2021

[4]Peter Mörters

1]Department of Physics, The Catholic University of Korea, 43 Jibong-ro,
Bucheon-si, 14662, Gyeonggi-do, Republic of Korea

2]Institut für Biologische Physik, Universität zu Köln, Zülpicher Str. 77,
50937 Köln, Germany

3]Laboratoire de Physique de l’École Normale Supérieure, ENS, Université PSL,
CNRS, Sorbonne Université, Université Paris Cité, F-75005 Paris, France

[4]Mathematisches Institut, Universität zu Köln, Weyertal 86–90, 50931 Köln,
Germany

# Branching with selection and mutation I:
Mutant fitness of Fréchet type

Su-Chan Park<EMAIL_ADDRESS>Joachim Krug<EMAIL_ADDRESS>Léo Touzo
<EMAIL_ADDRESS><EMAIL_ADDRESS>[  [  [  *

###### Abstract

We investigate two stochastic models of a growing population subject to
selection and mutation. In our models each individual carries a fitness which
determines its mean offspring number. Many of these offspring inherit their
parent’s fitness, but some are mutants and obtain a fitness randomly sampled
from a distribution in the domain of attraction of the Fréchet distribution.
We give a rigorous proof for the precise rate of superexponential growth of
these stochastic processes and support the argument by a heuristic and
numerical study of the mechanism underlying this growth.

###### keywords:

Branching process, random environment, population model, selection, mutation,
survival of the fittest, Fréchet distribution.

##  1 Introduction

While the theory of branching processes is undoubtedly one of the best
developed areas of probability theory, stochastic branching models that
incorporate effects of selection and mutation have only recently caught the
attention of mathematicians and physicists. This is despite the unquestionable
relevance of these effects to the evolution of populations in nature and in
the laboratory Park2010 ; Wiser2013 .

By contrast, deterministic high density models of a population undergoing
selection and mutation have been studied for quite some time. The model most
closely associated with our stochastic process is Kingman’s model Kingman1978
. This is a dynamical system on the space of probability measures describing
the fitness distribution of a population. After $t$ generations the fitness
distribution $p_{t}$ of the population is replaced by

$p_{t+1}(dx)=(1-\beta)\frac{x\,p_{t}(dx)}{\int x\,p_{t}(dx)}+\beta\mu(dx).$

Here a proportion $1-\beta$ of the new generation has been selected from the
current generation proportionally to their fitness and a proportion $\beta$
are mutants that get a new fitness, sampled independently from their past
using the mutant fitness distribution $\mu$. Note that this model is only
well-defined if the mean fitness remains finite and therefore requires moment
bounds for the mutant fitness distribution. Kingman’s model undergoes a
condensation phase transition, which is further studied in Dereich2013 .
Variants of the model have been considered in Yuan ; Yuan2020 ; Yuan2022 .

The study of individual based, stochastic models is much more recent. Park and
Krug Park2008  studied Kingman’s model for an unbounded fitness distribution
alongside a random model of fixed, finite population size. Despite its highly
simplified nature, the finite population model with an exponential
distribution of fitness effects is qualitatively consistent with the fitness
increase observed in Lenski’s long-term evolution experiment with bacteria
Sibani1998 ; Wiser2013 . A generalization of this model that includes the
response of the immune system to a population of pathogens was considered in
Bianconi2011 .

The first papers studying branching process models that express the selective
advantage of a fit individual in terms of its offspring distribution are
Dereich2017 , which deals with Weibull type fitness distributions and puts the
focus on the condensation phenomenon in that model, and Mailler2021  which
looks at the growth of the fittest family in the case of Gumbel type
distributions. Both papers are limited to bounded fitness distributions and
implicitly rely on the analogy to Kingman’s original model, though of course
the methods of study are entirely different in a stochastic setting. The
present paper initiates the study of branching processes with selection and
mutation for unbounded fitness distributions. We focus on the case of Fréchet
type fitness distributions where the mathematical challenge is linked to the
fact that the analogous Kingman model is ill-defined Park2008 .

The structure of the paper is as follows. In Sec. 2, we introduce the models
and state the main theorem. Sec. 3 explains the heuristics behind the formal
results and in Sec. 4 we present a rigorous proof of the theorem. Sec. 5
contains refined results for the empirical frequency distribution for one of
our models. These results are not yet accessible by a complete rigorous
mathematical analysis, so that we resort to a numerical and heuristic study
and a rigorous analysis of an approximating deterministic system. In Sec. 6 we
provide a short discussion that places our results into the context of
previous work and points to directions for future research.

##  2 Models and main result

We study two models of a population evolving in discrete generations. In both
models all individuals are assigned a fitness value, which is a positive real
number. As model parameters we fix a probability distribution $\mu$ on
$(0,\infty)$ from which the random fitness values $F$ are sampled, and a
mutation probability $\beta\in(0,1)$.

In both models we start from generation $t=0$ with a single individual111The
generalization to multiple individuals is straightforward. with fitness $f$.
Each individual in the population in generation $t\geq 0$ produces a Poisson
random number of offspring with mean given by its fitness. With probability
$1-\beta$ an offspring individual inherits its parent’s fitness and is added
to the population at generation $t+1$. Otherwise, with probability $\beta$, it
is a mutant. The two models differ in the fate of the mutants.

  * •

_Fittest mutant model (FMM):_ Every mutant is assigned a fitness sampled
independently from $\mu$. Only the fittest mutant (if there is one) is added
to the population at generation $t+1$. All other mutants die instantly.

  * •

_Multiple mutant model (MMM):_ Every mutant is assigned a fitness sampled
independently from $\mu$ and is added to the population at generation $t+1$.

We write $X(t)$ for the number of individuals in generation $t$ and study the
growth of the populations conditioned on the event of survival, i.e. when
$X(t)\neq 0$ for all times $t$. It is easy to see that the population size of
the MMM dominates the population size of the FMM at all times. Because the
growth is determined by the fittest mutants we expect both models to grow at
the same rate and to show this, it suffices to find an upper bound for the MMM
and a matching lower bound for the FMM.

Naturally, the rate of growth depends on the mutant fitness distribution
$\mu$. If $\mu$ is an unbounded distribution in both models individuals of
ever increasing fitness occur and hence the population will grow
superexponentially fast. By contrast, if $\mu$ is bounded we can only have
exponential growth. Indeed, if $\mu$ is continuous with essential supremum
one, then for a closely related continuous time model of immortal individuals,
it is shown in (Dereich2017, , Remark 1) that

$\lim_{t\to\infty}\frac{\log X(t)}{t}=\lambda^{*},$

where $\lambda^{*}\in[1-\beta,1)$ is the unique solution of the equation

$\beta\int\frac{\lambda^{*}}{\lambda^{*}-(1-\beta)x}\,\mu(dx)=1$

if $\beta\int\frac{1}{1-x}\,\mu(dx)\geq 1$, and otherwise
$\lambda^{*}:=1-\beta$. Further details on the long term growth of the process
depend on the classification of $\mu$ according to its membership in the max
domain of attraction of an extremal distribution. By the celebrated Fisher-
Tippett theorem there are three such universality classes, see for example
(Resnick, , Proposition 0.3). These are

  * •

the _Weibull class_ , which roughly occurs if $\mu$ is bounded with mass
decaying slowly near the essential supremum,

  * •

the _Gumbel class_ , which roughly occurs if the mass of $\mu$ is decaying
quickly near the essential supremum, which may be finite or infinite,

  * •

the _Fréchet class_ , which roughly occurs if $\mu$ is unbounded with mass
decaying slowly near infinity.

The assignment of mutant fitness distributions to extreme value universality
classes plays an important role in the interpretation of evolution experiments
Joyce2008 , and representatives of all three classes have been identified
empirically Das2022 .

In the present paper, we are mainly interested in the asymptotic behaviour of
the population size $X(t)$ in the so far unexplored case that $\mu$ belongs to
the Fréchet class (or, in short, is of _Fréchet type_). Precisely, this means
that the tail function

$G(x):=\mu((x,\infty))=\mathbb{P}(F>x)$

is regularly varying with index $-\alpha$ for some $\alpha>0$. In other words,
there exists a function $\ell\colon(0,\infty)\to\mathbb{R}$ which is slowly
varying at infinity such that $G(x)=x^{-\alpha}\ell(x)$. As in this case $\mu$
is an unbounded distribution, the process $(X(t)\colon t\geq 0)$ will grow
superexponentially fast on survival and therefore our discussion will focus on
the limiting quantity

$\displaystyle\nu=\lim_{t\rightarrow\infty}\frac{\log\log X(t)}{t}.$ (1)

Our main result is stated in the following theorem.

######  Theorem 1.

Given $\alpha>0$, let $T\in\mathbb{N}$ be the unique number such that

$\frac{(T-1)^{T}}{T^{T-1}}<\alpha\leq\frac{T^{T+1}}{(T+1)^{T}}$

and define

$\displaystyle\nu(\alpha):=\frac{1}{T}\log\frac{T}{\alpha}.$ (2)

Let $(X(t))_{t\geq 0}$ be the size of the population in either the FMM or the
MMM. Then, almost surely on survival,

$\lim_{t\to\infty}\frac{\log\log X(t)}{t}=\nu(\alpha).$

Before presenting the proof of the theorem in Sec. 4, in the next section we
motivate the expression (2).

##  3 Motivation of the main result

Here we explain the statement of Theorem 1 by a heuristic analysis of the FMM.
For convenience we take the fitness distribution to be of Pareto form,
$G(x)=x^{-\alpha}$ for $x\geq 1$ and $G(x)=1$ for $x<1$. Moreover, throughout
this section we assume that the initial fitness $f$ is so large that the
fluctuations induced by Poisson sampling are negligible at all times, which
implies that both the total population size and the sizes of subpopulations of
mutants are well approximated by their expectations. Denoting the fitness of
the mutant that is added to the population at generation $t$ by $W_{t}$, we
can then write

$\displaystyle
X(t)\approx(1-\beta)^{t}f^{t}+\sum_{i=1}^{t}(1-\beta)^{t-i}W_{i}^{t-i},$ (3)

where the factors $1-\beta$ account for the fact that (apart from the added
mutant) only the unmutated fraction of the population survives to the next
generation. For the same reason the total number $N_{t}$ of mutants produced
in generation $t$ (including the ones that die immediately) is approximately

$N_{t}\approx\frac{\beta}{1-\beta}X(t).$

Since the probability that the largest fitness $W_{t}$ among $N_{t}$
independent and identically distributed random variables with common
distribution $G$ is smaller than $x$ is $(1-x^{-\alpha})^{N_{t}}$, the random
variable $W_{t}$ can be sampled as

$W_{t}=\left(1-Z_{t}^{1/N_{t}}\right)^{-1/\alpha}\approx
X(t)^{1/\alpha}Y_{t},\quad
Y_{t}:=\left(\frac{1-\beta}{\beta}\log\frac{1}{Z_{t}}\right)^{-1/\alpha},$

where $Z_{t}$ is uniformly distributed in the interval $(0,1)$ and we have
approximated $Z_{t}^{1/N_{t}}\approx 1+(1/N_{t})\log Z_{t}$. Note that $Y_{t}$
does not depend on $X(t)$.

