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<div style="margin-left: 25em;"> | <div style="margin-left: 25em;"> | ||
<math>\sigma^2 = \mathcal{E}[(x- \mu)^2] =\int\limits_{-\infty}^{\infty} (x- \mu)^2 p(x)\, dx</math> | <math>\sigma^2 = \mathcal{E}[(x- \mu)^2] =\int\limits_{-\infty}^{\infty} (x- \mu)^2 p(x)\, dx</math> | ||
| + | </div> | ||
| + | |||
| + | For the multivariate normal density in ''d'' dimensions, f<sub>x</sub> is written as | ||
| + | |||
| + | <div style="margin-left: 25em;"> | ||
| + | <math>f_x = \frac{1}{(2 \pi)^ \frac{d}{2} |\boldsymbol{\Sigma}|^\frac{1}{2}} \exp [- \frac{1}{2} ('''x''' -\boldsymbol{\mu})^t\boldsymbol{\Sigma}^-1 ('''x''' -\boldsymbol{\mu})] </math> | ||
</div> | </div> | ||
Revision as of 17:37, 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 $
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 [- \frac{1}{2} ('''x''' -\boldsymbol{\mu})^t\boldsymbol{\Sigma}^-1 ('''x''' -\boldsymbol{\mu})] $
