stat341f11

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Notation

The following guidelines on notation were posted on the Wiki Course Note page for STAT 946. Add to them as necessary for consistent notation on this page.

Capital letters will be used to denote random variables and lower case letters denote observations for those random variables:

• [math]\displaystyle{ \{X_1,\ X_2,\ \dots,\ X_n\} }[/math] random variables
• [math]\displaystyle{ \{x_1,\ x_2,\ \dots,\ x_n\} }[/math] observations of the random variables

The joint probability mass function can be written as:

[math]\displaystyle{ P( X_1 = x_1, X_2 = x_2, \dots, X_n = x_n ) }[/math]

or as shorthand, we can write this as [math]\displaystyle{ p( x_1, x_2, \dots, x_n ) }[/math]. In these notes both types of notation will be used. We can also define a set of random variables [math]\displaystyle{ X_Q }[/math] where [math]\displaystyle{ Q }[/math] represents a set of subscripts.

Sampling - September 20, 2011

The meaning of sampling is to generate data points or numbers such that these data follow a certain distribution.
i.e. From [math]\displaystyle{ x \sim~f(x) }[/math] sample [math]\displaystyle{ \,x_{1}, x_{2}, ..., x_{1000} }[/math]

In practice, it maybe difficult to find the joint distribution of random variables. Through simulating the random variables, we can inference from the data.

Sampling from Uniform Distribution

Computers cannot generate random numbers as they are deterministic; however they can produce pseudo random numbers. Pseudo random numbers are numbers that mimic the properties of random numbers, but are generated using algorithms. Therefore, they are not truly random.

Multiplicative Congruential

• involves three parameters: integers [math]\displaystyle{ \,a, b, m }[/math], and an initial value [math]\displaystyle{ \,x_0 }[/math] we call the seed
• a sequence of integers is defined as

[math]\displaystyle{ x_{k+1} \equiv (ax_{k} + b) \mod{m} }[/math]

Example: [math]\displaystyle{ \,a=13, b=0, m=31, x_0=1 }[/math] creates a uniform histogram.

MATLAB code for generating 1000 random numbers using the multiplicative congruential method:

a = 13;
b = 0;
m = 31;
x(1) = 1;

for ii = 2:1000
    x(ii) = mod(a*x(ii-1)+b, m);
end

MATLAB code for displaying the values of x generated:

x

MATLAB code for plotting the histogram of x:

hist(x)

Facts about this algorithm:

• In this example, the first 30 terms in the sequence are a permutation of integers from 1 to 30 and then the sequence repeats itself.
• Values are between 0 and [math]\displaystyle{ m-1 }[/math], inclusive.
• Dividing the numbers by [math]\displaystyle{ m-1 }[/math] yields numbers in the interval [math]\displaystyle{ [0,1] }[/math].
• MATLAB's rand function once used this algorithm with [math]\displaystyle{ a = 7^5, b = 0, m = 2^{31}-1 }[/math], for reasons described in Park and Miller's 1988 paper "Random Number Generators: Good Ones are Hard to Find" (available online).
• Visual Basic's RND function also used this algorithm with [math]\displaystyle{ a = 1140671485, b = 12820163, m = 2^{24} }[/math] (Reference)

Inverse Transform Method

This is a basic method for sampling. Theoretically using this method we can generate sample numbers at random from any probability distribution once we know its cumulative distribution function (cdf).

Theorem

Take [math]\displaystyle{ U \sim~ \mathrm{Unif}[0, 1] }[/math] and let [math]\displaystyle{ X=F^{-1}(U) }[/math]. Then X has distribution function [math]\displaystyle{ F(\cdot) }[/math], where [math]\displaystyle{ F(x)=P(X \leq x) }[/math].

Let [math]\displaystyle{ F^{-1}( ) }[/math] denote the inverse of [math]\displaystyle{ F( ) }[/math]. Therefore [math]\displaystyle{ F(x)=u \implies x=F^{-1}(u) }[/math]

Proof

Recall that

[math]\displaystyle{ P(a \leq X\lt b)=\int_a^{b} f(x) dx }[/math]

[math]\displaystyle{ cdf=F(x)=P(X \leq x)=\int_{-\infty}^{x} f(x) dx }[/math]

Note that if [math]\displaystyle{ U \sim~ \mathrm{Unif}[0, 1] }[/math], we have [math]\displaystyle{ P(U \leq a)=a }[/math]

[math]\displaystyle{ \begin{align} P(F^{-1}(U) \leq x) &{}= P(F(F^{-1}(U)) \leq F(x)) \\ &{}= P(U \leq F(x)) \\ &{}= F(x) \end{align} }[/math]

Completing the proof.

Continuous Case

Generally it takes two steps to get random numbers using this method.

• Step 1. Draw [math]\displaystyle{ U \sim~ \mathrm{Unif}[0, 1] }[/math]
• Step 2. [math]\displaystyle{ X=F^{-1}(U) }[/math]

Example

Take the exponential distribution for example

[math]\displaystyle{ \,f(x)={\lambda}e^{-{\lambda}x} }[/math]

[math]\displaystyle{ \,F(x)=\int_0^x {\lambda}e^{-{\lambda}u} du=1-e^{-{\lambda}x} }[/math]

Let: [math]\displaystyle{ \,F(x)=y }[/math]

[math]\displaystyle{ \,y=1-e^{-{\lambda}x} }[/math]

[math]\displaystyle{ \,ln(1-y)={-{\lambda}x} }[/math]

[math]\displaystyle{ \,x=\frac{ln(1-y)}{-\lambda} }[/math]

[math]\displaystyle{ \,F^{-1}(x)=\frac{-ln(1-x)}{\lambda} }[/math]

Therefore, to get a exponential distribution from a uniform distribution takes 2 steps.

• Step 1. Draw [math]\displaystyle{ U \sim~ \mathrm{Unif}[0, 1] }[/math]
• Step 2. [math]\displaystyle{ x=\frac{-ln(1-U)}{\lambda} }[/math]

Note: If U~Unif[0, 1], then (1 - U) and U have the save distribution. This allows us to slightly simplify step 2 into an alternate form:

• Alternate Step 2. [math]\displaystyle{ x=\frac{-ln(U)}{\lambda} }[/math]

MATLAB code

for exponential distribution case,assuming [math]\displaystyle{ \lambda=0.5 }[/math]

for ii = 1:1000
    u = rand;
    x(ii) = -log(1-u)/0.5;
end
hist(x)

MATLAB result

Discrete Case - September 22, 2011

This same technique can be applied to the discrete case. Generate a discrete random variable [math]\displaystyle{ \,x }[/math] that has probability mass function [math]\displaystyle{ \,P(X=x_i)=P_i }[/math] where [math]\displaystyle{ \,x_0\lt x_1\lt x_2... }[/math] and [math]\displaystyle{ \,\sum_i P_i=1 }[/math]

• Step 1. Draw [math]\displaystyle{ u \sim~ \mathrm{Unif}[0, 1] }[/math]
• Step 2. [math]\displaystyle{ \,x=x_i }[/math] if [math]\displaystyle{ \,F(x_{i-1})\lt u \leq F(x_i) }[/math]

Example

Let x be a discrete random variable with the following probability mass function:

[math]\displaystyle{ \begin{align} P(X=0) = 0.3 \\ P(X=1) = 0.2 \\ P(X=2) = 0.5 \end{align} }[/math]

Given the pmf, we now need to find the cdf.

We have:

[math]\displaystyle{ F(x) = \begin{cases} 0 & x \lt 0 \\ 0.3 & 0 \leq x \lt 1 \\ 0.5 & 1 \leq x \lt 2 \\ 1 & 2 \leq x \end{cases} }[/math]

We can apply the inverse transform method to obtain our random numbers from this distribution.

