Probability Generating Functions, Poisson Processes, and Assorted Exam-Style Questions

The probability generating function for a discrete probability distribution is denoted P_N(t) for the random variable N, where Pr(N = n) = p_n. P_N(t) has the following form.

P_N(t) = p₀ + p₁t + p₂t² + ... + p_ntⁿ + ...

We can perform some derivative calculations for this functions.

P_N^'(t) = p₁ + 2p₂t + ... + np_nt^n-1 + ...

P_N^''(t) = 2*1p₂ + 2*3p₃t + 3*4p₄t² ... + n*(n-1)p_nt^n-2 + ...

We note that P_N^'(1) = p₁ + 2p₂ + 3p₃ + ... + np_n + ..., which by definition is the same as E[N]. Thus, P_N^'(1) = E[N].

We also note that P_N^''(1) = 2*1p₂ + 2*3p₃ + 3*4p₄ ... + n*(n-1)p_n + ...

We can add 0 = 0(-1)p₀ + 1*0p₁ to this quantity to get

P_N^''(1) = 0(-1)p₀ + 1*0p₁ + 2*1p₂ + 2*3p₃ + 3*4p₄ ... + n*(n-1)p_n + ...

This, by definition is E[N(N-1)] = E[N²] - E[N].

Thus, P_N^''(1) = E[N²] - E[N].

A Poisson random variable M with mean Λ is defined as follows (according to Daniel 2008):

50.1. M can only assume whole-number values.

50.2. Pr[M = k] = e^-ΛΛ^k/k!

50.3. E[M] = Var[M] = Λ

A Poisson process N, together with a rate function λ, can be described as follows (according to Daniel 2008):

50.4. N(0) = 0 and N is non-decreasing for whole-number values.

50.5. Any two non-overlapping time increments are independent with respect to N. What happens in one of those non-overlapping time increments does not affect what happens in the others.

50.6. "For all t ≥ 0 and h > 0, the increment N(t + h) - N(t) is a Poisson random variable with mean Λ = _t^t+h∫λ(z)dz."

50.7. The function m(t) = _t^t+h∫λ(z)dz is the mean value function or operational time.

50.8. Where λ is equal to a constant, N is a homogeneous poisson process.

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 S3L50-1. Similar to Question 23 from the Casualty Actuarial Society's Fall 2006 Exam 3.

The number of blue rhinoceroses that spontaneously grow wings on a given day follows a negative binomial distribution with mean 12 and variance 60. Find the probability that at least 13 but fewer than 15 blue rhinoceroses will grow wings today.

Solution S3L50-1. In a negative binomial distribution, E[M] = rβ and Var[M] = rβ(1+ β).

Thus, Var[M]/E[M] = (1+ β). In our case, 60/12 = (1+ β) = 5, so β = 4.

Since rβ = 12, it follows that r = 12/4 = 3.

Now we can use the formula P[M = k] = ([r(r + 1)...(r + k -1)]/k!)(β^k/(1+ β)^k+r).

We want to find P[M = 13] + P[M = 14].

P[M = 13] = [(3*4*5*6*7*8*9*10*11*12*13*14*15)/13!](4¹³/5¹⁶) = 2*14*15(4¹³/5¹⁶) = 0.1847179535.

P[M = 14] = [(3*4*5*6*7*8*9*10*11*12*13*14*15*16)/14!](4¹⁴/5¹⁷) = 2*15*16(4¹⁴/5¹⁷) = 0.168884986.

So our desired answer is 0.1847179535 + 0.168884986 = about 0.3536029395.

Problem S3L50-2. Similar to Question 25 from the Casualty Actuarial Society's Fall 2006 Exam 3.

You know the following about derivatives of the probability generating function for a discrete probability distribution:
P'(1) = 68

P''(1) = 19400.
Find Var(N), where N is the random variable for the distribution.

Solution S3L50-2. We know that P_N^'(1) = E[N] and P_N^''(1) = E[N²] - E[N].

Thus, 19400 = E[N²] - 68 and E[N²] = 19468.

Now we use the formula Var(N) = E[N²] - (E[N]²) = 19468 - 68² = Var(N) = 14844.

Problem S3L50-3. Similar to Question 27 from the Casualty Actuarial Society's Fall 2006 Exam 3. Sam the Speaker is always interrupted by Harold the Heckler when Sam gives a public speech. The length of the time period for which Sam can speak without being heckled is exponentially distributed with a mean of 5 minutes. Find the probability that, after the beginning of his speech, Sam will be heckled at least once within the first 3 minutes.

Solution S3L50-3. The survival function (in terms of minutes) for an exponential distribution with a mean of 5 minutes is s(x) = e^-x/5. Sam "survives" the first three minutes of the speech if he does not get heckled at all during that time. We want to find the complement of s(3), the probability that Sam gets heckled at least once. Thus, we want to find 1 - s(3) = 1 - e^-3/5 = about 0.4511883639.

Problem S3L50-4. Similar to Question 28 from the Casualty Actuarial Society's Fall 2006 Exam 3.

The number of flies that a flycatcher catches per hour follows a non-homogeneous process with the following rate function, with t in hours:

λ(t) = 56t + 3 for 0 ≤ t < 45.

λ(t) = 4t² - 34t + 4 for 45 ≤ t ≤ 90.

λ(t) = 845 for t > 90.

How many flies can the flycatcher be expected to catch between hours 17 and 100, inclusive?

Solution S3L50-4. We want to find Λ = ₁₇¹⁰⁰∫λ(z)dz =

Λ = ₁₇⁴⁵∫(56z + 3)dz + ₄₅⁹⁰∫(4z² - 34z + 4)dz + 845*10 (since, during the last 10 hours, the flycatcher can expect to catch 845 flies in each hour).

Thus, Λ = 48692 + 747405 + 8450 = 804,547 flies.

Problem S3L50-5. Similar to Question 26 from the Casualty Actuarial Society's Fall 2006 Exam 3.

Which of these statements about Poisson processes are true?

(a) What happens from time 3 to time 7 is independent of what happens from time 7 to time 16.

(b) What happens from time 3 to time 7 is independent of what happens from time 5 to time 13.

(c) The rate function for all Poisson processes is constant.

(d) For all Poisson random variables M, defined as M = N(t + h) - N(t), it is the case that

Var[M] = _t^t+h∫λ(z)dz.

Solution S3L50-5.

(a) is true, because [3, 7] and [7, 16] are not overlapping time intervals, even though they touch at an endpoint. So Condition 50.5 is met.

(b) is false, because [3, 7] and [5, 13] are overlapping time intervals, so independence of these intervals cannot be the case. Condition 50.5 is not met in this case.

(c) is false. Non-homogeneous Poisson processes have non-constant rate functions. We just saw an example of this in Problem S3L50-4.

(d) is true, because Var[M] = Λ by Condition 50.3, and Λ = _t^t+h∫λ(z)dz by Condition 50.6.

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