stat341 / CM 361

Computational Statistics and Data Analysis is a course offered at the University of Waterloo
Spring 2009
Instructor: Ali Ghodsi

Sampling (Generating random numbers)

Generating Random Numbers - May 12, 2009

Generating random numbers in a computational setting presents challenges. A good way to generate random numbers in computational statistics involves analyzing various distributions using computational methods. As a result, the probability distribution of each possible number appears to be uniform (pseudo-random). Outside a computational setting, presenting a uniform distribution is fairly easy (for example, rolling a fair die repetitively to produce a series of random numbers from 1 to 6).

We begin by considering the simplest case: the uniform distribution.

Multiplicative Congruential Method

One way to generate pseudo random numbers from the uniform distribution is using the Multiplicative Congruential Method. This involves three integer parameters a, b, and m, and a seed variable x₀. This method deterministically generates a sequence of numbers (based on the seed) with a seemingly random distribution (with some caveats). It proceeds as follows:

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

For example, with a = 13, b = 0, m = 31, x₀ = 1, we have:

[math]\displaystyle{ x_{i+1} = 13x_{i} \mod{31} }[/math]

So,

[math]\displaystyle{ \begin{align} x_{0} &{}= 1 \end{align} }[/math]

[math]\displaystyle{ \begin{align} x_{1} &{}= 13 \times 1 + 0 \mod{31} = 13 \\ \end{align} }[/math]

[math]\displaystyle{ \begin{align} x_{2} &{}= 13 \times 13 + 0 \mod{31} = 14 \\ \end{align} }[/math]

[math]\displaystyle{ \begin{align} x_{3} &{}= 13 \times 14 + 0 \mod{31} =27 \\ \end{align} }[/math]

etc.

The above generator of pseudorandom numbers is called a Mixed Congruential Generator or Linear Congruential Generator, as they involve both an additive and a muliplicative term. For correctly chosen values of a, b, and m, this method will generate a sequence of integers including all integers between 0 and m - 1. Scaling the output by dividing the terms of the resulting sequence by m - 1, we create a sequence of numbers between 0 and 1, which is similar to sampling from a uniform distribution.

Of course, not all values of a, b, and m will behave in this way, and will not be suitable for use in generating pseudo random numbers.

For example, with a = 3, b = 2, m = 4, x₀ = 1, we have:

[math]\displaystyle{ x_{i+1} = (3x_{i} + 2) \mod{4} }[/math]

So,

[math]\displaystyle{ \begin{align} x_{0} &{}= 1 \end{align} }[/math]

[math]\displaystyle{ \begin{align} x_{1} &{}= 3 \times 1 + 2 \mod{4} = 1 \\ \end{align} }[/math]

[math]\displaystyle{ \begin{align} x_{2} &{}= 3 \times 1 + 2 \mod{4} = 1 \\ \end{align} }[/math]

etc.

For an ideal situation, we want m to be a very large prime number, [math]\displaystyle{ x_{n}\not= 0 }[/math] for any n, and the period is equal to m-1. In practice, it has been found by a paper published in 1988 by Park and Miller, that a = 7⁵, b = 0, and m = 2³¹ - 1 = 2147483647 (the maximum size of a signed integer in a 32-bit system) are good values for the Multiplicative Congruential Method.

Java's Random class is based on a generator with a = 25214903917, b = 11, and m = 2⁴⁸<ref>http://java.sun.com/javase/6/docs/api/java/util/Random.html#next(int)</ref>. The class returns at most 32 leading bits from each x_i, so it is possible to get the same value twice in a row, (when x₀ = 18698324575379, for instance) without repeating it forever.

General Methods

Since the Multiplicative Congruential Method can only be used for the uniform distribution, other methods must be developed in order to generate pseudo random numbers from other distributions.

Inverse Transform Method

This method uses the fact that when a random sample from the uniform distribution is applied to the inverse of a cumulative density function (cdf) of some distribution, the result is a random sample of that distribution. This is shown by this theorem:

Theorem:

If [math]\displaystyle{ U \sim~ \mathrm{Unif}[0, 1] }[/math] is a random variable and [math]\displaystyle{ X = F^{-1}(U) }[/math], where F is continuous, monotonic, and is the cumulative density function (cdf) for some distribution, then the distribution of the random variable X is given by F(X).

Proof:

Recall that, if f is the pdf corresponding to F, then,

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

So F is monotonically increasing, since the probability that X is less than a greater number must be greater than the probability that X is less than a lesser number.

Note also that in the uniform distribution on [0, 1], we have for all a within [0, 1], [math]\displaystyle{ P(U \leq a) = a }[/math].

So,

[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.

Procedure (Continuous Case)

This method then gives us the following procedure for finding pseudo random numbers from a continuous distribution:

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

Example:

Suppose we want to draw a sample from [math]\displaystyle{ f(x) = \lambda e^{-\lambda x} }[/math] where [math]\displaystyle{ x \gt 0 }[/math] (the exponential distribution).

We need to first find [math]\displaystyle{ F(x) }[/math] and then its inverse, [math]\displaystyle{ F^{-1} }[/math].

[math]\displaystyle{ F(x) = \int^x_0 \theta e^{-\theta u} du = 1 - e^{-\theta x} }[/math]

[math]\displaystyle{ F^{-1}(x) = \frac{-\log(1-y)}{\theta} = \frac{-\log(u)}{\theta} }[/math]

Now we can generate our random sample [math]\displaystyle{ i=1\dots n }[/math] from [math]\displaystyle{ f(x) }[/math] by:

[math]\displaystyle{ 1)\ u_i \sim Unif[0, 1] }[/math]

[math]\displaystyle{ 2)\ x_i = \frac{-\log(u_i)}{\theta} }[/math]

The [math]\displaystyle{ x_i }[/math] are now a random sample from [math]\displaystyle{ f(x) }[/math].

