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Another probability theory question:

So, let $X_{1},\ldots,X_{n}$ bea martingale. Then, it appears that we have that for any $1 \leq a < b < c < d \leq n$, $\mathbb{E}((X_{d}-X_{c})(X_{b}-X_{a}))=0$.

I'm not sure how to argue this precisely. We have that $\mathbb{E}(X_{k+1}-X_{k}|X_{1},\ldots, X_{k})=0$ but how to conclude? Sorry if it is too elementary and I can't see it.

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    Condition on $X_1,...,X_b$ and see what you get when you use the tower rule.2012-03-07

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More generally, let $1 \leq a \leq b \leq n$ and $\mathcal{F}_k = \sigma (X_1, \cdots, X_k)$ be the filtration generated by $(X_k)$. If $Y$ is $\mathcal{F}_a$-measurable and $(X_b - X_a)Y$ is integrable, then we have $ \mathbb{E}[ (X_b - X_a) Y ] = 0. $ Indeed, by tower property of the conditional expectation, $ \mathbb{E}[ (X_b - X_a) Y ] = \mathbb{E}[ \mathbb{E}[ (X_b - X_a) Y | \mathcal{F}_a ] ]. $ Since $Y$ is $\mathcal{F}_a$-measurable, we can take it out and thus $ \begin{align*} \mathbb{E}[ (X_b - X_a) Y ] & = \mathbb{E}[ \mathbb{E}[ (X_b - X_a) Y | \mathcal{F}_a ] ] \\ & = \mathbb{E}[ \mathbb{E}[ X_b - X_a | \mathcal{F}_a ] Y ] \\ & = \mathbb{E}[ (\mathbb{E}[ X_b | \mathcal{F}_a ] - X_a) Y ]. \end{align*}$

But now we know that $\mathbb{E}[ X_b | \mathcal{F}_a ] = X_a$ by tower property again and induction as follows: $ \mathbb{E}[ X_b | \mathcal{F}_a ] = \mathbb{E}[ \mathbb{E}[ X_b | \mathcal{F}_{b-1} ] | \mathcal{F}_a ] = \mathbb{E}[ X_{b-1} | \mathcal{F}_a ] = \cdots = \mathbb{E}[ X_{a+1} | \mathcal{F}_a ] = \mathbb{E}[ X_a | \mathcal{F}_a ] = X_a. $

This completes the proof of the claim.