現在回頭再看看 Ito Formula 給我們的 Martingale 判別定理：
==============================
Theorem (Martingale PDE condition)
考慮 $t \in [0,T]$，若 $f \in \mathcal{C}^{1,2}(\mathbb{R}^+ \times \mathbb{R})$，且
\[
\frac{{\partial f}}{{\partial t}} + \frac{1}{2}\frac{{{\partial ^2}f}}{{\partial {x^2}}} =0
\]則 $X_t = f(t, B_t)$ 為一個 Local Martingale。
再者，若 ${\frac{{\partial f}}{{\partial x}}\left( {s,{B_s}} \right)}\in \mathcal{H}^2$，亦即
\[
E \left[\int_0^t \left ( {\frac{{\partial f}}{{\partial x}}\left( {s,{B_s}} \right)} \right)^2 ds \right] < \infty
\]則 $X_t$ 為一個 Martingale
==============================
現在再看個例子看看 Ito Formula 怎麼幫助我們獲得 Martingale
Example 1
令 $B_1(t), B_2(t), B_3 (t),...$ 互為獨立 Standard Brownian Motion。對 $k \in \mathbb{N}$ 定義函數 $g_k$ 與
\[
A_k(t) = \int_0^t g_k(B_1(s), B_2(s), ..., B_k(s))ds
\] 現在試求 $A_2$ 使得
\[
B_1(t)^2 B_2(t)^2 - A_2(t)
\]為 Martingale。 Hint: 利用 上述 PDE Martingale Condition。
Proof
我們要找
\[
{A_2}(t) = \int_0^t {{g_2}} ({B_1}(s),{B_2}(s))ds
\] 使得 $B_1(t)^2 B_2(t)^2 - A_2(t)$ 為 Martingale。
現在定義
\[f\left( {x,y} \right) = {x^2}{y^2}
\] 則我們首先計算其偏導數： ${f_x} = 2x{y^2}{,_{}}{f_{xx}} = 2{y^2}{,_{}}{f_y} = 2{x^2}y{,_{}}{f_{yy}} = 2{x^2}$ 現在由 Ito Formula：
\[
\begin{array}{l}
df\left( {{B_1}(t),{B_2}(t)} \right) = \frac{{\partial f}}{{\partial x}}\left( {{B_1}(t),{B_2}(t)} \right)d{B_1}(t)\\
\begin{array}{*{20}{c}}
{}&{}&{}
\end{array} + \frac{{\partial f}}{{\partial y}}\left( {{B_1}(t),{B_2}(t)} \right)d{B_2}(t) + \frac{1}{2}\left[ \begin{array}{l}
\frac{{{\partial ^2}f}}{{\partial {x^2}}}\left( {{B_1}(t),{B_2}(t)} \right)dt\\
+ \frac{{{\partial ^2}f}}{{\partial {y^2}}}\left( {{B_1}(t),{B_2}(t)} \right)dt\\
+ \frac{{{\partial ^2}f}}{{\partial x\partial y}}\left( {{B_1}(t),{B_2}(t)} \right)d\left\langle {{B_1},{B_2}} \right\rangle
\end{array} \right]\\
\Rightarrow df\left( {{B_1}(t),{B_2}(t)} \right) = \frac{{\partial f}}{{\partial x}}\left( {{B_1}(t),{B_2}(t)} \right)d{B_1}(t) + \frac{{\partial f}}{{\partial y}}\left( {{B_1}(t),{B_2}(t)} \right)d{B_2}(t)\\
\begin{array}{*{20}{c}}
{}&{}&{}&{}
\end{array} + \frac{1}{2}\left[ {\frac{{{\partial ^2}f}}{{\partial {x^2}}}\left( {{B_1}(t),{B_2}(t)} \right) + \frac{{{\partial ^2}f}}{{\partial {y^2}}}\left( {{B_1}(t),{B_2}(t)} \right)} \right]dt
