text:
\section{Introduction}
\subsection{Definition}\hspace{0pt}
Hawkes Process is a
time-homogeneous self-exciting locally-finite point process
on $\R$ with long-memory.  A realization of the process is a random locally-finite subset $S=S(\omega)$ of $\R$.
Any locally-finite point process is characterized by its intensity rate $r(t,\omega)$ defined as the intensity
of the point process at time $t$ conditioned on the past history of the process until time $t$, i.e. $r(t,\omega)$ can
be defined as
\begin{equation}
  r(t,\omega):=\lim_{\delta t \to 0} \frac{\P\bigg[\# \big( S(\omega)\cap [t,t+\delta t) \big) \geq 1 | {\cal
  F}_t\bigg]}{\delta t}
\end{equation}
where $\# \big( S(\omega)\cap [t,t+\delta t) \big)$ is the number of elements of the set $S(\omega)\cap
[t,t+\delta t)$ and ${\cal F}_t$ is the $\sigma-$field for $S_t(\omega):=S(\omega) \cap (-\infty,t)$ generated by
{\it elementary} events
$\{\omega: S_t(\omega)\cap I \neq \emptyset \}$
where $I$ varies over all intervals $I\subset (-\infty, t)$.

It is convenient to view the process as a random subset of $\Rp:=[0,\infty)$
and thus the  intensity  function $r(t,\omega)$ is defined for $t \in \Rp$.
Hawkes Process is defined by three $\Rp$-valued functions on $\Rp$.
Given these three functions $\lambda,h,g_0$ the
intensity rate is given by
  \begin{equation} \label{def:sigma} r(t,\omega)=\lambda \big(g_0(t)+\sum_{\tau \in S_t} h(t-\tau) \big) \end{equation}
where $g_0$ is some initial condition.

The above description is equivalent to the one  found in literature except for the term $g_0$.
Usually instead of starting from initial condition $g_0$, the process is
defined by giving $S_0$ a locally finite collection of points in $(-\infty,0)$. This set $S_0$ represents points of
process before time $0$. To connect to our setting $g_0$ can then be computed as
  \beq g_0(t)=\sum_{\tau \in S_0} h(t-\tau) \label{eq:old_form} \end{equation}
and the the two definitions are equivalent.  Defining process with $g_0$ as initial conditions also allow for
more general functions which are not of form \eqref{eq:old_form}.

This approach, besides providing this slight generalization of the model,
leads to a somewhat different perspective: Hawkes
process is a solution of a stochastic partial differential equation with jumps.
In fact Hawkes process corresponds to solution of one of the simplest such equations where
the state of the process at time $t$ is the random \textit {impulse function} $g_t(\cdot)$ that evolves by translation
in time $g_{t+\delta t}(\cdot)=g_t(\cdot +\delta t)$ with  jumps $g_t(\cdot)\to g_t(\cdot)+h(\cdot)$ occurring at rate
$\lambda(g_t(0))$.
There is a built-in invariance under time translation due to the form of \eqref{def:sigma}.
The evolution of the \textit{impulse function}  $g_t:\Rp \to \Rp$ is defined by:
  \begin{equation} g_t(s)=g_0(t+s)+\sum_{\tau\in S_t} h(t+s-\tau) \end{equation}
It is  $\F_t$-measurable and is a Markov process (it need not contain all the past information as is the case  when
$h(t)=e^{-t}$).
Given two times $s \ge t \ge 0$, the conditional distribution of the next $\tau \in S$ after time $t$
being greater than $s$ is given by
\begin{eqnarray}
  \P[\tau \ge s|{\cal F}_t]&:=&\P\big[ S \cap [s,t)=\emptyset, S \cap [t,\infty) \neq \emptyset | \F_t \big]\\
  &=&\exp \left[-\int_0^{s-t} \lambda(g_t(s))ds \right]
\end{eqnarray}
at which point $g_\tau(\cdot)$ jumps to $g_\tau(\cdot)+h(\cdot)$. Thus $g_t$ is by  itself a Markov process.

\subsection{Results}
The  two  main results of this paper, Theorems \ref{thrm:modulus} and \ref{thrm:jumps},  generalize the results of
Br\'{e}maud-Massouli\'{e} \cite{Bremaud} in different directions and prove that under certain conditions on
$\lambda$ and $h$, for a certain wide class $\mathcal C$ of initial conditions $g_0$, the distributions
of $g_t(\cdot)$ converge to a common limit.  Furthermore the limiting distribution is supported on $\mathcal C$. The
secondary new result, Theorem \ref{thrm:speed0}, is on speed of convergence.

In addition to these theorems we also observe several facts well known for attractive systems. These are
collected in Proposition \ref{prop:convergence}. The results are stated in sections \ref{sec:existence} and \ref{sec:uniqueness}.

\subsection {History and references to related work}
Point processes were first studied in \cite{Erlang} by Erlang in connection with queueing theory.
Hawkes process were first introduced in \cite{Hawkes} to study
self-exciting point processes.   See \cite{Cox,Liniger} for additional references.
The current work is the first one that covers cases where  $\lambda$  need not be Lipschitz and in fact can even have
jumps.  Large deviations  questions for  wide classes of   $\lambda$
and  $h$  have been studied by Zhu  in  \cite{Zhu1} and \cite{Zhu2};  for the special case of  linear $\lambda$
explicit  large deviation rates  and other limit theorems have been derived in  \cite{KZ}.

Currently Hawkes processes are used to model many phenomena ranging from queues and population growth
to mutations and  spread of infections; to defaults and jumps in financial markets; to neuroactivity and
social-networks; and finally even to modeling of artificial intelligence and creative thinking.
While most of the real world applications involve use of multi-dimensional Hawkes process, in order to keep the
presentation simple,  we limit ourselves to
the one-dimensional version. Possible generalizations  are outlined in section \ref{sec:gen}.

\subsection{Example} \label{ssec:example}
Simple example comes from population growth.
Consider a population that grows either by immigration or by birth generated by the current population. Immigrants are
treated as newborns when they arrive.   The immigration rate is $A$ and the birth rate for each individual is
$h(s)$, which depends only on  the current age $s$ of the individual. Then total growth rate of the population at time
$t$ is then given by
  \beq r(t)=A+\sum_{\tau \in S, \tau<t} h(t-\tau)\end{equation}  or if we normalize with $h$ by
$\int_0^\infty h(s) ds=1$, we replace this by
  \beq r(t)=A+B \sum_{\tau \in S, \tau<t} h(t-\tau)\end{equation}
since $B$ comes out of normalization.  So we see that this is a Hawkes process with $\lambda(z)=A+Bz$.  The Hawkes
processes with such linear $\lambda$ appear very often and are related to Galton-Watson trees which in turn provides an
easy way of studying many properties of Hawkes processes.

\vskip 0pt
\section{Formal Definitions} \label{sec:def}
\subsection{Formal Definition of a Hawkes Process}  \label{ssec:def}
\begin{definition} \label{P}
  \it{Hawkes process} is a random collection $S$ of
  points in $\Rp:=[0,\infty)$ characterized by triplet of functions $(\lambda,h,g_0)$ where the functions $\lambda,
  h,g_0: [0,\infty) \to [0,\infty)$. The conditional Poisson intensity at time $t$ is
  \beq r_t:=\lambda \left( g_0(t)+\sum_{0\le \tau<t, \tau \in S} h(t-\tau) \right) \end{equation} where the sum is over all
  previous points $\tau$ in $S_t:=S \cap [0,t)$. The function $g_0$ describes  the initial condition and the functions
  $\lambda$ and $h$ describe the evolution of the process.  We denote this Hawkes process by quadruple
  $(\P,\lambda,h,g)$.
\end{definition}
We will define a coupling of these different measures $\P_g$ by a canonical construction in the subsection
\ref{ssec:canonical}.

It is convenient to take as  state space  the space ${X }$ of locally integrable $\Rp$-valued functions on $\Rp$.
Let $\MM(X)$ be the space of probability distributions on ${X }$.
Then the process can be realized as an ${X }$-valued process.
Start at time $t$ from with initial condition $g_t \in {X }$ and consider the random evolution determined by two
components: deterministic
flow  $g_{t+\delta t}(s)=g_0(t+\delta t+s)$ up to stopping time $\tau$ at which point $g_\tau$ jumps
$g_{\tau+0}(x)=g_{\tau-0}(x)+h(x)$
with the
distribution of $\tau=\tau(g)$ given by
\begin{equation}
  \P_g[\tau \ge t+\delta t]=\exp\left[ -\int_0^{\delta t} \lambda(g_t(s)) ds \right].
\end{equation}
After time $\tau$, the process restarts in the sense that same procedure is to be repeated with $\tau$ as a new
starting time and $g_{\tau}$ as a new initial condition.
Each time stopping corresponds to a point in $S$.   The generator of the semi-group $T_t$ of the Markov process acting
on functions $F: {X } \to \Rp$  is given by:
  \beq \mathcal A:=\mathcal D+\lambda(g(0))(\theta_h-\mbox{Id}) \label{eq:generator}\end{equation}
where $\theta_h$ is a shift operator defined by $(\theta_h)F(g)=F(g+h)$ and $\mathcal D$ is the  derivative of
push-forward of time evolution defined by
  \beq \mathcal D F(g):=\lim_{\epsilon \to 0} \frac{F(\sigma_{\epsilon} g)-F(g)}{\epsilon} \end{equation}
where $\sigma_{\epsilon}$ is the time-shift operator defined by $\sigma_{\epsilon} g(s)=g(s+\epsilon)$.
Then this ${X }$-valued process $g_t$ satisfies
\begin{equation} \label{eq:g_t(s)}
  g_t(s)=g_0(t+s)+\sum_{\tau \in S_t} h(t+s-\tau).
\end{equation}

For any initial condition $g_0(\cdot)$, we have a  Markov
process $g_t(\cdot)$.
Then $\lambda(z_t)$ will be the intensity of the point process, where
  \begin{equation} z_t:=g_t(0)=g_0(t)+\sum_{\tau \in S_t} h(\tau -t). \end{equation}

It is sometimes more convenient to consider instead of $S_t$ the process
  \beq Q_t=(N_t,g_t) \end{equation}
where $N_t=\#(S_t)$ is the number of jumps from time $0$ to time $t$.  Then $Q_t$
can be viewed as a point process $N_t$  driven by the  Markov process  $g_t$.

\begin{remark}
  Without loss of generality we can assume $\int_0^{\infty} h(t) dt=1.$
  Indeed one can always achieve this if $\|h\|_1=\int h(t) dt<\infty$ by observing that triples
  $\big(\lambda(z),h(t),g(t)\big)$ and $\left(\lambda(z \|h\|_1),\frac{h(t)}{\|h\|_1}, \frac{g(t)}{\|h\|_1}\right)$
  produce the same Hawkes process. On the other hand if $\| h \|_1=\infty$ and $\inf_{z \in \Rp} \lambda(z)>0$
 then $\lim_{t \to \infty} g_t(0)=\infty$ a.s. and hence there can be  no  stationary version of the process.
\end{remark}

\subsection{Convergence notions} We equip the space $X$ with $L_1^{loc}$ metric
\beq \forall g,f \in X, \qquad d_X(g,f)=\sum_{i=1} \frac 1{2^n} \cdot \frac{\int_0^{n} |g(s)-f(s)|ds}{1+\int_0^{n} |g(s)-f(s)|ds} \end{equation}
The space $\MM(X)$  will then have the topology of  weak convergence  inherited from the metric space $X$.
Let $D(X)$ be the  space of $X$-valued functions on $[0,\infty)$ equipped with Skorohod $J_1$-metric $d$
and let $\G$ be the $\sigma$-field  generated by the
open sets in $({X }, d)$. Collection $\{\G_t\}$ is the corresponding filtration.
Let $g^*$ be the non-initial part of impulse function:
\begin{equation} \label{eq:g_t^*(s)}
  g_t^*(s)=\sum_{\tau \in S_t} h(t+s-\tau)
\end{equation}
and
\begin{eqnarray}
  \forall g, \tg \in D(X), \qquad d^*(g,\tg):=d(\mathcal S(g),\mathcal S(\tg))=d(g^*,\tg^*)
\end{eqnarray}
be the pull-back of $d$ under the map $\mathcal S:D(X)\to D(X)$, $\mathcal S: g \mapsto g^*$ defined according to \eqref{eq:g_t^*(s)} as
  \beq \big(\mathcal S(g)\big)_t(s)=g_t(s)-g_0(t+s) \end{equation}

Let $\mu_{f,s}$ be the distribution of $g_s$---impulse-function at time $s$---starting from initial condition
$f$.

The distribution of the whole process $g$ starting with initial condition $g_0$
 will be denoted by $\P_{g_0}$ and the associated expectations  will be denoted by $\E_{g_0}$.  Any $\mu \in \MM(X)$
can be viewed as a random initial condition and
  \beq \P_{\mu}:=\int \P_f \mu(df), \quad \E_{\mu}:=\int \E_f \mu(df), \end{equation}
denote the corresponding probability measure and expectation with respect to it.
Similarly $\P^*$ and $\E^*$ will be used as analogous expressions for the $g^*$.

In Theorems \ref{thrm:modulus} and \ref{thrm:jumps} we consider total variation $d_{TV}$
of the difference of $\P^*$ starting from two different initial conditions which can also be viewed as total variation of
the corresponding $\P$ but under a different $\sigma$-field $\G^*=\G \circ \mathcal S^{-1}$.

