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Merger agreed between Scotmid and Penrith Co-operatives
John Dalley, Chief Financial Officer of Scotmid, signs the paperwork with John Mills, Chief Executive Officer of Penrith Co-operative.
"This is a momentous decision for both societies."
Scotmid and Penrith Co-operative Societies are delighted to announce that their merger will proceed, after Penrith members voted in favour of the proposal on 28 August 2013 at a meeting of confirmation.
88 members of Penrith Co-operative attended the meeting in the Methodist Church in Wordsworth Street, Penrith, and voted overwhelmingly in support of the proposal to merge the societies, with Scotmid confirming the decision taken on the 13 August 2013.
Harry Cairney, Acting President of Scotmid, with Ralph Beaumont, President of Penrith.
Ralph Beaumont, President of Penrith Co-operative, thanked members for their support over the years for the Penrith society and looked forward to seeing the new enlarged society grow from strength to strength, and is confident that members will see the benefit of the coming together of these two great societies.
John Brodie, Chief Executive Officer of Scotmid Co-operative said "This is a momentous decision for both societies and one not taken lightly by the Boards of Penrith or Scotmid nor by the members, and together we can look forward, taking the best of both societies to serve the communities and extend co-operation to a wider geographic area."
Further details about the merger will be made available over the coming months.
Penrith members vote in support.
Scotmid Co-operative is Scotland's largest independent co-operative society and has been at the heart of local communities since 1859. Scotmid's latest financial figures (for the year ended 26th January, 2013) showed an operating profit of £6m and a turnover of £428m. The Society employs nearly 5000 staff in 350 retail and service outlets across Scotland, Northern Ireland and the North of England. This includes Scotmid Food, Semichem, The Fragrance House, funeral branches, post offices and a property division. The Society is owned, managed and governed in Scotland and as a co-operative is democratically controlled by its members. Further information on Scotmid can be found at; www.scotmid.coop; www.facebook.com/scotmid and on Twitter @scotmid
Penrith Co-operative is Cumbria's largest independent co-operative society and has been at the heart of local communities since 1890. Penrith's turnover for the year ended 7th January, 2013 was £13m. The Society employs 179 staff in nine retail outlets and Post Offices across Cumbria and County Durham. The Society which has its Head Office in Penrith is democratically controlled by its members. Further information on Penrith can be found at www.penrithco-op.co.uk
Scotmid Co-operative
© Scottish Midland Co-Operative Society Limited


text:
Room For One and Their Dreams
I returned to this spot for sunrise and sunset five or six times and didn't once think of getting into the hammock. Very few people did. This didn't strike me until just now. The scene was pretty, beautiful at times, but perhaps ostentatious in its singularity. The boldness of the idea that this hammock is for one person only and claiming it meant no one else could share it was too much for crowds on the beach, even me. Instead, no one reached for it, or only did so for the briefest moment. Collectively, everyone denied themselves and pressured each other to deny themselves this dream of a moment. The thought was, if everyone can't have it, no one can.
But, are our dreams really meant to be shared? Does one person really reach for the hands of others and collectively step in the direction of their dreams, or does that person lead and pull others in their wake? I feel that not sitting in that hammock is emblematic of the way we place our dreams just far enough out in front of those around us to be on display, but don't seize them as our peers look on because we have learned that behavior to be selfish. Stepping up to our dreams is stepping out of bounds, into the open, and beneath the scrutiny of those who think they don't belong to you. It takes a different kind of person to fold themselves into that space hanging between the posts. A person, maybe, who is different than me.
If I ever return to Gili Air I hope I can take that seat in that hammock, watch a sunrise or sunset and that I can feel ownership over the moment as I should. Until then, "Room for one and their dreams" will be my visualization and my manifestation.
"Room For One And Their Dreams" is a limited edition of 5 only
Standard: 24"x36" (0.9m)
Large: 35"x54" (1.4m)
Grand: 46"x72" (1.8m)
Commissions available
About the prints: I print on a premium substrate that is unparalleled in depth and luminosity. The material refracts light in such a way that light travels laterally through the paper as well as reflects to the audience creating a brilliant effect that many mistake to be rear illumination. Each print is then face mounted to acrylic and sandwiched to a protective backing. The artwork is then framed or a recessed mount is added to the back to float on the wall. I offer many framing options on this website or you can collect the artwork and have it framed at your preferred framer. I recommend Room for One to be framed in a narrow black or brown frame with three to five inches of white space. More arrangements are available. Liners are available in Black, White, Beige. See FRAMES in the menu above for more frames and liner options.

redpajama_set_name: RedPajamaCommonCrawl


Summer is the best part of the year to spend some time at the beach and soak in salt water and just enjoy the heat. However, this is not the very best times for our hair! It gets frizzy, dry, and unruly most of the time and gets it ready in time for the long and dry winter can be a huge task. Knowing the tricks on how to have a healthy and shiny looking hair is very important. Remember that our hair is our crowning glory and people can't help but notice it right away if it is unkempt. So here are the simple tricks to help you along the way.
What Causes Hair To Frizz?
Not eating a balanced diet.
You are not sleeping with satin or silk accessories causing your hair to rub roughly against the drying materials.
Not using enough hair conditioner and usually using the wrong kinds of hair products.
Your shampoo is drying out your hair.
The oils and moisturizers that you are using are not penetrating the hair shaft.
The medications that you are using can cause damage to your body and it affects the health of your hair.
There are so many suggestions on health and beauty magazines or blogs on how to properly care for our hair, but most of these are not really effective. It is important that you know how to take care of your hair to make it look healthy and shiny. Here are the easiest ways to help you achieve a healthier looking lock.
When you buy a shampoo, the most common instructions would be, "Lather, Rinse, Repeat," however, this is not applicable for everyone. If you want to achieve the healthiest looking locks, start by rinsing your hair in the shower. Rub a quarter-sized amount of shampoo into your roots and spend more time massaging your scalp if you have thick and long hair. There's no more need to scrub your ends or repeat the process.
Rinse your hair and condition it from the mid-length up to the ends and let it sit for a few minutes before you give it another rinse. Contrary to popular belief, giving your hair a final rinse of cold water will not leave it extra-shiny. According to the experts, a cold-water-rinse technique is just a tall tale. Just dry your hair with a squeeze using a towel. Remember not to rub it vigorously since it will cause breakage and will strip your hair off of its glorious shine.
Pollutants can easily get into your hair and it can make it look less shiny and duller. When we are using products onto our hair, the oils that we secret and the dirt that accumulates that we pick up every day get stuck in our hair. This is why we need an extra help when it comes to cleansing and clarifying shampoo is all that we need. This can remove all of the unwanted pollutants from our hair and will leave it looking fresh and shiny. However, this product can also strip our hair off with the good stuff if we use it too often. According to experts, using a clarifying shampoo on your hair once a week is more than enough.
Skip the shimmering oil and use Argan oil instead. The shimmering spray may be the easiest way to gloss it up but it can bring more harm to your hair. Most of the hairsprays have an ingredient called "mica" which can make your hair shine but it can also rough up the hair shaft and remove the natural gleam of your hair in long-term use. Hair experts these days recommend for their clients to use argan oil. These products will not only make your hair look shiny, but it can also improve its texture. Make sure that you start with a small amount especially if you have fine hair.
Vitamin E capsules can be bought even at the Whole Foods. Just snip off the ends of 10 -12 capsules and squeeze its contents into a bowl. Before you, shampoo, apply it to your hair and give it a little massage. Dr. Stafford Broumand, a cosmetic and plastic surgeon, Vitamin E can help increase the circulation of the scalp and by massaging your locks with the oil, your hair will feel stronger and will look healthier and shinier.
According to our grandmas, brushing your hair can make it look shinier and healthier. For better results, it is best that you use a 100% wild-boar brush because these stiff and rough-feeling brushes can draw natural oils from the roots of your hair down to its ends which can promote a natural gloss and shine. This can also help volumize fine hair.
Making sure that your hair gets the right nutrients and care that it needs regularly can help you achieve a healthier and shinier looking hair. Remember that when your hair is damaged, bringing back the shine and its health can be a challenge. The transformation does not happen overnight. You have to be patient and use the right products in order to bring back the life to your hair. Just follow the simple tricks listed above to help you achieve your hair goals easier. It won't be long until your hair will be back to its old self; healthy and shiny.
Great Depression Facts Copyright © 2019.


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text:
\section{Introduction}
Stepping stone model is a probability-measure-valued stochastic process
describing the evolution of relative frequencies
for different types of alleles in a large
population with geographical structure; see Shiga \cite{Shiga801, Shiga802, Shiga82}, Shiga and Uchiyama \cite{ShigaK}
for earlier work on the classical stepping stone model.
Without geographical structure, namely, the whole population lives in single colony, the model degenerates to the classical Fleming-Viot process  (cf. Fleming and Viot  \cite{FV_79}).
Notohara \cite{Notohara} and Handa \cite{Handa_2} introduced the dual process of the classical stepping stone model as a function-valued Markov process involving Kingman's coalescent respectively.

In the past three decades, more general coalescents have been proposed. For example, the $\Lambda$-coalescent (cf. Pitman \cite{Pitman99} and Sagitov \cite{SS_99}) is a coalescent
with possible multiple collisions and the $\Xi$-coalescent (cf. Sagitov \cite{SS_03} and Schweinsberg \cite{Jason-S}) is a coalescent with possible simultaneous multiple collisions.
It is then interesting to know whether there exists a generalized stepping stone model whose dual is
also a function-valued Markov process evolving in the same way as the dual of the classical stepping stone model but with Kingman's coalescent replaced by a more general
coalescent.
As far as we know, such kind of generalization mainly focus on the single colony model;
see Birkner et al. \cite{MJM}, Donnelly and Kurtz \cite{DK99b}, Li et al. \cite{LLXZ} and the references therein. However, there has been no breakthrough concerning generalization of the stepping stone model.

In this paper, we will consider replacing Kingman's coalescent in the classical stepping stone model with $\Xi$-coalescent and formulate a more generalized two dimensional probability-measure-valued stochastic process,
named as {\it the generalized stepping stone model with $\Xi$-resampling mechanism}. Intuitively, there is a large population of individuals living in two colonies. Each population undergoes reproduction that is described by a $\Xi$-coalescent with simultaneous multiple collisions, i.e., there are prolific individuals that can simultaneously give birth to children of large amounts comparable to the total population. At the same time, each individual is subject to independent mutation described by an operator $A$ and migration to the other colony at a certain rate.
Instead of considering the well-posedness of martingale problem for superprocess, the moment duality plays an explicit role in establishing the existence of the generalized stepping stone model. This is the most important highlight of our work.
An application of such approach has been introduced in  Evans \cite{Eva97}, where the author proved that subject to a weak duality condition, any system of coalescing Borel right processes gives rise to a Feller semigroup arising from the duality consideration.

Adapting this approach, we show that a function-valued Markov process involving $\Xi$-coalescent,
mutation operator $A$, and mutually geographical migration with rates $u_{\mathbbm{1}}$ and $u_{\mathbbm{2}}$, gives rise to a Feller semigroup, which is sufficient to determine the finite dimensional distributions of  the generalized stepping stone model with $\Xi$-resampling mechanism. Compared with \cite{Eva97}, our coalescing process has more general reproduction mechanism involving simultaneous multiple coalescent.
Further, we prove that
the generalized stepping stone model with $\Xi$-resampling mechanism has a unique invariant measure if the mutation process
allows a unique invariant measure.

The reversibility of a population genetic model is an important issue for statistical inference. Li et al. \cite{LSY99} proved the reversibility of the classical Fleming-Viot process with mutation operator of uniform type. Kermany et al. \cite{AXD} discussed the irreversibility of a two-island diffusion model.
Feng et al. \cite{FZ} proved the irreversibility for an interacting classical Fleming-Viot processes with mutation, selection and recombination.
Li et al. \cite{LLXZ} proved that the $\Xi$-Fleming-Viot process is not reversible except for the degenerate case in \cite{LSY99}.
In this paper we also discuss the reversibility for the generalized stepping stone model with $\Xi$-resampling mechanism. Assuming that the mutation operator is of uniform type, we verify that the process is not reversible.

The plan of the rest of the paper is as follows. In Section \ref{sec2}, we extend the
simultaneous multiple $\Xi$-coalescent to a geographically structured
model with mutation and mutual migration, i.e., the {\it $(\Xi,\text{A},u_{\mathbbm{1}},u_{\mathbbm{2}})$-coalescent process with geographical labels and migration} ({$(\Xi,\text{A},u_{\mathbbm{1}},u_{\mathbbm{2}})$-CPGM}),
which serves as the dual process and is described by a
continuous time function-valued Markov process. In Section \ref{sec3},
the generalized stepping stone model is specified by explicitly defining
a transition semigroup that has its associated ``moments" expressed as expectations for a system of  $(\Xi,\text{A},u_{\mathbbm{1}},u_{\mathbbm{2}})$-CPGMs.
Moreover, we discuss the stationary distribution and irreversibility of the generalized stepping stone model with $\Xi$-resampling mechanism in
Sections \ref{sec5} and \ref{sec6}, respectively.

Throughout the paper, we use $\mathbbm{P}\(\cdot\)$ (resp. $\mathbbm{Q}\(\cdot\)$)
to represent both the probability measure and the associated expectation on probability space $(\Omega,\mathcal{F},\mathbbm{P})$
(resp. $(\Omega^{*},\mathcal{F}^*,\mathbbm{Q})$).

\section{$(\Xi,\text{A},u_{\mathbbm{1}},u_{\mathbbm{2}})$-coalescent process with geographical labels and migration}\label{sec2}
In this section, we first give a short review on $\Xi$-coalescent, and then introduce the $\Xi$-coalescent process with
geographical labels and migration ($(\Xi,u_{\mathbbm{1}},u_{\mathbbm{2}})$-CPGM). Moreover, by adding the mutation operator, we construct a  function-valued Markov process,
named as the $\(\Xi,\text{A},u_{\mathbbm{1}},u_{\mathbbm{2}}\)$-CPGM. Its generator and a coupling property are derived at the end.

\subsection{$\Xi$-coalescent}
Put $[n]=\{1,\ldots,n\}$, $[n]_0=\{0,1,\ldots,n\}$, $[\infty]=\{1,2,\ldots\}$ and $[\infty]_0=\{0,1,2,\ldots\}$.
A {\it partition} of $D\subseteq[\infty]$ is a countable collection
$\pi=\{\pi_i, i=1,2,\ldots\}$ of disjoint blocks such that
$\cup_{i}\pi_i=D$ and $\min\pi_i<\min\pi_j$ for $i<j$.
Denote by $|\cdot|$ the cardinality of the corresponding set. Let
$\mathcal{P}_{D}$ be the collection of partitions for any $D\subseteq[\infty]$. In particular, we write
${\mathbf{1}}_{D}$ as the singleton partition for $D$. For example, ${\mathbf 1}_{[n]}=\{\{1\},\ldots,\{n\}\}$.

Given a partition $\pi\in\mathcal{P}_{D}$ with $|\pi|=n$ and
$\pi'\in\mathcal{P}_{[k]}$ with $n\leq k$, the {\it coagulation} of $\pi$ by $\pi'$, denoted by
$\text{Coag}(\pi,\pi')$, is defined as the following partition of
$D$,
\begin{equation}\label{coag}
\text{Coag}(\pi,\pi')=\pi''=\left\{\pi''_j:=\cup_{i\in \pi'_j}\pi_i, j=1,\ldots,|\pi'|\right\}.
\end{equation}

Given a partition $\pi$ with $|\pi|=n$ and a sequence of positive
integers $s,k_1,\ldots,k_r$ such that $k_i\geq 2, i=1,\ldots,r$
and $n=s+\sum_{i=1}^r k_i $, we say a partition $\pi''$ is obtained
by a {\it $(n;k_1,\ldots,k_r,s)$-collision} of $\pi$ if
$\pi''=\text{Coag}(\pi,\pi')$ for some partition $\pi'$ such that
\[\{|\pi'_i|: i=1,\ldots,|\pi'|
\}=\{k_1,\ldots,k_r,k_{r+1},\ldots,k_{r+s}\},\] where
$k_{r+1}=\cdots=k_{r+s}=1$, i.e., $\pi''$ is obtained by mergering the $n$
blocks of $\pi$ into $r+s$ blocks in which $s$ blocks remain
unchanged and the other $r$ blocks contain $k_1,\ldots,k_r$ blocks
from $\pi$, respectively.

