AP Statistics Curriculum 2007 MultivariateNormal

EBook - Multivariate Normal Distribution

The multivariate normal distribution, or multivariate Gaussian distribution, is a generalization of the univariate (one-dimensional) normal distribution to higher dimensions. A random vector is said to be multivariate normally distributed if every linear combination of its components has a univariate normal distribution. The multivariate normal distribution may be used to study different associations (e.g., correlations) between real-valued random variables.

Definition

In k-dimensions, a random vector \(X = (X_1, \cdots, X_k)\) is multivariate normally distributed if it satisfies any one of the following equivalent conditions (Gut, 2009):

• Every linear combination of its components Y = a₁X₁ + … + a_kX_k is normally distributed. In other words, for any constant vector \(a\in R^k\), the linear combination (which is univariate random variable) \(Y = a^TX = \sum_{i=1}^{k}{a_iX_i}\) has a univariate normal distribution.

• There exists a random ℓ-vector Z, whose components are independent normal random variables, a k-vector μ, and a k×ℓ matrix A, such that \(X = AZ + \mu\). Here ℓ is the rank of the variance-covariance matrix.

• There is a k-vector μ and a symmetric, nonnegative-definite k×k matrix Σ, such that the characteristic function of X is

\[ \varphi_X(u) = \exp\Big( iu^T\mu - \tfrac{1}{2} u^T\Sigma u \Big). \]

• When the support of X is the entire space R^k, there exists a k-vector μ and a symmetric positive-definite k×k variance-covariance matrix Σ, such that the probability density function of X can be expressed as

\[ f_X(x) = \frac{1}{ (2\pi)^{k/2}|\Sigma|^{1/2} } \exp\!\Big( {-\tfrac{1}{2}}(x-\mu)'\Sigma^{-1}(x-\mu) \Big) \], where |Σ| is the determinant of Σ, and where (2π)^k/2|Σ|^1/2 = |2πΣ|^1/2. This formulation reduces to the density of the univariate normal distribution if Σ is a scalar (i.e., a 1×1 matrix).

If the variance-covariance matrix is singular, the corresponding distribution has no density. An example of this case is the distribution of the vector of residual-errors in the ordinary least squares regression. Note also that the X_i are in general not independent; they can be seen as the result of applying the matrix A to a collection of independent Gaussian variables Z.

Bivariate (2D) case

In 2-dimensions, the nonsingular bi-variate Normal distribution with (\(k=rank(\Sigma) = 2\)), the probability density function of a (bivariate) vector (X,Y) is \[ f(x,y) = \frac{1}{2 \pi \sigma_x \sigma_y \sqrt{1-\rho^2}} \exp\left( -\frac{1}{2(1-\rho^2)}\left[ \frac{(x-\mu_x)^2}{\sigma_x^2} + \frac{(y-\mu_y)^2}{\sigma_y^2} - \frac{2\rho(x-\mu_x)(y-\mu_y)}{\sigma_x \sigma_y} \right] \right), \] where ρ is the correlation between X and Y. In this case, \[ \mu = \begin{pmatrix} \mu_x \\ \mu_y \end{pmatrix}, \quad \Sigma = \begin{pmatrix} \sigma_x^2 & \rho \sigma_x \sigma_y \\ \rho \sigma_x \sigma_y & \sigma_y^2 \end{pmatrix}. \]

In the bivariate case, the first equivalent condition for multivariate normality is less restrictive: it is sufficient to verify that countably many distinct linear combinations of X and Y are normal in order to conclude that the vector \( [ X, Y ] ^T\) is bivariate normal.

Properties

Normally distributed and independent

If X and Y are normally distributed and independent, this implies they are "jointly normally distributed", hence, the pair (X, Y) must have bivariate normal distribution. However, a pair of jointly normally distributed variables need not be independent - they could be correlated.

Two normally distributed random variables need not be jointly bivariate normal

The fact that two random variables X and Y both have a normal distribution does not imply that the pair (X, Y) has a joint normal distribution. A simple example is provided below:

Let X ~ N(0,1).

Let \(Y = \begin{cases} X,& |X| > 1.33,\\ -X,& |X| \leq 1.33.\end{cases}\)

Then, both X and Y are individually Normally distributed; however, the pair (X,Y) is not jointly bivariate Normal distributed (of course, the constant c=1.33 is not special, any other non-trivial constant also works).

Furthermore, as X and Y are not independent, the sum Z = X+Y is not guaranteed to be a (univariate) Normal variable. In this case, it's clear that Z is not Normal: \[Z = \begin{cases} 0,& |X| \leq 1.33,\\ 2X,& |X| > 1.33.\end{cases}\]

---

References

• Gut, A. (2009): An Intermediate Course in Probability, Springer 2009, chapter 5, ISBN 9781441901613.

---

• SOCR Home page: http://www.socr.ucla.edu

---

---

Translate this page:

(default)	Deutsch	Español	Français	Italiano	Português	日本語	България	الامارات العربية المتحدة	Suomi	इस भाषा में	Norge
한국어	中文	繁体中文	Русский	Nederlands	Ελληνικά	Hrvatska	Česká republika	Danmark	Polska	România	Sverige