Difference between revisions of "AP Statistics Curriculum 2007 MultivariateNormal"
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===Bivariate (2D) case=== | ===Bivariate (2D) case=== | ||
+ | : See the SOCR Bivariate Normal Distribution [[SOCR_BivariateNormal_JS_Activity| Activity]] and corresponding [http://socr.ucla.edu/htmls/HTML5/BivariateNormal/ Webapp]. | ||
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In 2-dimensions, the nonsingular bi-variate Normal distribution with (<math>k=rank(\Sigma) = 2</math>), the probability density function of a (bivariate) vector (X,Y) is | In 2-dimensions, the nonsingular bi-variate Normal distribution with (<math>k=rank(\Sigma) = 2</math>), the probability density function of a (bivariate) vector (X,Y) is | ||
: <math> | : <math> | ||
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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: | 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 X ~ N(0,1). | ||
− | : Let <math>Y = \begin{cases} X,& |X| > 1. | + | : Let <math>Y = \begin{cases} X,& |X| > 1.33,\\ |
− | -X,& |X| \leq 1. | + | -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: | ||
+ | : <math>Z = \begin{cases} 0,& |X| \leq 1.33,\\ | ||
+ | 2X,& |X| > 1.33.\end{cases}</math> | ||
− | + | ===Applications=== | |
+ | [[SOCR_EduMaterials_Activities_2D_PointSegmentation_EM_Mixture| This SOCR activity demonstrates the use of 2D Gaussian distribution, expectation maximization and mixture modeling for classification of points (objects) in 2D]]. | ||
===[[EBook_Problems_MultivariateNormal|Problems]]=== | ===[[EBook_Problems_MultivariateNormal|Problems]]=== | ||
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* SOCR Home page: http://www.socr.ucla.edu | * SOCR Home page: http://www.socr.ucla.edu | ||
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Latest revision as of 13:43, 3 March 2020
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 = a1X1 + … + akXk 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 Rk, 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 Xi 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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