3
$\begingroup$

I have noticed when reading (many) articles about Heston Calibration that not all (few actually) do care about the Feller condition. Below is a compilation of calibration results from some different authors (source):

enter image description here

If we calculate the Feller condition $2 \kappa \theta - \sigma^2$ for the above we can see that in all cases it is extremely close to 0. For row 1, 3 and 6 it is in fact negative.

So the Question is basically: In my own calibration code, should I use the Feller condition as a calibration constraint (e.g. as a penalty function) or should I skip the constraint since it doesn't always hold in the market?

Looking forward to your input!

Improve this question
$\endgroup$

2 Answers 2

Reset to default
5
$\begingroup$

You should not use the Feller condition as a constraint. In many cases its violation will be required for a good fit to the market data.

Improve this answer
$\endgroup$
2
  • $\begingroup$ Thanks for the comment @q.t.f. This is the kind of information I'm looking for. Have you read that in an article or do you know it from experience? Can it differ from market to market? I have seen a few articles incorporating the Feller condition as a constraint, and some do not. $\endgroup$
    – DoubleTrouble
    Jan 28, 2016 at 23:11
  • $\begingroup$ Recall Feller condition is a sufficient condition, but not necessary for positivity. $\endgroup$
    – Diego F Medina
    May 19, 2019 at 18:27
3
$\begingroup$

As q.t.f stated, you shouldn't pay too much attention to the Feller Condition since it is often violated in the Heston model, especially for options with more than a few weeks until maturity. However, you should make sure that your Characteristic Function stays continuous, else you'll end up with "wrong" prices. This is caused by the branch cut along the negativ real axis and can be avoided through a simple transformation. There are several formulations out there that prevent that from happening, a good and simple start is Albrecher et al. (2006).

Improve this answer
$\endgroup$
3
  • 1
    $\begingroup$ Thanks for the comment @Dede I'm using the characteristic function from Cui et al (2016) since it's much easier to differentiate. They also deal with the discontinuity problem! $\endgroup$
    – DoubleTrouble
    Sep 25, 2016 at 21:56
  • $\begingroup$ Yes, this one works as well, read it just recently. :) Do you also derive the analytical gradient? I haven't tried that yet, but their results sound promising. $\endgroup$
    – Andreas
    Sep 26, 2016 at 7:32
  • 1
    $\begingroup$ They derive the analytical gradient in the paper and I've tried implementing it. From a performance perspective it's amazing! $\endgroup$
    – DoubleTrouble
    Sep 26, 2016 at 8:17

Your Answer

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

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