Difference between revisions of "stat340s13"
(→Importance Sampling) 
m (Conversion script moved page Stat340s13 to stat340s13: Converting page titles to lowercase) 

(173 intermediate revisions by 47 users not shown)  
Line 42:  Line 42:  
=== Final ===  === Final ===  
−  Saturday August 10,2013  +  Saturday August 10,2013 from 7:30pm10:00pm 
=== TA(s): ===  === TA(s): ===  
Line 186:  Line 186:  
if y = ax + b, then <math>b:=y \mod a</math>. <br />  if y = ax + b, then <math>b:=y \mod a</math>. <br />  
−  '''  +  '''Example 1:'''<br /> 
<math>30 = 4 \cdot 7 + 2</math><br />  <math>30 = 4 \cdot 7 + 2</math><br />  
Line 201:  Line 201:  
<br />  <br />  
−  '''  +  '''Example 2:'''<br /> 
If <math>23 = 3 \cdot 6 + 5</math> <br />  If <math>23 = 3 \cdot 6 + 5</math> <br />  
Line 214:  Line 214:  
Then equivalently, <math>3 := 37\mod 40</math><br />  Then equivalently, <math>3 := 37\mod 40</math><br />  
+  
+  '''Example 3:'''<br />  
+  <math>77 = 3 \cdot 25 + 2</math><br />  
+  
+  <math>2 := 77\mod 3</math><br />  
+  <br />  
+  <math>25 = 25 \cdot 1 + 0</math><br />  
+  
+  <math>0: = 25\mod 25</math><br />  
+  <br />  
+  
+  
Line 221:  Line 233:  
==== Mixed Congruential Algorithm ====  ==== Mixed Congruential Algorithm ====  
−  We define the Linear Congruential Method to be <math>x_{k+1}=(ax_k + b) \mod m</math>, where <math>x_k, a, b, m \in \N, \;\text{with}\; a, m \neq 0</math>. Given a '''seed''' (i.e. an initial value <math>x_0 \in \N</math>), we can obtain values for <math>x_1, \, x_2, \, \cdots, x_n</math> inductively. The Multiplicative Congruential Method, invented by Berkeley professor D. H. Lehmer, may also refer to the special case where <math>b=0</math> and the Mixed Congruential Method is case where <math>b \neq 0</math> <br />  +  We define the Linear Congruential Method to be <math>x_{k+1}=(ax_k + b) \mod m</math>, where <math>x_k, a, b, m \in \N, \;\text{with}\; a, m \neq 0</math>. Given a '''seed''' (i.e. an initial value <math>x_0 \in \N</math>), we can obtain values for <math>x_1, \, x_2, \, \cdots, x_n</math> inductively. The Multiplicative Congruential Method, invented by Berkeley professor D. H. Lehmer, may also refer to the special case where <math>b=0</math> and the Mixed Congruential Method is case where <math>b \neq 0</math> <br />. Their title as "mixed" arises from the fact that it has both a multiplicative and additive term. 
An interesting fact about '''Linear Congruential Method''' is that it is one of the oldest and bestknown pseudo random number generator algorithms. It is very fast and requires minimal memory to retain state. However, this method should not be used for applications that require high randomness. They should not be used for Monte Carlo simulation and cryptographic applications. (Monte Carlo simulation will consider possibilities for every choice of consideration, and it shows the extreme possibilities. This method is not precise enough.)<br />  An interesting fact about '''Linear Congruential Method''' is that it is one of the oldest and bestknown pseudo random number generator algorithms. It is very fast and requires minimal memory to retain state. However, this method should not be used for applications that require high randomness. They should not be used for Monte Carlo simulation and cryptographic applications. (Monte Carlo simulation will consider possibilities for every choice of consideration, and it shows the extreme possibilities. This method is not precise enough.)<br />  
Line 384:  Line 396:  
'''Comments:'''<br />  '''Comments:'''<br />  
+  
+  Matlab code:  
+  a=5;  
+  b=7;  
+  m=200;  
+  x(1)=3;  
+  for ii=2:1000  
+  x(ii)=mod(a*x(ii1)+b,m);  
+  end  
+  size(x);  
+  hist(x)  
+  
+  
+  
Typically, it is good to choose <math>m</math> such that <math>m</math> is large, and <math>m</math> is prime. Careful selection of parameters '<math>a</math>' and '<math>b</math>' also helps generate relatively "random" output values, where it is harder to identify patterns. For example, when we used a composite (non prime) number such as 40 for <math>m</math>, our results were not satisfactory in producing an output resembling a uniform distribution.<br />  Typically, it is good to choose <math>m</math> such that <math>m</math> is large, and <math>m</math> is prime. Careful selection of parameters '<math>a</math>' and '<math>b</math>' also helps generate relatively "random" output values, where it is harder to identify patterns. For example, when we used a composite (non prime) number such as 40 for <math>m</math>, our results were not satisfactory in producing an output resembling a uniform distribution.<br />  
Line 435:  Line 461:  
</pre>  </pre>  
</div>  </div>  
+  Another algorithm for generating pseudo random numbers is the multiply with carry method. Its simplest form is similar to the linear congruential generator. They differs in that the parameter b changes in the MWC algorithm. It is as follows: <br>  
+  
+  1.) x<sub>k+1</sub> = ax<sub>k</sub> + b<sub>k</sub> mod m <br>  
+  2.) b<sub>k+1</sub> = floor((ax<sub>k</sub> + b<sub>k</sub>)/m) <br>  
+  3.) set k to k + 1 and go to step 1  
+  [http://www.javamex.com/tutorials/random_numbers/multiply_with_carry.shtml Source]  
=== Inverse Transform Method ===  === Inverse Transform Method ===  
Line 499:  Line 531:  
<pre style="fontsize:16px">  <pre style="fontsize:16px">  
>>u=rand(1,1000);  >>u=rand(1,1000);  
−  >>hist(u) #will generate a fairly uniform diagram  +  >>hist(u) # this will generate a fairly uniform diagram 
</pre>  </pre>  
[[File:ITM_example_hist(u).jpg300px]]  [[File:ITM_example_hist(u).jpg300px]]  
Line 535:  Line 567:  
Sol:  Sol:  
Let <math>y=x^5</math>, solve for x: <math>x=y^\frac {1}{5}</math>. Therefore, <math>F^{1} (x) = x^\frac {1}{5}</math><br />  Let <math>y=x^5</math>, solve for x: <math>x=y^\frac {1}{5}</math>. Therefore, <math>F^{1} (x) = x^\frac {1}{5}</math><br />  
−  Hence, to obtain a value of x from F(x), we first set u as an uniform distribution, then obtain the inverse function of F(x), and set  +  Hence, to obtain a value of x from F(x), we first set 'u' as an uniform distribution, then obtain the inverse function of F(x), and set 
<math>x= u^\frac{1}{5}</math><br /><br />  <math>x= u^\frac{1}{5}</math><br /><br />  
Line 600:  Line 632:  
To sample X with CDF F(x), <br />  To sample X with CDF F(x), <br />  
−  +  <math>1) U~ \sim~ Unif [0,1] </math>  
'''2) X = F<sup>1</sup>(u) '''<br />  '''2) X = F<sup>1</sup>(u) '''<br />  
Line 658:  Line 690:  
Note that after generating a random U, the value of X can be determined by finding the interval <math>[F(x_{j1}),F(x_{j})]</math> in which U lies. <br />  Note that after generating a random U, the value of X can be determined by finding the interval <math>[F(x_{j1}),F(x_{j})]</math> in which U lies. <br />  
+  
+  In summary:  
+  Generate a discrete r.v.x that has pmf:<br />  
+  P(X=xi)=Pi, x0<x1<x2<... <br />  
+  1. Draw U~U(0,1);<br />  
+  2. If F(x(i1))<U<F(xi), x=xi.<br />  
Line 1,180:  Line 1,218:  
== Class 4  Thursday, May 16 ==  == Class 4  Thursday, May 16 ==  
−  *When we want to find target distribution  +  
+  '''Goals'''<br>  
+  *When we want to find target distribution <math>f(x)</math>, we need to first find a proposal distribution <math>g(x)</math> that is easy to sample from. <br>  
*Relationship between the proposal distribution and target distribution is: <math> c \cdot g(x) \geq f(x) </math>, where c is constant. This means that the area of f(x) is under the area of <math> c \cdot g(x)</math>. <br>  *Relationship between the proposal distribution and target distribution is: <math> c \cdot g(x) \geq f(x) </math>, where c is constant. This means that the area of f(x) is under the area of <math> c \cdot g(x)</math>. <br>  
*Chance of acceptance is less if the distance between <math>f(x)</math> and <math> c \cdot g(x)</math> is big, and viceversa, we use <math> c </math> to keep <math> \frac {f(x)}{c \cdot g(x)} </math> below 1 (so <math>f(x) \leq c \cdot g(x)</math>). Therefore, we must find the constant <math> C </math> to achieve this.<br />  *Chance of acceptance is less if the distance between <math>f(x)</math> and <math> c \cdot g(x)</math> is big, and viceversa, we use <math> c </math> to keep <math> \frac {f(x)}{c \cdot g(x)} </math> below 1 (so <math>f(x) \leq c \cdot g(x)</math>). Therefore, we must find the constant <math> C </math> to achieve this.<br />  
*In other words, <math>C</math> is chosen to make sure <math> c \cdot g(x) \geq f(x) </math>. However, it will not make sense if <math>C</math> is simply chosen to be arbitrarily large. We need to choose <math>C</math> such that <math>c \cdot g(x)</math> fits <math>f(x)</math> as tightly as possible. This means that we must find the minimum c such that the area of f(x) is under the area of c*g(x). <br />  *In other words, <math>C</math> is chosen to make sure <math> c \cdot g(x) \geq f(x) </math>. However, it will not make sense if <math>C</math> is simply chosen to be arbitrarily large. We need to choose <math>C</math> such that <math>c \cdot g(x)</math> fits <math>f(x)</math> as tightly as possible. This means that we must find the minimum c such that the area of f(x) is under the area of c*g(x). <br />  
*The constant c cannot be a negative number.<br />  *The constant c cannot be a negative number.<br />  
+  
'''How to find C''':<br />  '''How to find C''':<br />  
+  
<math>\begin{align}  <math>\begin{align}  
&c \cdot g(x) \geq f(x)\\  &c \cdot g(x) \geq f(x)\\  
Line 1,192:  Line 1,234:  
&c= \max \left(\frac{f(x)}{g(x)}\right)  &c= \max \left(\frac{f(x)}{g(x)}\right)  
\end{align}</math><br>  \end{align}</math><br>  
+  
If <math>f</math> and <math> g </math> are continuous, we can find the extremum by taking the derivative and solve for <math>x_0</math> such that:<br/>  If <math>f</math> and <math> g </math> are continuous, we can find the extremum by taking the derivative and solve for <math>x_0</math> such that:<br/>  
<math> 0=\frac{d}{dx}\frac{f(x)}{g(x)}_{x=x_0}</math> <br/>  <math> 0=\frac{d}{dx}\frac{f(x)}{g(x)}_{x=x_0}</math> <br/>  
+  
Thus <math> c = \frac{f(x_0)}{g(x_0)} </math><br/>  Thus <math> c = \frac{f(x_0)}{g(x_0)} </math><br/>  
−  +  Note: This procedure is called the AcceptanceRejection Method.<br>  
−  The AcceptanceRejection method involves finding a distribution that we know how to sample from, g(x), and multiplying g(x) by a constant c so that <math>c \cdot g(x)</math> is always greater than or equal to f(x). Mathematically, we want <math> c \cdot g(x) \geq f(x) </math>.  +  
−  And it means, c has to be greater or equal to <math>\frac{f(x)}{g(x)}</math>. So the smallest possible c that satisfies the condition is the maximum value of <math>\frac{f(x)}{g(x)}</math><br/>.  +  '''The AcceptanceRejection method''' involves finding a distribution that we know how to sample from, g(x), and multiplying g(x) by a constant c so that <math>c \cdot g(x)</math> is always greater than or equal to f(x). Mathematically, we want <math> c \cdot g(x) \geq f(x) </math>. 
+  And it means, c has to be greater or equal to <math>\frac{f(x)}{g(x)}</math>. So the smallest possible c that satisfies the condition is the maximum value of <math>\frac{f(x)}{g(x)}</math><br/>.  
+  But in case of c being too large, the chance of acceptance of generated values will be small, thereby losing efficiency of the algorithm. Therefore, it is best to get the smallest possible c such that <math> c g(x) \geq f(x)</math>. <br>  
+  
+  '''Important points:'''<br>  
*For this method to be efficient, the constant c must be selected so that the rejection rate is low. (The efficiency for this method is <math>\left ( \frac{1}{c} \right )</math>)<br>  *For this method to be efficient, the constant c must be selected so that the rejection rate is low. (The efficiency for this method is <math>\left ( \frac{1}{c} \right )</math>)<br>  
*It is easy to show that the expected number of trials for an acceptance is <math> \frac{Total Number of Trials} {C} </math>. <br>  *It is easy to show that the expected number of trials for an acceptance is <math> \frac{Total Number of Trials} {C} </math>. <br>  
−  *recall the acceptance rate is 1/c. (Not rejection rate)  +  *recall the '''acceptance rate is 1/c'''. (Not rejection rate) 
:Let <math>X</math> be the number of trials for an acceptance, <math> X \sim~ Geo(\frac{1}{c})</math><br>  :Let <math>X</math> be the number of trials for an acceptance, <math> X \sim~ Geo(\frac{1}{c})</math><br>  
:<math>\mathbb{E}[X] = \frac{1}{\frac{1}{c}} = c </math>  :<math>\mathbb{E}[X] = \frac{1}{\frac{1}{c}} = c </math>  
*The number of trials needed to generate a sample size of <math>N</math> follows a negative binomial distribution. The expected number of trials needed is then <math>cN</math>.<br>  *The number of trials needed to generate a sample size of <math>N</math> follows a negative binomial distribution. The expected number of trials needed is then <math>cN</math>.<br>  
*So far, the only distribution we know how to sample from is the '''UNIFORM''' distribution. <br>  *So far, the only distribution we know how to sample from is the '''UNIFORM''' distribution. <br>  
+  
'''Procedure''': <br>  '''Procedure''': <br>  
+  
1. Choose <math>g(x)</math> (simple density function that we know how to sample, i.e. Uniform so far) <br>  1. Choose <math>g(x)</math> (simple density function that we know how to sample, i.e. Uniform so far) <br>  
−  The easiest case is  +  The easiest case is <math>U~ \sim~ Unif [0,1] </math>. However, in other cases we need to generate UNIF(a,b). We may need to perform a linear transformation on the <math>U~ \sim~ Unif [0,1] </math> variable. <br> 
2. Find a constant c such that :<math> c \cdot g(x) \geq f(x) </math>, otherwise return to step 1.  2. Find a constant c such that :<math> c \cdot g(x) \geq f(x) </math>, otherwise return to step 1.  
Line 1,218:  Line 1,268:  
#If <math>U \leq \frac{f(Y)}{c \cdot g(Y)}</math> then X=Y; else return to step 1 (This is not the way to find C. This is the general procedure.)  #If <math>U \leq \frac{f(Y)}{c \cdot g(Y)}</math> then X=Y; else return to step 1 (This is not the way to find C. This is the general procedure.)  
−  <hr><b>Example: Generate a random variable from the pdf</b><br>  +  <hr><b>Example: <br> 
+  
+  Generate a random variable from the pdf</b><br>  
<math> f(x) =  <math> f(x) =  
\begin{cases}  \begin{cases}  
Line 1,253:  Line 1,305:  
[[File:Beta(2,1)_example.jpg750x750px]]  [[File:Beta(2,1)_example.jpg750x750px]]  
−  Note: g follows uniform distribution, it only covers half of the graph which runs from 0 to 1 on yaxis. Thus we need to multiply by c to ensure that <math>c\cdot g</math> can cover entire f(x) area. In this case, c=2, so that makes g run from 0 to 2 on yaxis which covers f(x).  +  '''Note:''' g follows uniform distribution, it only covers half of the graph which runs from 0 to 1 on yaxis. Thus we need to multiply by c to ensure that <math>c\cdot g</math> can cover entire f(x) area. In this case, c=2, so that makes g run from 0 to 2 on yaxis which covers f(x). 
−  Comment:  +  '''Comment:'''<br> 
From the picture above, we could observe that the area under f(x)=2x is a half of the area under the pdf of UNIF(0,1). This is why in order to sample 1000 points of f(x), we need to sample approximately 2000 points in UNIF(0,1).  From the picture above, we could observe that the area under f(x)=2x is a half of the area under the pdf of UNIF(0,1). This is why in order to sample 1000 points of f(x), we need to sample approximately 2000 points in UNIF(0,1).  
And in general, if we want to sample n points from a distritubion with pdf f(x), we need to scan approximately <math>n\cdot c</math> points from the proposal distribution (g(x)) in total. <br>  And in general, if we want to sample n points from a distritubion with pdf f(x), we need to scan approximately <math>n\cdot c</math> points from the proposal distribution (g(x)) in total. <br>  
Line 1,266:  Line 1,318:  
</ol>  </ol>  
−  Note: In the above example, we sample 2 numbers. If second number (u) is less than or equal to first number (y), then accept x=y, if not then start all over.  +  '''Note:''' In the above example, we sample 2 numbers. If second number (u) is less than or equal to first number (y), then accept x=y, if not then start all over. 
<span style="fontweight:bold;color:green;">Matlab Code</span>  <span style="fontweight:bold;color:green;">Matlab Code</span>  
Line 1,373:  Line 1,425:  
To obtain a better proposing function <math>\begin{align}g(x)\end{align}</math>, we can first assume a new <math>\begin{align}q(x)\end{align}</math> and then solve for the normalizing constant by integrating.<br>  To obtain a better proposing function <math>\begin{align}g(x)\end{align}</math>, we can first assume a new <math>\begin{align}q(x)\end{align}</math> and then solve for the normalizing constant by integrating.<br>  
In the previous example, we first assume <math>\begin{align}q(x) = 3x\end{align}</math>. To find the normalizing constant, we need to solve <math>k *\sum 3x = 1</math> which gives us k = 2/3. So,<math>\begin{align}g(x) = k*q(x) = 2x\end{align}</math>.  In the previous example, we first assume <math>\begin{align}q(x) = 3x\end{align}</math>. To find the normalizing constant, we need to solve <math>k *\sum 3x = 1</math> which gives us k = 2/3. So,<math>\begin{align}g(x) = k*q(x) = 2x\end{align}</math>.  
−  +  
+  *Source: http://www.cs.bgu.ac.il/~mps042/acceptance.htm*  
'''Possible Limitations'''  '''Possible Limitations'''  
Line 1,650:  Line 1,703:  
* '''Example 3'''<br>  * '''Example 3'''<br>  
−  
−  
−  +  Suppose <math>\begin{align}p_{x} = e^{3}3^{x}/x! , x\geq 0\end{align}</math> (Poisson distribution)  
−  
−  
−  
−  +  '''First:''' Try the first few <math>\begin{align}p_{x}'s\end{align}</math>: 0.0498, 0.149, 0.224, 0.224, 0.168, 0.101, 0.0504, 0.0216, 0.0081, 0.0027 for <math>\begin{align} x = 0,1,2,3,4,5,6,7,8,9 \end{align}</math><br>  
−  
−  3  
−  Note: In this case, f(x)/g(x) is extremely difficult to differentiate so we were required to test points. If the function is  +  '''Proposed distribution:''' Use the geometric distribution for <math>\begin{align}g(x)\end{align}</math>;<br> 
+  
+  <math>\begin{align}g(x)=p(1p)^{x}\end{align}</math>, choose <math>\begin{align}p=0.25\end{align}</math><br>  
+  
+  Look at <math>\begin{align}p_{x}/g(x)\end{align}</math> for the first few numbers: 0.199 0.797 1.59 2.12 2.12 1.70 1.13 0.647 0.324 0.144 for <math>\begin{align} x = 0,1,2,3,4,5,6,7,8,9 \end{align}</math><br>  
+  
+  We want <math>\begin{align}c=max(p_{x}/g(x))\end{align}</math> which is approximately 2.12<br>  
+  
+  '''The general procedures to generate <math>\begin{align}p(x)\end{align}</math> is as follows:'''  
+  
+  1. Generate <math>\begin{align}U_{1} \sim~ U(0,1); U_{2} \sim~ U(0,1)\end{align}</math><br>  
+  
+  2. <math>\begin{align}j = \lfloor \frac{ln(U_{1})}{ln(.75)} \rfloor+1;\end{align}</math><br>  
+  
+  3. if <math>U_{2} < \frac{p_{j}}{cg(j)}</math>, set <math>\begin{align}X = x_{j}\end{align}</math>, else go to step 1.  
+  
+  Note: In this case, <math>\begin{align}f(x)/g(x)\end{align}</math> is extremely difficult to differentiate so we were required to test points. If the function is very easy to differentiate, we can calculate the max as if it were a continuous function then check the two surrounding points for which is the highest discrete value.  
+  
+  * Source: http://www.math.wsu.edu/faculty/genz/416/lect/l0446.pdf*  
*'''Example 4''' (Hypergeometric & Binomial)<br>  *'''Example 4''' (Hypergeometric & Binomial)<br>  
Line 1,741:  Line 1,805:  
The CDF of the Gamma distribution <math>Gamma(t,\lambda)</math> is(t denotes the shape, <math>\lambda</math> denotes the scale: <br>  The CDF of the Gamma distribution <math>Gamma(t,\lambda)</math> is(t denotes the shape, <math>\lambda</math> denotes the scale: <br>  
<math> F(x) = \int_0^{x} \frac{e^{y}y^{t1}}{(t1)!} \mathrm{d}y, \; \forall x \in (0,+\infty)</math>, where <math>t \in \N^+ \text{ and } \lambda \in (0,+\infty)</math>.<br>  <math> F(x) = \int_0^{x} \frac{e^{y}y^{t1}}{(t1)!} \mathrm{d}y, \; \forall x \in (0,+\infty)</math>, where <math>t \in \N^+ \text{ and } \lambda \in (0,+\infty)</math>.<br>  
+  
+  Note that the CDF of the Gamma distribution does not have a closed form.  
The gamma distribution is often used to model waiting times between a certain number of events. It can also be expressed as the sum of infinitely many independent and identically distributed exponential distributions. This distribution has two parameters: the number of exponential terms n, and the rate parameter <math>\lambda</math>. In this distribution there is the Gamma function, <math>\Gamma </math> which has some very useful properties. "Source: STAT 340 Spring 2010 Course Notes" <br/>  The gamma distribution is often used to model waiting times between a certain number of events. It can also be expressed as the sum of infinitely many independent and identically distributed exponential distributions. This distribution has two parameters: the number of exponential terms n, and the rate parameter <math>\lambda</math>. In this distribution there is the Gamma function, <math>\Gamma </math> which has some very useful properties. "Source: STAT 340 Spring 2010 Course Notes" <br/>  
Line 1,853:  Line 1,919:  
:<math>f(x) = \frac{1}{\sqrt{2\pi}}\, e^{ \frac{\scriptscriptstyle 1}{\scriptscriptstyle 2} x^2}</math>  :<math>f(x) = \frac{1}{\sqrt{2\pi}}\, e^{ \frac{\scriptscriptstyle 1}{\scriptscriptstyle 2} x^2}</math>  
−  *Warning : the General Normal distribution is  +  *Warning : the General Normal distribution is: 
−  :  
<table>  <table>  
<tr>  <tr>  
Line 1,906:  Line 1,971:  
Let <math> \theta </math> and R denote the Polar coordinate of the vector (X, Y)  Let <math> \theta </math> and R denote the Polar coordinate of the vector (X, Y)  
+  where <math> X = R \cdot \sin\theta </math> and <math> Y = R \cdot \cos \theta </math>  
[[File:rtheta.jpg]]  [[File:rtheta.jpg]]  
Line 1,922:  Line 1,988:  
We know that  We know that  
−  <math>  +  <math>R^{2}= X^{2}+Y^{2}</math> and <math> \tan(\theta) = \frac{y}{x} </math> where X and Y are two independent standard normal 
:<math>f(x) = \frac{1}{\sqrt{2\pi}}\, e^{ \frac{\scriptscriptstyle 1}{\scriptscriptstyle 2} x^2}</math>  :<math>f(x) = \frac{1}{\sqrt{2\pi}}\, e^{ \frac{\scriptscriptstyle 1}{\scriptscriptstyle 2} x^2}</math>  
:<math>f(y) = \frac{1}{\sqrt{2\pi}}\, e^{ \frac{\scriptscriptstyle 1}{\scriptscriptstyle 2} y^2}</math>  :<math>f(y) = \frac{1}{\sqrt{2\pi}}\, e^{ \frac{\scriptscriptstyle 1}{\scriptscriptstyle 2} y^2}</math>  
−  :<math>f(x,y) = \frac{1}{\sqrt{2\pi}}\, e^{ \frac{\scriptscriptstyle 1}{\scriptscriptstyle 2} x^2} * \frac{1}{\sqrt{2\pi}}\, e^{ \frac{\scriptscriptstyle 1}{\scriptscriptstyle 2} y^2}=\frac{1}{2\pi}\, e^{ \frac{\scriptscriptstyle 1}{\scriptscriptstyle 2} (x^2+y^2)} </math><br />  Since for independent distributions, their joint probability function is the multiplication of two independent probability functions  +  :<math>f(x,y) = \frac{1}{\sqrt{2\pi}}\, e^{ \frac{\scriptscriptstyle 1}{\scriptscriptstyle 2} x^2} * \frac{1}{\sqrt{2\pi}}\, e^{ \frac{\scriptscriptstyle 1}{\scriptscriptstyle 2} y^2}=\frac{1}{2\pi}\, e^{ \frac{\scriptscriptstyle 1}{\scriptscriptstyle 2} (x^2+y^2)} </math><br />  Since for independent distributions, their joint probability function is the multiplication of two independent probability functions. It can also be shown using 11 transformation that the joint distribution of R and θ is given by, 11 transformation:<br /> 
−  It can also be shown using 11 transformation that the joint distribution of R and θ is given by,  +  
−  11 transformation:<br />  +  
−  Let <math>d=R^2</math><br />  +  '''Let <math>d=R^2</math>'''<br /> 
+  
<math>x= \sqrt {d}\cos \theta </math>  <math>x= \sqrt {d}\cos \theta </math>  
<math>y= \sqrt {d}\sin \theta </math>  <math>y= \sqrt {d}\sin \theta </math>  
then  then  
<math>\left J\right = \left \dfrac {1} {2}d^{\frac {1} {2}}\cos \theta d^{\frac{1}{2}}\cos \theta +\sqrt {d}\sin \theta \dfrac {1} {2}d^{\frac{1}{2}}\sin \theta \right = \dfrac {1} {2}</math>  <math>\left J\right = \left \dfrac {1} {2}d^{\frac {1} {2}}\cos \theta d^{\frac{1}{2}}\cos \theta +\sqrt {d}\sin \theta \dfrac {1} {2}d^{\frac{1}{2}}\sin \theta \right = \dfrac {1} {2}</math>  
−  It can be shown that the  +  It can be shown that the joint density of <math> d /R^2</math> and <math> \theta </math> is: 
:<math>\begin{matrix} f(d,\theta) = \frac{1}{2}e^{\frac{d}{2}}*\frac{1}{2\pi},\quad d = R^2 \end{matrix},\quad for\quad 0\leq d<\infty\ and\quad 0\leq \theta\leq 2\pi </math>  :<math>\begin{matrix} f(d,\theta) = \frac{1}{2}e^{\frac{d}{2}}*\frac{1}{2\pi},\quad d = R^2 \end{matrix},\quad for\quad 0\leq d<\infty\ and\quad 0\leq \theta\leq 2\pi </math>  
Line 1,940:  Line 2,007:  
Note that <math> \begin{matrix}f(r,\theta)\end{matrix}</math> consists of two density functions, Exponential and Uniform, so assuming that r and <math>\theta</math> are independent  Note that <math> \begin{matrix}f(r,\theta)\end{matrix}</math> consists of two density functions, Exponential and Uniform, so assuming that r and <math>\theta</math> are independent  
<math> \begin{matrix} \Rightarrow d \sim~ Exp(1/2), \theta \sim~ Unif[0,2\pi] \end{matrix} </math>  <math> \begin{matrix} \Rightarrow d \sim~ Exp(1/2), \theta \sim~ Unif[0,2\pi] \end{matrix} </math>  
−  ::* <math> \begin{align} R^2 = x^2 + y^2 \end{align} </math>  +  ::* <math> \begin{align} R^2 = d = x^2 + y^2 \end{align} </math> 
::* <math> \tan(\theta) = \frac{y}{x} </math>  ::* <math> \tan(\theta) = \frac{y}{x} </math>  
<math>\begin{align} f(d) = Exp(1/2)=\frac{1}{2}e^{\frac{d}{2}}\ \end{align}</math>  <math>\begin{align} f(d) = Exp(1/2)=\frac{1}{2}e^{\frac{d}{2}}\ \end{align}</math>  
Line 1,946:  Line 2,013:  
<math>\begin{align} f(\theta) =\frac{1}{2\pi}\ \end{align}</math>  <math>\begin{align} f(\theta) =\frac{1}{2\pi}\ \end{align}</math>  
<br>  <br>  
+  
To sample from the normal distribution, we can generate a pair of independent standard normal X and Y by:<br />  To sample from the normal distribution, we can generate a pair of independent standard normal X and Y by:<br />  
+  
1) Generating their polar coordinates<br />  1) Generating their polar coordinates<br />  
2) Transforming back to rectangular (Cartesian) coordinates.<br />  2) Transforming back to rectangular (Cartesian) coordinates.<br />  
−  
−  Step 1: Generate <math>u_{1}</math> ~<math>Unif(0,1)</math>  +  '''Alternative Method of Generating Standard Normal Random Variables'''<br /> 
−  Step 2: Generate <math>Y_{1}</math> ~<math>Exp(1)</math>,<math>Y_{2}</math>~<math>Exp(2)</math>  +  
−  Step 3: If <math>Y_{2} \geq(Y_{1}1)^2/2</math>,set <math>V=Y1</math>,otherwise,go to step 1  +  Step 1: Generate <math>u_{1}</math> ~<math>Unif(0,1)</math><br /> 
−  Step 4: If <math>u_{1} \leq 1/2</math>,then <math>X=V</math>  +  Step 2: Generate <math>Y_{1}</math> ~<math>Exp(1)</math>,<math>Y_{2}</math>~<math>Exp(2)</math><br /> 
+  Step 3: If <math>Y_{2} \geq(Y_{1}1)^2/2</math>,set <math>V=Y1</math>,otherwise,go to step 1<br />  
+  Step 4: If <math>u_{1} \leq 1/2</math>,then <math>X=V</math><br />  
−  +  ===Expectation of a Standard Normal distribution===<br />  
−  The expectation of a standard normal distribution is 0  +  
−  :  +  The expectation of a standard normal distribution is 0<br /> 
+  
+  '''Proof:''' <br />  
:<math>\operatorname{E}[X]= \;\int_{\infty}^{\infty} x \frac{1}{\sqrt{2\pi}} e^{x^2/2} \, dx.</math>  :<math>\operatorname{E}[X]= \;\int_{\infty}^{\infty} x \frac{1}{\sqrt{2\pi}} e^{x^2/2} \, dx.</math>  
Line 1,968:  Line 2,040:  
:<math>=  \left[\phi(x)\right]_{\infty}^{\infty}</math>  :<math>=  \left[\phi(x)\right]_{\infty}^{\infty}</math>  
:<math>= 0</math><br />  :<math>= 0</math><br />  
−  
−  +  '''Note,''' more intuitively, because x is an odd function (f(x)+f(x)=0). Taking integral of x will give <math>x^2/2 </math> which is an even function (f(x)=f(x)). This is in relation to the symmetrical properties of the standard normal distribution. If support is from negative infinity to infinity, then the integral will return 0.<br />  
+  
+  
+  '''Procedure (BoxMuller Transformation Method):''' <br />  
+  
Pseudorandom approaches to generating normal random variables used to be limited. Inefficient methods such as inverse Gaussian function, sum of uniform random variables, and acceptancerejection were used. In 1958, a new method was proposed by George Box and Mervin Muller of Princeton University. This new technique was easy to use and also had the accuracy to the inverse transform sampling method that it grew more valuable as computers became more computationally astute. <br>  Pseudorandom approaches to generating normal random variables used to be limited. Inefficient methods such as inverse Gaussian function, sum of uniform random variables, and acceptancerejection were used. In 1958, a new method was proposed by George Box and Mervin Muller of Princeton University. This new technique was easy to use and also had the accuracy to the inverse transform sampling method that it grew more valuable as computers became more computationally astute. <br>  
The BoxMuller method takes a sample from a bivariate independent standard normal distribution, each component of which is thus a univariate standard normal. The algorithm is based on the following two properties of the bivariate independent standard normal distribution: <br>  The BoxMuller method takes a sample from a bivariate independent standard normal distribution, each component of which is thus a univariate standard normal. The algorithm is based on the following two properties of the bivariate independent standard normal distribution: <br>  
if <math>Z = (Z_{1}, Z_{2}</math>) has this distribution, then <br>  if <math>Z = (Z_{1}, Z_{2}</math>) has this distribution, then <br>  
+  
1.<math>R^2=Z_{1}^2+Z_{2}^2</math> is exponentially distributed with mean 2, i.e. <br>  1.<math>R^2=Z_{1}^2+Z_{2}^2</math> is exponentially distributed with mean 2, i.e. <br>  
<math>P(R^2 \leq x) = 1e^{x/2}</math>. <br>  <math>P(R^2 \leq x) = 1e^{x/2}</math>. <br>  
2.Given <math>R^2</math>, the point <math>(Z_{1},Z_{2}</math>) is uniformly distributed on the circle of radius R centered at the origin. <br>  2.Given <math>R^2</math>, the point <math>(Z_{1},Z_{2}</math>) is uniformly distributed on the circle of radius R centered at the origin. <br>  
We can use these properties to build the algorithm: <br>  We can use these properties to build the algorithm: <br>  
+  
1) Generate random number <math> \begin{align} U_1,U_2 \sim~ \mathrm{Unif}(0, 1) \end{align} </math> <br />  1) Generate random number <math> \begin{align} U_1,U_2 \sim~ \mathrm{Unif}(0, 1) \end{align} </math> <br />  
Line 1,996:  Line 2,073:  
−  Note: In steps 2 and 3, we are using a similar technique as that used in the inverse transform method. <br />  +  '''Note:''' In steps 2 and 3, we are using a similar technique as that used in the inverse transform method. <br /> 
The BoxMuller Transformation Method generates a pair of independent Standard Normal distributions, X and Y (Using the transformation of polar coordinates). <br />  The BoxMuller Transformation Method generates a pair of independent Standard Normal distributions, X and Y (Using the transformation of polar coordinates). <br />  
+  
If you want to generate a number of independent standard normal distributed numbers (more than two), you can run the BoxMuller method several times.<br/>  If you want to generate a number of independent standard normal distributed numbers (more than two), you can run the BoxMuller method several times.<br/>  
For example: <br />  For example: <br />  
Line 2,004:  Line 2,082:  
−  +  '''Matlab Code'''<br />  
+  
<pre style="fontsize:16px">  <pre style="fontsize:16px">  
>>close all  >>close all  
Line 2,019:  Line 2,098:  
>>hist(y)  >>hist(y)  
</pre>  </pre>  
+  <br>  
+  '''Remember''': For the above code to work the "." needs to be after the d to ensure that each element of d is raised to the power of 0.5.<br /> Otherwise matlab will raise the entire matrix to the power of 0.5."<br>  
−  +  '''Note:'''<br>the first graph is hist(tet) and it is a uniform distribution.<br>The second one is hist(d) and it is a exponential distribution.<br>The third one is hist(x) and it is a normal distribution.<br>The last one is hist(y) and it is also a normal distribution.  
−  
−  Note:<br>the first graph is hist(tet) and it is a uniform distribution.<br>The second one is hist(d) and it is a exponential distribution.<br>The third one is hist(x) and it is a normal distribution.<br>The last one is hist(y) and it is also a normal distribution.  
Attention:There is a "dot" between sqrt(d) and "*". It is because d and tet are vectors. <br>  Attention:There is a "dot" between sqrt(d) and "*". It is because d and tet are vectors. <br>  
Line 2,039:  Line 2,118:  
>>hist(x)  >>hist(x)  
>>hist(x+2)  >>hist(x+2)  
−  >>hist(x*2+2)  +  >>hist(x*2+2)<br> 
</pre>  </pre>  
−  +  <br>  
−  Note: randn is random sample from a standard normal distribution.<br />  +  '''Note:'''<br> 
−  +  1. randn is random sample from a standard normal distribution.<br />  
−  +  2. hist(x+2) will be centered at 2 instead of at 0. <br />  
+  3. hist(x*3+2) is also centered at 2. The mean doesn't change, but the variance of x*3+2 becomes nine times (3^2) the variance of x.<br />  
[[File:Normal_x.jpg300x300px]][[File:Normal_x+2.jpg300x300px]][[File:Normal(2x+2).jpg300px]]  [[File:Normal_x.jpg300x300px]][[File:Normal_x+2.jpg300x300px]][[File:Normal(2x+2).jpg300px]]  
<br />  <br />  
−  <b>Comment</b>: BoxMuller transformations are not computationally efficient. The reason for this is the need to compute sine and cosine functions. A way to get around this timeconsuming difficulty is by an indirect computation of the sine and cosine of a random angle (as opposed to a direct computation which generates U and then computes the sine and cosine of 2πU. <br />  +  <b>Comment</b>:<br /> 
+  BoxMuller transformations are not computationally efficient. The reason for this is the need to compute sine and cosine functions. A way to get around this timeconsuming difficulty is by an indirect computation of the sine and cosine of a random angle (as opposed to a direct computation which generates U and then computes the sine and cosine of 2πU. <br />  
+  
+  
'''Alternative Methods of generating normal distribution'''<br />  '''Alternative Methods of generating normal distribution'''<br />  
+  
1. Even though we cannot use inverse transform method, we can approximate this inverse using different functions.One method would be '''rational approximation'''.<br />  1. Even though we cannot use inverse transform method, we can approximate this inverse using different functions.One method would be '''rational approximation'''.<br />  
2.'''Central limit theorem''' : If we sum 12 independent U(0,1) distribution and subtract 6 (which is E(ui)*12)we will approximately get a standard normal distribution.<br />  2.'''Central limit theorem''' : If we sum 12 independent U(0,1) distribution and subtract 6 (which is E(ui)*12)we will approximately get a standard normal distribution.<br />  
Line 2,064:  Line 2,148:  
=== Proof of Box Muller Transformation ===  === Proof of Box Muller Transformation ===  
−  Definition:  +  '''Definition:'''<br /> 
A transformation which transforms from a '''twodimensional continuous uniform''' distribution to a '''twodimensional bivariate normal''' distribution (or complex normal distribution).  A transformation which transforms from a '''twodimensional continuous uniform''' distribution to a '''twodimensional bivariate normal''' distribution (or complex normal distribution).  
Line 2,348:  Line 2,432:  
Procedure:  Procedure:  
−  1) Generate U~Unif  +  1) Generate U~Unif (0, 1)<br> 
2) Set <math>x=F^{1}(u)</math><br>  2) Set <math>x=F^{1}(u)</math><br>  
3) X~f(x)<br>  3) X~f(x)<br>  
Line 2,354:  Line 2,438:  
'''Remark'''<br>  '''Remark'''<br>  
1) The preceding can be written algorithmically for discrete random variables as <br>  1) The preceding can be written algorithmically for discrete random variables as <br>  
−  Generate a random number U ~ U  +  Generate a random number U ~ U(0,1] <br> 
If U < p<sub>0</sub> set X = x<sub>0</sub> and stop <br>  If U < p<sub>0</sub> set X = x<sub>0</sub> and stop <br>  
If U < p<sub>0</sub> + p<sub>1</sub> set X = x<sub>1</sub> and stop <br>  If U < p<sub>0</sub> + p<sub>1</sub> set X = x<sub>1</sub> and stop <br>  
Line 2,385:  Line 2,469:  
Step1: Generate U~ U(0, 1)<br>  Step1: Generate U~ U(0, 1)<br>  
−  Step2: set <math>y=\, {\frac {1}{{\lambda_1 +\lambda_2}}} ln(u)</math><br>  +  
+  Step2: set <math>y=\, {\frac {1}{{\lambda_1 +\lambda_2}}} ln(1u)</math><br>  
+  
+  or set <math>y=\, {\frac {1} {{\lambda_1 +\lambda_2}}} ln(u)</math><br>  
+  Since it is a uniform distribution, therefore after generate a lot of times 1u and u are the same.  
+  
+  
+  * '''Matlab Code'''<br />  
+  <pre style="fontsize:16px">  
+  >> lambda1 = 1;  
+  >> lambda2 = 2;  
+  >> u = rand;  
+  >> y = log(u)/(lambda1 + lambda2)  
+  </pre>  
If we generalize this example from two independent particles to n independent particles we will have:<br>  If we generalize this example from two independent particles to n independent particles we will have:<br>  
Line 2,541:  Line 2,638:  
=== Example of Decomposition Method ===  === Example of Decomposition Method ===  
−  +  <math>F_x(x) = \frac {1}{3} x+\frac {1}{3} x^2+\frac {1}{3} x^3, 0\leq x\leq 1</math>  
−  +  Let <math>U =F_x(x) = \frac {1}{3} x+\frac {1}{3} x^2+\frac {1}{3} x^3</math>, solve for x.  
