# How the Ornstein–Uhlenbeck process can be considered as the continuous-time analogue of the discrete-time AR(1) process?

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!

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 $(k\Delta t)_{k\in \Bbb N_0}$.

• Very clear ... good answer. – Richard Oct 3 '14 at 8:09
• 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? – Richard Apr 1 '15 at 13:15
• @Richard seems to be the case – Ilya Apr 1 '15 at 17:57
• 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? – Bazman Sep 2 '15 at 15:25
• Amazing Answer thanks! :) – AIM_BLB Jun 17 '16 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

• Could you please suggest an article from where your answer is taken. Thanks! – Shanks Jan 4 at 14:45
• This is a great answer as it indicates how to discretise the continuous process properly so that it has the same stationary distribution. – James Griffin Mar 31 at 9:18

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.

• 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 – gota Oct 23 '15 at 14:38