2023

[1]Martial Longla

1]Department of mathematics, University of Mississippi, University Ave,
University, 38677, Mississippi, US
ORCID: 0000-0002-9432-0640 2]Department of mathematics, University of Maroua,
Cameroon

# Estimation problems for some perturbations of the independence copula.

<EMAIL_ADDRESS>Mous-Abou Hamadou<EMAIL_ADDRESS>[  [

###### Abstract

This work provides a study of parameter estimators based on functions of
Markov chains generated by some perturbations of the independence copula. We
provide asymptotic distributions of maximum likelihood estimators and
confidence intervals for copula parameters of several families of copulas
introduced in Longla (2023). Another set of moment-like estimators is proposed
along with a multivariate central limit theorem, that provides their
asymptotic distributions. We investigate the particular case of Markov chains
generated by sine copulas, sine-cosine copulas and the extended Farlie-Gumbel-
Morgenstern copula family. Some tests of independence are proposed. A
simulation study is provided for the three copula families of interest. This
simulation proposes a comparative study of the two introduced estimators and
the robust estimator of Longla and Peligrad (2021), showing advantages of the
proposed work.

###### keywords:

Estimation of copula parameters; Robust estimation; square integrable copulas;
Reversible Markov chains; Confidence interval for copula parameters;
Estimation of Long memory processes

###### pacs:

[

MSC Classification]62G08; 62M02; 60J35

##  1 Introduction

Over the past decades, many authors have worked on building copulas with
various properties such as prescribed diagonal section, known association
properties, perturbation and mixing. Some of these constructions can be found
in Nelsen (2006). Arakelian and Karlis (2014) studied mixtures of copulas,
that were investigated for mixing by Longla (2015). Longla (2014) constructed
copulas based on prescribed $\rho$-mixing properties obtained from extending
results of Beare (2010) on mixing for copula-based Markov chains. Longla and
al. (2022, 2022b) constructed copulas in general based on perturbations. In
these two works, two perturbation methods were considered. Our current work is
based on perturbations of the independence copula. These perturbations are of
the form $C_{D}(u,v)=C(u,v)+D(u,v)$, where $C(u,v)$ is the copula that is
undergoing the perturbation and $C_{D}(u,v)$ the perturbed copula. This method
of perturbation was presented in Komornik et al (2017), where authors mention
that it allows in the case of the independence copula to introduce some level
of dependence. Several other authors have considered extensions of existing
copula families via other methods, that are not the focus of this work, see
for instance, Morillas (2005), Aas et al (2009), Klement et al (2017),
Chesneau (2021), among others.

We should mention, that trigonometric copulas that were investigated in
Chesneau (2021), are very different from the perturbations that we have in
this paper. This can be seen on the level of the dependence structure and the
functional form of the copulas. This paper is concerned with the important
question of estimation, confidence intervals and testing for copula-based
Markov chains, that are generated by copula families of Longla (2023). These
families were introduced, but not studied for the mentioned questions. Their
mixing properties were examined, making it possible to explore central limit
theorems. We base our study of MLE on the work of Billingsley (1961) for
stationary Markov chains and their asymptotic distributions. Billingsley
(1961) has been used in the setup of Markov chains by several authors over the
years. Recently, it has been used by Hossain and Anam (2012) to study
modelling of rainfalls in Bangladesh via Markov chains and a logistic
regression model. Zhang et al. (2023) used it to study asymptotic normality of
estimators in an AR(1) Binomial model. Kuljus and Ranneby (2021) considered
maximum spacing as approximation of relative entropy for semi-markov processes
and continuous time Markov chains. A good resource for the use of copulas in
estimation problems is Chen and Fan (2006).

Our paper is dedicated to the analysis of three copula families. This analysis
relates mixing coefficients defined by Bradley (2007), in the context of
copulas, to the study of Markov chains. Copulas are functions that naturally
represent the effect of dependence in multivariate distributions. This fact is
due to Sklar’s theorem (see Sklar (1959)). Copulas are obtained by scalling
out the effect of marginal distributions from joint distributions. In the case
of continuous marginal distributions, most mixing coefficients of Markov
chains are not affected by the marginal distributions (see Longla (2015)).
Combining these facts with the special nature of the copula families of
interest of this paper, Longla (2023) showed that these copulas generate
$\psi$-mixing. It is worth mentioning that copula coefficients are eigen-
values of the copula operators. They are associated to the eigen-functions
$\varphi_{k}(x)$ present in the formula of the copula of Longla (2023).

The first family that we study has a density based on orthogonal cosine
functions. The second is based on orthogonal sine and cosine functions, and
the third family is based on Legendre polynomials. Orthogonality here is in
the sense of $L^{2}(0,1)$. This last family is an extension of the Farlie-
Gumbel-Morgenstern (FGM) copula family. It introduces a second dependence
coefficient $\lambda_{2}$. The FGM has been extensively studied in the
literature with several extensions (see Hurlimann (2017) or Ebaid and al.
(2022)).

In general, a multivariate central limit theorem is provided in our paper for
parameter estimators of copulas of Longla (2023). This central limit theorem
is based on Kipnis and Varadhan (1986). Kipnis and Varadhan (1986) showed that
for functions of a stationary reversible Markov chain with finite mean and
variance, when the variance of partial sums is asymptotically linear in $n$,
the central limit theorem holds.

A general theorem of Billingsley (1961) is used to establish the limiting
distribution of MLE. We show that they exist and are unique, when the true
parameters are inside the support region. Billingsley proved that under some
regularity conditions (which are checked for each of the copula families), the
MLE is asymptotically normal with variance provided by a form of Fisher
information. For practical reasons, given that the Fisher information can’t be
comptuted in closed form, an estimate was used. Computation time for this
procedure makes the MLE less attractive than the estimator that we propose.
Based on properties of the coefficients of the construction, our proposed
parameter estimators have closed form asymptotic variances.

###  1.1 Definitions and comments

We call copula in general a bivariate copula, which is defined as a bivariate
joint cumulative distibution function with uniform marginals on $[0,1]$. An
equivalent definition can be found in Nelsen (2006). A stationary Markov chain
can be represented by a copula and its stationary or marginal distribution.
The copula of a stationary Markov chain is that of any of its consecutive
variables. We say that a stationary sequence $(Y_{1},\cdots,Y_{n})$ with
finite mean $\mu$ and variance $\sigma^{2}$ satisfies the central limit
theorem if $\sqrt{n}(\bar{Y}-\mu)\to N(0,\sigma_{0}^{2})$, for some positive
number $\sigma^{2}_{0}$. Kipnis and Varadhan (1986) showed that when the
variance of partial sums of functions of the Markov chain satisfies
$\sigma_{f}^{2}=\lim_{n\to\infty}n^{-1}var(S_{n})$, where
$S_{n}=\sum_{k=1}^{n}f(X_{i})$, $Y=f(X)$, $\mu=\mathbb{E}(f(X))$, the central
limit theorem holds. This theorem is combined with the Cramér-Wold device to
prove a multivariate central limit theorem for parameter estimators in this
paper. In doing so, we use the $n$-fold product of copulas to compute the
limiting variance. The $n$-fold product of the copula uses partial derivatives
to find a new copula. The notation $C_{,i}(u,v)$ is for the derivative of
$C(u,v)$ with respect to the $i^{th}$ variable. The $n$-fold product of a
copula is defined in Nelsen (2006) (or by Darsow et al. (1992)) by
$C^{1}(u,v)=C(u,v),$

$\text{for}\quad
n>1,C^{n}(u,v)=C^{n-1}*C(u,v)=\int_{0}^{1}C^{n-1}_{,2}(u,t)C_{,1}(t,v)dt.$

Given that we are dealing with Markov chains, we recall that the fold product
for $n=2$ is the joint cumulative distribution of $(X_{1},X_{3})$ if
$(X_{1},X_{2},X_{3})$ is a stationary Markov chain with copula $C(u,v)$ and
the uniform marginal distribution (see Darsow et al. (1992)).

In general, the copula $C^{n}(u,v)$ is the joint distribution of
$(X_{0},X_{n})$ when $X_{0},\cdots,X_{n}$ is a stationary Markov chain
generated by $C(u,v)$ and the uniform distribution. These definitions can be
found in different forms in various papers, but are equivalent when the study
is concerned with absolutely continuous copulas and continuous marginal
distributions. This is because Sklar’s theorem guarantees uniqueness of the
representation of the joint distribution of a vector through its copula and
marginal distributions (see Sklar(1959)). Copulas simplify the study of mixing
properties of Markov chains due to the fact that for continuous random
variables, the marginal distributions are scaled out in the computation; the
formulas reduce to equivalent forms in terms of the uniform marginal
distributions. This is one of the advantages of dealing with copulas while
studying mixing.

Mixing coefficients, that we refer to, are defined in Bradley (2007), and were
used in Longla (2023) to establish properties of Markov chains for the copulas
that we are studying in this work. This work is not concerned by mixing in
general, but rather uses the that fact, that mixing was established in an
earlier work. This justifies the limitation of the background to the provided
references. The mixing rate allows convergence of the variance of partial sums
of functions of the Markov chains, and implies the needed ergodicity in the
Theorem of Kipnis and Varadhan (1986). This theorem on its turn is used to
establish central limit theorems for copula parameter estimators. In the
simulation part of this work, we use a result of Longla and Peligrad (2021) to
derive some estimators of copula parameter. These estimators are compared to
the MLE, and a second class of estimators, that is derived based on the fact,
that parameters are eigen-values of the operator induced by the copula of the
Markov chain on $L^{2}(0,1)$.

###  1.2 Structure of the paper and main results

We propose several estimators of copula parameters for the copulas created in
Longla (2023). Large sample study of these estimators is provided in general
via derivation of a multivariate central limit theorem. These results are
applied to 3 concrete families of copulas. We have conducted a comparative
simulation study in which 4 different estimators are used. For each of the
examples, we perform maximum likelihood estimation and show that the $R$ codes
for our proposed estimators take less time to run than the MLE and can be
preferable in large samples.

The rest of the paper is organized as follows. In section 2 we provide the
copula families of interest in this work along with some graphs of copula
densities and level curves. In section 3 we present parameter estimation
procedures for Markov chains generated by the copula families proposed by
Longla (2023). This section includes our proposed estimators, central limit
theorems for estimators and $\chi^{2}$ tests of independence of observations.
A multivariate central limit theorem is provided here as well. These
estimators and central limit theorems are results of this papers. They are
applied to each of the 3 examples that are proposed in Section 2. In section 4
we provide a simulation study of parameter estimators for the considered
copula families. For each of the proposed copulas, we prove existence of the
MLE, and derive its asymptotic distribution. A comparative study is provided
at the end of this section to show advantages of our proposed estimators. In
Section 5 we provide comments and conclusions of the work. An appendix of
proofs concludes the paper.

##  2 Copula families of interest

Copulas in general are defined as multivariate distributions with uniform
marginals on (0,1) (see Nelsen (2006), Durante and Sempi (2016) or Sklar
(1959) for more on the topic). Several authors have used this tool over the
years to model dependence in various applied fields of research. In this
paper, we are interested in copulas that were proposed by Longla (2023). Due
to their novelty, these copulas have not yet been considered for estimation
problems. There is no work that suggests how to use them, how to estimate
their parameters or test any related claims. Longla (2023) has shown that for
any sequences $\lambda_{k}$ and $\varphi_{k}(x)$ such that:

  * •

The sequence of $\lvert{}\lambda_{k}\lvert{}$ has a finite number of values or
converges to $0$.

  * •

The set $\\{\varphi_{k}(x),k\in\mathbb{N}\\}$ is an orthonormal basis of
$L^{2}(0,1)$.

The function

$c(u,v)=\sum_{k=1}^{\infty}\lambda_{k}\varphi_{k}(u)\varphi_{k}(v)$ (1)

is a copula density and the series converges pointwise and uniformly on
$(0,1)^{2}$, if

$1+\sum_{k=1}^{\infty}\lambda_{k}\alpha_{k}\geq
0,\quad\text{where}\quad\alpha_{k}=\begin{cases}\max\varphi^{2}_{k}&\text{if
}\quad\lambda_{k}<0\\\
\min\varphi_{k}\max\varphi_{k}&\text{if}\quad\lambda_{k}>0.\end{cases}$ (2)

Among examples of copula families of the form (1), that satisfy condition (2),
Longla (2023) introduced the following copulas.

  * •

The sine-cosine copulas as perturbations of $\Pi(u,v)$

Longla (2023) considered the trigonometric orthonormal basis of $L^{2}(0,1)$
that consists of functions: $\\{1,\sqrt{2}\cos 2\pi kx,\sqrt{2}\sin 2\pi
kx,k\in\mathbb{N^{*}}\\}$. He showed that

$\displaystyle
C(u,v)=uv+\frac{1}{2\pi^{2}}\sum_{k=1}^{\infty}\frac{\lambda_{k}}{k^{2}}\sin
2\pi ku\times\sin 2\pi kv+$
$\displaystyle+\frac{1}{2\pi^{2}}\sum_{k=1}^{\infty}\frac{\mu_{k}}{k^{2}}(1-\cos
2\pi ku)(1-\cos 2\pi kv)$ (3)

is a copula, when
$\displaystyle\sum_{k=1}^{\infty}(\lvert{}\lambda_{k}\lvert{}+\lvert{}\mu_{k}\lvert{})\leq
1/2$. This copula was called sine-cosine copula. It is a perturbation of the
independence copula.

Figure 1: Sine-cosine copula: $\lambda_{1}=0.14$, $\mu_{1}=-0.11$
$\lambda_{2}=0.13$ and $\mu_{2}=-0.12$

  * •

The sine copulas as perturbations of $\Pi(u,v)$

Sine copulas were obtained using the basis of $L^{2}(0,1)$ given by
$\\{1,\sqrt{2}\cos k\pi x,k>0,k\in\mathbb{N}\\}$. Their formula is

Figure 2: Sine copula with $\lambda_{1}=0.28$ and $\lambda_{2}=-0.15$

$C(u,v)=uv+\frac{2}{\pi^{2}}\sum_{k=1}^{\infty}\frac{\lambda_{k}}{k^{2}}\sin
k\pi u\sin k\pi
v,\quad\text{for}\quad\sum_{k=1}^{\infty}\lvert{}\lambda_{k}\lvert{}\leq 1/2.$
(4)

Any finite sum of terms from (4) that includes the first term $uv$ is a
copula. Sine copulas and Sine-cosine copulas are not part of the trigonometric
copulas of Chesneau (2021). These copulas are based on fundamental properties
of linear operators induced by copulas on the Hilbert space $L^{2}(0,1)$.
Their coefficients are eigen-values and the functions $\varphi_{k}(x)$ are
eigen-functions of the operator (see Longla (2023)).

