2
$\begingroup$

I am looking at a time-homogeneous local volatility model where

  • ATM implied volatility equals ATM local volatility: $\sigma_{imp}(S_0)=\sigma_{local}(S_0)$
  • ATM IV Skew = half of LV slope
  • In general $\sigma_{imp}(K) = \sqrt{\frac{\sigma_{local}^2(S_0)+\sigma_{local}^2(K)}{2}} $

LV is a function of spot and and is Calibrated to IV which is a function of strike (we are working with European call option in this case).

Claim: Delta for an ATM European Call Option in the LV model is given by: $$ \Delta_{LV} = \Delta_{BS} + \text{vega}_{BS}*\frac{\partial \sigma_{imp}(K)}{\partial K}\bigg\rvert_{K=S_0} $$ where $\Delta_{BS}$ and $\text{vega}_{BS}$ is the Black Scholes vega and Delta.

What is the proof for this claim?

Bassically I don't really know exactly how volatilites look in each of the delta term and that is why I can't construct a proof. So by explaining the model and the last equation thoroughly I will probably be able to reach the proof myself.

$\endgroup$
2
$\begingroup$

In a local volatility model, which is inhomogeneous in space, you'll end up with having that implied volatility of a vanilla option $(K,T)$ is a function of the spot price $S$, i.e. $$ \Sigma = \sigma(T,K,S) $$ As such when you compute the (total) derivative of the option price with respect to the spot price you'll have: \begin{align} \frac{d V}{d S} &= \left.\frac{\partial V}{\partial S}\right\vert_{\Sigma} + \left.\frac{\partial V}{\partial \Sigma}\right\vert_{S} \frac{\partial \sigma(T,K,S)}{\partial S} \\ &= \Delta_{BS}(S,\Sigma) + \nu_{BS}(S,\Sigma) \frac{\partial \sigma(T,K,S)}{\partial S} (S) \end{align} The last term should be how the implied volatility moves as the spot moves but the local volatility function is kept unchanged.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.