How can I estimate the Ornstein-Uhlenbeck paramters of some mean reverting data that I have on R?

I have mean reverting data (Difference of 2 stock prices, that I want to do pairs trading on). I want to simulate my own mean reverting data as similar as possible to the real data that I have.

The approach that I want to take is Least Squares Regression. Maximum Likelihood is too complicated.

Thank you all.

Similar to Juan Gil's answer but a bit differently I would say the following based on this:

The OU process $$dX_t = \kappa(\theta-X_t)dt + \sigma dW_t$$ can be (Euler-Maryuama discretization) discretized at times $n \Delta t,n=1,\ldots,\infty$ which gives with $t = k \Delta t$ $$X_{k+1} - X_k = \kappa \theta \Delta t -\kappa X_k \Delta t + \sigma (W_{k+1} - W_k),$$ rearranging and setting $\sigma (W_{k+1} - W_k) = \sigma \sqrt{\Delta t} \epsilon_k$ we get: $$X_{k+1} = \kappa \theta \Delta t - (\kappa \Delta t - 1) X_k + \sigma \sqrt{\Delta t} \epsilon_k.$$ So you can model an AR(1) process and then identify the parameters using the equation above.

Thinking about it again one can probably leave $X_{k+1} - X_k$ on the lhs and then one simply does a regression but I don't know exactly about the error terms in this case.

I have found this with R code, there an MLE approach is used. You find various solutions in this Stack Overflow question.

For a Ornstein-Uhlenbeck process, the maximum likelihood parameters are the ones from least squares regression.

If your process is:

$$dX=\kappa (\theta-X)dt+\sigma dW$$

you can do a linear regression in the form

$$\frac{dX}{dt}=a+bX+\epsilon$$

So your parameters will be:

$$\kappa=-b$$

$$\theta=-\frac{a}{b}$$

$$\sigma=std(\epsilon dt)$$

• After reading Richard's answer I've done a small edition in mine: each $\epsilon$ has to be multiplied by its $dt$ (especially if not all the $dt$ in your data are equal) to get the right volatility (the result would be in the units you are measuring $dt$) – Juan Ignacio Gil Apr 1 '15 at 15:36
• I know this is old, but doesn't Juan mean to divide each $\epsilon$ by $dt$, not times? Since each error is $N(0,\sigma^2 dt)$ we should be correcting for this by division, surely? – Patty Jul 24 '17 at 21:59