# Why the process $M_t = \sup_{0\leq s\leq t} W_s$ is not a markov process?

For the sake of completeness, I will define what is a Markov Process

Definition: An adapted, $$\mathcal{X}$$-valued stochastistic process $$X:\mathbb{R}_{+}\times \Omega \to \mathcal{X}$$ is a Markov process if, for all $$f\in \mathcal{B}_b(\mathcal{X})$$ and all $$0\leq s\leq t$$, $$\mathbb{E}[f(X_t)\mid \mathcal F_s]= \mathbb{E}[f(X_t)\mid X_s].$$

I recently started studying Markov processes and I am stuck in the following question

Problem: Let $$W_t$$ be the standard Brownian motion, show that the process $$M_t=\sup_{0\leq s\leq t}W_s$$ is not a Markov process.

The question does not actually say what is the filtration considered but I strongly think that is $$\mathcal{F}_t = \sigma(W_s,\ s\in[0,t])$$.

Generally, I post some ideas about what I have tried so far, but this time I am genuinely lost. I tried to use the equalities $$\mathbb{P}(M_t \geq a) = 2\mathbb{P}(W_t \geq a)$$ and $$\mathbb{P}(M_t \geq a) = \mathbb{P}(|W_t| \geq a)$$, but I failed miserably.

Can anyone help me?

• Usually when they say a Markov process they mean the filtration generated by the process itself. But funny is that when you want to prove something is a Markov process then you can prove it is Markov w.r.t. whatever filtration you like because it implies Markov w.r.t. the filtration generated by the process itself. But that does not go the other way around when you want to prove something is NOT Markov... But of course the exercise may have a different intention than what I'm used to. Oh btw I also want to see an answer! Oct 21, 2020 at 7:44

Here is a try: Fix $$s \leq t$$, then

\begin{align*} M_t &= \max \left\{ \sup_{r \leq s} B_r, \sup_{s < r \leq t} B_r \right\} \\ &= \max \left\{M_s, \sup_{r \leq t-s} (B_{r+s}-B_s)+B_s \right\}. \end{align*}

The restarted process $$W_r := B_{s+r}-B_s$$, $$r \geq 0$$, is again a Brownian motion. If we denote by $$M_t^W := \sup_{r \leq t} W_r$$ its running maximum, then we see that

$$M_t = \max\{M_s,M_{t-s}^W+B_s\}.$$

Since $$(W_t)_{t \geq 0}$$ is independent from $$\mathcal{F}_s$$, we find that

$$\mathbb{E}(M_t \mid \mathcal{F}_s) = g(M_s,B_s),\tag{1}$$

where

$$g(x,y) := \mathbb{E}( \max\{x,y+M_{t-s}^W\}).$$

The aim is to show that the function $$g(M_s,B_s)$$ cannot be measurable with respect to $$\sigma(M_s)$$ (intuivitely this is clear, but making it rigorous is not so easy). If we manage to show this, then it follows from $$(1)$$ that $$(M_t)_{t \geq 0}$$ is not Markovian (...because if it were Markovian then the left-hand side of $$(1)$$ would be $$\sigma(M_s)$$-measurable).

First we need to get our hands on $$g$$. To this end, we use the reflection principle. By definition,

$$g(x,y) = x \mathbb{P}(x>y+M_{t-s}^W) + \mathbb{E}((y+M_{t-s}^W) 1_{y+M_{t-s}^W \geq x}).$$

Using the fact that $$M_{t-s}^W$$ equals in distribution $$|W_{t-s}|$$, we see that

$$\mathbb{P}(x>y+M_{t-s}^W) = \mathbb{P}(|W_{t-s}| < x-y)$$

and

\begin{align*} \mathbb{E}(M_{t-s}^W 1_{y+M_{t-s}^W \geq x}) &= \mathbb{E}(|W_{t-s}| 1_{|W_{t-s}| \geq x-y}) \\ &= \sqrt{\frac{2}{\pi(t-s)}} \int_{x-y}^{\infty}z \exp \left(- \frac{z^2}{2(t-s)} \right) \, dz \\ &= \sqrt{\frac{2(t-s)}{\pi}} \exp \left(- \frac{(x-y)^2}{2(t-s)} \right). \end{align*}

