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=== Final ===
 
=== Final ===
Saturday August 10,2013 from7:30pm-10:00pm
+
Saturday August 10,2013 from 7:30pm-10:00pm
  
 
=== TA(s):  ===
 
=== TA(s):  ===
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=== Four Fundamental Problems ===
 
=== Four Fundamental Problems ===
 
<!-- br tag for spacing-->
 
<!-- br tag for spacing-->
1 Classification: Given an input object X, we have a function which will take in this input X and identify which 'class (Y)' it belongs to (Discrete Case) <br />
+
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) <br />
 
   <font size="3">i.e taking value from x, we could predict y.</font>
 
   <font size="3">i.e taking value from x, we could predict y.</font>
 
(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.) <br />
 
(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.) <br />
2 Regression: Same as classification but in the continuous case except y is non discrete. (Example of stock prices, height, weight, etc.) <br />
+
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.) <br />
 
(A simple practice might be investigating the hypothesis that higher levels of education cause higher levels of income.) <br />
 
(A simple practice might be investigating the hypothesis that higher levels of education cause higher levels of income.) <br />
 
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.) <br />
 
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.) <br />
4 Dimensionality Reduction (aka Feature extraction, Manifold learning): Used when we have a variable in high dimension space and we want to reduce the dimension <br />
+
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 <br />
  
 
=== Applications ===
 
=== Applications ===
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*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!
 
*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 (10% of final mark):'''
+
'''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.
 
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:<br/>
 
Example:<br/>
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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.
 
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 must first:
+
In order to perform a simulation study, we should:
<br\> 1 Use a computer to generate (pseudo) random numbers (rand in MATLAB).<br>
+
<br\> 1 Use a computer to generate (pseudo*) random numbers (rand in MATLAB).<br>
 
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).<br>
 
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).<br>
 
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)<br>
 
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)<br>
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(Source: Ross, Sheldon M., and Sheldon M. Ross. Simulation. San Diego: Academic, 1997. Print.)
 
(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 pseudo-random 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.
 
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.
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Generally, mod means taking the reminder after division by m.
 
Generally, mod means taking the reminder after division by m.
 
<br />
 
<br />
We say that n is congruent to r mod m if n = mq + r, where m is an integer. <br />
+
We say that n is congruent to r mod m if n = mq + r, where m is an integer.  
 +
Values are between 0 and m-1 <br />
 
if y = ax + b, then <math>b:=y \mod a</math>. <br />
 
if y = ax + b, then <math>b:=y \mod a</math>. <br />
  
'''For example:'''<br />
+
'''Example 1:'''<br />
  
 
<math>30 = 4 \cdot  7 + 2</math><br />
 
<math>30 = 4 \cdot  7 + 2</math><br />
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<br />
 
<br />
'''Another example:'''<br />
+
'''Example 2:'''<br />
  
 
If <math>23 = 3 \cdot  6 + 5</math> <br />
 
If <math>23 = 3 \cdot  6 + 5</math> <br />
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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 />
 +
 +
  
  
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==== 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 best-known 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 best-known 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 />
  
 +
[[File:Linear_Congruential_Statment.png‎|600px]] "Source: STAT 340 Spring 2010 Course Notes"
 +
 +
'''First consider the following algorithm'''<br />
 +
<math>x_{k+1}=x_{k} \mod m</math> <br />
  
 +
such that: if <math>x_{0}=5(mod 150)</math>, <math>x_{n}=3x_{n-1}</math>, find <math>x_{1},x_{8},x_{9}</math>. <br />
 +
<math>x_{n}=(3^n)*5(mod 150)</math> <br />
 +
<math>x_{1}=45,x_{8}=105,x_{9}=15</math> <br />
  
'''First consider the following algorithm'''<br />
 
<math>x_{k+1}=x_{k} \mod m</math>
 
  
  
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2. close all: closes all figures.<br />
 
2. close all: closes all figures.<br />
 
3. who: displays all defined variables.<br />
 
3. who: displays all defined variables.<br />
4. clc: clears screen.<br /><br />
+
4. clc: clears screen.<br />
5. ; : prevents the results from printing.<br /><br />
+
5. ; : prevents the results from printing.<br />
 +
6. disstool: displays a graphing tool.<br /><br />
  
 
<pre style="font-size:16px">
 
<pre style="font-size:16px">
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'''Comments:'''<br />
 
'''Comments:'''<br />
 +
 +
Matlab code:
 +
a=5;
 +
b=7;
 +
m=200;
 +
x(1)=3;
 +
for ii=2:1000
 +
x(ii)=mod(a*x(ii-1)+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 />
  
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</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 ===
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'''Proof of the theorem:'''<br />
 
'''Proof of the theorem:'''<br />
 
The generalized inverse satisfies the following: <br />
 
The generalized inverse satisfies the following: <br />
<math>\begin{align}
+
 
\forall u \in \left[0,1\right], \, x \in \R, \\
+
:<math>P(X\leq x)</math> <br />
&{} F^{-1}\left(u\right) \leq x &{} \\
+
<math>= P(F^{-1}(U)\leq x)</math> (since <math>X= F^{-1}(U)</math> by the inverse method)<br />
\Rightarrow &{} F\Big(F^{-1}\left(u\right)\Big) \leq F\left(x\right) &&{} F \text{ is non-decreasing} \\
+
<math>= P((F(F^{-1}(U))\leq F(x))</math>  (since <math>F </math> is monotonically increasing) <br />
\Rightarrow &{} F\Big(\inf \{y \in \R | F(y)\geq u \}\Big) \leq F\left(x\right) &&{} \text{by definition of } F^{-1} \\
+
<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>,<br />
\Rightarrow &{} \inf \{F(y) \in [0,1] | F(y)\geq u \} \leq F\left(x\right) &&{} F \text{ is right continuous and non-decreasing} \\
+
<math>= F(x) , \text{ where } 0 \leq F(x) \leq 1 </math>  <br />
\Rightarrow &{} u \leq F\left(x\right) &&{} \text{by definition of } \inf \\
+
 
\Rightarrow &{} x \in \{y \in \R | F(y) \geq u\} &&{} \\
+
This is the c.d.f. of X.  <br />
\Rightarrow &{} x \geq \inf \{y \in \R | F(y)\geq u \}\Big) &&{} \text{by definition of } \inf \\
+
<br />
\Rightarrow &{} x \geq F^{-1}(u) &&{} \text{by definition of } F^{-1} \\
 
\end{align}</math>
 
  
 
That is <math>F^{-1}\left(u\right) \leq x \Leftrightarrow u \leq F\left(x\right)</math><br />
 
That is <math>F^{-1}\left(u\right) \leq x \Leftrightarrow u \leq F\left(x\right)</math><br />
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Step 2: <math>  x=\frac{-ln(U)}{\lambda} </math> <br /><br />
 
Step 2: <math>  x=\frac{-ln(U)}{\lambda} </math> <br /><br />
  
 +
 +
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''':<br />
 
'''MatLab Code''':<br />
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<pre style="font-size:16px">
 
<pre style="font-size: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).jpg|300px]]
 
[[File:ITM_example_hist(u).jpg|300px]]
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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 />
  
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<math>x = 1-(1-u)^\frac {1}{\beta}</math><br />
 
<math>x = 1-(1-u)^\frac {1}{\beta}</math><br />
 
let β=3, use Matlab to construct N=1000 observations from Beta(1,3)<br />
 
let β=3, use Matlab to construct N=1000 observations from Beta(1,3)<br />
Matlab Code:<br />
+
'''MatLab Code''':<br />
>> u = rand(1,1000);<br />
+
 
x = 1-(1-u)^(1/3);<br />
+
<pre style="font-size:16px">
>> hist(x,50)<br />
+
>> u = rand(1,1000);
>> mean(x)<br />
+
x = 1-(1-u)^(1/3);
 +
>> hist(x,50)
 +
>> mean(x)
 +
</pre>
  
 
'''Example 5 - Estimating <math>\pi</math>''':<br/>
 
'''Example 5 - Estimating <math>\pi</math>''':<br/>
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== Class 3 - Tuesday, May 14 ==
 
== Class 3 - Tuesday, May 14 ==
 
=== Recall the Inverse Transform Method ===
 
=== Recall the Inverse Transform Method ===
 
+
Let U~Unif(0,1),then the random variable  X = F<sup>-1</sup>(u) has distribution F.  <br />
 
To sample X with CDF F(x), <br />
 
To sample X with CDF F(x), <br />
  
'''1) Draw u~U(0,1) '''<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 />
  
  
'''Proof''' <br />
 
First note that
 
<math>P(U\leq a)=a, \forall a\in[0,1]</math> <br />
 
  
:<math>P(X\leq x)</math> <br />
 
<math>= P(F^{-1}(U)\leq x)</math> (since <math>X= F^{-1}(U)</math> by the inverse method)<br />
 
<math>= P((F(F^{-1}(U))\leq F(x))</math>  (since <math>F </math> is monotonically increasing) <br />
 
<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>, this is explained further below)<br />
 
<math>= F(x) , \text{ where } 0 \leq F(x) \leq 1 </math>  <br />
 
  
This is the c.d.f. of X.  <br />
+
 
 
<br />
 
<br />
  
'''Note''': that the CDF of a U(a,b) random variable is:
+
'''Note''': CDF of a U(a,b) random variable is:
 
:<math>
 
:<math>
 
   F(x)= \begin{cases}
 
   F(x)= \begin{cases}
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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)  
 
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
+
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====
 
====Limitations of the Inverse Transform Method====
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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.
 
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 ===
 
=== Discrete Case ===
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Note that after generating a random U, the value of X can be determined by finding the interval <math>[F(x_{j-1}),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_{j-1}),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(i-1))<U<F(xi), x=xi.<br />
  
  
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Step 5: Go to step 3<br>
 
Step 5: Go to step 3<br>
 
*Note: These steps can be found in Simulation 5th Ed. by Sheldon Ross.
 
*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.<br>
+
*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.<br>
 
Step 1: Generate n uniform numbers U1 ... Un.<br>
 
Step 1: Generate n uniform numbers U1 ... Un.<br>
 
Step 2: X = <math>\sum U_i < = p</math> where P is the probability of success.
 
Step 2: X = <math>\sum U_i < = p</math> where P is the probability of success.
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<math>P(X=x_i) = \, p (1-p)^{x_{i}-1}</math>
 
<math>P(X=x_i) = \, p (1-p)^{x_{i}-1}</math>
 
We have CDF:
 
We have CDF:
<math>F(x)=P(X \leq x)=1-P(X>x) = 1-(1-p)^x</math>, P(X>x) means we get at least x failures before observe the first success.
+
<math>F(x)=P(X \leq x)=1-P(X>x) = 1-(1-p)^x</math>, P(X>x) means we get at least x failures before we observe the first success.
 
Now consider the inverse transform:
 
Now consider the inverse transform:
 
:<math>
 
:<math>
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'''Problems'''<br />
 
'''Problems'''<br />
1. We have to find <math> F^{-1} </math>
+
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).
2. For many distributions, such as Gaussian, it is too difficult to find the inverse of <math> F(x)</math>.<br>
+
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.
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 closed to the expected value 0.5.<br>
+
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.<br>
 
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.<br>
 
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.<br>
 
Example 3 uses inverse method to figure out the probability range of each random varible.
 
Example 3 uses inverse method to figure out the probability range of each random varible.
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3. Mixed continues discrete
 
3. Mixed continues discrete
  
'''Problems with Inverse-Transform Approach'''
 
 
1. must invert CDF, which may be different (numerical methods).
 
 
2. May not be the fastest or simplest approach for a given distribution.
 
  
 
'''Advantages of Inverse-Transform Method'''
 
'''Advantages of Inverse-Transform Method'''
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[[File:AR_Method.png]]
 
[[File:AR_Method.png]]
  
 
{{Cleanup|reason= Do not write <math>c*g(x)</math>. Instead write <math>c \times g(x)</math> or <math>\,c g(x)</math>
 
}}
 
  
 
The main logic behind the Acceptance-Rejection Method is that:<br>
 
The main logic behind the Acceptance-Rejection Method is that:<br>
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3. For each value of x, we accept and reject some points based on a probability, which will be discussed below.<br>
 
3. For each value of x, we accept and reject some points based on a probability, which will be discussed below.<br>
  
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> &forall;x. <br>
+
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> \,&forall;x. <br>
  
 
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.<br><br>
 
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.<br><br>
Line 993: Line 1,038:
 
3. Verify that <math>f(x)\leqslant c g(x)</math> at all the local maximums as well as the absolute maximums.<br>
 
3. Verify that <math>f(x)\leqslant c g(x)</math> at all the local maximums as well as the absolute maximums.<br>
 
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.<br>
 
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.<br>
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 p=P(U<= f(Y)/(cg(Y))) , P(N=n)=(1-p^(n-1))p ,n>=1.Thus on average the number of iterations required is given by E(N)=1/p
+
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)=(1-p(n-1))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 Acceptance-Rejection 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.  
 
c should be close to the maximum of f(x)/g(x), not just some arbitrarily picked large number. Otherwise, the Acceptance-Rejection 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.  
Line 1,173: Line 1,218:
  
 
== Class 4 - Thursday, May 16 ==  
 
== Class 4 - Thursday, May 16 ==  
*When we want to find a target distribution, denoted as <math>f(x)</math>, we need to first find a proposal distribution <math>g(x)</math>  that is easy to sample from. <br>  
+
 
*The relationship between the proposal distribution and target distribution is: <math> c \cdot g(x) \geq f(x) </math>, where c is a constant. This means that the area of f(x) is under the area of <math> c \cdot g(x)</math>. <br>
+
'''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>
 
*Chance of acceptance is less if the distance between <math>f(x)</math> and <math> c \cdot g(x)</math> is big, and vice-versa, 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 vice-versa, 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, a <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 1,185: 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/>
  
*The logic behind this:
+
Note: This procedure is called the Acceptance-Rejection Method.<br>
The Acceptance-Rejection 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/>. If c is 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>
+
'''The Acceptance-Rejection 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 UNIF(0,1). However, in other cases we need to generate UNIF(a,b). We may need to perform a linear transformation on the UNIF(0,1) variable. <br>
+
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 1,211: 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 1,246: Line 1,305:
 
[[File:Beta(2,1)_example.jpg|750x750px]]
 
[[File:Beta(2,1)_example.jpg|750x750px]]
  
Note: g follows uniform distribution, it only covers half of the graph which runs from 0 to 1 on y-axis. 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 runs from 0 to 2 on y-axis which covers f(x).
+
'''Note:''' g follows uniform distribution, it only covers half of the graph which runs from 0 to 1 on y-axis. 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 y-axis 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 1,259: 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="font-weight:bold;color:green;">Matlab Code</span>
 
<span style="font-weight:bold;color:green;">Matlab Code</span>
Line 1,309: Line 1,368:
  
 
<span style="font-weight:bold;colour:green;">Matlab Tip:</span>
 
<span style="font-weight:bold;colour:green;">Matlab Tip:</span>
Periods, ".",meaning "element-wise", 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 matrix dimensions must agree.
+
Periods, ".",meaning "element-wise", 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.
  
 
'''
 
'''
Line 1,339: Line 1,398:
 
=====Example of Acceptance-Rejection Method=====
 
=====Example of Acceptance-Rejection Method=====
  
<math> f(x) = 3x^2,  0<x<1 </math>
+
<math>\begin{align}
<math>g(x)=1,  0<x<1</math>
+
& f(x) = 3x^2,  0<x<1 \\
 +
\end{align}</math><br\>
 +
 
 +
<math>\begin{align}
 +
& g(x)=1,  0<x<1 \\
 +
\end{align}</math><br\>
  
 
<math>c = \max \frac{f(x)}{g(x)} = \max \frac{3x^2}{1} = 3 </math><br>
 
<math>c = \max \frac{f(x)}{g(x)} = \max \frac{3x^2}{1} = 3 </math><br>
Line 1,346: Line 1,410:
  
 
1. Generate two uniform numbers in the unit interval <math>U_1, U_2 \sim~ U(0,1)</math><br>
 
1. Generate two uniform numbers in the unit interval <math>U_1, U_2 \sim~ U(0,1)</math><br>
2. If <math>U_2 \leqslant {U_1}^2</math>, accept <math>U_1</math> as the random variable with pdf <math>f</math>, if not return to Step 1
+
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>g(x)=2x</math> for a more efficient algorithm
+
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>.
 
<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>g(x)</math>
+
Use the inverse method to sample from <math>\begin{align}g(x)\end{align}</math>
<math>G(x)=x^2</math>.
+
<math>\begin{align}G(x)=x^2\end{align}</math>.
Generate <math>U</math> from <math>U(0,1)</math> and set <math>x=sqrt(u)</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><br>
 
1. Generate two uniform numbers in the unit interval <math>U_1, U_2 \sim~ U(0,1)</math><br>
 
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
 
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 q(x) = c * g(x) is called an envelop or majoring function.<br>
+
*Note :the function <math>\begin{align}q(x) = c * g(x)\end{align}</math> is called an envelop or majoring function.<br>
To obtain a better proposing function g(x), we can first assume a new q(x) 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 q(x) = 3x. To find the normalizing constant, we need to solve k * <math>\sum 3x = 1</math> which gives us k = 2/3. So, g(x) = k*q(x) = 2x.
+
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 1,489: Line 1,554:
 
3) A constant c where <math>f(x)\leq c\cdot g(x)</math><br/>
 
3) A constant c where <math>f(x)\leq c\cdot g(x)</math><br/>
 
4) A uniform draw<br/>
 
4) A uniform draw<br/>
 
  
 
==== Interpretation of 'C' ====
 
==== Interpretation of 'C' ====
Line 1,498: Line 1,562:
 
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%.
 
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 bigger but close to one as much as possible so that <math>\tfrac{1}{c}</math> is smaller but close to one.
+
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 ==
 
== Class 5 - Tuesday, May 21 ==
Line 1,524: Line 1,602:
 
>>hist(x,30)                #30 is the number of bars
 
>>hist(x,30)                #30 is the number of bars
 
</pre>
 
</pre>
 +
 +
calculate process:
 +
<math>u_{1} <= \sqrt (1-(2u-1)^2) </math> <br>
 +
<math>(u_{1})^2 <=(1-(2u-1)^2) </math> <br>
 +
<math>(u_{1})^2 -1 <=(-(2u-1)^2) </math> <br>
 +
<math>1-(u_{1})^2 >=((2u-1)^2-1) </math> <br>
 +
  
 
MATLAB tips: hist(x,y) plots a histogram of variable x, where y is the number of bars in the graph.
 
MATLAB tips: hist(x,y) plots a histogram of variable x, where y is the number of bars in the graph.
Line 1,556: Line 1,641:
 
~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 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 acceptance-rejection method of pmf, the uniform probability is the same for all variables, and there 5 parameters(1,2,3,4,5), so g(x) is 0.2
+
the acceptance-rejection 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.
 
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.
<br />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 effect.
+
<br />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'''<br />
 
* '''Code for example 1'''<br />
Line 1,603: Line 1,688:
 
>>close all
 
>>close all
 
>>clear all
 
>>clear all
>>p=[.1 .3 .6];  
+
>>p=[.1 .3 .6];     %This a vector holding the values 
 
>>ii=1;
 
>>ii=1;
 
>>while ii < 1000
 
>>while ii < 1000
     y=unidrnd(3);
+
     y=unidrnd(3);   %generates random numbers for the discrete uniform distribution with maximum 3
     u=rand;
+
     u=rand;          
 
     if u<= p(y)/0.6
 
     if u<= p(y)/0.6
       x(ii)=y;
+
       x(ii)=y;    
       ii=ii+1;
+
       ii=ii+1;     %else ii=ii+1
 
     end
 
     end
 
   end
 
   end
Line 1,618: Line 1,703:
  
 
* '''Example 3'''<br>
 
* '''Example 3'''<br>
<math>p_{x}=e^{-3}3^{x}/x! , x>=0</math><br>(poisson distribution)
 
Try the first few p_{x}'s:  .0498 .149 .224 .224 .168 .101 .0504 .0216 .0081 .0027<br>
 
  
Use the geometric distribution for <math>g(x)</math>;<br>
+
Suppose <math>\begin{align}p_{x} = e^{-3}3^{x}/x! , x\geq 0\end{align}</math> (Poisson distribution)
<math>g(x)=p(1-p)^{x}</math>, choose p=0.25<br>
 
Look at <math>p_{x}/g(x)</math> for the first few numbers: .199 .797 1.59 2.12 2.12 1.70 1.13 .647 .324 .144<br>
 
We want <math>c=max(p_{x}/g(x))</math> which is approximately 2.12<br>
 
  
1. Generate <math>U_{1} \sim~ U(0,1); U_{2} \sim~ U(0,1)</math><br>
+
'''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>
2. <math>j = \lfloor \frac{ln(U_{1})}{ln(.75)} \rfloor+1;</math><br>
 
3. if <math>U_{2} < \frac{p_{j}}{cg(j)}</math>, set X = x<sub>j</sub>, else go to step 1.
 
  
Note: In this case, f(x)/g(x) is extremely difficult to differentiate so we were required to test points. If the function is easily differentiable, 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.
+
'''Proposed distribution:''' Use the geometric distribution for <math>\begin{align}g(x)\end{align}</math>;<br>
 +
 
 +
<math>\begin{align}g(x)=p(1-p)^{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/l04-46.pdf*
  
 
*'''Example 4''' (Hypergeometric & Binomial)<br>  
 
*'''Example 4''' (Hypergeometric & Binomial)<br>  
Line 1,710: Line 1,806:
 
<math> F(x) = \int_0^{x} \frac{e^{-y}y^{t-1}}{(t-1)!} \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^{t-1}}{(t-1)!} \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/>
  
 
Neither Inverse Transformation nor Acceptance-Rejection Method can be easily applied to Gamma distribution.
 
