現在我們來看看 Ito formula 可以幫助我們判別是否為 Martingale or Local Martingale。
現在考慮 $t \in [0,T]$，回憶雙變數的 Ito formula
\[
\small{
f(t,B_t) = f(0,B_0) + \int_0^t \frac{{\partial f}}{{\partial t}}\left( {s,{B_s}} \right)ds + \int_0^t \frac{{\partial f}}{{\partial x}}\left( {s,{B_s}} \right) dB_s + \frac{1}{2}\int_0^t \frac{{\partial^2 f}}{{\partial x^2}}\left( {s,{B_s}} \right)ds }
\]
現在將上式 $\int ds$ 項合併，可得
\[
\small{ f(t,{B_t}) = f(0,{B_0}) + \int_0^t {\left( {\frac{{\partial f}}{{\partial t}}\left( {s,{B_s}} \right) + \frac{1}{2}\frac{{{\partial ^2}f}}{{\partial {x^2}}}\left( {s,{B_s}} \right)} \right)} ds + \int_0^t {\frac{{\partial f}}{{\partial x}}} \left( {s,{B_s}} \right)d{B_s}}
\]觀察上式，如果 $\int ds$ 項為零，亦即
\[
{\frac{{\partial f}}{{\partial t}}\left( {s,{B_s}} \right) + \frac{1}{2}\frac{{{\partial ^2}f}}{{\partial {x^2}}}\left( {s,{B_s}} \right)}=0
\]，則 Ito formula剩餘的最後一項 為 Ito integral ，由我們剛剛提過的 $(\star)$ 可知，此 Ito integral
\[\int_0^t {\frac{{\partial f}}{{\partial x}}} \left( {s,{B_s}} \right)d{B_s}
\]為 Local martingale。 WHY!? 因為Ito formula假設 $f \in \mathcal{C}^{1,2}(\mathbb{R}^+ \times \mathbb{R})$，也就是說 Ito integral 項的積分變數 (對 $f$ 取一階偏導數) 為 ${\frac{{\partial f}}{{\partial x}}\left( {s,{B_s}} \right)} \in \mathcal{C}^1$，又因為 $t \in [0,T]$為compact domain，連續函數必定有界，也就是說積分變數是落在 $L_{LOC}^2$，亦即滿足下式
\[
\int_0^t \left ( {\frac{{\partial f}}{{\partial x}}\left( {s,{B_s}} \right)} \right)^2 ds < \infty
\]
另外，如果 Ito integral的 積分變數 ${\frac{{\partial f}}{{\partial x}}\left( {s,{B_s}} \right)} \in \mathcal{H}^2$，亦即滿足下式
\[
E \left [\int_0^t \left ( {\frac{{\partial f}}{{\partial x}}\left( {s,{B_s}} \right)} \right)^2 ds \right] < \infty
\]
，則我們可得到 Martingale。
我們現在將上述結果寫成下面這個定理：
Theorem (Martingale PDE condition)
考慮 $t \in [0,T]$，若 $f \in \mathcal{C}^{1,2}(\mathbb{R}^+ \times \mathbb{R})$，且
\[
\frac{{\partial f}}{{\partial t}} + \frac{1}{2}\frac{{{\partial ^2}f}}{{\partial {x^2}}} =0\ \ \ \ (1)
\]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
\[
E \left[\int_0^t \left ( {\frac{{\partial f}}{{\partial x}}\left( {s,{B_s}} \right)} \right)^2 ds \right] < \infty \ \ \ \ (2)
\]則 $X_t$ 為一個 Martingale
Proof
其實證明已經於前面討論寫完，但我們這邊把前述討論再稍作整理。
給定任意 $f \in \mathcal{C}^{1,2}(\mathbb{R}^+ \times \mathbb{R})$ 滿足雙變數 Ito formula：
\[
\small{ f(t,{B_t}) = f(0,{B_0}) + \int_0^t {\left( {\frac{{\partial f}}{{\partial t}}\left( {s,{B_s}} \right) + \frac{1}{2}\frac{{{\partial ^2}f}}{{\partial {x^2}}}\left( {s,{B_s}} \right)} \right)} ds + \int_0^t {\frac{{\partial f}}{{\partial x}}} \left( {s,{B_s}} \right)d{B_s}}
\]
由假設 $(1)$，上式變成
