第6講各種データ解析法

因子分析

因子得点の推定

因子負荷量を推定し，因子軸の回転により各因子の解釈ができたら，次は因子得点を推定する．

因子得点の推定量を $\hat{f}_{ki}$ とおき，各変量の線形式により推定することを考える．

{\hat{f}}_{k i} = \sum_{j = 1}^{p} β_{k j} z_{j i}, k = 1, \dots, m, i = 1, \dots, n

$\begin{equation} \hat{f}_{ki} = \sum_{j=1}^{p}\beta_{kj}z_{ji},~~~~k = 1, \ldots, m,~~~~i = 1, \ldots, n \end{equation}$

最小二乗法により，

β_{k j}

$\beta_{kj}$ を求める．

\begin{aligned} Q_{k} & = \sum_{i = 1}^{n} (f_{k i} - {\hat{f}}_{k i})^{2} \\ = \sum_{i = 1}^{n} {(f_{k i} - \sum_{j^{'} = 1}^{p} β_{k j^{'}} z_{j^{'} i})}^{2} \end{aligned}

$\begin{align} Q_k &= \sum_{i=1}^{n}(f_{ki} - \hat{f}_{ki})^2 \\ &= \sum_{i=1}^{n}\left(f_{ki} - \sum_{j'=1}^{p}\beta_{kj'}z_{j'i}\right)^2 \end{align}$

Q_{k}

$Q_k$ を

b e t a_{k j}

$beta_{kj}$ で偏微分して 0 とおくと，

\frac{\partial Q_{k}}{\partial β_{k j}} = 2 \sum_{i = 1}^{n} z_{j i} (f_{k i} - \sum_{j^{'} = 1}^{p} β_{k j^{'}} z_{j^{'} i}) = 0

$\begin{equation} \dfrac{\partial Q_k}{\partial \beta_{kj}} = 2 \sum_{i=1}^{n}z_{ji}\left(f_{ki} - \sum_{j'=1}^{p}\beta_{kj'}z_{j'i}\right) = 0 \end{equation}$

\sum_{i = 1}^{n} z_{j i} (f_{k i} - \sum_{j^{'} = 1}^{p} β_{k j^{'}} z_{j^{'} i}) = 0

$\begin{equation} \sum_{i=1}^{n}z_{ji}\left(f_{ki} - \sum_{j'=1}^{p}\beta_{kj'}z_{j'i}\right) = 0 \end{equation}$

\sum_{i = 1}^{n} z_{j i} f_{k i} - \sum_{i = 1}^{n} z_{j i} \sum_{j^{'} = 1}^{p} β_{k j^{'}} z_{j^{'} i} = 0

$\begin{equation} \sum_{i=1}^{n}z_{ji}f_{ki} - \sum_{i=1}^{n}z_{ji}\sum_{j'=1}^{p}\beta_{kj'}z_{j'i} = 0 \end{equation}$

\sum_{i = 1}^{n} z_{j i} \sum_{j^{'} = 1}^{p} β_{k j^{'}} z_{j^{'} i} = \sum_{i = 1}^{n} z_{j i} f_{k i}

$\begin{equation} \sum_{i=1}^{n}z_{ji}\sum_{j'=1}^{p}\beta_{kj'}z_{j'i} = \sum_{i=1}^{n}z_{ji}f_{ki} \end{equation}$

\sum_{j^{'} = 1}^{p} β_{k j^{'}} \sum_{i = 1}^{n} z_{j i} z_{j^{'} i} = \sum_{i = 1}^{n} z_{j i} f_{k i} \dots \dots (1)

$\begin{equation} \sum_{j'=1}^{p}\beta_{kj'}\sum_{i=1}^{n}z_{ji}z_{j'i} = \sum_{i=1}^{n}z_{ji}f_{ki}~~~~\cdots\cdots~(1) \end{equation}$

ここで，

z_{j i}

$z_{ji}$ と

z_{j^{'} i}

$z_{j'i}$ の相関係数を

r_{j j^{'}}

$r_{jj'}$ とおくと，

\begin{aligned} r_{j j^{'}} & = \frac{C o v [z_{j}, z_{j^{'}}]}{\sqrt{V [z_{j}]} \sqrt{V [z_{j^{'}}]}} \\ = C o v [z_{j}, z_{j^{'}}] \\ = E [z_{j} z_{j^{'}}] - E [z_{j}] E [z_{j^{'}}] \\ = E [z_{j} z_{j^{'}}] \\ = \frac{1}{n} \sum_{i = 1}^{n} z_{j i} z_{j^{'} i} \end{aligned}

$\begin{align} r_{jj'} &= \dfrac{Cov[z_{j}, z_{j'} ]}{\sqrt{V[z_{j}]}\sqrt{V[z_{j'}]}} \\ &= Cov[z_{j}, z_{j'} ] \\ &= E[z_{j} z_{j'} ] - E[z_{j}] E[z_{j'} ] \\ &= E[z_{j} z_{j'} ] \\ &= \frac{1}{n}\sum_{i=1}^{n}z_{ji}z_{j'i} \end{align}$

