text:
\section{\bf Introduction}
\label{sect-1}

Risk measures or premium calculation principles are functionals $\Pi: \mathcal{F} \to [0,+\infty]$ from the set $\mathcal{F}$ of all `loss' cumulative distribution functions (cdfs) $F$  to the extended non-negative real line $[0,+\infty]$ (e.g., Denuit et al. 2005). The corresponding random variables  $X \sim F$ are called `loss' random variables, and we assume them to be non-negative throughout this paper. We also assume that the variables $X$ follow, or are modeled with, continuous cdfs. Given a cdf $F$, the corresponding quantile function $Q:(0,1)\to [0,\infty )$ is defined by
\[
Q(t)=\inf\{x:\, F(x)\ge t\} .
\]
The quantile function $Q$ plays a pivotal role in defining numerous risk measures, and is a well known risk measure itself, called the value at risk and denoted by $\textrm{VaR}[t,F]$ (e.g., Denuit et al. 2005). Other illustrative examples of risk measures follow next.

\begin{example}\label{ex-1.1}\rm
Let $g:[0,1]\to [0,1]$ be a distortion function, that is, a non-decreasing function such that $g(0)=0$ and $g(1)=1$. The distortion risk measure (Denneberg, 1994; Wang, 1995, 1998) is defined by the formula
\[
\Pi_{d}[F]= \int_0^{\infty }g(1-F(x))dx,
\]
which can be rewritten in terms of the quantile function $Q$ as follows:
\begin{equation}
\Pi_{d}[F]=\int_{0}^{1} Q(s)d\Psi(s),
\label{pcp-dist}
\end{equation}
where $\Psi(s)=-g(1-s)$. Assuming that $g$ is left-continuous, the function $\Psi $ is right-continuous, and Note \ref{nt-1} below will clarify our reason for imposing this type of continuity. Hence, $\Pi_{d}[F]$ is an $L$-functional, a property that was highlighted and utilized by Jones and Zitikis (2003) and subsequently used by many researchers for developing statistical inferential results for distortion risk measures in light- and heavy-tailed settings.
\end{example}

\begin{note}\label{nt-1}\rm
To make this paper less cumbersome, we always write integrals as $\int_a^b g(x)dh(x)$ irrespectively of whether integrators $h$ are right- or left-continuous functions. What we have in mind behind such integrals is
$\int_{(a,b]} g(x)dh(x)$ in the case of right-continuous $h$ and $\int_{[a,b)} g(x)dh(x)$ in the case of left-continuous $h$. For example, the integral $\int_{0}^{1} Q(s)d\Psi(s)$ on the right-hand side of equation (\ref{pcp-dist}) means $\int_{(0,1]} Q(s)d\Psi(s)$ because $\Psi $ is right-continuous as noted below equation (\ref{pcp-dist}).
\end{note}

\begin{example}[Continuation of Example \ref{ex-1.1}]\label{ex-1.1b}\rm
Since $g$ is non-decreasing, the function $\Psi $ is also non-decreasing. In the special case when
\begin{equation}
\Psi(s)=-(1-s)^{1/\rho}
\label{psi-pht}
\end{equation}
with parameter $\rho \ge 1$, the distortion risk measure $\Pi_{d}[F]$ is called the proportional hazards transform and frequently denoted by $\textrm{PHT}[\rho,F]$. When
\begin{equation}
\Psi(s)={(s-t)_{+} \over 1-t }
\label{psi-cte}
\end{equation}
with parameter $t \in [0,1)$, then the risk measure $\Pi_{d}[F]$ is known as the tail value at risk and frequently denoted by  $\textrm{TVaR}[t,F]$. (We have used the classical notation $(s-t)_{+}$ for the positive part of $s-t$.) Since the cdf $F$ is continuous throughout this paper by assumption, the risk measure $\textrm{TVaR}[t,F]$ coincides with the conditional tail expectation  $\textrm{CTE}[t,F]=\mathbf{E}[X|X>Q(t)]$. For details on these and other risk measures, we refer to, e.g., Denuit et al. (2005) and references therein. This concludes Example \ref{ex-1.1}.
\end{example}

\begin{example}\label{ex-1.2a}\rm
The relative distortion risk measure (Wang, 1998) is given by the formula
\[
\Pi_{rd}[F]= {\Pi_{d}[F] \over \mathbf{E}[X]}.
\]
 The measure can be expressed (see equations of Example \ref{ex-1.1}) in terms of the quantile function $Q$ as follows:
\begin{equation}
\Pi_{rd}[F]={\int_{0}^{1} Q(s)d\Psi(s) \over \int_{0}^{1} Q(s)ds } .
\label{pcp-reldist}
\end{equation}
This is a ratio of $L$-functionals. With this note, we conclude Example \ref{ex-1.2a}.
\end{example}

\begin{example}\label{ex-1.2b}\rm
Let $w:[0,\infty )\to [0,\infty )$ be a non-decreasing function, called weight function. The weighted risk measure, or the weighted premium calculation principle (Furman and Zitikis, 2008), is given by the formula
\[
\Pi_{w}[F]={\frac{\mathbf{E}[Xw(X)] }{\mathbf{E}[w(X)]}}.
\]
For details on $\Pi_{w}[F]$ and its extensions, we refer to Furman and Zitikis (2009). With $U$ denoting a uniform on $[0,1]$ random variable, the variables $X$ and $Q(U)$ are equal in distribution, and so we can rewrite $\Pi_{w}[F]$ in terms of the quantile function $Q$ as follows:
\begin{equation}
\Pi_{w}[F]
= {\frac{\int_{0}^{1} H_{1}\circ Q(s)ds }{\int_{0}^{1} H_{2}\circ Q(s)ds}},
\label{pcp-weig}
\end{equation}
where $H_{1}(x)=xw(x)$ and $H_{2}(x)=w(x)$ with $H_{i}\circ Q(s)$
denoting the composition $H_{i}(Q(s))$ of the two functions $H_{i}$ and $Q$.
This concludes Example \ref{ex-1.2b}.
\end{example}

All of the aforementioned risk measures are special cases of the risk measure
\begin{equation}
\Pi_{r}[F]={\int_{0}^{1} H_{1}\circ Q(s)d\Psi_1(s) \over \int_{0}^{1} H_{2}\circ Q(s)d\Psi_2(s)  },
\label{gen-ratio}
\end{equation}
where $H_i:[0,\infty )\to [0,\infty ) $, $i=1,2$, are two non-decreasing and  left-continuous functions, and $\Psi_i:[0,1]\to \mathbf{R} $, $i=1,2$, are two non-decreasing and right-continuous functions. In addition to the aforementioned (absolute) PHT and CTE/TVaR risk measures, the ratio risk measure $\Pi_{r}[F]$ also includes the relative PHT risk measure $\int_0^{\infty }(1-F(x))^{1/\rho }dx / \mathbf{E}[X]$ and the relative CTE/TVaR risk measure $\mathbf{E}[X|X>Q(t)]/ \mathbf{E}[X]$. Many indices of economic inequality are also of form (\ref{gen-ratio}), as elucidated by Greselin et al. (2009). Another example of $\Pi_{r}[F]$ will follow after a note.

\begin{note}\rm
Some applications might lead to non-monotonic functions $H_{i}^{\circ}$. In this case we need to assume that the functions are of bounded variation, and to also require that each of the two non-decreasing components $H_{i}^{\ast}$ and $H_{i}^{\ast\ast}$ in the decomposition $H_{i}^{\circ}=H_{i}^{\ast}-H_{i}^{\ast\ast}$ satisfies the conditions to be imposed on $H_{i}$ in our following considerations. Same arguments apply to non-monotonic functions $\Psi_{i}$, assuming in particular that they are of bounded variation.
\end{note}

\begin{example}\label{ex-zenga}\rm
The risk measure $\Pi_{r}[F]$ includes, as a special case, the ratio
\[
R_F(p)
={\mathbf{E}[X|X \le Q(p)] \over \mathbf{E}[X|X > Q(p)] },
\]
which defines the Zenga curve $Z_F(p)$, $0\le p \le 1$, via the equation $Z_F(p)=1-R_F(p)$. We refer to the original papers by Zenga (1987, 2007) for interpretations and other details related to $Z_F$, and to Greselin et al. (2010) for statistical inferential results when (income) distributions are light-tailed. It is useful to rewrite $R_F(p)$ in terms of the aforementioned CTE/TVaR and PHT risk measures, which is accomplished by the equation
\[
\mathbf{E}[X|X \le Q(p)]=\bigg ( 1- {1\over p} \bigg )\mathbf{E}[X|X > Q(p)]+{1\over p}\mathbf{E}[X].
\]
Hence,
\[
R_F(p)=\mathcal{H}_p \big ( \mathbf{E}[X|X > Q(p)], \mathbf{E}[X] \big ),
\]
where the coupling function $\mathcal{H}_p:[0,\infty ) \times [0,\infty ) \to [0,\infty )$ is given by the formula
\[
\mathcal{H}_p(x,y)=1- {1\over p} +{1\over p}\bigg ( { y \over x } \bigg ).
\]
This concludes Example \ref{ex-zenga}.
\end{example}

Hence, the class of interesting and useful coupling functions spans beyond ratios $x/y$. This suggests considering a most general `coupled risk measure,' defined as follows: Let $\mathcal{H}:[0,\infty ) \times [0,\infty ) \to [0,\infty )$ be a `coupling' function, which couples two basic risk measures
\[
L_{i}[F]=\int_{0}^{1} H_{i}\circ Q(s)d\Psi_{i}(s), \quad i=1,2,
\]
into one
\[
\Pi[F]=\mathcal{H}(L_{1}[F],L_{2}[F]),
\]
which we call the coupled risk measure. Our task in this paper is to develop a statistical inferential theory for this risk measure when loss variables follow heavy-tailed distributions. The corresponding theory in the case of light-tailed distributions is already available in the literature, and we refer to Brazauskas et al. (2007, 2009) and references therein for details.

The rest of this paper is organized as follows. In Section \ref{sect-2} we introduce an empirical estimator, denoted by $\widehat{\Pi}_{n}$, of the coupled risk measure $\Pi[F]$ when losses are heavy-tailed. In Section \ref{sect-3a} we establish weak approximations and thus asymptotic normality of the estimator $\widehat{\Pi}_{n}$ under several sets of conditions. In Section \ref{sect-3b} we give illustrative examples of the aforementioned weak approximations. Proofs are given in Section \ref{sect-4}.

\section{\bf Constructing an estimator for $\Pi[F]$}
\label{sect-2}

The cdf $F$ is unknown, and thus the risk measure $\Pi[F]$ is unknown. Estimating $\Pi\lbrack F]$ crucially relies on estimating $L_{i}[F]$. To work out our initial intuition on the topic, we start with a brief discussion of what happens within the classical CLT-like framework, and why we need to depart from it.

It is natural to construct an estimator for $L_{i}[F]$ by simply replacing the (unknown) population cdf $F$ by its empirical counterpart $F_{n}$, which gives the weighted sum
\[
\widetilde{L}_{i,n}=\sum_{j=1}^{n}c_{i,j,n}  H_{i}(X_{j:n})
\]
of the (observable) random variables $H_{i}(X_{j:n})$, $1\le j \le n$,
with the coefficients
\[
c_{i,j,n}=\Psi_{i}\left(  \frac{j}{n}\right)-\Psi_{i}\left(  \frac{j-1}{n}\right) .
\]
It is known from the theory of $L$-statistics (e.g., Shorack and Wellner, 1986) that under some assumptions on $H_{i}$, $\Psi_{i}$, and $Q$, the following asymptotic-normality result holds:
\begin{equation}
\sqrt{n}\,(\widetilde{L}_{i,n}-L_{i}[F])\rightarrow_{d}\mathcal{N}(0,\sigma_{F}^{2}),
\label{L-stats0}
\end{equation}
provided that the variance
\[
\sigma_{F}^{2}=\int_{0}^{1}\int_{0}^{1}(H_{i}\circ Q)'(s)(H_{i}\circ
Q)'(t)(\min(s,t)-st)d\Psi_{i}(s)d\Psi_{i}(t)
\]
is finite. In the case of the CTE/TVaR risk measure, this scenario has been thoroughly investigated by Brazauskas et al. (2008), and for general $L$-type risk measures by Brazauskas et al. (2007, 2009). The finiteness of the variance $\sigma_{F}^{2}$ is, however, often violated by heavy-tailed distributions. Hence, we need to develop another approach for deriving statistical inferential results in the case of such distributions, and we shall do so next.

We note at the outset that a number of special cases that are covered by the coupled risk measure $\Pi[F]$ have been investigated in the literature within the heavy-tailed framework. For example, Peng (2001) has established a ground-breaking statistical inferential theory for the net premium. Necir et al. (2007), Necir and Meraghni (2009) have developed an analogous theory for the proportional hazards transform. Necir et al. (2010) have tackled the conditional tail expectation. Necir and Meraghni (2010) have devoted their research to general $L$-functionals. All of these risk measures are special cases of the above introduced coupled risk measure $\Pi[F]$, and thus our inferential theory developed in the following sections will cover all these special cases.