\[f(t,{B_t}) = f(0,{B_0}) + \int_0^t {\frac{{\partial f}}{{\partial x}}\left( {s,{B_s}} \right)} d{B_s}\]
由於 $f \in \mathcal{C}^{1,2}(\mathbb{R}^+ \times \mathbb{R}) \Rightarrow {\frac{{\partial f}}{{\partial x}}} \in \mathcal{C}^1$，故 ${\frac{{\partial f}}{{\partial x}}}$為連續函數，又因為 $t \in [0,T]$為compact domain，連續函數必定有界，也就是說積分變數 ${\frac{{\partial f}}{{\partial x}}}$ 是落在 $L_{LOC}^2$，亦即滿足下式
\[
\int_0^t \left ( {\frac{{\partial f}}{{\partial x}}\left( {s,{B_s}} \right)} \right)^2 ds < \infty
\]
故 Ito integral $\int_0^t {\frac{{\partial f}}{{\partial x}}\left( {s,{B_s}} \right)} d{B_s}$為一個Local Martingale。
另外如果假設 $(2)$ 成立；亦即
\[
E \left[\int_0^t \left ( {\frac{{\partial f}}{{\partial x}}\left( {s,{B_s}} \right)} \right)^2 ds \right] < \infty
\]則積分變數 ${\frac{{\partial f}}{{\partial x}}\left( {s,{B_s}} \right)}\in \mathcal{H}^2$，故
Ito integral $\int_0^t {\frac{{\partial f}}{{\partial x}}\left( {s,{B_s}} \right)} d{B_s}$為一個 Martingale。 $\square$
Example (Ruin Problem)
令 $B_t$ 為 標準布朗運動，定義一個隨機過程 $X_t$符合下式
\[
X_t := \mu t + \sigma B_t
\]
其中 $\mu \in \mathbb{R}$； $\sigma>0$；$A,B >0$。且定義停止時間
\[
\tau := \inf\{ t>0 : X_t = A \ or \ X_t = -B\}
\]計算 $P(X_{\tau} = A ) =?$
Comment:
上述隨機過程 $X_t := \mu t + \sigma B_t $ 一般稱之為 Arithmetic Brownian Motion。
Solution:
想法如下：
如果我們可以找到一個函數 $h(X_t)$ 使其為 Martingale (透過滿足 Martingale PDE condtion Theorem)，則利用 Martingale的性質我們知道
\[
E[h(X_0)] = E[h(X_{\tau})]
\]又因為 $E[h(X_{\tau})] = h(A)P(X_{\tau} = A) + h(-B) P(X_{\tau}=-B)$，故
\[
E[h(X_0)] = E[h(X_{\tau})] = h(A)P(X_{\tau} = A) + h(-B) P(X_{\tau}=-B)
\]又我們可以知道$h(A)$ 與 $h(-B)$，故即可解得 $P(X_{\tau} = A) $。
以下開始逐步求解：
為了要找出$h(X_t)$，我們令
$f(t,x):=h( \mu t + \sigma x)$，則 $f(t,B_t) = h( \mu t + \sigma B_t) = h(X_t)$
現在透過 Martingale PDE condition $(1)$：
\[
\frac{{\partial f}}{{\partial t}} + \frac{1}{2}\frac{{{\partial ^2}f}}{{\partial {x^2}}} =0 \ \ \ \ (**)
\]為了求解簡便起見，令 $z := \mu t + \sigma x $，則
\[\left\{ \begin{array}{l}
\frac{{\partial f}}{{\partial t}} = \frac{{\partial h}}{{\partial z}}\frac{{\partial z}}{{\partial t}} = \frac{{dh}}{{dz}}\mu \\
\frac{{\partial f}}{{\partial x}} = \frac{{\partial h}}{{\partial z}}\frac{{\partial z}}{{\partial x}} = \frac{{dh}}{{dz}}\sigma \\
\frac{{{\partial ^2}f}}{{\partial {x^2}}} = {\sigma ^2}\frac{{{d^2}h}}{{d{z^2}}}
\end{array} \right.\]則我們的 $(**)$ 變成
\[\begin{array}{l}
\frac{{dh}}{{dz}}\mu + \frac{1}{2}{\sigma ^2}\frac{{{d^2}h}}{{d{z^2}}} = 0\\
\Rightarrow h''\left( z \right) = \frac{{ - 2\mu }}{{{\sigma ^2}}}h'\left( z \right) \\
\Rightarrow \frac{{h''}}{{h'}}\left( z \right) = \frac{{ - 2\mu }}{{{\sigma ^2}}}\\
