The Negative Binomial Distribution and Some Associated Mixture Distributions: Practice Problems and Solutions

A negative binomial random variable is "a discrete random variable M with parameters

r > 0 and β > 0 for which

1. The only possible values are the non-negative integers 0, 1, 2, ..., etc.
2. P[M = k] = ([r(r + 1)...(r + k -1)]/k!)(β^k/(1+ β)^k+r)

3. E[M] = rβ

4. Var[M] = rβ(1+ β)" (Daniel, p. 11).

In a mixture distribution where β is itself a random variable with p.d.f. f(β) and with β ranging from a to b, the following formula applies.

P[M = k] = _a^b∫P[M = k│ β‌]*f(β)dβ

Sources: Broverman, Sam. Actuarial Exam Solutions - CAS Exam 3 - Fall 2006.

Daniel, James W. 2008. Poisson Processes (and Mixture Distributions).

Original Problems and Solutions from The Actuary's Free Study Guide

Problem S3L48-1. The number of great fleas sitting on the point of a needle follows a negative binomial distribution with r = 0.7 and β = 2. Find the probability that exactly 3 great fleas are sitting on the point of the needle.

Solution S3L48-1. We need to use the formula

P[M = k] = ([r(r + 1)...(r + k -1)]/k!)(β^k/(1+ β)^k+r), for k = 3, r = 0.7, and β = 2.

Thus, P[M = 3] = ((0.7)(1.7)(2.7)/3!)(2³/3^3.7) = about 0.0735361383.

Problem S3L48-2. The number of great fleas sitting on the point of a needle follows a negative binomial distribution with r = 0.7 and β = 2. Find the expected value of the number of great fleas that are sitting on the point of the needle.

Solution S3L48-2. We use the formula E[M] = rβ = 0.7*2 = 1.4 great fleas.

Problem S3L48-3. The number of great fleas sitting on the point of a needle follows a negative binomial distribution with r = 0.7 and β = 2. Find the variance of the number of great fleas that are sitting on the point of the needle.

Solution S3L48-3. We use the formula Var[M] = rβ(1+ β) = 0.7*2(1 + 2) = 4.2.

Problem S3L48-4. The number of magenta watermelons that an orange hippopotamus eats on a given day follows a negative binomial distribution with r = 4 and β being distributed uniformly on the interval between 1 and 5. Find the probability that this orange hippopotamus finds 2 magenta watermelons today. Set up the appropriate integral and then use any calculator to evaluate it.

Solution S3L48-4.

We first attempt to apply the formula

We need to use the formula

P[M = k] = ([r(r + 1)...(r + k -1)]/k!)(β^k/(1+ β)^k+r).

In this case, for k = 2 and r = 4, we have

P[M = 2 │ β‌] = (4*5/2!)(β²/(1+ β)⁶) = 10β²/(1+ β)⁶

β itself is a random variable with probability density function 1/(5-1) = ¼.

Thus, P[M = 2] = ₁⁵∫P[M = 2 │ β‌]*f(β)dβ = ₁⁵∫(1/4)(10β²/(1+ β)⁶)dβ =

₁⁵∫(5/2)(β²/(1+ β)⁶)dβ = about 0.0387088477.

Problem S3L48-5.

Similar to Question 18 from the Casualty Actuarial Society's Fall 2006 Exam 3.

The number of white cucumbers that a cucumber-hunter finds on a given day follows a negative binomial distribution with r = 2 and β being distributed uniformly on the interval between 3 and 6. Find the probability that this cucumber-hunter finds at least one white cucumber today.

Solution S3L48-5. We want to find P[M ≥ 1] = 1 - P[M = 0].

To find P[M = 0], we use the formula P[M = k] = ([r(r + 1)...(r + k -1)]/k!)(β^k/(1+ β)^k+r).

Here, k = 0 and r = 2, so [r(r + 1)...(r + k -1)] will be the smallest of r and (r + k - 1), which will be r + 0 - 1 = 2 - 1 = 1. Likewise, k! = 0! = 1, and β^k = β⁰ = 1.

So, we have P[M = 0│ β‌] = 1/(1+ β)^0+r = 1/(1+ β)².

β itself is a random variable with probability density function 1/(6-3) = 1/3.

P[M = 0] = ₃⁶∫P[M = 0 │ β‌]*f(β)dβ = ₃⁶∫(1/3)(1+ β)^-2*dβ = -(1/3)(1+ β)^-1│₃⁶ =

(1/3)(4)^-1-(1/3)(7)^-1= 1/12 - 1/21 = 1/28. Our desired answer is 1 - 1/28 = 27/28 = about 0.9642857143.

See other sections of The Actuary's Free Study Guide for Exam 3L.