# The ancestral selection graph for a $\Lambda$-asymmetric Moran model

Adrián González Casanova
adrian.gonzalez<EMAIL_ADDRESS>Noemi Kurt
<EMAIL_ADDRESS>José Luis Pérez<EMAIL_ADDRESS>Instituto
de Matematicas, Universidad Nacional Autonoma de Mexico (UNAM), Cuernavaca,
México and Department of Statistics, University of California at Berkeley
Institut für Mathematik, Johann Wolfgang Goethe-Universität, 60325 Frankfurt
am Main, Germany  Department of Probability and Statistics, Centro de
Investigación en Matemáticas A.C. Calle Jalisco s/n. C.P. 36240, Guanajuato,
Mexico

###### Abstract

Motivated by the question of the impact of selective advantage in populations
with skewed reproduction mechanims, we study a Moran model with selection. We
assume that there are two types of individuals, where the reproductive success
of one type is larger than the other. The higher reproductive success may stem
from either more frequent reproduction, or from larger numbers of offspring,
and is encoded in a measure $\Lambda$ for each of the two types.
$\Lambda$-reproduction here means that a whole fraction of the population is
replaced at a reproductive event. Our approach consists of constructing a
$\Lambda$-asymmetric Moran model in which individuals of the two populations
compete, rather than considering a Moran model for each population. Provided
the measure are ordered stochastically, we can couple them. This allows us to
construct the central object of this paper, the $\Lambda-$asymmetric ancestral
selection graph, leading to a pathwise duality of the forward in time
$\Lambda$-asymmetric Moran model with its ancestral process. We apply the
ancestral selection graph in order to obtain scaling limits of the forward and
backward processes, and note that the frequency process converges to the
solution of an SDE with discontinous paths. Finally, we derive a Griffiths
representation for the generator of the SDE and use it to find a semi-explicit
formula for the probability of fixation of the less beneficial of the two
types.

###### keywords:

Moran model, ancestral selection graph, duality, $\Lambda-$coalescent,
fixation probability

###### MSC:

[2020] 92D15, 60J28, 60J90

##  1 Introduction

There is a deep connection between the ancestry of a population and the
dynamics of its genetic configuration. Mathematical population genetics
exploits and formalises this connection, see for example [13] for an overview.
Its most classical instance is the moment duality relation between the block
counting process of the Kingman coalescent and the Wright Fisher diffusion,
which relates the past and future of a population that evolves on absence of
selection and in which the number of offsprings of a mother is of a smaller
order of magnitude than the population size.

This connection between the past and the future is ubiquitous and occurs in
many biologically inspired mathematical models. For example, the ancestry of
populations with skewed offspring distributions have been modelled by
$\Lambda$-coalescents or multiple merger coalescents [38, 41] different from
the Kingman coalescent, but which still have a moment dual, namely the
$\Lambda$-Fleming Viot process (see [5, 9, 10]). Skewed offspring
distributions occur in populations where the number of offspring of one mother
can be of the order of magnitude of the population size. It is believed that
they may be relevant for certain marine species such as the Atlantic cod,
where one mother may lay millions of eggs. See e.g. [1, 15] and references
therein for a more detailed discussion.

In the presence of weak selection, for example when one type of individuals
reproduces slightly faster than the others, it is cumbersome to formally
connect the ancestry and the forward in time changes in a population
configuration. Remarkably, there is a notion of potential ancestry that
overcomes this difficulty for populations in the universality class of the
Kingman coalescent: The celebrated ancestral selection graph (ASG) of Krone
and Neuhauser [31, 36]. Heuristically, in a population with only two types,
Krone and Neuhauser were able to describe the number of potential ancestors of
a sample as a branching coalescing process. Under the rule that an individual
is of selective type if at least one of its ancestors has selective advantage,
the forward in time propagation of types can be specified in terms of the
potential ancestry and the types at time zero. As a consequence, the frequency
of individuals with selective disadvantage is a Markov process, which is in
moment duality to the process of the number of potential ancestors (backward
in time). The graphical construction of the ancestral selection graph provides
a pathwise duality relation of the forward and backward processes, which for
example links fixation probabilities with the ancestral process, see e.g. [39,
30].

However, the classical ancestral selection graph couldn’t capture genetic
dynamics of a population of, say, cod in which a subpopulation is capable of
reproducing faster, as this leads to selective events individually affecting
large portions of the population. What is then a good model for populations
with skewed offspring distributions in the presence of selection? This was one
of the main motivating questions of work by Griffiths, Etheridge and Taylor
[12], see also Etheridge and Griffiths [11]. These authors showed that the
process of potential ancestors in this case should be a nonlinear branching
and coalescing process. They equipped their model with parent independent
mutation and found a duality in terms of the stationary distribution of the
forward process, in the case that the stationary distribution exists. They
also spelled out the reason for this mysterious nonlinear branching.

Furthermore, it was theorised by Gillespie [22, 23, 24] that subpopulations
with different reproduction mechanisms competing in a sequential sampling
experiment (for example, in the Lenski experiment [32, 26]) lead to different
macroscopic behaviours, which scale beyond the Wright-Fisher diffusion
universality class. A detailed mathematical analysis of these asymmetric
models was carried out in [7]. What does it mean that one subpopulation has
selective advantage over another if they have very different reproduction
behaviour? Can one compare the strategy of the cod with the strategy of the
rabbit?

The goal of the present paper is to revisit the motivating questions of [12]
and [22, 23, 24] to give a graphical representation of a Moran model with
$\Lambda$-reproduction in the presence of selection and asymmetry, following
the structure of the ancestral selection graph. Our representation provides a
pathwise duality between what we will call below the $\Lambda$-asymmetric
Moran model and its ancestral line counting process, from which the generator
duality can be derived. Contrary to Etheridge, Griffiths and Taylor we don’t
need to start from the invariant distribution in our construction and
therefore don’t need to include mutations in the model. Moreover, and probably
most importantly, our construction works without previously assigning types to
the individuals, since we can construct the ancestral selection graph
independently, in the spirit of Neuhauser and Krone.

Our construction is based on a coupling of the measures governing the
reproduction mechanisms of two subpopulations, which also has an
interpretation in terms of optimal transport theory. However, not every pair
of measures can be used to construct a $\Lambda$-ancestral selection graph,
and we provide explicit sufficient conditions. This is done by showing that a
pair of probability measures that is comparable with respect to the stochastic
partial order can be coupled to construct an ancestral selection graph. This
can be generalised to finite measures ordered in a similar manner.

Finally, we derive a representation inspired by Griffiths [27] for the
processes that arise as scaling limits of our $\Lambda$-Moran model with
selection. As an application we compute a recursion for the fixation
probabilities.

The paper is organised as follows. In the next section, we will define the
$\Lambda$-asymmetric Moran model and provide the generator of its frequency
process. In Section 3 we will state the central coupling lemma and construct
the ancestral selection graph. Using the construction of the ancestral graph,
we will consider the ancestral process and show our duality result in Section
4. A generalization of the $\Lambda$-asymmetric Moran to the case of finite
measures is provided in Section 5, as well as some remarks on the coupling.
Finally, we will discuss scaling limits in Section 6, and fixation
probabilites via Griffith’s representation in Section 7.

##  2 $\Lambda-$asymmetric Moran model

In this section, we define our main object of interest, the
$\Lambda$-asymmetric Moran model. It is related to the Moran model with
viability selection of Etheridge, Griffiths and Taylor [12], where for
simplicity we restrict ourselves to the case of two types and no mutation.
More importantly, we also require our process to be consistent and
exchangeable.

We consider a continuous time Moran model with fixed population size $N.$ The
two types will be denoted by $\oplus$ and $\ominus$, we write
$\tau(i,t)\in\\{\oplus,\ominus\\}$ for the type of individual
$i\in[N]=\\{1,...,N\\}$ at time $t\geq 0.$ Denote by $\mathcal{M}[0,1]$ the
set of finite measures and by $\mathcal{M}_{1}[0,1]$ the set of probability
measures on $[0,1]$ equipped with the Borel sigma field. Fix probability
measures $\Lambda^{\oplus},\Lambda^{\ominus}\in\mathcal{M}_{1}[0,1].$ These
measures will provide the reproduction rates and the strength of the selection
at a reproductive event, meaning that a fraction of the population, whose size
is determined by $\Lambda^{\oplus}$ or $\Lambda^{\ominus}$, will be replaced
by offspring of the reproducing individual. Note that the restriction to
probability measures is uniquely for the simplicity of the presentation. We
will explain in Subsection 3 below how the construction of the model easily
generalises to finite measures $\Lambda^{\oplus},\Lambda^{\ominus}$. We stress
that not only the construction of the model but all results presented below,
in particular the construction of the ancestral selection graph, also hold for
finite measures, with no changes to the proofs.

######  Definition 2.1 ($\Lambda$-asymmetric Moran model).

At rate 1, an individual chosen uniformly among the current population of size
$N$ reproduces. If at time $t$ individual $i$ reproduces and has type
$\tau(i,t)\in\\{\oplus,\ominus\\}$, then the selective strength of the
reproductive event is provided as a random variable $Y$ sampled independently
of everything else from the probability measure $\Lambda^{\tau(i,t)}$.
Conditional on $Y,$ each of the $N-1$ individuals in $[N]\setminus\\{i\\}$
participates in the reproduction event with probability $Y,$ meaning that the
participating individuals die and are replaced by offspring of the reproducing
individuals, carrying the type of individual $i.$

$\bm{\ominus}$$\bm{\ominus}$$\bm{\oplus}$$\bm{\ominus}$$\bm{\ominus}$$\bm{\ominus}$$\bm{\ominus}$$\bm{\oplus}$$\bm{\ominus}$$\bm{\ominus}$$\bm{\oplus}$$\bm{\oplus}$

Figure 1: A realisation of the $\Lambda$-asymmetric frequency process. Arrows
point from the reproducing individual to their offspring. In the first
reproductive event, a type $\oplus$ reproduces, in the second one a type
$\ominus$ individual. The role of selection in this construction will be
discussed later in Section 3, cf. also Figure 2.

Our model is closely related to the _Moran model with viability selection_
introduced in [12]. There, individuals reproduce at fixed rate $\lambda$ and
produce a number of offspring, among which only some of the children survive
to maturity. The probability of an individual of type $\tau$ to have $b$
mature offspring is given as $p_{\tau b}=\sum_{a=b}^{N-1}r_{a}v_{\tau ab},$
with $r_{a}$ the number of children, and $v_{\tau ab}$ the probability that
$b$ out of $a$ children survive to maturity. In particular $v_{\tau
ab}=\int_{[0,1]}\binom{a}{b}p^{b}(1-p)^{a-b}\nu_{\tau}(dp)$ for some
probability measure $\nu_{\tau}$ may be considered. This corresponds to our
model if we set $\Lambda^{\tau}=\lambda\nu_{\tau}$ and $r_{N-1}=1.$ Thus the
model of [12] is a priori more general, however, their methods require the
process to be in stationarity, and thus necessarily needs to incorporate
mutations. Our construction doesn’t need any stationarity assumption and works
for any initial conditions. Further, our construction always leads to a
consistent and exchangeable process. In the model of [12] this is not the case
if one takes for example $r_{a}=\delta_{k}$ for some fixed $k\in\mathbb{N}$
for all $a$, which would violate consistency. In view of [35], see also [28],
we expect our model to cover all consistent and exchangeable Moran models with
coordinated selection in terms of two finite $\Lambda$-measures.
Exchangeability plays a crucial role in the construction of the ancestral
selection graph for our model below. In our notation the $\Lambda-$asymmetric
Moran model doesn’t directly lead to an interpretation in terms of viability
selection. On the other hand, we can at least implicitly incorporate various
selective mechanisms in our finite measures $\Lambda^{\tau}.$ See also Remark
5.1 below for an interpretation of the $\Lambda$-measures in some special
cases.

The ancestral process of the $\Lambda$-asymmetric Moran model that will be
derived in Section 4 is similar to the one studied in [3], where the
probability of fixation for a related $\Lambda-$Moran model involving Beta
measures is calculated under stationarity assumptions. A notable difference is
the fact that the ancestral process of [3] has linear branching rates,which is
not in general the case for our model and for the model of [12]. A similar
process also appears for Moran models in a one-sided random environment, [8,
25]. In [2] an interesting construction of the ancestral selection graph for a
Moran model with frequency dependent selection is provided, leading to a
duality with respect to the same function as for our model, see Corollary 4.6.
Moran models with other types of $\Lambda$-reproduction undergoing selection
but with a different role of the $\Lambda$ measure have also been investigated
by Schweinsberg [42, 43] and studied further by Bah and Pardoux and Ged [4,
18].

