# Is the absolute value of Brownian motion a super martingale?Is it a sub martingale? Is it a Markov process?

I've just started to study random processes and I'm trying to solve the following problem:

Let $W(t)$ be a Brownian motion with filtration $F(t)$ generated by $W(t)$ (i.e., $F(t)=\sigma \left( W(s)\right)$, $s \in [0,t]$).

• Is the process $|W(t)|$ a sub martingale? A super martingale? Or neither?
• Is $|W(t)|$ a Markov process?

Unfortunately, I can't see how to do that using the definition of sub/super martingale.. Any help will be appreciated!

• Use Jensens inequality for the absolute value function. – Lutz Lehmann Feb 17 '14 at 0:19
• Conditional Jensen en.wikipedia.org/wiki/Conditional_expectation#Basic_properties – ir7 Feb 17 '14 at 0:23
• To rule out the other, being both a sub and super martingale implies being a martingale. What would $E[|W(t)|]$ be if $|W(t)|$ were a martingale? What would that imply? I would give a hint about the Markov property, but it depends on what definition you're using. – Chris Janjigian Feb 17 '14 at 0:37
• @Chris Janjigian Jensen inequality for conditional expectations yields $|E[W(t)|F(s)]| \le E[|W(t)||F(s)]$. Thus, since $W(t)$ is a martingale, the left side is equal to $|W(s)|$. So finally, I have the following inequality: $E[|W(t)||F(s)] \ge |W(s)|$ - which tells that |W(t)| is submartingale. Answering your question - if |W(t)| was martingale then $E[|W(t)||F(s)] = |W(s)|$. – mouse Feb 17 '14 at 1:41
• The hint was about the unconditional expectation $E[|W(t)|]$, not the conditional expectation. For the Markov property, notice that the filtration generated by $|W(t)|$ is a strict subset of the filtration generated by $W(t)$. Think about the tower rule. – Chris Janjigian Feb 17 '14 at 3:45

Concerning you first question, $|W|$ is a submartingale:

In general if $X$ is a martingale and $\phi: \mathbb R \rightarrow \mathbb R$ is a convex funtion such that $\mathbb E[\phi(X_t)] < \infty$, then a simple application of the conditional Jensen's inequality shows that $$\mathbb{E}\bigl[\phi(X_t)\mid \mathcal{F}_s \bigr] \geq \phi \Bigl(\mathbb{E}\bigl[X_t\mid \mathcal{F}_s \bigr] \Bigr) = \phi(X_s), \quad \text{a.s.}$$ Hence, the process $(\phi(X_t))_{t \geq 0}$ is a submartingale.

In your case $X=W$ and $\phi(x)=|x|$. It is easy to check that $\phi$ is convex and $E[|W_t|] < \infty$. Therefore $|W|$ is a submartingale.

Edit: the last part is incorrect as pointed out in the comments.

Concerning the Markov Property just keep in mind that $x \mapsto |x|$ is a measurable function (because it is continuous). Therefore the Markov property of $W$ implies the Markov property of $|W|$.

• It's not generally true that a measurable (or even continuous) function of a Markov process is a Markov process. For example, take simple random walk $X_n$ on $\mathbb{Z}$ and $Y_n = f(X_n)$ where $f(-2) = 2$ and $f(x) = x$ otherwise. Then $$1/2 = P(Y_3 = 3 \mid Y_2 = 2, Y_1 = 1) \ne P(Y_3 = 3 \mid Y_2 = 2, Y_1 = -1) = 0.$$ – Nate Eldredge Dec 14 '16 at 14:57
• The "submartingale" bit is good, but this justification of the Markov property is very bad. Try it with $f(x)= \max \{0,x\}$, for fun. – D. Thomine Dec 14 '16 at 15:02
• uuu, yeah thank you for pointing that out, I made a silly mistake – Cettt Dec 14 '16 at 15:08
• @NateEldredge It is my understanding that the function needs only to be measurable and bounded to imply the Markov property. Would it be allowable to say $|B_t| _=^d \sup_{0 \le s \le t} B_s$ sup of continuous function over finite interval is bounded, and sup B_s is measurable. So by the Markov property $|B_t|$ is Markov. I don't like it because it seems to be pulling at the ear with the wrong hand but I am weak at maths so cant see how to do it otherwise. – Novice May 30 '19 at 7:11
• @Novice: That really doesn't make any sense to me. I would suggest that you try to give a precise statement of the theorem you are trying to use, and then see whether its hypotheses are satisfied. – Nate Eldredge May 30 '19 at 13:59

