text:
\section{Introduction and the main result}

\subsection{Introduction}

It is known that the time until a birth and death process $X_{t}$
started at the origin reaches a certain level is distributed as a
sum of independent exponential variables whose parameters are the
negatives of the non-zero eigenvalues of the generator of the process
stopped at the given level (see Karlin~\cite{karlin1959coincidence}).
Diaconis and Miclo~\cite{diaconisMiclo2009quasi-stationarity} and
Swart~\cite{swart2010intertwining} gave a probabilistic proof of
this fact by finding a pure birth process $Y_{t}$ which reaches the
given level at the same time as $X_{t}$. The technique that Diaconis,
Miclo and Swart employ is called intertwining of Markov processes.
This technique was developed by Rogers and Pitman~\cite{rogersPitman1981markov},
Diaconis and Fill~\cite{diaconisFill90strong-stationary-times},
and Fill~\cite{fill92strong-stationary-duality}. It allows them
to add structure to the process $X_{t}$ such that it is initially
reluctant to be absorbed, but after each exponential time (which corresponds
to jump times of $Y_{t}$) it changes its behavior to be more and
more compelled to be absorbed. Since one-dimensional diffusions can
be obtained as limits of birth and death processes, it is interesting
to investigate whether this technique can be extended to the case
that $X_{t}$ is a diffusion. Since the general case is too difficult,
in this paper we consider the case that $X_{t}$ is the Wright-Fisher
diffusion with reflection at $1/2$, which has state-space $[1/2,1]$
and is absorbed at $1$. The generator and the semigroup of this diffusion
have the nice property that they map polynomials to polynomials of
the same order, which simplifies our proofs. We need that the diffusion
is reflected at $1/2$ for technical reasons; without it, one of our
proofs would not work (see Remark~\ref{rem:WF-fail}). We construct
an explosive pure birth process $Y_{t}$ such that $X_{t}$ is absorbed
at the same time as $Y_{t}$ explodes.

The idea of Diaconis, Miclo and Swart can be summarized as follows.
For a given transition semigroup $P_{t}$ of a birth and death process
$X_{t}$ on $\left\{ 0,\dots,n\right\} $ absorbed at $n$, Swart
finds a transition semigroup $Q_{t}$ of a pure birth process $Y_{t}$
on $\left\{ 0,\dots,n\right\} $ and a probability kernel $K$ which
satisfies
\begin{equation}
P_{t}K=KQ_{t}\qquad(t\ge0).\label{eq:intr-PK=00003DKQ}
\end{equation}
The algebraic relation~(\ref{eq:intr-PK=00003DKQ}) is called intertwining,
which gives the name to the intertwining of Markov processes. Swart
builds on an earlier work of Diaconis and Miclo, who found an intertwining
of the form $KP_{t}=Q_{t}K$. However, we focus on Swart's construction,
because our work on the Wright-Fisher diffusion is closer in spirit
to his. He uses a result proved by Fill~\cite{fill92strong-stationary-duality}
which says that if $X_{t}$ and $Y_{t}$ are Markov processes with
finite state-spaces related by~(\ref{eq:intr-PK=00003DKQ}), then
the two processes can be coupled (i.\,e.\ defined on the same probability
space) such that\footnote{Under certain conditions, Fill showed that this coupling can be extended
to countably infinite state-spaces. Fill built on earlier work of
Diaconis and Fill~\cite{diaconisFill90strong-stationary-times},
where an analogous result is proved for processes with discrete time.}
\begin{equation}
\mathrm{P}\left(Y_{t}=y|X_{u},0\leq u\leq t\right)=K\left(X_{t},y\right)\qquad\mathrm{a.s.}\quad(t\geq0).\label{eq:intr-coupling}
\end{equation}
Using (\ref{eq:intr-coupling}) and the fact that his kernel satisfies
\begin{equation}
K(x,n)=\boldsymbol{1}_{[x=n]}:=\begin{cases}
1, & \mathrm{if}\,x=n,\\
0, & \mathrm{otherwise,}
\end{cases}\label{eq:intr-K}
\end{equation}
Swart proves that $X_{t}$ and $Y_{t}$ can be coupled such that the
times of absorption of $X_{t}$ and $Y_{t}$ are a.\,s.\ the same.

In this paper, we derive analogue results for the case that $X_{t}$
is the Wright-Fisher diffusion with reflection at zero. We find an
explosive pure-birth process $Y_{t}$ on $\mbox{\ensuremath{\bar{\mathbb{N}}}:=}\mathbb{N}\cup\left\{ \infty\right\} :=\left\{ 0,1,\dots,\infty\right\} $
and a probability kernel $K$ from $[0,1]$ to $\bar{\mathbb{N}}$ satisfying
the intertwining relation~(\ref{eq:intr-PK=00003DKQ}) and
\[
K\left(x,\infty\right)=\boldsymbol{1}_{\left[x=1\right]}.
\]
We couple the two processes such that they satisfy~(\ref{eq:intr-coupling})
which allows us to conclude that the time of absorption of $X_{t}$
is a.\,s.\ equal to the time of explosion of $Y_{t}$. But since
the time of explosion of the pure birth process is the sum of independent
exponential variables whose intensities are the birth rates of $Y_{t}$,
this gives us a new proof of the distribution of the time to absorption
of $X_{t}$.\footnote{Just as with the birth and death processes, the distribution of the
time of absorption of the Wright-Fisher diffusion has long been known
(see e.\,g.\ Kent~\cite{kent1982spectral}), but our proof is new.}

\subsection{Intertwining of the Wright-Fisher diffusion}

Define
\[
\mathcal{D}(G)=\left\{ f\in\mathcal{C}^{2}[0,1];\,\frac{\partial}{\partial x}f(0)=0\right\}
\]
and
\begin{equation}
Gf(x)=\left(1-x^{2}\right)\frac{\partial^{2}}{\partial x^{2}}f(x),\qquad f\in\mathcal{D}(G),x\in[0,1].\label{eq:WF-generator}
\end{equation}
In the appendix we show that $G$ is closable and its closure generates
a Feller semigroup, which we denote $P_{t}$. We also show that the
associated Markov process, which we call the Wright-Fisher diffusion
with reflection at zero, has continuous sample paths. Note that the
generator of the Wright-Fisher diffusion is usually defined as
\begin{equation}
\frac{1}{2}x(1-x)\frac{\partial^{2}}{\partial x^{2}}f(x),\quad f\in\mathcal{C}^{2}[0,1],\,x\in[0,1]\label{eq:WF-=00005B0,1=00005D-generator}
\end{equation}
(see e.\,g.\ Liggett~\cite[Example 3.48]{liggett2010continuous}).
However, if $\tilde{X}_{t}$ is generated by~(\ref{eq:WF-=00005B0,1=00005D-generator}),
then $X_{t}=\left|2\tilde{X}_{2t}-1\right|$ is generated by~(\ref{eq:WF-generator}).

Define $H$ as the generator of an explosive pure birth process on
$\bar{\mathbb{N}}$ which jumps from $y$ to $y+1$ with the rate
\[
\lambda_{y}=\left(2y+1\right)\left(2y+2\right),\,y\in\mathbb{N}.
\]
That is, define $H$ as an operator from $\mathbb{R}^{\bar{\N}}$ to $\mathbb{R}^{\bar{\N}}$
by
\begin{eqnarray*}
Hf(y) & = & \lambda_{y}\left(f(y+1)-f(y)\right),\quad y\in\mathbb{N},\\
Hf(\infty) & = & 0,
\end{eqnarray*}
where $f$ is in $\mathbb{R}^{\bar{\N}}$. It is shown in the Appendix that the
restriction of $H$ to a suitable domain is the generator of a Feller
semigroup on $\mathcal{C}\left(\bar{\N}\right)$, which we denote by $Q_{t}$.
Define a probability kernel from $[0,1]$ to $\bar{\mathbb{N}}$ by
\begin{equation}
K(x,y)=\begin{cases}
\left(1-x^{2}\right)x^{2y}, & \mathrm{if\,}0\leq y<\infty,\\
\boldsymbol{1}_{[x=1]}, & \mathrm{if\,}y=\infty.
\end{cases}\label{eq:K-def}
\end{equation}
It can be shown that $K$ maps $\mathcal{C}\left(\bar{\mathbb{N}}\right)$ into $\mathcal{C}[0,1]$.
We claim that there is an intertwining relation:
\begin{thm}
\label{thm:PK=00003DKQ}We have
\begin{equation}
P_{t}K=KQ_{t},\qquad(t\geq0).\label{eq:PK=00003DKQ}
\end{equation}

\end{thm}
Using~(\ref{eq:PK=00003DKQ}), we are able to couple the two processes
in the spirit of~\cite{diaconisMiclo2009quasi-stationarity,fill92strong-stationary-duality,swart2010intertwining}.
We define the state-space $\boldsymbol{S}$ of the coupled process as the one-point
compactification of $[0,1]\times\mathbb{N}$, where we denote the point at
infinity by $(1,\infty)$. Using this notation, we can think of $\boldsymbol{S}$
as a subset of $[0,1]\times\bar{\N}$, but keep in mind that the topology
of $\boldsymbol{S}$ is not the one induced by $[0,1]\times\bar{\N}$. Note that
if the coupled process is to satisfy an analogue of~(\ref{eq:intr-coupling}),
then it would be natural to construct it on the space
\[
\left\{ (x,y)\in[0,1]\times\bar{\N};\,K(x,y)>0\right\} ,
\]
as was done for Markov processes with discrete state-spaces~\cite{diaconisFill90strong-stationary-times,fill92strong-stationary-duality}.
However, this space is not compact, so if we want to use the theory
of Feller semigroups we must compactify it (either implicitly or explicitly).
It turns out that $\boldsymbol{S}$ is the right compactification.

An analogous result to the following theorem was proved by Fill~\cite[Theorem 2]{fill92strong-stationary-duality}
for processes with discrete state-spaces and by Diaconis and Fill~\cite[Theorem 2.33]{diaconisFill90strong-stationary-times}
for processes with discrete time and space.
\begin{thm}
\label{thm:coupling}There exists a Feller process $\left(X_{t},Y_{t}\right)$
on $\boldsymbol{S}$ such that
\begin{equation}
\mathrm{E}\left(f\left(Y_{s+t}\right)|X_{u},Y_{u},0\leq u\leq s\right)=\left(Q_{t}f\right)\left(Y_{s}\right)\,\mathrm{a.s.}\label{eq:second-margin}
\end{equation}
for all $f\in\mathcal{C}\left(\bar{\mathbb{N}}\right)$ and $s,t\geq0$. Hence, $Y_{t}$
on its own is a pure birth process on $\bar{\mathbb{N}}$ with birth rates
$\lambda_{y}$. If the initial distribution satisfies
\begin{equation}
\pi_{0}^{(X,Y)}(A\times\{y\})=\int_{A}K(x,y)\pi_{0}^{X}(\mathrm{d} y),\label{eq:coupling-initial}
\end{equation}
where $\pi_{0}^{X}$ is an arbitrary probability measure on $[0,1]$,
then $X_{t}$ on its own is the Wright-Fisher diffusion with reflection
at zero with initial distribution $\pi_{0}^{X}$ and we have
\begin{equation}
\mathrm{P}\left(Y_{t}=y|X_{s},0\leq s\leq t\right)=K\left(X_{t},y\right)\quad\mathrm{a.s.}\label{eq:averaging}
\end{equation}
for all $y\in\bar{\mathbb{N}}$ and $t\geq0$.
\end{thm}
Note that if both $X_{t}$ and $Y_{t}$ start from zero, then~(\ref{eq:coupling-initial})
is satisfied. Using Theorem~\ref{thm:coupling} we can prove that
the time of absorption of the diffusion is a.\,s.\ equal to the
time of explosion of the pure birth process. Indeed, from~(\ref{eq:averaging})
we have
\[
\mathrm{P}\left(X_{t}\in A,Y_{t}\in B\right)=\mathrm{E}\left(\boldsymbol{1}_{A}\left(X_{t}\right)K\left(X_{t},B\right)\right).
\]
But since $K(x,\cdot)$ is concentrated on $\mathbb{N}$ for $x<1$ and on
$\{\infty\}$ for $x=1$, we have
\[
\mathrm{P}\left(X_{t}<1,Y_{t}=\infty\right)=\mathrm{P}\left(X_{t}=1,Y_{t}<\infty\right)=0.
\]

