Multivariate normal distribution

Properties

Moments

The kth-order moments of x are given by

${\displaystyle {\mu }_{1,\dots ,N}(\mathbf{x})\stackrel{\mathrm{d}\mathrm{e}\mathrm{f}}{=}{\mu }_{r_1,\dots ,r_N}(\mathbf{x})\stackrel{\mathrm{d}\mathrm{e}\mathrm{f}}{=}\mathrm{E}\left[\prod _{j=1}^N{X}_j^{r_j}\right]}$

where r₁ + r₂ + ⋯ + r_N = k.

The kth-order central moments are as follows

1. If k is odd, μ_{1, ..., N}(x − μ) = 0.
2. If k is even with k = 2λ, then $${\displaystyle {\mu }_{1,\dots ,2\lambda }(\mathbf{x}-\mathit{\mu })=\sum \left({\sigma }_{ij}{\sigma }_{k\ell }\cdots {\sigma }_{XZ}\right)}$$

where the sum is taken over all allocations of the set ${\displaystyle \left\{1,\dots ,2\lambda \right\}}$ into λ (unordered) pairs. That is, for a kth (= 2λ = 6) central moment, one sums the products of λ = 3 covariances (the expected value μ is taken to be 0 in the interests of parsimony):

${\displaystyle \begin{array}{cc}& \mathrm{E}[X_1{X}_2{X}_3{X}_4{X}_5{X}_6]\\ =& \mathrm{E}[X_1{X}_2]\mathrm{E}[X_3{X}_4]\mathrm{E}[X_5{X}_6]+\mathrm{E}[X_1{X}_2]\mathrm{E}[X_3{X}_5]\mathrm{E}[X_4{X}_6]+\mathrm{E}[X_1{X}_2]\mathrm{E}[X_3{X}_6]\mathrm{E}[X_4{X}_5]\\ & +\mathrm{E}[X_1{X}_3]\mathrm{E}[X_2{X}_4]\mathrm{E}[X_5{X}_6]+\mathrm{E}[X_1{X}_3]\mathrm{E}[X_2{X}_5]\mathrm{E}[X_4{X}_6]+\mathrm{E}[X_1{X}_3]\mathrm{E}[X_2{X}_6]\mathrm{E}[X_4{X}_5]\\ & +\mathrm{E}[X_1{X}_4]\mathrm{E}[X_2{X}_3]\mathrm{E}[X_5{X}_6]+\mathrm{E}[X_1{X}_4]\mathrm{E}[X_2{X}_5]\mathrm{E}[X_3{X}_6]+\mathrm{E}[X_1{X}_4]\mathrm{E}[X_2{X}_6]\mathrm{E}[X_3{X}_5]\\ & +\mathrm{E}[X_1{X}_5]\mathrm{E}[X_2{X}_3]\mathrm{E}[X_4{X}_6]+\mathrm{E}[X_1{X}_5]\mathrm{E}[X_2{X}_4]\mathrm{E}[X_3{X}_6]+\mathrm{E}[X_1{X}_5]\mathrm{E}[X_2{X}_6]\mathrm{E}[X_3{X}_4]\\ & +\mathrm{E}[X_1{X}_6]\mathrm{E}[X_2{X}_3]\mathrm{E}[X_4{X}_5]+\mathrm{E}[X_1{X}_6]\mathrm{E}[X_2{X}_4]\mathrm{E}[X_3{X}_5]+\mathrm{E}[X_1{X}_6]\mathrm{E}[X_2{X}_5]\mathrm{E}[X_3{X}_4].\end{array}}$

This yields ${\displaystyle \frac{(2\lambda -1)!}{2^{\lambda -1}(\lambda -1)!}}$ terms in the sum (15 in the above case), each being the product of λ (in this case 3) covariances. For fourth order moments (four variables) there are three terms. For sixth-order moments there are 3 × 5 = 15 terms, and for eighth-order moments there are 3 × 5 × 7 = 105 terms.

