The random variables $X$ and $Y$ have the joint probability density function $$ f_{X,Y}(x,y) = \left\{ \begin{array}{ccc} \frac{1}{y}, & 0 < x < y, & 0< y< 1 \\ 0, & \text{otherwise} & \end{array} \right. $$ Find the variance of $X$.
I need to find the variance of $x$. So far, I have found the mean and variance of $X$ conditional on $Y=y$ as:
$$ mean=\frac{y}{2}$$ and $$Variance=\frac{y^2}{12}$$
I know i must use $$V(X)=E(V(X|Y))+V(E(X|Y))$$, but i am getting the wrong answer.
You can do this by writing down the marginal $f_X$ instead of conditional distributions. $f_X(x)=\int_x^{1} \frac 1 y dy=-\log x$ for $0<x<1$. The first and the second moments can be computed by making the substitution $u=\log x$. You can try this and see if you get the right answer.