# copyofstat341

#### 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]U \sim~ \mathrm{Unif}[0, 1][/math] is a random variable and [math]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 where f is defined as 0 outside of its domain, then,

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

- [math]\int_1^{\infty} \frac{x^k}{x^2} dx[/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]P(U \leq a) = a[/math].

So,

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

**Example**:

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

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

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

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

Now we can generate our random sample [math]u_1\dots u_n[/math] from [math]F(x)[/math] by:

- [math]1)\ u_i \sim Unif[0, 1][/math]
- [math]2)\ x_i = \frac{-\log(u_i)}{\theta}[/math]

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

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

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

The major problem with this approach is that we have to find [math]F^{-1}[/math] and for many distributions it is too difficult (or impossible) to find the inverse of [math]F(x)[/math]. Further, for some distributions it is not even possible to find [math]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]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]x_0, \dots , x_n[/math] where:

- [math]\begin{align}P(X = x_i) &{}= p_i \end{align}[/math]
- [math]x_0 \leq x_1 \leq x_2 \dots \leq x_n[/math]
- [math]\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]U = 1[/math] is missed):

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

**Example** (from class):

Suppose we have the following discrete distribution:

- [math]\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] 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]u_0, \dots, u_n[/math] from [math]U \sim~ Unif[0, 1][/math]:

- [math] 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