本章导读 全书最后一章，给出随机微积分的两个应用。§16.1 Bessel 过程：16.1.1 Def 16.1 \(dX_t=\frac\alpha{X_t}dt+dB_t\) (16.1)；16.1.2 性质（命中概率 \(\varphi(x)\) 是鞅，由 Ito 公式 3 导出 ODE (16.3)，解得 (16.6)；Prop 16.1：\(\alpha\ge\frac12\) 时永不命中 0（\(T=\infty\)），\(\alpha<\frac12\) 时几乎必命中 0）；16.1.3 Prop 16.2：\(d\) 维标准 BM 的范数 \(|\mathbf B_t|\) 是参数 \(\alpha=\frac{d-1}2\) 的 Bessel 过程。§16.2 Black-Scholes-Merton 公式：16.2.1 一般 Feynman-Kac 公式（Thm 16.1：贴现期望 \(\varphi(t,x)=\mathbb E[e^{-\int_t^T r ds}F(X_T)\mid X_t=x]\) 满足 PDE \(\dot\varphi+\frac12\sigma^2\varphi''+m\varphi'-r\varphi=0\)）；16.2.2 欧式看涨期权（GBM (16.12)、风险中性 \(m=r\)、Black-Scholes 方程 (16.14)、BSM 公式 \(\varphi=N_0(d_1)x-N_0(d_2)Se^{-r(T-t)}\) (16.15)、\(d_1,d_2\) (16.16)/(16.17)、隐含波动率 Rmk 16.2；证明经热方程变换）。无图。

定义 16.1（Bessel 过程）/ Definition 16.1 (Bessel Process) 参数 \(\alpha\) 的 Bessel 过程是如下随机微分方程的解：\(dX_t=\dfrac\alpha{X_t}dt+dB_t\)，\(X_0=x_0>0\) (16.1)。A Bessel process with parameter \(\alpha\) is the solution to the SDE \(dX_t=\dfrac\alpha{X_t}dt+dB_t\), \(X_0=x_0>0\) (16.1).

命中概率 \(\varphi(x)\) 是鞅 / The hitting probability \(\varphi(x)\) is a martingale 研究 (16.1) 定义的 Bessel 过程 \(\{X_t\}\)，最终将证明它是 \(d\) 维标准 BM \(\{\mathbf B_t\}\) 的范数过程 \(X_t=|\mathbf B_t|\)。停时 \(T\equiv\inf\{t:X_t=0\}\)。设 \(x_0\in(r,R)\)（\(0Study the Bessel process \(\{X_t\}\) defined by (16.1); we will eventually show it is the norm process \(X_t=|\mathbf B_t|\) of a \(d\)-dimensional standard BM \(\{\mathbf B_t\}\). The stopping time \(T\equiv\inf\{t:X_t=0\}\). Let \(x_0\in(r,R)\) (\(0

