# Long edges in Galton-Watson trees

Sergey Bocharov111S.Bocharov: Department of Foundational Mathematics, Xian
Jiaotong-Liverpool University, Ren’Ai Road 111, Suzhou 215123, China, e-mail:
<EMAIL_ADDRESS>and Simon C. Harris222S.C.Harris: Department of
Statistics, University of Auckland, 38 Princes Street, Auckland, 1001, New
Zealand, e-mail<EMAIL_ADDRESS>
###### Abstract

In this article, we will establish a number of results concerning the limiting
behaviour of the longest edges in the genealogical tree generated by a
continuous-time Galton-Watson (GW) process. Separately, we consider the large
time behaviour of the longest pendant edges, the longest (strictly) interior
edges, and the longest of all the edges. These results extend the special case
of long pendant edges of birth-death processes established in Bocharov,
Harris, Kominek, Mooers, and Steel [1] .

##  1 Introduction

We consider a continuous-time Galton-Watson process with branching rate
$\beta>0$ and offspring distribution $(p_{k})_{k\geq 0}$ initiated by a single
particle. Such process may be constructed as follows. It starts with a single
particle at time $0$. This initial particle lives for a random time $T$ which
is exponentially distributed with rate parameter $\beta$. That is,
$\mathbb{P}(T>t)=\mathrm{e}^{-\beta t}$ and $\mathbb{E}T={1}/{\beta}$. At the
moment when the initial particle dies, it produces a random number $\xi$ of
new particles where $\mathbb{P}(\xi=k)=p_{k}$ for every integer $k\geq 0$.
Newly-born particles, independently of each other and of all the previous
history, replicate the initial particle’s behaviour, that is, they live for
random times with $Exp(\beta)$ distribution, produce random numbers of new
particles according to the offspring distribution $(p_{k})_{k\geq 0}$ at the
moment of their death, and so on. Graphically such a process is most naturally
represented by a tree in which each edge corresponds to the life span of a
particle, as in Figure 1. We use $\mathcal{T}_{t}$ to denote the entire
process (tree) evolved up to time $t$.

Figure 1: A typical realisation of $\mathcal{T}_{t}$.

In this article we shall study the limiting behaviour of the longest edges in
$\mathcal{T}_{t}$ as $t\to\infty$. We classify edges in $\mathcal{T}_{t}$ as
either pendant (edges corresponding to particles alive at time $t$) or
interior (all other edges). If one thinks of $\mathcal{T}_{t}$ as describing
some sort of evolutionary process then pendant edges would correspond to
extant species and interior edges would correspond to extinct species.

Throughout, we assume the generic offspring random variable $\xi$ has mean
$m:=\mathbb{E}(\xi)>1$ (the Galton-Watson process is _supercritical_ and has
some strictly positive chance of surviving forever) as well as the classical
condition that $\mathbb{E}(\xi\log\xi)<\infty$. (Note, these assumptions
permit cases where the offspring variance can be infinite.) Conditional on
survival of the population, for any real number $x$, we will show that, for
large times $t$, the number of _pendant edges_ longer than $(1-1/m)t+x$
converges in distribution to a mixture of Poisson random variables with
(random) parameter $e^{-m\beta x}M_{\infty}$, where the random variable
$M_{\infty}$ is the classical additive martingale limit for the GW process.
(Intuitively speaking, $M_{\infty}$ roughly measures the growth of the process
during its earlier stages, alternatively, $\log M_{\infty}$ can be thought of
as a random time delay for the process to get very large and start behaving
deterministically). As a corollary, we show the length of the $k^{th}$ longest
pendant edge at large times $t$ converges in distribution when centred around
$(1-1/m)t$ and we give an explicit representation for this limiting law. In
particular, the length of the longest pendant edge at time $t$ minus
$(1-1/m)t$ converges in the large time $t$ limit to a mixture of Gumbel
distributions.

Similarly, we will prove analogous results for the longest _interior edges_
and the longest of _all edges_. It turns out that long edges should always be
centred around size $(1-1/m)t$ and always converge to a mixture of Poisson
distributions, where only the mixture parameter varies according to whether
long pendant, long interior, or longest of all edges are being considered,
respectively. The analysis and proofs for the longest interior edges, and
similarly longest of all edges, turn out to be more delicate and require
considerably more involved calculations than the longest pendant edges case
(where there is much more independence). The _LlogL_ moment assumption is the
expected to be best possible for our results to hold in light of the natural
appearance of $M_{\infty}$ in the description of the limit and the classical
Kesten-Stigum theorem for GW processes. However, as this assumption permits
the offspring variance to be infinite in general, it necessitates an
additional careful truncation argument to allow application of a second moment
method involving use of the Paley-Zygmund inequality.

###  1.1 Preliminaries

We use $m$ to denote the mean of the offspring distribution:

$m:=\mathbb{E}\xi=\sum_{k\geq 0}kp_{k}$

As we shall see, this is the only quantity which determines the first-order
limiting behaviour of all the longest edges in the tree. We also use $v$ to
denote the second moment of the offspring distribution:

$v:=\mathbb{E}\big{[}\xi^{2}\big{]}=\sum_{k\geq 0}k^{2}p_{k}$

In this article it is always assumed that $m<\infty$, whereas $v$ may be
finite or infinite.

We denote the number of particles alive at time $t$ by $N_{t}$ and the event
of the population surviving forever by

$\\{\text{survival}\\}:=\bigcap_{t\geq 0}\big{\\{}N_{t}>0\big{\\}}.$

We let $\\{$extinction$\\}=\\{$survival$\\}^{c}$ be the event of the
population eventually becoming extinct.

It is well-known that the process

$M_{t}=\mathrm{e}^{-(m-1)\beta t}N_{t}\quad,t\geq 0$ (1.1)

is a (positive) martingale, which then has an almost sure limit
$M_{\infty}=\lim_{t\to\infty}M_{t}$.

