text:
\section{Introduction}

A stationary sequence $\{X(n)\}$ with finite variance is said to have \emph{long memory} or  \emph{long-range dependence}, if
\begin{equation}\label{eq:cov LRD}
\mathrm{Cov}[X(n),X(0)]\approx n^{2H-2}
\end{equation}
as $n\rightarrow\infty$, where $\approx$ means asymptotic equivalence up to a positive constant\footnote{In general, we can allow inserting a slowly varying factor (\citet{bingham:goldie:teugels:1989:regular}), e.g.,  a logarithmic function,   in the asymptotic relation (\ref{eq:cov LRD}), but we shall not do that for simplicity. }, the parameter  $H\in (1/2,1)$ is called the \emph{Hurst index}.  See, e.g., the recent monographs   \citet{giraitis:koul:surgailis:2009:large}, \citet{beran:2013:long}, \citet{samorodnitsky:2016:stochastic} and \citet{pipiras:taqqu:2017:long}  for more information on the notion long memory.

The study of the asymptotic behavior of partial sums of long-memory sequences has been of great interest in probability and statistics. In particular, complete results have been obtained for the following  class of models called \emph{Gaussian subordination}. Let $\{Y(n)\}$ be a stationary Gaussian process with long memory in the sense of (\ref{eq:cov LRD}), which we suppose to have mean $0$ and variance $1$ without loss of generality. Now consider the transformed process
\begin{equation}
X(n)=G(Y(n)),
\end{equation}
where $G(\cdot):\mathbb{R}\rightarrow\mathbb{R}$ is a function such that $\mathbb{E} X(n)^2=\mathbb{E} G(Y(n))^2<\infty$.  The goal is to establish limit theorems for the normalized sums  of $X(n)$, the transformed stationary process.

To develop the limit theorems, one has to use the notion of Hermite rank, which is
  an integer  attached to the function $G(\cdot)$. It is defined as follows.
Let $Z$ denote a standard Gaussian random variable  and let $\gamma(dx)=\phi(x)dx:=(2\pi)^{-1/2} e^{-x^2/2} dx$  denote its distribution.  The Hermite rank is associated with the orthogonal decomposition  of $G(\cdot)$ in the space
\[L^2(\gamma)=L^2(\phi)=\{G(\cdot):~\mathbb{E} G(Z)^2<\infty\}=\left\{G(\cdot):~\int_{\mathbb{R}} G(z)^2 \phi(z)dz<\infty\right\}
\]  into \emph{Hermite polynomials}, which are defined as $H_0(x)=1$, $H_1(x)=x$, $H_2(x)=x^2-1$, and more generally,
\begin{equation}\label{eq:Herm Poly}
H_m(x)=(-1)^m e^{x^2/2}\frac{d}{dx^m}e^{-x^2/2} \quad\text{ for }\quad m\ge 1.
\end{equation}
 Then $\{\frac{1}{\sqrt{m!}}H_m(\cdot), m\ge 0\}$ forms an orthonormal basis of $L^2(\gamma)$ (see \citet{pipiras:taqqu:2017:long}, Proposition 5.1.3).

\begin{Def}\label{Def:Herm rank}
  The  \emph{Hermite  rank}  $k$ of $G(\cdot)\in L^2(\gamma)$ is defined as
\begin{equation}\label{eq:def herm rank}
k=\inf\left\{m\ge 1: ~\mathbb{E} G(Z) H_m(Z)=\int_{\mathbb{R}}G(x)H_m(x)\gamma(dx)\neq 0 \right\},
\end{equation}
where $H_m(\cdot)$ is the $m$-th order Hermite polynomial.
Equivalently, $k$ is the order of the first nonzero coefficient in the $L^2(\gamma)$-expansion:
\begin{equation}\label{eq:def herm rank expan}
G(\cdot)-\mathbb{E} G(Z)=\sum_{m=k}^\infty c_m H_m(\cdot), \quad k\ge 1,
\end{equation}
that is, $c_m=0$ for $m<k$ and $c_k\neq 0$ where
\begin{equation}\label{eq:Herm Coeff}
c_m={\mathbb{E}[G(Z)H_m(Z)]\over m!} ~~\text{ for }~~m\ge 0.
\end{equation}
\end{Def}
The following celebrated results due to \citet{dobrushin:major:1979:non},  \citet{taqqu:1979:convergence} and \citet{breuer:major:1983:central} have been established involving the Hermite rank.
\begin{Thm}\label{Thm:nclt gaussian}
Suppose that $G(\cdot)\in L^2(\gamma)$ has Hermite rank $k\ge 1$.  Then the following conclusions hold.

\noindent $\bullet$   Central limit case: suppose that $1/2<H<1-\frac{1}{2k}$ (this implies $k\ge 2$).  Then $\{G(X(n))\}$ has short memory in the sense that   \[\sigma^2:=\sum_{n=-\infty}^\infty \mathrm{Cov}[G(X(n)),G(X(0))]\] converges absolutely and
\begin{equation}\label{eq:Herm limit}
\frac{1}{N^{1/2}}\sum_{n=1}^{[Nt]} \Big( G(X(n))-\mathbb{E} G(X(n)) \Big) \overset{f.d.d.}{\longrightarrow} \sigma B(t),\quad t\ge 0,
\end{equation}
where $\overset{f.d.d.}{\longrightarrow}$ denotes convergence of the finite-dimensional distributions and $B(t)$ is the standard Brownian motion.

\noindent $\bullet$ Non-central limit case:  suppose that $1-\frac{1}{2k}<H<1$.
$\{G(X(n))\}$ has long memory with Hurst index:
\begin{equation}\label{eq:H_G general}
H_G=(H-1)k+1\in \left(\frac{1}{2},1\right).
\end{equation}
Furthermore, as $N\rightarrow\infty$, we have\footnote{In fact,  we have weak convergence in the space $D[0,1]$  with uniform metric.}
\begin{equation}\label{eq:Herm limit2}
\frac{1}{N^{H_G}}\sum_{n=1}^{[Nt]} \Big( G(X(n))-\mathbb{E} G(X(n)) \Big) \overset{f.d.d.}{\longrightarrow} c Z_{H,k}(t),
\end{equation}
for some $c\neq 0$,  and
\begin{equation}\label{eq:Herm process}
Z_{H,k}(t)= \int_{\mathbb{R}^k}' \left[ \int_0^t \prod_{j=1}^k (s-x_j)_+^{H-3/2} ds\right]  B(dx_1)\ldots B(dx_k),
\end{equation}
is the so-called $k$-th order Hermite process,
where
 $\int_{\mathbb{R}^k}'   \cdot  \ B(dx_1)\ldots B(dx_k)$ denotes the $k$-tuple Wiener-It\^o integral with respect to the standard Brownian motion $B(\cdot)$ (\citet{major:2014:multiple}).
\end{Thm}

 \citet{bai:taqqu:2017:instability} point out  that despite  the probabilistic interest of Theorem \ref{Thm:nclt gaussian}, its  straightforward application  to large-sample statistical theories can be problematic. This is due to a   strong instability feature in the notion Hermite rank. This paper can be viewed as a technical companion to \citet{bai:taqqu:2017:instability} containing some mathematical characterization of such instability and related problems. In particular, the paper contains the following results:
\begin{enumerate}
\item The instability of the Hermite rank with respect to transformations including shift, scaling, etc.

\item A ``near higher order rank'' analysis of Theorem \ref{Thm:nclt gaussian} perturbed by a diminishing shift.

\item Coincidence of Hermite rank and power rank (\citet{ho:hsing:1997:limit}) in the Gaussian case.
\end{enumerate}
The third result is not closely related to the first two, but it is obtained as a direct byproduct of the analysis of the instability problems.
The  paper is organized as follows:
the   results described above are stated in Sections  \ref{Sec:affine}, \ref{Sec:near} and \ref{Sec:coin}.    Section \ref{Sec:aux} contain some auxiliary results involving
analytic function theory.
The theorems are proved in Section \ref{Sec:proof}.

Throughout the paper, $a_N\ll b_N$ means as $N\rightarrow\infty$, $a_N/b_N\rightarrow 0$, and $a_N\approx b_N$ means $a_N/b_N\rightarrow c$ for some positive constant $c>0$.
\section{How a transformation affects the Hermite rank}\label{Sec:affine}
First we state the results regarding the instability  of the Hermite rank  under transformation. Their relevance is explained in  Remark \ref{Rem:instab} which follows.  The proofs are given in Section \ref{Sec:proof}.

