【推导】Feynman-Kac To Black-Scholes-PDE

首先对Feynman-Kac theorem进行基本函数定义，推导并证明基本特性;

其次，展示五个版本的Black-Scholes-PDE推导，前三个采用了Feynman-Kac theorem，但是给的公式都不相同，后两个用的配无风险组合；

Feynman-Kac theorem

Cox-Ross-Robinsten 决定的 u and d

基于B站 Jerry Xu 视频的笔记

衍生品的 Probabilistic model

$$\mathbb{E}^\mathbb{Q}[e^{-r(T-t)}F(S_T)|\mathcal{F}_t]=u(t,x)$$

where $F(·):$ payoff, $u(t,x) :$ value at time t

衍生品的 Partial differential equation

$$\frac{\partial u}{\partial t}+rx\frac{\partial u}{\partial x}+\frac{1}{2}\sigma^2x^2\frac{\partial^2 u}{\partial x^2}=rx$$

Feynman-Kac theorem 将两者联系起来，只要两者知一，就能知道另一个

基本定义

Kronecker delta（克罗内克函数）

$$\delta_{ij}=\begin{cases}0&,i\ne j\\1&,i=j\end{cases}$$

股票价值

$$\overrightarrow{X}_t=(x_t^{(1)},x_t^{(2)},...,x_t^{(d)})^T\in\mathbb{R}^{d\times 1}$$ $$dX_t=\mu(t,X_t)dt+\sigma(t,X_t)dB_t$$ $$\begin{bmatrix} \mu_1\\ \mu_2\\...\\ \mu_d \end{bmatrix}_{d\times 1} + \begin{bmatrix} \sigma_{11}&\sigma_{12}&...&\sigma_{1m}\\ \sigma_{21}&\sigma_{22}&...&\sigma_{2m}\\ ...&...&...&...\\ \sigma_{d1}&\sigma_{d2}&...&\sigma_{dm}\\ \end{bmatrix}_{d\times m} \times \begin{bmatrix} dB_t^{(1)}\\ dB_t^{(2)}\\...\\ dB_t^{(m)} \end{bmatrix}_{m\times 1}$$

$B_t$: m-dimensioned BM [uncorrelated]

$$dB_idB_j=\begin{cases}dt&,i= j\\ \rho dt=0&,i\ne j\end{cases}$$

Also can be shown as

$$\begin{align} dx_t^{(i)}&=\mu_idt+\sum_{k=1}^m\sigma_{ik}dB_t^{(k)},\ i\in \{1,2,...,d\}\\ dx_t^{(j)}&=\mu_jdt+\sum_{l=1}^m\sigma_{jl}dB_t^{(l)},\ j\in \{1,2,...,d\} \end{align}$$

So we have

$$\begin{align} dx_t^{(i)}dx_t^{(j)}&=(\sum_{k=1}^m\sigma_{ik}dB_t^{(k)})(\sum_{l=1}^m\sigma_{jl}dB_t^{(l)})\\ &=\sum_{k=1}^m\sum_{l=1}^m\sigma_{ik}\sigma_{jl}dB_t^{(k)}dB_t^{(l)}\\ &=\sum_{k=1}^m\sum_{l=1}^m\sigma_{ik}\sigma_{jl}\delta_{kl}dt\\ &=\sum_{k=1}^m\sigma_{ik}\sigma_{jK}dt \end{align}$$

Generator

$$A=\sum_{i=1}^d\mu_i\frac{\partial}{\partial x_i}+\frac{1}{2}\sum_{i=1}^d\sum_{j=1}^d(\sigma \sigma^T)_{ij}\frac{\partial^2}{\partial x_i\partial x_j}$$

with

$$\begin{matrix}u(t,\underbrace{x_1,x_2,...,x_d})\\ \quad x_t^{(1)},x_t^{(2)},...,x_t^{(d)} \end{matrix}$$

