10
$\begingroup$

I've run up against a wall in reconciling two different definitions of the Ornstein-Uhlenbeck process, and would appreciate some help.

On the one hand, as discussed here, we can define an Ornstein-Uhlenbeck process as a Gaussian process with a kernel function of the form $k(x, y) \propto \exp(-\left\lVert x - y\right\rVert / \theta)$, which is clearly stationary.

On the other hand, we have the definition of the Ornstein-Uhlenbeck process as the solution to the stochastic differential equation $du(t) = \theta(\mu - u(t))+\sigma dW(t)$, which is given by $u(t)= u(0) \exp⁡(-\theta t)+\mu(1-\exp⁡(-\theta t) )+\sigma \exp⁡(-\theta t) \int_0^t \exp⁡(\theta\tau)dW(\tau)$. Using the Itô isometry, as shown here and here on SE, this process can be shown to have the covariance $k(x, y) = \frac{\sigma^2}{2\theta} \exp(-\theta (x+y)) (1 - \exp(2 \theta (x - y)))$.

Not only is this not equal to the previous definition, this is not even stationary. What gives? Are these two distinct definitions of what "Ornstein-Uhlenbeck process" means, or can these two processes be shown to be equal, possibly modulo certain assumptions?

As a bonus question, just because I've never seen anyone prove it - how do you prove that the solution to the above SDE is a Gaussian process (assuming it is)?

$\endgroup$

2 Answers 2

10
$\begingroup$

In order to have a stationary solution of the Stochastic Differential Equation (SDE), you have to start from a random initial value $u(0)$ at the fixed time $t=0$. This value must be drawn from the stationary distribution, here Gaussian.

An alternative is to consider that the process $u(t)$ "has started in the remote past". Although $u(t)$ is not differentiable, it is appealing to write the SDE as $$ \frac{\text{d} u(t)}{\text{dt}} = - \theta u(t) + \eta(t), $$ where $\eta(t)$ is a Gaussian white noise with variance $\sigma^2$, leading to the following representation of $u(t)$
$$ u(t) = \int_{-\infty}^t e^{-\theta (t -s)} \, \eta(s) \,\text{d}s $$
which can be made more rigorous by replacing $\eta(s)\text{d}s$ by $\sigma \text{d}W(s)$. Using this representation, it is easy to derive the covariance of $u(t)$. For $r \geq 0$ $$ \text{E}[u(t) u(t+r)] = \sigma^2 \, \text{E} \left[\int_{-\infty}^t\int_{-\infty}^{t +r} e^{-\theta (t -s)} \, e^{-\theta (t + r -s')} \, \text{d}W(s) \text{d}W(s') \right], $$ and we can exchange the expectation and the integral(s), the expectation $\text{E}[\text{d}W(s) \text{d}W(s')]$ being $0$ for $s \neq s'$ and $\text{d}s$ when $s= s'$. Then by simple integration, we find $$ \text{E}[u(t) u(t+r)] = \frac{\sigma^2}{ \, 2 \theta} \, e^{-\theta r} \quad (r \geq 0). $$

Most of the literature dedicated to Itô's calculus assumes that $W(0) = 0$; This can be quite misleading for statistical problems where there is usually no special time "$t=0$" corresponding to a deterministic value of the process. Some adaptations are needed to cope with continuous time linear filters.

$\endgroup$
6
  • $\begingroup$ Thanks for the answer! This makes perfect sense. Hell, from what I've seen, most of the introductory texts on the Itô calculus barely even admit the existence of integrals on any set but [0, T]. I get that extending it by shifting the process in question is trivial, but you'd think it'd at least be worth a sentence to talk about, eg, Itô integrals on (-infty,infty). Btw- the first time I saw the Itô integral, I was confused about how it seemed so analogous to the Riemann integral. That's very unintuitive, but is there a stochastic integral that instead takes after the Lebesgue integral? $\endgroup$ Nov 7, 2018 at 2:52
  • $\begingroup$ A SDE needs some kind of boundary conditions: usually an initial or a terminal condition, and these are in practice often stochastic rather than deterministic. I do not know if a different Stochastic Integral would help to get more flexibility. All in all, the same problem arises in discrete time: ARIMA and more general State Space models need initial conditions. These are not easy to cope with mathematically, although they involve only sums, not integrals. $\endgroup$
    – Yves
    Nov 7, 2018 at 8:28
  • 1
    $\begingroup$ Yes, we can use GPs without caring about SDEs. However, SDEs can provide interesting interpretation and also powerful algorithms. For instance a one-parameter/input GPs having a Matérn-Whitle with shape $(2p - 1)/2$ and $p>0$ integer is better regarded as a State-Space model, as it is the case for the OU corresponding to $p=1$. $\endgroup$
    – Yves
    Nov 8, 2018 at 7:29
  • 1
    $\begingroup$ Necroposting, but I find that very interesting, and would it be okay if I asked for just, like, a link to somewhere talking about an example of an algorithm that flows naturally from the intuitive link between these two? $\endgroup$ Dec 15, 2018 at 14:30
  • 1
    $\begingroup$ Maybe these slides by Simo Sârkkä's will help you, especially slides 23--28. $\endgroup$
    – Yves
    Dec 16, 2018 at 8:25
3
$\begingroup$

