Advanced High-School Mathematics

SECTION 6.2 Continuous Random Variables 363

of a geometric random variable. Just as we were able above to

transformrandinto an exponential random variable, we shall (ap-

proximately) transform the TI random integer variable “randInt”

into a geometric random variable.

First of all, the random integer generator has three inputs and has

the form randInt(nmin,nmax,N). The output consists of a se-

quence ofN randomly and uniformly distributed integers between

nmin andnmax. We shall, for convenience, take nmin = 1 and

setn=nmax. We letY be the geometric random variable (with

parameter p), and letX be a randomly-generated uniformly dis-

tributed integer 1≤X≤n. The goal is to find a functiongsuch

that Y = g(X). This will allow us to use the TI calculator to

generate samples of a geometric random variable (and therefore of

a negative binomial random variable).

Note first that

P(Y ≤k) =p+p(1−p)+p(1−p)^2 +···+p(1−p)k−^1 = 1−(1−p)k, k≥ 0

and that

P(X≤h) =

h

n

, 1 ≤h≤n.

At this point we see a potential problem in transforming from the

uniform variable to the geometric: the geometric random variable

has an infinite number of possible outcomes (with decreasing prob-

abilities) and the uniform random variable is an integer between 1

andn. Therefore, we would hope not to lose too much information

by allowingnto be reasonably large (n≈25 seems pretty good).

At any rate, we proceed in analogy with the analysis on page 353:

assuming thatY =g(X) we have