  * •

The extended Farlie - Gumbel - Morgenstern copulas or Legendre copulas.

They were obtained using shifted Legendre polynomials, orthonormal basis of
$L^{2}(0,1)$ for $\\{1,\varphi_{k},k>0,k\in\mathbb{N}\\}$. Legendre
polynomials are defined by Rodrigues’ formula (see Rodrigues (1816))

$P_{k}(x)=\frac{1}{2^{k}k!}\frac{d^{k}}{dx^{k}}(x^{2}-1)^{k}.$ (5)

Figure 3: Legendre copula with $\lambda_{1}=0.15$ and $\lambda_{2}=0.1$

The orthonormal basis of shifted Legendre polynomials is

$\varphi_{k}(x)=\sqrt{2k+1}P_{k}(2x-1).$ (6)

For $k>0,$ $\varphi_{k}(1)=\sqrt{2k+1}=\max\varphi_{k}$, for odd $k$,
$\min\varphi_{k}=-\sqrt{2k+1}=\varphi_{k}(0)$ and for even $k,$
$0>\min\varphi_{k}>-\sqrt{2k+1}$. This minimum is achieved at least at one
point. Therefore, when (1) contains $\varphi_{k}(x)$ alone, the range of
$\lambda_{k}$ is

$\frac{-1}{2k+1}\leq\lambda_{k}\leq\min(1,\frac{1}{(2k+1)\lvert{}\min
P_{k}(x)\lvert{}}).$

Using this basis in formula (1), we obtain what Longla (2023) called extended
FGM copula family for any sum of terms that contains the first two functions.
For more on Legendre polynomials, see Belousov (1962), Szegö (1975) and the
references therein.

##  3 Parameter estimation

The constructed copula families include the extended Farlie-Gumbel-Morgenstern
copula family. The FGM copula family has been extensively studied in recent
years. Earlier studies include Morgenstern (1956), Farlie (1960) and Gumbel
(1960). These authors discussed the bivariate FGM copula family. Johnson and
Kotz (1975) formulated the FGM copula as a multivariate distribution on
$[0,1]^{d}$ for any positive integer $d>1$. The multivariate FGM copula family
is just as valuable as the multivariate normal distribution thanks to its
simplicity. It has been applied to statistical modeling in various research
fields such as economics (see Patton (2006)), finance (see Cossette et al.
(2013)) and reliability engineering (see Navarro and Durante (2017) and
recently Ota and Kimura (2021)).

Suppose that we have a Markov chain generated by a copula and the uniform
distribution as stationary distribution. Assuming that this copula has the
representation (1), we have the following for estimators of the copula
coefficients.

######  Theorem 3.1.

For any copula-based Markov chain $(U_{0},U_{1},\cdots,U_{n})$ generated by
the copula (1) and the uniform marginal distribution,

$\hat{\lambda}_{k}=\frac{1}{n}\sum_{i=1}^{n}\varphi_{k}(U_{i})\varphi_{k}(U_{i-1})$
(7)

is an unbiased consistent estimator of $\hat{\lambda}_{k}$ provided that

$\sum\lvert{}\lambda_{z}\lvert{}(\int_{0}^{1}\varphi_{z}(x)\varphi_{k}^{2}(x)dx)^{2})<\infty\quad\text{and}\quad\int_{0}^{1}\varphi_{k}^{4}(x)dx<\infty.$

Moreover, $\sqrt{n}(\hat{\lambda}_{k}-\lambda_{k})\to N(0,\sigma^{2})$, where
the positve real number $\sigma^{2}$ is given by

$\sigma^{2}=(1-\lambda_{k}^{2}+\sum\lambda_{z}(\int_{0}^{1}\varphi_{z}(x)\varphi_{k}^{2}(x)dx)^{2})+$
$+2\lambda_{k}^{2}(\int_{0}^{1}\varphi^{4}_{k}(x)dx-1)+2\sum_{m=2}^{\infty}\lambda_{k}^{2}\sum\lambda^{m-1}_{z}(\int_{0}^{1}\varphi_{z}(x)\varphi_{k}^{2}(x)dx)^{2}.$

Note, that when $\varphi_{k}(x)$ are uniformly bounded, the conditions on
$\lambda_{k}$ reduce to $\sum\lvert\lambda_{k}\lvert<\infty,$ which is
satisfied under condition (2). Based Theorem (3.1), we can construct
confidence intervals for each of the values of $\lambda_{k}$ separately for
every case of $\varphi_{k}(x)$. This implies estimation of the variance as
shown in the examples below. But, to tackle any problem related to joint
parameters, we need to establish a result for the joint limiting distribution
of these estimators of the coefficients of our copula. These distributions can
be valuable in model selection for the number of parameters to include in the
estimator of the copula. We need to find the distribution of
$\Lambda=(\hat{\lambda}_{k_{1}},\cdots,\hat{\lambda}_{k_{s}})$ for any
sequence of increasing integers
$\hat{\lambda}_{k_{1}},\cdots,\hat{\lambda}_{k_{s}}$.

######  Theorem 3.2.

Assume $(U_{0},U_{1},\cdots,U_{n})$ is a copula-based Markov chain generated
by a copula of the form (1). Assume
$\int_{0}^{1}\varphi^{2}_{k_{i}}(x)\varphi^{2}_{k_{j}}(x)dx<\infty$,

$\sum_{z=1}^{\infty}\lvert{}\lambda_{z}\lvert{}(\int_{0}^{1}\varphi_{z}(x)\varphi_{k_{i}}(x)\varphi_{k_{j}}(x)dx)^{2}<\infty\quad\text{and}$
$\sum_{z=1}^{\infty}\lvert{}\lambda_{z}\lvert{}(\int_{0}^{1}\varphi_{z}(x)\varphi_{k_{i}}^{2}(x)dx)(\int_{0}^{1}\varphi_{z}(x)\varphi_{k_{j}}^{2}(x)dx)<\infty,$

then for some positive definite matrix $\Sigma$ we have

$\sqrt{n}(\hat{\Lambda}-\Lambda)\to N_{s}({\bf 0},\Sigma).$

######  Remark 1.

For an orthonormal basis of $L^{2}(0,1)$ made of uniformly bounded and
symmetric functions, the conditions of Theorem 3.2 are automatically satisfied
because the integrals are uniformly bounded and
$\sum_{i=1}^{\infty}\lvert{}\lambda_{i}\lvert{}<\infty$.

######  Proposition 3.3.

Under the assumptions of Theorem 3.2, the off-diagonal elements of the matrix
$\Sigma$ are given by

$\displaystyle\underset{k\neq j}{\Sigma_{k,j}}$ $\displaystyle=$
$\displaystyle\sum\limits_{z=1}^{s}\lambda_{z}(\int_{0}^{1}\varphi_{z}(x)\varphi_{k}(x)\varphi_{j}(x)dx)^{2}+\lambda_{k}\lambda_{j}\left(2(\int_{0}^{1}\varphi^{2}_{k}(x)\varphi_{j}^{2}(x)dx-1)\right.$
(8)
$\displaystyle\left.+2\sum\limits_{z=1}^{s}\sum\limits_{m=2}^{\infty}\lambda_{z}^{m-1}(\int_{0}^{1}\varphi_{z}(x)\varphi_{k}^{2}(x)dx)(\int_{0}^{1}\varphi_{z}(x)\varphi_{j}^{2}(x)dx)-1\right).$

###  3.1 Applications to some examples

For the classes of copulas given above, the Spearman’s $\rho(C)$ is estimated
by $\hat{\lambda}_{1}$.

  1. 1.

Sine-cosine copulas. Theorem 3.1 implies that for the copulas based on the
sine-cosine basis,

$\displaystyle\sigma^{2}_{k}=1+\frac{\lambda_{2k}}{2}+\frac{\lambda_{k}^{2}\lambda_{2k}}{1-\lambda_{2k}},\quad\text{for}\quad\hat{\lambda}_{k}=\frac{1}{n}\sum_{i=1}^{n}2\cos
2\pi kU_{i}\cos 2\pi kU_{i-1}$
$\text{and}\quad\sigma^{2}_{k}=1+\frac{\lambda_{2k}}{2}+\frac{\mu_{k}^{2}\lambda_{2k}}{1-\lambda_{2k}},\quad\text{for}\quad\hat{\mu}_{k}=\frac{1}{n}\sum_{i=1}^{n}2\sin
2\pi kU_{i}\sin 2\pi kU_{i-1}.$

It is also clear that for sine-cosine copulas, with $i\neq j$, $\Sigma_{i,j}$
is given by

$\varphi_{k_{i}}/\varphi_{k_{j}}$ | $\varphi_{1}(x)=\sqrt{2}\cos 2\pi k_{j}x$ | $\varphi_{2}(x)=\sqrt{2}\sin 2\pi k_{j}x$
---|---|---
$\varphi_{1}(x)$ | $\frac{1}{2}(\lambda_{k_{i}+k_{j}}+\lambda_{\lvert{}k_{i}-k_{j}\lvert{}})-\lambda_{k_{i}}\lambda_{k_{j}}$ | $\frac{1}{2}(\mu_{k_{i}+k_{j}}+\mu_{\lvert{}k_{i}-k_{j}\lvert{}})-\lambda_{k_{i}}\mu_{k_{j}}$
$\varphi_{2}(x)$ | $\frac{1}{2}(\mu_{k_{i}+k_{j}}+\mu_{\lvert{}k_{i}-k_{j}\lvert{}})-\mu_{k_{i}}\lambda_{k_{j}}$ | $\frac{1}{2}(\lambda_{k_{i}+k_{j}}+\lambda_{\lvert{}k_{i}-k_{j}\lvert{}})-\mu_{k_{i}}\mu_{k_{j}}$

Therefore, the multivariate central limit theorem holds with $\Sigma_{ij}$ as
entries of the covariance matrix $\Sigma$. A test of independence can be
formed based on this limit theorem. Under the hypothesis of independence of
observations agains that of a Markov chain with copula $C(u,v)$, all
coefficients are equal to $0$. The estimators are asymptotically independent
with $\sigma^{2}=1$. We can formulate this in the following proposition.

######  Proposition 3.4.

For any square integrable copula with representation in sine-cosine series,
under the assumption that all parameters are equal to zero, any vector of
estimators with finitely many components satisfies the $\sqrt{n}$-central
limit theorem with variance-covariance matrix $\Sigma=\mathbb{I}_{s}$, where
$s$ is the length of the vector and $\mathbb{I}_{s}$ is the identity matrix of
dimension $s$. Moreover, if a sequence $W_{n_{i}}$ takes on values of
$\hat{\lambda}_{k}$ or $\hat{\mu}_{k}$ for the Markov chain
$U_{0},\cdots,U_{n}$, then

$n\sum_{i=1}^{s}W_{n_{i}}^{2}\to\chi^{2}(s)\quad\text{as}\quad
n\to\infty\quad\text{and}\quad\frac{1}{\sqrt{2s}}(n\sum_{i=1}^{s}W_{n_{i}}^{2}-s)\to
N(0,1)$ $\text{as}\quad n\to\infty,\quad\text{then}\quad s\to\infty.$

  2. 2.

Sine copulas. Considering sine copulas, we have
$\displaystyle\sigma^{2}_{k}=1+\frac{\lambda_{2k}}{2}+\frac{\lambda_{k}^{2}\lambda_{2k}}{1-\lambda_{2k}}.$
Estimators are

$\hat{\lambda}_{k}=\frac{1}{n}\sum_{i=1}^{n}2\cos\pi kU_{i}\cos\pi kU_{i-1}.$

Therefore, Proposition 3.4 holds exactly with the same conditions.

##  4 Simulation study for the 3 examples

In this section, we compare these estimators to other known estimators such as
maximum likelihood estimator, the robust estimator of Longla and Peligrad
(2021) in two forms. The performance is studied on Markov chains simulated
using three different copulas from the families described above.

###  4.1  The specific copula examples

We consider 3 types of copulas with densities of the form (1). A sine copula,
a sine-cosine copula and a Legendre copula. The first two present several tops
and bottoms, while the second is an extension of the FGM copula family. This
third copula can be used to test departure from the FGM copula family in
modelling. They are defined as follows.

  1. 1.

Sine copulas

$C(u,v)=uv+\dfrac{2}{\pi^{2}}\lambda_{1}\sin(\pi u)\sin(\pi
v)+\dfrac{1}{2\pi^{2}}\lambda_{2}\sin(2\pi u)\sin(2\pi v),$ (9)

for $\lvert\lambda_{1}\lvert+\lvert\lambda_{2}\lvert\leq 0.5$ and density

$c(u,v)=1+2\lambda_{1}\cos(\pi u)\cos(\pi v)+2\lambda_{2}\cos(2\pi u)\cos(2\pi
v).$

  2. 2.

Sine-cosine copulas

$\displaystyle C(u,v)$ $\displaystyle=$ $\displaystyle
uv+\dfrac{1}{2\pi^{2}}(\lambda_{1}\sin(2\pi u)\sin(2\pi v)+\mu_{1}(1-\cos(2\pi
u))(1-\cos(2\pi v))$ $\displaystyle+\dfrac{1}{8\pi^{2}}(\lambda_{2}\sin(4\pi
u)\sin(4\pi v)+\mu_{2}(1-\cos(4\pi u))(1-\cos(4\pi v)),$

for
$\lvert\lambda_{1}\lvert+\lvert\lambda_{2}\lvert+\lvert\mu_{1}\lvert+\lvert\mu_{2}\lvert\leq
0.5$ and density

$\displaystyle c(u,v)$ $\displaystyle=$ $\displaystyle 1+2\lambda_{1}\cos(2\pi
u)\cos(2\pi v)+2\mu_{1}\sin(2\pi u)\sin(2\pi v)$ (10)
$\displaystyle+2\lambda_{2}\cos(4\pi u)\cos(4\pi v)+2\mu_{2}\sin(4\pi
u)\sin(4\pi v)).$

  3. 3.

