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I'm trying to apply the technique explained here for finding the expected value of the function of a random variable. Here's the exercise I tried it on:

"An auto-insurance policy has a deductible of 1 and a maximum claim payment of 5. Auto loss amounts follow an exponential distribution with mean 2. Calculate the expected value claim made for an auto loss."

One way to obtain the solution is by letting losses equal $X \sim \text{exp}(.5)$ and claim payments equal $Y = 0 \text{ for } 0 \leq X \leq 1, X-1 \text{ for } 1<X\leq 6, \text{ and } 5 \text{ for } X>6$. Then the pdf of X is $f(x) = .5e^{-.5x}$ and $$ \begin{align}E(Y) &= 0\int_{0}^{1}f(x)dx \text{ + } \int_{1}^{6} (x-1)f(x)dx \text{ + } 5\int_{6}^{\infty} f(x) dx =2(e^{-.5} - e^{-3}) \end{align}$$

which is the desired result. Using the approach outlined in the link, I tried setting up $E(Y)$ like this: $$E(Y) = 0\cdot P(0\leq X\leq 1) + E(X-1)\cdot P(1 \leq X \leq 6) + E(5) \cdot P(X>6)$$

The one problem I'm aware of is how I evaluated the expression $E(X-1)\cdot P(1 \leq X \leq 6)$. I tried to follow the example by assuming the memorylessness property of $X$ gave me $E(X-1) = E(X) = 2$, and that since $P(1 \leq X \leq 6) = e^{-.5} - e^{-3}$ the entire expression evaluates to $2(e^{-.5} - e^{-3})$. That's clearly wrong. I'd be grateful if someone could point out my mistake(s) and show me how to properly apply this technique.

Apologies in advance for the poor formatting -- double backslashes weren't working for me as line breaks in math mode!

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  • $\begingroup$ $E(X-1)=EX-1$. You can never have $E(X-1)=EX$. You seem to have a wrong idea of what memoryless property means. $\endgroup$ Commented Sep 27, 2024 at 22:47

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The middle term should be $E(X-1; 1\le X\le 6)$, which you correctly evaluated by computing $\int_1^6 (x-1) f(x)\,dx$.

The exponential distribution does have the property that for a fixed $b>0$, the conditional distribution of $X-b$, given that $X\ge b$, is exponential again, with the same rate. So the middle term can also be computed as $$ \eqalign{ E[(X-1)1_{\{1\le X\le 6\}}\mid| X\ge 1]\cdot P(X\ge 1) &=E[(X-1)1_{X-1\le 5}\mid| X\ge 1]\cdot e^{-.5}\cr &=E[X 1_{X \le 5} ]\cdot e^{-.5}\cr &=\int_0^5 x\cdot(.5)e^{-.5x}\,dx\cdot e^{-.5}\cr &=(2-7e^{-5/2})\cdot e^{-.5}\cr &=(2e^{-.5}-7e^{-3}).\cr } $$ Adding in the third term $5e^{-3}$ you get the expected result.

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  • $\begingroup$ Hi John -- Thanks for the answer! If I'm not mistaken, this "technique" is just an application of the total expectation theorem, stated in my text as E[X] = \sum_{i=1}^{n} P(A_{i})E[X|A_{i}], with A_{i}'s disjoint. But in the second and third terms above, it seems they overlap. I understand the motivation to use P(X>1) in combination with the function X-1, but doesn't that overlap with P(X>6)? I appreciate your help and patience! $\endgroup$ Commented Oct 2, 2024 at 14:50
  • $\begingroup$ The three terms correspond to the three non-overlapping events $\{X\le 1\}$, $\{1<X\le 6\}$, and $\{X>6\}$. $\endgroup$ Commented Oct 2, 2024 at 14:56
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    $\begingroup$ Oh, I think I see. The work you've shown rewrites the expectation for the event {1 < X <=6} in a way that allows you to exploit the memorylessness property of the distribution. Is that correct? $\endgroup$ Commented Oct 2, 2024 at 15:19
  • $\begingroup$ I'm sorry, now I'm totally confused. I initially found $E[Y]$ by summing the expectations of the three events that define $Y$ and obtaining $2(e^{-.5} - e^{-3})$ as the solution. However, the theorem of total expectation says that each event should be weighted by its respective probability, which I didn't do. So instead of $$E[Y] = E[0 | 0\leq X \leq 1] + E[X-1 | 1 < X \leq 6] + E[5 | X>6]$$ it should be $$E[Y] = E[0 | 0\leq X \leq 1]\cdot P(0 \leq X \leq 1) + E[X-1 | 1< X \leq 6] \cdot P(1 < X \leq 6)+ E[5 | X>6] \cdot P(X>6)$$ A completely different answer! Where am I going wrong? $\endgroup$ Commented Oct 2, 2024 at 18:05
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    $\begingroup$ I'm wondering if the answer is because each conditional pdf is supposed to be weighted by the probability of the conditioning event. So the reason my initial solution worked out is because $$E[X-1|1\leq X \leq 6]\cdot P(1\leq X \leq 6) ={\int_{1}^{6} \frac{f_{X}(x)}{P(1\leq X \leq 6)} dx}\cdot P(1 \leq X \leq 6) = \int_{1}^{6} f_{X}(x) dx$$ $\endgroup$ Commented Oct 2, 2024 at 18:32

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