Mean and variance of $\ln(u)$ Suppose $U$ follows $U(0.1)$.
1) find the mean and vairance of $\ln(u)$.
Question:
  I wish to confirm the 1st part of the proof. Are these steps correct?
CDF of $Y = P(Y \leq y) = P(\ln{U} \leq y)  = P ( U \geq e^{y}) 
= 1 – P(U < e^{y})  = 1 – f_{u}(e^{y})  
= 1 - e^{y}$
My concern is the reversal of the inequality by using the exponential base. Is this correct?
Thanks.
 A: You do not have to find the cdf of $\ln U$ to find the mean and variance of $\ln U$.
But since your question appears to be specifically about the cdf of $Y$, where $Y=\ln U$, we calculate that. 
let $-\infty\lt y \le 0$. Then $\Pr(Y\le y)=\Pr(\ln U\le y)=\Pr(U\le e^y)=e^y$. So for $-\infty\lt y\le 0$, we have $F_Y(y)=e^y$. If $y\gt 0$, we have $F_Y(y)=1$. And, for $y\lt 0$, we have $F_Y(y)=0$. Note that there was no cange in the direction of the inequality, since $\log x $ is an increasing funcion.  
Thus, differentiating, we find that the density of $Y$ is $e^y$ for $0\lt y\lt 1$, and $0$ elsewhere. 
But let's get back to the mean and variance of $Y$. We have
$$E(Y)=\int_0^1 \ln u\,du.$$
Integrating by parts, we find that
$$E(Y)=\int_0^1 \ln y\,dy=\left. (y\ln y-y\right|_0^1=-1.$$
For the variance, first find $E(Y^2)=\int_0^1 \ln^2\, u\, du$. This requires two integrations by parts (but we have essentially already done one of then). Then use the fact that $\operatorname{Var}(Y)=E(Y^2)-(E(Y))^2$.
A: 
Here I believe this should help you out. If not sorry
