Independent solution

How to solve this Law of Total Variance question

Setup

Setup

Let N be the annual count and let Lambda be its Poisson mean after the year type is known.

Pr⁡(Λ=20)=0.10,Pr⁡(Λ=15)=0.30,Pr⁡(Λ=10)=0.60\Pr(\Lambda=20)=0.10,\qquad \Pr(\Lambda=15)=0.30,\qquad \Pr(\Lambda=10)=0.60
E[N∣Λ]=Λ,Var⁡(N∣Λ)=ΛE[N\mid\Lambda]=\Lambda,\qquad \operatorname{Var}(N\mid\Lambda)=\Lambda

Model

Model

Separate ordinary Poisson variability within a year type from variability among the three conditional means.

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

Compute

Compute

Calculate the first two moments of the random Poisson mean and combine the two variance components.

E[Λ]=0.10(20)+0.30(15)+0.60(10)=12.50E[\Lambda]=0.10(20)+0.30(15)+0.60(10)=12.50
E[Λ2]=0.10(202)+0.30(152)+0.60(102)=167.50E[\Lambda^2]=0.10(20^2)+0.30(15^2)+0.60(10^2)=167.50
Var⁡(Λ)=167.50−12.502=11.25\operatorname{Var}(\Lambda)=167.50-12.50^2=11.25
Var⁡(N)=12.50+11.25=23.75\operatorname{Var}(N)=12.50+11.25=23.75

Answer

Answer

The unconditional annual count has variance 23.75.

23.75(E)\boxed{23.75\quad\text{(E)}}