Independent solution

How to solve this Compound Poisson Variance question

Setup

Setup

Write the aggregate loss as a random sum of independent severities. Both the severity variance and severity mean are needed because the number of losses is random.

E[N]=Var⁡(N)=4,E[L]=1000,Var⁡(L)=106E[N]=\operatorname{Var}(N)=4,\quad E[L]=1000,\quad \operatorname{Var}(L)=10^6

Model

Model

The law of total variance separates variation within individual losses from variation in the number of losses. For a Poisson count, this is equivalently the frequency mean multiplied by the severity second moment.

Var⁡(S)=E[N]Var⁡(L)+Var⁡(N)E[L]2\operatorname{Var}(S)=E[N]\operatorname{Var}(L)+\operatorname{Var}(N)E[L]^2

Compute

Compute

The severity second moment is two million. Multiplying by the Poisson mean of four gives an aggregate variance of eight million.

Var⁡(S)=4(106)+4(1000)2=8,000,000\operatorname{Var}(S)=4(10^6)+4(1000)^2=8{,}000{,}000

Answer

Answer

The total-loss variance is eight million.

8,000,000(C)\boxed{8{,}000{,}000\quad\text{(C)}}