- フィッシャーの判別分析
- 各群内の分散の導出
- \(\boldsymbol{\hat{\omega}}=\text{S}^{-1}(\overline{x}^{(1)}-\overline{x}^{(2)})\)が最適な射影であること
- 中点が\((\overline{x}^{(1)}-\overline{x}^{(2)})^{\text{T} }\text{S}^{-1}(\overline{x}^{(1)}+\overline{x}^{(2)})/2\)になること
- マハラノビス平方距離の式の導出
- マハラノビス平方距離を用いた\(D_2^2-D_1^2=2f(x)\)の導出
統計学実践ワークブックの行間埋め 第23章
\begin{eqnarray}
\frac{1}{n_j-1}\displaystyle \sum_{i=1}^{n_j} (y_i^{(j)}-\overline{y}^{(j)})^2
&=&
\frac{1}{n_j-1}\displaystyle \sum_{i=1}^{n_j} (\omega^{\text{T} }x_i^{(j)}-\omega^{\text{T} }\overline{x}^{(j)})^2 \\ \\
&=&
\frac{1}{n_j-1}\displaystyle \sum_{i=1}^{n_j} (\omega^{\text{T} }(x_i^{(j)}-\overline{x}^{(j)}))^2\\ \\
&=&
\frac{1}{n_j-1}\displaystyle \sum_{i=1}^{n_j} (\omega^{\text{T} }(x_i^{(j)}-\overline{x}^{(j)}))(\omega^{\text{T} }(x_i^{(j)}-\overline{x}^{(j)}))\\ \\
&=&
\frac{1}{n_j-1}\displaystyle \sum_{i=1}^{n_j} (\omega^{\text{T} }(x_i^{(j)}-\overline{x}^{(j)}))(\omega^{\text{T} }(x_i^{(j)}-\overline{x}^{(j)}))^{\text{T} }\\ \\
&=&
\frac{1}{n_j-1}\displaystyle \sum_{i=1}^{n_j} \omega^{\text{T} }(x_i^{(j)}-\overline{x}^{(j)}))(x_i^{(j)}-\overline{x}^{(j)})^{\text{T} }\omega\\ \\
&=&
\omega^{\text{T} }\left(\frac{1}{n_j-1}\displaystyle \sum_{i=1}^{n_j} (x_i^{(j)}-\overline{x}^{(j)}))(x_i^{(j)}-\overline{x}^{(j)})^{\text{T} }\right)\omega\\ \\
&=&
\omega^{\text{T} }\left(\frac{1}{n_j-1}\displaystyle \sum_{i=1}^{n_j} x_{iC}^{(j)}x_{iC}^{(j)\text{T} }\right)\omega\;\;\;...x_i^{(j)}-\overline{x}^{(j)}=x_{iC}^{(j)}とした\\ \\
&=&
\omega^{\text{T} }\left(\frac{1}{n_j-1}\displaystyle \sum_{i=1}^{n_j}
\left(
\begin{array}{cccc}
(x_{iC1}^{(j)})^2 & x_{iC1}^{(j)}x_{iC2}^{(j)} & \ldots & x_{iC1}^{(j)}x_{iCp}^{(j)} \\
x_{iC2}^{(j)}x_{iC1}^{(j)} & (x_{iC2}^{(j)})^2 & \ldots & x_{iC2}^{(j)}x_{iCp}^{(j)} \\
\vdots & \vdots & \ddots & \vdots \\
x_{iCp}^{(j)}x_{iC1}^{(j)} & x_{iCp}^{(j)}x_{iC2}^{(j)} & \ldots & (x_{iCp}^{(j)})^2
\end{array}
\right)
\right)\omega\\ \\
&=&
\omega^{\text{T} }\left(\frac{1}{n_j-1}\displaystyle
\left(
\begin{array}{cccc}
\sum_{i=1}^{n_j}(x_{iC1}^{(j)})^2 & \sum_{i=1}^{n_j}x_{iC1}^{(j)}x_{iC2}^{(j)} & \ldots & \sum_{i=1}^{n_j}x_{iC1}^{(j)}x_{iCp}^{(j)} \\
\sum_{i=1}^{n_j}x_{iC2}^{(j)}x_{iC1}^{(j)} & \sum_{i=1}^{n_j}(x_{iC2}^{(j)})^2 & \ldots & \sum_{i=1}^{n_j}x_{iC2}^{(j)}x_{iCp}^{(j)} \\
\vdots & \vdots & \ddots & \vdots \\
\sum_{i=1}^{n_j}x_{iCp}^{(j)}x_{iC1}^{(j)} & \sum_{i=1}^{n_j}x_{iCp}^{(j)}x_{iC2}^{(j)} & \ldots & \sum_{i=1}^{n_j}(x_{iCp}^{(j)})^2
\end{array}
\right)
\right)\omega\\ \\
&=&
\omega^{\text{T} }\left(\frac{1}{n_j-1}
\left(
\begin{array}{cccc}
x_{1C1}^{(j)} & x_{2C1}^{(j)} & \ldots & x_{pC1}^{(j)} \\
x_{1C2}^{(j)} & x_{2C2}^{(j)} & \ldots & x_{pC2}^{(j)} \\
\vdots & \vdots & \ddots & \vdots \\
x_{1Cp}^{(j)} & x_{2Cp}^{(j)} & \ldots & x_{pCp}^{(j)} \\