Pseudo Code for generating the random numbers:

Draw U ~ Unif[0,1] 
if U <= 0.3 
  return 0 
else if 0.3 < U <= 0.5 
  return 1
else if 0.5 < U <= 1 
  return 2

MATLAB code for generating 1000 random numbers in the discrete case:

for ii = 1:1000
  u = rand;
    
  if u <= 0.3
    x(ii) = 0;
  else if u <= 0.5
    x(ii) = 1;
  else
    x(ii) = 2;
  end
end

Pseudo code for the Discrete Case:

1. Draw U ~ Unif [0,1]

2. If [math]\displaystyle{ U \leq P_0 }[/math], deliver [math]\displaystyle{ X = x_0 }[/math]

3. Else if [math]\displaystyle{ U \leq P_0 + P_1 }[/math], deliver [math]\displaystyle{ X = x_1 }[/math]

4. Else If [math]\displaystyle{ U \leq P_0 +....+ P_k }[/math], deliver [math]\displaystyle{ X = x_k }[/math]

Limitations

Although this method is useful, it isn't practical in many cases since we can't always obtain [math]\displaystyle{ F }[/math] or [math]\displaystyle{ F^{-1} }[/math] as some functions are not integrable or invertible, and sometimes even [math]\displaystyle{ f(x) }[/math] itself cannot be obtained in closed form. Let's look at some examples:

• Continuous case

If we want to use this method to draw the pdf of normal distribution, we may find ourselves get stuck in finding its cdf. The simplest case of normal distribution is [math]\displaystyle{ f(x)=\frac{1}{\sqrt{2\pi}}e^{-\frac{x^2}{2}} }[/math], whose cdf is [math]\displaystyle{ F(x)=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{x}{e^{-\frac{u^2}{2}}}du }[/math]. This integral cannot be expressed in terms of elementary functions. So evaluating it and then finding the inverse is a very difficult task.

• Discrete case

It is easy for us to simulate when there are only a few values taken by the particular random variable, like the case above. And it is easy to simulate the binomial distribution [math]\displaystyle{ X \sim~ \mathrm{B}(n,p) }[/math] when the parameter n is not too large. But when n takes on values that are very large, say 50, it is hard to do so.

Acceptance/Rejection Method

The aforementioned difficulties of the inverse transform method motivates a sampling method that does not require analytically calculating cdf's and their inverses, which is the acceptance/rejection sampling method. Here, [math]\displaystyle{ f(x) }[/math] is approximated by another function, say [math]\displaystyle{ g(x) }[/math], with the idea being that [math]\displaystyle{ g(x) }[/math] is a "nicer" function to work with than [math]\displaystyle{ f(x) }[/math].

Suppose we assume the following:

1. There exists another distribution [math]\displaystyle{ g(x) }[/math] that is easier to work with and that you know how to sample from, and

2. There exists a constant c such that [math]\displaystyle{ f(x) \leq c \cdot g(x) }[/math] for all x

Under these assumptions, we can sample from [math]\displaystyle{ f(x) }[/math] by sampling from [math]\displaystyle{ g(x) }[/math]

General Idea

Looking at the image below we have graphed [math]\displaystyle{ c \cdot g(x) }[/math] and [math]\displaystyle{ f(x) }[/math].

Using the acceptance/rejection method we will accept some of the points from [math]\displaystyle{ g(x) }[/math] and reject some of the points from [math]\displaystyle{ g(x) }[/math]. The points that will be accepted from [math]\displaystyle{ g(x) }[/math] will have a distribution similar to [math]\displaystyle{ f(x) }[/math]. We can see from the image that the values around [math]\displaystyle{ x_1 }[/math] will be sampled more often under [math]\displaystyle{ c \cdot g(x) }[/math] than under [math]\displaystyle{ f(x) }[/math], so we will have to reject more samples taken at x₁. Around [math]\displaystyle{ x_2 }[/math] the number of samples that are drawn and the number of samples we need are much closer, so we accept more samples that we get at [math]\displaystyle{ x_2 }[/math]

Procedure

1. Draw y ~ g

2. Draw U ~ Unif [0,1]

3. If [math]\displaystyle{ U \leq \frac{f(y)}{c \cdot g(y)} }[/math] then x=y; else return to 1

Proof

Mathematically, we need to show that the sample points given that they are accepted have a distribution of f(x).

[math]\displaystyle{ \begin{align} P(y|accepted) &= \frac{P(y, accepted)}{P(accepted)} \\ &= \frac{P(accepted|y) P(y)}{P(accepted)}\end{align} }[/math] (Bayes' Rule)

[math]\displaystyle{ P(y) = g(y) }[/math]

[math]\displaystyle{ P(accepted|y) =P(u\leq \frac{f(y)}{c \cdot g(y)}) =\frac{f(y)}{c \cdot g(y)} }[/math],where u ~ Unif [0,1]

[math]\displaystyle{ P(accepted) = \sum P(accepted|y)\cdot P(y)=\int^{}_y \frac{f(y)}{c \cdot g(y)}g(y) dy=\int^{}_y \frac{f(y)}{c} dy=\frac{1}{c} \cdot\int^{}_y f(y) dy=\frac{1}{c} }[/math]

So,

[math]\displaystyle{ P(y|accepted) = \frac{ \frac {f(y)}{c \cdot g(y)} \cdot g(y)}{\frac{1}{c}} =f(y) }[/math]

Continuous Case

Example

Sample from Beta(2,1)

In general:

Beta([math]\displaystyle{ \alpha, \beta) = \frac{\Gamma (\alpha + \beta)}{\Gamma(\alpha)\Gamma(\beta)} }[/math] [math]\displaystyle{ x^{\alpha-1} }[/math] [math]\displaystyle{ (1-x)^{\beta-1} }[/math], [math]\displaystyle{ 0\lt x\lt 1 }[/math]

Note: [math]\displaystyle{ \Gamma(n) = (n-1)! }[/math] if n is a positive integer

[math]\displaystyle{ \begin{align} f(x) &= Beta(2,1) \\ &= \frac{\Gamma(3)}{\Gamma(2)\Gamma(1)} x^1(1-x)^0 \\ &= \frac{2!}{1! 0!}\cdot (1) x \\ &= 2x \end{align} }[/math]

We want to choose [math]\displaystyle{ g(x) }[/math] that is easy to sample from. So we choose [math]\displaystyle{ g(x) }[/math] to be uniform distribution.

We now want a constant c such that [math]\displaystyle{ f(x) \leq c \cdot g(x) }[/math] for all x from (0,1)

So,

[math]\displaystyle{ c \geq \frac{f(x)}{g(x)} }[/math], for all x from (0,1)

[math]\displaystyle{ \begin{align}c &\geq max (\frac {f(x)}{g(x)}, 0\lt x\lt 1) \\ &= max (\frac {2x}{1},0\lt x\lt 1) \\ &= 2 \end{align} }[/math]

Now that we have c =2,

1. Draw y ~ g(x) => Draw y ~ Unif [0,1]

2. Draw u ~ Unif [0,1]

3. if [math]\displaystyle{ u \leq \frac{2y}{2 \cdot 1} }[/math] then x=y; else return to 1

MATLAB code for generating 1000 samples following Beta(2,1):

close all
clear all
ii=1;
while ii < 1000
    y = rand;
    u = rand;
    
    if u <= y
        x(ii)=y;
        ii=ii+1;
    end
end
hist(x)

MATLAB result

Discrete Example

Generate random variables according to the p.m.f:

[math]\displaystyle{ \begin{align} P(Y=1) = 0.15 \\ P(Y=2) = 0.22 \\ P(Y=3) = 0.33 \\ P(Y=4) = 0.10 \\ P(Y=5) = 0.20 \end{align} }[/math]

find a g(y) discrete uniform distribution from 1 to 5

[math]\displaystyle{ c \geq \frac{P(y)}{g(y)} }[/math]
[math]\displaystyle{ c = \max \left(\frac{P(y)}{g(y)} \right) }[/math]
[math]\displaystyle{ c = \max \left(\frac{0.33}{0.2} \right) = 1.65 }[/math] Since P(Y=3) is the max of P(Y) and g(y) = 0.2 for all y.

1. Generate Y according to the discrete uniform between 1 - 5

2. U ~ unif[0,1]

3. If [math]\displaystyle{ U \leq \frac{P(y)}{1.65 \times 0.2} \leq \frac{P(y)}{0.33} }[/math], then x = y; else return to 1.

In MATLAB, the code would be:

   py = [0.15 0.22 0.33 0.1 0.2];
   ii =1;
   while ii <= 1000
       y = unidrnd(5);
       u = rand;
       if u <= py(y)/0.33
           x(ii) = y;
           ii = ii+1;
       end
   end
   hist(x);

MATLAB result

Limitations

Most of the time we have to sample many more points from g(x) before we can obtain an acceptable amount of samples from f(x), hence this method may not be computationally efficient. It depends on our choice of g(x). For example, in the example above to sample from Beta(2,1), we need roughly 2000 samples from g(X) to get 1000 acceptable samples of f(x).