\Rightarrow {\left( {\ln \left( {h'\left( z \right)} \right)} \right)^\prime } = \frac{{ - 2\mu }}{{{\sigma ^2}}}\\
\Rightarrow \ln \left( {h'\left( z \right)} \right) = \frac{{ - 2\mu }}{{{\sigma ^2}}}z + C\\
\Rightarrow h'\left( z \right) = {C_1}{e^{\frac{{ - 2\mu }}{{{\sigma ^2}}}z}}\\
\Rightarrow h\left( z \right) = {C_2}{e^{\frac{{ - 2\mu }}{{{\sigma ^2}}}z}} + {C_3}\\
\Rightarrow f\left( {t,x} \right) = h\left( {\mu t + \sigma x} \right) = {C_2}{e^{\frac{{ - 2\mu }}{{{\sigma ^2}}}\left( {\mu t + \sigma x} \right)}} + {C_3} \ \ \ \ (\star \star)
\end{array}
\]再來我們透過 Martingale PDE condition $(2)$，計算 $L^2$-norm
\[\begin{array}{l}
E\left[ {\int_0^t {{{\left( {\frac{{\partial f}}{{\partial x}}\left( {s,{B_s}} \right)} \right)}^2}} ds} \right] \\
\Rightarrow E\left[ {\int_0^t {{{\left( {\frac{{dh}}{{dz}}\sigma } \right)}^2}} ds} \right]= {\sigma ^2}E\left[ {{{\int_0^t {\left( {\frac{{dh}}{{dz}}} \right)} }^2}ds} \right]
\end{array}
\]由 $(\star \star)$，我們知道
\[
h\left( z \right) = {C_2}{e^{\frac{{ - 2\mu }}{{{\sigma ^2}}}\left( z \right)}} + {C_3} \Rightarrow \frac{{dh}}{{dz}} = {C_4}{e^{\frac{{ - 2\mu }}{{{\sigma ^2}}}\left( z \right)}}
\]，故我們可得到
\[
\begin{array}{l}
\Rightarrow E\left[ {\int_0^t {{{\left( {\frac{{dh}}{{dz}}\sigma } \right)}^2}} ds} \right] = {\sigma ^2}E\left[ {{{\int_0^t {\left( {{C_4}{e^{\frac{{ - 2\mu }}{{{\sigma ^2}}}\left( z \right)}}} \right)} }^2}ds} \right]\\
\Rightarrow {C_5}E\left[ {\int_0^t {{e^{\frac{{ - 4\mu }}{{{\sigma ^2}}}\left( {\mu s + \sigma {B_s}} \right)}}ds} } \right] = {C_5}E\left[ {\int_0^t {{e^{\frac{{ - 4\mu }}{{{\sigma ^2}}}\left( {\mu s} \right)}}{e^{\frac{{ - 4\mu }}{{{\sigma ^2}}}\left( {\sigma {B_s}} \right)}}ds} } \right]\\
\Rightarrow {C_5}E\left[ {\int_0^t {\underbrace {{e^{\frac{{ - 4\mu }}{{{\sigma ^2}}}\left( {\mu s} \right)}}}_{ \le 1}{e^{\frac{{ - 4\mu }}{\sigma }\left( {{B_s}} \right)}}ds} } \right] \le {C_5}E\left[ {\int_0^t {{e^{\frac{{ - 4\mu }}{\sigma }\left( {{B_s}} \right)}}ds} } \right]\\
\Rightarrow {C_5}E\left[ {\int_0^t {\underbrace {{e^{\frac{{ - 4\mu }}{{{\sigma ^2}}}\left( {\mu s} \right)}}}_{ \le 1}{e^{\frac{{ - 4\mu }}{\sigma }\left( {{B_s}} \right)}}ds} } \right] \le {C_5}\int_0^t {E\left[ {{e^{\frac{{ - 4\mu }}{\sigma }\left( {{B_s}} \right)}}} \right]ds}
\end{array}
\]注意到 ${E\left[ {{e^{\frac{{ - 4\mu }}{\sigma }\left( {{B_s}} \right)}}} \right]}$ 為 Gaussian Random Variable的 Moment Generating Function，亦即
\[
E\left[ {{e^{\frac{{ - 4\mu }}{\sigma }\left( {{B_s}} \right)}}} \right] = {e^{\frac{1}{2}{{\left( {\frac{{ - 4\mu }}{\sigma }} \right)}^2}s}}
