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 Ornstein-Uhlenbeck process 為例:
考慮如下SDE :
令 $\alpha, \sigma >0$,考慮如下 SDE
\[
dX_t = - \alpha X_t dt + \sigma dB_t
\]且給定初始條件 $X_0$ ,此 $X_0$ 與 $B_{\cdot}$ 互相獨立
Comment:
上述 SDE 稱為 Ornstein-Uhlenbeck process。
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 Ornstein-Uhlenbeck process 可知 $\mu(t, X_t) = - \alpha X_t, \ \sigma(t, X_t) = \sigma$
\[
|(-\alpha x) - (-\alpha y)|^2 + |\sigma - \sigma|^2 =(\alpha)^2 | x - y|^2
\]令 $K \geq \alpha^2$,則我們有如下關係:
\[
|(-\alpha x) - (-\alpha y)|^2 + |\sigma - \sigma|^2 \leq K | x - y|^2
\]亦即滿足 Lipschitz Condition。
接著我們檢驗 Growth Condition:
\[
|-\alpha x|^2 + |\sigma|^2 = \alpha^2 x^2 + \sigma^2
\]令 $K \geq \max\{\alpha^2, \sigma^2\}$ 則我們有
\[{\alpha ^2}{\left| x \right|^2} + {\sigma ^2} \le K\left( {1 + {{\left| x \right|}^2}} \right)
\]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 Ornstein-Uhlenbeck Process 解存在且唯一。
故現在開始求解:
由 Ornstein-Uhlenbeck Process 定義
\[
\begin{array}{l}
d{X_t} = - \alpha {X_t}dt + \sigma d{B_t}\\
\Rightarrow d{X_t} + \left( {\alpha dt} \right){X_t} = \sigma d{B_t} \ \ \ \ (*)
\end{array}
\]定義積分因子
\[{U_t}: = {e^{\int_0^t {\alpha dt} }} = {e^{\alpha t}}
\]將上式 $(*)$ 兩邊同乘積分因子,並且我們計算 (由 integration by part)
\[
d\left( {{X_t}{U_t}} \right) = {X_t}d{U_t} + {U_t}d{X_t} + d\left\langle {{X_t},{U_t}} \right\rangle \ \ \ \ (\star)
\]其中 $dU_t$ 與 cross variation: $d\left\langle {{X_t},{U_t}} \right\rangle $ 由 box calculation 計算如下:
\[\begin{array}{l}
d{U_t} = \alpha {e^{\alpha t}}dt = \alpha {U_t}dt\\
d\left\langle {{X_t},{U_t}} \right\rangle = d{X_t}d{U_t} = \left( { - \alpha {X_t}dt + \sigma d{B_t}} \right)\alpha {e^{\alpha t}}dt = 0
\end{array}
\],將上述結果代回 $(\star)$ 故我們可得
\[\begin{array}{l}
d\left( {{X_t}{U_t}} \right) = {X_t}\alpha {U_t}dt + {U_t}\left( { - \alpha {X_t}dt + \sigma d{B_t}} \right)\\
\Rightarrow d\left( {{X_t}{U_t}} \right) = {U_t}\sigma d{B_t}
\end{array}
\]現在將上式轉換回積分形式
\[\begin{array}{l}
\Rightarrow {X_t}{U_t} - {X_0}{U_0} = \sigma \int_0^t {{U_t}d{B_t}} = \sigma \int_0^t {{e^{\alpha s}}d{B_s}} \\
\Rightarrow {X_t}{e^{\alpha t}} = {X_0} + \sigma \int_0^t {{e^{\alpha s}}d{B_s}} \\
\Rightarrow {X_t} = {X_0}{e^{ - \alpha t}} + \sigma {e^{ - \alpha t}}\int_0^t {{e^{\alpha s}}d{B_s}}
\end{array}
\]至此我們找到了一個 Ornstein-Uhlenbeck process 的解,現在我們必須檢驗此解確實滿足 我們的 Ornstein-Uhlenbeck SDE,故我們計算 $dX_t$,步驟如下:
首先改寫\[\begin{array}{l}
{X_t} = {X_0}{e^{ - \alpha t}} + \sigma {e^{ - \alpha t}}\int_0^t {{e^{\alpha s}}d{B_s}} \\
\Rightarrow {X_t} = {e^{ - \alpha t}}\underbrace {\left[ {{X_0} + \sigma \int_0^t {{e^{\alpha s}}d{B_s}} } \right]}_{: = {V_t}}: = {e^{ - \alpha t}}{V_t}
\end{array}
\]現在 利用 Integration by part,可得
\[d{X_t} = d\left( {{e^{ - \alpha t}}{V_t}} \right) = {e^{ - \alpha t}}d{V_t} + {V_t}d\left( {{e^{ - \alpha t}}} \right) + d\left\langle {{U_t},{V_t}} \right\rangle
\]其中
\[\left\{ \begin{array}{l}
d{V_t} = \sigma {e^{\alpha t}}d{B_t}\\
d\left( {{e^{ - \alpha t}}} \right) = - \alpha {e^{ - \alpha t}}dt\\
d\left\langle {{U_t},{V_t}} \right\rangle = d{V_t} \cdot d{U_t} = 0
\end{array} \right.