In addition, in situations where a g(x) function is chosen and used, there can be a discrepancy between the functional behaviors of f(x) and g(x) that render this method unreliable. For example, given the normal distribution function as g(x) and a function of f(x) with a "fat" mid-section and "thin tails", this method becomes useless as more points near the two ends of f(x) will be rejected, resulting in a tedious and overwhelming number of sampling points having to be sampled due to the high rejection rate of such a method.

Sampling From Gamma and Normal Distribution - September 27, 2011

Sampling From Gamma

Gamma Distribution

The Gamma function is written as [math]\displaystyle{ X \sim~ Gamma (t, \lambda) }[/math]

[math]\displaystyle{ F(x) = \int_{0}^{\lambda x} \frac{e^{-y}y^{t-1}}{(t-1)!} dy }[/math]

If you have t samples of the exponential distribution,

[math]\displaystyle{ \begin{align} X_1 \sim~ Exp(\lambda)\\ \vdots \\ X_t \sim~ Exp(\lambda) \end{align} }[/math]

The sum of these t samples has a gamma distribution,

[math]\displaystyle{ X_1+X_2+ ... + X_t \sim~ Gamma (t, \lambda) }[/math]

[math]\displaystyle{ \sum_{i=1}^{t} X_i \sim~ Gamma (t, \lambda) }[/math] where [math]\displaystyle{ X_i \sim~Exp(\lambda) }[/math]

Method

We can sample the exponential distribution using the inverse transform method from previous class,

[math]\displaystyle{ \,f(x)={\lambda}e^{-{\lambda}x} }[/math]

[math]\displaystyle{ \,F^{-1}(u)=\frac{-ln(1-u)}{\lambda} }[/math]

[math]\displaystyle{ \,F^{-1}(u)=\frac{-ln(u)}{\lambda} }[/math]

1 - u is the same as x since [math]\displaystyle{ U \sim~ unif [0,1] }[/math]

[math]\displaystyle{ \begin{align} \frac{-ln(u_1)}{\lambda} - \frac{ln(u_2)}{\lambda} - ... - \frac{ln(u_t)}{\lambda} = x_1\\ \vdots \\ \frac{-ln(u_1)}{\lambda} - \frac{ln(u_2)}{\lambda} - ... - \frac{ln(u_t)}{\lambda} = x_t \end{align}  : }[/math]

[math]\displaystyle{ \frac {-\sum_{i=1}^{t} ln(u_i)}{\lambda} = x }[/math]

MATLAB code for a Gamma(1,3) is

x = sum(-log(rand(1000,3)),2); 
hist(x)

And the Histogram of X follows a Gamma distribution with long tail:

We can improve the quality of histogram by adding the number of bins we want, like hist(x, number_of_bins)

x = sum(-log(rand(20000,3)),2); 
hist(x,40)

File:untitled.jpg

R code for a Gamma(1,3) is

a<-apply(-log(matrix(runif(3000),nrow=1000)),1,sum);
hist(a);

And the histogram is

File:hist gamma.png

Here is another histogram of Gamma coding with R

a<-apply(-log(matrix(runif(3000),nrow=1000)),1,sum);
hist(a,freq=F);
lines(density(a),col="blue");
rug(jitter(a));

File:hist gamma 2.png

Sampling from Normal Distribution using Box-Muller Transform - September 29, 2011

Procedure

1. Generate [math]\displaystyle{ u_1 }[/math] and [math]\displaystyle{ u_2 }[/math], two values sampled from a uniform distribution between 0 and 1.
2. Set [math]\displaystyle{ R^2 = -2log(u_1) }[/math] so that [math]\displaystyle{ R^2 }[/math] is exponential with mean 1/2
   Set [math]\displaystyle{ \theta = 2*\pi*u_2 }[/math] so that [math]\displaystyle{ \theta }[/math] ~ Unif[0, 2[math]\displaystyle{ \pi }[/math]]
3. Set [math]\displaystyle{ X = R cos(\theta) }[/math]
   Set [math]\displaystyle{ Y = R sin(\theta) }[/math]

Justification

Suppose we have X ~ N(0, 1) and Y ~ N(0, 1) where X and Y are independent normal random variables. The relative probability density function of these two random variables using Cartesian coordinates is:

[math]\displaystyle{ f(X, Y) dxdy= f(X) f(Y) dxdy= \frac{1}{\sqrt{2\pi}}e^{-x^2/2} \frac{1}{\sqrt{2\pi}}e^{-y^2/2} dxdy= \frac{1}{2\pi}e^{-(x^2+y^2)/2}dxdy }[/math]

In polar coordinates [math]\displaystyle{ R^2 = x^2 + y^2 }[/math], so the relative probability density function of these two random variables using polar coordinates is:

[math]\displaystyle{ f(R, \theta) = \frac{1}{2\pi}e^{-R^2/2} }[/math]

If we have [math]\displaystyle{ R^2 }[/math] ~ exp(1/2) and [math]\displaystyle{ \theta }[/math] ~ unif[0, 2[math]\displaystyle{ \pi }[/math]] we get an equivalent relative probability density function. Notice that after the two on two transformation, a determinant of jocobian should be added according to the change of variable and rule of differential multiplication where

[math]\displaystyle{ |J|=\left|\frac{\partial(x,y)}{\partial(R,\theta)}\right|= \left|\begin{matrix}\frac{\partial x}{\partial R}&\frac{\partial x}{\partial \theta}\\\frac{\partial y}{\partial R}&\frac{\partial y}{\partial \theta}\end{matrix}\right|=R }[/math]

[math]\displaystyle{ f(X, Y) dxdy = f(R, \theta)|J|dRd\theta = \frac{1}{2\pi}e^{-R^2/2}R dRd\theta= \frac{1}{4\pi}e^{-\frac{s}{2}} dSd\theta }[/math]
where [math]\displaystyle{ S=R^2. }[/math]

Therefore we can generate a point in polar coordinates using the uniform and exponential distributions, then convert the point to Cartesian coordinates and the resulting X and Y values will be equivalent to samples generated from N(0, 1).

MATLAB code

In MatLab this algorithm can be implemented with the following code, which generates 20,000 samples from N(0, 1):

x = zeros(10000, 1);
y = zeros(10000, 1);
for ii = 1:10000
    u1 = rand;
    u2 = rand;
    R2 = -2 * log(u1);
    theta = 2 * pi * u2;
    x(ii) = sqrt(R2) * cos(theta);
    y(ii) = sqrt(R2) * sin(theta);
end
hist(x)

In one execution of this script, the following histogram for x was generated:

Non-Standard Normal Distributions

Example 1: Single-variate Normal

If X ~ Norm(0, 1) then (a + bX) has a normal distribution with a mean of [math]\displaystyle{ a }[/math] and a standard deviation of [math]\displaystyle{ b }[/math] (which is equivalent to a variance of [math]\displaystyle{ b^2 }[/math]). Using this information with the Box-Muller transform, we can generate values sampled from some random variable Y ~ Norm([math]\displaystyle{ a }[/math], [math]\displaystyle{ b^2 }[/math]) for arbitrary values of [math]\displaystyle{ a }[/math] and [math]\displaystyle{ b }[/math].

1. Generate a sample u from Norm(0, 1) using the Box-Muller transform.
2. Set v = a + bu.

The values for v generated in this way will be equivalent to sample from a Norm([math]\displaystyle{ a }[/math], [math]\displaystyle{ b^2 }[/math]) distribution. We can modify the MatLab code used in the last section to demonstrate this. We just need to add one line before we generate the histogram:

x = a + b * x;

For instance, this is the histogram generated when b = 15, a = 125:

500

Example 2: Multi-variate Normal

The Box-Muller method can be extended to higher dimensions to generate multivariate normals. The objects generated will be nx1 vectors, and their variance will be described by nxn covariance matrices.

