Worked solution · Exam P

SOA Exam P Sample Question #410 Solution

This Exam P sample reference tests Joint Cumulative Distribution Functions. This joint-CDF event combines a count restriction with an aggregate-loss restriction. Counts zero and one always qualify, while a count of two qualifies unless both severities equal two, giving 0.9575 and choice E.

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.

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Independent solution

How to solve this Joint Cumulative Distribution Functions question

Setup

The requested joint CDF includes outcomes with at most two events and aggregate severity at most three. Partition by the event count.

F(2,3)=\sum_{n=0}^{2}\Pr(X=n,\,Y\le3)

Model

With zero or one event, the aggregate cannot exceed three. With two events, the only disqualifying severity pair is two losses of size two.

\Pr(Y\le3\mid X=0)=\Pr(Y\le3\mid X=1)=1

\Pr(Y\le3\mid X=2)=1-(0.50)^2=0.75

Compute

Weight the conditional probabilities by the count probabilities.

F(2,3)=0.80+0.12+(0.05)(0.75)

F(2,3)=0.9575

Answer

The joint cumulative probability is 0.9575.

\boxed{0.9575\quad\text{(E)}}

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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 0.0250 is P(X=2,Y=3): two events occur and exactly one of their severities equals two. It omits all other outcomes in the cumulative event.
BThe value 0.0500 is P(X=2) and ignores both the aggregate restriction and the qualifying counts zero and one.
CThe value 0.0675 is the complement of 0.9325. It results from first evaluating the wrong event Y≤2 and then taking its complement.
DThe value 0.9325 equals 0.80+0.12+0.05(0.25), which imposes Y≤2 rather than Y≤3 when two events occur.

Original practice · fully worked

Original variant: nonuniform joint calibration scores

Normalized calibration scores U and V have joint density f(u,v)=(6/5)(u+v²) for 0≤u≤1 and 0≤v≤1, and zero elsewhere. Let G be their joint cumulative distribution function. Calculate G(0.80,0.50).

A
0.040
B
0.192
C
0.232
D
0.400
E
0.704

Answer and calculation

Integrate the nonconstant joint density over the CDF rectangle. The u and v-squared terms contribute 0.192 and 0.040, totaling 0.232 and selecting choice C.

Setup

The required event is the rectangle from the origin to u=0.80 and v=0.50. The scale factor normalizes the density over the full unit square.

\int_0^1\int_0^1\frac65(u+v^2)\,dv\,du=\frac65\left(\frac12+\frac13\right)=1

G(0.80,0.50)=\Pr(U\le0.80,\,V\le0.50)

Model

Integrate the density over the bounded rectangle and keep its two additive contributions separate.

G=\frac65\int_0^{0.80}\int_0^{0.50}(u+v^2)\,dv\,du

Compute

The u term and v term integrate directly.

\frac65\int_0^{0.80}\int_0^{0.50}u\,dv\,du=\frac65\frac{(0.50)(0.80)^2}{2}=0.192

\frac65\int_0^{0.80}\int_0^{0.50}v^2\,dv\,du=\frac65\frac{(0.80)(0.50)^3}{3}=0.040

G(0.80,0.50)=0.192+0.040=0.232

Answer

The joint cumulative probability is 0.232.

\boxed{0.232\quad\text{(C)}}

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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: nonuniform joint calibration scores

Answer and calculation

Setup

Model

Compute

Answer

Related Exam P solutions

Next steps for Exam P Multivariate Random Variables