\subsection{Discussion}

In addition to proving the a.\,s.\ equality of the time of absorption
and the time of explosion of the two processes, Theorem~\ref{thm:coupling}
shows more about the underlying structure. By inspection of the formula
for the coupled generator~(\ref{eq:coupled-generator-explicit})
below we see that conditionally on $Y_{t}=y\in\mathbb{N}$ we can interpret
$X_{t}$ as the Wright-Fisher diffusion with reflection at zero and
with additional drift, which at the point $X_{t}=x$ equals
\[
\frac{4y}{x}-4(y+1)x.
\]
It can be shown that the scale function $u(x)$ and the speed measure
$m(\mathrm{d} x)$ of this diffusion satisfy
\begin{eqnarray*}
u'(x) & = & \frac{1}{x^{4y}\left(1-x^{2}\right)^{2}},\\
m(\mathrm{d} x) & = & x^{4y}\left(1-x^{2}\right)\mathrm{d} x.
\end{eqnarray*}
Hence, by Mandl~\cite[pp.~24--25]{mandl1968analytical}, both boundaries
are entrance for $y>0$ and $0$ is a regular boundary while $1$
is an entrance boundary for $y=0$. In particular, the coupled process
lives on the set $\left\{ (x,y)\in[0,1]\times\bar{\N};\,K(x,y)>0\right\} $
as one might expect.

Moreover, we can see that there is an equilibrium point
\[
x_{y}=\sqrt{\frac{y}{y+1}}
\]
such that the drift is positive when $x<x_{y}$ and negative when
$x>x_{y}$. We can interpret this as, conditionally on $Y_{t}=y$,
$X_{t}$ is pushed toward the equilibrium point $x_{y}$. Obviously,
$x_{y}$ is monotonous in $y$ and goes from $0$ to $1$ as $y$
goes from $0$ to $\infty$. Thus, we can think of $X_{t}$ as being
initially reluctant to be absorbed, but later getting more and more
compelled to get absorbed.

In our paper we construct Markov processes from generators using the
Hille-Yosida theorem. We could also construct them as solutions to
Martingale problems or stochastic differential equations. However,
we chose the Hille-Yosida theorem for its simplicity. Theorem~\ref{thm:PK=00003DKQ}
extends results of Diaconis and Miclo~\cite{diaconisMiclo2009quasi-stationarity}
and Swart~\cite{swart2010intertwining}, who proved similar theorems
for birth and death processes. Theorem~\ref{thm:coupling} extends
results of Fill~\cite{fill92strong-stationary-duality}, who proved
similar result for Markov processes with continuous time and discrete
state-space, and of Diaconis and Fill~\cite{diaconisFill90strong-stationary-times},
who proved it for the case of discrete time and space. It remains
an open problem whether results like Theorems~\ref{thm:PK=00003DKQ}
and \ref{thm:coupling} hold for other diffusions than the modified
version of the Wright-Fisher diffusion we consider in our paper. It
seems that our proof of Theorem~\ref{thm:PK=00003DKQ} does not exploit
any peculiarity of the Wright-Fisher diffusion and we believe it could
be extended to other types of diffusions as well. On the other hand,
our proof of Theorem~\ref{thm:coupling} depends strongly on the
fact that the generator of the Wright-Fisher diffusion maps polynomials
to polynomials of the same order, and it seems that entirely different
proof techniques would be required for other diffusions.

\section{Proofs}

\subsection{Intertwining}

To prove Theorem~\ref{thm:PK=00003DKQ} we need to show that there
is an intertwining between semigroups $P_{t}$ and $Q_{t}$. The following
theorem says that we can show this by proving that there is an intertwining
between the generators. An analogous result for Markov processes with
discrete state-spaces was proved by Fill~\cite[Lemma 3]{fill92strong-stationary-duality}.
Although we use the following theorem only when $P_{t}$ and $Q_{t}$
are Feller semigroups and $K$ is a probability kernel, we are able
to prove it more generally.
\begin{thm}
\label{thm:Semigroup-intertwining}Let $L_{1},L_{2}$ be Banach spaces.
Let $P_{t}$ and $Q_{t}$ be strongly continuous contraction semigroups
defined on $L_{1},L_{2}$ and let $G$ and $H$ be their generators.
Let $K:L_{2}\rightarrow L_{1}$ be a continuous linear operator. Then
the following are equivalent:
\begin{enumerate}
\item \label{enu:SI-semigroup}For all $t\geq0$,
\begin{equation}
P_{t}K=KQ_{t}\label{eq:semigroup-intertwining}
\end{equation}
on $L_{2}$,
\item \label{enu:SI-generator}$K$ maps $\mathcal{D}(H)$ into $\mathcal{D}(G)$ and
\begin{equation}
GK=KH\label{eq:generator-intertwining}
\end{equation}
on $\mathcal{D}(H)$,
\item \label{enu:SI-core}There exists a core $D$ of $H$ (i.\,e.\ $D$
is a dense subspace of $\mathcal{D}(H)$ such that the closure of the restriction
of $H$ to $D$ is $H$) such that $K$ maps $D$ into $\mathcal{D}(G)$
and (\ref{eq:generator-intertwining}) holds on $D$.
\end{enumerate}
\end{thm}
\begin{proof}
To prove $\eqref{enu:SI-semigroup}\Rightarrow\eqref{enu:SI-generator}$,
fix $f\in\mathcal{D}(H)$. Then $\frac{1}{t}\left(Q_{t}f-f\right)$ converges
to $Hf$, so by the continuity of $K$, $\frac{1}{t}\left(KQ_{t}f-Kf\right)$
converges to $KHf$. By~(\ref{eq:semigroup-intertwining}), $\frac{1}{t}\left(P_{t}Kf-Kf\right)$
is also convergent, so $Kf$ is in $\mathcal{D}(G)$ and $GKf=KHf$.

In order to prove $\eqref{enu:SI-generator}\Rightarrow\eqref{enu:SI-semigroup}$,
fix $f\in\mathcal{D}(H)$ and define $u(t)=KQ_{t}f$. Since $Q_{t}f$ is
in $\mathcal{D}(H)$ by~\cite[Proposition 1.1.5]{EthierKurz}, $u(t)\in\mathcal{D}(G)$
for all $t\geq0$. By the continuity of $K$
\[
\frac{\mathrm{d}}{\mathrm{d} t}u(t)=K\frac{\mathrm{d}}{\mathrm{d} t}Q_{t}f=KHQ_{t}f=GKQ_{t}f=Gu(t).
\]
Since
\[
\frac{\mathrm{d}}{\mathrm{d} t}u(t)=KQ_{t}Hf,
\]
$Gu(t)=\frac{\mathrm{d}}{\mathrm{d} t}u(t)$ is continuous as a function of $t$.
By Proposition~1.3.4 in Ethier and Kurtz~\cite{EthierKurz}, $u(t)=P_{t}u(0)=P_{t}Kf$
which proves that (\ref{eq:semigroup-intertwining}) holds on $\mathcal{D}(H)$.
Since all operators involved in~(\ref{eq:semigroup-intertwining})
are continuous, the assertion now follows from the density of $\mathcal{D}(H)$
in $L_{2}$.

The implication $\eqref{enu:SI-generator}\Rightarrow\eqref{enu:SI-core}$
is trivial by taking $D=\mathcal{D}(H)$. To prove the converse, let $f$
be in $\mathcal{D}(H)$. Then there exist $f_{n}\in D$ such that $f_{n}\rightarrow f$
and $Hf_{n}\rightarrow Hf$. Since $K$ is continuous, $Kf_{n}\rightarrow Kf$
and $GKf_{n}=KHf_{n}\rightarrow KHf$, where we have used (\ref{eq:generator-intertwining})
for $f_{n}$. Since $G$ is a closed operator, $Kf$ is in $\mathcal{D}(G)$
and $GKf=KHf$.
\end{proof}
Theorem~\ref{thm:Semigroup-intertwining} shows that it suffices
to prove~(\ref{eq:generator-intertwining}) on a core of $H$. In
the Appendix it is shown that

\begin{equation}
D_{H}=\left\{ f\in\mathcal{C}\left(\bar{\mathbb{N}}\right);\,\exists y_{0}\in\mathbb{N}\,\mathrm{s.\,t.}\,\forall y>y_{0}\,f(y)=f(\infty)\right\} \label{eq:D_H}
\end{equation}
is a core of $H$. The following theorem verifies condition~\ref{enu:SI-core}
of Theorem~\ref{thm:Semigroup-intertwining}.
\begin{thm}
\label{thm:GK=00003DKH}$K$ maps $D_{H}$ into $\mathcal{D}(G)$ and
\[
GK=KH
\]
on $D_{H}$.\end{thm}
\begin{proof}
Fix $f$ in $D$ and let $y_{0}\in\mathbb{N}$ be such that $f(y)=f(\infty)$
for all $y\geq y_{0}$. Then for $x\in[0,1)$,
\begin{eqnarray}
Kf(x) & = & \left(1-x^{2}\right)\sum_{y=0}^{y_{0}-1}x^{2y}f(y)+x^{2y_{0}}f\left(y_{0}\right)\nonumber \\
 & = & f(0)+\sum_{y=1}^{y_{0}}x^{2y}\left(f(y)-f(y-1)\right).\label{eq:Kf}
\end{eqnarray}
As $x$ approaches $1$, $Kf(x)$ approaches $f\left(y_{0}\right)$.
Now
\[
Kf\left(1\right)=f(\infty)=f\left(y_{0}\right).
\]
Hence (\ref{eq:Kf}) holds also for $x=1$, and therefore $Kf$ is
in $\mathcal{C}^{\infty}[0,1]\subseteq\mathcal{C}^{2}(G)$. Moreover
\[
\frac{\partial}{\partial x}Kf(x)=\sum_{y=1}^{y_{0}}2yx^{2y-1}\left(f(y)-f(y-1)\right),
\]
hence
\[
\frac{\partial}{\partial x}Kf(0)=0.
\]
We have shown that $Kf$ is in $\mathcal{D}(G)$.