由 Ito 公式 3 导出 ODE 并求解 \(\varphi(x)\) / Deriving the ODE via Ito's Formula 3 and solving \(\varphi(x)\) 对 \(\varphi(X_t)\) 用 Ito 公式 3 (14.15)（此处 \(\dot\varphi=0\)）：\(d\varphi(X_t)=\frac12\varphi''(X_t)dt+\varphi'(X_t)dX_t=\frac12\varphi''(X_t)dt+\varphi'(X_t)\left(\frac\alpha{X_t}dt+dB_t\right)=\underbrace{\left(\frac12\varphi''(X_t)+\varphi'(X_t)\frac\alpha{X_t}\right)}_{\equiv\,\mu_t}dt+\varphi'(X_t)dB_t\)。由鞅性 \(\mu_t=0\)，得 ODE \(\frac12\varphi''(x)+\varphi'(x)\frac\alpha x=0\) (16.3)。令 \(g(x)=\varphi'(x)\)：\(\frac12 g'(x)+g(x)\frac\alpha x=0\Rightarrow g'(x)=-2\alpha\frac{g(x)}x\Rightarrow g(x)=C_1 x^{-2\alpha}\)。\(\alpha\neq\frac12\)：\(\varphi'(x)=C_1 x^{-2\alpha}\Rightarrow\varphi(x)=C_2 x^{-2\alpha+1}+C_3\)，边界 \(\varphi(r)=0,\varphi(R)=1\) 给 \(C_2=\frac1{R^{-2\alpha+1}-r^{-2\alpha+1}}\)、\(C_3=\frac{-r^{-2\alpha+1}}{R^{-2\alpha+1}-r^{-2\alpha+1}}\)，故 \(\varphi(x)=\frac{x^{-2\alpha+1}-r^{-2\alpha+1}}{R^{-2\alpha+1}-r^{-2\alpha+1}}\) (16.4)。\(\alpha=\frac12\)：\(\varphi'(x)=C_1 x^{-1}\Rightarrow\varphi(x)=C_1\ln x+C_2\)，边界给 \(C_1=\frac1{\ln R-\ln r}\)、\(C_2=\frac{-\ln r}{\ln R-\ln r}\)，故 \(\varphi(x)=\frac{\ln x-\ln r}{\ln R-\ln r}\) (16.5)。\(\blacksquare\)Apply Ito's Formula 3 (14.15) to \(\varphi(X_t)\) (here \(\dot\varphi=0\)): \(d\varphi(X_t)=\frac12\varphi''(X_t)dt+\varphi'(X_t)dX_t=\frac12\varphi''(X_t)dt+\varphi'(X_t)\left(\frac\alpha{X_t}dt+dB_t\right)=\underbrace{\left(\frac12\varphi''(X_t)+\varphi'(X_t)\frac\alpha{X_t}\right)}_{\equiv\,\mu_t}dt+\varphi'(X_t)dB_t\). By the martingale property \(\mu_t=0\), giving the ODE \(\frac12\varphi''(x)+\varphi'(x)\frac\alpha x=0\) (16.3). Let \(g(x)=\varphi'(x)\): \(\frac12 g'(x)+g(x)\frac\alpha x=0\Rightarrow g'(x)=-2\alpha\frac{g(x)}x\Rightarrow g(x)=C_1 x^{-2\alpha}\). \(\alpha\neq\frac12\): \(\varphi'(x)=C_1 x^{-2\alpha}\Rightarrow\varphi(x)=C_2 x^{-2\alpha+1}+C_3\); the boundaries \(\varphi(r)=0,\varphi(R)=1\) give \(C_2=\frac1{R^{-2\alpha+1}-r^{-2\alpha+1}}\), \(C_3=\frac{-r^{-2\alpha+1}}{R^{-2\alpha+1}-r^{-2\alpha+1}}\), so \(\varphi(x)=\frac{x^{-2\alpha+1}-r^{-2\alpha+1}}{R^{-2\alpha+1}-r^{-2\alpha+1}}\) (16.4). \(\alpha=\frac12\): \(\varphi'(x)=C_1 x^{-1}\Rightarrow\varphi(x)=C_1\ln x+C_2\); the boundaries give \(C_1=\frac1{\ln R-\ln r}\), \(C_2=\frac{-\ln r}{\ln R-\ln r}\), so \(\varphi(x)=\frac{\ln x-\ln r}{\ln R-\ln r}\) (16.5). \(\blacksquare\)

命中概率公式 (16.6) 与命题 16.1 / Hitting probability (16.6) and Proposition 16.1 综上，In summary,

$$\varphi(x)=\begin{cases}\dfrac{\ln x-\ln r}{\ln R-\ln r}&\text{if }\alpha=\tfrac12\\[2mm]\dfrac{x^{-2\alpha+1}-r^{-2\alpha+1}}{R^{-2\alpha+1}-r^{-2\alpha+1}}&\text{if }\alpha\neq\tfrac12\end{cases}\tag{16.6}$$

这也验证了 \(\varphi\) 关于 \(x\) 是 \(C^2\) 的。命题 16.1：设 \(T\equiv\inf\{t:X_t=0\}\) 是 (16.1) 的 Bessel 过程 \(\{X_t\}\) 的停时。若 \(\alpha\ge\frac12\)，则对任意 \(x_0>0\)，\(\mathbb P\{T=\infty\mid X_0=x_0\}=1\)；若 \(\alpha<\frac12\)，则对任意 \(x_0>0\)，\(\mathbb P\{T<\infty\mid X_0=x_0\}=1\)。which also verifies that \(\varphi\) is \(C^2\) in \(x\). Proposition 16.1: let \(T\equiv\inf\{t:X_t=0\}\) be the stopping time of the Bessel process \(\{X_t\}\) in (16.1). If \(\alpha\ge\frac12\), then for any \(x_0>0\), \(\mathbb P\{T=\infty\mid X_0=x_0\}=1\); if \(\alpha<\frac12\), then for any \(x_0>0\), \(\mathbb P\{T<\infty\mid X_0=x_0\}=1\).