Neither Inverse Transformation nor Acceptance-Rejection Method can be easily applied to Gamma distribution.
Line 1,820: 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 1,873: 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 1,889: Line 1,988:
 
We know that  
 
We know that  
  
:R<sup>2</sup>= X<sup>2</sup>+Y<sup>2</sup> and <math> \tan(\theta) = \frac{y}{x} </math> where X and Y are two independent standard normal
+
<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 1-1 transformation that the joint distribution of R and θ is given by, 1-1 transformation:<br />
It can also be shown using 1-1 transformation that the joint distribution of R and θ is given by,
+
 
1-1 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 pdf of <math> d </math> and <math> \theta </math> is:
+
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 1,907: 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 1,913: 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 />
  
Alternative Method of Generating Standard Normal Random Variables 
 
  
Step 1: Generate <math>u1~Unif(0,1)</math>
+
'''Alternative Method of Generating Standard Normal Random Variables'''<br /> 
Step 2: Generate <math>Y1~Exp(1),Y2~Exp(2)</math>
+
 
Step 3: If <math>Y2 \geq(Y1-1)^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>u1 \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<br />
  
==== Expectation of a Standard Normal distribution ====
+
'''Proof:''' <br />
The expectation of a standard normal distribution is 0
 
:Below is the proof:  
 
  
 
:<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 1,935: 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 />
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)). Note that 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 (Box-Muller Transformation Method):''' <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 (Box-Muller 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 acceptance-rejection 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 acceptance-rejection 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 Box-Muller 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 Box-Muller 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) = 1-e^{-x/2}</math>. <br>
 
<math>P(R^2 \leq x) = 1-e^{-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 1,963: 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 Box-Muller Transformation Method generates a pair of independent Standard Normal distributions, X and Y (Using the transformation of polar coordinates). <br />
 
The Box-Muller 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 Box-Muller method several times.<br/>
 
If you want to generate a number of independent standard normal distributed numbers (more than two), you can run the Box-Muller method several times.<br/>
 
For example: <br />
 
For example: <br />
Line 1,971: Line 2,082:
  
  
* '''Code'''<br />
+
'''Matlab Code'''<br />
 +
 
 
<pre style="font-size:16px">
 
<pre style="font-size:16px">
 
>>close all
 
>>close all
Line 1,986: 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>
  
"''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."
+
'''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 2,006: 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>
Note: hist(x+2) will be centered at 2 instead of at 0. <br />
+
1. randn is random sample from a standard normal distribution.<br />
      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 />
+
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.jpg|300x300px]][[File:Normal_x+2.jpg|300x300px]][[File:Normal(2x+2).jpg|300px]]
 
[[File:Normal_x.jpg|300x300px]][[File:Normal_x+2.jpg|300x300px]][[File:Normal(2x+2).jpg|300px]]
 
<br />
 
<br />
  
<b>Comment</b>: Box-Muller transformations are not computationally efficient. The reason for this is the need to compute sine and cosine functions. A way to get around this time-consuming 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 />
 +
Box-Muller transformations are not computationally efficient. The reason for this is the need to compute sine and cosine functions. A way to get around this time-consuming 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 2,031: Line 2,148:
 
=== Proof of Box Muller Transformation ===
 
=== Proof of Box Muller Transformation ===
  
Definition:
+
'''Definition:'''<br />
 
A transformation which transforms from a '''two-dimensional continuous uniform''' distribution to a '''two-dimensional bivariate normal''' distribution (or complex normal distribution).
 
A transformation which transforms from a '''two-dimensional continuous uniform''' distribution to a '''two-dimensional bivariate normal''' distribution (or complex normal distribution).
  
 
Let U<sub>1</sub> and U<sub>2</sub> be independent uniform (0,1) random variables. Then
 
Let U<sub>1</sub> and U<sub>2</sub> be independent uniform (0,1) random variables. Then
<math>X_{1} = (-2lnU)^0.5_{1}*cos(2\pi U_{2})</math>
+
<math>X_{1} = ((-2lnU_{1})^.5)*cos(2\pi U_{2})</math>
  
<math>X_{2} = (-2lnU)^0.5_{1}*sin(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.
 
are '''independent''' N(0,1) random variables.
  
Line 2,051: Line 2,168:
 
       u<sub>2</sub> = g<sub>2</sub> ^-1(x1,x2)
 
       u<sub>2</sub> = g<sub>2</sub> ^-1(x1,x2)
  
Inverting the above transformations, we have
+
Inverting the above transformation, we have
 
     u1 = exp^{-(x<sub>1</sub> ^2+ x<sub>2</sub> ^2)/2}
 
     u1 = exp^{-(x<sub>1</sub> ^2+ x<sub>2</sub> ^2)/2}
 
     u2 = (1/2pi)*tan^-1 (x<sub>2</sub>/x<sub>1</sub>)
 
     u2 = (1/2pi)*tan^-1 (x<sub>2</sub>/x<sub>1</sub>)
Line 2,294: Line 2,411:
  
  
</pre>
+
 
 
</div>
 
</div>
 
Note: We can also regard the Bernoulli Distribution as either a conditional distribution or <math>f(x)= p^{x}(1-p)^{(1-x)}</math>, x=0,1.
 
Note: We can also regard the Bernoulli Distribution as either a conditional distribution or <math>f(x)= p^{x}(1-p)^{(1-x)}</math>, x=0,1.
Line 2,315: Line 2,432:
 
Procedure:
 
Procedure:
  
1.Generate U~Unif [0, 1)<br>
+
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>
  
 
'''Remark'''<br>
 
'''Remark'''<br>
1. The preceding can be written algorithmically as
+
1) The preceding can be written algorithmically for discrete random variables as <br>
Generate a random number U
+
Generate a random number U ~ U(0,1] <br>
If U<<sub>p0</sub> set X=<sub>x0</sub> and stop
+
If U < p<sub>0</sub> set X = x<sub>0</sub> and stop <br>
If U<<sub>p0</sub>+<sub>p1</sub> set X=x1 and stop
+
If U < p<sub>0</sub> + p<sub>1</sub> set X = x<sub>1</sub> and stop <br>
...
+
... <br>
2. If the <sub>xi</sub>, i>=0, are ordered so that <sub>x0</sub><<sub>x1</sub><<sub>x2</sub><... and if we let F denote the distribution function of X, then X will equal <sub>xj</sub> if F(<sub>x(j-1)</sub>)<=U<F(<sub>xj</sub>)
+
2) If the x<sub>i</sub>, i>=0, are ordered so that x<sub>0</sub> < x<sub>1</sub> < x<sub>2</sub> <... and if we let F denote the distribution function of X, then X will equal x<sub>j</sub> if F(x<sub>j-1</sub>) <= U < F(x<sub>j</sub>)
  
 
'''Example 1'''<br>
 
'''Example 1'''<br>
Line 2,352: 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(1-u)</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 1-u and u are the same.
 +
 
 +
 
 +
* '''Matlab Code'''<br />
 +
<pre style="font-size: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 2,396: Line 2,526:
 
<math>U = 0.2</math> => <math>X = U^{1/20} = 0.923</math><br>
 
<math>U = 0.2</math> => <math>X = U^{1/20} = 0.923</math><br>
 
<br>
 
<br>
Observe from above that the values of X for n = 20 are close to 1, this is because we can view <math>X^n</sup></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.<br>
+
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.<br>
  
 
Recall that
 
Recall that
Line 2,508: Line 2,638:
 
=== Example of Decomposition Method ===
 
=== Example of Decomposition Method ===
  
F<sub>x</sub>(x) = 1/3*x+1/3*x<sup>2</sup>+1/3*x<sup>3</sup>, 0<= x<=1
+
<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 U =F<sub>x</sub>(x) = 1/3*x+1/3*x<sup>2</sup>+1/3*x<sup>3</sup>, solve for x.
+
Let <math>U =F_x(x) = \frac {1}{3} x+\frac {1}{3} x^2+\frac {1}{3} x^3</math>, solve for x.
  
P<sub>1</sub>=1/3, F<sub>x1</sub>(x)= x, P<sub>2</sub>=1/3,F<sub>x2</sub>(x)= x<sup>2</sup>,  
+
<math>P_1=\frac{1}{3}, F_{x1} (x)= x, P_2=\frac{1}{3},F_{x2} (x)= x^2,  
P<sub>3</sub>=1/3,F<sub>x3</sub>(x)= x<sup>3</sup>
+
P_3=\frac{1}{3},F_{x3} (x)= x^3</math>
  
 
'''Algorithm:'''
 
'''Algorithm:'''
  
Generate U ~ Unif [0,1)
+
Generate <math>\,U \sim Unif [0,1)</math>
  
Generate V~ Unif [0,1)
+
Generate <math>\,V \sim  Unif [0,1)</math>
  
if 0<u<1/3, x = v
+
if <math>0\leq u \leq \frac{1}{3}, x = v</math>
  
else if u<2/3, x = v<sup>1/2</sup>
+
else if <math>u \leq \frac{2}{3}, x = v^{\frac{1}{2}}</math>
  
else x = v<sup>1/3</sup><br>
+
else <math>x=v^{\frac{1}{3}}</math> <br>
  
  
Line 2,590: 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 2,598: Line 2,727:
 
(Basis of the Accept-Reject algorithm)
 
(Basis of the Accept-Reject 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.
+
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.<br />
 +
Inverse each part of partial CDF, the partial CDF is divided by the original CDF, partial range is uniform distribution.<br />
 +
More specific definition of the theorem can be found here.<ref>http://www.bus.emory.edu/breno/teaching/MCMC_GibbsHandouts.pdf</ref>
  
inverse each part of partial CDF, the partial CDF is divided by the original CDF, partial range is uniform distribution.
+
Matlab code:
 +
 
 +
<pre style="font-size:16px">
 +
close all
 +
clear all
 +
ii=1;
 +
while ii<1000
 +
u=rand
 +
y=R*(2*U-1)
 +
if (1-U^2)>=(2*u-1)^2
 +
x(ii)=y;
 +
ii=ii+1
 +
end
 +
</pre>
  
 
===Question 2===
 
===Question 2===
Line 2,643: Line 2,787:
 
===The Bernoulli distribution===
 
===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 if the event success, otherwise the probability is 1-p (we usually define a variate q, q= 1-p). The mean of Bernoulli is p, variance is p(1-p). 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 1-p is the probability of failure (we usually define a variate q, q= 1-p). The mean of Bernoulli is p and the variance is p(1-p). 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)=1-p, P(x=0)+P(x=1)=p+q=1
+
P(X=1)= p
 +
P(X=0)= q = 1-p
 +
Therefore, P(X=0) + P(X=1) = p + q = 1
  
 
'''Algorithm: '''
 
'''Algorithm: '''
  
1. Generate u~Unif(0,1) <br>
+
1) Generate <math>u\sim~Unif(0,1)</math> <br>
2. If u p, then x = 1 <br>
+
2) If <math>u \leq p</math>, then <math>x = 1 </math><br>
else x = 0 <br>
+
else <math>x = 0</math> <br>
 
The answer is: <br>
 
The answer is: <br>
when U≤p, x=1 <br>
+
when <math> U \leq p, x=1</math> <br>
when U>p, x=0<br>
+
when <math>U \geq p, x=0</math><br>
3.Repeat as necessary
+
3) Repeat as necessary
 +
 
 +
* '''Matlab Code'''<br />
 +
<pre style="font-size: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 2,764: Line 2,924:
  
 
P (X > x) = (1-p)<sup>x</sup>(because first x trials are not successful) <br/>
 
P (X > x) = (1-p)<sup>x</sup>(because first x trials are not successful) <br/>
 +
 +
NB: An advantage of using this method is that nothing is rejected. We accept all the points, and the method is more efficient. Also, this method is closer to the inverse transform method as nothing is being rejected. <br />
  
 
'''Proof''' <br/>
 
'''Proof''' <br/>
Line 2,978: Line 3,140:
 
=== Beta Distribution ===
 
=== Beta Distribution ===
 
The beta distribution is a continuous probability distribution. <br>
 
The beta distribution is a continuous probability distribution. <br>
 +
PDF:<math>\displaystyle \text{ } f(x) = \frac{\Gamma(\alpha+\beta)}{\Gamma(\alpha)\Gamma(\beta)}x^{\alpha-1}(1-x)^{\beta-1} </math><br>  where <math>0 \leq x \leq 1</math> and <math>\alpha</math>>0, <math>\beta</math>>0<br/>
 +
<div style = "align:left; background:#F5F5DC; font-size: 120%">
 +
Definition:
 +
In probability theory and statistics, the beta distribution is a family of continuous probability distributions defined on the interval [0, 1] parametrized by two positive shape parameters, denoted by α and β, that appear as exponents of the random variable and control the shape of the distribution.<br/.>
 +
More can be find in the link: <ref>http://en.wikipedia.org/wiki/Beta_distribution</ref>
 +
</div>
 +
 
There are two positive shape parameters in this distribution defined as alpha and beta: <br>
 
There are two positive shape parameters in this distribution defined as alpha and beta: <br>
-Both parameters greater than 0, and X within the interval [0,1]. <br>
+
-Both parameters are greater than 0, and X is within the interval [0,1]. <br>
 
-Alpha is used as exponents of the random variable. <br>
 
-Alpha is used as exponents of the random variable. <br>
 
-Beta is used to control the shape of the this distribution. We use the beta distribution to build the model of the behavior of random variables, which are limited to intervals of finite length. <br>
 
-Beta is used to control the shape of the this distribution. We use the beta distribution to build the model of the behavior of random variables, which are limited to intervals of finite length. <br>
Line 3,029: Line 3,198:
 
:<math>\displaystyle \text{f}(x) = \frac{\Gamma(\alpha+1)}{\Gamma(\alpha)\Gamma(1)}x^{\alpha-1}(1-x)^{1-1}=\alpha x^{\alpha-1}</math><br>
 
:<math>\displaystyle \text{f}(x) = \frac{\Gamma(\alpha+1)}{\Gamma(\alpha)\Gamma(1)}x^{\alpha-1}(1-x)^{1-1}=\alpha x^{\alpha-1}</math><br>
  
The CDF is <math>F(x) = x^{\alpha}</math> (using integration of <math>f(x)</math>)
+
By integrating <math>f(x)</math>, we find the CDF of X is <math>F(x) = x^{\alpha}</math>.
With CDF F(x) = x^α, if U have CDF, it is very easy to sample:
+
As <math>F(x)^{-1} = x^\frac {1}{\alpha}</math>, using the inverse transform method, <math> X = U^\frac {1}{\alpha} </math> with U ~ U[0,1].
y=x^α --> x=y^α --> inverseF(x)= x^(1/α)
 
U~U(0,1) --> x=u^(1/α)
 
Applying the inverse transform method with <math>y = x^\alpha \Rightarrow x = y^\frac {1}{\alpha}</math>
 
 
 
<math>F(x)^{-1} = y^\frac {1}{\alpha}</math>
 
 
 
between case 1 and case 2, when alpha and beta be different value, the beta distribution can simplify to other distribution.
 
  
 
'''Algorithm'''
 
'''Algorithm'''
Line 3,052: Line 3,214:
 
</pre>
 
</pre>
  
'''Case 3:'''<br\> To sample from beta in general. we use the property that <br\>
+
'''Case 3:'''<br\> To sample from beta in general, we use the property that <br\>
  
 
:if <math>Y_1</math> follows gamma <math>(\alpha,1)</math><br\>
 
:if <math>Y_1</math> follows gamma <math>(\alpha,1)</math><br\>
Line 3,190: Line 3,352:
 
An example of the 2-D case is given below:
 
An example of the 2-D case is given below:
  
<pre style='font-size:16px'>
+
<pre style='font-size:14px'>
 
 
 
>>a=[1 2];  
 
>>a=[1 2];  
 
>>b=[4 6];  
 
>>b=[4 6];  
Line 3,207: Line 3,368:
 
[[File:2d_ex.jpg|300px]]
 
[[File:2d_ex.jpg|300px]]
  
==== Code: ====
+
==== Matlab Code: ====
  
<pre style='font-size:16px'>
+
<pre style='font-size:14px'>
 
function x = urectangle (d,n,a,b)
 
function x = urectangle (d,n,a,b)
 
for ii = 1:d;
 
for ii = 1:d;
Line 3,216: Line 3,377:
 
     %keyboard                      #makes the function stop at this step so you can evaluate the variables
 
     %keyboard                      #makes the function stop at this step so you can evaluate the variables
 
end
 
end
 
  
 
>>x=urectangle(2, 100, 2, 5);
 
>>x=urectangle(2, 100, 2, 5);
Line 3,256: Line 3,416:
  
  
This is the picture of the example  
+
The following is a picture relating to the example
 +
 
 
[[File:Untitled.jpg]]
 
[[File:Untitled.jpg]]
  
matlab code:
+
Matlab code:
 
<pre style='font-size:16px'>
 
<pre style='font-size:16px'>
 
u = rand(d,n);
 
u = rand(d,n);
Line 3,284: Line 3,445:
 
==Class 10 - Thursday June 6th 2013 ==  
 
==Class 10 - Thursday June 6th 2013 ==  
 
MATLAB code for using Acceptance/Rejection Method to sample from a d-dimensional unit ball.
 
MATLAB code for using Acceptance/Rejection Method to sample from a d-dimensional unit ball.
 +
G: d-dimensional unit ball G
 +
W: d-dimensional Hypercube
  
 
<pre style='font-size:16px'>
 
<pre style='font-size:16px'>
1. U1~UNIF(0, 1)
+
1) U1~UNIF(0,1)
     U2~UNIF(0, 1)
+
     U2~UNIF(0,1)
 
     ...
 
     ...
     Ud~UNIF(0, 1)
+
     Ud~UNIF(0,1)
 +
2)  X1 = 1-2U1
 +
    X2 = 1-2U2
 +
    ...
 +
    Xd = 1-2Ud
 +
    R = sum(Xi^2)
 +
3)  If R<=1
 +
    X = (X1,X2,...,Xd),
 +
    else go to step 1
 
</pre>
 
</pre>
  
Line 3,327: Line 3,498:
 
z(:,ii) means all the numbers in the ii column starting from 1st column until the nth
 
z(:,ii) means all the numbers in the ii column starting from 1st column until the nth
 
column, which is the last one.
 
column, which is the last one.
 +
 +
higher dimension, less efficient and we need more data points
  
 
Save it with the name of the pattern.
 
Save it with the name of the pattern.
Line 3,399: Line 3,572:
 
[[File:3-dimensional unitball.jpg|400px]]
 
[[File:3-dimensional unitball.jpg|400px]]
  
Note that c increases exponentially as d increases, which will result in rejections of more points. So this method is not efficient for large values of d.
+
Note that c increases exponentially as d increases, which will result in a lower acceptance rate and more points being rejected. So this method is not efficient for large values of d.
  
 
In practice, when we need to vectorlize a high quality image or genes then d would have to be very large.  So AR method is not an efficient way to solve the problem.
 
In practice, when we need to vectorlize a high quality image or genes then d would have to be very large.  So AR method is not an efficient way to solve the problem.
Line 3,415: Line 3,588:
 
For example, for approximating value of <math>\pi</math>, when <math>d \text{(dimension)} =2</math>, the efficiency is around 0.7869; when <math>d=3</math>, the efficiency is around 0.5244; when <math>d=10</math>, the efficiency is around 0.0026: it is getting close to 0.
 
For example, for approximating value of <math>\pi</math>, when <math>d \text{(dimension)} =2</math>, the efficiency is around 0.7869; when <math>d=3</math>, the efficiency is around 0.5244; when <math>d=10</math>, the efficiency is around 0.0026: it is getting close to 0.
  
Thus, when we want to generate high dimension vectors, Acceptance-Rejection Method is not efficient to be used.
+
A 'C' value of 1 implies an acceptance rate of 100% (most efficient scenario) but as we sample from higher dimensions, 'C' usually gets larger. Thus, when we want to generate high dimension vectors, Acceptance-Rejection Method is not efficient to be used.
  
 
<span style="color:red;padding:0 auto;"><br>The end of midterm coverage</span>
 
<span style="color:red;padding:0 auto;"><br>The end of midterm coverage</span>
 
+
<div style="border:1px solid #cccccc;border-radius:10px;box-shadow: 0 5px 15px 1px rgba(0, 0, 0, 0.6), 0 0 200px 1px rgba(255, 255, 255, 0.5);padding:20px;margin:20px;background:#FFFFAD;">
 +
<h2 style="text-align:center;">Summary of vector acceptance-rejection sampling</h2>
 +
<p><b>Problem:</b> <math> f(x_1, x_2, ...x_n)</math> is difficult to sample from</p>
 +
<p><b>Plan:</b></p>
 +
Let W represent the sample space covered by <math> f(x_1, x_2, ...x_n)</math>
 +
<ol>
 +
<li>1.Draw <math>\vec{y}=y_1,y_2...y_n\sim~g()</math> where g has sample space G which is greater than W. g is a distribution that is easy to sample from (i.e. uniform)</li>
 +
<li>2.if <math>\vec{y} \subseteq W </math> then <math>\vec{x}=\vec{y} </math><br /> else go 1) </li>
 +
</ol>
 +
<p>x will have the desired distribution.</p>
 +
 
 +
</div>
 +
 
 
==== Stochastic Process ====
 
==== Stochastic Process ====
 
The basic idea of Stochastic Process (also called random process) is a collection of some random variables,  
 
The basic idea of Stochastic Process (also called random process) is a collection of some random variables,  
Line 3,425: 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 non-deterministic. This means that there is some indeterminacy in the final state, even if the initial condition is known.
+
A stochastic process is non-deterministic. 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 3,437: Line 3,622:
 
2. Markov Process- This is a stochastic process that satisfies the Markov property which can be understood as the memory-less 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 memory-less 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 memory-less 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 memory-less mouse in a maze, etc. <br>
 
   
 
   
 
Stochastic Process means even we get some conditions at the beginning, we just can guess some variables followed the first, but at the end the variable would be unpredictable.
 
  
 
=====Example=====
 
=====Example=====
Line 3,445: Line 3,628:
 
stochastic process always has state space and the index set to limit the range.
 
stochastic process always has state space and the index set to limit the range.
  
The state space is the set of cars , while <math>x_t</math> are sport cars.
+
The state space is the set of cars, while <math>x_t</math> are sport cars.
 +
 
 +
Births in a hospital occur randomly at an average rate
 +
 
 +
The number of cases of a disease in different towns
  
 
==== Poisson Process ====
 
==== Poisson Process ====
The Poisson process is a discrete counting process of number of occurrences over time.
+
[[File:Possionprocessidiagram.png‎]]
 +
 
 +
The Poisson process is a discrete counting process which counts the number of<br\>
 +
of events and the time that these occur in a given time interval.<br\>
 +
 
 +
e.g traffic accidents , arrival of emails. Emails arrive at random time <math>T_1, T_2 ... T_n</math> for example (2, 7, 3) is the number of emails received on day 1, day 2, day 3. This is a stochastic process and Poisson process with condition.
  
e.g traffic accidents , arrival of emails. Emails arrive at random time <math>T_1, T_2</math> ...
+
The probability of observing x events in a given interval is given by
 +
<math> P(X = x) = e^{-\lambda}* \lambda^x/ x! </math>
 +
where x = 0; 1; 2; 3; 4; ....
  