With $u(t,\overrightarrow{X}_t)$ and $r(t,\overrightarrow{X}_t)$, prove $M_t$ is a martingale

$$M_t=e^{-\int_0^trds}u-\int_0^te^{-\int_0^srdv}(\frac{\partial u}{\partial t}+Au-ru)ds$$

Proof：

Set $F_t=e^{-\int_0^trds}u$

$$\begin{align} dF_t&=[de^{-\int_0^trds}]u+e^{-\int_0^trds}du+0\\ &=-re^{-\int_0^trds}udt+e^{-\int_0^trds}[\frac{\partial u}{\partial t}dt+\sum_{i=1}^d\frac{\partial u}{\partial x_i}dx_t^{(i)}...\\ &...+\frac{1}{2}\sum_{i=1}^d\sum_{j=1}^d\frac{\partial^2 u}{\partial x_i\partial x_j}dx_t^{(i)}dx_t^{(j)}]\\ &=e^{-\int_0^trds}[-rudt+\frac{\partial u}{\partial t}dt+\sum_{i=1}^d\frac{\partial u}{\partial x_i}\mu_idt+\sum_{i=1}^d\frac{\partial u}{\partial x_i}\sum_{k=1}^m\sigma_{ik}dB_t^{(k)}...\\ &...+\frac{1}{2}\sum_{i=1}^d\sum_{j=1}^d\frac{\partial^2 u}{\partial x_i\partial x_j}\sum_{k=1}^m\sigma_{ik}\sigma_{jK}dt]\\ &=e^{-\int_0^trds}[-ru+\frac{\partial u}{\partial t}+\sum_{i=1}^d\frac{\partial u}{\partial x_i}\mu_i+\frac{1}{2}\sum_{i=1}^d\sum_{j=1}^d\frac{\partial^2 u}{\partial x_i\partial x_j}(\sigma \sigma^T)_{ij}]dt...\\ &...+e^{-\int_0^trds}(\nabla u)^T\sigma dB_t\\ &=e^{-\int_0^trds}[-ru+\frac{\partial u}{\partial t}+Au]dt+e^{-\int_0^trds}(\nabla u)^T\sigma dB_t \end{align}$$

So

$$\begin{align} dM_t&=dF_t-e^{-\int_0^trdv}(\frac{\partial u}{\partial t}+Au-ru)dt\\ &=e^{-\int_0^trds}(\nabla u)^T\sigma dB_t \end{align}$$

$\Rightarrow M_t$ is a Martingale

Similarly

Set $t\leq t’\leq T$ with $X_t^{t,x}=x$

$$\tilde{M}_{t'}=e^{-\int_t^{t'}rds}u-\int_t^{t'}e^{-\int_t^srdv}(\frac{\partial u}{\partial t}+Au-ru)ds$$

$\Rightarrow \tilde{M}_{t’}$ is also a Martingale

So $\mathbb{E}[\tilde{M}_T]=\mathbb{E}[\tilde{M}_t]=\tilde{M}_t=u(t,x)$

Feynman-Kac 证明

( Feynman-Kac Formula ) Let $X_t$ is an Ito diffusion given by

$$d\overrightarrow{X}_t=\mu(t,\overrightarrow{X}_t)dt+\sigma(t,\overrightarrow{X}_t)d\overrightarrow{B}_t$$

with generator A and initial condition $X_t^{t,x}=x$

Let $u(t,\overrightarrow{X}_t)$, $t\in\mathbb{R}^+$, $\overrightarrow{X}_t\in\mathbb{R}^{d\times 1}$ satisfy

$$\frac{\partial u}{\partial t}+Au-ru=0 \tag{1}$$

with final condition $u(T,x)=f(x)$

Then

$$\begin{align} u(t,x)&=\mathbb{E}[e^{-\int_t^Tr(s,x_s)ds}f(X_T)|\mathcal{F}_t]\\ &=\mathbb{E}[e^{-\int_t^Tr(s,x_s)ds}f(X_T)|X(t)=x]\\ &=\mathbb{E}[e^{-\int_t^Tr(s,X_s^{t,x})ds}f(X_T^{t,x})] \end{align}$$

