Binomial, Pareto, and Compound Probability Distributions: Practice Problems and Solutions

The following additional facts about the Pareto probability distribution are useful.

If random variable X follows a Pareto distribution with parameters θ and α and survival function s(x) = θ^α/(x + θ)^α, then the following is the case.

E[X] =θ/(α - 1)

E[X²] = 2θ²/[(α - 1)(α - 2)]

You may also be asked to work with truncated expected values for the Pareto distribution. These are denoted by E[X Λ K], which means the expected value of X, where X can assume any value less than or equal to K. The following holds under the Pareto distribution.

E[X Λ K] =[θ/(α - 1)][1 - [θ/(K + θ)]^{(α - 1)}]

Now we will briefly review the binomial probability distribution. If random variable N follows a binomial distribution with parameters m and q, then the following is the case.

Pr[N = n] = C(m, n)qⁿ(1 - q)^m-n

E[N] = mq

Var[N] = mq(1 - q)

Whenever we have a random variable S which follows a compound distribution, where some random variable N is chosen and then some different random variable X is chosen based on the choice on N, then the following is the case.

Var[S] = E[N]Var[X] + Var[N](E[X]²)

For instance, N could represent the frequency of losses and X could represent the severity of losses that occur. We will see another possible application of a compound distribution in one of the problems of this section.

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

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

Problem S3L51-1. The number of frogs on a lily pad follows a Pareto distribution with θ = 32 and α = 5. Find the expected value of the number of frogs on this lily pad.

Solution S3L51-1. We use the formula E[X] = θ/(α - 1) = 32/(5 - 1) = 32/4 = 8 frogs.

Problem S3L51-2. The number of lily pads in a pond follows a binomial distribution with m = 12400 and q = 0.24. Find the variance of the number of lily pads in this pond.

Solution S3L51-2. We use the formula Var[N] = mq(1 - q) = 12400*0.24*0.76 = Var[N] = 2261.76.

Problem S3L51-3. The number of lily pads in a pond follows a binomial distribution with m = 12400 and q = 0.24. The number of frogs on any lily pad follows a Pareto distribution with θ = 32 and α = 5. Find the variance of the total number of frogs that are sitting on lily pads throughout the pond.

Solution S3L51-3. We want to find Var[S], where S follows a compound distribution where the binomial distribution of N is considered first, followed by the Pareto distribution of X.

We already know from Solutions S3L51-1 and S3L51-2 that E[X] = 8, Var[N] = 2261.76.

We, moreover, find E[N] = mq = 12400*0.24 = 2976.

Then we find E[X²] = 2θ²/[(α - 1)(α - 2)] = 2*32²/(4*3) = 170.666666667

Thus, Var[X] = E[X²] - (E[X]²) = 170.666666667 - 64 = 106.666666667

Now we use the formula

Var[S] = E[N]Var[X] + Var[N](E[X]²) =

2976*106.666666667 +2261.76*64 = Var[S] = 462192.64.

Problem S3L51-4. Similar to Question 29 from the Casualty Actuarial Society's Fall 2006 Exam 3.

The frequency of losses in spaceship accidents (N) follows a binomial distribution with m = 21000 and q = 0.55. The severity of each loss (X) follows a Pareto distribution with θ = 2340 and α = 9. Find the standard deviation of the compound distribution of the random variable S, where S denotes the aggregate losses in spaceship accidents.

Solution S3L51-4. We first find E[N] = mq = 21000*0.55 =11550.

Then we find Var[N] = mq(1 - q) = 11550*0.45 = 5197.5.

Then we find E[X] = θ/(α - 1) = 2340/8 = 292.5

Then we find E[X²] = 2θ²/[(α - 1)(α - 2)] = 2*5475600/(8*7) = 195557.1429

Then we find Var[X] = E[X²] - (E[X]²) = 195557.1429 - 292.5² = 110000.8929

Now we find Var[S] = E[N]Var[X] + Var[N](E[X]²) =

11550*110000.8929 + 5197.5*292.5² = about 1715188922.

We find SD[S] = (1715188922)^1/2 = SD[S] = about 41414.83939.

Problem S3L51-5. Similar to Question 30 from the Casualty Actuarial Society's Fall 2006 Exam 3.

There are two insurance policies designed to insure against losses due to catastrophic cake splatterings. Policy C has no limits or deductibles and pays the full loss to the claimant. Policy D has a deductible of 800 Golden Hexagons (GH) and a limit of 7800 GH. That is, Policy D will pay all claims between 800 GH and 7800 GH. In year 0, losses due to catastrophic cake splatterings follow a Pareto Distribution with α = 5 and θ = 6000. Government officials are printing money like crazy in order to buy up all the cakes and throw them at the helpless populace. Therefore, the annual inflation rate is 56%. Find the absolute value of the difference between the expected losses under policies C and D in year 5.

Solution S3L51-5. As a result of inflation θ under the Pareto Distribution will increase from 6000 to 6000*1.56⁵ = 55433.74787. α will be unchanged.
Under Policy C, the expected loss in year 5 will simply be θ/(α - 1) = 55433.74787/4 = 13858.43697 GH.

Under Policy D, only the loss amounts between 800 and 7800 GH will be paid (a great deal for the insurance company due to the inflation that happened).

Thus, the expected loss under Policy D is E[X for X between 800 and 7800] =

E[X Λ 7800] - E[X Λ 800] =

[θ/(α - 1)][1 - [θ/(7800 + θ)]^{(α - 1)}] - [θ/(α - 1)][1 - [θ/(800 + θ)]^{(α - 1)}] =

[13858.43697][1 - [55433.74787/(7800 + 55433.74787)]⁴] -

[13858.43697][1 - [55433.74787/(800 + 55433.74787)]⁴] =

5673.496468 - 771.9492408 = 4901.547227.

The difference between the expected losses under these two policies is therefore

13858.43697 - 4901.547227 = about 8956.889739 GH.

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