# 2RV+HRV and Testing for Strong VS Full Dependence

Tiandong Wang
Shanghai Center for Mathematical Sciences, Fudan University
and
Sidney I. Resnick
School of Operations Research and Information Engineering, Cornell University
The first author gratefully acknowledges National Natural Science Foundation
of China Grant 12301660 and Science and Technology Commission of Shanghai
Municipality Grant 23JC1400700.

###### Abstract

Preferential attachment models of network growth are bivariate heavy tailed
models for in- and out-degree with limit measures which either concentrate on
a ray of positive slope from the origin or on all of the positive quadrant
depending on whether the model includes reciprocity or not. Concentration on
the ray is called full dependence. If there were a reliable way to distinguish
full dependence from not-full, we would have guidance about which model to
choose. This motivates investigating tests that distinguish between (i) full
dependence; (ii) strong dependence (support of the limit measure is a proper
subcone of the positive quadrant); (iii) weak dependence (limit measure
concentrates on positive quadrant). We give two test statistics, analyze their
asymptotically normal behavior under full and not-full dependence, and discuss
applicability using bootstrap methods applied to simulated and real data.

2RV+HRV and Testing for Strong VS Full Dependence

Keywords: Second order regular variation, hidden regular variation, hypothesis
test, asymptotic dependence.

##  1 Introduction

In multivariate heavy tail estimation, the support of the limit measure
provides information on the dependence structure of the random vector with the
heavy tail distribution ([28]). However, even in favorable circumstances in
$\mathbb{R}_{+}^{2}$, the positive quadrant in two dimensions, scatter or
diamond plots may have trouble distinguishing between

  * •

Full dependence where the limit measure concentrates on a ray of the form
$\\{(x,y)\in\mathbb{R}_{+}^{2}:y/x=m>0\\}$;

  * •

Strong dependence where  the support of the limit measure is a proper subcone
of $\mathbb{R}_{+}^{2}$ of the form
$\\{(x,y)\in\mathbb{R}_{+}^{2}:y/x\in[m_{l},m_{u}]\subsetneq[0,1]\\}$;

  * •

Weak dependence where the support of the limit measure is all of
$\mathbb{R}_{+}^{2}$; and

  * •

Asymptotic independence where the limit measure concentrates on the axes
$\mathbb{R}_{+}\times\\{0\\}\cup\\{0\\}\times\mathbb{R}_{+}$.

Estimation and visualization techniques that attempt to accurately distinguish
these cases encounter complications, the most glaring of which is the
requirement that data be thresholded according to the distance from the
origin. Plots can look rather different depending on the choice of threshold.
This is illustrated by diamond plots [11, 28] in Figure 1 of Exxon (XOM)
returns vs returns from Chevron (CVX) from January 04, 2016 to December 30,
2022. The data $\\{(x_{i},y_{i});1\leq i\leq 1761\\}$ is mapped to the
$L_{1}$-unit sphere via $(x,y)\mapsto(x,y)/(|x|+|y|)$ and then subsetted by
retaining only the points with $k$ largest values of $(|x|+|y|)$ where $k=100$
(left) or $k=500$ (right). Unsurprisingly, the two plots give different
impressions of where the limit measure concentrates.

Figure 1: Left: Diamond plot using the 100 points furthest from the origin.
Right: Diamond plot using 500 most remote points.

An automatic threshold selection technique is advocated in computer and
network science ([5, 38]) and implemented in [6] or [22]. This technique is
consistent ([2, 20]) and can increase ones’ comfort level with threshold
selection. However, this method offers no guarantee of best choice of
threshold and has the additional drawback of preventing tail estimators such
as Hill from being asymptotically normal ([20]). It would be desirable if
there were test statistics to guide us in choice of model from the bulleted
list above.

One reason for thinking about such problems was our interest in fitting
preferential attachment (PA) models of directed social networks to data
consisting of in- and out-degree of each node. The classical PA model of
directed edge growth ([25, 3]) when standardized to equal tail indices for
each component gives a heavy tail model with limit measure concentrating on
all of $\mathbb{R}_{+}^{2}$ ([35]). However, for reasonable ranges of model
parameters, these models do not correctly predict empirically observed values
for the reciprocity coefficient ([27]); this is discussed in [39, 40, 4, 41].
Adding the reciprocity feature to the theoretical model means predicted values
of the reciprocity coefficient can match empirical values. However, unlike the
classical model, this heavy tailed model with reciprocity has a limit measure
that concentrates on a ray. If there were a statistical test for full
dependence, it would provide guidance on whether one needs to add the
reciprocity feature to the model to obtain a satisfactory statistical fit for
social network data.

Of course network data or financial returns are not the same as iid
observations but this paper starts with the simple case and assumes all
observations come from a heavy tailed iid model by repeated sampling.

We give two test statistics which help distinguish full vs not-full asymptotic
dependence and show the statistics are asymptotically normal but with
different asymptotic variances, depending on the case. A somewhat novel aspect
of our approach is that hidden regular variation (HRV) arises naturally and is
employed in our proofs. The reason HRV is relevant is that to get asymptotic
normality with a constant centering for estimators of heavy tailed data
requires not only the regular variation assumption for the underlying
distribution but also second order regular variation (2RV) which controls
deviations between a finite sample mean and an asymptotic mean; this is
discussed at length in [15, 34, 17, 13, 31]. In the context of two dimensional
data, it is natural to expect we need a two-dimensional second order regular
variation assumption ([16, 14, 33, 12]) and this coupled with multivariate
regular variation with a limit measure concentrating on a proper subset of the
state space lead naturally to hidden regular variation. This is explained
further in Section 2.

We propose test statistics that offer guidance on appropriateness of the
different cases and give conditions under which the statistics are
asymptotically normal, paying attention to the centering and asymptotic
variance. The proposed hypothesis testing framework is  discussed further in
Section 5.

##  2 Multivariate, Hidden and Second Order Regular Variation

Here is a review of notation and concepts necessary for formulating and
proving the results. We particularize the concept of regular variation of
measures on a complete, separable metric space $\mathbb{X}$ for the case of
$\mathbb{X}=\mathbb{R}^{2}_{+}$ where visualization is easiest ([29, 24, 10,
26, 1]).

Suppose $\\{(X_{i},Y_{i});1\leq i\leq n\\}$ are iid random elements of
$\mathbb{R}_{+}^{2}$ and based on evidence provided by observing these
vectors, we need to analyze the asymptotic dependence structure of the
components. We do this in the context of regular variation of measures.

###  2.1 Multivariate regular variation.

We begin with the concept of $\mathbb{M}$-convergence.

######  Definition 2.1.

For a closed subcone $\mathbb{C}$ of $\mathbb{X}$, let
$\mathbb{M}(\mathbb{X}\setminus\mathbb{C})$ be the set of Borel measures on
$\mathbb{X}\setminus\mathbb{C}$ which are finite on sets bounded away from
$\mathbb{C}$, and $\mathcal{C}(\mathbb{X}\setminus\mathbb{C})$ be the set of
continuous, bounded, non-negative functions on $\mathbb{X}\setminus\mathbb{C}$
whose supports are bounded away from $\mathbb{C}$. Then for
$\mu_{n},\mu\in\mathbb{M}(\mathbb{X}\setminus\mathbb{C})$, we say
$\mu_{n}\to\mu$ in $\mathbb{M}(\mathbb{X}\setminus\mathbb{C})$, if $\int
f\mathrm{d}\mu_{n}\to\int f\mathrm{d}\mu$ for all
$f\in\mathcal{C}(\mathbb{X}\setminus\mathbb{C})$.

Without loss of generality ([29]), we can take functions in
$\mathcal{C}(\mathbb{X}\setminus\mathbb{C})$ to be uniformly continuous. The
modulus of continuity of a uniformly continuous function
$f:\mathbb{R}_{+}^{p}\mapsto\mathbb{R}_{+}$ is

$\Delta_{f}(\delta)=\sup\\{|f(\boldsymbol{x})-f(\boldsymbol{y})|:d(\boldsymbol{x},\boldsymbol{y})<\delta\\}$
(1)

where $d(\cdot,\cdot)$ is an appropriate metric on the domain of $f$. Uniform
continuity means $\lim_{\delta\to 0}\Delta_{f}(\delta)=0.$

Denote by $RV_{c}$, the class of regularly varying functions
$f:\mathbb{R}_{+}\mapsto\mathbb{R}_{+}$ with index $c\in\mathbb{R}$ and write
$f\in RV_{c}$. The formal definition of multivariate regular variation (MRV)
of distributions for the classical case $\mathbb{X}=\mathbb{R}_{+}^{2}$ and
$\mathbb{C}=\\{\boldsymbol{0}\\}$ is next.

######  Definition 2.2.

The distribution $\mathbb{P}[(X_{1},Y_{1})\in\cdot\,]$ of a random vector
$(X_{1},Y_{1})$ on $\mathbb{R}_{+}^{2}$, is (standard) regularly varying on
$\mathbb{R}_{+}^{2}\setminus\\{\boldsymbol{0}\\}$ with index $\alpha>0$ if
there exists some regularly varying scaling function
$b(t)\in\text{RV}_{1/\alpha}$ and a limit measure
$\eta(\cdot)\in\mathbb{M}(\mathbb{R}_{+}^{2}\setminus\\{\boldsymbol{0}\\})$
such that as $t\to\infty$,

$t\mathbb{P}[(X_{1},Y_{1})/b(t)\in\cdot\,]\rightarrow\eta(\cdot),\qquad\text{in
}\mathbb{M}(\mathbb{R}_{+}^{2}\setminus\\{\boldsymbol{0}\\}).$ (2)

It is convenient to write
$\mathbb{P}[(X_{1},Y_{1})\in\cdot\,]\in\text{MRV}(\alpha,b(t),\eta,\mathbb{R}_{+}^{2}\setminus\\{\boldsymbol{0}\\})$.

####  2.1.1 Cartesian to polar and back.

When analyzing the asymptotic dependence between components of a bivariate
random vector $\boldsymbol{Z}$ satisfying (2), it is often informative to make
a polar coordinate transform and consider the transformed points located on
the $L_{1}$-unit sphere

$\displaystyle(x,y)\mapsto\left(\frac{x}{x+y},\frac{y}{x+y}\right),$ (3)

after thresholding the data according to the $L_{1}$ norm. The plot of such
data is the (positive-quadrant) diamond plot and Figure 1 is the 4-quadrant
version.  In $\mathbb{R}_{+}^{2}$, the convenient version of the $L_{1}$-polar
coordinate transformation is
$T:\mathbb{R}_{+}^{2}\setminus\\{\boldsymbol{0}\\}\mapsto(\mathbb{R}_{+}\setminus\\{0\\})\times[0,1],$
defined by

$T(x,y)=\bigl{(}x+y,{\color[rgb]{0,0,0}x/(x+y)\bigr{)}=(r,\theta)}.$

The inverse transformation from polar to Cartesian coordinates is
$(r,\theta)\mapsto(r\theta,r(1-\theta)).$ The map $T$ disintegrates
$\eta(\cdot)$ into the product measure

$\eta\circ T^{-1}(\cdot)=\nu_{\alpha}\times S(\cdot),$

where $S(\cdot)$ can be taken to be a probability measure on $[0,1]$ called
the angular measure, and $\nu_{\alpha}(\cdot)$ is the Pareto measure with
$\nu_{\alpha}(x,\infty)=x^{-\alpha},\,x>0$.

###  2.2 Hidden regular variation.

Denote by $\mathbb{C}_{a,b}$ the subcone of $\mathbb{R}_{+}^{2}$ such that

$\mathbb{C}_{a,b}=\\{(x,y)\in\mathbb{R}_{+}^{2}:\theta=x/(x+y)\in[a,b]\\},\quad
0\leq a\leq b\leq 1.$

When the limit measure of regular variation $\eta(\cdot)$ concentrates on a
proper subcone $\mathbb{C}_{a,b}\subset\mathbb{X}=\mathbb{R}_{+}^{2}$ of the
full state space, we may improve estimates of probabilities in the complement
of the subcone, if there is a second hidden regular variation regime after
removing the subcone.

######  Definition 2.3.

The random vector $\boldsymbol{Z}$ in $\mathbb{R}^{2}_{+}$ has a distribution
that is regularly varying on $\mathbb{R}^{2}_{+}\setminus\\{\boldsymbol{0}\\}$
and has hidden regular variation on
$\mathbb{R}^{2}_{+}\setminus\mathbb{C}_{a,b}$ if there exist
$0<\alpha\leq\alpha_{0}$, scaling functions $b(t)\in RV_{1/\alpha}$ and
$b_{0}(t)\in RV_{1/\alpha_{0}}$ with $b(t)/b_{0}(t)\to\infty$ and limit
measures $\eta$ concentrating on $\mathbb{C}_{a,b}$ and another limit measure
$\eta_{0}$, such that

$\displaystyle\mathbb{P}(\boldsymbol{Z}\in\cdot)\in\text{MRV}(\alpha,b(t),\eta,\mathbb{R}_{+}^{2}\setminus\\{\boldsymbol{0}\\})\cap\text{MRV}(\alpha_{0},b_{0}(t),\eta_{0},\mathbb{R}_{+}^{2}\setminus\mathbb{C}_{a,b}).$
(4)

It is sometimes useful to characterize HRV is through the _generalized polar
coordinate transformation_ for $\mathbb{R}_{+}^{2}\setminus\mathbb{C}_{a,b}$,
assuming use of an associated metric $d(\cdot,\cdot)$ satisfying
$d(cx,cy)=cd(x,y)$ for scalars $c>0$. The metric $d(\cdot,\cdot)$ that we use
in practice is  a scaled $L_{1}$-metric. When using generalized polar
coordinates with respect to the forbidden zone $\mathbb{C}_{a,b}$, we define
$\aleph_{\mathbb{C}_{a,b}}:=\\{\boldsymbol{x}\in\mathbb{R}_{+}^{2}\setminus\mathbb{C}_{a,b}:d(\boldsymbol{x},\mathbb{C}_{a,b})=1\\}$,
the locus of points at distance 1 from $\mathbb{C}_{a,b}$. Then the
generalized polar coordinates are specified through the transformation,
$\text{GPOLAR}:\mathbb{R}_{+}^{2}\setminus\mathbb{C}_{a,b}\mapsto(0,\infty)\times\aleph_{\mathbb{C}_{a,b}}$
with

$\displaystyle\text{GPOLAR}(\boldsymbol{x})=\left(d(\boldsymbol{x},\mathbb{C}_{a,b}),\frac{\boldsymbol{x}}{d(\boldsymbol{x},\mathbb{C}_{a,b})}\right).$
(5)

For a probability measure $S_{0}(\cdot)$ on $\aleph_{\mathbb{C}_{a,b}}$, the
generalized polar coordinates allow re-writing the second part of (4) as

$t\mathbb{P}\left[\left(\frac{d(\boldsymbol{Z},\mathbb{C}_{a,b})}{b_{0}(t)},\frac{\boldsymbol{Z}}{d(\boldsymbol{Z},\mathbb{C}_{a,b})}\right)\in\cdot\right]\to(\nu_{\alpha_{0}}\times
S_{0})(\cdot)$

in
$\mathbb{M}\bigl{(}(\mathbb{R}_{+}\setminus\\{0\\})\times\aleph_{\mathbb{C}_{a,b}}\bigr{)}$.
In particular $\mathbb{P}[d(\boldsymbol{Z},\mathbb{C}_{a,b})>x]\in
RV_{-\alpha_{0}}$ is a lighter tail than $\mathbb{P}[\|\boldsymbol{Z}\|>x]\in
RV_{-\alpha}$. See [10] and [29] for details.

