I would like to estimate Ornstein–Uhlenbeck process' parameters via Kalman filter.

My process is the following one:


I'm interested in CIR process, too:

$\text{d}x_{t}=\alpha(\theta-x_{t})\text{d}t+\sigma x_{t}^{\beta}\text{d}W_{t}$

and my goal is to find the values of $\alpha$, $\theta$ and $\beta$ using Kalman filter over a state-space representation of the process.

How may I describe such a process in a form suitable to state-space representation and Kalman filter?

  • 1
    $\begingroup$ I tried to edit this for you but it hasn't gone through. there is no such term $x_{t}^{\beta}$ in a classical OU process. what you have up there is more like a CIR process. you say don't mind the diffusion though, so whatever. $\endgroup$
    – Vince
    May 11, 2013 at 19:39
  • $\begingroup$ CIR, you're right. $\endgroup$
    – Lisa Ann
    May 11, 2013 at 19:41
  • 1
    $\begingroup$ @Veeken I rejected your edit because changing an equation can fundamentally alter the question. If you suspect a formula is wrong, the best bet is to leave a comment as you've done here. All the same, thank you for pointing-out a suspicious post. $\endgroup$ May 11, 2013 at 21:17
  • $\begingroup$ @chrisaycock No problem with his correction, that was just my oversight to mix up UO with CIR formula. By the way, I decided to include both in my question :) $\endgroup$
    – Lisa Ann
    May 11, 2013 at 21:48
  • $\begingroup$ Well, you can find $\beta$ and $\sigma$ by using the quadratic variation of a process - no filtering is needed in such a case. W.r.t. $\alpha$ and $\theta$ you can use the UKF as Veeken suggested, or perhaps some particle filters. I am not an expert on filtering, but I'm pretty sure that both methods have their own advantages in your case. $\endgroup$
    – SBF
    May 21, 2013 at 11:52

2 Answers 2


This book goes through exactly this problem in quite detail (with C++ codes included). I've worked through it in the past, but can't sum it up off the top of my head.


this question can be quite straightforward or gnarly depending on whether you can observe measurements of $x_t$ directly or not. in the latter case, in general it will become a nonlinear system, and will require application of the extended kalman filter or its improvement, the unscented kalman filter.

On Edit: Now that the bounty has expired, let me answer by way of supplying this key reference (which i found as a result of this question*): "Estimating and Testing Exponential-Affine Term Structure Models by Kalman Filter" by Jin-Chuan Duan, Jean-Guy Simonato. Pg 13 of the paper gives the answer for the Vasicek model, page 15 for the CIR.

*Because of the results of Duffie and Kan (93), both these models lead to pricing equations for zero coupon bonds that are affine in the short rate, xt. As a result, one can use a linear Kalman filter to solve this, using zero coupon rate changes as the input. I originally thought one would definitely need the UKF.

  • 3
    $\begingroup$ I'm purging the comments and putting an end to this. QuantSE is about helping out people on things you know, and benefiting from other on things you don't know. There is no place here for negotiating reputation points. Had you given your answer, 5 upvotes would have given you the 50 points of the bounty. If you want additional points, ask good questions or solve unanswered questions. In general guys, comments boxes are no courtrooms. Please for this site's sake, stop arguing in comments, do it on the chat rooms if you really need to argue about something important. $\endgroup$
    – SRKX
    May 12, 2013 at 22:49
  • $\begingroup$ whatever happened to laissez faire? a friendly one liner to ask for a 'mini-bounty' that was heartily accepted by the OP is hardly a negotiation. I explained very well why i thought it was no big deal, but now that that has been purged, i have lost interest in posting here. I am sorry for any troubles, and in particular to those who have something against a guy who only wants to help out but have some fun while doing so. $\endgroup$
    – Vince
    May 13, 2013 at 3:29
  • $\begingroup$ again, i am fine with the rules, but when they are expressed to me in the way that was carried out, i see no reason to believe something equally distasteful won't be brought on to me in future while posting here. Best Regards and Wishes...ciao $\endgroup$
    – Vince
    May 13, 2013 at 3:30

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.