# The resolution and representation of time series in Banach space

Amie Albrecht  Konstantin Avrachenkov  Brendan K. Beare  John Boland  Massimo
Franchi  Phil Howlett  Scheduling and Control Group, Centre for Industrial and
Applied Mathematics, UniSA STEM, University of South Australia, Mawson Lakes
5095, Australia  INRIA, Sophia Antipolis, COSTNET CA15109, France  School of
Economics, University of Sydney, 2006, Australia  Department of Statistical
Sciences, Sapienza University of Rome, Rome, 00185, Italy  email:
<EMAIL_ADDRESS>url: https://orcid.org/0000-0001-7302-0369  email:
<EMAIL_ADDRESS>url: https://orcid.org/0000-0002-8124-8272
email<EMAIL_ADDRESS>url: https://orcid.org/0000-0001-9146-131X
email<EMAIL_ADDRESS>url: https://orcid.org/0000-0003-1132-7589
email<EMAIL_ADDRESS>url: https://orcid.org/0000-0002-3745-2233
email<EMAIL_ADDRESS>url: http://orcid.org/0000-0003-2382-8137

###### Abstract

We describe a new systematic procedure for resolution and representation of
unit root processes on Banach space. Each unit root process is a marginally
stable time series with a uniquely defined resolvent operator which is
singular at unity. The proposed method uses infinite-length Jordan chains to
find the key spectral projections which enable separation and solution of the
fundamental equations for the Laurent series coefficients. We use the
coefficients to find two distinct forms of the Granger-Johansen representation
for the time series—a natural form, expressed as the sum of a series of
negative powers and a series of nonnegative powers of backward differences, in
complementary subspaces defined by the key projections and justified by
additional assumptions on the resolvent—and an extended form requiring no
additional assumptions, which avoids the potentially divergent series of
nonnegative powers but preserves the natural spectral separation. The
representations remain valid when the resolvent has an isolated essential
singularity at unity.

###### keywords:

time series, operator pencils , generalized resolvent, fundamental equations

###### MSC:

[2020] 47A10 , 47B01 , 47N30 , 91B84

††journal: Journal of Mathematical Analysis and Applications

##  1 Introduction

A time series is a sequence of observed values indexed according to time in
order of occurrence. The observed data is often stochastic in nature but is
nevertheless used to make predictions about future values of the observed
variables. Common examples of time series include regular observations of
climate variables such as sunshine intensity and rainfall and regularly
updated values of various social science metrics such as stock market indices.
While the most basic time series are often sequences of real numbers it is now
commonplace to consider vector-valued variables where several real numbers are
observed at each time. More complex observations are also possible. Functional
data analysis [37] treats observations as values of some continuous function.
In such cases the corresponding observation at each time could be an infinite
sequence of coefficients—such as a Fourier series—taking values in some
abstract vector space. While the theoretical structures used to analyse time
series are well developed for vectors in finite-dimensional Euclidean space
the development of a theoretical basis for the resolution and representation
of time series in infinite-dimensional space is an active area of research [9,
10, 16, 17, 21, 23, 41, 45].

An autoregressive moving average ARMA$(p,q)$ process of order $(p,q)$ is a
time series where a linear combination of the current value
$\mbox{$\mbox{\boldmath$x$}$}(t)$ and the $p$ previous values
$\\{\mbox{$\mbox{\boldmath$x$}$}(s)\\}_{s=t-p}^{t-1}$ is modelled as a linear
combination of the current value $\mbox{$\mbox{\boldmath$n$}$}(t)$ and the $q$
previous values $\\{\mbox{$\mbox{\boldmath$n$}$}(s)\\}_{s=t-q}^{t-1}$ of a
white noise process. The order of an ARMA process can be determined from the
relevant autocorrelation functions. For further discussion see Box et al. [13]
and Brockwell and Davis [14]. Brockwell and Davis recommend using the Akaike
information criterion [1]. Alternatively one could use the closely related
Bayesian information criterion [42]. Estimation of the ARMA coefficients in
Banach space can be accomplished using a linear minimum norm regression. Model
fitting is not considered in this paper. Our main contribution is to propose
and demonstrate a new method for resolution and representation of an
ARMA$(p,q)$ unit root process on Banach space.

##  2 Contribution

We propose a systematic procedure for resolution and representation of an
ARMA$(p,q)$ unit root process taking values in Banach space.

  1. 1.

We apply a Maclaurin transform to an ARMA$(1,1)$ unit root process in order to
obtain a linear operator equation for the transformed time series. The
operator is defined by the characteristic linear pencil for the time series
and the corresponding resolvent operator is found by solving a doubly infinite
system of fundamental equations to find the Laurent series coefficients. The
resolvent operator is used to solve the linear operator equation and find a
resolution of the transformed time series.

  2. 2.

We use an inverse Maclaurin transform on the resolved transformed ARMA$(1,1)$
time series to recover the desired representation and hence derive two
different forms of a generalized Granger–Johansen representation [19, 26, 29,
35, 36].

We note that an ARMA$(p,q)$ process with bounded linear operator coefficients
that maps one Banach space onto another can be reduced to an equivalent
ARMA$(1,1)$ process with augmented bounded linear operator coefficients that
map one suitably defined Cartesian product Banach space onto another. We
outline this reduction in a special case later in the paper.

A time series is said to be stable if the resolvent $R(z)$ of the
characteristic polynomial is analytic on and inside the unit circle in the
complex plane111We use a Maclaurin transform to describe our proposed solution
because the linear pencil thus defined is the characteristic linear function
traditionally used in time series analysis. Many engineers would naturally use
a $Z$-transform for this purpose. The two transforms are entirely equivalent
under a $z\leftrightarrow 1/z$ correspondence. If a $Z$-transform is used then
the system will be stable if $S(z)=R(1/z)$ is analytic on and outside the unit
circle.. Our particular interest is the solution of marginally stable unit
root processes where $R(z)$ is analytic on a set

${\mathcal{V}}_{\,\epsilon}={\mathcal{U}}_{\,1+\epsilon}(0)\setminus\\{1\\}=\\{z\in{\mathbb{C}}\mid|z|<1+\epsilon\
\mbox{with}\ z\neq 1\\}$ (1)

for some $\epsilon\in{\mathbb{R}}$ with $\epsilon>0$. We will use the notation
${\mathcal{U}}_{\,r}(a)=\\{z\in{\mathbb{C}}\mid|z-a|<r\leq\infty\\}$ and
${\mathcal{U}}_{\,s,\,r}(a)=\\{z\in{\mathbb{C}}\mid 0\leq
s<|z-a|<r\leq\infty\\}$ for $a\in{\mathbb{C}}$ and $r,s\in[0,\infty]$
throughout the paper. We assume that the resolvent has an isolated singularity
at $z=1$ which may be a finite-order pole or an essential singularity. Our
proposed techniques are new and will include, extend and unify the current
methods used in the literature.

We will use recently established theory to construct the resolvent operator
[2, 4, 5] and so a key aim of our paper will be to explain the proposed method
of analysis, and to present a collection of suitable examples that show how
the new method can be applied to more general time series representation
problems on Banach space. Many of the current methods are restricted to
Euclidean space [25, 26, 29, 35, 36] or Hilbert space [9, 10] for pencils
where $R(z)$ has a finite-order pole at $z=1$ or where calculation of $R(z)$
depends on the existence of various orthogonal decompositions [16, 17, 21,
23]. Other methods [41] can only be applied in Banach space if certain direct
sum decompositions are assumed. We will show that many of these restrictions
can be removed by the systematic use of the natural non-orthogonal spectral
separation projections [2, 4, 5, 8, 24, 46] for the resolvent.

##  3 Structure of the paper

In Section 4 we review the relevant literature. The main results are described
briefly in Section 5. Section 6 is the main section. The Maclaurin transform
and the transformed equation are explained in Section 6.1 and the resolvent
operator for the characteristic linear pencil and the resolution of the
transformed equation are discussed in Section 6.2. The backward difference
operator is discussed in Section 6.3. A natural representation for the time
series is established in Section 6.4 by assuming that the resolvent is
analytic on a set ${\mathcal{V}}_{\epsilon}\cup{\mathcal{W}}_{\theta}$ for
some $\epsilon>0$ and some $\theta>0$ where

${\mathcal{W}}_{\theta}={\mathcal{U}}_{\,0,\,1+\theta}(1)=\\{z\in{\mathbb{C}}\mid
0<|z-1|<1+\theta\\}.$

The natural representation is defined in two distinct parts by the spectral
separation projections using negative and nonnegative powers of the backward
difference operator and can be interpreted as either a non-standard or
standard form of the Granger–Johansen representation. In Section 6.5 we show
that the natural representation obtained here for $I(1)$ and $I(2)$ processes
is closely aligned to the representations obtained by Beare and Seo [10]. The
natural representation is replaced in Section 6.6 by an extended form of the
representation that requires only the minimal assumption that the resolvent is
analytic on ${\mathcal{V}}_{\epsilon}$ for some $\epsilon>0$. The extended
representation no longer uses nonnegative powers of the backward difference
operator but nevertheless can still be interpreted as a generalized form of
the Granger–Johansen representation. In particular we show that the spectral
separation of the two distinct parts is preserved. In Section 7 we present an
informal discussion of the methodology used by Albrecht et al. [2, 4, 5] to
justify the representation of the resolvent operator. The main results are
illustrated for three specific applications in Section 8. In Section 9 we
discuss augmented linear pencils and reduction of order for $ARMA$ processes.
In Section 10 we draw some brief conclusions. The necessary background
material is outlined in Appendices A, B and C.

##  4 Literature review

The calculation of resolvent operators for linear pencils has a long history.
In particular we cite key papers by Stummel [46], Langenhop [32, 33], Bart and
Lay [8], Howlett et al. [28] and Albrecht et al. [2, 5]. See also [6, 18, 20,
22, 27, 31, 40, 43, 38, 47]. For a more extensive review of the literature on
inversion of linear pencils see [2, 3, 4, 5]. Additional information can be
found in the books by Gohberg et al. [24], Kato [30], Yosida [49], Campbell
and Meyer [15] and Avrachenkov et al. [7].

We will limit our detailed review of the time series literature to the most
relevant recent papers. For a full review of the literature on unit root
processes we refer to Beare and Seo [9, 10], Franchi and Paruolo [21, 23] and
Seo [41] and references therein. For a more general view and some applications
see Chang et al. [16, 17]. The paper by Spangenberg [45] does not consider
unit root processes but does discuss stationarity for ARMA processes with
values in Banach space and contains some interesting examples. See also the
text by Bosq [12] for the background theory of linear stochastic processes.
Our main interest is the Granger–Johansen representation—a multivariate
extension of the earlier Beveridge–Nelson representation [11, 36, 44]. The
papers by Hansen [25, 26] provide a nice introduction to the Granger–Johansen
representation. We quote from [25].

> The [Granger–Johansen] representation theorem states that a co-integrated
> vector autoregressive process can be decomposed into four components: a
> random walk, a stationary process, a deterministic part, and a term that
> depends on the initial conditions.

The paper by Beare and Seo [10] is the main focus of our review. Let
$(\Omega,\Sigma,\mu)$ be a probability space and let $H$ be a complex Hilbert
space with inner product $\langle\cdot,\cdot\rangle$. Beare and Seo consider
an autoregressive AR$(p)$ process where $p\in{\mathbb{N}}$ and
$\\{\mbox{$\mbox{\boldmath$x$}$}(\mbox{$\mbox{\boldmath$\omega$}$},t)\\}_{t\in{\mathbb{N}}}\in
H^{{\mathbb{N}}}$ is a random time series defined by an equation

$\mbox{$\mbox{\boldmath$x$}$}(t)=\mbox{$\sum_{j=1}^{p}$}\Phi_{j}\mbox{$\mbox{\boldmath$x$}$}(t-j)+\mbox{$\mbox{\boldmath$\epsilon$}$}(t)$

for $t\in{\mathbb{N}}$ and the initial values
$\mbox{$\mbox{\boldmath$x$}$}(-p+1),\ldots,\mbox{$\mbox{\boldmath$x$}$}(0)$.
The coefficients $\Phi_{j}\in{\mathcal{B}}(H)$ are compact bounded linear
operators for each $j=1,\ldots,p$ and
$\\{\mbox{$\mbox{\boldmath$\epsilon$}$}(\mbox{$\mbox{\boldmath$\omega$}$},t)\\}_{t\in{\mathbb{Z}}}\in
H^{\mathbb{Z}}$ is a time series of independent identically distributed
(i.i.d.) random variables with finite variance

$\sigma^{2}=\mbox{$\int_{\Omega}$}\|\mbox{$\mbox{\boldmath$\epsilon$}$}(\mbox{$\mbox{\boldmath$\omega$}$},t)\|^{2}d\mu(\mbox{$\mbox{\boldmath$\omega$}$})<\infty$

and zero mean

$\mbox{$\mbox{\boldmath$\mu$}$}=\mbox{$\int_{\Omega}$}\mbox{$\mbox{\boldmath$\epsilon$}$}(\mbox{$\mbox{\boldmath$\omega$}$},t)d\mu(\mbox{$\mbox{\boldmath$\omega$}$})=\mbox{$\mbox{\boldmath$0$}$}$

and with positive definite covariance $\Sigma\in{\mathcal{B}}(H)$ defined by

$\Sigma(\mbox{$\mbox{\boldmath$u$}$})=\mbox{$\int_{\Omega}$}\langle\mbox{$\mbox{\boldmath$u$}$},\mbox{$\mbox{\boldmath$\epsilon$}$}(\omega,t)\rangle\mbox{$\mbox{\boldmath$\epsilon$}$}(\omega,t)d\mu(\omega)$

for each fixed $\mbox{$\mbox{\boldmath$u$}$}\in H$ and each
$t\in{\mathbb{Z}}$. The characteristic function for the process is a
polynomial of degree $p$ defined by

$\Phi(z)=I-\mbox{$\sum_{j=1}^{p}$}\Phi_{j}z^{j}$

for all $z\in{\mathbb{C}}$. Beare and Seo assume that the operator polynomial
$\Phi(z)$ is a nonsingular mapping on the closed unit disk centred at $z=0$
except at $z=1$. It follows that the resolvent operator $\Phi(z)^{-1}$ is
analytic on the closed unit disk except for an isolated singularity at $z=1$.
The analytic Fredholm theorem [24, pp 203–204] shows that for some
$\epsilon>0$ the resolvent operator can be written in the form

$\Phi(z)^{-1}=\mbox{$\sum_{k\in{\mathbb{N}}-1-d}$}\Upsilon_{k}(z-1)^{k}$

for all $z\in{\mathcal{U}}_{\,0,\,\epsilon}(1)$ and some $d\in{\mathbb{N}}$,
where $\Upsilon_{k}\in{\mathcal{B}}(H)$ is a finite rank operator for each
$k=-d,\ldots,-1$ and $\Upsilon_{-d}\neq\mbox{\large$0$}$.

If $d=1$ let $H_{1}=\Phi(1)(H)$ denote the range space and let
$K_{1}=[\Phi(1)]^{-1}(\\{\mbox{$\mbox{\boldmath$0$}$}\\})$ denote the null
space of $\Phi(1)\in{\mathcal{B}}(H)$. Note that
$\mbox{dim}(H_{1}^{\perp})=\mbox{dim}(K_{1})<\infty$. Beare and Seo show [10,
Theorem 3.1] that $\Upsilon_{-1}(H)=K_{1}$ and that

$\mbox{$\mbox{\boldmath$x$}$}(t)=\mbox{$\mbox{\boldmath$z$}$}_{0}-\Upsilon_{-1}\left(\mbox{$\sum_{s=1}^{t}$}\mbox{$\mbox{\boldmath$\epsilon$}$}(s)\right)+\mbox{$\sum_{\ell\in{\mathbb{N}}-1}$}\Upsilon_{\ell}(\mbox{$\mbox{\boldmath$\epsilon$}$}(t-\ell))/\ell!$

for all $t\in{\mathbb{N}}$ where
$\mbox{$\mbox{\boldmath$z$}$}_{0}=\mbox{$\mbox{\boldmath$z$}$}_{0}(\mbox{$\mbox{\boldmath$\omega$}$})\in
H$ is a random variable with mean zero and square integrable norm. If
$\mbox{$\mbox{\boldmath$z$}$}_{0}\in K_{1}$ they conclude that for fixed
$\mbox{$\mbox{\boldmath$u$}$}\in H$ the time series
$\\{\langle\mbox{$\mbox{\boldmath$u$}$},\mbox{$\mbox{\boldmath$x$}$}(t)\rangle\\}_{t\in{\mathbb{N}}}$
is $I(0)$ if $\mbox{$\mbox{\boldmath$u$}$}\in K_{1}^{\perp}$ and is $I(1)$ if
$\mbox{$\mbox{\boldmath$u$}$}\notin K_{1}^{\perp}$.

If $d=2$ let $H_{2}=H_{1}+[\Phi^{\,\prime}(1)](K_{1})$ and
$K_{2}=[K_{1}^{\perp}+\Phi^{\,\prime}(1)^{*}(H_{1}^{\perp})]^{\perp}$. Beare
and Seo show [10, Theorem 4.1] that $\Upsilon_{-2}(H)=K_{2}$ has finite
dimension and that

$\mbox{$\mbox{\boldmath$x$}$}(t)=\mbox{$\mbox{\boldmath$z$}$}_{0}+\mbox{$\mbox{\boldmath$z$}$}_{1}t+\Upsilon_{-2}\left(\mbox{$\sum_{s=1}^{t}\sum_{r=1}^{s}$}\mbox{$\mbox{\boldmath$\epsilon$}$}(r)\right)-\Upsilon_{-1}\left(\mbox{$\sum_{s=1}^{t}$}\mbox{$\mbox{\boldmath$\epsilon$}$}(s)\right)+\mbox{$\sum_{\ell\in{\mathbb{N}}-1}$}\Upsilon_{\ell}(\mbox{$\mbox{\boldmath$\epsilon$}$}(t-\ell))/\ell!$

for all $t\in{\mathbb{N}}$ where
$\mbox{$\mbox{\boldmath$z$}$}_{\ell}=\mbox{$\mbox{\boldmath$z$}$}_{\ell}(\mbox{$\mbox{\boldmath$\omega$}$})\in
H$ are random variables with mean zero and square integrable norm for each
$\ell=1,2$. If $\mbox{$\mbox{\boldmath$z$}$}_{\ell}\in K_{2}$ for each
$\ell=1,2$ they conclude that for fixed $\mbox{$\mbox{\boldmath$u$}$}\in H$
the time series
$\\{\langle\mbox{$\mbox{\boldmath$u$}$},\mbox{$\mbox{\boldmath$x$}$}(t)\rangle\\}_{t\in{\mathbb{N}}}$
is either $I(0)$ or $I(1)$ if $\mbox{$\mbox{\boldmath$u$}$}\in K_{2}^{\perp}$,
and that the time series is $I(2)$ if $\mbox{$\mbox{\boldmath$u$}$}\notin
K_{2}^{\perp}$.

Beare and Seo define necessary and sufficient conditions for $d=1$, in which
case $\Phi(z)^{-1}$ has a first-order pole at $z=1$, and for $d=2$, in which
case $\Phi(z)^{-1}$ has a second-order pole at $z=1$. These conditions are
algebraically straightforward when $d=1$ and tolerably so when $d=2$. Franchi
and Paruolo [23] present a comprehensive analysis for unit root processes on
Hilbert space when $\Phi(z)^{-1}$ has a pole of order $d$ at $z=1$ for all
values $d\in{\mathbb{N}}$. The necessary and sufficient conditions given in
[23] for a pole of order $d$ are more complicated when $d>2$. A similar
orthogonal decomposition on Hilbert space was proposed by Howlett et al. [28]
and used to calculate the resolvent operator near a finite-order pole. This
calculation is also described in the book by Avrachenkov et al. [7, pp
270–285]. These methods were later extended, using Zorn’s lemma, by Albrecht
et al. [3] to find the required Laurent series expansion on Hilbert space when
the resolvent has an isolated essential singularity.

A recent paper by Seo [41] extends the results in [10] for $I(1)$ and $I(2)$
processes to Banach space under less restrictive assumptions but nevertheless
assumes that certain key subspaces can be complemented—an assumption that
cannot be guaranteed and would therefore require a posteriori validation in
Banach space.

The main advantages of our proposed method are (i) the method is valid for all
Banach spaces, (ii) the required subspaces are defined by the spectral
separation projections and hence are always complemented, (iii) both the key
subspaces and their complementary subspaces may be infinite dimensional, (iv)
the necessary and sufficient conditions that define the resolvent are simply
the elementary conditions that guarantee convergence of the Laurent series,
(v) a general method, described in A, has been proposed for solution of the
fundamental equations in separable Banach spaces using infinite-length
singular and regular Jordan chains, and (vi) the method can be applied when
the resolvent has an isolated essential singularity at $z=1$.

##  5 The main results

The main purpose of this paper is to establish a version of the
Granger–Johansen representation for marginally stable ARMA$(1,1)$
autoregressive moving average unit root processes on Banach space. A unit root
process is uniquely defined by a characteristic resolvent operator $R(z)$
which is necessarily analytic on a set of the form

$[\,{\mathcal{U}}_{\,1}(0)^{a}\setminus\\{1\\}\,]\cup{\mathcal{U}}_{\,0,\,\delta}(1)=[\\{z\in{\mathbb{C}}\mid|z|\leq
1\\}\setminus\\{1\\}\,]\cup\\{z\in{\mathbb{C}}\mid 0<|z-1|<\delta\\}$

for some $\delta>0$. In the above expression we have followed Yosida [49] and
written ${\mathcal{U}}_{\,1}(0)^{a}$ to denote the closure of
${\mathcal{U}}_{\,1}(0)$. In Proposition 1 and Theorem 1 we establish natural
preliminary forms of the Granger–Johansen representation using sums defined by
the negative and nonnegative powers of the backward difference operator under
the stronger assumption that $R(z)$ is analytic on
${\mathcal{V}}_{\epsilon}\cup{\mathcal{W}}_{\theta}=[\,{\mathcal{U}}_{\,1+\epsilon}(0)\setminus\\{1\\}]\cup{\mathcal{U}}_{\,0,\,1+\theta}(1)$
for some $\epsilon>0$ and $\theta>0$. The negative and nonnegative powers of
the backward difference operator generate terms in different complementary
subspaces defined by the spectral separation projection operators at $z=1$. In
Proposition 2 and Theorem 2 we establish extended forms of the
Granger–Johansen representation using only the minimal assumption that $R(z)$
is analytic on a set
${\mathcal{V}}_{\epsilon}=\left[\,{\mathcal{U}}_{\,1+\epsilon}(0)\setminus\\{1\\}\right]$
for some $\epsilon>0$. In this regard we will show that $R(z)$ is analytic on
a set ${\mathcal{V}}_{\epsilon}$ for some $\epsilon>0$ if and only if $R(z)$
is analytic on a set
$[\,{\mathcal{U}}_{\,1}(0)^{a}\setminus\\{1\\}\,]\cup{\mathcal{U}}_{\,0,\,\delta}(1)$
for some $\delta>0$. The extended forms do not make direct use of the
nonnegative powers of the backward difference operator because the relevant
sums are no longer guaranteed to converge. The convenient bipartite spectral
separation obtained in the natural forms is nevertheless preserved.

For matrix operators we know that $R(z)=A(z)^{-1}$ is analytic on
${\mathcal{V}}_{\epsilon}$ for some $\epsilon>0$ with a finite order pole at
$z=1$ if and only if $\det[A(z)]\neq 0$ for $z\in{\mathcal{V}}_{\epsilon}$ and
$\det[A(1)]=0$. For general linear operators the relationship between
$A(z)=C_{0}+C_{1}(z-1)$ and $R(z)=A(z)^{-1}$ is more complex and a more direct
approach is required. Albrecht et al [2] showed that the required Laurent
series expansion $R(z)=\sum_{j\in{\mathbb{Z}}}T_{j}(z-1)^{j}$ is well-defined
if and only if there exists a solution
$\\{T_{j}\\}_{j\in{\mathbb{Z}}}\subseteq{\mathcal{B}}(Y,X)$ to a doubly-
infinite set of fundamental equations defined by the coefficients
$C_{0},C_{1}\in{\mathcal{B}}(X,Y)$. This is essentially an implicit condition
on $A(z)$. In a subsequent paper Albrecht et al [5] proposed a general
solution procedure for these equations when $X,Y$ are separable Banach spaces.
The procedure is discussed informally in Section 8. The relevant theory is
outlined in A.

