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!
 A: 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 $|W|$.
A: 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. 