The covariances are then determined by replacing the terms of the list ${\displaystyle [1,\dots ,2\lambda ]}$ by the corresponding terms of the list consisting of r₁ ones, then r₂ twos, etc.. To illustrate this, examine the following 4th-order central moment case:

${\displaystyle \begin{array}{cc}\mathrm{E}\left[X_i^4\right]& =3{\sigma }_{ii}^2\\ \mathrm{E}\left[X_i^3{X}_j\right]& =3{\sigma }_{ii}{\sigma }_{ij}\\ \mathrm{E}\left[X_i^2{X}_j^2\right]& ={\sigma }_{ii}{\sigma }_{jj}+2{\sigma }_{ij}^2\\ \mathrm{E}\left[X_i^2{X}_j{X}_k\right]& ={\sigma }_{ii}{\sigma }_{jk}+2{\sigma }_{ij}{\sigma }_{ik}\\ \mathrm{E}\left[X_i{X}_j{X}_k{X}_n\right]& ={\sigma }_{ij}{\sigma }_{kn}+{\sigma }_{ik}{\sigma }_{jn}+{\sigma }_{in}{\sigma }_{jk}.\end{array}}$

where ${\displaystyle {\sigma }_{ij}}$ is the covariance of X_i and X_j. With the above method one first finds the general case for a kth moment with k different X variables, ${\displaystyle E\left[X_i{X}_j{X}_k{X}_n\right]}$, and then one simplifies this accordingly. For example, for ${\displaystyle \mathrm{E}[X_i^2{X}_k{X}_n]}$, one lets X_i = X_j and one uses the fact that ${\displaystyle {\sigma }_{ii}={\sigma }_i^2}$.

Functions of a normal vector

A quadratic form of a normal vector ${\displaystyle \mathit{x}}$, ${\displaystyle q(\mathit{x})={\mathit{x}}^{\prime }{\mathbf{Q}}_{\mathbf{2}}\mathit{x}+{{\mathit{q}}_{\mathbf{1}}}^{\prime }\mathit{x}+q_0}$ (where ${\displaystyle {\mathbf{Q}}_{\mathbf{2}}}$ is a matrix, ${\displaystyle {\mathit{q}}_{\mathbf{1}}}$ is a vector, and ${\displaystyle q_0}$ is a scalar), is a generalized chi-squared variable. The direction of a normal vector follows a projected normal distribution.

If ${\displaystyle f(\mathit{x})}$ is a general scalar-valued function of a normal vector, its probability density function, cumulative distribution function, and inverse cumulative distribution function can be computed with the numerical method of ray-tracing (Matlab code Archived 2025-02-20 at the Wayback Machine).

Likelihood function

If the mean and covariance matrix are known, the log likelihood of an observed vector ${\displaystyle \mathit{x}}$ is simply the log of the probability density function:

${\displaystyle \ln L(\mathit{x})=-\frac{1}{2}\left[\ln (|\mathbf{\Sigma }|)+(\mathit{x}-\mathit{\mu }{)}^{\prime }{\mathbf{\Sigma }}^{-1}(\mathit{x}-\mathit{\mu })+k\ln (2\pi )\right]}$,

The circularly symmetric version of the noncentral complex case, where ${\displaystyle \mathit{z}}$ is a vector of complex numbers, would be

${\displaystyle \ln L(\mathit{z})=-\ln (|\mathbf{\Sigma }|)-(\mathit{z}-\mathit{\mu }{)}^{†}{\mathbf{\Sigma }}^{-1}(\mathit{z}-\mathit{\mu })-k\ln (\pi )}$

i.e. with the conjugate transpose (indicated by ${\displaystyle †}$) replacing the normal transpose (indicated by ${\displaystyle {}^{\prime }}$). This is slightly different than in the real case, because the circularly symmetric version of the complex normal distribution has a slightly different form for the normalization constant.

A similar notation is used for multiple linear regression.

Since the log likelihood of a normal vector is a quadratic form of the normal vector, it is distributed as a generalized chi-squared variable.

Differential entropy

The differential entropy of the multivariate normal distribution is

$${\displaystyle \begin{array}{cc}h\left(f\right)& =-{\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}{\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}\cdots {\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}f(\mathbf{x})\ln f(\mathbf{x})d\mathbf{x}\\ & =\frac{1}{2}\ln \left|2\pi e\mathbf{\Sigma }\right|=\frac{k}{2}\left(1+\ln 2\pi \right)+\frac{1}{2}\ln \left|\mathbf{\Sigma }\right|,\end{array}}$$

where the bars denote the matrix determinant, k is the dimensionality of the vector space, and the result has units of nats.

