【推导】BS PDE To BS Model

基于 ChiuFai WONG 教授的FM2课件中从 BS-PDE 到 BS-MODEL 的推导过程；

带有个人理解；

预备知识

Black-Scholes partial differential equation

$$\frac{\partial V}{\partial t}+\frac{1}{2} \sigma^{2} S^{2} \frac{\partial^{2} V}{\partial S^{2}}+r S \frac{\partial V}{\partial S}-r V=0$$

with boundary conditions

$$C(S, T)=\max (S(T)-K, 0), \quad C(0, t)=0$$

$S(t)$ with Brownian motion $Z(t)$ has the following features

$$dS(t)=\alpha S(t)dt+\sigma S(t)dZ(t)$$ $$S(t)=S(0)e^{\sigma Z(t)+(\alpha -\frac12 \sigma^2)t}$$

Step 1 换元化简

In order to get rid of $S \frac{\partial C}{\partial S}$ and $S^{2} \frac{\partial^{2} C}{\partial S^{2}}$ in Black-Scholes equation, we set

$$S(t)=K e^{x}, \quad t=T-\frac{2 \tau}{\sigma^{2}}, \quad C(S, t)=K v(x, \tau)=K v\left(\ln \frac{S(t)}{K}, \frac{\sigma^{2}}{2}(T-t)\right)$$

这步主要排除偏导前 S 的影响, 同时为了配合 S 把 K 消除

So we have

$$\frac{\partial C}{\partial t}=-\frac{\sigma^{2} K}{2} \frac{\partial v}{\partial \tau}, \quad \frac{\partial C}{\partial S}=\frac{K}{S} \frac{\partial v}{\partial x}, \quad \frac{\partial^{2} C}{\partial S^{2}}=-\frac{K}{S^{2}} \frac{\partial v}{\partial x}+\frac{K}{S^{2}} \frac{\partial^{2} v}{\partial x^{2}}$$

Placing these expressions into the Black-Scholes partial differential equation and simplifying, we have

$$\frac{\partial^{2} v}{\partial x^{2}}+(k-1)\frac{\partial v}{\partial x}-kv=\frac{\partial v}{\partial \tau}, \quad -\infty<x<\infty, \quad 0\leq\tau\leq\frac{\sigma^2}2T$$

where $k=2\frac r{\sigma^{2}}$. The initial condition becomes

$$v(x, 0)=\frac{1}{K} C(S, T)=\max \left(e^{x}-1,0\right)$$

In order to eliminate $v$ and $\frac{\partial v}{\partial x}$, we set

$$v(x, \tau)=e^{a x+b \tau} u(x, \tau)$$

Computing the partials of $v$ in terms of $x$ and $\tau$ , we have

$$\begin{aligned} \frac{\partial v}{\partial \tau}&=b e^{a x+b \tau} u(x, \tau)+e^{a x+b \tau} \frac{\partial u}{\partial \tau} \\ \frac{\partial v}{\partial x}&=a e^{a x+b \tau} u(x, \tau)+e^{a x+b \tau} \frac{\partial u}{\partial x} \\ \frac{\partial^{2} v}{\partial x^{2}}&=a^{2} e^{a x+b \tau} u(x, \tau)+2 a e^{a x+b \tau} \frac{\partial u}{\partial x}+e^{a x+b \tau} \frac{\partial^{2} u}{\partial x^{2}} \end{aligned}$$

Placing these expressions into the partial differential equation

$$\begin{aligned} a^{2}u+2 a\frac{\partial u}{\partial x}+\frac{\partial^{2} u}{\partial x^{2}}+(k-1)(au+\frac{\partial u}{\partial x})-ku&=bu+\frac{\partial u}{\partial \tau} \\ \frac{\partial^{2} u}{\partial x^{2}}+(2a+k-1)\frac{\partial u}{\partial x}+(a^{2}+a(k-1)-b-k)u&=\frac{\partial u}{\partial \tau} \end{aligned}$$

