text:
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Posted on August 14, 2017 August 14, 2017 by seademon
Lack of Sense, Weak Arguments, To Ask For Probe Into Zahid's Wealth
"I am willing to be investigated to correct the perception of that person,"said Zahid Hamidi on 4 Aug 2017
Extraordinary wealth owned by a public official should be enough grounds for investigations into the person, said a group of people who call themselves lawyers, in an apparent reference to a claim made by U-Turn Mahathir that Deputy Prime Minister Ahmad Zahid Hamidi should be investigated by the MACC for having RM230 million when he was the UMNO Youth Chief.
Malaysian Incite reported that a so-called lawyer and "one-sided activist" called Ambiga said that the Malaysian Anti-Corruption Commission (MACC) could only ascertain the legitimacy of a person's funds through an investigation.
"Any sign of extraordinary wealth particularly when it concerns people in public office is enough of a basis to investigate. Unfortunately Dzulkifli Ahmad has shown that he is afraid to investigate people at the top," she was reported to have said, referring to the MACC chief commissioner.
Dzulkifli Ahmad stated that the MACC will only investigate a person's wealth if there is a basis for the allegation that there are elements of corruption and abuse of power in his wealth.
He said the amount of money one person has is not a reason for the commission to start investigating him.
"The issue is whether the wealth is accumulated through corruption and abuse of power…that basis must be there for the commission to investigate. People can be rich, but it is important that we look at the source of the their wealth because if it is through corruption and abuse of power, then it will become the MACC's responsibility to investigate," he said when asked whether the MACC will investigate Ahmad Zahid Hamidi on the excessive wealth he allegedly owned.
In an obviously choreographed act, U-Turn Mahathir claimed that Ahmad Zahid had RM230 million in his bank account back in 1996, while DAP subsequently lodged a police report on the allegation.
Firstly, Ambiga claims that any extraordinary amount of wealth particularly when it concerns people in the public office is an enough basis for the MACC to investigate.
Senility runs deep in the Opposition's ranks apparently. Zahid became the Member of Parliament for Teluk Intan in 1995 and subsequently the UMNO Youth Chief in October 1996, a position he held until he was arrested under the Internal Security Act for going against U-Turn Mahathir in 1998.
In response to U-Turn Mahathir's claim, Zahid offered to be investigated.
"I was a businessman, a CEO of a listed company and chairman of other companies," Zahid said of his past.
Zahid worked in banks such as OCBC, then became the Marketing Manager for Amanah Saham Nasional Berhad (ASNB). Other portfolios that he held then included Executive Director for Scandinavian Motors Sdn Bhd, the CEO for Kretam Holdings Berhad, Chairman of several companies such as Tekala Corporation Berhad, Seng Hup Berhad, Ramatex Berhad, and Bank Simpanan Nasional from 1995 until U-Turn Mahathir incarcerated him in Kamunting without trial in 1998.
Zahid was then the UMNO Youth Chief and a Member of Parliament. No law states that one has to give up his/her business practice in either post, and neither post is a public office as per the Constitution. Unless Zahid abused his position then as a Member of Parliament (a member of a public body as defined in the MACC Act) to obtain those posts, I see no reason for trying to implicate him on the alleged matter. How can remunerations from companies to a Director, or Shareholder, or Officer, derived from the businesses these companies are involved in, be a form of gratification obtained by a member of a public body?
Unlike in the case of Tokong Lim Guan Eng who clearly used his position to obtain gratification be it for himself or for his wife, Zahid did nothing of the above as alleged.
The MACC Act clearly states that in spite of any written law or rule of law to the contrary, a MACC officer of the rank of Commissioner and above who has reasonable reason to believe based on the investigation of an MACC officer into properties that are held as a result of an offence committed under the MACC Act, may by writing as the person being investigated to furnish the MACC with a list of the properties obtained through gratification.
What is there to ask for if (1) no offence has been committed, (2) no investigation has been conducted based on (1)? Request on mere speculation?
This is why Section 36 of the MACC Act is as such, to prevent a waste of taxpayers' money and the MACC's precious little time going on wild goose chase. You may refer to Section 3 of the said Act to know what it means by 'gratification' when referring to someone holding a public office or is a member of a public body. Therefore, how can there be any form of double standard committed by the MACC as alleged by Eric Paulsen of the Bar Council?
Both Eric Paulsen and Ambiga claim that the MACC is not courageous nor willing to investigate the 'big fish' such as "even Najib Razak if necessary."
But like I said at the beginning of this post, senility runs deep within the ranks of the Opposition. The MACC quizzed Najib Razak on the RM2.6 billion donation back in December 2015.
I wonder why don't Ambiga and Eric Paulsen read their own favourite news portal?
Anyway, before I end this posting, I would like to share what Section 23 (1) of the MACC Act says about the use of public office or position for 'gratification' :
Section 23 (1) of the MACC Act, 2009
If I may, I would like to join Ambiga's call to ask MACC to investigate the 'big fish' who, in the words of Ambiga herself, continue to loot the nation and if I may add, or also looted the nation.
Ambiga said those who loot the nation must be investigated
Ambiga and Eric Paulsen should now agree with me that a report was made by Pribumi's members on the extraordinary wealth accumulated by U-Turn Mahathir when he was definitely a 'member of a public body.'
PPBM Youth members lodged a report against U-Turn Mahathir over his extraordinary wealth
Therefore, in accordance with Section 23 (1) of the MACC Act, 2009, I challenge both Ambiga and Eric Paulsen to make a public call for an investigation to be made into U-Turn Mahathir's abuse of his office and position to obtain a contract through direct negotiation (more like direct command) from Telekom Malaysia to his son's company when he was still the Prime Minister. And I challenge the MACC to commence its investigation into the KELEPETOCRACY by U-Turn Mahathir.
KELEPETOCRACY, Mahathir-style
If anyone says this was done before the MACC Act, 2009 came into force, let it be known that in the Anti Corruption Act, 1997, the act of seeking gratification by someone holding an office was given via Section 15(1) of the Act.
So there have you. Investigate U-Turn Mahathir and see if he is going to try add this to his EVERYTHINGTUNHIDE 2.0 Q&No-A sessions.
CategoriesDaily Whatevers TagsAhmad Zahid Hamidi, Ambiga, Corruption, DAP, Dzulkifli Ahmad, Eric Paulsen, MACC, Mahathir Mohamad, malaysia, Malaysian Anti Corruption Commission, Malaysian Insight, OPCOM, politicians, Politics, SPRM, Telekom Malaysia, UMNO
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Without further ado, welcome to the second year of my blog. It's moderately impressive that I've managed to keep updating it for this long.
I know I haven't been posting a whole lot lately, but in fact, the reason for this is because I've settled into a pattern where I only write if I know I have a good idea. Some of my older posts on here seem cringe-y even now because of that, but I won't be deleting them out of shame. It's fun to see how someone progresses in their craft.
So yeah. I'll just be talking about my world on here and how nice it can be. And I'll still write about anime at some points. Go carry on with your day now.
← Have You Considered… Birdy the Mighty: Decode?


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redpajama_set_name: RedPajamaCommonCrawl


A new website has been launched with information on the upcoming Global Ministry Forums, which begin next month. The site is www.globalministryforums.com and contains dates, an agenda, venue addresses, directions, and other pertinent information.
In response to action taken at the 2018 Church of God International Assembly last summer, the Church of God will be conducting a series of Global Ministry Forums. Fifteen will take place in the United States, and eight at locations around the globe, including the Caribbean, Europe, Africa, Central America and the Philippines.
That open Ministry Forums be conducted globally to provide opportunity for deliberate and meaningful discussion, dialogue, questions/answers and time for spiritual insight regarding the importance and understanding of ministry ranks, qualifications, and women in leadership positions with attention upon the meaning and usage of the title "bishop." Following the forums, appropriate motion(s) be formulated by the International Executive Council specifically addressing the stated issues and brought to the 2020 International General Council.
"These forums have been designed to be informative and challenging and we need everyone's input as we move into a new decade," stated Church of God General Overseer Dr. Timothy M. Hill. Hill went on to explain that the forums are for ministers and laypersons and will offer a glimpse into where the potential lies for the future of the Church of God headed into the 2020s and beyond.
One of the most important aspects of the site is the Forums Survey, which everyone is encouraged to either peruse or complete on-line prior to attending. This survey is for everyone – ministers and laypersons – and may be completed whether or not a person will attend one of the forums. On-site completion of the survey will also be available at each location.
The first forum is set to take place on February 7, 2019 at the Living Waters Church in Fountain Valley, California. Although dates and locations have been confirmed, they are subject to change. Addresses and other contact information for each location is listed on the site, www.globalministryforums.com and will reflect any changes should they arise.


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Motley Fool Australia » Dividend Shares » 9 whopping dividend yields to pounce on right now!
9 whopping dividend yields to pounce on right now!
Christopher Georges | March 20, 2017 | More on: AX1 DDR FET MQG MTR RFG VLW WAM
Dividend shares continue to be very popular amongst investors and this is unsurprising when you consider term deposits and most inner city investment properties are offering net income returns of less than 2.5%.
With interest rates expected to remain at historically low levels for some time to come, I think dividend shares are still a great option for risk-tolerant investors who want to generate market-leading returns over the medium term.
With that in mind, here are nine high-yielding shares that income-seeking investors might want to take a closer look at in 2017:
Company Market Cap P/E Ratio Dividend Yield Grossed Up Yield 5 Year Total Return
Macquarie Group Ltd (ASX: MQG)
$30.2 billion 14.6 4.7% 5.7% 31.7%
Mantra Group Ltd (ASX: MTR)
$800 million 15.7 4.4% 6.3% N/A
RCG Corporation Ltd (ASX: RCG)
$569 million 14.1 5.6% 8.0% 30.5%
Dicker Data Ltd (ASX: DDR)
$336 million 13.1 7.8% 11.1% 47.4%
Villa World Ltd (ASX: VLW)
$273 million 7.9 7.5% 10.7% 29.0%
Retail Food Group Limited (ASX: RFG)
Folkestone Education Trust (ASX: FET)
WAM Capital Limited (ASX: WAM)
$1.5 billion 11.9 6.2% 8.9% 17.0%
Vanguard Australian Shares High Yield ETF (ASX: VHY) $736 million 20.2 5.6% 7.4% 10.7%
Importantly, all the shares listed above have either flagged higher dividends over the coming year, or are in a strong position to increase their dividends over the next 12-18 months.
Along with the individual shares above, I have also highlighted a high-yielding exchange traded fund (ETF) operated by Vanguard. This is a simple and convenient option for those investors who may be more risk-averse, or do not have the resources to research individual shares.
However, one point to note about this ETF, and others like it, is that it is dominated by a relatively small number of shares with a fairly large weighting towards the big four banks, Telstra Corporation Ltd (ASX: TLS) and Wesfarmers Ltd (ASX: WES). This means investors can expect to receive a fairly reliable dividend source, albeit at a pretty low growth rate.
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Q: "This video shows a heart transplant take place" or "taking place"? Which is the correct version of this sentence:

*

*This video shows a heart transplant take place.

*This video shows a heart transplant taking place.

I have a hunch that both are correct, but for the life of me I can't recall in grammatical terms why the first option is correct also.

A: The correct version is taking place, because in the video, the event is continually taking place.
To show "a heart surgery take place" is incorrect, though you might also say "a heart surgery that took place," because if there is a video of it, it clearly happened already.

