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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!

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1 Answer 1

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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.

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  • $\begingroup$ thanks for the answer! isn't that the same thing what I have done? (i.e. pulled out the same thing) $\endgroup$ Nov 22, 2014 at 17:00
  • $\begingroup$ @Epsilondelta No, because in the first equation there is a conditional expectation on the right-hand side (whereas, in your answer, you have $\mathbb{E}(Z_i-Z_{i-1})$). In this case the result is the same, but in general it might make a difference. $\endgroup$
    – saz
    Nov 22, 2014 at 21:17

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