Worked solution · Exam P

SOA Exam P Sample Question #246 Solution

This Exam P sample reference tests Poisson Distribution. Conditioning two independent Poisson counts on their sum makes X binomial with n=3 and p=1.7/4=0.425. Since X-Y=2X-3, the conditional variance is 4(3)(0.425)(0.575)=2.9325, choice C.

Independent analysis by ActuaryProof · reviewed July 31, 2026

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Independent solution

How to solve this Poisson Distribution question

Setup

Condition on the fixed combined count and use the Poisson splitting property.

X\mid(X+Y=3)\sim\operatorname{Binomial}\left(3,\frac{1.7}{1.7+2.3}\right)

p=0.425

Model

Under the condition, eliminate Y from the difference using the fixed total.

Y=3-X,\qquad X-Y=2X-3

Compute

A constant does not affect variance, while a scale factor is squared.

\operatorname{Var}(X-Y\mid X+Y=3)=4\operatorname{Var}(X\mid X+Y=3)

\operatorname{Var}(X-Y\mid X+Y=3)=4(3)(0.425)(0.575)=2.9325

Answer

The conditional variance rounds to 2.93.

\boxed{2.93\quad\text{(C)}}

Test the method on unseen problems

Probability Multivariate & Risk Models Practice Workbook

30 original multivariate 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.

AThe value 1.71 is 3(0.57), obtained by rounding the conditional share 0.575 down before reporting E[Y|X+Y=3]. It is a component mean, not the variance of the difference.
BThe value 1.77 can result from miscopying p(1-p)=0.244375 as 0.1475 before evaluating 4(3)p(1-p). Using the actual conditional binomial parameter gives 2.9325.
DThe value 3.14 is E[(X-Y)²|X+Y=3]=Var(X-Y|X+Y=3)+E[X-Y|X+Y=3]²=2.9325+0.2025=3.135. It is the second raw moment, not the variance.
EThe value 4.00 is the unconditional variance Var(X-Y)=Var(X)+Var(Y)=1.7+2.3. Conditioning on the combined count changes the variance.

Original practice · fully worked

Original variant: conditional covariance from an inferred split

Counts A and B from two independent event streams are Poisson with unknown positive means. Given that six events occurred in total, the conditional probability that all six came from stream A is 1/64. Calculate Cov(A,B) under this condition.

A
-3.000
B
-1.500
C
0
D
1.500
E
6.000

Answer and calculation

Conditioning on the total makes A binomial with six trials and an unknown split probability p. The all-A probability gives p=1/2, and because B=6-A, the conditional covariance is -Var(A)=-1.5, choice B.

Setup

Let p be stream A's share of the combined Poisson rate. Conditional splitting makes A binomial.

A\mid(A+B=6)\sim\operatorname{Binomial}(6,p)

Model

Use the probability that every conditioned event is allocated to stream A to recover p.

\Pr(A=6\mid A+B=6)=p^6=\frac1{64}

p=\frac12

Compute

Under the fixed total, B is a decreasing affine function of A.

B=6-A

\operatorname{Cov}(A,B\mid A+B=6)=\operatorname{Cov}(A,6-A)

=-\operatorname{Var}(A)=-6\left(\frac12\right)\left(\frac12\right)=-1.5

Answer

The conditional covariance is -1.500.

\boxed{-1.500\quad\text{(B)}}

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How to solve this Poisson Distribution question

Setup

Model

Compute

Answer

Probability Multivariate & Risk Models Practice Workbook

Why the other letters tempt

Original variant: conditional covariance from an inferred split

Answer and calculation

Setup

Model

Compute

Answer

Related Exam P solutions

Next steps for Exam P Multivariate Random Variables