We denote by

$X^{N}_{t}=\frac{1}{N}\sum_{i=1}^{N}1_{\\{\tau((t,i))=\ominus\\}},\qquad t\geq
0,$

the frequency at time $t$ of individuals of type $\ominus$, which will later
be the less fit type. Clearly the process $X^{N}=(X_{t}^{N})_{t\geq 0}$ is a
piecewise continuous Markov chain with state space
$S_{N}:=\\{0,1/N,...,(N-1)/N,1\\}$. Its transitions are given by

$x\mapsto\begin{cases}x+\frac{k}{N}&\text{ at rate
}x\int_{0}^{1}\binom{(1-x)N}{k}y^{k}(1-y)^{(1-x)N-k}\Lambda^{\ominus}(dy),\quad
k=1,...,(1-x)N\\\ x-\frac{k}{N}&\text{ at rate
}(1-x)\int_{0}^{1}\binom{xN}{k}y^{k}(1-y)^{xN-k}\Lambda^{\oplus}(dy),\quad
k=1,...,xN.\end{cases}$

Observe that atoms at 0 of $\Lambda^{\oplus}$ and $\Lambda^{\ominus}$ don’t
influence these rates. In words, upon a reproduction event of a type $\ominus$
individual, which happens at rate $x$, the strength $y$ of the reproduction is
determined according to the measure $\Lambda^{\ominus}$. Then, independently
with probability $y,$ each of the $N-1$ non-reproducing individuals dies and
is replaced by an offspring of type $\ominus$. With probability $1-y$ it
remains. Only a replacement of a type $\oplus$ individual by an offspring of
type $\ominus$ leads to an increase in type $\ominus$ individuals, thus there
are $(1-x)N$ individuals that may swap type from $\oplus$ to $\ominus$ at such
an event. The number of individuals switching type from $\oplus$ to $\ominus$
is thus binomial with success probability $y$ and $N(1-x)$ trials. A
reproduction of a type $\oplus$ individuals works vice versa. See Figure 1 for
an example.

Hence, the generator of the frequency process of the $\Lambda$-asymmetric
Moran model acts on bounded measurable functions $f:[0,1]\to[0,1]$ as

$\displaystyle\mathcal{B}^{N}f(x)$
$\displaystyle=x\mathbb{E}\left[f\left(x+\frac{1}{N}\text{Binom}(N(1-x),Y^{\ominus})\right)-f(x)\right]$
$\displaystyle+(1-x)\mathbb{E}\left[f\left(x-\frac{1}{N}\text{Binom}(Nx,Y^{\oplus})\right)-f(x)\right].$
(1)

Here, $x\in S_{N}$, and the expectation is taken with respect to the random
variables $Y^{\ominus}$ resp. $Y^{\oplus}$, which are distributed according to
the probability measures $\Lambda^{\ominus}$ resp. $\Lambda^{\oplus}$.

##  3 Stochastic ordering and the $\Lambda$-asymmetric ancestral selection
graph

The next aim is to construct an ancestral selection graph corresponding to the
$\Lambda$-asymmetric Moran model defined in the previous section. We will give
a graphical representation in terms of Poisson Point processes which forward
in time provides a construction of the $\Lambda$-asymmetric Moran model, and
backward in time gives the ancestry of a sample of such a population. Our
construction differs from the graphical representation of Etheridge, Griffiths
and Taylor in several ways. In particular, the construction works for any
assignment of types to the individuals, and it doesn’t need an invariant
distribution. It thus provides a pathwise construction of the duality found in
[12], see Corollary 4.6 below.

As a trade-off, we need the measures $\Lambda^{\ominus}$ and
$\Lambda^{\oplus}$ to be stochastically ordered. This assumption will allow us
to couple the two measures, a fact that will play a crucial role in the
construction. This order is readily interpreted in terms of the selective
advantage of the reproduction mechanisms: An individual of type $\oplus$ tends
to have more offspring at a selective event than an individual of type
$\ominus$. See also the discussion in Section 5 below. Moreover, it can easily
be generalised to finite measures.

We recall that two probability measures $\mu_{0}$ and $\mu_{1}$ on $[0,1]$ are
_stochastically ordered,_ we write $\mu_{0}\leq\mu_{1}$, if for every bounded
increasing function $f:[0,1]\to\mathbb{R}$ we have

$\int_{[0,1]}f(x)\mu_{0}(dx)\leq\int_{[0,1]}f(x)\mu_{1}(dx).$

This provides a partial order on the set of probability measures on $[0,1]$.
Note that by the usual measure theoretic approximation in terms of step
functions, this is equivalent to requiring $\mu_{0}[x,1]\leq\mu_{1}[x,1]$ for
every $x\in[0,1].$

The crucial step in our construction is the following coupling lemma. For
stochastically ordered measures $\Lambda^{\ominus}$ and $\Lambda^{\oplus}$, in
the construction of the $\Lambda-$asymmetric Moran model we can equivalently
consider a Moran model constructed from just one measure $\Lambda$ that
contains the information of a particular coupling of the pair
$(\Lambda^{\ominus},\Lambda^{\oplus}).$

######  Lemma 3.1 (Coupling lemma).

Let $\Delta=\\{(y,z)\in[0,1]^{2}:y+z\in[0,1]\\}$ and consider two probability
measures $\Lambda^{\oplus},\Lambda^{\ominus}$ on $[0,1]$ such that
$\Lambda^{\ominus}\leq\Lambda^{\oplus}$ in the stochastic partial order. Then
there exists a finite measure $\Lambda$ on $\Delta$ such that for all
$A,B\in\mathcal{B}([0,1]),$

$\Lambda^{\ominus}(A)=\Lambda(\\{(y,z):y\in A\\})\quad\text{ and
}\quad\Lambda^{\oplus}(B)=\Lambda(\\{(y,z):y+z\in B\\})$

hold.

###### Proof.

By stochastic ordering we know that
$\Lambda^{\ominus}[0,x]-\Lambda^{\oplus}[0,x]\geq 0$ for any $x\in[0,1]$.
Therefore by a version of Strassen’s Theorem (see (3) in [33]) there exists
two random variables $Z^{\ominus}$ and $Z^{\oplus}$ on a probability space
such that $Z^{\ominus}\sim\Lambda^{\ominus}(\cdot)$ and
$Z^{\oplus}\sim\Lambda^{\oplus}(\cdot),$ and $Z^{\ominus}\leq Z^{\oplus}$ a.s.
Here $\sim$ means that the random variable is distributed according to the
respective probability measure. Moreover, the joint distribution of
$(Z^{\ominus},Z^{\oplus})$, is given as

$\displaystyle\mathbb{P}\left((Z^{\ominus},Z^{\oplus})\in
A\right)=\mathbb{P}\left((F_{\ominus}^{-1}(U),F_{\oplus}^{-1}(U))\in
A\right),\qquad A\in\mathcal{B}([0,1]^{2}),$

where $F_{\ominus}$ (resp. $F_{\oplus}$) is the distribution function of the
random variable $Z^{\ominus}$ (resp. $Z^{\oplus}$) and $U$ is a unifom random
variable on $[0,1]$. Hence, by setting

$\displaystyle\Lambda(A):=\mathbb{P}\left((Z^{\ominus},(Z^{\oplus}-Z^{\ominus}))\in
A\right),\qquad\text{for any $A\in\mathcal{B}(\Delta)$,}$

we note that for any $B\in\mathcal{B}([0,1])$ we have
$\Lambda(B\times[0,1])=\mathbb{P}\left((Z^{\ominus},Z^{\oplus}-Z^{\ominus})\in
B\times[0,1]\right)=\mathbb{P}(Z^{\ominus}\in B)=\Lambda^{\ominus}(B)$ and
$\Lambda\left(\\{(y,z):y+z\in
A\\}\right)=\mathbb{P}\left(Z^{\ominus}+Z^{\oplus}-Z^{\ominus}\in
A\right)=\mathbb{P}\left(Z^{\oplus}\in A\right)=\Lambda^{\oplus}(A).$ ∎

By the coupling lemma, the probability measure $\rho$ on $[0,1]^{2}$ defined
by

$\displaystyle\rho(A\times B)=\Lambda(\\{(y,z):y\in A,y+z\in B\\}),\qquad
A,B\in\mathcal{B}([0,1]),$

is a monotone coupling of $\Lambda^{\ominus}$ and $\Lambda^{\oplus}$ (that is
$\rho\\{(y,z):y>z\\}=0$). We sometimes (by a slight abuse of notation) refer
to $\Lambda$ as the _selective coupling_ of $\Lambda^{\ominus}$ and
$\Lambda^{\oplus}$. We will usually apply it in the form

$\displaystyle\int_{\Delta}f(y)\Lambda(dy,dz)=\int_{[0,1]}f(y)\Lambda^{\ominus}(dy),\
\text{and}\
\int_{\Delta}f(y+z)\Lambda(dy,dz)=\int_{[0,1]}f(z)\Lambda^{\oplus}(dz)$ (2)

for measurable functions $f:[0,1]\mapsto[0,1]$. In Section 5 we explain how to
apply this Lemma in the case of finite measures $\Lambda^{\oplus}$ and
$\Lambda^{\ominus}$ and provide a small caveat on equation (2) in this case.
Moreover, we provide a brief discussion of uniqueness questions, and an
interpretation in terms of optimal transport.

$\bm{\ominus}$$\bm{\ominus}$$\bm{\oplus}$$\bm{\ominus}$$\bm{\ominus}$$\bm{\ominus}$$\bm{\ominus}$$\bm{\oplus}$$\bm{\ominus}$$\bm{\ominus}$$\bm{\oplus}$$\bm{\oplus}$

Figure 2: The realisation of the $\Lambda$-asymmetric frequency process from
Figure 1 now in terms of the coupling construction. Neutral arrows are black,
selective arrows grey. Individuals of type $\oplus$ may reproduce through any
arrow, individuals of type $\ominus$ only through neutral arrows. Therefore
some of the arrows and the last reproductive event weren’t present in Figure
1, where the measures applied to determine the reproduction depended on the
type of the reproducing individuals.

With this coupling, we will construct the $\Lambda$-asymmetric ancestral
selection graph. In order to make the construction more transparent, we first
provide an alternative description of the $\Lambda-$asymmetric Moran model,
using the stochastic partial order. Assume that the measures
$\Lambda^{\oplus}$ and $\Lambda^{\ominus}$ satisfy
$\Lambda^{\ominus}\leq\Lambda^{\oplus}$. Let $\Lambda$ be its selective
coupling in the sense of Lemma 3.1. This means that the random variable $Y$
that was sampled in Definition 2.1 according to either $\Lambda^{\oplus}$ or
$\Lambda^{\ominus},$ depending on the type of the reproducing individual, can
now be sampled instead as a pair $(Y,Z)$ according to $\Lambda$ from $\Delta,$
such that individuals are replaced with probability $Y$ if a type $\ominus$
individual reproduces, and individuals are replaced with probability $Y+Z$ if
a type $\oplus$ individual reproduces. The rates of the frequency process can
thus be rewritten as (cf. (2))

$x\mapsto\begin{cases}x+\frac{k}{N}&\text{ at rate
}x\int_{\Delta}\binom{(1-x)N}{k}y^{k}(1-y)^{(1-x)N-k}\Lambda(dy,dz),\quad
k=1,...,(1-x)N\\\ x-\frac{k}{N}&\text{ at rate
}(1-x)\int_{\Delta}\binom{xN}{k}(y+z)^{k}(1-(y+z))^{xN-k}\Lambda(dy,dz),\quad
k=1,...,Nx.\end{cases}$

The generator (2) becomes

$\displaystyle\mathcal{B}^{N}f(x)$
$\displaystyle=x\int_{\Delta}\mathbb{E}\left[f\left(x+\frac{1}{N}\text{Binom}(N(1-x),y)\right)-f(x)\right]\Lambda(dy,dz)$
$\displaystyle+(1-x)\int_{\Delta}\mathbb{E}\left[f\left(x-\frac{1}{N}\text{Binom}(Nx,y+z)\right)-f(x)\right]\Lambda(dy,dz).$
(3)

We are now ready to define the central object of this paper, which is the
ancestral selection graph for the $\Lambda-$asymmetric Moran model. We give a
construction in the spirit of Neuhauser and Krone in terms of a Poisson
process driven by the selective coupling $\Lambda.$

######  Definition 3.2 ($\Lambda-$asymmetric ancestral selection graph, ASG).

Consider a Poisson processes $M^{N}$ with values in
$\mathbb{R}_{+}\times[N]\times\Delta\times[0,1]^{N}$ and intensity measure
$dt\times dm\times\Lambda(dy,dz)\times du_{1}\times du_{2}...\times du_{N}$,
where $dm$ denotes the uniform measure on $[N]$. Each point
$(t,i)\in\mathbb{R}\times[N]$, represents the $i$-th individual alive at time
$t$.

  * 1.

We say that at time $t$ there is a _neutral arrow_ between $i$ and $j$ if
there is a point $(t,i,y,z,u_{1},u_{2},...,u_{N})\in M^{N}$ such that
$u_{j}\in[0,y]$.

  * 2.

We say that at time $t$ there is a _selective arrow_ between $i$ and $j$ if
there is a point $(t,i,y,z,u_{1},u_{2},...,u_{N})\in M^{N}$ such that
$u_{j}\in[y,y+z]$.

