Proof for ATM delta with Local col

Question

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.

Answer 1

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.