To proceed, we define $\omega_{t}$ as

$\displaystyle\omega_{t}:=\frac{\log X(t)}{\log f},$

which implies that $X(t)=f^{\omega_{t}}$ and $W_{t}\approx
Y_{t}f^{\omega_{t}/\alpha}$. Inserting these relations into (3) we obtain

$\displaystyle
f^{\omega_{t}}\approx(1-\beta)^{t}f^{t}+\sum_{i=1}^{t}(1-\beta)^{t-i}Y_{i}^{t-i}f^{(t-i)\omega_{i}/\alpha}.$
(4)

In the limit $f\to\infty$ the sum on the right hand side is dominated by the
term with the largest exponent. Correspondingly, the $\omega_{t}$ can be well
approximated by the solution $\chi_{t}$ of the recursion relation

$\displaystyle\chi_{t}=\max\left\\{t,\frac{t-1}{\alpha}\chi_{1},\frac{t-2}{\alpha}\chi_{2},\ldots,\frac{t-k}{\alpha}\chi_{k},\ldots,\frac{1}{\alpha}\chi_{t-1}\right\\}$
(5)

with $\chi_{1}=1$. We now argue that the $\chi_{t}$ grow at least
exponentially. Since for any $t_{0}\geq 1$ and any positive integer $m$

$\chi_{t_{0}+m}\geq\frac{m}{\alpha}\chi_{t_{0}},$

we have, for any $n\geq 1$

$\chi_{t_{0}+nm}\geq\left(\frac{m}{\alpha}\right)^{n}\chi_{t_{0}}.$

Correspondingly

$\displaystyle\lim_{t\rightarrow\infty}\frac{\log\chi_{t}}{t}=\lim_{n\rightarrow\infty}\frac{\log\chi_{t_{0}+nm}}{nm}\geq\frac{1}{m}\log\frac{m}{\alpha},$
(6)

where we have assumed that the limit is well-defined. Since (6) is valid for
any integer $m\geq 1$, an optimal lower-bound can be found by maximizing the
right hand side. As shown by Lemma 4 in Sec. 4, the maximizer over the
positive integers is precisely the function $\nu(\alpha)$ in Theorem 1. As the
population size depends exponentially on $\omega_{t}$ or $\chi_{t}$, the
heuristic argument makes it plausible that $\nu(\alpha)$ is a lower bound on
the double-exponential growth rate of $X(t)$. Remarkably, Theorem 1 states
that the bound is tight, and moreover applies also to the MMM. Informally this
implies that the population at time $t$ is dominated by the fittest mutant
that was generated at time $t-T$. As a consequence the mutant frequency
distribution changes periodically with period $T$ (see Sec. 5 for further
discussion).

In Fig. 1, we depict $\nu(\alpha)$ together with the numerical solution222The
direct numerical estimate of $\nu$ by extrapolating $\log\chi_{t}/t$ is
hampered by the fact that $\lim_{t\rightarrow\infty}\chi_{t}e^{-\nu t}$ does
not exist in general (see Sec. 5). The method used to obtain the data in Fig.
1 is explained at the end of Sec. 5.1. of the recursion relation (5). The fact
that $\nu(\alpha)$ is the exact exponential growth rate of $\chi_{t}$ is
proven rigorously in Lemma 6 in Sec. 4. In the inset of Fig. 1, we compare (2)
to an approximation obtained by treating $m$ in (6) as a continuous variable.
This yields

$\displaystyle\max\left\\{\frac{1}{x}\log\frac{x}{\alpha}\colon x\geq
1\right\\}=\begin{cases}1/(e\alpha),&\alpha e\geq 1,\\\ -\log\alpha,&\alpha
e<1.\end{cases}$ (7)

Although (7) is not exact, the relative error is less than 7% in all cases.

For $\alpha e<1$ the expressions (2) and (7) actually coincide. In this regime
of extremely heavy-tailed fitness distributions (more precisely, in the case
of $\alpha\leq 0.5$) selection becomes irrelevant, in the sense that the
double-exponential growth rate $\nu(\alpha)=\log(1/\alpha)$ persists in the
extreme case $\beta\to 1$ of the MMM, where all individuals are replaced by
mutants in each generation and the process becomes a classical Galton-Watson
process albeit with infinite mean, cf. Davies . In the case of the FMM, the
extreme case would stop the population from growing but the fitness $W_{t}$ of
the single individual present approximately satisfies $W_{t+1}\approx
W_{t}^{1/\alpha}$, which gives

$\lim_{t\rightarrow\infty}\frac{\log\log W_{t}}{t}=\log(1/\alpha).$

Figure 1:  Plots of $\nu$ vs $\alpha$. Solid line depicts (2) and symbols are
from numerical solution of the recursion relation (5). Inset: Plot of $(\nu
e\alpha)^{-1}-1$ vs. $\alpha$ with $\nu$ in (2). The error of the
approximation (7) is small and vanishes when $\alpha\leq 1/e$.

##  4 Proof of Theorem 1

###  4.1 Preparation for the proof

In this subsection we collect some tools that will be used in the proofs of
the lower and upper bounds in the estimate leading to Theorem 1. The lower
bound will be verified in Section 4.2 and the upper bound in Section 4.3.

For $t\in\mathbb{N}_{0}$ let $W_{t}$ be the fitness of the fittest of the
mutants in generation $t$ and $W_{t}=0$ if there are no mutants in generation
$t$. Our first observation is that under the weak assumption $G(x)>0$ for all
large $x$ (which always holds if $\mu$ is of Fréchet type) either the sequence
$(W_{t})$ is unbounded or the branching process dies out in finite time.
Heuristically speaking, on survival the accumulated number of mutants is
unbounded almost surely, which naturally entails unbounded largest fitness.

######  Lemma 2.

Almost surely on survival the sequence $(W_{t})$ is unbounded.

Proof: We first show that the branching process can be coupled to a sequence
$(\xi_{1},\ldots,\xi_{t})$ of independent Bernoulli variables with success
parameter $\beta$ and an independent sequence $(F_{1},\ldots,F_{t})$ of
independent fitnesses with distribution $\mu$ such that _on survival up to
generation $t$_ we have, for all $1\leq i\leq t$,

  * •

$\xi_{i}=1$ if there is at least one mutant in generation $i$, and

  * •

$W_{i}\geq F_{i}\xi_{i}$.

Indeed, once the random variables $(F_{1},\ldots,F_{t})$ and
$(\xi_{1},\ldots,\xi_{t})$ are generated with the given law we generate the
branching process as follows: Produce the offspring in the $n$th generation as
a Poisson distribution with the right parameter given by the previous
generation (possibly zero). If there is at least one offspring use $\xi_{n}$
to decide whether it is a mutant and if so give it fitness $F_{n}$. Then use
other newly sampled Bernoulli variables with parameter $\beta$ and fitnesses
to decide whether other variables are mutants and if they are decide their
fitness. Then surival implies $W_{n}\geq\xi_{n}F_{n}$ as required.

Now $N:=\sum_{i=1}^{t}\xi_{i}$ is binomially distributed with parameters
$\beta>0$ and $t\in\mathbb{N}$. We infer that, for any fixed $x>1$,

$\displaystyle\mathbb{P}(W_{i}\leq x~{}\text{ for all }i\leq t)$
$\displaystyle\leq\mathbb{P}(F_{i}\xi_{i}\leq x~{}\text{ for all }i\leq
t)+\mathbb{P}(\text{extinction})$ $\displaystyle=\sum_{i=0}^{t}{t\choose
i}\beta^{i}(1-\beta)^{t-i}\mathbb{P}(F\leq
x)^{i}+\mathbb{P}(\text{extinction})$ $\displaystyle=(\beta\mathbb{P}(F\leq
x)+(1-\beta))^{t}+\mathbb{P}(\text{extinction}).$

Since $\mathbb{P}(F\leq x)<1$ and $\beta>0$, we get

$\displaystyle\mathbb{P}(W_{i}\leq$ $\displaystyle\,x~{}\text{ for all
}i)=\lim_{t\uparrow\infty}\mathbb{P}(W_{i}\leq x~{}\text{ for all }i\leq t)$
$\displaystyle\leq\lim_{t\uparrow\infty}(\beta\mathbb{P}(F\leq
x)+(1-\beta))^{t}+\mathbb{P}(\text{extinction})=\mathbb{P}(\text{extinction}),$

hence $\mathbb{P}((W_{t})$ is unbounded$)=\mathbb{P}(\text{survival})$ as
claimed.

We next describe the distribution of $W_{t}$ given the process at time $t-1$.

######  Lemma 3.

Suppose that at generation $t-1$ there are $n$ individuals with fitness
$F_{1}$, $F_{2}$, …, $F_{n}$ and set $\mathcal{X}:=\sum_{i=1}^{n}F_{i}$. Then,
for all $x\geq 0$,

$\displaystyle\mathbb{P}(W_{t}>x)=1-e^{-\beta\mathcal{X}G(x)}.$

Proof: First fix a positive integer $n$ and suppose $W_{t}^{{}_{(n)}}$ is the
largest of $n$ independently sampled fitnesses and $W^{{}_{(0)}}_{t}=0$. Let
$\bar{G}(x)=1-G(x)$ and note that

$\mathbb{P}(W^{{}_{(n)}}_{t}>x)=1-\mathbb{P}(W^{{}_{(n)}}_{t}\leq
x)=1-\bar{G}(x)^{n}.$

Now let $N$ be the number of mutants in generation $t$, which is Poisson
distributed with mean $\beta\mathcal{X}$. Hence, for $x\geq 0$,

$\displaystyle\mathbb{P}(W_{t}>x)$
$\displaystyle=\sum_{n=0}^{\infty}\mathbb{P}(W_{t}^{{}_{(n)}}>x)\mathbb{P}(N=n)=\sum_{n=0}^{\infty}(1-\bar{G}(x)^{n})\mathbb{P}(N=n)$
$\displaystyle=1-\sum_{n=0}^{\infty}\bar{G}(x)^{n}\frac{(\beta\mathcal{X})^{n}}{n!}e^{-\beta\mathcal{X}}=1-e^{-\beta\mathcal{X}(1-\bar{G}(x))}.$

As $1-\bar{G}(x)=G(x)$ the proof is complete

The next two results concern the potential limit $\nu(\alpha)$. We first
characterise $\nu(\alpha)$ as a maximum and then as the growth rate in a
recursion relation. Note that the first result easily implies that
$\nu(\alpha)$ is decreasing, as well as continuous and positive.

######  Lemma 4.

We have

$\nu(\alpha)=\max\\{\tfrac{1}{m}\log(m/\alpha)\colon m\in\mathbb{N}\\}.$

In particular, for all $m\in\mathbb{N}$, we have

$\displaystyle m\leq\alpha e^{\nu(\alpha)m}\quad\text{ and }\quad T=\alpha
e^{\nu(\alpha)T}.$ (8)

Proof: First observe that $\alpha>m^{m+1}/(m+1)^{m}$ for $m<T$ and $\alpha\leq
m^{m+1}/(m+1)^{m}$ for $m\geq T$, where $m\in\mathbb{N}$. Since

$m(m+1)\left(\frac{1}{m}\log\frac{m}{\alpha}-\frac{1}{m+1}\log\frac{m+1}{\alpha}\right)=\log\frac{m^{m+1}}{(m+1)^{m}\alpha},$

we have the desired result.