###  2.3 Second order regular variation.

In one dimension, second order regular variation (2RV) controls deviation of
finite sample means from asymptotic means and allows a useful asymptotic
normality for estimators such as the Hill estimator. Our test statistics are
derived from two dimensional tail empirical measures and it is reasonable to
expect, therefore, that asymptotic normality requires two dimensional second
order regular variation conditions.

####  2.3.1 The second order condition.

There are several ways to state this condition which strengthens multivariate
regular variation. The first uses $\mathbb{M}$-convergence. We need a function
$A\in RV_{-\rho}$, $\rho>0$, and a signed measure $\chi(\cdot)$ which is not
identically 0 and is the difference of two measures in
$\mathbb{M}\bigl{(}(\mathbb{R}_{+}\setminus\\{0\\})\times[0,1]\bigr{)}$, such
that in
$\mathbb{M}\bigl{(}(\mathbb{R}_{+}\setminus\\{0\\})\times[0,1]\bigr{)}$,

$\displaystyle\frac{1}{A(t)}\Biggl{(}t\mathbb{P}\Bigl{(}(R/b(t),\Theta)\in\cdot\Bigr{)}-\nu_{\alpha}\times
S(\cdot)\Biggr{)}\to\chi(\cdot),$ (6)

meaning that evaluation of the signed measure on the left at a function
$f\in\mathcal{C}(\bigl{(}(\mathbb{R}_{+}\setminus\\{0\\})\times[0,1]\bigr{)}$
converges to the evaluation $\chi(f)$; or in symbols

$\frac{1}{A(t)}\Biggl{(}t\mathbb{E}f\bigl{(}R/b(t),\Theta\bigr{)}-\iint_{\mathbb{R}_{+}\setminus\\{0\\}\times[0,1]}f(r,\theta)\nu_{\alpha}(dr)S(d\theta)\Biggr{)}\to\chi(f).$
(7)

The second way to phrase condition (6) which looks more like convergence of
distribution functions is

$\frac{1}{A(t)}\Biggl{(}t\mathbb{P}\Bigl{(}\frac{R}{b(t)}>r,\Theta\leq\theta\Bigr{)}-r^{-\alpha}S[0,\theta]\Biggr{)}\to\chi\bigl{(}(r,\infty)\times[0,\theta]\bigr{)}$
(8)

locally uniform in $r\in(0,\infty)$ for each $\theta\in[0,1]$ where the limit
is specified before (6).

If $f_{1}(r)\in\mathbb{M}(\mathbb{R}_{+}\setminus\\{0\\})$, set
$f(r,\theta):=f_{1}(r)\theta\in\mathbb{M}\bigl{(}(\mathbb{R}_{+}\setminus\\{0\\})\times[0,1]\bigr{)}$
and inserting this into (7) gives

$\frac{1}{A(t)}\Biggl{(}t\mathbb{E}\Theta
f_{1}\bigl{(}R/b(t)\bigr{)}-\int_{[0,1]}\theta
S(d\theta)\nu_{\alpha}(f_{1})\Biggr{)}\to\int_{(0,\infty)\times[0,1]}\theta
f_{1}(r)\chi(dr,d\theta).$ (9)

or in convergence of signed measures formulation,

$\frac{1}{A(t)}\Biggl{(}t\mathbb{E}\Theta\epsilon_{R/b(t)}(\cdot)-\int_{[0,1]}\theta
S(d\theta)\nu_{\alpha}(\cdot)\Biggr{)}\to\iint_{(\cdot)\times[0,1]}\theta\chi(dr,d\theta).$
(10)

Note that (6) and (8) are formulated so they can be marginalized and therefore
the regularly varying distribution of $R$ is 2RV in one dimension. Also, the
second order condition allows (10) and (9), so controls the expectation of
$\Theta$ on the set where $R$ is large. If we set

$v(t)=E\Theta_{1}\boldsymbol{1}_{[R_{1}>t]},\quad\mu_{S}=\int_{[0,1]}\theta
S(d\theta),$

then (10) gives as $t\to\infty$,

$\frac{tv(b(t)x)-\mu_{S}x^{-\alpha}}{A(t)}\to
h(x):=\iint_{((x,\infty))\times[0,1]}\theta\chi(dr,d\theta).$ (11)

which leads to the more traditional form of the 2RV condition for $v(t)$,
namely

$\lim_{s\to\infty}\frac{\frac{v(sx)}{v(s)}-x^{-\alpha}}{A\circ
b^{\leftarrow}(s)}=h(x)/\mu_{S},$ (12)

where $A\circ b^{\leftarrow}\in RV_{-\rho\alpha}$ and the limit function
$h(x)$ must be of the form ([19, 31, 15]),

$h(x)=cx^{-\alpha}\Bigl{(}\frac{1-x^{-\rho\alpha}}{\rho\alpha}\Bigr{)},\quad
x>0,\,c\neq 0.$

####  2.3.2 2RV and HRV

We discuss why the second order condition (6) together with the assumption
$S([a,b])=1$ for $[a,b]\subsetneq[0,1]$ implies HRV. The essentials of the
argument in the context of asymptotic independence are in ([14, 33]).

######  Theorem 2.1.

Assume the 2RV condition (6) or (8) hold and $S([a,b])=1$ for
$[a,b]\subsetneq[0,1]$. Set $U(t)=t/A(t)\in RV_{1+\rho}$, so that
$U^{\leftarrow}(t)\in RV_{1/(1+\rho)}$ and therefore

$b_{0}(t):=b\circ U^{\leftarrow}(t)\in RV_{1/(\alpha(1+\rho))},\,\rho>0.$ (13)

Then provided $\chi(\cdot)$ is not identically 0 on
$(0,\infty)\times([0,1]\setminus[a,b])$,

$\displaystyle\mathbb{P}\left[\left(R,\Theta\right)\in\cdot\right]\in\text{MRV}(\alpha(1+\rho),b_{0}(t),\chi(\cdot),(\mathbb{R}_{+}\setminus\\{0\\})\times([0,1]\setminus[a,b])).$
(14)

###### Proof.

For $r>0$, $I\subset[0,1]$ such that $\chi(\partial((r,\infty)\times I))=0$,

$\displaystyle\chi((r,\infty)\times I)=$
$\displaystyle\lim_{t\to\infty}\frac{t\mathbb{P}[R/b(t)>r,\Theta\in
I]-r^{-\alpha}S(I)}{A(t)}$ and if $S(I)=0$, this is $\displaystyle=$
$\displaystyle\lim_{t\to\infty}\frac{t\mathbb{P}[R/b(t)>r,\Theta\in I]}{A(t)}$

Set $U(t):=t/A(t)$, $b_{0}:=b\circ U^{\leftarrow}$ and after a change of
variable, the proof of (14) is complete. ∎

##  3 Testing the existence of strong dependence

For strong convergence, we assume that $0\leq a\leq b\leq 1$ fixed, with
$[a,b]\subsetneq[0,1]$ and $S([a,b])=1.$ The condition
$\theta=x/(x+y)\in[a,b]$ translates to

$(x,y)\in\\{(u,v)\in\mathbb{R}_{+}^{2}:v/u\in[b^{-1}-1,a^{-1}-1]\\}=:\mathbb{C}_{a,b}.$

So the closed cone $\mathbb{C}_{a,b}$ is the set of first quadrant points
between the two rays $y=m_{u}x$ and $y=m_{l}x$, $x>0$, where the slopes are
$m_{u}=a^{-1}-1$, $m_{l}=b^{-1}-1$ and since $a\leq b$, we have $m_{u}\geq
m_{l}$. Define the scaled distance from $(x,y)\in\mathbb{R}_{+}^{2}$ to
$\mathbb{C}_{a,b}$ as

$d^{*}((x,y),\mathbb{C}_{a,b}):=\max\left\\{(b^{-1}-1)x-y,y-(a^{-1}-1)x,0\right\\}.$
(15)

Note

  1. 1.

when $(x,y)$ is above cone $\mathbb{C}_{a,b}$ so that $y/x>m_{u}$ and thus
$y>(a^{-1}-1)x$, $d^{*}((x,y),\mathbb{C}_{a,b})=y-(a^{-1}-1)x;$

  2. 2.

when $(x,y)$ is below the cone $\mathbb{C}_{a,b}$ so that $y/x<m_{l}$ and
$y<(b^{-1}-1)x$, $d^{*}((x,y),\mathbb{C}_{a,b})=(b^{-1}-1)x-y;$

  3. 3.

when $(x,y)\in\mathbb{C}_{a,b}$, $d^{*}((x,y),\mathbb{C}_{a,b})=0.$

  4. 4.

when $\mathbb{C}_{a,b}=\\{(x,y):y=(\theta_{0}^{-1}-1)x,x>0\\}$ because
$S\\{\theta_{0}\\}=1$, then
$d^{*}((x,y),\mathbb{C}_{a,b})=|(\theta_{0}^{-1}-1)x-y|.$

Using generalized polar coordinates, the HRV assumption on
$\mathbb{R}_{+}^{2}\setminus\mathbb{C}_{a,b}$ reads

$t\mathbb{P}\left(\left(\frac{d^{*}\bigl{(}(X,Y),\mathbb{C}_{a,b}\bigr{)}}{b_{0}(t)},\frac{(X,Y)}{d^{*}((X,Y),\mathbb{C}_{a,b})}\right)\in\cdot\right)\longrightarrow\nu_{\alpha_{0}}\times
S_{0}(\cdot)$

in
$\mathbb{M}((\mathbb{R}_{+}\setminus\\{0\\})\times\aleph_{\mathbb{C}_{a,b}})$
and in particular $P[d^{*}\bigl{(}(X,Y),\mathbb{C}_{a,b}\bigr{)}>x]\in
RV_{-\alpha_{0}}$ and assuming 2RV from the previous section,
$\alpha_{0}=\alpha(1+\rho)$.

Let $\\{(X_{i},Y_{i}):i\geq 1\\}$ be iid copies of $(X,Y)$, and set
$R_{i}:=X_{i}+Y_{i}$. We also define $(X_{i}^{*},Y_{i}^{*})$ to be the pair of
random variables such that $X_{i}^{*}+Y_{i}^{*}$ is the $i$-th largest order
statistic of $\\{R_{i}:1\leq i\leq n\\}$, i.e. $R_{(i)}$. Consider the
following hypotheses: for fixed and known $0<a\leq b<1$,

$\displaystyle H^{(1)}_{0}:\,S([a,b])=1,\qquad H^{(1)}_{a}:\,S([a,b])<1.$ (16)

In Theorem 3.1, we propose a test statistic for the test in (16).

######  Theorem 3.1.

Assume the 2RV condition (8) holds, $\alpha_{0}\equiv\alpha(1+\rho)>1$ and
$b_{0}(t)$ is defined in (13). Define the statistic

$\displaystyle D^{*}_{n}:=$
$\displaystyle\frac{1}{{k_{n}}}\sum_{i=1}^{k_{n}}\left(1+\frac{d^{*}\bigl{(}(X^{*}_{i},Y^{*}_{i}),\mathbb{C}_{a,b}\bigr{)}}{R_{(k_{n})}}\right)\log\frac{R_{(i)}}{R_{(k_{n})}}$
$\displaystyle=$ $\displaystyle
H_{k,n}+\frac{1}{{k_{n}}}\sum_{i=1}^{k_{n}}\left(\frac{d^{*}\bigl{(}(X^{*}_{i},Y^{*}_{i}),\mathbb{C}_{a,b}\bigr{)}}{R_{(k_{n})}}\right)\log\frac{R_{(i)}}{R_{(k_{n})}}$
(17)

where $H_{k_{n},n}$ is the Hill estimator of $1/\alpha$ applied to
$\\{R_{i},1\leq i\leq n\\}$ based on $k_{n}$ upper order statistics, and
$\\{k_{n}\\}$ is an intermediate sequence  (i.e. $k_{n}\to\infty$,
$n/k_{n}\to\infty,$ $n\to\infty$) satisfying

$\displaystyle\sqrt{k_{n}}\frac{b_{0}(n/k_{n})}{b(n/k_{n})}\to 0,\qquad
n\to\infty.$ (18)

Under $H^{(1)}_{0}$ as given in (16), we have

$\sqrt{k_{n}}(D^{*}_{n}-1/\alpha)\Rightarrow\frac{1}{\alpha}N(0,1).$ (19)

The proof of Theorem 3.1 is in Section 1 of the supplement; here we give some
remarks:

  1. 1.

Of course, $D_{n}^{*}$ depends on $a,b$ but this dependence is suppressed in
the notation. A consistent estimator of $a,b$ is suggested in Section 5.1.1
but for now we assume $a,b$ are fixed and known.

  2. 2.

Under $H^{(1)}_{0}$, for $(X_{i},Y_{i})$ such that $R_{i}$ is large, the
distance from $(X_{i},Y_{i})$ to $\mathbb{C}_{a,b}$ should be small with high
probability and therefore $D_{n}^{*}$ should be close to the Hill estimator
which is asymptotically normal. The proof of Theorem 3.1 shows that when
$S[a,b]=1$,

$\sqrt{k_{n}}(D_{n}^{*}-H_{k_{n},n})\Rightarrow 0,\quad(n\to\infty).$ (20)

  3. 3.

The condition $\alpha_{0}=\alpha(1+\rho)>1$ is mild as it is rare in practice
for tails to be so heavy that $\alpha<1$.

  4. 4.

The proof is based on asymptotic normality of the tail empirical measure. For
treatments explaining the need for the second order condition, see [34,
Section 9.1] or [15] and for background [36, 7, 16, 18, 21, 8, 9, 30].

  5. 5.

Theorem 3.1 suggests that for fixed $a,b$, we reject $H^{(1)}_{0}$ in (16) if
$|D_{n}^{*}-1/\alpha|>1.96/(\alpha/\sqrt{k_{n}})$. If we choose too wide an
interval $[a,b]\subsetneq[0,1]$, then the test statistic $D_{n}^{*}$ becomes
closer to $H_{k,n}$ as more data points are included in $\mathbb{C}_{a,b}$.
Failure to reject for the fixed interval means also that one fails to reject
for any bigger interval. So using only $D_{n}^{*}$, we cannot distinguish
whether the support of $S(\cdot)$ is in $[a,b]$ or a subset of $[a,b]$ and, in
particular, if we fail to reject $H_{0}^{(1)}$, it could be the support is
$\\{\theta_{0}\\}$ for some $\theta_{0}\in[a,b]$. Therefore, in the next
section, we give another test statistic that helps decide whether $(X,Y)$ is
asymptotically fully or strongly dependent.

  6. 6.