−  +  <math>P_1=\frac{1}{3}, F_{x1} (x)= x, P_2=\frac{1}{3},F_{x2} (x)= x^2,  
−  +  P_3=\frac{1}{3},F_{x3} (x)= x^3</math>  
'''Algorithm:'''  '''Algorithm:'''  
−  Generate U  +  Generate <math>\,U \sim Unif [0,1)</math> 
−  Generate V  +  Generate <math>\,V \sim Unif [0,1)</math> 
−  if 0  +  if <math>0\leq u \leq \frac{1}{3}, x = v</math> 
−  else if u  +  else if <math>u \leq \frac{2}{3}, x = v^{\frac{1}{2}}</math> 
−  else x = v  +  else <math>x=v^{\frac{1}{3}}</math> <br> 
Line 2,623:  Line 2,720:  
For More Details, please refer to http://www.stanford.edu/class/ee364b/notes/decomposition_notes.pdf  For More Details, please refer to http://www.stanford.edu/class/ee364b/notes/decomposition_notes.pdf  
−  
===Fundamental Theorem of Simulation===  ===Fundamental Theorem of Simulation===  
Line 2,635:  Line 2,731:  
More specific definition of the theorem can be found here.<ref>http://www.bus.emory.edu/breno/teaching/MCMC_GibbsHandouts.pdf</ref>  More specific definition of the theorem can be found here.<ref>http://www.bus.emory.edu/breno/teaching/MCMC_GibbsHandouts.pdf</ref>  
+  Matlab code:  
−  >  +  <pre style="fontsize:16px"> 
−  +  close all  
−  +  clear all  
−  +  ii=1;  
−  +  while ii<1000  
−  
−  
−  
−  
u=rand  u=rand  
y=R*(2*U1)  y=R*(2*U1)  
Line 2,651:  Line 2,744:  
ii=ii+1  ii=ii+1  
end  end  
+  </pre>  
===Question 2===  ===Question 2===  
Line 2,695:  Line 2,789:  
The Bernoulli distribution is a special case of the binomial distribution, where n = 1. X ~ Bin(1, p) has the same meaning as X ~ Ber(p), where p is the probability of success and 1p is the probability of failure (we usually define a variate q, q= 1p). The mean of Bernoulli is p and the variance is p(1p). Bin(n, p), is the distribution of the sum of n independent Bernoulli trials, Bernoulli(p), each with the same probability p, where 0<p<1. <br>  The Bernoulli distribution is a special case of the binomial distribution, where n = 1. X ~ Bin(1, p) has the same meaning as X ~ Ber(p), where p is the probability of success and 1p is the probability of failure (we usually define a variate q, q= 1p). The mean of Bernoulli is p and the variance is p(1p). Bin(n, p), is the distribution of the sum of n independent Bernoulli trials, Bernoulli(p), each with the same probability p, where 0<p<1. <br>  
For example, let X be the event that a coin toss results in a "head" with probability ''p'', then ''X~Bernoulli(p)''. <br>  For example, let X be the event that a coin toss results in a "head" with probability ''p'', then ''X~Bernoulli(p)''. <br>  
−  +  P(X=1)= p  
+  P(X=0)= q = 1p  
+  Therefore, P(X=0) + P(X=1) = p + q = 1  
'''Algorithm: '''  '''Algorithm: '''  
Line 2,706:  Line 2,802:  
when <math>U \geq p, x=0</math><br>  when <math>U \geq p, x=0</math><br>  
3) Repeat as necessary  3) Repeat as necessary  
+  
+  * '''Matlab Code'''<br />  
+  <pre style="fontsize:16px">  
+  >> p = 0.8 % an arbitrary probability for example  
+  >> for i = 1: 100  
+  >> u = rand;  
+  >> if u < p  
+  >> x(ii) = 1;  
+  >> else  
+  >> x(ii) = 0;  
+  >> end  
+  >> end  
+  >> hist(x)  
+  </pre>  
===The Binomial Distribution===  ===The Binomial Distribution===  
Line 3,500:  Line 3,610:  
'''Definition:''' In probability theory, a stochastic process /stoʊˈkæstɪk/, or sometimes random process (widely used) is a collection of random variables; this is often used to represent the evolution of some random value, or system, over time. This is the probabilistic counterpart to a deterministic process (or deterministic system). Instead of describing a process which can only evolve in one way (as in the case, for example, of solutions of an ordinary differential equation), in a stochastic or random process there is some indeterminacy: even if the initial condition (or starting point) is known, there are several (often infinitely many) directions in which the process may evolve. (from Wikipedia)  '''Definition:''' In probability theory, a stochastic process /stoʊˈkæstɪk/, or sometimes random process (widely used) is a collection of random variables; this is often used to represent the evolution of some random value, or system, over time. This is the probabilistic counterpart to a deterministic process (or deterministic system). Instead of describing a process which can only evolve in one way (as in the case, for example, of solutions of an ordinary differential equation), in a stochastic or random process there is some indeterminacy: even if the initial condition (or starting point) is known, there are several (often infinitely many) directions in which the process may evolve. (from Wikipedia)  
−  A stochastic process is nondeterministic. This means that  +  A stochastic process is nondeterministic. This means that even if we know the initial condition(state), and we know some possibilities of the states to follow, the exact value of the final state remains to be uncertain. 
We can illustrate this with an example of speech: if "I" is the first word in a sentence, the set of words that could follow would be limited (eg. like, want, am), and the same happens for the third word and so on. The words then have some probabilities among them such that each of them is a random variable, and the sentence would be a collection of random variables. <br>  We can illustrate this with an example of speech: if "I" is the first word in a sentence, the set of words that could follow would be limited (eg. like, want, am), and the same happens for the third word and so on. The words then have some probabilities among them such that each of them is a random variable, and the sentence would be a collection of random variables. <br>  
Line 3,512:  Line 3,622:  
2. Markov Process This is a stochastic process that satisfies the Markov property which can be understood as the memoryless property. The property states that the jump to a future state only depends on the current state of the process, and not of the process's history. This model is used to model random walks exhibited by particles, the health state of a life insurance policyholder, decision making by a memoryless mouse in a maze, etc. <br>  2. Markov Process This is a stochastic process that satisfies the Markov property which can be understood as the memoryless property. The property states that the jump to a future state only depends on the current state of the process, and not of the process's history. This model is used to model random walks exhibited by particles, the health state of a life insurance policyholder, decision making by a memoryless mouse in a maze, etc. <br>  
−  
−  
=====Example=====  =====Example=====  
Line 3,554:  Line 3,662:  
the rate parameter may change over time; such a process is called a nonhomogeneous Poisson process  the rate parameter may change over time; such a process is called a nonhomogeneous Poisson process  
−  ====  +  ==== Examples ==== 
<br />  <br />  
'''How to generate a multivariate normal with the builtin function "randn": (example)'''<br />  '''How to generate a multivariate normal with the builtin function "randn": (example)'''<br />  
Line 3,614:  Line 3,722:  
===Poisson Process===  ===Poisson Process===  
+  A Poisson Process is a stochastic approach to count number of events in a certain time period. <s>Strikethrough text</s>  
A discrete stochastic variable ''X'' is said to have a Poisson distribution with parameter ''λ'' > 0 if  A discrete stochastic variable ''X'' is said to have a Poisson distribution with parameter ''λ'' > 0 if  
:<math>\!f(n)= \frac{\lambda^n e^{\lambda}}{n!} \qquad n= 0,1,2,3,4,5,\ldots,</math>.  :<math>\!f(n)= \frac{\lambda^n e^{\lambda}}{n!} \qquad n= 0,1,2,3,4,5,\ldots,</math>.  
Line 3,637:  Line 3,746:  
'''Generate a Poisson Process'''<br />  '''Generate a Poisson Process'''<br />  
−  
−  
−  
1. set <math>T_{0}=0</math> and n=1<br/>  1. set <math>T_{0}=0</math> and n=1<br/>  
Line 3,721:  Line 3,827:  
</pre>  </pre>  
−  +  <br>  
The following plot is using TT = 50.<br>  The following plot is using TT = 50.<br>  
The number of points generated every time on average should be <math>\lambda</math> * TT. <br>  The number of points generated every time on average should be <math>\lambda</math> * TT. <br>  
The maximum value of the points should be TT. <br>  The maximum value of the points should be TT. <br>  
−  [[File:Poisson.jpg]]  +  [[File:Poisson.jpg]]<br> 
when TT be big, the plot of the graph will be linear, when we set the TT be 5 or small number, the plot graph looks like discrete distribution.  when TT be big, the plot of the graph will be linear, when we set the TT be 5 or small number, the plot graph looks like discrete distribution.  
Line 3,833:  Line 3,939:  
=== Examples of Transition Matrix ===  === Examples of Transition Matrix ===  
−  [[File:Mark13.png]]  +  [[File:Mark13.png]]<br> 
The picture is from http://www.google.ca/imgres?imgurl=http://academic.uprm.edu/wrolke/esma6789/graphs/mark13.png&imgrefurl=http://academic.uprm.edu/wrolke/esma6789/mark1.htm&h=274&w=406&sz=5&tbnid=6A8GGaxoPux9kM:&tbnh=83&tbnw=123&prev=/search%3Fq%3Dtransition%2Bmatrix%26tbm%3Disch%26tbo%3Du&zoom=1&q=transition+matrix&usg=__hZR1Cp6PbZ5PfnSjs2zU6LnCiI=&docid=PaQvi1F97P2urM&sa=X&ei=foTxUY3DBrMyQGvq4D4Cg&sqi=2&ved=0CDYQ9QEwAQ&dur=5515)  The picture is from http://www.google.ca/imgres?imgurl=http://academic.uprm.edu/wrolke/esma6789/graphs/mark13.png&imgrefurl=http://academic.uprm.edu/wrolke/esma6789/mark1.htm&h=274&w=406&sz=5&tbnid=6A8GGaxoPux9kM:&tbnh=83&tbnw=123&prev=/search%3Fq%3Dtransition%2Bmatrix%26tbm%3Disch%26tbo%3Du&zoom=1&q=transition+matrix&usg=__hZR1Cp6PbZ5PfnSjs2zU6LnCiI=&docid=PaQvi1F97P2urM&sa=X&ei=foTxUY3DBrMyQGvq4D4Cg&sqi=2&ved=0CDYQ9QEwAQ&dur=5515)  
Line 3,869:  Line 3,975:  
</div>  </div>  
−  <math>x_k+1= (ax_k+c) mod</math> <math>m</math><br />  +  <math>\begin{align}x_k+1= (ax_k+c) \mod m\end{align}</math><br /> 
+  
+  Where a, c, m and x<sub>1</sub> (the seed) are values we must chose before running the algorithm. While there is no set value for each, it is best for m to be large and prime. For example, Matlab uses a = 75,b = 0,m = 231 − 1.  
+  
+  '''Examples:'''<br>  
+  1. <math>\begin{align}X_{0} = 10 ,a = 2 , c = 1 , m = 13 \end{align}</math><br>  
+  
+  <math>\begin{align}X_{1} = 2 * 10 + 1\mod 13 = 8\end{align}</math><br>  
+  
+  <math>\begin{align}X_{2} = 2 * 8 + 1\mod 13 = 4\end{align}</math> ... and so on<br>  
−  
−  +  2. <math>\begin{align}X_{0} = 44 ,a = 13 , c = 17 , m = 211\end{align}</math><br>  
−  
−  +  <math>\begin{align}X_{1} = 13 * 44 + 17\mod 211 = 167\end{align}</math><br>  
−  
−  
−  +  <math>\begin{align}X_{2} = 13 * 167 + 17\mod 211 = 78\end{align}</math><br>  
−  +  
−  +  <math>\begin{align}X_{3} = 13 * 78 + 17\mod 211 = 187\end{align}</math> ... and so on<br>  
−  
−  
−  
=== Inverse Transformation Method ===  === Inverse Transformation Method ===  
Line 4,258:  Line 4,366:  
The vector <math>\underline{\mu_0}</math> is called the initial distribution. <br/>  The vector <math>\underline{\mu_0}</math> is called the initial distribution. <br/>  
−  <math>  +  <math> P^2~=P\cdot P </math> (as verified above) 
In general,  In general,  
−  <math>  +  <math> P^n~= \Pi_{i=1}^{n} P</math> (P multiplied n times)<br/> 
−  <math>\mu_n~=\mu_0  +  <math>\mu_n~=\mu_0 P^n</math><br/> 
where <math>\mu_0</math> is the initial distribution,  where <math>\mu_0</math> is the initial distribution,  
−  and <math>\mu_{m+n}~=\mu_m  +  and <math>\mu_{m+n}~=\mu_m P^n</math><br/> 
N can be negative, if P is invertible.  N can be negative, if P is invertible.  
Line 4,297:  Line 4,405:  
−  <math>\pi</math> is stationary distribution of the chain if <math>\pi</math>P = <math>\pi</math>  +  <math>\pi</math> is stationary distribution of the chain if <math>\pi</math>P = <math>\pi</math> In other words, a stationary distribution is when the markov process that have equal probability of moving to other states as its previous move. 
where <math>\pi</math> is a probability vector <math>\pi</math>=(<math>\pi</math><sub>i</sub>  <math>i \in X</math>) such that all the entries are nonnegative and sum to 1. It is the eigenvector in this case.  where <math>\pi</math> is a probability vector <math>\pi</math>=(<math>\pi</math><sub>i</sub>  <math>i \in X</math>) such that all the entries are nonnegative and sum to 1. It is the eigenvector in this case.  
Line 4,304:  Line 4,412:  
The above conditions are used to find the stationary distribution  The above conditions are used to find the stationary distribution  
+  In matlab, we could use <math>P^n</math> to find the stationary distribution.(n is usually larger than 100)<br/>  
+  
'''Comments:'''<br/>  '''Comments:'''<br/>  
Line 4,511:  Line 4,621:  
<math>\displaystyle \pi=(\frac{1}{3},\frac{4}{9}, \frac{2}{9})</math>  <math>\displaystyle \pi=(\frac{1}{3},\frac{4}{9}, \frac{2}{9})</math>  
−  <math>\displaystyle \lambda u=A u</math>  +  Note that <math>\displaystyle \pi=\pi p</math> looks similar to eigenvectors/values <math>\displaystyle \lambda vec{u}=A vec{u}</math> 
−  <math>\pi</math> can be considered as an eigenvector of P with eigenvalue = 1.  +  <math>\pi</math> can be considered as an eigenvector of P with eigenvalue = 1. But note that the vector <math>vec{u}</math> is a column vector and o we need to transform our <math>\pi</math> into a column vector. 
−  But the vector u  
−  <math>\pi</math><sup>T</sup>= P<sup>T</sup><math>\pi</math><sup>T</sup>  +  <math>=> \pi</math><sup>T</sup>= P<sup>T</sup><math>\pi</math><sup>T</sup><br/> 
Then <math>\pi</math><sup>T</sup> is an eigenvector of P<sup>T</sup> with eigenvalue = 1. <br />  Then <math>\pi</math><sup>T</sup> is an eigenvector of P<sup>T</sup> with eigenvalue = 1. <br />  
MatLab tips:[V D]=eig(A), where D is a diagonal matrix of eigenvalues and V is a matrix of eigenvectors of matrix A<br />  MatLab tips:[V D]=eig(A), where D is a diagonal matrix of eigenvalues and V is a matrix of eigenvectors of matrix A<br />  
==== MatLab Code ====  ==== MatLab Code ====  
+  <pre style='fontsize:14px'>  
+  
+  P = [1/3 1/3 1/3; 1/4 3/4 0; 1/2 0 1/2]  
−  =  +  pii = [1/3 4/9 2/9] 
−  
−  +  [vec val] = eig(P') %% P' is the transpose of matrix P  
−  +  
−  +  vec(:,1) = [0.5571 0.7428 0.3714] %% this is in column form  
−  
−  
−  +  a = vec(:,1)  
−  +  >> a =  
+  [0.5571 0.7428 0.3714]  
−  If the limiting distribution <math>\pi</math> exists, it must be equal to the stationary distribution.<br/>  +  %% a is in column form 
+  
+  %% Since we want this vector a to sum to 1, we have to scale it  
+  
+  b = a/sum(a)  
+  
+  >> b =  
+  [0.3333 0.4444 0.2222]  
+  
+  %% b is also in column form  
+  
+  %% Observe that b' = pii  
+  
+  </pre>  
+  </br>  
+  ==== Limiting distribution ====  
+  A Markov chain has limiting distribution <math>\pi</math> if  
+  
+  <math>\lim_{n\to \infty} P^n= \left[ {\begin{array}{ccc}  
+  \pi_1 \\  
+  \vdots \\  
+  \pi_n \\  
+  \end{array} } \right]</math>  
+  
+  That is <math>\pi_j=\lim[P^n]_{ij}</math> exists and is independent of i.<br/>  
+  
+  A Markov Chain is convergent if and only if its limiting distribution exists. <br/>  
+  
+  If the limiting distribution <math>\pi</math> exists, it must be equal to the stationary distribution.<br/>  
This convergence means that,in the long run(n to infinity),the probability of finding the <br/>  This convergence means that,in the long run(n to infinity),the probability of finding the <br/>  
Line 4,548:  Line 4,686:  
, find stationary distribution.<br/>  , find stationary distribution.<br/>  
We have:<br/>  We have:<br/>  
−  <math>0  +  <math>0\times \pi_0+0\times \pi_1+1\times \pi_2=\pi_0</math><br/> 
−  <math>1  +  <math>1\times \pi_0+0\times \pi_1+0\times \pi_2=\pi_1</math><br/> 
−  <math>0  +  <math>0\times \pi_0+1\times \pi_1+0\times \pi_2=\pi_2</math><br/> 
−  <math>\pi_0+\pi_1+\pi_2=1</math><br/>  +  <math>\,\pi_0+\pi_1+\pi_2=1</math><br/> 
this gives <math>\pi = \left [ \begin{matrix}  this gives <math>\pi = \left [ \begin{matrix}  
\frac{1}{3} & \frac{1}{3} & \frac{1}{3} \\[6pt]  \frac{1}{3} & \frac{1}{3} & \frac{1}{3} \\[6pt]  
Line 4,559:  Line 4,697:  
In general, there are chains with stationery distributions that don't converge, this means that they have stationary distribution but are not limiting.<br/>  In general, there are chains with stationery distributions that don't converge, this means that they have stationary distribution but are not limiting.<br/>  
+  === MatLab Code ===  
+  <pre style='fontsize:14px'>  
+  MATLAB  
+  >> P=[0, 1, 0;0, 0, 1; 1, 0, 0]  
−  +  P =  
−  +  0 1 0  
−  +  0 0 1  
−  +  1 0 0  
−  0  
−  0  
−  
−  +  >> pii=[1/3, 1/3, 1/3]  
−  
−  
−  
−  
−  
−  
−  
−  
−  
−  
−  
−  
−  
−  
+  pii =  
−  +  0.3333 0.3333 0.3333  
−  
−  +  >> pii*P  
−  +  ans =  
−  +  0.3333 0.3333 0.3333  
−  +  >> P^1000  
−  +  ans =  
−  +  0 1 0  
+  0 0 1  
+  1 0 0  
−  +  >> P^10000  
−  
−  >  
−  +  ans =  
−  0  +  0 1 0 
−  0  +  0 0 1 
−  0  +  1 0 0 
−  >> P^  +  >> P^10002 
ans =  ans =  
−  0  +  1 0 0 
−  0  +  0 1 0 
−  +  0 0 1  
−  >> P^  +  >> P^10003 
ans =  ans =  
−  0  +  0 1 0 
−  0  +  0 0 1 
−  0  +  1 0 0 
−  >> P^  +  >> %P^10000 = P^10003 
+  >> % This chain does not have limiting distribution, it has a stationary distribution.  
−  +  This chain does not converge, it has a cycle.  
+  </pre>  
−  +  The first condition of limiting distribution is satisfied; however, the second condition where <math>\pi</math><sub>j</sub> has to be independent of i (i.e. all rows of the matrix are the same) is not met.<br>  
−  +  This example shows the distinction between having a stationary distribution and convergence(having a limiting distribution).Note: <math>\pi=(1/3,1/3,1/3)</math> is the stationary distribution as <math>\pi=\pi*p</math>. However, upon repeatedly multiplying P by itself (repeating the step <math>P^n</math> as n goes to infinite) one will note that the results become a cycle (of period 3) of the same sequence of matrices. The chain has a stationary distribution, but does not converge to it. Thus, there is no limiting distribution.<br>  
−  
−  
−  +  '''Example:'''  
−  +  <math> P= \left [ \begin{matrix}  
+  \frac{4}{5} & \frac{1}{5} & 0 & 0 \\[6pt]  
+  \frac{1}{5} & \frac{4}{5} & 0 & 0 \\[6pt]  
+  0 & 0 & \frac{4}{5} & \frac{1}{5} \\[6pt]  
+  0 & 0 & \frac{1}{10} & \frac{9}{10} \\[6pt]  
+  \end{matrix} \right] </math>  
−  +  This chain converges but is not a limiting distribution as the rows are not the same and it doesn't converge to the stationary distribution.<br />  
−  +  <br />  
−  +  Double Stichastic Matrix: a double stichastic matrix is a matrix whose all colums sum to 1 and all rows sum to 1.<br />  
+  If a given transition matrix is a double stichastic matrix with n colums and n rows, then the stationary distribution matrix has all<br/>  
+  elements equals to 1/n.<br/>  
+  <br/>  
+  Example:<br/>  
+  For a stansition matrix <math> P= \left [ \begin{matrix}  
+  0 & \frac{1}{2} & \frac{1}{2} \\[6pt]  
+  \frac{1}{2} & 0 & \frac{1}{2} \\[6pt]  
+  \frac{1}{2} & \frac{1}{2} & 0 \\[6pt]  
+  \end{matrix} \right] </math>,<br/>  
+  We have:<br/>  
+  <math>0\times \pi_0+\frac{1}{2}\times \pi_1+\frac{1}{2}\times \pi_2=\pi_0</math><br/>  
+  <math>\frac{1}{2}\times \pi_0+0\times \pi_1+\frac{1}{2}\times \pi_2=\pi_1</math><br/>  
+  <math>\frac{1}{2}\times \pi_0+\frac{1}{2}\times \pi_1+0\times \pi_2=\pi_2</math><br/>  
+  <math>\pi_0+\pi_1+\pi_2=1</math><br/>  
+  The stationary distribution is <math>\pi = \left [ \begin{matrix}  
+  \frac{1}{3} & \frac{1}{3} & \frac{1}{3} \\[6pt]  
+  \end{matrix} \right] </math> <br/>  
−  >>  +  <span style="fontsize:20px;color:red">The following contents are problematic. Please correct it if possible.</span><br /> 
+  Suppose we're given that the limiting distribution <math> \pi </math> exists for stochastic matrix P, that is, <math> \pi = \pi \times P </math> <br>  
−  +  WLOG assume P is diagonalizable, (if not we can always consider the Jordan form and the computation below is exactly the same. <br>  
−  +  Let <math> P = U \Sigma U^{1} </math> be the eigenvalue decomposition of <math> P </math>, where <math>\Sigma = diag(\lambda_1,\ldots,\lambda_n) ; \lambda_i > \lambda_j, \forall i < j </math><br>  
−  
−  
+  Suppose <math> \pi^T = \sum a_i u_i </math> where <math> a_i \in \mathcal{R} </math> and <math> u_i </math> are eigenvectors of <math> P </math> for <math> i = 1\ldots n </math> <br>  
−  +  By definition: <math> \pi^k = \pi P = \pi P^k \implies \pi = \pi(U \Sigma U^{1}) (U \Sigma U^{1} ) \ldots (U \Sigma U^{1}) </math> <br>  
−  +  Therefore <math> \pi^k = \sum a_i \lambda_i^k u_i </math> since <math> <u_i , u_j> = 0, \forall i\neq j </math>. <br>  
−  
−  
−  >  +  Therefore <math> \lim_{k \rightarrow \infty} \pi^k = \lim_{k \rightarrow \infty} \lambda_i^k a_1 u_1 = u_1 </math> 
−  a  +  === MatLab Code === 
+  <pre style='fontsize:14px'>  
+  >> P=[1/3, 1/3, 1/3; 1/4, 3/4, 0; 1/2, 0, 1/2] % We input a matrix P. This is the same matrix as last class.  
−  +  P =  
−  
−  
−  +  0.3333 0.3333 0.3333  
+  0.2500 0.7500 0  
+  0.5000 0 0.5000  
+  
+  >> P^2  
ans =  ans =  
−  +  0.3611 0.3611 0.2778  
+  0.2708 0.6458 0.0833  
+  0.4167 0.1667 0.4167  
−  >>  +  >> P^3 
ans =  ans =  
−  0.  +  0.3495 0.3912 0.2593 
−  0.  +  0.2934 0.5747 0.1319 
−  0.  +  0.3889 0.2639 0.3472 
−  
−  +  >> P^10  
−  +  The example of code and an example of stand distribution, then the all the pi probability in the matrix are the same.  
−  
−  
−  
−  +  ans =  
−  +  0.3341 0.4419 0.2240  
+  0.3314 0.4507 0.2179  
+  0.3360 0.4358 0.2282  
−  The  +  >> P^100 % The stationary distribution is [0.3333 0.4444 0.2222] since values keep unchanged. 
−  +  ans =  
−  
−  
−  
−  +  0.3333 0.4444 0.2222  
−  +  0.3333 0.4444 0.2222  
−  +  0.3333 0.4444 0.2222  
−  
−  +  >> [vec val]=eigs(P') % We can find the eigenvalues and eigenvectors from the transpose of matrix P.  
−  
−  
−  >>  
−  +  vec =  
−  +  0.5571 0.2447 0.8121  
−  +  0.7428 0.7969 0.3324  
−  +  0.3714 0.5523 0.4797  
−  
−  +  val =  
−  0  +  1.0000 0 0 
+  0 0.6477 0  
+  0 0 0.0643  
−  >>  +  >> a=vec(:,1) % The eigenvectors can be mutiplied by (1) since λV=AV can be written as λ(V)=A(V) 
−  +  a =  
−  0.  +  0.5571 
+  0.7428  
+  0.3714  
−  >>  +  >> sum(a) 
ans =  ans =  
−  +  1.6713  
−  
−  
−  >>  +  >> a/sum(a) 
ans =  ans =  
−  +  0.3333  
−  +  0.4444  
−  +  0.2222  
−  
−  
−  
−  
−  
−  
−  
−  
−  
−  
−  
−  
−  
−  
−  
−  
−  
−  
−  
−  
−  
</pre>  </pre>  
−  +  This is <math>\pi_j = lim[p^n]_(ij)</math> exist and is independent of i  
−  
−  This  
Another example:  Another example:  
Line 4,801:  Line 4,921:  
'''Note:'''if there's a finite number N then every other state can be reached in N steps.  '''Note:'''if there's a finite number N then every other state can be reached in N steps.  
−  '''Note:'''Also note that a Ergodic chain is irreducible (all states communicate) and aperiodic (d = 1). An Ergodic chain is promised to have a stationary and limiting distribution.  +  '''Note:'''Also note that a Ergodic chain is irreducible (all states communicate) and aperiodic (d = 1). An Ergodic chain is promised to have a stationary and limiting distribution.<br/> 
+  '''Ergodicity:''' A state i is said to be ergodic if it is aperiodic and positive recurrent. In other words, a state i is ergodic if it is recurrent, has a period of 1 and it has finite mean recurrence time. If all states in an irreducible Markov chain are ergodic, then the chain is said to be ergodic.<br/>  
+  '''Some more:'''It can be shown that a finite state irreducible Markov chain is ergodic if it has an aperiodic state. A model has the ergodic property if there's a finite number N such that any state can be reached from any other state in exactly N steps. In case of a fully connected transition matrix where all transitions have a nonzero probability, this condition is fulfilled with N=1.<br/>  
Line 4,871:  Line 4,993:  
<math> \pi_0 = \frac{4}{19} </math> <br>  <math> \pi_0 = \frac{4}{19} </math> <br>  
<math> \pi = [\frac{4}{19}, \frac{15}{19}] </math> <br>  <math> \pi = [\frac{4}{19}, \frac{15}{19}] </math> <br>  
−  <math> \pi </math> is the long run distribution  +  <math> \pi </math> is the long run distribution, and this is also a limiting distribution. 
We can use the stationary distribution to compute the expected waiting time to return to state 'a' <br/>  We can use the stationary distribution to compute the expected waiting time to return to state 'a' <br/>  
Line 4,878:  Line 5,000:  
state 'a' given that we start at state 'a' is 19/4.<br/>  state 'a' given that we start at state 'a' is 19/4.<br/>  
−  definition of limiting distribution.  +  definition of limiting distribution: when the stationary distribution is convergent, it is a limiting distribution.<br/> 
remark：satisfied balance of <math>\pi_i P_{ij} = P_{ji} \pi_j</math>, so there is other way to calculate the step probability.  remark：satisfied balance of <math>\pi_i P_{ij} = P_{ji} \pi_j</math>, so there is other way to calculate the step probability.  
Line 5,003:  Line 5,125:  
<math>\pi_2 P_{2,3} = 4/9 \times 0 = 0,\, P_{3,2} \pi_3 = 0 \times 2/9 = 0 \Rightarrow \pi_2 P_{2,3} = P_{3,2} \pi_3</math><br>  <math>\pi_2 P_{2,3} = 4/9 \times 0 = 0,\, P_{3,2} \pi_3 = 0 \times 2/9 = 0 \Rightarrow \pi_2 P_{2,3} = P_{3,2} \pi_3</math><br>  
−  Remark：Detailed balance of <math> \pi_i  +  Remark：Detailed balance of <math> \pi_i \times Pij = Pji \times \pi_j</math> ， so there is other way to calculate the step probability<br /> 
<math>\pi</math> is stationary but is not limiting.  <math>\pi</math> is stationary but is not limiting.  
Detailed balance implies that <math>\pi</math> = <math>\pi</math> * P as shown in the proof and guarantees that <math>\pi</math> is stationary distribution.  Detailed balance implies that <math>\pi</math> = <math>\pi</math> * P as shown in the proof and guarantees that <math>\pi</math> is stationary distribution.  
Line 5,036:  Line 5,158:  
=== PageRank (http://en.wikipedia.org/wiki/PageRank) ===  === PageRank (http://en.wikipedia.org/wiki/PageRank) ===  
+  *PageRank is a probability distribution used to represent the likelihood that a person randomly clicking on links will arrive at any particular page. PageRank can be calculated for collections of documents of any size.  
*PageRank is a linkanalysis algorithm developed by and named after Larry Page from Google; used for measuring a website's importance, relevance and popularity.  *PageRank is a linkanalysis algorithm developed by and named after Larry Page from Google; used for measuring a website's importance, relevance and popularity.  
*PageRank is a graph containing web pages and their links to each other.  *PageRank is a graph containing web pages and their links to each other.  
Line 5,043:  Line 5,166:  
<br />'''The order of importance'''<br />  <br />'''The order of importance'''<br />  
−  1. A web page is important if many other pages point to it<br />  +  1. A web page is more important if many other pages point to it<br /> 
2. The more important a web page is, the more weight should be assigned to its outgoing links<br/ >  2. The more important a web page is, the more weight should be assigned to its outgoing links<br/ >  
3. If a webpage has many outgoing links, then its links have less value (ex: if a page links to everyone, like 411, it is not as important as pages that have incoming links)<br />  3. If a webpage has many outgoing links, then its links have less value (ex: if a page links to everyone, like 411, it is not as important as pages that have incoming links)<br />  
Line 5,074:  Line 5,197:  
<br />  <br />  
−  +  <math>C_j=</math> The number of outgoing links of page <math>j</math>:  
<math>C_j=\sum_i L_{ij}</math>  <math>C_j=\sum_i L_{ij}</math>  
(i.e. sum of entries in column j)<br />  (i.e. sum of entries in column j)<br />  
Line 5,085:  Line 5,208:  
<math>P_i=\sum_j L_{ij}</math> <br />(i.e. sum of entries in row i)  <math>P_i=\sum_j L_{ij}</math> <br />(i.e. sum of entries in row i)  
−  +  For each row of <math>L</math>, if there is a 1 in the third column, it means page three point to that page.  
+  
+  However, we should not define the rank of the page this way because links shouldn't be treated the same. The weight of the link is based on different factors. One of the factors is the importance of the page that link is coming from. For example, in this case, there are two links going to Page 4: one from Page 2 and one from Page 5. So far, both links have been treated equally with the same weight 1. But we must rerate the two links based on the importance of the pages they are coming from.  
A PageRank results from a mathematical algorithm based on the webgraph, created by all World Wide Web pages as nodes and hyperlinks as edges, taking into consideration authority hubs such as cnn.com or usa.gov. The rank value indicates an importance of a particular page. A hyperlink to a page counts as a vote of support. (This would be represented in our diagram as an arrow pointing towards the page. Hence in our example, Page 3 is the most important, since it has the most 'votes of support). The PageRank of a page is defined recursively and depends on the number and PageRank metric of all pages that link to it ("incoming links"). A page that is linked to by many pages with high PageRank receives a high rank itself. If there are no links to a web page, then there is no support for that page (In our example, this would be Page 1 and Page 5).  A PageRank results from a mathematical algorithm based on the webgraph, created by all World Wide Web pages as nodes and hyperlinks as edges, taking into consideration authority hubs such as cnn.com or usa.gov. The rank value indicates an importance of a particular page. A hyperlink to a page counts as a vote of support. (This would be represented in our diagram as an arrow pointing towards the page. Hence in our example, Page 3 is the most important, since it has the most 'votes of support). The PageRank of a page is defined recursively and depends on the number and PageRank metric of all pages that link to it ("incoming links"). A page that is linked to by many pages with high PageRank receives a high rank itself. If there are no links to a web page, then there is no support for that page (In our example, this would be Page 1 and Page 5).  
Line 5,120:  Line 5,245:  
<math>P_i= (1d) + d\cdot \sum_j \frac {L_{ji}P_j}{c_j}</math>  <math>P_i= (1d) + d\cdot \sum_j \frac {L_{ji}P_j}{c_j}</math>  
+  pi is the rank of a new created page(that no one knows about) is 0 since <math>L_ij</math> is 0 <br/>  
+  where 0 < d < 1 is constant (in original page rank algorithm d = 0.8), and <math>L_{ij}</math> is 1 if j has link to i, 0 otherwise.  
−  +  Note that the rank of a page is proportional to the number of its incoming links and inversely proportional to the number of its outgoing links.  
Interpretation of the formula:<br/>  Interpretation of the formula:<br/>  
Line 5,129:  Line 5,256:  
4) finally, we take a linear combination of the page rank obtained from above and a constant 1. This ensures that every page has a rank greater than zero.<br/>  4) finally, we take a linear combination of the page rank obtained from above and a constant 1. This ensures that every page has a rank greater than zero.<br/>  
5) d is the damping factor. It represents the probability a user, at any page, will continue clicking to another page.<br/>  5) d is the damping factor. It represents the probability a user, at any page, will continue clicking to another page.<br/>  
+  If there is no damping (i.e. d=1), then there are no assumed outgoing links for nodes with no links. However, if there is damping (e.g. d=0.8), then these nodes are assumed to have links to all pages in the web.  
Note that this is a system of N equations with N unknowns.<br/>  Note that this is a system of N equations with N unknowns.<br/>  
Line 5,146:  Line 5,274:  
0 & 0 & ... & c_N \end{matrix} } \right]</math>  0 & 0 & ... & c_N \end{matrix} } \right]</math>  
−  Then <math>P=~(1d)e+dLD^{1}P</math><br/> where e =[1 1 ....]<sup>T</sup> , i.e. a N by 1 vector.<br/>  +  Then <math>P=~(1d)e+dLD^{1}P</math>, P is an iegenvector of matrix A corresponding to an eigenvalue equal to 1.<br/> where e =[1 1 ....]<sup>T</sup> , i.e. a N by 1 vector.<br/> 
We assume that rank of all N pages sums to N. The sum of rank of all N pages can be any number, as long as the ranks have certain propotion. <br/>  We assume that rank of all N pages sums to N. The sum of rank of all N pages can be any number, as long as the ranks have certain propotion. <br/>  
i.e. e<sup>T</sup> P = N, then <math>~\frac{e^{T}P}{N} = 1</math>  i.e. e<sup>T</sup> P = N, then <math>~\frac{e^{T}P}{N} = 1</math>  
Line 5,170:  Line 5,298:  
<math>P=[(1d)~\frac{ee^T}{N}+dLD^{1}]P</math>  <math>P=[(1d)~\frac{ee^T}{N}+dLD^{1}]P</math>  
+  
+  <math>=> P=A*P</math>  
'''Explanation of an eigenvector'''  '''Explanation of an eigenvector'''  
Line 5,367:  Line 5,497:  
<math>1 \leftrightarrow 2 \rightarrow 3 \leftrightarrow 4 </math>  <math>1 \leftrightarrow 2 \rightarrow 3 \leftrightarrow 4 </math>  
−  
<br />  <br />  
<br />  <br />  
Line 5,375:  Line 5,504:  
1 & 0 & 0 & 0 \\  1 & 0 & 0 & 0 \\  
0 & 1 & 0 & 1 \\  0 & 1 & 0 & 1 \\  
−  0 & 0 & 1 & 0 \end{matrix} } \right]\;  +  0 & 0 & 1 & 0 \end{matrix} } \right]\;</math><br /> 
−  
−  
−  
−  
−  
−  
−  
−  
−  
−  
−  
−  
+  '''Matlab Code:'''<br>  
+  <pre style='fontsize:16px'>  
>> L=L= [0 1 0 0;1 0 0 0;0 1 0 1;0 0 1 0];  >> L=L= [0 1 0 0;1 0 0 0;0 1 0 1;0 0 1 0];  
>> C=sum(L);  >> C=sum(L);  
Line 5,394:  Line 5,513:  
>> d=0.8;  >> d=0.8;  
>> N=4;  >> N=4;  
−  >> A=(1d)*ones(N)/N+d*L*pinv(D)  +  >> A=(1d)*ones(N)/N+d*L*pinv(D); 
−  +  >> [vec val]=eigs(A);  
−  +  >> a=vec(:,1);  
−  
−  
−  
−  
−  
−  
−  >> [vec val]=eigs(A)  
−  
−  
−  
−  
−  
−  
−  
−  
−  
−  
−  
−  
−  
−  
−  
−  
−  
−  
−  >> a=vec(:,1)  
−  
−  
−  
−  
−  
−  
−  
−  
>> a=a/sum(a)  >> a=a/sum(a)  
+  a =  
+  0.1029 < Page 1  
+  0.1324 < Page 2  
+  0.3971 < Page 3  
+  0.3676 < Page 4  
−  +  % Therefore the PageRank for this matrix is: 3,4,2,1  
−  
−  
−  
−  
−  
</pre>  </pre>  
−  +  <br>  
−  
−  
==== Example 5 ====  ==== Example 5 ====  
Line 5,496:  Line 5,579:  
<br />  <br />  
−  Matlab Code<br />  +  '''Matlab Code:'''<br /> 
<pre style="fontsize:16px">  <pre style="fontsize:16px">  
−  >> d=0.8  +  >> d=0.8; 
+  >> L=[0 1 0 0 1;1 0 0 0 0;0 1 0 0 0;0 1 1 0 1;0 0 0 1 0];  
+  >> c=sum(L);  
+  >> D=diag(c);  
+  >> N=5;  
+  >> A=(1d)*ones(N)/N+d*L*pinv(D);  
+  >> [vec val]=eigs(A);  
+  >> a=vec(:,1);  
+  >> a=a/sum(a)  
+  a =  
+  0.1933 < Page 1  
+  0.1946 < Page 2  
+  0.0919 < Page 3  
+  0.2668 < Page 4  
+  0.2534 < Page 5  
−  +  % Therefore the PageRank for this matrix is: 4,5,2,1,3  
+  </pre>  
+  <br>  
−  +  == Class 17  Tuesday July 2nd 2013 ==  
+  === Markov Chain Monte Carlo (MCMC) ===  
−  +  ===Introduction===  
+  It is, in general, very difficult to simulate the value of a random vector X whose component random variables are dependent. We will present a powerful approach for generating a vector whose distribution is approximately that of X. This approach, called the Markov Chain Monte Carlo Methods, has the added significance of only requiring that the mass(or density) function of X be specified up to a multiplicative constant, and this, we will see, is of great importance in applications.  
+  (referenced by Sheldon M.Ross,Simulation)  
+  The basic idea used here is to generate a Markov Chain whose stationary distribution is the same as the target distribution.  
−  +  ====Definition:====  
+  Markov Chain  
+  A Markov Chain is a special form of stochastic process in which <math>\displaystyle X_t</math> depends only on <math> \displaystyle X_{t1}</math>.  
−  +  For example,  
−  +  :<math>\displaystyle f(X_1,...X_n)= f(X_1)f(X_2X_1)f(X_3X_2)...f(X_nX_{n1})</math>  
−  +  A random Walk is the best example of a Markov process  
−  
−  
−  >>  +  <br>'''Transition Probability:'''<br> 
+  The probability of going from one state to another state.  
