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\title{\bf Degree distribution of an inhomogeneous\\ random intersection graph}

% input author, affilliation, address and support information as follows;
% the address should include the country, and does not have to include
% the street address 

\author{Mindaugas  Bloznelis\thanks{Supported by the Research Council of Lithuania Grant MIP-053/2011.}\\
\small Faculty of mathematics and informatics\\[-0.8ex]
\small Vilnius University\\[-0.8ex] 
\small Vilnius, Lithuania\\
\small\tt mindaugas.bloznelis@mif.vu.lt\\
\and
Julius Damarackas\\
\small Faculty of mathematics and informatics\\[-0.8ex]
\small Vilnius University\\[-0.8ex]
\small Vilnius, Lithuania\\
\small\tt julius.damarackas@mif.stud.vu.lt
}

% \date{\dateline{submission date}{acceptance date}\\
% \small Mathematics Subject Classifications: comma separated list of
% MSC codes available from http://www.ams.org/mathscinet/freeTools.html}

\date{\dateline{Oct 8, 2012}{??? ?, 2013}\\
\small Mathematics Subject Classifications: 05C80, 05C07, 05C82}

\begin{document}

\maketitle

% E-JC papers must include an abstract. The abstract should consist of a
% succinct statement of background followed by a listing of the
% principal new results that are to be found in the paper. The abstract
% should be informative, clear, and as complete as possible. Phrases
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\begin{abstract}
We show the asymptotic degree distribution of the typical vertex of a 
sparse inhomogeneous random intersection graph. 

  % keywords are optional
  \bigskip\noindent \textbf{Keywords:} 
degree distribution; random graph; random intersection graph; power law
\end{abstract}

\section{Introduction}


 
Let $X_1,\dots, X_m, Y_1,\dots, Y_n$ be independent non-negative random variables such that
each $X_i$ has the  probability distribution $P_1$ and each $Y_j$ has the probability 
distribution
$P_2$. Given realized values $X=\{X_i\}_{i=1}^m$ and $Y=\{Y_j\}_{j=1}^n$
we define the random bipartite graph $H_{X,Y}$ with the bipartition 
$V=\{v_1,\dots, v_n\}$, $W=\{w_1,\dots, w_m\}$, 
where  edges $\{w_i,v_j\}$ are inserted 
with probabilities
$p_{ij}=\min\{1, X_iY_j(nm)^{-1/2}\}$ independently 
for each $\{i,j\}\in [m]\times [n]$.
The 
 {\it inhomogeneous} random intersection graph  $G(P_1,P_2, n,m)$ defines the adjacency 
relation on the  vertex 
set $V$: vertices $v',v''\in V$ are declared adjacent (denoted $v'\sim v''$) whenever
$v'$ and $v''$ have a common neighbour in $H_{X,Y}$. 

 The degree distribution of the typical vertex of the random  graph  $G(P_1,P_2, n,m)$ has 
been first considered  by
Shang \cite{Shang2010}.
The proof of the main result of  \cite{Shang2010} contains  gaps and the  result is 
incorrect in the regime where
 $m/n\to\beta\in (0,+\infty)$ as $m,n\to+\infty$. 
 We remark that this regime is of particular importance, because it leads to 
inhomogeneous graphs
 with the clustering property: the clustering coefficient 
$\PP(v_1\sim v_2|v_1\sim v_3, v_2\sim v_3)$ is bounded away from zero
 provided that $0<\beta<+\infty$ and 
  $\E X_1^3<\infty$, $\E Y_1^2<\infty$, see \cite{BloznelisKurauskas2012}. 
The aim of the present paper is to show the asymptotic degree distribution in the 
case where $m/n\to\beta$ for 
some $\beta\in [0,+\infty]$.


 We consider a sequence of graphs $\{G_n=G(P_1,P_2, n,m)\}$, 
where $m=m_n\to +\infty$ as $n\to \infty$, and
where $P_1, P_2$ do not depend on $n$.
We denote $a_k=\E X_1^k$, $b_k=\E Y_1^k$.
 By $d(v)=d_{G_n}(v)$ we denote the degree of a vertex $v$ in $G_n$ 
(the number of vertices adjacent to $v$ in $G_n$). 
We remark that for every $n$ the random variables
 $d_{G_n}(v_1),\dots, d_{G_n}(v_n)$ are identically distributed. 
In Theorem \ref{T1} below we show the asymptotic distribution of $d(v_1)$.  





\begin{theorem}
  \label{T1}
 Let $m,n\to\infty$. 

(i) Assume that $m/n\to 0$. Suppose that  that $\E X_1<\infty$. Then $\PP(d(v_1)=0)=1-o(1)$.




(ii) Assume that $m/n\to \beta$ for some $\beta\in (0,+\infty)$.
 Suppose  that $\E X_1^2<\infty$ and $\E Y_1<\infty$. 
Then  $d(v_1)$ converges in distribution to the random variable
\begin{equation}\label{d*1}
d_*=\sum_{j=1}^{\Lambda_1}\tau_j, 
\end{equation}
where $\tau_1,\tau_2,\dots$ are independent 
and identically distributed random variables independent of the random variable $\Lambda_1$.
They are distributed as follows. For $r=0,1,2,\dots$, we have
\begin{equation}\label{d*1++}
\PP(\tau_1=r)=\frac{r+1}{\E\Lambda_2}\PP(\Lambda_2=r+1)
\qquad
{\text{and}}
\qquad
\PP(\Lambda_i=r)=\E \,e^{-\lambda_i}\frac{\lambda_i^r}{r!},
\qquad
i=1,2.
\end{equation}
Here $\lambda_1=Y_1a_1\beta^{1/2}$ and $\lambda_2=X_1b_1\beta^{-1/2}$.


(iii) Assume that $m/n\to+\infty$.
 Suppose that  $\E X_1^2<\infty$ and $\E Y_1<\infty$. 
Then  $d(v_1)$ converges in distribution to a random variable
$\Lambda_3$ having the probability distribution
\begin{equation}\label{d*2}
\PP(\Lambda_3=r)=\E e^{-\lambda_3}\frac{\lambda_3^r}{r!},
\qquad
r=0,1,\dots.
\end{equation}
Here $\lambda_3=Y_1a_2b_1$.
\end{theorem}



\begin{remark}\label{Remark1}
The probability distributions $P_{\Lambda_i}$ of 
$\Lambda_i$, $i=1,2,3$, are Poisson mixtures. One way to
sample 
from the  distribution $P_{\Lambda_i}$  is to 
generate random variable  $\lambda_i$ and then, given $\lambda_i$, 
to generate Poisson random variable 
with the parameter $\lambda_i$. The realized value of the Poisson random variable 
has the  distribution $P_{\Lambda_i}$. 
\end{remark}

\begin{remark}\label{Remark2}
The asymptotic distributions (\ref{d*1}) and (\ref{d*2}) admit  
heavy tails.
In the case (ii) we obtain a power law asymptotic degree distribution 
(\ref{d*1}) provided that the heavier of the tails $t\to\PP(\tau_1>t)$ and $t\to\PP(\Lambda_1>t)$  has a power law, see, e.g., \cite{Foss}.
% least one 
%of $P_1$ and $P_2$ has a power law and $P_1(0),P_2(0)<1$.
Similarly, in the case (iii) 
we obtain a power law asymptotic degree distribution (\ref{d*2}) 
provided that  $P_2$ has a power law.  
 \end{remark}


\begin{remark}\label{Remark3}
 Since the second moment $a_2$ does not show up in (\ref{d*1}), (\ref{d*1++}) we expect that in the case 
(ii) the second moment condition $\E X_1^2<\infty$ is redundant and  could perhaps be replaced by the weaker first moment condition $\E X_1<\infty$.
\end{remark}

