[隨機分析] How to solve SDE practically (4) - Geometric Brownian Motion

這次要介紹的是如何直接求解隨機微分方程 (SDE)? 我們這邊將以 Geometric Brownian Motion 為例：

考慮如下 SDE:
\[
dX_t = \mu X_t dt + \sigma X_t dB_t
\] 其中 $\mu, \sigma$ 為固定常數滿足 $-\infty < \mu < \infty, \ \sigma >0$ ，且給定初始條件 $X_0$

Comment: 
1. 上述 SDE 稱為 Geometric Brownian Motion (GBM)。此 process 在財務上有重要的應用：EX: GBM 為股價波動的基本模型。

2. Compare to Arithmetic Brownian Motion (ABM):
\[
dX_t = \mu dt + \sigma dB_t
\]
Sol：
首先檢驗 Uniqueness 與 Existence ：

由 SDE 的 Uniqueness 與 Existence Theorem，
考慮 $t \in [0,T]$，SDE:
\[
dX_t = \mu(t,X_t)dt + \sigma(t,X_t) dB_t, \ X(0)=x_0
\]若其係數 $\mu, \sigma$滿足  Lipschitz condition
\[
|\mu(t,x) - \mu(t,y)|^2 + |\sigma(t,x) - \sigma(t,y)|^2 \leq K |x-y|^2
\]與 Growth condition
\[
|\mu(t,x)|^2 + |\sigma(t,x)|^2 \leq K(1 + |x|^2)
\]則我們的 SDE 解存在且唯一。

故此 我們首先檢驗 Lipschitz Condition : 觀察 GBM 可知 $\mu(t, X_t) =\mu X_t$, $\sigma(t,X_t) = \sigma X_t $，計算
\[
 |(\mu x) - (\mu y){|^2} + |(\sigma x) - (\sigma y){|^2} = \left( {{\mu ^2} + {\sigma ^2}} \right)|x - y{|^2}
\]令 $K \geq {{\mu ^2} + {\sigma ^2}}$，則我們有如下關係：
\[|(\mu x) - (\mu y){|^2} + |(\sigma x) - (\sigma y){|^2} \le K|x - y{|^2}
\]亦即滿足 Lipschitz Condition。

接著我們檢驗 Growth Condition：
\[|\mu x{|^2} + |\sigma x{|^2} = \left( {{\mu ^2} + {\sigma ^2}} \right){x^2}
\]故我們可選與 Lipschitz condition 相同的 $K$ 滿足 $K \geq {{\mu ^2} + {\sigma ^2}}$ ，則我們有
\[|\mu x{|^2} + |\sigma x{|^2} \le K\left( {{{\left| x \right|}^2}} \right) \le K\left( {1 + {{\left| x \right|}^2}} \right)
\]亦即滿足 Growth Condition。

由於 Lipschitz 與 Growth Condition皆滿足，故由 SDE 的 Uniqueness 與 Existence Theorem 我們可知 GBM 解存在且唯一。

故現在開始求解：
改寫 GBM 如下
\[\begin{array}{l}
d{X_t} = \mu {X_t}dt + \sigma {X_t}d{B_t}\\
 \Rightarrow d{X_t} - \left( {\mu dt + \sigma d{B_t}} \right){X_t} = 0
\end{array}
\]定義積分因子：
\[{U_t}: = {e^{ - \int_0^t {\left( {\mu dt + \sigma d{B_t}} \right) + \frac{1}{2}{\sigma ^2}t} }} = {e^{ - \mu t + \frac{1}{2}{\sigma ^2}t - \sigma {B_t}}}
\] 由 Integration by part，計算
\[d\left( {{X_t}{U_t}} \right) = {X_t}d{U_t} + {U_t}d{X_t} + d\left\langle {{X_t},{U_t}} \right\rangle \ \ \ \ (*)
\]其中
\[\left\{ \begin{array}{l}
d{U_t} = \left( { - \mu  + \frac{1}{2}{\sigma ^2}} \right){U_t}dt - \sigma {U_t}d{B_t} + \frac{1}{2}\sigma {U_t}dt\\
 \Rightarrow d{U_t} = {U_t}\left[ { - \mu dt + {\sigma ^2}dt - \sigma d{B_t}} \right]\\
d\left\langle {{X_t},{U_t}} \right\rangle  = d{X_t} \cdot d{U_t}\\
 = \left( {\mu {X_t}dt + \sigma {X_t}d{B_t}} \right)\left( {{U_t}\left[ { - \mu dt + {\sigma ^2}dt - \sigma d{B_t}} \right]} \right) =  - {\sigma ^2}{X_t}{U_t}dt
\end{array} \right.
\]注意到上述結果 $dU_t$ 為利用 Ito Formula 而得。

