Independent solution

How to solve this Poisson Distribution question

Setup

Setup

Let N have Poisson mean 3 and denote the positive-count event by A.

A={N>0},q=Pr⁡(A)=1−e−3A=\{N>0\},\qquad q=\Pr(A)=1-e^{-3}

Model

Model

Because multiplying either N or N squared by the indicator of A changes nothing, divide their unconditional moments by q.

E⁡[N∣A]=E⁡[N]q=3q\operatorname{E}[N\mid A]=\frac{\operatorname{E}[N]}{q}=\frac{3}{q}
E⁡[N2∣A]=E⁡[N2]q=3+32q=12q\operatorname{E}[N^2\mid A]=\frac{\operatorname{E}[N^2]}{q}=\frac{3+3^2}{q}=\frac{12}{q}

Compute

Compute

Subtract the square of the conditional mean from the conditional second moment.

q=0.9502129316q=0.9502129316
E⁡[N∣A]=3.1571870895,E⁡[N2∣A]=12.6287483579\operatorname{E}[N\mid A]=3.1571870895,\qquad \operatorname{E}[N^2\mid A]=12.6287483579
Var⁡(N∣A)=12q−(3q)2=2.6609180400\operatorname{Var}(N\mid A)=\frac{12}{q}-\left(\frac{3}{q}\right)^2=2.6609180400

Answer

Answer

The requested conditional variance rounds to 2.66.

2.66(C)\boxed{2.66\quad\text{(C)}}