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Understand normalized squared euclidean distance?

Posted 7 years ago

Hi everyone,

I try to understand the normalized squared euclidean distance formula from Wolfram documentation:

1/2*Norm[(u-Mean[u])-(v-Mean[v])]^2/(Norm[u-Mean[u]]^2+Norm[v-Mean[v]]^2)

I searched this formula on web but I did not found that anywhere. Someone can explain me how this formula is derived?

POSTED BY: Luca Zenesini
6 Replies

Note that it is a DistanceFunction option for ImageDistance. Maybe that helps some to see the context where it is used.

The relation to SquaredEuclideanDistance is:

NormalizedSquaredEuclideanDistance[x, y] ==
   (1/2) SquaredEuclideanDistance[x - Mean[x], y - Mean[y]]/
      (Norm[x - Mean[x]]^2 + Norm[y - Mean[y]]^2) 

So we see it is "normalized" "squared euclidean distance" between the "difference of each vector with its mean". I guess that was too long for a function name.. In any case the note under properties and relations ".. includes a squared Euclidean distance scaled by norms" makes little sense.

POSTED BY: george jefferson
Posted 7 years ago

Thanks for your reply, your definition of the formula has perfect sense for me, however I would like to understand whether and how such formula is derived (e.g. what is the meaning about 1/2 at the beggining of the formula?).

POSTED BY: Luca Zenesini

The 1/2 is just there such that the answer is bounded between 0 and 1, rather than 0 and 2.

POSTED BY: Sander Huisman
Posted 7 years ago

Ok you are right, I got confused by something trivial.

POSTED BY: Luca Zenesini

This is also asked elsewhere:

https://stats.stackexchange.com/questions/136232/definition-of-normalized-euclidean-distance

But I don't think an answer exists anywhere. I spent a fair amount of time looking for it myself. Ideally a function like this should directly point to a commonly known function that I can read about somewhere.

I've forwarded a report to the developer of this asking that clarification be added to the documentation about this function.

POSTED BY: Sean Clarke
Posted 7 years ago

I had already seen that answer some time ago but unfortunately its not enough exhaustive for me, however thanks for your report to the developer.

POSTED BY: Luca Zenesini
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