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Thread: Hellinger distance

  1. #1
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    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.
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  2. #2
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    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)?
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  3. #3
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    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 ?

    Thanks in advance. regards
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  4. #4
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    Re: Hellinger distance

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