##  6 Formulation and general resolution

Let $X,Y$ be Banach spaces. We assume that
$\\{\mbox{$\mbox{\boldmath$x$}$}(t)\\}_{t\in{\mathbb{N}}-1}:\Omega\rightarrow
X^{{\mathbb{N}}-1}$ is a random series whose value
$\\{\mbox{$\mbox{\boldmath$x$}$}(t)\\}_{t\in{\mathbb{N}}-1}(\mbox{$\mbox{\boldmath$\omega$}$})=\\{\mbox{$\mbox{\boldmath$x$}$}(\mbox{$\mbox{\boldmath$\omega$}$},t)\\}_{t\in{\mathbb{N}}-1}\in
X^{{\mathbb{N}}-1}$ depends on the outcome
$\mbox{$\mbox{\boldmath$\omega$}$}\in\Omega$ of some random process in a
probability space $(\Omega,\Sigma,\mu)$. Thus for each outcome of the process
the value of the series is simply defined as the corresponding series of
values. We include the possibility that
$\mbox{$\mbox{\boldmath$\omega$}$}=\\{\omega(t)\\}_{t\in{\mathbb{Z}}}\in{\mathbb{C}}^{\mathbb{Z}}$
where
$\mbox{$\mbox{\boldmath$x$}$}(\mbox{$\mbox{\boldmath$\omega$}$},t)=\mbox{$\mbox{\boldmath$x$}$}(\omega(t),t)$
for each $t\in{\mathbb{N}}-1$. If $A\in{\mathcal{B}}(X,Y)$ is a bounded linear
operator then we define
$[A\\{\mbox{$\mbox{\boldmath$x$}$}(t)\\}_{t\in{\mathbb{N}}-1}](\mbox{$\mbox{\boldmath$\omega$}$})=\\{A\mbox{$\mbox{\boldmath$x$}$}(\mbox{$\mbox{\boldmath$\omega$}$},t)\\}_{t\in{\mathbb{N}}-1}\in
Y^{{\mathbb{N}}-1}$ for each $\mbox{$\mbox{\boldmath$\omega$}$}\in\Omega$. We
will outline the solution procedure for an ARMA$(1,1)$ autoregressive moving
average process satisfying an equation in the form

$A_{0}\mbox{$\mbox{\boldmath$x$}$}(t)+A_{1}\mbox{$\mbox{\boldmath$x$}$}(t-1)=F_{0}\mbox{$\mbox{\boldmath$n$}$}(t)+F_{1}\mbox{$\mbox{\boldmath$n$}$}(t-1)$
(2)

for each $t\in{\mathbb{N}}$, where a linear combination of the current value
$\mbox{$\mbox{\boldmath$x$}$}(t)$ and the most recently observed value
$\mbox{$\mbox{\boldmath$x$}$}(t-1)$ of the time series is modelled as a linear
combination of the current value $\mbox{$\mbox{\boldmath$n$}$}(t)$ and the
most recently observed value $\mbox{$\mbox{\boldmath$n$}$}(t-1)$ of a strong
white noise process. The noise
$\\{\mbox{$\mbox{\boldmath$n$}$}(t)\\}_{t\in{\mathbb{Z}}}\in X^{\mathbb{Z}}$
is defined by a sequence of i.i.d. random vectors
$\mbox{$\mbox{\boldmath$n$}$}(t):\Omega\rightarrow X$ with value
$\mbox{$\mbox{\boldmath$n$}$}(\mbox{$\mbox{\boldmath$\omega$}$},t)\in X$ for
each $\mbox{$\mbox{\boldmath$\omega$}$}\in\Omega$ and all $t\in{\mathbb{Z}}$.
Thus we assume that $\mbox{$\mbox{\boldmath$n$}$}(t)$ is strictly stationary
with

$\sigma^{2}={\mathbb{E}}\left[\|\mbox{$\mbox{\boldmath$n$}$}\|^{2}\right]=\mbox{$\int_{\Omega}$}\|\mbox{$\mbox{\boldmath$n$}$}(\mbox{$\mbox{\boldmath$\omega$}$},t)\|^{2}d\mu(\mbox{$\mbox{\boldmath$\omega$}$})<\infty$

and

$\mbox{$\mbox{\boldmath$\mu$}$}={\mathbb{E}}\left[\mbox{$\mbox{\boldmath$n$}$}\right]=\mbox{$\int_{\Omega}$}\mbox{$\mbox{\boldmath$n$}$}(\mbox{$\mbox{\boldmath$\omega$}$},t)d\mu(\mbox{$\mbox{\boldmath$\omega$}$})=\mbox{$\mbox{\boldmath$0$}$}$

for each $t\in{\mathbb{Z}}$. Thus we say that the noise process
$\mbox{$\mbox{\boldmath$n$}$}(t)$ is $\mu$-square integrable with finite
variance and zero mean. Note that

${\mathbb{E}}\left[\|\mbox{$\mbox{\boldmath$n$}$}(t)\|\right]^{2}=\left(\mbox{$\int_{\Omega}$}\|\mbox{$\mbox{\boldmath$n$}$}(\mbox{$\mbox{\boldmath$\omega$}$},t)\|d\mu(\mbox{$\mbox{\boldmath$\omega$}$})\right)^{2}\leq\mbox{$\int_{\Omega}$}\|\mbox{$\mbox{\boldmath$n$}$}(\mbox{$\mbox{\boldmath$\omega$}$},t)\|^{2}d\mu(\mbox{$\mbox{\boldmath$\omega$}$})\cdot\mbox{$\int_{\Omega}$}d\mu(\mbox{$\mbox{\boldmath$\omega$}$})={\mathbb{E}}\left[\|\mbox{$\mbox{\boldmath$n$}$}(t)\|^{2}\right]<\infty$

implies that the noise process is also $\mu$-integrable. The observations
begin at $t=0$ and so it is necessary to impose an initial condition in the
form $\mbox{$\mbox{\boldmath$x$}$}(-1)=\mbox{$\mbox{\boldmath$c$}$}\in X$. The
operators $A_{0},A_{1},F_{0},F_{1}\in{\mathcal{B}}(X,Y)$ are bounded linear
operators. It is convenient to define a related time series
$\\{\mbox{$\mbox{\boldmath$g$}$}(t)\\}_{t\in{\mathbb{Z}}}=\\{\mbox{$\mbox{\boldmath$g$}$}(\mbox{$\mbox{\boldmath$\omega$}$},t)\\}_{t\in{\mathbb{Z}}}\in
Y^{\mathbb{Z}}$ by setting

$\mbox{$\mbox{\boldmath$g$}$}(t)=F_{0}\mbox{$\mbox{\boldmath$n$}$}(t)+F_{1}\mbox{$\mbox{\boldmath$n$}$}(t-1)$
(3)

for all $t\in{\mathbb{Z}}$. The series
$\\{\mbox{$\mbox{\boldmath$g$}$}(t)\\}_{t\in{\mathbb{Z}}}\in Y^{\mathbb{Z}}$
defined in (3) is a moving average MA$(1)$ process defined by a sequence of
identically distributed random vectors with finite variance and zero mean.

###### Remark 1

In our formulation the ARMA process has no deterministic driver. Thus our
representation is reduced to three characteristic components: a random walk, a
stationary process and a component that depends on the initial conditions.

###### Remark 2

A range of different assumptions is possible for the noise process. We make
the simplest possible assumption—that
$\\{\mbox{$\mbox{\boldmath$n$}$}(\mbox{$\mbox{\boldmath$\omega$}$},t)\\}_{t\in{\mathbb{Z}}}$
is a strong white noise process. In other words
$\\{\mbox{$\mbox{\boldmath$n$}$}(\mbox{$\mbox{\boldmath$\omega$}$},t)\\}_{t\in{\mathbb{Z}}}$
is a sequence of i.i.d. random vectors with finite variance and zero mean.
This assumption is sufficient to establish almost sure convergence of the
strictly stationary term in the standard form of the Granger–Johansen
representation. We do not use the square integrability as such and could
simply assume that the noise is i.i.d. and integrable with zero mean. Other
weaker assumptions are possible. Our representation remains valid if
$\\{\mbox{$\mbox{\boldmath$n$}$}(\mbox{$\mbox{\boldmath$\omega$}$},t)\\}_{t\in{\mathbb{Z}}}$
is a conventional weak white noise process. The condition that
$\|\mbox{$\mbox{\boldmath$n$}$}(\mbox{$\mbox{\boldmath$\omega$}$},t)\|$ be
integrable for all $t\in{\mathbb{Z}}$ could potentially be replaced by the
weaker condition that
$\log^{+}\|(A_{1}A_{0}^{-1}F_{0}+F_{1})\mbox{$\mbox{\boldmath$n$}$}(\mbox{$\mbox{\boldmath$\omega$}$},t)\|$
be integrable for all $t\in{\mathbb{Z}}$. See [45, Theorem 2.1] for discussion
of the relevant issues.

###  6.1 The Maclaurin transform and the transformed equation

The ARMA$(1,1)$ model can be formally solved using a Maclaurin transform—which
we shall refer to as an ${\mathcal{M}}$-transform. The
${\mathcal{\mathcal{M}}}$-transform
$\mbox{$\mbox{\boldmath$X$}$}(z)={\mathcal{M}}[\\{\mbox{$\mbox{\boldmath$x$}$}(t)\\}_{t\in{\mathbb{N}}-1}](z)$
of the time series
$\\{\mbox{$\mbox{\boldmath$x$}$}(t)\\}_{t\in{\mathbb{N}}-1}\in
X^{{\mathbb{N}}-1}$ is a random power series
$\mbox{$\mbox{\boldmath$X$}$}(z):\Omega\rightarrow X$ in the variable
$z\in{\mathbb{C}}$ with value

$\mbox{$\mbox{\boldmath$X$}$}(\mbox{$\mbox{\boldmath$\omega$}$},z)=\mbox{$\sum_{t\in{\mathbb{N}}-1}$}\mbox{$\mbox{\boldmath$x$}$}(\mbox{$\mbox{\boldmath$\omega$}$},t)z^{t}$
(4)

for each outcome $\mbox{$\mbox{\boldmath$\omega$}$}\in\Omega$ of the random
process. For a completely arbitrary time series we cannot decide a priori if
(4) will converge for any particular value of $z\in{\mathbb{C}}$ but we will
show that if the time series
$\\{\mbox{$\mbox{\boldmath$x$}$}(t)\\}_{t\in{\mathbb{N}}-1}$ can be modelled
as an ARMA$(1,1)$ process then the ${\mathcal{M}}$-transform can be used to
find a Granger–Johansen representation for the series. We can then recover a
retrospective answer to the convergence question for
$\mbox{$\mbox{\boldmath$X$}$}(z)$ and the corresponding representation for
$\\{\mbox{$\mbox{\boldmath$x$}$}(t)\\}_{t\in{\mathbb{N}}-1}$. If we apply an
${\mathcal{M}}$-transform to (2) we get

$A(z)\mbox{$\mbox{\boldmath$X$}$}(z)+A_{1}\,\mbox{$\mbox{\boldmath$c$}$}=\mbox{$\mbox{\boldmath$G$}$}(z)\iff
A(z)\mbox{$\mbox{\boldmath$X$}$}(z)=\mbox{$\mbox{\boldmath$G$}$}(z)-A_{1}\,\mbox{$\mbox{\boldmath$c$}$}$
(5)

where $\mbox{$\mbox{\boldmath$c$}$}=\mbox{$\mbox{\boldmath$x$}$}(-1)$, the
linear pencil $A(z)=A_{0}+A_{1}z$ is the characteristic function, and we have
defined
$\mbox{$\mbox{\boldmath$X$}$}(z)=\sum_{t\in{\mathbb{N}}-1}\mbox{$\mbox{\boldmath$x$}$}(t)z^{t}$
and used (3) to define
$\mbox{$\mbox{\boldmath$G$}$}(z)=\sum_{t\in{\mathbb{N}}-1}\mbox{$\mbox{\boldmath$g$}$}(t)z^{t}$.
Solution of (5) depends on the existence of a resolvent operator
$R(z)=A(z)^{-1}$ on some suitable region $z\in{\mathcal{U}}$ of the complex
plane.

###  6.2 The resolvent operator and the resolution

We are particularly interested in those marginally stable time series known as
unit root processes where the resolvent operator $R(z)=A(z)^{-1}$ is singular
at $z=1$ but is otherwise analytic on and inside the unit circle. To this end
we will assume that $R(z)$ is analytic on the set ${\mathcal{V}}_{\epsilon}$
defined in (1) for some $\epsilon>0$ and has a singularity at $z=1$ which may
be a pole of finite order or an isolated essential singularity. Hence $R(z)$
is analytic on the region ${\mathcal{U}}_{\,0,\,\epsilon}(1)$. We must
necessarily analyze the singularity at $z=1$ so we begin by writing

$A(z)=C_{0}+C_{1}(z-1)$ (6)

where $C_{0}=A_{0}+A_{1}\in{\mathcal{B}}(X,Y)$ and
$C_{1}=A_{1}\in{\mathcal{B}}(X,Y)$. Albrecht et al. [2] have shown that the
resolvent operator can be expressed as a Laurent series

$R(z)=\mbox{$\sum_{j\in{\mathbb{Z}}}$}T_{j}(z-1)^{j}$ (7)

for all $z\in{\mathcal{U}}_{\,0,\,\epsilon}(1)$ if and only if the
coefficients $\\{T_{j}\\}_{j\in{\mathbb{Z}}}$ satisfy a system of left and
right fundamental equations at $z=1$ given respectively by

$T_{j-1}C_{1}+T_{j}C_{0}=\left\\{\begin{array}[]{ll}I_{X}&\mbox{if}\ j=0\\\
\mbox{\large$0$}_{X}&\mbox{if}\ j\in{\mathbb{Z}},j\neq 0\end{array}\right.$
(8)

and

$C_{1}T_{j-1}+C_{0}T_{j}=\left\\{\begin{array}[]{ll}I_{Y}&\mbox{if}\ j=0\\\
\mbox{\large$0$}_{Y}&\mbox{if}\ j\in{\mathbb{Z}},j\neq 0\end{array}\right.$
(9)

and the coefficients $\\{T_{j}\\}_{j\in{\mathbb{Z}}}$ also satisfy the
magnitude constraints

$\lim_{k\rightarrow\infty}\|T_{-k}\|^{1/k}=0\quad\mbox{and}\quad\lim_{\ell\rightarrow\infty}\|T_{\ell}\|^{1/\ell}\leq
1/\epsilon.$ (10)

If (10) is satisfied then for each $\eta>0$ we can find a real constant
$c_{\eta}>0$ with $\|T_{-k}\|\leq c_{\eta}\eta^{k}$ for all $k\in{\mathbb{N}}$
and a real constant $d_{\epsilon}>0$ with $\|T_{\ell}\|\leq
d_{\epsilon}/\epsilon^{\ell}$ for all $\ell\in{\mathbb{N}}-1$.

If (8), (9) and (10) are all satisfied then

$T_{-k}=(-1)^{k-1}(T_{-1}C_{0})^{k-1}T_{-1}\in{\mathcal{B}}(Y,X)$ (11)

for all $k\in{\mathbb{N}}$ and

$T_{\ell}=(-1)^{\ell}(T_{0}C_{1})^{\ell}T_{0}\in{\mathcal{B}}(Y,X)$ (12)

for all $\ell\in{\mathbb{N}}-1$. Thus the solution to (8) and (9) on
${\mathcal{U}}_{\,0,\,\epsilon}(1)$ is completely determined by the basic
solution $\\{T_{-1},T_{0}\\}$ on ${\mathcal{U}}_{\,0,\,\epsilon}(1)$. The
operators $P=T_{-1}C_{1}\in{\mathcal{B}}(X)$ and
$P^{c}=I_{X}-P=T_{0}C_{0}\in{\mathcal{B}}(X)$ are complementary key
projections on $X$ and the operators $Q=C_{1}T_{-1}\in{\mathcal{B}}(Y)$ and
$Q^{c}=I_{Y}-Q=C_{0}T_{0}\in{\mathcal{B}}(Y)$ are corresponding complementary
key projections on $Y$. We also note that
$T_{-1}C_{i}T_{0}=T_{0}C_{i}T_{-1}=\mbox{\large$0$}_{Y,X}$ for each $i=0,1$.
We will write

$R(z)=R_{\mbox{\scriptsize sin}}(z)+R_{\mbox{\scriptsize reg}}(z)$ (13)

for all $z\in{\mathcal{U}}_{\,0,\,\epsilon}(1)$ where the singular part of the
Laurent series

$R_{\mbox{\scriptsize
sin}}(z)=[I_{X}(z-1)+T_{-1}C_{0}]^{-1}T_{-1}=\mbox{$\sum_{k\in{\mathbb{N}}}$}T_{-k}(z-1)^{-k}$
(14)

converges for all $z\in{\mathcal{U}}_{\,0,\,\infty}(1)$ and the regular part
of the Laurent series

$R_{\mbox{\scriptsize
reg}}(z)=[I_{X}+T_{0}C_{1}(z-1)]^{-1}T_{0}=\mbox{$\sum_{\ell\in{\mathbb{N}}-1}$}T_{\ell}(z-1)^{\ell}$
(15)

converges for all $z\in{\mathcal{U}}_{\,\epsilon}(1)$. See A for a formal
statement of the key results in Albrecht et al. [2]. Since the resolvent
operator satisfies the equation $R(z)A(z)=I_{X}$ for all
$z\in{\mathcal{U}}_{\,0,\,\epsilon}(1)$ we obtain a formal resolution of the
transformed equation (5) given by

$\mbox{$\mbox{\boldmath$X$}$}(z)=R(z)[\mbox{$\mbox{\boldmath$G$}$}(z)-A_{1}\mbox{$\mbox{\boldmath$c$}$}]=R_{\mbox{\scriptsize
sin}}(z)[\mbox{$\mbox{\boldmath$G$}$}(z)-A_{1}\mbox{$\mbox{\boldmath$c$}$}]+R_{\mbox{\scriptsize
reg}}(z)[\mbox{$\mbox{\boldmath$G$}$}(z)-A_{1}\mbox{$\mbox{\boldmath$c$}$}]$
(16)

for all $z\in{\mathcal{U}}_{\,0,\,\epsilon}(1)$. We will show that the formal
resolution in (16) leads naturally to an intuitive time series representation
using sums of negative and nonnegative powers of the backward difference
operator acting on a white noise process.

###  6.3 The backward difference operator

The backward difference operator222In the time series literature the backward
difference operator is often denoted by $\Delta$. In this paper we use the
more conventional mathematical notation and denote the backward difference
operator by $\nabla$. for a time series
$\\{\mbox{$\mbox{\boldmath$u$}$}(t)\\}_{t\in{\mathbb{Z}}}\in X$ is defined by

$\nabla\\{\mbox{$\mbox{\boldmath$u$}$}(t)\\}_{t\in{\mathbb{Z}}}=\\{\nabla\mbox{$\mbox{\boldmath$u$}$}(t)\\}_{t\in{\mathbb{Z}}}=\\{[\mbox{$\mbox{\boldmath$u$}$}(t)-\mbox{$\mbox{\boldmath$u$}$}(t-1)]\\}_{t\in{\mathbb{Z}}}.$

The backward difference operator is closely related to the lag operator

$L\\{\mbox{$\mbox{\boldmath$u$}$}(t)\\}_{t\in{\mathbb{Z}}}=\\{L\mbox{$\mbox{\boldmath$u$}$}(t)\\}_{t\in{\mathbb{Z}}}=\\{\mbox{$\mbox{\boldmath$u$}$}(t-1)\\}_{t\in{\mathbb{Z}}}.$

Indeed we have $\nabla=I_{X}-L$ where $I_{X}\in{\mathcal{B}}(X)$ is the
identity operator. We can define integer powers of the backward difference
operator by the formulæ

$\nabla^{-k}\mbox{$\mbox{\boldmath$u$}$}(t)=(I_{X}-L)^{-k}\mbox{$\mbox{\boldmath$u$}$}(t)=\mbox{$\sum_{s\in{\mathbb{N}}-1}$}\mbox{$\binom{s+k-1}{s}$}L^{s}\mbox{$\mbox{\boldmath$u$}$}(t)=\mbox{$\sum_{s\in{\mathbb{N}}-1}$}\mbox{$\binom{s+k-1}{s}$}\mbox{$\mbox{\boldmath$u$}$}(t-s)$
(17)

for all $k\in{\mathbb{N}}$, and

$\nabla^{\ell}\mbox{$\mbox{\boldmath$u$}$}(t)=(I_{X}-L)^{\ell}\mbox{$\mbox{\boldmath$u$}$}(t)=\mbox{$\sum_{s=0}^{\ell}$}\mbox{$\binom{\ell}{s}$}(-1)^{s}L^{s}\mbox{$\mbox{\boldmath$u$}$}(t)=\mbox{$\sum_{s=0}^{\ell}$}\mbox{$\binom{\ell}{s}$}(-1)^{s}\mbox{$\mbox{\boldmath$u$}$}(t-s)$
(18)

for all $\ell\in{\mathbb{N}}-1$. It is a matter of elementary algebra to show
that

${\mathcal{M}}[\\{\nabla^{j}\mbox{$\mbox{\boldmath$u$}$}_{+}(t)\\}_{t\in{\mathbb{Z}}}](z)=(1-z)^{j}\mbox{$\mbox{\boldmath$U$}$}(z)$

for each $j\in{\mathbb{Z}}$ where we have defined

$\mbox{$\mbox{\boldmath$u$}$}_{+}(t)=\left\\{\begin{array}[]{ll}\mbox{$\mbox{\boldmath$u$}$}(t)&\mbox{if}\
t\geq 0\\\ \mbox{$\mbox{\boldmath$0$}$}&\mbox{otherwise}.\end{array}\right.$
(19)

Since $\mbox{$\mbox{\boldmath$u$}$}_{+}(t)=0$ for $t<0$ it follows from (17)
and (18) that

$\nabla^{-k}\mbox{$\mbox{\boldmath$u$}$}_{+}(t)=\mbox{$\sum_{s\in{\mathbb{N}}-1}$}\mbox{$\binom{s+k-1}{s}$}\mbox{$\mbox{\boldmath$u$}$}_{+}(t-s)=\mbox{$\sum_{s=0}^{t}$}\mbox{$\binom{s+k-1}{s}$}\mbox{$\mbox{\boldmath$u$}$}(t-s)$
(20)

and

$\nabla^{\ell}\mbox{$\mbox{\boldmath$u$}$}_{+}(t)=\mbox{$\sum_{s=0}^{\ell}$}\mbox{$\binom{\ell}{s}$}(-1)^{s}\mbox{$\mbox{\boldmath$u$}$}_{+}(t-s)=\mbox{$\sum_{s=0}^{\min\\{\ell,t\\}}$}\mbox{$\binom{\ell}{s}$}(-1)^{s}\mbox{$\mbox{\boldmath$u$}$}(t-s).$
(21)

###  6.4 A natural representation of the time series

We assume that $R(z)$ is analytic on the set
${\mathcal{V}}_{\,\epsilon}\cup{\mathcal{W}}_{\theta}$ for some $\epsilon>0$
and some $\theta>0$.