 Random intersection graphs 
 have 
 attracted considerable attention in the recent literature, see, e.g., 
 \cite{Barbour2011,
 behrisch2007,
 Blackburn2009,
 Bradonjic2010,
 Britton2008,
 eschenauer2002,
 Rybarczyk2011}. 
 Starting with the paper  
  by Karo\'nski et al \cite{karonski1999}, see also \cite{Singer1995},
  where the  case  of degenerate distributions $P_1=P_{1n}$, $P_2=P_{2n}$ depending on $n$ was considered 
  (i.e., $\PP_{1n}(c_n)=\PP_{2n}(c_n)=1$,  for some $c_n>0$), 
 several more complex random intersection graph models were later introduced by Godehardt and Jaworski \cite{godehardt2001}, Spirakis et al. \cite{Spirakis2004}, Shang \cite{Shang2010}.  
 The asymptotic degree distribution  for various random intersection graph models was shown in  
  \cite{Bloznelis2008, Bloznelis2010, Bloznelis2010+, Bloznelis2011+,  Deijfen, JKS, JaworskiStark2008,
  Rybarczyk-degree2011, stark2004}.
 




\section{Proof}

Before the proof of Theorem \ref{T1} we introduce some notation and give two auxiliary lemmas.

The event that the edge $\{w_i, v_j\}$ is present in $H=H_{X,Y}$ is denoted $w_i\to v_j$. We denote
\begin{displaymath}
 {\mathbb I}_{ij}={\mathbb I}_{\{w_i\to v_j\}},
\qquad
{\mathbb I}_i={\mathbb I}_{i1},
\qquad
u_i=\sum_{2\le j\le n}{\mathbb I}_{ij},
\qquad
L=\sum_{i=1}^mu_i{\mathbb I}_i
\end{displaymath}
and remark that $u_i$ counts all neighbours of $w_i$ in $H$ 
belonging to the set $V\setminus \{v_1\}$. Denote
\begin{eqnarray}\nonumber
 &&
{\hat a}_k=m^{-1}\sum_{1\le i\le m}X^k_i,
\qquad
{\hat b}_k=n^{-1}\sum_{2\le j\le n}Y^k_j,
\\
\label{S-XY}
&&
\lambda_{ij}=\frac{X_iY_j}{\sqrt{mn}},
\qquad
 S_{XY}=\sum_{i=1}^m\sum_{j=2}^n\frac{X_i}{\sqrt{mn}}\lambda_{ij}\min\{1,\lambda_{ij}\},
\end{eqnarray}
and introduce the event ${\cal A}_1=\{\lambda_{i1}<1, \, 1\le i\le m\}$. We denote $\beta_n=m/n$.
By ${\PP}_1$ and $\E_1$ we denote the conditional probability and conditional expectation given $Y_1$.
By ${\tilde \PP}$ and ${\tilde \E}$ we denote the conditional probability 
and conditional expectation 
given $X,Y$.
By  $d_{TV}(\zeta,\xi)$ we denote the total variantion 
distance between the probability distributions
of
 random variables $\zeta$ and $\xi$. In the case where $\zeta,\xi$, $X, Y$ are 
defined on the same probability space we denote 
 by 
 ${\tilde d}_{TV}(\zeta,\xi)$ the total 
variation distance between the conditional distributions of $\zeta$ and $\xi$ given $X,Y$.

In the proof  below we use the following 
simple fact about the convergence of a sequence of random variables $\{\varkappa_n\}$: 
\begin{equation}\label{fact}
 \varkappa_n=o_P(1),
\quad
\E \sup_n\varkappa_n<\infty
\quad
\Rightarrow
\quad
\E\varkappa_n=o(1).
\end{equation}
We remark that the condition $\E \sup_n\varkappa_n<\infty$ (which assumes implicitly that
all $\varkappa_n$ are defined on the same probability space) can be replaced by a more 
restrictive condition that there exists a constant $c>0$ such that 
$|\varkappa_n|\le c$, for all $n$. In the latter case we do not need all $\varkappa_n$ to be defined on the same probability space.
In particular, given a sequence of bivariate random vectors $\{(\eta_n,\theta_n)\}$ such that, for  every $n$ and $m=m_n$ random variables $\eta_n$, $\theta_n$ and
 $\{X_i\}_{i=1}^m$, $\{Y_j\}_{j=1}^n$ are defined on the same probability space, we have 
\begin{displaymath}
{\tilde d}_{TV}(\eta_n,\theta_n)=o_P(1)
\
\Rightarrow
\
d_{TV}(\eta_n,\theta_n)\le \E{\tilde d}_{TV}(\eta_n,\theta_n) =o(1).
\end{displaymath}
Here and below all limits are taken as $n\to+\infty$ (if not stated otherwise).


\begin{lemma}
  \label{d-L}
   Assume that $\E X_1^2<\infty$ and $\E Y_1<\infty$.
We have as $n\to+\infty$
\begin{eqnarray}\label{d-L21}
&&
\PP(d(v_1)\not=L)=o(1),
\\
\label{d-L22}
&&
\PP({\cal A}_1)=1-o(1),
\\
\label{d-L23}
&&
S_{XY}=o_P(1),
\qquad
\E S_{XY}=o(1).
\end{eqnarray}

