text:
\section{Introduction}

A common feature of reinforced processes is the definition of transition probabilities in terms of a given \textit{reinforcement weight sequence} $(w_i)_{i\in\mathbb{N}\cup\{0\}}\in(\mathbb{R}_+)^{\mathbb{N}\cup\{0\}}$.
At any time $n$, if a reinforced process $X_n$ can jump to one of the $\ell$ states $x_1$, $x_2$, ..., $x_\ell$ that have already been visited respectively $I_1, I_2, \ldots, I_\ell$ times, then for each $1\leq k\leq \ell$ the probability that it chooses $x_k$ is
$$\mathbb{P}\left[\left.X_{n+1}=x_k\right|X_0, X_1,\dots,X_n\right]=\frac{w_{I_k}}{w_{I_1}+w_{I_2}+\dots+w_{I_\ell}}.$$

Since the reinforcement weight sequence takes values in $\mathbb{R}_+$, we can note that the above probability is in $(0,1)$, or equivalently, that it never takes the values $0$ or $1$. In this paper we generalize the set of possible reinforcement sequences by introducing for $i,j\in\mathbb{N}\cup\{0\}$ the possibility for $w_i$ to be ``infinitely larger'' than $w_j$, that is allowing the values $0$ and $1$ for the above probability.
In this framework, we prove a limit theorem for the {\RUP} with exponential reinforcement, introduced by Launay \cite{launay1}.

The simplest classical reinforced process consists of a single reinforced urn.
Without further mention we will assume here and later that the initial time is $0$, and that the urn is empty at time $0$.
Suppose that there are $C$ possible colors $\mathfrak{c}_1, \mathfrak{c}_2, \ldots, \mathfrak{c}_C$.
At time $n$, let there be $N_k\geq 0$ balls of color $\mathfrak{c}_k$, for each $k=1,\ldots, C$.  Then at time $n+1$ a new ball of color $\mathfrak{c}_k$ is added to the urn with probability
$$\frac{w_{N_k}}{\displaystyle \sum_{k^\prime=1}^C w_{N_{k^\prime}}}.$$
The first such process was introduced by P\'{o}lya in 1930 (\cite{polya}), and since it has been generalized in a number of ways, see for example the survey by Pemantle \cite{pemantle}.

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\noindent
{\footnotesize Figure 1:
Transition probabilities for a particular realization of a single $w$-urn at time 12.}
\setcounter{figure}{1}

\vspace{0.3cm}
\noindent

As already mention, the first author recently introduced a multiple urn model, called the {\RUP} (IUM), in which the urns interact through shared memory. Suppose that a number $U\in\mathbb{N}$ of urns is given, and let us fix the interaction parameter $p\in [0,1]$. The dynamics is given as follows:
at each time $n$, each urn draws independently a ball either from itself with probability $1-p$, or from all the urns combined with probability $p$. In other words, the higher the $p$ the more memory is shared between the urns. To be more precise, suppose that at time $n$, for each $k=1,\ldots, C$ and $u=1,\ldots, U$, there is $N^u_k$ balls of color $\mathfrak{c}_k$ in the urn number $u$. Then for each $u$, at time $n+1$ we add a new ball of color $\mathfrak{c}_k$ in the urn $u$ with probability
$$(1-p)\frac{w_{N_k^u}}{\displaystyle \sum_{k^\prime=1}^C w_{N_{k^\prime}^u}}+p\frac{w_{N_k^*}}{\displaystyle \sum_{k^\prime=1}^C w_{N_{k^\prime}^*}},$$
where $N_k^*=\sum_{1\leq v\leq U}N_{k^\prime}^v$ is the total number of balls of color $\mathfrak{c}_k$ in the configuration at time $n$.

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\noindent
{\footnotesize Figure 2:
Transition probabilities for a particular realization of the {{IUM}} with $U=3$ at time 12.}
\setcounter{figure}{2}

\vspace{0.3cm}
\noindent
For this model, it is proved in \cite{launay1} that if
$\liminf_{i\to\infty}w_{i+1}/w_{i}>1$, and in particular if $w_i=\rho^i$ with $\rho>1$, then
\begin{itemize}
\item if $p\geq 1/2$ all the $U$ urns eventually fixate on the same color ;
\item if $p < 1/2$ there exist two colors $\mathfrak{c}_1$ and $\mathfrak{c}_2$ such that a
certain number
of urns eventually draw a ball of color $\mathfrak{c}_1$ each time they draw out of all the urns combined and draw a ball of color $\mathfrak{c}_2$ otherwise, while all the
other urns fixate on $\mathfrak{c_1}$.
The number $N$ of urns that keep drawing balls of both colors is random and observes a deterministic bound
$N < U/(2-2p)$, almost surely (see \cite{launay1}).
\end{itemize}
An open question from \cite{launay1} was to determine the law of $N$.
The generalized reinforcement weights, will allow us to answer this question (see Theorem \ref{theorem} below) in the asymptotic regime $\rho \to \infty$, provided that $U=2$ or that $U$ is an odd number.

So let us now define the set of {\it generalized reinforcement weight sequence} as
$$\mathcal{W}:=\left(\mathbb{R}_+\times\mathbb{R}\right)^{\mathbb{N}\cup\{0\}}.$$
For $w=(w_i)_{i\in\mathbb{N}\cup\{0\}}\equiv (u_i,v_i)_{i\in\mathbb{N}\cup\{0\}}\in\mathcal{W}$ in the following, we shall use the abbreviated formal notation:
$$w_i=(u_i,v_i)=:u_i\infty^{v_i},~~\forall i\in\mathbb{N}\cup\{0\}.$$
The symbol $\infty$ in this notation is meant to recall the reader of the approximation
$\rho^i \approx \infty^i$ in the sense of Theorem \ref{T:approx} below.

If $a\in\mathbb{N}\cup\{0\}$ and $B=(b_1,b_2,\dots,b_\ell)\in \bigcup_{m\in\mathbb{N}}{\left(\mathbb{N}\cup\{0\}\right)}^m$ then we define
\begin{equation*}
\pi_{w}(a,B)=\pi_{w}(a,b_1,b_2,\dots,b_\ell)=\left\{
\begin{array}{ll}
0&\qquad \displaystyle \text{if}\quad v_a<\max_{1\leq k\leq \ell}{v_b};\\
\displaystyle \frac{u_a}{\displaystyle u_a+\sum_{1\leq k\leq \ell}u_{b_k}\, \delta_{v_a}(v_{b_k})}&\qquad \text{if}\quad \displaystyle v_a=\max_{1\leq k\leq \ell}{v_b};\\
1&\qquad \text{if}\quad \displaystyle v_a>\max_{1\leq k\leq \ell}{v_b}.
\end{array}
\right.
\end{equation*}
This quantity should be interpreted as the probability to draw a black ball in an urn that contains $a$ black balls and $b_k$ balls of color $\mathfrak{c}_k$ (different of black) for each $k\in\{1,\dots,\ell\}$. In particular the probability to draw a black ball among $a$ black and $b$ white balls is:
\begin{equation*}
\pi_{w}(a,b):=\left\{
\begin{array}{ll}
0&\qquad \text{if}\quad v_a<v_b;\\
\frac{u_a}{u_a+u_b}&\qquad \text{if}\quad v_a=v_b;\\
1&\qquad \text{if}\quad v_a>v_b.
\end{array}
\right.
\end{equation*}
When clear from the context we will simply write $\pi(a,b)=\pi_{w}(a,b)$.

\textbf{Remark} Note that a classical reinforcement weight sequence $(w_i)_{i\in\mathbb{N}\cup\{0\}}\in(\mathbb{R}_+)^{\mathbb{N}\cup\{0\}}$, could be identified with a generalized reinforcement weight sequence $(u_i\infty^{v_i})_{i\in\mathbb{N}\cup\{0\}}$ where $u_i=w_i$ and $v_i$ is constant.

For reasons of simplicity, we will consider from now on the above urn processes with only two colors, black and white. In the final section we make some remarks about the setting of three or more colors.

Let us first state some easy to derive facts on the behaviour of a single urn that corresponds to the above generalized reinforcement weight sequence.
\begin{lemme}
\label{L:single urn}
(a) If $v_i$ is increasing, the color of the first ball is chosen uniformly at random, and then all the balls drawn have the same color almost surely.\\
(b) If $v_i$ is decreasing, the color of the ball drawn at even times is chosen uniformly at random, and for each $k\in \mathbb{N}$ the color drawn at time $2k+1$ is almost surely different from that drawn at time $2k$.\\
(c) For each $i_0$ such that $v_{i_0}<v_i$ for all $i<i_0$, there is almost surely $i_0$ black balls and $i_0$ white balls in the urn at time $2i_0$.
\end{lemme}
\begin{proof}
Statements (a)--(b) are easy and left to the reader.
For (c) denote by $\tau_{i_0}$ the first (random) time at which one of the colors is drawn
exactly $i_0$ times.
Recall that the urn is empty at the initial time $0$.
Suppose WLOG that this color is black. Then clearly $\P(i_0\leq \tau_{i_0}\leq 2i_0-1) =1$ and the number of white balls drawn at time $\tau_{i_0}$ is precisely
$\tau_{i_0}-i_0 <i_0$.
Due to the above assumption on the sequence $(v_i)_i$, at time
$\tau_{i_0}$ the urn is bound to draw white, and
similarly at each time $\{\tau_{i_0}+1,\ldots, 2i_0-1\}$
the urn will draw white ball almost surely, which implies the stated claim.
\end{proof}

Davis \cite{davis} proved the well-known fact:
a single urn with a classical reinforcement weight sequence
$(w_i)_{i\in\mathbb{N}\cup\{0\}}\in (\mathbb{R}_+)^{\mathbb{N}\cup\{0\}}$ eventually fixates
on a single color if and only if $\sum_{i=0}^\infty w_i^{-1}<\infty$.
The following proposition extends this result in the setting of generalized reinforcement weight sequences.

\begin{prop} If the sequence $(v_i)_{i\in\mathbb{N}\cup\{0\}}$ attains its minimum $\underline{v}:=\min_{i\in\mathbb{N}\cup\{0\}}v_i$ and
$$\sum_{i=0}^\infty\frac{\indica{v_i=\underline{v}}}{u_i}<\infty,$$
then the urn eventually fixates on a single color almost surely.
Otherwise, both black and white balls are drawn infinitely often almost surely.
\end{prop}

\begin{proof}
First note that if $(v_i)_{i\in\mathbb{N}\cup\{0\}}$ does not attain its infimum,
then there are infinitely many indices $j\in\mathbb{N}$ such that $v_j<v_i$ for all $i<j$.
Then using Lemma \ref{L:single urn}(c), for each such $j$, there are $j$ balls of each color in the urn at time $2j$.
We can therefore conclude that almost surely, balls of both colors are drawn infinitely often.

