0
$\begingroup$

This question already has an answer here:

I've read quite a few explanations on this topic, liking this one the most: https://mccormickml.com/2014/07/22/mahalanobis-distance/

formula

But there is still one thing I don't understand.

I understand that the inverse of the covariance would deal with transformation of the (x-mu) to the standard gaussian. What I don't get is how is x^T infected by it?

To me, it seems that the we then take the dot product of elements 2 and 3 of the product, we transform only the (x-mu), and the first vector (x-mu)_transposed stays the same. So in the end we would get the dot product between the original vector and the transformed one.

$\endgroup$

marked as duplicate by whuber distributions Oct 4 at 13:12

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

1
$\begingroup$

Maybe your understanding is helped by rewriting a part of the formula as:

$(x-\mu)^\top S^{-1} (x-\mu) = (S^{-1/2} (x -\mu))^\top S^{-1/2} (x-\mu)$

Then you can see that both $x$ and $x^\top$ are transformed in the same way.

$\endgroup$

Not the answer you're looking for? Browse other questions tagged or ask your own question.