######  Remark 5.

Let $\alpha_{T}:=T^{T+1}/(T+1)^{T}$. The equality $m=\alpha e^{\nu(\alpha)m}$
holds iff ($m=T$) or ($m=T+1$ and $\alpha=\alpha_{T}$).

For the remainder of this subsection, we abbreviate $\nu:=\nu(\alpha)$.

######  Lemma 6.

For some positive sequence $(a_{n})$ we define inductively

$\chi_{t}:=\chi_{t}(\alpha,(a_{n})):=\max\big{\\{}a_{t},\tfrac{t-1}{\alpha}\chi_{1},\ldots,\tfrac{1}{\alpha}\chi_{t-1}\big{\\}}.$
(9)

Then, if $\displaystyle\lim_{n\to\infty}a_{n}e^{-\nu n}=0$, there are positive
constants $c$ and $c^{\prime}$ such that

$c^{\prime}e^{\nu t}\leq\chi_{t}\leq ce^{\nu t}\quad\text{ for all $t\geq
1$,}$

and therefore we have

$\lim_{t\to\infty}\frac{\log\chi_{t}}{t}=\nu.$

Proof: Abbreviate $\hat{\chi}_{t}:=\chi_{t}/c^{\prime}$ with
$c^{\prime}=e^{-\nu T}\min\\{\chi_{1},\chi_{2},\ldots,\chi_{T}\\}$. Obviously,
$\hat{\chi}_{t}\geq e^{\nu t}$ for $t\leq T$. Now assume $n\geq T$ and
$\hat{\chi}_{t}\geq e^{\nu t}$ for all $t\leq n$. By the assumption and (8),
we have

$\displaystyle\hat{\chi}_{n+1}\geq\frac{T}{\alpha}\hat{\chi}_{n+1-T}\geq\frac{T}{\alpha}e^{\nu(n+1-T)}=e^{\nu(n+1)}.$

Induction gives $\hat{\chi}_{t}\geq e^{\nu t}$ and hence $\chi_{t}\geq
c^{\prime}e^{\nu t}$ for all $t\geq 1$.

Now, choose a positive integer $n_{0}$ such that $a_{n}\leq e^{\nu n}$ for all
$n\geq n_{0}$. Let $\bar{\chi}_{t}=\chi_{t}/c$ with
$c=\max\\{1,\chi_{1},\chi_{2},\ldots,\chi_{n_{0}}\\}$. Obviously,
$\bar{\chi}_{t}\leq 1\leq e^{\nu t}$ for all $t\leq n_{0}$. Now let $n\geq
n_{0}$ and assume that $\bar{\chi}_{t}\leq e^{\nu t}$ for all $t\leq n$. Then,

$\displaystyle\bar{\chi}_{n+1}$
$\displaystyle=\max\left\\{\tfrac{a_{n+1}}{c},\tfrac{n}{\alpha}\bar{\chi}_{1},\tfrac{n-1}{\alpha}\bar{\chi}_{2},\ldots,\tfrac{n-k+1}{\alpha}\bar{\chi}_{k},\ldots,\tfrac{1}{\alpha}\bar{\chi}_{n}\right\\}$
$\displaystyle\leq\max\left\\{e^{\nu(n+1)},\tfrac{n}{\alpha}e^{\nu},\tfrac{n-1}{\alpha}e^{2\nu},\ldots\tfrac{n-k+1}{\alpha}e^{\nu
k},\ldots,\tfrac{1}{\alpha}e^{\nu n}\right\\}\leq e^{\nu(n+1)},$

where we have used (8). By induction, we have $\chi_{t}\leq ce^{\nu t}$ for
all $t\geq 1$.

For later reference we define

$\tilde{\chi}_{i}(t):=\begin{cases}-\infty,&\text{ if }i<0,\\\ a_{t},&\text{
if }i=0,\\\ (t-i)\chi_{i}/\alpha,&\text{ if }1\leq i\leq t-1.\end{cases}$ (10)

######  Lemma 7.

Define

$\displaystyle I_{t}:=\max\\{i<t\colon\chi_{t}=\tilde{\chi}_{i}(t)\\}.$ (11)

Then $t-I_{t}$ is bounded.

Proof: By Lemma 6 we have $c^{\prime}e^{\nu t}\leq\chi_{t}\leq ce^{\nu t}$ for
all $t$. Since there is $t_{0}$ such that $c^{\prime}e^{\nu t}>a_{t}$ for all
$t\geq t_{0}$, we can write, for $t>t_{0}$,

$\displaystyle\chi_{t}=\max\left\\{\tfrac{t-1}{\alpha}\chi_{1},\ldots,\tfrac{1}{\alpha}\chi_{t-1}\right\\}.$

Now it is enough to show that $t-I_{t}$ is bounded for $t>t_{0}$.

Note that, for $1\leq m\leq t-1$,

$\displaystyle ce^{\nu t}A(m)\geq\tfrac{m}{\alpha}\chi_{t-m}\geq
c^{\prime}e^{\nu t}A(m),$

where $A(m)=me^{-\nu m}/\alpha$ with $A(T)=1$. Since
$\lim_{m\rightarrow\infty}A(m)=0$, there is $m_{0}$ such that
$c^{\prime}>cA(m)$ and hence

$\tfrac{m}{\alpha}\chi_{t-m}<c^{\prime}e^{\nu t},\text{ for all $m\geq
m_{0}$.}$

As the right hand side is a lower bound of $\frac{T}{\alpha}\chi_{t-T}$ we get
that $t-I_{t}$ cannot be larger than $\max\\{m_{0},t_{0}\\}$, as desired.

######  Remark 8.

If we choose $T^{\prime}>\sup\\{t-I_{t}\colon t\in\mathbb{N}\\}$, then, for
all $t$,

$\displaystyle\chi_{t}=\max\\{a_{t},(t-1)\chi_{1}/\alpha,\ldots,\chi_{t-1}/\alpha\\}=\max\\{\tilde{\chi}_{t-T^{\prime}}(t),\ldots,\tilde{\chi}_{t-1}(t)\\}.$

In words, $\chi_{t}$ is completely determined by $\tilde{\chi}_{i}(t)$ for $i$
within the window $t-T^{\prime}\leq i\leq t-1$. This fact will play an
important role in the proof of Theorem 1.

We conclude the subsection with two estimates for classical Galton-Watson
processes.

######  Lemma 9.

Consider a supercritical Galton-Watson process $(\mathcal{X}_{t})_{t\geq 0}$
with Poisson offspring distribution with mean $\theta>1$, starting in
generation 0 with a single individual. Fix $0<x<1$ and an integer $n\geq 1$.
Then,

$\displaystyle\mathbb{P}\left(\mathcal{X}_{t}\geq x^{t}\theta^{t}\text{ for
all }1\leq t\leq n\right)\geq 1-n\frac{(1-x)^{-2}}{\theta-1}.$ (12)

Proof: First note that (see, e.g., H63 )

$\displaystyle\mathbb{E}(\mathcal{X}_{t})=\theta^{t},\quad\mathbb{V}(\mathcal{X}_{t})=\frac{\theta^{2t}-\theta^{t}}{\theta-1},$

and that

$\displaystyle\mathbb{P}\left(\mathcal{X}_{t}\geq x^{t}\theta^{t}\text{ for
all }1\leq t\leq n\right)\geq
1-\sum_{t=1}^{n}\mathbb{P}\left(\mathcal{X}_{t}\leq x^{t}\theta^{t}\right),$
(13)

where we have used the sub-additivity of probability measure. Using
Chebyshev’s inequality, we get

$\displaystyle\mathbb{P}$ $\displaystyle\left(\mathcal{X}_{t}\leq
x^{t}\theta^{t}\right)=\mathbb{P}\left(\theta^{t}-\mathcal{X}_{t}\geq\theta^{t}-x^{t}\theta^{t}\right)\leq\mathbb{P}\left(|\mathcal{X}_{t}-\theta^{t}|\geq\theta^{t}-x^{t}\theta^{t}\right)$
$\displaystyle\leq\left(1-x^{t}\right)^{-2}\frac{\mathbb{V}(\mathcal{X}_{t})}{\theta^{2t}}=\left(1-x^{t}\right)^{-2}\frac{(1-\theta^{-t})}{(\theta-1)}\leq\frac{(1-x)^{-2}}{\theta-1},$

which, along with (13), gives the claimed inequality.

######  Lemma 10.

For a Galton-Watson process $(\mathcal{X}_{t})$ with $\mathcal{X}_{0}=K_{0}$
and generation dependent offspring distribution $N_{t}$ with
$\mathbb{E}[N_{t}]\leq N$ for all $t$,

$\displaystyle\mathbb{P}\big{(}\mathcal{X}_{t}\leq KN^{t}B^{t}\text{ for all
}t\geq 1\big{)}\geq 1-\frac{K_{0}}{K(B-1)},$

for all $B>1$ and $K>0$.

Proof: By Markov’s inequality, we have

$\displaystyle\mathbb{P}(\mathcal{X}_{t}\geq
KN^{t}B^{t})\leq\frac{\mathbb{E}[\mathcal{X}_{t}]}{KN^{t}}B^{-t}.$

Since
$\mathbb{E}[\mathcal{X}_{t+1}|\mathcal{X}_{t}]=\mathcal{X}_{t}\mathbb{E}[N_{t}]$,
we have
$\mathbb{E}[\mathcal{X}_{t}]=K_{0}\prod_{i=0}^{t-1}\mathbb{E}[N_{i}]\leq
K_{0}N^{t}$, which gives

$\displaystyle\mathbb{P}(\mathcal{X}_{t}\geq
KN^{t}B^{t})\leq\frac{K_{0}}{KB^{t}}.$

Since

$\mathbb{P}(\mathcal{X}_{t}\leq KN^{t}B^{t}\text{ for all }t\geq 1)\geq
1-\sum_{t=1}^{\infty}\mathbb{P}(\mathcal{X}_{t}\geq KN^{t}B^{t}),$

a geometric sum gives the claimed inequality.