The {$\Xi$-coalescent} is a $\cP_{[\infty]}$-valued Markov process
 $\Pi_{[\infty]}=(\Pi_{[\infty]}(t))_{t\geq 0}$ starting from partition
$\Pi_{[\infty]}{(0)}\in \cP_{[\infty]}$ such that
 for any $D\subseteq[\infty]$, its
restriction to $D$, $\Pi_{D}=(\Pi_{D}(t))_{t\geq 0}$ is a Markov chain
and given that $\Pi_{D}(t)$ has $n$ blocks, each
$(n;k_1,\ldots,k_r;s)$-collision occurs at rate
$\la_{n;\,k_1,\ldots,k_r;\,s}$ with
\begin{equation*
\begin{split}
\la_{n;\,k_1,\ldots,k_r;\,s}
=\int_\Delta \sum_{\ell=0}^s\sum_{i_1\neq
\cdots\neq i_{r+\ell}}{s\choose{\ell}}x_{i_1}^{k_1}\ldots
x_{i_r}^{k_r}x_{i_{r+1}} \ldots x_{i_{r+\ell}}\left(1-\sum_{j=1}^\infty
x_j\right)^{s-\ell} \frac{\Xi(d\mathbf{x})}{\sum_{j=1}^\infty x_j^2},
\end{split}
\end{equation*}
where $\Xi$ is a finite measure on the infinite simplex
\[\Delta=\left\{\mathbf{x}=(x_1,x_2,\ldots): x_1\geq x_2\geq\cdots\geq 0, \,\sum_{i=1}^\infty x_i\leq 1\right\}.\]
There exists at least one $\pi^{'}\in\mathcal{P}_{[n]}$ that induces the $(n;k_1,\ldots,k_r;s)$-collision. We simply write $$\la_{\pi^{'}}:=\la_{n;\,k_1,\ldots,k_r;\,s}.$$
Given there are $n$ blocks, the total coalescence rate is
$$\la_{n}:=\sum_{\pi^{'}\in\mathcal{P}_{[n]}\setminus\mathbf{1}^{[n]}}\la_{\pi^{'}}.$$

\subsection{$(\Xi,u_{\mathbbm{1}},u_{\mathbbm{2}})$-CPGM}
Let $\mathbbm{S}=\{\mathbbm{1}, \mathbbm{2}\}$ be the labels of two colonies.
$\forall\, n\in [\infty]$, $\mathbbm{S}^n$ is the $n$-fold Cartesian product.
For any $\eta\in \cup_{n=1}^{\infty}\mathbbm{S}^n$ and $\mathbbm{i}\in\{\mathbbm{1,2}\}$,
denote by $|\eta|_{\mathbbm{i}}$
the number of coordinates in $\eta$ equal to $\mathbbm{i}$.
As a consequence we have $|\eta|=|\eta|_{\mathbbm{1}}+|\eta|_{\mathbbm{2}}$.

Given a partition $\pi=\{\pi_1,\pi_2,\ldots,\pi_n\}$  of $D\subseteq[\infty]$ with cardinality $n$ and a label vector   $\eta=\(\eta_1,\eta_2,\ldots,\eta_n\)\in\mathbbm{S}^n$,
a {\it $\eta$-labeled  partition} $\pi$ is defined as
$$\pi^{\eta}=\left\{\pi_1^{\eta_1},\pi_2^{\eta_2},\ldots,\pi_n^{\eta_n}\right\},~~~~~~~~~~~~~$$
where $\pi_{\ell}^{\eta_{\ell}}$ denotes a {\it $\eta_{\ell}$-labeled block}.
Set $\mathbbm{L}\({\pi}^{\eta}\)=\eta$ and  $\mathbbm{L}\({\pi_i}^{\eta_i}\)=\eta_i$ as the labels; $\mathbbm{B}\({\pi}^{\eta}\)=\pi$ and  $\mathbbm{B}\({\pi_i}^{\eta_i}\)=\pi_i$ as the collection of blocks.
For any $\mathbbm{i}\in\{\mathbbm{1,2}\}$, ${\pi^{\eta}}|_{\mathbbm{i}}$ is the
collection of blocks from $\pi^{\eta}$ with label $\mathbbm{i}$. As usual, the blocks in each collection are always ordered by their least elements.
For any $\pi^{'}\in\mathcal{P}_{[k]}$ with $k\geq |\eta|_{\mathbbm{i}}$, put  $$\text{Coag}^{\mathbbm{i}}\(\pi^{\eta},\pi^{'}\)=\text{Coag}\(\pi^{\eta}|_{\mathbbm{i}},\pi^{'}\)\cup\pi^{\eta}|_{\mathbbm{j}}\,\,\text{with}\,\,\mathbbm{j}
=\{\mathbbm{1,2}\}\setminus\mathbbm{i}$$
where Coag$\(\pi^{\eta}|_{\mathbbm{i}},\pi^{'}\)$ is the coagulation of $\pi^{\eta}|_{\mathbbm{i}}$ by $\pi^{'}$, obtained by (\ref{coag})
but keeping the labels unchanged.

\begin{definition}
The $(\Xi,u_{\mathbbm{1}},u_{\mathbbm{2}})$-CPGM
is a pure jump and labeled partition-valued Markov process.
Given any initial value,
the $\Xi$-coalescences independently take place among blocks with the same label.
Between coalescence times, with rate $u_{\mathbbm{i}}$, one of the blocks with label $\mathbbm{j}$ is randomly sampled with its label  replaced by label $\mathbbm{i}$, where $\{\mathbbm{i,\,j}\}=\{\mathbbm{1,\,2}\}$.
\end{definition}

\subsection{$\(\Xi,\text{A},u_{\mathbbm{1}},u_{\mathbbm{2}}\)\text{-CPGM}$}
Let $E=[0,1]$ be the state space. $\mathcal{E}=\mathscr{B}\(E\)$ is the Borel $\sigma$-algebra on $E$.
Set
\begin{equation*}
\begin{split}
&\mathcal{D}_0=\left\{E,\emptyset\right\};\\
&\mathcal{D}_n=\left\{\left[{i}/{2^n},{\(i+1\)}/{2^n}\right),\,i=0,1,\ldots,2^n-2,\right\}\cup{\big\{\left[{\(2^n-1\)}/{2^n},1\right]\big\} }\,\,\text{for}\,\,n\geq1.
\end{split}
\end{equation*}
Define $\mathcal{D}$ as the ring generated by
$\cup_{n=0}^{\infty}D_n$. Clearly, $\mathcal{D}$ is countable.
Let $\sigma\(\mathcal{D}\)$ be the $\sigma$-algebra generated by $\mathcal{D}$. Subsequently, $\mathcal{E}=\sigma\(\mathcal{D}\)$. Write
$$\mathcal{D}\times\mathcal{D}=\left\{C_i\times D_j |C_i,\,D_j\in\mathcal{D},\,i,j=1,2,\ldots\right\}.$$
One can see that $\sigma\(\mathcal{D}\times\mathcal{D}\)=\sigma\(\mathcal{E}\times\mathcal{E}\)=\mathscr{B}\(E^2\)$.
Let ${B}(E)$ be the real-valued bounded
functions on $E$ and ${C}(E)$ be the real-valued continuous functions on $E$.
Denote by ${M}_1(E)$ the collection of probability measures on $E$.
${B}(E^n)$, ${C}(E^n)$ and ${M}_1(E^n)$ are defined similarly on $E^n$.

The mutation operator $A$ is a jump-type Feller generator such that
\begin{equation}\label{eq:mu}
Af(x)=\frac{\theta}{2}\int_{E}\(f(y)-f(x)\)q(x,dy),
\end{equation}
where $f(x)\in B(E)$, $\theta>0$ is a constant,  and $q(x,\Gamma)$ is a probability transition function on $E\times \mathcal{E}$. Let $\(T_t\)_{t\geq0}$ be the Feller semigroup associated to $A$.
Denote by $A^{(n)}$ the linear operator on
$B\(E^n\)$ that generates the Feller semigroup $\(T^{(n)}_t\)_{t\geq
0}$ corresponding to $n$ independent copies of the processes
associated  to $(T_t)_{t\geq 0}$. Then
\begin{equation*}
\begin{split}
A^{\(n\)}f(x_1,\ldots,x_n)=\,&\sum_{k=1}^n\frac{\theta}{2}\int_{E}\left[f\(x_1,\ldots,x_{k-1},y,x_{k+1},\ldots,x_{n}\)\right.\\
&\quad\quad\quad\quad\left.-f\(x_1,\ldots,x_{k-1},x_k,x_{k+1},\ldots,x_{n}\)\right]q\(x_k,dy\).
\end{split}
\end{equation*}

Given any ${\eta}=\(\eta_1,\ldots,\eta_n\)\in
{\mathbbm{S}}^{n}$, $\mu=\(\mu_{\mathbbm{1}},\mu_{\mathbbm{2}}\)\in{M}_1(E^2)$ and $f\in B\(E^n\)$, we denote by
\begin{equation}\label{G_{f,u}}
\begin{split}
\mathcal{G}_{f,\,\eta}\(\mu\)&\,:=\,\mathcal{G}_{\mu}\(f,\eta\):=\,\left<\mu_{\eta},f\right>\\
&\,:=\,\int_{E}
\cdots\int_{E}f\(x_1,\ldots,x_n\)\mu_{{\eta}_1}\(dx_1\)\cdots\mu_{{\eta}_n}\(dx_n\).
\end{split}
\end{equation}
In particular, for $f\(x_1,\ldots,x_n\)=\prod_{k=1}^nf_k(x_k)$ with $f_k\in{B}\(E\)$, $k=1,\ldots,n$, we have
$$A^{\(n\)}f(x_1,\ldots,x_n)=\sum_{k=1}^n A f_k\(x_k\)\prod_{\ell\neq k}f_{\ell}\(x_{\ell}\)$$
and
\begin{equation*}\label{seq-fun}
\begin{split}
\mathcal{G}_{\mu}\(f,\eta\
\,=&\,\prod_{\{k|\eta_k=\mathbbm{1},\,k=1,\ldots,n\}}\<\mu_{\mathbbm{1}},f_k\>\prod_{\{k|\eta_k=\mathbbm{2},\,k=1,\ldots,n\}}\<\mu_{\mathbbm{2}},f_k\>\\
:=&\,\<\mu_{\mathbbm{1}}^{\otimes |\eta|_{\mathbbm{1}}},f^{\mathbbm{1}}\>\<\mu_{\mathbbm{2}}^{\otimes |\eta|_{\mathbbm{2}}},f^{\mathbbm{2}}\>
\end{split}
\end{equation*}
with
\[f^{\mathbbm{i}}=\otimes_{\{k|\eta_k=\mathbbm{i},\,k=1,\ldots,n\}}f_k\(x_k\)\,\,\text{for}\,\,\mathbbm{i}\in\{\mathbbm{1,2}\}.\]

We start to construct a $\cup_{m=1}^{\infty}{B}\(E^m\)\times \mathbbm{S}^m$-valued Markov process $\(Y\(t\),\eta\(t\)\)_{t\geq 0}$ on some probability space $(\Omega,\mathcal{F},\mathbbm{P})$.
Given initial value
$$\(Y(0),\eta\(0\)\)\in B\(E^n\)\times\mathbbm{S}^n,$$
we associate the process with a  $(\Xi,u_{\mathbbm{1}},u_{\mathbbm{2}})$-CPGM whose initial value is denoted by a singleton partition with label $\eta\(0\)$, i.e.,
${\mathbf{1}}_{[n]}^{\eta\(0\)}=\left\{\{1\}^{\eta\(0\)_1},\ldots,\{n\}^{{\eta}\(0\)_{n}}\right\}.$
$\(\eta\(t\)\)_{t>0}$ is defined as the label-valued process of the $(\Xi,u_{\mathbbm{1}},u_{\mathbbm{2}})$-CPGM. It is obvious that the value of $\eta\(t\)$ does not change until either coalescence or migration happens. Let$\{\tau_m\}_{m\geq1}$ be the sequence of jumping times of the associated $(\Xi,u_{\mathbbm{1}},u_{\mathbbm{2}})$-CPGM. Set $\tau_0=0$. $\(Y\(t\)\)_{t\geq 0}$ is formulated recursively. $\forall\,m\in[\infty]_0$,  define
$${Y}(t)=T^{\(|\eta(\tau_m)|\)}_{t-\tau_m}Y(\tau_m)\text{~~for any~~} t\in\left[\tau_m,
\tau_{m+1}\right).$$
At jumping time $\tau_{m+1}$, we need to introduce the operator concerning with the reduction of number of variables if coalescence happens.
For any $\mathbbm{i}\in\{\mathbbm{1,2}\}$ and $\pi\in\mathcal{P}_{[|\eta\(\tau_{m+1-}\)|_{\mathbbm{i}}]}\setminus{\mathbf 1}_{\left[|\eta\(\tau_{m+1-}\)|_{\mathbbm{i}}\right]}$, denote by
$$\pi\(\mathbbm{i}\)=\mathbbm{B}\(\text{Coag}^{\mathbbm{i}}\(\mathbf{1}_{[|\eta\(\tau_{m+1-}\)|]}^{\eta\(\tau_{m+1-}\)},\pi\)\)=\{\pi_{\ell},\ell=1,2,\ldots,|\pi\(\mathbbm{i}\)|\}.$$
A map $\Phi^{\mathbbm{i}}_{\pi}$ from ${B}\(E^{|\eta\(\tau_{m+1-}\)|}\)$ to ${B}\(E^{|\pi(\mathbbm{i})|}\)$ is defined as
$$\Phi^{\mathbbm{i}}_{\pi}g\(x_{1},x_2,\ldots,x_{|\eta\(\tau_{m+1-}\)|}\)=g\(x_{{i}_1},x_{{i}_2},\ldots,x_{{i}_{|\eta\(\tau_{m+1-}\)|}}\)$$
with $i_j=\ell$ for $i_j\in\pi_{\ell}$, $j\in\left[|\eta\(\tau_{m+1-}\)|\right]$.
Then we have
\begin{equation*}
\begin{split}
&\mathbb{P}\(Y\(\tau_{m+1}\)
=\Phi^\mathbbm{i}_{\pi}Y\(\tau_{m+1}-\)|\(\eta\(t\)\)_{0\leq t\leq \tau_{m+1}-}\)\\
&\quad
\ind_{\left\{|\eta(\tau_{m+1})|_{\mathbbm{i}}<|\eta(\tau_{m+1}-)|_{\mathbbm{i}},\,
\pi\in {\mathcal{P}_{\left[\left|\eta\(\tau_{m+1}-\)\right|_{\mathbbm{i}}\right]}\setminus
\mathbf{1}_{\left[\left|\eta\(\tau_{m+1}-\)\right|_{\mathbbm{i}}\right]}}\right\}}\,\,{\text{with}}\,\,\mathbbm{i}\in\{\mathbbm{1,2}\};\\
&\mathbb{P}\(Y\(\tau_{m+1}\)=Y\(\tau_{m+1}-\)|\(\eta\(t\)\)_{0\leq t\leq \tau_{m+1}-}\)=\ind_{\left\{|\eta(\tau_{m+1})|=|\eta(\tau_{m+1}-)|\right\}},
\end{split}
\end{equation*}
where $\ind_{\omega}$ denotes the indicator function of an
event $\omega$. The first equation is due to the coalescence within colony $\mathbbm{i}$; the second one is due to the migration between the two colonies.