+  :<math>p_{ij} = \Pr(X_{n}=j\mid X_{n1}= i). \,</math>  
−  +  <br>'''Transition Matrix:'''<br>  
+  For n states, transition matrix P is an <math>N \times N</math> matrix with entries <math>\displaystyle P_{ij}</math> as below:  
+  Markov Chain Monte Carlo (MCMC) methods are a class of algorithms for sampling from probability distributions based on constructing a Markov chain that has the desired distribution as its equilibrium distribution. The state of the chain after a large number of steps is then used as a sample of the desired distribution. The quality of the sample improves as a function of the number of steps. (http://en.wikipedia.org/wiki/Markov_chain_Monte_Carlo)</span>  
−  +  <a style="color:red" href="http://www.eecs.berkeley.edu/Pubs/TechRpts/2010/EECS2010165.pdf">some notes form UCb</a>  
−  +  '''One of the main purposes of MCMC''' : to simulate samples from a joint distribution where the joint random variables are dependent. In general, this is not easily sampled from. Other methods learned in class allow us to simulate i.i.d random variables, but not dependent variables . In this case, we could sample nonindependent random variables using a Markov Chain. Its Markov properties help to simplify the simulation process.  
−  
−  +  <b>Basic idea:</b> Given a probability distribution <math>\pi</math> on a set <math>\Omega</math>, we want to generate random elements of <math>\Omega</math> with distribution <math>\pi</math>. MCMC does that by constructing a Markov Chain with stationary distribution <math>\pi</math> and simulating the chain. After a large number of iterations, the Markov Chain will reach its stationary distribution. By sampling from the Markov chain for large amount of iterations, we are effectively sampling from the desired distribution as the Markov Chain would converge to its stationary distribution <br/>  
−  
−  
−  
−  
−  >  +  Idea: generate a Markov chain whose stationary distribution is the same as target distribution. <br/> 
−  
−  +  '''Notes'''  
−  +  # Regardless of the chosen starting point, the Markov Chain will converge to its stationary distribution (if it exists). However, the time taken for the chain to converge depends on its chosen starting point. Typically, the burnin period is longer if the chain is initialized with a value of low probability density.  
+  # Markov Chain Monte Carlo can be used for sampling from a distribution, estimating the distribution, and computing the mean and optimization (e.g. simulated annealing, more on that later).  
+  # Markov Chain Monte Carlo is used to sample using “local” information. It is used as a generic “problem solving technique” to solve decision/optimization/value problems, but is not necessarily very efficient.  
+  # MCMC methods do not suffer as badly from the "curse of dimensionality" that badly affects efficiency in the acceptancerejection method. This is because a point is always generated at each timestep according to the Markov Chain regardless of how many dimensions are introduced.  
+  # The goal when simulating with a Markov Chain is to create a chain with the same stationary distribution as the target distribution.  
+  # The MCMC method is usually used in continuous cases but a discrete example is given below.  
−  
−  +  '''Some properties of the stationary distribution <math>\pi</math>'''  
−  
−  
−  
−  
−  >>  +  <math>\pi</math> indicates the proportion of time the process spends in each of the states 1,2,...,n. Therefore <math>\pi</math> satisfies the following two inequalities: <br> 
−  +  # <math>\pi_j = \sum_{i=1}^{n}\pi_i P_{ij}</math> <br /> This is because <math>\pi_i</math> is the proportion of time the process spends in state i, and <math>P_{ij}</math> is the probability the process transition out of state i into state j. Therefore, <math>\pi_i p_{ij}</math> is the proportion of time it takes for the process to enter state j. Therefore, <math>\pi_j</math> is the sum of this probability over overall states i.  
+  #<math> \sum_{i=1}^{n}\pi_i= 1 </math> as <math>\pi</math> shows the proportion of time the chain is in each state. If we view it as the probability of the chain being in state i at time t for t sufficiently large, then it should sum to one as the chain must be in one of the states.  
−  +  ====Motivation example====  
+   Suppose we want to generate a random variable X according to distribution <math>\pi=(\pi_1, \pi_2, ... , \pi_m)</math> <br/>  
+  X can take m possible different values from <math>{1,2,3,\cdots, m}</math><br />  
+   We want to generate <math>\{X_t: t=0, 1, \cdots\}</math> according to <math>\pi</math><br />  
−  +  Suppose our example is of a bias die. <br/>  
−  +  Now we have m=6, <math>\pi=[0.1,0.1,0.1,0.2,0.3,0.2]</math>, <math>X \in [1,2,3,4,5,6]</math><br/>  
−  
−  
−  
−  +  Suppose <math>X_t=i</math>. Consider an arbitrary probability transition matrix Q with entry <math>q_{ij}</math> being the probability of moving to state j from state i. (<math>q_{ij}</math> can not be zero.) <br/>  
−  +  <math> \mathbf{Q} =  
−  +  \begin{bmatrix}  
−  +  q_{11} & q_{12} & \cdots & q_{1m} \\  
−  +  q_{21} & q_{22} & \cdots & q_{2m} \\  
−  +  \vdots & \vdots & \ddots & \vdots \\  
+  q_{m1} & q_{m2} & \cdots & q_{mm}  
+  \end{bmatrix}  
+  </math> <br/>  
−  +  We generate Y = j according to the ith row of Q. Note that the ith row of Q is a probability vector that shows the probability of moving to any state j from the current state i, i.e.<math>P(Y=j)=q_{ij}</math><br />  
−  +  In the following algorithm: <br>  
+  <math>q_{ij}</math> is the <math>ij^{th}</math> entry of matrix Q. It is the probability of Y=j given that <math>x_t = i</math>. <br/>  
+  <math>r_{ij}</math> is the probability of accepting Y as <math>x_{t+1}</math>. <br/>  
−  
−  
−  
−  
−  
−  +  '''How to get the acceptance probability?'''  
−  +  If <math>\pi </math> is the stationary distribution, then it must satisfy the detailed balance condition:<br/>  
−  +  If <math>\pi_i P_{ij}</math> = <math>\pi_j P_{ji}</math><br/>then <math>\pi </math> is the stationary distribution of the chain  
−  
−  
−  
−  >> a=  +  Since <math>P_{ij}</math> = <math>q_{ij} r_{ij}</math>, we have <math>\pi_i q_{ij} r_{ij}</math> = <math>\pi_j q_{ji} r_{ji}</math>.<br/> 
+  We want to find a general solution: <math>r_{ij} = a(i,j) \pi_j q_{ji}</math>, where a(i,j) = a(j,i).<br/>  
−  +  '''Recall'''  
+  <math>r_{ij}</math> is the probability of acceptance, thus it must be that <br/>  
−  +  1.<math>r_{ij} = a(i,j)</math> <math>\pi_j q_{ji} </math>≤1, then we get: <math>a(i,j) </math>≤ <math>1/(\pi_j q_{ji})</math>  
−  
−  
−  
−  
−  >> a  +  2. <math>r_{ji} = a(j,i) </math> <math>\pi_i q_{ij} </math> ≤ 1, then we get: <math>a(j,i)</math> ≤ <math>1/(\pi_i q_{ij})</math> 
−  a  +  So we choose a(i,j) as large as possible, but it needs to satisfy the two conditions above.<br/> 
−  +  <math>a(i,j) = \min \{\frac{1}{\pi_j q_{ji}},\frac{1}{\pi_i q_{ij}}\} </math><br/>  
−  
−  
−  
−  
−  </  
−  
−  =  +  Thus, <math> r_{ij} = \min \{\frac{\pi_j q_{ji}}{\pi_i q_{ij}}, 1\} </math><br/> 
−  
−  +  '''Note''':  
−  +  1 is the upper bound to make r<sub>ij</sub> a probability  
−  
−  
−  
−  
−  +  '''Algorithm:''' <br/>  
−  +  *<math> (*) P(Y=j) = q_{ij} </math>. <math>\frac{\pi_j q_{ji}}{\pi_i q_{ij}}</math> is a positive ratio.  
−  
−  <  +  *<math> r_{ij} = \min \{\frac{\pi_j q_{ji}}{\pi_i q_{ij}}, 1\} </math> <br/> 
−  +  *<math>  
−  +  x_{t+1} = \begin{cases}  
+  Y, & \text{with probability } r_{ij} \\  
+  x_t, & \text{otherwise} \end{cases} </math> <br/>  
+  * go back to the first step (*) <br/>  
−  <br>  +  We can compare this with the AcceptanceRejection model we learned before. <br/> 
−  +  * <math>U</math> ~ <math>Uniform(0,1)</math> <br/>  
−  +  * If <math>U < r_{ij}</math>, then accept. <br/>  
+  EXCEPT that a point is always generated at each timestep. <br>  
−  +  The algorithm generates a stochastic sequence that only depends on the last state, which is a Markov Chain.<br>  
−  +  ====Metropolis Algorithm====  
+  '''Proposition: ''' Metropolis works:  
−  <  +  The <math>P_{ij}</math>'s from Metropolis Algorithm satisfy detailed balance property w.r.t <math>\pi</math> . i.e. <math>\pi_i P_{ij} = \pi_j P_{ji}</math>. The new Markov Chain has a stationary distribution <math>\pi</math>. <br/> 
+  '''Remarks:''' <br/>  
+  1) We only need to know ratios of values of <math>\pi_i</math>'s.<br/>  
+  2) The MC might converge to <math>\pi</math> at varying speeds depending on the proposal distribution and the value the chain is initialized with<br/>  
−  
+  This algorithm generates <math>\{x_t: t=0,...,m\}</math>. <br/>  
+  In the long run, the marginal distribution of <math> x_t </math> is the stationary distribution <math>\underline{\Pi} </math><br>  
+  <math>\{x_t: t = 0, 1,...,m\}</math> is a Markov chain with probability transition matrix (PTM), P.<br>  
−  +  This is a Markov Chain since <math> x_{t+1} </math> only depends on <math> x_t </math>, where <br>  
+  <math> P_{ij}= \begin{cases}  
+  q_{ij} r_{ij}, & \text{if }i \neq j (q_{ij} \text{is the probability of generating j from i and } r_{ij} \text{ is the probiliity of accepting)}\\[6pt]  
+  1  \displaystyle\sum_{k \neq i} q_{ik} r_{ik}, & \text{if }i = j \end{cases} </math><br />  
−  +  <math>q_{ij}</math> is the probability of generating state j; <br/>  
−  +  <math> r_{ij}</math> is the probability of accepting state j as the next state. <br/>  
−  
−  
−  
−  
+  Therefore, the final probability of moving from state i to j when i does not equal to j is <math>q_{ij}*r_{ij}</math>. <br/>  
+  For the probability of moving from state i to state i, we deduct all the probabilities of moving from state i to any j that are not equal to i, therefore, we get the second probability.  
−  +  ===Proof of the proposition:===  
−  <math>\  +  A good way to think of the detailed balance equation is that they balance the probability from state i to state j with that from state j to state i. 
+  We need to show that the stationary distribition of the Markov Chain is <math>\underline{\Pi}</math>, i.e. <math>\displaystyle \underline{\Pi} = \underline{\Pi}P</math><br />  
+  <div style="textsize:20px">  
+  Recall<br/>  
+  If a Markov chain satisfies the detailed balance property, i.e. <math>\displaystyle \pi_i P_{ij} = \pi_j P_{ji} \, \forall i,j</math>, then <math>\underline{\Pi}</math> is the stationary distribution of the chain.<br /><br />  
+  </div>  
−  +  '''Proof:'''  
−  
−  +  WLOG, we can assume that <math>\frac{\pi_j q_{ji}}{\pi_i q_{ij}}<1</math><br/>  
−  
−  
−  
−  +  LHS:<br />  
−  +  <math>\pi_i P_{ij} = \pi_i q_{ij} r_{ij} = \pi_i q_{ij} \cdot \min(\frac{\pi_j q_{ji}}{\pi_i q_{ij}},1) = \cancel{\pi_i q_{ij}} \cdot \frac{\pi_j q_{ji}}{\cancel{\pi_i q_{ij}}} = \pi_j q_{ji}</math><br />  
−  +  RHS:<br />  
+  Note that by our assumption, since <math>\frac{\pi_j q_{ji}}{\pi_i q_{ij}}<1</math>, its reciprocal <math>\frac{\pi_i q_{ij}}{\pi_j q_{ji}} \geq 1</math><br />  
+  So <math>\displaystyle \pi_j P_{ji} = \pi_ j q_{ji} r_{ji} = \pi_ j q_{ji} \cdot \min(\frac{\pi_i q_{ij}}{\pi_j q_{ji}},1) = \pi_j q_{ji} \cdot 1 = \pi_ j q_{ji}</math><br />  
−  +  Hence LHS=RHS  
−  
−  
−  
−  
−  
−  
−  
+  If we assume that <math>\frac{\pi_j q_{ji}}{\pi_i q_{ij}}=1</math><br/> (essentially <math>\frac{\pi_j q_{ji}}{\pi_i q_{ij}}>=1</math>)<br/>  
−  +  LHS:<br />  
+  <math>\pi_i P_{ij} = \pi_i q_{ij} r_{ij} = \pi_i q_{ij} \cdot \min(\frac{\pi_j q_{ji}}{\pi_i q_{ij}},1) =\pi_i q_{ij} \cdot 1 = \pi_i q_{ij}</math><br />  
−  +  RHS:<br />  
−  <math>q_{ij}</math>  +  '''Note''' <br/> 
−  +  by our assumption, since <math>\frac{\pi_j q_{ji}}{\pi_i q_{ij}}\geq 1</math>, its reciprocal <math>\frac{\pi_i q_{ij}}{\pi_j q_{ji}} \leq 1 </math> <br />  
+  So <math>\displaystyle \pi_j P_{ji} = \pi_ j q_{ji} r_{ji} = \pi_ j q_{ji} \cdot \min(\frac{\pi_i q_{ij}}{\pi_j q_{ji}},1) = \cancel{\pi_j q_{ji}} \cdot \frac{\pi_i q_{ij}}{\cancel{\pi_j q_{ji}}} = \pi_i q_{ij}</math><br />  
−  +  Hence LHS=RHS which indicates <math>pi_i*P_{ij} = pi_j*P_{ji}</math><math>\square</math><br /><br />  
−  +  '''Note'''<br />  
−  +  1) If we instead assume <math>\displaystyle \frac{\pi_i q_{ij}}{\pi_j q_{ji}} \geq 1</math>, the proof is similar with LHS= RHS = <math> \pi_i q_{ij} </math> <br />  
−  +  2) If <math>\displaystyle i = j</math>, then detailed balance is satisfied trivially.<br />  
−  
−  +  since <math>{\pi_i q_{ij}}</math>, and <math>{\pi_j q_{ji}}</math> are smaller than one. so the above steps show the proof of <math>\frac{\pi_i q_{ij}}{\pi_j q_{ji}}<1</math>.  
−  <math>  
−  +  == Class 18  Thursday July 4th 2013 ==  
+  === Last class ===  
−  +  Recall: The Acceptance Probability,  
+  <math>r_{ij}=min(\frac {{\pi_j}q_{ji}}{{\pi_i}q_{ij}},1)</math> <br />  
−  +  1) <math>r_{ij}=\frac {{\pi_j}q_{ji}}{{\pi_i}q_{ij}}</math>, and <math>r_{ji}=1 </math>, (<math>\frac {{\pi_j}q_{ji}}{{\pi_i}q_{ij}} < 1</math>) <br />  
−  
−  +  2) <math>r_{ji}=\frac {{\pi_i}q_{ij}}{{\pi_j}q_{ji}}</math>, and <math> r{ij}=1 </math>, (<math>\frac {{\pi_j}q_{ji}}{{\pi_i}q_{ij}} \geq 1</math> ) <br />  
−  +  ===Example: Discrete Case===  
−  
−  +  Consider a biased die,  
−  +  <math>\pi</math>= [0.1, 0.1, 0.2, 0.4, 0.1, 0.1]  
−  +  We could use any <math>6 x 6 </math> matrix <math> \mathbf{Q} </math> as the proposal distribution <br>  
−  +  For the sake of simplicity ,using a discrete uniform distribution is the simplest. This is because all probabilities are equivalent, hence during the calculation of r, qxy and qyx will cancel each other out.  
−  
−  
−  
−  
−  +  <math> \mathbf{Q} =  
−  +  \begin{bmatrix}  
−  +  1/6 & 1/6 & \cdots & 1/6 \\  
−  +  1/6 & 1/6 & \cdots & 1/6 \\  
+  \vdots & \vdots & \ddots & \vdots \\  
+  1/6 & 1/6 & \cdots & 1/6  
+  \end{bmatrix}  
+  </math> <br/>  
−  
−  
−  '''  +  '''Algorithm''' <br> 
+  1. <math>x_t=5</math> (sample from the 5th row, although we can initialize the chain from anywhere within the support)<br />  
+  2. Y~Unif[1,2,...,6]<br />  
+  3. <math> r_{ij} = \min \{\frac{\pi_j q_{ji}}{\pi_i q_{ij}}, 1\} = \min \{\frac{\pi_j 1/6}{\pi_i 1/6}, 1\} = \min \{\frac{\pi_j}{\pi_i}, 1\}</math><br>  
+  Note: current state <math>i</math> is <math>X_t</math>, the candidate state <math>j</math> is <math>Y</math>. <br>  
+  Note: since <math>q_{ij}= q_{ji}</math> for all i and j, that is, the proposal distribution is symmetric, we have <math> r_{ij} = \min \{\frac{\pi_j}{\pi_i }, 1\} </math><br/>  
+  4. U~Unif(0,1)<br/>  
+  if <math>u \leq r_{ij}</math>, X<sub>t+1</sub>=Y<br />  
+  else X<sub>t+1</sub>=X<sub>t</sub><br />  
+  go back to 2<br>  
−  +  Notice how a point is always generated for X<sub>t+1</sub>, regardless of whether the candidate state Y is accepted <br>  
−  
−  
−  
+  '''Matlab'''  
+  <pre style="fontsize:14px">  
+  pii=[.1,.1,.2,.4,.1,.1];  
+  x(1)=5;  
+  for ii=2:1000  
+  Y=unidrnd(6); %%% Unidrnd(x) is a builtin function which generates a number between (0) and (x)  
+  r = min (pii(Y)/pii(x(ii1)), 1);  
+  u=rand;  
+  if u<r  
+  x(ii)=Y;  
+  else  
+  x(ii)=x(ii1);  
+  end  
+  end  
+  hist(x,6) %generate histogram displaying all 1000 points  
+  xx = x(501,end); %After 500, the chain will mix well and converge.  
+  hist(xx,6) % The result should be better.  
+  </pre>  
+  [[File:MH_example1.jpg300px]]  
−  
−  
−  
−  +  '''NOTE:''' Generally, we generate a large number of points (say, 1500) and throw away some of the points that were first generated(say, 500). Those first points are called the [[burnin period]]. A chain will converge to the limiting distribution eventually, but not immediately. The burnin period is that beginning period before the chain has converged to the desired distribution. By discarding those 500 points, our data set will be more representative of the desired limiting distribution; once the burnin period is over, we say that the chain "mixes well".  
−  
−  
−  
−  +  ===Alternate Example: Discrete Case===  
−  
−  
−  
−  +  Consider the weather. If it is sunny one day, there is a 5/7 chance it will be sunny the next. If it is rainy, there is a 5/8 chance it will be rainy the next.  
+  <math>\pi= [\pi_1 \ \pi_2] </math>  
−  +  Use a discrete uniform distribution as the proposal distribution, because it is the simplest.  
−  
−  
−  
−  
−  
−  +  <math> \mathbf{Q} =  
+  \begin{bmatrix}  
+  5/7 & 2/7 \\  
+  3/8 & 5/8\\  
+  
+  \end{bmatrix}  
+  </math> <br/>  
−  
−  
−  
−  +  '''Algorithm''' <br>  
−  +  1. Set initial chain state: <math>X_t=1</math> (i.e. sample from the 1st row, although we could also choose the 2nd row)<br />  
−  +  2. Sample from proposal distribution: Y~q(yx) = Unif[1,2]<br />  
+  3. <math> r_{ij} = \min \{\frac{\pi_j q_{ji}}{\pi_i q_{ij}}, 1\} = \min \{\frac{\pi_j 1/6}{\pi_i 1/6}, 1\} = \min \{\frac{\pi_j}{\pi_i}, 1\}</math><br>  
+  '''Note:''' Current state <math>i</math> is <math>X_t</math>, the candidate state <math>j</math> is <math>Y</math>. Since <math>q_{ij}= q_{ji}</math> for all i and j, that is, the proposal distribution is symmetric, we have <math> r_{ij} = \min \{\frac{\pi_j}{\pi_i }, 1\} </math>  
+  
+  4. U~Unif(0,1)<br>  
+  If <math>U \leq r_{ij}</math>, then<br>  
+  <math>X_t=Y</math><br>  
+  else<br />  
+  <math>X_{t+1}=X_t</math><br>  
+  end if<br />  
+  5. Go back to step 2<br>  
−  
−  +  '''Generalization of the above framework to the continuous case'''<br>  
−  +  In place of <math>\pi</math> use <math>f(x)</math>  
−  <math>  +  In place of r<sub>ij</sub> use <math>q(yx)</math> <br> 
+  In place of r<sub>ij</sub> use <math>r(x,y)</math> <br>  
+  Here, q(yx) is a friendly distribution that is easy to sample, usually a symmetric distribution will be preferable, such that <math>q(yx) = q(xy)</math> to simplify the computation for <math>r(x,y)</math>.  
−  
−  
−  
−  +  '''Remarks'''<br>  
+  1. The chain may not get to a stationary distribution if the # of steps generated are small. That is it will take a very large amount of steps to step through the whole support<br>  
+  2. The algorithm can be performed with a <math>\pi</math> that is not even a probability mass function, it merely needs to be proportional to the probability mass function we wish to sample from. This is useful as we do not need to calculate the normalization factor. <br>  
−  +  For example, if we are given <math>\pi^'=\pi\alpha=[5,10,11,2,100,1]</math>, we can normalize this vector by dividing the sum of all entries <math>s</math>.<br>  
+  However we notice that when calculating <math>r_{ij}</math>, <br>  
+  <math>\frac{\pi^'_j/s}{\pi^'_i/s}\times\frac{q_{ji}}{q_{ij}}=\frac{\pi^'_j}{\pi^'_i}\times\frac{q_{ji}}{q_{ij}}</math> <br>  
+  <math>s</math> cancels out in this case. Therefore it is not necessary to calculate the sum and normalize the vector.<br>  
−  +  This also applies to the continuous case,where we merely need <math> f(x) </math> to be proportional to the pdf of the distribution we wish to sample from. <br>  
−  
−  +  ===Metropolis–Hasting Algorithm===  
−  +  '''Definition''': <br>  
+  Metropolis–Hastings algorithm is a Markov chain Monte Carlo (MCMC) method for obtaining a sequence of random samples from a probability distribution for which direct sampling is difficult. The Metropolis–Hastings algorithm can draw samples from any probability distribution P(x), provided you can compute the value of a function f(x) which is proportional to the density of P. <br>  
−  
−  
−  
−  
−  +  '''Purpose''': <br>  
+  "The purpose of the MetropolisHastings Algorithm is to <b>generate a collection of states according to a desired distribution</b> <math>P(x)</math>. <math>P(x)</math> is chosen to be the stationary distribution of a Markov process, <math>\pi(x)</math>." <br>  
+  Source:(http://en.wikipedia.org/wiki/Metropolis%E2%80%93Hastings_algorithm)<br>  
−  +  MetropolisHastings is an algorithm for constructing a Markov chain with a given limiting probability distribution. In particular, we consider what happens if we apply the MetropolisHastings algorithm repeatedly to a “proposal” distribution which has already been updated.<br>  
−  
+  The algorithm was named after Nicholas Metropolis and W. K. Hastings who extended it to the more general case in 1970.<br>  
−  +  <math>q(yx)</math> is used instead of <math>qi,j</math>. In continuous case, we use these notation which means given state x, what's the probability of y.<br>  
−  <math>  
−  +  Note that the MetropolisHasting algorithm possess some advantageous properties. One of which is that this algorithm "can be used when \pi(x) is known up to the constant of proportionality". The second is that in this algorithm, "we do not require the conditional distribution, which, in contrast, is required for the Gibbs sampler. "  
−  +  Source:https://www.msu.edu/~blackj/Scan_2003_02_12/Chapter_11_Markov_Chain_Monte_Carlo_Methods.pdf  
−  
−  
−  
−  
−  
−  
−  
−  
+  '''Differences between the discrete and continuous case of the Markov Chain''':<br/>  
+  
+  1. <math>q(yx)</math> is used in continuous, instead of <math>q_{ij}</math> in discrete <br/>  
+  2. <math>r(x,y)</math> is used in continuous, instead of <math>r{ij}</math> in discrete <br/>  
+  3. <math>f</math> is used instead of <math>\pi</math> <br/>  
−  
−  
−  
−  
−  
−  
−  
−  
−  
−  
−  +  '''Build the Acceptance Ratio'''<br/>  
+  Before we consider the algorithm there are a couple general steps to follow to build the acceptance ratio:<br/>  
−  +  a) Find the distribution you wish to use to generate samples from<br/>  
−  <  +  b) Find a candidate distribution that fits the desired distribution, q(yx). (the proposed moves are independent of the current state)<br/> 
−  +  c) Build the acceptance ratio <math>\displaystyle \frac{f(y)q(xy)}{f(x)q(yx)}</math>  
−  
−  
−  
−  
−  
−  
−  
−  
−  
−  
−  
−  
−  
−  
−  </  
−  
−  
−  +  Assume that f(y) is the target distribution; Choose q(yx) such that it is a friendly distribution and easy to sample from.<br />  
+  '''Algorithm:'''<br />  
+  # Set <math>\displaystyle i = 0</math> and initialize the chain, i.e. <math>\displaystyle x_0 = s</math> where <math>\displaystyle s</math> is some state of the Markov Chain.  
+  # Sample <math>\displaystyle Y \sim q(yx)</math>  
+  # Set <math>\displaystyle r(x,y) = min(\frac{f(y)q(xy)}{f(x)q(yx)},1)</math>  
+  # Sample <math>\displaystyle u \sim \text{UNIF}(0,1)</math>  
+  # If <math>\displaystyle u \leq r(x,y), x_{i+1} = Y</math><br /> Else <math>\displaystyle x_{i+1} = x_i</math>  
+  # Increment i by 1 and go to Step 2, i.e. <math>\displaystyle i=i+1</math>  
−  +  <br> '''Note''': q(xy) is moving from y to x and q(yx) is moving from x to y.  
−  <  +  <br>We choose q(yx) so that it is simple to sample from. 
+  <br>Usually, we choose a normal distribution.  
−  +  NOTE2: The proposal q(yx) y depends on x (is conditional on x)the current state, this makes sense ,because it's a necessary condition for MC. So the proposal should depend on x (also their supports should match) e.g q(yx) ~ N( x, b<sup>2</sup>) here the proposal depends on x.  
+  If the next state is INDEPENDENT of the current state, then our proposal will not depend on x e.g. (A4 Q2, sampling from Beta(2,2) where the proposal was UNIF(0,1)which is independent of the current state. )  
−  +  However, it is important to remember that even if generating the proposed/candidate state does not depend on the current state, the chain is still a markov chain.  
−  +  
−  +  <br />  
−  +  Comparing with previous sampling methods we have learned, samples generated from MH algorithm are not independent of each other, since we accept future sample based on the current sample. Furthermore, unlike acceptance and rejection method, we are not going to reject any points in MetropolisHastings. In the equivalent of the "reject" case, we just leave the state unchanged. In other words, if we need a sample of 1000 points, we only need to generate the sample 1000 times.<br/>  
−  
−  
−  </  
+  <p style="fontsize:20px;color:red;">  
+  Remarks  
+  </p>  
+  ===='''Remark 1'''====  
+  <span style="textshadow: 0px 2px 3px 3399CC;marginright:1em;fontfamily: 'Nobile', Helvetica, Arial, sansserif;fontsize:16px;lineheight:25px;color:3399CC">  
+  A common choice for <math>q(yx)</math> is a normal distribution centered at x with standard deviation b. Y~<math>N(x,b^2)</math>  
+  In this case, <math> q(yx)</math> is symmetric.  
−  +  i.e.  
−  +  <math>q(yx)=q(xy)</math><br>  
−  +  (we want to sample q centered at the current state.)<br>  
−  +  <math>q(yx)=\frac{1}{\sqrt{2\pi}b}\,e^{ \frac{\scriptscriptstyle 1}{\scriptscriptstyle 2b^2} (yx)^2}</math>, (centered at x)<br>  
−  +  <math>q(xy)=\frac{1}{\sqrt{2\pi}b}\,e^{ \frac{\scriptscriptstyle 1}{\scriptscriptstyle 2b^2} (xy)^2}</math>,(centered at y)<br>  
−  +  <math>\Rightarrow (yx)^2=(xy)^2</math><br>  
+  so <math>~q(y \mid x)=q(x \mid y)</math> <br>  
+  In this case <math>\frac{q(x \mid y)}{q(y \mid x)}=1</math> and therefore <math> r(x,y)=\min \{\frac{f(y)}{f(x)}, 1\} </math> <br/><br />  
+  This is true for any symmetric q. In general if q(yx) is symmetric, then this algorithm is called Metropolis.<br/>  
+  When choosing function q, it makes sense to choose a distribution with the same support as the distribution you want to simulate. eg. If target is Beta, then can choose q ~ Uniform(0,1)<br>  
+  The chosen q is not necessarily symmetric. Depending on different target distribution, q can be uniform.</span>  
−  +  ===='''Remark 2'''====  
−  +  <span style="textshadow: 0px 2px 3px 3399CC;marginright:1em;fontfamily: 'Nobile', Helvetica, Arial, sansserif;fontsize:16px;lineheight:25px;color:3399CC">  
−  +  The value y is accepted if u<=<math>min\{\frac{f(y)}{f(x)},1\}</math>, so it is accepted with the probability <math>min\{\frac{f(y)}{f(x)},1\}</math>.<br/>  
−  +  Thus, if <math>f(y)>=f(x)</math>, then y is always accepted.<br/>  
−  +  The higher that value of the pdf is in the vicinity of a point <math>y_1</math> , the more likely it is that a random variable will take on values around <math>y_1</math>.<br/>  
−  +  Therefore,we would want a high probability of acceptance for points generated near <math>y_1</math>.<br>  
+  [[File:Diag1.png]]<br>  
+  '''Note''':<br/>  
+  If the proposal comes from a region with low density, we may or may not accept; however, we accept for sure if the proposal comes from a region with high density.<br>  
−  '''  +  ===='''Remark 3'''==== 
−  +  One strength of the MetropolisHastings algorithm is that normalizing constants, which are often quite difficult to determine, can be cancelled out in the ratio <math> r </math>. For example, consider the case where we want to sample from the beta distribution, which has the pdf:<br>  
−  +  （also notice that Metropolis Hastings is just a special case of Metropolis algorithm)  
−  
−  
+  <math>  
+  \begin{align}  
+  f(x;\alpha,\beta)& = \frac{1}{\mathrm{B}(\alpha,\beta)}\, x^{\alpha1}(1x)^{\beta1}\end{align}  
+  </math>  
−  '''  +  The beta function, ''B'', appears as a normalizing constant but it can be simplified by construction of the method. 
−  
−  
−  +  ====='''Example'''=====  
−  
−  
−  
−  +  <math>\,f(x)=\frac{1}{\pi^{2}}\frac{1}{1+x^{2}}</math>, where <math>\frac{1}{\pi^{2}} </math> is normalization factor and <math>\frac{1}{1+x^{2}} </math> is target distribution. <br>  
+  Then, we have <math>\,f(x)\propto\frac{1}{1+x^{2}}</math>.<br>  
+  And let us take <math>\,q(xy)=\frac{1}{\sqrt{2\pi}b}e^{\frac{1}{2b^{2}}(yx)^{2}}</math>.<br>  
+  Then <math>\,q(xy)</math> is symmetric since <math>\,(yx)^{2} = (xy)^{2}</math>.<br>  
+  Therefore Y can be simplified.  
−  
−  +  We get :  
−  
+  <math>\,\begin{align}  
+  \displaystyle r(x,y)  
+  & =min\left\{\frac{f(y)}{f(x)}\frac{q(xy)}{q(yx)},1\right\} \\  
+  & =min\left\{\frac{f(y)}{f(x)},1\right\} \\  
+  & =min\left\{ \frac{ \frac{1}{1+y^{2}} }{ \frac{1}{1+x^{2}} },1\right\}\\  
+  & =min\left\{ \frac{1+x^{2}}{1+y^{2}},1\right\}\\  
+  \end{align}  
+  </math>.  
−  +  <br/>  
−  +  <math>\pi=[0.1\,0.1\,...] </math> stands for probility;<br/>  
−  +  <math>\pi \propto [3\,2\, 10\, 100\, 1.5] </math> is not brobility, so we take:<br/>  
+  <math>\Rightarrow \pi=1/c \times [3\, 2\, 10\, 100\, 1.5]</math> is probility where<br/>  
+  <math>\Rightarrow c=3+2+10+100+1.5 </math><br/>  
+  <br/>  
+  <br/>  
+  In practice, if elements of <math>\pi</math> are functions or random variables, we need c to be the normalization factor, the summation/integration over all members of <math>\pi</math>. This is usually very difficult. Since we are taking ratios, with the MetropolisHasting algorithm, it is not necessary to do this.  
−  +  <br>  
−  +  For example, to find the relationship between weather temperature and humidity, we only have a proportional function instead of a probability function. To make it into a probability function, we need to compute c, which is really difficult. However, we don't need to compute c as it will be cancelled out during calculation of r.<br>  
−  
−  
−  
−  
−  
−  
−  
−  
−  
+  ======'''MATLAB'''======  
+  The Matlab code of the algorithm is the following :  
+  <pre style="fontsize:12px">  
+  clear all  
+  close all  
+  clc  
+  b=2;  
+  x(1)=0;  
+  for i=2:10000  
+  y=b*randn+x(i1);  
+  r=min((1+x(i1)^2)/(1+y^2),1);  
+  u=rand;  
+  if u<r  
+  x(i)=y;  
+  else  
+  x(i)=x(i1);  
+  end  
+  
+  end  
+  hist(x,100);  
+  %The Markov Chain usually takes some time to converge and this is known as the "burning time".  
+  </pre>  
+  [[File:MH_example2.jpg300px]]  
−  +  However, while the data does approximately fit the desired distribution, it takes some time until the chain gets to the stationary distribution. To generate a more accurate graph, we modify the code to ignore the initial points.<br>  
−  
−  +  '''MATLAB'''  
−  b)  +  <pre style="fontsize:16px"> 
−  +  b=2;  
+  x(1)=0;  
+  for ii=2:10500  
+  y=b*randn+x(ii1);  
+  r=min((1+x(ii1)^2)/(1+y^2),1);  
+  u=rand;  
+  if u<=r  
+  x(ii)=y;  
+  else  
+  x(ii)=x(ii1);  
+  end  
+  end  
+  xx=x(501:end) %we don't display the first 500 points because they don't show the limiting behaviour of the Markov Chain  
+  hist(xx,100)  
+  </pre>  
+  [[File:MH_Ex.jpg300px]]  
+  <br>  
+  '''If a function f(x) can only take values from <math>[0,\infty)</math>, but we need to use normal distribution as the candidate distribution, then we can use <math>q=\frac{2}{\sqrt{2\pi}}*exp(\frac{(yx)^2}{2})</math>, where y is from <math>[0,\infty)</math>. <br>(This is essentially the pdf of the absolute value of a normal distribution centered around x)'''<br><br>  
+  Example:<br>  
+  We want to sample from <math>exp(2), q(yx)~\sim~N(x,b^2)</math><br>  
+  <math>r=\frac{f(y)}{f(x)}=\frac{2*exp^(2y)}{2*exp^(2x)}=exp(2*(xy))</math><br>  
+  <math>r=min(exp(2*(xy)),1)</math><br>  
+  '''MATLAB'''  
+  <pre style="fontsize:16px">  
+  x(1)=0;  
+  for ii=2:100  
+  y=2*(randn*b+abs(x(ii1)))  
+  r=min(exp(2*(xy)),1);  
+  u=rand;  
+  if u<=r  
+  x(ii)=y;  
+  else  
+  x(ii)=x(ii1);  
+  end  
+  end  
+  </pre>  
+  <br>  
−  +  '''Definition of Burn in:'''  
−  '''  
−  
−  
−  
−  
−  
−  
−  
−  
−  
−  
−  
−  +  Typically in a MH Algorithm, a set of values generated at at the beginning of the sequence are "burned" (discarded) after which the chain is assumed to have converged to its target distribution. In the first example listed above, we "burned" the first 500 observations because we believe the chain has not quite reached our target distribution in the first 500 observations. 500 is not a set threshold, there is no right or wrong answer as to what is the exact number required for burnin. Theoretical calculation of the burnin is rather difficult, in the above mentioned example, we chose 500 based on experience and quite arbitrarily.  
−  
−  +  Burnin time can also be thought of as the time it takes for the chain to reach its stationary distribution. Therefore, in this case you will disregard everything uptil the burnin period because the chain is not stabilized yet.  
−  <  +  The Metropolis–Hasting Algorithm is started from an arbitrary initial value <math>x_0</math> and the algorithm is run for many iterations until this initial state is "forgotten". These samples, which are discarded, are known as ''burnin''. The remaining 
−  +  set of accepted values of <math>x</math> represent a sample from the distribution f(x).(http://en.wikipedia.org/wiki/Metropolis%E2%80%93Hastings_algorithm)<br/>  
−  +  Burnin time can also be thought of as the time it takes for the process to reach the stationary distribution pi. Suppose it takes 5 samples after which you reach the stationary distribution. You should disregard the first five samples and consider the remaining samples as representing your target distribution f(x). <br>  
−  +  
−  <  +  Several extensions have been proposed in the literature to speed up the convergence and reduce the so called “burnin” period. 
−  +  One common suggestion is to match the first few moments of q(yx) to f(x).  
−  
−  
−  +  '''Aside''': The algorithm works best if the candidate density q(yx) matches the shape of the target distribution f(x). If a normal distribution is used as a candidate distribution, the variance parameter b<sup>2</sup> has to be tuned during the burnin period. <br/>  
−  +  1. If b is chosen to be too small, the chain will mix slowly (smaller proposed move, the acceptance rate will be high and the chain will converge only slowly the f(x)).  
−  +  
−  +  2. If b is chosen to be too large, the acceptance rate will be low (larger proposed move and the chain will converge only slowly the f(x)).  
−  
−  
−  
−  
−  
−  
−  
−  
−  
−  
−  
−  
−  
−  
−  
−  
−  
−  +  '''Note''':  
+  The histogram looks much nicer if we reject the points within the burning time.<br>  
−  
−  
−  +  Example: Use MH method to generate sample from f(x)=2x  
−  +  0<x<1, 0 otherwise.  
−  f(x  
−  
−  +  1) Initialize the chain with <math>x_i</math> and set <math>i=0</math>  
−  =  +  2)<math>Y~\sim~q(yx_i)</math> 
+  where our proposal function would be uniform [0,1] since it matches our original ones support.  
+  =><math>Y~\sim~Unif[0,1]</math>  
−  <math>\  +  3)consider <math>\frac{f(y)}{f(x)}=\frac{y}{x}</math>, 
−  +  <math>r(x,y)=min (\frac{y}{x},1)</math> since q(yx<sub>i</sub>) and q(x<sub>i</sub>y) can be cancelled together.  
−  
−  
−  
+  4)<math>X_{i+1}=Y</math> with prob <math>r(x,y)</math>,  
+  <math>X_{i+1}=X_i</math>, otherwise  
−  +  5)<math>i=i+1</math>, go to 2  
−  <  +  <br> 
−  
−  
−  
−  
−  
−  
−  
−  +  Example form wikipedia  
−  
−  
−  
−  
−  
−  
−  +  ===Stepbystep instructions===  
−  <  +  Suppose the most recent value sampled is <math>x_t\,</math>. To follow the Metropolis–Hastings algorithm, we next draw a new proposal state <math>x'\,</math> with probability density <math>Q(x'\mid x_t)\,</math>, and calculate a value 
−  
−  +  :<math>  
−  +  a = a_1 a_2\,  
−  <  +  </math> 
−  
−  
−  
−  
−  
−  
−  
−  
−  
−  
−  
−  
−  
−  
−  
−  
−  
−  
−  </  
−  
−  +  where  
−  +  :<math>  
−  +  a_1 = \frac{P(x')}{P(x_t)} \,\!  
−  +  </math>  
−  
−  
−  
−  
−  
−  
−  
−  
−  
−  
−  
−  
−  
−  
−  
−  
−  
−  +  is the likelihood ratio between the proposed sample <math>x'\,</math> and the previous sample <math>x_t\,</math>, and  
−  
−  <math>  
−  
−  +  :<math>  
−  <  +  a_2 = \frac{Q(x_t \mid x')}{Q(x'\mid x_t)} 
−  +  </math>  
−  
−  
−  
−  
−  
−  
−  
−  x  
−  
−  
−  </  
−  
−  '''  +  is the ratio of the proposal density in two directions (from <math>x_t\,</math> to <math>x'\,</math> and ''vice versa''). 
+  This is equal to 1 if the proposal density is symmetric.  
