For a martingale $\{Z_n,n\geq1\}$, let $X_i=Z_i-Z_{i-1},i\geq1$, where $Z_0\equiv0$. Show that $Var(Z_n)=\sum^n_{i=1}Var(X_i)$.

Question

For a martingale $\{Z_n,n\geq1\}$, let $X_i=Z_i-Z_{i-1},i\geq1$, where $Z_0\equiv0$. Show that $$Var(Z_n)=\sum^n_{i=1}Var(X_i)$$

If $\{Z_n\}$ is a martingale, then $E[Z_{k+1}|\mathcal{F}_k]=Z_k$ for a filtration $\mathcal{F}_n$. I'm really not sure about how to start this problem. Could I get some hints about which properties of martingales to use? Thank you.

Answer 1

By definition, $Z_0=0$ and $Z_n = X_1 + \dots + X_n,~n=1,2,\dots$. By the Martingale property, $E[X_n |{\cal F}_{n-1}]=0$ for all $n=1,2,\dots$.

Now:

1. $\mbox{Var}(Z_n)=E[Z_n^2]$ for $n=0,1,2,\dots$.
2. $E[Z_0^2]=0$ and for $n=1,2,\dots$, we have the following formula, mentioned in an different, but equivalent form in the comment made by @user408858 \begin{align*} (*)\quad E[Z_n^2|{\cal F}_{n-1}] & =E[(Z_{n-1} +X_n)^2 |{\cal F}_{n-1}] \\ & = Z_{n-1}^2 +2E[X_n|{\cal F}_{n-1}] Z_{n-1} +X_n^2 \\ & = Z_{n-1}^2 + 0+ X_n^2. \end{align*} Taking expectations on both sides of $(*)$ gives $$\mbox{Var}(Z_n) = \mbox{Var}(Z_{n-1}) + E[X_n^2],$$ and so $$\mbox{Var}(Z_n) = \sum_{k=1}^n E[X_k^2].$$

Answer 2

$$ \begin{align} Var(X_i)&=Var(Z_i-Z_{i-1})\\ &=E[(Z_i-Z_{i-1}-E[Z_i-Z_{i-1}])^2]\\ &=E[(Z_i-Z_{i-1})^2]\\ &=E[Z_i^2-Z_{i-1}^2] \end{align} $$

Therefore

$$\sum_{i=1}^n Var(X_i) = \sum_{i=1}^n E[Z_i^2-Z_{i-1}^2] = E[Z_n^2]$$

Using $E[Z_n]=0$, it holds

$$\sum_{i=1}^n Var(X_i) = E[Z_n^2] - E[Z_n]^2 = Var(Z_n)$$