If $[a,b]=[0,1]$, then $\mathbb{C}_{a,b}=\mathbb{R}_{+}^{2}$ and
$d^{*}\bigl{(}(X^{*}_{i},Y^{*}_{i}),\mathbb{C}_{a,b}\bigr{)}=0$ so
$D_{n}^{*}=H_{k_{n},n}$ and (19) still holds  without any restriction on
$\\{k_{n}\\}$ beyond it being an intermediate sequence.

##  4 Full vs strong dependence

Consider the following hypothesis test:

$\displaystyle H^{(2)}_{0}:S(\\{\theta_{0}\\})=1\qquad
H^{(2)}_{a}:S([a,b])=1.$ (21)

where $\theta_{0}\in[a,b]$, and to capitalize on hidden regular variation
resulting from 2RV, we need the assumption that $[a,b]\subsetneq[0,1]$ is a
proper subset of $[0,1].$ Since $\theta_{0}\in[a,b]$ and $D_{n}^{*}$ given in
Theorem 3.1 is unable to distinguish between the two hypotheses in (21), we
now propose another test statistic. Let $\Theta_{i}^{*}$ be the concomitant of
$R_{(i)}$, and define

$\displaystyle
T_{n}:=\frac{\sum_{i=1}^{k_{n}}\Theta_{i}^{*}\log\frac{R_{(i)}}{R_{(k_{n})}}}{\sum_{i=1}^{k_{n}}\Theta_{i}^{*}}.$
(22)

The next results  recommend we distinguish between strong and full dependence
by assessing the asymptotic variance of $T_{n}$.  The methodology is discussed
more fully in the next Section 5. Under $H^{(2)}_{0}$ the asymptotic variance
of $T_{n}$ is $1/\alpha^{2}$ but under $H^{(2)}_{a}$ the asymptotic variance
is strictly greater than $1/\alpha^{2}$.

###  4.1 Full dependence

We begin by discussing a limit theorem that can aid in distinguishing between
full and strong dependence. This theorem is posed under the assumption
$H^{(2)}_{0}$ in (21) that full dependence holds with the limit angular
measure concentrating at a point $\theta_{0}\in(0,1)$. The proof machinery is
similar to that later in Theorem 4.2, with details included in the supplement.

######  Theorem 4.1.

Assume $H^{(2)}_{0}$ holds and the angular measure
$S(\cdot)=\epsilon_{\theta_{0}}(\cdot),\,\theta_{0}\in(0,1)$. Suppose the 2RV
condition in (6) holds with  $A(t)\in RV_{-\rho}$, $\rho>0$. Define
$b_{0}(t)$ as in (13) so $b_{0}(t)\in RV_{1/(\alpha(1+\rho))}$ and
$\alpha_{0}=\alpha(1+\rho)$. Let $\\{k_{n}\\}$ be an intermediate sequence
satisfying (18).Then for $W(\cdot)$ a standard Brownian Motion we have

$\displaystyle\sqrt{k_{n}}\left(T_{n}-\frac{1}{\alpha}\right)$
$\displaystyle\Rightarrow\frac{1}{\alpha}\left(\int_{0}^{1}W(s)\frac{\mathrm{d}s}{s}-W(1)\right){\color[rgb]{0,0,0}\stackrel{{\scriptstyle
d}}{{=}}\frac{1}{\alpha}W(1)\sim N(0,1/\alpha^{2}).}$ (23)

###  4.2 Strong dependence

Theorem 4.2 suggests identifying strong dependence in $H^{(2)}_{a}$ if the
asymptotic variance of $T_{n}$ is bigger than $1/\alpha^{2}$.

######  Theorem 4.2.

Consider the hypothesis test in (21)  with the assumption $H_{a}^{(2)}$, that
is, $S([a,b])=1.$ Suppose the 2RV condition in (6) holds with a limiting
signed measure $\chi(\cdot)$ and  $A(t)\in RV_{-\rho}$, $\rho>0$. Define
$b_{0}(t)$ as in (13), so $b_{0}(t)\in RV_{1/(\alpha(1+\rho))}$ and
$\alpha_{0}=\alpha(1+\rho)$. As before, $\\{k_{n}\\}$ is an intermediate
sequence satisfying (18). Define

$\displaystyle\mu$
$\displaystyle:=\int_{a}^{b}xS(\mathrm{d}x),\qquad\sigma^{2}:=\int_{a}^{b}(x-\mu)^{2}S(\mathrm{d}x),$

and under strong dependence assumption $H^{(2)}_{a}$, we have

$\displaystyle\sqrt{k_{n}}\left(T_{n}-\frac{1}{\alpha}\right)$
$\displaystyle\Rightarrow\frac{(1+\sigma^{2}/\mu^{2})^{1/2}}{\alpha}\left(\int_{0}^{1}\frac{W(s)}{s}\mathrm{d}s-W(1)\right)$
$\displaystyle\stackrel{{\scriptstyle
d}}{{=}}\frac{(1+\sigma^{2}/\mu^{2})^{1/2}}{\alpha}W(1)\sim
N\bigl{(}0,\frac{1}{\alpha^{2}}(1+\sigma^{2}/\mu^{2})\bigr{)}.$ (24)

The proof of Theorem 4.2 requires a functional central limit theorem for row
sums of a triangular array of $\mathbb{D}[0,1]$-functions ([32, Theorem 10.6])
that generalizes the sequential result of [23]. We give the formal proof for
results in Theorem 4.2 since it showcases the key proof steps for Theorem 4.1
as well.

###### Proof.

Proceed by steps.
(1) First, employ the functional central limit theorem given in Theorem 4.1 of
the supplement to show that in $D(0,1]$,

$\displaystyle\frac{\sqrt{k_{n}}}{(1+\sigma^{2}/\mu^{2})^{1/2}}\left(\frac{1}{\mu
k_{n}}\sum_{i=1}^{n}\Theta_{i}\epsilon_{R_{i}/b(n/k_{n})}(t^{-1/\alpha},\infty)-t\right)\Rightarrow
W(t),$ (25)

where $W(\cdot)$ is a standard Brownian motion.

We check all conditions in Theorem 4.1 of the supplement (details deferred to
Section 4 of the supplement), and draw the conclusion that in $D(0,1]$,

$\displaystyle\frac{(\mu^{2}+\sigma^{2})^{-1/2}}{\sqrt{k_{n}}}\sum_{i=1}^{n}\left(\Theta_{i}\epsilon_{\frac{R_{i}}{b(n/k_{n})}}(t^{-1/\alpha},\infty)-\mathbb{E}\bigl{(}\Theta_{i}\epsilon_{\frac{R_{i}}{b(n/k_{n})}}(t^{-1/\alpha},\infty)\bigr{)}\right)$
$\displaystyle\Rightarrow W(t).$ (26)

Note that by 2RV using (9) or (10) plus the marginalized version for the
distribution of $R_{1}$, we have

$\displaystyle\sqrt{k_{n}}\left(\frac{n}{k_{n}}\mathbb{E}\left(\Theta_{1}\boldsymbol{1}_{\\{R_{1}>b(n/k_{n})y\\}}\right)-\mu\frac{n}{k_{n}}\mathbb{P}(R_{1}>b(n/k_{n})y)\right)\to
0$ (27)

locally uniformly for $y>0$ and for $\\{k_{n}\\}$ satisfying (18). Combining
(27) with (26) then completes the proof of (25).
(2) Applying the composition map $(x(t),c)\mapsto x(ct)$ from
$D(0,\infty)\times(0,\infty)\mapsto D(0,\infty)$, we get in $D(0,\infty)$,

$\displaystyle\frac{\sqrt{k_{n}}}{(1+\sigma^{2}/\mu^{2})^{1/2}}\left(\frac{1}{\mu
k_{n}}\sum_{i=1}^{n}\Theta_{i}\epsilon_{R_{i}/R_{(k_{n})}}(y,\infty)-\left(y\frac{R_{(k_{n})}}{b(n/k_{n})}\right)^{-\alpha}\right)\Rightarrow
W(y^{-\alpha}).$ (28)

Repeating a similar argument as in Step 2 of the proof for Theorem 4.1
(details included in Section 3 of the supplement), we are able to justify the
application of

$x\mapsto\int_{1}^{\infty}\frac{x(s)}{s}\mathrm{d}s,$

which further leads to

$\displaystyle(\mu^{2}+\sigma^{2})^{-1/2}\sqrt{k_{n}}$
$\displaystyle\left(\frac{1}{\mu
k_{n}}\sum_{i=1}^{k_{n}}\Theta_{i}^{*}\log\frac{R_{(i)}}{R_{(k_{n})}}-\frac{1}{\alpha}\left(\frac{R_{(k_{n})}}{b(n/k_{n})}\right)^{-\alpha}\right)$
$\displaystyle\Rightarrow\frac{1}{\alpha}\int_{0}^{1}\frac{W(s)}{s}\mathrm{d}s.$
(29)

(3) We are left to justify the convergence of

$\displaystyle\sqrt{k_{n}}(T_{n}-1/\alpha)$
$\displaystyle-\sqrt{k_{n}}\left(\frac{1}{\mu
k_{n}}\sum_{i=1}^{k_{n}}\Theta_{i}^{*}\log\frac{R_{(i)}}{R_{(k_{n})}}-\frac{1}{\alpha}\left(\frac{R_{(k_{n})}}{b(n/k_{n})}\right)^{-\alpha}\right)$
$\displaystyle=\sqrt{k_{n}}\left(\left(\frac{1}{k_{n}}\sum_{i=1}^{k_{n}}\Theta_{i}^{*}\right)^{-1}-\frac{1}{\mu}\right)\frac{1}{k_{n}}\sum_{i=1}^{k_{n}}\Theta_{i}^{*}\log\frac{R_{(i)}}{R_{(k_{n})}}$
$\displaystyle+\frac{\sqrt{k_{n}}}{\alpha}\left(\left(\frac{R_{(k_{n})}}{b(n/k_{n})}\right)^{-\alpha}-1\right).$

Note that by (28),

$\displaystyle\frac{1}{k_{n}}\sum_{i=1}^{k_{n}}\Theta_{i}^{*}=\frac{1}{k_{n}}\sum_{i=1}^{n}\Theta_{i}\epsilon_{R_{i}/R_{(k_{n})}}(1,\infty)\stackrel{{\scriptstyle
p}}{{\longrightarrow}}\mu,$

and the convergence in (29) gives

$\displaystyle\frac{1}{k_{n}}\sum_{i=1}^{k_{n}}\Theta_{i}^{*}\log\frac{R_{(i)}}{R_{(k_{n})}}\stackrel{{\scriptstyle
p}}{{\longrightarrow}}\frac{\mu}{\alpha}.$

Therefore, it suffices to consider the convergence of

$\displaystyle\frac{\sqrt{k_{n}}}{\alpha}\left(\left(\frac{R_{(k_{n})}}{b(n/k_{n})}\right)^{-\alpha}-\frac{1}{\mu
k_{n}}\sum_{i=1}^{k_{n}}\Theta_{i}^{*}\right).$ (30)

To prove the convergence of (30), we first use Vervaat’s inversion ([37, 15,
34]) to obtain the convergence of the inverse of

$\eta_{n}(\cdot):=\frac{1}{\mu
k_{n}}\sum_{i=1}^{n}\Theta_{i}\epsilon_{R_{i}/b(n/k_{n})}\bigl{(}(\cdot)^{-1/\alpha},\infty\bigr{)}.$

Note that

$\displaystyle\eta_{n}^{\leftarrow}(t)$
$\displaystyle=\inf\left\\{s:\frac{1}{k_{n}}\sum_{i=1}^{n}\Theta_{i}\epsilon_{R_{i}/b(n/k_{n})}\bigl{(}s^{-1/\alpha},\infty\bigr{)}\geq\mu
t\right\\}$
$\displaystyle=\left(\sup\left\\{y:\frac{1}{k_{n}}\sum_{i=1}^{n}\Theta_{i}\epsilon_{R_{i}/b(n/k_{n})}\bigl{(}y,\infty\bigr{)}\geq\mu
t\right\\}\right)^{-\alpha}.$ (31)

Then with

$M_{n}(t):=\inf\left\\{m\geq
1:\frac{1}{k_{n}}\sum_{i=1}^{n}\Theta_{i}\epsilon_{R_{i}/b(n/k_{n})}\left(\frac{R_{(m)}}{b(n/k_{n})},\infty\right)\geq\mu
t\right\\},$

the inverse function in (31) becomes

$\displaystyle\eta_{n}^{\leftarrow}(t)=\left(\frac{R_{(M_{n}(t))}}{b(n/k_{n})}\right)^{-\alpha}.$

Applying Vervaat’s lemma ([37, 15, 34]) gives the joint convergence in
$D(0,1]\times D(0,1]$:

$\displaystyle\frac{\sqrt{k_{n}}}{(1+\sigma^{2}/\mu^{2})^{1/2}}\left(\eta_{n}(t)-t,\,\eta^{\leftarrow}_{n}(t)-t\right)\Rightarrow\left(W(t),-W(t)\right).$
(32)

Since for $t=\frac{1}{\mu k_{n}}\sum_{i=1}^{k_{n}}\Theta_{i}^{*}$,

$\displaystyle M_{n}\left(\frac{1}{\mu
k_{n}}\sum_{i=1}^{k_{n}}\Theta_{i}^{*}\right)$
$\displaystyle=\inf\left\\{m\geq
1:\frac{1}{k_{n}}\sum_{i=1}^{n}\Theta_{i}\epsilon_{R_{i}/b(n/k_{n})}\left(\frac{R_{(m)}}{b(n/k_{n})},\infty\right)\geq\frac{1}{k_{n}}\sum_{i=1}^{k_{n}}\Theta_{i}^{*}\right\\}=k_{n},$

then (32) gives

$\displaystyle\sqrt{k_{n}}\left(\left(\frac{R_{(k_{n})}}{b(n/k_{n})}\right)^{-\alpha}-\frac{1}{\mu
k_{n}}\sum_{i=1}^{k_{n}}\Theta_{i}^{*}\right)\Rightarrow-(1+\sigma^{2}/\mu^{2})^{1/2}W(1).$
(33)

Combining (33) with (29) shows that

$\sqrt{k_{n}}\left(T_{n}-1/\alpha\right)\Rightarrow\frac{(1+\sigma^{2}/\mu^{2})^{1/2}}{\alpha}\left(\int_{0}^{1}\frac{W(s)}{s}\mathrm{d}s-W(1)\right),$

thus verifying (24). ∎

##  5 Implementation of Testing

Applying the test statistics to data requires estimating a minimal length
interval $[a,b]$ containing the support of the angular measure. On the one
hand, choosing an unnecessarily wide interval $[a,b]$ leads $D_{n}^{*}$ to
conclude $S([a,b])=1$ but only shows the support is a subset of $[a,b]$. Also
making $[a,b]$ too wide may mean there are few points in
$[0,1]\setminus[a,b]$, so that even if the true support of $S$ is $[0,1]$, we
could falsely accept the existence of strong dependence. On the other hand,
fixing an excessively narrow interval $[a,b]$ may lead to $D_{n}^{*}$
inaccurately rejecting existence of strong dependence.

We begin with a method for estimating $a,b$ and then proceed to bootstrap
methods for implementing the tests. This is followed in Sections 5.2 and 5.3
by illustrations using simulated and real data.