From~(\ref{eq:Kf}) we have that for $x\in[0,1]$,
\begin{eqnarray*}
GKf(x) & = & \left(1-x^{2}\right)\sum_{y=1}^{y_{0}}2y(2y-1)x^{2y-2}\left(f(y+1)-f(y)\right)\\
 & = & \left(1-x^{2}\right)\sum_{y=0}^{y_{0}-1}\lambda_{y}x^{2y}\left(f(y+1)-f(y)\right).
\end{eqnarray*}
Now for $y<y_{0}$
\[
Hf(y)=\lambda_{y}\left(f(y+1)-f(y)\right),
\]
and for $y\geq y_{0}$,
\[
Hf(y)=0,
\]
hence for $x\in[-1,1]$,
\[
KHf(x)=\left(1-x^{2}\right)\sum_{y=0}^{y_{0}-1}\lambda_{y}x^{2y}\left(f(y+1)-f(y)\right).
\]

\end{proof}

\begin{proof}[Proof of Theorem~\ref{thm:PK=00003DKQ}.]
We use Theorem~\ref{thm:Semigroup-intertwining}. In the present
context, $L_{1}=\mathcal{C}[0,1]$ and $L_{2}=\mathcal{C}\left(\bar{\mathbb{N}}\right)$. Thus
we need to show that $K$ maps $\mathcal{C}\left(\bar{\N}\right)$ to $\mathcal{C}[0,1]$.
This is equivalent to saying that the measures $K(x,\cdot)$ are continuous
in $x$ with respect to the weak convergence. But this is easy to
prove, since $K(x,\cdot)$ is geometric distribution with success
parameter $1-x^{2}$ if $x<1$, and it is the degenerate distribution
$\delta_{\infty}$ if $x=1$. Theorem~\ref{thm:GK=00003DKH} verifies
condition~\ref{enu:SI-core} of Theorem~\ref{thm:Semigroup-intertwining}.
Theorem~\ref{thm:Semigroup-intertwining} shows that this is equivalent
to condition~\ref{enu:SI-semigroup}, and this is what we had to
prove.
\end{proof}

\subsection{Coupling}

In order to find a coupling of Theorem~\ref{thm:coupling}, recall
that $\boldsymbol{S}$ is the one-point compactification of $[0,1]\times\mathbb{N}$,
where $\left(1,\infty\right)$ denotes the point at infinity. It is
easy to see that $f:\boldsymbol{S}\rightarrow\mathbb{R}$ is continuous if and only if
$f\left(\cdot,y\right)$ is continuous for all $y\in\mathbb{N}$ and $f(x,y)\rightarrow f(1,\infty)$
as $y\rightarrow\infty$, uniformly in $x$. It is also easy to see
that
\[
\left\{ f\in\mathcal{C}(\boldsymbol{S});f(x,y)=f(1,\infty)\,\mathrm{for\,all}\,x\in[0,1]\,\mathrm{and}\,y>y_{0}\,\mathrm{for\,some}\,y_{0}\in\mathbb{N}\right\}
\]
is dense in $\mathcal{C}(\boldsymbol{S})$. Since even polynomials are dense in $\mathcal{C}[0,1]$
by the Stone-Weierstrass theorem, it follows that
\begin{equation}
\begin{aligned}\mathcal{D}\left(\boldsymbol{G}\right)= & \left\{ f\in\mathcal{C}(\boldsymbol{S});\exists y_{0}\in\mathbb{N}\,\mathrm{s.\,t.}\,f(x,y)=f(1,\infty)\,\mathrm{for\,all}\,x\in[0,1]\,\mathrm{and}\,y>y_{0},\right.\\
 & \left.f(\cdot,y)\,\mathrm{is\,an\,even\,polynomial\,for\,all}\,y\leq y_{0}\right\}
\end{aligned}
\label{eq:domain}
\end{equation}
is dense in $\mathcal{C}(\boldsymbol{S})$.

We now define an operator $\boldsymbol{G}$ with domain $\mathcal{D}(\boldsymbol{G})$
which we later prove generates a Feller process satisfying Theorem~\ref{thm:coupling}.
For the motivation of this definition, see section~\ref{sub:Generator-derivation}.
For $f\in\mathcal{D}(\boldsymbol{G})$ and $(x,y)\in(0,1)\times\mathbb{N}$ define
\begin{equation}
\boldsymbol{G}f(x,y)=\left(Hf\left(x,\cdot\right)\right)(y)+\frac{\left(Gf\left(\cdot,y\right)K\left(\cdot,y\right)\right)(x)-f(x,y)\left(GK\left(\cdot,y\right)\right)(x)}{K(x,y)}.\label{eq:coupled-generator-general}
\end{equation}
Here $\left(Gf\left(\cdot,y\right)K\left(\cdot,y\right)\right)$ denotes
the application of the operator $G$ to the product of $f\left(\cdot,y\right)$
and $K(\cdot,y)$, and $\left(GK\left(\cdot,y\right)\right)$ is the
application of the operator $G$ to $K(\cdot,y)$. In both cases,
$y$ is held fixed, so $f(\cdot,y)$ and $K(\cdot,y)$ are viewed
as functions of $x$ only. $\left(Hf\left(x,\cdot\right)\right)$
is interpreted similarly, but here $x$ is held fixed. Note that $f(\cdot,y)K(\cdot,y)$
and $K(\cdot,y)$ are even polynomials, hence they are in $\mathcal{D}(G)$.
Moreover, $f(x,\cdot)$ is in $D_{H}$ of~(\ref{eq:D_H}), which
is a core of $H$ as shown in the Appendix, hence $f(x,\cdot)$ is
in $\mathcal{D}(H)$. Finally, $K(x,y)>0$ since $x$ is in $(0,1)$ and
$y<\infty$. Therefore, all the expressions in~(\ref{eq:coupled-generator-general})
are well defined. After plugging in the definitions of $H$ and $G$,
we can get an explicit formula for $\boldsymbol{G}$.
\begin{lem}
\label{lem:coupled-generator-explicit}Let $f$ be in $\mathcal{D}(\boldsymbol{G})$
and $(x,y)\in(0,1)\times\mathbb{N}$. Let $\boldsymbol{G}f$ be defined by~(\ref{eq:coupled-generator-general}).
Then
\begin{equation}
\boldsymbol{G}f(x,y)=\lambda_{y}\left(f(x,y+1)-f(x,y)\right)+\left(1-x^{2}\right)\frac{\partial^{2}}{\partial x^{2}}f(x,y)+4\left[\frac{y}{x}-(y+1)x\right]\frac{\partial}{\partial x}f(x,y).\label{eq:coupled-generator-explicit}
\end{equation}
\end{lem}
\begin{proof}
Observe that for $(x,y)\in(0,1)\times\mathbb{N}$,
\begin{eqnarray*}
\left(Gf\left(\cdot,y\right)K\left(\cdot,y\right)\right)(x) & = & \left(1-x^{2}\right)\left(\frac{\partial^{2}}{\partial x^{2}}f(x,y)\right)K(x,y)\\
 &  & +2\left(1-x^{2}\right)\frac{\partial}{\partial x}f(x,y)\frac{\partial}{\partial x}K(x,y)\\
 &  & +\left(1-x^{2}\right)f(x,y)\frac{\partial^{2}}{\partial x^{2}}K(x,y).
\end{eqnarray*}
Hence
\begin{multline*}
\frac{\left(Gf\left(\cdot,y\right)K\left(\cdot,y\right)\right)(x)-f(x,y)\left(GK\left(\cdot,y\right)\right)(x)}{K(x,k)}\\
\begin{aligned}= & \left(1-x^{2}\right)\frac{\partial^{2}}{\partial x^{2}}f(x,y)+2\left(1-x^{2}\right)\frac{\frac{\partial}{\partial x}K(x,y)}{K(x,y)}\frac{\partial}{\partial x}f(x,y).\end{aligned}
\end{multline*}
Noting that
\begin{eqnarray*}
2\left(1-x^{2}\right)\frac{\frac{\partial}{\partial x}K(x,y)}{K(x,y)} & = & 2\frac{2yx^{2y-1}-(2y+2)x^{2y+1}}{y^{2k}}\\
 & = & \frac{4y}{x}-4(y+1)x,
\end{eqnarray*}
we get
\begin{multline}
\frac{\left(Gf\left(\cdot,y\right)K\left(\cdot,y\right)\right)(x)-f(x,y)\left(GK\left(\cdot,y\right)\right)(x)}{K(x,k)}\\
=\left(1-x^{2}\right)\frac{\partial^{2}}{\partial x^{2}}f(x,y)+4\left[\frac{y}{x}-(y+1)x\right]\frac{\partial}{\partial x}f(x,y).\label{eq:coupled-generator-part}
\end{multline}
Plugging~(\ref{eq:coupled-generator-part}) and the definition of
$H$ into~(\ref{eq:coupled-generator-general}), we get~(\ref{eq:coupled-generator-explicit}).
\end{proof}
Formula~(\ref{eq:coupled-generator-explicit}) is well defined even
for $x=1$. Moreover, since $f(x,y)$ is an even polynomial in $x$,
it follows that
\[
\lim_{x\downarrow0}\frac{y}{x}\frac{\partial}{\partial x}f(x,y)
\]
exists, hence we can define $\boldsymbol{G}f$ on $[0,1]\times\mathbb{N}$
by taking the limit. Observe that for $y>y_{0}$ (where $y_{0}$ is
as in~(\ref{eq:domain})), $\boldsymbol{G}f(x,y)=0$. Therefore,
if we define $\boldsymbol{G}f(1,\infty)=0$, then $\boldsymbol{G}f$
is in $\mathcal{D}\left(\boldsymbol{G}\right)\subseteq\mathcal{C}(\boldsymbol{S})$ and we can
view $\boldsymbol{G}:\mathcal{D}(\boldsymbol{G})\rightarrow\mathcal{C}(\boldsymbol{S})$ as a
linear operator.
\begin{thm}
\label{thm:generator}Operator $\boldsymbol{G}$ is closable and its
closure generates a Feller semigroup.
\end{thm}
In order to prove Theorem~\ref{thm:generator}, we use the following
corollary to the Hille-Yosida theorem.
\begin{prop}
\label{prop:Hille-Yosida-subspaces}Let $E$ be a compact metric space,
$\mathcal{D}(G)$ a subspace of $\mathcal{C}(E)$, and $G:\mathcal{D}(G)\rightarrow\mathcal{C}(E)$
a linear operator. Suppose that $1$ is in $\mathcal{D}(G)$ and $G1=0$,
$G$ satisfies the positive maximum principle, and there exist a sequence
$\left(L_{n}\right)_{n\in\mathbb{N}}$ of finite-dimensional subspaces of
$\mathcal{D}(G)$ such that $\bigcup_{n\in\mathbb{N}}L_{n}$ is dense in $\mathcal{C}(E)$
and $G:L_{n}\rightarrow L_{n}$. Then $G$ is closable and its closure
generates a Feller semigroup.\end{prop}
\begin{proof}
Lemma~4.2.1 in Ethier and Kurtz~\cite{EthierKurz} shows that $G$
is dissipative, and Proposition~1.3.5 in~\cite{EthierKurz} then
proves that $G$ is closable and its closure generates a strongly
continuous contraction semigroup. Finally, the fact that $G1=0$ proves
that $G$ is conservative.
\end{proof}

\begin{proof}[Proof of Theorem~\ref{thm:generator}]
Let us first prove that $\boldsymbol{G}$ satisfies the positive
maximum principle. Let $f\in D$ and $\left(x_{0},y_{0}\right)\in\boldsymbol{S}$
be such that
\[
\sup_{(x,y)\in\boldsymbol{S}}f(x,y)=f\left(x_{0},y_{0}\right)\geq0.
\]
First if $y_{0}=\infty$, then $\boldsymbol{G}f\left(\boldsymbol{z}_{0}\right)=0$
by definition. Second, let us assume that $\left(x_{0},y_{0}\right)\in[0,1]\times\mathbb{N}$.
Then we have $f\left(x_{0},y_{0}+1\right)-f\left(x_{0},y_{0}\right)\leq0$.
If $x_{0}\in(0,1)$, then $\frac{\partial}{\partial x}f\left(x_{0},y_{0}\right)=0$
and $\frac{\partial^{2}}{\partial x^{2}}f\left(x_{0},y_{0}\right)\leq0$,
hence $\boldsymbol{G}f\left(x_{0},y_{0}\right)\leq0$. If $x_{0}=1$,
then
\[
4\left[\frac{y_{0}}{x_{0}}-(y_{0}+1)x_{0}\right]\frac{\partial}{\partial x}f(x_{0},y_{0})\leq0
\]
and
\[
\left(1-x_{0}^{2}\right)\frac{\partial^{2}}{\partial x^{2}}f\left(x_{0},y_{0}\right)=0,
\]
so $\boldsymbol{G}f\left(x_{0},y_{0}\right)\leq0$. And if $x_{0}=0$,
then the second-order term of the polynomial $f\left(\cdot,y_{0}\right)$
must be non-positive, for otherwise $\left(0,y_{0}\right)$ could
not be a point of maximum. Hence
\[
\lim_{x\downarrow0}\frac{y_{0}}{x}\frac{\partial}{\partial x}f\left(x,y_{0}\right)\leq0
\]
and
\[
\left(1-x_{0}^{2}\right)\frac{\partial^{2}}{\partial x^{2}}f\left(x_{0},y_{0}\right)\leq0,
\]
so we again get that $\boldsymbol{G}f\left(x_{0},y_{0}\right)\leq0$.
We have shown that $\boldsymbol{G}$ satisfies the positive maximum
principle.

