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 $0=t_0<t_1<\cdots<t_n$. 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 $\mathbb{E}[B_t-B_{t-1} \vert B_{t-1}, \ldots,B_0]=\mathbb{E}[B_t-B_{t-1}]$.
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.
Is there any questions?