\end{lemma}



\begin{proof} Proof of (\ref{d-L21}).
We observe that the event ${\cal A}=\{d(v_1)\not=L\}$ occurs in the case where for some $2\le j\le n$
and some distinct $i_1,i_2\in [m]$ the event 
${\cal A}_{i_1,i_2,j}=
\{w_{i_1}\to v_1,\, w_{i_1}\to v_j,\, w_{i_2}\to v_1,\, w_{i_2}\to v_j\}$ occurs.
From the union bound and the inequality 
${\tilde \PP}({\cal A}_{i_1,i_2,j})\le m^{-2}n^{-2}X_{i_1}^2X_{i_2}^2Y_1^2Y_j^2$ we obtain
\begin{equation}\label{d-L1}
{\tilde \PP}({\cal A})
=
{\tilde \PP}
\left(\bigcup_{\{i_1, i_2\}\subset[m]}\bigcup_{2\le j\le n}{\cal A}_{i_1,i_2,j}\right)
\le n^{-1}{\hat b}_2Y_1^2 Q_X.
\end{equation}
Here $Q_X=m^{-2}\sum_{\{i_1,i_2\}\subset[m]}X_{i_1}^2X_{i_2}^2$. We note that
$Q_X$ and  $Y_1^2$ are stochastically bounded  and 
%We note that the product $Y_1^2Q_X$ is stochastically bounded and 
$n^{-1}{\hat b}_2=o_P(1)$ 
%converges to zero in probability
as $n\to+\infty$. Therefore, 
%the random variable 
${\tilde \PP}({\cal A})=o_P(1)$. 
%converges to zero in probability. 
Now (\ref{fact}) implies (\ref{d-L21}).
%This implies (\ref{d-L21}) since, for any $\varepsilon\in (0,1)$,
%\begin{displaymath}
%\PP({\cal A})=\E ({\tilde \PP}({\cal A}))
%\le 
%\varepsilon+\PP({\tilde \PP}({\cal A})>\varepsilon)
%=\varepsilon+o(1).
%\end{displaymath}

\medskip

Proof of (\ref{d-L22}).
Let ${\overline {\cal A}}_1$ denote the complement event to ${\cal A}_1$.
%Next we show that $\PP({\overline{\cal A}}_1)=o(1)$. 
%We obtain from 
We have, by 
the union bound  and Markov's inequality, 
%that
\begin{displaymath}
\PP_1({\overline {\cal A}}_1)\le \sum_{i\in [m]}\PP_1(\lambda_{i1}\ge 1)
\le 
(nm)^{-1}Y_1^2\sum_{i\in [m]}\E X_i^2=n^{-1}a_2Y_1^2.
\end{displaymath}
We conclude that $\PP_1({\overline {\cal A}}_1)=o_P(1)$, and now (\ref{fact}) implies $\PP({\overline {\cal A}}_1)=\E \PP_1({\overline {\cal A}}_1)=o(1)$.
%(\ref{d-L22}).
%Note that this implies (\ref{d-L22}) since, 
%for any $\varepsilon\in (0,1)$, we have
%\begin{displaymath}
% \PP({\overline{\cal A}}_1)
%=
%\E\PP_1({\overline {\cal A}}_1)
%\le 
%\varepsilon+\PP(\PP_1({\overline {\cal A}}_1)>\varepsilon)
%=
%\varepsilon+o(1).
%\end{displaymath}

\medskip

Proof of (\ref{d-L23}). Since the first bound of (\ref{d-L23}) follows from the second one, we only prove the latter.
%Now, we show that $S_{XY}=o_P(1)$.
Denote ${\hat X}_1=\max\{X_1, 1\}$ and ${\hat Y}_1=\max\{Y_1,1\}$.
We observe that $\E X_1^2<\infty$, $\E Y_1<\infty$ implies
\begin{displaymath}
\lim_{t\to+\infty}\E {\hat X}_1^2{\mathbb I}_{\{{\hat X}_1>t\}}=0,
\qquad
\lim_{t\to+\infty}\E {\hat Y}_1{\mathbb I}_{\{{\hat Y}_1>t\}}=0. 
\end{displaymath}
 Hence one can find
a strictly increasing function $\varphi:[1,+\infty)\to [0,+\infty)$ 
with 
$\lim_{t\to+\infty}\varphi(t)$ $=$ $+\infty$
such that
\begin{equation}\label{hat1}
\E {\hat X}_1^2\varphi({\hat X}_1)<\infty,
\qquad  
\E {\hat Y}_1\varphi({\hat Y}_1)<\infty.
\end{equation} 
We remark that one can find $\varphi$ which satisfies, in addition, the inequalities 
\begin{equation}\label{phi}
\varphi(t)<t
\qquad
{\text{and}}
\qquad
\varphi(st)
\le \varphi(t)\varphi(s), 
\qquad
\forall s,t\ge 1. 
\end{equation}
For this purpose we choose $\varphi$ of the form $\varphi(t)=e^{\psi(\ln (t))}$, where $\psi:[0,+\infty)\to [0,+\infty)$ is a
differentiable concave function, which grows slowly enough to satisfy (\ref{hat1}) and the first inequality of (\ref{phi}), and takes value $0$ at the origin. 
 We note that the second inequality of (\ref{phi})
follows from the concavity property of $\psi$. We remark, that (\ref{phi})
implies
\begin{equation}\label{phi1}
 t/(st)=s^{-1}\le 1/\varphi(s)\le\varphi(t)/\varphi(st),
\qquad
s,t\ge 1.
\end{equation}

Let ${\hat S}_{XY}$ be defined as in  (\ref{S-XY}) above, but with
$\lambda_{ij}$ replaced by ${\hat\lambda}_{ij}={\hat X}_i{\hat Y}_j/\sqrt{mn}$.
We note, that $S_{XY}\le {\hat S}_{XY}$. Furthermore, from the inequalities 
\begin{equation}\label{phi2}
 \min\{1,{\hat\lambda}_{ij}\}
\le
\min\Bigl\{1,\frac{\varphi({\hat X}_i{\hat Y}_j)}{\varphi(\sqrt{mn})}\Bigr\}
\le
\frac{\varphi({\hat X}_i{\hat Y}_j)}{\varphi(\sqrt{mn})}
\le 
\frac{\varphi({\hat X}_i)\varphi({\hat Y}_j)}{\varphi(\sqrt{mn})}
\end{equation}
we obtain 
$S_{XY}\le {\hat S}_{XY}\le S^*_{XY}/\varphi(\sqrt{nm})$, 
where 
\begin{displaymath}
S^*_{XY}
=
(mn)^{-1}
\left(\sum_{1\le i\le m}{\hat X}^2_i\varphi({\hat X}_i)\right)
\left(\sum_{2\le j\le n}{\hat Y}_j\varphi({\hat Y}_j)\right).
\end{displaymath}
We remark that (\ref{phi}) and (\ref{phi1}) and  imply the third and the first inequality of 
(\ref{phi2}), respectively. 

