Let $(X_n)_n$ be a sequence of square-integrable RVs and let $\mathcal{F}$ be given by $\mathcal{F}_n = \sigma (X_1, \ldots, X_n)$. Suppose that $S_n := \sum_{i=1}^n X_i$ is an $\mathcal{F}$-martingal. Show that
$$Cov[X_m, X_n] = \mathbb{E}[X_m X_n]$$
I know that by definition we have
$$Cov[X_m, X_n] = \mathbb{E}[(X_m- \mathbb{E}(X_m)) (X_n- \mathbb{E}[X_n])],$$
so I guess we should argue that $\mathbb{E}[X_m] = 0$ and $\mathbb{E}[X_n] = 0$, but I am not sure whether this is actually true. Could you please give me a hint?
It is true. The fact that $(S_n)$ is a martingale implies that $ES_{n+1}=E[E(S_{n+1}| S_n)]=ES_n$, so $EX_n=E(S_{n+1}-S_n)=0$ for all $n$.