\end{array}
\right)\left(
\begin{array}{cccc}
x_{1C1}^{(j)} & x_{1C2}^{(j)} & \ldots & x_{1Cp}^{(j)} \\
x_{2C1}^{(j)} & x_{2C2}^{(j)} & \ldots & x_{2Cp}^{(j)} \\
\vdots & \vdots & \ddots & \vdots \\
x_{pC1}^{(j)} & x_{pC2}^{(j)} & \ldots & x_{pCp}^{(j)} \\
\end{array}
\right)
\right)\omega\\ \\
&=&
\omega^{\text{T} }\left(\frac{1}{n_j-1}\text{X_C}^{(j)\text{T}}\text{X_C}^{(j)}\right)\omega\;\;\;...\left(
\begin{array}{cccc}
x_{1C1}^{(j)} & x_{1C2}^{(j)} & \ldots & x_{1Cp}^{(j)} \\
x_{2C1}^{(j)} & x_{2C2}^{(j)} & \ldots & x_{2Cp}^{(j)} \\
\vdots & \vdots & \ddots & \vdots \\
x_{pC1}^{(j)} & x_{pC2}^{(j)} & \ldots & x_{pCp}^{(j)} \\
\end{array}
\right)=\text{X_C}とした\\ \\
&=&
\omega^{\text{T} }\text{S_j}\omega\;\;\;...p.193より
\end{eqnarray}
\(\hat{\boldsymbol{\omega}}=\text{S}^{-1}(\overline{x}^{(1)}-\overline{x}^{(2)})\)を利用する。
\begin{eqnarray}
\frac{\hat{\omega} ^{\text{T} }\overline{x}^{(1)} +\hat{\omega} ^{\text{T} }\overline{x}^{(2)} }{2}
&=&
\frac{1}{2}\hat{\omega} ^{\text{T} }(\overline{x}^{(1)}+\overline{x}^{(2)}) \\ \\
&=&
\frac{1}{2}(\text{S}^{-1}(\overline{x}^{(1)}-\overline{x}^{(2)}))^{\text{T} }(\overline{x}^{(1)}+\overline{x}^{(2)}) \\ \\
&=&
\frac{1}{2}(\overline{x}^{(1)}-\overline{x}^{(2)})^{\text{T} }(\text{S}^{-1})^{\text{T} }(\overline{x}^{(1)}+\overline{x}^{(2)}) \\ \\
&=&
\frac{1}{2}(\overline{x}^{(1)}-\overline{x}^{(2)})^{\text{T} }(\text{S}^{\text{T} })^{-1}(\overline{x}^{(1)}+\overline{x}^{(2)}) \\ \\
&=&
\frac{1}{2}(\overline{x}^{(1)}-\overline{x}^{(2)})^{\text{T} }\text{S}^{-1}(\overline{x}^{(1)}+\overline{x}^{(2)}) \\ \\
\end{eqnarray}
\(\hat{\boldsymbol{\omega}}=\text{S}^{-1}(\overline{x}^{(1)}-\overline{x}^{(2)})\)を利用する。
\begin{eqnarray}
\lambda(\hat{\omega})
&=&
\frac{(\hat{\omega}^{\text{T} }(\overline{x}^{(1)}-\overline{x}^{(2)}))^2}{\hat{\omega}^{\text{T} }\text{S}\hat{\omega}} \\ \\
&=&
\frac{((\text{S}^{-1}(\overline{x}^{(1)}-\overline{x}^{(2)}))^{\text{T} }(\overline{x}^{(1)}-\overline{x}^{(2)}))^2}{(\text{S}^{-1}(\overline{x}^{(1)}-\overline{x}^{(2)}))^{\text{T} }\text{S}\text{S}^{-1}(\overline{x}^{(1)}-\overline{x}^{(2)})} \\ \\
&=&
\frac{((\overline{x}^{(1)}-\overline{x}^{(2)})^{\text{T} }(\text{S}^{-1})^{\text{T} }(\overline{x}^{(1)}-\overline{x}^{(2)}))^2}{(\overline{x}^{(1)}-\overline{x}^{(2)})^{\text{T} }(\text{S}^{-1})^{\text{T} }\text{S}\text{S}^{-1}(\overline{x}^{(1)}-\overline{x}^{(2)})} \\ \\
&=&
\frac{((\overline{x}^{(1)}-\overline{x}^{(2)})^{\text{T} }(\text{S}^{\text{T} })^{-1}(\overline{x}^{(1)}-\overline{x}^{(2)}))^2}{(\overline{x}^{(1)}-\overline{x}^{(2)})^{\text{T} }(\text{S}^{\text{T} })^{-1}(\overline{x}^{(1)}-\overline{x}^{(2)})} \\ \\
&=&
\frac{((\overline{x}^{(1)}-\overline{x}^{(2)})^{\text{T} }\text{S}^{-1}(\overline{x}^{(1)}-\overline{x}^{(2)}))^2}{(\overline{x}^{(1)}-\overline{x}^{(2)})^{\text{T} }\text{S}^{-1}(\overline{x}^{(1)}-\overline{x}^{(2)})} \\ \\
