Last time we discussed Random Number Generation and we found that random numbers need two properties:

Random numbers were generated by techniques such as the midsquare method.

Today we're going to generate random numbers by using distributions. Most of this information comes directly from Kobayashi.

One way to do this is by the transform method, page 241 Kobayashi.

Given:

The first thing we do is set:

Y = F( X ).

In this way, Y is defined on the range [0, 1].

If Y is a random variable uniformly distributed between 0 and 1 the variable X defined by:

X = F^-1( Y )

Note that this is the inverse function, not 1/F(Y).

Then X has cumulative distribution F(x):

P[ X £ x ] = P [ Y £ F(x) ] = F(x)

Let X be a random variable with exponential distribution:

F(x) = 1 - e^{-m x}, x ³ 0.

Set this equal to a random decimal number Y:

F( X ) = Y

1 - e^{-m X} = Y, or,

Y = 1 - e^{-m X}

Solve for X:

Y = 1 - e^{-m X}

e^{-m X} = 1 - Y

ln (e^{-m X}) = ln (1 - Y)

-m X = ln (1 - Y)

X = - { ln (1 - Y) }/m

So that:

X = F^-1(Y) = - { ln (1 - Y) }/m

Because Y is a random decimal number between 0 and 1, then (1 - Y) is also a random decimal number between 0 and 1. Therefore, the equation can be simplified as:

X = F^-1(1 - Y) = - { ln Y }/m

What's something we know about the natural log (ln)?

The natural log of decimal numbers less than 1 results in a negative decimal number.

So, these random numbers generated will be positive between 0 and 1.

So, we get (assume m = 3)

Let X be a random variable with Weibull distribution:

Set this equal to a random decimal number Y:

F( X ) = Y

Solve for X:

So that:

X = F^-1(Y) = -(a /e^b ) ln(1-Y) + v

Because Y is a random decimal number between 0 and 1, then (1 - Y) is also a random decimal number between 0 and 1. Therefore, the equation can be simplified as:

X = F^-1(1 - Y) = -(a /e^b ) ln(Y) + v

The rejection method, Kobayashi page 243, can be used when it is difficult to calculate the inverse function (as in the transform method above). In this method, let:

Let there be pairs of random decimal numbers (R₁, R₂) between 0 and 1. Then:

X₁ = a + (b - a)R₁

is a random number in [a, b]. Whenever we encounter a pair (R₁, R₂) that satisfies the relationship:

M× R₂ £ f(X₁)

we accept X₁ and reject otherwise.

The probability density function of accepted X₁'s will be f(x).

The number of trials before a successful pair is found is a random variable n with geometric distribution:

P[n] = r (1 - r )^n-1

with

r = 1/[M(b-a)]

and mean value n:

E[n] = 1/r = M(b-a)

So, the method is not appropriate for pdfs with large expected values.

Consider the following pdf:

Use the rejection technique to generate random variates according to this distribution.

So, a = 0, b = 1, and M = 2(1) = 2.

X₁ = a + (b - a)R₁

X₁ = 0 + (1 - 0)R₁

X₁ = R₁

is a random number in [0, 1]. Whenever we encounter a pair (R₁, R₂) that satisfies the relationship:

M× R₂ £ f(X₁)

2× R₂ £ 2× X₁

R₂ £ X₁

we accept X₁ and reject otherwise.

The probability density function of accepted X₁'s will be f(x).

The number of trials before a successful pair is found is a random variable n with geometric distribution:

P[n] = r (1 - r )^n-1

with

r = 1/[M(b-a)] = 1/[2(1-0)] = 1/2

P[n] = (1/2)(1 - (1/2))^n-1

P[n] = (1/2)(1/2)^n-1

P[n] = (1/2)ⁿ

and mean value n:

E[n] = 1/r = 2

The composition method, Kobayashi page 244, is another method to generate random numbers from a distribution. Let a distribution F(x) be a mixture of two distributions: F₁(x) and F₂(x):

F(x) = p× F₁(x) + (1 - p)× F₂(x)

We obtain a variate X by:

1. Generating a uniform variate U.
2. If U < p, then X = F₁(x), else X = F₂(x)

Example 4.4.7. Hyperexponential Distribution. Find a way to generate a sequence of variates that have the two-phase hyperexponential distribution:

where p + q = 1.

Since p + q = 1, 1 = p + q = p + (1 - p)

Now this is in the correct form.

So, we generate a random variate U and if U < p, then we set X equal to F₁(x), otherwise we use F₂(x).

This hyperexponential distribution has parameters (m ₁, m ₂, p). This distribution means that there are two types of service that may be requested, and are requested in proportions of p and q, respectively.