Independent solution

How to solve this Mixture Distributions question

Setup

Setup

Let T identify the policy class and let its class-specific Poisson mean be Lambda.

(Pr⁡(T=A),Pr⁡(T=B),Pr⁡(T=C))=(0.10,0.50,0.40)\left(\Pr(T=A),\Pr(T=B),\Pr(T=C)\right)=(0.10,0.50,0.40)
Λ=(1,2,10),X∣T∼Poisson⁡(Λ)\Lambda=(1,2,10),\qquad X\mid T\sim\operatorname{Poisson}(\Lambda)

Model

Model

Separate variation within a class from variation among the three class means.

Var⁡(X)=E⁡[Var⁡(X∣T)]+Var⁡(E⁡[X∣T])\operatorname{Var}(X)=\operatorname{E}[\operatorname{Var}(X\mid T)]+\operatorname{Var}(\operatorname{E}[X\mid T])
E⁡[Var⁡(X∣T)]=E⁡[Λ]\operatorname{E}[\operatorname{Var}(X\mid T)]=\operatorname{E}[\Lambda]

Compute

Compute

Evaluate the first two moments of the random class mean.

E⁡[Λ]=0.10(1)+0.50(2)+0.40(10)=5.10\operatorname{E}[\Lambda]=0.10(1)+0.50(2)+0.40(10)=5.10
E⁡[Λ2]=0.10(12)+0.50(22)+0.40(102)=42.10\operatorname{E}[\Lambda^2]=0.10(1^2)+0.50(2^2)+0.40(10^2)=42.10
Var⁡(Λ)=42.10−5.102=16.09\operatorname{Var}(\Lambda)=42.10-5.10^2=16.09
Var⁡(X)=5.10+16.09=21.19\operatorname{Var}(X)=5.10+16.09=21.19

Answer

Answer

The annual claim-count variance is 21.19.

21.19(C)\boxed{21.19\quad\text{(C)}}