The ancestral selection graph is then given by
$(\mathbb{R}_{+}\times[N],M^{N}).$

In the graphical representation of the ancestral selection graph, as usual,
individuals are represented by lines, and reproductive events are denoted by
arrows, where the arrow starts at the reproducing individual, and the tip
points to the line of (potential) offspring. Here we represent the reproducing
individuals by filled dots, the neutral arrows are black and the selective
arrows are grey, see Figure 2. Neutral and selective arrows may occur at the
same time. Observe that formally we can have both neutral and selective arrows
from $i$ to itself, but they won’t have an effect on the frequency process.
The ancestral selection graph may be constructed using any coupling that
satisfies the conditions of Lemma 3.1. In section 5 we show that the expected
number of selective arrows will be the same under any such coupling, while the
Strassen coupling applied in the proof of the coupling Lemma minimises the
variance of the number of these arrows.

From the graphical representation and the Poisson point process construction
we can derive the frequency process. Heuristically, at an arrival of the
Poisson process, the reproducing individual $i$ is chosen uniformly at random,
and the reproductive strength is determined by the measure $\Lambda.$ The
uniform values $u_{j}$ then determine whether individual $j$ participates in a
neutral or selective reproduction event. After introducing types, type
$\oplus$ will be propagated (that is, the individual at the tip ot the arrow
will be replaced) both through neutral and selective arrows, while type
$\ominus$ will be propagated only through neutral arrows. Hence we can
construct the frequency process from the ASG by distributing types at time
$t=0,$ and propagating them forward in time according to the arrows on the
ASG.

######  Proposition 3.3.

The frequency process constructed from the $\Lambda$-asymmetric ancestral
selection graph by propagating type $\oplus$ through both neutral and
selective arrows, and type $\ominus$ through neutral arrows only, is the same
as the frequency process of the $\Lambda$-asymmetric Moran model.

We will give a more formal definition of the frequency process and a proof of
this result in section 4 below.

##  4 Ancestral process, sampling function and duality

In this section we will formalize the connection between the
$\Lambda$-asymmetric Moran model and the ancestral selection graph from
Definition 3.2. We will also define the ancestral line counting process and
state its pathwise sampling duality with the frequency process.

We start by introducing suitable forward and backward filtrations. For any
subset of individuals $S\subset[N]$ and any time $T\in\mathbb{R}_{+}$ we will
write $S_{T}=\\{(T,i):i\in S\\}.$ We think of $S_{T}$ as a (random) sample of
individuals taken at time $T$. We will follow the construction detailed in
[25] and define the forward and backward filtrations of our ancestral
selection graph.

######  Definition 4.1.

Let $(\mathbb{R}_{+}\times[N],M^{N})$ be the $\Lambda$-asymmetric ancestral
selection graph, and let $S_{T}$ be a sample as defined above.

  * 1.

Let $\mathcal{F}^{T}_{t}$ denote the sigma algebra generated by $S_{T}$ and
the restriction of $M^{N}$ to $[T,T+t]$. Then
$\\{\mathcal{F}^{T}_{t}\\}_{t\geq 0}$ is the _forward filtration_.

  * 2.

Let $\mathcal{P}^{T}_{t}$ denote the sigma algebra generated by $S_{T}$ and
the restriction of $M^{N}$ to $[T,T-t]$. Then
$\\{\mathcal{P}^{T}_{t}\\}_{t\geq 0}$ is the _backward filtration_.

The ancestral selection graph is particularly useful as a construction that
can be used both forward and backward in time. In order to utilize this, we
introduce the notion of ancestral path in the ancestral selection graph, which
will go backward in time.

######  Definition 4.2.

Let $t,s\in\mathbb{R}_{+},i,j\in[N].$ An _ancestral path_ going from $(t+s,i)$
to $(t,j)$ is the graph of a càdlàg function $f:[t,t+s]\mapsto[N]$ such that:

  1. 1.

$f(t)=j$, $f(t+s)=i$.

  2. 2.

If $f(u^{-})=k\neq f(u)=l$ then there is an arrow (neutral or selective) in
the ancestral selection graph $M^{N}$ from $(u,k)$ to $(u,l)$.

  3. 3.

If there is an arrow from $(u,k)$ to $(u,l)$, then $f(u^{-})\neq l$.

We say that an individual $(t,j)$ is a _potential ancestor_ of $(t+s,i)$ if
there exists an ancestral path going from $(t+s,i)$ to $(t,j)$ and in this
case we write $(t,j)\sim(t+s,i)$.

Graphically, the concept of potential ancestor and of ancestral paths is
described in Figure 3.

***

$\bm{\oplus}$$\bm{\ominus}$$\bm{\ominus}$$\bm{\oplus}$

Figure 3: On the left: The ancestral line process of a sample of three
individuals (marked by an asterisk), using the realisation of the
$\Lambda$-asymmetric ancestral selection graph from Figure 2 and going
backward in time (right to left). Lines of potential ancestors are colored in
black. At a reproductive event, lines from individuals hit by a black
(neutral) arrow are alway merged with the reproducing line at the origin of
the arrow, where for those hit by a grey (selective) arrow, both lines are
continued. On the right: The ancestry of one individual. The type of this
individual is $\oplus$ if one potential ancestor has type $\oplus$, and
$\ominus$ otherwise.

Note that, as opposed to the Moran model, an individual in this model can have
many potential ancestors at any given time in its past. Moreover, recall that
we introduced neutral and selective arrows, which correspond to the measures
$\Lambda^{\oplus}$ and $\Lambda^{\ominus}$ in our original definition of the
$\Lambda$-asymmetric Moran model, where individuals of type $\oplus$ have a
selective advantage by reproducing according to $\Lambda^{\oplus}$ and
individuals of type $\ominus$ reproduce according to $\Lambda^{\ominus}$, with
$\Lambda^{\ominus}\leq\Lambda^{\oplus}.$ This means that if we assign type
$\ominus$ to all individuals of a sample $S_{T}$ taken at time $T$, and type
$\oplus$ to the individuals of $S_{T}^{c}$, then an individual $i$ at time
$T+s$ is of type $\oplus$ if and only if there exists an ancestral path from
$(T+s,i)$ to at least one individual in $S^{c}_{T}.$ In this case, we write
$\tau((s,i))=\oplus$. Otherwise, if all the potential ancestors of $(T+s,i)$
at time $T$ belong to $S_{T}$, we write $\tau((T+s,i))=\ominus$.

We therefore formally define the _frequency process in the
$\Lambda-$asymmetric ancestral selection graph_ by assigning types
$\ominus,\oplus$ to all individuals at time $T>0,$ and denote by
$X^{N,T}=\\{X^{N,T}_{t}:t\geq T\\}$ the frequency of type $\ominus$
individuals at time $t>T$. Thus,

$X^{N,T}_{t}=\frac{1}{N}\sum_{i=1}^{N}1_{\\{\tau((t,i))=\ominus\\}},\qquad
t\geq T,$

where $\tau(t,i)$ is constructed as explained above using the ancestral lines
of the ASG. By construction, it is straightforward to check that the process
$(X_{t}^{N,T})_{t\geq T}$ satisfies

$\displaystyle X^{N,T}_{t}$
$\displaystyle=X^{N,T}_{T}+\sum_{i=1}^{N}\int_{T}^{t}\int_{\Delta}\int_{[0,1]^{N}}1_{\\{\tau(s,i)=\ominus\\}}\frac{1}{N}\sum_{j=1}^{N}1_{\\{\tau(s,j)=\oplus,u_{j}\leq
y\\}}M^{N}(di,ds,dy,dz,du)$
$\displaystyle-\sum_{i=1}^{N}\int_{T}^{t}\int_{\Delta}\int_{[0,1]^{N}}1_{\\{\tau(s,i)=\oplus\\}}\frac{1}{N}\sum_{j=1}^{N}1_{\\{\tau(s,j)=\ominus,u_{j}\leq
y+z\\}}M^{N}(di,ds,dy,dz,du),\quad t\geq T.$ (4)

######  Proposition 4.3.

Fix $x\in[0,1]$ and assume that $X^{N,T}_{T}=x$, then the process $X^{N,T}$ is
a $\\{\mathcal{F}^{T}_{t}\\}_{t\geq 0}$ measurable continuous-time Markov
chain with values in $S_{N}$, and its infinitesimal generator
$\mathcal{B}^{N,T}$ is given for any $f\in\mathcal{C}^{2}([0,1])$ by

$\displaystyle\mathcal{B}^{N,T}f(x)$
$\displaystyle=x\int_{\Delta}\mathbb{E}\left[f\left(x+\frac{1}{N}\textup{Binom}(N(1-x),y)\right)-f(x)\right]\Lambda(dy,dz)$
$\displaystyle+(1-x)\int_{\Delta}\mathbb{E}\left[f\left(x-\frac{1}{N}\textup{Binom}(Nx,y+z)\right)-f(x)\right]\Lambda(dy,dz).$
(5)

###### Proof.

The first statement of the proposition follows from (4). Now, using (4) and an
application of Itô’s formula we obtain for any $f\in\mathcal{C}^{2}([0,1])$
and $t\geq T$

$\displaystyle f(X^{N,T}_{t})=f(X_{T}^{N,T})$
$\displaystyle+\sum_{i=1}^{N}\int_{T}^{t}\int_{\Delta}\int_{[0,1]^{N}}\Bigg{[}f\left(X_{s-}^{N,T}+1_{\\{\tau(s,i)=\ominus\\}}\frac{1}{N}\sum_{j=1}^{N}1_{\\{\tau(s,j)=\oplus,u_{j}\leq
y\\}}\right)-f(X_{s-}^{N,T})\Bigg{]}M^{N}(di,ds,dy,dz,du)$
$\displaystyle+\sum_{i=1}^{N}\int_{T}^{t}\int_{\Delta}\int_{[0,1]^{N}}\Bigg{[}f\left(X_{s-}^{N,T}-1_{\\{\tau(s,i)=\oplus\\}}\frac{1}{N}\sum_{j=1}^{N}1_{\\{\tau(s,j)=\ominus,u_{j}\leq
y+z\\}}\right)-f(X_{s-}^{N,T})\Bigg{]}M^{N}(di,ds,dy,dz,du).$

Taking expectations in the previous identity gives for $t\geq T$

$\displaystyle\mathbb{E}\left[f(X^{N,T}_{t})\right]=\mathbb{E}[f(X_{T}^{N,T})]$
$\displaystyle+\mathbb{E}\Bigg{[}\int_{T}^{t}\int_{\Delta}\int_{[0,1]^{N}}\frac{1}{N}\sum_{i=1}^{N}1_{\\{\tau(s,i)=\ominus\\}}\Bigg{[}f\left(X_{s-}^{N,T}+\frac{1}{N}\sum_{j=1}^{N}1_{\\{\tau(s,j)=\oplus,u_{j}\leq
y\\}}\right)-f(X_{s-}^{N,T})\Bigg{]}ds\Lambda(dy,dz)du$
$\displaystyle+\mathbb{E}\Bigg{[}\int_{T}^{t}\int_{\Delta}\int_{[0,1]^{N}}\frac{1}{N}\sum_{i=1}^{N}1_{\\{\tau(s,i)=\oplus\\}}\Bigg{[}f\left(X_{s-}^{N,T}-\frac{1}{N}\sum_{j=1}^{N}1_{\\{\tau(s,j)=\ominus,u_{j}\leq
y+z\\}}\right)-f(X_{s-}^{N,T})\Bigg{]}ds\Lambda(dy,dz)du$
$\displaystyle=\mathbb{E}[f(X_{T}^{N,T})]+\mathbb{E}\Bigg{[}\int_{T}^{t}\int_{\Delta}X_{s-}^{N,T}\Bigg{[}f\left(X_{s-}^{N,T}+\frac{1}{N}\textup{Binom}(N(1-X_{s-}^{N,T}),y)\right)-f(X_{s-}^{N,T})\Bigg{]}ds\Lambda(dy,dz)\Bigg{]}$
$\displaystyle+\mathbb{E}\Bigg{[}\int_{T}^{t}\int_{\Delta}\int_{[0,1]^{N}}(1-X_{s-}^{N,T})\Bigg{[}f\left(X_{s-}^{N,T}-\frac{1}{N}\textup{Binom}(NX_{s-}^{N,T},y+z)\right)-f(X_{s-}^{N,T})\Bigg{]}ds\Lambda(dy,dz)\Bigg{]}.$
(6)

Finally, by differentiating (4) and taking $t=T$ we obtain (4.3). ∎

###### Proof of Proposition 3.3.

The generator $\mathcal{B}^{N,T}$ has no true dependence on $T$, but only on
the cardinality of the sample $S_{T}$ of individuals of type $\ominus$, which
gives the initial frequency $|S_{T}|/N$ of individuals of type $\ominus$.
Comparing $\mathcal{B}^{N,T}$ and $\mathcal{B}^{N}$ shows that they are equal,
hence the processes they generate are equal in distribution. ∎

We may thus identify the two frequency processes $(X^{N}_{t})_{t\geq 0}$ and
$(X_{t}^{N,T})_{t\geq T}$ with $X^{N}_{0}=|S_{T}|/N,$ and call it the
frequency process associated with the $\Lambda$-asymmetric ancestral selection
graph.

One of the big advantages of the $\Lambda$-ancestral selection graph is that
it also allows for a $\\{\mathcal{P}^{T}_{t}\\}_{t\geq 0}$ measurable
ancestral process, the ancestral line counting process, which satisfies a
duality relation with the frequency process.