###  5.1 Methodology

####  5.1.1 Estimating $[a,b]$

We estimate $a,b$ as the minimizer of an objective function $g_{n}(a,b)$
subject to the constraint $0\leq a\leq b\leq 1$ where

$\displaystyle
g_{n}(a,b):=(b-a)+\sqrt{k_{n}}\left|D_{n}^{*}-H_{k_{n},n}\right|$ (34)

The first part of the objective function, $b-a$, favors a narrow interval
$[a,b]$ the second part requires a wide enough interval $[a,b]$ so that
$\left|D_{n}^{*}-H_{k_{n},n}\right|\approx 0$. Hence, by minimizing $g_{n}$,
we obtain an estimated interval $[\hat{a},\hat{b}]$ of reasonable length and
satisfying $\left|D_{n}^{*}-H_{k_{n},n}\right|\approx 0$. In practice, the
constrOptim function in R suffices for the minimization.

Theorem 5.1 gives the consistency of $\hat{a}$ and $\hat{b}$ for $\alpha>1$.

######  Theorem 5.1.

Suppose the support of $S$ is $[a,b]$, $\alpha>1$ and the intermediate
sequence $\\{k_{n}\\}$ satisfies (18). Let $\hat{a}$ and $\hat{b}$ be the
minimizer of (34). Then,

$\hat{a}\stackrel{{\scriptstyle
p}}{{\longrightarrow}}a,\qquad\hat{b}\stackrel{{\scriptstyle
p}}{{\longrightarrow}}b,$

as $n\to\infty$.

In fact, the consistency result in Theorem 5.1 also holds if we redefine

$g_{n}(s,t)={\color[rgb]{0,0,0}(t-s)}+\lambda\sqrt{k_{n}}\left|D_{n}^{*}-H_{k_{n},n}\right|,$

for some $\lambda>0$. In Section 5.2, we exemplify the value of this
flexibility.

###### Proof.

To shorten the proof and ease notation we make the simplifying assumption that
we know $b=1$. Define,

$\displaystyle\mathbb{C}_{s}=\mathbb{C}_{s,1}=\\{(x,y)\in R_{+}^{2}:s\leq
x/(x+y)\\},$ $\displaystyle
d^{*}_{s}\bigl{(}(x,y)\bigr{)}=d^{*}_{s,1}\bigl{(}(x,y)\bigr{)}=d^{*}\bigl{(}(x,y),\mathbb{C}_{s}\bigr{)}=\left\\{y-(s^{-1}-1)x\right\\}^{+},\qquad
0\leq s\leq 1.$ $\displaystyle
g_{n}(s)=g_{n}(s,1)=(1-s)+\frac{1}{\sqrt{k_{n}}}\sum_{i=1}^{k_{n}}\frac{d^{*}_{s}\bigl{(}(X_{i}^{*},Y_{i}^{*})\bigr{)}}{R_{(k_{n})}}\log\frac{R_{(i)}}{R_{(k_{n})}}.$

Note that $g_{n}(\cdot)$ is concave in $s$. We prove Theorem 5.1 by showing
that for some $\epsilon>0$, and
$\mathcal{I}_{\epsilon}:=[0,a-\epsilon]\cup[a+\epsilon,1],$

$\displaystyle\mathbb{P}\left(\inf_{s\in\mathcal{I}_{\epsilon}}(g_{n}(s)-g_{n}(a))>2\epsilon\right)\to
1,\quad(n\to\infty).$ (35)

Since $d^{*}_{s}\bigl{(}(x,y)\bigr{)}$ is increasing in $s$, Theorem 3.1
ensures that (see (20))

$\sup_{s\in[0,a]}\frac{1}{\sqrt{k_{n}}}\sum_{i=1}^{k_{n}}\frac{d^{*}_{s,t}\bigl{(}(X_{i}^{*},Y_{i}^{*})\bigr{)}}{R_{(k_{n})}}\log\frac{R_{(i)}}{R_{(k_{n})}}\stackrel{{\scriptstyle
p}}{{\longrightarrow}}0,$

and therefore,

$\displaystyle\lim_{n\to\infty}\mathbb{P}\left(\inf_{s\leq
a-\epsilon}(g_{n}(s)-g_{n}(a))>2\epsilon\right)=1.$ (36)

If $s\geq a+\epsilon,$ replace division by $\sqrt{k_{n}}$ with division by
$k_{n}$ and the definitions of $(R_{i},\Theta_{i})$ give

$\displaystyle\frac{1}{k_{n}}\sum_{i=1}^{k_{n}}\frac{d^{*}_{s}\bigl{(}(X_{i}^{*},Y_{i}^{*})\bigr{)}}{R_{(k_{n})}}\log\frac{R_{(i)}}{R_{(k_{n})}}=\frac{1}{k_{n}}\sum_{i=1}^{k_{n}}\left\\{1-s^{-1}\Theta^{*}_{i}\right\\}^{+}\frac{R_{(i)}}{R_{(k_{n})}}\log\frac{R_{(i)}}{R_{(k_{n})}},$
and since $R_{(i)}\geq R_{(k_{n})}$, this is bounded below by
$\displaystyle\geq\frac{1}{k_{n}}\sum_{i=1}^{k_{n}}\left\\{1-s^{-1}\Theta^{*}_{i}\right\\}^{+}\log\frac{R_{(i)}}{R_{(k_{n})}}.$
(37)

We show this converges in probability to a positive constant $L(a,s)>0$ and
thus

$\displaystyle\frac{1}{\sqrt{k_{n}}}$
$\displaystyle\sum_{i=1}^{k_{n}}\frac{d^{*}_{s}\bigl{(}(X_{i}^{*},Y_{i}^{*})\bigr{)}}{R_{(k_{n})}}\log\frac{R_{(i)}}{R_{(k_{n})}}\stackrel{{\scriptstyle
p}}{{\longrightarrow}}\infty.$ (38)

Combining (38) with (36) completes the proof of (35).

Returning to the expression in (37), write it as

$\iint_{\\{(r,\theta):r>1,\theta\in[0,1]\\}}\Bigl{(}1-\frac{\theta}{s}\Bigr{)}^{+}\log
r\frac{1}{k}\sum_{i=1}^{k_{n}}\epsilon_{(\Theta_{i}^{*},R_{(i)}/R_{(k)})}(d\theta,dr).$
(39)

On $[0,1]\times(1,\infty)$,

$\frac{1}{k}\sum_{i=1}^{k_{n}}\epsilon_{(\Theta_{i}^{*},R_{(i)}/R_{(k)})}(d\theta,dr)\to
S(d\theta)\nu_{\alpha}(dr)$

and the right side is a probability measure on $[0,1]\times(1,\infty)$, so
using familiar weak convergence arguments in (39) we get convergence to

$\displaystyle L(a,s):=$
$\displaystyle\iint_{\\{(r,\theta):r>1,\theta\in[0,1]\\}}\Bigl{(}1-\frac{\theta}{s}\Bigr{)}^{+}\log
rS(d\theta)\nu_{\alpha}(dr)$ and after some Fubini justified manipulations
this is $\displaystyle=$
$\displaystyle\frac{1}{\alpha}\int_{a}^{s}\Bigl{(}1-\frac{\theta}{s}\Bigr{)}S(d\theta).$

Remember $s>a+\epsilon$ and we verify $L(a,s)$ is positive. If not, $L(a,s)=0$
and $1-\theta/s=0$ or $\theta=s$ for almost all (with respect to $S(\cdot)$)
$\theta\in[a,,a+\epsilon]$ and this means $S[a,a+\epsilon]=0$, thus
contradicting $a$ being in the support of $S(\cdot)$. So $L(a,s)>0.$ ∎

####  5.1.2 Bootstrap methods

Formulating tests based on either Theorem 3.1 or 4.1 requires knowing the
values of $\alpha,a,b$, which, however, is unlikely to be true for real
datasets.  Substitution methods suggest replacing $\alpha$ with the
corresponding Hill estimator, $1/H_{k_{n},n}$ and investigating the effect on
the limit distribution but this will not work here due to (20). In the sequel,
we propose bootstrap methods to implement the proposed tests and try the
approach on simulated and real datasets. We do not formally justify the
bootstrap method–this is left for the future–but the numerical experiments
suggest its applicability.

Suppose we take $m_{n}\approx n/k_{n}$ so that $m_{n}/n\to 0$ and
$m_{n}\to\infty$. Let $\\{I_{1}(n),\ldots,I_{m_{n}}(n)\\}$ be iid discrete
uniform random variables on $\\{1,\ldots,n\\}$, independent from
$\\{(X_{i},Y_{i}):i\geq 1\\}$. We construct  a bootstrap resample of size
$m_{n}$ by

$(X_{I_{j}(n)},Y_{I_{j}(n)}),\qquad j=1,\ldots,m_{n}.$

Define $R^{\text{boot}}_{(i)}$ as the $i$-th largest order statistic among
$\\{R_{I_{j}(n)}\equiv X_{I_{j}(n)}+Y_{I_{j}(n)}:1\leq j\leq m_{n}\\}$, and
let $(X^{*}_{I_{i}(n)},Y^{*}_{I_{i}(n)})$ be the pair of random variables such
that $X^{*}_{I_{i}(n)}+Y^{*}_{I_{i}(n)}\equiv R^{\text{boot}}_{(i)}$.

(1) Test $H_{0}^{(1)}$. For the test in (16), we first solve (34)  using the
whole sample to estimate the support of the angular measure,
$[\hat{a},\hat{b}]$ from the sample. Then we obtain
$\mathbb{C}_{\hat{a},\hat{b}}:=\\{(x,y)\in\mathbb{R}_{+}^{2}:\hat{a}\leq
x/(x+y)\leq\hat{b}\\}$ and

$\displaystyle\widehat{D}^{*}_{m_{n}}=$
$\displaystyle\frac{1}{{k_{n}}}\sum_{i=1}^{k_{n}}\left(1+\frac{d^{*}\bigl{(}(X^{*}_{I_{j}(n)},Y^{*}_{I_{j}(n)}),{\mathbb{C}}_{\hat{a},\hat{b}}\bigr{)}}{R^{\text{boot}}_{(k_{m_{n}})}}\right)\log\frac{R^{\text{boot}}_{(i)}}{R^{\text{boot}}_{(k_{m_{n}})}}.$

Conditioning on the original sample, we presume from Theorem 3.1
$\sqrt{k_{m_{n}}}\left(\widehat{D}^{*}_{m_{n}}-H_{k_{n},n}\right)$ is
approximately normal with mean 0 and variance $H_{k_{n},n}^{2}$ for large $n$.
Therefore, we reject $H^{(1)}_{0}$ in (16) if

$\displaystyle\left|\widehat{D}^{*}_{m_{n}}-H_{k_{n},n}\right|>1.96\frac{H_{k_{n},n}}{\sqrt{k_{m_{n}}}}.$
(40)

In practice we would generate $B$ bootstrap samples and reject if more than 5%
satisfy (40).

(2) Full vs strong dependence. For the test in (21), generate $B$ bootstrap
resamples  indexed by $t=1,\dots,B$. For each $t$, let
$R^{\text{boot}}_{(i),t}$ be the $i$-largest order statistic in the $t$-th
resample; $\Theta^{*}_{i,t}$ is the corresponding concomitant. Compute the
corresponding test statistics for each resample,

$T_{m_{n}}^{(t)}=\frac{\sum_{i=1}^{k_{m_{n}}}\Theta_{i,t}^{*}\log\frac{R^{\text{boot}}_{(i),t}}{R^{\text{boot}}_{(k_{m_{n}}),t}}}{\sum_{i=1}^{k_{m_{n}}}\Theta_{i,t}^{*}},\qquad
t=1,\ldots,B.$

Based on Theorem 4.1, we presume under $H_{0}^{(2)}$ that conditional on the
original sample, $\sqrt{k_{m_{n}}}\left(T_{m_{n}}^{(t)}-H_{k_{n},n}\right)$ is
approximately normal with mean 0 and variance $H_{k_{n},n}^{2}$ for large $n$.
Using all $B$ resamples, we obtain the bootstrap estimate of the standard
error of $T_{n}$:

$SE_{\text{boot}}(m_{n}):=\left(\frac{1}{B-1}\sum_{t=1}^{B}\left(T_{m_{n}}^{(t)}-\bar{T}_{m_{n}}\right)^{2}\right)^{1/2},$

where $\bar{T}_{m_{n}}=\frac{1}{B}\sum_{t=1}^{B}T_{m_{n}}^{(t)}$. Then we
reject $H^{(2)}_{0}$ in (21) if

$k_{m_{n}}\frac{SE^{2}_{\text{boot}}(m_{n})}{H_{k_{n},n}^{2}}>\chi^{2}_{.95,B-1}/(B-1),$

where $\chi^{2}_{.95,B-1}$ denotes the 95% quantile of a chi-square
distribution with $B-1$ degrees of freedom.

(3) Strong vs weak dependence. When testing for strong vs weak dependence, we
rely on Theorem 4.2 and define
$\widetilde{\Theta}_{i}:=\Theta_{i}\boldsymbol{1}_{\\{\Theta_{i}\in[a,b]\\}}$,
$\widetilde{R}_{i}:=R_{i}\boldsymbol{1}_{\\{\Theta_{i}\in[a,b]\\}}$. Let
$\widetilde{\Theta}_{i}^{*}$ be the concomitant of $\widetilde{R}_{(i)}$, and
by assuming $0/0\equiv 1$ we define also

$\displaystyle\widetilde{T}_{n}:=\frac{\sum_{i=1}^{k_{n}}\widetilde{\Theta}_{i}^{*}\log\left(\frac{\widetilde{R}_{(i)}}{\widetilde{R}_{(k_{n})}}\vee
1\right)}{\sum_{i=1}^{k_{n}}\widetilde{\Theta}_{i}^{*}}.$ (41)

For $[a,b]\subsetneq[0,1]$, we want to test strong vs weak dependence, i.e.

