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The following is the problem that I am working on.

A loss random variable X has the following cdf: $$F(x)= 0 {\space} \text{if x<0}, .2+.3x {\space} \text{if $0 \le x<2$}, {\space}1 \text{if $x \ge2$}$$ An insurer will provide proportional insurance on this loss, covering fraction $\alpha$ of the loss ($0<\alpha<1$). $E[X]=0.5$. Find $\alpha$.

This was my approach.

The pdf is the derivative of the cdf, so I found $$f(x)=.3 {\space} \text{if $x\in(0,2)$}$$

Using this I did $$E[X]=\int_0^2 xf(x)dx = \int_0^2 .3xdx=.6$$

so $$\alpha=.3$$ but the answer says that it should be 0.5.

Can someone help me out?

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  • $\begingroup$ It's mixed so f(0) = 0.2, f(2) = 0.2. So $E[x] = 0 * 0.2 + 0.6 + 2 * 0.2 = 1$, where 0.6 is the part you computed. That should give you $\alpha = 0.5$. $\endgroup$ – Grid Apr 4 '14 at 19:07
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Note that at $x=2$, we have $0.2+0.3x=0.8$, which is not equal to $1$. What this means is that there is a "weight" of $0.2$ at $2$: Our random variable is neither continuous nor discrete.

To put it in loss terms, there is a probability of $0.2$ that the loss is $2$. That sort of thing happens if the payout is capped at $2$, perhaps because the value of the object is $2$. This is a common phenomenon in the insurance setting. In an accident, there may be a certain damage less than the value of the car. Or perhaps the car is "totalled."

The expected loss is the number you calculated, plus the contribution made from having a loss of $2$ with probability $0.2$. That contribution is $(2)(0.2)=0.4$.

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  • $\begingroup$ Ah, I see. The book used the formula $\int_0^\infty 1-F(x) dx$ so I wanted to know why what I am trying to do was not working. $\endgroup$ – hyg17 Apr 5 '14 at 0:47

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