Independent solution

How to solve this Exponential Distribution question

Setup

Setup

Let L be the loss amount and express the insurer's payment under an ordinary deductible of two.

L∼Exp⁡(1),P=(L−2)+L\sim\operatorname{Exp}(1),\qquad P=(L-2)_+

Model

Model

A payment occurs only when the loss crosses the deductible. Conditional on that event, exponential memorylessness makes the excess have the original mean of one.

Pr⁡(L>2)=e−2\Pr(L>2)=e^{-2}
E⁡[L−2∣L>2]=1\operatorname{E}[L-2\mid L>2]=1

Compute

Compute

Multiply the chance of a positive payment by its conditional mean amount.

E⁡[P]=Pr⁡(L>2)E⁡[L−2∣L>2]\operatorname{E}[P]=\Pr(L>2)\operatorname{E}[L-2\mid L>2]
E⁡[P]=e−2(1)=e−2=0.1353352832…\operatorname{E}[P]=e^{-2}(1)=e^{-2}=0.1353352832\ldots

Answer

Answer

The expected payment per loss is e to the power negative two.

e−2(A)\boxed{e^{-2}\quad\text{(A)}}