Furthermore let us  define $d_{TV,T}$ to be the  total variation after time $T$, i.e. the  total variation on the  sigma-algebra
$\G \circ \big(S|_{[T,\infty)} \big)^{-1}$ where operator $(S|_{[T,\infty})$ is defined by:
  \beq \big(\mathcal S|_{[T,\infty)}(g)\big)_t(s)=1_{t>T} \big(g_t(s)-g_0(t+s)\big) \end{equation}

\section{Tools} \label{sec:tools}
\subsection{Canonical Construction} \label{ssec:canonical}
One way to construct point processes is to start from a  canonical  Poisson point process $\P$.  To any measure $m$ one
can associate a Poisson  point process such  that for  any measurable set $A$, the  number of points in $A$ will be
random variable with  a Poisson distribution with mean  $m(A)$. In our case canonical  Poisson point process  will
correspond to the choice $m(A)=|A|$, the Lebesgue measure on $\Rp \times \Rp$.   Given $\lambda,h,g$ measure $\P$
induces a measure $\P_{g}^{\lambda,h}$ consistent with the definition above in the following way. If $S$ is our point
process then $\tau\in S$
if there is a point on the vertical line  ${\tau} \times [0,\lambda(z_t))$ of our plane $\Rp \times \Rp$. In the rest of
the paper we will denote $\P_g^{\lambda,h}$ by $\P_g$  because  $\lambda$ and $h$ would be the same unless stated
otherwise.

This induces a natural coupling of the collection of the measures $(\P_g)$ indexed by initial conditions $g$ that we
will call {\it canonical coupling}.  This coupling is no way unique but this particular coupling is maximal  when
$\lambda$ is  non-decreasing and we have two different initial conditions  one of which is strictly larger than the
other (by point-wise partial ordering).  The coupling is maximal in the sense that   the set of points common to both
of these two processes will  stochastically dominate the similar set of points in  any other coupling.

\subsection{Coupling / Stochastic Domination}
Here we present the canonical  coupling that we will  use.
\begin{tool*}[Stochastic Domination]
  Suppose $(S,\lambda,h,g_0)$ and $(\tS, \widetilde{\lambda}} \def\tilth{\widetilde{h}} \def\tiltg{\widetilde{g}} \def\tP{\widetilde{P},\tilth,\tiltg_0)$ are two Hawkes processes satisfying
  \beq \forall x \in \Rp, (h(x) \leq \tilth(x), g(x) \leq \tiltg(x)) \label{cond:order}\end{equation}
  \beq \forall x \leq y, \lambda(x) \leq \widetilde{\lambda}} \def\tilth{\widetilde{h}} \def\tiltg{\widetilde{g}} \def\tP{\widetilde{P}(y) \label{cond:midlambda}\end{equation}
  Then there exists a coupling such that  $S \ss \tS$.\\
  Note: $h, \tilth$ are not necessarily normalized.
\end{tool*}

\begin{proof}
  If $a(t)$, $b(t)$ are two intensities and if $a(t)\le b(t)$ for all $t$, one can couple the point processes,
  such that the two processes jump together with rate $a(t)$ and the second one jumps by itself at rate $b(t)-a(t)$.
  This can also be done if $\forall \omega \in \Omega, a(t,\omega) \leq b(t,\omega)$.  (In fact this is done naturally
  by our construction that is discussed in the previous subsection.)  Hence it remains to prove that
    \beq \lambda(g_t(0)) \leq \widetilde{\lambda}} \def\tilth{\widetilde{h}} \def\tiltg{\widetilde{g}} \def\tP{\widetilde{P}(\tiltg_t(0)) \label{eq:domination}.\end{equation}

  We know that up to the first jump in $\tS$ we have $g_t(0)=g_0(t), \tiltg_t(0)=\tiltg_0(t)$.  However we know from
  \eqref{cond:order} and \eqref{cond:midlambda} that
    \beq \lambda(g_0(t)) \leq \widetilde{\lambda}} \def\tilth{\widetilde{h}} \def\tiltg{\widetilde{g}} \def\tP{\widetilde{P}(\tiltg_0(t))\end{equation}
  which in turn implies that \eqref{eq:domination} holds up to first jump of $\tS$.
  Then we claim by induction that it holds for all times. Indeed at the time of jump the order of
  $g \leq \tiltg$ is preserved because there is no jump in $S$
  before first jump of $\tS$ and  the jump of  $\tiltg$ is always larger because $h \leq \tilth$.
\end{proof}

\subsection{Parent-Offspring Structure and the Branching representation}
We add the following parent-offspring structure to obtain
a random forest structure embedded in time. This will be  useful due to its connection to Galton-Watson trees (branching
processes).

We start with the Hawkes process corresponding to  functions $(\lambda,h,g_0)$ and let $S=\{\tau_1,\tau_2,...\}$, where
$\tau$ is an increasing sequence, i.e. $i<j \rightarrow \tau_i < \tau_j$.
Let us also denote
\beq \lambda_0(z):=\lambda(z)-\lambda(0) \end{equation}
Now to each $\tau_i$ we associate a randomly chosen parent element $p(\tau_i)$ from $S \cup \{-\infty\}$, where
$-\infty$ represents having no parent and being a root node.
Given a sequence $(\tau_1,\tau_2,...,\tau_i)$,   we define $\{p(\tau_i)\} $ to be  mutually independent with  distributions given by :
\begin{eqnarray}
   \P[p(\tau_i)=-\infty|\tau_1,...\tau_i]&=&
   \frac{\lambda(0)}{\lambda(z_{\tau_i})}+\frac{\lambda_0(z_{\tau_i})}{\lambda(z_{ \tau_i})}\frac{g(\tau_i)}{z_{\tau_i}}
\\ \P[p(\tau_i)=\tau_j|\tau_1,...\tau_i] &=&
    1_{j<i} \frac{\lambda_0(z_{\tau_i})}{\lambda(z_{ \tau_i})}\frac{h(\tau_i-\tau_j)}{z_{\tau_i}}
\end{eqnarray}
Note that $\{p(\tau_i)\}$ while   mutually independent are  not identically distributed.

Figure \ref{HawkesColor} provides a particular visualization of parent-offspring  structure that is consistent with the above description.
\begin{figure} \label{HawkesColor}
  \begin{center}
    \includegraphics[scale=0.5]{hawkesPC}
    \caption{For both $\lambda(z)=1+z, \quad h(t)=\frac{1}{(1+t)^2}$.  Each region corresponds to children of point of
    corresponding color}
  \end{center}
\end{figure}
We are specifically interested in linear case which provides us with dual description of the same process as described
in the following tool:

\begin{tool*}[Branching Process Equivalence]
  When $\lambda(z)=\bar{\lambda}} \def\bmu{\bar{\mu}} \def\bnu{\bar{\nu}(z)=A+Bz$ roots have rate $A+Bg(t)$ and each tree is a branching process with
  the number of branches having a Poisson distribution with mean $B$. The distribution of age $\tau_j$ of any parent  at
  the time $\tau$ of birth of any child  is then given by
    \beq \label{shape}  \P[ \tau-\tau_j >t | p(\tau)=\tau_j]=\int_t^{\infty} h(s) ds.\end{equation}
\end{tool*}
\begin{proof}
  By the above scheme we see that the intensity of roots is given by
  \beq \P[p(\tau_i)=-\infty]\lambda(z_{\tau_i})=\lambda(0)+Bz_{\tau_i}\frac{g(\tau_i)}{z_{\tau_i}}=A+Bg(\tau_i). \end{equation}
  Now each new point creates an area of size $B \|h\|_1=B$ and hence the number of children is Poisson$(B)$.  Now the
  shape of
  the area is given via $h$ which then proves \eqref{shape}.
\end{proof}
\vskip 0pt
\section{Existence} \label{sec:existence}
We begin with preliminary results regarding existence and ergodicity.
Let us consider the following hypothesis on Hawkes Process.

\begin{hyp}\label{hyp:existence}
  The function  $\lambda$ satisfies:
  \begin{equation} \label{cond:wd}
    \exists A,B \geq 0, \forall z \in \R, \lambda(z) \leq \bar{\lambda}} \def\bmu{\bar{\mu}} \def\bnu{\bar{\nu}(z):=A+Bz
  \end{equation}
\end{hyp}
which brings us to our first proposition:
\begin{proposition} \label{prop:convergence}
  Suppose Hawkes process satisfies Hypothesis~\ref{hyp:existence}. Then the following
  three statements hold:
  \begin{enumerate}[(i)]
  \item Hawkes process is well defined for all times.
  \item  When the Hypothesis \ref{hyp:existence} is satisfied with $B<1$, there exists an invariant distribution
    for process $g_t$.
  \item If in addition $\lambda$ is non-decreasing and we start from
    $0$ initial condition, i.e. $g_0=0$, then the distribution of $g_t$
    converges weakly to $\mu_0$, which being the unique minimal invariant measure is ergodic.
  \end{enumerate}
\end{proposition}

\begin{proof}[Proof of Proposition \ref{prop:convergence}]  (i) Consider Hawkes process with $\bar{\lambda}} \def\bmu{\bar{\mu}} \def\bnu{\bar{\nu}(z)=A+Bz$. By
  Galton-Watson representation it is well defined for all
  times. It can be coupled  with our  process  which it will dominate. Hence the dominated process is also  well defined
  for all times.

  (ii) If $B<1$, then the dominating process has  uniformly bounded  density; then so does the dominated process and we
  can get an invariant distribution $\mu_{g_0}$ by taking Cesaro limit along a subsequence
    \beq \label{eq:mu} \mu_{g_0}:=\lim_{t_k \to \infty} \frac 1{t_k} \int_0^{t_k} \mu_{s,g_0} ds \end{equation}
  where $\mu_{s,g_0}$ represents marginal distribution of the impulse-function at time $s$ given the initial condition
  $g_0$.

  (iii) The expected size of each tree is finite and hence we
  have bounded  density.  Since $\lambda$ is non-decreasing the process is order-preserving. Hence
    \begin{equation} \E_0\big[ f(g(t+s_1),\ldots, g(t+s_k)) \big] \end{equation}
  is non-decreasing in $t$ for any non-decreasing $f$ which implies convergence of $\mu_{s,0}$ to $\mu_0$.

  This gives us weak convergence.  We claim  that  invariant distribution $\mu_0$
  is  the unique minimal one, i.e. stochastically dominates any other invariant distribution.  To see this consider any
  other invariant $\widetilde{\mu}} \def\tpi{\widetilde{\pi}} \def\tS{\widetilde{S}} \def\tg{\widetilde{g}$ and choose initial condition chosen randomly  according  to  $\widetilde{\mu}} \def\tpi{\widetilde{\pi}} \def\tS{\widetilde{S}} \def\tg{\widetilde{g}$; but then we have
  point-wise domination initially and by coupling we see that  there is domination at all times and letting
  $t\to\infty$,  we see that $\widetilde{\mu}} \def\tpi{\widetilde{\pi}} \def\tS{\widetilde{S}} \def\tg{\widetilde{g}$ dominates $\mu$.  Hence $\mu$, being the unique   minimal  invariant measure, is
  extremal. Hence it  is ergodic.
\end{proof}

\section{Uniqueness} \label{sec:uniqueness}
\begin{definition} \label{stable}
  Let $\mu$ be an  invariant distribution supported on  a class of functions $\CC$, meaning that
    \beq \mu(\CC)=1 \end{equation}
  Then we say that the pair $(\lambda,h)$ is {\it
  $(\CC,\mu)$-stable} if starting from any initial condition $g_0$ in $\CC$ the distribution $\mu_{t,g_0}$ of the
  impulse
  function at time $t$ converges to $\mu$:
    \beq \lim_{t \to \infty} \mu_{t,g_0} \mathop{=}^d \mu \end{equation}
  where $\mathop{=}^d$ signifies that the limit is in the  sense of weak convergence.
\end{definition}

Now we turn to main results of this paper.  Here is our $2^{nd}$ hypothesis:

\begin{hyp}\label{hyp:H}
  The function $\lambda$ is non-decreasing and it satisfies
  \begin{equation} \label{cond:modulus}
      \sup_{x \in \Rp} \left( \lambda(x+s)-\lambda(x) \right)\le  \phi(s)
    \end{equation}
    for some concave non-decreasing $\phi$ satisfying
    \begin{equation} \label{cond:phiH}
      \int_{0}^{\infty} \phi(H(s))ds=C<\infty, \quad \mbox{where } H(s)=\int_s^\infty h(t)dt
    \end{equation}
\end{hyp}

\begin{theorem} \label{thrm:modulus}
  If Hawkes process satisfies both Hypothesis \ref{hyp:existence} with $B<1$ and Hypothesis \ref{hyp:H} then
  pair $(\lambda,h)$ is \it{$(\CC,\mu)$-stable} as in definition \ref{stable}
  with
  \begin{equation}
      \CC:=\left\{ g \in C(\Rp): \int_0^{\infty} \phi(g(s))ds< \infty \right\}
  \end{equation}
  and $\mu$ being $\mu_0$ from Proposition \ref{prop:convergence} part (iii).
\end{theorem}

\begin{remark}
  In this theorem we relax the Lipschitz condition with constant 1 on $\lambda$ that was imposed in \cite{Bremaud} (in a
  slightly different form since
  $\| h \|$ was not normalized).  When we do have this assumption the results of \cite{Bremaud} follow as a corollary of
  the above theorem \ref{thrm:modulus}.  In fact the results follow even under weaker assumption that $\lambda$ is
  Lipschitz with constant $L$ for some $L$:
  \beq \forall x,y \in \Rp |\lambda(x)-\lambda(y)|  \leq L |x-y| \end{equation}
\end{remark}

  Before we start with proof of theorem \ref{thrm:modulus} let us prove the following lemma:

\begin{lemma} \label{lemma:mean-field}
  If $\mu$ is a stationary distribution of the impulse function $g_t$ of a Hawkes process then
  \beq \E_{\mu}[g(s)]=\E_{\mu}[\lambda \big(g(0) \big)]H(s). \end{equation}
\end{lemma}

\begin{proof} By stationarity
  \begin{eqnarray}
    \nonumber \E_{\mu}[g(s)]  &=&\E_{\mu}\left[ \int_s^{\infty} \lambda\big(g(0))h(t) dt \right]
			    \\&=&\E_{\mu}[\lambda \big(g(0) \big)]\int_s^{\infty} h(t) dt
			    \\&=&\E_{\mu}[\lambda \big(g(0) \big)]H(s).\nonumber
  \end{eqnarray}
\end{proof}

\begin{proof}[Proof of Theorem \ref{thrm:modulus}:]
  We use a recurrence argument. It is enough to show that exists some class $\CC' \ss \CC$ such that:

  (i) $f \in \CC'$ implies that $\P^*_f$ and $\P^*_0$ have non-trivial overlap
      \beq I(\P^*_f, \P^*_0):=1-d_{TV}(\P^*_f, \P^*_0)\geq \delta >0 \end{equation}

  (ii) $\CC'$ is {\it recurrent with respect to $\CC$}: starting from any point in class $\CC$
    the impulse function will enter $\CC'$ at some time in the future, i.e.
      \beq \forall g_0 \in \CC, \P_{g_0}[\forall t \in \Rp, g_t \notin \CC' ]=0 \end{equation}

  We will split the proof into three steps.  In step (i) and (ii) we will show the above statements
  and in step (iii) we will complete  the proof using recurrence.  Step (i) in turn will suggest the choice of $\CC'$.