命题 16.1 证明 / Proof of Proposition 16.1 由重排 (16.6)。(1) \(\alpha=\frac12\)，令 \(r\downarrow0\)：\(\lim_{r\downarrow0}\varphi(x)=\lim_{r\downarrow0}\frac{\ln x-\ln r}{\ln R-\ln r}=\lim_{r\downarrow0}\frac{-\ln r}{-\ln r}=1\) (16.7)。(2) \(\alpha>\frac12\)，令 \(r\downarrow0\)（此时 \(-2\alpha+1<0\)，\(r^{-2\alpha+1}\to\infty\)）：\(\lim_{r\downarrow0}\varphi(x)=\lim_{r\downarrow0}\frac{x^{-2\alpha+1}-r^{-2\alpha+1}}{R^{-2\alpha+1}-r^{-2\alpha+1}}=\lim_{r\downarrow0}\frac{-r^{-2\alpha+1}}{-r^{-2\alpha+1}}=1\) (16.8)。合并 (16.7)、(16.8)：\(\alpha\ge\frac12\) 时对任意大 \(R\) 有 \(\lim_{r\downarrow0}\varphi(x)=1\)。回顾 \(\lim_{r\downarrow0}\varphi(x)\) 是先到 \(R\) 再到 \(0\) 的概率，故以概率 1，\(\{X_t\}\) 对任意大 \(R\) 都先到 \(R\) 再到 \(0\)，即以概率 1 \(T=\infty\)。(3) \(\alpha<\frac12\)（\(-2\alpha+1>0\)，\(r^{-2\alpha+1}\to0\)），令 \(r\downarrow0\)：\(\lim_{r\downarrow0}\varphi(x)=\frac{x^{-2\alpha+1}}{R^{-2\alpha+1}}=\left(\frac xR\right)^{-2\alpha+1}\)，于是 \(\lim_{R\uparrow\infty}\lim_{r\downarrow0}(1-\varphi(x))=1-\lim_{R\uparrow\infty}\left(\frac xR\right)^{-2\alpha+1}=1\)。即以概率 1，\(\{X_t\}\) 先到 \(0\) 再到 \(\infty\)，故 \(\mathbb P\{T<\infty\mid X_0=x_0\}=1\)。\(\blacksquare\)By rearranging (16.6). (1) \(\alpha=\frac12\), let \(r\downarrow0\): \(\lim_{r\downarrow0}\varphi(x)=\lim_{r\downarrow0}\frac{\ln x-\ln r}{\ln R-\ln r}=\lim_{r\downarrow0}\frac{-\ln r}{-\ln r}=1\) (16.7). (2) \(\alpha>\frac12\), let \(r\downarrow0\) (here \(-2\alpha+1<0\), so \(r^{-2\alpha+1}\to\infty\)): \(\lim_{r\downarrow0}\varphi(x)=\lim_{r\downarrow0}\frac{x^{-2\alpha+1}-r^{-2\alpha+1}}{R^{-2\alpha+1}-r^{-2\alpha+1}}=\lim_{r\downarrow0}\frac{-r^{-2\alpha+1}}{-r^{-2\alpha+1}}=1\) (16.8). Combining (16.7), (16.8): for \(\alpha\ge\frac12\), \(\lim_{r\downarrow0}\varphi(x)=1\) for arbitrarily large \(R\). Recalling \(\lim_{r\downarrow0}\varphi(x)\) is the probability of reaching \(R\) before 0, with probability 1 \(\{X_t\}\) reaches any arbitrarily large \(R\) before 0, i.e. with probability 1 \(T=\infty\). (3) \(\alpha<\frac12\) (\(-2\alpha+1>0\), \(r^{-2\alpha+1}\to0\)), let \(r\downarrow0\): \(\lim_{r\downarrow0}\varphi(x)=\frac{x^{-2\alpha+1}}{R^{-2\alpha+1}}=\left(\frac xR\right)^{-2\alpha+1}\), so \(\lim_{R\uparrow\infty}\lim_{r\downarrow0}(1-\varphi(x))=1-\lim_{R\uparrow\infty}\left(\frac xR\right)^{-2\alpha+1}=1\). So with probability 1 \(\{X_t\}\) reaches 0 before \(\infty\), hence \(\mathbb P\{T<\infty\mid X_0=x_0\}=1\). \(\blacksquare\)

