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PRML 演習問題 6.20(標準) 解答

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#問題

$(6.66)$ と $(6.67)$ の結果を確認せよ。

#方針

ガウス過程の予測分布は正規分布であり、その期待値と分散が

\begin{align}
m(\mathbf{x}_{N+1})
 &= \mathbf{k}^{\rm T} \mathbf{C}_N^{-1} \boldsymbol{\sf t}
\tag{6.66}\\

\sigma^2(\mathbf{x}_{N+1})
 &= c-\mathbf{k}^{\rm T} \mathbf{C}_N^{-1} \mathbf{k}
\tag{6.67}
\end{align}

であることを、同時分布

\begin{align}
p(\boldsymbol{\sf t}_{N+1})
&= \mathcal{N}(\boldsymbol{\sf t}_{N+1}|\mathbf{0}, \mathbf{C}_{N+1})
\tag{6.64}
\\

\mathbf{C}_{N+1}
 &= \left(
    \begin{array}{cc}
      \mathbf{C}_{N} & \mathbf{k} \\
      \mathbf{k}^{\rm T} & c
    \end{array}
  \right)
\tag{6.65}
\end{align}

から示す。

#解答

証明 以下の同時分布があるとき、

\begin{align}
p (\mathbf{x}) &=
\mathcal{N} (
  \mathbf{x} | \boldsymbol{\mu}, \boldsymbol{\Sigma}
) \\

\mathbf{x} =
\left( \begin{array}{c}
  \mathbf{x}_a \\ \mathbf{x}_b
\end{array} \right), \ \ 

\boldsymbol{\mu} &=
\left( \begin{array}{c}
  \boldsymbol{\mu}_a \\ \boldsymbol{\mu}_b
\end{array} \right), \ \ 

\boldsymbol{\Sigma} =
\left( \begin{array}{c}
  \boldsymbol{\Sigma}_{aa} & \boldsymbol{\Sigma}_{ab} \\
  \boldsymbol{\Sigma}_{ba} & \boldsymbol{\Sigma}_{bb}
\end{array} \right)
\tag{2.65-2.67}
\end{align}

条件付き分布は以下のように書けることが示されている。

\begin{align}
p (\mathbf{x}_a | \mathbf{x}_b) &=
\mathcal{N} (
  \mathbf{x}_a | \boldsymbol{\mu}_{a|b}, \boldsymbol{\Sigma}_{a|b}
) \\
\boldsymbol{\mu}_{a|b} &= 
\boldsymbol{\mu}_a + \boldsymbol{\Sigma}_{ab} \boldsymbol{\Sigma}_{bb}^{-1}
(\mathbf{x}_b - \boldsymbol{\mu}_b)
\tag{2.81}
\\
\boldsymbol{\Sigma}_{a|b} &=
\boldsymbol{\Sigma}_{aa}-
\boldsymbol{\Sigma}_{ab} \boldsymbol{\Sigma}_{bb}^{-1} \boldsymbol{\Sigma}_{ba}
\tag{2.82}
\\
\end{align}

これを $(6.64), (6.65)$ に当てはめると $(6.66), (6.67)$ が得られる。(証明終)

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