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A square-integrable linear shift with nonsummable physical dependence

Statement

Assume Countable Choice and Dependent Choice, and let $(ϵ_{t})_{t \in Z}$ be iid Rademacher variables ( $P (ϵ_{t} = 1) = P (ϵ_{t} = - 1) = 1 / 2$ ), and let $e (u) = 1$ in $H = L^{2} ([0, 1], d t; R)$ , so $∥ e ∥_{H} = 1$ . Let $N_{0}$ be the least $N \in N$ such that $\sum_{k > N} (k + 1)^{- 2} < 1.$ For $r \geq 1$ , recursively choose $N_{r}$ as the least integer greater than $N_{r - 1}$ such that $\sum_{k > N_{r}} (k + 1)^{- 2} < 2^{- 4 r} .$ Define a causal Borel filter by $f (x_{0}, x_{1}, \dots) = e \lim_{r \to \infty} \sum_{k = 0}^{N_{r}} \frac{x_{k}}{k + 1}$ when this subsequence converges, and define $f = 0$ otherwise. Then $X_{t} = f (ϵ_{t}, ϵ_{t - 1}, \dots)$ is centered and has finite second moment, but its physical-dependence coefficients satisfy $δ_{m}^{2} = 2 \sum_{k = m}^{\infty} \frac{1}{(k + 1)^{2}}, δ_{m} \geq \sqrt{\frac{2}{m + 1}} (m \geq 1) .$ Consequently $\sum_{m \geq 1} δ_{m} = \infty$ : the sufficient summability hypothesis in L2-m-approximable causal Bernoulli shifts does not apply to this shift. The example makes no claim that nonsummability alone precludes every possible central limit theorem.

Facts & Assumptions

Given: Countably many iid Rademacher innovations, the real Hilbert space $H = L^{2} ([0, 1], d t; R)$ , and the replacement convention in L2-m-approximable causal Bernoulli shifts.

[F1]

The countable product carrying iid Rademacher variables exists under Countable Choice and Dependent Choice (Countably many independent copies of a prescribed law exist, The Axiom of Countable Choice ( ${A C}_{ω}$ ), The axiom of dependent choice: a relation in which every element is related to something admits an $N$ -indexed chain).

[F2]

The scalar $L^{2}$ quotient is complete and its norm is induced by the integral inner product (The space $L^{p} (μ)$ as the quotient by null functions, $L^{2}$ with the integral pairing is a Hilbert space).

[F3]

Chebyshev's inequality bounds the probability of a deviation by its variance divided by the squared threshold, and the first Borel--Cantelli lemma makes events with summable probabilities occur only finitely often (Chebyshev's inequality for random variables, First Borel-Cantelli lemma for events).

[F4]

The stated two-point Rademacher law gives $E ϵ_{t} = 0$ and $E ϵ_{t}^{2} = 1$ by the expectation definition. Independence is the independence of the generated sigma-algebras (Independent random elements); expectations of products of integrable functions of distinct independent coordinates factor (Expectation of a nonnegative or integrable random variable, Expectations factor over finite products of independent random variables).

[F5]

Cauchy--Schwarz on a probability space gives $E ∣ Y ∣ \leq ∥ Y ∥_{2}$ ; thus $L^{2}$ convergence implies convergence of means and of $L^{2}$ norms (Cauchy-Schwarz for random variables).

[F6]

Since $x^{- 2}$ is decreasing, $\sum_{j = 2}^{N} j^{- 2} \leq \int_{1}^{N} x^{- 2} d x \leq 1$ , so $\sum_{j \geq 1} j^{- 2} < \infty$ . Since $x^{- 1 / 2}$ is decreasing, $\sum_{j = 2}^{N} j^{- 1 / 2} \geq \int_{2}^{N + 1} x^{- 1 / 2} d x \to \infty$ , so $\sum_{j \geq 2} j^{- 1 / 2} = \infty$ .

Proof

Given: The probability space and iid Rademacher innovations in the Statement, together with the independent replacement extension from L2-m-approximable causal Bernoulli shifts.