The following assumptions will be made in this article:

$m>1$ (A1)

and

$\mathbb{E}\big{[}\xi\log^{+}\xi\big{]}=\sum_{k\geq 1}k\log k\ p_{k}<\infty$
(A2)

Condition (A1) is necessary and sufficient for $\mathbb{P}($survival$)>0$ and
is known as the supercritical case. If condition (A1) is satisfied then the
celebrated Kesten-Stigum Theorem says that condition (A2) is necessary and
sufficient for $\mathbb{P}(M_{\infty}>0|$survival$)=1$ or, in other words, for
events $\\{M_{\infty}>0\\}$ and $\\{$survival$\\}$ to agree almost surely (see
e.g. [3]). This gives a fine approximation of the population growth:
$N_{t}\sim M_{\infty}\mathrm{e}^{(m-1)\beta t}$ as $t\to\infty$ a.s. on
survival (the words ‘a.s. on survival’ are commonly used to mean
‘$\mathbb{P}(\cdot|$survival$)=1$’). Let us remark that if condition
$\eqref{xlogx}$ is not satisfied then $\mathbb{P}(M_{\infty}=0)=1$ and
establishing the precise growth rate of $N_{t}$ is a difficult task.

###  1.2 Main results

We let

$\alpha^{\ast}:=1-\frac{1}{m}.$ (1.2)

As we shall soon see, all the longest edges in $\mathcal{T}_{t}$ are
approximately of length $\alpha^{\ast}t$ for $t$ large.

######  Definition 1.1.

For a length $l\geq 0$ and a time $t\geq 0$ we define $N_{t}^{l}$,
$\hat{N}_{t}^{l}$, and $\tilde{N}_{t}^{l}$ respectively to be the numbers of
pendant, interior, and all (counting both pendant and interior) edges in
$\mathcal{T}_{t}$ of length $\geq l$. Note, in what follows we shall often
take $l$ to be dependent on $t$.

For all the results stated below we assume that conditions (A1) and (A2) hold.
We let $\alpha^{\ast}$ be as in (1.2) and $M_{\infty}$ be the almost sure
limit of the martingale $M_{t}$ in (1.1).

######  Theorem 1.2.

Let $x\in\mathbb{R}$ be fixed, $t\geq 0$ variable and let us take

$l=l(t)=\alpha^{\ast}t+x.$

Then, under $\mathbb{P}(\cdot|$survival$)$,

$N_{t}^{l}\Rightarrow V_{x},\quad\hat{N}_{t}^{l}\Rightarrow\hat{V}_{x}\ \text{
and }\tilde{N}_{t}^{l}\Rightarrow\tilde{V}_{x}$ (1.3)

as $t\to\infty$, where $V_{x}$, $\hat{V}_{x}$ and $\tilde{V}_{x}$ are mixtures
of Poisson distributions with respective rates $\mathrm{e}^{-m\beta
x}M_{\infty}$, $\frac{1}{m-1}\mathrm{e}^{-m\beta x}M_{\infty}$ and
$\frac{m}{m-1}\mathrm{e}^{-m\beta x}M_{\infty}$ conditional on $M_{\infty}>0$.
To be precise, for $k=0$, $1$, $2$, …,

$\mathbb{P}\big{(}V_{x}=k\big{|}\text{survival}\big{)}=\mathbb{E}\Big{[}\frac{1}{k!}\Big{(}\mathrm{e}^{-m\beta
x}M_{\infty}\Big{)}^{k}\exp\big{\\{}-\mathrm{e}^{-m\beta
x}M_{\infty}\big{\\}}\Big{|}M_{\infty}>0\Big{]},$ (1.4)
$\mathbb{P}\big{(}\hat{V}_{x}=k\big{|}\text{survival}\big{)}=\mathbb{E}\Big{[}\frac{1}{k!}\Big{(}\frac{\mathrm{e}^{-m\beta
x}M_{\infty}}{m-1}\Big{)}^{k}\exp\Big{\\{}-\frac{\mathrm{e}^{-m\beta
x}M_{\infty}}{m-1}\Big{\\}}\Big{|}M_{\infty}>0\Big{]},$ (1.5)

and

$\mathbb{P}\big{(}\tilde{V}_{x}=k\big{|}\text{survival}\big{)}=\mathbb{E}\Big{[}\frac{1}{k!}\Big{(}\frac{m\mathrm{e}^{-m\beta
x}M_{\infty}}{m-1}\Big{)}^{k}\exp\Big{\\{}-\frac{m\mathrm{e}^{-m\beta
x}M_{\infty}}{m-1}\Big{\\}}\Big{|}M_{\infty}>0\Big{]}.$ (1.6)

Let us comment that more can be shown to be true. For example, if we take
$l_{1}=\alpha^{\ast}t+x$ and $l_{2}=\alpha^{\ast}t+y$ for $x<y$ then
$(N_{t}^{l_{1}},N_{t}^{l_{2}})\Rightarrow(V_{x},V_{y})$, where
$(V_{x},V_{y}-V_{x})$ is a mixture of pair independent Poisson distributions
with rates $\mathrm{e}^{-m\beta x}M_{\infty}$ and $(\mathrm{e}^{-m\beta
x}-\mathrm{e}^{-m\beta y})M_{\infty}$ conditional on $M_{\infty}>0$.

From Theorem 1.2 one can now recover the limiting distributions of all longest
edges in $\mathcal{T}_{t}$ (shifted by $\alpha^{\ast}t$).

######  Definition 1.3.

For $k=1$, $2$, $3$, … let $L_{t}^{(k)}$, $\hat{L}_{t}^{(k)}$ and
$\tilde{L}_{t}^{(k)}$ be the lengths of $k$th longest edge among pendant,
interior and all edges in $\mathcal{T}_{t}$ respectively.

