LemmaStatement: AI-adaptedProof: AI-generatedPipeline-generatedjudge pass (gpt-6-luna)

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Asymptotic squared-norm normalization of a tapered window

Statement

Let $w$ satisfy the taper conditions in Tapered block bootstrap. As the positive integer $b \to \infty$ , $w^{2}$ is Riemann integrable on $[0, 1]$ , and

$\frac{1}{b} \sum_{j = 1}^{b} w ⁣ {(\frac{j - \frac{1}{2}}{b})}^{2} ⟶ I_{w} : = \int_{0}^{1} w (u)^{2} d u > 0.$

Consequently, $W_{b} > 0$ for all sufficiently large $b$ and $a_{b} = \sqrt{b} / W_{b} \to I_{w}^{- 1 / 2}$ .

Facts & Assumptions

Given: The bounded, symmetric window $w$ , nondecreasing on $[0, 1 / 2]$ , and its midpoint weights from Tapered block bootstrap.

[F1]

Tapered block bootstrap: $w$ takes values in $[0, 1]$ , is symmetric around $1 / 2$ , is nondecreasing on $[0, 1 / 2]$ , and is positive on a neighbourhood of $1 / 2$ ; $W_{b}^{2} = \sum_{j = 1}^{b} w ((j - \frac{1}{2}) / b)^{2}$ .

[F2]

A monotone function on $[a, b]$ is Riemann integrable: for the uniform partition into $N$ parts the upper minus lower sum telescopes to $∣ f (b) - f (a) ∣ (b - a) / ι (N)$ : A bounded monotone function on a compact interval is Riemann integrable. On the uniform partition into $N$ parts, its upper-minus-lower Darboux sum is the endpoint variation times the interval length divided by $N$ .

[F3]

For every $ε > 0$ in a complete ordered field there is a natural $n \geq 1$ with $1 / n < ε$ : For every $ε > 0$ there is an integer $N \geq 1$ with $1 / N < ε$ ; since $1 / N$ decreases with $N$ , this implies $1 / N \to 0$ .

[F4]

Riemann's criterion: a bounded $f$ on $[a, b]$ is Darboux integrable if and only if for every real $ε > 0$ there is a partition $P$ with $U (f, P) - L (f, P) < ε$ : A bounded function on a compact interval is Riemann integrable if every $ε > 0$ admits a partition whose upper-minus-lower sum is less than $ε$ .

[F5]

The Darboux and Riemann definitions agree: a bounded $f$ on $[a, b]$ is Darboux integrable with integral $I$ if and only if for every real $ε > 0$ there is a real $δ > 0$ such that $∣ S (f, P, ξ) - I ∣ < ε$ for every tagged partition of mesh below $δ$ : If a bounded function is Riemann integrable, then for every $ε > 0$ all tagged Riemann sums on sufficiently fine partitions differ from its integral by less than $ε$ .

[F6]

The lower and upper Darboux integrals of a bounded $f$ on $[a, b]$ as $\sup_{P} L (f, P)$ and $\inf_{P} U (f, P)$ , Darboux integrability as their equality, and the notation $\int_{a}^{b} f$ and For bounded $f$ on $[a, b]$ and a partition $P$ : the infimum $m_{i}$ and supremum $M_{i}$ of $f$ on the $i$ -th subinterval, and the lower and upper Darboux sums $L (f, P) = \sum_{i} m_{i} Δ_{i}$ and $U (f, P) = \sum_{i} M_{i} Δ_{i}$ : The Darboux integral of an integrable function is the supremum of its lower sums and the infimum of its upper sums; in particular each lower sum is at most the integral.

[F7]

Continuity and derivatives of positive-base real powers: The positive-base real power $x \mapsto x^{- 1 / 2}$ is continuous on $(0, \infty)$ .

Proof

Given: The window and weights in the Statement.

1.1F1F2F3F4givenchoosealgebra

Put $g = w^{2}$ . On $[0, 1 / 2]$ , $g$ is bounded and nondecreasing; symmetry makes it bounded and nonincreasing on $[1 / 2, 1]$ . For each $N \geq 1$ , combine the uniform $N$ -part partitions of these two half-intervals. By [F2], the Darboux gap on each half is at most $1 / (2 N)$ , since $0 \leq g \leq 1$ ; the gap on the combined partition is their sum and is at most $1 / N$ . By [F3], choose $N$ with $1 / N < ε$ for each $ε > 0$ ; [F4] therefore gives Riemann integrability of $g$ on $[0, 1]$ .

1.2F1F6givenchoosealgebra

Choose $0 < δ < 1 / 2$ so that $[1 / 2 - δ, 1 / 2 + δ]$ lies inside the neighbourhood where $w$ is positive, and put $c = w (1 / 2 - δ) > 0$ . Monotonicity on the left half and symmetry imply $g (u) \geq c^{2}$ throughout this central interval, while $g \geq 0$ on $[0, 1]$ . The partition with endpoints $0, 1 / 2 - δ, 1 / 2 + δ, 1$ therefore has lower sum at least $2 δ c^{2}$ . By [F6], $I_{w} \geq 2 δ c^{2} > 0$ .

2.1F3F5givenstep 1.1algebra

Let $I_{w} = \int_{0}^{1} g (u) d u$ . The uniform partition into $b$ parts with midpoint tags has mesh $1 / b \to 0$ , and its Riemann sum is exactly $W_{b}^{2} / b$ . By [F3], [F5] and step 1.1, $W_{b}^{2} / b \to I_{w}$ .

3.1F7step 2.1step 1.2algebra∎

Since $W_{b}^{2} / b \to I_{w} > 0$ , it is positive and bounded away from zero for all sufficiently large $b$ . For those $b$ , $a_{b} = (W_{b}^{2} / b)^{- 1 / 2}$ , so [F7] gives $a_{b} \to I_{w}^{- 1 / 2}$ .

Depends on

Used by

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Sources

D. Pilavakis, E. Paparoditis, and T. Sapatinas, Moving block and tapered block bootstrap for functional time series with an application to the K-sample mean problem, Bernoulli 25 (2019), Section 2.3, Assumption 2 (standard reference, not scraped)