\]將上述結果帶入 $dX_t$ 中可得
\[\begin{array}{l}
d{X_t} = {e^{ - \alpha t}}d{V_t} + {V_t}d\left( {{e^{ - \alpha t}}} \right) + d\left\langle {{U_t},{V_t}} \right\rangle \\
\Rightarrow d{X_t} = {e^{ - \alpha t}}\sigma {e^{\alpha t}}d{B_t} - \alpha {V_t}{e^{ - \alpha t}}dt\\
\Rightarrow d{X_t} = \sigma d{B_t} - \alpha {V_t}{e^{ - \alpha t}}dt
\end{array}
\] 又因為 ${X_t} = {e^{ - \alpha t}}{V_t} \Rightarrow {e^{\alpha t}}{X_t} = {V_t}$,將此代入上式我們得到
\[\begin{array}{l}
d{X_t} = \sigma d{B_t} - \alpha {V_t}{e^{ - \alpha t}}dt\\
\Rightarrow d{X_t} = \sigma d{B_t} - \alpha {e^{\alpha t}}{X_t}{e^{ - \alpha t}}dt\\
\Rightarrow d{X_t} = \sigma d{B_t} - \alpha {X_t}dt
\end{array}
\]上式即為 Ornstein-Uhlenbeck process , 亦即 $X_t = {X_0}{e^{ - \alpha t}} + \sigma {e^{ - \alpha t}}\int_0^t {{e^{\alpha s}}d{B_s}} $ 確實為 Ornstein-Uhlenbeck process 的 (唯一)解。 $\square$
Comments:
1. 注意到 Ornstein-Uhlenbeck process 的解:
\[
X_t = {X_0}{e^{ - \alpha t}} + \sigma {e^{ - \alpha t}}\int_0^t {{e^{\alpha s}}d{B_s}}
\] 現在觀察上式的第二項 (Ito integral term),亦即 $\sigma {e^{ - \alpha t}}\int_0^t {{e^{\alpha s}}d{B_s}} $ 此項的 積分變數為 $f(s) := {{e^{\alpha s}}}$ 亦即不是隨機過程 (為確定 (determinstic) 函數)。故由 FACT: 可知 對任意非隨機連續函數 $f(t)$,其 Ito integral 為一個 Gaussian Process with zero mean 且 variance 為 $\int_0^t f(s)^2 ds$。故我們可知此項為 Gaussian Process:亦即
\[\begin{array}{l}
\sigma {e^{ - \alpha t}}\int_0^t {{e^{\alpha s}}d{B_s}} \sim N\left( {0,{\sigma ^2}{e^{ - 2\alpha t}}\left( {\frac{1}{{2\alpha }}\left( {{e^{2\alpha t}} - 1} \right)} \right)} \right)\\
\Rightarrow \sigma {e^{ - \alpha t}}\int_0^t {{e^{\alpha s}}d{B_s}} \sim N\left( {0,\frac{{{\sigma ^2}}}{{2\alpha }}\left( { 1- {e^{- 2\alpha t}}} \right)} \right)
\end{array}
\]
2. 再者,如果當 $t \rightarrow \infty$,此解
\[{{X_t} = \underbrace {{X_0}{e^{ - \alpha t}}}_{ \to 0} + \underbrace {\sigma {e^{ - \alpha t}}\int_0^t {{e^{\alpha s}}d{B_s}} }_{ \to N\left( {0,\frac{{{\sigma ^2}}}{{2\alpha }}} \right)}}
\]上述極限為 Convergence in distribution。且注意到由於 $X_0$ 與 $B_{\cdot}$ 互相獨立,故兩個獨立的 Gaussian random variable 相加,亦即:
\[
{X_t} \to N\left( {0 + 0,\frac{{{\sigma ^2}}}{{2\alpha }}} \right) = N\left( {0,\frac{{{\sigma ^2}}}{{2\alpha }}} \right)
\] in distribution
考慮如下SDE :
令 $\alpha, \sigma >0$,考慮如下 SDE
\[
dX_t = - \alpha X_t dt + \sigma dB_t
\]且給定初始條件 $X_0$ ,此 $X_0$ 與 $B_{\cdot}$ 互相獨立
Comment:
上述 SDE 稱為 Ornstein-Uhlenbeck process。
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 Ornstein-Uhlenbeck process 可知 $\mu(t, X_t) = - \alpha X_t, \ \sigma(t, X_t) = \sigma$
\[
|(-\alpha x) - (-\alpha y)|^2 + |\sigma - \sigma|^2 =(\alpha)^2 | x - y|^2
\]令 $K \geq \alpha^2$,則我們有如下關係:
