Worked solution · Exam P

SOA Exam P Sample Question #626 Solution

This Exam P sample reference tests Poisson Distribution. For a Poisson variable, the variance equals its mean and the zero-count probability is exponential in that mean. Taking the logarithm of the stated probability ratio gives a variance difference of ln(2)=0.69315, so choice A 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 Poisson Distribution question

Setup

Let the two Poisson means be lambda_1 and lambda_2. Their variances equal those same parameters.

V_1=\lambda_1,\qquad V_2=\lambda_2

Model

A Poisson count is zero with probability e raised to the negative mean. Express the given zero-count ratio with that identity.

e^{-\lambda_1}=\frac{1}{2}e^{-\lambda_2}

Compute

Divide the exponentials and take natural logarithms to isolate the parameter difference.

e^{-(\lambda_1-\lambda_2)}=\frac{1}{2}

\lambda_1-\lambda_2=\ln 2=0.6931471806\ldots

Answer

The same difference applies to the two Poisson variances.

\boxed{V_1-V_2=0.693\quad\text{(A)}}

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.

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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.

BThis exceeds ln(2) because it effectively replaces the exact one-half probability ratio by a smaller rounded ratio before taking the logarithm. The ratio is given exactly and should not be perturbed.
CThis treats the zero-count relation too much like a linear probability difference. Poisson zero probabilities are exponential in the means, so a logarithm is required.
DThis can arise from taking a logarithm after incorrectly adding a probability adjustment to the one-half ratio. The algebra requires division of the two zero-count probabilities.
EThis is another inflated parameter difference produced by altering the exact ratio. Once the exponential terms are divided, the result is fixed uniquely at the natural logarithm of two.

Original practice · fully worked

Original variant: triaged sensor alerts

A monitoring station receives a Poisson number of sensor alerts with mean 6 per shift. Each alert is independently classified as urgent with probability 0.40; all other alerts are routine. Let U and R be the urgent and routine counts. Calculate Var(2U-R).

A
2.4
B
6.0
C
8.4
D
13.2
E
36.0

Answer and calculation

Poisson thinning makes U and R independent Poisson counts with means 2.4 and 3.6. Therefore Var(2U-R)=4(2.4)+3.6=13.2, which is choice D.

Setup

Split the mean total count between the urgent and routine categories.

E[U]=6(0.40)=2.4

E[R]=6(0.60)=3.6

Model

Independent thinning of a Poisson count produces independent Poisson category counts, each with variance equal to its mean.

U\sim\operatorname{Poisson}(2.4),\qquad R\sim\operatorname{Poisson}(3.6)

\operatorname{Cov}(U,R)=0

Compute

Square the coefficients when taking the variance of the linear combination.

\operatorname{Var}(2U-R)=2^2\operatorname{Var}(U)+(-1)^2\operatorname{Var}(R)

\operatorname{Var}(2U-R)=4(2.4)+3.6=13.2

Answer

The variance of the weighted alert score is 13.2.

\boxed{13.2\quad\text{(D)}}

Keep practicing

Next steps for Exam P Univariate Random Variables

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How to solve this Poisson Distribution question

Setup

Model

Compute

Answer

Probability Univariate Distributions Practice Workbook

Why the other letters tempt

Original variant: triaged sensor alerts

Answer and calculation

Setup

Model

Compute

Answer

Related Exam P solutions

Next steps for Exam P Univariate Random Variables