To show the Markov property there is a direct approach.

Let $$u \in \mathcal{B}_b(\mathbb{R})$$, and $$s,t \ge 0$$. Then by the independent and stationary increment properties of BM:

\begin{align*} \mathbb{E}[u(|B_{t+s}|) | \mathcal{F_s}] &= \mathbb{E}[u(|B_{t+s}-B_s +B_s|)| \mathcal{F_s}]\\ &= \mathbb{E}u(|B_{t+s}-B_s +y|)\bigg|_{y=B_s}\\ &= \mathbb{E}u(|B_t +y|)\bigg|_{y=B_s} \end{align*}

By symmetry of BM we get that, for $$g_t(z)$$ normal denisty,

\begin{align*} \mathbb{E}[u(|B_{t+s}|) | \mathcal{F_s}] &= \frac{1}{2} \left[\mathbb{E}u(|B_t +y|)\bigg|_{y=B_s} + \mathbb{E}u(|B_t -y|)\bigg|_{y=B_s} \right]\\ &= \frac{1}{2} \left[ \int_{-\infty}^{+\infty}\left(u(|z+y|) + u(|z-y|)\right)g_t(z)dz \right]\bigg|_{y=B_s}\\ &= \frac{1}{2} \left[ \int_{-\infty}^{+\infty}u(|z|)(g_t(z+y) + g_t(z-y))dz \right]\bigg|_{y=B_s}\\ &= \int_{0}^{+\infty}u(|z|)(g_t(z+y) + g_t(z-y))dz \bigg|_{y=B_s} \end{align*}

The integral is independent of $$s$$. Hence $$\mathbb{E}[u(|B_{t+s}|) | \mathcal{F_s}]$$ is a function of $$|B_s|$$.

Thus

$$\begin{equation*} \mathbb{E}[u(|B_{t+s}|) | \mathcal{F_s}] = g(B_s) \end{equation*}$$

for some Borel function $$g$$.

It follows, by the tower property for conditional expectations, that

\begin{align*} \mathbb{E}[u(|B_{t+s}|) | X_s] &= \mathbb{E}\left[ \mathbb{E}[u(|B_{t+s}|) | \mathcal{F_s}|X_s\right]\\ &= \mathbb{E}[g(B_s)|X_s]\\ &=g(B_s). \end{align*}

Hence we conclude that

$$\begin{equation*} \mathbb{E}[u(|B_{t+s}|) | \mathcal{F_s}] = \mathbb{E}[u(|B_{t+s}|) | X_s]. \end{equation*}$$

Thus $$|B_t|$$ is Markov.

• There are some mismatched parentheses that make this hard to read. – Nate Eldredge May 30 '19 at 14:00
• Thanks Nate for the remark. Can you let me know if this is now correct. Best wishes. – Novice Jun 1 '19 at 16:45
• The first two lines still look wrong, with the parentheses mismatched around the absolute values. – Nate Eldredge Jun 1 '19 at 18:22
• why can we have $$\mathbb{E}\Big[h(|W_{s+t}-W_{s}+W_{s}|)\ \Big|\ \mathcal{F}_{s}\Big]=\mathbb{E}h(|W_{s+t}-W_{s}+y|)\Big|_{y=W_{s}}?$$ – JacobsonRadical Jun 6 '20 at 19:28