Worked solution · Exam P

SOA Exam P Sample Question #574 Solution

This Exam P sample reference tests Joint Distributions. Only the row totals are needed for the strawberry marginal. The readable rows have masses 0.25, 0.33, and 0.24, so normalization fixes the final row at 0.18. These masses give E[S]=2.35 and E[S²]=6.61, hence variance 1.0875 and 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.

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Independent solution

How to solve this Joint Distributions question

Setup

Sum each readable row of the joint table to obtain the first three marginal probabilities.

\Pr(S=1)=0.25,\qquad \Pr(S=2)=0.33,\qquad \Pr(S=3)=0.24

Model

The unread entries all lie in the same final row, so their individual allocation is irrelevant. Total probability determines that row's combined mass.

\Pr(S=4)=1-(0.25+0.33+0.24)=0.18

Compute

Calculate the first two moments of the resulting marginal distribution.

E[S]=1(0.25)+2(0.33)+3(0.24)+4(0.18)=2.35

E[S^2]=1(0.25)+4(0.33)+9(0.24)+16(0.18)=6.61

\operatorname{Var}(S)=6.61-2.35^2=1.0875

Answer

The variance rounds to 1.09.

\boxed{1.09\quad\text{(D)}}

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.

AUnderstating the final-row contribution to E[S²] by 0.06 gives 6.55-2.35²=1.0275, which rounds to 1.03.
BUsing the incorrect mean 2.36 while keeping E[S²]=6.61 gives 6.61-2.36²=1.0404.
CUnderstating E[S²] by 0.02 gives 6.59-2.35²=1.0675, which rounds to 1.07.
EThis assumes the allocation among the unread cells is needed. For a function of S alone, only their combined row mass matters, and normalization supplies it.

Original practice · fully worked

Original variant: recovering a missing joint cell

Two discrete variables U and V take values U∈{0,1} and V∈{0,2}. Their joint probabilities are P(U=0,V=0)=0.20, P(U=0,V=2)=0.30, P(U=1,V=0)=x, and P(U=1,V=2)=0.50-x. Given Cov(U,V)=0.10, determine x.

A
0.000
B
0.050
C
0.100
D
0.200
E
0.450

Answer and calculation

The table gives E[U]=0.5, E[V]=1.6-2x, and E[UV]=1-2x. Therefore Cov(U,V)=0.2-x; equating this to 0.10 gives x=0.10 and choice C.

Setup

Compute the marginal mean of U and express the other required moments in terms of the unknown cell.

E[U]=0.50

E[V]=2(0.30+0.50-x)=1.6-2x

E[UV]=2(0.50-x)=1-2x

Model

Use the covariance identity with the moment expressions from the table.

\operatorname{Cov}(U,V)=E[UV]-E[U]E[V]

Compute

Simplify the covariance and solve the supplied covariance equation.

\operatorname{Cov}(U,V)=1-2x-0.5(1.6-2x)=0.2-x

0.2-x=0.10

x=0.10

Answer

The missing joint probability is 0.10.

\boxed{x=0.10\quad\text{(C)}}

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How to solve this Joint Distributions question

Setup

Model

Compute

Answer

Probability Multivariate & Risk Models Practice Workbook

Why the other letters tempt

Original variant: recovering a missing joint cell

Answer and calculation

Setup

Model

Compute

Answer

Related Exam P solutions

Next steps for Exam P Multivariate Random Variables