# Hellinger distance

• Oct 7th 2012, 12:20 PM
ghali
Hellinger distance
Hi all,
I would appreciate if someone can help me to prove the following proposition.

Let$\displaystyle$P$$and \displaystyle Q$$ be arbitrary distributions over $\displaystyle$\Omega = \{w_1,w_2,... ,w_n\}.let $\displaystyle ||x|| = || (x_1, x_2, ..., x_n) || = \sqrt{\varSigma_{i=1}^n {x_i}^2$ denote the usual Euclidean norm. how we can prove that the Hellinger distance $\displaystyle$h(P,Q)$$is the norm of vector with \displaystyle i$$th coordinate equal to the difference of the square roots of the probabilities of seeing $\displaystyle$w_i$$according to \displaystyle P$$ and
$\displaystyle$Q$$normalized by dividing by \displaystyle \sqrt{2}$$. that is prove $\displaystyle h(P,Q) = || \sqrt{\frac {P(w_1)}{2} - \frac {Q(w_1)}{2}},... ,\sqrt{\frac {P(w_n)}{2} - \frac {Q(w_n)}{2}}||.$

• Oct 8th 2012, 01:19 AM
chiro
Re: Hellinger distance
Hey ghali.

Do you have a distributional definition of this? Like for example a conditional distribution like P(W=w|P=p)?
• Oct 8th 2012, 02:43 AM
ghali
Re: Hellinger distance
Hi chiro,
Thank you very much for your reply. Actually we do not have conditional distribution like P(W=w|P=p). dose it make the proof hard ?