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I'm a biologist trying to understand math. It would help me out no end if somebody could clarify my misunderstandings. Question 1 is causing me the most grief.

The image below is taken from the paper 'Fast Gauss Transforms based on a High Order Singular Value Decomposition for Nonlinear Filtering' by Mittelman and Miller, 2007. My questions refer to Equation 6, which shows a weighted sum of Gaussian kernels.

1) In the expressions for $s$ and $\xi$, why isn't $\Lambda^{1/2}$ inverted, i.e. $\Lambda^{-1/2}$? My reasoning is this: if $\Sigma = V^T \Lambda V$ then $\Sigma^{-1} = V^T \Lambda^{-1} V = V^T \Lambda^{-1/2} \Lambda^{-1/2} V $ (note that V is orthogonal so that $V^{-1} = V^T$). The transformation of $x$ and $f(x)$ in $s$ and $\xi$, respectively, looks to me like a whitening transformation (specifically Mahalanobis whitening), where a vector of random variables with a known covariance matrix is transformed into a set of new variables whose covariance is the identity matrix, meaning that they are uncorrelated and each have variance 1. However, this would require us to transform $x$ and $f(x)$ by pre-multiplying them with $\Lambda^{-1/2} V$, as derived above, not $\Lambda^{1/2} V$, to normalize the distance (i.e., compute the Mahalonobis distance)

2) Why is $\begin{vmatrix} \Sigma \end{vmatrix} ^{1/2}$ not included in the normalizing term $\kappa$ given that it forms a part of the normalization constant for a multivariate gaussian?

3) Is the factorization of the covariance matrix simply eigendecomposition? I normally see this written as $\Sigma = V \Lambda V^T$ not $\Sigma = V^T \Lambda V$ so I wasn't sure.

Thank you so much in advance!


excerpt from Mittelman and Miller, 2007

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  • $\begingroup$ Hi and welcome to the site, please learn MathJax typesetting. It is similar to the math typesetting in for example LaTeX. $\endgroup$ – mathreadler Feb 17 '18 at 22:13
  • $\begingroup$ Isn't a downvote a little harsh so early? This is rather good a first question compared to many other I've seen. $\endgroup$ – mathreadler Feb 17 '18 at 22:18
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    $\begingroup$ for (3) since $\Sigma$ is symmetric, $V$ or $V^T$ doesnt matter since they are inverses of eachother $\endgroup$ – dunno Feb 17 '18 at 22:40
  • $\begingroup$ This page is hard to understand on its own. We are missing the description of what is going on, including equation (1). Anyway, suppose the state noise in (1) is large (or small). Then Λ would be large (or small). It appears that this will make the probability in (6) small (or large). Reversing 1/2 to -1/2 would invert that relationship. Which way makes more sense, given the meanings of the quantities involved? $\endgroup$ – Matt Dec 11 '18 at 0:11

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