Worked solution · Exam P

SOA Exam P Sample Question #589 Solution

This Exam P sample reference tests Compound Poisson Distribution. This is a compound-Poisson expectation with an ordinary deductible. The expected payment per event is the tail integral from 2 to 8, equal to 1.125, so multiplying by the expected count 12 gives 13.5 and choice B is correct.

Independent analysis by ActuaryProof · reviewed July 30, 2026

Official wording withheld. Read the problem in the official SOA sample-question PDF. This page provides only our independently written analysis.

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Independent solution

How to solve this Compound Poisson Distribution question

Setup

Let Y be the payment from one event. Under a deductible of 2, Y is the positive part of the loss above 2.

Y=(X-2)_+

Model

Integrating the stated density gives a simple survival function on the loss support. The tail-integral identity then gives the mean payment.

\Pr(X>x)=\frac{(8-x)^2}{64},\qquad 0\le x\le 8

E[Y]=\int_2^8\Pr(X>x)\,dx

Compute

Evaluate the per-event mean, then apply the compound-sum mean formula with expected event count 12.

E[Y]=\int_2^8\frac{(8-x)^2}{64}\,dx=\frac{6^3}{3(64)}=1.125

E\!\left[\sum_{i=1}^{N}Y_i\right]=E[N]E[Y]=12(1.125)=13.5

Answer

The expected aggregate payment is 13.5.

\boxed{13.5\quad\text{(B)}}

Test the method on unseen problems

Probability Univariate Distributions Practice Workbook

30 original univariate random variables problems with complete solutions and wrong-choice analysis. Written for Exam P, not copied from the SOA set. One-time $19 PDF.

See the cover and sample pages

Error diagnosis

Why the other letters tempt

These notes identify the calculation error associated with each wrong letter when that error is reproducible.

AThis uses E[(X-2)+]=E[X]-2, giving 12(8/3-2)=8, and incorrectly permits negative payments below the deductible.
CThis can arise from integrating the excess against an incorrectly normalized tail rather than the survival function implied by the density.
DThis overstates the conditional excess and then applies it to every event instead of weighting by the chance that the deductible is exceeded.
EThis is 12E[X]=12(8/3)=32 and ignores the deductible entirely.

Original practice · fully worked

Original variant: pop-up service expenses

The number of technical-support incidents during a pop-up event is Poisson with mean 5. Each incident costs 2 units with probability 0.6 and 5 units with probability 0.4, independently of all other incidents and of the count. Calculate the variance of the total incident cost.

A
10.8
B
16.0
C
51.2
D
62.0
E
256.0

Answer and calculation

A single incident has second moment 0.6(2²)+0.4(5²)=12.4. A compound Poisson sum has variance λ E[Y²], so the aggregate variance is 5(12.4)=62 and choice D is correct.

Setup

Let Y denote one incident cost and S the random sum of costs.

S=\sum_{i=1}^{N}Y_i,\qquad N\sim\operatorname{Poisson}(5)

Model

For a compound Poisson sum, the aggregate variance is the Poisson mean times the severity second moment.

\operatorname{Var}(S)=5E[Y^2]

Compute

Calculate the discrete severity second moment and multiply by the frequency mean.

E[Y^2]=0.6(2^2)+0.4(5^2)=12.4

\operatorname{Var}(S)=5(12.4)=62

Answer

The total incident cost has variance 62.

\boxed{62.0\quad\text{(D)}}

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Next steps for Exam P Univariate Random Variables

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How to solve this Compound Poisson Distribution question

Setup

Model

Compute

Answer

Probability Univariate Distributions Practice Workbook

Why the other letters tempt

Original variant: pop-up service expenses

Answer and calculation

Setup

Model

Compute

Answer

Related Exam P solutions

Next steps for Exam P Univariate Random Variables