Conditional expectation of a sum Let $U_1, U_2,...,U_n$ be a sequence of independent random variables such that for every $i$:
$P(U_i=1)=P(U_i=-1)=\frac{1}{2}$
And we define $X_n=\sum_{i=1}^{n}U_i$.
So for $m\geq n$, what is $E(U_n|X_m)$?
I'm having troubles with this, since for some m, some x can not even be received.
For example $m=4$ and $x=3$.
So I'm not sure how to separate for different cases
 A: Take $m=3$, for example. By the evident symmetry,
$$
E[U_1|X_3]=E[U_2|X_3]=E[U_3|X_3].
$$
Add up the $E[U_k|X_3]$:
$$
\eqalign{
3E[U_1|X_3]
&=E[U_1|X_3]+E[U_2|X_3]+E[U_3|X_3]\cr
&=E[U_1+U_2+U_3|X_3]\cr
&=E[X_3|X_3]=X_3.\cr
}
$$
A: I will give a try for this question. By the definition of conditional expectation
\begin{align}
E(U_n|X_m = s) =& P(U_n = 1|X_m = s) - P(U_n = -1|X_m = s)\\
=& P(U_n = 1|X_m = s) - (1- P(U_n = 1|X_m = s))\\
=& 2P(U_n = 1|X_m = s) -1 
\end{align}
and by the Bayes rule,
\begin{align}
P(U_n = 1|X_m = s) =& \frac{P(U_n = 1,X_m = s)}{P(X_m =s)}\\
=& \frac{P(X_m = s|U_n = 1)P(U_n = 1)}{P(X_m =s)}\\
=& \frac{1}{2}\cdot\frac{P(X_{m-1} = s-1)}{P(X_m =s)}
\end{align}
So the problem reduces to figure out the probability of $P(X_m =s)$. To proceed with computation, we need to discuss whether $m$ is even or odd. Say $m$ is even, then $s$ has to be even, so let's assume $m = 2k$ and $s = 2l$. Here $ k\geq 1$ is an integer and $l$ could take values as integers between $-k$ and $k$. To make $X_{2s} = 2l$, there are $k+l$ $U_n$'s taking 1 and the others taking -1. Hence
\begin{equation}
P(X_m =s) = P(X_{2s} =2l) = \frac{C_{2k}^{k+l}}{2^{2k}}
\end{equation}
We could do similar analysis for the odd number $m-1$, this will give
\begin{equation}
P(X_{m-1} =s-1) = P(X_{2s-1} =2l-1) = \frac{C_{2k-1}^{k+l-1}}{2^{2k-1}}
\end{equation}
Plug into the formula for $P(U_n = 1|X_m = s)$ yields
\begin{equation}
P(U_n = 1|X_m = s) = \frac{1}{2}\cdot\frac{2^{2k}}{C_{2k}^{k+l}} \cdot \frac{C_{2k-1}^{k+l-1}}{2^{2k-1}} = \frac{k+l}{2k}
\end{equation}
This further implies
\begin{equation}
E(U_n|X_m = s) = 2\cdot \frac{k+l}{2k} -1 = \frac{2l}{2k} = \frac{s}{m}
\end{equation}
for $m$ even, and $s$ even between $-m$ and $m$. The case for $m$ odd could be obtained similarly. This means $E(U_n|X_m) = X_m/m$.
