Sum of r.v’s are martingales then $E[X_iX_j]=0$

conditional-expectationmartingalesprobability theory

The exercise I am about to ask is from a basic probability course, that is, we did not learn all these $\sigma$-algebra, filtration to explain martingales. We used the following definition:

A sequence of random variables $(Z_n)_{n\geq 0}$ is a martingale w.r.t $(X_n)_{n\geq 0}$ if the following holds for all $n\geq0$. (a) $Z_n$ is a function of $X_0, X_1, \cdots X_n$, (b) $\mathbb{E}[|Z_n|]<\infty$, and (c)
$$\mathbb{E}[Z_{n+1}|X_0, X_1, \cdots X_n]=Z_n$$

With this given, I want to solve the following exercise:

Let $X_0, X_1, \cdots $ be a sequence of random variables, and let $S_i=\sum_{j=1}^i X_j$. Show that if $S_0, S_1, \cdots $ is a martingale w.r.t $X_0, X_1, \cdots $, then for all $i\neq j$, $\mathbb{E}[X_iX_j]=0$.

(This is Exercise 13.2 in the book Probability and Computing by Michael Mitzenmacher and Eli Upfal
.) By the way, I don't know whether the expression of $S_i$ is a typo or not; the sequence starts from $0$, but the sum starts from $1$. I guess it doesn't matter that much(correct me if I'm wrong).

What I have done:

If $(S_n)_n$ is a martingale w.r.t $(X_n)_n$, then
$$\mathbb{E}[S_{n+1}|X_0, X_1, \cdots X_n]=\mathbb{E}[\sum_{j=1}^{n+1}X_j|X_0, \cdots X_n]=\sum_{j=1}^n X_j+\mathbb{E}[X_{n+1}|X_0, \cdots X_n]=S_n$$
so we must have $\mathbb{E}[X_{n+1}|X_0, \cdots X_n]=0$ for all $n\geq 0$. Using the law of total expectation, we can furthermore see that $\mathbb{E}[X_{n+1}]=0$ for all $n\geq 0$.

How should I proceed now? I think I should use the tower properties, but it seems difficult to deduce a product of two random variables from the given form of conditional expectation. Any help would be appreciated. Thank you in advance.

Best Answer

Hint: $\mathbb{E}(X_{n+1} X_i) = \mathbb{E}(\mathbb{E}(X_{n+1}X_i|\sigma(X_1,...,X_n))$ Where $1\leq i\leq n$.

Then, $$\mathbb{E}(\mathbb{E}(X_{n+1}X_i|\sigma(X_1,...,X_n))=\mathbb{E}(X_{i}\mathbb{E}(X_{n+1}|\sigma(X_1,...,X_n)))=\mathbb{E}(X_i \cdot 0)= 0$$ (since you proved in the question $\mathbb{E}(X_{n+1}|\sigma(X_1,...,X_n))=0$).

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