When X,Y independent identically distributed random variable what about E[X|Z] where Z = X-Y

Question

$X,Y$ are iid random variable and $Z = X + Y$. To find $E[X|Z]$ I follow this procedure.

$E[Z|Z] = Z \implies E[X+Y|Z] = Z \implies E[X|Z] + E[Y|Z] = Z \implies 2E[X|Z] = Z (since X,Y iid E[X|Z] = E[Y|Z]) \implies E[X|Z] = Z/2$

for $Z = X-Y $ the procedure gives $E[Z|Z] = E[X-Y|Z] = E[X|Z] - E[Y|Z] \implies E[Z|Z] = 0 \implies Z=0$ ,which obviously cannot be true.

Please let me know where the confusion is or is there anything wrong in this approach

Answer 1

When $X, Y$ are iid, we have by symmetry $E[X|X+Y]=E[Y|X+Y]$. However, $E[X|X-Y]=E[Y|X-Y]$ is not true.

In addition if you assume that the distribution is symmetric about zero, then $X, -Y$ are iid, and so $$E[X\,|\,X-Y]\,\,=\,\,E[X\,|\,X+(-Y)]\,\,=\,\,E[(-Y)\,|\,X+(-Y)]\,\,=\,\,-E[Y\,|\,X-Y].$$

Answer 2

In you second case, we do not have necessarly $E(X|Z)=E(Y|Z)$, because $Z=X-Y$ and the negative sign prevents e a total symmetry of the problem. However, we would expect $E(X|Y-X)=E(Y|X-Y)$