######  Definition 4.4 (Ancestral process).

Fix a sample $S_{T}$ of individuals at some time $T\in\mathbb{R}$. The
_ancestral line counting process_ $(A^{N,T}_{t})_{0\leq t\leq T}$ of a sample
taken at time $T$ is given by

$A^{N,T}_{t}=\sum_{i=1}^{N}1_{\\{(T-t,i)\sim(T,j),\text{ for some }(T,j)\in
S_{T}\\}},\qquad t\in[0,T].$

Hence by definition $A_{t}^{N,T}$ counts the number of potential ancestors
living at time $t\in[0,T]$ of individuals from the sample $S_{T}$.

######  Proposition 4.5.

The process $(A^{N,T}_{t})_{0\leq t\leq T}$ is a $\\{\mathcal{P}_{t}\\}_{t\geq
0}$ measurable continuous-time Markov chain with values in $[N]$ starting at
$A_{0}^{N,T}=|S_{T}|$ and transition rates

$n\mapsto\begin{cases}n-k&\text{ at rate
}\frac{n}{N}\int_{\Delta}\binom{n-1}{k}y^{k}(1-y)^{n-1-k}\Lambda(dy,dz)\\\
&\hskip
36.98866pt+(1-\frac{n}{N})\int_{\Delta}\binom{n}{k+1}y^{k+1}(1-y)^{n-k-1}\Lambda(dy,dz),\quad
k=1,...,n-1\\\ n+1&\text{ at rate
}(1-\frac{n}{N})\int_{\Delta}[(1-y)^{n}-(1-y-z)^{n}]\Lambda(dy,dz).\end{cases}$

Since again the process doesn’t depend on $T$ but only on $|S_{T}|,$ we drop
the $T$ from the notation and call $(A^{N}_{t})_{t\geq 0}$ the ancestral line
counting process or process of potential ancestors. For probability measures
(and more generally when $\|\Lambda^{\oplus}\|=\|\Lambda^{\ominus}\|$) we can
apply (2) and rewrite the rates in terms of the original measures:

$n\mapsto\begin{cases}n-k&\text{ at rate
}\frac{n}{N}\int_{[0,1]}\binom{n-1}{k}y^{k}(1-y)^{n-1-k}\Lambda^{\ominus}(dy)\\\
&\hskip
36.98866pt+(1-\frac{n}{N})\int_{[0,1]}\binom{n}{k+1}y^{k+1}(1-y)^{n-k-1}\Lambda^{\ominus}(dy),\quad
k=1,...,n-1,\\\ n+1&\text{ at rate
}(1-\frac{n}{N})\left[\int_{[0,1]}(1-y)^{n}\Lambda^{\ominus}(dy)-\int_{[0,1]}(1-z)^{n}\Lambda^{\oplus}(dz)\right].\end{cases}$

Hence we can write the dual block counting process and the duality also in
terms of the original measures. In the block counting process, the effect of
the selective advantage of $\Lambda^{\oplus}$ is reduced to the question of
whether there is at least one selective arrow coming from outside the sample
at a time where there is no neutral arrow occurring. However, we would like to
stress that for the construction of the ancestral selection graph and
therefore also for the _pathwise_ construction of the duality the existence of
a monotone coupling is crucial.

###### Proof.

Measurability is clear, as well as the initial value. In order to understand
the rates, we consider the graphical representation and go backward from
sampling time $T.$ If we have currently $n$ potential ancestors, then the next
event backward in time is one of the following three cases (cf. Figure 4):

  1. 1.

From one of the $n$ potential ancestors, $k$ neutral arrows emerge and hit $k$
of the remaining $n-1$ lines of potential ancestors. In that case, the $k$
lines coalesce with the reproducing line, reducing the number of potential
ancestors by $k.$ According to the construction of the ancestral selection
graph, this happens at rate
$\frac{n}{N}\int_{\Delta}\binom{n-1}{k}y^{k}(1-y)^{n-1-k}\Lambda(dy,dz),$
leading to the first line in the above transition rates. There might be
selective arrows at the same event, but these don’t change the number of
potential ancestors, since in that case both lines continue to be potential
ancestors.

  2. 2.

From one of the $N-n$ lines that don’t currently belong to the set of
potential ancestors there are neutral arrows to $k\geq 2$ out of the $n$
current potential ancestors. In that case those $k$ lines merge, but the line
of the reproducing individual is added to the set of potential ancestors. This
happens at rate
$(1-\frac{n}{N})\int_{\Delta}\binom{n}{k}y^{k}(1-y)^{n-k}\Lambda(dy,dz).$

  3. 3.

From one of the $N-n$ lines outside the current set of potential ancestors
there is a selective arrow to at least one of the $n$ individuals from the
current set of ancestors, while at the same time there are no neutral arrows
to any of the $n$ current ancestors in the sample. In that case, the
reproducing line is added to the set of potential ancestors, while all the
other lines remain. This happens at rate
$(1-\frac{n}{N})\int_{\Delta}[(1-y)^{n}-(1-y-z)^{n}]\Lambda(dy,dz),$ since
$(1-y)^{n}-(1-y-z)^{n}$ is the probability that there are no neutral arrows to
individuals of the ancestal process, while there is at least one selective
arrow.

There are no further events in the graphical construction that could change
the number of potential ancestors. ∎

Figure 4: The three possible types of transitions of the ancestral process. On
the left, the reproducing individuals belongs to the current sample(thick
black lines), there is one neutral (black) and two selctive (grey) arrows. The
line of the individual at the tip of the neutral arrow is discarded, resp.
merges with the reproducing line, while the lines hit by a selective arrow
remain, as well as those lines not affected by the reproduction event. In the
middle, the reproducing individuals doesn’t belong to the current sample,
there are two neutral and one selctive arrows. The line of the reproducing
individual is thus added, the lines of the individuals at the end of a neutral
arrow are discarded, resp. merge with the reproducing line. On the right, only
selective arrows are present, where the incoming lines are kept, but the
reproducing line is added. We therefore see a branching.

For $x\in S_{N}$, $n\in[N]$, and $t\geq 0$ we define the _sampling function_
$S_{t}(x,n)$ as follows. Let $\\{u_{i}\\}_{i\in[n]}$ be a uniformly sampled
subset of $[N]$ of size $n$, and let $S_{t}(x,n)$ be the probability that all
individuals in the sample $S_{t}=\\{(u_{i},t)\\}$ are of type $\ominus$
conditional on the initial frequency of type $\ominus$ individuals at time $0$
is $x$. Note that for $t=0$

$S_{0}(x,n)=\prod_{i=1}^{n}\frac{Nx+1-i}{N+1-i}=\frac{\binom{Nx}{n}}{\binom{N}{n}}.$
(7)

Observe that for fixed $n$ and large $N$ is of order $x^{n}+O(1/N),$ a fact
that we will use in Proposition 6.5 below.

Now, using the graphical representation, for any $t>0$ we can write
$S_{t}(x,n)$ in terms of $S_{0}(x,n)$ and $A^{N}_{t}$ started at
$A_{0}^{N}=n$. Recall that an individual at time $t$ has type $\ominus$ if and
only if all its potential ancestors at time $0$ have type $\ominus$. Thus the
probability that $n$ individuals sampled at time $t$ is given in terms of the
number of potential ancestors of these individuals, and we obtain

$S_{t}(x,n)=\mathbb{E}[S_{0}(x,A^{N}_{t})\,|\,A_{0}^{N}=n]=\mathbb{E}_{n}[S_{0}(x,A^{N}_{t})].$

Here, $\mathbb{E}$ denotes the expectation with respect to the law of the
ancestral selection graph, and $\mathbb{E}_{n}$ the expectation with respect
to the Markov chain $(A_{t}^{N})_{t\geq 0}$ started at $n.$ Similarly, we can
write $S_{t}(x,n)$ in terms of $S_{0}(x,n)$ and $X^{N}_{t}$ started at
$X_{0}^{N}=x$ as

$S_{t}(x,n)=\mathbb{E}[S_{0}(X^{N}_{t},n)\,|\,X_{0}=x]=\mathbb{E}_{x}[S_{0}(X^{N}_{t},n)].$

Hence we have shown

######  Corollary 4.6.

The processes $(X_{t}^{N})_{t\geq 0}$ and $(A_{t}^{N})_{t\geq 0}$ are dual
with respect to the duality function $S_{0},$ that is,

$\mathbb{E}_{x}[S_{0}(X^{N}_{t},n)]=\mathbb{E}_{n}[S_{0}(x,A^{N}_{t})]\quad\forall
t\geq 0,x\in S_{N},n\in\mathbb{N}.$

Duality with respect to the sampling function $S_{0}$ is a classical concept
which goes back to Cannings and Gladstien [6, 19, 20, 21] and was studied
extensively by Möhle [34]. The same type of duality in a similar model appears
in [2].

##  5 Generalisations of the model and some remarks on the coupling

In this section we show how the construction can be extended to finite
measures, which also extends the interpretation of selective advantage encoded
in the stochastic order. Further, we provide some comments concerning the
question of uniqueness of the coupling.

### Extension to finite measures

Assume now that $\Lambda^{\oplus}$ and $\Lambda^{\ominus}$ are finite measures
on $[0,1]$ satisfying

$\displaystyle\Lambda^{\ominus}[x,1]\leq\Lambda^{\oplus}[x,1]$ (8)

for all $x\in[0,1].$ We thus write again by analogy to the stochastic order
$\Lambda^{\ominus}\leq\Lambda^{\oplus}.$ Note that also in this case this is a
partial order on the set of finite measures. Write $\|\Lambda^{\oplus}\|$ and
$\|\Lambda^{\ominus}\|$ for their respective total masses. Now define

$\mu^{\oplus}:=\frac{\Lambda^{\oplus}}{\|\Lambda^{\oplus}\|}\qquad\text{and}\qquad\mu^{\ominus}:=\frac{\Lambda^{\ominus}}{\|\Lambda^{\oplus}\|}+c\delta_{0},$
(9)

where $c=1-\frac{\|\Lambda^{\ominus}\|}{\|\Lambda^{\oplus}\|}$ and
$\delta_{0}$ denotes the Dirac measure at 0. Then clearly $\mu^{\ominus}$ and
$\mu^{\oplus}$ are probability measures, and we have
$\mu^{\ominus}\leq\mu^{\oplus}$ in the stochastic order if and only if
$\Lambda^{\ominus}\leq\Lambda^{\oplus}$ in the sense of (8).

We can define the $\Lambda$-asymmetric Moran model for finite measures
$\Lambda^{\ominus},\Lambda^{\oplus}$ exactly as in Definition 2.1. The rates
of the frequency process then become

$x\mapsto\begin{cases}x+\frac{k}{N}&\text{ at rate
}x\|\Lambda^{\oplus}\|\int_{0}^{1}\binom{(1-x)N}{k}y^{k}(1-y)^{(1-x)N-k}\mu^{\ominus}(dy),\quad
k=1,...,(1-x)N\\\ x-\frac{k}{N}&\text{ at rate
}(1-x)\|\Lambda^{\oplus}\|\int_{0}^{1}\binom{xN}{k}y^{k}(1-y)^{xN-k}\mu^{\oplus}(dy),\quad
k=1,...,xN,\end{cases}$

since an atom at 0 doesn’t give a contribution. Thus the $\Lambda$-Moran model
with finite measures $\Lambda^{\oplus},\Lambda^{\ominus}$ is a simple linear
time change of the Model with probability measures
$\mu^{\oplus},\mu^{\ominus}$ according to the above transformation.
Reproductive events now happen at rate $\|\Lambda^{\oplus}\|.$ The ancestral
selection graph can be constructed by using the selective coupling $\mu$ of
$\mu^{\ominus},\mu^{\oplus}$ and then running the Poisson point process of
Definition 3.2 with the coupling measure $\|\Lambda^{\oplus}\|\cdot\mu$ in the
place of $\Lambda$. Hence all results in this paper also hold for finite
measures provided they are ordered in the above sense. One should simply be
slightly careful when applying the identity (2), since whenever $c\neq 0,$
that is, $\|\Lambda^{\ominus}\|<\|\Lambda^{\oplus}\|,$ this only holds for
functions $f:[0,1]\to\mathbb{R}$ such that $f(0)=0.$ The measure $\Lambda$ via
(2) determines $\Lambda^{\oplus}$ and $\Lambda^{\ominus}$ only up to possible
atoms at 0.

######  Remark 5.1.

When considering finite measures, we see in particular that
$\Lambda^{\ominus}\leq\Lambda^{\oplus}$ in the following two prototypical
cases of selective mechanisms:

  1. 1.

(Faster reproduction) if $\Lambda^{\oplus}=(1+\alpha)\Lambda^{\ominus}$ for
some $\alpha>0$.

  2. 2.

(Bigger reproductive events) There exists a function $s:[0,1]\mapsto[0,1]$
such that $s(x)-x\geq 0$ and $\Lambda^{\ominus}(s(A))=\Lambda^{\oplus}(A)$.

