3
$\begingroup$

The following is an exercise from Rice's Mathematical Statistics and Data Analysis:

This problem shows one way to generate discrete random variables from a uniform random number generator. Suppose that F is the cdf of an integer-valued random variable; let U be uniform on [0, 1]. Define a random variable Y = k if F(k − 1) < U ≤ F(k). Show that Y has cdf F.

I know how to prove this (by writing F's cdf as a sum of cdf of uniform distribution, which then eventually cancels / simplifies nicely to give F_Y(x) = F(x)) but I'm having trouble developing an intuition as to what this method means and how it works. I am not looking for a solution to the exercise / a proof. I want to understand why this method works at a conceptual level.

I'm not sure what this method is called so I couldn't search for other resources on this.

$\endgroup$
1

2 Answers 2

2
$\begingroup$

I'll try to give a working example. Let's say you want to create a RV from the following discrete distribution: $$p_X(x)=\begin{cases} 0.3, &x=1\\0.2, &x=2\\0.1,&x=3\\0.4,&x=4 \end{cases}$$

One simple way to simulate this distribution is the following intuitive if-else block:

u = rand() // uniform 0-1 RV
if u < 0.3
   x = 1   // we'll be here with 0.3 probability
else if u < 0.5
   x = 2   // we'll be here with 0.5 - 0.3 = 0.2 probability
else if u < 0.6
   x = 3   // we'll be here with 0.6 - 0.5 = 0.1 probability 
else
   x = 4   // we'll be here with 1 - 0.6 = 0.4 probability
end

which is the same procedure described analytically.

$\endgroup$
1
  • 1
    $\begingroup$ Just saw this. (+1) Nice to show if...then...else code. $\endgroup$
    – BruceET
    Apr 16, 2020 at 9:22
1
$\begingroup$

Whether for generating values from a continuous or discrete random variable, the method is sometimes called the 'inverse CDF method' or the 'quantile method'. I will illustrate the discrete case.

Suppose we want to simulate $X \sim \mathsf{Binom}(n=2,p=1/2).$ Its CDF table can be found in R as follows:

x = 0:2; cdf = pbinom(x, 2, 1/2);  cbind(x, cdf)
     x  cdf
[1,] 0 0.25
[2,] 1 0.75
[3,] 2 1.00

Suppose you want to simulate $X \sim \mathsf{Binom}(2,1/2)$ using a sample from $U \sim \mathsf{Unif}(0,1).$ Then you could get

  • $P(X = 0) = 1/4$ by using values of $U$ in the interval $(0,0.25)$ because $P(0 \le U < 0.25) = 1/4;$

  • $P(X = 1) = 1/2 $ by using values of $U$ in $(0.25, 0.75)$, and

  • $P(X = 2) = 1/4 $ by using values of $U$ in $(0.75, 1.00).$

The graph on the left below shows the CDF of $X \sim \mathsf{Binom}(2,1/2)$ and on the right its inverse CDF (quantile function). For reference, heavy lines on both graphs show $P(X=1) = 1/2.$

enter image description here

In R, if I want generate 7 observations from $\mathsf{Binom}(2, 1/2),$ I can do it by using rbinom or by using runif with qbinom; by using the same seed both times in the example below, I get exactly the same sample each way.

set.seed(123); rbinom(7, 2, 1/2)
[1] 1 2 1 2 2 0 1
set.seed(123); qbinom(runif(7), 2, 1/2)
[1] 1 2 1 2 2 0 1

Thus, it is reasonable to assume that the R function rbinom implements the quantile method as illustrated above. (But for $p > .5,$ R uses a slightly different method.)

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.