命题 16.2 / Proposition 16.2 设 \(\{\mathbf B_t\}\) 是标准 \(d\) 维布朗运动，定义 \(\{X_t\}\) 为其（欧氏）范数过程 \(X_t=|\mathbf B_t|\)（\(\forall t\)）。则 \(\{X_t\}\) 满足 \(dX_t=\dfrac\alpha{X_t}dt+dW_t\)，其中 \(\{W_t\}\) 是标准一维布朗运动、\(\alpha=\dfrac{d-1}2\)。此时称 \(\{X_t\}\) 为 \(d\) 维 Bessel 过程。证明：由 \(d\) 维 BM 退出概率 (9.4) 与 Bessel 过程退出概率 (16.6) 在 \(\alpha=\frac{d-1}2\) 时的等价性直接得到——\(X_t=|\mathbf B_t|\) 满足 (9.4)，等价于满足 \(\alpha=\frac{d-1}2\) 的 (16.6)，故 \(X_t\) 解 \(dX_t=\frac\alpha{X_t}dt+dW_t\)（\(\alpha=\frac{d-1}2\)）。\(\blacksquare\)Let \(\{\mathbf B_t\}\) be a standard \(d\)-dimensional Brownian motion, and define \(\{X_t\}\) as its (Euclidean) norm process \(X_t=|\mathbf B_t|\) (for all \(t\)). Then \(\{X_t\}\) satisfies \(dX_t=\dfrac\alpha{X_t}dt+dW_t\), where \(\{W_t\}\) is a standard one-dimensional Brownian motion and \(\alpha=\dfrac{d-1}2\). In this case \(\{X_t\}\) is called a \(d\)-dimensional Bessel process. Proof: directly from the equivalence between the \(d\)-dimensional BM exit probability (9.4) and the Bessel-process exit probability (16.6) when \(\alpha=\frac{d-1}2\) — \(X_t=|\mathbf B_t|\) satisfies (9.4), equivalent to satisfying (16.6) with \(\alpha=\frac{d-1}2\), so \(X_t\) solves \(dX_t=\frac\alpha{X_t}dt+dW_t\) with \(\alpha=\frac{d-1}2\). \(\blacksquare\)

贴现期望 \(\varphi(t,x)\) 的设定 / Setup of the discounted expectation \(\varphi(t,x)\) 设 \(\{X_t\}\) 是标的股票价格，满足 SDE \(dX_t=m(t,X_t)dt+\sigma(t,X_t)dB_t\)，\(X_0=x_0\)。\(F(X_T)\) 是基于未来时刻 \(T\) 股价 \(X_T\) 的收益。\(r(t,X_t)\) 是连续贴现率，使得日期 \(t\) 处收益 1 的时刻 0 价值为 \(e^{-\int_0^t r(s,X_s)ds}\)。定义贴现因子 \(R_t=R_0 e^{\int_0^t r(s,X_s)ds}\)（\(R_0=1\)）。\(\varphi(t,x)\) 是时刻 \(t\le T\)、给定 \(X_t=x\) 时收益的期望值：\(\varphi(t,x)=\mathbb E\!\left[\dfrac{R_t}{R_T}F(X_T)\mid X_t=x\right]=\mathbb E\!\left[e^{-\int_t^T r(s,X_s)ds}F(X_T)\mid X_t=x\right]\)。Let \(\{X_t\}\) be the price of an underlying stock satisfying the SDE \(dX_t=m(t,X_t)dt+\sigma(t,X_t)dB_t\), \(X_0=x_0\). \(F(X_T)\) is a payoff at future time \(T\) based on the stock price \(X_T\). \(r(t,X_t)\) is the continuous discount rate such that the time-0 value of a payoff of 1 at date \(t\) is \(e^{-\int_0^t r(s,X_s)ds}\). Define the discount factor \(R_t=R_0 e^{\int_0^t r(s,X_s)ds}\) (\(R_0=1\)). \(\varphi(t,x)\) is the expected value of the payoff at time \(t\le T\) given \(X_t=x\): \(\varphi(t,x)=\mathbb E\!\left[\dfrac{R_t}{R_T}F(X_T)\mid X_t=x\right]=\mathbb E\!\left[e^{-\int_t^T r(s,X_s)ds}F(X_T)\mid X_t=x\right]\).

