This Exam P sample reference tests Exponential Distribution. The mean of a payment limited at 1000 is the integral of the loss survival function up to that limit. Substitution of beta=400/ln(2) gives 475.064, which rounds to choice D.
These notes identify the calculation error associated with each wrong letter when that error is reproducible.
AThis computes E[X 1(X≤1000)]=beta-(1000+beta) exp(−1000/beta)=298.29 and omits the capped 1000 payments from larger losses.
BThis drops the division by ln(2), treats 400 as the mean, and obtains 400(1-exp(−1000/400))=367.17.
CThis treats the numerator 400 as the unlimited mean and then ignores the effect of the reimbursement limit.
EThis reports the unlimited exponential mean 400/ln(2)=577.08 instead of the mean after applying the cap.
Original practice · fully worked
Original variant: variance after an attachment point
A weather-repair cost X, measured in thousand-dollar units, follows an exponential model with mean 4. A relief plan pays max(X-4,0). Calculate the variance of the payment.
A 1.4715
B 5.8861
C 9.6068
D 11.7721
E 16.0000
Variant answer in brief
The payment has first moment 4/e and second moment 32/e. Their difference E[Y²]-E[Y]² is 9.6068, so choice C is correct.
Setup
Setup
Let Y be the excess payment and note the probability that the attachment point is exceeded.
Y=(X−4)+,Pr(X>4)=e−1
Model
Model
Exponential memorylessness gives an exponential excess with mean four whenever a positive payment occurs.
E[Y]=e−1(4)
E[Y2]=e−1(2(42))
Compute
Compute
Subtract the squared first moment from the second moment.
Var(Y)=e32−(e4)2
Var(Y)=9.606777586
Answer
Answer
The payment variance is approximately 9.6068 square thousand-dollar units.
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