Legendre copulas
If $\varphi_{k}\in\\{\sqrt{3}(2x-1),~{}\sqrt{5}(6x^{2}-6x+1)\\}$ and
$3|\lambda_{1}|+5|\lambda_{2}|\leq 1$, we obtain the following Legendre
copula:

$C(u,v)=1+3\lambda_{1}(u^{2}-v)(v^{2}-v)+5\lambda_{2}(2u^{3}-3u^{2}+u)(2v^{3}-3v^{2}+v),$
(11)

with density

$c(u,v)=1+3\lambda_{1}(2u-1)(2v-1)+5\lambda_{2}(6u^{2}-6u+1)(6v^{2}-6v+1).$
(12)

Each of these copulas is a perturbation of the independence copula and can be
used to seek some level of dependence in the data. We generate Markov chains
using copulas from these families for known values of parameters. Then, we use
them to build confidence intervals, and compute coverage probabilities based
on the asymptotic distrubtions of estimators that are proposed.

###  4.2 Algorithm for simulation of the Markov chains

As shown in Darsow et al. (1992), the transition kernel for a copula-based
Markov chain is given by the derivative with respect to the first variable of
the copula by $P(x,[0,y])=C_{,1}(x,y)$. Therefore, to generate a realization
of a Markov chain $(U_{0},\cdots,U_{n})$ using each of the three considered
copulas and the uniform marginal distribution, we use the following algorithm.

  1. 1.

generate $U_{0}\sim Unif(0,1)$;

  2. 2.

$U[1]=U_{0}$;

  3. 3.

For $2\leq i\leq n+1$:

a). generate $w\sim Unif(0,1)$;

b). $U[i]$ is the unique root on $[0,1]$ of the equation
$C_{,1}(U[i-1],U[i])-w=0$.

For each of the used copulas, the last step uses the $R$ function $Uniroot$.
For n=999, we have obtained the following graphs, displaying time series
copula 1, copula 2 and copula 3 respectively for sine copula, sine-cosine
copula and Legendre copula.

Figure 4: Data generated under each of the copula for n=999.

###  4.3 Study of the proposed estimator for our examples

Assume a Markov chain $(U_{0},\cdots,U_{n})$ is generated using a copula of
the form (1). The estimator of $\lambda_{k}$ is given by

$\hat{\lambda}_{k}=\dfrac{1}{n}\sum\limits_{i=1}^{n}\varphi_{k}(U_{i})\varphi_{k}(U_{i-1}),$

and $\hat{\Lambda}=(\hat{\lambda}_{1},...,\hat{\lambda}_{s})$, satisfies
$\sqrt{n}(\hat{\Lambda}-\Lambda)\longrightarrow{N}\left({\bf
0},\Sigma\right),$ where $\Sigma$ is the variance-covariance matrix defined in
the Proposition 4.1 below.

######  Proposition 4.1.

Copulas of the examples are constructed by means of uniformly bounded
functions $\varphi_{k}(x)$ and the sum of absolute values of parameters is
bounded. Therefore, Theorem 3.2 holds with the following $\Sigma$.

  1. 1.

For the sine copula of formula (9) with
$\hat{\Lambda}=(\hat{\lambda}_{1},\hat{\lambda}_{2})$:

$\Sigma=\left(\begin{matrix}1+\dfrac{\lambda_{2}}{2}+\dfrac{\lambda_{1}^{2}\lambda_{2}}{1-\lambda_{2}}&~{}~{}\lambda_{1}\left(\dfrac{1}{2}-\lambda_{2}\right)\\\
~{}\\\ \lambda_{1}\left(\dfrac{1}{2}-\lambda_{2}\right)&1\end{matrix}\right).$
(13)

  2. 2.

For the Legendre copula of formula (11) with
$\hat{\Lambda}=(\hat{\lambda}_{1},\hat{\lambda}_{2})$:

$\Sigma=\left(\begin{matrix}1+\dfrac{4}{5}\lambda_{2}+\lambda_{1}^{2}\dfrac{3+5\lambda_{2}}{5(1-\lambda_{2})}&~{}~{}\dfrac{4}{5}\lambda_{1}+\lambda_{1}\lambda_{2}\dfrac{1+7\lambda_{2}}{7(1-\lambda_{2})}\\\
~{}\\\
\dfrac{4}{5}\lambda_{1}+\lambda_{1}\lambda_{2}\dfrac{1+7\lambda_{2}}{7(1-\lambda_{2})}&~{}~{}1+\dfrac{20}{49}\lambda_{2}+\lambda_{2}^{2}\dfrac{63-23\lambda_{2}}{49(1-\lambda_{2})}\end{matrix}\right).$
(14)

  3. 3.

For the sine-cosine copula with density (10) and
$\hat{\Lambda}=(\hat{\lambda}_{1},\hat{\lambda}_{2},\hat{\mu}_{1},\hat{\mu}_{2})$:

$\left(\begin{matrix}1+\dfrac{\lambda_{2}}{2}+\dfrac{\lambda_{1}^{2}\lambda_{2}}{1-\lambda_{2}}&~{}~{}\lambda_{1}\left(\dfrac{1}{2}-\lambda_{2}\right)&~{}~{}\dfrac{\mu_{2}}{2}-2\lambda_{1}\mu_{1}&~{}~{}\dfrac{\mu_{1}}{2}-\lambda_{1}\mu_{2}\\\
~{}\\\
\lambda_{1}\left(\dfrac{1}{2}-\lambda_{2}\right)&~{}~{}1&~{}~{}\dfrac{\mu_{1}}{2}-\lambda_{2}\mu_{1}&~{}~{}-2\lambda_{2}\mu_{2}\\\
~{}\\\
\dfrac{\mu_{2}}{2}-2\lambda_{1}\mu_{1}&~{}~{}\dfrac{\mu_{1}}{2}-\lambda_{2}\mu_{1}&~{}~{}1+\dfrac{\lambda_{2}}{2}+\dfrac{\mu_{1}^{2}\lambda_{2}}{1-\lambda_{2}}&~{}~{}\dfrac{\lambda_{1}}{2}-\mu_{1}\mu_{2}\\\
~{}\\\
\dfrac{\mu_{1}}{2}-\lambda_{1}\mu_{2}&~{}~{}-2\lambda_{2}\mu_{2}&~{}~{}\dfrac{\lambda_{1}}{2}-\mu_{1}\mu_{2}&~{}~{}1\end{matrix}\right).$
(15)

Proposition 4.1 shows that the estimators are asymptotically independent if
$\lambda_{1}=0$ or $\lambda_{2}=0.5$ for the sine copula. This is true for the
Legendre copula if $\lambda_{1}=0$. These matrices allow us to construct
confidence intervals for single copula parameters. Any test of hypotheses on
these parameters can use the obtained multivariate central limit theorem.

######  Remark 2.

Under the assumptions of Proposition 4.1, the individual confidence interval
for each of the parameters is given by

$\left(\hat{\lambda}_{k}-z_{\frac{\alpha}{2}}\frac{\hat{\sigma}_{k}}{\sqrt{n}},\quad\hat{\lambda}_{k}+z_{\frac{\alpha}{2}}\frac{\hat{\sigma}_{k}}{\sqrt{n}}\right)\text{
and
}\left(\hat{\mu}_{k}-z_{\frac{\alpha}{2}}\frac{\hat{\tau}_{k}}{\sqrt{n}},\quad\hat{\mu}_{k}+z_{\frac{\alpha}{2}}\frac{\hat{\tau}_{k}}{\sqrt{n}}\right),$
(16) $\text{Moreover,}\quad
Q=n(\hat{\Lambda}-\Lambda)^{\prime}\Sigma^{-1}(\hat{\Lambda}-\Lambda)\underset{n\rightarrow\infty}{\longrightarrow}\chi^{2}_{s},\quad\text{and}$
$\displaystyle
C=\left\\{\Lambda\in\mathbb{R}^{s}:\hat{Q}<\chi^{2}_{\alpha}(s)\right\\}=\left\\{\Lambda\in\mathbb{R}^{s}:n(\hat{\Lambda}-\Lambda)^{\prime}\hat{\Sigma}^{-1}(\hat{\Lambda}-\Lambda)<\chi_{\alpha}^{2}(s),\right\\}$
(17)

is the $(1-\alpha)100\%$ confidence region for the vector $\Lambda$,
$\chi^{2}_{\alpha}(s)$ is given by
$P(\chi^{2}(s)>\chi^{2}_{\alpha}(s))=\alpha$, $\hat{\sigma}_{k}$
($\hat{\tau}_{k}$) is the estimate of $\sigma_{k}$ ($\tau_{k}$) obtained by
replacing $\lambda_{k}$ and $\mu_{k}$ by their estimates in the matrix
$\Sigma$. The last comment is true thanks to consistency of the estimators.

###  4.4 The Maximum likelihood estimator (MLE)

For the Markov chain $(U_{0},\cdots,U_{n})$ generated by each of the
considered copulas and the uniform marginal distribution, the log-likelihood
function is:

  1. 1.

For the sine copula:

$\ell(\Lambda)=\sum\limits_{i=1}^{n}\ln(1+2\lambda_{1}\cos(\pi U_{i})\cos(\pi
U_{i-1})+2\lambda_{2}\cos(2\pi U_{i})\cos(2\pi U_{i-1})),$

and the MLE of $\Lambda=(\lambda_{1},\lambda_{2})$ is given by:

$\hat{\Lambda}^{ML}=\underset{\lvert\lambda_{1}\lvert+\lvert\lambda_{2}\lvert\leq.5}{Argmax}\ell(\Lambda).$
(18)

  2. 2.

For the sine-cosine copula:

$\displaystyle\ell(\Lambda)=\sum\limits_{i=1}^{n}\ln(1+2\lambda_{1}\cos(2\pi
U_{i})\cos(2\pi U_{i-1})+2\mu_{1}\sin(2\pi U_{i})\sin(2\pi U_{i-1})$
$\displaystyle+2\lambda_{2}\cos(4\pi U_{i})\cos(4\pi
U_{i-1})+2\mu_{2}\sin(4\pi U_{i})\sin(4\pi U_{i-1})),$

and the MLE of $\Lambda=(\lambda_{1},\lambda_{2},\mu_{1},\mu_{2})$ is given
by:

$\hat{\Lambda}^{ML}=\underset{\lvert\lambda_{1}\lvert+\lvert\lambda_{2}\lvert+\lvert\mu_{1}\lvert+\lvert\mu_{2}\lvert\leq.5}{Argmax}\ell(\Lambda).$
(19)

  3. 3.

For the Legendre copula:

$\ell=\sum\limits_{i=1}^{n}\ln(1+3\lambda_{1}(2U_{i}-1)(2U_{i-1}-1)+5\lambda_{2}(6U_{i}^{2}-6U_{i}+1)(6U_{i-1}^{2}-6U_{i-1}+1)),$

and the MLE of $\Lambda$ is given by:

$\hat{\Lambda}^{ML}=\underset{3\lvert\lambda_{1}\lvert+5\lvert\lambda_{2}\lvert\leq
1}{Argmax}\ell(\Lambda).$ (20)

Billingsley (1961) proved asymptotic likelihood theory for the MLE of
parameters of a Markov process ($\sqrt{n}(\hat{\Lambda}^{ML}-\Lambda)\to
N({\bf 0},\Sigma^{-1})$) under the following regularity conditions.

  1. 1.

The Markov chain is ergodic and not necessarity stationary.

  2. 2.

For $u\in[0,1]$, the set of $v$ for which $c(u,v)>0$ does not depend on
$\Lambda$.

  3. 3.

For $1\leq i,j,k\leq s$, where $s$ is the length of the vector $\Lambda$,

$\frac{\partial}{\partial\lambda_{i}}\ln
c(u,v),\frac{\partial^{2}}{\partial\lambda_{i}\lambda_{j}}\ln c(u,v)\text{ and
}\frac{\partial^{3}}{\partial\lambda_{i}\lambda_{j}\lambda_{k}}\ln c(u,v)$

exist and are continuous on the set of all considered $\Lambda$.

  4. 4.

For $1\leq i,j,k\leq s$, $A^{\prime}$ an open subset of the support of the
vector $\Lambda$:
$\mathbb{E}_{\Lambda}\left[\left(\frac{\partial}{\partial\lambda_{i}}\ln
c(u,v)\right)^{2}\right]<\infty$ and
$\mathbb{E}_{\Lambda}\left[\underset{\Lambda\in
A^{\prime}}{\sup}\lvert\frac{\partial^{3}}{\partial\lambda_{i}\lambda_{j}\lambda_{k}}\ln
c(u,v)\lvert\right]<\infty$.

  5. 5.

$\sigma_{i,j}=\mathbb{E}_{\Lambda}\left[\frac{\partial}{\partial\lambda_{i}}\ln
c(u,v)\frac{\partial}{\partial\lambda_{j}}\ln c(u,v)\right]$ (21)

form a non-singular matrix $\Sigma$.

######  Theorem 4.2.

For any Markov chain generated by the considered copulas and the uniform
distribution on $[0,1]$, there exist a consistent MLE $\Lambda^{ML}$ such that

$\sqrt{n}(\Lambda^{ML}-\Lambda)\to N({\bf 0},\Sigma^{-1})$

and the $(1-\alpha)100\%$ confidence intervals for $\lambda_{k}$ (or $\mu_{k}$
for the sine-cosine copula) is

$\left({\lambda}^{ML}_{k}-z_{\dfrac{\alpha}{2}}\sqrt{\left[I_{n}^{-1}({\Lambda}^{ML})\right]_{k,k}}~{},~{}{\lambda}^{ML}_{k}+z_{\dfrac{\alpha}{2}}\sqrt{[I_{n}^{-1}({\Lambda}^{ML})]_{k,k}}\right).$
(22)

######  Remark 3.

In each of the cases, we have the following.

1\. For the sine copula:

$I_{n}(\Lambda)=\left(\sum_{i=1}^{n}\frac{4\cos(k\pi U_{i})\cos(k\pi
U_{i-1})\cos(j\pi U_{i})\cos(j\pi
U_{i-1})}{c^{2}(U_{i},U_{i-1})}\right)_{1\leq k,j\leq 2}$

where $c(u,v)$ is the density of the cosine copula.

