text:
\section{Introduction}
\label{sec:intro}
The QED (Quality and Efficiency Driven) regime for many-server systems combines large capacity with high utilization while maintaining satisfactory system performance.
In the QED regime the arrival rate $\lambda$ and the number of servers $s$ are scaled in such a way that while they both increase towards infinity, the traffic intensity $\rho=\rho_s=\lambda/\mu$ (assuming service rate $\mu/s$ per server) approaches one and
\begin{equation}\label{scaling}
(1-\rho_s)\sqrt{s}\rightarrow \beta, \quad \beta\in(-\infty,\infty).
\end{equation}
Halfin and Whitt \cite{halfin1981htl} introduced the QED regime for the $GI/M/s$ system. Under the scaling \eqref{scaling}, assuming $\beta>0$, the stationary probability of delay was shown to converge to a non-degenerate limit, bounded away from both zero and one. Limit theorems for other, more general systems were obtained in \cite{garnett2002designing, janssen2011refining, jelenkovic2004heavy, maglaras2004diffusion,mandelbaum2008queues,reed2009g}, and for all those cases, the limiting probability of delay remains in the interval $(0,1)$. In fact, not only the probability of delay, but several other performance characteristics or objective functions are shown to behave (near) optimally in the QED regime (see \cite{borst2004dimensioning, gans2003telephone}).

Associated with the near optimal behaviour is the fact that the process $Q_s(\cdot)$ that counts the total number of customers in the system at any time exhibits relatively small fluctuations. Halfin and Whitt \cite{halfin1981htl} showed for the $GI/M/s$ system that under \eqref{scaling} a sequence of standardized processes $X_s(\cdot)$, with  $X_s(t):=(Q_s(t)-s)/\sqrt{s}$, converges as $s\to\infty$ to a diffusion process $X(\cdot)$. This diffusion process behaves like a Brownian motion with drift above zero and like an Ornstein--Uhlenbeck process below zero. The interpretation of this scaled process is as follows: In case, $X_s(t)>0$, it represents the scaled number of customers waiting for service, whereas in case $X_s(t)<0$, it represents the scaled number of idle servers. This result shows that the natural scale that emerges is of the order $\sqrt{s}$: Specifically, both the queue length and the number of idle servers in the system are of the order~$\sqrt{s}$.

Due to its favorable behavior, the many-server QED regime has been a major focal point of applied probability and stochastic operations research in the past 30 years. The many extensions of the Halfin--Whitt exposition in  \cite{halfin1981htl} have lead to theoretical advances in the areas of stochastic-process limits and asymptotic dimensioning. From an operational point of view, the QED regime has found many applications in the planning, analysis and optimization of queueing, inventory and service systems (see for example \cite{armony2004customer}).

In this paper we explore the presence of a BRAVO effect in the QED regime. BRAVO is short for Balancing Reduces Asymptotic Variance of Outputs. This again would be a favourable property of QED, this time for the departure process
$N_{\rm dep}(0,\cdot]$, where $N_{\rm dep}(0,t]$ counts the number of serviced customers during the time interval $(0,t]$. The study of departure or output processes of queues has a long tradition, see for example the classic surveys \cite{daley1976queueing} and \cite{disney1985queueing}, yet only recently the surprising phenomenon of the BRAVO effect has been reported. The BRAVO effect is captured in terms of the asymptotic ratio of the variance and the mean of the departure process
\eqan{\label{limrat}
\mathcal{D} := \lim_{t\to\infty} \frac{\mathop{\rm var}\nolimits \big( N_{\rm dep}(0,t] \big)}{{\rm E} \big(N_{\rm dep}(0,t]\big)}.
}
For Poisson processes $\mathcal{D}=1$, and more generally for renewal processes, $\mathcal{D}$ equals the ratio of the variance of the renewal lifetime and the square of the mean lifetime. Thus it is initially surprising that for $M/M/1/K$ systems, $\mathcal{D}$ is minimized when the arrival rate $\lambda$ is equal to the service rate $\mu$ with a minimum equal to $\frac{2}{3} + o_K(1)$, where $o_K(1)$ is a term that vanishes as $K \to \infty$. This was shown in \cite{NazarathyWeiss0336}. Further, when $K=\infty$, it is well known that $\mathcal{D}=1$ whenever $\lambda \neq \mu$, yet it was shown in \cite{al2011asymptotic} that in the critical case that $\lambda=\mu$, $\mathcal{D} = 2(1-2/\pi) \neq \frac 23$. The work in \cite{al2011asymptotic} goes further, generalizing this $M/M/1$ result to $GI/GI/1$ systems and even multi-server $GI/GI/s$ systems with a finite bounded number of servers $s$. Hence, by BRAVO we mean that $\mathcal{D}< 1$ when $\rho=1$. For overviews of BRAVO results we refer to \cite{daley2011revisiting} and \cite{nazarathy2011variance}.

In this paper we study the BRAVO effect in the QED regime by equipping the
many-server $M/M/s$ system with a finite waiting capacity $K$. In order to create a finite-capacity effect in the QED regime that is neither dominant nor negligible, it is plausible to assume that $K \approx \eta\sqrt{s}$, because the natural scale of the queue length is $\sqrt{s}$. More precisely, we study a sequence of systems in which both $K$ and $s$ grow in such a way that
\begin{equation}
\label{scaling2}
\frac{K}{\sqrt{s}}\rightarrow \eta
\end{equation}
for some positive $\eta$.
A similar threshold $K\approx\eta\sqrt{s}$ in the context of many-server systems in the QED regime has been considered in \cite{armony2004customer, janssenScaled,massey2005asymptotically,whitt2004diffusion,whitt2005heavy}.
Hence, in addition to the parameter $\beta$ in \eqref{scaling} describing the scaled shortfall from one in the system capacity, our system includes a parameter $\eta$ describing the relative buffer size. Our result on the BRAVO effect in the QED regime is in terms of
\eqan{\label{double}
\mathcal{D}_{\beta, \eta} :=\lim_{s,K\to\infty} \lim_{t\to\infty} \frac{\mathop{\rm var}\nolimits \big( N_{\rm dep}(0,t] \big)}{{\rm E} \big(N_{\rm dep}(0,t]\big)},
}
where the outer limit is taken under the constraints at \eqref{scaling} and \eqref{scaling2}.

Our analysis that leads to explicit representations for
 $\mathcal{D}_{\beta,\eta}$ is based on the following general result for output processes of birth--death processes.  Consider a finite, irreducible, birth--death process $Q(\cdot)$ on $\{0,1,\ldots,J\}$ with birth rates $\lambda_0,\lambda_1,\ldots,\lambda_{J-1}$ and death rates $\mu_1,\mu_2,\ldots,\mu_J$.  Let $\{\pi_i\}$ denote the stationary distribution for $Q(\cdot)$ with cumulative distribution $P_i = \sum_{j=0}^i \pi_j$, and let
 $N_{\rm dep}(0,t]$ denote the number of deaths in $(0,t]$.
 Denote the departure rate by
\[
\lambda^* := \lim_{t \to \infty} \frac{{\rm E}(N_{\rm dep}(0,t])}{t} = \sum_{j=1}^J \mu_j \pi_j ,
\]
and write
\[
\Lambda^*_i= \frac{ \sum_{j=1}^i \mu_j \pi_j }{\lambda^*}
\]
for what are cumulative probabilities.
Then we know from \cite[Theorem 1]{NazarathyWeiss0336} that
\begin{equation}
\label{eq:newR1}
\mathcal{D}_{\pi} :=
\lim_{t \to \infty} \frac{\mathop{\rm var}\nolimits\big(N_{\rm dep}(0,t]\big)}{{\rm E}(N_{\rm dep}(0,t])}
=
1 - 2 \sum_{i=0}^J (P_i - \Lambda^*_i)\Big(1 - \frac{\lambda^*}{\pi_i \lambda_i}  (P_i - \Lambda^*_i) \Big).
\end{equation}
Note that \eqref{eq:newR1} is in a slightly different form from that appearing in \cite{NazarathyWeiss0336}; the translation between the two forms is immediate.

In this paper we use  \eqref{eq:newR1} so as to obtain explicit expressions for $\mathcal{D}_{\beta,\eta}$ in the case of $M/M/s/K$ systems.  To do so we use the fact that in birth--death processes with $\lambda_i \equiv \lambda$, we have
\[
\lambda^*=\lambda (1- \pi_J), \qquad  \Lambda^*_i = \frac{P_{i-1}}{(1-\pi_J)},
\qquad
P_i - \Lambda^*_i = \frac{\pi_i - \pi_J P_i}{1-\pi_J},
\]
and after basic manipulation, \eqref{eq:newR1} can be represented as
\begin{equation}
\label{mainn}
\mathcal{D}_{\pi}
= 1 - 2 \frac{\pi_J}{1-\pi_J} \sum_{i=0}^{J} P_i \Big(1 - \pi_J \frac{P_i}{\pi_i}\Big).
\end{equation}

In the case of the $M/M/s/K$ system, this elegant form proves amenable to manipulation and asymptotics under the QED regime, yielding our desired explicit formulae for $\mathcal{D}_{\beta,\eta}$ defined in \eqref{double}. Carrying out these asymptotics is the main contribution of the current paper. A further virtue of the form \eqref{mainn}, which is of independent interest, is that it demonstrates that (for the case $\lambda_i=\lambda$),
\begin{equation}
\label{eq:LowBoundHalf}
\mathcal{D}_\pi \ge \frac{1/2 - \pi_J}{1-\pi_J}.
\end{equation}
To see this observe that the function $x\mapsto x(1-x)$ is maximized at $x=1/4$, so that
\[
\frac{1}{2}(1-\mathcal{D}_\pi)(1-\pi_J) = \sum_{i=0}^J \pi_i   \frac{ \pi_JP_i}{\pi_i} \Big(1- \frac{ \pi_JP_i}{\pi_i}\Big)
\le \frac{1}{4}.
\]
The lower bound \eqref{eq:LowBoundHalf} implies that as long as $\pi_J \to 0$, as in the $M/M/s/K$ system, $\mathcal{D}_{\beta,\eta} \ge 1/2$. Our explicit expressions for $\mathcal{D}_{\beta,\eta}$ for $\rho \equiv 1$ in fact establish that $\mathcal{D}_{0,\eta}$ is in the range $(0.6, \frac{2}{3})$ with the exact value depending on $\eta$. Hence for QED systems, the magnitude of the BRAVO effect is not exactly the same as for single server systems, but it is in a similar range. Similar results are found for QED systems with non-zero $\beta$ for which $|\beta|$ is not too big.

The remainder of this  paper is structured as follows. In Section \ref{sec:bravoForQED} we give our main theorem on the BRAVO effect for the $M/M/s/K$ system, which presents an expression for the asymptotic ratio $\mathcal{D}_{\beta,\eta}$ for both the case $\rho \equiv 1$ (i.e.~$\beta=0$) and for the case $\beta \neq 0$ (i.e.~$\rho \approx 1-\beta/\sqrt{s}$).
The proofs for these two cases are presented in Sections \ref{proofthm1} and \ref{proofthm2}, respectively. We present some conclusions and ideas for future work in Section~\ref{sec:conclusion}. The appendix contains some needed asymptotic properties of Poisson probabilities.

\section{Main result}
\label{sec:bravoForQED}

We now state the main theorem of this paper, which identifies the BRAVO effect in many-server QED systems in terms of the asymptotic output ratio.
Let $\Phi$ and $\phi$ denote the distribution and density of a standard
normal random variable.

\begin{theorem}
\label{thm1}
Let $s,K\to\infty$ in such a way that ${K}/{\sqrt{s}}\rightarrow \eta$
for some finite positive $\eta$.