Kullback–Leibler divergence

The Kullback–Leibler divergence from ${\displaystyle {\mathcal{N}}_1({\mathit{\mu }}_1,{\mathbf{\Sigma }}_1)}$ to ${\displaystyle {\mathcal{N}}_0({\mathit{\mu }}_0,{\mathbf{\Sigma }}_0)}$, for non-singular matrices Σ₁ and Σ₀, is:

${\displaystyle D_{\text{KL}}({\mathcal{N}}_0\parallel {\mathcal{N}}_1)=\frac{1}{2}\left\{\mathrm{tr}\left({\mathbf{\Sigma }}_1^{-1}{\mathbf{\Sigma }}_0\right)+{\left({\mathit{\mu }}_1-{\mathit{\mu }}_0\right)}^{\mathrm{T}}{\mathbf{\Sigma }}_1^{-1}({\mathit{\mu }}_1-{\mathit{\mu }}_0)-k+\ln \frac{|{\mathbf{\Sigma }}_1|}{|{\mathbf{\Sigma }}_0|}\right\},}$

where ${\displaystyle |\cdot |}$ denotes the matrix determinant, ${\displaystyle tr(\cdot )}$ is the trace, ${\displaystyle \ln (\cdot )}$ is the natural logarithm and ${\displaystyle k}$ is the dimension of the vector space.

The logarithm must be taken to base e since the two terms following the logarithm are themselves base-e logarithms of expressions that are either factors of the density function or otherwise arise naturally. The equation therefore gives a result measured in nats. Dividing the entire expression above by log_e 2 yields the divergence in bits.

When ${\displaystyle {\mathit{\mu }}_1={\mathit{\mu }}_0}$,

${\displaystyle D_{\text{KL}}({\mathcal{N}}_0\parallel {\mathcal{N}}_1)=\frac{1}{2}\left\{\mathrm{tr}\left({\mathbf{\Sigma }}_1^{-1}{\mathbf{\Sigma }}_0\right)-k+\ln \frac{|{\mathbf{\Sigma }}_1|}{|{\mathbf{\Sigma }}_0|}\right\}.}$

Mutual information

The mutual information of two multivariate normal distribution is a special case of the Kullback–Leibler divergence in which ${\displaystyle P}$ is the full ${\displaystyle k}$ dimensional multivariate distribution and ${\displaystyle Q}$ is the product of the ${\displaystyle k_1}$ and ${\displaystyle k_2}$ dimensional marginal distributions ${\displaystyle X}$ and ${\displaystyle Y}$, such that ${\displaystyle k_1+k_2=k}$. The mutual information between ${\displaystyle X}$ and ${\displaystyle Y}$ is given by:

${\displaystyle I(\mathit{X},\mathit{Y})=\frac{1}{2}\ln \left(\frac{det({\mathrm{\Sigma }}_X)det({\mathrm{\Sigma }}_Y)}{det(\mathrm{\Sigma })}\right),}$

where

${\displaystyle \mathrm{\Sigma }=\left[\begin{array}{cc}{\mathrm{\Sigma }}_X& {\mathrm{\Sigma }}_{XY}\\ {\mathrm{\Sigma }}_{XY}& {\mathrm{\Sigma }}_Y\end{array}\right].}$

If ${\displaystyle Q}$ is product of ${\displaystyle k}$ one-dimensional normal distributions, then in the notation of the Kullback–Leibler divergence section of this article, ${\displaystyle {\mathbf{\Sigma }}_1}$ is a diagonal matrix with the diagonal entries of ${\displaystyle {\mathbf{\Sigma }}_0}$, and ${\displaystyle {\mathit{\mu }}_1={\mathit{\mu }}_0}$. The resulting formula for mutual information is:

${\displaystyle I(\mathit{X})=-\frac{1}{2}\ln |{\mathit{\rho }}_0|,}$

where ${\displaystyle {\mathit{\rho }}_0}$ is the correlation matrix constructed from ${\displaystyle {\mathbf{\Sigma }}_0}$.

