Hilbert-Valued CLT for Functional Time Series

For $h\in \mathbb{Z}$ put $C_h=\mathrm{Cov}(Y_0,Y_h)$. These covariance terms are finite by [F8]. Strict stationarity gives $C_{-h}=C_h^T$, and $m$-dependence gives $C_h=0$ when $\mathrm{\mid }h\mathrm{\mid }>m$. Expanding the finite variance of the sum yields $\frac{1}{n}\mathrm{Cov}\text{\hspace{0.17em}⁣}\left({\sum }_{t=1}^n{Y}_t\right)={\sum }_{\mathrm{\mid }h\mathrm{\mid }\le m}{(1-\frac{\mathrm{\mid }h\mathrm{\mid }}{n})}_{+}{C}_h⟶{\mathrm{\Sigma }}_m\mathrm{.}$ Thus ${\mathrm{\Sigma }}_m$ is symmetric, and for every $a\in {\mathbb{R}}^d$, $a^T{\mathrm{\Sigma }}_{m}a$ is the limit of nonnegative variances divided by $n$, so it is nonnegative. Hence $N_d(0,{\mathrm{\Sigma }}_m)$ exists by [F5].

Fix $a\in {\mathbb{R}}^d$ and let $Z_t=a^T{Y}_t$ and ${\sigma }^2=a^T{\mathrm{\Sigma }}_{m}a$. If ${\sigma }^2=0$, step 1.1 gives $\mathbb{E}\mathrm{\mid }{n}^{-1\mathrm{/}2}{\sum }_{t=1}^n{Z}_t{\mathrm{\mid }}^2\to 0$; [F6] and then [F3] give convergence in distribution to zero. Suppose ${\sigma }^2>0$. For $K\ge 1$ define $Z_t^{(K)}=\max (-K,\min (K,Z_t))-\mathbb{E}[\max (-K,\min (K,Z_0))]$. This is centered, stationary, $m$-dependent and bounded by $2K$; [F7] gives $\mathrm{\parallel }{Z}_0-Z_0^{(K)}{\mathrm{\parallel }}_2\to 0$. Write ${\sigma }_K^2={\lim }_n{n}^{-1}\mathrm{Var}({\sum }_{t=1}^n{Z}_t^{(K)})={\sum }_{\mathrm{\mid }h\mathrm{\mid }\le m}\mathrm{Cov}(Z_0^{(K)},Z_h^{(K)})$. Finite summation and [F8] give ${\sigma }_K^2\to {\sigma }^2$, so ${\sigma }_K^2>0$ for all sufficiently large $K$.

Fix such a $K$, put ${\mathrm{\ell }}_n=\lfloor n^{1\mathrm{/}4}\rfloor$, and retain $q_n=\lfloor n\mathrm{/}({\mathrm{\ell }}_n+m)\rfloor$ blocks of length ${\mathrm{\ell }}_n$, with $m$ omitted indices between successive blocks. Write $b$ for the last index in a retained block. The next block starts at $b+m+1$, so $m$-dependence makes its sigma-algebra independent of the sigma-algebra generated by all earlier blocks; induction gives row-wise independence of the block sums $B_{n,r}$. If ${\gamma }_h^{(K)}=\mathrm{Cov}(Z_0^{(K)},Z_h^{(K)})$, stationarity gives $\mathrm{Var}(B_{n,r})={\mathrm{\ell }}_n{\gamma }_0^{(K)}+2{\sum }_{h=1}^m({\mathrm{\ell }}_n-h){\gamma }_h^{(K)},\frac{s_n^2}{n}:=\frac{q_n\mathrm{Var}(B_{n,1})}{n}⟶{\sigma }_K^2\mathrm{.}$ Here $q_n{\mathrm{\ell }}_n\mathrm{/}n\to 1$ and $q_n\mathrm{/}n\to 0$. Also $\mathrm{\mid }{B}_{n,r}\mathrm{\mid }\le 2K{\mathrm{\ell }}_n$ and $s_n\mathrm{/}\sqrt{n}\to {\sigma }_K>0$. Thus $q_n\ge 1$ and $s_n>0$ for all sufficiently large $n$; on this tail, for each $\epsilon >0$ every block is eventually bounded by $\epsilon s_n$, so the Lindeberg sum is zero. Applying [F1] to this tail of rows, $s_n^{-1}{\sum }_r{B}_{n,r}\Rightarrow N(0,1)$, and [F2] gives $n^{-1\mathrm{/}2}{\sum }_r{B}_{n,r}\Rightarrow N(0,{\sigma }_K^2)$.

