這次要介紹的是如何直接求解隨機微分方程 (SDE)?
考慮如下SDE (Brownian Bridge):
定義 $t \in [0 ,T], T<1$
\[
dX_t = \frac{-X_t}{1-t}dt + dB_t \ \ \ \ X_0 =0 \ \ \ \ (*)
\]
Comment:
此SDE 稱作 Brownian Bridge,亦即 其行為類似一個 Bridge 連接兩端點為 $0$ ,在時間 $t=0$時值為 $0$,最後到 時間 $t=t$ 亦回到 $0$,如下圖所示:
Sol:
1. Check the existence and uniqueness
由Existence 與 Uniqueness Theorem得知,給定一 SDE具有下列形式:\[
dX_t = \mu(t,X_t)dt + \sigma(t,X_t) dB_t, \ X(0)=x_0 \ \ \ \ (*)
\]若我們要有唯一解,則其係數必須滿足 Lipschitz condtion 與 Growth conditon,故我們在求解之前先確認此兩個條件滿足。
首先檢驗 Lipschitz conditon
\[
|\mu(t,x) - \mu(t,y)|^2 + |\sigma(t,x) - \sigma(t,y)|^2 \leq K |x-y|^2
\]由上式左邊可得
\[{ \Rightarrow {{\left| {\left( {\frac{{ - {X_t}}}{{1 - t}}} \right) - \left( {\frac{{ - {Y_t}}}{{1 - t}}} \right)} \right|}^2} + |1 - 1{|^2} = {{\left( {\frac{1}{{1 - t}}} \right)}^2}{{\left| {{X_t} - {Y_t}} \right|}^2}}
\]故令 $K = {\left( {\frac{1}{{1 - t}}} \right)^2}$ 即可滿足Lipschitz condtion。
再者檢驗 Growth condition
\[
|\mu(t,x)|^2 + |\sigma(t,x)|^2 \leq K(1 + |x|^2)
\]故由上式左邊可得
\[ \Rightarrow {\left| {\left( {\frac{{ - {X_t}}}{{1 - t}}} \right)} \right|^2} + |1{|^2} = {\left( {\frac{1}{{1 - t}}} \right)^2}{\left| {{X_t}} \right|^2} + 1 \le {\left( {\frac{1}{{1 - t}}} \right)^2}(1 + |X_t{|^2})
\]故取 $K={\left( {\frac{1}{{1 - t}}} \right)^2}$ 即可滿足Growth condition。
由於係數滿足Lipschitz condtion 與 Growth conditon,故SDE $(*)$有唯一解,
2. Solve the SDE by using integration factor method
現在我們可以開始進行求解。想法為利用ODE中的積分因子法求解。步驟如下
首先改寫上式如下
\[ \Rightarrow d{X_t} + \frac{{dt}}{{1 - t}}{X_t} = d{B_t}
\]定義積分因子 (積分因子的定法請參見此篇 :[微分方程] Solving a linear ODE by using Integration factor method )
\[{U_t}: = {e^{\int_0^t {\frac{1}{{1 - s}}ds} }} = {e^{ - \ln \left( {1 - t} \right)}}
\]現在觀察 $d\left( {{X_t}{U_t}} \right)$,由integration by part (Ito formula):
\[d(XU) = XdU + UdX + d\left\langle {X,U} \right\rangle
\] 我們可得
\[\begin{array}{l}
\Rightarrow d\left( {{X_t}{U_t}} \right) = Xd\left( {{e^{ - \ln \left( {1 - t} \right)}}} \right) + {e^{ - \ln \left( {1 - t} \right)}}\left( {\frac{{ - {X_t}}}{{1 - t}}dt + d{B_t}\;} \right) \\
\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ + \left( {\frac{{ - {X_t}}}{{1 - t}}dt + d{B_t}\;} \right)d\left( {{e^{ - \ln \left( {1 - t} \right)}}} \right)\\
\Rightarrow d\left( {{X_t}{U_t}} \right) = X{e^{ - \ln \left( {1 - t} \right)}}\left( {\frac{1}{{1 - t}}} \right)dt + {e^{ - \ln \left( {1 - t} \right)}}\left( {\frac{{ - {X_t}}}{{1 - t}}dt + d{B_t}\;} \right)\\
\begin{array}{*{20}{c}}
{}&{}&{}&{}&{}&{}
\end{array} + \left( {\frac{{ - {X_t}}}{{1 - t}}dt + d{B_t}\;} \right){e^{ - \ln \left( {1 - t} \right)}}\left( {\frac{1}{{1 - t}}} \right)dt\\
\Rightarrow d\left( {{X_t} \cdot {U_t}} \right) = {e^{ - \ln \left( {1 - t} \right)}}\left[ {d{B_t}\;} \right]
\end{array}\]故對上式兩邊同取積分,我們可得
\[\begin{array}{l}
{X_t} \cdot {U_t} = \int_0^t {{e^{ - \ln \left( {1 - s} \right)}}d{B_s}} \\
\Rightarrow {X_t} \cdot \left( {{e^{ - \ln \left( {1 - t} \right)}}} \right) = \int_0^t {\frac{1}{{1 - s}}d{B_s}} \\
\Rightarrow {X_t} = {e^{\ln \left( {1 - t} \right)}}\int_0^t {\frac{1}{{1 - s}}d{B_s}}\\
{ \Rightarrow {X_t} = \left( {1 - t} \right)\int_0^t {\frac{1}{{1 - s}}d{B_s}} }, \ \forall t \in [0,1)
\end{array}
\]上式即為Brownian Bridge $(*)$的解。