######  Corollary 1.4.

For all $k=1$, $2$, $3$, …, under $\mathbb{P}(\cdot|$survival$)$,

$L_{t}^{(k)}-\alpha^{\ast}t\Rightarrow
W^{(k)},\quad\hat{L}_{t}^{(k)}-\alpha^{\ast}t\Rightarrow\hat{W}^{(k)}\ \text{
and }\tilde{L}_{t}^{(k)}-\alpha^{\ast}t\Rightarrow\tilde{W}^{(k)},$ (1.7)

where

$\mathbb{P}\big{(}W^{(k)}\leq
x|\text{survival}\big{)}=\mathbb{E}\Big{[}\sum_{j=0}^{k-1}\frac{1}{j!}\Big{(}\mathrm{e}^{-m\beta
x}M_{\infty}\Big{)}^{j}\exp\big{\\{}\\!-\\!\mathrm{e}^{-m\beta
x}M_{\infty}\big{\\}}\Big{|}M_{\infty}>0\Big{]},$
$\mathbb{P}\big{(}\hat{W}^{(k)}\leq
x|\text{survival}\big{)}=\mathbb{E}\Big{[}\sum_{j=0}^{k-1}\frac{1}{j!}\Big{(}\frac{\mathrm{e}^{-m\beta
x}M_{\infty}}{m-1}\Big{)}^{j}\exp\Big{\\{}-\frac{\mathrm{e}^{-m\beta
x}M_{\infty}}{m-1}\Big{\\}}\Big{|}M_{\infty}>0\Big{]}$

and

$\mathbb{P}\big{(}\tilde{W}^{(k)}\leq
x|\text{survival}\big{)}=\mathbb{E}\Big{[}\sum_{j=0}^{k-1}\frac{1}{j!}\Big{(}\frac{m\mathrm{e}^{-m\beta
x}M_{\infty}}{m-1}\Big{)}^{j}\exp\Big{\\{}-\frac{m\mathrm{e}^{-m\beta
x}M_{\infty}}{m-1}\Big{\\}}\Big{|}M_{\infty}>0\Big{]}$

In particular, the limiting distributions of the longest pendant, interior and
any edge centred around $\alpha^{\ast}t$ are mixtures of Gumbel distributions:

$\mathbb{P}\big{(}W^{(1)}\leq
x|\text{survival}\big{)}=\mathbb{E}\Big{[}\exp\big{\\{}\\!-\\!\mathrm{e}^{-m\beta
x}M_{\infty}\big{\\}}\Big{|}M_{\infty}>0\Big{]}$

(and similarly for $\hat{W}^{(1)}$ and $\tilde{W}^{(1)}$).

###### Proof of Corollary 1.4.

We note that

$\mathbb{P}\big{(}L_{t}^{(k)}>\alpha^{\ast}t+x\big{|}\text{survival}\big{)}=\mathbb{P}\big{(}L_{t}^{(k)}\geq\alpha^{\ast}t+x\big{|}\text{survival}\big{)}=\mathbb{P}\big{(}N_{t}^{l}\geq
k\big{|}\text{survival}\big{)},$

where $l=\alpha^{\ast}t+x$. Hence, combining this observation with Theorem
1.2,

$\mathbb{P}(L_{t}^{(k)}-\alpha^{\ast}t\leq
x|\text{survival})=\mathbb{P}\big{(}N_{t}^{l}<k\big{|}\text{survival}\big{)}\to\mathbb{P}\big{(}V_{x}<k\big{|}\text{survival}\big{)}$

as $t\to\infty$. The proofs for $\hat{L}_{t}^{(k)}$ and $\tilde{L}_{t}^{(k)}$
are essentially the same. ∎

######  Remark 1.5.

Let us note that conditioning on the process surviving forever may be easily
replaced with conditioning on survival up to time $t$ whenever it would make
sense since $\mathbb{P}(N_{t}>0)\searrow\mathbb{P}($survival$)$ as
$t\to\infty$. For example, $\mathbb{P}(V_{x}=k|$survival$)$ in Theorem 1.2 may
be replaced with $\lim_{t\to\infty}\mathbb{P}(N_{t}^{l}=k|N_{t}>0)$ and
likewise for interior and all edges. Similarly, $\mathbb{P}(W^{(k)}\leq x|$
survival $)$ in Corollary 1.4 may be replaced with
$\lim_{t\to\infty}\mathbb{P}(L_{t}^{(k)}-\alpha^{\ast}t\leq x|N_{t}>0)$ and
likewise for interior and all edges.

It is also easy to see (after dividing (1.7) by $t$ and using some standard
results about convergence of random variables) that
$\frac{L_{t}^{(k)}}{t}\to\alpha^{\ast}$ in
$\mathbb{P}(\cdot|$survival$)$-probability. Or, for the same reason as above,
$\lim_{t\to\infty}\mathbb{P}\big{(}\big{|}\frac{L_{t}^{(k)}}{t}-\alpha^{\ast}\big{|}>\epsilon\big{|}N_{t}>0\big{)}=0$
for all $\epsilon>0$ and likewise for interior and pendant edges. In fact, a
stronger result is true.

######  Theorem 1.6.

For $k=1$, $2$, $3$, …,

$\frac{L_{t}^{(k)}}{t}\to\alpha^{\ast},\quad\frac{\hat{L}_{t}^{(k)}}{t}\to\alpha^{\ast}\
\text{ and }\frac{\tilde{L}_{t}^{(k)}}{t}\to\alpha^{\ast}$ (1.8)

as $t\to\infty$ a.s. on survival.

####  1.2.1 Special case: Long edges in the Birth-Death process

We can specialise the above results about long edges in general Galton-Watson
results to the Birth-Death process. The Birth-Death process is not only an
important model for many applications, but it also permits further explicit
calculations and simplified expressions. Previously, only the behaviour of
long pendant edges for Birth-Death processes had been established in [1].

######  Example 1.7.

Consider a supercritical birth-death process with birth rate $\lambda$ and
death rate $\mu$ ($\lambda>\mu$). This is a special case of a Galton-Watson
process with branching rate $\beta=\lambda+\mu$ and the offspring distribution
$p_{0}=\frac{\mu}{\lambda+\mu}$, $p_{2}=\frac{\lambda}{\lambda+\mu}$.

In this case $m=\frac{2\lambda}{\lambda+\mu}$ and so

$\alpha^{\ast}=1-\frac{\lambda+\mu}{2\lambda}=\frac{\lambda-\mu}{2\lambda}=\frac{1}{2}\big{(}1-\frac{\mu}{\lambda}\big{)}.$

Moreover, in this case $M_{\infty}$ has $Exp(1-\frac{\mu}{\lambda})$
distribution conditional on survival (see e.g. [2]).

