[隨機分析] Uniqueness and Existence theorem for S.D.E. (4) - The Existence of Solution

延續前篇。繼續逐步完成 Existence 的證明：
現在我們有Picard iteration 與 如下 LEMMA

==============================
Iteration Scheme (Picard Iteration):
給定 $X_t^{(0)}:=x_0$，與下列跌代過程：
$$X_t^{(n+1)} = x_0 + \int_0^t \mu(s, X_s^{(n)})ds + \int_0^t \sigma(s,X_s^{(n)})dBs \ \ \ \ (*)$$ ==============================

===============================
Lemma 
若 $\mu, \sigma$ 滿足Lipschitz condtion：
$$| \mu(t,x) - \mu(t,y)|^2 + |\sigma(t,x) - \sigma(t,y)|^2 \leq K |x-y|^2$$ 則存在一個常數 $C$ 使得 由Picard iteration所定義的隨機過程 $X_t^{(n)}$滿足下列不等式：
$$E \left[ \displaystyle \sup_{0 \leq s \leq t} |X_s^{(n+1)} - X_s^{(n)}|^2 \right ] \leq C \int_0^t E \left[ | X_s^{(n)} - X_s^{(n-1)}|^2 \right ]$$ ===============================

有了這個Lemma 我們便可以開始證明Picard iteration 的跌代過程sequence $X_t^{(n)} \rightarrow X_t$ almost surely。(why we need almost surely?) 因為事實上我們需要證明SDE的解為連續，為了要保持連續性，我們需要almost sure uniformly convergence。故想法如下：

注意到為了要證明 uniform convergence，我們需要先製造出Cauchy Sequence 再透過Completeness 推得 Picard iteration 產生的 $X_t^{(n)} \rightarrow X_t$ almost surely & uniformly。

現在令
$$g_n(t) := E[\sup_{0\leq s \leq t} | X_s^{(n+1)} - X_s^{(n)}|^2]$$
則Lemma告訴我們
$$\begin{array}{l} E\left[ {\mathop {\sup }\limits_{0 \le s \le t} |X_s^{(n + 1)} - X_s^{(n)}{|^2}} \right] \le C\int_0^t E \left[ {|X_s^{(n)} - X_s^{(n - 1)}{|^2}} \right]\\  \Rightarrow {g_n}(t) \le C\int_0^t {{g_{n - 1}}(s)ds} \ \ \ \ (\star) \end{array}$$ 注意到對$t \in [0,T]$，存在一個常數$M$ 使得 $g_0(t) \leq M$ (WHY?)


；利用 $(\star)$我們可以計算下一步跌代的上界為 ${g_1}(t) \le MCt$，且by induction，可以推得對第$n$步跌代的上界為
$$\Rightarrow {g_n}(t) \le M{C^n}\frac{{{t^n}}}{{n!}}$$ 由Markov inequality: $P\left( {X \ge a} \right) \le \frac{{E\left[ {{X^r}} \right]}}{{{a^r}}}, \ r > 0$可知
$$\begin{array}{l} P\left( {\mathop {\sup }\limits_{0 \le s \le t} |X_s^{(n + 1)} - X_s^{(n)}| \ge {2^{ - n}}} \right) \le \frac{{E\left[ {\mathop {\sup }\limits_{0 \le s \le t} |X_s^{(n + 1)} - X_s^{(n)}{|^2}} \right]}}{{{{\left( {{2^{ - n}}} \right)}^2}}}\\  \Rightarrow P\left( {\mathop {\sup }\limits_{0 \le s \le t} |X_s^{(n + 1)} - X_s^{(n)}| \ge {2^{ - n}}} \right) \le \left( {\frac{1}{{{2^{ - 2n}}}}} \right)\left( {M{C^n}\frac{{{t^n}}}{{n!}}} \right)\\  \Rightarrow P\left( {\mathop {\sup }\limits_{0 \le s \le t} |X_s^{(n + 1)} - X_s^{(n)}| \ge {2^{ - n}}} \right) \le \frac{{M{C^n}{T^n}{2^{2n}}}}{{n!}} \end{array}$$ 上述機率為summable (亦即加到無窮大不會爆掉，因為 分母項的 $n!$ dominated 分子項，故當 $n$ 夠大的時候，上式收斂到0)。

現在我們手上有summable的機率，則由 Borel-Cantelli Lemma 給予我們Almost surely limit。
----
Borel Cantelli Lemma
若 $\{A_i \}$ 為任意事件sequence，則
$$\displaystyle \sum_{i=1}^\infty P(\{ A_i \}) < \infty \Rightarrow P \left (\displaystyle \sum_{i=1}^\infty 1_{\{ A_i \}} <\infty \right) =1$$ ----