考慮如下SDE (Brownian Bridge):
定義 $t \in [0 ,T], T<1$
\[
dX_t = \frac{-X_t}{1-t}dt + dB_t \ \ \ \ X_0 =0 \ \ \ \ (*)
\]
Comment:
此SDE 稱作 Brownian Bridge,亦即 其行為類似一個 Bridge 連接兩端點為 $0$ ,在時間 $t=0$時值為 $0$,最後到 時間 $t=t$ 亦回到 $0$,如下圖所示:
Sol:
1. Check the existence and uniqueness
由Existence 與 Uniqueness Theorem得知,給定一 SDE具有下列形式:\[
dX_t = \mu(t,X_t)dt + \sigma(t,X_t) dB_t, \ X(0)=x_0 \ \ \ \ (*)
\]若我們要有唯一解,則其係數必須滿足 Lipschitz condtion 與 Growth conditon,故我們在求解之前先確認此兩個條件滿足。
首先檢驗 Lipschitz conditon
\[
|\mu(t,x) - \mu(t,y)|^2 + |\sigma(t,x) - \sigma(t,y)|^2 \leq K |x-y|^2
\]由上式左邊可得
\[{ \Rightarrow {{\left| {\left( {\frac{{ - {X_t}}}{{1 - t}}} \right) - \left( {\frac{{ - {Y_t}}}{{1 - t}}} \right)} \right|}^2} + |1 - 1{|^2} = {{\left( {\frac{1}{{1 - t}}} \right)}^2}{{\left| {{X_t} - {Y_t}} \right|}^2}}
\]故令 $K = {\left( {\frac{1}{{1 - t}}} \right)^2}$ 即可滿足Lipschitz condtion。
再者檢驗 Growth condition
\[
|\mu(t,x)|^2 + |\sigma(t,x)|^2 \leq K(1 + |x|^2)
\]故由上式左邊可得
\[ \Rightarrow {\left| {\left( {\frac{{ - {X_t}}}{{1 - t}}} \right)} \right|^2} + |1{|^2} = {\left( {\frac{1}{{1 - t}}} \right)^2}{\left| {{X_t}} \right|^2} + 1 \le {\left( {\frac{1}{{1 - t}}} \right)^2}(1 + |X_t{|^2})
\]故取 $K={\left( {\frac{1}{{1 - t}}} \right)^2}$ 即可滿足Growth condition。
由於係數滿足Lipschitz condtion 與 Growth conditon,故SDE $(*)$有唯一解,
2. Solve the SDE by using integration factor method
現在我們可以開始進行求解。想法為利用ODE中的積分因子法求解。步驟如下
首先改寫上式如下
\[ \Rightarrow d{X_t} + \frac{{dt}}{{1 - t}}{X_t} = d{B_t}
\]定義積分因子 (積分因子的定法請參見此篇 :[微分方程] Solving a linear ODE by using Integration factor method )
\[{U_t}: = {e^{\int_0^t {\frac{1}{{1 - s}}ds} }} = {e^{ - \ln \left( {1 - t} \right)}}
\]現在觀察 $d\left( {{X_t}{U_t}} \right)$,由integration by part (Ito formula):
\[d(XU) = XdU + UdX + d\left\langle {X,U} \right\rangle
\] 我們可得
\[\begin{array}{l}
\Rightarrow d\left( {{X_t}{U_t}} \right) = Xd\left( {{e^{ - \ln \left( {1 - t} \right)}}} \right) + {e^{ - \ln \left( {1 - t} \right)}}\left( {\frac{{ - {X_t}}}{{1 - t}}dt + d{B_t}\;} \right) \\
\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ + \left( {\frac{{ - {X_t}}}{{1 - t}}dt + d{B_t}\;} \right)d\left( {{e^{ - \ln \left( {1 - t} \right)}}} \right)\\
\Rightarrow d\left( {{X_t}{U_t}} \right) = X{e^{ - \ln \left( {1 - t} \right)}}\left( {\frac{1}{{1 - t}}} \right)dt + {e^{ - \ln \left( {1 - t} \right)}}\left( {\frac{{ - {X_t}}}{{1 - t}}dt + d{B_t}\;} \right)\\
\begin{array}{*{20}{c}}
{}&{}&{}&{}&{}&{}
\end{array} + \left( {\frac{{ - {X_t}}}{{1 - t}}dt + d{B_t}\;} \right){e^{ - \ln \left( {1 - t} \right)}}\left( {\frac{1}{{1 - t}}} \right)dt\\
\Rightarrow d\left( {{X_t} \cdot {U_t}} \right) = {e^{ - \ln \left( {1 - t} \right)}}\left[ {d{B_t}\;} \right]
\end{array}\]故對上式兩邊同取積分,我們可得
\[\begin{array}{l}
{X_t} \cdot {U_t} = \int_0^t {{e^{ - \ln \left( {1 - s} \right)}}d{B_s}} \\
\Rightarrow {X_t} \cdot \left( {{e^{ - \ln \left( {1 - t} \right)}}} \right) = \int_0^t {\frac{1}{{1 - s}}d{B_s}} \\
\Rightarrow {X_t} = {e^{\ln \left( {1 - t} \right)}}\int_0^t {\frac{1}{{1 - s}}d{B_s}}\\
{ \Rightarrow {X_t} = \left( {1 - t} \right)\int_0^t {\frac{1}{{1 - s}}d{B_s}} }, \ \forall t \in [0,1)
\end{array}
\]上式即為Brownian Bridge $(*)$的解。
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