Hellinger distance

Remarks

The value does not depend on the choice of $\lambda$. Expanding the square gives $H^2(P, Q) = 1 - \int \sqrt{p q} \, d\lambda$, so $0 \le H \le 1$. The map $H$ is a metric on the set of probability measures on $(\mathcal{X}, \mathcal{F})$. Some authors omit the factor $\frac{1}{2}$, which multiplies $H$ by $\sqrt{2}$. With the convention above, the total variation distance $\delta$ satisfies $H^2(P, Q) \le \delta(P, Q) \le \sqrt{2} \, H(P, Q)$.

Example: on $\set{0, 1}$ with counting measure, let $P$ give mass $\frac{1}{2}$ to each point and let $Q$ be the point mass at $1$. Then $\int \sqrt{pq} = \sqrt{0 \cdot \tfrac{1}{2}} + \sqrt{1 \cdot \tfrac{1}{2}} = 1/\sqrt{2}$, so $H^2(P, Q) = 1 - 1/\sqrt{2}$.

Non-example: the quantity $\frac{1}{2} \int (p - q)^2 \, d\lambda$, without square roots, is not the Hellinger distance. It changes with $\lambda$: replacing $\lambda$ by $2\lambda$ halves $p$ and $q$ and divides the quantity by $2$.