The $\cup_{m=1}^{\infty}{B}\(E^m\)\times {\mathbbm{S}}^m$-valued Markov process $\(Y\(t\),\eta\(t\)\)_{t\geq 0}$ constructed above is called the {\it  $\(\Xi,\text{A},u_{\mathbbm{1}},u_{\mathbbm{2}}\)\text{-CPGM}$}.

\begin{proposition}\label{ge-pro}
Given $\(Y(0),\eta(0)\)=\(f,\eta\)\in\cup_{m=1}^{\infty}{B}\(E^m\)\times {\mathbbm{S}}^m$, $\forall$ $\mu\in M_1\(E^2\)$, the generator $\mathcal{L}$ of the $\(\Xi,\text{A},u_{\mathbbm{1}},u_{\mathbbm{2}}\)$-CPGM $\(Y\(t\),\eta\(t\)\)_{t\geq0}$ is given by
{\rm
\begin{equation}\label{generator-dual}
\begin{split}
&\mathcal{L}\mathcal{G}_{\mu}\(f,\eta\)\\
&=\mathcal{G}_{\mu}\(A^{\(|\eta|\)}f,\eta\)+\sum_{\mathbbm{i}\in\{\mathbbm{1,2}\}}\sum_{\pi\in\mathcal{P}_{\left[|\eta|_{\mathbbm{i}}\right]}\setminus\mathbf{1}^{\left[|\eta|_{\mathbbm{i}}\right]}}
\la_{\pi}\(\mathcal{G}_{\mu}\(\Phi_{\pi}^{\mathbbm{i}}f,{\beta}_{\pi}^{\mathbbm{i}}\(\eta\)\)-\mathcal{G}_{\mu}\(f,\eta\)\)\\
&\qquad+\sum_{\{\mathbbm{i,\,j}\}=\{\mathbbm{1,2}\}}\sum_{\{k|\eta_k={\mathbbm{j}},\,k=1,\ldots,|\eta|\}}u_{\mathbbm{i}}
\(\mathcal{G}_{\mu}\(f,{\gamma}_{k,\,\mathbbm{i}}\(\eta\)\)-\mathcal{G}_{\mu}\(f,\eta\)\),
\end{split}
\end{equation}
}
where
{\rm
\begin{eqnarray*}
\begin{split}
&\be^{\mathbbm{i}}_{\pi}\(\eta\)=\mathbbm{L}\({\text{Coag}}^{\mathbbm{i}}\({\mathbf 1}_{[|\eta|]}^{\eta},\pi\)\)\text{\,for any\,}\pi\in \mathcal{P}_{\left[|\eta|_{\mathbbm{i}}\right]}\setminus{\mathbf 1}_{\left[|\eta|_{\mathbbm{i}}\right]};\\
&\ga_{k,\,\mathbbm{i}}\(\eta\)=\(\eta_1,\ldots,\eta_{k-1},\mathbbm{i},\eta_{k+1},\ldots,\eta_{|\eta|}\),\,\, 1\leq k\leq |\eta|.
\end{split}
\end{eqnarray*}
}

\end{proposition}

\begin{proof}
The result is easily obtained from the construction of the $\(\Xi,\text{A},u_{\mathbbm{1}},u_{\mathbbm{2}}\)$-CPGM.
\end{proof}

Next we present a coupling property for the $(\Xi,\text{A}, u_{\mathbbm{1}},u_{\mathbbm{2}})$-CPGM. It implies that the impact of a  $\(\Xi,\text{A},u_{\mathbbm{1}},u_{\mathbbm{2}}\)$-CPGM is the same as that of superposition of countable infinitely many  $\(\Xi,\text{A},u_{\mathbbm{1}},u_{\mathbbm{2}}\)$-CPGMs as long as they are all driven by a  $\(\Xi,u_{\mathbbm{1}},u_{\mathbbm{2}}\)$-CPGM
and the superposition of initial values coincides with that of  the single process.

\begin{proposition}\label{coupling property}
Let $\(Y_i\(t\),\eta\(t\)\)_{t\geq 0},\, i\in[\infty]$ be a sequence of $\(\Xi,\text{A},u_{\mathbbm{1}},u_{\mathbbm{2}}\)$-CPGMs that are driven by a  $(\Xi,u_{\mathbbm{1}},u_{\mathbbm{2}})$-CPGM.
Given any $\mu\in{M}_1(E^2)$, $\forall$ $t\geq0$, if
\begin{equation*}
\sum_{i=1}^{\infty}\mid\mathcal{G}_{\mu}\(Y_i\(t\),\eta\(t\)\)\mid<\infty,
\end{equation*}
one can see that
\begin{equation*}
\sum_{i=1}^{\infty}\mathcal{G}_{\mu}\(Y_i\(t\),\eta\(t\)\)=\mathcal{G}_{\mu}\(Y\(t\),\eta\(t\)\)
\end{equation*}
where $\(Y\(t\),\eta\(t\)\)$ is also a $\(\Xi,\text{A},u_{\mathbbm{1}},u_{\mathbbm{2}}\)$-CPGM driven by the same $(\Xi,u_{\mathbbm{1}},u_{\mathbbm{2}})$-CPGM with initial value $Y\(0\)=\sum_{i=1}^{\infty}Y_i\(0\)$.
\end{proposition}

\begin{proof}
For any $s\in[0,\tau_1)$, one can see that
\begin{equation*}
\sum_{i=1}^{\infty}Y_i\(s\)=\sum_{i=1}^{\infty}T_s^{\otimes |\eta\(0\)|}Y_i\(0\)=T_s^{\otimes |\eta\(0\)|}Y\(0\)=Y\(s\).
\end{equation*}
Further, if $\tau_1$ is a migration time, then
\begin{equation*}
\sum_{i=1}^{\infty}Y_i\(\tau_1\)=\sum_{i=1}^{\infty}Y_i\(\tau_1-\)=Y\(\tau_1-\)=Y\(\tau_1\);
\end{equation*}
otherwise, $\tau_1$ is a coalescence time. Since the deduction of variables in $Y_i\(\tau_1-\)$, $i\in[\infty]$ and $Y\(\tau_1-\)$ is driven by the same coalescence event. Subsequently, for any $\Phi^\mathbbm{i}_{\pi}$ with $\mathbbm{i}\in\{\mathbbm{1,2}\}$ and $\pi\in  \mathcal{P}_{\left[|\eta\(\tau_1-\)|_{\mathbbm{i}}\right]}\setminus{\mathbf 1}_{\left[|\eta\(\tau_1-\)|_{\mathbbm{i}}\right]}$, we still have
\begin{equation*}
\sum_{i=1}^{\infty}Y_i\(\tau_1\)=\sum_{i=1}^{\infty}\Phi^{\mathbbm{i}}_{\pi}Y_i\(\tau_1-\)
=\Phi^{\mathbbm{i}}_{\pi}\(\sum_{i=1}^{\infty}Y_i\(\tau_1-\)\)=\Phi^{\mathbbm{i}}_{\pi}Y\(\tau_1-\)=Y\(\tau_1\).
\end{equation*}
By mathematical induction on the jumping times, one can see that $\sum_{i=1}^{\infty}Y_i\(t\)=Y\(t\)$ holds for any $t\geq 0$.
Then the desired result follows from (\ref{G_{f,u}}).

\end{proof}

\section{Generalized stepping stone model with $\Xi$-resampling mechanism}\label{sec3}
In this section, we formulate {\it the generalized stepping stone model with $\Xi$-resampling mechanism} to describe the gene frequencies for a population model with two colonies.
We start with introducing the multidimensional Hausdorff moment problem
Then, subject to moment duality, the transition semigroup of the generalized stepping stone model with $\Xi$-resampling mechanism is uniquely defined from its dual $\(\Xi,\text{A},u_{\mathbbm{1}},u_{\mathbbm{2}}\)$-CPGM. As a consequence, the transition semigroup with initial value determines the finite dimensional distributions of the generalized stepping stone model.

\subsection{Multidimensional Hausdorff moment problem}
The lemma below provides some necessary and sufficient conditions for the solution to multidimensional Hausdorff moment problem.

\begin{lemma}[cf. Proposition 4.6.11 in Berg et al. \cite{Berg}]\label{Haus}
Given any $k\in[\infty]$,  for a function $\psi:[\infty]_0^k\rightarrow\mathbb{R}$, the following conditions are equivalent:
\begin{enumerate}[(i)]
\item $\psi$ is completely monotone;
\item
$$\sum_{\mathbf{0}\leq \mathbf{p}\leq \mathbf{n}}\(-1\)^{\parallel\mathbf{p}\parallel}{\mathbf{n}\choose\mathbf{p}}
\psi\(\mathbf{m}+\mathbf{p}\)\geq0\text{~~~~~~for all~~}\mathbf{n},\mathbf{m}\in[\infty]_0^k,$$
where {$\bf{\leq}$} denotes the usual coordinatewise partial order on $[\infty]_0^{k}$,
$\mathbf{n}=\(n_1,\ldots,n_k\)$, $\mathbf{m}=\(m_1,\ldots,m_{k}\)$, $\mathbf{p}=\(p_1,\ldots,p_k\)$,
$\parallel\mathbf{p}\parallel=p_1+\cdots+p_k$  and
\[{\mathbf{n}\choose\mathbf{p}}=\prod_{i=1}^{k}{{n_i\choose p_i}};\]
\item There exists $\mu\in{M}_{+}\([0,1]^k\)$ such that
$$\psi(\mathbf{n})=\int_{[0,1]^k}\mathbf{x}^{\mathbf{n}}d\mu\({\mathbf{x}}\),~~~~{\mathbf{n}\in[\infty]_0^{k}},$$
where ${M}_{+}\(X\)$ is the collection of Radon measures on $X$.
\end{enumerate}
\end{lemma}

\subsection{Generalized stepping stone model with $\Xi$-resampling mechanism}
Denote by
$\(C\(M_1\(E^2\)\),\|\cdot\|\)$ the Banach space of continuous functions on $M_1\(E^2\)$
with norm $\|\mathbbm{F}\|=\sup_{\mu\in M_1\(E^2\)}|\mathbbm{F}\(\mu\)|$.
In order to formulate a measure-valued stochastic process, we need to introduce some subspace of $C\(M_1\(E^2\)\)$ serving as the core for the generator.
Denote by $C_p\(M_1\(E^2\)\)$  the linear span of monomials of the form
\begin{equation*}
\mathbb{F}_{f,\,\eta,\,n}\(\mu\)=\int_{E}\cdots\int_{E}f\(x_1,\ldots,x_n\)
\prod_{i=1}^n\mu_{\eta_i}\(dx_i\)
\end{equation*}
with $n\in[\infty]$, $f\(x_1,\ldots,x_n\)=f_1\(x_1\)\cdots f_n\(x_n\)\in C\(E\)^{n}$
 and $\eta=\(\eta_1,\ldots,\eta_n\)\in\mathbbm{S}^n$. It follows from the Stone-Weierstrass Theorem
that ${C}_p({M}_1\(E^2\))$ is dense in ${C}({M}_1\(E^2\))$.
Let $B_{b,\,p}({M}_1\(E^2\))$ be the collection of those $\mathbb{F}_{f,\,\eta,\,n}\(\mu\)$ with $n\in[\infty]$, $f\in{B}\(E^n\)$
and $\eta\in\mathbbm{S}^n$. Denote by $\bar{B}_{b,\,p}\({M}_1\(E^2\)\)$ the closure of ${B}_{b,\,p}\({M}_1\(E^2\)\)$. $B\(M_1\(E^2\)\)$ is the collection of bounded functions on $M_1\(E^2\)$. Clearly, we have ${C}_p({M}_1\(E^2\))\subseteq B_{b,\,p}({M}_1\(E^2\))\subseteq\bar{B}_{b,\,p}\({M}_1\(E^2\)\)\subseteq B\(M_1\(E^2\)\)$.
Throughtout this paper, we make the following assumptions:
\begin{enumerate}[(I)]
\item A finite measure $\Xi$ on
$\Delta$ characterizes the simultaneous multiple coalescent;
\item The mutation operator $A$ is a jump-type Feller generator given by (\ref{eq:mu});
\item The migration rates $u_{\mathbbm 1}$ and $u_{\mathbbm 2}$ are positive.
\end{enumerate}

\begin{proposition}\label{cor1}
For each $\mu=\(\mu_{\mathbbm{1}},\mu_{\mathbbm{2}}\)\in M_1\(E^2\)$ and $t\geq0$,
there exists a $M_1\(E^2\)$-valued random variable $V_{\mu,t}$ on some complete probability space $\(\Omega^{*},\mathcal{F}^{*},\mathbbm{Q}\)$
such that for all $k$, $\ell\in[\infty]$ with $k+\ell>0$,
$\mathbf{n_1}=(n_{11}, \ldots, n_{1k})\in\left[\infty\right]_0^k$, $\mathbf{n_2}=(n_{21}, \ldots, n_{2\ell})\in\left[\infty\right]_0^{\ell}$,
$\{C_i\times D_j\}_{i\in [k],\,j\in[\ell]}\subseteq\mathcal{D}\times\mathcal{D}$, we have
{\rm
\begin{equation*}
\begin{split}
&\mathbbm{Q}\left[V_{\mu,\,t}\(\otimes_{i=1}^k\ind_{C_i}^{\otimes n_{1i}}\otimes_{i=1}^{\ell}\ind_{D_i}^{\otimes n_{2i}}\)\right]
=\mathbbm{P}\left[\mathcal{G}_{\mu}\(Y_{\mathbf{n_1,\,n_2}}\(t\),\eta_{\mathbf{n_1,\,n_2}}\(t\)\)\right],
\end{split}
\end{equation*}
}
where $\(Y_{\mathbf{n_1,\,n_2}}\(t\),\eta_{\mathbf{n_1,\,n_2}}\(t\)\)_{t\geq0}$ is a $\(\Xi,\text{A},u_{\mathbbm{1}},u_{\mathbbm{2}}\)$-CPGM with initial value
{\rm$$ Y_{\mathbf{n_1,\,n_2}}\(0\)=\otimes_{i=1}^k\ind_{C_i}^{\otimes n_{1i}}\otimes_{i=1}^{\ell}\ind_{D_i}^{\otimes n_{2i}}\,\,\,\,\text{and}\,\,\,\,\eta_{\mathbf{n_1,\,n_2}}\(0\)=\{\underbrace{\mathbbm{1},\ldots,\mathbbm{1}}_{n_1},\underbrace{\mathbbm{2},\ldots,\mathbbm{2}}_{n_2}\}.$$
}

\end{proposition}

\begin{proof}
Given that $Y_{\mathbf{n_1,\,n_2}}\(0\)$ is in the form of tensor product of one variable functions, one can see that $Y_{\mathbf{n_1,\,n_2}}\(t\)$ is always in a similar form of one variable functions from the construction of the $\(\Xi,\text{A},u_{\mathbbm{1}},u_{\mathbbm{2}}\)$-CPGM.  Subsequently, given any $t\geq 0$, we may  assume that
\begin{equation*}
\begin{split}
Y_{\mathbf{n_1,\,n_2}}\(t\)&=g_1\(x_1\)g_2\(x_2\)\ldots g_{|\eta_{\mathbf{n_1,\,n_2}}\(t\)|}\(x_{|\eta_{\mathbf{n_1,\,n_2}}\(t\)|}\)\\
:&= Y_{\mathbf{n_1}}^{\mathbbm{1}}\(t\)\otimes Y_{\mathbf{n_2}}^{\mathbbm{2}}\(t\)
\end{split}
\end{equation*}
with
$$Y_{\mathbf{n_1}}^{\mathbbm{1}}\(t\)=\otimes_{\{i|\eta_{\mathbf{n_1,\,n_2}}\(t\)_i=\mathbbm{1},\,i=1,\ldots,|\eta_{\mathbf{n_1,\,n_2}}\(t\)|\}}g_i\(x_i\);$$
$$Y_{\mathbf{n_2}}^{\mathbbm{2}}\(t\)=\otimes_{\{i|\eta_{\mathbf{n_1,\,n_2}}\(t\)_i=\mathbbm{2},\,i=1,\ldots,|\eta_{\mathbf{n_1,\,n_2}}\(t\)|\}}g_i\(x_i\).$$
For simplicity of notation, let's write
$$\al_{\mathbf{n_1}}^{\mathbbm{1}}(t)=|\eta_{\mathbf{n_1,\,n_2}}(t)|_{\mathbbm{1}};\,\, \al_{\mathbf{n_2}}^{\mathbbm{2}}(t)=|\eta_{\mathbf{n_1,\,n_2}}(t)|_{\mathbbm{2}}$$
and set $\|\mathbf{v}\|=v_{1}+\cdots+v_{k}$ for any $\mathbf{v}=\(v_{1},\ldots,v_{k}\)\in[\infty]_0^k$.
Define
\begin{eqnarray}\label{F_{n,m}}
\begin{split}
F(\mathbf{n_1},\,\mathbf{n_2}):&=\mathbbm{P}\left[\mathcal{G}_{\mu}\(Y_{\mathbf{n_1,\,n_2}}(t),\,\eta_{\mathbf{n_1,\,n_2}}(t)\)\right]\\
&=\mathbbm{P}\left[\<\mu_{\mathbbm{1}}^{\otimes\al_{\mathbf{n_1}}^{\mathbbm{1}}(t)},
Y_{\mathbf{n_1}}^{{\mathbbm{1}}}(t)\>\<{\mu_{\mathbbm{2}}}^{\otimes\al_{\mathbf{n_2}}^{\mathbbm{2}}(t)},
Y_{\mathbf{n_2}}^{\mathbbm{2}}(t)\>\right].
\end{split}
\end{eqnarray}

