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Covariate shift gives a joint-law density

Statement

Assume AC. Let $P_{S}, P_{T}$ be source and target joint laws with input marginals $μ_{S}, μ_{T}$ and specified conditional label kernels $K_{S}, K_{T}$ as in the definition of source/target distributions. Suppose $μ_{T} ≪ μ_{S}$ and there is a $μ_{T}$ -null set outside of which $K_{S} (x, \cdot) = K_{T} (x, \cdot)$ . Let $w$ be any nonnegative measurable representative of $d μ_{T} / d μ_{S}$ . Then for every product-measurable set $E \subseteq X \times Y$ , $P_{T} (E) = \int_{E} w (x) P_{S} (d (x, y)) .$ Consequently $P_{T} ≪ P_{S}$ and $w \circ π_{X}$ is a representative of the likelihood-ratio density of $P_{T}$ with respect to $P_{S}$ .

Statement references

Sugiyama, §3.1, book pp. 930–931: the density-based covariate-shift identity that this lemma establishes for general measurable spaces using supplied conditional kernels.

Facts & Assumptions

Given: AC; source and target joint probability laws on the common product measurable space; their input marginals and conditional label kernels; $μ_{T} ≪ μ_{S}$ ; and a nonnegative representative $w$ of $d μ_{T} / d μ_{S}$ .

[A1]

The Axiom of Choice: AC is assumed for the Radon-Nikodym density selection used below.

[F1]

Source and target joint laws, risks, and covariate shift: for every nonnegative product-measurable $g$ , $\int g d P_{D} = \int_{X} \int_{Y} g (x, y) K_{D} (x, d y) μ_{D} (d x)$ .

[F2]

Likelihood densities and Likelihood ratio as a Radon–Nikodym derivative: for the two-law experiment with parameter set ${S, T}$ dominated by $μ_{S}$ , AC supplies a finite-valued nonnegative representative $w_{0}$ of the Radon–Nikodym density class of $μ_{T}$ with respect to $μ_{S}$ ; every other representative $w$ of that class agrees with $w_{0}$ $μ_{S}$ -almost everywhere. Since these are probability measures, the constant sequence $X_{n} = X$ is a common finite-measure exhaustion.

[F3]

Likelihood-ratio change of measure: if $μ_{T} ≪ μ_{S}$ , then $\int q d μ_{T} = \int q w d μ_{S}$ for every nonnegative measurable $q$ .

[F4]

The measure with density $f$ relative to $μ$ : a nonnegative measurable density $r$ relative to a measure $P$ defines the measure $r d P$ by $E \mapsto \int_{E} r d P$ .

[F5]

A measurable function between measurable spaces: a map is measurable when preimages of measurable sets are measurable.

[F6]

Closure properties of measurable functions used by the integral: if $f \geq 0$ is measurable and $E$ is measurable, then $f 1_{E}$ is measurable.

[F7]

Pi-systems and The product sigma-algebra and its finite iterates: measurable rectangles form a $π$ -system generating $Σ_{X} \otimes Σ_{Y}$ .

[F8]

Finite measures agreeing on a generating pi-system and on the whole space are equal: finite measures with equal total mass that agree on a generating $π$ -system agree on the generated sigma-algebra.

[F9]

Source and target joint laws, risks, and covariate shift: under covariate shift, the specified kernels agree as probability measures outside a $μ_{T}$ -null set.

[F10]

Measure kernel and probability kernel: for each measurable $B \subseteq Y$ , $x \mapsto K_{D} (x, B)$ is measurable.

[F11]

Measure kernel and probability kernel: a probability kernel satisfies $K_{D} (x, Y) = 1$ for every $x$ .

[F12]

A nonnegative integral over a null set vanishes: the integral of a nonnegative measurable function over a measurable null set is zero.

[F13]

Monotonicity and nonnegative homogeneity of the nonnegative integral: multiplication by a finite nonnegative scalar is homogeneous for the nonnegative integral; the zero scalar case gives integral zero.