&=&
(\overline{x}^{(1)}-\overline{x}^{(2)})^{\text{T} }\text{S}^{-1}(\overline{x}^{(1)}-\overline{x}^{(2)}) \;\;\;...スカラーの計算のため約分できる\\ \\
\end{eqnarray}
スカラーの転置は元の値と同じになることを用いる。(\(x^{\text{T}}=x\))
\begin{eqnarray}
D_2^2-D_1^2
&=&
(\overline{x}-\overline{x}^{(2)})^{\text{T} }\text{S}^{-1}(\overline{x}-\overline{x}^{(2)})&&-(\overline{x}-\overline{x}^{(1)})^{\text{T} }\text{S}^{-1}(\overline{x}-\overline{x}^{(1)}) \\ \\
&=&
\overline{x}^{\text{T} }\text{S}^{-1}\overline{x}-\overline{x}^{\text{T} }\text{S}^{-1}\overline{x}^{(2)}-\overline{x}^{(2)\text{T} }\text{S}^{-1}\overline{x}+\overline{x}^{(2)\text{T} }\text{S}^{-1}\overline{x}^{(2)}&&-\overline{x}^{\text{T} }\text{S}^{-1}\overline{x}+\overline{x}^{\text{T} }\text{S}^{-1}\overline{x}^{(1)}+\overline{x}^{(1)\text{T} }\text{S}^{-1}\overline{x}-\overline{x}^{(1)\text{T} }\text{S}^{-1}\overline{x}^{(1)} \\ \\
&=&
\overline{x}^{\text{T} }\text{S}^{-1}\overline{x}-\overline{x}^{\text{T} }\text{S}^{-1}\overline{x}&&+\overline{x}^{\text{T} }\text{S}^{-1}(\overline{x}^{(1)}-\overline{x}^{(2)})+(\overline{x}^{(1)\text{T} }-\overline{x}^{(2)\text{T} })\text{S}^{-1}\overline{x}&&+\overline{x}^{(2)\text{T} }\text{S}^{-1}\overline{x}^{(2)}-\overline{x}^{(1)\text{T} }\text{S}^{-1}\overline{x}^{(1)} \\ \\
&=&
0&&+\overline{x}^{\text{T} }\hat{\omega}+((\text{S}^{-1})^{\text{T}}(\overline{x}^{(1)}-\overline{x}^{(2)}))^{\text{T}}\overline{x}&&-\overline{x}^{(1)\text{T} }\text{S}^{-1}\overline{x}^{(1)}+\overline{x}^{(2)\text{T} }\text{S}^{-1}\overline{x}^{(2)}+(\overline{x}^{(1)\text{T} }\text{S}^{-1}\overline{x}^{(2)}-\overline{x}^{(1)\text{T} }\text{S}^{-1}\overline{x}^{(2)}) \\ \\
&=&
\overline{x}^{\text{T} }\hat{\omega}+(\text{S}^{-1}(\overline{x}^{(1)}-\overline{x}^{(2)}))^{\text{T}}\overline{x}&&-\overline{x}^{(1)\text{T} }\text{S}^{-1}\overline{x}^{(1)}+\overline{x}^{(2)\text{T} }\text{S}^{-1}\overline{x}^{(2)}+(\overline{x}^{(1)\text{T} }\text{S}^{-1}\overline{x}^{(2)}-\overline{x}^{(2)\text{T} }\text{S}^{-1}\overline{x}^{(1)}) \\ \\
&=&
\overline{x}^{\text{T} }\hat{\omega}+\hat{\omega}^{\text{T} }\overline{x}&&-\overline{x}^{(1)\text{T} }\text{S}^{-1}(\overline{x}^{(1)}-\overline{x}^{(2)})-\overline{x}^{(2)\text{T} }\text{S}^{-1}(\overline{x}^{(1)}-\overline{x}^{(2)}) \\ \\
&=&
(\overline{x}^{\text{T} }\hat{\omega})^{\text{T} }+\hat{\omega}^{\text{T} }\overline{x}&&-(\overline{x}^{(1)\text{T} }+\overline{x}^{(2)\text{T} })\text{S}^{-1}(\overline{x}^{(1)}-\overline{x}^{(2)}) \\ \\
&=&
2\hat{\omega}^{\text{T} }\overline{x}&&-(\overline{x}^{(1)\text{T} }+\overline{x}^{(2)\text{T} })\text{S}^{-1}(\overline{x}^{(1)}-\overline{x}^{(2)}) \\ \\
&=&
2f(x)
\end{eqnarray}