Percent loss covered on average with a deductible

Question

Suppose you have auto insurance with a deductible of \$200, and with no restriction on maximal payment. The probability of a loss is 0.10, and suppose that the distribution of the loss is exponential with a mean of \$1000.

What percentage of the loss does the insurance cover on average? The answer in the back of the book is 57.4%, but I am not sure how to reproduce it. This is how I thought of doing the problem:

Say that \$1000 = 1 unit of money (for ease of calculation), so that the deductible $d = 0.20$

The payment function may be described as $$r(X) = \begin{cases} 0, \quad \quad x \leq 0.20 \\ x-0.2, \quad x > 0.20 \end{cases} $$

where $x$ is the realization of the stochastic variable $$X = \begin{cases} \xi, \quad p = 0.10 \\ 0, \quad p = 0.90 \end{cases} $$

Here, $\xi$ is the exponentially distributed loss. Since we are taking \$1000 to be a single unit of money, this just has a mean of 1.

Since we are interested in the percentage of the loss covered, we are assuming that a loss has occurred and that that loss is larger than the deductible. So, the expected value of the loss under this condition is $$E[\xi | \xi > 0.20] = 0.2 + E[\xi] = 0.2 + 1 = 1.2$$ which is just a result of the shifted exponential distribution. In this case, the insurance will cover 1 unit of money, or \$1000, and so we have 83% coverage on average when there is a loss. (1 divided by 1.2).

What is the mistake?

Answer 1

The error comes from the statement in bold:

Since we are interested in the percentage of the loss covered, we are assuming that a loss has occurred and that that loss is larger than the deductible.

This is a mistaken interpretation of the question. Instead, the question is implying that when a loss happens, some of these losses are below the deductible, and therefore not covered. In such a case, the percentage of loss covered is $0\%$.

Consequently, the calculation of the expected proportion of loss covered should be $$\operatorname{E}\left[\frac{(X - 0.2)_+}{X}\right],$$ where $(X - 0.2)_+ = \max(0, X - 0.2)$ is the claim payment and $$X \sim \operatorname{Exponential}(1)$$ is the ground-up loss given that a loss occurs. Note we do not need $\xi$ nor do we need the information that a loss occurs with probability $0.1$, because the first part of your statement "we are assuming a loss has occurred" is in fact correct.

Then the calculation proceeds as $$\begin{align*} \operatorname{E}\left[\frac{(X - 0.2)_+}{X}\right] &= \operatorname{E}[0 \mid X \le 0.2]\Pr[X \le 0.2] + \operatorname{E}\left[1 - \frac{1}{5X} \,\bigg| \; X > 0.2 \right]\Pr[X > 0.2] \\ &= e^{-1/5} - \int_{x=0.2}^\infty \frac{1}{5x} e^{-x} \, dx \\ &\approx 0.574201. \end{align*}$$

Answer 2

I don't see a basis in the problem statement for conditioning on $\xi\gt0.20$. Quite to the contrary, it seems to me that if you want to know the average percentage of the loss covered by your insurance, the fact that it covers $0\%$ if the loss is below the deductible seems like a very relevant piece of information to include in the calculation.

Without the conditioning, you get

$$E[r(X)]=\mathrm e^{-0.2}\cdot1=\mathrm e^{-0.2}\approx81.9\%\;,$$

slightly less than your value under conditioning. This, however, is not the average percentage of loss covered; it's the average loss covered, expressed as a percentage of the average loss incurred. The average percentage of loss covered is

$$ \left(1-\mathrm e^{-0.2}\right)\cdot0+\mathrm e^{-0.2}\int_0^\infty\frac x{x+0.2}\,\mathrm e^{-x}\mathrm dx=0.2\operatorname{Ei}(-0.2)+\mathrm e^{-0.2}\approx57.4\%\;, $$

where $\operatorname{Ei}$ denotes the exponential integral.