One can then easily check that $V_{x}$, $\hat{V}_{x}$, and $\tilde{V}_{x}$
from Theorem 1.2 follow Geometric distribution on $\\{0,1,2,\cdots\\}$ with
parameters $\frac{1}{1+\frac{\lambda}{\lambda-\mu}\mathrm{e}^{-m\beta x}}$,
$\frac{1}{1+\frac{\lambda}{\lambda-\mu}\frac{\mathrm{e}^{-m\beta x}}{m-1}}$,
and $\frac{1}{1+\frac{\lambda}{\lambda-\mu}\frac{m\mathrm{e}^{-m\beta
x}}{m-1}}$, respectively.

In addition, the cumulative distribution functions of $W^{(k)}$,
$\hat{W}^{(k)}$ and $\tilde{W}^{(k)}$ from Corollary 1.4 become:

$\mathbb{P}\big{(}W^{(k)}\leq
x|\text{survival}\big{)}=1-\frac{1}{\big{(}1+\frac{\lambda-\mu}{\lambda}\mathrm{e}^{m\beta
x}\big{)}^{k}},$ $\mathbb{P}\big{(}\hat{W}^{(k)}\leq
x|\text{survival}\big{)}=1-\frac{1}{\big{(}1+\frac{\lambda-\mu}{\lambda}(m-1)\mathrm{e}^{m\beta
x}\big{)}^{k}}$

and

$\mathbb{P}\big{(}\tilde{W}^{(k)}\leq
x|\text{survival}\big{)}=1-\frac{1}{\big{(}1+\frac{\lambda-\mu}{\lambda}\frac{m-1}{m}\mathrm{e}^{m\beta
x}\big{)}^{k}}.$

In particular, $W^{(1)}$, $\hat{W}^{(1)}$ and $\tilde{W}^{(1)}$ all follow
Logistic distributions with the same scale parameter $\frac{1}{m\beta}$ but
different location parameters
$\frac{1}{m\beta}\log(\frac{\lambda}{\lambda-\mu})$,
$\frac{1}{m\beta}\log(\frac{1}{m-1}\cdot\frac{\lambda}{\lambda-\mu})$, and
$\frac{1}{m\beta}\log(\frac{m}{m-1}\cdot\frac{\lambda}{\lambda-\mu})$,
respectively.

####  1.2.2 Structure of this article

The rest of the article is organised as follows. A simple proof of Theorem 1.2
for pendant edges will be presented in Section 2. The proof of Theorem 1.2 for
interior and all edges will be given in Section 3 with all the necessary heavy
calculations presented in Section 4 separately. Finally, Theorem 1.6 will be
proved in Section 5.

##  2 Proof of Theorem 1.2 for pendant edges.

Recall that $N_{t}^{l}$ is the number of pendant edges of length $\geq l$ at
time $t$.

###### Proof of Theorem 1.2 for pendant edges.

We observe that for $t>l$

$N_{t}^{l}=\sum_{k=1}^{N_{t-l}}\mathbf{1}_{A_{k}},$ (2.1)

where

$A_{k}=\big{\\{}k\text{th particle alive at time }t-l\text{ does not die in
the time interval }[t-l,t]\big{\\}}.$

Events $A_{k}$, $k\in N_{t-l}$ are independent under
$\mathbb{P}(\cdot|\mathcal{F}_{t-l})$ with

$\mathbb{P}(A_{k}|\mathcal{F}_{t-l})=\mathrm{e}^{-\beta l}.$

We define $m_{t}^{l}:=\mathbb{E}N_{t}^{l}$ and
$v_{t}^{l}=\mathbb{E}\big{(}N_{t}^{l}\big{)}^{2}$. If we then recall the facts
(see e.g. [2]) that

$\mathbb{E}N_{\tau}=\mathrm{e}^{\beta(m-1)\tau}$ (2.2)

and, assuming $v<\infty$,

$\mathbb{E}\big{[}N_{\tau}(N_{\tau}-1)\big{]}=\frac{v-m}{m-1}\Big{(}\mathrm{e}^{2\beta(m-1)\tau}-\mathrm{e}^{\beta(m-1)\tau}\Big{)}$
(2.3)

for any $\tau\geq 0$ then it follows from (2.1) that

$m_{t}^{l}=\mathrm{e}^{-\beta l}\mathbb{E}N_{t-l}=\mathrm{e}^{-\beta
l}\cdot\mathrm{e}^{\beta(m-1)(t-l)}=\mathrm{e}^{-\beta(ml-(m-1)t)}$ (2.4)

and, assuming $v<\infty$,

$\displaystyle v_{t}^{l}=$
$\displaystyle\mathbb{E}\Big{[}\sum_{k=1}^{N_{t-l}}\mathbf{1}_{A_{k}}+\sum_{\begin{subarray}{c}i,j=1\\\
i\neq j\end{subarray}}^{N_{t-l}}\mathbf{1}_{A_{i}}\mathbf{1}_{A_{j}}\Big{]}$
(2.5) $\displaystyle=$ $\displaystyle\mathrm{e}^{-\beta
l}\mathbb{E}N_{t-l}+\mathrm{e}^{-2\beta
l}\mathbb{E}\big{[}N_{t-l}(N_{t-l}-1)\big{]}$ $\displaystyle=$
$\displaystyle\mathrm{e}^{-\beta
l}\cdot\mathrm{e}^{\beta(m-1)(t-l)}+\frac{v-m}{m-1}\mathrm{e}^{-2\beta
l}\Big{(}\mathrm{e}^{2\beta(m-1)(t-l)}-\mathrm{e}^{\beta(m-1)(t-l)}\Big{)}$
$\displaystyle=$
$\displaystyle\mathrm{e}^{-\beta(ml-(m-1)t)}\Big{[}1+\frac{v-m}{m-1}\mathrm{e}^{-\beta(ml-(m-1)t)}-\frac{v-m}{m-1}\mathrm{e}^{-\beta
l}\Big{]}$ (2.6)

Moreover, the probability generating function of $N_{t}^{l}$ satisfies

$\mathbb{E}\Big{[}\theta^{N_{t}^{l}}\Big{]}=\mathbb{E}\Big{[}\theta^{\sum_{k=1}^{N_{t-l}}\mathbf{1}_{A_{k}}}\Big{]}=\mathbb{E}\Big{[}\prod_{k=1}^{N_{t-l}}\theta^{\mathbf{1}_{A_{k}}}\Big{]}=\mathbb{E}\Big{[}\Big{(}\theta\mathbb{P}(A_{k})+1-\mathbb{P}(A_{k})\Big{)}^{N_{t-l}}\Big{]}=\mathbb{E}\Big{[}\Big{(}1+(\theta-1)\mathrm{e}^{-\beta
l}\Big{)}^{N_{t-l}}\Big{]},$

