SOA Exam P Sample Question #589. Official problem wording is withheld; follow the source link to read it.

SOA Exam P Sample Question #589 — Verified Solution

Independent solution

How to solve this Compound Poisson Distribution question

Answer in brief

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.

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)}}

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

Variant answer in brief

A single incident has second moment 0.6(2^2)+0.4(5^2)=12.4. A compound Poisson sum has variance lambda E[Y^2], 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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ActuaryProof. “SOA Exam P Sample Question #589 — Verified Solution.” Verified 2026-07-30.

Source reference: official SOA sample-question PDF. Last reviewed 2026-07-30.