# A question on implied volatility surface

Let a stock price process be $(S_t)_{t\geq 0}$ and let $(K, T)\longrightarrow \sigma^*(K,T)$ be the volatility surface corresponding to vanilla options on the stock. What is, for any time $T$ the implied distribution density of $S_T$ corresponding to strike $K$? Also, what is $\mathbb P(S_T=K)$ as a function of the call price, for maturity $T$ and strike $K$? And what is its expression as a function of $\sigma^*$ and its derivatives?

Let $$q (S) := \frac{d\mathbb {Q}(S_T \leq S)}{dS}$$ denote the probability density function of the stock price at time $T>0$ under the risk-neutral measure.

By definition, the price of a European call then writes \begin{align} C (K,T) &= P (0,T) E_0^{\mathbb {Q}}[(S_T-K)^+] \\ &= P (0,T) \int_K^\infty (S - K) q (S) dS \end{align} with $P (0,T)$ the relevant discount factor.

Compute the second derivative of the last equality with respect to $K$ to obtain what is known as the Breeden-Litzenberger identity, $\forall T>0$:

$$q (S_T=K) = \frac {1}{P (0,T)} \frac {\partial^2 C }{\partial K^2} (K,T)$$

$$C(K,T) = BS (P (0,T), F (0,T), \sigma^*(K,T), K, T)$$
where $BS (.)$ represents the Black-Scholes analytical formula, then work out the expression of the second derivative $\frac {\partial^2 C }{\partial K^2} (K,T)$ as a function of $\sigma^*(T,K)$ using standard calculus (chain rule).