{\rm(a)} Let $\rho=1$. Then
\eqan{
\mathcal{D}_{0,\eta}=\frac23-L(\eta),
}
where
\eqan{L(\eta)=
\frac{\left(2-\frac{ \pi }{2}\right) \eta+ \sqrt{2 \pi } (1-\log 2-\frac{\pi}{12})}{ \left(\eta+\sqrt{\frac{\pi }{2}}\right)^3}.
}

{\rm(b)} Let $\rho=1-\beta/\sqrt{s}$ for finite $\beta\ne 0$.
Then
\eqan{\label{exacttt}
\mathcal{D}_{\beta,\eta}
= 1 - \frac{2\beta^2{\rm e}^{-\beta\eta}h^2(\eta,\beta)}{\phi(\beta)}f(\eta,\beta)
   +g(\eta,\beta),
 }
where
\eqan{
h(\eta,\beta)=\frac{1}{1-{\rm e}^{-\beta\eta}+\frac{\beta\Phi(\beta)}{\phi(\beta)}},
}
\eqan{
f(\eta,\beta)  = \int_{-\beta }^\infty
  \Big(1 - \beta {\rm e}^{-\beta\eta}h(\eta,\beta) \frac{\Phi(-u)}{\phi(u)}\Big)
\,\Phi(-u)\,{\rm d} u,
 }
and, with $h=h(\eta,\beta)$,
\eqan{
  g(\eta,\beta)
=
2 {\rm e}^{-\beta\eta}h(1+{\rm e}^{-\beta\eta}h)\Big(1-\beta\eta-{\rm e}^{-\beta\eta}+(1-2\beta\eta{\rm e}^{-\beta\eta}-{\rm e}^{-2\beta\eta})h\Big).
 }
\end{theorem}

We start by discussing Theorem \ref{thm1}(a). Figure~\ref{fig:rho1} displays $\mathcal{D}_{0,\eta}$ as a function of $\eta$.
Observe that as $\eta \to \infty$ which includes the case of a fixed finite number of servers (with a large, but finite buffer), we have $\mathcal{D}_{0,\eta} \to {2 \over 3}$. Further, for $\eta= 0$ we have
\eqan{
\mathcal{D}_{0,\eta} = 1-\frac{4(1-\log2)}{\pi} \approx 0.6093.
}
It is also easy to verify that as a function of $\eta$, $\mathcal{D}_{0,\eta}$ has a unique global minimum at
\eqan{
\eta =
\frac{
\sqrt{2 \pi}(\log8-2)
}{
4-\pi
}
\approx
0.232,
}
yielding $\inf _\eta \mathcal{D}_{0,\eta} \approx 0.6018$.
\begin{figure}
\begin{center}
\includegraphics[width=10cm]{DforRho1FofEta.pdf}
\end{center}
\caption{{\small The asymptotic value of $\mathcal{D}_{0,\eta}$ as a function of $\eta$ when $\rho \equiv 1.$}
\label{fig:rho1}}
\end{figure}
We now turn to Theorem \ref{thm1}(b). Figure \ref{fig:rhon1} displays $\mathcal{D}_{\beta,\eta}$ as a function of $\beta$ and $\eta$. The figure suggests that, for fixed $\eta$,
\begin{equation*}
\lim_{\beta\to\infty}\mathcal{D}_{\beta,\eta}=\lim_{\beta\to -\infty}\mathcal{D}_{\beta,\eta}=1
\end{equation*}
and indeed this follows from the exact expression in \eqref{exacttt}. In fact, this can also be explained using the following heuristic reasoning.
As $\beta\to \infty$, the system becomes lightly loaded, and the process $Q_s(\cdot)$  behaves as an infinite-server system, which is reversible and therefore has an asymptotic output ratio equal to one. Also, as $\beta\to -\infty$, the system becomes increasingly overloaded, so that the process $Q_s(\cdot)$ behaves like a single-server system in which all servers work all the time. Such a single-server process is again a reversible birth--death process, which has an asymptotic output ratio equal to one. For any finite $\beta$, the behaviour of the process $Q_s(\cdot)$ resembles a mixture of an infinite-server system and a single-server system, and it is this alternation between two different stable systems that may explain the BRAVO effect. This effect is most pronounced for  values of $\beta$ close to zero.

\begin{figure}
\begin{center}
\includegraphics[width=12.5cm]{genDetaBetaCurves.pdf
\end{center}
\caption{{\small The asymptotic value of $\mathcal{D}_{\beta,\eta}$ for various $\eta$ values.
\label{fig:rhon1}}}
\end{figure}

We now present some further arguments about why the scaling relations in \eqref{scaling} and \eqref{scaling2} are the precise scalings needed to create the BRAVO effect. We first show that the scaling relation in fact leads to a well-defined stochastic-process limit of the entire queue-length process. Then we establish, using Gaussian approximations for the Poisson distribution,
non-degenerate limits for the stationary distribution.

Let ``$\Rightarrow$'' denote weak convergence in the space $D[0,\infty)$ or convergence in distribution. The following result is proved in \cite[Thm.~1.2]{pang2007martingale}.

\begin{proposition}[Weak convergence to a diffusion process]\label{prop111}
Assume \eqref{scaling} and \eqref{scaling2}.  If $X_s(0)\Rightarrow X(0)\in\mathbb{R}$, then for every $t\geq 0$, as $s\to \infty$,
$
X_s(t)\Rightarrow X(t)\wedge\eta,
$
where the limit $X(\cdot)$ is the diffusion process with infinitesimal drift $ -\beta$ when $x> 0$ and $ -\beta-x$ when $x<0$, and constant infinitesimal variance $2$.
\end{proposition}

With minor abuse of notation, let $X$, $X_s$ and $Q_s$ without time arguments denote  stationary random variables.
One of the signature features of the QED regime is that, due to economies of scale, the stationary probability of delay $\mathbb{P}(Q_s\geq s)=\sum_{i=s}^{K}\pi_i$ converges to a limit that is neither zero nor one. This feature continues to exist for our model with the finite-capacity scaling. For convenience we state it formally below, showing the relation of $h(\eta,\beta)$ to the probability of delay (the result appears at equation (4.7) of the unpublished manuscript \cite{massey2005asymptotically}):

\begin{proposition}[Probability of delay in QED regime]\label{prop2}
Assume \eqref{scaling} and \eqref{scaling2}.  Then
\begin{equation}\label{qedresult}
\lim_{s\rightarrow\infty}\mathbb{P}(Q_s\geq s)=\frac{1-{\rm e}^{-\beta\eta}}{1-{\rm e}^{-\beta\eta}+ \frac{\beta\Phi(\beta)}{\phi(\beta)}}=(1-{\rm e}^{-\beta\eta})h(\eta,\beta).
\end{equation}
\end{proposition}

It can be shown that the right-hand side of \eqref{qedresult} corresponds to the probability $\mathbb{P}(X>0)$ that the stationary diffusion process is positive (see \cite{browne1995piecewise}), as suggested by Proposition~\ref{prop111}. We choose, however, to give a direct derivation, starting from the exact expression for $\mathbb{P}(Q_s\geq s)$. We do so in order to give insight into the crucial role played by Gaussian approximations of the Poisson distribution; approximations constitute an important ingredient for the proof of the main theorem.

Recall the representation for the stationary distribution $\{\pi_i\}$
of the number in the many-server queueing system $M/M/s/K$, arrival rate $\lambda$ and service rate $\mu/s$ per
server,
\eqan{ \label{ssm}
\pi_i =
\begin{cases}
\frac{(s\lambda/\mu)^i}{i!}\,\pi_0 & \mbox{for} ~~ i=0,1, \ldots,s,\\
(\lambda/\mu)^{i-s}\pi_s & \mbox{for} ~~ i=s,s+1,\ldots,s+K,
\end{cases}
:=
\begin{cases}
\frac{{(s\rho)}^i}{i!}\,\pi_0,\\
\rho^{i-s} \pi_s.
\end{cases}
 }
For finite $\kappa>0$, let $\{\varpi_i(\kappa) = {\rm e}^{-\kappa}\kappa^i/i!\,, i=0,1,\ldots\}$ denote the Poisson distribution with mean $\kappa$. Observe that for $i=0,\ldots,s$,
\eqan{
\pi_i = \pi_0{\rm e}^{s\rho} \varpi_i(s \rho) := b_s(\rho)\varpi_i(s\rho),
 }
where $b_s(\rho) := \pi_0{\rm e}^{s\rho}$.

The stationary distribution $\{\pi_i\}$ has three parameters: $\rho$, $K$ and $s$, but they must satisfy the constraint
\eqan{
1 = \sum_{i=0}^{s+K} \pi_i
= \Big(\sum_{i=0}^s + \sum_{i=s+1}^{s+K} \Big)\pi_i.
 }
Thus when $\lambda=\mu$, i.e.\ $\rho=1$, the latter sum equals $K \pi_s$, and the former sum equals
\eqan{
b_s(1) \sum_{i=0}^s \varpi_i(s)
=
b_s(1)\,\Big(\frac12+\frac{\psi_{ss}}{\sqrt{s}}\Big)
 }
by the central limit property for the Poisson distribution (see Lemma~\ref{lemma:pois2}) for constants $\psi_{ss}$ satisfying $\sup_s |\psi_{ss}| < \infty$. Using Stirling's formula,
 \eqan{
 \frac{\pi_s}{\pi_0} = \frac{\varpi_s(s)}{\varpi_0(s)}
 = \frac{s^s}{s!}
= \frac{s^s}{(s/{\rm e})^s \,\sqrt{2\pi s}\,}\Big(1+\frac{\vartheta_s}{12s}\Big)
= [1+O(s^{-1})]{\rm e}^s/\sqrt{2\pi s},
 }
for some $\vartheta_s$ for which $|\vartheta_s|<1$.
So $b_s(1) = \pi_0{\rm e}^s = \pi_s\sqrt{2\pi s}\,[1 + \vartheta_s/12s]$, and
\eqan{
1 = \pi_s\bigg[\Big(\frac12 + \frac{\psi_{ss}}{\sqrt{2\pi s}}\Big)
  \frac{\sqrt{2\pi s} }{1 + \vartheta_s/12s} + K\bigg].
  } Then
\eqan{
1 = \pi_s \sqrt{s}\Big(\sqrt{\pi/2} + \frac{K}{\sqrt{s}} + o(1)\Big),
 }
hence, using \eqref{scaling2},
\begin{equation}
\label{eq:pisRootS}
\pi_s\sqrt{s} \to \frac{1}{\sqrt{\pi/2}+\eta}.
\end{equation}
Also, $b_s(1) = \pi_s \sqrt{s}\,\sqrt{2\pi}\,(1+\vartheta_s/12s) \to \sqrt{2\pi}/(\sqrt{\pi/2}+\eta) =: b_\infty$ as $s \to \infty$.

More generally, for Theorem~\ref{thm1} we examine $M/M/s/K$ systems
for which $\rho_s = \lambda/\mu = 1 - \beta/\sqrt{s}$ and $s,K\to\infty$
as at \eqref{scaling2}.  The sums of terms $\pi_i$ over $i\le s$ and $i>s$ now equal
\eqan{
\label{eq:geoSum1}
b_s(\rho)        \sum_{i=0}^s \varpi_i(s\rho)
= b_s(\rho)       \Big[\Phi\Big(\frac{s-s\rho}{\sqrt{s\rho}\,}\Big)
+ O(s^{-{1\over2}})\Big]\quad
 \hbox{ and } \qquad \pi_s \sum_{i=1}^{K} \rho^i
  = \pi_s\frac{\rho(\rho^K -1)}{\rho-1}
 }
respectively. Here we have used  the central limit property of a Poisson distribution with mean $ s \rho$  as in Lemma~\ref{lemma:pois2}.
Substitution for $\rho$ gives, correct to terms that are $O(1/\sqrt{s})$,  $\pi_0{\rm e}^{s\rho} \Phi(\beta)$ and $\sqrt{s}\,\pi_s \big(1 - {\rm e}^{-\beta\eta}\,\big)/\beta$ when $s,K \to\infty$ as in \eqref{scaling2}.  The local asymptotic normality of Poisson probabilities (cf.~Lemma~\ref{lemma:pois1}) implies
\eqan{
b_s(\rho)       = \pi_s\cdot\frac{\pi_0}{\pi_s}\cdot {\rm e}^{s\rho}
\,=\, \frac{\pi_s}{{\rm e}^{-s\rho} (s\rho)^s\big/s!}
\,=\,   \frac{\sqrt{s\rho}\,\pi_s } {\phi\big( (s-s\rho)/\sqrt{s\rho}\,\big)}
\big(1+O(s^{-{1\over2}})\big)
\,=\,   \frac{\sqrt{s}\,\pi_s}{\phi(\beta)}\big(1+O(s^{-{1\over2}})\big).
 }
The terms on the left-hand sides of the relations in \eqref{eq:geoSum1} add to 1, so that
\eqan{\label{220}
1 + o(1) = \sqrt{s}\,\pi_s\Big(\frac{\Phi(\beta)}{\phi(\beta)}
+ \frac{1 - {\rm e}^{-\beta\eta} } {\beta}\Big)
 =: \frac{\sqrt{s}\,\pi_s} {\beta h(\eta,\beta) }
 }
and $\sqrt{s}\,\pi_s \to \beta h(\eta,\beta)$.   Also,
$\lim_{s\to\infty} b_s(\rho) = \lim_{s\to\infty} \pi_0{\rm e}^{s\rho} = \beta h(\eta,\beta)/\phi(\beta)$.
Proposition \ref{prop2} follows from combining \eqref{eq:geoSum1} and \eqref{220}.

