Double conditional expectation of martingale Assume $X$ a martingale. A common exercise is showing that every martingale has uncorrelated increments. That is, with $a < b < c< d$, 
$$ Cov( X_a - X_b, X_c - X_d) = 0 $$
While there are a few straightforward ways of showing this result, I was curious whether we could reach the same result through noting that, say, $X_a = \mathbb{E}(X_n | \mathcal{F}_a)$, for some $n$ such, that $\mathcal{F}_a \subset \mathcal{F}_n$.
This way, 
$$ Cov( X_a - X_b, X_c - X_d) = Cov( \mathbb{E}(X_n | \mathcal{F}_a) - \mathbb{E}(X_n | \mathcal{F}_b), \mathbb{E}(X_n | \mathcal{F}_c) - \mathbb{E}(X_n | \mathcal{F}_d))  $$
And since $\mathbb{E}(\mathbb{E}(X_n | \mathcal{F}_c) - \mathbb{E}(X_n | \mathcal{F}_d))=0$, this should reduce to four sums, where I'm guessing every member should equate to zero. 
Namely,
$$ \mathbb{E}(\mathbb{E}(X_n | \mathcal{F}_i)\mathbb{E}(X_n | \mathcal{F}_j)) = 0$$
would seem to hold for every $i \neq j$ such, that $\mathcal{F}_{i,j} \subset \mathcal{F}_n$


*

*Is this true, or am I messing things ups?

*How should we be dealing with double expectation of a multiplication of moments? It's not clear that the two expectations are independent (since $\mathcal{F}_i \subset \mathcal{F}_j$ might hold, or the other way round). 
 A: No, the expectation of the product does, in general, not equal zero.

Lemma: Let $(X_n)_{n \in \mathbb{N}}$ be a square-integrable martingale. Then $$\mathbb{E}(X_i X_j) = \mathbb{E}(X_{\min\{i,j\}}^2) \tag{1}$$ for any $i,j \in \mathbb{N}$.

Since $\mathbb{E}(X_n \mid \mathcal{F}_i) = X_i$ for all $i \leq n$, we can $(1)$ rewrite as
$$\mathbb{E} \big[ \mathbb{E}(X_n \mid \mathcal{F}_i) \mathbb{E}(X_n \mid \mathcal{F}_j) \big] = \mathbb{E} \big[ \mathbb{E}(X_n \mid \mathcal{F}_{\min\{i,j\}})^2 \big]$$
for $i,j \leq n$. Hence, the expression equals zero if, and only if, $ \mathbb{E}(X_n \mid \mathcal{F}_{\min\{i,j\}})=0$ almost surely (i.e. $X_{\min\{i,j\}}=0$ almost surely). However, it is possible to use the above lemma to prove that $X$ has uncorrelated increments, see below.
Proof of the lemma: The assertion is obvious for $i=j$. Now if $i,j \in \mathbb{N}$ are such that $i<j$, then $$\mathbb{E}(X_j \mid \mathcal{F}_i) = X_i,$$ and therefore it follows from the tower property that
$$\mathbb{E}(X_i X_j) = \mathbb{E} \big[ \mathbb{E}(X_i X_j \mid \mathcal{F}_i \big] = \mathbb{E} \big[ X_i \mathbb{E}(X_j \mid \mathcal{F}_i \big] = \mathbb{E}(X_i^2)$$ which proves the assertion.

Corollary Any square integrable martingale $(X_n)_{n \in \mathbb{N}}$ has uncorrelated increments.

Proof: Since a martingale has constant expectation, we have
$$\text{cov}(X_a -X_b, X_c-X_d) = \mathbb{E}((X_a-X_b)(X_c-X_d))$$
for any $a<b<c<d$. Hence,
$$\text{cov}(X_a -X_b, X_c-X_d) = \mathbb{E}(X_a X_c - X_a X_d - X_b X_c+ X_b X_d).$$
Applying the above lemma (four times), we get
$$\text{cov}(X_a -X_b, X_c-X_d) = \mathbb{E}(X_a^2) - \mathbb{E}(X_a^2) - \mathbb{E}(X_b^2)+ \mathbb{E}(X_b^2)=0.$$ 
