Plug-in long-run variance for a stable AR mean

Stability means that $\varphi$ has no zero in $\mathrm{\mid }z\mathrm{\mid }\le 1$. In particular $\varphi (1)=1-{\sum }_j{\varphi }_j\mathrm{\ne }0$. The finite augmented Gram matrix gives a finite second moment for each lagged coordinate, and strict stationarity gives the same finite second moment and mean at every time. Strict stationarity also makes the bivariate laws shift-invariant, so the covariance depends only on lag. Thus [F5] gives weak stationarity and verifies the solution premise of [F1]. The causal AR coefficients are then the Taylor coefficients of $1\mathrm{/}\varphi (z)$ and decay exponentially. Their sum converges absolutely; evaluating the power-series identity $\varphi (z)\mathrm{\Psi }(z)=1$ at $z=1$ gives $\mathrm{\Psi }=1\mathrm{/}\varphi (1)=1\mathrm{/}(1-{\sum }_j{\varphi }_j)$. Since ${\sigma }^2>0$ and $\varphi (1)$ is finite and nonzero, ${\sigma }^2{\mathrm{\Psi }}^2>0$.

By [F2], $U_n:=\sqrt{n}({\widehat{\beta }}_n-\beta )$ converges in distribution to a finite-dimensional normal law, so [F6] makes $U_n$ bounded in probability. Fix $\delta ,\eta >0$ and choose $M>0$ with ${\mathrm{lim\hspace{0.17em}sup}}_n\mathbb{P}(\mathrm{\parallel }{U}_n{\mathrm{\parallel }}_2>M)\le \eta$. For all sufficiently large $n$, $\sqrt{n}\delta >M$, so $\mathbb{P}(\mathrm{\parallel }{\widehat{\beta }}_n-\beta {\mathrm{\parallel }}_2>\delta )\le \mathbb{P}(\mathrm{\parallel }{U}_n{\mathrm{\parallel }}_2>M)$. Its limiting upper probability is at most $\eta$; letting $\eta \downarrow 0$ proves ${\widehat{\beta }}_n{\to }_{\mathbb{P}}\beta$, and hence ${\widehat{\varphi }}_n{\to }_{\mathbb{P}}\varphi$ because the slope subvector norm is at most the full-vector norm. Supplier [F2] also gives ${\widehat{\sigma }}_n^2{\to }_{\mathbb{P}}{\sigma }^2$.

Put $d_0:=1-{\sum }_j{\varphi }_j\mathrm{\ne }0$ and $d_n:=1-{\sum }_j{\widehat{\varphi }}_{n,j}$. Step 1.2 and finite coordinate pairing in [F3] give $({\widehat{\sigma }}_n^2,{\widehat{\varphi }}_n){\to }_{\mathbb{P}}({\sigma }^2,\varphi )$ and $d_n{\to }_{\mathbb{P}}{d}_0$. The everywhere-continuous map $g(s,v):=s\mathrm{/}\max \{(1-{\sum }_j{v}_j{)}^2,d_0^2\mathrm{/}4\}$ therefore satisfies $g({\widehat{\sigma }}_n^2,{\widehat{\varphi }}_n){\to }_{\mathbb{P}}{\sigma }^2\mathrm{/}{d}_0^2$ by [F3]. On $\{\mathrm{\mid }{d}_n\mathrm{\mid }\ge \mathrm{\mid }{d}_0\mathrm{\mid }\mathrm{/}2\}$ this map equals the displayed estimator ${\widehat{\tau }}_n^2$. The complement has probability at most $\mathbb{P}(\mathrm{\mid }{d}_n-d_0\mathrm{\mid }\ge \mathrm{\mid }{d}_0\mathrm{\mid }\mathrm{/}2)\to 0$. For every $\epsilon >0$, the estimator's deviation probability is bounded by that complement probability plus $\mathbb{P}(\mathrm{\mid }g({\widehat{\sigma }}_n^2,{\widehat{\varphi }}_n)-{\sigma }^2\mathrm{/}{d}_0^2\mathrm{\mid }>\epsilon )$, so it tends to zero. This proves consistency including the zero-denominator fallback.

By step 1.1, the limiting long-run variance is ${\sigma }^2{\mathrm{\Psi }}^2>0$; step 2.1 proves that ${\widehat{\tau }}_n^2$ is a nonnegative consistent estimate of it. The exponential coefficient bound in step 1.1 gives ${\sum }_j\sqrt{j+1}\mathrm{\mid }{\psi }_j\mathrm{\mid }<\mathrm{\infty }$, so all additional hypotheses of [F4] hold; its AC branch is supplied by [A1]. The stable linear-process CLT gives the stated standardized mean limit and interval coverage. This uses the fitted OLS innovation variance and fixed-order slope estimate; it makes no claim for a data-selected order or an unspecified HAC estimator.