###  4.2 Proof of the lower bound

In this subsection we show that, for given $\alpha>0$ and all
$\alpha^{\prime}>\alpha$, we have

$\displaystyle\mathbb{P}\Big{(}\liminf_{t\rightarrow\infty}\frac{\log\log
X(t)}{t}\geq\nu(\alpha^{\prime})\,\Big{|}\,\text{survival}\Big{)}=1.$ (14)

In both models at each generation $s$ the lineage originating from the mutant
with fitness $W_{s}$ dominates a version $(\hat{X}_{t}(f))_{t\geq s}$ of the
same model starting in generation $s$ with a single individual of fitness
$f=W_{s}$. If there is at least one $s$ such that

$\liminf_{t\rightarrow\infty}\frac{\log\log\hat{X}_{t}(W_{s})}{t}\geq\nu(\alpha^{\prime})$

then (14) is proved. As $(W_{t})$ is unbounded almost surely on survival it
therefore suffices to show that

$\displaystyle\lim_{f\rightarrow\infty}\mathbb{P}\left(\liminf_{t\rightarrow\infty}\frac{\log\log\hat{X}_{t}(f)}{t}\geq\nu(\alpha^{\prime})\right)=1.$
(15)

As $(\hat{X}_{t}(x))$ can be coupled to an FMM $(S_{t}(x))$ with the same
initial condition such that $\hat{X}_{t}(x)\geq S_{t}(x)$ for all $t\geq 0$,
the result follows by combining Lemma 6 with the following statement.

######  Lemma 11.

Fix $0<\varepsilon<1/2$ and let

$E(f):=\big{\\{}S_{t}(f)\geq(1-\beta)^{t}f^{\chi_{t}^{\prime}}\text{ for all
}t\geq 1\big{\\}},$

where $\chi_{t}^{\prime}:=\chi_{t}(\alpha^{\prime},(\frac{n}{2}))$ with
$\alpha^{\prime}:=\alpha/(1-2\varepsilon)$. Then

$\lim_{f\rightarrow\infty}\mathbb{P}(E(f))=1.$

Proof: We define $m_{0}:=f$, $n_{0}:=f^{1/2}$, and ($i\geq 1$)

$\displaystyle m_{i}:=f^{(1-\varepsilon)\chi_{i}^{\prime}/\alpha},\quad
n_{i}:=f^{(1-2\varepsilon)\chi_{i}^{\prime}/\alpha}=f^{\chi_{i}^{\prime}/\alpha^{\prime}}.$

For later reference, we also define $U_{i}:=n_{i}/m_{i}$ for all $i\geq 0$.

Set $\varepsilon_{0}=\varepsilon/(2-2\varepsilon)$. By our assumption on
$\mu$, there is $f_{0}$ such that

$G(x)\geq x^{-\alpha(1+\varepsilon_{0})}\text{ for all $x>f_{0}$.}$

Since we are only interested in the limit as $f\to\infty$, we may assume that
$f$ is so large that $(1-\beta)m_{0}\geq 2$, $(1-\beta)m_{1}\geq 2$,
$U_{0}<1/2$, $U_{1}<1/2$, and $m_{1}>f_{0}$. Notice that by assumption,

$G(m_{1})\geq G(m_{i})\geq
m_{i}^{-\alpha(1+\varepsilon_{0})}=f^{-(1-\varepsilon/2)\chi_{i}^{\prime}}\text{
for all $i\geq 1$.}$

For $\chi^{\prime}_{t}$, we choose $T^{\prime}$ as in Remark 8. By $N_{i,t}$
we denote the number of individuals with fitness $W_{i}$ at generation $t$.
Define events

$\displaystyle A_{i}:=\\{N_{i,t}\geq(1-\beta)^{t-i}n_{i}^{t-i}\text{ for all
}i<t\leq i+T^{\prime}\\},\quad B_{i}:=\\{W_{i}\geq m_{i}\\}.$

Let $D_{-1}$ be the certain event and, for $i\in\mathbb{N}_{0}$,

$D_{i}:=D_{i-1}\cap A_{i}\cap B_{i},\quad A:=\bigcap_{i=0}^{\infty}D_{i}.$

Now observe that

$\displaystyle\mathbb{P}(A)=\lim_{i\rightarrow\infty}\mathbb{P}(D_{i}),\quad\mathbb{P}(D_{i})=\mathbb{P}(A_{i}|B_{i}\cap
D_{i-1})\mathbb{P}(B_{i}|D_{i-1})\mathbb{P}(D_{i-1}).$

By Lemma 9 we have

$\displaystyle\mathbb{P}(A_{i}|B_{i}\cap D_{i-1})\geq
1-\frac{T^{\prime}(1-n_{i}/W_{i})^{-2}}{(1-\beta)W_{i}-1}\geq
1-\frac{T^{\prime}(1-U_{i})^{-2}}{(1-\beta)m_{i}-1},$

where we have used (12).

To proceed, we find the $\mathcal{X}$ in Lemma 3 on the event $D_{n-1}$ as

$\displaystyle\mathcal{X}\geq\sum_{i=0}^{t-1}N_{i,t-1}W_{i}\geq(1-\beta)^{t}\sum_{i=t-T^{\prime}-1}^{t-1}f^{\tilde{\chi}_{i}^{\prime}(t)}\geq(1-\beta)^{t}f^{\chi_{t}^{\prime}},$

where we have used $W_{i}\geq m_{i}\geq n_{i}$ and
$\tilde{\chi}_{i}^{\prime}(t)$ as in (10) for parameters $\alpha^{\prime}$ and
$a_{s}=s/2$. Using Lemma 3 with $G(m_{i})\geq
f^{-(1-\varepsilon/2)\chi_{i}^{\prime}}$, we have

$\displaystyle\mathbb{P}(B_{i}|D_{i-1})\geq
1-\exp\left(-\beta(1-\beta)^{i}f^{\varepsilon\chi_{i}^{\prime}/2}\right).$

Now we define

$\displaystyle
b_{i}:=\frac{T^{\prime}(1-U_{i})^{-2}}{(1-\beta)m_{i}-1}+(1-\delta_{i,0})\exp\left(-\beta(1-\beta)^{i}f^{\varepsilon\chi_{i}^{\prime}/2}\right),\quad\phi(f):=\sum_{i=0}^{\infty}b_{i},$

where $\delta_{ij}$ is the Kronecker delta symbol. Trivially, we have
$\lim_{f\rightarrow\infty}b_{i}=0$ for all $i\geq 0$. Since, for sufficiently
large $f$, $b_{s}$ for fixed $s$ is a bounded and decreasing function of $f$
and since Lemma 6 gives

$\lim_{s\rightarrow\infty}b_{s}2^{s}=0,$

there is $s_{0}$ such that $|b_{s}|<2^{-s}$ for all $s>s_{0}$ and for all
assumed value of $f$. Therefore, the series defining $\phi(f)$ converges
uniformly for sufficiently large $f$ and $\lim_{f\rightarrow\infty}\phi(f)=0$.
Therefore, for sufficiently large $f$, we get

$\displaystyle\mathbb{P}(A)\geq\prod_{i=0}^{\infty}(1-b_{i})\geq
1-\phi(f),\quad\lim_{f\rightarrow\infty}\mathbb{P}(A)=1,$

where we have used $(1-x)(1-y)\geq 1-x-y$ for $x,y\geq 0$. As, on the event
$A$,

$\displaystyle
S_{t}(f)\geq\sum_{i=t-T^{\prime}}^{t}N_{i,t}\geq(1-\beta)^{t}f^{\chi_{t}^{\prime}}$

where we have assumed $N_{i,t}=0$ for $i<0$, we see that $A\subset E(f)$ and
the proof is completed.

In fact, Lemma 11 and its proof are applicable to the MMM verbatim, except
that $S_{t}$ is replaced by $\hat{X}_{t}$. If we are interested in the proof
only for the MMM, we actually do not need to introduce $S_{t}$.

###  4.3 Proof of the upper bound

In this subsection we show that, for given $\alpha>0$ and all
$\alpha^{\prime}<\alpha$, we have for the MMM denoted by $(M_{t})$, or
$(M_{t}(x))$ if in the initial generation there is a single individual with
fixed fitness $x$, that

$\displaystyle\mathbb{P}\Big{(}\limsup_{t\rightarrow\infty}\frac{\log\log
M_{t}}{t}\leq\nu(\alpha^{\prime})\Big{)}=1.$ (16)

In case of extinction the upper bound holds by convention. One can construct
two processes with initial fitness $x\leq y$ on the same probability space
such that $M_{t}(x)\leq M_{t}(y)$ for all $t$. Indeed, this can be done as
follows. First construct $(M_{t}(y))$ and look at its genealogical tree
truncated after the first mutant in every line of descent from the root.
Removing any individual in that tree together with all its offspring from
$(M_{t}(y))$ independently with probability $x/y$ we obtain $(M_{t}(x))$.

We now construct an MMM with special initial conditions. Fix $\varepsilon>0$.
For given $\alpha$, let $\delta=(1+2\varepsilon)/(1+3\varepsilon)$,
$\alpha^{\prime}=\alpha/(1+3\varepsilon)$,
$\nu^{\prime}=\nu(\alpha^{\prime})$, $T=T(\alpha^{\prime})$, and

$\Delta_{0}=\frac{\varepsilon}{\nu^{\prime}(1+3\varepsilon)},$

which is equivalent to $\nu^{\prime}\Delta_{0}+\delta=1$. We choose $\Delta$
such that $0<\Delta\leq\Delta_{0}$ and

$\displaystyle\hat{n}:=\frac{T}{\Delta}$

is an integer. We define, for a given $f>0$,

$\displaystyle\chi_{t}^{\prime}:=e^{\nu^{\prime}t},\quad\kappa_{n,t}:=e^{\nu^{\prime}(t-T+n\Delta)},\quad
g_{n,t}:=f^{\kappa_{n,t}/\alpha},\quad
h_{n,t}:=f^{(1+\varepsilon)\kappa_{n,t}/\alpha}.$

We consider the MMM $(\tilde{M}_{t}(f))_{t\geq T-1}$ starting in generation
$T-1$ with an initial condition such that there are $T$ different mutant
classes with fitness $g_{\hat{n},m}$ for $0\leq m\leq T-1$ and the number of
individuals with fitness $g_{\hat{n},m}$ is
$\lfloor(g_{\hat{n},m})^{T-m-1}\rfloor$. We only consider $f$ sufficiently
large so that $(1-\beta)f>2$ and $(1-\beta)f^{\varepsilon/\alpha}>2$.

Now assume that we have proved, for all $\alpha^{\prime}<\alpha$,

$\displaystyle\lim_{f\rightarrow\infty}\mathbb{P}\Big{(}\limsup_{t\rightarrow\infty}\frac{\log\log\tilde{M}_{t}(f)}{t}\leq\nu(\alpha^{\prime})\Big{)}=1.$
(17)

Given an arbitrary $f>0$ and $\varepsilon>0$ pick $f_{\varepsilon}$ such that
the probability above exceeds $1-\varepsilon$ and the smallest fitness in the
initial condition of $\tilde{M}_{t}(f_{\varepsilon})$ is larger than $f$. Then
(17) guarantees that

$\mathbb{P}\Big{(}\limsup_{t\rightarrow\infty}\frac{\log\log
M_{t}(f)}{t}\leq\nu(\alpha^{\prime})\Big{)}>1-\varepsilon,$

which proves (16). So it is enough to prove (17). Once (17) is proved, we use
the natural coupling such that $S_{t}\leq M_{t}$ for all $t$. Then, almost
surely on survival,

$\displaystyle\limsup_{t\rightarrow\infty}\frac{\log\log
S_{t}}{t}\leq\limsup_{t\rightarrow\infty}\frac{\log\log
M_{t}}{t}\leq\nu(\alpha),$

which completes the proof of Theorem 1.