よって，

\sum_{i = 1}^{n} z_{j i} z_{j^{'} i} = n r_{j j^{'}} \dots \dots (2)

$\begin{equation} \sum_{i=1}^{n}z_{ji}z_{j'i} = n r_{jj'}~~~~\cdots\cdots~(2) \end{equation}$

また，

\begin{aligned} \sum_{i = 1}^{n} z_{j i} f_{k i} & = \sum_{i = 1}^{n} (\sum_{k^{'} = 1}^{m} a_{j k^{'}} f_{k^{'} i} + e_{j i}) f_{k i} \\ = \sum_{i = 1}^{n} (\sum_{k^{'} = 1}^{m} a_{j k^{'}} f_{k i} f_{k^{'} i} + f_{k i} e_{j i}) \\ = \sum_{i = 1}^{n} \sum_{k^{'} = 1}^{m} a_{j k^{'}} f_{k i} f_{k^{'} i} + \sum_{i = 1}^{n} f_{k i} e_{j i} \\ = \sum_{k^{'} = 1}^{m} a_{j k^{'}} \sum_{i = 1}^{n} f_{k i} f_{k^{'} i} + \sum_{i = 1}^{n} f_{k i} e_{j i} \dots \dots (3) \end{aligned}

$\begin{align} \sum_{i=1}^{n}z_{ji}f_{ki} &= \sum_{i=1}^{n}\left(\sum_{k'=1}^{m}a_{jk'}f_{k'i} + e_{ji}\right)f_{ki} \\ &= \sum_{i=1}^{n}\left(\sum_{k'=1}^{m}a_{jk'}f_{ki}f_{k'i} + f_{ki}e_{ji}\right) \\ &= \sum_{i=1}^{n}\sum_{k'=1}^{m}a_{jk'}f_{ki}f_{k'i} + \sum_{i=1}^{n}f_{ki}e_{ji} \\ &= \sum_{k'=1}^{m}a_{jk'}\sum_{i=1}^{n}f_{ki}f_{k'i} + \sum_{i=1}^{n}f_{ki}e_{ji}~~~~\cdots\cdots~(3) \\ \end{align}$

k \neq k^{'}

$k \ne k'$ に対して

C o v [f_{k}, f_{k^{'}}] = 0

$Cov[f_{k}, f_{k'}] = 0$ であるので，

\begin{aligned} C o v [f_{k}, f_{k^{'}}] & = E [f_{k} f_{k^{'}}] - E [f_{k}] E [f_{k^{'}}] \\ = E [f_{k} f_{k^{'}}] \\ = \frac{1}{n} \sum_{i = 1}^{n} f_{k i} f_{k^{'} i} \end{aligned}

$\begin{align} Cov[f_{k}, f_{k'}] &= E[f_{k} f_{k'}] - E[f_{k}] E[f_{k'}] \\ &= E[f_{k} f_{k'}] \\ &= \frac{1}{n}\sum_{i=1}^{n}f_{ki}f_{k'i} \\ \end{align}$

∴ \sum_{i = 1}^{n} f_{k i} f_{k^{'} i} = 0

$\begin{equation} \therefore~\sum_{i=1}^{n}f_{ki}f_{k'i} = 0 \end{equation}$

k = k^{'}

$k = k'$ に対し，

C o v [f_{k}, f_{k^{'}}] = V [f_{k}] = 1

$Cov[f_{k}, f_{k'}] = V[f_{k}] = 1$ であるので，

C o v [f_{k}, f_{k^{'}}] = \frac{1}{n} \sum_{i = 1}^{n} f_{k i} f_{k^{'} i} = 1

$\begin{equation} Cov[f_{k}, f_{k'}] = \frac{1}{n}\sum_{i=1}^{n}f_{ki}f_{k'i} = 1 \end{equation}$

∴ \sum_{i = 1}^{n} f_{k i} f_{k^{'} i} = n

$\begin{equation} \therefore~\sum_{i=1}^{n}f_{ki}f_{k'i} = n \end{equation}$

両条件を合わせると，

\sum_{i = 1}^{n} f_{k i} f_{k^{'} i} = δ_{k k^{'}} n \dots \dots (4)

$\begin{equation} \sum_{i=1}^{n}f_{ki}f_{k'i} = \delta_{k k'}n~~~~\cdots\cdots~(4) \end{equation}$

また，

f_{k}

$f_{k}$ と

e_{j}

$e_{j}$ は無相関であるので，

\sum_{i = 1}^{n} f_{k i} e_{j i} = 0 \dots \dots (5)

$\begin{equation} \sum_{i=1}^{n}f_{ki}e_{ji} = 0~~~~\cdots\cdots~(5) \end{equation}$