This example can be illustrated in Matlab using the code below. Generate [math]\displaystyle{ u_i }[/math], calculate [math]\displaystyle{ x_i }[/math] using the above formula and letting [math]\displaystyle{ \theta=1 }[/math], plot the histogram of [math]\displaystyle{ x_i }[/math]'s for [math]\displaystyle{ i=1,...,100,000 }[/math].

u=rand(1,100000);
x=-log(1-u)/1;
hist(x)

The histogram shows exponential pattern as expected.

The major problem with this approach is that we have to find [math]\displaystyle{ F^{-1} }[/math] and for many distributions it is too difficult (or impossible) to find the inverse of [math]\displaystyle{ F(x) }[/math]. Further, for some distributions it is not even possible to find [math]\displaystyle{ F(x) }[/math] (i.e. a closed form expression for the distribution function, or otherwise; even if the closed form expression exists, it's usually difficult to find [math]\displaystyle{ F^{-1} }[/math]).

Procedure (Discrete Case)

The above method can be easily adapted to work on discrete distributions as well.

In general in the discrete case, we have [math]\displaystyle{ x_0, \dots , x_n }[/math] where:

[math]\displaystyle{ \begin{align}P(X = x_i) &{}= p_i \end{align} }[/math]

[math]\displaystyle{ x_0 \leq x_1 \leq x_2 \dots \leq x_n }[/math]

[math]\displaystyle{ \sum p_i = 1 }[/math]

Thus we can define the following method to find pseudo random numbers in the discrete case (note that the less-than signs from class have been changed to less-than-or-equal-to signs by me, since otherwise the case of [math]\displaystyle{ U = 1 }[/math] is missed):

• Step 1: Draw [math]\displaystyle{ U~ \sim~ Unif [0,1] }[/math].
• Step 2:
  • If [math]\displaystyle{ U \lt p_0 }[/math], return [math]\displaystyle{ X = x_0 }[/math]
  • If [math]\displaystyle{ p_0 \leq U \lt p_0 + p_1 }[/math], return [math]\displaystyle{ X = x_1 }[/math]
  • ...
  • In general, if [math]\displaystyle{ p_0+ p_1 + \dots + p_{k-1} \leq U \lt p_0 + \dots + p_k }[/math], return [math]\displaystyle{ X = x_k }[/math]

Example (from class):

Suppose we have the following discrete distribution:

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

The cumulative density function (cdf) for this distribution is then:

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

Then we can generate numbers from this distribution like this, given [math]\displaystyle{ u_0, \dots, u_n }[/math] from [math]\displaystyle{ U \sim~ Unif[0, 1] }[/math]:

[math]\displaystyle{ x_i = \begin{cases} 0, & \text{if } u_i \leq 0.3 \\ 1, & \text{if } 0.3 \lt u_i \leq 0.5 \\ 2, & \text{if } 0.5 \lt u_i \leq 1 \end{cases} }[/math]

This example can be illustrated in Matlab using the code below:

p=[0.3,0.2,0.5];
for i=1:1000;
  u=rand;
  if u <= p(1)
    x(i)=0;
  elseif u < sum(p(1,2))
    x(i)=1;
  else
    x(i)=2;
  end
end

Acceptance-Rejection Sampling - May 14, 2009

Today, we continue the discussion on sampling (generating random numbers) from general distributions with the Acceptance/Rejection Method.

Acceptance/Rejection Method

Suppose we wish to sample from a target distribution [math]\displaystyle{ f(x) }[/math] that is difficult or impossible to sample from directly. Suppose also that we have a proposal distribution [math]\displaystyle{ g(x) }[/math] from which we have a reasonable method of sampling (e.g. the uniform distribution). Then, if there is a constant [math]\displaystyle{ c \ |\ c \cdot g(x) \geq f(x)\ \forall x }[/math], accepting samples drawn in successions from [math]\displaystyle{ c \cdot g(x) }[/math] with ratio [math]\displaystyle{ \frac {f(x)}{c \cdot g(x)} }[/math] close to 1 will yield a sample that follows the target distribution [math]\displaystyle{ f(x) }[/math]; on the other hand we would reject the samples if the ratio is not close to 1.

The following graph shows the pdf of [math]\displaystyle{ f(x) }[/math] (target distribution) and [math]\displaystyle{ c \cdot g(x) }[/math] (proposal distribution)

File:fxcgx.JPG

At x=7; sampling from [math]\displaystyle{ c \cdot g(x) }[/math] will yield a sample that follows the target distribution [math]\displaystyle{ f(x) }[/math]

At x=9; we will reject samples according to the ratio [math]\displaystyle{ \frac {f(x)}{c \cdot g(x)} }[/math] after sampling from [math]\displaystyle{ c \cdot g(x) }[/math]

Proof

Note the following:

• [math]\displaystyle{ Pr(X|accept) = \frac{Pr(accept|X) \times Pr(X)}{Pr(accept)} }[/math] (Bayes' theorem)
• [math]\displaystyle{ Pr(accept|X) = \frac{f(x)}{c \cdot g(x)} }[/math]
• [math]\displaystyle{ Pr(X) = g(x)\frac{}{} }[/math]

So, [math]\displaystyle{ Pr(accept) = \int^{}_x Pr(accept|X) \times Pr(X) dx }[/math] [math]\displaystyle{ = \int^{}_x \frac{f(x)}{c \cdot g(x)} g(x) dx }[/math] [math]\displaystyle{ = \frac{1}{c} \int^{}_x f(x) dx }[/math] [math]\displaystyle{ = \frac{1}{c} }[/math]

Therefore, [math]\displaystyle{ Pr(X|accept) = \frac{\frac{f(x)}{c\ \cdot g(x)} \times g(x)}{\frac{1}{c}} = f(x) }[/math] as required.