######  Lemma 12.

Let $Z_{t}$ be the number of _non-mutated_ descendants at generation $t$ of
$\mathcal{X}$ individuals with fitness in a bounded interval $I$ with right
endpoint $b$ at generation $m$ of an MMM. Assume $\mathcal{X}\leq K$. Then,
for all $B>1$,

$\mathbb{P}(Z_{t}\leq K(1-\beta)^{t-m}b^{t-m}B^{t-m}\text{ for all }t\geq
m+1)\geq 1-\frac{1}{B-1}.$

Proof: As the mean number of non-mutated offspring of an individual is bounded
by $(1-\beta)b$ we get the result by applying Lemma 10.

######  Lemma 13.

Suppose at generation $t-1$ of an MMM the population consists of $n$
individuals with fitness $F_{1},\ldots,F_{n}$. Let

$\displaystyle Y_{t}:=\beta\sum_{i=1}^{n}F_{i}$ (18)

and let $Z$ be the number of mutants in generation $t$ with fitness in the
interval $(a,b]$. Then, with $p:=\mu((a,b])$, we have

$\displaystyle\mathbb{P}(Z>K)=e^{-Y_{t}p}\sum_{n=K+1}^{\infty}\frac{(Y_{t}p)^{n}}{n!}.$

Proof: Observe that $Z$ is Poisson distributed with mean $Y_{t}p$.

######  Remark 14.

Using Markov’s inequality, we get

$\displaystyle\mathbb{P}(Z>K)\leq\frac{Y_{t}p}{K},$ (19)

which is useful when $K\gg Y_{t}p$. By Chebyshev’s inequality, for $K>Y_{t}$,

$\displaystyle\mathbb{P}(Z>K)\leq\mathbb{P}(|Z-Y_{t}p|\geq
K-Y_{t}p)\leq\frac{Y_{t}p}{(K-Y_{t}p)^{2}}\leq\frac{Y_{t}}{(K-Y_{t})^{2}},$
(20)

which is useful when $(K-Y_{t})^{2}\gg Y_{t}$. For $K=0$, we will use

$\displaystyle\mathbb{P}(Z=0)=e^{-Y_{t}p}\geq 1-Y_{t}p\geq 1-Y_{t}G(a).$ (21)

We denote the number of non-mutated descendants of initial individuals with
fitness $g_{\hat{n},m}$ at generation $t\geq T-1$ by $N_{m,T-1,t}$ and define

$N_{T-1,t}:=\sum_{m=0}^{T-1}N_{m,T-1,t}.$

The number of mutants that appear at generation $t\geq T$ with fitness in the
interval $(h_{n-1,t},h_{n,t}]$ is denoted by $N_{n,t,t}$ for $0\leq
n\leq\hat{n}+1$, where we have assumed $h_{-1,t}:=0$ and
$h_{\hat{n}+1,t}:=\infty$. Typically, $N_{\hat{n}+1,t,t}$ will be zero. The
number of non-mutated descendants of $N_{n,m,m}$ at generation $t>m$ is
denoted by $N_{n,m,t}$. For $t\geq m\geq T$ define

$\displaystyle N_{m,t}:=\sum_{n=0}^{\hat{n}+1}N_{n,m,t},$

which gives

$\displaystyle\tilde{M}_{t}(f)=\sum_{m=T-1}^{t}N_{m,t}.$

Let $(\theta_{t})_{t\geq T-1}$ be a sequence satisfying
$\theta_{T-1}=\theta_{T}=T$ and, for $t>T$,

$\theta_{t}=\hat{n}+\sum_{m=T-1}^{t-1}\theta_{m}.$

Since $\theta_{t+1}-\theta_{t}=\theta_{t}$ for $t\geq T+1$, we have

$\displaystyle\theta_{t}=\begin{cases}2^{t-T-1}(2T+\hat{n}),&\text{ for
}t>T\\\ T,&\text{ for }t\leq T.\end{cases}$ (22)

######  Lemma 15.

For $T\leq x\leq m<t$ ($t,m$ are integers and $x$ is real), we have

$\displaystyle
e^{\nu^{\prime}m}+\tfrac{t-m}{\alpha^{\prime}}e^{\nu^{\prime}(m-T)}\leq
e^{\nu^{\prime}t},$ (23) $\displaystyle
e^{\nu^{\prime}m}-e^{\nu^{\prime}(x-\Delta)}+\delta\tfrac{t-m}{\alpha^{\prime}}e^{\nu^{\prime}x}\leq
e^{\nu^{\prime}t}.$ (24)

Proof: Using (8), we have

$\displaystyle
e^{\nu^{\prime}m}\left(1+\tfrac{t-m}{\alpha^{\prime}}e^{-\nu^{\prime}T}\right)=e^{\nu^{\prime}m}\tfrac{T+t-m}{T}\leq
e^{\nu^{\prime}t}e^{\nu^{\prime}T}\tfrac{\alpha^{\prime}}{T}=e^{\nu^{\prime}t},$

which proves (23). If $\delta(t-m)-\alpha^{\prime}e^{-\nu^{\prime}\Delta}$ is
negative, then (24) is trivially valid. If
$\delta(t-m)-\alpha^{\prime}e^{-\nu^{\prime}\Delta}$ is positive, then the
left hand side of (24) has maximum at $x=m$. Therefore, it is enough to prove
(24) only for $x=m$. Plugging $x=m$, we have

$\displaystyle
e^{\nu^{\prime}m}-e^{\nu^{\prime}(m-\Delta)}+\delta\tfrac{t-m}{\alpha^{\prime}}e^{\nu^{\prime}m}$
$\displaystyle=e^{\nu^{\prime}m}(1-e^{-\nu^{\prime}\Delta})+\delta\tfrac{t-m}{\alpha^{\prime}}e^{\nu^{\prime}m}$
$\displaystyle\leq e^{\nu^{\prime}t}\left(\nu^{\prime}\Delta+\delta\right)\leq
e^{\nu^{\prime}t},$

where we have used $1-e^{-x}\leq x$, $e^{\nu^{\prime}m}\leq
e^{\nu^{\prime}t}$, and $t-m\leq\alpha^{\prime}e^{\nu^{\prime}(t-m)}$.

######  Lemma 16.

Let $E(f):=\\{\tilde{M}_{t}(f)\leq\theta_{t+1}f^{\chi^{\prime}_{t}}$ for all
$t\geq T\\}$. Then

$\lim_{f\rightarrow\infty}\mathbb{P}(E(f))=1,$

which implies (17).

Proof: Set $\varepsilon_{0}=\varepsilon/(2+2\varepsilon)$. By our assumption
on $\mu$ there is $f_{0}$ such that $G(f)\leq f^{-\alpha(1-\varepsilon_{0})}$
for all $f>f_{0}$. Now we assume $h_{0,T}=f^{(1+\varepsilon)/\alpha}>f_{0}$,
which gives

$\displaystyle G(h_{n,t})\leq
f^{-(1+\varepsilon/2)\kappa_{n,t}},\quad\text{for all }0\leq
n\leq\hat{n}\text{ and }t\geq T.$ (25)

Let

$A_{n,T-1}:=\left\\{N_{n,T-1,t}\leq\left\lfloor(g_{\hat{n},n})^{T-n-1}\right\rfloor(1-\beta)^{t-T+1}f^{(t-T+1)\chi^{\prime}_{n}/\alpha^{\prime}}\mbox{
for all }t\geq T\right\\}$

and define $A_{T-1}:=\bigcap_{n=0}^{T-1}A_{n,T-1}$. By Lemma 12 with
$K=\lfloor(g_{\hat{n},n})^{T-n-1}\rfloor$,
$b=g_{\hat{n},n}=f^{\chi_{n}^{\prime}/\alpha}$, and
$bB=f^{\chi_{n}^{\prime}/\alpha^{\prime}}=f^{(1+3\varepsilon)\chi_{n}^{\prime}/\alpha}$,
we have

$\displaystyle
1-\mathbb{P}(A_{n,T-1})\leq\left(f^{3\varepsilon\chi_{n}^{\prime}/\alpha}-1\right)^{-1},$
$\displaystyle\quad\mathbb{P}(A_{T-1})\geq 1-b_{T-1},$ $\displaystyle
b_{T-1}:=\sum_{n=0}^{T-1}\left(f^{3\varepsilon\chi_{n}^{\prime}/\alpha}-1\right)^{-1}.$

Since $me^{-\nu^{\prime}m}\leq\alpha^{\prime}$ and $g_{\hat{n},n}\leq
f^{\chi^{\prime}_{n}/\alpha^{\prime}}$, we have

$N_{n,T-1,t}\leq f^{\chi_{n}^{\prime}(t-n)/\alpha^{\prime}}\leq
f^{\chi^{\prime}_{t}}\quad\text{for all }n,t.$

Therefore,

$\displaystyle\sum_{n=0}^{T-1}N_{n,T-1,t}\leq
Tf^{\chi^{\prime}_{t}},\quad\text{for all }t\geq T,$ (26)

on the event $A_{T-1}$. Let

$A_{n,m,t}:=\left\\{\begin{array}[]{ll}\big{\\{}N_{0,m,t}\leq\theta_{m}f^{\chi_{m}^{\prime}}(1-\beta)^{t-m}f^{(t-m)\kappa_{0,m}/\alpha^{\prime}}\big{\\}},&\quad\text{
for }n=0,\\\\[5.69054pt]
\big{\\{}N_{n,m,t}\leq\frac{f^{\chi_{n}^{\prime}}}{(g_{n-1,m})^{\alpha}}(1-\beta)^{t-m}f^{\delta(t-m)\kappa_{n,m}/\alpha^{\prime}}\big{\\}},&\quad\text{
for }1\leq n\leq\hat{n},\\\\[8.53581pt]
\big{\\{}N_{\hat{n}+1,m,t}=0\big{\\}}&\quad\text{ for
}n=\hat{n}+1.\end{array}\right.$

Note that $A_{n,t,t}$ has information on the empirical distribution of
mutants’ fitness that appear at generation $t$. We define

$\displaystyle\tilde{A}_{n,m}=\bigcap_{t=m+1}^{\infty}A_{n,m,t},\quad
A_{n,m}:=A_{n,m,m}\cap\tilde{A}_{n,m},\quad
A_{m}:=\bigcap_{n=0}^{\hat{n}+1}A_{n,m},$ $\displaystyle
D_{T-1}:=A_{T-1},\quad D_{n}:=D_{n-1}\cap A_{n},\quad
A(f):=\bigcap_{n=T-1}^{\infty}D_{n}.$

By Lemma 15, we have on the event $A_{m}$, that for all $t\geq m$,

$\displaystyle\sum_{n=0}^{\hat{n}+1}N_{n,m,t}\leq(\theta_{m}+\hat{n})f^{\chi^{\prime}_{t}},$

and, in turn,

$\displaystyle\tilde{M}_{t}(f)\leq\left(\hat{n}+\sum_{m=T-1}^{t}\theta_{m}\right)f^{\chi^{\prime}_{t}}=\theta_{t+1}f^{\chi^{\prime}_{t}}$

on the event $A(f)$. Therefore, $A(f)\subset E(f)$ and the proof is complete
if we show

$\displaystyle\lim_{f\rightarrow\infty}\mathbb{P}(A(f))=1.$

Now we investigate $\mathbb{P}(A_{m}|D_{m-1})$. First note that

$\displaystyle 1-\mathbb{P}(A_{m}|D_{m-1})$
$\displaystyle\leq\sum_{n=0}^{\hat{n}+1}\left[1-\mathbb{P}(A_{n,m}|D_{m-1})\right],$
$\displaystyle\mathbb{P}(A_{n,m}|D_{m-1})$
$\displaystyle=\mathbb{P}(\tilde{A}_{n,m}|A_{n,m,m}\cap
D_{m-1})\mathbb{P}(A_{n,m,m}|D_{m-1}),$

and, on the event $D_{m-1}$,

$\displaystyle Y_{m}\leq\beta\theta_{m}f^{\chi^{\prime}_{m}},$ (27)

where $Y_{m}$ is defined in (18).

