# AP Statistics Curriculum 2007 MultivariateNormal

## Contents

## 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*=*a*_{1}*X*_{1}+ … +*a*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._{k}X_{k}

- 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: \[Z = \begin{cases} 0,& |X| \leq 1.33,\\ 2X,& |X| > 1.33.\end{cases}\]

### Applications

### Problems

### References

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

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

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