\]故，
\[
E\left[ {\int_0^t {{{\left( {\frac{{\partial f}}{{\partial x}}\left( {s,{B_s}} \right)} \right)}^2}} ds} \right] \leq {C_5}\int_0^t {{e^{\frac{1}{2}{{\left( {\frac{{ - 4\mu }}{\sigma }} \right)}^2}s}}ds} < \infty
\]，至此我們知道
\[
f\left( {t,{B_t}} \right) = h\left( {\mu t + \sigma {B_t}} \right) = h(X_t)= {C_2}{e^{\frac{{ - 2\mu }}{{{\sigma ^2}}}\left( {\mu t + \sigma {B_t}} \right)}} + {C_3}
\] 為Martingale。
由於 $ h(X_t) $ 為Martingale $\Rightarrow$ $h(X_{t \wedge \tau})$亦為 Martingale。
\[
E[h(X_0)] = E[h(X_{t \wedge \tau})]
\]且因為$h(X_t)$有界 (bounded by A or -B)；亦即
\[
|h(X_{t \wedge \tau}) | \leq \max_{-B \leq x \leq A} h(x)
\]
，故由Dominated Convergence Theorem，我們知道當 $t \rightarrow \infty$，
\[
E[h(X_0)] = E[h(X_{t \wedge \tau})] \rightarrow E[h(X_{\tau})]
\]也就是說
\[
E[h(X_0)] = E[h(X_{\tau})] \ \ \ \ (**)
\]
且我們知道 $E[h(X_{\tau})] = h(A) P(X_{\tau} =A) + h(-B) P(X_{\tau} = -B)$，又
$X_0 = 0$ 故 $ h (X_0) = h(0) \Rightarrow E[h(X_0)] = E[h(0)] =h(0) $
現在我們可以求解 $(**)$如下
\[
E[h(X_0)] = E[h(X_{\tau})] \Rightarrow h(0) =h(A) P(X_{\tau} =A) + h(-B) P(X_{\tau} = -B)
\]
因為我們要求 $P(X_{\tau}=A)$故令邊界條件 $h(A) =1, h(-B)=0$；故上式改寫
\[
E[h(X_0)] = E[h(X_{\tau})] \Rightarrow h(0) =1 \cdot P(X_{\tau} =A) \ \ \ \ (\star \star)
\]且由先前計算得到的
\[
h(z) = {C_2}{e^{\frac{{ - 2\mu }}{{{\sigma ^2}}}\left( z \right)}} + {C_3}
\]
透過邊界條件 $h(A) =1, h(-B)=0$，我們可解 $C_2, C_3$如下
\[\begin{array}{l}
h(z) = \frac{1}{{{e^{\frac{{ - 2\mu }}{{{\sigma ^2}}}A}} - {e^{\frac{{2\mu }}{{{\sigma ^2}}}B}}}}{e^{\frac{{ - 2\mu }}{{{\sigma ^2}}}\left( z \right)}} + \frac{{ - {e^{\frac{{2\mu }}{{{\sigma ^2}}}B}}}}{{{e^{\frac{{ - 2\mu }}{{{\sigma ^2}}}A}} - {e^{\frac{{2\mu }}{{{\sigma ^2}}}B}}}} = \frac{{{e^{\frac{{ - 2\mu }}{{{\sigma ^2}}}\left( z \right)}} - {e^{\frac{{2\mu }}{{{\sigma ^2}}}B}}}}{{{e^{\frac{{ - 2\mu }}{{{\sigma ^2}}}A}} - {e^{\frac{{2\mu }}{{{\sigma ^2}}}B}}}}\\
\Rightarrow h(0) = \frac{{1 - {e^{\frac{{2\mu }}{{{\sigma ^2}}}B}}}}{{{e^{\frac{{ - 2\mu }}{{{\sigma ^2}}}A}} - {e^{\frac{{2\mu }}{{{\sigma ^2}}}B}}}}
\end{array}\]
現在比較上式與 $(\star \star)$，我們得到
\[\begin{array}{l}
h(0) = 1 \cdot P({X_\tau } = A)\\
\Rightarrow \;\frac{{1 - {e^{\frac{{2\mu }}{{{\sigma ^2}}}B}}}}{{{e^{\frac{{ - 2\mu }}{{{\sigma ^2}}}A}} - {e^{\frac{{2\mu }}{{{\sigma ^2}}}B}}}} = P({X_\tau } = A)
\end{array}\]
即為所求。 $\square$
ref:
J. M. Steele, Stochastic Calculus and Financial Applications, Springer
B. Øksendal, Stochastic Differential Equations: An Introduction with Applications 6th, Springer
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