By Lemma \ref{Haus}, in order to show that (\ref{F_{n,m}}) is the joint moment of some measure, it suffices to verify that
\begin{eqnarray*}
\begin{split}
&I:\,=\sum_{\mathbf{0}\leq\mathbf{p}\leq\mathbf{r}}\sum_{\mathbf{0}\leq\mathbf{q}\leq
\mathbf{h}}(-1)^{\|\mathbf{p}\|+\|\mathbf{q}\|}  {\mathbf{r}\choose\mathbf{p}}{\mathbf{h}\choose\mathbf{q}}
F(\mathbf{n_1}+\mathbf{p},\,
\mathbf{n_2}+\mathbf{q})\geq 0
\end{split}
\end{eqnarray*}
for any $\mathbf{r}=(r_1, \ldots, r_k)\in[\infty]_0^k$ and $\mathbf{h}=\(h_1,\ldots,h_{\ell}\)\in[\infty]_0^{\ell}$,
$\mathbf{p}=(p_1, \ldots, p_k)$ with $p_i\leq r_i$ and
$\mathbf{q}=\(q_1,\ldots,q_{\ell}\)$ with $q_i\leq h_i$.
It follows from (\ref{F_{n,m}}) that
\begin{eqnarray*}
\begin{split}
I=\sum_{\mathbf{0}\leq\mathbf{p}\leq\mathbf{r}}\sum_{\mathbf{0}\leq\mathbf{q}\leq\mathbf{h}}{\mathbf{r}\choose\mathbf{p}}{\mathbf{h}\choose\mathbf{q}}
(-1)^{\|\mathbf{p}\|+\|\mathbf{q}\|}\mathbbm{P}\left[\<\mu_{\mathbbm{1}}^{\otimes\al^{\mathbbm{1}}_{\mathbf{n_1}+\mathbf{p}}(t)},
Y_{\mathbf{n_1}+\mathbf{p}}^{\mathbbm{1}}(t)\>\<\mu_{\mathbbm{2}}^{\otimes\al^{\mathbbm{2}}_{\mathbf{n_2}+\mathbf{q}}(t)},
Y_{\mathbf{n_2}+\mathbf{q}}^{\mathbbm{2}}(t)\>\right]
\end{split}
\end{eqnarray*}
with initial values
$$Y_{\mathbf{n_1}+\mathbf{p}}^{\mathbbm{1}}(0)=\otimes_{i=1}^k\ind_{C_i}^{\otimes
n_{1i}+p_i}; \, Y_{\mathbf{n_2}+\mathbf{q}}^{\mathbbm{2}}(0)=\otimes_{i=1}^{\ell}\ind_{D_i}^{\otimes
n_{2i}+q_i};$$
$$\al^{\mathbbm{1}}_{\mathbf{n_1}+\mathbf{p}}(0)=\|\mathbf{n_1}\|+\|\mathbf{p}\|; \,
\al^{\mathbbm{2}}_{\mathbf{n_2}+\mathbf{q}}(0)=\|\mathbf{n_2}\|+\|\mathbf{q}\|.$$
For $\mathbf{r}$ and $\mathbf{h}$, let's write
\[\mathscr{C}_\mathbf{r}:=\{c=c_1\times\cdots\times c_k: c_i\subset [r_i]_0,\,
1\leq i\leq k\};\]
\[{\mathscr{C}}_\mathbf{h}:=\{\tilde{c}=\tilde{c}_1\times\cdots\times \tilde{c}_{\ell}: \tilde{c}_i\subset [h_i]_0,\,
1\leq i\leq {\ell}\}.\]
Put $\mathbf{c}=(|c_1|,\ldots,|c_k|)\in[\infty]_0^k$
and $\tilde{\mathbf{c}}=(|\tilde c_1|,\ldots,|\tilde
c_{\ell}|)\in[\infty]_0^{\ell}$.
Then
\begin{eqnarray*}
\begin{split}
I&\,=\sum_{\scriptstyle\mathbf{0}\leq\mathbf{p}\leq\mathbf{r}}\sum_{\scriptstyle\mathbf{0}\leq\mathbf{q}\leq\mathbf{h}}\sum_{\scriptstyle c\in
{\mathscr{C}}_\mathbf{r},\atop \scriptstyle |c_i|=p_i,\,i\in[k]} \sum_{\scriptstyle \tilde c\in
{\mathscr{C}}_\mathbf{h},\atop \scriptstyle |\tilde c_i|=q_i,\,
i\in[\ell]}(-1)^{\|\mathbf{p}\|+\|\mathbf{q}\|}\\
&\qquad\times\mathbbm{P}\left[\<\mu_{\mathbbm{1}}^{\otimes\al^{\mathbbm{1}}_{\mathbf{n_1+c}}(t)},
Y_{\mathbf{n_1+c}}^{\mathbbm{1}}(t)\>\<\mu_{\mathbbm{2}}^{\otimes\al^{\mathbbm{2}}_{\mathbf{n_2}+\tilde{\mathbf{c}}}(t)},
Y_{\mathbf{n_2}+\tilde{\mathbf{c}}}^{\mathbbm{2}}(t)\>\right]\\
&\,=\sum_{c\in {\mathscr{C}}_{\mathbf{r}}}\sum_{\tilde c\in {\mathscr{C}}_{\mathbf{h}}}(-1)^{\|\mathbf
c\|+\|\tilde{\mathbf c}\|}\mathbbm{P}\left[\<\mu_{\mathbbm{1}}^{\otimes\al^{\mathbbm{1}}_{\mathbf{n_1+c}}(t)},
Y_{\mathbf{n_1+c}}^{\mathbbm{1}}(t)\>\<\mu_{\mathbbm{2}}^{\otimes\al^{\mathbbm{2}}_{\mathbf{n_2}+\tilde{\mathbf{c}}}(t)},
Y_{\mathbf{n_2}+\tilde{\mathbf{c}}}^{\mathbbm{2}}(t)\>\right]\\
\end{split}
\end{eqnarray*}
with initial values
$$Y_{\mathbf{n_1+c}}^{\mathbbm{1}}(0)=\otimes_{i=1}^k\(\otimes_{j=1}^{n_{1i}}\ind_{C_i}\otimes_{j=n_{1i}+1}^{n_{1i}+r_i}h_{ij}\);\,\,
Y_{\mathbf{n_2}+\tilde{\mathbf
c}}^{\mathbbm{2}}(0)=\otimes_{i=1}^{\ell}\(\otimes_{j=1}^{n_{2i}}\ind_{D_i}\otimes_{j=n_{2i}+1}^{n_{2i}+h_i}g_{ij}\);$$
$$\al^{\mathbbm{1}}_{\mathbf{n_1+c}}(0)=\|\mathbf{n_1}\|+\|\mathbf c\|;\,\,
\al^{\mathbbm{2}}_{\mathbf{n_2}+\tilde{\mathbf{c}}}(0)=\|\mathbf{n_2}\|+\|\tilde{\mathbf c}\|;$$
where
\begin{equation*}
\begin{aligned}
h_{ij}=\left\{\begin{array}{lll}
\ind_{C_i}\,\,&\mbox{if\,\,} j-n_{1i}\in c_i;\\
\ind      &\text{if\,\,} j-n_{1i}\not\in c_i;
\end{array}
\right.\\
\end{aligned}
\begin{aligned}
\quad\quad
g_{ij}=\left\{\begin{array}{lll}
\ind_{D_i} &\mbox{if\,\,} j-n_{2i}\in \tilde c_i;\\
       \ind       &\text{if\,\,} j-n_{2i}\not\in \tilde c_i.\\
\end{array}
\right.
\end{aligned}
\end{equation*}
Therefore,
\begin{eqnarray*}
\begin{split}
I&\,=\mathbbm{P}\left[\<\mu_{\mathbbm{1}}^{\otimes\al^{\mathbbm{1}}_{\mathbf {n_1}+\mathbf r}(t)}, Y_{\mathbf
{n_1}+\mathbf r}^{\mathbbm{1}}(t)\>\<\mu_{\mathbbm{2}}^{\otimes\al^{\mathbbm{2}}_{\mathbf{n_2}+\mathbf h}(t)},
Y_{\mathbf{n_2}+\mathbf h}^{\mathbbm{2}}(t)\>\right]\\
&\,=\mathbbm{P}\left[\mathcal{G}_{\mu}\(Y_{\mathbf{n_1}+\mathbf r,\,\mathbf{n_2}+\mathbf h}(t), \eta_{\mathbf{n_1}+\mathbf r,\,\mathbf{n_2}+\mathbf
h}(t)\)\right]
\,\geq 0
\end{split}
\end{eqnarray*}
with initial values
$$Y_{\mathbf{n_1}+\mathbf
r}^{\mathbbm{1}}(0)=\otimes_{i=1}^k\left[\otimes_{j=1}^{n_{1i}}\ind_{C_i}\otimes_{j=n_{1i}+1}^{n_{1i}+r_i}(\ind-\ind_{C_i})\right];$$
$$Y_{\mathbf{n_2}+\mathbf
h}^{\mathbbm{2}}(0)=\otimes_{i=1}^{\ell}\left[\otimes_{j=1}^{n_{2i}}\ind_{D_i}\otimes_{j=n_{2i}+1}^{n_{2i}+h_i}(\ind-\ind_{D_i})\right];$$
$$\al^{\mathbbm{1}}_{\mathbf{n_1}+\mathbf r}(0)= \|\mathbf{n_1}\|+\|\mathbf r\|;\,\,\al^{\mathbbm{2}}_{\mathbf{n_2}+\mathbf h}(0)= \|\mathbf{n_2}\|+\|\mathbf h\|.$$

Consequently, for any $\mathbf{n_1}\in[\infty]_0^k$ and
$\mathbf{n_2}\in[\infty]_0^{\ell}$ with $k+\ell>0$, there exist $[0, 1]$-valued random
variables $W_1, \ldots, W_k$ and $Z_1, \ldots, Z_{\ell}$ on some complete probability space $(\Omega^*, \mathcal{F}^*, \mathbbm{Q})$ such that
\begin{eqnarray}\label{double sequence}
\mathbbm{Q}\left[\prod_{i=1}^k W_i^{n_{1i}}\prod_{j=1}^{\ell} Z_j^{n_{2j}}\right]=F(\mathbf{n_1},\,
\mathbf{n_2}).
\end{eqnarray}
Define $V_{\mu,\,t}$ as a $M_1\(E^2\)$-valued random variable satisfying
{\rm
\begin{equation}\label{eq:V}
\begin{split}
&V_{\mu,\,t}\(\otimes_{i=1}^k\ind_{C_i}^{\otimes n_{1i}}\otimes_{i=1}^{\ell}\ind_{D_i}^{\otimes n_{2i}}\)
=\prod_{i=1}^k W_i^{n_{1i}}\prod_{j=1}^{\ell} Z_j^{n_{2j}}.
\end{split}
\end{equation}
}
Combining with (\ref{F_{n,m}}), (\ref{double sequence}) and (\ref{eq:V}), we obtain the result.

\end{proof}

\begin{proposition}\label{cor0}
For each $\mu\in M_1\(E^2\)$ and $t\geq0$, the $M_1\(E^2\)$-valued random variable $V_{\mu,\,t}$
has the following properties:
$$V_{\mu,\,t}\(\(C_1\cup C_2\)\times D\)=V_{\mu,\,t}\({C_1\times D}\)+V_{\mu,\,t}\({C_2\times D}\);$$
$$V_{\mu,\,t}\(C\times \(D_1\cup D_2\)\)=V_{\mu,\,t}\(C\times D_1\)+V_{\mu,\,t}\(C\times D_2\),$$
where $C_1,\,C_2,\,D_1,\,D_2,\,C,\,D\in\mathcal{D}$ with $C_1\cap C_2=D_1\cap D_2=\emptyset$. Moreover, we have $V_{\mu,\,t}\(\emptyset\times\emptyset\)=0$ and $V_{\mu,\,t}\(E\times E\)=1.$
\end{proposition}

\begin{proof}
Note that
\begin{equation*}
\begin{split}
&\left[V_{\mu,\,t}\(\(C_1\cup C_2\)\times D\)-V_{\mu,\,t}\({C_1\times D}\)-V_{\mu,\,t}\({C_2\times D}\)\right]^2\\
&\,=V_{\mu,\,t}^2\(\(C_1\cup C_2\)\times D\)+V_{\mu,\,t}^2\({C_1\times D}\)+V_{\mu,\,t}^2\({C_2\times D}\)\\
&\qquad-2V_{\mu,\,t}\(\(C_1\cup C_2\)\times D\)\times V_{\mu,\,t}\({C_1\times D}\)\\
&\qquad-2V_{\mu,\,t}\(\(C_1\cup C_2\)\times D\)\times V_{\mu,\,t}\({C_2\times D}\)\\
&\qquad+V_{\mu,\,t}\({C_1\times D}\)\times V_{\mu,\,t}\({C_2\times D}\)\\
&\qquad+V_{\mu,\,t}\({C_2\times D}\)\times V_{\mu,\,t}\({C_1\times D}\).
\end{split}
\end{equation*}
By the coupling property, we can associate these terms above with seven $\(\Xi,\text{A},u_{\mathbbm{1}},u_{\mathbbm{2}}\)$-CPGMs
$\(Y_i\(t\),\eta\(t\)\)_{t\geq 0}$, $i=1,\ldots,7$ driven by a $\(\Xi,u_{\mathbbm{1}},u_{\mathbbm{2}}\)$-CPGM.
Put $\eta\(0\)=\{\mathbbm{1,1,2,2}\}$ and
\begin{equation*}
\begin{split}
&Y_1\(0\)=\ind_{C_1\cup C_2}\(x_1\)\ind_{C_1\cup C_2}\(x_2\)\ind_{D}\(y_1\)\ind_{D}\(y_2\);\\
&Y_2\(0\)=\ind_{C_1}\(x_1\)\ind_{C_1}\(x_2\)\ind_{D}\(y_1\)\ind_{D}\(y_2\);\\
&Y_3\(0\)=\ind_{C_2}\(x_1\)\ind_{C_2}\(x_2\)\ind_{D}\(y_1\)\ind_{D}\(y_2\);\\
&Y_4\(0\)=-2\ind_{C_1\cup C_2}\(x_1\)\ind_{C_1}\(x_2\)\ind_{D}\(y_1\)\ind_{D}\(y_2\);\\
&Y_5\(0\)=-2\ind_{C_1\cup C_2}\(x_1\)\ind_{C_2}\(x_2\)\ind_{D}\(y_1\)\ind_{D}\(y_2\);\\
&Y_6\(0\)=\ind_{C_1}\(x_1\)\ind_{C_2}\(x_2\)\ind_{D}\(y_1\)\ind_{D}\(y_2\);\\
&Y_7\(0\)=\ind_{C_2}\(x_1\)\ind_{C_1}\(x_2\)\ind_{D}\(y_1\)\ind_{D}\(y_2\).
\end{split}
\end{equation*}
It follows from
Proposition \ref{cor1} that
\begin{equation*}
\begin{split}
&\mathbbm{Q}\left[V_{\mu,\,t}\(\(C_1\cup C_2\)\times D\)-V_{\mu,\,t}\({C_1\times D}\)-V_{\mu,\,t}\({C_2\times D}\)\right]^2\\
&\,=\sum_{i=1}^7\mathbbm{P}\left[\mathcal{G}_{\mu}\(Y_i\(t\),\eta\(t\)\)\right]\\
&\,=\mathbbm{P}\left[\mathcal{G}_{\mu}\({Y}\(t\),\eta\(t\)\)\right]
\end{split}
\end{equation*}
where the last equality follows from  Proposition \ref{coupling property}, and $\({Y}\(t\),\eta\(t\)\)_{t\geq 0}$ is  also a $\(\Xi,\text{A},u_{\mathbbm{1}},u_{\mathbbm{2}}\)$-CPGMs driven by the same $(\Xi,u_{\mathbbm{1}},u_{\mathbbm{2}})$-CPGM  with
$$Y\(0\)=\sum_{i=1}^7Y_i\(0\)=0.$$ As a consequence, $\mathbbm{P}\left[\mathcal{G}_{\mu}\({Y}\(t\),\eta\(t\)\)\right]=0$ and
 we have
\begin{equation*}\label{1}
V_{\mu,\,t}\(\(C_1\cup C_2\)\times D\)=V_{\mu,\,t}\({C_1\times D}\)+V_{\mu,\,t}\({C_2\times D}\).
\end{equation*}
Similarly, one can show that
\begin{equation*}\label{2}
V_{\mu,\,t}\(C\times \(D_1\cup D_2\)\)=V_{\mu,\,t}\(C\times D_1\)+V_{\mu,\,t}\(C\times D_2\).
\end{equation*}
Choosing $Y_{\(1\),\,\(1\)}\(0\)=\ind_{\emptyset}\otimes\ind_{\emptyset}$ and $Y_{\(1\),\,\(1\)}\(0\)=\ind_{E}\otimes\ind_{E}$ in  (\ref{F_{n,m}}),
we obtain
\begin{equation}\label{eq:normal}
\mathbbm{Q}\left[V_{\mu,\,t}\(\emptyset\times\emptyset\)\right]=0\,\,\text{and}\,\,\mathbbm{Q}\left[V_{\mu,\,t}\(E\times E\)\right]=1.
\end{equation}
It follows from (\ref{eq:V}) that $0\leq V_{\mu, t}(C\times D)\leq1$  holds for any $C\times D\in\mathcal{D}\times\mathcal{D}$. Therefore,
\begin{equation*}
V_{\mu,\,t}\(\emptyset\times\emptyset\)=0\,\text{and}\,V_{\mu,\,t}\(E\times E\)=1.
\end{equation*}

