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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.

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up vote 1 down vote accepted

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 $X$.

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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.

So, your answer is:

$$ 100 + 300 ln(\frac{1}{.05}) = 998.71 \approx 999 $$

This "shortcut" is for a very specific kind of problem, but it isn't hard to memorize, and you will be glad you did if you are lucky enough to get this question on your test.

One should also note that a "benefit limit" does not change the percentile value at all with an exponential distribution. Use the closed form, smile, and go on to the next question!

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.

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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.

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