6
$\begingroup$

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?

$\endgroup$
0

1 Answer 1

3
$\begingroup$

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.