Consider the term $\mbox{$\mbox{\boldmath$X$}$}_{\mbox{\scriptsize
sin}}(z)=R_{\mbox{\scriptsize
sin}}(z)[\mbox{$\mbox{\boldmath$G$}$}(z)-C_{1}\mbox{$\mbox{\boldmath$c$}$}]$.
We have

$\displaystyle\mbox{$\mbox{\boldmath$X$}$}_{\mbox{\scriptsize sin}}(z)$
$\displaystyle=$
$\displaystyle\mbox{$\sum_{k\in{\mathbb{N}}}$}T_{-k}(z-1)^{-k}\left[\mbox{$\mbox{\boldmath$G$}$}(z)-C_{1}\mbox{$\mbox{\boldmath$c$}$}\right]$
(22) $\displaystyle=$
$\displaystyle(-1)\mbox{$\sum_{k\in{\mathbb{N}}}$}(T_{-1}C_{0})^{k-1}T_{-1}(1-z)^{-k}\left[\mbox{$\mbox{\boldmath$G$}$}(z)-C_{1}\mbox{$\mbox{\boldmath$c$}$}\right]$
$\displaystyle=$
$\displaystyle(-1)\mbox{$\sum_{k\in{\mathbb{N}}}$}(T_{-1}C_{0})^{k-1}T_{-1}\,{\mathcal{M}}\left[\\{\nabla^{-k}(\mbox{$\mbox{\boldmath$g$}$}-C_{1}\mbox{$\mbox{\boldmath$c$}$}\,\delta)_{+}(t)\\}_{t\in{\mathbb{Z}}}\right](z)$

where the time series $\\{\delta(t)\\}_{t\in{\mathbb{Z}}}$ is defined by
$\delta(t)=0$ if $t\neq 0$ and $\delta(0)=1$. Now we can apply (20) to deduce
that

$\displaystyle\mbox{$\mbox{\boldmath$X$}$}_{\mbox{\scriptsize sin}}(z)$
$\displaystyle=$
$\displaystyle(-1)\mbox{$\sum_{k\in{\mathbb{N}}}$}(T_{-1}C_{0})^{k-1}T_{-1}\,{\mathcal{M}}\left[\\{\mbox{$\sum_{s=0}^{t}\binom{s+k-1}{s}$}(\mbox{$\mbox{\boldmath$g$}$}-C_{1}\mbox{$\mbox{\boldmath$c$}$}\,\delta)(t-s)\\}_{t\in{\mathbb{Z}}}\right](z)$
(23) $\displaystyle=$
$\displaystyle{\mathcal{M}}\left[\\{(-1)\mbox{$\sum_{k\in{\mathbb{N}}}$}(T_{-1}C_{0})^{k-1}T_{-1}\,\mbox{$\sum_{s=0}^{t}\binom{s+k-1}{s}$}(\mbox{$\mbox{\boldmath$g$}$}-C_{1}\mbox{$\mbox{\boldmath$c$}$}\,\delta)(t-s)\\}_{t\in{\mathbb{Z}}}\right](z)$
$\displaystyle=$
$\displaystyle{\mathcal{M}}\left[\\{\mbox{$\sum_{k\in{\mathbb{N}}}$}(-1)^{k}T_{-k}\nabla^{-k}\mbox{$\mbox{\boldmath$g$}$}_{+}(t)\\}_{t\in{\mathbb{Z}}}\right](z)$
$\displaystyle\hskip
85.35826pt+{\mathcal{M}}\left[\\{\mbox{$\sum_{k\in{\mathbb{N}}}\binom{t+k-1}{t}$}(T_{-1}C_{0})^{k-1}T_{-1}C_{1}\mbox{$\mbox{\boldmath$c$}$}\\}_{t\in{\mathbb{Z}}}\right](z).$

We justify convergence of (23) as follows. We have assumed that $R(z)$ is
analytic on ${\mathcal{V}}_{\,\epsilon}$ for some $\epsilon>0$ and so
$R_{\mbox{\scriptsize sin}}(z)=\sum_{k\in{\mathbb{N}}}T_{-k}(z-1)^{-k}$ is
analytic for $z\in{\mathcal{U}}_{\,0,\infty}(1)$. Hence, for each $\eta>0$, we
can find $c_{\eta}>0$ such that

$\|T_{-k}\|=\|(T_{-1}C_{0})^{k-1}T_{-1}\|\leq c_{\eta}\eta^{k}$

for all $k\in{\mathbb{N}}$. Therefore

$\displaystyle{\mathbb{E}}\left[\|\mbox{$\sum_{k\in{\mathbb{N}}}$}(-1)^{k-1}T_{-k}\nabla^{-k}\mbox{$\mbox{\boldmath$g$}$}_{+}(t)\|\right]$
$\displaystyle=$
$\displaystyle{\mathbb{E}}\left[\,\|\mbox{$\sum_{k\in{\mathbb{N}}}$}(T_{-1}C_{0})^{k-1}T_{-1}\,\mbox{$\sum_{s=0}^{t}\binom{s+k-1}{s}$}\mbox{$\mbox{\boldmath$g$}$}(t-s)\|\right]$
$\displaystyle\leq$
$\displaystyle{\mathbb{E}}\left[\|\mbox{$\mbox{\boldmath$g$}$}\|\right]\cdot\mbox{$\sum_{k\in{\mathbb{N}}}$}\|(T_{-1}C_{0})^{k-1}T_{-1}\|\cdot\mbox{$\sum_{s=0}^{t}\binom{s+k-1}{s}$}$
$\displaystyle\leq$ $\displaystyle
c_{\eta}{\mathbb{E}}\left[\|\mbox{$\mbox{\boldmath$g$}$}\|\right]\cdot\mbox{$\sum_{k\in{\mathbb{N}}}\binom{t+k}{t}$}\eta^{k}$
$\displaystyle=$ $\displaystyle
c_{\eta}{\mathbb{E}}\left[\|\mbox{$\mbox{\boldmath$g$}$}\|\right]\cdot\left[1/(1-\eta)^{t+1}-1\right]$

is well defined and finite. Furthermore

$\mbox{$\sum_{k\in{\mathbb{N}}}\binom{t+k-1}{t}$}(T_{-1}C_{0})^{k-1}T_{-1}C_{1}\mbox{$\mbox{\boldmath$c$}$}=(I_{X}-T_{-1}C_{0})^{-t-1}T_{-1}C_{1}\mbox{$\mbox{\boldmath$c$}$}$

is also well defined. The function $R_{\mbox{\scriptsize
sin}}(z)=[I_{X}(z-1)+T_{-1}C_{0}]^{-1}T_{-1}$ is analytic for all $z\neq 1$
and hence is analytic at $z=0$ with

$R_{\mbox{\scriptsize sin}}(z)=\mbox{$\sum_{s\in{\mathbb{N}}-1}$}U_{s}z^{s}$
(24)

for all $z\in{\mathcal{U}}_{\,1}(0)$ where the coefficients
$\\{U_{s}\\}_{s\in{\mathbb{N}}-1}$ are given by the formula

$\displaystyle U_{s}$ $\displaystyle=$
$\displaystyle(1/s!)\left[d^{s}R_{\mbox{\scriptsize
sin}}(z)/dz^{s}\right]_{z=0}$ (25) $\displaystyle=$
$\displaystyle(1/s!)\left[d^{s}[I_{X}(z-1)+T_{-1}C_{0}]^{-1}T_{-1}/dz^{s}\right]_{z=0}$
$\displaystyle=$ $\displaystyle(-1)(I_{X}-T_{-1}C_{0})^{-s-1}T_{-1}$

for each $s\in{\mathbb{N}}-1$. If we define

$\displaystyle\mbox{$\mbox{\boldmath$x$}$}_{\mbox{\scriptsize sin,+}}(t)$
$\displaystyle=$
$\displaystyle\mbox{$\sum_{k\in{\mathbb{N}}}$}(-1)^{k}T_{-k}\nabla^{-k}\mbox{$\mbox{\boldmath$g$}$}_{+}(t)+(I_{X}-T_{-1}C_{0})^{-t-1}T_{-1}C_{1}\mbox{$\mbox{\boldmath$c$}$}$
(26) $\displaystyle=$
$\displaystyle\mbox{$\sum_{k\in{\mathbb{N}}}$}(-1)^{k}T_{-k}\nabla^{-k}\mbox{$\mbox{\boldmath$g$}$}_{+}(t)-U_{t}C_{1}\mbox{$\mbox{\boldmath$c$}$}$

for $t\in{\mathbb{N}}-1$ then we have shown that
${\mathcal{M}}\left[\\{\mbox{$\mbox{\boldmath$x$}$}_{\mbox{\scriptsize
sin,+}}(t)\\}_{t\in{\mathbb{Z}}}\right](z)=\mbox{$\mbox{\boldmath$X$}$}_{\mbox{\scriptsize
sin}}(z)$.

Now consider the term $\mbox{$\mbox{\boldmath$X$}$}_{\mbox{\scriptsize
reg}}(z)=R_{\mbox{\scriptsize
reg}}(z)[\mbox{$\mbox{\boldmath$G$}$}(z)-C_{1}\mbox{$\mbox{\boldmath$c$}$}]$.
We have

$\displaystyle\mbox{$\mbox{\boldmath$X$}$}_{\mbox{\scriptsize reg}}(z)$
$\displaystyle=$
$\displaystyle\mbox{$\sum_{\ell\in{\mathbb{N}}-1}$}T_{\ell}(z-1)^{\ell}[\mbox{$\mbox{\boldmath$G$}$}(z)-C_{1}\mbox{$\mbox{\boldmath$c$}$}]$
(27) $\displaystyle=$
$\displaystyle\mbox{$\sum_{\ell\in{\mathbb{N}}-1}$}(T_{0}C_{1})^{\ell}T_{0}(1-z)^{\ell}\left[\mbox{$\mbox{\boldmath$G$}$}(z)-C_{1}\mbox{$\mbox{\boldmath$c$}$}\right]$
$\displaystyle=$
$\displaystyle\mbox{$\sum_{\ell\in{\mathbb{N}}-1}$}(T_{0}C_{1})^{\ell}T_{0}\,{\mathcal{M}}\left[\\{\nabla^{\ell}(\mbox{$\mbox{\boldmath$g$}$}-C_{1}\mbox{$\mbox{\boldmath$c$}$}\,\delta)_{+}(t)\\}_{t\in{\mathbb{Z}}}\right](z).$

By applying (21) we can see that

$\displaystyle\mbox{$\mbox{\boldmath$X$}$}_{\mbox{\scriptsize reg}}(z)$
$\displaystyle=$
$\displaystyle\mbox{$\sum_{\ell\in{\mathbb{N}}-1}$}(T_{0}C_{1})^{\ell}T_{0}\,{\mathcal{M}}\left[\\{\mbox{$\sum_{s=0}^{\min\\{\ell,t\\}}\binom{\ell}{s}$}(-1)^{s}(\mbox{$\mbox{\boldmath$g$}$}-C_{1}\mbox{$\mbox{\boldmath$c$}$}\,\delta)(t-s)\\}_{t\in{\mathbb{Z}}}\right](z)$
(28) $\displaystyle=$
$\displaystyle{\mathcal{M}}\left[\\{\mbox{$\sum_{\ell\in{\mathbb{N}}-1}$}(T_{0}C_{1})^{\ell}T_{0}\,\mbox{$\sum_{s=0}^{\min\\{\ell,t\\}}\binom{\ell}{s}$}(-1)^{s}(\mbox{$\mbox{\boldmath$g$}$}-C_{1}\mbox{$\mbox{\boldmath$c$}$}\,\delta)(t-s)\\}_{t\in{\mathbb{Z}}}\right](z)$
$\displaystyle=$
$\displaystyle{\mathcal{M}}\left[\\{\mbox{$\sum_{\ell\in{\mathbb{N}}-1}$}(-1)^{\ell}T_{\ell}\nabla^{\ell}\mbox{$\mbox{\boldmath$g$}$}_{+}(t)\\}_{t\in{\mathbb{Z}}}\right](z)$
$\displaystyle\hskip
28.45274pt-\,{\mathcal{M}}\left[\\{\mbox{$\sum_{s=0}^{t}$}(-1)^{s}\left[\mbox{$\sum_{\ell\in{\mathbb{N}}-1+t}\binom{\ell}{s}$}(T_{0}C_{1})^{\ell+1}\mbox{$\mbox{\boldmath$c$}$}\right]\delta(t-s)\\}_{t\in{\mathbb{Z}}}\right](z)$
$\displaystyle=$
$\displaystyle{\mathcal{M}}\left[\\{\mbox{$\sum_{\ell\in{\mathbb{N}}-1}$}(-1)^{\ell}T_{\ell}\nabla^{\ell}\mbox{$\mbox{\boldmath$g$}$}_{+}(t)\\}_{t\in{\mathbb{Z}}}\right](z)$
$\displaystyle\hskip
85.35826pt-\,{\mathcal{M}}\left[\\{(-1)^{t}\mbox{$\sum_{\ell\in{\mathbb{N}}-1+t}\binom{\ell}{t}$}(T_{0}C_{1})^{\ell+1}\mbox{$\mbox{\boldmath$c$}$}\\}_{t\in{\mathbb{Z}}}\right](z)$

We can justify convergence of (28) in the following way. We have assumed that
$R(z)$ is analytic on ${\mathcal{W}}_{\theta}$ for some $\theta>0$ and so
$R_{\mbox{\scriptsize
reg}}(z)=\sum_{\ell\in{\mathbb{N}}-1}T_{\ell}(z-1)^{\ell}$ is analytic for all
$z\in{\mathcal{U}}_{\,1+\theta}(1)$. Hence we can find $d_{\theta}>0$ such
that

$\|T_{\ell}\|=\|(T_{0}C_{1})^{\ell}T_{0}\|\leq d_{\theta}/(1+\theta)^{\ell}$

for all $\ell\in{\mathbb{N}}-1$. It follows that

$\displaystyle{\mathbb{E}}\left[\|\mbox{$\sum_{\ell\in{\mathbb{N}}-1}$}(-1)^{\ell}T_{\ell}\nabla^{\ell}\mbox{$\mbox{\boldmath$g$}$}_{+}(t)\|\right]$
$\displaystyle=$
$\displaystyle{\mathbb{E}}\left[\|\mbox{$\sum_{\ell\in{\mathbb{N}}-1}(T_{0}C_{1})^{\ell}T_{0}\,\sum_{s=0}^{\min\\{\ell,t\\}}\binom{\ell}{s}$}(-1)^{s}\mbox{$\mbox{\boldmath$g$}$}(t-s)\|\right]$
$\displaystyle\leq$ $\displaystyle
d_{\theta}\,{\mathbb{E}}\left[\|\mbox{$\mbox{\boldmath$g$}$}\|\right]\mbox{$\sum_{s=0}^{t}\sum_{\ell\in{\mathbb{N}}-1+s}\binom{\ell}{s}$}[1/(1+\theta)]^{\ell}$
$\displaystyle=$ $\displaystyle
d_{\theta}\,{\mathbb{E}}\left[\|\mbox{$\mbox{\boldmath$g$}$}\|\right]\,\mbox{$\sum_{s=0}^{t}\sum_{m\in{\mathbb{N}}-1}\binom{m+s}{s}$}\left[1/(1+\theta)\right]^{m+s}$
$\displaystyle=$ $\displaystyle
d_{\theta}\,{\mathbb{E}}\left[\|\mbox{$\mbox{\boldmath$g$}$}\|\right]\,(1+\theta)\,\mbox{$\sum_{s=0}^{t}$}(1/\theta)^{s+1}$

is well defined and finite for all $t\in{\mathbb{N}}-1$. A similar argument
shows that

$\mbox{$\sum_{\ell\in{\mathbb{N}}-1+t}\binom{\ell}{t}$}(T_{0}C_{1})^{\ell+1}\mbox{$\mbox{\boldmath$c$}$}=(I_{X}-T_{0}C_{1})^{-t-1}(T_{0}C_{1})^{t+1}\mbox{$\mbox{\boldmath$c$}$}$

is also well defined. The function $R_{\mbox{\scriptsize
reg}}(z)=[I_{x}+T_{0}C_{1}(z-1)]^{-1}T_{0}$ is analytic for
$z\in{\mathcal{U}}_{\,1+\theta}(1)$ and hence is analytic at $z=0$ with

$R_{\mbox{\scriptsize reg}}(z)=\mbox{$\sum_{s\in{\mathbb{N}}-1}$}V_{s}z^{s}$
(29)

for all $z\in{\mathcal{U}}_{\,\theta}(0)$ where the coefficients
$\\{V_{s}\\}_{s\in{\mathbb{N}}-1}$ are given by the formula

$\displaystyle V_{s}$ $\displaystyle=$
$\displaystyle(1/s!)\left[d^{s}R_{\mbox{\scriptsize
reg}}(z)/dz^{s}\right]_{z=0}$ (30) $\displaystyle=$
$\displaystyle(1/s!)\left[d^{s}[I_{X}+T_{0}C_{1}]^{-1}T_{0}/dz^{s}\right]_{z=0}$
$\displaystyle=$
$\displaystyle(-1)^{s}(I_{X}-T_{0}C_{1})^{-s-1}(T_{0}C_{1})^{s}T_{0}$

for each $s\in{\mathbb{N}}-1$. If we set

$\displaystyle\mbox{$\mbox{\boldmath$x$}$}_{\mbox{\scriptsize reg,+}}(t)$
$\displaystyle=$
$\displaystyle\mbox{$\sum_{\ell\in{\mathbb{N}}-1}$}(-1)^{\ell}T_{\ell}\nabla^{\ell}\mbox{$\mbox{\boldmath$g$}$}_{+}(t)+(-1)^{t+1}(I_{X}-T_{0}C_{1})^{-t-1}(T_{0}C_{1})^{t+1}\mbox{$\mbox{\boldmath$c$}$}$
(31) $\displaystyle=$
$\displaystyle\mbox{$\sum_{\ell\in{\mathbb{N}}-1}$}(-1)^{\ell}T_{\ell}\nabla^{\ell}\mbox{$\mbox{\boldmath$g$}$}_{+}(t)-V_{t}C_{1}\mbox{$\mbox{\boldmath$c$}$}$

for all $t\in{\mathbb{N}}-1$ then
${\mathcal{M}}\left[\\{\mbox{$\mbox{\boldmath$x$}$}_{\mbox{\scriptsize reg,
+}}(t)\\}_{t\in{\mathbb{Z}}}\right](z)=\mbox{$\mbox{\boldmath$X$}$}_{\mbox{\scriptsize
reg}}(z)$. We can now combine (26) and (31) to state a non-standard natural
form of the Granger–Johansen representation.

###### Proposition 1

If $R(z)$ is analytic on the set
${\mathcal{V}}_{\epsilon}\cup{\mathcal{W}}_{\theta}$ for some $\epsilon>0$ and
$\theta>0$ and if
$\mbox{$\mbox{\boldmath$x$}$}(-1)=\mbox{$\mbox{\boldmath$c$}$}$ then we can
write

$\mbox{$\mbox{\boldmath$x$}$}(t)=\mbox{$\sum_{k\in{\mathbb{N}}}$}(-1)^{k}\,T_{-k}\nabla^{-k}\mbox{$\mbox{\boldmath$g$}$}_{+}(t)-U_{t}C_{1}\mbox{$\mbox{\boldmath$c$}$}+\mbox{$\sum_{\ell\in{\mathbb{N}}-1}$}(-1)^{\ell}T_{\ell}\nabla^{\ell}\mbox{$\mbox{\boldmath$g$}$}_{+}(t)-V_{t}C_{1}\mbox{$\mbox{\boldmath$c$}$}$
(32)

for all $t\in{\mathbb{N}}-1$. $\hfill\Box$

Proposition 1 shows that $x(t)$ may be written as the sum of a stochastic
trend depending only on negative powers of the backward difference operator,
deterministic components $(-1)U_{t}C_{1}\mbox{$\mbox{\boldmath$c$}$}$ and
$(-1)V_{t}C_{1}\mbox{$\mbox{\boldmath$c$}$}$ which do not depend on
$\mbox{$\mbox{\boldmath$g$}$}_{+}$, and the term

$\mbox{$\mbox{\boldmath$h$}$}_{t}(t)=\mbox{$\sum_{\ell\in{\mathbb{N}}-1}$}(-1)^{\ell}T_{\ell}\nabla^{\ell}\mbox{$\mbox{\boldmath$g$}$}_{+}(t)=\mbox{$\sum_{s=0}^{t}\sum_{\ell\in{\mathbb{N}}-1+s}\binom{\ell}{s}$}(-1)^{\ell+s}T_{\ell}\,\mbox{$\mbox{\boldmath$g$}$}(t-s)$
(33)

The expression (32) is non-standard because
$\mbox{$\mbox{\boldmath$h$}$}_{t}(t)$ is only asymptotically stationary—rather
than strictly stationary. However one advantage is that
$\mbox{$\mbox{\boldmath$h$}$}_{t}(t)$ is a posteriori observable because
$\mbox{$\mbox{\boldmath$g$}$}(t)=A_{0}\mbox{$\mbox{\boldmath$x$}$}(t)+A_{1}\mbox{$\mbox{\boldmath$x$}$}(t-1)$
can be recovered from the observed values $\mbox{$\mbox{\boldmath$x$}$}(t)$
for $t\in{\mathbb{N}}-1$ and the initial value
$\mbox{$\mbox{\boldmath$x$}$}(-1)$ if the coefficients $\\{A_{0},A_{1}\\}$ of
the model are known. This assumes, in essence, that the model fitting was
successful [1, 13, 14, 42]. We will now show that (32) can be converted into
standard form. Define

$\mbox{$\mbox{\boldmath$h$}$}_{\infty}(t)=\mbox{$\sum_{\ell\in{\mathbb{N}}-1}$}(-1)^{\ell}T_{\ell}\nabla^{\ell}\mbox{$\mbox{\boldmath$g$}$}(t)=\mbox{$\sum_{s\in{\mathbb{N}}-1}\sum_{\ell\in{\mathbb{N}}-1+s}\binom{\ell}{s}$}(-1)^{\ell+s}T_{\ell}\,\mbox{$\mbox{\boldmath$g$}$}(t-s)$
(34)

and write
$\mbox{$\mbox{\boldmath$h$}$}_{t}(t)=\mbox{$\mbox{\boldmath$h$}$}_{\infty}(t)-\mbox{$\sum_{\ell\in{\mathbb{N}}-1}$}(-1)^{\ell}T_{\ell}\nabla^{\ell}[\mbox{$\mbox{\boldmath$g$}$}-\mbox{$\mbox{\boldmath$g$}$}_{+}](t)$
for all $t\in{\mathbb{N}}-1$. Now

$\displaystyle\mbox{$\sum_{\ell\in{\mathbb{N}}-1}$}(-1)^{\ell}T_{\ell}\nabla^{\ell}[\mbox{$\mbox{\boldmath$g$}$}-\mbox{$\mbox{\boldmath$g$}$}_{+}](t)$
$\displaystyle=$
$\displaystyle\mbox{$\sum_{s\in{\mathbb{N}}-1}\sum_{\ell\in{\mathbb{N}}-1+s}\binom{\ell}{s}$}(-1)^{s}(T_{0}C_{1})^{\ell}T_{0}\,[\mbox{$\mbox{\boldmath$g$}$}-\mbox{$\mbox{\boldmath$g$}$}_{+}](t-s)$
$\displaystyle=$
$\displaystyle\mbox{$\sum_{s\in{\mathbb{N}}+t}\sum_{\ell\in{\mathbb{N}}-1+s}\binom{\ell}{s}$}(-1)^{s}(T_{0}C_{1})^{\ell}T_{0}\,\mbox{$\mbox{\boldmath$g$}$}(t-s)$
$\displaystyle=$
$\displaystyle\mbox{$\sum_{r\in{\mathbb{N}}}\sum_{m\in{\mathbb{N}}-1}\binom{m+r+t}{r+t}$}(-1)^{m}(T_{0}C_{1})^{m+r+t}T_{0}\,\mbox{$\mbox{\boldmath$g$}$}(-r)$
$\displaystyle=$
$\displaystyle\mbox{$\sum_{r\in{\mathbb{N}}}$}(-1)^{r+t}(I_{X}-T_{0}C_{1})^{-r-t-1}(T_{0}C_{1})^{r+t}T_{0}\,\mbox{$\mbox{\boldmath$g$}$}(-r)$
$\displaystyle=$
$\displaystyle(-1)^{t}(I_{X}-T_{0}C_{1})^{-t-1}(T_{0}C_{1})^{t+1}\mbox{$\mbox{\boldmath$k$}$}$
$\displaystyle=$ $\displaystyle V_{t}C_{1}\,\mbox{$\mbox{\boldmath$k$}$}$

for all $t\in{\mathbb{N}}-1$ where we have defined

$\mbox{$\mbox{\boldmath$k$}$}=\mbox{$\sum_{r\in{\mathbb{N}}}(-1)^{r}$}(I_{X}-T_{0}C_{1})^{-r}(T_{0}C_{1})^{r-1}T_{0}\,\mbox{$\mbox{\boldmath$g$}$}(-r)\in
P^{c}(X).$ (35)

The significance of the expressions in (34) for
$\mbox{$\mbox{\boldmath$h$}$}_{\infty}(t)$ and (35) for $k$ is the dependence
on the function $\mbox{$\mbox{\boldmath$g$}$}(t)$ which is strictly
stationary. Thus, in each case, the entire expression is strictly stationary.
By comparison the expression in (33) for $\mbox{$\mbox{\boldmath$h$}$}_{t}(t)$
depends on the truncated function $\mbox{$\mbox{\boldmath$g$}$}_{+}(t)$ which
is not strictly stationary. We can now state a natural form of the standard
Granger–Johansen representation.

