协方差矩阵公式推导

     已知n维随机变量\(\vec{X}=(X_{1},X_{2},...,X_{n})\)的协方差矩阵为\(C = \begin{bmatrix}c_{11} & c_{12} & ... & c_{1n} \\c_{21} & c_{22} & ... & c_{2n} \\. & .& &.\\. & .& &.\\. & .& &.\\c_{n1} & c_{n2} &...&c_{nn}\end{bmatrix} \)，其中\(c_{ij} = E\big\{[X_{i}-E(X_{i})][X_{j}-E(X_{j})]\big\}\)。那么，能否将协方差矩阵写成向量形式呢？即写成和一维随机向量的方差公式类似的形式：\(D(X) = E(X-E(X))^{2} \)。

     我们来证明一下，设\(X_{i}\)的样本量为m，由协方差矩阵的公式，可知：

\(C = \frac{1}{m}\begin{bmatrix} (X_{1}-E(X_{1}))^{T}(X_{1}-E(X_{1})) & ... & (X_{1}-E(X_{1}))^{T}(X_{n}-E(X_{n})) \\ (X_{2}-E(X_{2}))^{T}(X_{1}-E(X_{1})) & ... & (X_{2}-E(X_{2}))^{T}(X_{n}-E(X_{n})) \\. &. & .\\. &. & .\\. &. & .\\ (X_{n}-E(X_{n}))^{T}(X_{1}-E(X_{1})) & ... & (X_{n}-E(X_{n}))^{T}(X_{n}-E(X_{n})) \end{bmatrix} \)

\(   = \frac{1}{m}\begin{bmatrix} (X_{1}-E(X_{1}))^{T} \\ (X_{2}-E(X_{2}))^{T} \\ .\\. \\. \\ (X_{n}-E(X_{n}))^{T} \end{bmatrix} \begin{bmatrix} X_{1}-E(X_{1}), & X_{2}-E(X_{2}), &...,& X_{n}-E(X_{n}) \end{bmatrix} \)

\(   = \frac{1}{m}(\vec{X}-E(\vec{X}))^{T}(\vec{X}-E(\vec{X})) \)

\(   = E((\vec{X}-E(\vec{X}))^{T}(\vec{X}-E(\vec{X}))) \)

     推导结果与一维随机向量的方差公式是一致的，所以，我们的猜测是正确的，证毕。

 

转载于:https://www.cnblogs.com/thinkers-dym/p/4492817.html