也就是說事件
$$\{ {\mathop {\sup }\limits_{0 \le s \le t} |X_s^{(n + 1)} - X_s^{(n)}| \ge {2^{ - n}}} \}$$ 只會發生有限次。故我們得到 對$n$夠大的時候，我們有
$${\mathop {\sup }\limits_{0 \le s \le t} |X_s^{(n + 1)} - X_s^{(n)}| < {2^{ - n}}} \ \text{almost surely (a.s.)}$$ 亦即對$n$夠大的時候，我們有 almost surely 的 Cauchy sequence of functions。
$$||X_s^{(n + 1)} - X_s^{(n)}|| < {2^{ - n}} \ a.s.$$ 由 Completeness 與 FACT: Cauchy sequence of functions 必為 uniform convergence，我們可知極限存在且連續性被保證。我們稱此almost sure continuous limit 為 $X_t$亦即
$$X_t^{\left( n \right)} \to {X_t} \ \ a.s.$$

接著如果我們可以證明極限 $X_t$ 是 $L^2$ boundedness；則我們就可得到
$$X_t^{(n)} \rightarrow X_t \in L^2(dP), \ \forall \ t\in [0,T]$$
故現在觀察
$$\begin{array}{l} E\left[ {{{\left( {X_t^{\left( {n + 1} \right)} \to X_t^{\left( n \right)}} \right)}^2}} \right] \le {g_n}(t) \le \frac{{M{C^n}{T^n}{2^{2n}}}}{{n!}}\\  \Rightarrow {\left( {E\left[ {{{\left( {X_t^{\left( {n + 1} \right)} \to X_t^{\left( n \right)}} \right)}^2}} \right]} \right)^{1/2}} \le {\left( {\frac{{M{C^n}{T^n}{2^{2n}}}}{{n!}}} \right)^{1/2}}\\  \Rightarrow {\left\| {X_t^{\left( n \right)} \to {X_t}} \right\|_{{L^2}\left( {dP} \right)}} \le {\left( {\frac{{M{C^n}{T^n}{2^{2n}}}}{{n!}}} \right)^{1/2}} \end{array}$$
也就是說$X_t^{(n)}$ 為 Cauchy sequence in $L^2$。又由先前證明我們知道$X_t^{n)}$ 有 almost sure 極限。
$$\left\{ \begin{array}{l} X_t^{\left( n \right)} \to {X_t}\begin{array}{*{20}{c}} {} \end{array}{\rm{in}}\begin{array}{*{20}{c}} {} \end{array}{L^2}\\ X_t^{\left( n \right)} \to {X_t}\begin{array}{*{20}{c}} {} \end{array}{\rm{a}}{\rm{.s}}{\rm{.}} \end{array} \right.$$ 由極限的唯一性我們知道此兩個極限 ($L^2$ limit 與 $a.s$ limit) 必須相等。

