Verified solution · Exam P

SOA Exam P Sample Question #444 Solution

This Exam P sample reference tests Joint Distributions. Only the three states with Y=1 contribute to the expected ratio. Their masses are 3/18, 2/18, and 1/18 for X=1,2,3, producing 3/18+(2/18)/2+(1/18)/3=13/54=0.2407 and choice C.

Independent analysis by ActuaryProof · reviewed August 4, 2026

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Independent solution

How to solve this Joint Distributions question

Setup

Evaluate the supplied joint mass formula on the row where Y equals one. The numerator then simplifies to 4-X.

p(x,1)=\frac{-2x-4+x+8}{18}=\frac{4-x}{18}

p(1,1)=\frac3{18},\quad p(2,1)=\frac2{18},\quad p(3,1)=\frac1{18}

Model

Apply the expectation-of-a-function formula. Every state with Y=0 has function value zero.

\operatorname{E}\!\left[\frac{Y}{X}\right]=\sum_{x=1}^{3}\sum_{y=0}^{1}\frac{y}{x}p(x,y)

\operatorname{E}\!\left[\frac{Y}{X}\right]=\sum_{x=1}^{3}\frac1x p(x,1)

Compute

Substitute the three nonzero masses and combine the fractions.

\operatorname{E}\!\left[\frac{Y}{X}\right]=\frac3{18}+\frac12\frac2{18}+\frac13\frac1{18}

\operatorname{E}\!\left[\frac{Y}{X}\right]=\frac9{54}+\frac3{54}+\frac1{54}=\frac{13}{54}=0.2407407407\ldots

Answer

The expected ratio rounds to 0.241.

\boxed{0.241\quad\text{(C)}}

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Error diagnosis

Why the other letters tempt

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

AAssigning the same mass 1/18 to all three Y=1 states gives (1+1/2+1/3)/18=11/108=0.10185, near 0.102. The actual masses are 3/18, 2/18, and 1/18.
BThis is the ratio of expectations, E[Y]/E[X]=(1/3)/(5/3)=0.200. In general, E[Y/X] is not E[Y]/E[X].
DThe three exact nonzero contributions are 0.16667, 0.05556, and 0.01852, totaling 0.24074. A value of 0.306 necessarily overcounts at least one joint state.
EThis is E[1/X]=1(1/2)+(1/2)(1/3)+(1/3)(1/6)=13/18=0.7222. It omits the factor Y that zeroes the entire Y=0 row.

Original practice · fully worked

Original variant: nonlinear load under a random intensity

A controller selects intensity X=1 with probability 0.40 and X=3 with probability 0.60. Conditional on X=x, the number Y of incoming signals is Poisson distributed with mean x. A load score is C=XY+Y(Y-1). Calculate E[C].

A
2.2
B
4.4
C
5.8
D
11.6
E
13.8

Variant answer in brief

Given X=x, Poisson moments give E[Y|X=x]=x and E[Y(Y-1)|X=x]=x². Thus E[C|X=x]=2x²; averaging over X gives 2[0.40(1)+0.60(9)]=11.6, choice D.

Setup

Condition on the selected intensity and write the first two factorial moments of the resulting Poisson count.

Y\mid X=x\sim\operatorname{Poisson}(x)

\operatorname{E}[Y\mid X=x]=x

\operatorname{E}[Y(Y-1)\mid X=x]=x^2

Model

For fixed x, apply conditional expectation to both terms of the nonlinear score.

\operatorname{E}[C\mid X=x]=x\operatorname{E}[Y\mid X=x]+\operatorname{E}[Y(Y-1)\mid X=x]

\operatorname{E}[C\mid X=x]=x^2+x^2=2x^2

Compute

Average the two conditional load scores using the intensity probabilities.

\operatorname{E}[C]=0.40[2(1)^2]+0.60[2(3)^2]

\operatorname{E}[C]=0.8+10.8=11.6

Answer

The expected load score is 11.6.

\boxed{11.6\quad\text{(D)}}

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ActuaryProof. “SOA Exam P Sample Question #444 — Verified Solution.” Verified 2026-07-31.

Source reference: official SOA sample-question PDF. Last reviewed 2026-07-31.

How to solve this Joint Distributions question

Setup

Model

Compute

Answer

All 718 Exam P sample solutions in syllabus order

Why the other letters tempt

Original variant: nonlinear load under a random intensity

Variant answer in brief

Setup

Model

Compute

Answer

Related Exam P solutions

Turn this solution into a complete study sequence

Cite this solution