$\theta\in(-1,1)$. If we now take

$l=l(t)=\alpha^{\ast}t+x=\big{(}1-\frac{1}{m}\big{)}t+x$

for some fixed $x$ and a variable $t$ then

$\displaystyle\mathbb{E}\Big{[}\Big{(}1+(\theta-1)\mathrm{e}^{-\beta
l}\Big{)}^{N_{t-l}}\Big{]}=$
$\displaystyle\mathbb{E}\Big{[}\Big{(}\big{(}1+(\theta-1)\mathrm{e}^{-\beta
l}\big{)}^{\mathrm{e}^{\beta(m-1)(t-l)}}\Big{)}^{\mathrm{e}^{-\beta(m-1)(t-l)}N_{t-l}}\Big{]}$
$\displaystyle=$
$\displaystyle\mathbb{E}\Big{[}\Big{(}\big{(}1+(\theta-1)\mathrm{e}^{-\beta
l}\big{)}^{\mathrm{e}^{\beta(m-1)(t-l)}}\Big{)}^{M_{t-l}}\Big{]}$
$\displaystyle=$
$\displaystyle\mathbb{E}\Big{[}\Big{(}\big{(}1+(\theta-1)\mathrm{e}^{-m\beta
x}\mathrm{e}^{-\beta(m-1)(t-l)}\big{)}^{\mathrm{e}^{\beta(m-1)(t-l)}}\Big{)}^{M_{t-l}}\Big{]}$
$\displaystyle\to$
$\displaystyle\mathbb{E}\Big{[}\exp\big{\\{}(\theta-1)\mathrm{e}^{-m\beta
x}M_{\infty}\big{\\}}\Big{]}$

as $t\to\infty$ using the facts that $\mathrm{e}^{\beta(m-1)(t-l)}\to\infty$
as $t\to\infty$, $(1+\frac{c}{\tau})^{\tau}\to\mathrm{e}^{c}$ as
$\tau\to\infty$, $M_{t-l}\to M_{\infty}$ as $t\to\infty$ a.s. on survival and
bounded convergence theorem. It also follows that

$\displaystyle\mathbb{E}\big{[}\theta^{N_{t}^{l}}\big{|}\text{survival}\big{]}=$
$\displaystyle\frac{\mathbb{E}\big{[}\theta^{N_{t}^{l}}\big{]}-\mathbb{E}\big{[}\theta^{N_{t}^{l}}\mathbf{1}_{\\{extinction\\}}\big{]}}{\mathbb{P}(survival)}$
$\displaystyle\to$
$\displaystyle\frac{\mathbb{E}\big{[}\exp\\{(\theta-1)\mathrm{e}^{-m\beta
x}M_{\infty}\\}\big{]}-\mathbb{P}(extinction)}{\mathbb{P}(survival)}$
$\displaystyle=$
$\displaystyle\frac{\mathbb{E}\big{[}\exp\\{(\theta-1)\mathrm{e}^{-m\beta
x}M_{\infty}\\}\big{]}-\mathbb{P}(M_{\infty}=0)}{\mathbb{P}(M_{\infty}>0)}$
$\displaystyle=$
$\displaystyle\mathbb{E}\Big{[}\exp\big{\\{}(\theta-1)\mathrm{e}^{-m\beta
x}M_{\infty}\big{\\}}\Big{|}M_{\infty}>0\Big{]}$

recalling that events $\\{$extinction$\\}$ and $\\{M_{\infty}>0\\}$ are almost
surely equal. This establishes (1.3) for $N_{t}^{l}$ since
$\exp\big{\\{}(\theta-1)\mathrm{e}^{-m\beta x}M_{\infty}\big{\\}}$ is the
probability generating function of a Poisson random variable with parameter
$\mathrm{e}^{-m\beta x}M_{\infty}$. ∎

##  3 Proof of Theorem 1.2 for interior and all edges.

Recall that $\hat{N}_{t}^{l}$ and $\tilde{N}_{t}^{l}$ are the numbers of
interior and all edges respectively of length $\geq l$ in the tree
$\mathcal{T}_{t}$.

######  Proposition 3.1.

Let $\hat{m}_{t}^{l}=\mathbb{E}\hat{N}_{t}^{l}$ and
$\tilde{m}_{t}^{l}=\mathbb{E}\tilde{N}_{t}^{l}$. Then

$\hat{m}_{t}^{l}=\left\\{\begin{array}[]{rl}0&\text{if }t\leq l\\\
\frac{1}{m-1}\mathrm{e}^{-\beta
l}\big{(}\mathrm{e}^{\beta(m-1)(t-l)}-1\big{)}&\text{if
}t>l\end{array}\right.$ (3.1)

and

$\tilde{m}_{t}^{l}=\left\\{\begin{array}[]{rl}0&\text{if }t\leq l\\\
\frac{1}{m-1}\mathrm{e}^{-\beta
l}\big{(}m\mathrm{e}^{\beta(m-1)(t-l)}-1\big{)}&\text{if
}t>l\end{array}\right.$ (3.2)

###### Proof.

Trivially $\hat{m}_{t}^{l}=\tilde{m}_{t}^{l}=0$ when $t\leq l$ because edges
in $\mathcal{T}_{t}$ cannot be longer than $t$. For the rest of the proof let
$l$ be fixed and let $t>l$. If $T$ is the branching time of the initial
particle (so that $\mathbb{P}(T>t)=\mathrm{e}^{-\beta t}$) and $\xi$ is the
number of children produced by the initial particle then

$\displaystyle\hat{N}_{t}^{l}=$ $\displaystyle
0\cdot\mathbf{1}_{\\{T>t\\}}+\Big{[}1+\sum_{k=1}^{\xi}\hat{N}_{t-T}^{l}(k)\Big{]}\mathbf{1}_{\\{l<T\leq
t\\}}+\Big{[}0+\sum_{k=1}^{\xi}\hat{N}_{t-T}^{l}(k)\Big{]}\mathbf{1}_{\\{0\leq
T\leq l\\}}$ $\displaystyle=$ $\displaystyle\mathbf{1}_{\\{l<T\leq
t\\}}+\mathbf{1}_{\\{0\leq T\leq t\\}}\sum_{k=1}^{\xi}\hat{N}_{t-T}^{l}(k),$
(3.3)

where $\hat{N_{\tau}^{l}}(k)$ is the number of interior edges of length $\geq
l$ in the subtree initiated by $k$th child of the initial particle and which
has evolved for a period of time $\tau$ after it has been initiated.