A: Handcuffed by the question, I have to say 'This video shows a heart transplant taking place.'It's correct but more elegant to say 'this video shows a heart transplant operation'. Sentences with cumbersome tenses and involved clauses beginning with 'that', 'which', 'what' and so on, I try to avoid in the interests of brevity, clarity and holding readers' interest.

redpajama_set_name: RedPajamaStackExchange


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\section{Introduction}

The F-KPP equation
\begin{equation}
 \label{KPP
\begin{array}{l}
\displaystyle {\partial_t u}= \frac{1}{2}  \displaystyle {\partial^2_{x} u} + u^2 -u, \quad x \in \mathbb{R},\,
t>0,\\
\\
\displaystyle u(0,x)= \displaystyle u_0(x), \quad x \in \mathbb{R}.
\end{array}
\end{equation}
was first introduced as a central model of front propagation for reaction-diffusion phenomena.
Both Fisher and Kolmogorov, Petrovskii and Piskunov \cite{Fisher, KPP} proved independently
the existence of an infinite number of traveling wave solutions.
In 1975, McKean established a first link with a microscopic model (a particle system),
showing an exact connection with Branching Brownian motion (BBM) \cite{MK}.
This sprouted a large effort of research focusing on defining corresponding microscopic models
for front propagation both in the physics and mathematics literature.
The seminal papers \cite{BD2,BD} set the basic questions on the influence of microscopic effects on the front propagation properties.
Three types of models were then considered: adding noise to the deterministic equation \cite{BDMM, MMQ, MMQ2}, introducing a cut-off \cite{BenDep, DP} or defining truly microscopic models via interacting particles systems conserving the total number of particles \cite{BDMM2,DR,M,deMasi}.
This allows to study the effects produced by the finite size population on the front propagation.
We focus here on the third direction of research.

Brunet and Derrida \cite{BD} highlighted that in contrast to the macroscopic equations, which admit infinitely many traveling waves (characterized by their velocities), microscopic models should select a unique velocity.
Defining finite population particle systems, of size $N$, with branching and selection (in discrete time), they showed using heuristic arguments and simulations that the asymptotic speed of a particle system has a deviation of order $(\log N)^{-2}$ from the macroscopic speed. This sheds light on the effects of the finite size population, as the corresponding speed converges much slower than expected.
This was rigorously proved in \cite{BG} for $N$-Branching Random Walks ($N$-BRW), a system in which particles branch, perform random walks and undergo a selection mechanism.

Several similar particle systems involving diffusive movements and sequential steps of branching and selection have been considered in this context.
In \cite{DR} the authors consider a model closely related to $N-$BRW and prove that the distribution function of the empirical measure converges, for fixed times, to a free boundary integro-differential equation in the F-KPP class, as well as a speed selection phenomenon.
The same conclusions were conjectured to hold in \cite{GJ2} for the $N-$Branching Brownian Motion ($N$-BBM), introduced and studied in \cite{M}. These conclusions have been recently settled in \cite{BBP18, deMasi}. More on this macroscopic equation can be found in \cite{BBD17}.

In the present article, we define and study a finite-population particle system in continuous time, following closely the dynamics proposed by Brunet and Derrida in \cite{BD}.
In our setting $N$ particles diffuse independently according to one dimensional Brownian motion, branch and get selected, with the important feature that particles are paired at each branching event.
At these times, the particle on the pair with largest position  branches into two and the one (in the pair) with smallest position is eliminated from the system.
This corresponds to the branching-selection mechanism.

We first prove that the cumulative distribution function of the empirical measure of this system converges to a solution of the F-KPP equation as the number of particles goes to infinity. As a consequence, we are able to prove that the system selects the minimal speed of the macroscopic model as $N$ goes to infinity.
In order to do that, we first show the existence of a stationary regime for the system seen from its leftmost particle
and the existence of an asymptotic speed for fixed $N$, using classical arguments. Combining this with the hydrodynamic limit previously obtained, we get a lower bound for the speed.
The upper bound can be obtained by constructing our process as a pruning of $N$ independent BBMs.

Some particular features of our setting, in contrast with previously mentioned works are
(i) the proofs do not depend on explicit computations but rather on a careful analysis of the system (for instance, the limiting speed is obtained through comparison with F-KPP solutions and not through Legendre transforms),
(ii) we provide explicit bounds for the decorrelation between particles and hence we can quantify the propagation of chaos,
(iii) we believe that the analysis is robust enough to be extended to more general diffusions and branching-selection mechanisms (see \cite{GSL}) and
(iv) the behavior of the system as $N$ goes to infinity is given exactly by the solution of the F-KPP an not by approximations.

The rest of the paper is organized as follows.
In Section \ref{sec:mod}, we introduce the model and state the main results.
The proof of propagation of chaos and convergence towards the F-KPP is given in Section \ref{sec:proof1}.
Finally in Section \ref{sec:proof3}, we prove the existence and properties of the velocities for fixed $N$ and then proceed to prove the {\em selection principle}: as $N$ goes to infinity, the velocities converge to the minimal velocity of the F-KPP equation.

\section{Model and main results}\label{sec:mod}
We first describe the dynamics in an informal way.
The system starts with $N$ particles located in the real line, each of which performs independently a standard Brownian motion and carries a Poisson clock with rate one.
At Poissonian times, the particle chooses another particle uniformly among the other ones. If the choosen particle has larger position, it jumps on top of it. Throughout the article we will refer indistinctly to this interaction as a \emph{jump} of the smaller particle of the pair to the position of the larger one or as a \emph{branching} of the larger particle with killing of the smaller one. More precisely $(\xi_t)_{t\ge0}$ is a Markov process in $\mathbb{R}^N$ with generator given by
\begin{equation}
\label{generator}
\L f(\xi)=\frac{1}{2} \sum_{i=1}^{N} \partial^2_{x_i} f(\xi) + \frac{1}{2 (N-1)} \sum_{i=1}^{N} \sum_{j \neq i}
\left(f(\theta_{ij}(\xi))-f(\xi)\right),
\end{equation}
for $f \in {C}^2_0(\mathbb{R}^N)$, the set of twice differentiable functions with compact support. Here
\begin{equation}
\label{teta}
\theta_{ij}(\xi)(k)=
\begin{cases}
\max\{\xi(j), \xi(i)\} & \text{ if $k\in \{i,j\}$},\\
\xi(k) & \text{ otherwise}.
\end{cases}
\end{equation}
Namely, given two labels $i,j$, $\theta_{ij}$ replaces the position of the particle with smaller position with the position of the larger one.

We denote by $\xig{}{t}=(\xi_t(1), \ldots, \xi_t(N))$ the process at time $t$ and define the  \emph{empirical measure} associated to $\xig{}{t}$ by
\begin{equation*}
\mu^N_t:= \frac{1}{N} \sum_{i=1}^{N} \delta_{\xi_t(i)}, \qquad t>0.
\end{equation*}
The \emph{cumulative distribution function} of $\mu^N_t$ is given by,
\begin{equation*}
F_N(x,t):= \mu^N_t((-\infty,x])	, \qquad x\in \mathbb{R},t>0.
\end{equation*}
For simplicity, we omit the dependence of $\mu^N_t$ and $F_N(x,t)$ on ${\xi}_t$ if not necessary.

\bigskip
\noindent
{\bf Notation.}
In what follows, we denote by $\mathcal{P}$ the space of probability measures on $\mathbb{R}$ endowed with the weak topology. For a polish metric space $S$, $D([0,T], S)$ is the space of c\`adl\`ag functions from $[0,T]$ to $S$ endowed with the Skorokhod topology.
We assume that all our processes are defined in a probability space $(\Omega, \mathcal F, P)$ and denote with $E$, the expectation with respect to $P$.
When needed, we use $P_\mu$ (respectively $P_{\xig{}{0}}$) whenever the initial condition is distributed at random according to the probability measure $\mu$ (respectively $\delta_{\xig{}{0}}$), with the same convention for expectation.
The $\sigma$-algebra generated by the process up to time $t$ is denoted by $\mathcal{F}_t$.
For simplicity, we write $u_N(x,t):=E_{\xig{}{0}}[F_N(x,t)]$ and $\| \cdot \| = \| \cdot \|_\infty$ for the infinity norm of a real valued function.

\subsection{Hydrodynamic limit}
The following theorem provides a link between the aforementioned model and the solutions of the F-KPP equation,
through the hydrodynamic limit.

\bigskip

\begin{theorem}
\label{mresult}
Let $u_0$ be a distribution function in $\mathbb{R}$
such that $\lim_{N\to\infty} \|F_N(\cdot,0)-u_0\| = 0$ in probability and $u$ the solution to \eqref{KPP}.
Then, for any $t>0$ we have
\begin{equation}
\label{convergence}
\lim_{N\to\infty}\|F_N(\cdot,t) - u(\cdot,t)\| = 0 \qquad \text{in probability}.
\end{equation}
\end{theorem}

{We can also state a path-wise version of this result.}

\begin{corollary}
\label{path}
In the setting of Theorem \ref{mresult}, let $\mu_t$ be the deterministic measure in $\mathbb{R}$ given by $\mu_t((-\infty,x])=u(x,t)$. Then,
\begin{equation*}
\lim_{N\to\infty}(\mu^N_t)_{t\ge 0}  = (\mu_t)_{t \ge 0},
\end{equation*}
weakly in $D([0,+\infty),\mathcal P)$, in probability.
\end{corollary}

\subsection{Speed selection} For $\xi \in \mathbb{R}^N$, denote $\xi[i]$ the $i$-th order statistic of $\xi$, i.e.
\[
 \xi[1]=\min_{1\le j \le N}\xi(j), \qquad \xi[i+1] = \min_{j\colon \xi(j)\ge\xi[i]} \xi(j).
\]
We use the label of the particles to break ties. The position of the $i$-th particle as seen from the minimum is given by $\etap{i}{t}=\osp{i+1}{t}-\osp{1}{t}$ and the process as seen from the minimum is defined by
\begin{equation}
\label{leftmost
\pfl=(\etap{1}{t},\ldots,\etap{N-1}{t}).
\end{equation}
\begin{theorem}
\label{finiteN}
Let $N \geq 1$. Then,
\begin{enumerate}
\item
The process $(\pfl)_{t \geq 0}$ has a unique stationary distribution $\nu^N$, which is absolutely continuous with respect to the $N$-dimensional Lebesgue measure.
\item
For any initial distribution $\rho^N_0$
$$\lim_{t\to \infty}\| P_{\rho^N_0}(\pfl \in \cdot)-\nu^N(\cdot)\|_{TV}= 0.$$
\item
There exists $v_N \geq 0$ such that
\begin{equation}
\label{velocityE
\lim_{t \to \infty} \frac{\osp{1}{t}}{t}=\lim_{t \to \infty} \frac{\osp{N}{t}}{t}=v_N, \quad \text{almost surely and in $L^1$.}
\end{equation}
\item $v_{N+1}\ge v_N$ and
\begin{equation}
\label{minimal}
\lim_{N \to \infty} v_N=\sqrt{2}.
\end{equation}
\end{enumerate}
\end{theorem}

{\bf Remark.}
Equation \eqref{minimal} can be thought as a \emph{selection principle} in the sense that though the macroscopic equation admits infinitely many travelling waves with different velocities, for each $N$ and for \emph{any} initial distribution the microscopic system has a \emph{unique} velocity which converges to the \emph{minimal velocity} of the macroscopic equation, when $N$ goes to infinity.
This kind of result was first proved in \cite{BCdMFLS}.

\section{Hydrodynamic limit}\label{sec:proof1}
We begin now with the proof of Theorem \ref{mresult}.
The main idea is to show that $u_N$ verifies, in the limit as $N$ goes to infinity, the F-KPP equation and that the variances $V_{\xig{}{0}}[F_N(x,t)]$ converge to zero, which boils down to prove asymptotic decorrelation of the particles (propagation of chaos).
The following graphical construction will turn to be instrumental for this purpose.

