1
$\begingroup$

I am calculating Hellinger distance for different vectors. I initially assumed that the value returned by it in in the range of 0 to 1.

However for the following two vectors I received Hellinger score as 1.0488088481701514, which is > 1.

vector_1 = [0.0,0.5,0.7]
vector_2 = [1.0,0.0,0.0]

Now, I am curious to know the range of Hellinger distance values. Please explain me why that value exceeded 1.

$\endgroup$
  • 1
    $\begingroup$ Which implementation of Hellinger distance? Did you implement it yourself or are you using a package? $\endgroup$ – Mephy Sep 11 '17 at 13:51
1
$\begingroup$

It is bounded by unity, but your first vector does not encode a probability mass function, since 0.5 + 0.7 > 1.0. If the 0.7 had been 0.5 or the 0.5 a 0.3, the distance would have been 1.0 since the distributions are maximally separated, having no overlap.

$\endgroup$
  • $\begingroup$ Can you please let me know how to adjust my data according to probability mass function? Will it be solved if I normalize my feature vectors before calculating hellinger distance? $\endgroup$ – Smith Volka Sep 11 '17 at 23:39
  • 1
    $\begingroup$ The Hellinger distance applies to probability distributions. First make sure you have one. It does not make sense for arbitrary vectors. If normalizing your vector can be reasonably interpreted as producing a pmf then go ahead. $\endgroup$ – Emre Sep 12 '17 at 0:19
  • $\begingroup$ Thanks! What do you mean by producing a pmf? $\endgroup$ – Smith Volka Sep 12 '17 at 0:20
  • 1
    $\begingroup$ Probability mass function. Are you dealing with probabilities? $\endgroup$ – Emre Sep 12 '17 at 0:29
  • $\begingroup$ Yes, I actually have probabilities. However, as you have said, their sum is not equal to 1. Is there any better solution I can do instead of normalising? $\endgroup$ – Smith Volka Sep 12 '17 at 0:31

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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