In the first case, in particular
$\|\Lambda^{\oplus}\|=(1+\alpha)\|\Lambda^{\ominus}\|,$ hence we are in the
situation of the classical ancestral selection graph by Krone and Neuhauser
[31, 36] where the type with a selective advantage reproduces faster than the
other. The second case is satisfied for example if
$\Lambda^{\ominus}=\delta_{a}$ and $\Lambda^{\oplus}=\delta_{b},$ with $0\leq
a\leq b\leq 1$ where $\Lambda^{\ominus}$ generates reproductive events of size
$a$, and $\Lambda^{\oplus}$ of size $b.$

### Non-uniqueness of the coupling

We now turn to a discussion of the coupling from Lemma 3.1. A natural question
to ask is whether the coupling $\Lambda$ we use in our construction of the
ancestral selection graph is unique. As we saw, there is always some freedom
in adding atoms at 0. However, also without atoms at 0 the coupling is not
unique, as can be seen from a simple example.

######  Example 5.2.

Consider
$\Lambda^{\ominus}(dx)=\frac{1}{2}(\delta_{1/4}(dx)+\delta_{1/2}(dx))$ and
$\Lambda^{\oplus}(dx)=\frac{1}{3}(\delta_{1/2}(dx)+\delta_{3/4}(dx)+\delta_{1}(dx)).$
Clearly these are probability measures on $[0,1]$ satisfying
$\Lambda^{\ominus}\leq\Lambda^{\oplus}$. Let

$\displaystyle\Lambda_{1}(dy,dz)=\frac{1}{6}\sum_{i=1}^{2}\sum_{j=2}^{4}\delta_{(i/4,(j-i)/4)}(dy,dz),$

and

$\displaystyle\Lambda_{2}(dy,dz)$
$\displaystyle=\frac{1}{4}\delta_{(1/4,1/4)}(dy,dz)+\frac{1}{12}\delta_{(1/2,0)}(dy,dz)+\frac{1}{8}(\delta_{(1/4,1/2)}(dy,dz)+\delta_{(1/4,3/4)}(dy,dz))$
$\displaystyle+\frac{5}{24}(\delta_{(1/2,1/4)}(dy,dz)+\delta_{(1/2,1/2)}(dy,dz)).$

It is straightforward to check that $\Lambda_{1}$ and $\Lambda_{2}$ are two
couplings of $\Lambda^{\ominus}$ and $\Lambda^{\oplus}$ in the sense of Lemma
3.1.

In the ancestral selection graph, we can use any coupling $\Lambda$ satisfying
the conditions of Lemma 3.1. By (2) this always leads to the same frequency
process, and at least as long as $\|\Lambda^{\oplus}\|=\|\Lambda^{\ominus}\|$,
also the rates of the ancestral line counting process remain the same. The
choice of the specific coupling is therefore not important on the level of
these processes. However, the fact that at least one selective coupling exists
is of course crucial in the construction of the ASG, and therefore pathwise
construction of the duality. It is also important in Section 7.

In spite of this non-uniqueness, Strassen’s theorem as used in the proof of
Lemma 3.1 provides us with a characterisation of the coupling $\rho$
introduced after Lemma 3.1. In fact, as we will argue, it minimizes the
variance of the number of potential ancestors in the construction of the
$\Lambda$-asymmetric ancestral graph given in Definition 3.2 in the case in
which $\Lambda^{\ominus}$ and $\Lambda^{\oplus}$ are both probability
measures. To be more precise, denote for $n\in\mathbb{N}$.

$c(y,z):=(z-y)^{2},\qquad y,z\in[0,1]^{2},$

and consider the following optimal transport problem, consisting in finding a
probability measure $\gamma^{*}$ on $[0,1]^{2}$ such that the following
infimum is achieved

$\displaystyle
V(n,\Lambda^{\ominus},\Lambda^{\oplus}):=\inf\left\\{\int_{[0,1]^{2}}c(y,z)\gamma(dy,dz):\gamma\in\Gamma(\Lambda^{\ominus},\Lambda^{\oplus})\right\\},$
(10)

where $\Gamma(\Lambda^{\ominus},\Lambda^{\oplus})$ is the set of probability
measures on $[0,1]^{2}$ with marginals $\Lambda^{\ominus}$, $\Lambda^{\oplus}$
on $[0,1]$. We first note that by construction

$\displaystyle\int_{\Delta}z^{2}\Lambda(dy,dz)$
$\displaystyle=\int_{[0,1]^{2}}\left(z-y\right)^{2}\rho(dy,dz)=\mathbb{E}\left[c(Z^{\ominus},Z^{\oplus})\right],$
(11)

where $Z^{\ominus},Z^{\oplus}$, given in the proof of Lemma 3.1, satisfy that
$(Z^{\ominus},Z^{\oplus})=(F_{\ominus}^{-1}(U),F_{\oplus}^{-1}(U))$.

On the other hand, the function $c$ is symmetric and satisfies the ‘Monge”
conditions, i.e. $c$ is continuous and satisfies that for $y^{\prime}\geq y$
and $z^{\prime}\geq z$

$\displaystyle
c(y^{\prime},z^{\prime})-c(y,z^{\prime})-c(y^{\prime},z)+c(y,z)\leq 0.$

Hence, as in Section 7.1 in [40] we have that the solution to the optimization
problem given in (10) is given by a measure $\gamma^{*}$ with distribution
$F^{*}(y,z):=\min\\{F_{\ominus}(y),F_{\oplus}(z)\\}$ for $y,z\in[0,1]^{2}$, or
in terms of random variables, $\gamma^{*}$ is the distribution function of the
random vector $(F_{\ominus}^{-1}(U),F_{\oplus}^{-1}(U))$, where $U$ a uniform
random variable. Comparing with (11) we see that $\gamma^{*}=\rho$ and
therefore the measure $\rho$ from our coupling is the solution to the
minimisation problem (10) i.e.

$V(n,\mu_{1},\mu_{2})=\int_{[0,1]^{2}}\left(z-y\right)^{2}\rho(dy,dz),$

for each $n\in\mathbb{N}$. To arrive at the interpretation in terms of
selective arrows, first observe that by definition of the ancestral selection
graph, the number of selective arrows for any coupling $\Lambda$ is given by

$\int_{\Delta}z\Lambda(dy,dz)=\int_{[0,1]^{2}}\left(z-y\right)\rho(dy,dz),$

which is a constant, since by (2)

$\int_{\Delta}z\Lambda(dy,dz)=\int_{[0,1]}(y+z)\Lambda^{\oplus}(dz)-\int_{[0,1]}y\Lambda^{\ominus}(dy):=C.$

Therefore the expected number of selective arrows associated with any coupling
in the class $\Gamma(\Lambda^{\ominus},\Lambda^{\oplus})$ is constant and
equal to $C$. On the other hand, the variance of the number of selective
arrows is given by

$\int_{\Delta}z^{2}\Lambda(dy,dz)-C^{2}=\int_{[0,1]^{2}}\left(z-y\right)^{2}\rho(dy,dz)-C^{2}.$

Hence minimising $\int_{[0,1]^{2}}\left(z-y\right)^{2}\gamma(dy,dz)$ in the
class $\Gamma(\Lambda^{\ominus},\Lambda^{\oplus})$ also minimises this
variance.

##  6 Scaling Limits

In this section we will study the scaling limit of the frequency process
associated to the $\Lambda$-asymmetric ancestral selection graph introduced in
Section 3. We start by showing that the limiting object is well defined, which
is the content of the next result. We defer the proof to A.

######  Proposition 6.1.

Assume that $\Lambda$ is a measure on
$\Delta=\\{(y,z)\in[0,1]^{2}:y+z\in[0,1]\\}$ such that

$\displaystyle\int_{\Delta}(y^{2}+z)\Lambda(dy,dz)<\infty.$ (12)

Then, for any $x\in[0,1]$ there exists a unique strong solution
$Y=(Y_{t})_{t\geq 0}$ to the following stochastic differential equation

$\displaystyle Y_{t}$
$\displaystyle=x+\int_{0}^{t}\int_{0}^{1}\int_{\Delta}\left(y(1-Y_{s-})1_{\\{u\leq
Y_{s-}\\}}-(y+z)Y_{s-}1_{\\{u\geq
Y_{s-}\\}}\right)1_{\\{Y_{s-}\in[0,1]\\}}\tilde{N}(ds,du,dy,dz)$
$\displaystyle-\int_{\Delta}z\Lambda(dy,dz)\int_{0}^{t}(1-Y_{s-})Y_{s-}1_{\\{Y_{s-}\in[0,1]\\}}ds,$
(13)

where $N$ is a Poisson random measure on $(0,\infty)\times[0,1]\times\Delta$
with intensity measure $dt\times du\times\Lambda(dy,dz)$ and
$\tilde{N}(ds,du,dy,dz):=N(ds,du,dy,dz)-dsdu\Lambda(dy,dz)$ denotes the
compensated random measure associated to $N$. We refer to the process $Y$ as
the _$\Lambda$ -asymmetric frequency process_.

######  Remark 6.2.

Note that if the stronger condition $\int_{\Delta}(y+z)\Lambda(dy,dz)<\infty$,
holds, then the process $Y$, solution to the SDE given in (6.1), can be
described as the solution to the simpler SDE given by

$\displaystyle
Y_{t}=x+\int_{0}^{t}\int_{\Delta}\int_{0}^{1}\left(y(1-Y_{s-})1_{\\{u\leq
Y_{s-}\\}}-(y+z)Y_{s-}1_{\\{u\geq
Y_{s-}\\}}\right)1_{\\{Y_{s-}\in[0,1]\\}}N(ds,du,dy,dz).$

In the following result we derive the convergence to the process $Y$, of the
frequency processes associated to the $\Lambda$-ancestral selection graph
introduced in Section 3. In doing so we need to take some care in dealing with
the possible singularity at 0 of the measure $\Lambda$. To this end, we will
apply a truncation procedure at 0, and study a sequence of frequency processes
$Y^{N}$ that correspond to the truncated $\Lambda$-measures.

Fix a measure $\Lambda$ on $\Delta$ such that
$\int_{\Delta}(y^{2}+z)\Lambda(dy,dz)<\infty$, an initial frequency
$x\in[0,1]$ and a number $\alpha\in(0,1/2)$. For each $N\in\mathbb{N}$ we
define the truncated frequency process $Y^{N}=(Y_{t}^{N})_{t\geq 0}$ with
initial value $Y^{N}_{0}=\lfloor xN\rfloor/N$ as the frequency process of an
asymmetric $\Lambda$-Moran model with reproduction mechanism
$\Lambda^{(N)}(A):=\int_{\Delta^{N}}1_{A}(y,z)\Lambda(dy,dz)$, where
$\Delta^{N}:=\\{(y,z)\in\Delta:y^{2}>1/N^{\alpha}\\}$.

Note that the measure $\Lambda^{(N)}$ is a finite measure on $\Delta$, indeed
by the definition of $\Delta^{N}$ we have

$\Lambda^{(N)}(\Delta)=\Lambda(\Delta^{N})\leq
N^{\alpha}\int_{\Delta}(y^{2}+z)\Lambda(dy,dz)<\infty.$

Hence $Y^{N}$ is well defined. We will now show the convergence of the
sequence of process $(Y^{N})_{N\in\mathbb{N}}$ as the total size of the
population $N$ grows to infinity to the $\Lambda$-asymmetric frequency process
$Y$ defined in Proposition 6.1. The proof is deferred to B.

######  Proposition 6.3.

The sequence $(Y^{N})_{N\in\mathbb{N}}$ converges weakly in
$\mathbb{D}([0,T],[0,1])$ to a limit $Y$, where $Y$ is the
$\Lambda$-asymmetric frequency process given in (6.1).

Let $A^{N}:=(A^{N}_{t})_{t>0}$ denote the ancestral process associated to the
frequency process $Y^{N}$ defined in Proposition 6.3. In the next result we
show the convergence of the sequence of processes $(A^{N})_{N\in\mathbb{N}}$
as the total size of the population $N$ grows to infinity.