\subsection{Graphical construction}
\label{graphicalconstruction}
We construct the process as a deterministic function of $N$ independent Brownian motions and an homogeneous Poisson process (for the jumps).\\
To each particle $i$, we associate a marked Poisson process $\omega^i$ and an independent standard Brownian motion $(B^i_t)_{t\geq 0}$.
The process $\omega^i$ is defined on $\mathbb{R} \times (\{1,\ldots,N\} \backslash \{i\})$ with intensity measure $dt d\beta_i$, where $\beta_i$ is the uniform distribution on $\{1,\ldots,N\} \backslash \{i\}$. The processes $(\omega^i, (B^i_t)_{t \geq 0})_{1 \leq i \leq N}$ are independent.
Consider the superposition  $\omega=\bigcup \omega^i {\times \{i\}}  $ and sort the marks in order of appearance.
We denote this sequence with $(\tau_k,j_k,i_k)_{k \in \mathbb{N}}$ where the first coordinate $\tau$ is the {\em time mark}, the second one is the {\em partner mark} chosen according to $\beta_i$ and the third one is the {\em label mark} (the label mark is $i$ if it occurs at $\omega^i$). We define $\xig{}{t}$ inductively as follows.
\begin{itemize}
\item
At time $0$, the configuration is $ \xig{}{0}$.
\item
Assume $ \xig{}{\tau_k}$ is defined. For $t \in (\tau_k, \tau_{k+1}]$ we define
\begin{equation*}
\xip{l}{t}=
\begin{cases}
\xig{}{t^-}(j_{k+1}) & \text{ if $t=\tau_{k+1}$, $l=i_{k+1}$ and $\xig{}{t^-}(i_{k+1}) < \xig{}{t^-}(j_{k+1})$}\\
\xip{l}{\tau_{k}} + B^l_t - B^l_{\tau_k}  & \text{ otherwise.}
\end{cases}
\end{equation*}
\end{itemize}

It is straightforward to check that $( \xig{}{t})_{t \geq 0}$, as constructed above is Markov, with generator given by \eqref{generator}.

For each particle $i$ and a time $t>0$, we construct backwards a set of (labels of) particles that we call {\em ancestors}.
The important feature of this set is that the position of the particle is measurable with respect to the $\sigma$-algebra generated by the Brownian motions and Poisson processes \emph{attached to the ancestors up to time $t$}.
For any two particles, conditioned on the event that their {\em clan of ancestors} do not intersect, they are independent.
More details below.

\subsection{The clan of ancestors}
Given a set of labels $A \subseteq \{1,\ldots,N\}$ we denote by $\omega^A[0,s]$ the superposition of all the points of $\omega^j$ with $j \in A$ restricted to the interval $[0,s]$.
We add $(0,0,0)$ to this set for convenience. That is, $\omega^A[0,s]=\bigcup_{j\in A} \omega^j([0,s]) \bigcup \{(0,0,0)\}$.
For each $1\le i \le N$, we define inductively a sequence of times and sets as follows.
\begin{itemize}
\item Let $s_1$ be the largest time mark in $\omega^{\{i\}}[0,t]$ and $j_1$ its partner mark. If $s_1=0$ stop and define $\psi^i_t=\{i\}$. Otherwise define $A^i_1=\{i,j_1\}$.
\item
Suppose $s_k, A^i_k$ are defined.
Then, let
$s_{k+1}$ be the largest time in $\omega^{A^i_k}[0,s_k]$ and $j_{k+1}$ its partner mark. If $s_{k+1}=0$ stop and define $\psi^i_t=A^i_k$. Otherwise define $A^i_{k+1}=A^i_k \cup \{j_{k+1}\}$.
\end{itemize}
As $N$ is finite, this procedure finishes after a finite number of steps  $n^ i_t$  almost surely.

The next lemma gives a bound on the probability of intersection of two clans of ancestors which allows us to control the two-particles correlation.

\begin{lemma}
\label{influence
For $i \ne j$  and $t>0$,
\begin{equation*}
P_{{{\mu^N_0}}}(\psi^i_t\cap \psi^j_t \neq \varnothing) \leq \frac{e^{t}-1}{N-1}.
\end{equation*}
\end{lemma}

In the event $\{\psi^i_t \cap \psi^j_t= \varnothing\}$ the only possible dependence between $\xi^i_t$ and $\xi^j_t$ is due to the initial condition. As a consequence, for deterministic initial conditions, we obtain the following bound.

\begin{lemma} For $t\ge 0$ and $x,y \in \mathbb{R}$, we have
\label{Lemma:decorrelation}
\begin{equation*}
\sup_{ \xig{}{} \in \mathbb{R}^N} \left|\sum_{i,j=1}^{N} E_{ \xig{}{}}[\I{\xip{i}{t} \leq x}\I{\xip{j}{t} \leq y}]-E_{ \xig{}{}}[\I{\xip{i}{t}\leq x}]E_{\xig{}{}}[\I{\xip{j}{t}\leq y}] \right| \leq 2N e^{t}.
\end{equation*}
\end{lemma}
The proof of this lemma is similar to the one in \cite[Section 3]{AFG} once we have Lemma \ref{influence}.
The only minor modification is to consider the events $\{\xip{i}{t} \leq x, \xip{j}{t} \leq y\}$ instead of $\{\xip{i}{t}=x, \xip{j}{t}=y\}$.
For the proof of Lemma \ref{influence} we also refer the reader to \cite[Lemma 2.1]{AFG}.

The following lemma allows to control perturbations of the F-KPP equation.

\begin{lemma}
\label{COROCP}
Let $u$ be a solution of \eqref{KPP} with initial datum $u_0$ and $w$ a smooth bounded function in $\mathbb{R}\times [0,T]$ such that
\[
  \partial_t w = \frac12\partial^2_{x} w + w^2 - w + g, \qquad w(x,0)=w_0(x).
\]
Then
\begin{equation*}
 \| u(\cdot, t) - w(\cdot, t)\| \le \left (\|u_0 - w_0\| + \int_0^t e^{-s} \|g(\cdot, s)\| \, ds\right ) e^{{C}t} ,
\end{equation*}
{with $C=\max\{\|g\|T + \|w_0\|,1\}$.}
\end{lemma}

\begin{proof}
Call $\underline z=w-u$ and
\[
\overline z(x,t)= \left (\|u_0 - w_0\| + \int_0^t e^{-s} \|g(\cdot, s)\| \, ds\right ) e^{Ct}.
\]
By the maximum principle for parabolic equations in unbounded domains \cite[Proposition 52.4]{QS} we get $\|w\| \le C$ and hence, this function verifies
\begin{align*}
\partial_t \overline z =&\, \frac12\partial^2_{x} \overline z + C\overline z + \|g(\cdot,t)\|\\
\ge & \, \frac12\partial^2_{x} \overline z + (w+u-1) \overline z + \|g(\cdot,t)\|,
\end{align*}
with $\overline z(x,0)=\|u_0 - w_0\|$, while for $\underline z$ we have
\begin{align*}
\partial_t \underline z =&\, \frac12\partial^2_{x} \underline z + w^2-u^2 -  \underline z + g\\
= & \, \frac12\partial^2_{x} \underline z + (w+u-1) \underline z + g.
\end{align*}
Hence $e=\underline z - \overline z$ verifies
\[
\partial_t e  \le \frac12\partial^2_{x} e + (w+u-1)e.\\
\]
With $e(x,0)=w_0(x)-u_0(x) - \|u_0 - w_0\| \le 0$. Since $u$ and $w$ are bounded, using again the maximum principle we obtain $e(x,t)\le 0$ for all $(x,t) \in \mathbb{R}\times[0,T]$. Proceeding in the same way with $-\overline z - \underline z$ we get that for all $(x,t) \in \mathbb{R}\times[0,T]$, $-\overline z(x,t) \le  w(x,t) - u(x,t) \le \overline z(x,t)$, which conludes the proof.
\end{proof}

We are ready to prove Theorem \ref{mresult}.

\begin{proof}[Proof of Theorem \ref{mresult}]
Throughout this proof, we denote $E_{\xig{}{0}}, V_{ \xig{}{0}}$ and $P_{ \xig{}{0}}$ briefly by $E,V$ and $P$, respectively. We also use $\mathcal{G}(x,t;z)=\int_{-\infty}^x (2\pi t)^{-1/2}e^{-(y-z)^2/2t}\, dy$ for the probability that a Brownian motion started at $z$ is less than $x$ at time $t$.
The proof is divided into two steps.

\bigskip
\noindent
{\em Step 1.} Derivation of the equation for $u_N(x,t)$.

\medskip
Let us first observe that
\begin{equation}
\label{EP}
u_N(x,t)=\frac{ 1}{N} \sum_{i=1}^N P(\xip{i}{t} \leq x).
\end{equation}
We will focus on each term of the previous sum separately.
Fix $i$ and let $\tau$ be the {last time-mark before time $t$ of $\omega^i$}.
If there is no such a mark, we set $\tau=0$.
Notice that particle $i$ performs a Brownian motion in $(\tau,t)$ and that $\tau$ is the minimum between $t$ and an exponential random variable with mean $1$.
Thus, by conditioning on $\tau$ we get
\begin{align*}
P(\xip{i}{t}\leq x)
&= e^{-t} \mathcal{G}(x,t;\xip{i}{0}) +
\int_{0}^{t} e^{-s} E[\mathcal{G}(x,t-s;\xip{i}{s})| \tau=s] ds.
\end{align*}
Observe that for any $z$, $\mathcal{G}(\cdot, \cdot;z)$ verifies the heat equation.
Hence, for every $1\le i \le N$, the function $q_i(x,t):=P(\xip{i}{t}\leq x)$ verifies,
\begin{equation}
\label{Efep}
\partial_t q_i= \frac12\partial^2_{x} q_i - q_i + P(\xip{i}{t}\leq x |\tau=t),
\end{equation}
We proceed to examine the last term in \eqref{Efep}.
Let $J_{ij}$ be the event that the partner mark is $j$.
Conditioning on $J_{ij}$ we get
\begin{align}
\nonumber
P(\xip{i}{t}\leq x |\tau=t)&=
\sum_{j \neq i} P \left(\max(\xip{i}{t^-},\xip{j}{t^-})\leq x |\tau=t,J_{ij} \right) P(J_{ij}|\tau=t)\\
\nonumber &= \frac{1}{N-1} E \left ( \sum_{j \neq i} \I{\xip{i}{t}\leq x}\I{\xip{j}{t}\leq x} \right )\\
\label{vt}
&=\frac{1}{N-1} E \left ( \sum_{j=1}^N \I{\xip{i}{t}\leq x}\I{\xip{j}{t}\leq x} - \I{\xip{i}{t}\leq x}\right ).
\end{align}
Since $F_N^2(x,t)=\frac{1}{N^2} \sum_{i,j=1}^{N} \I{\xip{i}{t}\leq x}\I{\xip{j}{t}\leq x}$, combining \eqref{EP},\eqref{Efep} and \eqref{vt} we get
\begin{align}
\nonumber
\partial_t u_N &= \partial^2_{x} u_N - \tfrac{N}{N-1} E[F_N (1-F_N)]\\
\label{KPPwe}
&= \partial^2_{x} u_N - u_N (1-u_N) + \tfrac{N}{N-1} V(F_N(x,t)) + \tfrac{1}{N-1} (u_N (1-u_N)).
\end{align}

Next we prove that solutions of  \eqref{KPPwe} converge to solutions of \eqref{KPP} when $N$ tends to infinity.

\bigskip

\noindent
{\em Step 2.} Control of the variance and stability of the F-KPP.