Taking the expectation of the above equation and using the Markov property
gives

$\displaystyle\hat{m}_{t}^{l}=$ $\displaystyle(\mathrm{e}^{-\beta
l}-\mathrm{e}^{-\beta
t})+\mathbb{E}[\xi]\int_{0}^{t}\hat{m}_{t-s}^{l}\beta\mathrm{e}^{-\beta
s}\mathrm{d}s$ $\displaystyle=$ $\displaystyle(\mathrm{e}^{-\beta
l}-\mathrm{e}^{-\beta t})+m\beta\mathrm{e}^{-\beta
t}\int_{0}^{t}\hat{m}_{u}^{l}\mathrm{e}^{\beta u}\mathrm{d}u.$

This translates to the differential equation

$\left\\{\begin{array}[]{rl}(\hat{m}_{t}^{l})^{\prime}&=\beta(m-1)\hat{m}_{t}^{l}+\beta\mathrm{e}^{-\beta
l}\quad,\ t>l\\\ \hat{m}_{l+}^{l}&=0\end{array}\right.$

(Continuity and differentiability of $\hat{m}_{t}^{l}$ follow from the
integral equation.) Multiplying the differential equation by the integrating
factor $\mathrm{e}^{-\beta(m-1)t}$ gives

$(\hat{m}_{t}^{l})^{\prime}\mathrm{e}^{-\beta(m-1)t}-\beta(m-1)\mathrm{e}^{-\beta(m-1)t}\hat{m}_{t}^{l}=\beta\mathrm{e}^{-\beta
l}\mathrm{e}^{-\beta(m-1)t}$

Integrating this equation over the time interval $[l,t]$ and noting that the
left hand side can be written as
$\frac{\mathrm{d}}{\mathrm{d}t}\big{(}\hat{m}_{t}^{l}\mathrm{e}^{-\beta(m-1)t}\big{)}$
we get

$\Big{[}\hat{m}_{s}^{l}\mathrm{e}^{-\beta(m-1)s}\Big{]}_{l}^{t}=\frac{\mathrm{e}^{-\beta
l}}{m-1}\int_{l}^{t}\beta(m-1)\mathrm{e}^{-\beta(m-1)s}\mathrm{d}s$

and hence using the initial condition $\hat{m}_{l+}^{l}=0$ we get

$\hat{m}_{t}^{l}\mathrm{e}^{-\beta(m-1)t}=\frac{\mathrm{e}^{-\beta
l}}{m-1}\Big{(}\mathrm{e}^{-\beta(m-1)l}-\mathrm{e}^{-\beta(m-1)t}\Big{)}$

which yields

$\hat{m}_{t}^{l}=\frac{1}{m-1}\mathrm{e}^{-\beta
l}\big{(}\mathrm{e}^{\beta(m-1)(t-l)}-1\big{)}$

for $t>l$. Then using the fact that
$\tilde{N}^{l}_{t}=N^{l}_{t}+\hat{N}^{l}_{t}$ and recalling the expression for
$m_{t}^{l}=\mathbb{E}N_{t}^{l}$ from (2.4) we get

$\tilde{m}_{t}^{l}=m_{t}^{l}+\hat{m}_{t}^{l}=\frac{1}{m-1}\mathrm{e}^{-\beta
l}\big{(}m\mathrm{e}^{\beta(m-1)(t-l)}-1\big{)}$

for $t>l$. ∎

One may now see that if $l=l(t)$ is taken to depend on $t$ in any way such
that

$t-l\to\infty$ (3.4)

as $t\to\infty$ then

$\hat{m}_{t}^{l}\sim\frac{1}{m-1}\mathrm{e}^{-\beta
l}\cdot\mathrm{e}^{\beta(m-1)(t-l)}=\frac{1}{m-1}\mathrm{e}^{-\beta(ml-(m-1)t)}$

and

$\tilde{m}_{t}^{l}\sim\frac{m}{m-1}\mathrm{e}^{-\beta
l}\cdot\mathrm{e}^{\beta(m-1)(t-l)}=\frac{m}{m-1}\mathrm{e}^{-\beta(ml-(m-1)t)}$

as $t\to\infty$. Thus, if in addition to (3.4) we assume that $l=l(t)$ is such
that

$ml-(m-1)t\to\infty\quad\text{ as }t\to\infty$ (3.5)

(for example, if we take $l=\alpha t$, $\alpha\in(\alpha^{\ast},1)$) then
$\hat{m}_{t}^{l}$ and $\tilde{m}_{t}^{l}$ will converge to $0$, whereas if we
assume that

$ml-(m-1)t\to-\infty\quad\text{ as }t\to\infty$ (3.6)

(for example, if we take $l=\alpha t$, $\alpha\in(0,\alpha^{\ast}$) then
$\hat{m}_{t}^{l}$ and $\tilde{m}_{t}^{l}$ will diverge to $\infty$.

In fact, under conditions (3.4) and (3.5) on $l=l(t)$, the decay rates of
$\hat{m}_{t}^{l}$ and $\tilde{m}_{t}^{l}$ will give the decay rates of
$\mathbb{P}(\hat{N}_{t}^{l}>0)$ and $\mathbb{P}(\tilde{N}_{t}^{l}>0)$ as is
stated in the following proposition.