Thus, our task now is to modify the classical estimator $\widetilde{L}_{i,n}$ in such a way that it would work in the heavy-tailed setting. For this, keeping in mind that high quantiles are estimated differently from the intermediate ones of the (observable) random variables $H_{i}(X_{1}),\dots,H_{i}(X_{n})$, we introduce integers $k=k_{n}$, which depend on $n$ and are such that
\begin{equation}
k\rightarrow\infty \quad \textrm{and} \quad k/n\rightarrow0,
\label{con-00e}
\end{equation}
with further assumptions specified later in this paper. Next we write the decomposition $L_{i}[F]=L_{i,n}(1)+L_{i,n}(2)$ with
\[
L_{i,n}(1)=\int_{0}^{1-k/n}H_{i}\circ Q(s)d\Psi_{i}(s)
\]
and
\[
L_{i,n}(2)=\int_{1-k/n}^{1}H_{i}\circ Q(s)d\Psi_{i}(s).
\]
Finally, we replace the quantile function $Q$ in the definition of $L_{i,n}(1)$ by the
classical non-parametric estimator $Q_{n}$, and then replace $H_{i}\circ Q$ in the definition of $L_{i,n}(2)$ by any of the many high-quantile estimators available in the literature (see, e.g., Beirlant et al., 2004; Castillo et al. 2005; and references therein), which we denote by $\widehat{H_{i}\circ Q}$. Throughout the present paper we work with the Weissman (1978) estimator
\[
\widehat{H_{i}\circ Q}(s)=\left(  \frac{k}{n}\right)  ^{\widehat{\gamma}_{i}}H_{i}(X_{n-k:n})(1-s)^{-\widehat{\gamma}_{i}},\quad s\in\big (1-k/n,1\big ),
\]
where $\widehat{\gamma}_{i}$ is the Hill (1975) estimator of the tail index $\gamma_{i}\in(1/2,1)$ defined by the formula
\[
\widehat{\gamma}_{i}={\frac{1}{k}}\sum_{j=1}^{k}
\log \bigg ( {H_{i}(X_{n-j+1:n}) \over H_{i}(X_{n-k:n})} \bigg ).
\]
In summary, we have arrived at the estimator
\begin{equation}
\widehat{L}_{i,n}=\widehat{L}_{i,n}(1)+\widehat{L}_{i,n}(2),
\label{decomp-emp}
\end{equation}
where
\[
\widehat{L}_{i,n}(1)   =\sum_{j=1}^{n-k}c_{i,j,n}  H_{i}\left(X_{j:n}\right)
\]
and
\[
\widehat{L}_{i,n}(2) = c_{i,\vartriangle,n} H_{i}(X_{n-k:n})
\]
with
\[
c_{i,\vartriangle,n}=\left(  \frac{k}{n}\right)
^{\widehat{\gamma}_{i}}\int_{1-k/n}^{1}(1-s)^{-\widehat{\gamma}_{i}}d\Psi
_{i}(s).
\]
Note the similarity between $\widehat{L}_{i,n}(1)$ and the `classical' estimator $\widetilde{L}_{i,n}$, but $\widehat{L}_{i,n}(2)$ and $\widetilde{L}_{i,n}$ are quite different. Replacing $L_{1}[F]$ and $L_{2}[F]$ in $\mathcal{H}(L_{1}[F],L_{2}[F])$ by the above constructed $\widehat{L}_{1,n}$ and $\widehat{L}_{2,n}$, we obtain the estimator
\[
\widehat{\Pi}_{n}=\mathcal{H}(\widehat{L}_{1,n},\widehat{L}_{2,n})
\]
of $\Pi[F]$. In the next section we shall establish the asymptotic distribution of $\widehat{\Pi}_{n}$.

For developing statistical inferential results (e.g., confidence intervals and hypothesis tests) for $\Pi[F]$, we need to derive asymptotic distributions of $\widehat{L}_{1,n}$ and $\widehat{L}_{2,n}$, which we do in the next section.
The following regular variation condition plays a decisive role in the derivations. Namely, let the function $H_{i}\circ Q$ be regularly varying at $1$ with index $-\gamma_{i}<0$, that is,
\begin{equation}
\lim_{\epsilon\downarrow0}{\frac{H_{i}\circ Q(1-\epsilon s)}{H_{i}\circ
Q(1-\epsilon)}}=s^{-\gamma_{i}}
\label{tag-2u}
\end{equation}
for every $s>0$. This is the so-called first-order condition, which is sufficient  for proving consistency of the estimator $\widehat{\Pi}_{n}$. For establishing the asymptotic distribution of the estimator, we need a second-order condition, which specifies the rate of convergence in the first-order condition. Namely, assume that, for every $s>0$,
\begin{equation}
\lim_{\epsilon\downarrow0}{\frac{1}{A_{i}(1/\epsilon)}}\left( {\frac {H_{i}\circ Q(1-\epsilon s)}{H_{i}\circ Q(1-\epsilon)}}-s^{-\gamma_{i}}\right)  =s^{-\gamma_{i}}\frac{s^{-\omega_{i}}-1}{\omega_{i}} ,
\label{tag-2A}
\end{equation}
where $\omega_{i}\leq0$ is the so-called second-order parameter and
$x\mapsto A_{i}(x)$ is a function that does not change its
sign for all sufficiently large $x$ and converges to $0$ when $x\uparrow \infty $.
When $\omega_{i}=0$, then the ratio on the right-hand side of equation
(\ref{tag-2A}) is interpreted as $\log s$.

\section{\bf Establishing the asymptotic distribution of $\widehat{\Pi}_{n}$}
\label{sect-3a}

We assume that the first partial derivatives of the functional $\mathcal{H}$ are continuous at the point $(L_{1}[F],L_{2}[F])$, and not both of them are zero at the point. By $\mathcal{H}^{(1)}_{x}(L_{1}[F],L_{2}[F])$ and $\mathcal{H}^{(1)}_{y}(L_{1}[F],L_{2}[F])$ we denote the first partial
derivatives  of $\mathcal{H}(x,y)$ with respect to $x$ and $y$, respectively, both evaluated at the point $(x,y)=(L_{1}[F],L_{2}[F])$. The following lemma plays a fundamental role in our considerations.

\begin{lemma}\label{lemma-funda}
Assume that there are standard Brownian bridges $B_{n}$ defined on a possibly different probability space than the original one, and also assume that, for each $i=1$ and $i=2$, there are constants $b_{i,n}$ and $\lambda_{i}$, and linear functionals $\ell_{i,n}$ defined on the space of all functions such that
\begin{equation}
b_{i,n}\big(\widehat{L}_{i,n}-L_{i}[F]\big)
=\lambda_{i}+\ell_{i,n}(B_{n})+o_{\mathbf{P}}(1)
\label{approxim-1}
\end{equation}
when $n\rightarrow\infty$. Furthermore, let the constants $b_{i,n}$ be such that
\begin{equation}
{\frac{b_{2,n}}{b_{1,n}+b_{2,n}}}\rightarrow\delta
\label{approxim-2}
\end{equation}
for some $\delta\in [0,1]$. Then, when $n\rightarrow\infty$,
\begin{equation}
{\frac{b_{1,n}b_{2,n}}{b_{1,n}+b_{2,n}}}\big(\widehat{\Pi}_{n}-\Pi[F]\big)
=\lambda+\ell_n(B_{n})+o_{\mathbf{P}}(1)
\label{approxim-1a}
\end{equation}
with the bias term
\[
\lambda=\delta\mathcal{H}^{(1)}_{x}(L_{1}[F],L_{2}[F])\lambda
_{1}+(1-\delta)\mathcal{H}^{(1)}_{y}(L_{1}[F],L_{2}[F])\lambda_{2}%
\]
and the linear functional
\[
\ell_n=\delta\mathcal{H}^{(1)}_{x}(L_{1}[F],L_{2}[F])\ell_{1,n}
+(1-\delta)\mathcal{H}^{(1)}_{y}(L_{1}[F],L_{2}[F])\ell_{2,n}.
\]
\end{lemma}

Lemma \ref{lemma-funda} reduces our task of deriving the asymptotic distribution of the coupled risk measure $\Pi[F]$ to establishing statement (\ref{approxim-1}) for both $i=1$ and $i=2$. Recall that $L_{i}=L_{i,n}(1)+L_{i,n}(2)$ and $\widehat{L}_{i,n}=\widehat{L}_{i,n}(1)+\widehat{L}_{i,n}(2)$. These two equations further reduce our task to showing that, with the above defined constants $b_{i,n}$ and Brownian bridges $B_{n}$, for each $\Delta =1$ and $\Delta =2$ there are constants $\lambda_{i}(\Delta)$ and linear functionals $\ell_{i,n}(\Delta;\bullet)$ such that, when $n\rightarrow\infty$,
\begin{equation}
b_{i,n}\big(\widehat{L}_{i,n}(\Delta)-L_{i,n}(\Delta)\big)
=\lambda_{i}(\Delta)+\ell_{i,n}(\Delta;B_{n})+o_{\mathbf{P}}(1).
\label{approxim-4}
\end{equation}
In view of this assumption, the earlier noted constants $\lambda_{i}$ and linear functionals $\ell_{i}$ are the sums
\begin{equation}
\lambda_{i}=\lambda_{i}(1)+\lambda_{i}(2)
\label{approxim-4con}
\end{equation}
and
\begin{equation}
\ell_{i,n}=\ell_{i,n}(1;\bullet)+\ell_{i,n}(2;\bullet).
\label{approxim-4lin}
\end{equation}
In various generalities, statement (\ref{approxim-4}) when $\Delta =1$ will be established in Theorems \ref{theor-1ab} and \ref{theor-1a} below, whereas Theorem \ref{theor-1b} and Corollary \ref{utheor-1b} will deal with the case $\Delta=2$.

\begin{note}\label{probab-space}\rm
In all of the following results, the Brownian bridges $B_{n}$ and the probability space on which they are defined can be chosen same, and this is a crucial property as it allows us to combine various asymptotic results into one.
\end{note}

In what follows we use the notation
\[
D_{i,n}=H_{i}\circ Q(1-k/n)\bigg({k \over n}\bigg )^{\gamma_{i}}\int_{1-k/n}^{1}(1-s)^{-\gamma
_{i}}d\Psi_{i}(s)
\]
and assume that, when $n\rightarrow\infty$,
\begin{equation}
\sqrt{{n\over k}} ~ D_{i,n}\rightarrow\infty .
\label{con-0e}
\end{equation}

The next theorem establishes statement (\ref{approxim-4}) when $\Delta =1$ under a very weak assumption on the function $F\circ H_{i}^{-1}$, assuming only that it is continuous, but this generality is achieved at the expense of requiring $\Psi_{i}(s)=s$ for all $s\in [0,1]$, which is indeed a restriction though luckily satisfied by a number of important examples.

\begin{theorem}\label{theor-1ab}
Let the function $F\circ H_{i}^{-1}$ be continuous and $\Psi_{i}(s)=s$ for all $s\in [0,1]$. Then there are Brownian bridges $B_{n}$ (on a possibly different probability space) such that
\begin{equation}
\frac{\sqrt{n}(\widehat{L}_{i,n}(1)-L_{i,n}(1)}{(n/k)^{1/2}D_{i,n}}=\ell
_{i,n}(1;B_{n})+o_{\mathbf{P}}(1)
\label{equa-10cc1}
\end{equation}
with
\[
\ell_{i,n}(1;B_{n})={\frac{-1}{(n/k)^{1/2}D_{i,n}}}\int_{0}^{1-k/n}%
B_{n}(s)dH_{i}\circ Q(s).
\]
\end{theorem}

The following theorem allows for a general class of functions $\Psi_{i}$ but imposes a requirement that $H_{i}\circ Q$ is continuously differentiable.

\begin{theorem}\label{theor-1a}
Let $H_{i}\circ Q$ be continuously differentiable on $[0,1)$. Furthermore, let $\sqrt{n}\,(\Psi_{i}(1/n)-\Psi_{i}(0))=O(1)$ when $n\rightarrow\infty$, and let $s\rightarrow\int_{s}^{1}(1-t)^{-\gamma_{i}}d\Psi
_{i}(t)$ be regularly varying at $1$ with an index $\kappa\in(0,1/2)$.  Then there
are Brownian bridges $B_{n}$ (on a possibly different probability space) such that statement (\ref{equa-10cc1}) holds with
\[
\ell_{i,n}(1;B_{n})={\frac{-1}{(n/k)^{1/2}D_{i,n}}}\int_{0}^{1-k/n}(H_{i}\circ
Q)'(s)B_{n}(s)d\Psi_{i}(s).
\]
\end{theorem}

The previous two theorems deal with low and moderate quantiles. The next theorem and its corollary deal with high quantiles.

\begin{theorem}\label{theor-1b}
Let the following three limits exist and be finite:
\begin{equation}
b_{i}=\lim_{n\rightarrow\infty}\sqrt{k}\,A_{i}(n/k),
\label{con-0a}
\end{equation}
\begin{equation}
c_{i}=\lim_{n\rightarrow\infty}{\frac{\int_{0}^{1}s^{-\gamma_{i}}\log
(s)d\Psi_{i}(1-ks/n)}{\int_{0}^{1}s^{-\gamma_{i}}d\Psi_{i}(1-ks/n)}},
\label{con-0b}
\end{equation}
\begin{equation}
d_{i}=\lim_{n\rightarrow\infty}\frac{\int_{0}^{1}s^{-\gamma_{i}}(s^{-\omega_{i}%
}-1)d\Psi_{i}\left(  1-ks/n\right)  }{\int_{0}^{1}s^{-\gamma_{i}}d\Psi
_{i}\left(  1-ks/n\right)  }.
\label{con-0c}
\end{equation}
Furthermore, assume that, for some $\delta>0$ and when $n\rightarrow\infty$,
\begin{equation}
{\frac{\int_{0}^{1}s^{-\gamma_{i}-\omega_{i}-\delta}d\Psi_{i}(1-ks/n)}{\int
_{0}^{1}s^{-\gamma_{i}}d\Psi_{i}(1-ks/n)}}=O(1).
\label{equa-10c}
\end{equation}
Then there are Brownian bridges $B_{n}$ (on a possibly different probability space) such that
\begin{equation}
\frac{\sqrt{n}(\widehat{L}_{i,n}(2)-L_{i,n}(2))}{(n/k)^{1/2}D_{i,n}}
={\frac{-b_{i}d_{i}}{\omega_{i}}}+\ell_{i,n}(2;B_{n})+o_{\mathbf{P}}(1),
\label{equa-10cc2}
\end{equation}
where
\[
\ell_{i,n}(2;B_{n})=-\gamma_{i}(1+c_{i})\sqrt{{\frac{n}{k}}}\,B_{n}(1-k/n)
+c_{i}\gamma_{i}\sqrt{{\frac{n}{k}}}\,\int_{1-k/n}^{1}\frac{B_{n}(s)}{1-s}ds.
\]
\end{theorem}

The following corollary is a special case of Theorem \ref{theor-1b} when $\Psi_{i}(s)=s$.

\begin{corollary}\label{utheor-1b}
Let $\Psi_{i}(s)=s$ for all $s\in [0,1]$, and let the limit $b_i$ defined in (\ref{con-0a}) exist and be finite. Furthermore, let $\gamma_{i}+\omega_{i}<1$. Then there are Brownian bridges $B_{n}$ (on a possibly different probability space) such that
\begin{equation}
\frac{\sqrt{n}(\widehat{L}_{i,n}(2)-L_{i,n}(2))}{(n/k)^{1/2}D_{i,n}}%
={\frac{-b_{i}d_{i}}{\omega_{i}}}+\ell_{i,n}(2;B_{n})+o_{\mathbf{P}}(1),
\label{uequa-10cc2}
\end{equation}
where $c_{i}$ in the definition of $\ell_{i,n}(2;B_{n})$ and the constant $d_{i}$ are given by the formulas
\[
c_{i}={-1 \over 1-\gamma_{i}}
\quad \textrm{and} \quad
d_{i}={\omega_{i} \over 1-\gamma_{i}-\omega_{i}}.
\]
\end{corollary}

