Independent solution

How to solve this Poisson Distribution question

Answer in brief

The Poisson mean-variance identity turns the two supplied relationships into lambda_X minus lambda_Y equal to 3 and lambda_X equal to twice lambda_Y, so lambda_Y is 3. Therefore E[Y squared] equals lambda_Y squared plus lambda_Y, or 12, matching choice D.

Setup

Setup

Assign rates to the two Poisson variables and use the fact that each rate is both its mean and its variance.

X∼Poisson⁡(λX),Y∼Poisson⁡(λY)X\sim\operatorname{Poisson}(\lambda_X),\quad Y\sim\operatorname{Poisson}(\lambda_Y)
E[X]=Var⁡(X)=λX,E[Y]=Var⁡(Y)=λY\mathbb{E}[X]=\operatorname{Var}(X)=\lambda_X,\quad \mathbb{E}[Y]=\operatorname{Var}(Y)=\lambda_Y

Model

Model

Translate the mean-difference and variance-ratio information into two equations for the rates.

λX−λY=3\lambda_X-\lambda_Y=3
λX=2λY\lambda_X=2\lambda_Y

Compute

Compute

Solve for the target rate, then apply the Poisson second-moment identity.

2λY−λY=3⟹λY=32\lambda_Y-\lambda_Y=3\quad\Longrightarrow\quad\lambda_Y=3
E[Y2]=Var⁡(Y)+E[Y]2\mathbb{E}[Y^2]=\operatorname{Var}(Y)+\mathbb{E}[Y]^2
E[Y2]=3+32=12\mathbb{E}[Y^2]=3+3^2=12

Answer

Answer

The second raw moment is 12, the value listed under choice D.

E[Y2]=12(D)\boxed{\mathbb{E}[Y^2]=12\quad\text{(D)}}