In the bivariate case the expression for the mutual information is:

${\displaystyle I(x;y)=-\frac{1}{2}\ln (1-{\rho }^2).}$

Joint normality

Normally distributed and independent

If ${\displaystyle X}$ and ${\displaystyle Y}$ are normally distributed and independent, this implies they are "jointly normally distributed", i.e., the pair ${\displaystyle (X,Y)}$ must have multivariate normal distribution. However, a pair of jointly normally distributed variables need not be independent (would only be so if uncorrelated, ${\displaystyle \rho =0}$ ).

Two normally distributed random variables need not be jointly bivariate normal

The fact that two random variables ${\displaystyle X}$ and ${\displaystyle Y}$ both have a normal distribution does not imply that the pair ${\displaystyle (X,Y)}$ has a joint normal distribution. A simple example is one in which ${\displaystyle X}$ has a normal distribution with expected value 0 and variance 1, and ${\displaystyle Y=X}$ if ${\displaystyle |X|>c}$ and ${\displaystyle Y=-X}$ if ${\displaystyle |X|<c}$, where ${\displaystyle c>0}$. There are similar counterexamples for more than two random variables. In general, they sum to a mixture model.

Correlations and independence

In general, random variables may be uncorrelated but statistically dependent. But if a random vector has a multivariate normal distribution then any two or more of its components that are uncorrelated are independent. This implies that any two or more of its components that are pairwise independent are independent. But, as pointed out just above, it is not true that two random variables that are (separately, marginally) normally distributed and uncorrelated are independent.

Conditional distributions

If N-dimensional x is partitioned as follows

${\displaystyle \mathbf{x}=\left[\begin{array}{c}{\mathbf{x}}_1\\ {\mathbf{x}}_2\end{array}\right]\text{ with sizes }\left[\begin{array}{c}q\times 1\\ (N-q)\times 1\end{array}\right]}$

and accordingly μ and Σ are partitioned as follows

${\displaystyle \mathit{\mu }=\left[\begin{array}{c}{\mathit{\mu }}_1\\ {\mathit{\mu }}_2\end{array}\right]\text{ with sizes }\left[\begin{array}{c}q\times 1\\ (N-q)\times 1\end{array}\right]}$

${\displaystyle \mathbf{\Sigma }=\left[\begin{array}{cc}{\mathbf{\Sigma }}_{11}& {\mathbf{\Sigma }}_{12}\\ {\mathbf{\Sigma }}_{21}& {\mathbf{\Sigma }}_{22}\end{array}\right]\text{ with sizes }\left[\begin{array}{cc}q\times q& q\times (N-q)\\ (N-q)\times q& (N-q)\times (N-q)\end{array}\right]}$

then the distribution of x₁ conditional on x₂ = a is multivariate normal (x₁ | x₂ = a) ~ N(μ, Σ) where

${\displaystyle \overline{\mathit{\mu }}={\mathit{\mu }}_1+{\mathbf{\Sigma }}_{12}{\mathbf{\Sigma }}_{22}^{-1}\left(\mathbf{a}-{\mathit{\mu }}_2\right)}$

and covariance matrix

${\displaystyle \overline{\mathbf{\Sigma }}={\mathbf{\Sigma }}_{11}-{\mathbf{\Sigma }}_{12}{\mathbf{\Sigma }}_{22}^{-1}{\mathbf{\Sigma }}_{21}.}$

Here ${\displaystyle {\mathbf{\Sigma }}_{22}^{-1}}$ is the generalized inverse of ${\displaystyle {\mathbf{\Sigma }}_{22}}$. The matrix ${\displaystyle \overline{\mathbf{\Sigma }}}$ is the Schur complement of Σ₂₂ in Σ. That is, the equation above is equivalent to inverting the overall covariance matrix, dropping the rows and columns corresponding to the variables being conditioned upon, and inverting back to get the conditional covariance matrix.

Note that knowing that x₂ = a alters the variance, though the new variance does not depend on the specific value of a; perhaps more surprisingly, the mean is shifted by ${\displaystyle {\mathbf{\Sigma }}_{12}{\mathbf{\Sigma }}_{22}^{-1}\left(\mathbf{a}-{\mathit{\mu }}_2\right)}$; compare this with the situation of not knowing the value of a, in which case x₁ would have distribution ${\displaystyle {\mathcal{N}}_q\left({\mathit{\mu }}_1,{\mathbf{\Sigma }}_{11}\right)}$.

