Call Strike Derivatives and Risk-Neutral Probabilities

Summary

The document corrects a proposed relationship between a call option's strike derivative and the probability that the terminal stock price is below the strike. Under risk-neutral pricing, the first derivative of the call price with respect to strike is the negative discounted probability that the terminal price exceeds the strike, assuming a density and the stated pricing setup. The discount factor matters, and the event in the corrected identity is the upper-tail event rather than the lower-tail event in the question.

The derivation writes the call value as a discounted integral of its payoff over terminal prices above the strike, then differentiates with respect to strike. Differentiating again relates the second strike derivative to the discounted terminal-price density at the strike. These results connect option prices across strikes to risk-neutral tail probabilities and densities. The explanation assumes the relevant density and differentiability; it does not address atoms, market quote noise, or extracting stable distributions from discrete option prices.

Key ideas

• The first strike derivative of a call price gives a discounted risk-neutral upper-tail probability with a negative sign.
• The probability concerns the terminal stock price exceeding the strike under the stated setup.
• The second strike derivative recovers the discounted density at that strike.
• The derivation assumes a terminal-price density and differentiable call prices.

Tags

# probability that the stock price is below the strike price


# probability that the stock price is below the strike price












How can I prove that under the risk-neutral probability:

$\mathbb{P}[S_{t}<K]=-\frac{\partial{C}}{\partial{K}}(K,T)$

where

$S_{t}$ is the stock price, K is the strike price, C is the call option price

Thank you !

## Answer by Gordon (score 4, accepted)

https://quant.stackexchange.com/a/17650

Your posting has an error, that is, the identity should be \begin{align*} -P(0, T) \mathbb{P}(S_T > K) = \frac{\partial C}{\partial K}. \end{align*} The derivation below is based on this assumption. We denote by $f(x)$ the density function for $S_T$. Then \begin{align*} \mathbb{P}(S_T > K) = \int_K^{\infty} f(x) dx, \end{align*} and \begin{align*} C(K, T) &= P(0, T) \mathbb{E}\big((S_T-K)^+ \big)\\ &=P(0, T) \int_K^{\infty}(x-K) f(x) dx\\ &= P(0, T)\bigg[\int_K^{\infty} x f(x) dx - K \int_K^{\infty}f(x) dx \bigg]. \end{align*} Then, \begin{align*} \frac{\partial C}{\partial K} = -P(0, T)\int_K^{\infty} f(x) dx. \end{align*} That is, \begin{align*} -P(0, T) \mathbb{P}(S_T > K) = \frac{\partial C}{\partial K}. \end{align*}

We can additionally obtain that \begin{align*} P(0, T) f(K) = \frac{\partial^2 C}{\partial K^2}. \end{align*}

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.