I'm trying to understand the Markov property for Brownian motions in full generality. The textbook I'm following states it like this:

Recall that we have a family of measures $P_x, x \in \mathbb{R}$ on $(C,\mathcal{C})$ so that under $P_x, B_t(\omega) = \omega(t)$ is a Brownian motion starting at $x$. For $s \geq 0$, we define the shift transformation $\theta_s: C \to C$ by $$(\theta_s \omega)(t) = \omega(s+t) \quad \textrm{for} \quad t \geq 0$$ Theorem 8.2.1: If $s \geq 0$ and $Y$ is bounded and $\mathcal{C}$-measurable, then for all $x \in \mathbb{R}$ $$E_x(Y \circ \theta_s | \mathcal{F}_s^+) = E_{B_s}Y$$ where the right-hand side is the function $\phi(x) = E_xY$ evaluated at $x = B_s$.

I think I understand the general idea, but there are still 2 gaps which make me think I'm fundamentally missing something here:

1) I'm confused by the $x$ subscript on $E_x$ in the left hand side. I understand that this is expectation with respect to $P_x$, but then shouldn't this also appear on the right side somewhere?

2) I'm trying to figure out what function $Y: C[0,\infty) \to \mathbb{R}$ (as in the theorem statement above) would allow me to state this Markov property in terms of a simpler function $f: \mathbb{R} \to \mathbb{R}$, involving for example $f(B_{t+s} - B_s)$. I think this is something like $Y(\omega(\cdot)) = \omega(t)$ but I'm pretty shaky on this.

  • $\begingroup$ by any chance, the textbook you're using is that written by Liggett? $\endgroup$
    – user190080
    Jul 24, 2015 at 20:08
  • $\begingroup$ @user190080 Durrett. Somewhat frustrating because he often leaves notation unexplained and makes big leaps. But it's possible that I'm just dumb. $\endgroup$
    – gogurt
    Jul 24, 2015 at 20:34

1 Answer 1

  1. In some sense, there is an $x$ on the right-hand side because the equality $$\mathbb{E}_x(Y \circ \theta_s \mid \mathcal{F}_s^+)(\omega) = \mathbb{E}_{B_s(\omega)}(Y)$$ holds only $\mathbb{P}_x$-almost surely.
  2. For a function $f: \mathbb{R} \to \mathbb{R}$ the Markov property reads $$\mathbb{E}_x (f(B_{t+s}) \mid \mathcal{F}_s^+) = \mathbb{E}_{B_s} f(B_{t}), \qquad t \geq 0.$$ This implies in particular $$\mathbb{E}_x (f(B_{t+s}) \mid \mathcal{F}_s^+) = \mathbb{E}_{x}(f(B_{t+s}) \mid B_s);$$ this means that the future does not depend on the past, but only on the current position of the process.
  • $\begingroup$ Thanks for this. But 2 follow-up questions: so in 1) I'm still a bit confused. Where is the $\omega$ that the $\mathbb{P}^x$-a.s. refers to? And in 2), so am I correct in saying that the function here is $Y(\omega(\cdot)) = \omega(t)$? $\endgroup$
    – gogurt
    Jul 24, 2015 at 20:33
  • $\begingroup$ @gogurt 1) See my edited answer. 2) If you choose $Y(\omega(\cdot)) = \omega(t)$,then the Markov property states $$\mathbb{E}_x (B_{t+s} \mid \mathcal{F}_s^+) = \mathbb{E}_{B_s}(B_t).$$ If you want to get the identity stated in the second part of my answer, then you have to choose $Y(\omega(\cdot)) := f(\omega(t))$. $\endgroup$
    – saz
    Jul 25, 2015 at 5:10
  • $\begingroup$ awesome. thanks for your help! $\endgroup$
    – gogurt
    Jul 26, 2015 at 14:04

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