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Multivariate normal distribution - Partitioned vectors

Let X be a Kx1 multivariate normal random vector with mean mu and covariance matrix V. In this lecture we present some useful facts about partitionings of X, that is, about subdivisions of X into two sub-vectors $X_{a}$ and $X_{b}$ such that[eq1]where $X_{a}$ and $X_{b}$ have dimensions $K_{a}	imes 1$ and $K_{b}	imes 1$ respectively and $K_{a}+K_{b}=K$.

Table of Contents

Notation

In what follows, we will denote by:

Given this notation, it follows that[eq7]and[eq8]

Normality of the sub-vectors

The following proposition states a first (trivial) fact about the two sub-vectors.

Proposition Both $X_{a}$ and $X_{b}$ have a multivariate normal distribution, i.e.,[eq9]

Proof

The random vector $X_{a}$ can be written as a linear transformation of X:[eq10]where A is a $K_{a}	imes K$ matrix whose entries are either zero or one. Thus, $X_{a}$ has a multivariate normal distribution, because it is a linear transformation of the multivariate normal random vector X and multivariate normality is preserved by linear transformations (see the lecture entitled Linear combinations of normal random variables). By the same token, also $X_{b}$ has a multivariate normal distribution, because it can be written as a linear transformation of X:[eq11]where $B$ is a $K_{b}	imes K$ matrix whose entries are either zero or one.

This, of course, implies that any sub-vector of X is multivariate normal when X is multivariate normal.

Independence of the sub-vectors

The following proposition states a necessary and sufficient condition for the independence of the two sub-vectors.

Proposition $X_{a}$ and $X_{b}$ are independent if and only if $V_{ab}=0$.

Proof

$X_{a}$ and $X_{b}$ are independent if and only if their joint moment generating function is equal to the product of their individual moment generating functions (see the lecture entitled Joint moment generating function). Since $X_{a}$ is multivariate normal, its joint moment generating function is[eq12]The joint moment generating function of $X_{b}$ is[eq13]The joint moment generating function of $X_{a}$ and $X_{b}$, which is just the joint moment generating function of X, is[eq14]from which it is obvious that [eq15] if and only if $V_{ab}=0$.

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