 
-Let <math>N_t</math> denote the number of arrivals within the time interval <math>(0,t]</math><br\>
 
-Let <math>N_t</math> denote the number of arrivals within the time interval <math>(0,t]</math><br\>
Line 3,466: Line 3,660:
 
E[N<sub>t</sub>] = <math>\lambda t</math> and Var[N<sub>t</sub>] = <math>\lambda t</math>
 
E[N<sub>t</sub>] = <math>\lambda t</math> and Var[N<sub>t</sub>] = <math>\lambda t</math>
  
==== ====
+
the rate parameter may change over time; such a process is called a non-homogeneous Poisson process
 +
 
 +
==== Examples ====
 
<br />
 
<br />
 
'''How to generate a multivariate normal with the built-in function "randn": (example)'''<br />
 
'''How to generate a multivariate normal with the built-in function "randn": (example)'''<br />
Line 3,478: Line 3,674:
 
                       %matrix to 1*n matrix;
 
                       %matrix to 1*n matrix;
 
</pre>
 
</pre>
 +
For example, if we use mu = [2 5], we would get <br/>
 +
<math> = \left[ \begin{array}{ccc}
 +
3.8214 & 0.3447 \\
 +
6.3097 & 5.6157 \end{array} \right]</math>
 +
  
and if we want to use box-muller to generate a multivariate normal, we could use the code in lecture 6:
+
If we want to use box-muller to generate a multivariate normal, we could use the code in lecture 6:
 
<pre style='font-size:16px'>
 
<pre style='font-size:16px'>
 
d = length(mu);
 
d = length(mu);
Line 3,513: Line 3,714:
  
 
(The definition of CLT is from http://en.wikipedia.org/wiki/Central_limit_theorem)
 
(The definition of CLT is from http://en.wikipedia.org/wiki/Central_limit_theorem)
 +
 +
<math> \lim_{n \to \infty} P*[{\frac{X_1 + ... + X_n -n*\mu}{\sigma*\surd n}} < x] = \Phi (x)</math>
  
 
==Class 11 - Tuesday,June 11, 2013==
 
==Class 11 - Tuesday,June 11, 2013==
Line 3,519: 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>Strike-through 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,\ldots,</math>.
+
:<math>\!f(n)= \frac{\lambda^n e^{-\lambda}}{n!}  \qquad n= 0,1,2,3,4,5,\ldots,</math>.
  
 
<math>\{X_t:t\in T\}</math>  where <math>\ X_t </math> is state space and T is index set.
 
<math>\{X_t:t\in T\}</math>  where <math>\ X_t </math> is state space and T is index set.
Line 3,530: Line 3,734:
 
(c) '''Individuality:'''  for a sufficiently short time period of length h, the probability of 2 or more events occurring in the interval is close to 0, or formally <math>\mathcal{O}(h)</math><br>
 
(c) '''Individuality:'''  for a sufficiently short time period of length h, the probability of 2 or more events occurring in the interval is close to 0, or formally <math>\mathcal{O}(h)</math><br>
  
 
+
NOTE: it is very important to note that the time between the occurrence of consecutive events (in a Poisson Process) is exponentially distributed with the same parameter as that in the Poisson distribution. This characteristic is used when trying to simulate a Poisson Process.
'''Notation'''<br>
 
N<sub>t</sub> denotes the number of arrivals up to t, i.e.(0,t] <br>
 
N(a,b] = N<sub>b</sub> - N<sub>a</sub> denotes the number of arrivals in I(a, b]. <br>
 
 
 
  
 
For a small interval (t,t+h], where h is small<br>
 
For a small interval (t,t+h], where h is small<br>
1. The number of arrivals in this interval is independent of the number of arrivals up to t(N<sub>t</sub>)<br>
+
1. The number of arrivals up to time t(N<sub>t</sub>) is independent of the number of arrival in the interval<br>
 
2. <math> P (N(t,t+h)=1|N_{t} ) = P (N(t,t+h)=1) =\frac{e^{-\lambda h} (\lambda h)^1}{1!} =e^{-\lambda h} {\lambda h} \approx \lambda h </math> since <math>e^{-\lambda h} \approx 1</math> when h is small.<br>
 
2. <math> P (N(t,t+h)=1|N_{t} ) = P (N(t,t+h)=1) =\frac{e^{-\lambda h} (\lambda h)^1}{1!} =e^{-\lambda h} {\lambda h} \approx \lambda h </math> since <math>e^{-\lambda h} \approx 1</math> when h is small.<br>
  
Line 3,547: Line 3,747:
 
'''Generate a Poisson Process'''<br />
 
'''Generate a Poisson Process'''<br />
  
<math>U_n \sim U(0,1)</math><br>
+
1. set <math>T_{0}=0</math> and n=1<br/>
<math>T_n-T_{n-1}=-\frac {1}{\lambda} log(U_n)</math><br>
 
  
1. set T<sub>0</sub>=0 and n=1<br />
+
2. <math>U_{n} \sim~ U(0,1)</math><br />
  
2. U<sub>n</sub>~ U(0,1)<br />
+
3. <math>T_{n} = T_{n-1}-\frac {1}{\lambda} log (U_{n})  </math> (declare an arrival)<br />
  
3. T<sub>n</sub> = T<sub>n-1</sub> <math> -\frac {1}{\lambda} </math> log (U<sub>n</sub>) (declare an arrival)<br />
+
4. if <math>T_{n} \gneq T</math> stop<br />
 
 
4. if T<sub>n</sub>>T stop<br />
 
 
&nbsp;&nbsp;&nbsp;&nbsp;else<br />
 
&nbsp;&nbsp;&nbsp;&nbsp;else<br />
 
&nbsp;&nbsp;&nbsp;&nbsp;n=n+1 go to step 2<br />
 
&nbsp;&nbsp;&nbsp;&nbsp;n=n+1 go to step 2<br />
Line 3,630: 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 3,656: Line 3,853:
 
*Technology: The Google link analysis algorithm "PageRank"<br />
 
*Technology: The Google link analysis algorithm "PageRank"<br />
  
 +
 +
'''Definition''' An irreducible Markov Chain is said to be aperiodic if for some n <math>\ge 0 </math> and some state j.<br />
 +
<math> P*(X_n=j | X_0 =j) > 0 </math>    and    <math>  P*(X_{n+1} | X_0=j) > 0 </math> <br />
 +
 +
It can be shown that if the Markov Chain is irreducible and aperiodic then, <br />
 +
<math> \pi_j = \lim_{n -> \infty} P*(X_n = j) for j=1...N </math> <br />
 +
Source: From Simulation textbook <br />
  
 
Product Rule (Stochastic Process):<br />
 
Product Rule (Stochastic Process):<br />
Line 3,735: 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=__hZR-1Cp6PbZ5PfnSjs2zU6LnCiI=&docid=PaQvi1F97P2urM&sa=X&ei=foTxUY3DB-rMyQGvq4D4Cg&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=__hZR-1Cp6PbZ5PfnSjs2zU6LnCiI=&docid=PaQvi1F97P2urM&sa=X&ei=foTxUY3DB-rMyQGvq4D4Cg&sqi=2&ved=0CDYQ9QEwAQ&dur=5515)
  
Line 3,766: Line 3,970:
  
 
=== Multiplicative Congruential Algorithm ===
 
=== Multiplicative Congruential Algorithm ===
x<sub>k+1</sub>= (ax<sub>k</sub>+c) mod m
+
<div style="border:1px solid red">
 +
A Linear Congruential Generator (LCG) yields a sequence of randomized numbers calculated with a linear equation. The method represents one of the oldest and best-known pseudorandom number generator algorithms.[1] The theory behind them is easy to understand, and they are easily implemented and fast, especially on computer hardware which can provide modulo arithmetic by storage-bit truncation.<br>
 +
from wikipedia
 +
</div>
 +
 
 +
<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.
  
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.
+
'''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>
  
Examples:
+
<math>\begin{align}X_{2} = 2 * 8  + 1\mod 13 = 4\end{align}</math> ... and so on<br>
      X<sub>0</sub> = 10 ,a = 2 , c = 1 , m = 13
+
 
 +
 
 +
2. <math>\begin{align}X_{0} = 44 ,a = 13 , c = 17 , m = 211\end{align}</math><br>
 
        
 
        
          X<sub>1</sub> = 2 * 10 + mod 13 = 8
+
<math>\begin{align}X_{1} = 13 * 44 + 17\mod 211 = 167\end{align}</math><br>
          X<sub>2</sub> = 2 * 8 + mod 13 = 4
+
 
          ... and so on
+
<math>\begin{align}X_{2} = 13 * 167 + 17\mod 211 = 78\end{align}</math><br>
  
      X<sub>0</sub> = 44 ,a = 13 , c = 17 , m = 211
+
<math>\begin{align}X_{3} = 13 * 78 + 17\mod 211 = 187\end{align}</math> ... and so on<br>
     
 
          X<sub>1</sub> = 13 * 44 + 17  mod 211 = 167
 
          X<sub>2</sub> = 13 * 167 + 17 mod 211 = 78
 
          X<sub>3</sub> = 13 * 78  + 17  mod 211 = 187
 
          ... and so on
 
  
 
=== Inverse Transformation Method ===
 
=== Inverse Transformation Method ===
Line 3,839: Line 4,050:
  
 
Models the waiting time until the first success.<br>
 
Models the waiting time until the first success.<br>
 +
<math>X\sim~Exp(\lambda)</math> <br />
  
X~Exp<math>(\lambda) </math><br>
+
<math>f(x) = \lambda e^{-\lambda x} \, , x>0 </math><br/>
<math> f (x) = \lambda e^{-\lambda x}</math> , <math>x>0 </math><br/>
 
  
1. U~Unif(0,1)
+
<math>1.\, U\sim~U(0,1)</math>
 
+
<br />
2. The inverse of exponential function is x = <math>\frac{-1}{\lambda} log(U)</math>
+
<math>2.\, x = \frac{-1}{\lambda} log(U)</math>
  
 
===Normal===
 
===Normal===
Line 3,863: Line 4,074:
 
In the multivariate case,<br/ >
 
In the multivariate case,<br/ >
 
<math>\underline{Z}\sim N(\underline{0},I)\rightarrow  \underline{X} \sim N(\underline{\mu},\Sigma)</math> <br/ >
 
<math>\underline{Z}\sim N(\underline{0},I)\rightarrow  \underline{X} \sim N(\underline{\mu},\Sigma)</math> <br/ >
<math>\underline{X} = \underline{\mu} +\Sigma ^{1/2} \underline{Z}</math>
+
<math>\underline{X} = \underline{\mu} +\Sigma ^{1/2} \underline{Z}</math><br/>
 +
Note: <math>\Sigma^{1/2}</math> can be obtained from Cholesky decomposition (chol(A) in MATLAB), which is guaranteed to exist, as  <math>\Sigma</math> is positive semi-definite.
  
 
=== Gamma ===
 
=== Gamma ===
Line 3,877: Line 4,089:
 
<math>=\frac {-1}{\lambda}\log(\prod_{j=1}^{t} U_j)</math>
 
<math>=\frac {-1}{\lambda}\log(\prod_{j=1}^{t} U_j)</math>
  
This is a property of gamma distribution.
+
This is a special property of gamma distribution.
  
 
=== Bernoulli ===
 
=== Bernoulli ===
Line 3,883: Line 4,095:
 
A Bernoulli random variable can only take two possible values: 0 and 1. 1 represents "success" and 0 represents "failure." If p is the probability of success, we have pdf
 
A Bernoulli random variable can only take two possible values: 0 and 1. 1 represents "success" and 0 represents "failure." If p is the probability of success, we have pdf
  
<math> f(x)= p^x (1-p)^{1-x}, x=0,1 </math><br>
+
<math> f(x)= p^x (1-p)^{1-x},\,  x=0,1 </math><br>
  
 
To generate a Bernoulli random variable we use the following procedure:
 
To generate a Bernoulli random variable we use the following procedure:
  
sample u~U(0,1)<br>
+
<math> 1. U\sim~U(0,1)</math><br>
if u <= p, then x=1<br>
+
<math> 2. if\, u <= p, then\, x=1\,</math><br />  
else x=0<br>
+
<math> else\, x=0</math><br/>
 
where 1 stands for success and 0 stands for failure.<br>
 
where 1 stands for success and 0 stands for failure.<br>
  
Line 3,896: Line 4,108:
 
The sum of n independent Bernoulli trials
 
The sum of n independent Bernoulli trials
 
<br\>
 
<br\>
X~ Bin(n,p)<br/>
+
<math> X\sim~ Bin(n,p)</math><br/>
1. U1, U2, ... Un ~ U(0,1)<br/>
+
1.<math> U1, U2, ... Un \sim~U(0,1)</math><br/>
 
2. <math> X= \sum^{n}_{1} I(U_i \leq p) </math> ,where <math>I(U_i \leq p)</math> is an indicator for a successful trial.<br/>
 
2. <math> X= \sum^{n}_{1} I(U_i \leq p) </math> ,where <math>I(U_i \leq p)</math> is an indicator for a successful trial.<br/>
 
Return to 1<br/>
 
Return to 1<br/>
  
I is an indicator variable if for U <= P, then I(U<=P)=1; else I(U>P)=0.
+
I is an indicator variable if for <math>U \leq P,\, then\, I(U\leq P)=1;\, else I(U>P)=0.</math>
  
 
Repeat this N times if you need N samples.
 
Repeat this N times if you need N samples.
Line 3,915: Line 4,127:
 
simulate this binomial distribution.
 
simulate this binomial distribution.
  
1) Generate <math> U_1....U_{10} </math> ~ <math> U(0,1) </math>  <br>
+
1) Generate <math>U_1....U_{10} \sim~ U(0,1) </math>  <br>
 
2) <math> X= \sum^{10}_{1} I(U_i \leq \frac{1}{6}) </math> <br>
 
2) <math> X= \sum^{10}_{1} I(U_i \leq \frac{1}{6}) </math> <br>
3)Return to one.
+
3)Return to 1)
  
 
=== Beta Distribution ===
 
=== Beta Distribution ===
Line 3,978: Line 4,190:
 
If X~Unif (0,1), Y= floor(5U)-2 = [5U]-2 -> Y~ DU[-2,2]  
 
If X~Unif (0,1), Y= floor(5U)-2 = [5U]-2 -> Y~ DU[-2,2]  
 
<br>
 
<br>
 +
 +
There is also another intuitive method:<br>
 +
1. Draw U ~ U(0,1)<br>
 +
2. i = 1, Pi = 1 - (1 - P)^i. <br>
 +
3. If u <= Pi = 1 - (1 - P)^i, set X = i.
 +
Else, i = i + 1. <br>
  
 
===Poisson===
 
===Poisson===
Line 4,019: Line 4,237:
 
<br>N-Step Transition Matrix: a matrix <math> P_n </math> whose elements are the probability of moving from state i to state j in n steps. <br/>
 
<br>N-Step Transition Matrix: a matrix <math> P_n </math> whose elements are the probability of moving from state i to state j in n steps. <br/>
 
<math>P_n (i,j)=Pr⁡(X_{m+n}=j|X_m=i)</math> <br/>
 
<math>P_n (i,j)=Pr⁡(X_{m+n}=j|X_m=i)</math> <br/>
 +
 +
Explanation: (with an example) Suppose there 10 states { 1, 2, ..., 10}, and suppose you are on state 2, then P<sub>8</sub>(2, 5) represent the probability of moving from state 2 to state 5 in 8 steps.
  
 
One-step transition probability:<br/>
 
One-step transition probability:<br/>
Line 4,053: Line 4,273:
  
 
<math>P_2 = P_1 P_1 </math><br\>
 
<math>P_2 = P_1 P_1 </math><br\>
 +
 +
<math>P_3 = P_1 P_2 </math><br\>
 +
 +
<math>P_n = P_1 P_(n-1) </math><br\>
  
 
<math>P_n = P_1^n </math><br\>
 
<math>P_n = P_1^n </math><br\>
Line 4,095: Line 4,319:
 
Note: <math>P_2 = P_1\times P_1; P_n = P^n</math><br />
 
Note: <math>P_2 = P_1\times P_1; P_n = P^n</math><br />
 
The equation above is a special case of the Chapman-Kolmogorov equations.<br />
 
The equation above is a special case of the Chapman-Kolmogorov equations.<br />
It is true because of the Markov property or<br />
+
It is true because of the Markov property or the memoryless property of Markov chains, where the probabilities of going forward to the next state <br />
the memoryless property of Markov chains, where the probabilities of going forward to the next state <br />
 
 
only depends on your current state, not your previous states. By intuition, we can multiply the 1-step transition <br />
 
only depends on your current state, not your previous states. By intuition, we can multiply the 1-step transition <br />
 
matrix n-times to get a n-step transition matrix.<br />
 
matrix n-times to get a n-step transition matrix.<br />
Line 4,102: Line 4,325:
 
Example: We can see how <math>P_n = P^n</math> from the following:
 
Example: We can see how <math>P_n = P^n</math> from the following:
 
<br/>
 
<br/>
<math>\mu_1=\mu_0\cdot P</math> <br/>
+
<math>\vec{\mu_1}=\vec{\mu_0}\cdot P</math> <br/>
<math>\mu_2=\mu_1\cdot P</math> <br/>
+
<math>\vec{\mu_2}=\vec{\mu_1}\cdot P</math> <br/>
<math>\mu_3=\mu_2\cdot P</math> <br/>
+
<math>\vec{\mu_3}=\vec{\mu_2}\cdot P</math> <br/>
 
Therefore,  
 
Therefore,  
 
<br/>
 
<br/>
<math>\mu_3=\mu_0\cdot P^3
+
<math>\vec{\mu_3}=\vec{\mu_0}\cdot P^3
 
</math> <br/>
 
</math> <br/>
  
 
<math>P_n(i,j)</math> is called n-steps transition probability. <br>
 
<math>P_n(i,j)</math> is called n-steps transition probability. <br>
<math>\mu_0 </math> is called the '''initial distribution'''. <br>
+
<math>\vec{\mu_0} </math> is called the '''initial distribution'''. <br>
<math>\mu_n = \mu_0* P^n </math> <br />
+
<math>\vec{\mu_n} = \vec{\mu_0}* P^n </math> <br />
  
 
Example with Markov Chain:
 
Example with Markov Chain:
Line 4,143: 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> P_2~=P_1 P_1 </math> (as verified above)  
+
<math> P^2~=P\cdot P </math> (as verified above)  
  
 
In general,
 
In general,
<math> P_n~=(P_1)^n </math> **Note that <math>P_1</math> is equal to the matrix P <br/>
+
<math> P^n~= \Pi_{i=1}^{n} P</math> (P multiplied n times)<br/>
<math>\mu_n~=\mu_0 P_n</math><br/>
+
<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 P_n</math><br/>
+
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 4,182: 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 4,189: 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 4,200: Line 4,425:
 
Ans: There is not clear way to find that out  
 
Ans: There is not clear way to find that out  
  
How do you increase the time it takes to reach the steady state.
+
How do you increase the time it takes to reach the steady state?
Ans: Make the probabilities of transition much smaller, to reach from state 0 to state 1 and visevera p=0.005. And make the probabilities of staying in the same state extremely high. To stay in state 0 or state 1 p=0.995, then the matrix is in a "sticky state"
+
Ans: Make the probabilities of transition much smaller, to reach from state 0 to state 1 and vice-versa p=0.005. And make the probabilities of staying in the same state extremely high. To stay in state 0 or state 1 p=0.995, then the matrix is in a "sticky state"
 +
 
 +
 
 +
EXAMPLE : Random Walk on the cycle S={0,1,2}
 +
 
 +
<math>P^2 = \left[ \begin{array}{ccc}
 +
2pq & q^2 & p^2 \\
 +
p^2 & 2pq & q^2 \\
 +
q^2 & p^2 & 2pq \end{array} \right]</math>
 +
 
 +
Suppose<br/>
 +
<math>P(x_0=0)=\frac{1}{4}</math><br/>
 +
<math>P(x_0=1)=\frac{1}{2}</math><br/>
 +
<math>P(x_0=2)=\frac{1}{4}</math><br/>
 +
Thus<br/>
 +
<math>\pi_0 = \left[ \begin{array}{c} \frac{1}{4} \\ \frac{1}{2} \\ \frac{1}{4} \end{array} \right]</math><br/>
 +
so<br/>
 +
<math>\,\pi^2 = \pi_0 * P^2 </math>
 +
<math>= \left[ \begin{array}{c} \frac{1}{4} \\ \frac{1}{2} \\ \frac{1}{4} \end{array} \right] * \left[ \begin{array}{ccc}
 +
2pq & q^2 & p^2 \\
 +
p^2 & 2pq & q^2 \\
 +
q^2 & p^2 & 2pq \end{array} \right]</math>
 +
<math>= \left[ \begin{array}{c} \frac{1}{2}pq + \frac{1}{2}p^2+\frac{1}{4}q^2 \\ \frac{1}{4}q^2+pq+\frac{1}{4}p^2 \\ \frac{1}{4}p^2+\frac{1}{2}q^2+\frac{1}{2}pq\end{array} \right]</math>
  
 
==== MatLab Code ====
 
==== MatLab Code ====
Line 4,374: 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 here needs to be a column vector. So we need to transform <math>\pi</math> into a column vector.
 
