\( \newcommand{\matr}[1] {\mathbf{#1}} \newcommand{\vertbar} {\rule[-1ex]{0.5pt}{2.5ex}} \newcommand{\horzbar} {\rule[.5ex]{2.5ex}{0.5pt}} \)
deepdream of
          a sidewalk
Show Answer
\( \newcommand{\cat}[1] {\mathrm{#1}} \newcommand{\catobj}[1] {\operatorname{Obj}(\mathrm{#1})} \newcommand{\cathom}[1] {\operatorname{Hom}_{\cat{#1}}} \newcommand{\multiBetaReduction}[0] {\twoheadrightarrow_{\beta}} \newcommand{\betaReduction}[0] {\rightarrow_{\beta}} \newcommand{\betaEq}[0] {=_{\beta}} \newcommand{\string}[1] {\texttt{"}\mathtt{#1}\texttt{"}} \newcommand{\symbolq}[1] {\texttt{`}\mathtt{#1}\texttt{'}} \)
Math and science::INF ML AI

Covariance matrix

Let \( X \) and \( Y \) be two random variables. The covariance between \( X \) and \( Y \) is defined as:

\[\begin{aligned} Cov[X,Y] &:= E[(X-E[X])(Y-E[Y])] \\ &= [...] \end{aligned} \]

Let the vector \( Z \) be defined like so: \( Z := \begin{bmatrix} X \\ Y\end{bmatrix} \). Thus, \( Z \) is a vector of random variables.

The covariance matrix for \( Z \) is defined as:

\[ \begin{aligned} Cov[Z] &:= E[(Z - E[Z])(Z - E[Z])^T] \\ &= [...] \\ \end{aligned} \]

Where the expectation is an elementwise operation. The covariance matrix is a result of a matrix multiplication of two vector-like matrices, which produces a 2x2 matrix. (Yes, it is valid!).