######  Proposition 3.2.

Let $l=l(t)$ depend on $t$ in any way such that $t-l\to\infty$ and
$ml-(m-1)t\to\infty$ as $t\to\infty$ (e.g., one may have $l=\alpha t+x$ for
some $\alpha\in(\alpha^{\ast},1)$ and $x\in\mathbb{R}$). Then

$\mathbb{P}\big{(}\hat{N}_{t}^{l}>0\big{)}\sim\hat{m}_{t}^{l}\sim\frac{1}{m-1}\mathrm{e}^{-\beta(ml-(m-1)t)}=\frac{1}{m-1}\mathrm{e}^{-\beta
m(l-\alpha^{\ast}t)}$ (3.7)

and

$\mathbb{P}\big{(}\tilde{N}_{t}^{l}>0\big{)}\sim\tilde{m}_{t}^{l}\sim\frac{m}{m-1}\mathrm{e}^{-\beta(ml-(m-1)t)}=\frac{m}{m-1}\mathrm{e}^{-\beta
m(l-\alpha^{\ast}t)}$ (3.8)

as $t\to\infty$.

The proof of Proposition 3.2, due to its length, will be given separately in
Section 4. Assuming validity of (3.7) and (3.8) we shall now proceed with the
proof of Theorem 1.2 for interior and all edges.

###### Proof of Theorem 1.2 for interior and all edges.

Let us fix $x\in\mathbb{R}$. For a variable $t\geq 0$ we take

$l=l(t)=\alpha^{\ast}t+x.$

We need to show that $\hat{N}_{t}^{l}$ and $\tilde{N}_{t}^{l}$ converge in
distribution to certain mixed Poisson distributions given by (1.5) and (1.6).
We shall do this by proving that the probability generating functions of
$\hat{N}_{t}^{l}$ and $\tilde{N}_{t}^{l}$ satisfy

$\mathbb{E}\big{[}\theta^{\hat{N}_{t}^{l}}\big{|}\text{survival}\big{]}\to\mathbb{E}\Big{[}\exp\big{\\{}(\theta-1)\frac{\mathrm{e}^{-m\beta
x}M_{\infty}}{m-1}\big{\\}}\Big{|}M_{\infty}>0\Big{]}$ (3.9)

and

$\mathbb{E}\big{[}\theta^{\tilde{N}_{t}^{l}}\big{|}\text{survival}\big{]}\to\mathbb{E}\Big{[}\exp\big{\\{}(\theta-1)\frac{m\mathrm{e}^{-m\beta
x}M_{\infty}}{m-1}\big{\\}}\Big{|}M_{\infty}>0\Big{]}$ (3.10)

as $t\to\infty$ for all $\theta\in(-1,1)$.

Let us take a small $\delta>0$ ($\delta<\frac{\alpha^{\ast}}{m}$ will
suffice). Then for all $t$ sufficiently large,

$\sum_{k=1}^{N_{\delta
t}}\hat{N}_{(1-\delta)t}^{l}(k)\leq\hat{N}_{t}^{l}\leq\sum_{k=1}^{N_{\delta
t}}\mathbf{1}_{B_{k}}+\sum_{k=1}^{N_{\delta t}}\hat{N}_{(1-\delta)t}^{l}(k)$
(3.11)

and

$\sum_{k=1}^{N_{\delta
t}}\tilde{N}_{(1-\delta)t}^{l}(k)\leq\tilde{N}_{t}^{l}\leq\sum_{k=1}^{N_{\delta
t}}\mathbf{1}_{B_{k}}+\sum_{k=1}^{N_{\delta
t}}\tilde{N}_{(1-\delta)t}^{l}(k),$ (3.12)

where for $k\in\\{1,2,\cdots,N_{\delta t}\\}$, $\hat{N}_{(1-\delta)t}^{l}(k)$
and $\tilde{N}_{(1-\delta)t}^{l}(k)$ are the numbers of interior and all edges
of length $\geq l$ in subtrees $\mathcal{T}_{(1-\delta)t}(k)$ initiated by
$k$th particle at time $\delta t$ and

$B_{k}=\big{\\{}k\text{th particle alive at time }\delta t\text{ does not die
in the time interval }[\delta t,l]\big{\\}}.$

Since $\mathbb{P}(B_{k}|\mathcal{F}_{\delta t})=\mathrm{e}^{-\beta(l-\delta
t)}$, $k=1$, …, $N_{\delta t}$, it follows that

$\mathbb{P}\Big{(}\sum_{k=1}^{N_{\delta
t}}\mathbf{1}_{B_{k}}>0\Big{|}\mathcal{F}_{\delta
t}\Big{)}\leq\mathbb{E}\Big{[}\sum_{k=1}^{N_{\delta
t}}\mathbf{1}_{B_{k}}\big{|}\mathcal{F}_{\delta
t}\Big{]}=\mathrm{e}^{-\beta(l-\delta t)}N_{\delta
t}=\mathrm{e}^{-\beta(l-m\delta t)}M_{\delta t}\to 0$ (3.13)

$\mathbb{P}$-a.s. as $t\to\infty$.

Let us also consider

$\hat{\varphi}^{l}_{(1-\delta)t}(\theta)=\mathbb{E}\Big{[}\theta^{\hat{N}^{l}_{(1-\delta)t}}\Big{]}=\sum_{k=0}^{\infty}\theta^{k}\mathbb{P}\big{(}\hat{N}^{l}_{(1-\delta)t}=k\big{)}$

for $\theta\in(-1,1)$. We may write it as

$\displaystyle\hat{\varphi}^{l}_{(1-\delta)t}(\theta)=$
$\displaystyle\mathbb{P}\big{(}\hat{N}^{l}_{(1-\delta)t}=0\big{)}+\theta\mathbb{P}\big{(}\hat{N}^{l}_{(1-\delta)t}>0\big{)}+\sum_{k=2}^{\infty}(\theta^{k}-\theta)\mathbb{P}\big{(}\hat{N}^{l}_{(1-\delta)t}=k\big{)}$
$\displaystyle=$ $\displaystyle
1+(\theta-1)\mathbb{P}\big{(}\hat{N}^{l}_{(1-\delta)t}>0\big{)}+\sum_{k=2}^{\infty}(\theta^{k}-\theta)\mathbb{P}\big{(}\hat{N}^{l}_{(1-\delta)t}=k\big{)}.$