On the contrary, if $(v_i)_{i\in\mathbb{N}\cup\{0\}}$ attains its minimum, denote by $\alpha(0)$, $\alpha(1)$, $\alpha(2)$... the (finite or infinite) sequence of subscripts such that $v_{\alpha(i)}=\underline{v}:=\min_{j\in\mathbb{N}\cup\{0\}}v_j$, for each $i$.
Let us suppose first that this sequence is infinite.
We can then use the classical time line argument of Rubin.
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\noindent
{\footnotesize Figure 3:
Rubin's time line technique}
\setcounter{figure}{3}

\vspace{0.3cm}
\noindent
The time interval indicated by $\mathcal{E}(i)$ (or $\mathcal{E}'(i)$) in the figure has  length distributed as Exponential (rate $u_{\alpha(i)}$) random variable, for each $i\in \mathbb{N}$. Moreover, all the Exponential variables are mutually independent.
Note that by Lemma \ref{L:single urn}(c), at time $2\alpha(0)$ there are almost surely $\alpha(0)$ black balls and $\alpha(0)$ white balls in the urn.
Then one can construct a realization of the urn process from the time-lines just as in Davis \cite{davis} (see e.g.~also \cite{attracom}) using the fact that if the current count of one color is at $i$ such that $v_i=\underline{v}$, and of the other color is at $j$ such that $v_j>\underline{v}$, then the urn ``deterministically'' draws the latter color on this time step. The time-lines are used to decide the next draw on steps where both counts are contained in $(\alpha(i))_{i\geq 0}$, that is when the draw is not deterministic.

If on the other hand, the sequence $\alpha(i)$ defined above is finite, let $\alpha^*$ be the maximum of $\alpha(i)$. Then the urn will find itself in the situation analogous to that of Lemma \ref{L:single urn}(a) after finitely many steps, in the sense that
the color for which the count becomes first higher than $\alpha^*$ will be drawn forever after.
\end{proof}

Let us denote by $\mathfrak{U}:=\{\mathfrak{b},\mathfrak{w}\}^\mathbb{N}$ the set of all possible evolutions of an urn. If $\mathfrak{u}\in\mathfrak{U}$, then for $i\in\mathbb{N}$, $\mathfrak{u}(i)$ equals $\mathfrak{b}$ (resp. $\mathfrak{w}$) if the ball drawn at time $i$ is black (resp. white). Then $\mathfrak{U}^2$ is the set of all possible joint evolutions of the two urns. Let $F\subset \mathfrak{U}^2$ be the set of all evolutions of two urns such that both of them fixate on the same color, or more precisely;
$$F:=\{(\mathfrak{u},\mathfrak{v})\in\mathfrak{U}^2,~ \text{ such that }~\exists n_0,\forall n\geq n_0, \mathfrak{u}(n)=\mathfrak{v}(n)=\mathfrak{u}(n_0)\}$$

Let us now state the first main result of this paper.
From now on we consider the Generalized Interacting Urn Model (GIUM) with
two urns (that is, $U=2$). In the final section we address extensions to three or more urns.
\begin{theorem}\label{theorem} If $w_i=\infty^i$ (or equivalently $w_i=(1,i)$)
and if the interaction parameter $p\leq 1/2$, then
$$\mathbb{P}\left[F\right]=\frac{1-p+C(p)(3p-2)}{(1-p)(2-\lambda_{-}(p))}+C(p)$$
where
$$C(p)=-\frac{2 (1-p)^2 (1+p)}{2 p^3-6p^2+9p-4}\text{~~~and~~~}\lambda_{-}(p)=\frac{1-\sqrt{1-p^2\left(1-\frac{p}{2}\right)^2}}{2\left(1-p/2\right)^2}.$$
\end{theorem}

\textbf{Remark} The reinforcement weight sequence $w_i=\infty^i$ could be replaced
by any $w_i=u_i\infty^{v_i}$ with $u_i=1$ and $(v_i)_i$ increasing.
The law of the GIUM is trivially equal to that of GIUM with $w_i=\infty^i$.
Indeed, this law is entirely determined by the family $(\pi_w(i,j))_{i,j})$ and when $(v_i)_i$ is increasing, we have $\pi_w(i,j)=\indica{i>j}+\indica{i=j}/2$.

Note that if $p>1/2$, we already know (Theorem 3.2 from \cite{launay1}) that the two urns fixate on the same color almost surely. Therefore, the graph of
$$p\mapsto \mathbb{P}\left[F\right]$$
is as follows:

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\noindent
{\footnotesize Figure 4: Fixation probability plot}
\setcounter{figure}{4}

\vspace{0.3cm}
\noindent
	\end{center}

Before proving Theorem \ref{theorem} in Section \ref{S:proofs}, we explain
our main motivation for obtaining this result, which concerns the limiting regime for the Interacting Urn with exponential reinforcement, where $\rho \to \infty$.

For any given $\rho>1$ set $w(\rho)=(\rho^i)_{i\in\mathbb{N}\cup\{0\}}$, let and $w(\infty)=(\infty^i)_{i\in\mathbb{N}\cup\{0\}}$. Denote by $\P_{\rho},\mathbb{E}_\rho$ (resp.~$\P_{\infty},E_{\infty}$) the probability laws and expectations induced by the reinforcement weight $w(\rho)$ (resp.~$w(\infty)$) on $\mathfrak{U}^2$.
In fact, in Section \ref{S:approx} we will enrich the filtration to also include the i.i.d.~Bernoulli coin flips which tell us from which drawing pool (the urn itself or the two urns combined) are the draws made, and we will call the resulting laws again $\P_\rho,\P_\infty$. This should however not confuse the reader since the projection of the enriched laws to $\mathfrak{U}^2$ gives precisely the above defined laws.
\begin{theorem}\label{T:approx}
We have
$$\lim_{\rho\to\infty}\P_{\rho}\left[F\right]=\P_{\infty}\left[F\right],$$
where the right-hand side equals the expression given in Theorem \ref{theorem}.
\end{theorem}

Note that $F$ is a countable set, so we have
$$\P_{\rho}\left[F\right]=\P_{\rho}\left[\bigcup_{(\mathfrak{u}, \mathfrak{v})\in F}\{(\mathfrak{u},\mathfrak{v})\}\right]=\sum_{(\mathfrak{u},\mathfrak{v})\in F}\P_{\rho}\left[(\mathfrak{u},\mathfrak{v})\right].$$
For any $(\mathfrak{u},\mathfrak{v})\in F$, it is relatively easy to show that
$$\lim_{\rho\to\infty}\P_{\rho}\left[(\mathfrak{u},\mathfrak{v})\right]=\P_{\infty}\left[(\mathfrak{u},\mathfrak{v})\right].$$
However, this convergence is not sufficient to prove the convergence of the series, and the most concise rigorous argument that we could find is perhaps surprisingly long. Since we use some elements of the proof of Theorem \ref{theorem} in the proof of Theorem \ref{T:approx}, this is postponed to Section \ref{S:approx}.

\section{Proofs of Theorems \ref{theorem} and \ref{T:approx}}
\label{S:proofs}
Our first observation is that for all $i,j\in\mathbb{N}$, we have $\pi(i,j)=\pi(i-1,j-1)$. This means that if an urn contains both black and white balls, one can withdraw one black ball and one white ball from this urn without changing the law of the next ball to be drawn. In other words, if an urn contains $B$ black balls and $W$ white balls with $B\geq W$, its behaviour is the same as the behaviour of an urn that contains $B-W$ black balls and no white balls.

This remark allows us to simplify the set of possible configurations for two interacting urns. Let us consider two urns, containing respectively $B_1$ and $B_2$ black balls and $W_1$ and $W_2$ white balls. We can classify the possible configurations into three cases:

\textbf{Configuration 1.} $B_1+B_2=W_1+W_2$. It also means that $B_1-W_1=W_2-B_2$, and according to the previous remark, this configuration is equivalent to a configuration in which one urn contains $|B_1-W_1|=|B_2-W_2|$ black balls and the other as many white balls. We denote this configuration by $C_1(|B_1-W_1|)$.

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\noindent
{\footnotesize Figure 5:
Several examples of two urns in configuration $C_1(4)$.}
\setcounter{figure}{5}

\vspace{0.3cm}
\noindent
	\end{center}
The configuration in which both urns are empty is denoted by $C_1(0)$.

\textbf{Configuration 2.} $B_1+B_2 \neq W_1+W_2$ and $(B_2-W_2)(B_1-W_1)\leq 0$. This means that there is a global majority color in the two urns, but one of the two urns has a majority of the other color. Suppose without loss of generality that $B_1+B_2> W_1+W_2$, $B_1>W_1$ and $B_2\leq W_2$. Note that starting from this configuration, all the balls drawn from the urn 1 will be black almost surely. Therefore, the law of the next ball to be drawn starting from this configuration only depends on the difference $W_2-B_2$. We denote this configuration by $C_2(W_2-B_2)$.
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\noindent
{\footnotesize Figure 6:
Several examples of two urns in configuration $C_2(2)$.}
\setcounter{figure}{6}

\vspace{0.3cm}
\noindent
	\end{center}

\textbf{Configuration 3.} $B_1+B_2\neq W_1+W_2$, and $(B_2-W_2)(B_1-W_1)>0$. Here the global majority color is the same as the majority color in each of the two urns. Starting from this configuration, all the balls drawn will have the majority color, almost surely. We denote this configuration by $C_3$.