\medskip
For $\varepsilon>0$ we have,
\begin{align}
\label{cheby}
\nonumber
P_{\rho^N_0}(|F_N(x,t)& -u(x,t)|> \varepsilon)
= \int P_{\xig{}{0}} (|F_N(x,t)-u(x,t)|> \varepsilon) \ \rho^N_0(d \xig{}{0})\\
& \nonumber
\leq \int P_{ \xig{}{0}}(|F_N(x,t)-u_N(x,t)|> \tfrac{\varepsilon}{2}) +
P_{ \xig{}{0}}(|u_N(x,t)-u(x,t)|> \tfrac{\varepsilon}{2}) \ \rho^N_0(d \xig{}{0})\\
&\leq \int \frac{4}{\varepsilon^2} V_{\xig{}{0}}(F_N(x,t)) + P_{\xig{}{0}}(\|u_N(\cdot,t)-u(\cdot,t)\|> \tfrac{\varepsilon}{2}) \ \rho^N_0(d \xig{}{0}).
\end{align}
By Lemma \ref{Lemma:decorrelation}, we get that
\begin{equation}
\label{v-d}
\sup_{ \xig{}{0}} V_{ \xig{}{0}}(F_N(x,t)) \leq \tfrac{2}{N} e^T.
\end{equation}
For the second term in the integral in \eqref{cheby} we apply {Lemma} \ref{COROCP} with $w=u_N$ and $g=\tfrac{N}{N-1} V(F_N(x,t)) + \tfrac{1}{N-1} (u_N (1-u_N))$ to obtain
\begin{align*}
\nonumber
\|u_N(\cdot,t)-u(\cdot,t)\| &\leq \left(\|F_N(\cdot, 0) - u_0(\cdot)\| + \int_0^t e^{-s} \|g(\cdot, s)\| \, ds\right) e^{Ct}\\
& \leq \left(\|F_N(\cdot, 0) - u_0(\cdot)\| + \tfrac{1}{N-1} (1+ 2 e^{T}) \right ) e^{CT}.
\end{align*}
In the last inequality we use that $0 \leq u_N \leq 1$ and \eqref{v-d}.
Taking $N$ such that ${C<2}$ and $\tfrac{1}{N-1} (1+ 2 e^{T}){e^{2T}}<\tfrac{\varepsilon}{4}$ and using the two previous observations we can bound \eqref{cheby} by
\begin{equation}
\label{finalcheby}
\frac{8}{N \varepsilon^2} e^T + \rho_0^N(\| F_N(\cdot, 0) - u_0(\cdot) \|> \tfrac{\varepsilon}{4} e^{-CT}),
\end{equation}
and hence, due to our assumption on $\rho^N_0$, for every $x\in \mathbb{R}$ and $t>0$, $F_N(x,t) \to u(x,t)$ in probability. Since $u(\cdot,t)$ is a continuous distribution function, the convergence holds uniformly \cite[Remark 5.28]{GeorgiiSto}, which proves \eqref{convergence}.

Note also that the upper bound \eqref{finalcheby} is independent of $x$ and $t$, hence the convergence in probability holds uniformly in $[0,T]$.
\end{proof}

{\em Proof of Corollary \ref{path}.}
The main ingredient to obtain Corollary \ref{path} is to show that for every $T>0$, the sequence $(\mu^N_t)_{0\le t\le T}$ is tight in $D([0,T],\mathcal P)$.
Afterwards, Theorem \ref{mresult} implies that there is a unique limit point and hence the desired result.
\bigskip

\noindent
{\em Tightness.} \
To prove tightness of $(\mu^N_t)_{0 \leq t \leq T}$ it is enough to show that for any continuous and bounded function $\varphi$ on $\mathbb{R}$ the sequence of real-valued processes $(\langle \mu^N_t,\varphi\rangle)_{0 \leq t \leq T}$, with  $\langle \mu^N_t,\varphi\rangle=\int_{-\infty}^{+\infty} \varphi(x) \mu^N_t(dx)$, is tight in $D([0,T], \mathbb{R})$ \cite[Theorem 16.16]{kallenberg}.

According to Aldous criterion \cite[Theorem 16.11  and Lemma 16.12]{kallenberg} the following two conditions are enough to get tightness for the previous sequence:
\begin{enumerate}
\item
For every $t \in [0,T]\cap\mathbb{Q}$ and every $\varepsilon>0$, there is an $L>0$ such that
$$\sup_{N>0} P_{}(|\langle \mu^N_t, \varphi \rangle|>L)\leq \varepsilon.$$
\item
Let $\mathfrak T^N_T$ be the collection of stopping times with respect to the natural filtration associated to $\langle\mu^N_t,\varphi\rangle$ that are almost surely bounded by $T$. For every $\varepsilon>0$
$$\lim_{r \to 0} \limsup_{N \to \infty}\sup_{\stackrel{\kappa \in \mathfrak T^N_T}{s<r}} P_{}\Big(|\langle \mu^N_{(\kappa+s)\wedge T}, \varphi \rangle - \langle \mu^N_\kappa, \varphi \rangle|>\varepsilon \Big)=0.$$
\end{enumerate}

The first condition follows immediately by taking $L> \|\varphi\|$. For the second one, we decompose $\varphi$ as the sum of two functions, with one of them uniformly continuous.
Let $\chi_x$ be the indicator function of the compact interval $[-x,x]$.
We have that
$$\langle \mu^N_t, \varphi \rangle= \langle \mu^N_t, \varphi \chi_x \rangle + \langle \mu^N_t, \varphi (1-\chi_x) \rangle.$$
We hence obtain
\begin{equation}
\label{ineq:comp}
\begin{split}
P(|\langle \mu^N_{(\kappa+s)\wedge T}, \varphi  \rangle - \langle \mu^N_\kappa, \varphi \rangle|>\varepsilon) &\le P_{}(|\langle \mu^N_{(\kappa+s)\wedge T}, \varphi \chi_x  \rangle - \langle \mu^N_\kappa, \varphi  \chi_x \rangle|>\varepsilon/2)\\
& + P_{}(|\langle \mu^N_{(\kappa+s)\wedge T}, (1-\chi_x)  \rangle - \langle \mu^N_\kappa, (1-\chi_x) \rangle|>\varepsilon/(2 \|\varphi\| )).
\end{split}
\end{equation}

We first deal with the first term of the r.h.s. of the previous inequality.
Fix $N>0, \varepsilon>0$ and $\kappa \in \mathfrak T^N_T$ and let $\mathcal{N}^i$ be the total number of marks of $\omega^i[0,T]$, with $(\tau^i_k)_{1 \leq k \leq \mathcal{N}^i}$ its time-marks.

We abbreviate $\varphi \chi_x$ to $\psi$.
Using the uniform continuity of $\psi$ in a compact interval, we get that there exists $a>0$ such that
$$\sup_{\stackrel{z,z' \in [-x,x]}{|z- z'| \le a}} |\psi(z)-\psi(z')| \le \varepsilon/12.$$
We thus have that
$$|\psi(z)-\psi(z')| \leq 2 \|\psi\| \I{|z-z'|>a} + \varepsilon/12.$$

We {now} rewrite $|\langle \mu^N_{(\kappa+s) \wedge T}, \psi \rangle - \langle \mu^N_\kappa, \psi \rangle|$ as
\begin{equation*}
\begin{split}
\Bigg|
\tfrac{1}{N} \sum_{i=1}^{N} \Bigg[
& \psi(\xip{i}{(\kappa+s)\wedge T})-\psi(\xip{i}{\tau^i_{\mathcal{N}^i}}) + \psi(\xip{i}{{\tau^{i}_1}-})-\psi(\xip{i}{\kappa}) \\
&  + \sum_{k=2}^{\mathcal{N}^i} \psi(\xip{i}{{\tau^i_{k}}-}) - \psi(\xip{i}{\tau^{i}_{k-1}})
+ \sum_{k=1}^{\mathcal{N}^i} \psi(\xip{i}{\tau^{i}_{k}}) - \psi(\xip{i}{{\tau^i_{k}}-}) \Bigg]
\Bigg|,
\end{split}
\end{equation*}
which can be bounded above by
\begin{equation*}
\begin{split}
\tfrac{1}{N} \sum_{i=1}^{N} \bigg[ 2 \|\psi\| \bigg(
\I{|\xip{i}{(\kappa+s)\wedge T}-\xip{i}{\tau^i_{\mathcal{N}^i}}|>a}
+
\I{|\xip{i}{{\tau^{i}_1}-}-\xip{i}{\kappa}|>a} \bigg) + \frac{\varepsilon}{6} \bigg]
+
\frac{4 \|\psi\| }{N} \mathcal{N}_s,
\end{split}
\end{equation*}
being $\mathcal{N}_s$ the total number of jumps taking place in the interval $[\kappa, \kappa +s]$,
Then,
\begin{equation*}
\begin{split}
P_{}(|\langle \mu^N_{(\kappa+s) \wedge T},& \psi \rangle - \langle \mu^N_\kappa, \psi \rangle|> \tfrac{\varepsilon}{2})  \leq
P_{}(\tfrac{4}{N} \|\psi\| \mathcal{N}_s > \tfrac{\varepsilon}{6}) \\
& + P_{}\bigg(
\tfrac{2 \|\psi\|}{N} \sum_{i=1}^{N}
\I{|\xip{i}{(\kappa+s)\wedge T}-\xip{i}{\tau^i_{\mathcal{N}^i}}|>a}
+
\I{|\xip{i}{{\tau^{i}_1}-}-\xip{i}{\kappa}|>a} > \tfrac{\varepsilon}{6}
\bigg)
\\
& \leq \frac{12}{N \varepsilon} \|\psi\| E(\mathcal{N}_s)
+
\frac{12}{\varepsilon} \|\psi\| \sup_{0 \leq r \leq s} P (|B_r|>a) \\
& \leq \frac{12}{\varepsilon} s \|\psi\| + \frac{12}{\varepsilon} \|\psi\|  (2-2 \Phi(\tfrac{a}{\sqrt{s}})),
\end{split}
\end{equation*}
where $\Phi$ is the standard Gaussian distribution function.
In the third line we applied Markov's inequality and use the fact that the displacement of the particles before (after) the first (last) jump is given by independent Brownian motions. Then we use that $\mathcal{N}$ is a Poisson process.

The second term in (\ref{ineq:comp}) can be controlled using {\em Step 2} above and the fact that the solution of the F-KPP equation is uniformly continuous in compact space-time intervals. Hence the second condition is fulfilled as well.

\medskip

\noindent
{\em Uniqueness of limit points.}
Let $(\mu^{N_k}_t)_{0 \leq t \leq T}$ be any convergent subsequence and $(\mu_t)_{0 \leq t \leq T}$ its limit.
Its time marginals $\mu_t$ are characterized by their distribution functions $F_{\mu_t}$.
Due to Theorem \ref{mresult} we have that $F_{\mu_t}=u(\cdot, t)$.
This, together with \cite[Theorem 13.1]{Bi99} provide us the desired conclusion.

\section{Speed selection}\label{sec:proof3}
In this section, we prove Theorem \ref{finiteN}.
The proofs for fixed $N$ are similar to the ones in \cite{DR,BG}.
We include them for the reader's convenience but we point to those references for the details.
However, the proof of convergence of the velocities $v_N \nearrow \sqrt{2}$ as $N\to \infty$ requires a completely different strategy and is based on the hydrodynamic limit, Theorem \ref{mresult}.

Parts {\em (1)} and {\em (2)} of Theorem \ref{finiteN} are obtained by proving positive Harris recurrence for the process $(\pfl)_{t \geq 0}$ and part {\em (3)} by means of the subadditive ergodic theorem.
The monotonicity of the velocities {\em (4)}, is proved by coupling two process with different number of particles but comparable initial conditions. With some abuse of notation we will denote $(\eta_k)_{k \geq 0}$ a process constructed in this section which coincides in law with $(\eta_t)_{t\ge0}$ observed at jump times, although the construction is different.

In what follows we omit the dependence of some of the variables on $N$.
Consider a Poisson process $(T_k)_{k \geq 1}$ with rate {$\frac{N}{2}$} and an i.i.d. family $(V_k)_{k \geq 1}$ with $V_k=(V_k^1,V_k^2)\sim \mathcal{U}\{O_N\}$, where $O_N=\{(i,j) \colon \ 1\leq i<j \leq N \}$.
Let $({B}^N_s)_{s\geq 0}$ be an $N-$dimensional standard Brownian motion.
For $\xig{}{} \in \mathbb{R}^N$ we denote $\sigma(\xig{}{})=(\xig{}{}[1],\ldots, \xig{}{}[N])$.