$\displaystyle H^{(3)}_{0}:\text{Support of
$S(\cdot)=[a,b]$}\qquad\text{v.s.}\qquad H^{(3)}_{a}:\text{Support of
$S(\cdot)=[0,1]$}.$ (42)

Under $H^{(3)}_{0}$, $\widetilde{T}_{n}$ must have the same asymptotic
distribution as $T_{n}$. Here we apply the bootstrap method again to test
whether $T_{n}$ and $\widetilde{T}_{n}$ have the same asymptotic variance.
Again estimate $[\hat{a},\hat{b}]$ from (34). To obtain the $t$-th resample,
we generate $\\{I_{1,t}(n),\ldots,I_{m_{n},t}(n)\\}$ iid discrete uniform
random variables on $\\{1,\ldots,n\\}$, and compute

$\widetilde{\Theta}_{i,t}:=\Theta_{I_{i,t}(n)}\boldsymbol{1}_{\\{\Theta_{I_{i,t}(n)}\in[\hat{a},\hat{b}]\\}},\qquad\widetilde{T}_{m_{n}}^{(t)}=\frac{\sum_{i=1}^{k_{m_{n}}}\widetilde{\Theta}_{i,t}^{*}\log\left(\frac{\widetilde{R}_{(i),t}}{\widetilde{R}_{(k_{m_{n}}),t}}\vee
1\right)}{\sum_{i=1}^{k_{m_{n}}}\widetilde{\Theta}_{i,t}^{*}}.$

We repeat the bootstrap resampling scheme twice to obtain
$T_{m_{n}}^{(1)},\ldots,T_{m_{n}}^{(B)}$,
$\widetilde{T}_{m_{n}}^{(1)},\ldots,\widetilde{T}_{m_{n}}^{(B)}$, and reject
$H^{(3)}_{0}$ if

$\frac{\frac{1}{B-1}\sum_{t=1}^{B}\left(T_{m_{n}}^{(t)}-\bar{T}_{m_{n}}\right)^{2}}{\frac{1}{B-1}\sum_{s=1}^{B}\left(\widetilde{T}_{m_{n}}^{(t)}-\bar{\widetilde{T}}_{m_{n}}\right)^{2}}>F_{0.975,B-1,B-1}\quad\text{or}\quad<F_{0.025,B-1,B-1},$

where
$\bar{\widetilde{T}}_{m_{n}}=\frac{1}{B}\sum_{t=1}^{B}\widetilde{T}_{m_{n}}^{(t)}$
and $F_{p,B-1,B-1}$ denotes the $100p\%$-percentile of an $F$-distribution
with numerator and denominator degrees of freedom both equal to $B-1$.

###  5.2 Simulation study

Example 1. Consider a simulated data example as below. Set $a=0.25$, $b=0.75$.
Suppose $R_{1}\sim\text{Pareto}(2)$, $R_{2}\sim\text{Pareto}(4)$,
$Z\sim\text{Beta}(0.05,0.1)$,
$\Theta_{2}\sim\text{Unif}([0,1]\setminus[a,b])$, and
$B\sim\text{Bernoulli}(0.5)$. Assume the random variables are all independent,
and let $\Theta_{1}:=a+(b-a)Z$. Now define the vector $(X,Y)$ as

$\displaystyle X$ $\displaystyle:=BR_{1}\Theta_{1}+(1-B)R_{2}\Theta_{2}$
$\displaystyle Y$
$\displaystyle:=BR_{1}(1-\Theta_{1})+(1-B)R_{2}(1-\Theta_{2}).$

By construction, $(X,Y)$ is MRV on
$\mathbb{R}^{2}_{+}\setminus\\{\boldsymbol{0}\\}$ with tail parameter
$\alpha=2$. The second order condition (6) also holds since

$\frac{1}{t^{-1}}\left(t\mathbb{P}\left[\left(\frac{R}{b(t)},\Theta\right)\in\cdot\right]-p\nu_{2}\times\mathbb{P}[\Theta_{1}\in\cdot]\right)\to(1-p)\nu_{4}\times\mathbb{P}[\Theta_{2}\in\cdot].$

Furthermore, for

$\mathbb{C}_{a,b}=\\{(x,y)\in\mathbb{R}^{2}_{+}:y/3\leq x\leq 3y\\},$

the vector $(X,Y)$ has HRV on $\mathbb{R}^{2}_{+}\setminus\mathbb{C}_{a,b}$
with tail parameter $\alpha_{0}=4$.

We generate $n=30{,}000$ iid samples from the distribution of $(X,Y)$. The
scatter plot is in the left panel of Figure 2 and illustrates the dependence
structure. Thresholding with $k_{n}=100$ yields the histogram of angles in the
right panel of Figure 2; this also describes the dependence structure of
$(X,Y)$ and based on high values of $|x|+|y|$, shows that the support of the
angular measure is $[0.25,0.75]$. Based on this sample, the Hill estimate with
$k_{n}=100$ is $H_{k_{n},n}=0.473$.

Figure 2: Simulated data example 1. Left: Scatter plot of $30{,}000$ data
points. Histogram of $\theta_{1}$ with $k_{n}=100$.

To estimate $[a,b]$, we solve the optimization problem in (34)  using the
constrOptim function in R, and Table 1 shows the estimated $[\hat{a},\hat{b}]$
for different choices of the tuning parameter $\lambda$. We see that for a
bimodal histogram as in the right panel of Figure 2, solving (34) provides
good estimates for $a,b$, invariant to different choices of the tuning
parameter.

$\lambda\sqrt{k_{n}}$ | $\sqrt{k_{n}}$ | $2\sqrt{k_{n}}$ | $2^{2}\sqrt{k_{n}}$ | $2^{3}\sqrt{k_{n}}$ | $2^{4}\sqrt{k_{n}}$
---|---|---|---|---|---
$[\hat{a},\hat{b}]$ | $[0.25,0.75]$ | $[0.25,0.75]$ | $[0.250,0.750]$ | $[0.25,0.75]$ | $[0.25,0.75]$
Table 1: Estimated $[\hat{a},\hat{b}]$ for different choices of the tuning
parameter $\lambda$ when $[a,b]=[0.25,0.75]$.

Next, we set $m_{n}=500$, $k_{m_{n}}=25$, and generate $B=2{,}000$ bootstrap
resamples to test

$H^{(1)}_{0}:\,S([0.25,0.75])=1,\qquad H^{(1)}_{a}:\,S([0.25,0.75])<1.$

For each bootstrap resample, we compute the corresponding
$\widehat{D}_{m_{n}}^{*}$ and use (40) to decide whether to reject
$H^{(1)}_{0}$ or not. Among the 2,000 bootstrap resamples generated, the
rejection rate is 0.045, indicating we shall accept $H^{(1)}_{0}$. In
addition, applying the boostrap testing procedure for $H_{0}^{(3)}$, we have

$\displaystyle\frac{\frac{1}{B-1}\sum_{t=1}^{B}\left(T_{m_{n}}^{(t)}-\bar{T}_{m_{n}}\right)^{2}}{\frac{1}{B-1}\sum_{t=1}^{B}\left(\widetilde{T}_{m_{n}}^{(t)}-\bar{\widetilde{T}}_{m_{n}}\right)^{2}}$
$\displaystyle=1.077\in[0.916,1.092]$
$\displaystyle=[F_{0.025,1999,1999},F_{0.975,1999,1999}].$

This confirms the existence of strong dependence on
$\mathbb{R}_{+}^{2}\setminus\mathbb{C}_{a,b}$. To check further the existence
of full dependence, we compute $SE_{\text{boot}}(m_{n})=0.546$ based on the
bootstrap resamples, which gives

$k_{m_{n}}\frac{SE^{2}_{b}(m_{n})}{H_{k_{n},n}^{2}}=1.336>\chi^{2}_{.95,1999}/1999=1.053.$

Hence, we reject the full dependence hypothesis in $H_{0}^{(2)}$.

Example 2. In fact, whether to reject $H_{0}^{(2)}$ also depends on the
variability of $\Theta_{1}$. For instance, suppose instead the random variable
$Z$ in the previous example follows a $\text{Beta}(1,2)$ distribution, and all
other assumptions remain identical to Example 1. The scatter plot and the
histogram of angles are given in Figure 3.

Figure 3: Simulated data example 2. Left: Scatter plot of $30{,}000$ data
points.Histogram of $\theta_{1}$ with $k_{n}=100$.

We again estimate $[a,b]$ by solving the optimization problem in (34), and
Table 2 shows the estimated $[\hat{a},\hat{b}]$ for different choices of the
tuning parameter $\lambda$. We see that here the estimation procedure provides
an accurate estimate for $\hat{a}$ across all chosen values of $\lambda$, but
estimated values of $\hat{b}$ are all smaller than $b=0.75$. Overall,
$\lambda=2^{4}$ provides the most accurate estimates.

$\lambda\sqrt{k_{n}}$ | $\sqrt{k_{n}}$ | $2\sqrt{k_{n}}$ | $2^{2}\sqrt{k_{n}}$ | $2^{3}\sqrt{k_{n}}$ | $2^{4}\sqrt{k_{n}}$
---|---|---|---|---|---
$[\hat{a},\hat{b}]$ | $[0.251,0.554]$ | $[0.251,0.596]$ | $[0.251,0.597]$ | $[0.251,0.654]$ | $[0.251,0.670]$
Table 2: Estimated $[\hat{a},\hat{b}]$ for different choices of the tuning
parameter $\lambda$ when $[a,b]=[0.25,0.75]$.

For this simulated dataset, we now proceed with the true values of $a$ and
$b$. Following the testing procedure as in the previous example, we see that
out of $2{,}000$ bootstrap resamples, the overall rejection rate for
$H^{(1)}_{0}$ is 0.0465, and the boostrap method for testing $H_{0}^{(3)}$
returns a test statistic

$\displaystyle\frac{\frac{1}{B-1}\sum_{t=1}^{B}\left(T_{m_{n}}^{(t)}-\bar{T}_{m_{n}}\right)^{2}}{\frac{1}{B-1}\sum_{t=1}^{B}\left(\widetilde{T}_{m_{n}}^{(t)}-\bar{\widetilde{T}}_{m_{n}}\right)^{2}}$
$\displaystyle=1.035\in[0.916,1.092]$
$\displaystyle=[F_{0.025,1999,1999},F_{0.975,1999,1999}],$

confirming the existence of strong dependence. However, these bootstrap
resamples give $SE_{\text{boot}}(m_{n})=0.508$, thus giving

$k_{m_{n}}\frac{SE^{2}_{\text{boot}}(m_{n})}{H_{k_{n},n}^{2}}=0.939<\chi^{2}_{.95,1999}/1999=1.053.$

This makes us fail to reject the full dependence hypothesis in $H_{0}^{(2)}$.
In fact, since $\Theta_{1}$ follows a Beta$(1,2)$ distribution, then we have
$\mu=0.417$ and $\sigma^{2}=0.014$, which leads to
$(\sigma^{2}/\mu^{2}+1)^{1/2}=1.039<1.053$. Hence, the small variation in the
underlying distribution of $\Theta_{1}$ may lead to false acceptance of
$H_{0}^{(2)}$. If replacing $[a,b]$ with the estimated
$[\hat{a},\hat{b}]=[0.251,0.670]$, we obtain the same conclusion.

So for this example, we fail to reject all three null hypotheses,
$H_{0}^{(i)}$, $i=1,2,3$, and the third failure to reject is in error. One way
to fix this is to check the proportion of resamples among $t=1,\ldots,B$,
which have

$\displaystyle\left|T_{m_{n}}^{(t)}-H_{k_{n},n}\right|>1.96\frac{H_{k_{n},n}}{\sqrt{k_{m_{n}}}},$
(43)

so that we will reject $H_{0}^{(2)}$. Here the reject rate among the $2{,}000$
resamples is 0.059, which suggests we should reject the full dependence
hypothesis; this brings us to the correct decision for this (simulated) data.

###  5.3 Real data examples

Figure 4: Acf plots for the log returns of every-other-day stock prices.

We now consider the application of the bootstrap method to real data. We
download the daily adjusted stock prices of Chevron (CVX), Exxon (XOM) and
Apple (AAPL) during the time period from January 04, 2016 to December 30,
2022. To remove the possible serial dependence of stock returns, we compute
the log returns of these three stocks using their every-other-day prices. The
acf plots in Figure 4 show little serial dependence for all three stocks. This
leads to a reduced dataset of $n=880$ observations for each stock.

####  5.3.1 CVX vs XOM

Figure 5: CVX vs XOM. Left: Scatter plot of CVX and XOM returns. Right:
Histogram of angles (absolute returns of CVX) with $k_{n}=100$.

In the left panel of Figure 5, we present the scatter plot of the log returns
of CVX and XOM. To understand the dependence structure between absolute log
returns of CVX and XOM, we also graph the histogram of $|x|/(|x|+|y|)$ in the
right panel of Figure 5, where the threshold is chosen as $k_{n}=100$.

The corresponding Hill estimate gives
$\hat{\alpha}=1/H_{k_{n},n}=1/0.342=2.926$. By setting $\lambda=4$, we obtain
estimates $\hat{a}=0.336$ and $\hat{b}=0.853$ (estimates remain the same for
$\lambda\geq 4$) as well as

$\frac{1}{k_{n}}\sum_{i=1}^{k_{n}}\Theta_{i}^{*}=0.482\equiv\widehat{\theta}_{0}.$

Then we generate 2,000 bootstrap resamples with $m_{n}=200$ and $k_{m_{n}}=20$
to test

$H^{(1)}_{0}:\,S([0.336,0.853])=1,\qquad H^{(1)}_{a}:\,S([0.336,0.853])<1.$

For each bootstrap resample, we compute the corresponding test statistic
$\widehat{D}_{k_{m_{n}}}^{*}$, and see that only 2.1% of the 2000 bootstrap
trials reject $H^{(1)}_{0}$. In addition, consider strong vs weak dependence
(i.e. $H^{(3)}_{0}$ vs $H^{(3)}_{a}$), and calculate

$\displaystyle\frac{\frac{1}{B-1}\sum_{t=1}^{B}\left(T_{m_{n}}^{(t)}-\bar{T}_{m_{n}}\right)^{2}}{\frac{1}{B-1}\sum_{t=1}^{B}\left(\widetilde{T}_{m_{n}}^{(t)}-\bar{\widetilde{T}}_{m_{n}}\right)^{2}}$
$\displaystyle=1.072\in[0.916,1.092].$

Therefore, we accept the existence of strong dependence and conclude
$S([0.336,0.853])=1$.

To distinguish between full and strong dependence, we obtain
$SE_{\text{boot}}(m_{n})=0.317$, which leads to

$k_{m_{n}}\frac{SE^{2}_{\text{boot}}(m_{n})}{H_{k_{n},n}^{2}}=0.861<\chi^{2}_{.95,1999}/1999=1.053.$

So we fail to reject the hypothesis of full dependence, i.e.
$H^{(2)}_{0}:S(\\{0.482\\})=1$. Here even if we consider the rejection rate
using the criterion in (43), only 3.35% of the 2000 bootstrap trials rejects
$H^{(2)}_{0}$. Hence, we conclude that the absolute returns of CVX and XOM
show full asymptotic dependence.

####  5.3.2 CVX vs AAPL

Figure 6: CVX vs AAPL. Left: Scatter plot of CVX and AAPL returns. Right:
Histogram of angles (absolute returns of CVX) with $k_{n}=100$.

Next, we inspect the dependence structure between absolute returns of CVX and
AAPL. Based on analyses using Hill plot (not shown), we choose $k_{n}=100$ and
estimate the marginal tail indices $\hat{\alpha}=1/0.294=3.401$. We give the
scatter plot and the histogram of $|x_{1}|/(|x_{1}|+|x_{2}|)$ in the left and
right panels Figure 6, respectively.