Define
\begin{eqnarray*}
L_{n} & = & \left\{ f\in\mathcal{C}(\boldsymbol{S});\,f(\cdot,y)\,\mathrm{is\,an\,even\,polynomial\,of\,degree\,at\,most}\,2n\,\mathrm{for\,all}\,y\leq n,\right.\\
 &  & \left.f(\cdot,y)=f(1,\infty)\,\mathrm{for\,all}\,y>n\right\} .
\end{eqnarray*}
It is easy to see that $\boldsymbol{G}:L_{n}\rightarrow L_{n}$ and
$\bigcup_{n\in\mathbb{N}}L_{n}=\mathcal{D}(\boldsymbol{G})$ is dense in $\mathcal{C}[0,1]$.
Finally, $\boldsymbol{G}1=0$, so by Proposition~\ref{prop:Hille-Yosida-subspaces},
$\boldsymbol{G}$ is closable and its closure generates a Feller semigroup.\end{proof}
\begin{rem}
\label{rem:WF-fail}The proof of Theorem~\ref{thm:generator} is
the only place where our argument fails for the Wright-Fisher diffusion
without reflection at zero. Indeed, we could take $P_{t}$ to be the
semigroup of the diffusion on the whole interval $[-1,1]$, that is,
$P_{t}$ would be generated by
\begin{equation}
Gf(x)=\left(1-x^{2}\right)\frac{\partial^{2}}{\partial x^{2}}f(x),\qquad f\in\mathcal{C}^{2}[-1,1],x\in[-1,1].\label{eq:WF-=00005B-1,1=00005D-generator}
\end{equation}
We could now extend kernel $K$ to be from $[-1,1]$ to $\bar{\mathbb{N}}$
using the same formula~(\ref{eq:K-def}). Our proof of Theorem~\ref{thm:PK=00003DKQ}
would still work. We could define $\boldsymbol{G}$ by~(\ref{eq:coupled-generator-general})
where $G$ would now be defined by~(\ref{eq:WF-=00005B-1,1=00005D-generator}).
But now we could not take $\mathcal{D}(\boldsymbol{G})$ to be functions
such that $f(\cdot,y)$ are even polynomials, because they are not
dense in $\mathcal{C}[-1,1]$. But if we allowed all polynomials, then for
$y>0$ we could not extend $\left(\boldsymbol{G}f\right)(\cdot,y)$
to a continuous function on $[-1,1]$ because of the term $\frac{y}{x}\frac{\partial}{\partial x}f(x,y)$
in~(\ref{eq:coupled-generator-explicit}). The deeper reason for
this problem is that $K(0,y)=0$ for $y>0$, so if the process $\left(X_{t},Y_{t}\right)$
satisfies~(\ref{eq:averaging}), then after $Y_{t}$ departs from
zero, $X_{t}$ is no longer allowed to cross zero, so the behavior
of the diffusion on $[-1,0]$ and $[0,1]$ are independent. To overcome
this problem, we could define
\[
\boldsymbol{S}=[-1,1]\times\{0\}\cup\left[-1,0^{-}\right]\times\{1,2,\dots\}\cup\left[0^{+},1\right]\times\{1,2\dots\}\cup\{-1,1\}\times\{\infty\},
\]
where we think of $0^{-}$ and $0^{+}$ as two different points. Now
we could take $\mathcal{D}(\boldsymbol{G})$ to be functions such that $f(\cdot,0)$
is a polynomial, $f(\cdot,y)$ is an even polynomial with possibly
different coefficients on $\left[-1,0^{-}\right]$ and on $\left[0^{+},1\right]$
and from some $y_{0}$, $f(x,y)$ equals either $f(-1,\infty)$ or
$f(1,\infty)$ depending on whether $x$ is in $\left[-1,0^{-}\right]$
or $\left[0^{+},1\right]$. This set is dense in $\mathcal{C}(\boldsymbol{S})$. Then,
however, $\left(Gf\right)(\cdot,0)$ could be discontinuous at $x=0$
because of the term $\lambda_{y}\left(f(x,y+1)-f(x,y)\right)$ in~(\ref{eq:coupled-generator-explicit}).
To get around this problem, we decided to work with the Wright-Fisher
diffusion with reflection at zero.
\end{rem}
In order to prove Theorem~\ref{thm:coupling}, we need the following
theorem due to Rogers and Pittman~\cite{rogersPitman1981markov}.
\begin{thm}
\label{thm:Markov-function}Let $\left( \boldsymbol{S}, \pmb{\mathscr{S}} \right)$
and $\left(S,\mathscr{S}\right)$ be measurable spaces and let $\phi:\boldsymbol{S}\rightarrow S$
be a measurable transformation. Let $\Lambda$ be a probability kernel
from $S$ to $\boldsymbol{S}$ and define a probability kernel from
$\boldsymbol{S}$ to $S$ by
\[
\Phi f=f\circ\phi.
\]
Let $\boldsymbol{X}_{t}$ be a continuous-time Markov process with
state space $\left( \boldsymbol{S}, \pmb{\mathscr{S}} \right)$, transition
semigroup $\boldsymbol{P}_{t}$ and initial distribution $\boldsymbol{\pi}_{0}=\pi_{0}\Lambda$,
for some distribution $\pi_{0}$ on $S$. Suppose further:
\begin{enumerate}
\item $\Lambda\Phi=I$, the identity kernel on $S$,
\item for each $t\geq0$ the probability kernel $P{}_{t}:=\Lambda\boldsymbol{P}_{t}\Phi$
from $S$ to $S$ satisfies
\begin{equation}
\Lambda\boldsymbol{P}_{t}=P{}_{t}\Lambda.\label{eq:LP=00003DQL}
\end{equation}

\end{enumerate}
Then $P{}_{t}$ is a transition semigroup on $S$, $\phi\circ\boldsymbol{X}_{t}$
is Markov with transition semigroup $P_{t}$ and the initial distribution
$\pi_{0}$ and
\[
\mathrm{P}\left(\boldsymbol{X}_{t}\in A|\phi\circ\boldsymbol{X}_{s},0\leq s\leq t\right)=\Lambda\left(\phi\circ\boldsymbol{X}_{t},A\right)
\]
a.\,s.\ for all $t\geq0$ and $A \in \pmb{\mathscr{S}}$.\end{thm}
\begin{proof}
Rogers and Pittman~\cite[Theorem 2]{rogersPitman1981markov} proved
this for the case that $\pi_{0}=\delta_{y}$ for some $y\in S$. The
general case follows by integration with respect to $\pi_{0}$.
\end{proof}

\begin{proof}[Proof of Theorem~\ref{thm:coupling}.]

It is intuitively clear from the form of $\boldsymbol{G}$ that $Y_{t}$
on its own is generated by $H$, but here we give a short formal proof.
Note that for $f\in\mathcal{C}\left(\bar{\mathbb{N}}\right)$ and $s,t\geq0$,
\[
\mathrm{E}\left(f\left(Y_{s+t}\right)|X_{u},Y_{u},0\leq u\leq s\right)=\left(\boldsymbol{P}_{t}\Psi f\right)\left(X_{s},Y_{s}\right),
\]
where $\boldsymbol{P}_{t}$ is the semigroup generated by $\boldsymbol{G}$
and $\Psi$ is a kernel given by $\Psi f=f\circ\psi$ where $\psi(x,y)=y$.
Hence, in order to prove~(\ref{eq:second-margin}) we need to show
that
\begin{equation}
\boldsymbol{P}_{t}\Psi f=\Psi Q_{t}f\label{eq:second-margin-semigroup}
\end{equation}
for all $f\in\mathcal{C}\left(\bar{\mathbb{N}}\right)$. By Theorem~\ref{thm:Semigroup-intertwining}
it suffices to prove that there exists a core $D_{H}$ of $H$ such
that $\Psi$ maps $D_{H}$ into $\mathcal{D}(\boldsymbol{G})$ and
\begin{equation}
\boldsymbol{G}\Psi=\Psi H\label{eq:second-margin-generator-intertwining}
\end{equation}
on $D_{H}$. It is shown in Lemma~\ref{lem:H-core} that
\[
D_{H}=\left\{ f\in\mathcal{C}\left(\bar{\mathbb{N}}\right);\,\exists y_{0}\,\mathrm{s.t.}\,f(y)=f(\infty)\,\mathrm{for\,all}\,y>y_{0}\right\}
\]
is a core of $H$ and it is easy to see that for $f\in D_{H}$, $\Psi f$
is in $\mathcal{D}(\boldsymbol{G})$. Moreover, for $f\in D_{H}$ we have
$\left(Gf(y)K(\cdot,y)\right)(x)=f(y)\left(GK(\cdot,y)\right)(x)$,
so from~(\ref{eq:coupled-generator-general}) it is easy to see that~(\ref{eq:second-margin-generator-intertwining})
holds.

In order to prove the claims about $X_{t}$, we will use Theorem~\ref{thm:Markov-function}.
In the present setting, $S=[0,1]$ and $\phi(x,y)=x$. Define a probability
kernel from $[0,1]$ to $\boldsymbol{S}$ by
\begin{eqnarray*}
\Lambda(x,A\times\{y\}) & = & \delta_{x}(A)K(x,y)\\
\Lambda(x,\left(1,\infty\right)) & = & K(x,\infty)
\end{eqnarray*}
where $x$ is in $[0,1]$, $A$ is in $\mathcal{B}[0,1]$ and $y\in\mathbb{N}$. In
other words,
\begin{equation}
\Lambda f(x)=\sum_{0\leq y\leq\infty}K(x,y)f(x,y)\label{eq:lf}
\end{equation}
for $f\in\mathcal{C}(\boldsymbol{S})$ and $x\in[0,1]$. Observe that $\pi_{0}^{X}\Lambda=\boldsymbol{\pi}_{0}^{(X,Y)}$.
Also observe that for $f\in\mathcal{C}[0,1]$ and $x\in[0,1]$ we have
\[
\Lambda\Phi f(x)=\sum_{0\leq y\leq\infty}K(x,y)f(x)=f(x),
\]
hence
\begin{equation}
\Lambda\Phi=I.\label{eq:LPhi=00003DI}
\end{equation}

Let us now prove that
\begin{equation}
\Lambda\boldsymbol{P}_{t}=P_{t}\Lambda.\label{eq:LP=00003DPL}
\end{equation}
By Theorem~\ref{thm:Semigroup-intertwining}, it suffices to prove
that $\Lambda$ maps $\mathcal{D}(\boldsymbol{G})$ into $\mathcal{C}^{2}[0,1]$ and
\begin{equation}
\Lambda\boldsymbol{G}=G\Lambda\label{eq:coupled-intertwining}
\end{equation}
on $\mathcal{D}(\boldsymbol{G})$. Let $f$ be in $\mathcal{D}(\boldsymbol{G})$.
Then there is $y_{0}$ such that $f(x,y)=f(1,\infty)$ for $y>y_{0}$.
Since we know that $\Lambda\boldsymbol{G}1=G\Lambda1=0$, we may without
loss of generality assume that $f(\infty)=0$. Then
\[
\Lambda f(x)=\sum_{y=0}^{y_{0}}K(x,y)f(x,y),
\]
which is a polynomial, hence in $\mathcal{C}^{2}[0,1]$.