######  Proposition 6.4.

The sequence $(A^{N})_{N\in\mathbb{N}}$ converges weakly in
$\mathbb{D}([0,T],\mathbb{N})$ to a limit $A=(A_{t})_{t\geq 0}$, where $A$ is
a continuous-time Markov chain with values in $\mathbb{N}$, whose generator
has the following transition rates

$m\mapsto\begin{cases}m-k+1&\text{ at rate
}\int_{\Delta}\binom{m}{k}y^{k}(1-y)^{m-k}\Lambda(dy,dz),\quad k=2,...,m\\\
m+1&\text{ at rate
}\int_{\Delta}[(1-y)^{m}-(1-y-z)^{m}]\Lambda(dy,dz).\end{cases}$

###### Proof.

Fix $m\in\mathbb{N}$. Using that $|(1-y)^{m}-(1-y-z)^{m}|\leq mz$, for
$y,z\in\Delta$ we obtain that (12) together with dominated convergence gives

$\displaystyle\lim_{N\to\infty}\left(1-\frac{m}{N}\right)\int_{\Delta^{N}}\left[(1-y)^{m}-(1-y-z)^{m}\right]\Lambda(dy,dz)=\int_{\Delta}\left[(1-y)^{m}-(1-y-z)^{m}\right]\Lambda(dy,dz).$

On the other hand, for $k\geq 2$, by (12) and dominated convergence

$\displaystyle\lim_{N\to\infty}\left(1-\frac{m}{N}\right)\int_{\Delta^{N}}\binom{m}{k}y^{k}(1-y)^{m-k}\Lambda(dy,dz)=\int_{\Delta}\binom{m}{k}y^{k}(1-y)^{m-k}\Lambda(dy,dz).$

By the definiton of $\Delta^{N}$ we note that dominated convergence gives

$\displaystyle\lim_{N\to\infty}\frac{m}{N}\int_{\Delta^{N}}my(1-y)^{m-1}\Lambda(dy,dz)\leq\lim_{N\to\infty}m^{2}\frac{1}{N^{1-\alpha/2}}\int_{\Delta^{N}}y^{2}\Lambda(dy,dz)=0.$

Similarly, for $k\geq 2$, proceeding as in the previous identities we get

$\displaystyle\lim_{N\to\infty}\frac{m}{N}\int_{\Delta^{N}}\binom{m}{k}y^{k}(1-y)^{m-k}\Lambda(dy,dz)=0.$

Therefore, by Problem 3(ii) in [14] we obtain the result. ∎

As expected we show in the next result that the processes $Y$ and $A$ are
moment duals. To this end, we note that the infinitesimal generator
$\mathcal{A}$ of the process $A$ is given for any
$f:\mathbb{N}\mapsto\mathbb{N}$ by

$\displaystyle\mathcal{A}f(n)$
$\displaystyle=\sum_{k=1}^{n}(f(n-k+1)-f(n))\int_{\Delta}\binom{n}{k}y^{k}(1-y)^{n-k}\Lambda(dy,dz)$
$\displaystyle+(f(n+1)-f(n))\int_{\Delta}[(1-y)^{n}-(1-y-z)^{n}]\Lambda(dy,dz).$
(14)

######  Proposition 6.5.

Consider the $\Lambda$-asymmetric frequency process $Y$ defined in Proposition
6.1 and the ancestral process $A$ with rates given in Proposition 6.4. Then
the processes $Y$ and $A$ are moment duals, i.e. for any $x\in[0,1]$ and
$n\in\mathbb{N}$

$\mathbb{E}_{x}\left[Y_{t}^{n}\right]=\mathbb{E}_{n}\left[x^{A_{t}}\right],\qquad\text{for
all $t\geq 0$}.$

###### Proof.

We will consider $\mathbb{N}$ endowed with the discrete topology and
$\mathbb{N}\times[0,1]$ with the product topology. We denote for every fixed
$x\in[0,1]$, $H(n,x)=x^{n}$, which is bounded and continuous. In addition, for
every fixed $k\in\mathbb{N}$, $H(k,x)=x^{k}$ is continuous. Therefore, we
conclude that $H:\mathbb{N}\times[0,1]\mapsto[0,1]$ is continuous.

We observe that $H(\cdot,n)$ is a polynomial in $[0,1]$ for fixed
$n\in\mathbb{N}$. This fact clearly implies that
$H(\cdot,n)\in\mathcal{C}^{2}([0,1])$ and hence it lies in the domain of the
generator $\mathcal{B}$ of the $\Lambda$-asymmetric frequency process $Y$ as
in (43). Therefore, the process

$H(n,Y_{t})-\int_{0}^{t}\mathcal{B}H(n,Y_{s})ds,$

is a martingale.

Additionally, we have that for fixed $x\in[0,1]$ the function $H(\cdot,x)$
lies in the domain of the generator $\mathcal{A}$, which implies that the
process

$H(A_{t},x)-\int_{0}^{t}\mathcal{A}H(A_{s},x)ds,$

is also a martingale.

We will compute $\mathcal{B}H(n,x)$ for $x\in[0,1]$ and $n\in\mathbb{N}$, then
using (43)

$\displaystyle\mathcal{B}H(n,x)=\int_{\Delta}\left[x(x+y(1-x))^{n}+(1-x)(x-(y+z)x)^{n}-x^{n}\right]\Lambda(dy,dz),\
x\in[0,1],n\in\mathbb{N}.$ (15)

We note that

$\displaystyle
x(x+y(1-x))^{n}=x\sum_{k=0}^{n}\binom{n}{k}y^{k}x^{n-k}(1-y)^{n-k}=x^{n+1}(1-y)^{n}+\sum_{k=1}^{n}\binom{n}{k}y^{k}x^{n-k+1}(1-y)^{n-k}.$

Additionaly,

$\displaystyle(1-x)(x-(y+z)x)^{n}=(x^{n}-x^{n+1})(1-y-z)^{n},$

and finally

$\displaystyle
x^{n}=x^{n}(1-y)^{n}+x^{n}\sum_{k=1}^{n}\binom{n}{k}y^{k}(1-y)^{n-k}.$

Hence, using the previous identities gives

$\displaystyle x(x+y(1-x))^{n}+(1-x)(x-(y+z)x)^{n}-x^{n}$
$\displaystyle=\sum_{k=2}^{n}\binom{n}{k}y^{k}(1-y)^{n-k}(x^{n-k+1}-x^{n})$
$\displaystyle+(x^{n+1}-x^{n})\left[(1-y)^{n}-(1-y-z)^{n}\right].$ (16)

Using (6) in (15) we obtain

$\displaystyle\mathcal{B}H(n,x)$
$\displaystyle=\sum_{k=2}^{n}(x^{n-k+1}-x^{n})\int_{\Delta}\binom{n}{k}y^{k}(1-y)^{n-k}\Lambda(dy,dz)$
$\displaystyle+(x^{n+1}-x^{n})\int_{\Delta}\left[(1-y)^{n}-(1-y-z)^{n}\right]\Lambda(dy,dz)=\mathcal{A}H(n,x).$

Finally an application of Proposition 5 in [7] gives the result. ∎

##  7 Griffiths representation and the probability of fixation

We motivate our next result by recalling a result for the two-type
$\Lambda$-Fleming Viot process, which is defined as the solution to the SDE

$\displaystyle Z_{t}$
$\displaystyle=y+\int_{0}^{t}\int_{0}^{1}\int_{\Delta}\left(y(1-Z_{s-})1_{\\{u\leq
Z_{s-}\\}}-yZ_{s-}1_{\\{u\geq
Z_{s-}\\}}\right)1_{\\{Z_{s-}\in[0,1]\\}}\tilde{N}(ds,du,dy)$
$\displaystyle-\int_{0}^{t}s(1-Z_{s-})Z_{s-}1_{\\{Z_{s-}\in[0,1]\\}}ds,\quad
t\geq 0.$

Independently and using different techniques, Foucart [16] and Griffiths [27]
obtained explicit conditions for the two-type $\Lambda$-Fleming Viot process
to be absorbed at 0 almost surely, or to be absorbed in either of the
boundaries $\\{0,1\\}$ with positive probability. In particular, they observed
that for every $\Lambda$ it is always possible to find a small enough
selection parameter $s$ such that
$\mathbb{P}(\lim_{t\rightarrow\infty}Z_{t}=1)\mathbb{P}(\lim_{t\rightarrow\infty}Z_{t}=0)>0$.
Foucart uses a duality technique which relies on the observation that the
$\Lambda$-Fleming Viot process can be absorbed at $1$ if and only if its dual,
that we denote $(F_{t})_{t\geq 0}$, is positive recurrent. The process
$(F_{t})_{t\geq 0}$ coalesces like the block counting process of the
$\Lambda$-coalescent and branches at rate $ns$. In our model, the dual process
is given by the ancestral line counting process $(A_{t})_{t\geq 0}$. It
coalesces at the same rate and branches at rate
$\int_{0}^{1}\int_{0}^{1}\left[(1-y)^{n-1}-(1-y-z)^{n-1}\right]\Lambda(dy,dz)$
which is smaller than $ns$ for all $n>n(s)$, for some $n(s)\gg 1$, which
exists for every $s$. This implies that $(F_{t})_{t\geq 0}$ reflected in
$n(s)$ stochastically dominates $(A_{t})_{t\geq 0}$. In turn this implies that
$(A_{t})_{t\geq 0}$ is positive recurrent. A duality argument allows to
conclude that the limiting frequency process $(Y_{t})_{t\geq 0}$ in equation
(6.1) fullfils
$\mathbb{P}(\lim_{t\rightarrow\infty}Y_{t}=1)\mathbb{P}(\lim_{t\rightarrow\infty}Y_{t}=0)>0$.

Here, instead of formalising the previous argument, we will exploit Griffiths’
technique in order to compute a semi-explicit expression for the probability
of fixation of type $\ominus$ individuals in a $\Lambda$-asymmetric frequency
process. To this end, we will start by obtaining an alternative expression for
its generator $\mathcal{B}$, given in (43), following the ideas of Theorem 1
in [27]. This is given in the next result. We define the measure
$\tilde{\Lambda}$ on the simplex $\Delta$ by
$\tilde{\Lambda}(dy,dz):=(y^{2}+z)\Lambda(dy,dz).$

######  Proposition 7.1.

Let $\mathcal{B}$ be the infinitesimal generator of a $\Lambda$-asymmetric
frequency process as in (43). Let $U,V,Y,Z$ be independent random variables,
such that $V$ has a uniform distribution on $[0,1]$, $U$ has density $2u$ with
respect to the Lebesgue measure, and $(Y,Z)$ is distributed according to
$\tilde{\Lambda}/\|\tilde{\Lambda}\|$. Then, for any
$f\in\mathcal{C}^{2}([0,1])$,

$\displaystyle\mathcal{B}f(x)=\frac{\|\tilde{\Lambda}\|}{2}x(1-x)\mathbb{E}\left[f^{\prime\prime}(x(1-UY)+UYV)\frac{Y^{2}}{Z+Y^{2}}-2f^{\prime}(x-xY-
xZV)\frac{Z}{Z+Y^{2}}\right].$

###### Proof.

Using that $V$ has a uniform distribution on $[0,1]$ we obtain that

$\displaystyle\mathbb{E}\left[f^{\prime\prime}(x(1-UY)+UYV)\frac{Y^{2}}{Z+Y^{2}}\right]$
$\displaystyle=\mathbb{E}\left[\frac{Y^{2}}{Z+Y^{2}}\int_{0}^{1}f^{\prime\prime}(x(1-UY)+UYv)dv\right]$
$\displaystyle=\mathbb{E}\left[\frac{Y}{Z+Y^{2}}\left[f^{\prime}(x(1-UY)+UY)-f^{\prime}(x(1-UY))\right]\frac{1}{U}\right]$
$\displaystyle=\mathbb{E}\left[\frac{2}{Z+Y^{2}}\left(\frac{f(x(1-Y)+Y)-f(x)}{1-x}+\frac{f(x(1-Y))-f(x)}{x}\right)\right]$
$\displaystyle=\frac{2}{\|\tilde{\Lambda}\|}\int_{\Delta}\left[\frac{f(x(1-y)+y)-f(x)}{1-x}+\frac{f(x(1-y))-f(x)}{x}\right]\frac{\tilde{\Lambda}(dy,dz)}{z+y^{2}}$
(17)

where in the third equality we used that $U$ has density given by $2u$ on
$[0,1]$.

In a similar way, we obtain

$\displaystyle\mathbb{E}\left[\frac{Z}{Z+Y^{2}}f^{\prime}(x-xY-xZV)\right]$
$\displaystyle=-\mathbb{E}\left[\frac{Z}{Z+Y^{2}}\left(f(x-xZ-
xY)-f(x-xY)\right)\frac{1}{xZ}\right]$
$\displaystyle=-\frac{1}{\|\tilde{\Lambda}\|}\int_{\Delta}\left[\frac{f(x-xy-
xz)-f(x-xy)}{x}\right]\frac{\tilde{\Lambda}(dy,dz)}{z+y^{2}}.$ (18)

Therefore, using (7) and (7) we obtain for $x\in[0,1]$

$\displaystyle\frac{\|\tilde{\Lambda}\|}{2}$ $\displaystyle
x(1-x)\mathbb{E}\left[f^{\prime\prime}(x(1-UY)+UYV)\frac{Y^{2}}{Z+Y^{2}}-2f^{\prime}(x-xZV)\frac{Z}{Z+Y^{2}}\right]$
$\displaystyle=\int_{\Delta}\left[x(f(x(1-y)+y)-f(x))+(1-x)(f(x(1-y))-f(x))\right]\frac{\tilde{\Lambda}(dy,dz)}{z+y^{2}}$
$\displaystyle+\int_{\Delta}(1-x)\left[f(x(1-y-z))-f(x(1-y))\right]\frac{\tilde{\Lambda}(dy,dz)}{z+y^{2}}$
$\displaystyle=\int_{\Delta}\left[x(f(x(1-y)+y)-f(x))+(1-x)(f(x(1-y-z))-f(x))\right]\frac{\tilde{\Lambda}(dy,dz)}{z+y^{2}}$
$\displaystyle=\mathcal{B}f(x).$

∎

Denote by $p(x)$ for $x\in[0,1]$ the probability of fixation of type $\ominus$
in the $\Lambda$-asymmetric frequency process, where $x$ denotes the initial
frequency of type $\ominus$ individuals. Since $p$ is a harmonic function for
the generator $\mathcal{B}$, we see from the previous proposition that it
satisfies for $x\in(0,1)$

$\displaystyle\mathbb{E}\left[p^{\prime\prime}(x(1-UY)+UYV)\frac{Y^{2}}{Z+Y^{2}}-2p^{\prime}(x-xY-
xZV)\frac{Z}{Z+Y^{2}}\right]=0,$ (19)

with boundary conditions given by $p(0)=0$ and $p(1)=1$.

