For a martingale $(Z_n)_{n\in \mathbb N}$ define $X_i=Z_i-Z_{i-1}$ with $Z_0=0$
Show:
$$Var(Z_n)=\sum_{i=1}^nVar(X_i)$$
My attempt:
We can write $Z_n=\sum_{i=1}^nX_i$, so we actually just have to show that $Cov(X_i,X_j)=0$ for all $i\neq j$
Let $i>j$ we have:
$Cov(X_i,X_j)=Cov(Z_{i}-Z_{i-1},Z_j-Z_{j-1})=\mathbb E((Z_i-Z_{i-1})(Z_j-Z_{j-1}))-\mathbb E(Z_i-Z_{i-1})\mathbb E(Z_j-Z_{j-1})$
The second addend is zero because $\mathbb E(Z_i-Z_{i-1})\mathbb E(Z_j-Z_{j-1})=(\mathbb E(Z_i)-\mathbb E(Z_{i-1}))(\mathbb E(Z_j-Z_{j-1})=0$
(Because $n\mapsto \mathbb E(Z_n)$ is constant for a martingale $Z_n$)
For the first addend:
We know that $Z_n$ is $\mathfrak{F}_n-adapted$, where $\mathfrak{F}_n$ is monotonically-nested.
We condition on $\mathfrak{F}_j$:
$\mathbb E((Z_i-Z_{i-1})(Z_j-Z_{j-1}))=\mathbb E(\mathbb E((Z_i-Z_{i-1})(Z_j-Z_{j-1}))$|$\mathfrak{F}_j)$
By the property of conditional expectation which is called "pull out whats known") we obtain:
$\mathbb E((Z_j-Z_{j-1})\mathbb E(Z_i-Z_{i-1})$|$\mathfrak{F}_j)=\mathbb E((Z_j-Z_{j-1})(\mathbb E(Z_i)-\mathbb E(Z_{i-1}))$|$\mathfrak{F}_j)$
(We "know" $Z_j$ and $Z_{j-1})$
Which is zero, again because $n\mapsto \mathbb E(Z_n)$ is constant for a martingale $Z_n$
The assertion follows.
I hope someone can go trough it and verify for me if everything is correct..especially when I condition on something.
General question: Is there a rule on which you should condition when you want to prove things like that?
Thanks in advance!
Looks to me as you "pulled out" a bit too much: Pull out gives
$$\mathbb{E}\big[ (Z_i-Z_{i-1}) \cdot (Z_j-Z_{j-1}) \mid \mathcal{F}_j \big] = (Z_j-Z_{j-1}) \cdot \mathbb{E}(Z_i-Z_{i-1} \mid \mathcal{F}_j).$$
Note that there is still the conditional expectation on the right-hand side. Now the martingale property gives
$$\mathbb{E}(Z_i-Z_{i-1} \mid \mathcal{F}_j) = \mathbb{E}(Z_i \mid \mathcal{F}_j) - \mathbb{E}(Z_{i-1} \mid \mathcal{F}_j) = Z_j-Z_j =0$$
since $i>j$. The remaining part of your proof is fine.