Since the two urns are initially empty, they stay for some time in Configuration 1 and then jump to Configuration 2 (or possibly directly to Configuration 3). Once in Configuration 2, they can either stay in Configuration 2 forever, or jump to Configuration 3, and stay there forever after.
A typical evolution of this process is schematically depicted by the following figure:

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\noindent
{\footnotesize Figure 7:
The three phases of the evolution.}
\setcounter{figure}{7}

\vspace{0.3cm}
\noindent
	\end{center}

\subsection{Proof of Theorem \ref{theorem}}
For $\ell\geq 0$ we denote by
$$q_\ell\equiv q_\ell(p):=\mathbb{P}\left[\text{The two urns fixate on the same color starting from configuration }C_1(\ell)\right],$$
and
$$r_\ell\equiv r_\ell(p):=\mathbb{P}\left[\text{The two urns fixate on the same color starting from configuration }C_2(\ell)\right].$$

As already noted, starting from configuration $C_3$, the probability that the two urns fixate on the same color equals 1.
We proceed by observing that, due to the Markov-like property, some simple relations connect these different probabilities.

Starting from $C_1(0)$, each urn independently draws a white or black ball with probability 1/2. Therefore with probability 1/2, the two urns draw the same color and immediately (at time $1$) enter the configuration $C_3$, and with probability 1/2 the two urns choose different colors and jump to configuration $C_1(1)$ at time $1$. This implies
\begin{equation}
\label{eqn_p0}q_0=\frac{1}{2}+\frac{1}{2}q_1.
\end{equation}
For $\ell\geq 1$, starting from $C_1(\ell)$, the urns can jump to $C_1(\ell-1)$, $C_1(\ell+1)$ or $C_2(\ell-1)$, and this decision is made independently form the past. More precisely:
\begin{itemize}
\item They jump to configuration $C_1(\ell-1)$ if and only if each urn draws a ball of the color which is not its majority color. In order that this happens, for each of the urns the drawing must be done in the two urns combined (with probability $p$) and the minority color of the urns combined must be chosen (with probability $1/2$). So the probability of this move equals $(p/2)^2$.
\item They jump to configuration $C_1(\ell+1)$ if and only if each urn draws a ball of its majority color. This happens with probability $((1-p)+p/2)^2=(1-p/2)^2$.
\item They jump to configuration $C_2(\ell-1)$ if the two urns draw a ball of the same color. This happens with the remaining probability $p(1-p/2)$.
\end{itemize}
This implies the following sequence of relations:
\begin{equation}\label{eqn_p1}
q_\ell=\left(\frac{p}{2}\right)^2 q_{\ell-1}+\left(1-\frac{p}{2}\right)^2 q_{\ell+1}+p(1-\frac{p}{2})r_{\ell-1}, \ \ell\geq 1.
\end{equation}

Starting from configuration $C_2(0)$, the urns either jump to configuration $C_3$ with probability $(1+p)/2$, or to configuration $C_2(1)$ with probability $(1-p)/2$.
We conclude that
\begin{equation}\label{eqn_r1}
r_0= \frac{1+p}{2}+\frac{1-p}{2}r_1.
\end{equation}

Similarly, for $\ell\geq 1$, starting from configuration $C_2(\ell)$, the urns either jump to configuration $C_2(\ell-1)$ with probability $p$, or to configuration $C_2(\ell-1)$ with probability $1-p$, again independently of the past. This translates into
\begin{equation}\label{eqn_r2}
r_\ell = p r_{\ell-1} + (1-p)r_{\ell+1}.
\end{equation}
In fact, the above reasoning shows that if $Y_n$ is an integer-valued random variable such that at time $n$, the urns are in configuration $C_2(Y_n)$, then $(Y_{k \wedge \tau_0}, \,k\geq n)$ is the simple random walk with drift $1-2p$ killed at
$\tau_0:=\inf\{j\geq n:\, Y_j=0\}$.
It is well known (recall that $p<1/2$) that
$\P(\tau_0<\infty\,|\, {\mathcal F}_n)=p/(1-p)$ on $\{Y_n=1\}$, yielding
$r_1=\frac{p}{1-p}r_0$, and
together with (\ref{eqn_r1}) this  gives
$r_0=\frac{1+p}{2-p}.$
Now one can deduce by induction from (\ref{eqn_r2}) that
$$r_\ell=\frac{1+p}{2-p}\left(\frac{p}{1-p}\right)^\ell,\ \ell\geq 0.$$
We then introduce these values into (\ref{eqn_p1}) to obtain
\begin{equation}\label{eqn_p2}
q_\ell=\left(\frac{p}{2}\right)^2 q_{\ell-1}+\left(1-\frac{p}{2}\right)^2 q_{\ell+1}+\frac{p(1+p)}{2}\left(\frac{p}{1-p}\right)^{\ell-1}, \ \ell\geq 1.
\end{equation}
Define a function
$$f_p(x):=\sum_{\ell=0}^{\infty}q_\ell(p)\frac{x^\ell}{\ell !}.$$
Differentiating both sides and applying (\ref{eqn_p2}) we get
\begin{eqnarray*}
f_p^{\prime}(x)&=&\sum_{\ell=0}^{\infty}q_\ell(p)\frac{x^{\ell-1}}{(\ell-1) !}\\
&=&\sum_{\ell=1}^{\infty}\left(\left(\frac{p}{2}\right)^2 q_{\ell-1}+\left(1-\frac{p}{2}\right)^2 q_{\ell+1}+\frac{p(1+p)}{2}\left(\frac{p}{1-p}\right)^{\ell-1}\right)\frac{x^{\ell-1}}{(\ell-1) !}\\
&=&\left(\frac{p}{2}\right)^2\sum_{\ell=0}^{\infty}q_\ell(p)\frac{x^{\ell}}{\ell !}+\left(1-\frac{p}{2}\right)^2\sum_{\ell=0}^{\infty}q_{\ell+2}(p)\frac{x^{\ell}}{\ell !}+\frac{p(1+p)}{2}\sum_{\ell=0}^{\infty}\frac{\left(\frac{px}{1-p}\right)^{\ell}}{\ell !}\\
&=&\left(\frac{p}{2}\right)^2f_p(x)+\left(1-\frac{p}{2}\right)^2f_p^{\prime\prime}(x)+\frac{p(1+p)}{2}\exp\left(\frac{px}{1-p}\right).\\
\end{eqnarray*}
So $f_p$ is a solution of the ODE
\begin{equation}\label{ODE}
\left(1-\frac{p}{2}\right)^2y^{\prime\prime}-y^{\prime}+\left(\frac{p}{2}\right)^2y+\frac{p(1+p)}{2}\exp\left(\frac{px}{1-p}\right)=0.
\end{equation}
The characteristic polynomial of the homogeneous ODE is
$$\left(1-\frac{p}{2}\right)^2\lambda^2-\lambda+\left(\frac{p}{2}\right)^2=0,$$
and its two solutions are
\begin{equation}\label{Elambda}
\lambda_{\pm}(p)=\frac{1\pm\sqrt{1-4\left(\frac{p}{2}\right)^2\left(1-\frac{p}{2}\right)^2}}{2\left(1-p/2\right)^2}=\frac{1\pm\sqrt{1-p^2\left(1-\frac{p}{2}\right)^2}}{2\left(1-p/2\right)^2}.
\end{equation}
So the general solution of the homogeneous equation is of the form
$$f_p^1(x)=A(p)\exp{(\lambda_{-}(p)x)}+B(p)\exp{(\lambda_{+}(p)x)}.$$
Let us now search for a particular solution to (\ref{ODE}) of the form $f_p^2(x)=C(p)\exp(xp/(1-p))$. Plugging it into (\ref{ODE}), we obtain
\begin{equation}\label{EC}
C(p)=-\frac{2 (1-p)^2 (1+p)}{2 p^3-6p^2+9p-4}\ .
\end{equation}
Hence the general solution to (\ref{ODE}) is of the form
$$f_p(x)=f_p^1(x)+f_p^2(x)=A(p)\exp{(\lambda_{-}(p)x)}+B(p)\exp{(\lambda_{+}(p)x)}+C(p)\exp\left(\frac{p}{1-p}x\right).$$
In order to find $A(p)$ and $B(p)$, note that since $q_\ell\leq 1$ for all $\ell\geq 0$, we have
$$f_p(x)=\sum_{\ell=0}^{\infty}q_\ell(p)\frac{x^\ell}{\ell !}\leq \sum_{\ell=0}^{\infty}\frac{x^\ell}{\ell !}=\exp(x).$$
So $f_p(x)=o(\exp{(\lambda_{+}(p)x)})$ since $\lambda_+(p)>1$ as can be easily checked.
Therefore we deduce that $B(p)=0$.
We can now find $A(p)$ from the following linear system:
$$\left\{\begin{array}{l}
q_0=f_p(0)=A(p)+C(p)\\
q_1=f_p^{\prime}(0)=A(p)\lambda_{-}(p)+C(p)\frac{p}{1-p}
\end{array}\right. .$$
Recalling (\ref{eqn_p0}) we then obtain
$$A(p)+C(p)=\frac{1}{2}+\frac{1}{2}\left(A(p)\lambda_{-}(p)+C(p)\frac{p}{1-p}\right),$$
and deduce that
$$A(p)=\frac{1-p+C(p)(3p-2)}{(1-p)(2-\lambda_{-}(p))}.$$
Therefore
$$q_0(p)=A(p)+C(p)=\frac{1-p+C(p)(3p-2)}{(1-p)(2-\lambda_{-}(p))}+C(p),$$
where $C(p)$ and $\lambda_{-}(p)$ are as in (\ref{EC}) and (\ref{Elambda}), respectively, concluding the proof of Theorem \ref{theorem}.

One can check in particular that $C(0)=1/2$, $C(1/2)=1$, $A(0)=A(1/2)=0$, and so
$$q_0(0)=1/2\text{~~~and~~~}q_0(1/2)=1.$$
Indeed, when $p=0$ the two urns are independent and the probability that they fixate on the same color is $1/2$, while $p=1/2$ is the critical phase of the IUM, for which we already knew that the two urns fixate on the same color, almost surely (see Theorem 3.5 in \cite{launay1}).

