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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 $Var[X_t]=Var[\sum_{i\leq t}\Delta_j]$=$\sum_{i,j}Cov(\Delta_i,\Delta_{j})=\sum_{i,j}Var(\Delta_j)+2\sum_{i<j\leq t}Cov(\Delta_i,\Delta_j)$.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.

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  • $\begingroup$ 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 $X_t$ $\endgroup$ – Babado May 23 at 1:53

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