# Making Sense of Independent Increments Proving The Martingale Property

I've got a question about a proof about using the independent increment property of a stochastic process when it comes to showing the Martingale Property of the said process.

Suppose $$\{B_t\}_{t\geq0}$$ is a standard Brownian motion with some $$\sigma\in(0,\infty)$$ sampled at the times $$0=t_0. That is we are working in discret time. We are to show the Martingale property for this process holds, and that arguments goes as follows (in my textbook):

$$\mathbb{E}[B_t \vert B_{t-1}, \ldots,B_0] = \mathbb{E}[B_t - B_{t-1} + B_{t-1}\vert B_{t-1}, \ldots,B_0] =^* \mathbb{E}[B_t-B_{t-1}]+ \mathbb{E}[B_{t-1} \vert B_{t-1}, \ldots,B_0] = B_{t-1}$$

My question is about the * part. The argument is that Brownian motion has indepdent increments. But how do we make sense of saying that $$B_t-B_{t-1}$$ is independent of every single one of the $$B_{t-1}, \ldots, B_0$$. Why can we just throw away the conditioning by claiming "independent increments"?

Note, I understand that the unconditional expectation is $$0$$ and that the other conditional will be $$B_{t-1}$$. My question is solely about how $$\mathbb{E}[B_t-B_{t-1} \vert B_{t-1}, \ldots,B_0]=\mathbb{E}[B_t-B_{t-1}]$$.

• Your version of martingale property looks like you work in discrete time. Maybe you meant $E(B_t | B_{t_n}, \ldots, B_{t_0})$ with $t_n > t_{n-1} > \ldots > t_0$ instead of $E(B_t | B_{t-1}, \ldots, B_{0})$? Dec 22, 2020 at 22:02
• Yes sorry, I forgot to add that it is the Brownian motion sampled at discrete time steps. So we are working in discrete time. Dec 22, 2020 at 22:11
• Which textbook are you using? Dec 22, 2020 at 22:18

Let us consider vector $$(B_t, B_{t-1}, \ldots, B_0)$$. It has normal distribution (a propery of Brownian motion). Consider the vector $$Z = (B_t - B_{t-1}, B_{t-1}, B_{t-2}, \ldots, B_1, B_0)$$. It's a linear transformation of $$(B_t, B_{t-1}, \ldots, B_0)$$ and hence is has normal distribution too. We know that $$cov(B_{i}, B_j) = \min(i,j)$$. Hence the covariance matrix $$\Sigma$$ of vector $$Z$$ is such that $$\Sigma_{1i} = \Sigma_{i1} = 0$$ for $$i \ge 2$$. We know the form of characteristic function of $$Z$$, because $$Z = (Z_t, \ldots, Z_0)$$ is normal. It's easy to see that $$Ee^{i (\vec{s}, \vec{Z})} = Ee^{i s_t Z_t} \cdot Ee^{i (\sum_{k=0}^{t-1} s_k Z_k)}$$. Hence $$Z_t$$ is independent of $$(Z_{t-1}, \ldots, Z_0)$$. Q.e.d.
• Alright. But I was thinking more of why it is okay to say $B_t-B_{t-1}$ is independent of $B_{t-1}, B_{t-2}, \ldots B_0$. Perhaps it's because we can say $B_t-B_{t-1}$ should be independent of $B_{t-1}- B_{0}$ and $B_{t-2}- B_{0}$ and so on. But I'm not sure if this is actually correct. Dec 22, 2020 at 22:28