[隨機分析] How to solve SDE practically (5) - Stochastic Exponential

Again，這是這系列的第五篇，這次來看看一個稍微不一樣的例子

Example 1: Stochastic Exponential for Standard Brownian Motion
令 $\lambda \in \mathbb{R}$，考慮如下 Stochastic Process:
\[
X_t = 1 + \lambda \int_0^t X_s dB_s \ \ \ \ (*)
\]試求解上式。

Comment:
回憶在微分方程，
\[
g(t) = 1 + \lambda \int_0^t g(s) ds
\]我們有 Unique 的解且具備 Exponential 形式： $g(t) := e^{\lambda t}$ 。故可以想見式 $(*)$ 應該也有此 Exponential 的解的樣子。

Solution
首先將 $(*)$ 改寫成微分形式：
\[\begin{array}{l}
{X_t} = 1 + \lambda \int_0^t {{X_s}} d{B_s}\\
 \Rightarrow d{X_t} = \lambda {X_t}d{B_t}
\end{array}
\] 這邊我們直接求解 (這邊略過此SDE的 唯一性 與 存在性 檢驗，直接求解。)
 進一步改寫 $d{X_t} - \lambda {X_t}d{B_t} = 0$ 並且定義積分因子 (透過一點 trial and error)：
\[{U_t}: = {e^{ - \lambda \int_0^t {d{B_s}}  + \frac{1}{2}{\lambda ^2}t}} = {e^{ - \lambda {B_t} + \frac{1}{2}{\lambda ^2}t}}
\]利用 Integration by part (Product rule) 計算 $d(X_t U_t)$：
\[
 d\left( {{X_t}{U_t}} \right) = {X_t}d\left( {{U_t}} \right) + {U_t}d\left( {{X_t}} \right) + d\left\langle {{U_t},{X_t}} \right\rangle  \ \ \ \ (\star)
\]其中
\[\left\{ \begin{array}{l}
d\left( {{U_t}} \right) = \frac{1}{2}{\lambda ^2}{U_t}dt - \lambda {U_t}d{B_t} + \frac{1}{2}{\lambda ^2}{U_t}dt\\
d\left\langle {{U_t},{X_t}} \right\rangle  = d{U_t} \cdot d{X_t} = \left( {\frac{1}{2}{\lambda ^2}{U_t}dt - \lambda {U_t}d{B_t} + \frac{1}{2}{\lambda ^2}{U_t}dt} \right)\lambda {X_t}d{B_t}\\
\begin{array}{*{20}{c}}
{}&{}&{}&{}
\end{array} =  - {\lambda ^2}{U_t}{X_t}dt
\end{array} \right.
\] 將上式代入 $(\star)$，我們得到
\[\begin{array}{l}
d\left( {{X_t}{U_t}} \right) = {X_t}\left( {{\lambda ^2}{U_t}dt - \lambda {U_t}d{B_t}} \right) + {U_t}\lambda {X_t}d{B_t} - {\lambda ^2}{U_t}{X_t}dt\\
 \Rightarrow d\left( {{X_t}{U_t}} \right) = 0
\end{array}
\] 現在將上式轉換回積分形式：
\[\begin{array}{l}
{X_t}{U_t} - {X_0}{U_0} = 0\\
 \Rightarrow {X_t}{U_t} = {X_0}\\
 \Rightarrow {X_t}{e^{ - \lambda {B_t} + \frac{1}{2}{\lambda ^2}t}} = {X_0}\\
 \Rightarrow {X_t} = {X_0}{e^{\lambda {B_t} - \frac{1}{2}{\lambda ^2}t}}\\
 \Rightarrow {X_t} = {e^{\lambda {B_t} - \frac{1}{2}{\lambda ^2}t}}\\
\end{array}
\] (注意到 $X_0=1, U_0=1$)；現在我們用積分因子法找到了一組解 $X_t$，我們必須回頭測試此解確實滿足 SDE，利用 Ito Formula，計算
\[d{X_t} = \left( { - \frac{1}{2}{\lambda ^2}} \right){X_t}dt + \lambda {X_t}d{B_t} + \frac{1}{2}{\lambda ^2}{X_t}dt\]故
\[ \Rightarrow d{X_t} = \lambda {X_t}d{B_t}
\]亦即此為我們的SDE。故
\[
{X_t} = {e^{\lambda {B_t} - \frac{1}{2}{\lambda ^2}t}}
\]確實為 SDE的(唯一)解。此解被稱為 Stochastic Exponential。 $\square$

現在我們更進一步考慮如何拓展上述結果：
如果我們把 Standard Brownian Motion 替換成 Standard Process 上述結果是否仍能成立呢? 答案是肯定的，我們用下面這個例子說明 Stochastic Exponential 對 standard process 的情況：

