A martingale with uncorrelated components

Question

Let $\zeta_{i}$ be a sequence of random variables such that the partial sums \begin{align*} X_{n}=\zeta_{0}+\zeta_{1}+\cdots+\zeta_{n},\quad n\geq 1, \end{align*} determine a martingale. Show that the summands are mutually uncorrelated, i.e. that $\mathbb {E}(\zeta_{i}\zeta_{j})=\mathbb {E}(\zeta_{i})\mathbb {E}(\zeta_{j})$ for $i\neq j$. I'm seeking for some hints to start with since I don't quite see what's the connection between the given condition and the conclusion. Thanks.

Answer 1

I guess that $X_0$ should also be defined as well, otherwise there would be no reason for $\zeta_0$ and $\zeta_1$ to be uncorrelated. (Let for example n=1 and $\zeta_1=\zeta_0$ be integrable correlated random variables. The $1$-step process $X$ is then clearly a martingale as it must only satisfy integrability.)

Let $(\mathcal{F}_i)_{0\leqslant i \leqslant n}$ be the corresponding filtration. First notice that $\mathbb{E}(\zeta_i)=0$ for $i\geqslant 1$, as, using tower property of conditional expectation and the Martingale property of $X_i$, $$\mathbb{E}(\zeta_i)=\mathbb{E}(X_i-X_{i-1})=\mathbb{E}(\mathbb{E}(X_i-X_{i-1}|\mathcal{F}_{i-1}))=0.$$

W.l.o.g. let $i<j$. Now the result directly follows as $$ \mathbb{E}(\zeta_i \zeta_j)=\mathbb{E}(\zeta_i (X_j-X_{j-1}))=\mathbb{E}(\mathbb{E}(\zeta_i (X_j-X_{j-1})|\mathbb{F}_i))=\mathbb{E}(\zeta_i \mathbb{E}(X_j-X_{j-1}|\mathbb{F}_i))=\mathbb{E}(\zeta_i (\mathbb{E}(X_j|\mathcal{F}_i)-\mathbb{E}(X_{j-1}|\mathcal{F}_i)))=\mathbb{E}(\zeta_i(X_i-X_i))=0.$$