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.