Worked solution · Exam P

SOA Exam P Sample Question #337 Solution

This Exam P sample reference tests Exponential Distribution. This is the expected excess of an exponential loss above a deductible. The probability of reaching the deductible is exp(−3/10), and memorylessness leaves mean excess 10, so expected payment is 10 exp(−0.3)≈7.41 and choice D.

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 Exponential Distribution question

Setup

Represent the insurer payment as the positive excess above the deductible.

Y=(X-3)_+,\qquad X\sim\operatorname{Exponential}\left(\text{mean }10\right)

Model

A positive payment requires survival beyond three, and the exponential residual loss then has its original mean.

\Pr(X>3)=e^{-3/10}

\operatorname{E}[X-3\mid X>3]=10

Compute

Multiply the payment probability by the conditional mean payment.

\operatorname{E}[Y]=e^{-3/10}(10)=7.408182207

Answer

The expected claim payment is approximately 7.41.

\boxed{7.41\quad\text{(D)}}

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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 2.59 is 10(1-exp(−0.3)), the expected amount of loss limited to the deductible. It is the retained layer, not the insurer's excess payment.
BThe value 5.19 is approximately twice the limited-loss amount 10(1-exp(−0.3)), double-counting a component that is not the claim payment.
CThe value 7.00 subtracts the deductible from the unconditional mean, 10-3, and ignores the zero payment when the loss does not exceed three.
EThe value 9.63 is E[X 1{X>3}]=13 exp(−0.3), which includes the full loss whenever a claim occurs instead of removing the deductible.

Original practice · fully worked

Original variant: repair payment under a severity mixture

A repair is minor with probability 0.30 and major with probability 0.70. Conditional on a minor repair, loss X is exponential with mean 4 units; conditional on a major repair, X is exponential with mean 10 units. A service plan pays the amount above a deductible of 5 units. Calculate the expected plan payment.

A
0.344
B
4.246
C
4.457
D
4.590
E
8.200

Answer and calculation

For an exponential loss with mean μ, expected payment above five is μe to the power (−5/μ). Weighting the minor and major values by 0.30 and 0.70 gives 0.30(4 exp(−5/4))+0.70(10 exp(−1/2))≈4.590 and choice D.

Setup

Use the exponential stop-loss expectation separately within each repair class.

\operatorname{E}[(X-5)_+\mid M]=4e^{-5/4}

\operatorname{E}[(X-5)_+\mid J]=10e^{-5/10}

Model

Average the two conditional payment means with their repair-class probabilities.

\operatorname{E}[Y]=0.30\left(4e^{-5/4}\right)+0.70\left(10e^{-1/2}\right)

Compute

Evaluate the minor and major contributions and add them.

0.30\left(4e^{-5/4}\right)=0.3438057562

0.70\left(10e^{-1/2}\right)=4.245714618

\operatorname{E}[Y]=4.589520374

Answer

The expected service-plan payment is approximately 4.590.

\boxed{4.590\quad\text{(D)}}

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How to solve this Exponential Distribution question

Setup

Model

Compute

Answer

Probability Univariate Distributions Practice Workbook

Why the other letters tempt

Original variant: repair payment under a severity mixture

Answer and calculation

Setup

Model

Compute

Answer

Related Exam P solutions

Next steps for Exam P Univariate Random Variables