\section{Proof of Theorem \ref{thm1}: the case $\rho=1$}\label{proofthm1}
 Since $\pi_i=\pi_s$ for $i\ge s$, $P_i = 1 - (K+s-i)\pi_s$. Write \eqref{mainn} as
\eqan{
\label{eq:newRdecompose}
(1-\pi_J)(1 - \mathcal{D}_{\beta,\eta}) =  2 \pi_s \sum_{i=0}^J \frac{\pi_J}{\pi_s}\,P_i
  \Big(1 - \frac{\pi_J}{\pi_i}\,P_i\Big) \,=\,
   2\pi_s \bigg(\sum_{i=0}^{s-1} + \sum_{i=s}^{s+K}\bigg)\big(\cdots\big),
 }
in which the last sum, for which $\pi_s=\pi_i=\pi_J$, equals
\begin{align*}
  2\pi_s \sum_{i'=0}^{K} \Big(1 - [K-i']\pi_s\Big)\,
   (K-i')\pi_s
 \,&=\, K(K+1)\pi^2_s - \frac{K(K+1)(2K+1)\pi_s^3}{3}\,,\\
 &\to
   \frac{\eta^2}{(\sqrt{\pi/2}+\eta)^2} - \frac{\frac{2}{3}\eta^3}{(\sqrt{\pi/2}+\eta)^3}
  \,=\, \frac{\eta^2 \sqrt{\pi/2}+\frac{1}{3}\eta^3}{(\sqrt{\pi/2}+\eta)^3}
\end{align*}
when $s,K\to\infty$ as at \eqref{scaling2}, using also \eqref{eq:pisRootS}.

It thus remains to consider the sum
\begin{equation}
\label{eq:mainSum1}
S_s := 2\sum_{i=0}^{s-1} \Big(1 - \frac{\pi_s}{\pi_i}\,P_i\Big)\pi_s P_i,
\end{equation}
which in terms of Poisson probabilities $\varpi_i := \varpi_i(s)$ and the
multiplier $b_s:=b_s(1)$ becomes
\begin{equation}
\label{eq:mainSum2}
S_s=
2 b_s^2
 \sum_{i=1}^s
\Big(1 - b_s\varpi_s \frac{\Pi_{s-i}(s)}{\varpi_{s-i}}\Big)
\varpi_s \Pi_{s-i}(s).
\end{equation}
Now both components of this sum for $S_s$ have finite limits (see below), and
$b_s\to b_\infty$ (see also below \eqref{eq:pisRootS}), finite and positive; we use this limit in examining the expression at \eqref{eq:mainSum2}.

Substitute from Lemmas \ref{lemma:pois1} and \ref{lemma:pois2} for the Poisson probabilities and assume for the moment that conditions for uniform convergence are met. Then
\eqan{
S_s
 \approx
2b_\infty^2 \sum_{i=1}^s \,\Big(1 - b_\infty {\rm e}^{{1\over2} x_{s i}^2} \Phi(-x_{s i})\Big) \frac{1}{\sqrt{2\pi}\,\sqrt{s}\,}
  \Phi(-x_{s i}),
 }
where $x_{si} = i/\sqrt{s}\,$. Recognize that this sum is an approximation to the Riemann integral
\begin{equation}
\label{eq:propInt1}
\frac{2b_\infty^2}{\sqrt{2\pi}} \int_0^{\sqrt{s}}
   \Big[1 - b_\infty\, {\rm e}^{{1\over2} u^2} \Phi(-u)\Big]
  \,\Phi(-u) \,{\rm d} u
\end{equation}
based on the $s$ intervals determined by the $s+1$ points $\{i/\sqrt{s}\,\}$, $i=0,\ldots,s$.  These integrals certainly converge as $s\to\infty$, because the (improper) Riemann integral is finite (its integrand is non-negative and dominated by $\Phi(-u)$ whose integral on $\bbbR_+$ equals $1/\sqrt{2\pi}\,$).

To prove that the improper integral based on \eqref{eq:propInt1} is indeed equal to  $\lim_{s\to\infty} S_s$, there are essentially two tasks: one relates to the finiteness of the limit of the finite sum, and the other to the convergence of the summands to the function involving the normal distribution function $\Phi$ and its density $\phi$.  These two tasks overlap in the sense that the part of the argument in the latter requiring uniform convergence follows from truncating the infinite sum.  It is convenient to rewrite the sum for $S_s$ as an integral, namely
\eqan{
\label{eq:mainInt1}
S_s =  \int_{\bbbR_+} g_s(x)\,\nu_s({\rm d} x),
 }
where for each positive integer $s$, $\nu_s$ is a purely atomic measure on $\bbbR_+$ with support set $\{x_{si}\}:=\{i/\sqrt{s}:i=0,\ldots,s-1\}$ and mass $2 b_s^2 \varpi_s(s)\, \Pi_{s-i}(s)$ at $i/\sqrt{s}$, and $g_s(x)$ is a right-continuous simple function defined on $\bbbR_+$ equal to $1- b_s \Pi_s(s)$ at 0 and with upward jumps at each point $i/\sqrt{s}$, $i=1,\ldots,s$, where $g_s(x_{si}+0)= 1 -  b_s     (\varpi_s/\varpi_{s-i})\,\Pi_{s-i}(s)$.

We first show that $\Pi_{s-i}(s)/\varpi_{s-i}(s)$ is a monotone sequence (in $i$), and that its analogue ${\rm e}^{{1\over2} x^2} \Phi(-x)$ is monotonic in $x$.
For the former, recall that the partial sum $\Pi_i(s)$ is a tail integral of a gamma density function (e.g.\ Johnson {\sl et al.}\ (1993), equation (4.108)), so
\eqan{\label{4999}
\frac{\Pi_i(s)}{\varpi_i(s)}
  = \frac{{\rm e}^s i!}{s^i} \int_s^\infty {\rm e}^{-u} \frac{u^i}{i!} \,{\rm d} u
  \,=\, \int_0^\infty {\rm e}^{-v} \Big(1 + \frac{v}{s}\Big)^i\,{\rm d} v,
 }
which is monotonic increasing in $i$, hence $1 - b_s \Pi_s(s) \le 1 - b_s \varpi_s \Pi_{s-i}(s)/\varpi_{s-i}(s) \le 1 - b_s \varpi_s$ for $0\le i \le s$. This monotonicity implies that each function $g_s$ has all jumps upwards as asserted.

For the size of the jumps we evaluate (omitting here the common argument $s$)
\eqan{
\label{eq:omegaRatios}
\varpi_s\bigg[ \frac{\Pi_{i+1}}{\varpi_{i+1}}
   - \frac{\Pi_i}{\varpi_i}\bigg]
  = \varpi_s\,\frac{\varpi_i\big[\Pi_i + \varpi_{i+1}\big] - \varpi_{i+1}\Pi_i}
  {\varpi_{i+1}\,\varpi_i}
  \,=\, \varpi_s\Big[1 - \Big(1 - \frac{i+1}{s}\Big)\frac{\Pi_i}{\varpi_i}\Big],
 }
so the jumps in $g_s(x)$ are bounded uniformly in $x$ by $b_s \varpi_s(s) \le b_s  \big[1 + 1/12s\big]\big/\sqrt{2\pi\,s}\,\le B/\sqrt{s}$, for a constant $B$, uniformly in $s$.

For the putative limit function ${\rm e}^{{1\over2} x^2}\Phi(-x)$ in the integrand at \eqref{eq:propInt1}, we have for $x\in\bbbR_+$,
\eqan{
\sqrt{2\pi}\, {\rm e}^{{1\over2} x^2} \Phi(-x) = \int_x^\infty {\rm e}^{{1\over2} x^2} {\rm e}^{-{1\over2} v^2}\,{\rm d} v = \int_0^\infty {\rm e}^{-{1\over2} w^2 - w x}\,{\rm d} w\,,
 }
which is monotonic in $x$, so $1 - b_\infty \le 1 - b_\infty {\rm e}^{{1\over2} x^2} \Phi(-x) \le 1$ for $0\le x \to\infty$.  Thus, omitting the multiplier $b_\infty^2/\sqrt{\pi/2}$, the improper integral based on \eqref{eq:propInt1} is dominated by $\int_0^\infty \Phi(-u)\,{\rm d} u= 1/\sqrt{2\pi}\,,$ which is finite. Hence, given $\epsilon>0$, there exists $U_\epsilon$ such that
$$\int_0^{U_\epsilon} \Phi(-u)\big[1 - b_\infty {\rm e}^{{1\over2} u^2} \Phi(-u)\big]\,{\rm d} u$$
 is within $\epsilon$ of the similar integral but over $\bbbR_+$.

We turn to the measures $\nu_s$.  Consider the sums $S_{sj} := \varpi_s \sum_{i=0}^j  \Pi_i(s)$. Using again the integral representation for $\Pi_i(s)$ as in \eqref{4999},
\begin{align}
S_{si} &= \varpi_s \sum_{j=0}^i \int_s^\infty {\rm e}^{-u} \frac{u^j}{j!}\,{\rm d} u \nonumber
  \,=\,\varpi_s \int_s^\infty {\rm d} u \int_u^\infty {\rm e}^{-v} \frac{v^i}{i!}\,{\rm d} v
\\ \nonumber
&= \varpi_s \int_s^\infty {\rm e}^{-v}\frac{v^i}{i!}(v-s)\,{\rm d} v
= \varpi_s\big[(i+1)\Pi_{i+1}(s) - s \Pi_i(s)\big]
\\ \nonumber
&= \varpi_s \big[(i+1)\varpi_{i+1}(s) - (s - i-1)\Pi_i(s)\big]
= \varpi_s \big[s\varpi_i(s) - (s - i-1)\Pi_i(s)\big]
\\
&= s \,\varpi_s \varpi_i
\Big[1 - \Big(1 - \frac{i+1}{s}\Big)\frac{\Pi_i(s)}{\varpi_i}\Big]
\label{eq:brokenSum}
= s\varpi_s^2 \frac{\varpi_i}{\varpi_s}
\Big[1 - \Big(1 - \frac{i+1}{s}\Big)\frac{\Pi_i(s)}{\varpi_i}\Big]\,.
\end{align}
In particular, for $i=s-1$, $S_{s,s-1} = s\varpi_s\varpi_{s-1} = s\varpi_s^2$, ${}= \sum_{i=1}^s \varpi_s \Pi_{s-i}(s) = \nu_s(\bbbR_+)/2b_s^2$.

From Stirling's formula (see e.g.\ Appendix A) we know that $s\varpi_s^2 = [1 + \vartheta_s/12s]^2/2\pi$ for some $|\vartheta_s| \le 1$, so $\sup_s \nu_s(\bbbR_+)<\infty$ uniformly in $s$. Therefore, given $\epsilon > 0$, for every $s$ there exists $j_s$ such that $S_{s,j_s} > S_{s,s-1}-\epsilon$.