We have noted above that the Brownian bridges $B_{n}$ and the probability space are same in all our results. This allows us to pair, for example, Theorem \ref{theor-1ab} with Corollary \ref{theor-1b}, and Theorem \ref{theor-1a} with Theorem \ref{theor-1b}. Two results of this pairing are given next.

\begin{corollary}\label{cor-1}
Assume that one of the following two requirements is satisfied:
\begin{enumerate}
  \item $\Psi_i(x)=s$ for all $s\in [0,1]$, and let the conditions of Theorem \ref{theor-1ab} and Corollary \ref{theor-1b} be satisfied.
  \item The conditions of Theorems \ref{theor-1a} and \ref{theor-1b} are satisfied.
\end{enumerate}
Then there are Brownian bridges $B_{n}$ (on a possibly different probability space) such that
\begin{equation}
\frac{\sqrt{n}(\widehat{L}_{i,n}-L_{i})}{(n/k)^{1/2}D_{i,n}}={\frac
{-b_{i}d_{i}}{\omega_{i}}}+\ell_{i,n}(B_{n})+o_{\mathbf{P}}(1), \label{equa-10fin}
\end{equation}
where $\ell_{i,n}=\ell_{i,n}(1;\bullet )+\ell_{i,n}(2;\bullet)$.
\end{corollary}

Corollary \ref{cor-1} together with Lemma \ref{lemma-funda} imply the following theorem, which we view as the main result of the present paper.

\begin{theorem}\label{th-main-2}
Assume that when $i=1$, and also when $i=2$, one of the following two requirements is satisfied:
\begin{enumerate}
  \item $\Psi_i(x)=s$ for all $s\in [0,1]$, and let the conditions of Theorem \ref{theor-1ab} and Corollary \ref{theor-1b} be satisfied.
  \item The conditions of Theorems \ref{theor-1a} and \ref{theor-1b} are satisfied.
\end{enumerate}
Furthermore, let there exist a constant $\delta\in [0,1]$ such that, when $n\rightarrow\infty$,
\begin{equation}
{\frac{D_{1,n}}{D_{1,n}+D_{2,n}}}\rightarrow\delta.
\label{approxim-5a}
\end{equation}
Then there are Brownian bridges $B_{n}$ (on a possibly different probability space) such that
\begin{equation}
\frac{\sqrt{n}\big(\widehat{\Pi}_{n}-\Pi[F]\big)}{(n/k)^{1/2}(D_{1,n}+D_{2,n})}
=\lambda+\ell_n(B_{n})+o_{\mathbf{P}}(1)
\label{approxim-5b}
\end{equation}
with the bias term
\[
\lambda=\delta\mathcal{H}^{(1)}_{x}(L_{1}[F],L_{2}[F]){\frac{-b_{1}d_{1}}{\omega_{1}}}
+(1-\delta)\mathcal{H}^{(1)}_{y}(L_{1}[F],L_{2}[F]){\frac{-b_{2}d_{2}}{\omega_{2}}}
\]
and the leading Gaussian term
\[
\ell_n=\delta\mathcal{H}^{(1)}_{x}(L_{1}[F],L_{2}[F])\ell_{1,n}
+(1-\delta)\mathcal{H}^{(1)}_{y}(L_{1}[F],L_{2}[F])\ell_{2,n},
\]
where $\ell_{i,n}=\ell_{i,n}(1;\bullet )+\ell_{i,n}(2;\bullet)$.
\end{theorem}

We can use Theorem \ref{th-main-2} to derive asymptotic distributions of various special cases of $\widehat{\Pi}_{n}$ and thus obtain statistical inferential results for the corresponding $\Pi[F]$. Naturally, one of the most challenging tasks that inevitably arises is obtaining explicit expressions for $\ell_n(B_{n})$. To illustrate how this can be done, in the next section we present detailed examples tackling the PHT and CTE/TVaR risk measures.

\section{\bf Brownian bridge approximations: illustrative examples}
\label{sect-3b}

Our illustrative examples in Section \ref{sect-1} were based on three risk measures: the proportional hazards transform $\textrm{PHT}[\rho,F]$, the conditional tail expectation $\textrm{CTE}[t,F]$, and the net premium $\mathbf{E}[X]$, which can be viewed, for example, as the proportional hazards transform $\textrm{PHT}[\rho,F]$ with the parameter $\rho =1$. Hence, in order to develop a statistical inferential theory for $\Pi[F]=\mathcal{H}(L_{1}[F],L_{2}[F])$ with $L_{1}[F]$ and $L_{2}[F]$ being two of the aforementioned three risk measures, we need to establish asymptotic representations (\ref{equa-10fin}) for the proportional hazards transform (PHT) and the conditional tail expectation (CTE). This we do next. Since we work individually with each of the two risk measures, we shall drop the subindex $i$ from the rest of this section.

\subsection*{PHT}

In the case of the proportional hazards transform $\textrm{PHT}[\rho,F]$, we have   $\Psi(s)=-(1-s)^{1/\rho}$ and $H(x)=x$ (Example \ref{ex-1.1}). Using decomposition (\ref{decomp-emp}), the estimator $\widehat{PHT}_n$ can be expressed as
\[
\widehat{PHT}_n= \sum_{j=1}^{n-k}
\bigg\{ \bigg(  1-\frac{j-1}{n}\bigg)  ^{1/\rho}-\bigg(  1-\frac{j}{n}\bigg)  ^{1/\rho}\bigg \}X_{j:n}
+\bigg( \frac{k}{n}\bigg)^{1/\rho}\frac{X_{n-k:n}}{1-\rho\widehat{\gamma}}
\]
with
\begin{equation}
\widehat{\gamma}={1\over k}\sum_{j=1}^{k}
\log\bigg ( {X_{n-j+1:n} \over X_{n-k:n}} \bigg ).
\label{hill-1}
\end{equation}
Brownian bridge approximations for $\widehat{PHT}_n$ can be established using Corollary \ref{cor-1} under the conditions of Theorems \ref{theor-1a} and \ref{theor-1b}. For this, we first calculate
\[
D_{n}=\bigg( \frac{k}{n}\bigg)^{1/\rho}\frac{Q(1-k/n)}{1-\rho\gamma}
\]
and easily check that $(n/k)^{1/2}D_{n}\to \infty $ when $n\to \infty $.  Assuming that the quantile function $Q$ is continuously differentiable on $[0,1)$,  with other conditions of Theorem \ref{theor-1a} satisfied automatically, we have that
\begin{equation}
\ell_{n}(1;B_{n})={\frac{-(1/\rho-\gamma)}{(k/n)^{1/\rho -1/2}Q(1-k/n)}}\int
_{0}^{1-k/n}(1-s)^{1/\rho-1}B_{n}(s)dQ(s).
\label{ell-pht-1}
\end{equation}
Assume that $\lim_{n\rightarrow\infty}\sqrt{k}\,A_{1}(n/k)=0$. Limits (\ref{con-0b}) and (\ref{con-0c}) are
\begin{equation}
c={-1 \over 1/\rho-\gamma}
\quad \textrm{and} \quad
d={\omega \over 1/\rho-\gamma-\omega},
\label{lim-cANDd}
\end{equation}
provided that $1/\rho-\gamma-\omega>0$. The ratio on the right hand side of equation (\ref{equa-10c}) is equal to $( 1/\rho -\gamma )/( 1/\rho -\gamma-\omega-\delta) $ under the assumption that $\delta>0$ is so small that $\delta<1/\rho-\gamma-\omega$. (We can always choose such $\delta>0$ because $1/\rho-\gamma-\omega>0$ by assumption.)  We can now conclude from Theorem \ref{theor-1b} that
\begin{equation}
\ell_{n}(2;B_{n})=-\gamma \bigg ( 1- {1 \over 1/\rho-\gamma}\bigg ) \sqrt{{\frac{n}{k}}}\,B_{n}(1-k/n)
-\gamma{ 1 \over 1/\rho-\gamma}\sqrt{{\frac{n}{k}}}\,\int_{1-k/n}^{1}\frac{B_{n}(s)}{1-s}ds.
\label{ell-pht-2}
\end{equation}
Hence, $\ell_{n}(B_{n})$ is the sum of the right-hand sides of equations (\ref{ell-pht-1}) and (\ref{ell-pht-2}). The sum is asymptotically Gaussian with the mean zero and the variance (cf.\, Necir and Meraghni, 2009)
\begin{equation}
\frac{\gamma^{2}(  \gamma^{2}\rho^{2}-2\gamma^{2}\rho^{3}+4\gamma\rho^{2}-2\gamma\rho+\rho^{2}-2\rho+1)  }{(\gamma\rho-1)^{2}}+\frac{2\gamma^{2}(\rho+\gamma\rho-1)  }{\rho+2\gamma\rho-2}.
\label{var-1}
\end{equation}

\subsection*{An interlude}

We note that it is not the variance expression (\ref{var-1}) that is important for us --  unless we consider the proportional hazards transform $\textrm{PHT}[\rho,F]$ as a stand alone risk measure -- but the above derived expression for $\ell_{n}(B_{n})$ as it allows us to combine the expression with the corresponding one of the other risk measure making up the coupled risk measure under consideration. This explains our interest in deriving expressions of $\ell_{n}(B_{n})$ for various risk measures. For example, if the second risk measure is the mean, which is the net premium, then an expression for the corresponding $\ell_{n}(B_{n})$ follows from the formulas of the previous subsection by setting $\rho =1$. If the second risk measure is the CTE/TVaR, then we refer to our next subsection.

\subsection*{CTE/TVaR}

In the case of the conditional tail expectation $\textrm{CTE}[t,F]$, we have   $\Psi(s)=(s-t)_{+} / (1-t)$ and $H(x)=x$ (Example \ref{ex-1.1}). Using decomposition (\ref{decomp-emp}), we express the estimator $\widehat{CTE}_n$ by the formula
\[
\widehat{CTE}_n=\dfrac{1}{1-t}\sum_{j=1}^{n-k} \bigg ( \left(  \frac{j}{n}-t\right)_{+}-\left(  \frac{j-1}{n}-t\right)_{+}  \bigg )  X_{j:n}+\frac{kX_{n-k:n}}{n\left(  1-t\right)  \left(  1-\widehat{\gamma}\right)  }
\]
for any fixed $0<t<1$ and for all sufficiently large $n$, where $\widehat{\gamma}$ is same as in (\ref{hill-1}). Assume that $\lim_{n\rightarrow\infty}\sqrt{k}\,A_{1}(n/k)=0$. Limits (\ref{con-0b}) and (\ref{con-0c}) are
\[
c={-1 \over 1-\gamma}
\quad \textrm{and} \quad
d={\omega \over 1-\gamma-\omega},
\]
provided that $\gamma+\omega<1$. (Same as in (\ref{lim-cANDd}) but with $\rho=1$.) The ratio on the right hand side of equation (\ref{equa-10c}) is equal to $1/( 1 -\gamma-\omega-\delta) $, assuming that $\delta>0$ is so small that $\delta<1-\gamma-\omega$. From Theorems \ref{theor-1a} and \ref{theor-1b} we have Brownian bridges $B_{n}$ such that
\begin{equation}
\ell_{n}(1;B_{n})=\frac{-(1-\gamma )}{(k/n)^{1/2}Q(1-k/n)} \int_{0}^{1-k/n}B_{n}(s)dQ(s)
\label{stat-1}
\end{equation}
and
\begin{equation}
\ell_{n}(2;B_{n})={\gamma^2 \over 1-\gamma }
 \sqrt{{n\over k}}\,B_{n}(1-k/n)
-{\gamma \over 1-\gamma }
\sqrt{{n\over k}}\,\int_{1-k/n}^{1}\frac{B_{n}(s)}{1-s}ds.
\label{stat-2}
\end{equation}
Hence, $\ell_{n}(B_{n})$ is equal to the sum of the right-hand sides of equations (\ref{stat-1}) and (\ref{stat-2}). This sum is asymptotically Gaussian with the mean zero and the variance (cf.\, Necir et al., 2010)
\begin{equation}
{\frac{\gamma^{4}}{(1-\gamma)^{2}(2\gamma-1)}}.
\label{var-2}
\end{equation}
We highlight again that it is not the latter variance but the above derived formula for $\ell_{n}(B_{n})$ that is of our primary interest in the present paper.

\subsection*{Useful formulas}

Calculating the second moments or, equivalently, the variances of the functionals  $\ell_{n}(B_{n})$ in our above examples reduces to calculating the second and mixed moments of the following three random variables:
\begin{align*}
W_{1,n}&={1\over (k/n)^{1/\rho -1/2}Q(1-k/n)}\int_{0}^{1-k/n}(1-s)^{1/\rho-1}B_{n}(s)dQ(s),
\\
W_{2,n}&= {\sqrt{{n\over k}}\,B_{n}(1-k/n)},
\\
W_{3,n}&= {\sqrt{{n\over k}}\,\int_{1-k/n}^{1}\frac{B_{n}(s)}{1-s}ds}.
\end{align*}
We next give formulas that make such calculations straightforward. Namely, when $n\to \infty $, we have the limits:
\begin{align*}
& \mathbf{E}[W_{1,n}^{2}]\rightarrow {\gamma^2 \over (1/\rho-\gamma-1)(1/\rho-\gamma-1/2)},
\\
& \mathbf{E}[W_{2,n}^{2}]\rightarrow 1,
\\
& \mathbf{E}[W_{3,n}^{2}]\rightarrow 2,
\\
& \mathbf{E}[W_{1,n}W_{2,n}]\rightarrow {\gamma \rho \over 1/\rho-\gamma-1},
\\
& \mathbf{E}[W_{1,n}W_{3,n}]\rightarrow {-\gamma  \over 1/\rho-\gamma-1},
\\
& \mathbf{E}[W_{2,n}W_{3,n}]\rightarrow 1.
\end{align*}
Using these limits, we can now easily check, for example, that $\mathbf{E}[\ell_{n}^2(B_{n})]$ is asymptotically (when $n\to \infty $) equal to quantity (\ref{var-1}) in the PHT case and to (\ref{var-2}) in the CTE/TVaR case.