\end{array}
\] 注意到上式中的 Cross Variation term: $ d\left\langle {{B_1},{B_2}} \right\rangle = 0$；另外為了形式簡潔起見，我們把 函數中的 時間 $t$ 先移除不寫，則上式變成：
\[\begin{array}{l}
df\left( {{B_1},{B_2}} \right) = 2{B_1}{B_2}^2d{B_1} + 2{B_1}^2{B_2}d{B_2} + \frac{1}{2}\left[ {2{B_2}^2dt + 2{B_1}^2dt} \right]\\
\Rightarrow df\left( {{B_1},{B_2}} \right) = 2{B_1}{B_2}^2d{B_1} + 2{B_1}^2{B_2}d{B_2} + {B_2}^2dt + {B_1}^2dt
\end{array}
\] 現在將其轉換回 積分形式：
\[\begin{array}{l}
\underbrace {f\left( {{B_1},{B_2}} \right)}_{ = {B_1}{{(t)}^2}{B_2}{{(t)}^2}} = 2\int_0^t {{B_1}{B_2}^2d{B_1}} + 2\int_0^t {{B_1}^2{B_2}d{B_2}} + \int_0^t {\left( {{B_2}^2 + {B_1}^2} \right)dt} \\
\Rightarrow {B_1}^2{B_2}^2 - \int_0^t {\left( {{B_2}^2 + {B_1}^2} \right)dt} = 2\int_0^t {{B_1}{B_2}^2d{B_1}} + 2\int_0^t {{B_1}^2{B_2}d{B_2}}
\end{array}
\]注意到上式等號右邊為 Local Martingale。現在檢驗 積分變數 是否落在 $\cal{H}^2$ 中，如果是的話，我們即得到 Martingale：故
\[\begin{array}{l}
E\left[ {\int_0^T {{{\left( {{B_1}{B_2}^2} \right)}^2}ds} } \right] = E\left[ {\int_0^T {{B_1}^2{B_2}^4ds} } \right] = \int_0^T {E\left[ {{B_1}^2{B_2}^4} \right]ds} \\
= \int_0^T {E\left[ {{B_1}^2} \right]E\left[ {{B_2}^4} \right]ds} = \int_0^T {s\left( {3{s^2}} \right)ds} = 3\int_0^T {{s^3}ds} = \frac{3}{4}{T^4} < \infty
\end{array}\] 故可知 ${{B_1}{B_2}^2} \in \cal{H}^2$
同理可證 ${{B_1}^2{B_2}} \in \cal{H}^2$，故我們得到
\[{B_1}{(t)^2}{B_2}{(t)^2} - \int_0^t {\left( {{B_2}^2 + {B_1}^2} \right)dt} \]為 Martingale。亦即
\[
{A_2}(t) = \int_0^t {\left( {{B_2}^2 + {B_1}^2} \right)dt} \ \ \ \ \square
\]
現在我們接續前述的例子，我們看看 $k=3$ 的情況：
Example 2
令 $B_1(t), B_2(t), B_3 (t),...$ 互為獨立 Standard Brownian Motion。對 $k \in \mathbb{N}$ 定義函數 $g_k$ 與
\[
A_k(t) = \int_0^t g_k(B_1(s), B_2(s), ..., B_k(s))ds
\] 現在試求 $A_3$ 使得
\[
B_1(t)^2 B_2(t)^2 B_3(t)^2 - A_3(t)
\]為 Martingale。 Hint: 利用 上述 PDE Martingale Condition。
Proof
如前例，令 $f\left( {x,y,z} \right) = {x^2}{y^2}{z^2}{,_{}} $ ，
\[\left\{ \begin{array}{l}
{f_x} = 2x{y^2}{z^2},\begin{array}{*{20}{c}}
{}
\end{array}{f_{xx}} = 2{y^2}{z^2},\\
{f_y} = 2{x^2}y{z^2},\begin{array}{*{20}{c}}
{}
\end{array}{f_{yy}} = 2{x^2}{z^2},\\
{f_z} = 2{x^2}{y^2}z,\begin{array}{*{20}{c}}
{}
\end{array}{f_{zz}} = 2{x^2}{y^2},
\end{array} \right.
