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I read in a paper that the squared Hellinger distance between two densities $f$ and $g$

$H^2(f,g)=\frac{1}{2}\int \left(\sqrt{f(x)}-\sqrt{g(x)}\right)^2 dx$

is not a metric. I wonder if there is a nice counterexample showing this.

Thanks in advance.

1 Answers 1

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Here is an approach to create examples: You would like to contradict the triangle inequality, as the other properties of a metric are satisfied, so you are looking for densities $f,g,h$ such that $2(H^2(f,g) + H^2(g,h) < 2H^2(f,h)$, or equivalently by definition $\int f + g - 2\sqrt{fg} + \int g + h - 2\sqrt{gh} < \int f + h - 2 \sqrt{fh},$ which is equivalent to $\int 2g - 2(\sqrt{fg} + \sqrt{gh}) < \int -2\sqrt{fh}.$ Now you can construct examples using any densities $f,g,h$ satisfying the following: $fh = 0$ $\sqrt{fg} + \sqrt{gh} > g.$

For instance consider the domain $[0,1]$ and the densities $g(x) \equiv 1 $ $f(x) = 2 \mathcal I_{[0,1/2]}(x)$ $h(x) = 2 \mathcal I_{[1/2,1]}(x).$

$\mathcal I_A$ denotes the indicator function of a set $A$, being $1$ at points of $A$ and $0$ otherwise, so $\sqrt{fg} + \sqrt{gh} \equiv \sqrt{2} > 1 \equiv g$.

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    Yes it is. I do not see any obvious easy example, but you can construct examples out of the above one. Consider densities $f_n, g_n, h_n$ supported on $\mathbb R$ which converge to $f,g,h$ respectively. Then for big $n$ things will work.2012-12-19