By~(\ref{eq:coupled-generator-general}) we have for $x\in(0,1)$
and $y\leq y_{0}$ that
\begin{equation}
K(x,y)\left(\boldsymbol{G}f\right)(x,y)=K(x,y)\left(Hf\left(\cdot,x\right)\right)(y)+G\left(f\left(\cdot,y\right)K\left(\cdot,y\right)\right)(x)-f(x,y)\left(GK\left(\cdot,y\right)\right)(x).\label{eq:KG}
\end{equation}
Since both sides of the equality are continuous in $x$, the equality
also holds for all $x\in[0,1]$. Using~(\ref{eq:lf}), (\ref{eq:KG})
and noting that $\left(\boldsymbol{G}f\right)(x,y)=0$ for $y>y_{0}$
by~(\ref{eq:coupled-generator-general}) and $\boldsymbol{G}f(1,\infty)=0$
by definition, we get
\[
\Lambda\boldsymbol{G}f(x)=\sum_{y=0}^{y_{0}}K(x,y)\left(Hf\left(x,\cdot\right)\right)(y)+\sum_{y=0}^{y_{0}}G\left(f\left(\cdot,y\right)K\left(\cdot,y\right)\right)(x)-\sum_{y=0}^{y_{0}}f(x,y)\left(GK\left(\cdot,y\right)\right)(x).
\]
The second term is just $G\Lambda f$, since $f(x,y)=0$ for $y>y_{0}$.
The first term can be rewritten as
\[
\sum_{y=0}^{y_{0}}K(x,y)\sum_{z=0}^{y_{0}}H(y,z)f(x,z),
\]
where we have again used that $f(x,z)=0$ for $z>y_{0}$. The last
term can be written as
\begin{eqnarray*}
\sum_{y=0}^{y_{0}}f(x,y)\left(GK\boldsymbol{1}_{\{y\}}\right)(x) & = & \sum_{y=0}^{y_{0}}f(x,y)\left(KH\boldsymbol{1}_{\{y\}}\right)(x)\\
 & = & \sum_{y=0}^{y_{0}}f(x,y)\sum_{z=0}^{y_{0}}K(x,z)H(z,y)
\end{eqnarray*}
where in the first equality we have used Theorem~\ref{thm:GK=00003DKH}
and in the second equality we have used that $H$ is an upper triangular
matrix. Therefore, $\Lambda\boldsymbol{G}f=G\Lambda f$.

Finally, from (\ref{eq:LPhi=00003DI}) and (\ref{eq:LP=00003DPL})
we get that $P_{t}=\Lambda\boldsymbol{P}_{t}\Phi$. Thus, we have
verified all requirements of Theorem~\ref{thm:Markov-function}.
It follows that $X_{t}$ is the Wright-Fisher diffusion with reflection
at zero with the initial distribution $\pi_{0}^{X}$ and
\begin{eqnarray*}
\mathrm{P}\left(Y_{t}=y|X_{s},0\leq s\leq t\right) & = & \Lambda\left(X_{t},[0,1]\times\{y\}\right)\\
 & = & K\left(X_{t},y\right)\,a.s.
\end{eqnarray*}
for $y\in\bar{\mathbb{N}}$.
\end{proof}

\subsection{\label{sub:Generator-derivation}Derivation of the generator for
the coupled process}

In this section we show how formula~(\ref{eq:coupled-generator-general})
for the generator of the coupled process can be derived. Strictly
speaking, this derivation is not necessary since (\ref{eq:coupled-generator-general})
can be taken as a definition (and we therefore choose to make this
derivation informal for the sake of brevity). However, we believe
that this derivation can provide insight into the problem. We use
the technique of Diaconis and Fill~\cite{diaconisFill90strong-stationary-times}
who derived an analogous result for Markov processes with discrete
space and time. Fill~\cite{fill92strong-stationary-duality} then
extended the result to Markov processes with continuous time and discrete
space. In Fill's setting, $P_{t}$ and $Q_{t}$ are transition semigroups
of Markov processes with discrete state-spaces $\mathcal{S}_{1}$
and $\mathcal{S}_{2}$. For a fixed $t>0$, he defines a probability
kernel on
\[
\mathcal{S}=\left\{ (x,y)\in\mathcal{S}_{1}\times\mathcal{S}_{2};\,K(x,y)>0\right\} .
\]
by
\begin{equation}
\boldsymbol{P}^{(t)}f=Q_{t}\boldsymbol{1}_{\left[P_{t}K\neq0\right]}\frac{P_{t}fK}{P_{t}K}.\label{eq:P^t-discrete-ananlogy-compact}
\end{equation}
In~(\ref{eq:P^t-discrete-ananlogy-compact}), $P_{t}fK$ denotes
the application of semigroup $P_{t}$ to the product of functions
$f$ and $K$ (here, $K$ is not viewed as a kernel but simply as
a function of $x$ and $y$). Although $P_{t}$ normally acts on functions
of $x$, we make it act on functions of $x$ and $y$ by fixing $y$.
Similarly, $P_{t}K$ denotes the application of $P_{t}$ to $K$ (this
can alternatively be interpreted as the composition of the two kernels).
Then we take the pointwise division of $P_{t}fK$ and $P_{t}K$, which
we define to be zero if the denominator is zero. We then get $\boldsymbol{P}^{(t)}f$
as the application of $Q_{t}$ to this function. This time, $x$ is
considered fixed when we apply $Q_{t}$.

It turns out that $\boldsymbol{P}^{(t)}$ does not satisfy the Chapman-Kolmogorov
equations and hence cannot be used to construct the coupled process
directly. However, Fill proves that there exists a generator $\boldsymbol{G}$
on $\mathcal{S}$ (for which he gives an explicit formula) such that
\begin{equation}
\frac{\boldsymbol{P}^{(t)}-I}{t}\rightarrow\boldsymbol{G}\label{eq:Fill-generator}
\end{equation}
as $t\downarrow0$. He then shows the bivariate Markov process associated
with $\boldsymbol{G}$ (with suitable initial distribution) has the
desired properties, i.\,e. its margins on their own are processes
with transition semigroups $P_{t}$ and $Q_{t}$ and satisfy~(\ref{eq:intr-coupling}).

Now we return back to our setting where $P_{t}$ is the semigroup
of the Wright-Fisher diffusion with reflection at zero and $Q_{t}$
is the semigroup of an explosive pure birth process. Note that~(\ref{eq:P^t-discrete-ananlogy-compact})
is not a suitable definition in this case, since $P_{t}$ operates
on continuous functions, but the indicator in (\ref{eq:P^t-discrete-ananlogy-compact})
can introduce discontinuity. To get around this problem, it can be
proved that $\frac{P_{t}fK}{P_{t}K}$ can be extended to a continuous
function (Hudec~\cite[Theorem 3.10]{hudec2016Cascades} proves this
for a slightly different kernel $K$, but his proof can easily be
adapted to our setting). Now we can define $\boldsymbol{P}^{(t)}$
by

\begin{equation}
\boldsymbol{P}^{(t)}f=Q_{t}\frac{P_{t}fK}{P_{t}K}.\label{eq:approximation-of-coupled-semigroup}
\end{equation}
Observe that $P_{t}fK\rightarrow fK$, $P_{t}K\rightarrow K$ and
$\frac{P_{t}fK}{P_{t}K}\rightarrow\frac{fK}{K}=f$ as $t\downarrow0$
(provided we choose $\left(x,y\right)$ such that $K(x,y)>0$). Hence,
\[
\frac{1}{t}\left(\boldsymbol{P}^{(t)}f-f\right)=\frac{1}{t}\left(Q_{t}-1\right)\frac{P_{t}fK}{P_{t}K}+\frac{\frac{1}{t}\left(\left(P_{t}-1\right)fK\right)}{P_{t}K}-f\frac{\frac{1}{t}\left(\left(P_{t}-1\right)K\right)}{P_{t}K}
\]
is expected to converge to
\[
Hf+\frac{GfK}{K}-f\frac{GK}{K}.
\]

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text:
Klaus Gysi (Neukölln, 1912. március 3. – Berlin, 1999. március 6.) német politikus, diplomata és újságíró. A második világháborúban a francia ellenállás tagja volt, utána az NDK-ban lett politikus. 1966 és 1973 közt kulturális miniszter, 1973 és 1978 közt nagykövet Olaszországban, a Vatikánban és Máltán, 1979 és 1988 közt pedig vallásügyi miniszter volt a Német Szocialista Egységpárt színeiben. Az újjáegyesítés után a PDS tagja lett. Fia, Gregor Gysi is politikus.

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1999-ben elhunyt személyek

redpajama_set_name: RedPajamaWikipedia


After 3 months of living in an RV on the coast, the Justins find an apartment in Santa Monica, much to the dismay of Glasses.
Filmed and edited this one during a day off from drawing last week. I've got a few more Justins in the pipe right now, including a mock 'informercial' for the book of blogs I've published through Blurb (we're talking Slap-chop meets the Juicer meets these guys), and a 'pre-going-out-on-the-town' short where Flattie educates Glasses in the art of picking-up chicks, dancing and overall kanoodling.
Thanks for watching if you watched.


text:
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#define crypto_shorthash_siphash24_BYTES 8U
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size_t crypto_shorthash_siphash24_bytes(void);

#define crypto_shorthash_siphash24_KEYBYTES 16U
SODIUM_EXPORT
size_t crypto_shorthash_siphash24_keybytes(void);

SODIUM_EXPORT
int crypto_shorthash_siphash24(unsigned char *out, const unsigned char *in,
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#ifdef __cplusplus
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#endif

