I've read the following paper of Gatheral and Jacquier https://arxiv.org/pdf/1204.0646.pdf about volatility surfaces. I'm thinking about the SSVI surface. Is there any motivation why they choose exactly this parameterization? Moreover I had a look at this https://www.imperial.ac.uk/media/imperial-college/research-centres-and-groups/cfm-imperial-institute-of-quantitative-finance/events/distinguished-lectures/Gatheral-2nd-Lecture.pdf. On page 17 he writes that: This smile is completely defined given three observables. The ATM volatility and ATM skew are obvious choices for two of them. The most obvious choice for the third observable in equity markets would be the asymptotic slope for negative and in FX markets and interest rate markets, perhaps the ATM curvature of the smile might be more appropriate. Does these mean I need this data for calibrating SSVI? And then when Implementing he does this in a quite complicated way, it seems to me. Is there an easier method? And if I fix $\theta=\theta_t$ and plot $\omega(k,\theta)$ I should have a smile? Am I right? Because I only get a line. For $\phi$ I used the heston like function. Can maybe anybody tell me values for $\rho$ and $\lambda$ to get a curve, so I can see this is possible with this parameterization. Thanks in advance!

Update Something with my update doesn't work. So I have given these data: k=[ -0.0193, -0.0070, 0.0040, 0.0150, 0.0260] implied volatility= [0.0366, 0.0331, 0.0320, 0.0329, 0.0344] t=0.25. So total implied variance=0.25*(implied volatility)^2. By interpolation I can find $\theta_t=3.0000e-04$ Now my program gives me $\rho=-0.9993, \lambda=3.8324$. If I plot this it looks like the file attached. Can anybody tell me where I made a mistake?enter image description here


1 Answer 1


In short no you don't need to estimate the values for these quantities to calibrate the model. Instead what you want to do is to perform a least squares optimization on the model parameters against market volatilities, this gives you the calibrated surface to the market prices and if you want to output the model quantities you mentioned then you can.

So you want to minimise the error between

$$\sigma_{model} = \sqrt{w(k,\theta_t;t)/t}$$ and $$\sigma_{market}(k,t)$$ as

$$\sum_{k, t} |\sigma_{model}(k, \theta_t;t) - \sigma_{market}(k,t)|^2 < \epsilon$$

for some $\epsilon >0$. You will either have quotes from the market in terms of prices or volatilites, if you have prices then you must use a numerical method such as Newton-Raphson to obtain the market volatilities. During the optimisation procedure you will need to enforce the constraints mentioned in the paper.

  • $\begingroup$ I tried to do this. But instead of a curve I only get a line. My try now was to begin with minimizing the market volatilites agains the model volatilities. Then I used the calibrated parameters to determine the jumpwing parametrization and this lead to an better approximation, but its quite complicated. And is my understanding right, that $\omega(k,\theta_t)$ and the raw parametrization $\omega(k,\chi_R)$ or the jumpwing parametrization describe the same smile? $\endgroup$
    – P.G.
    Jan 7, 2019 at 22:41
  • $\begingroup$ I’m a little unsure how you get a flat smile, could you post some plots? You can just work directly with the Heston parametrisation in the SSVI rather than in one of the reparametrisations of the SVI model. Your $\theta_t$ Should be the total ATM BS variance right? This should be an input rather than something calibrated to. $\endgroup$ Jan 8, 2019 at 20:03
  • $\begingroup$ I have updatet my question and attached a plot $\endgroup$
    – P.G.
    Jan 8, 2019 at 23:20
  • $\begingroup$ I noticed when you posted the update on my post you had something like $1.0e-3 * 0.26...$ type "total volatilities" but your new update on your post has different numbers, the original ones seemed more like equity style implied vols at level around 0.26 but I was unsure where the other multiplicative factor came from. Because your $\theta_t$ is so low, your optimiser is finding a solution with $~\rho=-1.0$ which gives a straight line. $\endgroup$ Jan 9, 2019 at 11:07
  • $\begingroup$ As an example if I interpolate to find the implied vol corresponding to $k=0$ as $\sigma_{ATM}=0.26$, I would expect my total ATM variance $\theta_{0.25}=0.25*0.26*0.26=0.0169$. If you try plotting with sensible params ($\rho=-0.7, \lambda = 0.5$), you should get a nice smile. Let me know if this solves the problem. $\endgroup$ Jan 9, 2019 at 11:11

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