   Recall that $x$ is an \emph{accumulation point} or \emph{limit point} of a set $E\subset\mathbb{R}^p$,  if every neighborhood of $x$ contains an infinite number of elements of $E$.  The so-called \emph{derived set} $E'$ consists of all the accumulation points of $E$. For example, if $E=\{0\}\cup [1,2)$, then $ E'=[1,2]$.

\begin{Thm}[Instability with respect to  shift]\label{Thm:shift}
Suppose that the measurable function $G(\cdot):\mathbb{R}\rightarrow\mathbb{R}$ is  not a.e.\ constant  and $G(\cdot\pm M)\in L^2(\gamma)$ for some $M>0$. Then $G(\cdot +x)\in L^2$ for all $x\in[-M,M]$, and the  interval $(-M,M)$ contains no accumulation point of the set
 \[
 E=\{x\in (-M,M):  ~ G(\cdot+x) \text{ has Hermite rank   }  \ge 2\},
 \]
namely, $E'\cap (-M,M)=\emptyset$.
\end{Thm}

\begin{Thm}[Instability with respect to  scaling]\label{Thm:scale}
Suppose that the measurable function $G:\mathbb{R}\rightarrow \mathbb{R}$ is not a.e.\ symmetric  and    $G(\cdot\times M)\in L^2(\gamma)$ for some   $M>1$.    Then $G(\cdot \times y)\in L^2$ for all $y\in (0,M]$, and the interval $(0,M)$ contains no accumulation point of the set
 \[
 E=\{y\in (0,M):  ~ G(\cdot\times y) \text{ has Hermite rank   } \ge  2\},\]
namely,  $E'\cap (0,M)=\emptyset$.
\end{Thm}
\begin{Rem}\label{Rem:sym}
The requirement of $G$ being non-symmetric in Theorem \ref{Thm:scale} is essential because any non-zero symmetric function $f\in L^2(\gamma)$ has Hermite rank  $2$ since due to the symmetry of the measure $\gamma$
\[
\int_{\mathbb{R}} f(z)H_1(z) \gamma(dz)=\int_{\mathbb{R}} f(z)z \gamma(dz)=0,
\] while
\[
\int_{\mathbb{R}} f(z) (z^2-1) \gamma(dz)=
\int_{\mathbb{R}} f(z) H_2(z) \gamma(dz)=2\int_0^\infty f(z)z^2 \gamma(dz)>0
\] due to symmetry of $\gamma$.
\end{Rem}
\begin{Rem}\label{Rem:instab}
The preceding two theorems point to the \emph{instability of the Hermite rank}. They imply, for example, that the particular values $x=0$ in Theorem \ref{Thm:shift} or $y=1$ in Theorem \ref{Thm:scale} cannot be an accumulation point of the set $E$ with  Hermite rank greater or equal to $2$. Hence some neighborhood of $x=0$ or $y=1$ contains      at most one point ($x=0$ or $y=1$) with Hermite  rank $k\ge 2$. This means that a slight  level shift or scale change (when the original transformation is non-symmetric) will force the   rank to change from perhaps being $\ge 2$  to being  $1$. In the case of shift, this is stated as Theorem 2.14 in \citet{bai:taqqu:2017:instability}. Note that the rank $2$ of a non-constant symmetric function in Remark \ref{Rem:sym}  is still unstable with respect to a level shift by Theorem \ref{Thm:shift}.
\end{Rem}

In addition, one can consider the       shift and scaling  joint together, namely, deal with an affine transformation.
\begin{Thm}[Instability with respect to  affine transformation]\label{Thm:joint}
Suppose that the measurable $G(\cdot)$ is not constant a.e. and $G(\pm M_1+\cdot\times M_2)\in L^2(\gamma)$. Then the  set \[
E=\{(x,y)\in (-M_1,M_1)\times (0,M_2):  ~ G(x+ \cdot \times y ) \text{ has Hermite rank   }\ge   2\}\]
has Hausdorff dimension (see \citet{falconer:2004:fractal}, Section 2.2) not exceeding $1$. So in particular,  $E$ has $0$ two-dimensional Lebesgue measure.
\end{Thm}

To illustrate Theorem \ref{Thm:joint}, consider the function $G(z)=z^2=1+H_2(z)$ which has Hermite rank $2$. Suppose that it is perturbed by an affine transformation involving a shift $x$ and a scale $y$ and becomes
\[(x+zy)^2=x^2+2xyz+y^2z^2= x^2+y^2 + (2xy)z+y^2(z^2-1)=x^2+y^2 + (2xy)H_1(z)+y^2H_2(z).
\]
After centering,    the centered function in $z$ becomes $(2xy)H_1(z)+y^2H_2(z)$, which has Hermite rank $1$  if and only if $x\neq 0$ (it no longer has Hermite rank $2$ as $G(z)$ does) . The  set $E$ in Theorem \ref{Thm:joint} corresponds to the one-dimensional line $\{(x,y):~x=0\}$ which has Lebesgue measure zero.

 Finally, we can formulate an abstract result about the instability with respect to  general nonlinear transformations.
\begin{Thm}\label{Thm:nonlinear per}
Suppose that $G\in L^2(\gamma)$  and let $F_\theta$ be a family of transformations parameterized by  $\theta\in D\subset \mathbb{R}^p$, where $D$ is an open region containing the origin $\mathbf{0}=(0,\ldots,0)$ and let $F_\mathbf{0}$ be the identity transformation. Suppose that $G$ is perturbed and becomes $G\circ F_\theta$.

 Assume that  $G\circ F_\theta\in L^2(\phi)$ for all $\theta\in D$, $Z$ is a standard normal random variable, and
let
\[
U(\theta):= \mathbb{E} [  (G\circ F_\theta)(Z)\times H_1( Z)]=\mathbb{E} [ (G\circ F_\theta)(Z)\times Z  ].
\]
be the first coefficient of the Hermite expansion of $G\circ F_\theta$, and
  assume that the following properties hold:
\begin{enumerate}[(a)]
\item  $U(\theta)$ is real analytic in $\theta\in D$;
\item   If $U(\theta)=0$ holds for all $\theta\in D$, then $G$ is   constant a.e..
\end{enumerate}
 Then either $G$ is constant a.e., or if not, then the set
 \[E=\{\theta\in D: ~G\circ F_\theta \text{ has Hermite rank }\ge 2\}=\{\theta\in D:  ~U(\theta)=0\}\] has Hausdorff dimension  not exceeding $p-1$. In particular,  $E$ has zero $p$-dimensional Lebesgue measure.
\end{Thm}

\begin{Rem}
Here is a more informative description of the set $E$ in   Theorem \ref{Thm:joint} and \ref{Thm:nonlinear per} above.  First, $E$ is called the zero set of $U$ if it consists of all $\theta\in D$ such that $U(\theta)=0$. Second,       the zero set $E=\{\theta\in D:~U(\theta)=0\}$  of a (multivariate) real analytic function $U$ is called a real analytic variety. According to Lojasiewicz's Structure Theorem (Theorem 6.3.3 of \citet{krantz:parks:2002:primer}),  $E\subset \mathbb{R}^p$ can be expressed as a union of   submanifolds of dimensions $0,1,\ldots,p-1$. We state here the results in terms of the Hausdorff dimension   for simplicity.
\end{Rem}

\section{Being near a higher order Hermite rank}\label{Sec:near}
 In the so-called near integration analysis of unit root (e.g., \citet{phillips:1987:towards}), one studies the asymptotic behavior of an autoregressive model when the autoregression coefficient  tends to $1$ (unit root) as the sample size increases. Such analysis sheds lights on the situation where the coefficient is close to but not exactly $1$.
Here in a similar spirit, we carry out a ``near higher order rank''  analysis  for Theorem \ref{Thm:shift} regarding the limit theorem for the sum $\sum_{n=1}^N G(Y(n)+x_N)$, where the shift $x_N\rightarrow 0$  as the number of summands $N\rightarrow\infty$. The result is given in the following theorem,  where  $c,c_i$'s denote   constants whose value can change from line to line.
\begin{Thm}\label{Thm:near higher order}
Let $\{Y(n)\}$ be a  standardized stationary Gaussian process whose covariance satisfies (\ref{eq:cov LRD}) with Hurst index $H\in (1/2,1)$. Suppose that the function $G(\cdot)$ is not constant a.e.,     $G(\cdot\pm M)\in L^2(\gamma)$ for some $M>0$, and has an Hermite rank $k\ge 1$.  Set
\begin{equation}\label{e:s1}
S_N(x_N)=\sum_{n=1}^N\Big[G(Y(n)+x_N)-\mathbb{E} G(Y(n)+x_N)\Big],
\end{equation}
where $x_N\rightarrow 0$.

\noindent $\bullet$  Central limit case: suppose that $1/2<H<1-{1\over {2k}} $ (if $k=1$ this case does not exist).
Then as $N\rightarrow\infty$, $N^{(1/2-H)/(k-1)}\rightarrow 0$ and
\begin{flalign*}
    \begin{array}{L@{\quad}l@{{}\overset{f.d.d.}{\longrightarrow}{}}l}
    (a)    if  $ x_N\ll N^{(1/2-H)/(k-1)}$:   &  N^{-1/2}S_{[Nt]}(x_N)   &  c B(t);   \\
     (b)  if  $x_N\approx N^{(1/2-H)/(k-1)}$:        &N^{-1/2}S_{[Nt]}(x_N)   & c_1 Z_{H,1} (t) + c_2 B (t); \\
     (c)  if  $ N^{(1/2-H)/(k-1)}\ll x_N \rightarrow 0$: & N^{-H} x_N^{1-k} S_{[Nt]}(x_N)   &  c  Z_{H,1} (t);
    \end{array}
\end{flalign*}
where the fractional Brownian motion $Z_{H,1}(t)$ and the Brownian motion $B(t)$ are independent.

