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Projection CLT for a two-sample U-statistic

Statement

Assume AC and the setup of Two-sample order-r U-statistic and first projections, $r$ is fixed, $m / N \to p \in (0, 1)$ , $θ (P, Q; h) = 0$ , and $0 < E_{P} [g_{X}^{2}] = σ_{X}^{2} < \infty$ and $0 < E_{Q} [g_{Y}^{2}] = σ_{Y}^{2} < \infty$ . If the fixed-order kernel has finite second moments under all mixed assignments, then

$\sqrt{m} U_{m, n} \Rightarrow N ⁣ (0, r^{2} (σ_{X}^{2} + \frac{p}{1 - p} σ_{Y}^{2})) .$

Facts & Assumptions

Given: Independent IID samples from $P$ and $Q$ , fixed $r$ , $m / N \to p \in (0, 1)$ , a zero parameter, and positive finite first-projection variances.

[F1]

Two-sample order-r U-statistic and first projections defines $U_{m, n}$ and its centered projections $g_{X}, g_{Y}$ .

[F2]

Second-moment order of the degenerate remainder gives $\sqrt{N} R_{m, n} \to 0$ in $L^{2}$ under the mixed second-moment premises.

[F3]

Lindeberg-Levy iid central limit theorem: under AC, each IID centered real sample mean with positive finite variance has its normalized normal limit.

[F4]

Characteristic functions under affine maps and independent sums, Characteristic function of a normal law, and Characteristic function criterion for weak convergence give the independent-sum normal limit.

[F5]

Slutsky's theorem for real random variables transfers a weak limit after adding a term converging to zero in probability.

[A1]

AC is assumed exactly for the CLT and characteristic-function suppliers in [F3]–[F4]. The Axiom of Choice

Proof

Given: The assumptions in the Statement. Prove the centered normal limit at scale $\sqrt{m}$ .

1.1F1F2givenalgebra

By [F1] and [F2], $\sqrt{m} U_{m, n} = r \frac{1}{\sqrt{m}} \sum_{i = 1}^{m} g_{X} (X_{i}) + r \sqrt{\frac{m}{n}} (\frac{1}{\sqrt{n}} \sum_{j = 1}^{n} g_{Y} (Y_{j})) + \sqrt{m} R_{m, n},$ and the final term tends to zero in probability because $m / N$ is bounded and $E [(\sqrt{N} R_{m, n})^{2}] \to 0$ .

1.2F1F3A1given

The summands $g_{X} (X_{i})$ are IID, centered, and have variance $σ_{X}^{2}$ ; [F3] gives $m^{- 1 / 2} \sum_{i} g_{X} (X_{i}) \Rightarrow N (0, σ_{X}^{2})$ . The $g_{Y} (Y_{j})$ are IID, centered, and have variance $σ_{Y}^{2}$ , so the same result gives $n^{- 1 / 2} \sum_{j} g_{Y} (Y_{j}) \Rightarrow N (0, σ_{Y}^{2})$ . The two sums are independent because the two samples are independent.

2.1F1F4step 1.1step 1.2algebra

Since $\sqrt{m / n} \to \sqrt{p / (1 - p)}$ , the characteristic function of the sum of the first two terms in step 1.1 is the product of the two independent characteristic functions. By [F4] this product converges to the characteristic function of a centered normal with variance $r^{2} (σ_{X}^{2} + (p / (1 - p)) σ_{Y}^{2})$ ; the positivity assumptions make that variance nonzero.

3.1F5step 1.1step 2.1∎

The first two terms in step 1.1 therefore have the stated weak limit. The last term converges to zero in probability by step 1.1, so [F5] gives the claimed limit for $\sqrt{m} U_{m, n}$ . All divisions are by positive finite variances, and no claim is made when either first-projection variance is zero.

Depends on

Used by

Studentized permutation validity under a weak two-sample parameter null Theorem

Dependency tree · two levels

52 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

E. Y. Chung and J. P. Romano, Asymptotically Valid and Exact Permutation Tests Based on Two-Sample U-Statistics, Stanford Technical Report 2011-09, Remark 3.1 (standard reference, not scraped)
Péter Major, Some Results About U-statistics, in On the Estimation of Multiple Random Integrals and U-Statistics, Lecture Notes in Mathematics 2079, Theorem 9.1 (standard reference, not scraped)