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

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.

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 $X$.

• Ah yes that is a perfectly clear explanation. Thanks! – TheHopefulActuary Dec 3 '12 at 19:14

There is a very excellent way to solve this problem, which is both fast, and has a low risk of error (which is ideal for the acutary exam). It makes use of 2 facts...

1. The exponential distribution has a closed form for all percentiles. It is $\frac{1}{\lambda}ln(\frac{1}{1-p_n})$. In this case, the 95th percentile is calculated as $300 ln(\frac{1}{.05})$.

2. In the case of a deductible, with the exponential distribution, one can calculate the percentile, and then just add the deductible to that answer. The sages of this forum can explain why that works... I believe it might have something to do with the "memoryless" property of the exponential distribution.

$$100 + 300 ln(\frac{1}{.05}) = 998.71 \approx 999$$
Finally, there is also a closed form for calculating the expected value of an exponential with a deductible. That is $$E[X] = \frac{1}{\lambda}e^{-\lambda D}$$ This formula applies to the "per loss average," that is, the mean of all the insurer's payments including those that are 0 because they are below the deductible.
The "per payment average" $$E[x-d|x>d] = E[x] = \frac {1}{\lambda}$$ Again, unchanged due to the memoryless property.