\medskip
\noindent $\bullet$  Non-central limit case:  suppose that $1-{1\over {2k}}<H<1$.
 Let $H_G$ be
as in (\ref{eq:H_G general}).
Then as $N\rightarrow\infty$, $N^{H-1}\rightarrow 0$, and
\begin{flalign*}
 \begin{array}{L@{\quad}l@{{}\overset{f.d.d.}{\longrightarrow}{}}l}
    (a)    if  $ 0<x_N\ll N^{H-1}$:   &  N^{-H_G}S_{[Nt]}(x_N) & c Z_{H,k} (t)  ;   \\
     (b)   if  $x_N\approx N ^{H-1}$: & N^{-H_G}S_{[Nt]}(x_N)&\sum_{m=1}^k  c_m Z_{H,m} (t); \\
     (c)  if  $N^{H-1}\ll x_N\rightarrow 0$:& N^{-H} x_N^{1-k} S_{[Nt]}(x_N)  & c  Z_{H,1} (t);
    \end{array}
\end{flalign*}
where the Hermite processes $Z_{H,m}(t)$'s  are defined through the same Brownian motion in (\ref{eq:Herm process}).
\end{Thm}

The terms "central limit case" and "non-central limit case" in the preceding theorem refer to the
 terminology used in Theorem \ref{Thm:nclt gaussian} and thus to the type of limits one obtains when there is no shift, that is when $x_N=0$.
\begin{Rem}
Theorem \ref {Thm:near higher order} has some interesting implications.
In the central limit case,     the critical order $N^{(1/2-H)/(k-1)}$  depends negatively on $H$. The larger $H$, the smaller the critical order. Thus the larger $H$ (but below $1-1/(2k)$) is, the more easily the effect of a higher-order rank   in the limit theorem gets surpassed by a shift perturbation. In  the non-central limit case, however, the relation is reversed since the critical order $N^{H-1}$ depends positively on $H$. Note that in the non-central limit case,  the order $N^{H-1}$ determining the border of the regimes  does not depend on the rank $k$
\end{Rem}

In Theorem \ref{Thm:near higher order non-center} below, we  provide a result  on  the \emph{non-centered} sum $\sum_{n=1}^N  G(Y(n)+x_N)$, assuming that $\mathbb{E} G(Y(n))=0$. It is expected that if $x_N$ tends to $0$ too slowly, then a deterministic trend will appear in the limit.

\begin{Thm}\label{Thm:near higher order non-center}
Under the assumptions of Theorem \ref{Thm:near higher order}, if in addition $\mathbb{E} G(Y(n))=0$ and $S_N$ is replaced by
\begin{equation}\label{e:s2}
\widetilde{S}_N(x_N)=\sum_{n=1}^N  G(Y(n)+x_N),
\end{equation}
then the following conclusions hold:

\noindent $\bullet$  Central limit case:    suppose that  $H<1-{1\over {2k}} $.
Then as $N\rightarrow\infty$:
\begin{flalign*}
 \begin{array}{L@{\quad}l@{{}\overset{f.d.d.}{\longrightarrow}{}}l}
    (a) if $x_N\ll N^{-1/(2k)}$: & N^{-1/2}\widetilde{S}_{[Nt]}(x_N) & c B(t)\\
 (b) if $x_N\approx N^{1-1/(2k)}$:& N^{-1/2}\widetilde{S}_{[Nt]}(x_N)& c_1t+ c_2 B(t);   \\
     (c) if  $N^{1-1/(2k)}\ll x_N\rightarrow 0$: & N^{-1}x_N^{-k}\widetilde{S}_{[Nt]}(x_N)& ct ;
    \end{array}
\end{flalign*}
where $Z_{H,k}(t)$ and $B(t)$ are independent.

\medskip
\noindent $\bullet$  Non-central limit case:    suppose that  $H>1-{1\over {2k}}$.
 Let $H_G$ be
as in (\ref{eq:H_G general}).
Then as $N\rightarrow\infty$:
\begin{flalign*}
 \begin{array}{L@{\quad}l@{{}\overset{f.d.d.}{\longrightarrow}{}}l}
    (a) if  $0<x_N\ll N^{H-1}$:  &N^{-H_G}\widetilde{S}_{[Nt]}(x_N) & c Z_{H,k} (t)  ;\\
 (b) if  $x_N\approx N ^{H-1}$: & N^{-H_G}\widetilde{S}_{[Nt]}(x_N) & c_0t+   \sum_{m=1}^k  c_m Z_{H,m} (t)  ;   \\
     (c) if  $N^{H-1}\ll x_N\rightarrow 0$:   & N^{-1} x_N^{-k} \widetilde{S}_{[Nt]}(x_N) & ct  ;
    \end{array}
\end{flalign*}
 where the Hermite processes $Z_{H,m}(t)$'s  are defined through the same Brownian motion in (\ref{eq:Herm process}).
\end{Thm}

In the next theorem the argument $x_N$ in Theorem \ref{Thm:near higher order non-center} is replaced by a subtracted sample mean.
\begin{Thm}\label{Thm:near higher order centering}
Under the assumptions of Theorem \ref{Thm:near higher order non-center},   assume in addition that $G(\cdot)$ is a polynomial. Let $\bar{Y}_N=N^{-1}\sum_{n=1}^N Y(n)$. Then the following conclusions hold:

\noindent $\bullet$  Central limit case:    suppose that  $H<1-{1\over {2k}} $. Then as $N\rightarrow\infty$,
\[
N^{-1/2} \widetilde{S}_{[Nt]}(-\bar{Y}_N)\overset{f.d.d.}{\longrightarrow} cB(t)
\]
\medskip
\noindent $\bullet$  Non-central limit case:    suppose that  $H>1-{1\over {2k}}$.
 Let $H_G$ be
as in (\ref{eq:H_G general}).
Then as $N\rightarrow\infty$,
\[
N^{-H_G}\widetilde{S}_{[Nt]}(-\bar{Y}_N)\overset{f.d.d.}{\longrightarrow}  \sum_{m=1}^k c_m Z_{H,m}(t)
\]
where   the Hermite processes $Z_{H,m}(t)$ 's are defined through the same Brownian motion in (\ref{eq:Herm process}).
\end{Thm}
\begin{Rem}
As an example, one may take $G(z)=z^2$ in (\ref{e:s2}), which leads to the sample variance as considered in  \citet{hosking:1996:asymptotic} and \citet{dehling:taqqu:1991:bivariate}. See also Section 3.1 of \citet{bai:taqqu:2017:instability}. Comparing Theorem \ref{Thm:near higher order centering} with Theorem \ref{Thm:nclt gaussian}, centering $\{Y(n)\}$ by subtracting the sample mean may not affect the fluctuation order of the sum, but  can change the limit distribution.
\end{Rem}

The proofs of Theorem \ref{Thm:shift}, \ref{Thm:scale}, \ref{Thm:joint} and Theorems \ref{Thm:near higher order}, \ref{Thm:near higher order non-center} and
  \ref{Thm:near higher order centering} can  be found in   Section \ref{Sec:proof}, and are all based on the analysis of analytic functions.

\section{On the coincidence of Hermite and power ranks}\label{Sec:coin}

The Hermite rank has been defined in (\ref{eq:def herm rank}). We now define the \emph{power rank} which is used in the approach of \citet{ho:hsing:1997:limit} for limit theorems for transformations of long-memory  moving-average processes.
Given a function $G(\cdot)$ and a random variable $Y$ satisfying $\mathbb{E} G(Y)^2<\infty$, let
\begin{equation}\label{eq:G_infty}
G_\infty(y)=\mathbb{E}  G (Y+y)
\end{equation}
given that the expectation exists  and suppose that $G_\infty(\cdot)$ has   derivatives of  order sufficiently high. The power rank of $G(\cdot)$ with respect to $Y$ is defined as
\begin{equation}\label{eq:power rank}
\inf\{m \ge 1:~ G_\infty^{(m)}(0)\neq 0  \},
\end{equation}
where $G_\infty^{(m)}(y)$ denotes the $m$-th derivative of $G_\infty(y)$.
The following fact  was stated in \citet{ho:hsing:1997:limit} without a detailed proof. It was proved by \citet{levy-leduc:taqqu:2013:long} in the case where $G(\cdot)$ is a polynomial.
\begin{Pro}\label{Pro:coincide}
Suppose that $G(\cdot)\in L^2(\gamma)$. Then
the power rank in (\ref{eq:power rank}) coincides with the Hermite rank \eqref{eq:def herm rank} if $Y$ is Gaussian.
\end{Pro}
\begin{proof
Proposition \ref{Pro:coincide} is a direct consequence of (\ref{eq:ana expan G_infty}) below, namely,
\[ \sum_{m=0}^\infty \frac{G_\infty^{(m)}(0)}{m!} x^m =\sum_{m=0}^\infty \frac{\mathbb{E}[G(Z)H_m(Z)]}{m!} x^m,
\]
\end{proof}

\section{Auxiliary results}\label{Sec:aux}
We prove here some auxiliary results involving the Weierstrass transform and analytic function theory.

