共分散行列 (variance-covariance matrix)

$m$ 個の確率変数 ${\displaystyle X_1}$ ， ${\displaystyle X_2}$ ， ･･･， ${\displaystyle X_m}$ ， $n$ 個の確率変数 ${\displaystyle Y_1}$ ， ${\displaystyle Y_2}$ ， ･･･， ${\displaystyle Y_n}$ があり，確率ベクトルを

${\displaystyle X=\left(\begin{array}{c}{X}_1\\ X_2\\ ⋮\\ X_m\end{array}\right)}$ ， ${\displaystyle Y=\left(\begin{array}{c}{Y}_1\\ Y_2\\ ⋮\\ Y_n\end{array}\right)}$

とすると，共分散行列 ${\displaystyle \text{Cov}\left(X,Y\right)}$ は

${\displaystyle \text{Cov}\left(X,Y\right)=E\left(\left(X-{\mu }_X\right){}^t\left(Y-{\mu }_Y\right)\right)}$

ただし

${\displaystyle {\mu }_X=\left(\begin{array}{c}E\left(X_1\right)\\ E\left(X_2\right)\\ ⋮\\ E\left(X_m\right)\end{array}\right)=\left(\begin{array}{c}{\mu }_{X1}\\ {\mu }_{X2}\\ ⋮\\ {\mu }_{Xm}\end{array}\right)}$ ， ${\displaystyle {\mu }_Y=\left(\begin{array}{c}E\left(Y_1\right)\\ E\left(Y_2\right)\\ ⋮\\ E\left(Y_n\right)\end{array}\right)=\left(\begin{array}{c}{\mu }_{Y1}\\ {\mu }_{Y2}\\ ⋮\\ {\mu }_{Yn}\end{array}\right)}$

と定義される． ${\displaystyle \text{Cov}\left(X,Y\right)}$ は以下のように書き換えられる．

■ ${\displaystyle \text{Cov}\left(X,Y\right)}$ の性質

${\displaystyle X}$は$m$ 次元確率ベクトル ， ${\displaystyle Y}$ は $n$ 次元確率ベクトル， ${\displaystyle A}$は $m$ 次正方行列， ${\displaystyle B}$ は $n$ 次正方行列， ${\displaystyle a}$は$m$次元定数ベクトル ， ${\displaystyle b}$は$n$次元定数ベクトル とする

• ${\displaystyle \text{Cov}\left(a+X,b+Y\right)}$ ${\displaystyle =\text{Cov}\left(X,Y\right)}$

  ●証明

  ${\displaystyle \text{Cov}\left(a+X,b+Y\right)}$ ${\displaystyle =\text{Cov}\left(\left(\begin{array}{c}{a}_1+X_1\\ a_2+X_2\\ ⋮\\ a_m+X_m\end{array}\right),\left(\begin{array}{c}{b}_1+Y_1\\ b_2+Y_2\\ ⋮\\ b_n+Y_n\end{array}\right)\right)}$

  ${\displaystyle =\left(\begin{array}{cccc}\text{Cov}\left(a_1+X_1,b_1+Y_1\right)& \text{Cov}\left(a_1+X_1,b_2+Y_2\right)& \cdots & \text{Cov}\left(a_1+X_1,b_n+Y_n\right)\\ \text{Cov}\left(a_2+X_2,b_1+Y_1\right)& \text{Cov}\left(a_2+X_2,b_2+Y_2\right)& \cdots & \text{Cov}\left(a_2+X_2,b_n+Y_n\right)\\ ⋮& ⋮& \ddots & ⋮\\ \text{Cov}\left(a_m+X_m,b_1+Y_1\right)& \text{Cov}\left(a_m+X_m,b_2+Y_2\right)& \cdots & \text{Cov}\left(a_m+X_m,b_n+Y_n\right)\end{array}\right)}$

  ${\displaystyle \text{Cov}\left(c+U,d+V\right)=\text{Cov}\left(U,V\right)}$ の性質より

  ${\displaystyle =\left(\begin{array}{cccc}\text{Cov}\left(X_1,Y_1\right)& \text{Cov}\left(X_1,Y_2\right)& \cdots & \text{Cov}\left(X_1,Y_n\right)\\ \text{Cov}\left(X_2,Y_1\right)& \text{Cov}\left(X_2,Y_2\right)& \cdots & \text{Cov}\left(X_2,Y_n\right)\\ ⋮& ⋮& \ddots & ⋮\\ \text{Cov}\left(X_m,Y_1\right)& \text{Cov}\left(X_m,Y_2\right)& \cdots & \text{Cov}\left(X_m,Y_n\right)\end{array}\right)}$

  ${\displaystyle =\text{Cov}\left(X,Y\right)}$

• ${\displaystyle \text{Cov}\left(AX,BY\right)}$ ${\displaystyle =A\text{Cov}\left(X,Y\right){}^{t}B}$

  ●証明

  ${\displaystyle \text{Cov}\left(AX,BY\right)}$ ${\displaystyle =E\left(\left\{A\left(X-{\mu }_X\right)\right\}{}^t\left\{B\left(Y-{\mu }_Y\right)\right\}\right)}$

  ただし， ${\displaystyle {\mu }_X=E\left(X\right)}$ ， ${\displaystyle {\mu }_Y=E\left(Y\right)}$

  ${\displaystyle =E\left(A\left(X-{\mu }_X\right){}^t\left(Y-{\mu }_Y\right){}^{t}B\right)}$

  ${\displaystyle E\left(X\right)}$ の性質を用いて式変形をすると

  ${\displaystyle =AE\left(\left(X-{\mu }_X\right){}^t\left(Y-{\mu }_Y\right){}^{t}B\right)}$

  ${\displaystyle =AE\left(\left(X-{\mu }_X\right){}^t\left(Y-{\mu }_Y\right)\right){}^{t}B}$

  ${\displaystyle =A\text{Cov}\left(X,Y\right){}^{t}B}$

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　最終更新日： 2026年7月11日