\hfill{\Box}

\begin{proposition}
\label{prop:weakProp1}
The measures $\{\nu_s\}$ introduced at \eqref{eq:mainInt1}  converge weakly to the limit measure $\nu_\infty$ for which
\eqan{
\label{eq:nuProp}
\nu_\infty([0,x]) = \frac{2}{(\sqrt{\pi/2}+\eta)^2}\int_0^x \Phi(-v)\,{\rm d} v.
 }

\end{proposition}

{\noindent\sl Remark}: Because the limit function at \eqref{eq:nuProp} is continuous, $\sup_{x\in\bbbR_+} |\nu_s([0,x]) - \nu_\infty([0,x])| \,\to\,0$ for $s\to\infty$, i.e.\ we have not merely weak convergence but uniformity of that convergence on the domain of definition (e.g.\ Problem 5 in III.\S1 in \cite{shiryayev1984probability}).

\begin{proof} (of Proposition \ref{prop:weakProp1})
Let $f : \bbbR_+ \mapsto {\Bbb{R}}$ be a bounded continuous function; without loss of generality suppose $0 \le f(x)\le C_f$ for $x\in\bbbR_+$ and finite $C_f$. From our definitions,
\eqan{
\label{eq:intAndSum1}
\int_{\bbbR_+} f(x) \,\nu_s({\rm d} x)  = 2b_s^2 \sum_{i=0}^s f(x_{si})\,\varpi_s \Pi_{s-i}(s),
 }
where $x_{si} = i/\sqrt{s}$ as earlier. Because $f$ is bounded and $S_{si} \le \nu_s(\bbbR_+)/2b_s^2 \to 1/(2\pi)$ as $s\to\infty$, there exists $\{j_s\}$ such that the integral at \eqref{eq:intAndSum1}, written as $\sum_{i=0}^{j_s-1} + \sum_{i=j_s}^s$, is approximated arbitrarily closely for $s$ large enough by the former of these sums; for example $j_s = s^{5/8}$.  Then using (A.6),
\begin{align}
\nonumber
\sum_{i=0}^{j_s-1} f(x_{si})\,\nu_s(\{i/\sqrt{s}\,\})
  &= 2b_s^2\sum_{i=0}^{j_s-1} f(x_{si})\,\varpi_s
  \Big[\Phi(-x_{si}) + \frac{\psi_{si}}{\sqrt{s}}\,\Big]
\\
\label{eq:bOmegaSum2}
&= 2b_s^2 \sum_{i=0}^{j_s-1} f(x_{si})\,\Phi(-x_{si}) \varpi_s
  + 2b_s^2 \sum_{i=0}^{j_s-1} f(x_{si})\,\frac{\psi_{si}}{\sqrt{s}} \,\varpi_s,
\end{align}
where the constants $\psi_{si}$ are bounded uniformly in $i$ and $s$, $|\psi_{si}| \le C_\psi$ say.
The last sum at \eqref{eq:bOmegaSum2} is bounded by $j_s C_f C_\psi/s \sqrt{2\pi}\,$, and when $j_s = s^{5/8}$, this bound is arbitrarily small for $s$ sufficiently large.  So the left-hand side of \eqref{eq:intAndSum1} is arbitrarily close to the sum at \eqref{eq:bOmegaSum2} involving $\Phi(\cdot)$.  Each of these sums is an approximation to the Riemann integral $2 b_\infty^2 \int_0^{s^{1/8}} f(v) \,\Phi(-v) \,{\rm d} v$, which in turn is arbitrarily close (for large enough $s$) to the improper integral over $\bbbR_+$.
\end{proof}

This weak convergence property of $\{\nu_s\}$ is not directly applicable to the integral at \eqref{eq:mainInt1} in which the functions $g_s$ are not continuous; but they are monotonic and bounded, and have bounded increments with $g_s(x+1/\sqrt{s})- g_s(x) \le B/\sqrt{s}$ (see around \eqref{eq:omegaRatios}).  We show that
 $g_\infty(x) := \lim_{s\to\infty}g_s(x)$ exists and equals the bounded continuous function $1 - b_\infty {\rm e}^{{1\over2} x^2} \Phi(-x)$ for $x\in\bbbR_+$. To this end, consider $x\in[0,U_\epsilon]$, where $U_\epsilon$ is as defined below \eqref{eq:omegaRatios}, and write
\eqan{
g_s(x) = 1 - b_s(1) \frac{\Phi\big(-i_s(x)\big/\sqrt{s}\,\big) +
    \psi_{s,i_s(x)}/\sqrt{s} }{\phi\big(i_s(x)\big/\sqrt{s}\,\big)
  \big[1 + O(s^{-1/8  })\big]\big/\phi(0)}\,,
 }
where $i_s(x) = \min\big(\lfloor x\sqrt{s}\,\rfloor,\,s\big)$. For given $x$, $i_s(x)/\sqrt{s} \to x$ as $s\to\infty$, and thus $g_s(x) \to g_\infty(x)$ as defined.  Further, because $\phi(i_s(x)/\sqrt{s}\,)$ is bounded away from 0 on $[0,U_\epsilon]$, this convergence satisfies
$\sup_{x\in [0,U_\epsilon]} |g_s(x) - g_\infty(x)| < B_g/\sqrt{s}$ for
some finite constant $B_g$.  We can now write
\begin{align}
 \int_{\bbbR_+} g_s(x)\,\nu_s({\rm d} x)
 &= \int_{(U_\eps,\infty)} g_s(x) \nu_s({\rm d} x)
   + \int_{[0,U_\eps]} [g_s(x) - g_{\infty(x)}]\,\nu_s({\rm d} x) \nonumber
 \\
&\qquad+ \int_{[0,U_\eps]} g_\infty(x)\,[\nu_s({\rm d} x) - \nu_\infty({\rm d} x)]
   - \int_{(U_\eps,\infty)} g_\infty(x)\,\nu_\infty({\rm d} x) \nonumber
\\ \nonumber
&\qquad+ \int_{\bbbR_+} g_{\infty}(x)\,\nu_\infty({\rm d} x)
\\
\label{eq:5terms}
&:= T_1 + T_2 + T_3 - T_4 + \int_{\bbbR_+} g_{\infty}(x)\,\nu_{\infty}({\rm d} x).
\end{align}
Each of the terms $T_1,\ldots,T_4$ here can be made smaller than any given positive $\epsilon$, first by the choice of $U_\eps$ for $T_4$, then by choice of $s$ for $T_3$ and $T_1$ by appealing to weak convergence, and finally by choice of $s$ for $T_2$ by the uniform convergence of $g_s(x)$ in $[0,U_\eps]$. \hfill{\Box}

It remains to demonstrate that the improper integral in \eqref{eq:propInt1} has the value as asserted.  We do this
in the proof of the next lemma.
\begin{lemma}\label{lemm6}
\[
I_+:=\int_0^\infty \Phi(-u)[1 - {\rm e}^{{1\over2} u^2} \Phi(-u)]\,{\rm d} u = \sqrt{2 \pi} (1 - \log \sqrt{2}).
\]
\end{lemma}
\begin{proof}
Write $\int_0^\infty \Phi(-u)\,{\rm d} u - I_+
  = \int_0^\infty {\rm e}^{{1\over2} u^2} \,{\rm d} u (2\pi)^{-1}
  \int_u^\infty {\rm e}^{-{1\over2} v^2} \,{\rm d} v \int_u^\infty {\rm e}^{-{1\over2} w^2}\,{\rm d} w$,
noting that the left-hand side equals $1/\sqrt{2\pi} - I_+ =: J_0$ say.
Use polar coordinates
 $(v,w) = (r\cos\theta, r\sin\theta)$ so that
$ {\rm d} v\,{\rm d} w \mapsto r\,{\rm d} r \,{\rm d}\theta $ and multiply by $2\pi$, i.e.\
\begin{align*}
2\pi J_0 :\!&=
\int_0^\infty {\rm e}^{{1\over2} u^2}\,{\rm d} u
  \int_u^\infty {\rm e}^{-{1\over2} v^2}\,{\rm d} v
  \int_u^\infty {\rm e}^{-{1\over2} w^2}\,{\rm d} w \\
&=
\int_0^\infty {\rm e}^{{1\over2} u^2}\,{\rm d} u
  \int_{u\sqrt{2}}^\infty \Big({\textstyle\half}\pi - 2 \arcsin \frac{u}{r}\Big)
  {\rm e}^{-{1\over2} r^2} r\,{\rm d} r \\
&=
0 + 2 \int_0^\infty {\rm e}^{{1\over2} u^2}\,{\rm d} u
  \int_{u\sqrt{2}}^\infty
\Big(- \frac{u}{r^2}\cdot \frac{1}{\sqrt{1 - u^2/r^2}}  \Big)
\big( - {\rm e}^{-{1\over2} r^2} \big)\,{\rm d} r \qquad\hbox{(integration by parts)},\\
&=
2 \int_0^\infty {\rm e}^{-{1\over2} z^2 \cos 2\theta } z\,{\rm d} z
  \int_0^{\mathop{\rm{artan}}\nolimits(1/\sqrt{2}\,)}  \frac{ z \sin\theta }{ z \cos\theta
  \sqrt{z^2 \cos 2\theta} }\,{\rm d}\theta
\qquad\big((r,u)=(z\cos\theta,z\sin\theta)\big),\\
&=
2 \int_0^{\mathop{\rm{artan}}\nolimits(1/\sqrt{2}\,)}
  \frac{\tan\theta\,{\rm d}\theta}{\sqrt{\cos 2\theta}}
  \int_0^\infty {\rm e}^{-{1\over2} z^2 \cos 2\theta}\,{\rm d} z
   \qquad\hbox{(Fubini's theorem)},\\
  &= 2 \int_0^{\mathop{\rm{artan}}\nolimits(1/\sqrt{2}\,)}
  \frac{\tan\theta\,{\rm d}\theta}{\cos 2\theta}
  \int_0^\infty {\rm e}^{-{1\over2} y^2}\,{\rm d} y
  \qquad(y = z\sqrt{\cos 2\theta}\,),\\
&=
\sqrt{2\pi}\int_0^{\mathop{\rm{artan}}\nolimits(1/\sqrt{2}\,)}
  \frac{\tan \theta \,{\rm d}\theta}{\cos 2\theta}
  \qquad\hbox{(integration)},\\
&=
\sqrt{2\pi} \int_0^{1/\sqrt{2}} \frac{ t \,{\rm d} t}{1-t^2}
  \qquad(t=\tan\theta),\\
&=
\, \sqrt{2\pi} \int_0^{1\over2}
  \frac{{1\over2} \,{\rm d} u}{1-u}
 \,=\, {\textstyle\half} \sqrt{2\pi} \log \big(1/{\textstyle\half}\big) \ =  \
\sqrt{2\pi}\,\log\sqrt{2}\,.
\end{align*}
Recalling the definition of $J_0$ yields the expression for $I_+$.
\end{proof}

\section{Proof of Theorem \ref{thm1}: the case $\rho=1-\beta/\sqrt{s}$}
\label{proofthm2}

We now have $\rho=1-\beta/\sqrt{s}$ for finite $\beta\ne 0$. The proof is similar to the approach and methods used in the case $\rho=1$ above; the particular algebraic expressions differ.  Start by breaking the sum at \eqref{eq:newRdecompose} into the two sums as before. Write the first of these as
\eqan{
\label{eq:sumInt2}
2 \sum_{i=1}^s
   \Big(1 - \frac{\pi_J}{\pi_s}\,\varpi_s(s\rho)\,
   \frac{\Pi_{s-i}(s\rho)}{\varpi_{s-i}(s\rho)}\Big)
\pi_J \Pi_{s-i}(s \rho) \,=\, \int_{\bbbR_+} g_s(x)\,\nu_s({\rm d} x),
 }
where the measure $\nu_s$ is atomic as before but now has mass $ 2b^2_s(\rho) \varpi_s(s\rho)\,\rho^K \Pi_{s-i}(s\rho)$ located at $x_{si}=i/\sqrt{s}$ for $i=0,\ldots,s-1$, and the function $g_s$ is a right-continuous simple function defined on $\bbbR_+$ equal to $1- b_s(\rho)\rho^K \Pi_s(s\rho)$ at 0 and has jumps at each point $x_{si}$, $(i=1,\ldots,s)$, where $g_s(x_{si}+0)=1 - b_s(\rho)\rho^K \Pi_{s-i}(s\rho)\big/ [\varpi_{s-i}(s\rho)/\varpi_s(s\rho)]$,  so that $g_s(\cdot)$ is monotonic.