用 Ito 与乘积公式推导 Feynman-Kac PDE / Deriving the Feynman-Kac PDE via Ito and the product rule 设 \(\varphi(t,x)\) 关于 \(t\) 是 \(C^1\)、关于 \(x\) 是 \(C^2\)。由 Ito 公式 2 (14.15)：\(d\varphi(t,X_t)=\left(\dot\varphi+\frac12\sigma^2\varphi''+\varphi'm\right)dt+\varphi'\sigma\,dB_t\) (16.9)。定义 \(M_t=\mathbb E\!\left[\frac1{R_T}F(X_T)\mid\mathcal F_t\right]\)（时刻 0 收益在时刻 \(t\) 的评估值），\(M_T=\frac1{R_T}F(X_T)\)，\(M_t=\mathbb E[M_T\mid\mathcal F_t]\) 由塔性质是鞅。改写 \(M_t=\frac1{R_t}\mathbb E\!\left[\frac{R_t}{R_T}F(X_T)\mid\mathcal F_t\right]=\frac1{R_t}\varphi(t,X_t)\) (16.10)（马氏性 + \(X_t\in\mathcal F_t\)）。又 \(d\frac1{R_t}=-r(t,X_t)\frac1{R_t}dt\) (16.11)（无 \(dB_t\) 项）。由乘积公式 (14.21)（协变差项为 0，因 \(\frac1{R_t}\) 无扩散）：\(dM_t=\frac1{R_t}d\varphi(t,X_t)+\varphi(t,X_t)d\frac1{R_t}=\frac1{R_t}\underbrace{\left(\dot\varphi+\frac12\sigma^2\varphi''+\varphi'm-\varphi r\right)}_{\equiv\,\mu_t}dt+\frac1{R_t}\varphi'\sigma\,dB_t\)。由 \(M_t\) 鞅性 \(\mu_t=0\)，得 Feynman-Kac PDE \(\dot\varphi+\frac12\sigma^2\varphi''+\varphi'm-\varphi r=0\)。\(\blacksquare\)Let \(\varphi(t,x)\) be \(C^1\) in \(t\) and \(C^2\) in \(x\). By Ito's Formula 2 (14.15): \(d\varphi(t,X_t)=\left(\dot\varphi+\frac12\sigma^2\varphi''+\varphi'm\right)dt+\varphi'\sigma\,dB_t\) (16.9). Define \(M_t=\mathbb E\!\left[\frac1{R_T}F(X_T)\mid\mathcal F_t\right]\) (the time-0 payoff evaluated at time \(t\)), with \(M_T=\frac1{R_T}F(X_T)\) and \(M_t=\mathbb E[M_T\mid\mathcal F_t]\) a martingale by the tower property. Rewrite \(M_t=\frac1{R_t}\mathbb E\!\left[\frac{R_t}{R_T}F(X_T)\mid\mathcal F_t\right]=\frac1{R_t}\varphi(t,X_t)\) (16.10) (Markov property + \(X_t\in\mathcal F_t\)). Also \(d\frac1{R_t}=-r(t,X_t)\frac1{R_t}dt\) (16.11) (no \(dB_t\) term). By the product rule (14.21) (the covariation term is 0 since \(\frac1{R_t}\) has no diffusion): \(dM_t=\frac1{R_t}d\varphi(t,X_t)+\varphi(t,X_t)d\frac1{R_t}=\frac1{R_t}\underbrace{\left(\dot\varphi+\frac12\sigma^2\varphi''+\varphi'm-\varphi r\right)}_{\equiv\,\mu_t}dt+\frac1{R_t}\varphi'\sigma\,dB_t\). By the martingale property of \(M_t\), \(\mu_t=0\), giving the Feynman-Kac PDE \(\dot\varphi+\frac12\sigma^2\varphi''+\varphi'm-\varphi r=0\). \(\blacksquare\)

