SOA/CAS Exam P Question (from previous exam (Nov. '09)): Finding percentiles

Question

So this question comes from SOA/CAS Exam P of November 2009, I'm not sure why the solution is the way it is. The question says this:

An insurance company sells an auto insurance policy that covers losses incurred by a policyholder, subject to a deductible of 100. Loses incurred follow an exponential distribution with mean 300. What is the $95^{th}$ percentile of actual losses that exceed the deductible?

I know that $\lambda=\frac1\mu$ where $\mu$ is the mean. So $\lambda=\frac{1}{300}$. But here is where it got crazy (to me at least). This is what the solution says:

The probability that a claim exceeds the $100$ deductible is:$$1-F(100)=1-\left(1-e^{-\frac{100}{300}}\right)=e^{-\frac13}=0.7165$$ The $95^{th}$ percentile of losses that exceed the deductible is the $96.42^{th}$ percentile of all losses, since:$$1-0.05\times0.7165=0.9642$$ This is found as: $$\begin{align}0.9642=F(x')=1-e^{\frac{-x'}{300}}\\ \Rightarrow x'=-300ln(.0358)=999\end{align}$$

So what I'm confused by is 1) Why the cumulative distribution function was used. And 2) How the $2$nd line was come up with in the solution. What's with this 96.42th percentile? Otherwise I know why they solved for $x'$. If you read this whole thing, I give you much thanks, I know it's long.

Answer 1

Let $X$ represent the loss incurred by a policy holder. Note that $X$ "ignores" the deductible. The $95$'th percentile of $X$ is the value $\alpha$ such that $95\%$ of all losses is at most $\alpha$. Note that many of these losses might be less than the deductible; thus, this would not give the $95$'th percentile of losses that exceed the deductible.

The percentage of losses that exceed the deductible is $71.65$. This was computed in 1), where the cumulative distribution was used simply because it was appropriate: $$P[X>100]=1-P[X\le 100]=1-F(100).$$ The reason this calculation is made is because now we can find the required percentile:

We wish to find the $95$'th percentile, call it $\beta$, of the losses that exceed $100$. Let $f$ be the density of $X$. Note that the area under the graph of $f$ and over the interval $[100,\beta]$ is $95$ percent of $.7165$. You should be able to see now that the area underneath the graph of $f$ to the right of $\beta$ is $(.05\cdot 71.65)$ percent of the entire area underneath the graph of $f$. This leads to what was done "on the second line" and to finding the $96.42$'nd percentile of $X$.

Answer 2

1) Percentiles divide the cumulative frequency into hundredths. However, here frequencies are continuous, hence the use of CDF (which represents the area). See the definition of percentile.

2) The reason why we would use 96.42 is because it is the closest to 95 on the table. We could use 95, but then we would need to integrate and the result would be almost identical. The result would differ only in the 4th decimal place or something like that.