To check that $\nu_s$ has finite total mass, and that there is a bound on $\nu_s(\bbbR_+)$ that is uniform in $s$, note that $\lim_{s,K\to\infty} b_s^2(\rho) \rho^K$ is finite for finite $\beta$ and equal to $[\beta h(\eta,\beta)/\phi(\beta)]^2(1-{\rm e}^{-\beta\eta})$ (see around (2.7)); for the rest, mimic the
calculation leading to \eqref{eq:brokenSum}, now with $\rho=1-\beta/\sqrt{s}\,,$ in computing
\begin{align*}
S_{si} := \varpi_s(s\rho) \sum_{j=0}^i \Pi_j(s\rho)=s\rho\,\varpi_s(s\rho) \varpi_i(s\rho)
\bigg[1 - \Big(1 - \frac{i+1}{s\rho}\Big)
   \frac{\Pi_i(s\rho)}{\varpi_i(s\rho)}\bigg]
   \end{align*}
so that
\begin{align*}
 S_{s,s-1} = s\rho\,\varpi_s(s\rho)\,\varpi_{s-1}(s\rho)
\bigg[1 - \Big(1 - \frac{1}{\rho}\Big)
   \frac{\Pi_{s-1}(s\rho)}{\varpi_{s-1}(s\rho)}\bigg]
= s\, \varpi_s^2(s\rho)\bigg[1 + \frac{\beta }{\rho\sqrt{s}\,}
   \frac{\Pi_{s-1}(s\rho)}{\varpi_{s-1}(s\rho)}\bigg],
\end{align*}
which for given (finite) $\beta$ is bounded in $s$.  Thus, $\sup_s \nu_s(\bbbR_+) <\infty$.  Note that
\eqan{
\nu_s(\bbbR_+) = 2 b^2_s(\rho) \rho^K S_{s,s-1}  \,\to\, \frac{\beta^2h^2(\eta,\beta)\,(1-{\rm e}^{-\beta\eta})}{\pi\,\phi^2(\beta)}\, \big(1 + \beta\sqrt{\pi/2}\,\big) \qquad (s\to\infty).
 }

From Lemma~\ref{lemma:pois2}, for $\rho = 1 - \beta/\sqrt{s}$,  $0\le i\le s$ and $s$ large,
\eqan{
\Pi_{s-i}(s\rho)
  - \frac{\psi_{si}}{\sqrt{s\rho}\,}
= \Phi\Big(\frac{s-i-s\rho}{\sqrt{s\rho\,}}\Big)
 &= \Phi\big(- x_{si} + \beta +O(s^{-{1\over2}})\big)\nonumber\\
 &= \Phi\big(-x_{si} + \beta
\big) +O(s^{-{1\over2}}),
 }
where $o(\cdot)$ in the last term is uniform in $i$ and $s$. Hence
\eqan{
\label{eq:2ndPoissRel}
\Pi_{s-i}(s\rho) =
  \Phi\big(- x_{si} + \beta\big)
  + \tilde\psi_{si}\big/\sqrt{s}\,,
 }
where by Lemma~\ref{lemma:pois2} $\tilde\psi_{si}$ is uniformly bounded in $i$ and $s$.

For $x\in\bbbR_+$ define $i_s(x) = \lfloor x\sqrt{s} \rfloor$ so that $i_s(x)/\sqrt{s} \to x$ as $s\to\infty$, and
\begin{align*}
g_s(x) &= 1 -  \frac{b_s(\rho)\rho^K\,\Pi_{s-i_s(x)}(s\rho)}
 {\varpi_{s-i_s(x)}(s\rho)\big/ \varpi_s(s\rho)}
 \\
  &= 1 - b_s(\rho)
\frac{\rho^K\big[\Phi(-x_{s,i_s(x)}+\beta)
  + \tilde\psi_{s,i_s(x)}/\sqrt{s}\,\big] }
  {\phi(-x_{s,i_s(x)} + \beta)/\phi(\beta)\,[1 + O(s^{-{1\over2}})]}\,.
\end{align*}
Then for given $x$ and $s\to\infty$, the right-hand side $\to g_\infty(x)$ defined by
\eqan{\label{45}
g_\infty(x) := 1 - \beta h(\eta,\beta)\,\frac{(1-{\rm e}^{-\beta\eta}) \Phi(-x+\beta)}
   {\phi(-x+\beta)}
 }
at a uniform rate of convergence that is $O(s^{-{1\over2}})$ for $x$ on compact sets like $U_\eps$ where the denominator above is bounded away from 0, $b_\infty(\rho) := \lim_{s\to\infty} b_s(\rho) = \beta h(\eta,\beta)/\phi(\beta)$ as below (2.7), and
\eqan{
\lim_{s\to\infty}
   \big(1 - \beta/\sqrt{s}\,\big)^{\sqrt{s}\eta}
  = {\rm e}^{-\beta\eta}.
 }

\begin{proposition}
\label{prop:weakProp2}
The measures $\{\nu_s\}$ defined below \eqref{eq:sumInt2} converge weakly to the limit measure $\nu_\infty$ for which
\eqan{
\label{eq:nuProp2}
\nu_\infty([0,x]) =
  \frac{2 [\beta h(\eta,\beta)]^2 \,(1-{\rm e}^{-\beta\eta})}{\phi(\beta)}
\int_{(-\beta,x-\beta]} \Phi(-v)\,{\rm d} v.
 }
\end{proposition}

\begin{proof}
The same argument as used in establishing Proposition~\ref{prop:weakProp1} holds,  subject to using \eqref{eq:2ndPoissRel} in place of \eqref{eq:piRel3} from Lemma~\ref{lemma:pois2}. The analogue of \eqref{eq:bOmegaSum2} for $\beta\ne 0$, for bounded continuous $f$, is
\eqan{
\sum_{i=0}^{j_s-1} f(x_{si}) \,\nu_s\big(\{i/\sqrt{s}\,\}\big)
   = \sum_{i=0}^{j_s-1} f(x_{si})\,2 b^2_s(\rho)\rho^K \varpi_s(s\rho)
  \Big[\Phi(-x_{si} + \beta) + \frac{\tilde\psi_{si}}{\sqrt{s}\,}\Big].
 }
Each term involving some $\tilde\psi_{si}$ is at most $O(1/s)$, so when $j_s< O(s)$ their sum is negligibly small for large $s$. Since $\varpi_s(s\rho) =       {\rm e}^{-{1\over2}\beta^2}/\sqrt{2\pi s}\,[1 + O(s^{-{1\over2}})]$ the other term is an approximation to the Riemann integral
\eqan{
2\big(\lim_{s\to\infty} b^2_s(\rho) \rho^K\big)\phi(\beta)
\int_0^{j_s/\sqrt{s}}  f(u)\,\Phi(-u+\beta)\,{\rm d} u,
 }
from which we deduce \eqref{eq:nuProp2}.
\end{proof}

An argument similar to that leading to \eqref{eq:5terms} shows that the integrals at the right of \eqref{eq:sumInt2} converge to the limit $\int_{\bbbR_+} g_\infty(x)\, \nu_\infty({\rm d} x)$.

It remains to consider the analogue of the last sum at \eqref{eq:newRdecompose} for which $i\ge s$.  As earlier, calculate $P_i^{}$ via its tail which now reads
\[
P_i^{} = 1 - \sum_{j=i+1}^J \pi_j
  \,=\, 1 - \pi_J\sum_{j=i+1}^J \rho^{j-J}
 \,=\, 1 - \rho^{i+1-J} \frac{\pi_J(\rho^{J-i}  - 1)}{\rho-1}
 = 1 + \frac{\rho\pi_J(1 - \rho^{-(J-i)})}{1-\rho}\,,
\]
so
\begin{align}
\Big(\,1 - \frac{\pi_J}{\pi_i}P_i^{}\Big)P_i
 &= \Big(1 - \rho^{J-i} P_i^{}\Big)P_i \nonumber
\\
&= \Big(1 + \frac{\rho\pi_J(1 - \rho^{-(J-i)})} {1 - \rho}\Big)
   \bigg[1 - \rho^{J-i} \Big(1 + \frac{\rho\pi_J(1 - \rho^{-(J-i)})}{1 - \rho}
\Big)\bigg] \nonumber
\\
&= \Big(1 + \frac{\rho \pi_J(1-\rho^{-(J-i)}}{1-\rho}\Big)(1-\rho^{J-i})
  \Big(1 + \frac{\rho\pi_J}{1-\rho}\Big) \nonumber
\\
&= \Big(1 + \frac{\rho\pi_J}{1-\rho}\Big)
  \bigg[1 - \rho^{J-i} + \frac{\rho\pi_J\big(2 - \rho^{J-i} - \rho^{-(J-i)}\big)}
    {1-\rho} \bigg]\,.
\end{align}
Hence, for the second part of the sum of \eqref{eq:newRdecompose} for the case $\rho\neq 1$ and $i\geq s$ we have
\begin{align}
2\pi_J\sum_{i=s}^J \Big(1 - \frac{\pi_J}{\pi_i}P_i^{}\Big)P_i \nonumber
  &=  2\pi_J\Big(1 + \frac{\rho\pi_J}{1-\rho}\Big)
  \bigg[(K+1)\Big(1 + \frac{2\rho\pi_J}{1-\rho}\Big)
\\
  &
  - \frac{\rho^K (\rho^{-K-1}-1)} {\rho^{-1}-1} \nonumber
\Big(1 + \frac{\rho\pi_J}{1-\rho}\Big)
  - \frac{\rho^{-K} (\rho^{K+1} - 1)}{\rho-1} \cdot\frac{\rho\pi_J}{1-\rho}
\bigg]
\\
\label{eq:bigSecondSum}
=  2\pi_J\Big(1 + \frac{\rho\pi_J}{1-\rho}\Big)
  &\bigg[\Big(K+1-\frac{1-\rho^{K+1}}{1-\rho}\Big)
     \Big(1 + \frac{\rho\pi_J}{1-\rho}\Big)
  + \Big(K+1 - \frac{1-\rho^{K+1}}{\rho^K(1-\rho)}\Big)
 \frac{\rho\pi_J}{1-\rho}
\bigg].\qquad\quad
 \end{align}
 For $\rho = 1 - \beta/\sqrt{s}$ and $s,K\to\infty$ as at \eqref{scaling2}
\eqan{
\frac{\rho \pi_J}{1-\rho} = \frac{\tsqrt{s}\,\pi_s \rho^{K+1}}{\beta}
\approx \frac{\sqrt{s}\,\pi_s (1-{\rm e}^{-\beta\eta}) }{\beta}\,,
 }
so the limit of the right-hand side of \eqref{eq:bigSecondSum} is the same as the limit of
\begin{align}
&-2\pi_s \rho^K\Big(1 + \frac{\sqrt{s}\,\pi_s\,(1-{\rm e}^{-\beta\eta})}{\beta}\Big)
  \bigg[ \Big(K - \frac{\sqrt{s}(1 - (1-{\rm e}^{-\beta\eta}))}{\beta}\Big)
\Big(1 + \frac{\sqrt{s}\,\pi_s\,(1-{\rm e}^{-\beta\eta})}{\beta}\Big)
\nonumber \\ \nonumber
&\quad + \Big(K - \frac{\sqrt{s}(1 - (1-{\rm e}^{-\beta\eta}))}{(1-{\rm e}^{-\beta\eta})\,\beta} \Big)
  \frac{\sqrt{s}\,\pi_s\,(1-{\rm e}^{-\beta\eta})}{\beta}
\bigg]
\\ \nonumber
 &\to -2\beta h(\eta,\beta)\,(1-{\rm e}^{-\beta\eta})
  \Big(1 + \frac{\beta h(\eta,\beta)\,(1-{\rm e}^{-\beta\eta})}{\beta}\Big)
  \bigg[\eta - \frac{1-(1-{\rm e}^{-\beta\eta})}{\beta}\Big)
   \Big(1 + \frac{\beta h(\eta,\beta)\,(1-{\rm e}^{-\beta\eta})}{\beta}\Big)
   \\ \nonumber
 &\quad + \Big(\eta
  - \frac{1-(1-{\rm e}^{-\beta\eta})}{\beta\,(1-{\rm e}^{-\beta\eta})}\Big)
  \frac{\beta h(\eta,\beta)\,(1-{\rm e}^{-\beta\eta})}{\beta}\bigg]
\\
 &= -2\widetilde{C}\bigg(1 + \frac{\widetilde{C}}{\beta}\bigg) \bigg[
 \bigg(\eta - \frac{1-(1-{\rm e}^{-\beta\eta})}{\beta}\bigg)
  \bigg(1 + \frac{\widetilde{C}}{\beta}\bigg)
 +\bigg(\eta
 - \frac{1-(1-{\rm e}^{-\beta\eta})}{\beta (1-{\rm e}^{-\beta\eta})}\bigg)
  \frac{\widetilde{C}}{\beta}\bigg],\label{413}
\end{align}
where $\widetilde{C}=\beta h(\eta,\beta)\,(1-{\rm e}^{-\beta\eta})$. Combining \eqref{45} and \eqref{eq:nuProp2}, and \eqref{413}, as required for the sum at \eqref{eq:newRdecompose}, now yields
\begin{equation}\label{intttt}
1-\mathcal{D}_{\beta,\eta}
=  2\lim_{s\to\infty}\tsqrt{s}\,\pi_s \int_{-\beta}^\infty
   (1-{\rm e}^{-\beta\eta})
  \Big(1 - \frac{(1-{\rm e}^{-\beta\eta})\,\phi(\beta)\,\Phi(-u)}{\phi(u)}
   \Big)\Phi(-u)\,{\rm d} u + g(\eta,\beta).
 \end{equation}
We should like at this point to evaluate the integral at \eqref{intttt}.
For this purpose we have mimicked the steps followed in establishing the
expression for $I_+$ in Lemma~\ref{lemm6}, but have succeeded only
in reaching the expressions in \eqref{111} and \eqref{222} below in which final
quadratures are unresolved.  Inspection of \eqref{intttt} shows that we must evaluate,
for both positive and negative $\beta$,
 $$ J_\beta :=  \int_{\beta}^\infty \frac{[\Phi(-u)]^2}{\phi(u)}
  \,{\rm d} u \,=\, \int_{\beta}^\infty {\rm e}^{{1\over2} u^2}\,{\rm d} u
  \int_u^\infty \int_u^\infty \frac{{\rm e}^{-{1\over2}(v^2+w^2)}}{2\pi}\,{\rm d} v \,{\rm d} w.
$$
By following steps similar to those in the proof of Lemma~\ref{lemm6} we deduced first
that for $\beta>0$,
\eqan{\label{111}J_\beta  =
- \frac{\Phi(-|\beta|\,)}{\sqrt{2\pi}}\,\log 2
  + \int_1^\infty \frac{\beta {\rm e}^{-{1\over2} \beta^2 x^2}}{2\pi} \,
   \log\Big(1 + \frac{1}{x^2}\Big)\,{\rm d} x.
   }
For $\beta<0$, integration takes place (cf.\ second line of the evaluation of
$J_0$ in the proof of Lemma~\ref{lemm6}) over $0<r<\infty$, with angular coordinate having arc-lengths $2\pi$
for $r<|\beta|$, $2\pi - 4\,\arccos(|\beta|/r) = 4\,\arcsin(|\beta|/r)$ for
$|\beta|<r<|\beta|\sqrt{2}$, and ${1\over2}\pi + 2\,\arcsin(|\beta|/r)$
for $r>|\beta|\sqrt{2}$.  Ultimately this yields
\begin{align}\label{222}
J_{-|\beta|}
  = J_{|\beta|} - J_0 + \int_0^{|\beta|} {\rm e}^{{1\over2} u^2}\,{\rm d} u
  &+ \frac{4\log 2}{\sqrt{2\pi}}\Big[\Phi(|\beta|\sqrt{2}\,) - {1\over2}\Big]\cr
  &- \int_0^{\sqrt{2}} \frac{4|\beta|{\rm e}^{-{1\over2}\beta^2x^2}}{2\pi}
\,\log\Big(1 + \frac{1}{x^2}\Big)\,{\rm d} x.
\end{align}