Finally, the bound $\E S_{XY}=o(1)$ follows from the inequality 
$S_{XY}\le S^*_{XY}/\varphi(\sqrt{nm})$
and the fact  that $\E S^*_{XY}$ remains bounded as $n,m\to+\infty$, 
see  (\ref{hat1}).
\end{proof}


























%*****************************************************************************************************************************







In the proof of Theorem \ref{T1} we use the following inequality 
 referred to as LeCam's inequality, see e.g., \cite{Steele}.
%

\begin{lemma}
  \label{LeCamLemma}
   Let $S={\mathbb I}_1+ {\mathbb I}_2+\dots+ {\mathbb I}_n$ be the sum of independent random indicators
with probabilities $\PP({\mathbb I}_i=1)=p_i$. Let $\Lambda$ be Poisson random variable with mean $p_1+\dots+p_n$. The total variation
distance between the distributions $P_S$ of $P_{\Lambda}$ of $S$ and $\Lambda$
\begin{equation}\label{LeCam}
\sup_{A\subset \{0,1,2\dots \}}|\PP(S\in A)-\PP(\Lambda\in A)|= \frac{1}{2} \sum_{k\ge 0}|\PP(S=k)-\PP(\Lambda=k)|
\le
\sum_{1\le i\le n}p_i^2.
\end{equation}
\end{lemma}








































%*****************************************************************************************************************************




































\begin{proof}[Proof of Theorem~\ref{T1}] 


{\it The case (i).} Since $\PP(d(v_1)\le L)=1$ it suffices to show that $\PP(L>\varepsilon)=o(1)$, for any $\varepsilon\in (0,1)$.
For this purpose we write $\PP(L>\varepsilon)=\E\PP_1(L>\varepsilon)$ and prove  that $\PP_1(L>\varepsilon)=o_P(1)$, see  (\ref{fact}).
Here we estimate $\PP_1(L>\varepsilon)$ using the union bound and Markov's inequality,  
\begin{displaymath}
\PP_1(L>\varepsilon)
\le 
\sum_{1\le k\le m}\PP_1({\mathbb I}_k=1)
\le
\E_1\sum_{1\le k\le m}\lambda_{k1}
=
\sqrt{m/n}Y_1\E X_1
=
o_P(1).
\end{displaymath}

{\it The case (ii)}. 
Before the proof we introduce some notation. Given $X,Y$, let 
\begin{equation}\label{sequences_xi}
\{\Delta_{lk}\}_{k=1}^m,
 \quad
 l=1,2,3,
 \qquad
 \{\xi_{rk}\}_{k=1}^m,
\quad
r=1,2,3,4,
\quad
{\text{and}}
\quad
\{\xi'_{3k}\}_{k=1}^m,
\quad
\{\xi'_{4k}\}_{k=1}^m,
\end{equation} 
and $\{\eta_k\}_{k=1}^m$ be sequences of conditionally independent (within each sequence) 
Poisson random variables with mean values
\begin{eqnarray}\nonumber
&&
{\tilde \E}\xi_{1k}=\sum_{2\le j\le n} p_{kj},
\qquad
{\tilde \E}\xi_{2k}=\sum_{2\le j\le n} \lambda_{kj},
\qquad
{\tilde \E}\xi_{3k}=\frac{1}{\sqrt{\beta_n}}b_1X_k,
\qquad
{\tilde \E}\xi_{4k}=\frac{1}{\sqrt{\beta}}b_1X_k,
\\
\nonumber
&&
{\tilde \E}\Delta_{1k}=\sum_{2\le j\le n} (\lambda_{kj}-p_{kj}),
\qquad
{\tilde \E}\Delta_{2k}=\frac{1}{\sqrt{\beta_n}}\delta_2 X_k,
\qquad
{\tilde \E}\Delta_{3k}=\frac{1}{\sqrt{\beta_n}}\delta_3 X_k,
\\
\nonumber
&&
{\tilde \E}\xi'_{3k}=\beta_n^{-1/2}bX_k,
\qquad
{\tilde \E}\xi'_{4k}=\beta' b_1 X_k,
\qquad
{\tilde \E}\eta_{k}=\lambda_{k1}.
\end{eqnarray}
Here we denote $b:=\min\{{\hat b}_1, b_1\}$, $\delta_{2}:= {\hat b}_1-b$, $\delta_{3}:=b_1-b$ and 
$\beta':=\min\{\beta^{-1/2},\beta_n^{-1/2}\}$.
We recall that ${\tilde \E}$ denotes the conditional expectation given $X,Y$.
We assume that, given $X,Y$,
the sequences $\{\xi'_{3k}\}_{k=1}^m$, $\{\Delta_{2k}\}_{k=1}^m$, $\{\Delta_{3k}\}_{k=1}^m$ are conditionally independent, the sequences
$\{{\mathbb I}_{k}\}_{k=1}^m$, $\{\xi_{3k}\}_{k=1}^m$ are conditionally independent,
the sequences $\{\xi_{1k}\}_{k=1}^m$, $\{\Delta_{1k}\}_{k=1}^m$ are conditionally independent, and
 the sequence $\{\eta_k\}_{k=1}^m$ is conditionally independent of the sequences (\ref{sequences_xi}).
  We assume, in addition, that for $1\le k\le m$
\begin{displaymath}
\xi_{2k}:=\xi_{1k}+\Delta_{1k},
\qquad
\xi_{3k}:=\xi_{4k}'+(\beta_n^{-1/2}-\beta') b_1 X_k,
\qquad
\xi_{4k}:=\xi_{4k}'+(\beta^{-1/2}-\beta') b_1 X_k.
\end{displaymath}
% almost surely (one may assume that $\xi_{2k}$ is defined as the sum $\xi_{1k}+\Delta_{1k}$). 
Furthermore, we suppose that, given $X,Y_1$,
the sequence $\{\eta_k\}_{k=1}^m$ is conditionally independent of the subgraph of $H_{X,Y}$ spanned by the vertices $\{v_2,\dots, v_n\}\cup W$.

\medskip

Next we introduce several random variables which are intermediate between $L$ and $d_*$:
\begin{eqnarray}\nonumber
&&
L_0=\sum_{1\le k\le m}\eta_ku_k,
\qquad
L_r=\sum_{1\le k\le m}\eta_k\xi_{rk},
\qquad
r=1,2,3,4,
\\
\nonumber
&&
L'_2=\sum_{1\le k\le m}\eta_k(\xi'_{3k}+\Delta_{2k}), 
\qquad
L'_3=\sum_{1\le k\le m}\eta_k(\xi'_{3k}+\Delta_{3k}).
\end{eqnarray}

\medskip

Finally, we denote
\begin{eqnarray}\nonumber
 &&
f_\tau(t)=\E e^{it\tau_1},
\qquad
{\bar f}_{\tau}(t)=\sum_{r\ge 0}e^{itr}{\bar p}_r, 
\qquad
{\bar p}_r
=
{\bar \lambda}^{-1}\sum_{1\le k\le m}\lambda_{k1}{\mathbb I}_{\{\xi_{4k}=r\}},
\qquad
{\bar \lambda}=\sum_{k=1}^m\lambda_{k1},
\\
\nonumber
&&
\delta=({\bar f}_\tau(t)-1){\bar \lambda}-(f_{\tau}(t)-1)\lambda_1,
\qquad
f(t)=\E_1e^{itd_*}, 
\qquad
{\bar f}(t)={\bar\E}e^{it L_4}.
\end{eqnarray}
Here ${\bar \E}$ denotes the conditional expectation given $X,Y$ 
and $\xi_{41},\dots, \xi_{4m}$. We recall that $\E_1$ denotes 
the conditional expectation given $Y_1$.

