This Exam P sample reference tests Exponential Stop-Loss Moments. The deductible leaves 90% of the original exponential mean. Its payment variance is 0.99 times the original variance, a 1% reduction, choice A.
How to solve this Exponential Stop-Loss Moments question
Setup
Setup
Let the payment be the amount of loss above the deductible. The stated expected-payment reduction implies that a loss exceeds the deductible with probability 0.9.
E[X]=μ,Var(X)=μ2,q=Pr(X>d)=0.9
Model
Model
By the memoryless property, a positive payment has the same exponential mean and second moment as the original loss. Weighting those moments by the 0.9 chance of a positive payment gives the displayed unconditional moments.
E[Y]=qμ=0.9μ,E[Y2]=2qμ2=1.8μ2
Compute
Compute
Subtracting the squared mean from the second moment leaves 99% of the original variance. The variance reduction is therefore 1%.
Var(Y)=1.8μ2−(0.9μ)2=0.99μ2
Answer
Answer
The variance falls by one percent.
1%(A)
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CThe value 10% is the probability of no payment. The variance reduction is the square of that probability, not the probability itself.
DThe value 20% doubles the no-payment probability and omits the squared-mean correction in the variance calculation.
Original practice · fully worked
Original variant: variance reduction from a service deductible
A repair cost is exponentially distributed. A deductible is selected so that the expected payment above the deductible is 80% of the original expected cost. Determine the percentage reduction in the payment variance.
A 1%
B 4%
C 8%
D 16%
E 20%
Variant answer in brief
With survival probability 0.8, the payment variance is (1.6-0.64) times the original variance, or 0.96, so the reduction is 4%, choice B.
Setup
Setup
Let the payment be the amount of loss above the deductible. An expected payment equal to 80% of the original mean implies a deductible-exceedance probability of 0.8.
q=0.8,E[Y]=0.8μ,E[Y2]=1.6μ2
Model
Model
Conditional on a positive payment, memorylessness preserves the original exponential excess distribution. Weighting its first two moments by 0.8 gives the displayed payment moments.
Var(Y)=1.6μ2−(0.8μ)2
Compute
Compute
The payment variance is 96% of the original variance, so the reduction is 4%.
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