\section{Conclusion}
\label{sec:conclusion}

The analysis presented in this paper both answers questions and raises
further points that need clarification. Recall that in the setup in \cite{NazarathyWeiss0336}, Nazarathy and Weiss observed that for $M/M/1/K$ systems with $K \to \infty$, $\mathcal{D}_\pi \to 1$ for $0<\rho<\infty$ except when $\rho=1$, for which they found
$\mathcal{D}_\pi \to {2\over 3}$.  A similar effect occurs in multi-server $M/M/s/K$ systems. This is illustrated in Figure~\ref{fig:preLim}.

\begin{figure}[h]
\begin{center}
\includegraphics[width=12cm]{prelimitDfigure.pdf}
\end{center}
\caption{{\small M/M/$s/K$ systems with $K=\sqrt{s}$ for increasing values of $s$. $\mathcal{D}_\pi$ as a function of $\rho$. The black dot is $\mathcal{D}_{0,1}$ corresponding to $s \to \infty$ with $\eta=1$.
\label{fig:preLim}}}
\end{figure}

Our theorem shows that in the multi-server case, this discontinuity is merely a matter
of scale: setting $\beta = (1-\rho)\sqrt{s}$ and considering
 $$

 \mathcal{D}_\beta =
 \lim_{t\to\infty}
 \frac{\mathop{\rm var}\nolimits\big(N_{\rm dep}(0,t]\big)}{ {\rm E} \big( N_{\rm dep}(0,t]\big)}
\Bigg|_{\rho=1-\beta/\sqrt{s}}
\quad ,$$
the limit function  is no longer discontinuous but reflects the
asymptotic variance over a much finer range of values as in Figure~\ref{fig:rhon1} of Section~\ref{sec:bravoForQED}.

In turn, this emphasizes that, for the systems that we have considered, there
exists some
subtle interplay between the service facility on the one hand and customers
on the other. In terms of customers there may be a `deficit' (so some servers are idle)
or a `surfeit' (so some customers are waiting), and when
the system is `balanced', meaning that $\rho\approx 1$, the occasions
when the system is either empty or full are rare (i.e.\ both $\pi_0$ and
$\pi_J$ are `small') so that mostly periods with above or below average
net arrivals are balanced by periods of above or below average productive
service, and `consequently' the variability of throughput of the system
as measured by the output $N_{\rm dep}$ is to some extent `smoothed'.
This intuitive explanation of the BRAVO effect is one we have not been able
to translate into a mathematical explanation.

In other work concurrently under preparation, we have examined the
many-server  $M/M/s$ system with reneging to see whether it too exhibits the
BRAVO effect.
In studying a system with reneging, the departure process $N_{\rm dep}(0,\cdot\,]$ no longer
consists precisely of the deaths in the birth--death process $Q_s(\cdot)$ but only
of a subset of those deaths.  Consequently the second equality at (1.5) is
no longer available as an expression for $\mathcal{D}_\pi$, but an independent
computation, with the advantage from an expository viewpoint that it relies
on more primitive results, yields a substitute expression
\begin{equation}
\label{eq:Rfuture}
1 - 2 \sum_{i=0}^J (P_i - \Lambda^*_i)\Big(q_{i+1} - \frac{\lambda^*}{\pi_i \lambda_i}  (P_i - \Lambda^*_i) \Big),
\end{equation}
where $q_{i}$ is the probability of having a death at state $i$ of $Q_s(\cdot)$ count as an increment of $N_{\rm dep}(0,\cdot]$. A complete derivation of
this result has been deferred from this paper, in which context the present work should have been more self-contained.

The same substitute expression
can also cope with a sequence of systems with balking, where we expect the
BRAVO effect may still be observed, but whether such a system must be finite,
with a buffer
in place as well as the balking mechanism, remains to be discovered.

\vspace{10pt}
{\bf Acknowledgment}: DJD's work done while an Honorary Professorial Associate at the University
of Melbourne. JvL is supported by a Starting Grant of the European Research Council. YN's work done in part while at the Department of Mathematics at the Swinburne University of Technology. YN is supported by Australian Research Council (ARC) grants DP130100156 and DE130100291.

