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?
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1$\begingroup$ 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). $\endgroup$– ArrigoNov 3, 2014 at 14:20
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2 Answers
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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.
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1$\begingroup$ 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? $\endgroup$– cgaoNov 3, 2014 at 14:53
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$\begingroup$ 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 $\endgroup$– PaulNov 3, 2014 at 16:17
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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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1$\begingroup$ There is weak stationarity too ... then you only consider the first 2 moments. $\endgroup$– Richi WaNov 6, 2014 at 8:17