An interesting fact derived in order to prove this result, is that the random vectors ${\displaystyle {\mathbf{x}}_2}$ and ${\displaystyle {\mathbf{y}}_1={\mathbf{x}}_1-{\mathbf{\Sigma }}_{12}{\mathbf{\Sigma }}_{22}^{-1}{\mathbf{x}}_2}$ are independent.

The matrix Σ₁₂Σ₂₂⁻¹ is known as the matrix of regression coefficients.

Bivariate case

In the bivariate case where x is partitioned into ${\displaystyle X_1}$ and ${\displaystyle X_2}$, the conditional distribution of ${\displaystyle X_1}$ given ${\displaystyle X_2}$ is

${\displaystyle X_1\mid X_2=a\sim \mathcal{N}\left({\mu }_1+\frac{{\sigma }_1}{{\sigma }_2}\rho (a-{\mu }_2),(1-{\rho }^2){\sigma }_1^2\right)}$

where ${\displaystyle \rho =\frac{{\sigma }_{12}}{{\sigma }_1{\sigma }_2}}$ is the correlation coefficient between ${\displaystyle X_1}$ and ${\displaystyle X_2}$.

Bivariate conditional expectation, centered case with unit variances

${\displaystyle \left(\begin{array}{c}{X}_1\\ X_2\end{array}\right)\sim \mathcal{N}\left(\left(\begin{array}{c}0\\ 0\end{array}\right),\left(\begin{array}{cc}1& \rho \\ \rho & 1\end{array}\right)\right)}$

The conditional expectation of X₁ given X₂ is

${\displaystyle \mathrm{E}(X_1\mid X_2=x_2)=\rho x_2}$

and the conditional variance is

${\displaystyle \mathrm{var}(X_1\mid X_2=x_2)=1-{\rho }^2;}$

thus the conditional variance does not depend on x₂.

The conditional expectation of X₁ given that X₂ is smaller/bigger than z is:

${\displaystyle \mathrm{E}(X_1\mid X_2<z)=-\rho \frac{\phi (z)}{\mathrm{\Phi }(z)},}$

${\displaystyle \mathrm{E}(X_1\mid X_2>z)=\rho \frac{\phi (z)}{(1-\mathrm{\Phi }(z))},}$

where the final ratio here is called the inverse Mills ratio.

Proof: the last two results are obtained using the result ${\displaystyle \mathrm{E}(X_1\mid X_2=x_2)=\rho x_2}$, so that

${\displaystyle \mathrm{E}(X_1\mid X_2<z)=\rho E(X_2\mid X_2<z)}$ and then using the properties of the expectation of a truncated normal distribution.

Marginal distributions

To obtain the marginal distribution over a subset of multivariate normal random variables, one only needs to drop the irrelevant variables (the variables that one wants to marginalize out) from the mean vector and the covariance matrix. The proof for this follows from the definitions of multivariate normal distributions and linear algebra.

Example

Let X = [X₁, X₂, X₃] be multivariate normal random variables with mean vector μ = [μ₁, μ₂, μ₃] and covariance matrix Σ (standard parametrization for multivariate normal distributions). Then the joint distribution of X′ = [X₁, X₃] is multivariate normal with mean vector μ′ = [μ₁, μ₃] and covariance matrix ${\displaystyle {\mathbf{\Sigma }}^{\prime }=\left[\begin{array}{cc}{\mathbf{\Sigma }}_{11}& {\mathbf{\Sigma }}_{13}\\ {\mathbf{\Sigma }}_{31}& {\mathbf{\Sigma }}_{33}\end{array}\right]}$.

Affine transformation

If Y = c + BX is an affine transformation of ${\displaystyle \mathbf{X}\sim \mathcal{N}(\mathit{\mu },\mathbf{\Sigma }),}$ where c is an ${\displaystyle M\times 1}$ vector of constants and B is a constant ${\displaystyle M\times N}$ matrix, then Y has a multivariate normal distribution with expected value c + Bμ and variance BΣB^T i.e., ${\displaystyle \mathbf{Y}\sim \mathcal{N}\left(\mathbf{c}+\mathbf{B}\mathit{\mu },\mathbf{B}\mathbf{\Sigma }{\mathbf{B}}^{\mathrm{T}}\right)}$. In particular, any subset of the X_i has a marginal distribution that is also multivariate normal. To see this, consider the following example: to extract the subset (X₁, X₂, X₄)^T, use

${\displaystyle \mathbf{B}=\left[\begin{array}{ccccccc}1& 0& 0& 0& 0& \dots & 0\\ 0& 1& 0& 0& 0& \dots & 0\\ 0& 0& 0& 1& 0& \dots & 0\end{array}\right]}$

which extracts the desired elements directly.