\end{proof}

\begin{theorem}
Given any $\mu\in M_1\(E^2\)$, there exists a $M_1\(E^2\)$-valued Markov process $\(\mu\(t\)\)_{t\geq0}$ on some complete probability space $\(\Omega^*,\mathcal{F}^{*},\mathbbm{Q}\)$ with transition semigroup $\left(Q_t\right)_{t\geq0}$ on $\bar{B}_{b,\,p}\(M_1\(E^2\)\)\subseteq B\(M_1\(E^2\)\)$.
In particular, for any $\mathbb{F}_{f,\,\eta\,,n}\in B_{b,\,p}\(M_1\(E^2\)\)\subseteq B\(M_1\(E^2\)\)$ with $f\in B\(E^n\)$
and $\eta\in\mathbbm{S}^n$,
\begin{equation}\label{eq-dual-2}
\begin{split}
Q_t\mathbb{F}_{f,\,\eta,\,n}\(\mu\)&=\int_{M_1\(E^2\)}\mathbb{F}_{f,\,\eta,\,n}\(v\)Q_t\(\mu,dv\)\\
&=\mathbbm{P}_{(f,\,\eta)}\left[\<{\mu}_{\eta(t)}, Y(t)\>\right]\\
&=\mathbbm{P}_{(f,\,\eta)}\mathbb{F}_{Y(t),\,\eta(t),\,|\eta(t)|}(\mu),
\end{split}
\end{equation}
where $\(Y\(t\),\eta\(t\)\)_{t\geq 0}$ is a $\(\Xi,\text{A},u_{\mathbbm{1}},u_{\mathbbm{2}}\)$-CPGM with initial value $\(Y\(0\),\eta\(0\)\)=\(f,\eta\)$.
The $M_1\(E^2\)$-valued Markov process $\(\mu\(t\)\)_{t\geq 0}$ is  {\it the generalized stepping stone model with $\Xi$-resampling mechanism.}

\end{theorem}

\begin{proof}
Initial value and transition semigroup are sufficient to determine the finite dimensional distributions of a $M_1\(E^2\)$-valued Markov process. We carry out the proof in two steps.

{\bf (i) Existence of the measure.} We show that for each $\mu\in M_1\(E^2\)$ and $t\geq0$, there
exists a random measure $\bar{V}_{\mu,\,t}\in M_1\(E^2\)$ with distribution function  denoted by $p_t\(\mu,\cdot\)$.

For any $f\in C\(E^2\)$, let $\(f_n\)_{n\geq 1}$ be a sequence of step functions with
\begin{equation*}
\begin{split}
f_n\(x,y\)=&\sum_{i=0}^{2^n-2}\sum_{j=0}^{2^n-2}f\(\frac{i}{2^n},\frac{j}{2^n}\)
\ind_{\left[\frac{i}{2^n},\,\frac{i+1}{2^n}\right)\times \left[\frac{j}{2^n},\,\frac{j+1}{2^n}\right)}\(x,y\)\\
&\quad+\sum_{j=0}^{2^n-2}f\(\frac{2^n-1}{2^n},\frac{j}{2^n}\)
\ind_{\left[\frac{2^n-1}{2^n},\,1\right]\times \left[\frac{j}{2^n},\,\frac{j+1}{2^n}\right)}\(x,y\)\\
&\quad+\sum_{i=0}^{2^n-2}f\(\frac{i}{2^n},\frac{2^n-1}{2^n}\)
\ind_{\left[\frac{i}{2^n},\,\frac{i+1}{2^n}\right)\times \left[\frac{2^n-1}{2^n},\,1\right]}\(x,y\)\\
&\quad+f\(\frac{2^n-1}{2^n},\frac{2^n-1}{2^n}\)\ind_{\left[\frac{2^n-1}{2^n},\,1\right]\times \left[\frac{2^n-1}{2^n},\,1\right]}\(x,y\).
\end{split}
\end{equation*}
Clearly, $\(f_n\)_{n\geq 1}$ uniformly converges to $f$. Define $\Lambda\,:\,C\(E^2\)\rightarrow\mathbbm{R}$
such that
\begin{equation*}
\begin{split}
\Lambda\(f\):=\lim_{n\rightarrow\infty}\Lambda\(f_n\):=\lim_{n\rightarrow\infty}\left[I_1\(n\)+I_2\(n\)+I_3\(n\)+I_4\(n\)\right]
\end{split}
\end{equation*}
with
\begin{equation*}
\begin{split}
I_1\(n\)=&\sum_{i=0}^{2^n-2}\sum_{j=0}^{2^n-2}f\(\frac{i}{2^n},\frac{j}{2^n}\)
V_{\mu,\,t}\(\left[\frac{i}{2^n},\,\frac{i+1}{2^n}\right)\times \left[\frac{j}{2^n},\,\frac{j+1}{2^n}\right)\);\\
I_2\(n\)=&\sum_{j=0}^{2^n-2}f\(\frac{2^n-1}{2^n},\frac{j}{2^n}\)
V_{\mu,\,t}\(\left[\frac{2^n-1}{2^n},\,1\right]\times \left[\frac{j}{2^n},\,\frac{j+1}{2^n}\right)\);\\
I_3\(n\)=&\sum_{i=0}^{2^n-2}f\(\frac{i}{2^n},\frac{2^n-1}{2^n}\)
V_{\mu,\,t}\(\left[\frac{i}{2^n},\,\frac{i+1}{2^n}\right)\times \left[\frac{2^n-1}{2^n},\,1\right]\);\\
I_4\(n\)=&f\(\frac{2^n-1}{2^n},\frac{2^n-1}{2^n}\)V_{\mu,\,t}\(\left[\frac{2^n-1}{2^n},\,1\right]\times \left[\frac{2^n-1}{2^n},\,1\right]\).
\end{split}
\end{equation*}

Combining the uniform continuity of function $f$ with Proposition \ref{cor0}, one can show that for any positive integer $m$,
$$|\Lambda\(f_{n+m}\)-\Lambda\(f_{n}\)|\leq \epsilon$$
holds for all $n$ large enough, namely, $\(\Lambda\(f_{n}\)\)_{n\geq 1}$ is a Cauchy sequence. The limit of the sequence exists and  $\Lambda$ is well-defined on $C\(E^2\)$.
The linear property of $\Lambda$ can be easily obtained from its definition. For any $f,\,g\in C\(E^2\)$ with
$$\| f-g\|=\sup_{\(x,\,y\)\in E^2}\mid f\(x,\,y\)-g\(x,\,y\)\mid\leq \epsilon,$$
we can choose the value of $n$ large enough and obtain
\begin{equation*}
\begin{split}
&\mid\Lambda\(f\)-\Lambda\(g\)\mid\\
&\leq\mid\Lambda\(f\)-\Lambda\(f_n\)\mid+\mid\Lambda\(f_n\)-\Lambda\(g_n\)\mid+\mid\Lambda\(g_n\)-\Lambda\(g\)\mid\\
&\leq3\epsilon.
\end{split}
\end{equation*}
In conclusion, $\Lambda:\,C\(E^2\)\rightarrow\mathbbm{R}$ is a continuous linear functional satisfying $\Lambda\(f\)\geq 0$ whenever $f\geq 0$.
By Riesz representation theorem, there exits a unique random probability measure $\bar{V}_{\mu,\,t}$
on $\sigma\(\mathcal{E}\times\mathcal{E}\)$
such that
$$\Lambda\(f\)=\int_{E^2}f d\bar{V}_{\mu,\,t}.$$
Define $p_t\(\mu,\cdot\)$ as the distribution of $\bar{V}_{\mu,\,t}$
for any $t\geq 0$.

{\bf (ii) Existence of semigroup $\(Q_t\)_{t\geq 0}$ on $\bar{B}_{b,\,p}\(M_1\(E^2\)\)$.} Denote by
\begin{equation}\label{eq:Q_t}
Q_t\mathbbm{F}\(\mu\)=\int_{M_1\(E^2\)}\mathbbm{F}\(\nu\)p_t\(\mu,d\,\nu\)
\end{equation}
for any $\mathbbm{F}\in B\(M_1\(E^2\)\)$. We first verify that there exists enough $\mathbbm{F}$ such that the value of (\ref{eq:Q_t}) can be
represented from the moments of the dual  $\(\Xi,\text{A},u_{\mathbbm{1}},u_{\mathbbm{2}}\)$-CPGMs. Then we discuss the semigroup property.

(a) {\it Dual representation.}
For any $n\in[\infty]$,
$\mathcal{D}^n$ is a $\pi$-system on $E^{n}$.
Let $\mathcal{H}$ be the collection of functions  $f\in {B}\(E^{n}\)$ with $n\in[\infty]$ satisfyin
\begin{equation*}\label{eq-dual-4}
\int_{M_1\(E^2\)}\mathbb{F}_{f,\,\eta,\,n}\(v\)p_t\(\mu,dv\)=\mathbbm{P}_{(f,\,\eta)}\left[\<{\mu}_{\eta(t)}, Y(t)\>\right],
\end{equation*}
where $({Y}(t), \eta(t))_{t\geq0}$ is the dual
$\(\Xi,\text{A},u_{\mathbbm{1}},u_{\mathbbm{2}}\)$-CPGM
with initial value $\(f,\eta\)$ satisfying $\eta\in{\mathbbm{S}}^n$.
It follows from (\ref{double sequence}) that
$$\otimes_{i=1}^k\ind_{C_i}^{\otimes n_{1i}}\otimes_{j=1}^{\ell}\ind_{D_j}^{\otimes n_{2j}}\in\mathcal{H}.$$
By the linearity of expectation and integration we also have
$ah+bg\in\mathcal{H}$ for any $h,g\in\mathcal{H}$ and $a,b\in\mathbbm{R}$.
It is obvious that
$\ind_{E^n}\in\mathcal{H}$. If $h_m\in\mathcal{H},m\geq1,$ $0\leq h_m\uparrow h$ and $h$ is bounded,
let $(Y_m(s), \eta(s))_{s\leq t}$ and $(Y(s), \eta(s))_{s\leq t}$ be the associated dual processes driven by a $(\Xi,u_{\mathbbm{1}},u_{\mathbbm{2}})$-CPGM with initial values $(h_m, \eta)$ and $(h, \eta)$, respectively.  Observing that $Y_m(t)\uparrow Y(t)$ as $m\rightarrow\infty$, then by the dominated convergence theorem, we have
\begin{equation*}
\begin{split}
\int_{M_1\(E^2\)}\mathbb{F}_{h,\,\eta,\,n}\(v\)p_t\(\mu,dv\)
&=\lim_{m\rightarrow\infty} \int_{M_1\(E^2\)}\mathbb{F}_{h_m,\,\eta,\,n}\(v\)p_t\(\mu,dv\)  \\
&=\lim_{m\rightarrow\infty} \mathbbm{P}_{(h_m,\,\eta)}\left[\langle \mu_{\eta(t)}, Y_m(t)\rangle \right] \\
&=\mathbbm{P}_{(h,\,\eta)}\left[\<{\mu}_{\eta(t)}, Y(t)\>\right].
\end{split}
\end{equation*}
Subsequently,  $h\in\mathcal{H}$.
By (\ref{double sequence}) we know that
{\rm $\ind_{C}\in\mathcal{H}$} for any $ C\in\mathcal{D}^n$. Applying
the  monotone class theorem, $\mathcal{H}$ contains all of the $\sigma\(\mathcal{D}^n\)$-measurable real-valued functions and
therefore, (\ref{eq-dual-2})
holds for any $\mathbb{F}_{f,\,\eta,\,n}\in {B_{b,\,p}}\(M_1\(E^2\)\)$

(b) {\it Semigroup property.}
 By the Markov property of the dual $\(\Xi,\text{A},u_{\mathbbm{1}},u_{\mathbbm{2}}\)$-CPGM $(Y(s),\eta(s))_{s\geq 0}$, one can see that given any
$\mathbbm{F}_{f,\,\eta,\,n}\in B_{b,\,p}\(M_1\(E^2\)\)$ and $\mu\in M_1\(E^2\)$, $\forall\,s,\,t\geq 0$,
\begin{equation*}
\begin{split}
Q_sQ_t\mathbbm{F}_{f,\,\eta,\,n}(\mu)
&=\mathbbm{P}_{(f,\,\eta)}\left[Q_s\mathbbm{F}_{Y(t),\,\eta\(t\),\,|\eta\(t\)|}(\mu)\right]\\
&=\mathbbm{P}_{(f,\,\eta)}\left[\mathbbm{F}_{Y(s+t),\,\eta\(s+t\),\,|\eta\(s+t\)|}(\mu)\right]\\
&=Q_{s+t} \mathbbm{F}_{f,\,\eta,\,n}(\mu).
\end{split}
\end{equation*}
For general $\mathbbm{F}\in \bar{B}_{b,\,p}\(M_1\(E^2\)\)$, there exists an approximating sequence
$\{\mathbbm{F}_{k}, k\geq1\}\subseteq B_{b,\,p}\(M_1\(E^2\)\)$ such that
$$\lim_{k\rightarrow\infty}\|\mathbbm{F}_{k}-\mathbbm{F}\|=\lim_{k\rightarrow\infty}\sup_{\mu\in M_1\(E^2\)}\|\mathbbm{F}_{k}\(\mu\)-\mathbbm{F}\(\mu\)\|=0.$$
For any $t\geq 0$,
we have
\begin{equation*}
\begin{split}
|Q_t\mathbbm{F}\(\mu\)-Q_t\mathbbm{F}_{k}\(\mu\)|
&\,=\int_{M_1\(E^2\)}\(\mathbbm{F}\(\nu\)-\mathbbm{F}_{k}\(\nu\)\)p_t\(\mu,d\,\nu\)\\
&\,\leq\|\mathbbm{F}-\mathbbm{F}_{k}\|\rightarrow 0\text{~~as~~}k\rightarrow\infty.
\end{split}
\end{equation*}
That is to say
$Q_t\mathbbm{F}\(\mu\)=\lim_{k\rightarrow\infty}Q_t\mathbbm{F}_{k}\(\mu\).$
Therefore, for any $s,\,t\geq 0$ and $\mathbbm{F}\in \bar{B}_{b,\,p}\(M_1\(E^2\)\)$, we have
\begin{equation*}
\begin{split}
Q_sQ_t\mathbbm{F}(\mu)
&\,=\lim_{k\rightarrow\infty}Q_sQ_t\mathbbm{F}_{k}(\mu)=\lim_{k\rightarrow\infty}Q_{s+t}\mathbbm{F}_{k}(\mu)
=Q_{s+t} \mathbbm{F}(\mu).
\end{split}
\end{equation*}
The semigroup property then follows.