The omitted gaps and terminal remainder contain $r_n\le m{q}_n+{\mathrm{\ell }}_n+m=O(n\mathrm{/}{\mathrm{\ell }}_n+{\mathrm{\ell }}_n)=o(n)$ indices. For a centered $m$-dependent sequence bounded by $2K$, at most $(2m+1)r_n$ ordered pairs of omitted terms can have nonzero covariance, each covariance having absolute value at most $4K^2$. Therefore the omitted sum divided by $\sqrt{n}$ tends to zero in $L^2$, hence in probability by [F6]. By [F3], $n^{-1\mathrm{/}2}{\sum }_{t=1}^n{Z}_t^{(K)}\Rightarrow N(0,{\sigma }_K^2)$.

The centered $m$-dependent truncation errors $W_t^{(K)}=Z_t-Z_t^{(K)}$ satisfy $\mathbb{E}{\mid n^{-1\mathrm{/}2}{\sum }_{t=1}^n{W}_t^{(K)}\mid }^2\le (2m+1)\mathbb{E}\mathrm{\mid }{W}_0^{(K)}{\mathrm{\mid }}^2⟶0(K\to \mathrm{\infty }),$ by expanding the variance and bounding each of its at most $2m+1$ nonzero lag terms with [F8]. For fixed real $u$, $\mathrm{\mid }{e}^{iux}-e^{iuy}\mathrm{\mid }\le \mathrm{\mid }u\mathrm{\mid }\mathrm{\mid }x-y\mathrm{\mid }$; thus the characteristic function of $n^{-1\mathrm{/}2}\sum Z_t$ differs from that of its $K$-truncation by at most $\mathrm{\mid }u\mathrm{\mid }\sqrt{(2m+1)\mathbb{E}\mathrm{\mid }{W}_0^{(K)}{\mathrm{\mid }}^2}$. First taking $n\to \mathrm{\infty }$ using step 4.1 and then $K\to \mathrm{\infty }$ using ${\sigma }_K^2\to {\sigma }^2$ proves convergence to $e^{-u^2{\sigma }^2\mathrm{/}2}$. By [F4], $n^{-1\mathrm{/}2}\sum Z_t\Rightarrow N(0,{\sigma }^2)$. This also covers every projection with positive variance; the zero-variance case was handled in step 2.1.

Steps 2.1 and 5.1 show for every $a\in {\mathbb{R}}^d$ that $a^T{n}^{-1\mathrm{/}2}{\sum }_{t=1}^n{Y}_t\Rightarrow N(0,a^T{\mathrm{\Sigma }}_{m}a)$. These are precisely the projections of the law $N_d(0,{\mathrm{\Sigma }}_m)$ established in step 1.1, so [F5] gives the asserted vector convergence, including singular and zero covariance.

By [F1], choose an orthonormal eigenbasis $(e_j{)}_{j\in J}$ of $\mathrm{\Gamma }$ with eigenvalues ${\lambda }_j\ge 0$. This basis is finite or countable: take a dense sequence $(h_k)$ in $H$ and assign to each unit basis vector the least index $k$ with $\mathrm{\parallel }{e}_j-h_k{\mathrm{\parallel }}_H<1\mathrm{/}4$. Distinct orthogonal unit vectors are separated by $\sqrt{2}$, so the assignment is injective. By [F2], ${\sum }_{j\in J}{\lambda }_j=\mathrm{tr}(\mathrm{\Gamma })<\mathrm{\infty }$.

By [A1] and [F3], take independent standard normal variables $(Z_j{)}_{j\in J}$ on a common probability space. Define $T={\sum }_{j\in J}{\lambda }_j{Z}_j^2$ as a nonnegative extended random variable. Tonelli [F5] gives $\mathbb{E}T={\sum }_j{\lambda }_j\mathbb{E}{Z}_j^2=\mathrm{tr}(\mathrm{\Gamma })<\mathrm{\infty }$, so $T<\mathrm{\infty }$ almost surely: if $\mathbb{P}(T=\mathrm{\infty })>0$, then $\mathbb{E}T\ge M\mathbb{P}(T=\mathrm{\infty })$ for every $M>0$, a contradiction.