定理 16.1（Feynman-Kac 公式）/ Theorem 16.1 (Feynman-Kac Formula) 设 \(\{X_t\}\) 满足 \(dX_t=m(t,X_t)dt+\sigma(t,X_t)dB_t\)（\(X_0=x_0\)），\(r(t,x)\ge0\) 是连续贴现率。\(F(X_T)\) 是基于标的 \(X_T\) 的时刻 \(T\) 收益，\(\mathbb E[|F(X_T)|]<\infty\)。设 \(\varphi(t,x)=\mathbb E\!\left[e^{-\int_t^T r(s,X_s)ds}F(X_T)\mid X_t=x\right]\)。则 \(\varphi(t,x)\) 满足偏微分方程Let \(\{X_t\}\) satisfy \(dX_t=m(t,X_t)dt+\sigma(t,X_t)dB_t\) (\(X_0=x_0\)), with \(r(t,x)\ge0\) a continuous discount rate. \(F(X_T)\) is a payoff at time \(T\) as a function of the underlying \(X_T\) with \(\mathbb E[|F(X_T)|]<\infty\). Let \(\varphi(t,x)=\mathbb E\!\left[e^{-\int_t^T r(s,X_s)ds}F(X_T)\mid X_t=x\right]\). Then \(\varphi(t,x)\) satisfies the PDE

$$\dot\varphi(t,x)+\frac12\sigma^2(t,x)\varphi''(t,x)+\varphi'(t,x)m(t,x)-\varphi(t,x)r(t,x)=0,\qquad 0\le t\le T,$$

其中 \(\dot\varphi=\frac{\partial\varphi}{\partial t}\)、\(\varphi'=\frac{\partial\varphi}{\partial x}\)、\(\varphi''=\frac{\partial^2\varphi}{\partial x^2}\)，且满足终端条件 \(\varphi(T,x)=F(x)\)。where \(\dot\varphi=\frac{\partial\varphi}{\partial t}\), \(\varphi'=\frac{\partial\varphi}{\partial x}\), \(\varphi''=\frac{\partial^2\varphi}{\partial x^2}\), with the terminal condition \(\varphi(T,x)=F(x)\).

设定：几何布朗运动与风险中性测度 / Setup: GBM and the risk-neutral measure 设 \(\{X_t\}\) 是标的股价，服从几何布朗运动 \(dX_t=mX_t dt+\sigma X_t dB_t\) (16.12)。在风险中性概率测度下 \(m=r\)（注 16.1：客观测度 \(\mathbf P\) 下漂移 \(m\) 不等于无风险利率 \(r\)，但在风险中性测度 \(\mathbf Q\) 下二者相等——这正是 \(\mathbf Q\) 的定义）。欧式看涨期权：行权价 \(S\)、行权日 \(T\)、收益 \(F(X_T)=\max\{X_T-S\}\equiv(X_T-S)_+\)。无风险贴现率 \(r(t)\) 使日期 \(T\) 收益 1 的时刻 \(t\) 价值为 \(e^{-\int_t^T r(s)ds}\)；下设 \(r(t)=r\) 为常数。\(\varphi(t,x)=\mathbb E_{\mathbf Q}\!\left[e^{-r(T-t)}F(X_T)\mid X_t=x\right]\)。Let \(\{X_t\}\) be the underlying stock price following the geometric Brownian motion \(dX_t=mX_t dt+\sigma X_t dB_t\) (16.12). Under the risk-neutral probability measure, \(m=r\) (Remark 16.1: under the objective measure \(\mathbf P\) the drift \(m\) does not equal the risk-free rate \(r\), but they are equal under the risk-neutral measure \(\mathbf Q\) — which is precisely the definition of \(\mathbf Q\)). European call option: exercise price \(S\), exercise date \(T\), payoff \(F(X_T)=\max\{X_T-S\}\equiv(X_T-S)_+\). The risk-free discount rate \(r(t)\) makes the time-\(t\) value of a payoff of 1 at date \(T\) equal to \(e^{-\int_t^T r(s)ds}\); below assume \(r(t)=r\) constant. \(\varphi(t,x)=\mathbb E_{\mathbf Q}\!\left[e^{-r(T-t)}F(X_T)\mid X_t=x\right]\).

