Independent solution

How to solve this Uniform Distribution question

Setup

Setup

Let d be the fixed deductible and write the per-loss payment as the positive excess over d.

Y=(X−d)+Y=(X-d)_+
X∼Uniform⁡(0,1)X\sim\operatorname{Uniform}(0,1)

Model

Model

Integrate the payment over losses above the deductible and use the supplied expected payment to recover the remaining support length.

E⁡[Y]=∫d1(x−d) dx=(1−d)22=0.245\operatorname{E}[Y]=\int_d^1(x-d)\,dx=\frac{(1-d)^2}{2}=0.245
(1−d)2=0.49(1-d)^2=0.49

Compute

Compute

Take the nonnegative root, compute the second payment moment, and subtract the squared mean.

1−d=0.701-d=0.70
E⁡[Y2]=∫d1(x−d)2 dx=0.7033=0.1143333333\operatorname{E}[Y^2]=\int_d^1(x-d)^2\,dx=\frac{0.70^3}{3}=0.1143333333
Var⁡(Y)=0.1143333333−(0.245)2=0.0543083333\operatorname{Var}(Y)=0.1143333333-(0.245)^2=0.0543083333

Answer

Answer

The payment variance rounds to 0.054.

0.054(B)\boxed{0.054\quad\text{(B)}}