[math]\displaystyle{ \mathbf{z} = N(\mathbf{u}, \Sigma) }[/math] defines the n by 1 vector [math]\displaystyle{ \mathbf{z} }[/math] such that:

• [math]\displaystyle{ u_i }[/math] is the average of [math]\displaystyle{ z_i }[/math]
• [math]\displaystyle{ \!\Sigma_{ii} }[/math] is the variance of [math]\displaystyle{ z_i }[/math]
• [math]\displaystyle{ \!\Sigma_{ij} }[/math] is the co-variance of [math]\displaystyle{ z_i }[/math] and [math]\displaystyle{ z_j }[/math]

If [math]\displaystyle{ z_1, z_2, ..., z_d }[/math] are normal variables with mean 0 and variance 1, then the vector [math]\displaystyle{ (z_1, z_2,..., z_d) }[/math] has mean 0 and variance [math]\displaystyle{ \!I }[/math], where 0 is the zero vector and [math]\displaystyle{ \!I }[/math] is the identity matrix. This fact suggests that the method for generating a multivariate normal is to generate each component individually as single normal variables.

The mean and the covariance matrix of a multivariate normal distribution can be adjusted in ways analogous to the single variable case. If [math]\displaystyle{ \mathbf{z} \sim N(0,I) }[/math], then [math]\displaystyle{ \Sigma^{1/2}\mathbf{z}+\mu \sim N(\mu,\Sigma) }[/math]. Note here that the covariance matrix is symmetric and nonnegative, so its square root should always exist.

We can compute [math]\displaystyle{ \mathbf{z} }[/math] in the following way:

1. Generate an n by 1 vector [math]\displaystyle{ \mathbf{x} = \begin{bmatrix}x_{1} & x_{2} & ... & x_{n}\end{bmatrix} }[/math] where [math]\displaystyle{ x_{i} }[/math] ~ Norm(0, 1) using the Box-Muller transform.
2. Calculate [math]\displaystyle{ \!\Sigma^{1/2} }[/math] using singular value decomposition.
3. Set [math]\displaystyle{ \mathbf{z} = \Sigma^{1/2} \mathbf{x} + \mathbf{u} }[/math].

The following MatLab code provides an example, where a scatter plot of 10000 random points is generated. In this case x and y have a co-variance of 0.9 - a very strong positive correlation.

x = zeros(10000, 1);
y = zeros(10000, 1);
for ii = 1:10000
    u1 = rand;
    u2 = rand;
    R2 = -2 * log(u1);
    theta = 2 * pi * u2;
    x(ii) = sqrt(R2) * cos(theta);
    y(ii) = sqrt(R2) * sin(theta);
end

E = [1, 0.9; 0.9, 1];
[u s v] = svd(E);
root_E = u * (s ^ (1 / 2));

z = (root_E * [x y]);
z(:,1) = z(:,1) + 5;
z(:,2) = z(:,2) + -8;

scatter(z(:,1), z(:,2))

This code generated the following scatter plot:

File:scatter covar.jpg

In Matlab, we can also use the function "sqrtm()" to calculate Square root of a matrix directly. Here is an example:

E = [1, 0.9; 0.9, 1];
sqrtm(E);

R code for a multivariate normal distribution:

n=10000;
r2<--2*log(runif(n));
theta<-2*pi*(runif(n));
x<-sqrt(r2)*cos(theta);

y<-sqrt(r2)*sin(theta);
a<-matrix(c(x,y),nrow=n,byrow=F);
e<-matrix(c(1,.9,09,1),nrow=2,byrow=T);
svde<-svd(e);
root_e<-svde$u %*% diag(svde$d)^1/2;
z<-t(root_e %*%t(a));
z[,1]=z[,1]+5;
z[,2]=z[,2]+ -8;
par(pch=19);
plot(z,col=rgb(1,0,0,alpha=0.06))

File:m normal.png

Remarks

MATLAB's randn function uses the ziggurat method to generate normal distributed samples. It is an efficient rejection method based on covering the probability density function with a set of horizontal rectangles so as to obtain points within each rectangle. It is reported that a 800 MHz Pentium III laptop can generate over 10 million random numbers from normal distribution in less than one second. (Reference)

Sampling From Binomial Distributions

In order to generate a sample x from X ~ Bin(n, p), we can follow the following procedure:

1. Generate n uniform random numbers sampled from Unif[0, 1]: [math]\displaystyle{ u_1, u_2, ..., u_n }[/math].

2. Set x to be the total number of cases where [math]\displaystyle{ u_i \lt = p }[/math] for all [math]\displaystyle{ 1 \lt = i \lt = n }[/math].

In MatLab this can be coded with a single line. The following generates a sample from X ~ Bin(n, p):

>> sum(rand(n, 1) <= p, 1)

Bayesian Inference and Frequentist Inference - October 4, 2011

Bayesian inference vs Frequentist inference

The Bayesian method has become popular in the last few decades as simulation and computer technology makes it more applicable. For more information about its history and application, please refer to http://en.wikipedia.org/wiki/Bayesian_inference. As for frequentists, please refer to http://en.wikipedia.org/wiki/Frequentist_inference.

Example

Consider: A person drinks a cup of coffee on a specific day.

Frequentist: There is no explanation to this situation. It is essentially meaningless since it has only occurred once. Therefore, it is not a probability.
Bayesian: Probability is not just about the frequent occurrences but it is what you believe about this probability.

Example of face identification

Take the face as input x. And the person as output y. The person can be either Ali or Tom. If it is Ali, y=1. Otherwise, y=0. We can divide the picture into 100*100 pixels and then list them into a 10,000*1 column vector which is x.

If you are a frequentist, you would compare Pr(X=x|y=1) with Pr(X=x|y=0) and see which one is higher. But if you are a Bayesianist, you would compare Pr(y=1|X=x) with Pr(y=0|X=x).

Summary of differences between two schools

• Frequentist: Probability refers to limiting relative frequency. (objective)
• Bayesian: Probability describes degree of belief not frequency. (subjective)

e.g. The probability that you drank a cup of tea on May 20, 2001 is 0.62 does not refer to any frequency.

---

• Frequentist: Parameters are fixed, unknown constants.
• Bayesian: Parameters are random variables and we can make probabilistic statement about them.

---

• Frequentist: Statistical procedures should have long run frequency probabilities.

e.g. a 95% confidence interval should trap true value of the parameter for at least 95% of limited frequency

• Bayesian: It makes inferences about [math]\displaystyle{ \theta }[/math] by producing a prbability distribution for [math]\displaystyle{ \theta }[/math]. Inference (e.g. point estimation) will be extracted from this distribution.

Bayesian inference

Bayesian inference is usually carried out in the following way:

1. Choose a prior probability density function of [math]\displaystyle{ \theta }[/math] which is [math]\displaystyle{ f(\theta) }[/math]. This is our belief about [math]\displaystyle{ \theta }[/math] before we see any data.

2. Choose a statistical model [math]\displaystyle{ f(x|\theta) }[/math] that reflects our beliefs about X.

3. After observing data [math]\displaystyle{ x_1,...,x_n }[/math], we update our beliefs and calculate the posterior probability.

[math]\displaystyle{ f(\theta|x) = \frac{f(\theta,x)}{f(x)}=\frac{f(x|\theta) \cdot f(\theta)}{f(x)}=\frac{f(x|\theta) \cdot f(\theta)}{\int^{}_\theta f(x|\theta) \cdot f(\theta) d\theta} }[/math], where [math]\displaystyle{ f(\theta|x) }[/math] is the posterior probability, [math]\displaystyle{ f(\theta) }[/math] is the prior probability, [math]\displaystyle{ f(x|\theta) }[/math] is the likelihood of observing X=x given [math]\displaystyle{ \theta }[/math] and f(x) is the marginal probability of X=x.

If we have i.i.d. observations [math]\displaystyle{ x_1,...,x_n }[/math], we can replace [math]\displaystyle{ f(x|\theta) }[/math] with [math]\displaystyle{ f({x_1,...,x_n}|\theta)=\prod_{i=1}^n f(x_i|\theta) }[/math] because of independency.

We denote [math]\displaystyle{ f({x_1,...,x_n}|\theta) }[/math] as [math]\displaystyle{ L_n(\theta) }[/math] which is called likelihood. And we use [math]\displaystyle{ x^n }[/math] to denote [math]\displaystyle{ (x_1,...,x_n) }[/math].

[math]\displaystyle{ f(\theta|x^n) = \frac{f(x^n|\theta) \cdot f(\theta)}{f(x^n)}=\frac{f(x^n|\theta) \cdot f(\theta)}{\int^{}_\theta f(x^n|\theta) \cdot f(\theta) d\theta} }[/math] , where [math]\displaystyle{ \int^{}_\theta f(x^n|\theta) \cdot f(\theta) d\theta }[/math] is a constant [math]\displaystyle{ c_n }[/math]. So [math]\displaystyle{ f(\theta|x^n) \propto f(x^n|\theta) \cdot f(\theta) }[/math]. The posterior probability is proportional to the likelihood times prior probability.