  Step (i): First observe that by applying Jensen's inequality and concavity of $\phi$ we obtain:
    \begin{eqnarray}
      I(\P^*_f, \P^*_0)&=&\E_{0}\left[ \exp \left(-\int_0^{\infty} |\lambda(z_t+f(t))-\lambda(z_t)| dt \right) \right]
      \\&\geq& \exp \left(-\E_{0}\left[\int_0^{\infty} |\lambda(z_t+f(t))-\lambda(z_t)| dt \right]\right)
      \\&\geq& \exp \left(-\E_{0}\left[\int_0^{\infty} \phi(f(t)) dt \right] \right)
      \\&=& \exp \left(-\int_0^{\infty} \phi(f(t)) dt \right)
    \end{eqnarray}

  Step (ii): Step (i) suggests that  we take
    \beq \label{CC'} \CC':=\left\{ g \in C(\Rp): \int_0^{\infty} \phi(g(s))ds< K \right\} \end{equation}
    whereas for $\CC'$ to satisfy the recurrence condition we will make  a suitable choice of  $K$.  We know that for
    any finite time $t$ process which started from $\CC$ will remain in $\CC$ since $\phi$ is convex.  Now
    consider starting from $\CC$ but replacing  $\lambda$  by with $\bar{\lambda}} \def\bmu{\bar{\mu}} \def\bnu{\bar{\nu}$ in the definition  of our process.
    Then, because $\phi$ is nondecreasing   by applying stochastic domination   we get
    \begin{eqnarray} \label{eq:theorem3start}
      \E^{\lambda}_{g} \left[\int_0^{\infty} \phi(g_t(s))ds \right] &\leq&  \E^{\bar{\lambda}} \def\bmu{\bar{\mu}} \def\bnu{\bar{\nu}}_{g} \left[\int_0^{\infty}
      \phi(g_t(s))ds \right].
    \end{eqnarray}
    We now take  $\limsup_{t \to \infty}$ on  both sides of \eqref{eq:theorem3start} and  by proposition
    \ref{prop:convergence}(iii),   the
    right hand side converges to a limit from any initial $g$.

    \begin{eqnarray}
      \overline{\lim_{t \to \infty}} \E^{\lambda}_{g} \left[\int_0^{\infty} \phi(g_t(s))ds \right]
      &\hspace{-10pt} \leq&  \lim_{t \to \infty} \E^{\bar{\lambda}} \def\bmu{\bar{\mu}} \def\bnu{\bar{\nu}}_{g} \left[\int_0^{\infty} \phi(g_t(s))ds \right] \\
      &\hspace{-10pt} \leq&  \lim_{t \to \infty} \E^{\bar{\lambda}} \def\bmu{\bar{\mu}} \def\bnu{\bar{\nu}}_{0} \left[\int_0^{\infty} \phi(g_t(s))ds \right] \\
      \label{eq:theorem3end2}&\hspace{-10pt} \leq&  \lim_{t \to \infty} \int_0^{\infty} \phi(
      \E^{\bar{\lambda}} \def\bmu{\bar{\mu}} \def\bnu{\bar{\nu}}_{0}[\lambda(g_t(0))]H(s))ds
    \end{eqnarray}
    where the line \eqref{eq:theorem3end2} follows from Lemma \ref{lemma:mean-field}.
    Recall that part of assumption of hypothesis 2 is that $\phi$ is concave and increasing. While  concavity  implies
    that $\phi(cx) \leq c \phi(x)$ for $c \geq 1$, monotonicity implies  that $\phi(cx) \leq \phi(x)$ for $c<1$.
    Combining these two we get $\phi(cx) \leq \max(c,1) \phi(x)$. Applied to \eqref{eq:theorem3end2} with
    $c=\E^{\bar{\lambda}} \def\bmu{\bar{\mu}} \def\bnu{\bar{\nu}}_{0}[\lambda(g_t(0))]$ we obtain:

    \begin{eqnarray}
      \label{eq:theorem3end} \label{eq:K}
      \overline{\lim_{t \to \infty}} \E^{\lambda}_{g} \left[\int_0^{\infty} \phi(g_t(s))ds \right]
      &\hspace{-10pt} \leq& \hspace{-5pt} C  \max( \lim_{t \to \infty} \E^{\bar{\lambda}} \def\bmu{\bar{\mu}} \def\bnu{\bar{\nu}}_{0}[\lambda(g_t(0))],1)
    \end{eqnarray}
    where $C=\int_0^{\infty} \phi(H(s))ds<\infty$ by Hypothesis \ref{hyp:H}.
    Hence \eqref{eq:K} is finite since  $\lim_{t \to \infty} \E^{\bar{\lambda}} \def\bmu{\bar{\mu}} \def\bnu{\bar{\nu}}_{0}[\lambda(g_t(0))]<\infty$
    by branching representation. Hence $\mu(\CC)=1$ and in \eqref{CC'} it is enough to set
    \begin{eqnarray}
      K=\max( \lim_{t \to \infty} \E^{\bar{\lambda}} \def\bmu{\bar{\mu}} \def\bnu{\bar{\nu}}_{0}[\lambda(g_t(0))],1) \times \big[ \int_0^{\infty} \phi(H(s))ds\big].
    \end{eqnarray}

  Step (iii): Hence we return almost surely  to the class $\CC'$.  Now starting from any initial condition in $\CC'$,
    \begin{eqnarray} \label{stepiiibound}
      I(\P^*_f, \P^*_0)&\geq&1-\exp \left(-K \right)
    \end{eqnarray}
    and result follows from the following schematic representation
      \beq \label{eq:collapsed} g \rightarrow \CC  \rightleftarrows \CC' \rightarrow \mu \end{equation}
    where all arrows represent transitions with uniformly positive probability
    which implies the convergence to $\mu$.

    Let us describe this in detail. Define two sequences of alternating  stopping times as follows:
    Let stopping time ${\varsigma}_1$ be the first time that $g_t$ is in $\CC'$.
    Then we try to couple the process $S$ with $S^{(1)}$ where $S^{(1)}$ is Hawkes process with same pair $(\lambda,h)$
    but that starts at time $t$ with $g_t=0$; we use canonical coupling.
    Then by step (i) the coupling is successful with probability at least $\delta>0$.
    If it is not then there is a stopping time $\upsilon_1 \in S \setminus S^{(1)}$, in this case we
    restart the procedure defining ${\varsigma}_i, \upsilon_i$ by the following recursive definition
    \begin{eqnarray}
      {\varsigma}_1		&=&\inf \{t: g_t \in \CC'\} \nonumber \\
      \forall i \geq 1, \upsilon_i 	&=&\inf \{t>{\varsigma}_i: t \in S \setminus S^{(k)}\} \label{eq:stimes}\\
      \forall i >1,	{\varsigma}_i		&=&\inf \{t>\upsilon_{i-1}: g_t \in \CC' \} \nonumber
    \end{eqnarray}
    where $S^{(k)}$ is Hawkes Process that starts at time ${\varsigma}_k$ with condition $g_t=0$.
    Hence we want to show that almost surely for some $i$, ${\varsigma}_i=\infty$ which is immediate from step (i):

    \begin{eqnarray}
      \P[\forall i, {\varsigma}_i<\infty]
      &=&\E[\prod_{i=1}^{\infty} I(\P^*_{g_{\tau_i}}, \P^*_0)]\\
      &\leq& \prod_{i=1}^{\infty}  \big( 1-e^{-K} \big)=0.
    \end{eqnarray}
\end{proof}

Finally we turn to the last theorem where $\lambda$ can have jumps.

\begin{hyp} \label{hyp:jumps}
  (i) Function $\lambda$ is non-decreasing and satisfies:
    \begin{equation} \label{cond:jumps:l}
      \mbox{ $\lambda(0)>0$, $\lambda \leq \bar{\lambda}} \def\bmu{\bar{\mu}} \def\bnu{\bar{\nu}$, $B<1$}
    \end{equation}
  (ii) Function $h$ is convex and satisfies:
    \begin{equation} \label{cond:jumps:h}
      \mbox{$\|h\|_{\infty}<\infty$, $\log |h'(x)|=o(x)$},
      \int_0^{\infty} t h(t) dt<\infty
    \end{equation}
\end{hyp}

\begin{remark}
  These assumptions allow large tails in $h$ which was the aim of this study, in
  particular $h(x)=\frac{p}{(1+x)^{p+1}}$ where $p>0$.
\end{remark}
\begin{remark}
  These assumptions can be weakened to allow piece-wise decreasing $h$ as long as $h^{-1}(y)$ is finite for any $y \in
  \Rp$.
\end{remark}

\begin{theorem} \label{thrm:jumps}
  If Hawkes process satisfies Hypothesis \ref{hyp:existence} with $B<1$ and
  Hypothesis \ref{hyp:jumps} then pair $(\lambda,h)$ is \it{$(\CC,\mu)$-stable} as in definition \ref{stable}
  with
  \begin{equation} \label{def:Class2}
    \CC:=\left\{ g \in C(\Rp):  \int_0^{\infty} t g(t)dt<\infty \right\}
  \end{equation}
  and $\mu$ being $\mu_0$ from Proposition \ref{prop:convergence} part (iii).
\end{theorem}

\begin{definition} \label{def:nu}
  Let $\nu$ denote distribution of $g(0)$ under $\mu$.
\end{definition}

Next we state estimates on the tail of $\nu$ as well as on the probability density of $\nu$.
We will prove the following two  Lemmas after we complete  the proof of Theorem \ref{thrm:jumps}.

\begin{lemma}  \label{lemma:total} Under Hypothesis
    \ref{thrm:jumps} the following statements hold:
    \begin{equation} \label{eq:exp_mu}
      \exists \theta>0, \E_{\mu}[e^{\theta z_t}]<\infty,
    \end{equation}
    In particular $\nu$ has exponential tail: there exist constants   $0< c_1<\infty$ and $\theta_1>0$  such that for
    all  $z\ge 0$,
    \begin{equation}
      \nu[z,\infty) \leq c_1 e^{-\theta_1 z},
    \end{equation}
\end{lemma}

\begin{lemma} \label{lemma:local} Under Hypothesis \ref{hyp:jumps}, there exists a constant $c_1'$ such that for any invariant distribution
$\mu$, the associated $\nu$ (distribution of $g(0)$ under $\mu$),
  has a density $\psi(z)$ that satisfies
  \begin{equation}
    \psi(z) \leq c'_1 \lambda(z) \nu[z+h(0), \infty)
  \end{equation}
  Furthermore there exist constants  $c_2,\theta_2$ such that
  \begin{equation}
    \psi(z) \leq c_2 e^{-\theta_2 z}
  \end{equation}
\end{lemma}

\begin{proof}[Proof of Theorem \ref{thrm:jumps}:]
  Unlike proof of Theorem \ref{thrm:jumps} the comparison is done between initial conditions $f$ and $g+f$ where $g$ is
  randomly chosen with distribution $\mu_0$. We denote  by $\mu^\ast_f$ the distribution of $g+f$, i.e. $\mu_0$
  shifted    by $f$, i.e. $\mu^\ast_f(A):=\mu_0 \big(\{g-f: g \in A\} \big)$.
  The idea of the  proof remains the same as that of  Theorem \ref{thrm:jumps} while the calculations
  of \eqref{eq:theorem3start}-\eqref{eq:theorem3end} are redone as follows.

