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