On $\{T<\mathrm{\infty }\}$ the partial sums $G_N={\sum }_{j\in J,j\le N}\sqrt{{\lambda }_j}{Z}_j{e}_j$ are Cauchy in $H$, since orthonormality gives $\mathrm{\parallel }{G}_M-G_N{\mathrm{\parallel }}_H^2={\sum }_{N<j\le M}{\lambda }_j{Z}_j^2$. Let $G$ be their limit there, and set $G=0$ on the measurable null complement. The convergence set is measurable by the Cauchy criterion. For $F=\mathrm{\varnothing }$ the preimage is empty. For nonempty closed $F\subseteq H$, the set $\{G\in F\}$ is the union of $\{T<\mathrm{\infty },\mathrm{dist}(G_N,F)\to 0\}$ and, when $0\in F$, $\{T=\mathrm{\infty }\}$; the first set is measurable because each distance map is continuous and the second is measurable. Thus $G$ is Borel. To see strong measurability, let $(h_k)$ be a dense sequence in $H$ and, for each $N$, map $x$ to the first point among $h_1,\dots ,h_N$ minimizing distance to $x$. The finitely many tie-broken cells are Borel, so these are finite-valued Borel maps; their distances from $x$ tend to zero by density. Composing them with $G$ gives measurable simple maps converging pointwise to $G$, as required by [F6]. On $\{T<\mathrm{\infty }\}$, $\mathrm{\parallel }G-G_N{\mathrm{\parallel }}_H^2={\sum }_{j>N}{\lambda }_j{Z}_j^2$; [F5] gives $\mathbb{E}\mathrm{\parallel }G-G_N{\mathrm{\parallel }}_H^2={\sum }_{j>N}{\lambda }_j\to 0$ and $\mathbb{E}\mathrm{\parallel }G{\mathrm{\parallel }}_H^2={\sum }_j{\lambda }_j=\mathrm{tr}(\mathrm{\Gamma })<\mathrm{\infty }$.

Fix $h\in H$. The finite sum $⟨G_N,h{⟩}_H$ is a sum of independent centered normal variables, so its characteristic function is $\exp (-t^2{\sum }_{j\le N}{\lambda }_j⟨e_j,h{⟩}_H^2\mathrm{/}2)$ by [F3] and [F4]. The series of variances increases to ${\sum }_j{\lambda }_j⟨e_j,h{⟩}_H^2=⟨\mathrm{\Gamma }h,h{⟩}_H$, and $G_N\to G$ in $L^2$ by step 2.1. Since $\mathrm{\mid }{e}^{itx}-e^{ity}\mathrm{\mid }\le \mathrm{\mid }t\mathrm{\mid }\mathrm{\mid }x-y\mathrm{\mid }$, the characteristic functions converge to $\exp (-t^2⟨\mathrm{\Gamma }h,h{⟩}_H\mathrm{/}2)$; [F3]–[F4] identify $⟨G,h{⟩}_H$ as the centered normal with that variance, including variance zero.

For $h,k\in H$, the finite-sum covariances are $\mathbb{E}[⟨G_N,h{⟩}_H⟨G_N,k{⟩}_H]={\sum }_{j\le N}{\lambda }_j⟨h,e_j{⟩}_H⟨e_j,k{⟩}_H$ by independence, centering and [F3]–[F4]. The $L^2$ convergence from step 2.1 and [F7] pass the left side to the limit, while the eigen-expansion of $\mathrm{\Gamma }$ makes the right side tend to $⟨\mathrm{\Gamma }h,k{⟩}_H$. Hence $G$ has the asserted covariance.