 \medskip


Let us prove the statement (ii).  In view of 
(\ref{d-L21}) the random variables $d(v_1)$ and $L$ have the same asymptotic distribution 
(if any).
We shall show  that  $L$ converges in distribution to (\ref{d*1}).
We proceed in two steps. 
Firstly, we   approximate the distribution of $L$ by that of $L_4$ 
using LeCam's inequality, 
see Lemma \ref{LeCamLemma},
and  a coupling argument.  
Secondly, we prove the convergence of the characteristic functions 
$\E e^{itL_4}\to\E e^{itd_*}$. Here and below $i=\sqrt{-1}$ denotes the imaginary unit.





































%XXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXX











{\it Step 1.} Let us show that $L$ and $L_4$ 
have the same asymptotic probability distribution (if any). To this aim we prove that 
\begin{eqnarray}\label{DeltaT}
&&
d_{TV}(L, L_0)=o(1),
\qquad
d_{TV}(L_0,L_1)=o(1),
\\
\label{L-L}
&&
\E|L_1-L_2|=o(1),
\qquad
L'_2-L'_3=o_P(1),
\qquad
\E|L_3-L_4|=o(1),
\end{eqnarray}
and observe that $L_2$ (respectively $L_3$) has the same distribution as 
$L_2'$ (respectively $L_3'$).


%....................................




\quad% 
Let us prove the first bound of (\ref{DeltaT}).
In view of (\ref{fact}) it suffices to show that
${\tilde d}_{TV}(L_0,L)=o_P(1)$. In order to prove the latter bound we apply the inequality
\begin{equation}\label{TV1}
 {\tilde d}_{TV}(L_0, L){\mathbb I}_{{\cal A}_1}\le n^{-1}Y_1^2{\hat a}_2
\end{equation} 
shown below. 
We remark that (\ref{TV1}) implies 
\begin{displaymath}
 {\tilde d}_{TV}(L_0,L)
 \le 
 {\tilde d}_{TV}(L_0,L) {\mathbb I}_{{\cal A}_1}+{\mathbb I}_{{\overline{\cal A}}_1}
 \le
 n^{-1}Y_1^2{\hat a}_2+{\mathbb I}_{{\overline{\cal A}}_1}=o_P(1).
\end{displaymath}
Here $n^{-1}Y_1^2{\hat a}_2=o_P(1)$, because $Y_1^2{\hat a}_2$ is stochastically bounded. 
Furthermore,  the bound
${\mathbb I}_{{\overline{\cal A}}_1}=o_P(1)$
follows from (\ref{d-L22}).
It remains to prove (\ref{TV1}).
% from LeCam's inequality, see Lemma \ref{LeCamLemma} below. 
We denote  
${\bar L}_k=\sum_{l=1}^k{\mathbb I}_lu_l$ $+$ $\sum_{l=k+1}^m\eta_lu_l$ and write, 
by the triangle inequality, 
%of the total variation distance,
\begin{displaymath}
 {\tilde d}_{TV}(L_0,L)\le \sum_{k=1}^m{\tilde d}_{TV}({\bar L}_{k-1}, {\bar L}_k).
\end{displaymath}
Then we estimate 
${\tilde d}_{TV}({\bar L}_{k-1}, {\bar L}_k)
\le 
{\tilde d}_{TV}(\eta_k,{\mathbb I}_k)\le (nm)^{-1}Y_1^2X_k^2$.
Here the first inequality follows from the properties of the total variation distance. 
The second inequality follows from Lemma \ref{LeCamLemma} and the fact 
that on the event ${\cal A}_1$ we have
$p_{k1}=\lambda_{k1}$.


\medskip

\quad%
 Let us prove the second bound of (\ref{DeltaT}).
In view of (\ref{fact}) it suffices to  show that
  ${\tilde d}_{TV}(L_0,L_1)=o_P(1)$. We denote  
$L^*_k=\sum_{l=1}^k\eta_lu_l+\sum_{l=k+1}^m\eta_l\xi_{1l}$ and write, 
by the triangle inequality, 
%of the total variation distance,
\begin{equation}\label{dTV4}
 {\tilde d}_{TV}(L_0,L_1)\le \sum_{k=1}^m{\tilde d}_{TV}(L^*_{k-1}, L^*_k).
\end{equation}
Here
\begin{equation}\label{dTV41}
{\tilde d}_{TV}(L^*_{k-1}, L^*_k)\le {\tilde d}_{TV}(\eta_ku_k,\eta_k\xi_{1k})
\le 
{\tilde \PP}(\eta_k\not=0){\tilde d}_{TV}(u_k,\xi_{1k}).
\end{equation}
Now, invoking the inequalities  
${\tilde \PP}(\eta_k\not=0)=1-e^{-\lambda_{k1}}\le \lambda_{k1}$
and ${\tilde d}_{TV}(u_k,\xi_{1k})\le \sum_{j=2}^np^2_{kj}$, we obtain the desired bound from (\ref{d-L23}),
(\ref{dTV4}) and (\ref{dTV41}) 
\begin{displaymath}
 {\tilde d}_{TV}(L_0,L_1)\le \sum_{k=1}^m\lambda_{k1}\sum_{j=2}^np^2_{kj} 
\le 
Y_1S_{XY}=o_P(1).
\end{displaymath}



\medskip

\quad%
Let us prove the first  bound of (\ref{L-L}).
We observe that 
\begin{displaymath}
 |L_2-L_1|=L_2-L_1
=
\sum_{1\le k\le m}\eta_k\Delta_{1k}
\end{displaymath}
and 
\begin{displaymath}
{\tilde \E}\sum_{1\le k\le m}\eta_k\Delta_{1k}
=
\sum_{1\le k\le m}\lambda_{k1}\sum_{2\le j\le n}
(\lambda_{kj}-1){\mathbb I}_{\{\lambda_{kj}>1\}}
\le Y_1S_{XY}. 
\end{displaymath}
Hence, we obtain $\E|L_2-L_1|\le \E Y_1\E S_{XY}=o(1)$, see
 (\ref{d-L23}).