現在我們開始利用 $L^2$ limit 來幫助我們計算
$${\left\| {\int_0^t {\sigma (s,X_s^{(n)})} d{B_s} - \int_0^t {\sigma (s,{X_s})} d{B_s}} \right\|_{{L_2}}} \\ = E{\left[ {{{\left( {\int_0^t {\left[ {\sigma (s,X_s^{(n)}) - \sigma (s,{X_s})} \right]} d{B_s}} \right)}^2}} \right]^{1/2}}$$ 由Ito isometry可知
$$\Rightarrow E{\left[ {\int_0^t {{{\left[ {\sigma (s,X_s^{(n)}) - \sigma (s,{X_s})} \right]}^2}} ds} \right]^{1/2}} \ \ \ \ (\star \star)$$
再由Lipschitz condition 與 三角不等式 我們可得
$$\begin{array}{l}  \Rightarrow (\star \star) \le {K^{1/2}}E{\left[ {\int_0^t {{{\left| {X_s^{(n)} - {X_s}} \right|}^2}} ds} \right]^{1/2}}\begin{array}{*{20}{c}} {} \end{array}\left( {{\rm{Lipschitz}}{\rm{.}}} \right)\\ \begin{array}{*{20}{c}} {} \end{array}\begin{array}{*{20}{c}} {} \end{array}\begin{array}{*{20}{c}} {} \end{array}\begin{array}{*{20}{c}} {} \end{array}\begin{array}{*{20}{c}} {} \end{array}\begin{array}{*{20}{c}} {} \end{array}\begin{array}{*{20}{c}} {} \end{array}\begin{array}{*{20}{c}} {} \end{array}\begin{array}{*{20}{c}} {} \end{array}\begin{array}{*{20}{c}} {} \end{array}\begin{array}{*{20}{c}} {} \end{array}\begin{array}{*{20}{c}} {} \end{array}\begin{array}{*{20}{c}} {} \end{array}\begin{array}{*{20}{c}} {} \end{array}\begin{array}{*{20}{c}} {} \end{array}\begin{array}{*{20}{c}} {} \end{array}\begin{array}{*{20}{c}} {} \end{array}\begin{array}{*{20}{c}} {} \end{array}\begin{array}{*{20}{c}} {} \end{array}\begin{array}{*{20}{c}} {} \end{array}\begin{array}{*{20}{c}} {} \end{array}\begin{array}{*{20}{c}} {} \end{array}\begin{array}{*{20}{c}} {} \end{array}\begin{array}{*{20}{c}} {} \end{array}\begin{array}{*{20}{c}} {} \end{array}\begin{array}{*{20}{c}} {} \end{array} \le {K^{1/2}}\sum\limits_{m = n}^\infty  {{{\left\{ {E\left[ {\int_0^t {{{\left| {X_s^{m + 1} - {X_s}^{\left( m \right)}} \right|}^2}} ds} \right]} \right\}}^{1/2}}} \\ \begin{array}{*{20}{c}} {} \end{array}\begin{array}{*{20}{c}} {} \end{array}\begin{array}{*{20}{c}} {} \end{array}\begin{array}{*{20}{c}} {} \end{array}\begin{array}{*{20}{c}} {} \end{array}\begin{array}{*{20}{c}} {} \end{array}\begin{array}{*{20}{c}} {} \end{array}\begin{array}{*{20}{c}} {} \end{array}\begin{array}{*{20}{c}} {} \end{array}\begin{array}{*{20}{c}} {} \end{array}\begin{array}{*{20}{c}} {} \end{array}\begin{array}{*{20}{c}} {} \end{array}\begin{array}{*{20}{c}} {} \end{array}\begin{array}{*{20}{c}} {} \end{array}\begin{array}{*{20}{c}} {} \end{array}\begin{array}{*{20}{c}} {} \end{array}\begin{array}{*{20}{c}} {} \end{array}\begin{array}{*{20}{c}} {} \end{array}\begin{array}{*{20}{c}} {} \end{array}\begin{array}{*{20}{c}} {} \end{array}\begin{array}{*{20}{c}} {} \end{array}\begin{array}{*{20}{c}} {} \end{array}\begin{array}{*{20}{c}} {} \end{array}\begin{array}{*{20}{c}} {} \end{array}\begin{array}{*{20}{c}} {} \end{array}\begin{array}{*{20}{c}} {} \end{array} \le {K^{1/2}}\sum\limits_{m = n}^\infty  {{{\left( {\int_0^t {{g_m}\left( s \right)ds} } \right)}^{1/2}}}  \to 0\begin{array}{*{20}{c}} {} \end{array}. \left( {{\rm{\Delta  - ineq}}{\rm{.}}} \right) \end{array}$$ 最後式成立是因為由Lemma的結果得知其收斂。
同樣的對 $n \rightarrow \infty$
$$\int_0^t {{{\left[ {\mu (s,X_s^{(n)}) - \mu (s,{X_s})} \right]}^2}} ds \to 0$$
故我們現在有
$$\left\{ \begin{array}{l} \int_0^t {\mu (s,X_s^{(n)})} ds \to \int_0^t {\mu (s,{X_s})} ds,\begin{array}{*{20}{c}} {} \end{array}in\begin{array}{*{20}{c}} {} \end{array}{L^2}\\ \int_0^t {\sigma (s,X_s^{(n)})} ds \to \int_0^t {\sigma (s,{X_s})} ds,\begin{array}{*{20}{c}} {} \end{array}in\begin{array}{*{20}{c}} {} \end{array}{L^2} \end{array} \right.$$ 現在我們可以檢驗 $X_t$ 確實為SDE的解：
令$n \rightarrow \infty$，觀察 Picard iteration
$$X_t^{(n+1)} = x_0 + \int_0^t \mu(s, X_s^{(n)})ds + \int_0^t \sigma(s,X_s^{(n)})dBs \ \ \ \ (*)$$ 上式的左邊，我們知道
$$X_t^{(n)} \rightarrow X_t \ \text{uniformly on $[0,T]$}$$ 至於上式右方，我們由$L^2$ convergence 表示 存在一個 almost convergent 的子序列 (subsequence)；故對任意固定 $t \in [0,T]$我們有一個subsequence $\{n_k\}$使得
$$\left\{ \begin{array}{l} \int_0^t {\mu (s,X_s^{({n_k})})} ds \to \int_0^t {\mu (s,{X_s})} ds,\begin{array}{*{20}{c}} {} \end{array}a.s.\begin{array}{*{20}{c}} {} \end{array}\forall t \in \left[ {0,T} \right] \cap \mathbb{Q} \\ \int_0^t {\sigma (s,X_s^{({n_k})})} dB_s \to \int_0^t {\sigma (s,{X_s})} dB_s,\begin{array}{*{20}{c}} {} \end{array}a.s.\begin{array}{*{20}{c}} {} \end{array}\forall t \in \left[ {0,T} \right] \cap \mathbb{Q} \end{array} \right.$$ 因此，若我們取 $n_k \rightarrow \infty$，則可得到
$${X_t} = {x_0} + \int_0^t {\mu (s,{X_s})} ds + \int_0^t {\sigma (s,{X_s})} dB_s$$ 對 $\forall t \in [0,T] \cap \mathbb{Q}$。
最後，因為上式左右兩邊為連續，故對 $\forall t \in [0,T] \cap \mathbb{Q}$ 成立的結果可以被拓展到 $\forall t \in [0,T]$ 至此證明完畢。 $\square$