We begin with $\mathbb{P}(A_{\hat{n}+1,m}|D_{m-1})$, which clearly equals
$\mathbb{P}(A_{\hat{n}+1,m,m}|D_{m-1})$. Using (21) with $a=h_{\hat{n},m}$ and
$Y_{m}G(a)\leq\theta_{m}f^{-\varepsilon\chi^{\prime}_{m}/2}$, we obtain

$\displaystyle\mathbb{P}(A_{\hat{n}+1,m}|D_{m-1})\geq 1-b_{\hat{n}+1,m},\quad
b_{\hat{n}+1,m}:=\theta_{m}f^{-\varepsilon\chi^{\prime}_{m}/2}.$

Now we consider $\mathbb{P}(A_{0,m}|D_{m-1})$. Using (20) with
$K=\theta_{m}f^{\chi_{m}^{\prime}}$ and (27), we have

$\displaystyle\mathbb{P}(A_{0,m,m}|D_{m-1})\geq
1-\frac{\beta}{(1-\beta)^{2}\theta_{m}}f^{-\chi^{\prime}_{m}}.$

Using Lemma 12 with
$B=f^{\kappa_{0,m}/\alpha^{\prime}}/h_{0,m}=f^{2\varepsilon\kappa_{0,m}/\alpha}$,
we have

$\displaystyle\mathbb{P}(\tilde{A}_{0,m}|A_{0,m,m}\cap D_{m-1})\geq
1-\left(f^{2\varepsilon\kappa_{0,m}/\alpha}-1\right)^{-1}.$

Therefore, defining

$b_{0,m}:=\frac{\beta}{(1-\beta)^{2}\theta_{m}}f^{-\chi^{\prime}_{m}}+\left(f^{2\varepsilon\kappa_{0,m}/\alpha}-1\right)^{-1},$

we have $\mathbb{P}(A_{0,m}|D_{m-1})\geq 1-b_{0,m}$.

Finally, we move on to $\mathbb{P}(A_{n,m}|D_{m-1})$ for $1\leq n\leq\hat{n}$.
Using (19) with $K=f^{\chi_{m}^{\prime}-\kappa_{n-1,m}}$, $a=h_{n-1,m}$, and
$Y_{m}G(a)\leq\theta_{m}f^{\chi_{m}^{\prime}-(1+\varepsilon/2)\kappa_{n-1,m}}$,
we have

$\displaystyle\mathbb{P}(A_{n,m,m}|D_{m-1})\geq
1-\theta_{m}f^{-\varepsilon\kappa_{n-1,m}/2}.$

Using Lemma 12 with
$B=f^{\delta\kappa_{n,m}/\alpha^{\prime}}/h_{n,m}=f^{\varepsilon\kappa_{n,m}/\alpha}$,
we have

$\displaystyle\mathbb{P}(\tilde{A}_{n,m}|A_{n,m,m}\cap D_{m-1})\geq
1-\left(f^{\varepsilon\kappa_{n,m}/\alpha}-1\right)^{-1}.$

Therefore, defining

$b_{n,m}:=\theta_{m}f^{-\varepsilon\kappa_{n-1,m}/2}+\left(f^{\varepsilon\kappa_{n,m}/\alpha}-1\right)^{-1},$

we have $\mathbb{P}(A_{n,m}|D_{m-1})\geq 1-b_{n,m}.$ We define

$b_{m}:=\sum_{n=0}^{\hat{n}+1}b_{n,m}\text{ for }m\geq T,\quad\text{ and
}\phi(f):=\sum_{m=T-1}^{\infty}b_{m}.$

Recall that we have assumed $(1-\beta)f>2$ and
$(1-\beta)f^{\varepsilon/\alpha}>2$. Since $b_{m}$ for given $m$ is a bounded
function of $f$ which is decreasing to zero and

$\lim_{m\rightarrow\infty}b_{m}2^{m}=0,$

there is $m_{0}$ such that $|b_{m}|<2^{-m}$ for all $m>m_{0}$. Hence the
series defining $\phi(f)$ converges uniformly for sufficiently large $f$ and,
accordingly, $\lim_{f\rightarrow\infty}\phi(f)=0$. Therefore, for sufficiently
large $f$,

$\displaystyle\mathbb{P}(A(f))\geq\prod_{m=T-1}^{\infty}(1-b_{m})\geq
1-\phi(f),$

and $\displaystyle\lim_{f\rightarrow\infty}\mathbb{P}(A(f))=1$, which
completes the proof.

##  5 Empirical frequency distributions

Apart from the fact that the population is dominated by a single mutant class
at all times, the proof of the double-exponential growth rate $\nu$ presented
in Sec. 4 does not give any insight into the structure of the population.
However, since the solution $\chi_{t}$ of the recursion relation (9) correctly
describes the asymptotic growth of $X(t)$, it provides a natural starting
point for addressing this question at least on a heuristic level. In this
section, we analyze the recursion relation in more depth to understand the
demographic structure in the long time limit, which turns out to display a
rather rich behaviour.

###  5.1 Numerical solution of the recursion relation

To characterize the empirical frequency distribution we introduce the
following quantities:

$\displaystyle J_{i}(t)$ $\displaystyle:=\frac{\log W_{i}}{\log
W_{t}}\approx\frac{\chi_{i}}{\chi_{t}},\;\;P(t):=\frac{\log X(t+1)-\log
X(t)}{\log W_{t}}\approx\alpha\frac{\chi_{t+1}-\chi_{t}}{\chi_{t}},$
$\displaystyle R_{i}(t)$ $\displaystyle:=\frac{\log W_{i}^{t-i}-\log
X(t)}{\log X(t)}=\frac{t-i}{\alpha}J_{i}(t)-1,$

where the second approximate relations in the definitions of $J_{i}(t)$ and
$P(t)$ become equalities in the formal deterministic limit $f\to\infty$ (see
Sec. 3). The ratio $J_{i}(t)\in[0,1]$ compares the log-fitness of the mutant
class born at time $i$ to the log-fitness of the current fittest mutant. Since
$X(t+1)\approx(1-\beta)X(t)\bar{F}_{t}$ with $\bar{F}_{t}$ denoting the mean
fitness of the population at generation $t$, $P(t)$ quantifies the mean
fitness at generation $t$ on the same scale. The decomposition in Eq. (3)
shows that the fraction of the population in mutant class $i$ at time $t$ is
proportional to $W_{i}^{t-i}$, and therefore $R_{i}(t)\in[-1,0]$ serves as
proxy of the (logarithmic) frequency distribution at generation $t$ over
mutant classes $i$.

Figure 2:  (a) Plots of $R_{i}(t)$ vs. $J_{i}(t)$ for $\alpha=1$ at
generations $t=16$, $17$, $\ldots$, $24$. The vertical line at each panel
indicates the location of the mean fitness $P(t)$. (b) Data collapse plots of
$R_{i}(t)$ vs. $J_{i}(t)$ for each column of (a).

In Fig. 2(a), we plot $R_{i}(t)$ against $J_{i}(t)$ for nine consecutive
generations, obtained by numerically solving the recursion relation (5) for
$\alpha=1$. The salient feature is the periodic behavior of the frequency
distribution with period $3$; note that $T=3$ for $\alpha=1$. To illustrate
the accuracy of the periodic behaviour, we present data-collapse plots in Fig.
2(b). In most regions of $J_{i}$, the collapse looks perfect (since the number
of mutant classes increases with the number of generations, the frequency
distributions at different times cannot be identical). A rigorous proof of the
periodicity will be given in Sec. 5.2.

The periodicity was taken into account in the numerical estimates of $\nu$
reported in Fig. 1. Rather than monitoring $\log\chi_{t}/t$, which converges
very slowly, we computed the quantity

$\displaystyle\hat{\nu}(t):=\frac{1}{T}\log\frac{\chi_{t+T}}{\chi_{t}},$

which approaches a constant in a relatively short time.

###  5.2 Periodicity of $\chi_{t}e^{-\nu t}$

By Lemma 6, we know that

$\displaystyle c_{t}:=\chi_{t}e^{-\nu t}$

is bounded away from zero and infinity. Now we show that $c_{t}$ is not only
bounded, but eventually becomes periodic.

######  Proposition 17.

For any sequence $(a_{n})$ in the recursion relation (9), there is a $t_{1}$
such that $c_{t}=c_{t+T}$ for all $t\geq t_{1}$.