\subsection{Proof of Theorem \ref{T:approx}}
\label{S:approx}
In most of this argument we do not fix some $\rho\in (1,\infty]$ but rather consider the whole family of laws $\P_\rho$, $\rho\in (1,\infty]$.
Denote by $S_k$ the state of the two (generalized) interacting urns at time $k$.
The reader should think of $S_k$ as a ``struct'' that contains the count of black and white balls for each of the urns at time $k$, as well as the information on the drawing pools used at time $k-1$ (for the draws made to update the state at time $k$ from the state at time $k-1$).
Denote by ${\mathcal F}=({\mathcal F}_k)_{k \geq 0}$ the filtration generated by the process $S$, that is,
${\mathcal F}_k=\sigma\{S_n:n =0,\ldots,k\}$.
Define furthermore $\sigma_1:=0$,
$$
\sigma_2:=\inf\{k \geq 0: S_k \in \cup_{i=1}^\infty C_2(i)\},\
\sigma_3:=\inf\{k \geq 0: S_k \in C_3\}.
$$
Clearly, all the $\sigma_i$ are stopping times with respect to ${\mathcal F}$.
As depicted in Figure 7, we have $\P_\infty(\sigma_1 \leq \sigma_2 \wedge \sigma_3 \leq  \sigma_3)=1$.
As argued in the course of the proof of Theorem \ref{theorem}, we also have that
if $S_k \in C_2(\ell)$ for some $k$ and $\ell\geq 1$, then
$\P_\infty(S_{k+1} \in \{C_2(\ell-1),C_2(\ell+1)\})=1$.
Similarly, if $S_k \in C_2(0)$ for some $k$ then
$ \P_\infty(S_{k+1} \in \{C_2(1),C_3\})=1$, and if $S_k \in C_3$ then
$\P_\infty(S_{k+1} \in C_3)=1$. Therefore
\[
F= \{\sigma_3< \infty\},\ \P_\infty-\mbox{almost surely}.
\]
Our main problem in proving the current result is related to the fact that most of the above statements fail
to hold under $\P_\rho$ if $\rho<\infty$.

Let $\bar{A}$ be following event: there exist some $m \in \mathbb{N}$ and an urn $u$ such that the color of the ball drawn for $u$ at time $m$ is the minority color in the drawing pool  prescribed by the corresponding Bernoulli($p$) flip (if $1$ this pool is the two urns combined, otherwise this pool is the urn $u$ itself). Such a draw is asymptotically impossible, or equivalently $\P_\infty(\bar{A})=0$, and henceforth we will refer to it as ``AI-draw''. Let $A$ be the complement of $\bar{A}$.
Our strategy is to show the following two points.
\begin{lemme}
\label{L:two pts}
(i) $\lim_{\rho\to\infty}\P_\rho[F \cap A]=\P_\infty[F \cap A]=\P_\infty[F]$,\\
(ii) $\P_\rho[F\cap \bar{A}]=O(1/\rho)$.
\end{lemme}
Let $\tau:= \inf\{k \geq 0: S_k$ is a result of an AI-draw$\}$, with the usual convention that $\tau=\infty$ if such a draw is never made.
Then is it clear that $\tau$ is a stopping time with respect to ${\mathcal F}$ (for each $\rho\in (1,\infty)$).

Our interest in $\tau$ comes from the observation that
the state of the two urns changes according to the same general pattern (depicted in Figure 7) under $\P_\rho$ and under $\P_\infty$ until the time of the first AI-draw, where the transition probabilities depend on $\rho$.
More precisely, we have $\sigma_1=\sigma_1\wedge \tau \leq \tau \wedge \sigma_2
\wedge \sigma_3 \leq \tau \wedge \sigma_3$, and also if $S_k \in C_2(\ell)$ for some $k<\tau$ and $\ell\geq 1$, then either
$\{\tau=k+1\}$ or $\{\tau > k+1, S_{k+1} \in \{C_2(\ell-1),C_2(\ell+1)\}\}$ happen
$\P_\rho$-almost surely.
Furthermore, if $S_k \in C_2(0)$ for some $k<\tau$, then
$\P_\rho(\tau=k+1) + \P_\rho(\tau > k+1, S_{k+1} \in \{C_2(1),C_3\})=1$
and if $S_k \in C_3$ for some $k<\tau$, then
$\P_\rho(\tau=k+1) + \P_\rho(\tau > k+1, S_{k+1} \in C_3)=1$.
Therefore, $F\cap A=\{\sigma_3<\infty, \tau=\infty\}$ and more importantly $F\cap \bar{A}=\{\sigma_3<\infty, \tau<\infty\}$, $\P_\rho$-almost surely, for any $\rho\in(1,\infty]$.

In order to show part (i) of Lemma \ref{L:two pts}, we note that the probability of any path (in the countable set of infinite paths) contributing to $F\cap A$
is a product of finitely many terms of the form $c\times\frac{\rho^a}{\rho^a+\rho^b}$ with
$c\in \{1,p,1-p\}$ and
$a\geq b$ (otherwise, there would exist an AI-draw along the path)
and infinitely many terms of the form $p\times\frac{\rho^a}{\rho^a+\rho^b} +
(1-p)\times\frac{\rho^c}{\rho^c+\rho^d}$, where again $a/b \wedge c/d\geq 1$.
Therefore for each infinite path $\pi\in F\cap A$,
$$\rho \mapsto \P_{\rho}(\pi)$$
increases to $\P_{\infty}(\pi)$ and so by the monotone convergence theorem, we have $\lim_{\rho\to\infty}\P_\rho(F\cap A)=\P_\infty(F\cap A)=\P_\infty(F)$, where the last equality is due to
$\P_\infty(\bar{A})=0$.

Arguing part (ii) of Lemma \ref{L:two pts} turns out to be more tricky.
First note that $\P_\rho(F\cap \bar{A})=$
\[
\P_\rho(\sigma_3<\infty, \tau<\infty)\leq \P_\rho(\tau \leq \sigma_2 \wedge \sigma_3,\sigma_3<\infty) + \P_\rho(\sigma_2< \tau \leq \sigma_3<\infty) + \P_\rho(\sigma_3< \tau <\infty).
\]
The final term above is the simplest one to study so we consider it first.
Indeed, at time $\sigma_3$
the count of the global majority color for all the pools (each of the urns, and therefore the urns combined) is at least by one higher than that for the minority color.
Moreover,
as long as the two urns remain in state $C_3$ (which is certainly true until time $\tau$), the count of the majority color for all the pools (each of the urns, and therefore the urns combined) increases by at least one in each step.
Therefore, by nested conditioning with respect to ${\mathcal F}_{\sigma_3+i}$, we have
\begin{eqnarray*}
\P_\rho(\sigma_3< \tau <\infty)&=&
\sum_{i=0}^\infty \P_\rho(\sigma_3<\infty,\sigma_3+i<\tau,\sigma_3+i+1= \tau)\\
&\leq& \sum_{i=0}^\infty \P_\rho(\sigma_3<\infty,\sigma_3+i<\tau) \cdot \frac{1}{1+\rho^{i+1}}\\
&\leq& \sum_{i=0}^\infty \frac{1}{1+\rho^{i+1}}= O\left(\frac{1}{\rho} \right).
\end{eqnarray*}

Note that $\sigma_2 \wedge \sigma_3$ is a finite (even stochastically bounded by a
geometric, see the proof of Lemma \ref{L:unibd on tau} below) random variable under $\P_\rho$, for each $\rho$.
We split the event $\{\tau\leq \sigma_2 \wedge \sigma_3\}$ according to the state of the system at time $\tau-1$ and the value of $\tau$. Note that AI-draw is impossible from state $C_1(0)$, for any $\rho\in \mathbb{R}_+$. Also note that $\sigma_2<\sigma_3$ (in words, the two urns enter $\cup_{\ell \geq 0}C_2(\ell)$ at the exit time from $\cup_i C_1(i)$) unless
$S_{\sigma_2 \wedge \sigma_3 -1}\in C_1(0)$ in which case the two urns jump directly to $C_3$.
Thus
\begin{eqnarray*}
\P_\rho(\tau\leq \sigma_2 \wedge \sigma_3,\sigma_3<\infty)&=&
\sum_{i=1}^\infty \P_\rho(S_{\tau-1}\in C_1(i), \tau\leq \sigma_2, \sigma_3<\infty)\\
&\leq& \sum_{i=1}^\infty \sum_{k=0}^\infty \P_\rho(\tau>k-1, S_{k-1}\in C_1(i),
\tau=k,\sigma_3<\infty)\\
&\leq& \sum_{i=1}^\infty \sum_{k=0}^\infty
\frac{1}{1+\rho^i} \P_\rho(\tau>k-1, S_{k-1}\in C_1(i),\,\sigma_3<\infty)\\
&\leq& \sum_{i=1}^\infty \frac{1}{1+\rho^i} \mathbb{E}(\#\{\mbox{steps in }C_1(i)
\mbox{ before }\tau\};\,\sigma_3<\infty)\\
&\leq& O\left(\frac{1}{\rho} \right) \mathbb{E}(\#\{\mbox{steps in }\cup_i C_1(i)
 \mbox{ before }\tau\};\,\sigma_3<\infty),
\end{eqnarray*}
where the inequality in the third line is again obtained by nested conditioning with respect to ${\mathcal F}_{k-1}$ and the fact that on $\{S_{k-1}\in C_1(i)\}$ the difference in the counts of majority and minority colors is equal to $i$.

In an analogous fashion (we leave the details to the reader) one can obtain
$$
\P_\rho(\sigma_2< \tau \leq \sigma_3<\infty) \leq
O\left(\frac{1}{\rho} \right) \mathbb{E}(\#\{\mbox{steps in }\cup_i C_2(i)
 \mbox{ before }\tau\};\,\sigma_3<\infty).
$$

\begin{lemme}
\label{L:unibd on tau}
Fix $p\in (0,1/2)$.
There exist $\rho_0<\infty$ such that for each $\rho > \rho_0$ we have
$$
\mathbb{E}_\rho(\tau \wedge \sigma_3 \cdot \indica{\sigma_3<\infty}) < C,
$$
where $C$ is a finite constant which depends only on $p$.
\end{lemme}

Since, due to the above calculations,
$$
\P_\rho(F\cap \bar{A})=O\left(\frac{1}{\rho} \right) (1+ \mathbb{E}(\#\{\mbox{steps in }\cup_{j=1,2}\cup_i C_j(i) \mbox{ before }\tau\};\,\sigma_3<\infty),
$$
and the expectation on the right-hand side above is precisely
$\mathbb{E}_\rho(\tau \wedge \sigma_3 \cdot \indica{\sigma_3<\infty})$, we obtained the required claim, and this completes the proof of Lemma \ref{L:two pts}, and in turn the proof of Theorem \ref{T:approx}.