1.1givenF1F6constructalgebra

By [F6], the coefficient series $\sum_{k \geq 0} (k + 1)^{- 2}$ converges and its tails tend to zero. Hence the set defining $N_{0}$ is nonempty, so its least element exists. For $r \geq 1$ , once $N_{r - 1}$ is chosen, the tail tends to zero, so some integer $N > N_{r - 1}$ satisfies the required bound $\sum_{k > N} (k + 1)^{- 2} < 2^{- 4 r}$ ; thus the least such $N_{r}$ exists. For a deterministic sign sequence $x = (x_{k})_{k \geq 0}$ , put $s_{N} (x) = \sum_{k = 0}^{N} x_{k} / (k + 1)$ . Each $s_{N}$ is product-measurable. The set on which $(s_{N_{r}})_{r \geq 0}$ is Cauchy is Borel, because the Cauchy condition is a countable combination of inequalities between these measurable real functions. On that set their limit is measurable, and extending it by zero on the complement is product-measurable. Thus $f$ is a total Borel map into the one-dimensional subspace $R e \subset H$ .

2.1F2F3F4F5F6step 1.1choosealgebra

On the given probability space carrying the iid innovations, the partial sums $S_{N} = \sum_{k = 0}^{N} ϵ_{- k} / (k + 1)$ are Cauchy in $L^{2}$ , since for $M > N$ independence, centering and [F4] give $E ∣ S_{M} - S_{N} ∣^{2} = \sum_{k = N + 1}^{M} (k + 1)^{- 2} ⟶ 0.$ By [F2] they converge in $L^{2}$ to a random variable $S$ . Each $S_{N}$ has mean zero and squared norm $∥ S_{N} ∥_{2}^{2} = \sum_{k = 0}^{N} (k + 1)^{- 2}$ ; [F5] therefore gives $E S = 0$ and $E S^{2} = ∥ S ∥_{2}^{2} = \sum_{k = 0}^{\infty} (k + 1)^{- 2} < \infty$ by [F6]. Choose increasing $N_{r}$ with $\sum_{k > N_{r}} (k + 1)^{- 2} < 2^{- 4 r}$ . Taking $M \to \infty$ in the finite tail-variance identity gives $E ∣ S_{N_{r}} - S ∣^{2} = \sum_{k > N_{r}} (k + 1)^{- 2}$ . Thus [F3] gives $P (∣ S_{N_{r}} - S ∣ > 2^{- r}) \leq 2^{2 r} E ∣ S_{N_{r}} - S ∣^{2} < 2^{- 2 r} .$ These probabilities are summable, so [F3] implies $S_{N_{r}} \to S$ almost surely. Hence the defining subsequence filter equals $S e$ almost surely. In particular $X_{0}$ is centered and $E ∥ X_{0} ∥_{H}^{2} = E S^{2} = \sum_{k = 0}^{\infty} (k + 1)^{- 2} < \infty .$ The same argument applies to every shifted iid input sequence.

3.1F1F2F4step 2.1algebra

For fixed $m \geq 1$ , the mixed sequence used by $X_{0, m}$ is itself iid Rademacher, so its filter is the corresponding $L^{2}$ limit. Subtracting its partial sums from those for $X_{0}$ cancels the terms $k < m$ and leaves $X_{0} - X_{0, m} = e \sum_{k = m}^{\infty} \frac{ϵ_{- k} - ϵ_{0, m, - k}^{'}}{k + 1} in L^{2} .$ The summands are independent, centered, and have variance $2 / (k + 1)^{2}$ by [F4]; finite-tail variance computation followed by $L^{2}$ convergence therefore gives $δ_{m}^{2} = 2 \sum_{k = m}^{\infty} (k + 1)^{- 2}$ .

4.1F6step 2.1step 3.1algebra∎

Since $x \mapsto x^{- 2}$ is decreasing on $(0, \infty)$ , $\sum_{k = m}^{\infty} (k + 1)^{- 2} \geq \int_{m + 1}^{\infty} x^{- 2} d x = \frac{1}{m + 1} .$ Thus $δ_{m} \geq \sqrt{2 / (m + 1)}$ , and the decreasing-function integral comparison gives $\sum_{m \geq 1} δ_{m} \geq \sqrt{2} \sum_{j \geq 2} j^{- 1 / 2} \geq \sqrt{2} \int_{2}^{\infty} x^{- 1 / 2} d x = + \infty$ by [F6]. The moment condition holds by step 2.1, but the required dependence-coefficient summability fails.

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