###### Theorem 1

If $R(z)$ is analytic on the set
${\mathcal{V}}_{\epsilon}\cup{\mathcal{W}}_{\theta}$ for some $\epsilon>0$ and
$\theta>0$ and if
$\mbox{$\mbox{\boldmath$x$}$}(-1)=\mbox{$\mbox{\boldmath$c$}$}$ then we can
write

$\mbox{$\mbox{\boldmath$x$}$}(t)=\mbox{$\sum_{k\in{\mathbb{N}}}$}(-1)^{k}\,T_{-k}\nabla^{-k}\mbox{$\mbox{\boldmath$g$}$}_{+}(t)-U_{t}C_{1}\mbox{$\mbox{\boldmath$c$}$}+\mbox{$\sum_{\ell\in{\mathbb{N}}-1}$}(-1)^{\ell}T_{\ell}\nabla^{\ell}\mbox{$\mbox{\boldmath$g$}$}(t)-V_{t}C_{1}\mbox{$\mbox{\boldmath$c$}$}-\mbox{$\mbox{\boldmath$k$}$}(t)$
(36)

where we have defined
$\mbox{$\mbox{\boldmath$k$}$}(t)=V_{t}C_{1}\mbox{$\mbox{\boldmath$k$}$}$ for
all $t\in{\mathbb{N}}-1$. $\hfill\Box$

In the natural form (36) of the standard Granger–Johansen representation the
term $\mbox{$\mbox{\boldmath$h$}$}_{\infty}(t)$ defined in (34) is strictly
stationary but non-observable because $\mbox{$\mbox{\boldmath$g$}$}(t-s)$ is
only observed for $s\leq t$. In practice one would need to check the observed
properties of $\mbox{$\mbox{\boldmath$h$}$}_{t}(t)$ to determine whether the
unobservable $\mbox{$\mbox{\boldmath$h$}$}_{\infty}(t)$ is truly strictly
stationary. The natural form of the Granger–Johansen representation shows that
the key terms derived from the negative and nonnegative powers of the backward
difference operator lie in complementary subspaces defined by the spectral
separation projection $P=T_{-1}C_{1}$.

###### Corollary 1

Let $P=T_{-1}C_{1}\in{\mathcal{B}}(X)$ and
$P^{c}=I_{X}-P=T_{0}C_{0}\in{\mathcal{B}}(X)$ denote the spectral separation
projections centred at $z=1$ on $X$. If $R(z)$ is analytic on the set
${\mathcal{V}}_{\epsilon}\cup{\mathcal{W}}_{\theta}$ for some $\epsilon>0$ and
$\theta>0$ and if
$\mbox{$\mbox{\boldmath$x$}$}(-1)=\mbox{$\mbox{\boldmath$c$}$}$ then we have

$\mbox{$\mbox{\boldmath$x$}$}_{\mbox{\scriptsize\rm
sin}}(t)=\mbox{$\sum_{k\in{\mathbb{N}}}$}(-1)^{k}\,T_{-k}\nabla^{-k}P\mbox{$\mbox{\boldmath$g$}$}_{+}(t)-U_{t}T_{-1}C_{1}\mbox{$\mbox{\boldmath$c$}$}\in
P(X)$ (37)

and

$\mbox{$\mbox{\boldmath$x$}$}_{\mbox{\scriptsize\rm
reg}}(t)=\mbox{$\sum_{\ell\in{\mathbb{N}}-1}$}(-1)^{\ell}T_{\ell}\nabla^{\ell}P^{c}\mbox{$\mbox{\boldmath$g$}$}(t)-V_{t}C_{1}\mbox{$\mbox{\boldmath$c$}$}-\mbox{$\mbox{\boldmath$k$}$}(t)\in
P^{c}(X)$ (38)

for all $t\in{\mathbb{N}}-1$. $\hfill\Box$

Proof. We have
$T_{-k}\mbox{$\mbox{\boldmath$g$}$}(t)=(-1)(T_{-1}C_{0})^{k-1}T_{-1}(C_{1}T_{-1}\mbox{$\mbox{\boldmath$g$}$}(t))=T_{-k}P\mbox{$\mbox{\boldmath$g$}$}(t)$
for each $k\in{\mathbb{N}}$ and
$T_{\ell}\mbox{$\mbox{\boldmath$g$}$}(t)=(-1)^{\ell}(T_{0}C_{1})^{\ell}T_{0}(C_{0}T_{0}\mbox{$\mbox{\boldmath$g$}$}(t))=T_{\ell}P^{c}\mbox{$\mbox{\boldmath$g$}$}(t)$
for each $\ell\in{\mathbb{N}}-1$. $\hfill\Box$

We know that $\|(T_{-1}C_{0})^{k}\|^{1/k}\rightarrow 0$ and that
$\|(C_{0}T_{-1})^{k}\|^{1/k}\rightarrow 0$ as $k\rightarrow\infty$. We have
the following related result.

###### Lemma 1

The operators $T_{-1}C_{0}\in{\mathcal{B}}(X)$ and
$C_{0}T_{-1}\in{\mathcal{B}}(Y)$ are either both nilpotent of index $d+1$ for
some finite $d\in{\mathbb{N}}$ or both quasi-nilpotent. $\hfill\Box$

Proof. If $(T_{-1}C_{0})^{k}\neq\mbox{\large$0$}_{X}$ then
$(T_{-1}C_{0})^{k}(T_{-1}C_{1}+T_{0}C_{0})\neq\mbox{\large$0$}_{X}$. Since
$T_{-1}C_{0}T_{0}=\mbox{\large$0$}_{Y,X}$ we know that
$T_{-1}(C_{0}T_{-1})^{k}C_{1}\neq\mbox{\large$0$}_{X}$. Therefore
$(C_{0}T_{-1})^{k}\neq\mbox{\large$0$}_{Y}$. A similar argument shows that the
reverse implication is also true. In the first case
$T_{-k}=\mbox{\large$0$}_{Y,X}$ for all $k\geq d+1$ and the resolvent $R(z)$
has a pole of order $d$ at $z=1$. In the second case
$T_{-k}\neq\mbox{\large$0$}_{Y,X}$ for all $k\in{\mathbb{N}}$ and the
resolvent $R(z)$ has an isolated essential singularity at $z=1$. $\hfill\Box$

###### Corollary 2

Suppose $R(z)$ is analytic on the set
${\mathcal{V}}_{\epsilon}\cup{\mathcal{W}}_{\theta}$ for some $\epsilon>0$ and
$\theta>0$ and that
$\mbox{$\mbox{\boldmath$x$}$}(-1)=\mbox{$\mbox{\boldmath$c$}$}$. If $R(z)$ has
a pole of order $d$ at $z=1$ then $T_{-k-1}=\mbox{\large$0$}_{Y,X}$ for $k\geq
d+1$ and $T_{-d}\neq\mbox{\large$0$}_{X,Y}$ and we have

$\mbox{$\mbox{\boldmath$x$}$}(t)=\mbox{$\sum_{k=1}^{d}$}(-1)^{k}\,T_{-k}\nabla^{-k}\mbox{$\mbox{\boldmath$g$}$}_{+}(t)-U_{t}C_{1}\mbox{$\mbox{\boldmath$c$}$}+\mbox{$\sum_{\ell\in{\mathbb{N}}-1}$}(-1)^{\ell}T_{\ell}\nabla^{\ell}\mbox{$\mbox{\boldmath$g$}$}(t)-V_{t}C_{1}\mbox{$\mbox{\boldmath$c$}$}-\mbox{$\mbox{\boldmath$k$}$}(t)$
(39)

for all $t\in{\mathbb{N}}-1$. We say that the time series is integrated of
order $d$ and we write
$\\{\mbox{$\mbox{\boldmath$x$}$}_{+}(t)\\}_{t\in{\mathbb{Z}}}$ is $I(d)$.
$\hfill\Box$

###  6.5 Special cases

Beare and Seo [10] consider specific results for $I(d)$ processes when
$d=1,2$. We have the following remarks. We suppose $R(z)$ is analytic on the
set ${\mathcal{V}}_{\epsilon}\cup{\mathcal{W}}_{\theta}$ for some $\epsilon>0$
and $\theta>0$ and that
$\mbox{$\mbox{\boldmath$x$}$}(-1)=\mbox{$\mbox{\boldmath$c$}$}$. From a purely
algebraic viewpoint it is more convenient to refer the following arguments to
the non-standard representation in Proposition 1. When $d=1$ we know that
$T_{-k}=\mbox{\large$0$}$ for $k>1$ and so
$T_{-2}=T_{-1}C_{0}T_{-1}=\mbox{\large$0$}$. Therefore the standard series
expansion gives

$U_{t}=(-1)[I_{X}-T_{-1}C_{0}]^{-1}T_{-1}=(-1)[T_{-1}+(T_{-1}C_{0})T_{-1}+\cdots]=-T_{-1}.$

Hence (32) becomes

$\mbox{$\mbox{\boldmath$x$}$}(t)=(-1)T_{-1}\nabla^{-1}\mbox{$\mbox{\boldmath$g$}$}_{+}(t)+T_{-1}C_{1}\mbox{$\mbox{\boldmath$c$}$}+\mbox{$\sum_{\ell\in{\mathbb{N}}-1}$}(-1)^{\ell}T_{\ell}\nabla^{\ell}\mbox{$\mbox{\boldmath$g$}$}_{+}(t)-V_{t}C_{1}\mbox{$\mbox{\boldmath$c$}$}$
(40)

for all $t\in{\mathbb{N}}-1$. In the above expression
$\nabla^{-1}\mbox{$\mbox{\boldmath$g$}$}_{+}(t)=\sum_{s=0}^{t}\mbox{$\mbox{\boldmath$g$}$}(s)$
for all $t\in{\mathbb{N}}-1$.

###### Remark 3

Let $\mbox{$\mbox{\boldmath$f$}$}\in X^{*}$ be a bounded linear functional on
$X$. We follow Luenberger [34, Sections 5.6, 5.7, pp 115–118] and use the
scalar product notation
$\langle\mbox{$\mbox{\boldmath$x$}$},\mbox{$\mbox{\boldmath$f$}$}\rangle=\mbox{$\mbox{\boldmath$f$}$}(\mbox{$\mbox{\boldmath$x$}$})$
to denote the value of $f$ at the point $\mbox{$\mbox{\boldmath$x$}$}\in X$.
For each $S\subseteq X$ the orthogonal complement is

$S^{\perp}=\\{\mbox{$\mbox{\boldmath$f$}$}\in
X^{*}\mid\langle\mbox{$\mbox{\boldmath$x$}$},\mbox{$\mbox{\boldmath$f$}$}\rangle=0\
\mbox{for all}\ \mbox{$\mbox{\boldmath$x$}$}\in S\\}\subseteq X^{*}.$

We know from Theorem 4 in A that

$Y=C_{1}T_{-1}(Y)\oplus C_{0}T_{0}(Y)=Q(Y)\oplus
Q^{c}(Y)=Y_{\mbox{\rm\scriptsize sin}}\oplus Y_{\mbox{\rm\scriptsize reg}}.$

Since $T_{-1}C_{1}(X_{\mbox{\rm\scriptsize
sin}})=P[P(X)]=P(X)=X_{\mbox{\rm\scriptsize sin}}$ and
$T_{-1}C_{0}(X_{\mbox{\rm\scriptsize
reg}})=T_{-1}C_{0}[T_{0}C_{0}(X)]=\\{\mbox{$\mbox{\boldmath$0$}$}\\}$ it
follows that

$T_{-1}(Y)=T_{-1}(Y_{\mbox{\rm\scriptsize sin}})\oplus
T_{-1}(Y_{\mbox{\rm\scriptsize reg}})=T_{-1}C_{1}(X_{\mbox{\rm\scriptsize
sin}})+T_{-1}C_{0}(X_{\mbox{\rm\scriptsize reg}})=X_{\mbox{\rm\scriptsize
sin}}.$

Now consider the time series
$\\{\langle\mbox{$\mbox{\boldmath$x$}$}(t),\mbox{$\mbox{\boldmath$f$}$}\rangle\\}_{t\in{\mathbb{N}}-1}$
formed by the action of $f$ on
$\\{\mbox{$\mbox{\boldmath$x$}$}(t)\\}_{t\in{\mathbb{N}}-1}$. If
$\mbox{$\mbox{\boldmath$x$}$}(-1)=\mbox{$\mbox{\boldmath$c$}$}\in T^{-1}(Y)$
then
$T_{0}C_{1}\mbox{$\mbox{\boldmath$c$}$}=T_{0}C_{1}T_{-1}\mbox{$\mbox{\boldmath$d$}$}$
for some $\mbox{$\mbox{\boldmath$d$}$}\in Y$ and so
$T_{0}C_{1}\mbox{$\mbox{\boldmath$c$}$}=\mbox{$\mbox{\boldmath$0$}$}$.
Therefore
$V_{t}C_{1}\mbox{$\mbox{\boldmath$c$}$}=\mbox{$\mbox{\boldmath$0$}$}$ for all
$t\in{\mathbb{N}}-1$. We also have

$\displaystyle\mbox{$\sum_{\ell\in{\mathbb{N}}-1}$}(-1)^{\ell}T_{\ell}\nabla^{\ell}\mbox{$\mbox{\boldmath$g$}$}_{+}(t)$
$\displaystyle=$
$\displaystyle\mbox{$\sum_{\ell\in{\mathbb{N}}-1}$}(-1)^{\ell}T_{\ell}\left[\mbox{$\sum_{s=0}^{\min\\{\ell,t\\}}\binom{\ell}{s}$}(-1)^{s}\mbox{$\mbox{\boldmath$g$}$}(t-s)\right]$
$\displaystyle=$
$\displaystyle\mbox{$\sum_{s=0}^{t}$}\left[\rule{0.0pt}{11.38092pt}\mbox{$\sum_{\ell\in{\mathbb{N}}-1+s}(-1)^{\ell+s}T_{\ell}\binom{\ell}{s}$}\right]\mbox{$\mbox{\boldmath$g$}$}(t-s)$
$\displaystyle=$
$\displaystyle\mbox{$\sum_{s=0}^{t}$}(-1)^{s}\left[\rule{0.0pt}{11.38092pt}\mbox{$\sum_{m\in{\mathbb{N}}-1}\binom{s+m}{s}(T_{0}C_{1})^{m}$}\right](T_{0}C_{1})^{s}T_{0}\,\mbox{$\mbox{\boldmath$g$}$}(t-s)$
$\displaystyle=$
$\displaystyle\mbox{$\sum_{s=0}^{t}$}(-1)^{s}(I_{X}-T_{0}C_{1})^{-s-1}(T_{0}C_{1})^{s}T_{0}\,\mbox{$\mbox{\boldmath$g$}$}(t-s)$
$\displaystyle=$
$\displaystyle\rule{0.0pt}{11.38092pt}\mbox{$\sum_{s=0}^{t}$}V_{s}\,\mbox{$\mbox{\boldmath$g$}$}(t-s)$

and so (40) reduces to

$\mbox{$\mbox{\boldmath$x$}$}(t)=T_{-1}\left[C_{1}\mbox{$\mbox{\boldmath$c$}$}-\mbox{$\sum_{s=0}^{t}$}\,\mbox{$\mbox{\boldmath$g$}$}(t-s)\right]+\mbox{$\sum_{s=0}^{t}$}V_{s}\mbox{$\mbox{\boldmath$g$}$}(t-s)$
(41)

for all $t\in{\mathbb{N}}-1$. Now when $\mbox{$\mbox{\boldmath$f$}$}\in
X_{\mbox{\rm\scriptsize sin}}^{\perp}=T_{-1}(Y)^{\perp}\subseteq X^{*}$ we can
see that

$\langle\mbox{$\mbox{\boldmath$x$}$}(t),\mbox{$\mbox{\boldmath$f$}$}\rangle=\mbox{$\sum_{s=0}^{t}$}V_{s}\,\langle\mbox{$\mbox{\boldmath$g$}$}(t-s),\mbox{$\mbox{\boldmath$f$}$}\rangle$
(42)

for all $t\in{\mathbb{N}}-1$ and so
$\\{\langle\mbox{$\mbox{\boldmath$x$}$}(t),\mbox{$\mbox{\boldmath$f$}$}\rangle\\}_{t\in{\mathbb{N}}-1}$
is $I(0)$. Hence $\mbox{$\mbox{\boldmath$x$}$}(t)$ is co-integrated. In this
regard we recall that $R_{\mbox{\rm\scriptsize
reg}}(z)=\sum_{s\in{\mathbb{N}}-1}V_{s}z^{s}$ for all
$z\in{\mathcal{U}}_{\,\theta}(0)$ and that $R_{\mbox{\rm\scriptsize
reg}}(z)=\sum_{\ell\in{\mathbb{N}}-1}T_{\ell}(z-1)^{\ell}$ for all
$z\in{\mathcal{U}}_{\,1+\theta}(1)$. Thus we know that
$R_{\mbox{\rm\scriptsize reg}}(1)=T_{0}$ is nonsingular on the subspace
$P^{c}(X)$. Therefore $R_{\mbox{\rm\scriptsize
reg}}(1)=T_{0}\neq\mbox{\large$0$}$ provided $P^{c}\neq\mbox{\large$0$}$. Thus
(41) and (42) provide a natural extension of the results obtained by Beare and
Seo. $\hfill\Box$

When $d=2$ we know that
$T_{-k}=(-1)^{k-1}(T_{-1}C_{0})^{k-1}T_{-1}=\mbox{\large$0$}$ for $k>2$ and so
the standard series expansion gives

$U_{t}C_{1}\mbox{$\mbox{\boldmath$c$}$}=(-1)(I_{X}-T_{-1}C_{0})^{-t-1}T_{-1}C_{1}\mbox{$\mbox{\boldmath$c$}$}=(-1)[T_{-1}C_{1}\mbox{$\mbox{\boldmath$c$}$}+(t+1)T_{-1}C_{0}T_{-1}C_{1}\mbox{$\mbox{\boldmath$c$}$}]$

for all $t\in{\mathbb{N}}-1$. Therefore (32) becomes

$\displaystyle\mbox{$\mbox{\boldmath$x$}$}(t)$ $\displaystyle=$ $\displaystyle
T_{-2}\nabla^{-2}\mbox{$\mbox{\boldmath$g$}$}_{+}(t)-T_{-1}\nabla^{-1}\mbox{$\mbox{\boldmath$g$}$}_{+}(t)$
(43) $\displaystyle\hskip
14.22636pt+T_{-1}C_{1}\mbox{$\mbox{\boldmath$c$}$}-(t+1)T_{-2}C_{1}\mbox{$\mbox{\boldmath$c$}$}+\mbox{$\sum_{s=0}^{t}$}V_{s}\mbox{$\mbox{\boldmath$g$}$}(t-s)-V_{t}C_{1}\mbox{$\mbox{\boldmath$c$}$}$

for all $t\in{\mathbb{N}}-1$. In the above expression
$\nabla^{-1}\mbox{$\mbox{\boldmath$g$}$}_{+}(t)=\sum_{s=0}^{t}\,\mbox{$\mbox{\boldmath$g$}$}(t-s)$
and
$\nabla^{-2}\mbox{$\mbox{\boldmath$g$}$}_{+}(t)=\sum_{s=0}^{t}\,(s+1)\mbox{$\mbox{\boldmath$g$}$}(t-s)$.

###### Remark 4

If $\mbox{$\mbox{\boldmath$x$}$}(-1)=\mbox{$\mbox{\boldmath$c$}$}\in
T_{-2}(Y)$ then
$\mbox{$\mbox{\boldmath$c$}$}=T_{-2}\mbox{$\mbox{\boldmath$d$}$}$ for some
$\mbox{$\mbox{\boldmath$d$}$}\in Y$. Since $T_{0}C_{1}T_{-2}=\mbox{\large$0$}$
it follows that
$T_{0}C_{1}\mbox{$\mbox{\boldmath$c$}$}=\mbox{$\mbox{\boldmath$0$}$}$. We also
have
$T_{-1}C_{1}\mbox{$\mbox{\boldmath$c$}$}=(T_{-1}C_{1}+T_{0}C_{0})T_{-2}\mbox{$\mbox{\boldmath$d$}$}=T_{-2}\mbox{$\mbox{\boldmath$d$}$}$.
Therefore (43) reduces to

$\mbox{$\mbox{\boldmath$x$}$}(t)=T_{-2}\nabla^{-2}\mbox{$\mbox{\boldmath$g$}$}_{+}(t)-T_{-1}\nabla^{-1}\mbox{$\mbox{\boldmath$g$}$}_{+}(t)+T_{-2}\mbox{$\mbox{\boldmath$d$}$}+(t+1)T_{-2}C_{0}T_{-2}\mbox{$\mbox{\boldmath$d$}$}+\mbox{$\sum_{s=0}^{t}$}V_{s}\,\mbox{$\mbox{\boldmath$g$}$}(t-s).$
(44)

If we also have $\mbox{$\mbox{\boldmath$f$}$}\in T_{-2}(Y)^{\perp}$ then (44)
implies

$\langle\mbox{$\mbox{\boldmath$x$}$}(t),\mbox{$\mbox{\boldmath$f$}$}\rangle=-T_{-1}\mbox{$\sum_{s=0}^{t}$}\,\langle\mbox{$\mbox{\boldmath$g$}$}(t-s),\mbox{$\mbox{\boldmath$f$}$}\rangle+\mbox{$\sum_{s=0}^{t}$}V_{s}\,\langle\mbox{$\mbox{\boldmath$g$}$}(t-s),\mbox{$\mbox{\boldmath$f$}$}\rangle$

for all $t\in{\mathbb{N}}-1$. Thus
$\\{\langle\mbox{$\mbox{\boldmath$x$}$}(t),\mbox{$\mbox{\boldmath$f$}$}\rangle\\}_{t\in{\mathbb{N}}-1}$
is $I(1)$ if $\mbox{$\mbox{\boldmath$f$}$}\notin T_{-1}(Y)^{\perp}$ and is
$I(0)$ if $\mbox{$\mbox{\boldmath$f$}$}\in T_{-1}(Y)^{\perp}$. Therefore
$\mbox{$\mbox{\boldmath$x$}$}(t)$ is co-integrated. Once again we note that
$R_{\mbox{\rm\scriptsize reg}}(1)=T_{0}\neq\mbox{\large$0$}$ provided
$P^{c}\neq\mbox{\large$0$}$. Note that
$T_{-2}(Y)=T_{-1}C_{0}T_{-1}(Y)\subseteq T_{-1}(Y)$ implies
$T_{-1}(Y)^{\perp}\subseteq T_{-2}(Y)^{\perp}$. If
$\mbox{$\mbox{\boldmath$f$}$}\notin T_{-2}(Y)^{\perp}$ then
$\\{\langle\mbox{$\mbox{\boldmath$x$}$}(t),\mbox{$\mbox{\boldmath$f$}$}\rangle\\}_{t\in{\mathbb{N}}-1}$
is $I(2)$. $\hfill\Box$

###  6.6 The extended representation

Now suppose only that
$\\{\mbox{$\mbox{\boldmath$x$}$}(t)\\}_{t\in{\mathbb{Z}}}$ is a well-defined
unit root process. Therefore $R(z)$ is analytic for
$z\in[\,{\mathcal{U}}_{\,1}(0)^{a}\setminus\\{1\\}\,]\cup{\mathcal{U}}_{\,0,\,\delta}(1)$
for some $\delta>0$. We have the following result.

###### Lemma 2

The resolvent operator $R(z)$ is analytic on a set
$[\,{\mathcal{U}}_{\,1}(0)^{a}\setminus\\{1\\}\,]\cup{\mathcal{U}}_{\,0,\,\delta}(1)$
for some $\delta>0$ if and only if $R(z)$ is analytic on a set
${\mathcal{V}}_{\epsilon}={\mathcal{U}}_{1+\epsilon}(0)\setminus\\{1\\}$ for
some $\epsilon>0$. $\hfill\Box$

Proof. Suppose $R(z)$ is analytic on ${\mathcal{V}}_{\,\epsilon}$ for some
$\epsilon>0$. Since

$[\,{\mathcal{U}}_{\,1}(0)^{a}\setminus\\{1\\}\,]\cup{\mathcal{U}}_{\,0,\,\epsilon}(1)\subseteq{\mathcal{V}}_{\epsilon}$

it follows that $R(z)$ is analytic on
$[\,{\mathcal{U}}_{\,1}(0)^{a}\setminus\\{1\\}\,]\cup{\mathcal{U}}_{\,0,\,\epsilon}(1)$.