\smallskip
\noindent
{\em Proof of Lemma \ref{L:unibd on tau}.}
Note that
$$
\mathbb{E}_\rho(\tau \wedge \sigma_3 \cdot \indica{\sigma_3<\infty}) \leq
\mathbb{E}_\rho(\tau \wedge \sigma_2 \wedge \sigma_3) + \mathbb{E}_\rho((\tau \wedge \sigma_3 - \tau \wedge \sigma_2 \wedge \sigma_3) \,\indica{\sigma_3<\infty}).
$$
Again, bounding the first expectation on the right-hand side above is simpler. For this, note that on $\{k < \sigma_2 \wedge \sigma_3\}$ the two urns have a probability uniformly bounded from below by $2(p/2)^2=p^2/2>0$ for $\sigma_2$ to happen at step $k$, since for this it is sufficient that the draws for both urns are made from the two urns combined, and that the same (equally likely) color is chosen for both.
Therefore, as already mentioned, the random variable $\sigma_2 \wedge \sigma_3$ is stochastically bounded by a geometric random variable with success probability $p^2/2$. We conclude that
$\tau \wedge \sigma_2 \wedge \sigma_3$ has expectation bounded by $2/p^2$ under any $\P_\rho$, for $\rho\in (1,\infty]$.

To bound
$ \mathbb{E}_\rho((\tau \wedge \sigma_3 - \tau \wedge \sigma_2 \wedge \sigma_3) \,\indica{\sigma_3<\infty})$, we use the fact that on $\{\sigma_2<\sigma_3\}$
(WLOG we can and will assume that this happens), under $\P_\rho$, the steps of the two urn model during the time interval $[\tau \wedge \sigma_2, \tau \wedge \sigma_3]$
go through the classes $C_2(i)$ according to the pattern illustrated by Figure 7.
More precisely, it is easily seen that
\begin{equation}
\label{EenterC3}
\begin{array}{ll}
\P_{\rho}(\sigma_3=k+1\,|\, {\mathcal F}_k) \indica{\tau > k,\, S_k \in C_2(0)}
\!\!\! &
=\P_{\rho}(S_{k+1} \in C_3 \,|\, {\mathcal F}_k) \indica{\tau > k,\, S_k \in C_2(0)}\\
& \geq \frac{1}{4}
\indica{\tau > k, S_k \in C_2(0)},
\end{array}
\end{equation}
uniformly in $k$, and moreover that for any $\ell \geq 1$ and any $k\in\mathbb{N}$.
on $\{\tau > k,\, S_k \in C_2(\ell)\}$
\begin{equation}
\label{EstepsC2}
\begin{array}{ll}
\P_{\rho}(S_{k+1} \in C_2(\ell+1),\tau>k+1 \,|\, {\mathcal F}_k) =
(1-p) (1+O_{k,\ell+1}(\rho)),\\
\P_{\rho}(S_{k+1} \in C_2(\ell-1),\tau>k+1 \,|\, {\mathcal F}_k) =
p (1+O_{k,\ell-1}(\rho)),
\end{array}
\end{equation}
where for each $k$ and all $\ell\in \mathbb{N}$, $O_{k,\ell+1}(\rho),O_{k,\ell-1}(\rho)$
are ${\mathcal F}_k$-measurable random variables satisfying the following important
property: there exists some $C<\infty$ such that
\[
\P_\rho(O_{k,\ell-1}(\rho)\leq C/\rho)=\P_\rho(O_{k,\ell+1}(\rho)\leq C/\rho)=1, \ \forall k,\ell\in \mathbb{N}, \ \rho\in (1,\infty).
\]

Now fix some $\rho \in (1,\infty)$.
On $\{\sigma_2 < \tau \wedge \tau_3\}$, the two urns enter the set of configurations $\cup_\ell C_2(\ell)$
via $C_2(I)$, where $I$ is such that $S_{\sigma_2}\in C_2(I)$, $\P_\rho$-almost surely.
Due to the reasoning used in bounding $\mathbb{E}(\tau \wedge \sigma_2 \wedge \sigma_3)$, $I$ is stochastically bounded by a geometric random variable with success probability $p^2/2$.
Recall the process $Y_n$ defined in the paragraph which contains relation
(\ref{eqn_r2}).
Note that the same definition for $Y$ makes sense under $\P_\rho$ (at least) until time $\tau$, and recall the stopping time $\tau_0=\inf\{j\geq n:\, Y_{j\wedge \tau}=0\}$.

Due to (\ref{EenterC3}), there are at most geometric (with success probability $1/4$) many excursions of $Y_{\cdot \wedge \tau}$ above $0$ before $\sigma_3$.
The first excursion starts at $I$ and ends at $0$, while all the others start and end at $0$.
Due to (\ref{EstepsC2}), all the excursions but (possibly) the first one
are ``identically distributed up to an error of order
$(1+C/\rho)^{\text{excursion length}}$''.
By this we mean that for a fixed excursion path ${\mathbf{e}}$ away from $0$, the probability
that the $k$th excursion above takes value ${\mathbf{e}}$ equals
$p({\mathbf{e}})x (1+g_k(\rho,{\mathbf{e}}))^{|{\mathbf{e}}|}$
where $p({\mathbf{e}})$ depends only on the steps in ${\mathbf{e}}$, $|{\mathbf{e}}|$ is the number of steps in
${\mathbf{e}}$, and $|g_k(\rho,{\mathbf{e}})|\leq C/\rho$ for all $k,\rho,{\mathbf{e}}$. In the sequel, we will explain this even more precisely.
Due to Wald identity, we now know that for bounding
$\mathbb{E}_\rho((\tau \wedge \sigma_3 - \tau \wedge \sigma_2 \wedge \sigma_3) \,\indica{\sigma_3<\infty})$, it suffices to show that
both
\begin{equation}
\label{Etau from i}
\sum_i (1- p^2/2)^{i-1}\mathbb{E}_\rho(\tau_0;\tau_0<\infty|Y_0=i)<\infty
\end{equation}
and $\mathbb{E}(\tau_0;\tau_0<\infty|Y_0=0)<\infty$.
The excursion starting at $i$ and ending at $0$ can again be split into
$i$ almost identically distributed excursions (due to (\ref{EstepsC2})),
where the $j$th excursion starts at $i-j+1$ and ends at the time of
the first visit to level $i-j$, $j=1,\ldots,i$.
Denote by $\tau_{j}$ the time of the first visit to level $j$.
Therefore, bounding $\mathbb{E}_\rho(\tau_0;\tau_0<\infty|Y_0=i)$ for $i\geq 1$ amounts to bounding
$\sum_{j=1}^i \mathbb{E}_\rho(\tau_{i-j}-\tau_{i-j+1};\tau_{i-j}<\infty|Y_0=i)$.

Now due to (\ref{EstepsC2}), we know that
\[
\mathbb{E}_\rho(\tau_{i-j}-\tau_{i-j+1};\tau_{i-j}<\infty|Y_0=i)
\leq \mathbb{E}(\tau_0'\cdot (1+C/\rho)^{\tau_0'};\tau_0'<\infty|Y_0'=1),
\ 1\leq j \leq i,
\]
where $Y'$ is now the simple asymmetric random walk with $\P(\Delta Y' =1)=1-p=1-\P(\Delta Y' =-1)$ (as it is under $\P_\infty$) and $\tau_0'$ is its corresponding $\tau_0$.
With a finite upper bound $B$ for the expectation on the right-hand side in the last display, we would have
$\mathbb{E}_\rho(\tau_0;\tau_0<\infty|Y_0=i)\leq i \times B$ and this would make the sum in (\ref{Etau from i}) bounded by a multiple of $B$, that would depend only on $p$.
Furthermore, since trivially $\mathbb{E}_\rho(\tau_0;\tau_0<\infty|Y_0=0)=
1 + \mathbb{E}_\rho(\tau_0;\tau_0<\infty|Y_0=1)$ (the walk $Y$ is reflected at $0$),
one sees that a uniform (in large $\rho$) upper bound on
$$\mathbb{E}(\tau_0 \cdot (1+C/\rho)^{\tau_0'};\tau_0'<\infty|Y_0'=1)$$
is sufficient to conclude the argument.

The next step is to condition on $\tau_0'<\infty$.
It is well-known (and easy to check) that the law of $Y'$
(simple asymmetric random walk with bias $1-2p>0$), conditioned on $\tau_0'<\infty$,
becomes the law of the simple asymmetric random walk $Y''$ with bias $2p-1<0$,
where $\P(\Delta Y'' =1)=p=1-\P(\Delta Y'' =-1)$.
Due to the well-known correspondence between the excursion of $Y''$
and the subcritical Galton-Watson branching (of geometric offspring distribution
with success probability $(1-p)$ and mean $1/(1-p)-1<1$),
one can easily see that there exists $\lambda>1$ such that
\[
\mathbb{E}(\lambda^{\tau_0''}) = \mathbb{E}(\lambda^{2 Z})<\infty,
\]
where $Z$ is the total population size of the above Galton-Watson process,
started from a single individual.
Denote by $Z_m$ the total size od the first $m$ generations. Then,
by conditioning on the size of the first generation, one obtains that
\[
\mathbb{E}(\nu^{Z_m})= \nu \cdot (1-p) \cdot \frac{1}{1-p\cdot \mathbb{E}(\nu^{Z_{m-1}})}, \ m\geq 1, \, \nu>0.
\]
Consider
$$
g= \nu \cdot (1-p) \cdot \frac{1}{1-p\cdot g},
$$
and note that
if $\nu_0\equiv\nu_0(p)=(4p(1-p))^{-1}>1$ (recall that $p<1/2$) one can solve the above equation for all $\nu\in[0, \nu_0]$ to obtain a functional solution $g_p(\nu)$, continuously increasing in $\nu$.
Since clearly for $\nu\in (1,\nu_0]$ we have $\mathbb{E}(\nu^{Z_0})=\nu< g_p(\nu)$
and since $x<g_p(\nu)$ implies $(1-p) \cdot \frac{1}{1-p\cdot x}<g_p(\nu)$, one concludes that
$\sup_m \mathbb{E}(\nu^{Z_m}) = \lim_m \mathbb{E}(\nu^{Z_m})$ is finite for $\nu\in [0,\nu_0]$, and furthermore that $g_p(\nu)$ must equal $\mathbb{E}(\nu^Z)$.

