Conditional Expectations Given Sum of I.I.D. Given that $X_1,...Xn$ are all identical independent random variables.
$\mathbb{E}(X_1|\sum_{k=1}^{n}X_k)$ = ?
I am unsure how to proceed on this one. I know the default relation: $\mathbb{E}(X|Y)$ = $\mathbb{E}(X*I_{[Y=y]})\over\mathbb{P}(Y=y))$, where I is an indicator function.
Intuitively, I believe the answer should be the sum of the random variables divided by how many random variables or the average of the sum.

For the late comers for this question. I think clear solution of the problem might be as follows: 
Let $Y = \sum_{i=1}^n X_i$, by using the property of conditional expectation $\mathbb{E}[g(Y)|Y] = g(Y)$ (i.e.$\mathbb{E}[Y|Y] = Y$), 
we can write 
$$ Y = \mathbb{E}[X_1 + X_2 + \cdots + X_n |Y] = \mathbb{E}[X_1|Y]+ \mathbb{E}[X_2|Y]+ \cdots + \mathbb{E}[X_n|Y] = n\mathbb{E}[X_1|Y],$$ 
since $X_i$'s are i.i.d., then 
$$\mathbb{E}[X_1|Y] = \frac{Y}{n}.$$
