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Let $S$ be the covariance matrix of some data set. $S^{-1}$ is the inverse covariance matrix, also called the precision matrix.

Question: In practice, then, what does $S^{-1}x$ mean for a data point $x$?

Question 2: What does $x^TS^{-1}x$ mean, then?

Do these have attractive interpretations?

ttnphns
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Lepidopterist
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    Question (2) is answered at http://stats.stackexchange.com/questions/62092 (on the Mahalanobis distance). As to the first question, is there a particular context in which you have seen $S^{-1}x$ used that would demand an interpretation? – whuber Oct 03 '14 at 21:41
  • Thanks, that's interesting. Maybe the first question comes from nowhere, I remember seeing these somewhere and wanted to ask about them. Though I clearly remember that I saw the second expression *without* the square root, which I suppose indicates that someone considered the squared Mahalanobis distance. I don't know if that's strange or not. – Lepidopterist Oct 03 '14 at 23:01
  • It's actually pretty natural. Often squared distances are easier to deal with than the distances themselves: there are no square roots and in either case the relative order of the distances is the same as that of their squares. I'm not challenging the applicability or interest of the first question, but it would be nice to know of at least one context in which the expression $S^{-1}x$ occurs in a meaningful way. – whuber Oct 03 '14 at 23:04
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    It's a bit more natural to think in terms of $S^{-1/2} x$, because then if $X$ has mean $0$ and covariance matrix $S$ we have $\mbox{Var}(S^{-1/2} X) = S^{-1/2} S S^{-1/2} = \mathbf I$ where $\mathbf I$ is the identity. So multiplication by $S^{-1/2}$ is akin to dividing by the standard deviation for scalar random variables. – guy Oct 04 '14 at 02:46

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