Choose measurable representatives of $x$ and $y$. By [F1]–[F2], ${\int }_{I^2}\mathrm{\mid }x(u)y(v){\mathrm{\mid }}^{2}d(dt\times dt)=({\int }_I\mathrm{\mid }x(u){\mathrm{\mid }}^{2}du)({\int }_I\mathrm{\mid }y(v){\mathrm{\mid }}^{2}dv)<\mathrm{\infty }\mathrm{.}$ Changing either representative changes the product only on $(N\times I)\cup (I\times M)$ for null sets $N,M\subset I$; [F1] gives this union product measure zero. Hence $x⊠y$ is a well-defined class in $K$ and the displayed integral proves its norm identity.

Choose Borel representatives of $k$ and $f$. The function $(u,v)\mapsto k(u,v)f(v)$ is product-measurable and belongs to $L^1(dt\times dt)$, since Cauchy–Schwarz on $I^2$ gives ${\int }_{I^2}\mathrm{\mid }k(u,v)f(v)\mathrm{\mid }dudv\le \mathrm{\parallel }k{\mathrm{\parallel }}_K{({\int }_{I^2}\mathrm{\mid }f(v){\mathrm{\mid }}^{2}dudv)}^{1\mathrm{/}2}=\mathrm{\parallel }k{\mathrm{\parallel }}_K\mathrm{\parallel }f{\mathrm{\parallel }}_H\mathrm{.}$ By [F3] its section integrals are defined almost everywhere and have a measurable representative. For almost every $u$, sectionwise Cauchy–Schwarz and [F2] give $\mathrm{\mid }(T_{k}f)(u){\mathrm{\mid }}^2\le \mathrm{\parallel }f{\mathrm{\parallel }}_H^2{\int }_I\mathrm{\mid }k(u,v){\mathrm{\mid }}^{2}dv,\mathrm{\parallel }{T}_{k}f{\mathrm{\parallel }}_H^2\le \mathrm{\parallel }k{\mathrm{\parallel }}_K^2\mathrm{\parallel }f{\mathrm{\parallel }}_H^2\mathrm{.}$ If $f$ is changed on a $dt$-null set, its section integral is unchanged for every $u$ whose $k(u,\cdot )$ is square-integrable. If $k$ is changed on a $(dt\times dt)$-null set, the difference times $f(v)$ is in $L^1$ by the same Cauchy–Schwarz estimate, so [F3] makes its section integral zero for almost every $u$. Thus $T_k$ is well-defined on the quotients. Linearity follows from linearity of the section integral, and the second estimate proves boundedness and $\mathrm{\parallel }{T}_k\mathrm{\parallel }\le \mathrm{\parallel }k{\mathrm{\parallel }}_K$.

For $k=x⊠y$ and almost every $u$, the section is integrable by step 1.2 and $(T_{x⊠y}f)(u)=x(u){\int }_{I}y(v)f(v)dv=x(u)⟨f,y{⟩}_H=((x\otimes y)f)(u)\mathrm{.}$ Thus $T_{x⊠y}=x\otimes y$ as operators on $H$. This includes $x=0$ or $y=0$, when both operators are zero.

For $x,y,g,h\in H$, [F3] applies to $x(u)y(v)g(u)h(v)$ because Cauchy–Schwarz on $I^2$ bounds its absolute integral by $\mathrm{\parallel }x{\mathrm{\parallel }}_H\mathrm{\parallel }g{\mathrm{\parallel }}_H\mathrm{\parallel }y{\mathrm{\parallel }}_H\mathrm{\parallel }h{\mathrm{\parallel }}_H$. Fubini then factors the integral into $⟨x,g{⟩}_H⟨y,h{⟩}_H$, proving the stated inner-product identity.

Fix $x\in H$ and $N$. Orthonormality gives $⟨x-P_{N}x,e_j{⟩}_H=0$ for each $j\le N$, so $x-P_{N}x$ is orthogonal to both $P_{N}x$ and every vector in the first-$N$ span. By [F1], $\mathrm{\parallel }x{\mathrm{\parallel }}_H^2=\mathrm{\parallel }{P}_{N}x{\mathrm{\parallel }}_H^2+\mathrm{\parallel }x-P_{N}x{\mathrm{\parallel }}_H^2,\mathrm{\parallel }{P}_{N}x{\mathrm{\parallel }}_H^2={\sum }_{k\le N}\mathrm{\mid }⟨x,e_k{⟩}_H{\mathrm{\mid }}^2\mathrm{.}$ For any $y$ in that span, $P_{N}x-y$ also lies in the span and is orthogonal to $x-P_{N}x$, hence $\mathrm{\parallel }x-y{\mathrm{\parallel }}_H^2=\mathrm{\parallel }x-P_{N}x{\mathrm{\parallel }}_H^2+\mathrm{\parallel }{P}_{N}x-y{\mathrm{\parallel }}_H^2\ge \mathrm{\parallel }x-P_{N}x{\mathrm{\parallel }}_H^2\mathrm{.}$ Thus $P_{N}x$ is a closest vector to $x$ in the first-$N$ span.

