Well, question is in the title. Assume I have two different models (for example a local volatility model and a stochastic volatility model such as a Dupire model and a SABR model for example) and I am looking at European options only. Let's assume that both models have a parameter set in which they produce the exact same volatility smile, e.g. for a give strike, they agree on a European option price. For simplicity, let's really focus on EOs here.

Does that also imply their sensitivities coincide? I know that a LV delta and for example a SABR delta can both differ for an option but I can't really find much information about whether that comes with a different surface (so there is SOME dynamic which would be different) or not.

My thinking so far tells me that no, they must be equal. Assume I take a FD estimator of any greek. As the prices produced from both models coincide, both are also the same and thus in the limit the same quantity.

I'm currently thinking about the greeks of straddles or strangles or other option strategies. For those, I have option prices quoted and I also have different volatilities per component. I would technically just add their BS greeks naively if I wanted to have the "overall delta exposure" of the product but I know that of course skew plays a role.

Am I overthinking this? Would you in practice really go an say "here, these three models fit the market perfectly and all their greeks here are slightly different so pick what's interesting for you" if they were in fact different?

  • 4
    $\begingroup$ To each stoch vol model there is a local vol model with the same terminal distribution. So the vanilla surface will be the same, but the stoch vol model greeks will be different than its local vol counterpart. The difference comes from the implied (forward) dynamics which will be different in the two models. $\endgroup$
    – user34971
    Jul 24, 2022 at 13:39
  • $\begingroup$ What is the right choice then if both models fit perfectly and I only have "terminal value exposure" (e.g. european-style options)? That would also imply different hedges. Also, do you know where I can find a proof to that statement? Really curious. $\endgroup$
    – freistil90
    Jul 24, 2022 at 14:01

1 Answer 1


Too long for a comment, so:

As mentioned above, and as I think you already know, to each SV model there corresponds a LV model with the same marginals. Both will therefore generate the same vanilla implied volatility surface.

The question is will the sensitivities be the same under the LV and SV models. In general the answer is no, and you asked is there a proof of this statement.

I have not seen a general proof of the statement, but I'll give you a plausibility argument restricted to so-called SV homogeneous models (which includes Heston, SABR with beta=1, and other popular SV models)

Many stochastic vol models are homogeneous of degree 1 in spot and strike. Meaning, if $C_{SV}(S_t,K,T)$ is the stochastic volatility vanilla option price, then for $\lambda \in \mathbb R$, $$ C_{SV}(\lambda S_t,\lambda K,T) = \lambda C_{SV} (S_t, K,T) $$

The consequence of this is, by differentiating both sides wrt $\lambda$ and setting $\lambda = 1$, $$ C_{SV} (S_t, K,T) = S \partial_S C_{SV} (S_t, K,T) + K \partial_K C_{SV} (S_t, K,T) $$ Let $C_{BS} (S_t,K,I(K),T)$ be the Black-Scholes call price formula such that $$ C_{BS} (S_t,K,I(K),T) = C_{SV} (S_t, K,T) $$ then you see that for homogeneous models the delta of the option can be read off the smile (i.e. `model-independent'): \begin{align} S \partial_S C_{SV} (S_t, K,T) &= C_{BS} (S_t,K,I,T) - K \partial_K C_{SV} (S_t, K,T) \\ &= C_{BS} (S_t,K,I,T) - K \left\{ \partial_K C_{BS} (S_t,I(K),K,T) \right. \\ &\quad \left. + \partial_K I(K) \partial_I C_{BS} (S_t,I(K),K,T) \right\} \end{align}

Local volatility models are not homogeneous in general, and so their delta will be different than the SV model delta derived above. But how do you know LV models are not homogeneous you might say. For that let's look at the simple case $T-t \ll 1$, and let the SV model be $$ dS_t = \sigma_t S_t dW_t $$ and the corresponding LV model $$ dS_t = \sigma(S_t,t)S_t dW_t $$ Then, for the SV model we can write for $T-t \ll 1$ $$ E_t \left[(S_T - K)_+\right] = E_t \left[(S_t + \sigma_t S_t (W_T-W_t) - K)_+\right] $$ which is clearly homogeneous of degree 1. For the LV model though $$ E_t \left[(S_T - K)_+\right] = E_t \left[(S_t + \sigma_t(S_t,t) S_t (W_T-W_t) - K)_+\right] $$ is not in general homogeneous because of the dependence of the instantaneous volatility on spot.

As to which delta you should use. I don't think there is a definitive answer to that. I mean in the SV model the instantaneous vol is a separate (possibly) correlated process, whereas in the LV model the instantaneous vol is driven by the spot price. So in one there is Vega risk, in the other there isn't really / strictly speaking.

The homogeneity argument can be applied also to gamma. But Vega and vanna and Volga in LV versus SV are trickier concepts which I won't speak about here / now.

Hope this helps.

  • $\begingroup$ Thanks for this and I can follow. To the "meat" of the question, which delta to use, beyond "mathematical niceness", why is it realistic that a market model is homogeneous? $\endgroup$
    – freistil90
    Jul 24, 2022 at 15:31
  • 2
    $\begingroup$ As said, there is no right or wrong delta coming from an arb free model. It's a choice you / your desk makes. $\endgroup$
    – user34971
    Jul 24, 2022 at 15:36
  • 1
    $\begingroup$ Industry standard is LSV with rough vol gaining some traction. $\endgroup$
    – user34971
    Jul 24, 2022 at 15:36
  • $\begingroup$ Hm. Maybe I'm asking the wrong question - if there is a delta for each model and my interpretation of delta is in the end the 'hedge ratio' (e.g. I need this and that many unit of the underlying today to "delta-hedge" my position), does delta actually determine this? In the BS world, it does because it is the term that shows up in the replicating portfolio in that spot. Now if all these models fit my volatility surface and I have no way to distinguish based on the instruments - which one is the correct hedge to make my position "delta-neutral"? Especially like above... for EO structures? $\endgroup$
    – freistil90
    Jul 24, 2022 at 15:44
  • $\begingroup$ Again, pricing fine but I think I'm suddenly a bit unsure about what greeks actually mean if there are multiple models that "just fit" and there's no good way to distinguish between them. I always had thought that for EO structures, local vol is "just fine", since I only care about the terminal distribution. I just thought about greeks being potentially completely different and having no way to decide what works better even for the "easy" instruments. Barriers, cliquets etc., yeah here we know that SV is better than LV because of FWD vol. But how does it look like for "trivial" strats? $\endgroup$
    – freistil90
    Jul 24, 2022 at 15:52

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