Worked solution · Exam P

SOA Exam P Sample Question #719 Solution

This Exam P sample reference tests Poisson Distribution. The Poisson mean-variance identity turns the two supplied relationships into lambda_X minus lambda_Y equal to 3 and lambda_X equal to twice lambda_Y, so lambda_Y is 3. Therefore E[Y squared] equals lambda_Y squared plus lambda_Y, or 12, matching choice D.

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.

Read the solution Open official source

Independent solution

How to solve this Poisson Distribution question

Setup

Assign rates to the two Poisson variables and use the fact that each rate is both its mean and its variance.

X\sim\operatorname{Poisson}(\lambda_X),\quad Y\sim\operatorname{Poisson}(\lambda_Y)

\mathbb{E}[X]=\operatorname{Var}(X)=\lambda_X,\quad \mathbb{E}[Y]=\operatorname{Var}(Y)=\lambda_Y

Model

Translate the mean-difference and variance-ratio information into two equations for the rates.

\lambda_X-\lambda_Y=3

\lambda_X=2\lambda_Y

Compute

Solve for the target rate, then apply the Poisson second-moment identity.

2\lambda_Y-\lambda_Y=3\quad\Longrightarrow\quad\lambda_Y=3

\mathbb{E}[Y^2]=\operatorname{Var}(Y)+\mathbb{E}[Y]^2

\mathbb{E}[Y^2]=3+3^2=12

Answer

The second raw moment is 12, the value listed under choice D.

\boxed{\mathbb{E}[Y^2]=12\quad\text{(D)}}

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.

AThe value 3 is E[Y] and Var(Y); it omits the squared-mean term needed for the second raw moment.
BThe value 6 results from using 2 lambda_Y, which doubles the variance but is not the identity for E[Y squared].
CThe value 9 is lambda_Y squared, equivalently E[Y(Y-1)], and leaves out the additional lambda_Y term.
EThe value 15 comes from lambda_Y squared plus 2 lambda_Y, so the mean contribution has been counted twice.

Original practice · fully worked

Original variant: cryogenic-detector count

A cryogenic detector's number N of spurious pulses during a test window follows a Poisson distribution. Calibration establishes that the factorial second moment E[N(N − 1)] equals 16. Calculate E[(N + 1)²].

A
16
B
20
C
21
D
25
E
29

Answer and calculation

For a Poisson variable, E[N(N − 1)] equals λ squared, so the nonnegative rate is 4. Expanding the requested square and using E[N squared] = 20 gives 29, making E correct.

Setup

Use the Poisson factorial second-moment identity to identify the rate.

\mathbb{E}[N(N-1)]=\lambda^2=16

\lambda=4

Model

Convert the factorial moment to the ordinary second moment.

\mathbb{E}[N^2]=\mathbb{E}[N(N-1)]+\mathbb{E}[N]

\mathbb{E}[N^2]=16+4=20

Compute

Expand the shifted square and take expectations term by term.

(N+1)^2=N^2+2N+1

\mathbb{E}[(N+1)^2]=20+2(4)+1=29

Answer

The shifted count has second moment 29.

\boxed{\mathbb{E}[(N+1)^2]=29\quad\text{(E)}}

Keep practicing

Next steps for Exam P Univariate Random Variables

Work through the related free solutions for this topic, review the formulas you used, or keep every current Exam P solution in one searchable PDF.

More Exam P Univariate Random Variables practice problems Free Exam P formula sheet Exam P study manual PDF — $49

How to solve this Poisson Distribution question

Setup

Model

Compute

Answer

Probability Univariate Distributions Practice Workbook

Why the other letters tempt

Original variant: cryogenic-detector count

Answer and calculation

Setup

Model

Compute

Answer

Related Exam P solutions

Next steps for Exam P Univariate Random Variables