\section{\bf Proofs}
\label{sect-4}

We begin with a proof of Theorem \ref{theor-1a}, which is then followed by a proof of Theorem \ref{theor-1ab}. This sequence of proofs is natural since Theorem \ref{theor-1ab} assumes $\Psi_{i}(s)=s$ and thus allows us to employ an additional technical tool, called the Vervaat process, which enables us to relax the differentiability of $F\circ H_{i}^{-1}$ to only continuity.

\begin{proof}[\textbf{Proof of Theorem \ref{theor-1a}}]

We first show that
\begin{equation}
{\frac{\sqrt{n}}{(n/k)^{1/2}D_{i,n}}}\int_{0}^{1/n}\left(  H_{i}\circ
Q_{n}(s)-H_{i}\circ Q(s)\right)  d\Psi_{i}(s)=o_{\mathbf{P}}(1).
\label{stat-1a}
\end{equation}
Since $Q_{n}(s)=X_{1:n}$ when $s\in (0,1/n]$ and
$X_{1:n}=O_{\mathbf{p}}(1)$, and since the function $H_{i}$ is non-decreasing,
we have that $\sup_{s\in (0,1/n]}H_{i}\circ Q_{n}(s)=O_{\mathbf{p}}(1) $. Consequently,  the quantity $\sqrt{n}\int_{0}^{1/n}H_{i}\circ Q_{n}(s)ds$ is of the order
$O_{\mathbf{p}}( 1/\sqrt{n})  $ and thus, in particular, converges to zero in probability. Furthermore, since the function $H_{i}\circ Q$ is non-decreasing, we have that
\begin{equation}
\sqrt{n}\int_{0}^{1/n}H_{i}\circ Q(s)d\Psi_{i}(s)
\leq \sqrt{n}(\Psi_{i}(1/n) -\Psi_{i}(0) ) H_{i}\circ Q(1/n) .
\label{stat-1ak}
\end{equation}
By assumption, $\sqrt{n}(\Psi_{i}(1/n) -\Psi_{i}(0) )$ is bounded, and so the left-hand side of inequality (\ref{stat-1ak}) is asymptotically bounded. This proves statement (\ref{stat-1a}) because $(n/k)^{1/2}D_{i,n}\rightarrow \infty $ when $n\rightarrow\infty$ by assumption (\ref{con-0e}).

In view of statement (\ref{stat-1a}), the proof of Theorem \ref{theor-1a} reduces to showing that
\begin{equation}
{\frac{\sqrt{n}}{(n/k)^{1/2}D_{i,n}}}\int_{1/n}^{1-k/n}\left(  H_{i}\circ
Q_{n}(s)-H_{i}\circ Q(s)\right)  d\Psi_{i}(s)=\ell_{i,n}(1;B_{n})+o_{\mathbf{P}}(1). \label{stat-1new}
\end{equation}
Before doing so, we introduce additional notation. Let $E_{n}^{-1}$ denote the uniform quantile function based on the uniform on $[0,1]$ random variables $F(X_{1}),\dots,F(X_{n})$ (recall that the cdf $F$ is continuous by assumption).
Hence, $Q_{n}(s)=Q(E_{n}^{-1}(s))$. With the notation
\begin{equation}
\vartheta_{n}(s)=E_{n}^{-1}(s)-s ,
\label{quant-process}
\end{equation}
and using the assumed differentiability of the function $H_{i}\circ Q$, we have that
\begin{align}
& {\frac{\sqrt{n}}{(n/k)^{1/2}D_{i,n}}} \int_{1/n}^{1-k/n}\left(  H_{i}\circ
Q(s+\vartheta_{n}(s))-H_{i}\circ Q(s)\right)  d\Psi_{i}(s)
\notag
\\
&={\frac{\sqrt{n}}{(n/k)^{1/2}D_{i,n}}}\int_{1/n}^{1-k/n}\mathbf{E}_{\tau
}\left[  (H_{i}\circ Q)'(s+\tau\vartheta_{n}(s))\right]  \vartheta
_{n}(s)d\Psi_{i}(s)
\notag
\\
&= {\frac{\sqrt{n}}{(n/k)^{1/2}D_{i,n}}}\int_{1/n}^{1-k/n}\Big (\mathbf{E}_{\tau
}\left[  (H_{i}\circ Q)'(s+\tau\vartheta_{n}(s))\right]
-(H_{i}\circ Q)'(s)\Big )\vartheta_{n}(s)d\Psi_{i}(s)
\notag
\\
&\quad +{\frac{\sqrt{n}}{(n/k)^{1/2}D_{i,n}}}\int_{1/n}^{1-k/n}(H_{i}\circ
Q)'(s)\vartheta_{n}(s)d\Psi_{i}(s)
\label{stat-1b}
\end{align}
where $\tau$ is a uniform on $[0,1]$ random variable, independent of all
other random variables, and $\mathbf{E}_{\tau}$ denotes the expectation with
respect to $\tau$, with all other random variables being fixed.

We shall next prove that
\begin{equation}
{\frac{\sqrt{n}}{(n/k)^{1/2}D_{i,n}}}\int_{1/n}^{1-k/n}\Big (\mathbf{E}_{\tau
}\left[  (H_{i}\circ Q)'(s+\tau\vartheta_{n}(s))\right]
-(H_{i}\circ Q)'(s)\Big )\vartheta_{n}(s)d\Psi_{i}(s)=o_{\mathbf{P}}(1).
\label{stat-1d}
\end{equation}
For this we first fix any $\epsilon\in(0,1)$ and establish statement (\ref{stat-1d}) with the integral $\int_{1/n}^{1-k/n}$ replaced by $\int_{1/n}^{1-\epsilon}$. The function $(H_{i}\circ Q)'$ is uniformly continuous on the interval $[0,1-\epsilon]$ and the process $\vartheta_{n}$ converges to $0$ uniformly on $[0,1]$. Moreover, $\sup_{0<s<1}|\sqrt{n}\,\vartheta_{n}(s)|=O_{\mathbf{P}}(1)$. Since $(n/k)^{1/2}D_{i,n}\rightarrow\infty$, we therefore have that
\begin{equation}
{\frac{\sqrt{n}}{(n/k)^{1/2}D_{i,n}}}\int_{1/n}^{1-\epsilon}|\vartheta
_{n}(s)|d\Psi_{i}(s)=O_{\mathbf{P}}(1).
\label{stat-1d4}
\end{equation}
To complete the proof of statement (\ref{stat-1d}), we are left to
verify statement (\ref{stat-1d}) with the integral $\int_{1/n}^{1-k/n}$ replaced by $\int_{1-\epsilon}^{1-k/n}$, that is, we need to prove that
\begin{equation}
{\frac{\sqrt{n}}{(n/k)^{1/2}D_{i,n}}}\int_{1-\epsilon}^{1-k/n}\Big (\mathbf{E}%
_{\tau}\left[  (H_{i}\circ Q)'(s+\tau\vartheta_{n}(s))\right]
-(H_{i}\circ Q)'(s)\Big )\vartheta_{n}(s)d\Psi_{i}(s)=o_{\mathbf{P}}(1).
\label{stat-1d2}
\end{equation}
To this end, we first use the fact that $\sup_{1-\epsilon<s<1-k/n}{|\sqrt
{n}\,\vartheta_{n}(s)|/(1-s)^{1/2}}=O_{\mathbf{P}}(1)$, which follows from
statement (2.2) on page 40 of Cs\"{o}rg\H{o} et al. (1986) and the proof of statement (2.39) on page 49 of Cs\"{o}rg\H{o} et al. (1986). Then we use the fact that
\begin{equation}
\sup_{1-\epsilon\leq s\leq1-k/n}\bigg |{\frac{\mathbf{E}_{\tau}\left[
(H_{i}\circ Q)'(s+\tau\vartheta_{n}(s))\right]  }{(H_{i}\circ
Q)'(s)}}-1\bigg |=o_{\mathbf{P}}(1), \label{stat-1d6}%
\end{equation}
which is a consequence of Lemma 3 in Necir and Meraghni (2009). Hence, statement (\ref{stat-1d2})
holds if
\begin{equation}
{\frac{1}{(n/k)^{1/2}D_{i,n}}}\int_{1-\epsilon}^{1-k/n}(H_{i}\circ
Q)'(s)^{1/2}d\Psi_{i}(s)=O(1). \label{stat-1d8}%
\end{equation}
To verify statement (\ref{stat-1d8}), we use the fact that
$(1-s)(H_{i}\circ Q)'(s)/H_{i}\circ Q(s)$ is asymptotically bounded when
$s\uparrow1$. This allows us to replace the derivative $(H_{i}\circ
Q)'(s)$ by $(1-s)H_{i}\circ Q(s)$ and reduces the proof of statement (\ref{stat-1d8}) to showing that, when $n\rightarrow \infty $,
\begin{equation}
{\frac{1}{a(1-k/n)b(1-k/n)}}\int_{1-\epsilon}^{1-k/n}a(s)db(s)=O(1),
\label{stat-3b0}
\end{equation}
where $a(s)=H_{i}\circ Q(s)(1-s)^{\gamma_{i}-1/2}$
and $b(s)=\int_{s}^{1}(1-t)^{-\gamma_{i}}d\Psi_{i}(t)$. We shall establish statement (\ref{stat-3b0}) in Lemma \ref{lemma4.1} below, taking now the validity of the statement for granted. This concludes the proof of statement (\ref{stat-1d8}).

In summary, we have prove that
\begin{multline}
{\frac{\sqrt{n}}{(n/k)^{1/2}D_{i,n}}}\int_{1/n}^{1-k/n}\left(  H_{i}\circ
Q_{n}(s)-H_{i}\circ Q(s)\right)  d\Psi_{i}(s)\\
={\frac{\sqrt{n}}{(n/k)^{1/2}D_{i,n}}}\int_{1/n}^{1-k/n}(H_{i}\circ
Q)'(s)\vartheta_{n}(s)d\Psi_{i}(s)+o_{\mathbf{P}}(1).
\label{stat-1c}
\end{multline}
Our next task is to establish a weak approximation for the main term on the right-hand side of equation (\ref{stat-1c}). Namely, we shall show that
\begin{multline}
{\frac{\sqrt{n}}{(n/k)^{1/2}D_{i,n}}}\int_{1/n}^{1-k/n}(H_{i}\circ Q)^{\prime
}(s)\vartheta_{n}(s)d\Psi_{i}(s)\\
={\frac{-1}{(n/k)^{1/2}D_{i,n}}}\int_{1/n}^{1-k/n}(H_{i}\circ Q)^{\prime
}(s)B_{n}(s)d\Psi_{i}(s)+o_{\mathbf{P}}(1),
\label{stat-9e}
\end{multline}
where, defined on a possibly different probability space, $B_{n}$ are Brownian bridges such that (see statement (2.2) on page 40 of Cs\"{o}rg\H{o} et al., 1986) for
any $0\leq\nu<1/2$ we have
\begin{equation}
\sup_{1/n\leq s\leq(n-1)/n}n^{\nu}{\frac{|\sqrt{n}\,\vartheta_{n}%
(s)+B_{n}(s)|}{s^{1/2-\nu}(1-s)^{1/2-\nu}}}=O_{\mathbf{P}}(1)
\label{stat-cchm}
\end{equation}
when $n\rightarrow\infty$.

\begin{note}\rm
The need to change the original probability space into a new one does not affect our results. In fact, we can start working on the probability space of Cs\"{o}rg\H{o} et al. (1986) right at the very outset since all our results are `in probability.'
\end{note}

In view of (\ref{stat-cchm}), statement (\ref{stat-9e}) follows if
\begin{equation}
{\frac{n^{-\nu_{i}}}{(n/k)^{1/2}D_{i,n}}}\int_{1/n}^{1-k/n}(H_{i}\circ
Q)'(s)^{1/2-\nu_{i}}d\Psi_{i}(s)\rightarrow0,
\label{stat-9a}
\end{equation}
where $\nu_{i}\in [0,1/2)$ can be any, even though our arguments below will require
$\nu_{i}$ to be strictly positive, though no matter how small it could be.

To prove statement (\ref{stat-9a}), we employ the earlier noted fact that $(1-s)(H_{i}\circ Q)'(s)/H_{i}\circ Q(s)=O(1)$ when $s\uparrow1$, and reduce the proof to showing that
\begin{equation}
{\frac{1}{k^{\nu_{i}}}}\bigg ({\frac{\int_{1/n}^{1-k/n}H_{i}\circ
Q(s)(1-s)^{\gamma_{i}-1/2-\nu_{i}}(1-s)^{-\gamma_{i}}d\Psi_{i}(s)}{H_{i}\circ
Q(1-k/n)(k/n)^{\gamma_{i}-1/2-\nu_{i}}\int_{1-k/n}^{1}(1-s)^{-\gamma_{i}}%
d\Psi_{i}(s)}}\bigg )\rightarrow0.
\label{stat-10a}
\end{equation}
In view of statement (\ref{stat-3b0}), but now with the function $a(s)=H_{i}%
\circ Q(s)(1-s)^{\gamma_{i}-1/2-\nu_{i}}$ (see Lemma \ref{lemma4.1} below), we have that the ratio in the parentheses is asymptotically bounded. Since $\nu_{i}>0$ and  $k=k_{n}\rightarrow\infty$, statement (\ref{stat-10a}) holds and thus statement (\ref{stat-9a}) follows.

