This Exam P sample reference tests Loss Models. Normalizing the conditional severity probabilities gives K=60/137. Weighting the three payments above the deductible and then the claim frequency gives 0.03139, choice A.
First normalize the conditional severity probabilities, which are proportional to the reciprocals of the five possible severities.
1=Kn=1∑5n1,K=13760
Model
Model
Only severities three, four, and five produce payment above the deductible of two. Weight each excess amount by its normalized conditional probability.
E[(N−2)+∣loss]=K(31+42+53)
Compute
Compute
The expected payment per loss is then multiplied by the annual claim probability 0.05, giving approximately 0.0313869.
E[Y]=0.0513760(31+21+53)=0.0313869
Answer
Answer
The expected annual payment is approximately 0.031, selecting choice A.
0.031(A)
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Original variant: compound expected payment with a discrete deductible
The annual number of device losses is Poisson with mean 0.08. Individual severities independently equal 2, 4, 6, or 8 with probability proportional to severity. A policy pays the amount above a deductible of 3 for every loss. Find the expected aggregate annual payment.
A 0.160
B 0.248
C 0.310
D 0.800
E 3.100
Variant answer in brief
Severity weights normalize by 20. Expected payment per loss is 3.1, and the compound-Poisson mean is 0.08×3.1=0.248, choice B.
Setup
Setup
Normalize the four severity probabilities, which are proportional to severity and therefore have total weight 20.
Pr(X=x∣L)=x/(2+4+6+8)=x/20,x∈{2,4,6,8}
Model
Model
For each severity, apply the deductible of three and weight the positive excess by the normalized severity probability. The expected payment per loss is 3.1.
E[(X−3)+∣L]=200(2)+1(4)+3(6)+5(8)=3.1
Compute
Compute
A compound-Poisson mean equals the frequency mean times the mean payment per loss. Multiplying 0.08 by 3.1 gives 0.248.
E[S]=E[M]E[Y]=0.08(3.1)=0.248
Answer
Answer
The expected aggregate annual payment is 0.248, selecting choice B.
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