Let $Z$ denote throughout a standard Gaussian random variable and let $\phi=d\gamma/dx=(2\pi)^{-1/2}e^{-x^2/2}$ denote its density.  Define  the function
\begin{equation}\label{eq:G_infty}
 G_\infty(x)= \mathbb{E}  G (Z+x)=\int_{\mathbb{R}}{G}(z+x)\phi(z)dz=\int_{\mathbb{R}}{G}(z)\phi(z-x)dz,
\end{equation}
whenever the integrability holds.
The function $G$   may not be smooth, but the function $G_\infty$ is   smooth due to the convolution with the smooth $\phi(z)$.  $G_\infty$ is called the \emph{Weierstrass transform} of $G$ (see \citet{hirschman:widder:2005:convolution}, Chapter VIII).

First we state some preliminary   facts.
Recall that a function $f:\mathbb{R}^p\rightarrow \mathbb{R}$ is real analytic over an open domain $D\subset\mathbb{R}^p$, if at every point $y=(y_1,\ldots,y_p)\in D$, there exists a  neighborhood $B\subset D$ of $y$, such that
\[
f(x_1,\ldots,x_p)=\sum_{i_1,\ldots,i_p=1}^\infty a_{i_1,\ldots,i_p}  (x_1-y_1)^{i_1}\ldots (x_p-y_p)^{i_p}
\]
for some coefficient $a(i_1,\ldots,i_k)$, where the series converges absolutely in $B$.
It is well-known that $f$ is  infinitely differentiable, and common elementary operations including composition, affine transform, multiplication preserve  analyticity. See, e.g., \citet{krantz:parks:2002:primer}, Chapter 1.
\begin{Lem}\label{Lem:shift scale still int}~
\begin{enumerate}[(a)]
\item
If $g(\cdot+ a)$, $g(\cdot+ b) \in L^1(\phi)$, $a<b$, then
\begin{enumerate}
\item[(a1)]
$g(\cdot + x)\in L^1(\phi)$ for any $a<x<b$;
\item[(a2)]
 the Weierstrass transform $\mathbb{E} g(Z+x)$ is a real analytic function in $x\in (a,b)$.
\end{enumerate}
\item If $g(\cdot \times c)\in L^1(\phi)$, then $g(\cdot \times y)\in L^1(\phi)$ for any $0<y<c$.
\end{enumerate}
\end{Lem}
\begin{proof}

(a) The first statement (a1) follows from the fact that $\phi(z-x)\le  A\phi(z- a)+ B \phi(z- b)$ for some sufficiently large constants $A,B>0$ by exploring the shape  of $\phi$.   To obtain  statement (a2), first as in item 2.2 of \citet{hirschman:widder:2005:convolution} Chapter VIII, the Weierstrass transform $(\mathcal{W} g)(x):=\mathbb{E} g(Z+x)=(g\ast \phi)(x)$ can be expressed by  a bilateral Laplace transform  $(\mathcal{L}g)(x):=\int_{\mathbb{R}}e^{-zx}g(z)dz$  as
\begin{equation}\label{eq:laplace}
(\mathcal{W}g)(x)=\frac{1}{\sqrt{2\pi}}\int_{\mathbb{R}} f(z)  e^{-(x-z)^2/2}dz=\frac{e^{-x^2/2}}{\sqrt{2\pi}} \int_{\mathbb{R}}   e^{(-x)(-z)} [f(z) e^{-z^2/2}]dz=  \frac{e^{-x^2/2}}{\sqrt{2\pi}} (\mathcal{L} g)(-x),
\end{equation}
where $g(z)=f(z) e^{-z^2/2}$. Note that $ e^{-x^2/2}$ is analytic.
Then the analyticity   of $\mathcal{W} g$  follows from the analyticity of the bilateral Laplace transform  $\mathcal{L} g$. See \citet{widder:2010:laplace} Chapter VI, p.140. The statement
(b) follows from $\phi(x/y)\le C \phi(x/c)$ for some sufficiently large constant $C>0$, because $\int_{\mathbb{R}} |g(xc)|\phi(x)dx= c^{-1}\int_{\mathbb{R}} |g(x)|\phi(x/c)dx<\infty$
\end{proof}

\begin{Lem}\label{Lem:analytic}
Suppose $G\in L^2(\phi)$. Then $\mathbb{E}  |G (Z+x)|<\infty$, namely, $G(x+\cdot)\in L^1(\phi)$, for all $x\in\mathbb{R}$. Furthermore,  $G_\infty(x)$ in (\ref{eq:G_infty}) admits the analytic expansion
\begin{equation}\label{eq:ana expan G_infty}
G_\infty(x)=\sum_{m=0}^\infty \frac{G_\infty^{(m)}(0)}{m!} x^m =\sum_{m=0}^\infty c_m x^m,
\end{equation}
where $G_\infty^{(m)}$ is the $m$-th derivative of $G_\infty$ and \[c_m=\frac{\mathbb{E}[G(Z)H_m(Z)]}{m!}\] are the coefficients of the Hermite expansion   in (\ref{eq:Herm Coeff}).

If in addition, $G(\cdot \pm M)\in L^2(\phi)$, $M>0$, then
\begin{equation}\label{eq:G^m(x)}
 G_\infty^{(m)}(x)  =  \mathbb{E} [G(Z+x) H_m(Z)],\qquad |x|<M,  \quad m=0,1,2,\ldots.
\end{equation}
\end{Lem}
\begin{proof}
Let $H_m(x)$'s be the Hermite polynomials defined in \eqref{eq:Herm Poly}. Then  by the Cauchy-Schwartz inequality and Corollary 5.1.2 of \citet{pipiras:taqqu:2017:long},   one has for any $x\in \mathbb{R}$ that
\begin{equation}\label{eq:abs sum}
\sum_{m=1}^\infty \frac{|x|^m}{m!} \int_{\mathbb{R}}|{G}(z) H_m(z)|\phi(z) dz \le \sum_{m=1}^\infty \frac{|x|^m}{m!}  \|{G}\|_{L^2(\phi)} \|H_m\|_{L^2(\phi)}=\|{G}\|_{L^2(\phi)}\sum_{m=1}^\infty \frac{|x|^m}{\sqrt{m!}}  <\infty
\end{equation}
since $\|H_m\|_{L^2(\phi)}=\sqrt{m!}$.
By Proposition 1.4.2 of  \citet{nourdin:peccati:2012:normal},  one has
\[
\sum_{m=0}^\infty \frac{x^m}{m!}H_m(z)=\exp( -(z-x)^2/2 +z^2/2) .
\]
So
\[
\sum_{m=0}^\infty \frac{ x^m}{m!} \int_{\mathbb{R}} {G}(z) H_m(z) \phi(z) dz =\int_{\mathbb{R}}{G}(z)   e^{ -(z-x)^2/2 +z^2/2} \phi(z)  dz=\int_{\mathbb{R}}{G}(z)   \phi(z-x)  dz=G_\infty(x),
\]
where the change of the order between the sum and the integral in the first equality  can be justified by (\ref{eq:abs sum}) and Fubini's Theorem.
This shows that $G_\infty$   has the desired expansion (\ref{eq:ana expan G_infty}).  Note that the same computation as above with $G$ replaced by $|G|$   shows that $\mathbb{E}  |G (Z+x)|=\int_{\mathbb{R}}|{G}(z)|   \phi(z-x)  dz<\infty$ by (\ref{eq:abs sum}).