  
<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='font-size:14px'>
  
==== Limiting distribution ====
+
P = [1/3 1/3 1/3; 1/4 3/4 0; 1/2 0 1/2]
A Markov chain has limiting distribution <math>\pi</math> if
 
  
<math>\lim_{n\to \infty} P^n= \left[ {\begin{array}{ccc}
+
pii = [1/3 4/9 2/9]
\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/>
+
[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 Markov Chain is convergent if and only if its limiting distribution exists. <br/>
+
a = -vec(:,1)
 +
 
 +
>> a =
 +
[0.5571 0.7428 0.3714]   
 +
 
 +
%% 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/>
 
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/>
 +
Markov chain in state j is approximately <math>\pi_j</math> no matter in which state <br/>
 +
the chain began at time 0. <br/>
  
 
'''Example:'''
 
'''Example:'''
Line 4,407: Line 4,686:
 
, find stationary distribution.<br/>
 
, find stationary distribution.<br/>
 
We have:<br/>
 
We have:<br/>
<math>0*\pi_0+0*\pi_1+1*\pi_2=\pi_0</math><br/>
+
<math>0\times \pi_0+0\times \pi_1+1\times \pi_2=\pi_0</math><br/>
<math>1*\pi_0+0*\pi_1+0*\pi_2=\pi_1</math><br/>
+
<math>1\times \pi_0+0\times \pi_1+0\times \pi_2=\pi_1</math><br/>
<math>0*\pi_0+1*\pi_1+0*\pi_2=\pi_2</math><br/>
+
<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 4,418: 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='font-size:14px'>
 +
MATLAB
 +
>> P=[0, 1, 0;0, 0, 1; 1, 0, 0]
 +
 +
P =
  
'''Example:'''
+
    0    1    0
 +
    0    0    1
 +
    1    0    0
 +
 
 +
>> pii=[1/3, 1/3, 1/3]
 +
 
 +
pii =
  
<math> P= \left [ \begin{matrix}
+
    0.3333    0.3333    0.3333
\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 />
+
>> pii*P
<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/>
 
The stationary distribution is <math>\pi = \left [ \begin{matrix}
 
\frac{1}{3} & \frac{1}{3} & \frac{1}{3} \\[6pt]
 
\end{matrix} \right] </math> <br/>
 
  
 +
ans =
  
<span style="font-size:20px;color:red">The following contents are problematic. Please correct it if possible.</span><br />
+
    0.3333    0.3333    0.3333
Suppose we're given that the limiting distribution <math> \pi </math> exists for  stochastic matrix P, that is, <math> \pi = \pi * 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>
+
>> P^1000
  
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>
+
ans =
  
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>
+
    0    1     0
 +
    0    0    1
 +
    1    0    0
  
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>
+
>> P^10000
  
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>
+
ans =
  
Therefore <math> \lim_{k \rightarrow \infty} \pi^k = \lim_{k \rightarrow \infty}  \lambda_i^k * a_1 * u_1 = u_1 </math>
+
    0    1    0
 +
    0    0    1
 +
    1    0    0
  
=== MatLab Code ===
+
>> P^10002
<pre style='font-size: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 =
+
ans =
  
     0.3333    0.3333    0.3333
+
    1     0     0
     0.2500    0.7500        0
+
    0     1     0
     0.5000        0    0.5000
+
    0     0    1
  
>> P^2
+
>> P^10003
  
 
ans =
 
ans =
  
     0.3611    0.3611    0.2778
+
    0    1     0
     0.2708    0.6458    0.0833
+
    0     0     1
     0.4167    0.1667    0.4167
+
    1     0    0
  
>> P^3
+
>> %P^10000 = P^10003
 +
>> % This chain does not have limiting distribution, it has a stationary distribution. 
  
ans =
+
This chain does not converge, it has a cycle.
 +
</pre>
  
    0.3495    0.3912    0.2593
+
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>
    0.2934    0.5747    0.1319
 
    0.3889    0.2639    0.3472
 
  
>> P^10
+
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>
  
the example of code and an example of stand distribution, then the all the pi probability in the matrix are the same.
+
'''Example:'''
  
ans =
+
<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>
  
    0.3341    0.4419    0.2240
+
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 />
    0.3314    0.4507    0.2179
+
<br />
    0.3360    0.4358    0.2282
+
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/>
  
>> P^100                                            % The stationary distribution is [0.3333 0.4444 0.2222]  since values keep unchanged.
 
  
ans =
+
<span style="font-size: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>
  
    0.3333    0.4444    0.2222
+
WLOG assume P is diagonalizable, (if not we can always consider the Jordan form and the computation below is exactly the same. <br>
    0.3333    0.4444    0.2222
 
    0.3333    0.4444    0.2222
 
  
 +
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>
  
>> [vec val]=eigs(P')                              % We can find the eigenvalues and eigenvectors from the transpose of matrix P.
+
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>
  
vec =
+
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>
  
  -0.5571    0.2447    0.8121
+
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>
  -0.7428  -0.7969  -0.3324
 
  -0.3714    0.5523  -0.4797
 
  
 +
Therefore <math> \lim_{k \rightarrow \infty} \pi^k = \lim_{k \rightarrow \infty}  \lambda_i^k  a_1  u_1 = u_1 </math>
  
val =
+
=== MatLab Code ===
 +
<pre style='font-size: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. 
  
    1.0000        0        0
+
P =
        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)
+
    0.3333    0.3333    0.3333
 +
    0.2500    0.7500        0
 +
    0.5000        0    0.5000
 +
 
 +
>> P^2
 +
 
 +
ans =
 +
 
 +
    0.3611    0.3611    0.2778
 +
    0.2708    0.6458    0.0833
 +
    0.4167    0.1667    0.4167
 +
 
 +
>> P^3
 +
 
 +
ans =
 +
 
 +
    0.3495    0.3912    0.2593
 +
    0.2934    0.5747    0.1319
 +
    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
 +
 
 +
>> 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 =
 +
 
 +
    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 =
 
a =
Line 4,543: Line 4,891:
 
</pre>
 
</pre>
  
This is <math>\pi_j = lim[p^n]_(ij)</math> exist and is independent of i  
+
This is <math>\pi_j = lim[p^n]_(ij)</math> exist and is independent of i
  
Example: Find the stationary distribution of P= <math>\left[ {\begin{array}{ccc}
+
Another example:
0 & 1 & 0 \\
+
 
0 & 0 & 1 \\
+
 
1 & 0 & 0 \end{array} } \right]</math>
+
Find the stationary distribution of P= <math>\left[ {\begin{array}{ccc}
 +
0.5 & 0 & 0 \\
 +
1 & 0 & 0.5 \\
 +
0 & 1 & 0.5 \end{array} } \right]</math>
  
 
<math>\pi=\pi~P</math><br>
 
<math>\pi=\pi~P</math><br>
Line 4,556: Line 4,907:
 
The system of equations is:
 
The system of equations is:
  
0*<math>\pi</math><sub>0</sub>+0*<math>\pi</math><sub>1</sub>+1*<math>\pi</math><sub>2</sub> = <math>\pi</math><sub>0</sub> => <math>\pi</math><sub>2</sub> = <math>\pi</math><sub>0</sub><br>  
+
<math>0.5\pi_0+1\pi_1+0\pi_2= \pi_0=> 2\pi_1 = \pi_0</math><br>  
1*<math>\pi</math><sub>0</sub>+0*<math>\pi</math><sub>1</sub>+0*<math>\pi</math><sub>2</sub> = <math>\pi</math><sub>1</sub> => <math>\pi</math><sub>1</sub> = <math>\pi</math><sub>0</sub><br>  
+
<math>0\pi_0+0\pi_1+1\pi_2= \pi_1 => \pi_1=\pi_2</math><br>  
0*<math>\pi</math><sub>0</sub>+1*<math>\pi</math><sub>1</sub>+0*<math>\pi</math><sub>2</sub> = <math>\pi</math><sub>2</sub> <br>
+
<math>0\pi_0+0.5\pi_1+0.5\pi_2 = \pi_2 => \pi_1 = \pi_2</math><br>
<math>\pi</math><sub>0</sub>+<math>\pi</math><sub>1</sub>+<math>\pi</math><sub>2</sub> = 1<br>
+
<math>\pi_0+\pi_1+\pi_2 = 1</math><br>
  
<math>\pi</math><sub>0</sub>+<math>\pi</math><sub>0</sub>+<math>\pi</math><sub>0</sub> = 3<math>\pi</math><sub>0</sub> = 1, which gives <math>\pi</math><sub>0</sub> = 1/3 <br>
+
<math>2\pi_1+\pi_1+\pi_1 = 4\pi_1 = 1</math>, which gives <math>\pi_1=\frac {1}{4}</math> <br>
Also, <math>\pi</math><sub>1</sub> = <math>\pi</math><sub>2</sub> = 1/3 <br>
+
Also, <math>\pi_1 = \pi_2 = \frac {1}{4}</math> <br>
So, <math>\pi</math> = <math>[\frac{1}{3}, \frac{1}{3}, \frac{1}{3}]</math> <br>
+
So, <math>\pi = [\frac{1}{2}, \frac{1}{4}, \frac{1}{4}]</math> <br>
  
when the p matrix is a standard matrix, then all the probabilities of pi are the same in the matrix.
+
=== Ergodic Chain ===
  
=== MatLab Code ===
+
A Markov chain is called an ergodic chain if it is possible to go from every state to every state (not necessarily in one move). For instance, note that we can claim a Markov chain is ergodic if it is possible to somehow start at any state i and end at any state j in the matrix. We could have a chain with states 0, 1, 2, 3, 4 where it is not possible to go from state 0 to state 4 in just one step. However, it may be possible to go from 0 to 1, then from 1 to 2, then from 2 to 3, and finally 3 to 4 so we can claim that it is possible to go from 0 to 4 and this would satisfy a requirement of an ergodic chain. The example below will further explain this concept.
<pre style='font-size:14px'>
 
MATLAB
 
>> P=[0, 1, 0;0, 0, 1; 1, 0, 0]
 
  
P =
+
'''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.<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 non-zero probability, this condition is fulfilled with N=1.<br/>
  
    0    1    0
 
    0    0    1
 
    1    0    0
 
  
>> pii=[1/3, 1/3, 1/3]
+
====Example====
 +
<math> P= \left[ \begin{matrix}
 +
\frac{1}{3} \; & \frac{1}{3} \; & \frac{1}{3} \\ \\
 +
\frac{1}{4} \; & \frac{3}{4} \; & 0 \\ \\
 +
\frac{1}{2} \; & 0 \; & \frac{1}{2}
 +
\end{matrix} \right] </math><br />
  
pii =
 
  
    0.3333    0.3333    0.3333
+
<math> \pi=\left[ \begin{matrix}
 +
\frac{1}{3} & \frac{4}{9} & \frac{2}{9}
 +
\end{matrix} \right] </math><br />
  
>> pii*P
 
  
ans =
+
There are three states in this example.
  
    0.3333    0.3333    0.3333
+
[[File:ab.png]]
  
>> P^1000
+
In this case, state a can go to state a, b, or c; state b can go to state a, b, or c; and state c can go to state a, b, or c so it is possible to go from every state to every state. (Although state b cannot directly go into c in one move, it must go to a, and then to c.).
  
ans =
+
A k-by-k matrix indicates that the chain has k states.
  
    0    1    0
+
- Ergodic Markov chains are irreducible.
    0    0    1
 
    1    0    0
 
  
>> P^10000
+
- A Markov chain is called a '''regular''' chain if some power of the transition matrix has only positive elements.<br />
 +
*Any transition matrix that has no zeros determines a regular Markov chain
 +
*However, it is possible for a regular Markov chain to have a transition matrix that has zeros.
 +
<br />
 +
For example, recall the matrix of the Land of Oz
  
ans =
+
<math>P = \left[ \begin{matrix}
 +
& R & N & S \\
 +
R & 1/2 & 1/4 & 1/4 \\
 +
N & 1/2 & 0 & 1/2 \\
 +
S & 1/4 & 1/4 & 1/2 \\
 +
\end{matrix} \right]</math><br />
  
    0    1    0
+
=== Theorem ===
    0    0    1
+
An ergodic Markov chain has a unique stationary distribution <math>\pi</math>. The limiting distribution exists and is equal to <math>\pi</math><br/>
    1    0    0
+
Note: Ergodic Markov Chain is irreducible, aperiodic and positive recurrent.
  
>> P^10002
+
Example: Consider the markov chain of <math>\left[\begin{matrix}0 & 1 \\ 1 & 0\end{matrix}\right]</math>, the stationary distribution is obtained by solving <math>\pi P = \pi</math>, getting <math>\pi=[0.5, 0.5]</math>, but from the assignment we know that it does not converge, ie. there is no limiting distribution, because the Markov chain is not aperiodic and cycle repeats <math>P^2=\left[\begin{matrix}1 & 0 \\ 0 & 1\end{matrix}\right]</math> and <math>P^3=\left[\begin{matrix}0 & 1 \\ 1 & 0\end{matrix}\right]</math>
  
ans =
+
'''Another Example'''
  
    1    0    0
+
<math>P=\left[ {\begin{array}{ccc}
    0    1     0
+
\frac{1}{4} & \frac{3}{4} \\[6pt]
    0    0    1
+
\frac{1}{5} & \frac{4}{5} \end{array} } \right]</math> <br>
  
>> P^10003
 
  
ans =
 
  
    0    1    0
+
[[File:Untitled*.jpg]]
    0    0    1
 
    1    0    0
 
  
>> %P^10000 = P^10003
+
This matrix means that there are two points in the space, let's call them a and b<br/>
>> % This chain does not have limiting distribution, it has a stationary distribution. 
+
Starting from a, the probability of staying in a is 1/4 <br/>
 +
Starting from a, the probability of going from a to b is 3/4 <br/>
 +
Starting from b, the probability of going from b to a is 1/5 <br/>
 +
Starting from b, the probability of staying in b is 4/5 <br/>
  
This chain does not converge, it has a cycle.
+
Solve the equation <math> \pi = \pi P </math> <br>
</pre>
+
<math> \pi_0 = .25 \pi_0 + .2 \pi_1 </math> <br>
 
+
<math> \pi_1 = .75 \pi_0 + .8 \pi_1 </math> <br>
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>
+
<math> \pi_0 + \pi_1 = 1 </math> <br>
 
+
Solving this system of equations we get: <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>
+
<math> \pi_0 = \frac{4}{15} \pi_1 </math> <br>
 +
<math> \pi_1 = \frac{15}{19} </math> <br>
 +
<math> \pi_0 = \frac{4}{19} </math> <br>
 +
<math> \pi = [\frac{4}{19}, \frac{15}{19}] </math> <br>
 +
<math> \pi </math> is the long run distribution, and this is also a limiting distribution.
  
Another example:
+
We can use the stationary distribution to compute the expected waiting time to return to state 'a' <br/>
 +
given that we start at state 'a' and so on.. Formula for this will be : <math> E[T_{i,i}]=\frac{1}{\pi_i}</math><br/>
 +
In the example above this will mean that that expected waiting time for the markov process to return to<br/>
 +
state 'a' given that we start at state 'a' is 19/4.<br/>
  
 +
definition of limiting distribution: when the stationary distribution is convergent, it is a limiting distribution.<br/>
  
Find the stationary distribution of P= <math>\left[ {\begin{array}{ccc}
+
remark:satisfied balance of <math>\pi_i P_{ij} = P_{ji} \pi_j</math>, so there is other way to calculate the step probability.
0.5 & 0 & 0 \\
 
1 & 0 & 0.5 \\
 
0 & 1 & 0.5 \end{array} } \right]</math>
 
  
<math>\pi=\pi~P</math><br>
+
=== MatLab Code ===
 +
In the following, P is the transition matrix. eye(n) refers to the n by n Identity matrix. L is the Laplacian matrix, L = (I - P). The Laplacian matrix will have at least 1 zero Eigenvalue. For every 0 in the diagonal, there is a component. If there is exactly 1 zero Eigenvalue, then the matrix is connected and has only 1 component. The number of zeros in the Laplacian matrix is the number of parts in your graph/process. If there is more than one zero on the diagonal of this matrix, means there is a disconnect in the graph.
  
<math>\pi=</math> [<math>\pi</math><sub>0</sub>, <math>\pi</math><sub>1</sub>, <math>\pi</math><sub>2</sub>]<br>
 
  
The system of equations is:
+
<pre style='font-size:14px'>
 +
>> P=[1/3, 1/3, 1/3; 1/4, 3/4, 0; 1/2, 0, 1/2]
  
<math>0.5\pi_0+1\pi_1+0\pi_2= \pi_0=> 2\pi_1 = \pi_0</math><br>
+
P =
<math>0\pi_0+0\pi_1+1\pi_2= \pi_1 => \pi_1=\pi_2</math><br>
 
<math>0\pi_0+0.5\pi_1+0.5\pi_2 = \pi_2 => \pi_1 = \pi_2</math><br>
 
<math>\pi_0+\pi_1+\pi_2 = 1</math><br>
 
  
<math>2\pi_1+\pi_1+\pi_1 = 4\pi_1 = 1</math>, which gives <math>\pi_1=\frac {1}{4}</math> <br>
+
    0.3333    0.3333    0.3333
Also, <math>\pi_1 = \pi_2 = \frac {1}{4}</math> <br>
+
    0.2500    0.7500        0
So, <math>\pi = [\frac{1}{2}, \frac{1}{4}, \frac{1}{4}]</math> <br>
+
    0.5000        0    0.5000
  
=== Ergodic Chain ===
+
>> eye(3) %%returns 3x3 identity matrix
  
A Markov chain is called an ergodic chain if it is possible to go from every state to every state (not necessarily in one move). For instance, note that we can claim a Markov chain is ergodic if it is possible to somehow start at any state i and end at any state j in the matrix. We could have a chain with states 0, 1, 2, 3, 4 where it is not possible to go from state 0 to state 4 in just one step. However, it may be possible to go from 0 to 1, then from 1 to 2, then from 2 to 3, and finally 3 to 4 so we can claim that it is possible to go from 0 to 4 and this would satisfy a requirement of an ergodic chain. The example below will further explain this concept.
+
ans =
  
'''Note:'''if there's a finite number N then every other state can be reached in N steps.
+
    1    0    0
'''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 distribution.
+
    0    1    0
 +
    0    0    1
  
 +
>> L=(eye(3)-P) 
  
====Example====
+
L =
<math> P= \left[ \begin{matrix}
 
\frac{1}{3} \; & \frac{1}{3} \; & \frac{1}{3} \\ \\
 
\frac{1}{4} \; & \frac{3}{4} \; & 0 \\ \\
 
\frac{1}{2} \; & 0 \; & \frac{1}{2}
 
\end{matrix} \right] </math><br />
 
  
 +
    0.6667  -0.3333  -0.3333
 +
  -0.2500    0.2500        0
 +
  -0.5000        0    0.5000
  
<math> \pi=\left[ \begin{matrix}
+
>> [vec val]=eigs(L)
\frac{1}{3} & \frac{4}{9} & \frac{2}{9}
 
\end{matrix} \right] </math><br />
 
 
 
  
There are three states in this example.
+
vec =
  
[[File:ab.png]]
+
  -0.7295    0.2329    0.5774
 +
    0.2239  -0.5690    0.5774
 +
    0.6463    0.7887    0.5774
  
In this case, state a can go to state a, b, or c; state b can go to state a, b, or c; and state c can go to state a, b, or c so it is possible to go from every state to every state. (Although state b cannot directly go into c in one move, it must go to a, and then to c.).
 
  
A k-by-k matrix indicates that the chain has k states.
+
val =
  
- Ergodic Markov chains are irreducible.
+
    1.0643        0        0
 +
        0    0.3523        0
 +
        0        0  -0.0000
  
- A Markov chain is called a '''regular''' chain if some power of the transition matrix has only positive elements.<br />
+
%% Only one value of zero on the diagonal means the chain is connected
*Any transition matrix that has no zeros determines a regular Markov chain
 
*However, it is possible for a regular Markov chain to have a transition matrix that has zeros.
 
<br />
 
For example, recall the matrix of the Land of Oz
 
  
<math>P = \left[ \begin{matrix}
+
>> P=[0.8, 0.2, 0, 0;0.2, 0.8, 0, 0; 0, 0, 0.8, 0.2; 0, 0, 0.1, 0.9]
& R & N & S \\
 
R & 1/2 & 1/4 & 1/4 \\
 
N & 1/2 & 0 & 1/2 \\
 
S & 1/4 & 1/4 & 1/2 \\
 
\end{matrix} \right]</math><br />
 
  
=== Theorem ===
+
P =
An ergodic Markov chain has a unique stationary distribution <math>\pi</math>. The limiting distribution exists and is equal to <math>\pi</math><br/>
 
Note: Ergodic Markov Chain is irreducible, aperiodic and positive recurrent.
 
  
Example: Consider the markov chain of <math>\left[\begin{matrix}0 & 1 \\ 1 & 0\end{matrix}\right]</math>, the stationary distribution is obtained by solving <math>\pi P = \pi</math>, getting <math>\pi=[0.5, 0.5]</math>, but from the assignment we know that it does not converge, ie. there is no limiting distribution, because the Markov chain is not aperiodic and cycle repeats <math>P^2=\left[\begin{matrix}1 & 0 \\ 0 & 1\end{matrix}\right]</math> and <math>P^3=\left[\begin{matrix}0 & 1 \\ 1 & 0\end{matrix}\right]</math>
+
    0.8000    0.2000        0         0
 +
    0.2000    0.8000        0        0
 +
        0        0    0.8000    0.2000
 +
        0         0   0.1000    0.9000
  
'''Another Example'''
+
>> eye(4)
  
<math>P=\left[ {\begin{array}{ccc}
+
ans =
\frac{1}{4} & \frac{3}{4} \\[6pt]
 
\frac{1}{5} & \frac{4}{5} \end{array} } \right]</math> <br>
 
  
 +
    1    0    0    0
 +
    0    1    0    0
 +
    0    0    1    0
 +
    0    0    0    1
  
 +
>> L=(eye(4)-P)
  
[[File:Untitled*.jpg]]
+
L =
  
This matrix means that there are two points in the space, let's call them a and b<br/>
+
    0.2000  -0.2000        0        0
Starting from a, the probability of staying in a is 1/4 <br/>
+
  -0.2000    0.2000        0        0
Starting from a, the probability of going from a to b is 3/4 <br/>
+
        0        0    0.2000  -0.2000
Starting from b, the probability of going from b to a is 1/5 <br/>
+
        0        0  -0.1000    0.1000
Starting from b, the probability of staying in b is 4/5 <br/>
 
  
Solve the equation <math> \pi = \pi P </math> <br>
+
>> [vec val]=eigs(L)
<math> \pi_0 = .25 \pi_0 + .2 \pi_1 </math> <br>
 
<math> \pi_1 = .75 \pi_0 + .8 \pi_1 </math> <br>
 
<math> \pi_0 + \pi_1 = 1 </math> <br>
 
Solving this system of equations we get: <br>
 
<math> \pi_0 = \frac{4}{15} \pi_1 </math> <br>
 
<math> \pi_1 = \frac{15}{19} </math> <br>
 
<math> \pi_0 = \frac{4}{19} </math> <br>
 
<math> \pi = [\frac{4}{19}, \frac{15}{19}] </math> <br>
 
<math> \pi </math> is the long run distribution
 
  
We can use the stationary distribution to compute the expected waiting time to return to state 'a' <br/>
+
vec =
given that we start at state 'a' and so on.. Formula for this will be : <math> E[T_{i,i}]=\frac{1}{\pi_i}</math><br/>
 
In the example above this will mean that that expected waiting time for the markov process to return to<br/>
 
state 'a' given that we start at state 'a' is 19/4.<br/>
 
  
definition of limiting distribution.
+
    0.7071        0    0.7071        0
 +
  -0.7071        0    0.7071        0
 +
        0    0.8944        0    0.7071
 +
        0  -0.4472        0    0.7071
  
remark:satisfied balance of <math>\pi_i P_{ij} = P_{ji} \pi_j</math>, so there is other way to calculate the step probability.
 
  
=== MatLab Code ===
+
val =
In the following, P is the transition matrix. eye(n) refers to the n by n Identity matrix. L is the Laplacian matrix, L = (I - P). The Laplacian matrix will have at least 1 zero Eigenvalue. For every 0 in the diagonal, there is a component. If there is exactly 1 zero Eigenvalue, then the matrix is connected and has only 1 component. The number of zeros in the Laplacian matrix is the number of parts in your graph/process. If there is more than one zero on the diagonal of this matrix, means there is a disconnect in the graph.
 
  
 +
    0.4000        0        0        0
 +
        0    0.3000        0        0
 +
        0        0  -0.0000        0
 +
        0        0        0  -0.0000
  
<pre style='font-size:14px'>
+
%% Two values of zero on the diagonal means there are two 'islands' of chains
>> P=[1/3, 1/3, 1/3; 1/4, 3/4, 0; 1/2, 0, 1/2]
 
  
P =
+
</pre>
  
    0.3333    0.3333    0.3333
+
<math>\Pi</math> satisfies detailed balance if <math>\Pi_i P_{ij}=P_{ji} \Pi_j</math>. Detailed balance guarantees that <math>\Pi</math> is stationary distribution.<br />
    0.2500    0.7500        0
 
    0.5000        0    0.5000
 
  
>> eye(3) %%returns 3x3 identity matrix
+
'''Adjacency matrix''' - a matrix <math>A</math> that dictates which states are connected and way of portraying which vertices in the matrix are adjacent. Two vertices are adjacent if there exists a path between them of length 1.If we compute <math>A^2</math>, we can know which states are connected with paths of length 2.<br />
  
ans =
+
A '''Markov chain''' is called an irreducible chain if it is possible to go from every state to every state (not necessary in one more).<br />
 +
Theorem: An '''ergodic''' Markov chain has a unique stationary distribution <math>\pi</math>. The limiting distribution exists and is equal to <math>\pi</math>. <br />
  
    1    0    0
 
    0    1    0
 
    0    0    1
 
  
>> L=(eye(3)-P)  
+
Markov process satisfies detailed balance  if and only if it is a '''reversible''' Markov process
 +
where P is the matrix of Markov transition.<br />
  
L =
+
Satisfying the detailed balance condition guarantees that <math>\pi</math> is stationary distributed.
  
