Random vectors and matrices

deejayosamu·2026년 1월 18일

다변량통계

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Suppose XX and YY are vector
covariance matrix 설명

  • Linearity and bilinearity

    • E(aX+bY)=aE(X)+bE(Y), a,bRE(aX+bY)=aE(X)+bE(Y),\space \forall a,b \in \mathbb{R}
    • Cov(X+Y,Z)=Cov(X,Z)+Cov(Y,Z)Cov(X+Y,Z)=Cov(X,Z)+Cov(Y,Z)
    • Cov(X,Y+Z)=Cov(X,Y)+Cov(X,Z)Cov(X,Y+Z)=Cov(X,Y)+Cov(X,Z)
    • For any matrices AA and BB, {E(AX)=AE(X)E(XB)=E(X)BCov(AX,Y)=ACov(X,Y)Cov(X,BY)=Cov(X,Y)BT\left\{\begin{matrix} E(AX)=AE(X)\\ E(XB)=E(X)B\\ Cov(AX,Y)=ACov(X,Y)\\ Cov(X,BY)=Cov(X,Y)B^T \end{matrix}\right.
  • Variance matrix

    • Var(AX)=AVar(X)AT=Cov(AX,AX)Var(AX)=A Var(X) A^T=Cov(AX,AX)
    • Var(X+b)=Var(X)Var(X+b)=Var(X)
    • Var(X+Y)=Var(X)+Cov(X,Y)+Cov(Y,X)+Var(Y) (CovT(X,Y)=Cov(Y,X))Var(X+Y)=Var(X)+Cov(X,Y)+Cov(Y,X)+Var(Y) \space (Cov^T(X,Y)=Cov(Y,X))
    • Var(X)Var(X) is singular <=> a(0)Rp\exists a(\neq0) \in \mathbb{R}^p s.t. P(aT(XE(X))=0)=1P(a^T(X-E(X))=0)=1
    • Correlation matrix
      Corr(X)=Cov(Xi,Xj)Var(Xi)Var(Xj)=diag(1Var(Xi))Var(X)diag(1Var(Xj))Corr(X)=\frac{Cov(X_i,X_j)}{\sqrt{Var(X_i)} \sqrt{Var(X_j)}}=diag(\frac{1}{\sqrt{Var(X_i)}}) Var(X) diag(\frac{1}{\sqrt{Var(X_j)}})
  • Canonical Correlation
    For X1RkX_1 \in \mathbb{R}^k and X2RmX_2 \in \mathbb{R}^m, we have maxa,b,c,d Corr(cTX2+d,aTX1+d)=max1imλi\underset{a,b,c,d}{max} \space Corr(c^T X_2+d,a^T X_1+d )=\sqrt{\underset{1 \leq i \leq m}{max} \lambda_i}
    where λ1,,λm\lambda_1, \cdots, \lambda_m are eigenvalues of Σ221/2Σ21Σ111Σ12Σ221/2\Sigma_{22}^{-1/2} \Sigma_{21} \Sigma_{11}^{-1} \Sigma_{12} \Sigma_{22}^{-1/2}
    pf)
    can._cor.

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