The formula (\ref{eq:G^m(x)}) follows from
\[
 G_\infty^{(m)}(x)=\frac{d}{dx^m}\int_{\mathbb{R}} G(z) \phi(z-x)dz =\int_{\mathbb{R}} G(z) \frac{d}{dx^m}\phi(z-x)dz= \int G(z)H_m(z-x)\phi(z-x)dx
\]
by (\ref{eq:Herm Poly}), where the differentiation under the integral can be justified by the Dominated Convergence Theorem and the Mean Value Theorem as in the proof of Lemma \ref{Lem:biv ana} below.
\end{proof}

\begin{Lem}[Uniqueness of  Weierstrass transform]\label{Lem:Gaussian vanish}
If $f\in L^1(\phi)$ and the convolution $f\ast \phi=0$ a.e., then $f=0$ a.e..
\end{Lem}
\begin{proof}
First, express as in (\ref{eq:laplace}) the Weierstrass transform  by a bilateral Laplace transform. Then the conclusion follows from the uniqueness of the bilateral Laplace transform. See \citet{widder:2010:laplace} Chapter VI, Theorem 6b.
\end{proof}

\begin{Lem}\label{Lem:scale ana}
Suppose that $G(\cdot\times M)\in L^2(\phi)$ for some $M\in (1,+\infty)$. Then $\mathbb{E} |G(yZ)Z|<\infty$ and
\[F(y):=\mathbb{E} G(yZ)Z
\] is an analytic function in  $ y\in (0,M)$.
\end{Lem}
\begin{proof}
Since $\phi(t/y)\le c \phi (t/M)$ for some constant $c>0$,
\[
\mathbb{E} |G(yZ)Z|=\int_\mathbb{R} |G(y  z)z|\phi(z)dz=\frac{1}{y^2}\int_\mathbb{R} |G(t)t|\phi(t/y)dt \le C\int_\mathbb{R} |G(M  z)z|\phi(z)dz,
\]
and hence by Cauchy-Schwartz,
\[
\mathbb{E} |G(yZ)Z|\le  C \|G(\cdot \times M)\|_{L^2(\phi)} <\infty
\]
for some constant $C>0$   depending only on $y$.
 Next
applying the change of variable $u=(yz)^2$ on $z\ge 0$ and on $z<0$, we have
\begin{equation}\label{eq:H scale}
F(y)=\mathbb{E} G(yZ)Z= \int_{\mathbb{R}}  zG(yz) \frac{e^{-z^2/2}}{\sqrt{2\pi}} dz = \frac{1}{2y^2\sqrt{2\pi}} \left[ \int_{0}^{\infty} [G(u^{1/2})-G(-u^{1/2})] \exp[-u  (2y^2)^{-1}] du   \right].
\end{equation}
The   integral  in (\ref{eq:H scale}) is a   Laplace transform  evaluated  at $(2y^2)^{-1}$.  The Laplace transform is real analytic over the the open half interval where the integrability holds (\citet{widder:2010:laplace} Chapter II, Theorem 5a). The  function $y^{-2}$ is also analytic in $y\in (0,M)$. The conclusion then  follows since elementary operations preserve analyticity.
\end{proof}

\begin{Lem}\label{Lem:biv ana}
Let $D=(-M_1,M_1)\times (0,M_2)$, $M_1>0$, $M_2>1$. Suppose that $g$ is a measurable function such that $ g(\pm M_1+ \cdot\times M_2)   \in L^2(\phi)$. Then the function
\begin{equation*}
f(x,y):=    \int_{\mathbb{R}} g(u)  \exp\left[  -\frac{(u-x)^2}{2 y^2} \right]du
\end{equation*}
is    real analytic over $(x,y)\in D$.
\end{Lem}
\begin{proof}
The assumption implies that
\begin{equation}\label{eq:in L2}
g(x+\cdot\times y )   \in L^2(\phi)
\end{equation}  for all $(x,y)\in D$ by Lemma \ref{Lem:shift scale still int}.
By Theorem 1 of \citet{siciak:1970:characterization},  one needs to show that
\begin{enumerate}[(1)]
\item   $f$ is   $C^\infty$ (infinitely smooth);
\item $f$ is univariately analytic along any direction at an arbitrary point $(x_0,y_0)\in D$.
\end{enumerate}

For  part (1),  we need to check that the partial derivatives of all orders of $f(x,y)$ exist and are continuous. First, the   derivatives with respect to $x$ and $y$ can be taken under the integral sign  by applying the Dominated Convergence Theorem and the Mean Value Theorem  with the help of the following facts which are justified afterwards:
\begin{enumerate}
\item For any integers $i,j \ge 0$,
\[\frac{\partial^{i+j}}{\partial x^i \partial y^j} \exp( -\frac{(u-x)^2}{2 y^2}) =Q(u,x,y^{-1})\exp( -\frac{(u-x)^2}{2 y^2})\]
for some trivariate polynomial $Q(x_1,x_2,x_2)$.
\item  For any $x_0\in \mathbb{R} $ and  small $\delta>0$, there exists a polynomial $Q^*$ with non-negative coefficient  such that
 \[\sup_{|x-x_0|\le \delta}|Q(u,x,y^{-1})|\le Q^*(|u|,|y|^{-1});\]
  A similar bound holds for $\sup_{|y-y_0|\le \delta}|Q(u,x,y^{-1})|$.
\item If $\delta$ is chosen sufficiently small so that $\exp(-\frac{4\delta^2}{y^2})\ge 1/2,$
 then
 \[\sup_{|x-x_0|\le \delta}\exp( -\frac{(u-x)^2}{2 y^2})\le \exp( -\frac{(u-x_0+\delta)^2}{2 y^2})+\exp( -\frac{(u-x_0-\delta)^2}{2 y^2}),\] and for any $\delta>0$, \[\sup_{|y-y_0|\le \delta}\exp( -\frac{(u-x+\delta)^2}{2 y^2})=\exp( -\frac{(u-x+\delta)^2}{2 (y_0+\delta)^2}).\]
\item  For any integer $n\ge 0$ and polynomial $Q$, \[\int |g(u)| Q(|u|) \exp( -\frac{(u-x)^2}{2 y^2}) du<\infty\]  for all $(x,y)\in D$.
\end{enumerate}

Item 1 can be verified easily by induction and the elementary differentiation rules.

To see Item 2, write the polynomial as $Q(u,x,v)= a_0(x)+a_1(x)u+a_2(x)v+\ldots $. So $|Q(u,x,v)|\le |a_0(x)|+|a_1(x)u|+|a_2(x)v|+\ldots $.
Note that each coefficient $|a_i(x)|$ is a continuous function. Over a  compact interval $[x_0-\delta,x_0+\delta]$, each $|a_i(x)|$  is bounded   by a positive constant.  Replacing $|a_i(x)|$'s by these constants yields the conclusion.

For Item 3, the maximum of the function $\exp( -\frac{(u-x)^2}{2 y^2})$ in $x$ over the interval $[x_0-\delta,x_0+\delta]$ can   be either $1$ if  $u \in [x_0-\delta,x_0+\delta]$, or $\exp( -\frac{(u-x_0\pm \delta)^2}{2 y^2})$ otherwise. In the case where $u \in [x_0-\delta,x_0+\delta]$,   by the choice of $\delta$ described in Item 3 and  $|u-x_0\pm \delta|\le 2 \delta $, we have  \[\exp( -\frac{(u-x_0+\delta)^2}{2 y^2})+\exp( -\frac{(u-x_0-\delta)^2}{2 y^2})\ge1/2+1/2\ge  1;\]
 in the case where $u \notin [x_0-\delta,x_0+\delta]$, e.g., if $u<x_0-\delta$, then one has $\exp( -\frac{(u-x_0)^2}{2 y^2})\le \exp( -\frac{(u-x_0+\delta)^2}{2 y^2})$ by monotonicity. The second inequality follows easily from monotonicity.

To obtain Item 4, by Cauchy-Schwartz and (\ref{eq:in L2}), we have
\[
\int |g(u)| Q(|u|) \exp( -\frac{(u-x)^2}{2 y^2}) du\le y(2\pi)^{1/2} [\mathbb{E} g(x+yZ)^2]^{1/2} [\mathbb{E} Q(|x+yZ|)^2 ]^{1/2}<\infty
\]

Continuity of $f(x,y)$ can as well be verified  using Dominated Convergence and the facts 2$\sim$4 above.

