已知n維隨機變量\(\vec{X}=(X_{1},X_{2},...,X_{n})\)的協方差矩陣為\(C = \begin{bmatrix}c_{11} & c_{12} & ... & c_{1n} \\c_{21} & c_{22} & ... & c_{2n} \\. & .& &.\\. & .& &.\\. & .& &.\\c_{n1} & c_{n2} &...&c_{nn}\end{bmatrix} \)，其中\(c_{ij} = E\big\{[X_{i}-E(X_{i})][X_{j}-E(X_{j})]\big\}\)。那么，能否將協方差矩陣寫成向量形式呢？即寫成和一維隨機向量的方差公式類似的形式：\(D(X) = E(X-E(X))^{2} \)。
我們來證明一下，設\(X_{i}\)的樣本量為m，由協方差矩陣的公式，可知：
\(C = \frac{1}{m}\begin{bmatrix} (X_{1}-E(X_{1}))^{T}(X_{1}-E(X_{1})) & ... & (X_{1}-E(X_{1}))^{T}(X_{n}-E(X_{n})) \\ (X_{2}-E(X_{2}))^{T}(X_{1}-E(X_{1})) & ... & (X_{2}-E(X_{2}))^{T}(X_{n}-E(X_{n})) \\. &. & .\\. &. & .\\. &. & .\\ (X_{n}-E(X_{n}))^{T}(X_{1}-E(X_{1})) & ... & (X_{n}-E(X_{n}))^{T}(X_{n}-E(X_{n})) \end{bmatrix} \)
\( = \frac{1}{m}\begin{bmatrix} (X_{1}-E(X_{1}))^{T} \\ (X_{2}-E(X_{2}))^{T} \\ .\\. \\. \\ (X_{n}-E(X_{n}))^{T} \end{bmatrix} \begin{bmatrix} X_{1}-E(X_{1}), & X_{2}-E(X_{2}), &...,& X_{n}-E(X_{n}) \end{bmatrix} \)
\( = \frac{1}{m}(\vec{X}-E(\vec{X}))^{T}(\vec{X}-E(\vec{X})) \)
\( = E((\vec{X}-E(\vec{X}))^{T}(\vec{X}-E(\vec{X}))) \)
推導結果與一維隨機向量的方差公式是一致的，所以，我們的猜測是正確的，證畢。