Black-Scholes 方程 (16.14) 与 BSM 公式 (16.15) / The Black-Scholes equation and the BSM formula 把 \(m(t,X_t)=mX_t\)、\(\sigma(t,X_t)=\sigma X_t\)、\(r(t,X_t)=r(t)\) 代入 Feynman-Kac（定理 16.1）：\(\dot\varphi+\frac12\sigma^2 x^2\varphi''+\varphi'mx-\varphi r=0\) (16.13)。代入 \(r(t)=r\)、\(m=r\) 得 Black-Scholes 方程 \(\dot\varphi=\varphi r-rx\varphi'-\frac12\sigma^2 x^2\varphi''\) (16.14)。代入 \(F(X_T)=(X_T-S)_+\)，得行权日 \(T\)、行权价 \(S\)、时刻 \(t\) 股价 \(x\) 的欧式看涨期权时刻 \(t\) 价格为Plug \(m(t,X_t)=mX_t\), \(\sigma(t,X_t)=\sigma X_t\), \(r(t,X_t)=r(t)\) into Feynman-Kac (Theorem 16.1): \(\dot\varphi+\frac12\sigma^2 x^2\varphi''+\varphi'mx-\varphi r=0\) (16.13). Plugging \(r(t)=r\), \(m=r\) gives the Black-Scholes equation \(\dot\varphi=\varphi r-rx\varphi'-\frac12\sigma^2 x^2\varphi''\) (16.14). Plugging \(F(X_T)=(X_T-S)_+\), the time-\(t\) price of a European call with exercise date \(T\), exercise price \(S\), and time-\(t\) stock price \(x\) is

$$\varphi(t,x)=N_0(d_1)\,x-N_0(d_2)\,Se^{-r(T-t)},\tag{16.15}$$

其中 \(d_1=\dfrac1{\sigma\sqrt{T-t}}\left[\ln\dfrac xS+\left(r+\dfrac{\sigma^2}2\right)(T-t)\right]\) (16.16)、\(d_2=d_1-\sigma\sqrt{T-t}\) (16.17)，\(N_0(\cdot)\) 是 \(\mathcal N(0,1)\) 的累积分布函数。注 16.2：若观测到欧式看涨期权的市场价格，可由 (16.15)–(16.17) 反解出标的资产的隐含波动率 \(\sigma\)。where \(d_1=\dfrac1{\sigma\sqrt{T-t}}\left[\ln\dfrac xS+\left(r+\dfrac{\sigma^2}2\right)(T-t)\right]\) (16.16), \(d_2=d_1-\sigma\sqrt{T-t}\) (16.17), and \(N_0(\cdot)\) is the CDF of \(\mathcal N(0,1)\). Remark 16.2: if we observe the market price of a European call, we can obtain the implied volatility \(\sigma\) of the underlying by inverting (16.15)–(16.17).

