continuous distribution with change of variable I'm trying to do this question:
If $X$ is a continuous random variable with a mean of 2 and a variance of 4, find the mean and variance of $Y$, where $Y=\log{X}$.
I know how to find the expectation and variance from a pdf, but I don't understand how to do this from a completely different variable.
 A: First of all, X must also be positive for the log-transform to work. I will assume that is true. In general, there is no simple formula for converting the moments of a random variable to the moments of its transform. If you have the pdf (i.e, $f$) of X, then you can calculate the moments of Y by recognizing:
$f_Y(Y=y) = f_X(X=e^y)$ Therefore, you substitute $e^y$ for $x$ in $f_x$ to get $f_y$. However, this substution will effectively compress $f_X$ in the horizontal direction (here the degree of compression increases exponentially with y, since it will "accelerate" through the domain of $f_X$ at an increasing rate), which means that it wont integrate to 1 after substitution. You need to divide by the jacobian, which indicates how much the transform "stretches" the distribution, to restore its integral to 1 (i.e., $f_Y \space y = f_X\space dx$ or $f_Y = f_X \frac{dx}{dy}$). The jacobian for this transformation is $\frac{df_y}{dx}|_{e^y}=\frac{d}{dx} log(x)|_{e^y}=\frac{1}{x}|_{e^y}=e^{-y}$. Therefore, $f_Y(y) = e^{y}f_X(e^y)=f_X \frac{dx}{dy}$.
Since you said you know how to compute moments from pdfs, I'll leave the rest to you.
