Independent solution

How to solve this Mixed Poisson Distribution question

Setup

Setup

Conditional on rate Lambda, N is Poisson; the rate itself is uniform from 0 to 3.

N∣Λ∼Poisson⁡(Λ)N\mid\Lambda\sim\operatorname{Poisson}(\Lambda)
Λ∼Unif⁡(0,3)\Lambda\sim\operatorname{Unif}(0,3)

Model

Model

The law of total variance separates ordinary Poisson variation within a fixed rate from variation of the random rate across observations.

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])

Compute

Compute

The uniform rate has mean 1.5 and variance 0.75. Adding these two mixed-Poisson components gives variance 2.25.

E[Λ]=1.5E[\Lambda]=1.5
Var⁡(Λ)=3212=0.75\operatorname{Var}(\Lambda)=\frac{3^2}{12}=0.75
Var⁡(N)=1.5+0.75=2.25\operatorname{Var}(N)=1.5+0.75=2.25

Answer

Answer

Therefore Var(N)=2.25, corresponding to choice E.

2.25(E)\boxed{2.25\quad\text{(E)}}