Let $\epsilon >0$. By density choose a finite linear combination $y$ of the $e_k$ with $\mathrm{\parallel }x-y{\mathrm{\parallel }}_H<\epsilon$, and choose $N$ at least as large as every index used in $y$. Then $y$ lies in the first-$N$ span, so step 1.1 gives $\mathrm{\parallel }x-P_{N}x{\mathrm{\parallel }}_H\le \mathrm{\parallel }x-y{\mathrm{\parallel }}_H<\epsilon$. Hence $P_{N}x\to x$; the first identity in step 1.1 then gives the asserted Parseval limit. For a finite family, $P_{N}x=x$ once all its vectors are included.

For each $k$, the coordinate variables $U_k=⟨U,e_k{⟩}_H$ and $V_k=⟨V,e_k{⟩}_H$ are measurable, centered by [F1], and independent by [F2]. Their product is integrable by Cauchy--Schwarz, so [F3] gives $\mathbb{E}[U_k{V}_k]=\mathbb{E}[U_k]\mathbb{E}[V_k]=0$. Every finite sum $⟨P_{N}U,P_{N}V{⟩}_H={\sum }_{k\le N}{U}_k{V}_k$ therefore has expectation zero.

By [F4], $P_{N}U\to U$ and $P_{N}V\to V$ almost surely (indeed for every vector in $H$); continuity of the inner product gives $⟨P_{N}U,P_{N}V{⟩}_H\to ⟨U,V{⟩}_H$. Since projections are contractions, $\mathrm{\mid }⟨P_{N}U,P_{N}V{⟩}_H\mathrm{\mid }\le \mathrm{\parallel }U{\mathrm{\parallel }}_H\mathrm{\parallel }V{\mathrm{\parallel }}_H,$ and this majorant is integrable by [F5]. Dominated convergence [F5] and step 1.1 now yield $\mathbb{E}⟨U,V{⟩}_H=0$.

Put $\sigma =(\mathbb{E}\mathrm{\parallel }{X}_0{\mathrm{\parallel }}_H^2{)}^{1\mathrm{/}2}$. For each $j\ge 0$, the map $(x,y)\mapsto x\otimes y$ is continuous into $S_1(H)$, since [F2] and the triangle inequality give $\mathrm{\parallel }x\otimes y-x^{\mathrm{\prime }}\otimes y^{\mathrm{\prime }}{\mathrm{\parallel }}_1\le \mathrm{\parallel }x-x^{\mathrm{\prime }}{\mathrm{\parallel }}_H\mathrm{\parallel }y{\mathrm{\parallel }}_H+\mathrm{\parallel }{x}^{\mathrm{\prime }}{\mathrm{\parallel }}_H\mathrm{\parallel }y-y^{\mathrm{\prime }}{\mathrm{\parallel }}_H\mathrm{.}$ Likewise $(x,y)\mapsto x⊠y$ is continuous into $K$ by [F5]. Strong simple approximants to $X_0$ and $X_j$ therefore make both tensor-valued maps strongly measurable. Their norms have expectations at most $(\mathbb{E}\mathrm{\parallel }{X}_0{\mathrm{\parallel }}^2{)}^{1\mathrm{/}2}(\mathbb{E}\mathrm{\parallel }{X}_j{\mathrm{\parallel }}^2{)}^{1\mathrm{/}2}={\sigma }^2$ by Cauchy–Schwarz and stationarity.