We construct a discrete time auxiliary Markov process ${\zeta}_k$ as follows. Given $\zeta_0$, define
\begin{equation}
\label{coupling}
{\zeta}_{k+1}:= \sigma\big(\theta_{{V}_k}(\sigma({\zeta}_{k} + {B}^N_{T_{k+1}}-{B}^N_{T_k} ))\big)
\end{equation}
with $\theta_{ij}$ defined in \eqref{teta} and assuming $T_0=0$. In words, between jump times, particles evolve according to an $N-$dimensional Brownian motion.
At jump times, two ranks $V=(V^1,V^2)$ are chosen uniformly in $O_N$ and the particle with rank $V^1$ jumps to the position of the particle with rank $V^2$.
Afterwards, particles are sorted in increasing order. We omit the trajectories between jumps, so that $\zeta_k$ is the state of the system at the $k-$th jump of $(\xi_t)$.

The law of $({\zeta}_k)_{k \geq 0}$ is given by the order statistics of a process with generator \eqref{generator} at jump times as can be seen by direct computation.
This process is a deterministic function of the initial condition $\zeta_0$, the displacements $D_{k}:={B}^N_{T_{k}}-{B}^N_{T_{k-1}}$ and the jump marks $(V_k)_{k \geq 1}$. We emphasize this by writing
\begin{equation}
\label{det.function}
 (\zeta_k)_{k\ge 0}= \Upsilon((V_k)_{k\ge 1},(D_k)_{k\ge 1},\zeta_0).
\end{equation}

As in \eqref{leftmost}, let $(\eta_k)_{k \geq 0}$ be the process $(\zeta_{k})_{k \geq 0}$ as seen from the leftmost particle and
$\Omega_N:=\{(z_1, \ldots,z_{N-1}) \in \mathbb{R}^{N-1} : 0 \leq z_1 \leq \ldots \leq  z_{N-1} \}$ its state space.
Any set $R \subseteq \Omega_N$ is embedded in $\mathbb{R}^N$ by defining $R^0=\{0\}\times R$. We now turn to prove each of the statements stated in Theorem \ref{finiteN}.

\subsection{Positive Harris recurrence.}
To prove Positive Harris recurrence it is sufficient (see for instance \cite{asmussen2003}) to show that there exist a set $R\subseteq \Omega_N$ such that
\begin{enumerate}[(i)]
\item
\label{C1}
$E_\eta(\tau_R) < \infty$, for all {$\eta \in \Omega_N$}, where $\tau_R=\inf\{k\geq0 \ : \ \eta_k \in R\} $.
\item
\label{C2}
There exists a probability measure $q$ on $R$, $\lambda>0$ and $r \in \mathbb{N}$ so that \\
$P_{\eta_0}(\eta_r \in \tilde{R}) \geq \lambda q(\tilde{R})$ for all {$\eta_0 \in R$} and all $\tilde{R} \subseteq R$.

\end{enumerate}
Take
$$R=\{ \eta \in \Omega_N \ : \ \etap{i+1}{}-\etap{i}{} \in (0,N) \text{ \ for \ } i=1, \ldots, N-1\}.$$
We claim that any initial condition $\eta \in \Omega_N$ can be driven by the dynamics into the set $R$ in exactly {$N-1$} steps with a positive probability uniformly bounded below away from zero. In fact that is the case if (a) each particle does not displace more than ${1}/{2}$ between jumps (due to Brownian movement) and
(b) the particles chosen by $(V_k)_{k \geq 1}$ to perform the jumps are always the first and the last one (hence the first particle jumps to the position of the last one). In this event we have that $\eta_{N-1} \in R$.

The probability of this event can be bounded below uniformly on the initial condition by
\begin{equation}
\label{ub}
P_\eta(\eta_{N-1} \in R) \geq {\left ( {a_N}P(V_1=(1,N))  \right )^{N-1}}= \left( \frac{{2} a_N}{N(N-1)} \right )^{N-1},
\end{equation}
with $a_N>0$ being the probability that a Brownian motion started at the origin does not leave $(-1/2,1/2)$ before a time given by an independent exponential random variable with parameter $N-1$.

This implies \eqref{C1}.
To prove \eqref{C2} note that if {$\tilde{R}\subseteq R$}, then
$$\tilde{R}=\{\eta \in \Omega_N \ : \ \eta(i+1)-\eta(i) \in \tilde{R}_i \subseteq (0,N), \quad i=1, \ldots, N-1\}.$$
Observe also that the displacement of the particles between jumps has the law of the order statistics for $N$ independent Brownian motions, whose density is uniformly bounded below for any initial condition {$\eta_0 \in R$} by a positive constant $c$.
Taking $q$ as the normalized Lebesgue measure restricted to $R$, $r=1$ and $\lambda=c$ we get that \eqref{C2} holds.
It is easy to check, by means of \eqref{ub} and the strong Markov property, that $\sup_\eta E_\eta(\tau_R) < \infty$ and hence $(\eta_k)_{k \geq 0}$ is positive Harris recurrent, which implies that a unique invariant measure exists and is finite.
The fact that $\nu^N$ is absolutely continuous follows from the fact that the distribution of the process is absolutely continuous for every positive time, for every initial distribution.

\subsection{Convergence to equilibrium.}
To prove {\em (2)} it suffices to show the result for the subsequences $(\eta_{N m + j})_{m \geq 0}$, with $0\leq j < N$.
Furthermore, due to the Markov property, we only need to consider the case $j=0$.
The inequality obtained in \eqref{ub} (which is independent of $\eta$) implies that $\sup_\eta P_\eta(\tau_R>t)<1$ for some $t>0$.
The result follows by applying Theorem $4.1 (ii)$ in \cite{AN78} to the subsequence $(\eta_{N m})_{m \geq 0}$.

\medskip

\subsection{Existence of a velocity.} We prove that both limits in \eqref{velocityE} exist almost surely and in $L^1$ and that they are nonrandom as in \cite{BG}.
The argument makes use of the subadditive ergodic theorem \cite{Durrett1991}.
First observe that due to the monotonicity property of the coupling introduced in \eqref{coupling} and translation invariance, it is sufficient to prove the result when all the particles start at the origin. \\
We construct several copies of our process simultaneously.
The index $n$ refers to the $n$-th copy and the index $k$ refers to time.
Given the sequences $(D_k)_{k \geq 1}$ for the displacements and $(V_k)_{k \geq 1}$ for the jumps we construct each copy of the process in the following manner:
\begin{equation*}
\zeta_{n,0}=\xi_0, \qquad (\zeta_{n,k})_{k\geq 1}=\Upsilon((V_{r})_{r \geq n}, (D_{r})_{r \geq n},\zeta_0), \qquad n\geq 0.
\end{equation*}
Here $\Upsilon$ refers to the construction introduced in \eqref{det.function}.
Let us stress the fact that the $n$-th copy of the process does not make use of the first $n$ elements of the sequences $(V_{r})_{r \geq 1}, (D_{r})_{r \geq 1}$.
Thus, for $d \in \mathbb{N}$,  $(\zeta_{dm,d})_{m\geq 1}$ is an i.i.d. sequence and the distribution of $(\zeta_{n,k})_{k \geq 1}$ does not depend on $n$.

A key observation is that if we run the process up to time $k$, next we put all the particles at the position of $\zeta_k[N]$, and finally we run it for another $l$ steps, we obtain a configuration that dominates the one obtained by running the process $k+l$ steps.
In other words,
\begin{equation*}
\zeta_{0,n+k} [N] \leq \{\zeta_{n,k} + \zeta_{0,n}[N]\}[N].
\end{equation*}
Now, define $\gamma_{n,k}=\zeta_{n,k-n}$ for $0\leq n\leq k$.
Taking into account the aforementioned considerations it is simple to verify that the sequence $(\gamma_{n,k}[N])_{n,k \geq 1}$ fulfills all the conditions of the subadditive ergodic theorem.
Thus, $\lim_{k\to \infty} {\gamma_{0,k}[N]}/{k}$ exists almost surely and in $L^1$, and it is nonrandom.
Finally, observe that $(\gamma_{0,k})_{k \geq 0}$ and $(\zeta_{k})_{k \geq 0}$ coincide in law and hence the result holds also for $\lim_{k \to \infty} {\zeta_k[N]}/{k}$ and $\lim_{t\to \infty} \xi_t[N]/t$.
The last limit holds since $(\zeta_k[N])_{k \geq 0}$ and $(\xi_{\tau_k}[N])_{k \geq 0}$ have the same distribution and the maximum displacements of $(\xi_t[N])_{t \geq 0}$ between jump times are tight (and hence converge to zero when divided by $t$). The existence of the limit $\lim_{t\to\infty}\xi_t[1]/t$ can be obtained in a similar way.

Finally observe that {\em (1)} and {\em (2)} of Theorem \ref{finiteN}, imply that $\etap{N-1}{t}$ converges in distribution, as $t \to \infty$ to an almost surely finite random variable $X$.
Therefore, ${\etap{N-1}{t}}/{t}=({\xi}_t[N]-{\xi}_t[1])/{t}$ converges in distribution to $0$ and thus, also in probability. This finishes the proof of {\em (3)}.

\subsection{Velocity.}
To prove the monotonicity stated in {\em (4)} we use a construction similar to the one in \eqref{coupling} to couple two systems, one with $N$ particles, denoted by $(\zeta^N_k)_{k \geq 0}$, and another with $N+1$ particles denoted by $(\zeta^{N+1}_k)_{k \geq 0}$.
{This coupling is similar to the one used in \cite{BG, M, DR} to prove the same monotonicity result.
The main difference is that in those models the rate at which each pair of particles produce a branching-selection event is independent of $N$, while in our case it is $1/(N-1)$}.

For the ease of the reading, we first describe the coupling informally. Let us suppose that the initial conditions are ordered in the following sense
\begin{equation}
\label{strong.order}
\zeta^{N+1}_0[i+1] \ge \zeta^{N}_0[i], \quad i=1, \ldots, N.
\end{equation}
We start by \emph{matching} the order statistics of both systems. Each $\zeta^N[i], i=1, \ldots, N$  is paired with $\zeta^{N+1}[i+1]$. {The particle at $\zeta^{N+1}[1]$} is not coupled.
We re-{match} the particles of the two processes just before and inmediately  after every time a jump takes place.
We do not re-match between jumps, but during these periods the Brownian displacements are coupled among both systems according to the match.
More precisely, after each jump we use the same Brownian motion to drive the particles at positions $\zeta^{N+1}[i+1]$ and $\zeta^{N}[i]$ , for every $1\le i \le N.$
This guarantees that the pairs matched by the coupling keep their relative order.
In particular, each $i-$th order statistic of the $N-$system has at least $N-i$ particles of the $(N+1)-$system to the right.

Given that the rate at which each  pair of order statistics produce a jump in a system with $N$ particles is $\tfrac{1}{N-1}$ (and hence depends on $N$), we couple the jumps as follows.
Each time there is a jump from position $\zeta^N[i]$ to position $\zeta^N[j]$ with $i<j,$ we enforce the same jump from $\zeta^{N+1}[i+1]$ to $\zeta^{N+1}[j+1]$ with probability $\tfrac{N-1}{N}$ while, with probability $\tfrac{1}{N}$, the jump occurs from $\zeta^{N+1}[1]$ to $\zeta^{N+1}[j+1]$.
Note that in order to obtain the correct rates for the $(N+1)-$system, extra jumps from $\zeta^{N+1}[1]$ to the other particles should be included.
Observe that {while this procedure gives the correct rates, neither of these jumps alter the ordering \eqref{strong.order}.