\] 利用 Ito Formula 我們可得
\[\begin{array}{l}
df\left( {{B_1},{B_2},{B_3}} \right) = \frac{{\partial f}}{{\partial x}}\left( {{B_1},{B_2},{B_3}} \right)d{B_1} + \frac{{\partial f}}{{\partial y}}\left( {{B_1},{B_2},{B_3}} \right)d{B_2}\\
\begin{array}{*{20}{c}}
{}&{}&{}&{}
\end{array} + \frac{{\partial f}}{{\partial z}}\left( {{B_1},{B_2},{B_3}} \right)d{B_3} + \frac{1}{2}\left[ \begin{array}{l}
\frac{{{\partial ^2}f}}{{\partial {x^2}}}\left( {{B_1},{B_2},{B_3}} \right)dt\\
\begin{array}{*{20}{c}}
{}
\end{array} + \frac{{{\partial ^2}f}}{{\partial {y^2}}}\left( {{B_1},{B_2},{B_3}} \right)dt\\
\begin{array}{*{20}{c}}
{}
\end{array} + \frac{{{\partial ^2}f}}{{\partial {z^2}}}\left( {{B_1},{B_2},{B_3}} \right)dt
\end{array} \right]\\
\Rightarrow df\left( {{B_1},{B_2},{B_3}} \right) = \frac{{\partial f}}{{\partial x}}\left( {{B_1},{B_2},{B_3}} \right)d{B_1}\\
\begin{array}{*{20}{c}}
{}&{}&{}&{}
\end{array} + \frac{{\partial f}}{{\partial y}}\left( {{B_1},{B_2},{B_3}} \right)d{B_2} + \frac{{\partial f}}{{\partial z}}\left( {{B_1},{B_2},{B_3}} \right)d{B_3}\\
\begin{array}{*{20}{c}}
{}&{}&{}&{}
\end{array} + \frac{1}{2}\left[ {\frac{{{\partial ^2}f}}{{\partial {x^2}}}\left( {{B_1},{B_2},{B_3}} \right) + \frac{{{\partial ^2}f}}{{\partial {y^2}}}\left( {{B_1},{B_2},{B_3}} \right) + \frac{{{\partial ^2}f}}{{\partial {z^2}}}\left( {{B_1},{B_2},{B_3}} \right)} \right]dt
\end{array}
\] 故
\[\begin{array}{l}
df\left( {{B_1},{B_2},{B_3}} \right) = 2{B_1}{B_2}^2{B_3}^2d{B_1} + 2{B_1}^2{B_2}{B_3}^2d{B_2} + 2{B_1}^2{B_2}^2{B_3}d{B_3}\\
\begin{array}{*{20}{c}}
{}&{}&{}&{}
\end{array} + \frac{1}{2}\left[ {2{B_2}^2{B_3}^2 + 2{B_1}^2{B_3}^2 + 2{B_1}^2{B_2}^2} \right]dt
\end{array}
\]現在轉換回積分形式：
\[\begin{array}{l}
\Rightarrow \underbrace {f\left( {{B_1},{B_2},{B_3}} \right)}_{ = {B_1}^2{B_2}^2{B_3}^2} = 2\left[ \begin{array}{l}
\int_0^t {{B_1}{B_2}^2{B_3}^2d{B_1}} + \int_0^t {{B_1}^2{B_2}{B_3}^2d{B_2}} \\
\begin{array}{*{20}{c}}
{}&{}
\end{array} + \int_0^t {{B_1}^2{B_2}^2{B_3}d{B_3}}
\end{array} \right]\\
\begin{array}{*{20}{c}}
{}&{}&{}&{}&{}&{}&{}&{}
\end{array} + \int_0^t {\left( {{B_2}^2{B_3}^2 + {B_1}^2{B_3}^2 + {B_1}^2{B_2}^2} \right)} ds\\
\Rightarrow {B_1}^2{B_2}^2{B_3}^2 - \int_0^t {\left( {{B_2}^2{B_3}^2 + {B_1}^2{B_3}^2 + {B_1}^2{B_2}^2} \right)} ds\\
\begin{array}{*{20}{c}}
{}&{}&{}&{}&{}&{}&{}&{}
\end{array} = 2\left[ \begin{array}{l}
