Expected Value and Variance of transformed Random variable I am trying to find the expected value and variance of $Y_i=\ln(X_i)$ for $X$ is uniformly distributed between $1$ and $3$. I believe that $E(Y_i)=(\ln3)/2$ and $\operatorname{Var}(x)=(\ln3)^2/12$. Could someone please confirm this to be true? Isn't $Y$ uniformly distributed on $[0,\ln(3)]$ now?
 A: No.  By a change of variables (an application of the chain rule):
$$\begin{align}
 y=\ln x &\iff x=e^y
\\[2ex]
f_Y(y) & = f_X(e^y) \left\lvert\frac{\mathrm d e^y}{\mathrm d y}\right\rvert
\\[1ex] & = \frac{e^y}2 \;\mathbf 1_{y\in [\ln 1;\ln 3]}
\end{align}$$
This is not a uniform distribution.  Which you should expect by looking at a sketch of $y=\ln x$.  Equally placed points along the $x$-axis do not map to equally spaced points along the $y$-axis.
Buy you can now determine the mean and variance.  Use:
$$\begin{align}
\mathsf E(Y) & = \tfrac 1 2 \int_{0}^{\ln 3} ye^y\operatorname d y
\\[2ex]
\mathsf E(Y^2) & = \tfrac 1 2 \int_{0}^{\ln 3} y^2e^y\operatorname d y
\\[2ex]
\mathsf {Var}(Y) & = \mathsf E(Y^2)-\mathsf E(Y) ^2
\end{align}$$
A: If $X$ is uniformly distributed between $1$ and $3$, then $Y=\ln X$ is always between $\ln 1$ and $\ln 3$, but $Y=\ln X$ is not uniformly distributed. 
If $0\le y\le \ln 3$ then
$$
F_Y(y) = \Pr(Y\le y) = \Pr(\ln X\le y) = \Pr(X\le e^y) = \frac{e^y-1}{3-1}.
$$
The derivative of that is the density function, and it is not constant between $0$ and $\ln 3$.  In order for such a transformation to transform a uniform distribution to another uniform distribution, the graph of the transforming function would have to be a straight line, and the graph of $\ln$ is not a straight line.
