統計学実践ワークブックの行間埋め　第26章

\begin{eqnarray} \|\boldsymbol{x}_i-\boldsymbol{x}_j\|^2 &=& (\boldsymbol{x}_i-\boldsymbol{x}_j)^{\text{T} }(\boldsymbol{x}_i-\boldsymbol{x}_j)&& \\ \\ &=& \boldsymbol{x}_i^{\text{T} }\boldsymbol{x}_i-\boldsymbol{x}_j^{\text{T} }\boldsymbol{x}_i -\boldsymbol{x}_i^{\text{T} }\boldsymbol{x}_j+\boldsymbol{x}_j^{\text{T} }\boldsymbol{x}_j& \\ \\ &=& \| \boldsymbol{x}_i\|^2+\|\boldsymbol{x}_j\|^2-(\boldsymbol{x}_j^{\text{T} }\boldsymbol{x}_i)^{\text{T}} -\boldsymbol{x}_i^{\text{T} }\boldsymbol{x}_j&&...\text{スカラーであるため、転置をとっても同じ値になる} \\ \\ &=& \| \boldsymbol{x}_i\|^2+\|\boldsymbol{x}_j\|^2-\boldsymbol{x}_i^{\text{T} }\boldsymbol{x}_j -\boldsymbol{x}_i^{\text{T} }\boldsymbol{x}_j&& \\ \\ &=& \| \boldsymbol{x}_i\|^2+\|\boldsymbol{x}_j\|^2-2\boldsymbol{x}_i^{\text{T} }\boldsymbol{x}_j&& \\ \\ \end{eqnarray}

ここでは\(\displaystyle \sum_{i=1}^nd_{ij}^2\)の計算について示す。 \begin{eqnarray} \displaystyle \sum_{i=1}^nd_{ij}^2 &=& \displaystyle \sum_{i=1}^n \| \boldsymbol{x}_i\|^2+\|\boldsymbol{x}_j\|^2-2\boldsymbol{x}_i^{\text{T} }\boldsymbol{x}_j&& \\ \\ &=& a+n\|\boldsymbol{x}_j\|^2-2\displaystyle \sum_{i=1}^n \boldsymbol{x}_i^{\text{T} }\boldsymbol{x}_j&& \\ \\ &=& a+n\|\boldsymbol{x}_j\|^2-2\left(\displaystyle \sum_{i=1}^n \boldsymbol{x}_i^{\text{T} } \right)\boldsymbol{x}_j&& \\ \\ &=& a+n\|\boldsymbol{x}_j\|^2-2 \;\boldsymbol{0} ^{\text{T} }\boldsymbol{x}_j&&...p.232より\displaystyle \sum_{i=1}^n \boldsymbol{x}_i=0 \\ \\ &=& a+n\|\boldsymbol{x}_j\|^2 \end{eqnarray}

\begin{eqnarray} \displaystyle \sum_{i=1}^n\sum_{j=1}^nd_{ij}^2 &=& \displaystyle \sum_{i=1}^n\sum_{j=1}^n \| \boldsymbol{x}_i\|^2+\|\boldsymbol{x}_j\|^2-2\boldsymbol{x}_i^{\text{T} }\boldsymbol{x}_j&& \\ \\ &=& \displaystyle \sum_{i=1}^n\left(\sum_{j=1}^n \| \boldsymbol{x}_i\|^2+\|\boldsymbol{x}_j\|^2-2\boldsymbol{x}_i^{\text{T} }\boldsymbol{x}_j\right)&& \\ \\ &=& \displaystyle \sum_{i=1}^n\left( a+n\|\boldsymbol{x}_j\|^2 \right)&& \\ \\ &=& an+\displaystyle \sum_{i=1}^n\left( n\|\boldsymbol{x}_j\|^2 \right)&& \\ \\ &=& an+n\displaystyle \sum_{i=1}^n\|\boldsymbol{x}_j\|^2 && \\ \\ &=& an+na && \\ \\ &=& 2na \end{eqnarray}