2\. For the sine-cosine copula: $\displaystyle
I_{n}(\Lambda)=\left(a_{j,k},\right)_{1\leq j,k\leq 4}\quad\text{where}$

$\displaystyle
a_{1,1}=\sum_{i=1}^{n}\dfrac{4C^{2}_{1}(U_{i})C^{2}_{1}(U_{i-1})}{c^{2}(U_{i},U_{i-1})}$
$\displaystyle\mathchar 24635\relax\;$ $\displaystyle
a_{1,2}=\sum_{i=1}^{n}\dfrac{4C_{1}(U_{i})S_{1}(U_{i})C_{1}(U_{i-1})S_{1}(U_{i-1})}{c^{2}(U_{i},U_{i-1})}\mathchar
24635\relax\;$ $\displaystyle
a_{1,3}=\sum\limits_{i=1}^{n}\dfrac{4C_{1}(U_{i})C_{2}(U_{i})C_{1}(U_{i-1})C_{2}(U_{i-1})}{c^{2}(U_{i},U_{i-1})}$
$\displaystyle
a_{1,4}=\sum\limits_{i=1}^{n}\dfrac{4C_{1}(U_{i})S_{2}(U_{i})C_{1}(U_{i-1})S_{2}(U_{i-1})}{c^{2}(U_{i},U_{i-1})}\mathchar
24635\relax\;$ $\displaystyle
a_{2,2}=\sum\limits_{i=1}^{n}\dfrac{4S^{2}_{1}(U_{i})S^{2}_{1}(U_{i-1})}{c^{2}(U_{i},U_{i-1})}$
$\displaystyle\mathchar 24635\relax\;$ $\displaystyle
a_{2,3}=\sum\limits_{i=1}^{n}\dfrac{4C_{2}(U_{i})S_{1}(U_{i})C_{2}(U_{i-1})S_{1}(U_{i-1})}{c^{2}(U_{i},U_{i-1})}\mathchar
24635\relax\;$ $\displaystyle
a_{2,4}=\sum\limits_{i=1}^{n}\dfrac{4S_{2}(U_{i})S_{1}(U_{i})S_{2}(U_{i-1})S_{1}(U_{i-1})}{c^{2}(U_{i},U_{i-1})}$
$\displaystyle\mathchar 24635\relax\;$ $\displaystyle
a_{3,3}=\sum\limits_{i=1}^{n}\dfrac{4C^{2}_{2}(U_{i})C^{2}_{2}(U_{i-1})}{c^{2}(U_{i},U_{i-1})}\mathchar
24635\relax\;$ $\displaystyle
a_{3,4}=\sum\limits_{i=1}^{n}\dfrac{4C_{2}(U_{i})S_{2}(U_{i})C_{2}(U_{i-1})S_{2}(U_{i-1})}{c^{2}(U_{i},U_{i-1})}$
$\displaystyle
a_{4,4}=\sum\limits_{i=1}^{n}\dfrac{4S_{2}(U_{i})S_{1}(U_{i})S_{2}(U_{i-1})S_{1}(U_{i-1})}{c^{2}(U_{i},U_{i-1})},\text{
$C_{k}(x)=\cos(2\pi kx),~{}S_{k}(x)=\sin(2\pi kx)$}.$

3\. For the Legendre copula: $\displaystyle
I_{n}(\Lambda)=\left(a_{j,k}\right)_{1\leq j,k\leq 2}$ where

$a_{1,1}=\sum\limits_{i=1}^{n}\dfrac{9L^{2}(i)L^{2}(i-1)}{c^{2}(U_{i},U_{i-1})}\mathchar
24635\relax\;~{}a_{2,2}=\sum\limits_{i=1}^{n}\dfrac{25P^{2}(i)P^{2}(i-1)}{c^{2}(U_{i},U_{i-1})}\quad\text{and}$
$a_{1,2}=\sum\limits_{i=1}^{n}\dfrac{15L(i)L(i-1)P(i)P(i-1)}{c^{2}(U_{i},U_{i-1})},\quad\text{
$L(i)=2U_{i}-1\mathchar 24635\relax\;P(i)=6U_{i}^{2}-6U_{i}+1$.}$

###  4.5 The robust estimator of Longla and Peligrad (2021)

The robust estimator defined in Longla and Peligrad (2021) uses an independent
random sample to reduce the effect of the dependence. It makes use of kernel
estimators to derive a central limit theorem that doesn’t require knowledge of
the variance of partial sums. For each of the three considered Markov chains,
we consider the sequence $Y_{k}=(Y_{k1},\cdots,Y_{kn})$, where

$Y_{ki}=\varphi_{k}(U_{i})\varphi_{k}(U_{i-1}),\quad\mu_{Y_{k}}=\mathbb{E}Y_{ki}=\lambda_{k}.$
(23)

This sequence is a function of the stationary Markov chain
$(\xi_{0},\cdots,\xi_{n-1})$, with $\xi_{i}=(U_{i},U_{i+1})$. According to
Longla and Peligrad (2021) and considering the Gaussian kernel, the robust
estimator of $\mu_{Y_{k}}$, is

$\tilde{\lambda}_{k}=\dfrac{1}{nh_{n}}\sum\limits_{i=1}^{n}Y_{ki}\exp\left(-0.5(\dfrac{X_{i}}{h_{n}})^{2}\right),$
(24)

where $X_{i}\sim\mathcal{N}(0,1)$ is the independently generated random
sample, the bandwidths sequence is
$h_{n}=\left[\dfrac{\overline{Y_{k}^{2}}}{\overline{{Y}_{k}}^{2}n\sqrt{2}}\right]^{1/5}$
and estimators of means of $Y_{k,i}$ and $Y^{2}_{k,i}$ are given by
$\overline{{Y}_{k}}=\dfrac{1}{n}\sum\limits_{i=1}^{n}Y_{k,i}$ and
$\overline{{Y}_{k}^{2}}=\dfrac{1}{n}\sum\limits_{i=1}^{n}Y_{k,i}^{2}$, leading
to the $(1-\alpha)100\%$ confidence intervals

$\displaystyle\left(\tilde{\lambda}_{k}\sqrt{1+h_{n}^{2}}-z_{\alpha/2}\left(\dfrac{\overline{{Y}_{k}^{2}}}{nh_{n}\sqrt{2}}\right)^{1/2},~{}\tilde{\lambda}_{k}\sqrt{1+h_{n}^{2}}+z_{\alpha/2}\left(\dfrac{\overline{{Y}_{k}^{2}}}{nh_{n}\sqrt{2}}\right)^{1/2}\right).$
(25)

######  Remark 4.

Given that the means of $Y_{k,i}$ and $Y^{2}_{k,i}$ can also be computed as
follows

$\hat{\mathbb{E}}[Y_{k,i}]=\hat{\lambda}_{k}\text{ and
}\hat{\mathbb{E}}[Y^{2}_{k,i}]=\hat{\mathbb{E}}(\varphi^{2}_{k}(U_{0})\varphi^{2}_{k}(U_{1}))=1+\sum_{z=1}^{s}\hat{\lambda}_{z}(\int_{0}^{1}\varphi_{z}(x)\varphi_{k}^{2}(x)dx)^{2},$

the bandwidths sequence is provided by
$\hat{h}_{n,k}=\left[\dfrac{\hat{\mathbb{E}}[Y^{2}_{k,i}]}{\hat{\mathbb{E}}^{2}[Y_{k,i}]n\sqrt{2}}\right]^{1/5}$,
and the estimator of $\lambda_{k}$ becomes

$\bar{\lambda}_{k}=\dfrac{1}{n\hat{h}_{n,k}}\sum\limits_{i=1}^{n}Y_{k,i}\exp\left(-0.5(\dfrac{X_{i}}{\hat{h}_{n,k}})^{2}\right)$
(26)

which leads to the following $(1-\alpha)100\%$ confidence interval

$\left(\bar{\lambda}_{k}\sqrt{1+\hat{h}_{n,k}^{2}}-z_{\alpha/2}\left(\dfrac{\hat{\mathbb{E}}[Y^{2}_{k,i}]}{n\hat{h}_{n,k}\sqrt{2}}\right)^{1/2},~{}\bar{\lambda}_{k}\sqrt{1+\hat{h}_{n,k}^{2}}+z_{\alpha/2}\left(\dfrac{\hat{\mathbb{E}}[Y^{2}_{k,i}]}{n\hat{h}_{n,k}\sqrt{2}}\right)^{1/2}\right)$
(27)

where

  1. 1.

For the sine copula
$\hat{\mathbb{E}}[Y^{2}_{1,i}]=1+\frac{\hat{\lambda}_{2}}{2}$,
$\hat{\mathbb{E}}[Y^{2}_{2,i}]=1$,
$\hat{h}_{n,1}=\left[\frac{2+\hat{\lambda}_{2}}{2n\sqrt{2}\hat{\lambda}_{1}^{2}}\right]^{1/5}$
and
$\hat{h}_{n,2}=\left[\frac{1}{n\sqrt{2}\hat{\lambda}_{2}^{2}}\right]^{1/5}$.

  2. 2.

For the sine-cosine copula
$\hat{\mathbb{E}}[Y^{2}_{1,i}]=1+\frac{\hat{\lambda}_{2}}{2}$,
$\hat{\mathbb{E}}[Y^{2}_{2,i}]=1$,
$\hat{\mathbb{E}}[Y^{2}_{3,i}]=1+\frac{\hat{\lambda}_{2}}{2}$,
$\hat{h}_{n,4}=\left[\frac{1}{n\sqrt{2}\hat{\mu}_{2}^{2}}\right]^{1/5}$,

$\hat{\mathbb{E}}[Y^{2}_{4,i}]=1$,
$\hat{h}_{n,1}=\left[\frac{2+\hat{\lambda}_{2}}{2n\sqrt{2}\hat{\lambda}_{1}^{2}}\right]^{1/5}$,
$\hat{h}_{n,2}=\left[\frac{1}{n\sqrt{2}\hat{\lambda}_{2}^{2}}\right]^{1/5}$,
$\hat{h}_{n,3}=\left[\frac{2+\hat{\lambda}_{2}}{2n\sqrt{2}\hat{\mu}_{1}^{2}}\right]^{1/5}$.

  3. 3.

For the Legendre copula
$\hat{\mathbb{E}}[Y^{2}_{1,i}]=1+\frac{4}{5}\hat{\lambda}_{2}$,
$\hat{\mathbb{E}}[Y^{2}_{2,i}]=1+\frac{20}{49}\hat{\lambda}_{2}$,
$\hat{h}_{n,1}=\left[\frac{5+4\hat{\lambda}_{2}}{5n\sqrt{2}\hat{\lambda}_{1}^{2}}\right]^{1/5}$
and
$\hat{h}_{n,2}=\left[\frac{49+20\hat{\lambda}_{2}}{49n\sqrt{2}\hat{\lambda}_{1}^{2}}\right]^{1/5}$.

###  4.6 Data simulation and analysis of the examples

We consider copulas with the following description for generation of Markov
chains using the given algorithm.

  1. 1.

For the sine copula, $\lambda_{1}=0.28$ and $\lambda_{2}=-0.15$;

  2. 2.

For the sine-cosine copula,
$\lambda_{1}=0.14,~{}\lambda_{2}=0.13,~{}\mu_{1}=-0.11,~{}\mu_{2}=-0.12$;

  3. 3.

For the Legendre copula, $\lambda_{1}=0.15$ and $\lambda_{2}=0.1$.

The obtained data is then used to estimate parameters and construct confidence
intervals for the three cases studied above. We also provide comparisons with
the MLE, the estimators $\hat{\Lambda}$ and $\tilde{\Lambda}$. The obtained
confidence intervals are based on formulas (16), (22) and (25). For the MLE,
we have implemented an algorithm on $R$ using the level of significancy
$\alpha=.05$. The graphs are obtained using Mathlab.

####  4.6.1 Graphs of the log-likelihood functions

######  Remark 5.

Comments on the log-likelihood graphs

  1. 1.

The log-likelihood function of the sine copula in figure 5 is obtained using
the Markov generated with parameter $\Lambda=(0.14,-0.8)$;

  2. 2.

For the sine-cosine copula, we use the Markov chain generated with parameter
$\Lambda=(0.07,0.065,-0.055,-0.06)$. In order to obtain the representation in
figure 6, we’ve fixed $\mu_{1}$ and $\mu_{2}$ in the likelihood function;

  3. 3.

The log-likelihood function of the Legendre copula shown in Figure 7 uses the
Markov chain generated with parameter $\Lambda=(0.075,0.05)$.

In each of the cases, the contour plot of the log-likelihood function exhibits
evident uniqueness of the MLE. This is shown by spectral colors of the level
curves stretching towards a single interior point. This observation provides
reassurance regarding reliability of the MLE and confirms the theoretical
insights.

Figure 5: Log-likelihood function under thesine copula

Figure 6: Log-likelihood function under the sine-cosine copula

Figure 7: Log-likelihood function under the Legendre copula

####  4.6.2 Analysis of simulation results

For every given sample size, a data set has been simulated, representing a
Markov chain generated by the considered copula and the uniform marginal
distribution. Computations have been made for each of the copulas in $R$.
After running 100 times each of these simulations, we have counted the number
of intervals including the true value of the parameter to get the coverage
probability, wich should represent $95$ out of the $100$ constructed intervals
per case. We also calculated the mean length of the confidence interval.
Conclusions of the simulation study are given in the remark below. In these
tables, we use CP for coverage probability and CIML for confidence interval’s
mean length.

Sine-cosine copula for
$\lambda_{1}=0.14,~{}\lambda_{2}=0.13,~{}\mu_{1}=-0.11,~{}\mu_{2}=-0.12$
---
Estimator | n=4999 | n=9999 | n=19999
CP | CIML | CP | CIML | CP | CIML
$\hat{\lambda}_{1}=0.1504$ | 97 | 0.0523 | 97 | 0.0369 | 95 | 0.0256
$\hat{\lambda}_{1}^{ML}=0.1349$ | 98 | 0.0543 | 98 | 0.0361 | 98 | 0.0255
$\tilde{\lambda}_{1}=0.1278$ | 98 | 0.0517 | 93 | 0.0525 | 95 | 0.0407
$\bar{\lambda}_{1}=0.1280$ | 96 | 0.0549 | 95 | 0.0560 | 93 | 0.0415
$\hat{\lambda}_{2}=0.1339$ | 98 | 0.0511 | 97 | 0.0380 | 98 | 0.0272
$\hat{\lambda}_{2}^{ML}=0.1315$ | 95 | 0.0497 | 98 | 0.0362 | 97 | 0.0253
$\tilde{\lambda}_{2}$=0.1281 | 94 | 0.0509 | 92 | 0.0515 | 94 | 0.0398
$\bar{\lambda}_{2}$=0.1130 | 90 | 0.0529 | 0.575 | 99 | 94 | 0.0413
$\hat{\mu_{1}}=-0.1039$ | 96 | 0.0732 | 91 | 0.0347 | 93 | 0.0250
${\hat{\mu}_{1}^{ML}}=-0.1023$ | 95 | 0.0694 | 95 | 0.0360 | 93 | 0.0249
$\tilde{\mu}_{1}$=0.0945 | 92 | 0.0665 | 92 | 0.0506 | 92 | 0.0384
$\bar{\mu}_{1}=-0.1067$ | 93 | 0.0702 | 96 | 0.0522 | 96 | 0.0396
$\hat{\mu_{2}}=-0.1223$ | 99 | 0.0744 | 94 | 0.0368 | 90 | 0.0249
$\hat{\mu}_{2}^{ML}=-0.1233$ | 99 | 0.0683 | 95 | 0.0359 | 93 | 0.0248
$\tilde{\mu}_{2}=-0.1042$ | 95 | 0.0662 | 95 | 0.0530 | 93 | 0.0395
$\bar{\mu}_{2}=-0.1294$ | 93 | 0.0678 | 93 | 0.0513 | 93 | 0.0392
Table 1: Estimates, coverage probabilities and mean lengths.

