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

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  • $\begingroup$ 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 $E(B_t | B_{t-1}, \ldots, B_{0})$? $\endgroup$
    – Botnakov N.
    Dec 22, 2020 at 22:02
  • $\begingroup$ Yes sorry, I forgot to add that it is the Brownian motion sampled at discrete time steps. So we are working in discrete time. $\endgroup$
    – Sinbad The Sailor
    Dec 22, 2020 at 22:11
  • $\begingroup$ Which textbook are you using? $\endgroup$
    – Yuval Peres
    Dec 22, 2020 at 22:18

1 Answer 1

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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?

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  • $\begingroup$ 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. $\endgroup$
    – Sinbad The Sailor
    Dec 22, 2020 at 22:28
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    $\begingroup$ @SinbadTheSailor, this argument is not correct: in general case pairwise independence is not enough for mutual independence. That's why I used another approach: method of characteristic funcitons. When we have normal distributions, then in some sense pairwise independence is enough for mutual independence. $\endgroup$
    – Botnakov N.
    Dec 22, 2020 at 22:30

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