  \begin{eqnarray}
    I(\P^*_f, \P^*_{\mu_f^*})&=&\E_{\mu_0}\left[ \exp \left(-\int_0^{\infty}
    \lambda \big(z_t+f(t) \big)-\lambda(z_t) dt \right) \right]
    \\&\geq& \exp \left(-\E_{\mu_0}\left[\int_0^{\infty} \lambda \big(z_t+f(t) \big)-\lambda(z_t) dt \right]\right)
  \end{eqnarray}
  by Jensen's Inequality.  Since  $\mu_0$  is invariant we can replace $z_t$ by $z_0$:
  \begin{eqnarray}
    E_1&:=&\E_{\mu_0}\left[\int_0^{\infty} \lambda \big(z_t+f(t) \big)-\lambda(z_t) dt \right]
    \\&=& \E_{\mu_0}\left[\int_0^{\infty} \lambda \big(z_0+f(t) \big)-\lambda(z_0) dt \right]
    \\&=& \mathop{\mathop{\int \int \int}_{z,x,t \in \Rp}}_{z \leq x \leq z+f(t) } \psi(z) d\lambda(x)  dz dt
    \\&=& \mathop{\mathop{\int \int \int}_{z,x,t \in \Rp}}_{x-f(t) \leq z \leq x } \psi(z) dz d\lambda(x)  dt
  \end{eqnarray}
  We can now apply the estimate on $\psi(z)$ from  Lemma \ref{lemma:local} and obtain
  \begin{eqnarray}
    E_1&\leq& \mathop{\mathop{\int \int \int}_{z,x,t \in \Rp}}_{x-f(t) \leq z \leq x }  c_2 e^{-\theta_2 z} dz
d\lambda(x)  dt
    \\&\leq& \mathop{\mathop{\int \int \int}_{z,x,t \in \Rp}}_{-f(t) \leq z \leq 0 }  c_2 e^{-\theta_2 x} e^{-\theta_2 z} dz d\lambda(x)  dt
    \\&\leq& \mathop{\mathop{\int \int \int}_{z,x,t \in \Rp}}_{-f(t) \leq z \leq 0 }
	      c_2 e^{-\theta_2 x} e^{\theta_2 f(t)} dz d\lambda(x)  dt d\lambda(x)  dt
    \\&=&  c_2 \|f e^{\theta_2 f} \|_{L_1} \mathop{\int}_{x \in \Rp} e^{-\theta_2 x} d\lambda(x)
    \\&\leq&  c_2 \|f\|_{L_1} e^{\theta_2 \|f\|_{L_{\infty}}} \mathop{\int}_{x \in \Rp} e^{-\theta_2 x} d\lambda(x)
    \\&<&\infty
  \end{eqnarray}
  where the last line follows from Hypothesis \ref{hyp:jumps} since $f \in L_1 \cap L_{\infty}$; last term  is integrated by parts:
  \begin{eqnarray}
    \mathop{\int}_{x \in \Rp} e^{-\theta_2 x} d\lambda(x)=\lambda(0)+
    \mathop{\int}_{x \in \Rp} \theta_2e^{-\theta_2 x} \lambda(x)dx<\infty
  \end{eqnarray}

  The rest of the proof follows analogously to Theorem \ref{thrm:modulus}.
\end{proof}

\subsection{Lemmas for Theorem \ref{thrm:jumps}}

The two lemmas in this section provide estimates on the
density as well as the tail  probabilities of distribution $\nu$.
Let us first prove the tail estimate.
\begin{proof}[Proof of Lemma \ref{lemma:total}:]
  Observe first that $\nu[z,\infty] \leq \bnu[z,\infty]$ by stochastic domination. We can now  use Galton-Watson representation in the
  following way.
  Consider a  tree  $T$  viewed as a population starting with one individual born at time $0$.
  Each individual born at time $s$  gives  birth to new individuals at  rate   $Bh(t-s)$. Since $B<1$ and $\int h(s) ds=1$, the size of the
  tree $T$ is finite. Consider the quantities:
  \begin{eqnarray}
    \label{def:Lambda_T} \Lambda_{\theta}(t)&:=&{\mbox{ln}}} \def\lg {{\mbox{log}}} \def\eps {{\epsilon}\, \E [\exp( \theta \sum_{\tau \in T \atop \tau\le t}h(t-\tau))]\\
    \label{def:Lambda} \Lambda_{\theta}&:=&{\mbox{ln}}} \def\lg {{\mbox{log}}} \def\eps {{\epsilon} \,\E_{\mu}[\exp(\theta z_t)]
  \end{eqnarray}

  If we ignore time,  we have a Galton-Watson tree with the number of branches having a Poisson distribution with
  parameter $B<1$.
  Then it is known that (see \cite{Nakayama} for example), the total size $Z$ of the population has an exponential moment, i.e.
  \beq \E [e^{\theta Z}]<\infty \end{equation}
  for some $\theta_0>0$. In particular since $\|h\|_\infty<\infty$, exists $\theta=\frac{\theta_0}{\|h\|_\infty}$
  \beq \E \big[\exp( \theta \sum_{\tau \in T\atop\tau\le t}h(t-\tau) \big]\le \E[\exp(\theta \|h\|_\infty Z)]<\infty \end{equation}
  By dominated convergence theorem
  \beq \lim_{t\to\infty} \Lambda_{\theta}(t)=0\end{equation}

  Observe that for linear $\bar{\lambda}} \def\bmu{\bar{\mu}} \def\bnu{\bar{\nu}(z)=A+Bz$
  \beq \exp[\sum_{\tau\in T\atop\tau\le t} \theta h(t-\tau)-\theta h(t)-B\int_0^t [e^{\Lambda_\theta(t-s)}-1]h(s)ds] \end{equation}
  is a martingale. Therefore $\Lambda_\theta(t)$ satisfies
    \beq \label{eq:Lambda_T} 	\Lambda_{\theta}(t)=\theta h(t)+B\int_0^t (e^{\Lambda_{\theta}(t-s)}-1)h(s)ds \end{equation}
  Similarly
    \beq \label{eq:Lambda} 	\Lambda_{\theta }=A \int_0^{\infty} (e^{\Lambda_{\theta}(s)}-1)ds<\infty \end{equation}
  as we add the contributions of  $e^{\Lambda_T(s)}-1$ from  trees at $-s$ that are created at rate $A$.  It only remains
  to show that right hand side of \eqref{eq:Lambda_T} is finite.

  Integrate \eqref{eq:Lambda_T} with respect to $t$ from $0$ to $\infty$.
  \begin{eqnarray}
    \int_0^{\infty} \Lambda_\theta(t)  &\le& \theta +B \int_0^{\infty} dt\int_0^t  (e^{\Lambda_{\theta}(t-s)}-1)h(s)dsdt\\
    &=& \theta + B\int\int_{0\le s\le t<\infty} (e^{\Lambda_{\theta}(t-s)}-1) h(s)dtds\\
    &\le &\theta+ B\int_0^{\infty}  (e^{\Lambda_{\theta}(t)}-1)dt
  \end{eqnarray}
  Since $\Lambda_\theta(t)\to 0$ as $t\to\infty$, given $\delta>0$, we can bound $ (e^{\Lambda_{\theta}(t)}-1)$ by
  $(1+\delta)\Lambda_\theta(t)$
  for sufficiently large $t$,  providing us with an estimate:
    \beq \int_0^{\infty} \Lambda_\theta(t) dt\le \theta+C_\delta+B(1+\delta)\int_0^{\infty} \Lambda_\theta(t) dt \end{equation}
  We choose $\delta>0$ so that $B(1+\delta)<1$. This proves that $\int_0^\infty \Lambda_\theta(t) dt$ as well as $\int_0^\infty
  (e^{\Lambda_{\theta}(t)}-1)dt$
  are finite.
\end{proof}

Having obtained exponential tail  estimates we now turn to  the proof of the density estimate:
\begin{proof}[Proof of lemma \ref{lemma:local}:] Let $L(s,t,A)$ be  time spent in $A$ by $z_t:=g_t(0)$
  between time $s$ and $t$ starting from arbitrary initial condition, i.e.
  \begin{equation}
    L(s, t,A)=\int_s^t 1_{z_s \in A}ds.
  \end{equation}
  Let $I=[z,z+\epsilon]$, $\tau_0=0$ and $\tau_i, i>0$ be $i^{th}$ jump after $0$, that is
    \begin{equation}\tau_i=\inf\{t>\tau_{i-1}: t \in S\}. \end{equation}
  Furthermore let
    \begin{equation} n_t=\inf\{s>t: s \in S\}.  \end{equation}
  be the first jump after time $t$. We note that we can calculate $\nu(I)$ as

    \begin{eqnarray} \nu[I]&=&\lim_{t \to \infty} \frac{\E_{\mu}\left[L(0,t,I)\right]}t\end{eqnarray}
  We consider first the numerator.
    \begin{eqnarray} L(0,t,I)&\leq&\sum_{i} L(\tau_i,\tau_{i+1},I) 1_{\tau_i<t}\end{eqnarray}
  Taking expectations and using the properties of  conditional expectations :
  \begin{eqnarray}
    \E[L(0,t,I)] 	&\leq&	\E[\sum_{i} \E[L(\tau_i,\tau_{i+1},I)|\F_{\tau_i-}] 1_{\tau_i<t}]
    \\			&=&	\E[\sum_{i} F(g_{\tau_i-}) 1_{\tau_i<t}]
  \end{eqnarray}
  where function $F:X \to \Rp$ is defined as
    \begin{eqnarray} F(g)=\E_g[L(0,\tau_1,I)] \end{eqnarray}
  We   finally have
  \begin{eqnarray} \label{eq:LandF}
    \E[L(0,t,I)] \leq \E[\int_0^t F(g_{t-}) dN_t]
  \end{eqnarray}
  which we can now view  as  a stochastic integral  of a predictable process with respect to  $dN_t$: using \eqref{eq:LandF} and the
  stationarity of $g(t)$ we obtain
  \begin{eqnarray}
    \nu[I]&=&\lim_{t \to \infty} \frac{\E_{\mu}\left[L(0,t,I)\right]}t
    \\&\leq&\lim_{t \to \infty} \frac{\E_{\mu}\left[ \int_0^t F(g_{t-}) dN_t \right] }t
    \\&\leq&\lim_{t \to \infty} \frac{\E_{\mu}\left[ \int_0^t F(g_{t-}+h) \lambda(g_t) dt \right] }t
    \\&\leq&\lim_{t \to \infty} \frac{\E_{\mu}\left[ \int_0^t F(g_{0}+h) \lambda(g_0) dt \right] }t
    \\&=&\int_{g \in X} \E_{g+h}\left[L(0,\tau_1,I)\right] \lambda(g(0))  d\mu(g)
    \\&\leq& \int_{g \in X} \E_{g+h}\left[\frac{\epsilon }{-h'(\tau_1)}\right] 1_{g(0)+h(0)>z} \lambda(g(0))  d\mu(g) \label{eq:product}
  \end{eqnarray}
  The  last equation follows from condition \eqref{cond:jumps:h}. By convexity,  $-h'$ is decreasing and
  $L(0,\tau_1,I) \leq \frac{\epsilon}{g'(\tau_{1})}\leq \frac{\epsilon}{h'(\tau_{1})}$ since between two jumps $g(t)$  can cross
  $I$ only once. Because $\lambda(g(0))\ge A>0$,  we have the estimate  $\P[\tau_1>t]\leq e^{-At}$.
  In view of  condition  \eqref{cond:jumps:h} we obtain
    \begin{eqnarray} \E_{g}\left[\frac{\epsilon}{-h'(\tau_1)}\right] &\geq& \epsilon C_1 \end{eqnarray}
  for some constant  constant $C_1>0$. Then we continue from \eqref{eq:product}
  \begin{eqnarray}
    \nu[I]&\leq&	\int_g \E_{g+h}\left[\frac{\epsilon }{-h'(\tau_1)}\right] 1_{g(0)+h(0)>z}\lambda(g(0))  d\mu(g)
    \\	&\leq& 	\epsilon C \int_g 1_{g(0)+h(0)>z}\lambda(g(0))  d\mu(g)
    \\	&=& 	\epsilon C \int_{x>z+h(0)} \lambda(x)  d\nu(x)
    \\	&=& 	\epsilon c'_1 \lambda(z)  \nu[z+h(0),\infty)
  \end{eqnarray}
  where the last  inequality follows the fact that $d\nu(x)$ has exponential tail.  Hence we have
  \begin{equation} \label{eq:local_final}
    \nu[I]\leq \epsilon c'_1 \lambda(z) \nu[z+h(0),\infty)
  \end{equation}
  from which we first observe that there is density $\psi$ since then $\nu[I] \leq c'_1 \lambda(z) \epsilon$ and then taking limit as
  $\epsilon \to 0$ in \eqref{eq:local_final} we obtain the estimate on the density.
\end{proof}

\section{Speed of Convergence}

In general, estimating  the speed of convergence by the methods  of Theorem \ref{thrm:modulus} and \ref{thrm:jumps}
appears to be difficult. To overcome  this we need to strengthen  Hypothesis \ref{hyp:H} to the following:
\begin{hyp} \label{hyp:H2}
  The function $\lambda$ is non-decreasing and it satisfies
  \begin{equation} \label{cond:modulus2}
    \sup_{x \in \Rp} \left( \lambda(x+s)-\lambda(x) \right)\le  \phi(s)
  \end{equation}
  for some $\phi$ that satisfies
  \beq \label{eq:phi(h)} \tilde B:=B \int_0^{\infty} \phi(h(t)) dt<1. \end{equation}
    $\tilde h(t)=\frac{B \phi(h(t))}{\tilde B}$ that satisfies:
  \beq \label{eq:iterative} \limsup_{t \to \infty}
    \left( \frac{\sum_{n=1}^{\infty} \tilde B^n n \int_{x>\frac{t}{2n}} {\tilde h}(x) dx}{\int_{x>t} {\tilde h}(x) dx} \right)=:C_4<\infty
  \end{equation}
    In addition ${\tilde g(t)}=\phi(g(t))$ also satisfies $\|\tilde g\|_{1}=\int_0^{\infty}{ \tilde g}(t)dt<\infty$
\end{hyp}
\begin{remark}
  Condition \eqref{eq:iterative} can be replaced by other conditions as can  be seen from the proof.  The  important
  feature is that all nicely decaying functions like  ${\tilde h}(x)=\frac p{(1+x)^{p+1}},p>0$ satisfy  Hypothesis  \ref{hyp:H2}, while
  the following $h$ does not: $h(x)=\sum_{i=0}^{\infty} 2^{-2i-1} 1_{2^{i}<x<2^{i+1}}$ where $1_{2^{i} \leq x<2^{i+1}}$ is
  the indicator function that is $1$ if $x \in [2^i,2^{i+1})$ and 0 otherwise.
\end{remark}

\begin{theorem} \label{thrm:speed0}
  If Hawkes process satisfies Hypothesis \ref{hyp:existence} with $B<1$, and Hypothesis \ref{hyp:H2} then:
  \begin{eqnarray} \label{eq:exact}
    d_{TV,t}(\P_g, \P_0) \leq B \int_t^{\infty} [(I- BQ_{\tilde h})^{-1}{\tilde g}] (s)ds
  \end{eqnarray}
  where $Q_{\tilde h}$ stands for the  convolution operator $Q_{\tilde h}g={\tilde h}\ast g$.

