I have seen the Markov Property stated in a variety of ways. I'm trying find the equivalence between the two statements below, but I'm having a tough time:

The Markov Property of Brownian Motion started at $a \in \mathbb{R}$ is stated as: $$\mathbf{E}^a[B_{t+s}|\mathcal{F}_s] = \mathbf{E}^{B_s}[B_t]$$ where the notation $\mathbf{E}^a$ means expectation with respect to $\mathbf{P}^a$.

And separately:

Fix $t_0 \in T \subset \mathbb{R}^{+}$ and consider a process: $$B'_t(\omega) = B_{t_0+t}(\omega) - B_{t_0}(\omega)$$ Then $(B'_t)_{t \geq 0}$ is a Brownian motion and independant of $\mathcal{F}_{t_0}$. This is to say for any $A\subset \mathbb{R}^n, M \subset \mathcal{F}_{t_0},$ and any $0 \leq t_0 \leq t_1 \leq ... \leq t_k$: $$\textbf{P}(\{B'_{t_1},...,B'_{t_k}\} \in A \cap M) = \textbf{P}(\{B'_{t_1},...,B'_{t_k}\} \in A) \textbf{P}(M) = \textbf{P}(\{B_{t_1},...,B_{t_k}\} \in A) \textbf{P}(M)$$

I think this post is relevant, but I haven't able to get enough out of it: (Elementary) Markov property of the Brownian motion

I have a similar question about the strong Markov property. Perhaps if someone offers a hint/suggestion/reference I'll try this myself before asking.


1 Answer 1


I can help with one way.

Note that if $(B_t)_{t\geq 0}$ is a $(\mathcal{F}_t)$-Brownian motion, such that among other things we have independent and stationary increments. I.e for any $t_0\geq 0$; $(B'_t)_{t\geq 0}$ is independent of $\mathcal{F}_{t_0}$, where $ B'_t = B'_{t+t_0} - B'_{t_0} $ for any $t\geq 0$. Then we have that \begin{align} E^a(B_{t+s}|\mathcal{F}_s) &= E^a(B_{t+s}-B_s+B_s |\mathcal{F}_s) \\ &= E^a(B_{t+s}-B_s|\mathcal{F}_s) +B_s \quad \quad \{\text{$B_s$ is $\mathcal{F}_s$-measurable}\} \\ &= E^a(B'_t |\mathcal{F}_s) +B_s \\ &= E^a(B'_t) +B_s \quad \quad \{\text{$(B'_t)_{t\geq 0}$ is independent of $\mathcal{F}_s$}\} \\ &= E^0(B_t) +B_s \\ &= E^{B_s}(B_t) \end{align}


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