######  Remark 6 (On simulations).

The following summarizes the results of the conducted simulations.

  1. 1.

In Table 1, the estimates are provided for a sample size of $10000$. For Table
2, we utilize a sample size of $5000$, while for Table 3, a sample size of
$20000$ is considered for the estimates.

  2. 2.

Through $R$’s constrOptim function, the MLE is obtained. This function is
initialized at $(0,0)$ for the sine and Legendre copulas and at $(0,0,0,0)$
for the sine-cosine copula. We successfully optimized the likelihood function
while maintaining the linear inequality constraints thanks to the function’s
flexibility.

  3. 3.

For each type of copula, the proposed estimator $\hat{\Lambda}$ challenges the
MLE by demonstrating its effectiveness in providing reliable parameter
estimates in situations involving small sample sizes. Furthermore, it exhibits
an increased level of precision as the sample size grows larger.

Legendre copula for $\lambda_{1}=0.15,~{}\lambda_{2}=0.1$
---
Estimator | n=4999 | n=9999 | n=19999
CP | CIML | CP | CIML | CP | CIML
$\hat{\lambda}_{1}$=0.1511 | 93 | 0.0522 | 94 | 0.0387 | 94 | 0.0274
$\hat{\lambda}_{1}^{ML}$=0.1512 | 96 | 0.0535 | 97 | 0.0383 | 96 | 0.0268
$\tilde{\lambda}_{1}$=0.1309 | 92 | 0.0737 | 97 | 0.0586 | 91 | 0.0418
$\bar{\lambda}_{1}$= 0.1555 | 92 | 0.0733 | 95 | 0.0572 | 93 | 0.0426
$\hat{\lambda}_{2}$=0.1101 | 88 | 0.0502 | 89 | 0.0359 | 92 | 0.0234
$\hat{\lambda}_{2}^{ML}$=0.1081 | 99 | 0.0539 | 99 | 0.0381 | 91 | 0.0248
$\tilde{\lambda}_{2}$=0.1042 | 90 | 0.0665 | 91 | 0.0509 | 88 | 0.0373
$\bar{\lambda}_{2}$=0.1042 | 90 | 0.0600 | 91 | 0.0476 | 93 | 0.0375
Table 2: Estimates, coverage probabilities and mean lengths.

  4. 4.

One significant advantage of $\hat{\Lambda}$ over $\Lambda^{ML}$ is that the
latter necessitates more computational time due to the absence of a closed of
Fisher information matrix for the MLE and simple form of the variance for and
estimators of our proposed method for considered copulas. For each of the
cases $\hat{\Lambda}$ has an explicit expression for both the estimate and the
confidence interval, resulting in reduced computation time and reduced errors
for the algorithm.

Sine copula for $\lambda_{1}=0.28,~{}\lambda_{2}=-0.15$
---
Estimator | n=4999 | n=9999 | n=19999
CP | CIML | CP | CIML | CP | CIML
$\hat{\lambda}_{1}$=0.2998 | 94 | 0.0417 | 98 | 0.0241 | 97 | 0.0127
$\hat{\lambda}_{1}^{ML}=0.2991$ | 95 | 0.0298 | 96 | 0.0084 | 97 | 0.0013
$\tilde{\lambda}_{1}=0.2784$ | 91 | 0.0662 | 97 | 0.0455 | 91 | 0.0290
$\bar{\lambda}_{1}=0.3078$ | 97 | 0.0699 | 98 | 0.0491 | 97 | 0.0334
$\hat{\lambda}_{2}$ =$-0.1354$ | 97 | 0.0235 | 95 | 0.0058 | 96 | 0.0003
$\hat{\lambda}_{2}^{ML}=0.1438$ | 97 | 0.0044 | 94 | 0.0003 | 95 | 0.0002
$\tilde{\lambda}_{2}=-0.1213$ | 98 | 0.0356 | 94 | 0.0202 | 96 | 0.0071
$\bar{\lambda}_{2}=-0.1320$ | 97 | 0.0383 | 93 | 0.0236 | 95 | 0.0053
Table 3: Estimates, coverage probabilities and mean lengths.

  5. 5.

In Table 3, the coverage probabilities of the MLE are unexpectedly high for
small sample sizes, given the specified significance level. This can be
interpreted as an instance of over-coverage, indicating that the MLE provides
intervals that are wider than necessary to achieve the desired level of
confidence.

  6. 6.

The alternative estimator $\bar{\Lambda}$, in comparison to $\tilde{\Lambda}$,
offers improved point estimates and coverage probabilities for each copula. It
demonstrates enhanced accuracy in estimating the true parameter values and
achieves more reliable coverage probabilities within the considered framework.

##  5 Conclusions and remarks

This paper studies copula-based Markov chains with uniform marginals. These
Markov chains are based, in general, on copulas of Longla (2023). This is the
first study of properties of estimators of parameters of these copulas. A
multivariate central limit theorem is provided for copula parameters
estimators. These copulas have densities of the general form (1), which eases
the investigation of asymptotic properties via mixing. This central limit
theorem has been applied to examples of sine copulas, sine-cosine copulas and
Legendre copulas introduced in Longla (2023). The proposed estimators of
parameters have been compared to the robust estimators of Longla and Peligrad
(2021). This theorem has been used, to show that our estimators perform better
than the MLE for sample sizes above $5000$.

The simulation study that has been performed using $R$, shows that it takes
more time to run the procedures of MLE than to run their equivalents for our
proposed estimators. Moreover, the possibility of having several points that
indicate the maximum of the likelihood functions, and the difficulty to obtain
the true variance of the MLE, are indicators that our estimators are better
tools. Note that it takes long iterations to approximate integrals that lead
to the variance-covariance matrix of the asymptotic distribution of the MLE,
while the variance-covariance matrix of our estimator is know in closed form.
It is also good to note that on large samples, coverage probabilities also
indicate that the MLE is not better.

One of the main takeaways of this work is the provided theoretical proof of
large sample properties of our estimators and the MLE for Markov chains
generated by the copulas with densities of the form (1). This opens a path to
several other questions that can be investigated for these copula families.
Future research on the topic includes the use of these theorems to
investigates tests of equality of two or more bivariate distributions based on
realizations of Markov chains that each of them generates. We also consider
working on general estimation problems, when the functions $\varphi_{k}(x)$
have a parameter. All these issues can also be included in problems with a
general marginal distribution. The marginal distribution can be considered
with parametric or non-parametetric form. All these questions are interesting
and not obvious, and are among our topics of current research.

Another extension that can be considered here is the work on such questions
for a perturbation of a copula other than the independence copula. The
theoretical work on this last topic is a little more complicated, because
orthogonality becomes more complex.

## Statements and declarations

All authors certify that they have no affiliations with or involvement in any
organization or entity with any financial interest or non-financial interest
in the subject matter or materials discussed in this manuscript. They have no
competing interest and have no funding to disclose.

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##  6 Appendix of proofs of major results

###  6.1 Proof of Theorem 3.1

Assume $U_{0},U_{1},\cdots,U_{n}$ is a copula-based Markov chain generated by
a copula of the form (1). The stationary distribution of the Markov chain is
uniform on $(0,1)$. By construction, $\mathbb{E}(\varphi_{j}(U_{i}))=0$,
$\mathbb{E}(\varphi_{j}(U_{i})\varphi_{l}(U_{i}))=\delta_{il}$ for any
$U_{i}$. Here $\delta_{il}=1$ when $i=l$ and $\delta_{il}=0$ when $i\neq l$.
It follows that $\mathbb{E}(\hat{\lambda}_{k})=\lambda_{k}$. Moreover,

$var(\hat{\lambda}_{k})=\frac{1}{n^{2}}\sum_{i=1}^{n}\sum_{j=1}^{n}cov(\varphi_{k}(U_{i})\varphi_{k}(U_{i-1}),\varphi_{k}(U_{j})\varphi_{k}(U_{j-1}))$
$=\frac{1}{n^{2}}\sum_{i=1}^{n}\sum_{j=1}^{n}\mathbb{E}(\varphi_{k}(U_{i})\varphi_{k}(U_{i-1})\varphi_{k}(U_{j})\varphi_{k}(U_{j-1}))-\lambda_{k}^{2}.$

Using stationarity and reversibility, we obtain

$var(\hat{\lambda}_{k})=\frac{1}{n}\mathbb{E}(\varphi^{2}_{k}(U_{1})\varphi^{2}_{k}(U_{0}))+\frac{2(n-1)}{n^{2}}\mathbb{E}(\varphi_{k}(U_{0})\varphi^{2}_{k}(U_{1})\varphi_{k}(U_{2}))+$
$+\frac{1}{n^{2}}\sum_{s=2}^{n-1}2(n-s)\mathbb{E}(\varphi_{k}(U_{0})\varphi_{k}(U_{1})\varphi_{k}(U_{s})\varphi_{k}(U_{s+1}))=\lambda_{k}^{2}.$

To end these computations, we need the three joint distributions of
$(U_{0},U_{1},U_{2})$ and $(U_{0},U_{1},U_{s},U_{s+1})$ for $s\geq 2$. The
following Lemma can be easily established using the definitions.

######  Lemma 1.

If $(U_{0},U_{1},\cdots,U_{n})$ is a copula-based Markov chain generated by
the copula with density (1), then the following holds:

  1. 1.

The density of $(U_{0},U_{1},U_{2})$ is $\displaystyle
c(u_{0},u_{1},u_{2})=c(u_{0},u_{1})c(u_{1},u_{2})=$

$=1+\sum\lambda_{m}\varphi_{m}(u_{0})\varphi_{m}(u_{1})+\sum\lambda_{n}\varphi_{z}(u_{1})\varphi_{z}(u_{2})$
$+\sum\sum\lambda_{m}\lambda_{z}\varphi_{m}(u_{0})\varphi_{m}(u_{1})\varphi_{z}(u_{1})\varphi_{z}(u_{2}).$

  2. 2.

$\displaystyle\mathbb{E}(\varphi_{k}(U_{0})\varphi^{2}_{k}(U_{1})\varphi_{k}(U_{2}))=\lambda_{k}^{2}\int_{0}^{1}\varphi^{4}_{k}(x)dx$.

  3. 3.

$\displaystyle\mathbb{E}(\varphi^{2}_{k}(U_{1})\varphi^{2}_{k}(U_{0}))=1+\sum\lambda_{z}(\int_{0}^{1}\varphi_{z}(x)\varphi_{k}^{2}(x)dx)^{2}.$

  4. 4.

The joint density of $(U_{0},U_{1},U_{s},U_{s+1})$ is

$\displaystyle
c(u_{0},u_{1},u_{s},u_{s+1})=c(u_{0},u_{1})(1+\sum\lambda_{z}^{s-1}\varphi_{z}(u_{1})\varphi_{z}(u_{s}))c(u_{s},u_{s+1}).$

  5. 5.

$\mathbb{E}(\varphi_{k}(U_{0})\varphi_{k}(U_{1})\varphi_{k}(U_{s})\varphi_{k}(U_{s+1})\lvert{}U_{0},U_{1},U_{s})=\lambda_{k}\varphi_{k}(U_{0})\varphi_{k}(U_{1})\varphi^{2}_{k}(U_{s}).$

  6. 6.

The joint density of $(U_{0},U_{1},U_{s})$ is

$\displaystyle
c(u_{0},u_{1},u_{s},u_{s+1})=c(u_{0},u_{1})(1+\sum\lambda_{z}^{s-1}\varphi_{z}(u_{1})\varphi_{z}(u_{s})).$

  7. 7.

$\mathbb{E}(\varphi_{k}(U_{0})\varphi_{k}(U_{1})\varphi_{k}(U_{s})\varphi_{k}(U_{s+1}))=\lambda_{k}\mathbb{E}(\varphi_{k}(U_{0})\varphi_{k}(U_{1})\varphi^{2}_{k}(U_{s}))$

$=\lambda_{k}^{2}\mathbb{E}(\varphi^{2}_{k}(U_{1})\varphi^{2}_{k}(U_{s})).$

  8. 8.