  In turn this implies that tail of $d_{TV,t}(\P_g, \P_0)$ cannot be much worse than that of $\phi(h(t))$ or $\phi(g(t))$;
  more precisely:
  \begin{eqnarray} \label{eq:tail}
    \limsup_{t \to \infty} \frac{d_{TV,t}(\P_g, \P_0)}{\int_{\frac t2}^{\infty} \phi(g(s))+\phi(h(s)) ds}< \infty
  \end{eqnarray}
\end{theorem}
\begin{proof}
  Consider the canonical coupling of  the measures $\P_g$ and $\P_0$, i.e.  a pair of processes $(g,g^{(0)})$
  which start from initial conditions $g_0=g$ and $g^{(0)}_0=0$ respectively.
  Let $\tri g=g-g^{(0)}$ be the difference of impulse functions, $\tri S=S \setminus S^{(0)}$
  be the difference in sets and the last point in $\tri S$ being:
    \beq L:=\max \{s \in \Rp: s \in \tri S\} \end{equation}

  Then observe that the concavity of $\phi$ along with  $\phi(0)=0$ implies subadditivity of $\phi$, i.e. $\phi(a+b) \leq \phi(a)+\phi(b)$.
  It now follows from  \eqref{cond:modulus2} that
  \begin{eqnarray}
    \lambda(g_t(s))&-&\lambda(g_t^0(s))
    \\&\leq& \phi \big(\tri (g_t(s)) \big)
    \\ &=&\phi(g_0(t+s)+\sum_{\tau \in \tri S\atop  \tau<t} h(t+s-\tau))\\
    &\leq& \phi(g_0(t+s))+\sum_{\tau \in \tri S\atop \tau<t} \phi(h(t+s-\tau))
  \end{eqnarray}

  By setting $s=0$, we see  that the point process   with rate $\widetilde r_t$ at time $t$ given by
  \begin{eqnarray}
    \widetilde r_t:=\phi(g_0(t))+\sum_{\tau \in \tri S\atop \tau<t} \phi(h(t-\tau))
  \end{eqnarray}
  dominates stochastically the process $\tri  S$.  Consider the process $\tri D$ with the  following description: one begins with roots that
  are created with rate $\phi(g(t))$ at time $t$ and then each root generates a Galton-Watson tree with birthrate  at time $t$ along  any
  branch born  at time $s$ given by   $\phi(h(t-s))$.  Let the last point of $\tri D$   be:
    \beq L_D=\max \{s \in \Rp: s \in \tri D\} \end{equation}

  Notice that
    \beq d_{TV,t}(\P_g, \P_0)=\P[L>t] \leq \P[L_D>t] \leq \E[\#(\tri D \cup [t,\infty))] \label{eq:rec:ineq} \end{equation}
  where the first inequality follows from $L \leq L_D$ because $\tri S \subset \tri D$ and the second
  from  $L_D \in \tri D$. So we will be estimating  $\E[\#(\tri D \cup [t,\infty))]$ which turns out to
  be much simpler to work with.
  If we define
    \beq l( A):= \E[\#(\tri D \cap A)] \end{equation}
  then:
    \begin{eqnarray}
    \int_0^{\infty} e^{\theta t} l(dt)= \E[\sum_{\tau \in \tri D} e^{\theta \tau}]
    \end{eqnarray}
  Next we compute $\E[\sum_{\tau \in \tri D} e^{\theta \tau}]$ and from that can obtain  either an exact formula or estimate  the tail
behavior. Observe that we have the recurrence relation:
  \beq \label{eq:rec:tree}
    \E[\sum_{\tau \in T_{\phi}} e^{\theta(\tau-\rho)}]= 1+{\tilde B}\cdot \left(\int_0^{\infty} {\tilde B}{\tilde h}(t) e^{\theta t} dt
    \right) \E[\sum_{\tau \in T_{\phi}} e^{\theta(\tau-\rho)}]
  \end{equation}
  where $1$ comes from the root; ${\tilde B}{\tilde h}(t)$ is the rate of birth of the (direct) children of the root at time $\rho+t$;
  $e^{\theta t}$ comes from time shift and finally due to the fact that each child generates independently the same random tree we multiply
  $\E[\sum_{\tau \in T_{\phi}} e^{\theta(\tau-\rho)}]$. This in turn implies that
  \beq
  \E[\sum_{\tau \in T_{\phi}} e^{\theta(\tau-\rho)}]=
  \frac{1}{1-{\tilde B} \cdot \int_0^{\infty} {\tilde h}(t)e^{\theta t} dt}
  \end{equation}

  Now similarly we let $\rho$ be an arbitrary root and $T_{\rho}$
  be its tree we  obtain from  \eqref{eq:rec:tree}:
  \begin{eqnarray} \label{eq:rec:final0}
  \E[\sum_{\tau \in \tri D} e^{\theta \tau}]&=&
  {\tilde B} \cdot \left(\int_0^{\infty} {\tilde g}(t) e^{\theta t} dt \right) \E[\sum_{\tau \in T_{\phi}} e^{\theta(\tau-\rho)}]
  \\\label{eq:rec:final} &=& \frac{\tilde B \cdot \left(\int_0^{\infty} {\tilde g}(t) e^{\theta t} dt \right) }{1-\tilde B \cdot
\int_0^{\infty} {\tilde h}(t) e^{\theta
  t} dt}
  \end{eqnarray}
  where the recurrence relation  in \eqref{eq:rec:final0} is similar to \eqref{eq:rec:tree}: there is no $1$ here since there is no initial
birth at time  $0$;
  the birth rate rate is ${\tilde B}{\tilde g}(t))$ and the rest of the calculation  is the same.

  Finally when combining \eqref{eq:rec:ineq}, \eqref{eq:rec:tree} and \eqref{eq:rec:final}
  we obtain the exact formula \eqref{eq:exact}.  Observe that if
  $f_1,f_2,...f_k$  are probability density functions and $a_i \geq 0, i \in \{1,2,...k\}, \sum_{i=1}^{k} a_i=1$, then
  \begin{eqnarray}
  \int_{s=t}^{\infty} (f_1*f_2*...*f_k)(s)(ds) \nonumber
    &=&\hspace{-10pt}\mathop{\int \cdots \int}_{\sum_{i=1}^k s_i>t} f_1(s_1)...f_k(s_2) ds_1...ds_k
  \\&\leq&\sum_{i=1}^n \mathop{\int \cdots \int}_{s_i>a_i t} f_i(s_i)ds_1...ds_k \label{eq:conv_split}
  \\&\leq&\sum_{i=1}^n \mathop{\int}_{s_i>a_i t} f_i(s_i)ds_i \nonumber
  \end{eqnarray}
  where second line follows from the fact that ${\sum_{i=1}^k s_i>t}$ implies that for some $i$, $s_i>a_it$.
  (Another way to see this is via probability interpretation if random variables $X_i$ are independent and have distribution $f_i$
  then $\P[\sum_{i=1}^{k}X_i>t] \leq \P\big[X_i>a_it \mbox{ for some }i \in \{1,2,...,k\} \big]=\sum_{i=1}^{k} \P[X_i>a_it]$ where the
  first and last expressions correspond to those of \eqref{eq:conv_split}). Then
  \begin{eqnarray}
  \hspace{15pt} d_{TV,t}(\P_g, \P_0)
  \hspace{-5pt}&=&\int_t^{\infty} {\tilde g}* \left(\frac {1}{1-{\tilde B} \hat {\tilde h}}\right)^{\vee}(s) ds
  \\&=&\|\tg\|_1\sum_{n=0}^{\infty} {\tilde B}^n \int_t^{\infty} \big(\frac{\tg}{\|\tg\|_1}*\underbrace{{\tilde h}*\cdots*{\tilde
h}}_{\mbox{n times}}\big)(s) ds
  \\&\leq&\|\tg\|_1 \sum_{n=0}^{\infty} {\tilde B}^n \left( \int_{s> \frac t2} \frac{\tg(s)}{\|\tg\|_1} ds+n \int_{s>\frac t{2n}} {\tilde
h}(s) ds \right)
  \\&\leq& \frac{1}{1-{\tilde B}} \int_{s> \frac t2} \tg(s) ds+\|\tg\|_1\sum_{n=0}^{\infty} {\tilde B}^n n \int_{s>\frac t{2n}} {\tilde
h}(s) ds
  \end{eqnarray}
  using \eqref{eq:conv_split} with
  $k=n+1, \ f_1=\frac{\tg}{\|\tg\|_1}, \ f_2=f_3=\cdots=f_{k+1}={\tilde h}, \ a_1=\frac 12,\\  a_2=a_3=\cdots=a_{k+1}=\frac 1{2n}$.
  Finally:
  \begin{eqnarray}
  &&\limsup_{t \to \infty} \frac{d_{TV,t}(\P_g, \P_0)}{\int_{\frac t2}^{\infty} \tg(s)+{\tilde h}(s)ds}
  \\&\leq&\frac{\frac{1}{1-{\tilde B}} \int_{s> \frac t2} \tg(s) ds+\|\tg\|_1\sum_{n=0}^{\infty} {\tilde B}^n n \int_{s>\frac t{2n}} {\tilde
h}(s) ds}
  {\int_{\frac t2}^{\infty} \tg(s)+\phi(h(s))ds}
  \\&\leq&\frac 1{1-{\tilde B}}+\|\tg\|_1C_4<\infty
  \end{eqnarray}
\end{proof}

\section{Generalizations} \label{sec:gen}
\subsection{Multi-type models} \label{ssec:multi-type}
This section suggests two generalizations: first making Hawkes multi-type and second replacing $h$ by a general measure.
The proofs will be  simple modification of the ones we have presented here.

Consider that points are now of finitely many types and the set of type is $E=\{1,...,d\}$.
Now the process is specified by analogues triple $(h,\lambda,g)$:
\begin{itemize}
 \item $h_{ij}(t): E \times E \times \R \to \Rp$ where $i,j \in E$, $\| h_{ij} \|_{L_1}=\frac 1d$
 \item $\lambda_i: E \times \mathbb R_+^{d^2} \to \Rp
 \item $g_{ij}(t): E \times E \times \R \to \Rp$.
\end{itemize}
Then for all $e \in E$ the rate at time $t$ of type $e$ is given by
\beq r_e(t)=\lambda_e(z(t)) \end{equation}
where $z_{ij}(t)$ is a matrix given by
\beq z_{ij}=g_{ij}(t)+\sum_{i} \sum_{\tau \in S_i(t)} h_{ij}(t-\tau) \end{equation}
whre $S_i(t)$ are all the point of type $i$ before time $t$.
When $h$ is a measure one extends the definition of the can extend functionals $\int_a^b r_e(s) ds$ if exist
such $k^{e}_{ij}$ for each $i,j,e \in E$ that satisfy:
\begin{equation} \label{eq:limit}
\forall e \in, \exists k^{e}_{ij} \in \mathbb R_+^{d^2}, \lim_{c \to \infty} \frac{\lambda_e(cz)}{c}=\sum_{i,j}
z_{i,j}k^{e}_{ij}.
\end{equation}
As the analogue of Hypothesis \ref{hyp:existence} we now have:
 \begin{equation} \label{cond:existance1}
\exists k \in \mathbb R_+^{d^3}, \forall e \in E, \lambda_e (z) \leq \bar{\lambda}} \def\bmu{\bar{\mu}} \def\bnu{\bar{\nu}_e(z):=c_e+\sum_{i,j} z_{i,j}k^{e}_{ij}
\end{equation}

As a generalization of Propositoin \ref{prop:convergence} we get:

\begin{corollary1} Suppose condition \eqref{cond:existance1} is satisfied and if $h$ is a measure
then condition \eqref{eq:limit} is satisfied.  Then the following statements hold:\\
(i) If $\forall e\in E$, $\lambda_e$ is non-decreasing then Hawkes process is well defined\\
(ii) If in addition matrix \beq m_{i,e}=\sum_{j \in E} k^{e}_{ij} \end{equation} indexed by $i,e \in E$
has spectrum supported on unit disk $\{ c \in \mathbb C: |c| < 1 \}$ then invariant measure exists
and is unique minimal invariant measure and hence ergodic.
\end{corollary1}
\subsection{Stationary initial conditions as a generalization}  The proofs of this paper can also accommodate additional
levels of complexity
where one adds a stationary signal in the following sense.  Suppose you add another bounded stationary process $p(t)$ to the rate function
before passing through $\lambda$ so that now rate $r_t$ is defined as:
\begin{equation}
 r_t:=\lambda \left( g_0(t)+p(t)+\sum_{\tau \in S_t} h(t-\tau) \right)
\end{equation}
The results of this paper will still hold. One can also add another signal bounded signal $q(t)$ outside of lambda where rate will be:
\begin{equation}
 r_t:=\lambda \left( g_0(t)+p(t)+\sum_{\tau \in S_t} h(t-\tau) \right)+q(t)
\end{equation}
One useful application of this is a Hawkes-like linear process where the roots of the trees have some non-Poisson but
stationary
behavior while the trees are just like in Hawkes process.
\section*{Acknowledgements}
I would like to thank my adviser S. R. Srinivasa Varadhan for giving this topic to me and Lingjiong
Zhu.  He not only gave us this rich and interesting topic, but also provided guidance in
motivation, formulation of results and clear view of the subject.  I also thank Lingjiong Zhu for
helpful discussions and review of my paper.

