Independent solution

How to solve this Random-Sum Variance question

Setup

Setup

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)=0.0576E[N]=0.75,\quad \operatorname{Var}(N)=0.5625,\quad E[U]=0.24,\quad \operatorname{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\operatorname{Var}(S)=E[N]\operatorname{Var}(U)+\operatorname{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\operatorname{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)\boxed{0.0756\quad\text{(B)}}