\[
|(-\alpha x) - (-\alpha y)|^2 + |\sigma - \sigma|^2 \leq K | x - y|^2
\]亦即滿足 Lipschitz Condition。
接著我們檢驗 Growth Condition:
\[
|-\alpha x|^2 + |\sigma|^2 = \alpha^2 x^2 + \sigma^2
\]令 $K \geq \max\{\alpha^2, \sigma^2\}$ 則我們有
\[{\alpha ^2}{\left| x \right|^2} + {\sigma ^2} \le K\left( {1 + {{\left| x \right|}^2}} \right)
\]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 Ornstein-Uhlenbeck Process 解存在且唯一。
故現在開始求解:
由 Ornstein-Uhlenbeck Process 定義
\[
\begin{array}{l}
d{X_t} = - \alpha {X_t}dt + \sigma d{B_t}\\
\Rightarrow d{X_t} + \left( {\alpha dt} \right){X_t} = \sigma d{B_t} \ \ \ \ (*)
\end{array}
\]定義積分因子
\[{U_t}: = {e^{\int_0^t {\alpha dt} }} = {e^{\alpha t}}
\]將上式 $(*)$ 兩邊同乘積分因子,並且我們計算 (由 integration by part)
\[
d\left( {{X_t}{U_t}} \right) = {X_t}d{U_t} + {U_t}d{X_t} + d\left\langle {{X_t},{U_t}} \right\rangle \ \ \ \ (\star)
\]其中 $dU_t$ 與 cross variation: $d\left\langle {{X_t},{U_t}} \right\rangle $ 由 box calculation 計算如下:
\[\begin{array}{l}
d{U_t} = \alpha {e^{\alpha t}}dt = \alpha {U_t}dt\\
d\left\langle {{X_t},{U_t}} \right\rangle = d{X_t}d{U_t} = \left( { - \alpha {X_t}dt + \sigma d{B_t}} \right)\alpha {e^{\alpha t}}dt = 0
\end{array}
\],將上述結果代回 $(\star)$ 故我們可得
\[\begin{array}{l}
d\left( {{X_t}{U_t}} \right) = {X_t}\alpha {U_t}dt + {U_t}\left( { - \alpha {X_t}dt + \sigma d{B_t}} \right)\\
\Rightarrow d\left( {{X_t}{U_t}} \right) = {U_t}\sigma d{B_t}
\end{array}
\]現在將上式轉換回積分形式
\[\begin{array}{l}
\Rightarrow {X_t}{U_t} - {X_0}{U_0} = \sigma \int_0^t {{U_t}d{B_t}} = \sigma \int_0^t {{e^{\alpha s}}d{B_s}} \\
\Rightarrow {X_t}{e^{\alpha t}} = {X_0} + \sigma \int_0^t {{e^{\alpha s}}d{B_s}} \\
\Rightarrow {X_t} = {X_0}{e^{ - \alpha t}} + \sigma {e^{ - \alpha t}}\int_0^t {{e^{\alpha s}}d{B_s}}
\end{array}
\]至此我們找到了一個 Ornstein-Uhlenbeck process 的解,現在我們必須檢驗此解確實滿足 我們的 Ornstein-Uhlenbeck SDE,故我們計算 $dX_t$,步驟如下:
首先改寫\[\begin{array}{l}
{X_t} = {X_0}{e^{ - \alpha t}} + \sigma {e^{ - \alpha t}}\int_0^t {{e^{\alpha s}}d{B_s}} \\
\Rightarrow {X_t} = {e^{ - \alpha t}}\underbrace {\left[ {{X_0} + \sigma \int_0^t {{e^{\alpha s}}d{B_s}} } \right]}_{: = {V_t}}: = {e^{ - \alpha t}}{V_t}
\end{array}
\]現在 利用 Integration by part,可得
\[d{X_t} = d\left( {{e^{ - \alpha t}}{V_t}} \right) = {e^{ - \alpha t}}d{V_t} + {V_t}d\left( {{e^{ - \alpha t}}} \right) + d\left\langle {{U_t},{V_t}} \right\rangle
\]其中
\[\left\{ \begin{array}{l}
d{V_t} = \sigma {e^{\alpha t}}d{B_t}\\
d\left( {{e^{ - \alpha t}}} \right) = - \alpha {e^{ - \alpha t}}dt\\
d\left\langle {{U_t},{V_t}} \right\rangle = d{V_t} \cdot d{U_t} = 0
\end{array} \right.