[math]\displaystyle{ E(\theta)=\int^{}_\theta \theta \cdot f(\theta|x^n) d\theta }[/math] which is the posterior mean of [math]\displaystyle{ \theta }[/math].

Let [math]\displaystyle{ \tilde{\theta}=(\theta_1,...,\theta_d)^T }[/math], then [math]\displaystyle{ f(\theta_1|x^n) = \int^{} \int^{} \dots \int^{}f(\theta|X)d\theta_2d\theta_3 \dots d\theta_d }[/math] and [math]\displaystyle{ E(\theta_1)=\int^{}\theta_1 \cdot f(\theta_1|x^n) d\theta_1 }[/math]

Example 1: Estimating parameters of a univariate Gaussian distribution

Suppose X follows a univariate Gaussian distribution (i.e. a Normal distribution) with parameters [math]\displaystyle{ \mu }[/math] and [math]\displaystyle{ \sigma^2 }[/math].

(a) For Frequentists:

[math]\displaystyle{ f(x|\theta)= \frac{1}{\sqrt{2\pi}\sigma} \cdot e^{-\frac{1}{2}(\frac{x-\mu}{\sigma})^2} }[/math]

[math]\displaystyle{ L_n(\theta)= \prod_{i=1}^n \frac{1}{\sqrt{2\pi}\sigma} \cdot e^{-\frac{1}{2}(\frac{x_i-\mu}{\sigma})^2} }[/math]

[math]\displaystyle{ \ln L_n(\theta) = l(\theta) = \sum_{i=1}^n -\frac{1}{2}\ln 2\pi-\ln \sigma-\frac{1}{2}(\frac{x_i-\mu}{\sigma})^2 }[/math]

To get the maximum likelihood estimator of [math]\displaystyle{ \mu }[/math] (mle), we find the [math]\displaystyle{ \hat{\mu} }[/math] which maximizes [math]\displaystyle{ L_n(\theta) }[/math]:

[math]\displaystyle{ \frac{\partial l(\theta)}{\partial \mu}= \sum_{i=1}^n \frac{1}{\sigma}(\frac{x_i-\mu}{\sigma})=0 \Rightarrow \sum_{i=1}^n x_i = n\mu \Rightarrow \hat{\mu}_{mle}=\bar{x} }[/math]

(b) For Bayesians:

[math]\displaystyle{ f(\theta|x) \propto f(x|\theta) \cdot f(\theta) }[/math]

We assume that the mean of the above normal distribution is itself distributed normally with mean [math]\displaystyle{ \mu_0 }[/math] and variance [math]\displaystyle{ \Gamma }[/math].

Suppose [math]\displaystyle{ \mu }[/math] ~ N([math]\displaystyle{ \mu_0, \Gamma^2 }[/math]),

so [math]\displaystyle{ f(\mu) = \frac{1}{\sqrt{2\pi}\Gamma} \cdot e^{-\frac{1}{2}(\frac{\mu-\mu_0}{\Gamma})^2} }[/math]

[math]\displaystyle{ f(\mu|x) = \frac{1}{\sqrt{2\pi}\tilde{\sigma}} \cdot e^{-\frac{1}{2}(\frac{\mu-\tilde{\mu}}{\tilde{\sigma}})^2} }[/math]

[math]\displaystyle{ \tilde{\mu} = \frac{\frac{n}{\sigma^2}}{\frac{n}{\sigma^2}+\frac{1}{\Gamma^2}}\bar{x}+\frac{\frac{1}{\Gamma^2}}{\frac{n}{\sigma^2}+\frac{1}{\Gamma^2}}\mu_0 }[/math], where [math]\displaystyle{ \tilde{\mu} }[/math] is the estimator of [math]\displaystyle{ \mu }[/math].

• If prior belief about [math]\displaystyle{ \mu_0 }[/math] is strong, then [math]\displaystyle{ \Gamma }[/math] is small and [math]\displaystyle{ \frac{1}{\Gamma^2} }[/math] is large. [math]\displaystyle{ \tilde{\mu} }[/math] is close to [math]\displaystyle{ \mu_0 }[/math] and the observations will not affect too much. On the contrary, if prior belief about [math]\displaystyle{ \mu_0 }[/math] is weak, [math]\displaystyle{ \Gamma }[/math] is large and [math]\displaystyle{ \frac{1}{\Gamma^2} }[/math] is small. [math]\displaystyle{ \tilde{\mu} }[/math] depends more on observations.(This is intuitive, when our original belief is reliable, then the sample is not important in improving the result; when the belief is not reliable, then we depend a lot on the sample.)

• When the sample is large (i.e. n [math]\displaystyle{ \to \infty }[/math]), [math]\displaystyle{ \tilde{\mu} \to \bar{x} }[/math] and the impact of prior belief about [math]\displaystyle{ \mu }[/math] is weakened.

Basic Monte Carlo Integration - October 6th, 2011

Three integration methods would be taught in this course:

• Basic Monte Carlo Integration
• Importance Sampling
• Markov Chain Monte Carlo (MCMC)

The first, and most basic, method of numerical integration we will see is Monte Carlo Integration. We use this to solve an integral of the form: [math]\displaystyle{ I = \int_{a}^{b} h(x) dx }[/math]

Note the following derivation: [math]\displaystyle{ I = \int_{a}^{b} h(x)dx = \int_{a}^{b} h(x)((b-a)/(b-a)) dx = \int_{a}^{b} (h(x)(b-a))(1/(b-a))dx = \int_{a}^{b} w(x)f(x)dx = E[w(x)] }[/math] ~ [math]\displaystyle{ (1/n) \sum_{i=1}^{n} w(x) }[/math]

Where w(x) = h(x)(b-a) and f(x) is the probability density function of a uniform random variable on the interval [a,b]. The expectation, with respect to the distribution of f, of w is taken from n samples of x.

General Procedure

i) Draw n samples x_i ~ uniform[a,b]

ii) Compute w(x_i)

        [math]\displaystyle{  \hat{I} = 1/n \sum_{i=1}^{n} w(x }[/math]i[math]\displaystyle{  ) }[/math]

By the strong law of large numbers I^{^} converges to I as n approaches [math]\displaystyle{ \infty }[/math]. Because of this, we can compute all sorts of useful information, such as variance, standard error, and confidence intervals.

Standard Error: [math]\displaystyle{ SE = Standard Deviation / \sqrt{n} }[/math]

Variance: [math]\displaystyle{ V = (\sum_{i=1}^{n} (w(x)-I)^2)/(n-1) }[/math]

Confidence Interval: [math]\displaystyle{ I +/- t \_(\alpha/2) SE }[/math]

Example: Uniform Distribution

Consider the integral, [math]\displaystyle{ \int_{0}^{1} x^3dx }[/math], which is easily solved through standard analytical integration methods, and is equal to .25. Now, let us check this answer with a numerical approximation using Monte Carlo Integration.

We generate a 1 by 10000 vector of uniform (on the interval [0,1]) random variables and call that vector 'u'. We see that our 'w' in this case is [math]\displaystyle{ x^3 }[/math], so we set [math]\displaystyle{ w = u^3 }[/math]. Our I^{^} is equal to the mean of w.

In Matlab, we can solve this integration problem with the following code:

u = rand(1,10000);
w = u.^3;
mean(w)
  ans = 0.2475

Note the '.' after 'u' in the second line of code, indicating that each entry in the matrix is cubed. Also, our approximation is close to the actual value of .25. Now let's try to get an even better approximation by generating more sample points.

u= rand(1,100000);
w= u.^3;
mean(w)
  ans = .2503

We see that when the number of sample points is increased, our approximation improves, as one would expect.

Generalization

Up to this point we have seen how to numerically approximate an integral when the distribution of f is uniform. Now we will see how to generalize this to other distributions.

[math]\displaystyle{ I = \int h(x)f(x)dx }[/math]

If f is a distribution function (pdf), then [math]\displaystyle{ I }[/math] can be estimated as E_f[h(x)]. This means taking the expectation of h with respect to the distribution of f. Our previous example is the case where f is the uniform distribution between [a,b].