Consequently,

\begin{align*} g(x,y) &= x \mathbb{P}(|W_{t-s}|

Writing $$\mathbb{P}(|W_{t-s}| we see that $$g(x,y) = x+h(x-y) \tag{2}$$ for some continuous function $$h$$. More precisely, $$h(r) := - r \mathbb{P}(|W_{t-s}| \geq r) + \sqrt{\frac{2(t-s)}{\pi}} \exp \left(- \frac{r^2}{2(t-s)} \right), \qquad r \geq 0.$$

Pick disjoint intervals $$[a,b]$$ and $$[c,d]$$ such that $$h^{-1}([a,b])$$ and $$h^{-1}([c,d])$$ have positive Lebesgue measure.

Finally, we are ready to check that $$g(M_s,B_s)$$ cannot be $$\sigma(M_s)$$-measurable. Suppose, to the contrary, that it were $$\sigma(M_s)$$-measurable. Then it is immediate from $$(2)$$that $$h(M_s-B_s)$$ is also $$\sigma(M_s)$$-measurable. Consequently, there would be a Borel set, say $$A$$, such that

$$\{h(M_s-B_s) \in [a,b]\} = \{M_s \in A\}. \tag{3}$$

Since $$M_s-B_s$$ has a strictly positive density on $$(0,\infty)$$, we have, by our choice of $$[a,b]$$,

$$\mathbb{P}(M_s \in A)>0,$$

and so $$A$$ has strictly positive Lebesgue measure. Moreover, the fact that $$(M_s,B_s)$$ has a strictly positive density (on its support) implies that $$(M_s,M_s-B_s)$$ has a strictly positive density (on its support). Since $$A$$ and $$h^{-1}([c,d])$$ have positive Lebesgue measure, we obtain that

$$0 < \mathbb{P}(M_s \in A, M_s-B_s \in h^{-1}([c,d])) = \mathbb{P}(M_s \in A,h(M_s-B_s) \in [c,d]). \tag{4}$$

On the other hand, $$(3)$$ and the disjointness of the intervals $$[a,b]$$ and $$[c,d]$$ shows that

$$\mathbb{P}(M_s \in A,h(M_s-B_s) \in [c,d]) = \mathbb{P}(h(M_s-B_s) \in [a,b], h(M_s-B_s) \in [c,d])=0,$$

which contradicts $$(4)$$.

Remark: Using a reasoning very similar to that at the beginning of this answer, it is possible to show that the two-dimensional process $$(M_t,B_t)_{t \geq 0}$$ is Markovian. By the way, also $$M_t-B_t$$ is Markovian.

• Wow, it seems like there should be an easier way to show this. Here is the analogous result for a simple random walk: math.stackexchange.com/questions/683123/… Oct 28, 2020 at 21:09
• @Math1000 Yeah, I saw that answer as well but in the discrete case it's much simpler to get to a contradiction because the supremum (and also the random walk) take discrete values; in particular, $\mathbb{P}(M_n=x)>0$ for suitable $x \in \mathbb{N}$. In contrast, for BM, we have $\mathbb{P}(M_t = x)=0$ for any $x$. I tried different approaches but I always ended up having trouble with exceptional null sets. Feel free to come up with a more elegant proof.
– saz
Oct 28, 2020 at 21:15
• To give you some idea how I would have liked to argue (perhaps you can fix it) starting from (2): If $g(M_s,B_s)$ were $\sigma(M_s)$-measurable, then $h(M_s-B_s) = u(M_s)$ for some mapping $u$. At first I thought that this already gives a contradiction, but it doesn't seem to be that easy. In the discrete case, you can explicitly construct paths where things go wrong... but for Brownian motion it's more difficult because $\mathbb{P}(M_s=x)=0$ for all $x$ (if there were $x$ with $\mathbb{P}(M_s=x)>0$, then $h(x-B_s)=u(x)$ on a set of positive probability would immediately give a contradiction.)
– saz
Oct 28, 2020 at 21:27