QuoteOriginally posted by: QuantupletCuch, not sure about the last identity, what about if [$]\rho=0[$], to me the relationship [$]\eta(1+\vert\rho\vert) \le 2[$] then implies [$]\eta \le 2[$].Yyang, I tried running an SSVI calibration on your data this morning. Because the [$]w(k,t)[$] numbers were so small (I guess), the fit was horrible. I therefore introduced an offset of +0.5 (without loss of generality) to all your [$]w(k,t)[$] figures, ran the calibration routine on this modified data set, and shifted the results back of -0.5 to obtain the fit to the original data. That way, the fit is very decent!My final parameters (for the +0.5 shifted data) are : [$] \theta = 0.5002 [$][$] \rho = -7.0294e-4 [$][$] \eta = 0.8150 [$][$] \gamma = 0.5901 [$]Take these parameters and shift the resulting SSVI total variance smile of -0.5 to see how it fits the original data.(Seems like I was not too far off with the initial guess of [$]\rho_0=0, \eta_0=0.5[$] and [$]\gamma_0=0.5[$] that I recommended you yesterday!).Still, I should investigate further this issue, is it machine precision related or SSVI that does not behave well when [$]\theta \rightarrow 0[$] or simply something I did wrong in my implementation?PS: I did not manage to attach an image of the fit. I can add it if you find it useful... and if you explain me how to!Here's something interesting. For Raw SVI and Natural SVI, the best approach for now is scaling moneyness and total implied variance to an appropriate level(my scaler is [$]\frac{2}{sigma_x}[$]) and just use a tolerance of [$]1e-6[$]; while for SSVI, the best approach for now is shifting total implied variance by an appropriate amount(using Quantuplet's 0.5, and using his p0 guesses) and use a lower tolerance(my [$]tol = 1e-20[$]).Here's my optimal params for SSVI:[$]\theta = 0.50020759 \\\rho = -0.03887581 \\\eta = 0.72415426 \\\gamma = 0.32759533 \\[$]which is very close to Quantuplet's results

QuoteOriginally posted by: AntonioQuantuplet, thank you for your comment.Indeed, the power law seems to have nicer properties.As far as I have seen, though, if you take the SPX, for maturities not too small, the Heston function phi is also pretty good.PS: I know the remark on Page 14, as I am the other author.....Nice to meet you! I now realise that my comment must have sounded pretty stupid :)Truth is, I did not push my tests very far. As a matter of fact, I simply observed that on a several occasions the power law gave incredibly good results while those of the heston-like function were decent but not great. There might be a bug on my side though, never investigated that, since I already found an excellent candidat. But as you point out, i must say that even the power law function is too rigid when you look at very short dated options.In practice I tend to use ssvi for single names / "illiquid" underlyings for which the smiles are not too convex and the bid/ask spreads are "large". When it comes to indices with small bid/ask spreads I instead use it to obtain an initial guess (arb free and smooth), and then fine tune the fit of each smile using svi (plus penalising for non smoothness and arb, which is more of an art than a science if you ask me).

QuoteOriginally posted by: AntonioQuantuplet, thank you for your comment.Indeed, the power law seems to have nicer properties.As far as I have seen, though, if you take the SPX, for maturities not too small, the Heston function phi is also pretty good.PS: I know the remark on Page 14, as I am the other author.....loooooooooooooool, hi author, I chose the the modified power law parameterization because in Remark 4.4, you stated clearly the non static arbitrage condition, while heston like parameterization only stated the restriction that satisfies condition 3 in corollary 4.1.

QuoteOriginally posted by: AntonioQuantuplet, thank you for your comment.Indeed, the power law seems to have nicer properties.As far as I have seen, though, if you take the SPX, for maturities not too small, the Heston function phi is also pretty good.PS: I know the remark on Page 14, as I am the other author..... Now I understand that the heston like function is in range[$](0, \frac{1}{2})[$], so condition 4 in corollary 4.1 stands once condition 3 stands.Another thing I would like to know is that, in you mind, what kind of vol is SSVI most suitable for? The more noisy live vol surface or the more stable settle vol surface? Also from my implementation, the heston like parameterization's results depends a lot on the shift amount on total implied variance. I have tested shifts from 0 to 200, and the model performance increase monotonically. What's perplexing to me is that [$]\mu_{TIV}[$] determines optimzation result, not [$]\sigma_{TIV}[$]. Magnifying [$]\sigma_{TIV}[$] helped, but over-magnifying destroys the optimization result. So this lead me wondering, how dependent are the 'appropreiate' [$]{\mu_{TIV}}_{shift}[$] and [$]{\sigma_{TIV}}_{mult}[$] on data? Does this mean it's possible we have to adjust [$]{\mu_{TIV}}_{shift}[$] and [$]{\sigma_{TIV}}_{mult}[$] every time encountering new data? In other words, this cannot be fully automated?

QuoteOriginally posted by: yyangQuoteOriginally posted by: CuchulainnQuoteOriginally posted by: yyangQuoteOriginally posted by: AntonioDear Cuchulainn, Quantuplet,Jim G. does indeed use R.I personally use Python/IPython, with the L-BFGS-B constrained optimisation algorithm (http://docs.scipy.org/doc/scipy-0.15.1/ ... fgs_b.html).So far, it has been working pretty nicely.Best,Doesn't L-BFGS-B only take bounds? How would u deal with [$]\eta (1+|\rho|) \leq 2[$]? as Quantuplet indicatedFirst thoughts,Looking at DE we use a population of possible vector solutions and we find the fittest one. DE finds the global minimum for _any_ seed. These are 2 very attractive properties.Quotethe python implementation of DE only accepts constant boundaries, and does not handle variable dependent boundaries. DE uses box constraints. Intuitively, a nonlinear constraint as above could be converted to one in a box. Since DE is robust, rough-and-ready boxes are probably OK I reckon. QuoteI did try DE once, but not for such kinds of problems, how has it been working for you ?I di it for the usual tests and implied Vol. It has also been done for Heston etc.Would like to do it for SSVI.//BTW it can be use to solve NL systems f(x) = 0 by converting it to a least-squares problem which is kind of cool.[$]\eta (1+|\rho|) \leq 2[$] ==> [$]\eta \leq 1[$]??For your last question, [$]\eta (1+|\rho|) \leq 2[$] ==> [$]\eta \leq \frac{2}{1 + |\rho|}[$]?? Quantuplet gave an example.Correct; I modify my formula indeed [$]\eta \leq 2[$] (thx Qt as well)

QuoteOriginally posted by: QuantupletPS: I did not manage to attach an image of the fit. I can add it if you find it useful... and if you explain me how to!If you mean .. you have to put image in a zip file and download/attach zip file.

Thanks Cuch, I'll keep that in mind for next time

QuoteOriginally posted by: CuchulainnQuoteOriginally posted by: QuantupletPS: I did not manage to attach an image of the fit. I can add it if you find it useful... and if you explain me how to!If you mean .. you have to put image in a zip file and download/attach zip file.lol, noted

QuoteOriginally posted by: yyangQuoteOriginally posted by: CuchulainnQuoteOriginally posted by: QuantupletPS: I did not manage to attach an image of the fit. I can add it if you find it useful... and if you explain me how to!If you mean .. you have to put image in a zip file and download/attach zip file.lol, notedYeah, this is a hi-tech skill know to a few Wilmotters