Procedure for the General Case

i) Draw n samples from f

ii) Compute h(x_i)

iii) [math]\displaystyle{ \hat{I} = 1/n \sum_{i=1}^{n} h(x }[/math]_i[math]\displaystyle{ ) }[/math]

Example: Exponential Distribution

Find [math]\displaystyle{ E[\sqrt{x}] }[/math] for [math]\displaystyle{ f = e^{-x} }[/math], which is the exponential distribution with mean 1.

[math]\displaystyle{ I = \int_{0}^{\infty} \sqrt{x} e^{-x}dx }[/math]

We can see that we must draw samples from f, the exponential distribution.

To find a numerical solution using Monte Carlo Integration we see that:

u= rand(1,10000) X= -log(u) h= [math]\displaystyle{ \sqrt{x} }[/math] I= mean(h)

To implement this procedure in Matlab, use the following code:

u = rand(1,10000);
X = -log(u);
h = x.^.5;
mean(h)
  ans = .8841

An easy way to check whether your approximation is correct is to use the built in Matlab function 'quadl' which takes a function and bounds for the integral and returns a solution for the definite integral of that function. For this specific example, we can enter:

f = @(x) sqrt(x).*exp(-x);
% quadl runs into computational problems when the upper bound is "inf" or an extremely large number, 
% so choose just a moderately large number.
quadl(f,0,100)
ans =
  0.8862

From the above result, we see that our approximation was quite close.

Example: Normal Distribution

Let [math]\displaystyle{ f(x) = (1/(2 \pi)^{1/2}) e^{(-x^2)/2} }[/math]. Compute the cumulative distribution function at some point x.

[math]\displaystyle{ F(x)= \int_{-\infty}^{x} f(s)ds = \int_{-\infty}^{x}(1)(1/(2 \pi)^{1/2}) e^{(-s^2)/2}ds }[/math]. The (1) is inserted to illustrate that our h(x) will be the constant function 1, and our f(x) is the normal distribution. To take into account the upper bound of integration, x, any values sampled that are greater than x will be set to zero.

This is the Matlab code for solving F(2):

u = randn(1,10000)
h = u < 2;
mean(h)
  ans = .9756

We generate a 1 by 10000 vector of standard normal random variables and we return a value of 1 if u is less than 2, and 0 otherwise.

We can also build the function F(x) in matlab in the following way:

function F(x)
u=rand(1,1000000);
h=u<x;
mean(h)

Example: Binomial Distribution

In this example we will see the Bayesian Inference for 2 Binomial Distributions.

Let [math]\displaystyle{ X ~ Bin(n,p) }[/math] and [math]\displaystyle{ Y ~ Bin(m,q) }[/math], and let [math]\displaystyle{ \delta = p-q }[/math].

Therefore, [math]\displaystyle{ \delta = x/n - y/m }[/math] which is the frequentist approach.

Bayesian wants [math]\displaystyle{ f(p,q|x,y) = f(x,y|p,q)f(p,q)/f(x,y) }[/math]

Thus, [math]\displaystyle{ f(p,q|x,y) \alpha f(x,y|p,q)f(p,q) }[/math]. Now we assume that [math]\displaystyle{ f(p,q) = f(p)f(q) = 1 }[/math] and f(p) and f(q) are uniform.

Therefore, [math]\displaystyle{ f(p,q|x,y) \alpha p^x(1-p)^{n-x}q^y(1-q)^{m-y} }[/math].

[math]\displaystyle{ E[\delta] = \int_{0}^{1} \int_{0}^{1} (p-q)f(p,q|x,y)dxdy }[/math].

As you can see this is much tougher than the frequentist approach.

Importance Sampling and Basic Monte Carlo Integration - October 11th, 2011

Example: Binomial Distribution (Continued)

Suppose we are given two independent Binomial Distributions [math]\displaystyle{ \displaystyle X \sim Bin(n, p_1) }[/math], [math]\displaystyle{ \displaystyle Y \sim Bin(m, p_2) }[/math]. We would like to give an Monte Carlo estimate of [math]\displaystyle{ \displaystyle \delta = p_1 - p_2 }[/math]

Frequentist approach:

[math]\displaystyle{ \displaystyle \hat{p_1} = \frac{X}{n} }[/math] ; [math]\displaystyle{ \displaystyle \hat{p_2} = \frac{Y}{m} }[/math]

[math]\displaystyle{ \displaystyle \hat{\delta} = \hat{p_1} - \hat{p_2} = \frac{X}{n} - \frac{Y}{m} }[/math]

Bayesian approach to compute the expected value of [math]\displaystyle{ \displaystyle \delta }[/math]:

[math]\displaystyle{ \displaystyle E(\delta) = \int\int(p_1-p_2) f(p_1,p_2|X,Y)\,dp_1dp_2 }[/math]

Assume that [math]\displaystyle{ \displaystyle n = 100, m = 100, p_1 = 0.5, p_2 = 0.8 }[/math] and the sample size is 1000.
MATLAB code of the above example:

n = 100;
m = 100;
p_1 = 0.5;
p_2 = 0.8;
p1 = mean(rand(n,1000)<p_1);
p2 = mean(rand(m,1000)<p_2);
delta = p2 - p1;
hist(delta)
mean(delta)

In one execution of the code, the mean of delta was 0.3017. The histogram of delta generated was:

Through Monte Carlo simulation, we can obtain an empirical distribution of delta and carry out inference on the data obtained, such as computing the mean, maximum, variance, standard deviation and the standard error of delta.

Importance Sampling

Motivation

Consider the integral [math]\displaystyle{ \displaystyle I = \int h(x)f(x)\,dx }[/math]

According to basic Monte Carlo Integration, if we can sample from the probability density function [math]\displaystyle{ \displaystyle f(x) }[/math] and feed the samples of [math]\displaystyle{ \displaystyle f(x) }[/math] back to [math]\displaystyle{ \displaystyle h(x) }[/math], [math]\displaystyle{ \displaystyle I }[/math] can be estimated as an average of [math]\displaystyle{ \displaystyle h(x) }[/math] ( i.e. [math]\displaystyle{ \hat{I} = \frac{1}{n} \sum_{i=1}^{n} h(x_i) }[/math] )
However, the Monte Carlo method works when we know how to sample from [math]\displaystyle{ \displaystyle f(x) }[/math]. In the case where it is difficult to sample from [math]\displaystyle{ \displaystyle f(x) }[/math], importance sampling is a technique that we can apply. Importance Sampling relies on another function [math]\displaystyle{ \displaystyle g(x) }[/math] which we know how to sample from.

The above integral can be rewritten as follow:
[math]\displaystyle{ \begin{align} \displaystyle I & = \int h(x)f(x)\,dx \\ & = \int h(x)f(x)\frac{g(x)}{g(x)}\,dx \\ & = \int \frac{h(x)f(x)}{g(x)}g(x)\,dx \\ & = \int y(x)g(x)\,dx \\ & = E_g(y(x)) \\ \end{align} }[/math]
[math]\displaystyle{ where \ y(x) = \frac{h(x)f(x)}{g(x)} }[/math]

The integral can thus be simulated as [math]\displaystyle{ \displaystyle \hat{I} = \frac{1}{n} \sum_{i=1}^{n} Y_i \ , \ where \ Y_i = \frac{h(x_i)f(x_i)}{g(x_i)} }[/math]

Procedure

Suppose we know how to sample from [math]\displaystyle{ \displaystyle g(x) }[/math]

1. Choose a suitable [math]\displaystyle{ \displaystyle g(x) }[/math] and draw n samples [math]\displaystyle{ x_1,x_2....,x_n \sim~ g(x) }[/math]
2. Set [math]\displaystyle{ Y_i =\frac{h(x_i)f(x_i)}{g(x_i)} }[/math]
3. Compute [math]\displaystyle{ \hat{I} = \frac{1}{n}\sum_{i=1}^{n} Y_i }[/math]
   

By the Law of large numbers, [math]\displaystyle{ \displaystyle \hat{I} \rightarrow I }[/math] provided that the sample size n is large enough.