Procedure (Continuous Case)

• Choose [math]\displaystyle{ g(x) }[/math] (a density function that is simple to sample from)
• Find a constant c such that :[math]\displaystyle{ c \cdot g(x) \geq f(x) }[/math]

1. Let [math]\displaystyle{ Y \sim~ g(y) }[/math]
2. Let [math]\displaystyle{ U \sim~ Unif [0,1] }[/math]
3. If [math]\displaystyle{ U \leq \frac{f(x)}{c \cdot g(x)} }[/math] then X=Y; else reject and go to step 1

Example:

Suppose we want to sample from Beta(2,1), for [math]\displaystyle{ 0 \leq x \leq 1 }[/math]. Recall:

[math]\displaystyle{ Beta(2,1) = \frac{\Gamma (2+1)}{\Gamma (2) \Gamma(1)} \times x^1(1-x)^0 = \frac{2!}{1!0!} \times x = 2x }[/math]

• Choose [math]\displaystyle{ g(x) \sim~ Unif [0,1] }[/math]
• Find c: c = 2 (see notes below)

1. Let [math]\displaystyle{ Y \sim~ Unif [0,1] }[/math]
2. Let [math]\displaystyle{ U \sim~ Unif [0,1] }[/math]
3. If [math]\displaystyle{ U \leq \frac{2Y}{2} = Y }[/math], then X=Y; else go to step 1

[math]\displaystyle{ c }[/math] was chosen to be 2 in this example since [math]\displaystyle{ \max \left(\frac{f(x)}{g(x)}\right) }[/math] in this example is 2. This condition is important since it helps us in finding a suitable [math]\displaystyle{ c }[/math] to apply the Acceptance/Rejection Method.

In MATLAB, the code that demonstrates the result of this example is:

   j = 1;
       while i < 1000
           y = rand;
           u = rand;
           if u <= y
               x(j) = y;
               j = j + 1;
               i = i + 1;
           end
       end
       hist(x);
       

The histogram produced here should follow the target distribution, [math]\displaystyle{ f(x) = 2x }[/math], for the interval [math]\displaystyle{ 0 \leq x \leq 1 }[/math].

The histogram shows the PDF of a Beta(2,1) distribution as expected.

The Discrete Case

The Acceptance/Rejection Method can be extended for discrete target distributions. The difference compared to the continuous case is that the proposal distribution [math]\displaystyle{ g(x) }[/math] must also be discrete distribution. For our purposes, we can consider g(x) to be a discrete uniform distribution on the set of values that X may take on in the target distribution.

Example

Suppose we want to sample from a distribution with the following probability mass function (pmf):

P(X=1) = 0.15
P(X=2) = 0.55
P(X=3) = 0.20
P(X=4) = 0.10 

• Choose [math]\displaystyle{ g(x) }[/math] to be the discrete uniform distribution on the set [math]\displaystyle{ \{1,2,3,4\} }[/math]
• Find c: [math]\displaystyle{ c = \max \left(\frac{f(x)}{g(x)} \right)= 0.55/0.25 = 2.2 }[/math]

1. Generate [math]\displaystyle{ Y \sim~ Unif \{1,2,3,4\} }[/math]
2. Let [math]\displaystyle{ U \sim~ Unif [0,1] }[/math]
3. If [math]\displaystyle{ U \leq \frac{f(x)}{2.2 \times 0.25} }[/math], then X=Y; else go to step 1

Limitations

If the proposed distribution is very different from the target distribution, we may have to reject a large number of points before a good sample size of the target distribution can be established. It may also be difficult to find such [math]\displaystyle{ g(x) }[/math] that satisfies all the conditions of the procedure.

We will now begin to discuss sampling from specific distributions.

Special Technique for sampling from Gamma Distribution

Recall that the cdf of the Gamma distribution is:

[math]\displaystyle{ F(x) = \int_0^{\lambda x} \frac{e^{-y}y^{t-1}}{(t-1)!} dy }[/math]

If we wish to sample from this distribution, it will be difficult for both the Inverse Method (difficulty in computing the inverse function) and the Acceptance/Rejection Method.

Additive Property of Gamma Distribution

Recall that if [math]\displaystyle{ X_1, \dots, X_t }[/math] are independent exponential distributions with mean [math]\displaystyle{ \lambda }[/math] (in other words, [math]\displaystyle{ X_i\sim~ Exp (\lambda) }[/math]), then [math]\displaystyle{ \Sigma_{i=1}^t X_i \sim~ Gamma (t, \lambda) }[/math]

It appears that if we want to sample from the Gamma distribution, we can consider sampling from t independent exponential distributions with mean [math]\displaystyle{ \lambda }[/math] (using the Inverse Method) and add them up. Details will be discussed in the next lecture.

Techniques for Normal and Gamma Sampling - May 19, 2009

We have examined two general techniques for sampling from distributions. However, for certain distributions more practical methods exist. We will now look at two cases,
Gamma distributions and Normal distributions, where such practical methods exist.

Gamma Distribution

Given the additive property of the gamma distribution,

If [math]\displaystyle{ X_1, \dots, X_t }[/math] are independent random variables with [math]\displaystyle{ X_i\sim~ Exp (\lambda) }[/math] then,

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

We can use the Inverse Transform Method and sample from independent uniform distributions seen before to generate a sample following a Gamma distribution.

Procedure

1. Sample independently from a uniform distribution [math]\displaystyle{ t }[/math] times, giving [math]\displaystyle{ u_1,\dots,u_t }[/math]
2. Sample independently from an exponential distribution [math]\displaystyle{ t }[/math] times, giving [math]\displaystyle{ x_1,\dots,x_t }[/math] such that,
   [math]\displaystyle{ \begin{align} x_1 \sim~ Exp(\lambda)\\ \vdots \\ x_t \sim~ Exp(\lambda) \end{align} }[/math]
   
   Using the Inverse Transform Method,
   [math]\displaystyle{ \begin{align} x_i = -\frac {1}{\lambda}\log(u_i) \end{align} }[/math]
3. Using the additive property,
   [math]\displaystyle{ \begin{align} X &{}= x_1 + x_2 + \dots + x_t \\ X &{}= -\frac {1}{\lambda}\log(u_1) - \frac {1}{\lambda}\log(u_2) \dots - \frac {1}{\lambda}\log(u_t) \\ X &{}= -\frac {1}{\lambda}\log(\prod_{i=1}^{t}u_i) \sim~ Gamma (t, \lambda) \end{align} }[/math]

