SOA Exam P Sample Question #630. Official problem wording is withheld; follow the source link to read it.

SOA Exam P Sample Question #630 — Verified Solution

Independent solution

How to solve this Exponential Distribution question

Answer in brief

This is an expected excess-loss calculation for an exponential severity above an ordinary deductible. The payment expectation equals the survival probability at two times the mean residual loss, giving e^{-2}, so choice A is correct.

Setup

Let L be the loss amount and express the insurer's payment under an ordinary deductible of two.

L\sim\operatorname{Exp}(1),\qquad P=(L-2)_+

Model

A payment occurs only when the loss crosses the deductible. Conditional on that event, exponential memorylessness makes the excess have the original mean of one.

\Pr(L>2)=e^{-2}

\operatorname{E}[L-2\mid L>2]=1

Compute

Multiply the chance of a positive payment by its conditional mean amount.

\operatorname{E}[P]=\Pr(L>2)\operatorname{E}[L-2\mid L>2]

\operatorname{E}[P]=e^{-2}(1)=e^{-2}=0.1353352832\ldots

Answer

The expected payment per loss is e to the power negative two.

\boxed{e^{-2}\quad\text{(A)}}

Error diagnosis

Why the other letters tempt

These notes identify the calculation error associated with each wrong letter when that error is reproducible.

BMultiplying the deductible amount two by the probability of exceeding it gives 2e^{-2}; the payment is the excess above the deductible, not the deductible itself.
CThis uses survival beyond one rather than beyond two, effectively replacing the stated deductible by the distribution's mean.
DThe quantity 3e^{-2} is E[L times the indicator that L exceeds 2], which includes the retained first two units and therefore does not subtract the deductible.
EThis is the mean loss before applying any deductible and consequently overstates the insurer's expected payment.

Original practice · fully worked

Original variant: capped equipment-repair layer

Equipment damage severity L is exponentially distributed with mean 2 repair units. A service contract pays the portion of a loss lying between 1 and 4 units: it pays nothing below 1, L-1 between 1 and 4, and a maximum of 3 units above 4. Calculate the expected contract payment per incident.

A
0.1353
B
0.4712
C
0.6065
D
0.9424
E
1.2131

Variant answer in brief

The capped layer can be valued by integrating its payment tail from zero to three. This gives 2(e^{-1/2}-e^{-2})=0.942391, so choice D is correct.

Setup

Represent the contract as a three-unit layer attaching when the severity reaches one.

L\sim\operatorname{Exp}(1/2)

W=\min\!\left((L-1)_+,3\right)

Model

For a nonnegative capped payment, integrate the probability that the payment exceeds each level from zero up to its cap.

\operatorname{E}[W]=\int_0^3\Pr(W>t)\,dt

\Pr(W>t)=\Pr(L>1+t)=e^{-(1+t)/2}

Compute

Evaluate the payment-tail integral across the full three-unit layer.

\operatorname{E}[W]=\int_0^3 e^{-(1+t)/2}\,dt

\operatorname{E}[W]=2\left(e^{-1/2}-e^{-2}\right)=0.9423907530\ldots

Answer

The expected capped-layer payment is approximately 0.9424 repair units.

\boxed{0.9424\quad\text{(D)}}

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ActuaryProof. “SOA Exam P Sample Question #630 — Verified Solution.” Verified 2026-07-30.

Source reference: official SOA sample-question PDF. Last reviewed 2026-07-30.