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Small L² log-density perturbations can remain far in Hellinger distance

Statement

For integers $k \geq 1$ , on $([0, 1], B, Lebesgue)$ let $p_{0} (x) = 1$ , $A_{k} = [0, k^{- 4}]$ , $w_{k} = k 1_{A_{k}}$ , and $p_{k} = e^{w_{k}} / \int_{0}^{1} e^{w_{k}} d x$ . Then $∥ w_{k} ∥_{L^{2}} \to 0$ but $h (P_{k}, P_{0}) \to \sqrt{2}$ . Thus the $L^{2}$ norm of a log-density perturbation need not control the induced Hellinger distance.

Facts & Assumptions

Given: The stated finite-measure space and functions.

[F1]

Root-L2 Hellinger distance between density laws: $h^{2} (P_{k}, P_{0}) = \int_{0}^{1} (\sqrt{p_{k}} - 1)^{2} d x = 2 - 2 \int_{0}^{1} \sqrt{p_{k}} d x$ .

Verification

1.1givenalgebra

Since $∣ A_{k} ∣ = k^{- 4}$ , $∥ w_{k} ∥_{2}^{2} = k^{2} ∣ A_{k} ∣ = k^{- 2} \to 0$ . The normalizer is $Z_{k} = 1 - k^{- 4} + e^{k} k^{- 4}$ , so $p_{k}$ is a positive unit-mass density.

1.2F1givenalgebra

The Hellinger affinity is $\int_{0}^{1} \sqrt{p_{k}} d x = \frac{1 - k^{- 4} + k^{- 4} e^{k / 2}}{\sqrt{Z_{k}}} .$ Since $Z_{k} \geq e^{k} / k^{4}$ , this is at most $k^{2} e^{- k / 2} + k^{- 2}$ . The exponential series gives $e^{k / 2} \geq (k / 2)^{3} / 3! = k^{3} / 48$ , hence $k^{2} e^{- k / 2} \leq 48 / k \to 0$ .

2.1F1step 1.2algebra∎

By [F1], $h^{2} (P_{k}, P_{0}) \to 2$ , and nonnegativity of Hellinger distance gives $h (P_{k}, P_{0}) \to \sqrt{2}$ .

Depends on

Used by

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Sources

A. W. van der Vaart and J. H. van Zanten, Rates of contraction of posterior distributions based on Gaussian process priors, §3.1 (standard reference, not scraped)
Bas J. K. Kleijn, Aad W. van der Vaart and Harry van Zanten, Lectures on Nonparametric Bayesian Statistics, §12.2 (standard reference, not scraped)