Another corollary is that the distribution of Z = b · X, where b is a constant vector with the same number of elements as X and the dot indicates the dot product, is univariate Gaussian with ${\displaystyle Z\sim \mathcal{N}\left(\mathbf{b}\cdot \mathit{\mu },{\mathbf{b}}^{\mathrm{T}}\mathbf{\Sigma }\mathbf{b}\right)}$. This result follows by using

${\displaystyle \mathbf{B}=\left[\begin{array}{cccc}{b}_1& b_2& \dots & b_n\end{array}\right]={\mathbf{b}}^{\mathrm{T}}.}$

Observe how the positive-definiteness of Σ implies that the variance of the dot product must be positive.

An affine transformation of X such as 2X is not the same as the sum of two independent realisations of X.

Geometric interpretation

The equidensity contours of a non-singular multivariate normal distribution are ellipsoids (i.e. affine transformations of hyperspheres) centered at the mean. Hence the multivariate normal distribution is an example of the class of elliptical distributions. The directions of the principal axes of the ellipsoids are given by the eigenvectors of the covariance matrix ${\displaystyle \mathbf{\Sigma }}$. The squared relative lengths of the principal axes are given by the corresponding eigenvalues.

If Σ = UΛU^T = UΛ^1/2(UΛ^1/2)^T is an eigendecomposition where the columns of U are unit eigenvectors and Λ is a diagonal matrix of the eigenvalues, then we have

${\displaystyle \mathbf{X}\sim \mathcal{N}(\mathit{\mu },\mathbf{\Sigma })⟺\mathbf{X}\sim \mathit{\mu }+\mathbf{U}{\mathbf{\Lambda }}^{1/2}\mathcal{N}(0,\mathbf{I})⟺\mathbf{X}\sim \mathit{\mu }+\mathbf{U}\mathcal{N}(0,\mathbf{\Lambda }).}$

Moreover, U can be chosen to be a rotation matrix, as inverting an axis does not have any effect on N(0, Λ), but inverting a column changes the sign of U's determinant. The distribution N(μ, Σ) is in effect N(0, I) scaled by Λ^1/2, rotated by U and translated by μ.

Conversely, any choice of μ, full rank matrix U, and positive diagonal entries Λ_i yields a non-singular multivariate normal distribution. If any Λ_i is zero and U is square, the resulting covariance matrix UΛU^T is singular. Geometrically this means that every contour ellipsoid is infinitely thin and has zero volume in n-dimensional space, as at least one of the principal axes has length of zero; this is the degenerate case.

"The radius around the true mean in a bivariate normal random variable, re-written in polar coordinates (radius and angle), follows a Hoyt distribution."

In one dimension the probability of finding a sample of the normal distribution in the interval ${\displaystyle \mu \pm \sigma }$ is approximately 68.27%, but in higher dimensions the probability of finding a sample in the region of the standard deviation ellipse is lower.

Statistical inference

Parameter estimation

The derivation of the maximum-likelihood estimator of the covariance matrix of a multivariate normal distribution is straightforward.

In short, the probability density function (pdf) of a multivariate normal is

${\displaystyle f(\mathbf{x})=\frac{1}{\sqrt{(2\pi {)}^k|\mathbf{\Sigma }|}}\exp \left(-\frac{1}{2}(\mathbf{x}-\mathit{\mu }{)}^{\mathrm{T}}{\mathbf{\Sigma }}^{-1}(\mathbf{x}-\mathit{\mu })\right)}$

and the ML estimator of the covariance matrix from a sample of n observations is

${\displaystyle \hat{\mathbf{\Sigma }}=\frac{1}{n}\sum _{i=1}^n({\mathbf{x}}_i-\overline{\mathbf{x}})({\mathbf{x}}_i-\overline{\mathbf{x}}{)}^{\mathrm{T}}}$

which is simply the sample covariance matrix. This is a biased estimator whose expectation is

