這次要介紹的是如何直接求解隨機微分方程 (SDE)?
Example 1
考慮如下SDE ：
定義 $t \in [0 , 1)$
\[
dX_t = \frac{-2 X_t}{1-t}dt + \sqrt{2t (1-t)} \ dB_t \ \ \ \ X_0 =0 \ \ \ \ (*)
\]
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，此性質留給讀者確認。計算後可得上述SDE係數滿足Lipschitz condtion 與 Growth conditon，故SDE $(*)$有唯一解，細節在此不贅述
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SDE如下：
\[\begin{array}{l}
d{X_t} = \frac{{ - 2{X_t}}}{{1 - t}}dt + \sqrt {2t(1 - t)} \;d{B_t}\;\;\;\;{X_0} = 0\;\;\;\;(*)\\
\Rightarrow d{X_t} + \frac{{2dt}}{{1 - t}}{X_t} = \sqrt {2t(1 - t)} \;d{B_t}\;
\end{array}
\]定義積分因子
\[{U_t}: = {e^{\int_0^t {\frac{2}{{1 - s}}ds} }} = {e^{2\int_0^t {\frac{1}{{1 - s}}ds} }} = \frac{1}{{{{\left( {1 - t} \right)}^2}}} \\
\Rightarrow d{U_t} = {e^{2\int_0^t {\frac{1}{{1 - s}}ds} }}\left( {\frac{2}{{1 - t}}} \right)dt
\]接著計算
\[\begin{array}{l}
\Rightarrow d\left( {{X_t}{U_t}} \right) = {X_t}{e^{2\int_0^t {\frac{1}{{1 - s}}ds} }}\left( {\frac{2}{{1 - t}}} \right)dt\\
\begin{array}{*{20}{c}}
{}&{}&{}&{}
\end{array} + {e^{2\int_0^t {\frac{1}{{1 - s}}ds} }}\left( {\frac{{ - 2{X_t}}}{{1 - t}}dt + \sqrt {2t(1 - t)} \;d{B_t}} \right)\\
\begin{array}{*{20}{c}}
{}&{}&{}&{}
\end{array} + \left( {{e^{2\int_0^t {\frac{1}{{1 - s}}ds} }}\left( {\frac{2}{{1 - t}}} \right)dt} \right)\left( {\frac{{ - 2{X_t}}}{{1 - t}}dt + \sqrt {2t(1 - t)} \;d{B_t}} \right)
\end{array}
\]故
\[\begin{array}{l}
\Rightarrow d\left( {{X_t}{U_t}} \right) = {e^{2\int_0^t {\frac{1}{{1 - s}}ds} }}\left[ {\sqrt {2t(1 - t)} \;d{B_t}} \right] = \frac{1}{{{{\left( {1 - t} \right)}^2}}}\sqrt {2t(1 - t)} \;d{B_t}\\
\Rightarrow \int_0^t d \left( {{X_t}{U_t}} \right) = \int_0^t {\frac{1}{{{{\left( {1 - s} \right)}^2}}}\sqrt {2s(1 - s)} \;d{B_s}} \\
\Rightarrow {X_t}{U_t} - {X_0}{U_0} = \int_0^t {\frac{1}{{{{\left( {1 - s} \right)}^2}}}\sqrt {2s(1 - s)} \;d{B_s}} \\
\Rightarrow {X_t}{U_t} = \int_0^t {\frac{1}{{{{\left( {1 - s} \right)}^2}}}\sqrt {2s(1 - s)} \;d{B_s}} \\
\Rightarrow {X_t} = {\left( {1 - t} \right)^2}\int_0^t {\frac{1}{{{{\left( {1 - s} \right)}^2}}}\sqrt {2s(1 - s)} \;d{B_s}}
\end{array}
\]解得上式之後，我們需回頭檢驗此解是否確實為SDE的解：
故計算
\[\begin{array}{l}