Now suppose $R(z)$ is analytic on
$[\,{\mathcal{U}}_{\,1}(0)^{a}\setminus\\{1\\}\,]\cup{\mathcal{U}}_{\,0,\,\delta}(1)$
for some $\delta>0$. Define the compact set $C=\\{z\in{\mathbb{C}}\mid|z|=1\
\mbox{and}\ z\notin{\mathcal{U}}_{\,\delta/2}(1)\\}$. The resolvent $R(z)$ is
analytic on $C$ and so for each $w\in C$ we can find $\epsilon(w)>0$ such that
$R(z)$ is analytic on ${\mathcal{U}}_{\,\epsilon(w)}(w)$. Therefore

$C\subseteq\mbox{$\bigcup_{w\in C}$}\,{\mathcal{U}}_{\,\epsilon(w)}(w).$

Since $C$ is compact there exists a finite sub-covering. Thus we have a finite
collection of neighbourhoods
$\\{{\mathcal{U}}_{\epsilon_{i}}(w_{i})\\}_{i=1}^{n}$ with $w_{i}\in C$ and
$\epsilon_{i}>0$ for each $i=1,\ldots,n$ such that

$C\subseteq\mbox{$\bigcup_{i=1}^{n}$}\,{\mathcal{U}}_{\,\epsilon_{i}}(w_{i}).$

Thus there is some $\epsilon\in(0,\delta/2)$ such that

$N(C,\epsilon)=\\{z\in{\mathbb{C}}\mid|z-w|<\epsilon\ \mbox{for some}\ w\in
C\\}\subseteq\mbox{$\bigcup_{i=1}^{n}$}\,{\mathcal{U}}_{\,\epsilon_{i}}(w_{i}).$

Therefore $R(z)$ is analytic for $z\in N(C,\epsilon)$. Hence $R(z)$ is
analytic for

$z\in{\mathcal{V}}_{\,\epsilon}\subseteq
N(C,\epsilon)\cup[\,{\mathcal{U}}_{\,1}(0)^{a}\setminus\\{1\\}\,]\cup{\mathcal{U}}_{\,0,\,\delta}(1).$

This completes the proof. $\hfill\Box$.

Lemma 2 shows that if
$\\{\mbox{$\mbox{\boldmath$x$}$}(t)\\}_{t\in{\mathbb{Z}}}$ is a well-defined
unit root process then there is some $\epsilon>0$ such that $R(z)$ is analytic
for $z\in{\mathcal{V}}_{\epsilon}$. It is still true that the function
$R_{\mbox{\scriptsize sin}}(z)$ is analytic for all $z\neq 1$ and hence is
analytic at $z=0$ with

$R_{\mbox{\scriptsize sin}}(z)=\mbox{$\sum_{s\in{\mathbb{N}}-1}$}U_{s}z^{s}$

for all $z\in{\mathcal{U}}_{\,1}(0)$ where the coefficients
$\\{U_{s}\\}_{s\in{\mathbb{N}}-1}$ are given by the formula

$U_{s}=(-1)(I_{X}-T_{-1}C_{0})^{-s-1}T_{-1}$

for each $s\in{\mathbb{N}}-1$. The resolvent $R(z)=(A_{0}+A_{1}z)^{-1}$ is
also analytic for $z\in{\mathcal{U}}_{\,1}(0)$ and so $A_{0}^{-1}$ is well
defined. Hence we can use a Neumann expansion to write

$R(z)=\mbox{$\sum_{s\in{\mathbb{N}}-1}$}R_{s}z^{s}$

for all $z\in{\mathcal{U}}_{\,1}(0)$ where

$R_{s}=(-1)^{s}(A_{0}^{-1}A_{1})^{s}A_{0}^{-1}$ (45)

for all $s\in{\mathbb{N}}-1$. Since $R(z)$ is singular at $z=1$ it follows
that
$\lim_{s\rightarrow\infty}\|R_{s}\|^{1/s}=\lim_{s\rightarrow\infty}\|(A_{0}^{-1}A_{1})^{s}\|^{1/s}=1$.
We also know that

$R_{\mbox{\scriptsize reg}}(z)=R(z)-R_{\mbox{\scriptsize sin}}(z)$ (46)

for $z\in{\mathcal{U}}_{\,\epsilon}(1)$. Since $R(z)$ and
$R_{\mbox{\scriptsize sin}}(z)$ are analytic for
$z\in{\mathcal{V}}_{\epsilon}$ and since $R_{\mbox{\scriptsize reg}}(z)$ is
analytic at $z=1$ we can use (46) to extend $R_{\mbox{\scriptsize reg}}(z)$ to
an analytic function on ${\mathcal{U}}_{\,1+\epsilon}(0)$. Therefore

$R_{\mbox{\scriptsize reg}}(z)=\mbox{$\sum_{s\in{\mathbb{N}}-1}$}Q_{s}z^{s}$
(47)

for $z\in{\mathcal{U}}_{\,1+\epsilon}(0)$ where $Q_{s}\in{\mathcal{B}}(Y,X)$
and $\lim_{s\rightarrow\infty}\|Q_{s}\|^{1/s}\leq 1/(1+\epsilon)$. Now (46)
gives

$Q_{s}=R_{s}-U_{s}=(-1)^{s}(A_{0}^{-1}A_{1})^{s}A_{0}^{-1}+(I_{X}-T_{-1}C_{0})^{-s-1}T_{-1}$
(48)

for each $s\in{\mathbb{N}}-1$. We have $C_{0}=A_{0}+A_{1}$ and so we can
calculate $\\{Q_{s}\\}_{s\in{\mathbb{N}}-1}$ from a knowledge of the
coefficients $A_{0}$, $A_{1}$ and $T_{-1}$.

The inequality $\lim_{s\rightarrow\infty}\|Q_{s}\|^{1/s}\leq 1/(1+\epsilon)$
means that we can find a real constant $c>0$ such that $\|Q_{s}\|\leq
c/(1+\epsilon)^{s}$ for all $s\in{\mathbb{N}}-1$. If we write $\zeta=z-1$ we
have

$\displaystyle R_{\mbox{\scriptsize reg}}(z)$ $\displaystyle=$
$\displaystyle\mbox{$\sum_{s\in{\mathbb{N}}-1}$}Q_{s}(1+\zeta)^{s}$ (49)
$\displaystyle=$
$\displaystyle\mbox{$\sum_{s\in{\mathbb{N}}-1}$}Q_{r}\left[\,\mbox{$\sum_{\ell=0}^{s}$}\mbox{$\binom{s}{\ell}$}\zeta^{\ell}\,\right]$
$\displaystyle=$
$\displaystyle\mbox{$\sum_{\ell\in{\mathbb{N}}-1}$}\left[\,\mbox{$\sum_{s\in{\mathbb{N}}-1+\ell}$}\mbox{$\binom{s}{\ell}$}Q_{s}\,\right]\zeta^{\ell}$
$\displaystyle=$
$\displaystyle\mbox{$\sum_{\ell\in{\mathbb{N}}-1}$}\left[\,\mbox{$\sum_{r\in{\mathbb{N}}-1}$}\mbox{$\binom{\ell+r}{\ell}$}Q_{\ell+r}\,\right](z-1)^{\ell}$

for all $z\in{\mathcal{U}}_{\,\epsilon}(1)$. It follows from (49) that as well
as the formula $T_{\ell}=(-1)^{\ell}(T_{0}C_{1})^{\ell}T_{0}$ for each
$\ell\in{\mathbb{N}}-1$ we can also calculate
$\\{T_{\ell}\\}_{\ell\in{\mathbb{N}}-1}$ indirectly using the formula

$T_{\ell}=\mbox{$\sum_{r\in{\mathbb{N}}-1}$}\mbox{$\binom{\ell+r}{\ell}$}Q_{\ell+r}$
(50)

for each $\ell\in{\mathbb{N}}-1$. Since $\|Q_{\ell+r}\|\leq
c/(1+\epsilon)^{\ell+r}$ it follows that (50) converges absolutely with
$\|T_{\ell}\|\leq c(1+\epsilon)/\epsilon^{\ell+1}$. Before continuing we will
illustrate our considerations with an elementary matrix example.

###### Example 1

Let $\epsilon\in(0,1)\subseteq{\mathbb{R}}$ and define

$A_{0}=\left[\begin{array}[]{cc}1&-\epsilon\\\
1&1\end{array}\right]\quad\mbox{and}\quad
A_{1}=\left[\begin{array}[]{cc}-1&0\\\ -1&-1\end{array}\right].$

Therefore

$A(z)=\left[\begin{array}[]{cc}1-z&-\epsilon\\\ 1-z&1-z\end{array}\right]$

for all $z\in{\mathbb{C}}$ and

$R(z)=(1+\epsilon-z)^{-1}\left[\begin{array}[]{cc}1&\epsilon(1-z)^{-1}\\\
-1&1\end{array}\right]=(1+\epsilon-z)^{-1}\left[\begin{array}[]{cc}1&-1\\\
-1&1\end{array}\right]+(1-z)^{-1}\left[\begin{array}[]{cc}0&1\\\
0&0\end{array}\right]$

for all $z\in{\mathbb{C}}$ with $z\notin\\{1,1+\epsilon\\}$. If we wish to
find expansions centred at $z=1$ then we have $A(z)=C_{0}+C_{1}(z-1)$ where

$C_{0}=\left[\begin{array}[]{cc}0&-\epsilon\\\
0&0\end{array}\right]\quad\mbox{and}\quad
C_{1}=\left[\begin{array}[]{cc}-1&0\\\ -1&-1\end{array}\right]$

and $R(z)=T_{-1}(z-1)^{-1}+\sum_{\ell\in{\mathbb{N}}-1}T_{\ell}(z-1)^{\ell}$
where

$T_{-1}=\left[\begin{array}[]{cc}0&-1\\\
0&0\end{array}\right]\quad\mbox{and}\quad
T_{\ell}=\epsilon^{-(\ell+1)}\left[\begin{array}[]{cc}1&-1\\\
-1&1\end{array}\right]=\epsilon^{-\ell}\left[\begin{array}[]{cc}\epsilon^{-1}&-\epsilon^{-1}\\\
-\epsilon^{-1}&\epsilon^{-1}\end{array}\right]=\epsilon^{-\ell}T_{0}$

for each $\ell\in{\mathbb{N}}-1$. Since
$\|T_{\ell}\|_{2}=2\epsilon^{-(\ell+1)}$ it follows that the Laurent series
for $R(z)$ centred at $z=1$ converges for
$z\in{\mathcal{U}}_{\,0,\,\epsilon}(1)$. The spectral separation projections
are

$P=T_{-1}C_{1}=\left[\begin{array}[]{cc}1&1\\\
0&0\end{array}\right]\quad\mbox{and}\quad
Q=C_{1}T_{-1}=\left[\begin{array}[]{cc}0&1\\\ 0&1\end{array}\right].$

We note that $PT_{-1}=T_{-1}\Rightarrow T_{-1}\in P(X)$ and
$P^{c}T_{\ell}=(I-P)T_{\ell}=T_{\ell}\Rightarrow T_{\ell}\in P^{c}(X)$ for
each $\ell\in{\mathbb{N}}-1$. Although the singular part of the resolvent can
be written directly in the form

$R_{\mbox{\scriptsize
sin}}(z)=T_{-1}(z-1)^{-1}=\mbox{$\sum_{s\in{\mathbb{N}}-1}$}(-1)T_{-1}z^{s}\in
P(X)$

as a legitimate Maclaurin transform the same direct approach for the regular
part gives

$\displaystyle R_{\mbox{\scriptsize reg}}(z)$ $\displaystyle=$
$\displaystyle\mbox{$\sum_{\ell\in{\mathbb{N}}-1}$}\epsilon^{-\ell}T_{0}(z-1)^{\ell}$
$\displaystyle=$
$\displaystyle\mbox{$\sum_{\ell\in{\mathbb{N}}-1}$}(-1)^{\ell}\epsilon^{-\ell}T_{0}\,\mbox{$\sum_{r=0}^{\ell}\binom{\ell}{r}$}z^{r}$
$\displaystyle=$
$\displaystyle\mbox{$\sum_{r\in{\mathbb{N}}-1}$}\left[\mbox{$\sum_{\ell\in{\mathbb{N}}+r-1}\binom{\ell}{r}$}(-1)^{\ell}\epsilon^{-\ell}\right]T_{0}\,z^{r}$
$\displaystyle=$
$\displaystyle\mbox{$\sum_{r\in{\mathbb{N}}-1}$}\left[\mbox{$\sum_{m\in{\mathbb{N}}-1}\binom{m+r}{r}$}(-1)^{m}\epsilon^{-m}\right]T_{0}\,(-1)^{r}(z/\epsilon)^{r}$

which is not a legitimate Maclaurin transform because each coefficient of
$z^{r}$ involves the sum of a divergent series. Thus we must take a different
approach. We can write

$\displaystyle R_{\mbox{\scriptsize reg}}(z)$ $\displaystyle=$ $\displaystyle
R(z)-R_{\mbox{\scriptsize sin}}(z)$ $\displaystyle=$
$\displaystyle(1+\epsilon-z)^{-1}\left[\begin{array}[]{cc}1&\epsilon(1-z)^{-1}\\\
-1&1\end{array}\right]-(1-z)^{-1}\left[\begin{array}[]{cc}0&1\\\
0&0\end{array}\right]$ $\displaystyle=$
$\displaystyle(1+\epsilon-z)^{-1}\left[\begin{array}[]{cc}1&-1\\\
-1&1\end{array}\right]$ $\displaystyle=$
$\displaystyle\mbox{$\sum_{s\in{\mathbb{N}}-1}$}\epsilon(1+\epsilon)^{-(s+1)}T_{0}\,z^{s}\in
P^{c}(X)$

for all $z\in{\mathcal{U}}_{\,1+\epsilon}(0)$. $\hfill\Box$

It is important to note that in Example 1 we have $Q_{s}\in P^{c}(X)$ for all
$s\in{\mathbb{N}}-1$. We will now argue that this is true in general.

###### Lemma 3

Let

$Q(z)=\left\\{\begin{array}[]{ll}R(z)-R_{\mbox{\rm\scriptsize
sin}}(z)&\mbox{for}\ z\in{\mathcal{U}}_{\,1+\epsilon}(0)\setminus\\{1\\}\\\
R_{\mbox{\rm\scriptsize reg}}(z)&\mbox{for}\ z=1\end{array}\right.$

and suppose that the Maclaurin series representation is given by
$Q(z)=\sum_{s\in{\mathbb{N}}-1}Q_{s}z^{s}$ for all
$z\in{\mathcal{U}}_{\,1+\epsilon}(0)$. Then $Q_{s}\in{\mathcal{B}}(P^{c}(X))$
for all $s\in{\mathbb{N}}-1$. $\hfill\Box$

Proof. Let $P=T_{-1}C_{1}$ and let
$PQ(z)=\sum_{s\in{\mathbb{N}}-1}PQ_{s}z^{s}$ for
$z\in{\mathcal{U}}_{\,1+\epsilon}(0)$ denote the projection of $Q(z)$ onto the
subspace $P(X)$. We have $PT_{\ell}=\mbox{\large$0$}$ for all
$\ell\in{\mathbb{N}}-1$ and so

$PQ(z)=P[R(z)-R_{\mbox{\scriptsize sin}}(z)]=PR_{\mbox{\scriptsize
reg}}(z)=\mbox{$\sum_{\ell\in{\mathbb{N}}-1}$}PT_{\ell}(z-1)^{\ell}=\mbox{\large$0$}$

for all $z\in{\mathcal{U}}_{\,\epsilon}(1)$. Since
$PQ(z)\equiv\mbox{\large$0$}$ on a non-trivial open set it follows by analytic
continuation that $PQ(z)\equiv\mbox{\large$0$}$ for all
$z\in{\mathcal{U}}_{\,1+\epsilon}(0)$ and hence that $PQ_{s}=\mbox{\large$0$}$
for all $s\in{\mathbb{N}}-1$. Thus we deduce that
$Q_{s}=P^{c}Q_{s}\in{\mathcal{B}}(P^{c}(X))$ for all $s\in{\mathbb{N}}-1$.
$\hfill\Box$

We can now find a more general representation for
$\\{\mbox{$\mbox{\boldmath$x$}$}_{\mbox{\scriptsize
reg,+}}(t)\\}_{t\in{\mathbb{Z}}}$. We have

$\displaystyle\mbox{$\mbox{\boldmath$X$}$}_{\mbox{\scriptsize reg}}(z)$
$\displaystyle=$ $\displaystyle R_{\mbox{\scriptsize
reg}}(z)[G(z)-C_{1}\mbox{$\mbox{\boldmath$c$}$}]$ (53) $\displaystyle=$
$\displaystyle\mbox{$\sum_{s\in{\mathbb{N}}-1}$}Q_{s}z^{s}\,\mbox{$\sum_{r\in{\mathbb{N}}-1}$}(\mbox{$\mbox{\boldmath$g$}$}-C_{1}\mbox{$\mbox{\boldmath$c$}$}\,\delta)_{+}(r)z^{r}$
$\displaystyle=$
$\displaystyle\mbox{$\sum_{t\in{\mathbb{N}}-1}$}\left[\mbox{$\sum_{s\in{\mathbb{N}}-1}$}Q_{s}\,(\mbox{$\mbox{\boldmath$g$}$}-C_{1}\mbox{$\mbox{\boldmath$c$}$}\,\delta)_{+}(t-s)\right]z^{t}$
$\displaystyle=$
$\displaystyle\mbox{$\sum_{t\in{\mathbb{N}}-1}$}\left[\mbox{$\sum_{s=0}^{t}$}Q_{s}\,\mbox{$\mbox{\boldmath$g$}$}(t-s)-Q_{t}\,C_{1}\mbox{$\mbox{\boldmath$c$}$}\right]z^{t}$
$\displaystyle=$
$\displaystyle{\mathcal{M}}\left[\\{\mbox{$\sum_{s=0}^{t}$}Q_{s}\,\mbox{$\mbox{\boldmath$g$}$}(t-s)\\}_{t\in{\mathbb{N}}-1}-\\{Q_{t}\,C_{1}\mbox{$\mbox{\boldmath$c$}$}\\}_{t\in{\mathbb{N}}-1}\right](z)$

for all $z\in{\mathcal{U}}_{\,1+\epsilon}(0)$. Therefore

$\mbox{$\mbox{\boldmath$x$}$}_{\mbox{\scriptsize
reg}}(t)=\mbox{$\sum_{s=0}^{t}$}Q_{s}\,\mbox{$\mbox{\boldmath$g$}$}(t-s)-Q_{t}\,C_{1}\mbox{$\mbox{\boldmath$c$}$}$

for all $t\in{\mathbb{N}}-1$. It is now possible to state a more general form
of the non-standard Granger–Johansen representation.

###### Proposition 2

If $R(z)$ is analytic on the set ${\mathcal{V}}_{\epsilon}$ for some
$\epsilon>0$ then we can write

$\mbox{$\mbox{\boldmath$x$}$}(t)=\mbox{$\sum_{k\in{\mathbb{N}}}$}(-1)^{k}\,T_{-k}\nabla^{-k}\mbox{$\mbox{\boldmath$g$}$}_{+}(t)-U_{t}C_{1}\mbox{$\mbox{\boldmath$c$}$}+\,\mbox{$\sum_{s=0}^{t}$}Q_{s}\,\mbox{$\mbox{\boldmath$g$}$}(t-s)-Q_{t}\,C_{1}\mbox{$\mbox{\boldmath$c$}$}$
(54)

for all $t\in{\mathbb{N}}-1$. $\hfill\Box$

The term

$\widehat{\mbox{$\mbox{\boldmath$h$}$}}_{t}(t)=\mbox{$\sum_{s=0}^{t}$}Q_{s}\,\mbox{$\mbox{\boldmath$g$}$}(t-s)$
(55)

is observable but is only asymptotically stationary. If we define

$\widehat{\mbox{$\mbox{\boldmath$h$}$}}_{\infty}(t)=\mbox{$\sum_{s\in{\mathbb{N}}-1}$}Q_{s}\mbox{$\mbox{\boldmath$g$}$}(t-s)$
(56)

for all $t\in{\mathbb{N}}-1$ then we have

$\widehat{\mbox{$\mbox{\boldmath$h$}$}}_{t}(t)=\widehat{\mbox{$\mbox{\boldmath$h$}$}}_{\infty}(t)-\mbox{$\sum_{s\in{\mathbb{N}}+t}$}Q_{s}\mbox{$\mbox{\boldmath$g$}$}(t-s)=\widehat{\mbox{$\mbox{\boldmath$h$}$}}_{\infty}(t)-\mbox{$\sum_{r\in{\mathbb{N}}}$}Q_{r+t}\mbox{$\mbox{\boldmath$g$}$}(-r)$

for all $t\in{\mathbb{N}}-1$. The extended term
$\widehat{\mbox{$\mbox{\boldmath$h$}$}}_{\infty}(t)$ is strictly stationary
but is no longer a posteriori observable.

The original relationships (2) and (3) show that

$A_{0}\mbox{$\mbox{\boldmath$x$}$}(t)+A_{1}\mbox{$\mbox{\boldmath$x$}$}(t-1)=\mbox{$\mbox{\boldmath$g$}$}(t)$

for all $t\in{\mathbb{N}}$. Therefore

$\mbox{$\mbox{\boldmath$x$}$}(t)=A_{0}^{-1}\mbox{$\mbox{\boldmath$g$}$}(t)-A_{0}^{-1}A_{1}\mbox{$\mbox{\boldmath$x$}$}(t-1)$

for all $t\in{\mathbb{N}}$. If we extend this relationship to all
$t\in{\mathbb{Z}}$ then repeated application gives

$\mbox{$\mbox{\boldmath$x$}$}(-1)=\mbox{$\sum_{r=1}^{n}$}(-1)^{r-1}(A_{0}^{-1}A_{1})^{r-1}A_{0}^{-1}\mbox{$\mbox{\boldmath$g$}$}(-r)+(-1)^{n}(A_{0}^{-1}A_{1})^{n}\mbox{$\mbox{\boldmath$x$}$}(-n-1)$
(57)

for each $n\in{\mathbb{N}}$. We can also write the original relationship in
the form

$(C_{0}-C_{1})\mbox{$\mbox{\boldmath$x$}$}(t)+C_{1}\mbox{$\mbox{\boldmath$x$}$}(t-1)=\mbox{$\mbox{\boldmath$g$}$}(t)$

for all $t\in{\mathbb{N}}$. Thus we have

$T_{-1}C_{0}\mbox{$\mbox{\boldmath$x$}$}(t)-P\mbox{$\mbox{\boldmath$x$}$}(t)+P\mbox{$\mbox{\boldmath$x$}$}(t-1)=T_{-1}\mbox{$\mbox{\boldmath$g$}$}(t)$

where we have written $P=T_{-1}C_{1}$. Since
$T_{-1}C_{0}T_{0}=\mbox{\large$0$}$ and $T_{-1}C_{1}+T_{0}C_{0}=I_{X}$ it
follows that
$P\mbox{$\mbox{\boldmath$x$}$}(t)-T_{-1}C_{0}\mbox{$\mbox{\boldmath$x$}$}(t)=(I_{X}-T_{-1}C_{0})P\mbox{$\mbox{\boldmath$x$}$}(t)$.
Some elementary algebra now shows that

$P\mbox{$\mbox{\boldmath$x$}$}(t)=(-1)(I_{X}-T_{-1}C_{0})^{-1}T_{-1}\mbox{$\mbox{\boldmath$g$}$}(t)+(I_{X}-T_{-1}C_{0})^{-1}P\mbox{$\mbox{\boldmath$x$}$}(t-1)$

for all $t\in{\mathbb{N}}$. If we extend this relationship to all
$t\in{\mathbb{Z}}$ then repeated application gives

$P\mbox{$\mbox{\boldmath$x$}$}(-1)=\mbox{$\sum_{r=1}^{n}$}(-1)(I_{X}-T_{-1}C_{0})^{-r}T_{-1}\mbox{$\mbox{\boldmath$g$}$}(-r)+(I_{X}-T_{-1}C_{0})^{-n}P\mbox{$\mbox{\boldmath$x$}$}(-n-1)$
(58)

for each $n\in{\mathbb{N}}$. By applying (57) and (58) and using (25) and (45)
we deduce that