\medskip

\quad%
Let us prove the second  bound of (\ref{L-L}).  It suffices to show that ${\tilde \E}|L'_2-L'_3|=o_P(1)$.
We have 
\begin{equation}\label{L2-L3}
 {\tilde \E}|L'_2-L'_3|
=
(\delta_2+\delta_3)Y_1{\hat a}_2=|{\hat b}_1-b_1|Y_1{\hat a}_2=o_P(1).
\end{equation}
In the last step we used the fact that $Y_1{\hat a}_2=O_P(1)$ and ${\hat b}_1-b_1=o_P(1)$.
%,by the law of large numbers.  
%Finally, we show that (\ref{L2-L3}) implies
%the bound $|L'_2-L'_3|=o_P(1)$. Denoting, for short, $H=|L_2'-L_3'|$ and 
%$h={\tilde \E}{\mathbb I}_{\{H\ge \varepsilon\}}$ we write, for $\varepsilon\in (0,1)$,
%\begin{equation}\label{HHHH}
% \PP(H\ge\varepsilon)
%=
%\E h
%=
%\E h
%({\mathbb I}_{\{{\tilde \E}H\ge\varepsilon^2\}}
%+
%{\mathbb I}_{\{{\tilde \E}H<\varepsilon^2\}}).
%\end{equation}
%Using the simple inequality $h\le 1$ and the inequality,
%$h\le \varepsilon^{-1}{\tilde \E}H$, which follows from Markov's inequality, we obtain
%\begin{eqnarray}\nonumber
% \E h
%{\mathbb I}_{\{{\tilde \E}H\ge\varepsilon^2\}}
%&
%\le 
%&
%\E 
%{\mathbb I}_{\{{\tilde \E}H\ge\varepsilon^2\}}=\PP({\tilde \E}H\ge\varepsilon^2)=o(1),
%\\
%\nonumber
%\E h {\mathbb I}_{\{{\tilde \E}H<\varepsilon^2\}}
%&
%\le 
%&
%\E (\varepsilon^{-1}{\tilde \E}H){\mathbb I}_{\{{\tilde \E}H<\varepsilon^2\}}
%< 
%\varepsilon.
%\end{eqnarray}
%Invoking these inequalities in (\ref{HHHH}) we obtain $\PP(H\ge \varepsilon)<\varepsilon+o(1)$.
%Hence $H=o_P(1)$. 


\quad%
The third bound of (\ref{L-L}) is straightforward
\begin{displaymath}
 \E|L_3-L_4|
\le 
\Bigl|\frac{1}{\sqrt{\beta_n}}-\frac{1}{\sqrt{\beta}}\Bigr|
b_1\sum_{1\le k\le m}\E\eta_k X_k
=
\Bigl|1-\frac{\sqrt{\beta_n}}{\sqrt{\beta}}\Bigr|
b_1^2a_2=o(1).
\end{displaymath}











%XXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXX














































{\it Step 2.} Here we show that for every real $t$ we have $\E \Delta(t)=o(1)$, where
$\Delta(t)$ $=$ $e^{itL_4}-e^{itd_*}$. To this aim we prove that 
for any realized value $Y_1$ 
there exists a positive constant 
$c=c(t,Y_1)$ such that for every $0<\varepsilon<0.5$ 
we have
\begin{equation}\label{epsilon}
 \limsup_{n,m\to+\infty}|\E(\Delta(t)|Y_1)|<c\, \varepsilon.
\end{equation}
Clearly, (\ref{epsilon}) implies $\E(\Delta(t)| Y_1)=o(1)$.
 This fact together with the simple inequality
 $|\Delta(t)|\le 2$  yields the desired bound $\E\Delta(t)=o(1)$, 
by Lebesgue's dominated convergence theorem.


\bigskip



\quad%
We fix $0<\varepsilon<0.5$ and prove (\ref{epsilon}). 
Introduce  event
${\cal D}=\{|{\hat a}_1-a_1|<\varepsilon\min\{1, a_1\}\}$ and 
let ${\overline{\cal D}}$ denote the complement event.  
Furthermore, select a number 
$T>1/\varepsilon$  such that $\PP(\tau_1\ge T)< \varepsilon$. By $c_1, c_2, \dots$ 
we denote positive numbers which do not depend on $n,m$.

We observe that, given $Y_1$, the conditional distribution of $d_*$ is the 
compound Poisson distribution
with the characteristic function $f(t)=e^{\lambda_1(f_{\tau}(t)-1)}$. Similarly,
given $X,Y$ 
and $\xi_{41},\dots, \xi_{4m}$, 
the conditional distribution of $L_4$ is the 
compound Poisson distribution
with the characteristic function ${\bar f}(t)=e^{{\bar \lambda}({\bar f}_{\tau}(t)-1)}$.
In the proof of (\ref{epsilon}) we exploit the convergence
${\bar\lambda}\to\lambda_1$ and ${\bar f}_{\tau}(t)\to f_{\tau}(t)$. In what follows 
we assume that $m,n$ are so 
large that $\beta\le 2\beta_n\le 4\beta$.

\quad%
Let us show  (\ref{epsilon}). For this purpose 
we write 
\begin{displaymath}
\E(\Delta(t)|Y_1)=\E_1\Delta(t)=I_{1}+I_{2},
\qquad
I_{1}=\E_1\Delta(t){\mathbb I}_{\cal D},
\qquad
I_2=\E_1\Delta(t){\mathbb I}_{\overline{\cal D}}        
\end{displaymath}
and estimate $I_1$ and $I_2$. We have 
$|I_2|\le 2\PP_1({\overline {\cal D}})=2\PP({\overline {\cal D}})=o(1)$, 
by the law of large numbers.

\quad%
Now we show that $I_{1}\le c\,\varepsilon+o(1)$. Combining
 the identity $\E_1\Delta(t)=\E_1f(t)(e^{\delta}-1)$
with the inequalities $|f(t)|\le 1$ and $|e^{s}-1|\le |s|e^{|s|}$
we obtain
\begin{equation}\label{2013-05-08}
 |I_1|\le \E_1|\delta|e^{|\delta|}{\mathbb I}_{\cal D}
\le 
c_1\E_1|\delta|{\mathbb I}_{\cal D}.
\end{equation}
Here we estimated $e^{|\delta|}\le e^{8\lambda_1}=:c_1$ using the inequalities
 \begin{displaymath}
 |\delta|\le 2{\bar \lambda}+2\lambda_1,
\qquad
{\bar \lambda}=Y_1\beta_n^{1/2}{\hat a}_1\le 3\lambda_1. 
 \end{displaymath}
We remark that the last  inequality holds for 
$\beta_n\le 2\beta$ provided that the event ${\cal D}$ occurs.
We shall show below that
\begin{equation}\label{2013-05-08+1}
\E_1|\delta|{\mathbb I}_{\cal D}\le (c_2+\lambda_1c_3+\lambda_1c_4)\varepsilon+o(1). 
\end{equation}
This inequality
together with (\ref{2013-05-08}) yields the desired upper bound for $I_1$. 