    0.6667  -0.3333  -0.3333
+
<math> \pi </math> satisfies detailed balance if <math> \pi_i P_{ij} = P_{ji} \pi_j </math> <br>
  -0.2500    0.2500        0
+
which is the same as the Markov process equation.
  -0.5000        0    0.5000
 
  
>> [vec val]=eigs(L)
+
Example in the class:
 +
<math>P= \left[ {\begin{array}{ccc}
 +
\frac{1}{3} & \frac{1}{3} & \frac{1}{3} \\[6pt]
 +
\frac{1}{4} & \frac{3}{4} & 0 \\[6pt]
 +
\frac{1}{2} & 0 & \frac{1}{2} \end{array} } \right]</math>
  
vec =
+
and  <math>\pi=(\frac{1}{3},\frac{4}{9}, \frac{2}{9})</math>
  
  -0.7295    0.2329    0.5774
+
<math>\pi_1 P_{1,2} = 1/3 \times 1/3 = 1/9,\, P_{2,1} \pi_2 = 1/4 \times 4/9 = 1/9 \Rightarrow \pi_1 P_{1,2} = P_{2,1} \pi_2 </math><br>
    0.2239  -0.5690    0.5774
 
    0.6463    0.7887    0.5774
 
  
 +
<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 \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.
 +
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.
  
val =
+
== Class 15 - Tuesday June 25th 2013 ==
 +
=== Announcement ===
 +
Note to all students, the first half of today's lecture will cover the midterm's solution; however please do not post the solution on the Wikicoursenote.<br />
  
    1.0643        0        0
+
====Detailed balance====
        0    0.3523        0
+
<div style="border:2px solid black">
        0        0  -0.0000
+
<b>Definition (from wikipedia)</b>
 +
The principle of detailed balance is formulated for kinetic systems which are decomposed into elementary processes (collisions, or steps, or elementary reactions): At equilibrium, each elementary process should be equilibrated by its reverse process.
 +
</div>
 +
Let <math>P</math> be the transition probability matrix of a Markov chain. If there exists a distribution vector <math>\pi</math> such that <math>\pi_i \cdot P_{ij}=P_{ji} \cdot \pi_j, \; \forall i,j</math>, then the Markov chain is said to have '''detailed balance'''. A detailed balanced Markov chain must have <math>\pi</math> given above as a stationary distribution, that is <math>\pi=\pi P</math>, where <math>\pi</math> is a 1 by n matrix and P is a n by n matrix.<br>
  
%% Only one value of zero on the diagonal means the chain is connected
 
  
>> P=[0.8, 0.2, 0, 0;0.2, 0.8, 0, 0; 0, 0, 0.8, 0.2; 0, 0, 0.1, 0.9]
+
need to remember:
 +
'''Proof:''' <br>
 +
<math>\; [\pi P]_j = \sum_i \pi_i P_{ij} =\sum_i P_{ji}\pi_j =\pi_j\sum_i P_{ji} =\pi_j  ,\forall j</math>
  
P =
+
:Note: Since <math>\pi_j</math> is a sum of column j and we can do this proof for every element in matrix P; in general, we can prove <math>\pi=\pi P</math> <br>
  
    0.8000    0.2000        0        0
+
Hence <math>\pi</math> is always a stationary distribution of <math>P(X_{n+1}=j|X_n=i)</math>, for every n.  
    0.2000    0.8000        0        0
 
        0        0    0.8000    0.2000
 
        0        0    0.1000    0.9000
 
  
>> eye(4)
+
In other terms, <math> P_{ij} = P(X_n = j| X_{n-1} = i) </math>, where <math>\pi_j</math> is the equilibrium probability of being in state j and <math>\pi_i</math> is the equilibrium probability of being in state i. <math>P(X_{n-1} = i) = \pi_i</math> is equivalent to <math>P(X_{n-1} = i,  Xn = j)</math> being symmetric in i and j.
  
ans =
+
Keep in mind that the detailed balance is a sufficient but not required condition for a distribution to be stationary.
 +
i.e. A distribution satisfying the detailed balance is stationary, but a stationary distribution does not necessarily satisfy the detailed balance.
  
    1    0    0    0
+
In the stationary distribution <math>\pi=\pi P</math>, in the proof the sum of the p is equal 1 so the <math>\pi P=\pi</math>.
    0    1    0    0
 
    0    0    1    0
 
    0    0    0    1
 
  
>> L=(eye(4)-P)
+
=== PageRank (http://en.wikipedia.org/wiki/PageRank) ===
  
L =
+
*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 link-analysis 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.
 +
*Many social media sites use this (such as Facebook and Twitter)
 +
*It can also be used to find criminals (ie. theives, hackers, terrorists, etc.) by finding out the links.
 +
This is what made Google the search engine of choice over Yahoo, Bing, etc.- What made Google's search engine a huge success is not its search function, but rather the algorithm it used to rank the pages. (Ex. If we come up with 100 million search results, how do you list them by relevance and importance so the users can easily find what they are looking for. Most users will not go past the first 3 or so search pages to find what they are looking for. It is this ability to rank pages that allow Google to remain more popular than Yahoo, Bink, AskJeeves, etc.). It should be noted that after using the PageRank algorithm, Google uses other processes to filter results.<br/>
  
    0.2000  -0.2000        0        0
+
<br />'''The order of importance'''<br />
  -0.2000    0.2000        0        0
+
1. A web page is more important if many other pages point to it<br />
        0        0    0.2000  -0.2000
+
2. The more important a web page is, the more weight should be assigned to its outgoing links<br/ >
        0        0  -0.1000    0.1000
+
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 />
  
>> [vec val]=eigs(L)
+
<br />
 +
[[File:diagram.jpg]]
 +
<math>L=
 +
\left[ {\begin{matrix}
 +
0 & 0 & 0 & 0 & 0 \\
 +
1 & 0 & 0 & 0 & 0 \\
 +
1 & 1 & 0 & 1 & 0 \\
 +
0 & 1 & 0 & 0 & 1 \\
 +
0 & 0 & 0 & 0 & 0 \end{matrix} } \right]</math>
  
vec =
+
The first row indicates who gives a link to 1. As shown in the diagram, nothing gives a link to 1, and thus it is all zero.
 +
The second row indicates who gives a link to 2. As shown in the diagram, only 1 gives a link to 2, and thus column 1 is a 1 for row 2, and the rest are all zero.
  
    0.7071        0    0.7071        0
+
ie: According to the above example <br/ >
  -0.7071        0    0.7071        0
+
Page 3 is the most important since it has the most links pointing to it, therefore more weigh should be placed on its outgoing links.<br/ >
        0    0.8944        0    0.7071
+
Page 4 comes after page 3 since it has the second most links pointing to it<br/ >
        0  -0.4472        0    0.7071
+
Page 2 comes after page 4 since it has the third most links pointing to it<br/>
 +
Page 1 and page 5 are the least important since no links point to them<br/ >
 +
As page 1 and page 2 have the most outgoing links, then their links have less value compared to the other pages. <br/ >
  
 +
:<math>
 +
Lij = \begin{cases}
 +
1, & \text{if j has a link to i} \\
 +
0, & \text{otherwise}
 +
\end{cases}</math>
  
val =
+
<br />
 +
<math>C_j=</math> The number of outgoing links of page <math>j</math>:
 +
<math>C_j=\sum_i L_{ij}</math>
 +
(i.e. sum of entries in column j)<br />
 +
<br />
 +
<math>P_j</math> is the rank of page <math>j</math>.<br />
 +
Suppose we have <math>N</math> pages, <math>P</math> is a vector containing ranks of all pages.<br />
 +
- <math>P</math> is a <math>N \times 1</math> vector.
  
    0.4000        0        0        0
+
- <math>P_i</math> counts the number of incoming links of page <math>i</math>
        0    0.3000        0        0
+
<math>P_i=\sum_j L_{ij}</math> <br />(i.e. sum of entries in row i)
        0        0  -0.0000        0
 
        0        0        0  -0.0000
 
  
%% Two values of zero on the diagonal means there are two 'islands' of chains
+
For each row of <math>L</math>, if there is a 1 in the third column, it means page three point to that page.
  
</pre>
+
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.
  
<math>\Pi</math> satisfies detailed balance if <math>\Pi_i P_{ij}=P_{ji} \Pi_j</math>. Detailed balance guarantees that <math>\Pi</math> is stationary distribution.<br />
+
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).
 +
(source:http://en.wikipedia.org/wiki/PageRank#Description)
  
'''Adjacency matrix''' - a matrix <math>A</math> that dictates which states are connected and way of portraying which vertices in the matrix are adjacent. If we compute <math>A^2</math>, we can know which states are connected with paths of length 2.<br />
+
For those interested in PageRank, here is the original paper by Google co-founders Brin and Page: http://infolab.stanford.edu/pub/papers/google.pdf
  
A '''Markov chain''' is called an irreducible chain if it is possible to go from every state to every state (not necessary in one more).<br />
+
=== Example of Page Rank Application in Real Life ===
Theorem: An '''ergodic''' Markov chain has a unique stationary distribution <math>\pi</math>. The limiting distribution exists and is equal to <math>\pi</math>. <br />
 
  
 +
'''Page Rank checker'''
 +
- This is a free service to check Google™ page rank instantly via online PR checker or by adding a PageRank checking button to the web pages.
 +
  <font size="3">(http://www.prchecker.info/check_page_rank.php)</font>
  
Markov process satisfies detailed balance  if and only if it is a '''reversible''' Markov process
 
where P is the matrix of  Markov transition.<br />
 
  
Satisfying the detailed balance condition guarantees that <math>\pi</math> is stationary distributed.
+
GoogleMatrix G = d * [ (Hyperlink Matrix H) + (Dangling Nodes Matrix A) ] + ((1-d)/N) * (NxN Matrix U of all 1's)
  
<math> \pi </math> satisfies detailed balance if <math> \pi_i P_{ij} = P_{ji} \pi_j </math> <br>
 
which is the same as the Markov process equation.
 
  
Example in the class:
+
[[File:Google matrix.png]]
<math>P= \left[ {\begin{array}{ccc}
 
\frac{1}{3} & \frac{1}{3} & \frac{1}{3} \\[6pt]
 
\frac{1}{4} & \frac{3}{4} & 0 \\[6pt]
 
\frac{1}{2} & 0 & \frac{1}{2} \end{array} } \right]</math>
 
  
and  <math>\pi=(\frac{1}{3},\frac{4}{9}, \frac{2}{9})</math>
 
  
<math>\pi_1 P_{1,2} = 1/3 \times 1/3 = 1/9,\, P_{2,1} \pi_2 = 1/4 \times 4/9 = 1/9 \Rightarrow \pi_1 P_{1,2} = P_{2,1} \pi_2 </math><br>
+
(source: https://googledrive.com/host/0B2GQktu-wcTiaWw5OFVqT1k3bDA/)
  
<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>
+
== Class 16 - Thursday June 27th 2013 ==
Remark:Detailed balance of <math> \pi_i * Pij = Pji * \pi_j</math> , so there is other way to calculate the step probability<br />
 
<math>\pi</math> is stationary but is not limiting.
 
Detailed balance guarantees that <math>\pi</math> is stationary distribution.
 
  
== Class 15 - Tuesday June 25th 2013 ==
+
=== Page Rank ===
=== Announcement ===
+
*<math>
Note to all students, the first half of today's lecture will cover the midterm's solution; however please do not post the solution on the Wikicoursenote.<br />
+
L_{ij} = \begin{cases}
 +
1, & \text{if j has a link to i }  \\
 +
0, & \text{otherwise} \end{cases} </math> <br/>
  
====Detailed balance====
+
*<math>C_j</math>: number of outgoing links for page j, where <math>c_j=\sum_i L_{ij}</math>
  
Let <math>P</math> be the transition probability matrix of a Markov chain. If there exists a distribution vector <math>\pi</math> such that <math>\pi_i \cdot P_{ij}=P_{ji} \cdot \pi_j, \; \forall i,j</math>, then the Markov chain is said to have '''detailed balance'''. A detailed balanced Markov chain must have <math>\pi</math> given above as a stationary distribution, that is <math>\pi=\pi P</math>, where <math>\pi</math> is a 1 by n matrix and P is a n by n matrix.<br>
+
P is N by 1 vector contains rank of all N pages; for page i, the rank is <math>P_i</math>
  
 +
<math>P_i= (1-d) + 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.
  
'''Proof:''' <br>
+
Interpretation of the formula:<br/>
<math>\; [\pi P]_j = \sum_i \pi_i P_{ij} =\sum_i P_{ji}\pi_j =\pi_j\sum_i P_{ji} =\pi_j  ,\forall j</math>
+
1) sum of L<sub>ij</sub> is the total number of incoming links<br/>
 +
2) the above sum is weighted by page rank of the pages that contain the link to i (P<sub>j</sub>) i.e. if a high-rank page points to page i, then this link carries more weight than links from lower-rank pages.<br/>
 +
3) the sum is then weighted by the inverse of the number of outgoing links from the pages that contain links to i (c<sub>j</sub>). i.e. if a page has more outgoing links than other pages then its links carry less weight.<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/>
 +
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: Since <math>\pi_j</math> is a sum of column j and we can do this proof for every element in matrix P; in general, we can prove <math>\pi=\pi P</math> <br>
+
<math>c_j</math> is the number of outgoing links, less outgoing links means more important.<br/>
  
Hence <math>\pi</math> is always a stationary distribution of <math>P(X_{n+1}=j|X_n=i)</math>, for every n.
 
  
In other terms, <math> P_{ij} = P(X_n = j| X_{n-1} = i) </math>, where <math>\pi_j</math> is the equilibrium probability of being in state j and <math>\pi_i</math> is the equilibrium probability of being in state i. <math>P(X_{n-1} = i) = \pi_i</math> is equivalent to <math>P(X_{n-1} = i,  Xn = j)</math> being symmetric in i and j.
+
Let D be a diagonal N by N matrix such that <math> D_{ii}</math> = <math>c_i</math>
  
Keep in mind that the detailed balance is a sufficient but not required condition for a distribution to be stationary.
+
Note: Ranks are arbitrary, all we want to know is the order. That is, we want to know how important the page rank relative to the other pages and are not interested in the value of the page rank.  
i.e. A distribution satisfying the detailed balance is stationary, but a stationary distribution does not necessarily satisfy the detailed balance.
 
  
In the stationary distribution <math>\pi=\pi P</math>, in the proof the sum of the p is equal 1 so the <math>\pi P=\pi</math>.
+
<math>D=
 +
\left[ {\begin{matrix}
 +
c_1 & 0 & ... & 0  \\
 +
0 & c_2 & ...  &  0  \\
 +
0 & 0 & ... &  0 \\
 +
0 & 0 & ... & c_N \end{matrix} } \right]</math>
  
=== PageRank (http://en.wikipedia.org/wiki/PageRank) ===
+
Then <math>P=~(1-d)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/>
 +
i.e. e<sup>T</sup> P = N, then <math>~\frac{e^{T}P}{N} = 1</math>
  
*PageRank is a link-analysis 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.
 
*Many social media sites use this (such as Facebook and Twitter)
 
*It can also be used to find criminals (ie. theives, hackers, terrorists, etc.) by finding out the links.
 
This is what made Google the search engine of choice over Yahoo, Bing, etc.- What made Google's search engine a huge success is not its search function, but rather the algorithm it used to rank the pages. (Ex. If we come up with 100 million search results, how do you list them by relevance and importance so the users can easily find what they are looking for. Most users will not go past the first 3 or so search pages to find what they are looking for. It is this ability to rank pages that allow Google to remain more popular than Yahoo, Bink, AskJeeves, etc.). It should be noted that after using the PageRank algorithm, Google uses other processes to filter results.<br/>
 
  
<br />'''The order of importance'''<br />
+
D<sup>-1</sup> will be:
1. A web page is 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/ >
 
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 />
 
  
<br />
+
D<sup>-1</sup><math>=  
[[File:diagram.jpg]]
 
<math>L=  
 
 
\left[ {\begin{matrix}
 
\left[ {\begin{matrix}
0 & 0 & 0 & 0 & 0 \\
+
\frac {1}{c_1} & 0 & ... & 0 \\
1 & 0 & 0 & 0 & 0 \\
+
0 & \frac {1}{c_2} & ...  & 0 \\
1 & 1 & 0 & 1 & 0 \\
+
0 & 0 & ... & 0 \\
0 & 1 & 0 & 0 & 1 \\
+
0 & 0 & ... & \frac {1}{c_N} \end{matrix} } \right]</math>
0 & 0 & 0 & 0 & 0 \end{matrix} } \right]</math>
 
  
The first row indicates who gives a link to 1. As shown in the diagram, nothing gives a link to 1, and thus it is all zero.
+
<math>P=~(1-d)e+dLD^{-1}P</math>  where <math>e=\begin{bmatrix}
The second row indicates who gives a link to 2. As shown in the diagram, only 1 gives a link to 2, and thus column 1 is a 1 for row 2, and the rest are all zero.
+
1\\
 +
1\\
 +
...\\
 +
1
 +
\end{bmatrix}</math>
  
ie: According to the above example <br/ >
+
<math>P=(1-d)~\frac{ee^{T}P}{N}+dLD^{-1}P</math>
Page 3 is the most important since it has the most links pointing to it, therefore more weigh should be placed on its outgoing links.<br/ >
 
Page 4 comes after page 3 since it has the second most links pointing to it<br/ >
 
Page 2 comes after page 4 since it has the third most links pointing to it<br/>
 
Page 1 and page 5 are the least important since no links point to them<br/ >
 
<math>As page 1</math> and page 2 has the most outgoing links, then their links have less value compared to the other pages. <br/ >
 
  
:<math>
+
<math>P=[(1-d)~\frac{ee^T}{N}+dLD^{-1}]P</math>
Lij = \begin{cases}
 
1, & \text{if j has a link to i} \\
 
0, & \text{otherwise}
 
\end{cases}</math>
 
  
<br />
+
<math>=> P=A*P</math>
C<sub>j</sub> The number of outgoing links of page <math>j</math>:
 
<math>C_j=\sum_i L_{ij}</math>
 
(i.e. sum of entries in column j)<br />
 
<br />
 
<math>P_j</math> is the rank of page <math>j</math>.<br />
 
Suppose we have <math>N</math> pages, <math>P</math> is a vector containing ranks of all pages.<br />
 
- <math>P</math> is a <math>N \times 1</math> vector.
 
  
- <math>P_i</math> counts the number of incoming links of page <math>i</math>
+
'''Explanation of an eigenvector'''
<math>P_i=\sum_j L_{ij}</math> <br />(i.e. sum of entries in row i)
 
  
for each row, if there is a 1 in the third column, it means page three point to that page.
+
An eigenvector is a non-zero vector '''v''' such that when multiplied by a square matrix, A, the result is a scalar times the vector '''v''' itself. <br>
 +
That is, A*v = c*v. Where c is the eigenvalue of A corresponding to the eigenvector v. In our case of Page Rank, the eigenvalue c=1. <br>
  
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).
+
We obtain that <math>P=AP</math> where <math>A=(1-d)~\frac{ee^T}{N}+dLD^{-1}</math><br/>
(source:http://en.wikipedia.org/wiki/PageRank#Description)
+
Thus, <math>P</math> is an eigenvector of <math>P</math> correspond to an eigen value equals 1.<br/>
  
For those interested in PageRank, here is the original paper by Google co-founders Brin and Page: http://infolab.stanford.edu/pub/papers/google.pdf
 
  
=== Example of Page Rank Application in Real Life ===
+
Since,
 +
L is a N*N matrix,
 +
D<sup>-1</sup> is a N*N matrix,
 +
P is a N*1 matrix <br/>
 +
Then as a result, <math>LD^{-1}P</math> is a N*1 matrix. <br/>
  
'''Page Rank checker'''
+
N is a N*N matrix, d is a constant between 0 and 1.
- This is a free service to check Google™ page rank instantly via online PR checker or by adding a PageRank checking button to the web pages.
 
  <font size="3">(http://www.prchecker.info/check_page_rank.php)</font>
 
  
 +
'''P=AP'''<br />
 +
P is an eigenvector of A with corresponding eigenvalue equal to 1.<br>
 +
'''P<sup>T</sup>=P<sup>T</sup>A<sup>T</sup><br>'''
 +
Notice that all entries in A are non-negative and each row sums to 1. Hence A satisfies the definition of a transition probability matrix.<br>
 +
P<sup>T</sup> is the stationary distribution of a Markov Chain with transition probability matrix A<sup>T</sup>.
  
GoogleMatrix G = d * [ (Hyperlink Matrix H) + (Dangling Nodes Matrix A) ] + ((1-d)/N) * (NxN Matrix U of all 1's)
+
We can consider A to be the matrix describing all possible movements following links on the internet, and P<sup>t</sup> as the probability of being on any given webpage if we were on the internet long enough.
  
 +
Definition of rank page and proof it steps by steps, it shows with 3 n*n matrix and and one n*1 matrix and a constant d between 0 to 1.
 +
p is the stationary distribution so p=Ap.
  