We conclude the proof of Theorem \ref{theor-1a} by noting that the
integral on the right-hand side of equation (\ref{stat-9e}) can be made into
$\ell_{i,n}(1;B_{n})$. For this we need to check that
\[
{\frac{-1 }{(n/k)^{1/2}D_{i,n} }}\int_{0}^{1/n} (H_{i}\circ Q)'(s)
B_{n}(s) d\Psi_{i}(s) = o_{\mathbf{P}}(1),
\]
but latter statement holds because the integral is asymptotically bounded in probability and $(n/k)^{1/2}D_{i,n}\to\infty$ when $n\to\infty$. This concludes the proof of Theorem \ref{theor-1a}.
\end{proof}

\begin{lemma}\label{lemma4.1}
Statement (\ref{stat-3b0}) holds with the functions $a(s)=H_{i}\circ Q(s)(1-s)^{\gamma_{i}-1/2-\nu_i}$ and $b(s)=\int_{s}^{1}(1-t)^{-\gamma_{i}}d\Psi_{i}(t)$ assuming that the parameter
$\nu_{i}$ is either zero or (strictly) positive but sufficiently small.
\end{lemma}

\begin{proof}
To prove the lemma, we show that the limit
\begin{equation}
\lim_{\epsilon\downarrow0}{1\over \alpha(\epsilon)\beta(\epsilon)} \int_{\epsilon}^{1/2} \alpha
(s)d\beta(s)
\label{statem-this}
\end{equation}
is finite, where $\alpha(s)=H_{i}\circ Q(1-s) s^{\gamma_{i}-1/2-\nu_i}$ and $\beta(s)=\int_{1-s}^{1} (1-t)^{-\gamma_{i}}d\Psi_{i}(t)$. This reformulation of the problem helps us to connect the current proof with that of Lemma 1 in Necir and Meraghni (2010). Since $H_{i}\circ Q$ is differentiable and regularly varying at $1$ with the index $-\gamma_{i} <0$, the function $s\mapsto \alpha(s)$ is differentiable and regularly
varying at $0$ with the index $-\gamma_{i} + (\gamma_{i}-1/2-\nu_i)$.
Since the function $s\mapsto b(s)$ is regularly varying at $1$ with the index $\kappa> 0$ by an assumption of Theorem \ref{theor-1a}, the function $s\mapsto \beta(s)$ is regularly varying at $0$ with the index $\kappa$.
Consequently, the product function $s\mapsto \alpha(s)\beta(s)$ is regularly
varying at $0$ with the index $\kappa-1/2-\nu_i$, which is strictly negative for all sufficiently small $\nu_i$, and thus the product function converges to $+\infty$ when $s\downarrow0$. This fact and integration by parts formula imply that limit (\ref{statem-this}) is finite when the limit
\begin{equation}
\lim_{\epsilon\downarrow 0}{\frac{1 }{\alpha(\epsilon)\beta(\epsilon)}} \int_{\epsilon}^{1/2}\alpha'(s)\beta(s)ds
\label{statem-this-2}
\end{equation}
is such. Since $\epsilon \alpha'(\epsilon)=(-1/2)\alpha(\epsilon)(1+o(1))$ when $\epsilon\downarrow0$ by Karamata's representation (e.g., Proposition
B.1.9 (11) on page 367 of de Hann and Ferreira, 2006), limit (\ref{statem-this-2}) is finite when the limit
\begin{equation}
\lim_{\epsilon\downarrow0}{\frac{1 }{\epsilon\alpha'(\epsilon)\beta(\epsilon)}} \int_{\epsilon}^{1/2}\alpha'(s)\beta(s)ds
\label{stat-3b0ii}
\end{equation}
is such. The function $s\mapsto \alpha'(s)\beta(s)$ is regularly varying at $0$ with the index $(-1/2)-1+\kappa<0$, and thus Theorem B.1.5 on page 363 of de Haan and Ferreira (2006) implies the finiteness of limit (\ref{stat-3b0ii}). This proves the asymptotic boundedness of the right-hand side of equation (\ref{stat-3b0ii}) and  completes the proof of Lemma \ref{lemma4.1}.
\end{proof}

\begin{proof}[\textbf{Proof of Theorem \ref{theor-1ab}}]

We have
\begin{align}
\frac{\sqrt{n}(\widehat{L}_{i,n}(1)-L_{i,n}(1)}{(n/k)^{1/2}D_{i,n}}  &
={\frac{\sqrt{n}}{(n/k)^{1/2}D_{i,n}}}\int_{0}^{1-k/n}\left(  H_{i}\circ
Q_{n}(s)-H_{i}\circ Q(s)\right)  ds.\nonumber\\
&  ={\frac{\sqrt{n}}{(n/k)^{1/2}D_{i,n}}}\int_{1/n}^{1-k/n}\left(  H_{i}\circ
Q_{n}(s)-H_{i}\circ Q(s)\right)  ds+o_{\mathbf{P}}(1), \label{eq-start-1}%
\end{align}
where the second equality follows from statement (\ref{stat-1a}). Our
next step is based on the Vervaat process
\[
V_{n}(t)=\int_{0}^{t}\left(  H_{i}\circ Q_{n}(s)-H_{i}\circ Q(s)\right)
ds+\int_{-\infty}^{H_{i}\circ Q(t)}\left(  F_{n}\circ H_{i}^{-1}(x)-F\circ
H_{i}^{-1}(x)\right)  dx.
\]
We write the integral on the right-hand side of equation (\ref{eq-start-1})
as follows:
\begin{align}
\int_{1/n}^{1-k/n}
&\left(  H_{i}\circ Q_{n}(s)-H_{i}\circ Q(s)\right)  ds
\notag
\\
&=-\int_{H_{i}\circ Q(1/n)}^{H_{i}\circ Q(1-k/n)}\left(  F_{n}\circ H_{i}%
^{-1}(x)-F\circ H_{i}^{-1}(x)\right)  dx+V_{n}(1-k/n)-V_{n}(1/n)
\notag
\\
&=-\int_{H_{i}\circ Q(1/n)}^{H_{i}\circ Q(1-k/n)}e_{n}(F\circ H_{i}^{-1}(x))dx+V_{n}(1-k/n)-V_{n}(1/n),
\label{eq-0}
\end{align}
where $e_{n}(t)=\sqrt{n}\,(F_{n}\circ Q(t)-t)$ is the uniform empirical process.
It is known (see Zitikis, 1998; Davydov and Zitikis, 2003,
2004) that $V_{n}(t)$ is non-negative and does not exceed $-(F_{n}\circ
H_{i}^{-1}\circ H_{i}\circ Q(t))-t)(H_{i}\circ Q_{n}(t)-H_{i}\circ Q(t))$.
Since $F\circ H_{i}^{-1}$ is continuous by assumption, we therefore have the bound
\begin{equation}
\sqrt{n}\,V_{n}(t)\leq|e_{n}(t)||H_{i}\circ Q_{n}(t)-H_{i}\circ Q(t)|.
\label{eq-1}
\end{equation}
Consequently,
\begin{equation}
\frac{\sqrt{n}(\widehat{L}_{i,n}(1)-L_{i,n}(1)}{(n/k)^{1/2}D_{i,n}}%
=\frac{-1}{(n/k)^{1/2}D_{i,n}}\int_{H_{i}\circ Q(1/n)}^{H_{i}\circ Q(1-k/n)}e_{n}(F\circ H_{i}^{-1}(x))dx+O_{\mathbf{P}}(r_{n}^{*}+r_{n}^{**}),
\label{eq-4}
\end{equation}
where
\[
r_{n}^{*}={\frac{|e_{n}(1-k/n)||H_{i}\circ Q_{n}(1-k/n)-H_{i}\circ Q(1-k/n)|}{(n/k)^{1/2}D_{i,n}}}
\]
and
\[
r_{n}^{**}={\frac{|e_{n}(1/n)||H_{i}\circ Q_{n}(1/n)-H_{i}\circ Q(1/n)|}{(n/k)^{1/2} D_{i,n}}}.
\]
We have $r_{n}^{**}=o_{\mathbf{P}}(1)$ because
$|e_{n}(1/n)||H_{i}\circ Q_{n}(1/n)-H_{i}\circ Q(1/n)|=O_{\mathbf{P}}(1)$ and
$(n/k)^{1/2}D_{i,n}\rightarrow\infty$. To show that $r_{n}^{*}=o_{\mathbf{P}}(1)$, we shall next check that
\begin{equation}
\frac{|e_{n}(1-k/n)|}{(k/n)^{1/2}}\bigg |{\frac{H_{i}\circ Q_{n}(1-k/n)}%
{H_{i}\circ Q(1-k/n)}}-1\bigg |=o_{\mathbf{P}}(1). \label{eq-10}%
\end{equation}
For this, we write the bound
\begin{equation}
\frac{|e_{n}(1-k/n)|}{(k/n)^{1/2}}\leq\frac{|e_{n}(1-k/n)-B_{n}(1-k/n)|}%
{(k/n)^{1/2}}+\frac{|B_{n}(1-k/n)|}{(k/n)^{1/2}}. \label{eq-12}%
\end{equation}
The first summand on the right-hand side of bound (\ref{eq-12}) is of the
order $O_{\mathbf{P}}(1)$ due to Corollary 2.1 on page 48 of Cs\"{o}rg\H{o}
et al. (1986), which states that on an appropriate probability space and for any $0\leq\nu<1/4$, we have that
\begin{equation}
\sup_{1/n\leq s\leq1-1/n}n^{\nu}{\frac{|e_{n}(s)-B_{n}(s)|}{s^{1/2-\nu
}(1-s)^{1/2-\nu}}}=O_{\mathbf{P}}(1).
\label{str-approx}
\end{equation}
Statement (\ref{str-approx}) with $\nu=0$ implies that the first summand on the right-hand side of bound (\ref{eq-12}) is of the order $O_{\mathbf{P}}(1)$.

\begin{note}\rm
Compare statements (\ref{str-approx}) and (\ref{stat-cchm}): both of them use
same Brownian bridges $B_{n}$ and probability spaces as specified in
Cs\"{o}rg\H{o} et al. (1986).
\end{note}

The second summand on the right-hand side of bound (\ref{eq-12}) is also of the order $O_{\mathbf{P}}(1)$ due to a statement on page 49 of Cs\"{o}rg\H{o} et al. (1986): see the displayed bound there just below statement (2.39). Hence, statement (\ref{eq-10}) follows from
\begin{equation}
{\frac{H_{i}\circ Q_{n}(1-k/n)}{H_{i}\circ Q(1-k/n)}}=1+o_{\mathbf{P}}(1),
\label{eq-13}
\end{equation}
which we shall establish in Lemma \ref{lemma4.2} below.
In summary, we have proved that
\begin{equation}
\frac{\sqrt{n}(\widehat{L}_{i,n}(1)-L_{i,n}(1)}{(n/k)^{1/2}D_{i,n}}
=\frac{-1}{(n/k)^{1/2}D_{i,n}}\int_{H_{i}\circ Q(1/n)}^{H_{i}\circ Q(1-k/n)}e_{n}(F\circ H_{i}^{-1}(x))dx+o_{\mathbf{P}}(1).
\label{eq-4oo}
\end{equation}

We next replace the empirical process $e_{n}$ n the right-hand side of equation
(\ref{eq-4oo}) by an appropriate Brownian bridge $B_{n}$ with an error term of the order $o_{\mathbf{P}}(1)$. Statement (\ref{str-approx})
implies that the replacement of $e_{n}$ by $B_{n}$ is possible with an error  $o_{\mathbf{P}}(1)$, provided that the quantity
\[
\frac{1}{n^{\nu}(n/k)^{1/2}D_{i,n}}\int_{H_{i}\circ Q(1/n)}^{H_{i}\circ Q(1-k/n)}F\circ H_{i}%
^{-1}(x)^{1/2-\nu}(1-F\circ H_{i}^{-1}(x))^{1/2-\nu}dx
\]
converges to $0$ when $n\rightarrow\infty$. To show this, we first change the
variable of integration, then integrate by parts, and in this way reduce our task to
showing that, when $n\rightarrow\infty$,
\begin{equation}
\frac{1}{n^{\nu}(n/k)^{1/2}D_{i,n}}s^{1/2-\nu}(1-s)^{1/2-\nu}H_{i}\circ Q(s)|_{1/n}^{1-k/n} \to 0
\label{eq-re1}
\end{equation}
and
\begin{equation}
\frac{1}{n^{\nu}(n/k)^{1/2}D_{i,n}}\int_{1/n}^{1-k/n}s^{-1/2-\nu}
(1-s)^{-1/2-\nu}H_{i}\circ Q(s)ds \to 0.
\label{eq-re2}
\end{equation}

To prove statement (\ref{eq-re1}), we first recall that we are
currently dealing with the case $\Psi_i(s)=s$, and then replace $D_{i,n}$ by
$(k/n)H_{i}\circ Q(1-k/n)$ since, up to a constant, the two quantities are asymptotically equivalent. We have that
\begin{multline}
\frac{1}{n^{\nu}(k/n)^{1/2}H_{i}\circ Q(1-k/n)}s^{1/2-\nu}(1-s)^{1/2-\nu}H_{i}\circ Q(s)|_{1/n}^{1-k/n}
\\
=O\bigg (\frac{1}{k^{\nu}}\bigg )+O\bigg (\frac{H_{i}\circ Q(1/n)}{n^{1/2}(n/k)^{1/2}D_{i,n}}\bigg ),
\label{eq-8}
\end{multline}
which converges to $0$ because $k=k_{n}\rightarrow\infty$ and $(n/k)^{1/2}%
D_{i,n}\rightarrow\infty$ when $n\rightarrow\infty$.

To prove statement (\ref{eq-re2}), we use the notation
$\phi(u)=H_{i}\circ Q(1-u)/u^{1/2+\nu}$ and write
\begin{equation}
\frac{1}{n^{\nu}(k/n)^{1/2}H_{i}\circ Q(1-k/n)}
\int_{1/n}^{1-k/n}(1-s)^{-1/2-\nu}H_{i}\circ Q(s)ds
={\frac{1}{k^{\nu}}}{\frac{\int_{k/n}^{1-1/n}\phi(s)ds}{(k/n)\phi(k/n)}}.
\label{eq-9}
\end{equation}
The right-hand side converges to $0$ when $n\rightarrow\infty$ as seen from
Result 1 in Appendix of Necir and Meraghni (2009).

