Converting between continuous and discrete-time stochastic processes I'm reading through Dixit and Pindyck's Investment under Uncertainty, where I found the following passage. First, they introduce the Ornstein-Uhlenbeck process
$$
 dx = \eta (x - \bar{x})dt + \sigma dz\ ,
$$
and then claim that the above equation is the "continuous-time version of the first-order autoregressive process in discrete time. Specifically, [the above equation] is the limiting case as $\Delta t \to 0$ of the following AR(1) process:
$$
 x_t - x_{t-1} = \bar{x}(1 - e^{-\eta}) + (e^{-\eta} - 1) x_{t-1} + \epsilon_t\ ,
$$
where $\epsilon_t$ is normally distributed with mean zero and standard deviation $\sigma_\epsilon$, and
$$
 \sigma_\epsilon^2 = \frac{\sigma^2}{2 \eta}(1 - e^{-2\eta})\ ."
$$
It is not at all obvious to me how they arrive at this result: would someone be kind enough to the fill in the details? My naive approach would have been to simply replace the $d$'s by $\Delta$'s, but in doing so I would have missed out on a lot.
I am also curious as to how well such a discrete-time writing of a continuous-time Ito diffusion generalises; that is, is it true that I can write down a discrete time formula for an arbitrary Ito diffusion
$$
 dx = a(x) dt + b(x) dz\ ?
$$
What would be the AR(1) version of Brownian motion
$$
 dx = \mu dt + \sigma dz\ ?
$$
 A: We start with the OU/Vasicek SDE:
$$dX_t=\alpha(\mu-X_t)dt+\sigma dW_t$$
Standard Ito calculus yields:
$$d(X_te^{\alpha t})=\alpha \mu e^{\alpha t}dt+\sigma e^{\alpha t}dW_t$$
Now we obtain the following by integrating from $t$ to $t+\Delta t$
$$\begin{aligned}&X_{t+\Delta t} e^{\alpha (t+\Delta t)}-X_te^{\alpha t}=\alpha \mu \int_t^{t+\Delta t}e^{\alpha s}ds+\sigma \int_t^{t+\Delta t}e^{\alpha s}dW_s\\
\implies &X_{t+\Delta t} e^{\alpha (t+\Delta t)}-X_te^{\alpha t}=\mu(e^{\alpha (t+\Delta t)}-e^{\alpha t})+\sigma \int_t^{t+\Delta t}e^{\alpha s}dW_s\\
\implies &X_{t+\Delta t}-X_te^{-\alpha \Delta t}=\mu(1-e^{-\alpha \Delta t})+\sigma \int_t^{t+\Delta t}e^{-\alpha(t+\Delta t- s)}dW_s\\
\implies &X_{t+\Delta t}-X_t=X_t(e^{-\alpha \Delta t}-1)+\mu(1-e^{-\alpha \Delta t})+\sigma \int_t^{t+\Delta t}e^{-\alpha(t+\Delta t- s)}dW_s\end{aligned}$$
The RHS integral is a Gaussian rv with mean zero and variance $\frac{\sigma^2}{2\alpha}(1-e^{-2\alpha \Delta t})$, as desired. Choose $\Delta t=1$ to obtain the AR(1) process.