We note that

$\displaystyle\mathbb{E}\left[p^{\prime\prime}(x(1-UY)+UYV)\frac{Y^{2}}{Z+Y^{2}}\right]=\mathbb{E}\left[\frac{p^{\prime}(x(1-W)+W)-p^{\prime}(x(1-W))}{W}\frac{Y^{2}}{Z+Y^{2}}\right],$
(20)

with $W:=UY$.

In order to find the solution to (19), following [27], we will consider
polynomials of the form

$\displaystyle h_{n}(x)=\sum_{r=0}^{n}a_{n,r}x^{r},$ (21)

with $h_{0}(x)=1$.

Then, we will prove that we can take a choice of coefficients
$\\{a_{n,r}\\}_{0\leq r\leq n}$ such that

$\displaystyle\mathbb{E}\left[\frac{h_{n}(x(1-W)+W)-h_{n}(x(1-W))}{W}\frac{Y^{2}}{Z+Y^{2}}\right]=n\mathbb{E}\left[h_{n-1}\left(x(1-Y-ZV)\right)\frac{Z}{Z+Y^{2}}\right].$
(22)

with the choice

$a_{n,n}=\prod_{i=1}^{n-1}\frac{\mathbb{E}\left[(1-Y-ZV)^{i}\displaystyle\frac{Z}{Z+Y^{2}}\right]}{\mathbb{E}\left[(1-W)^{i}\displaystyle\frac{Y^{2}}{Z+Y^{2}}\right]},\qquad
n=0,1\dots$ (23)

for the diagonal elements. Indeed, by (21)

$\displaystyle\mathbb{E}\left[\frac{h_{n}(x(1-W)+W)-h_{n}(x(1-W))}{W}\frac{Y^{2}}{Z+Y^{2}}\right]$
$\displaystyle=\mathbb{E}\left[\frac{Y^{2}}{Z+Y^{2}}\sum_{r=0}^{n}\frac{a_{n,r}}{W}\left[(x(1-W)+W)^{r}-(x(1-W))^{r}\right]\right]$
$\displaystyle=\sum_{r=1}^{n}a_{n,r}\sum_{j=0}^{r-1}\binom{r}{j}\mathbb{E}\left[(1-W)^{j}W^{r-j-1}\frac{Y^{2}}{Z+Y^{2}}\right]x^{j}$
$\displaystyle=\sum_{j=0}^{n-1}\sum_{r=j+1}^{n}\binom{r}{j}\mathbb{E}\left[(1-W)^{j}W^{r-j-1}\frac{Y^{2}}{Z+Y^{2}}\right]a_{n,r}x^{j}.$

Hence in order for (22) to hold we need that

$\displaystyle\sum_{j=0}^{n-1}\sum_{r=j+1}^{n}\binom{r}{j}\mathbb{E}\left[(1-W)^{j}W^{r-j-1}\frac{Y^{2}}{Z+Y^{2}}\right]a_{n,r}x^{j}=\sum_{j=0}^{n-1}na_{n-1,j}\mathbb{E}\left[(1-Y-ZV)^{j}\frac{Z}{Z+Y^{2}}\right]x^{j},$

or equivalently that

$\displaystyle
a_{n-1,j}=\sum_{r=j+1}^{n}\binom{r}{j}\frac{\mathbb{E}\left[(1-W)^{j}W^{r-j-1}\displaystyle\frac{Y^{2}}{Z+Y^{2}}\right]}{n\mathbb{E}\left[(1-Y-ZV)^{j}\displaystyle\frac{Z}{Z+Y^{2}}\right]}a_{n,r}.$
(24)

Following the discussion in [27], we can determine the coefficients
$\\{a_{n,j}\\}_{j=0}^{n-1}$ of the polynomial $h_{n}$, from the coefficients
$\\{a_{n-1,j}\\}_{j=0}^{n}$ of the polynomial $h_{n-1}$. Indeed, we can take
$a_{n,n}$ as in (23) and then, using (24), recursively obtain $a_{n,j}$ for
$j=n-1,\dots,0$. We notice that the coefficient $a_{n,0}$ can be chosen
arbitrarly, and will be specified later in the construction of the probability
of fixation $p$.

Now let us return to the probability of fixation $p$, where for the first
derivative $p^{\prime}(x)$ we make the ansatz

$\displaystyle p^{\prime}(x)=A\sum_{n=1}^{\infty}2^{n}c_{n}h_{n-1}(x),$ (25)

with $p(1)=1$ and $p(0)=0$. Using (25) in (19), together with (20) and (22)
gives

$\displaystyle\mathbb{E}\left[\frac{p^{\prime}(x(1-W)+W)-p^{\prime}(x(1-W))}{W}\frac{Y^{2}}{Z+Y^{2}}-2p^{\prime}(x(1-Y-ZV))\frac{Z}{Z+Y^{2}}\right]$
$\displaystyle=A\sum_{n=1}^{\infty}2^{n}c_{n}(n-1)\mathbb{E}\left[h_{n-2}\left(x(1-Y-ZV)\right)\frac{Z}{Z+Y^{2}}\right]-2A\sum_{n=1}^{\infty}2^{n}c_{n}\mathbb{E}\left[h_{n-1}\left(x(1-Y-ZV)\right)\frac{Z}{Z+Y^{2}}\right].$

By chosing $c_{n}=\frac{1}{(n-1)!},$ for $n\in\mathbb{N}$, we obtain

$\displaystyle\mathbb{E}\left[\frac{p^{\prime}(x(1-W)+W)-p^{\prime}(x(1-W))}{W}\frac{Y^{2}}{Z+Y^{2}}-2p^{\prime}(x(1-Y-ZV))\frac{Z}{Z+Y^{2}}\right]$
$\displaystyle=A\sum_{n=2}^{\infty}\frac{2^{n}}{(n-2)!}\mathbb{E}\left[h_{n-2}\left(x(1-Y-ZV)\right)\frac{Z}{Z+Y^{2}}\right]-A\sum_{n=1}^{\infty}\frac{2^{n+1}}{(n-1)!}\mathbb{E}\left[h_{n-1}\left(x(1-Y-ZV)\right)\frac{Z}{Z+Y^{2}}\right]=0.$

Hence, $p$ is a solution to (19). Now, by integrating (25) we obtain that

$\displaystyle
p(x)=A\sum_{n=1}^{\infty}\int_{0}^{x}\frac{2^{n}}{(n-1)!}h_{n-1}(u)du.$

So if we choose $\\{h_{n}(0)\\}_{n\geq 1}$ such that
$\int_{0}^{1}nh_{n-1}(u)du=1,$ then by the fact that $p(1)=1$ and $p(0)=0$ we
obtain that

$\displaystyle
1=p(1)-p(0)=A\sum_{n=1}^{\infty}\frac{2^{n}}{n!}\int_{0}^{1}nh_{n-1}(u)du=A(e^{2}-1),$

and hence $A=(e^{2}-1)^{-1}$.

So putting the pieces together we have that

$p(x)=(e^{2}-1)^{-1}\sum_{n=1}^{\infty}\frac{2^{n}}{n!}H_{n}(x),$

where $H_{n}(x):=\int_{0}^{x}nh_{n-1}(u)du$, and $\\{h_{n}\\}_{n\geq 1}$
satisfies (22).

The previous discussion leads to the following main result of this section.

######  Proposition 7.2.

The fixation probability of type $\ominus$ individuals is given by

$p(x)=(e^{2}-1)^{-1}\sum_{n=1}^{\infty}\frac{2^{n}}{n!}H_{n}(x),\qquad
x\in[0,1],$

where the polynomials $\\{H_{n}\\}_{n=0}^{\infty}$ are given by

$H_{n}(x)=\int_{0}^{x}nh_{n-1}(u)du,\qquad x\in[0,1],$

with $\\{h_{n}\\}_{n=0}^{\infty}$ given in (21), (23), and (24) and the
constants $\\{h_{n}(0)\\}_{n=0}^{\infty}$ chosen so that
$\int_{0}^{1}nh_{n-1}=1.$

##  Appendix A Proof of Proposition 6.1

Let us denote for $(x,y,z,u)\in[0,1]\times\Delta\times[0,1]$

$g(x,y,z,u):=y(1-x)1_{\\{u<x\\}}-(y+z)x1_{\\{u\geq x\\}}.$

For fixed $(y,z,u)\in\Delta\times[0,1]$ we have

$x+g(x,y,z,u)=\begin{cases}y+x(1-y)&\mbox{if }u<x,\\\ x(1-(y+z))&\mbox{if
}u\geq x.\end{cases}$ (26)

First, we note that for fixed $(y,z,u)\in\Delta\times[0,1]$

$\displaystyle 0\leq x+g(x,y,z,u)=x+y(1-x)1_{\\{u<x\\}}-(y+z)x1_{\\{u\geq
x\\}}\leq 1.$

Hence, by a modification of Proposition 2.1 in [17] (see also Corollary 6.2 in
[37]) we obtain that $\mathbb{P}\left(Y_{t}\in[0,1]\ \text{for all $t\geq
0$}\right)=1$.

On the other hand, (26) implies that for fixed $(y,z,u)\in\Delta\times[0,1]$
the mapping $x\mapsto x+g(x,y,z,u)$ is non-decreasing for $x\in[0,1]$.

Denote for $x\in[0,1]$

$b(x):=\left(\int_{\Delta}z\Lambda(dy,dz)\right)x(1-x).$

Then, for $x_{1},x_{2}\in[0,1]$

$\displaystyle|b(x_{1})-b(x_{2})|=|x_{2}(1-x_{2})-x_{1}(1-x_{1})|\int_{\Delta}z\Lambda(dy,dz)\leq
2\int_{\Delta}z\Lambda(dy,dz)|x_{2}-x_{1}|.$ (27)

Similarly, for $x_{1},x_{2}\in[0,1]$

$\displaystyle|g(x_{1},y,z,u)-g(x_{2},y,z,u)|$
$\displaystyle=y(x_{2}-x_{1})1_{\\{u\leq x_{1}\wedge
x_{2}\\}}-(y(1-x_{2})+(y+z)x_{1})1_{\\{x_{1}\leq u\leq x_{2}\\}}$
$\displaystyle+(y(1-x_{1})+(y+z)x_{2})1_{\\{x_{2}\leq u\leq
x_{1}\\}}-(y+z)(x_{1}-x_{2})1_{\\{x_{1}\vee x_{2}\leq u\\}}.$ (28)

Using (A) we obtain for $x_{1},x_{2}\in[0,1]$

$\displaystyle\int_{\Delta}\int_{0}^{1}$
$\displaystyle|g(x_{1},y,z,u)-g(x_{2},y,z,u)|^{2}du\Lambda(dy,dz)$
$\displaystyle\leq\int_{\Delta}\int_{0}^{1}\Big{[}y^{2}|x_{2}-x_{1}|^{2}1_{\\{u\leq
x_{1}\wedge x_{2}\\}}+(y(1-x_{2})+(y+z)x_{1})^{2}1_{\\{x_{1}\leq u\leq
x_{2}\\}}$ $\displaystyle+(y(1-x_{1})+(y+z)x_{2})^{2}1_{\\{x_{2}\leq u\leq
x_{1}\\}}+(y+z)^{2}|x_{1}-x_{2}|^{2}1_{\\{x_{1}\vee x_{2}\leq
u\\}}\Big{]}du\Lambda(dy,dz)$ $\displaystyle\leq
6|x_{2}-x_{1}|\int_{\Delta}(y^{2}+z)\Lambda(dy,dz).$ (29)

Finally we note that for $x\in[0,1]$

$\displaystyle|b(x)|^{2}=\left(\int_{\Delta}z\Lambda(dy,dz)\right)^{2}x^{2}(1-x)^{2}\leq\left(\int_{\Delta}z\Lambda(dy,dz)\right)^{2}x^{2},$
(30)

and

$\displaystyle\int_{\Delta}\int_{0}^{1}|g(x,y,z,u)|^{2}du\Lambda(dy,dz)$
$\displaystyle=\int_{\Delta}\int_{0}^{1}\left[y^{2}(1-x)^{2}1_{\\{u<x\\}}+(y+z)^{2}x^{2}1_{\\{u\geq
x\\}}\right]du\Lambda(dy,dz)$
$\displaystyle\leq\int_{\Delta}\left[y^{2}(1-x)^{2}x+x^{2}(y+z)^{2}(1-x)\right]\Lambda(dy,dz)$
$\displaystyle\leq(1+x^{2})\int_{\Delta}(y^{2}+z)\Lambda(dy,dz).$ (31)

Therefore, there exists $K>0$ such that for $x\in[0,1]$

$\displaystyle|b(x)|^{2}+\int_{\Delta}\int_{0}^{1}|g(x,y,z,u)|^{2}du\Lambda(dy,dz)\leq
K(1+|x|^{2}).$ (32)

Hence using (27), (A) and (A) together with the fact that the mapping
$x\mapsto x+g(x,y,z,u)$ is non-decreasing for $x\in[0,1]$, we obtain by a
slight modification of Theorem 5.1 in [37] that there exists a unique strong
solution to (6.1).

