Suppose $f(x,y) = (x+y)\mathbb{1}_{1 \geq x \geq 0}\mathbb{1}_{1 \geq y \geq 0}$. Find $\mathbb{E}(X|Y=y)$ and density of $V = \mathbb{E}(X|Y)$.
The first one is rather easy:
$$ f_y(y) = \int_0^1(x+y)\mathbb{1}_{1 \geq y \geq 0}dx = (\frac{1}{2}+y)\mathbb{1}_{1 \geq y \geq 0} $$
$$ f_{X|Y}(x, y) = \frac{(x+y)\mathbb{1}_{1 \geq x \geq 0}\mathbb{1}_{1 \geq y \geq 0}}{(\frac{1}{2}+y)\mathbb{1}_{1 \geq y \geq 0}} = \frac{(x+y)\mathbb{1}_{1 \geq x \geq 0}}{(\frac{1}{2}+y)} $$
$$ \mathbb{E}(X|Y=y) = \int_{\mathbb{R}}x\frac{(x+y)\mathbb{1}_{1 \geq x \geq 0}}{(\frac{1}{2}+y)}dx = \frac{1}{\frac{1}{2}+y} \int_0^1x(x+y) = \frac{\frac{1}{3}+\frac{y}{2}}{\frac{1}{2}+y} $$
so
$$ \mathbb{E}(X|Y)=\frac{\frac{1}{3}+\frac{Y}{2}}{\frac{1}{2}+Y} $$
How can I find the density of $V$?
I tried:
$$ \frac{\frac{1}{3}+\frac{Y}{2}}{\frac{1}{2}+Y} = \frac{1}{2} \frac{\frac{4}{6}+Y}{\frac{1}{2}+Y} = \frac{1}{2}\left(1 + \frac{\frac{1}{6}}{\frac{1}{2}+Y}\right) = \frac{1}{2}\left(1 + \frac{1}{3+6Y}\right) = \frac{1}{2}+\frac{1}{6+12Y} $$
and then
$$ F_Z(z) = \mathbb{P}(Z \leq z) = \mathbb{P}(\frac{1}{2}+\frac{1}{6+12Y} \leq z) = \\ = \mathbb{P}(\frac{1}{6+12Y} \leq z - \frac{1}{2}) = 1 - \mathbb{P}(6+12Y \leq \frac{2}{2z-1}) = \\ = 1 - \mathbb{P}(Y \leq \frac{2}{24z-12} - \frac{1}{2}) $$ this is already becoming too weird to be true. Is that a correct attempt and the answer is just $$ f_V(z) = -(\frac{2}{24z-12} - \frac{1}{2})'f_Y\left(\frac{2}{24z-12} - \frac{1}{2}\right) $$ or did I make a mistake somewhere?