[[File:Google matrix.png]]
+
=== Damping Factor "d" ===
  
 +
The PageRank assumes that any imaginary user who is randomly clicking on links will eventually stop clicking. The probability, at any step, that the person will keep on clicking is a damping factor, <math>d</math>. After many studies, the approximation of <math>d</math> is 0.85. Other values for <math>d</math> have been used in class and may appear on assignments and exams.<br/>
  
(source: https://googledrive.com/host/0B2GQktu-wcTiaWw5OFVqT1k3bDA/)
+
In addition, <math>d</math> is a vector of ranks that are arbitrary. For example the rank can be [1 3 2], or [10 30 20], or [0.1 0.3 0.2]. All three of these examples are relative/equivalent since they are ranks, we could even have [1 10 3]. Therefore, <math>d</math> must have a relative rank.<br/>
  
== Class 16 - Thursday June 27th 2013 ==
+
So <math>P_1 + P_2 + \cdots + P_n=N</math> <br/>
 +
Which is equivalent to:
 +
<math>e^{T}P= [1 \cdots 1] [P_1 \cdots P_n]^T </math> <br/>
 +
Where <math>[1 \cdots 1]</math> is a 1 scalar vector and <math>[P_1 \cdots P_n]^T</math> is a rank vector. <br/>
 +
So <math>e^{T}P=N -> (e^{T}P)/N = 1 </math>
  
=== Page Rank ===
+
===Examples===
<math>L_{ij}</math> equals 1 if j has a link to i, and equals 0 otherwise. <br>
+
<span style="background:#F5F5DC">
<math>C_j</math> :The number of outgoing links for page j, where <math>c_j=\sum_i L_{ij}</math>  
 
  
P is N by 1 vector contains rank of all N pages; for page i, the rank is <math>P_i</math>
+
==== Example 1 ====
  
<math>P_i= (1-d) + d\cdot \sum_j \frac {L_{ji}P_j}{c_j}</math>
 
  
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.
+
[[File:eg1.jpg]]
 +
<br />
 +
</span>
 +
<math>L=
 +
\left[ {\begin{matrix}
 +
0 & 0 & 1  \\
 +
1 & 0 & 0  \\
 +
0 & 1 & 0 \end{matrix} } \right]\;c=  
 +
\left[ {\begin{matrix}
 +
1 & 1 & 1 \end{matrix} } \right]\;D=
 +
\left[ {\begin{matrix}
 +
1 & 0 & 0  \\
 +
0 & 1 & 0  \\
 +
0 & 0 & 1 \end{matrix} } \right]</math>
  
Interpretation of the formula:<br/>
+
<pre style='font-size:14px'>
1) sum of L<sub>ij</sub> is the total number of incoming links<br/>
 
2) the above sum is weighted by page rank of the pages that contain the link to i (P<sub>j</sub>) i.e. if a high-rank page points to page i, then this link carries more weight than links from lower-rank pages.<br/>
 
3) the sum is then weighted by the inverse of the number of outgoing links from the pages that contain links to i (c<sub>j</sub>). i.e. if a page has more outgoing links than other pages then its links carry less weight.<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/>
 
  
Note that this is a system of N equations with N unknowns.<br/>
+
MATLAB Code
  
<math>c_j</math> is the number of outgoing links, less outgoing links means more important.<br/>
+
d=0.8
 +
N=3
 +
A=(1-d)*ones(N)/N+d*L*pinv(D) #pinv: Moore-Penrose inverse (pseudoinverse) of symbolic matrix
 +
We use the pinv(D) function [pseudo-inverse] instead of the inv(D) function because in
 +
the case of a non-invertible matrix, it would not crash the program.
 +
[vec val]=eigs(A) (eigen-decomposition)
 +
a=-vec(:,1) (find the eigenvector equals to 1)
 +
a=a/sum(a) (normalize a)
 +
or to show that A transpose is a stationary transition matrix
 +
(transpose(A))^200 will be the same as a=a/sum(a)
 +
</pre>
  
 +
'''NOTE:''' Changing the value of d, does not change the ranking order of the pages.
  
Let D be a diagonal N by N matrix such that <math> D_{ii}</math> = <math>c_i</math>
+
By looking at each entry after normalizing a, we can tell the ranking order of each page.<br>
 +
<span style="background:#F5F5DC">
  
Note: Ranks are arbitrary, all we want to know is the order. That is, we want to know how important the page rank relative to the other pages and are not interested in the value of the page rank.  
+
c = [1 1 1] since there are 3 pages, each page is one way recurrent to each other and there is only one outgoing for each page. Hence, D is a 3x3 standard diagonal matrix.
  
<math>D=  
+
==== Example 2 ====
\left[ {\begin{matrix}
 
c_1 & 0 & ... & 0  \\
 
0 & c_2 & ...  &  0  \\
 
0 & 0 & ... &  0 \\
 
0 & 0 & ... & c_N \end{matrix} } \right]</math>
 
  
Then <math>P=~(1-d)e+dLD^{-1}P</math><br/> where e =[1 1 ....]<sup>T</sup> , i.e. a N by 1 vector.<br/>
+
[[File:Screen_shot_2013-07-02_at_3.43.04_AM.png]]
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>
 
  
  
D<sup>-1</sup> will be:  
+
<math>L=
 
+
\left[ {\begin{matrix}
D<sup>-1</sup><math>=  
+
0 & 0 & 1  \\
 +
1 & 0 & \\
 +
0 & 1 & 0 \end{matrix} } \right]\;
 +
c=
 +
\left[ {\begin{matrix}
 +
1 & 1 & 2 \end{matrix} } \right]\;
 +
D=  
 
\left[ {\begin{matrix}
 
\left[ {\begin{matrix}
\frac {1}{c_1} & 0 & ... & 0  \\
+
1 & 0 & 0  \\
0 & \frac {1}{c_2} & ...  & 0  \\
+
0 & 1 & 0  \\
0 & 0 & ... &  0 \\
+
0 & 0 & 2 \end{matrix} } \right]</math>
0 & 0 & ... & \frac {1}{c_N} \end{matrix} } \right]</math>
 
  
<math>P=~(1-d)e+dLD^{-1}P</math>  where <math>e=\begin{bmatrix}
+
<pre style='font-size:14px'>
1\\
 
1\\
 
...\\
 
1
 
\end{bmatrix}</math>
 
  
<math>P=(1-d)~\frac{ee^{T}P}{N}+dLD^{-1}P</math>
+
Matlab code
  
<math>P=[(1-d)~\frac{ee^T}{N}+dLD^{-1}]P</math>
+
>> L=[0 0 1;1 0 1;0 1 0];
 +
>> C=sum(L);
 +
>> D=diag(C);
 +
>> d=0.8;
 +
>> N=3;
 +
>> A=(1-d)*ones(N)/N+d*L*pinv(D);
 +
>> [vec val]=eigs(A)
  
'''Explanation of an eigenvector'''
+
vec =
  
An eigenvector is a non-zero vector '''v''' such that when multiplied by a square matrix, A, the result is a scalar times the vector '''v''' itself. <br>
+
  -0.3707            -0.3536 + 0.3536i  -0.3536 - 0.3536i
That is, A*v = c*v. Where c is the eigenvalue of A corresponding to the eigenvector v. In our case of Page Rank, the eigenvalue c=1. <br>
+
  -0.6672            -0.3536 - 0.3536i  -0.3536 + 0.3536i
 +
  -0.6461            0.7071            0.7071         
  
P=AP
 
  
 +
val =
  
N is a N*N matrix,
+
  1.0000                  0                  0         
L is a N*N matrix,
+
        0            -0.4000 - 0.4000i        0         
D<sup>-1</sup> is a N*N matrix,
+
        0                  0            -0.4000 + 0.4000i
P is a N*1 matrix
 
d is a constant between 0 and 1
 
  
'''P=AP'''<br />
+
>> a=-vec(:,1)
P is an eigenvector of A with corresponding eigenvalue equal to 1.<br>
 
'''P<sup>T</sup>=P<sup>T</sup>A<sup>T</sup><br>'''
 
Notice that all entries in A are non-negative and each row sums to 1. Hence A satisfies the definition of a transition probability matrix.<br>
 
P<sup>T</sup> is the stationary distribution of a Markov Chain with transition probability matrix A<sup>T</sup>.
 
  
We can consider A to be the matrix describing all possible movements following links on the internet, and P<sup>t</sup> as the probability of being on any given webpage if we were on the internet long enough.
+
a =
  
Definition of rank page and proof it steps by steps, it shows with 3 n*n matrix and and one n*1 matrix and a constant d between 0 to 1.
+
    0.3707
p is the stationary distribution so p=Ap.
+
    0.6672
 +
    0.6461
  
=== Damping Factor "d" ===
+
>> a=a/sum(a)
  
The PageRank assumes that any imaginary user who is randomly clicking on links will eventually stop clicking. The probability, at any step, that the person will keep on clicking is a damping factor, "d". After many studies, the approximation of "d" is 0.85. Other values for "d" have been used in class and may appear on assignments and exams.
+
a =
  
===Examples===
+
    0.2201
<span style="background:#F5F5DC">
+
    0.3962
 +
    0.3836
 +
</pre>
 +
'''NOTE:''' Page 2 is the most important page because it has 2 incomings. Similarly, page 3 is more important than page 1 because page 3 has the incoming result from page 2.
 +
 
 +
This example is similar to the first example, but here, page 3 can go back to page 2, so the matrix of the outgoing matrix, the third column of the D matrix is 3 in the third row. And we use the code to calculate the p=Ap. Therefore 2, 3, 1 is the order of importance.
  
==== Example 1 ====
+
==== Example 3 ====
  
 +
[[File:eg 3.jpg]]<br>
  
[[File:eg1.jpg]]
 
<br />
 
</span>
 
 
<math>L=  
 
<math>L=  
 
\left[ {\begin{matrix}
 
\left[ {\begin{matrix}
0 & 0 & 1 \\
+
0 & 1 & 0  \\
1 & 0 & 0 \\
+
1 & 0 & 1 \\
0 & 1 & 0 \end{matrix} } \right]\;c=  
+
0 & 1 & 0 \end{matrix} } \right]\;
 +
c=  
 
\left[ {\begin{matrix}
 
\left[ {\begin{matrix}
1 & 1 & 1 \end{matrix} } \right]\;D=  
+
1 & 2 & 1 \end{matrix} } \right]\;
 +
D=  
 
\left[ {\begin{matrix}
 
\left[ {\begin{matrix}
 
1 & 0 & 0  \\
 
1 & 0 & 0  \\
0 & 1 & 0  \\
+
0 & 2 & 0  \\
 
0 & 0 & 1 \end{matrix} } \right]</math>
 
0 & 0 & 1 \end{matrix} } \right]</math>
  
<pre style='font-size:14px'>
+
<math>d=0.8</math><br>
 +
<math>N=3</math><br>
  
MATLAB Code
 
  
d=0.8
+
</span>
N=3
+
this example is the second page have 2 outgoings.
A=(1-d)*ones(N)/N+d*L*pinv(D) #pinv: Moore-Penrose inverse (pseudoinverse) of symbolic matrix
 
We use the pinv(D) function [pseudo-inverse] instead of the inv(D) function because in
 
the case of a non-invertible matrix, it would not crash the program.
 
[vec val]=eigs(A) (eigen-decomposition)
 
a=-vec(:,1) (find the eigenvector equals to 1)
 
a=a/sum(a) (normalize a)
 
or to show that A transpose is a stationary transition matrix
 
(transpose(A))^200 will be the same as a=a/sum(a)
 
</pre>
 
  
'''NOTE:''' Changing the value of d, does not change the ranking order of the pages.
 
  
By looking at each entry after normalizing a, we can tell the ranking order of each page.<br>
 
<span style="background:#F5F5DC">
 
  
c = [1 1 1] since there are 3 pages, each page is one way recurrent to each other and there is only one outgoing for each page. Hence, D is a 3x3 standard diagonal matrix.
+
Another Example:
 
 
==== Example 2 ====
 
 
 
[[File:Screen_shot_2013-07-02_at_3.43.04_AM.png]]
 
  
 +
Consider: 1 -> ,<-2 ->3
  
 
<math>L=  
 
<math>L=  
 
\left[ {\begin{matrix}
 
\left[ {\begin{matrix}
0 & 0 & 1 \\
+
0 & 1 & 0  \\
1 & 0 & 1 \\
+
1 & 0 & 0 \\
 
0 & 1 & 0 \end{matrix} } \right]\;
 
0 & 1 & 0 \end{matrix} } \right]\;
 
c=  
 
c=  
 
\left[ {\begin{matrix}
 
\left[ {\begin{matrix}
1 & 1 & 2 \end{matrix} } \right]\;
+
1 & 1 & 1 \end{matrix} } \right]\;
 
D=  
 
D=  
 
\left[ {\begin{matrix}
 
\left[ {\begin{matrix}
 
1 & 0 & 0  \\
 
1 & 0 & 0  \\
 
0 & 1 & 0  \\
 
0 & 1 & 0  \\
0 & 0 & 2 \end{matrix} } \right]</math>
+
0 & 0 & 1 \end{matrix} } \right]</math>
  
<pre style='font-size:14px'>
+
==== Example 4 ====
  
Matlab code
+
<math>1 \leftrightarrow 2 \rightarrow 3 \leftrightarrow 4 </math>
 +
<br />
 +
<br />
 +
<math>L=
 +
\left[ {\begin{matrix}
 +
0 & 1 & 0 & 0 \\
 +
1 & 0 & 0 & 0 \\
 +
0 & 1 & 0 & 1 \\
 +
0 & 0 & 1 & 0 \end{matrix} } \right]\;</math><br />
  
>> L=[0 0 1;1 0 1;0 1 0];
+
'''Matlab Code:'''<br>
 +
<pre style='font-size:16px'>
 +
>> L=L= [0 1 0 0;1 0 0 0;0 1 0 1;0 0 1 0];
 
>> C=sum(L);
 
>> C=sum(L);
 
>> D=diag(C);
 
>> D=diag(C);
 
>> d=0.8;
 
>> d=0.8;
>> N=3;
+
>> N=4;
 
>> A=(1-d)*ones(N)/N+d*L*pinv(D);
 
>> A=(1-d)*ones(N)/N+d*L*pinv(D);
>> [vec val]=eigs(A)
+
>> [vec val]=eigs(A);
 +
>> a=vec(:,1);
 +
>> a=a/sum(a)
 +
    a =
 +
        0.1029 <- Page 1
 +
        0.1324 <- Page 2
 +
        0.3971 <- Page 3
 +
        0.3676 <- Page 4
  
vec =
+
        % Therefore the PageRank for this matrix is: 3,4,2,1
 +
</pre>
 +
<br>
  
  -0.3707            -0.3536 + 0.3536i  -0.3536 - 0.3536i
+
==== Example 5 ====
  -0.6672            -0.3536 - 0.3536i  -0.3536 + 0.3536i
 
  -0.6461            0.7071            0.7071         
 
  
 +
<math>L=
 +
\left[ {\begin{matrix}
 +
0 & 1 & 0 & 1 \\
 +
1 & 0 & 1 & 1 \\
 +
1 & 0 & 0 & 1 \\
 +
1 & 0 & 0 & 0 \end{matrix} } \right]</math>
  
val =
+
<math>c=  
 +
\left[ {\begin{matrix}
 +
3 & 1 & 1 & 3 \end{matrix} } \right]</math>
  
  1.0000                  0                 0        
+
<math>D=
        0           -0.4000 - 0.4000i        0         
+
\left[ {\begin{matrix}
        0                  0           -0.4000 + 0.4000i
+
3 & 0 & 0 & 0 \\
 +
0 & 1 & 0 & 0 \\
 +
0 & 0 & 1 & 0 \\
 +
0 & 0 & 0 & 3 \end{matrix} } \right]</math>
  
>> a=-vec(:,1)
+
<pre style='font-size:14px'>
  
a =
+
Matlab code
 
 
    0.3707
 
    0.6672
 
    0.6461
 
  
 +
>> L= [0 1 0 1; 1 0 1 1; 1 0 0 1;1 0 0 0];
 +
>> d = 0.8;
 +
>> N = 4;
 +
>> C = sum(L);
 +
>> D = diag(C);
 +
>> A=(1-d)*ones(N)/N+d*L*pinv(D);
 +
>> [vec val]=eigs(A);
 +
>> a=vec(:,1);
 
>> a=a/sum(a)
 
>> a=a/sum(a)
  
 
a =
 
a =
  
     0.2201
+
     0.3492
     0.3962
+
     0.3263
     0.3836
+
     0.1813
 +
    0.1431
 
</pre>
 
</pre>
'''NOTE:''' Page 2 is the most important page because it has 2 incomings. Similarly, page 3 is more important than page 1 because page 3 has the incoming result from page 2.
 
 
This example is similar to the first example, but here, page 3 can go back to page 2, so the matrix of the outgoing matrix, the third column of the D matrix is 3 in the third row. And we use the code to calculate the p=Ap.
 
 
==== Example 3 ====
 
 
[[File:eg 3.jpg]]<br>
 
  
 +
==== Example 6 ====
 
<math>L=  
 
<math>L=  
 
\left[ {\begin{matrix}
 
\left[ {\begin{matrix}
0 & 1 & 0 \\
+
0 & 1 & 0 & 0 & 1\\
1 & 0 & 1  \\
+
1 & 0 & 0 & 0 & 0\\
0 & 1 & 0 \end{matrix} } \right]\;
+
0 & 1 & 0 & 0 & 0\\
c=
+
0 & 1 & 1 & 0 & 1\\
\left[ {\begin{matrix}
+
0 & 0 & 0 & 1 & 0 \end{matrix} } \right]</math>
1 & 2 & 1 \end{matrix} } \right]\;
+
<br />
D=
 
\left[ {\begin{matrix}
 
1 & 0 & 0 \\
 
0 & 2 & 0 \\
 
0 & 0 & 1 \end{matrix} } \right]</math>
 
  
<math>d=0.8</math><br>
+
'''Matlab Code:'''<br />
<math>N=3</math><br>
+
<pre style="font-size:16px">
 +
>> 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=(1-d)*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>
  
</span>
+
== Class 17 - Tuesday July 2nd 2013 ==
this example is the second page have 2 outgoings.
+
=== 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_{t-1}</math>.
  
Another Example:
+
For example,
 +
:<math>\displaystyle f(X_1,...X_n)= f(X_1)f(X_2|X_1)f(X_3|X_2)...f(X_n|X_{n-1})</math>
 +
A random Walk is the best example  of a Markov process
  
Consider: 1 -> ,<-2 ->3
+
<br>'''Transition Probability:'''<br>
 +
The probability of going from one state to another state.
 +
:<math>p_{ij} = \Pr(X_{n}=j\mid X_{n-1}= i). \,</math>
  
L= [0 1 0; 1 0 0; 0 1 0]; c=[1,1,1]; D= [1 0 0; 0 1 0; 0 0 1]
+
<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>
  
==== Example 4 ====
+
<a style="color:red" href="http://www.eecs.berkeley.edu/Pubs/TechRpts/2010/EECS-2010-165.pdf">some notes form UCb</a>
  
<math>1 \leftrightarrow 2 \rightarrow 3 \leftrightarrow 4 </math>
+
'''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 non-independent random variables using a Markov Chain. Its Markov properties help to simplify the simulation process.
<br />
 
<br />
 
<br />
 
<math>L=
 
\left[ {\begin{matrix}
 
0 & 1 & 0 & 0 \\
 
1 & 0 & 0 & 0 \\
 
0 & 1 & 0 & 1 \\
 
0 & 0 & 1 & 0 \end{matrix} } \right]\;
 
c=
 
\left[ {\begin{matrix}
 
1 & 2 & 1 & 1 \end{matrix} } \right]\;
 
D=
 
\left[ {\begin{matrix}
 
1 & 0 & 0 & 0 \\
 
0 & 2 & 0 & 0 \\
 
0 & 0 & 1 & 0  \\
 
0 & 0 & 0 & 1 \end{matrix} } \right]</math><br />
 
  
Matlab code
 
<pre style='font-size:14px'>
 
  
>> L=L= [0 1 0 0;1 0 0 0;0 1 0 1;0 0 1 0];
+
<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/>
>> C=sum(L);
 
>> D=diag(C);
 
>> d=0.8;
 
>> N=4;
 
>> A=(1-d)*ones(N)/N+d*L*pinv(D)
 
  
A =
+
Idea: generate a Markov chain whose stationary distribution is the same as target distribution. <br/>
  
    0.0500    0.4500    0.0500    0.0500
 
    0.8500    0.0500    0.0500    0.0500
 
    0.0500    0.4500    0.0500    0.8500
 
    0.0500    0.0500    0.8500    0.0500
 
  
>> [vec val]=eigs(A)
+
'''Notes'''
  
vec =
+
# 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 burn-in 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 acceptance-rejection method. This is because a point is always generated at each time-step 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.
  
    0.1817  -0.0000  -0.4082    0.4082
 
    0.2336    0.0000    0.5774    0.5774
 
    0.7009  -0.7071    0.4082  -0.4082
 
    0.6490    0.7071  -0.5774  -0.5774
 
  
 +
'''Some properties of the stationary distribution <math>\pi</math>'''
  
val =
+
<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>
  
    1.0000        0        0        0
+
# <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.
        0  -0.8000        0        0
+
#<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.
        0        0  -0.5657        0
 
        0        0        0    0.5657
 
  
>> a=vec(:,1)
+
====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 />
  
>> a=vec(:,1)
+
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/>
  
a =
+
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/>
  
    0.1817
+
<math> \mathbf{Q} =
    0.2336
+
\begin{bmatrix}
    0.7009
+
q_{11} & q_{12} & \cdots & q_{1m} \\
    0.6490
+
q_{21} & q_{22} & \cdots & q_{2m} \\
 +
\vdots & \vdots & \ddots & \vdots \\
 +
q_{m1} & q_{m2} & \cdots & q_{mm}
 +
\end{bmatrix}
 +
</math> <br/>
  
>> a=a/sum(a)
 
  
a =
+
We generate Y = j according to the i-th row of Q. Note that the i-th 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 />
  
    0.1029
+
In the following algorithm: <br>
    0.1324
+
<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/>
    0.3971
+
<math>r_{ij}</math> is the probability of accepting Y as <math>x_{t+1}</math>. <br/>
    0.3676
 
</pre>
 
'''NOTE:''' The ranking of each page is as follows: Page 3, Page 4, Page 2 and Page 1. Page 3 is the highest since it has the most incoming links. All of the other pages only have one incoming link but since Page 3, highest ranked page, links to Page 4, Page 4 is the second highest ranked. Lastly, since Page 2 links into Page 3 it is the next highest rank.
 
  
Page 2 has 2 outgoing links. Pages with the same incoming links can be ranked closest to the highest ranked page. If the highest page P1 is incoming into a page P2,  then P2 is ranked second, and so on.
 
  
==== Example 5 ====
+
'''How to get the acceptance probability?'''
  
