What is the risk neutral expectiation of an option price given a move in spot?

Lets say we have a volatility surface for the SPX at time t with spot S. We consequently know the price of some call option at maturity T with strike K. What is the risk neutral expectation of the option price at t+1day, if spot moves 1% ( E[option price | S*1.01] ) ?

• Be more specific. What is '1' here, infinitesimal / very small, or 1 month, 1 year? Apr 13, 2023 at 13:54
• Lets say 1 day. Apr 13, 2023 at 14:07
• Why not just use the delta and gamma then? Apr 13, 2023 at 14:30
• Because volatility also moves. Apr 13, 2023 at 15:29
• See answer below. Apr 13, 2023 at 15:32

Your question is unclear / lacks relevant detail, but I suspect what you're really asking is what happens to the vol change when the spot change is given.

Assume, as an example, the following dynamics: \begin{align} dS_t &= \sigma_t S_t dW_t \\ d\sigma_t &= \alpha \sigma_t ( \rho dW_t + \sqrt{1-\rho^2} dZ_t) \end{align} with $$dW dZ = 0$$.

You'd like to calculate $$E_t[ C(S_{t+dt},\sigma_{t+dt},K) | dS_t = c]$$. Now $$C(S_{t+dt},\sigma_{t+dt},K) = C(S_t,\sigma_t, K) + dC(S_t,\sigma_t,K)$$ with $$dC(S_t,\sigma_t,K) = \Delta dS_t + \nu d\sigma_t$$ with $$\Delta$$ the delta of the option and $$\nu$$ the vega.

Given $$dS_t = c$$ then \begin{align} dS_t &= c\\ d\sigma_t &= \frac{\rho \alpha}{S_t} dS_t + (\cdot) dZ_t \\ & = \frac{\rho \alpha}{S_t} c + (\cdot) dZ_t \end{align} Thus \begin{align} E_t[ C(S_{t+dt},\sigma_{t+dt},K) |dS_t = c] &= C(S_t,\sigma_t,K) + c \Delta + \frac{\rho \alpha}{S_t} c \nu + E[ (\cdot) dZ_t] \\ &= C(S_t,\sigma_t,K) + c \Delta + \frac{\rho \alpha}{S_t} c \nu \end{align}

• The problem now becomes: What is the risk neutral estimate of the spot-vol correlation Apr 13, 2023 at 15:33
• That's a different question but it's well known that correlation is proportional to the ATM slope. Apr 13, 2023 at 15:35
• Well, you framed your answer in a way that left one variable to which we do not know the risk neutral estimate: the spot vol correlation, which should indeed be somewhat proportional to the skew. Apr 13, 2023 at 15:45
• Yeah ok, and you're welcome by the way. Apr 13, 2023 at 15:49
• Didn't mean to sound smug, sorry if was the case, thanks for taking your time to reply. Apr 13, 2023 at 16:11