d{X_t} = d\left( {{{\left( {1 - t} \right)}^2}{V_t}} \right) = {\left( {1 - t} \right)^2}d{V_t} + {V_t}d{\left( {1 - t} \right)^2} + {\left\langle {V,{{\left( {1 - t} \right)}^2}} \right\rangle _t}\\
\Rightarrow d{X_t} = {\left( {1 - t} \right)^2}\left( {\frac{1}{{{{\left( {1 - t} \right)}^2}}}\sqrt {2t(1 - t)} \;d{B_t}} \right)\\
\begin{array}{*{20}{c}}
{}&{}&{}&{}
\end{array} - 2\left( {1 - t} \right)dt\left( {\int_0^t {\frac{1}{{{{\left( {1 - s} \right)}^2}}}\sqrt {2s(1 - s)} \;d{B_s}} } \right)\\
\Rightarrow d{X_t} = \sqrt {2t(1 - t)} \;d{B_t} - 2\left( {1 - t} \right)dt\left( {\frac{{{X_t}}}{{{{\left( {1 - t} \right)}^2}}}} \right)\\
\Rightarrow d{X_t} = \sqrt {2t(1 - t)} \;d{B_t} - 2\frac{{{X_t}}}{{\left( {1 - t} \right)}}dt
\end{array}\]
Example 2
考慮如下SDE ：
\[
dY_t = rdt + \alpha Y_t dB_t \ \ \ \ (*)
\]給定 初始條件 $Y_0$ 且與 標準布朗運動 $B_t$ 互相獨立。
Sol：
此題較為特殊，如果選積分因子
\[{U_t}: = {e^{ - \alpha {B_t}}}
\]會失敗，故需做一點trial and error 修正：
\[{U_t}: = {e^{ - \alpha {B_t} + \frac{1}{2}{\alpha ^2}t}}
\]且注意到上式積分因子為隨機，故 $dU_t$ 需用Ito formula來幫助我們
\[\begin{array}{l}
d{U_t} = \frac{1}{2}{\alpha ^2}{e^{ - \alpha {B_t} + \frac{1}{2}{\alpha ^2}t}}dt + {e^{ - \alpha {B_t} + \frac{1}{2}{\alpha ^2}t}}\left( { - \alpha } \right)d{B_t} + \frac{1}{2}{\alpha ^2}{e^{ - \alpha {B_t} + \frac{1}{2}{\alpha ^2}t}}dt\\
\Rightarrow d{U_t} = {e^{ - \alpha {B_t} + \frac{1}{2}{\alpha ^2}t}}\left[ {{\alpha ^2}dt - \alpha d{B_t}} \right]
\end{array}
\]現在我們將SDE兩邊同乘積分因子，並計算
\[\begin{array}{l}
\Rightarrow d\left( {{Y_t}{U_t}} \right) = {Y_t}d\left( {{U_t}} \right) + {U_t}d\left( {{Y_t}} \right) + d{\left\langle {Y,U} \right\rangle _t}\\
\Rightarrow d\left( {{Y_t}{U_t}} \right) = {Y_t}\left( {{e^{ - \alpha {B_t} + \frac{1}{2}{\alpha ^2}t}}\left[ {{\alpha ^2}dt - \alpha d{B_t}} \right]} \right)\\
\begin{array}{*{20}{c}}
{}&{}&{}&{}
\end{array} + {e^{ - \alpha {B_t} + \frac{1}{2}{\alpha ^2}t}}\left( {rdt + \alpha {Y_t}d{B_t}} \right)\\
\begin{array}{*{20}{c}}
{}&{}&{}&{}
\end{array} + \left( {rdt + \alpha {Y_t}d{B_t}} \right)\left( {{e^{ - \alpha {B_t} + \frac{1}{2}{\alpha ^2}t}}\left[ {{\alpha ^2}dt - \alpha d{B_t}} \right]} \right)\\
\Rightarrow d\left( {{Y_t}{U_t}} \right) = {e^{ - \alpha {B_t} + \frac{1}{2}{\alpha ^2}t}}rdt
