【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
質問
岡野原大輔さんの著書、拡散モデル データ生成技術の数理を読んでて, p70の導出がわかりませんでした.
条件付き確率の導出方法について, どなたかご教示頂けないでしょうか？
確率微分方程式が与えられているとする.
\begin{equation} dx=f(t)xdt+g(t)dw \end{equation}
このときの, 一般解を求めよ.
\begin{equation} dx - f(t)xdt = g(t)dw(t) \end{equation}
両辺に$\exp\left(-\int_{0}^{t}f(t)dt\right)$をかける.
\begin{equation} \exp\left(-\int_{0}^{t}f(t)dt\right)dx -\exp\left(-\int_{0}^{t}f(t)dt\right)f(t)xdt =\exp\left(-\int_{0}^{t}f(t)dt\right)g(t)dw(t) \end{equation}
伊藤の公式より,
\begin{equation}
d(\exp\left(-\int_{0}^{t}f(t)dt\right)x) = \exp\left(-\int_{0}^{t}f(t)dt\right)dx
-\exp\left(-\int_{0}^{t}f(t)dt\right)f(t)xdt
\end{equation}
よって,
\begin{equation}
d(\exp\left(-\int_{0}^{t}f(t)dt\right)x)
=\exp\left(-\int_{0}^{t}f(t)dt\right)g(t)dw(t)
\end{equation}
両辺を積分すると,
\begin{equation}
\exp\left(-\int_{0}^{t}f(t)dt\right)x-x_0
=\int_{0}^{t}\exp\left(-\int_{0}^{t}f(t)dt\right)g(t)dw(t)
\end{equation}
よって,
\begin{equation}
x
=x(0)\exp\left(\int_{0}^{t}f(t)dt\right)
+\exp\left(\int_{0}^{t}f(t)dt\right)\int_{0}^{t}\exp\left(-\int_{0}^{t}f(t)dt\right)g(t)dw(t)
\end{equation}
確率微分方程式が与えられているとする.
\begin{equation} dx=f(t)xdt+g(t)dw \end{equation}
このとき, 条件付き確率は,
\begin{equation}
p(x(t)|x(0))=\mathcal{N}(s(t)x(0),s(t)^2\sigma(t)^2I)
\end{equation}
ただし,
\begin{eqnarray}
s(t)&=&\exp\left(\int_{0}^{t}f(\xi)d\xi\right)\\
\sigma(t)&=&\sqrt{\int_{0}^{t}\dfrac{g(\xi)^2}{s(\xi)^2}d\xi}
\end{eqnarray}
である.
簡単のため, 1次元の場合について考える.
\begin{eqnarray} \mathbb{E}\lbrack x(t)\rbrack &=& \mathbb{E}\Bigl\lbrack x(0)\exp\left(\int_{0}^{t}f(t)dt\right)\Bigr\rbrack +\mathbb{E}\Bigl\lbrack \exp\left(\int_{0}^{t}f(t)dt\right)\int_{0}^{t}\exp\left(-\int_{0}^{t}f(t)dt\right)g(t)dw(t)\Bigr\rbrack\\ &=& x(0)\exp\left(\int_{0}^{t}f(t)dt\right) \ \ \ \left(\because \mathbb{E}\Bigl\lbrack \int h(t)dw(t)\Bigr\rbrack =0\right) \end{eqnarray}
\begin{eqnarray} \mathbb{E}\lbrack x^2(t)\rbrack &=& \mathbb{E}\Bigl\lbrack \left(x(0)\exp\left(\int_{0}^{t}f(t)dt\right)\right)^2 \Bigr\rbrack +\mathbb{E}\Bigl\lbrack \left( \exp\left(\int_{0}^{t}f(t)dt\right)\int_{0}^{t}\exp\left(-\int_{0}^{t}f(t)dt\right)g(t)dw(t) \right)^2 \Bigr\rbrack \\ &&\left(\because \mathbb{E}\Bigl\lbrack \int h(t)dw(t)\Bigr\rbrack =0\right) \\ &=& \mathbb{E}\Bigl\lbrack \left(x(0)\exp\left(\int_{0}^{t}f(t)dt\right)\right)^2 \Bigr\rbrack +\mathbb{E}\Bigl\lbrack \left( \exp\left(\int_{0}^{t}f(t)dt\right)\right)^2 \int_{0}^{t}\left(\exp\left(-\int_{0}^{t}f(t)dt\right)g(t)\right)^2dt \Bigr\rbrack\\ &&(\because 伊藤積分の等長性)\\ &=& \left(x(0)\exp\left(\int_{0}^{t}f(t)dt\right)\right)^2 + \left( \exp\left(\int_{0}^{t}f(t)dt\right)\right)^2 \int_{0}^{t}\left(\exp\left(-\int_{0}^{t}f(t)dt\right)g(t)\right)^2dt \end{eqnarray}
\begin{eqnarray} \mathbb{E}\lbrack \left(x(t)-\mathbb{E}[x(t)]\right)^2\rbrack &=& \mathbb{E}\lbrack x^2(t)\rbrack-\mathbb{E}\lbrack x(t)\rbrack^2\\ &=& \left( \exp\left(\int_{0}^{t}f(t)dt\right)\right)^2 \int_{0}^{t}\left(\exp\left(-\int_{0}^{t}f(t)dt\right)g(t)\right)^2dt \end{eqnarray}
