SOA Exam P Sample Question #626. Official problem wording is withheld; follow the source link to read it.

SOA Exam P Sample Question #626 — Verified Solution

Independent solution

How to solve this Poisson Distribution question

Answer in brief

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.

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)}}

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

Variant answer in brief

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)}}

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ActuaryProof. “SOA Exam P Sample Question #626 — Verified Solution.” Verified 2026-07-30.

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