# 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?

• Stationary means that the process does not depend on a specific time instant, but only on a time interval. For example, a process for a stock price is stationary if the probability density of the price over a year is tied to the time interval (i.e. 1 year) and not on a specific year (2013 or 2014 or any other year). Nov 3 '14 at 14:20

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.

• Thanks @Paul for the comments. But for OU process, its discrete form (explicit) will be $$r_{t_2} - r_{t_1} = (\alpha - \beta r_{t_1})(t_2 - t_1) + \sigma \epsilon_{t_2-t_1}$$, which seems to be dependent on value of $r_{t_1}$. So the mean of $r_{t_2} - r_{t_1}$: $$E[r_{t_2} - r_{t_1}]$$ is not purely a function of $(t_2-t_1)$. Did I misunderstand something?
– cgao
Nov 3 '14 at 14:53
• Solving the recurrence $m_i+\beta(t_i-t_{i-1})m_{i-1}= \alpha(t_i-t_{i-1})$ where $m_i:=\mathbb E[r_{t_i}]$ you may find the right answer
– Paul
Nov 3 '14 at 16:17

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.

• There is weak stationarity too ... then you only consider the first 2 moments.
– Ric
Nov 6 '14 at 8:17