Finite-dependent vector central limit theorem

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.