Proof: In this proof, $k$ and $k^{\prime}$ are exclusively used as integers in
the range $1\leq k,k^{\prime}\leq T$. Since $\chi_{t+T}\geq e^{\nu T}\chi_{t}$
(see Sec. 3), the sequence $(c_{k+nT})_{n}$ is nondecreasing and bounded.
Consequently,

$\displaystyle C_{k}:=\lim_{n\rightarrow\infty}c_{k+nT}$ (28)

is well defined. Note that $\max\\{C_{k}:1\leq k\leq T\\}$ becomes the optimal
upper bound in Lemma 6. If $n$ satisfies $nT>T^{\prime}$ with
$T^{\prime}>\max\\{t-I_{t}\\}$ (see Remark 8), then we have

$\displaystyle c_{k+nT}$
$\displaystyle=e^{-\nu(k+nT)}\max\left\\{\frac{k+nT-t^{\prime}}{\alpha}\chi_{t^{\prime}}:k+nT-T^{\prime}\leq
t^{\prime}<k+nT\right\\}$ $\displaystyle=\max\left\\{\frac{s}{\alpha}e^{-\nu
s}c_{k+nT-s}:1\leq s\leq T^{\prime}\right\\}.$ (29)

Taking $n$ to infinity, we get

$\displaystyle C_{k}=\max\left\\{\frac{s}{\alpha}e^{-\nu s}C_{k-s}:1\leq s\leq
T^{\prime}\right\\},$ (30)

and, by definition, $C_{k+mT}=C_{k}$ for any integer $m$. Since $Te^{-\nu
T}/\alpha=1$ and $C_{k-T}=C_{k}$, we can rewrite (30) as

$\displaystyle C_{k}=\max\left\\{\frac{s}{\alpha}e^{-\nu s}C_{k-s}:1\leq s\leq
T+1\right\\}.$

Comparing terms with $s=T-1,T,T+1$ for any $k$, we have

$\displaystyle\frac{T-1}{T}e^{\nu}C_{k+1}=\frac{T-1}{\alpha}e^{-\nu(T-1)}C_{k+1}\leq
C_{k},$
$\displaystyle\frac{T+1}{T}e^{-\nu}C_{k-1}=\frac{T+1}{\alpha}e^{-\nu(T+1)}C_{k-1}\leq
C_{k},$

which gives

$\displaystyle\frac{T+1}{T}\leq\frac{C_{k+1}e^{\nu}}{C_{k}}\leq\frac{T}{T-1},$

for all $k$. Let $\varphi_{k}=C_{k}e^{\nu}/C_{k-1}$ with
$\varphi_{1}=\varphi_{T+1}=C_{1}e^{\nu}/C_{T}$. Then,

$C_{k}=C_{1}e^{-\nu(k-1)}\prod_{j=2}^{k}\varphi_{j},$

for $k>1$. Setting $k=T+1$ and considering $C_{T+1}=C_{1}$, we have

$\prod_{j=1}^{T}\varphi_{j}=e^{\nu T}=\frac{T}{\alpha}.$

To sum up, $C_{k}$ takes the form

$\displaystyle C_{k}=C_{0}e^{-\nu k}\prod_{j=1}^{k}\varphi_{j},$

where $C_{0}$ is a positive constant (note that
$\chi_{t}(\alpha,(C_{0}a_{n}))=C_{0}\chi_{t}(\alpha,(a_{n}))$) and
$\varphi_{k}$ satisfies

$\displaystyle\frac{T+1}{T}\leq\varphi_{k}\leq\frac{T}{T-1},\quad\prod_{j=1}^{T}\varphi_{j}=\frac{T}{\alpha}=e^{\nu
T}.$ (31)

If $\alpha=\alpha_{T}$ (see Remark 5), then $e^{\nu}=(T+1)/T$ and the only
possible value of $\varphi_{k}$ is $\varphi_{k}=e^{\nu}$ for all $k$ because
of (31).

To simplify (29) for large $n$, we use the following observation. For
$p\in\mathbb{N}$ with $X:=1/T$ and $C_{k-(T\pm p)}=C_{k\mp p}$, we have

$\displaystyle\frac{T-p}{\alpha}e^{-\nu(T-p)}C_{k+p}$
$\displaystyle=\frac{T-p}{T}e^{\nu
p}\frac{C_{k+p}}{C_{k}}C_{k}=C_{k}\frac{T-p}{T}\prod_{j=1}^{p}e^{\nu}\frac{C_{k+j}}{C_{k+j-1}}$
$\displaystyle\leq
C_{k}\frac{T-p}{T}\left(\frac{T}{T-1}\right)^{p}=C_{k}\frac{1-pX}{(1-X)^{p}},$

and

$\displaystyle\frac{T+p}{\alpha}e^{-\nu(T+p)}C_{k-p}$
$\displaystyle=\frac{T+p}{T}e^{-\nu
p}\frac{C_{k-p}}{C_{k}}C_{k}=C_{k}\frac{T+p}{T}\prod_{j=1}^{p}\frac{C_{k-j}}{e^{\nu}C_{k-j+1}}$
$\displaystyle\leq
C_{k}\frac{T+p}{T}\left(\frac{T}{T+1}\right)^{p}=C_{k}\frac{1+pX}{(1+X)^{p}}.$

Since $\sup_{p\geq 2}(1+pX)/(1+X)^{p}<1$ for all nonzero $X$ not smaller than
$-1$, relating $s$ and $p$ by $p=|s-T|$ we can choose $\varepsilon>0$ such
that $C_{k}-\varepsilon>se^{-\nu s}C_{k-s}/\alpha$ for all $s$ with $|s-T|>1$
and for all $k$. By (28), for this $\varepsilon$, there is an integer $m_{0}$
such that $C_{k}-\varepsilon<c_{k+nT}\leq C_{k}$ for all $n\geq m_{0}$ and for
all $k$. If $n>m_{0}$, then

$c_{k+nT}>C_{k}-\varepsilon>se^{-\nu s}C_{k-s}/\alpha\geq se^{-\nu
s}c_{k-s+nT}/\alpha,$

for all $s$ with $|s-T|>1$, which reduces (29) to

$\displaystyle
c_{k+(n+1)T}=\max\left\\{c_{k+nT},\tfrac{T+1}{T}e^{-\nu}c_{k-1+nT},\tfrac{T-1}{T}e^{\nu}c_{k+1+nT}\right\\}.$
(32)

In the following, $n$ is assumed so large that (32) is valid for all $k$.
Defining $\delta_{k,n}:=$$1-c_{k+nT}/C_{k}$ with the convention
$\delta_{k+mT,n}:=\delta_{k,n+m}$ for integer $m$ and using the definition of
$\varphi_{k}$, we can write

$\displaystyle\delta_{k,n+1}=\min\left\\{\delta_{k,n},1-\tfrac{T+1}{T\varphi_{k}}(1-\delta_{k-1,n}),1-\tfrac{T-1}{T}\varphi_{k+1}(1-\delta_{k+1,n})\right\\}.$

As $c_{k+nT}\rightarrow C_{k}$, we have $\delta_{k,n}\rightarrow 0$ as
$n\rightarrow\infty$. Accordingly, if $(T+1)/(T\varphi_{k})<1$, then the term
with $(T+1)/(T\varphi_{k})$ cannot be a minimum for large $n$. The same
argument is applicable to the term with $(T-1)\varphi_{k+1}/T$.

If $\alpha=\alpha_{T}$, then $\varphi_{k}=(T+1)/T$ for all $k$ and,
accordingly, we have $\delta_{k,n+1}=\min\\{\delta_{k,n},\delta_{k-1,n}\\}$
for all $k$ and for all sufficiently large $n$. If there is $m$ and
$k^{\prime}$ such that $\delta_{k^{\prime},m}=0$, then
$\delta_{k^{\prime},n}=0$ for all $n\geq m$ and $\delta_{k^{\prime}+1,m+1}=0$,
which again gives $\delta_{k^{\prime}+2,m+2}=0$ and so on. Therefore, we have
$\delta_{k,n}=0$ for all $n>m+T$ and all $k$. Hence, to complete the proof for
this case, we need to elicit a contradiction if $\delta_{k,n}$ is strictly
positive for all $n$ and for all $k$. Since $\delta_{k,n}$ is a nonincreasing
sequence of $n$, we have

$\displaystyle\delta_{k,n+s}$
$\displaystyle=\min\\{\delta_{k,n+s-1},\delta_{k-1,n+s-1}\\}=\min\\{\delta_{k,n+s-2},\delta_{k-1,n+s-2},\delta_{k-1,n+s-1}\\}$
$\displaystyle=\min\\{\delta_{k,n+s-2},\delta_{k-1,n+s-1}\\}=\min\\{\delta_{k,n},\delta_{k-1,n+s-1}\\},$

for all $s\in\mathbb{N}$. Since $\delta_{k-1,n+s-1}$ should approach zero
monotonically as $s\rightarrow\infty$, there should be $s_{0}$ such that
$\delta_{k-1,n+s-1}<\delta_{k,n}$ for all $k$ and for all $s>s_{0}$.
Therefore, we get
$\delta_{k,n+s+T}=\delta_{k-1,n+s+T-1}=\delta_{k-2,n+s+T-2}=\delta_{k,n+s}$
for all $s>s_{0}$. Since $\delta$ cannot increase, we conclude that
$\delta_{k,n}$ is a constant for all sufficiently large $n$. If $\delta_{k,n}$
is strictly positive for all $n$ as assumed, $C_{k}$ cannot be a limit and we
arrive at a contradiction. Therefore, there is $t_{1}$ such that
$c_{t+T}=c_{t}$ for all $t\geq t_{1}$ in this case.

If $\alpha\neq\alpha_{T}$, then there is at least one $\varphi_{k}$ such that
$(T+1)/T<\varphi_{k}$. If $\varepsilon$ also satisfies
$\varepsilon/C_{k}<1-(T+1)/(T\varphi_{k})$, then we can write

$\displaystyle\delta_{k,n+1}=\min\left\\{\delta_{k,n},1-\tfrac{T-1}{T}\varphi_{k+1}(1-\delta_{k+1,n})\right\\},$

for all $n>m_{0}$. If $\varphi_{k+1}<T/(T-1)$, then $\delta_{k,n}$ will
eventually be smaller than $1-(T-1)\varphi_{k+1}/T$ and we have
$\delta_{k,n+1}=\delta_{k,n}=0$ for all large $n$. On the other hand, if
$\varphi_{k+1}=T/(T-1)>(T+1)/T$, we have

$\displaystyle\delta_{k,n+1}$
$\displaystyle=\min\left\\{\delta_{k,n},\delta_{k+1,n}\right\\},$
$\displaystyle\delta_{k+1,n+1}$
$\displaystyle=\min\left\\{\delta_{k+1,n},1-\tfrac{T-1}{T}\varphi_{k+2}(1-\delta_{k+2,n})\right\\}.$
(33)

Since it is impossible for all $\varphi_{k}$ to be $T/(T-1)$, there exists
$k^{\prime}$ such that $\varphi_{k+i}=T/(T-1)$ for $1\leq i\leq k^{\prime}$
and $\varphi_{k+k^{\prime}+1}<T/(T-1)$. Therefore,

$\delta_{k+k^{\prime},n+1}=\delta_{k+k^{\prime},n},$

for all sufficiently large $n$. Once $\delta_{k+k^{\prime},m}=0$, then
$\delta_{k+i,n}=0$ for all $0\leq i\leq k^{\prime}$ and for all $n>m+T$ by
(33). If $\varphi_{k+k^{\prime}+1}>(T+1)/T$, we can repeat the above
procedure. If $\varphi_{k+k^{\prime}+1}=(T+1)/T$, we have

$\delta_{k+k^{\prime}+1,n+1}=\min\left\\{\delta_{k+k^{\prime}+1,n},\delta_{k+k^{\prime},n},1-\tfrac{T-1}{T}\varphi_{k+k^{\prime}+2}(1-\delta_{k+k^{\prime}+2,n})\right\\}=0$

for all sufficiently large $n$. Hence, the proof is complete.

