Conditional Expectation Question from Grimmett & Stirzaker 1000 exercises in probability Let $X$ and $Y$ be independent discrete random variables, let $z$ be a constant real number. 
Why is $\Bbb E(X \mid  X + Y = z)$ not equal to $\Bbb E(X \mid X = z - Y)$? Is ${X+Y=z}$ not the same event as ${X=z-Y}$?


The whole question is, what is wrong with the following: Let X and Y be (independent) random variables with mean μ, (then) E(X | X + Y = z) = E(X | X = z − Y) = E(z − Y) = z − μ.

The trouble is not with $E(X \mid X + Y = z) = E(X \mid X = z − Y )$, always true since, as you noted, $[X+Y=z]=[X=z-Y]$, but with your assertion that $E(X \mid X = z − Y ) = E(z − Y )$. Why would this identity be true, even with $X$ and $Y$ independent? 
To get a grasp of the problem, consider the case when $X$ and $Y$ are i.i.d. Bernoulli uniform on $\{0,1\}$ and $z=2$. Then $[X=z-Y]=[X=Y=1]$ hence $E(X \mid X = z − Y )=1\ne\frac32=E(z-Y)$.
