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<math>\mu = \mathcal{E}[x] =\int\limits_{-\infty}^{\infty} xp(x)\, dx</math>         
 
<math>\mu = \mathcal{E}[x] =\int\limits_{-\infty}^{\infty} xp(x)\, dx</math>         
 +
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and where the expected squared deviation or ''variance'' is
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<math>\sigma^2 = \mathcal{E}[(x- \mu)^2] =\int\limits_{-\infty}^{\infty} (x- \mu)^2 p(x)\, dx</math>       
 
</div>
 
</div>

Revision as of 17:13, 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 $

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Sees the importance of signal filtering in medical imaging

Dhruv Lamba, BSEE2010