Risk-neutral pricing: $C = e^{-r(T-t)}\,\mathbb E[\max(S_T-K,0)]$

Proved

BlackScholesModel.callPrice_eq_risk_neutral_expectation

by Lucas · Oct 9, 2026 · Mathlib 0df444a (Lean v4.33.1)

black-scholesmathematical-financeoption-pricing

Let $K>0$ , $\sigma>0$ , $S>0$ , $t<T$ , $r\in\mathbb R$ and $\tau=T-t$ . Let $Z$ be a standard normal random variable and let

S_T=S\exp\Big(\big(r-\tfrac{\sigma^2}2\big)\tau+\sigma\sqrt\tau\,Z\Big)

be the terminal stock price under the risk-neutral measure (geometric Brownian motion with drift $r$ ). Then the Black–Scholes call price is the discounted expected payoff:

C(S,t)=e^{-r\tau}\,\mathbb E\big[\max(S_T-K,0)\big].

Formalization Note The expectation is the integral with respect to the standard Gaussian measure on $\mathbb R$ (gaussianReal 0 1) of $z\mapsto\max(S_T(z)-K,0)$ .

Preamble

import Mathlib
import Definitions.Def_BlackScholesModel_Core

open Real MeasureTheory ProbabilityTheory Filter Topology
open scoped ContDiff

Formal statement

namespace BlackScholesModel

theorem callPrice_eq_risk_neutral_expectation (K r σ T S t : ℝ) (hK : 0 < K) (hσ : 0 < σ) (hS : 0 < S) (ht : t < T) :
    callPrice K r σ T S t
      = Real.exp (-r * (T - t))
          * ∫ z, max (terminalPrice S r σ (T - t) z - K) 0 ∂(gaussianReal 0 1) := by sorry

end BlackScholesModel

Source

Wikipedia, "Black–Scholes model" (snapshot supplied as Black–Scholes_model.pdf, 22 pp.), https://en.wikipedia.org/wiki/Black%E2%80%93Scholes_model; p. 6 (Derivations: the option price is the expected value of the discounted payoff under the risk-neutral measure; risk-neutral stock price)

Read-back

What the Lean code literally says, in plain math · Aristotle (Harmonic)

Non-blind read-back. This read-back was written by the same agent that drafted the Lean statement (at the explicit instruction of the proposal owner), not by an independent blind auditor. It is not independent testimony; reviewers should compare it against the Lean code themselves.

Here $K,r,\sigma,T,S,t$ are arbitrary real numbers subject to the hypotheses $K>0$ , $\sigma>0$ , $S>0$ and $t<T$ (so $\tau:=T-t>0$ ); $r$ is unrestricted (it may be zero or negative). $N$ , $N'$ , $d_\pm$ , $C$ , $P$ are the functions of the companion definition file (normal distribution function given by its integral, normal density in closed form, $d_\pm=d_\pm(S,K,r,\sigma,T-t)$ , call price $C(S,t)=N(d_+)S-N(d_-)Ke^{-r(T-t)}$ , put price $P(S,t)=N(-d_-)Ke^{-r(T-t)}-N(-d_+)S$ ).

Let $\gamma$ be the standard Gaussian probability measure on $\mathbb R$ (mean $0$ , variance $1$ ) and $S_T(z)=S\exp\big((r-\sigma^2/2)(T-t)+\sigma\sqrt{T-t}\,z\big)$ . The theorem asserts

C(S,t)=e^{-r(T-t)}\int_{\mathbb R}\max\big(S_T(z)-K,\,0\big)\,d\gamma(z),

where the integral is the Bochner integral (which the formal system sets to $0$ if the integrand is not integrable).

Human review

Endorsed by Shuze Chen · Oct 10, 2026
Confirmed by the moderator at approval.
Endorsed by Lucas · Oct 10, 2026
Confirmed by the mission captain (proposal self-audit).

View graph

Risk-neutral pricing: C=e−r(T−t) E[max⁡(ST−K,0)]C = e^{-r(T-t)}\,\mathbb E[\max(S_T-K,0)]C=e−r(T−t)E[max(ST​−K,0)]

Risk-neutral pricing: $C = e^{-r(T-t)}\,\mathbb E[\max(S_T-K,0)]$