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Let $\{X_k\}_{k\in[0,T]}$ be a stochastic process adapted to the filtered sigma algebra $\sigma_k$.

I wonder, does the following equal zero? $\Bbb{E}[X_t-X_s|\sigma_s]$. I can show that it is true easily if $X$ is a Brownian motion process, but I wonder if it is true outside of such a circumstance. I do take still that intervals of $X$ are independent, but I lose the condition that $X_t-X_s\sim \mathcal{N}(0,t-s)$.

So we have $\Bbb{E}[X_t-X_s|\sigma_s]=\Bbb{E}[X_t-X_s]$

Jason
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  • The expression equals $0$ iff $(X_t){t \in [0,T]}$ is a martingale with respct to $(\mathcal{F}_t){t \in [0,T]}$. – saz Apr 08 '17 at 14:57

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