 To check  claim (2), set $x(t)=x_0+at$, $y=y_0+bt$ where $(a,b)\in \mathbb{R}^2$ satisfying $a^2+b^2=1$.
Suppose that  $(x(t),y(t))\in D$
when  $|t|<\delta$ , where  $\delta>0$ will be adjusted smaller if necessary  later. We want to check the real analyticity of
\begin{equation}\label{eq:f(t)}
f(t):=f(x(t),y(t))=\int_{\mathbb{R}} g(u)  \exp\left[  -\frac{(u-x_0-at)^2}{2 (y_0+bt)^2} \right]du.
\end{equation}
at $t=0$.
Note that the function
\[h(u,z):= \exp(A(u,z)):=\exp\left[  -\frac{(u-x_0-az)^2}{2 (y_0+bz)^2} \right]
\] is  complex analytic in $z\in B_\delta:=\{z\in \mathbb{C}: |z|<\delta\}$ for each fixed $u\in \mathbb{R}$ since     $y_0+bz\neq 0$.   Next, by choosing $\delta$ small enough, one can ensure that for some  constant $c,\mu>0$  sufficiently large  and  $\sigma\in (0,M_2)$, such that
\begin{align}
\mathrm{Re}[A(u,z)]& =  -  \frac{\mathrm{Re}[ (u-x_0-az)^2(y_0+b\bar{z})^2 ]}{2| y_0+bz|^4 }= \frac{- y_0^2 u^2  +\mathrm{Re}[\ldots] }{2| y_0 +bz |^4 } = \frac{-  u^2  +\mathrm{Re}[\ldots] }{2| y_0^{1/2}+bz/y_0^{1/2}|^4 }\label{eq:Re}.
\end{align}
Since $y_0\in (0,M_2)$ and $|z|<\delta$ where $\delta$ is sufficiently small, there exists $\sigma\in (0,M_2)$, so that\begin{equation}\label{eq:den bound}
| y_0^{1/2}+bz/y_0^{1/2}|^4\ge \sigma^2
\end{equation} for all $|z|<\delta$. On the other hand,  the bracket $[\ldots]$ in (\ref{eq:Re}) is of the form
\[
c_1(z,\bar{z},x_0)u+c_0(z,\bar{z},x_0),
\]
where $c_1$ and $c_0$ are polynomials. For all $|z|<\delta$, one has  for some large $\mu,c>0$ that
\begin{equation}\label{eq:Re bound}
\mathrm{Re}[\ldots]\le  |c_1(z,\bar{z},x_0)||u|+|c_0(z,\bar{z},x_0)|\le 2\mu |u|-\mu^2  + 2\sigma^2 c
\end{equation}
Combining (\ref{eq:Re}), (\ref{eq:den bound}) and (\ref{eq:Re bound}), one has
\[
\mathrm{Re}[A(u,z)]\le  \frac{-   u ^2 + 2\mu |u|-\mu^2}{2\sigma^2}+c
\]
Hence for some constant $C>0$,
\begin{equation}\label{eq:bound}
|h(u,z)| = \exp(\mathrm{Re}[A(u,z)]             )\le C[  \exp \left( -  (u^2+\mu)^2/(2\sigma^2)\right) +\exp\left(-  (u^2-\mu)^2/(2\sigma^2)\right) ]  ,\quad z\in B_\delta.
\end{equation}
In   $\int_{\mathbb{R}}| g(u) h(u,z)| du$,
replace $|h(u,z)|$ by the preceding bound (\ref{eq:bound}), and note that $g(\pm\mu+\cdot\times \sigma)   \in L^1(\phi)$ for any $\mu\in \mathbb{R}$ and $\sigma\in (0,M)$   by Lemma \ref{Lem:shift scale still int} (b) and Lemma \ref{Lem:analytic}. Then one deduces that
\[
\sup_{z\in B_\delta}\int_{\mathbb{R}}| g(u) h(u,z)| du<\infty.
\]
  This  fact with Fubini's Theorem justifies the following order change of integrals:
\[
\oint_{\Delta}f(z)dz=\oint_{\Delta}dz\int_{\mathbb{R}} g(u)h(u,z)du  = \int_{\mathbb{R}} g(u) du\oint_{\Delta} h(u,z)  dz =0,
\]
where $\Delta$ is a closed triangle within $B_\delta$, and the last equality is due to  Cauchy's theorem since $ h(u,z)$ is complex analytic in $z\in B_\delta$. Then the complex analyticity of $f(z)$ over ${B}_\delta$ follows from Morera's Theorem (Theorem 10.17 of \citet{rudin:1987:real}). Restricting $B_\delta$ to $\mathrm{Im}(z)=0$ yields the real analyticity.
\end{proof}

\section{Proof of the theorems}\label{Sec:proof}
We prove here Theorems \ref{Thm:shift}, \ref{Thm:scale}, \ref{Thm:joint} involving the instability of the Hermite rank and
and Theorems \ref{Thm:near higher order}, \ref{Thm:near higher order non-center} and   \ref{Thm:near higher order centering} involving "near higher order rank".

\begin{proof}[Proof of Theorem \ref{Thm:shift}] Suppose that the function G is not a.e.\ constant.
Define $E=\{x\in (-M,M):~ G(\cdot+x) \text{ has Hermite rank} \ge 2 \}$. Note first that we have
$E= \{x\in (-M,M): ~G_\infty^{(1)}(x)=0\}$ in view of Lemma \ref{Lem:analytic} applied to $G(\cdot+x)$. Indeed, since the Hermite rank of $G(\cdot+x)$ is greater than $1$, we have $G_\infty^{(1)}(x)=\mathbb{E} G(Z+x)H_1(Z)=0$.
Now suppose by contradiction that $E$ has an accumulation point in $(-M,M)$.
Since $G_\infty$ is analytic by Lemma \ref{Lem:analytic}, so is the derivative $G_\infty^{(1)}(x)$. By \citet{rudin:1976:principles} Theorem 8.5,  $G_\infty^{(1)}$ is identically zero on $(-M,M)$ and hence $G_\infty$ is a constant on $(-M,M)$. But by (\ref{eq:G_infty}), $G_\infty=G\ast \phi$. By Lemma \ref{Lem:Gaussian vanish} and linearity, ${G}$ is a.e.\ a constant as well  which contradicts the theorem assumption.
\end{proof}

\begin{proof}[Proof of Theorem \ref{Thm:scale}]
Suppose that the function G is not a.e.\ symmetric.
Now the set with Hermite rank $\ge 2$  is $E=\{y\in (0,M): ~F(y):=\mathbb{E} G(yZ)Z=0\}$.
 The proof is similar to that of Theorem \ref{Thm:shift} above. Suppose by contradiction that $E$ has an accumulation point in $(0,M)$. Since  $F$ is analytic on $(0,M)$ by Lemma \ref{Lem:scale ana}, so $F(y)$ is identically zero on $(0,M)$.
Then we apply  the uniqueness of the Laplace transform  (\citet{widder:2010:laplace}  Chapter II, Theorem 6.3)
to relation (\ref{eq:H scale}) to conclude that $G(z)=G(-z)$ a.e., which contradicts   the assumption.
\end{proof}

\begin{proof}[Proof of Theorem \ref{Thm:joint}]
Define
\[F(x,y)=\mathbb{E}[ Z G(x+yZ)].
\] As before $E=\{(x,y)\in D: ~F(x,y)=0 \}$.
By a change of variable $z= ( u-x)/y$ we can write
\begin{equation}\label{eq:F(x,y)}
F(x,y)= \frac{1}{y^2\sqrt{2\pi}} \int_{\mathbb{R}}( u-x)G(u) \exp\left[  -\frac{(u-x)^2}{2 y^2} \right] du.
\end{equation}
Then $F(x,y)$ is a bivariate real analytic function by Lemma \ref{Lem:biv ana} since
$ (\pm M_1 + M_2 u) G(\pm M_1+M_2 u )\in L^2(\phi)$ by Cauchy-Schwartz.  If $F(x,y)\equiv 0$, which implies $F(x,1)\equiv 0$, then the Hermite rank of $G(\cdot+x)$ is greater than $1$ for all $x\in (-M_1,M_1)$, which contradicts Theorem \ref{Thm:shift}.
 So $F(x,y)$ is not identically zero.
The claimed properties of $E$, the zero set of the analytic $F$, then follow from \citet{mityagin:2015:zero} (see also \citet{krantz:parks:2002:primer}, Section 4.1).

\end{proof}

\begin{proof}[Proof of Theorem \ref{Thm:nonlinear per}]
If $G$ is not a constant a.e., by assumption  (b), the analytic $U(\theta)$ is not identically zero on $D$. So taking into account assumption (a),  the zero set $E$ has the claimed properties by \citet{mityagin:2015:zero}.
\end{proof}

\begin{proof}[Proof of Theorem \ref{Thm:near higher order}]

First by the Hermite expansion  (\ref{eq:def herm rank expan}) we can write:
\begin{equation}\label{e:G2}
G(Z+x_N)=\sum_{m=0}^\infty c_m(x_N)  H_m(Z),
\end{equation}
 We set as in (\ref{eq:G_infty}) that $G_\infty(x_N)=\mathbb{E} G(Z+x_N)$, where $Z\sim N(0,1)$. In view of Lemma \ref{Lem:analytic}, we have
\[
c_m(x_N)=\frac{1}{m!}\mathbb{E}  G(Z+x_N)H_m(Z) =\frac{1}{m!} G_\infty^{(m)}(x_N).
\]
and
\begin{equation}\label{eq:H_infty}
F_\infty(x_N):= \mathbb{E} G(Z+x_N)^2=\sum_{m=0}^\infty c_m(x_N)^2 \mathbb{E} H_m(Z)^2=\sum_{m=0}^\infty c_m(x_N)^2m!.
\end{equation}

