# Conditional expectation of increments of stochastic process [closed]

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 $$\mathcal{F}_0$$).

• What is the book? What page / section is the proof from? Commented Oct 31, 2023 at 19:42
• "Stochastic Finance" by Nicolas Privault (doi.org/10.1201/b16359), it's on page 112 Commented Oct 31, 2023 at 20:11

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 $$\mathbb{E}[X|\mathcal{F_t}] = \mathbb{E}[X]$$.
$$\mathcal{F}_s$$ 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 $$E(A|\sigma(B)) = E(A)$$.