Weak stationarity of a process modulus Is it true that if a continuous-time stochastic process $X_t$ is weakly stationary then $|X_t|$ is also weakly stationary?
 A: For simplicity, assume discrete-time process $\{X_t\}_{t\in\mathbb Z}$. A counterexample is the following. One can easily finds random variables $Y$ and $Z$, such that $E(Y) = E(Z), E|Y|\neq E|Z|$ and $E(Y^2)=E(Z^2)$ (*). Then, let $X_{2n}$ be i.i.d. copies of $Y$, and $X_{2n-1}$ be i.i.d. copies of $Z$, and let $X_{2n}$ and $X_{2m-1}$ for all $n,m\in\mathbb Z$ be independent as well.
Now construct an example such that (*) holds. Consider $Y = \epsilon U$, where $U$ is the uniform random variable on $[0,1]$, $P(\epsilon = 1)=P(\epsilon=-1)=1/2$ and $\epsilon$, $U$ are independent. Then, $E Y^2 = 1/3, EY = 0$ and $E|Y| = 1/2$. Take $Z = \epsilon' \delta_{1/\sqrt 3}$. $EZ^2 = 1/3, EZ = 0$, but $E|Z| = 1/\sqrt{3}$.
Now for the continuous-time counterexample. Let $U$ be uniform on $[0,1]$, and let $\tilde X$ denote the discrete one defined above. Define
$X_t = \tilde X_{\lfloor t+U \rfloor}$.
Then, for $|t-s|<1$, $E(|X_tX_s|) = (t-s)E|X_0|^2$.
A: A simple counterexample is the following.  Let $m(t)$ be a non-constant function that satisfies $m(t) \ge 2$ for all $t$, and let $p(t)=1/m(t)^2$.  Now take $X_t$ at each point in time to be $m(t)$ with probability $p(t)$, $-m(t)$ with probability $p(t)$, and $0$ with probability $1-2p(t)$.  Clearly $E[X_t] = 0$ for all $t$.  Also $E[X_t^2] = 2p(t)m(t)^2 = 2$; and $E[X_t X_{t+h}]=0$ for $h \neq 0$.  Therefore $X_t$ is weakly stationary.  On the other hand, $E[|X_t|] = 2p(t)m(t) = 2/m(t)$, which is not constant; so $|X_t|$ is not weakly stationary.