Remarks: One can think of [math]\displaystyle{ \frac{f(x)}{g(x)} }[/math] as a weight to [math]\displaystyle{ \displaystyle h(x) }[/math] in the computation of [math]\displaystyle{ \hat{I} }[/math]

[math]\displaystyle{ \displaystyle i.e. \ \hat{I} = \frac{1}{n}\sum_{i=1}^{n} Y_i = \frac{1}{n}\sum_{i=1}^{n} (\frac{f(x_i)}{g(x_i)})h(x_i) }[/math]

Therefore, [math]\displaystyle{ \displaystyle \hat{I} }[/math] is a weighted average of [math]\displaystyle{ \displaystyle h(x_i) }[/math]

Problem

If [math]\displaystyle{ \displaystyle g(x) }[/math] is not chosen appropriately, then the variance of the estimate [math]\displaystyle{ \hat{I} }[/math] may be very large. Here we actually face a similar problem with Rejection-Acceptance Approach. Consider the second moment of [math]\displaystyle{ \displaystyle I }[/math]:

[math]\displaystyle{ \begin{align} \displaystyle I & = E_g((y(x))^2) \\ & = \int (\frac{h(x)f(x)}{g(x)})^2 g(x) dx \\ & = \int \frac{h^2(x) f^2(x)}{g^2(x)} g(x) dx \\ & = \int \frac{h^2(x)f^2(x)}{g(x)} dx \\ \end{align} }[/math]

When [math]\displaystyle{ \displaystyle g(x) }[/math] is very small, then the above integral could be infinitely large, hence the variance can be very large when g is not chosen appropriately. This occurs when [math]\displaystyle{ \displaystyle g(x) }[/math] has a thinner tail than [math]\displaystyle{ \displaystyle f(x) }[/math] such that the quantity [math]\displaystyle{ \displaystyle \frac{h^2(x)f^2(x)}{g(x)} }[/math] is large.

Remarks:

1. We can actually compute the form of [math]\displaystyle{ \displaystyle g(x) }[/math] to have optimal variance.
Mathematically, it is to find [math]\displaystyle{ \displaystyle g(x) }[/math] subject to [math]\displaystyle{ \displaystyle \min_g [\ E_g([y(x)]^2) - (E_g[y(x)])^2\ ] }[/math]
It can be shown that the optimal [math]\displaystyle{ \displaystyle g(x) }[/math] is [math]\displaystyle{ \displaystyle \frac{|h(x)|f(x)}{\int_{-\infty}^{\infty}|h(s)|f(s)ds} }[/math]. Using the optimal [math]\displaystyle{ \displaystyle g(x) }[/math] will minimize the variance of estimation in Importance Sampling. This is of theoretical interest but not useful in practice. As we can see, if we can actually show the expression of g(x), we must first have the value of the integration---which is what we want in the first place.

2. In practice, we shall choose [math]\displaystyle{ \displaystyle g(x) }[/math] which has similar shape as [math]\displaystyle{ \displaystyle f(x) }[/math] but with a thicker tail than [math]\displaystyle{ \displaystyle f(x) }[/math] in order to avoid the problem mentioned above.

Example

Estimate [math]\displaystyle{ \displaystyle I = Pr(Z\gt 3),\ where\ Z \sim N(0,1) }[/math]

Method 1: Basic Monte Carlo

[math]\displaystyle{ \begin{align} Pr(Z\gt 3) & = \int^\infty_3 f(x)\,dx \\ & = \int^\infty_{-\infty} h(x)f(x)\,dx \end{align} }[/math]
[math]\displaystyle{ where \ h(x) = \begin{cases} 0, & \text{if } x \le 3 \\ 1, & \text{if } x \gt 3 \end{cases} }[/math] [math]\displaystyle{ \ ,\ f(x) = \frac{1}{\sqrt{2\pi}}e^{-\frac{1}{2}x^2} }[/math]

MATLAB code to compute [math]\displaystyle{ \displaystyle I }[/math] from 100 samples of standard normal distribution:

h = randn(100,1) > 3;
I = mean(h)

In one execution of the code, it returns a value of 0 for [math]\displaystyle{ \displaystyle I }[/math], which differs significantly from the true value of [math]\displaystyle{ \displaystyle I \approx 0.0013 }[/math]. The problem of using Basic Monte Carlo in this example is that [math]\displaystyle{ \displaystyle Pr(Z\gt 3) }[/math] has a small value, and hence many points sampled from the standard normal distribution will be wasted. Therefore, although Basic Monte Carlo is a feasible method to compute [math]\displaystyle{ \displaystyle I }[/math], it gives a poor estimation.

Method 2: Importance Sampling

[math]\displaystyle{ \displaystyle I = Pr(Z\gt 3)= \int^\infty_3 f(x)\,dx }[/math]

To apply importance sampling, we have to choose a [math]\displaystyle{ \displaystyle g(x) }[/math] which we will sample from. In this example, we can choose [math]\displaystyle{ \displaystyle g(x) }[/math] to be the probability density function of exponential distribution, normal distribution with mean 0 and variance greater than 1 or normal distribution with mean greater than 0 and variance 1 etc.. For the following, we take [math]\displaystyle{ \displaystyle g(x) }[/math] to be the pdf of [math]\displaystyle{ \displaystyle N(4,1) }[/math].

Procedure:

1. Draw n samples [math]\displaystyle{ x_1,x_2....,x_n \sim~ g(x) }[/math]
2. Calculate [math]\displaystyle{ \begin{align} \frac{f(x)}{g(x)} & = \frac{ \frac{1}{\sqrt{2\pi}}e^{-\frac{1}{2}x^2} }{ \frac{1}{\sqrt{2\pi}}e^{-\frac{1}{2}(x-4)^2} } \\ & = e^{8-4x} \end{align} }[/math]
   
3. Set [math]\displaystyle{ Y_i = h(x_i)e^{8-4x_i}\ with\ h(x) = \begin{cases} 0, & \text{if } x \le 3 \\ 1, & \text{if } x \gt 3 \end{cases} }[/math]
   
4. Compute [math]\displaystyle{ \hat{Y} = \frac{1}{n}\sum_{i=1}^{n} Y_i }[/math]
   

The above procedure from 100 samples of [math]\displaystyle{ \displaystyle g(x) }[/math]can be implemented in MATLAB as follow:

for ii = 1:100
  x = randn + 4 ;
  h = x > 3 ;
  y(ii) = h * exp(8-4*x) ;
end
mean(y)

In one execution of the code, it returns a value of 0.001271 for [math]\displaystyle{ \hat{Y} }[/math], which is much closer to the true value of [math]\displaystyle{ \displaystyle I \approx 0.0013 }[/math]. From many executions of the code, the variance of basic monte carlo is approximately 150 times that of importance sampling. This demonstrates that this method can provide a better estimate than the Basic Monte Carlo method.

Importance Sampling with Normalized Weight and Markov Chain Monte Carlo - October 13th, 2011

Importance Sampling with Normalized Weight

Recall that we can think of [math]\displaystyle{ \displaystyle b(x) = \frac{f(x)}{g(x)} }[/math] as a weight applied to the samples [math]\displaystyle{ \displaystyle h(x) }[/math]. If the form of [math]\displaystyle{ \displaystyle f(x) }[/math] is known only up to a constant, we can use an alternate, normalized form of the weight, [math]\displaystyle{ \displaystyle b^*(x) }[/math]. (This situation arises in Bayesian inference.) Importance sampling with normalized or standard weight is also called indirect importance sampling.

We derive the normalized weight as follows:
[math]\displaystyle{ \begin{align} \displaystyle I & = \int h(x)f(x)\,dx \\ &= \int h(x)\frac{f(x)}{g(x)}g(x)\,dx \\ &= \frac{\int h(x)\frac{f(x)}{g(x)}g(x)\,dx}{\int f(x) dx} \\ &= \frac{\int h(x)\frac{f(x)}{g(x)}g(x)\,dx}{\int \frac{f(x)}{g(x)}g(x) dx} \\ &= \frac{\int h(x)b(x)g(x)\,dx}{\int\ b(x)g(x) dx} \end{align} }[/math]

[math]\displaystyle{ \hat{I}= \frac{\sum_{i=1}^{n} h(x_i)b(x_i)}{\sum_{i=1}^{n} b(x_i)} }[/math]

Then, the normalized weight is [math]\displaystyle{ b^*(x) = \displaystyle \frac{b(x_i)}{\sum_{i=1}^{n} b(x_i)} }[/math]

Markov Chain Monte Carlo

We still want to solve [math]\displaystyle{ I = \displaystyle\int^\ h(x)f(x)\,dx }[/math]

Stochastic Process

A stochastic process [math]\displaystyle{ \{ x_t : t \in T \} }[/math] is a collection of random variables. Variables [math]\displaystyle{ \displaystyle x_t }[/math] take values in some set [math]\displaystyle{ \displaystyle X }[/math] called the space set. The set [math]\displaystyle{ \displaystyle T }[/math] is called the index set.

