Find expectation of $X_1$ given $\sum_{k=1}^n X_k=j$

Question

$X_1,X_2,...$ are iid non-negative random variables, distributed as the counting random variable $X$ having PMF $$p_X(k)=\mathbb{P}(X=k), \quad k=0,1,2,...$$

Show that $$\mathbb{E}\left(X_1 \,\bigg| \sum_{k=1}^nX_k=j\right)={j \over n}, \quad j,n,\in\mathbb{N}.$$

Intuitively, because all $X_k$ are iid, given the sum is $j$, a single $X_1$ is expected to be the average $j/n$. Without knowing the specific distribution of $X_k$, how should I write down this intuition formally?

Answer 1

Let $i \in \{1,...,n\}$ and note that $E [ X_i | \sum_k X_k = j ] = E [ X_1 | \sum_k X_k = j ]$.

Since $E [ \sum_i X_i | \sum_k X_k = j ] = j$ and $E [ \sum_i X_i | \sum_k X_k = j ] = n E [ X_1 | \sum_k X_k = j ]$, we have the desired result.