This procedure can be illustrated in Matlab using the code below with [math]\displaystyle{ t = 5, \lambda = 1 }[/math] :

U = rand(10000,5);
X = sum( -log(U), 2);
hist(X)

Normal Distribution

The cdf for the Standard Normal distribution is:

[math]\displaystyle{ F(Z) = \int_{-\infty}^{Z}\frac{1}{\sqrt{2\pi}}e^{-x^2/2}dx }[/math]

We can see that the normal distribution is difficult to sample from using the general methods seen so far, eg. the inverse is not easy to obtain from F(Z); we may be able to use the Acceptance-Rejection method, but there are still better ways to sample from a Standard Normal Distribution.

=====Box-Muller Method===== [Add a picture WikiSysop 19:25, 1 June 2009 (UTC)]

"Diagram of the Box Muller transform, which transforms uniformly distributed value pairs to normally distributed value pairs." [Box-Muller Transform, Wikipedia]

The Box-Muller or Polar method uses an approach where we have one space that is easy to sample in, and another with the desired distribution under a transformation. If we know such a transformation for the Standard Normal, then all we have to do is transform our easy sample and obtain a sample from the Standard Normal distribution.

Properties of Polar and Cartesian Coordinates

If x and y are points on the Cartesian plane, r is the length of the radius from a point in the polar plane to the pole, and θ is the angle formed with the polar axis then,

• [math]\displaystyle{ \begin{align} r^2 = x^2 + y^2 \end{align} }[/math]
• [math]\displaystyle{ \tan(\theta) = \frac{y}{x} }[/math]

• [math]\displaystyle{ \begin{align} x = r \cos(\theta) \end{align} }[/math]
• [math]\displaystyle{ \begin{align} y = r \sin(\theta) \end{align} }[/math]

Let X and Y be independent random variables with a standard normal distribution,

[math]\displaystyle{ X \sim~ N(0,1) }[/math]

[math]\displaystyle{ Y \sim~ N(0,1) }[/math]

also, let [math]\displaystyle{ r }[/math] and [math]\displaystyle{ \theta }[/math] be the polar coordinates of (x,y). Then the joint distribution of independent x and y is given by,

[math]\displaystyle{ \begin{align} f(x,y) = f(x)f(y) &{}= \frac{1}{\sqrt{2\pi}}e^{-\frac{x^2}{2}}\frac{1}{\sqrt{2\pi}}e^{-\frac{y^2}{2}} \\ &{}=\frac{1}{2\pi}e^{-\frac{x^2+y^2}{2}} \end{align} }[/math]

It can also be shown that the joint distribution of r and θ is given by,

[math]\displaystyle{ \begin{matrix} f(r,\theta) = \frac{1}{2}e^{-\frac{d}{2}}*\frac{1}{2\pi},\quad d = r^2 \end{matrix} }[/math]

Note that [math]\displaystyle{ \begin{matrix}f(r,\theta)\end{matrix} }[/math] consists of two density functions, Exponential and Uniform, so assuming that r and [math]\displaystyle{ \theta }[/math] are independent [math]\displaystyle{ \begin{matrix} \Rightarrow d \sim~ Exp(1/2), \theta \sim~ Unif[0,2\pi] \end{matrix} }[/math]

Procedure for using Box-Muller Method

1. Sample independently from a uniform distribution twice, giving [math]\displaystyle{ \begin{align} u_1,u_2 \sim~ \mathrm{Unif}(0, 1) \end{align} }[/math]
2. Generate polar coordinates using the exponential distribution of d and uniform distribution of θ,
   [math]\displaystyle{ \begin{align} d = -2\log(u_1),& \quad r = \sqrt{d} \\ & \quad \theta = 2\pi u_2 \end{align} }[/math]
3. Transform r and θ back to x and y,
   [math]\displaystyle{ \begin{align} x = r\cos(\theta) \\ y = r\sin(\theta) \end{align} }[/math]

Notice that the Box-Muller Method generates a pair of independent Standard Normal distributions, x and y.

This procedure can be illustrated in Matlab using the code below:

u1 = rand(5000,1);
u2 = rand(5000,1);

d = -2*log(u1);
theta = 2*pi*u2;

x = d.^(1/2).*cos(theta);
y = d.^(1/2).*sin(theta);

figure(1);

subplot(2,1,1);
hist(x);
title('X');
subplot(2,1,2);
hist(y);
title('Y');

Also, we can confirm that d and theta are indeed exponential and uniform random variables, respectively, in Matlab by:

subplot(2,1,1);
hist(d);
title('d follows an exponential distribution');
subplot(2,1,2);
hist(theta);
title('theta follows a uniform distribution over [0, 2*pi]');

Useful Properties (Single and Multivariate)

Box-Muller can be used to sample a standard normal distribution. However, there are many properties of Normal distributions that allow us to use the samples from Box-Muller method to sample any Normal distribution in general.

