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The following example is taken from the book "Introduction to Probability Models" of Sheldon M. Ross (Chapter 5, example 5.4).

The dollar amount of damage involved in an automobile accident is an exponential random variable with mean 1000. Of this, the insurance company only pays that amount exceeding (the deductible amount of) 400. Find the expected value and the standard deviation of the amount the insurance company pays per accident."

In the solution, the author states that:

By the lack of memory property of the exponential, it follows that if a damage amount exceeds 400, then the amount by which it exceeds it is exponential with mean 1000.

After reading several implications of this property, I easily map such statement to something like: if you have been waiting for 400s without seeing the bus, then the expected time until the next bus is always 1000s. (Please correct me if I'm wrong)

In case I've understood well, what makes me confuse is this next equation:

$$ E[Y|I=1] = 1000 $$

where:

$X$: the dollar amount of damage resulting from an accident

$Y=(X-400)^+$: the amount paid by the insurance company (where $a^+$ is $a$ if $a>0$ and 0 if $a<=0$).

$I = 1*(X > 400) + 0*(X<=400)$

I don't get why that equality holds given the memoryless property. Straightforwardly, I think with respect to 400 subtraction, it should be something like: $E[Y|I] = 1000 - 400 = 600$ (or some other value). Can anyone give me an explanation about this?

In case you are not clear about my description, please refer to this link with example 5.4.

hoangtuansu
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  • $(X-400)^+$ is the excess on top of the $400$ that is already accounted for. Upon conditioning that $X>400$, the average of this excess is still $1000$; this is just another way of saying what you said back up at the top. This is a slightly strange property of the exponential distribution which, strictly speaking, is a flaw in its use in many of the contexts in which it is used. – Ian Aug 30 '16 at 19:38
  • To see why this is a flaw, think about this example of the waiting time for a bus. Suppose it is being modeled as Exp(1) hours. It is probably not going to take more than 24 hours for the bus to come. Yet the exponential distribution model says that once you've waited 24 hours, you should still expect to wait 1 more hour. – Ian Aug 30 '16 at 19:39

2 Answers2

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Comment: The link didn't work on my computer. Here is a simulation of the situation as I understand it from your description.

A million accidents are simulated using R statistical software. Those costs below \$ 400 are set to $0.$ The sample mean and SD (including the $0$ costs) are shown. The histogram shows the simulated distribution very accurately, except that the bin $(0,100)$ contains only $0$'s. This a very long-tailed distribution; the maximum payout in a million simulated accidents is almost \$ 14,500.

x = rexp(10^6, 1/1000)  # R uses rate (reciprocal of mean)
x[x<=400] = 0           # adjust for deductible
mean(x)
## 936.641              # aprx avg payment/accident
sd(x)
## 1047.132             # aprx SD payment/accident
mean(x==0)              
## 0.329995             # aprx fraction with 0 payment
pexp(400, 1/1000)
## 0.32968              # exact theoretical fraction with 0 payment
max(x)
## 14474.76             # max payment in a million sim accidents

You did not give Ross's answer or yours. If my interpretation is right, results should agree to at least two significant digits.

enter image description here

Note: The bus example is technically correct. If it does not seem reasonable, that is because waiting times for buses are not actually exponentially distributed in real life. The exponential distribution is so easy to use (because of the no-memory property and because of it's simple mathematical form) that there is a temptation to use it in circumstances where it is not appropriate. (You may find a lot ot textbook examples of that type; perhaps fewer in Ross's book than in most.) I would have to see extensive actuarial evidence to believe that payouts on car accidents are truly exponential. Totaling a new Mercedes would cost someone a lot more than \$14,500.

BruceET
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  • Your answer is useful to verify the results. I reproduce your code with Matlab and add one more line: x[x>400] = x - 400 (due to the deducible amount) and I get the same result in Ross's solution. Thanks. – hoangtuansu Sep 01 '16 at 13:43
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Indeed. We are looking at: $$\begin{align}\mathsf E(Y \mid I=1) ~=~& \mathsf E(Y\mid X>400) &:~& I=\big[ X>400\big] \text{ an indicator function} \\ ~=~& \mathsf E(X-400\mid X>400) &:~& Y=(X-400)^+ = \max(X-400, 0) \\ =~& \mathsf E(X\mid X>400) - 400 &:~& \text{Linearity of Expectation} \\ =~& (400+\mathsf E(X))-400 &:~& \mathsf E(X\mid X>400)=400+\mathsf E(X) \tag{1} \\ =~& \mathsf E(X) &:~& \text{algebraic cancelation} \end{align}$$

$(1)$ is a consequence of the memoryless property since $X$ has an exponential distribution.   The distribution of damage above 400 is the same as the distribution of the damage above $0$.   So the expected total damage, when given that it is more than 400, is 400 plus the expected damage above 400.

This should lead on to : $$\mathsf E(Y) ~=~ \mathsf E(X)~\mathsf P(X>400)$$

Via the Law of Total Expectation and that $\mathsf E(Y\mid I=0)~=~0$.

Graham Kemp
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