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))
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.