Properties of Normal distributions

• [math]\displaystyle{ \begin{align} \text{If } & X = \mu + \sigma Z, & Z \sim~ N(0,1) \\ &\text{then } X \sim~ N(\mu,\sigma ^2) \end{align} }[/math]

• [math]\displaystyle{ \begin{align} \text{If } & \vec{Z} = (Z_1,\dots,Z_d)^T, & Z_1,\dots,Z_d \sim~ N(0,1) \\ &\text{then } \vec{Z} \sim~ N(\vec{0},I) \end{align} }[/math]

• [math]\displaystyle{ \begin{align} \text{If } & \vec{X} = \vec{\mu} + \Sigma^{1/2} \vec{Z}, & \vec{Z} \sim~ N(\vec{0},I) \\ &\text{then } \vec{X} \sim~ N(\vec{\mu},\Sigma) \end{align} }[/math]

These properties can be illustrated through the following example in Matlab using the code below:

Example: For a Multivariate Normal distribution [math]\displaystyle{ u=\begin{bmatrix} -2 \\ 3 \end{bmatrix} }[/math] and [math]\displaystyle{ \Sigma=\begin{bmatrix} 1&0.5\\ 0.5&1\end{bmatrix} }[/math]

u = [-2; 3];
sigma = [ 1 1/2; 1/2 1];

r = randn(15000,2);
ss = chol(sigma);

X = ones(15000,1)*u' + r*ss;
plot(X(:,1),X(:,2), '.');

Note: In the example above, we had to generate the square root of [math]\displaystyle{ \Sigma }[/math] using the Cholesky decomposition,

ss = chol(sigma);

which gives [math]\displaystyle{ ss=\begin{bmatrix} 1&0.5\\ 0&0.8660\end{bmatrix} }[/math]. Matlab also has the sqrtm function:

ss = sqrtm(sigma);

which gives a different matrix, [math]\displaystyle{ ss=\begin{bmatrix} 0.9659&0.2588\\ 0.2588&0.9659\end{bmatrix} }[/math], but the plot looks about the same (X has the same distribution).

Bayesian and Frequentist Schools of Thought - May 21, 2009

In this lecture we will continue to discuss sampling from specific distributions , introduce Monte Carlo Integration, and also talk about the differences between the Bayesian and Frequentist views on probability, along with references to Bayesian Inference.

Binomial Distribution

A Binomial distribution [math]\displaystyle{ X \sim~ Bin(n,p) }[/math] is the sum of [math]\displaystyle{ n }[/math] independent Bernoulli trials, each with probability of success [math]\displaystyle{ p }[/math]. For each trial we generate an independent uniform random variable: [math]\displaystyle{ U_1, \ldots, U_n \sim~ Unif(0,1) }[/math]. Then X is the number of times that [math]\displaystyle{ U_i \leq p }[/math]. In this case if n is large enough, by the central limit theorem, the Normal distribution can be used to approximate a Binomial distribution.

Sampling from Binomial distribution in Matlab is done using the following code:

n=3;
p=0.5;
trials=1000;
X=sum((rand(trials,n))'<=p);
hist(X)

Where the histogram is a Binomial distribution, and for higher [math]\displaystyle{ n }[/math], it would resemble a Normal distribution.

Monte Carlo Integration

Monte Carlo Integration is a numerical method of approximating the evaluation of integrals using random numbers generated from simulations. In this course we will mainly look at three methods for approximating integrals:

1. Basic Monte Carlo Integration
2. Importance Sampling
3. Markov Chain Monte Carlo (MCMC)

Bayesian VS Frequentists

During the history of statistics, two major schools of thought emerged along the way and have been locked in an on-going struggle in trying to determine which one has the correct view on probability. These two schools are known as the Bayesian and Frequentist schools of thought. Both the Bayesians and the Frequentists holds a different philosophical view on what defines probability. Below are some fundamental differences between the Bayesian and Frequentist schools of thought:

Frequentist

• Probability is objective and refers to the limit of an event's relative frequency in a large number of trials. For example, a coin with a 50% probability of heads will turn up heads 50% of the time.
• Parameters are all fixed and unknown constants.
• Any statistical process only has interpretations based on limited frequencies. For example, a 95% C.I. of a given parameter will contain the true value of the parameter 95% of the time.

Bayesian

• Probability is subjective and can be applied to single events based on degree of confidence or beliefs. For example, Bayesian can refer to tomorrow's weather as having 50% of rain, whereas this would not make sense to a Frequentist because tomorrow is just one unique event, and cannot be referred to as a relative frequency in a large number of trials.
• Parameters are random variables that has a given distribution, and other probability statements can be made about them.
• Probability has a distribution over the parameters, and point estimates are usually done by either taking the mode or the mean of the distribution.

Bayesian Inference

Example: If we have a screen that only displays single digits from 0 to 9, and this screen is split into a 4x5 matrix of pixels, then all together the 20 pixels that make up the screen can be referred to as [math]\displaystyle{ \vec{X} }[/math], which is our data, and the parameter of the data for this case, which we will refer to as [math]\displaystyle{ \theta }[/math], would be a discrete random variable that can take on the values of 0 to 9. In this example, a Bayesian would be interested in finding [math]\displaystyle{ Pr(\theta=a|\vec{X}=\vec{x}) }[/math], whereas a Frequentist would be more interested in finding [math]\displaystyle{ Pr(\vec{X}=\vec{x}|\theta=a) }[/math]

Bayes' Rule

[math]\displaystyle{ f(\theta|X) = \frac{f(X | \theta)\, f(\theta)}{f(X)}. }[/math]

Note: In this case [math]\displaystyle{ f (\theta|X) }[/math] is referred to as posterior, [math]\displaystyle{ f (X | \theta) }[/math] as likelihood, [math]\displaystyle{ f (\theta) }[/math] as prior, and [math]\displaystyle{ f (X) }[/math] as the marginal, where [math]\displaystyle{ \theta }[/math] is the parameter and [math]\displaystyle{ X }[/math] is the observed variable.

Procedure in Bayesian Inference

• First choose a probability distribution as the prior, which represents our beliefs about the parameters.
• Then choose a probability distribution for the likelihood, which represents our beliefs about the data.
• Lastly compute the posterior, which represents an update of our beliefs about the parameters after having observed the data.

As mentioned before, for a Bayesian, finding point estimates usually involves finding the mode or the mean of the parameter's distribution.

