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

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}