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Positive kernel second moment does not imply nondegenerate first projections

Statement refuted

For a two-sample order- $(r, r)$ kernel, positivity of its second moment under every P/Q assignment implies that both first-projection variances are positive.

Facts & Assumptions

Given: The Rademacher law, the kernel above, and the definitions in Two-sample order-r U-statistic and first projections.

[F1]

Two-sample order-r U-statistic and first projections: the parameter is $E h$ and the first projections are conditional section expectations centered by $E h$ .

[F2]

Second-moment order of the degenerate remainder: if both first projections vanish, then $E [U_{m, n}^{2}] = O (N^{- 2})$ for this fixed-order kernel.

[F3]

Moments, variance, and covariance on a probability space: variance is the expected squared centered deviation.

Counterexample

Let $P = Q$ be the Rademacher law, assigning probability $1 / 2$ to each of $- 1$ and $1$ . Take $r = 2$ and

$h (x_{1}, x_{2}; y_{1}, y_{2}) = x_{1} x_{2} - y_{1} y_{2} .$

Given: Independent Rademacher observations in both groups and the specified kernel. Verify the null, moments, and first projections.

1.1F1givenalgebra

The kernel is symmetric within each block and changes sign when the blocks are exchanged. Since each Rademacher variable has mean zero, [F1] gives $θ = E [X_{1} X_{2}] - E [Y_{1} Y_{2}] = 0$ .

1.2givenF3algebra

The products $A = X_{1} X_{2}$ and $B = Y_{1} Y_{2}$ are independent Rademacher variables. Hence $E [h^{2}] = E [(A - B)^{2}] = E [A^{2}] + E [B^{2}] - 2 E [A B] = 2$ . Because $P = Q$ , this same strictly positive finite second moment holds for every assignment of the four kernel arguments.

1.3F1givenalgebra

For every fixed $x$ , $E [h (x, X_{2}; Y_{1}, Y_{2})] = x E [X_{2}] - E [Y_{1}] E [Y_{2}] = 0$ ; for every fixed $y$ , $E [h (X_{1}, X_{2}; y, Y_{2})] = E [X_{1}] E [X_{2}] - y E [Y_{2}] = 0$ . Thus $g_{X} = g_{Y} = 0$ and both first-projection variances vanish.

2.1F2step 1.3algebra∎

By [F2], $E [U_{m, n}^{2}] = O (N^{- 2})$ , so $\sqrt{m} U_{m, n} \to 0$ in $L^{2}$ when $m / N \to p \in (0, 1)$ . The positive kernel second moment therefore does not provide a nondegenerate normal pivot; the explicit positive first-projection assumptions in the asymptotic theorem cannot be dropped.

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