Wikipedia says

The Ornstein–Uhlenbeck process can also be considered as the continuous-time analogue of the discrete-time AR(1) process.

I was wondering how the Ornstein–Uhlenbeck process can be considered as the continuous-time analogue of the discrete-time AR(1) process?

An Ornstein–Uhlenbeck process, $x_t$, satisfies the following stochastic differential equation: $$ dx_t = \theta (\mu-x_t)\,dt + \sigma\, dW_t $$ where $\theta > 0, \mu$ and $\sigma > 0$ are parameters and $W_t$ denotes the Wiener process.

The $AR(p)$ model, i.e. an autoregressive model of order $p$, is defined as $$ X_t = c + \sum_{i=1}^p \varphi_i X_{t-i}+ \varepsilon_t \, $$ where $\varphi_1, \ldots, \varphi_p$ are the parameters of the model, $c$ is a constant, and $\varepsilon_t$ is white noise.

Thanks and regards!


3 Answers 3


In case $p=1$ you have $$ x_{k+1} = c+a x_k + b\varepsilon_k $$ so that if you put $c = \theta\mu\Delta t$, $a = -\theta\Delta t$ and $b = \sigma\sqrt{\Delta t}$ you get $$ x_{k+1} = \theta(\mu - x_k)\Delta t + \sigma \varepsilon_k\sqrt{\Delta t} $$ which is exactly an Euler-Maryuama discretization of OU at times $(k\Delta t)_{k\in \Bbb N_0}$.

  • $\begingroup$ Very clear ... good answer. $\endgroup$
    – Richi W
    Oct 3, 2014 at 8:09
  • 3
    $\begingroup$ I remembered your answer and posted it a bit differently answering another question here. Don't you need the approximation $dX_t \approx X_{k+1}-X_k$ on the left hand side and then you put the $X_k$ on the rhs and get another paramter as the AR coefficient? $\endgroup$
    – Richi W
    Apr 1, 2015 at 13:15
  • $\begingroup$ @Richard seems to be the case $\endgroup$
    – Ilya
    Apr 1, 2015 at 17:57
  • 1
    $\begingroup$ I am a little confused by the interpretation of the result above. If a=-theta*Delta t then theta=-a/(Delta t). In this case if you have a positive AR parameter such as 0.9 it will give rise to a negative theta say -0.9 (if Delta t=1 Delta t will always be positive). In this case if you get an observation above the mean (mu-x_k) will be negative and this value will be multiplied by a negative value meaning that the drift term will add to an observation that is already above the mean? Is this correct? $\endgroup$
    – Bazman
    Sep 2, 2015 at 15:25
  • $\begingroup$ Amazing Answer thanks! :) $\endgroup$
    – AIM_BLB
    Jun 17, 2016 at 15:26

Rather than using an approximation (like Euler-Maryuama), one can just sample* the continuous-time solution:

$$x(t)=x(0)\,e^{-\theta t}+\mu \,(1-e^{-\theta t})+\sigma \int _{0}^{t}e^{-\theta (t-s)}\,dW_{s}.\, $$

Writing $X_k=x(k\,\Delta t)$, the sampled process is described by the AR(1) process: $$ X_{k+1} = c + \varphi \,X_k + \varepsilon_k $$ where: \begin{align} X_0&=x(0),\\ \varphi &= e^{-\theta \Delta t},\\ c &= (1-\varphi)\mu, \\ \epsilon_k &\sim \mathcal{N}\Big(0,\frac{1}{2\theta}\sigma^2\big(1-e^{-2 \theta \Delta t}\big)\Big). \end{align}

* I mean sample, not sample

  • 2
    $\begingroup$ Could you please suggest an article from where your answer is taken. Thanks! $\endgroup$
    – Shanks
    Jan 4, 2019 at 14:45
  • $\begingroup$ This is a great answer as it indicates how to discretise the continuous process properly so that it has the same stationary distribution. $\endgroup$ Mar 31, 2019 at 9:18
  • $\begingroup$ This works if one has a known close form solution for the SDE of the continuous-time process; but in most cases one does not. So, how do you "sample" then? $\endgroup$
    – Confounded
    Jun 2, 2020 at 13:01
  • 2
    $\begingroup$ @Confounded Sorry, I am not following you... The OP asked about the OU process, whose general form is $dx_t = \theta (\mu-x_t)\,dt + \sigma\, dW_t$. For this process, I provided the corresponding AR(1) in my answer. If you are asking about general SDEs (not OUs), then my answer does not apply but, in general, there is no corresponding discrete-time AR(1) process either. $\endgroup$
    – Luca Citi
    Jun 4, 2020 at 18:24
  • $\begingroup$ @Shanks you asked for the article reference right. This comes directly from the wikipedia article--or at least the same derivation is basically there as well, in the "formal solution" section. en.wikipedia.org/wiki/Ornstein%E2%80%93Uhlenbeck_process $\endgroup$
    – krishnab
    Dec 20, 2020 at 5:49

Should be a comment on Ilya's answer but I don't have enough reputation --

Although the other answer does show how the AR process is equivalent to the OU process, keep in mind that the Euler-Maruyama discretization is just an approximation. In order to derive the exact relationship between the two models, you would have to integrate the OU process from time t to t+1, and then derive the various relationships between the parameters. Luckily there is a closed form solution to the OU SDE, so this is not too difficult.

EDIT: you could also use the Milstein Discretization, which is a more accurate approximation because it includes the convexity term from Ito's lemma. But it is still easier than the exact solution.

  • 3
    $\begingroup$ Paul, I don't think in this case the Milstein Discretization and the Euler-Maruyama discretization yield a different result. Because the diffusion term does not have a state dependence $\endgroup$
    – gota
    Oct 23, 2015 at 14:38

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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