Worked solution · Exam P

SOA Exam P Sample Question #652 Solution

This Exam P sample reference tests Marginal Distributions. Summing the joint mass over the second coordinate gives marginal probabilities 0.1, 0.3, and 0.6 at the positive values of X. The resulting first and second moments are 2.5 and 6.7, so the variance is 0.45 and choice A.

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 Marginal Distributions question

Setup

Marginalize the supplied joint mass function over every allowed value of the second coordinate.

p_X(x)=\sum_{y=0}^{x}\frac{x+2y}{40},\qquad x=0,1,2,3

Model

The finite arithmetic sum reduces to a simple marginal mass function.

\sum_{y=0}^{x}(x+2y)=x(x+1)+x(x+1)=2x(x+1)

p_X(x)=\frac{x(x+1)}{20}

Compute

Calculate both raw moments from the marginal probabilities.

\Pr(X=1)=0.1,\qquad \Pr(X=2)=0.3,\qquad \Pr(X=3)=0.6

\operatorname{E}[X]=1(0.1)+2(0.3)+3(0.6)=2.5

\operatorname{E}[X^2]=1^2(0.1)+2^2(0.3)+3^2(0.6)=6.7

\operatorname{Var}(X)=6.7-(2.5)^2=0.45

Answer

The marginal variance is 0.45.

\boxed{\operatorname{Var}(X)=0.45\quad\text{(A)}}

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.

BThis is the raw second moment 6.70 with the decimal point shifted; a variance must subtract the squared mean.
CThis candidate fails the direct mean-centered sum and is too large once the mass concentrated near X = 3 is centered at 2.5.
DThis remains on the scale of the raw second moment because it subtracts only a small correction instead of the full squared mean 6.25.
EThis is E[X²] = 6.70, not Var(X).

Original practice · fully worked

Original variant: inspection tags and manual reviews

A quality team records the number A of priority tags on a production lot and whether a manual review B is requested. Their joint probabilities are P(A=0,B=0)=0.10, P(A=0,B=1)=0.05, P(A=1,B=0)=0.20, P(A=1,B=1)=0.25, P(A=2,B=0)=0.15, and P(A=2,B=1)=0.25. Calculate the variance of A.

A
0.4000
B
0.4500
C
0.4875
D
1.2500
E
2.0500

Answer and calculation

Marginalizing over the review indicator gives probabilities 0.15, 0.45, and 0.40 for A = 0, 1, and 2. The first two raw moments are 1.25 and 2.05, so the variance is 0.4875 and choice C.

Setup

Add the two joint cells in each row to obtain the distribution of the tag count.

\Pr(A=0)=0.10+0.05=0.15

\Pr(A=1)=0.20+0.25=0.45

\Pr(A=2)=0.15+0.25=0.40

Model

Use the first and second raw moments of this three-point marginal distribution.

\operatorname{Var}(A)=\operatorname{E}[A^2]-\operatorname{E}[A]^2

Compute

Evaluate the moments and subtract the squared mean.

\operatorname{E}[A]=0(0.15)+1(0.45)+2(0.40)=1.25

\operatorname{E}[A^2]=0^2(0.15)+1^2(0.45)+2^2(0.40)=2.05

\operatorname{Var}(A)=2.05-1.25^2=0.4875

Answer

The priority-tag count has variance 0.4875.

\boxed{0.4875\quad\text{(C)}}

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How to solve this Marginal Distributions question

Setup

Model

Compute

Answer

Probability Multivariate & Risk Models Practice Workbook

Why the other letters tempt

Original variant: inspection tags and manual reviews

Answer and calculation

Setup

Model

Compute

Answer

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Next steps for Exam P Multivariate Random Variables