redpajama_set_name: RedPajamaArXiv


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After some initial uncertainty regarding the status of Cameron Biedscheid, Notre Dame granted Biedschied permission to transfer. As we noted earlier this week, it has been a rough week for the Irish who lost Jerian Grant to an academic issue over the weekend. Biedscheid, who took a redshirt just before the season started, was a top-30 recruit coming into school, but did not have a particularly impressive freshman season. News of Biedscheid's intention to transfer leaked over the weekend, but then questions arose as to when and whether Notre Dame would grant the transfer request.
Some might consider it a little early to start debating the best conference in the country, but in some ways it might be the second best time to judge conferences since the remainder of the regular season will essentially be conference match-ups (the best time being after the season). As such, Matt Norlander's analysis of the top conferences in the country using the RPI and KenPom rankings is somewhat instructive. Most of the findings are not too surprising for those of us that follow the sport regularly, but it is interesting to see just how bad some conferences have been (like the Mountain West and SEC) and how disappointing some other ones have been (ACC).
With Saturday's massive intrastate match-up between Louisville and Kentucky looming, Sporting News took a look at the top 10 players from Rick Pitino and John Calipari (fair warning: they are both slideshows). Given their well-traveled coaching paths neither list is composed exclusively of Louisville and Kentucky players, but it is interesting although not surprising to see how heavily weighted Pitino's list is towards Kentucky. Having said that it will be interesting to see how Pitino's legacy in particular is defined in terms of his team association when he hangs it up. Given the arc of his career we suspect he will be someone like Larry Brown without a firm affiliation to one school.
When we saw the headline that Reggie Johnson was transferring from Miami we were initially very confused. It turns out that it was not about the rather large former center for the Hurricanes. Instead it was referring to the 6'1″ sophomore guard from Miami (Ohio). This Reggie Johnson is the second-leading scorer for his team at 11.8 points per game and states that he is looking for "great chemistry, a winning program, a place where I can excel and reach my full potential and of course a great education." Although the other Reggie Johnson was probably a more desirable target this one should have a fair number of suitors with nearly two and a half years of eligibility remaining.
We have spent a fair amount of time on Twitter criticizing the breaks that teams (particularly professional ones) get from the government. It should not come as a surprise that other organizations get these benefits too, but the size of the benefits–particularly for a company like ESPN–may surprise you. According to The New York Times, ESPN has received almost $260 million in state tax breaks and credits over the past 12 years. We won't get into the economic or political discussions regarding the reasoning behind this decision or the implications it has on the state government and economy, but we suspect that it might become a big debate whenever the next economic crisis hits Connecticut.
It wasn't always easy for Miami to get to this year's Sweet Sixteen, but it was certainly never easy for Marquette to make the same journey. Ultra-tight games and impressive comebacks highlighted the Golden Eagles' close wins over Davidson and Butler, while Miami romped over Pacific and held on against Illinois. Say what you will about Buzz Williams' team, but improbably this team has figured out how to handle the big gut-check moments. During the regular season, Marquette played four overtime games and won three of them. I'm honestly uncertain if "clutchness" is a real phenomenon, but the Golden Eagles give my doubts doubts.
Big news came for Miami when it was announced that Reggie Johnson would not be traveling with the team for tonight's match-up, but it's unclear just how significant this news actually is. Johnson missed a huge chunk of the season with an injury and was out of shape and often ineffectual when he finally returned to the court. Against the smallish Golden Eagles, how much would the lumbering Johnson actually have played anyway? In Miami's statement game in the ACC championship against small ball North Carolina, the center played all of three minutes. His time instead went to Rion Brown and Julian Gamble, a pair of players who supercharge the Hurricanes' offense and defense. It feels unlikely that Johnson would have (or should have) played all that much against Marquette. It's not that this news doesn't have a big impact on Miami's title chances, but for the purposes of this match-up, it doesn't feel particularly significant for those who have watched this team closely.
After clinching the ACC Tournament title with three straight double-digit wins, Miami was riding high. The media present in Greensboro — including myself — were riding a similar high after watching one of the best offensive basketball games of the year. It felt like a slap in the face to watch as the seeds were announced and the Hurricanes weren't rewarded for their efforts with a #1 seed. It was the first time in history that the consensus ACC champion failed to earn a spot on top of a region.
Matt Patton is an ACC microsite writer. He filed this report from the ACC Tournament championship game between North Carolina and Miami in Greensboro this afternoon.
An ACC school from outside the state of North Carolina has only won the regular season title outright eight times since 1954 (most recently, Maryland won it in 2002). An ACC school from outside the state of North Carolina has only won the ACC Tournament 10 times since 1954 (most recently, Florida State won it last year). Miami is the first team to ever do both in the same year. An unbalanced conference schedule has cheapened the regular season title in recent years, but that's shouldn't throw any shade on what the Hurricanes accomplished this year. The Miami win along with Florida State's ACC Championship last season marks the second time in the league's history that consecutive ACC Tournaments were won by schools outside of North Carolina (the 1984 and 1985 tournaments were won by Maryland and Georgia Tech, respectively).
And the Hurricanes won their final two games in front of very hostile crowds. They won because of tremendous coaching from Jim Larranaga — whose lineup changes proved instrumental in games against North Carolina and Boston College. They won because Shane Larkin was the best player in an ACC Tournament full of outstanding performances (Olivier Hanlan, Durand Scott and Dez Wells all went for over 30 points in a game). They won because experience doesn't get rattled. In short, they won because they were the best team on the floor.
Against North Carolina, Miami found itself in a different position than usual. The Tar Heels and their new and improved smaller lineup came out firing. With just over 10 minutes to go in the first half, North Carolina led, 18-13. Miami proceeded to score on its next nine possessions and 13 of 16 of the final possessions of the half (over that time they missed three shots). North Carolina only scored on eight of 15 possessions, but PJ Hairston hit four threes and Marcus Paige added another to keep the game within a possession at the half. It was the best overall 10-minute offensive stretch I've seen this year. Both teams moved the ball to find open shots and both teams knocked down nearly every shot available. At one point the lead changed hands on eight straight possessions.
We are in the final stretch before conference tournaments tip off and there are several key match-ups this week that will help to determine not only postseason seeds but regular season titles. Let's not waste any time and get to the breakdowns!
How does Tom Izzo stop Indiana's Victor Olidipo from lighting up the Spartans again?
It doesn't get any bigger than this game this week, as the Hoosiers and the Spartans battle for sole possession of first place in the Big Ten. Michigan State faces a gauntlet of games as it goes up against Indiana at home, Ohio State and Michigan on the road, and then back home against Wisconsin. It is highly unlikely that they will get through the next four unscathed, so this first game at home is crucial if they want to win a regular season Big Ten crown. In the teams' previous game this season, IU defeated MSU 75-70 in Bloomington. Tom Izzo's squad turned the ball over too many times, couldn't hit their two-point shots, and didn't get to the free throw line enough. Since then, the Spartans have won five straight and are looking better and better each game. They are being led by point guard Keith Appling. Appling is averaging 16.4 points per game in their current five-game winning streak, since going 1-of-4 with three points in the loss to IU. Appling will once again be a key factor as the Spartans will need his scoring and play-making abilities. Izzo will also need his defense to figure out a way to stop Indiana's Victor Olidipo. Olidipo torched the Spartans last time out for 21 points on 9-of-12 shooting. If Indiana is allowed to go 19-of-32 again from inside the paint, Michigan State will struggle to find a way to win. The Spartans need a better defensive effort on the interior and better rebounding if they are going to overtake the Hoosiers for the Big Ten lead.
While Miami remains undefeated in the ACC and is aiming for a #1 seed in the NCAA Tourney, they face two tough tests the remainder of the season against Virginia at home and against Duke on the road. The Cavaliers have been anything but road warriors in the ACC (2-4), but they present a tough match-up because of their ability to shoot the ball, especially from three. Keep a close on Virginia's Joe Harris and Akil Mitchell. If they are to knock off the Hurricanes, these two players must have very big games. If UVA is still without 6'11" Mike Tobey (mononucleosis), scoring on the interior will be tough against Miami's 6'11" Kenny Kadji and 6'10" Reggie Johnson. The three-point shot is a significant part of the Cavaliers' offense, so pay close attention to their effort early from beyond the arc. If they can stay in the game with some made threes, they will have a shot to win it in the end. However, the Hurricanes play lockdown perimeter defense too. This is a tall task for the Cavaliers, but it will still be an interesting match-up nonetheless.
It was a good weekend for basketball fans in the ACC though fans of individual teams may be feeling more ambivalent. Of the six games this weekend, all but one of them had a final margin of four points or less. In fact, if North Carolina State hadn't managed a four-point overtime win, all but one of the games this weekend would have had a final score where the outcome was within one possession. Even the single game that wasn't close had its own fascinating subplot and March ramifications. It was a good weekend for close games, but did we learn anything?
Duke Is Fouling Too Much. On the season, Duke has been fouling opponents at a nice and low rate, posting and opponent free throw attempted to field goal attempted ratio of 30.5%. In conference, however, Duke has jumped in this metric to 36.4%. Amazingly, like many of Duke's problems, this rise can probably be accredited to the injury of Ryan Kelly. Kelly's replacements foul at a very high rate while still allowing a conference-worst 49.5% shooting from inside the arc. Physical defense that prevents easy buckets can sometimes be used to excuse high fouling rates, but Duke's interior is offering up the worst of both worlds. On Saturday, Duke forced 26 turnovers against Maryland and shot nearly 50% and the Terrapins still won.
I LOVED…. my friend's text response to my joking suggestion that Dickie V might not survive five overtimes at his age during the Louisville-Notre Dame game Saturday night: "At one point in the third OT he said on air, 'Dan I gotta go to the bathroom. Seriously I'm not kidding,' and then he was gone from the broadcast for like five minutes." Priceless.
I LOVED…. Indiana going from an are-you-kidding-me loss to Illinois, complete with a late-game collapse and unforgivable defensive sequence on the buzzer-beater, to looking like a can't-miss Final Four team by completely dominating Ohio State on the road. Of course, what good college team hasn't pulled this Jekyll and Hyde routine at least once this season? If Victor Oladipo and Cody Zeller are on their A-game, though, this Hoosiers team is awfully tough to beat.
I LOVED…. Tim Hardaway Jr.'s stroke. It's been a while since I've seen someone pure so many consecutive threes as he did against Ohio State on Tuesday. As LeBron pointed out in a tweet, it's pretty because the form is so consistent each time he rises up. His clutch marksmanship should have gotten the Wolverines a win at Wisconsin on Saturday, and it could be part of some fun nail-biters in March, too.
I LOVED…. the cat-and-mouse ending to the Kansas-Oklahoma game. Bill Self made the absolute right call switching to the triangle-and-two defense, which got the Jayhawks back into the game in the final minutes. He dared the Sooners to beat them from the perimeter, and they did just that by knocking down multiple three-balls in the closing minutes. Perfect strategy shift, but better response.
What if Lefty Driesell was given the Duke job?
What if Adolph Rupp had taken over for Duke in the mid-1970's?
The first question is fascinating. Driesell built Maryland, but Duke already had a history of success (only five teams beat the Blue Devils to the 1,000 win mark). Could he have taken the Blue Devils to similar heights (and lows)? Just how different would Duke's program be today if the (aptly described) "mercurial" Driesell ushered in the modern era instead of Coach K. Also, what would have happened to Mike Krzyzewski? Similar butterfly effects happen if Rupp takes over. The article also has historical anecdotes about the dominance of the Durham YMCA in the 1920's. Seriously, give it a read.
ESPN: Well, the inevitable has arrived. Despite not receiving bids from Madison Square Garden or the Barclays Center in New York City, "because of the league's changing membership," those two arenas will still be in the running for the 2016-2021 ACC Tournaments. The move makes sense, but it has the potential to be a major flop too. The atmosphere at the ACC Tournament the past few years hasn't been the same. The declining excitement is largely thanks to an increase in noncompetitive teams, the addition of Thursday and an expanding geographic footprint. Moving the tournament to New York could exacerbate the issues if the league continues to aim for a balanced allotment of tickets.
ACC Sports Journal: The ACC is slowly rebuilding. Almost all programs appear to be moving in the right direction, though there are still plenty of questions surrounding almost all of the new coaches: Can Jim Larranaga and Steve Donahue recruit at the ACC level consistently? Can Brian Gregory and Brad Brownell break through to the next level? And can Jeff Bzdelik and Donahue pull their teams out of the cellar? The next couple of seasons are critical to the success of the ACC going forward because coaching stability is a huge factor in sustained success.
Raleigh News & Observer: NC State took a gut-punch against Miami without junior guard Lorenzo Brown. The Wolfpack controlled for most of the game, but a late Miami run and some costly errors from CJ Leslie (missed foul shots, turnovers, and dumb fouls) gave the Hurricanes the chance to win. But two stories more important than Reggie Johnson's buzzer-beating tip are starting to show through the game. For one, Miami is a solid two games ahead of Duke in the loss column (everyone else has three or more losses). That's a very, very good place to be going into the second half of conference play. Second, Tyler Lewis finally started showing why he was a McDonald's All-American. Lewis ran NC State's offense very well against the best defense in the ACC, and he didn't look nearly as lost on defense. He still needs some work, but developing Lewis is crucial in the long run.
Atlanta Journal-Constitution: Georgia Tech was a different team Sunday than the one that got smacked in Charlottesville (to be fair the home-road splits are looking fairly dramatic for Virginia too). The Yellow Jackets looked like they might be due for a repeat of their last game with the Cavaliers as they went into the half down by nine. Brian Gregory said after the loss that his team needed to learn how to finish. Well, the second time around they did just that. Georgia Tech held Virginia to six points in the final 9:40 of the game. The Yellow Jackets were the first ACC team to drop 60 on Virginia. Good luck ranking the middle and bottom of the ACC this season. It's a train-wreck, though it's a train-wreck played at a higher level than last year.


Cambridge, Mass. — Engineers from Harvard University have designed and demonstrated ice-free nanostructured materials that literally repel water droplets before they even have the chance to freeze.
The finding, reported online in ACS Nano on November 9, could lead to a new way to keep airplane wings, buildings, powerlines, and even entire highways free of ice during the worst winter weather. Moreover, integrating anti-ice technology right into a material is more efficient and sustainable than conventional solutions like chemical sprays, salt, and heating.
A team led by Joanna Aizenberg, Amy Smith Berylson Professor of Materials Science at the Harvard School of Engineering and Applied Sciences (SEAS) and a Core Member of the Wyss Institute for Biologically Inspired Engineering at Harvard, focused on preventing rather than fighting ice buildup.
For initial inspiration, the researchers turned to some elegant solutions seen in nature. For example, mosquitos can defog their eyes, and water striders can keep their legs dry thanks to an array of tiny bristles that repel droplets by reducing the surface area each one encounters.
To gain a detailed understanding of the process, the researchers watched high-speed videos of supercooled droplets hitting surfaces that were modeled after those found in nature. They saw that when a cold droplet hits the nanostructured surface, it first spreads out, but then the process runs in reverse: the droplet retracts to a spherical shape and bounces back off the surface before ever having a chance to freeze.
By contrast, on a smooth surface without the structured properties, a droplet remains spread out and eventually freezes.
"We fabricated surfaces with various geometries and feature sizes—bristles, blades, and interconnected patterns such as honeycombs and bricks—to test and understand parameters critical for optimization," says Lidiya Mishchenko, a graduate student in Aizenberg's lab and first author of the paper.
The use of such precisely engineered materials enabled the researchers to model the dynamic behavior of impacting droplets at an amazing level of detail, leading them to create a better design for ice-preventing materials.
Another important benefit of testing a wide variety of structures, Mishchenko adds, was that it allowed the team to optimize for pressure-stability. They discovered that the structures composed of interconnected patterns were ideally suited for stable, liquid-repelling surfaces that can withstand high-impact droplet collisions, such as those encountered in driving rain or by planes in flight.
The nanostructured materials prevent the formation of ice even down to temperatures as low as –25 to –30 degrees Celsius. Below that, due to the reduced contact area that prevents the droplets from fully wetting the surface, any ice that forms does not adhere well and is much easier to remove than the stubborn sheets that can form on flat surfaces.
In comparison with traditional ice prevention or removal methods like salting or heating, the nanostructured materials approach is efficient, non-toxic, and environmentally friendly. Further, when chemicals are used to de-ice a plane, for example, they can be washed away into the environment and their disposal must be carefully monitored. Similarly, salt on roads can lead to corrosion and run-off problems in local water sources.
The researchers anticipate that with their improved understanding of the ice forming process, a new type of coating integrated directly into a variety of materials could soon be developed and commercialized.
In addition to Aizenberg, who is also the Susan S. and Kenneth L. Wallach Professor at the Radcliffe Institute for Advanced Study and a Professor of Chemistry and Chemical Biology at Harvard, and Mishchenko, the co-authors of the paper included Benjamin Hatton and Vaibhav Bahadur, both at SEAS and Wyss, and Ashley Taylor and Tom Krupenkin, both at the University of Wisconsin-Madison.
The researchers acknowledge L. Stirling and A. Grinthal for their valuable contribution and funding from DARPA (Award Number HR0011-08-C-0114); the Wyss Institute for Biologically Inspired Engineering at Harvard University; and the U.S. Department of Homeland Security (DHS) Scholarship and Fellowship Program.
Note: High-resolution images and videos are available upon request.


text:
Between 2011
and 2014, the
Lab traveled
the Lab?
A mobile laboratory
focused on urban life
The Lab opened in
New York, Berlin, and Mumbai
Happening?
about the BMW
Guggenheim Lab
What Is the Lab?
Mumbai Lab Team
Mumbai Lab Collaborators
Mumbai Program Consultants
Berlin Lab Team
José Gómez-Márquez
Carlo Ratti
Corinne Rose
Rachel Smith
Berlin Lab Collaborators
Berlin Program Consultants
New York Lab Team
New York Lab Collaborators
New York Program Consultants
Curatorial Team
Former Curator
Lab | Log
Lab Learnings
José Gómez-Márquez is interested in innovative uses of technology to empower citizens in the process of actively making. Following a DIY philosophy, Gómez-Márquez's programs will focus on the tangible aspects of building your own city and hacking space through the use of tools for rapid prototyping and creative construction.
José Gómez-Márquez is the program director for the Innovations in International Health (IIH) initiative at the Massachusetts Institute of Technology (MIT) and cofounder of LDTC+Labs, a design and strategy consultancy for international development technology. Among the projects under his technology practice at IIH is the Aerovax Drug Delivery System, a device for mass delivery of inhalable drugs and vaccines to remote populations. His IIH invention portfolio also includes SafePilot, a wireless cane for the blind, and most recently, the X out TB program, which aims to increase tuberculosis-therapy adherence in developing countries using novel diagnostics and mobile technology. Recently, the group has developed the MEDIKit, a series of design building blocks that empower doctors and nurses in developing countries to invent medical technologies. The lab's work has been profiled in Discover, Wired, and The Economist. Gómez-Márquez is also an instructor of MIT's D-Lab: Health, a course on designing global health technologies.
Gómez-Márquez serves on the European Union's Science Against Poverty Initiative Task Force and has participated as an expert advisor on the U.S. President's Council of Advisors on Science and Technology. He is a three-time MIT IDEAS Competition winner, with two Lemelson Awards for International Technology. In 2009 Technology Review included Gómez-Márquez on its "TR35" list of innovators under 35 and named him Humanitarian of the Year. In 2011 he was selected as a TEDGlobal Fellow. Gómez-Márquez arrived in the United States from his native Honduras on a Rotary scholarship in 1997. After working in institutional investments and international development, he studied at the Worcester Polytechnic Institute, where he focused on policy research studies covering international technology transfer and small-team innovation. He currently lives in Cambridge, Massachusetts.
Photo: TEDGlobal 2011
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...minus the chocolate, of course! I am only slightly concerned that nobody has asked me what kind of cupcakes I do want, or who should be invited to the party. I hope you all got invites!
Happy birthday sweet adorable Foz!!!!!! I hope you get lots of pressies and cupcakes!
Happy Birthday, Foz! Enjoy those (non-chocolate) cupcakes!
happy birthday to Foz. maybe birthday biscuits are a better choice. Have fun!
Happy birthday, little Foz! I have a recipe for doggy biscuits that you might like.
YAY FOZZIE'S BIRTHDAY! Yet another reason to eat carbs!!


