Independent solution

How to solve this Poisson Distribution question

Setup

Setup

Express the annual payment through the event count above the two-event retention.

Y=1000(N−2)+,N∼Poisson⁡(1)Y=1000(N-2)_+,\qquad N\sim\operatorname{Poisson}(1)

Model

Model

Use the identity that a count equals its capped portion through two plus its excess above two.

(N−2)+=N−min⁡(N,2)(N-2)_+=N-\min(N,2)
E⁡[min⁡(N,2)]=Pr⁡(N≥1)+Pr⁡(N≥2)\operatorname{E}[\min(N,2)]=\Pr(N\ge1)+\Pr(N\ge2)

Compute

Compute

Evaluate the two Poisson tails and subtract the retained portion from the mean count.

Pr⁡(N≥1)=1−e−1,Pr⁡(N≥2)=1−2e−1\Pr(N\ge1)=1-e^{-1},\qquad \Pr(N\ge2)=1-2e^{-1}
E⁡[(N−2)+]=1−(2−3e−1)=−1+3e−1=0.1036383\operatorname{E}[(N-2)_+]=1-\left(2-3e^{-1}\right)=-1+3e^{-1}=0.1036383
E⁡[Y]=1000(0.1036383)=103.6383\operatorname{E}[Y]=1000(0.1036383)=103.6383

Answer

Answer

The expected annual payment rounds to 104.

104(B)\boxed{104\quad\text{(B)}}