This Exam P sample reference tests Random-Sum Variance. The accident count has mean 0.75 and variance 0.5625; a retained loss has mean 0.24 and variance 0.0576, producing aggregate variance 0.0756, choice B.
Let the total retained loss be the sum of the retained amounts from a random number of accidents. The accident count is independent of the retained loss from each accident.
E[N]=0.75,Var(N)=0.5625,E[U]=0.24,Var(U)=0.0576
Model
Model
The random-sum variance formula adds two sources of uncertainty: variation in retained severity at a fixed count and variation in the count acting on the mean retained severity.
Var(S)=E[N]Var(U)+Var(N)E[U]2
Compute
Compute
The severity contribution is 0.0432 and the frequency contribution is 0.0324. Adding them gives 0.0756.
Var(S)=0.75(0.0576)+0.5625(0.24)2=0.0756
Answer
Answer
The variance of total retained loss is 0.0756.
0.0756(B)
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AThe value 0.0432 includes only the expected count multiplied by retained-severity variance. It omits the count-variance contribution of 0.0324.
Original practice · fully worked
Original variant: uncovered repair cost from machine faults
Five independent machine faults each occur with probability 0.20. A fault cost is exponentially distributed with mean 2 units, and the owner retains 40% of each cost. Determine the variance of total retained cost.
A 0.512
B 0.960
C 1.152
D 1.280
E 1.600
Variant answer in brief
The fault count has mean 1 and variance 0.8. Retained severity has mean 0.8 and variance 0.64, so total variance is 0.64+0.8(0.64)=1.152, choice C.
Setup
Setup
Let the total retained cost be the sum of retained costs from the five possible independent faults. The fault count has mean one and variance 0.8.
E[N]=1,Var(N)=0.8,E[U]=0.8,Var(U)=0.64
Model
Model
Apply the random-sum variance decomposition to combine retained-severity variation with variation in the number of faults.
Var(S)=1(0.64)+0.8(0.8)2
Compute
Compute
The severity contribution is 0.64 and the frequency contribution is 0.512. Their sum is 1.152.
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