\int_0^t {{B_1}{B_2}^2{B_3}^2d{B_1}} + \int_0^t {{B_1}^2{B_2}{B_3}^2d{B_2}} \\
\begin{array}{*{20}{c}}
{}&{}
\end{array} + \int_0^t {{B_1}^2{B_2}^2{B_3}d{B_3}}
\end{array} \right]
\end{array}\] 故上式等號右邊 為 Local Martingale。(讀者可自行驗證其 $\int (\cdot) dB$ 積分的 積分變數全部都落在 $\cal{H}^2$)，故為 Martingale，亦即
\[{B_1}^2{B_2}^2{B_3}^2 - \int_0^t {\left( {{B_2}^2{B_3}^2 + {B_1}^2{B_3}^2 + {B_1}^2{B_2}^2} \right)} ds\]為 Martingale。
且
\[{A_3}(t) = \int_0^t {\left( {{B_2}^2{B_3}^2 + {B_1}^2{B_3}^2 + {B_1}^2{B_2}^2} \right)} ds\]
==============================
Theorem (Martingale PDE condition)
考慮 $t \in [0,T]$，若 $f \in \mathcal{C}^{1,2}(\mathbb{R}^+ \times \mathbb{R})$，且
\[
\frac{{\partial f}}{{\partial t}} + \frac{1}{2}\frac{{{\partial ^2}f}}{{\partial {x^2}}} =0
\]則 $X_t = f(t, B_t)$ 為一個 Local Martingale。
再者，若 ${\frac{{\partial f}}{{\partial x}}\left( {s,{B_s}} \right)}\in \mathcal{H}^2$，亦即
\[
E \left[\int_0^t \left ( {\frac{{\partial f}}{{\partial x}}\left( {s,{B_s}} \right)} \right)^2 ds \right] < \infty
\]則 $X_t$ 為一個 Martingale
==============================
現在再看個例子看看 Ito Formula 怎麼幫助我們獲得 Martingale
Example 1
令 $B_1(t), B_2(t), B_3 (t),...$ 互為獨立 Standard Brownian Motion。對 $k \in \mathbb{N}$ 定義函數 $g_k$ 與
\[
A_k(t) = \int_0^t g_k(B_1(s), B_2(s), ..., B_k(s))ds
\] 現在試求 $A_2$ 使得
\[
B_1(t)^2 B_2(t)^2 - A_2(t)
\]為 Martingale。 Hint: 利用 上述 PDE Martingale Condition。
Proof
我們要找
\[
{A_2}(t) = \int_0^t {{g_2}} ({B_1}(s),{B_2}(s))ds
\] 使得 $B_1(t)^2 B_2(t)^2 - A_2(t)$ 為 Martingale。
現在定義
\[f\left( {x,y} \right) = {x^2}{y^2}
\] 則我們首先計算其偏導數： ${f_x} = 2x{y^2}{,_{}}{f_{xx}} = 2{y^2}{,_{}}{f_y} = 2{x^2}y{,_{}}{f_{yy}} = 2{x^2}$ 現在由 Ito Formula：
\[
\begin{array}{l}
df\left( {{B_1}(t),{B_2}(t)} \right) = \frac{{\partial f}}{{\partial x}}\left( {{B_1}(t),{B_2}(t)} \right)d{B_1}(t)\\
\begin{array}{*{20}{c}}
{}&{}&{}
\end{array} + \frac{{\partial f}}{{\partial y}}\left( {{B_1}(t),{B_2}(t)} \right)d{B_2}(t) + \frac{1}{2}\left[ \begin{array}{l}
\frac{{{\partial ^2}f}}{{\partial {x^2}}}\left( {{B_1}(t),{B_2}(t)} \right)dt\\
+ \frac{{{\partial ^2}f}}{{\partial {y^2}}}\left( {{B_1}(t),{B_2}(t)} \right)dt\\
+ \frac{{{\partial ^2}f}}{{\partial x\partial y}}\left( {{B_1}(t),{B_2}(t)} \right)d\left\langle {{B_1},{B_2}} \right\rangle
\end{array} \right]\\
\Rightarrow df\left( {{B_1}(t),{B_2}(t)} \right) = \frac{{\partial f}}{{\partial x}}\left( {{B_1}(t),{B_2}(t)} \right)d{B_1}(t) + \frac{{\partial f}}{{\partial y}}\left( {{B_1}(t),{B_2}(t)} \right)d{B_2}(t)\\