This research is supported by NSF DMS-0904701, DMS-1208334 and DARPA grants.

redpajama_set_name: RedPajamaArXiv


text:
Q: AlertView issue in iOS 8 I added a UIVIew in AlertView as follow:
UIView *alertSubView = [[UIView alloc]initWithFrame:CGRectMake(0, 0, 260, 55)];
        [alertSubView addSubview:textField];
        [alertSubView addSubview:charLeftLabel];

        [alert setValue:alertSubView forKey:@"accessoryView"];

Problem
UIView* alertSubView = [alertView valueForKey:@"accessoryView"];

App is crash when I try to get that subview in Alertview Delegate Method.
It is working fine in iOS7 but there is issue with iOS 8 only
Error Terminating app due to uncaught exception 'NSUnknownKeyException', reason: '[ valueForUndefinedKey:]: this class is not key value coding-compliant for the key accessoryView.

A: Alert View is deprecated. You should use the UIAlertController. This is just a suggestion.
For example as your aspect :
UIAlertController *alertController = [UIAlertController
                                      alertControllerWithTitle:@"Alert Controller"
                                      message:@"Alert Message"
                                      preferredStyle:UIAlertControllerStyleAlert];

UIViewController *viewController = [[UIViewController alloc]init];
[viewController.view setBackgroundColor:[UIColor blueColor]];

UILabel *lbl = [[UILabel alloc]initWithFrame:CGRectMake(10, 8, 250, 30)];
lbl.text = @"This is a label";
lbl.textAlignment = NSTextAlignmentCenter;
lbl.textColor = [UIColor whiteColor];
[viewController.view addSubview:lbl];

UITextField *tf = [[UITextField alloc]initWithFrame:CGRectMake(10, 35, 250, 30)];
tf.borderStyle = UITextBorderStyleRoundedRect;
tf.placeholder = @"Enter your name";
[viewController.view addSubview:tf];

[alertController setValue:viewController forKey:@"contentViewController"];

UIAlertAction *cancelAction = [UIAlertAction
                               actionWithTitle:@"Cancel"
                               style:UIAlertActionStyleCancel
                               handler:^(UIAlertAction *action)
                               {
                                   NSLog(@"Cancel action");
                               }];

UIAlertAction *okAction = [UIAlertAction
                           actionWithTitle:@"OK"
                           style:UIAlertActionStyleDefault
                           handler:^(UIAlertAction *action)
                           {
                               NSLog(@"OK action");

                               NSLog(@"Text Value : %@",tf.text);
                           }];

[alertController addAction:cancelAction];
[alertController addAction:okAction];

dispatch_async(dispatch_get_main_queue(), ^{
    [self presentViewController:alertController animated:YES completion:nil];
});

Hope it helps you.

A: You can make this same thing in swift 2.0 like following code:
let alert = UIAlertController(title: "Alert Controller", message: "Alert Message", preferredStyle: UIAlertControllerStyle.Alert)

let cancelAction = UIAlertAction(
    title: "Cancel",
    style: UIAlertActionStyle.Destructive) { (action) in

}

let confirmAction = UIAlertAction(
    title: "OK", style: UIAlertActionStyle.Default) { (action) in

}

alert.addAction(confirmAction)
alert.addAction(cancelAction)

let VC = UIViewController()
VC.view.backgroundColor = UIColor.blackColor()

let lbl =  UILabel()
lbl.frame =  CGRectMake(10, 8, 250, 30)
lbl.text = "this is a label"
lbl.textAlignment = NSTextAlignment.Center
lbl.textColor = UIColor.whiteColor()
VC.view .addSubview(lbl)

let txt = UITextField()
txt.frame =  CGRectMake(10, 35, 250, 30)
txt.borderStyle = UITextBorderStyle.RoundedRect
txt.placeholder = "enter text"
VC.view .addSubview(txt)

alert.setValue(VC, forKey: "contentViewController")

self.presentViewController(alert, animated: true, completion: nil)

Download Demo from Github: https://github.com/nitingohel/NGAlertViewController-Swift2.0

A: Try this code :
UIAlertView *alert = [[UIAlertView alloc] initWithTitle:@"TEST" message:@"subview" delegate:nil cancelButtonTitle:@"NO" otherButtonTitles:@"YES", nil];
UIView *view = [[UIView alloc] initWithFrame:CGRectMake(0, 0, 260, 55)];
[view addSubview:textField];
[view addSubview:charLeftLabel];
[alert setValue:view forKey:@"accessoryView"];
[alert show];

A: UIAlertView *alert = [[UIAlertView alloc] initWithTitle:@"TEST" message:@"subview" delegate:self cancelButtonTitle:@"NO" otherButtonTitles:@"YES", nil];
UIView *view = [[UIView alloc] initWithFrame:CGRectMake(0, 0, 260, 55)];
[view addSubview:textField];
[view addSubview:charLeftLabel];
[alert setValue:view forKey:@"accessoryView"];
[alert show];

redpajama_set_name: RedPajamaStackExchange


text:
<?php

namespace Faker\Provider\ng_NG;

class Address extends \Faker\Provider\Address
{
    protected static $postcode = array('#####', '## ###');

    private static $county = array(
        'Abuja', 'Abia', 'Adamawa', 'Akwa Ibom', 'Anambra',
        'Bauchi', 'Bayelsa', 'Benue', 'Borno',
        'Cross River',
        'Delta',
        'Ebonyi', 'Edo', 'Ekiti', 'Enugu', 'Gombe',
        'Imo', 'Jigawa',
        'Kaduna', 'Kano', 'Katsina', 'Kebbi', 'Kogi', 'Kwara',
        'Lagos',
        'Nasarawa', 'Niger',
        'Ogun', 'Ondo', 'Osun', 'Oyo',
        'Plateau',
        'Rivers',
        'Sokoto',
        'Taraba',
        'Yobe',
        'Zamfara'
    );

    protected static $country = array(
        'Afghanistan', 'Albania', 'Algeria', 'American Samoa', 'Andorra', 'Angola', 'Anguilla', 'Antarctica (the territory South of 60 deg S)', 'Antigua and Barbuda', 'Argentina', 'Armenia', 'Aruba', 'Australia', 'Austria', 'Azerbaijan',
        'Bahamas', 'Bahrain', 'Bangladesh', 'Barbados', 'Belarus', 'Belgium', 'Belize', 'Benin', 'Bermuda', 'Bhutan', 'Bolivia', 'Bosnia and Herzegovina', 'Botswana', 'Bouvet Island (Bouvetoya)', 'Brazil', 'British Indian Ocean Territory (Chagos Archipelago)', 'British Virgin Islands', 'Brunei Darussalam', 'Bulgaria', 'Burkina Faso', 'Burundi',
        'Cambodia', 'Cameroon', 'Canada', 'Cape Verde', 'Cayman Islands', 'Central African Republic', 'Chad', 'Chile', 'China', 'Christmas Island', 'Cocos (Keeling) Islands', 'Colombia', 'Comoros', 'Congo', 'Congo', 'Cook Islands', 'Costa Rica', 'Cote d\'Ivoire', 'Croatia', 'Cuba', 'Cyprus', 'Czech Republic',
        'Denmark', 'Djibouti', 'Dominica', 'Dominican Republic',
        'Ecuador', 'Egypt', 'El Salvador', 'Equatorial Guinea', 'Eritrea', 'Estonia', 'Ethiopia',
        'Faroe Islands', 'Falkland Islands (Malvinas)', 'Fiji', 'Finland', 'France', 'French Guiana', 'French Polynesia', 'French Southern Territories',
        'Gabon', 'Gambia', 'Georgia', 'Germany', 'Ghana', 'Gibraltar', 'Greece', 'Greenland', 'Grenada', 'Guadeloupe', 'Guam', 'Guatemala', 'Guernsey', 'Guinea', 'Guinea-Bissau', 'Guyana',
        'Haiti', 'Heard Island and McDonald Islands', 'Holy See (Vatican City State)', 'Honduras', 'Hong Kong', 'Hungary',
        'Iceland', 'India', 'Indonesia', 'Iran', 'Iraq', 'Ireland', 'Isle of Man', 'Israel', 'Italy',
        'Jamaica', 'Japan', 'Jersey', 'Jordan',
        'Kazakhstan', 'Kenya', 'Kiribati', 'Korea', 'Korea', 'Kuwait', 'Kyrgyz Republic',
        'Lao People\'s Democratic Republic', 'Latvia', 'Lebanon', 'Lesotho', 'Liberia', 'Libyan Arab Jamahiriya', 'Liechtenstein', 'Lithuania', 'Luxembourg',
        'Macao', 'Macedonia', 'Madagascar', 'Malawi', 'Malaysia', 'Maldives', 'Mali', 'Malta', 'Marshall Islands', 'Martinique', 'Mauritania', 'Mauritius', 'Mayotte', 'Mexico', 'Micronesia', 'Moldova', 'Monaco', 'Mongolia', 'Montenegro', 'Montserrat', 'Morocco', 'Mozambique', 'Myanmar',
        'Namibia', 'Nauru', 'Nepal', 'Netherlands Antilles', 'Netherlands', 'New Caledonia', 'New Zealand', 'Nicaragua', 'Niger', 'Nigeria', 'Niue', 'Norfolk Island', 'Northern Mariana Islands', 'Norway',
        'Oman',
        'Pakistan', 'Palau', 'Palestinian Territories', 'Panama', 'Papua New Guinea', 'Paraguay', 'Peru', 'Philippines', 'Pitcairn Islands', 'Poland', 'Portugal', 'Puerto Rico',
        'Qatar',
        'Reunion', 'Romania', 'Russian Federation', 'Rwanda',
        'Saint Barthelemy', 'Saint Helena', 'Saint Kitts and Nevis', 'Saint Lucia', 'Saint Martin', 'Saint Pierre and Miquelon', 'Saint Vincent and the Grenadines', 'Samoa', 'San Marino', 'Sao Tome and Principe', 'Saudi Arabia', 'Senegal', 'Serbia', 'Seychelles', 'Sierra Leone', 'Singapore', 'Slovakia (Slovak Republic)', 'Slovenia', 'Solomon Islands', 'Somalia', 'South Africa', 'South Georgia and the South Sandwich Islands', 'Spain', 'Sri Lanka', 'Sudan', 'Suriname', 'Svalbard & Jan Mayen Islands', 'Swaziland', 'Sweden', 'Switzerland', 'Syrian Arab Republic',
        'Taiwan', 'Tajikistan', 'Tanzania', 'Thailand', 'Timor-Leste', 'Togo', 'Tokelau', 'Tonga', 'Trinidad and Tobago', 'Tunisia', 'Turkey', 'Turkmenistan', 'Turks and Caicos Islands', 'Tuvalu',
        'Uganda', 'Ukraine', 'United Arab Emirates', 'United Kingdom', 'United States of America', 'United States Minor Outlying Islands', 'United States Virgin Islands', 'Uruguay', 'Uzbekistan',
        'Vanuatu', 'Venezuela', 'Vietnam',
        'Wallis and Futuna', 'Western Sahara',
        'Yemen',
        'Zambia', 'Zimbabwe'
    );

    private static $regions = array(
        'Abakaliki', 'Abeokuta', 'Akure', 'Asaba', 'Awka',
        'Bauchi',
        'Calabar', 'City',
        'Damaturu', 'Dutse',
        'Ekiti', 'Enugu',
        'Gombe', 'Gusau',
        'Ibadan', 'Ikeja', 'Ilorin',
        'Jalingo', 'Jos',
        'Kaduna', 'Kano', 'Katsina', 'Kebbi',
        'Lafia', 'Lokoja',
        'Maiduguri', 'Makurdi', 'Minna',
        'Oshogbo', 'Owerri',
        'Port-Harcourt',
        'Sokoto',
        'Umuahia', 'Uyo',
        'Yenagoa', 'Yola'
    );

    /**
     * Randomly returns a Nigerian state or county.
     *
     * @example 'Lagos'
     *
     * @return string
     */
    public static function county()
    {
        return static::randomElement(static::$county);
    }

    /**
     * Randomly returns a Nigerian region of a state.
     *
     * @example 'Ikeja'
     *
     * @return string
     */
    public static function region()
    {
        return static::randomElement(static::$regions);
    }
}

redpajama_set_name: RedPajamaGithub


I really don't want to write right now because I just don't want to do it. However when you don't want to do something then maybe you should. I'd rather be watching youtube videos or playing video games but really I need to stay on point.
One of my weak points is constancy.
You see I can create youtube videos for my channel or write a decent blog post but I'm not consistent. I tend to create everyday then I just stop for no reason.
I give myself excuses. I have writers block, people don't like my videos, I don't get enough views, ect. I have a million excuses for myself none of them are good. Last Spring I was making videos 2-13 times per week, then… nothing. There's a lot of talk on the internet about "how to become a youtuber" and truthfully a lot of it's negative. People say the market is saturated and it's too hard to make videos and not worth it. They say you'll never make money at it and all kinds of other stuff. Well… yes it's difficult… very difficult, but not impossible.
Don"t let the negative talk and people and ideas take you down. They'll do everything in their power to keep you at their level. Make friends with other bloggers and content creators, get into their mindset. If you take a few steps today then take a few more tomorrow. The journey is tough. but not impossible. Anything worth achieving in life is not easy.
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Conductive hearing loss occurs when there is an issue with the outer or middle ear, which interferes with sounds passing through to the inner ear. It can be caused by a build up of earwax, an ear infection, a perforated ear drum, a build up of fluid, or otosclerosis (an abnormal bone growth in the middle ear).
Sensorineural hearing loss occcurs when there is damage to the inner ear (hearing organ, cochlea, or auditory nerve), meaning that it is unable to accurately transmit information to the brain. Sensorineural hearing loss is permanent.
It can be caused by the natural ageing process, excessive noise exposure, or certain types of medications.
Mixed hearing loss occurs when both Conductive and Sensorineural hearing loss are present. The sensorineural component of the hearing loss is permanent, whereas the conductive component can be either permanent or temporary.
EG: You may have sensorineural hearing loss because you work in an environment exposed to loud noises, and you may also have fluid in your middle ear.
© 2019 Newcastle Hearing Centre.