$\displaystyle(-1)R_{t}C_{1}\mbox{$\mbox{\boldmath$x$}$}(-1)+U_{t}C_{1}\mbox{$\mbox{\boldmath$x$}$}(-1)$
(59) $\displaystyle=$
$\displaystyle(-1)^{t+1}(A_{0}^{-1}A_{1})^{t+1}A_{1}\mbox{$\mbox{\boldmath$x$}$}(-1)-(I_{X}-T_{-1}C_{0})^{-t-1}P\mbox{$\mbox{\boldmath$x$}$}(-1)$
$\displaystyle=$
$\displaystyle\mbox{$\sum_{r=1}^{n}$}(-1)^{r+t}(A_{0}^{-1}A_{1})^{r+t}A_{0}^{-1}\mbox{$\mbox{\boldmath$g$}$}(-r)+(-1)^{n+t+1}(A_{0}^{-1}A_{1})^{n+t}A_{0}^{-1}A_{1}\mbox{$\mbox{\boldmath$x$}$}(-n-1)$
$\displaystyle\hskip
14.22636pt+\mbox{$\sum_{r=1}^{n}$}(I_{X}-T_{-1}C_{0})^{-r-t-1}T_{-1}\mbox{$\mbox{\boldmath$g$}$}(-r)-(I_{X}-T_{-1}C_{0})^{-n-t-1}P\mbox{$\mbox{\boldmath$x$}$}(-n-1)$
$\displaystyle=$
$\displaystyle\mbox{$\sum_{r=1}^{n}$}Q_{r+t}\mbox{$\mbox{\boldmath$g$}$}(-r)-R_{n+t}C_{1}\mbox{$\mbox{\boldmath$x$}$}(-n-1)+U_{n+t}C_{1}\mbox{$\mbox{\boldmath$x$}$}(-n-1)$

for all $t\in{\mathbb{N}}-1$ and all $n\in{\mathbb{N}}$. Rearranging and
writing $\mbox{$\mbox{\boldmath$x$}$}(-1)=\mbox{$\mbox{\boldmath$c$}$}$ gives

$\displaystyle\mbox{$\sum_{r=1}^{n}Q_{t+r}$}\mbox{$\mbox{\boldmath$g$}$}(-r)$
$\displaystyle=$
$\displaystyle\left[R_{n+t}C_{1}\mbox{$\mbox{\boldmath$x$}$}(-n-1)-U_{n+t}C_{1}\mbox{$\mbox{\boldmath$x$}$}(-n-1)\right]-\left[R_{t}C_{1}\mbox{$\mbox{\boldmath$c$}$}-U_{t}C_{1}\mbox{$\mbox{\boldmath$c$}$}\right]$
(60) $\displaystyle=$ $\displaystyle
Q_{n+t}C_{1}\mbox{$\mbox{\boldmath$x$}$}(-n-1)-Q_{t}C_{1}\mbox{$\mbox{\boldmath$c$}$}$

for all $t\in{\mathbb{N}}-1$. We have

${\mathbb{E}}\left[\,\|\mbox{$\sum_{r=1}^{n}$}Q_{t+r}\mbox{$\mbox{\boldmath$g$}$}(-r)\|\right]\leq
c\,{\mathbb{E}}[\,\|\mbox{$\mbox{\boldmath$g$}$}\|\,]\mbox{$\sum_{r\in{\mathbb{N}}}$}1/(1+\epsilon)^{t+r}=c\,{\mathbb{E}}[\,\|\mbox{$\mbox{\boldmath$g$}$}\|\,]/[\epsilon(1+\epsilon)^{t}]\rightarrow
0$

as $t\rightarrow\infty$. Hence (60) shows that we can define a
$\mu$-integrable time series
$\\{\mbox{$\mbox{\boldmath$k$}$}(t)\\}_{t\in{\mathbb{N}}-1}$ by setting

$\mbox{$\mbox{\boldmath$k$}$}(t)=\lim_{n\rightarrow\infty}\left[Q_{n+t}C_{1}\mbox{$\mbox{\boldmath$x$}$}(-n-1)-Q_{t}C_{1}\mbox{$\mbox{\boldmath$c$}$}\right]$
(61)

for each $t\in{\mathbb{N}}-1$. Thus we can now write

$\mbox{$\sum_{r\in{\mathbb{N}}}Q_{t+r}$}\mbox{$\mbox{\boldmath$g$}$}(-r)=\mbox{$\mbox{\boldmath$k$}$}(t)$
(62)

for all $t\in{\mathbb{N}}-1$. Therefore

$\widehat{\mbox{$\mbox{\boldmath$h$}$}}_{t}(t)=\widehat{\mbox{$\mbox{\boldmath$h$}$}}_{\infty}(t)-\mbox{$\mbox{\boldmath$k$}$}(t)$
(63)

for all $t\in{\mathbb{N}}-1$. This relationship allows us to state a general
form of the standard Granger–Johansen representation. This is our main result.

###### Theorem 2

If $R(z)$ is analytic on the set ${\mathcal{V}}_{\epsilon}$ for some
$\epsilon>0$ and if
$\mbox{$\mbox{\boldmath$x$}$}(-1)=\mbox{$\mbox{\boldmath$c$}$}$ then we can
write

$\mbox{$\mbox{\boldmath$x$}$}(t)=\mbox{$\sum_{k\in{\mathbb{N}}}$}(-1)^{k}\,T_{-k}\nabla^{-k}\mbox{$\mbox{\boldmath$g$}$}_{+}(t)-U_{t}C_{1}\mbox{$\mbox{\boldmath$c$}$}+\mbox{$\sum_{s\in{\mathbb{N}}-1}$}Q_{s}\,\mbox{$\mbox{\boldmath$g$}$}(t-s)-Q_{t}\,C_{1}\mbox{$\mbox{\boldmath$c$}$}-\mbox{$\mbox{\boldmath$k$}$}(t)$
(64)

for all $t\in{\mathbb{N}}-1$. $\hfill\Box$

The extended representation in Theorem 2 is more robust than the natural
representation in Theorem 1 and while we can no longer use the coefficients
$\\{T_{\ell}\\}_{\ell\in{\mathbb{N}}-1}$ in the extended representation the
spectral separation of the singular and regular parts is nevertheless
preserved. In each case we have
$\mbox{$\mbox{\boldmath$X$}$}_{\mbox{\scriptsize sin}}(z)\in P(X)$ and
$\mbox{$\mbox{\boldmath$X$}$}_{\mbox{\scriptsize reg}}(z)\in P^{c}(X)$.

##  7 The resolvent operator for a linear pencil on Banach space

The methods described in this paper depend on a systematic methodology for
calculation of the resolvent operator for a linear pencil on Banach space. The
basic methodology was introduced earlier in Section 6.2. In this section we
pick up where that earlier discussion left off. Albrecht et al. [2, 4, 5] show
that the resolvent operator $R(z)=(C_{0}+C_{1}(z-1))^{-1}$ exists and is
defined by the Laurent series $R(z)=\sum_{j\in{\mathbb{Z}}}T_{j}(z-1)^{j}$ if
and only if the coefficients $\\{T_{j}\\}_{j\in{\mathbb{Z}}}$ satisfy the left
and right fundamental equations (8), (9) and the magnitude constraint (10) in
which case the coefficients are defined by (11) and (12). It follows from (10)
and (11) that

$\lim_{k\rightarrow\infty}\|T_{-k}\|^{1/k}=0\iff\lim_{k\rightarrow\infty}\|(T_{-1}C_{0})^{k}\|^{1/k}=0\iff\lim_{k\rightarrow\infty}\|(C_{0}T_{-1})^{k}\|^{1/k}=0.$

We will show that the operators $T_{-1}C_{0}\in{\mathcal{B}}(X)$ and
$C_{0}T_{-1}\in{\mathcal{B}}(Y)$ are either both nilpotent of index $d+1$ for
some finite $d\in{\mathbb{N}}$ or both quasi-nilpotent. We argue as follows.
If $(T_{-1}C_{0})^{k}\neq\mbox{\large$0$}_{X}$ then
$(T_{-1}C_{0})^{k}(T_{-1}C_{1}+T_{0}C_{0})\neq\mbox{\large$0$}_{X}$. Since
$T_{-1}C_{0}T_{0}=\mbox{\large$0$}_{Y,X}$ we know that
$T_{-1}(C_{0}T_{-1})^{k}C_{1}\neq\mbox{\large$0$}_{X}$. Therefore
$(C_{0}T_{-1})^{k}\neq\mbox{\large$0$}_{Y}$. A similar argument shows that the
reverse implication is also true. In the first case
$T_{-k}=\mbox{\large$0$}_{Y,X}$ for all $k\geq d+1$ and the resolvent $R(z)$
has a pole of order $d$ at $z=1$. In the second case
$T_{-k}\neq\mbox{\large$0$}_{Y,X}$ for all $k\in{\mathbb{N}}$ and the
resolvent $R(z)$ has an isolated essential singularity at $z=1$.

We can find the resolvent $R(z)$ by finding a basic solution to the
fundamental equations. The task is much easier if we begin by finding the key
spectral separation projections $P=T_{-1}C_{1}\in{\mathcal{B}}(X)$ and
$Q=C_{1}T_{-1}\in{\mathcal{B}}(Y)$.

###  7.1 Calculation of the key spectral separation projections

Albrecht et al. [5] have used infinite-length Jordan chains to find the
projections $P\in{\mathcal{B}}(X)$ and $Q\in{\mathcal{B}}(Y)$. The set
$J_{\mbox{\scriptsize\rm sin}}\subseteq X^{\mathbb{N}}$ of all infinite-length
singular Jordan chains for the pencil $A(z)$ on
${\mathcal{U}}_{\,s,\,\infty}(1)$ is the set of sequences
$\\{\mbox{$\mbox{\boldmath$x$}$}_{-n}\\}_{n\in{\mathbb{N}}}\in X^{\mathbb{N}}$
such that
$C_{0}\mbox{$\mbox{\boldmath$x$}$}_{-n}+C_{1}\mbox{$\mbox{\boldmath$x$}$}_{-n-1}=\mbox{$\mbox{\boldmath$0$}$}_{Y}$
for all $n\in{\mathbb{N}}$ and
$\lim_{n\rightarrow\infty}\|\mbox{$\mbox{\boldmath$x$}$}_{-n}\|^{1/n}\leq s$.
The subspace

$X_{\mbox{\scriptsize\rm
sin}}=\\{\mbox{$\mbox{\boldmath$x$}$}_{-1}\mid\\{\mbox{$\mbox{\boldmath$x$}$}_{-n}\\}_{n\in{\mathbb{N}}}\in
J_{\mbox{\scriptsize\rm sin}}\\}$

is the generating set for $J_{\mbox{\scriptsize\rm sin}}$. The set
$J_{\mbox{\scriptsize\rm reg}}\subseteq X^{\mathbb{N}}$ of all infinite-length
regular Jordan chains for the pencil $A(z)$ on ${\mathcal{U}}_{\,0,\,r}(1)$ is
the set of sequences
$\\{\mbox{$\mbox{\boldmath$x$}$}_{n}\\}_{n\in{\mathbb{N}}}\in X^{\mathbb{N}}$
such that
$C_{0}\mbox{$\mbox{\boldmath$x$}$}_{n+1}+C_{1}\mbox{$\mbox{\boldmath$x$}$}_{n}=\mbox{$\mbox{\boldmath$0$}$}_{Y}$
for all $n\in{\mathbb{N}}$ and
$\lim_{n\rightarrow\infty}\|\mbox{$\mbox{\boldmath$x$}$}_{n}\|^{1/n}\leq 1/r$.
The subspace

$X_{\mbox{\scriptsize\rm
reg}}=\\{\mbox{$\mbox{\boldmath$x$}$}_{1}\mid\\{\mbox{$\mbox{\boldmath$x$}$}_{n}\\}_{n\in{\mathbb{N}}}\in
J_{\mbox{\scriptsize\rm reg}}\\}$

is the generating set for $J_{\mbox{\scriptsize\rm reg}}$. Albrecht et al. [5]
show that $X_{\mbox{\scriptsize\rm sin}}=P(X)$ and $X_{\mbox{\scriptsize\rm
reg}}=P^{c}(X)=(I_{X}-P)(X)$ and that $X=X_{\mbox{\scriptsize\rm sin}}\oplus
X_{\mbox{\scriptsize\rm reg}}\cong X_{\mbox{\scriptsize\rm sin}}\times
X_{\mbox{\scriptsize\rm reg}}$. They also show that $Y_{\mbox{\scriptsize\rm
sin}}=Q(Y)=C_{0}(X_{\mbox{\scriptsize\rm sin}})$ and $Y_{\mbox{\scriptsize\rm
reg}}=Q^{c}(Y)=(I_{Y}-Q)(Y)=C_{1}(X_{\mbox{\scriptsize\rm reg}})$ and that
$Y=Y_{\mbox{\scriptsize\rm sin}}\oplus Y_{\mbox{\scriptsize\rm reg}}\cong
Y_{\mbox{\scriptsize\rm sin}}\times Y_{\mbox{\scriptsize\rm reg}}$.

The spectral separation projections $P\in{\mathcal{B}}(X)$ and
$Q\in{\mathcal{B}}(Y)$ can be found by calculating general solutions to the
defining equations for the Jordan chains subject to the limit requirements on
the chains determined by the annular region ${\mathcal{U}}_{\,s,\,r}(1)$. This
allows us to find the spaces $X_{\mbox{\scriptsize\rm sin}}$ and
$X_{\mbox{\scriptsize\rm reg}}$ in the domain space and the corresponding
spaces $Y_{\mbox{\scriptsize\rm sin}}$ and $Y_{\mbox{\scriptsize\rm reg}}$ in
the range space and hence to find the key projections.

For matrix pencils the full generality of the Jordan chain methodology is not
necessary. This is also true for some operator pencils on infinite-dimensional
spaces. We will consider a set of specific examples in Section 8 to illustrate
various options.

###  7.2 Separation of the fundamental equations

The key projections $P=T_{-1}C_{1}\in{\mathcal{B}}(X)$ and
$Q=C_{1}T_{-1}\in{\mathcal{B}}(Y)$ allow us to separate each of (8) and (9)
into two singly infinite sets of equations—one for the singular part of the
resolvent and the other for the regular part. We prefer to write
$M=P(X)=X_{\mbox{\scriptsize\rm sin}}$ and
$M^{c}=P^{c}(X)=X_{\mbox{\scriptsize\rm reg}}$ so that $X=M\times M^{c}$ and
$N=Q(Y)=Y_{\mbox{\scriptsize\rm sin}}$ and
$N^{c}=Q^{c}(Y)=Y_{\mbox{\scriptsize\rm reg}}$ so that $Y=N\times N^{c}$. Now
we define restricted operators ${\mathfrak{C}}_{i}\in{\mathcal{B}}(M,N)$,
${\mathfrak{C}}_{i}^{c}\in{\mathcal{B}}(M^{c},N^{c})$ for each $i=0,1$,
${\mathfrak{T}}_{-k}\in{\mathcal{B}}(N,M)$ for each $k\in{\mathbb{N}}$ and
${\mathfrak{T}}_{\ell}\in{\mathcal{B}}(N^{c},M^{c})$ for each
$\ell\in{\mathbb{N}}-1$ by setting

$C_{i}\cong\left[\begin{array}[]{c}Q\\\
Q^{c}\end{array}\right]C_{i}\left[\begin{array}[]{cc}P&P^{c}\end{array}\right]=\left[\begin{array}[]{cc}QC_{i}P&QC_{i}P^{c}\\\
Q^{c}C_{i}P&Q^{c}C_{i}P^{c}\end{array}\right]=\left[\begin{array}[]{cc}{\mathfrak{C}}_{i}&\mbox{\large$0$}_{M^{c},N}\\\
\mbox{\large$0$}_{M,N^{c}}&{\mathfrak{C}}_{i}^{c}\end{array}\right],$
$T_{-k}\cong\left[\begin{array}[]{c}P\\\
P^{c}\end{array}\right]T_{-k}\left[\begin{array}[]{cc}Q&Q^{c}\end{array}\right]=\left[\begin{array}[]{cc}PT_{-k}Q&PT_{-k}Q^{c}\\\
P^{c}T_{-k}Q&P^{c}T_{-k}Q^{c}\end{array}\right]=\left[\begin{array}[]{cc}{\mathfrak{T}}_{-k}&\mbox{\large$0$}_{N^{c},M}\\\
\mbox{\large$0$}_{N,M^{c}}&\mbox{\large$0$}_{N^{c},M^{c}}\end{array}\right]$

and

$T_{\ell}\cong\left[\begin{array}[]{c}P\\\
P^{c}\end{array}\right]T_{\ell}\left[\begin{array}[]{cc}Q&Q^{c}\end{array}\right]=\left[\begin{array}[]{cc}PT_{\ell}Q&PT_{\ell}Q^{c}\\\
P^{c}T_{\ell}Q&P^{c}T_{\ell}Q^{c}\end{array}\right]=\left[\begin{array}[]{cc}\mbox{\large$0$}_{N,M}&\mbox{\large$0$}_{N^{c},M}\\\
\mbox{\large$0$}_{N,M^{c}}&{\mathfrak{T}}_{\ell}^{c}\end{array}\right].$

If we restrict our attention to $M$ the left fundamental equations (8) become
a singly infinite set of left fundamental equations for the singular part of
the resolvent given by

$\begin{array}[]{rcl}{\mathfrak{T}}_{-1}{\mathfrak{C}}_{1}&=&{\mathfrak{I}}_{M}\\\
{\mathfrak{T}}_{-k-1}{\mathfrak{C}}_{1}+{\mathfrak{T}}_{-k}{\mathfrak{C}}_{0}&=&\mbox{\large$0$}_{M}\quad\mbox{for}\
k\in{\mathbb{N}}\end{array}$ (65)

where ${\mathfrak{I}}_{M}\in{\mathcal{B}}(M)$ denotes the identity operator on
$M$. If we restrict our attention to $M^{c}$ the left fundamental equations
(8) become a singly infinite set of left fundamental equations for the regular
part of the resolvent given by

$\begin{array}[]{rcl}{\mathfrak{T}}_{0}^{c}{\mathfrak{C}}_{0}^{c}&=&{\mathfrak{I}}_{M^{c}}\\\
{\mathfrak{T}}_{\ell-1}^{c}{\mathfrak{C}}_{1}^{c}+{\mathfrak{T}}_{\ell}^{c}{\mathfrak{C}}_{0}^{c}&=&\mbox{\large$0$}_{M^{c}}\quad\mbox{for}\
\ell\in{\mathbb{N}}\end{array}$ (66)

where ${\mathfrak{I}}_{M^{c}}\in{\mathcal{B}}(M^{c})$ denotes the identity
operator on $M^{c}$. The systems (65) and (66) are completely separate. If we
restrict our attention to $N$ the right fundamental equations (9) become a
singly infinite set of right fundamental equations for the singular part of
the resolvent given by

$\begin{array}[]{rcl}{\mathfrak{C}}_{1}{\mathfrak{T}}_{-1}&=&{\mathfrak{I}}_{N}\\\
{\mathfrak{C}}_{1}{\mathfrak{T}}_{-k-1}+{\mathfrak{C}}_{0}{\mathfrak{T}}_{-k}&=&\mbox{\large$0$}_{N}\quad\mbox{for}\
k\in{\mathbb{N}}\end{array}$ (67)

where ${\mathfrak{I}}_{N}\in{\mathcal{B}}(N)$ denotes the identity operator on
$N$. If we restrict our attention to $N^{c}$ the right fundamental equations
(9) become a singly infinite set of right fundamental equations for the
regular part of the resolvent given by

$\begin{array}[]{rcl}{\mathfrak{C}}_{0}^{c}{\mathfrak{T}}_{0}^{c}&=&{\mathfrak{I}}_{N^{c}}\\\
{\mathfrak{C}}_{1}^{c}{\mathfrak{T}}_{\ell-1}^{c}+{\mathfrak{C}}_{0}^{c}{\mathfrak{T}}_{\ell}^{c}&=&\mbox{\large$0$}_{N^{c}}\quad\mbox{for}\
\ell\in{\mathbb{N}}\end{array}$ (68)

where ${\mathfrak{I}}_{N^{c}}\in{\mathcal{B}}(N^{c})$ denotes the identity
operator on $N^{c}$. The systems (67) and (68) are completely separate. In
infinite-dimensional space the analysis depends on both the left and right
sets of fundamental equations. We need both
${\mathfrak{T}}_{-1}{\mathfrak{C}}_{1}={\mathfrak{I}}_{M}$ and
${\mathfrak{C}}_{1}{\mathfrak{T}}_{-1}={\mathfrak{I}}_{N}$ to deduce that
${\mathfrak{T}}_{-1}={\mathfrak{C}}_{1}^{-1}\in{\mathcal{B}}(N,M)$ is the
uniquely-defined inverse of ${\mathfrak{C}}_{1}$. Similarly we need both
${\mathfrak{T}}_{0}^{c}{\mathfrak{C}}_{0}^{c}={\mathfrak{I}}_{M^{c}}$ and
${\mathfrak{C}}_{0}^{c}{\mathfrak{T}}_{0}^{c}={\mathfrak{I}}_{N^{c}}$ to
deduce that
${\mathfrak{T}}_{0}^{c}=[{\mathfrak{C}}_{0}^{c}]^{-1}\in{\mathcal{B}}(N^{c},M^{c})$
is the uniquely-defined inverse of ${\mathfrak{C}}_{0}^{c}$.

The existence of ${\mathfrak{C}}_{1}^{-1}$ and $[{\mathfrak{C}}_{0}^{c}]^{-1}$
means that $M$ is isomorphic to $N$ and that $M^{c}$ is isomorphic to $N^{c}$.
Thus $X=M\times M^{c}$ and $Y=N\times N^{c}$ are isomorphic. Nevertheless
there are situations where we may wish to regard these isomorphic spaces as
different. See [2, 28] and [7, pp 282–285] for some specific instances.

###  7.3 Solution of the fundamental equations

The systems (65) and (67) have a unique solution
$\\{{\mathfrak{T}}_{-k}\\}_{k\in{\mathbb{N}}}$ where

${\mathfrak{T}}_{-k}=(-1)^{k-1}({\mathfrak{C}}_{1}^{-1}{\mathfrak{C}}_{0})^{k-1}{\mathfrak{C}}_{1}^{-1}$
(69)

for each $k\in{\mathbb{N}}$ and the systems (66) and (68) have a unique
solution $\\{{\mathfrak{T}}_{\ell}\\}_{\ell\in{\mathbb{N}}-1}$ where

${\mathfrak{T}}_{\ell}^{c}=(-1)^{\ell}([{\mathfrak{C}}_{0}^{c}]^{-1}{\mathfrak{C}}_{1}^{c})^{\ell}[{\mathfrak{C}}_{0}^{c}]^{-1}$
(70)

for each $\ell\in{\mathbb{N}}-1$. The formulæ (69) and (70) validate (11) and
(12). It follows that if we can find the key spectral separation projections
$P\in{\mathcal{B}}(X)$ and $Q\in{\mathcal{B}}(Y)$ on
${\mathcal{U}}_{\,s,\,r}(1)$ then it is a simple matter to define the
resolvent operator by the Laurent series
$R(z)=\sum_{j\in{\mathbb{Z}}}T_{j}(z-1)^{j}$ for all
$z\in{\mathcal{U}}_{\,s,\,r}(1)$.

##  8 Some specific applications

We consider three specific examples to demonstrate our proposed method for
calculation of the resolvent operator. Example 2 is taken from Seo [41,
Example 4.1, pp 13–14 and Example 4.3, p 15]. Example 3 concerns a Volterra
integral operator and amplifies the discussions in Seo [41, version 4, Example
4.2, p 14] and Spangenberg [45, Example 3.4, p 135]. Example 4 is a discrete
time analogue with added noise of a model proposed by Albrecht et al. [4,
Section 6, pp 170–173]. In each case $X,Y$ are separable Banach spaces and
$A(z)\in{\mathcal{B}}(X,Y)$ is a linear operator pencil. Our primary task is
to find the resolvent operator $R(z)=A(z)^{-1}\in{\mathcal{B}}(Y,X)$ on an
annular region ${\mathcal{U}}_{\,0,\,r}(1)$ for some $r>0$. In Examples 2 and
4 we make explicit use of Jordan chains to find the key projection operators
and hence solve the fundamental equations. In Example 3, where
$P^{c}=\mbox{\large$0$}_{X}\iff Q^{c}=\mbox{\large$0$}_{Y}$, we show that
overt use of Jordan chains can be avoided by using a Neumann expansion
instead. In this case the regular part of the Laurent series is absent. Our
aim is to demonstrate that the existing Hilbert space methods—which invariably
depend on a knowledge of advanced functional analysis—can be replaced by
systematic procedures that are more intuitive and more general.

