Hilbert-Valued CLT for Functional Time Series — Examples

By [F6], the coefficient series ${\sum }_{k\ge 0}(k+1{)}^{-2}$ converges and its tails tend to zero. Hence the set defining $N_0$ is nonempty, so its least element exists. For $r\ge 1$, once $N_{r-1}$ is chosen, the tail tends to zero, so some integer $N>N_{r-1}$ satisfies the required bound ${\sum }_{k>N}(k+1{)}^{-2}<2^{-4r}$; thus the least such $N_r$ exists. For a deterministic sign sequence $x=(x_k{)}_{k\ge 0}$, put $s_N(x)={\sum }_{k=0}^N{x}_k\mathrm{/}(k+1)$. Each $s_N$ is product-measurable. The set on which $(s_{N_r}{)}_{r\ge 0}$ 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 $\mathbb{R}e\subset H$.

On the given probability space carrying the iid innovations, the partial sums $S_N={\sum }_{k=0}^N{ϵ}_{-k}\mathrm{/}(k+1)$ are Cauchy in $L^2$, since for $M>N$ independence, centering and [F4] give $\mathbb{E}\mathrm{\mid }{S}_M-S_N{\mathrm{\mid }}^2={\sum }_{k=N+1}^M(k+1{)}^{-2}⟶0.$ By [F2] they converge in $L^2$ to a random variable $S$. Each $S_N$ has mean zero and squared norm $\mathrm{\parallel }{S}_N{\mathrm{\parallel }}_2^2={\sum }_{k=0}^N(k+1{)}^{-2}$; [F5] therefore gives $\mathbb{E}S=0$ and $\mathbb{E}{S}^2=\mathrm{\parallel }S{\mathrm{\parallel }}_2^2={\sum }_{k=0}^{\mathrm{\infty }}(k+1{)}^{-2}<\mathrm{\infty }$ by [F6]. Choose increasing $N_r$ with ${\sum }_{k>N_r}(k+1{)}^{-2}<2^{-4r}$. Taking $M\to \mathrm{\infty }$ in the finite tail-variance identity gives $\mathbb{E}\mathrm{\mid }{S}_{N_r}-S{\mathrm{\mid }}^2={\sum }_{k>N_r}(k+1{)}^{-2}$. Thus [F3] gives $P(\mathrm{\mid }{S}_{N_r}-S\mathrm{\mid }>2^{-r})\le 2^{2r}\mathbb{E}\mathrm{\mid }{S}_{N_r}-S{\mathrm{\mid }}^2<2^{-2r}\mathrm{.}$ These probabilities are summable, so [F3] implies $S_{N_r}\to S$ almost surely. Hence the defining subsequence filter equals $Se$ almost surely. In particular $X_0$ is centered and $\mathbb{E}\mathrm{\parallel }{X}_0{\mathrm{\parallel }}_H^2=\mathbb{E}{S}^2={\sum }_{k=0}^{\mathrm{\infty }}(k+1{)}^{-2}<\mathrm{\infty }\mathrm{.}$ The same argument applies to every shifted iid input sequence.

For fixed $m\ge 1$, 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 $X_0-X_{0,m}=e{\sum }_{k=m}^{\mathrm{\infty }}\frac{{ϵ}_{-k}-{ϵ}_{0,m,-k}^{\mathrm{\prime }}}{k+1}\text{in }{L}^2\mathrm{.}$ The summands are independent, centered, and have variance $2\mathrm{/}(k+1{)}^2$ by [F4]; finite-tail variance computation followed by $L^2$ convergence therefore gives ${\delta }_m^2=2{\sum }_{k=m}^{\mathrm{\infty }}(k+1{)}^{-2}$.

Since $x\mapsto x^{-2}$ is decreasing on $(0,\mathrm{\infty })$, ${\sum }_{k=m}^{\mathrm{\infty }}(k+1{)}^{-2}\ge {\int }_{m+1}^{\mathrm{\infty }}{x}^{-2}dx=\frac{1}{m+1}\mathrm{.}$ Thus ${\delta }_m\ge \sqrt{2\mathrm{/}(m+1)}$, and the decreasing-function integral comparison gives ${\sum }_{m\ge 1}{\delta }_m\ge \sqrt{2}{\sum }_{j\ge 2}{j}^{-1\mathrm{/}2}\ge \sqrt{2}{\int }_2^{\mathrm{\infty }}{x}^{-1\mathrm{/}2}dx=+\mathrm{\infty }$ by [F6]. The moment condition holds by step 2.1, but the required dependence-coefficient summability fails.

The vector $e$ has norm one, so $X_t$ has mean zero and $\mathbb{E}\mathrm{\parallel }{X}_t{\mathrm{\parallel }}_H^2=\mathbb{E}({ϵ}_t+2{ϵ}_{t-1}{)}^2=5$ by [F2]. For $m=1$, the replacement changes only ${ϵ}_{-1}$ in $X_0$, so $X_0-X_{0,1}=2({ϵ}_{-1}-{ϵ}_{0,1,-1}^{\mathrm{\prime }})e$ and [F2] gives ${\delta }_1^2=4(1+1)=8$. For $m\ge 2$, both innovations on which $X_0$ depends are retained, so $X_{0,m}=X_0$ and ${\delta }_m=0$. Hence ${\sum }_{m\ge 1}{\delta }_m=2\sqrt{2}<\mathrm{\infty }$.