\end{array}
\]轉換回積分型態：
\[\begin{array}{l}
{Y_t}{U_t} - {Y_0}{U_0} = r\int_0^t {{e^{ - \alpha {B_s} + \frac{1}{2}{\alpha ^2}s}}d} s\\
\Rightarrow {Y_t}{U_t} = {Y_0} + r\int_0^t {{e^{ - \alpha {B_s} + \frac{1}{2}{\alpha ^2}s}}d} s\\
\Rightarrow {Y_t} = {Y_0}{e^{\alpha {B_t} - \frac{1}{2}{\alpha ^2}t}} + r{e^{\alpha {B_t} - \frac{1}{2}{\alpha ^2}t}}\int_0^t {{e^{ - \alpha {B_s} + \frac{1}{2}{\alpha ^2}s}}d} s\\
\Rightarrow {Y_t} = {Y_0}{e^{\alpha {B_t} - \frac{1}{2}{\alpha ^2}t}} + r\int_0^t {{e^{\alpha \left( {{B_t} - {B_s}} \right) - \frac{1}{2}{\alpha ^2}\left( {t - s} \right)}}d} s
\end{array}
\]
現在回頭檢驗此解是否滿足SDE:
首先將我們的解改寫定義如下：
\[{Y_t} = {e^{\alpha {B_t} - \frac{1}{2}{\alpha ^2}t}}\left( {{Y_0} + r\int_0^t {{e^{ - \alpha {B_s} + \frac{1}{2}{\alpha ^2}s}}d} s} \right): = {e^{\alpha {B_t} - \frac{1}{2}{\alpha ^2}t}}\left( {{V_t}} \right)
\]計算
\[d\left( {{Y_t}} \right) = d\left( {{e^{\alpha {B_t} - \frac{1}{2}{\alpha ^2}t}}{V_t}} \right) = {e^{\alpha {B_t} - \frac{1}{2}{\alpha ^2}t}}d{V_t} + {V_t}d\left( {{e^{\alpha {B_t} - \frac{1}{2}{\alpha ^2}t}}} \right) + {\left\langle {V,{e^{\alpha {B_t} - \frac{1}{2}{\alpha ^2}t}}} \right\rangle _t}
\]其中\[V_t^{} = {Y_0} + r\int_0^t {{e^{ - \alpha {B_s} + \frac{1}{2}{\alpha ^2}s}}d} s \Rightarrow d{V_t} = r{e^{ - \alpha {B_t} + \frac{1}{2}{\alpha ^2}t}}dt\]
且\[d\left( {{e^{\alpha {B_t} - \frac{1}{2}{\alpha ^2}t}}} \right): = {e^{\alpha {B_t} - \frac{1}{2}{\alpha ^2}t}}\alpha d{B_t}
\]上式由Ito formula推得 (因為是隨機，需用Ito formula幫助我們計算)
故
\[\begin{array}{l}
\Rightarrow d\left( {{Y_t}} \right) = {e^{\alpha {B_t} - \frac{1}{2}{\alpha ^2}t}}\left[ {r{e^{ - \alpha {B_t} + \frac{1}{2}{\alpha ^2}t}}dt + \alpha \left( {{Y_0} + r\int_0^t {{e^{ - \alpha {B_s} + \frac{1}{2}{\alpha ^2}s}}d} s} \right)d{B_t}} \right]\\
\Rightarrow d\left( {{Y_t}} \right) = {e^{\alpha {B_t} - \frac{1}{2}{\alpha ^2}t}}\left[ {r{e^{ - \alpha {B_t} + \frac{1}{2}{\alpha ^2}t}}dt + \alpha \left( {{e^{ - \alpha {B_t} + \frac{1}{2}{\alpha ^2}t}}{Y_t}} \right)d{B_t}} \right]\\
\Rightarrow d\left( {{Y_t}} \right) = \left[ {rdt + \alpha {Y_t}d{B_t}} \right]
\end{array}
\]亦即我們的解確實滿足SDE。至此計算完畢。
Example 1
