Independent solution

How to solve this Uniform Distribution question

Setup

Setup

Let X be the uniform loss and let Y be the payment limited to the unknown maximum.

X∼Uniform⁡(0,2000)X\sim\operatorname{Uniform}(0,2000)
Y=min⁡(X,m)Y=\min(X,m)

Model

Model

Split the expectation between losses below the maximum and losses that reach the maximum.

E⁡[Y]=∫0mx2000 dx+mPr⁡(X>m)\operatorname{E}[Y]=\int_0^m\frac{x}{2000}\,dx+m\Pr(X>m)
E⁡[Y]=m24000+m(1−m2000)=m−m24000\operatorname{E}[Y]=\frac{m^2}{4000}+m\left(1-\frac{m}{2000}\right)=m-\frac{m^2}{4000}

Compute

Compute

Set the limited expectation equal to the supplied value and solve the quadratic.

m−m24000=910m-\frac{m^2}{4000}=910
m2−4000m+3640000=0m^2-4000m+3640000=0
m=1400 or 2600m=1400\ \text{or}\ 2600

Answer

Answer

A policy maximum above 2000 would be outside the stated loss range, leaving 1400 as the admissible solution.

m=1400(C)\boxed{m=1400\quad\text{(C)}}