\quad%
Let us prove (\ref{2013-05-08+1}).
We write
\begin{displaymath}
\delta
=
({\bar f}_{\tau}(t)-1)({\bar\lambda}-\lambda_1)+({\bar f}_{\tau}(t)-f_{\tau}(t))\lambda_1
\end{displaymath}
and estimate
\begin{displaymath}
 |\delta|\le 2|{\bar\lambda}-\lambda_1|+\lambda_1|{\bar f}_{\tau}(t)-f_{\tau}(t)|.
\end{displaymath}
Now, from the inequalities $|{\hat a}_1-a_1|<\varepsilon$ and $\beta_n\le 2\beta$ we obtain
\begin{displaymath}
 |{\bar\lambda}-\lambda_1|
\le 
Y_1|{\hat a}_1-a|\beta_n^{1/2}
+Y_1a_1|\beta_n^{1/2}-\beta^{1/2}|
\le 
2Y_1\beta^{1/2}\varepsilon+o(1).
\end{displaymath}
Hence, $\E_1|{\bar\lambda}-\lambda_1|{\mathbb I}_{\cal D}\le c_2\varepsilon+o(1)$,
where $c_2=2Y_1\beta^{1/2}$.
We complete the proof of (\ref{2013-05-08+1}) by showing that
\begin{equation}\label{2013-05-06}
 \E_1|{\bar f}_{\tau}(t)-f_{\tau}(t)|{\mathbb I}_{\cal D}\le (c_3+c_4)\varepsilon+o(1).
\end{equation}


In order to prove (\ref{2013-05-06}) we  split 
\begin{displaymath}
 {\bar f}_{\tau}(t)-f_{\tau}(t)=\sum_{r\ge 0}e^{itr}({\bar p}_r-p_r)=R_1-R_2+R_3,
\end{displaymath}
and estimate separately the terms
\begin{displaymath}
 R_1=\sum_{r\ge T}e^{itr}{\bar p}_r,
\qquad
R_2=\sum_{r\ge T}e^{itr}p_r,
\qquad 
R_3=\sum_{0\le r<T}e^{itr}({\bar p}_r-p_r).
\end{displaymath}
Here we denote
$p_r=\PP(\tau_1=r)$. The upper bound for $R_2$ follows by the choice of $T$ 
\begin{displaymath}
 |R_2|\le \sum_{r\ge T}p_r=\PP(\tau_1\ge T)< \varepsilon.
\end{displaymath}
Next,
combining the identity
${\bar p}_r=({\hat a}_1m)^{-1}\sum_{1\le k\le m}X_k{\mathbb I}_{\{\xi_{4k}=r\}}$
with the inequalities
\begin{eqnarray}
 |R_1|
\le 
({\hat a}_1m)^{-1}
\sum_{r\ge T}\sum_{1\le k\le m}X_k{\mathbb I}_{\{\xi_{4k}=r\}}
=
({\hat a}_1m)^{-1}\sum_{1\le k\le m}X_k{\mathbb I}_{\{\xi_{4k}\ge T\}}
\end{eqnarray}
and ${\hat a}_1^{-1}{\mathbb I}_{\cal D}\le 2a_1^{-1}$, we estimate
\begin{equation}\nonumber
 \E_1|R_1|{\mathbb I}_{\cal D}\le 2a_1^{-1}\E_1(X_k{\mathbb I}_{\{\xi_{4k}\ge T\}})
\le 
2a_1^{-1}T^{-1}\E_1(X_k\xi_{4k})
=
2a_1^{-1}a_2b_1\beta^{-1/2}\varepsilon.
\end{equation}
Hence, we obtain $ \E_1|R_1|{\mathbb I}_{\cal D}\le c_4\varepsilon$, 
where $c_4=2a_1^{-1}a_2b_1\beta^{-1/2}$.

\quad%
Now we estimate $R_3$. 
We denote $p'_r=({\hat a}_1/a_1){\bar p}_r$
and observe that the inequality $|{\hat a}_1-a|\le \varepsilon a_1$ implies
$|{\hat a}_1a_1^{-1}-1|\le \varepsilon$ and
\begin{displaymath}
 |\sum_{0\le r\le T}e^{itr}({\bar p}_r- p'_r)|\le \varepsilon\sum_{0\le r\le T}
{\bar p}_r\le \varepsilon.
\end{displaymath}
In the last inequality we use the fact that the probabilities  $\{{\bar p}_r\}_{r\ge 0}$ 
sum up to $1$. It follows now that
\begin{displaymath}
 |R_3|{\mathbb I}_{\cal D}\le \varepsilon+\sum_{0\le r\le T}|p'_r-p_r|.
\end{displaymath}
Furthermore, observing that 
$\E_1p'_r=a^{-1}\E_1X_k{\mathbb I}_{\{\xi_{4k}=r\}}=p_r$, for $1\le k\le m$,
 we obtain 
\begin{displaymath}
 \E_1|p'_r-p_r|^2
= 
m^{-1}\E_1|a^{-1}X_k{\mathbb I}_{\{\xi_{4k}=r\}}-p_r|^2\le m^{-1}a_1^{-2}\E X_k^2.
\end{displaymath}
Hence, $\E_1|p'_r-p_r|=O(m^{-1/2})$. We conclude that
\begin{displaymath}
\E_1 |R_3|{\mathbb I}_{\cal D}
\le 
\varepsilon+O(|T|m^{-1/2})=\varepsilon+o(1),
\end{displaymath}
thus completing the proof of (\ref{2013-05-06}).































%XXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXX











































{\it The case (iii)}.  
Before the proof we introduce several random variables. 
Given $\varepsilon\in (0,1)$, 
denote
\begin{displaymath}
{\mathbb I}'_{k}={\mathbb I}_{\{X_k\beta_n^{-1/2}b_1<\varepsilon\}},
\quad
\gamma_{k}=X_k\beta_n^{-1/2}b_1{\mathbb I}'_{k},
\quad
1\le k\le m,
\qquad
{\text{and}}
\qquad
\gamma=\sum_{1\le k\le m}\lambda_{k1}\gamma_{k}.
\end{displaymath}
 Given $X,Y$, 
let ${\tilde {\mathbb I}}_{1},\dots, {\tilde {\mathbb I}}_{m}$ be conditionally independent
Bernoulli random variables with success probabilities 
\begin{displaymath}
{\tilde \PP}({\tilde {\mathbb I}}_{k}=1)=1-{\tilde \PP}({\tilde {\mathbb I}}_{k}=0)=\gamma_{k}.
\end{displaymath}
We assume that, given $X,Y$, 
the sequences 
$\{{\mathbb I}_k\}_{k=1}^m$, 
$\{{\tilde {\mathbb I}}_{k}\}_{k=1}^m$  and 
$\{\xi_{3 k}\}_{k=1}^m$ (see (\ref{sequences_xi}))
are conditionally independent. Furthermore, introduce random variables
\begin{displaymath}
L_5=\sum_{1\le k\le m}{\mathbb I}_k\xi_{3 k},
\qquad
L_6=\sum_{1\le k\le m}{\mathbb I}_k{\mathbb I}'_{k}\xi_{3 k},
\qquad
L_7=\sum_{1\le k\le m}{\mathbb I}_k{\tilde {\mathbb I}}_{k}
\end{displaymath}
and let $L_8$ be a random variable with the distribution $\PP(L_8=r)=\E e^{-\gamma}\gamma^r/r!$, for $r=0,1,\dots$.
We remark that the probability distribution  of $L_8$ is a Poisson mixture, i.e., 
the Poisson distribution with random parameter $\gamma$.