\]將上述結果帶入 $dX_t$ 中可得
\[\begin{array}{l}
d{X_t} = {e^{ - \alpha t}}d{V_t} + {V_t}d\left( {{e^{ - \alpha t}}} \right) + d\left\langle {{U_t},{V_t}} \right\rangle \\
\Rightarrow d{X_t} = {e^{ - \alpha t}}\sigma {e^{\alpha t}}d{B_t} - \alpha {V_t}{e^{ - \alpha t}}dt\\
\Rightarrow d{X_t} = \sigma d{B_t} - \alpha {V_t}{e^{ - \alpha t}}dt
\end{array}
\] 又因為 ${X_t} = {e^{ - \alpha t}}{V_t} \Rightarrow {e^{\alpha t}}{X_t} = {V_t}$,將此代入上式我們得到
\[\begin{array}{l}
d{X_t} = \sigma d{B_t} - \alpha {V_t}{e^{ - \alpha t}}dt\\
\Rightarrow d{X_t} = \sigma d{B_t} - \alpha {e^{\alpha t}}{X_t}{e^{ - \alpha t}}dt\\
\Rightarrow d{X_t} = \sigma d{B_t} - \alpha {X_t}dt
\end{array}
\]上式即為 Ornstein-Uhlenbeck process , 亦即 $X_t = {X_0}{e^{ - \alpha t}} + \sigma {e^{ - \alpha t}}\int_0^t {{e^{\alpha s}}d{B_s}} $ 確實為 Ornstein-Uhlenbeck process 的 (唯一)解。 $\square$
Comments:
1. 注意到 Ornstein-Uhlenbeck process 的解:
\[
X_t = {X_0}{e^{ - \alpha t}} + \sigma {e^{ - \alpha t}}\int_0^t {{e^{\alpha s}}d{B_s}}
\] 現在觀察上式的第二項 (Ito integral term),亦即 $\sigma {e^{ - \alpha t}}\int_0^t {{e^{\alpha s}}d{B_s}} $ 此項的 積分變數為 $f(s) := {{e^{\alpha s}}}$ 亦即不是隨機過程 (為確定 (determinstic) 函數)。故由 FACT: 可知 對任意非隨機連續函數 $f(t)$,其 Ito integral 為一個 Gaussian Process with zero mean 且 variance 為 $\int_0^t f(s)^2 ds$。故我們可知此項為 Gaussian Process:亦即
\[\begin{array}{l}
\sigma {e^{ - \alpha t}}\int_0^t {{e^{\alpha s}}d{B_s}} \sim N\left( {0,{\sigma ^2}{e^{ - 2\alpha t}}\left( {\frac{1}{{2\alpha }}\left( {{e^{2\alpha t}} - 1} \right)} \right)} \right)\\
\Rightarrow \sigma {e^{ - \alpha t}}\int_0^t {{e^{\alpha s}}d{B_s}} \sim N\left( {0,\frac{{{\sigma ^2}}}{{2\alpha }}\left( { 1- {e^{- 2\alpha t}}} \right)} \right)
\end{array}
\]
2. 再者,如果當 $t \rightarrow \infty$,此解
\[{{X_t} = \underbrace {{X_0}{e^{ - \alpha t}}}_{ \to 0} + \underbrace {\sigma {e^{ - \alpha t}}\int_0^t {{e^{\alpha s}}d{B_s}} }_{ \to N\left( {0,\frac{{{\sigma ^2}}}{{2\alpha }}} \right)}}
\]上述極限為 Convergence in distribution。且注意到由於 $X_0$ 與 $B_{\cdot}$ 互相獨立,故兩個獨立的 Gaussian random variable 相加,亦即:
\[
{X_t} \to N\left( {0 + 0,\frac{{{\sigma ^2}}}{{2\alpha }}} \right) = N\left( {0,\frac{{{\sigma ^2}}}{{2\alpha }}} \right)
\] in distribution
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