In summary, we have proved that
\begin{align}
\frac{\sqrt{n}(\widehat{L}_{i,n}(1)-L_{i,n}(1)}{(n/k)^{1/2}D_{i,n}}
&=\frac{-1}{(n/k)^{1/2}D_{i,n}}\int_{H_{i}\circ Q(1/n)}^{H_{i}\circ Q(1-k/n)}B_{n}\circ F\circ H_{i}^{-1}(x)dx
+o_{\mathbf{P}}(1)
\notag
\\
&  ={\frac{-1}{(n/k)^{1/2}D_{i,n}}}\int_{1/n}^{1-k/n}B_{n}(s)dH_{i}\circ
Q(s)+o_{\mathbf{P}}(1)
\notag
\\
&  ={\frac{-1}{(n/k)^{1/2}D_{i,n}}}\int_{0}^{1-k/n}B_{n}(s)dH_{i}\circ
Q(s)+o_{\mathbf{P}}(1).
\label{eq-6aa}
\end{align}
Upon noticing that the right-hand side of equation (\ref{eq-6aa}) is equal to $\ell
_{i,n}(1;B_{n})+o_{\mathbf{P}}(1)$, we complete the proof of Theorem \ref{theor-1ab}.
\end{proof}

\begin{lemma}\label{lemma4.2}
Statement (\ref{eq-13}) holds.
\end{lemma}

\begin{proof}
The distribution of $H_{i}\circ Q_{n}(1-k/n) $ is same as that
of $H_{i}\circ Q\left(  E_{n}^{-1}(1-k/n) \right)  $, where $E_{n}^{-1}$ is
the uniform empirical quantile function. Furthermore, the two processes $\{
1-E_{n}^{-1}( 1-s) ,\ 0\leq s\leq1\}$ and $\{ E_{n}^{-1}( s) ,\ 0\leq
s\leq1\}$ are equal in distribution. Hence, statement (\ref{eq-13})
is equivalent to
\begin{equation}
{\frac{H_{i}\circ Q\left(  1-E_{n}^{-1}(k/n)\right)  }{H_{i}\circ Q(1-k/n)}}
=1+o_{\mathbf{P}}(1).
\label{eq-13a}
\end{equation}
From the Glivenko-Cantelli theorem, $E_{n}^{-1}(k/n)-k/n\to0$ almost surely.
This also implies that $E_{n}^{-1}(k/n)\to0$ since $k/n\to0$. Moreover, by
Theorem 0 and Remark 1 of Wellner (1978), we have
$\sup_{1/n\leq s\leq1}s^{-1}\left\vert E_{n}^{-1}(s)-s\right\vert =o_{\mathbf{P}}(1)$. Hence,
\begin{equation}
(n/k)E_{n}^{-1}(k/n)=1+o_{\mathbf{P}}(1).
\label{eq-13b}
\end{equation}
Since $s\mapsto H_{i}\circ Q\left(  1-s\right)  $ is slowly varying at zero,
using Potter's inequality (see, e.g., the $5^{\text{th}}$ assertion of
Proposition B.1.9 on page 367 of de Haan and Ferreira (2006)) we have
\begin{equation}
\frac{H_{i}\circ Q\left(  1-E_{n}^{-1}(k/n)\right)  }{H_{i}\circ Q(1-k/n)
}=\left(  1+o_{\mathbf{P}}(1)\right)  \left(  (n/k)E_{n}^{-1}(k/n)\right)
^{-\gamma_{i}\pm\theta}
\label{eq-13c}
\end{equation}
for any $\theta\in(0,\gamma_{i} )$. In view of (\ref{eq-13b}), the right-hand
side of equation (\ref{eq-13c}) is equal to $1+o_{\mathbf{P}}(1)$. This
implies statement (\ref{eq-13a}) and thus, in turn, statement (\ref{eq-13}).
\end{proof}

\begin{proof}[\textbf{Proof of Theorem \ref{theor-1b}}]

Let $Y$ be a random variable with the cdf $G(z)=1-1/z$, $z\geq1$. Furthermore, let $\mathbb{U}_{i}(z)=H_{i}\circ Q(1-1/z)$. Since $G(Y)$ is uniform on the interval
$[0,1]$, the random variables $H_{i}(X)$ and $\mathbb{U}_{i}(Y)$ are
equal in distribution. We have the representations
\[
\widehat{L}_{i,n}(2) =\mathbb{U}_{i}(Y_{n-k:n}) \left(  \frac{k}%
{n}\right)  ^{\widehat{\gamma}_{i}} \int_{1-k/n}^{1}(1-s)^{-\widehat{\gamma
}_{i}}d\Psi_{i}(s)
\]
and
\[
D_{i,n} =\mathbb{U}_{i}(n/k)\left(  \frac{k}{n}\right)  ^{\gamma_{i}}
\int_{1-k/n}^{1} (1-s)^{-\gamma_{i}}d\Psi_{i}(s).
\]
Consequently,
\begin{align}
\frac{\sqrt{n}\,( \widehat{L}_{i,n}(2)-L_{i,n}(2)) }{(n/k)^{1/2}D_{i,n}}
= &
\sqrt{k}\,\left(  \frac{\mathbb{U}_{i}(Y_{n-k:n}) }{\mathbb{U}_{i}(n/k)
}-{\frac{\int_{0}^{1}H_{i}\circ Q(1-ks/n) d\Psi
_{i}(1-ks/n) }{H_{i}\circ Q(1-k/n) \int_{0}^{1} s^{-\gamma_{i}}d\Psi
_{i}(1-ks/n)}} \right)
\notag
\\
 =  &
\sqrt{k}\,\left(  \frac{\mathbb{U}_{i}(Y_{n-k:n}) }{\mathbb{U}_{i}(n/k)
}-\left(  \frac{Y_{n-k:n}}{n/k}\right)  ^{\gamma_{i}}\right)  {\frac{\left(
k/n\right)  ^{\widehat{\gamma}_{i}} \int_{1-k/n}^{1}(1-s)^{-\widehat
{\gamma}_{i}}d\Psi_{i}(s) }{\left(  k/n\right)  ^{\gamma_{i}}
\int_{1-k/n}^{1} (1-s)^{-\gamma_{i}}d\Psi_{i}(s) }}\nonumber\\
&  +\sqrt{k}\,\left(  \left(  \frac{Y_{n-k:n}}{n/k}\right)  ^{\gamma_{i}%
}-1\right)  {\frac{\left(  k/n\right)  ^{\widehat{\gamma}_{i}}
\int_{1-k/n}^{1}(1-s)^{-\widehat{\gamma}_{i}}d\Psi_{i}(s) }{(k/n)  ^{\gamma_{i}} \int_{1-k/n}^{1} (1-s)^{-\gamma_{i}}d\Psi_{i}(s)
}}\nonumber\\
&  + \sqrt{k}\, \left(  {\frac{\left(  k/n\right)  ^{\widehat{\gamma
}_{i}} \int_{1-k/n}^{1}(1-s)^{-\widehat{\gamma}_{i}}d\Psi_{i}(s) }{\left(
k/n\right)  ^{\gamma_{i}} \int_{1-k/n}^{1} (1-s)^{-\gamma_{i}}%
d\Psi_{i}(s) }} -1 \right) \nonumber\\
&  +\sqrt{k}\, \left(  1 -{\frac{\int_{0}^{1}H_{i}\circ Q(1-ks/n) d\Psi
_{i}(1-ks/n) }{H_{i}\circ Q(1-k/n) \int_{0}^{1} s^{-\gamma_{i}}d\Psi
_{i}(1-ks/n)}} \right)  .
\label{equa-1}
\end{align}
We shall next investigate the asymptotic behaviour of the four terms on the right-hand
side of equation (\ref{equa-1}).

\subsubsection*{First term on the RHS of equation (\ref{equa-1})}

We shall show that this term is of the order $o_{\mathbf{P}}(1)$, which follows from the two statements:
\begin{equation}
\sqrt{k}\left(  {\frac{\mathbb{U}_{i}(Y_{n-k:n})}{\mathbb{U}_{i}(n/k)}}-\left(  \frac{Y_{n-k:n}}{n/k}\right)^{\gamma_{i}}\right)  =o_{\mathbf{P}}(1)
\label{equa-10}
\end{equation}
and
\begin{equation}
{\frac{\left(  k/n\right)  ^{\widehat{\gamma}_{i}}\int_{1-k/n}%
^{1}(1-s)^{-\widehat{\gamma}_{i}}d\Psi_{i}(s)}{\left(  k/n\right)
^{\gamma_{i}}\int_{1-k/n}^{1}(1-s)^{-\gamma_{i}}d\Psi_{i}(s)}}=1+o_{\mathbf{P}}(1).
\label{equa-2}
\end{equation}
To prove statement (\ref{equa-10}), we note that since $H_{i}\circ Q$ is
regularly varying at $1$ with the index $(-\gamma_{i})$, the statement can be verified using analogous arguments as those used for proving  statement $T_{n1}=o_{\mathbf{P}}(1)$ by Necir et al. (2007). To check
statement (\ref{equa-2}), we note that we shall later establish a
non-degenerate limiting distribution of the third term on the right-hand side
of equation (\ref{equa-1}), which is a stronger statement than (\ref{equa-2}). With these notes we conclude our proof that the first term on the right-hand side of equation (\ref{equa-1}) is of the order $o_{\mathbf{P}}(1)$.

\subsubsection*{Second term on the RHS of equation (\ref{equa-1})}

Following the arguments on page 158 of Necir et al. (2007) and making notational adjustments, we have that
\begin{equation}
\sqrt{k}\,\left(  \left(  \frac{Y_{n-k:n}}{n/k}\right)  ^{\gamma_{i}%
}-1\right)  =\gamma_{i}\sqrt{{\frac{n}{k}}}\,\sqrt{n}\,\vartheta
_{n}(1-k/n)+o_{\mathbf{P}}(1), \label{equa-12-0}%
\end{equation}
where the (un-normalized) uniform quantile process $\vartheta_{n}$ is defined by equation (\ref{quant-process}). Utilizing weak approximation results of Cs\"{o}rg\H{o}
et al. (1986), we have Brownian bridges $B_n$ on an appropriate probability space such that
\begin{equation}
\sqrt{k}\,\left(  \left(  \frac{Y_{n-k:n}}{n/k}\right)  ^{\gamma_{i}%
}-1\right)  =-\gamma_{i}\sqrt{{\frac{n}{k}}}\,B_{n}(1-k/n)+o_{\mathbf{P}}(1).
\label{equa-12}
\end{equation}
This and also statement (\ref{equa-2}) imply that the second term on the right-hand side of equation (\ref{equa-1}) has the same asymptotic behaviour as the right-hand side of equation (\ref{equa-12}).

\subsubsection*{Third term on the RHS of equation (\ref{equa-1})}

We rewrite the term as follows:
\begin{align}
\sqrt{k}\,\left(  {\frac{\left(  k/n\right)  ^{\widehat{\gamma}_{i}%
}\int_{1-k/n}^{1}(1-s)^{-\widehat{\gamma}_{i}}d\Psi_{i}(s)}{\left( k/n\right)  ^{\gamma_{i}}\int_{1-k/n}^{1}(1-s)^{-\gamma_{i}}d\Psi_{i}(s)}}-1\right)
&  =\sqrt{k}\,\left(  {\frac{\int_{0}^{1}s^{-\widehat{\gamma}_{i}}
d\Psi_{i}(1-ks/n)}{\int_{0}^{1}s^{-\gamma_{i}}d\Psi_{i}(1-ks/n)}}-1\right)
\nonumber
\\
&  =\sqrt{k}\,{\frac{\int_{0}^{1}(  s^{-\widehat{\gamma}_{i}}-s^{-\gamma_{i}})  d\Psi_{i}(1-ks/n)}{\int_{0}^{1}s^{-\gamma_{i}}d\Psi_{i}(1-ks/n)}}.
\label{equa-12uu}
\end{align}
Theorems 2.3 and 2.4 of Cs\"{o}rg\H{o} and Mason (1985) imply that there are Brownian bridges $B_{n}$ on a probability space specified by Cs\"{o}rg\H{o} et al. (1986) such that
\begin{equation}
{\frac{\sqrt{k}\left(  \widehat{\gamma}_{i}-\gamma_{i}\right)  }{\gamma_{i}}}
=\sqrt{{\frac{n}{k}}}\,B_{n}(1-k/n)-\sqrt{{\frac{n}{k}}}\,
\int_{1-k/n}^{1}\frac{B_{n}(s)}{1-s}ds+o_{\mathbf{P}}(1).
\label{weak-approx-1}
\end{equation}
Hence, in particular,  $\widehat{\gamma}$ is a consistent estimator of $\gamma$, and so we have
\begin{multline}
\int_{0}^{1}\left(  s^{-\widehat{\gamma}_{i}}-s^{-\gamma_{i}}\right)
d\Psi_{i}(1-ks/n)=-\left(  \widehat{\gamma}_{i}-\gamma_{i}\right)  \int
_{0}^{1}s^{-\gamma_{i}}\log(s)d\Psi_{i}(1-ks/n)\\
+o_{\mathbf{P}}(1)\left\vert \widehat{\gamma}_{i}-\gamma_{i}\right\vert
\int_{0}^{1}s^{-\gamma_{i}-\delta}d\Psi_{i}(1-ks/n),
\label{equa-3}
\end{multline}
where $\delta>0$ can be any fixed number. Furthermore, in view of statement (\ref{weak-approx-1}) we have that $\sqrt{k}(\widehat{\gamma}_{i}-\gamma)=O_{\mathbf{P}}(1)$. This result,
assumptions (\ref{con-0b}) and (\ref{equa-10c}), and asymptotic representation  (\ref{equa-3}) imply that the right-hand side of equation (\ref{equa-12uu}) is equal to $-c_{i}\sqrt{k}(\widehat{\gamma}_{i}-\gamma_{i})+o_{\mathbf{P}}(1)$. Utilizing result (\ref{weak-approx-1}) in its full generality, we conclude that
\begin{multline}
\sqrt{k}\,\left(  {\frac{(k/n)^{\widehat{\gamma}_{i}}
\int_{1-k/n}^{1}(1-s)^{-\widehat{\gamma}_{i}}d\Psi_{i}(s)}{(k/n)^{\gamma_{i}}\int_{1-k/n}^{1}(1-s)^{-\gamma_{i}}d\Psi_{i}(s)}}-1\right)  =-\gamma_{i}c_{i}\sqrt{{\frac{n}{k}}}\,B_{n}(1-k/n)
\\
+\gamma_{i}c_{i}\sqrt{{\frac{n}{k}}}\,\int_{1-k/n}^{1}\frac{B_{n}(s)}{1-s}ds
+o_{\mathbf{P}}(1).
\label{equa-15}
\end{multline}
This determines the asymptotic behaviour of the third term on the right-hand side of equation (\ref{equa-1}) and also establishes the earlier noted statement (\ref{equa-2}).