Now for the above $\nu_0$ we pick $\lambda\in (1,\sqrt{\nu_0})$, and then
$\rho_0$ such that $1+C/\rho_0 < \lambda$.
Then there exists $n_0$ such that for all $\rho \geq \rho_0$
$$n (1+C/\rho)^n \leq n(1+C/\rho_0)^n \leq \lambda^n, \ \forall n\geq n_0,$$
and so we finally have that for all $\rho \geq \rho_0$
$$
\mathbb{E}(\tau_0'' \cdot (1+C/\rho)^{\tau_0''}) \leq n_0 \lambda^{n_0} + \mathbb{E}(\lambda^{\tau_0''})
= \lambda^{n_0} + G_p(\lambda^2)< \infty.
$$

\section{Concluding remarks}
The following remarks lead to natural open questions.

\begin{enumerate}
\item
Suppose that $\rho=\infty$.
In the case of Interacting Urn Models with more than two urns, one can easily derive the law of the number of non-conformist urns (that is the number of urns that do not fixate on the majority color, see \cite{launay1}) when the number $U$ of urns is odd.
Indeed, if $U$ is an odd number then the majority color in the $U$ urns combined  is necessarily decided at the very first step. Then all the urns that have chosen the majority color immediately fixate and and all the urns that chose the minority color are in configuration $C_2(1)$, and each eventually fixates on the majority color with probability
$$r_1(p)=\frac{p(1+p)}{(1-p)(2-p)},$$ independently of the others.

If the number $U$ of urns is even and strictly bigger than two, then we can still write the probability of fixation as the countable sum of all the urn paths until the majority color is decided but it seems more difficult to figure out an explicit formula for this expression (as the one given by Theorem \ref{theorem}).

\item
Suppose that $\rho=\infty$.
If there are more than two colors and $U=2$, then generalizing our results is trivial.
Indeed, if there are $C>2$ colors to start with, then the two urns draw the same color at the first step with probability $1/C$ and they draw two different colors with the remaining probability $(C-1)/C$. In the former case the model starts from configuration $C_3$ and the two urns fixate on the chosen color. In the latter case, the model starts in configuration $C_1(1)$ and the two urns fixate on the same color with probability $q_1(p)$.
If there are more than two urns, then the we cannot conclude much from the results of this paper.
\item
Theorem \ref{T:approx} can also be generalized for $U\geq 2$ and $C\geq 2$. In fact, the proof that $\lim_{\rho\to\infty}\P_\rho[A]=\P_\infty[A]$ is exactly the same (recall that $A$ is the set of all asymptotically possible urn paths). The proof that $\lim_{\rho\to\infty}\P_{\rho}[\bar{A}]=0$ needs to be adapted but can be done in the same fashion. The idea is that if an urn path is in $\bar{A}$ then it has an AI-draw, and its probability is at least $\rho$ times smaller than the same urn path without this AI-draw. Thus $\P_{\rho}[\bar{A}]=O(\rho^{-1})$ which proves the limit.
\end{enumerate}

\bibliographystyle{plain}

redpajama_set_name: RedPajamaArXiv


taleA corrupt ATF agent becomes obsessed with capturing a serial killer who kills and mutilates his victims in the most unusual way.is an unflinching tale about two psychopaths who get their kicks and exact their revenge the backwoods way. The locales in this film don't take kindly to the law and when they have a score to settle they use theon their enemies. What is a, let's just say that it is something that you are not soon to erase from your mind after you have seen it. To go into any more detail about what exactly ais would lessen the impact of this sadistic torture / revenge device.


text:
EnglandLondonHackney
Living in The London Borough of Hackney
An enviable choice of night life venues, creative spirit and a strong sense of community brings young working professionals to this culturally diverse borough in droves. So what's it really like to live in Hackney?
"…where high-end luxury fashion meets the edgy innovation of BOX PARK, and Michelin-starred restaurants sit alongside street-food tents, Hackney is where you'll find the culinary stars of tomorrow…"
Moving to Hackney
One of the capital's fastest-growing and most exciting areas, Hackney attracts people from all over the world with its creativity and vibrant urban life. It's where high-end luxury fashion meets the edgy innovation of BOX PARK, and Michelin-starred restaurants sit alongside street-food tents where you'll find the culinary stars of tomorrow. Galleries, start-up businesses and creative offices also line the streets of Shoreditch and Hoxton, while exclusive member's club Shoreditch House offers a place for the cultural elite and fashion stars of London to mingle. It's this kaleidoscopic mix of innovative business, alternative entertainment and hip restaurants that makes Hackney so appealing to young professionals and affluent students.
However, on the other side of the coin, Hackney is at the forefront of London's current controversial wave of gentrification, with long-term residents rapidly becoming priced out of the borough to make way for new property developments and city planners seeking to take advantage of the borough's cool, creative reputation. This is a borough of contrasts, where council estates sit alongside designer outlet stores, and 41% of local children live in property while city workers buy up million-pound properties on the streets of Shoreditch. Hackney's highest scoring Locality Reality postcode is N4 1HJ, hitting the 9.5 mark on Rowley Gardens.
House Prices in Hackney
Due to the lively lifestyle the London Borough of Hackney offers its residents, the area is becoming a sought-after residential location, particularly with young professionals, and new builds are popping up everywhere to meet the rising demand. New apartment blocks are a common sight; however, more historic terraced houses can still be found throughout the borough. Hackney has seen enormous amounts of regeneration over the last few decades, and as a result properties here are now £600,000 on average, slightly more than the London average, making Hackney the 12th most expensive borough in the capital. The most expensive areas in the borough tend to be those closest to Central London, such as Shoreditch and Hoxton, with better bargains to be found in areas including Clapton and Homerton.
"Busy, eclectic Hackney has become one of London's most vibrant cultural hotspots, with regular festivals and events held throughout the borough"
Travelling from Hackney
Although there are very few Underground stations in the London Borough of Hackney, the area makes up for this with strong rail connections on the Overground and a vast network of bus routes. The sole Underground station can be found at Manor House on the border of the Borough of Haringey, and offers access to the Piccadilly Line. For many commuters, the best route into Central London is on the Overground via Highbury and Islington station where connections to the Victoria Line can be made, or alternatively via one of the many direct buses.
Restaurants in Hackney
The dining scene in this borough now rivals that of Soho and Covent Garden, with many innovative new restaurants now choosing to set up here rather than in Central London. Eateries for all tastes and budgets can be found, ranging from the Michelin-starred Clove Club and Lyle's, to Irani-cafe inspired Indian food at Dishoom and enormous pizza slices at Voodoo Ray's in BoxPark. In addition, the borough is also home to a range of street food sites and festivals, including Street Feast's Dinerama, Night Tales, and Urban Food Fest, while Brick Lane is home to some of the best curries in the capital.
Hackney is equally well known for its vibrant night life, with some of the capital's coolest bars and clubs including the Queen of Hoxton, XOYO, The Nest and Oslo, as well as a seemingly endless number of pop-up bars, cocktail lounges and traditional pubs.
Shopping opportunities in Hackney are also exceptionally good, particularly for those seeking up-and-coming designers or vintage finds. Brick Lane and the surrounding streets are a hub for vintage clothing, with popular shops including the enormous Beyond Retro, Rockit and Absolute Vintage, whilst BOXPARK is packed with small designers and Commercial Street is home to big brands such as Urban Outfitters and All Saints.
Hackney residents who prefer home-cooking to the restaurants of Shoreditch are well-served by good supermarkets throughout the borough, including large branches of Tesco, Lidl and Sainsbury's. The diversity of the borough has also led to a selection of excellent international supermarkets, such as Longdan Express on Hackney Road and Akdeniz International Supermarket on Well Street.
There are a wide range of fitness facilities available in Hackney, ranging from branches of popular gym chains such as Soho Gym, PureGym and Fitness4Less to council-run leisure centres such as Britannia Leisure Centre, Clissold Leisure Centre and Kings Hall Leisure Centre.
As a result of the young professional crowd, there are also a range of high-end fitness studios to be found, including those dedicated to pilates and yoga.
Busy, eclectic Hackney has become one of London's most vibrant cultural hotspots, with regular festivals and events held throughout the borough, a number of brilliant museums, and two independent art-house cinemas – not to mention the theatres and music venues.
Top museums in the area include the Hackney Museum, Geffrye Museum and V&A Museum of Childhood, while film fans are well provided for with both the Hackney Picturehouse and Rio Cinema. Hackney Empire is the ultimate venue in the borough for a range of entertainment from dance performances to stand-up comedy, and the Arcola Theatre is one of the best spots for established and emerging theatre outside the West End.
Finally, if you're looking for something a bit different, check out the schedule at venues such as the Old Truman Brewery, which regularly hosts events such as the London Coffee Festival, D&AD Festival and the Foodies Festival.
"Schools in Hackney have improved hugely over the last few decades, and as of 2016 students in the borough achieved better GCSE results than the national average"
Schools in Hackney have improved hugely over the last few decades, and as of 2016 students in the borough achieved better GCSE results than the national average. The best secondary schools in Hackney include Cardinal Pole RC School and Yesodey Hatorah Senior Girls School in the state sector, and Beis Chinuch Lebonos Girls School in the independent sector. As for primary schools in Hackney, Orchard Primary School, Queensbridge Primary School and Parkwood Primary School are among the best in the borough, with over 95% of pupils reaching national English and Maths targets.
Whilst crime levels in Hackney are still higher than the London average, resting around 105 crimes per 1000 people as of 2016, they are gradually decreasingly and have seen a fall of over 30% since 2003.
For a borough so associated with urban areas such as Shoreditch and Dalston, there are surprisingly good selection of green spaces to choose from, particularly in the northern end of the borough. There are a total of 58 parks in Hackney, with popular spots including Clissold Park and Abney Park in Stoke Newington, and Hackney Downs in Lower Clapton, where you'll find a range of sports facilities such as an athletics track and tennis courts.
Once a modest Tudor town, dotted with grand palaces and spread with leafy hunting grounds, the London Borough of Hackney has since evolved to take over the scene as London's hub of urban cool.
Best area to live in London?

