Independent solution

How to solve this Insurance Payment Variables question

Answer in brief

The density corresponds to a Pareto tail S(x)=(1000/x)^3 above 1000. The expected positive excess over 1100 is the integral of this survival function, equal to 1000^3 divided by 2(1100)^2, or 413.2231, so choice C is correct.

Setup

Setup

Integrate the density once to express the loss survival function above its lower endpoint.

S(x)=Pr⁡(X>x)=(1000x)3,x≥1000S(x)=\Pr(X>x)=\left(\frac{1000}{x}\right)^3,\qquad x\ge1000

Model

Model

For an ordinary deductible, the expected positive excess equals the tail integral beginning at the deductible.

E[(X−1100)+]=∫1100∞S(x) dxE[(X-1100)_+]=\int_{1100}^{\infty}S(x)\,dx

Compute

Compute

Evaluate the power-tail integral.

E[(X−1100)+]=10003∫1100∞x−3 dxE[(X-1100)_+]=1000^3\int_{1100}^{\infty}x^{-3}\,dx
E[(X−1100)+]=100032(1100)2=413.2231405…E[(X-1100)_+]=\frac{1000^3}{2(1100)^2}=413.2231405\ldots

Answer

Answer

The expected payout rounds to the listed value 413.

413(C)\boxed{413\quad\text{(C)}}