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Given a suitable filtered space, if $B=\{B_t: t \in [0, \infty)\} $ is a Brownian motion and we define $M_t = \sup_{0\le s \le t} B_s, $ on a suitable filtered space, then it is known that the process $M =\{M_t: t \in [0, \infty)\} $ is not a Markov process.

However, I have seen in a book a statement about the fact that $\{(B_t, M_t): t \in [0, \infty)\} $ IS a Markov process, instead.

I am aware about how to find the joinst distribution of $(B_t, M_t), $ but I do not seem to have a working idea about how to prove that the defined bivariate entity is a Markov process.

I would really appreciate if someone could provide any insight. Thank you in advance for your kindness

Maurice

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    $\begingroup$ I believe using the following: $S_t \stackrel{d}= \lvert B_t \rvert $ will be helpful. $\endgroup$
    – oliverjones
    Aug 23, 2022 at 19:36
  • $\begingroup$ Yes, thank you. That is what I had thought as well. But now I need to prove something like $E[f(S_t, B_t)| {\cal F}_v] = g(S_v, B_v) $ for some Borel measurable functions $f, g. $ And, while with martingales I can find my way around problems, with Markov processes, I am very clumsy, and this one has two components as well... Any chance you could at least get me started on the path to how to set up proving the desired result? $\endgroup$
    – Maurice
    Aug 23, 2022 at 20:23
  • $\begingroup$ You could show $\mathbb{E}[f(B_{t+h},S_{t+h}) \mid \mathcal{F}_t] = \mathbb{E}[f(B_{t+h},S_{t+h}) \mid (B_t,S_t)]$ for any Borel function $f$. Coupled with my previous comment that should be enough to get you there. $\endgroup$
    – oliverjones
    Aug 23, 2022 at 22:28

1 Answer 1

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I could not really follow up on the hints of using $|B_t|. $ However, while I am quite "green" in the matter of Markov processes, I am better at using conditional expectations. So, here is my argument that uses the Freezing Lemma, sometimes called even the Independence Lemma (I always wondered what it was useful for and finally I have my answer, albeit 15 years or more after I studied it!).... Here it goes:

Let $f: R^2 \mapsto R $ be a measurable and bounded function. Then, for $0 \le s \le t: $ \begin{equation*}\begin{split} E[f(B_t, M_t) \mid {\cal F}_s] &= E[f(B_s + (B_t-B_s), M_s \vee \sup_{s \le u \le t} B_u) \mid {\cal F}_s]\\ &= E[f(B_s+ (B_t-B_s), M_s \vee \{B_s + \sup_{s\le u\le t} (B_u-B_s)\}) \mid {\cal F}_s]. \end{split}\end{equation*} At this point, we should recognize that $M_s $ and $B_s $ are ${\cal F}_s$-measurable and that $B_t-B_s $ and $B_s + \sup_{s\le u\le t} (B_u-B_s) $ are instead independent of $ {\cal F}_s $ and an application of the Freezing Lemma yields that $$ E[f(B_t, M_t) \mid {\cal F}_s] = g(B_s, M_s) $$ where $$ g(x, y) := E[f(x+ y \vee \sup_{s\le u \le t} (B_u -B_s), x+ (B_t-B_s))]. $$

Hence $(B, M) $ is a Markov process.

I wonder if, by trying to prove the results using transition kernels could use the the fact that $M_t $ and $|B_t| $ have the same distribution

PS... On an unrelated issue... what happened to guys like Did, Saz and other mathematicians of high stature that used to have an asnwer to almost anything? Are they no longer associated with the website?

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