Example 2: Stochastic Exponential for Standard Process
定義 $Y_t$ 為 Standard process 亦即
\[
Y_t := Y_0 + \int_0^t a(\omega, s) ds + \int_0^t b(\omega,s) dB_s, \ \ 0 \leq t \leq T
\]其中 $a(\cdot, \cdot)$ 與 $b(\cdot, \cdot)$ 為 adapted, measurable processes 且滿足以下積分條件：
\[
\int_0^T |a(\omega,s)|ds + \int_0^T |b(\omega,s)|^2 ds < \infty \ \text{almost surely}
\] 現在試求 $Z_t$ 滿足
\[
Z_t = 1 + \lambda \int_0^t Z_s dY_s
\]
Solution
在求解之前我們可以猜測此SDE對應的解的形式應該仍保留 Stochastic Exponential 的樣子。不過我們為了統一性，我們仍使用積分因子法求解：

首先改寫上述 $Z_t$ 為 微分形式
\[\begin{array}{l}
{Z_t} = 1 + \lambda \int_0^t {{Z_s}} d{Y_s}\\
 \Rightarrow d{Z_t} = \lambda {Z_t}d{Y_t}\\
 \Rightarrow d{Z_t} - \lambda {Z_t}d{Y_t} = 0
\end{array}\]
定義積分因子如下：
\[{U_t}: = {e^{ - \lambda \int_0^t {d{Y_s}}  + \frac{1}{2}{\lambda ^2}{{\left\langle Y \right\rangle }_t}}} = {e^{ - \lambda {Y_t} + \frac{1}{2}{\lambda ^2}{{\left\langle Y \right\rangle }_t}}}
\]利用 Integration by part 計算
\[d\left( {{Z_t}{U_t}} \right) = {Z_t}d\left( {{U_t}} \right) + {U_t}d\left( {{Z_t}} \right) + d\left\langle {{U_t},{Z_t}} \right\rangle
\]其中
\[\left\{ \begin{array}{l}
d\left( {{U_t}} \right) =  - \lambda {U_t}d{Y_t} + \frac{1}{2}{\lambda ^2}{U_t}d{\left\langle Y \right\rangle _t} + \frac{1}{2}{\lambda ^2}{U_t}d{\left\langle Y \right\rangle _t}\\
\begin{array}{*{20}{c}}
{}&{}&{}&{}
\end{array} =  - \lambda {U_t}d{Y_t} + {\lambda ^2}{U_t}d{\left\langle Y \right\rangle _t}\\
d{Z_t} = \lambda {Z_t}d{Y_t}\\
d\left\langle {{U_t},{Z_t}} \right\rangle  = d{U_t}d{Z_t} = \left( { - \lambda {U_t}d{Y_t} + {\lambda ^2}{U_t}d{{\left\langle Y \right\rangle }_t}} \right)\lambda {Z_t}d{Y_t}\\
\begin{array}{*{20}{c}}
{}&{}&{}&{}
\end{array} =  - {\lambda ^2}{U_t}{Z_t}d{Y_t}d{Y_t} =  - {\lambda ^2}{U_t}{Z_t}d{\left\langle Y \right\rangle _t}
\end{array} \right.
\] 將上式代入 $d(Z_t U_t)$ 可得
\[\begin{array}{l}
d\left( {{Z_t}{U_t}} \right) = {Z_t}d\left( {{U_t}} \right) + {U_t}d\left( {{Z_t}} \right) + d\left\langle {{U_t},{Z_t}} \right\rangle \\
 \Rightarrow d\left( {{Z_t}{U_t}} \right) = {Z_t}\left( { - \lambda {U_t}d{Y_t} + {\lambda ^2}{U_t}d{{\left\langle Y \right\rangle }_t}} \right) + {U_t}\lambda {Z_t}d{Y_t} - {\lambda ^2}{U_t}{Z_t}d{\left\langle Y \right\rangle _t}\\
 \Rightarrow d\left( {{Z_t}{U_t}} \right) = 0
\end{array}
\]現在轉換為積分形式
\[\begin{array}{l}
{Z_t}{U_t} - {Z_0}{U_0} = 0\\
 \Rightarrow {Z_t}{U_t} = {Z_0}\\
 \Rightarrow {Z_t}{e^{ - \lambda {Y_t} + \frac{1}{2}{\lambda ^2}{{\left\langle Y \right\rangle }_t}}} = 1\\
 \Rightarrow {Z_t} = {e^{\lambda {Y_t} - \frac{1}{2}{\lambda ^2}{{\left\langle Y \right\rangle }_t}}}
\end{array}
\] 故我們求得一個對 $Z_t$ 的解，現在對其進行驗證測試其確實滿足SDE：利用 Ito Formula 計算 $dZ_t$ 可得
\[\begin{array}{l}
d{Z_t} = \lambda {Z_t}d{Y_t} - \frac{1}{2}{\lambda ^2}{Z_t}d{\left\langle Y \right\rangle _t} + \frac{1}{2}{\lambda ^2}{Z_t}d{\left\langle Y \right\rangle _t}\\
 \Rightarrow d{X_t} = \lambda {Z_t}dY
\end{array}\]亦即 此解
\[
{Z_t} = {e^{\lambda {Y_t} - \frac{1}{2}{\lambda ^2}{{\left\langle Y \right\rangle }_t}}}
\]確實滿足SDE，故為此即為 SDE的 解。