\begin{eqnarray} &&d_{ij}^2&=&\| \boldsymbol{x}_i\|^2+\|\boldsymbol{x}_j\|^2-2\boldsymbol{x}_i^{\text{T} }\boldsymbol{x}_j \\ \\ &\Leftrightarrow& 2\boldsymbol{x}_i^{\text{T} }\boldsymbol{x}_j&=&-d_{ij}^2+&\| \boldsymbol{x}_i\|^2&+&\|\boldsymbol{x}_j\|^2& \\ \\ && &=&-d_{ij}^2+&\frac{1}{n}\left( \displaystyle \sum_{j=1}^n d_{ij}^2 -a\right)&+&\frac{1}{n}\left( \sum_{i=1}^n d_{ij}^2 -a\right)& \\ \\ && &=&-d_{ij}^2+&\frac{1}{n}\left( \displaystyle \sum_{j=1}^n d_{ij}^2 \right)&+&\frac{1}{n}\left( \sum_{i=1}^n d_{ij}^2 \right)&-&\frac{2a}{n}& \\ \\ && &=&-d_{ij}^2+&\frac{1}{n}\left( \displaystyle \sum_{j=1}^n d_{ij}^2 \right)&+&\frac{1}{n}\left( \sum_{i=1}^n d_{ij}^2 \right)&-&\frac{1}{n}\left( \frac{1}{n}\sum_{i=1}^n\sum_{j=1}^n d_{ij}^2 \right)&\;&...2na=\sum_{i=1}^n\sum_{j=1}^nd_{ij}^2より& \\ \\ &\Leftrightarrow& \boldsymbol{x}_i^{\text{T} }\boldsymbol{x}_j&=&-\frac{1}{2}(d_{ij}^2+&\frac{1}{n}\left( \displaystyle \sum_{j=1}^n d_{ij}^2 \right)&+&\frac{1}{n}\left( \sum_{i=1}^n d_{ij}^2 \right)&-&\frac{1}{n}\left( \frac{1}{n}\sum_{i=1}^n\sum_{j=1}^n d_{ij}^2 \right)&) \\ \\ \end{eqnarray}

