# $Var (X_n)= Var(X_0) +\sum_{k=0}^{n} Var (X_{k}-X_{k-1})$, Martingale Variance

i'm already diving through the martingales properties and try to prove the following problem:

Let $$X_{n}, Y_{n}$$ two martingales squared integrable respect to the filtred probability space $$(\Omega, \mathcal{F},(\mathcal{F}_n)_{0\leq n}, \mathbb{P}),$$ prove:

$$Var (X_n)= Var(X_0) +\sum_{k=0}^{n} Var (X_{k}-X_{k-1})$$

$$X_0$$ is independent from $$(X_k-X_{k})$$

I tried to begin asumming that $$X_0=0$$ and for $$0\leq t, \Delta_T=X_t-X_{t-1}$$ then we can write $$X_t$$ as $$\sum_{i=1}^{t}\Delta$$, where $$\mathbb{E}(\Delta|\Delta_1 \cdots \Delta_{j-1})=\mathbb{E}(X_j-X_{j-1}|\mathcal{F})=0$$. In other words, $$\Delta_i$$ is uncorrelated with $$\Delta_1 \cdots \Delta_{j-1}$$ This is ok to prove the independence or i have to go to the definition on independent increments?

My attempt: If we suppose we want $$Var(S_n)=\mathbb{E}[(X_t-\mathbb{E}(X_t))^2]=\mathbb{E}(X_t)^2$$ (Recall where assuming $$X_t=0$$, so $$\mathbb{E}(X_t)=0$$ by the martingale property). Then we have $$Var[X_t]=Var[\sum_{i\leq t}\Delta_j]$$=$$\sum_{i,j}Cov(\Delta_i,\Delta_{j})=\sum_{i,j}Var(\Delta_j)+2\sum_{i.Now then $$Var(\Delta_{i})=\mathbb{E}[\Delta_{i}^2]$$ but we have $$\mathbb{E}(\Delta_{i})=0$$, then $$Cov(\Delta_i,\Delta_j)=\mathbb{E}(\Delta_i,\Delta_j)$$ by the tower property of conditional expectation over $$\Delta_i$$, $$\mathbb{E}[\mathbb{E}[\Delta_j|\mathcal{F_{j-1}}]|\Delta_j]=0$$, it follows in this case $$Var[X_t]=\sum_{i\leq t}\mathbb{E}(\Delta_i^2)$$

As you see i didn't get $$\sum_{k=0}^{n} Var (X_{k}-X_{k-1})$$, i'm thinking that the correct aproach of this is using the fact that $$Var(X)= E(Var(X|Y))+Var(E(X|Y))$$

Anyone could help me out? And even more to specify what really means the variance of a martingale, or is just analogue to random variables, thanks.

• Your notation is not clear in some places... so I didn't quite understand your question. And when you mention the variatnce of a martingale, you are usually refering to the variance of $X_t$ – Babado May 23 at 1:53