Suppose $X_1, X_2, \dots, $ be i.i.d. with finite mean and assume $S_n=\sum_{i=1}^n X_i$. Compute $$\mathbb{E}(X_1\mid S_n=a).$$
I know that $$\mathbb{E}(X_1\mid S_n)=\frac{S_n}{n}.$$ Does it work?
Suppose $X_1, X_2, \dots, $ be i.i.d. with finite mean and assume $S_n=\sum_{i=1}^n X_i$. Compute $$\mathbb{E}(X_1\mid S_n=a).$$
I know that $$\mathbb{E}(X_1\mid S_n)=\frac{S_n}{n}.$$ Does it work?
Copyright © 2021 JogjaFile Inc.
Answer is correct. Let me add in more details.
$$\sum_{i=1}^n \mathbb{E}(X_i|S_n) = \mathbb{E}(\sum_{i=1}^nX_i|S_n)= \mathbb{E}(S_n|S_n)=S_n$$
by IID, we have $\mathbb{E}(X_i|S_n)=\mathbb{E}(X_1|S_n)$,
Hence $$n \mathbb{E}(X_1|S_n)=S_n$$
$$ \mathbb{E}(X_1|S_n)=\frac{S_n}n$$