考慮如下SDE ：
定義 $t \in [0 , 1)$
\[
dX_t = \frac{-2 X_t}{1-t}dt + \sqrt{2t (1-t)} \ dB_t \ \ \ \ X_0 =0 \ \ \ \ (*)
\]
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，此性質留給讀者確認。計算後可得上述SDE係數滿足Lipschitz condtion 與 Growth conditon，故SDE $(*)$有唯一解，細節在此不贅述
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SDE如下：
\[\begin{array}{l}
d{X_t} = \frac{{ - 2{X_t}}}{{1 - t}}dt + \sqrt {2t(1 - t)} \;d{B_t}\;\;\;\;{X_0} = 0\;\;\;\;(*)\\
\Rightarrow d{X_t} + \frac{{2dt}}{{1 - t}}{X_t} = \sqrt {2t(1 - t)} \;d{B_t}\;
\end{array}
\]定義積分因子
\[{U_t}: = {e^{\int_0^t {\frac{2}{{1 - s}}ds} }} = {e^{2\int_0^t {\frac{1}{{1 - s}}ds} }} = \frac{1}{{{{\left( {1 - t} \right)}^2}}} \\
\Rightarrow d{U_t} = {e^{2\int_0^t {\frac{1}{{1 - s}}ds} }}\left( {\frac{2}{{1 - t}}} \right)dt
\]接著計算
\[\begin{array}{l}
\Rightarrow d\left( {{X_t}{U_t}} \right) = {X_t}{e^{2\int_0^t {\frac{1}{{1 - s}}ds} }}\left( {\frac{2}{{1 - t}}} \right)dt\\
\begin{array}{*{20}{c}}
{}&{}&{}&{}
\end{array} + {e^{2\int_0^t {\frac{1}{{1 - s}}ds} }}\left( {\frac{{ - 2{X_t}}}{{1 - t}}dt + \sqrt {2t(1 - t)} \;d{B_t}} \right)\\
\begin{array}{*{20}{c}}
{}&{}&{}&{}
\end{array} + \left( {{e^{2\int_0^t {\frac{1}{{1 - s}}ds} }}\left( {\frac{2}{{1 - t}}} \right)dt} \right)\left( {\frac{{ - 2{X_t}}}{{1 - t}}dt + \sqrt {2t(1 - t)} \;d{B_t}} \right)
\end{array}
\]故
\[\begin{array}{l}
\Rightarrow d\left( {{X_t}{U_t}} \right) = {e^{2\int_0^t {\frac{1}{{1 - s}}ds} }}\left[ {\sqrt {2t(1 - t)} \;d{B_t}} \right] = \frac{1}{{{{\left( {1 - t} \right)}^2}}}\sqrt {2t(1 - t)} \;d{B_t}\\
\Rightarrow \int_0^t d \left( {{X_t}{U_t}} \right) = \int_0^t {\frac{1}{{{{\left( {1 - s} \right)}^2}}}\sqrt {2s(1 - s)} \;d{B_s}} \\
\Rightarrow {X_t}{U_t} - {X_0}{U_0} = \int_0^t {\frac{1}{{{{\left( {1 - s} \right)}^2}}}\sqrt {2s(1 - s)} \;d{B_s}} \\
\Rightarrow {X_t}{U_t} = \int_0^t {\frac{1}{{{{\left( {1 - s} \right)}^2}}}\sqrt {2s(1 - s)} \;d{B_s}} \\
\Rightarrow {X_t} = {\left( {1 - t} \right)^2}\int_0^t {\frac{1}{{{{\left( {1 - s} \right)}^2}}}\sqrt {2s(1 - s)} \;d{B_s}}
\end{array}
\]解得上式之後，我們需回頭檢驗此解是否確實為SDE的解：
故計算
\[\begin{array}{l}