So we can obtain an equation by choosing

$$\begin{aligned} &b=a^2+(k-1)a-k,\quad 2a=1-k\\ \Rightarrow\quad &a=-\frac12(k-1)=\frac{\sigma^2-2r}{2\sigma^2}, \quad b=-\frac14(k+1)^2=-(\frac{\sigma^2+2r}{2\sigma^2})^2\\ \Rightarrow\quad & v(x, \tau)=e^{-\frac{1}{2}(k-1) x-\frac{1}{4}(k+1)^{2} \tau} u(x, \tau) \end{aligned}$$

We then have the heat equation

$$\frac{\partial u}{\partial \tau}=\frac{\partial^{2} u}{\partial x^{2}}, \quad-\infty<x<\infty, \quad 0 \leq \tau \leq \frac{\sigma^{2}}{2} T$$

with initial condition

$$u(x, 0)=e^{-\alpha x} v(x, 0)=\max \left(e^{\frac{1}{2}(k+1) x}-e^{\frac{1}{2}(k-1) x}, 0\right)$$

Step 2 使用Laplace变换的方法求热传导公式

知识点回顾 Fourier Transform

Define

$$U(x, s)=\mathfrak{L}_{s}(u(x, \tau)):=\int_{0}^{\infty} e^{-s \tau} u(x, \tau) d \tau$$

where $\mathfrak{L}_{s}$​ is the Laplace operator with parameter $s$. Then, we have

$$sU(x, s)-u(x, 0)=U_{x x}(x, s)$$

Its characteristic equation is $r^2-s=0$. The roots are $r_1=\sqrt{s},r_2=-\sqrt{s}$

The homogeneous solution is

$$U_{h}(x, s)=c_{1} e^{x \sqrt{s}}+c_{2} e^{-x \sqrt{s}}$$

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Before calculating particular solution, we need Definition 4.12 and Theorem 4.13

Definition 4.12

Define error function $\operatorname{erf}(x)=\frac{1}{\sqrt{\pi}} \int_{-x}^{x} e^{-t^{2}} d t=\frac{2}{\sqrt{\pi}} \int_{0}^{x} e^{-t^{2}} d t$ and complementary error function $\operatorname{erfc}(x)=1-\operatorname{erf}(x)=\frac{2}{\sqrt{\pi}} \int_{x}^{\infty} e^{-t^{2}} d t$

Theorem 4.13

Suppose $\frac{\partial u}{\partial \tau}=\frac{\partial^{2} u}{\partial x^{2}}$ with $u(x, 0)=u(\infty, \tau)=0, u(0, \tau)=1$ . Then we have

1. $u(x, \tau)=\frac{2}{\sqrt{\pi}} \int_{\frac{x}{2 \sqrt{\tau}}}^{\infty} e^{-t^{2}} d t$ .
2. $\mathfrak{L}_{s}\left(\frac{1}{2 \sqrt{\pi \tau}} e^{-\frac{x^{2}}{4 \tau}}\right)=\frac{e^{-x \sqrt{s}}}{2 \sqrt{s}} \text { for } x>0$ .

proof 1

Let $\xi=\frac{x}{\sqrt{\tau}}$ such that $u(x, \tau)=U(\xi)$

此处的 $U$ 与之前的不相同，这里表示的是以 $\xi$ 为变量的 $u$ 的另外一个形式

$$\begin{aligned} &\frac{\partial u}{\partial \tau}=U^{\prime}(\xi) \frac{d \xi}{d \tau}=U^{\prime}(\xi)\left(-\frac{1}{2} x \tau^{-\frac{3}{2}}\right)=-\frac{1}{2 \tau} \xi U^{\prime}(\xi) \\ &\frac{\partial u}{\partial x}=U^{\prime}(\xi) \frac{d \xi}{d x}=U^{\prime}(\xi) \frac{1}{\sqrt{\tau}} \\ &\frac{\partial^{2} u}{\partial x^{2}}=U^{\prime \prime}(\xi) \frac{d \xi}{d x} \frac{1}{\sqrt{\tau}}=U^{\prime \prime}(\xi) \frac{1}{\tau} \end{aligned}$$ $$\frac{\partial^{2} u}{\partial x^{2}}=\frac{\partial u}{\partial \tau} \Rightarrow U^{\prime \prime}(\xi) \frac{1}{\tau}=-\frac{1}{2 \tau} \xi U^{\prime}(\xi) \Rightarrow \frac{U^{\prime \prime}(\xi)}{U^{\prime}(\xi)}=-\frac{1}{2} \xi \Rightarrow \int \frac{U^{\prime \prime}(\xi)}{U^{\prime}(\xi)} d \xi=\int-\frac{1}{2} \xi d \xi$$ $$\begin{aligned} &U^{\prime}(\xi)=C e^{-\frac{\xi^{2}}{4}} \\ &U(\xi)=C \int_{0}^{\xi} e^{-\frac{s^{2}}{4}} d s+D \\ &u(x, 0)=u(\infty, \tau)=0, u(0, \tau)=1 \Rightarrow U(0)=1, U(\infty)=0\\ \end{aligned}$$