\bigskip
We now provide a precise construction.
Let $(T_k)_{k \geq 1}$ a Poisson process with rate $\lambda^N+\lambda^+ $, with $\lambda^N:=\frac{N}{2}$ and $\lambda^+=1-\frac{1}{N-1}$.
Let $(V_k)_{k \geq 1}$, with $V \sim \mathcal{U}\{O_N\}$ as in \eqref{coupling}.
We decompose $B^{N+1}_s=(B^{1}_s,B^{N}_s)$ where $(B^{N}_s)_{s\geq0}$ is a $N$-dimensional Brownian motion.
Finally, let $(W_k)_{k\geq 1}$ and $(C_k)_{k\geq 1}$ be two Bernoulli processes with success probability $\frac{\lambda^N}{\lambda^N + \lambda^+}$ and $\frac{N-1}{N}$ respectively, and $(U^+_k)_{k\geq 1}$ i.i.d. random variables with $U^+ \sim \mathcal{U}\{2,\ldots,N+1\}$.
Given $\zeta^{N}_0,\zeta^{N+1}_0$ (and using the previously defined notations $\sigma$ and $\theta$), we construct both processes as follows:
$$
\zeta^{N}_{k+1}=
\begin{cases}
\sigma\big(\theta_{V_k}(\sigma(\zeta^{N}_{k} + {B}^{N}_{T_{k+1}}-{B}^N_{T_k} ))\big),  & \text{if $W_k=1$,}\\
\zeta^{N}_{k} + {B}^{N}_{T_{k+1}}-{B}^N_{T_k},  & \text{if $W_k=0$,}
\end{cases}
$$
$$
\zeta^{N+1}_{k+1}=
\begin{cases}
\sigma\big(\theta_{V_k+(1,1)}(\sigma(\zeta^{N+1}_{k} + {B}^{N+1}_{T_{k+1}}-{B}^{N+1}_{T_k} ))\big),  & \text{if $W_k=1, \ C_k=1$,}\\
\sigma\big(\theta_{(1,V_k(2)+1)}(\sigma(\zeta^{N+1}_{k} + {B}^{N+1}_{T_{k+1}}-{B}^{N+1}_{T_k} ))\big),  & \text{if $W_k=1, \ C_k=0$,}\\
\sigma\big(\theta_{(1,j)}(\sigma(\zeta^{N+1}_{k} + {B}^{N+1}_{T_{k+1}}-{B}^{N+1}_{T_k} ))\big),  & \text{if $W_k=0, \ U^+_k=j, \ j\ge 2$}.
\end{cases}
$$
It is easy to check that with this construction $$\zeta^{N+1}_k[i+1]>\zeta^{N}_k[i], \ \forall \ k\geq 1, \ \forall \ i=1, \ldots, N,$$ which implies the claimed monotonicity $v_{N+1} \ge v_N$.

\bigskip

To prove {\eqref{minimal}} we provide a lower bound for the speed of the empirical mean and an upper bound for the speed of the rightmost particle. The result will follow from the fact that both bounds coincide as $N\to \infty$. For the lower bound, we make use of the monotonicity of the spacings of $\xig{}{t}$, which is established in the next lemma.
\begin{lemma}
\label{mono}
Let $(\xi_t)_{t \geq 0}$, $(\tilde\xi_t)_{t \geq 0}$ be two processes with generator \eqref{generator} and initial conditions $\xig{}{0}$, $\tilde\xi_0$ respectively such that
$$\osp{i+1}{0}-\osp{i}{0} \leq_{st} \tilde\xi_0[i+1]-\tilde\xi_0[i], \quad i=1, \ldots, N-1.$$
Then,
$$\osp{i+1}{t}-\osp{i}{t} \leq_{st} \tilde\xi_t[i+1]-\tilde\xi_t[i], \quad i=1, \ldots, N-1, \quad \forall \ t \geq 0.$$
\end{lemma}
Here $\leq_{st}$ denotes stochastic domination.
\begin{proof}
Let us first observe that the result holds for the spacings of $N$ independent Brownian motions (see \cite[Theorem 4.1]{R2000}), which is the law of the particles between jump times. Next, note that if the jumps for both processes are coupled as in \eqref{coupling} the domination of spacings is preserved after each jump. Proceeding inductively we obtain the stochastic domination for all times.
\end{proof}

\noindent
{\emph{Lower bound.}} Let $u^0$ be the solution of the F-KPP equation with initial condition given by the Heavyside function and $w_{\sqrt{2}}(x)$ the minimal velocity traveling wave for \eqref{KPP}. Let $m(t)$ be the median of $u^0(\cdot,t)$. The following facts hold \cite{B83}.
\begin{enumerate}
\item[(i)] $\|u^0(\cdot + m(t),t) - w_{\sqrt{2}}\| \to 0$ as $t\to \infty$.

\item[(ii)] $\sqrt{2}= \int_{-\infty}^{+\infty} w_{\sqrt{2}}(x) (1-w_{\sqrt{2}}(x)) \ dx.$
\end{enumerate}
Our strategy consists in comparing the velocity of the empirical mean of the system in equilibrium as seen from the leftmost particle and the system with initial condition given by the Heavyside function.
Afterwards, Theorem \ref{mresult} will provide us the link to use (i)-(ii). This idea appears in \cite{BCdMFLS}.

We denote by $\bar{\nu}^N$ the distribution obtained by placing particle labeled 1 at the origin and the remaining ones according to $\nu^N$. Let $m^N_t=\sum_{i=1}^N\xi_t(i)$ be the empirical mean and observe that
\begin{equation*}
E_{\bar{\nu}^N}[m^N_t]=
\int_{0}^{+\infty} E_{\bar{\nu}^N}[1-F_N(x,t)] \ dx - \int_{-\infty}^{0} E_{\bar{\nu}^N}[F_N(x,t)] \ dx,
\end{equation*}
and hence differentiating under the integral sign we get
\begin{align}
\nonumber
\partial_t E_{\bar{\nu}^N}[m^N_t]&=
-\int_{-\infty}^{+\infty} \partial_t E_{\bar{\nu}^N}[F_N(x,t)] \ dx \\
\label{whatwewant}
&=\tfrac{N}{N-1} \int_{-\infty}^{+\infty}  E_{\bar{\nu}^N}[F_N(x,t)(1-F_N(x,t))] \ dx.
\end{align}
In the last equality we make use of \eqref{KPPwe}.
Let us notice that
$$\int_{-\infty}^{+\infty}F_N(x,t)(1-F_N(x,t))dx=\sum_{k=1}^{N-1} \frac{k (N-k)}{N^2} (\osp{k+1}{t}-\osp{k}{t}),$$
and hence it is monotone in the spacings of the system.\\
We bound \eqref{whatwewant} from below as follows
\begin{align}
\label{linkKPP}
\nonumber \int_{-\infty}^{+\infty} E_{\bar{\nu}^N}[F_N(x,t)(1-F_N(x,t))] \ dx
& \geq
\int_{-\infty}^{+\infty} E_{0}[F_N(x,t)(1-F_N(x,t))] \ dx\\
& \geq
\int_{m(t)-b}^{m(t)+b} E_{0}[F_N(x,t)(1-F_N(x,t))] \ dx,
\end{align}
for all $0<b< \infty$.
Now, \eqref{linkKPP} can be written as
\begin{align}
\label{vd}
&\int_{m(t)-b}^{m(t)+b} E_{0}[F_N(x,t)(1-F_N(x,t))]\ dx
- \int_{m(t)-b}^{m(t)+b} u^0(1-u^0)(x,t) \ dx\\
\label{kpptf}
+ & \int_{m(t)-b}^{m(t)+b} u^0(1-u^0)(x,t) \ dx.
\end{align}
Due to (i)-(ii) we can fix $b$ and $t$ big enough to make \eqref{kpptf} as close to $\sqrt 2$ as desired. For those values of $b$ and $t$, the term \eqref{vd} converge to zero as $N\to \infty$. Here we use that, due to Theorem \ref{mresult}, $\|F_N(1-F_N)(\cdot,t) - u^0(1-u^0)(\cdot, t) \|$ converges to zero in probability (and hence in $L^1$) as $N\to\infty$. So,
\begin{equation}
\label{liminf.KPP}
\liminf_{N\to\infty} \int_{-\infty}^{+\infty} E_{\bar{\nu}^N}[F_N(x,t)(1-F_N(x,t))] \ dx \ge \sqrt{2}.
\end{equation}
We claim that $\partial_t E_{\bar{\nu}^N}[m^N_t]=v_N$.
Indeed, as $\nu^N$ is the equilibrium  distribution for the system seen from the leftmost particle, a renewal argument readily implies that $(\osp{1}{t})_{t \geq 0}$ has stationary increments and thus
\begin{equation*}
E_{\bar{\nu}^N}[m^N_t]=E_{\bar{\nu}^N}[m^N_t - \osp{1}{t}] + E_{\bar{\nu}^N}[\osp{1}{t}]=c_1+c_2 t,
\end{equation*}
with $c_1$ and $c_2$ depending on $N$.
The limit in \eqref{velocityE} forces $c_2$ to equal $v_N$ which implies our claim. Combining this with \eqref{whatwewant} and \eqref{liminf.KPP} we conclude that
$$\liminf_{n \to \infty} v_N\geq \sqrt{2}.$$
\noindent
\emph{Upper bound.}
For this bound, we embed the process into $N$ independent BBMs. Recall that the maximum $M_t$ of a BBM at time $t$ verifies
\begin{equation}
\label{maximumBBM}
\lim_{t \to \infty} \tfrac{M_t}{t}=\sqrt{2} \quad  \text{a.s.}
\end{equation}

We provide a last alternative construction of the process.
Instead of using a Poisson clock of rate $1$ for the jumps of each particle, we use a Poisson clock of rate $\tfrac{k-1}{N-1}$ for each rank $k=1, \ldots,N$.
Whenever the $k$-th clock rings the particle whose rank is $k$ branches into two particles.
At that time, a particle is chosen uniformly at random among those particles with rank less than $k$ and is eliminated from the system. Between jumps, particles evolve as independent Brownian motions.
It is clear that (i) the process constructed in this way has generator \eqref{generator} and (ii) each particle branches at a rate which is less or equal than one. Hence we can couple this process with a BBM by adding extra branchings to achive rate one.

More precisely, let us construct a coupling between a process with generator \eqref{generator} and $N$ independent BBM systems. We will still denote it $(\xi_t)_{t \geq 0}$.
Let $X_1(t),\ldots,X_N(t), X_{N+1}(t), \ldots$ denote the positions of all the particles of $N$ independent BBM starting at $\xip{1}{0}, \ldots, \xip{N}{0}$, labeled according to its appearence time.
Let $(\tau_i)_{i \geq 1}$ be the branching times (in increasing order) and $(l_i)_{i \geq 1}$ the corresponding labels of the branching particles.
Let $(U_i)_{i \geq 1}$ and $(D_i^k)_{k \leq N,i \geq 1}$ be independent random variables with $U_i$ uniform in $[0,1]$, $i\ge 1$ and $D^k_i$ uniform in $\{1, \ldots , k-1\}$ for $i\ge 1$, $k\le N$.
Finally, let $\alpha_k=\tfrac{k-1}{N-1}, \ k=2, \ldots,N$. We construct a process $(L^1_t, \ldots, L^N_t)$ on the set of labels $\mathbb{N}$.
We will use these labels to determine the particles of the BBM that are going to be used to construct the coupling.
For $j=1, \ldots, N$, we define
$$
R^j_t= \text{rank of $X_{L^j_t}(t)$ in the set $\{ X_{L^1_t}(t), \ldots, X_{L^N_t}(t) \}$}
\subseteq
\{1, \ldots, N \}.
$$

We build $(L^1_t, \ldots, L^N_t)$ inductively as follows:
\begin{itemize}
\item
At time zero $(L^1_0, \ldots, L^N_0)=(1,\ldots,N)$.

