Let X and Y be random variables such that:
f$_Y$(y)=2y for 0$\leq$y$\leq$1 , and for Y=y, X~U[0,y]
find var(Y | X=0.5). answer by book: 1:48
I managed to find (unless i did a mistake) that the PDF of X is f$_X$(x)=2-2x, and the conditional density of Y|X is 2 when X=0.5. And so I get that E(Y|X=0.5)=1 and E(Y$^2$|X=0.5)=2:3 which means that I got a result that is not only wrong but negative...
thanks in advance
HINT
Given that $Y \in [0, 1]$, intuitively it makes no sense that $E[Y \mid X=0.5]$ would take the extreme value of $1$.
I think your $f_X(x)$ is correct though, so you made a mistake after that.
The formula for density of $Y\mid X$ is not enough... what is the support (i.e. set of possible values) for $Y$ when $X=0.5$?
Can you finish from here or do you need another hint?