Vrftjkry

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:
$$dx=theta(bar{x}-x)dt+sigma dW tag{O-U}$$
$$x_{t+1}=thetabar{x}+(1-theta)x_t +sigmaepsilon_ttag{AR(1)}$$
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
$$dln z_t=-(1-rho) ln z_t +sigma sqrt{(1-rho^2)}dW_t$$
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

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

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

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

Thus, we can write (*) as

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

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

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

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

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.