I have been researching volatility smoothing techniques and risk-neutral pdf. I noticed one interesting post in Does the risk neutral pdf that is derived using Litzenberger-Breeden Method correspond to gamma and it's integral correspond to delta? Kevin, the answerer of the post, pointed out a property to derive spot delta based on dual delta and spot gamma based on dual gamma.

I am wondering is there a similar way to derive vega, theta, and rho in a (almost) model-free approach? Any suggestions and comments are highly appreciated!

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    $\begingroup$ Kevin qualifies his answer by writing (in a comment) "if we knew the right Gamma". I did not fully grasp the meaning of this, and I wish I understood it better. Can someone clarify? Thanks. $\endgroup$
    – nbbo2
    Commented Jul 27, 2023 at 1:56
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    $\begingroup$ @nbbo2 Sory for the misleading statement. This surely isn't one of my best answers. Assuming homogeneity, I derived that the Breeden-Litzenberger PDF is proportional to gamma and OP asked if we could calculate the risk-neutral density from gamma. Well, yes we could, if the assumptions are true and if we knew what gamma is. But estimating gamma in a (mostly) model-free fashion essentially boils down to a butterfly spread (ie, the Breeden Litzenberger formula). So that doesn't help too much, I'm afraid. $\endgroup$
    – Kevin
    Commented Jul 27, 2023 at 4:22
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    $\begingroup$ Regarding the new question here, I'm not sure we can calculate vega in a model-free way. In fact, I don't think "vega" makes sense in a model-free sense. Like what is vega in the Heston model? The derivative wrt spot volatility? Or to both, spot volatility and the long run average volatility (ie, a parallel shift in the entire vol curve)? I think different authors give different definitions. There's somewhere a question where Frido commented on this (if I remember correctly). Given that, if the definition of vega depends on the model, I don't think it can be calculated model-free. $\endgroup$
    – Kevin
    Commented Jul 27, 2023 at 4:25
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    $\begingroup$ Theta can be estimated in a model-free sense as a finite difference if we have options with multiple expiries and same moneyness. So that should work. I'm not sure about rho again as it seems to depend on how interest rates are modelled (if at all). So the definition of rho could be model-dependent (like vega). $\endgroup$
    – Kevin
    Commented Jul 27, 2023 at 4:27
  • $\begingroup$ Although theta is nonlinear in time, meaning unless you have daily expiries, you will get a misleading answer when applying finite difference, especially for short term options. See here for some details on theta in the BSM model. $\endgroup$
    – AKdemy
    Commented Jul 27, 2023 at 6:31

2 Answers 2


No, vega cannot be derived in a model-free manner. The reason for this is because in contrast to delta and gamma, there are multiple definitions of vega, and an even deeper underlying reason may be that spot volatility is not an observable. The consequence of this is that 'vega' starts to depend on the quoting mechanism; i.e. do you express the market price of an option in terms of a local vol model, a stochastic vol model, a jump diffusion, or Black Scholes with an implied volatility? All these different quoting mechanisms will lead to a different concept and value of 'vega'.

As an example; suppose you are able to fit a pure jump model to the market price of options for a particular time to maturity. What is vega in a pure jump model? However you can (always) translate your pure jump model prices to Black-Scholes prices with implied vols that depend on strike. Then you do have vega. Hope this example makes it clearer.

Regarding theta: here too there is no model-free quantity. Recall that in SV models the theta of an option balances the gamma, vega, vanna, volga. As only gamma potentially is model-free, you cannot hope to have model-free theta.

As regards delta and gamma: Here under some circumstances (for the class of so-called homogeneous stochastic vol models) there is a model free definition, and this is (partly) because the spot price is an observable and therefore has to occur in any function you use to quote the market price by.

Last remark: as option market prices can always be expressed in terms of Black-Scholes prices with strike dependent IV, BS vega is always well-defined.

  • $\begingroup$ Thank you so much for the detailed explanations! The facts that "spot price is an observable" and "theta and volatility are not" answer my question! Much appreciated!! $\endgroup$
    – Frank
    Commented Jul 27, 2023 at 7:49

Calculating vega in a model free manner is an oxymoron.

Vega can be seen as the way to jump from one model's price to another.

For example, a model which specifies vol as 10 bps and a model which specifies vol as 11 bps have price difference equal to vega at 10 bps (roughly).

If we're saying vega is independent of models (='what vol is'), then prices must be linear in vol. Which doesn't work out because when vol is infinite, price is finite.


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