Independent solution

How to solve this Uniform Distribution question

Setup

Setup

Scale the claim payment above the deductible and record the variance of the original uniform loss.

C=(X−b/2)+C=(X-b/2)_+
Var⁡(X)=b212\operatorname{Var}(X)=\frac{b^2}{12}

Model

Model

Integrate the first two payment moments only over losses that exceed the deductible.

E⁡[C]=1b∫b/2b(x−b/2) dx\operatorname{E}[C]=\frac{1}{b}\int_{b/2}^{b}(x-b/2)\,dx
E⁡[C2]=1b∫b/2b(x−b/2)2 dx\operatorname{E}[C^2]=\frac{1}{b}\int_{b/2}^{b}(x-b/2)^2\,dx

Compute

Compute

Form the payment variance and cancel the common scale when taking the ratio.

E⁡[C]=b8,E⁡[C2]=b224\operatorname{E}[C]=\frac{b}{8},\qquad \operatorname{E}[C^2]=\frac{b^2}{24}
Var⁡(C)=b224−(b8)2=5b2192\operatorname{Var}(C)=\frac{b^2}{24}-\left(\frac{b}{8}\right)^2=\frac{5b^2}{192}
Var⁡(C)Var⁡(X)=5b2/192b2/12=516\frac{\operatorname{Var}(C)}{\operatorname{Var}(X)}=\frac{5b^2/192}{b^2/12}=\frac{5}{16}

Answer

Answer

The variance ratio is 5 to 16.

5:16(D)\boxed{5:16\quad\text{(D)}}