$\mathbb{E}(\varphi_{k}(U_{0})\varphi_{k}(U_{1})\varphi_{k}(U_{s})\varphi_{k}(U_{s+1}))=\lambda_{k}^{2}(1+\sum\lambda^{s-1}_{z}(\int_{0}^{1}\varphi_{z}(x)\varphi_{k}^{2}(x)dx)^{2}).$

Based on Lemma 1, we can conclude that

$var(\hat{\lambda}_{k})=\frac{1}{n}(1-\lambda_{k}^{2}+\sum\lambda_{z}(\int_{0}^{1}\varphi_{z}(x)\varphi_{k}^{2}(x)dx)^{2})+\frac{2(n-1)\lambda_{k}^{2}}{n^{2}}(\int_{0}^{1}\varphi^{4}_{k}(x)dx-1)$
$+\sum_{s=2}^{n-1}\frac{2(n-s)\lambda_{k}^{2}}{n^{2}}(\sum\lambda^{s-1}_{z}(\int_{0}^{1}\varphi_{z}(x)\varphi_{k}^{2}(x)dx)^{2}).$
(28)

Therefore, knowing that
$\sum\lvert{}\lambda_{z}\lvert{}(\int_{0}^{1}\varphi_{z}(x)\varphi_{k}^{2}(x)dx)^{2})=K<\infty$
and $\int_{0}^{1}\varphi_{k}^{4}(x)dx<\infty$, we have

$\lim_{n\to\infty}var(\hat{\lambda}_{k})=\lim_{n\to\infty}\frac{2\lambda_{k}^{2}}{n^{2}}\sum_{s=3}^{n-1}(n-s)(\sum\lambda^{s-1}_{z}(\int_{0}^{1}\varphi_{z}(x)\varphi_{k}^{2}(x)dx)^{2}).$
(29)

Provided that $\underline{\lambda}=\sup_{z}\lvert{}\lambda_{z}\lvert{}<1$ , we
obtain

$\lim_{n\to\infty}var(\hat{\lambda}_{k})\leq\lim_{n\to\infty}\frac{2\lambda_{k}^{2}}{n^{2}}\sum_{s=3}^{n-1}(n-s)\underline{\lambda}^{s-2}(\sum\lvert{}\lambda_{z}\lvert{}(\int_{0}^{1}\varphi_{z}(x)\varphi_{k}^{2}(x)dx)^{2}),$
$\lim_{n\to\infty}var(\hat{\lambda}_{k})\leq\lim_{n\to\infty}\frac{2K\lambda_{k}^{2}}{n^{2}}\sum_{s=3}^{n-1}(n-s)\underline{\lambda}^{s-2}.$

Moreover, the last sum can be written as

$\sum_{s=3}^{n-1}(n-s)\underline{\lambda}^{s-2}=(n\sum_{s=3}^{n-1}\underline{\lambda}^{s-2}-\frac{1}{\underline{\lambda}}\frac{d}{d\underline{\lambda}}(\sum_{s=3}^{n-1}\underline{\lambda}^{s}))=$
$\frac{n\underline{\lambda}(1-\underline{\lambda}^{n-2})}{1-\underline{\lambda}}-\frac{1}{\underline{\lambda}}\frac{d}{d\underline{\lambda}}\frac{\underline{\lambda}^{2}(1-\underline{\lambda}^{n-2})}{1-\underline{\lambda}}=\frac{n(\underline{\lambda}-\underline{\lambda}^{n-1})-(2-n\underline{\lambda}^{n-2})}{1-\underline{\lambda}}+\frac{\underline{\lambda}^{2}-\underline{\lambda}^{n}}{(1-\underline{\lambda})^{2}}.$

If $\sup_{z}\lvert{}\lambda_{z}\lvert{}<1$, then as $n>2$ this sum behaves
like $\displaystyle n\frac{\underline{\lambda}}{1-\underline{\lambda}}.$
Therefore $\lim_{n\to\infty}var(\hat{\lambda}_{k})=0$. This implies, that
under the conditions of the statement, $\hat{\lambda}_{k}$ is a consistent
estimator of $\lambda_{k}$. Moroever, from this proof, it follows that
$nvar(\hat{\lambda}_{k})\to\sigma^{2}$ with $0<\sigma^{2}<\infty$. We can also
establish easily that $Y_{i}=(X_{i-1},X_{i}),i=1,\cdots,n$ forms a reversible
Markov chain. The variables
$Z_{i}=\varphi_{k}(U_{i})\varphi_{k}(U_{i-1}):=f(Y_{i})$ satisfy
$n^{-1}var(\sum Z_{i})\to\sigma^{2}$ because $\sum Z_{i}=n\hat{\lambda}_{k}$.
Therefore, the Kipnis and Varadhan (1986) central limit theorem holds for
$\hat{\lambda}_{k}$.

###  6.2 Proof of Theorem 3.2

Assume $(U_{0},U_{1},\cdots,U_{n})$ is a copula-based Markov chain generated
by a copula of the form (1). It is already clear that each of the
$\hat{\lambda}_{k}$ is a consistent estimator of $\lambda_{k}$. To prove this
theorem, all we need to do now is to show by the Cramér-Wold device that for
every ${\bf t}=(t_{1},\cdots,t_{s})^{\prime}$, $\sqrt{n}({\bf
t}^{\prime}\hat{\Lambda}-{\bf t}^{\prime}\Lambda)\to N({\bf 0},{\bf
t}^{\prime}\Sigma{\bf t})$. And by the Kipnis and Varadhan central limit
theorem for reversible Markov chains, to complete this proof, we need to show
that $nvar(t^{\prime}\hat{\lambda})\to t^{\prime}\Sigma t$ as $n\to\infty$.
But it is easy to realize that

$nvar({\bf
t}^{\prime}\hat{\Lambda})=\sum_{j=1}^{s}\sum_{i=1}^{s}t_{j}t_{i}ncov(\hat{\lambda}_{k_{i}},\hat{\lambda}_{k_{j}}).$

Therefore, to prove the central limit theorem, it is enough to show that
$ncov(\hat{\lambda}_{k_{i}},\hat{\lambda}_{k_{j}})\to\Sigma_{ij}$ as
$n\to\infty$. As we have already shown this convergence for $i=j$, it is
enough to conduct the proof for $i\neq j$. But,

$ncov(\hat{\lambda}_{k_{i}},\hat{\lambda}_{k_{j}})=\frac{1}{n}\sum_{p=1}^{n}\sum_{q=1}^{n}cov(\varphi_{k_{i}}(U_{p-1})\varphi_{k_{i}}(U_{p}),\varphi_{k_{j}}(U_{q-1})\varphi_{k_{j}}(U_{q})).$

To compute this covariance, we consider $4$ separate cases
$(p=q,p=q-1,q=p-1,\lvert{}p-q\lvert{}=m>1)$. The cases $p=q-1$ and $q=p-1$ are
equivalent due to symmetry.

  1. 1.

For $p=q$ and $i\neq j$, using stationarity we have

$cov(\varphi_{k_{i}}(U_{p-1})\varphi_{k_{i}}(U_{p}),\varphi_{k_{j}}(U_{q-1})\varphi_{k_{j}}(U_{q}))=$
$\mathbb{E}(\varphi_{k_{i}}(U_{0})\varphi_{k_{i}}(U_{1})\varphi_{k_{j}}(U_{0})\varphi_{k_{j}}(U_{1}))-\lambda_{k_{j}}\lambda_{k_{i}}=$
$=\sum_{z=1}^{\infty}\lambda_{z}(\int_{0}^{1}\varphi_{z}(x)\varphi_{k_{i}}(x)\varphi_{k_{j}}(x)dx)^{2}-\lambda_{k_{j}}\lambda_{k_{i}}.$

  2. 2.

For $q=p-1$ and $i\neq j$, using stationarity we have

$cov(\varphi_{k_{i}}(U_{p-1})\varphi_{k_{i}}(U_{p}),\varphi_{k_{j}}(U_{p})\varphi_{k_{j}}(U_{p+1}))=$
$\mathbb{E}(\varphi_{k_{i}}(U_{0})\varphi_{k_{i}}(U_{1})\varphi_{k_{j}}(U_{1})\varphi_{k_{j}}(U_{2}))-\lambda_{k_{j}}\lambda_{k_{i}}=$
$=\lambda_{k_{j}}\lambda_{k_{i}}(\int_{0}^{1}\varphi^{2}_{k_{i}}(x)\varphi^{2}_{k_{j}}(x)dx-1).$

  3. 3.

For $\lvert{}p-q\lvert{}=m\geq 2$ and $j\neq j$, using stationary we have

$cov(\varphi_{k_{i}}(U_{p-1})\varphi_{k_{i}}(U_{p}),\varphi_{k_{j}}(U_{p+m-1})\varphi_{k_{j}}(U_{p+m}))=$
$\mathbb{E}(\varphi_{k_{i}}(U_{0})\varphi_{k_{i}}(U_{1})\varphi_{k_{j}}(U_{m})\varphi_{k_{j}}(U_{m+1}))-\lambda_{k_{j}}\lambda_{k_{i}}=$
$\lambda_{k_{j}}\lambda_{k_{i}}(\mathbb{E}(\varphi^{2}_{k_{i}}(U_{1})\varphi^{2}_{k_{j}}(U_{m}))-1)=$
$=\lambda_{k_{j}}\lambda_{k_{i}}\sum_{z=1}^{\infty}\lambda_{z}^{m-1}(\int_{0}^{1}\varphi_{z}(x)\varphi_{k_{i}}^{2}(x)dx)(\int_{0}^{1}\varphi_{z}(x)\varphi_{k_{j}}^{2}(x)dx).$

Similarly to Haggstrom and Rosenthal (2007), it follows that

$\lim_{n\to\infty}ncov(\hat{\lambda}_{k_{i}},\hat{\lambda}_{k_{j}})=cov(\varphi_{k_{i}}(U_{0})\varphi_{k_{i}}(U_{1}),\varphi_{k_{j}}(U_{0})\varphi_{k_{j}}(U_{1}))+$
$2\sum_{m=1}^{\infty}cov(\varphi_{k_{i}}(U_{0})\varphi_{k_{i}}(U_{1}),\varphi_{k_{j}}(U_{m})\varphi_{k_{j}}(U_{m+1})).$

Knowing that $c(u,v)$ is square integrable, we can establish easily that for
$m\leq 3$ each of the terms of the series is in absolute value smaller than
$\underline{\lambda}^{m-3}K$, where $\underline{\lambda}$ is the suppremum of
$\lvert{}\lambda_{z}\lvert{}$ over $z$ and $K$ is a constant that depends only
on $k_{i},k_{j}$. This implies that the series with sum starting at $m=3$ is
convergent (using geometric series to end the argument as
$0<\underline{\lambda}<1$). Moreover the term with $m=1$ is finite under any
conditions and the term with $m=2$ is finite if and only if
$\sum_{z=1}^{\infty}\lambda_{z}(\int_{0}^{1}\varphi_{z}(x)\varphi_{k_{i}}^{2}(x)dx)(\int_{0}^{1}\varphi_{z}(x)\varphi_{k_{j}}^{2}(x)dx)<\infty.$
On the other hand,
$cov(\varphi_{k_{i}}(U_{0})\varphi_{k_{i}}(U_{1}),\varphi_{k_{j}}(U_{0})\varphi_{k_{j}}(U_{1}))$
is finite if and only if

$\sum_{z=1}^{\infty}\lambda_{z}(\int_{0}^{1}\varphi_{z}(x)\varphi_{k_{i}}(x)\varphi_{k_{j}}(x)dx)^{2}<\infty.$

###  6.3 Proof of Proposition 3.3

Following the proof of Theorem 3.2 and Haggstorm amd Rosenthal (2007), we have

$\displaystyle\underset{k\neq j}{\Sigma_{k,j}}$ $\displaystyle=$
$\displaystyle\underset{n\rightarrow\infty}{\lim~{}}ncov(\hat{\lambda}_{k},\hat{\lambda}_{j})$
(30) $\displaystyle=$ $\displaystyle
cov(\varphi_{k}(U_{0})\varphi_{k}(U_{1}),\varphi_{j}(U_{0})\varphi_{j}(U_{1}))+2cov(\varphi_{k}(U_{0})\varphi_{k}(U_{1}),\varphi_{j}(U_{1})\varphi_{j}(U_{2}))$
$\displaystyle+2\sum\limits_{m=2}^{\infty}cov(\varphi_{k}(U_{0})\varphi_{k}(U_{1}),\varphi_{k}(U_{m})\varphi_{k}(U_{m+1}).$

Using the obtained expected values, we derive the following.

$\displaystyle
cov(\varphi_{k}(U_{0})\varphi_{k}(U_{1}),\varphi_{j}(U_{0})\varphi_{j}(U_{1}))$
$\displaystyle=$
$\displaystyle\mathbb{E}\left(\varphi_{k}(U_{0})\varphi_{k}(U_{1})\varphi_{j}(U_{0})\varphi_{j}(U_{1})\right)-\lambda_{k}\lambda_{j}$
$\displaystyle=$
$\displaystyle\sum\limits_{z=1}^{s}\lambda_{z}\left(\int_{0}^{1}\varphi_{z}(x)\varphi_{k}(x)\varphi_{j}(x)dx\right)^{2}-\lambda_{k}\lambda_{j}.$
$\displaystyle
cov(\varphi_{k}(U_{0})\varphi_{k}(U_{1}),\varphi_{j}(U_{1})\varphi_{j}(U_{2}))$
$\displaystyle=$
$\displaystyle\mathbb{E}\left(\varphi_{k}(U_{0})\varphi_{k}(U_{1})\varphi_{j}(U_{1})\varphi_{j}(U_{2})\right)-\lambda_{k}\lambda_{j}$
$\displaystyle=$
$\displaystyle\lambda_{k}\lambda_{j}\left(\int_{0}^{1}\varphi^{2}_{k}(x)\varphi_{j}^{2}(x)-1\right).$
$\displaystyle
cov(\varphi_{k}(U_{0})\varphi_{k}(U_{1}),\varphi_{k}(U_{m})\varphi_{k}(U_{m+1})=$
$\displaystyle=\mathbb{E}\left(\varphi_{k}(U_{0})\varphi_{k}(U_{1})\varphi_{j}(U_{m})\varphi_{j}(U_{m+1})\right)-\lambda_{k}\lambda_{j}=$
$\displaystyle=\lambda_{k}\lambda_{j}\sum\limits_{z=1}^{\infty}\lambda_{z}^{m-1}\left(\int_{0}^{1}\varphi_{z}(x)\varphi_{k}^{2}(x)dx\right)\left(\int_{0}^{1}\varphi_{z}(x)\varphi_{j}^{2}(x)dx\right).$

Therefore, formula (30) becomes

$\displaystyle\underset{k\neq j}{\Sigma_{k,j}}$ $\displaystyle=$
$\displaystyle\sum\limits_{z=1}^{s}\lambda_{z}(\int_{0}^{1}\varphi_{z}(x)\varphi_{k}(x)\varphi_{j}(x)dx)^{2}+2\lambda_{k}\lambda_{j}(\int_{0}^{1}\varphi^{2}_{k}(x)\varphi_{j}^{2}(x)dx-1)$
$\displaystyle+2\lambda_{k}\lambda_{j}\sum\limits_{z=1}^{s}\sum\limits_{m=2}^{\infty}\lambda_{z}^{m-1}(\int_{0}^{1}\varphi_{z}(x)\varphi_{k}^{2}(x)dx)(\int_{0}^{1}\varphi_{z}(x)\varphi_{j}^{2}(x)dx)-\lambda_{k}\lambda_{j}.$

