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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!

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2 Answers 2

up vote 8 down vote accepted

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 exactly an Euler-Maryuama discretization of OU at times $(k\Delta t)_{k\in \Bbb N_0}$.

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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 at 13:15
@Richard seems to be the case – Ilya Apr 1 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 at 15:25

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.

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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 – Nuno Calaim Oct 23 at 14:38

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