\end{proof}

\begin{proposition}
 $(Q_t)_{t\geq 0}$ is a Feller semigroup on $C(M_1(E^2))$.
\end{proposition}

\begin{proof}
We need to verify the required properties point by point.

(a) {\it Semigroup property.} Since $C\(M_1\(E^2\)\)\subseteq \bar{B}_{b,\,p}\(M_1\(E^2\)\)$, the restriction of $\(Q_t\)_{t\geq 0}$
on $C\(M_1\(E^2\)\)$ satisfies the semigroup property.

(b) {\it Strongly continuous property.}
Given $\mathbbm{F}_{f,\,\eta,\,n}\in C_{p}\(M_1\(E^2\)\)$, let's recall that $\tau_1$ is the first jumping time of the dual $\(\Xi,\text{A},u_{\mathbbm{1}},u_{\mathbbm{2}}\)$-CPGM with initial value $\(f,\eta\)$. Thus $\tau_1\sim \exp\(\lambda\)$ with
$\lambda=|\eta|_{\mathbbm{2}}u_{\mathbbm{1}}+|\eta|_{\mathbbm{1}}u_{\mathbbm{2}}+\lambda_{|\eta|_{\mathbbm{2}}}+\lambda_{|\eta|_{\mathbbm{1}}}$.
The probability  of $\{\tau_1<t\}$
(that is either coalescence or migration has occurred by time $t$) converges to $0$ as $t\rightarrow 0$. Then
\begin{equation*}\label{eq-expand}
\begin{split}
&\lim_{t\rightarrow 0}| Q_t\mathbbm{F}_{f,\,\eta,\,n}(\mu)-\mathbbm{F}_{f,\,\eta,\,n}(\mu)|=\lim_{t\rightarrow 0}|e^{-\lambda t}\mathcal{G}_{\mu}\(T_t^{\(n\)}f,\eta\)-\mathcal{G}_{\mu}\(f,\eta\)|=0.
\end{split}
\end{equation*}
For general $\mathbbm{F}\in C\(M_1\(E^2\)\)$, we reapply the approximating sequence
$\{\mathbbm{F}_{k}, k\geq1\}\subseteq C_{p}\(M_1\(E^2\)\)$ to get
\begin{equation*}
\begin{split}
&\lim_{t\rightarrow 0}| Q_t\mathbbm{F}(\mu)-\mathbbm{F}(\mu)|\\
&\leq\lim_{k\rightarrow\infty}\lim_{t\rightarrow 0}\(|Q_t\mathbbm{F}(\mu)-Q_t\mathbbm{F}_{k}(\mu)|+|Q_t\mathbbm{F}_{k}(\mu)-\mathbbm{F}_{k}(\mu)|
+|\mathbbm{F}_{k}(\mu)-\mathbbm{F}(\mu)|\)\\
&=0\end{split}
\end{equation*}
Therefore $Q_t$ is strongly continuous.

(c) {\it Contraction operator.}
\begin{equation*}
\begin{split}
\| Q_t\mathbbm{F}\|=&\sup_{\mu}|Q_t\mathbbm{F}\(\mu\)|
=\sup_{\mu}\left|\int_{M_1\(E^2\)}\mathbbm{F}\(\nu\)p_t\(\mu,d\nu\)\right|
\leq \| \mathbbm{F}\|.
\end{split}
\end{equation*}
Consequently, $Q_t$ is a contraction operator.

(d) {\it $Q_t\,:\,C\(M_1\(E^2\)\)\rightarrow C\(M_1\(E^2\)\)$.} For any  $\mathbbm{F}\in C\(M_1\(E^2\)\)$,
$$Q_t\mathbbm{F}\(\mu\)=\lim_{k\rightarrow\infty}Q_t\mathbbm{F}_{k}\(\mu\),$$
we know that $Q_t\mathbbm{F}\(\mu\)$ is a limit point of a sequence belonging to $C_p\(M_1\(E^2\)\)$.
Thus, $Q_t\mathbbm{F}\(\mu\)\in C\(M_1\(E^2\)\)=\overline{C_p\(M_1\(E^2\)\)}$.

Combining with (a)-(d), one can see that $\(Q_t\)_{t\geq 0}$  is a Feller semigroup.
\end{proof}

Let $\mathcal{L}^{*}$ be the generator of the generalized stepping stone model with $\Xi$-resampling mechanism $(\mu(t))_{t\geq 0}$. For any $\mathbb{F}_{f,\,\eta\,,n}\in B_{b,\,p}\(M_1\(E^2\)\)$ with $f\in B\(E^n\)$
and $\eta\in\mathbbm{S}^n$,  one can see that
$$\mathcal{L}^{*}\mathbb{F}_{f,\,\eta,\,n}\(\mu\)=\lim_{t\rightarrow 0}\frac{Q_t\mathbb{F}_{f,\,\eta,\,n}\(\mu\)-\mathbb{F}_{f,\,\eta,\,n}\(\mu\)}{t}, \quad \mu\in M_1\(E^2\)$$
whenever the limit exists. What follows is a proposition to represent  $\mathcal{L}^{*}$ from the generator of the dual $\(\Xi,\text{A},u_{\mathbbm{1}},u_{\mathbbm{2}}\)$-CPGM. Recall $\mathcal{G}_{\mu} $ and $\mathcal{L}$ defined in (\ref{G_{f,u}}) and (\ref{generator-dual}), respectively.

\begin{proposition}
The generator $\mathcal{L}^{*}$ of the generalized stepping stone model with $\Xi$-resampling mechanism $(\mu(t))_{t\geq 0}$ with initial value $\mu\(0\)=\mu$ is of the form below
\begin{equation}\label{generator}
\begin{split}
&\mathcal{L}^{*}\mathbb{F}_{f,\,\eta,\,n}\(\mu\)
=\mathcal{L}\mathcal{G}_{\mu}\(f,\eta\)
\end{split}
\end{equation}
where $\mathbb{F}_{f,\,\eta\,,n}\in B_{b,\,p}\(M_1\(E^2\)\)$ with $\(f,\,\eta\)\in B\(E^n\)\times \mathbbm{S}^n$.
\end{proposition}
\begin{proof}
Applying (\ref{eq-dual-2}), one can see that
\begin{equation*}
\begin{split}
\mathcal{L}^{*}\mathbb{F}_{f,\,\eta,\,n}\(\mu\)=&\lim_{t\rightarrow0}\frac{Q_t\mathbb{F}_{f,\,\eta,\,n}\(\mu\)-\mathbb{F}_{f,\,\eta,\,n}\(\mu\)}{t}\\
=&\lim_{t\rightarrow0}\frac{\mathbbm{P}\left[\mathcal{G}_{\mu}\(Y\(t\),\eta\(t\)\)\right]-\mathcal{G}_{\mu}\(f,\eta\)}{t}\\
=&\mathcal{L}\mathcal{G}_{\mu}\(f,\eta\).
\end{split}
\end{equation*}

\end{proof}

\section{Stationary distribution of the generalized stepping stone model}\label{sec5}
In this section we discuss the limit stationary distribution for the generalized stepping stone model with $\Xi$-resampling mechanism under the condition below.

\noindent{\bf Condition A:} The mutation operator
$A$ generates an irreducible semigroup $\(T_t\)_{t\geq 0}$ and $\tilde{\pi}$ is the unique invariant measure in ${M}_1(E)$ such that $T^{*}_t\nu\rightarrow\tilde{\pi}$ weakly for any $\nu\in{M}_1\(E\)$
as $t\rightarrow\infty$, where $T_t^{*}$ is the adjoint for $T_t$.

\begin{theorem}
Under Condition A, the generalized stepping stone model with $\Xi$-resampling mechanism $\({\mu}(t)\)_{t\geq 0}$ has a unique
stationary distribution $\Pi\in {M}_1({M}_1(E^2))$ such that
\begin{equation}\label{stationary distribution}
\int_{{M}_1(E^2)}\<\mu_{\xi}, f\>\Pi(d{\mu})=
\<\tilde{\pi}, f\>,
\end{equation} for any $\xi\in \mathbbm{S}$ and $f\in B\(E\)$.
\end{theorem}

\begin{proof}
The result of the current theorem for the $\Xi$-Fleming-Viot process and the classical stepping stone model was  proved
in  \cite{LLXZ} and \cite{Handa_2}, respectively.
Our proof is an adaption  of their work.

We first show the existence of a stationary distribution for the generalized stepping stone model.
As $\(T_t^*, T_t^*\)\mu\rightarrow\(\tilde{\pi},\tilde{\pi}\)$, the family $\{\(T_t^*,T_t^*\)\mu:\ t\ge
0\}$ is pre-compact, and hence, tight in ${M}_1(E^2)$.
Thus, for any
$\ep>0$, there exists a compact subset $K_\ep$ of $E^2$ such that
$\(T_t^*,T_t^*\)\mu\(K^c_\ep\)<\ep$ for all $t\ge 0$. Let
\[\KK_\ep=\left\{\mu\in {{M}}_1(E^2):\;\mu(K^c_{\ep k^{-1}2^{-k}})\le
k^{-1},\;\;\forall\ k\ge 1\right\}.\] For any $\de>0$, choose  $k\ge 1$
be such that $k^{-1}<\de$. Then for all $\mu\in\KK_\ep$, we have
\[\mu(K^c_{\ep k^{-1}2^{-k}})\le
k^{-1}<\de,\] and hence, $\KK_\ep$ is tight in ${M}_1(E^2)$. Then
$\KK_\ep$ is a pre-compact subset of ${M}_1(E^2)$.
Note that
\begin{eqnarray*}
t^{-1}\int^t_0\mathbbm{Q}\left[\mu^{-1}\(s\)\right]ds(\KK^c_\ep):&=&t^{-1}\int^t_0\mathbbm{Q}\left[\mu\(s\)\in\KK^c_\ep\right]ds
\\&=&t^{-1}\int^t_0\mathbbm{Q}\left[\exists\
k\ge
1,\;\;\mu\(s\)(K^c_{\ep k^{-1}2^{-k}})>k^{-1}\right]ds\\
&\le&t^{-1}\int^t_0\sum^\infty_{k=1}k\mathbbm{Q}\left[ \mu\(s\)(K^c_{\ep k^{-1}2^{-k}})\right]ds\\
&=&t^{-1}\int^t_0\sum^\infty_{k=1}k \(T_s^*, T_s^*\)\mu(K^c_{\ep k^{-1}2^{-k}})ds\\
&<&t^{-1}\int^t_0\sum^\infty_{k=1}k \ep k^{-1}2^{-k}ds=\ep.
\end{eqnarray*}
Thus, the family  $\left\{t^{-1}\int^t_0\mathbbm{Q}[\mu^{-1}\(s\)]ds:\;t\ge
0\right\}$ is tight, and hence, pre-compact in ${M}_1({M}_1(E^2))$. Let
$\Pi$ be a limit point. Then there exists a sequence $(t_n)$ such
that $t_n\uparrow \infty$ and
\[\lim_{n\goto\infty }t_n^{-1}\int^{t_n}_0\mathbbm{Q}[\mu^{-1}\(s\)]ds=\Pi.\]
For any $r\ge 0$,
\begin{eqnarray*}
\Pi [\mu^{-1}\(r\)]&=&\lim_{n\to\infty}t_n^{-1}\int^{t_n}_0\mathbbm{Q}[\mu^{-1}\(s\)]\circ[\mu^{-1}\(r\)]
ds\\
&=&\lim_{n\to\infty}t_n^{-1}\int^{t_n+r}_r\mathbbm{Q}[\mu^{-1}\(s\)]ds\\
&=&\Pi.
\end{eqnarray*}
Namely, $\Pi$ is an invariant measure of the stochastic process
$\(\mu\(t\)\)_{t\geq 0}$.

It follows from the construction of the dual $\(\Xi,\text{A},u_{\mathbbm{1}},u_{\mathbbm{2}}\)$-CPGM that
$|\eta(t)|$ is non-increasing with respect to $t$ and
$\lim_{t\rightarrow\infty}|\eta(t)|=1$. Denote by $\tau=\inf\{t\geq0: |\eta\(t\)|=1\}$.
Note that $\tau<\infty$ a.s. and $\(Y\(t\), \eta\(t\)\)\in {B}(E)\times {\mathbbm{S}}$ for any $t\geq\tau$. One can see that
\begin{equation}\label{limit state distribution}
\begin{split}
\int_{M_1\(E^2\)}\left<\mu_{\eta},f\right>\Pi\(d\mu\)
=&\lim_{t\rightarrow\infty}\mathbbm{Q}_{\Pi}\left<\mu\(t\)_{\eta},f\right>\\
=&\lim_{t\rightarrow\infty}\int_{M_1\(E^2\)}\mathbbm{P}_{\(f,\,\eta\)}\left<\mu_{\eta\(t\)},Y\(t\)\right>\Pi\(d\mu\)\\
=&\lim_{t\rightarrow\infty}\int_{M_1\(E^2\)}\mathbbm{P}_{\(f,\,\eta\)}\left[\left<\mu_{\eta\(t\)},Y\(t\)\right>{\ind_{\{\tau\leq t\}}}\right]\Pi\(d\mu\)\\
=&\lim_{t\rightarrow\infty}\int_{M_1\(E^2\)}\mathbbm{P}_{\(f,\,\eta\)}\mathbbm{P}_{\(Y\(\tau\),\,\eta\(\tau\)\)}\left[\left<\mu_{\eta\(t-\tau\)},Y\(t-\tau\)\right>{\ind_{\{\tau\leq t\}}}\right]\Pi\(d\mu\)\\
=&\lim_{t\rightarrow\infty}\int_{M_1\(E^2\)}\mathbbm{P}_{\(f,\,\eta\)}\mathbbm{P}_{\(Y\(\tau\),\,\eta\(\tau\)\)}\left<\mu_{\eta\(t\)},Y\(t\)\right>\Pi\(d\mu\)\\
=&\lim_{t\rightarrow\infty}\int_{M_1\(E^2\)}\mathbbm{P}_{\(f,\,\eta\)}\left<\mu_{\eta\(t+\tau\)},T_tY\(\tau\)\right>\Pi\(d\mu\)\\
=&\lim_{t\rightarrow\infty}\int_{M_1\(E^2\)}\mathbbm{P}_{\(f,\,\eta\)}\left<T_t^{*}\mu_{\eta\(t+\tau\)},Y\(\tau\)\right>\Pi\(d\mu\)\\
=&\mathbbm{P}_{\(f,\,\eta\)}\left[\left<\tilde{\pi},Y\(\tau\)\right>\right].
\end{split}
\end{equation}
Thus, (\ref{stationary distribution}) follows from (\ref{limit state distribution}) with
$n=1$.
\end{proof}

\section{Reversibility of the generalized stepping stone model}\label{sec6}
In this section we discuss the reversibility of the generalized stepping stone model with $\Xi$-resampling mechanism.
In view of the well-known results in the literature, we make the following
assumption on the mutation operator.