redpajama_set_name: RedPajamaCommonCrawl


text:
← Concerned MRA warns women that feminism will ruin their lives by making them bitchy
Sorry, Men's Rights Activists, Feminists Actually Do Care About Girls in Africa →
Red Piller: Become an Alpha Sex God by Imitating a Cult Leader and Convicted Rapist
Aravindan Balakrishnan: Convicted serial rapist and abuser, Red Pill role model?
The We Hunted the Mammoth Pledge Drive is almost done! If you haven't already, please support this blog by donating through the PayPal button below. Thanks!
Aravindan Balakrishnan was a Maoist cult leader who used psychological torture and sexual violence to keep a small group of women under his control for decades in a hellish "commune" in South London.
Earlier this month, he was convicted of multiple counts of rape and assorted other crimes — only a tiny fraction of the violence and abuse he presumably inflicted upon his female followers/captives since the late 1970s.
Over in the Red Pill subreddit, the regulars are looking at his sordid history of abuse for tips on how to become more like this "dark triad" master who had "a harem of 10 women … worshipping him as a god."
It's a central tenet of Red Pill thought that women swoon and "tingle' when faced with men displaying so-called "dark triad" traits — that is, narcissism, Machiavellianism and psychopathy. Red Pillers who aren't already "dark triad" men to begin with often try to emulate these not-so-nice guys.
Now they're looking, at least semi-seriously, to the abusive monster known as Comrade Bala for dating tips.
As the Red Piller Redditor no_face sees it, Comrade Bala's story makes clear that "dark triad" traits matter a lot more than looks — and even lifting — when it comes to the ladies.
Take a look at him. He is a short, bald, bespectacled 75 year old Indian guy. I'm not a physical fitness expert, but I'm wagering money that this dude does not lift. After you read this, you have lost the right to complain about your looks. He doesn't even have a lot of money either.
How did he manage this? Oh, did he tap into that Indian mystical mumbo-jumbo? No! This guy gets laid by preaching … gasp …. communism! At redpill, we know that the validity of the actual stuff he preached does not matter in the slightest. Its all about his intense dark triad characteristics.
And these he apparently had in spades. No_face, evidently an aspiring Dark Traid-er himself, quotes extensively, and sort-of, kind-of approvingly, from a Daily Telegraph article detailing Bala's abuses. [You may wish to stop reading at this point; it only gets worse from here.]
Even the detailed and explicit description of Bala's sexual abuse of his "followers" doesn't seem to phase him. No_Face suggests that the rapes Bala has now been convicted of were probably just instances of "rough sex."
And after quoting a passage in the Telegraph article noting that the victims of Bala's sexual assaults "were ordered to swallow when he ejaculated as it was the 'the elixir of life,'"no_face suggested that they probably wanted it.
I'm guessing the saying "woman gladly swallows for an alpha" has some bearing here. Think about this while you guys are emptying elixir into old socks.
So edgy, these Red Pillers.
No_face sums up the "Lessons Learned' from Bala's disgusting story:
Raw [Dark Triad] charisma can overcome almost any physical limitation.
Eventually your plates will crash, some even after 47 years.
You can literally be a god to women.
Your women can and will turn against you, even if you were their fucking god for most of their lifetime.
Apparently it's news to Red Pillers that even the most demoralized and beaten-down abuse victims may come to see their abusers as something less than a godlike figures.
Some of the Red Pill regulars responded to the story with jokes.
"Hey fellas, I'm just saying – a little alcohol and dancing close to rhythmic music also works," wrote one.
Of course, most of the jokes still managed to suggest a certain admiration for Comrade Bala's "achievements."
"Is it safe to say that Balla was a balla?" asked one. "Filthy dirty old man, I like him ☺" wrote another.
But others saw Bala's story as just more evidence that men need to become ever more ruthless towards women. Eugenics2015 declared:
This is what i've been saying forever, men have the power to control womens minds if you do it right.
As troghin2 saw it,
That dude either is insane or a genius at human interaction.
For a few, the story was proof that Asian men, allegedly despised by all the hot babes of the world, can only succeed with women by becoming abusive assholes. As imsfncy2 wrote,
This article teaches that an indian guy must be dark triad to make up for his ethnic disadvantage unlike other white guys that can just rely on social status and beta game.
The game IS rigged for everyone, especially for asian guys, so only the dark triad guy of any race thrives.
At this point, the Red Pill subreddit is basically a school for abusers, with occasional jokes.
Posted in: alpha males, creepy, do you even lift, emotional abuse, empathy deficit, entitled babies, men who should not ever be with women ever, misogyny, PUA, rape, rape culture, red pill, reddit, sexual abuse, sexual exploitation, taking pleasure in women's pain
Wait. Which one was the reddit troll comment and which was the YouTube troll comment?
nparker says:
I see the shortest one as the most common kind of Youtube troll, but actually they both fit the bill, don't they?
Orion says:
Seriously, it looks like TRP has completely crossed the line from misogynistic rants into being an advice board for abusers (if it wasn't already).
It's more insidious than that. It's not an advice board for abusers, it's a board where abusers can go for advice. There are posts that don't seem outright hateful at first, if often laughably mis-guided. Usually the abusers will turn up in the comments, but not always. Occaisonally you have entire threads that seem innocuous. There's thread on the front page about MBTI personality testing. The original poster is for it. Some agree that it's useful, some question the science behind it, some compare it to cold reading, etc. Nobody in the thread mentions women or gender at all, except insofar as they assert that certain "red pill" habits are easier to practice for some personalities than others.
David tends to "cherry-pick" the very worst stuff from his sources, and people who only know the various subreddits and 4Chan boards from reading WHTM often assume that the whole site is like the stuff he posts. It's not.
Let me be clear about this: I'm not actually criticizing David here; I'm not saying that he's trying to deceive, or even that he's sloppy or sensationalistic. I think he has a very good reason for doing what he does. It's *important* to dredge up the worst of the worst and make sure that we get a chance to look at it closely. It's easy for the worst stuff to drop out of sight when looking at a group as a whole, or to excuse a group by saying that *most* of what they post isn't outright criminal instead of condemning them for allowing that content at all. Highlighting the most terrible parts of these terrible sites is work worth doing.
Still, if you're worried about these sites it's worth familiarizing yourself with the more "moderate" content that's less likely to show up in a WHTM piece. In my opinion it's important to be aware of the "moderates" because they make these sites more, not less, dangerous.
TRP isn't 10 abusers sitting around swapping rape tips. It's 1 person saying something smart, 1 person saying something stupid but not explicitly bigoted, 5 people on misogynist tirades, and 3 abusers swapping rape tips in the corner (or yelling them at the crowd) while everyone else conspicuously ignores them.
Which is, honestly, even creepier.
mockingbird says:
I must apologise to Julotte with all my heart. Sorry, Julotte. But you are unoriginal.
At least Walter's trolling is so over the top that it produces some genuine WTF moments.
And then he discusses comics and anime.
Julotte, et al – Bring a shtick to the table or something.
AGuy423 says:
Did anyone bother to see how many comments, and how much "support" this thread actually had on reddit?
Redsilkphoenix says:
Reading about stuff like this make me think that C. S. Lewis was onto more than he realized when he wrote The Screwtape Letters and the sequel story Screwtape Proposes a Toast. The demons in those stories try to bring humans to Hell so they can eventually eat their souls (and any demon who fails to bring their assigned human target to Hell gets eaten in their place – Lewis based this version of Hell off an extremely toxic workplace).
The stronger the negative emotions are in a human, the more of a delicacy it is for the demons. So the demons work to encourage said emotions while the human is alive so they get the best possible treat when the human dies.
There are times when I think a lot of the poison going around the manosphere is the result of these demons prepping the living for eventual inclusion in an afterlife casserole of some kind. Makes reading about it all a bit more endurable, anyway. Sometimes.
And sometimes, not so much.
Virtually Out of Touch says:
Bull. All he has to be is hot. My super good looking cousins are never without girlfriends and they are really "nice guys".
I've also got some assholes in my family and try as they might, they don't score.
Roosh V Forum members agree: The real victim in the Stanford rape case is Brock Turner June 10, 2016
A little gender experiment confirms that Reddit is full of douchebags April 22, 2012
GirlWritesWhat on "The Necessity of Domestic Violence": "I don't really find too much [that's] seriously ethically questionable." August 16, 2012
Harry Potter pushes "willing cuckoldry as a moral imperative," weirdo Nazoid pickup artist explains April 11, 2017
Women with self-esteem: A grave threat to modern man January 22, 2013


…Linda Johansson, head of human resources at Kanthal®, about the new Kanthal® Student Award.
As head of HR, what is your main take on the Kanthal® Student Award?
"It's all about finding great people with innovative ideas that can further develop Kanthal®. The prize is aimed at students, as fresh new ideas from young people benefit the company in finding new angles on what we do.
I'm a student thinking about future employers. Why would I choose Kanthal®?
"If you like the idea of contributing to a better world, I think you would love working at Kanthal®. We provide the industry with sustainable heating solutions and medical wire that save lives and contribute to a higher quality of life for many people.
"You will also feel very welcome. Kanthal® is a company with about 1,200 employees globally. It's a perfect place to work in an international environment with great opportunities for development, yet with the feeling of being part of a family. Decision chains are short, so you will easily be able to try out your ideas.
What is the culture of Kanthal® – what's the vibe there?
"Kanthal® is driven by diversity, both geographically and cognitively. We are genuinely open to new influences and the positive effects that come from that. New, clever minds that want to contribute to this will feel at home here. The company is not that small, but there is still a familiarity among those who work here.


Steve Bailey: Who will hold their nerve for the run-in?
Don't panic. As disappointing as Albion's late defeat to Newcastle was, it won't be the game that defines the season.
Ask anyone involved at the club and they'd have bitten your hand off to be in the position Albion are with 12 games to go and what is, on paper, a favourable run-in.
Albion are already on 71 points from 34 matches, so are on course for 90-plus points which should be more than enough to secure a top-two finish, although there's no doubt more twists and turns to come.
Can Brighton bounce back quickly from Tuesday? How will Newcastle do with back-to-back games at Huddersfield and Reading? Can Huddersfield continue their outstanding run? Who will keep their bottle when the pressure really mounts?
As Newcastle boss Rafa Benitez said before and in the aftermath of Tuesday's match, April is the key month of the season.
Albion play just one team currently in the top six in their run-in, when they travel to Elland Road to play Leeds later this month.
The only other teams they have to play in the top half are Derby County and Norwich City.
I know there's the argument you don't want to play teams fighting relegation but just look at Brighton's record against the bottom eight sides this season, nine wins out of nine.
Albion still have to play seven teams in the bottom eight as well, beginning with a trip to seventh-from-bottom Nottingham Forest on Saturday and then away to rock bottom Rotherham – a potential banana skin – on Tuesday.
Win both of these upcoming matches and everything will be rosy for Seagulls fans again heading into the last ten games of the season.
Huddersfield, like Albion, have just one team in the top six left to play when they welcome Newcastle on Saturday.
Newcastle have the toughest run-in of the top three. As well as Huddersfield, they also have to come up against Reading, Sheffield Wednesday and Leeds in the top six.
Add Fulham, Preston, Cardiff and Barnsley, all in the top half, to the mix and the Toon Army have the trickiest games to come. But with the quality in their squad, you'd still think they've got more than enough to cope and be celebrating an immediate return to the Premier League in May.
Now is the time for the Albion players to re-focus and re-group for the last quarter of the season. On another day, Newcastle's fluke equaliser on Tuesday flies wide or over the bar, the Albion hang on to win and are now four points clear at the top.
The manner of the defeat will no doubt hurt the players but they have the resilience and team spirit to quickly bounce back – and they have to. Chris Hughton will be reminding the players what they have achieved so far this season and the big prize that is on offer at the end of the season if they can hold their nerve.
I'm still confident the Seagulls will be in the top two come the end of the season but can't look past Newcastle for the title.