BSM 公式证明（经热方程变换，条理梗概）/ Proof of the BSM formula (via a heat-equation transformation — outline) 从 Black-Scholes 方程 \(\dot\varphi=\varphi r-mx\varphi'-\frac12\sigma^2 x^2\varphi''\)（终端 \(\varphi(T,x)=(x-S)_+\)）出发。(1) 换元：令 \(\tau=T-t\)、\(y=\ln x\)，则 \(\frac{\partial\varphi}{\partial\tau}=\frac12\sigma^2\frac{\partial^2\varphi}{\partial y^2}+\left(m-\frac12\sigma^2\right)\frac{\partial\varphi}{\partial y}-r\varphi\) (16.21)，记 \(A=\frac12\sigma^2\)、\(B=m-\frac12\sigma^2\)、\(C=-r\)。(2) 化为热方程：令 \(h=e^{\alpha y+\beta\tau}\varphi\) (16.22)，可任意取 \(\alpha,\beta\)；选 \(\alpha=\frac B{2A}\)、\(\beta=\frac{B^2}{4A}-C\) 消去一阶项与零阶项，得标准热方程 \(\frac{\partial h}{\partial\tau}=A\frac{\partial^2 h}{\partial y^2}\) (16.28)。(3) 解热方程：\(h=\frac1{\sqrt{4\pi A\tau}}\int_{-\infty}^\infty g(z)e^{-\frac{(y-z)^2}{4A\tau}}dz\) (16.29)（猜测并验证，(16.30)、(16.31) 验证），其中 \(g(y)=e^{\alpha y}f(y)\)、\(f(y)=\max\{e^y-S,0\}\) 来自终端条件（\(\lim_{\tau\to0}h(\tau,y)=g(y)\) (16.32)）。(4) 回代：经 Gaussian 积分把 \(\varphi=e^{-\alpha y-\beta\tau}h\) 写成两部分 (16.34)，并用引理 16.1（若 \(N_1\) 是 \(\mathcal N(\mu,\sigma^2)\) 的 CDF、\(N_0\) 是 \(\mathcal N(0,1)\) 的 CDF，则 \(N_1(q)=N_0\!\left(\frac{q-\mu}\sigma\right)\)）：Part 1 \(=xe^{m(T-t)}N_0(d_1)\) (16.35)、Part 2 \(=N_0(d_2)\) (16.36)。代入 \(m=r\)：\(\varphi(t,x)=e^{-r(T-t)}xe^{r(T-t)}N_0(d_1)-Se^{-r(T-t)}N_0(d_2)=N_0(d_1)x-N_0(d_2)Se^{-r(T-t)}\)，其中 \(d_2=\frac{\ln\frac xS+(m-\frac12\sigma^2)(T-t)}{\sigma\sqrt{T-t}}=d_1-\sigma\sqrt{T-t}\)，证毕。\(\blacksquare\)Start from the Black-Scholes equation \(\dot\varphi=\varphi r-mx\varphi'-\frac12\sigma^2 x^2\varphi''\) (terminal \(\varphi(T,x)=(x-S)_+\)). (1) Change of variables: let \(\tau=T-t\), \(y=\ln x\); then \(\frac{\partial\varphi}{\partial\tau}=\frac12\sigma^2\frac{\partial^2\varphi}{\partial y^2}+\left(m-\frac12\sigma^2\right)\frac{\partial\varphi}{\partial y}-r\varphi\) (16.21), writing \(A=\frac12\sigma^2\), \(B=m-\frac12\sigma^2\), \(C=-r\). (2) To the heat equation: let \(h=e^{\alpha y+\beta\tau}\varphi\) (16.22), with \(\alpha,\beta\) free; choosing \(\alpha=\frac B{2A}\), \(\beta=\frac{B^2}{4A}-C\) kills the first- and zeroth-order terms, giving the standard heat equation \(\frac{\partial h}{\partial\tau}=A\frac{\partial^2 h}{\partial y^2}\) (16.28). (3) Solve the heat equation: \(h=\frac1{\sqrt{4\pi A\tau}}\int_{-\infty}^\infty g(z)e^{-\frac{(y-z)^2}{4A\tau}}dz\) (16.29) (guess and verify via (16.30), (16.31)), where \(g(y)=e^{\alpha y}f(y)\), \(f(y)=\max\{e^y-S,0\}\) from the terminal condition (\(\lim_{\tau\to0}h(\tau,y)=g(y)\) (16.32)). (4) Back-substitute: through Gaussian integrals, write \(\varphi=e^{-\alpha y-\beta\tau}h\) as two parts (16.34), and using Lemma 16.1 (if \(N_1\) is the CDF of \(\mathcal N(\mu,\sigma^2)\) and \(N_0\) the CDF of \(\mathcal N(0,1)\), then \(N_1(q)=N_0\!\left(\frac{q-\mu}\sigma\right)\)): Part 1 \(=xe^{m(T-t)}N_0(d_1)\) (16.35), Part 2 \(=N_0(d_2)\) (16.36). With \(m=r\): \(\varphi(t,x)=e^{-r(T-t)}xe^{r(T-t)}N_0(d_1)-Se^{-r(T-t)}N_0(d_2)=N_0(d_1)x-N_0(d_2)Se^{-r(T-t)}\), where \(d_2=\frac{\ln\frac xS+(m-\frac12\sigma^2)(T-t)}{\sigma\sqrt{T-t}}=d_1-\sigma\sqrt{T-t}\), completing the proof. \(\blacksquare\)