We set $\lambda=4$ and obtain the estimated $[\hat{a},\hat{b}]=[0.182,0.928]$
($\lambda>4$ gives too wide an interval). When testing

$H^{(1)}_{0}:\,S([0.182,0.928])=1\qquad\text{vs}\qquad
H^{(1)}_{a}:\,S([0.182,0.928])<1,$

we compute $\widehat{D}_{m_{n}}^{*}$ for each of the $2{,}000$ resamples and
only 3.5% of them rejects $H^{(1)}_{0}$. Note that the estimated support
$[\hat{a},\hat{b}]$ is already quite wide, then the low rejection rate can be
a result of having too wide a support of $S$. So we next generate two sets of
2,000 bootstrap resamples with $m_{n}=200$ and $k_{m_{n}}=20$ to test

$H^{(3)}_{0}:\,S([0.182,0.928])=1\qquad\text{vs}\qquad
H^{(3)}_{a}:\,S([0,1])=1.$

This gives a test statistic

$\displaystyle\frac{\frac{1}{B-1}\sum_{t=1}^{B}\left(T_{m_{n}}^{(t)}-\bar{T}_{m_{n}}\right)^{2}}{\frac{1}{B-1}\sum_{t=1}^{B}\left(\widetilde{T}_{m_{n}}^{(t)}-\bar{\widetilde{T}}_{m_{n}}\right)^{2}}$
$\displaystyle=0.814\notin[0.916,1.092],$

indicating the existence of weak dependence. Hence, we end up with the
conclusion that considering the absolute returns of CVX and AAPL, the support
of the angular measure is likely to be $[0,1]$.

##  6 Supplementary Material

This supplement contains detailed proofs on the main theorems. Section 6.1
proves Theorem 3.1, and Section 6.2 proves asymptotic normality of the test
statistic $T_{n}$ under the null hypothesis of full dependence. Section 6.3
checks conditions of the functional central limit theorem in Theorem 10.6 of
[32].

###  6.1 Proof of Theorem 3.1

We start by showing that for intermediate sequence $\\{k_{n}\\}$ satisfying
(18),

$\displaystyle W_{n}(y):=$
$\displaystyle\sqrt{k_{n}}\left(\frac{1}{{k_{n}}}\sum_{i=1}^{n}\left(1+\frac{d^{*}\bigl{(}(X_{i},Y_{i}),\mathbb{C}_{a,b}\bigr{)}}{R_{(k_{n})}}\right)\boldsymbol{1}_{\left\\{\frac{R_{i}}{b(n/k_{n})}>y\right\\}}-y^{-\alpha}\right)$
$\displaystyle\Rightarrow$ $\displaystyle W(y^{-\alpha}),$ (44)

in $D(0,\infty)$, where $W(\cdot)$ is a standard Brownian motion.

We begin by showing that the sequence of processes in the variable $y$ satisfy

$\displaystyle\frac{1}{\sqrt{k_{n}}}\sum_{i=1}^{n}\frac{d^{*}\bigl{(}(X_{i},Y_{i}),\mathbb{C}_{a,b}\bigr{)}}{b(n/k_{n})}\boldsymbol{1}_{\left\\{\frac{R_{i}}{b(n/k_{n})}>y\right\\}}\Rightarrow
0,\quad\text{ in }D(0,\infty),$ (45)

and it is here that HRV and assumption $[a,b]\subsetneq[0,1]$ is used. We have
that

$\displaystyle\frac{1}{k_{n}}\sum_{i=1}^{n}\frac{d^{*}\bigl{(}(X_{i},Y_{i}),\mathbb{C}_{a,b}\bigr{)}}{b(n/k_{n})}\boldsymbol{1}_{\left\\{\frac{R_{i}}{b(n/k_{n})}>y\right\\}}$
$\displaystyle=\frac{b_{0}(n/k_{n})}{b(n/k_{n})}\frac{1}{k_{n}}\sum_{i=1}^{n}\frac{d^{*}\bigl{(}(X_{i},Y_{i}),\mathbb{C}_{a,b}\bigr{)}}{b_{0}(n/k_{n})}\boldsymbol{1}_{\left\\{\frac{R_{i}}{b(n/k_{n})}>y\right\\}}$
$\displaystyle\leq\frac{b_{0}(n/k_{n})}{b(n/k_{n})}\frac{1}{k_{n}}\sum_{i=1}^{n}\frac{d^{*}\bigl{(}(X_{i},Y_{i}),\mathbb{C}_{a,b}\bigr{)}}{b_{0}(n/k_{n})}\left(\boldsymbol{1}_{\left\\{\frac{R_{i}}{b(n/k_{n})}>y,d^{*}\bigl{(}(X_{i},Y_{i}),\mathbb{C}_{a,b}\bigr{)}>b_{0}(n/k_{n})\epsilon\right\\}}\right.$
$\displaystyle\left.\qquad+\boldsymbol{1}_{\left\\{\frac{R_{i}}{b(n/k_{n})}>y,d^{*}\bigl{(}(X_{i},Y_{i}),\mathbb{C}_{a,b}\bigr{)}\leq
b_{0}(n/k_{n})\epsilon\right\\}}\right)$
$\displaystyle\leq\frac{b_{0}(n/k_{n})}{b(n/k_{n})}\left(\frac{1}{k_{n}}\sum_{i=1}^{n}\frac{d^{*}\bigl{(}(X_{i},Y_{i}),\mathbb{C}_{a,b}\bigr{)}}{b_{0}(n/k_{n})}\boldsymbol{1}_{\left\\{\frac{d^{*}\bigl{(}(X_{i},Y_{i}),\mathbb{C}_{a,b}\bigr{)}}{b_{0}(n/k_{n})}>\epsilon\right\\}}+\frac{\epsilon}{k_{n}}\sum_{i=1}^{n}\boldsymbol{1}_{\left\\{\frac{R_{i}}{b(n/k_{n})}>y\right\\}}\right)$
$\displaystyle=A+B.$

To handle $B$ observe for each fixed $y>0$, that the monotone function in $y$,

$\frac{1}{k_{n}}\sum_{i=1}^{n}\boldsymbol{1}_{\left\\{\frac{R_{i}}{b(n/k_{n})}>y\right\\}}\Rightarrow
y^{-\alpha_{0}}$

using, for example [34, Theorem 5.3(ii), p. 139]. Therefore, when $k_{n}$
satisfies (18), we have $\sqrt{k}_{n}B\Rightarrow 0$ in $D(0,\infty)$.

For $A$ we claim

$\frac{1}{k_{n}}\sum_{i=1}^{n}\frac{d^{*}\bigl{(}(X_{i},Y_{i}),\mathbb{C}_{a,b}\bigr{)}}{b_{0}(n/k_{n})}\boldsymbol{1}_{\left\\{\frac{d^{*}\bigl{(}(X_{i},Y_{i}),\mathbb{C}_{a,b}\bigr{)}}{b_{0}(n/k_{n})}>\epsilon\right\\}}=O_{p}(1),$

since for any $M>0$

$\displaystyle\mathbb{P}\Biggl{[}\frac{1}{k_{n}}\sum_{i=1}^{n}$
$\displaystyle\frac{d^{*}\bigl{(}(X_{i},Y_{i}),\mathbb{C}_{a,b}\bigr{)}}{b_{0}(n/k_{n})}\boldsymbol{1}_{\left\\{\frac{d^{*}\bigl{(}(X_{i},Y_{i}),\mathbb{C}_{a,b}\bigr{)}}{b_{0}(n/k_{n})}>\epsilon\right\\}}>M\Biggr{]}$
$\displaystyle\leq$
$\displaystyle\frac{1}{M}\frac{n}{k_{n}}\mathbb{E}\left(\frac{d^{*}\bigl{(}(X_{1},Y_{1}),\mathbb{C}_{a,b}\bigr{)}}{b_{0}(n/k_{n})}\boldsymbol{1}_{\left\\{\frac{d^{*}\bigl{(}(X_{1},Y_{1}),\mathbb{C}_{a,b}\bigr{)}}{b_{0}(n/k_{n})}>\epsilon\right\\}}\right)$
and because $\alpha_{0}>1$ we may apply Karamata’s theorem on integration to
get convergence, as $n\to\infty$ to $\displaystyle\to$
$\displaystyle\frac{1}{M}\int_{\epsilon}^{\infty}x\nu_{\alpha_{0}}(dx)<\infty.$

Therefore $\sqrt{k}_{n}A\Rightarrow 0$ in $D(0,\infty)$. This proves (44).

Since $R_{(k_{n})}/b(n/k_{n})\stackrel{{\scriptstyle p}}{{\longrightarrow}}1$
(eg. [34, p. 82]), we also have

$\frac{1}{\sqrt{k_{n}}}\sum_{i=1}^{n}\frac{d^{*}\bigl{(}(X_{i},Y_{i}),\mathbb{C}_{a,b}\bigr{)}}{R_{(k_{n})}}\boldsymbol{1}_{\left\\{\frac{R_{i}}{b(n/k_{n})}>y\right\\}}\Rightarrow
0,\quad\text{ in }D(0,\infty).$

So to prove (44) it remains to verify

$\sqrt{k_{n}}\left(\frac{1}{{k_{n}}}\sum_{i=1}^{n}\boldsymbol{1}_{\left\\{\frac{R_{i}}{b(n/k_{n})}>y\right\\}}-y^{-\alpha}\right)\Rightarrow
W(y^{-\alpha}),$ (46)

in $D(0,\infty)$. The regular variation of $P[R_{1}>x]$ implies ([34, Theorem
9.1, p. 292] or [15]) that

$\sqrt{k_{n}}\left(\frac{1}{{k_{n}}}\sum_{i=1}^{n}\boldsymbol{1}_{\left\\{\frac{R_{i}}{b(n/k_{n})}>y\right\\}}-\frac{n}{k_{n}}\mathbb{P}[R_{1}/b(n/k)>y]\right)\Rightarrow
W(y^{-\alpha}),$

in $D(0,\infty)$ and the 2RV assumption  in (8) marginalized to the
distribution of $R_{1}$ and the choice of $k_{n}$ in (18) imply

$\sqrt{k_{n}}\left(\frac{n}{k_{n}}\mathbb{P}\left(\frac{R_{1}}{b(n/k_{n})}>y\right)-y^{-\alpha}\right)\to
0,$

locally uniformly in $y$. This gives (46) and completes the proof of (44).

Apply the composition map $(x(t),c)\mapsto x(ct)$ from
$D(0,\infty)\times(0,\infty)\mapsto D(0,\infty)$ to (44) in the form

$\Bigl{(}W_{n}(y),\frac{R_{(k_{n})}}{b(n/k)}\Bigr{)}\mapsto
W_{n}(\frac{R_{(k_{n})}}{b(n/k)}y)\Rightarrow W(y^{-\alpha})$

to get

$\displaystyle\sqrt{k_{n}}\Biggl{(}$
$\displaystyle\frac{1}{{k_{n}}}\sum_{i=1}^{n}\left(1+\frac{d^{*}\bigl{(}(X_{i},Y_{i}),\mathbb{C}_{a,b}\bigr{)}}{R_{(k_{n})}}\right)\boldsymbol{1}_{\left\\{\frac{R_{i}}{R_{(k_{n})}}>y\right\\}}-{\color[rgb]{0,0,0}\bigl{(}\frac{R_{(k_{n})}}{b(n/k)}y\bigr{)}^{-\alpha}}\Biggr{)}$
$\displaystyle\Rightarrow W(y^{-\alpha}).$ (47)

Couple this with a Vervaat inversion of (46) ([15, p. 357] or [34, p. 57]).
The inversion yields in $D(0,\infty)$

$\sqrt{k_{n}}\Biggl{(}\Bigl{(}\frac{R_{([kt])}}{b(n/k_{n})}\Bigr{)}^{-\alpha}-t\Biggr{)}\Rightarrow-W(t),\quad\text{
in }D(0,\infty),$ (48)

and the convergence is joint with the one in (6.1). Combining (6.1) with (48)
gives

$\displaystyle\sqrt{k_{n}}\Bigl{(}$
$\displaystyle\frac{1}{{k_{n}}}\sum_{i=1}^{n}\left(1+\frac{d^{*}\bigl{(}(X_{i},Y_{i}),\mathbb{C}_{a,b}\bigr{)}}{R_{(k_{n})}}\right)\boldsymbol{1}_{\left\\{\frac{R_{i}}{R_{(k_{n})}}>y\right\\}}-y^{-\alpha}\Bigr{)}$
$\displaystyle\Rightarrow W(y^{-\alpha})-y^{-\alpha}W(1).$ (49)

Finally apply the mapping

$x\mapsto\int_{1}^{\infty}\frac{x(s)}{s}\mathrm{d}s$

to (6.1)  using justifications similar to [34, Section 9.1]. This yields the
asymptotic normality of $D^{*}_{n}$ under the null hypothesis in
$H^{(1)}_{0}$,

$\displaystyle\sqrt{k_{n}}$
$\displaystyle\left(\frac{1}{{k_{n}}}\sum_{i=1}^{n}\left(1+\frac{d^{*}\bigl{(}(X^{*}_{i},Y^{*}_{i}),\mathbb{C}_{a,b}\bigr{)}}{R_{(k_{n})}}\right)\log\frac{R_{(i)}}{R_{(k_{n})}}-\frac{1}{\alpha}\right)$
$\displaystyle\Rightarrow\frac{1}{\alpha}\left(\int_{0}^{1}\frac{W(s)}{s}\mathrm{d}s-W(1)\right)\stackrel{{\scriptstyle
d}}{{=}}\frac{1}{\alpha}W(1)\sim N(0,1/\alpha^{2}).$

###  6.2 Proof of Theorem 4.1

The proof proceeds in a series of steps.

  1. 1.

To prove the convergence in Eq.(23) of main document under $H^{(2)}_{0}$, we
first show that in $D(0,\infty)$,

$\displaystyle\tilde{W}_{n}(y):=\frac{\sqrt{k_{n}}}{\theta_{0}}\left(\frac{1}{k_{n}}\sum_{i=1}^{n}\Theta_{i}\epsilon_{R_{i}/b(n/k_{n})}(y,\infty)-\theta_{0}y^{-\alpha}\right)\Rightarrow
W(y^{-\alpha}).$ (50)

The LHS of (50)  can be decomposed as

$\displaystyle\frac{1}{\theta_{0}}\frac{1}{\sqrt{k_{n}}}\sum_{i=1}^{n}$
$\displaystyle\left(\Theta_{i}-\theta_{0}\right)\epsilon_{R_{i}/b(n/k_{n})}(y,\infty)$
$\displaystyle+\sqrt{k_{n}}\left(\frac{1}{k_{n}}\sum_{i=1}^{n}\epsilon_{R_{i}/b(n/k_{n})}(y,\infty)-y^{-\alpha}\right)=:B1+B2.$
(51)

From full dependence, we have
$\mathbb{C}_{a,b}\equiv\\{(x,y)\in\mathbb{R}_{+}^{2}:y=(1/\theta_{0}-1)x\\}$.
Remember $\Theta_{i}=X_{i}/R_{i}=X_{i}/(X_{i}+Y_{i})$ and then $|B1|$ is
bounded by

$\displaystyle\frac{1}{\theta_{0}}\frac{1}{\sqrt{k_{n}}}$
$\displaystyle\sum_{i=1}^{n}\Bigl{|}\frac{X_{i}}{R_{i}}-\theta_{0}\Bigr{|}\epsilon_{R_{i}/b(n/k_{n})}(y,\infty)$
$\displaystyle=$
$\displaystyle\frac{1}{\theta_{0}}\frac{1}{\sqrt{k_{n}}}\sum_{i=1}^{n}\theta_{0}\frac{|Y_{i}-X_{i}(\theta_{0}^{-1}-1)|}{R_{i}}\boldsymbol{1}_{\\{R_{i}>b(n/k_{n})y\\}}$
$\displaystyle=$
$\displaystyle\frac{1}{\sqrt{k_{n}}}\sum_{i=1}^{n}\frac{d^{*}((X_{i},Y_{i}),\mathbb{C}_{a,b})}{b(n/k_{n})y}\boldsymbol{1}_{\\{R_{i}>b(n/k_{n})y\\}}\Rightarrow
0$ (52)

in $D(0,\infty)$,  since (45) is still applicable. As in (46), the second term
$B2$ in (1) converges weakly  in $D(0,\infty)$ to $W(y^{-\alpha})$ under the
2RV condition for $\mathbb{P}[R_{1}>x]$, thus completing the proof of (50).