\subsubsection*{Fourth term on the RHS of equation (\ref{equa-1})}

We rewrite the term as follows:
\begin{multline}
\sqrt{k}\,\left(  1-{\frac{\int_{0}^{1}H_{i}\circ Q(1-ks/n)d\Psi_{i}%
(1-ks/n)}{H_{i}\circ Q(1-k/n)\int_{0}^{1}s^{-\gamma_{i}}d\Psi_{i}(1-ks/n)}%
}\right) \\
={\frac{-\sqrt{k}}{\int_{0}^{1}s^{-\gamma_{i}}d\Psi_{i}(1-ks/n)}}
\int_{0}^{1}\left(  {\frac{H_{i}\circ Q(1-ks/n)}{H_{i}\circ
Q(1-k/n)}}-s^{-\gamma_{i}}\right)  d\Psi_{i}(1-ks/n).
\label{equa-10a}
\end{multline}
From Theorem 2.3.9 on page 48 of de Haan and Ferreira (2006) we have a function $\widetilde{A}_{i}$ with the property $\widetilde{A}_{i}(x)\sim A_{i}(x) $ when $x\rightarrow \infty $ such that for any $\delta>0$ there exists $\epsilon_{0,i}\in (0,1) $ such that, whenever $\epsilon s\leq \epsilon_{0,i}$,
\begin{equation}
\left\vert {\frac{1}{\widetilde{A}_{i}(1/\epsilon)}}\left(  {\frac{H_{i}\circ
Q(1-\epsilon s)}{H_{i}\circ Q(1-\epsilon)}}-s^{-\gamma_{i}}\right)  -s^{-\gamma_{i}}%
\frac{s^{-\omega_{i}}-1}{\omega_{i}}\right\vert \leq\delta s^{-\gamma_{i}-\omega
_{i}-\delta}.
\label{equa-11z}
\end{equation}
Upon recalling condition (\ref{con-0a}), we have $\sqrt{k}\,\widetilde{A}_{i}(n/k)  \rightarrow b_{i}$. Inequality (\ref{equa-11z}) implies that for all sufficiently large $n$ the right-hand side of equation (\ref{equa-10a}) is equal to
\begin{equation}
\bigg ( {\frac{-b_{i}}{\omega_{i}}} \bigg ) {\frac{\int_{0}^{1}s^{-\gamma_{i}}(s^{-\omega_{i}%
}-1)d\Psi_{i}(1-ks/n)}{\int_{0}^{1}s^{-\gamma_{i}}d\Psi_{i}(1-ks/n)}}+o(1)
\frac{\int_{0}^{1}s^{-\gamma_{i}-\omega
_{i}-\delta}d\Psi_{i}(1-ks/n)}{\int_{0}^{1}s^{-\gamma_{i}}d\Psi_{i}(1-ks/n)}.
\label{equa-11d}
\end{equation}
In view of assumptions (\ref{con-0c}) and (\ref{equa-10c}), we conclude that quantity (\ref{equa-11d}) and thus the fourth term on the right-hand side of equation (\ref{equa-1}) are of the order $-b_{i}d_{i}/\omega_{i}+o(1)$.

Collecting the above established asymptotic properties of the four terms on
the right-hand side of equation (\ref{equa-1}), we finish the proof of
Theorem \ref{theor-1b}.
\end{proof}

\section*{\bf Acknowledgments}

The research has been partially supported by
the Natural Sciences and Engineering Research Council (NSERC) of Canada.

\section*{\bf References}
\def\hangindent=\parindent\noindent{\hangindent=\parindent\noindent}

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52, 255--264.

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\textit{Empirical Processes with Applications to Statistics.}
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\textit{Insurance: Mathematics and Economics} 17, 43--54.

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{Zenga M.} (1987).
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{Zenga, M.} (2007).
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\end{document}

redpajama_set_name: RedPajamaArXiv


Why make your Pet a New Member of the Skype Community anyway?
Let others know you are a proud owner of the cutest pet in the world.
Meet other happy pet-lovers and their pets around Skype.
To get you updated with what your pets are doing (if they're okay), while you're out or perhaps busy at work all throughout the day.
In order to find an ideal pet sitter.
Keep in touch with your pet sitter and give your instructions from time to time.
Take cute videos and pictures with your pet you can flaunt around social networking sites.
So you won't miss them even if you're on a month long vacation.
Start a new form of pet and pet owner bonding.
Preserve beautiful moments with your pet you can look back from time to time.
In the Tools menu at the uppermost part of your Skype screen, click Options. Then, a new window will appear on your screen.
Hit the Take a Video Snapshot button you will see just below your displayed image. This will enable another window to display a larger view.
Then, at the lowermost part of the new window, click the button of the Take a Video Snapshot option. After which, you may choose to adjust your picture's size and crop the exact angle you want to save.
To save the photo, click Save Snapshot.
On your Skype screen's left panel, choose a contact you wish to share your photo.
Choose the Send file option, then a new window will appear.
Select the photo you want to send, click it, and click the "Open" button. You may need to wait a few minutes for your contact to accept the file you've sent.
These are just some things you can do with Skype to feature your pet. You may also start a video or audio call to your pet sitter to keep you updated with your pet, or a group video call to introduce your pet to other pet owners.
PS: Get Supertintin.com's Skype Video Call Recorder to record every happy moment with your pet in Skype. Grab it now!
This entry was posted in Skype How To's, Skype Tips and Tricks and tagged Skype Community. Bookmark the permalink.


text:
Q: Ionic2 sqlitePlugin is not defined How to evade this error: VM18193:27 Unable to open database ReferenceError: sqlitePlugin is not defined(…)
    setTimeout(function() {
      let db = new SQLite();
      db.openDatabase({
        name: "data.db",
        location: "default"
      }).then(() => {
        db.executeSql("CREATE TABLE IF NOT EXISTS people (id INTEGER PRIMARY KEY AUTOINCREMENT, firstname TEXT, lastname TEXT)", {}).then((data) => {
          console.log("TABLE CREATED: ", data);

        }, (error) => {
          console.error("Unable to execute sql", error);
        })
      }, (error) => {
        console.error("Unable to open database", error);
      });
    }, 2000);

How can i execute some query?

if(SqlSettingsService.openDb){
      this.db = SqlSettingsService.getDB();
      this.db.executeSql("CREATE TABLE IF NOT EXISTS people (id INTEGER PRIMARY KEY AUTOINCREMENT, firstname TEXT, lastname TEXT", {}).then

instead i get error.
console:

SqlSettingsService() starts
VM21750:27 Unhandled Promise rejection: Cannot read property 'executeSql' of null ; Zone: <root> ; Task: Promise.then ; Value: TypeError: Cannot read property 'executeSql' of null(…) TypeError: Cannot read property 'executeSql' of null

A: The plugin sqlLite does not work in a browser, you can use Websql for the browser instead (with compatible browsers, including Chrome and Opera as far as I know).
Transactions written for Websql are compatible with transactions written for SQLite.
Here is a service I have done to manage the Connection to the DB and make it work regardless if the program runs in a browser or a on real device:
import { Injectable } from '@angular/core';
import { SQLite } from 'ionic-native';
import { Platform } from 'ionic-angular';
import { Storage } from '@ionic/storage';

@Injectable()
export class SqlSettingsService {

    private db: any = null;
    private isOpened: boolean = false;

    constructor() {
        console.log('SqlSettingsService() starts');
    }

    public getDB(){
        return this.db;
    }

    public openDb = (platform:Platform,winSer:any):Promise<any> => {
        console.log('SqlSettingsService() opend DB starts');
        let p:Promise<any>;
        if(!this.isOpened){
            this.isOpened = true;

            if(platform.is('core')){
                this.db = winSer.window.openDatabase("ionic2BrowserDev","1.0","",5*1024*1024);
                p = new Promise(function(resolve,reject){resolve('websql success')});
            } else {
                this.db = new SQLite();
                p = this.db.openDatabase({
                    name: 'data.db',
                    location: 'default' // the location field is required
                }).then(
                    ()=>{console.log("SqlSettingsService open db successful")},
                    (err)=>{console.error(err)}
                );
            }

        } else {
             p = new Promise(function(resolve,reject){
                resolve('db already opened');
             });
        }
        return p;

    }

    public closeDb = () => {
        this.isOpened = false;
        return this.db.close();
    }

}

winSer is another service to access the window Object that I use in my app.component.ts when I call openDB() on SqlSettingsService. It's just this:
import { Injectable } from '@angular/core';

@Injectable()
export class WindowService {
  public window = window;
}

To execute query:
    [SqlSettingsService-instance].openDb();
[SqlSettingsSevice-instance].getDB().transaction(
            function(tx){
                tx.executeSql([your sql],[bracket values you want to pass],success,error);
                function success(tx,rs){
                     console.log("success exec sql: ")
                     console.info(rs);

                }
                function error(tx,error){
                     console.log('execSqlCustom error ' + error.message + " for tx " + tx);

                }
            });

redpajama_set_name: RedPajamaStackExchange


text:
16/01/2021 09:42:11pm
Malawi churches add voice to calls for an end to sexual violence, GBV
December 11, 2020 Watipaso Mzungu – Nyasa Times 1 Comment
Christian mother bodies, the Episcopal Conference of Malawi (ECM), Malawi Council of Churches (MCC), and Evangelical Association of Malawi (EAM), have added their voice to calls for an end to sexual and gender-based violence (SGBV) in the country.
Fr Saindi co-signed the statement
In a statement issued as the world winds up the observance of the 16 days of activism a GBV, the bodies say they have noted with concern the alarming increase in cases of rape and defilement in the country that is threatening lives and rights of innocent women, men, girls and boys.
The statement has been signed by the Reverend Father Henry Saindi, who is ECM General Secretary, and his MCC and EAM counterparts, Bishop Gilford Matonga and Reverend Francis Mkandawire, respectively.
The faith leaders says even more disturbing are delays in response and support mechanisms to both the survivors and victims of these barbaric acts against humanity.
"These acts leave lifelong physical and psychological scars in the life of those affected and undermine their dignity and confidence," they say.
They bodies state that as Christian mother bodies, they strongly hold the belief that although we are physically different and possess unique functions for human procreation and sustenance, we are all created in the image of God (Genesis 1 verse 27).
"In this regard, we are all equal before the Creator. We further believe in the supremacy of human dignity regardless of sex, age, race and/or nationality. The recent developments are therefore in direct conflict with God's creation and are tantamount to assault on the dignity of life at the core. This is unacceptable and should be stopped by all means. The Scriptures command us to love others just as we love ourselves and do unto them as we would have loved them do unto us (Mathews 7 Verse 12; Mark 12 Verse 31)," they preach.
The faith leaders have since appealed to their fellow religious leaders, as moral advocates in society, to seriously teach and campaign against rape, defilement and all forms of gender based injustices against women and children.
They say various platforms within various places of worship and outside including community gatherings should be used to fight against these malpractices.
"We should take it as our responsibility to ensure all religious institutions support survivors and victims of gender injustice by protecting and providing safe space and other essential materials to survivors of GBV, providing counseling and guidance where necessary. We should promote reporting of all cases for redress without fear or favour until the survivors as well as victims get justice," they emphasize.
They have also called for collaboration among stakeholders committed to the same cause, including relevant Government Departments, other Non-Governmental Organisations and development partners in promoting positive masculinities in fighting against gender injustices through among others mobilisation of the men's guild and youths on the same.
In the same vein, the Christian mother bodies have asked traditional leaders to stop and get rid of all harmful cultural practices and rituals that predispose women and children to gender injustices, adopt appropriate community enforcement mechanisms to ensure complete compliance within the community in fighting gender injustice and ensure enforcement of by-laws particularly those to do with child and teen marriages which fuel SGBV.
"Chiefs should also take a leaders role in raising awareness on gender and encourage your subjects to report cases of Sexual GBV to relevant authorities. On the other hand, the government and other duty bearers should provide sufficient funding and relevant support to fight gender injustices, especially fund the Malawi Human Rights Commission for the Implementation of the Gender Equality Act.
"Parliament should provide protected funding for promotion of gender justice and fighting inequalities while law enforcers and security agents must ensure that the law is applied and enforced without bias or favour. There are reports of GBV cases being mishandled by the law enforcement agencies under the influence of corruption. This is a double blow to survivors and victims of gender injustice and must be stopped," continues the statement.
To the media, the faith leaders have asked journalists to aggressively disseminate information on gender injustice of any form, using all media outlets for the people to hear and change their mindset.
The churches call upon all Malawians of goodwill to pray earnestly for God's divine intervention that Gender Based Violence may be completely uprooted in our country.
defilementgender-based-violenceMalawi Council of Churches
Great Angels Choir drops 5th album 'Ufumu Umodzi Suukilani'
'Slightly unwell' Archbishop Ziyaye misses priestly ordination: Hospitalised in Namibia
Khekhi
Za abortion ndi him on sexuality bwezi atatokota kale kale
After spending a period of six solid years without releasing an album, the country's local leading gospel music group Great...

redpajama_set_name: RedPajamaCommonCrawl


As the impact of my Mother's death lessens, I can now put into perspective the way my body reacted to the grief at losing my Mother to cancer last year. Some have likened the grief process to accelerated ageing, with changes to metabolism, increased muscle stiffness, joint pains, loss of vitality, and a general state of 'foggy' head. I have experienced all of this over the last seven months. When you are going through the grieving process, though, it's just all part of the coping mechanism; not to pay too much attention to the individual effects, and rather concentrate on becoming a whole person again, even though you have lost a loved one, an important part of yourself.
I have written previously about how my grief started before Mom died, because her illness was so extensive, we knew there was no hope. I don't know if this helped or hindered my recovery. I don't suppose anyone can tell. It affected my sleep, I suffered weight loss, aches and pains, and felt hollow inside. After Mom died, there was a brief period of well being, mainly relief because Mom was no longer suffering, and that it was all over. I wasn't thinking about the future, just getting through each day. It wasn't a conscious decision; again, it's just part of the process. The mind taking over, in order to protect the body.
The legalities of death and probate hit you at a vulnerable time, and I found that many of the companies who should have specific staff to deal with probate and help families are woefully ill equipped to do so. Banks, building societies and HMRC are the worst. If I had had more energy at the time, I would have created a scene – perhaps I still will. And for me, it was the legal and financial stuff that set me back. It made me angry, and that was unhelpful. It wasn't anyone's fault that Mom died. I couldn't assign any blame, but those companies made it so much worse.
It's tempting to want to put a time frame, or deadline, on grief. I knew I shouldn't but psychologically Christmas, Mom and Dad's wedding anniversary, Dad's birthday and Mother's Day all came and went with a slightly numb 'let's not cry' feeling. We planned a wonderful holiday to a tropical paradise. I rested and felt restored, but still came home feeling so tired. So tired in fact that those around me thought I was seriously ill. Two more weeks past, and slowly but surely I was able to exercise for longer, and more efficiently too. But the aches and pains have been slow to recede. Arthritic flare-ups have impeded by recovery, ligaments in my feet have been inflamed too. It's not been all bad. Not every day is painful. It seems to be when something else causes a small stress in every day life, that my resilience is low, and my body tells me so.
It's been quite difficult to find information on this topic. Most of what is out there is written from a complementary therapy point of view. It seems a little recognised topic on the part of the medical people. I have already questioned the GP's wisdom about just leaving someone to grieve after a significant bereavement. I'm not suggesting that drugs are required, simply that more attention needs to be paid to these physiological, as opposed to psychological, symptoms of grief.
Spring days, warm sunshine, wonderful colours are all contributing to the healing process. Human beings are after all part of the natural world. I am fortunate to be surrounded by nature, renewal and re-birth, living as I do in the countryside. So if like me, you are grieving the loss of a loved one, be kind to yourself. Understand that it's not all in the mind. Your body will be taking a battering too.