d{X_t} = d\left( {{{\left( {1 - t} \right)}^2}{V_t}} \right) = {\left( {1 - t} \right)^2}d{V_t} + {V_t}d{\left( {1 - t} \right)^2} + {\left\langle {V,{{\left( {1 - t} \right)}^2}} \right\rangle _t}\\
\Rightarrow d{X_t} = {\left( {1 - t} \right)^2}\left( {\frac{1}{{{{\left( {1 - t} \right)}^2}}}\sqrt {2t(1 - t)} \;d{B_t}} \right)\\
\begin{array}{*{20}{c}}
{}&{}&{}&{}
\end{array} - 2\left( {1 - t} \right)dt\left( {\int_0^t {\frac{1}{{{{\left( {1 - s} \right)}^2}}}\sqrt {2s(1 - s)} \;d{B_s}} } \right)\\
\Rightarrow d{X_t} = \sqrt {2t(1 - t)} \;d{B_t} - 2\left( {1 - t} \right)dt\left( {\frac{{{X_t}}}{{{{\left( {1 - t} \right)}^2}}}} \right)\\
\Rightarrow d{X_t} = \sqrt {2t(1 - t)} \;d{B_t} - 2\frac{{{X_t}}}{{\left( {1 - t} \right)}}dt
\end{array}\]
Example 2
考慮如下SDE ：
\[
dY_t = rdt + \alpha Y_t dB_t \ \ \ \ (*)
\]給定 初始條件 $Y_0$ 且與 標準布朗運動 $B_t$ 互相獨立。
Sol：
此題較為特殊，如果選積分因子
\[{U_t}: = {e^{ - \alpha {B_t}}}
\]會失敗，故需做一點trial and error 修正：
\[{U_t}: = {e^{ - \alpha {B_t} + \frac{1}{2}{\alpha ^2}t}}
\]且注意到上式積分因子為隨機，故 $dU_t$ 需用Ito formula來幫助我們
\[\begin{array}{l}
d{U_t} = \frac{1}{2}{\alpha ^2}{e^{ - \alpha {B_t} + \frac{1}{2}{\alpha ^2}t}}dt + {e^{ - \alpha {B_t} + \frac{1}{2}{\alpha ^2}t}}\left( { - \alpha } \right)d{B_t} + \frac{1}{2}{\alpha ^2}{e^{ - \alpha {B_t} + \frac{1}{2}{\alpha ^2}t}}dt\\
\Rightarrow d{U_t} = {e^{ - \alpha {B_t} + \frac{1}{2}{\alpha ^2}t}}\left[ {{\alpha ^2}dt - \alpha d{B_t}} \right]
\end{array}
\]現在我們將SDE兩邊同乘積分因子，並計算
\[\begin{array}{l}
\Rightarrow d\left( {{Y_t}{U_t}} \right) = {Y_t}d\left( {{U_t}} \right) + {U_t}d\left( {{Y_t}} \right) + d{\left\langle {Y,U} \right\rangle _t}\\
\Rightarrow d\left( {{Y_t}{U_t}} \right) = {Y_t}\left( {{e^{ - \alpha {B_t} + \frac{1}{2}{\alpha ^2}t}}\left[ {{\alpha ^2}dt - \alpha d{B_t}} \right]} \right)\\
\begin{array}{*{20}{c}}
{}&{}&{}&{}
\end{array} + {e^{ - \alpha {B_t} + \frac{1}{2}{\alpha ^2}t}}\left( {rdt + \alpha {Y_t}d{B_t}} \right)\\
\begin{array}{*{20}{c}}
{}&{}&{}&{}
\end{array} + \left( {rdt + \alpha {Y_t}d{B_t}} \right)\left( {{e^{ - \alpha {B_t} + \frac{1}{2}{\alpha ^2}t}}\left[ {{\alpha ^2}dt - \alpha d{B_t}} \right]} \right)\\
\Rightarrow d\left( {{Y_t}{U_t}} \right) = {e^{ - \alpha {B_t} + \frac{1}{2}{\alpha ^2}t}}rdt
\end{array}
\]轉換回積分型態：
