Robert_V._Hogg,_Joseph_W._McKean,_Allen_T._Craig

3.1. The Binomial and Related Distributions 159

Theorem 3.1.1.LetX 1 ,X 2 ,...,Xmbe independent random variables such that

Xihas binomialb(ni,p)distribution, fori=1, 2 ,...,m.LetY=

∑m

i=1Xi.Then

Yhas a binomialb(

∑m
i=1ni,p)distribution.
Proof:The mgf ofXiisMXi(t)=(1−p+pet)ni. By independence it follows from
Theorem 2.6.1 that

MY(t)=

∏m

i=1

(1−p+pet)ni=(1−p+pet)

Pm

i=1ni.

Hence,Yhas a binomialb(

∑m
i=1ni,p) distribution.
For the remainder of this section, we discuss some important distributions that
are related to the binomial distribution.

3.1.1 Negative Binomial and Geometric Distributions

Consider a sequence of independent Bernoulli trials with constant probabilitypof

success. Let the random variableY denote the total number of failures in this

sequence before therth success, that is,Y+r is equal to the number of trials

necessary to produce exactlyrsuccesses with the last trial as a success. Herer

is a fixed positive integer. To determine the pmf ofY,letybe an element of

{y:y=0, 1 , 2 ,...}. Then, since the trials are independent,P(Y =y)isequal

to the product of the probability of obtaining exactlyr−1 successes in the first

y+r−1 trials times the probabilitypof a success on the (y+r)th trial. Thus the

pmf ofYis

pY(y)=

{ (y+r− 1

r− 1

)

pr(1−p)y y=0, 1 , 2 ,...

0elsewhere.

(3.1.3)

A distribution with a pmf of the formpY(y) is called anegative binomial dis-
tributionand any suchpY(y) is called a negative binomial pmf. The distribution
derives its name from the fact thatpY(y) is a general term in the expansion of
pr[1−(1−p)]−r. It is left as an exercise to show that the mgf of this distribution
isM(t)=pr[1−(1−p)et]−r,fort<−log(1−p). The R call to computeP(y≤y)
ispnbinom(y,r,p).

Example 3.1.6.Suppose the probability that a person has blood type B is 0.12.

In order to conduct a study concerning people with blood type B, patients are

sampled independently of one another until 10 are obtained who have blood type

B. Determine the probability that at most 30 patients have to have their blood type

determined. LetYhave a negative binomial distribution withp=0.12 andr= 10.

Then, the desired probability is

P(Y≤20) =

∑^20

j=0

(

j+9

9

)

0. 12100. 88 j.

Its computation in R ispnbinom(20,10,0.12) = 0.0019.