Recovering Risk-Neutral Stock Density from Call Prices and Implied Volatility
Summary
The note explains how to infer the risk-neutral distribution of a stock price at a fixed maturity from European call prices. Starting from the discounted expected payoff formula, it identifies the second derivative of call price with respect to strike, divided by the discount factor, as the density at that strike. This is the Breeden-Litzenberger relationship and also addresses the question of probability mass at a single strike, subject to the usual smooth-density interpretation.
To connect the result to an implied volatility surface, the note proposes substituting the Black-Scholes call formula using strike- and maturity-dependent implied volatility, then differentiating twice by strike with the chain rule. It does not work through that differentiation or give a final expanded expression. The result is risk-neutral, not a direct forecast of real-world probabilities, and its use requires suitable smoothness and reliable option prices across strikes.
Key ideas
- The second strike derivative of a call price, adjusted for discounting, gives the risk-neutral density at that strike.
- The density relationship follows by differentiating the discounted expected call payoff twice with respect to strike.
- An implied volatility surface can be substituted into the Black-Scholes formula before applying the same derivatives.
- The document outlines the chain-rule step but does not provide the expanded implied-volatility expression.
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Full text
# A question on implied volatility surface
# 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?
## Answer by Quantuple (score 5, accepted)
https://quant.stackexchange.com/a/25211
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, $\forall T>0$:
$$q (S_T=K) = \frac {1}{P (0,T)} \frac {\partial^2 C }{\partial K^2} (K,T) $$
which answers your two first questions.
To further express the pdf as a function of the implied volatility smile, just use the fact that
$$ C(K,T) = BS (P (0,T), F (0,T), \sigma^*(K,T), K, T) $$
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).Shown in full with attribution under the source's licence. Licence: CC BY-SA 4.0 (Stack Exchange)
This summary was written by Stratmill's research agent from the original; it is not a copy of the source.