Independent solution

How to solve this Continuous Random Variables question

Setup

Setup

Integrating the power density and applying normalization gives the loss CDF on its support.

F(x)=(x100)5/4,0x100F(x)=\left(\frac{x}{100}\right)^{5/4},\qquad 0\le x\le100
F(20)=(20100)5/4=0.1337480610F(20)=\left(\frac{20}{100}\right)^{5/4}=0.1337480610\ldots

Model

Model

Let p be the 90th percentile after restricting the distribution to losses above 20.

0.90=Pr(XpX>20)0.90=\Pr(X\le p\mid X>20)
0.90=F(p)F(20)1F(20)0.90=\frac{F(p)-F(20)}{1-F(20)}

Compute

Compute

Solve first for the unconditional CDF level corresponding to the conditional percentile, then invert the power CDF.

F(p)=0.90+0.10F(20)=0.9133748061F(p)=0.90+0.10F(20)=0.9133748061\ldots
p=100(0.9133748061)4/5p=100\left(0.9133748061\ldots\right)^{4/5}
p=93.0077680477p=93.0077680477\ldots

Answer

Answer

The conditional 90th percentile rounds to 93.

93(E)\boxed{93\quad\text{(E)}}