ExampleConstruction: AI-generatedVerification: AI-generatedPipeline-generatedjudge pass (gpt-6-luna)

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A likelihood tilt can be a finite measure of mass above one

Example

On $X = {0, 1}$ with counting measure, let $p_{0} = (1 / 1000, 999 / 1000)$ , $p_{a} = (10^{- 6}, 999999 / 10^{6})$ , and $p_{b} = (1, 0)$ , where each pair lists the values at $0$ and $1$ . Give $a$ and $b$ prior probability $1 / 2$ each. Then $a$ is the unique KL minimizer and the tilt at $b$ has mass $1000$ .

Facts & Assumptions

Given: The finite density model and prior in the Example.

[A1]

Assume AC for the imported directional KL convention The Axiom of Choice. No additional choice is used.

[F1]

KL projection sets and relative KL neighborhoods: a KL projection minimizes $K (θ) = K L (P_{0} ∥ P_{θ})$ .

[F2]

Likelihood-tilted measures at a KL projection: for selected $θ^{*} = a$ , $Q_{b} (X) = E_{P_{0}} [p_{b} / p_{a}]$ .

[F3]

Likelihood tilts are sigma-finite measures: the likelihood tilt is a sigma-finite measure, whether or not its total mass is one.

Verification

1.1A1F1algebra

The density $p_{a}$ is positive at both sample points, so $K (a) = K L (P_{0} ∥ P_{a}) < \infty$ . At $b$ , $p_{b} (1) = 0$ while $p_{0} (1) > 0$ , so $P_{0} ≪̸ P_{b}$ and $K (b) = + \infty$ by [F1]. Therefore $a$ is the unique minimizer.

2.1F2F3algebra∎

Since $p_{b} (0) / p_{a} (0) = 10^{6}$ and $p_{b} (1) / p_{a} (1) = 0$ , $Q_{b} (X) = \frac{1}{1000} \cdot 10^{6} + \frac{999}{1000} \cdot 0 = 1000 > 1.$ The tilt is a finite nonprobability measure by [F2] and [F3].

Depends on

Used by

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Sources

B. J. K. Kleijn and A. W. van der Vaart, The Bernstein-Von-Mises Theorem under Misspecification, §3.2 (standard reference, not scraped)