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Let $S_n = \sum_{i=1}^n X_i$ be the symmetric random walk. Compute the conditional expectation $E(X_i \mid \mathcal{G}_{n+1})$ for $1 \leq i \leq n$ with $\mathcal{G}_{n} = \sigma(S_{n},S_{n+1},\dots)$.

I know that $S_n$ is a martingale and that I could perhaps use symmetry in terms of $E(X_i \mid \mathcal{G}_{n+1}) = E(X_n \mid \mathcal{G}_{n+1})$ but I still don't know how to compute it.

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    This is only (maybe) the first part of the answer: $$ E(X_i \mid \mathcal G_{n+1})=E(X_i \mid S_{n+1}, X_{n+2}, X_{n+3},\ldots) = E(X_i \mid S_{n+1}) $$ – 2017-01-10

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Hint: $$S_{n+1} = \mathbb{E}(S_{n+1} \mid \mathcal{G}_{n+1}) = \sum_{i=1}^{n+1} \mathbb{E}(X_i \mid \mathcal{G}_{n+1}).$$ Now use the the fact that the random variables are independent and identically distributed to conclude that $$\mathbb{E}(X_i \mid \mathcal{G}_{n+1}) = \frac{S_{n+1}}{n+1}.$$

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    I don't see why $S_n$ is $\mathcal{G}_{n+1}$ measurable. – 2017-01-10
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    It must be a typo, the answer is $\frac{S_{n+1}}{n+1}$ – 2017-01-10
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    @Cettt Ah, sorry, my mistake. – 2017-01-10
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    But the question should be asking for $$E(X_i\mid\mathcal G_n)\quad (i\leqslant n)$$ which makes more sense, and then this answer would be ok... :-) – 2017-01-10
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    @Did Yeah, that was actually what I had in mind. – 2017-01-10