Index: The Book of Statistical ProofsGeneral Theorems ▷ Probability theory ▷ Covariance ▷ Symmetry

Theorem: Each covariance matrix is symmetric:

$\label{eq:covmat-symm} \Sigma_{XX}^\mathrm{T} = \Sigma_{XX} \; .$

Proof: The covariance matrix of a random vector $X$ is defined as

$\label{eq:covmat} \Sigma_{XX} = \begin{bmatrix} \mathrm{Cov}(X_1,X_1) & \ldots & \mathrm{Cov}(X_1,X_n) \\ \vdots & \ddots & \vdots \\ \mathrm{Cov}(X_n,X_1) & \ldots & \mathrm{Cov}(X_n,X_n) \end{bmatrix} \; .$

A symmetric matrix is a matrix whose transpose is equal to itself. The transpose of $\Sigma_{XX}$ is

$\label{eq:covmat-trans} \Sigma_{XX}^\mathrm{T} = \begin{bmatrix} \mathrm{Cov}(X_1,X_1) & \ldots & \mathrm{Cov}(X_n,X_1) \\ \vdots & \ddots & \vdots \\ \mathrm{Cov}(X_1,X_n) & \ldots & \mathrm{Cov}(X_n,X_n) \end{bmatrix} \; .$

Because the covariance is a symmetric function, i.e. $\mathrm{Cov}(X,Y) = \mathrm{Cov}(Y,X)$, this matrix is equal to

$\label{eq:covmat-symm-qed} \Sigma_{XX}^\mathrm{T} = \begin{bmatrix} \mathrm{Cov}(X_1,X_1) & \ldots & \mathrm{Cov}(X_1,X_n) \\ \vdots & \ddots & \vdots \\ \mathrm{Cov}(X_n,X_1) & \ldots & \mathrm{Cov}(X_n,X_n) \end{bmatrix}$

which is equivalent to our original definition in \eqref{eq:covmat}.

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Metadata: ID: P350 | shortcut: covmat-symm | author: JoramSoch | date: 2022-09-26, 10:54.