Hello Every I am just wondering to know if I want to calibrate Heston model what kind of market price I should apply. i.e. Options based on S&P 500 ? And where I can get these data. Many thanks Max

To what market data you want to calibrate will depend on the optionsyou are planning to price. If those options (maybe a bunch of non-liquidexotic options) happen to have S&P500 as the underlying, then you willindeed want to calibrate the model to liquid European vanilla options onS&P500.I’m not an expert on where to get data. I want to suggest, however, anarticle by Schoutens et al. called “A Perfect Calibration! Now What?”. Youwill find this article on scholar.google.com.In this article the authors give an example of market data. They alsocalibrate the Heston model to that market data, and show the resultingfive parameters they get (without, however, mentioning what methodthey used to calibrate the model).I was pleased to find out that by using a global minimization methodcalled Differential Evolution, I was able to get exactly the same parametersas they did (and even better, if I let the algorithm run long enough).What always bothered me, though, was that those parametersturned out to land “infinitely” close to the boundary of the surfacesigma^2 > 2*kappa*theta. This is certainly not a good propertyof a model, is it? I mean, this inequality is supposed to prevent zerovolatilities, thus preventing the impossible. The minimum point shouldtherefore not lie on this boundary, I believe. But I still got the same resultsas the authors of the abovementioned article did, which makes me believethat both me and the authors are making a serious error somewhere.Can someone shed a light on what the issue might be?

What can I do if I only have american options for calibration?Europan option can be computed with analytical formulas for the Heston model, but what can I do when I have only american ones?

QuoteOriginally posted by: skyrmionWhat can I do if I only have american options for calibration?Europan option can be computed with analytical formulas for the Heston model, but what can I do when I have only american ones?Then you would have to, I guess, price each individual optionusing Monte Carlo or finite difference methods, instead of theanalytical formula. Hence, the calibration process would takemuch longer.

Thank you Y0da! I have PM you my specific questions. Do you know how to guess the initial values for these five parameters? Thanks

QuoteOriginally posted by: Y0daWhat always bothered me, though, was that those parametersturned out to land “infinitely” close to the boundary of the surfacesigma^2 > 2*kappa*theta. This is certainly not a good propertyof a model, is it? I mean, this inequality is supposed to prevent zerovolatilities, thus preventing the impossible. The minimum point shouldtherefore not lie on this boundary, I believe. But I still got the same resultsas the authors of the abovementioned article did, which makes me believethat both me and the authors are making a serious error somewhere.Can someone shed a light on what the issue might be?Observing zero values of the instanteneous variance is not a bad property of the model if you are not confusing the instanteneous variance with the realized variance that mostly determines the price of vanilla options. On short time scale, the instanteneous variance can be approximated by the realized variance realized during small time interval (i-1, i): v(i)=(ln[(S(i)/S(i-1)])^2/AnnualizedTime(i-1,i). I don't see any problems if the latter quantity turns out to be close to zero, you might also inspect the historical time series to conform that small values of v(i) are not that impossible. By calibrating the Heston model to SPX options, the Feller condition you stated is never satisfied.Unfortunately, the problem with the Heston model is deep indeed. When the Feller condition is not satisfied you have to respect the boundary for v(i) at zero which is very problematic with PDE and MC methods, and makes the Heston model very hard to deal with numerically. Also there is a lot of problems with evaluation of "analytical" integrals.Fortunately, there is a stochastic volatility model which is much more robust than Heston model and which is relatively easy to deal with numerically.

QuoteFortunately, there is a stochastic volatility model which is much more robust than Heston model and which is relatively easy to deal with numerically.and that would be...?

QuoteOriginally posted by: davidrisingQuoteFortunately, there is a stochastic volatility model which is much more robust than Heston model and which is relatively easy to deal with numerically.and that would be...?Hehe exactly. And that would be what model?

NIG? VG?

QuoteOriginally posted by: AVtNIG? VG?Hardly a Stoch Vol Model....