\medskip

\quad%
We note that by (\ref{DeltaT}) and the first two bounds of (\ref{L-L}), the random variables $L$ and $L_3$ 
have the same asymptotic distribution (if any). Now we prove 
that $L_3$ converges in distribution to $\Lambda_3$. For this purpose we show  that
for any $\varepsilon\in (0,1)$
\begin{eqnarray}\label{TL1}
&&
d_{TV}(L_3, L_5)=o(1),
\qquad
\E|L_5-L_6|=o(1),
\\
\label{TL2}
&&
d_{TV}(L_6, L_7)\le a_2b_1^2 \varepsilon,
\qquad
d_{TV}(L_7,L_8)=o(1),
\\
\label{TL3}
&&
\E e^{itL_8}-\E e^{it\Lambda_3}=o(1).
\end{eqnarray}

\quad%
Let us prove (\ref{TL1}). The first bound of (\ref{TL1}) 
is obtained in the same way as the first bound of (\ref{DeltaT}). To show the second 
bound  of (\ref{TL1}) we write
\begin{displaymath}
{\tilde \E}|L_5-L_6|
=
\sum_{1\le k\le m}(1-{\mathbb I}_k'){\tilde \E}{\mathbb I}_{k}{\tilde \E}\xi_{3k}
\le
Y_1b_1m^{-1}\sum_{1\le k\le m}(1-{\mathbb I}_k')X_k^2
\end{displaymath}
and obtain
\begin{displaymath}
\E |L_5-L_6|=\E{\tilde  \E}|L_5-L_6|
\le 
b_1^2\E X_1^2{\mathbb I}_{\{X_1\beta_n^{-1/2}b_1\ge \varepsilon \}}=o(1). 
\end{displaymath}
We note that the right hand side tends to zero since $\beta_n\to+\infty$.

\medskip

\quad%
Let us prove the first inequality of (\ref{TL2}).
Proceeding as in (\ref{dTV4}), (\ref{dTV41}) and using the 
identity ${\tilde {\mathbb I}}_k={\tilde {\mathbb I}}_k{\mathbb I}_k'$  we write
\begin{displaymath}
 {\tilde d}_{TV}(L_6, L_7)
\le 
\sum_{1\le k\le m}
{\mathbb I}_k'
{\tilde \PP}({\mathbb I}_k\not=0)
{\tilde d}_{TV}(\xi_{3k}, {\tilde {\mathbb I}}_k).
\end{displaymath}
Next, we estimate 
${\mathbb I}_k'{\tilde d}_{TV}(\xi_{3k}, {\tilde {\mathbb I}}_k)\le \gamma_k^2$, by
LeCam's inequality (\ref{LeCam}), and invoke the inequality
${\tilde \PP}({\mathbb I}_k\not=0)\le \lambda_{k1}$. 
We obtain
\begin{displaymath}
 {\tilde d}_{TV}(L_6, L_7)
\le 
\sum_{1\le k\le m}
{\mathbb I}_k'\lambda_{k1}\gamma_k^2
\le 
\varepsilon\sum_{1\le k\le m}
{\mathbb I}_k'\lambda_{k1}\gamma_k
\le 
\varepsilon Y_1b_1{\hat a}_2.
\end{displaymath}
Here we estimated $\gamma_k^2\le \varepsilon\gamma_k$. Now the inequalities
$d_{TV}(L_6,L_7)$ $\le$ $\E{\tilde d}_{TV}(L_6,L_7)$ $\le$  $a_2b_1^2\varepsilon$
imply the first relation of (\ref{TL2}).

\medskip

\quad%
Let us prove the second relation of (\ref{TL2}).
In view of  
(\ref{fact}) it suffices to show that ${\tilde d}_{TV}(L_7,L_8)=o_P(1)$.
For this purpose we write 
\begin{displaymath}
{\tilde d}_{TV}(L_7,L_8)\le 
{\mathbb I}_{{\cal A}_1}{\tilde d}_{TV}(L_7,L_8)+{\mathbb I}_{{\overline {\cal A}}_1},
\end{displaymath}
 where
${\mathbb I}_{{\overline {\cal A}}_1}=o_P(1)$, see (\ref{d-L22}), and estimate 
using  LeCam's inequality (\ref{LeCam})
\begin{displaymath}
{\mathbb I}_{{\cal A}_1}{\tilde d}_{TV}(L_7,L_8)
\le 
{\mathbb I}_{{\cal A}_1}\sum_{1\le k\le m}{\tilde \PP}^2({\mathbb I}_k{\tilde {\mathbb I}}_k=1){\mathbb I}_k'
= 
{\mathbb I}_{{\cal A}_1}\sum_{1\le k\le m}\lambda_{k1}^2\gamma_k^2
\le
b_1^2Y_1^2m^{-1}{\hat a}_4=o_P(1). 
\end{displaymath}
In the last step we used the fact that $\E X_1^2<\infty$ implies $m^{-1}{\hat a}_4=o_P(1)$.

\quad%
Finally, we show (\ref{TL3}). We write
${\tilde \E} e^{itL_8}=e^{\gamma(e^{it}-1)}$ and observe that 
\begin{equation}\label{Y1gamma}
 Y_1b_1a_2-\gamma=o_P(1).
\end{equation}
Furthermore, since for any real $t$ the function $z\to e^{z(e^{it}-1)}$ is bounded and uniformly continuous for $z\ge 0$, 
we conclude that (\ref{Y1gamma}) implies the convergence 
\begin{displaymath}
 \E e^{itL_8}=\E e^{\gamma(e^{it}-1)}\to\E e^{Y_1b_1a_2(e^{it}-1)}=\E e^{it\Lambda_3}.
\end{displaymath}
It remains to prove (\ref{Y1gamma}). 
We write $Y_1b_1a_2-\gamma=Y_1b_1(a_2-{\hat a}_2)+Y_1b_1{\hat a}_2-\gamma$ and 
note that
$a_2-{\hat a}_2=o_P(1)$, by the law of large numbers, and
\begin{displaymath}
 0\le \E (Y_1b_1{\hat a}_2-\gamma)
=
b_1^2\E X_1^2{\mathbb I}_{\{X_1\beta_n^{-1/2}b_1\ge \varepsilon\}}=o(1).
\end{displaymath}















\end{proof}




\bigskip
 
\subsection*{Acknowledgements}
  We thank the anonymous referee for remarks and
suggestions that have improved the presentation.

  

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\end{document}



