Why Zero Correlation Sensitivity Does Not Imply Zero Implied-Volatility Cross Derivative

# Dependence of implied volatility on spot-vol correlation


# Dependence of implied volatility on spot-vol correlation












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.

## Answer by bhutes (score 1)

https://quant.stackexchange.com/a/45480

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

Consider,

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

So,

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

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 $\rho =0$.