Proof：

we have

$$\mathbb{E}[\tilde{M}_{T}]=\mathbb{E}[e^{-\int_t^{T}r(S,X_s^{t,x})ds}u(T,X_T^{t,x})-\int_t^{T}e^{-\int_t^sr(v,X_v^{t,x})dv}(\frac{\partial u}{\partial t}+Au-ru)(s,X_s^{t,x})ds]$$

Under condition (1)

$$\mathbb{E}[\tilde{M}_{T}]=\mathbb{E}[e^{-\int_t^{T}r(S,X_s^{t,x})ds}u(T,X_T^{t,x})]$$

and final condition

$$\mathbb{E}[\tilde{M}_{T}]=\mathbb{E}[e^{-\int_t^{T}r(S,X_s^{t,x})ds}f(X_T^{t,x})] \tag{2}$$

总结：凡是满足以上的所有条件的，都能写成期望公式(2)的格式，因此两种形式可以互通

Black-Scholes-PDE推导

1. 三个采用Feynman-Kac theorem的推导

【第一版-B站版】

根据B站杨维强老师的讲解，将Feynman-Kac theorem比作了一个连接SDE与PDE的桥梁

Recall：

$$SDE: dX_t = \mu(X_t)dt + \sigma(X_t)dB_t \ , \ X_0 = x$$

Feynman-Kac theorem 使用概率方法去解出PDE：

$$u(t,x) = E^X[e^{-\int_0^t\lambda(X_s)ds}f(X_t)]$$

其中：

$$PDE: u_t = \mu(x)u_x+\frac{1}{2}\sigma^2(x)u_{xx}-\lambda(x)u \ , \ u(0,x)=f(x)$$

有些资料会以另一个方式写(相当于做一个时间变换)：

$$PDE：u_t + \mu(x)u_x+\frac{1}{2}\sigma^2(x)u_{xx}-\lambda(x)u = 0 \ , \ u(T,x)=f(x)$$

由原有方程：

$$dX_t = rX_tdt + \sigma X_tdB_t\\ V(t,x) = E_Q[e^{-r(T-t)}h(X_T)|X_t=x]$$

当设 $\mu(x)=rx,\sigma(x)=\sigma x,\lambda(x)=r$ , 得到PDE：

$$rV = V_t + rxV_x+\frac{1}{2}\sigma^2x^2V_{xx} \ , \ V(T,x)=h(x)$$

【第二版-格拉版】

区别的地方是采用BM构建Process：

$$\begin{align} V(t,x) &= E_Q[e^{-r(T-t)}h(X_T)|X_t=x]\\ &\Downarrow\\ V(t,x) &= E_Q[e^{-r(T-t)}h(B(T))|B(t)=f(t,x)] = u(t,f(t,x)) \end{align}$$

其中$B(t)$的式子由：

$$\begin{align} X_t &= X_0e^{(r-\frac{\sigma^2}{2})t+\sigma B(t) }= x\\ B(t) &= \frac{ln(\frac{x}{X_0})-(r-\frac{\sigma^2}{2})t}{\sigma}=f(t,x) \end{align}$$

By the Feynman-Kac theorem with $\mu(x)=0,\sigma(x)=1$：

$$ru = u_t+\frac{1}{2}u_{ff}$$

为了求解，得以下三个式子：

$$\begin{align} V_t &= u_t + u_ff_t\\ V_x &= u_ff_x\\ V_{xx} &= u_{ff}(f_x)^2 + u_ff_{xx} \end{align}$$

