【推导】Black Scholes Model

基于格拉ECO5009课件；

省略了关于测度变换的部分，直接求解；

是比较早的笔记，完整的可以看进阶版；

前情提要

• stock price 基于GBM：

$$dS = \mu Sdt + \sigma SdW$$

• Ito’s lemma：

$$df = (\frac{\partial f}{\partial S}\mu S + \frac{\partial f}{\partial t} + \frac{1}{2}\frac{\partial^2 f}{\partial S^2}\sigma^2S^2)dt + \frac{\partial f}{\partial S}\sigma SdW$$

推导

• let f = lnS, we have：

$$\frac{\partial f}{\partial S} = \frac{1}{S}, \frac{\partial^2 f}{\partial S^2} = \frac{1}{S^2}, \frac{\partial f}{\partial t} = 0$$

• so ：

$$dlnS_t = (\mu - \frac{\sigma^2}{2})dt +\sigma dW_t$$

• Integrating equation from time 0 to t and taking the exponential：

$$S_t = S_0exp((\mu - \frac{\sigma^2}{2})t +\sigma W_t)$$

• In the distribution $W_t \sim \epsilon \sqrt{t}$

$$S_t = S_0exp((\mu - \frac{\sigma^2}{2})t +\sigma \epsilon \sqrt{t})$$

• With the risk-neutral valuation, where $\tau = T-t$：

$$c = e^{-r \tau}\widetilde{E}_t((S_T-K)^+)$$

• 代入$S_T$（老师写的是$-\sigma \epsilon \sqrt{\tau}$ ,但说不影响EXP），$y \sim N(0,1)$

$$\begin{align} c &= e^{-r \tau}\widetilde{E}_t((S_texp((r - \frac{\sigma^2}{2})\tau +\sigma \sqrt{\tau}y)-K)^+)\\ &(对y求积分)\\ &= \int_{-\infty}^{\infty} \frac{1}{\sqrt{2\pi}}e^{-\frac{1}{2}y^2} e^{-r \tau}(S_texp((r - \frac{\sigma^2}{2})\tau +\sigma \sqrt{\tau}y)-K)^+dy \end{align}$$

• 分析，存在某一个点使$()^+$部分从正数到0

$$\begin{align} S_texp((r - \frac{\sigma^2}{2})\tau +\sigma \sqrt{\tau}y) &> K\\ y &> (ln\frac{K}{S_t}-(r - \frac{\sigma^2}{2})\tau)\frac{1}{\sigma \sqrt{\tau}}\\ y &< (ln\frac{S_t}{K}+(r - \frac{\sigma^2}{2})\tau)\frac{1}{\sigma \sqrt{\tau}}=d2 \end{align}$$

• 代入前一个式子：

$$\begin{align} c &= \int_{-\infty}^{d2} \frac{1}{\sqrt{2\pi}}e^{-\frac{1}{2}y^2} e^{-r \tau}(S_texp((r - \frac{\sigma^2}{2})\tau +\sigma \sqrt{\tau}y)-K)dy\\ &= \int_{-\infty}^{d2} \frac{1}{\sqrt{2\pi}}e^{-\frac{1}{2}y^2} e^{-r \tau}(S_texp(r - \frac{\sigma^2}{2})\tau +\sigma \sqrt{\tau}y)dy - \int_{-\infty}^{d2} \frac{1}{\sqrt{2\pi}}e^{-\frac{1}{2}y^2} e^{-r \tau}Kdy\\ &= \frac{1}{\sqrt{2\pi}}S_t\int_{-\infty}^{d2} e^{-r \tau}e^{-\frac{1}{2}y^2} e^{(r - \frac{\sigma^2}{2})\tau +\sigma \sqrt{\tau}y}dy - e^{-r \tau}KN(d2)\\ &= \frac{1}{\sqrt{2\pi}}S_t\int_{-\infty}^{d2}e^{-\frac{1}{2}(y-\sigma \sqrt{\tau})^2}dy - e^{-r \tau}KN(d2)\\ \end{align}$$

• 第一部分去掉$S_T$相当于：

$$\begin{align} N(y-\sigma \sqrt{\tau}<d2) &= N(y<d2+\sigma \sqrt{\tau}) = N(d1)\\ d1 &= (ln\frac{S_t}{K}+(r + \frac{\sigma^2}{2})\tau)\frac{1}{\sigma \sqrt{\tau}} \end{align}$$

• 得到最终结果：

$$c = S_tN(d1)- e^{-r \tau}KN(d2)\\$$

MATLAB公式

[Call,Put] = blsprice(Price,Strike,Rate,Time,Volatility)