Find $\mathbb E(X | X + 2Y)$, where $X$ and $Y$ ~ $U[0,1]$ Let $X$ and $Y$ two independent random variables with unifrom distribution $U[0,1]$. Need to find $\mathbb E(X | X + 2Y)$. Is my calculation below correct?
$\mathbb E(X | X + 2Y) = \mathbb E(X|X) + \mathbb E(X | 2Y) = X + 2\mathbb E(X|Y) = X + 2*\frac{1}{2} = X +1$
 A: $$\mathbb{E}[X|X+2Y]+\mathbb{E}[Y|X+2Y]+\mathbb{E}[Y|X+2Y]=\mathbb{E}[X+2Y|X+2Y]=X+2Y$$
Being
$$X \perp\!\!\!\!\!\! \perp Y$$
they are exchangeable too, thus
$$\mathbb{E}[X|X+2Y]=\frac{X+2Y}{3}$$

Alternative solution:
Looking at the following drawing
and setting $U=X$ and $Z=X+2Y$

*

*If $0<z<1$ $X$ is uniform in $(0;z)$ with expectation $\frac{z}{2}$


*If $1<z<2$ $X$ is uniform in $(0;1)$ with expectation $\frac{1}{2}$


*If $2<z<3$ $X$ is uniform in $(z-2;1)$ with expectation $\frac{z-1}{2}$
Thus the expected value is
$$\frac{z}{2}\cdot\frac{1}{4}+\frac{1}{2}\cdot\frac{1}{2}+\frac{z-1}{2}\cdot\frac{1}{4}=\frac{2z+1}{8}$$
A: Maybe that example can hint you :
\begin{align*}
\mathbb E[X|X+Y]+\mathbb E[Y|X+Y]=\mathbb E[X+2Y|X+Y]=X+Y
\end{align*}
But by symmetry $\mathbb E[X|X+Y]=\mathbb E[Y|X+Y]$ which means that $$\mathbb E[X|X+Y]=\frac{X+Y}{2}$$
A: Let $z=x+2y$.  Draw a graph with $x$ on the horizontal and $z$ on the vertical.  Find the region in the plane that can be reached by $(x,z)$.  It should be a parallelogram.
Draw horizontal lines through the region, finding the intervals of allowed $x$ values for each $z$ value.
The midpoint of an interval is the expected $x$-value for that $z$-value.
