The squared Hellinger distance is, for two densities $f(x)$ and $g(x)$,

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

Using that $\int f(x)dx=\int g(x)dx=1$, we may write this as

$$HD^2(f,g)=1-\int \sqrt{f(x)g(x)}dx$$

The Wikipedia entry invokes the Cauchy-Schwarz inequality to show that $$0\leq HD^2(f,g)\leq1$$ My question: Is that necessary?

The first display shows $HD^2(f,g)$ to be an integral over the nonnegative function $\left[\sqrt{f(x)}-\sqrt{g(x)}\right]^2$, which will be nonnegative, too, so that $HD^2(f,g)\geq0$. The second display subtracts $\sqrt{f(x)g(x)}$ from one, i.e., the square root of the product of two densities, which are nonnegative, so that the integrand is again nonnegative, so that $HD^2(f,g)\leq1$.

What, if anything, is wrong with that reasoning?


1 Answer 1


Your derivation is correct. You don't need Cauchy-Schwarz here, unless you are looking for a simple application of it for illustrative purposes.


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