Since $G$ has Hermite rank   $k$, in view of Lemma \ref{Lem:analytic} we have $G_\infty^{(1)}(0)=\ldots=G_\infty^{k-1}(0)=0\neq G_\infty^{k}(0)$. By the analytic expansion,
\[
G_\infty(x_N)=\frac{G_\infty^{(k)}(0)}{k!} x_N^k+O(x_N^{k+1})
\]
and
\[
\quad G_\infty^{(m)}(x_N) =\frac{d^mG}{dx^m}  (x_N)  =\frac{1}{(k-m)!} g_k x_N^{k-m} +O(x_N^{k-m+1}), \quad m=1,\ldots,k ,
\]
where
\[g_k=G_\infty^{(k)}(0).\]
 Then
\begin{align}\label{eq:S_{[Nt]}}
S_{[Nt]}(x_N)&= \sum_{n=1}^{[Nt]}\Big[G(Y(n)+x_N)-\mathbb{E} G(Y(n)+x_N)\Big]\notag\\
&=A_N(t)+B_N(t)+C_N(t)+D_N(t) +H.O.T.=A_N(t)+B_N(t)+R_N(t) +H.O.T.
\end{align}
In (\ref{eq:S_{[Nt]}}),
\begin{align}
&A_N(t)= (x_N^{k-1}N^{H})  \frac{g_k}{(k-1)!} \left[\frac{1}{N^H}\sum_{n=1}^{[Nt]}  Y(n) \right], ~~ B_N(t)=\sum_{m=2}^{k-1} (x_N^{k-m} N^{H[m]}) \frac{g_k}{(k-m)!}  \left[\frac{1}{N^{H[m]}}\sum_{n=1}^{[Nt]}  H_k(Y(n))\right]\notag\\
& C_N(t)=  N^{H[k]} \frac{g_k }{k!}\left[ \frac{1}{N^{H[k]}} \sum_{n=1}^{[Nt]}  H_k (Y(n)) \right],~~ D_N(t)=\sum_{m=k+1}^\infty c_m(x_N)  \sum_{n=1}^{[Nt]}   H_m(Y(n)),~~R_N(t)=C_N(t)+D_N(t),\label{eq:A-D}
\end{align}
where
\begin{equation}\label{eq:H(m)}
H[m]=\max\Big((H-1)m+1,~ 1/2\Big),
\end{equation}
 $\sum_{m=a}^b\ldots$ is understood as zero if $a>b$, and
  $H.O.T.$ stands for a higher-order term that will be asymptotically negligible as $N\rightarrow\infty$ after suitable normalization. In  (\ref{eq:A-D}), the expressions within a bracket  $[ ~\cdot ~]$ converge  in distribution and  their variances also converge. See, e.g., \citet{pipiras:taqqu:2017:long}, Chapter 5. The term $A_N(t)$ involves the first Hermite polynomial since $H_1(Y)=Y$; the term $B_N(t)$ involves the second up to the $(k-1)$th Hermite polynomial; the term $C_N(t)$ involves the $k$th Hermite polynomial; the term $D_N(t)$  involves all the Hermite polynomials of orders higher than $k$. $R_N(t)$ involves the terms $m\ge k$ in (\ref{e:G2}).

 We assume  for simplicity that $H\neq 1/(2m)$, $m=1,\ldots,k$. Otherwise the proof undergoes a slight modification involving an logarithmic factor.
 We have to consider the behavior of $ \sum_{n=1}^{[Nt]}   H_m(Y(n))$ for all $ m\geq 1 $.
  Let
\begin{equation}\label{eq:k_0}
k_0=\sup\{m\in  \mathbb{Z}_+:~ H>  1-1/(2m)\}.
\end{equation}

\noindent
In view of (\ref{eq:k_0}), when $m>k_0$ the central limit theorem in Theorem \ref{Thm:nclt gaussian} holds and when $m<k_0$ it is the non-central limit theorem that holds. Alternatively,
the central limit theorem  holds when
$H<1-1/(2k)$ and the non-central limit theorem holds when $H>1-1/(2k)$.

\bigskip
\noindent $\bullet$ Consider first  the central limit case where $H<1-1/(2k)$.

 In this case, $A_N(t)$ is associated with the order $x_N^{k-1} N^{H}$ and  $R_N(t)$ is   associated with the order $N^{1/2}$.
We now focus on $B_N(t)$ and claim that it is asymptotically negligible compared with either $A_N(t)$ or $R_N(t)$, namely,
\begin{equation}\label{eq:red clt}
x_N^{k-m} N^{H[m]}\ll \max( x_N^{k-1} N^{H},  N^{1/2}),
\quad m=2,\ldots,k-1.
\end{equation}
Indeed, when $m>k_0$, this is obvious since $H[m]=1/2$; Suppose now $m\le k_0$. In this case, $H[m]=(H-1)m+1$. However, in cases (a) and (b), where $x_N\ll$ or $\approx N^{(1/2-H)/(k-1)}$, the factor $x_N^{k-m}N^{H[m]}$ has an exponent bounded by
\begin{equation}
\frac{(1/2-H)(k-m)}{k-1}+(H-1)m+1=\frac{(m-1)k}{k-1} H +\frac{k+m-2km}{2(k-1)}+1<\frac{1}{2},
\end{equation}
where the last inequality can be obtained by plugging in the inequality $H<1-1/(2k)$ and some elementary computation. Hence in cases (a) and (b),  the relation (\ref{eq:red clt}) is proved.
In the case (c) where  $x_N\gg  N^{(1/2-H)/(k-1)}$,  it can be checked that $x_N^{k-m} N^{(H-1)m+1}\ll  x_N^{k-1} N^{H} $ using the inequality $\frac{1/2-H}{k-1}>H-1$, and the latter inequality follows from $H<1-1/(2k)$.
We have thus proved (\ref{eq:red clt}) and hence $B_N(t)$ is asymptotically negligible. We are left with $A_N(t)$ and $R_N(t)$.

The different asymptotic regimes  in the central limit case in Theorem \ref{Thm:near higher order} come about when comparing the order $x_N^{k-1} N^H$ of $A_N(t)$ with the order $N^{1/2}$ of $R_N(t)=C_N(t)+D_N(t)$. But we also have to deal with the convergence of the processes.
By \citet{bai:taqqu:2013:1-multivariate},  we have the joint convergence
\begin{equation}\label{eq:CLT joint}
 \left(\frac{1}{N^{H}}\sum_{n=1}^{[Nt]} Y(n), \frac{1}{N^{1/2}}\sum_{n=1}^{[Nt]}  \sum_{m=k}^\infty \frac{g_m}{m!} H_m(Y(n)) \right) \overset{f.d.d.}{\longrightarrow}  \left(c_1 Z_{H,1}(t), c_2 B(t) \right)
\end{equation}
where the fractional Brownian motion $Z_{H,1}(t)$ and the Brownian motion $B(t)$ are independent. From (\ref{eq:CLT joint}) we set
\[
R_N'(t)=\sum_{n=1}^{[Nt]}  \sum_{m=k}^\infty \frac{g_m}{m!} H_m(Y(n)),
\]
which is $R_N(t)$ with $x_N=0$.

We cannot use (\ref{eq:CLT joint}) directly because we have $R_N(t)$ instead of $R_N'(t)$. We thus need to compare them first.
 Since $Y(n)$ is standardized, we have  $|\gamma_Y(n)|^{m}\le |\gamma_Y(n)|^{k}$ if $m\ge k$, where $\gamma_Y(n)=\mathrm{Cov}[Y(n),Y(0)]$. Thus  by a computation similar to those on p.299 of \citet{pipiras:taqqu:2017:long} using the orthogonality of the Hermite polynomials,
\begin{align}\label{eq:diff D B}
 \mathbb{E} |R_N(t)-R_N'(t)|^2 &=  \sum_{m>K}  m! (c_m(x_N) -c_m(0))^2 \sum_{|n|<[Nt]} ([Nt]-|n|) \gamma_Y(n)^m\notag\\
 & \le (Nt)\left(\sum_{n=-\infty}^\infty  |\gamma_Y(n)|^k\right)  \sum_{m=k}^\infty  m! (c_m(x_N)-c_m(0))^2.
\end{align}
Note that  for  a $K>k$,
\begin{align}\label{eq:near int 1}
&\sum_{m=k}^\infty  m! (c_m(x_N)-c_m(0))^2\le \sum_{m=k}^K  m! (c_m(x_N)-c_m(0))^2+ 2\sum_{m>K}  m! ( c_m(x_N)^2+ c_m(0)^2).
\end{align}
Recall $F_\infty(x)=\sum_{m=0}^\infty m! c_m(x)^2$  in (\ref{eq:H_infty}), which is analytic in view of Lemma \ref{Lem:shift scale still int}.
Hence
\[
\sum_{m>K}  m!  c_m(x_N)^2 = F_\infty(x_N)-F_\infty(0)+F_\infty(0)- \sum_{m=0}^K  m!   c_m(x_N)^2.
\]
As $N\rightarrow\infty$, by the continuity of $F_\infty$ and $c_m(x)=G^{(m)}_\infty(x)/m!$, we have
\begin{equation}\label{eq:near int 2}
\sum_{m>K}  m!  c_m(x_N)^2
\rightarrow  \sum_{m>K} m! c_m(0)^2.
\end{equation}
Take the limit $N\rightarrow\infty$ in  (\ref{eq:near int 1}) using (\ref{eq:near int 2}) and the continuity of $c_m$,  the right-hand side of (\ref{eq:near int 1}) is only left with the term $2\sum_{m>K} m! c_m(0)^2$, which tends to $0$  as $K\rightarrow\infty$. Combining this with (\ref{eq:diff D B}), we have
\begin{equation}\label{eq:approx clt}
\lim_{N\rightarrow\infty} N^{-1} \mathbb{E} | R_N(t)-R_N'(t)|^2= 0.
\end{equation}
It thus follows from (\ref{eq:CLT joint}) that
\[
\left(\frac{1}{N^{H}}\sum_{n=1}^{[Nt]} Y(n), \frac{1}{N^{1/2}}\sum_{n=1}^{[Nt]}  R_N(t) \right) \overset{f.d.d.}{\longrightarrow}  \left(c_1 Z_{H,1}(t), c_2 B(t) \right).
\]
In case (a), we have $x_N^{k-1} N^{H}\ll N^{1/2}$ and thus as $N\rightarrow\infty$, $A_N(t)$ is negligible compared to $R_N(t)$ and thus we get $N^{-1/2}R_N(t)\overset{f.d.d.}{\longrightarrow} c_2 B(t)$. In case (c), $x_N^{k-1}N^H \gg N^{1/2}$ and  the limit is $c_1 Z_{H,1}(t)$. In case (b), where $x_N^{k-1}N^H \approx N^{1/2}$, both limits appear.
Finally, note that $N^{-H} x_N^{1-k}  A_N(t)$ is proportional to the first component on the left-hand side  of (\ref{eq:CLT joint}), while by (\ref{eq:approx clt}), the term $N^{-1/2}R_N(t)$  can be asymptotically replaced by $N^{-1/2}R_N'(t)$, which is  the second component  on the left-hand side of (\ref{eq:CLT joint}). The rest of the proof in the central limit case can be carried out easily.