###### Example 2

Let $X=Y=c_{0}$ be the Banach space of all sequences
$\mbox{$\mbox{\boldmath$x$}$}=\\{x_{n}\\}_{n\in{\mathbb{N}}}\subseteq{\mathbb{C}}^{\mathbb{N}}$
with $\lim_{n\rightarrow\infty}x_{n}=0$ and norm defined by
$\|\mbox{$\mbox{\boldmath$x$}$}\|_{\infty}=\max_{n\in{\mathbb{N}}}|x_{n}|$.
The space $c_{0}$ is a closed separable subspace of $\ell^{\infty}$ with the
standard basis $\\{\mbox{$\mbox{\boldmath$e$}$}_{j}\\}_{j\in{\mathbb{N}}}$.
Let $\lambda\in{\mathbb{R}}$ with $\lambda\in(0,1)$ and define
$A(z)=A_{0}+A_{1}z\in{\mathcal{B}}(X,Y)$ where $A_{0}=I$ is the identity
mapping and

$A_{1}=(-1)\left[\begin{array}[]{ccccc}1&1&0&0&\cdots\\\ 0&1&0&0&\cdots\\\
0&0&\lambda&0&\cdots\\\ 0&0&0&\lambda^{2}&\cdots\\\
\vdots&\vdots&\vdots&\vdots&\ddots\end{array}\right].$

Near $z=1$ we can write $A(z)=A_{0}+A_{1}z=C_{0}+C_{1}(z-1)$ where

$C_{0}=I+A_{1}=\left[\begin{array}[]{ccccc}0&-1&0&0&\cdots\\\
0&0&0&0&\cdots\\\ 0&0&1-\lambda&0&\cdots\\\ 0&0&0&1-\lambda^{2}&\cdots\\\
\vdots&\vdots&\vdots&\vdots&\ddots\end{array}\right]$

and $C_{1}=A_{1}$. We wish to find a Laurent expansion for the resolvent
operator $R(z)=A(z)^{-1}\in{\mathcal{B}}(Y,X)$ in the form of a Laurent series
$R(z)=\sum_{j\in{\mathbb{Z}}}T_{j}(z-1)^{j}$ which converges on some region
${\mathcal{U}}_{\,0,\,\rho}(1)=\\{z\in{\mathbb{C}}\mid 0<|z-1|<\rho\\}$. Thus
we need to find $\\{T_{j}\\}_{j\in{\mathbb{Z}}}\in{\mathcal{B}}(Y,X)$
satisfying the left fundamental equations (8) and the right fundamental
equations (9). It is sufficient to find the basic solution
$\\{T_{-1},T_{0}\\}\in{\mathcal{B}}(Y,X)$ on some region
${\mathcal{U}}_{\,0,\,\rho}(1)$.

We begin by finding a general form for the infinite-length singular Jordan
chains $\\{\mbox{$\mbox{\boldmath$x$}$}_{-n}\\}_{n\in{\mathbb{N}}}$ satisfying

$C_{0}\mbox{$\mbox{\boldmath$x$}$}_{-n}+C_{1}\mbox{$\mbox{\boldmath$x$}$}_{-n-1}=\mbox{$\mbox{\boldmath$0$}$}$

for all $n\in{\mathbb{N}}$ with
$\|\mbox{$\mbox{\boldmath$x$}$}_{-n}\|_{\infty}^{1/n}\rightarrow 0$ as
$n\rightarrow\infty$. We have

$\displaystyle-x_{-n,2}-x_{-n-1,1}-x_{-n-1,2}$ $\displaystyle=$ $\displaystyle
0$ $\displaystyle-x_{-n-1,2}$ $\displaystyle=$ $\displaystyle 0$
$\displaystyle(1-\lambda)x_{-n,3}-\lambda x_{-n-1,3}$ $\displaystyle=$
$\displaystyle 0$ $\displaystyle(1-\lambda^{2})x_{-n,4}-\lambda^{2}x_{-n-1,4}$
$\displaystyle=$ $\displaystyle 0$ $\displaystyle\vdots$ $\displaystyle\vdots$

from which we deduce

$\displaystyle x_{-n-1,1}$ $\displaystyle=$ $\displaystyle-x_{-n,2}$
$\displaystyle x_{-n-1,2}$ $\displaystyle=$ $\displaystyle 0$ $\displaystyle
x_{-n-1,3}$ $\displaystyle=$ $\displaystyle(1/\lambda-1)x_{-n,3}$
$\displaystyle x_{-n-1,4}$ $\displaystyle=$
$\displaystyle(1/\lambda^{2}-1)x_{-n,4}$ $\displaystyle\vdots$
$\displaystyle\vdots$

and so on. If we let
$\mbox{$\mbox{\boldmath$x$}$}_{-1}=\mbox{$\mbox{\boldmath$t$}$}=\\{t_{j}\\}_{j\in{\mathbb{N}}}\in
c_{0}$ then we have $x_{-n,2}=0$ for $n\geq 2$ and $x_{-n,1}=-x_{-n+1,2}=0$
for $n\geq 3$. In general we have $x_{-n,j}=(1/\lambda^{j-2}-1)^{n-1}t_{j}$
for all $n\in{\mathbb{N}}$ and all $j\geq 3$. If $t_{j}\neq 0$ then

$\|\mbox{$\mbox{\boldmath$x$}$}_{-n}\|_{\infty}^{1/n}\geq|1/\lambda^{j-2}-1|^{1-1/n}|t_{j}|^{1/n}\rightarrow|1/\lambda^{j-2}-1|$

as $n\rightarrow\infty$. Since $\lambda\neq 1$ it is not possible to have
$\lim_{n\rightarrow\infty}\|\mbox{$\mbox{\boldmath$x$}$}_{-n}\|_{\infty}^{1/n}=0$.
Consequently we must choose $t_{j}=0$ for $j\geq 3$. Therefore the subspace
$X_{\mbox{\scriptsize\rm sin}}\subseteq c_{0}$ generated by the infinite-
length singular Jordan chains is the closed subspace
${\mathcal{S}}(\\{\mbox{$\mbox{\boldmath$e$}$}_{1},\mbox{$\mbox{\boldmath$e$}$}_{2}\\})$
generated by $\mbox{$\mbox{\boldmath$e$}$}_{1}$ and
$\mbox{$\mbox{\boldmath$e$}$}_{2}$.

Our next step is to find a general form for the infinite-length regular Jordan
chains $\\{\mbox{$\mbox{\boldmath$x$}$}_{n}\\}_{n\in{\mathbb{N}}}$ satisfying

$C_{0}\mbox{$\mbox{\boldmath$x$}$}_{n+1}+C_{1}\mbox{$\mbox{\boldmath$x$}$}_{n}=\mbox{$\mbox{\boldmath$0$}$}$

for all $n\in{\mathbb{N}}$ with
$\|\mbox{$\mbox{\boldmath$x$}$}_{n}\|_{\infty}^{1/n}\rightarrow\rho$ for some
$\rho>0$. We have

$\displaystyle-x_{n+1,2}-x_{n,1}-x_{n,2}$ $\displaystyle=$ $\displaystyle 0$
$\displaystyle-x_{n,2}$ $\displaystyle=$ $\displaystyle 0$
$\displaystyle(1-\lambda)x_{n+1,3}-\lambda x_{n,3}$ $\displaystyle=$
$\displaystyle 0$ $\displaystyle(1-\lambda^{2})x_{n+1,4}-\lambda^{2}x_{n,4}$
$\displaystyle=$ $\displaystyle 0$ $\displaystyle\vdots$ $\displaystyle\vdots$

from which we deduce

$\displaystyle x_{n+1,2}$ $\displaystyle=$ $\displaystyle-x_{n,1}$
$\displaystyle x_{n,2}$ $\displaystyle=$ $\displaystyle 0$ $\displaystyle
x_{n+1,3}$ $\displaystyle=$ $\displaystyle[1/(1-\lambda)-1]x_{n,3}$
$\displaystyle x_{n+1,4}$ $\displaystyle=$
$\displaystyle[1/(1-\lambda^{2})-1]x_{n,4}$ $\displaystyle\vdots$
$\displaystyle\vdots$

and so on. If we set
$\mbox{$\mbox{\boldmath$x$}$}_{1}=\mbox{$\mbox{\boldmath$s$}$}=\\{s_{j}\\}_{j\in{\mathbb{N}}}$
then $x_{1,2}=s_{2}=0$. We also have $x_{2,2}=0$ and $x_{2,2}=-x_{1,1}=s_{1}$.
Therefore $s_{1}=0$ as well. The remaining equations show that
$x_{n,j}=[1/(1-\lambda^{j-2})-1]^{n-1}s_{j}$ for all $n\in{\mathbb{N}}$ and
all $j\geq 3$. If we choose
$\mbox{$\mbox{\boldmath$x$}$}_{1}=\mbox{$\mbox{\boldmath$e$}$}_{k}$ for some
$k\geq 3$ then
$\mbox{$\mbox{\boldmath$x$}$}_{n}=[1/(1-\lambda^{k-2})-1]^{n-1}\mbox{$\mbox{\boldmath$e$}$}_{k}$
and hence

$\|\mbox{$\mbox{\boldmath$x$}$}_{n}\|_{\infty}^{1/n}=[1/(1-\lambda^{k-2})-1]^{1-1/n}\rightarrow[1/(1-\lambda^{k-2})-1]=\lambda^{k-2}/(1-\lambda^{k-2})\leq\lambda/(1-\lambda)$

as $n\rightarrow\infty$. It follows that if we define the subspace
$X_{\mbox{\scriptsize\rm reg}}\subset c_{0}$ as the closed subspace
${\mathcal{S}}(\\{\mbox{$\mbox{\boldmath$e$}$}_{3},\mbox{$\mbox{\boldmath$e$}$}_{4},\ldots\\})$
generated by $\\{\mbox{$\mbox{\boldmath$e$}$}_{n}\\}_{n\in{\mathbb{N}}+2}$
then $X=X_{\mbox{\scriptsize\rm sin}}\oplus X_{\mbox{\scriptsize\rm reg}}\cong
X_{\mbox{\scriptsize\rm sin}}\times X_{\mbox{\scriptsize\rm reg}}$ is the
desired spectral decomposition of the domain space for the region
${\mathcal{U}}_{\,0,\,\rho}$ where $\rho=1/\lambda-1$. The corresponding
decomposition for $Y=c_{0}$ is defined by $Y_{\mbox{\scriptsize\rm
sin}}=C_{1}(X_{\mbox{\scriptsize\rm sin}})$ and $Y_{\mbox{\scriptsize\rm
reg}}=C_{0}(X_{\mbox{\scriptsize\rm reg}})$. We have

$C_{1}\mbox{$\mbox{\boldmath$e$}$}_{1}=-\mbox{$\mbox{\boldmath$e$}$}_{1}\quad\mbox{and}\quad
C_{1}\mbox{$\mbox{\boldmath$e$}$}_{2}=-\mbox{$\mbox{\boldmath$e$}$}_{1}-\mbox{$\mbox{\boldmath$e$}$}_{2}$

which means that $Y_{\mbox{\scriptsize\rm sin}}\subseteq c_{0}$ is the
subspace
${\mathcal{S}}(\\{\mbox{$\mbox{\boldmath$e$}$}_{1},\mbox{$\mbox{\boldmath$e$}$}_{2}\\})$
and

$C_{0}\mbox{$\mbox{\boldmath$e$}$}_{j}=(1-\lambda^{j-2})\mbox{$\mbox{\boldmath$e$}$}_{j}$

for each $j\geq 3$. Therefore $Y_{\mbox{\scriptsize\rm reg}}\subseteq c_{0}$
is the subspace
${\mathcal{S}}(\\{\mbox{$\mbox{\boldmath$e$}$}_{3},\mbox{$\mbox{\boldmath$e$}$}_{4},\ldots\\})$
and $Y=Y_{\mbox{\scriptsize\rm sin}}\oplus Y_{\mbox{\scriptsize\rm reg}}\cong
Y_{\mbox{\scriptsize\rm sin}}\times Y_{\mbox{\scriptsize\rm reg}}$ is the
corresponding spectral decomposition of the range space on
${\mathcal{U}}_{\,0,\,\rho}(1)$.

The corresponding key projections are clearly

$P=Q=\left[\begin{array}[]{ccccc}1&0&0&0&\cdots\\\ 0&1&0&0&\cdots\\\
0&0&0&0&\cdots\\\ 0&0&0&0&\cdots\\\
\vdots&\vdots&\vdots&\vdots&\ddots\end{array}\right].$

The basic solution is found by solving the equations $T_{-1}C_{0}=P$ and
$C_{0}T_{-1}=Q$ to find $T_{-1}$ and the equations $T_{0}C_{0}=P^{c}=I-P$ and
$C_{0}T_{0}=Q^{c}=I-Q$ to find $T_{0}$. Some elementary algebra gives gives

$T_{-1}=\left[\begin{array}[]{cc|ccc}-1&1&0&0&\cdots\\\ 0&-1&0&0&\cdots\\\
\hline\cr 0&0&0&0&\cdots\\\ 0&0&0&0&\cdots\\\
\vdots&\vdots&\vdots&\vdots&\ddots\end{array}\right]\quad\mbox{and}\quad
T_{0}=\left[\begin{array}[]{cc|ccc}0&0&0&0&\cdots\\\ 0&0&0&0&\cdots\\\
\hline\cr 0&0&1/(1-\lambda)&0&\cdots\\\ 0&0&0&1/(1-\lambda^{2})&\cdots\\\
\vdots&\vdots&\vdots&\vdots&\ddots\end{array}\right]$

Although the algebra is simply Gaussian elimination we note that calculation
of $T_{0}$ requires consideration of both $T_{0}C_{0}=P^{c}$ and
$C_{0}T_{0}=Q^{c}$. Now we can calculate the entire Laurent series with

$T_{-k}=(-1)^{k-1}(T_{-1}C_{0})^{k-1}T_{-1}$

for all $k\in{\mathbb{N}}$ which gives

$T_{-2}=(-1)T_{-1}C_{0}T_{-1}=\left[\begin{array}[]{cc|ccc}0&1&0&0&\cdots\\\
0&0&0&0&\cdots\\\ \hline\cr 0&0&0&0&\cdots\\\ 0&0&0&0&\cdots\\\
\vdots&\vdots&\vdots&\vdots&\ddots\end{array}\right]$

and $T_{-k}=\mbox{\large$0$}$ for $k\geq 3$. Therefore $R(z)$ has a pole of
order $2$ at $z=1$. We can also calculate

$T_{\ell}=(-1)^{\ell}(T_{0}C_{1})^{\ell}T_{0}=\left[\begin{array}[]{cc|ccc}0&0&0&0&\cdots\\\
0&0&0&0&\cdots\\\ \hline\cr 0&0&1/(1-\lambda)^{\ell}&0&\cdots\\\
0&0&0&1/(1-\lambda^{2})^{\ell}&\cdots\\\
\vdots&\vdots&\vdots&\vdots&\ddots\end{array}\right]$

for each $\ell\in{\mathbb{N}}-1$. The singular part of the resolvent is

$R_{\mbox{\scriptsize\rm
sin}}(z)=\left[\begin{array}[]{cc|ccc}-1/(z-1)&-1/(z-1)^{2}+1/(z-1)&0&0&\cdots\\\
0&-1/(z-1)&0&0&\cdots\\\ \hline\cr 0&0&0&0&\cdots\\\ 0&0&0&0&\cdots\\\
\vdots&\vdots&\vdots&\vdots&\ddots\end{array}\right]$

and the regular part is

$R_{\mbox{\scriptsize\rm
reg}}(z)=\left[\begin{array}[]{cc|ccc}0&0&0&0&\cdots\\\ 0&0&0&0&\cdots\\\
\hline\cr 0&0&(1-\lambda)/[1-\lambda-(z-1)]&0&\cdots\\\
0&0&0&(1-\lambda^{2})/[1-\lambda^{2}-(z-1)]&\cdots\\\
\vdots&\vdots&\vdots&\vdots&\ddots\end{array}\right]$

and the Laurent series converges for $z\in{\mathcal{U}}_{\,0,\,\rho}(1)$ where
$\rho=1/(1-\lambda)$. Our series solution relies only on intuitive use of
elementary algebra and is identical to the solution found by Seo [41] on the
same example. $\hfill\Box$

###### Example 3

Let $X=Y=L^{2}([0,1])$ be the Hilbert space of all complex-valued square-
integrable functions
$\mbox{$\mbox{\boldmath$x$}$}:[0,1]\rightarrow{\mathbb{C}}$. Let
$V\in{\mathcal{B}}(X,Y)$ be the Volterra integral operator defined by

$[V\mbox{$\mbox{\boldmath$x$}$}](t)=\mbox{$\int_{[0,t]}$}\mbox{$\mbox{\boldmath$x$}$}(s)ds$

for all $t\in[0,1]$. We consider the operator pencil
$A(z)=I-z(I-V)=V-(z-1)(I-V)$. We will show that the resolvent $R(z)=A(z)^{-1}$
has an isolated essential singularity at $z=1$. We will prove that $V$ is
quasi-nilpotent but not nilpotent. That is we will show that
$\|V^{n}\|^{1/n}\rightarrow 0$ as $n\rightarrow\infty$ but that
$V^{n}\neq\mbox{\large$0$}$ for all $n\in{\mathbb{N}}$. Clearly
$V^{n}\mbox{$\mbox{\boldmath$1$}$}=\mbox{$\mbox{\boldmath$t$}$}^{n}/n!$ where
we have defined $\mbox{$\mbox{\boldmath$1$}$}(t)=1$ and
$\mbox{$\mbox{\boldmath$t$}$}^{n}(t)=t^{n}$ for all $t\in[0,1]$ and all
$n\in{\mathbb{N}}$. Hence $V^{n}\neq\mbox{\large$0$}$ for all
$n\in{\mathbb{N}}$. The Cauchy–Schwartz inequality shows that

$|V\mbox{$\mbox{\boldmath$x$}$}(t)|\leq\left[\mbox{$\int_{[0,t]}$}|\mbox{$\mbox{\boldmath$x$}$}(s)|^{2}ds\right]^{1/2}\left[\mbox{$\int_{[0,t]}$}1^{2}ds\right]^{1/2}\leq\|\mbox{$\mbox{\boldmath$x$}$}\|\cdot
t^{1/2}\leq\|\mbox{$\mbox{\boldmath$x$}$}\|$

for all $t\in[0,1]$. Now it follows
$|V^{n}\mbox{$\mbox{\boldmath$x$}$}(t)|\leq\|\mbox{$\mbox{\boldmath$x$}$}\|t^{n-1}/(n-1)!\leq\|\mbox{$\mbox{\boldmath$x$}$}\|/(n-1)!$
for all $t\in[0,1]$ and all $n\in{\mathbb{N}}$. Consequently $\|V^{n}\|\leq
1/(n-1)!$ for all $n\in{\mathbb{N}}$. Choose $p\in{\mathbb{N}}$. An elementary
argument shows that $(n-1)!>p^{n}$ for $n\in{\mathbb{N}}$ sufficiently large.
Therefore we can find $m=m(p)\in{\mathbb{N}}$ such that $\|V^{n}\|^{1/n}<1/p$
for all $n\geq m$. Since $p\in{\mathbb{N}}$ was chosen arbitrarily it follows
that $\lim_{n\rightarrow\infty}\|V^{n}\|^{1/n}=0$.

The adjoint operator $V^{*}\in{\mathcal{B}}(Y,X)$ is defined by

$\langle
V^{*}\mbox{$\mbox{\boldmath$y$}$},\mbox{$\mbox{\boldmath$x$}$}\rangle=\langle\mbox{$\mbox{\boldmath$y$}$},V\mbox{$\mbox{\boldmath$x$}$}\rangle=\mbox{$\int_{[0,1]}$}\mbox{$\mbox{\boldmath$y$}$}(t)\left[\mbox{$\int_{[0,t]}$}\overline{\mbox{$\mbox{\boldmath$x$}$}}(s)ds\right]dt=\mbox{$\int_{[0,1]}$}\left[\mbox{$\int_{[s,1]}$}\mbox{$\mbox{\boldmath$y$}$}(t)dt\right]\overline{\mbox{$\mbox{\boldmath$x$}$}}(s)ds$

for all $\mbox{$\mbox{\boldmath$x$}$}\in X$ and
$\mbox{$\mbox{\boldmath$y$}$}\in Y$. Therefore

$V^{*}\mbox{$\mbox{\boldmath$y$}$}(s)=\mbox{$\int_{[s,1]}$}\mbox{$\mbox{\boldmath$y$}$}(t)dt$

for all $s\in[0,1]$ and so

$V^{*}V\mbox{$\mbox{\boldmath$x$}$}(s)=\mbox{$\int_{[s,1]}$}\left[\mbox{$\int_{[0,t]}$}\mbox{$\mbox{\boldmath$x$}$}(r)dr\right]ds$

for all $\mbox{$\mbox{\boldmath$x$}$}\in X$. Since $V^{*}V$ is self-adjoint
and compact it has a discrete spectrum on the non-negative real axis [49, pp
282–286]. Suppose $\mbox{$\mbox{\boldmath$x$}$}\in X$ is an eigenvector
corresponding to a real eigenvalue $\lambda\geq 0$. Then we must have

$\mbox{$\int_{[s,1]}$}\left[\mbox{$\int_{[0,t]}$}\mbox{$\mbox{\boldmath$x$}$}(r)dr\right]ds=\lambda\mbox{$\mbox{\boldmath$x$}$}(s)$

and differentiation shows that

$(-1)\mbox{$\mbox{\boldmath$x$}$}(s)=\lambda\mbox{$\mbox{\boldmath$x$}$}^{\,\prime\prime}(s)$

for all $s\in[0,1]$. Therefore

$\mbox{$\mbox{\boldmath$x$}$}(s)=c\cos(s/\sqrt{\lambda})+d\sin(s/\sqrt{\lambda}).$

Direct calculation gives

$\displaystyle\mbox{$\int_{[s,1]}$}\left[\mbox{$\int_{[0,t]}$}\mbox{$\mbox{\boldmath$x$}$}(r)dr\right]dt$
$\displaystyle=$
$\displaystyle\lambda[c\cos(s/\sqrt{\lambda})+d\sin(s/\sqrt{\lambda})]$
$\displaystyle\hskip
56.9055pt-[c\cos(1/\sqrt{\lambda})+d\sin(1/\sqrt{\lambda})]+d\sqrt{\lambda}[1-s]$

and so we must have $d=0$ and $\cos(1/\sqrt{\lambda})=0$. Thus the eigenvalues
are

$\lambda_{n}=\frac{4}{(2n+1)^{2}\pi^{2}}$

and the corresponding eigenvectors $\mbox{$\mbox{\boldmath$x$}$}_{n}\in X$ are
defined by

$\mbox{$\mbox{\boldmath$x$}$}_{n}(s)=\cos(s/\sqrt{\lambda_{n}})$

for all $s\in[0,1]$ and each $n\in{\mathbb{N}}-1$. Therefore
$\|V^{*}V\|=\max_{n\in{\mathbb{N}}-1}\lambda_{n}=4/\pi^{2}$ which implies
$\|V\|=2/\pi<1$. Hence we can apply the Neumann expansion to see that

$(I-V)^{-1}=I+V+V^{2}+\cdots$

is well defined with
$\|(I-V)^{-1}\|\leq\sum_{k\in{\mathbb{N}}-1}(2/\pi)^{k}=\pi/(\pi-2)$. By once
more applying the Neumann expansion we have

$\displaystyle R(z)$ $\displaystyle=$
$\displaystyle\left[V-(z-1)(I-V)\right]^{-1}$ $\displaystyle=$
$\displaystyle-(I-V)^{-1}\left[I-V(I-V)^{-1}/(z-1)\right]^{-1}/(z-1)$
$\displaystyle=$
$\displaystyle-(I-V)^{-1}\left[I+V(I-V)^{-1}/(z-1)+V^{2}(I-V)^{-2}/(z-1)^{2}+\cdots\right]/(z-1)$
$\displaystyle=$
$\displaystyle-\left[(I-V)^{-1}/(z-1)+V(I-V)^{-2}/(z-1)^{2}+V^{2}(I-V)^{-3}/(z-1)^{3}+\cdots\right]$

for all $z\neq 1$ because
$\|V^{n}(I-V)^{-n}\|^{1/n}\leq\|V^{n}\|^{1/n}\|(I-V)^{-n}\|^{1/n}=\|V^{n}\|^{1/n}\cdot\pi/(\pi-2)\rightarrow
0$ as $n\rightarrow\infty$. Thus $V(I-V)^{-1}$ is quasi-nilpotent.