#endif

redpajama_set_name: RedPajamaGithub


text:
New ACC art instructor makes space for learning | News, Sports, Jobs
News Photo by Crystal Nelson
Alpena Community College Art Instructor Brian Schorn is pictured Wednesday hanging art in the Olin H. Joynton Fine Arts Center.
ALPENA — Alpena Community College Art Instructor Brian Schorn has big plans for the college's fine arts program.
Schorn, an Alpena native, who officially began working as faculty member on Monday, is settling into his office, developing a new art curriculum, and reimagining how the space in the Olin H. Joynton Fine Arts Center on the college campus will be used.
"I'm looking forward to a fresh start, really, in everything," he said.
Schorn plans to teach photography and digital imaging in the fall and is busy revamping the building's computer lab for when the fall semester begins on Aug. 30. He said the state-of-the-art computer lab includes new iMac computers, new scanners, printers, and a 3-D printer.
The art program's curriculum is also being redeveloped to include five courses students will be able to transfer to other colleges and universities under the Michigan Transfer Agreement — art history I, art history II, drawing, 2-D design, and 3-D design.
"These five classes will then allow students to automatically know that their credits will transfer from here to another college within the state of Michigan," he said. "That's been a challenge, because students take classes and then they end up having to take them over at another college, and it causes a lot of grief and is unnecessary."
The college will also begin offering a concentration in graphic design in 2022, which will align more closely with the trades industry.
"This allows the art program to sort of fit in with the rest of the trades and, essentially, allow us to offer a different type of trade," he said. "That's going to be a two-year program that will have to be developed with four to five new courses in graphic design, which I will teach."
"I really wanted to do that because it's difficult for students who take art, particularly fine art, to find a career, even though they do exist, but to add graphic design allows for a lot more options for jobs for people who are interested in the arts. So this makes it similar to the concrete program, the nursing program, the electrical engineering program, in that they learn a skill that is needed in the world, and it provides jobs easily."
Schorn also has plans for other classrooms and spaces in the building. He has cleaned out the building's gallery, which he said has never been used as a gallery, and wants to offer student and faculty exhibitions as well as art from nationally recognized artists.
A new space for a student lounge was also created in the building and space for vending machines if students need a drink or a snack.
Posted in: Graphic DesignTagged : ACC,art,instructor,Jobs,learning,News,Space,Sports
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Social Media's Dial-Up Ancestor: The Bulletin Board System

redpajama_set_name: RedPajamaCommonCrawl


Many epidemiological studies have associated exposure to traffic-derived air pollution from motor vehicles, air toxins from industry and other environmental toxins by pregnant women with measures of poor birth outcomes including preeclampsia (PE), preterm birth (PTB) and intra-uterine growth restriction (IUGR). Preeclampsia, preterm labor and IUGR, collectively known as ischemic placental disease, are strongly correlated with infant morbidity and a host of adult diseases ranging from coronary artery disease to cancer. Although it is widely believed that the pathophysiological mechanisms leading to complications of ischemic placental disease and placental insufficiency have similar biological origins, starting as early as defective placental implantation, to date there are no predictive studies that prospectively examine placental structure and/or function. This may be related to existing technologies, routinely employed in clinical decision-making, such as ultrasonography and certain biomarkers, lacking precision in the first trimester, and adequate sensitivity and specificity in the second to third trimesters. Thus, assessment and prediction of normalcy versus aberrancy of placental function are lacking. Therefore, the overarching objective of this proposal is to develop and evaluate a suit of cutting-edge multi-parametric magnetic resonance imaging (mp-MRI) technologies (primary predictor), and translate these novel placental imaging modalities to assessing the impact of environmental pollution exposure on prediction of the composite of placental aberrancy/insufficiency and related outcomes (PE, PTB and IUGR) (primary outcome).
Our central hypothesis is that chronic exposure to high rates of environmental pollution, independent of socio-economic status (SES), increases the risk of placental insufficiency due to early gestational development of adverse placental structure/function as detected by mp-MRI technological advances.
AIM 1: Develop non-contrast free-breathing placental mp-MRI techniques that can non-invasively characterize the in vivo placental perfusion and microstructure through all trimesters.
Develop a free-breathing multi-delay pCASL sequence using background-suppressed localized-volume 3D GRASE imaging for quantitative measurement of placental perfusion within a 5-10 min scan (1.5x1.5x2 mm3 resolution).
Develop a free-breathing high-resolution golden-angle 3D stack-of-radial sequence with self-navigation and retrospective motion compensation using compressed sensing for mapping placental microstructure within a 5-10 min scan (1x1x1 mm3 isotropic resolution).
AIM 2: Evaluate placental mp-MRI by longitudinal comparison with uterine artery Doppler ultrasound through gestation as a predictive marker of placental insufficiency.
Conduct a 3-year prospective cohort study of pregnant women, performing newly developed mp-MRI (AIM1) in the 1st and 2nd trimesters in parallel to ultrasound measurement of uterine artery Doppler, to assess mp-MRI prediction capability versus ultrasound.
Collect detailed pregnancy outcome data of prospective cohort, including diagnosis, neonatal outcome, and placenta for pathological and histological analysis to ascertain risk factors for placental insufficiency as a response to environmental exposures (AIM 3).
AIM 3: Perform in-depth analyses in recruited subjects to determine associations of environmental pollution exposures and both placental mp-MRI measurements and concomitant pregnancy outcomes.
Perform in-depth analyses to measures of modeled environmental pollution exposure in our recruited subjects to assess associations with mp-MRI measures of placental structure and perfusion throughout gestation, taking into account multiple confounding factors such as maternal age at delivery, race/ethnicity, and SES.
Correlate mp-MRI measures of placental structure and perfusion throughout gestation with pregnancy outcomes, post-parturient assessment of placental weight, pathology, and neonatal outcomes. These studies will establish the impact of environmental pollution on placental structure and function, and related pregnancy outcomes, with obstetric and neonatal practices embracing surveillance tactics based on the residential location of pregnant women and their offspring. In addition, the proposed research will establish the potential for early recognition of placental insufficiency, thereby staging future strategies of preventive and interventional therapies targeting reversal before encountering detrimental consequences.


L&C gets roughly 2500-2600 applicants, of which they accept 1000.
Of those 1000, roughly 220-250 attend.
Mean GPA and LSAT is 3.49 and 161.
For comparison, L&C's closest competitor, UofO law receives roughly 2000 applicants, of which they accept 850-900. Of those roughly 160-190 attend. However, UofO does not hold a part-time or evening program which could explain the discrepency. UofO also pays closer attention to in-state residents, and their mean GPA and LSAT is lower than L&C.
KraftEZ wrote: L&C gets roughly 2500-2600 applicants, of which they accept 1000.
I wish they would YP. I bet they could jump into the 40's or 50's if they tried this and other strategies.
My checker has updated to 'complete' three different time - 1/26, 2/26, and 3/2. As the most recent update was the third, i assume that it means decision rendered. That said, I am not able to log into the student page yet.
barga24 wrote: My checker has updated to 'complete' three different time - 1/26, 2/26, and 3/2. As the most recent update was the third, i assume that it means decision rendered. That said, I am not able to log into the student page yet.
At least in my case, I couldn't log onto the site until about a week after my last status update. Also, even though most of us seem to have had three updates before a decision was rendered, I don't think it necessarily means much. Keep trying to access the admitted students site and check your mailbox. Good luck!
Can the current student who posted here (about helping on ASD) tell us whether it's true that for most classes you have to read about 12 cases per night? An alumnus of the school (2005) mentioned that figure to me and it kind of freaked me out. How could you keep up with that in 4-5 classes at a time?
It's untrue. Unless that includes reading for 2-3 classes total and includes note cases as well as the primary cases. Also, it depends on the textbook and subject. In our torts book the cases were often edited to half a page. In Constitutional Law we had 19 pages for two cases.
Status update changed yesterday (3/4) and I just got the email stating that "the admissions committee has completed its review of your application, but has not yet made a decision. Your file is still active and we will continue to keep it under consideration. In late April or early May, we will have a clearer idea of how many seats are still available, if any. If we have additional space in the class, we will do a second review of your file and we may extend an offer of admission to you at that time. If you are not admitted after a second review, your file may be denied or added to a waitlist. "
What exactly is the difference between this and the waitlist? Anyone have any other insight or advice to offer? Who else is in this position? What have you heard?
zaw wrote: Status update changed yesterday (3/4) and I just got the email stating that "the admissions committee has completed its review of your application, but has not yet made a decision. Your file is still active and we will continue to keep it under consideration. In late April or early May, we will have a clearer idea of how many seats are still available, if any. If we have additional space in the class, we will do a second review of your file and we may extend an offer of admission to you at that time. If you are not admitted after a second review, your file may be denied or added to a waitlist. "
I just got this too. Weird. Well I might withdraw. Late April is a long time to wait.
I got this today too, my first reponse from any school. Hoping for some better news over the next few weeks.
Got that email today too. Though, I'll be waiting on all of the other schools till then too, and I'd really like to go to L&C.
For those who are interested-I got my financial aid award letter today from Lewis and Clark. I filled out my FAFSA very early on, but it was still nice to get an award letter so early on. Now I know how much debt I am going to accrue over three years!
heckyes wrote: For those who are interested-I got my financial aid award letter today from Lewis and Clark. I filled out my FAFSA very early on, but it was still nice to get an award letter so early on. Now I know how much debt I am going to accrue over three years!
Received it today as well. Unfortunately, it has put into writing the sad fact that L&C isn't giving money to keep it a competitive option for me. I shall be withdrawing soon.
After 3 weeks of forcing the issue, I finally got into the admitted students site 5 minutes ago. No mail confirmation or anything has come yet. But I assume this isn't a glitch.
My 3rd "complete" update on status checker was 2/25 (i.e., last Thursday).
ozarkhack wrote: I reckon I'm in.
also, was that your third update?
I had this situation with UGA. I called them and they said sending along an LOCI or another recommendation is completely fine. I guess the way to think about it is, "can this hurt?" and I think the answer is no. As long as you are courteous, write your letter, and otherwise don't make a pest of yourself, I'm sure it won't hurt and could even help!
I ended up sending a LOCI (kind of casually, in the form of an email) and informed admissions through that letter that a supervisor would be sending along an additional recommendation on my behalf.
Got a large envelope in the mail today, I'm in! L & C is one of my top choices so I'm definitely psyched. I got the letter about running out of scholarship money and that they will reevaluate in late April, that'll probably be the deciding factor.
I read that some of you got travel vouchers. There was nothing in my acceptance packet to that end; was this where you found it? Do you think there's a chance they'll give me one if I contact admissions?
What e-mail address did you send the LOCI to?
"At this time we do not have a formal wait list established. Your file has been initially reviewed by the committee and no final decision has been made so it is basically on "hold". After our April 15th 1st seat deposit deadline, the committee will assess whether there is additional space in the class at that time. If so, they will do second reviews on applicants with your status and may admit some folks then. If it appears there is no space available, we will be in touch asking if you are interested in being added to our wait list as we will be establishing an official wait list at that point. Final decisions will hinge on space availability in the class after mid April."
See the bolded, I didn't send one to L&C, I was accepted there. I was just trying to say that in general, its probably not a bad idea to do so if you're on any kind of waitlist.


One internet surfing "hobby" of mine is wikitravel. Oh boy, do I love wikitravel – it's almost as good as being there. Well, not quite, but it's a time waster. The user-written "travel guides" cover more than just the ordinary and are loaded with great "personal experience"-driven data that is at the same time free and invaluable.
Quite often, though, I find checking out places like Labrador or Sudan the most fascinating. Yes, I imagine those two places have never been mentioned in the same sentence. While I have an affinity for the remote (explaining Labrador), when it comes to wikitravel surfing, I love to read the user-crafted warnings for extremely dangerous locales imagining a completely naive adventurer-to-be trying to plan a trip to South Ossetia. Now, who in their right mind needs to travel to South Ossetia? Needless to say, wikitravel is the only place one may find a travel guide for the place. They even give tips on how to deal with captors in the case of being kidnapped – now that is well thought out!
The North Pole is an obvious choice – considering it might be one of the most experience. Heck, any "cruise ship" plan running north – whether it's to the pole or through the Northwest Passages is one of the most expensive trips a person can sign up for. $20,000+ is the norm, with $10,000 or so likely getting you a trip around Northern Labrador and maybe to the first reaches of Nunavut if you're lucky. I haven't even gotten into the high Russian artic, which is all but impossible for even the richest tourists.
Then there are places where it can be relatively "easy" to get in, but getting around is just not (or nearly not) possible. Count Myanmar on that list – where most of the country is hands off to tourists. North Korea is another one. Put on the nicest of nice faces, and you are – for all intents and purposes – quite safe in the country. Try to venture outside of the general tourist area and you could be in big trouble.
War torn or politically ravaged countries are an easy bunch. Is it worth listing them all? Some are incredibly obvious. Some aren't so. Some aren't even recognized countries yet (South Ossetia being one). If you have any desire to travel to a danger zone it might be helpful to check out Robert Young Pelton's site Come Back Alive. Mr. Pelton has visited most of the places a lot of us wouldn't even spend a second considering – and, well, has come back alive to tell about it. He also has set up travel itineraries for those looking for a truly unique, and dangerous, vacation.
Cuba – while beautiful and old-worldly in nature – is, of course, blocked for travel by Americans (unless you have a valid reason, of which journalism is one, but I don't think you'll see me try to use that to get there anytime soon).
Bolivian visas not only cost quite a bit (around $130 last I checked) – but if you want to get there to explore the Yungas Road, you better be extra careful as this is by far the world's most physically dangerous road. Of course any remote road could be considered dangerous – this includes the Trans-Taiga in Northern Quebec and the Canning Stock Route in Western Australia.
La Rinconada in Peru is a difficult voyage due to the town's extreme height and lack of easy access. It's also devoid of any real tourism and mercury poisoning is a real problem for the mining town – the highest settlement in the world.
And then there is Antarctica – which may very well be the nadir of every adventurer's life. It's the "lost" continent, so to speak, inhospitable for sure, but still home to thousands of scientists year round. That said, traveling to Antarctica is akin to traveling into space – it's a frontier of sorts for travelers who really want to stand out from the crowd. And it's rather doable with total trip costs in the $7,000-10,000 range for a voyage (including flights to the southernmost of South America).
I've always been in love with the remote and while I don't have the cash to shuttle myself around like Richard Branson (who has been planning $200,000 space travel packages), I have a couple of these on my list – namely Labrador and especially Antarctica. That, or a visit to the arguably most remotely inhabited place on earth – Tristan de Cuna. When I get there, you will most certainly know.