The lag-zero covariance is $5(e\otimes e)$. At lag one, the only nonzero product in $({ϵ}_0+2{ϵ}_{-1})({ϵ}_1+2{ϵ}_0)$ is $2{ϵ}_0^2$, so ${\mathbb{E}}_B[X_0\otimes X_1]=2(e\otimes e)$. Reversing the factors gives the same operator. At every lag $\mathrm{\mid }j\mathrm{\mid }\ge 2$ the innovation sets are disjoint, so independence and centering make the covariance zero. Therefore the long-run covariance sum is $(5+2+2)(e\otimes e)=9(e\otimes e)$.

Fix a cut $a\in \mathbb{Z}$. The variables $({ϵ}_j{)}_{j\in \mathbb{Z}}$ are independent by the iid assumption, so their generated sigma-algebras form an independent family by Independent random elements and Independent sigma-algebras and independent events. Each $X_s$ is measurable with respect to $\sigma ({ϵ}_s,{ϵ}_{s-1})$. Thus $\sigma (X_s:s\le a)\subseteq \sigma ({ϵ}_j:j\le a)$, while $\sigma (X_s:s\ge a+2)\subseteq \sigma ({ϵ}_j:j\ge a+1)$. By [F4] the two source sigma-algebras are independent; their sub-sigma-algebras generated by the process groups inherit independence. The process is therefore $1$-dependent under the definition in L2-m-approximable causal Bernoulli shifts.

The moment, centering and summability conditions were verified in step 1.1, and step 1.2 gives the covariance required by [F3]. Applying [F3] proves the asserted sample-sum convergence; the Gaussian limit is allowed to have finite rank, as it does here. This conclusion concerns the $H$-valued sample sums.

For every input sequence with $\mathrm{\mid }{ϵ}_t\mathrm{\mid }=1$, ${\sum }_{k\ge 0}\mathrm{\mid }{2}^{-k}{ϵ}_{t-k}\mathrm{\mid }\le 2$, so the series converges absolutely and uniformly over the product space. Its partial sums are product-measurable and bounded by $2$, so their pointwise limit is a Borel causal map. Finite partial-sum variances are ${\sum }_{k\le N}{4}^{-k}$ by [F2]; the uniform bound and dominated convergence give centering and $\mathbb{E}\mathrm{\parallel }{X}_0{\mathrm{\parallel }}_H^2={\sum }_{k\ge 0}{4}^{-k}=4\mathrm{/}3<\mathrm{\infty }$. For the replacement, the first $m$ terms cancel; finite tails are independent centered differences with variance $2{\sum }_{k=m}^N{4}^{-k}$ and are bounded in absolute value by $2{\sum }_{k=m}^{\mathrm{\infty }}{2}^{-k}$. Dominated convergence [F2] passes these variances to the infinite tail, giving ${\delta }_m^2=2{\sum }_{k=m}^{\mathrm{\infty }}{4}^{-k}=\frac{8}{3}{4}^{-m}\mathrm{.}$ Thus ${\sum }_{m\ge 1}{\delta }_m=\sqrt{8\mathrm{/}3}{\sum }_{m\ge 1}{2}^{-m}<\mathrm{\infty }$.

For $j\ge 0$, truncate both linear sums at index $N$. Their product is bounded by $4$ and converges pointwise to $X_0{X}_j$, so dominated convergence [F2] passes expectations to the limit. The finite sums share exactly innovations with indices $-k=j-(k+j)$; their coefficient products are $2^{-k}{2}^{-(k+j)}$, while all other products have zero expectation by [F2]. The finite shared-coefficient sums tend to ${\sum }_{k\ge 0}{2}^{-k}{2}^{-(k+j)}$, giving ${\gamma }_j=\mathbb{E}⟨X_0,X_j{⟩}_H=\frac{4}{3}{2}^{-j}\mathrm{.}$ Symmetry gives ${\gamma }_{-j}={\gamma }_j$. Therefore the scalar long-run variance is ${\gamma }_0+2{\sum }_{j\ge 1}{\gamma }_j=\frac{4}{3}+\frac{8}{3}{\sum }_{j\ge 1}{2}^{-j}=4,$ and the operator is $4(e\otimes e)$.

Step 1.1 gives centering, a finite second moment and summable physical-dependence errors; step 2.1 gives the long-run covariance. Applying [F3] yields convergence of the normalized $H$-valued sums to a centered Gaussian with covariance $4(e\otimes e)$.

Since $\sin (0)=\sin (\pi )=0$, both even and odd paths satisfy $w_n(1)=0$. Thus their terminal laws are all the point mass ${\delta }_0$ and in particular converge weakly to ${\delta }_0$ in $H$.

Define $\mathrm{\Phi }:\mathcal{C}\to \mathbb{R}$ by $\mathrm{\Phi }(w)=\min \{1,{\sup }_{0\le t\le 1}\mathrm{\parallel }w(t){\mathrm{\parallel }}_H\}$. The supremum norm inequality gives $\mathrm{\mid }{\sup }_t\mathrm{\parallel }w(t){\mathrm{\parallel }}_H-{\sup }_t\mathrm{\parallel }v(t){\mathrm{\parallel }}_H\mathrm{\mid }\le {\sup }_t\mathrm{\parallel }w(t)-v(t){\mathrm{\parallel }}_H=d(w,v),$ and truncation by $x\mapsto \min (1,x)$ is 1-Lipschitz on $[0,\mathrm{\infty })$. Hence $\mathrm{\Phi }$ is bounded and continuous. It equals $0$ on every even path and $1$ on every odd path, so its expectations under the path laws alternate between $0$ and $1$ and do not converge. By [F1], those path laws cannot converge weakly.

The example has convergence only after applying the terminal evaluation map; the bounded continuous path functional in step 1.2 detects the missing path-space convergence. Therefore endpoint convergence alone supplies no functional CLT conclusion, consistently with the scope in [F2].