Detail from the Codex Zouche-Nuttall, a pre-Columbian (1200-1521) screenfold manuscript book painted on deer skin, depicting the life of the Mixtec ruler Eight Deer Jaguar-Claw (right).
The Mixtec of Mexico have long known how to predict changes in the weather from the behaviour of birds and plants. When they hear the first grave and sad notes of the chicucu bird, for example, they know that it heralds the end of the rainy season. As long as the chicucu sings, farmers know that the dry season will be prolonged, and that the time has not yet come to plant their beans and pumpkins.
To determine the right time to plant maize, the farmers rely on the juniper tree. If they see that the leaves are beginning to turn to dust, it means that the rainy season will be late, and that, if necessary, they should delay planting until mid-June, or even later.
Ndejama means "change", while cuia has many meanings – the expression can be used to designate the changes of the month, the year, or the season. The phrase ini zaza designates one of the four types of heat (ini) distinguished in the Mixtec language: lo´o (mild), keva kandeinio (bearable), kini (great), zaza (unbearable). The frequency with which we now hear of ini zaza – which used to be an exceptional phenomenon – has been increasing. That is why farmers now associate this phrase with climate change. What scientists today define as "climate variability", the Mixtec have always called ndejama cuia.
The Mixtec have been in tune with nature for centuries. Today, as in olden times, they interpret the signs that nature sends them to make the right decisions. Their language contains a great variety of information that could provide solutions to the current problems raised by climate change.
In 2019, UNESCO's LINKS programme on Local and Indigenous Knowledge Systems will publish a glossary of Mixtec expressions and terms related to climate phenomena in the Mixteca Baja region, to facilitate discussions between experts and the holders of traditional knowledge.
Read the UNESCO publication Indigenous and Local Knowledge about Pollination, UNESCO, 2015.
Mexican linguist Aracely Torres Morales is the daughter of palm farmers in the Mixteca Baja region. The youngest of eight children, she was the only one who was able to finish school. In 2018, she was part of the UNESCO Local and Indigenous Knowledge Systems (LINKS) programme, contributing her expertise on the Mixtec language and its relationship to the environment.


text:
Q: PHP Code on Separate Server From Apache? This is something I've never seen done, and I'm not turning up in my research, but my boss is interested in the idea.  We are looking at some load balancing options, and wonder if it is possible to have apache and php installed on multiple servers, managed by a load balancer, but have all the actual php code on one server, with the various apache servers pointing to the one central code base?

A: For instance NFS mounts are certainly possible, but I wouldn't recommend it. A lot of the advantage of loadbalancing is lost, and you're reintroducing a single point of failure. When syncing code, and rsync cronjob can handle itself very nicely, or a manual rsync on deployment can be done.
What is the reason you would want this central code base? I'm about 99% sure there is a better solution then a single server dishing out code.

A: I believe it is possible. To add to Wrikken's answer, I can imagine NFS could be a good choice.  However, there are some drawbacks and caveats.  For one, when Apache tries access files on an NFS share that has gone away (connection dropped, host failed, etc) very bad things happen.  Apache locks up, and continues to try to retrieve the file.  The processes attempting to access the share, for whatever reason, do not die, and it is necessary to re-boot the server.
If you do wind up doing this, I would recommend an opcode cache, such as APC.  APC will cache the pre-processed php locally, and eliminate round trips to your storage.  Just be prepared to clear the opcode cache whenever you update your application/

A: PHP has to run under something to act as a web processor, Apache is the most popular. I've done NFS mounts across servers without problem. Chances are if NFS is down, the network is down. But it doesn't take long to do an rsync across servers to replicate files, and is really a better idea.
I'm not sure what your content is like, but you can separate static files like javascript, css and images so they are on their own server. lighttpd is a good, light weight web server for things like this. Then you end up with a "dedicated" php server. You don't even need a load balancer for this setup.
Keep in mind that PHP stores sessions on the local file system. So if you are using sessions, you need to make sure users always return to the same server. Otherwise you need to do something like store sessions in memcache.


text:
La  de missiles anti-aériens est une brigade de missiles sol-air des forces terrestres russes. Faisant partie de la  de la Garde, la brigade est basée à Koursk.

Formée à partir d'un régiment anti-aérien en 1967 en Arménie, la brigade a été transférée en Allemagne de l'Est peu après sa formation et a fait partie du Groupement des forces armées soviétiques en Allemagne pendant le reste de la guerre froide. Après le retrait des troupes soviétiques d'Allemagne en 1992, la brigade a été transférée à Koursk.

Une équipe d'enquête conjointe de l'Australie, la Belgique, la Malaisie, des Pays-Bas et de l'Ukraine a conclu que la brigade a fourni l'unité qui a abattu le vol Malaysia Airlines 17 qui a tué 298 personnes.

Historique

Guerre froide
Le 268 régiment de missiles anti-aériens est formé en 1961 à Artik avec la  de la Garde, équipée du SA-2. Le , il devient la  de missiles anti-aériens, avec les 677, 679 et 682 bataillons de missiles anti-aériens séparés. À cette époque, la brigade a reçu le système de missile SA-4 Kroug. Du  au , la brigade est transférée en Allemagne de l'Est. La brigade est devenue une partie de la  blindée de la Garde à Altenbourg. Le 677 bataillon était basé à Altenbourg, le 679 à Mersebourg et le 682 à Zeitz. Le , la  est transférée à la  de la Garde mais est réintègre la  blindée de la Garde en 1984. En novembre 1986, le 1578 bataillon séparé anti-missile a été formé à Oschatz après que le 677 soit déplacé à Wittemberg. À cette époque, la brigade est rééquipée du système de missiles SA-11 Buk. En 1992, la brigade est transférée à Koursk et est devenue une partie de la 20 armée de la Garde.

Service des forces terrestres russes

Entre 1994 et 1998, la  est stationnée à Goudaouta dans le cadre de la mission russe de maintien de la paix en Abkhazie. La brigade a reçu à plusieurs reprises le prix de la meilleure brigade de défense aérienne de la  de la Garde et du district militaire de Moscou pour ses performances lors des exercices entre 2005 et 2008. La brigade a participé au défilé du jour de la victoire à Moscou en 2008 et aux exercices "Zapad-2009", "Zapad-2011" et "Zapad-2014".

Destruction du vol 17 de Malaysia Airlines

Bellingcat et Correctiv ont montré que la brigade a fourni l'unité de lanceur de montage de transporteur et de radar (TELAR) qui a abattu le Vol Malaysia Airlines 17 le . Selon l'enquête de janvier 2016 de Bellingcat, les soldats de la brigade ont été impliqués dans l'abattage de l'avion.

Le , la joint investigation team (JIT) a conclu que le vol MH17 avait été abattu à l'aide d'un système de missiles Buk appartenant à la  de missiles antiaériens.

Commandants
La brigade a été commandée par les officiers suivants :

 Colonel I.P. Popov (1967  1973)
 Colonel A.S. Samoïlenko (1973  1977)
 Colonel I.S. Doudtchenko (1977  1983)
 Colonel Iou.A. Iartchak (1983  1985)
 Colonel Iou.V. Filkov (1985  1991)
 Colonel I.P. Kondrachov (1991  1992)
 Colonel G.N. Tchernov (1992  2004)
 Colonel Alexandre Viktorovitch Donets (2005  2009)
 Colonel VV Koukouchkine (2009  2011)
 Colonel A. Iou. Zolotov (2011  2013)
 Colonel Sergueï Borissovitch Moutchkaïev (2013  2020)
 Colonel Viatcheslav Borissovitch Vlassov (depuis 2020)

Notes et références

Unité ou formation militaire créée en 1967
Brigade de l'armée de terre russe

redpajama_set_name: RedPajamaWikipedia


As a homeowner you're faced with a huge list of things to get done, one which can be air conditioner replacement. Sure, there's a lot to get done around the house, so when it's time to take on one of the bigger tasks, make certain you've done your research so you can lessen the impact it has on your wallet.
Because you live in Hattiesburg, you could qualify for utility rebates and other local incentive programs when you get a high-efficiency air conditioner from Lennox®. These rebates are tailored to your specific location, so rebates will be different, but chances are, there are savings to be had if you make sure to take the time to do a bit of research.
Savvy homeowners looking for air conditioner replacement can save hundreds, and even thousands, of dollars when getting an energy efficient air conditioner this summer. Between the federal credits and those available in Hattiesburg, you can boost your comfort while saving money. Something to keep in mind as you're doing your research is that local incentives can change pretty fast and without notice, so utilize them when they are available so you can save the most possible.
If you decide to go the solar route, you may be eligible for even greater rebates as well. Adding a solar air conditioner like the SunSource® Home Energy System means you're eligible for the federal renewable-energy tax credit through 2016. That's just the tip of the iceberg for solar bonuses too, so be sure to keep that choice open as you shop around for the ideal option.
Considering air conditioner replacement doesn't just mean you're getting a new system, but you're also gaining the benefits of updated technology. The lifespan of the average air conditioner is around 10-15 years, and the advancements made in energy efficiency alone can make the investment worth it. You'll receive the benefits month after month too, with greater comfort and lower energy costs month after month.
If you want to find out more about the variety of savings available to you in Hattiesburg, give the team at Oak Grove Heating & Air Conditioning a call or check with your local utility company to make certain you're getting the most savings possible.


text:
#pragma once
#include <aws/pinpoint/Pinpoint_EXPORTS.h>
#include <aws/pinpoint/model/JourneyResponse.h>
#include <utility>

namespace Aws
{
template<typename RESULT_TYPE>
class AmazonWebServiceResult;

namespace Utils
{
namespace Json
{
  class JsonValue;
} // namespace Json
} // namespace Utils
namespace Pinpoint
{
namespace Model
{
  class AWS_PINPOINT_API UpdateJourneyResult
  {
  public:
    UpdateJourneyResult();
    UpdateJourneyResult(const Aws::AmazonWebServiceResult<Aws::Utils::Json::JsonValue>& result);
    UpdateJourneyResult& operator=(const Aws::AmazonWebServiceResult<Aws::Utils::Json::JsonValue>& result);

    inline const JourneyResponse& GetJourneyResponse() const{ return m_journeyResponse; }

    inline void SetJourneyResponse(const JourneyResponse& value) { m_journeyResponse = value; }

    inline void SetJourneyResponse(JourneyResponse&& value) { m_journeyResponse = std::move(value); }

    inline UpdateJourneyResult& WithJourneyResponse(const JourneyResponse& value) { SetJourneyResponse(value); return *this;}

    inline UpdateJourneyResult& WithJourneyResponse(JourneyResponse&& value) { SetJourneyResponse(std::move(value)); return *this;}

  private:

    JourneyResponse m_journeyResponse;
  };

} // namespace Model
} // namespace Pinpoint
} // namespace Aws


Authors: Hoffmann, Alex C., Stein, Louis E.
The subject of this book is centrifugal gas separation devices, namely cyclones used for gas purification, solids collection and for demisting. The optimization of cyclone performance for any given task is an often-sought goal but is seldom achieved in practice. Understanding cyclone performance as a function of a cyclone's size, geometry, feed properties, feed flow rates and the system of which it is a part, is essential if one wishes to successfully design, operate, troubleshoot or predict cyclone performance. The information presented in this expanded book edition will help to develop this understanding.