If you are someone who lives in Texas, you know that this place is not going to be very cold and that the weather there can be really hot. Maybe you are someone who lives in a very cold country and if you do and if you would really want to visit a place where the weather is pretty hot, you should really go and visit Texas as the place there is really hot. Because there are so many people who live in Texas that do not really like the hot weather, they are really looking for good solutions for these things. There are so many ways that you can stay cool when you are visiting or if you are a resident of Texas and we are going to be looking at some of these things here in this article that we have for you today. We hope that you will learn a lot from this article that is before you today.
There is one really great way that you can try out when you are that someone who really wants to stay cool in the state of Texas. Texas is a really hot places and if you are outdoors so much, you can really start sweating and dripping sweat all over because of the hot weather. This is why there are so many places in Texas that are air conditioned and really cooled well and if you would like to get these AC systems, you can go and get them. What are you waiting for? Get these air conditioners for your home and have them installed at your place today and you will really not regret it as they can really help you stay comfortable and happy in your own house in Texas.
If it is summer in Texas, it is really toing to be hotter than ever before and if your AC system is broken down, you really need to get a good repair service that can fix and repair your broken down AC system for you. Getting an AC repair service to help get your AC systems repaired can really help you and benefit you a whole lot indeed so never hesitate to go and hire them for your own good needs. Your AC might have broken down as you use it too much especially during those really hot Texan months.


Areas of Body Affected: Skin, circulatory, muscles and joints, respiratory, immune, nervous systems.
Properties of Oil: analgesic, antiseptic, antispasmodic, antiviral, balsamic, decongestant, deodorant, diuretic, expectorant, parasiticide, stimulant.
Physical Conditions: Cuts, sores, ulcerations, muscle/joint pain, sports-related injuries, asthma, bronchitis, congestion, coughs, respiratory infections, colds, flu, headaches, exhaustion, sciatica.
Holistic Uses: In Chinese medicine, tonic for Lung-Qi.
Exceptions: Not compatible with homeopathy. Non-toxic, non-irritant when diluted, non-sensitizing. Do not take internally.
Blends Well With: This oil blends well with citrus oils.
Eucalyptus Peppermint is pale yellow to colorless in color. Its aroma is fresh with a spicy-minty aroma.
Process of Extraction: Steam distillation from leaves and stems.


text:
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CELEBRITY INSIDER – THE LATEST CELEBRITY NEWS, CELEBRITY GOSSIP, HOLLYWOOD CELEBRITY NEWS, CELEBRITY ENTERTAINMENT NEWS. CELEBRITY INSIDER is an independent network of reporters and contributors with access to insider celebrity reports, unbiased news and other information related to celebrities in Hollywood, TV shows, movies, music, fashion,...
Who are the top 10 most famous celebrities in the world?
Taylor Swift. Rihanna. Selena Gomez. Kim Kardashian. Kylie Jenner. Kendall Jenner. Justin Bieber. Miley Cyrus.
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Mila Kunis And Ashton Kutcher Shock The Internet By Revealing Their Family's Unusual Showering Habits! Valerie Bertinelli And Demi Lovato Spotted On The Set Of Their New Show 'Hungry!'
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Harry Potter's Devon Murray, Dancing With the Stars' Witney Carson and more celebrities have welcomed babies in 2021. The actor's girlfriend, Shannon McCaffrey Quinn, gave birth to their son on...
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Nikki Glaser Pays Tribute To 'The World's Friend' Bob Saget With New Song Bob Saget died on Jan. 9 at the age of 65. Jordan Cashmyer, '16 And Pregnant Star,' Dead at …
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7 hours ago · Piers Morgan has taken aim at Prince Harry and Meghan Markle in his latest column. It comes after the Duke of Sussex is reportedly set to launch legal action, saying it's "unsafe" for him and his ...


Re: Skipping the proxy cache based on a cookie?
OK finally figured it out I think.
Seems it may be to do with the scope of the user set variables (not sure) but after moving adding proxy_no_cache and proxy_cache_bypass to the same location instead of the centrally called ones, It seems to be working as expected.
Skipping the proxy cache based on a cookie?


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text:
Musicians under fire in France and Ethiopia
Hamé after trial
Photo: Daniel Brown
For the last six years one of France's leading underground rap groups, La Rumeur, has been locked in a legal battle against current French President Nicolas Sarkozy. On 3 June, the fourth trial took place, pitting one of the band members, Hamé, against the Ministry of the Interior. World Tracks covered the court case.
World Tracks: Musicians under fire in France and Ethiopia
listen	25 minutes 6 seconds
19/06/2008 by Daniel Brown
Hamé and defence lawyer Dominique Tricaud
This week's World Tracks focuses on censorship and attacks against musicians. We first hear from Tiken Jah Fakoly, the Ivorian who won an award in 2008 for his defence of freedom of speech.
It was bequeathed by Freemuse, the Danish-based association which, for the past decade, has monitored human rights abuses against musicians, including censorship.
Characteristically, Fakoly dedicated his award and some of the prize money to Moussa Kaka, the RFI journalist imprisoned for the past nine months in his native Niger.
It was awarded to the reggaeman as a result, says the NGO, of his tireless work for freedom of musical expression in West Africa. The idol of millions of Africans, especially the youth, Fakoly has been banned, censored and exiled after he denounced political corruption in his native Côte d'Ivoire. In his songs, he also condemns the French legacy of colonialism in Africa.
This legacy also reverberated into France in early June. That's when a trial of one of France's most popular underground rap-bands, La Rumeur, debated over subjects like the massacre in Paris of over 400 French Algerians who, on 17 October 1961, were killed after peacefully demonstrating against the war.
For the past six years La Rumeur rapper Mohammed Bourokba, better known as Hamé, has been prosecuted by the French government for defamation against the national police. It followed an article he published in a fanzine that he and La Rumeur published to accompany their debut album L'Ombre sur la Mesure. In it, Hamé states that France's Interior Minister will never report on the "hundreds of our brothers, who were killed by the police without the assassins ever being worried".
To corroborate his claim, Hamé invited several expert witnesses for his defence. One, rap specialist Gregory Protche, insisted that La Rumeur and Hamé are perpetuating an old tradition of anti-establishment music. These include such revered figures as Georges Brassens and Léo Ferré. And all La Rumeur is doing is taking this tradition into the rough ghettoes where they come from.
32-year-old Hamé is a rare mixture of street savvy and sophistication. The son of an illiterate farmer from Algeria, he survived the impoverished suburban ghettoes called "cité" here and is currently studying for his doctorate at New York University. His 227 line article in the La Rumeur fanzine was a dense critique of French society and the police force that according to Amnesty international is one of the most notorious in Europe.
After the four-hour trial on June 3 he remained defiant: "Our arguments have a true legitimacy," he said, "They should be heard and understood. I wrote the article three years before the explosion in the cite ghettoes in 2005."
"Nowadays, this kind of police violence is becoming commonplace. And it's getting more and more worrying. In the last week, there were three murders – there's a feeling of impunity on the ground. The people responsible for this violence remain untroubled."
The singer continued: "The debate during this trial went to the heart of the matter. All that I wrote in that fanzine six years ago has been amplified nowadays. And we did our best to add testimonies that support my analysis."
"When a policeman kills, he must be tried. If he abuses his authority, he must be punished. And the same justice must be used to chastise him, in the same way as any citizen. This is the very cement of our society. It's just so obvious."
For defence lawyer Dominique Tricaud, the court hearing in Versailles was fundamentally different to court cases in 2003, 2004 and 2006: "It's become a trial of the police," he stated outside the courtroom. "The accuracy of what Hamé wrote six years ago has been underscored by the 2005 rioting. So he cannot be indicted. All that he wrote has been proven right on the ground."
The veteran lawyer remains untroubled about the persistance of the government. "I'm optimistic that he will again be discharged. It's been a political trial from the start. It's not my fault if the signature at the bottom of the complaint is that of Nicolas Sarkozy. Unfortunately for him, however, every time politics enters a courtroom, justice triumphs".
The trial is finally set to end on 23 September when the French court will give its decision. One thing seems certain: win or lose, the finances of Hamé and La Rumeur have been stretched to the limit by a defence that has so far cost them 20,000 euros. And the French system is such that it will be extremely difficult for them to recuperate that from the Sarkozy government.
World Tracks
Moving on up with Angélique Kidjo and Bibi Tanga and the Selenites
Today's guests have chosen to live out of Africa but celebrate its native languages and defend their right to move and groove with the times.
World Tracks: Angélique Kidjo, Bibi Tanga and the Selenites visit RFI's studio 136
Nneka: hip-hop till you drop
Nigerian-German hip-hop/soul artist Nneka has come a long way since she first spoke to Worldtracks in 2007. With the release of her fourth album Concrete Jungle and a guest appearance on the David Letterman show in the US, she's quickly becoming one to watch.
World Tracks: Nneka
Sounds from the River Niger
Salif Keita's latest album seeks to fight discrimination against his fellow albinos, and Mamane Barka, the last master of the biram, battles to save the traditional instrument.
World Tracks: Festival du Niger
Electro sounds with a beating heart
Well-established French electro-pop duo Air have returned with a groovy and uptempo production, while innovative newcomers Sourya have used a Nintendo DS to pack more emotion into their debut album than a dancefloor can normally withstand.
World Tracks: Air/Sourya
Sculpting music with the ondes martenot
Many musicians have been inspired by the ondes martenot. The electronic keyboard can produce a huge range of fascinating sounds but the instrument itself remains somewhat marginal and misunderstood. Christine Ott is determined to change that.
World Tracks: Christine Ott
Lhasa: a unique voice
The Mexican-American singer Lhasa de Sela died on 1 January 2010, aged just 37. In tribute to her remarkable voice and force we rebroadcast a Worldtracks from 2003 which followed the release of her second album The Living Road.
World Tracks: A tribute to Lhasa de Sela
Breton festival puts unknowns on stage
French festival Les Transmusicales is renowned for its eclectic line-up of relatively unknown bands from all over the world. But some acts have gone on to bigger things, including Nirvana, Dizzee Rascal and Justice.
World Tracks: Les Transmusicales de Rennes
Pink Martini, Inna Modja: festive tonic
The latest album from Pink Martini, Splendour in the Grass, pulls out all the stops. And Inna Modja is a newcomer who dared go knock on Salif Keita's door!
World Tracks: Pink Martini, Inna Modja
Folk meets blues meets Africa
New Yorkers Bethany & Rufus join forces with African percussionists, while Langhorne Slim delivers his own pared-down, acoustic love songs with disarming sincerity.
World Tracks: Bethany & Rufus Roots Quartet and Langhorne Slim
Abraham Inc and Faren Kahn - the revival of Klezmer
Klezmer is one of the most nomadic forms of world music and no Jewish wedding would be complete without it! World Tracks talks to David Krakauer about the revival of Klezmer and his mission to keep it out of the museum.
World tracks: the revival of Klezmer


