Independent solution

How to solve this Transformations of Random Variables question

Setup

Setup

Let X be the modeled loss and Y the insurer payment after applying the ordinary deductible.

X∼Unif⁡(0,15000)X\sim\operatorname{Unif}(0,15000)
Y=(X−1000)+=max⁡(X−1000,0)Y=(X-1000)_+=\max(X-1000,0)

Model

Model

The payment has a point mass at zero, so calculate both raw moments from the original loss density rather than treating Y as an ordinary uniform variable.

E[Yk]=115000∫100015000(x−1000)k dx,k=1,2\mathbb{E}[Y^k]=\frac{1}{15000}\int_{1000}^{15000}(x-1000)^k\,dx,\qquad k=1,2

Compute

Compute

Evaluate the two integrals and center the second moment.

E[Y]=1400022(15000)=6533.3333…\mathbb{E}[Y]=\frac{14000^2}{2(15000)}=6533.3333\ldots
E[Y2]=1400033(15000)=60977777.7778…\mathbb{E}[Y^2]=\frac{14000^3}{3(15000)}=60977777.7778\ldots
Var⁡(Y)=E[Y2]−E[Y]2=18293333.3333…\operatorname{Var}(Y)=\mathbb{E}[Y^2]-\mathbb{E}[Y]^2=18293333.3333\ldots
SD⁡(Y)=18293333.3333…=4277.0706…\operatorname{SD}(Y)=\sqrt{18293333.3333\ldots}=4277.0706\ldots

Answer

Answer

Rounding to the precision of the choices gives a payment standard deviation of 4277.

4277(B)\boxed{4277\quad\text{(B)}}