0

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.

Centar 15
  • 3
  • 1

1 Answers1

2

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.

nope
  • 604
  • 3
  • 8
  • Thank you so much! I've combined your answer with to figure everything out: https://www.visiondummy.com/2014/04/geometric-interpretation-covariance-matrix/ – Centar 15 Oct 04 '19 at 13:20