If $X$ and $Y$ are i.i.d. random variables with $Binomial(n, p)$ distribution, calculate $E(X|X+Y)$.
Attempt:
To calculate the conditional expectation, obviously you first need to determine the conditional probability funtion. Let us denote $Z = X + Y$. Then $Z \sim Binomial(2n, p)$. Now we need to find $P_{X|Z}(x|z)$. By the definition of conditional probability, that is:
$$ \frac{P_{X, Z}(x, z)}{P_Z(z)} $$
The denominator is known. I'm having problem with determining the joint distribution, i.e., the numerator. Notice that $X$ and $Z$ are not independent (because $Z = X + Y$ so $Z$ depends on $X$), therefore the joint distribution is not the product of the marginals.
Anyway, I still tried to come up with the joint pmf, and I came up with this:
$$ P(X = x, Z = z) = P(\{X = x\} \cap \{Z = z\}) = P(X = x)P(Z = z)\mathbf{1} $$
Where $\mathbf{1}$ is the indicator random variable that equals $1$ if $x$ and $z$ are in $\{0, 1, 2, 3, ...\}$
Official Answer
$$ \frac{X + Y}{2} $$
A possible approach though not the most elegant. We first compute the distribution of $X\mid X+Y=m$ for $m=0,\dotsc, 2n$. To this end note that by the definition of conditional probability $$ P(X=k\mid X+Y=m)=\frac{P(X=k, X+Y=m)}{P(X+Y=m)}=\frac{P(X=k)P(Y=m-k)}{P(X+Y=m)}\tag{0} $$ where we used the equality of events $(X=k, X+Y=m)=(X=k, Y=m-k)$ and the independence of $X$ and $Y$ in the final equality. After a bit of simplification one finds that $$ P(X=k\mid X+Y=m)=\frac{\dbinom{n}{k}\dbinom{n}{m-k}}{\dbinom{2n}{m}}; \quad (k=0,\dotsc, m) $$ It follows that $$ E(X\mid X+Y=m)=\binom{2n}{m}^{-1}\sum_{k=0}^m k\binom{n}{k}\binom{n}{m-k}\tag{1}. $$ We claim that $$ \sum_{k=0}^m k\binom{n}{k}\binom{n}{m-k}=\frac{m}{2}\binom{2n}{m}\tag{2} $$ To do this we recognize that the LHS of $(2)$ is the coefficient of $x^m$ in the product of $F(x)=nx(1+x)^{n-1}=\sum_{k=0}^nk\binom{n}{k}x^k$ and $G(x)=(1+x)^n=\sum_{k=0}^n\binom{n}{k}x^k$ (see cauchy product). But $$ F(x)G(x)=nx(1+x)^{2n-1} $$ whence by the binomial theorem the coefficient of $x^m$ in the product is $$n\binom{2n-1}{m-1}=\frac{n(2n-1)(2n-2)\dotsb(2n-1-m)}{(m-1)!}=m\frac{2 n(2n-1)(2n-2)\dotsb(2n-1-m)}{2m(m-1)!}$$ which equals $\frac{m}{2}\binom{2n}{m}$ as desired.
(1) and $(2)$ together imply that $E(X\mid X+Y=m)=\frac{m}{2}$ as desired.
Addendum. As pointed out in the comments. The above work is unnecessary. Note that $$ E(X\mid X+Y)+E(Y\mid X+Y)=E(X+Y\mid X+Y)=X+Y $$ and because $X$ equals in distribution to $Y$ and $X$ is independent of $Y$, $$ E(X\mid X+Y)=E(Y\mid X+Y) $$ from which the result follows.