Methods

• Mode: [math]\displaystyle{ \theta = \arg\max_{\theta} f(\theta|X) \gets }[/math] value of [math]\displaystyle{ \theta }[/math] that maximizes [math]\displaystyle{ f(\theta|X) }[/math]
• Mean: [math]\displaystyle{ \bar\theta = \int^{}_\theta \theta \cdot f(\theta|X)d\theta }[/math]

If it is the case that [math]\displaystyle{ \theta }[/math] is high-dimensional, and we are only interested in one of the components of [math]\displaystyle{ \theta }[/math], for example, we want [math]\displaystyle{ \theta_1 }[/math] from [math]\displaystyle{ \vec{\theta}=(\theta_1,\dots,\theta_n) }[/math], then we would have to calculate the integral: [math]\displaystyle{ \int^{} \int^{} \dots \int^{}f(\theta|X)d\theta_2d\theta_3 \dots d\theta_n }[/math]

This sort of calculation is usually very difficult or not feasible to compute, and thus we would need to do it by simulation.

Note:

1. [math]\displaystyle{ f(x)=\int^{}_\theta f(X | \theta)f(\theta) d\theta }[/math] is not a function of [math]\displaystyle{ \theta }[/math], and is called the Normalization Factor
2. Therefore, since f(x) is like a constant, the posterior is proportional to the likelihood times the prior: [math]\displaystyle{ f(\theta|X)\propto f(X | \theta)f(\theta) }[/math]

Monte Carlo Integration - May 26, 2009

Today's lecture completes the discussion on the Frequentists and Bayesian schools of thought and introduces Basic Monte Carlo Integration.

Frequentist vs Bayesian Example - Estimating Parameters

Estimating parameters of a univariate Gaussian:

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

Frequentist Approach

Let [math]\displaystyle{ X^N }[/math] denote [math]\displaystyle{ (x_1, x_2, ..., x_n) }[/math]. Using the Maximum Likelihood Estimation approach for estimating parameters we get:

[math]\displaystyle{ L(X^N; \theta) = \prod_{i=1}^N \frac{1}{\sqrt{2\pi}\sigma}e^{-\frac{1}{2}(\frac{x_i- \mu} {\sigma})^2} }[/math]

[math]\displaystyle{ l(X^N; \theta) = \sum_{i=1}^N -\frac{1}{2}log (2\pi) - log(\sigma) - \frac{1}{2} \left(\frac{x_i- \mu}{\sigma}\right)^2 }[/math]

[math]\displaystyle{ \frac{dl}{d\mu} = \displaystyle\sum_{i=1}^N(x_i-\mu) }[/math]

Setting [math]\displaystyle{ \frac{dl}{d\mu} = 0 }[/math] we get

[math]\displaystyle{ \displaystyle\sum_{i=1}^Nx_i = \displaystyle\sum_{i=1}^N\mu }[/math]

[math]\displaystyle{ \displaystyle\sum_{i=1}^Nx_i = N\mu \rightarrow \mu = \frac{1}{N}\displaystyle\sum_{i=1}^Nx_i }[/math]

Bayesian Approach

Assuming the prior is Gaussian:

[math]\displaystyle{ P(\theta) = \frac{1}{\sqrt{2\pi}\tau}e^{-\frac{1}{2}(\frac{x-\mu_0}{\tau})^2} }[/math]

[math]\displaystyle{ f(\theta|x) \propto \frac{1}{\sqrt{2\pi}\sigma}e^{-\frac{1}{2}(\frac{x_i-\mu}{\sigma})^2} * \frac{1}{\sqrt{2\pi}\tau}e^{-\frac{1}{2}(\frac{x-\mu_0}{\tau})^2} }[/math]

By completing the square we conclude that the posterior is Gaussian as well:

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

Where

[math]\displaystyle{ \tilde{\mu} = \frac{\frac{N}{\sigma^2}}{{\frac{N}{\sigma^2}}+\frac{1}{\tau^2}}\bar{x} + \frac{\frac{1}{\tau^2}}{{\frac{N}{\sigma^2}}+\frac{1}{\tau^2}}\mu_0 }[/math]

The expectation from the posterior is different from the MLE method. Note that [math]\displaystyle{ \displaystyle\lim_{N\to\infty}\tilde{\mu} = \bar{x} }[/math]. Also note that when [math]\displaystyle{ N = 0 }[/math] we get [math]\displaystyle{ \tilde{\mu} = \mu_0 }[/math].

Basic Monte Carlo Integration

We start with a simple example:

[math]\displaystyle{ I = \displaystyle\int_a^b h(x)\,dx }[/math]

[math]\displaystyle{ = \displaystyle\int_a^b w(x)f(x)\,dx }[/math]

where

[math]\displaystyle{ w(x) = h(x)(b-a) }[/math]

[math]\displaystyle{ f(x) = \frac{1}{b-a} \rightarrow }[/math] the p.d.f. is Unif[math]\displaystyle{ (a,b) }[/math]

[math]\displaystyle{ \hat{I} = E_f[w(x)] = \frac{1}{N}\displaystyle\sum_{i=1}^Nw(x_i) }[/math]

If [math]\displaystyle{ x_i \sim~ Unif(a,b) }[/math] then by the Law of Large Numbers [math]\displaystyle{ \frac{1}{N}\displaystyle\sum_{i=1}^Nw(x_i) \rightarrow \displaystyle\int w(x)f(x)\,dx = E_f[w(x)] }[/math]

Process

Given [math]\displaystyle{ I = \displaystyle\int^b_ah(x)\,dx }[/math]

1. [math]\displaystyle{ \begin{matrix} w(x) = h(x)(b-a)\end{matrix} }[/math]
2. [math]\displaystyle{ \begin{matrix} x_1, x_2, ..., x_n \sim UNIF(a,b)\end{matrix} }[/math]
3. [math]\displaystyle{ \hat{I} = \frac{1}{N}\displaystyle\sum_{i=1}^Nw(x_i) }[/math]

From this we can compute other statistics, such as

1. [math]\displaystyle{ SE=\frac{s}{\sqrt{N}} }[/math] where [math]\displaystyle{ s=\frac{\sum_{i=1}^{N}(Y_i-\hat{I})^2 }{N-1} }[/math] with [math]\displaystyle{ \begin{matrix}Y_i=w(i)\end{matrix} }[/math]
2. [math]\displaystyle{ \begin{matrix} 1-\alpha \end{matrix} }[/math] CI's can be estimated as [math]\displaystyle{ \hat{I}\pm Z_\frac{\alpha}{2}*SE }[/math]