The National Insurance Scheme (NIS) is looking to implement a slew of new measures to combat fraud and improve the services it offers.
The list of new measures includes one which is referred to as the photographic verification of life system. This will require pensioners to submit a photograph every six months via phone application or webcam to facilitate the processing of life certificates.
Chairman of the NIS Dr. Surendra Persaud made this disclosure at a recent event to celebrate the Scheme's 49th anniversary. It is believed that this system will prevent fraud, as sometimes relatives of persons who have died continue to collect the pension.
"Guyanese are pretty much at every corner of this globe and if you're going to ask some Guyanese to walk into this building or another to verify that they're alive, that's not sustainable," Dr. Persaud said.
The Scheme has 51,178 pensioners as at the end of August 2018. For the period of January to August 2018, the Scheme also processed 82,934 medical claims.
The state-run NIS continues to see escalating expenses contributing to the Scheme running a deficit every year. This year, it is projected to record a deficit of $854.9M.
The Scheme is looking to upgrade its system to make payments through the commercial banks and allow employer to also make remittances online.
"I envision a day, that if you're going to interact with the Scheme, it will be unnecessary for you to step foot into one of the buildings. We anticipate that where it comes to the payments for employers, they will be able to make those contributions online and eventually all of your benefits should go exactly into the bank of your choice," Dr Persaud stated.
Citizens often times raise concerns about the time taken to conduct business with the NIS.
Dr. Persaud said the Scheme will also be putting in place a system through which citizens can make appointments to meet with its officers and a list of requirements will be provided in order to decrease waiting time. This decision was made at its last board meeting.


text:
Celebrity Tidings
Home » World News » Two kids, 4 and 6, killed in freak car strike while playing mini-golf with their parents
Two kids, 4 and 6, killed in freak car strike while playing mini-golf with their parents
12/05/2020 World News
TWO young kids were killed when they were hit by a car while playing mini-golf with their parents in northwest Florida.
Police said a Chevy pickup left the road, crashed into the golf course in Panama City before striking the four-year-old boy and six-year-old girl.
WJHG reports Panama City Beach officials said the children were hit by a vehicle while playing at the Coconut Creek Family Fun Park mini-golf course with their parents.
It happened about 3.30pm yesterday.
One of the children died at the scene, while the other was taken to a nearby hospital and died later.
According to Panama City Beach police officials, the driver of the Chevy Silverado that struck them has been identified as Scott Donaldson from Panama City Beach.
It is unclear what caused the driver to veer off the westbound carriageway crash into the mini-golf course.
Police say the investigation has been handed over to the traffic homicide unit.
Debbie Ward, public information officer at the city of Panama City Beach, said charges were pending, reports the Panama City News Herald.
She confirmed the family was from Louisville, Kentucky.
4, 6, and, f, In, Kids, killed, two
Priyanka Chopra Gets Candid About Racist Bullying in High School
Priyanka Chopra is opening up about a tough time in her past. The actress, who spoke about her life in upcoming memoir Unfinished out on February 9, detailed her experience of being Indian while attending an American high school as a teenager. PHOTOS: Check out the latest pics of Priyanka Chopra She reveals in the book that she left the
Donald Trump Left a Note for Joe Biden in the White House & It Is Now a Viral Meme
It was discovered that Donald Trump left a note for President Joe Biden in the White House shortly before he left the morning of the 2021 Inauguration on Wednesday (January 20). Well, the contents of the note have and likely never will be made public. However, that did not stop anyone from guessing the content of the note in question.
Kailyn Lowry takes a lot of flak on social media for the decisions she makes in her personal life. Kail likes to say she's been unlucky with men, but her haters believe all of her suffering is self-created, and she's a victim of her own bad decisions. The truth, of course, is probably somewhere in the middle. But if the
David Beckham's PT shares recipe for the 'best' healthy pancakes
Chris Wallace Has Some Thoughts About Biden's Inaugural Address
Here's Why You're Seeing So Much Purple On Inauguration Day
Danny Masterson's Attorney Enters Not Guilty Plea to Rape Charges
Ray caught at last as Abi gets evidence to expose his crimes in Corrie?
Melania Trump's 'clenched teeth' expose ex-FLOTUS fury at husband's last day in office


The mineral or gemstone chrysoberyl, also known as cymophane or golden beryl, is an aluminate of beryllium with the formula BeAl2O4. Its crystals form in an orthorhombic crystal system[?] with contact and penetration twins common, often repeated forming rosetted structures. Hardness 8.5, refractive index 1.74-1.75, Dispersion 0.015, Double refraction 0.009, optic sign[?] positive, specific gravity 3.5-3.84, distinct cleavage, conchoidal to uneven fracture, vitreous lustre, and colored various shades of green and sometimes yellow. Chrysoberyl is transparent to translucent, sometimes opalescent[?]. The word chrysoberyl is derived from the Greek words meaning golden and beryl.
Chrysoberyl occurs in granitic rocks, pegmatites and mica schists; often it is found in alluvial deposits. It has also been found in contact metamorphic deposits of dolomitic marble with corundum, and in fluorine skarns. Most chrysoberyl is recovered from river sands and gravels.
The alexandrite variety displays pleochroism. Alexandrite from the Ural mountains in Russia is green by daylight and red by transmitted light. Other varieties of alexandrite may be yellowish or pink in daylight and yellow-orange or green by transmitted light. It results from small scale replacement of alumina by chromic oxide, which is responsible for alexandrite's characteristic green to red color change. Alexandrite was first discovered in 1831 in an emerald mining region of the Ural Mountains in the Soviet Union, it is said on the very same day that the Russian czar Alexander II came of age. It was named "alexandrite" in his honor by the mineralogist Adolf Nordenskjold. It is an interesting coincidence that the Russian national colors are green and red.
Translucent yellowish chatoyant[?] chyroberyl is called cymophane or cat's eye. Cymophane has its derivation also from the Greek words meaning wave and appearance, in reference to the opalescence sometimes exhibited. In this variety, microscopic tubelike cavities or needlelike inclusions of rutile occur in an orientation parallel to the c-axis producing a chatoyant effect visible as a single ray of light passing across the crystal. This effect is best seen in gemstones cut in cabochon[?] form perpendicular to the c-axis. The color in yellow chrysoberyl is due to Fe+3 impurities.
Although other minerals such as tourmaline, scapolite[?], corundum, spinel and quartz can form "cat's eye" stones similar in appearance to cymophane, the jewelry industry designates these stones as "quartz cat's eyes", or "ruby cat's eyes" and only chrysoberyl can be referred to as "cat's eye" with no other designation.


text:
"My life was not useless; I gave important truths to the world, and it was only for want of understanding that they were disregarded. I have been ahead of my time."
— Robert Owen, Deathbed statement (November 1858), in response to a church minister who asked if he regretted wasting his life on fruitless projects; as quoted in Harold Hill : A People's History http://www.haroldhill.org/section_two/section_two_page_one.htm (2004).
Robert Owen29
Welsh social reformer 1771 - 1858
"As I advanced in my career in librarianship, I have been a woman in a man's world. However, this issue has not been an important factor in my thinking."
— Henriette Avram American computer programmer and system analyst. She developed the MARC formatting used in libraries 1919 - 2006
World, Thinking
"' He doesn't understand. The navy is my life. My only life, now. The war will not last for ever, but until it ends I shall stand in the line-of-battle as I have been entrusted to!'"
— Douglas Reeman British author 1924 - 2017
For My Country's Freedom, Cap 13 "Loneliness"
Life, War, Understanding
"I think journalism was the most important thing I could have done with my life. I just can't think of anything I could have done with my life that would have been more meaningful."
— John Seigenthaler American journalist, writer, and political figure 1927 - 2014
Reported in his Tennessean's obituary; quoted in "John Seigenthaler dies at 86" http://www.poynter.org/2014/john-seigenthaler-dies-at-86/258597/ by Andrew Beaujon, poynter.org (11 July 2014)
Life, Thinking
"My life was in crisis. All my values were becoming meaningless. I was discovering that my chosen profession was empty, foolish, as useless as—as playing chess."
— Robert Silverberg American speculative fiction writer and editor 1935
Short fiction, Schwartz Between the Galaxies (1974)
"Love has always been the most important business in my life; I should say the only one."
— Stendhal French writer 1783 - 1842
L'amour a toujours été pour moi la plus grande des affaires ou plutôt la seule. La Vie d'Henri Brulard (1890) Variant translation: Love has always been the most important business in my life, or rather the only one.
Love, Life, Business
"I gave you my life, you gave me my life.
Like a gush of wind in my hair."
— Yoko Ono Japanese artist, author, and peace activist 1933
Context: I gave you my life, you gave me my life. Like a gush of wind in my hair. Why do we forget what's been said And play the game of life with our hearts? "Walking On Thin Ice" on Season of Glass (1981).
"I have descended to the depths of suffering, and it seems to me that after that one cannot help being raised up. But I have spent the last three nights weeping. Pity! Pity! Remembrance is the true imperishable life... I should like to live my life over again, to record it, to admit my weaknesses; no, this is useless; I have sinned, I have suffered, I have atoned for it. I could write only a bad novel by relating what has been related a thousand times."
— Berthe Morisot painter from France 1841 - 1895
1881 - 1895, Quote from Berthe's Morisot's letter, 1887 - after the death of her husband Eugène Manet
Life, Suffering, Time, Help
"My life is very simple. I like simplicity and for my time to be my own, so that I have the freedom to devote the majority of it to Islam. That is the foundation from which I keep everything in perspective. I also want to let people know about Islam, how Islam can be a way of life. I want them to really understand its richness and its beauty and to see that Islam is for everybody. For me, this is the most important thing. From this, everything else falls into place."
— Hakeem Olajuwon Nigerian–American basketball player 1963
Sourced Quotes, Slam dunk - interview with basketball player Hakeem Olajuwon - Interview, Feb, 1994 by Spike Lee.
Life, People, Beauty, Way
"The only time I see the truth is when I cross my eyes."
— Louise Erdrich, The Last Report on the Miracles at Little No Horse
Truth, Time
"My life is lived, and I have played
The part that Fortune gave."
— John Conington British classical scholar 1825 - 1869
Translations, The Aeneid of Virgil (1866), Book IV, p. 138
Life, Parting
"All my life I've known better than to depend on the experts. How could I have been so stupid, to let them go ahead?"
— John F. Kennedy 35th president of the United States of America 1917 - 1963
Attributed, Conversation with Theodore C. Sorensen concerning the Bay of Pigs Invasion; as quoted in Sorensen's Kennedy (1965), p. 309.
Life, Stupidity, Dependency
"I was taken to see the World Exhibition in Moscow [in 1880]. I found it all very boring. But when I came to a section devoted to art – there were only paintings, and this was the first time of my life [Jawlensky was 16 years old] I had seen paintings – I was so deeply affected that it was a case of Saul becoming Paul. It was the turning point of my life. Even since then art has been my ideal, my holy of holiest, that for which my entire soul and my entire self yearn."
— Alexej von Jawlensky Russian painter 1864 - 1941
1936 - 1941, Life Memories' (1938), p. 20 from: 'Lebenserinnerungen', 1938
Life, World, Art, Soul
"Everybody wants me to show my vagina to the world. And the truth is, I don't have to."
— Lady Gaga American singer, songwriter, and actress 1986
Interview for Maxim Magazine.
Truth, World
"Writing — I exist only for that. It's the most important thing in my life."
— Taylor Caldwell Novelist 1900 - 1985
1970s-, Context: About half of my published novels were written before I was published. So I didn't write a book every two years, as some people think. Writing — I exist only for that. It's the most important thing in my life. It's not apart from me. I have no other interests, except cooking. I don't belong to any organizations, clubs — I don't go to lunches. This is my life, the most important thing — far more important than anything else I do. It has to be that way, otherwise you're just a hobbyist. Now, a painter needs only to know the technique of his painting, and he has to have a tremendous emotional response to it. Musicians, sculptors — the same way. But they don't have to know about everything. A writer does.<!-- He has to do a tremendous amount of reading, too. I'd rather go without food, sleep, even cigarettes, than go without books. I read at least three of four books a week, plus all kinds of publications, some very weird. I like to know what's going on, what people think. I read the far left, the far right, and in between, to see what people are doing and saying.
"We've been here before, haven't we? Last time you were starving, I gave you my blood. It was a little homoerotic, maybe, but I'm secure in my sexuality."
— Cassandra Clare, book City of Heavenly Fire
The Mortal Instruments, City of Heavenly Fire (2014), Context: "Why didn't you say something?" "Because, it's not like there's animals I can feed on here." "There's us." "I don't want to feed on my friends' blood." "Why not? We've been here before, haven't we? Last time you were starving, I gave you my blood. It was a little homoerotic, maybe, but I'm secure in my sexuality." Jace Herondale and Simon Lewis, pg. 509
"All my life, I've always wanted to be somebody, but I see now I should have been more specific."
— Jane Wagner Playwright, actress 1935
The Search for Signs of Intelligent Life in the Universe (1985), "Chrissy"