導(dǎo)航

• Source of Randomness And Ito SDE
• SDE解的存在性和唯一性
• 等價SDE
• 參考資料

Source of Randomness And Ito SDE

考慮如下常微分方程
$$\{\begin{array}{l}dX(t)=a(t,X(t))dt\\ X(0)=x_0\end{array}$$
隨機(jī)性來源分析
(1) 初值隨機(jī)化
$$\{\begin{array}{l}d{X}_t=a(t,X_t)dt\\ X_0(w)=Y(w)\end{array}$$
(2) 隨機(jī)噪聲
$$d{X}_t=a(t,X_t)dt+b(t,X_t)d{B}_t$$
兩邊積分可以得到Ito-SDE
$$X_t=X_0+\underset{Riemann.Integral}{{\int }_0^{t}a(s,X_s)ds}+\underset{Ito.Integral}{{\int }_0^{t}b(s,X_s)d{B}_s}$$

SDE解的存在性和唯一性

Strong Solution
(1) $X_t$ is adapted $B_t$ i.e. $X_t$ is a function of $B_s，s\le t$.
(2) 以下兩個積分是well defined
$$\begin{gathered} {\int }_0^{t}a(s,X_s)ds \\ {\int }_0^{t}b(s,X_s)d{B}_s \end{gathered}$$
(3) $X$ is a function of the underlying sample path and of $a(t,x)$ and $b(t,x)$.

定理(強(qiáng)解存在唯一性)：
設(shè)初值為$X_0$，滿足以下兩個條件：
1.$\mathbb{E}[X_0^2]<\infty$.
2.$X_0$與$B_t$相互獨(dú)立，對于任意的$x,y\in \mathbb{R}$有
$a(t,x),b(t,x)$是連續(xù)的.
$|a(t,x)?a(t,y)|+|b(t,x)?b(t,y)|\le K|x?y|,?K\in [0,T]$. (Lipschitz Condition)
那么Ito-SDE有唯一的強(qiáng)解$X_t\in [0,T]$.
例：
線性SDE
$$X_t=X_0+{\int }_0^t(c_1{X}_s+c_2)ds+{\int }_0^t({\sigma }_1{X}_s+{\sigma }_2)d{B}_s$$
其中
$$\begin{gathered} a(t,x)=c_{1}x+c_2 \\ b(t,x)={\sigma }_{1}x+{\sigma }_2 \end{gathered}$$
驗(yàn)證Lipschitz條件
$$\begin{gathered} |a(t,x)?a(t,y)|+|b(t,x)?b(t,y)|=|c_1||x?y|+|{\sigma }_1||x?y|\le K|x?y| \\ K\ge |c_1|+|{\sigma }_1| \end{gathered}$$

等價SDE

轉(zhuǎn)換定理(Transformation Formula)：假設(shè)Ito-SDE
$$d{X}_t=a(t,X_t)dt+b(t,X_t)d{B}_t$$
將Ito積分轉(zhuǎn)為Stratonovich積分
$${\int }_0^{T}f(t,X_t)°d{B}_t={\int }_0^{T}f(t,X_t)d{B}_t+\frac{1}{2}{\int }_0^{T}b(t,X_t)f_x(t,X_t)dt$$
例 1：
設(shè)Ito-SDE
$$d{X}_t=a(t,X_t)dt+b(t,X_t)d{B}_t\text{\hspace{1em}(1)}$$
轉(zhuǎn)為Stratonovich-SDE
$$d{X}_t=a(t,X_t)dt+b(t,X_t)°d{B}_t\text{\hspace{1em}(2)}$$
根據(jù)轉(zhuǎn)換定理：
$${\int }_0^{T}f(t,X_t)°d{B}_t={\int }_0^{T}f(t,X_t)d{B}_t+\frac{1}{2}{\int }_0^{T}b(t,X_t)f_x(t,X_t)dt$$
不妨令$f(t,X_t)=b(t,X_t)$，可以得到
$$b(t,X_t)°d{B}_t=b(t,X_t)d{B}_t+\frac{1}{2}b(t,X_t)b_x(t,X_t)dt$$
代入Ito-SDE形式
$$\begin{array}{rl}d{X}_t& =a(t,X_t)dt+b(t,X_t)d{B}_t\\ & =a(t,X_t)dt+b(t,X_t)°d{B}_t?\frac{1}{2}b(t,X_t)b_x(t,X_t)dt\\ & =\underset{a}{(a(t,X_t)?\frac{1}{2}b(t,X_t)b_x(t,X_t))}dt+\underset{b}{b(t,X_t)}°d{B}_t\end{array}$$
可以得到
$$\{\begin{array}{l}a=a(t,X_t)?\frac{1}{2}b(t,X_t)b_x(t,X_t)\\ b=b(t,X_t)\end{array}$$
方程$(1)$和$(2)$是等價SDE.

例 2：
Ito-SDE
$$d{X}_t=\frac{1}{2}f(X_t)f^{\prime }(X_t)dt+f(X_t)d{B}_t$$
可知
$$\{\begin{array}{l}a(t,x)=\frac{1}{2}f(X_t)f^{\prime }(X_t)\\ b(t,x)=f(X_t)\end{array}$$
根據(jù)轉(zhuǎn)換定理轉(zhuǎn)為Stratonovich-SDE
$$\{\begin{array}{l}a(t,x)=0\\ b(t,x)=f(X_t)\end{array}$$
等價SDE為
$$d{X}_t=f(X_t)°d{B}_t$$
求解analogue:
$$dk(t)=f(k(t))dc(t)$$
移項(xiàng)得到
$$\int \frac{dl(t)}{f(l(t))}=dc(t)$$
例 3：
$$d{X}_t=\frac{1}{2}n{X}_t^{2n?1}dt+X_t^{n}d{B}_t$$
轉(zhuǎn)為等價SDE
$$d{X}_t=f(X_t)°d{B}_t=X_t^n°d{B}_t$$
令$k(t)=X_t,c(t)=B_t$，得到analogue-ODE:
$$dk(t)=k(t{)}^{n}dc(t)$$
在區(qū)間$[0,t]$上解之得
$$\begin{gathered} {\int }_{l(0)}^{l(t)}\frac{1}{k(t{)}^n}dk(t)=c(t)?c(0) \\ \frac{1}{1?n}[k(t){]}^{1?n}?\frac{1}{1?n}[k(0){]}^{1?n}=c(t)?c(0) \end{gathered}$$
令$X_t=k(t),B_t=c(t)$，得到
$$\begin{gathered} X_t^{1?n}?X_0^{1?n}=(1?n)B_t \\ {X}_t=[X_0^{1?n}+(1?n)B_t{]}^{\frac{1}{1?n}} \end{gathered}$$

參考資料

stochastic differential equation(SDE) part1 Jerry Xu

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