Worked solution · Exam P

SOA Exam P Sample Question #173 Solution

This Exam P sample reference tests Conditional Variance. On the conditioning diagonal, the masses for N=0,1,2 are 0.10, 0.18, and 0.12, totaling 0.40. The normalized moments are 1.05 and 1.65, so the variance is 1.65-1.05²=0.5475, choice B.

Independent analysis by ActuaryProof · reviewed July 31, 2026

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Independent solution

How to solve this Conditional Variance question

Setup

Keep only joint cells whose two counts sum to two.

(N,S)\in\{(0,2),(1,1),(2,0)\}

0.10+0.18+0.12=0.40

Model

Normalize the three retained masses to obtain the conditional distribution of N.

\Pr(N=0\mid N+S=2)=0.25

\Pr(N=1\mid N+S=2)=0.45

\Pr(N=2\mid N+S=2)=0.30

Compute

Calculate the first two conditional moments and subtract the squared mean.

E[N\mid N+S=2]=0.45+2(0.30)=1.05

E[N^2\mid N+S=2]=0.45+4(0.30)=1.65

\operatorname{Var}(N\mid N+S=2)=1.65-(1.05)^2=0.5475

Answer

The conditional variance rounds to 0.55.

\boxed{0.55\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 0.48 is the rounded unnormalized calculation 0.66-(0.42)²=0.4836. Conditional moments require dividing the diagonal masses by 0.40 first.
CThe value 0.67 is the variance 2/3 of a uniform distribution on {0,1,2}. The three conditional probabilities are not equal.
DA variance of 0.91 with mean 1.05 would require second moment 2.0125, but the normalized table gives only 1.65.
EA variance of 1.25 exceeds the verified second central moment from the three-point conditional distribution; it is incompatible with the normalized masses.

Original practice · fully worked

Original variant: variance after continuous tail conditioning

A response measure X has density f(x)=2x/9 on 0<x<3 and zero elsewhere. Given that X exceeds 1, calculate the conditional variance of X.

A
0.1111
B
0.3056
C
2.1667
D
4.4444
E
5.0000

Answer and calculation

The tail probability is 8/9. Its normalized first and second moments are 13/6 and 5, so the conditional variance is 5-(13/6)²=11/36=0.3056, choice B.

Setup

Integrate the density to find the probability of the conditioning event.

F(x)=\frac{x^2}{9}

\Pr(X>1)=1-\frac19=\frac89

Model

Write the two tail moments divided by the tail probability.

E[X\mid X>1]=\frac{\int_1^3x(2x/9)\,dx}{8/9}

E[X^2\mid X>1]=\frac{\int_1^3x^2(2x/9)\,dx}{8/9}

Compute

Evaluate the integrals and form the variance.

E[X\mid X>1]=\frac{13}{6}

E[X^2\mid X>1]=5

\operatorname{Var}(X\mid X>1)=5-\left(\frac{13}{6}\right)^2=\frac{11}{36}

Answer

The conditional variance is approximately 0.3056.

\boxed{0.3056\quad\text{(B)}}

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How to solve this Conditional Variance question

Setup

Model

Compute

Answer

Probability Multivariate & Risk Models Practice Workbook

Why the other letters tempt

Original variant: variance after continuous tail conditioning

Answer and calculation

Setup

Model

Compute

Answer

Related Exam P solutions

Next steps for Exam P Multivariate Random Variables