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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

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  • $\begingroup$ Analogous to the advice of "plot your data" is "generate all possibilities 'by hand'" for a small example (or two or three). With $m=4$, there are 16 sequences. $X_m$ can take on values -4, -2, 0, 2, and 4 so there are 5 conditional expectations to find. $\endgroup$
    – JimB
    May 28 at 15:38
  • $\begingroup$ And I assume when you wrote $x=3$ you meant $n=3$. $\endgroup$
    – JimB
    May 28 at 15:55
  • $\begingroup$ One may also assume $X_0=0$, right? $\endgroup$ May 28 at 15:57

2 Answers 2

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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 $E[U_k|X_3]$: $$ \eqalign{ 3E[U_1|X_3] &=E[U_1|X_3]+E[U_2|X_3]+E[U_3|X_3]\cr &=E[U_1+U_2+U_3|X_3]\cr &=E[X_3|X_3]=X_3.\cr } $$

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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 $E(U_n|X_m) = X_m/m$.

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