Take the 2-minute tour ×
Quantitative Finance Stack Exchange is a question and answer site for finance professionals and academics. It's 100% free, no registration required.

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?

share|improve this question
"variance periodically oscillates over time"...that doesn't sound time homogeneous to me –  quasi Dec 9 '13 at 21:58
Check out my example. –  Hans Dec 9 '13 at 22:23
gotcha, get it now –  quasi Dec 9 '13 at 23:19
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 –  quasi Dec 9 '13 at 23:22
This does not work. The variance of the mean-reverting Ornstein-Uhlenbeck process strictly increases over time. –  Hans Dec 10 '13 at 3:04

1 Answer 1

up vote 2 down vote accepted

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 $t-s=0$.

share|improve this answer
why is $t-a=0$? Also the autocovariance is to be a function of the time difference not necessarily a constant. –  Hans Dec 7 '13 at 21:40
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. –  Richard Dec 7 '13 at 22:16
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. –  Hans Dec 9 '13 at 4:24
No problem - when you post a new question then I would be happy to help - if I can. –  Richard Dec 9 '13 at 8:34
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. –  Richard Dec 11 '13 at 9:37

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.