依照之前的$f(t,x)$公式

$$\begin{align} f &= \frac{ln(\frac{x}{X_0})-(r-\frac{\sigma^2}{2})t}{\sigma}\\ f_t &= -\frac{1}{\sigma}(r-\frac{\sigma^2}{2})\\ f_x &= \frac{1}{\sigma x} \ (lnax = lna + lnx)\\ f_{xx} &= -\frac{1}{\sigma x^2} \end{align}$$

代入前一个式子并换位

$$\begin{align} V_t + \frac{1}{\sigma}(r-\frac{\sigma^2}{2})u_f &= u_t \\ \sigma xV_x &= u_f\\ \sigma^2 x^2V_{xx} + \frac{\sigma^2 x^2}{\sigma x^2}u_f &= u_{ff} \end{align}$$

代入最开始的式子

$$\begin{align} rV &= V_t +(r-\frac{\sigma^2}{2})xV_x +\frac{1}{2}(\sigma^2 x^2V_{xx} + \sigma^2 xV_x)\\ &= V_t +rxV_x-\frac{\sigma^2 x}{2}V_x +\frac{\sigma^2 x^2}{2}V_{xx} + \frac{\sigma^2 x}{2}V_x\\ &= V_t +rxV_x +\frac{\sigma^2 x^2}{2}V_{xx} \end{align}$$

【第三版-参考书】

$$\begin{align} V(t,x) &= E_Q[e^{-r(T-t)}V(X_T)|X_t=x]\\ g(t,x) &= e^{-rt}V(t,x) \end{align}$$

By the Feynman-Kac theorem with time in [0,T]：

$$g_t + rxg_x+\frac{1}{2}\sigma^2 x^2g_{xx}= 0$$

对g求导：

$$\begin{align} g_t &= -re^{rt}V + e^{-rt}V_t\\ g_x &= e^{-rt}V_x\\ g_{xx} &= e^{-rt}V_{xx} \end{align}$$

代入：

$$\begin{align} -re^{rt}V + e^{-rt}V_t + rxe^{-rt}V_x+\frac{1}{2}\sigma^2 x^2e^{-rt}V_{xx}= 0\\ -rV + V_t + rxV_x+\frac{1}{2}\sigma^2 x^2V_{xx}= 0\\ \end{align}$$

2. 对冲版推导

【第一版-Delta 对冲】

做一个Delta对冲的投资组合：

$$\Pi = C(X,t) - \Delta X \ , \ \Delta = \frac{\partial C}{\partial X}$$

Ito formula：

$$\begin{align} d\Pi &= \frac{\partial C}{\partial t}dt + \frac{\partial C}{\partial X}dX + \frac{1}{2}\sigma^2X^2\frac{\partial^2C}{\partial X^2}dt - \Delta dX\\ &= \frac{\partial C}{\partial t}dt + \frac{1}{2}\sigma^2X^2\frac{\partial^2C}{\partial X^2}dt + (\frac{\partial C}{\partial X} - \Delta)dX\\ &= \frac{\partial C}{\partial t}dt + \frac{1}{2}\sigma^2X^2\frac{\partial^2C}{\partial X^2}dt \end{align}$$

It should be equal to the same return as if we invested in riskless interest-bearing account：

$$d\Pi = r\Pi dt = r(C(X,t) - \Delta X)dt = r(C(X,t) - X \frac{\partial C}{\partial X})dt$$

because of no arbitrage：

$$\frac{\partial C}{\partial t}dt + \frac{1}{2}\sigma^2X^2\frac{\partial^2C}{\partial X^2}dt = r(C(X,t) - X \frac{\partial C}{\partial X})dt$$

so that：

$$\frac{\partial C}{\partial t} + rX \frac{\partial C}{\partial X} + \frac{1}{2}\sigma^2X^2\frac{\partial^2C}{\partial X^2} = rC$$