\item
Assume $(L^1_{\tau_{i-1}}, \ldots, L^N_{\tau_{i-1}})$ is known.
For $t \in (\tau_{i-1}, \tau_{i}]$ we define:
\begin{equation*}
L^k_t=
\begin{cases}
N + i  & \text{ if $t=\tau_i$, \quad $l_i=L^j_t, \ j \neq  k$ , \quad $U_i < \alpha_{R^{j}_t}$, \quad $D^{R^{j}_t}_i= R^k_t$} \\
L^k_{t^-}  & \text{ otherwise,}
\end{cases}
\end{equation*}
$k=1 , \ldots, N$.
\end{itemize}

We leave to the reader to verify that
\begin{equation*}
(\xi_t(1), \dots, \xi_t(N)):=(X_{L^1_t}(t), \ldots, X_{L^N_t}(t)), \qquad t \geq 0
\end{equation*}
has generator \ref{generator}. With this coupling we have that
\begin{equation}
\label{ENBBM}
\osp{N}{t} \leq \max(M^1_t, \ldots, M^N_t), \qquad \forall \ t \geq 0,
\end{equation}
with $M^j_t$ the maximum of the $j$-th BBM.
Combining \eqref{maximumBBM} with \eqref{ENBBM} we conclude
$$\limsup_{t \to \infty} \tfrac{\osp{N}{t}}{t}\leq \sqrt{2}.$$

\bigskip

The proof is concluded by observing that
$$\osp{1}{t} \leq m^N_t \leq \osp{N}{t}, \qquad \forall \ t \geq 0.$$

\section{Acknowledgments}
We thank Nahuel Soprano-Loto for pointing out an error in a preliminary version of the manuscript. The authors are partially supported by grants UBACYT 20020160100147BA and PICT 2015-3154.

\bibliographystyle{plain}

redpajama_set_name: RedPajamaArXiv


text:
David "Dave" Richmond (født 1940 i Brighton Sussex, England) er en engelsk bassist og komponist.
Richmond er nok mest kendt fra sit samarbejde med Manfred Mann og dennes gruppe. Han var også session musiker i 1970´erne for feks. Cliff Richard, The Shadows, Olivia Newton-John, Elton John, Cilla Black, Serge Gainsbourg, Jane Birkin, Dusty Springfield etc.
Richmond dannede gruppen Collage (1973) med Brian Bennett og Alan Hawkshaw som lavede en lp Misty. Selvom han i starten af sin karriere primært var elbassist, spiller han også kontrabas. Han spiller i dag stadig som sessionbassist bla med den tyske slagermusiker Bert Kaempfert´s bigband, og i andre forskellige jazz og rock sammenhænge.

Udvalgte Plader

Solo lp´er

 Mainly Bass - 1983
 Thought Patterns - 1979 med Clive Hicks
 Tolls of the Trade - 1980 med Harold Fisher

Som sideman

 Manfred Mann - As is - 1966
 Manfred Mann - The Soul of Mann - 1967
 Marvin, Welch & Farrar - Marvin, Welch & Farrar - 1971
 Marvin, Welch & Farrar - Second Opinion - 1971
 Cliff Richard - Cliff Goes East - 1972
 Collage/Misty - 1973 trio med Brian Bennett og Alan Hawkshaw
 Moody - The Gentle Rain - 1973
 Brian Bennett - Aim High (Brian Bennett at KPM 1973-1976) - 2005
 The Shadows - Shades of Rock - 1970 (med på enkelte numre).
 The Shadows - Specs Appeal - 1975
 Hank Marvin - Guitar Syndicate - 1977
 Elton John - Elton John - 1970
 Elton John - Chronicles - 2005
 Olivia Newton-John - Olivia - 1998
 Bert Kaempfert - Celebrating a Legend - 2004
 Serge Gainsbourg - London Paris - 1963-1971 - 2016

Kilde

 Om Dave Richmond på www.allmusic.com
 Om Dave Richmond på www.jazz4now.co.uk

Bassister fra England
Komponister fra England

redpajama_set_name: RedPajamaWikipedia


Mind the Gap is about a young woman, Elle Peterssen who was attacked at a subway station and now she finds herself disembodied and wandering around the spirit-world of the hospital where her potentially body resides. There are a couple levels of dramatic action here. Peterssen not only has to come to grips with finding herself near-dead and in a new world with its own rules but has lost her memory, most importantly the memory of who her attacker is. Also, in the embodied world, Peterssen's friends and family console each other and argue about the terrible situation they find themselves in. There's a lot going on here and I worry that it might be too much for the comic book medium.
Mind the Gap tries to tell an ambitious amount of story in a comic book. The comic book medium is set up for success in terms of the potential to tell any story or even function strictly on a high (A)rt level. The medium has visuals that can show characterization and two dimensional beauty but it also has text to allow for dialogue, exposition and any other wonder than can come from the written word. There should be no story that cannot be told through comic books. However, McCann and Mind the Gap may have found a story that might very well be too complex for the limited time and physical space of comic books.
Mind the Gap lives in two different worlds, one that is new to the reader and needs to be defined as well as nearly a dozen characters that all need their accounting. There are supernatural aspects as well as a murder mystery that unravels as the book progresses. McCann has taken on a lot of details to be managed in a medium that is limited by visual and temporal space – the turn of a page is a moment of transition. While McCann does a great job of not breaking the natural rules that come with the medium, the book becomes cluttered and at times lacks any real propulsion of story. While it might have been more successful as a TV show where acting can drive necessarily dull or even just flat moments of detail and character definition, Mind the Gap becomes thick and plodding as a comic book.
This entry was posted in Columns and tagged adrian alphona, alternative comics, arif prianto, beny maulana, damien lucchese, dave lanphear, IMAGE COMICS, jim mccann, mind the gap, rob levin, rodin esquejo, sonia oback. Bookmark the permalink. Follow any comments here with the RSS feed for this post. Post a comment or leave a trackback: Trackback URL.


text:
All the gigs of my life: Gig 76 – The Darkness, Wednesday, October 22, 2003, Hammersmith Apollo
By Scruffy on May 7, 2020 • ( Leave a comment )
My one and only gig from The Darkness, who came to save us all from tedium with their guitar solos and hair-flinging rock majesty.
Looking back at 2003, it feels as though the musical landscape was brimming with a diverse array of glittering delights. Pop and indie bands both still had a strong hold on the charts, dance music was as thrilling as ever, and the occasional weird mavericks were still given their time to shine.
And no mavericks were quite as extraordinary as the Darkness. I'd bought their album, "Permission To Land", in July of 2003, and I wrote a mini-review in my diary at the end of that month:
'I mean LOOK at them! The hair of Aerosmith, the moustache of Queen, the pout of T-Rex and the wardrobe of KISS! You just don't do that in the zeroes!* I have to admit, it was purely the thought of seeing them strutting about in zebra-striped catsuits and flailing perms that made me buy a ticket for their October gig. So having done that, I had to get their album of course. And bloody hell! Forget twelve years of indie training and luxuriate in the scribbling slashery of the solos, screeches of "geetar!" and manly falsetto. You gotta love a band whose party anthem includes the line "needlework on Tuesday!" And their ballads are lovely, as is the way with these sorts of bands. I get the impression they're a shade past their early twenties too, which is kinda cheering.'
(*Note: I have always called the first decade of this century 'the zeroes'. Because it's the name of a Bowie song, and thus way cooler than 'the noughties', which is just silly.)
But then when the time came, as with my Kosheen gig the night before this one, I found myself in a period of life where I was far too wrapped up in my new geeky, hobbit and wizard themed interests to write more than a couple of words in my diary about the gig itself. At least this time the words I wrote were positive ones: 'pure rawk thrills!' But not a lot to go on for a blog post.
I really have only one actual memory of the night: that I had a seat up on the balcony, and the moment the band came on stage, every single person in the crowd stood up as if propelled by some mysterious rock'n'roll force. And we stayed that way throughout the whole night, lifted by the sheer energy and melodic power of their already impressive catalogue of songs.
But here comes my geeky 2003 livejournal to the rescue again! And, according to this entry entitled "I believe in a thing called RAWK", it turns out my memory is nearly accurate:
'Well, after last's night's disappointing gig experience, I wasn't really in the mood for going out again, but I'm so glad I did. The Darkness certainly restored my faith in live music. I guess I'm a rawk girl at heart. The atmosphere was brilliant – I was on the balcony and quite close to the back, and every single person was on their feet before the band took the stage. I've never seen that from a balcony crowd before. They're a band who really know how to entertain, be a spectacle, as well as having great tunes. Ah, the catsuits, the falsetto, the strutting, the fiddly geetar solos, 'twas all brilliant. Definitely a band I want to see up close at some point. If it was a great gig from the back of the balcony, it must have been utterly storming for those crushed against the barrier.
Justin: gimme a D!
Crowd: D!!!!
Justin: gimme an Arkness!
Crowd: ARKNESS!!!!!
Gotta love the lunacy.'
As with Kosheen, I've been doing a little research to find out what they're up to now – and it turns out they're still recording and touring and in fact, I just missed seeing them on their latest UK tour in December. As fate would have it however, the last gig I went to before lockdown started included a Darkness song, with the Lounge Kittens' majestic cover of "Love Is Only A Feeling" at their Union Chapel gig.
The thing that hits me the most upon revisiting the Darkness's moment in the spotlight is just how fun and vibrant and varied the chart music scene was back in 2003, and how generic it feels now. Where are the hair-flinging mavericks, the boys and girls in spandex and mascara, the bands and singers who are not afraid to look ridiculous in the name of Pop Glory? In the current pop landscape, there's a touch of it in Harry Styles, but that's about it. Even Lady Gaga seems to have watered down many of the elements that made her so shockingly exciting a decade ago.
So now I know that the Darkness are still striding on with their idiosyncratic rawk fabulousness, I find myself definitely of the same mind as my 2003 self – I want to see them again. What with lockdown and the fact that they've already played London recently, who knows when that chance will come. But until then I've got their back catalogue to catch up on, and when the time comes I'll be there, up close to the stage, to scream and fling my hair along with their flamboyant rawk majesty once more.
Tagged as: the darkness
All the gigs of my life: Gig 75 – Kosheen, Tuesday, October 21, 2003, Shepherd's Bush Empire
All the gigs of my life: Gig 77 – David Bowie, Monday, November 17, 2003, MEN Arena, Manchester


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If you have been in a motorcycle accident, you should get an attorney. The first thing to look for is someone who handles these types of cases, because motorcycle accidents bring with them a lot of baggage, and the attorney will have to be able frame the case in the strongest way.
You will also want an attorney who understands the medicine and the injuries, so that they can be a strong advocate for the person, either in the pre-litigation stage with the insurance company or in front of a judge or jury.
The Most Difficult Aspects Of Motorcycle Accident Cases for an Attorney
The difficulty for the attorney it's the same as for the client. It's necessary to overcome the biases that follow motorcyclists, such as the one that says they are reckless just because they are riding a motorcycle. That makes it necessary to frame the case so that the injured motorcyclist is a responsible person.
No matter what, the key is in overcoming the biases of an 80-year old woman on your jury who believes that the bigger the car, the safer you are, and that motorcyclists who are injured deserve it because they shouldn't ride motorcycles.
It's also necessary and difficult to remove the emotional aspects of the case and get the client to focus. Sometimes it's necessary to explain to the client that the other side sees the case in a very non-emotional way.
To the insurance company, the case is a business decision to them, which is difficult for the injured person to understand because their injury is very personal, but they have to be able to see it that way. They have to be able to look at the pros and cons, as well as the risks and the benefits of going forward to trial or settlement.
The lawyer's job is never to force a client to resolve a case. The lawyer's job is to gather all the information, make the strongest case possible and counsel their client on what they believe is the best course of action. However, ultimately, the case is always the client's and what to do is always the client's decision.
Important Things to Remember About Motorcycle Accidents
Those who have been involved in a motorcycle accident should be checked out and treated by a doctor before anything else. After that, they should speak to a lawyer. They should also take plenty of photos of the accident scene, as well as the motorcycle, the other vehicles involved and themselves, preferably within the first 72 hours of the accident.
By doing that, you've preserved the evidence that you might need to have a successful claim.
For more information on Hiring a Lawyer, please call (219) 874-4878 today to schedule a free initial consultation. Get the information and legal answers you're seeking.