${\displaystyle E\left[\hat{\mathbf{\Sigma }}\right]=\frac{n-1}{n}\mathbf{\Sigma }.}$

An unbiased sample covariance is

${\displaystyle \hat{\mathbf{\Sigma }}=\frac{1}{n-1}\sum _{i=1}^n({\mathbf{x}}_i-\overline{\mathbf{x}})({\mathbf{x}}_i-\overline{\mathbf{x}}{)}^{\mathrm{T}}=\frac{1}{n-1}\left[X^{\prime }\left(I-\frac{1}{n}\cdot J\right)X\right]}$ (matrix form; ${\displaystyle I}$ is the ${\displaystyle K\times K}$ identity matrix, J is a ${\displaystyle K\times K}$ matrix of ones; the term in parentheses is thus the ${\displaystyle K\times K}$ centering matrix)

The Fisher information matrix for estimating the parameters of a multivariate normal distribution has a closed form expression. This can be used, for example, to compute the Cramér–Rao bound for parameter estimation in this setting. See Fisher information for more details.

Bayesian inference

In Bayesian statistics, the conjugate prior of the mean vector is another multivariate normal distribution, and the conjugate prior of the covariance matrix is an inverse-Wishart distribution ${\displaystyle {\mathcal{W}}^{-1}}$ . Suppose then that n observations have been made

${\displaystyle \mathbf{X}=\{{\mathbf{x}}_1,\dots ,{\mathbf{x}}_n\}\sim \mathcal{N}(\mathit{\mu },\mathbf{\Sigma })}$

and that a conjugate prior has been assigned, where

${\displaystyle p(\mathit{\mu },\mathbf{\Sigma })=p(\mathit{\mu }\mid \mathbf{\Sigma })p(\mathbf{\Sigma }),}$

where

${\displaystyle p(\mathit{\mu }\mid \mathbf{\Sigma })\sim \mathcal{N}({\mathit{\mu }}_0,m^{-1}\mathbf{\Sigma }),}$

and

${\displaystyle p(\mathbf{\Sigma })\sim {\mathcal{W}}^{-1}(\mathbf{\Psi },n_0).}$

Then

${\displaystyle \begin{array}{rcl}p(\mathit{\mu }\mid \mathbf{\Sigma },\mathbf{X})& \sim & \mathcal{N}\left(\frac{n\overline{\mathbf{x}}+m{\mathit{\mu }}_0}{n+m},\frac{1}{n+m}\mathbf{\Sigma }\right),\\ p(\mathbf{\Sigma }\mid \mathbf{X})& \sim & {\mathcal{W}}^{-1}\left(\mathbf{\Psi }+n\mathbf{S}+\frac{nm}{n+m}(\overline{\mathbf{x}}-{\mathit{\mu }}_0)(\overline{\mathbf{x}}-{\mathit{\mu }}_0{)}^{\prime },n+n_0\right),\end{array}}$

where

${\displaystyle \begin{array}{cc}\overline{\mathbf{x}}& =\frac{1}{n}\sum _{i=1}^n{\mathbf{x}}_i,\\ \mathbf{S}& =\frac{1}{n}\sum _{i=1}^n({\mathbf{x}}_i-\overline{\mathbf{x}})({\mathbf{x}}_i-\overline{\mathbf{x}}{)}^{\prime }.\end{array}}$

Multivariate normality tests

Multivariate normality tests check a given set of data for similarity to the multivariate normal distribution. The null hypothesis is that the data set is similar to the normal distribution, therefore a sufficiently small p-value indicates non-normal data. Multivariate normality tests include the Cox–Small test and Smith and Jain's adaptation of the Friedman–Rafsky test created by Larry Rafsky and Jerome Friedman.