(4) と (5) を (3) に代入すると，

\begin{aligned} \sum_{i = 1}^{n} z_{j i} f_{k i} & = \sum_{k^{'} = 1}^{m} a_{j k^{'}} \sum_{i = 1}^{n} f_{k i} f_{k^{'} i} + \sum_{i = 1}^{n} f_{k i} e_{j i} \\ = \sum_{k^{'} = 1}^{m} a_{j k^{'}} \cdot δ_{k k^{'}} n \\ = a_{j k} n \dots \dots (6) \end{aligned}

$\begin{align} \sum_{i=1}^{n}z_{ji}f_{ki} &= \sum_{k'=1}^{m}a_{jk'}\sum_{i=1}^{n}f_{ki}f_{k'i} + \sum_{i=1}^{n}f_{ki}e_{ji} \\ &= \sum_{k'=1}^{m}a_{jk'} \cdot \delta_{k k'}n \\ &=a_{jk}n~~~~\cdots\cdots~(6) \end{align}$

(2) と (6) を (1) に代入すると，

\sum_{j^{'} = 1}^{p} β_{k j^{'}} \cdot n r_{j j^{'}} = a_{j k} n

$\begin{equation} \sum_{j'=1}^{p}\beta_{kj'} \cdot n r_{jj'} = a_{jk}n \end{equation}$

∴ \sum_{j^{'} = 1}^{p} β_{k j^{'}} r_{j j^{'}} = a_{j k}

$\begin{equation} \therefore~\sum_{j'=1}^{p}\beta_{kj'}r_{jj'} = a_{jk} \end{equation}$

これより，

k = 1, \dots, m

$k = 1, \ldots, m$ に対して，

{\begin{matrix} β_{k 1} r_{11} + \dots + β_{k p} r_{1 p} = a_{1 k} \\ ⋮ \\ β_{k 1} r_{p 1} + \dots + β_{k p} r_{p p} = a_{p k} \end{matrix}

$\begin{equation} \left\{ \begin{array}{c} \beta_{k1}r_{11} + \cdots + \beta_{kp}r_{1p} = a_{1k} \\ \vdots \\ \beta_{k1}r_{p1} + \cdots + \beta_{kp}r_{pp} = a_{pk} \end{array} \right. \end{equation}$

行列で表現すると，

(\begin{matrix} r_{11} & \dots & r_{1 p} \\ ⋮ & ⋱ & ⋮ \\ r_{p 1} & \dots & r_{p p} \end{matrix}) (\begin{matrix} β_{k 1} \\ ⋮ \\ β_{k p} \end{matrix}) = (\begin{matrix} a_{1 k} \\ ⋮ \\ a_{p k} \end{matrix})

$\begin{equation} \begin{pmatrix} r_{11} & \cdots & r_{1p} \\ \vdots & \ddots & \vdots \\ r_{p1} & \cdots & r_{pp} \end{pmatrix} \begin{pmatrix} \beta_{k1} \\ \vdots \\ \beta_{kp} \end{pmatrix} = \begin{pmatrix} a_{1k} \\ \vdots \\ a_{pk} \end{pmatrix} \end{equation}$

\begin{aligned} (\begin{matrix} β_{k 1} \\ ⋮ \\ β_{k p} \end{matrix}) & = {(\begin{matrix} r_{11} & \dots & r_{1 p} \\ ⋮ & ⋱ & ⋮ \\ r_{p 1} & \dots & r_{p p} \end{matrix})}^{- 1} (\begin{matrix} a_{1 k} \\ ⋮ \\ a_{p k} \end{matrix}) \\ = (\begin{matrix} r^{11} & \dots & r^{1 p} \\ ⋮ & ⋱ & ⋮ \\ r^{p 1} & \dots & r^{p p} \end{matrix}) (\begin{matrix} a_{1 k} \\ ⋮ \\ a_{p k} \end{matrix}) \end{aligned}

$\begin{align} \begin{pmatrix} \beta_{k1} \\ \vdots \\ \beta_{kp} \end{pmatrix} &= \begin{pmatrix} r_{11} & \cdots & r_{1p} \\ \vdots & \ddots & \vdots \\ r_{p1} & \cdots & r_{pp} \end{pmatrix}^{-1} \begin{pmatrix} a_{1k} \\ \vdots \\ a_{pk} \end{pmatrix} \\ &= \begin{pmatrix} r^{11} & \cdots & r^{1p} \\ \vdots & \ddots & \vdots \\ r^{p1} & \cdots & r^{pp} \end{pmatrix} \begin{pmatrix} a_{1k} \\ \vdots \\ a_{pk} \end{pmatrix} \end{align}$

ただし，

r^{j j^{'}}

$r^{jj'}$ は相関行列

R = (r_{j j^{'}})

$R = (r_{jj'})$ の逆行列の要素

(j, j^{'})

$(j,~j')$ とする．これより，

β_{k j} = \sum_{j^{'} = 1}^{p} a_{j^{'} k} r^{j j^{'}}

$\begin{equation} \beta_{kj} = \sum_{j'=1}^{p}a_{j'k}r^{jj'} \end{equation}$

第6講 各種データ解析法

因子分析

因子得点の推定

第6講各種データ解析法