+  Then the new state <math>\displaystyle x_{t+1}</math> is chosen according to the following rules.  
−  +  :<math>  
+  \begin{matrix}  
+  \mbox{If } a \geq 1: & \\  
+  & x_{t+1} = x',  
+  \end{matrix}  
+  </math>  
+  :<math>  
+  \begin{matrix}  
+  \mbox{else} & \\  
+  & x_{t+1} = \left\{  
+  \begin{array}{lr}  
+  x' & \mbox{ with probability }a \\  
+  x_t & \mbox{ with probability }1a.  
+  \end{array}  
+  \right.  
+  \end{matrix}  
+  </math>  
−  +  The Markov chain is started from an arbitrary initial value <math>\displaystyle x_0</math> and the algorithm is run for many iterations until this initial state is "forgotten".  
+  These samples, which are discarded, are known as ''burnin''. The remaining set of accepted values of <math>x</math> represent a sample from the distribution <math>P(x)</math>.  
−  The  +  The algorithm works best if the proposal density matches the shape of the target distribution <math>\displaystyle P(x)</math> from which direct sampling is difficult, that is <math>Q(x'\mid x_t) \approx P(x') \,\!</math>. 
−  +  If a Gaussian proposal density <math>\displaystyle Q</math> is used the variance parameter <math>\displaystyle \sigma^2</math> has to be tuned during the burnin period.  
+  This is usually done by calculating the ''acceptance rate'', which is the fraction of proposed samples that is accepted in a window of the last <math>\displaystyle N</math> samples.  
+  The desired acceptance rate depends on the target distribution, however it has been shown theoretically that the ideal acceptance rate for a one dimensional Gaussian distribution is approx 50%, decreasing to approx 23% for an <math>\displaystyle N</math>dimensional Gaussian target distribution.<ref name=Roberts/>  
−  +  If <math>\displaystyle \sigma^2</math> is too small the chain will ''mix slowly'' (i.e., the acceptance rate will be high but successive samples will move around the space slowly and the chain will converge only slowly to <math>\displaystyle P(x)</math>). On the other hand,  
−  +  if <math>\displaystyle \sigma^2</math> is too large the acceptance rate will be very low because the proposals are likely to land in regions of much lower probability density, so <math>\displaystyle a_1</math> will be very small and again the chain will converge very slowly.  
−  
−  
−  '''  +  == Class 19  Tuesday July 9th 2013 == 
+  '''Recall: Metropolis–Hasting Algorithm'''  
−  1.  +  1) <math>X_i</math> = State of chain at time i. Set <math>X_0</math> = 0<br> 
+  2) Generate proposal distribution: Y ~ q(yx) <br>  
+  3) Set <math>\,r=min[\frac{f(y)}{f(x)}\,\frac{q(xy)}{q(yx)}\,,1]</math><br>  
+  4) Generate U ~ U(0,1)<br>  
+  If <math>U<r</math>, then<br>  
+  <math>X_{i+1} = Y</math> % i.e. we accept Y as the next point in the Markov Chain <br>  
+  else <br>  
+  <math>X_{i+1}</math> = <math>X_i</math><br>  
+  End if<br>  
+  5) Set i = i + 1. Return to Step 2. <br>  
−  
+  Why can we use this algorithm to generate a Markov Chain?<br>  
+  <math>\,Y</math>~<math>\,q(yx)</math> satisfies the Markov Property, as the current state does not depend on previous trials. Note that Y does not '''''have''''' to depend on X<sub>t1</sub>; the Markov Property is satisfied as long as Y is not dependent on X<sub>0</sub>, X<sub>1</sub>,..., X<sub>t2</sub>. Thus, time t will not affect the choice of state.<br>  
−  
−  
+  ==='''Choosing b: 3 cases'''===  
+  If y and x have the same domain, say R, we could use normal distribution to model <math>q(yx)</math>. <math>q(xy)~normal(y,b^2), and q(yx)~normal(x,b^2)</math>.  
+  In the continuous case of MCMC, <math>q(yx)</math> is the probability of observing y, given you are observing x. We normally assume <math>q(yx)</math> ~ N(x,b^2). A reasonable choice of b is important to ensure the MC does indeed converges to the target distribution f. If b is too small it is not possible to explore the whole support because the jumps are small. If b is large than the probability of accepting the proposed state y is small, and it is very likely that we reject the possibilities of leaving the current state, hence the chain will keep on producing the initial state of the Markov chain.  
−  +  To be precise, we are discussing the choice of variance for the proposal distribution.Large b simply implies larger variance for our choice of proposal distribution (Gaussian) in this case. Therefore, many points will be rejected and we will generate same points many times since there are many points that have been rejected.<br>  
−  
−  +  In this example, <math>q(yx)=N(x, b^2)</math><br>  
−  +  Demonstrated as follows, the choice of b will be significant in determining the quality of the Metropolis algorithm. <br>  
−  
−  
−  +  This parameter affects the probability of accepting the candidate states, and the algorithm will not perform well if the acceptance probability is too large or too small, it also affects the size of the "jump" between the sampled <math>Y</math> and the previous state x<sub>i+1</sub>, as a larger variance implies a larger such "jump".<br>  
−  <math>  
−  +  If the jump is too large, we will have to repeat the previous stage; thus, we will repeat the same point for many times.<br>  
−  <  
−  +  '''MATLAB b=2, b= 0.2, b=20 '''  
+  <pre style="fontsize:12px">  
+  clear all  
+  close all  
+  clc  
+  b=2 % b=0.2 b=20;  
+  x(1)=0;  
+  for i=2:10000  
+  y=b*randn+x(i1);  
+  r=min((1+x(i1)^2)/(1+y^2),1);  
+  u=rand;  
+  if u<r  
+  x(i)=y;  
+  else  
+  x(i)=x(i1);  
+  end  
+  
+  end  
+  figure(1);  
+  hist(x(5000:end,100));  
+  figure(2);  
+  plot(x(5000:end));  
+  %The Markov Chain usually takes some time to converge and this is known as the "burning time"  
+  %Therefore, we don't display the first 5000 points because they don't show the limiting behaviour of the Markov Chain  
−  <  +  generate the Markov Chain with 10000 random variable, using a large b and a small b. 
+  </pre>  
−  +  b tells where the next point is going to be. The appropriate b is supposed to explore all the support area.  
−  +  f(x) is the stationary distribution list of the chain in MH. We generating y using q(yx) and accept it with respect to r.  
−  +  ===='''b too small====  
+  If <math>b = 0.02</math>, the chain takes small steps so the chain doesn't explore enough of sample space.  
−  +  If <math>b = 20</math>, jumps are very unlikely to be accepted; i.e. <math> y </math> is rejected as <math> u> r </math> and <math> Xt+1 = Xt</math>.  
−  +  i.e <math>\frac {f(y)}{f(x)}</math> and consequent <math> r </math> is very small and very unlikely that <math> u < r </math>, so the current value will be repeated.  
−  </math>  
−  +  ==='''Detailed Balance Holds for MetropolisHasting'''===  
−  +  In metropolishasting, we generate y using q(yx) and accept it with probability r, where <br>  
−  
−  <  
−  +  <math>r(x,y) = min\left\{\frac{f(y)}{f(x)}\frac{q(xy)}{q(yx)},1\right\} = min\left\{\frac{f(y)}{f(x)},1\right\}</math><br>  
−  +  Without loss of generality we assume <math>\frac{f(y)}{f(x)}\frac{q(xy)}{q(yx)} > 1</math><br>  
−  
−  </math>  
−  +  Then r(x,y) (probability of accepting y given we are currently in x) is <br>  
−  
−  
−  +  <math>r(x,y) = min\left\{\frac{f(y)}{f(x)}\frac{q(xy)}{q(yx)},1\right\} = \frac{f(y)}{f(x)}\frac{q(xy)}{q(yx)}</math><br>  
−  \  
−  \  
−  
−  
−  
−  
−  \  
−  
−  
−  
−  
−  
−  
−  
−  
−  </math>  
−  +  Now suppose that the current state is y and we are generating x; the probability of accepting x given that we are currently in state y is <br>  
−  
−  +  <math>r(x,y) = min\left\{\frac{f(x)}{f(y)}\frac{q(yx)}{q(xy)},1\right\} = 1 </math><br>  
−  
−  
−  
−  +  This is because <math>\frac{f(y)}{f(x)}\frac{q(xy)}{q(yx)} < 1 </math> and its reverse <math>\frac{f(x)}{f(y)}\frac{q(yx)}{q(xy)} > 1 </math>. Then <math>r(x,y) = 1</math>.<br>  
−  +  We are interested in the probability of moving from from x to y in the Markov Chain generated by MH algorithm: <br>  
+  P(yx) depends on two probabilities:  
+  1. Probability of generating y, and<br>  
+  2. Probability of accepting y. <br>  
−  ==  +  <math>P(yx) = q(yx)*r(x,y) = q(yx)*{\frac{f(y)}{f(x)}\frac{q(xy)}{q(yx)}} = \frac{f(y)*q(xy)}{f(x)} </math> <br> 
−  
−  +  The probability of moving to x given the current state is y:  
−  
−  
−  
−  
−  
−  
−  
−  
−  
−  
+  <math>P(xy) = q(xy)*r(y,x) = q(xy)</math><br>  
−  +  So does detailed balance hold for MH? <br>  
−  <math>  +  If it holds we should have <math>f(x)*P(yx) = f(y)*P(xy)</math>.<br> 
+  Lefthand side: <br>  
−  +  <math>f(x)*P(yx) = f(x)*{\frac{f(y)*q(xy)}{f(x)}} = f(y)*q(xy)</math><br>  
−  
−  +  Righthand side: <br>  
−  +  <math>f(y)*P(xy) = f(y)*q(xy)</math><br>  
−  +  Thus LHS and RHS are equal and the detailed balance holds for MH algorithm. <br>  
+  Therefore, f(x) is the stationary distribution of the chain.<br>  
−  +  == Class 20  Thursday July 11th 2013 ==  
+  === Simulated annealing ===  
+  <br />  
+  '''Definition:''' Simulated annealing (SA) is a generic probabilistic metaheuristic for the global optimization problem of locating a good approximation to the global optimum of a given function in a large search space. It is often used when the search space is discrete (e.g., all tours that visit a given set of cities). <br />  
+  (http://en.wikipedia.org/wiki/Simulated_annealing) <br />  
+  "Simulated annealing is a popular algorithm in simulation for minimizing functions." (from textbook)<br />  
−  +  Simulated annealing is developed to solve the traveling salesman problem: finding the optimal path to travel all the cities needed<br/>  
−  +  It is called "Simulated annealing" because it mimics the process undergone by misplaced atoms in a metal when<br />  
−  +  its heated and then slowly cooled.<br />  
−  +  (http://mathworld.wolfram.com/SimulatedAnnealing.html)<br />  
−  
−  
−  
−  
−  
−  
−  
−  
−  
−  
−  
−  
−  
−  
−  
−  
−  
−  
−  
−  
−  
−  +  It is a probabilistic method proposed in Kirkpatrick, Gelett and Vecchi (1983) and Cerny (1985) for finding the global minimum of a function that may have multiple local minimums.<br />  
−  </  +  (http://www.mit.edu/~dbertsim/papers/Optimization/Simulated%20annealing.pdf)<br /> 
−  +  Simulated annealing was developed as an approach for finding the minimum of complex functions <br />  
+  with multiple peaks; where standard hillclimbing approaches may trap the algorithm at a less that optimal peak.<br />  
−  +  Suppose we generated a point <math> x </math> by an existing algorithm, and we would like to get a "better" point. <br>  
−  +  (eg. If we have generated a local min of a function and we want the global min) <br>  
−  
−  
−  
−  
−  
−  
−  
−  
−  
−  
−  
−  
−  
−  
−  
−  
−  
−  
−  
−  
−  
−  
−  
−  
−  
−  
−  
−  
−  
−  
−  
−  
−  
−  
−  
−  
−  
−  
−  
−  
−  
−  
−  
−  
−  
−  
−  
−  
−  
−  
−  
−  
−  
−  
−  
−  Suppose we generated a point <math> x </math> by an existing algorithm, and we would like to get a "better" point. <br>  
−  (eg. If we have generated a local min of a function and we want the global min) <br>  
Then we would use simulated annealing as a method to "perturb" <math> x </math> to obtain a better solution. <br>  Then we would use simulated annealing as a method to "perturb" <math> x </math> to obtain a better solution. <br>  
Line 6,347:  Line 6,362:  
<b>4.</b> <math>r = \min\{\frac{f(y)}{f(x)},1\}</math><br />  <b>4.</b> <math>r = \min\{\frac{f(y)}{f(x)},1\}</math><br />  
<b>5.</b> U ~ U(0,1)<br />  <b>5.</b> U ~ U(0,1)<br />  
−  <b>6.</b> If U < r, <math>X_{t+1}=  +  <b>6.</b> If U < r, <math>X_{t+1}=y</math> <br/> 
else, <math>X_{t+1}=X_t</math><br/>  else, <math>X_{t+1}=X_t</math><br/>  
<b>7.</b> end decrease T, and let i=i+1. Go back to 3. (This is where the difference lies between SA and MH. <br />  <b>7.</b> end decrease T, and let i=i+1. Go back to 3. (This is where the difference lies between SA and MH. <br />  
Line 6,371:  Line 6,386:  
Assume T is large <br />  Assume T is large <br />  
1. h(y) < h(x), e<sup>(h(x)h(y))/T </sup> > 1, then r = 1, y will always be accepted.<br />  1. h(y) < h(x), e<sup>(h(x)h(y))/T </sup> > 1, then r = 1, y will always be accepted.<br />  
−  2. h(y) > h(x), e<sup>(h(x)h(y))/T </sup>< 1, then r < 1, y will be accepted with probability r.<br />  +  2. h(y) > h(x), e<sup>(h(x)h(y))/T </sup>< 1, then r < 1, y will be accepted with probability r. '''Remark:'''this will help to scape from local minimum, because the algorithm prevents it from reaching and staying in the local minimum forever. <br /> 
Assume T is small<br />  Assume T is small<br />  
1. h(y) < h(x), then r = 1, y will always be accepted.<br />  1. h(y) < h(x), then r = 1, y will always be accepted.<br />  
Line 6,463:  Line 6,478:  
<h3> <span class="mwheadline" id="Motivation:_Simulated_Annealing_and_the_Travelling_Salesman_Problem"> Motivation: Simulated Annealing and the Travelling Salesman Problem </span></h3>  <h3> <span class="mwheadline" id="Motivation:_Simulated_Annealing_and_the_Travelling_Salesman_Problem"> Motivation: Simulated Annealing and the Travelling Salesman Problem </span></h3>  
<p>The Travelling Salesman Problem asks: <br />  <p>The Travelling Salesman Problem asks: <br />  
−  Given n numbers of cities and the distances between each pair of cities, what is the shortest possible route that visits each city exactly once and returns to the original city?  +  Given n numbers of cities and the distances between each pair of cities, what is the shortest possible route that visits each city exactly once and returns to the original city? By letting two permutations if one results from an interchange of two of the coordinates of the other, we can use simulated annealing to approximate the best path. 
<p>[[File:Salesman n5.png350px]]  <p>[[File:Salesman n5.png350px]]  
</p>  </p>  
Line 6,488:  Line 6,503:  
=== Gibbs Sampling===  === Gibbs Sampling===  
'''Definition'''<br>  '''Definition'''<br>  
−  Gibbs sampling is a  +  In statistics and in statistical physics, Gibbs sampling or a Gibbs sampler is a Markov chain Monte Carlo (MCMC) algorithm for obtaining a sequence of observations which are approximately from a specified multivariate probability distribution (i.e. from the joint probability distribution of two or more random variables), when direct sampling is difficult.<br/> 
+  (http://en.wikipedia.org/wiki/Gibbs_sampling)<br/>  
−  *  +  The Gibbs sampling method was originally developed by Geman and Geman [1984]. It was later brought into mainstream statistics by Gelfand and Smith [1990] and Gelfand, et al. [1990]<br/> 
+  Source: https://www.msu.edu/~blackj/Scan_2003_02_12/Chapter_11_Markov_Chain_Monte_Carlo_Methods.pdf<br/>  
+  
+  Gibbs sampling is a general method for probabilistic inference which is often used when dealing with incomplete information. However, generality comes at some computational cost, and for many applications including those involving missing information, there are often alternative methods that have been proven to be more efficient in practice. For example, say we want to sample from a joint distribution <math>p(x_1,...,x_k)</math> (i.e. a posterior distribution). If we knew the full conditional distributions for each parameter (i.e. <math>p(x_ix_1,x_2,...,x_{i1},x_{i+1},...,x_k)</math>), we can use the Gibbs sampler to sample from these conditional distributions. <br>  
+  
+  When utilizing the Gibbs sampler, the candidate state is always accepted as the next state of the chain.(from text book)<br/>  
+  
+  *Another Markov Chain Monte Carlo (MCMC) method (first MCMC method introduced in this course is the MH Algorithm) <br/>  
*a special case of MetropolisHastings sampling where the random value is always accepted, i.e. as long as a point is proposed, it is accepted. <br/>  *a special case of MetropolisHastings sampling where the random value is always accepted, i.e. as long as a point is proposed, it is accepted. <br/>  
* useful and make it simple and easier for sampling a ddimensional random vector <math>\vec{x} = (x_1, x_2,...,x_d)</math><br />  * useful and make it simple and easier for sampling a ddimensional random vector <math>\vec{x} = (x_1, x_2,...,x_d)</math><br />  
−  * then the observations of ddimensional random vectors <math>{\vec{x_1}, \vec{x_2}, ... , \vec{x_n}}</math> form a ddimensional Markov Chain and the joint density <math>f(x_1, x_2, ... , x_d)</math> is an invariant distribution for the chain.  +  * then the observations of ddimensional random vectors <math>{\vec{x_1}, \vec{x_2}, ... , \vec{x_n}}</math> form a ddimensional Markov Chain and the joint density <math>f(x_1, x_2, ... , x_d)</math> is an invariant distribution for the chain. i.e. for sampling multivariate distributions.<br /> 
−  i.e. for sampling multivariate distributions.<br />  
* useful if sampling from conditional pdf, since they are easier to sample, in comparison to the joint distribution.<br/>  * useful if sampling from conditional pdf, since they are easier to sample, in comparison to the joint distribution.<br/>  
*Definition of univariate conditional distribution: all the random variables are fixed except for one; we need to use n such univariate conditional distributions to simulate n random variables.  *Definition of univariate conditional distribution: all the random variables are fixed except for one; we need to use n such univariate conditional distributions to simulate n random variables.  
'''Difference between Gibbs Sampling & MH'''<br>  '''Difference between Gibbs Sampling & MH'''<br>  
−  Gibbs Sampling generates new value based on the conditional distribution of other components.<br/>  +  Gibbs Sampling generates new value based on the conditional distribution of other components (unlike MH, which does not require conditional distribution).<br/> 
eg. We are given the following about <math> f(x_1,x_2) , f(x_1x_2),f(x_2x_1) </math><br/>  eg. We are given the following about <math> f(x_1,x_2) , f(x_1x_2),f(x_2x_1) </math><br/>  
1. let <math>x^*_1 \sim f(x_1x_2)</math><br/>  1. let <math>x^*_1 \sim f(x_1x_2)</math><br/>  
−  2. <math>x^*_2 \sim f(x^*_1  +  2. <math>x^*_2 \sim f(x_2x^*_1)</math><br/> 
3. substitute <math>x^*_2</math> back into first step and repeat the process. <br/>  3. substitute <math>x^*_2</math> back into first step and repeat the process. <br/>  
+  Also, for Gibbs sampling, we will "always accept a candidate point", unlike MH<br/>  
+  Source: https://www.msu.edu/~blackj/Scan_2003_02_12/Chapter_11_Markov_Chain_Monte_Carlo_Methods.pdf<br/>  
+  
+  <div style = "align:left; background:#F5F5DC; fontsize: 120%">  
+  '''Gibbs Sampling as a special form of the Metropolis Hastings algorithm'''<br>  
+  
+  The Gibbs Sampler is simply a case of the Metropolis Hastings algorithm<br>  
+  
+  here, the proposal distribution is <math>q(YX)=f(X^jX^*_i, i\neq j)=\frac{f(Y)}{f(X_i, i\neq j)}</math> for <math>X=(X_1,...,X_n)</math>, <br>  
+  which is simply the conditional distribution of each element conditional on all the other elements in the vector. <br>  
+  similarly <math>q(XY)=f(XY^*_i, i\neq j)=\frac{f(X)}{f(Y_i, i\neq j)}</math><br>  
+  notice that <math>(Y_i, i\neq j)</math> and <math>(X_i, i\neq j)</math> are identically distributed. <br>  
+  
+  the distribution we wish to simulate from is <math>p(X) = f(X) </math>  
+  also, <math>p(Y) = f(Y) </math>  
+  
+  Hence, the acceptance ratio in the MetropolisHastings algorithm is: <br>  
+  <math>r(x,y) = min\left\{\frac{f(x)}{f(y)}\frac{q(yx)}{q(xy)},1\right\} = min\left\{\frac{f(x)}{f(y)}\frac{f(y)}{f(x)},1\right\} = 1 </math><br>  
+  so the new point will always be accepted, and no points are rejected and the Gibbs Sampler is an efficient algorithm in that aspect. <br>  
+  </div>  
<b>Advantages </b><ref>  <b>Advantages </b><ref>  
Line 6,531:  Line 6,573:  
Let's suppose that we are interested in sampling from the posterior p(xy), where x is a vector of three parameters, x1, x2, x3. <br\>  Let's suppose that we are interested in sampling from the posterior p(xy), where x is a vector of three parameters, x1, x2, x3. <br\>  
The steps to a Gibbs Sampler are:<br\>  The steps to a Gibbs Sampler are:<br\>  
−  1. Pick a vector of starting value x(0).<br\>  +  1. Pick a vector of starting value x(0). Any x(0) will converge eventually, but it can be chosen to take fewer iterations<br\> 
2. Start with any x(order does not matter, but I will start with x1 for convenience). Draw a value x1(1)from the full conditional p(x1x2(0),x3(0),y)<br\>  2. Start with any x(order does not matter, but I will start with x1 for convenience). Draw a value x1(1)from the full conditional p(x1x2(0),x3(0),y)<br\>  
3. Draw a value x2(1) from the full conditional p(x2x1(1),x3(0),y). Note that we must use the updated value of x1(1).<br\>  3. Draw a value x2(1) from the full conditional p(x2x1(1),x3(0),y). Note that we must use the updated value of x1(1).<br\>  
Line 6,541:  Line 6,583:  
'''The Basic idea:''' <br>  '''The Basic idea:''' <br>  
−  The distinguishing feature of Gibbs sampling is that the underlying Markov chain is constructed  +  The distinguishing feature of Gibbs sampling is that the underlying Markov chain is constructed from a sequence of conditional distributions. The essential idea is updating one part of the previous element while keeping the other parts fixed  it is useful in many instances where the state variable is a random variable taking values in a general space, not just in R<sup>n</sup>. (Simulation and the Monte Carlo Method, Reuven Y. Rubinstein) 
'''Note:''' <br>  '''Note:''' <br>  
−  1.Other optimizing algorithms introduced such as Simulated Annealing settles on a minimum eventually,which means that if we generate enough observations and plot them in a time series plot, the plot will eventually flatten at the optimal value.  +  1.Other optimizing algorithms introduced such as Simulated Annealing settles on a minimum eventually,which means that if we generate enough observations and plot them in a time series plot, the plot will eventually flatten at the optimal value.<br\> 
−  2.For Gibbs Sampling however, when convergence is achieved, instead of staying at the optimal value, the Gibbs Sampler continues to wonder through the target distribution (i.e. will not stay at the optimal point) forever.<br>  +  2.For Gibbs Sampling however, when convergence is achieved, instead of staying at the optimal value, the Gibbs Sampler continues to wonder through the target distribution (i.e. will not stay at the optimal point) forever.<br\> 
−  '''Special Example'''<br>  +  '''Special Example'''<br\> 
<pre>  <pre>  
function gibbs2(n, thin)  function gibbs2(n, thin)  
Line 6,568:  Line 6,610:  
7.6084020137786865  7.6084020137786865  
</pre>  </pre>  
+  
+  '''Theoretical Example''' <br/>  
+  
+  Gibbs Sampler Application (Inspired by Example 10b in the Ross Simulation (4th Edition Textbook))  
+  
+  Suppose we are a truck driver who randomly puts n basketballs into a 3D storage cube sized so that each edge of the cube is 300cm in length. The basket balls are spherical and have a radius of 25cm each.  
+  
+  Because the basketballs have a radius of 25cm, the centre of each basketball must be at least 50cm away from the centre of another basketball. That is to say, if two basketballs are touching (as close together as possible) their centres will be 50cm apart.  
+  
+  Clearly the distribution of n basketballs will need to conditioned on the fact that no basketball is placed so that its centre is closer than 50cm to another basketball.  
+  
+  This gives:  
+  
+  Beta = P{the centre of no two basketballs are within 50cm of each other}  
+  
+  That is to say, the placement of basketballs is conditioned on the fact that two balls cannot overlap.  
+  
+  This distribution of n balls can be modelled using the Gibbs sampler.  
+  
+  1. Start with n basketballs positioned in the cube so that no two centres are within 50cm of each other<br />  
+  2. Generate a random number U and let I = floor(n*U) + 1<br />  
+  3. Generate another random point <math>X_k</math> in the storage box.<br />  
+  4. If <math>X_k</math> is not within 50cm of any other point, excluding point <math>X_I</math>: <br />  
+  then replace <math>X_I</math> by this new point. <br />  
+  Otherwise: return to step 3.<br />  
+  
+  After many iterations, the set of n points will approximate the distribution.  
+  
+  
'''Example1''' <br/>  '''Example1''' <br/>  
We want to sample from a target joint distribution f(x<sub>1</sub>, x<sub>2</sub>), which is not easy to sample from but the conditional pdfs f(x<sub>1</sub>x<sub>2</sub>) & f(x<sub>2</sub>x<sub>1</sub>) are very easy to sample from. We can find the stationary distribution (target distribution) using Gibbs sampling: <br/>  We want to sample from a target joint distribution f(x<sub>1</sub>, x<sub>2</sub>), which is not easy to sample from but the conditional pdfs f(x<sub>1</sub>x<sub>2</sub>) & f(x<sub>2</sub>x<sub>1</sub>) are very easy to sample from. We can find the stationary distribution (target distribution) using Gibbs sampling: <br/>  
Line 6,740:  Line 6,811:  
ezsurf(exp(x given ~ 2 ...)/i) this gives a n dimensional plot  ezsurf(exp(x given ~ 2 ...)/i) this gives a n dimensional plot  
</pre>  </pre>  
+  
+  ezsurf(fun) creates a graph of fun(x,y) using the surf function. fun is plotted over the default domain: 2π < x < 2π, 2π < y < 2π.  
+  http://www.mathworks.com/help/matlab/ref/ezsurf.html<br>  
'''Example:''' ezsurf((x+y)^2+(xy)^3)<br/>  '''Example:''' ezsurf((x+y)^2+(xy)^3)<br/>  
Line 6,748:  Line 6,822:  
'''Example:''' Face recognition <br />  '''Example:''' Face recognition <br />  
−  X is a greyscale image of the  +  X is a greyscale image of the person and Y is the person.<br /> 
Here,We have a 100 x 100 grid where each cell is a number from 0 to 255 representing the darkness of the cell (from white to black).<br>  Here,We have a 100 x 100 grid where each cell is a number from 0 to 255 representing the darkness of the cell (from white to black).<br>  
Let x be a vector of length 100*100=10,000 and y be a vector with each element being a picture of a person's face.<br>  Let x be a vector of length 100*100=10,000 and y be a vector with each element being a picture of a person's face.<br>  
Line 6,754:  Line 6,828:  
<br />[[Frequentist approach]]<br />  <br />[[Frequentist approach]]<br />  
−  *A frequentist would say X is a random variable and  +  *A frequentist would say X is a random variable and Y is not, so they would use Pr{xy} (given that y is Tom, how likely is it that x is an image of Tom?). 
<math>\displaystyle P(XY)</math>, y is person and x is how likely the picture is of this person. Here, y is known. <br/>  <math>\displaystyle P(XY)</math>, y is person and x is how likely the picture is of this person. Here, y is known. <br/>  
Line 6,801:  Line 6,875:  
'''Definition'''<br/>  '''Definition'''<br/>  
*Variance reduction is a procedure used to increase the precision of the estimates that can be obtained for a given number of iterations. Every output random variable from the simulation is associated with a variance which limits the precision of the simulation results. <br/>  *Variance reduction is a procedure used to increase the precision of the estimates that can be obtained for a given number of iterations. Every output random variable from the simulation is associated with a variance which limits the precision of the simulation results. <br/>  
−  *In order to make a simulation statistically efficient,(i.e. to obtain a greater precision and smaller confidence intervals for the output random variable of interest),variance reduction techniques can be used. The main ones are: Common random numbers, antithetic variates, control variates, importance sampling and stratified sampling), We will only be learning one of the methods  importance sampling. http://en.wikipedia.org/wiki/Variance_reduction  +  *In order to make a simulation statistically efficient,(i.e. to obtain a greater precision and smaller confidence intervals for the output random variable of interest),variance reduction techniques can be used. The main ones are: Common random numbers, antithetic variates, control variates, importance sampling and stratified sampling), We will only be learning one of the methods  importance sampling. Importance sampling is used to generate more statistically significant points rather than generating those points that do not have any value, such as generating in the middle of the bell curve rather than at the tail end of the bell curve. http://en.wikipedia.org/wiki/Variance_reduction 
*<br />It can be seen that the integral <math>\displaystyle\int \frac{h(x)f(x)}{g(x)}g(x)\,dx = \int \frac{f(x)}{g(x)}h(x) g(x)\,dx</math> is just <math> \displaystyle E_g(h(x)) \rightarrow</math>the expectation of h(x) with respect to g(x), where <math>\displaystyle \frac{f(x)}{g(x)} </math> is a weight <math>\displaystyle\beta(x)</math>. In the case where <math>\displaystyle f > g</math>, a greater weight for <math>\displaystyle\beta(x)</math> will be assigned. <br /><ref>  *<br />It can be seen that the integral <math>\displaystyle\int \frac{h(x)f(x)}{g(x)}g(x)\,dx = \int \frac{f(x)}{g(x)}h(x) g(x)\,dx</math> is just <math> \displaystyle E_g(h(x)) \rightarrow</math>the expectation of h(x) with respect to g(x), where <math>\displaystyle \frac{f(x)}{g(x)} </math> is a weight <math>\displaystyle\beta(x)</math>. In the case where <math>\displaystyle f > g</math>, a greater weight for <math>\displaystyle\beta(x)</math> will be assigned. <br /><ref>  
http://wikicoursenote.com/wiki/Stat341#Importance_Sampling_2  http://wikicoursenote.com/wiki/Stat341#Importance_Sampling_2  
</ref>  </ref>  
+  *Variance reduction uses the fact that the variance of a finite integral is zero. <br/>  
−  We  +  We would like to use simulation for this algorithm. We can use Monte Carlo Integration framework from previous classes. 
<math>E_f [h(x)] = \int h(x)f(x) dx</math>. The motivation is that a lot of integrals need to be calculated. <br/><br/>  <math>E_f [h(x)] = \int h(x)f(x) dx</math>. The motivation is that a lot of integrals need to be calculated. <br/><br/>  
−  +  '''Some addition knowledge:''' <br/>  
+  Common Random Numbers: The common random numbers variance reduction technique is a popular and useful variance reduction technique which applies when we are comparing at least two alternative configurations (of a system) instead of investigating a single configuration. <br/> <br/>  
'''Case 1 Basic Monte Carlo Integration''' <br/>  '''Case 1 Basic Monte Carlo Integration''' <br/>  
'''Idea:'''Evaluating an integral means calculating the area under the desired curve f(x).<br/>  '''Idea:'''Evaluating an integral means calculating the area under the desired curve f(x).<br/>  
Line 6,825:  Line 6,901:  
This is referred to as '''Monte Carlo Integration.''' http://web.mit.edu/~wingated/www/introductions/mcmcgibbsintro.pdf  This is referred to as '''Monte Carlo Integration.''' http://web.mit.edu/~wingated/www/introductions/mcmcgibbsintro.pdf  
−  Suppose we have integral of this form<br />  +  Suppose we have an integral of this form<br /> 
<math>I = \int_a ^b h(x)dx =\int_a ^b h(x) (\frac {ba}{ba} )dx =\int_a ^b h(x)(ba) (\frac {1}{ba} )dx =\int_a ^b w(x) f(x)dx </math><br />  <math>I = \int_a ^b h(x)dx =\int_a ^b h(x) (\frac {ba}{ba} )dx =\int_a ^b h(x)(ba) (\frac {1}{ba} )dx =\int_a ^b w(x) f(x)dx </math><br />  
Line 6,903:  Line 6,979:  
Consider <math>I = \int_0 ^1 e^x dx</math><br />  Consider <math>I = \int_0 ^1 e^x dx</math><br />  
−  The exact answer is e  1.  +  The exact answer is (e^1  e^0) = 2.718281828  1 = 1.718281828 
Comparing to the simulation, the matlab code is as follows:  Comparing to the simulation, the matlab code is as follows:  
Line 6,917:  Line 6,993:  
1.7178  1.7178  
</pre>  </pre>  
−  The answer 1.7178 is really  +  The answer 1.7178 is really close enough to the exact answer e  1 = 1.71828182846. The accuracy will increase if n is larger, for example n=100000000. 
<br/>  <br/>  
Line 7,033:  Line 7,109:  
== Class 23, Tuesday July 23 ==  == Class 23, Tuesday July 23 ==  
===Importance Sampling===  ===Importance Sampling===  
+  Start with  
+  <math>I = \int^{b}_{a} f(x)\,dx </math><br/> = <math>\int f(x)*(ba) * \frac{1}{(ba)}\,dx </math><br/>  
+  <math>\widehat{I} = \frac{1}{n} \sum_{i = 1}^{n} w({x_{i})}</math> where <math>w_{i} ~\sim Unif(a,b)</math>  
+  
Recall the definition of crude Monte Carlo Integration: <br/>  Recall the definition of crude Monte Carlo Integration: <br/>  
−  <math>E[  +  <math>E[h(X)]=\int f(x)h(x)\,dx</math><br/> 
If <math>x~\sim U(0,1)</math> and hence <math>\,f(x)=1</math>, then we have the basis of other variance reduction techniques. Now we consider what happens if X is not uniformly distributed.  If <math>x~\sim U(0,1)</math> and hence <math>\,f(x)=1</math>, then we have the basis of other variance reduction techniques. Now we consider what happens if X is not uniformly distributed.  
−  In the control variate case, we change the formula b adding and subtracting a known function h(x): basically, by adding zero to the integral,  +  In the control variate case, we change the formula b adding and subtracting a known function h(x): basically, by adding zero to the integral, keeping it unbiased and allowing us to have an easier time of solving it. In importance sampling, we will instead multiply by 1. The known function in this case will be g(x), which is selected under a few assumptions. 
There are cases where another distribution gives a better fit to integral to approximate, and results in a more accurate estimate; importance sampling is useful here.  There are cases where another distribution gives a better fit to integral to approximate, and results in a more accurate estimate; importance sampling is useful here.  
Line 7,047:  Line 7,127:  
 rare event is the event when you sample from its distribution, you rarely get an satisfied sample.<br/>   rare event is the event when you sample from its distribution, you rarely get an satisfied sample.<br/>  
<br/>  <br/>  
+  *Importance sampling can solve the cases listed above. It makes use of some functions that are easier to sample from. <br/>  
*Importance sampling is a variance reduction technique that can be used in the Monte Carlo method. Although it is not exactly like a Markov Chain Monte Carlo (MCMC) algorithm, it also approximately samples a vector where the mass function is specified up to some constant.<br/>  *Importance sampling is a variance reduction technique that can be used in the Monte Carlo method. Although it is not exactly like a Markov Chain Monte Carlo (MCMC) algorithm, it also approximately samples a vector where the mass function is specified up to some constant.<br/>  
*The idea behind importance sampling is that, certain values of the input random variables in a simulation have more impact on the parameter being estimated than the others. If these "important" values are emphasized by being sampled more frequently, then the estimator variance can be reduced.<br/>  *The idea behind importance sampling is that, certain values of the input random variables in a simulation have more impact on the parameter being estimated than the others. If these "important" values are emphasized by being sampled more frequently, then the estimator variance can be reduced.<br/>  
*Hence, the basic methodology in importance sampling is to choose a distribution which "encourages" the important values. This use of "biased" distributions will result in a biased estimator if it is applied directly in the simulation. (http://en.wikipedia.org/wiki/Importance_sampling)<br/>  *Hence, the basic methodology in importance sampling is to choose a distribution which "encourages" the important values. This use of "biased" distributions will result in a biased estimator if it is applied directly in the simulation. (http://en.wikipedia.org/wiki/Importance_sampling)<br/>  
*However, the simulation outputs are weighted to correct for the use of the biased distribution, and this ensures that the new importance sampling estimator is unbiased. (http://en.wikipedia.org/wiki/Importance_sampling)  *However, the simulation outputs are weighted to correct for the use of the biased distribution, and this ensures that the new importance sampling estimator is unbiased. (http://en.wikipedia.org/wiki/Importance_sampling)  
−  Example:<br/>  +  
+  '''Example''':<br/>  
+  
* Bit Error Rate on a channel.<br/>  * Bit Error Rate on a channel.<br/>  
The bit error rate (BER) is the number of bit errors over the total number of bits during a specific time. BER has no unit associated to it. BER is often written as a percentage. <br/>  The bit error rate (BER) is the number of bit errors over the total number of bits during a specific time. BER has no unit associated to it. BER is often written as a percentage. <br/>  
Line 7,063:  Line 7,146:  
As n approaches infinity, <math>\hat{I}</math> approaches <math>{I}</math>  As n approaches infinity, <math>\hat{I}</math> approaches <math>{I}</math>  
−  w(x) is called the  +  '''Note''': <br/> 
−  A good importance function will be large when the integrand is large and small otherwise.<br/>  +  
−  This is the importance sampling estimator of <math>I</math>, and is unbiased. That is, the estimation procedure is to generate i.i.d. samples from <math>g(x)</math>, and for each sample which exceeds the upper bound of the integral  +  Even though the uniform distribution sampling method only works for a definite integral, you can use still uniform distribution sampling method for I in the case of indefinite integral  this can be done by manipulating the function to adjust the integral range, such that the integral becomes definite. 
+  
+  w(x) is called the Importance Function. <br/>  
+  *A good importance function will be large when the integrand is large and small otherwise.<br/>  
+  
+  This is the importance sampling estimator of <math>I</math>, and is unbiased. That is, the estimation procedure is to generate i.i.d. samples from <math>g(x)</math>, and for each sample which exceeds the upper bound of the integral, the estimate is incremented by the weight W, evaluated at the sample value. The results are averaged over N trials. <br/>  
http://en.wikipedia.org/wiki/Importance_sampling <br/>  http://en.wikipedia.org/wiki/Importance_sampling <br/>  
Line 7,079:  Line 7,167:  
−  '''  +  
+  '''Example 1:'''<br/>  
<math>I=\int^{1}_{\infty} f(x)\,dx</math>, where <math>\displaystyle f(x) \sim N(0,1)</math><br/>  <math>I=\int^{1}_{\infty} f(x)\,dx</math>, where <math>\displaystyle f(x) \sim N(0,1)</math><br/>  
Define  Define  
Line 7,093:  Line 7,182:  
Therefore,<math>\widehat{I} = \frac{1}{n} \sum_{i = 1}^{n} h({x_{i})}</math> where <math>x_{i} ~\sim N(0,1)</math>  Therefore,<math>\widehat{I} = \frac{1}{n} \sum_{i = 1}^{n} h({x_{i})}</math> where <math>x_{i} ~\sim N(0,1)</math>  
which gives <math>\widehat{I}= \frac{\text{number of observations }<= 1}{n}</math><br/>  which gives <math>\widehat{I}= \frac{\text{number of observations }<= 1}{n}</math><br/>  
−  <br  +  <br> 
+  '''Note''': <br/>  
+  h(x) is acting as an indicator variable which follows a Bernoulli distribution with p = P(x<=1).<br/>  
+  h(x) is used to count the points greater than 1.  