Towards the end of our South Africa safari trip, Bwana Mbogo disappeared. We had come back from our afternoon hunt, and had gone to the room to wash our faces. When he was done, he left to go to the lodge for his evening drink and I was to join him when I was done with my cleanup. When I arrived at the lodge bar, he wasn't there. I searched and searched, and finally I found that he'd been eaten by a hippo!
Bwana Mbogo got eaten by the hippo!
Thankfully, there were several professional hunters who were experienced in dealing with hippos, and we saved Bwana Mbogo!
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Home>Celebrities>Actor>Shrinivas Pokale
Shrinivas Pokale
Shrinivas Pokale Bio :
Born : 25 June 2010
Age : 8 Year
Birthplace : Mumbai
Education : N/A
Shrinivas Pokale is a child actor who hails from Mumbai who was born and brought up at this place. He is the student of fourth standard who will be doing his debut with a Marathi Movie called Naal, which is releasing in November 2018. He is a young and cute boy who is barely 9 years old but when he comes on stage makes people simply love him. He has played the character of Chaitanya who lives in a small village in Maharashtra over the banks of a river. His parents' roles have been played by Nagraj Popatrao Manjule and Devika Daftardar in the film.
The family is a happy to go family who takes care of the kid who remains happy and make everyone happy and feel cool in the company of others. One day he leaves the home with other boy, which makes his parents worried. So, what goes is interesting to catch. After the release of the trailer, the people in the B Town have gone ga-ga over the performance of this young boy. Many of the people in the M Town have been seeing his incredible and bright future in the coming time when he finally grows up adult.
There is not much discussed about his personal life and family, so you need to remain in touch with us and our portal to get the latest about him and others in the M Town. Stay tuned and do comment about the kid and his upcoming movie, which has become the talk of the time with his incredible performance. The film is directed by the debutant director Sudhakar Reddy who is opening his innings with this movie in M Town.
Previous Baazaar 1st Weekend Box Office Collection
Next Naal Trailer Review – worthy story, competitive performances and picturesque locations make the film a delightful watch
Aditya Durve
Saorabh Rajnish Choughule
The Punjabi Music industry is more vibrant when we compare with the other music industry …


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Holwell Sports' cup tie at Quorn re-arranged...again
Holwell Sports will be hoping for third time lucky after Tuesday's re-arranged Westerby Cup tie at Quorn fell victim to the cold snap.
The Farley Way playing surface failed a 4pm pitch inspection, and the quarter-final has been provisionally rescheduled for Tuesday, February 24.
Holwell were also out of luck in the league as Saturday's football programme was also badly hit by snow and ice. Their East Midlands Counties League match at Aylestone Park was called off and has been re-arranged for Saturday, April 18.
Sports will hope for better on Saturday when they travel to Kimberley Miners Welfare (ko 3pm).
The local derby at Welby Road between Holwell Reserves and Asfordby Amateurs in Leics Senior League was also postponed as was Melton Mowbray's match at Kirby Muxloe Reserves.


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/*
 * This code was generated by https://github.com/googleapis/google-api-java-client-services/
 * Modify at your own risk.
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package com.google.api.services.androidenterprise.model;

/**
 * Represents the list of app restrictions available to be pre-configured for the product.
 *
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 This is the Java data model class that specifies how to parse/serialize into the JSON that is
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 * https://developers.google.com/api-client-library/java/google-http-java-client/json
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 * @author Google, Inc.
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Austin Cindric gets first NASCAR Xfinity Series victory in duel at The Glen
August 3, 2019 Reid Spencer, NASCAR Wire Service Main Page, NASCAR News, NASCAR Xfinity Series, Top Stories 0
WATKINS GLEN, NEW YORK - AUGUST 03: Austin Cindric, driver of the #22 MoneyLion Ford, takes the checkered flag to win the NASCAR Xfinity Series Zippo 200 at The Glen at Watkins Glen International on August 03, 2019 in Watkins Glen, New York. (Photo by Sean Gardner/Getty Images)
Opportunity knocked for Austin Cindric after a mechanical failure knocked Kyle Busch out of Saturday's Zippo 200 at Watkins Glen International.
And after Cindric and road course ace AJ Allmendinger traded knocks on the final two laps of the NASCAR Xfinity Series race, Cindric—on fresher tires—emerged the winner for the first time in his career.
Following post-race inspection, Allmendinger's No. 10 Kaulig Racing Chevrolet was disqualified for a rear height violation, dropping Allmendinger to last place in the running order. But fans left the track with the memory of an intense battle over the closing laps.
When NASCAR called the sixth caution on Lap 69 of 82, Brian Wilson, crew chief on Cindric's No. 22 Team Penske Ford, made a courageous call that later proved decisive. Wilson brought Cindric from the lead to pit road for fresh tires, dropping the 20-year-old driver to ninth in the running order for a restart on Lap 73.
An immediate caution for a pile-up in Turn 1 left Cindric in sixth for the next restart on Lap 76. When Christopher Bell was knocked sideways in Turn 2, and Justin Allgaier and Tyler Reddick lost momentum while fighting for the second spot, Cindric charged around the outside into second place and chased Allmendinger.
It took three laps for Cindric to trim Allmendinger's advantage from 1.714 seconds to .521 seconds, and from that point, the game was on. On Lap 81, Cindric nudged Allmendinger up the track in the carousel and took the lead. Allmendinger returned the favor approaching Turn 7 and regained the top spot as Cindric was forced wide.
But Allmendinger entered Turn 7 too wide and Cindric drove back underneath to lead Lap 81. On the final circuit he pulled away to win by 1.168 seconds and clinched a spot in the Xfinity Series Playoffs.
Cindric and Wilson had discussed the possibility of a late pit stop before the race.
"I've been on the other side of it," Cindric said. "(Bell) had been pretty good all day, so it was going to be hard to hold him off depending on which lane he had on the restart."
The only thing that went wrong was Cindric's celebratory burnout. But it was helpful that pole winner Kyle Busch had critical issues during the race itself.
"I was kind of bummed," Cindric said. "I broke the clutch out of it trying to do a burnout, so my guys are going to have to do a little extra work and I'll have to buy them an even bigger dinner. We talked before the weekend, and we knew something would have to go wrong with (Kyle Busch), but I'm so blessed to be able to be here."
After winning the first stage and pitting thereafter, Busch had just passed Ryan Blaney for the lead entering the inner loop when the upper control arm on the left front of his No. 18 Toyota broke. Busch retired from the race and opened the door for Cindric.
"As soon as KB went out, everybody's eyes opened up, and it was like, 'OK, here we go,'" Allmendinger said. "When you take tires like that, it's all about getting lucky. If you get a restart where you get by a chunk of cars, it makes that strategy work—and he did it.
"Congrats to Austin. He went in there and nudged me, and that was fair. I nudged him—you race how you get raced—but that's what racing's all about. He deserved it. He was on it the whole race."
With the Allmendinger disqualification, Bell inherited the runner-up spot, followed by Allgaier, who traded hard knocks with Ross Chastain, eliminating Chastain from the race after hard contact with the barrier in the carousel. Blaney and Reddick ran fourth and fifth.
NASCAR Xfinity Series Race Results Zippo 200 at The Glen
Watkins Glen, New York
1. (2) Austin Cindric, Ford, 82.
2. (4) Christopher Bell, Toyota, 82.
3. (9) Justin Allgaier, Chevrolet, 82.
4. (3) Ryan Blaney(i), Ford, 82.
5. (5) Tyler Reddick, Chevrolet, 82.
6. (6) Chase Briscoe #, Ford, 82.
7. (13) Cole Custer, Ford, 82.
8. (14) Michael Annett, Chevrolet, 82.
9. (18) Noah Gragson #, Chevrolet, 82.
10. (8) Ryan Preece(i), Chevrolet, 82.
11. (11) Jeremy Clements, Chevrolet, 82.
12. (20) John Hunter Nemechek #, Chevrolet, 82.
13. (23) Scott Heckert, Chevrolet, 82.
14. (12) Justin Haley #, Chevrolet, 82.
15. (15) Gray Gaulding, Chevrolet, 82.
16. (21) Alex Labbe, Chevrolet, 82.
17. (10) Brandon Jones, Toyota, 82.
18. (27) Brandon Brown #, Chevrolet, 82.
19. (30) Josh Williams, Chevrolet, 82.
20. (28) Stephen Leicht, Chevrolet, 82.
21. (26) Cody Ware, Chevrolet, 82.
22. (24) Ray Black II, Chevrolet, 82.
23. (31) Stanton Barrett(i), Toyota, 82.
24. (29) Garrett Smithley, Chevrolet, 80.
25. (34) Dan Corcoran(i), Chevrolet, 75.
26. (16) Josh Bilicki, Chevrolet, Accident, 72.
27. (33) BJ McLeod, Chevrolet, DVP, 67.
28. (22) Tommy Joe Martins, Toyota, Accident, 65.
29. (32) David Starr, Chevrolet, Fuel Pump, 57.
30. (17) Ryan Sieg, Chevrolet, Accident, 47.
31. (1) Kyle Busch(i), Toyota, Suspension, 35.
32. (35) Chad Finchum, Toyota, Brakes, 30.
33. (19) Ross Chastain(i), Chevrolet, Accident, 19.
34. (36) Joey Gase, Toyota, Transmission, 9.
35. (25) JJ Yeley(i), Chevrolet, Overheating, 6.
36. (37) Vinnie Miller, Chevrolet, Ignition, 0.
37. (7) AJ Allmendinger, Chevrolet, 82.
Average Speed of Race Winner: 88.611 mph.
Time of Race: 2 Hrs., 16 Mins, 2 Secs. Margin of Victory: 1.168 Seconds.
Caution Flags: 7 for 16 laps.
Lead Changes: 9 among 6 drivers.
Lap Leaders: K. Busch(i) 1-21;C. Bell 22-23;R. Blaney(i) 24-36;A. Allmendinger 37-50;J. Allgaier 51-63;A. Cindric 64-69;A. Allmendinger 70-71;C. Bell 72;A. Allmendinger 73-80;A. Cindric 81-82.
Leaders Summary (Driver, Times Lead, Laps Led): AJ Allmendinger 3 times for 24 laps; Kyle Busch(i) 1 time for 21 laps; Ryan Blaney(i) 1 time for 13 laps; Justin Allgaier 1 time for 13 laps; Austin Cindric 2 times for 8 laps; Christopher Bell 2 times for 3 laps.
Stage #1 Top Ten: 18,20,51,11,12,8,9,39,08,23
Stage #2 Top Ten: 12,8,22,98,23,19,7,11,20,39
Written by: Reid Spencer, NASCAR Wire Service, August 3, 2019
Austin Cindric
Denny Hamlin aka "Mr. Glue."
Chase Elliott leads Hendrick sweep of Watkins Glen front row