\begin{eqnarray} -2\text{X}\text{X}^{\text{T}}&=& -2\left( \begin{array}{cccc} \boldsymbol{x}_1^{\text{T}} \\ \boldsymbol{x}_2^{\text{T}} \\ \vdots \\ \boldsymbol{x}_n^{\text{T}} \end{array} \right) \left(\boldsymbol{x}_1,\boldsymbol{x}_2,\ldots,\boldsymbol{x}_n\right) \\ \\ &=& -2\left( \begin{array}{cccc} \boldsymbol{x}_1^{\text{T}}\boldsymbol{x}_1 & \boldsymbol{x}_1^{\text{T}}\boldsymbol{x}_2 & \ldots & \boldsymbol{x}_1^{\text{T}}\boldsymbol{x}_n \\ \boldsymbol{x}_2^{\text{T}}\boldsymbol{x}_1 & \boldsymbol{x}_2^{\text{T}}\boldsymbol{x}_2 & \ldots & \boldsymbol{x}_2^{\text{T}}\boldsymbol{x}_n \\ \vdots & \vdots & \ddots & \vdots \\ \boldsymbol{x}_n^{\text{T}}\boldsymbol{x}_1 & \boldsymbol{x}_n^{\text{T}}\boldsymbol{x}_2 & \ldots & \boldsymbol{x}_n^{\text{T}}\boldsymbol{x}_n \end{array} \right) \\ \\ &=& \left( \begin{array}{cccc} d_{11}^2 & d_{12}^2 & \ldots & d_{1n}^2 \\ d_{21}^2 & d_{22}^2 & \ldots & d_{2n}^2 \\ \vdots & \vdots & \ddots & \vdots \\ d_{n1}^2 & d_{n2}^2 & \ldots & d_{nn}^2 \end{array} \right) -\frac{1}{n}\displaystyle \sum_{i=1}^n \left( \begin{array}{cccc} d_{i1}^2 & d_{i2}^2 & \ldots & d_{in}^2 \\ d_{i1}^2 & d_{i2}^2 & \ldots & d_{in}^2 \\ \vdots & \vdots & \ddots & \vdots \\ d_{i1}^2 & d_{i2}^2 & \ldots & d_{in}^2 \\ \end{array} \right) -\frac{1}{n}\displaystyle \sum_{j=1}^n \left( \begin{array}{cccc} d_{1j}^2 & d_{1j}^2 & \ldots & d_{1j}^2 \\ d_{2j}^2 & d_{2j}^2 & \ldots & d_{2j}^2 \\ \vdots & \vdots & \ddots & \vdots \\ d_{nj}^2 & d_{nj}^2 & \ldots & d_{nj}^2 \\ \end{array} \right) +\frac{1}{n^2}\displaystyle \sum_{i=1}^n\sum_{j=1}^n d_{ij} \left( \begin{array}{cccc} 1 & 1 & \ldots & 1 \\ 1 & 1 & \ldots & 1 \\ \vdots & \vdots & \ddots & \vdots \\ 1 & 1 & \ldots & 1 \\ \end{array} \right) \\ \\ &=& \text{D} -\frac{1}{n} \left( \begin{array}{cccc} 1 & 1 & \ldots & 1 \\ 1 & 1 & \ldots & 1 \\ \vdots & \vdots & \ddots & \vdots \\ 1 & 1 & \ldots & 1 \\ \end{array} \right) \text{D} -\frac{1}{n} \text{D} \left( \begin{array}{cccc} 1 & 1 & \ldots & 1 \\ 1 & 1 & \ldots & 1 \\ \vdots & \vdots & \ddots & \vdots \\ 1 & 1 & \ldots & 1 \\ \end{array} \right) +\frac{1}{n^2} \left( \begin{array}{cccc} 1 & 1 & \ldots & 1 \\ 1 & 1 & \ldots & 1 \\ \vdots & \vdots & \ddots & \vdots \\ 1 & 1 & \ldots & 1 \\ \end{array} \right) \text{D} \left( \begin{array}{cccc} 1 & 1 & \ldots & 1 \\ 1 & 1 & \ldots & 1 \\ \vdots & \vdots & \ddots & \vdots \\ 1 & 1 & \ldots & 1 \\ \end{array} \right) \\ \\ &=& \text{D} -\frac{1}{n} \text{J}_n \text{D} -\frac{1}{n} \text{D} \text{J}_n +\frac{1}{n^2} \text{J}_n\text{D}\text{J}_n \\ \\ &=& \text{I}_n\text{D}\text{I}_n -\frac{1}{n} \text{J}_n \text{D}\text{I}_n -\frac{1}{n}\text{I}_n \text{D} \text{J}_n +\frac{1}{n^2} \text{J}_n\text{D}\text{J}_n \\ \\ &=& (\text{I}_n-\frac{1}{n}\text{J}_n) \text{D}(\text{I}_n-\frac{1}{n}\text{J}_n) \\ \\ \Leftrightarrow\text{X}\text{X}^{\text{T}}&=&-\frac{1}{2}(\text{I}_n-\frac{1}{n}\text{J}_n) \text{D}(\text{I}_n-\frac{1}{n}\text{J}_n) \end{eqnarray}

\(\boldsymbol{x}\)について計算する。ここで、 \begin{eqnarray} E[\boldsymbol{x}]=\boldsymbol{0} \end{eqnarray} とする。そうすることで、 \begin{eqnarray} E[\boldsymbol{a}^{\text{T}}\boldsymbol{x}]=\boldsymbol{0} \end{eqnarray} となる。分散を求めるにあたって、\(\text{X}=(\boldsymbol{x}_1,\boldsymbol{x}_2,\ldots,\boldsymbol{x}_n)^{\text{T}}\)とすると、 \begin{eqnarray} S_{xx}=\frac{1}{n-1}\text{X}^{\text{T} }\text{X} \end{eqnarray} である(p.193より)。期待値が0なので、\(X=X_C\)と見なせ、\(\boldsymbol{a}^{\text{T}}\boldsymbol{x}\)の分散を計算すると \begin{eqnarray} \frac{1}{n-1}(\text{X}\boldsymbol{a})^{\text{T} }(\text{X}\boldsymbol{a}) &=& \boldsymbol{a}^{\text{T} }\frac{1}{n-1}\text{X}^{\text{T} }\text{X}\boldsymbol{a} \\ \\ &=& \boldsymbol{a}^{\text{T} }S_{xx}\boldsymbol{a} \end{eqnarray} 同様の計算が\(\boldsymbol{y}\)についても行えるため、\(\boldsymbol{b}^{\text{T} }\boldsymbol{y}\)の分散は\(\boldsymbol{b}^{\text{T} }S_{yy}\boldsymbol{b}\)