−  Consider <math>I= \int h(x)f(x)\,dx </math> again. Importance sampling is used to overcome the following two cases:  +  
−  <br />*cases we don't know how to sample from f(x), because f(x) is a complicated distribution  +  '''Consider <math>I= \int h(x)f(x)\,dx </math> again.''' <br/> 
−  <br />*cases in which h(x) corresponds to a rare event over f (e.g. less than 3 in a standard normal distribution)  +  
−  <br />  +  Importance sampling is used to overcome the following two cases: <br/> 
+  *cases we don't know how to sample from f(x), because f(x) is a complicated distribution. <br/>  
+  *cases in which h(x) corresponds to a rare event over f (e.g. less than 3 in a standard normal distribution). <br/>  
+   In the second case, using the basic method without importance sampling will result in high variability in the simulated results (which goes against the purpose of variance reduction) <br/>  
+  
+  
<math>\begin{align}  <math>\begin{align}  
Line 7,120:  Line 7,217:  
Note that <math>\hat{I}</math> is an unbiased estimator for <math>I</math> as <math>\ E_x(\hat{I}) = E_x(\frac{1}{n} \sum_{i = 1}^{n} w(X_i)) = \frac{1}{n} \sum_{i = 1}^{n} E_x(\frac{h(X_i)f(X_i)}{g(X_i)}) = \frac{1}{n} \sum_{i = 1}^{n} \int \frac{h(x)f(x)}{g(x)}g(x)dx = \frac{1}{n} \sum_{i = 1}^{n} I = I</math>  Note that <math>\hat{I}</math> is an unbiased estimator for <math>I</math> as <math>\ E_x(\hat{I}) = E_x(\frac{1}{n} \sum_{i = 1}^{n} w(X_i)) = \frac{1}{n} \sum_{i = 1}^{n} E_x(\frac{h(X_i)f(X_i)}{g(X_i)}) = \frac{1}{n} \sum_{i = 1}^{n} \int \frac{h(x)f(x)}{g(x)}g(x)dx = \frac{1}{n} \sum_{i = 1}^{n} I = I</math>  
−  [[Problem:]]The variance of <math> \widehat{I}</math> could be very large with bad choice of g. <br/>  +  ''''''[[Problem:]]''''''The variance of <math> \widehat{I}</math> could be very large with bad choice of g. <br/> 
−  '''Advice 1''':Choose g such that g has thicker tails compare to f.  +  
+  '''Advice 1''': <br/>  
+  Choose g such that g has thicker tails compare to f. <br/>  
In general, if over a set A, g is small but f is large, then f(x)/g(x) could be large. ie: the variance could be large. (the values for which h(x) is exceedingly small) <br/>  In general, if over a set A, g is small but f is large, then f(x)/g(x) could be large. ie: the variance could be large. (the values for which h(x) is exceedingly small) <br/>  
−  '''Advice 2'''  +  '''Advice 2'''： 
−  In general, it is better to choose g such that it is similar to f in terms of shape, but has thicker tails.  +  Choose g to have similar shape with f. <br/> 
+  In general, it is better to choose g such that it is similar to f in terms of shape, but has thicker tails. <br/>  
<br><br>  <br><br>  
−  <b>Procedure</b><br>  +  <b>'''Procedure'''</b><br> 
+  
1. Sample <math> x_{1}, x_{2}, ..., x_{n} ~\sim g(x) </math> <br /><br />  1. Sample <math> x_{1}, x_{2}, ..., x_{n} ~\sim g(x) </math> <br /><br />  
2. <math>\widehat{I} = \frac{1}{n} \sum_{i = 1}^{n} w({x_{i})}</math> where <math> w(x_i) = \frac{h(x_i)f(x_i)}{g(x_i)} </math> for <math>i=1\dots n</math><br />  2. <math>\widehat{I} = \frac{1}{n} \sum_{i = 1}^{n} w({x_{i})}</math> where <math> w(x_i) = \frac{h(x_i)f(x_i)}{g(x_i)} </math> for <math>i=1\dots n</math><br />  
+  
+  
Line 7,149:  Line 7,252:  
which gives <math>\widehat{I}= \frac{\text{number of observations }< 3}{n}</math><br/>  which gives <math>\widehat{I}= \frac{\text{number of observations }< 3}{n}</math><br/>  
−  Matlab Code: <br/>  +  
+  
+  '''Matlab Code:''' <br/>  
+  
<pre style="fontsize:16px">  <pre style="fontsize:16px">  
−  
n = 200;  n = 200;  
x = randn(1,n);  x = randn(1,n);  
Line 7,158:  Line 7,263:  
>> mean(I)  >> mean(I)  
>> var(I) % to calculate the variance of the estimates  >> var(I) % to calculate the variance of the estimates  
−  
</pre>  </pre>  
<br>  <br>  
+  '''Comments on Example 2''':<br/>  
−  
*Since observations less than 3 are a relatively rare event, this method will give us a relatively high variance. <br/>  *Since observations less than 3 are a relatively rare event, this method will give us a relatively high variance. <br/>  
*To illustrate this, suppose we sample 100 points each time for many times, we will be getting mostly 0's and some 1's and occasionally 2's. This data has large variances.<br/>  *To illustrate this, suppose we sample 100 points each time for many times, we will be getting mostly 0's and some 1's and occasionally 2's. This data has large variances.<br/>  
+  
'''Note''' ： h(x) is counting the number of observations that are less than 3.  '''Note''' ： h(x) is counting the number of observations that are less than 3.  
+  
+  
'''Remarks'''：  '''Remarks'''：  
*We can actually compute the form of <math>\displaystyle g(x)</math> to have optimal variance. <br>Mathematically, it is to find <math>\displaystyle g(x)</math> subject to <math>\displaystyle \min_g [\ E_g([y(x)]^2)  (E_g[y(x)])^2\ ]</math><br>  *We can actually compute the form of <math>\displaystyle g(x)</math> to have optimal variance. <br>Mathematically, it is to find <math>\displaystyle g(x)</math> subject to <math>\displaystyle \min_g [\ E_g([y(x)]^2)  (E_g[y(x)])^2\ ]</math><br>  
+  
It can be shown that the optimal <math>\displaystyle g(x)</math> is <math>\displaystyle {h(x)f(x)}</math>. Using the optimal <math>\displaystyle g(x)</math> will minimize the variance of estimation in Importance Sampling. This is of theoretical interest but not useful in practice. As we can see, if we can actually show the expression of g(x), we must first have the value of the integrationwhich is what we want in the first place. <br/>  It can be shown that the optimal <math>\displaystyle g(x)</math> is <math>\displaystyle {h(x)f(x)}</math>. Using the optimal <math>\displaystyle g(x)</math> will minimize the variance of estimation in Importance Sampling. This is of theoretical interest but not useful in practice. As we can see, if we can actually show the expression of g(x), we must first have the value of the integrationwhich is what we want in the first place. <br/>  
−  +  In practice, we shall choose <math>\displaystyle g(x)</math> which has similar shape as <math>\displaystyle f(x)</math> but with a thicker tail than <math>\displaystyle f(x)</math> in order to avoid the problem mentioned above.<br>  
+  
+  The case when <math> g(x) </math> is important it should have the same support. If <math> g(x) </math> does not have the same support then it may not be able to sample from <math> f </math> like before. Also, if <math> g(x) </math> is not a good choice then it increases the variance very badly. <br/>  
−  
−  +  '''Note:'''  
+  Normalized imporatance sampling is biased, but it is asymptotically unbiased.<br/>  
+  
<math>I=\int h(x)f(x)dx</math> <br>  <math>I=\int h(x)f(x)dx</math> <br>  
<math>I=\int \frac{h(x)f(x)}{g(x)} g(x) dx </math><br>  <math>I=\int \frac{h(x)f(x)}{g(x)} g(x) dx </math><br>  
Line 7,190:  Line 7,301:  
[[File:IMP ex part 2.png600px]] <br \>  [[File:IMP ex part 2.png600px]] <br \>  
Source: STAT 340 Spring 2010 Course Notes <br>  Source: STAT 340 Spring 2010 Course Notes <br>  
+  
+  
'''Example:'''<br>  '''Example:'''<br>  
Line 7,203:  Line 7,316:  
Actual value of the integral is 1 <br>  Actual value of the integral is 1 <br>  
−  Matlab Code:<br>  +  '''Matlab Code''':<br> 
+  
<pre style="fontsize:16px">  <pre style="fontsize:16px">  
>> clear all  >> clear all  
Line 7,209:  Line 7,323:  
>> n=1000;  >> n=1000;  
>> u=rand(1,n);  >> u=rand(1,n);  
−  >> x=log(u);  +  >> x=log(u); % Generates number from exponential distribution using inverse transformation method 
>> w=(1./(1+x).^2).*exp(x);  >> w=(1./(1+x).^2).*exp(x);  
>> sum(w)/n  >> sum(w)/n  
Line 7,219:  Line 7,333:  
'''Another Method'''<br />  '''Another Method'''<br />  
−  +  By changing the variable so that the bounds is (0,1), we can apply the Unif(0,1) method: <br />  
Let <math>y= \frac{1}{x+1}, dy= \frac{1}{(x+1)^2}dx =y^2dx</math><br />  Let <math>y= \frac{1}{x+1}, dy= \frac{1}{(x+1)^2}dx =y^2dx</math><br />  
Line 7,225:  Line 7,339:  
We can express the integral as <br />  We can express the integral as <br />  
<math>\int^{1}_{0} \frac {1}{y^2} y^2 dy =\int^{1}_{0} 1 dy </math><br />  <math>\int^{1}_{0} \frac {1}{y^2} y^2 dy =\int^{1}_{0} 1 dy </math><br />  
−  which we recognise that it is just a <math>Unif(0,1)</math> and the result follows.  +  which we recognise that it is just a <math>Unif(0,1)</math> and the result follows. <br /> 
+  <br />  
+  '''The following are general forms for the change of variable method for different cases'''  
+  :<math>  
+  \int_a^{+\infty}f(x) \, dx =\int_0^1 f\left(a + \frac{u}{1u}\right) \frac{du}{(1u)^2} </math>  
+  
+  :<math>  
+  \int_{\infty}^a f(x) \, dx = \int_0^1 f\left(a  \frac{1u}{u}\right) \frac{du}{u^2}</math>  
−  ===Problem of Importance Sampling===  +  :<math> 
+  \int_{\infty}^{+\infty} f(x) \, dx = \int_{1}^{+1} f\left( \frac{u}{1u^2} \right) \frac{1+u^2}{(1u^2)^2} \, du,  
+  </math>  
+  Source: Wikipedia Numerical Integration  
+  
+  <math>Insert formula here</math>===Problem of Importance Sampling===  
The variance of <math>\hat{I}</math> '''could be very large''' (infinitely large) with a bad choice of <math>g</math> <br>  The variance of <math>\hat{I}</math> '''could be very large''' (infinitely large) with a bad choice of <math>g</math> <br>  
−  
<math>\displaystyle Var(w) = E(w^2)  (E(w))^2 </math> <br>  <math>\displaystyle Var(w) = E(w^2)  (E(w))^2 </math> <br>  
<math> \begin{align}  <math> \begin{align}  
−  E(w^2) &= \int (\frac{h(x)f(x)}{g(x)})^2 g(x) dx  +  E(w^2) &= \int (\frac{h(x)f(x)}{g(x)})^2 g(x) dx , where w = (\frac{h(x)f(x)}{g(x)})\\ 
&= \int (\frac{h^2(x)f^2(x)}{g^2(x)}) g(x) dx \\  &= \int (\frac{h^2(x)f^2(x)}{g^2(x)}) g(x) dx \\  
&= \int (\frac{h^2(x)f^2(x)}{g(x)}) dx  &= \int (\frac{h^2(x)f^2(x)}{g(x)}) dx  
Line 7,389:  Line 7,514:  
==Class 24, Thursday, July 25, 2013==  ==Class 24, Thursday, July 25, 2013==  
===Importance Sampling===  ===Importance Sampling===  
−  Importance Sampling is the most fundamental variance reduction technique  +  Importance Sampling is the most fundamental variance reduction technique and usually leads to a dramatic variance reduction. <br /> 
−  Importance sampling involves choosing a sampling distribution that  +  Importance sampling involves choosing a sampling distribution that favour important samples*.(Simulation and the Monte Carlo Method, Reuven Y. Rubinstein) <br /> 
+  
+  * Here "favour important samples" implies encouraging the occurrence of the desired event or part of the desired event. For instance, if the event of interest is rare (probability of occurrence is close to zero), we "favour important samples" by choosing a sampling distribution such that the event has higher probability of occurrence.  
Definition of importance sampling from Wikipedia:<br>  Definition of importance sampling from Wikipedia:<br>  
Line 7,401:  Line 7,528:  
If g(x) is another probability density function, <br />  If g(x) is another probability density function, <br />  
+  note: in summary, a good importance sampling function g(x) should satisfies:<br />  
+  
+  1. g(x) > 0 whenever f(x)not equal to 0<br />  
+  2. g(x) should be equal or close to the absolute value of f(x)<br />  
+  3. easy to simulate values from g(x)<br />  
+  4. easy to compute the density of g(x)<br />  
+  
+  original source is here<br /> http://ib.berkeley.edu/labs/slatkin/eriq/classes/guest_lect/mc_lecture_notes.pdf<br />  
then we have: <br />  then we have: <br />  
Line 7,408:  Line 7,543:  
In order to estimate I we have:<br/>  In order to estimate I we have:<br/>  
−  <math>\widehat{I}=\frac{1}{n}\sum_{i=1}^{n}w(x)</math> and <math>g^{*}(x) = \frac{h(x)f(x)}{\int h(x)f(x)dx}</math>  +  <math>\widehat{I}=\frac{1}{n}\sum_{i=1}^{n}w(x)</math> and <math>g^{*}(x) = \frac{h(x)f(x)}{\int h(x)f(x)dx}</math>, where <math> h(x)>=0 </math> for all x <br> 
+  
+  Higher values of n correspond to values of <math>\widehat{I}</math> closer to <math>{I}</math>, which approaches <math>\widehat{I}</math> as n approaches infinity.  
'''Note:''' g(x) should be chosen carefully. It should be easy to sample from. Also, since this method is for minimizing the variance, g(x) should be chosen in a manner such that the variance is minimized. g*(x) is the distribution that minimizes the variance.  '''Note:''' g(x) should be chosen carefully. It should be easy to sample from. Also, since this method is for minimizing the variance, g(x) should be chosen in a manner such that the variance is minimized. g*(x) is the distribution that minimizes the variance.  
Line 7,419:  Line 7,556:  
:<math>\, Var(I) = Var(\frac{1}{n} \sum_{i = 1}^{n} w({x_{i})})= Var(w)/n </math> <br>  :<math>\, Var(I) = Var(\frac{1}{n} \sum_{i = 1}^{n} w({x_{i})})= Var(w)/n </math> <br>  
+  
+  Note:  
+  This expression has equivalent to the summation of all variances of W, because W’s are independent, hence covariance terms are zero.  
Line 7,434:  Line 7,574:  
:'''Note''' ： No matter what g is, the second term is always constant with respect to g at <math> I^2</math>.<br>  :'''Note''' ： No matter what g is, the second term is always constant with respect to g at <math> I^2</math>.<br>  
+  since <math> I^2 </math> is constant with respect to g, if we want to minimize the variance, we only need to consider the first term.<br>  
So, we need to minimize the first term.<br />  So, we need to minimize the first term.<br />  
Line 7,464:  Line 7,605:  
Therefore, at <math>\,g(x), Var(w)=I^2 I^2=0</math><br />  Therefore, at <math>\,g(x), Var(w)=I^2 I^2=0</math><br />  
'''Note that although this proof uses the assumption of h(x) ≥ 0, the result still holds for functions h(x) that are not always nonnegative (however, the variance will not be 0)<br/>  '''Note that although this proof uses the assumption of h(x) ≥ 0, the result still holds for functions h(x) that are not always nonnegative (however, the variance will not be 0)<br/>  
−  More specifically, since <math>\int h(x)f(x) dx \geq \int h(x)f(x) dx = I</math> where <math>h(x)</math> can be negative, so <math>\, Var(w) \geq I^2I^2 = 0 </math>, and as a result Var(w) will always be nonnegative.  +  More specifically, since <math>\int h(x)f(x) dx \geq \int h(x)f(x) dx = I</math> where <math>h(x)</math> can be negative, so <math>\, Var(w) \geq I^2I^2 = 0 </math>, and as a result Var(w) will always be nonnegative.<br/> 
+  '''  
+  Remark: Since <math> I^2 </math> is constant of g, we only consider minimizing the first term to minimize the variance.  
'''  '''  
−  === Normalized Importance  +  === Normalized Importance Sampling === 
<math>I= \frac{\int h(x)f(x) dx}{\int f(x) dx}</math> since f(x) is a pdf, and the integral is just equal to 1<br />  <math>I= \frac{\int h(x)f(x) dx}{\int f(x) dx}</math> since f(x) is a pdf, and the integral is just equal to 1<br />  
Line 7,482:  Line 7,625:  
<math>\beta_i^*= \biggl[ \frac{\beta_1}{\beta_1+...+\beta_n}, \frac{\beta_2}{\beta_1+...+\beta_n}, ... , \frac{\beta_n}{\beta_1+...+\beta_n} \biggr]</math><br />  <math>\beta_i^*= \biggl[ \frac{\beta_1}{\beta_1+...+\beta_n}, \frac{\beta_2}{\beta_1+...+\beta_n}, ... , \frac{\beta_n}{\beta_1+...+\beta_n} \biggr]</math><br />  
+  <br/>  
+  Note: Above is not included in exam.  
:'''Note:'''  :'''Note:'''  
Line 7,488:  Line 7,633:  
:*Normalized Importance Sampling however, performs worse than regular importance sampling as we are approximating the normalizing constant  :*Normalized Importance Sampling however, performs worse than regular importance sampling as we are approximating the normalizing constant  
−  [http://www.youtube.com/watch?v=gYvlnu5AAzE Here is video explaining normalized importance sampling sightly differently.]  +  [http://www.youtube.com/watch?v=gYvlnu5AAzE Here is a video explaining normalized importance sampling sightly differently.] 
===Final Exam Review===  ===Final Exam Review===  
+  
+  '''Summary of Final Exam Topics:'''  
+  ''PreMidterm:''  
+  • Multiplicative Congruential Algorithm  
+  • Inverse Transform Method  
+  • Acceptance Rejection Method  
+  • Multivariate Random Variable Generation  
+  • Vector Acceptance Rejection Method  
+  ''PostMidterm:''  
+  • Poisson Process  
+  • Markov Chains (MC)  
+  • Page Rank (MC Application)  
+  • Markov Chain Monte Carlo (MCMC)  
+  • MetropolisHasting Algorithm (MCMC Application)  
+  • Simulated Annealing (MCMC Application)  
+  • Gibbs Sampling (MetropolisHasting adaptation)  
+  • Monte Carlo Integration  
+  • Importance Sampling  
+  
+  
+  Only review of material not covered on the midterm, the final will be cumulative<br/>  
+  For review of material covered on the midterm refer to class 12  June 13th.  
'''Stochastic Processes''' (we learned Poisson Process and Markov Chain).  '''Stochastic Processes''' (we learned Poisson Process and Markov Chain).  
Line 7,498:  Line 7,665:  
*The two most important stochastic processes we looked at in this term are Poission Process and Markov Chain.<br/>  *The two most important stochastic processes we looked at in this term are Poission Process and Markov Chain.<br/>  
−  +  ===Poisson Process=== (useful for counting number of arrivals): <br/>  
 two assumptions: <br/>   two assumptions: <br/>  
# the number of arrivals in nonoverlapping intervals are independent <br/>  # the number of arrivals in nonoverlapping intervals are independent <br/>  
Line 7,530:  Line 7,697:  
'''Markov Chain''': <br/>  '''Markov Chain''': <br/>  
Recall that:<br />  Recall that:<br />  
−  +  *A Markov Chain is a discrete random process which transits from one state to another. The number of states in a Markov Chain can be finite or countable.<br />  
−  +  
−  <math>Pr(X_t=x_tX_{t1}=x_{t1},..., X_1=x_1)= Pr(X_t=x_tX_{t1}=x_{t1})</math>. In other words, the current state only depends on the previous state and no other prior states. This property is also called the "Markov property".<br />  +  *A Markov Chain has the Memoriless Property: 
+  <math>Pr(X_t=x_tX_{t1}=x_{t1},..., X_1=x_1)= Pr(X_t=x_tX_{t1}=x_{t1})</math>.  
+  <br>In other words, the current state only depends on the previous state and no other prior states. This property is also called the "Markov property".<br />  
−  +  *The possible values of X<sub>i</sub> are called the "state space" of the chain.  
−  +  *Transition probability P<sub>ij</sub> = Pr {x<sub>t+1</sub>=j  x<sub>t</sub> = i} = P(i,j)<br/>  
−  +  *Transition matrix P = [P<sub>11</sub> ... P<sub>1n</sub> ; ... ; P<sub>n1</sub> ... P<sub>nn</sub>]. where P<sub> ij</sub> >= 0, row sum = 1 <br/>  
−  +  *Nstep transition matrix P<sub>n(i,j)</sub> = Pr {x<sub>t+n</sub>=j  x<sub>t</sub> = i}, P<sub>n</sub> = P<sup>n</sup> <br/>  
−  +  *Marginal distribution:<math>\mu_1~ = \mu_0P</math> <br>  
In general, <math>\mu_n~ = \mu_0P^n</math><br />  In general, <math>\mu_n~ = \mu_0P^n</math><br />  
+  where <math> \mu_0</math> is initial dust.<br>  
−  +  *'''Stationary distribution''': <math>\pi</math> = <math>\pi</math> P <br/>  
−  +  There are three conditions to calculate Stationary Distribution<br/>  
−  +  1. <math>\mu_1~ = \mu_0P</math> <br>  
+  2. sum of <math>\pi</math>= 1<br>  
+  3. <math>\pi</math> is greater than 0<br>  
+  *'''Limiting distribution''':  
<math>\lim_{n\to \infty} P^n= \left[ {\begin{array}{ccc}  <math>\lim_{n\to \infty} P^n= \left[ {\begin{array}{ccc}  
\pi_1 \\  \pi_1 \\  
Line 7,575:  Line 7,748:  
<math>P_i= (1d) + d\cdot \sum_j \frac {L_{ij}P_j}{c_j}</math>, where 0 < d < 1 is constant <br/>  <math>P_i= (1d) + d\cdot \sum_j \frac {L_{ij}P_j}{c_j}</math>, where 0 < d < 1 is constant <br/>  
where <math> L_{ij} </math> is 1 if j has link to i, and 0 otherwise;  where <math> L_{ij} </math> is 1 if j has link to i, and 0 otherwise;  
−  <math> C_j = \sum_i L_{  +  <math> C_j = \sum_i L_{ij} </math> <br/> 
+  Note: we solved this using systems of equations or eigenvalues and eigenvectors  
'''Matrix form:''' <br/>  '''Matrix form:''' <br/>  
Line 7,585:  Line 7,759:  
'''Markov Chain Monte Carlo (MCMC)''': <br/>  '''Markov Chain Monte Carlo (MCMC)''': <br/>  
−  Recall that MCMC is a special form of stochastic process  +  Recall that MCMC is a special form of stochastic process where X<sub>t</sub> depends only on X<sub>t1</sub><br/><br/> 
+  The two applications of MCMC are Metropolis–Hasting algorithm and Simulated Annealing.<br/>  
+  
 ''''''Metropolis–Hasting Algorithm'''''': <br/>   ''''''Metropolis–Hasting Algorithm'''''': <br/>  
If we have target distribution f, which we want to sample from, then<br/>  If we have target distribution f, which we want to sample from, then<br/>  
Line 7,613:  Line 7,789:  
1) Set T to be a large number, Set i = 0, <math>X_{t}</math> = 0<br />  1) Set T to be a large number, Set i = 0, <math>X_{t}</math> = 0<br />  
−  2) <math>Y</math>~<math>  +  2) <math>Y</math>~<math>Q(YX)</math><br/> 
−  3) <math>r(x,y) = \min\{\frac{f(y)}{f(x)},1\}</math><br />  +  3) <math>r(x,y) = \min\{\frac{f(y)}{f(x)},1\}</math><br /> Since q(.) is symetric 
4) <math>U</math>~<math>U(0,1)</math><br/>  4) <math>U</math>~<math>U(0,1)</math><br/>  
−  5) If <math>U<r</math>, <math>X_{t+1}=Y</math>; else <math>X_{t+1}=X_t</math><br/>  +  5) If <math>U<r</math>, <math>X_{t+1}=Y</math>; else <math>X_{t+1}=X_t</math><br/> 
6) Decrease T. Return to Step 2.<br/><br/>  6) Decrease T. Return to Step 2.<br/><br/>  
+  
+  * note: popular candidates for Q(YX) are uniform distribution and normal distribution.(symmetric)  
Line 7,651:  Line 7,829:  
 '''Proof of Simulated Annealing Algorithm (convergence):''' <br/>   '''Proof of Simulated Annealing Algorithm (convergence):''' <br/>  
−  Detailed Balance: <math>f(x) P(yx) = f(y) P(xy)</math><br/>  +  Detailed Balance: <math>\,f(x) P(yx) = f(y) P(xy)</math><br/> 
Since q(yx) is symmetric > q(yx)=q(xy)<br/>  Since q(yx) is symmetric > q(yx)=q(xy)<br/>  
Line 7,661:  Line 7,839:  
2) <math>\frac {f(y)}{f(x)}>1</math> <br>  2) <math>\frac {f(y)}{f(x)}>1</math> <br>  
−  => <math>r(x,y) = 1</math><br>  +  => <math>\, r(x,y) = 1</math><br> 
<math>\begin{align}  <math>\begin{align}  
Line 7,671:  Line 7,849:  
<math>\begin{align}  <math>\begin{align}  
\text{RHS} & = f(y)P(xy)= f(y)q(xy)r(y,x) \\  \text{RHS} & = f(y)P(xy)= f(y)q(xy)r(y,x) \\  
−  & =f(y)q(xy)  +  & =f(y)q(xy)\times 1 \\ 
& =f(y)q(xy) = \text{LHS}  & =f(y)q(xy) = \text{LHS}  
\end{align}</math><br>  \end{align}</math><br>  
Line 7,678:  Line 7,856:  
'''Gibbs Sampling''':<br>  '''Gibbs Sampling''':<br>  
The most widely used version of the MetropolisHastings algorithm is the Gibbs sampler. <br />  The most widely used version of the MetropolisHastings algorithm is the Gibbs sampler. <br />  
−  This sampling method is useful when dealing with multivariable  +  This sampling method is useful when dealing with multivariable distributions.<br> 
<math>f(x_1, x_2, ..., x_d)</math><br />  <math>f(x_1, x_2, ..., x_d)</math><br />  
Line 7,726:  Line 7,904:  
*This method is one of the Monte Carlo methods that numerically computes definite integrals. <br/>  *This method is one of the Monte Carlo methods that numerically computes definite integrals. <br/>  
−  *The above integral can be rewritten as  +  *The above integral can be rewritten as following:<br> 
−  <math>I = \int_a^b h(x)dx = \int_a^b h(x) \frac{ba}{ba} dx = \int_a^b \frac{h(x)}{ba} (ba) dx </math><br/>  +  <math>I = \int_a^b h(x)dx = \int_a^b h(x) \frac{ba}{ba} dx = \int_a^b \frac{h(x)}{ba} (ba) dx </math> where <math> U(a,b) = 1/(ba) </math> <br/> 
Line 7,736:  Line 7,914:  
<math>I = \int h(x)f(x)dx</math><br />  <math>I = \int h(x)f(x)dx</math><br />  
−  <math>\hat{I} = \frac{1}{n} \sum _{i=1}^n h(x_i) , x_i \sim f</math>  +  <math>\hat{I} = \frac{1}{n} \sum _{i=1}^n h(x_i) , \text{where} \ x_i \sim f</math> 
+  
+  ===Importance Sampling===  
+  Importance Sampling is a useful technique for variance reduction.<br />  
+  
+  Using importance sampling, we have:<br/>  
+  
+  <math>I=\int_{a}^{b}f(x)dx = \int_{a}^{b}f(x)(ba) \times \frac{1}{ba}dx</math> <br />  
+  
+  If g(x) is another probability density function, <br />  
+  
+  <math>I = \int h(x)f(x)\,dx =\int\frac{h(x)f(x)}{g(x)}\times g(x)\,dx</math>, where <math>w(x) = \frac{h(x)f(x)}{g(x)}</math><br />  
+  
+  To approximate I,<br/>  
+  
+  <math>\widehat{I}=\frac{1}{n}\sum_{i=1}^{n}w(x)</math> and <math>g^{*}(x) = \frac{h(x)f(x)}{\int h(x)f(x)dx}</math>, where <math> h(x)>=0 </math> for all x <br>  
+  
+  '''Note:''' g(x) should be chosen carefully so that its distribution would minimize the variance. 
Latest revision as of 09:46, 30 August 2017
If you use ideas, plots, text, code and other intellectual property developed by someone else in your `wikicoursenote' contribution , you have to cite the original source. If you copy a sentence or a paragraph from work done by someone else, in addition to citing the original source you have to use quotation marks to identify the scope of the copied material. Evidence of copying or plagiarism will cause a failing mark in the course.
Example of citing the original source
Assumptions Underlying Principal Component Analysis can be found here<ref>http://support.sas.com/publishing/pubcat/chaps/55129.pdf</ref>
Contents
 1 Important Notes
 2 Introduction, Class 1  Tuesday, May 7
 3 Class 2  Thursday, May 9
 4 Class 3  Tuesday, May 14
 5 Summary of Inverse Transform Method
 6 Class 4  Thursday, May 16
 7 Class 5  Tuesday, May 21
 8 AcceptanceRejection Method
 9 Class 6  Thursday, May 23
 10 Class 7  Tuesday, May 28
 11 Class 8  Thursday, May 30, 2013
 12 Class 9  Tuesday, June 4, 2013
 13 Class 10  Thursday June 6th 2013
 14 Summary of vector acceptancerejection sampling
 15 Class 11  Tuesday，June 11, 2013
 16 Class 12  Thursday，June 13, 2013
 16.1 Midterm Review
 16.2 Multiplicative Congruential Algorithm
 16.3 Inverse Transformation Method
 16.4 AcceptanceRejection Method
 16.5 Multivariate
 16.6 Vector A/R Method
 16.7 Common distribution
 16.8 Exponential
 16.9 Normal
 16.10 Gamma
 16.11 Bernoulli
 16.12 Binomial
 16.13 Beta Distribution
 16.14 Geometric
 16.15 Poisson
 17 Class 13  Tuesday June 18th 2013
 18 Class 14  Thursday June 20th 2013
 19 Class 15  Tuesday June 25th 2013
 20 Class 16  Thursday June 27th 2013
 21 Class 17  Tuesday July 2nd 2013
 22 Class 18  Thursday July 4th 2013
 23 Class 19  Tuesday July 9th 2013
 24 Class 20  Thursday July 11th 2013
 25 Class 21  Tuesday July 16, 2013
 26 Class 22, Thursday, July 18, 2013
 27 Class 23, Tuesday July 23
 28 Class 24, Thursday, July 25, 2013
Important Notes
To make distinction between the material covered in class and additional material that you have add to the course, use the following convention. For anything that is not covered in the lecture write:
In the news recently was a story that captures some of the ideas behind PCA. Over the past two years, Scott Golder and Michael Macy, researchers from Cornell University, collected 509 million Twitter messages from 2.4 million users in 84 different countries. The data they used were words collected at various times of day and they classified the data into two different categories: positive emotion words and negative emotion words. Then, they were able to study this new data to evaluate subjects' moods at different times of day, while the subjects were in different parts of the world. They found that the subjects generally exhibited positive emotions in the mornings and late evenings, and negative emotions midday. They were able to "project their data onto a smaller dimensional space" using PCS. Their paper, "Diurnal and Seasonal Mood Vary with Work, Sleep, and Daylength Across Diverse Cultures," is available in the journal Science.<ref>http://www.pcworld.com/article/240831/twitter_analysis_reveals_global_human_moodiness.html</ref>.
Assumptions Underlying Principal Component Analysis can be found here<ref>http://support.sas.com/publishing/pubcat/chaps/55129.pdf</ref>
Introduction, Class 1  Tuesday, May 7
Course Instructor: Ali Ghodsi
Lecture:
001: T/Th 8:309:50am MC1085
002: T/Th 1:002:20pm DC1351
Tutorial:
2:303:20pm Mon M3 1006
Office Hours:
Friday at 10am, M3 4208
Midterm
Monday June 17,2013 from 2:30pm3:20pm
Final
Saturday August 10,2013 from 7:30pm10:00pm
TA(s):
TA  Day  Time  Location 

Lu Cheng  Monday  3:305:30 pm  M3 3108, space 2 
Han ShengSun  Tuesday  4:006:00 pm  M3 3108, space 2 
Yizhou Fang  Wednesday  1:003:00 pm  M3 3108, space 1 
Huan Cheng  Thursday  3:005:00 pm  M3 3111, space 1 
Wu Lin  Friday  11:001:00 pm  M3 3108, space 1 
Four Fundamental Problems
1 Classification: Given input object X, we have a function which will take this input X and identify which 'class (Y)' it belongs to (Discrete Case)
i.e taking value from x, we could predict y.
(For example, if you have 40 images of oranges and 60 images of apples (represented by x), you can estimate a function that takes the images and states what type of fruit it is  note Y is discrete in this case.)
2 Regression: Same as classification but in the continuous case except y is non discrete. Results from regression are often used for prediction,forecasting and etc. (Example of stock prices, height, weight, etc.)
(A simple practice might be investigating the hypothesis that higher levels of education cause higher levels of income.)
3 Clustering: Use common features of objects in same class or group to form clusters.(in this case, x is given, y is unknown; For example, clustering by provinces to measure average height of Canadian men.)
4 Dimensionality Reduction (also known as Feature extraction, Manifold learning): Used when we have a variable in high dimension space and we want to reduce the dimension
Applications
Most useful when structure of the task is not well understood but can be characterized by a dataset with strong statistical regularity
Examples:
 Computer Vision, Computer Graphics, Finance (fraud detection), Machine Learning
 Search and recommendation (eg. Google, Amazon)
 Automatic speech recognition, speaker verification
 Text parsing
 Face identification
 Tracking objects in video
 Financial prediction(e.g. credit cards)
 Fraud detection
 Medical diagnosis
Course Information
Prerequisite: (One of CS 116, 126/124, 134, 136, 138, 145, SYDE 221/322) and (STAT 230 with a grade of at least 60% or STAT 240) and (STAT 231 or 241)
Antirequisite: CM 361/STAT 341, CS 437, 457
General Information
 No required textbook
 Recommended: "Simulation" by Sheldon M. Ross
 Computing parts of the course will be done in Matlab, but prior knowledge of Matlab is not essential (will have a tutorial on it)
 First midterm will be held on Monday, June 17 from 2:30 to 3:30
 Announcements and assignments will be posted on Learn.
 Other course material on: http://wikicoursenote.com/wiki/
 Log on to both Learn and wikicoursenote frequently.
 Email all questions and concerns to UWStat340@gmail.com. Do not use your personal email address! Do not email instructor or TAs about the class directly to their personal accounts!
Wikicourse note (complete at least 12 contributions to get 10% of final mark):
When applying for an account in the wikicourse note, please use the quest account as your login name while the uwaterloo email as the registered email. This is important as the quest id will be used to identify the students who make the contributions.
Example:
User: questid
Email: questid@uwaterloo.ca
After the student has made the account request, do wait for several hours before students can login into the account using the passwords stated in the email. During the first login, students will be ask to create a new password for their account.
As a technical/editorial contributor: Make contributions within 1 week and do not copy the notes on the blackboard.
All contributions are now considered general contributions you must contribute to 50% of lectures for full marks
 A general contribution can be correctional (fixing mistakes) or technical (expanding content, adding examples, etc.) but at least half of your contributions should be technical for full marks.
Do not submit copyrighted work without permission, cite original sources. Each time you make a contribution, check mark the table. Marks are calculated on an honour system, although there will be random verifications. If you are caught claiming to contribute but have not, you will not be credited.
Wikicoursenote contribution form : https://docs.google.com/forms/d/1Sgq0uDztDvtcS5JoBMtWziwH96DrBz2JiURvHPNdxs/viewform
 you can submit your contributions multiple times.
 you will be able to edit the response right after submitting
 send email to make changes to an old response : uwstat340@gmail.com
Tentative Topics
 Random variable and stochastic process generation
 DiscreteEvent Systems
 Variance reduction
 Markov Chain Monte Carlo
Class 2  Thursday, May 9
Generating Random Numbers
Introduction
Simulation is the imitation of a process or system over time. Computational power has introduced the possibility of using simulation study to analyze models used to describe a situation.
In order to perform a simulation study, we should:
<br\> 1 Use a computer to generate (pseudo*) random numbers (rand in MATLAB).
2 Use these numbers to generate values of random variable from distributions: for example, set a variable in terms of uniform u ~ U(0,1).
3 Using the concept of discrete events, we show how the random variables can be used to generate the behavior of a stochastic model over time. (Note: A stochastic model is the opposite of deterministic model, where there are several directions the process can evolve to)
4 After continually generating the behavior of the system, we can obtain estimators and other quantities of interest.
The building block of a simulation study is the ability to generate a random number. This random number is a value from a random variable distributed uniformly on (0,1). There are many different methods of generating a random number:
Physical Method: Roulette wheel, lottery balls, dice rolling, card shuffling etc.
Numerically/Arithmetically: Use of a computer to successively generate pseudorandom numbers. The
sequence of numbers can appear to be random; however they are deterministically calculated with an
equation which defines pseudorandom.
(Source: Ross, Sheldon M., and Sheldon M. Ross. Simulation. San Diego: Academic, 1997. Print.)
 We use the prefix pseudo because computer generates random numbers based on algorithms, which suggests that generated numbers are not truly random. Therefore pseudorandom numbers is used.
In general, a deterministic model produces specific results given certain inputs by the model user, contrasting with a stochastic model which encapsulates randomness and probabilistic events.
A computer cannot generate truly random numbers because computers can only run algorithms, which are deterministic in nature. They can, however, generate Pseudo Random Numbers
Pseudo Random Numbers are the numbers that seem random but are actually determined by a relative set of original values. It is a chain of numbers preset by a formula or an algorithm, and the value jump from one to the next, making it look like a series of independent random events. The flaw of this method is that, eventually the chain returns to its initial position and pattern starts to repeat, but if we make the number set large enough we can prevent the numbers from repeating too early. Although the pseudo random numbers are deterministic, these numbers have a sequence of value and all of them have the appearances of being independent uniform random variables. Being deterministic, pseudo random numbers are valuable and beneficial due to the ease to generate and manipulate.
When people repeat the test many times, the results will be the closed express values, which make the trials look deterministic. However, for each trial, the result is random. So, it looks like pseudo random numbers.
Mod
Let [math]n \in \N[/math] and [math]m \in \N^+[/math], then by Division Algorithm,
[math]\exists q, \, r \in \N \;\text{with}\; 0\leq r \lt m, \; \text{s.t.}\; n = mq+r[/math],
where [math]q[/math] is called the quotient and [math]r[/math] the remainder. Hence we can define a binary function
[math]\mod : \N \times \N^+ \rightarrow \N [/math] given by [math]r:=n \mod m[/math] which returns the remainder after division by m.
Generally, mod means taking the reminder after division by m.
We say that n is congruent to r mod m if n = mq + r, where m is an integer.
Values are between 0 and m1
if y = ax + b, then [math]b:=y \mod a[/math].
Example 1:
[math]30 = 4 \cdot 7 + 2[/math]
[math]2 := 30\mod 7[/math]
[math]25 = 8 \cdot 3 + 1[/math]
[math]1: = 25\mod 3[/math]
[math]3=5\cdot (1)+2[/math]
[math]2:=3\mod 5[/math]
Example 2:
If [math]23 = 3 \cdot 6 + 5[/math]
Then equivalently, [math]5 := 23\mod 6[/math]
If [math]31 = 31 \cdot 1[/math]
Then equivalently, [math]0 := 31\mod 31[/math]
If [math]37 = 40\cdot (1)+ 3[/math]
Then equivalently, [math]3 := 37\mod 40[/math]
Example 3:
[math]77 = 3 \cdot 25 + 2[/math]
[math]2 := 77\mod 3[/math]
[math]25 = 25 \cdot 1 + 0[/math]
[math]0: = 25\mod 25[/math]
Note: [math]\mod[/math] here is different from the modulo congruence relation in [math]\Z_m[/math], which is an equivalence relation instead of a function.
The modulo operation is useful for determining if an integer divided by another integer produces a nonzero remainder. But both integers should satisfy [math]n = mq + r[/math], where [math]m[/math], [math]r[/math], [math]q[/math], and [math]n[/math] are all integers, and [math]r[/math] is smaller than [math]m[/math]. The above rules also satisfy when any of [math]m[/math], [math]r[/math], [math]q[/math], and [math]n[/math] is negative integer, see the third example.
Mixed Congruential Algorithm
We define the Linear Congruential Method to be [math]x_{k+1}=(ax_k + b) \mod m[/math], where [math]x_k, a, b, m \in \N, \;\text{with}\; a, m \neq 0[/math]. Given a seed (i.e. an initial value [math]x_0 \in \N[/math]), we can obtain values for [math]x_1, \, x_2, \, \cdots, x_n[/math] inductively. The Multiplicative Congruential Method, invented by Berkeley professor D. H. Lehmer, may also refer to the special case where [math]b=0[/math] and the Mixed Congruential Method is case where [math]b \neq 0[/math]
. Their title as "mixed" arises from the fact that it has both a multiplicative and additive term.
An interesting fact about Linear Congruential Method is that it is one of the oldest and bestknown pseudo random number generator algorithms. It is very fast and requires minimal memory to retain state. However, this method should not be used for applications that require high randomness. They should not be used for Monte Carlo simulation and cryptographic applications. (Monte Carlo simulation will consider possibilities for every choice of consideration, and it shows the extreme possibilities. This method is not precise enough.)
