Conditional expectation of increments of stochastic process [closed]

Question

I have come across the following result in my book on stochastic finance and I have trouble understanding the proof.

On a filtered probability space with filtration $(\mathcal{F}_t)_{t \in \mathbb{R}^+}$, any integrable stochastic process $(X_t)_{t \in \mathbb{R}^+}$ with centred and independent increments is a martingale.

The proof goes as follows.

For $0 \leq s \leq t$, we have \begin{align} E[X_t \vert \mathcal{F}_s] &= E[X_t - X_s + X_s \vert \mathcal{F}_s]\\ &= E[X_t - X_s \vert \mathcal{F}_s] + E[X_s \vert \mathcal{F}_s]\\ &= E[X_t - X_s] + X_s\\ &= X_s \text{.} \end{align}

In the third equality, I don't understand why $E[X_t - X_s \vert \mathcal{F}_s] = E[X_t - X_s]$. The only explanation I can find is that it has to do with the fact that the increments are independent but I don't see how it is related exactly. I had come to understand that $E[Y_t] = E[Y_t \vert \mathcal{F}_0]$ for any stochastic process $(Y_t)_{t \in \mathbb{R}^+}$, but here it is $E[X_t - X_s] = E[X_t - X_s \vert \mathcal{F}_s]$ (so conditional on $\mathcal{F}_s$ instead of $\mathcal{F}_0$).

Answer 1

Since the process has independent increments, the increment $X_t - X_s$ is independent of $X_s - X_0$. So, your estimate of $X_t - X_s$, based on information learned by observing the process upto time $s$, $\mathbb{E}[X_t - X_s|\mathcal{F}_s]$, is as good as having no information at all aka $\mathbb{E}[X_t - X_s]$. This is a well-known property of conditional expectation.

If $X$ is independent of $\mathcal{F}_t$, symbolically written as $X \perp \mathcal{F_t}$, then $\mathbb{E}[X|\mathcal{F_t}] = \mathbb{E}[X]$.

Answer 2

Generally, this proof is learned in the context of Brownian motion, but it works the same with processes that have independent increments.

$\mathcal{F}_s$ is called the filtration and contains all the information of the events up to time $s$. And so if you are computing a probability or expectation that is independent of the past, you "remove" the past (the conditional).

It's the same idea as, if $A$ and $B$ are independent then $P(A|B) = P(A)$ and $E(A|\sigma(B)) = E(A)$.