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text:
Testing Rules from Amateurs,
"Fixing" Brains, Public Education,
Teacher Shortage, Your Tax Dollars at Work,
Accountability Double Standards,
Retention in Grade
LEAVE THE TESTING ANALYSIS TO THE EXPERTS
MI: When Legislators Don't Understand Testing
When I was teaching, I administered individual diagnostic tests to students. The instructions for every one of the tests reminded me that the test was to be used for its intended purpose. No conclusions unrelated to that intended purpose were considered valid.
Standardized achievement tests, however, are frequently (at this point, more frequently than not, I would wager) used for making conclusions unrelated to their intended purpose.
The reason? Legislators and politicians have taken over the responsibility of choosing how to evaluate children...and, for the most part, they don't know what they're doing.
Tests should not be given for any purpose other than that for which it was intended. To do so, as most states are doing, is invalid, irresponsible, and a form of educational malpractice.
In this post we learn of Michigan legislators who consider a bill which requires teachers to "pass the SAT" before earning a teaching certificate. If that sounds odd to you, it's because you cannot "pass" the SAT. Nor can you "fail" it. It's not a spelling test, or a final exam.
Pass the SAT? What does that even mean? The SAT gives you a score, which as I told my students every year, is neither "good" nor "bad" until the college you're applying to says so. I talk to someone on line with ties to the testing and data biz and she absolutely hates it when people talk about passing or failing test. And yet, here we are, demonstrating once again that civilians (even elected ones) don't understand that tests are produced for very specific purposes and can't just be swapped to whatever purpose you like as if all tests are fundamentally the same. And instead of seeing some rich source of nuanced data that can be carefully decoded for a wealth of information, these citizens just see a thing that you either pass or fail. No more nuance or richness than a light switch.
And these are the people who legislate how tests must be used and what rewards and punishments will be doled out because of them. Yes, one of the biggest problems with modern ed reform is that it's amateur hour in education. Knowing what the heck you're talking about-- that's the test that people in power keep failing.
BILLIONAIRES WANT TO "FIX" BRAINS RATHER THAN ADDRESS POVERTY
Billionaires Want Poor Children's Brains to Work Better
Gates and his billionaire friends are determined to find the cause of low achievement anywhere but with poverty (just like DeVos, and other NRA shills, look for the answer to gun violence anywhere but with the actual guns). The billionaires are afraid that the solutions might cost money (see The Schools Chicago's Students Deserve).
They want to fund research in executive functioning and why students who live in poverty have such trouble. How about if they start with these reports of actual research already done...
How Poverty Gets Under the Skin: The Effects of Deprivation on Blood, the Brain, and the Body
Poverty Affects The Brain, Causes Lower Test Scores: Study
Poverty Shrinks Brains from Birth
Poverty and Potential: Out-of-School Factors and School Success
EPA scientists find black communities disproportionately hit by pollution
The U.S. does not have an education problem. It has a poverty problem.
...the billionaires reason that not only can executive malfunctioning cause substantial classroom learning problems and school failure, it also can adversely affect socio-economic status, physical health, drug problems, and criminal convictions in adulthood. Consequently, if teachers of poor students know how to improve executive function, their students will do well academically and reap future "real-world benefits." For Gates, who is always looking for "the next big thing," this can be it in education.
Most people looking at this reasoning would likely think, "If executive functioning is poorer in poor children, why not eliminate the apparent cause of the deficiency, i.e., poverty?" Not so for the billionaires.
THE WAR AGAINST PUBLIC EDUCATION AND PUBLIC EDUCATORS
Our Schools Are Not Failing; Our Policy Makers Are : Raleigh's Amorphous Way of Measuring Schools
With the exception of "class size caps" the words "North Carolina" in the following quote (and its source blog post) can be replaced with "Indiana" (or any number of other states).
And when you are the North Carolina General Assembly that is trying to privatize the public school system, you undertake a series of actions that weaken public schools such as school performance grades aligned with achievement, intentionally not fully fund schools, create class size caps with no funding of new classrooms, and throw millions of dollars into vouchers.
You try and disenchant the teaching profession by removing due-process rights and graduate degree pay from new teachers to a point where state education programs have experienced a significant drop in candidates.
And yet public schools are still doing the job.
PAYING FOR EDUCATION: THE TEACHER SHORTAGE
Fact Sheet: Yes, Increase the Salaries of All Teachers
Indiana and other states need to do something to reverse the growing teacher shortage. The number of students enrolled in teacher education programs in Indiana in 2015-16 has dropped by half since 2010-11. In 2010-11 there were 13,493 students enrolled in teacher training programs. That number was 6,813 in 2015-16.
For the last few decades public school teachers have been made the scapegoat for the failure of students to achieve.
The state government under Mitch Daniels began the punishment of teachers in 2011. Since then...
collective bargaining rights for teachers have been restricted.
the state began what is now the largest private/parochial school voucher program in the nation, and increased funding for privately owned and operated charter schools.
the state passed a property tax cap amendment to the constitution, and shifted state funding of public education to the state legislature.
teachers have lost tenure (due process) and seniority protections.
the importance of experience and education level as a factor in teacher salaries has been reduced.
accountability measures requiring teacher evaluations to be based on student test scores despite lack of validity have been instituted.
A raise in teacher pay is only the first step towards restoring the teaching profession.
Note that the legislature, policy makers, and politicians are not held accountable for societal issues leading to lowered achievement such as funding, class size, and the effects of poverty.
The annual pay for teachers fell sharply from 1995 to 2015 in relation to the annual pay of similar workers. According to the Economic Policy Institute, public school teachers are paid less than other comparable workers in every state, and they earn 11 percent less on average, when accounting for nonwage benefits. This calculation is based on comparable weekly wages [emphasis added].
MONEY LAUNDERING FOR SCHOOL "CHOICE"
FL Schools Using Taxpayer Money to Teach Ridiculous Lies
Should tax dollars be used to fund schools which teach that "dinosaurs and humans lived together, that God's intervention prevented Catholics from dominating North America and that slaves who 'knew Christ' were better off than free men who did not."
This report from Florida discusses what's taught in private schools using textbooks from Abeka, BJE Press, and Accelerated Christian Education (ACE). Some of Indiana's parochial schools use the same books.
Where is your educational tax dollar going?
The constitutional issues here are rather complex. There are two arguments that can be made here on either side. On the one hand, giving taxpayer money to religious entities seems like a clear violation of the Establishment Clause, especially when it's used to teach things that advocate very sectarian ideas, something the government is clearly forbidden from doing.
On the other hand, the voucher is not aimed specifically at religious schools. Parents get a voucher and can use it to send their kids to any kind of school, religious or secular. The fact that the money is "laundered" through parental choice does make a difference constitutionally because it's akin to someone getting public assistance and then using a portion of it to tithe at church, or buy some religious product or service. The government is not funding the religious activity directly, so that does mitigate, at least to some degree, the Establishment Clause problem.
Either way, we can be appalled by the fact that our tax dollars are used to promote vile and dishonest ideas like this.
THE DOUBLE STANDARD IN SCHOOL ACCOUNTABILITY
'Wild West of education'
Where is the accountability for all non-public schools which receive state tax dollars? You know that if a public school was avoiding accountability the "reformers" in the state would be all over them. Yet accountability somehow doesn't seem to matter when it comes to F rated charter or voucher schools.
Indiana grades schools with an A-F system, and according to the state grades, IVS is a failing school. In fact, all virtual charter schools in Indiana received F grades from the state in both 2016 and 2017, according to the State Board of Education's recent report. Any one of them could be closed by its authorizer, only to be replaced by yet another virtual school.
As Cavazos' recent explorations of the peculiar origins of the new Indiana Agriculture and Technology School show, Indiana is the Wild West of education. There are few rules for virtual schools to follow, but lots of money to be made.
This past session, our legislators killed three bills regarding accountability for charter school authorizing, even though Gov. Eric Holcomb and State Superintendent Jennifer McCormick called for improved accountability in virtual charter schools.
LA FINALLY ACCEPTS YEARS OF RESEARCH INTO RETENTION
Louisiana ends policy that held thousands of students back a grade or more
Being forced to choose Social Promotion or Retention is a false dichotomy. It doesn't have to be either one or the other. Investing in education and providing students the help they need (not just what they can afford), is the answer. Not every child will succeed...but many, many more children won't fail.
"But then when I got the numbers for New Orleans and for Louisiana – and you know a lot of Louisiana was not affected by Katrina – New Orleans was a little bit worse but Louisiana was still really bad on retention," she says. "And as I talked to more people it was clear that it was an effect of standardized testing."
Reckdahl recently wrote about overage students in Louisiana and investigated the impacts of retention for The Hechinger Report. So many students have been held back due to mandatory retention that in 2017 the Louisiana State Legislature decided to end it. Now, schools offer summer classes, online classes and help from specialized teachers as alternatives for students who don't pass the LEAP test.
Reckdahl says there's one big takeaway from the state's "experiment" with retention.
"It's not enough to scare a kid into performing," she says. "You can't just say I'm going to hold you back."
🚌🎓🏫
Posted by Stu at 9:55 AM
Labels: Accountability, Article Medleys, Choice, poverty, Public Ed, retention, TeacherShortage, Testing, vouchers
Hey, we could use some consistency, here!
2018 Medley #15: Separation of Children from Paren...
Father's Day 2018: A Reminder to Read Aloud to You...
What Being a Teacher Has Taught Me
Privatization – Still Failing After All These Year...
2018 Medley #13: Investing in Children

redpajama_set_name: RedPajamaCommonCrawl


We had so much fun for Tripp's second birthday! I can't believe what a big boy I have these days! His birthday party fell on Cinco de Mayo so OF COURSE we had a little fiesta!
Instead of stressing over food we decided to cater in Rosa's and it was amazing and a WONDERFUL price point! Pro tip: Make sure to have lots of fruit and pouches if you have a lot of kiddos attending. They loved snacking on that while parents enjoyed chips, salsa and guacamole (and some of the kiddos loved that too)!
I had so much fun decorating with all the pinatas and mini hats! I also went simple with mini cupcakes and purchased themed cupcake toppers. The party favors were little maracas. The kiddos thought they were super fun to play with and they served as good decoration (win-win)!
We had inflatable pools and water toys out side to play with and I couldn't leave the little kids table with no decoration! I also made sure to have plenty of chalk for them to draw on the back porch too! I have linked everything below that we enjoyed at the party (including the toys we had out for kids to play with).


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text:
Pantinia darwini är en insektsart som beskrevs av William Edward China 1962. Pantinia darwini ingår i släktet Pantinia och familjen Peloridiidae. Inga underarter finns listade i Catalogue of Life.

Källor

Externa länkar

Halvvingar
darwini


text:
Q: Exact sequence of finitely generated abelian groups Let $$0\rightarrow G_{1}\xrightarrow{f_{1}} G_{2}\xrightarrow{f_{2}} G_{3}\rightarrow 0$$ is a short exact sequence of finitely generated abelian groups. We call $\overline{G_{i}}$ the quotient of $G_{i}$ by its torsion. Then, we consider :
$$0\rightarrow \bar{G_{1}}\xrightarrow{\bar{f_{1}}} \bar{G_{2}}\xrightarrow{\bar{f_{2}}} \bar{G_{3}}\rightarrow 0.$$

Prove that $\bar{f_{2}}$ is surjective and that $Ker(\bar{f_{2}})/Im(\bar{f_{1}})$ is a torsion group.

My answer is (I'm not sure, that's why it looks like questions) : Is it true that $\bar{f_{2}}$ is surjective since $f_{2}$ is ?
Is it true that $Ker(\bar{f_{2}})/Im(\bar{f_{1}})$ is a torsion group since $Im(f_{1})=Ker(f_{2})$ ?
Thank you.

A: Writing $G_i = \mathbb{Z}^{r_i} \oplus T_i,$ where $r_i \geq 0$ is an integer and $T_i$ is a torsion group, we have the following exact sequence:
$$
0 \rightarrow \mathbb{Z}^{r_1} \oplus T_1 \xrightarrow{f_1} \mathbb{Z}^{r_2} \oplus T_2 \xrightarrow{f_2} \mathbb{Z}^{r_3} \oplus T_3 \rightarrow 0.
$$
From this, taking quotient by trosion subgroups, we get the following complex:
$$
0 \rightarrow \mathbb{Z}^{r_1} \xrightarrow{\bar{f_1}} \mathbb{Z}^{r_2} \xrightarrow{\bar{f_2}} \mathbb{Z}^{r_3} \rightarrow 0.
$$
This is not exact in general. First we want to show that $\bar{f_2}$ is surjective. This follows from the fact that $f_2$ is surjective, and $f_2(t) \in T_3$ for every $t \in T_2.$ Now we want to show that $\text{ker}{\bar{f_2}}/ \text{Im}{\bar{f_1}}$ is a torsion group. Let $x \in \text{ker}{\bar{f_2}}.$ Then $\bar{f_2}(x) = 0 \Rightarrow f_2(x) \in T_3.$ But $T_3$ is a torsion group. So there exists $n \in \mathbb{N},$ such that $f_2(nx) = nf_2(x) = 0.$ Hence there is $y \in G_1$ such that $f_1(y) = nx.$ Note that $y \notin T_1.$ So $nx = 0$ in $\text{ker}{\bar{f_2}}/ \text{Im}{\bar{f_1}}.$ This shows that, every element of $\text{ker}{\bar{f_2}}/ \text{Im}{\bar{f_1}}$ is a torsion element.


May, 1861, saw this high 30-acre bluff overlooking Lafayette quickly become an induction center for enlistees in the Civil War until its end in 1865. Ample water, good drainage and access to the railroad at its west end served the purpose well, however rough its shelters. Regiments were organized here including the 10th, 15th, 20th, 40th, 72nd (a part of Wilder's famous "Lighning Brigade") and the 86th, as well as the 10th, 11th and 16th Light Artillery Batteries attached to other units.