Combine (50) with

$\frac{R_{(k_{n})}}{b(n/k_{n})}\stackrel{{\scriptstyle
p}}{{\longrightarrow}}1,$

to get joint convergence in $D(0,\infty)\times\mathbb{R}_{+}$. Applying the
composition map $(x(t),c)\mapsto x(ct)$ from
$D(0,\infty)\times(0,\infty)\mapsto D(0,\infty)$, we get in $D(0,\infty)$,

$\displaystyle\frac{\sqrt{k_{n}}}{\theta_{0}}\left(\frac{1}{k_{n}}\sum_{i=1}^{n}\Theta_{i}\epsilon_{R_{i}/R_{(k_{n})}}(y,\infty)-\theta_{0}\left(\frac{R_{(k_{n})}}{b(n/k_{n})}y\right)^{-\alpha}\right)\Rightarrow
W(y^{-\alpha}).$ (53)

Apply Vervaat inversion again as in (48) and (49), we again conclude

$\sqrt{k_{n}}\left(\frac{1}{k_{n}\theta_{0}}\sum_{i=1}^{n}\Theta_{i}\epsilon_{R_{i}/R_{(k_{n})}}(y,\infty)-y^{-\alpha}\right)\Rightarrow
W(y^{-\alpha})-y^{-\alpha}W(1).$ (54)

  2. 2.

Next, we need to justify application of the map

$x\mapsto\int_{1}^{\infty}\frac{x(s)}{s}\mathrm{d}s$

to (54), which, if justified, leads to

$\sqrt{k_{n}}\left(\frac{1}{k_{n}\theta_{0}}\sum_{i=1}^{k_{n}}\Theta^{*}_{i}\log\frac{R_{(i)}}{R_{(k_{n})}}-\frac{1}{\alpha}\right)\Rightarrow\int_{1}^{\infty}\frac{W(y^{-\alpha})}{y}\mathrm{d}y-\frac{1}{\alpha}W(1)\sim\frac{1}{\alpha}N(0,1).$
(55)

The proof is similar to the one given in Proposition 9.1 of [34], and we defer
details to Section 6.2.1.

  3. 3.

From (55),

$\frac{1}{k_{n}}\sum_{i=1}^{k_{n}}\Theta^{*}_{i}\log\frac{R_{(i)}}{R_{(k_{n})}}\stackrel{{\scriptstyle
p}}{{\longrightarrow}}\frac{\theta_{0}}{\alpha},$ (56)

which suggests comparing

$\displaystyle\sqrt{k_{n}}\Bigl{(}T_{n}$
$\displaystyle-\frac{1}{\theta_{0}}\frac{1}{k_{n}}\sum_{i=1}^{k_{n}}\Theta^{*}_{i}\log\frac{R_{(i)}}{R_{(k_{n})}}\Bigr{)}$
(57) $\displaystyle=$
$\displaystyle\sqrt{k_{n}}\Bigl{(}\frac{1}{\frac{1}{k_{n}}\sum_{i=1}^{k_{n}}\Theta^{*}_{i}}-\frac{1}{\theta_{0}}\Bigr{)}\frac{1}{k_{n}}\sum_{i=1}^{k_{n}}\Theta^{*}_{i}\log\frac{R_{(i)}}{R_{(k_{n})}}$
and applying (56), this is $\displaystyle=$
$\displaystyle\sqrt{k_{n}}\Bigl{(}\frac{1}{\frac{1}{k_{n}}\sum_{i=1}^{k_{n}}\Theta^{*}_{i}}-\frac{1}{\theta_{0}}\Bigr{)}O_{p}(1)=\sqrt{k_{n}}\Biggl{(}\frac{\theta_{0}-\frac{1}{k_{n}}\sum_{i=1}^{k_{n}}\Theta^{*}_{i}}{\frac{1}{k_{n}}\sum_{i=1}^{k_{n}}\Theta^{*}_{i}\theta_{0}}\Biggr{)}O_{p}(1).$
(58)

Now

$\frac{1}{k_{n}}\sum_{i=1}^{n}\epsilon_{(\Theta_{i},R_{i}/R_{(k)})}\Rightarrow
S\times\nu_{\alpha}=\epsilon_{\theta_{0}}\times\nu_{\alpha}$

in $\mathbb{M}([0,1]\times\mathbb{R}_{+}\setminus\\{0\\})$ (eg. [34, p. 180])
and so

$\displaystyle\int_{[0,1]\times(1,\infty)}\theta$
$\displaystyle\frac{1}{k_{n}}\sum_{i=1}^{n}\epsilon_{(\Theta_{i},R_{i}/R_{(k)})}(d\theta,dr)=\frac{1}{k_{n}}\sum_{i=1}^{n}\Theta_{i}\boldsymbol{1}_{[R_{i}>R_{(k_{n})}]}$
$\displaystyle=$
$\displaystyle\frac{1}{k_{n}}\sum_{i=1}^{k_{n}}\Theta_{i}^{*}\Rightarrow\int_{[0,1]\times(1,\infty)}\theta\epsilon_{\theta_{0}}\times\nu_{\alpha}(d\theta,dr)=\theta_{0}.$

Thus the denominator of (58) is also $O_{p}(1)$. A successful comparison in
(57) has the difference converging to $0$ and so it remains to show

$\frac{1}{\sqrt{k_{n}}}\Bigl{|}k_{n}\theta_{0}-\sum_{i=1}^{k_{n}}\Theta^{*}_{i}\Bigr{|}\Rightarrow
0.$ (59)

Since under $H^{(2)}_{0}$,

$\displaystyle\frac{1}{\sqrt{k_{n}}}\Bigl{|}\sum_{i=1}^{k_{n}}(\Theta_{i}^{*}$
$\displaystyle-\theta_{0})\Bigr{|}=\frac{1}{\sqrt{k_{n}}}\left|\sum_{i=1}^{n}(\Theta_{i}-\theta_{0})\boldsymbol{1}_{\\{R_{i}\geq
R_{(k_{n})}\\}}\right|$
$\displaystyle\leq\frac{1}{\sqrt{k_{n}}}\sum_{i=1}^{n}\frac{d^{*}((X_{i},Y_{i}),\mathbb{C}_{a,b})}{R_{(k_{n})}}\boldsymbol{1}_{\\{R_{i}\geq
R_{(k_{n})}\\}}\stackrel{{\scriptstyle p}}{{\longrightarrow}}0,$

which can be seen as in the proof of (45) by replacement of $R_{(k_{n})}$ by
$b(n/k_{n})$ at the cost of $1\pm\epsilon$ for any $\epsilon>0$ with high
probability. This confirms (59) and thus proves convergence to 0 in (57). In
turn, this coupled with (55) proves the theorem.

####  6.2.1 Details for Step 2 in the proof of Theorem 4.1

The proof of (55) requires justifying the application of the mapping

$x\mapsto\int_{1}^{\infty}x(s)\frac{\mathrm{d}s}{s},$

and applying this mapping to (53) leads to

$\displaystyle\frac{\sqrt{k_{n}}}{\theta_{0}}\left(\frac{1}{k_{n}}\sum_{i=1}^{k_{n}}\Theta^{*}_{i}\log\frac{R_{(i)}}{R_{(k_{n})}}-\frac{\theta_{0}}{\alpha}\left(\frac{R_{(k_{n})}}{b(n/k_{n})}\right)^{-\alpha}\right)\Rightarrow\int_{1}^{\infty}\frac{W(y^{-\alpha})}{y}\mathrm{d}y.$
(60)

For $M$ large, applying the map

$x\mapsto\int_{1}^{M}\frac{x(s)}{s}\mathrm{d}s$

to (53) gives

$\displaystyle\frac{\sqrt{k_{n}}}{\theta_{0}}$
$\displaystyle\left(\int_{1}^{M}\frac{1}{k_{n}}\sum_{i=1}^{n}\Theta_{i}\epsilon_{R_{i}/R_{(k_{n})}}(s,\infty)\frac{\mathrm{d}s}{s}-\left(\frac{R_{(k_{n})}}{b(n/k_{n})}\right)^{-\alpha}\theta_{0}\int_{1}^{M}s^{-\alpha-1}\mathrm{d}s\right)$
$\displaystyle\Rightarrow\int_{1}^{M}W(s^{-\alpha})\frac{\mathrm{d}s}{s}.$

As $M\to\infty$,

$\int_{1}^{M}W(s^{-\alpha})\frac{\mathrm{d}s}{s}\to\int_{1}^{\infty}\frac{W(s^{-\alpha})}{s}\mathrm{d}s.$

Hence, it remains to verify that for any $\delta>0$,

$\displaystyle\lim_{M\to\infty}\limsup_{n\to\infty}\,$
$\displaystyle\mathbb{P}\left(\sqrt{k_{n}}\left|\int_{M}^{\infty}\frac{1}{k_{n}}\sum_{i=1}^{n}\Theta_{i}\epsilon_{R_{i}/R_{(k_{n})}}(s,\infty)\frac{\mathrm{d}s}{s}\right.\right.$
$\displaystyle\left.\left.\qquad-\theta_{0}\left(\frac{R_{(k_{n})}}{b(n/k_{n})}\right)^{-\alpha}\int_{M}^{\infty}s^{-\alpha-1}\mathrm{d}s\right|>\delta\right)=0$
(61)

Rewrite the probability in (6.2.1) as

$\displaystyle\mathbb{P}$
$\displaystyle\left(\sqrt{k_{n}}\left|\int_{M}^{\infty}\left(\frac{1}{k_{n}}\sum_{i=1}^{n}\Theta_{i}\epsilon_{R_{i}/R_{(k_{n})}}(s,\infty)-\theta_{0}\left(\frac{R_{(k_{n})}}{b(n/k_{n})}s\right)^{-\alpha}\right)\frac{\mathrm{d}s}{s}\right|>\delta\right)$
$\displaystyle\leq\mathbb{P}\left(\sqrt{k_{n}}\int_{M}^{\infty}\left|\frac{1}{k_{n}}\sum_{i=1}^{n}\Theta_{i}\epsilon_{R_{i}/R_{(k_{n})}}(s,\infty)-\theta_{0}\left(\frac{R_{(k_{n})}}{b(n/k_{n})}s\right)^{-\alpha}\right|\frac{\mathrm{d}s}{s}>\delta\right)$
$\displaystyle=\mathbb{P}\left(\sqrt{k_{n}}\int_{MR_{(k_{n})}/b(n/k_{n})}^{\infty}\left|\frac{1}{k_{n}}\sum_{i=1}^{n}\Theta_{i}\epsilon_{R_{i}/b(n/k_{n})}(s,\infty)-\theta_{0}s^{-\alpha}\right|\frac{\mathrm{d}s}{s}>\delta\right)$
$\displaystyle\leq\mathbb{P}\left(\sqrt{k_{n}}\int_{M(1-\eta)}^{\infty}\left|\frac{1}{k_{n}}\sum_{i=1}^{n}\Theta_{i}\epsilon_{R_{i}/b(n/k_{n})}(s,\infty)-\theta_{0}s^{-\alpha}\right|\frac{\mathrm{d}s}{s}>\delta\right)$
$\displaystyle\qquad\qquad+\mathbb{P}\left(\left|\frac{R_{(k_{n})}}{b(n/k_{n})}-1\right|>\eta\right),$

for $\eta>0$. Since $R_{(k_{n})}/b(n/k_{n})\stackrel{{\scriptstyle
p}}{{\longrightarrow}}1$, it suffices to consider

$\displaystyle\mathbb{P}$
$\displaystyle\left(\sqrt{k_{n}}\int_{M(1-\eta)}^{\infty}\left|\frac{1}{k_{n}}\sum_{i=1}^{n}\Theta_{i}\epsilon_{R_{i}/b(n/k_{n})}(s,\infty)-\theta_{0}s^{-\alpha}\right|\frac{\mathrm{d}s}{s}>\delta\right)$
$\displaystyle\leq\frac{k_{n}}{\delta^{2}}\mathbb{E}\left[\int_{M(1-\eta)}^{\infty}\left(\frac{1}{k_{n}}\sum_{i=1}^{n}\Theta_{i}\epsilon_{R_{i}/b(n/k_{n})}(s,\infty)-\theta_{0}s^{-\alpha}\right)^{2}\frac{\mathrm{d}s}{s}\right].$

Write what is inside the square by centering the random term to get

$\displaystyle\frac{1}{k_{n}}$
$\displaystyle\sum_{i=1}^{n}\Theta_{i}\epsilon_{R_{i}/b(n/k_{n})}(s,\infty)-\frac{n}{k_{n}}\mathbb{E}\left(\Theta_{1}\boldsymbol{1}_{\\{R_{1}>b(n/k_{n})s\\}}\right)$
$\displaystyle+$
$\displaystyle\frac{n}{k_{n}}\mathbb{E}\left(\Theta_{1}\boldsymbol{1}_{\\{R_{1}>b(n/k_{n})s\\}}\right)-\theta_{0}s^{-\alpha}$

and we get the probability in (6.2.1) bounded by

$\displaystyle\leq\frac{k_{n}}{\delta^{2}}\int_{M(1-\eta)}^{\infty}\frac{n}{k_{n}^{2}}\text{Var}\left(\Theta_{1}\boldsymbol{1}_{\\{R_{1}>b(n/k_{n})s\\}}\right)\frac{\mathrm{d}s}{s}$
$\displaystyle\quad+\frac{k_{n}}{\delta^{2}}\int_{M(1-\eta)}^{\infty}\left(\frac{n}{k_{n}}\mathbb{E}\left(\Theta_{1}\boldsymbol{1}_{\\{R_{1}>b(n/k_{n})s\\}}\right)-\theta_{0}s^{-\alpha}\right)^{2}\frac{\mathrm{d}s}{s}$
$\displaystyle=:I_{n}+II_{n}$

Since $\Theta_{1}\leq 1$ a.s., the term $I_{n}$ is bounded by

$\frac{1}{\delta^{2}}\int_{M(1-\eta)}^{\infty}\frac{n}{k_{n}}\mathbb{E}\left(\Theta_{1}\boldsymbol{1}_{\\{R_{1}>b(n/k_{n})s\\}}\right)^{{\color[rgb]{0,0,0}2}}\frac{\mathrm{d}s}{s}\leq\frac{1}{\delta^{2}}\int_{M(1-\eta)}^{\infty}\frac{n}{k_{n}}\mathbb{P}(\\{R_{1}>b(n/k_{n})s)\frac{\mathrm{d}s}{s}.$

