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Is it true that if a continuous-time stochastic process $X_t$ is weakly stationary then $|X_t|$ is also weakly stationary?

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@Raskolnikov, how exactly it follows? We need to show that $E|X_tX_{t+h}|$ does not depend on $t$, and this should follow from the fact that $EX_tX_{t+h}$ does not depend on $t$. I am probably missing some trick, because I do not understand how to exploit the facts you suggested. – mpiktas Jan 14 '11 at 13:14
@Raskolnikov, I've understood the trick, but I do not think it works. We have $|E|X_t|-E|X_{t+h}||=|E(|X_t|-|X_{t+h}|)|\le E||X_t|-|X_{t+h}||\le E|X_t-X_{t+h}|$ which is not helpful. – mpiktas Jan 14 '11 at 13:37
Yes, you are right, I made a mistake from there on. Somebody posted a counterexample, but it was a discrete process, now the post is gone. Maybe we can construct a similar continuous time counterexample. The idea was to use two different discrete time processes with same means but different means of absolute values and then build a process that combines both by switching between them. I think this can be generalized to a continuous process. – Raskolnikov Jan 14 '11 at 13:44
up vote 3 down vote accepted

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 $|t-s|<1$, $E(|X_tX_s|) = (t-s)E|X_0|^2$.

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Oh, you adapted your reply. The definition on the wikipage is for a continuous time process though. But I guess it's not too hard to adapt your example by having a switch between continuous time processes at regular time intervals. – Raskolnikov Jan 14 '11 at 13:49
Yea. Actually, I missed a condition to check in (*) in the previous post. – Morning Jan 14 '11 at 13:51
+1, nice example. In your condition (*) there should be $Z$ instead of $X$. – mpiktas Jan 14 '11 at 14:02
@mpiktas: Corrected. Thanks. – Morning Jan 14 '11 at 14:44

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.

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