Independent solution

How to solve this Exponential Distribution question

Answer in brief

For an exponential loss, the expected excess above a deductible equals the mean times the survival probability at that deductible. Applying that identity gives a mean of approximately 1790.190, which matches choice B.

Setup

Setup

Represent the insurer's payment as the positive part of the loss after the deductible.

d=0.4β,Y=(Xd)+d=0.4\beta,\qquad Y=(X-d)_+

Model

Model

Use the exponential survival function and memoryless property to factor the expected payment into the chance of crossing the deductible and the mean residual loss.

E[Y]=Pr(X>d)E[XdX>d]\mathbb{E}[Y]=\Pr(X>d)\,\mathbb{E}[X-d\mid X>d]
Pr(X>d)=ed/β=e0.4,E[XdX>d]=β\Pr(X>d)=e^{-d/\beta}=e^{-0.4},\qquad \mathbb{E}[X-d\mid X>d]=\beta

Compute

Compute

Equate the stop-loss expectation to the supplied expected benefit and solve for the exponential mean.

1200=βe0.41200=\beta e^{-0.4}
β=1200e0.4=1790.189637\beta=1200e^{0.4}=1790.189637\ldots

Answer

Answer

The nearest listed amount is 1790.

β1790.190(B)\boxed{\beta\approx 1790.190\quad\text{(B)}}