Recall $\int_{0}^{\infty} e^{-\frac{s^{2}}{4}} d s=\sqrt{\pi}$ . We have $D=1, \quad C=-\frac{1}{\sqrt{\pi}}$

$$U(\xi)=\frac{1}{\sqrt{\pi}} \int_{\xi}^{\infty} e^{-\frac{s^{2}}{4}} d s$$ $$u(x, \tau)=\frac{1}{\sqrt{\pi}} \int_{\frac{x}{\sqrt{\tau}}}^{\infty} e^{-\frac{s^{2}}{4}} d s \quad \stackrel{t=\frac{s}{2}}{=} \frac{2}{\sqrt{\pi}} \int_{\frac{x}{2 \sqrt{\tau}}}^{\infty} e^{-t^{2}} d t=\operatorname{erfc}\left(\frac{x}{2 \sqrt{\tau}}\right)$$

proof 2

We have found that

$$U(x, s)=c_{1} e^{x \sqrt{s}}+c_{2} e^{-x \sqrt{s}}$$ $$\begin{aligned} &U(\infty, s)=\mathfrak{L}_{s}(u(\infty, \tau))=0 \Rightarrow c_{1}=0 \\ &U(0, s)=\mathfrak{L}_{s}(u(0, \tau))=\int_{0}^{\infty} e^{-s \tau} d \tau=\frac{1}{s} \Rightarrow c_{2}=\frac{1}{s} \end{aligned}$$

So we have both $u(x,\tau)$ and $U(x,s)$

$$\frac{e^{-x \sqrt{s}}}{s}=U(x, s)=\mathfrak{L}_{s}\left(\operatorname{erfc}\left(\frac{x}{2 \sqrt{\tau}}\right)\right)$$

Since $\mathfrak{L}_{s}\left(f^{\prime}(\tau)\right)=s \mathfrak{L}_{s}(f(\tau))-\lim _{\tau \rightarrow 0^{+}} f(\tau)$,

$$\begin{aligned} \mathfrak{L}_{s}\left(\frac{d}{d \tau} \operatorname{erf} c\left(\frac{x}{2 \sqrt{\tau}}\right)\right)&=s \mathfrak{L}_{s}\left(\operatorname{erfc}\left(\frac{x}{2 \sqrt{\tau}}\right)\right)-\operatorname{erfc}(\infty) \quad \text { if } x>0 \\ \mathfrak{L}_{s}\left(\left(\frac{x}{2}\right)\left(-\frac{1}{2}\right)\left(\frac{1}{\sqrt{\tau^{3}}}\right) \frac{-2}{\sqrt{\pi}} e^{-\frac{x^{2}}{4 \tau}}\right)&=e^{-x \sqrt{s}}+0 \\ \mathfrak{L}_{s}\left(\frac{-x \tau}{2 \sqrt{\pi \tau^{3}}} e^{-\frac{x^{2}}{4 \tau}}\right)=\frac{d}{d s} \mathfrak{L}_{s}\left(\frac{x}{2 \sqrt{\pi \tau^{3}}} e^{-\frac{x^{2}}{4 \tau}}\right)&=\frac{d}{d s} e^{-x \sqrt{s}}=\frac{-x e^{-x \sqrt{s}}}{2 \sqrt{s}}\end{aligned}$$

第三行则是考虑到了此类变换的求导性质

• $f^{(n)}(\tau)\leftrightarrow(s)^ng(s)$
• $(-i\tau)^nf(\tau)\leftrightarrow g^{(n)}(s)$