\noindent{\bf Condition B:} The mutation operator
$A$ is of the uniform type, i.e.,
$$Af(x)=\frac{\theta}{2}\int_E\(f(y)-f(x)\)\nu_0(dy)$$
for some $\theta>0$, $\nu_0\in M_1{\(E\)}$ and $f\in B\(E\)$.

\begin{definition}
The generalized stepping stone model with $\Xi$-resampling mechanism $({\mu}(t))_{t\geq0}=(\mu_{\mathbbm{1}}(t),\mu_{\mathbbm{2}}(t))_{t\geq0}$ with stationary
distribution $\Pi$ is {\it reversible} if  for any
nonnegative integers $n, m, p$ and $q$, we have
\begin{equation}\label{eq-rev}
\begin{split}
&\mathbbm{Q}_{\Pi}\left[\(\<{{\mu_{\mathbbm{1}}}(t)}^{\otimes n}, f^{{\mathbbm{1}}}\>\<{{\mu_{\mathbbm{2}}}(t)}^{\otimes
m},
f^{{\mathbbm{2}}}\>\)\mathcal{L}^{*}\(\<{{\mu_{\mathbbm{1}}}(t)}^{\otimes p}, g^{{\mathbbm{1}}}\>\<{{\mu_{\mathbbm{2}}}(t)}^{\otimes q}, g^{{\mathbbm{2}}}\>\)\right]\\
&=\mathbbm{Q}_{\Pi}\left[\(\<{\mu_{\mathbbm{1}}(t)}^{\otimes p}, g^{{\mathbbm{1}}}\>\<{{\mu_{\mathbbm{2}}}(t)}^{\otimes
q}, g^{{\mathbbm{2}}}\>\)\mathcal{L}^{*}\(\<{{\mu_{\mathbbm{1}}}(t)}^{\otimes n}, f^{{\mathbbm{1}}}\>\<{{\mu_{\mathbbm{2}}}(t)}^{\otimes
m}, f^{{\mathbbm{2}}}\>\)\right],
\end{split}
\end{equation}
where $f^{{\mathbbm{1}}}\in {B}(E^n)$, $f^{{\mathbbm{2}}}\in {B}(E^m)$, $g^{{\mathbbm{1}}}\in
{B}(E^{p})$, $g^{{\mathbbm{2}}}\in {B}(E^{q})$ and $\mathcal{L}^{*}$ is the generator given by $(\text{\ref{generator}})$.
Without loss of generality, we denote by $\<{\mu_{\mathbbm{i}}(t)}^{\otimes 0}, \cdot\>=1$ with $\mathbbm{i}\in\{\mathbbm{1,2}\}$ and $\mathcal{L}^{*}1=0$.
\end{definition}
\begin{theorem}\label{reversibility theorem}
Under Condition B, the generalized stepping stone model with $\Xi$-resampling mechanism is not reversible.
\end{theorem}
\begin{proof}
The proof is based on computations  of moments and joint moments of different orders. The software {\it Maple} is used to
carry out symbolic calculations. We sketch the main steps and refer to Appendix for more details.

Let $E^{*}$ be any subset of $E$ with $0<\nu_0(E^{*})=\alpha<1$.  We start with $(p,q)=(0,0)$. By choosing different values of $n$, $m$,
$f^{\mathbbm{1}}$ and $f^{\mathbbm{2}}$ in (\ref{eq-rev}), expressions for the moments and joint moments
\begin{equation*}
\begin{split}
&\mathbbm{Q}\left[{\mu_{\mathbbm{1}}}(E^{*})\right],\,\,\mathbbm{Q}\left[{\mu_{\mathbbm{2}}}(E^{*})\right],\,\,\mathbbm{Q}\left[{\mu_{\mathbbm{1}}}^2(E^{*})\right],\,\,\mathbbm{Q}\left[{\mu_{\mathbbm{1}}}(E^{*}){\mu_{\mathbbm{2}}}(E^{*})\right],\,\,\mathbbm{Q}\left[{\mu_{\mathbbm{2}}}^2(E^{*})\right],\\
&\mathbbm{Q}\left[{\mu_{\mathbbm{1}}}^3(E^{*})\right],\,\,\mathbbm{Q}\left[{\mu_{\mathbbm{1}}}^2(E^{*}){\mu_{\mathbbm{2}}}(E^{*})\right],\,\,\mathbbm{Q}\left[{\mu_{\mathbbm{1}}}(E^{*}){\mu_{\mathbbm{2}}}^2(E^{*})\right],\,\,\mathbbm{Q}\left[{\mu_{\mathbbm{2}}}^3(E^{*})\right],\\
&\mathbbm{Q}\left[{\mu_{\mathbbm{1}}}^4(E^{*})\right],\,\,\mathbbm{Q}\left[{\mu_{\mathbbm{1}}}^3(E^{*}){\mu_{\mathbbm{2}}}(E^{*})\right],\,\,\mathbbm{Q}\left[{\mu_{\mathbbm{1}}}^2(E^{*}){\mu_{\mathbbm{2}}}^2(E^{*})\right],
\mathbbm{Q}\left[{\mu_{\mathbbm{1}}}(E^{*}){\mu_{\mathbbm{2}}}^3(E^{*})\right],\,\,\mathbbm{Q}\left[{\mu_{\mathbbm{2}}}^4(E^{*})\right]
\end{split}
\end{equation*}
are all available.

We then consider the case that $(p,q)\neq(0,0)$.
For  $$(n, m)=(1, 0),\,(p, q)=(0,1)\,\text{and}\,f^{\mathbbm{1}}=g^{\mathbbm{2}}=\ind_{E^{*}},$$ by (\ref{eq-rev})
we have that the condition  ${u_{\mathbbm 1}}={u_{\mathbbm 2}}$ is necessary for the process being reversible.
Further substituting  $$(n, m)=(2, 0),\,(p, q)=(0,1)\,\text{and}\,f^{\mathbbm{1}}=\ind_{E^{*}\times E^{*}},\,g^{\mathbbm{2}}=\ind_{E^{*}}$$ in (\ref{eq-rev}), we have that
the condition  $\al=1/2$ is necessary for the process being reversible.
Given $\al=1/2$ and ${u_{\mathbbm 1}}={u_{\mathbbm 2}}$, we choose
$$(n, m)=(1, 1),\,(p, q)=(2,0),\,f^{\mathbbm{1}}=f^{\mathbbm{2}}=\ind_{E^{*}},\,g^{\mathbbm{1}}=\ind_{E^{*}\times E^{*}}$$ and
$$(n, m)=(2, 1),\,(p, q)=(1,0),\,f^{\mathbbm{1}}=\ind_{E^{*}\times E^{*}},\,f^{\mathbbm{2}}=g^{\mathbbm{1}}=\ind_{E^{*}}$$ in (\ref{eq-rev}), respectively. Note that the two equations can not  be true at the same time and we reach  a contradiction.
Thus, the process is not reversible.
\end{proof}

\begin{appendices}
\section{The detailed calculations in the proof of Theorem \ref{reversibility theorem}}
For simplicity of notation, we write
\begin{equation*}
\begin{split}
&a_2=\la_{2;2;0},\,\,\,\,\,\,a_{21}=\la_{3;2;1},\,\,\,\,\,\,a_3=\la_{3;3;0},\\
&a_{211}=\la_{4;2;2},\,\,a_{22}=\la_{4;2,2;0},\,\,a_{31}=\la_{4;3;1}, \,\,a_4=\la_{4;4;0}.
\end{split}
\end{equation*}
Let $E^{*}$ be  a subset of the type space $E$ with $\nu_0\(E^{*}\)=\al>0$. For any positive integers  $n$ and $m$, the joint moments are defined as $$M_{n,m}=\mathbbm{Q}\left[{\mu_{\mathbbm{1}}}^n\(E^{*}\){\mu_{\mathbbm{2}}}^m\(E^{*}\)\right].$$

We begin with the discussion for the first moments, i.e., $n+m=1$. \\
For $(n,m)=(1,0),\,(p,q)=(0,0)$ and $f^{{\mathbbm{1}}}=\ind_{E^{*}}$ in (\ref{eq-rev}), we have
\begin{equation}\label{E_{11}}
\left(\frac{\theta}{2}+u_{\mathbbm 2}\right)M_{1,0}-u_{\mathbbm 2}M_{0,1}=\frac{\theta\al}{2}.
\end{equation}
For $(n,m)=(0,1),\,(p,q)=(0,0)$ and $f^{{\mathbbm{2}}}=\ind_{E^{*}}$ in (\ref{eq-rev}), we have
\begin{equation}\label{E_{12}}
-u_{\mathbbm 1}M_{1,0}+\left(\frac{\theta}{2}+u_{\mathbbm 1}\right)M_{0,1}=\frac{\theta\al}{2}.
\end{equation}
By (\ref{E_{11}}) and (\ref{E_{12}}) we have  $M_{1,0}=M_{0,1}=\al.$

Then we consider the second moments, i.e., $n+m=2$.\\
For $(n,m)=(2,0),\,(p,q)=(0,0)$ and $f^{{\mathbbm{1}}}=\ind_{E^{*}\times E^{*}}$ in (\ref{eq-rev}), we have
\begin{equation}\label{E_{21}}
\left( \theta+2\,u_{\mathbbm 2}+a_
{{2}} \right) M_{{2,0}}-2\,u_{\mathbbm 2}M_{{1,1}}=\theta\,{\alpha}^{2}+a_{{2}}\,\alpha.
\end{equation}
For $(n,m)=(0,2),\,(p,q)=(0,0)$ and $f^{{\mathbbm{2}}}=\ind_{E^{*}\times E^{*}}$ in (\ref{eq-rev}), we have
\begin{equation}\label{E_{22}}
-2\,u_{\mathbbm 1}M_{{1,1}}+\left( \theta+2\,u_{\mathbbm 1}+a_
{{2}} \right) M_{{0,2}}=\theta\,{\alpha}^{2}+a_{{2}}\,\alpha.
\end{equation}
For $(n,m)=(1,1),\,(p,q)=(0,0)$ and $f^{{\mathbbm{1}}}=f^{{\mathbbm{2}}}=\ind_{E^{*}}$ in (\ref{eq-rev}), we have
\begin{equation}\label{E_{23}}
-u_{\mathbbm 1}M_{
{2,0}}+ \left( \theta+u_{\mathbbm 2}+u_{\mathbbm 1} \right) M_{{1,1}}-u_{\mathbbm 2}M_{{0,2}}=\theta\,{\alpha}^{2}.
\end{equation}
(\ref{E_{21}})-(\ref{E_{23}}) constitute a system of linear equations. Note that the coefficient matrix of
$M_{2,0}$, $M_{1,1}$, $M_{0,2}$ is invertible. Thus, the solution is unique.
Substituting $(n,m)=(1,0),\,(p,q)=(0,1)$ and $f^{{\mathbbm{1}}}=\ind_{E^{*}},\,g^{{\mathbbm{2}}}=\ind_{E^{*}}$ in (\ref{eq-rev}), we have the reversible equation
\begin{equation}\label{S_{1}}
\left( u_{\mathbbm 1}-u_{\mathbbm 2} \right) M_{{1,1}}-u_{\mathbbm 1}M_{{2,0}}+u_{\mathbbm 2}M_{{0,2}}=0.
\end{equation}
Replacing the second moments, the numerator of (\ref{S_{1}}) equals to $0$, i.e.,
\begin{equation*}
{{\alpha\,\theta\,a_{{2}} \left(u_{\mathbbm 1}-u_{\mathbbm 2} \right)  \left( \theta+2\,u_{\mathbbm 1}
+a_{{2}}+2\,u_{\mathbbm 2} \right)  \left( \alpha-1 \right) }}=0.
\end{equation*}
Thus, a necessary condition for this process being reversible is
that ${u_{\mathbbm 1}}={u_{\mathbbm 2}}$.

Now we continue to consider the third moments, i.e., $m+n=3$.\\
For $(n,m)=(3,0),\,(p,q)=(0,0)$ and $f^{{\mathbbm{1}}}=\ind_{E^{*}\times E^{*}\times E^{*}}$ in (\ref{eq-rev}), we have
\begin{equation}\label{E_{31}}
\left( 3\,a_{{21}}+a_{{3}}+\frac32\,\theta+3\,u_{\mathbbm 2} \right) M_{{3,0}}-3\,u_{\mathbbm 2}M_{
{2,1}}=\left( 3\,a_{{21}}+\frac32\,\alpha\,\theta \right) M_{{2,0}}+a_{{3}}\alpha.
\end{equation}
For $(n,m)=(2,1),\,(p,q)=(0,0)$ and $f^{{\mathbbm{1}}}=\ind_{E^{*}\times E^{*}},\,f^{{\mathbbm{2}}}=\ind_{E^{*}}$ in (\ref{eq-rev}), we have
\begin{equation}\label{E_{32}}
-u_{\mathbbm 1}M_{{3,0}}+\left( a_{{2}}+\frac32\,\theta+2\,u_{\mathbbm 2}+u_{\mathbbm 1} \right) M_{{2,1}}-2\,u_{\mathbbm 2}M_{{1,2
}}=\left( \alpha\,\theta+a_{{2}} \right) M_{{1,1}}+\frac{\theta\,\alpha}{2}\,M
_{{2,0}}.
\end{equation}
For $(n,m)=(1,2),\,(p,q)=(0,0)$ and $f^{{\mathbbm{1}}}=\ind_{E^{*}},\,f^{{\mathbbm{2}}}=\ind_{E^{*}\times E^{*}}$ in (\ref{eq-rev}), we have
\begin{equation}\label{E_{33}}
-2
\,u_{\mathbbm 1}M_{{2,1}}+\left( a_2+\frac32\,\theta+u_{\mathbbm 2}+2\,u_{\mathbbm 1} \right) M_{{1,2}}-u_{\mathbbm 2}M_{{0,3}}=\left( \alpha\,\theta+a_{{2}} \right) M_
{{1,1}}+\frac{\theta\,\alpha}{2}\,M_{{0,2}}.
\end{equation}
For $(n,m)=(0,3),\,(p,q)=(0,0)$ and $f^{{\mathbbm{2}}}=\ind_{E^{*}\times E^{*}\times E^{*}}$ in (\ref{eq-rev}), we have
\begin{equation}\label{E_{34}}
-3\,u_{\mathbbm 1}M_{{
1,2}}+\left( 3\,a_{{21}}+a_{{3}}+\frac32\,\theta+3\,u_{\mathbbm 1}\right) M_{{0,3}}=\left( 3\,a_{{21}}+\frac32\,\alpha\,\theta \right) M_{{0,2}}+a_{{3}}\alpha.
\end{equation}
(\ref{E_{31}})-(\ref{E_{34}}) constitute a system of linear equations and the coefficient matrix of $M_{3,0},\,M_{2,1}$, $M_{1,2},\,M_{0,3}$ is invertible. Thus, the system of equations has a unique solution.
Substituting $(n,m)=(2,0),\,(p,q)=(0,1)$ and $f^{{\mathbbm{1}}}=\ind_{E^{*}\times E^{*}},\,g^{{\mathbbm{2}}}=\ind_{E^{*}}$ in (\ref{eq-rev}), we have
\begin{equation}\label{T_{1}}
\left( \alpha\,\theta+a_{{2}} \right) M_{{1,1}}-\frac{\theta\,\alpha}{2}\,M
_{{2,0}}- \left( 2\,u_{\mathbbm 2}+a_{{2}}+\frac{\theta}{2}-u_{\mathbbm 1} \right) M_{{2,1}}-u_{\mathbbm 1}M_{{3,0}}+
2\,u_{\mathbbm 2}M_{{1,2}}=0.
\end{equation}
Note that $a_2=a_{21}+a_3$. Substituting the expressions of the moments, we have the numerator of (\ref{T_{1}}) equal to $0$, i.e.,
\begin{equation*}
4\,\alpha\,{\theta}^{2}u_{\mathbbm 1} \left( \alpha-1 \right)  \left( 3\,{a_{{21}}}
^{2}+{a_{{3}}}^{2}+8\,a_{{3}}u+2\,a_{{3}}\theta+4\,a_{{3}}a_{{21}}
 \right)  \left( -1+2\,\alpha \right)=0.
\end{equation*}
So far we conclude that the process is not reversible except the case when there exists a subset $E^{*}\subseteq E$ with $\alpha=\nu_0\(E^{*}\)=1/2$.