Example 1

Find [math]\displaystyle{ E[\sqrt{x}] }[/math] for [math]\displaystyle{ \begin{matrix} f(x) = e^{-x}\end{matrix} }[/math]

1. We need to draw from [math]\displaystyle{ \begin{matrix} f(x) = e^{-x}\end{matrix} }[/math]
2. [math]\displaystyle{ \hat{I} = \frac{1}{N}\displaystyle\sum_{i=1}^Nw(x_i) }[/math]

This example can be illustrated in Matlab using the code below:

u=rand(100,1)
x=-log(u)
h= x.* .5
mean(h)
%The value obtained using the Monte Carlo method
F = @ (x) sqrt (x). * exp(-x)
quad(F,0,50)
%The value of the real function using Matlab

Example 2 Find [math]\displaystyle{ I = \displaystyle\int^1_0h(x)\,dx, h(x) = x^3 }[/math]

1. [math]\displaystyle{ I = x^4/4 = 1/4 }[/math]
2. [math]\displaystyle{ W(x) = x^3*(1-0) }[/math]
3. [math]\displaystyle{ Xi \sim~Unif(0,1) }[/math]
4. [math]\displaystyle{ \hat{I} = \frac{1}{N}\displaystyle\sum_{i=1}^Nw(x_i^3) }[/math]

This example can be illustrated in Matlab using the code below:

x = rand (1000)
mean(x^3)

Example 3 To estimate an infinite integral such as [math]\displaystyle{ I = \displaystyle\int^\infty_5 h(x)\,dx, h(x) = 3e^{-x} }[/math]

1. Substitute in [math]\displaystyle{ y=\frac{1}{x-5+1} =\gt dy=\frac{1}{(x-4)^2}dx =\gt dy=-y^2dx }[/math]
2. [math]\displaystyle{ I = \displaystyle\int^1_0 \frac{3e^{-(\frac{1}{y}+4)}}{-y^2}\,dy }[/math]
3. [math]\displaystyle{ w(y) = \frac{3e^{-(\frac{1}{y}+4)}}{-y^2}(1-0) }[/math]
4. [math]\displaystyle{ Y_i \sim~Unif(0,1) }[/math]
5. [math]\displaystyle{ \hat{I} = \frac{1}{N}\displaystyle\sum_{i=1}^Nw(\frac{3e^{-(\frac{1}{y_i}+4)}}{-y_i^2}) }[/math]

Importance Sampling and Monte Carlo Simulation - May 28, 2009

During this lecture we covered two more examples of Monte Carlo simulation, finishing that topic, and begun talking about Importance Sampling.

Binomial Probability Monte Carlo Simulations

Example 1:

You are given two independent Binomial distributions with probabilities [math]\displaystyle{ \displaystyle p_1\text{, }p_2 }[/math]. Using a Monte Carlo simulation, approximate the value of [math]\displaystyle{ \displaystyle \delta }[/math], where [math]\displaystyle{ \displaystyle \delta = p_1 - p_2 }[/math].

[math]\displaystyle{ \displaystyle X \sim BIN(n, p_1) }[/math]; [math]\displaystyle{ \displaystyle Y \sim BIN(n, p_2) }[/math]; [math]\displaystyle{ \displaystyle \delta = p_1 - p_2 }[/math]

So [math]\displaystyle{ \displaystyle f(p_1, p_2 | x,y) = \frac{f(x, y|p_1, p_2)*f(p_1,p_2)}{f(x,y)} }[/math] where [math]\displaystyle{ \displaystyle f(x,y) }[/math] is a flat distribution and the expected value of [math]\displaystyle{ \displaystyle \delta }[/math] is as follows:

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

Since X, Y are independent, we can split the conditional probability distribution:

[math]\displaystyle{ \displaystyle f(p_1,p_2|X,Y) \propto f(p_1|X)f(p_2|Y) }[/math]

We need to find conditional distribution functions for [math]\displaystyle{ \displaystyle p_1, p_2 }[/math] to draw samples from. In order to get a distribution for the probability 'p' of a Binomial, we have to divide the Binomial distribution by n. This new distribution has the same shape as the original, but is scaled. A Beta distribution is a suitable approximation. Let

[math]\displaystyle{ \displaystyle f(p_1 | X) \sim \text{Beta}(x+1, n-x+1) }[/math] and [math]\displaystyle{ \displaystyle f(p_2 | Y) \sim \text{Beta}(y+1, n-y+1) }[/math], where

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

Process:

1. Draw samples for [math]\displaystyle{ \displaystyle p_1 }[/math] and [math]\displaystyle{ \displaystyle p_2 }[/math]: [math]\displaystyle{ \displaystyle (p_1,p_2)^{(1)} }[/math], [math]\displaystyle{ \displaystyle (p_1,p_2)^{(2)} }[/math], ..., [math]\displaystyle{ \displaystyle (p_1,p_2)^{(n)} }[/math];
2. Compute [math]\displaystyle{ \displaystyle \delta = p_1 - p_2 }[/math] in order to get n values for [math]\displaystyle{ \displaystyle \delta }[/math];
3. [math]\displaystyle{ \displaystyle \hat{\delta}=\frac{\displaystyle\sum_{\forall i}\delta^{(i)}}{N} }[/math].
   
   

Matlab Code:

The Matlab code for recreating the above example is as follows:

n=100; %number of trials for X
m=100; %number of trials for Y
x=80; %number of successes for X trials
y=60; %number of successes for y trials
p1=betarnd(x+1, n-x+1, 1, 1000);
p2=betarnd(y+1, m-y+1, 1, 1000);
delta=p1-p2;
mean(delta);

The mean in this example is given by 0.1938.