"Source: STAT 340 Spring 2010 Course Notes"
First consider the following algorithm
[math]x_{k+1}=x_{k} \mod m[/math]
such that: if [math]x_{0}=5(mod 150)[/math], [math]x_{n}=3x_{n1}[/math], find [math]x_{1},x_{8},x_{9}[/math].
[math]x_{n}=(3^n)*5(mod 150)[/math]
[math]x_{1}=45,x_{8}=105,x_{9}=15[/math]
Example
[math]\text{Let }x_{0}=10,\,m=3[/math]
 [math]\begin{align} x_{1} &{}= 10 &{}\mod{3} = 1 \\ x_{2} &{}= 1 &{}\mod{3} = 1 \\ x_{3} &{}= 1 &{}\mod{3} =1 \\ \end{align}[/math]
[math]\ldots[/math]
Excluding [math]x_{0}[/math], this example generates a series of ones. In general, excluding [math]x_{0}[/math], the algorithm above will always generate a series of the same number less than M. Hence, it has a period of 1. The period can be described as the length of a sequence before it repeats. We want a large period with a sequence that is random looking. We can modify this algorithm to form the Multiplicative Congruential Algorithm.
[math]x_{k+1}=(a \cdot x_{k} + b) \mod m [/math](a little tip: [math](a \cdot b)\mod c = (a\mod c)\cdot(b\mod c))[/math]
Example
[math]\text{Let }a=2,\, b=1, \, m=3, \, x_{0} = 10[/math]
[math]\begin{align}
\text{Step 1: } 0&{}=(2\cdot 10 + 1) &{}\mod 3 \\
\text{Step 2: } 1&{}=(2\cdot 0 + 1) &{}\mod 3 \\
\text{Step 3: } 0&{}=(2\cdot 1 + 1) &{}\mod 3 \\
\end{align}[/math]
[math]\ldots[/math]
This example generates a sequence with a repeating cycle of two integers.
(If we choose the numbers properly, we could get a sequence of "random" numbers. How do we find the value of [math]a,b,[/math] and [math]m[/math]? At the very least [math]m[/math] should be a very large, preferably prime number. The larger [math]m[/math] is, the higher the possibility to get a sequence of "random" numbers. This is easier to solve in Matlab. In Matlab, the command rand() generates random numbers which are uniformly distributed on the interval (0,1)). Matlab uses [math]a=7^5, b=0, m=2^{31}1[/math] – recommended in a 1988 paper, "Random Number Generators: Good Ones Are Hard To Find" by Stephen K. Park and Keith W. Miller (Important part is that [math]m[/math] should be large and prime)
Note: [math]\frac {x_{n+1}}{m1}[/math] is an approximation to the value of a U(0,1) random variable.
MatLab Instruction for Multiplicative Congruential Algorithm:
Before you start, you need to clear all existing defined variables and operations:
>>clear all >>close all
>>a=17 >>b=3 >>m=31 >>x=5 >>mod(a*x+b,m) ans=26 >>x=mod(a*x+b,m)
(Note:
1. Keep repeating this command over and over again and you will get random numbers – this is how the command rand works in a computer.
2. There is a function in MATLAB called RAND to generate a random number between 0 and 1.
For example, in MATLAB, we can use rand(1,1000) to generate 1000's numbers between 0 and 1. This is essentially a vector with 1 row, 1000 columns, with each entry a random number between 0 and 1.
3. If we would like to generate 1000 or more numbers, we could use a for loop
(Note on MATLAB commands:
1. clear all: clears all variables.
2. close all: closes all figures.
3. who: displays all defined variables.
4. clc: clears screen.
5. ; : prevents the results from printing.
6. disstool: displays a graphing tool.
>>a=13 >>b=0 >>m=31 >>x(1)=1 >>for ii=2:1000 x(ii)=mod(a*x(ii1)+b,m); end >>size(x) ans=1 1000 >>hist(x)
(Note: The semicolon after the x(ii)=mod(a*x(ii1)+b,m) ensures that Matlab will not print the entire vector of x. It will instead calculate it internally and you will be able to work with it. Adding the semicolon to the end of this line reduces the run time significantly.)
This algorithm involves three integer parameters [math]a, b,[/math] and [math]m[/math] and an initial value, [math]x_0[/math] called the seed. A sequence of numbers is defined by [math]x_{k+1} = ax_k+ b \mod m[/math].
Note: For some bad [math]a[/math] and [math]b[/math], the histogram may not look uniformly distributed.
Note: In MATLAB, hist(x) will generate a graph representing the distribution. Use this function after you run the code to check the real sample distribution.
Example: [math]a=13, b=0, m=31[/math]
The first 30 numbers in the sequence are a permutation of integers from 1 to 30, and then the sequence repeats itself so it is important to choose [math]m[/math] large to decrease the probability of each number repeating itself too early. Values are between [math]0[/math] and [math]m1[/math]. If the values are normalized by dividing by [math]m1[/math], then the results are approximately numbers uniformly distributed in the interval [0,1]. There is only a finite number of values (30 possible values in this case). In MATLAB, you can use function "hist(x)" to see if it looks uniformly distributed. We saw that the values between 030 had the same frequency in the histogram, so we can conclude that they are uniformly distributed.
If [math]x_0=1[/math], then
 [math]x_{k+1} = 13x_{k}\mod{31}[/math]
So,
 [math]\begin{align} x_{0} &{}= 1 \\ x_{1} &{}= 13 \times 1 + 0 &{}\mod{31} = 13 \\ x_{2} &{}= 13 \times 13 + 0 &{}\mod{31} = 14 \\ x_{3} &{}= 13 \times 14 + 0 &{}\mod{31} =27 \\ \end{align}[/math]
etc.
For example, with [math]a = 3, b = 2, m = 4, x_0 = 1[/math], we have:
 [math]x_{k+1} = (3x_{k} + 2)\mod{4}[/math]
So,
 [math]\begin{align}
x_{0} &{}= 1 \\
x_{1} &{}= 3 \times 1 + 2 \mod{4} = 1 \\
x_{2} &{}= 3 \times 1 + 2 \mod{4} = 1 \\
\end{align}[/math]
Another Example, a =3, b =2, m = 5, x_0=1 etc.
FAQ:
1.Why is it 1 to 30 instead of 0 to 30 in the example above?
[math]b = 0[/math] so in order to have [math]x_k[/math] equal to 0, [math]x_{k1}[/math] must be 0 (since [math]a=13[/math] is relatively prime to 31). However, the seed is 1. Hence, we will never observe 0 in the sequence.
Alternatively, {0} and {1,2,...,30} are two orbits of the left multiplication by 13 in the group [math]\Z_{31}[/math].
2.Will the number 31 ever appear?Is there a probability that a number never appears?
The number 31 will never appear. When you perform the operation [math]\mod m[/math], the largest possible answer that you could receive is [math]m1[/math]. Whether or not a particular number in the range from 0 to [math]m  1[/math] appears in the above algorithm will be dependent on the values chosen for [math]a, b[/math] and [math]m[/math].
Examples:[From Textbook]
[math]\text{If }x_0=3 \text{ and } x_n=(5x_{n1}+7)\mod 200[/math], [math]\text{find }x_1,\cdots,x_{10}[/math].
Solution:
[math]\begin{align}
x_1 &{}= (5 \times 3+7) &{}\mod{200} &{}= 22 \\
x_2 &{}= 117 &{}\mod{200} &{}= 117 \\
x_3 &{}= 592 &{}\mod{200} &{}= 192 \\
x_4 &{}= 2967 &{}\mod{200} &{}= 167 \\
x_5 &{}= 14842 &{}\mod{200} &{}= 42 \\
x_6 &{}= 74217 &{}\mod{200} &{}= 17 \\
x_7 &{}= 371092 &{}\mod{200} &{}= 92 \\
x_8 &{}= 1855467 &{}\mod{200} &{}= 67 \\
x_9 &{}= 9277342 &{}\mod{200} &{}= 142 \\
x_{10} &{}= 46386717 &{}\mod{200} &{}= 117 \\
\end{align}[/math]
Comments:
Matlab code: a=5; b=7; m=200; x(1)=3; for ii=2:1000 x(ii)=mod(a*x(ii1)+b,m); end size(x); hist(x)
Typically, it is good to choose [math]m[/math] such that [math]m[/math] is large, and [math]m[/math] is prime. Careful selection of parameters '[math]a[/math]' and '[math]b[/math]' also helps generate relatively "random" output values, where it is harder to identify patterns. For example, when we used a composite (non prime) number such as 40 for [math]m[/math], our results were not satisfactory in producing an output resembling a uniform distribution.
The computed values are between 0 and [math]m1[/math]. If the values are normalized by dividing by [math]m1[/math], their result is numbers uniformly distributed on the interval [math]\left[0,1\right][/math] (similar to computing from uniform distribution).
From the example shown above, if we want to create a large group of random numbers, it is better to have large, prime [math]m[/math] so that the generated random values will not repeat after several iterations. Note: the period for this example is 8: from '[math]x_2[/math]' to '[math]x_9[/math]'.
There has been a research on how to choose uniform sequence. Many programs give you the options to choose the seed. Sometimes the seed is chosen by CPU.
Theorem (extra knowledge)
Let c be a nonzero constant. Then for any seed x0, and LCG will have largest max. period if and only if
(i) m and c are coprime;
(ii) (a1) is divisible by all prime factor of m;
(iii) if and only if m is divisible by 4, then a1 is also divisible by 4.
We want our LCG to have a large cycle. We call a cycle with m element the maximal period. We can make it bigger by making m big and prime. Recall:any number you can think of can be broken into a factor of prime Define coprime:Two numbers X and Y, are coprime if they do not share any prime factors.
Example:
Xn=(15Xn1 + 4) mod 7
(i) m=7 c=4 > coprime;
(ii) a1=14 and a1 is divisible by 7;
(iii) dose not apply.
(The extra knowledge stops here)
In this part, I learned how to use R code to figure out the relationship between two integers division, and their remainder. And when we use R to calculate R with random variables for a range such as(1:1000),the graph of distribution is like uniform distribution.
Summary of Multiplicative Congruential Algorithm
Problem: generate Pseudo Random Numbers.
Plan:
 find integer: a b m(large prime) x_{0}(the seed) .
 [math]x_{k+1}=(ax_{k}+b)[/math]mod m
Matlab Instruction:
>>clear all >>close all >>a=17 >>b=3 >>m=31 >>x=5 >>mod(a*x+b,m) ans=26 >>x=mod(a*x+b,m)
Another algorithm for generating pseudo random numbers is the multiply with carry method. Its simplest form is similar to the linear congruential generator. They differs in that the parameter b changes in the MWC algorithm. It is as follows:
1.) x_{k+1} = ax_{k} + b_{k} mod m
2.) b_{k+1} = floor((ax_{k} + b_{k})/m)
3.) set k to k + 1 and go to step 1
Source
Inverse Transform Method
Now that we know how to generate random numbers, we use these values to sample form distributions such as exponential. However, to easily use this method, the probability distribution consumed must have a cumulative distribution function (cdf) [math]F[/math] with a tractable (that is, easily found) inverse [math]F^{1}[/math].
Theorem:
If we want to generate the value of a discrete random variable X, we must generate a random number U, uniformly distributed over (0,1).
Let [math]F:\R \rightarrow \left[0,1\right][/math] be a cdf. If [math]U \sim U\left[0,1\right][/math], then the random variable given by [math]X:=F^{1}\left(U\right)[/math]
follows the distribution function [math]F\left(\cdot\right)[/math],
where [math]F^{1}\left(u\right):=\inf F^{1}\big(\left[u,+\infty\right)\big) = \inf\{x\in\R  F\left(x\right) \geq u\}[/math] is the generalized inverse.
Note: [math]F[/math] need not be invertible everywhere on the real line, but if it is, then the generalized inverse is the same as the inverse in the usual case. We only need it to be invertible on the range of F(x), [0,1].
Proof of the theorem:
The generalized inverse satisfies the following:
 [math]P(X\leq x)[/math]
[math]= P(F^{1}(U)\leq x)[/math] (since [math]X= F^{1}(U)[/math] by the inverse method)
[math]= P((F(F^{1}(U))\leq F(x))[/math] (since [math]F [/math] is monotonically increasing)
[math]= P(U\leq F(x)) [/math] (since [math] P(U\leq a)= a[/math] for [math]U \sim U(0,1), a \in [0,1][/math],
[math]= F(x) , \text{ where } 0 \leq F(x) \leq 1 [/math]
This is the c.d.f. of X.
That is [math]F^{1}\left(u\right) \leq x \Leftrightarrow u \leq F\left(x\right)[/math]
Finally, [math]P(X \leq x) = P(F^{1}(U) \leq x) = P(U \leq F(x)) = F(x)[/math], since [math]U[/math] is uniform on the unit interval.
This completes the proof.
Therefore, in order to generate a random variable X~F, it can generate U according to U(0,1) and then make the transformation x=[math] F^{1}(U) [/math]
Note that we can apply the inverse on both sides in the proof of the inverse transform only if the pdf of X is monotonic. A monotonic function is one that is either increasing for all x, or decreasing for all x. Of course, this holds true for all CDFs, since they are monotonic by definition.
In short, what the theorem tells us is that we can use a random number [math] U from U(0,1) [/math] to randomly sample a point on the CDF of X, then apply the inverse of the CDF to map the given probability to its domain, which gives us the random variable X.
Example 1  Exponential: [math] f(x) = \lambda e^{\lambda x}[/math]
Calculate the CDF:
[math] F(x)= \int_0^x f(t) dt = \int_0^x \lambda e ^{\lambda t}\ dt[/math]
[math] = \frac{\lambda}{\lambda}\, e^{\lambda t}\,  \underset{0}{x} [/math]
[math] = e^{\lambda x} + e^0 =1  e^{ \lambda x} [/math]
Solve the inverse:
[math] y=1e^{ \lambda x} \Rightarrow 1y=e^{ \lambda x} \Rightarrow x=\frac {ln(1y)}{\lambda}[/math]
[math] y=\frac {ln(1x)}{\lambda} \Rightarrow F^{1}(x)=\frac {ln(1x)}{\lambda}[/math]
Note that 1 − U is also uniform on (0, 1) and thus −log(1 − U) has the same distribution as −logU.
Steps:
Step 1: Draw U ~U[0,1];
Step 2: [math] x=\frac{ln(U)}{\lambda} [/math]
EXAMPLE 2 Normal distribution
G(y)=P[Y<=y)
=P[sqr (y) < z < sqr (y)) =integrate from sqr(z) to Sqr(z) 1/sqr(2pi) e ^(z^2/2) dz = 2 integrate from 0 to sqr(y) 1/sqr(2pi) e ^(z^2/2) dz
its the cdf of Y=z^2
pdf g(y)= G'(y) pdf pf x^2 (1)
MatLab Code:
>>u=rand(1,1000); >>hist(u) # this will generate a fairly uniform diagram
#let λ=2 in this example; however, you can make another value for λ >>x=(log(1u))/2; >>size(x) #1000 in size >>figure >>hist(x) #exponential
Example 2  Continuous Distribution:
[math] f(x) = \dfrac {\lambda } {2}e^{\lambda \left x\theta \right } for \infty \lt X \lt \infty , \lambda \gt 0 [/math]
Calculate the CDF:
[math] F(x)= \frac{1}{2} e^{\lambda (\theta  x)} , for \ x \le \theta [/math]
[math] F(x) = 1  \frac{1}{2} e^{\lambda (x  \theta)}, for \ x \gt \theta [/math]
Solve for the inverse:
[math]F^{1}(x)= \theta + ln(2y)/\lambda, for \ 0 \le y \le 0.5[/math]
[math]F^{1}(x)= \theta  ln(2(1y))/\lambda, for \ 0.5 \lt y \le 1[/math]
Algorithm:
Steps:
Step 1: Draw U ~ U[0, 1];
Step 2: Compute [math]X = F^1(U)[/math] i.e. [math]X = \theta + \frac {1}{\lambda} ln(2U)[/math] for U < 0.5 else [math]X = \theta \frac {1}{\lambda} ln(2(1U))[/math]
Example 3  [math]F(x) = x^5[/math]:
Given a CDF of X: [math]F(x) = x^5[/math], transform U~U[0,1].
Sol:
Let [math]y=x^5[/math], solve for x: [math]x=y^\frac {1}{5}[/math]. Therefore, [math]F^{1} (x) = x^\frac {1}{5}[/math]
Hence, to obtain a value of x from F(x), we first set 'u' as an uniform distribution, then obtain the inverse function of F(x), and set
[math]x= u^\frac{1}{5}[/math]
Algorithm:
Steps:
Step 1: Draw U ~ rand[0, 1];
Step 2: X=U^(1/5);
Example 4  BETA(1,β):
Given u~U[0,1], generate x from BETA(1,β)
Solution:
[math]F(x)= 1(1x)^\beta[/math],
[math]u= 1(1x)^\beta[/math]
Solve for x:
[math](1x)^\beta = 1u[/math],
[math]1x = (1u)^\frac {1}{\beta}[/math],
[math]x = 1(1u)^\frac {1}{\beta}[/math]
let β=3, use Matlab to construct N=1000 observations from Beta(1,3)
MatLab Code:
>> u = rand(1,1000); x = 1(1u)^(1/3); >> hist(x,50) >> mean(x)
Example 5  Estimating [math]\pi[/math]:
Let's use rand() and Monte Carlo Method to estimate [math]\pi[/math]
N= total number of points
N_{c} = total number of points inside the circle
Prob[(x,y) lies in the circle=[math]\frac {Area(circle)}{Area(square)}[/math]
If we take square of size 2, circle will have area =[math]\pi (\frac {2}{2})^2 =\pi[/math].
Thus [math]\pi= 4(\frac {N_c}{N})[/math]
For example, UNIF(a,b)
[math]y = F(x) = (x  a)/ (b  a) [/math] [math]x = (b  a ) * y + a[/math] [math]X = a + ( b  a) * U[/math]
where U is UNIF(0,1)
Limitations:
1. This method is flawed since not all functions are invertible or monotonic: generalized inverse is hard to work on.
2. It may be impractical since some CDF's and/or integrals are not easy to compute such as Gaussian distribution.
We learned how to prove the transformation from cdf to inverse cdf,and use the uniform distribution to obtain a value of x from F(x). We can also use uniform distribution in inverse method to determine other distributions. The probability of getting a point for a circle over the triangle is a closed uniform distribution, each point in the circle and over the triangle is almost the same. Then, we can look at the graph to determine what kind of distribution the graph resembles.
Probability Distribution Function Tool in MATLAB
disttool #shows different distributions
This command allows users to explore different types of distribution and see how the changes affect the parameters on the plot of either a CDF or PDF.
change the value of mu and sigma can change the graph skew side.
Class 3  Tuesday, May 14
Recall the Inverse Transform Method
Let U~Unif(0,1),then the random variable X = F^{1}(u) has distribution F.
To sample X with CDF F(x),
[math]1) U~ \sim~ Unif [0,1] [/math]
2) X = F^{1}(u)
Note: CDF of a U(a,b) random variable is:
 [math] F(x)= \begin{cases} 0 & \text{for }x \lt a \\[8pt] \frac{xa}{ba} & \text{for }a \le x \lt b \\[8pt] 1 & \text{for }x \ge b \end{cases} [/math]
Thus, for [math] U [/math] ~ [math]U(0,1) [/math], we have [math]P(U\leq 1) = 1[/math] and [math]P(U\leq 1/2) = 1/2[/math].
More generally, we see that [math]P(U\leq a) = a[/math].
For this reason, we had [math]P(U\leq F(x)) = F(x)[/math].
Reminder:
This is only for uniform distribution [math] U~ \sim~ Unif [0,1] [/math]
[math] P (U \le 1) = 1 [/math]
[math] P (U \le 0.5) = 0.5 [/math]
[math] P (U \le a) = a [/math]
Note that on a single point there is no mass probability (i.e. [math]u[/math] <= 0.5, is the same as [math] u [/math] < 0.5) More formally, this is saying that [math] P(X = x) = F(x) \lim_{s \to x^}F(x)[/math] , which equals zero for any continuous random variable
Limitations of the Inverse Transform Method
Though this method is very easy to use and apply, it does have a major disadvantage/limitation:
 We need to find the inverse cdf [math] F^{1}(\cdot) [/math]. In some cases the inverse function does not exist, or is difficult to find because it requires a closed form expression for F(x).
For example, it is too difficult to find the inverse cdf of the Gaussian distribution, so we must find another method to sample from the Gaussian distribution.
In conclusion, we need to find another way of sampling from more complicated distributions
Discrete Case
The same technique can be used for discrete case. We want to generate a discrete random variable x, that has probability mass function:
 [math]\begin{align}P(X = x_i) &{}= p_i \end{align}[/math]
 [math]x_0 \leq x_1 \leq x_2 \dots \leq x_n[/math]
 [math]\sum p_i = 1[/math]
Algorithm for applying Inverse Transformation Method in Discrete Case (Procedure):
1. Define a probability mass function for [math]x_{i}[/math] where i = 1,....,k. Note: k could grow infinitely.
2. Generate a uniform random number U, [math] U~ \sim~ Unif [0,1] [/math]
3. If [math]U\leq p_{o}[/math], deliver [math]X = x_{o}[/math]
4. Else, if [math]U\leq p_{o} + p_{1} [/math], deliver [math]X = x_{1}[/math]
5. Repeat the process again till we reached to [math]U\leq p_{o} + p_{1} + ......+ p_{k}[/math], deliver [math]X = x_{k}[/math]
Note that after generating a random U, the value of X can be determined by finding the interval [math][F(x_{j1}),F(x_{j})][/math] in which U lies.
In summary:
Generate a discrete r.v.x that has pmf:
P(X=xi)=Pi, x0<x1<x2<...
1. Draw U~U(0,1);
2. If F(x(i1))<U<F(xi), x=xi.
Example 3.0:
Generate a random variable from the following probability function:
x  2  1  0  1  2 
f(x)  0.1  0.5  0.07  0.03  0.3 
Answer:
1. Gen U~U(0,1)
2. If U < 0.5 then output 1
else if U < 0.8 then output 2
else if U < 0.9 then output 2
else if U < 0.97 then output 0 else output 1
Example 3.1 (from class): (Coin Flipping Example)
We want to simulate a coin flip. We have U~U(0,1) and X = 0 or X = 1.
We can define the U function so that:
If [math]U\leq 0.5[/math], then X = 0
and if [math]0.5 \lt U\leq 1[/math], then X =1.
This allows the probability of Heads occurring to be 0.5 and is a good generator of a random coin flip.
[math] U~ \sim~ Unif [0,1] [/math]
 [math]\begin{align} P(X = 0) &{}= 0.5\\ P(X = 1) &{}= 0.5\\ \end{align}[/math]
The answer is:
 [math] x = \begin{cases} 0, & \text{if } U\leq 0.5 \\ 1, & \text{if } 0.5 \lt U \leq 1 \end{cases}[/math]
 Code
>>for ii=1:1000 u=rand; if u<0.5 x(ii)=0; else x(ii)=1; end end >>hist(x)
Note: The role of semicolon in Matlab: Matlab will not print out the results if the line ends in a semicolon and vice versa.
Example 3.2 (From class):
Suppose we have the following discrete distribution:
 [math]\begin{align} P(X = 0) &{}= 0.3 \\ P(X = 1) &{}= 0.2 \\ P(X = 2) &{}= 0.5 \end{align}[/math]
The cumulative distribution function (cdf) for this distribution is then:
 [math] F(x) = \begin{cases} 0, & \text{if } x \lt 0 \\ 0.3, & \text{if } x \lt 1 \\ 0.5, & \text{if } x \lt 2 \\ 1, & \text{if } x \ge 2 \end{cases}[/math]
Then we can generate numbers from this distribution like this, given [math]U \sim~ Unif[0, 1][/math]:
 [math] x = \begin{cases} 0, & \text{if } U\leq 0.3 \\ 1, & \text{if } 0.3 \lt U \leq 0.5 \\ 2, & \text{if } 0.5 \lt U\leq 1 \end{cases}[/math]
"Procedure"
1. Draw U~u (0,1)
2. if U<=0.3 deliver x=0
3. else if 0.3<U<=0.5 deliver x=1
4. else 0.5<U<=1 deliver x=2
Can you find a faster way to run this algorithm? Consider:
 [math] x = \begin{cases} 2, & \text{if } U\leq 0.5 \\ 1, & \text{if } 0.5 \lt U \leq 0.7 \\ 0, & \text{if } 0.7 \lt U\leq 1 \end{cases}[/math]
The logic for this is that U is most likely to fall into the largest range. Thus by putting the largest range (in this case x >= 0.5) we can improve the run time of this algorithm. Could this algorithm be improved further using the same logic?
 Code (as shown in class)
Use Editor window to edit the code
>>close all >>clear all >>for ii=1:1000 u=rand; if u<=0.3 x(ii)=0; elseif u<=0.5 x(ii)=1; else x(ii)=2; end end >>size(x) >>hist(x)
The algorithm above generates a vector (1,1000) containing 0's ,1's and 2's in differing proportions. Due to the criteria for accepting 0, 1 or 2 into the vector we get proportions of 0,1 &2 that correspond to their respective probabilities. So plotting the histogram (frequency of 0,1&2) doesn't give us the pmf but a frequency histogram that shows the proportions of each, which looks identical to the pmf.
Example 3.3: Generating a random variable from pdf
 [math] f_{x}(x) = \begin{cases} 2x, & \text{if } 0\leq x \leq 1 \\ 0, & \text{if } otherwise \end{cases}[/math]
 [math] F_{x}(x) = \begin{cases} 0, & \text{if } x \lt 0 \\ \int_{0}^{x}2sds = x^{2}, & \text{if } 0\leq x \leq 1 \\ 1, & \text{if } x \gt 1 \end{cases}[/math]
 [math]\begin{align} U = x^{2}, X = F^{1}x(U)= U^{\frac{1}{2}}\end{align}[/math]
Example 3.4: Generating a Bernoulli random variable
 [math]\begin{align} P(X = 1) = p, P(X = 0) = 1  p\end{align}[/math]
 [math] F(x) = \begin{cases} 1p, & \text{if } x \lt 1 \\ 1, & \text{if } x \ge 1 \end{cases}[/math]
1. Draw [math] U~ \sim~ Unif [0,1] [/math]
2. [math]
X = \begin{cases}
0, & \text{if } 0 \lt U \lt 1p \\
1, & \text{if } 1p \le U \lt 1
\end{cases}[/math]
Example 3.5: Generating Binomial(n,p) Random Variable
[math] use p\left( x=i+1\right) =\dfrac {ni} {i+1}\dfrac {p} {1p}p\left( x=i\right) [/math]
Step 1: Generate a random number [math]U[/math].
Step 2: [math]c = \frac {p}{(1p)}[/math], [math]i = 0[/math], [math]pr = (1p)^n[/math], [math]F = pr[/math]
Step 3: If U<F, set X = i and stop,
Step 4: [math] pr = \, {\frac {c(ni)}{(i+1)}} {pr}, F = F +pr, i = i+1[/math]
Step 5: Go to step 3
 Note: These steps can be found in Simulation 5th Ed. by Sheldon Ross.
 Note: Another method by seeing the Binomial as a sum of n independent Bernoulli random variables, U1, ..., Un. Then set X equal to the number of Ui that are less than or equal to p. To use this method, n random numbers are needed and n comparisons need to be done. On the other hand, the inverse transformation method is simpler because only one random variable needs to be generated and it makes 1 + np comparisons.
Step 1: Generate n uniform numbers U1 ... Un.
Step 2: X = [math]\sum U_i \lt = p[/math] where P is the probability of success.
Example 3.6: Generating a Poisson random variable
"Let X ~ Poi(u). Write an algorithm to generate X. The PDF of a poisson is:
 [math]\begin{align} f(x) = \frac {\, e^{u} u^x}{x!} \end{align}[/math]
We know that
 [math]\begin{align} P_{x+1} = \frac {\, e^{u} u^{x+1}}{(x+1)!} \end{align}[/math]
The ratio is [math]\begin{align} \frac {P_{x+1}}{P_x} = ... = \frac {u}{{x+1}} \end{align}[/math] Therefore, [math]\begin{align} P_{x+1} = \, {\frac {u}{x+1}} P_x\end{align}[/math]
Algorithm:
1) Generate U ~ U(0,1)
2) [math]\begin{align} X = 0 \end{align}[/math]
[math]\begin{align} F = P(X = 0) = e^{u}*u^0/{0!} = e^{u} = p \end{align}[/math]
3) If U<F, output x
Else, [math]\begin{align} p = (u/(x+1))^p \end{align}[/math]
[math]\begin{align} F = F + p \end{align}[/math]
[math]\begin{align} x = x + 1 \end{align}[/math]
4) Go to 1"
Acknowledgements: This is an example from Stat 340 Winter 2013
Example 3.7: Generating Geometric Distribution:
Consider Geo(p) where p is the probability of success, and define random variable X such that X is the total number of trials required to achieve the first success. x=1,2,3..... We have pmf: [math]P(X=x_i) = \, p (1p)^{x_{i}1}[/math] We have CDF: [math]F(x)=P(X \leq x)=1P(X\gt x) = 1(1p)^x[/math], P(X>x) means we get at least x failures before we observe the first success. Now consider the inverse transform:
 [math] x = \begin{cases} 1, & \text{if } U\leq p \\ 2, & \text{if } p \lt U \leq 1(1p)^2 \\ 3, & \text{if } 1(1p)^2 \lt U\leq 1(1p)^3 \\ .... k, & \text{if } 1(1p)^{k1} \lt U\leq 1(1p)^k .... \end{cases}[/math]
Note: Unlike the continuous case, the discrete inversetransform method can always be used for any discrete distribution (but it may not be the most efficient approach)
General Procedure
1. Draw U ~ U(0,1)
2. If [math]U \leq P_{0}[/math] deliver [math]x = x_{0}[/math]
3. Else if [math]U \leq P_{0} + P_{1}[/math] deliver [math]x = x_{1}[/math]
4. Else if [math]U \leq P_{0} + P_{1} + P_{2} [/math] deliver [math]x = x_{2}[/math]
...
Else if [math]U \leq P_{0} + ... + P_{k} [/math] deliver [math]x = x_{k}[/math]
===Inverse Transform Algorithm for Generating a Binomial(n,p) Random Variable(from textbook)===
step 1: Generate a random number U
step 2: c=p/(1p),i=0, pr=(1p)^{n}, F=pr.
step 3: If U<F, set X=i and stop.
step 4: pr =[c(ni)/(i+1)]pr, F=F+pr, i=i+1.
step 5: Go to step 3.
Problems
Though this method is very easy to use and apply, it does have a major disadvantage/limitation:
We need to find the inverse cdf F^{1}(\cdot) . In some cases the inverse function does not exist, or is difficult to find because it requires a closed form expression for F(x).
For example, it is too difficult to find the inverse cdf of the Gaussian distribution, so we must find another method to sample from the Gaussian distribution.
In conclusion, we need to find another way of sampling from more complicated distributions
Flipping a coin is a discrete case of uniform distribution, and the code below shows an example of flipping a coin 1000 times; the result is close to the expected value 0.5.
Example 2, as another discrete distribution, shows that we can sample from parts like 0,1 and 2, and the probability of each part or each trial is the same.
Example 3 uses inverse method to figure out the probability range of each random varible.
Summary of Inverse Transform Method
Problem:generate types of distribution.
Plan:
Continuous case:
 find CDF F
 find the inverse F^{1}
 Generate a list of uniformly distributed number {x}
 {F^{1}(x)} is what we want
Matlab Instruction
>>u=rand(1,1000); >>hist(u) >>x=(log(1u))/2; >>size(x) >>figure >>hist(x)
Discrete case:
 generate a list of uniformly distributed number {u}
 d_{i}=x_{i} if[math] X=x_i, [/math] if [math] F(x_{i1})\lt U\leq F(x_i) [/math]
 {d_{i}=x_{i}} is what we want
Matlab Instruction
>>for ii=1:1000 u=rand; if u<0.5 x(ii)=0; else x(ii)=1; end end >>hist(x)
Generalized InverseTransform Method
Valid for any CDF F(x): return X=min{x:F(x)[math]\leq[/math] U}, where U~U(0,1)
1. Continues, possibly with flat spots (i.e. not strictly increasing)
2. Discrete
3. Mixed continues discrete
Advantages of InverseTransform Method
Inverse transform method preserves monotonicity and correlation
which helps in
1. Variance reduction methods ...
2. Generating truncated distributions ...
3. Order statistics ...
AcceptanceRejection Method
Although the inverse transformation method does allow us to change our uniform distribution, it has two limits;
 Not all functions have inverse functions (ie, the range of x and y have limit and do not fix the inverse functions)
 For some distributions, such as Gaussian, it is too difficult to find the inverse
To generate random samples for these functions, we will use different methods, such as the AcceptanceRejection Method. This method is more efficient than the inverse transform method. The basic idea is to find an alternative probability distribution with density function f(x);
Suppose we want to draw random sample from a target density function f(x), x∈S_{x}, where S_{x} is the support of f(x). If we can find some constant c(≥1) (In practice, we prefer c as close to 1 as possible) and a density function g(x) having the same support S_{x} so that f(x)≤cg(x), ∀x∈S_{x}, then we can apply the procedure for AcceptanceRejection Method. Typically we choose a density function that we already know how to sample from for g(x).
The main logic behind the AcceptanceRejection Method is that:
1. We want to generate sample points from an unknown distribution, say f(x).
2. We use [math]\,cg(x)[/math] to generate points so that we have more points than f(x) could ever generate for all x. (where c is a constant, and g(x) is a known distribution)
3. For each value of x, we accept and reject some points based on a probability, which will be discussed below.
Note: If the red line was only g(x) as opposed to [math]\,c g(x)[/math] (i.e. c=1), then [math]g(x) \geq f(x)[/math] for all values of x if and only if g and f are the same functions. This is because the sum of pdf of g(x)=1 and the sum of pdf of f(x)=1, hence, [math]g(x) \ngeqq f(x)[/math] \,∀x.
Also remember that [math]\,c g(x)[/math] always generates higher probability than what we need. Thus we need an approach of getting the proper probabilities.
c must be chosen so that [math]f(x)\leqslant c g(x)[/math] for all value of x. c can only equal 1 when f and g have the same distribution. Otherwise:
Either use a software package to test if [math]f(x)\leqslant c g(x)[/math] for an arbitrarily chosen c > 0, or:
1. Find first and second derivatives of f(x) and g(x).
2. Identify and classify all local and absolute maximums and minimums, using the First and Second Derivative Tests, as well as all inflection points.
3. Verify that [math]f(x)\leqslant c g(x)[/math] at all the local maximums as well as the absolute maximums.
4. Verify that [math]f(x)\leqslant c g(x)[/math] at the tail ends by calculating [math]\lim_{x \to +\infty} \frac{f(x)}{\, c g(x)}[/math] and [math]\lim_{x \to \infty} \frac{f(x)}{\, c g(x)}[/math] and seeing that they are both < 1. Use of L'Hopital's Rule should make this easy, since both f and g are p.d.f's, resulting in both of them approaching 0.
5.Efficiency: the number of times N that steps 1 and 2 need to be called(also the number of iterations needed to successfully generate X) is a random variable and has a geometric distribution with success probability [math]p=P(U \leq f(Y)/(cg(Y)))[/math] , [math]P(N=n)=(1p(n1))p ,n \geq 1[/math].Thus on average the number of iterations required is given by [math] E(N)=\frac{1} p[/math]
c should be close to the maximum of f(x)/g(x), not just some arbitrarily picked large number. Otherwise, the AcceptanceRejection method will have more rejections (since our probability [math]f(x)\leqslant c g(x)[/math] will be close to zero). This will render our algorithm inefficient.
The expected number of iterations of the algorithm required with an X is c.
Note:
1. Value around x_{1} will be sampled more often under cg(x) than under f(x).There will be more samples than we actually need, if [math]\frac{f(y)}{\, c g(y)}[/math] is small, the acceptancerejection technique will need to be done to these points to get the accurate amount.In the region above x_{1}, we should accept less and reject more.
2. Value around x_{2}: number of sample that are drawn and the number we need are much closer. So in the region above x_{2}, we accept more. As a result, g(x) and f(x) are comparable.
3. The constant c is needed because we need to adjust the height of g(x) to ensure that it is above f(x). Besides that, it is best to keep the number of rejected varieties small for maximum efficiency.
Another way to understand why the the acceptance probability is [math]\frac{f(y)}{\, c g(y)}[/math], is by thinking of areas. From the graph above, we see that the target function in under the proposed function c g(y). Therefore, [math]\frac{f(y)}{\, c g(y)}[/math] is the proportion or the area under c g(y) that also contains f(y). Therefore we say we accept sample points for which u is less then [math]\frac{f(y)}{\, c g(y)}[/math] because then the sample points are guaranteed to fall under the area of c g(y) that contains f(y).
There are 2 cases that are possible:
Sample of points is more than enough, [math]c g(x) \geq f(x) [/math]
Similar or the same amount of points, [math]c g(x) \geq f(x) [/math]
There is 1 case that is not possible:
Less than enough points, such that [math] g(x) [/math] is greater than [math] f [/math], [math]g(x) \geq f(x)[/math]
Procedure
 Draw Y~g(.)
 Draw U~u(0,1) (Note: U and Y are independent)
 If [math]u\leq \frac{f(y)}{cg(y)}[/math] (which is [math]P(acceptedy)[/math]) then x=y, else return to Step 1
Note: Recall [math]P(U\leq a)=a[/math]. Thus by comparing u and [math]\frac{f(y)}{\, c g(y)}[/math], we can get a probability of accepting y at these points. For instance, at some points that cg(x) is much larger than f(x), the probability of accepting x=y is quite small.
ie. At X_{1}, low probability to accept the point since f(x) is much smaller than cg(x).
At X_{2}, high probability to accept the point. [math]P(U\leq a)=a[/math] in Uniform Distribution.
Note: Since U is the variable for uniform distribution between 0 and 1. It equals to 1 for all. The condition depends on the constant c. so the condition changes to [math]c\leq \frac{f(y)}{g(y)}[/math]
introduce the relationship of cg(x)and f(x),and prove why they have that relationship and where we can use this rule to reject some cases.
and learn how to see the graph to find the accurate point to reject or accept the ragion above the random variable x.
for the example, x1 is bad point and x2 is good point to estimate the rejection and acceptance
Some notes on the constant C
1. C is chosen such that [math] c g(y)\geq f(y)[/math], that is,[math] c g(y)[/math] will always dominate [math]f(y)[/math]. Because of this,
C will always be greater than or equal to one and will only equal to one if and only if the proposal distribution and the target distribution are the same. It is normally best to choose C such that the absolute maxima of both [math] c g(y)[/math] and [math] f(y)[/math] are the same.
2. [math] \frac {1}{C} [/math] is the area of [math] F(y)[/math] over the area of [math] c G(y)[/math] and is the acceptance rate of the points generated. For example, if [math] \frac {1}{C} = 0.7[/math] then on average, 70 percent of all points generated are accepted.
3. C is the average number of times Y is generated from g .
Theorem
Let [math]f: \R \rightarrow [0,+\infty][/math] be a welldefined pdf, and [math]\displaystyle Y[/math] be a random variable with pdf [math]g: \R \rightarrow [0,+\infty][/math] such that [math]\exists c \in \R^+[/math] with [math]f \leq c \cdot g[/math]. If [math]\displaystyle U \sim~ U(0,1)[/math] is independent of [math]\displaystyle Y[/math], then the random variable defined as [math]X := Y \vert U \leq \frac{f(Y)}{c \cdot g(Y)}[/math] has pdf [math]\displaystyle f[/math], and the condition [math]U \leq \frac{f(Y)}{c \cdot g(Y)}[/math] is denoted by "Accepted".
Proof
Recall the conditional probability formulas:
[math]\begin{align}
P(AB)=\frac{P(A \cap B)}{P(B)}, \text{ or }P(AB)=\frac{P(BA)P(A)}{P(B)} \text{ for pmf}
\end{align}[/math]
[math]P(yaccepted)=f(y)=\frac{P(acceptedy)P(y)}{P(accepted)}[/math]
based on the concept from procedurestep1:
[math]P(y)=g(y)[/math]
[math]P(acceptedy)=\frac{f(y)}{cg(y)}[/math]
(the larger the value is, the larger the chance it will be selected)
[math]
\begin{align}
P(accepted)&=\int_y\ P(acceptedy)P(y)\\
&=\int_y\ \frac{f(s)}{cg(s)}g(s)ds\\
&=\frac{1}{c} \int_y\ f(s) ds\\
&=\frac{1}{c}
\end{align}[/math]
Therefore:
[math]\begin{align}
P(x)&=P(yaccepted)\\
&=\frac{\frac{f(y)}{cg(y)}g(y)}{1/c}\\
&=\frac{\frac{f(y)}{c}}{1/c}\\
&=f(y)\end{align}[/math]
Here is an alternative introduction of AcceptanceRejection Method
Comments:
AcceptanceRejection Method is not good for all cases. The limitation with this method is that sometimes many points will be rejected. One obvious disadvantage is that it could be very hard to pick the [math]g(y)[/math] and the constant [math]c[/math] in some cases. We have to pick the SMALLEST C such that [math]cg(x) \leq f(x)[/math] else the the algorithm will not be efficient. This is because [math]f(x)/cg(x)[/math] will become smaller and probability [math]u \leq f(x)/cg(x)[/math] will go down and many points will be rejected making the algorithm inefficient.