\begin{array}{*{20}{c}}
{}&{}&{}&{}
\end{array} + \frac{1}{2}\left[ {\frac{{{\partial ^2}f}}{{\partial {x^2}}}\left( {{B_1}(t),{B_2}(t)} \right) + \frac{{{\partial ^2}f}}{{\partial {y^2}}}\left( {{B_1}(t),{B_2}(t)} \right)} \right]dt
\end{array}
\] 注意到上式中的 Cross Variation term: $ d\left\langle {{B_1},{B_2}} \right\rangle = 0$；另外為了形式簡潔起見，我們把 函數中的 時間 $t$ 先移除不寫，則上式變成：
\[\begin{array}{l}
df\left( {{B_1},{B_2}} \right) = 2{B_1}{B_2}^2d{B_1} + 2{B_1}^2{B_2}d{B_2} + \frac{1}{2}\left[ {2{B_2}^2dt + 2{B_1}^2dt} \right]\\
\Rightarrow df\left( {{B_1},{B_2}} \right) = 2{B_1}{B_2}^2d{B_1} + 2{B_1}^2{B_2}d{B_2} + {B_2}^2dt + {B_1}^2dt
\end{array}
\] 現在將其轉換回 積分形式：
\[\begin{array}{l}
\underbrace {f\left( {{B_1},{B_2}} \right)}_{ = {B_1}{{(t)}^2}{B_2}{{(t)}^2}} = 2\int_0^t {{B_1}{B_2}^2d{B_1}} + 2\int_0^t {{B_1}^2{B_2}d{B_2}} + \int_0^t {\left( {{B_2}^2 + {B_1}^2} \right)dt} \\
\Rightarrow {B_1}^2{B_2}^2 - \int_0^t {\left( {{B_2}^2 + {B_1}^2} \right)dt} = 2\int_0^t {{B_1}{B_2}^2d{B_1}} + 2\int_0^t {{B_1}^2{B_2}d{B_2}}
\end{array}
\]注意到上式等號右邊為 Local Martingale。現在檢驗 積分變數 是否落在 $\cal{H}^2$ 中，如果是的話，我們即得到 Martingale：故
\[\begin{array}{l}
E\left[ {\int_0^T {{{\left( {{B_1}{B_2}^2} \right)}^2}ds} } \right] = E\left[ {\int_0^T {{B_1}^2{B_2}^4ds} } \right] = \int_0^T {E\left[ {{B_1}^2{B_2}^4} \right]ds} \\
= \int_0^T {E\left[ {{B_1}^2} \right]E\left[ {{B_2}^4} \right]ds} = \int_0^T {s\left( {3{s^2}} \right)ds} = 3\int_0^T {{s^3}ds} = \frac{3}{4}{T^4} < \infty
\end{array}\] 故可知 ${{B_1}{B_2}^2} \in \cal{H}^2$
同理可證 ${{B_1}^2{B_2}} \in \cal{H}^2$，故我們得到
\[{B_1}{(t)^2}{B_2}{(t)^2} - \int_0^t {\left( {{B_2}^2 + {B_1}^2} \right)dt} \]為 Martingale。亦即
\[
{A_2}(t) = \int_0^t {\left( {{B_2}^2 + {B_1}^2} \right)dt} \ \ \ \ \square
\]
現在我們接續前述的例子，我們看看 $k=3$ 的情況：
Example 2
令 $B_1(t), B_2(t), B_3 (t),...$ 互為獨立 Standard Brownian Motion。對 $k \in \mathbb{N}$ 定義函數 $g_k$ 與
\[
A_k(t) = \int_0^t g_k(B_1(s), B_2(s), ..., B_k(s))ds
\] 現在試求 $A_3$ 使得
\[
B_1(t)^2 B_2(t)^2 B_3(t)^2 - A_3(t)
\]為 Martingale。 Hint: 利用 上述 PDE Martingale Condition。
Proof
如前例，令 $f\left( {x,y,z} \right) = {x^2}{y^2}{z^2}{,_{}} $ ，
\[\left\{ \begin{array}{l}
{f_x} = 2x{y^2}{z^2},\begin{array}{*{20}{c}}
{}
\end{array}{f_{xx}} = 2{y^2}{z^2},\\
{f_y} = 2{x^2}y{z^2},\begin{array}{*{20}{c}}
{}
\end{array}{f_{yy}} = 2{x^2}{z^2},\\
{f_z} = 2{x^2}{y^2}z,\begin{array}{*{20}{c}}
{}
\end{array}{f_{zz}} = 2{x^2}{y^2},
\end{array} \right.