Mardia's test

Mardia's test is based on multivariate extensions of skewness and kurtosis measures. For a sample {x₁, ..., x_n} of k-dimensional vectors we compute

${\displaystyle \begin{array}{cc}& \hat{\mathbf{\Sigma }}=\frac{1}{n}\sum _{j=1}^n\left({\mathbf{x}}_j-\overline{\mathbf{x}}\right){\left({\mathbf{x}}_j-\overline{\mathbf{x}}\right)}^{\mathrm{T}}\\ & A=\frac{1}{6n}\sum _{i=1}^n\sum _{j=1}^n{\left[({\mathbf{x}}_i-\overline{\mathbf{x}}{)}^{\mathrm{T}}{\hat{\mathbf{\Sigma }}}^{-1}({\mathbf{x}}_j-\overline{\mathbf{x}})\right]}^3\\ & B=\sqrt{\frac{n}{8k(k+2)}}\left\{\frac{1}{n}\sum _{i=1}^n{\left[({\mathbf{x}}_i-\overline{\mathbf{x}}{)}^{\mathrm{T}}{\hat{\mathbf{\Sigma }}}^{-1}({\mathbf{x}}_i-\overline{\mathbf{x}})\right]}^2-k(k+2)\right\}\end{array}}$

Under the null hypothesis of multivariate normality, the statistic A will have approximately a chi-squared distribution with ⁠1/6⁠⋅k(k + 1)(k + 2) degrees of freedom, and B will be approximately standard normal N(0,1).

Mardia's kurtosis statistic is skewed and converges very slowly to the limiting normal distribution. For medium size samples ${\displaystyle (50\le n<400)}$, the parameters of the asymptotic distribution of the kurtosis statistic are modified For small sample tests (${\displaystyle n<50}$) empirical critical values are used. Tables of critical values for both statistics are given by Rencher for k = 2, 3, 4.

Mardia's tests are affine invariant but not consistent. For example, the multivariate skewness test is not consistent against symmetric non-normal alternatives.

BHEP test

The BHEP test computes the norm of the difference between the empirical characteristic function and the theoretical characteristic function of the normal distribution. Calculation of the norm is performed in the L²(μ) space of square-integrable functions with respect to the Gaussian weighting function ${\displaystyle {\mu }_{\beta }(\mathbf{t})=(2\pi {\beta }^2{)}^{-k/2}{e}^{-|\mathbf{t}{|}^2/(2{\beta }^2)}}$. The test statistic is

${\displaystyle \begin{array}{cc}{T}_{\beta }& ={\int }_{{\mathbb{R}}^k}{\left|\frac{1}{n}\sum _{j=1}^n{e}^{i{\mathbf{t}}^{\mathrm{T}}{\hat{\mathbf{\Sigma }}}^{-1/2}({\mathbf{x}}_j-\overline{\mathbf{x})}}-e^{-|\mathbf{t}{|}^2/2}\right|}^2{\mathit{\mu }}_{\beta }(\mathbf{t})d\mathbf{t}\\ & =\frac{1}{n^2}\sum _{i,j=1}^n{e}^{-\frac{{\beta }^2}{2}({\mathbf{x}}_i-{\mathbf{x}}_j{)}^{\mathrm{T}}{\hat{\mathbf{\Sigma }}}^{-1}({\mathbf{x}}_i-{\mathbf{x}}_j)}-\frac{2}{n(1+{\beta }^2{)}^{k/2}}\sum _{i=1}^n{e}^{-\frac{{\beta }^2}{2(1+{\beta }^2)}({\mathbf{x}}_i-\overline{\mathbf{x}}{)}^{\mathrm{T}}{\hat{\mathbf{\Sigma }}}^{-1}({\mathbf{x}}_i-\overline{\mathbf{x}})}+\frac{1}{(1+2{\beta }^2{)}^{k/2}}\end{array}}$

The limiting distribution of this test statistic is a weighted sum of chi-squared random variables.

A detailed survey of these and other test procedures is available.

Computational methods

Drawing values from the distribution

A widely used method for drawing (sampling) a random vector x from the N-dimensional multivariate normal distribution with mean vector μ and covariance matrix Σ works as follows:

1. Find any real matrix A such that AA^T = Σ. When Σ is positive-definite, the Cholesky decomposition is typically used because it is widely available, computationally efficient, and well known. If a rank-revealing (pivoted) Cholesky decomposition such as LAPACK's dpstrf() is available, it can be used in the general positive-semidefinite case as well. A slower general alternative is to use the matrix A = UΛ^1/2 obtained from a spectral decomposition Σ = UΛU⁻¹ of Σ.
2. Let z = (z₁, ..., z_N)^T be a vector whose components are N independent standard normal variates (which can be generated, for example, by using the Box–Muller transform).
3. Let x be μ + Az. This has the desired distribution due to the affine transformation property.