Let $U$ be a uniform distribution on $[0,1]$
1) Find the distribution function of $V = -log(U)$ (where log is the natural log)
2) Find $E(V)$
What I got:
1) $F_V(x) = P(V<x) = P(-log(U) < x) = P(log(U) > -x) = 1 - P(U < e^{-x}) = 1-e^{-x}$
2) confuses me, I know it's to do with the distribution of an exponential but doesn't the domain $[0,1]$ effect this in any way?
Hint for part 2:
$$E(g(X)) = \int_{-\infty}^{\infty} g(x) f_{X}(x) dx$$
In particular,
$$E(-\log(U)) = \int_{0}^{1} -log(x) dx $$