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The unlimited severity distribution for claim amounts under an auto liability insurance policy is given by the cumulative distribution:

$$ F(x) = 1 - 0.8e^{-0.02x}-0.2e^{-0.001x} , x \geq 0$$

The insurance policy pays amounts up to a limit of $1000$ per claim.

Calculate the expected payment under this policy for one claim.

The answer only calculates $E(X)$ where $x$ is less than $1000$, however with a policy limit of $1000$, shouldn't $1000[1-F(1000)]$ also be added to the expected payment?

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You have a typo in your CDF: the function you stated is not nondecreasing for all $x \ge 0$. Most likely, you should have a $-0.2$ instead of $+0.2$.

I believe there is some misunderstanding of what exactly the solution calculated. I am guessing that the solution wrote $${\rm E}[X \wedge L] = \int_{x=0}^L S_X(x) \, dx,$$ where $L = 1000$ and $S_X(x) = 1-F_X(x)$. In this case, the calculation is correct.

To understand why, recall that for a continuous random variable $X$ with nonnegative support, $$\begin{align*} {\rm E}[X \wedge L] &= \int_{x=0}^\infty \min(x,L) f_X(x) \, dx \\ &= \int_{x=0}^L x f_X(x) \, dx + L \int_{x=L}^\infty f_X(x) \, dx \\ &= \int_{x=0}^L \int_{y=0}^x 1 \, dy \cdot f_X(x) \, dx + L S_X(L) \\ &= \int_{y=0}^L \int_{x=y}^L f_X(x) \, dx \, dy + L S_X(L) \\ &= \int_{y=0}^L S_X(y) - S_X(L) \, dy + L S_X(L) \\ &= \int_{y=0}^L S_X(y) \, dy - L S_X(L) + L S_X(L) \\ &= \int_{y=0}^L S_X(y) \, dy. \end{align*}$$ Thus, the limited expected value is simply the integral of the survival function on $[0,L]$. It already takes into account the remaining portion of the expectation above $L$. Your thinking is probably along the lines of the law of total probability $${\rm E}[X \wedge L] = {\rm E}[X \mid X \le L] \Pr[X \le L] + {\rm E}[X \mid X > L]\Pr[X > L],$$ which is also correct, but the first expectation on the RHS is not equal to the integral of the survival function. It is $${\rm E}[X \mid X \le L] = \frac{1}{F_X(L)} \int_{x = 0}^L x f_X(x) \, dx.$$ Indeed, if you wish, it is an instructive exercise to calculate the density $f_X(x)$ and compute the limited expected value explicitly using $X \wedge 1000 = \min(X,1000)$.

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  • $\begingroup$ I know that I'm a cat and you're a dog, but I think we could still be friends! Could you help explain to me how to calculate the expectation of r.v.'s of a mixed probability distribution in this problem? math.stackexchange.com/questions/2524217/… I'd be forever thankful! $\endgroup$ – ALannister Nov 17 '17 at 5:33

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