Worked solution · Exam P

SOA Exam P Sample Question #239 Solution

This Exam P sample reference tests Joint Cumulative Distribution Functions. The factorized joint CDF gives marginal masses P(X=3)=0.5³(1-0.5)=0.0625 and P(Y=3)=0.2³(1-0.2)=0.0064. Their product is 0.00040, choice B.

Independent analysis by ActuaryProof · reviewed July 31, 2026

Official wording withheld. Read the problem in the official SOA sample-question PDF. This page provides only our independently written analysis.

Read the solution Open official source

Independent solution

How to solve this Joint Cumulative Distribution Functions question

Setup

Read the two factors of the joint CDF as marginal CDFs. Their product shows that the counts are independent.

F_X(x)=1-0.5^{x+1},\qquad F_Y(y)=1-0.2^{y+1}

Model

Obtain each point mass by differencing its marginal CDF at adjacent integers.

\Pr(X=3)=F_X(3)-F_X(2)=0.5^3-0.5^4=0.0625

\Pr(Y=3)=F_Y(3)-F_Y(2)=0.2^3-0.2^4=0.0064

Compute

Multiply the two marginal masses.

\Pr(X=3,Y=3)=(0.0625)(0.0064)=0.0004

Answer

The exact joint point probability is 0.00040.

\boxed{0.00040\quad\text{(B)}}

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.

See the cover and sample pages

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 0.00004 is P(X=4,Y=4)=(0.03125)(0.00128). It shifts both CDF differences one count too far.
CUsing F(3,1)=0.9000 as the added corner in the four-term difference gives 0.9360-0.8736-0.9300+0.9000=0.0324, near 0.03244. The correct lower-left corner is F(2,2).
DThe value 0.06800 is F(3,3)-F(2,2)=0.9360-0.8680. A diagonal difference includes two boundary strips, not just the single cell.
EThe value 0.12440 is [F(3,3)-F(2,3)]+[F(3,2)-F(2,2)]=0.0624+0.0620. It adds two nested X=3 strips instead of differencing them to isolate Y=3.

Original practice · fully worked

Original variant: covariance recovered from a binary joint CDF

Binary alarm indicators U and V take values 0 or 1. Their joint CDF G satisfies G(0,1)=0.55, G(1,0)=0.40, and G(0,0)=0.25. Calculate Cov(U,V).

A
0.0300
B
0.2500
C
0.3000
D
0.4500
E
0.6000

Answer and calculation

The CDF values give P(U=1)=0.45 and P(V=1)=0.60. Inclusion-exclusion gives P(U=1,V=1)=0.30, so Cov(U,V)=0.30-(0.45)(0.60)=0.03, choice A.

Setup

Because the indicators are binary, two boundary CDF values are marginal zero probabilities.

\Pr(U=0)=G(0,1)=0.55,\qquad \Pr(V=0)=G(1,0)=0.40

E[U]=0.45,\qquad E[V]=0.60

Model

Use the lower-left CDF corner to recover the probability that both indicators equal one.

\Pr(U=1,V=1)=1-\Pr(U=0)-\Pr(V=0)+\Pr(U=0,V=0)

Compute

Evaluate the joint one-cell and subtract the product of the marginal means.

E[UV]=1-0.55-0.40+0.25=0.30

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

\operatorname{Cov}(U,V)=0.30-(0.45)(0.60)=0.03

Answer

The covariance of the two alarm indicators is 0.03.

\boxed{0.0300\quad\text{(A)}}

Keep practicing

Next steps for Exam P Multivariate Random Variables

Work through the related free solutions for this topic, review the formulas you used, or keep every current Exam P solution in one searchable PDF.

More Exam P Multivariate Random Variables practice problems Free Exam P formula sheet Exam P study manual PDF — $49

How to solve this Joint Cumulative Distribution Functions question

Setup

Model

Compute

Answer

Probability Multivariate & Risk Models Practice Workbook

Why the other letters tempt

Original variant: covariance recovered from a binary joint CDF

Answer and calculation

Setup

Model

Compute

Answer

Related Exam P solutions

Next steps for Exam P Multivariate Random Variables