# AP Statistics Curriculum 2007 MultivariateNormal

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-X,& |X| \leq 1.33.\end{cases}</math> | -X,& |X| \leq 1.33.\end{cases}</math> | ||

- | + | 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: | 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: |

## Revision as of 02:50, 25 January 2011

## 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 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 , the linear combination (which is univariate random variable) 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*=*A**Z*+ μ. 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

- 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

- , 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* = *r**a**n**k*(Σ) = 2), the probability density function of a (bivariate) vector (X,Y) is

where *ρ* is the correlation between *X* and *Y*. In this case,

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

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:

### 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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