\] 利用 Ito Formula 我們可得
\[\begin{array}{l}
df\left( {{B_1},{B_2},{B_3}} \right) = \frac{{\partial f}}{{\partial x}}\left( {{B_1},{B_2},{B_3}} \right)d{B_1} + \frac{{\partial f}}{{\partial y}}\left( {{B_1},{B_2},{B_3}} \right)d{B_2}\\
\begin{array}{*{20}{c}}
{}&{}&{}&{}
\end{array} + \frac{{\partial f}}{{\partial z}}\left( {{B_1},{B_2},{B_3}} \right)d{B_3} + \frac{1}{2}\left[ \begin{array}{l}
\frac{{{\partial ^2}f}}{{\partial {x^2}}}\left( {{B_1},{B_2},{B_3}} \right)dt\\
\begin{array}{*{20}{c}}
{}
\end{array} + \frac{{{\partial ^2}f}}{{\partial {y^2}}}\left( {{B_1},{B_2},{B_3}} \right)dt\\
\begin{array}{*{20}{c}}
{}
\end{array} + \frac{{{\partial ^2}f}}{{\partial {z^2}}}\left( {{B_1},{B_2},{B_3}} \right)dt
\end{array} \right]\\
\Rightarrow df\left( {{B_1},{B_2},{B_3}} \right) = \frac{{\partial f}}{{\partial x}}\left( {{B_1},{B_2},{B_3}} \right)d{B_1}\\
\begin{array}{*{20}{c}}
{}&{}&{}&{}
\end{array} + \frac{{\partial f}}{{\partial y}}\left( {{B_1},{B_2},{B_3}} \right)d{B_2} + \frac{{\partial f}}{{\partial z}}\left( {{B_1},{B_2},{B_3}} \right)d{B_3}\\
\begin{array}{*{20}{c}}
{}&{}&{}&{}
\end{array} + \frac{1}{2}\left[ {\frac{{{\partial ^2}f}}{{\partial {x^2}}}\left( {{B_1},{B_2},{B_3}} \right) + \frac{{{\partial ^2}f}}{{\partial {y^2}}}\left( {{B_1},{B_2},{B_3}} \right) + \frac{{{\partial ^2}f}}{{\partial {z^2}}}\left( {{B_1},{B_2},{B_3}} \right)} \right]dt
\end{array}
\] 故
\[\begin{array}{l}
df\left( {{B_1},{B_2},{B_3}} \right) = 2{B_1}{B_2}^2{B_3}^2d{B_1} + 2{B_1}^2{B_2}{B_3}^2d{B_2} + 2{B_1}^2{B_2}^2{B_3}d{B_3}\\
\begin{array}{*{20}{c}}
{}&{}&{}&{}
\end{array} + \frac{1}{2}\left[ {2{B_2}^2{B_3}^2 + 2{B_1}^2{B_3}^2 + 2{B_1}^2{B_2}^2} \right]dt
\end{array}
\]現在轉換回積分形式：
\[\begin{array}{l}
\Rightarrow \underbrace {f\left( {{B_1},{B_2},{B_3}} \right)}_{ = {B_1}^2{B_2}^2{B_3}^2} = 2\left[ \begin{array}{l}
\int_0^t {{B_1}{B_2}^2{B_3}^2d{B_1}} + \int_0^t {{B_1}^2{B_2}{B_3}^2d{B_2}} \\
\begin{array}{*{20}{c}}
{}&{}
\end{array} + \int_0^t {{B_1}^2{B_2}^2{B_3}d{B_3}}
\end{array} \right]\\
\begin{array}{*{20}{c}}
{}&{}&{}&{}&{}&{}&{}&{}
\end{array} + \int_0^t {\left( {{B_2}^2{B_3}^2 + {B_1}^2{B_3}^2 + {B_1}^2{B_2}^2} \right)} ds\\
\Rightarrow {B_1}^2{B_2}^2{B_3}^2 - \int_0^t {\left( {{B_2}^2{B_3}^2 + {B_1}^2{B_3}^2 + {B_1}^2{B_2}^2} \right)} ds\\
\begin{array}{*{20}{c}}
{}&{}&{}&{}&{}&{}&{}&{}
\end{array} = 2\left[ \begin{array}{l}
\int_0^t {{B_1}{B_2}^2{B_3}^2d{B_1}} + \int_0^t {{B_1}^2{B_2}{B_3}^2d{B_2}} \\
\begin{array}{*{20}{c}}
{}&{}
\end{array} + \int_0^t {{B_1}^2{B_2}^2{B_3}d{B_3}}
\end{array} \right]
\end{array}\] 故上式等號右邊 為 Local Martingale。(讀者可自行驗證其 $\int (\cdot) dB$ 積分的 積分變數全部都落在 $\cal{H}^2$)，故為 Martingale，亦即
\[{B_1}^2{B_2}^2{B_3}^2 - \int_0^t {\left( {{B_2}^2{B_3}^2 + {B_1}^2{B_3}^2 + {B_1}^2{B_2}^2} \right)} ds\]為 Martingale。
且
\[{A_3}(t) = \int_0^t {\left( {{B_2}^2{B_3}^2 + {B_1}^2{B_3}^2 + {B_1}^2{B_2}^2} \right)} ds\]
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