###  5.3 Non-uniqueness of periodic solutions

Proposition 17 and its proof have shown the general periodic solutions for
$t>t_{1}$ to be of the form

$\displaystyle\chi_{t}=\chi_{t_{1}}\prod_{k=t_{1}+1}^{t}\varphi_{k},\quad
c_{t}=c_{t_{1}}e^{-\nu(t-t_{1})}\prod_{k=t_{1}+1}^{t}\varphi_{k},$

where the $\varphi_{k}$ satisfy $\varphi_{T+k}=\varphi_{k}$ and (31). Since
$(T+1)/T\leq e^{\nu}<T/(T-1)$, setting $\varphi_{i}=e^{\nu}$ for all $i$
satisfies (31), which gives the constant sequence $c_{t}=c_{t_{1}}$. We refer
to this solution as the homogeneous state. Recall that the homogeneous state
is the unique possibility for $\alpha=\alpha_{T}$, as shown right after (31).
By constructing an appropriate sequence $(a_{n})$, we now show that any set
$\\{\varphi_{k}\\}$ that satisfies the conditions (31) can give rise to a
periodic solution $c_{t}$. Therefore, the periodic solution $c_{t}$ is not
unique and can vary substantially with $(a_{n})$ unless $\alpha=\alpha_{T}$ or
$T=1$.

######  Proposition 18.

Let

$\displaystyle a_{t}=\max\left\\{\frac{T-i+t-1}{\alpha}\psi_{i}\colon 0\leq
i<T\right\\},\quad\psi_{i}:=\prod_{j=1}^{i}\varphi_{j},$ (34)

with the convention $\prod_{j=1}^{0}=1$. Then

$\displaystyle\chi_{t}=\psi_{t+T-1}:=\prod_{j=1}^{t+T-1}\varphi_{j},$ (35)

where the $\varphi_{j}$ are as in (31) with periodicity
$\varphi_{T+j}=\varphi_{j}$.

Proof: To find $a_{1}$, we observe that for $0\leq i<T-1$

$\displaystyle\frac{T-i-1}{\alpha}\psi_{i+1}$
$\displaystyle=\frac{T-i}{\alpha}\psi_{i}\frac{T-i-1}{T-i}\varphi_{i+1}$
$\displaystyle\leq\frac{T-i}{\alpha}\psi_{i}\frac{(T-i-1)T}{(T-i)(T-1)}=\frac{T-i}{\alpha}\psi_{i}\frac{T^{2}-(i+1)T}{T^{2}-(i+1)+i}\leq\frac{T-i}{\alpha}\psi_{i},$

where we have used $\varphi_{i}\leq T/(T-1)$. Therefore, we get

$\chi_{1}=a_{1}=\frac{T}{\alpha}=\prod_{j=1}^{T}\varphi_{j}=\psi_{T},$

which is (35) for $t=1$. Note that this $\chi_{1}$ is trivially valid for
$T=1$.

Now assume (35) is valid up to $t=n$. Then,

$\chi_{n+1}=\max\left\\{\frac{T-i+n}{\alpha}\psi_{i}:0\leq i\leq
n+T-1\right\\}.$

For $i\leq n$, we have

$\displaystyle\frac{T-i+n}{\alpha}\psi_{i}$
$\displaystyle=\frac{T-i+n+1}{\alpha}\psi_{i-1}\frac{T-i+n}{T-i+n+1}\varphi_{i}$
$\displaystyle\geq\frac{T-i+n+1}{\alpha}\psi_{i-1}\frac{(T-i+n)(T+1)}{(T-i+n+1)T}$
$\displaystyle=\frac{T-i+n+1}{\alpha}\psi_{i-1}\frac{T^{2}+(n+1-i)T+n-i}{T^{2}+(n+1-i)T}$
$\displaystyle\geq\frac{T-i+n+1}{\alpha}\psi_{i-1},$

and for $n+T-1>i\geq n$ and $T>1$, we have

$\displaystyle\frac{T-i+n}{\alpha}\psi_{i}$
$\displaystyle=\frac{T-i+n-1}{\alpha}\psi_{i+1}\frac{T-i+n}{T-i+n-1}\frac{1}{\varphi_{i+1}}$
$\displaystyle\geq\frac{T-i+n-1}{\alpha}\psi_{i+1}\frac{(T-i+n)(T-1)}{(T-i+n-1)T}$
$\displaystyle=\frac{T-i+n-1}{\alpha}\psi_{i+1}\frac{T^{2}+(n-1-i)T+i-n}{T^{2}+(n-1-i)T}$
$\displaystyle\geq\frac{T-i+n-1}{\alpha}\psi_{i+1}.$

Therefore, we have

$\chi_{n+1}=\frac{T}{\alpha}\psi_{n}=\psi_{n}\prod_{j=n+1}^{T+n}\varphi_{j}=\prod_{j=1}^{T+n}\varphi_{j}.$

Induction completes the proof.

For a realization of (34) in the branching process, consider an initial
condition such that there are $T$ different mutant classes with fitness
$f_{i}:=f^{\psi_{i}/\alpha}$ ($0\leq i<T$) and the number $N_{i}$ of
individuals with fitness $f_{i}$ is

$N_{i}=\left\lfloor f^{(T-i-1)\psi_{i}/\alpha}\right\rfloor.$

Notice that this initial condition with $\varphi_{j}=e^{\nu^{\prime}}$
together with a shift in time was used in the proof of Lemma 16. In the limit
$f\rightarrow\infty$ as in Sec. 3, $X(t)$ is well approximated by
$f^{\chi_{t}}$ with $a_{t}$ in (34).

In the original branching process, the sequence $(a_{n})$ depends both on the
initial condition and the stochastic evolution in the early time regime before
the deterministic approximation through the recursion relation (9) becomes
valid. To see this, we recall from Sec. 3 how the recursion relation arises
from the stochastic process. Since on survival the total population size as
well as the largest fitness increases indefinitely, there should be a
generation $t_{0}$ such that $X(t_{0}+1)>K$ for any preassigned $K$. Let
$W_{0}$ be the largest fitness at generation $t_{0}$, define $Y=X(t_{0}+1)$
and introduce a shifted time variable $t^{\prime}=t-t_{0}$. If $K$ is
extremely large, $X(t^{\prime})$ can be well approximated as

$\displaystyle X(1)=Y\approx Y^{\chi_{1}},\quad W_{1}\approx
Y^{\chi_{1}/\alpha},$ $\displaystyle
X(2)=Y^{a_{2}}+Y^{\chi_{1}/\alpha}+1\approx Y^{\chi_{2}},\quad W_{2}\approx
Y^{\chi_{2}/\alpha},$ $\displaystyle
X(3)=Y^{a_{3}}+Y^{2\chi_{1}/\alpha}+Y^{\chi_{2}/\alpha}+1\approx
Y^{\chi_{3}},\quad W_{3}\approx Y^{\chi_{3}/\alpha}$

where $Y^{a_{n}}$ is the population size of all mutant classes that appeared
prior to generation $t_{0}$. Since $Y^{a_{n}}\leq YW_{0}^{n}$, we naturally
have $\lim_{n\rightarrow\infty}a_{n}e^{-\nu n}=0$, and $(a_{n})$ is a
permissible sequence that can be entered into the recursion relation (9).

###  5.4 Empirical frequency distribution for large $\alpha$

Whereas the preceding subsection has shown that the empirical frequency
distribution at long times is generally non-universal, we will now argue that
it nevertheless has a well-defined limit for $\alpha\rightarrow\infty$. Let us
begin with the homogeneous state. In this case,

$\displaystyle J_{i}(t)=e^{-\nu(t-i)},\quad P(t)\equiv\alpha(e^{\nu}-1),$
$\displaystyle
R_{i}(t)=\frac{(t-i)}{\alpha}J_{i}(t)-1=-\frac{1}{\nu\alpha}J_{i}(t)\log
J_{i}(t)-1.$

Since $\nu\alpha\rightarrow 1/e$ as $\alpha\rightarrow\infty$, the homogeneous
state for all sufficiently large $\alpha$ is well described by

$\displaystyle R_{i}\approx-eJ_{i}\log J_{i}-1,$ (36)

and the mean log fitness converges to $P=\frac{1}{e}$.

Moreover, since

$\frac{T}{T-1}-\frac{T+1}{T}=\frac{1}{(T-1)T}=O(T^{-2})$

and $T/\alpha\rightarrow e$ as $\alpha\rightarrow\infty$, in this limit all
periodic solutions that satisfy the constraints (31) become close to
homogeneous, $\varphi_{i}=e^{\nu}+O(\alpha^{-2})$. Therefore, we conjecture
that the empirical distribution on survival has (36) as a limit distribution
for $\alpha\rightarrow\infty$. As an illustration, in Fig. 3 we compare (36)
to numerical solutions of the recursion relation for $\alpha=3$, 4, 5, 6. The
numerical data are hardly distinguishable from (36) already for $\alpha=5$.

Figure 3:  Plot of $R_{i}$ vs $J_{i}$ at $t=500$ and 501 for $\alpha=3$, 4, 5,
6. For comparison, the asymptotic prediction (36) is depicted as a red solid
curve. For $\alpha=3$ and 4, the changes of the empirical distributions
between generations 500 and 501 are still visible, but the distributions
become indistinguishable from the asymptotic form (36) for $\alpha\geq 5$.

##  6 Summary and Discussion

In this article we have provided a detailed characterization of the
superexponential population growth in two closely related stochastic models of
evolution. To the best of our knowledge, this is the first rigorous analysis
of a branching process with selection and random mutations drawn from an
unbounded fitness distribution. A remarkable feature of the models considered
here is the emergence of an integer-valued time scale $T$ which depends
(discontinuously) on the index $\alpha$ of the underlying Fréchet
distribution. A partial understanding of the resulting periodic behaviour of
the population structure was achieved in a deterministic approximation.
Further work on this problem is needed, addressing in particular how the
stochastic initial phase of the process determines the non-universal aspects
of the asymptotic population distribution.

It is instructive to compare our findings for the branching process to the
earlier analysis of a stochastic fixed finite population version of Kingman’s
model in Park2008 . In both cases the long-time behavior is dominated by, and
can quantitatively understood in terms of extremal mutation events in the
past. However, in the fixed finite population model the likelihood of
generating mutants that exceed the current population fitness declines with
time, and the dynamics reduces to a modified record process, where the
takeover of the population by a fit mutant is instantaneous compared to the
waiting time for the next fitter mutant. As a consequence, the population at
time $t$ is dominated by a mutant that arose at a time of order $t$ in the
past. By contrast, in the branching process with Fréchet-type distributions,
the declining probability of exceeding the current fitness is compensated by
the rapid growth of the population in such a way that the time lag since the
birth of the currently dominant mutant takes on a fixed value $T$.

It is reasonable to expect that the growth of the population fitness in the
branching process is intermediate between that of the fixed finite population
model and the deterministic infinite population model. For Fréchet type
fitness distributions the deterministic model is ill-defined, but the analysis
of the fixed finite population model predicts a polynomial increase of the
fitness with exponent $1/\alpha$ Park2008 , which is indeed much slower than
the superexponential growth in the branching process. For unbounded Gumbel
type distributions the growth law of the fitness is known for infinite as well
as for finite populations Kingman1978 ; Park2008 . The corresponding behaviour
of the branching process will be addressed in future work.

Acknowledgments

S-CP acknowledges the support by the National Research Foundation of Korea
(NRF) grant funded by the Korea government (MSIT) (Grant No. 2020R1F1A1077065)
and by the Catholic University of Korea, research fund 2021. JK and PM were
supported by the German Excellence Initiative through the UoC Forum “Classical
and Quantum Dynamics of Interacting Particle Systems”.

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