I initially believed that the mean parameter had a part to play as, in this link Rasmussen & Williams, 2006, Appendix B, equation B.27, the OU SDE is written as:

$$ dX_t = - a_0 X(t)dt + b_0 dW_t $$

whereas the traditional equation is:

$$ dX_t = \theta \mu dt - \theta X_tdt + \sigma dW_t $$

But this is not the case, the covariance is not dependant on the parameter $\mu$ as Billy pointed out.

We'd expect the moments not to depend on location in a stationary process. The OU isn't stationary even when the parameter $\mu$ is 0, evident in Wikipedia's solution, given below. This in fact, tends to the RW solution:

$$Cov(X_t, X_s) = \frac{\sigma^2}{2\theta} \left( e^{-\theta|t-s|} - e^{-\theta(t+s)} \right) \rightarrow \frac{\sigma^2}{2\theta} e^{-\theta|t-s|} $$

... as t and/or s increase. So, as the process gets farther and farther from its initial condition, it tends to a stationary process.

I can only speculate that they did not include the full non-stationary solution and provided the limiting case or perhaps the method of obtaining the covariance from the power spectrum assumes stationarity perhaps, I do not know.


As for the proof of the the OU process being a GP, it's possible to solve the Fokker-Planck equation - you end up with the normal density (see the Wikipedia link). Hence, any at any finite set of points $\bf t$, $X$ will have a multivariate normal distribution, which is the definition of a GP.

$\endgroup$
6
  • 2
    $\begingroup$ Your guess is correct, if you solve the equation, you end up with a normal distribution. And then you find the kernel function by determining $cov(x_i, x_j)$, as they also do on the wikipedia page. You now have a multivariate normal distribution, and a function to determine the covariance between any two variables thereof. Gaussian Process. $\endgroup$
    – dwcoder
    Sep 21, 2018 at 6:09
  • $\begingroup$ But when you compute the covariance, you're actually subtracting off the mean deterministic drift - so I can't see why this makes any difference. I'm going through the calculation again and still winding up with a nonstationary process at the end, even if I assume x0 = 0 and mu = 0 - in fact, precisely the same covariance. $\endgroup$ Sep 21, 2018 at 14:06
  • 1
    $\begingroup$ It appears that you are right. The kernel mentioned in R&W seems to be some sort of a steady-state solution, so if either of $s$ or $t$ go to infinity, the wikipedia kernel tends to the R&W kernel. I will update my answer. $\endgroup$
    – adityar
    Sep 21, 2018 at 15:13
  • $\begingroup$ Ah, that makes sense! I feel like they could put more emphasis on this when discussing GPs - perhaps a little note when discussing the process with stationary covariance that this isn't exactly the OU process, which is more commonly defined as the solution to the stochastic differential equation above, but rather a limit of the process as the index parameter goes to infinity. So intuitively, the "actual" OU process is a sum of the stationary exponential kernel, plus an exponential kernel that just penalizes parameters with large norm. $\endgroup$ Sep 22, 2018 at 14:40
  • 1
    $\begingroup$ Yeah, true. I think that the extra term penalises how far you are (in a time sense) from the initial starting point, not necessarily the parameters per say. So, for example, you might start the OU process with an initial condition of like 100 when the mean is zero, so the initial observations will be correlated with the initial condition (this is when the t+s are small) but once you're far enough away, i.e. t+s large, the initial condition has no bearing on the process, and the process behaves as a stationary one $\endgroup$
    – adityar
    Sep 22, 2018 at 15:36

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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