A 95% confidence interval for [math]\displaystyle{ \delta }[/math] is represented by the interval between the 2.5% and 97.5% quantiles which covers 95% of the probability distribution. In Matlab, this can be calculated as follows:

q1=quantile(delta,0.025);
q2=quantile(delta,0.975);

The interval is approximately [math]\displaystyle{ 95% CI \approx (0.06606, 0.32204) }[/math]

The histogram of delta is:

Note: In this case, we can also find [math]\displaystyle{ E(\delta) }[/math] analytically since [math]\displaystyle{ E(\delta) = E(p_1 - p_2) = E(p_1) - E(p_2) = \frac{x+1}{n+2} - \frac{y+1}{m+2} \approx 0.1961 }[/math]. Compare this with the maximum likelihood estimate for [math]\displaystyle{ \delta }[/math]: [math]\displaystyle{ \frac{x}{n} - \frac{y}{m} = 0.2 }[/math].

Example 2:

Bayesian Inference for Dose Response

We conduct an experiment by giving rats one of ten possible doses of a drug, where each subsequent dose is more lethal than the previous one:

[math]\displaystyle{ \displaystyle x_1\lt x_2\lt ...\lt x_{10} }[/math]

For each dose [math]\displaystyle{ \displaystyle x_i }[/math] we test n rats and observe [math]\displaystyle{ \displaystyle Y_i }[/math], the number of rats that survive. Therefore,

[math]\displaystyle{ \displaystyle Y_i \sim~ BIN(n, p_i) }[/math]
.

We can assume that the probability of death grows with the concentration of drug given, i.e. [math]\displaystyle{ \displaystyle p_1\lt p_2\lt ...\lt p_{10} }[/math]. Estimate the dose at which the animals have at least 50% chance of dying.

Let [math]\displaystyle{ \displaystyle \delta=x_j }[/math] where [math]\displaystyle{ \displaystyle j=min\{i|p_i\geq0.5\} }[/math]

We are interested in [math]\displaystyle{ \displaystyle \delta }[/math] since any higher concentrations are known to have a higher death rate.

Solving this analytically is difficult:

[math]\displaystyle{ \displaystyle \delta = g(p_1, p_2, ..., p_{10}) }[/math] where g is an unknown function

[math]\displaystyle{ \displaystyle \hat{\delta} = \int \int..\int_A \delta f(p_1,p_2,...,p_{10}|Y_1,Y_2,...,Y_{10})\,dp_1dp_2...dp_{10} }[/math]

where [math]\displaystyle{ \displaystyle A=\{(p_1,p_2,...,p_{10})|p_1\leq p_2\leq ...\leq p_{10} \} }[/math]

Process: Monte Carlo
We assume that

1. Draw [math]\displaystyle{ \displaystyle p_i \sim~ BETA(y_i+1, n-y_i+1) }[/math]
2. Keep sample only if it satisfies [math]\displaystyle{ \displaystyle p_1\leq p_2\leq ...\leq p_{10} }[/math], otherwise discard and try again.
3. Compute [math]\displaystyle{ \displaystyle \delta }[/math] by finding the first [math]\displaystyle{ \displaystyle p_i }[/math] sample with over 50% deaths.
4. Repeat process n times to get n estimates for [math]\displaystyle{ \displaystyle \delta_1, \delta_2, ..., \delta_N }[/math].
5. [math]\displaystyle{ \displaystyle \bar{\delta} = \frac{\displaystyle\sum_{\forall i} \delta_i}{N} }[/math].

Importance Sampling

In statistics, Importance Sampling is a

In [math]\displaystyle{ I = \displaystyle\int h(x)f(x)\,dx }[/math], Monte Carlo simulation can be used only if it easy to sample from f(x). Otherwise, another method must be applied. If sampling from f(x) is difficult but there exists a probability distribution function g(x) which is easy to sample from, then [math]\displaystyle{ I }[/math] can be written as

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

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

[math]\displaystyle{ = \displaystyle E_g(w(x)) \rightarrow }[/math]the expectation of w(x) with respect to g(x)

[math]\displaystyle{ = \frac{\displaystyle\sum_{i=1}^{N} w(x_i)}{N} }[/math] where [math]\displaystyle{ \displaystyle w(x) = \frac{h(x)f(x)}{g(x)} }[/math]

Process

1. Choose [math]\displaystyle{ \displaystyle g(x) }[/math] such that it's easy to sample from.
2. Compute [math]\displaystyle{ \displaystyle w(x)=\frac{h(x)f(x)}{g(x)} }[/math]
3. [math]\displaystyle{ \displaystyle \hat{I} = \frac{\displaystyle\sum_{i=1}^{N} w(x_i)}{N} }[/math]
   
   

"Weighted" average

The term "importance sampling" is used to describe this method because a higher 'importance' or 'weighting' is given to the values sampled from [math]\displaystyle{ \displaystyle g(x) }[/math] that are closer to [math]\displaystyle{ \displaystyle f(x) }[/math], the original distribution we would ideally like to sample from (but cannot because it is too difficult).

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

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

[math]\displaystyle{ =\displaystyle \int \frac{f(x)}{g(x)}E_g(h(x))\,dx }[/math] which is the same thing as saying that we are applying a regular Monte Carlo Simulation method to [math]\displaystyle{ \displaystyle\int h(x)g(x)\,dx }[/math], taking each result from this process and weighting the more accurate ones (i.e. the ones for which [math]\displaystyle{ \displaystyle \frac{f(x)}{g(x)} }[/math] is high) higher.

One can view [math]\displaystyle{ \frac{f(x)}{g(x)}\ = B(x) }[/math] as a weight.

Then [math]\displaystyle{ \displaystyle \hat{I} = \frac{\displaystyle\sum_{i=1}^{N} w(x_i)}{N} = \frac{\displaystyle\sum_{i=1}^{N} B(x_i)*h(x_i)}{N} }[/math]

i.e. we are computing a weighted sum of [math]\displaystyle{ h(x_i) }[/math] instead of a sum