This is an edit to the previous question, on stationary process, which was answered by Richard below.

Let $x_t$ be a zero mean, time homogeneous Markovian process over time $t$ starting from $x_0=0$. What are the examples of $x_t$ where the variance at $t$ does not increase over $t$?

1) In discrete time and discrete state, the followig is a very simple example where the variance periodically oscillates over time.

$$x_{t+1} = \eta(1-|x_t|),\, x_0=0;\, \eta\in\{-1,1\},\mbox{ with probability of } \frac{1}{2} \mbox{ on each value of }\eta.$$

2) In continuous time, but discontinuous path setting, is the following jump diffusion process a correct example?

$$dx_t = -\alpha x_t dt+dz_t+ y\eta dN_t,\, x_0 = 0,$$ where $\alpha\gg 0$, $z_t$ is the standard brownian motion with mean $0$ and standard deviation $t$, $N_t$ is the Poisson process with frequency $0<\lambda\ll 1$, $\eta$ takes on values $-1$ or $1$ with $0.5$ probability each, $z_{t_1}$, $N_{t_2}$ and $\eta$ are independent of each other at arbitrary $t_1$ and $t_2$, and constant $y\gg 1$.

On second thought, this is not a correct example. One can solve this equation and one will find the variance of this process is the sum of the variance from $dz_t$ and that from $dN_t$ due their independence. We will have to make the jumps negatively correlated to $z_t$.

A better setup is to shift $x_t$ beyond a barrier directly back to the $x=0$ line. So the process resides on the topology of two cylinders touched along a longitude. However, it seems to me, even this set up with $x_t$ being either a standard Browniam motion or mean reverting one without any jump process still has its variance increasing with time.

Therefore, I am still without a valid example in this setup.

3) What are the examples for continuous path? I suspect it is not possible. Can anyone prove this if it is indeed impossible?

  • $\begingroup$ "variance periodically oscillates over time"...that doesn't sound time homogeneous to me $\endgroup$
    – quasi
    Dec 9 '13 at 21:58
  • $\begingroup$ Check out my example. $\endgroup$
    – Hans
    Dec 9 '13 at 22:23
  • $\begingroup$ gotcha, get it now $\endgroup$
    – quasi
    Dec 9 '13 at 23:19
  • $\begingroup$ en.wikipedia.org/wiki/Ornstein%E2%80%93Uhlenbeck_process; start it from its steady state distribution. note this has mean-reverting behavior, similar to your example. oops, doesn't start at 0 though $\endgroup$
    – quasi
    Dec 9 '13 at 23:22
  • $\begingroup$ This does not work. The variance of the mean-reverting Ornstein-Uhlenbeck process strictly increases over time. $\endgroup$
    – Hans
    Dec 10 '13 at 3:04

This is the answer to the first version of the question which asked whether a stationary process has an increasing variance over time.

No the definition of (weakly) stationary (http://en.wikipedia.org/wiki/Stationary_process) is that the variance is the same for each point in time.

In the literature it is often dealt with the covariance function. For a stationary time series, the covariance between $X_t$ and $X_s$ only depends on the time span $|t-s|$. For the varianace of $X_t$ we have $t-s=0$.

  • $\begingroup$ why is $t-a=0$? Also the autocovariance is to be a function of the time difference not necessarily a constant. $\endgroup$
    – Hans
    Dec 7 '13 at 21:40
  • $\begingroup$ If we want to apply the formulation of covariance to the simple variance case, then $t=s$ and thus $t-s=0$. The ACF is a function of the time difference, true. But this is zero for the variance. A random walk e.g. is not stationary. The variance increases with the square-root of time. An white noise on the other hand is stationary - the variance at each point in time is the same. $\endgroup$
    – Ric
    Dec 7 '13 at 22:16
  • $\begingroup$ You are right, Richard. Thank you. I am a bit embarrassed by my ignorance betrayed by this question. I need to edit the question to bring out what I am really after. $\endgroup$
    – Hans
    Dec 9 '13 at 4:24
  • 1
    $\begingroup$ No problem - when you post a new question then I would be happy to help - if I can. $\endgroup$
    – Ric
    Dec 9 '13 at 8:34
  • 1
    $\begingroup$ Yes, I have seen it but I don't have time at the moment. Why did you rephrase this question and did not post an new one? I can only delete my answer ... your question is a hard one and I don't have the time to think about it at the moment, sorry. $\endgroup$
    – Ric
    Dec 11 '13 at 9:37

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