This Exam P sample reference tests Compound Poisson Distribution. This is a compound-Poisson expectation with an ordinary deductible. The expected payment per event is the tail integral from 2 to 8, equal to 1.125, so multiplying by the expected count 12 gives 13.5 and choice B is correct.
These notes identify the calculation error associated with each wrong letter when that error is reproducible.
AThis uses E[(X-2)+]=E[X]-2, giving 12(8/3-2)=8, and incorrectly permits negative payments below the deductible.
CThis can arise from integrating the excess against an incorrectly normalized tail rather than the survival function implied by the density.
DThis overstates the conditional excess and then applies it to every event instead of weighting by the chance that the deductible is exceeded.
EThis is 12E[X]=12(8/3)=32 and ignores the deductible entirely.
Original practice · fully worked
Original variant: pop-up service expenses
The number of technical-support incidents during a pop-up event is Poisson with mean 5. Each incident costs 2 units with probability 0.6 and 5 units with probability 0.4, independently of all other incidents and of the count. Calculate the variance of the total incident cost.
A 10.8
B 16.0
C 51.2
D 62.0
E 256.0
Variant answer in brief
A single incident has second moment 0.6(2²)+0.4(5²)=12.4. A compound Poisson sum has variance λ E[Y²], so the aggregate variance is 5(12.4)=62 and choice D is correct.
Setup
Setup
Let Y denote one incident cost and S the random sum of costs.
S=i=1∑NYi,N∼Poisson(5)
Model
Model
For a compound Poisson sum, the aggregate variance is the Poisson mean times the severity second moment.
Var(S)=5E[Y2]
Compute
Compute
Calculate the discrete severity second moment and multiply by the frequency mean.
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