\medskip

\noindent $\bullet$ Now we consider the non-central limit case  where $H>1-\frac{1}{2k}$.
  We have to study, as before, the behavior of $ \sum_{n=1}^{[Nt]}   H_m(Y(n))$ for all $m\geq 1$.

  We first consider the terms $A_N(t)$, $B_N(t)$ and $C_N(t)$ which involve $m \leq k$.
In this case, we have the following relation: any $n,m \in \mathbb{Z}_+$,
\[x_N^{k-m}N^{(H-1)m+1} \ll x_N^{k-n} N^{(H-1)n+1} \iff    x_N\ll N^{H-1},\quad \text{which holds with $\ll$ replaced by $\approx$ or $\gg$ as well.}
\]
Hence  in case (a),   the term $A_N(t)$ in (\ref{eq:S_{[Nt]}}) contributes; in case (b), the terms $A_N(t)$, $B_N(t)$ and $C_N(t)$ all contribute; in case (c),  the term  $C_N(t)$ contributes.

We shall show below that the term $D_N(t)$, which involves $m>k$, is  negligible.
Set
$\gamma_Y(n)=\mathrm{Cov}(Y(n),Y(0))\approx n^{2H-2}$.  By a similar computation as in (\ref{eq:diff D B}),
\begin{equation}\label{eq:E D_N^2}
\mathbb{E} D_N(t)^2 = \sum_{m>k}  m! c_m(x_N)^2 \sum_{|n|<[Nt]} ([Nt]-|n|) \gamma_Y(n)^m .
\end{equation}
For an arbitrarily small $\epsilon>0$,  we have  for $m>k$ that
\begin{equation}\label{eq:bound nclt}
\left|\sum_{|n|<[Nt]} ([Nt]-|n|) \gamma_Y(n)^m \right|\le  c_1 (Nt) \sum_{|n|<[Nt]}  n^{(2H-2)(k+1)}  \le c_2 (Nt)^{2H[k+1]},
\end{equation}
for some constants $c_1,c_2>0$.
where $H[m]$ is as in (\ref{eq:H(m)}). Since $H[k+1] <H[k]$ when $H>1-1/(2k)$,  in view of (\ref{eq:E D_N^2}),  (\ref{eq:bound nclt}) and (\ref{eq:H_infty}),
\[
\mathbb{E} D_N(t)^2\le  c_2 (Nt)^{2H[k+1]} F_\infty(x_N)\ll N^{2H[k]} \approx \mathbb{E} C_N(t)^2.
\]
So the order of $D_N(t)$ is always dominated by that of $C_N(t)$. So we only need to focus on $A_N(t)$, $B_N(t)$ and $C_N(t)$. Then  the rest of the proof can be carried out  using the following consequence of \citet{bai:taqqu:2013:1-multivariate}:
\begin{equation}\label{eq:NCLT joint}
\left(\frac{1}{N^{(H-1)m+1}}\sum_{n=1}^{[Nt]}     H_m(Y(n)), ~m=1,\ldots,k \right) \overset{f.d.d.}{\longrightarrow}  \Big(c_m Z_{H, m}(t), ~ m=1,\ldots,k \Big)
\end{equation}
where the Hermite processes $Z_{H,m}(t)$'s are defined through the same Brownian motion in (\ref{eq:Herm process}).

\end{proof}

\begin{proof}[Proof of Theorem \ref{Thm:near higher order non-center}]
The proof is done  as the proof of Theorem \ref{Thm:near higher order} based on (\ref{eq:CLT joint}) and (\ref{eq:NCLT joint}), and  we thus only provide an outline.
By Taylor expansion of $G_\infty$ in (\ref{eq:ana expan G_infty}), and since   $\mathbb{E} G(Y(n))=g_0=G_\infty(0)=0$ and   the Hermite rank is $k$, we have
\[
\mathbb{E} G(Y(n)+x_N)=G_\infty(x_N)=\frac{g_k}{k!}  x_N^k+O(x_N^{k+1}).
\]
Adding the leading term above to (\ref{eq:S_{[Nt]}}), we have
\begin{align}\label{eq:sum no center}
\widetilde{S}_{[Nt]}(x_N)=Z_N(t)+A_N(t)+B_N(t)+C_N(t)+D_N(t) +H.O.T.=Z_N(t)+A_N(t)+B_N(t)+R_N(t) +H.O.T.
\end{align}
where
\[
Z_N(t)=\frac{g_k }{k!} (x_N^k [Nt]) +H.O.T.
\]

In the central limit case where $H<1-1/(2k)$,   the term $B_N(t)$  is negligible compared to $A_N(t)$ and $R_N(t)$ as in the proof of Theorem \ref{Thm:near higher order} above.
We thus compare the orders $x_N^k N$ of $Z_N(t)$, $x_N^{k-1} N^H$ of $A_N(t)$ and $N^{1/2}$ of $R_N(t)$. Indeed, if $x_N\ll N^{-1/(2k)}$, then $x_N^{k}N\ll N^{1/2}$ and $x_N^{k-1} N^H \ll N^{1/2}$, so only $R_N(t)$ contributes; if $x_N\approx N^{-1/(2k)}$, then $x_N^{k}N\approx N^{1/2}$ and $x_N^{k}N\ll N^{1/2}$, so  $Z_N(t)$ and  $R_N(t)$ contribute; if  $x_N\gg N^{-1/(2k)}$, then $x_N^{k}N\gg N^{1/2}$ and $x_N^{k}N\gg x_N^{k-1} N^H$, so only $Z_N(t)$ contributes.

A similar analysis can be carried out in the non-central limit case. As in the proof of Theorem \ref{Thm:near higher order},  the term $B_N(t)$ only contributes in the case $x_N\approx N^{H-1}$.  The asymptotic regimes in the other cases are determined by   comparing the orders $x_N^k N$ of $Z_N(t)$, $x_N^{k-1} N^H$ of $A_N(t)$ and $N^{(H-1)k+1}$ of $C_N(t)$.
\end{proof}

\begin{proof}[Proof of Theorem \ref{Thm:near higher order centering}]
The proof is again similar to those of Theorem \ref{Thm:near higher order} and \ref{Thm:near higher order non-center}, and we thus only provide an outline.
Since $G(\cdot)$ is a polynomial, in (\ref{eq:A-D}) the $\infty$ in the sum defining $D_N(t)$ is replaced by a finite integer. This is important since the arguments leading to (\ref{eq:approx clt}) can no longer be applied.

The key is to note that $-\bar{Y}_N$ is associated with the order $N^{H-1}$, which replaces $x_N$ in all the terms in (\ref{eq:sum no center}). More precisely,
in the central limit case where $H<1-1/(2k)$,  comparing the order $N^{k(H-1)}$ of $Z_N(t)$,  the order $N^{(k-m)(H-1)+H[m]}$ of $A_N(t)$ and $B_N(t)$, $0<m<k$,  and the order $N^{1/2}$ of $R_N(t)$, where $H[m]$ is as in (\ref{eq:H(m)}), one can find that $R_N(t)$ is the contributing term. The conclusion then follows from (\ref{eq:CLT joint})  and the Slutsky Lemma using the continuity of the coefficients $c_m(\cdot)$. In the non-central limit case where $H>1-1/(2k)$, we compare the order $N^{k(H-1)}$ of $Z_N(t)$,  the order $N^{k (H-1)+1}$ of $A_N(t),B_N(t),C_N(t)$, $0<m\le k$,       and the order $N^{H[k+1]}$ of $D_N(t)$. In this case, the terms $A_N(t)$ and $B_N(t)$ contribute.

\end{proof}

\bigskip
\textbf{Acknowledgment.} This work was partially supported by the NSF grant  DMS-1309009  at Boston University.

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