I have the following general SV model:

$$ dS = \sigma S dW_S $$ $$ d\sigma = a(\sigma,t) dt + b (\sigma, t) dW_\sigma $$ $$ dW_S dW_\sigma = \rho dt $$ where $a , b$ are deterministic functions of $\sigma$ and $t$ only, and $\rho$ is constant.

My question is the following:

Suppose that for any value of the spot, at any time before maturity of a vanilla call option, there is a strike where the sensitivity of the implied volatility to correlation is zero, that is $$ \Sigma_\rho = 0 $$ where subscripts denotes partial derivative, and where the implied volatility is of course defined as follows: $$ C^{BS} (S,t,K,T;\Sigma) = C^{SV} (S,t,K,T;\sigma) $$ where the subscript "BS" means Black-Scholes price, and "SV" means stochastic vol model price.

What can we then say about $$ \Sigma_{S \sigma} = ? $$ My conjecture is that the second order derivative above will be zero at the strike where the sensitivity of the implied volatility to correlation is zero. But I cannot prove it precisely.

The hand-waving argument is as follows. Since $\Sigma$ is stochastic, $$ d\Sigma = \Sigma_t dt + \Sigma_S dS + \Sigma_\sigma d\sigma + \frac{1}{2} \Sigma_{S S} (dS)^2 + \frac{1}{2} \Sigma_{\sigma \sigma} ( d\sigma)^2 + \Sigma_{S \sigma} dS d\sigma $$

The term involving $dS d\sigma$ above will contain $\rho$, and intuition suggests that $\Sigma$ would be independent of $\rho$ if $\Sigma_{S \sigma} = 0$, but of course this is not a hard-proof.

I would be more than satisfied to restrict the question to the case where $\rho = 0$ to start with, i.e. a symmetric smile. [Needless to say a symmetric smile doesn't mean there is no sensitivity to correlation.]

Any help appreciated. This is a research question by the way, so not expecting a full answer, but some ideas would be great.


1 Answer 1


Please see, if the below serves as a counter-example -


$\Sigma= \rho S\sigma - \rho K S\sigma^2 +S\sigma$


$\Sigma_\rho = S\sigma - K S \sigma^2$

There exists $K$, such that $K= \frac 1 \sigma$ where $\Sigma_\rho = 0$.

Evaluating $\Sigma_{S \sigma}$ below -

$\Sigma_S= \rho \sigma - \rho K \sigma^2 +\sigma $

$\Sigma_{S \sigma} = \rho - 2 \rho K \sigma +1 $

Here, $\Sigma_{S\sigma} \neq 0 $ at $K=\frac 1 \sigma $ and $\rho =0$.

  • $\begingroup$ Thanks, I wish things were that easy :) At least two things not right with your counter-example: 1.) The smile would be flat if $\rho=0$ which is not possible in a SV model. 2.) The dimensions / units are incorrect, e.g. $K$ is dimensionless (or has units in dollars if you wish), so it can never equal $1/\sigma$ which has different units. $\endgroup$
    – user34971
    May 6, 2019 at 7:16
  • $\begingroup$ 1) $\rho=0$ is not required by the counter-example. It works at any $\rho$ (rather it works, even if $\rho$ is zero. 2). I was looking at these as general multi-variate functions with dimensionless variables. Anyway, I can brood over this some more. And wait for other replies to educate myself. Thanks for your comment. $\endgroup$
    – bhutes
    May 6, 2019 at 7:25
  • $\begingroup$ Sure. The main problem with trying to find specific counter-examples in a functional form such as the one you gave is that then you have to find at least one SV model that generates a smile exactly given by that particular functional form. It wouldn't be the route I'd follow. $\endgroup$
    – user34971
    May 6, 2019 at 7:33
  • $\begingroup$ Intuitively, I think the conjecture seems approximately correct. Both expressions are close to zero for ATM options I believe. Sabr model experience shows that ATM options have no sensitivity to correlation. Meanwhile the second expression is the sensitivity of delta to changes in vol. but delta of ATM options is approximately 50 regardless of vol. $\endgroup$
    – dm63
    Jun 5, 2019 at 10:11

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