<math>L=
+
If <math>\pi </math> is the stationary distribution, then it must satisfy the detailed balance condition:<br/>
\left[ {\begin{matrix}
+
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
0 & 1 & 0 & 1 \\
 
1 & 0 & 1 & 1 \\
 
1 & 0 & 0 & 1 \\
 
1 & 0 & 0 & 0 \end{matrix} } \right]</math>
 
  
<math>c=  
+
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/>
\left[ {\begin{matrix}
+
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/>  
3 & 1 & 1 & 3 \end{matrix} } \right]</math>
 
  
<math>D=
+
'''Recall'''
\left[ {\begin{matrix}
+
<math>r_{ij}</math> is the probability of acceptance, thus it must be that <br/>
3 & 0 & 0 & 0 \\
 
0 & 1 & 0 & 0 \\
 
0 & 0 & 1 & 0  \\
 
0 & 0 & 0 & 3 \end{matrix} } \right]</math>
 
  
<pre style='font-size:14px'>
+
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>
  
Matlab code
+
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>
  
>> L= [0 1 0 1; 1 0 1 1; 1 0 0 1;1 0 0 0];
+
So we choose a(i,j) as large as possible, but it needs to satisfy the two conditions above.<br/>
>> d = 0.8;
 
>> N = 4;
 
>> C = sum(L);
 
>> D = diag(C);
 
>> A=(1-d)*ones(N)/N+d*L*pinv(D);
 
>> [vec val]=eigs(A);
 
>> a=vec(:,1);
 
>> a=a/sum(a)
 
  
a =
+
<math>a(i,j) = \min \{\frac{1}{\pi_j q_{ji}},\frac{1}{\pi_i q_{ij}}\} </math><br/>
  
    0.3492
+
Thus, <math> r_{ij} = \min \{\frac{\pi_j q_{ji}}{\pi_i q_{ij}}, 1\} </math><br/>
    0.3263
 
    0.1813
 
    0.1431
 
</pre>
 
  
==== Example 6 ====
+
'''Note''':
<math>L=
+
1 is the upper bound to make r<sub>ij</sub> a probability
\left[ {\begin{matrix}
 
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 \end{matrix} } \right]</math>
 
<br />
 
  
Matlab Code<br />
 
<pre style="font-size:16px">
 
>> d=0.8
 
  
d =
+
'''Algorithm:'''  <br/>
 +
*<math> (*) P(Y=j) = q_{ij} </math>. <math>\frac{\pi_j q_{ji}}{\pi_i q_{ij}}</math> is a positive ratio.
  
    0.8000
+
*<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/>
  
>> 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]
+
We can compare this with the Acceptance-Rejection 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 time-step. <br>
  
L =
+
The algorithm generates a stochastic sequence that only depends on the last state, which is a Markov Chain.<br>
  
    0    1    0    0    1
+
====Metropolis Algorithm====
    1    0    0    0    0
 
    0    1    0    0    0
 
    0    1    1    0    1
 
    0    0    0    1    0
 
  
>> c=sum(L)
+
'''Proposition: ''' Metropolis works:
  
c =
+
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/>
  
    1    3    1    1    2
 
  
>> D=diag(c)
+
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>
  
D =
+
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 />
  
    1    0    0    0    0
+
<math>q_{ij}</math> is the probability of generating state j; <br/>
    0    3    0    0    0
+
<math> r_{ij}</math> is the probability of accepting state j as the next state. <br/>
    0    0    1    0    0
 
    0    0    0    1    0
 
    0    0    0    0    2
 
  
>> N=5
+
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.
  
N =
+
===Proof of the proposition:===
  
    5
+
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="text-size: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>
  
>> A=(1-d)*ones(N)/N+d*L*pinv(D)
+
'''Proof:'''
  
A =
+
WLOG, we can assume that <math>\frac{\pi_j q_{ji}}{\pi_i q_{ij}}<1</math><br/>
  
    0.0400    0.3067    0.0400    0.0400    0.4400
+
LHS:<br />
    0.8400    0.0400    0.0400    0.0400    0.0400
+
<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 />
    0.0400    0.3067    0.0400    0.0400    0.0400
 
    0.0400    0.3067    0.8400    0.0400    0.4400
 
    0.0400    0.0400    0.0400    0.8400    0.0400
 
  
>> [vec val]=eigs(A)
+
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 />
  
vec =
+
Hence LHS=RHS
  
  Columns 1 through 4
+
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/>
  
  -0.4129            0.4845 + 0.1032i  0.4845 - 0.1032i  -0.0089 + 0.2973i
+
LHS:<br />
  -0.4158            -0.6586            -0.6586            -0.5005 + 0.2232i
+
<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 />
  -0.1963            0.2854 - 0.0608i  0.2854 + 0.0608i  -0.2570 - 0.2173i
 
  -0.5700            0.1302 + 0.2612i  0.1302 - 0.2612i  0.1462 - 0.3032i
 
  -0.5415            -0.2416 - 0.3036i -0.2416 + 0.3036i  0.6202         
 
  
  Column 5
+
RHS:<br />
 +
'''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 />
  
  -0.0089 - 0.2973i
+
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 />
  -0.5005 - 0.2232i
 
  -0.2570 + 0.2173i
 
  0.1462 + 0.3032i
 
  0.6202         
 
  
 +
Hence LHS=RHS which indicates <math>pi_i*P_{ij} = pi_j*P_{ji}</math><math>\square</math><br /><br />
  
val =
+
'''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 />
  
  Columns 1 through 4
+
2) If <math>\displaystyle i = j</math>, then detailed balance is satisfied trivially.<br />
  
  1.0000                  0                  0                  0         
+
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>.
        0            -0.5886 - 0.1253i        0                  0         
 
        0                  0            -0.5886 + 0.1253i        0         
 
        0                  0                  0            0.1886 - 0.3911i
 
        0                  0                  0                  0         
 
  
  Column 5
+
== Class 18 - Thursday July 4th 2013 ==
 +
=== Last class ===
  
        0         
+
Recall: The Acceptance Probability,
        0         
+
<math>r_{ij}=min(\frac {{\pi_j}q_{ji}}{{\pi_i}q_{ij}},1)</math> <br />
        0         
 
        0         
 
  0.1886 + 0.3911i
 
  
>> a=-vec(:,1)
+
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 />
  
a =
 
  
     0.4129
+
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 />
    0.4158
 
    0.1963
 
    0.5700
 
    0.5415
 
  
>> a=a/sum(a)
+
===Example: Discrete Case===
  
a =
 
  
    0.1933
+
Consider a biased die,
    0.1946
+
<math>\pi</math>= [0.1, 0.1, 0.2, 0.4, 0.1, 0.1]
    0.0919
 
    0.2668 % (the most important)
 
    0.2534
 
</pre>
 
For the matrix, the rank is: page 4, page 5, page 2, page 1, page 3.<br />
 
  
== Class 17 - Tuesday July 2nd 2013 ==
+
We could use any <math>6 x 6 </math> matrix <math> \mathbf{Q} </math> as the proposal distribution <br>
=== Markov Chain Monte Carlo (MCMC) ===
+
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.
Idea: generate a Markov Chain whose stationary distribution is the same as the target distribution.
 
 
 
Motivation example<br />
 
-We would like to sample from random variable X according to distribution <math>\pi=(\pi_1, \pi_2,  ...  , \pi_m)</math> <br/>
 
X can take m different values from <math>X \in [1,2,3,...,m]</math> <br />
 
-Suppose X can take values from: {Xt, t=0, 1, ....., n } according to <math>\pi</math><br />
 
{X0, X1, ....., Xn}<br />
 
 
 
 
 
example:
 
M=6<br/>
 
<math> \pi=[0.1,0.1,0.1,0.2,0.3,0.2]</math><br/>
 
<math>X \in [1,2,3,4,5,6]</math><br/>
 
 
 
- Suppose Xt=i<br />
 
consider an arbitrary transition matrix Q with entry q<sub>ij</sub>. <br/>
 
  
 
<math> \mathbf{Q} =  
 
<math> \mathbf{Q} =  
 
  \begin{bmatrix}
 
  \begin{bmatrix}
  q_{11} & q_{12} & \cdots & q_{1m} \\
+
  1/6 & 1/6 & \cdots & 1/6 \\
  q_{21} & q_{22} & \cdots & q_{2m} \\
+
  1/6 & 1/6 & \cdots & 1/6 \\
 
  \vdots & \vdots & \ddots & \vdots \\
 
  \vdots & \vdots & \ddots & \vdots \\
  q_{m1} & q_{m2} & \cdots & q_{mm}
+
  1/6 & 1/6 & \cdots & 1/6
 
  \end{bmatrix}
 
  \end{bmatrix}
 
</math> <br/>
 
</math> <br/>
  
q<sub>ij</sub> is the probability of moving to any state j from current state i. <br/>
 
Generate Y according to i-th row of matrix Q.
 
i.e.P(Y=j)=q<sub>ij</sub><br />
 
r<sub>ij</sub> is the probabiliy of accepting j. <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>
  
'''Algorithm:'''  <br/>
+
Notice how a point is always generated for X<sub>t+1</sub>, regardless of whether the candidate state Y is accepted <br>
*<math> (*) P(Y=j) = q_{ij} </math> <br/>
 
*<math> r_{ij} = min (\frac{\pi_j q_{ji}}{\pi_i q_{ij}}, 1) </math>,  Notice that the reason why we take minimum is because we know that <math>\frac{\pi_j q_{ji}}{\pi_i q_{ij}} >0</math>, and may greater than 1. Since r is a probability, it is no more than 1, so we take the minimum.<br/>
 
* <math>
 
x_{t+1} = \begin{cases}
 
Y, & \text{w/p}:  r_{ij} \\
 
x_t, & \text{otherwise} \end{cases} </math> <br/>
 
*go to (*)  <br/>
 
 
 
This algorithm generates <math> {x_t, t=0,...,n} </math>. In the long run, the marginal distribution of <math> x_t </math> is <math>\underline{\Pi} </math><br/>
 
<math> {{x_t, t = 0, 1,...,n}}</math> is a Markov chain with probability transition matrix P.
 
This is a Markov Chain since <math> x_t+1 </math> only depends on
 
<math> x_t
 
\text{Where }P_{ij}= \begin{cases}
 
q_{ij} r_{ij}, & \text{if }i \neq j \\
 
1 - \sum_{j} (q_{ij} r_{ij}), & \text{if i = j} \end{cases} </math>
 
  
Detailed Balance:
+
'''Matlab'''
 +
<pre style="font-size:14px">
 +
pii=[.1,.1,.2,.4,.1,.1];
 +
x(1)=5;
 +
for ii=2:1000
 +
  Y=unidrnd(6);                %%% Unidrnd(x) is a built-in function which generates a number between (0) and (x)
 +
  r = min (pii(Y)/pii(x(ii-1)), 1);
 +
  u=rand;
 +
  if u<r
 +
    x(ii)=Y;
 +
  else
 +
    x(ii)=x(ii-1);
 +
  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.jpg|300px]]
  
if <math>\pi_i P_{ij} = P_{ji} \pi_j</math>, then <math>\underline\pi</math> is the stationary distribution of this markov chain.
 
  
LHS:
+
'''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 [[burn-in period]]. A chain will converge to the limiting distribution eventually, but not immediately. The burn-in 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 burn-in period is over, we say that the chain "mixes well".
<math>\pi_i P_{ij} = \pi_i q_{ij} r_{ij} = \pi_i q_{ij} \frac{\pi_j q_{ij}}{\pi_i q_{ij}} = \pi_i q_{ji} </math> <br/>
 
RHS:
 
<math>\ P_{ji} \pi_j=q_{ji} r_{ji} \pi_i </math> <br/>
 
note: we assume <math>r_{ij}</math> is smaller than 1, then <math>r_{ji}</math> should be equal to 1.
 
  
<math>q_{ij}</math> is the chance of generating j from i, but may not be accepted, so we consider the chance of accepting j as the next step, which is <math>r_{ij}</math>.
+
===Alternate Example: Discrete Case===
If P<sub>ij</sub> is not zero for any i,j, then the chain is ergodic.
 
  
'''Example''':
+
 
\pi=[0.1 0.1 0.2 0.2 0.2 0.1]<br/>
+
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} =  
 
<math> \mathbf{Q} =  
 
  \begin{bmatrix}
 
  \begin{bmatrix}
  1/6 & 1/6 & 1/6 & 1/6 & 1/6 & 1/6 \\
+
  5/7 & 2/7 \\
1/6 & 1/6 & 1/6 & 1/6 & 1/6 & 1/6 \\
+
  3/8 & 5/8\\
1/6 & 1/6 & 1/6 & 1/6 & 1/6 & 1/6 \\
+
   
  1/6 & 1/6 & 1/6 & 1/6 & 1/6 & 1/6 \\
 
  1/6 & 1/6 & 1/6 & 1/6 & 1/6 & 1/6 \\
 
1/6 & 1/6 & 1/6 & 1/6 & 1/6 & 1/6 \\
 
 
  \end{bmatrix}
 
  \end{bmatrix}
 
</math> <br/>
 
</math> <br/>
  
Y~unif[1,2,3,4,5,6]<br/>
 
Suppose that the current state of i is X_t</math> = 4<br/>
 
'''Algorithm''':<br/>
 
1. Initialize the chain, X_t</math> = 4<br/>
 
2. Draw Y~unif[1,2,3,4,5,6]<br/>
 
<math> r_{ij} = min (\frac{\pi_j q_{ji}}{\pi_i q_{ij}}, 1) = min (\frac{\pi_j}{\pi_i}, 1) </math> </br>
 
3. Sample u~unif(0,1)<br/>
 
4. If u <= r_{ij}, x_{t+1} = y <br/>
 
5. Else x_{t+1} = x_{t} <br/>
 
6. Go back to 2 <br/>
 
  
== Class 17 - Tuesday July 2nd 2013 ==
 
=== Markov Chain Monte Carlo (MCMC) ===
 
  
===Introduction===
+
'''Algorithm''' <br>
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 important in applications.
+
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 />
(referenced by Sheldon M.Ross,Simulation)
+
2. Sample from proposal distribution: Y~q(y|x) = 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>
  
====Definition:====
+
4. U~Unif(0,1)<br>
Markov Chain
+
  If  <math>U \leq r_{ij}</math>, then<br>
A Markov Chain is a special form of stochastic process in which <math>\displaystyle X_t</math> depends only on <math> \displaystyle X_{t-1}</math>.
+
        <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>
  
For example,
 
:<math>\displaystyle f(X_1,...X_n)= f(X_1)f(X_2|X_1)f(X_3|X_2)...f(X_n|X_{n-1})</math>
 
A random Walk is the best example  of a Markov process
 
  
<br>'''Transition Probability:'''<br>
+
'''Generalization of the above framework to the continuous case'''<br>
The probability of going from one state to another state.
 
:<math>p_{ij} = \Pr(X_{n}=j\mid X_{n-1}= i). \,</math>
 
  
<br>'''Transition Matrix:'''<br>
+
In place of <math>\pi</math> use <math>f(x)</math>
For n states, transition matrix P is an <math>N \times N</math> matrix with entries <math>\displaystyle P_{ij}</math> as below:
+
In place of r<sub>ij</sub> use <math>q(y|x)</math> <br>
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>
+
In place of r<sub>ij</sub> use <math>r(x,y)</math> <br>
 +
Here, q(y|x) is a friendly distribution that is easy to sample, usually a symmetric distribution will be preferable, such that <math>q(y|x) = q(x|y)</math> to simplify the computation for <math>r(x,y)</math>.
  
<a style="color:red" href="http://www.eecs.berkeley.edu/Pubs/TechRpts/2010/EECS-2010-165.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 non-independent random variables using a Markov Chain. Its Markov properties help to simplify the simulation process.
+
'''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>
  
<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/>  
+
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>
  
Idea: generate a Markov chain whose stationary distribution is the same as target distribution. <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>
  
'''Note''' <br/>
 
1) 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 burn-in period is longer if the chain is initialized with a value of low probability density<br/>
 
  
2) 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). <br>
 
  
3) 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.<br/>
+
'''Purpose''': <br>
 +
"The purpose of the Metropolis-Hastings 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>
  
4) MCMC methods do not suffer as badly from the "curse of dimensionality" that badly affects efficiency in the acceptance-rejection method. This is because a point is always generated at each time-step according to the Markov Chain regardless of how many dimensions are introduced.<br>
 
  
5) The goal when simulating with a Markov Chain is to create a chain with the same stationary distribution as the target distribution.<br/>
+
Metropolis-Hastings is an algorithm for constructing a Markov chain with a given limiting probability distribution. In particular, we consider what happens if we apply the Metropolis-Hastings algorithm repeatedly to a “proposal” distribution which has already been updated.<br>
  
6) The MCMC method is usually used in continuous cases but a discrete example is given below.<br />
 
  
 +
The algorithm was named after Nicholas Metropolis and W. K. Hastings who extended it to the more general case in 1970.<br>
  
'''Some properties of the stationary distribution <math>\pi</math>'''
+
<math>q(y|x)</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>\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>
+
Note that the Metropolis-Hasting 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
  
1. <math>\pi_j</math> = <math>\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. <br>
 
  
2)<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. <br>
 
  
====Motivation example====
+
'''Differences between the discrete and continuous case of the Markov Chain''':<br/>
- 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/>
+
1. <math>q(y|x)</math> is used in continuous, instead of <math>q_{ij}</math> in discrete <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/>
+
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/>
  
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} =
+
'''Build the Acceptance Ratio'''<br/>
\begin{bmatrix}
+
Before we consider the algorithm there are a couple general steps to follow to build the acceptance ratio:<br/>
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/>
 
  
 +
a) Find the distribution you wish to use to generate samples from<br/>
 +
b) Find a candidate distribution that fits the desired distribution, q(y|x). (the proposed moves are independent of the current state)<br/>
 +
c) Build the acceptance ratio <math>\displaystyle \frac{f(y)q(x|y)}{f(x)q(y|x)}</math>
  
We generate Y = j according to the i-th row of Q. Note that the i-th 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/>
 
  
 +
Assume that f(y) is the target distribution; Choose q(y|x) such that it is a friendly distribution and easy to sample from.<br />
 +
'''Algorithm:'''<br />
  
'''How to get the acceptance probability?'''
+
# 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(y|x)</math>
 +
# Set <math>\displaystyle r(x,y) = min(\frac{f(y)q(x|y)}{f(x)q(y|x)},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>
  
If <math>\pi </math> is the stationary distribution, then it must satisfy the detailed balance condition:<br/>
+
<br> '''Note''': q(x|y) is moving from y to x and q(y|x) is moving from x to y.
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
+
<br>We choose q(y|x) so that it is simple to sample from.
 +
<br>Usually, we choose a normal distribution.
  
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/>
+
NOTE2: The proposal q(y|x) 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(y|x) ~ N( x, b<sup>2</sup>) here the proposal depends on x.  
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/>
+
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. )
  
'''Recall'''
+
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.
<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>
+
<br />
 +
Comparing with previous sampling methods we have learned, samples generated from M-H 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 Metropolis-Hastings. 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/>
  
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>
+
<p style="font-size:20px;color:red;">
 +
Remarks
 +
</p>
 +
===='''Remark 1'''====
 +
<span style="text-shadow: 0px 2px 3px 3399CC;margin-right:1em;font-family: 'Nobile', Helvetica, Arial, sans-serif;font-size:16px;line-height:25px;color:3399CC">
 +
A common choice for <math>q(y|x)</math> is a normal distribution centered at x with standard deviation b. Y~<math>N(x,b^2)</math>
  
So we choose a(i,j) as large as possible, but it needs to satisfy the two conditions above.<br/>
+
In this case, <math> q(y|x)</math> is symmetric.
  
<math>a(i,j) = \min \{\frac{1}{\pi_j q_{ji}},\frac{1}{\pi_i q_{ij}}\} </math><br/>
+
i.e.
 +
<math>q(y|x)=q(x|y)</math><br>
 +
(we want to sample q centered at the current state.)<br>
 +
<math>q(y|x)=\frac{1}{\sqrt{2\pi}b}\,e^{- \frac{\scriptscriptstyle 1}{\scriptscriptstyle 2b^2} (y-x)^2}</math>, (centered at x)<br>
 +
<math>q(x|y)=\frac{1}{\sqrt{2\pi}b}\,e^{- \frac{\scriptscriptstyle 1}{\scriptscriptstyle 2b^2} (x-y)^2}</math>,(centered at y)<br>
 +
<math>\Rightarrow (y-x)^2=(x-y)^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(y|x) 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>
  
Thus, <math> r_{ij} = \min \{\frac{\pi_j q_{ji}}{\pi_i q_{ij}}, 1\} </math><br/>
+
===='''Remark 2'''====
 
+
<span style="text-shadow: 0px 2px 3px 3399CC;margin-right:1em;font-family: 'Nobile', Helvetica, Arial, sans-serif;font-size:16px;line-height:25px;color:3399CC">
'''Note''':
+
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/>
1 is the upper bound to make r<sub>ij</sub> a probability
+
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>
  
'''Algorithm:''' <br/>
+
===='''Remark 3'''====
*<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/>
+
One strength of the Metropolis-Hastings 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>
*<math>
+
(also notice that Metropolis Hastings is just a special case of Metropolis algorithm)
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/>
 
  
We can compare this with the Acceptance-Rejection model we learned before. <br/>
+
<math>
* <math>U</math> ~ <math>Uniform(0,1)</math> <br/>
+
\begin{align}
* If <math>U < r_{ij}</math>, then accept. <br/>
+
f(x;\alpha,\beta)& = \frac{1}{\mathrm{B}(\alpha,\beta)}\, x^{\alpha-1}(1-x)^{\beta-1}\end{align}
EXCEPT that a point is always generated at each time-step. <br>
+
</math>
  
The algorithm generates a stochastic sequence that only depends on the last state, which is a Markov Chain.<br>
+
The beta function, ''B'', appears as a normalizing constant but it can be simplified by construction of the method.
  
====Metropolis Algorithm====
+
====='''Example'''=====
  
'''Proposition: ''' Metropolis works:
+
<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(x|y)=\frac{1}{\sqrt{2\pi}b}e^{-\frac{1}{2b^{2}}(y-x)^{2}}</math>.<br>
 +
Then <math>\,q(x|y)</math> is symmetric since <math>\,(y-x)^{2} = (x-y)^{2}</math>.<br>
 +
Therefore Y can be simplified.
  
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/>
 
  
 +
We get :
  
This algorithm generates <math>\{x_t:  t=0,...,m\}</math>. <br/>
+
<math>\,\begin{align}
In the long run, the marginal distribution of <math> x_t </math> is the stationary distribution <math>\underline{\Pi} </math><br>
+
\displaystyle r(x,y)
<math>\{x_t: t = 0, 1,...,m\}</math> is a Markov chain with probability transition matrix (PTM), P.<br>
+
& =min\left\{\frac{f(y)}{f(x)}\frac{q(x|y)}{q(y|x)},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>.
  