Fix $j\ge 1$ and put $Z_j=X_{j,j-1}$. For any $h,g\in H$, the scalar maps $x\mapsto ⟨x,g{⟩}_H$ and $y\mapsto ⟨h,y{⟩}_H$ are continuous. By [F1] and [F4] their values at $X_0$ and $Z_j$ are independent, and the first has mean zero. The bounded linear functional $L_{h,g}(T)=⟨Th,g{⟩}_H$ on $S_1(H)$ commutes with Bochner integration by [F2]–[F3], so $L_{h,g}({\mathbb{E}}_B[X_0\otimes Z_j])=\mathbb{E}[⟨h,Z_j{⟩}_H⟨X_0,g{⟩}_H]=0.$ For each $h$, taking $g={\mathbb{E}}_B[X_0\otimes Z_j]h$ shows that this vector has zero squared norm, so ${\mathbb{E}}_B[X_0\otimes Z_j]=0$; reversing the tensor factors gives ${\mathbb{E}}_B[Z_j\otimes X_0]=0$ as well. For the kernel expectation, pair ${\mathbb{E}}_B[X_0⊠Z_j]$ with any product kernel $g⊠h$. By [F3] and [F5], the scalar pairing is $\mathbb{E}[⟨X_0,g{⟩}_H⟨Z_j,h{⟩}_H]$, which is zero by independence and centering. Such product kernels span a dense subspace of $K$ by [F5], so ${\mathbb{E}}_B[X_0⊠Z_j]=0$; similarly ${\mathbb{E}}_B[Z_j⊠X_0]=0$.

By step 1.2, stationarity, and [F1], for every $j\ge 1$, $\begin{array}{rl}{\displaystyle \mathrm{\parallel }{\mathbb{E}}_B[X_0\otimes X_j]{\mathrm{\parallel }}_1}& {\displaystyle =\mathrm{\parallel }{\mathbb{E}}_B[X_0\otimes (X_j-Z_j)]{\mathrm{\parallel }}_1}\\ {\displaystyle }& {\displaystyle \le \mathbb{E}[\mathrm{\parallel }{X}_0{\mathrm{\parallel }}_H\mathrm{\parallel }{X}_j-Z_j{\mathrm{\parallel }}_H]\le \sigma {\delta }_{j-1}\mathrm{.}}\end{array}$ The same bound holds for ${\mathbb{E}}_B[X_j\otimes X_0]$. It also holds for $\mathrm{\parallel }{\kappa }_j{\mathrm{\parallel }}_K$ and $\mathrm{\parallel }{\tilde{\kappa }}_j{\mathrm{\parallel }}_K$ by [F3] and [F5]. Since ${\delta }_0\le 2\sigma$ and ${\sum }_{m\ge 1}{\delta }_m<\mathrm{\infty }$, both series of trace-class terms and product-kernel terms are absolutely summable. The Banach property of $S_1(H)$ in [F2] and the Hilbert completeness of $K$ in [A1], [F5] give their norm limits.

For $n\ge 1$, set $S_n={\sum }_{t=1}^n{X}_t$ and ${\mathrm{\Gamma }}_n=n^{-1}{\mathbb{E}}_B[S_n\otimes S_n]$. Step 1.1 makes this Bochner expectation well-defined. It is positive and self-adjoint because for every $h\in H$, $⟨{\mathrm{\Gamma }}_{n}h,h{⟩}_H=n^{-1}\mathbb{E}⟨S_n,h{⟩}_H^2\ge 0.$ Expanding the finite sum and using stationarity gives ${\mathrm{\Gamma }}_n={\mathbb{E}}_B[X_0\otimes X_0]+{\sum }_{j=1}^{n-1}(1-\frac{j}{n})({\mathbb{E}}_B[X_0\otimes X_j]+{\mathbb{E}}_B[X_j\otimes X_0])\mathrm{.}$ Absolute summability from step 2.1 implies ${\mathrm{\Gamma }}_n\to \mathrm{\Gamma }$ in trace norm: for any fixed finite lag cutoff the coefficients tend to $1$, and the remaining trace-norm tail is uniformly bounded by the tail of the absolutely convergent series. Since $\mathrm{\parallel }T\mathrm{\parallel }\le \mathrm{\parallel }T{\mathrm{\parallel }}_1$, positivity and self-adjointness pass to the limit. Thus $\mathrm{\Gamma }$ is positive self-adjoint and trace class.