\begin{eqnarray} E[\boldsymbol{x}]&=&\boldsymbol{0} \\ \\ E[\boldsymbol{y}]&=&\boldsymbol{0} \end{eqnarray} とする。そうすることで、 \begin{eqnarray} E[\boldsymbol{a}^{\text{T}}\boldsymbol{x}]=\boldsymbol{0} \\ \\ E[\boldsymbol{b}^{\text{T}}\boldsymbol{y}]=\boldsymbol{0} \end{eqnarray} となる。共分散を求めるにあたって、 \begin{eqnarray} \text{X}=(\boldsymbol{x}_1,\boldsymbol{x}_2,\ldots,\boldsymbol{x}_n)^{\text{T}} \\ \\ \text{Y}=(\boldsymbol{y}_1,\boldsymbol{y}_2,\ldots,\boldsymbol{y}_n)^{\text{T}} \end{eqnarray} とする。これらの共分散を計算すると \begin{eqnarray} S_{xy}=\frac{1}{n-1}\text{X}^{\text{T} }\text{Y} \end{eqnarray} であるから、これを利用して、 \begin{eqnarray} \frac{1}{n-1}(\text{X}\boldsymbol{a})^{\text{T} }(\text{Y}\boldsymbol{b}) &=& \boldsymbol{a}^{\text{T} }\frac{1}{n-1}\text{X}^{\text{T} }\text{Y}\boldsymbol{b} \\ \\ &=& \boldsymbol{a}^{\text{T} }S_{xy}\boldsymbol{b} \end{eqnarray} が得られる。

\begin{eqnarray} &&\left\{ \begin{array}{l} S_{xy}\boldsymbol{b}-\lambda S_{xx}\boldsymbol{a}=0 \\ S_{xy}^{\text{T} }\boldsymbol{a}-\lambda S_{yy}\boldsymbol{b}=0 \end{array} \right.\\ \\ &\Leftrightarrow& \left\{ \begin{array}{l} S_{xy}\boldsymbol{b}-\lambda S_{xx}\boldsymbol{a}=0 \\ \frac{1}{\lambda}S_{yy}^{-1}S_{xy}^{\text{T} }\boldsymbol{a}=\boldsymbol{b} \end{array} \right.\\ \\ &\Rightarrow& S_{xy}(\frac{1}{\lambda}S_{yy}^{-1}S_{xy}^{\text{T} }\boldsymbol{a})-\lambda S_{xx}\boldsymbol{a}=0 \\ \\ &\Rightarrow& S_{xy}S_{yy}^{-1}S_{xy}^{\text{T} }\boldsymbol{a}=\lambda^2 S_{xx}\boldsymbol{a} \\ \\ \end{eqnarray}