The salary statistics and trend of engineering technicians, except drafters in the industry of alumina and aluminum production and processing are shown in the following tables and chart. In Table 3 we compare engineering technician, except drafters salaries in different industries within the manufacturing sector.
Table 1 shows the average annual salaries for engineering technicians, except drafters in the industry of alumina and aluminum production and processing. The salaries are shown in 5 percentile scales. The average annual salary for the 90th percentile (the top 10 percent of the highest paid) engineering technicians, except drafters is $77,190. The median annual salary (50th percentile) is $50,910. The average annual salary for the bottom 10 percent engineering technicians, except drafters is $36,050.
The table and chart below show the trend of the median salary of engineering technicians, except drafters in the industry of Alumina and Aluminum Production and Processing from 2012 to 2017.


Jelle Smet has done a screencast on his NSCAweb addon for Nagios. Check it out!
NSCAweb part 1: Introduction from Jelle Smet on Vimeo.
Many of our clients are Managed Service Providers (MSPs). The challenges they face when monitoring their clients' networks vary greatly, but a common problem they face is how to monitor remote Windows machines that are located behind a firewall.
A simple solution to this challenge involves configuring the NSClient++ agent to act as a passive agent which reports it status to Nagios. We've written a document that describes the steps needed to configure passive monitoring of Windows boxes.
Passive checks are extremely useful for integrating information from external applications and agents. They're also the primary method of monitoring remote machines that are behind firewalls, proxies, and on private networks.
In an effort to simplify the setup of new remote agents and applications with Nagios XI, we developed a method for easily capturing and configuring new passive host and service check results.
Category: Configuration, Nagios XI and Passive Checks. Closed Tags: passive, XI.


On the Sonesta Sun Goddess Cruise Ship, enjoy an authentic Nile adventure that create memories to last a lifetime. Cruise the Nile for 4 or 6 nights from Aswan to Luxor, as the Sun Goddess travels from the Valley of Kings, to the mummified crocodiles of Kom Ombo. This elegant Nile cruise ship features cabins with broad panoramic views, bathrooms with full-size bathtubs and all modern conveniences. Enjoy a swim in the outdoor pool, experience the exciting nightlife in the ship's discotheque or a relaxing massage treatment at the spa. Deluxe amenities on the Sun Goddess Cruise Ship are delivered with the uncompromising personal service that is unmistakably Sonesta.
If you would like more information on the Sun Goddess or would like to discuss a cruise, please call one of our Cruise Experts on 0800 008 6677.


text:
Staff Profile: Cecial Patrick, Sponsored Programs Office
After 35 years at VCU, Cecial Patrick, grants research coordinator in the Sponsored Programs Office, will retire on January 1, 2020.
Cecial began working for VCU in November of 1984. She was originally hired by Lamont Kier, Ph.D., to work in the Department of Medicinal Chemistry as the office service assistant. "Dr. Kier was very sincere and compassionate," says Cecial. "The position required someone to operate the IBM Displaywriter System 6580, which was a microcomputer-based word processing machine with a floppy unit. It was large and had a lot of parts."
Besides the Department of Medicinal Chemistry, Cecial also spent time in the Training and Technical Assistance Center and the Department of Biology. And for the past five years, Cecial worked in the dean's office as a fiscal technician, grants specialist and grant research coordinator where she manages financial support, verified sponsored awards, and monitored deficit reports and monthly expense reports. Her proudest accomplishment in her professional career is her mastery of all the financial programs used by VCU. "I had to learn to use so many different systems starting with Financial Record Systems in the 1980s, Banner 8, Banner 9, eVA, and now, RealSource," Cecial explains. "Switching from system to system taught me to be prepared for change in the blink of an eye."
Among her many favorite memories from her time at VCU, Cecial will always treasure watching her daughter graduate in 2016 with her B.S. in psychology. "I will never forget her smile while walking across the stage to get her degree." She'll also miss the training opportunities that are offered and the campus environment as a whole. "I enjoy meeting new students and watching them progress throughout their time at VCU," says Cecial.
What's next for Cecial? "I'm going to start a new chapter in my life. I plan to spend time with family, travel and enjoy life," says Cecial. As she says goodbye to VCU, Cecial is guided by the quote, "Your present circumstances don't determine where you can go; they merely determine where you start," by Nido Qubein.
Cecial can be reached at patrickcecial@gmail.com.


The cut flowers industry is an important sector for several countries in East Africa, chiefly Kenya and Ethiopia. Kenya is the top exporter of roses to the European market, and both Kenya and Ethiopia are eyeing expansion into new markets, especially to the U.S. because of its duty-free status under the African Growth and Opportunity Act (AGOA).
From June 14- 15, the Hub is supporting eight Kenyan and Ethiopian flower exporters to exhibit at the International Floriculture Trade Expo in Chicago, IL through a grant to Kenya Flower Council and Ethiopian Horticulture Exporters Association. The International Floriculture Exposition (IFE) has provided the exhibitors with an opportunity to meet with the highest-volume buyers from the floral industry, connecting them with companies that offer an array of products and services.
The Hub's participation at the expo contributes to its objective of facilitating a 40 percent increase in exports from the East Africa region to the U.S under AGOA.
The IFE Expo boasts over 60,000 square feet of floral exhibits, featuring anything a floral category buyer might need for their department from cut flowers and potted plants to giftware and accessories. It provides an opportunity for exhibitors to meet hundreds of vendors face-to-face.


Coming to theaters this Easter!
See Dwight & Forge in Theaters Nationally!
Forge alumni Sam Bhatt is behind the upcoming animated feature film "The Pilgrim's Progress" which will be playing in theatres all around the country on April 18th and April 20th. Based on John Bunyan's epic masterpiece, The Pilgrim's Progress helps families understand that the spiritual journey is the most important journey in life, as well as the importance of sticking to the path, even when the journey gets difficult.
Invite your family, friends and church community to this amazing event! Also consider hanging up the movie poster on the back of this newsletter at your local church to help spread the word! Then get there early to see Forge's newest mission trailer and get excited about what God is doing in and through this ministry. Amen!
Find a theater near you that is showing the upcoming animated feature film The Pilgrim's Progress, and purchase your tickets now!
Purchase Tickets at Your Local Theater!


text:
Men's Ice Hockey Alumnus Mike Seward '15 Earns NCAA's Walter Byers Postgraduate Scholarship
Mike Seward '15 has been awarded the NCAA"s prestigious Walter Byers Postgraduate Scholarship.
INDIANAPOLIS, Ind. – Former men's hockey ice player Mike Seward has been awarded the prestigious Walter Byers Postgraduate Scholarship by the NCAA, the organization announced Tuesday (May 8). Seward, who graduated from Harvard magna cum laude in 2015 as an organismic and evolutionary biology concentrator, and is currently enrolled at Harvard Medical School, will receive a $24,000 grant towards his studies. He plans to pursue a career in clinical medicine that includes research and an active role in health policy.
Seward had previously earned an NCAA Postgraduate Scholarship last summer.
Established in 1988, the Walter Byers Postgraduate Scholarship program each year awards grants to one male and one female recipient. The grants can be renewed for a second year. Recipients chosen by the NCAA Walter Byers Scholarship Committee are recognized as combining the best elements of mind and body to achieve national distinction for their achievements and to be future leaders in their chosen field of career service.
The following is an excerpt from the NCAA's announcement:
At Harvard, Seward realized good fuel could give him an edge. "I found that when I paid more attention to eating nutrient-rich foods, I noticed a considerable change in how I felt and functioned on and off the ice," he said.
Since then, he has turned that desire to maximize his training into a passion for improving health and wellness. During his junior year, Seward developed traffic-light nutrition labels for a Harvard cafe. Green labels indicated nutrient-rich food; yellow cautioned eat less often; and red meant eat sparingly. He expanded the program to 12 campus cafeterias for his senior thesis, publishing the results in the American Journal of Public Health.
Seward also had a reputation on his team for being a natural leader and a disciplined worker. "His food choices in the dining halls and work ethic in the gym rubbed off on his teammates," said Ted Donato, The Robert D. Ziff '88 Head Coach for Harvard Men's Ice Hockey. "He became the go-to off-ice expert, frequently answering questions about the best breakfast before games or tips on the team strength and conditioning testing."
Among his various service activities, Seward co-chaired the Harvard Student-Athlete Advisory Committee, and he volunteered in Vietnam one summer, teaching physics to eighth- and ninth-grade students, often using sports to explore trajectory or acceleration.


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RussianOperaWorkshop.com | Ghena on Tatyana in Eugene Onegin
Tchaikovsky on 'Tatyana's Letter Scene'
Tchaikovsky writes to Nadezhda von Meck that Tatyana is not just a provincial little girl that falls for the man of the world. She is a dreamy and impressionable young woman in search of her ideal love. If no one matches Tatyana's ideal, she remains disappointed but serene. With the appearance of one man (Onegin) who stands out, Tatyana's imagines that her ideal has come and her intense passion is revealed
Commenting on composer/pianist Sergei Taneyev's mistaken remark, Tchaikovsky explains that Tatyana falls in love with Onegin instantly. Before Onegin even appears on the scene, Tatyana is in love with her ideal of a romantic hero, instilling these qualities into Onegin at first sight.
Brother's Criticism
Apparently, Modest Tchaikovsky, composer's brother, criticized Onegin for potentially lack of scenic effect and action. This annoyed Pyotr, but he didn't care. He was immersed in the opera, in love with Tatyana's image, and enchanted by Pushkin's verse.