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| − | where '''x''' is a ''d''-component column vector, '''μ''' is the ''d''-component mean vector, '''Σ''' is the ''d''-by-''d'' covariance matrix, and '''|Σ| and '''Σ<sup>-1</sup>''' are its determinant and inverse respectively. Also, ('''x -&mu''')<sup>t</sup> denotes the transpose of ('''x -&mu'''). | + | where '''x''' is a ''d''-component column vector, '''μ''' is the ''d''-component mean vector, '''Σ''' is the ''d''-by-''d'' covariance matrix, and '''|Σ|''' and '''Σ<sup>-1</sup>''' are its determinant and inverse respectively. Also, ('''x -&mu''')<sup>t</sup> denotes the transpose of ('''x -&mu'''). |
and | and | ||
Revision as of 17:45, 4 April 2013
Discriminant Functions For The Normal Density
Lets begin with the continuous univariate normal or Gaussian density.
$ f_x = \frac{1}{\sqrt{2 \pi} \sigma} \exp \left [- \frac{1}{2} \left ( \frac{x - \mu}{\sigma} \right)^2 \right ] $
for which the expected value of x is
$ \mu = \mathcal{E}[x] =\int\limits_{-\infty}^{\infty} xp(x)\, dx $
and where the expected squared deviation or variance is
$ \sigma^2 = \mathcal{E}[(x- \mu)^2] =\int\limits_{-\infty}^{\infty} (x- \mu)^2 p(x)\, dx $
The univariate normal density is completely specified by two parameters; its mean μ and variance σ2. The function fx can be written as N(μ,σ) which says that x is distributed normally with mean μ and variance σ2.
For the multivariate normal density in d dimensions, fx is written as
$ f_x = \frac{1}{(2 \pi)^ \frac{d}{2} |\boldsymbol{\Sigma}|^\frac{1}{2}} \exp \left [- \frac{1}{2} (\mathbf{x} -\boldsymbol{\mu})^t\boldsymbol{\Sigma}^{-1} (\mathbf{x} -\boldsymbol{\mu}) \right] $
where x is a d-component column vector, μ is the d-component mean vector, Σ is the d-by-d covariance matrix, and |Σ| and Σ-1 are its determinant and inverse respectively. Also, (x -&mu)t denotes the transpose of (x -&mu).
and
$ \boldsymbol{\Sigma} = \mathcal{E} \left [(\mathbf{x} - \boldsymbol{\mu})(\mathbf{x} - \boldsymbol{\mu})^t \right] = \int(\mathbf{x} - \boldsymbol{\mu})(\mathbf{x} - \boldsymbol{\mu})^t p(\mathbf{x})\, dx $