Note: When [math]f(y)[/math] is very different than [math]g(y)[/math], it is less likely that the point will be accepted as the ratio above would be very small and it will be difficult for [math]U[/math] to be less than this small value.
An example would be when the target function ([math]f[/math]) has a spike or several spikes in its domain  this would force the known distribution ([math]g[/math]) to have density at least as large as the spikes, making the value of [math]c[/math] larger than desired. As a result, the algorithm would be highly inefficient.
AcceptanceRejection Method
Example 1 (discrete case)
We wish to generate X~Bi(2,0.5), assuming that we cannot generate this directly.
We use a discrete distribution DU[0,2] to approximate this.
[math]f(x)=Pr(X=x)=2Cx×(0.5)^2\,[/math]
[math]x[/math]  0  1  2 
[math]f(x)[/math]  1/4  1/2  1/4 
[math]g(x)[/math]  1/3  1/3  1/3 
[math]c=f(x)/g(x)[/math]  3/4  3/2  3/4 
[math]f(x)/(cg(x))[/math]  1/2  1  1/2 
Since we need [math]c \geq f(x)/g(x)[/math]
We need [math]c=3/2[/math]
Therefore, the algorithm is:
1. Generate [math]u,v~U(0,1)[/math]
2. Set [math]y= \lfloor 3*u \rfloor[/math] (This is using uniform distribution to generate DU[0,2]
3. If [math](y=0)[/math] and [math](v\lt \tfrac{1}{2}), output=0[/math]
If [math](y=2) [/math] and [math](v\lt \tfrac{1}{2}), output=2 [/math]
Else if [math]y=1, output=1[/math]
An elaboration of “c”
c is the expected number of times the code runs to output 1 random variable. Remember that when [math]u \lt \tfrac{f(x)}{cg(x)}[/math] is not satisfied, we need to go over the code again.
Proof
Let [math]f(x)[/math] be the function we wish to generate from, but we cannot use inverse transform method to generate directly.
Let [math]g(x)[/math] be the helper function
Let [math]kg(x)\gt =f(x)[/math]
Since we need to generate y from [math]g(x)[/math],
[math]Pr(select y)=g(y)[/math]
[math]Pr(output yselected y)=Pr(u\lt f(y)/(cg(y)))= f(y)/(cg(y))[/math] (Since u~Unif(0,1))
[math]Pr(output y)=Pr(output y1selected y1)Pr(select y1)+ Pr(output y2selected y2)Pr(select y2)+…+ Pr(output ynselected yn)Pr(select yn)=1/c[/math]
Consider that we are asking for expected time for the first success, it is a geometric distribution with probability of success=1/c
Therefore, [math]E(X)=1/(1/c))=c[/math]
Acknowledgements: Some materials have been borrowed from notes from Stat340 in Winter 2013.
Use the conditional probability to proof if the probability is accepted, then the result is closed pdf of the original one. the example shows how to choose the c for the two function [math]g(x)[/math] and [math]f(x)[/math].
Example of AcceptanceRejection Method
Generating a random variable having p.d.f.
[math]\displaystyle f(x) = 20x(1  x)^3, 0\lt x \lt 1 [/math]
Since this random variable (which is beta with parameters (2,4)) is concentrated in the interval (0, 1), let us consider the acceptancerejection method with
[math]\displaystyle g(x) = 1,0\lt x\lt 1[/math]
To determine the constant c such that f(x)/g(x) <= c, we use calculus to determine the maximum value of
[math]\displaystyle f(x)/g(x) = 20x(1  x)^3 [/math]
Differentiation of this quantity yields
[math]\displaystyle d/dx[f(x)/g(x)]=20*[(1x)^33x(1x)^2][/math]
Setting this equal to 0 shows that the maximal value is attained when x = 1/4,
and thus,
[math]\displaystyle f(x)/g(x)\lt = 20*(1/4)*(3/4)^3=135/64=c [/math]
Hence,
[math]\displaystyle f(x)/cg(x)=(256/27)*(x*(1x)^3)[/math]
and thus the simulation procedure is as follows:
1) Generate two random numbers U1 and U2 .
2) If U_{2}<(256/27)*U_{1}*(1U_{1})^{3}, set X=U_{1}, and stop Otherwise return to step 1). The average number of times that step 1) will be performed is c = 135/64.
(The above example is from http://www.cs.bgu.ac.il/~mps042/acceptance.htm, example 2.)
use the derivative to proof the accepetancerejection method, find the local maximum of f(x)/g(x). and we can calculate the best constant c.
Another Example of AcceptanceRejection Method
Generate a random variable from:
[math]\displaystyle f(x)=3*x^2, 0\lt x\lt 1 [/math]
Assume g(x) to be uniform over interval (0,1), where 0< x <1
Therefore:
[math]\displaystyle c = max(f(x)/(g(x)))= 3[/math]
the best constant c is the max(f(x)/(cg(x))) and the c make the area above the f(x) and below the g(x) to be small.
because g(.) is uniform so the g(x) is 1. max(g(x)) is 1
[math]\displaystyle f(x)/(cg(x))= x^2[/math]
Acknowledgement: this is example 1 from http://www.cs.bgu.ac.il/~mps042/acceptance.htm
Class 4  Thursday, May 16
Goals
 When we want to find target distribution [math]f(x)[/math], we need to first find a proposal distribution [math]g(x)[/math] that is easy to sample from.
 Relationship between the proposal distribution and target distribution is: [math] c \cdot g(x) \geq f(x) [/math], where c is constant. This means that the area of f(x) is under the area of [math] c \cdot g(x)[/math].
 Chance of acceptance is less if the distance between [math]f(x)[/math] and [math] c \cdot g(x)[/math] is big, and viceversa, we use [math] c [/math] to keep [math] \frac {f(x)}{c \cdot g(x)} [/math] below 1 (so [math]f(x) \leq c \cdot g(x)[/math]). Therefore, we must find the constant [math] C [/math] to achieve this.
 In other words, [math]C[/math] is chosen to make sure [math] c \cdot g(x) \geq f(x) [/math]. However, it will not make sense if [math]C[/math] is simply chosen to be arbitrarily large. We need to choose [math]C[/math] such that [math]c \cdot g(x)[/math] fits [math]f(x)[/math] as tightly as possible. This means that we must find the minimum c such that the area of f(x) is under the area of c*g(x).
 The constant c cannot be a negative number.
How to find C:
[math]\begin{align}
&c \cdot g(x) \geq f(x)\\
&c\geq \frac{f(x)}{g(x)} \\
&c= \max \left(\frac{f(x)}{g(x)}\right)
\end{align}[/math]
If [math]f[/math] and [math] g [/math] are continuous, we can find the extremum by taking the derivative and solve for [math]x_0[/math] such that:
[math] 0=\frac{d}{dx}\frac{f(x)}{g(x)}_{x=x_0}[/math]
Thus [math] c = \frac{f(x_0)}{g(x_0)} [/math]
Note: This procedure is called the AcceptanceRejection Method.
The AcceptanceRejection method involves finding a distribution that we know how to sample from, g(x), and multiplying g(x) by a constant c so that [math]c \cdot g(x)[/math] is always greater than or equal to f(x). Mathematically, we want [math] c \cdot g(x) \geq f(x) [/math].
And it means, c has to be greater or equal to [math]\frac{f(x)}{g(x)}[/math]. So the smallest possible c that satisfies the condition is the maximum value of [math]\frac{f(x)}{g(x)}[/math]
.
But in case of c being too large, the chance of acceptance of generated values will be small, thereby losing efficiency of the algorithm. Therefore, it is best to get the smallest possible c such that [math] c g(x) \geq f(x)[/math].
Important points:
 For this method to be efficient, the constant c must be selected so that the rejection rate is low. (The efficiency for this method is [math]\left ( \frac{1}{c} \right )[/math])
 It is easy to show that the expected number of trials for an acceptance is [math] \frac{Total Number of Trials} {C} [/math].
 recall the acceptance rate is 1/c. (Not rejection rate)
 Let [math]X[/math] be the number of trials for an acceptance, [math] X \sim~ Geo(\frac{1}{c})[/math]
 [math]\mathbb{E}[X] = \frac{1}{\frac{1}{c}} = c [/math]
 The number of trials needed to generate a sample size of [math]N[/math] follows a negative binomial distribution. The expected number of trials needed is then [math]cN[/math].
 So far, the only distribution we know how to sample from is the UNIFORM distribution.
Procedure:
1. Choose [math]g(x)[/math] (simple density function that we know how to sample, i.e. Uniform so far)
The easiest case is [math]U~ \sim~ Unif [0,1] [/math]. However, in other cases we need to generate UNIF(a,b). We may need to perform a linear transformation on the [math]U~ \sim~ Unif [0,1] [/math] variable.
2. Find a constant c such that :[math] c \cdot g(x) \geq f(x) [/math], otherwise return to step 1.
Recall the general procedure of AcceptanceRejection Method
 Let [math]Y \sim~ g(y)[/math]
 Let [math]U \sim~ Unif [0,1] [/math]
 If [math]U \leq \frac{f(Y)}{c \cdot g(Y)}[/math] then X=Y; else return to step 1 (This is not the way to find C. This is the general procedure.)
Example:
Generate a random variable from the pdf
[math] f(x) =
\begin{cases}
2x, & \mbox{if }0 \leqslant x \leqslant 1 \\
0, & \mbox{otherwise}
\end{cases} [/math]
We can note that this is a special case of Beta(2,1), where,
[math]beta(a,b)=\frac{\Gamma(a+b)}{\Gamma(a)\Gamma(b)}x^{(a1)}(1x)^{(b1)}[/math]
Where Γ (n) = (n  1)! if n is positive integer
[math]Gamma(z)=\int _{0}^{\infty }t^{z1}e^{t}dt[/math]
Aside: Beta function
In mathematics, the beta function, also called the Euler integral of the first kind, is a special function defined by
[math]B(x,y)=\int_0^1 \! {t^{(x1)}}{(1t)^{(y1)}}\,dt[/math]
[math]beta(2,1)= \frac{\Gamma(3)}{(\Gamma(2)\Gamma(1))}x^1 (1x)^0 = 2x[/math]
[math]g=u(0,1)[/math]
[math]y=g[/math]
[math]f(x)\leq c\cdot g(x)[/math]
[math]c\geq \frac{f(x)}{g(x)}[/math]
[math]c = \max \frac{f(x)}{g(x)} [/math]
[math]c = \max \frac{2x}{1}, 0 \leq x \leq 1[/math]
Taking x = 1 gives the highest possible c, which is c=2
Note that c is a scalar greater than 1.
cg(x) is proposal dist, and f(x) is target dist.
Note: g follows uniform distribution, it only covers half of the graph which runs from 0 to 1 on yaxis. Thus we need to multiply by c to ensure that [math]c\cdot g[/math] can cover entire f(x) area. In this case, c=2, so that makes g run from 0 to 2 on yaxis which covers f(x).
Comment:
From the picture above, we could observe that the area under f(x)=2x is a half of the area under the pdf of UNIF(0,1). This is why in order to sample 1000 points of f(x), we need to sample approximately 2000 points in UNIF(0,1).
And in general, if we want to sample n points from a distritubion with pdf f(x), we need to scan approximately [math]n\cdot c[/math] points from the proposal distribution (g(x)) in total.
Step
 Draw y~U(0,1)
 Draw u~U(0,1)
 if [math]u \leq \frac{(2\cdot y)}{(2\cdot 1)}, u \leq y,[/math] then [math] x=y[/math]
 Else go to Step 1
Note: In the above example, we sample 2 numbers. If second number (u) is less than or equal to first number (y), then accept x=y, if not then start all over.
Matlab Code
>>close all >>clear all >>ii=1; # ii:numbers that are accepted >>jj=1; # jj:numbers that are generated >>while ii<1000 y=rand; u=rand; jj=jj+1; if u<=y x(ii)=y; ii=ii+1; end end >>hist(x) # It is a histogram >>jj jj = 2024 # should be around 2000
 *Note: The reason that a for loop is not used is that we need to continue the looping until we get 1000 successful samples. We will reject some samples during the process and therefore do not know the number of y we are going to generate.
 *Note2: In this example, we used c=2, which means we accept half of the points we generate on average. Generally speaking, 1/c would be the probability of acceptance, and an indicator of the efficiency of your chosen proposal distribution and algorithm.
 *Note3: We use while instead of for when looping because we do not know how many iterations are required to generate 1000 successful samples. We can view this as a negative binomial distribution so while the expected number of iterations required is n * c, it will likely deviate from this amount. We expect 2000 in this case.
 *Note4: If c=1, we will accept all points, which is the ideal situation. However, this is essentially impossible because if c = 1 then our distributions f(x) and g(x) must be identical, so we will have to be satisfied with as close to 1 as possible.
Use Inverse Method for this Example
 [math]F(x)=\int_0^x \! 2s\,ds={x^2}0={x^2}[/math]
 [math]y=x^2[/math]
 [math]x=\sqrt y[/math]
 [math] F^{1}\left (\, x \, \right) =\sqrt x[/math]
 Procedure
 1: Draw [math] U~ \sim~ Unif [0,1] [/math]
 2: [math] x=F^{1}\left (\, u\, \right) =\sqrt u[/math]
Matlab Code
>>u=rand(1,1000); >>x=u.^0.5; >>hist(x)
Matlab Tip: Periods, ".",meaning "elementwise", are used to describe the operation you want performed on each element of a vector. In the above example, to take the square root of every element in U, the notation U.^0.5 is used. However if you want to take the square root of the entire matrix U the period, "." would be excluded. i.e. Let matrix B=U^0.5, then [math]B^T*B=U[/math]. For example if we have a two 1 X 3 matrices and we want to find out their product; using "." in the code will give us their product. However, if we don't use ".", it will just give us an error. For example, a =[1 2 3] b=[2 3 4] are vectors, a.*b=[2 6 12], but a*b does not work since the matrix dimensions must agree.
Example for AR method:
Given [math] f(x)= \frac{3}{4} (1x^2), 1 \leq x \leq 1 [/math], use AR method to generate random number
Let g=U(1,1) and g(x)=1/2
let y ~ f, [math] cg(x)\geq f(x), c\frac{1}{2} \geq \frac{3}{4} (1x^2) /1, c=max 2\cdot\frac{3}{4} (1x^2) = 3/2 [/math]
The process:
 1: Draw U1 ~ U(0,1)
 2: Draw U2 ~ U(0,1)
 3: let [math] y = U1*2  1 [/math]
 4: if [math]U2 \leq \frac { \frac{3}{4} * (1y^2)} { \frac{3}{4}} = {1y^2}[/math], then x=y, note that (3/4(1y^2)/(3/4) is getting from f(y) / (cg(y)) )
 5: else: return to step 1
Example of AcceptanceRejection Method
[math]\begin{align} & f(x) = 3x^2, 0\lt x\lt 1 \\ \end{align}[/math]<br\>
[math]\begin{align} & g(x)=1, 0\lt x\lt 1 \\ \end{align}[/math]<br\>
[math]c = \max \frac{f(x)}{g(x)} = \max \frac{3x^2}{1} = 3 [/math]
[math]\frac{f(x)}{c \cdot g(x)} = x^2[/math]
1. Generate two uniform numbers in the unit interval [math]U_1, U_2 \sim~ U(0,1)[/math]
2. If [math]U_2 \leqslant {U_1}^2[/math], accept [math]\begin{align}U_1\end{align}[/math] as the random variable with pdf [math]\begin{align}f\end{align}[/math], if not return to Step 1
We can also use [math]\begin{align}g(x)=2x\end{align}[/math] for a more efficient algorithm
[math]c = \max \frac{f(x)}{g(x)} = \max \frac {3x^2}{2x} = \frac {3x}{2} [/math]. Use the inverse method to sample from [math]\begin{align}g(x)\end{align}[/math] [math]\begin{align}G(x)=x^2\end{align}[/math]. Generate [math]\begin{align}U\end{align}[/math] from [math]\begin{align}U(0,1)\end{align}[/math] and set [math]\begin{align}x=sqrt(u)\end{align}[/math]
1. Generate two uniform numbers in the unit interval [math]U_1, U_2 \sim~ U(0,1)[/math]
2. If [math]U_2 \leq \frac{3\sqrt{U_1}}{2}[/math], accept [math]U_1[/math] as the random variable with pdf [math]f[/math], if not return to Step 1
 Note :the function [math]\begin{align}q(x) = c * g(x)\end{align}[/math] is called an envelop or majoring function.
To obtain a better proposing function [math]\begin{align}g(x)\end{align}[/math], we can first assume a new [math]\begin{align}q(x)\end{align}[/math] and then solve for the normalizing constant by integrating.
In the previous example, we first assume [math]\begin{align}q(x) = 3x\end{align}[/math]. To find the normalizing constant, we need to solve [math]k *\sum 3x = 1[/math] which gives us k = 2/3. So,[math]\begin{align}g(x) = k*q(x) = 2x\end{align}[/math].
Possible Limitations
This method could be computationally inefficient depending on the rejection rate. We may have to sample many points before
we get the 1000 accepted points. In the example we did in class relating the [math]f(x)=2x[/math],
we had to sample around 2070 points before we finally accepted 1000 sample points.
If the form of the proposal distribution, g, is very different from target distribution, f, then c is very large and the algorithm is not computationally efficient.
Acceptance  Rejection Method Application on Normal Distribution
[math]X \sim∼ N(\mu,\sigma^2), \text{ or } X = \sigma Z + \mu, Z \sim~ N(0,1) [/math]
[math]\vert Z \vert[/math] has probability density function of
f(x) = (2/[math]\sqrt{2\pi}[/math]) e^{x2/2}
g(x) = e^{x}
Take h(x) = f(x)/g(x) and solve for h'(x) = 0 to find x so that h(x) is maximum.
Hence x=1 maximizes h(x) => c = [math]\sqrt{2e/\pi}[/math]
Thus f(y)/cg(y) = e^{(y1)2/2}
learn how to use code to calculate the c between f(x) and g(x).
How to transform [math]U(0,1)[/math] to [math]U(a, b)[/math]
1. Draw U from [math]U(0,1)[/math]
2. Take [math]Y=(ba)U+a[/math]
3. Now Y follows [math]U(a,b)[/math]
Example: Generate a random variable z from the Semicircular density [math]f(x)= \frac{2}{\pi R^2} \sqrt{R^2x^2}, R\leq x\leq R[/math].
> Proposal distribution: UNIF(R, R)
> We know how to generate using [math] U \sim UNIF (0,1) [/math] Let [math] Y= 2RUR=R(2U1)[/math], therefore Y follows [math]U(R,R)[/math]
> In order to maximize the function we must maximize the top and minimize the bottom.
Now, we need to find c:
Since c=max[f(x)/g(x)], where
[math]f(x)= \frac{2}{\pi R^2} \sqrt{R^2x^2}[/math], [math]g(x)=\frac{1}{2R}[/math], [math]R\leq x\leq R[/math]
Thus, we have to maximize R^2x^2.
=> When x=0, it will be maximized.
Therefore, c=4/pi. * Note: This also means that the probability of accepting a point is [math]\pi/4[/math].
We will accept the points with limit f(x)/[cg(x)]. Since [math]\frac{f(y)}{cg(y)}=\frac{\frac{2}{\pi R^{2}} \sqrt{R^{2}y^{2}}}{\frac{4}{\pi} \frac{1}{2R}}=\frac{\frac{2}{\pi R^{2}} \sqrt{R^{2}R^{2}(2U1)^{2}}}{\frac{2}{\pi R}}[/math]
 Note: Y= R(2U1)
We can also get Y= R(2U1) by using the formula y = a+(ba)*u, to transform U~(0,1) to U~(a,b). Letting a=R and b=R, and substituting it in the formula y = a+(ba)*u, we get Y= R(2U1).
Thus, [math]\frac{f(y)}{cg(y)}=\sqrt{1(2U1)^{2}}[/math] * this also means the probability we can accept points
The algorithm to generate random variable x is then:
1. Draw [math]\ U[/math] from [math]\ U(0,1)[/math]
2. Draw [math]\ U_{1}[/math] from [math]\ U(0,1)[/math]
3. If [math]U_{1} \leq \sqrt{1(2U1)^2}, set x = U_{1}[/math]
else return to step 1.
The condition is
[math] U_{1} \leq \sqrt{(1(2U1)^2)}[/math]
[math]\ U_{1}^2 \leq 1  (2U 1)^2[/math]
[math]\ U_{1}^2  1 \leq (2U  1)^2[/math]
[math]\ 1  U_{1}^2 \geq (2U  1)^2[/math]
One more example about AR method
(In this example, we will see how to determine the value of c when c is a function with unknown parameters instead of a value)
Let [math]f(x)=x×e^{x}, x \gt 0 [/math]
Use [math]g(x)=a×e^{a×x}[/math] to generate random variable
Solution: First of all, we need to find c
[math]cg(x)\gt =f(x)[/math]
[math]c\gt =\frac{f(x)}{g(x)}[/math]
[math]\frac{f(x)}{g(x)}=\frac{x}{a} * e^{(1a)x}[/math]
take derivative with respect to x, and set it to 0 to get the maximum,
[math]\frac{1}{a} * e^{(1a)x}  \frac{x}{a} * e^{(1a)x} * (1a) = 0 [/math]
[math]x=\frac {1}{1a}[/math]
[math]\frac {f(x)}{g(x)} = \frac {e^{1}}{a*(1a)} [/math]
[math]\frac {f(0)}{g(0)} = 0[/math]
[math]\frac {f(\infty)}{g(\infty)} = 0[/math]
therefore, [math]c= \frac {e^{1}}{a*(1a)}[/math]
In order to minimize c, we need to find the appropriate a
Take derivative with respect to a and set it to be zero,
We could get [math]a= \frac {1}{2}[/math]
[math]c=\frac{4}{e}[/math]
Procedure:
1. Generate u v ~unif(0,1)
2. Generate y from g, since g is exponential with rate 2, let y=0.5*ln(u)
3. If [math]v\lt \frac{f(y)}{c\cdot g(y)}[/math], output y
Else, go to 1
Acknowledgements: The example above is from Stat 340 Winter 2013 notes.
Summary of how to find the value of c
Let [math]h(x) = \frac {f(x)}{g(x)}[/math], and then we have the following:
1. First, take derivative of h(x) with respect to x, get x_{1};
2. Plug x_{1} into h(x) and get the value(or a function) of c, denote as c_{1};
3. Check the endpoints of x and sub the endpoints into h(x);
4. (if c_{1} is a value, then we can ignore this step) Since we want the smallest value of c such that [math]f(x) \leq c\cdot g(x)[/math] for all x, we want the unknown parameter that minimizes c.
So we take derivative of c_{1} with respect to the unknown parameter (ie k=unknown parameter) to get the value of k.
Then we submit k to get the value of c_{1}. (Double check that [math]c_1 \geq 1[/math]
5. Pick the maximum value of h(x) to be the value of c.
For the two examples above, we need to generate the probability function to uniform distribution, and figure out [math]c=max\frac {f(y)}{g(y)} [/math]. If [math]v\lt \frac {f(y)}{c\cdot g(y)}[/math], output y.
Summary of when to use the Accept Rejection Method
1) When the calculation of inverse cdf cannot to be computed or is too difficult to compute.
2) When f(x) can be evaluated to at least one of the normalizing constant.
3) A constant c where [math]f(x)\leq c\cdot g(x)[/math]
4) A uniform draw
Interpretation of 'C'
We can use the value of c to calculate the acceptance rate by [math]\tfrac{1}{c}[/math].
For instance, assume c=1.5, then we can tell that 66.7% of the points will be accepted ([math]\tfrac{1}{1.5} = 0.667[/math]). We can also call the efficiency of the method is 66.7%.
Likewise, if the minimum value of possible values for C is [math]\tfrac{4}{3}[/math], [math]1/ \tfrac{4}{3}[/math] of the generated random variables will be accepted. Thus the efficient of the algorithm is 75%.
In order to ensure the algorithm is as efficient as possible, the 'C' value should be as close to one as possible, such that [math]\tfrac{1}{c}[/math] approaches 1 => 100% acceptance rate.
>> close All
>> clear All
>> i=1
>> j=0;
>> while ii<1000
y=rand
u=rand
if u<=y;
x(ii)=y
ii=ii+1
end
end
Class 5  Tuesday, May 21
Recall the example in the last lecture. The following code will generate a random variable required by the question.
 Code
>>close all >>clear all >>ii=1; >>R=1; #Note: that R is a constant in which we can change i.e. if we changed R=4 then we would have a density between 4 and 4 >>while ii<1000 u1 = rand; u2 = rand; y = R*(2*u21); if (1u1^2)>=(2*u21)^2 x(ii) = y; ii = ii + 1; #Note: for beginner programmers that this step increases the ii value for next time through the while loop end end >>hist(x,20) # 20 is the number of bars >>hist(x,30) #30 is the number of bars
calculate process:
[math]u_{1} \lt = \sqrt (1(2u1)^2) [/math]
[math](u_{1})^2 \lt =(1(2u1)^2) [/math]
[math](u_{1})^2 1 \lt =((2u1)^2) [/math]
[math]1(u_{1})^2 \gt =((2u1)^21) [/math]
MATLAB tips: hist(x,y) plots a histogram of variable x, where y is the number of bars in the graph.
Discrete Examples
 Example 1
Generate random variable [math]X[/math] according to p.m.f
[math]\begin{align}
P(x &=1) &&=0.15 \\
P(x &=2) &&=0.25 \\
P(x &=3) &&=0.3 \\
P(x &=4) &&=0.1 \\
P(x &=5) &&=0.2 \\
\end{align}[/math]
The discrete case is analogous to the continuous case. Suppose we want to generate an X that is a discrete random variable with pmf f(x)=P(X=x). Suppose also that we use the discrete uniform distribution as our target distribution, then [math] g(x)= P(X=x) =0.2 [/math] for all X.
The following algorithm then yields our X:
Step 1 Draw discrete uniform distribution of 1, 2, 3, 4 and 5, [math]Y \sim~ g[/math].
Step 2 Draw [math]U \sim~ U(0,1)[/math].
Step 3 If [math]U \leq \frac{f(Y)}{c \cdot g(Y)}[/math], then X = Y ;
Else return to Step 1.
C can be found by maximizing the ratio :[math] \frac{f(x)}{g(x)} [/math]. To do this, we want to maximize [math] f(x) [/math] and minimize [math] g(x) [/math].
 [math]c = max \frac{f(x)}{g(x)} = \frac {0.3}{0.2} = 1.5 [/math]
Note: In this case [math]f(x)=P(X=x)=0.3[/math] (highest probability from the discrete probabilities in the question)
 [math]\frac{p(x)}{cg(x)} = \frac{p(x)}{1.5*0.2} = \frac{p(x)}{0.3} [/math]
Note: The U is independent from y in Step 2 and 3 above. ~The constant c is a indicator of rejection rate or efficiency of the algorithm. It can represent the average number of trials of the algorithm. Thus, a higher c would mean that the algorithm is comparatively inefficient.
the acceptancerejection method of pmf, the uniform probability is the same for all variables, and there are 5 parameters(1,2,3,4,5), so g(x) is 0.2
Remember that we always want to choose [math] cg [/math] to be equal to or greater than [math] f [/math], but as close as possible.
limitations: If the form of the proposal dist g is very different from target dist f, then c is very large and the algorithm is not computatively efficient.
 Code for example 1
>>close all >>clear all >>p=[.15 .25 .3 .1 .2]; %This a vector holding the values >>ii=1; >>while ii < 1000 y=unidrnd(5); %generates random numbers for the discrete uniform u=rand; distribution with maximum 5. if u<= p(y)/0.3 x(ii)=y; ii=ii+1; end end >>hist(x)
unidrnd(k) draws from the discrete uniform distribution of integers [math]1,2,3,...,k[/math] If this function is not built in to your MATLAB then we can do simple transformation on the rand(k) function to make it work like the unidrnd(k) function.
The acceptance rate is [math]\frac {1}{c}[/math], so the lower the c, the more efficient the algorithm. Theoretically, c equals 1 is the best case because all samples would be accepted; however it would only be true when the proposal and target distributions are exactly the same, which would never happen in practice.
For example, if c = 1.5, the acceptance rate would be [math]\frac {1}{1.5}=\frac {2}{3}[/math]. Thus, in order to generate 1000 random values, on average, a total of 1500 iterations would be required.
A histogram to show 1000 random values of f(x), more random value make the probability close to the express probability value.
 Example 2
p(x=1)=0.1
p(x=2)=0.3
p(x=3)=0.6
Let g be the uniform distribution of 1, 2, or 3
g(x)= 1/3
[math]c=max(\tfrac{p_{x}}{g(x)})=0.6/(\tfrac{1}{3})=1.8[/math]
Hence [math]\tfrac{p(x)}{cg(x)} = p(x)/(1.8 (\tfrac{1}{3}))= \tfrac{p(x)}{0.6}[/math]
1,y~g
2,u~U(0,1)
3, If [math]U \leq \frac{f(y)}{cg(y)}[/math], set x = y. Else go to 1.
 Code for example 2
>>close all >>clear all >>p=[.1 .3 .6]; %This a vector holding the values >>ii=1; >>while ii < 1000 y=unidrnd(3); %generates random numbers for the discrete uniform distribution with maximum 3 u=rand; if u<= p(y)/0.6 x(ii)=y; ii=ii+1; %else ii=ii+1 end end >>hist(x)
 Example 3
Suppose [math]\begin{align}p_{x} = e^{3}3^{x}/x! , x\geq 0\end{align}[/math] (Poisson distribution)
First: Try the first few [math]\begin{align}p_{x}'s\end{align}[/math]: 0.0498, 0.149, 0.224, 0.224, 0.168, 0.101, 0.0504, 0.0216, 0.0081, 0.0027 for [math]\begin{align} x = 0,1,2,3,4,5,6,7,8,9 \end{align}[/math]
Proposed distribution: Use the geometric distribution for [math]\begin{align}g(x)\end{align}[/math];
[math]\begin{align}g(x)=p(1p)^{x}\end{align}[/math], choose [math]\begin{align}p=0.25\end{align}[/math]
Look at [math]\begin{align}p_{x}/g(x)\end{align}[/math] for the first few numbers: 0.199 0.797 1.59 2.12 2.12 1.70 1.13 0.647 0.324 0.144 for [math]\begin{align} x = 0,1,2,3,4,5,6,7,8,9 \end{align}[/math]
We want [math]\begin{align}c=max(p_{x}/g(x))\end{align}[/math] which is approximately 2.12
The general procedures to generate [math]\begin{align}p(x)\end{align}[/math] is as follows:
1. Generate [math]\begin{align}U_{1} \sim~ U(0,1); U_{2} \sim~ U(0,1)\end{align}[/math]
2. [math]\begin{align}j = \lfloor \frac{ln(U_{1})}{ln(.75)} \rfloor+1;\end{align}[/math]
3. if [math]U_{2} \lt \frac{p_{j}}{cg(j)}[/math], set [math]\begin{align}X = x_{j}\end{align}[/math], else go to step 1.
Note: In this case, [math]\begin{align}f(x)/g(x)\end{align}[/math] is extremely difficult to differentiate so we were required to test points. If the function is very easy to differentiate, we can calculate the max as if it were a continuous function then check the two surrounding points for which is the highest discrete value.
 Example 4 (Hypergeometric & Binomial)
Suppose we are given f(x) such that it is hypergeometically distributed, given 10 white balls, 5 red balls, and select 3 balls, let X be the number of red ball selected, without replacement.
Choose g(x) such that it is binomial distribution, Bin(3, 1/3). Find the rejection constant, c
Solution:
For hypergeometric: [math]P(X=0) =\binom{10}{3}/\binom{15}{3} =0.2637, P(x=1)=\binom{10}{2} * \binom{5}{1} /\binom{15}{3}=0.4945, P(X=2)=\binom{10}{1} * \binom{5}{2} /\binom{15}{3}=0.2198,[/math]
[math]P(X=3)=\binom{5}{3}/\binom{15}{3}= 0.02198[/math]
For Binomial g(x): P(X=0) = (2/3)^3=0.2963; P(X=1)= 3*(1/3)*(2/3)^2 = 0.4444, P(X=2)=3*(1/3)^2*(2/3)=0.2222, P(X=3)=(1/3)^3=0.03704
Find the value of f/g for each X
X=0: 0.8898; X=1: 1.1127; X=2: 0.9891; X=3: 0.5934
Choose the maximum which is c=1.1127
Looking for the max f(x) is 0.4945 and the max g(x) is 0.4444, so we can calculate the max c is 1.1127. But for the graph, this c is not the best because it does not cover all the point of f(x), so we need to move the c*g(x) graph to cover all f(x), and decreasing the rejection ratio.
Limitation: If the shape of the proposed distribution g is very different from the target distribution f, then the rejection rate will be high (High c value). Computationally, the algorithm is always right; however it is inefficient and requires many iterations.
Here is an example:
In the above example, we need to move c*g(x) to the peak of f to cover the whole f. Thus c will be very large and 1/c will be small.
The higher the rejection rate, more points will be rejected.
More on rejection/acceptance rate: 1/c is the acceptance rate. As c decreases (note: the minimum value of c is 1), the acceptance rate increases. In our last example, 1/c=1/1.5≈66.67%. Around 67% of points generated will be accepted.
which brings the acceptance rate low which leads to very time consuming sampling
AcceptanceRejection Method
Problem: The CDF is not invertible or it is difficult to find the inverse.
Plan:
 Draw y~g(.)
 Draw u~Unif(0,1)
 If [math]u\leq \frac{f(y)}{cg(y)}[/math]then set x=y. Else return to Step 1
x will have the desired distribution.
Matlab Example
close all clear all ii=1; R=1; while ii<1000 u1 = rand; u2 = rand; y = R*(2*u21); if (1u1^2)>=(2*u21)^2 x(ii) = y; ii = ii + 1; end end hist(x,20)
Recall that,
Suppose we have an efficient method for simulating a random variable having probability mass function {q(j),j>=0}. We can use this as the basis for simulating from the distribution having mass function {p(j),j>=0} by first simulating a random variable Y having mass function {q(j)} and then accepting this simulated value with a probability proportional to p(Y)/q(Y).
Specifically, let c be a constant such that p(j)/q(j)<=c for all j such that p(j)>0
We now have the following technique, called the acceptancerejection method, for simulating a random variable X having mass function p(j)=P{X=j}.
Sampling from commonly used distributions
Please note that this is not a general technique as is that of acceptancerejection sampling. Later, we will generalize the distributions for multidimensional purposes.
 Gamma
The CDF of the Gamma distribution [math]Gamma(t,\lambda)[/math] is(t denotes the shape, [math]\lambda[/math] denotes the scale:
[math] F(x) = \int_0^{x} \frac{e^{y}y^{t1}}{(t1)!} \mathrm{d}y, \; \forall x \in (0,+\infty)[/math], where [math]t \in \N^+ \text{ and } \lambda \in (0,+\infty)[/math].
Note that the CDF of the Gamma distribution does not have a closed form.
The gamma distribution is often used to model waiting times between a certain number of events. It can also be expressed as the sum of infinitely many independent and identically distributed exponential distributions. This distribution has two parameters: the number of exponential terms n, and the rate parameter [math]\lambda[/math]. In this distribution there is the Gamma function, [math]\Gamma [/math] which has some very useful properties. "Source: STAT 340 Spring 2010 Course Notes"
Neither Inverse Transformation nor AcceptanceRejection Method can be easily applied to Gamma distribution. However, we can use additive property of Gamma distribution to generate random variables.
 Additive Property
If [math]X_1, \dots, X_t[/math] are independent exponential distributions with hazard rate [math] \lambda [/math] (in other words, [math] X_i\sim~ Exp (\lambda) [/math][math], Exp (\lambda)= Gamma (1, \lambda)), then \Sigma_{i=1}^t X_i \sim~ Gamma (t, \lambda) [/math]
Side notes: if [math] X_i\sim~ Gamma(a,\lambda)[/math] and [math] Y_i\sim~ Gamma(B,\lambda)[/math] are independent gamma distributions, then [math]\frac{X}{X+Y}[/math] has a distribution of [math] Beta(a,B). [/math]
If we want to sample from the Gamma distribution, we can consider sampling from [math]t[/math] independent exponential distributions using the Inverse Method for each [math] X_i[/math] and add them up. Note that this only works the specific set of gamma distributions where t is a positive integer.
According to this property, a random variable that follows Gamma distribution is the sum of i.i.d (independent and identically distributed) exponential random variables. Now we want to generate 1000 values of [math]Gamma(20,10)[/math] random variables, so we need to obtain the value of each one by adding 20 values of [math]X_i \sim~ Exp(10)[/math]. To achieve this, we generate a 20by1000 matrix whose entries follow [math]Exp(10)[/math] and add the rows together.
[math] x_1 \sim~Exp(\lambda)[/math]
[math]x_2 \sim~Exp(\lambda)[/math]
...
[math]x_t \sim~Exp(\lambda)[/math]
[math]x_1+x_2+...+x_t~[/math]
>>l=1 >>urand(1,1000); >>x=(1/l)*log(u); >>hist(x) >>rand
 Procedure
 Sample independently from a uniform distribution [math]t[/math] times, giving [math] U_1,\dots,U_t \sim~ U(0,1)[/math]
 Use the Inverse Transform Method, [math] X_i = \frac {1}{\lambda}\log(1U_i)[/math], giving [math] X_1,\dots,X_t \sim~Exp(\lambda)[/math]
 Use the additive property,[math] X = \Sigma_{i=1}^t X_i \sim~ Gamma (t, \lambda) [/math]
 Note for Procedure
 If [math]U\sim~U(0,1)[/math], then [math]U[/math] and [math]1U[/math] will have the same distribution (both follows [math]U(0,1)[/math])
 This is because the range for [math]1U[/math] is still [math](0,1)[/math], and their densities are identical over this range.
 Let [math]Y=1U[/math], [math]Pr(Y\lt =y)=Pr(1U\lt =y)=Pr(U\gt =1y)=1Pr(U\lt =1y)=1(1y)=y[/math], thus [math]1U\sim~U(0,1)[/math]
 Code
>>close all >>clear all >>lambda = 1; >>u = rand(20, 1000); Note: this command generate a 20x1000 matrix (which means we generate 1000 number for each X_i with t=20); all the elements are generated by rand >>x = (1/lambda)*log(1u); Note: log(1u) is essentially the same as log(u) only if u~U(0,1) >>xx = sum(x) Note: sum(x) will sum all elements in the same column. size(xx) can help you to verify >>size(sum(x)) Note: see the size of x if we forget it (the answer is 20 1000) >>hist(x(1:)) Note: the graph of the first exponential distribution >>hist(xx)
size(x) and size(u) are both 20*1000 matrix. Since if u~unif(0, 1), u and 1  u have the same distribution, we can substitute 1u with u to simply the equation. Alternatively, the following command will do the same thing with the previous commands.
 Code
>>close all >>clear all >>lambda = 1; >>xx = sum((1/lambda)*log(rand(20, 1000))); ''This is simple way to put the code in one line. Here we can use either log(u) or log(1u) since U~U(0,1); >>hist(xx)
In the matrix rand(20,1000) means 20 row with 1000 numbers for each. use the code to show the generalize the distributions for multidimensional purposes in different cases, such as sum xi (each xi not equal xj), and they are independent, or matrix. Finally, we can see the conclusion is shown by the histogram.
Other Sampling Method: Box Muller
 From cartesian to polar coordinates
[math] R=\sqrt{x_{1}^2+x_{2}^2}= x_{2}/sin(\theta)= x_{1}/cos(\theta)[/math]
[math] tan(\theta)=x_{2}/x_{1} \rightarrow \theta=tan^{1}(x_{2}/x_{1})[/math]
 BoxMuller Transformation:
It is a transformation that consumes two continuous uniform random variables [math] X \sim U(0,1), Y \sim U(0,1) [/math] and outputs a bivariate normal random variable with [math] Z_1\sim N(0,1), Z_2\sim N(0,1). [/math]
Matlab
If X is a matrix,
 X(1,:) returns the first row
 X(:,1) returns the first column
 X(i,j) returns the (i,j)th entry
 sum(X,1) or sum(X) is a summation of the rows of X. The output is a row vector of the sums of each column.
 sum(X,2) is a summation of the columns of X, returning a vector.
 rand(r,c) will generate uniformly distributed random numbers in r rows and c columns.
 The dot operator (.), when placed before a function, such as +,,^, *, and many others specifies to apply that function to every element of a vector or a matrix. For example, to add a constant c to elements of a matrix A, do A.+c as opposed to simply A+c. The dot operator is not required for functions that can only take a number as their input (such as log).
 Matlab processes loops very slow, while it is fast with matrices and vectors, so it is preferable to use the dot operator to and matrices of random numbers than loops if it is possible.
Class 6  Thursday, May 23
Announcement
1. On the day of each lecture, students from the morning section can only contribute the first half of the lecture (i.e. 8:30  9:10 am), so that the second half can be saved for the ones from the afternoon section. After the day of lecture, students are free to contribute anything.