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{eqnarray} \mathbb{E}\lbrack x(t)\rbrack &=& \mathbb{E}\Bigl\lbrack x(0)\exp\left(\int_{0}^{t}f(t)dt\right)\Bigr\rbrack +\mathbb{E}\Bigl\lbrack \exp\left(\int_{0}^{t}f(t)dt\right)\int_{0}^{t}\exp\left(-\int_{0}^{t}f(t)dt\right)g(t)dw(t)\Bigr\rbrack\\ &=& x(0)\exp\left(\int_{0}^{t}f(t)dt\right) \ \ \ \left(\because \mathbb{E}\Bigl\lbrack \int h(t)dw(t)\Bigr\rbrack =0\right) \end{eqnarray}
\begin{eqnarray} \mathbb{E}\lbrack x(t)x^T(t)\rbrack &=& \mathbb{E}\Bigl\lbrack \left(\exp\left(\int_{0}^{t}f(t)dt\right)\right)^2x(0)x(0)^T \Bigr\rbrack\\ &+& \mathbb{E}\Bigl\lbrack \left(\exp\left(\int_{0}^{t}f(t)dt\right)\right)^2x(0)\left(\int_{0}^{t}\exp\left(-\int_{0}^{t}f(t)dt\right)g(t)dw(t)\right) ^T\Bigr\rbrack\\ &+& \mathbb{E}\Bigl\lbrack \left(\exp\left(\int_{0}^{t}f(t)dt\right)\right)^2\left(\int_{0}^{t}\exp\left(-\int_{0}^{t}f(t)dt\right)g(t)dw(t)\right)x(0)^T(t) \Bigr\rbrack\\ &+& \mathbb{E}\Bigl\lbrack \left(\exp\left(\int_{0}^{t}f(t)dt\right)\right)^2\left(\int_{0}^{t}\exp\left(-\int_{0}^{t}f(t)dt\right)g(t)dw(t)\right)\left(\int_{0}^{t}\exp\left(-\int_{0}^{t}f(t)dt\right)g(t)dw(t)\right)^T \Bigr\rbrack\\ &=& \left(\exp\left(\int_{0}^{t}f(t)dt\right)\right)^2 x(0)x(0)^T \\ &+& \left(\exp\left(\int_{0}^{t}f(t)dt\right)\right)^2 \mathbb{E}\Bigl\lbrack x(0)\left(\int_{0}^{t}\exp\left(-\int_{0}^{t}f(t)dt\right)g(t)dw(t)\right)^T \Bigr\rbrack\\ &+& \left(\exp\left(\int_{0}^{t}f(t)dt\right)\right)^2 \mathbb{E}\Bigl\lbrack \left(\int_{0}^{t}\exp\left(-\int_{0}^{t}f(t)dt\right)g(t)dw(t)\right)x(0)^T \Bigr\rbrack\\ &+& \left(\exp\left(\int_{0}^{t}f(t)dt\right)\right)^2 \mathbb{E}\Bigl\lbrack \left(\int_{0}^{t}\exp\left(-\int_{0}^{t}f(t)dt\right)g(t)dw(t)\right)\left(\int_{0}^{t}\exp\left(-\int_{0}^{t}f(t)dt\right)g(t)dw(t)\right)^T \Bigr\rbrack\\ &=& \exp\left(\int_{0}^{t}f(t)dt\right)^2 x(0)x(0)^T + \exp\left(\int_{0}^{t}f(t)dt\right)^2 \mathbb{E}\Bigl\lbrack \left(\int_{0}^{t}\exp\left(-\int_{0}^{t}f(t)dt\right)^2g^2(t)dt \right)I \Bigr\rbrack\\ &=& \exp\left(\int_{0}^{t}f(t)dt\right)^2 x(0)x(0)^T + \exp\left(\int_{0}^{t}f(t)dt\right)^2 \left(\int_{0}^{t}\exp\left(-\int_{0}^{t}f(t)dt\right)^2g^2(t)dt \right)I \end{eqnarray}
\begin{eqnarray} \mathbb{E}\Bigl\lbrack \left(x(t)-\mathbb{E}\lbrack x(t)\rbrack\right) \left(x(t)-\mathbb{E}\lbrack x(t)\rbrack\right)^T \Bigr\rbrack &=& \mathbb{E}\Bigl\lbrack x(t)x(t)^T \Bigr\rbrack - \mathbb{E}\Bigl\lbrack x(t) \Bigr\rbrack \mathbb{E}\Bigl\lbrack x(t)^T \Bigr\rbrack - \mathbb{E}\Bigl\lbrack x(t) \Bigr\rbrack \mathbb{E}\Bigl\lbrack x(t)^T \Bigr\rbrack + \mathbb{E}\Bigl\lbrack x(t) \Bigr\rbrack \mathbb{E}\Bigl\lbrack x(t)^T \Bigr\rbrack\\ &=& \mathbb{E}\Bigl\lbrack x(t)x(t)^T \Bigr\rbrack - \mathbb{E}\Bigl\lbrack x(t) \Bigr\rbrack \mathbb{E}\Bigl\lbrack x(t) \Bigr\rbrack^T\\ &=& \exp\left(\int_{0}^{t}f(t)dt\right)^2 \left(\int_{0}^{t}\exp\left(-\int_{0}^{t}f(t)dt\right)^2g^2(t)dt \right)I \end{eqnarray}
本当は, コルモゴロフの前向き方程式とかコルモゴロフの後ろ向き方程式から条件付き確率を求めたほうがいいんでしょうが, 導出方法がわかりませんでした...