Variance & Joint Density Function $X$ and $Y$ have joint density given by $$f_{XY}(x,y)=\begin{cases}2,& 0≤x≤y≤1 \\0,& \text{elsewhere}\end{cases}$$ a) Find $\text{Var}(Y|X=x_0)$. 
b) What is the answer if $x_0$ is not in the interval $[0,1]$? 
So I know that if the $x_0$ is not in the interval, then the answer is $0$, right? I need help finding the variance.

It suffice to compute
$$
f_{Y|X}(y|x)=\frac{f_{X,Y}(x,y)}{f_{X}(x)}=\frac{2*\chi{0\le x\le y\le 1}}{\int^{1}_{x}2dy *0\le x\le y\le 1}=\frac{1}{1-x}*\chi{0\le x\le y\le 1}
$$
Therefore as a function of $y$, $f_{Y|X}(y|x)$ is well defined on $0\le x\le y\le 1$ except at $y=1,x=1$. Excluding this case we may compute that
$$
Var(Y|X=x_{0})=\frac{(1-x_0)^{2}}{12}, x_{0}\not=1
$$
and for $x_{0}\not \in [0,1]$ the conditional distribution itself is not properly defined. 