We have

$$\mathfrak{L}_{s}\left(\frac{1}{2 \sqrt{\pi \tau}} e^{-\frac{x^{2}}{4 \tau}}\right)=\frac{e^{-x \sqrt{s}}}{2 \sqrt{s}}$$

同时除 $x$ 并且消掉 $\tau$ 后得到

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Let us come back to particular solution.

$$y_{1}(x)=e^{x \sqrt{s}} \text { and } y_{2}(x)=e^{-x \sqrt{s}}$$

The Wronskian of $y_1(x)$ and $y_2(x)$ is

$$W\left(y_{1}, y_{2}\right)=y_{1} y_{2}^{\prime}-y_{1}^{\prime} y_{2}=-\sqrt{s} e^{x \sqrt{s}} e^{-x \sqrt{s}}-\sqrt{s} e^{x \sqrt{s}} e^{-x \sqrt{s}}=-2 \sqrt{s}$$

Recall that

$$U_{x x}(x, s)-sU(x, s)=-u(x, 0)$$

Then, the particular solution is

$$U_{p}(x, s)=u_{1}(x) y_{1}(x)+u_{2}(x) y_{2}(x)=u_{1}(x) e^{x \sqrt{s}}+u_{2}(x) e^{-x \sqrt{s}}$$

with the computating of Nonhomogeneous Equations

$$u_{1}^{\prime}(x)=\frac{y_{2}(x) u(x, 0)}{W\left(y_{1}, y_{2}\right)(x)}=-\frac{e^{-x \sqrt{s}} u(x, 0)}{2 \sqrt{s}} \text { and } u_{2}^{\prime}(x)=-\frac{y_{1}(x) u(x, 0)}{W\left(y_{1}, y_{2}\right)(x)}=\frac{e^{x \sqrt{s}} u(x, 0)}{2 \sqrt{s}} \text {. }$$

So

$$\begin{aligned} &u_{1}(x)=-\int \frac{e^{-x \sqrt{s}} u(x, 0)}{2 \sqrt{s}} d x=-\int_{\infty}^{x} \frac{e^{-\xi \sqrt{s}} u(\xi, 0)}{2 \sqrt{s}} d \xi=\int_{x}^{\infty} \frac{e^{-\xi \sqrt{s}} u(\xi, 0)}{2 \sqrt{s}} d \xi\\ &u_{2}(x)=\int \frac{e^{x \sqrt{s}} u(x, 0)}{2 \sqrt{s}} d x=\int_{-\infty}^{x} \frac{e^{\xi \sqrt{s}} u(\xi, 0)}{2 \sqrt{s}} d \xi \end{aligned}$$

11th ODE教材中写到，对于求积的下限 $t_0$ “ t0 is any conveniently chosen point in interval I ”

$$\begin{aligned} U_{p}(x, s) &=u_{1}(x) e^{x \sqrt{s}}+u_{2}(x) e^{-x \sqrt{s}} \\ &=e^{x \sqrt{s}} \int_{x}^{\infty} \frac{e^{-\xi \sqrt{s}} u(\xi, 0)}{2 \sqrt{s}} d \xi+e^{-x \sqrt{s}} \int_{-\infty}^{x} \frac{e^{\xi \sqrt{s}} u(\xi, 0)}{2 \sqrt{s}} d \xi \\ &=\int_{x}^{\infty} \frac{e^{-(\xi-x) \sqrt{s}} u(\xi, 0)}{2 \sqrt{s}} d \xi+\int_{-\infty}^{x} \frac{e^{-(x-\xi) \sqrt{s}} u(\xi, 0)}{2 \sqrt{s}} d \xi \\ &=\int_{-\infty}^{\infty} \frac{e^{-|\xi-x| \sqrt{s}} u(\xi, 0)}{2 \sqrt{s}} d \xi \end{aligned}$$

Thus, to get $u(x,\tau)$ it has to be done inverse Laplace operation

$$\begin{aligned} u(x, \tau) &=\mathfrak{L}_{s}^{-1}(U(x, s)) \\ &=\mathfrak{L}_{s}^{-1}\left(\int_{-\infty}^{\infty} \frac{e^{-|\xi-x| \sqrt{s}} u(\xi, 0)}{2 \sqrt{s}} d \xi\right) \\ &=\int_{-\infty}^{\infty} \mathfrak{L}_{s}^{-1}\left(\frac{e^{-|\xi-x| \sqrt{s}}}{2 \sqrt{s}}\right) u(\xi, 0) d \xi \\ &=\frac{1}{2 \sqrt{\pi \tau}} \int_{-\infty}^{\infty} e^{-\frac{(\xi-x)^{2}}{4 \tau}} u(\xi, 0) d \xi \end{aligned}$$