\[\begin{array}{l}
{Y_t}{U_t} - {Y_0}{U_0} = r\int_0^t {{e^{ - \alpha {B_s} + \frac{1}{2}{\alpha ^2}s}}d} s\\
\Rightarrow {Y_t}{U_t} = {Y_0} + r\int_0^t {{e^{ - \alpha {B_s} + \frac{1}{2}{\alpha ^2}s}}d} s\\
\Rightarrow {Y_t} = {Y_0}{e^{\alpha {B_t} - \frac{1}{2}{\alpha ^2}t}} + r{e^{\alpha {B_t} - \frac{1}{2}{\alpha ^2}t}}\int_0^t {{e^{ - \alpha {B_s} + \frac{1}{2}{\alpha ^2}s}}d} s\\
\Rightarrow {Y_t} = {Y_0}{e^{\alpha {B_t} - \frac{1}{2}{\alpha ^2}t}} + r\int_0^t {{e^{\alpha \left( {{B_t} - {B_s}} \right) - \frac{1}{2}{\alpha ^2}\left( {t - s} \right)}}d} s
\end{array}
\]
現在回頭檢驗此解是否滿足SDE:
首先將我們的解改寫定義如下：
\[{Y_t} = {e^{\alpha {B_t} - \frac{1}{2}{\alpha ^2}t}}\left( {{Y_0} + r\int_0^t {{e^{ - \alpha {B_s} + \frac{1}{2}{\alpha ^2}s}}d} s} \right): = {e^{\alpha {B_t} - \frac{1}{2}{\alpha ^2}t}}\left( {{V_t}} \right)
\]計算
\[d\left( {{Y_t}} \right) = d\left( {{e^{\alpha {B_t} - \frac{1}{2}{\alpha ^2}t}}{V_t}} \right) = {e^{\alpha {B_t} - \frac{1}{2}{\alpha ^2}t}}d{V_t} + {V_t}d\left( {{e^{\alpha {B_t} - \frac{1}{2}{\alpha ^2}t}}} \right) + {\left\langle {V,{e^{\alpha {B_t} - \frac{1}{2}{\alpha ^2}t}}} \right\rangle _t}
\]其中\[V_t^{} = {Y_0} + r\int_0^t {{e^{ - \alpha {B_s} + \frac{1}{2}{\alpha ^2}s}}d} s \Rightarrow d{V_t} = r{e^{ - \alpha {B_t} + \frac{1}{2}{\alpha ^2}t}}dt\]
且\[d\left( {{e^{\alpha {B_t} - \frac{1}{2}{\alpha ^2}t}}} \right): = {e^{\alpha {B_t} - \frac{1}{2}{\alpha ^2}t}}\alpha d{B_t}
\]上式由Ito formula推得 (因為是隨機，需用Ito formula幫助我們計算)
故
\[\begin{array}{l}
\Rightarrow d\left( {{Y_t}} \right) = {e^{\alpha {B_t} - \frac{1}{2}{\alpha ^2}t}}\left[ {r{e^{ - \alpha {B_t} + \frac{1}{2}{\alpha ^2}t}}dt + \alpha \left( {{Y_0} + r\int_0^t {{e^{ - \alpha {B_s} + \frac{1}{2}{\alpha ^2}s}}d} s} \right)d{B_t}} \right]\\
\Rightarrow d\left( {{Y_t}} \right) = {e^{\alpha {B_t} - \frac{1}{2}{\alpha ^2}t}}\left[ {r{e^{ - \alpha {B_t} + \frac{1}{2}{\alpha ^2}t}}dt + \alpha \left( {{e^{ - \alpha {B_t} + \frac{1}{2}{\alpha ^2}t}}{Y_t}} \right)d{B_t}} \right]\\
\Rightarrow d\left( {{Y_t}} \right) = \left[ {rdt + \alpha {Y_t}d{B_t}} \right]
\end{array}
\]亦即我們的解確實滿足SDE。至此計算完畢。
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