Dr. Neal, WKU
MATH 329 The Negative Binomial
Distribution
A negative binomial random variable, denoted by X~nb(r,p), is a generalization of the
geometric random variable. Suppose we have probability p>0 of success on every
attempt and we make a sequence of independent attempts. Then
X
counts the number
of attempts needed to obtain the
r
th success, for some designated integer
r≥1
. For
r=1
, then X~geo(p). Because its takes at least
r
attempts to obtain
r
successes, the
range of X~nb(r,p) is {
r
,
r+1
,
r+2
, . . . }.
Probability Distribution Function
We again let q=1−p be the probability of a failure on each attempt. To have
X
equal
k
, for some
k≥r
, we must have exactly
r
successes in
k
attempts with the last success
being on the
k
th (i.e., final) attempt. And we must choose
r−1
of the preceding
k−1
attempts to have the other successes. So there are k−1
r−1
ways to have a sequence of
k
attempts with exactly
r
successes and with the last success being on the last attempt.
And each such individual sequence occurs with probability p
rqk−r
. Thus, the
probability of having the
r
th success on the
k
th attempt, for
k≥r
, is given by
P(X=k)=k−1
r−1
prqk−r.
There is no closed-form formula for the cumulative probability P(X≤k) or for
computing probabilities such as P(j≤X≤k). In each case, the individual probabilities
must be summed, or we must use a calculator/computer command:
P(X≤k) = i=r
k
∑P(X=i)
= i−1
r−1
prqi−r
i=r
k
∑
and P(j≤X≤k) = i−1
r−1
prqi−r
i=j
k
∑
.
Expected Value, Variance, and Standard Deviation
We can easily derive the expected value and variance of X~nb(r,p) by writing
X
as a
sum of independent geometric random variables. First, let X
1
~ geo(p) be the number
of attempts needed for the first success. Then for
2≤i≤r
, let Xi ~ geo(p) be the
additional number of attempts after the (i−1)st success until the
i
th success.
Then the Xi are independent of each other (due to independent attempts) and the
total number of attempts needed for
r
successes is given by
X
1+X2+... +Xr=X~nb(r,p)
. Moreover, E[Xi]=1/ p and Var(Xi)=q/p
2
for
1≤i≤r
. By the linearity of expected value, we then have
E[X]=E[X1+X2+.. .+Xr]=E[X1]+E[X2]+... +E[Xr]
=1/ p+1/ p+... +1/ p=r
p.