Step 3 将系数代回

It is convenient to make the change of variable $x_{1}=\frac{\xi-x}{\sqrt{2 \tau}}$， so that

$$\begin{aligned} u(x, \tau) &=\frac{1}{\sqrt{2 \pi}} \int_{-\infty}^{\infty} u\left(x_{1} \sqrt{2 \tau}+x, 0\right) e^{-\frac{x_{1}^{2}}{2}} d x_{1} \\ &=\underbrace{\frac{1}{\sqrt{2 \pi}} \int_{-\frac{x}{\sqrt{2 \tau}}}^{\infty} e^{\frac{1}{2}(k+1)\left(x+x_{1} \sqrt{2 \tau}\right)} e^{-\frac{x_{1}^{2}}{2}} d x_{1}}_{I_{1}}-\underbrace{\frac{1}{\sqrt{2 \pi}} \int_{-\frac{x}{\sqrt{2 \tau}}}^{\infty} e^{\frac{1}{2}(k-1)\left(x+x_{1} \sqrt{2 \tau}\right)} e^{-\frac{x_{1}^{2}}{2}} d x_{1}}_{I_{2}} \end{aligned}$$ $$\begin{aligned} I_{1} &=\frac{1}{\sqrt{2 \pi}} \int_{-\frac{x}{\sqrt{2 \tau}}}^{\infty} e^{\frac{1}{2}(k+1)\left(x+x_{1} \sqrt{2 \tau}\right)} e^{-\frac{x_1^{2}}{2}} d x_{1} \\ &=\frac{e^{\frac{1}{2}(k+1) x}}{\sqrt{2 \pi}} \int_{-\frac{x}{\sqrt{2 \tau}}}^{\infty} e^{\frac{1}{4}(k+1)^{2} \tau} e^{-\frac{1}{2}\left(x_{1}-\frac{1}{2}(k+1) \sqrt{2 \tau}\right)^{2}} d x_{1} \\ &=\frac{e^{\frac{1}{2}(k+1) x+\frac{1}{4}(k+1)^{2} \tau}}{\sqrt{2 \pi}} \int_{-\frac{x}{\sqrt{2 \tau}}-\frac{1}{2}(k+1) \sqrt{2 \tau}}^{\infty} e^{-\frac{1}{2} \rho^{2}} d \rho \quad\left[\rho=x_{1}-\frac{1}{2}(k+1) \sqrt{2 \tau}\right] \\ &=e^{\frac{1}{2}(k+1) x+\frac{1}{4}(k+1)^{2} \tau} N\left(d_{1}\right) \end{aligned}$$

Similarly, $I_{2}=e^{\frac{1}{2}(k-1) x+\frac{1}{4}(k-1)^{2} \tau} N\left(d_{2}\right)$

$$\begin{aligned} v(x, \tau) &=e^{\alpha x+\beta t} u(x, \tau) \\ &=e^{-\frac{1}{2}(k-1) x-\frac{1}{4}(k+1)^{2} \tau}\left(e^{\frac{1}{2}(k+1) x+\frac{1}{4}(k+1)^{2} \tau} N\left(d_{1}\right)-e^{\frac{1}{2}(k-1) x+\frac{1}{4}(k-1)^{2} \tau} N\left(d_{2}\right)\right) \\ &=e^{x} N\left(d_{1}\right)-e^{-k \tau} N\left(d_{2}\right) \\ &=\frac{S(t)}{K} N\left(d_{1}\right)-e^{-\frac{2 r}{\sigma^{2}} \frac{\sigma^{2}}{2}(T-t)} N\left(d_{2}\right) \\ &=\frac{S(t)}{K} N\left(d_{1}\right)-e^{-r(T-t)} N\left(d_{2}\right) \end{aligned}$$

额外资料-求解 Black-Scholes 偏微分方程