Multivariate normal distribution

Definition

Let $k \ge 1$ and let $X = (X_1, \dots, X_k)^{\mathsf{T}}$ be a random vector in $\mathbb{R}^k$. Then $X$ is multivariate normal if there exist $\mu \in \mathbb{R}^k$, an integer $\ell \ge 1$ and a real $k \times \ell$ matrix $A$ with this property. There is a vector $Z = (Z_1, \dots, Z_\ell)^{\mathsf{T}}$ of independent $\mathcal{N}(0, 1)$ random variables with $$X = AZ + \mu.$$ Then the law of $X$ depends only on $\mu$ and on $\Sigma = AA^{\mathsf{T}}$. It is written $\mathcal{N}_k(\mu, \Sigma)$.

Remarks

Here $\mu$ is the mean vector, with $\mu_i = E[X_i]$, and $\Sigma$ is the covariance matrix, with $\Sigma_{ij} = \operatorname{Cov}(X_i, X_j)$. An equivalent condition: $X$ is multivariate normal if and only if $a^{\mathsf{T}} X$ has a normal distribution for each $a \in \mathbb{R}^k$. Here a constant counts as normal with variance $0$. If $\Sigma$ is positive definite, $X$ has the density $$f(x) = \frac{1}{\sqrt{(2\pi)^k \det \Sigma}} \exp\left(-\tfrac{1}{2} (x - \mu)^{\mathsf{T}} \Sigma^{-1} (x - \mu)\right), \quad x \in \mathbb{R}^k.$$ The components of a multivariate normal vector are independent if and only if they are uncorrelated, that is, if $\Sigma$ is diagonal.

Explanations

L1 human

A multivariate normal vector is a list of normal random variables that may be linked to each other. The covariance matrix says how strongly each pair of components moves together. In two dimensions, the density looks like a hill with elliptical contour lines.