現在將上述結果代入 $(*)$ 可得
\[\begin{array}{l}
 \Rightarrow d\left( {{X_t}{U_t}} \right) = {X_t}{U_t}\left[ { - \mu dt + {\sigma ^2}dt - \sigma d{B_t}} \right]\\
\begin{array}{*{20}{c}}
{}&{}&{}&{}&{}&{}
\end{array} + {U_t}\left( {\mu {X_t}dt + \sigma {X_t}d{B_t}} \right) - {\sigma ^2}{X_t}{U_t}dt\\
 \Rightarrow d\left( {{X_t}{U_t}} \right) = 0
\end{array}
\]現在將其轉換回積分形式：
\[\begin{array}{l}
d\left( {{X_t}{U_t}} \right) = 0\\
 \Rightarrow {X_t}{U_t} - {X_0}{U_0} = 0\\
 \Rightarrow {X_t}{U_t} = {X_0}{U_0} = {X_0}\\
 \Rightarrow {X_t}{e^{ - \mu t + \frac{1}{2}{\sigma ^2}t - \sigma {B_t}}} = {X_0}\\
 \Rightarrow {X_t} = {X_0}{e^{\mu t - \frac{1}{2}{\sigma ^2}t + \sigma {B_t}}}
\end{array}
\] 故我們得到一個 GBM 的解，現在我們檢驗此解確實滿足 GBM：
定義
\[
{X_t} = {X_0}\underbrace {{e^{\mu t - \frac{1}{2}{\sigma ^2}t + \sigma {B_t}}}}_{: = {V_t}}: = {X_0}{V_t}
\]利用 integration by part：
\[d{X_t} = d\left( {{X_0}{V_t}} \right) = {X_0}d\left( {{V_t}} \right) + {V_t}d\left( {{X_0}} \right) + d\left\langle {{X_0},{V_t}} \right\rangle  \ \ \ \ (\star)
\]其中
\[\left\{ \begin{array}{l}
d{V_t} = {V_t}\left[ {\left( {\mu  - \frac{1}{2}{\sigma ^2}} \right)dt + \sigma d{B_t} + \frac{1}{2}{\sigma ^2}dt} \right] = {V_t}\left[ {\mu dt + \sigma d{B_t}} \right]\\
d\left( {{X_0}} \right) = 0\\
d\left\langle {{X_0},{V_t}} \right\rangle  = d{X_0}d{V_t} = 0\\
\end{array} \right.
\]將上述結果代入 $\star$ 可得
\[\begin{array}{l}
d{X_t} = {X_0}d\left( {{V_t}} \right) + {V_t}d\left( {{X_0}} \right) + d\left\langle {{X_0},{V_t}} \right\rangle \\
 \Rightarrow d{X_t} = {X_0}{V_t}\left[ {\mu dt + \sigma d{B_t}} \right] = {X_0}{e^{\mu t - \frac{1}{2}{\sigma ^2}t + \sigma {B_t}}}\left[ {\mu dt + \sigma d{B_t}} \right]
\end{array}
\] 又因為我們之前定義 $X_t := X_0 V_t$，故 ${X_t}{e^{ - \mu t + \frac{1}{2}{\sigma ^2}t - \sigma {B_t}}} = {X_0}$，將此結果也代入上式中的 $X_0$ 我們得到
\[\begin{array}{l}
d{X_t} = {X_t}{e^{ - \mu t + \frac{1}{2}{\sigma ^2}t - \sigma {B_t}}}{e^{\mu t - \frac{1}{2}{\sigma ^2}t + \sigma {B_t}}}\left[ {\mu dt + \sigma d{B_t}} \right]\\
 \Rightarrow d{X_t} = \mu {X_t}dt + \sigma {X_t}d{B_t}
\end{array}
\]此為 GBM ，故 ${X_t} = {X_0}{e^{\mu t - \frac{1}{2}{\sigma ^2}t + \sigma {B_t}}}$ 確實為 GBM 的 (唯一) 解。 $\square$

Comments:
關於上述的 GBM 的解可寫作下列形式
\[{X_t} = {X_0}{e^{\mu t - \frac{1}{2}{\sigma ^2}t + \sigma {B_t}}} = {X_0}{e^{t\left( {\mu  - \frac{1}{2}{\sigma ^2} + \sigma \frac{{{B_t}}}{t}} \right)}}
\]注意到上式中的 $\frac{B_t}{t} \rightarrow 0 \ \text{almost surely}$ 當 $t \rightarrow \infty$ ，(此結果可利用 Strong Law of Large number 證明，在此略過)，故現在觀察上式，當 $t \rightarrow \infty$ 我們得到以下的結果：
\[{X_t} = \left\{ \begin{array}{l}
\infty ,\begin{array}{*{20}{c}}
{}&{}
\end{array}\text{if}\begin{array}{*{20}{c}}
{}
\end{array}\mu  > \frac{1}{2}{\sigma ^2}\\
0,\begin{array}{*{20}{c}}
{}&{}
\end{array}\text{if}\begin{array}{*{20}{c}}
{}
\end{array}\mu  < \frac{1}{2}{\sigma ^2}\\
{X_0}{e^{\sigma {B_t}}},\begin{array}{*{20}{c}}
{}&{}
\end{array}\text{if}\begin{array}{*{20}{c}}
{}
\end{array}\mu  = \frac{1}{2}{\sigma ^2}
\end{array} \right.
\]