\(S_{xy}S_{yy}^{-1}S_{xy}^{\text{T} }\boldsymbol{a}=\lambda^2 S_{xx}\boldsymbol{a}\)に\(\boldsymbol{c}=S_{xx}^{\frac{1}{2} }\boldsymbol{a}\)を代入する。 \begin{eqnarray} &&&S_{xy}S_{yy}^{-1}S_{xy}^{\text{T} }&\boldsymbol{a}&=&\lambda^2 S_{xx}\boldsymbol{a} \\ \\ &\Leftrightarrow& S_{xx}^{-\frac{1}{2}}&S_{xy}S_{yy}^{-1}S_{xy}^{\text{T} }&(S_{xx}^{-\frac{1}{2}}S_{xx}^{\frac{1}{2}})\boldsymbol{a}&=&S_{xx}^{-\frac{1}{2}}\lambda^2S_{xx}\boldsymbol{a} \\ \\ &\Leftrightarrow& S_{xx}^{-\frac{1}{2}}&S_{xy}S_{yy}^{-1}S_{xy}^{\text{T} }&S_{xx}^{-\frac{1}{2}}S_{xx}^{\frac{1}{2}}\boldsymbol{a}&=&\lambda^2S_{xx}^{\frac{1}{2}}\boldsymbol{a} \\ \\ &\Leftrightarrow& S_{xx}^{-\frac{1}{2}}&S_{xy}S_{yy}^{-1}S_{xy}^{\text{T} }&S_{xx}^{-\frac{1}{2}}\boldsymbol{c}&=&\lambda^2\boldsymbol{c} \\ \\ \end{eqnarray}

p.236より、\( \|\boldsymbol{c}\|^2=1 \)であることを用いる。\(\boldsymbol{a}^{\text{T}}S_{xx}\boldsymbol{a}\)を求めると \begin{eqnarray} \boldsymbol{a}^{\text{T}}S_{xx}\boldsymbol{a} &=& &(S_{xx}^{-\frac{1}{2}}\boldsymbol{c})^{\text{T} }&S_{xx}S_{xx}^{-\frac{1}{2}}\boldsymbol{c} \\ \\ &=& &\boldsymbol{c}^{\text{T} }(S_{xx}^{-\frac{1}{2}})^{\text{T} }&S_{xx}^{\frac{1}{2}}\boldsymbol{c} \\ \\ &=& &\boldsymbol{c}^{\text{T} }(S_{xx}^{\text{T} })^{-\frac{1}{2}}&S_{xx}^{\frac{1}{2}}\boldsymbol{c} \\ \\ &=& &\boldsymbol{c}^{\text{T} }S_{xx}^{-\frac{1}{2}}&S_{xx}^{\frac{1}{2}}\boldsymbol{c} \\ \\ &=& &\boldsymbol{c}^{\text{T} }\boldsymbol{c} \\ \\ &=& &\|\boldsymbol{c}\|^2 \\ \\ &=& &1 \\ \\ \end{eqnarray} が得られる。次に\(\boldsymbol{b}^{\text{T}}S_{yy}\boldsymbol{b}\)を求める。 \begin{eqnarray} \boldsymbol{b}^{\text{T}}S_{yy}\boldsymbol{b} &=& (\frac{1}{\sqrt{\eta_1}}S_{yy}^{-1}S_{xy}^{\text{T} }\boldsymbol{a})^{\text{T}}&S_{yy}&\frac{1}{\sqrt{\eta_1}}S_{yy}^{-1}S_{xy}^{\text{T} }\boldsymbol{a} \\ \\ &=& \frac{1}{\eta_1}(S_{yy}^{-1}S_{xy}^{\text{T} }\boldsymbol{a})^{\text{T}}&S_{yy}&S_{yy}^{-1}S_{xy}^{\text{T} }\boldsymbol{a} \\ \\ &=& \frac{1}{\eta_1}\boldsymbol{a}^{\text{T}}S_{xy}(S_{yy}^{-1})^{\text{T}}&S_{yy}S_{yy}^{-1}&S_{xy}^{\text{T} }\boldsymbol{a} \\ \\ &=& \frac{1}{\eta_1}\boldsymbol{a}^{\text{T}}S_{xy}(S_{yy}^{\text{T}})^{-1}&1&S_{xy}^{\text{T} }\boldsymbol{a} \\ \\ &=& \frac{1}{\eta_1}\boldsymbol{a}^{\text{T}}S_{xy}S_{yy}^{-1}&1&S_{xy}^{\text{T} }\boldsymbol{a} \\ \\ &=& \frac{1}{\eta_1}\boldsymbol{a}^{\text{T}}S_{xy}&S_{yy}^{-1}&S_{xy}^{\text{T} }\boldsymbol{a} \\ \\ &=& \frac{1}{\eta_1}(S_{xx}^{-\frac{1}{2}}\boldsymbol{c})^{\text{T}}S_{xy}&S_{yy}^{-1}&S_{xy}^{\text{T} }S_{xx}^{-\frac{1}{2}}\boldsymbol{c} \\ \\ &=& \frac{1}{\eta_1}\boldsymbol{c}^{\text{T}}(S_{xx}^{-\frac{1}{2}})^{\text{T}}S_{xy}&S_{yy}^{-1}&S_{xy}^{\text{T} }S_{xx}^{-\frac{1}{2}}\boldsymbol{c} \\ \\ &=& \frac{1}{\eta_1}\boldsymbol{c}^{\text{T}}(S_{xx}^{\text{T}})^{-\frac{1}{2}}S_{xy}&S_{yy}^{-1}&S_{xy}^{\text{T} }S_{xx}^{-\frac{1}{2}}\boldsymbol{c} \\ \\ &=& \frac{1}{\eta_1}\boldsymbol{c}^{\text{T}}S_{xx}^{-\frac{1}{2}}S_{xy}&S_{yy}^{-1}&S_{xy}^{\text{T} }S_{xx}^{-\frac{1}{2}}\boldsymbol{c} \\ \\ &=& \frac{1}{\eta_1}&\boldsymbol{c}^{\text{T}}&(S_{xx}^{-\frac{1}{2}}S_{xy}S_{yy}^{-1}S_{xy}^{\text{T} }S_{xx}^{-\frac{1}{2}}\boldsymbol{c}) \\ \\ &=& \frac{1}{\eta_1}&\boldsymbol{c}^{\text{T}}&\lambda^2\boldsymbol{c}&S_{xx}^{-\frac{1}{2}}&S_{xy}S_{yy}^{-1}S_{xy}^{\text{T} }&S_{xx}^{-\frac{1}{2}}\boldsymbol{c}&=&\lambda^2\boldsymbol{c}より \\ \\ &=& \frac{1}{\eta_1}&\boldsymbol{c}^{\text{T}}&\eta_1\boldsymbol{c} \\ \\ &=& &\boldsymbol{c}^{\text{T} }\boldsymbol{c} \\ \\ &=& &\|\boldsymbol{c}\|^2 \\ \\ &=& &1 \\ \\ \end{eqnarray}

