I have been stuck for hours now trying to figure out the following question: Say that $X$ and $Y$ are independent variables. $X$ is uniformly distributed over $(0,2)$, while $Y$ is exponentially distributed with mean $1$. Let $W = \min\{X,Y\}$. What is the $E[W]$?
My solution so far goes like this:
$$F_W(w) = P(W > t) = P(X>t,Y>t) = P(X>t)P(Y>t)$$ $$=\Biggl(\int_t^2 \frac{1}{2} \, dx\Biggl)\Biggl(\int_t^\infty e^{-y} \, dy\Biggl)$$ $$=e^{-t}-\frac{1}{2}te^{-t}$$
I then take the derivative of this to find the density function of W which ends up equaling:
$$-e^{-t}-\frac{1}{2}e^{-t}+\frac{1}{2}te^{-t}$$
In order to find the expected value, I then attempt to integrate like so:
$$\int t \Biggl(-e^{-t}-\frac{1}{2}e^{-t}+\frac{1}{2}te^{-t}\Biggl)dt$$
The issue is that this seems to come out with a very complex answer that I don't think is correct. Is there a formula I'm not picking up on or a simpler way to compute these problems then the method I am using?
First, $\Pr(W > w) = 1-F_W(w)$.
Second, $$\int_0^2(1-F_W(w))\,dw = \left[w(1-F_W(w))\right]_0^2+\int_0^2 wf_W(w) \, dw = 0 + E[W]=E[W].$$ So you do not need to take the derivative of $F_W(w)$ at all, nor you need to compute an integral that involves a quadratic expression.