For the case of $\al=1/2$, we need higher moments to reach a contradiction. \\
For $(n,m)=(4,0),\,(p,q)=(0,0)$ and $f^{{\mathbbm{1}}}=\ind_{E^{*}\times E^{*}\times E^{*}\times E^{*}}$ in (\ref{eq-rev}), we have
\begin{equation}\label{E_{41}}
\begin{split}
& \left( 4\,u_{\mathbbm 2}+6\,a_{{
211}}+3\,a_{{22}}+4\,a_{{31}}+a_{{4}}+2\,\theta \right) M_{{4,0}}-4\,u_{\mathbbm 2}M
_{{31}}\\
&\,\,=\left( 2\,\theta\,\alpha+6\,a_{{211}} \right) M_{{3,0}}+ \left( 3\,a_{
{22}}+4\,a_{{31}} \right) M_{{2,0}}+a_{{4}}\,\alpha.
\end{split}
\end{equation}
For $(n,m)=(3,1),\,(p,q)=(0,0)$ and $f^{{\mathbbm{1}}}=\ind_{E^{*}\times E^{*}\times E^{*}},\,f^{{\mathbbm{2}}}=\ind_{E^{*}}$ in (\ref{eq-rev}), we have
\begin{equation}\label{E_{42}}
\begin{split}
&-u_{\mathbbm 1}M_{{4,0}}+ \left( a_{{3}
}+3\,a_{{21}}+2\,\theta+3\,u_{\mathbbm 2}+u_{\mathbbm 1} \right) M_{{3,1}}-3\,u_{\mathbbm 2}M_{{2,2}}\\
&\,\,= \left( 3\,a_{{21}}+\frac{3\,\theta\,\alpha}{2} \right) M_{{2,1}}+a_{{3}}M_{{1,1
}}+\frac{\,\theta\,\alpha}{2}\,M_{{3,0}}.
\end{split}
\end{equation}
For $(n,m)=(2,2),\,(p,q)=(0,0)$ and $f^{{\mathbbm{1}}}=\ind_{E^{*}\times E^{*}},\,f^{{\mathbbm{2}}}=\ind_{E^{*}\times E^{*}}$ in (\ref{eq-rev}), we have
\begin{equation}\label{E_{43}}
\begin{split}
&-2\,u_{\mathbbm 1}M_{{3,1}}+ \left( 2\,a_{{2}}
+2\,\theta+2\,u_{\mathbbm 2}+2\,u_{\mathbbm 1} \right) M_{{2,2}}-2\,u_{\mathbbm 2}M_{{1,3}}\\
&\,\,=\left( \theta\,\alpha+a_{{2}} \right) M_{{1,2}}+ \left( \theta\,\alpha
+a_{{2}} \right) M_{{2,1}}.
\end{split}
\end{equation}
For $(n,m)=(1,3),\,(p,q)=(0,0)$ and $f^{{\mathbbm{1}}}=\ind_{ E^{*}},\,f^{{\mathbbm{2}}}=\ind_{E^{*}\times E^{*}\times E^{*}}$ in (\ref{eq-rev}), we have
\begin{equation}\label{E_{44}}
 \begin{split}
&-3\,u_{\mathbbm 1}M_{{2,2}}+ \left( 2\,
\theta+u_{\mathbbm 2}+3\,u_{\mathbbm 1}+3\,a_{{21}}+a_{{3}} \right) M_{{1,3}}-u_{\mathbbm 2}M_{{0,4}}\\
 &\,\,=\frac{\theta\,\alpha}{2}\,M_{{0,3}}+ \left( 3\,a_{{21}}+\frac{3\,\theta\,\alpha}{2}
 \right) M_{{1,2}}+a_{{3}}M_{{1,1}}.
 \end{split}
\end{equation}
For $(n,m)=(0,4),\,(p,q)=(0,0)$ and $f^{{\mathbbm{2}}}=\ind_{E^{*}\times E^{*}\times E^{*}\times E^{*}}$ in (\ref{eq-rev}), we have
\begin{equation}\label{E_{45}}
\begin{split}
 &-4\,u_{\mathbbm 1}M_{{
1,3}}+ \left( 6\,a_{{211}}+3
\,a_{{22}}+4\,a_{{31}}+a_{{4}}+2\,\theta+4\,u_{\mathbbm 1} \right) M_{{0,4}}\\
&\,\,=\left( 2\,\theta\,\alpha+6\,a_{{211}} \right) M_{{0,3}}+ \left( 3\,a_{{
22}}+4\,a_{{31}} \right) M_{{0,2}}+a_{{4}}\,\alpha.
\end{split}
\end{equation}
We solve for $M_{4,0},\,M_{3,1},\,M_{2£¬2},\,M_{1,3},\,M_{0,4}$ uniquely from (\ref{E_{41}})-(\ref{E_{45}}) which is also a system of linear equations with invertible coefficient matrix.
Substituting $(n,m)=(1,1),\,(p,q)=(2,0)$ and $f^{{\mathbbm{1}}}=f^{{\mathbbm{2}}}=\ind_{E^{*}},\,g^{{\mathbbm{1}}}=\ind_{E^{*}\times E^{*}}$, we have
\begin{equation}\label{F_{1}}
\begin{split}
\frac{\,\theta\,\alpha}{2}\,M_{{3,0}}-\(a_{{2}}+\frac{\theta\,
\alpha}{2}\)\,M_{{2,1}}-u_{\mathbbm 2}M_{{2,2}}+u_{\mathbbm 1}M_{{4,0}}+\(a_2-u_{\mathbbm 1}+u_{\mathbbm 2}\)M_{{3,1}}=0.
\end{split}
\end{equation}
Substituting $(n,m)=(2,1),\,(p,q)=(1,0)$ and $f^{{\mathbbm{1}}}=\ind_{E^{*}\times E^{*}},\,f^{{\mathbbm{2}}}=g^{{\mathbbm{1}}}=\ind_{E^{*}}$, we have
\begin{equation}\label{F_{2}}
\begin{split}
 \left( a_{{2}}+\frac{\theta\,\alpha}{2} \right) M_{{2,1}}+\frac{\theta\,
\alpha}{2}\,M_{{3,0}}- \left( \theta+u_{\mathbbm 2}+u_{\mathbbm 1}+a_{{2}} \right) M_{{3,1}}+u_{\mathbbm 2}M_{{2,2}}
+u_{\mathbbm 1}M_{{4,0}}=0.
\end{split}
\end{equation}
Substituting all the joint moments, the numerator of $$(\ref{F_{1}})-2\times(\ref{F_{2}})=0$$  is equivalent to
\begin{equation}\label{eq-contra}
\begin{split}
& \left( \theta+4\,u_{\mathbbm{1}} \right) \theta\,u_{\mathbbm{1}}a_{{3}} \left( -8\,a_{{4}}u_{\mathbbm{1}}-4\,a_
{{4}}a_{{3}}-4\,\theta\,a_{{4}}+8\,{a_{{3}}}^{2}+22\,a_{{3}}u_{\mathbbm{1}}+11\,a_{{
3}}\theta+8\,{a_{{21}}}^{2}+16\,a_{{21}}a_{{3}}\right.\\
&\left.\,\,\,\,\,\,\,\,\,\,\,\,+12\,a_{{211}}a_{{3}}-4
\,a_{{4}}a_{{21}}+10\,a_{{21}}u_{\mathbbm{1}}+5\,a_{{21}}\theta+3\,a_{{211}}\theta+6
\,a_{{211}}u_{\mathbbm{1}}  +12\,a_{{211}}a_{{21}}\right)=0.
\end{split}
\end{equation}
The term in the second parentheses  is always positive because
$-8a_4u_{\mathbbm{1}}+22a_3u_{\mathbbm{1}}\geq0$, $-4a_4a_3+8a_3^2\geq0$, $-4\theta a_4+11a_3\theta\geq0$
and
$-4a_4a_{21}+16a_{21}a_{3}\geq0.$
Since $u_{\mathbbm{1}}>0$ and $\theta>0$, a necessary condition for  (\ref{eq-contra}) is that $a_3=0$. By the consistent condition of coalescent rates
\begin{equation*
\begin{cases}
a_2=a_{21}+a_3;\\
a_3=a_{31}+a_4;\\
a_{21}=a_{211}+a_{22}+a_{31},
\end{cases}
\end{equation*}
$a_3=0$ implies that  $a_4=a_{31}=a_{22}=0$ and $a_{2}=a_{21}=a_{211}=:a>0$. Substituting these values into the numerator of (\ref{F_{2}}), we get
$4a^3=0$ which contradicts $a>0$. Therefore, the process is not reversible.

A Maple note for all the calculations carried out in this paper is available upon request.
\end{appendices}

\renewcommand{\bibname}{References}
\bibliographystyle{plainnat}

redpajama_set_name: RedPajamaArXiv


I'm back, now with more parentheses!
1. Almost no book I own can be easily summed up in a few words. Go ahead, try it, I'll wait.
Maybe I'm just picky, but I think describing most of my books — particularly since I read a lot of sci-fi and fantasy — in a few words either makes them sound terrible, or doesn't properly explain the book. I feel like you want the answer to be, "Ponies!" or "The Franco-Prussian War!" Actually, I may just make that second one my stock answer.
No it is not and no I did not. Which isn't to say that I don't occasionally read a romance novel (ideally a Regency, even more ideally with a breeches role. You find me a Georgette Heyer with a girl pretending to be a boy and hilarious consequences and I am a happy camper.), but I get tired of every book I read being assumed to be a romance. This happened most recently with Robert Heinlein's Friday, and I didn't have the energy to explain that it's about a bioengineered super spy finding her identity in the world, so I settled for a withering look.
3. I'M READING! Interrupting me while reading, much like interrupting me while watching a new episode of Buffy: The Vampire Slayer (back when that was a thing) may cause me to get stabby. (My brother knows this well, as when Buffy first started airing, that hour of TV time was sacrosanct and any interruption would be met with, "No talk. Buffy." A repetition would probably have led to flung items, but my family never risked it.) Why does everyone assume I'm just dying to put down my book and tell them why it's great? Why don't we understand that yanking me out of the story suddenly results in squishy whiplashed brain, making it even more impossible for me to give you a decent answer?
Get your own book! If you must know more about mine, stop playing Candy Crush for two seconds and look up the title. Read a synopsis by someone who WASN'T blindsided by having to edit a 200 page book down to on sentence instantly.
Damn kids on my lawn.
I don't think everyone else is like that. I'm like you: if I'm deeply into something, any distraction is unwelcome and causes my brain to grind gears for almost a minute as I try to grasp what is happening. The folks who have interrupted me all look surprised – "What is his problem?" So I think other people can effortlessly change mental tracks. We think more deeply, and change tracks much more slowly.
This entry was posted on July 13, 2013 by naralesser and tagged Manifestos again?, NERD!, owls, ranty mcranterson, terrible ideas, this won't end well.


During the summer, and for those who do not have the fortune of being able to go on vacation, the swimming pool kid in the garden at home can be an excellent choice of relaxation and fun for adults and children. With balls: It consists of making two teams with the children and it will be necessary to have two beach balls of the inflatables. The game is that each child carries the ball from one end of the pool to the other, but not with his hands but pushing it with his nose.
Win the game that first arrives. Using several balloons in the swimming pool kid, the idea is for the children to take the balloons and try to sink them to hide them under their body, this is fun for the little ones who, even if they can not, want to hide many balloons. With toys: It is necessary to divide the pool with a rope or net and on each side several floating objects are placed.
The children are divided into two groups occupying each part of the swimming pool kid. Set a time of 10 minutes to remove the objects that are on your side and leave it on the side of your opponents, who should also do the same, trying to prevent your space is filled with toys. At the end of the agreed time, a count is made and the group that has fewer objects on its side will be the winner.


text:
Q: display: inline-block vs display: table (no display: table-cell) I have used both of the display properties whenever I wanted the width of an element to be the same as its content.
But is one better than the other? I'm referring to accessibility, browser compatibility, responsiveness, etc.
Note: I am asking because I'm planing to use only one of these in my new web site. I just don't know which one is better, if any.

A: Ultimately, it depends on the use case:

*

*display: inline-block will create an inline-block element

*display: table will create a table element

Here they are in use:

span.mySpan {
  background-color: red;
}

  A span element.
  a display: table element.
  Another span element.

  A span element.
  a display: inline-block element.
  Another span element.

As can be seen, the results are very different. The table element positions itself on a new line, and causes the next element to be on a new line as well. The inline-block element positions itself inline with it's sibling elements.
In many cases, the above differences will be enough to choose one or the other.
If not, let's continue...
There are some cases when display: table is useful:

*

*Horizontal and vertical centering of elements

*Equal height elements

However, browsers can produce inconsistent results when not implemented correctly so you should always couple display: table with the standard table markup (using rows and cells):

.table {
  display: table;
}

.table-row {
  display: table-row:
}

.table-cell {
  display: table-cell;
  background-color: #eaeaea;
  padding: 10px;
}

      Content

      Content

      Content

This becomes pretty tedious. And with modern CSS we can accomplish the same using display: flex, with a simpler HTML structure and less CSS:

.flex {
  display: flex;
}

.flex-cell {
  background-color: #eaeaea;
  padding: 10px;
}

    Content

    Content

    Content

Honestly, I can't think of many times I would need to decide between display: inline-block and display: table as they produce such different results. However, if I were on the fence I'd follow this decision tree:

*

*Do I need to make a table? Use a true

 element

*Do I need equal height/width elements, and/or vertical centering? Use a display: flex element

*Otherwise, use the appropriate HTML element (display: inline-block)

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text:
Velká francouzská válka je označení používané pro sérii konfliktů v letech 1792 až 1815, známých jako francouzské revoluční a napoleonské války. Bojovala v ní Francie proti většině Evropy.

Průběh
Roku 1789 začala Velká francouzská revoluce a roku 1792 začala válka první koalice jako odpověď na popravu francouzského krále Ludvíka XVI. a jeho manželky, královny Marie Antoinetty. Tím začaly revoluční války. První koalice se rozpadla roku 1797 po italském tažení vedeném generálem Bonapartem. Brzy poté zahájil Bonaparte tažení do Egypta, ale nepovedlo se mu dosáhnout rozhodného vítězství a nakonec se musel vrátit, protože v Evropě mezitím vznikla druhá koalice, která pod velením ruského maršála Suvorova porážela revoluční vojska v Itálii.

Napoleon 14. června 1800 porazil koalici u Marenga a postupoval do Rakouska, které raději uzavřelo mír; ten nakonec uzavřely i ostatní členové koalice. Roku 1803 vpadl Napoleon do Hanoverska a tím zahájil napoleonské války. Po jeho korunovaci císařem (2. prosince 1804) vznikla třetí koalice, jež se však po bitvách u Ulmu a Slavkova rozpadla. Napoleon poté zvítězil nad čtvrtou koalicí, ale válka ve Španělsku skončila neúspěchem a ve válce páté koalice prohrál bitvu u Aspern. Po bitvě u Aspern Napoleon nakonec zvítězil u Wagramu, ale to již bylo jasné, že není neporazitelný.

Tažení do Ruska roku 1812 skončilo zničením jeho půlmilionové armády a následným vznikem šesté koalice. Ta mu uštědřila zdrcující porážku u Lipska a 30. března 1814 obsadila Paříž. Napoleon musel abdikovat a odejít do exilu na ostrově Elba. 1. března 1815 z Elby utekl a později se vylodil v zátoce San Juan u Cannes. Francouzský král Ludvík XVIII. musel prchnout, ale Napoleon zůstal císařem jen sto dní, neboť 18. června 1815 mu nově vzniklá sedmá koalice uštědřila zdrcující porážku u Waterloo. Napoleon nyní byl deportován na ostrov Svatá Helena, kde roku 1821 zemřel. Poválečné uspořádání Evropy bylo dohodnuto na Vídeňském kongresu.

Ve francouzských revolučních válkých zahynulo 1 100 000 vojáků a v napoleonských 3 350 000 až 6 500 000. Šlo až do první světové války o největší ozbrojený konflikt historie. Bojovalo se v severní a jižní Africe, jižní a Severní Americe, Karibské oblasti, na Blízkém východě, v Indii a většině Atlantského a Indické oceánu.

Odkazy

Literatura
 Kolektiv autorů: Války a válečníci
 Lev Tolstoj: Vojna a mír
 P. A. Žilin: Kutuzov
 Josef Hotmar: Dobrodružství Velké revoluce 1789–1799
 Geofrey Regan: Guinnesova kniha válečných omylů

Související články
 Moderní dějiny
 Velká francouzská revoluce
 Napoleon Bonaparte
 Vídeňský kongres
 Francouzské revoluční války
 Napoleonské války

Války Francie
Války Ruského impéria
Války Spojeného království
Novověká Evropa
Francouzské revoluční války
Války Habsburské monarchie
Války Španělska
Války Nizozemska
Války Portugalska
Války Saska
Války Bavorska

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Lunch Club is a well established ministry of East Beach Church which has served the community fo many years. A two course meal is provided free to anyone who wants to come and enjoy good food and good company. (Some love to give something towards the meal and so a box is provided at the entrance of the church for donations.) The meal is cooked and served by Christian volunteers from East Beach Church and other local churches.
To book a place, just come along on a Sunday to our morning service at 10am and put your name on the form on the notice board, or ring the church office and give your name and contact number. Spaces are limited.