
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 $\displaystyle h(P,Q) =  \sqrt{\frac {P(w_1)}{2}  \frac {Q(w_1)}{2}},... ,\sqrt{\frac {P(w_n)}{2}  \frac {Q(w_n)}{2}}.$
Thanks in advance.

Re: Hellinger distance
Hey ghali.
Do you have a distributional definition of this? Like for example a conditional distribution like P(W=wP=p)?

Re: Hellinger distance
Hi chiro,
Thank you very much for your reply. Actually we do not have conditional distribution like P(W=wP=p). dose it make the proof hard ?
Thanks in advance. regards

Re: Hellinger distance
Hint: What is the length of a unit vector plus a unit vector when both are orthogonal?