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

Multivariate Gaussian distribution

This card derives the general multivariate normal distribution from the standard multivariate normal distribution.

Standard multivariate Gaussian/normal distribution

Let \( (\Omega, \mathrm{F}, \mathbb{P}) \) be a probability space. Let \( X : \Omega \to \mathbb{R}^K \) be a continuous random vector. \( X \) is said to have a standard multivariate normal distribution iff its joint probability density function is:

[\[ f_X(x) = \quad ? \] ]

As a vector of random variables

\( X \) can be considered to be a vector of independent random variables, each having a standard normal distribution. The proof of this formulation on the reverse side.

General multivariate

The general multivariate normal distribution is best understood as being the distribution that results from applying a linear transformation to a random variable having a multivariate standard normal distribution.

General multivariate normal distribution

Let \( (\Omega, \mathrm{F}, \mathbb{P}) \) be a probability space, and let \( Z : \Omega \to \mathrm{R}^K \) be a random vector with a multivariate standard normal distribution. Then let \( X = \mu + \Sigma Z \) be another random vector. \( X \) has a distribution \( f_X : \mathbb{R}^K \to \mathbb{R} \) which is a transformed version of \( Z \)'s distribution, \( f_Z : \mathbb{R}^K \to \mathbb{R} \):

[\[ \begin{alignat*}{3}f_X(x) &= \frac{1}{\text{rescaling factor} } \; &&f_z(\text{z in terms of x}) \\&= \frac{1}{\text{what?} } &&f_z(\Sigma^{-1}(x - \mu)) \\ &= \frac{1}{\text{what?} } &&(\frac{1}{\sqrt{2\pi} })^K e^{\text{what?} } \\ \end{alignat*}\]]