【拡散モデル】確率微分方程式と拡散過程における条件付き確率について

$$dx-f(t)xdt=g(t)dw(t)$$

両辺に$\exp (-{\int }_0^{t}f(t)dt)$をかける.

$$\exp (-{\int }_0^{t}f(t)dt)dx-\exp (-{\int }_0^{t}f(t)dt)f(t)xdt=\exp (-{\int }_0^{t}f(t)dt)g(t)dw(t)$$

伊藤の公式より,
$$d(\exp (-{\int }_0^{t}f(t)dt)x)=\exp (-{\int }_0^{t}f(t)dt)dx-\exp (-{\int }_0^{t}f(t)dt)f(t)xdt$$

よって,
$$d(\exp (-{\int }_0^{t}f(t)dt)x)=\exp (-{\int }_0^{t}f(t)dt)g(t)dw(t)$$

両辺を積分すると,
$$\exp (-{\int }_0^{t}f(t)dt)x-x_0={\int }_0^t\exp (-{\int }_0^{t}f(t)dt)g(t)dw(t)$$

よって,
$$x=x(0)\exp ({\int }_0^{t}f(t)dt)+\exp ({\int }_0^{t}f(t)dt){\int }_0^t\exp (-{\int }_0^{t}f(t)dt)g(t)dw(t)$$

簡単のため, 1次元の場合について考える.

$$\begin{array}{rcl}\mathbb{E}[x(t)]& =& \mathbb{E}[x(0)\exp ({\int }_0^{t}f(t)dt)]+\mathbb{E}[\exp ({\int }_0^{t}f(t)dt){\int }_0^t\exp (-{\int }_0^{t}f(t)dt)g(t)dw(t)]\\ & =& x(0)\exp ({\int }_0^{t}f(t)dt)(\because \mathbb{E}[\int h(t)dw(t)]=0)\end{array}$$

$$\begin{array}{rcl}\mathbb{E}[x^2(t)]& =& \mathbb{E}[{(x(0)\exp ({\int }_0^{t}f(t)dt))}^2]+\mathbb{E}[{(\exp ({\int }_0^{t}f(t)dt){\int }_0^t\exp (-{\int }_0^{t}f(t)dt)g(t)dw(t))}^2]\\ & & (\because \mathbb{E}[\int h(t)dw(t)]=0)\\ & =& \mathbb{E}[{(x(0)\exp ({\int }_0^{t}f(t)dt))}^2]+\mathbb{E}[{(\exp ({\int }_0^{t}f(t)dt))}^2{\int }_0^t{(\exp (-{\int }_0^{t}f(t)dt)g(t))}^{2}dt]\\ & & (\because 伊藤積分の等長性)\\ & =& {(x(0)\exp ({\int }_0^{t}f(t)dt))}^2+{(\exp ({\int }_0^{t}f(t)dt))}^2{\int }_0^t{(\exp (-{\int }_0^{t}f(t)dt)g(t))}^{2}dt\end{array}$$

$$\begin{array}{rcl}\mathbb{E}[{(x(t)-\mathbb{E}[x(t)])}^2]& =& \mathbb{E}[x^2(t)]-\mathbb{E}[x(t){]}^2\\ & =& {(\exp ({\int }_0^{t}f(t)dt))}^2{\int }_0^t{(\exp (-{\int }_0^{t}f(t)dt)g(t))}^{2}dt\end{array}$$

多次元の場合はわからなかったので, どなたか補足お願いします.
一応, 私なりに多次元の場合を証明したので下記に示します.

$$\begin{array}{rcl}\mathbb{E}[x(t)]& =& \mathbb{E}[x(0)\exp ({\int }_0^{t}f(t)dt)]+\mathbb{E}[\exp ({\int }_0^{t}f(t)dt){\int }_0^t\exp (-{\int }_0^{t}f(t)dt)g(t)dw(t)]\\ & =& x(0)\exp ({\int }_0^{t}f(t)dt)(\because \mathbb{E}[\int h(t)dw(t)]=0)\end{array}$$

$$\begin{array}{rcl}\mathbb{E}[x(t)x^T(t)]& =& \mathbb{E}[{(\exp ({\int }_0^{t}f(t)dt))}^{2}x(0)x(0{)}^T]\\ & +& \mathbb{E}[{(\exp ({\int }_0^{t}f(t)dt))}^{2}x(0){({\int }_0^t\exp (-{\int }_0^{t}f(t)dt)g(t)dw(t))}^T]\\ & +& \mathbb{E}[{(\exp ({\int }_0^{t}f(t)dt))}^2({\int }_0^t\exp (-{\int }_0^{t}f(t)dt)g(t)dw(t))x(0{)}^T(t)]\\ & +& \mathbb{E}[{(\exp ({\int }_0^{t}f(t)dt))}^2({\int }_0^t\exp (-{\int }_0^{t}f(t)dt)g(t)dw(t)){({\int }_0^t\exp (-{\int }_0^{t}f(t)dt)g(t)dw(t))}^T]\\ & =& {(\exp ({\int }_0^{t}f(t)dt))}^{2}x(0)x(0{)}^T\\ & +& {(\exp ({\int }_0^{t}f(t)dt))}^2\mathbb{E}[x(0){({\int }_0^t\exp (-{\int }_0^{t}f(t)dt)g(t)dw(t))}^T]\\ & +& {(\exp ({\int }_0^{t}f(t)dt))}^2\mathbb{E}[({\int }_0^t\exp (-{\int }_0^{t}f(t)dt)g(t)dw(t))x(0{)}^T]\\ & +& {(\exp ({\int }_0^{t}f(t)dt))}^2\mathbb{E}[({\int }_0^t\exp (-{\int }_0^{t}f(t)dt)g(t)dw(t)){({\int }_0^t\exp (-{\int }_0^{t}f(t)dt)g(t)dw(t))}^T]\\ & =& \exp {({\int }_0^{t}f(t)dt)}^{2}x(0)x(0{)}^T+\exp {({\int }_0^{t}f(t)dt)}^2\mathbb{E}[({\int }_0^t\exp {(-{\int }_0^{t}f(t)dt)}^2{g}^2(t)dt)I]\\ & =& \exp {({\int }_0^{t}f(t)dt)}^{2}x(0)x(0{)}^T+\exp {({\int }_0^{t}f(t)dt)}^2({\int }_0^t\exp {(-{\int }_0^{t}f(t)dt)}^2{g}^2(t)dt)I\end{array}$$

$$\begin{array}{rcl}\mathbb{E}[(x(t)-\mathbb{E}[x(t)]){(x(t)-\mathbb{E}[x(t)])}^T]& =& \mathbb{E}[x(t)x(t{)}^T]-\mathbb{E}[x(t)]\mathbb{E}[x(t{)}^T]-\mathbb{E}[x(t)]\mathbb{E}[x(t{)}^T]+\mathbb{E}[x(t)]\mathbb{E}[x(t{)}^T]\\ & =& \mathbb{E}[x(t)x(t{)}^T]-\mathbb{E}[x(t)]\mathbb{E}[x(t){]}^T\\ & =& \exp {({\int }_0^{t}f(t)dt)}^2({\int }_0^t\exp {(-{\int }_0^{t}f(t)dt)}^2{g}^2(t)dt)I\end{array}$$