If X and Y are two iid random variables, why $E[X|X+2Y] = E[Y|X + 2Y]$?

Question

If X and Y are two iid random variables, why $E[X|X+2Y] = E[Y|X + 2Y]$?

I can see that $E[X|X+Y] = E[Y|X + Y]$, but if there is a coefficient, why it's true?

Answer 1

Who told you that? It's wrong.

Take for instance $X,Y\in\{0,1\}$. The map $(x,y)\in\{0,1\}\times\{0,1\}\mapsto x+2y\in\{0,1,2,3\}$ being a bijection, we have $\sigma(X+2Y)=\sigma(X,Y)$, hence $\mathbb E[X\vert X+2Y]=X$ and $\mathbb E[Y\vert X+2Y]=Y$.

Answer 2

Here is another counter-example.

Let $X, Y \sim \mathcal{N}(0, 1)$ be i.i.d. standard normal variables. Since $X+2Y$ and $2X-Y$ are jointly normal and uncorrelated, they are independent. This means that

\begin{align*} \mathbf{E}[X+2Y \mid X+2Y] &= X+2Y, \\ \mathbf{E}[2X-Y \mid X+2Y] &= \mathbf{E}[2X-Y] = 0. \end{align*}

Solving this for $\mathbf{E}[X \mid X+2Y]$ and $\mathbf{E}[Y \mid X+2Y]$, we get

$$ \mathbf{E}[X \mid X+2Y] = \frac{X+2Y}{5}, \qquad \mathbf{E}[Y \mid X+2Y] = \frac{2(X+2Y)}{5}. $$