Download MATH 382 The Negative Binomial Distribution ∑ P( X = i) = ∑ ∑

Dr. Neal, WKU
MATH 382
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
 k − 1
 ways to have a sequence of k
attempts to have the other successes. So there are 
 r − 1
attempts with exactly r successes and with the last success being on the last attempt.
And each such individual sequence occurs with probability p r q k−r . Thus, the
probability of having the r th success on the k th attempt, for k ≥ r , is given by
 k − 1 r k −r
p q .
P(X = k ) = 
 r − 1
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:
k
k  i − 1
k  i − 1
 pr q i−r and P( j ≤ X ≤ k) = ∑ 
 r i−r .
P(X ≤ k) = ∑ P( X = i) = ∑ 
 r − 1 p q
r − 1
i =r
i =r
i=j
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 X1 ~ 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 the sum
2
X1 + X 2 + . . . + Xr = X ~ nb(r, p) . Moreover, E[ Xi ] = 1 / p and Var( Xi ) = q / p for
1 ≤ i ≤ r . By the linearity of expected value, we then have
E[ X ] = E[ X1 + X2 + . . . + X r ] = E[X1 ] + E[ X2 ] + . . . + E[ Xr ]
r
=1/ p +1/ p +. .. +1/ p = .
p
Dr. Neal, WKU
Then by independence of the Xi , we have
Var( X) = Var(X1 + X 2 + . . . + Xr ) = Var( X1 ) + Var(X 2 ) + . . . + Var( Xr )
rq
= q / p 2 + q / p2 + . . . + q / p2 = 2 .
p
And for X ~ nb(r, p) , the standard deviation is given by σ X =
rq
.
p
Mode
The most likely number of attempts needed to attain r successes, i.e., the mode, is always
r − q 
r−q
 . However, a
given by the largest integer k such that k ≤
, denoted by k = 
 p 
p
r−q
negative binomial distribution will be bi-modal when k =
is an integer. In this
p
case, this value of k and the previous integer k − 1 will be the two modes, except in the
case of r = 1 . When r = 1 , then k = 1 is the only mode. The expression for the mode
can be derived with the same argument used for deriving the mode of a binomial
distribution.
Example. Suppose we have a 40% chance of winning a bet and we play until we win 5
times.
(a) What is the most likely number of attempts needed for 5 wins and the what is the
probability of needing exactly this number of attempts?
(b) What are the average number and the standard deviation of the number of attempts
needed to win 5 times?
(c) What is the probability that it takes at least 10 attempts to achieve 5 wins?
Solution. Here X ~ nb(5, 0.40) . (a) The most likely number of attempts needed for 5
 5 − 0.6   5 − 0.6 
wins is k = 
. So 10 and 11 are both the most likely number
 0.4  =  0. 4  =  11
of attempts needed with
 9
P(X = 10) =   0. 45 0.65 ≈ 0.100329 and
 4
 10 
P(X = 11) =   0.4 5 0.66 ≈ 0.100329 .
4 
(b) The average number of attempts needed is E[ X ] = 5 / 0.4 = 12. 5 with a standard
5 × 0.6
deviation of σ X =
≈ 4.33 . (c) The probability that it takes at least 10 attempts
0. 4
is given by
P ( X ≥ 10 ) = 1 − P(5 ≤ X ≤ 9)
 4
5
0  5
5
1  6
5
2  7
5
3  8
5
4
= 1 −   0. 4 0.6 −   0. 4 0.6 −   0.4 0.6 −   0. 4 0.6 −   0. 4 0. 6
4
4
4
4
4
≈ 0.7334
Dr. Neal, WKU
Exercises
1. Suppose you have a 20% chance of winning a bet and you play until you win 4 times.
(a) What is the most likely number of attempts needed for 4 wins and the what is the
probability of needing exactly this number of attempts?
(b) What are the average number and the standard deviation of the number of attempts
needed to win 4 times?
(c) Compute P (20 ≤ X ≤ 22) .
2. Derive the expression for the mode of X ~ nb(r, p) .