##  Appendix B Proof of Proposition 6.3

We recall that the infinitesimal generator $\mathcal{B}^{N}$ of the process
$Y^{N}$ is given for any $f\in\mathcal{C}^{2}([0,1])$ by

$\displaystyle\mathcal{B}^{N}f(x)=\int_{\Delta^{N}}\Bigg{\\{}\frac{\lfloor
xN\rfloor}{N}\mathbb{E}\Bigg{[}f$ $\displaystyle\left(\frac{\lfloor
xN\rfloor}{N}+\frac{1}{N}B_{N}\right)-f\left(\frac{\lfloor
xN\rfloor}{N}\right)\Bigg{]}$ $\displaystyle+\left(1-\frac{\lfloor
xN\rfloor}{N}\right)\mathbb{E}\Bigg{[}f\left(\frac{\lfloor
xN\rfloor}{N}-\frac{1}{N}\tilde{B}_{N}\right)-f\left(\frac{\lfloor
xN\rfloor}{N}\right)\Bigg{]}\Bigg{\\}}\Lambda(dy,dz),$

where $B_{N}:=\text{Binom}\left(N\left(1-\frac{\lfloor
xN\rfloor}{N}\right);y\right)$ and $\tilde{B}_{N}:=\text{Binom}\left(\lfloor
xN\rfloor;y+z\right)$.

Now, for any $f\in\mathcal{C}^{2}([0,1])$ and $x\in[0,1]$ we have by Taylor’s
Theorem that

$\displaystyle f\left(\frac{\lfloor xN\rfloor}{N}+\frac{1}{N}B_{N}\right)$
$\displaystyle-f\left(\frac{\lfloor xN\rfloor}{N}\right)-(f(x+(1-x)y)-f(x))$
$\displaystyle=R\left(\frac{\lfloor xN\rfloor}{N},\frac{\lfloor
xN\rfloor}{N}+\frac{1}{N}B_{N}\right)-R(x,x+(1-x)y).$ (33)

where $R(u,v):=\int_{u}^{v}f^{\prime}(t)dt$.

Now, we note that

$\displaystyle R\Bigg{(}\frac{\lfloor xN\rfloor}{N},\frac{\lfloor
xN\rfloor}{N}+\frac{1}{N}B_{N}\Bigg{)}-R(x,x+(1-x)y)$
$\displaystyle\leq\int_{\frac{\lfloor
xN\rfloor}{N}}^{x}f^{\prime}(t)dt+\int_{x+(1-x)y}^{\frac{\lfloor
xN\rfloor}{N}+\frac{1}{N}B_{N}}f^{\prime}(t)dt.$ (34)

Therefore, using (34)

$\displaystyle\Bigg{|}R\Bigg{(}\frac{\lfloor xN\rfloor}{N},\frac{\lfloor
xN\rfloor}{N}+\frac{1}{N}B_{N}\Bigg{)}$ $\displaystyle-R(x,x+(1-x)y)\Bigg{|}$
$\displaystyle\leq\|f^{\prime}\|_{[0,1]}\left|x-\frac{\lfloor
xN\rfloor}{N}\right|+\|f^{\prime}\|_{[0,1]}\left|\frac{\lfloor
xN\rfloor}{N}+\frac{1}{N}B_{N}-x-(1-x)y\right|$
$\displaystyle\leq\|f^{\prime}\|_{[0,1]}\frac{1}{N}+\|f^{\prime}\|_{[0,1]}\left[\frac{1}{N}(1-y)+\left|\frac{1}{N}B_{N}-\left(1-\frac{\lfloor
xN\rfloor}{N}\right)y\right|\right],$ (35)

where $\|g\|_{[0,1]}:=\sup_{x\in[0,1]}|g(x)|$ for any
$g\in\mathcal{C}^{2}([0,1])$.

On the other hand, by the Cauchy-Schwarz inequality

$\displaystyle\mathbb{E}\left[\left|\frac{1}{N}B_{N}-\left(1-\frac{\lfloor
xN\rfloor}{N}\right)y\right|\right]\leq\left(\mathbb{E}\left[\left|\frac{1}{N}B_{N}-\left(1-\frac{\lfloor
xN\rfloor}{N}\right)y\right|^{2}\right]\right)^{1/2}=\left(\frac{1}{N}\left(1-\frac{\lfloor
xN\rfloor}{N}\right)y(1-y)\right)^{1/2},$

Hence, by taking expectations on (B), we can find a constant $C_{f}>0$ only
dependent on $f$ such that

$\displaystyle\mathbb{E}\Bigg{[}\Bigg{|}R\Bigg{(}\frac{\lfloor
xN\rfloor}{N},\frac{\lfloor
xN\rfloor}{N}+\frac{1}{N}B_{N}\Bigg{)}-R(x,x+(1-x)y)\Bigg{|}\Bigg{]}=C_{f}\left(\frac{1}{N}+\frac{1}{N^{1/2}}y^{1/2}\right).$
(36)

Hence, by (B) together with (36) there exists a constant $C_{f}>0$ only
dependent on $f$ such that

$\displaystyle\Bigg{|}f\left(\frac{\lfloor
xN\rfloor}{N}+\frac{1}{N}B_{N}\right)$ $\displaystyle-f\left(\frac{\lfloor
xN\rfloor}{N}\right)-(f(x+(1-x)y)-f(x))\Bigg{|}\leq
C_{f}\left(\frac{1}{N}+\frac{1}{N^{1/2}}y^{1/2}\right).$ (37)

Very similarly we can deal with the other term in the generator involving
$\tilde{B}_{N}$, and after the same kind of calculations as before we arrive
at

$\displaystyle\Bigg{|}f\left(\frac{\lfloor
xN\rfloor}{N}-\frac{1}{N}\tilde{B}_{N}\right)-f\left(\frac{\lfloor
xN\rfloor}{N}\right)-(f(x-x(y+z))-f(x))\Bigg{|}\leq\tilde{C}_{f}\left(\frac{1}{N}+\frac{1}{N^{1/2}}(y+z)^{1/2}\right).$
(38)

Using (37) together with (38) we obtain, for $x\in[0,1]$,

$\displaystyle\sup_{x\in[0,1]}\Bigg{|}\int_{\Delta^{N}}$
$\displaystyle\Bigg{\\{}x\mathbb{E}\Bigg{[}f\left(\frac{\lfloor
xN\rfloor}{N}+\frac{1}{N}B_{N}\right)-f\left(\frac{\lfloor
xN\rfloor}{N}\right)\Bigg{]}$ $\displaystyle\hskip
42.67912pt+(1-x)\mathbb{E}\Bigg{[}f\left(\frac{\lfloor
xN\rfloor}{N}-\frac{1}{N}\tilde{B}_{N}\right)-f\left(\frac{\lfloor
xN\rfloor}{N}\right)\Bigg{]}\Bigg{\\}}\Lambda(dy,dz)$
$\displaystyle-\int_{\Delta}\Bigg{\\{}xf(x+y(1-x))+(1-x)f(x-(y+z)x)-f(x)\Bigg{\\}}\Lambda(dy,dz)\Bigg{|}$
$\displaystyle\leq
2(C_{f}+\tilde{C}_{f})\frac{1}{N^{1/2}}\int_{\Delta^{N}}\Lambda(dy,dz)$
$\displaystyle+\sup_{x\in[0,1]}\int_{\Delta\backslash\Delta^{N}}\left|xf(x+y(1-x))+(1-x)f(x-(y+z)x)-f(x)\right|\Lambda(dy,dz).$
(39)

Expanding the integrand in the right-hand side of (B) gives

$\displaystyle|xf(x+y(1-x))$
$\displaystyle+(1-x)f(x-(y+z)x)-f(x)|=\Bigg{|}x\int_{x}^{x+y(1-x)}\frac{f^{\prime\prime}(t)}{2}(x+y(1-x)-t)dt$
$\displaystyle+(1-x)\int_{x}^{x-(y+z)x}\frac{f^{\prime\prime}(t)}{2}(x-(y+z)x-t)dt-x(1-x)zf^{\prime}(x)\Bigg{|}$
$\displaystyle\leq\frac{\|f^{\prime\prime}\|_{[0,1]}}{4}\left(x(1-x)^{2}y^{2}+(1-x)(y+z)^{2}x^{2}\right)+\|f^{\prime}\|_{[0,1]}x(1-x)z\leq
K_{f}(y^{2}+z),$ (40)

where $K_{f}$ is positive constant only dependent on $f$. Hence, using (B) in
(B)

$\displaystyle\lim_{n\to\infty}\sup_{x\in[0,1]}\Bigg{|}\int_{\Delta^{N}}\Bigg{\\{}x\mathbb{E}\Bigg{[}f$
$\displaystyle\left(\frac{\lfloor
xN\rfloor}{N}+\frac{1}{N}B_{N}\right)-f\left(\frac{\lfloor
xN\rfloor}{N}\right)\Bigg{]}$
$\displaystyle+(1-x)\mathbb{E}\Bigg{[}f\left(\frac{\lfloor
xN\rfloor}{N}-\frac{1}{N}\tilde{B}_{N}\right)-f\left(\frac{\lfloor
xN\rfloor}{N}\right)\Bigg{]}\Bigg{\\}}\Lambda(dy,dz)$
$\displaystyle-\int_{\Delta}\Bigg{\\{}xf(x+y(1-x))+(1-x)f(x-(y+z)x)-f(x)\Bigg{\\}}\Lambda(dy,dz)\Bigg{|}$
$\displaystyle\leq\lim_{n\to\infty}\left[(C_{f}+\tilde{C}_{f})\frac{1}{N^{1/2-\alpha}}\int_{\Delta}(y^{2}+z)\Lambda(dy,dz)+K_{f}\int_{\Delta\backslash\Delta^{N}}(y^{2}+z)\Lambda(dy,dz)\right]$
$\displaystyle=0.$ (41)

Finally, we note that

$\displaystyle\lim_{N\to\infty}$
$\displaystyle\sup_{x\in[0,1]}\Bigg{|}\int_{\Delta^{N}}\Bigg{\\{}\frac{\lfloor
xN\rfloor}{N}\mathbb{E}\Bigg{[}f\left(\frac{\lfloor
xN\rfloor}{N}+\frac{1}{N}B_{N}\right)-f\left(\frac{\lfloor
xN\rfloor}{N}\right)\Bigg{]}$ $\displaystyle+\left(1-\frac{\lfloor
xN\rfloor}{N}\right)\mathbb{E}\Bigg{[}f\left(\frac{\lfloor
xN\rfloor}{N}-\frac{1}{N}\tilde{B}_{N}\right)-f\left(\frac{\lfloor
xN\rfloor}{N}\right)\Bigg{]}\Bigg{\\}}\Lambda(dy,dz)$
$\displaystyle-\int_{\Delta^{N}}\Bigg{\\{}x\mathbb{E}\Bigg{[}f\left(\frac{\lfloor
xN\rfloor}{N}+\frac{1}{N}B_{N}\right)-f\left(\frac{\lfloor
xN\rfloor}{N}\right)\Bigg{]}$ $\displaystyle\hskip
42.67912pt+(1-x)\mathbb{E}\Bigg{[}f\left(\frac{\lfloor
xN\rfloor}{N}-\frac{1}{N}\tilde{B}_{N}\right)-f\left(\frac{\lfloor
xN\rfloor}{N}\right)\Bigg{]}\Bigg{\\}}\Lambda(dy,dz)\Bigg{|}$
$\displaystyle\leq\lim_{N\to\infty}2\|f\|_{[0,1]}\frac{1}{N^{1-\alpha}}\int_{\Delta}(y^{2}+z)\Lambda(dy,dz)=0.$
(42)

Following the steps of Proposition 3 in [7] we have that the infinitesimal
generator $\mathcal{B}$ of the $\Lambda$-asymmetric frequency process $Y$ is
given for any $f\in\mathcal{C}^{2}([0,1])$ and $x\in[0,1]$ by

$\displaystyle\mathcal{B}f(x)=\int_{\Delta}\left[xf(x+y(1-x))+(1-x)f(x-(y+z)x)-f(x)\right]\Lambda(dy,dz),\qquad
x\in[0,1].$ (43)

Hence, by (B) together with (B) we obtain that
$\mathcal{B}^{N}f\to\mathcal{B}f$ as $N\to\infty$ uniformly in $[0,1]$.
Therefore, Theorem 17.25 in [29] implies that $Y^{N}\to Y$ as $N\to\infty$
weakly in the space $\mathbb{D}([0,T],[0,1])$.

### Acknowledgements

The authors thank two anonymous referees for extremely helpful and
constructive comments, that immensely improved the presentation. AGC was
supported by the grant PAPIIT UNAM IN101722 “Nuevas aplicaciones de la
dualidad de momentos y de la construccion Lookdown”, and acknowledge support
from the Hausdorff Research Institute for Mathematics in Bonn where he made a
3 month research visit in the summer of 2022. The authors would like to thank
Fernanda Lopez, who wrote her Master thesis at UNAM on a simplified version of
the model discussed in this manuscript.

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