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I am trying to understand the concept of strong Markov property quoted from Wikipedia:

Suppose that $X=(X_t:t\geq 0)$ is a stochastic process on a probability space $(\Omega,\mathcal{F},\mathbb{P})$ with natural filtration $\{\mathcal{F}\}_{t\geq 0}$. Then $X$ is said to have the strong Markov property if, for each stopping time $\tau$, conditioned on the event $\{\tau < \infty\}$, the process $X_{\tau + \cdot}$ (which maybe needs to be defined) is independent from $\mathcal{F}_{\tau}:=\{A \in \mathcal{F}: \tau \cap A \in \mathcal{F}_t ,\, \ t \geq 0\}$ and $X_{\tau + t} − X_{\tau}$ has the same distribution as $X_t$ for each $t \geq 0$.

Here are some questions that make me stuck:

  1. In $\mathcal{F}_{\tau}:=\{A \in \mathcal{F}: \tau \cap A \in \mathcal{F}_t ,\, \ t \geq 0\} $, what does $\tau \cap A $ mean? $\tau$ is a stopping time and therefore a random variable and $A$ is a $\mathcal{F}$-measurable subset, but what does $\tau \cap A$ mean?
  2. How is the process $X_{\tau + \cdot}$ defined from the process $X_{\cdot}$ ? Is it the translated version of the latter by $\tau$?
  3. How is the conditional independence between a process, such as $X_{\tau + \cdot}$, and the sigma algebra, such as $\mathcal{F}_{\tau}$, given an event, such as $\{\tau < \infty\}$, defined?

    Related question, is independence between a random variable and a sigma algebra defined as independence between the sigma algebra of the random variable and the sigma algebra?

  4. Is "$X_{\tau+ t} − X_{\tau}$ has the same distribution as $X_t$ for each $t \geq 0$" also conditional on the event $\{\tau < \infty\}$?

Thanks and regards!

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    $\begingroup$ After reading that I think, really, you should be learning about the strong Markov property from somewhere other than Wikipedia. $\endgroup$ Mar 9, 2011 at 23:00
  • $\begingroup$ @George: I tried to look up in several books, many of which are just talking about the theorem of strong Markov property for Brownian motion. One (Lamperti's book) uses some notation that I don't understand yet. Maybe I will sometime but who knows. So I very much appreciate some nice and clear references to clear things up. $\endgroup$
    – Tim
    Mar 9, 2011 at 23:14
  • $\begingroup$ I don't have time to write up a proper answer now, but I do have a definition of the strong Markov property on my blog (but it assumes a good knowledge of measure-theoretic probability theory) almostsure.wordpress.com/2010/07/19/… $\endgroup$ Mar 9, 2011 at 23:50
  • $\begingroup$ And in that Wikipedia page, they don't say what notation they are using for the bits you labeled (1) and (2), and it looks weird. (4) is just plain wrong. $\endgroup$ Mar 9, 2011 at 23:54
  • $\begingroup$ @George: Thanks for your blog! In the third paragraph there, "Recall that the law of an inhomogeneous Markov process $X$ is described by a transition function ${\{P_t\}_{t\ge0}}$ on some measurable space ${(E,\mathcal{E})}$." Do you mean "inhomogeneous" or "homogenenous"? $\endgroup$
    – Tim
    Mar 10, 2011 at 10:37

2 Answers 2

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Here is a less garbled version of the Wikipedia definition. (Use TheBridge's correction for the definition of ${\cal F}_\tau$.) The post-$\tau$ process $X_{\tau+\cdot}$ is defined on the event $\{\tau<\infty\}$ by $$ X_{\tau+t}(\omega) = X_{\tau(\omega)+t}(\omega),\qquad t\ge 0, $$ for $\omega\in\{\tau<\infty\}$. One way to state the strong Markov property is this: The conditional distribution of $X_{\tau+\cdot}$ given ${\cal F}_\tau$ is (a.s.) equal to the conditional distribution of $X_{\tau+\cdot}$ given $\sigma\{X_\tau\}$, on the event $\{\tau<\infty\}$. More precisely, $$ P[ X_{\tau+t}\in B|{\cal F}_\tau] = P[ X_{\tau+t}\in B|X_\tau],\qquad \hbox{almost surely on }\{\tau<\infty\}, $$ for all $t\ge 0$, and all measurable subsets $B$ of the state space of $X$.

This is equivalent to the statement that $X_{\tau+\cdot}$ and ${\cal F}_\tau$ are conditionally independent, given $X_\tau$: $$ P[ F\cap \{X_{\tau+t}\in B\}|X_\tau] = P[ F|X_\tau]\cdot P[X_{\tau+t}\in B|X_\tau],\qquad \hbox{almost surely on }\{\tau<\infty\}, $$

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  • $\begingroup$ The condition you wrote after your "More precisely" characterizes the conditional distribution of the random variable $X_{\tau+t}$ conditionally on $\mathcal{F}_\tau$, for every $t\ge0$. This is strictly weaker than the property of the conditional distribution of the process $X_{\tau+\cdot}$ conditionally on $\mathcal{F}_\tau$ that is required for $X$ to be strong Markov. $\endgroup$
    – Did
    Mar 10, 2011 at 20:56
  • $\begingroup$ Good point! That really should be $$ P[ G|{\cal F}_\tau] = P[ G|X_\tau],\qquad \hbox{almost surely on }\{\tau<\infty\},\leqno(*) $$ for all events $G$ in the $\sigma$-field generated by $X_{\tau+\cdot}$, and similarly for the conditional independence form of the property. On the other hand, if the weaker form found in my first post holds {\it for all\/} stopping times, then so does the stronger form ($*$). $\endgroup$
    – JohnD
    Mar 10, 2011 at 21:34
  • $\begingroup$ D Sure, by the chain rule for conditional probabilities. But my point is that, since the OP asks for clear and explicit definitions and proofs about a notion he/she is trying to understand in depth, you might want to add to your post an expanded version of the remark you made in the last sentence of your comment. $\endgroup$
    – Did
    Mar 11, 2011 at 6:07
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For the first one the notation is wrong it should I think $A\cap\{\tau\le t\}\in\mathcal{F}_t$ instead of $A\cap\tau$.

And for the fourth point, look at George Lowther comments below, that fully address the problematic.

Regard

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  • $\begingroup$ No, the fourth point is saying that it has identical increments, which is true for Levy processes, but not part of the strong Markov property. $\endgroup$ Mar 9, 2011 at 23:58
  • $\begingroup$ In fact, the fourth point on its own implies that it is a Levy process starting from zero. $\endgroup$ Mar 10, 2011 at 0:03
  • $\begingroup$ You are right how I missed that I edit my post accordingly $\endgroup$
    – TheBridge
    Mar 10, 2011 at 10:53
  • $\begingroup$ I'm guessing, but I think the person who wrote that Wikipedia entry copied a statement from somewhere which was only dealing with Levy processes, and also made some typos. $\endgroup$ Mar 10, 2011 at 12:18

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