This is a Markov Chain since <math> x_{t+1} </math> only depends on <math> x_t </math>, where <br>
+
<br/>
<math> P_{ij}= \begin{cases}
+
<math>\pi=[0.1\,0.1\,...] </math> stands for probility;<br/>
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]
+
<math>\pi \propto [3\,2\, 10\, 100\, 1.5] </math> is not brobility, so we take:<br/>
1 - \displaystyle\sum_{k \neq i} q_{ik} r_{ik}, & \text{if }i = j \end{cases} </math><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/>
  
<math>q_{ij}</math> is the probability of generating state j; <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 Metropolis-Hasting algorithm, it is not necessary to do this.  
<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/>
+
<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.
+
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>
  
===Proof of the proposition:===
+
======'''MATLAB'''======
 +
The Matlab code of the algorithm is the following :
 +
<pre style="font-size:12px">
 +
clear all
 +
close all
 +
clc
 +
b=2;
 +
x(1)=0;
 +
for i=2:10000
 +
    y=b*randn+x(i-1);
 +
    r=min((1+x(i-1)^2)/(1+y^2),1);
 +
    u=rand;
 +
    if u<r
 +
        x(i)=y;
 +
    else
 +
        x(i)=x(i-1);
 +
    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.jpg|300px]]
  
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.
+
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>
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="text-size: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:'''
+
'''MATLAB'''
 
+
<pre style="font-size:16px">
WLOG, we can assume that <math>\frac{\pi_j q_{ji}}{\pi_i q_{ij}}<1</math><br/>
+
b=2;
 +
x(1)=0;
 +
for ii=2:10500
 +
y=b*randn+x(ii-1);
 +
r=min((1+x(ii-1)^2)/(1+y^2),1);
 +
u=rand;
 +
if u<=r
 +
x(ii)=y;
 +
else
 +
x(ii)=x(ii-1);
 +
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.jpg|300px]]
 +
<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{-(y-x)^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>
  
LHS:<br />
+
Example:<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 />
+
We want to sample from <math>exp(2), q(y|x)~\sim~N(x,b^2)</math><br>
 +
<math>r=\frac{f(y)}{f(x)}=\frac{2*exp^(-2y)}{2*exp^(-2x)}=exp(2*(x-y))</math><br>
 +
<math>r=min(exp(2*(x-y)),1)</math><br>
  
RHS:<br />
+
'''MATLAB'''
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 />
+
<pre style="font-size:16px">
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 />
+
x(1)=0;
 +
for ii=2:100
 +
y=2*(randn*b+abs(x(ii-1)))
 +
r=min(exp(2*(x-y)),1);
 +
u=rand;
 +
if u<=r
 +
x(ii)=y;
 +
else
 +
x(ii)=x(ii-1);
 +
end
 +
end
 +
</pre>
 +
<br>
  
Hence LHS=RHS
+
'''Definition of Burn in:'''
  
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/>
+
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 burn-in. Theoretical calculation of the burn-in is rather difficult, in the above mentioned example, we chose 500 based on experience and quite arbitrarily. 
  
LHS:<br />
+
Burn-in 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 burn-in period because the chain is not stabilized yet.
<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 />
+
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 ''burn-in''. The remaining
'''Note''' <br/>
+
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/>
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 />
+
 
 +
Burn-in 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 “burn-in” period.
 +
One common suggestion is to match the first few moments of q(y|x) to f(x).
  
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 />
+
'''Aside''': The algorithm works best if the candidate density q(y|x) 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 burn-in period. <br/>
  
Hence LHS=RHS <math>\square</math><br /><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)).
  
'''Note'''<br />
+
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)).
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>.
 
  
== Class 18 - Thursday July 4th 2013 ==
+
'''Note''':
=== Last class ===
+
The histogram looks much nicer if we reject the points within the burning time.<br>
  
Recall : The Acceptance Probability <math>r_{ij}</math>
 
<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 />
+
Example: Use M-H method to generate sample from f(x)=2x
 +
0<x<1, 0 otherwise.
  
 +
1) Initialize the chain with <math>x_i</math> and set <math>i=0</math>
  
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 />
+
2)<math>Y~\sim~q(y|x_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>
  
===Example: Discrete Case===
+
3)consider <math>\frac{f(y)}{f(x)}=\frac{y}{x}</math>,
 +
<math>r(x,y)=min (\frac{y}{x},1)</math> since q(y|x<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
  
Consider a biased die
+
5)<math>i=i+1</math>, go to 2
<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>
+
<br>
For the sake of simplicity ,using a discrete uniform distribution is the simplest.
 
  
<math> \mathbf{Q} =
+
Example form wikipedia
\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/>
 
  
 +
===Step-by-step 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
  
'''Algorithm''' <br>
+
:<math>
1. <math>x_t=5</math> (sample from the 5th row, although we can initialize the chain from anywhere within the support)<br />
+
a = a_1 a_2\,
2. Y~Unif[1,2,...,6]<br />
+
</math>
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>
 
  
4. U~Unif(0,1)<br />
+
where
  if <math>u \leq r_{ij}</math>,<br />X<sub>t+1</sub>=Y<br />
 
  else<br />
 
  X<sub>t+1</sub>=X<sub>t</sub><br />
 
  end if<br />
 
  go 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>
+
:<math>
 +
a_1 = \frac{P(x')}{P(x_t)} \,\!
 +
</math>
  
'''Matlab'''
+
is the likelihood ratio between the proposed sample <math>x'\,</math> and the previous sample <math>x_t\,</math>, and
<pre style="font-size:14px">
 
pii=[.1,.1,.2,.4,.1,.1];
 
x(1)=5;
 
for ii=2:1000
 
  Y=unidrnd(6);                %%% Unidrnd(x) is a built-in function which generates a number between (0) and (x)
 
  r = min (pii(Y)/pii(x(ii-1)), 1);
 
  u=rand;
 
  if u<r
 
    x(ii)=Y;
 
  else
 
    x(ii)=x(ii-1);
 
  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.jpg|300px]]
 
  
 +
:<math>
 +
a_2 = \frac{Q(x_t \mid x')}{Q(x'\mid x_t)}
 +
</math>
  
'''NOTE:''' Generally we generate a large number of points (say, 1500) and throw away the first points (say, 500). Those first points are called the [[burn-in period]]. Since the chain is said to converge in the long run, the burn-in period is where the chain is converging toward the limiting distribution, but has not converged yet; by discarding those 500 points, our data set will be more representative of the desired limiting distribution, once the burn-in period is over, we say that the chain "mixes well".
+
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.
  
===Alternate Example: Discrete Case===
+
:<math>
 
+
\begin{matrix}
 
+
\mbox{If } a \geq 1: &  \\
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.
+
& x_{t+1} = x',
<math>\pi</math>= [pi1 pi2]
+
\end{matrix}
 
+
</math>
Use a discrete uniform distribution as the proposal distribution, because it is the simplest.
+
:<math>
 +
\begin{matrix}
 +
\mbox{else} & \\
 +
& x_{t+1} = \left\{
 +
                  \begin{array}{lr}
 +
                      x' & \mbox{ with probability }a \\
 +
                      x_t & \mbox{ with probability }1-a.
 +
                  \end{array}
 +
            \right.
 +
\end{matrix}
 +
</math>
  
<math> \mathbf{Q} =
+
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".  
\begin{bmatrix}
+
These samples, which are discarded, are known as ''burn-in''. The remaining set of accepted values of <math>x</math> represent a sample from the distribution <math>P(x)</math>.
5/7 & 2/7 \\
 
1/8 & 5/8\\
 
   
 
\end{bmatrix}
 
</math> <br/>
 
  
 +
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 burn-in 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.
  
'''Algorithm''' <br>
+
== Class 19 - Tuesday July 9th 2013 ==
1. <math>x_t=1</math> (sample from the 1st row, although we could also choose the second)<br />
+
'''Recall: Metropolis–Hasting Algorithm'''
2. Y~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>. <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>
 
  
4. U~Unif(0,1)<br />
+
1) <math>X_i</math> = State of chain at time i. Set <math>X_0</math> = 0<br>
   if <math>u \leq r_{ij}</math>,<br />X<sub>t+1</sub>=Y<br />
+
2) Generate proposal distribution: Y ~ q(y|x) <br>
   else<br />
+
3) Set <math>\,r=min[\frac{f(y)}{f(x)}\,\frac{q(x|y)}{q(y|x)}\,,1]</math><br>
  X<sub>t+1</sub>=X<sub>t</sub><br />
+
4) Generate U ~ U(0,1)<br>
   end if<br />
+
   If <math>U<r</math>, then<br>
  go to (2)<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>
  
  
'''Generalization of the above framework to the continuous case'''<br>
+
Why can we use this algorithm to generate a Markov Chain?<br>
  
In place of <math>\pi</math> use <math>f(x)</math>
+
<math>\,Y</math>~<math>\,q(y|x)</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>t-1</sub>; the Markov Property is satisfied as long as Y is not dependent on  X<sub>0</sub>, X<sub>1</sub>,..., X<sub>t-2</sub>. Thus, time t will not affect the choice of state.<br>
In place of r<sub>ij</sub> use <math>q(y|x)</math> <br>
 
In place of r<sub>ij</sub> use <math>r(x,y)</math> <br>
 
Here, q(y|x) is a friendly distribution that is easy to sample, usually a symmetric distribution will be preferable, such that <math>q(y|x) = q(x|y)</math> to simplify the computation for <math>r(x,y)</math>.
 
  
  
'''Remarks'''<br>
+
==='''Choosing b: 3 cases'''===
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>
+
If y and x have the same domain, say R, we could use normal distribution to model <math>q(y|x)</math>. <math>q(x|y)~normal(y,b^2), and q(y|x)~normal(x,b^2)</math>.
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>
+
In the continuous case of MCMC, <math>q(y|x)</math> is the probability of observing y, given you are observing x. We normally assume <math>q(y|x)</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.  
  
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>
+
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>
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>
+
In this example, <math>q(y|x)=N(x, b^2)</math><br>
  
===Metropolis–Hasting Algorithm===
+
Demonstrated as follows, the choice of b will be significant in determining the quality of the Metropolis algorithm. <br>
  
'''Definition''': <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>
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. <br>
 
  
 +
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>
  
'''Purpose''': <br>
+
'''MATLAB b=2, b= 0.2, b=20 '''
"The purpose of the Metropolis-Hastings 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>
+
<pre style="font-size:12px">
Source:(http://en.wikipedia.org/wiki/Metropolis%E2%80%93Hastings_algorithm)<br>
+
clear all
 +
close all
 +
clc
 +
b=2 % b=0.2 b=20;
 +
x(1)=0;
 +
for i=2:10000
 +
    y=b*randn+x(i-1);
 +
    r=min((1+x(i-1)^2)/(1+y^2),1);
 +
    u=rand;
 +
    if u<r
 +
        x(i)=y;
 +
    else
 +
        x(i)=x(i-1);
 +
    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.
  
Metropolis-Hastings is an algorithm for constructing a Markov chain with a given limiting probability distribution. In particular, we consider what happens if we apply the Metropolis-Hastings algorithm repeatedly to a “proposal” distribution which has already been updated.<br>
+
f(x) is the stationary distribution list of the chain in MH. We generating y using q(y|x) 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.
  
The algorithm was named after Nicholas Metropolis and W. K. Hastings who extended it to the more general case in 1970.<br>
+
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.
  
 +
==='''Detailed Balance Holds for Metropolis-Hasting'''===
  
'''Differences between the discrete and continuous case of the Markov Chain''':<br/>
+
In metropolis-hasting, we generate y using q(y|x) and accept it with probability r, where <br>
  
1. <math>q(y|x)</math> is used in continuous, instead of <math>q_{ij}</math> in discrete <br/>
+
<math>r(x,y) = min\left\{\frac{f(y)}{f(x)}\frac{q(x|y)}{q(y|x)},1\right\} = min\left\{\frac{f(y)}{f(x)},1\right\}</math><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/>
 
  
 +
Without loss of generality we assume <math>\frac{f(y)}{f(x)}\frac{q(x|y)}{q(y|x)} > 1</math><br>
  
'''Build the Acceptance Ratio'''<br/>
+
Then r(x,y) (probability of accepting y given we are currently in x) is <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/>
+
<math>r(x,y) = min\left\{\frac{f(y)}{f(x)}\frac{q(x|y)}{q(y|x)},1\right\} = \frac{f(y)}{f(x)}\frac{q(x|y)}{q(y|x)}</math><br>
b) Find a candidate distribution that fits the desired distribution, q(y|x). (the proposed moves are independent of the current state)<br/>
 
c) Build the acceptance ratio <math>\displaystyle \frac{f(y)q(x|y)}{f(x)q(y|x)}</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(y|x)}{q(x|y)},1\right\} = 1 </math><br>
  
Assume that f(y) is the target distribution; Choose q(y|x) such that it is a friendly distribution and easy to sample from.<br />
+
This is because <math>\frac{f(y)}{f(x)}\frac{q(x|y)}{q(y|x)} < 1 </math> and its reverse <math>\frac{f(x)}{f(y)}\frac{q(y|x)}{q(x|y)} > 1 </math>. Then <math>r(x,y) = 1</math>.<br>
'''Algorithm:'''<br />
+
We are interested in the probability of moving from from x to y in the Markov Chain generated by MH algorithm: <br>
 +
P(y|x) depends on two probabilities:
 +
1. Probability of generating y, and<br>
 +
2. Probability of accepting y. <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.
+
<math>P(y|x) = q(y|x)*r(x,y) = q(y|x)*{\frac{f(y)}{f(x)}\frac{q(x|y)}{q(y|x)}} = \frac{f(y)*q(x|y)}{f(x)} </math> <br>
# Sample <math>\displaystyle Y \sim q(y|x)</math>
 
# Set <math>\displaystyle r(x,y) = min(\frac{f(y)q(x|y)}{f(x)q(y|x)},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(x|y) is moving from y to x and q(y|x) is moving from x to y.
+
The probability of moving to x given the current state is y:
<br>We choose q(y|x) so that it is simple to sample from.
 
<br>Usually, we choose a normal distribution.
 
  
NOTE2: The proposal q(y|x) 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(y|x) ~ N( x, b<sup>2</sup>) here the proposal depends on x.
+
<math>P(x|y) = q(x|y)*r(y,x) = q(x|y)</math><br>
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. )
 
  
<br />
+
So does detailed balance hold for MH? <br>
Comparing with previous sampling methods we have learned, samples generated from M-H 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 Metropolis-Hastings. 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="font-size:20px;color:red;">
+
If it holds we should have <math>f(x)*P(y|x) = f(y)*P(x|y)</math>.<br>
Remarks
 
</p>
 
===='''Remark 1'''====
 
<span style="text-shadow: 0px 2px 3px 3399CC;margin-right:1em;font-family: 'Nobile', Helvetica, Arial, sans-serif;font-size:16px;line-height:25px;color:3399CC">
 
A common choice for q(y|x) is a normal distribution centered at x with standard deviation b. q(y|x)=N(x,b<sup>2</sup>)
 
  
i.e.
+
Left-hand side: <br>
<math>q(y|x)=q(x|y)</math><br>
 
(we want to sample q centered at the current state.)<br>
 
<math>q(y|x)=\frac{1}{\sqrt{2\pi}b}\,e^{- \frac{\scriptscriptstyle 1}{\scriptscriptstyle 2b^2} (y-x)^2}</math>, (centered at x)<br>
 
<math>q(x|y)=\frac{1}{\sqrt{2\pi}b}\,e^{- \frac{\scriptscriptstyle 1}{\scriptscriptstyle 2b^2} (x-y)^2}</math>,(centered at y)<br>
 
<math>\Rightarrow (y-x)^2=(x-y)^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(y|x) 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. Beta ---> 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'''====
+
<math>f(x)*P(y|x) = f(x)*{\frac{f(y)*q(x|y)}{f(x)}} = f(y)*q(x|y)</math><br>
<span style="text-shadow: 0px 2px 3px 3399CC;margin-right:1em;font-family: 'Nobile', Helvetica, Arial, sans-serif;font-size:16px;line-height: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/>
+
Right-hand side: <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'''====
+
<math>f(y)*P(x|y) = f(y)*q(x|y)</math><br>
  
One strength of the Metropolis-Hastings 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>
+
Thus LHS and RHS are equal and the detailed balance holds for MH algorithm. <br>
(also notice that Metropolis Hastings is just a special case of Metropolis algorithm)
+
Therefore, f(x) is the stationary distribution of the chain.<br>
  
<math>
+
== Class 20 - Thursday July 11th 2013 ==
\begin{align}
+
=== Simulated annealing ===
f(x;\alpha,\beta)& = \frac{1}{\mathrm{B}(\alpha,\beta)}\, x^{\alpha-1}(1-x)^{\beta-1}\end{align}
+
<br />
</math>
+
'''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 />
  
The beta function, ''B'', appears as a normalizing constant but it can be simplified by construction of the method.
+
Simulated annealing is developed to solve the traveling salesman problem: finding the optimal path to travel all the cities needed<br/>
  
====='''Example'''=====
+
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 />
  
<math>\,f(x)=\frac{1}{\pi^{2}}\frac{1}{1+x^{2}}</math><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 />
Then, we have <math>\,f(x)\propto\frac{1}{1+x^{2}}</math>.<br>
+
(http://www.mit.edu/~dbertsim/papers/Optimization/Simulated%20annealing.pdf)<br />
And let us take <math>\,q(x|y)=\frac{1}{\sqrt{2\pi}b}e^{-\frac{1}{2b^{2}}(y-x)^{2}}</math>.<br>
 
Then <math>\,q(x|y)</math> is symmetric since <math>\,(y-x)^{2} = (x-y)^{2}</math>.<br>
 
Therefore Y can be simplified.
 
  
 +
Simulated annealing was developed as an approach for finding the minimum of complex functions <br />
 +
with multiple peaks; where standard hill-climbing approaches may trap the algorithm at a less that optimal peak.<br />
  
We get :
+
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>
 +
 +
Suppose we would like to min <math> h(x)</math>, for any arbitrary constant <math> T > 0</math>, this problem is equivalent to  max <math>e^{-h(x)/T}</math><br />
 +
Note that the exponential function is monotonic. <br />
 +
Consider f proportional  to  e<sup>-h(x)/T</sup>, sample of this distribution when T is small and
 +
close to the optimal point of h(x). Based on this observation, SA algorithm is introduced as :<br />
 +
<b>1.</b> Set T to be a large number<br />
 +
<b>2.</b> Initialize the chain: set <math>\,X_{t}  (ie.  i=0, x_0=s)</math><br />
 +
<b>3.</b> <math>\,y</math>~<math>\,q(y|x)</math><br/>
 +
(q should be symmetric)<br />
 +
<b>4.</b> <math>r = \min\{\frac{f(y)}{f(x)},1\}</math><br />
 +
<b>5.</b> U ~ U(0,1)<br />
 +
<b>6.</b> If U < r, <math>X_{t+1}=y</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 />
 +
(repeat the procedure until T is very small)<br/>
 +
<br/>
 +
<b>Note</b>: q(y|x) does not have to be symmetric. If q is non-symmetric, then the original MH formula is used.<br />
 +
 
 +
The significance of T <br />
 +
Initially we set T to be large when initializing the chain so as to explore the entire sample space and to avoid the possibility of getting stuck/trapped in one region of the sample space. Then we gradually start decreasing T so as to get closer and closer to the actual solution. 
 +
 
 +
Notice that we have:
 +
    <math> r = \min\{\frac{f(y)}{f(x)},1\} </math><br/>
 +
    <math> = \min\{\frac{e^{\frac{-h(y)}{T}}}{e^{\frac{-h(x)}{T}}},1\} </math>  <br/>
 +
    <math> = \min\{e^{\frac{h(x)-h(y)}{T}},1\} </math><br/>
 +
 
 +
Reasons we start with a large T but not a small T at the beginning:<br />
 +
 
 +
<ul><li>A point in the tail when T is small would be rejected <br />
 +
</li><li>Chances that we reject points get larger as we move from large T to small T <br />
 +
</li><li>Large T helps get to the mode of maximum value<br />
 +
</li></ul>
  
<math>\,\begin{align}
+
Assume T is large <br />
\displaystyle r(x,y)  
+
1. h(y) < h(x), e<sup>(h(x)-h(y))/T </sup> > 1, then r = 1, y will always be accepted.<br />
& =min\left\{\frac{f(y)}{f(x)}\frac{q(x|y)}{q(y|x)},1\right\} \\
+
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 />
& =min\left\{\frac{f(y)}{f(x)},1\right\} \\
+
Assume T is small<br />
& =min\left\{ \frac{ \frac{1}{1+y^{2}} }{ \frac{1}{1+x^{2}} },1\right\}\\
+
1. h(y) < h(x), then r = 1, y will always be accepted.<br />
& =min\left\{ \frac{1+x^{2}}{1+y^{2}},1\right\}\\
+
2. h(y) > h(x), e<sup>(h(x)-h(y))/T </sup> approaches to 0, then r goes to 0 and y will almost never be accepted.
\end{align}
 
</math>.
 
  
<br/>
+
<p><br /> All in all, choose a large T to start off with in order for a higher chance that the points can explore. <br />
<math>\pi=[0.1\,0.1\,...] </math><br/>
 
<math>\pi \propto [3\,2\, 10\, 100\, 1.5] </math><br/>
 
<math>\Rightarrow \pi=1/c \times [3\, 2\, 10\, 100\, 1.5]</math><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 Metropolis-Hasting algorithm, it is not necessary to do this.  
+
'''Note''': The variable T is known in practice as the "Temperature", thus the higher T is, the more variability there is in terms of the expansion and contraction of materials. The term "Annealing" follows from here, as annealing is the process of heating materials and allowing them to cool slowly.<br />
  
<br>
+
Asymptotically this algorithm is guaranteed to generate the global optimal answer, however in practice, we never sample forever and this may not happen.
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'''======
+
</p><p><br />
The Matlab code of the algorithm is the following :
+
</p><p>Example: Consider <math>h(x)=3x^2</math>, 0&lt;x&lt;1
 +
</p><p><br />1) Set T to be large, for example, T=100<br />
 +
<br />2) Initialize the chain<br />
 +
<br />3) Set <math>q(y|x)~\sim~Unif[0,1]</math><br />
 +
<br />4) <math>r=min(exp(\frac{(3x^2-3y^2)}{100}),1)</math><br />
 +
<br />5) <math>U~\sim~U[0,1]</math><br />
 +
<br />6) If <i>U</i> &lt; <i>r</i> then <i>X</i><sub><i>t</i> + 1</sub> = <i>y</i> <br>
 +
<i>e</i><i>l</i><i>s</i><i>e</i>,<i>X</i><sub><i>t</i> + 1</sub> = <i>x</i><sub><i>t</i></sub><br />
 +
<br />7) Decrease T, go back to 3<br />
 +
</p>
 +
<div style="border:1px red solid">
 +
<p><b>MATLAB </b>
 +
</p>
 
<pre style="font-size:12px">
 
<pre style="font-size:12px">
clear all
+
Syms x
close all
+
Ezplot('(x-3)^2',[-6,12])
clc
+
Ezplot('exp(-((x-3)^2))', [-6, 12])
b=2;
 
x(1)=0;
 
for i=2:10000
 
    y=b*randn+x(i-1);
 
    r=min((1+x(i-1)^2)/(1+y^2),1);
 
    u=rand;
 
    if u<r
 
        x(i)=y;
 
    else
 
        x(i)=x(i-1);
 
    end
 
   
 
end
 
hist(x,100);
 
%The Markov Chain usually takes some time to converge and this is known as the "burning time".
 
 
</pre>
 
</pre>
[[File:MH_example2.jpg|300px]]
 
  
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>
+
[[File:Snip2013.png|350px]]
 +
 
 +
[[File:Snip20131.png|350px]]
 +
 
 +
[[File:STAT_340.JPG]]
 +
http://www.wolframalpha.com/input/?i=graph+exp%28-%28x-3%29%5E2%2F10%29
 +
<b>MATLAB </b>
  
'''MATLAB'''