笨潔旻畫蝴蘭

這篇文章是當初剛接觸期權時整理的心得筆記，需要微積分及微分方程的背景知識。由於文章用到的字句都是簡單的句子以及專有名詞，所以就不翻譯成中文了。

reference: https://ocw.mit.edu/courses/sloan-school-of-management/15-070j-advanced-stochastic-processes-fall-2013/index.htm

Definition 1. A random process $I_t$ is an Ito Process if and only if
$$dI_t=\mu(t,I_t)dt+\sigma(t,I_t)dB_t$$
,where $B_t$ is the standard Brownian motion process, that is $B_t\sim N(0,t)$. Moreover,
$$\left(\int_0^t (dB_\alpha)^2=t=\int_0^td\alpha\right)\Rightarrow \left((dB_t)^2=dt\right)$$

Lemma 2. （Ito's Lemma） Given a standard Brownian motion process $B_t$ and twice-differentiable scalar function $f(B_t)$, by intermediate value theorem,
$$\begin{align}f(B_{t+dt})&=f(B_t)+\frac{df(B_t)}{dB_t}(B_{t+dt}-B_t)+\frac{1}{2}\frac{d^2f(B_t)}{dB_t^2}(B_{t+dt}-B_t)^2\\
\rightarrow f(B_t)&=f(B_0)+\int_0^t \frac{df(B_\alpha)}{dB_\alpha}dB_\alpha + \frac{1}{2}\int_0^t \frac{d^2f(B_\beta)}{dB_\beta^2}dB_\beta^2\\
&=f(B_0)+\int_0^t \frac{df(B_\alpha)}{dB_\alpha}dB_\alpha + \frac{1}{2}\int_0^t \frac{d^2f(B_\beta)}{dB_\beta^2}d\beta\end{align}$$

• The additional term dt arises because Brownian motion B is not differentiable and instead has quadratic variation.

Example 3. Given a standard Brownian motion process $B_t$ and a function $S(t,B_t)$ which satisfies
$$dS(t,B_t)=\mu S(t,B_t)dt+\sigma S(t,B_t)dB_t$$
, where $\mu$ and $\sigma$ are scalars. Then,
$$\begin{align}dS(t,B_t)&=\frac{\partial S(t,B_t)}{\partial t}dt+\frac{\partial S(t,B_t)}{\partial B_t}dB_t+\frac{1}{2}\frac{\partial^2 S(t,B_t)}{\partial B_t^2}dt\\
&=\left(\frac{\partial S(t,B_t)}{\partial t}+\frac{1}{2}\frac{\partial^2 S(t,B_t)}{\partial B_t^2}\right)dt+\frac{\partial S(t,B_t)}{\partial B_t}dB_t\end{align}$$
. Compare these equations, we have
$$\begin{cases}\frac{\partial S(t,B_t)}{\partial t}+\frac{1}{2}\frac{\partial^2 S(t,B_t)}{\partial B_t^2}=\mu S(t,B_t)\\
\frac{\partial S(t,B_t)}{\partial B_t}=\sigma S(t,B_t)\end{cases}$$
. Therefore,
$$S(t,B_t)= S(0,B_0)e^{\sigma B_t}e^{(\mu-\frac{1}{2}\sigma^2)t}$$

Theorem 4. （Black-Scholes Model）Let $u(t)$ be the unit step function, then
$$\begin{align}&E\left\{e^{-rt}\left[S(t,B_t)-L\right]u\left[S(t,B_t)-L\right]\right\}\\
=&\int_{-\infty}^\infty e^{-rt}\left[S_0 e^{\sigma \alpha}e^{(\mu-\frac{1}{2}\sigma^2)t}-L\right]u\left[S_0 e^{\sigma \alpha}e^{(\mu-\frac{1}{2}\sigma^2)t}-L\right]f_{B_t}(\alpha)d\alpha\end{align}$$
. Let
$$i=-\frac{\ln\frac{L}{S_0}-\left(\mu-\frac{1}{2}\sigma^2\right)t}{\sigma\sqrt{t}}=\frac{\left(\mu-\frac{1}{2}\sigma^2\right)t+\ln\frac{S_0}{L}}{\sigma\sqrt{t}}$$
, then
$$\begin{align}&\int_{-\infty}^\infty e^{-rt}\left[S_0 e^{\sigma \alpha}e^{(\mu-\frac{1}{2}\sigma^2)t}-L\right]u\left[S_0 e^{\sigma \alpha}e^{(\mu-\frac{1}{2}\sigma^2)t}-L\right]f_{B_t}(\alpha)d\alpha\\
=&\int_{-i}^\infty e^{-rt}\left[S_0 e^{\sigma\sqrt{t} \beta}e^{(\mu-\frac{1}{2}\sigma^2)t}-L\right]f_{N(0,1)}(\beta)d\beta\\
=&\int_{-i}^\infty e^{-rt}\left[S_0 e^{\sigma\sqrt{t} \beta}e^{(\mu-\frac{1}{2}\sigma^2)t}\right]f_{N(0,1)}(\beta)d\beta-\int_{-i}^\infty e^{-rt}Lf_{N(0,1)}(\beta)d\beta\\
=&S_0 e^{(\mu-r)t}\int_{-i}^\infty e^{-\frac{1}{2}\sigma^2 t+\sigma\sqrt{t}\beta}\frac{1}{\sqrt{2\pi}}e^{-\frac{\beta^2}{2}}d\beta-Le^{-rt}\int_{-\infty}^i f_{N(0,1)}(\beta)d\beta\\
=&S_0 e^{(\mu-r)t}\int_{-i}^\infty \frac{1}{\sqrt{2\pi}}e^{-\frac{1}{2}(\sigma\sqrt{t}-\beta)^2}d\beta-Le^{-rt}N(i)\\
=&S_0 e^{(\mu-r)t}\int_{-i-\sigma\sqrt{t}}^\infty \frac{1}{\sqrt{2\pi}}e^{-\frac{\gamma^2}{2}}d\gamma-Le^{-rt}N(i)\\
=&S_0 e^{(\mu-r)t}\int_{-\infty}^{i+\sigma\sqrt{t}} \frac{1}{\sqrt{2\pi}}e^{-\frac{\gamma^2}{2}}d\gamma-Le^{-rt}N(i)\\
=&S_0 e^{(\mu-r)t}N(i+\sigma\sqrt{t})-Le^{-rt}N(i)\end{align}$$