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?
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 $(2)(0.2)=0.4$.