Before the availability of high-speed internet, the world of entertainment for consumers was pretty bleak. Other than a small selection of channels, we had very little choice available to us in terms of movie and TV entertainment. We were also limited when it came to music, as all we had was the radio and access to any CDs and records we could afford to buy.
Fortunately, this has all changed over recent years. With the availability of superfast broadband speeds coupled with a huge choice of online entertainment, people have been able to really broaden their horizons. This is something that has enables us to explore new cultures, explore different genres, and develop new tastes in terms of entertainment.
We all have different tastes and preferences when it comes to entertainment. Being forced to watch or listen to whatever happens to be on at the time is therefore not ideal. Thankfully, modern technology now enables us to enjoy entertainment on our own terms. You can now watch movies from any genre and from countries around the world at the touch of a button. The same goes for TV shows and even music.
There are many free movie sites that offer access to a huge database of movies and shows. This list on Techwhack provides details of some of the top free sites you can access. Not only is the movie and TV entertainment on these sites completely free to access, but you will also find an incredible choice of content with something to suit everyone. This makes it easier to explore new genres, as it won't cost you anything and you will find plenty to select from. Better still, you can watch the content at a time that suits you rather than having to abide by the schedules of TV network operators. Of course, there are also subscriptions sites that you can sign up to, such as Amazon Prime and Netflix. However, many people don't bother with paid sites because of the sheer number of free sites that are available.
Music is another area where we have been able to diversify in terms of what we listen to. You can go online and access a huge range of free music streaming and download sites. This means that you can download and listen to your favorite songs and artists as well as new ones. You can broaden your musical tastes, enjoy music from other parts of the world, and even create playlists to suit every mood or occasion.
Of course, another important consideration is the flexibility that online entertainment provides us with. With so many people now using mobile devices, it is only natural that many want to access entertainment while on the go. With online movies, shows, and music, this is exactly what you can do. As long as you have access to the internet, you can enjoy watching content or listening to music on your device no matter where you are.


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Q: Distribution Fitting with multiple changepoints I have a time series of a random variable x, which randomly moves in and out of three broader states- "generally increasing", "generally decreasing", "more or less staying still". The distribution of x in each of these 3 states is gaussian with different means and variances.
Given the above time series of x, is there a way to iteratively fit the 3 distributions without knowing when the multiple changes happen? Preferably without using hidden markov models?

A: The changepoint package in the R software provides the option for doing the above.  To demonstrate with a toy example:
library(changepoint)
set.seed(1)
x=c(rnorm(50,0,1),rnorm(50,5,3),rnorm(50,10,1),rnorm(50,3,10))
out=cpt.meanvar(x,method="PELT")
cpts(out) # gives 50,100,150 as changepoints

The above uses the cpt.meanvar function as you want to identify a change in mean and variance.  The default is to assume a Normal distribution so you are good there.  The default is to identify 1 change so we use method='PELT' to tell the function to identify multiple changes uses the PELT algorithm (exact minimization of the objective function in approximately linear time).
There are many things you can do with the output, cpts(out) gives you the changepoint locations but you can also plot:
plot(out,cpt.width=3)

Here i've changed the width of the changepoint lines (means) to be thicker than the default (1), you can also use all your normal plotting arguments.
We have used the default penalty MBIC in calculating the segmentation above but if your data are not independent within each segment you need to change this.  There is the CROPS penalty which gives you a range of segmentations between two penalty values:
out1=cpt.meanvar(x,method='PELT',penalty='CROPS',pen.value=c(10,500))
plot(out1,diagnostic=TRUE)

You need to make sure that your pen.value upper limit gives a segmentation with 0 changepoints for the diagnostic to make sense.  In the diagnostic plot you get the number of changepoints against the negative-log-likelihood. The idea is that when true changes are added the negative-log-likelihood goes down a large amount, when you add false changes then the negative-log-likelihood show a small improvement in fit.  For this simulation it is clear that the appropriate number of changes is 3 but for real data it can be less clear.  Once the number of changes has been decided you can plot that number using:
plot(out1,ncpts=3,cpt.width=3)


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Pro Basketball|Fultz Records 2nd Triple-Double, Magic Beat Lakers 119-118
https://nyti.ms/36TpJ4t
Fultz Records 2nd Triple-Double, Magic Beat Lakers 119-118
LOS ANGELES — Markelle Fultz hit two big layups in the final minute while scoring 21 points during his second career triple-double, and the Orlando Magic snapped the Los Angeles Lakers' nine-game winning streak with a 119-118 victory Wednesday night.
Fultz added 11 rebounds and 10 assists, while Aaron Gordon had 21 points and a late go-ahead basket as the Magic became the first sub-.500 team to beat the Western Conference-leading Lakers this season despite blowing a 21-point lead in the second half.
Wes Iwundu scored a career-high 19 points, and Nikola Vucevic scored seven of his 19 points in the fourth quarter of the Magic's eighth win in 12 games.
LeBron James had 19 points and matched his career high with 19 assists before missing a tying 3-pointer with 3 seconds left for the Lakers (33-8), who hit the halfway point of their regular season with their first loss since Christmas.
Quinn Cook scored a season-high 22 points, including a 3-pointer at the buzzer, while Troy Daniels added a season-high 17 for the Lakers. Los Angeles finally lost in its fourth straight game without injured superstar Anthony Davis.
Kentavious Caldwell-Pope also scored 17 points for the Lakers in their final home game before a five-game road trip while the Grammy Awards take over Staples Center.
HEAT 106, SPURS 100
MIAMI (AP) — Kendrick Nunn scored 33 points, Goran Dragic added 17 and Miami held on to beat San Antonio and move back into the No. 2 spot in the Eastern Conference.
Bam Adebayo — cut from USA Basketball's national team by Spurs coach Gregg Popovich over the summer — had 14 points, 13 rebounds and seven assists for the Heat. Miami improved to 18-1 at home, one win shy of matching last season's home win total.
Jimmy Butler scored 12 and Duncan Robinson added 11 for Miami.
DeMar DeRozan scored 30 points on 12-of-14 shooting for San Antonio, while Patty Mills added 21 for the Spurs.
San Antonio had won three of its last four games, all those wins coming against top Eastern Conference teams Milwaukee, Boston and Toronto.
TRAIL BLAZERS 117, ROCKETS 107
HOUSTON (AP) — Damian Lillard scored 25 points and CJ McCollum added 24 as Portland withstood a late surge by Houston.
Russell Westbrook had a triple-double with 31 points, 12 assists and 11 rebounds on a night when James Harden scored a season-low 13 points. Houston lost consecutive games for the first time since a season-high three-game skid from Nov. 20-24, and has dropped three of its last four.
Carmelo Anthony added 18 points with 12 rebounds for Portland in his first game against Houston since the Rockets parted ways with him after just 10 games last season.
76ERS 117, NETS 106
PHILADELPHIA (AP) — Tobias Harris scored 34 points, including two big baskets in the final two minutes that proved to be the difference, as Philadelphia topped Brooklyn.
Ben Simmons had 20 points and 11 assists for the 76ers, who improved to 19-2 at home.
Brooklyn has lost nine of 11 overall and six straight away from home. The Nets were led by Spencer Dinwiddie's 26 points and eight assists. Jarret Allen had 17 points.
Josh Richardson had 15 points and Furkan Korkmaz added 15 off the bench. Al Horford had 14 points and eight rebounds for the Sixers.
PISTONS 113, CELTICS 103
BOSTON (AP) — Sekou Doumbouya scored a season-high 24 points and Detroit snapped a three-game losing streak with a win over Boston.
Markieff Morris matched his season best with 23 points. Derrick Rose added 22 points and Sviatoslav Mykhailiuk had a season-high 21. The Pistons shot a season-best 60.3% from the field.
Andre Drummond added 13 points and 13 rebounds to post his 34th double-double of the season.
Gordon Hayward had 25 points and seven rebounds, Jaylen Brown finished with 24 points and 12 boards and Kemba Walker scored 19 points on 7-of-19 shooting and 2 for 11 from 3-point range.
BULLS 115, WIZARDS 106
CHICAGO (AP) — Zach LaVine had 30 points, seven rebounds and seven assists, and banged-up Chicago beat Bradley Beal and Washington.
Chicago placed seven players in double figures in its second win in its last nine games. Thaddeus Young had 18 points and eight boards, and Tomas Satoransky scored 18 points against his former team.
Beal scored 23 points on 8-for-13 shooting after he was listed as questionable because of right shoulder soreness.
Jordan McRae had 16 points for Washington, which has dropped three of four. Isaiah Thomas and Davis Bertans scored 12 points apiece.
RAPTORS 130, THUNDER 121
OKLAHOMA CITY (AP) — Norman Powell scored 23 points, and Toronto held off a furious rally and beat Oklahoma City.
Pascal Siakam and OG Anunoby each added 21 points for the Raptors, who shot a season-high 61.2% from the field.
Dennis Schroder scored 25 points, Danilo Gallinari scored 23 and Shai Gilgeous-Alexader added 21 for the Thunder, who cut a 30-point deficit to three before Toronto held on.
PACERS 104, TIMBERWOLVES 99
MINNEAPOLIS (AP) — Domantas Sabonis matched a season high with 29 points and added 13 rebounds and six assists to lead Indiana over Minnesota.
Malcolm Brogdon added 21 points and Doug McDermott scored 14 off the bench for the Pacers, who have won three straight.
Jarrett Culver led Minnesota with 17 points and Gorgui Dieng had 15 points and 11 boards. The Timberwolves have lost three in a row.
Minnesota big man Karl-Anthony Towns missed his 15th straight game with a left knee sprain. A combination of several players matched up on Sabonis struggled early to keep him in check in Towns' absence.
NUGGETS 100, HORNETS 86
DENVER (AP) — Rookie reserve Michael Porter Jr. scored 19 points and recent call-up P.J. Dozier had a big Denver debut filling in for injured point guard Jamal Murray in the second half of a win over Charlotte.
Murray exited the game with 1:32 remaining in the second quarter after rolling his left ankle.
Dozier finished with 12 points in 13 minutes, including a stretch in the second half where he scored nine straight for the Nuggets.
Terry Rozier scored 20 points for the Hornets, who wrapped up a winless four-game trip.
MAVERICKS 127, KINGS 123
SACRAMENTO, Calif. (AP) — Luka Doncic had 25 points, 15 rebounds and a career-high 17 assists for his NBA-leading 12th triple-double, and Dallas held off a late run to beat Sacramento.
Seth Curry scored 21 points off the bench as Dallas won its third straight despite the absence of Kristaps Porzingis, who missed his ninth consecutive game. Dwight Powell added 17 points and nine rebounds, and Tim Hardaway Jr. scored 18.
De'Aaron Fox had 27 points and 12 assists for Sacramento. Harrison Barnes and Buddy Hield scored 25 apiece.


text:
Union Twp. student in custody after bomb squad finds explosives
By Post-Tribune staff
A Union Township Middle School student is accused of bringing two improvised explosive devices to school in a backpack Thursday.
Authorities said the Porter County Bomb Squad discovered the devices and safely removed them. An unidentified juvenile was taken into custody and is facing a felony charge of possession of a destructive device.
School administrators and a school resource officer relocated students from classrooms in proximity to the backpack about 12:30 p.m. and secured the area, according to a release from the Porter County Sheriff's Department.
The school, at 599 West 300 North next to Wheeler High School, houses students in grades 6-8.
— Carole Carlson
Chicago Sky in WNBA free agency: Candace Parker and Courtney Vandersloot are leaving. What will Allie Quigley decide?