Independent solution

How to solve this Uniform Distribution question

Setup

Setup

Let X be uniform on (0,1000) and let d be the deductible. The required expected excess payment is one quarter of the original mean 500, namely 125.

XUnif(0,1000)X\sim\operatorname{Unif}(0,1000)
E[X]=500,E[(Xd)+]=125E[X]=500,\qquad E[(X-d)_+]=125

Model

Model

For d between 0 and 1000, integrate the excess x-d over the portion of the uniform support above d.

E[(Xd)+]=11000d1000(xd)dx=(1000d)22000E[(X-d)_+]=\frac1{1000}\int_d^{1000}(x-d)\,dx=\frac{(1000-d)^2}{2000}

Compute

Compute

Equating the stop-loss mean to 125 gives a remaining support width of 500. Since 1000-d is nonnegative, the admissible solution is d=500.

(1000d)22000=125\frac{(1000-d)^2}{2000}=125
1000d=5001000-d=500
d=500d=500

Answer

Answer

The deductible is therefore 500, corresponding to choice C.

500(C)\boxed{500\quad\text{(C)}}