Verified 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 August 4, 2026

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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)}}

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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 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

Variant answer in brief

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)}}

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ActuaryProof. “SOA Exam P Sample Question #239 — Verified Solution.” Verified 2026-07-31.

Source reference: official SOA sample-question PDF. Last reviewed 2026-07-31.

How to solve this Joint Cumulative Distribution Functions question

Setup

Model

Compute

Answer

All 718 Exam P sample solutions in syllabus order

Why the other letters tempt

Original variant: covariance recovered from a binary joint CDF

Variant answer in brief

Setup

Model

Compute

Answer

Related Exam P solutions

Turn this solution into a complete study sequence

Cite this solution