【第二版-Ito 公式】

建立一个期权的复制组合

• $\Delta(t)$：持有股票数

• $X(t)$：portfolio 价值

• 股票价值增量：$dS(t)=\alpha S(t)dt+\sigma S(t)dW(t)$

• portfolio 价值增量：超额收益+无风险收益+风险

  $$\begin{align} dX(t)&=\Delta(t)dS(t)+r(X(t)-\Delta(t)S(t))dt\\ &=\Delta(t)\alpha S(t)dt+\Delta(t)\sigma S(t)dW(t)+rX(t)dt-r\Delta(t)S(t)dt\\ &=(\alpha-r)\Delta(t)S(t)dt+rX(t)dt+\Delta(t)\sigma S(t)dW(t) \end{align}$$

股票价值的折现过程

$$\begin{align} d(e^{-rt}S(t))&=S(t)de^{-rt}+e^{-rt}dS(t)+dS(t)de^{-rt}\\ &=-re^{-rt}S(t)dt+e^{-rt}(\alpha S(t)dt+\sigma S(t)dW(t))\\ &=(\alpha-r)e^{-rt}S(t)dt+e^{-rt}\sigma S(t)dW(t) \end{align}$$

组合价值的折现过程

$$\begin{align} d(e^{-rt}X(t))&=X(t)de^{-rt}+e^{-rt}dX(t)+dX(t)de^{-rt}\\ &=-re^{-rt}X(t)dt+e^{-rt}(\alpha-r)\Delta(t)S(t)dt...\\ &...+e^{-rt}rX(t)dt+e^{-rt}\Delta(t)\sigma S(t)dW(t)\\ &=\Delta(t)(e^{-rt}(\alpha-r)S(t)dt+e^{-rt}\sigma S(t)dW(t))\\ &=\Delta(t)d(e^{-rt}S(t)) \end{align}$$

• $C(t,x)$：Option price

当复制成立时

$$d(e^{-rt}C(t,S(t)))=d(e^{-rt}X(t))$$

先准备

$$\begin{align} d(C(t,S(t)))&=C_tdt+C_xdS(t)+\frac12C_{xx}dS(t)dS(t)\\ &=C_tdt+C_x(\alpha S(t)dt+\sigma S(t)dW(t))...\\ &...+\frac12C_{xx}\sigma^2S(t)^2dt\\ &=(C_t+C_x\alpha S(t)+\frac12C_{xx}\sigma^2S(t)^2)dt+C_x\sigma S(t)dW(t) \end{align}$$

代入

$$\begin{align} d(e^{-rt}C(t,S(t)))&=Cde^{-rt}+e^{-rt}dC\\ &=-re^{-rt}Cdt+e^{-rt}C_x\sigma S(t)dW(t)...\\ &...+e^{-rt}(C_t+C_x\alpha S(t)+\frac12C_{xx}\sigma^2S(t)^2)dt\\ &=e^{-rt}(C_t+C_x\alpha S(t)+\frac12C_{xx}\sigma^2S(t)^2-rC)dt...\\ &...+e^{-rt}C_x\sigma S(t)dW(t) \end{align}$$

与组合价值的折现过程对比

$$\Delta(t)(e^{-rt}(\alpha-r)S(t)dt+e^{-rt}\sigma S(t)dW(t))$$

得出

$$\Delta(t)=C_x$$ $$rC=C_t+\frac12C_{xx}\sigma^2S(t)^2+rC_xS(t)$$

【附加资料】

Black-Scholes equation is written in the form ：

$$\frac{\partial C}{\partial t} + bX \frac{\partial C}{\partial X} + \frac{1}{2}\sigma^2X^2\frac{\partial^2C}{\partial X^2} - rC = 0$$

where：

Asset with on dividend	$b = r$
Asset with dividend $D$	$b = r - D$
Foreign currency with return $r_f$	$b = r - r_f$
Commodity with storage cost $q$	$b = r + q$
Futures	$b = 0$

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Reference

1. 金融数学课程

2. 孙健 《金融衍生品定价模型—数理金融引论 》中国经济出版社