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The hellinger distance for a univariate distribution is

$$ \ H(x) = 1 - \int \sqrt {f(x)g(x)}\; dx $$

I wish to use it for a bivariate distribution, by extending it to this form

$$ \ H(x) = 1 - \int\sqrt {f(x, y)g(x, y)} \;dx dy $$

Can someone comment on whether this is still a valid Hellinger distance?

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  • $\begingroup$ By "$H(x)$" do you perhaps mean $H(f,g)$? If not, then what is "$x$"? BTW, it seems to me that the definition of Hellinger distance makes no reference to the dimension. $\endgroup$ – whuber Mar 20 '13 at 15:27
  • $\begingroup$ Oops, you're right it should be H(f,g) $\endgroup$ – gamerx Mar 20 '13 at 16:31
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It's definitely valid. In fact, Hellinger distance between two measure $P$ and $Q$, which are dominated by the same (sigma-finite) measure $\lambda$, is defined as $$ H^2(P,Q) = \frac{1}{2}\int\left( \sqrt{\frac{dP}{d\lambda}} - \sqrt{\frac{dQ}{d\lambda}} \right)^2 \, d\lambda \, . $$ Therefore, you're considering the special case when $\lambda$ is two-dimensional Lebesgue measure, and the densities $f$ and $g$ are versions of the Radon-Nikodym derivatives $dP/d\lambda$ and $dQ/d\lambda$.

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  • $\begingroup$ Zen, are you aware of an implementation of the Hellinger distance for the N >= 2 cases? $\endgroup$ – mjs Dec 30 '19 at 17:17

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