[F14]

Source and target joint laws, risks, and covariate shift: the input marginal satisfies $μ_{D} (A) = P_{D} (A \times Y)$ for every $A \in Σ_{X}$ .

Proof

1.1A1F1F2F3F4F5F7F11F13given

Let $π_{X} (x, y) = x$ . For each $A \in Σ_{X}$ , $π_{X}^{- 1} (A) = A \times Y \in Σ_{X} \otimes Σ_{Y}$ , so $π_{X}$ is measurable by [F5] and [F7]. For every Borel $C \subseteq [0, + \infty]$ , $(w_{0} \circ π_{X})^{- 1} (C) = π_{X}^{- 1} (w_{0}^{- 1} (C))$ is product-measurable. Define $ν_{0} : = (w_{0} \circ π_{X}) d P_{S}$ ; [F4] makes $ν_{0}$ a measure. Since $K_{S} (x, Y) = 1$ by [F11], [F1] and [F13] give $ν_{0} (X \times Y) = \int w_{0} d μ_{S}$ . Applying [F3] to $q = 1$ yields $\int w_{0} d μ_{S} = \int 1 d μ_{T} = 1$ . Thus $ν_{0}$ and $P_{T}$ are finite measures of equal total mass.

1.2F1F3F5F6F7F9F10F13step 1.1given

For $A \in Σ_{X}$ and $B \in Σ_{Y}$ , the function $g (x, y) = 1_{A \times B} (x, y) w_{0} (x)$ is nonnegative and product-measurable by [F5]–[F7]. By [F10], $x \mapsto K_{S} (x, B)$ is measurable, and by [F6] the function $q (x) = 1_{A} (x) K_{S} (x, B)$ is measurable. Since $w_{0} (x)$ is finite, [F13] pulls this scalar through the section integral in [F1]. Applying [F3] to $q$ gives $ν_{0} (A \times B) = \int_{A} w_{0} (x) K_{S} (x, B) μ_{S} (d x) = \int_{A} K_{S} (x, B) μ_{T} (d x) = \int_{A} K_{T} (x, B) μ_{T} (d x) = P_{T} (A \times B) .$ The middle equality uses the $μ_{T}$ -almost-everywhere equality of the specified kernels, and the last equality uses their target disintegration [F1].

2.1F2F4F7F8step 1.1step 1.2

The measurable rectangles form the generating $π$ -system in [F7], so [F8] and steps 1.1–1.2 give $ν_{0} = P_{T}$ on the entire product sigma-algebra. Since [F2] identifies $w_{0}$ as a representative of $d μ_{T} / d μ_{S}$ , the setwise equality makes $w_{0} \circ π_{X}$ a Radon–Nikodym representative of $P_{T}$ with respect to $P_{S}$ .

3.1F2F4F5F7F12F14step 2.1given∎

Let $w$ be the arbitrary representative in the Statement. The same preimage argument as in step 1.1, using [F5] and [F7], shows $w \circ π_{X}$ is measurable. By [F2], there is a measurable $N \in Σ_{X}$ with $μ_{S} (N) = 0$ outside of which $w = w_{0}$ . By [F14], $P_{S} (N \times Y) = 0$ . For each measurable $E \subseteq X \times Y$ , the two density integrands agree on $E ∖ (N \times Y)$ , and both integrals over $E \cap (N \times Y)$ vanish by [F12]. Hence $\int_{E} w (x) d P_{S} = \int_{E} w_{0} (x) d P_{S} = P_{T} (E)$ . In particular, if $P_{S} (E) = 0$ , [F12] gives $P_{T} (E) = 0$ , so $P_{T} ≪ P_{S}$ .

Depends on

Used by

Importance weighting under covariate shift Theorem

Dependency tree · two levels

53 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

Masashi Sugiyama, Learning under Non-stationarity: Covariate Shift Adaptation by Importance Weighting, Handbook of Computational Statistics, 2nd ed., Chapter 31 (2012), §3.1 (standard reference, not scraped)