3

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 $$ dx = \mu dt + \sigma dz\ ? $$

Anthony
  • 833
  • 5
  • 17

1 Answers1

3

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.

Snoop
  • 18,347
  • Thanks for the answer! Would you care to comment on the other two questions? Namely, is the above a "trick" that happens to work for this process, or can something similar be done for an arbitrary stochastic process? Also, for a Brownian motion with drift, is it true that the discrete-time version is simply $\Delta x_t = \epsilon_t$, where $\epsilon_t \sim N(\mu, \sigma^2)$? – Anthony Jan 06 '23 at 20:00
  • 1
    @Anthony arguments like the above apply when we can isolate terms nicely, especially elements like the rhs integral term, which becomes a 'nice' error term. This is the exception rather than the norm. For a Brownian motion with drift, we obtain $X_{t+\Delta t}-X_t=\mu\Delta t + \sigma (W_{t+\Delta t}-W_t)$. If you set $\Delta t =1$, you obtain the claim. – Snoop Jan 09 '23 at 05:43