Show $D_{n} : = X_n - X_{n-1}, n \in \mathbb{N}$ are pairwise uncorrelated, where $X_n$ is a square integrable martingale

Question

Assume that $(X_n)_{n \in \mathbb{N}}$ is a square integrable martingale w.r.t. the filtration $\mathcal({F}_n)$.

Let $$D_{n} : = X_n - X_{n-1}, n \in \mathbb{N}$$

Show that $D_n$ are pairwise uncorrelated.

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I'm struggling pretty much with this problem above. What I have already found out is that for $X_n$ is a square integrable martingale and an example for it ( i.e. $E(X_{n}^{2}) < \infty, n \in \mathbb{N_{0}}$

$D_n$ are pairwise uncorrelated (i.e. $\mathbb{E}(D_nD_m) = 0$ for all $n,m \in \mathbb{N} $ with $ m \neq n$ Should I use the conditional expectations of integrable random variables forms? It doesn't seem to be very reasonable.

I would really appreciate any kind of help a lot.

Answer 1

Hint: For $s< n$ you have $$ E[X_n X_s \mid \mathcal F_s ] = X_s E[X_n \mid \mathcal F_s]= X_s^2$$ since $X_s $ is $\mathcal F_s$ measurable and $X_sX_n$ is integrable. By taking expectation on both sides, you get that $E[X_nX_s]=E[X_s^2]$. This should be enough to calculate $E[D_nD_m]$.

Edit: For $s <n$ \begin{align*} E[D_nD_s]&= E[(X_n-X_{n-1})(X_s-X_{s-1})] \\ &= E[X_nX_s]- E[X_nX_{s-1}] -E[X_{n-1}X_s] + E[X_{n-1}X_{s-1}] \\ &=E[X_s^2]-E[X_{s-1}^2]-E[X_s^2]+E[X_{s-1}^2]=0 \end{align*}