Expected value of $E(X-Y|X+Y)$ if $X \sim Y$

Question

If we have two discret random values with $X \sim Y$, prove that $$E(X-Y|X+Y)=0$$ My first try was to do this $$E(X-Y) =E(X)-E(Y)=0$$ In other hand $$E(X-Y)=\sum_{z} E(X-Y|X+Y=z)P(X+Y=z)$$ I don't know what to do.

Answer 1

Being $X\sim Y$ it is also $X|(X+Y)\sim Y|(X+Y)$

thus

$$\mathbb{E}[X-Y|X+Y]=\mathbb{E}[X|X+Y]-\mathbb{E}[Y|X+Y]=0$$

Answer 2

I don't believe the statement to be necessarily true unless $\ X\ $ and $\ Y\ $ are independent.

Suppose the joint distribution of $\ X\ $ and $\ Y\ $ is given by the following array, the entry in row $\ i\ $ and column $\ j\ $ being $\ P(X=i,Y=j)\ $. $$ \begin{array}{c|ccc} &1&2&3\\ \hline 1&\frac{1}{20}&\frac{2}{20}&\frac{3}{20}\\ 2&\frac{5}{20}&\frac{2}{20}&\frac{1}{20}\\3&0&\frac{4}{20}&\frac{2}{20} \end{array} $$ The distributions of $\ X\ $ and $\ Y\ $ are the same, with probability mass vector $\ \big(\frac{3}{10},\frac{2}{5},\frac{3}{10}\big)\ $. However, \begin{align} E(X-Y|X+Y=3)&=\frac{3}{7}\\ E(X-Y|X+Y=4)&={-}\frac{6}{5}\\ E(X-Y|X+Y=5)&=\frac{3}{5}\ . \end{align}