Stationarity of Ornstein–Uhlenbeck Processes and Time Shifts

Summary

The document addresses confusion about how an Ornstein–Uhlenbeck (OU) process can be stationary when its mean and variance may change over time. One response points to a distributional condition involving increments over time intervals of equal length and suggests examining mean and variance as functions of the interval length. This can help distinguish time dependence in a process’s parameters or transient behavior from statistical invariance under shifts in time.

A second response gives the standard definition of stationarity: the joint distributions of observations should be invariant when all observation times are shifted together. It recommends comparing that condition for Brownian motion and the OU process. The discussion is brief and presents competing formulations; in particular, invariance of increment distributions alone describes stationary increments and does not establish strict stationarity of the process levels. Whether an OU process is stationary also depends on its initialization.

Key ideas

• Strict stationarity requires joint distributions of process levels to remain invariant under time shifts.
• Stationary increments require increment distributions to depend on interval length rather than absolute time.
• Stationarity of increments alone does not prove that a process is strictly stationary.
• An OU process’s stationarity depends on its initialization, a condition the short discussion does not develop.

Tags

# Why is OU process stationary?


# Why is OU process stationary?












The mean and variance of Ornstein–Uhlenbeck (OU) process have time dependence (exponentially decay in time). So they are not constant in time. How can it to be stationary?

## Answer by Paul (score 2)

https://quant.stackexchange.com/a/15267

I think you misunderstood the definition. Be stationary does not mean not depend of the time as you can check here. (Sorry for putting an wikipedia link here as I suppose you may have read it)

Another way to think is that the law any increment of the process is given by a same function of the difference of time. More precisely $\forall ~t_2\geq t_1,$ :

$$\mathcal L \left\{X_{t_2}-X_{t_1}\right\}= \Gamma(t_2-t_1)$$

In particular in the case of a stationary Gaussian process whose law as you know is well determined by its mean and variance, the above condition can be expressed by

$$\mathbb E \left[X_{t_2}-X_{t_1}\right]= m(t_2-t_1)$$

$$\text{Var} \left[X_{t_2}-X_{t_1}\right]= v(t_2-t_1)$$

which is the case for OU.

## Answer by Drew (score 0)

https://quant.stackexchange.com/a/15272

Im not sure its very Paul clear. By your definition, a Brownian Motion is stationary. In fact, for a stochastic process, stationarity is defined as statistically invariant under translations.

Try calculating this for the Brownian Motion and OU Process:

$\forall A \in \mathbb{R}^N$

$Pr\{X_1, ..., X_n \in A\} = Pr\{X_{1+h}, ..., X_{n+h} \in A\}$

If those are equal, then the process is stationary.

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This summary was written by Stratmill's research agent from the original; it is not a copy of the source.