By Karamata’s theorem, the right side converges as $n\to\infty$ to

$\frac{1}{\delta^{2}}\int_{M(1-\eta)}^{\infty}s^{-\alpha-1}\mathrm{d}s=\frac{1}{\alpha\delta^{2}}(M(1-\eta))^{-\alpha}\stackrel{{\scriptstyle
M\to\infty}}{{\longrightarrow}}0.$

For $II_{n}$, with
$v(t)=\mathbb{E}(\Theta_{1}\boldsymbol{1}_{\\{R_{1}>t\\}})$, we notice that

$\displaystyle\frac{\theta_{0}^{2}}{A^{2}(n/k_{n})}\left(\frac{n}{k_{n}\theta_{0}}\mathbb{E}\left(\Theta_{1}\boldsymbol{1}_{\\{R_{1}>b(n/k_{n})s\\}}\right)-s^{-\alpha}\right)^{2}$
$\displaystyle=\frac{\theta_{0}^{2}}{A^{2}(n/k_{n})}\left(\frac{n}{k_{n}\theta_{0}}\mathbb{E}\left(\Theta_{1}\boldsymbol{1}_{\\{R_{1}>b(n/k_{n})s\\}}\right)-\frac{v(b(n/k_{n})s)}{v(b(n/k_{n}))}+\frac{v(b(n/k_{n})s)}{v(b(n/k_{n}))}-s^{-\alpha}\right)^{2}$
$\displaystyle\leq\left|\frac{\theta_{0}}{A(n/k_{n})}\left(\frac{n}{k_{n}\theta_{0}}\mathbb{E}\left(\Theta_{1}\boldsymbol{1}_{\\{R_{1}>b(n/k_{n})s\\}}\right)-\frac{v(b(n/k_{n})s)}{v(b(n/k_{n}))}\right)\right|^{2}$
$\displaystyle\quad+\left|\frac{\theta_{0}}{A(n/k_{n})}\left(\frac{v(b(n/k_{n})s)}{v(b(n/k_{n}))}-s^{-\alpha}\right)\right|^{2}$
$\displaystyle\quad+2\left|\frac{\theta_{0}}{A(n/k_{n})}\left(\frac{n}{k_{n}\theta_{0}}\mathbb{E}\left(\Theta_{1}\boldsymbol{1}_{\\{R_{1}>b(n/k_{n})s\\}}\right)-\frac{v(b(n/k_{n})s)}{v(b(n/k_{n}))}\right)\right|$
$\displaystyle\qquad\times\left|\frac{\theta_{0}}{A(n/k_{n})}\left(\frac{v(b(n/k_{n})s)}{v(b(n/k_{n}))}-s^{-\alpha}\right)\right|.$

By [15, Theorem 2.3.9], for any $\epsilon>0$, $\delta\in(0,\alpha(1+\rho))$,
there exists $A_{0}(n/k_{n})\sim A(n/k_{n})$ as $n\to\infty$ and $n_{0}\equiv
n_{0}(\epsilon,\delta)$ such that for all $b(n/k_{n}),b(n/k_{n})s\geq n_{0}$,

$\displaystyle\left|\frac{1}{A_{0}(n/k_{n})}\left(\frac{v(b(n/k_{n})s)}{v(b(n/k_{n}))}-s^{-\alpha}\right)-s^{-\alpha}\frac{1-s^{-\alpha\rho}}{\alpha\rho}\right|\leq\epsilon
s^{-\alpha(1+\rho)}\max\\{s^{-\delta},s^{\delta}\\},$

so that

$\displaystyle\left|\frac{1}{A_{0}(n/k_{n})}\left(\frac{v(b(n/k_{n})s)}{v(b(n/k_{n}))}-s^{-\alpha}\right)\right|$
$\displaystyle\leq
s^{-\alpha}\left(\left|\frac{1-s^{-\alpha\rho}}{\alpha\rho}\right|+\epsilon
s^{-\alpha\rho}\max\\{s^{-\delta},s^{\delta}\\}\right).$

Hence,

$\displaystyle\int_{M(1-\eta)}^{\infty}$
$\displaystyle\left|\frac{\theta_{0}}{A_{0}(n/k_{n})}\left(\frac{v(b(n/k_{n})s)}{v(b(n/k_{n}))}-s^{-\alpha}\right)\right|^{2}\frac{\mathrm{d}s}{s}$
$\displaystyle\leq\int_{M(1-\eta)}^{\infty}s^{-2\alpha}\left(\left|\frac{1-s^{-\alpha\rho}}{\alpha\rho}\right|+\epsilon
s^{-\alpha\rho}\max\\{s^{-\delta},s^{\delta}\\}\right)^{2}\frac{\mathrm{d}s}{s}<\infty,$

which further implies

$\displaystyle\int_{M(1-\eta)}^{\infty}$
$\displaystyle\left|\frac{\theta_{0}}{A(n/k_{n})}\left(\frac{v(b(n/k_{n})s)}{v(b(n/k_{n}))}-s^{-\alpha}\right)\right|^{2}\frac{\mathrm{d}s}{s}<\infty,$

as $A_{0}(n/k_{n})\sim A(n/k_{n})$. In addition, since

$\displaystyle\frac{\theta_{0}}{A(n/k_{n})}$
$\displaystyle\left(\frac{n}{k_{n}\theta_{0}}\mathbb{E}\left(\Theta_{1}\boldsymbol{1}_{\\{R_{1}>b(n/k_{n})s\\}}\right)-\frac{v(b(n/k_{n})s)}{v(b(n/k_{n}))}\right)$
$\displaystyle=\frac{v(b(n/k_{n})s)}{v(b(n/k_{n}))\theta_{0}}\frac{1}{A(n/k_{n})}\left(\frac{n}{k_{n}}v(b(n/k_{n}))-\theta_{0}\right),$

then

$\displaystyle\int_{M(1-\eta)}^{\infty}$
$\displaystyle\left|\frac{\theta_{0}}{A(n/k_{n})}\left(\frac{n}{k_{n}\theta_{0}}\mathbb{E}\left(\Theta_{1}\boldsymbol{1}_{\\{R_{1}>b(n/k_{n})s\\}}\right)-\frac{v(b(n/k_{n})s)}{v(b(n/k_{n}))}\right)\right|^{2}\frac{\mathrm{d}s}{s}$
$\displaystyle=\left|\frac{1}{A(n/k_{n})}\left(\frac{n}{k_{n}}v(b(n/k_{n}))-\theta_{0}\right)\right|^{2}\int_{M(1-\eta)}^{\infty}\frac{v^{2}(b(n/k_{n})s)}{\theta_{0}^{2}v^{2}(b(n/k_{n}))}\frac{\mathrm{d}s}{s}$
$\displaystyle\to\left(\frac{1}{\theta_{0}}\int\int_{(1,\infty)\times[0,1]}\theta\chi(\mathrm{d}r,\mathrm{d}\theta)\right)^{2}\frac{(M(1-\eta))^{-2\alpha}}{2\alpha}<\infty.$

Since under the condition in (18), the intermediate sequence $\\{k_{n}\\}$
also satisfies $\sqrt{k_{n}}A(n/k_{n})\to 0$ as $n\to\infty$. We then see that

$\displaystyle k_{n}$
$\displaystyle\int_{M(1-\eta)}^{\infty}\left|\left(\frac{n}{k_{n}}\mathbb{E}\left(\Theta_{1}\boldsymbol{1}_{\\{R_{1}>b(n/k_{n})s\\}}\right)-\theta_{0}\frac{v(b(n/k_{n})s)}{v(b(n/k_{n}))}\right)\right|^{2}\frac{\mathrm{d}s}{s}$
$\displaystyle=$ $\displaystyle(\sqrt{k_{n}}A(n/k_{n}))^{2}$
$\displaystyle\quad\times\int_{M(1-\eta)}^{\infty}\left|\frac{\theta_{0}}{A(n/k_{n})}\left(\frac{n}{k_{n}\theta_{0}}\mathbb{E}\left(\Theta_{1}\boldsymbol{1}_{\\{R_{1}>b(n/k_{n})s\\}}\right)-\frac{v(b(n/k_{n})s)}{v(b(n/k_{n}))}\right)\right|^{2}\frac{\mathrm{d}s}{s}\to
0,$ and similarly, $\displaystyle k_{n}$
$\displaystyle\int_{M(1-\eta)}^{\infty}\left|{\theta_{0}}\left(\frac{v(b(n/k_{n})s)}{v(b(n/k_{n}))}-s^{-\alpha}\right)\right|^{2}\frac{\mathrm{d}s}{s}\to
0.$

So we conclude that $II_{n}\to 0$. This justifies (6.2.1), thus completing the
proof of (60).

Recall (54), and applying the mapping

$(x,y)\mapsto\left(\int_{1}^{\infty}x(s)\frac{\mathrm{d}s}{s},y\right)$

to (54) gives

$\displaystyle\frac{\sqrt{k_{n}}}{\theta_{0}}\left(\frac{1}{k_{n}}\sum_{i=1}^{k_{n}}\Theta^{*}_{i}\log\frac{R_{(i)}}{R_{(k_{n})}}-\frac{\theta_{0}}{\alpha}\right)\Rightarrow\int_{1}^{\infty}\frac{W(y^{-\alpha})}{y}\mathrm{d}y-\frac{1}{\alpha}W(1).$
(62)

###  6.3 Check conditions of the functional central limit theorem

We now present the functional central limit theorem given in [32, Theorem
10.6].

######  Theorem 6.1.

Consider the triangular array $\\{f_{n,i}(t):t\in T\\}$ with envelop function
$F_{n,i}$, independent within each row. Suppose also that $\\{f_{n,i}\\}$
satisfy

  1. (i)

$\\{f_{n,i}\\}$ are manageable;

  2. (ii)

For $X_{n}(t)=\sum_{i}(f_{n,i}(t)-\mathbb{E}(f_{n,i}(t)))$,
$H(s,t)=\lim_{n\to\infty}\mathbb{E}[X_{n}(t)X_{n}(s)]$ exists for every
$s,t\in T$;

  3. (iii)

The envelope function satisfies
$\limsup_{n\to\infty}\mathbb{E}(F^{2}_{n,i})<\infty$, and

$\sum_{i}\mathbb{E}\left(F^{2}_{n,i}\boldsymbol{1}_{\\{F_{n,i}>\eta\\}}\right)\to
0,$

for each $\eta>0$;

  4. (iv)

Let
$\rho_{n}(s,t)=\left(\sum_{i}\mathbb{E}\left(f_{n,i}(t)-f_{n,i}(s)\right)^{2}\right)^{1/2}$,
then the limit $\rho(s,t)=\lim_{n\to\infty}\rho_{n}(s,t)$ is well-defined, and
for deterministic sequences $\\{s_{n}\\}$, $\\{t_{n}\\}$, if
$\rho(s_{n},t_{n})\to 0$, then $\rho_{n}(s_{n},t_{n})\to 0$.

Then $X_{n}$ converges to a Gaussian process with zero mean and covariance
given by $H$.

To align with the statement in Theorem 6.1, we define

$f_{n,i}(t):=\frac{(\mu^{2}+\sigma^{2})^{-1/2}}{k_{n}}\Theta_{i}\epsilon_{R_{i}/b(n/k_{n})}(t^{-1/\alpha},\infty),\qquad
t\in(0,1],$

and the envelope function

$F_{n,i}:=\frac{(\mu^{2}+\sigma^{2})^{-1/2}}{k_{n}}\epsilon_{R_{i}/b(n/k_{n})}(1,\infty).$

By Definition 7.9 of [32], we see that $\\{f_{n,i}\\}$ are manageable. Also,
with

$X_{n}(t)=\sum_{i=1}^{n}\left(f_{n,i}(t)-\mathbb{E}[f_{n,i}(t)]\right),$

we have

$\displaystyle\mathbb{E}$
$\displaystyle\left(X_{n}(t)X_{n}(s)\right)=\frac{(\mu^{2}+\sigma^{2})^{-1}}{k_{n}}\sum_{i=1}^{n}\mathbb{E}\left(\Theta_{i}^{2}\epsilon_{R_{i}/b(n/k_{n})}\bigl{(}(t\wedge
s)^{-1/\alpha},\infty\bigr{)}\right)$
$\displaystyle-\frac{(\mu^{2}+\sigma^{2})^{-1}}{k_{n}}\sum_{i=1}^{n}\mathbb{E}\left(\Theta_{i}\epsilon_{R_{i}/b(n/k_{n})}\bigl{(}t^{-1/\alpha},\infty\bigr{)}\right)\mathbb{E}\left(\Theta_{i}\epsilon_{R_{i}/b(n/k_{n})}\bigl{(}s^{-1/\alpha},\infty\bigr{)}\right)$
$\displaystyle\longrightarrow t\wedge s,\qquad n\to\infty.$

For the envelope function $F_{n,i}$, we have

$\displaystyle\limsup_{n\to\infty}\sum_{i=1}^{n}\mathbb{E}(F^{2}_{n,i})\to(\mu^{2}+\sigma^{2})^{-1/2},$

and for $\delta>0$,

$\displaystyle\sum_{i=1}^{n}\mathbb{E}(F^{2+\delta}_{n,i})$
$\displaystyle=(\mu^{2}+\sigma^{2})^{-1/2}\frac{n}{k_{n}^{2+\delta}}\mathbb{P}\left(\frac{R_{1}}{b(n/k_{n})}>1\right)\to
0,$

which further implies for each $\eta>0$,

$\displaystyle\sum_{i=1}^{n}\mathbb{E}(F^{2}_{n,i}\boldsymbol{1}_{\\{F_{n,i}>\eta\\}})\to
0.$

Assume $t_{1}>t_{2}$, then for

$\rho_{n}(t_{1},t_{2}):=\left(\sum_{i=1}^{n}\mathbb{E}\left(f_{n,i}(t_{1})-f_{n,i}(t_{2})\right)^{2}\right)^{1/2},$

we have

$\displaystyle\rho_{n}(t_{1},t_{2})$
$\displaystyle=(\mu^{2}+\sigma^{2})^{-1/2}\left(\frac{n}{k_{n}}\mathbb{E}\left(\Theta_{1}\epsilon_{R_{1}/b(n/k_{n})}(t_{1}^{-1/\alpha},t_{2}^{-1/\alpha}\right)\right)^{1/2}$
$\displaystyle\longrightarrow(\mu^{2}+\sigma^{2})^{-1/2}(t_{1}-t_{2}).$

Therefore, all conditions in Theorem 6.1 are satisfied, which gives the
conclusion that in $D(0,1]$,

$\displaystyle X_{n}(t)$
$\displaystyle=\frac{(\mu^{2}+\sigma^{2})^{-1/2}}{\sqrt{k_{n}}}\sum_{i=1}^{n}\left(\Theta_{i}\epsilon_{\frac{R_{i}}{b(n/k_{n})}}(t^{-1/\alpha},\infty)-\mathbb{E}\bigl{(}\Theta_{i}\epsilon_{\frac{R_{i}}{b(n/k_{n})}}(t^{-1/\alpha},\infty)\bigr{)}\right)$
$\displaystyle\Rightarrow W(t).$

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