###  6.4 Proof of Proposition 4.1

Using the provided theorems for each of the copula families, we get:

1\. The sine copula, $\varphi_{1}(x)=\sqrt{2}\cos\pi x$ and
$\varphi_{2}(x)=\sqrt{2}\cos 2\pi x$. It follows that

$\displaystyle\sigma_{1}^{2}$ $\displaystyle=$ $\displaystyle
1-\lambda_{1}^{2}+\lambda_{1}(\int_{0}^{1}(\sqrt{2}\cos\pi
x)^{3}dx)^{2}+\lambda_{2}(\int_{0}^{1}\sqrt{2}\cos 2\pi x(\sqrt{2}\cos\pi
x)^{2}dx)^{2}+$ (31)
$\displaystyle+2\lambda_{1}^{2}(\int_{0}^{1}(\sqrt{2}\cos\pi
x)^{4}dx-1)+2\lambda_{1}^{2}\sum\limits_{m=2}^{\infty}\lambda_{1}^{m-1}(\int_{0}^{1}(\sqrt{2}\cos\pi
x)^{3}dx)^{2}$
$\displaystyle+2\lambda_{1}^{2}\sum\limits_{m=2}^{\infty}\lambda_{2}^{m-1}(\int_{0}^{1}\sqrt{2}\cos
2\pi x(\sqrt{2}\cos\pi x)^{2}dx)^{2}.$

Given that $\displaystyle\int_{0}^{1}(\sqrt{2}\cos\pi
x)^{3}dx)=0,\int_{0}^{1}\sqrt{2}\cos 2\pi x(\sqrt{2}\cos\pi
x)^{2}dx)=\dfrac{1}{\sqrt{2}}$ and

$\int_{0}^{1}4\cos^{4}\pi xdx-1=\dfrac{1}{2},\quad\text{ we
obtain}\quad\sigma_{1}^{2}=1+\dfrac{\lambda_{2}}{2}+\lambda_{1}^{2}\dfrac{\lambda_{2}}{1-\lambda_{2}}.$
$\text{Similarly,}\quad\int_{0}^{1}(\sqrt{2}\cos 2\pi
x)^{3}dx=0,\int_{0}^{1}\sqrt{2}\cos\pi x(\sqrt{2}\cos 2\pi x)^{2}dx)=0$

and $\int_{0}^{1}(\sqrt{2}\cos 2\pi x)^{4}dx-1=\dfrac{1}{2}$ imply
$\sigma^{2}_{2}=1,$
$\displaystyle\Sigma_{1,2}=\Sigma_{2,1}=\lambda_{1}\left(\dfrac{1}{2}-\lambda_{2}\right).$

2\. The Sine-cosine copula. We have

$\varphi_{1}(x)=\cos 2\pi x\mathchar 24635\relax\;~{}~{}\varphi_{2}(x)=\cos
4\pi x\mathchar 24635\relax\;~{}~{}\varphi_{3}(x)=\sin 2\pi x\text{ and
}\varphi_{4}(x)=\sin 4\pi x.$

Via simple computations, we get

$\displaystyle\int_{0}^{1}(\sqrt{2})^{3}\cos^{3}2\pi
xdx=0,\int_{0}^{1}(\sqrt{2})^{3}\cos^{2}2\pi x\cos 4\pi
xdx=\dfrac{1}{\sqrt{2}},$ $\displaystyle\int_{0}^{1}(\sqrt{2})^{3}\cos^{2}2\pi
x\sin 2\pi xdx=0,\int_{0}^{1}2\sqrt{2}\cos^{2}2\pi x\sin 4\pi x=0,$
$\displaystyle\text{ and }\int_{0}^{1}4\cos^{4}2\pi
xdx=\dfrac{3}{2}.\quad\text{Using these integrals leads to}$
$\Sigma_{1,1}=\sigma^{2}_{1}=1-\lambda_{1}^{2}+\dfrac{\lambda_{2}}{2}+\lambda_{1}^{2}\lambda_{1}^{2}+\dfrac{\lambda_{2}}{1-\lambda_{2}}=1+\dfrac{\lambda_{2}}{2}+\lambda_{1}^{2}\dfrac{\lambda_{2}}{1-\lambda_{2}}.$

$\displaystyle\bullet~{}\Sigma_{2,2}=\sigma_{2}^{2}=1-\lambda_{2}^{2}+\lambda_{1}(\int_{0}^{1}2\sqrt{2}\cos
2\pi x\cos^{2}4\pi xdx)^{2}+$
$\displaystyle\lambda_{2}(\int_{0}^{1}2\sqrt{2}\cos^{3}4\pi
xdx)^{2}+\mu_{1}(\int_{0}^{1}2\sqrt{2}\cos^{2}4\pi x\sin 2\pi xdx)^{2}+$
$\displaystyle\mu_{2}(\int_{0}^{1}2\sqrt{2}\cos^{2}4\pi x\sin 4\pi
xdx)^{2}+2\lambda_{2}^{2}(\int_{0}^{1}4\cos^{4}4\pi xdx-1)+$ $\displaystyle
2\lambda_{2}^{2}\sum\limits_{m=2}^{\infty}\lambda_{1}^{m-1}(\int_{0}^{1}2\sqrt{2}\cos^{2}4\pi
x\cos 2\pi xdx)^{2}+$ $\displaystyle
2\lambda_{2}^{2}\left(\sum\limits_{m=2}^{\infty}\lambda_{2}^{m-1}(2\sqrt{2}\int_{0}^{1}\cos^{3}4\pi
xdx)^{2}\right)+$
$\displaystyle+2\lambda_{2}^{2}\left(\sum\limits_{m=2}^{\infty}\mu_{1}^{m-1}(\int_{0}^{1}2\sqrt{2}\sin
2\pi x\cos^{2}4\pi xdx)^{2}\right)+$ $\displaystyle
2\lambda_{2}^{2}\sum\limits_{m=2}^{\infty}\mu_{2}^{m-1}(\int_{0}^{1}2\sqrt{2}\sin
4\pi x\cos^{2}4\pi xdx)^{2}.$

Knowing $\displaystyle\int_{0}^{1}2\sqrt{2}\cos^{3}4\pi
xdx=\int_{0}^{1}2\sqrt{2}\cos 2\pi x\cos^{2}4\pi xdx=0$,
$\displaystyle\int_{0}^{1}2\sqrt{2}\cos^{2}4\pi x\sin 2\pi
xdx=\int_{0}^{1}2\sqrt{2}\cos^{2}4\pi x\sin 4\pi xdx=0,$
$\text{and}\quad\displaystyle\int_{0}^{1}2\cos^{2}4\pi xdx-1=\dfrac{1}{2},$ we
derive $\Sigma_{2,2}=\sigma_{2}^{2}=1-\lambda_{2}^{2}+\lambda_{2}^{2}=1.$

$\displaystyle\bullet~{}\Sigma_{3,3}=\sigma_{3}^{2}=1-\mu_{1}^{2}+\lambda_{1}(\int_{0}^{1}\sqrt{2}\cos
2\pi x(\sqrt{2}\sin 2\pi x)^{2}dx)^{2}+$
$\displaystyle\lambda_{2}(\int_{0}^{1}\sqrt{2}\cos 4\pi x(\sqrt{2}\sin 2\pi
x)^{2}dx)^{2}+\mu_{1}(\int_{0}^{1}(\sqrt{2}\sin 2\pi x)^{3}dx)^{2}+$
$\displaystyle\mu_{2}(\int_{0}^{1}(\sqrt{2}\sin 2\pi x)^{2}\sqrt{2}\sin 4\pi
xdx)^{2}+2\mu_{1}^{2}(\int_{0}^{1}(\sqrt{2}\sin 2\pi x)^{4}dx-1)+$
$\displaystyle
2\mu_{1}^{2}\sum\limits_{m=2}^{\infty}\lambda_{1}^{m-1}(\int_{0}^{1}(\sqrt{2}\sin
2\pi x)^{2}\sqrt{2}\cos 2\pi xdx)^{2}+$ $\displaystyle
2\mu_{1}^{2}\sum\limits_{m=2}^{\infty}\lambda_{2}^{m-1}(\int_{0}^{1}(\sqrt{2}\sin
2\pi x)^{2}\sqrt{2}\cos 4\pi xdx)^{2}+$ $\displaystyle
2\mu_{1}^{2}\sum\limits_{m=2}^{\infty}\mu_{1}^{m-1}(\int_{0}^{1}(\sqrt{2}\sin
2\pi x)^{3}dx)^{2}+$ $\displaystyle
2\mu_{1}^{2}\sum\limits_{m=2}^{\infty}\mu_{2}^{m-1}(\int_{0}^{1}\sqrt{2}\sin
4\pi x(\sqrt{2}\sin 2\pi x)^{2}dx)^{2}$

$\displaystyle\text{and}\quad\int_{0}^{1}2\sqrt{2}\sin^{3}2\pi
xdx=0,\int_{0}^{1}2\sqrt{2}\cos 2\pi x\sin^{2}2\pi xdx=0,$
$\displaystyle\int_{0}^{1}2\sqrt{2}\sin^{2}2\pi x\cos 4\pi
xdx=-\dfrac{1}{\sqrt{2}},$ $\displaystyle\int_{0}^{1}2\sqrt{2}\sin^{2}2\pi
x\sin 4\pi xdx=0,\int_{0}^{1}4\sin 2\pi xdx-1=\dfrac{1}{2}.$

$\displaystyle\text{Thus},\quad\Sigma_{3,3}=\sigma_{3}^{2}$ $\displaystyle=$
$\displaystyle
1-\mu_{1}^{2}+\dfrac{\lambda_{2}}{2}+\mu_{1}^{2}+\mu_{1}^{2}\dfrac{\lambda_{2}}{1-\lambda_{2}}=1+\dfrac{\lambda_{2}}{2}+\mu_{1}^{2}\dfrac{\lambda_{2}}{1-\lambda_{2}}.$

$\displaystyle\bullet\Sigma_{4,4}=\sigma_{4}^{2}=1-\mu_{2}^{2}+\lambda_{1}(\int_{0}^{1}\sqrt{2}\cos
2\pi x\cdot(\sqrt{2}\sin 4\pi x)^{2}dx)^{2}+$
$\displaystyle\lambda_{2}(\int_{0}^{1}\sqrt{2}\cos 4\pi x\cdot(\sqrt{2}\sin
4\pi x)^{2}dx)^{2}+\mu_{1}(\int_{0}^{1}\sqrt{2}\sin 2\pi x\cdot(\sqrt{2}\sin
4\pi x)^{2}dx)^{2}+$ $\displaystyle\mu_{2}(\int_{0}^{1}(\sqrt{2}\sin 4\pi
x)^{3}dx)^{2}+2\mu_{2}^{2}(\int_{0}^{1}(\sqrt{2}\sin 4\pi x)^{4}dx-1)+$
$\displaystyle
2\mu_{2}^{2}\sum\limits_{m=2}^{\infty}\lambda_{1}^{m-1}(\int_{0}^{1}(\sqrt{2}\sin
4\pi x)^{2}\cdot\sqrt{2}\cos 2\pi xdx)^{2}+$ $\displaystyle
2\mu_{2}^{2}\sum\limits_{m=2}^{\infty}\lambda_{2}^{m-1}(\int_{0}^{1}(\sqrt{2}\sin
4\pi x)^{2}\cdot\sqrt{2}\cos 4\pi xdx)^{2}+$ $\displaystyle
2\mu_{2}^{2}\sum\limits_{m=2}^{\infty}\mu_{1}^{m-1}(\int_{0}^{1}\sqrt{2}\sin
2\pi x\cdot(\sqrt{2}\sin 4\pi x)^{2}dx)^{2}+$ $\displaystyle
2\mu_{2}^{2}\sum\limits_{m=2}^{\infty}\mu_{2}^{m-1}(\int_{0}^{1}(\sqrt{2}\sin
4\pi x)^{3}dx)^{2},$

$\displaystyle\text{and}\int_{0}^{1}\left(\sqrt{2}\sin 4\pi
x\right)^{3}dx=0,\int_{0}^{1}\left(\sqrt{2}\cos 2\pi
x\right)\left(\sqrt{2}\sin 4\pi x\right)^{2}dx=0,$
$\displaystyle\int_{0}^{1}\left(\sqrt{2}\sin 4\pi
x\right)^{2}\left(\sqrt{2}\cos 4\pi x\right)dx=0,$
$\displaystyle\int_{0}^{1}\left(\sqrt{2}\sin 2\pi x\right)\left(\sqrt{2}\sin
4\pi x\right)^{2}dx=0,\int_{0}^{1}\left(\sqrt{2}\sin 4\pi
x\right)^{4}dx-1=\dfrac{1}{2}.$
$\text{Therefore,}\quad\Sigma_{4,4}=\sigma_{4}^{2}=1-\mu_{2}^{2}+\mu_{2}^{2}=1.$

$\displaystyle\bullet~{}\Sigma_{1,2}$ $\displaystyle=$
$\displaystyle\sum\limits_{z=1}^{4}\lambda_{z}(\int_{0}^{1}\varphi_{z}(x)\varphi_{1}(x)\varphi_{2}(x)dx)^{2}+\lambda_{1}\lambda_{2}\left(2(\int_{0}^{1}\varphi^{2}_{1}(x)\varphi_{2}^{2}(x)dx-1)\right.$
$\displaystyle\left.+2\sum\limits_{z=1}^{4}\sum\limits_{m=2}^{\infty}\lambda_{z}^{m-1}(\int_{0}^{1}\varphi_{z}(x)\varphi_{1}^{2}(x)dx)(\int_{0}^{1}\varphi_{z}(x)\varphi_{2}^{2}(x)dx)-1\right).$

Moreover, just as for the other cases, we have

$\displaystyle\int_{0}^{1}\varphi^{2}_{1}(x)\varphi_{2}(x)dx=\dfrac{1}{\sqrt{2}},\int_{0}^{1}\varphi^{2}_{2}(x)\varphi_{1}(x)dx=\int_{0}^{1}\varphi_{3}(x)\varphi_{1}(x)\varphi_{2}(x)dx=0,$
$\displaystyle\int_{0}^{1}\varphi_{4}(x)\varphi_{1}(x)\varphi_{2}(x)dx=\int_{0}^{1}\varphi^{2}_{1}(x)\varphi^{2}_{2}(x)dx-1=\int_{0}^{1}\varphi^{2}_{1}(x)\varphi_{3}(x)dx=0,$
$\displaystyle\int_{0}^{1}\varphi^{2}_{1}(x)\varphi_{4}(x)dx=\int_{0}^{1}\varphi^{2}_{2}(x)\varphi_{3}(x)dx=0,\text{
and }\int_{0}^{1}\varphi^{2}_{2}(x)\varphi_{4}(x)dx=0.$