Standard Normal distribution
If X ~ N(0,1) i.e. Standard Normal Distribution  then its p.d.f. is of the form
 [math]f(x) = \frac{1}{\sqrt{2\pi}}\, e^{ \frac{\scriptscriptstyle 1}{\scriptscriptstyle 2} x^2}[/math]
 Warning : the General Normal distribution is:
[math] f(x) = \frac{1}{\sigma\sqrt{2\pi}} e^{ \frac{(x\mu)^2}{2\sigma^2} } [/math] 
which is almost useless in this course

where [math] \mu [/math] is the mean or expectation of the distribution and [math] \sigma [/math] is standard deviation
 N(0,1) is standard normal. [math] \mu [/math] =0 and [math] \sigma [/math]=1
Let X and Y be independent standard normal.
Let [math] \theta [/math] and R denote the Polar coordinate of the vector (X, Y) where [math] X = R \cdot \sin\theta [/math] and [math] Y = R \cdot \cos \theta [/math]
Note: R must satisfy two properties:
 1. Be a positive number (as it is a length)
 2. It must be from a distribution that has more data points closer to the origin so that as we go further from the origin, less points are generated (the two options are Chisquared and Exponential distribution)
The form of the joint distribution of R and [math]\theta[/math] will show that the best choice for distribution of R^{2} is exponential.
We cannot use the Inverse Transformation Method since F(x) does not have a closed form solution. So we will use joint probability function of two independent standard normal random variables and polar coordinates to simulate the distribution:
We know that
[math]R^{2}= X^{2}+Y^{2}[/math] and [math] \tan(\theta) = \frac{y}{x} [/math] where X and Y are two independent standard normal
 [math]f(x) = \frac{1}{\sqrt{2\pi}}\, e^{ \frac{\scriptscriptstyle 1}{\scriptscriptstyle 2} x^2}[/math]
 [math]f(y) = \frac{1}{\sqrt{2\pi}}\, e^{ \frac{\scriptscriptstyle 1}{\scriptscriptstyle 2} y^2}[/math]
 [math]f(x,y) = \frac{1}{\sqrt{2\pi}}\, e^{ \frac{\scriptscriptstyle 1}{\scriptscriptstyle 2} x^2} * \frac{1}{\sqrt{2\pi}}\, e^{ \frac{\scriptscriptstyle 1}{\scriptscriptstyle 2} y^2}=\frac{1}{2\pi}\, e^{ \frac{\scriptscriptstyle 1}{\scriptscriptstyle 2} (x^2+y^2)} [/math]
 Since for independent distributions, their joint probability function is the multiplication of two independent probability functions. It can also be shown using 11 transformation that the joint distribution of R and θ is given by, 11 transformation:
Let [math]d=R^2[/math]
[math]x= \sqrt {d}\cos \theta [/math] [math]y= \sqrt {d}\sin \theta [/math]
then [math]\left J\right = \left \dfrac {1} {2}d^{\frac {1} {2}}\cos \theta d^{\frac{1}{2}}\cos \theta +\sqrt {d}\sin \theta \dfrac {1} {2}d^{\frac{1}{2}}\sin \theta \right = \dfrac {1} {2}[/math] It can be shown that the joint density of [math] d /R^2[/math] and [math] \theta [/math] is:
 [math]\begin{matrix} f(d,\theta) = \frac{1}{2}e^{\frac{d}{2}}*\frac{1}{2\pi},\quad d = R^2 \end{matrix},\quad for\quad 0\leq d\lt \infty\ and\quad 0\leq \theta\leq 2\pi [/math]
Note that [math] \begin{matrix}f(r,\theta)\end{matrix}[/math] consists of two density functions, Exponential and Uniform, so assuming that r and [math]\theta[/math] are independent [math] \begin{matrix} \Rightarrow d \sim~ Exp(1/2), \theta \sim~ Unif[0,2\pi] \end{matrix} [/math]
 [math] \begin{align} R^2 = d = x^2 + y^2 \end{align} [/math]
 [math] \tan(\theta) = \frac{y}{x} [/math]
[math]\begin{align} f(d) = Exp(1/2)=\frac{1}{2}e^{\frac{d}{2}}\ \end{align}[/math]
[math]\begin{align} f(\theta) =\frac{1}{2\pi}\ \end{align}[/math]
To sample from the normal distribution, we can generate a pair of independent standard normal X and Y by:
1) Generating their polar coordinates
2) Transforming back to rectangular (Cartesian) coordinates.
Alternative Method of Generating Standard Normal Random Variables
Step 1: Generate [math]u_{1}[/math] ~[math]Unif(0,1)[/math]
Step 2: Generate [math]Y_{1}[/math] ~[math]Exp(1)[/math],[math]Y_{2}[/math]~[math]Exp(2)[/math]
Step 3: If [math]Y_{2} \geq(Y_{1}1)^2/2[/math],set [math]V=Y1[/math],otherwise,go to step 1
Step 4: If [math]u_{1} \leq 1/2[/math],then [math]X=V[/math]
===Expectation of a Standard Normal distribution===
The expectation of a standard normal distribution is 0
Proof:
 [math]\operatorname{E}[X]= \;\int_{\infty}^{\infty} x \frac{1}{\sqrt{2\pi}} e^{x^2/2} \, dx.[/math]
 [math]\phi(x) = \frac{1}{\sqrt{2\pi}}\, e^{ \frac{\scriptscriptstyle 1}{\scriptscriptstyle 2} x^2}.[/math]
 [math]=\;\int_{\infty}^{\infty} x \phi(x), dx.[/math]
 Since the first derivative ϕ′(x) is −xϕ(x)
 [math]=\;\  \int_{\infty}^{\infty} \phi'(x), dx.[/math]
 [math]=  \left[\phi(x)\right]_{\infty}^{\infty}[/math]
 [math]= 0[/math]
Note, more intuitively, because x is an odd function (f(x)+f(x)=0). Taking integral of x will give [math]x^2/2 [/math] which is an even function (f(x)=f(x)). This is in relation to the symmetrical properties of the standard normal distribution. If support is from negative infinity to infinity, then the integral will return 0.
Procedure (BoxMuller Transformation Method):
Pseudorandom approaches to generating normal random variables used to be limited. Inefficient methods such as inverse Gaussian function, sum of uniform random variables, and acceptancerejection were used. In 1958, a new method was proposed by George Box and Mervin Muller of Princeton University. This new technique was easy to use and also had the accuracy to the inverse transform sampling method that it grew more valuable as computers became more computationally astute.
The BoxMuller method takes a sample from a bivariate independent standard normal distribution, each component of which is thus a univariate standard normal. The algorithm is based on the following two properties of the bivariate independent standard normal distribution:
if [math]Z = (Z_{1}, Z_{2}[/math]) has this distribution, then
1.[math]R^2=Z_{1}^2+Z_{2}^2[/math] is exponentially distributed with mean 2, i.e.
[math]P(R^2 \leq x) = 1e^{x/2}[/math].
2.Given [math]R^2[/math], the point [math](Z_{1},Z_{2}[/math]) is uniformly distributed on the circle of radius R centered at the origin.
We can use these properties to build the algorithm:
1) Generate random number [math] \begin{align} U_1,U_2 \sim~ \mathrm{Unif}(0, 1) \end{align} [/math]
2) Generate polar coordinates using the exponential distribution of d and uniform distribution of θ,
[math] \begin{align} R^2 = d = 2\log(U_1), & \quad r = \sqrt{d} \\ & \quad \theta = 2\pi U_2 \end{align} [/math]
[math] \begin{matrix} \ R^2 \sim~ Exp(1/2), \theta \sim~ Unif[0,2\pi] \end{matrix} [/math]
Note: If U~unif(0,1), then ln(1U)=ln(U)
3) Transform polar coordinates (i.e. R and θ) back to Cartesian coordinates (i.e. X and Y),
[math] \begin{align} x = R\cos(\theta) \\ y = R\sin(\theta) \end{align} [/math]
.
Alternatively,
[math] x =\cos(2\pi U_2)\sqrt{2\ln U_1}\, [/math] and
[math] y =\sin(2\pi U_2)\sqrt{2\ln U_1}\, [/math]
Note: In steps 2 and 3, we are using a similar technique as that used in the inverse transform method.
The BoxMuller Transformation Method generates a pair of independent Standard Normal distributions, X and Y (Using the transformation of polar coordinates).
If you want to generate a number of independent standard normal distributed numbers (more than two), you can run the BoxMuller method several times.
For example:
If you want 8 independent standard normal distributed numbers, then run the BoxMuller methods 4 times (8/2 times).
If you want 9 independent standard normal distributed numbers, then run the BoxMuller methods 5 times (10/2 times), and then delete one.
Matlab Code
>>close all >>clear all >>u1=rand(1,1000); >>u2=rand(1,1000); >>d=2*log(u1); >>tet=2*pi*u2; >>x=d.^0.5.*cos(tet); >>y=d.^0.5.*sin(tet); >>hist(tet) >>hist(d) >>hist(x) >>hist(y)
Remember: For the above code to work the "." needs to be after the d to ensure that each element of d is raised to the power of 0.5.
Otherwise matlab will raise the entire matrix to the power of 0.5."
Note:
the first graph is hist(tet) and it is a uniform distribution.
The second one is hist(d) and it is a exponential distribution.
The third one is hist(x) and it is a normal distribution.
The last one is hist(y) and it is also a normal distribution.
Attention:There is a "dot" between sqrt(d) and "*". It is because d and tet are vectors.
As seen in the histograms above, X and Y generated from this procedure have a standard normal distribution.
 Code
>>close all >>clear all >>x=randn(1,1000); >>hist(x) >>hist(x+2) >>hist(x*2+2)<br>
Note:
1. randn is random sample from a standard normal distribution.
2. hist(x+2) will be centered at 2 instead of at 0.
3. hist(x*3+2) is also centered at 2. The mean doesn't change, but the variance of x*3+2 becomes nine times (3^2) the variance of x.
Comment:
BoxMuller transformations are not computationally efficient. The reason for this is the need to compute sine and cosine functions. A way to get around this timeconsuming difficulty is by an indirect computation of the sine and cosine of a random angle (as opposed to a direct computation which generates U and then computes the sine and cosine of 2πU.
Alternative Methods of generating normal distribution
1. Even though we cannot use inverse transform method, we can approximate this inverse using different functions.One method would be rational approximation.
2.Central limit theorem : If we sum 12 independent U(0,1) distribution and subtract 6 (which is E(ui)*12)we will approximately get a standard normal distribution.
3. Ziggurat algorithm which is known to be faster than BoxMuller transformation and a version of this algorithm is used for the randn function in matlab.
If Z~N(0,1) and X= μ +Zσ then X~[math] N(\mu, \sigma^2)[/math]
If Z_{1}, Z_{2}... Z_{d} are independent identically distributed N(0,1), then Z=(Z_{1},Z_{2}...Z_{d})^{T} ~N(0, I_{d}), where 0 is the zero vector and I_{d} is the identity matrix.
For the histogram, the constant is the parameter that affect the center of the graph.
Proof of Box Muller Transformation
Definition:
A transformation which transforms from a twodimensional continuous uniform distribution to a twodimensional bivariate normal distribution (or complex normal distribution).
Let U_{1} and U_{2} be independent uniform (0,1) random variables. Then [math]X_{1} = ((2lnU_{1})^.5)*cos(2\pi U_{2})[/math]
[math]X_{2} = (2lnU_{1})^0.5*sin(2\pi U_{2})[/math] are independent N(0,1) random variables.
This is a standard transformation problem. The joint distribution is given by
f(x1 ,x2) = f_{u1}, _{u2}(g1^− 1(x1,x2),g2^− 1(x1,x2)) *  J 
where J is the Jacobian of the transformation,
J = ∂u_{1}/∂x_{1},∂u_{1}/∂x_{2} ∂u_{2}/∂x_{1},∂u_{2}/∂x_{2}
where
u_{1} = g_{1} ^1(x1,x2) u_{2} = g_{2} ^1(x1,x2)
Inverting the above transformation, we have
u1 = exp^{(x_{1} ^2+ x_{2} ^2)/2} u2 = (1/2pi)*tan^1 (x_{2}/x_{1})
Finally we get
f(x1,x2) = {exp^((x1^2+x2^2)/2)}/2pi
which factors into two standard normal pdfs.
(The quote is from http://mathworld.wolfram.com/BoxMullerTransformation.html)
(The proof is from http://www.math.nyu.edu/faculty/goodman/teaching/MonteCarlo2005/notes/GaussianSampling.pdf)
General Normal distributions
General normal distribution is a special version of the standard normal distribution. The domain of the general normal distribution is affected by the standard deviation and translated by the mean value.
 The pdf of the general normal distribution is
[math] f(x) = \frac{1}{\sigma\sqrt{2\pi}} e^{ \frac{(x\mu)^2}{2\sigma^2} } [/math] 
which is almost useless in this course

where [math] \mu [/math] is the mean or expectation of the distribution and [math] \sigma [/math] is standard deviation
The probability density must be scaled by 1/sigma so that the integral is still 1.(Acknowledge: https://en.wikipedia.org/wiki/Normal_distribution) The special case of the normal distribution is standard normal distribution, which the variance is 1 and the mean is zero. If X is a general normal deviate, then [math] Z=\dfrac{X  (\mu)}{\sigma} [/math] will have a standard normal distribution.
If Z ~ N(0,1), and we want [math]X [/math]~[math] N(\mu, \sigma^2)[/math], then [math]X = \mu + \sigma * Z[/math] Since [math]E(x) = \mu +\sigma*0 = \mu [/math] and [math]Var(x) = 0 +\sigma^2*1[/math]
If [math]Z_1,...Z_d[/math] ~ N(0,1) and are independent then [math]Z = (Z_1,..Z_d)^{T} [/math]~ [math]N(0,I_d)[/math] ie.
 Code
>>close all >>clear all >>z1=randn(1,1000); <generate variable from standard normal distribution >>z2=randn(1,1000); >>z=[z1;z2]; <produce a vector >>plot(z(1,:),z(2,:),'.')
If Z~N(0,Id) and X= [math]\underline{\mu} + \Sigma^{\frac{1}{2}} \,Z [/math] then [math]\underline{X}[/math] ~[math]N(\underline{\mu},\Sigma)[/math]
NonStandard Normal Distributions
Example 1: Singlevariate Normal
If X ~ Norm(0, 1) then (a + bX) has a normal distribution with a mean of [math]\displaystyle a[/math] and a standard deviation of [math]\displaystyle b[/math] (which is equivalent to a variance of [math]\displaystyle b^2[/math]). Using this information with the BoxMuller transform, we can generate values sampled from some random variable [math]\displaystyle Y\sim N(a,b^2) [/math] for arbitrary values of [math]\displaystyle a,b[/math].
 Generate a sample u from Norm(0, 1) using the BoxMuller transform.
 Set v = a + bu.
The values for v generated in this way will be equivalent to sample from a [math]\displaystyle N(a, b^2)[/math]distribution. We can modify the MatLab code used in the last section to demonstrate this. We just need to add one line before we generate the histogram:
v = a + b * x;
For instance, this is the histogram generated when b = 15, a = 125:
Example 2: Multivariate Normal
The BoxMuller method can be extended to higher dimensions to generate multivariate normals. The objects generated will be nx1 vectors, and their variance will be described by nxn covariance matrices.
[math]\mathbf{z} = N(\mathbf{u}, \Sigma)[/math] defines the n by 1 vector [math]\mathbf{z}[/math] such that:
 [math]\displaystyle u_i[/math] is the average of [math]\displaystyle z_i[/math]
 [math]\!\Sigma_{ii}[/math] is the variance of [math]\displaystyle z_i[/math]
 [math]\!\Sigma_{ij}[/math] is the covariance of [math]\displaystyle z_i[/math] and [math]\displaystyle z_j[/math]
If [math]\displaystyle z_1, z_2, ..., z_d[/math] are normal variables with mean 0 and variance 1, then the vector [math]\displaystyle (z_1, z_2,..., z_d) [/math] has mean 0 and variance [math]\!I[/math], where 0 is the zero vector and [math]\!I[/math] is the identity matrix. This fact suggests that the method for generating a multivariate normal is to generate each component individually as single normal variables.
The mean and the covariance matrix of a multivariate normal distribution can be adjusted in ways analogous to the single variable case. If [math]\mathbf{z} \sim N(0,I)[/math], then [math]\Sigma^{1/2}\mathbf{z}+\mu \sim N(\mu,\Sigma)[/math]. Note here that the covariance matrix is symmetric and nonnegative, so its square root should always exist.
We can compute [math]\mathbf{z}[/math] in the following way:
 Generate an n by 1 vector [math]\mathbf{x} = \begin{bmatrix}x_{1} & x_{2} & ... & x_{n}\end{bmatrix}[/math] where [math]x_{i}[/math] ~ Norm(0, 1) using the BoxMuller transform.
 Calculate [math]\!\Sigma^{1/2}[/math] using singular value decomposition.
 Set [math]\mathbf{z} = \Sigma^{1/2} \mathbf{x} + \mathbf{u}[/math].
The following MatLab code provides an example, where a scatter plot of 10000 random points is generated. In this case x and y have a covariance of 0.9  a very strong positive correlation.
x = zeros(10000, 1); y = zeros(10000, 1); for ii = 1:10000 u1 = rand; u2 = rand; R2 = 2 * log(u1); theta = 2 * pi * u2; x(ii) = sqrt(R2) * cos(theta); y(ii) = sqrt(R2) * sin(theta); end E = [1, 0.9; 0.9, 1]; [u s v] = svd(E); root_E = u * (s ^ (1 / 2)) * u'; z = (root_E * [x y]'); z(1,:) = z(1,:) + 0; z(2,:) = z(2,:) + 3; scatter(z(1,:), z(2,:))
Note: The svd command computes the matrix singular value decomposition.
[u,s,v] = svd(E) produces a diagonal matrix s of the same dimension as E, with nonnegative diagonal elements in decreasing order, and unitary matrices u and v so that E = u*s*v'.
This code generated the following scatter plot:
In Matlab, we can also use the function "sqrtm()" or "chol()" (Cholesky Decomposition) to calculate square root of a matrix directly. Note that the resulting root matrices may be different but this does materially affect the simulation. Here is an example:
E = [1, 0.9; 0.9, 1]; r1 = sqrtm(E); r2 = chol(E);
R code for a multivariate normal distribution:
n=10000; r2<2*log(runif(n)); theta<2*pi*(runif(n)); x<sqrt(r2)*cos(theta); y<sqrt(r2)*sin(theta); a<matrix(c(x,y),nrow=n,byrow=F); e<matrix(c(1,.9,09,1),nrow=2,byrow=T); svde<svd(e); root_e<svde$u %*% diag(svde$d)^1/2; z<t(root_e %*%t(a)); z[,1]=z[,1]+5; z[,2]=z[,2]+ 8; par(pch=19); plot(z,col=rgb(1,0,0,alpha=0.06))
Bernoulli Distribution
The Bernoulli distribution is a discrete probability distribution, which usually describes an event that only has two possible results, i.e. success or failure (x=0 or 1). If the event succeed, we usually take value 1 with success probability p, and take value 0 with failure probability q = 1  p.
P ( x = 0) = q = 1  p
P ( x = 1) = p
P ( x = 0) + P (x = 1) = p + q = 1
If X~Ber(p), its pdf is of the form [math]f(x)= p^{x}(1p)^{(1x)}[/math], x=0,1
P is the success probability.
The Bernoulli distribution is a special case of binomial distribution, where the variate x only has two outcomes; so that the Bernoulli also can use the probability density function of the binomial distribution with the variate x taking values 0 and 1.
The most famous example for the Bernoulli Distribution would be the "Flip Coin" question, which has only two possible outcomes(Success or Failure) with the same probabilities of 0.5
Let x1,x2 denote the lifetime of 2 independent particles, x1~exp([math]\lambda[/math]), x2~exp([math]\lambda[/math]) we are interested in y=min(x1,x2)
Procedure: To simulate the event of flipping a coin, let P be the probability of flipping head and X = 1 and 0 represent flipping head and tail respectively: 1. Draw U ~ Uniform(0,1) 2. If U <= P X = 1 Else X = 0 3. Repeat as necessary
An intuitive way to think of this is in the coin flip example we discussed in a previous lecture. In this example we set p = 1/2 and this allows for 50% of points to be heads or tails.
 Code to Generate Bernoulli(p = 0.3)
i = 1; while (i <=1000) u =rand(); p = 0.3; if (u <= p) x(i) = 1; else x(i) = 0; end i = i + 1; end hist(x)
However, we know that if [math]\begin{align} X_i \sim Bernoulli(p) \end{align}[/math] where each [math]\begin{align} X_i \end{align}[/math] is independent,
[math]U = \sum_{i=1}^{n} X_i \sim Binomial(n,p)[/math]
So we can sample from binomial distribution using this property.
Note: We can consider Binomial distribution as the sum of n, independent, Bernoulli distributions
 Code to Generate Binomial(n = 20,p = 0.7)
p = 0.7; n = 20; for k=1:5000 i = 1; for i=1:n u=rand(); if (u <= p) y(i) = 1; else y(i) = 0; end end x(k) = sum(y==1); end hist(x)
Note: We can also regard the Bernoulli Distribution as either a conditional distribution or [math]f(x)= p^{x}(1p)^{(1x)}[/math], x=0,1.
Comments on Matlab: When doing operations on vectors, always put a dot before the operator if you want the operation to be done to every element in the vector. example: Let V be a vector with dimension 2*4 and you want each element multiply by 3.
The Matlab code is 3.*V
some examples for using code to generate distribution.
Class 7  Tuesday, May 28
Universality of the Uniform Distribution/Inverse Method
The inverse method is universal in the sense that we can potentially sample from any distribution where we can find the inverse of the cumulative distribution function.
Procedure:
1) Generate U~Unif (0, 1)
2) Set [math]x=F^{1}(u)[/math]
3) X~f(x)
Remark
1) The preceding can be written algorithmically for discrete random variables as
Generate a random number U ~ U(0,1]
If U < p_{0} set X = x_{0} and stop
If U < p_{0} + p_{1} set X = x_{1} and stop
...
2) If the x_{i}, i>=0, are ordered so that x_{0} < x_{1} < x_{2} <... and if we let F denote the distribution function of X, then X will equal x_{j} if F(x_{j1}) <= U < F(x_{j})
Example 1
Let [math]X[/math]_{1},[math]X[/math]_{2} denote the lifetime of two independent particles:
[math]X[/math]_{1}~exp([math]\lambda[/math]_{1})
[math]X[/math]_{2}~exp([math]\lambda[/math]_{2})
We are interested in [math]y=min(X[/math]_{1}[math],X[/math]_{2}[math])[/math]
Design an algorithm based on the InverseTransform Method to generate samples according to [math]f[/math]_{y}[math](y)[/math]
Solution:
x_{1}~exp([math]\lambda_1[/math])
x_{2}~exp([math]\lambda_2[/math])
[math]f_{x(x)}=\lambda e^{\lambda x},x\geq0 [/math]
[math]F_X(x)=1e^{\lambda x}, x\geq 0[/math]
[math]1F_Y(y) = P(Y\gt y)[/math] = P(min(X_{1},X_{2}) > y) = [math]\, P((X_1)\gt y) P((X_2)\gt y) = e^{\, (\lambda_1 + \lambda_2) y}[/math]
[math]F_Y(y)=1e^{\, (\lambda_1 + \lambda_2) y}, y\geq 0[/math]
[math]U=1e^{\, (\lambda_1 + \lambda_2) y}[/math] => [math]y=\, {\frac {1}{{\lambda_1 +\lambda_2}}} ln(1u)[/math]
Procedure:
Step1: Generate U~ U(0, 1)
Step2: set [math]y=\, {\frac {1}{{\lambda_1 +\lambda_2}}} ln(1u)[/math]
or set [math]y=\, {\frac {1} {{\lambda_1 +\lambda_2}}} ln(u)[/math]
Since it is a uniform distribution, therefore after generate a lot of times 1u and u are the same.
 Matlab Code
>> lambda1 = 1; >> lambda2 = 2; >> u = rand; >> y = log(u)/(lambda1 + lambda2)
If we generalize this example from two independent particles to n independent particles we will have:
[math]X[/math]_{1}~exp([math]\lambda[/math]_{1})
[math]X[/math]_{2}~exp([math]\lambda[/math]_{2})
...
[math]X[/math]_{n}~exp([math]\lambda[/math]_{n})
.
And the algorithm using the inversetransform method as follows:
step1: Generate U~U(0,1)
Step2: [math]y=\, {\frac {1}{{ \sum\lambda_i}}} ln(1u)[/math]
Example 2
Consider U~Unif[0,1)
[math]X=\, a (1\sqrt{1u})[/math],
where a>0 and a is a real number
What is the distribution of X?
Solution:
We can find a form for the cumulative distribution function of X by isolating U as U~Unif[0,1) will take values from the range of F(X)uniformly. It then remains to differentiate the resulting form by X to obtain the probability density function.
[math]X=\, a (1\sqrt{1u})[/math]
=>[math]1\frac {x}{a}=\sqrt{1u}[/math]
=>[math]u=1(1\frac {x}{a})^2[/math]
=>[math]u=\, {\frac {x}{a}} (2\frac {x}{a})[/math]
[math]f(x)=\frac {dF(x)}{dx}=\frac {2}{a}\frac {2x}{a^2}=\, \frac {2}{a} (1\frac {x}{a})[/math]
Example 3
Suppose F_{X}(x) = x^{n}, 0 ≤ x ≤ 1, n ∈ N > 0. Generate values from X.
Solution:
1. Generate [math]U ~\sim~ Unif[0, 1)[/math]
2. Set [math]X = U^{1/n}[/math]
For example, when [math]n = 20[/math],
[math]U = 0.6[/math] => [math]X = U^{1/20} = 0.974[/math]
[math]U = 0.5 =\gt [/math] [math]X = U^{1/20} = 0.966[/math]
[math]U = 0.2[/math] => [math]X = U^{1/20} = 0.923[/math]
Observe from above that the values of X for n = 20 are close to 1, this is because we can view [math]X^n[/math] as the maximum of n independent random variables [math]X,[/math] [math]X~\sim~Unif(0,1)[/math] and is much likely to be close to 1 as n increases. This is because when n is large the exponent tends towards 0. This observation is the motivation for method 2 below.
Recall that
If Y = max (X_{1}, X_{2}, ... , X_{n}), where X_{1}, X_{2}, ... , X_{n} are independent,
F_{Y}(y) = P(Y ≤ y) = P(max (X_{1}, X_{2}, ... , X_{n}) ≤ y) = P(X_{1} ≤ y, X_{2} ≤ y, ... , X_{n} ≤ y) = F_{x1}(y) F_{x2}(y) ... F_{xn}(y)
Similarly if [math] Y = min(X_1,\ldots,X_n)[/math] then the cdf of [math]Y[/math] is [math]F_Y = 1 [/math][math]\prod[/math][math](1 F_{X_i})[/math]
Method 1: Following the above result we can see that in this example, F_{X} = x^{n} is the cumulative distribution function of the max of n uniform random variables between 0 and 1 (since for U~Unif(0, 1), F_{U}(x) =
Method 2: generate X by having a sample of n independent U~Unif(0, 1) and take the max of the n samples to be x. However, the solution given above using inversetransform method only requires generating one uniform random number instead of n of them, so it is a more efficient method.
Generate the Y = max (X1, X2, ... , Xn), Y = min (X1, X2, ... , Xn), pdf and cdf, but (xi and xj are independent) i,j=1,2,3,4,5.....
Example 4 (New)
Now, we are having an similar example as example 1 just doing the maximum way.
Let X_{1},X_{2} denote the lifetime of two independent particles:
[math]\, X_1, X_2 \sim exp(\lambda)[/math]
We are interested in Z=max(X_{1},X_{2})
Design an algorithm based on the InverseTransform Method to generate samples according to f_{Z}(z)
[math]\, F_Z(z)=P[Z\lt =z] = F_{X_1}(z) \cdot F_{X_2}(z) = (1e^{\lambda z})^2[/math]
[math] \text{thus } F^{1}(z) = \frac{1}{\lambda}\log(1\sqrt z)[/math]
To sample Z:
[math]\, \text{Step 1: Generate } U \sim U[0,1)[/math]
[math]\, \text{Step 2: Let } Z = \frac{1}{\lambda}\log(1\sqrt U)[/math], therefore we can generate random variable of Z.
Discrete Case:
u~unif(0,1)
x < 0, S < P_{0}
while u < S
x < x + 1
S < S + P_{0}
Return x
Decomposition Method
The CDF, F, is a composition if [math]F_{X}(x)[/math] can be written as:
[math]F_{X}(x) = \sum_{i=1}^{n}p_{i}F_{X_{i}}(x)[/math] where
1) p_{i} > 0
2) [math]\sum_{i=1}^{n}[/math]p_{i} = 1.
3) [math]F_{X_{i}}(x)[/math] is a CDF
The general algorithm to generate random variables from a composition CDF is:
1) Generate U,V ~ [math] Unif(0,1)[/math]
2) If U < p_{1}, V = [math]F_{X_{1}}(x)[/math]^{1}
3) Else if U < p_{1} + p_{2}, V = [math]F_{X_{2}}(x)[/math]^{1}
4) Repeat from Step 1 (if N randomly generated variables needed, repeat N times)
Explanation
Each random variable that is a part of X contributes [math]p_{i} F_{X_{i}}(x)[/math] to [math]F_{X}(x)[/math] every time.
From a sampling point of view, that is equivalent to contributing [math]F_{X_{i}}(x)[/math] [math]p_{i}[/math] of the time. The logic of this is similar to that of the AcceptReject Method, but instead of rejecting a value depending on the value u takes, we instead decide which distribution to sample it from.
Simplified Version
1) Generate [math]u \sim Unif(0,1)[/math]
2) Set [math] X=0, s=P_0[/math]
3) While [math] u \gt s, [/math]
set [math] X = X+1[/math] and [math] s=s+P_x [/math]
4) Return [math] X [/math]
Examples of Decomposition Method
Example 1
[math]f(x) = \frac{5}{12}(1+(x1)^4) 0\leq x\leq 2[/math]
[math]f(x) = \frac{5}{12}+\frac{5}{12}(x1)^4 = \frac{5}{6} (\frac{1}{2})+\frac {1}{6}(\frac{5}{2})(x1))^4[/math]
Let[math]f_{x_1}= \frac{1}{2}[/math] and [math]f_{x_2} = \frac {5}{2}(x1)^4[/math]
Algorithm:
Generate U~Unif(0,1)
If [math]0\lt u\lt \frac {5}{6}[/math], then we sample from f_{x1}
Else if [math]\frac{5}{6}\lt u\lt 1[/math], we sample from f_{x2}
We can find the inverse CDF of f_{x2} and utilize the Inverse Transform Method in order to sample from f_{x2}
Sampling from f_{x1} is more straightforward since it is uniform over the interval (0,2)
divided f(x) to two pdf of x1 and x2, with uniform distribution, of two range of uniform.
Example 2
[math]f(x)=\frac{1}{4}e^{x}+2x+\frac{1}{12}, \quad 0\leq x \leq 3 [/math]
We can rewrite f(x) as [math]f(x)=(\frac{1}{4}) e^{x}+(\frac{2}{4}) 4x+(\frac{1}{4}) \frac{1}{3}[/math]
Let f_{x1} = [math]e^{x}[/math], f_{x2} = 4x, and f_{x3} = [math]\frac{1}{3}[/math]
Generate U~Unif(0,1)
If [math]0\lt u\lt \frac{1}{4}[/math], we sample from f_{x1}
If [math]\frac{1}{4}\leq u \lt \frac{3}{4}[/math], we sample from f_{x2}
Else if [math]\frac{3}{4} \leq u \lt 1[/math], we sample from f_{x3}
We can find the inverse CDFs of f_{x1} and f_{x2} and utilize the Inverse Transform Method in order to sample from f_{x1} and f_{x2}
We find F_{x1} = [math] 1e^{x}[/math] and F_{x2} = [math]2x^{2}[/math]
We find the inverses are [math] X = ln(1u)[/math] for F_{x1} and [math] X = \sqrt{\frac{U}{2}}[/math] for F_{x2}
Sampling from f_{x3} is more straightforward since it is uniform over the interval (0,3)
In general, to write an efficient algorithm for:
[math]F_{X}(x) = p_{1}F_{X_{1}}(x) + p_{2}F_{X_{2}}(x) + ... + p_{n}F_{X_{n}}(x)[/math]
We would first calculate [math] {q_i} = \sum_{j=1}^i p_j, \forall i = 1,\dots, n[/math]
Then Generate [math] U \sim~ Unif(0,1) [/math]
If [math] U \lt q_1 [/math] sample from [math] f_1 [/math]
else if [math] u\lt q_i [/math] sample from [math] f_i [/math] for [math] 1 \lt i \lt n [/math]
else sample from [math] f_n [/math]
when we divided the pdf of different range of f(x1) f(x2) and f(x3), and generate all of them and inverse, U~U(0,1)
Example of Decomposition Method
[math]F_x(x) = \frac {1}{3} x+\frac {1}{3} x^2+\frac {1}{3} x^3, 0\leq x\leq 1[/math]
Let [math]U =F_x(x) = \frac {1}{3} x+\frac {1}{3} x^2+\frac {1}{3} x^3[/math], solve for x.
[math]P_1=\frac{1}{3}, F_{x1} (x)= x, P_2=\frac{1}{3},F_{x2} (x)= x^2, P_3=\frac{1}{3},F_{x3} (x)= x^3[/math]
Algorithm:
Generate [math]\,U \sim Unif [0,1)[/math]
Generate [math]\,V \sim Unif [0,1)[/math]
if [math]0\leq u \leq \frac{1}{3}, x = v[/math]
else if [math]u \leq \frac{2}{3}, x = v^{\frac{1}{2}}[/math]
else [math]x=v^{\frac{1}{3}}[/math]
Matlab Code:
u=rand # U is v=rand if u<1/3 x=v elseif u<2/3 x=sqrt(v) else x=v^(1/3) end
=== Example of Decomposition Method(new) ===
F_{x}(x) = 1/2*x+1/2*x^{2}, 0<= x<=1
let U =F_{x}(x) = 1/2*x+1/2*x^{2}, solve for x.
P_{1}=1/2, F_{x1}(x)= x, P_{2}=1/2,F_{x2}(x)= x^{2},
Algorithm:
Generate U ~ Unif [0,1)
Generate V~ Unif [0,1)
if 0<u<1/2, x = v
else x = v^{1/2}
Matlab Code:
u=rand v=rand if u<1/2 x=v else x=sqrt(v) end
Extra Knowledge about Decomposition Method
There are different types and applications of Decomposition Method
1. Primal decomposition
2. Dual decomposition
3. Decomposition with constraints
4. More general decomposition structures
5. Rate control
6. Single commodity network ﬂow
For More Details, please refer to http://www.stanford.edu/class/ee364b/notes/decomposition_notes.pdf
Fundamental Theorem of Simulation
Consider two shapes, A and B, where B is a subshape (subset) of A. We want to sample uniformly from inside the shape B. Then we can sample uniformly inside of A, and throw away all samples outside of B, and this will leave us with a uniform sample from within B. (Basis of the AcceptReject algorithm)
The advantage of this method is that we can sample a unknown distribution from a easy distribution. The disadvantage of this method is that it may need to reject many points, which is inefficient.
Inverse each part of partial CDF, the partial CDF is divided by the original CDF, partial range is uniform distribution.
More specific definition of the theorem can be found here.<ref>http://www.bus.emory.edu/breno/teaching/MCMC_GibbsHandouts.pdf</ref>
Matlab code:
close all clear all ii=1; while ii<1000 u=rand y=R*(2*U1) if (1U^2)>=(2*u1)^2 x(ii)=y; ii=ii+1 end
Question 2
Use Acceptance and Rejection Method to sample from [math]f_X(x)=b*x^n*(1x)^n[/math] , [math]n\gt 0[/math], [math]0\lt x\lt 1[/math]
Solution: This is a beta distribution, Beta ~[math]\int _{0}^{1}b*x^{n}*(1x)^{n}dx = 1[/math]
U_{1~Unif[0,1) }
U_{2~Unif[0,1)
}
fx=[math] bx^{1/2}(1x)^{1/2} \lt = bx^{1/2}\sqrt2 ,0\lt =x\lt =1/2 [/math]
The beta distribution maximized at 0.5 with value [math](1/4)^n[/math].
So, [math]c=b*(1/4)^n[/math]
Algorithm:
1.Draw [math]U_1[/math] from [math]U(0, 1)[/math]. [math] U_2[/math] from [math]U(0, 1)[/math]
2.If [math]U_2\lt =b*(U_1)^n*(1(U_1))^n/b*(1/4)^n=(4*(U_1)*(1(U_1)))^n[/math]
then X=U_1 Else return to step 1.
Discrete Case:
Most discrete random variables do not have a closed form inverse CDF. Also, its CDF [math]F:X \rightarrow [0,1][/math] is not necessarily onto. This means that not every point in the interval [math] [0,1] [/math] has a preimage in the support set of X through the CDF function.
Let [math]X[/math] be a discrete random variable where [math]a \leq X \leq b[/math] and [math]a,b \in \mathbb{Z}[/math] .
To sample from [math]X[/math], we use the partition method below:
[math]\, \text{Step 1: Generate u from } U \sim Unif[0,1][/math]
[math]\, \text{Step 2: Set } x=a, s=P(X=a)[/math]
[math]\, \text{Step 3: While } u\gt s, x=x+1, s=s+P(X=x)[/math]
[math]\, \text{Step 4: Return } x[/math]
Class 8  Thursday, May 30, 2013
In this lecture, we will discuss algorithms to generate 3 wellknown distributions: Binomial, Geometric and Poisson. For each of these distributions, we will first state its general understanding, probability mass function, expectation and variance. Then, we will derive one or more algorithms to sample from each of these distributions, and implement the algorithms on Matlab.
The Bernoulli distribution
The Bernoulli distribution is a special case of the binomial distribution, where n = 1. X ~ Bin(1, p) has the same meaning as X ~ Ber(p), where p is the probability of success and 1p is the probability of failure (we usually define a variate q, q= 1p). The mean of Bernoulli is p and the variance is p(1p). Bin(n, p), is the distribution of the sum of n independent Bernoulli trials, Bernoulli(p), each with the same probability p, where 0<p<1.
For example, let X be the event that a coin toss results in a "head" with probability p, then X~Bernoulli(p).
P(X=1)= p
P(X=0)= q = 1p
Therefore, P(X=0) + P(X=1) = p + q = 1
Algorithm:
1) Generate [math]u\sim~Unif(0,1)[/math]
2) If [math]u \leq p[/math], then [math]x = 1 [/math]
else [math]x = 0[/math]
The answer is:
when [math] U \leq p, x=1[/math]
when [math]U \geq p, x=0[/math]
3) Repeat as necessary
 Matlab Code
>> p = 0.8 % an arbitrary probability for example >> for i = 1: 100 >> u = rand; >> if u < p >> x(ii) = 1; >> else >> x(ii) = 0; >> end >> end >> hist(x)
The Binomial Distribution
In general, if the random variable X follows the binomial distribution with parameters n and p, we write X ~ Bin(n, p). (Acknowledge: https://en.wikipedia.org/wiki/Binomial_distribution) If X ~ B(n, p), then its pmf is of form:
f(x)=(nCx) p^{x}(1p)^{(nx)}, x=0,1,...n
Or f(x) = [math](n!/x!(nx)!)[/math] p^{x}(1p)^{(nx)}, x=0,1,...n
Mean (x) = E(x) = [math] np [/math]
Variance = [math] np(1p) [/math]
Generate n uniform random number [math]U_1,...,U_n[/math] and let X be the number of [math]U_i[/math] that are less than or equal to p.
The logic behind this algorithm is that the Binomial Distribution is simply a Bernoulli Trial, with a probability of success of p, repeated n times. Thus, we can sample from the distribution by sampling from n Bernoulli. The sum of these n bernoulli trials will represent one binomial sampling. Thus, in the below example, we are sampling 1000 realizations from 20 Bernoulli random variables. By summing up the rows of the 20 by 1000 matrix that is produced, we are summing up the 20 bernoulli outcomes to produce one binomial sampling. We have 1000 rows, which means we have realizations from 1000 binomial random variables when this sum is done (the output of the sum is a 1 by 1000 sized vector).
To continue with the previous example, let X be the number of heads in a series of n independent coin tosses  where for each toss, the probability of coming up with a head is p  then X~Bin(n, p).
MATLAB tips: to get a pdf f(x), we can use code binornd(N,P). N means number of trials and p is the probability of success. a=[2 3 4],if set a<3, will produce a=[1 0 0]. If you set "a == 3", it will produce [0 1 0]. If a=[2 6 9 10], if set a<4, will produce a=[1 0 0 0], because only the first element (2) is less than 4, meanwhile the rest are greater. So we can use this to get the number which is less than p.
Algorithm for Bernoulli is given as above
Code
>>a=[3 5 8]; >>a<5 ans= 1 0 0 >>rand(20,1000) >>rand(20,1000)<0.4 >>A = sum(rand(20,1000)<0.4) #sum of raws ~ Bin(20 , 0.3) >>hist(A) >>mean(A) Note: `1` in the above code means sum the matrix by column >>sum(sum(rand(20,1000)<0.4)>8)/1000 This is an estimate of Pr[A>8].
remark: a=[2 3 4],if set a<3, will produce a=[1 0 0]. If you set "a == 3", it will produce [0 1 0]. using code to find some value what i want to get from the matrix. It`s useful to defi