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Let $\{X_n:n \geq 1\}$ be a sequence of random variables. Let $S_n = \sum_{i=1}^{n} (X_i-E(X_i|X_1,...,X_{i-1})).$ Show that $\{S_n\}$ is a martingale.

This is what I have so far:

To show it's martingale then : $E(S_{n+1}|F_n)$

=$E(S_{n+1}|S_1,S_2,..,S_n) \rightarrow E(S_{n+1}-S_n|X_1,X_2,..,X_n) =$

$\rightarrow E(X_{n+1}-E(X_n+1|X_1,...,X_n])$ am I going in the right path?

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I will write $\mathcal{F}_n = \sigma(X_1,...,X_n)$. Then you're given that $S_n = \sum_{i=1}^{n} X_i-E(X_i \mid \mathcal{F}_{i-1})$. We now have

\begin{align} E(S_{n+1} \mid \mathcal{F}_n) &= E\Big(\sum_{i=1}^{n+1} (X_i-E(X_i \mid \mathcal{F}_{i-1})) \mid \mathcal{F}_n \Big) \\ &= \sum_{i=1}^{n+1} E(X_i \mid \mathcal{F}_n) - E( E(X_i \mid \mathcal{F}_{i-1}) \mid \mathcal{F}_n) \\ &= \sum_{i=1}^n \Big[ E(X_i \mid \mathcal{F}_n) - E( E(X_i \mid \mathcal{F}_{i-1}) \mid \mathcal{F}_n) \Big] + E(X_{n+1} \mid \mathcal{F}_n) - E( E(X_{n+1} \mid \mathcal{F}_n) \mid \mathcal{F}_n) \\ &= \sum_{i=1}^n X_i - E(X_i \mid \mathcal{F}_{i-1}) = S_n. \end{align}

Do you see how to justify each step?

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  • $\begingroup$ Yeah, I knew that the $E(X_i)=X_i$ but I didn't know how to approach the rest thank you. $\endgroup$ – Killercamin May 7 '17 at 21:54
  • $\begingroup$ It's not true that $E(X_i)=X_i$. That would mean that each $X_i$ is constant. $\endgroup$ – grndl May 7 '17 at 21:56
  • $\begingroup$ Sorry was getting confused with another example, but I do understand the steps now, once I saw my notes and your explanation I was able to follow along. $\endgroup$ – Killercamin May 7 '17 at 21:58

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