I have the AR(1) process of the following form:

$$\ln z_{t+1}=\rho \ln z_t+\sigma \sqrt{(1-\rho^2)}\epsilon_t$$

And need to find its continous time corresponding Ornstein-Uhlenbeckprocess.

I have the info that an AR(1) can be translated to an Ornstein-Uhlenbeck as follows: $$dx=\theta(\bar{x}-x)dt+\sigma dW \tag{O-U}$$ $$x_{t+1}=\theta\bar{x}+(1-\theta)x_t +\sigma\epsilon_t\tag{AR(1)}$$ but fail to see how to apply this to this specific case. I have looked at other answers considering the translation between AR(1) and OU, but again do not clearly see it for this specific form.

Any help is highly appreciated :-)

edit: Using the answer below and a linear approximation of $e^{-\theta \Delta}\approx1-\theta \Delta$ and setting $\Delta=1$ I get $$d\ln z_t=-(1-\rho) \ln z_t +\sigma \sqrt{(1-\rho^2)}dW_t$$ which corresponds to the approximation as given by (AR(1))


1 Answer 1


We can solve

$$dx=\theta(\bar{x}-x)ds+\sigma dW_s $$

by multiplying by an integrating factor $e^{\theta s}$ and integrating over $[t,t+\Delta]$ to obtain

$$e^{\theta(t+\Delta)}x(t + \Delta) - e^{\theta t}x(t) = \bar{x} ( e^{\theta(t + \Delta)}- e^{\theta t}) + \sigma\int_t^{t + \Delta} e^{\theta s}\, dW_s$$

Rearranging we get,

$$\tag{*}x(t + \Delta) =\bar{x}(1 - e^{-\theta \Delta}) + e^{-\theta \Delta}x(t) + \underbrace{\sigma e^{-\theta \Delta}\int_t^{t + \Delta} e^{\theta(s- t)}\, dW_s}_{I(t,\Delta)}$$

The moments of the stochastic integral on the RHS are

$$\mathbb{E}\left(I(t, \Delta) \right) = 0, \\ var\left(I(t, \Delta) \right)= \sigma^2 e^{-2\theta \Delta}\int_t^{t + \Delta} e^{2\theta(s- t)}\, ds = \frac{\sigma^2(1 - e^{-2\theta \Delta})}{2 \theta}$$

Thus, we can write (*) as

$$x(t + \Delta) =\bar{x}(1 - e^{-\theta \Delta}) + e^{-\theta \Delta}x(t) + \sigma\sqrt{\frac{1 - e^{-2\theta \Delta}}{2\theta}}\xi,$$

where $\xi \sim N(0,1)$ is a standard normal random variable.

Making the association with your AR(1) process we have

$$x(t+\Delta) \iff \ln z_{t+1} \quad x(t) \iff \ln z_t, \\ \bar{x} = 0, \quad \rho = e^{-\theta \Delta}, \quad \sigma\sqrt{1- \rho^2} \iff \sigma\sqrt{\frac{1 - e^{-2\theta \Delta}}{2\theta}}$$

So the continuous time process is

$$d \ln z_t = \frac{\ln \rho}{\Delta}\ln z_t \,dt + \sigma \sqrt{\frac{-\ln \rho}{\Delta}} \, dW_t$$

where $\Delta$ is the time interval between sampled observations.

  • $\begingroup$ just for notational consistency, how would you get rid of $\Delta$ so that I only have "dt" terms? $\endgroup$
    – user469216
    Commented Nov 14, 2018 at 22:08
  • $\begingroup$ It's just a matter of using a dimensionless time variable $\hat{t} = t/\Delta$ $\endgroup$
    – RRL
    Commented Nov 14, 2018 at 22:11
  • 1
    $\begingroup$ Or I could have integrated the SDE over the interval $[t,t+1]$ and it goes away. $\endgroup$
    – RRL
    Commented Nov 14, 2018 at 22:12
  • 1
    $\begingroup$ So I left out the $dt$ in the answer by mistake -- I now added it.Clearly you can see how $\Delta$ can be absorbed into the time variable. Also the Wiener process is $\mathcal{O}(dt)$ which is why the $\sqrt{\Delta}$ shows up as a coefficient. Short answer -- just set $\Delta = 1$. $\endgroup$
    – RRL
    Commented Nov 14, 2018 at 22:16
  • 1
    $\begingroup$ The OU and AR(1) you wrote with the same parameters $\theta$ and $\sigma$ appearing in both is only correct if the spacing between discrete steps is the same as the time unit for $t$, e.g. 1 year. $\endgroup$
    – RRL
    Commented Nov 14, 2018 at 23:25

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