$\text{Therefore},\quad\Sigma_{1,2}=\Sigma_{2,1}=\dfrac{\lambda_{1}}{2}-\lambda_{1}\lambda_{2}=\lambda_{1}\left(\dfrac{1}{2}-\lambda_{2}\right).$

$\displaystyle\bullet~{}\Sigma_{2,3}$ $\displaystyle=$
$\displaystyle\sum\limits_{z=1}^{4}\lambda_{z}(\int_{0}^{1}\varphi_{z}(x)\varphi_{3}(x)\varphi_{2}(x)dx)^{2}+\lambda_{2}\mu_{1}\left(2(\int_{0}^{1}\varphi^{2}_{3}(x)\varphi_{2}^{2}(x)dx-1)\right.$
$\displaystyle\left.+2\sum\limits_{z=1}^{4}\sum\limits_{m=2}^{\infty}\lambda_{z}^{m-1}(\int_{0}^{1}\varphi_{z}(x)\varphi_{3}^{2}(x)dx)(\int_{0}^{1}\varphi_{z}(x)\varphi_{2}^{2}(x)dx)-1\right).$

Moreover,

$\displaystyle\int_{0}^{1}\varphi^{2}_{2}(x)\varphi_{3}(x)dx=\int_{0}^{1}\varphi_{4}(x)\varphi_{3}(x)\varphi_{2}(x)dx=\int_{0}^{1}\varphi^{2}_{3}(x)\varphi^{2}_{2}(x)dx=0,$
$\displaystyle\int_{0}^{1}\varphi^{2}_{3}(x)\varphi_{1}(x)dx=\int_{0}^{1}\varphi^{2}_{3}(x)\varphi_{4}(x)dx=0,\int_{0}^{1}\varphi^{2}_{3}(x)\varphi_{2}(x)dx=-\dfrac{1}{\sqrt{2}}.$

$\text{Theorefore,}\quad\Sigma_{2,3}=\Sigma_{3,2}=\dfrac{\mu_{1}}{2}-\lambda_{2}\mu_{1}=\mu_{1}\left(\dfrac{1}{2}-\lambda_{2}\right).$

$\displaystyle\bullet~{}\Sigma_{1,3}=\sum\limits_{z=1}^{4}\lambda_{z}(\int_{0}^{1}\varphi_{z}(x)\varphi_{3}(x)\varphi_{1}(x)dx)^{2}+\lambda_{1}\mu_{1}\left(2(\int_{0}^{1}\varphi^{2}_{3}(x)\varphi_{1}^{2}(x)dx-1)\right.$
$\displaystyle\left.+2\sum\limits_{z=1}^{4}\sum\limits_{m=2}^{\infty}\lambda_{z}^{m-1}(\int_{0}^{1}\varphi_{z}(x)\varphi_{3}^{2}(x)dx)(\int_{0}^{1}\varphi_{z}(x)\varphi_{1}^{2}(x)dx)-1\right).$

$\text{Since},\int_{0}^{1}\varphi^{2}_{3}(x)\varphi^{2}_{1}(x)dx-1=-\dfrac{1}{2},\int_{0}^{1}\varphi_{4}(x)\varphi_{3}(x)\varphi_{1}(x)dx=\dfrac{1}{\sqrt{2}},$
$\text{it follows
that}\quad\Sigma_{1,3}=\Sigma_{3,1}=\dfrac{\mu_{2}}{2}-2\lambda_{1}\mu_{1}.$

$\displaystyle\bullet~{}\Sigma_{1,4}=\sum\limits_{z=1}^{4}\lambda_{z}(\int_{0}^{1}\varphi_{z}(x)\varphi_{4}(x)\varphi_{1}(x)dx)^{2}+\lambda_{1}\mu_{2}\left(2(\int_{0}^{1}\varphi^{2}_{4}(x)\varphi_{1}^{2}(x)dx-1)\right.$
$\displaystyle\left.+2\sum\limits_{z=1}^{4}\sum\limits_{m=2}^{\infty}\lambda_{z}^{m-1}(\int_{0}^{1}\varphi_{z}(x)\varphi_{4}^{2}(x)dx)(\int_{0}^{1}\varphi_{z}(x)\varphi_{1}^{2}(x)dx)-1\right).$

$\displaystyle\text{
Moreover,}\int_{0}^{1}\varphi^{2}_{4}(x)\varphi^{2}_{1}(x)dx-1=\int_{0}^{1}\varphi^{2}_{4}(x)\varphi_{1}(x)dx=0,$
$\displaystyle\int_{0}^{1}\varphi_{4}(x)\varphi_{1}(x)\varphi_{3}(x)dx=\dfrac{1}{\sqrt{2}}.\quad\text{Thus,}\quad\Sigma_{1,4}=\Sigma_{4,1}=\dfrac{\mu_{1}}{2}-\lambda_{1}\mu_{2}.$
$\displaystyle\bullet~{}\Sigma_{2,4}$ $\displaystyle=$
$\displaystyle\sum\limits_{z=1}^{4}\lambda_{z}(\int_{0}^{1}\varphi_{z}(x)\varphi_{4}(x)\varphi_{2}(x)dx)^{2}+\lambda_{2}\mu_{2}\left(2(\int_{0}^{1}\varphi^{2}_{4}(x)\varphi_{2}^{2}(x)dx-1)\right.$
$\displaystyle\left.+2\sum\limits_{z=1}^{4}\sum\limits_{m=2}^{\infty}\lambda_{z}^{m-1}(\int_{0}^{1}\varphi_{z}(x)\varphi_{4}^{2}(x)dx)(\int_{0}^{1}\varphi_{z}(x)\varphi_{2}^{2}(x)dx)-1\right).$
$\displaystyle\text{So,}\int_{0}^{1}\varphi^{2}_{4}(x)\varphi^{2}_{2}(x)dx=-\dfrac{1}{2},\int_{0}^{1}\varphi^{2}_{4}(x)\varphi_{2}(x)dx=0,\Sigma_{2,4}=\Sigma_{4,2}=-2\lambda_{2}\mu_{2}.$

$\displaystyle\bullet~{}\Sigma_{3,4}$ $\displaystyle=$
$\displaystyle\sum\limits_{z=1}^{4}\lambda_{z}(\int_{0}^{1}\varphi_{z}(x)\varphi_{4}(x)\varphi_{3}(x)dx)^{2}+\mu_{1}\mu_{2}\left(2(\int_{0}^{1}\varphi^{2}_{4}(x)\varphi_{3}^{2}(x)dx-1)\right.$
$\displaystyle\left.+2\sum\limits_{z=1}^{4}\sum\limits_{m=2}^{\infty}\lambda_{z}^{m-1}(\int_{0}^{1}\varphi_{z}(x)\varphi_{4}^{2}(x)dx)(\int_{0}^{1}\varphi_{z}(x)\varphi_{3}^{2}(x)dx)-1\right).$
$\displaystyle\text{But,}\quad\int_{0}^{1}\varphi^{2}_{4}(x)\varphi^{2}_{3}(x)dx-1=\int_{0}^{1}\varphi^{2}_{4}(x)\varphi_{3}(x)dx=0,$
$\displaystyle\int_{0}^{1}\varphi^{2}_{4}(x)\varphi_{2}(x)dx=0,\int_{0}^{1}\varphi_{4}(x)\varphi_{1}(x)\varphi_{3}(x)dx=\dfrac{1}{\sqrt{2}}.$
$\text{Thus,}\quad\Sigma_{2,4}=\Sigma_{4,2}=\dfrac{\lambda_{1}}{2}-\mu_{1}\mu_{2}.$

For the Legendre copula:

$\displaystyle\int_{0}^{1}\varphi^{3}_{1}(x)dx)=\int_{0}^{1}(\sqrt{3}(2x-1))^{3}dx=0,$
$\displaystyle\int_{0}^{1}\varphi_{2}(x)\varphi_{1}^{2}(x)dx)=\int_{0}^{1}\sqrt{5}(6x^{2}-6x+1)(\sqrt{3}(2x-1))^{2}dx=\dfrac{2}{\sqrt{5}},$
$\displaystyle\int_{0}^{1}\varphi^{4}_{1}(x)dx-1=\int_{0}^{1}(\sqrt{3}(2x-1))^{4}dx-1=\dfrac{4}{5}.$
$\text{Thus,}\quad\sigma_{1}^{2}=1-\lambda_{1}^{2}+\dfrac{4}{5}\lambda_{2}+\dfrac{8}{5}\lambda_{1}^{2}+\dfrac{8}{5}\lambda_{1}^{2}\sum\limits_{m=2}^{\infty}\lambda_{2}^{m-1}=1+\dfrac{4}{5}\lambda_{2}+\lambda_{1}^{2}\dfrac{3+5\lambda_{2}}{5(1-\lambda_{2})}.$

$\displaystyle\text{Similarly,}\quad\int_{0}^{1}\varphi^{3}_{2}(x)dx)=\int_{0}^{1}(\sqrt{5}(6x^{2}-6x+1))^{3}dx=\dfrac{2\sqrt{5}}{7},$
$\displaystyle\int_{0}^{1}\varphi_{1}(x)\varphi_{2}^{2}(x)dx)=\int_{0}^{1}5\sqrt{3}(6x^{2}-6x+1)^{2}(2x-1)dx=0,$
$\displaystyle\int_{0}^{1}\varphi^{4}_{2}(x)dx-1=\int_{0}^{1}(\sqrt{5}(6x^{2}-6x+1))^{4}dx-1=\dfrac{8}{7},\quad\text{implies}$

$\displaystyle\sigma_{2}^{2}$ $\displaystyle=$ $\displaystyle
1-\lambda_{2}^{2}+\dfrac{20}{49}\lambda_{2}+\dfrac{16}{7}\lambda_{2}^{2}+\dfrac{40}{49}\lambda_{2}^{2}\sum\limits_{m=2}^{\infty}\lambda_{2}^{m-1}=1+\dfrac{20}{49}\lambda_{2}+\lambda_{2}^{2}\dfrac{63-23\lambda_{2}}{49(1-\lambda_{2})}.$

The correlation in this case is obtained as

$\displaystyle\Sigma_{1,2}$ $\displaystyle=$
$\displaystyle\sum\limits_{z=1}^{2}\lambda_{z}(\int_{0}^{1}\varphi_{z}(x)\varphi_{1}(x)\varphi_{2}(x)dx)^{2}+\lambda_{1}\lambda_{2}\left(2(\int_{0}^{1}\varphi^{2}_{1}(x)\varphi_{2}^{2}(x)dx-1)\right.$
$\displaystyle\left.+2\sum\limits_{z=1}^{2}\sum\limits_{m=2}^{\infty}\lambda_{z}^{m-1}(\int_{0}^{1}\varphi_{z}(x)\varphi_{1}^{2}(x)dx)(\int_{0}^{1}\varphi_{z}(x)\varphi_{2}^{2}(x)dx)-1\right).$

$\displaystyle\quad\text{But,}\int_{0}^{1}\varphi_{2}(x)\varphi_{1}^{2}(x)dx)=\dfrac{2}{\sqrt{5}},\int_{0}^{1}\varphi_{1}^{2}(x)\varphi^{2}_{2}(x)dx-1=\dfrac{4}{7}.$

$\displaystyle\text{So,}\quad\Sigma_{1,2}$ $\displaystyle=$
$\displaystyle\dfrac{4}{5}\lambda_{1}+\lambda_{1}\lambda_{2}\left(\dfrac{8}{7}-1+\dfrac{8}{7}\dfrac{\lambda_{2}}{(1-\lambda_{2})}\right)=\dfrac{4}{5}\lambda_{1}+\lambda_{1}\lambda_{2}\dfrac{1+7\lambda_{2}}{7(1-\lambda_{2})}.$

###  6.5 Proof of Theorem 4.2

Verification of conditions

  1. 1.

Considering the parameters space, the density of each of the copulas is
$c(u,v)\leq M<2$ inside the parameter space. Therefore, according to Longla et
al. (2022c), the Markov chains generated by such copulas are $\psi-$mixing.
Since $\psi$-mixing implies ergodicity, the first condition is satisfied.

  2. 2.

The second condition is satisfied as well since $c(u,v)>0$ on
$[0,1]\times[0,1]$ for any set of parameters not on the boundary of the
parameter set.

  3. 3.

Since $c(u,v)>0$ and is continuous on $[0,1]^{2}$, all partial derivatives
mentioned in the third condition exist and are continuous on $[0,1]^{2}$ when
$\Lambda$ is not on the boundary of the set.

  4. 4.

Since there is a constant $M<c(u,v)<2$ on $[0,1]^{2}$ when $\Lambda$ is an
interior point, then

$\mathbb{E}_{\Lambda}\left[\left(\frac{\partial}{\partial\lambda_{i}}\ln
c(u,v)\right)^{2}\right]<\infty.$

Via simple calculations, we establish that when the parameters are strictly
inside the region, we have for some $K$ not depending on $\Lambda$,

$\lvert\frac{\partial^{3}}{\partial\lambda_{i}\lambda_{j}\lambda_{k}}\ln
c(u,v)\lvert\leq
K\quad\text{and}\quad\mathbb{E}_{\Lambda}\left[\underset{\Lambda\in
A^{\prime}}{\sup}\lvert\frac{\partial^{3}}{\partial\lambda_{i}\lambda_{j}\lambda_{k}}\ln
c(u,v)\lvert\right]<\infty.$

  5. 5.

Since entries of the Fisher information matrix $\Sigma(\Lambda)$ cannot be
computed in closed form, an estimation procedure is used in Matlab to
approximate it. It appears to have a strictly positive determinant for each of
the cases, implying that it is inversible. Considering the extra regularity
condition, $n\Sigma$ is aproximated as an average by $I_{n}(\Lambda)$ because
$I_{n}/n$ converges in probability to $\Sigma$ thanks to ergodicity (see Sun
et al (2020)). Here,

$I_{n}(\Lambda)=-\left(\frac{\partial^{2}}{\partial\lambda_{k}\lambda_{j}}~{}\ell(\Lambda)\right)_{1\leq
k,j\leq s}.$

This completes the proof of the following statement. $\square$