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The gym measurements feature allows you to track and trend a member's measurements very easily. Usually the reason why someone joins a gym or membership facility is for them to to meet a certain goal or get better at something. With the measurements software feature, your gym or membership facility has the ability to track the progress of your members goals easily and efficiently. The software does not have limits on what you can track. It allows you to add as many categories as you need. For example, the software allows you to track categories like weight loss, weight gain, body mass index and blood pressure. These are just a few common categories, but you can add whatever your gym or membership facility needs to track.
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The Team Australian Customer Service outbound communication solution is one that starts with customer CRM databases, and highly sophisticated SMS communication software. Scripts and SMS messages are written to achieve the required outcomes jointly between client or agency and TACS with calls or SMS made by a dedicated trained team at the Team Australian Customer Service Sydney call centre.
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Tag Archives: Harry Potter And The Deathly Hallows: Part I
The Harry Potter Retrospective – The Films
Posted on July 20, 2011 by mildconcern
You may remember that last month we spent two nights at the BFI IMAX watching all seven Harry Potter films over the course of two nights. (Thanks BFI IMAX!) We finished our journey through the franchise on Monday night as we watched and scored Harry Potter and the Deathly Hallows: Part 2. What follows is a run down of all eight films, written using the increasingly brief and incoherent notes we made at the time. Spoilers lie ahead.
Harry Potter and the Philosopher's Stone
In which Harry Potter learns he is a wizard, goes to Hogwarts School of Witchcraft and Wizardry and defeats a teacher harbouring the evil wizard Voldemort at the back of his head.
We start the franchise with an over-long film with terrible acting, odd prosthetics and scenes bordering on pantomime. From the initial scenes with the Dursleys playing out as a knockabout comedy to the final showdown in which a man completely disintegrates, Christopher Columbus produced a completely uneven film which relies mostly on reaction shots for laughs. Horrible acting from the kids is made up for by sheer cuteness and ultimately the film is a bit better than you remember. 6/10
The series continues with Harry hearing voices and writing in a diary which writes back. Turns out that pesky diary was Voldemort again.
As the kids seem to have experienced a growth spurt since the previous film they are less cute and their acting has improved slightly to compensate. Early scenes at the Burrows with the Weasley parents are great but even Julie Walters can't make exposition work properly. Jason Isaacs and Kenneth Branagh are pretty awesome but Christopher Columbus again fails to make anything remarkable happen. With students being attacked (but surviving) the series begins its journey into becoming "dark". 6.5/10
Posted in Features | Tagged alan rickman, Alfonso Cuarón, Christopher Columbus, Daniel Radcliffe, david yates, emma watson, Gary Oldman, Harry Potter, Harry Potter and the Deathly Hallows: Part 2, Harry Potter And The Deathly Hallows: Part I, Imelda Staunton, Jason Isaacs, julie walters, Kenneth Branagh, Robert Patterson, rupert grint | 1 Comment
2010's Best 20 Films
Posted on January 4, 2011 by mildconcern
Bank holidays are gone for a while so it's time to look back at 2010, and then maybe look forward at 2011. I messed about with this list a lot, made some graphics, lost them and decided to simply list the films without too much flair.
These may not be the top 20 films I've seen this year but are the top 20 that have been released in cinemas. I think I may have already seen some of the best of 2011.
20. Harry Potter And The Deathly Hallows: Part I
This slot is more of a token than a definitive "Top 20 Film" as Deathly Hallows deserves to be highlighted for making the Potter series credible at last. It's the film equivalent of Doctor Who, another British series to finally finding its feet in 2010. The kids can act and the plot is finally given time to unfold. Lovely.
19. Greenberg
Ben Stiller in an understated role in a film that bridged the gap between mumblecore and mainstream should get a bit of respect. It's not going to blow you away but will certainly help heal the wounds left by the Focker franchise.
18. Piranha
The only 3D film to make it onto the list, because it didn't take itself seriously and brought plenty of laughs, gore and boobs to the big screen. A sequel is on its way and I hope James Cameron gets invited to the premiere.
17. World's Greatest Dad
I can't help but love the unseen underdog, especially one featuring a comic actor giving a subdued performance. Try to see this, the darkest comedy of 2010. A must for anyone curious to see the genie from Aladdin in the buff.
16. The Kids Are All Right
It's a comedy, it's a drama, it's a truly modern family! From laugh out loud funny to cry into your sleeve devastating, this is the film so many others want to be.
15. Eccentricities of a Blonde Haired Girl
Drawn in by an intriguing review, we were delighted by a hilarious and eccentric film about one young man's love for the girl he sees from his office window. Directed in a unique fashion by a 100 year old Portuguese film-maker, you won't see anything like it again.
14. Four Lions
A black comedy from Chris Morris was always going to be brilliant, though could possibly have been better. Still proud to see a British film this funny, brave and with no romance in sight.
13. Monsters
A subtle sci-fi with amazing effects on a shoe-string budget, and from the UK. Slightly shaky dialogue and inexperienced actors are easily masked by the sheer beauty of every shot, with the film climaxing visually in its final scene.
12. Buried
A wonderfully tense thriller all set in a tiny coffin. Ryan Reynolds proves himself as a proper actor and the ninety minutes fly by. It's enough to make Phone Booth look like a good idea. Bonus points for having a great poster.
11. Kick Ass
After the first viewing this could have been in the top three films of the year, but a second viewing reminded me of the annoying beginning and the less than perfect lead. Still a brilliant black comedy and awesome showcase for Chloe Moretz. Shame Scott Pilgrim came along and showed how much better a comic book film could be.
10. Please Give
I may regret giving this film such a high position but it was surprisingly enjoyable and genuinely emotional. Not a film for everyone but worth a look if you're into more subtle comedies that don't have happy endings.
Almost forgotten but a re-watch over Christmas brought this early gem back to my attention. Lots of good performances, an unpredictable plot and a little bit of heartbreak thrown in. If only Clooney hadn't gone on to do The American.
8. Catfish
The most surprising film of the year and for a moment the most terrifying. If you still haven't seen it and have managed to remain unspoiled then get watching, you'll never add someone on Facebook again.
7. Black Dynamite
Finally another decent parody, burying the like of Date Movie simply by actually being funny. Were it not for a few familiar faces this could have come straight from the era it is parodying as the aesthetic is pitch perfect.
6. Let Me In
The highest horror on the list is a truly classy example of the genre. Not relying on fake scares, nudity or gore, we instead have a character study filled with tragedy, heart and intrigue. Chloe Moretz puts in her second performance on this list, a totally different but equally well crafted character. Arguably better than the original, this is a tale of love and horror.
5. Another Year
The most genuine film of the lot, so easily relatable and filled with characters you feel you could meet out here in the real world. Lesley Manville comes out of nowhere with a powerhouse of a performance. Heartbreaking and hilarious, a bit of a theme in 2010.
Lower down the list than I expected, and I'm the one who put the list together. Inception is as good as an action film can get as the usual set pieces are presented in wonderful new ways, the audience is made to think and small indie actors get their chance at the mainstream.
3. Whip It
Ranking so high for the sheer delight the film brings and the way it was overlooked by most people. The ugly poster hid away what is this year's ultimate feel good film. Ellen Page is doing something right.
2. The Social Network
Aaron Sorkin wrote the tightest script to be released in 2010 and was lucky to have it so well taken care of. Eisenberg and Garfield made the world take notice and even Timberlake managed to impress. Not so much a tale of how Facebook came to be but a look at friendship and the price of success.
1. Scott Pilgrim Vs. The World
No surprise here considering the sheer number of times we've had to use the "Scott Pilgrim" tag this year. Almost perfect and equally enjoyable at every repeat viewing. Never before has a film been so carefully crafted, every cut planned years in advance and every eye movement accompanied by a subtle noise. And how this is not long-listed for best original song I have no idea. Edgar Wright has shown just what he can do when given the freedom to run wild and the cinema audiences of the world have shown why we don't get many good films as the weakest of romantic comedies could beat it at the box office. For me 2010 will always be the year Scott Pilgrim came out and I'll be re-watching long after 3D has it's next revival.
So there you have it, my top 20 picks. Do you agree?
Posted in Features | Tagged another year, black dynamite, buried, Catfish, eccentricities of a blonde-haired girl, four lions, greenberg, Harry Potter And The Deathly Hallows: Part I, Inception, Kick Ass, let me in, monsters, piranha, please give, Scott Pilgrim Vs. The World, the kids are all right, The Social Network, up in the air, whip it, world's greatest dad | 2 Comments


This. is. my. favorite. festival. ever.
Comfest (Community Festival) is an annual holiday festival in Columbus that promotes human and equal rights for all, environmentalism and group participation in building communities. It is a three-day festival that takes place in Goodale Park.
Have I mentioned that this is my favorite festival ever?
This year, like last, I shared a tent with Adornments by Karla and Kathryn of Arteest. This year we also had Amy from Painted House and More and Melanie from Mel's Clay. We had two side-by-side spots, which we turned into one large joint space, and proceeded to have the single most fun weekend of the year.
Comfest is also my biggest show of the year, running three days and getting thousands upon thousands of people attending. But the best part is always the PEOPLE. People after people: meeting new ones who like my work, seeing old friends, and having a beer with them all! There is nothing that beats that feeling. Spending warm summer days outside, talking about art, having drinks and delicious food, and meeting hundreds of people who love Columbus just as much as I do... it truly is my favorite weekend of the year. I can't wait to do it all again!


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text:
Athletic Giving
TRITON GEAR
Thanksgiving Classic
Dalayna
Sampton
Experience: 1 Year
Among the top players in UC San Diego program history, Dalayna Sampton was added to the coaching staff as an assistant coach prior to the 2018-19 season. Sampton primarily works with the Triton post players, while also having a significant role in handling social media, community service outreach, scouting and recruiting.
Sampton had a remarkable two-year career with the Tritons after transferring from the University of Hawai'i. She was named California Collegiate Athletic Association (CCAA) Newcomer of the Year in her first season in La Jolla, before earning CCAA Player of the Year honors as a senior. In addition, Sampton received Women's Basketball Coaches Association (WBCA) All-America Honorable Mention, Division 2 Conference Commissioners Association (CCA) All-America Honorable Mention and D2CCA All-West Region First-Team accolades. She was also selected to the NCAA West Region All-Tournament Team as a senior and was one of 12 UC San Diego Athletes of the Year. In the 2017-18 season, Sampton averaged 14.6 points per game on 49.7 percent shooting from the field and ranked fifth in the nation in total rebounds (359) with 10.9 per contest. She scored a career-high 30 points with 11 rebounds in the final game of her UC San Diego career against Montana State Billings in the NCAA West Region semifinals.
A native of Vallejo, Sampton attended Florin High School in Sacramento and graduated in 2014.
Interview: Dalayna Sampton
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