In this example explicit solution of the fundamental equations can be
suppressed because $A(z)=C_{0}+C_{1}(z-1)$ where $C_{0}=V$ and
$C_{1}=(-1)(I-V)$ is invertible. It follows that $T_{-1}=C_{1}^{-1}$ and that
the key spectral separation projection is $P=T_{-1}C_{1}=I$ with $P^{c}=0$.
Thus the Laurent series for $R(z)$ at $z=1$ contains only a singular part and
no regular part. The solution to the fundamental equations is given by
$T_{-k}=(-1)^{k-1}T_{-1}(C_{0}T_{-1})^{k-1}=(-1)(I-V)^{-1}[V(I-V)^{-1}]^{k-1}$
for all $k\in{\mathbb{N}}$ with $T_{\ell}=0$ for all $\ell\in{\mathbb{N}}-1$.
Thus $R(z)=\sum_{k\in{\mathbb{N}}}T_{-k}/(z-1)^{k}$ for all
$z\in{\mathcal{U}}_{\,0,\,\infty}(1)$. $\hfill\Box$

###### Example 4

Let $\\{\lambda_{n}\\}\in{\mathbb{R}}$ be a sequence of positive numbers with
$\sigma=\sum_{n\in{\mathbb{N}}}\lambda_{n}<1$. Let
$\\{f_{n}\\}:{\mathbb{Z}}\rightarrow{\mathbb{R}}$ for each $n\in{\mathbb{N}}$
and $g:{\mathbb{Z}}\rightarrow{\mathbb{R}}$ be real-valued functions with
$f_{n}(t)=0$ for $t\leq 0$ and with $g(0)=1$ and $g(t)=0$ for $t<0$ and
suppose that

$f_{n}(t+1)-f_{n}(t)-\lambda_{n}[f_{n+1}(t)+g(t)]=0$

for each $n\in{\mathbb{N}}$ and

$g(t+1)-(1-\sigma)g(t)=\epsilon w(t)$

for all $t\in{\mathbb{N}}-1$ where $\\{w(t)\\}_{t\in{\mathbb{Z}}}$ is a random
process with ${\mathbb{P}}[w(t)=0]=p$ and ${\mathbb{P}}[w(t)=1]=1-p$ for each
$t\in{\mathbb{Z}}$ and some $p\in(0,1)$. We could regard $g(t)$ as an
investment fund with an initial endowment that is progressively distributed
over time to feed an infinite hierarchy of subsidiary assets
$\\{f_{n}(t)\\}_{n\in{\mathbb{N}}}$, each of which contribute, in turn, to the
incremental growth of the next lowest asset in the hierarchy. The total return
for the investor is the lowest ranked asset $f_{1}(t)$ which is the only one
that can be realized. The noise term means that the investment fund receives a
regular random supplement.

An ${\mathcal{M}}$-transform gives

$\displaystyle(1-z)F_{n}(z)-\lambda_{n}z[F_{n+1}(z)+G(z)]$ $\displaystyle=$
$\displaystyle 0$ (71) $\displaystyle\left[1-(1-\sigma)z\right]G(z)$
$\displaystyle=$ $\displaystyle\epsilon zW(z)+1$ (72)

for each $n\in{\mathbb{N}}$ where we have used the notation

$F(z)={\mathcal{M}}[\\{f(t)\\}_{t\in{\mathbb{N}}-1}](z)=\mbox{$\sum_{t\in{\mathbb{N}}-1}$}f(t)z^{t}$

to denote the ${\mathcal{M}}$-transform of a function
$\\{f(t)\\}_{t\in{\mathbb{N}}-1}$. In this case it is convenient to define
$\zeta=1/z$ and write $F_{n}(z)=H_{n}(\zeta)$, $G(z)=K(\zeta)$ and
$W(z)=Y(\zeta)$. Now (71) and (72) become

$\displaystyle(\zeta-1)H_{n}(\zeta)-\lambda_{n}[H_{n+1}(\zeta)+K(\zeta)]$
$\displaystyle=$ $\displaystyle 0$ (73)
$\displaystyle\left[\zeta-1+\sigma\right]K(\zeta)$ $\displaystyle=$
$\displaystyle\epsilon Y(\zeta)+\zeta$ (74)

for each $n\in{\mathbb{N}}$. If we assume we can find
$\\{m_{n}\\}_{n\in{\mathbb{N}}}>0$ with $\sum_{n\in{\mathbb{N}}}m_{n}<\infty$
and $r>0$ such that $|H_{n}(\zeta)|\leq m_{n}$ for $0<|\zeta-1|<r$ then we can
define vectors

$\mbox{$\mbox{\boldmath$\lambda$}$}=\left[\begin{array}[]{c}\lambda_{1}\\\
\lambda_{2}\\\ \lambda_{3}\\\
\vdots\end{array}\right]\in\ell^{1}\quad\mbox{and}\quad\mbox{$\mbox{\boldmath$H$}$}(\zeta)=\left[\begin{array}[]{c}H_{1}(\zeta)\\\
H_{2}(\zeta)\\\ \vdots\end{array}\right]\in\ell^{1}$

for $0<|\zeta-1|<r$. Now we can write the equations (73) and (74) as a vector
equation

$\left[\begin{array}[]{c|c}(\zeta-1)I-U&-\mbox{$\mbox{\boldmath$\lambda$}$}\\\
\hline\cr\mbox{$\mbox{\boldmath$0$}$}^{*}&\zeta-1+\sigma\end{array}\right]\left[\begin{array}[]{c}\mbox{$\mbox{\boldmath$H$}$}(\zeta)\\\
\hline\cr
K(\zeta)\end{array}\right]=\left[\begin{array}[]{c}\mbox{$\mbox{\boldmath$0$}$}\\\
\hline\cr(1-\sigma)[Y(\zeta)-1]\end{array}\right]$ (75)

or equivalently
$A(\zeta)\mbox{$\mbox{\boldmath$H$}$}(\zeta)=\mbox{$\mbox{\boldmath$L$}$}(\zeta)$
where
$A(\zeta)=C_{0}+C_{1}(\zeta-1)\in{\mathcal{B}}(\ell^{1}\times{\mathbb{C}})$,

$C_{0}=\left[\begin{array}[]{c|c}-U&-\mbox{$\mbox{\boldmath$\lambda$}$}\\\
\hline\cr\mbox{$\mbox{\boldmath$0$}$}^{*}&\sigma\end{array}\right],\quad
C_{1}=\left[\begin{array}[]{c|c}I&\mbox{$\mbox{\boldmath$0$}$}\\\
\hline\cr\mbox{$\mbox{\boldmath$0$}$}^{*}&1\end{array}\right]\in{\mathcal{B}}(\ell^{1}\times{\mathbb{C}}),$
$\mbox{$\mbox{\boldmath$L$}$}(\zeta)=\left[\begin{array}[]{c}\mbox{$\mbox{\boldmath$0$}$}\\\
\hline\cr(1-\sigma)[Y(\zeta)-1]\end{array}\right]\in\ell^{1}\times{\mathbb{C}}$

and where

$U=\left[\begin{array}[]{ccccc}\mbox{$\mbox{\boldmath$0$}$}&\lambda_{1}\mbox{$\mbox{\boldmath$e$}$}_{1}&\lambda_{2}\mbox{$\mbox{\boldmath$e$}$}_{2}&\lambda_{3}\mbox{$\mbox{\boldmath$e$}$}_{3}&\cdots\end{array}\right]\in{\mathcal{B}}(\ell^{1}).$

The set $X_{\mbox{\scriptsize\rm sin}}=\\{\mbox{$\mbox{\boldmath$x$}$}_{-1}\in
X\mid\,\mbox{there exists
$\\{\mbox{$\mbox{\boldmath$x$}$}_{-n}\\}_{n\in{\mathbb{N}}}$ with}\
C_{0}\mbox{$\mbox{\boldmath$x$}$}_{-n}+C_{1}\mbox{$\mbox{\boldmath$x$}$}_{-n-1}=\mbox{$\mbox{\boldmath$0$}$}\
\mbox{for all}\ n\in{\mathbb{N}}\ \mbox{and}\
\lim_{n\rightarrow\infty}\|\mbox{$\mbox{\boldmath$x$}$}_{-n}\|^{1/n}=0\\}$ is
the generating set for the infinite-length singular Jordan chains on
${\mathcal{U}}_{\,0,\,\infty}(1)$ for the linear pencil $A(z)$. Albrecht et
al. [5] have shown that $X_{\mbox{\scriptsize\rm sin}}=P(X)$ where
$P=T_{-1}C_{1}$ is the key spectral separation projection operator on
${\mathcal{U}}_{\,0,\infty}(1)$. To find $X_{\mbox{\scriptsize\rm sin}}$ we
consider the equation

$C_{0}\mbox{$\mbox{\boldmath$x$}$}_{-n}+C_{1}\mbox{$\mbox{\boldmath$x$}$}_{-n-1}=\mbox{$\mbox{\boldmath$0$}$}\quad\Rightarrow\quad\mbox{$\mbox{\boldmath$x$}$}_{-n-1}=(-1)^{n}C_{0}^{n}\mbox{$\mbox{\boldmath$x$}$}_{-1}$

for each $n\in{\mathbb{N}}$. Some elementary algebra shows that

$C_{0}^{n}=\left[\begin{array}[]{c|c}(-1)^{n}U^{n}&[\,\sum_{k=0}^{n-1}\sigma^{k}(-1)^{n-k}U^{n-1-k}\,]\mbox{$\mbox{\boldmath$\lambda$}$}\\\
\hline\cr\mbox{$\mbox{\boldmath$0$}$}^{*}&\sigma^{n}\end{array}\right].$

Now we have

$U^{2}=\left[\begin{array}[]{cccccc}\mbox{$\mbox{\boldmath$0$}$}&\mbox{$\mbox{\boldmath$0$}$}&\lambda_{1}\lambda_{2}\mbox{$\mbox{\boldmath$e$}$}_{1}&\lambda_{2}\lambda_{3}\mbox{$\mbox{\boldmath$e$}$}_{2}&\lambda_{3}\lambda_{4}\mbox{$\mbox{\boldmath$e$}$}_{3}&\cdots\end{array}\right],$
$U^{3}=\left[\begin{array}[]{ccccccc}\mbox{$\mbox{\boldmath$0$}$}&\mbox{$\mbox{\boldmath$0$}$}&\mbox{$\mbox{\boldmath$0$}$}&\lambda_{1}\lambda_{2}\lambda_{3}\mbox{$\mbox{\boldmath$e$}$}_{1}&\lambda_{2}\lambda_{3}\lambda_{4}\mbox{$\mbox{\boldmath$e$}$}_{2}&\lambda_{3}\lambda_{4}\lambda_{5}\mbox{$\mbox{\boldmath$e$}$}_{3}&\cdots\end{array}\right],$

and so on. It is clear that the first $n$ columns of $U^{n}$ are zero columns.
Consequently
$U^{n}\mbox{$\mbox{\boldmath$e$}$}_{j}=\mbox{$\mbox{\boldmath$0$}$}$ for
$n>j$. It follows that

$\mbox{$\mbox{\boldmath$x$}$}_{-1}=\left[\begin{array}[]{c}\mbox{$\mbox{\boldmath$e$}$}_{j}\\\
\hline\cr
0\end{array}\right]\quad\Longrightarrow\quad\mbox{$\mbox{\boldmath$x$}$}_{-n}=(-1)^{n-1}C_{0}^{n-1}\mbox{$\mbox{\boldmath$x$}$}_{-1}=\left[\begin{array}[]{c}\mbox{$\mbox{\boldmath$0$}$}\\\
\hline\cr 0\end{array}\right]$

for all $n>j+1$. Therefore $\|\mbox{$\mbox{\boldmath$x$}$}_{-n}\|_{1}^{1/n}=0$
for $n>j+1$ and hence
$\ell^{1}={\mathcal{S}}(\\{\mbox{$\mbox{\boldmath$e$}$}_{1},\mbox{$\mbox{\boldmath$e$}$}_{2},\ldots\\})\subseteq
X_{\mbox{\scriptsize\rm sin}}$. We can also see that

$\mbox{$\mbox{\boldmath$x$}$}_{-1}=\left[\begin{array}[]{c}\mbox{$\mbox{\boldmath$0$}$}\\\
\hline\cr 1\end{array}\right]\ \Rightarrow\
\mbox{$\mbox{\boldmath$x$}$}_{-n-1}=\left[\begin{array}[]{c}[\,\sum_{k=0}^{n-1}\sigma^{k}(-1)^{k+1}U^{n-1-k}\,]_{\rule{0.0pt}{11.38109pt}}\mbox{$\mbox{\boldmath$\lambda$}$}\\\
\hline\cr(-1)^{n}\sigma^{n}\end{array}\right]$

from which it follows that
$\|\mbox{$\mbox{\boldmath$x$}$}_{-n}\|_{1}\geq\sigma^{n-1}\Rightarrow\|\mbox{$\mbox{\boldmath$x$}$}_{-n}\|_{1}^{1/n}=\sigma^{1-1/n}\rightarrow\sigma>0$
as $n\rightarrow\infty$. Therefore

$\left[\begin{array}[]{c}\mbox{$\mbox{\boldmath$0$}$}\\\ \hline\cr
1\end{array}\right]\notin X_{\mbox{\scriptsize\rm sin}}.$

Hence we have shown that $X_{\mbox{\scriptsize\rm
sin}}=\ell^{1}\subseteq\ell^{1}\times{\mathbb{C}}$.

The set $X_{\mbox{\scriptsize\rm reg}}=\\{\mbox{$\mbox{\boldmath$x$}$}_{1}\in
X\mid\ \mbox{there exists
$\\{\mbox{$\mbox{\boldmath$x$}$}_{n}\\}_{n\in{\mathbb{N}}}$ with}\
C_{0}\mbox{$\mbox{\boldmath$x$}$}_{n+1}+C_{1}\mbox{$\mbox{\boldmath$x$}$}_{n}=\mbox{$\mbox{\boldmath$0$}$}\
\mbox{for all}\ n\in{\mathbb{N}}\ \mbox{and}\
\lim_{n\rightarrow\infty}\|\mbox{$\mbox{\boldmath$x$}$}_{n}\|^{1/n}\leq\rho\\}$
for some $\rho>0$ is the generating set for the infinite-length regular Jordan
chains on some set $\,{\mathcal{U}}_{\,\rho}(1)$. Albrecht et al. [5] have
shown that $X_{\mbox{\scriptsize\rm reg}}=P^{c}(X)$ where
$P^{c}=I_{X}-P=T_{0}C_{0}$ is the complementary key spectral projection on
${\mathcal{U}}_{\,\rho}(1)$. To find $X_{\mbox{\scriptsize\rm reg}}$ we
consider the equation

$C_{0}\mbox{$\mbox{\boldmath$x$}$}_{n+1}+C_{1}\mbox{$\mbox{\boldmath$x$}$}_{n}=\mbox{$\mbox{\boldmath$0$}$}\quad\Rightarrow\quad(-1)^{n-1}C_{0}^{n-1}\mbox{$\mbox{\boldmath$x$}$}_{n}=\mbox{$\mbox{\boldmath$x$}$}_{1}$

for all $n\in{\mathbb{N}}$. Therefore we must have

$\lim_{n\rightarrow\infty}(-1)^{n-1}C_{0}^{n-1}\mbox{$\mbox{\boldmath$x$}$}_{n}=\mbox{$\mbox{\boldmath$x$}$}_{1}.$

We have already seen that

$(-1)^{n-1}C_{0}^{n-1}\left[\begin{array}[]{c}\mbox{$\mbox{\boldmath$e$}$}_{j}\\\
\hline\cr 0\end{array}\right]=\mbox{$\mbox{\boldmath$0$}$}$

for $n>j+1$ and so we might expect to find

$\mbox{$\mbox{\boldmath$x$}$}_{1}=\lim_{n\rightarrow\infty}(-1)^{n-1}C_{0}^{n-1}\left[\begin{array}[]{c}\mbox{$\mbox{\boldmath$0$}$}\\\
\hline\cr\alpha_{n}\end{array}\right]=\lim_{n\rightarrow\infty}\alpha_{n}(-1)^{n-1}C_{0}^{n-1}\left[\begin{array}[]{c}\mbox{$\mbox{\boldmath$0$}$}\\\
\hline\cr 1\end{array}\right]$

for some suitably chosen sequence
$\\{\alpha_{n}\\}_{n\in{\mathbb{N}}}\in{\mathbb{R}}$ with $\alpha_{n}\neq 0$
for all $n\in{\mathbb{N}}$. Our earlier calculations showed us that

$\displaystyle(-1)^{n}C_{0}^{n}/\sigma^{n}\left[\begin{array}[]{c}\mbox{$\mbox{\boldmath$0$}$}\\\
\hline\cr 1\end{array}\right]$ $\displaystyle=$
$\displaystyle\left[\begin{array}[]{c}[\,\sum_{k=0}^{n-1}(-1/\sigma)^{n-k}U^{n-1-k}\,]\mbox{$\mbox{\boldmath$\lambda$}$}\\\
\hline\cr 1\end{array}\right]$ $\displaystyle\rightarrow$
$\displaystyle\left[\begin{array}[]{c}-(\sigma
I+U)^{-1}\mbox{$\mbox{\boldmath$\lambda$}$}\\\ \hline\cr
1\end{array}\right]=\mbox{$\mbox{\boldmath$v$}$}$

as $n\rightarrow\infty$. Therefore
$(C_{0}/\sigma)\mbox{$\mbox{\boldmath$v$}$}=\mbox{$\mbox{\boldmath$v$}$}\iff
C_{0}\mbox{$\mbox{\boldmath$v$}$}=\sigma\mbox{$\mbox{\boldmath$v$}$}$. If we
choose $\mbox{$\mbox{\boldmath$x$}$}_{1}=\mbox{$\mbox{\boldmath$v$}$}$ then
$\mbox{$\mbox{\boldmath$x$}$}_{n}=(-1)^{n-1}C_{0}^{n-1}\mbox{$\mbox{\boldmath$v$}$}=(-1)^{n-1}\sigma^{n-1}\mbox{$\mbox{\boldmath$v$}$}$
for all $n\in{\mathbb{N}}$. Thus
$\|\mbox{$\mbox{\boldmath$x$}$}_{n}\|_{1}=\sigma^{n-1}\|\mbox{$\mbox{\boldmath$v$}$}\|_{1}$
and hence
$\|\mbox{$\mbox{\boldmath$x$}$}_{n}\|_{1}^{1/n}=\sigma^{1-1/n}\|\mbox{$\mbox{\boldmath$v$}$}\|_{1}^{1/n}\rightarrow\sigma$
as $n\rightarrow\infty$. It follows that $\mbox{$\mbox{\boldmath$v$}$}\in
X_{\mbox{\scriptsize\rm reg}}$. Since this is the only possible solution we
have shown that $X_{\mbox{\scriptsize\rm
reg}}=\\{\mbox{$\mbox{\boldmath$x$}$}\in
X\mid\mbox{$\mbox{\boldmath$x$}$}=\alpha\mbox{$\mbox{\boldmath$v$}$}\
\mbox{for some}\ \alpha\in{\mathbb{C}}\\}$. We have also shown that
$\|\mbox{$\mbox{\boldmath$x$}$}_{n}\|^{1/n}\rightarrow\sigma$ for every
infinite-length regular Jordan chain. Hence $\rho=\sigma$.

Therefore

$P^{c}=\left[\begin{array}[]{c|c}\mbox{\large$0$}&-(\sigma
I+U)^{-1}\mbox{$\mbox{\boldmath$\lambda$}$}\\\
\hline\cr\mbox{$\mbox{\boldmath$0$}$}^{*}&1\end{array}\right]$

and

$P=I-P^{c}=\left[\begin{array}[]{c|c}I&(\sigma
I+U)^{-1}\mbox{$\mbox{\boldmath$\lambda$}$}\\\
\hline\cr\mbox{$\mbox{\boldmath$0$}$}^{*}&0\end{array}\right]$

are the key spectral separation projections on
${\mathcal{U}}_{\,0,\,\sigma}(1)={\mathcal{U}}_{\,\sigma}(1)\cap{\mathcal{U}}_{\,0,\,\infty}(1)$.
For each

$\mbox{$\mbox{\boldmath$x$}$}=\left[\begin{array}[]{c}\mbox{$\mbox{\boldmath$\alpha$}$}\\\
\hline\cr\beta\end{array}\right]\in X$

we can write
$\mbox{$\mbox{\boldmath$x$}$}=P\mbox{$\mbox{\boldmath$x$}$}+P^{c}\mbox{$\mbox{\boldmath$x$}$}$
to find the unique decomposition

$\left[\begin{array}[]{c}\mbox{$\mbox{\boldmath$\alpha$}$}\\\
\hline\cr\beta\end{array}\right]=\left[\begin{array}[]{c}\mbox{$\mbox{\boldmath$\alpha$}$}+\beta(\sigma
I+U)^{-1}\mbox{$\mbox{\boldmath$\lambda$}$}\\\ \hline\cr
0\end{array}\right]+\left[\begin{array}[]{c}-\beta(\sigma
I+U)^{-1}\mbox{$\mbox{\boldmath$\lambda$}$}\\\
\hline\cr\beta\end{array}\right]\in X_{\mbox{\scriptsize\rm sin}}\oplus
X_{\mbox{\scriptsize\rm reg}}.$

Since $C_{1}=I$ and $T_{-1}C_{1}=P$ we deduce that $T_{-1}=P$. Thus we have

$T_{-1}=\left[\begin{array}[]{c|c}I&(\sigma
I+U)^{-1}\mbox{$\mbox{\boldmath$\lambda$}$}\\\
\hline\cr\mbox{$\mbox{\boldmath$0$}$}^{*}&0\end{array}\right].$

We also have

$T_{0}C_{0}=P^{c}\iff\left[\begin{array}[]{c|c}T_{0,11}&\mbox{$\mbox{\boldmath$t$}$}\\\
\hline\cr\mbox{$\mbox{\boldmath$s$}$}^{*}&\gamma\end{array}\right]\left[\begin{array}[]{c|c}-U&-\mbox{$\mbox{\boldmath$\lambda$}$}\\\
\hline\cr\mbox{$\mbox{\boldmath$0$}$}^{*}&\sigma\end{array}\right]=\left[\begin{array}[]{c|c}\mbox{\large$0$}&-(\sigma
I+U)^{-1}\mbox{$\mbox{\boldmath$\lambda$}$}\\\
\hline\cr\mbox{$\mbox{\boldmath$0$}$}^{*}&1\end{array}\right]$

which implies $T_{0,11}U=\mbox{\large$0$}$,
$-T_{0,11}\mbox{$\mbox{\boldmath$\lambda$}$}+\sigma\mbox{$\mbox{\boldmath$t$}$}=-(\sigma
I+U)^{-1}\mbox{$\mbox{\boldmath$\lambda$}$}$,
$\mbox{$\mbox{\boldmath$s$}$}^{*}U=\mbox{$\mbox{\boldmath$0$}$}^{*}$, and
$-\mbox{$\mbox{\boldmath$s$}$}^{*}\mbox{$\mbox{\boldmath$\lambda$}$}+\gamma\sigma=1$.
Since
$U=[\mbox{$\mbox{\boldmath$0$}$},\lambda_{1}\mbox{$\mbox{\boldmath$e$}$}_{1},\lambda_{2}\mbox{$\mbox{\boldmath$e$}$}_{2},\dots]$
it follows that $T_{0,11}U=\mbox{\large$0$}$ and
$\mbox{$\mbox{\boldmath$s$}$}^{*}U=\mbox{$\mbox{\boldmath$0$}$}^{*}$ implies
$T_{0,11}=\mbox{\large$0$}$ and
$\mbox{$\mbox{\boldmath$s$}$}=\mbox{$\mbox{\boldmath$0$}$}$. Therefore

$T_{0}=\left[\begin{array}[]{c|c}\mbox{\large$0$}&-(\sigma
I+U)^{-1}\mbox{$\mbox{\boldmath$\lambda$}$}/\sigma\\\
\hline\cr\mbox{$\mbox{\boldmath$0$}$}^{*}&1/\sigma\end{array}\right].$

We can now calculate

$T_{-k}=(-1)^{k-1}(T_{-1}C_{0})^{k-1}T_{-1}=\left[\begin{array}[]{c|c}U^{k-1}&U^{k-1}(\sigma
I+U)^{-1}\mbox{$\mbox{\boldmath$\lambda$}$}\\\
\hline\cr\mbox{$\mbox{\boldmath$0$}$}^{*}&0\end{array}\right]$

for each $k\in{\mathbb{N}}$ and

$T_{\ell}=(-1)^{\ell}(T_{0}C_{1})^{\ell}T_{0}=(-1)^{\ell}\left[\begin{array}[]{c|c}\mbox{\large$0$}&-(\sigma
I+U)^{-1}\mbox{$\mbox{\boldmath$\lambda$}$}/\sigma^{\ell+1}\\\
\hline\cr\mbox{$\mbox{\boldmath$0$}$}^{*}&1/\sigma^{\ell+1}\end{array}\right]$

for each $\ell\in{\mathbb{N}}-1$. An elementary series argument shows that the
singular part of the resolvent is

$R_{\mbox{\scriptsize\rm