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A weak law for exchangeable arrays close to uncorrelated reference rows

Statement

For each integer $N \geq 2$ , let $(X_{N, i}, U_{N, i})_{i = 1}^{N}$ be a jointly exchangeable row of real random pairs: its law is invariant when the same permutation is applied to every index. Assume that $U_{N, 1}, \dots, U_{N, N}$ are pairwise uncorrelated, all have the law of a fixed square-integrable random variable $U$ , and $∥ X_{N, 1} - U_{N, 1} ∥_{2} : = (E ∣ X_{N, 1} - U_{N, 1} ∣^{2})^{1 / 2} ⟶ 0.$ Here $Cov (V, W) : = E [(V - E V) (W - E W)]$ . Then the empirical mean ${\overset{ˉ}{X}}_{N} = N^{- 1} \sum_{i = 1}^{N} X_{N, i}$ converges to $E U$ in $L^{2}$ , meaning $E ∣ {\overset{ˉ}{X}}_{N} - E U ∣^{2} \to 0$ , and hence in probability.

Facts & Assumptions

Given: The exchangeable rows and moment hypotheses in the Statement.

[A1]

Joint exchangeability means that the joint law of the paired row is invariant under every permutation of its $N$ indices.

[A2]

The variables $U_{N, i}$ all have the law of $U$ , with $E U^{2} < \infty$ , and $Cov (U_{N, 1}, U_{N, 2}) = 0$ .

[A3]

$∥ X_{N, 1} - U_{N, 1} ∥_{2} \to 0$ as $N \to \infty$ .

[F1]

For square-integrable random variables, $∣ E Z ∣ \leq (E ∣ Z ∣^{2})^{1 / 2}$ and $E ∣ Z W ∣ \leq ∥ Z ∥_{2} ∥ W ∥_{2}$ . Holder's inequality for random variables

Proof

Given: The hypotheses in the Statement.

1.1A1A2A3F1givenalgebra

By [A1], all $X_{N, i}$ have the same distribution and every distinct pair has the same covariance. By [A3] and [F1], $E X_{N, 1} \to E U$ ; the same $L^{2}$ convergence and $E U^{2} < \infty$ give $\sup_{N} ∥ X_{N, 1} ∥_{2} < \infty$ .

2.1A1A2A3F1givenstep 1.1algebra

Write $X_{i} = X_{N, i}$ and $U_{i} = U_{N, i}$ . The identity $X_{1} X_{2} - U_{1} U_{2} = (X_{1} - U_{1}) X_{2} + U_{1} (X_{2} - U_{2})$ and [F1] give $∣ Cov (X_{1}, X_{2}) - Cov (U_{1}, U_{2}) ∣ \leq ∥ X_{1} - U_{1} ∥_{2} ∥ X_{2} ∥_{2} + ∥ U_{1} ∥_{2} ∥ X_{2} - U_{2} ∥_{2} + ∣ E X_{1} E X_{2} - E U_{1} E U_{2} ∣ .$ By exchangeability the two $L^{2}$ differences have the same norm and both means converge as in step 1.1. Thus [A2]–[A3] imply $Cov (X_{N, 1}, X_{N, 2}) \to 0$ .

3.1A1step 1.1step 2.1algebra∎

Expanding the variance and using [A1] yields $Var ({\overset{ˉ}{X}}_{N}) = N^{- 1} Var (X_{N, 1}) + (1 - N^{- 1}) Cov (X_{N, 1}, X_{N, 2}) ⟶ 0,$ because the first variance is uniformly bounded by step 1.1 and the second term tends to zero by step 2.1. Also $E {\overset{ˉ}{X}}_{N} = E X_{N, 1} \to E U$ by step 1.1. Hence $E ∣ {\overset{ˉ}{X}}_{N} - E U ∣^{2} = Var ({\overset{ˉ}{X}}_{N}) + ∣ E {\overset{ˉ}{X}}_{N} - E U ∣^{2} \to 0$ .

Depends on

Holder's inequality for random variables

Used by

Dependency tree · two levels

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Sources

Jan Mandel, Loren Cobb, and Jonathan D. Beezley, On the Convergence of the Ensemble Kalman Filter, §2, Lemma 2 (standard reference, not scraped)