\begin{eqnarray} \boldsymbol{a}^{\text{T}}S_{xy}\boldsymbol{b} &=& (S_{xx}^{-\frac{1}{2}}\boldsymbol{c})^{\text{T}}S_{xy}\frac{1}{\sqrt{\eta_1}}S_{yy}^{-1}S_{xy}^{\text{T} }\boldsymbol{a} \\ \\ &=& \frac{1}{\sqrt{\eta_1}}\boldsymbol{c}^{\text{T}}(S_{xx}^{-\frac{1}{2}})^{\text{T}}S_{xy}S_{yy}^{-1}S_{xy}^{\text{T} }(S_{xx}^{-\frac{1}{2}}\boldsymbol{c}) \\ \\ &=& \frac{1}{\sqrt{\eta_1}}\boldsymbol{c}^{\text{T}}(S_{xx}^{\text{T}})^{-\frac{1}{2}}S_{xy}S_{yy}^{-1}S_{xy}^{\text{T} }S_{xx}^{-\frac{1}{2}}\boldsymbol{c} \\ \\ &=& \frac{1}{\sqrt{\eta_1}}\boldsymbol{c}^{\text{T}}S_{xx}^{-\frac{1}{2}}S_{xy}S_{yy}^{-1}S_{xy}^{\text{T} }S_{xx}^{-\frac{1}{2}}\boldsymbol{c} \\ \\ &=& \frac{1}{\sqrt{\eta_1}}\boldsymbol{c}^{\text{T}}\lambda^2\boldsymbol{c} \\ \\ &=& \frac{1}{\sqrt{\eta_1}}\boldsymbol{c}^{\text{T}}\eta\boldsymbol{c} \\ \\ &=& \sqrt{\eta_1}\boldsymbol{c}^{\text{T}}\boldsymbol{c} \\ \\ &=& \sqrt{\eta_1} \\ \\ \end{eqnarray}