Markov Chain

A Markov Chain is a stochastic process for which the distribution of [math]\displaystyle{ \displaystyle x_t }[/math] depends only on [math]\displaystyle{ \displaystyle x_t-1 }[/math]. It is a random process characterized as being memoryless; meaning that the next occurrence of a defined event is only dependent on the current event and not on the sequence of events that preceded it.

  Formal Definition: The process [math]\displaystyle{  \{ x_t : t \in T \} }[/math] is a Markov Chain if [math]\displaystyle{ \displaystyle Pr(x_t|x_0, x_1,..., x_{t-1})= Pr(x_t|x_{t-1}) }[/math] for all [math]\displaystyle{  \{t \in T \} }[/math] and for all [math]\displaystyle{  \{x \in X \} }[/math]

For a Markov Chain, [math]\displaystyle{ \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]

Real Life Example:
When going for an interview, the employer only looks at your highest education achieved. The employer would not look at the past educations received (elementary school, high school etc.) because the employer believes that the highest education achieved summarizes your previous educations. Therefore, anything before your most recent previous education is irrelevant. In other word, we can say that[math]\displaystyle{ x_t }[/math]is regarded as the summary of [math]\displaystyle{ x_{t-1},...,x_2,x_1 }[/math], so when we need to determine [math]\displaystyle{ x_{t+1} }[/math], we only need to pay attention in [math]\displaystyle{ x_{t} }[/math].

Transition Probabilities

A Transition Probability is the probability of jumping from one state to another state.

  Formal Definition: We call [math]\displaystyle{ \displaystyle P_{ij} = Pr(x_{t+1}=j|x_t=i) }[/math] the transition probability.

The matrix P whose (i,j) element is [math]\displaystyle{ \displaystyle P_{ij} }[/math] is called the Transition Matrix.

Properties of P:

1) [math]\displaystyle{ \displaystyle P_{ij} \gt = 0 }[/math]

2) [math]\displaystyle{ \displaystyle \sum_{\forall i}P_{ij} = 1 }[/math]

Random Walk

Example: Start at one point and flip a coin where [math]\displaystyle{ \displaystyle Pr(H)=p }[/math] and [math]\displaystyle{ \displaystyle Pr(T)=1-p=q }[/math]. Take one step right if heads and one step left if tails. If at an endpoint, stay there. The transition matrix is [math]\displaystyle{ P=\left(\begin{matrix}1&0&0&\dots&\dots&0\\ q&0&p&0&\dots&0\\ 0&q&0&p&\dots&0\\ \vdots&\vdots&\vdots&\ddots&\vdots&\vdots\\ \vdots&\vdots&\vdots&\vdots&\ddots&\vdots\\ 0&0&\dots&\dots&\dots&1 \end{matrix}\right) }[/math]

Let [math]\displaystyle{ \displaystyle P_n }[/math] be the matrix such that its (i,j) element is [math]\displaystyle{ \displaystyle P_{ij}(n) }[/math]. This is called n-step probability.

[math]\displaystyle{ \displaystyle P_n = P^n }[/math]

[math]\displaystyle{ \displaystyle P_1 = P }[/math]

[math]\displaystyle{ \displaystyle P_2 = P^2 }[/math]

Markov Chain Properties and Page Rank - October 18th, 2011

Transition Matrix: [math]\displaystyle{ P(i,j) = Pr(x_{t+1} = j | x_t = i) }[/math]

n-step transition matrix [math]\displaystyle{ P_n(i,j) = Pr(x_{m+n}=j|x_m = i) }[/math]

[math]\displaystyle{ P_n = P^n }[/math]

Marginal distribution of a Markov Chain

We represent state at at time t as a vector.

[math]\displaystyle{ \mu_t = (\mu_t(1) \; \mu_t(2) ... \mu_t(n)) }[/math]

Consider this Markov Chain:

[math]\displaystyle{ \mu_t = (A \; B) }[/math], where A is the probability of being in state a and B is the probability of being in state b at time t.

For example if [math]\displaystyle{ \mu_t = (0.1 \; 0.9) }[/math], we have a 10% chance of being in state a, and a 90% chance of being in state b at time t.

Suppose we run this Markov chain many times, and record the state at each step.

In this example, we run 4 trials, up until t=5.

t	Trial 1	Trial 2	Trial 3	Trial 4	Observed [math]\displaystyle{ \mu }[/math]
1	a	b	b	a	(0.5, 0.5)
2	b	a	a	a	(0.75, 0.25)
3	a	a	b	a	(0.75, 0.25)
4	b	b	a	b	(0.25, 0.75)
5	b	b	b	a	(0.25, 0.75)

Imagine simulating the chain many times.

If we collect all the outcomes at time t from all the chains, the histogram of this data would look like [math]\displaystyle{ \mu_t }[/math].

We can find the marginal probabilities as [math]\displaystyle{ \mu_n = \mu_0 P^n }[/math]

Markov Chain Definitions:

Stationary Distribution:

Let [math]\displaystyle{ \Pi(\Pi_i, \forall i \in x) }[/math] be a vector of non-negative numbers that sum to 1. ([math]\displaystyle{ \Pi }[/math] is a pmf)

If [math]\displaystyle{ \Pi = \Pi P }[/math], [math]\displaystyle{ \Pi }[/math] is a stationary distribution, also known as an invariant distribution.

Limiting Distribution:

A chain has limiting distribution \pi if P^n = column [\pi \pi \pi]

That is, \pi_j = lim[P^n]_ij as n approaches infinity exists and is independent of i.

Detailed Balance:

If \Pi has the detailed balance property, then \Pi_iP_ij = P_ji\Pi_j

If \Pi has the detailed balance property, then it is stationary.

This means that if \Pi_iP_ij = P_ji\Pi_j, then \Pi = \Pi P

Proof:

Multiply out \Pi P. = \sum over all i, \Pi_i P_ij \sum over all i, \P_ji \pi_j, by the fact that \pi has detailed balance Bring out the \Pi_j, the rest must sum to 1. = \Pi_j

Find the transition matrix of a Markov Chain: Insert Example Here:

P = [.2 .8; .6 .4]

% starting state mu = [.1 .9]

% at time = 1 mu = mu * P

Notice that this example converges, if you repeatedly run: mu = mu * P

It converges to \mu = (0.4286, 0.5714)

Note: There are chains with stationary distributions, that don't converge (The chain might not naturally reach the stationary distribution, and isn't limiting)

An example of this is: P = [0 1 0; 0 0 1; 1 0 0] mu = [1/3 1/3 1/3] mu * P % mu * P == mu, thus there is a stationary distribution % P^n doesn't converge though

Intuition

Used by Google to rank web pages.

When you enter a query, there are a set of webpages which are matched by this query, but this set of pages must be ordered.

We represent webpages by a set of nodes, where web links are represented as edges.

If many pages point to a page, that page should be important.

Links from important pages should be more highly valued.

If a page links to tons of other pages, each link should be weighted less heavily.

Modelling

We need to model this set of links:

L_ij =

1 if j points to i

0 otherwise

c_j = \sum over all i of L_ij, this is the number of outgoing links from page j

Insert example here:

0 0 1 0
1 0 0 0
1 1 0 1 
0 0 0 0

c =

2 1 1 1

The rank of a page is P_i.

P_i = (1-d) + d\sum ((L_ij/c_j) P_j)

We divide by the number of outgoing links, and sum all incoming.

We don't want to completely ignore pages with no incoming links, which is why we add (1-d).

L is a matrix of L_11 to L_NN D = the matrix of zeroes, with c on the diagonal

Then P = (1-d)e + dLD^-1P

Where e is = [1 1 ... 1]

To simplify the problem, we let e^t P = N.

This means that the average importance of all pages on the internet is 1.

Then P = (1-d)(ee^tP)/N + dLD^-1P

P = [(1-d)(ee^t)/N + dLD^-1] * P

Lets call [(1-d)(ee^t)/N + dLD^-1] = A.

Then P = AP

Then we know that P is an eigenvector of A with eigenvalue 1.

So we can find the ranks of all pages by finding an eigenvector of A.

In reality this is too expensive, so Google uses a method to approximate an eigenvector with eigenvalue 1.