Then recalling from Proposition 3.2 that

$\mathbb{P}\big{(}\hat{N}^{l}_{(1-\delta)t}>0\big{)}\sim\mathbb{E}\hat{N}^{l}_{(1-\delta)t}=\hat{m}^{l}_{(1-\delta)t}\sim\frac{1}{m-1}\mathrm{e}^{-\beta(ml-(m-1)(1-\delta)t)}=\frac{\mathrm{e}^{-\beta
mx}}{m-1}\mathrm{e}^{-\beta(m-1)\delta t}$

as $t\to\infty$ and using the fact that $|\theta^{k}-\theta|\leq 2$ we see
that

$\Big{|}\sum_{k=2}^{\infty}(\theta^{k}-\theta)\mathbb{P}\big{(}\hat{N}^{l}_{(1-\delta)t}=k\big{)}\Big{|}\leq
2\mathbb{P}\big{(}\hat{N}^{l}_{(1-\delta)t}>1\big{)}\leq
2\Big{(}\mathbb{E}\hat{N}^{l}_{(1-\delta)t}-\mathbb{P}\big{(}\hat{N}^{l}_{(1-\delta)t}>0\big{)}\Big{)}=o\big{(}\mathrm{e}^{-\beta(m-1)\delta
t}\big{)}$

as $t\to\infty$ and therefore

$\displaystyle\hat{\varphi}^{l}_{(1-\delta)t}(\theta)=$ $\displaystyle
1+(\theta-1)\mathbb{P}\big{(}\hat{N}^{l}_{(1-\delta)t}>0\big{)}+o\big{(}\mathrm{e}^{-\beta(m-1)\delta
t}\big{)}$ $\displaystyle=$ $\displaystyle
1+(\theta-1)\frac{\mathrm{e}^{-\beta mx}}{m-1}\mathrm{e}^{-\beta(m-1)\delta
t}+o\big{(}\mathrm{e}^{-\beta(m-1)\delta t}\big{)}$ (3.14)

Then using (3.11), (3.13), the facts that
$\mathbb{E}\big{[}\theta^{\sum_{k=1}^{N_{\delta
t}}\hat{N}_{(1-\delta)t}^{l}(k)}\big{|}\mathcal{F}_{\delta
t}\big{]}=\big{(}\hat{\varphi}^{l}_{(1-\delta)t}(\theta)\big{)}^{N_{\delta
t}}$,

$\\{N_{\delta t}>0\\}\searrow\\{$survival$\\}$ and
$\mathbb{P}(M_{\infty}>0|$survival$)=1$ and bounded convergence we see that

$\displaystyle\mathbb{E}\Big{[}\theta^{\hat{N}^{l}_{t}}\big{|}\text{survival}\Big{]}=$
$\displaystyle\mathbb{E}\Big{[}\theta^{\hat{N}^{l}_{t}}\big{|}N_{\delta
t}>0\Big{]}+o(1)$ $\displaystyle=$
$\displaystyle\mathbb{E}\Big{[}\theta^{\sum_{k=1}^{N_{\delta
t}}\hat{N}_{(1-\delta)t}^{l}(k)}\big{|}N_{\delta t}>0\Big{]}+o(1)$
$\displaystyle=$
$\displaystyle\mathbb{E}\Big{[}\Big{(}\hat{\varphi}^{l}_{(1-\delta)t}(\theta)\Big{)}^{N_{\delta
t}}\big{|}N_{\delta t}>0\Big{]}+o(1)$ $\displaystyle=$
$\displaystyle\mathbb{E}\Big{[}\Big{(}\hat{\varphi}^{l}_{(1-\delta)t}(\theta)\Big{)}^{N_{\delta
t}}\big{|}\text{survival}\Big{]}+o(1)$ $\displaystyle=$
$\displaystyle\mathbb{E}\Big{[}\Big{(}1+(\theta-1)\frac{\mathrm{e}^{-\beta
mx}}{m-1}\mathrm{e}^{-\beta(m-1)\delta
t}+o\big{(}\mathrm{e}^{-\beta(m-1)\delta
t}\big{)}\Big{)}^{\mathrm{e}^{\beta(m-1)\delta t}M_{\delta
t}}\big{|}M_{\infty}>0\Big{]}+o(1)$ $\displaystyle\to$
$\displaystyle\mathbb{E}\Big{[}\exp\Big{\\{}(\theta-1)\frac{\mathrm{e}^{-m\beta
x}}{m-1}M_{\infty}\Big{\\}}\big{|}M_{\infty}>0\Big{]}$

as $t\to\infty$, which proves (3.9). Convergence in (3.10) is proved in
exactly the same way. ∎

##  4 Proof of Proposition 3.2

Recall that in the previous section we have calculated
$\hat{m}_{t}^{l}=\mathbb{E}\hat{N}_{t}^{l}$ and
$\tilde{m}_{t}^{l}=\mathbb{E}\tilde{N}_{t}^{l}$, which was relatively easy.
Let us now consider $\hat{v}_{t}^{l}=\mathbb{E}[(\hat{N}_{t}^{l})^{2}]$ and
$\tilde{v}_{t}^{l}=\mathbb{E}[(\tilde{N}_{t}^{l})^{2}]$. Finding the exact
expressions for both of these quantities is quite possible (as will be
apparent from the calculations below) but they are cumbersome. Instead, we
will find simple upper bounds on $\hat{v}_{t}^{l}$ and $\tilde{v}_{t}^{l}$.
These upper bounds will be used to tell us that if $v<\infty$ then under
conditions (3.4) and (3.5) on $l$, $\hat{m}_{t}^{l}\sim\hat{v}_{t}^{l}$ and
$\tilde{m}_{t}^{l}\sim\tilde{v}_{t}^{l}$ as $t\to\infty$. This together with a
suitable truncation of the branching process $\mathcal{T}_{t}$, $t\geq 0$ will
enable us to prove Proposition 3.2.

######  Proposition 4.1.

Let us suppose that $v=\mathbb{E}[\xi^{2}]=\sum_{k\geq 0}k^{2}p_{k}<\infty$.
If $t\geq l$ then

$\hat{v}_{t}^{l}\leq\frac{1}{m-1}\mathrm{e}^{-\beta(ml-(m-1)t)}\Big{[}1+2\beta
m(t-l)\mathrm{e}^{-\beta
ml}+\frac{v-m}{(m-1)^{2}}\mathrm{e}^{-\beta(ml-(m-1)t)}\Big{]}$ (4.1)

and

$\tilde{v}_{t}^{l}\leq\frac{m}{m-1}\mathrm{e}^{-\beta(ml-(m-1)t)}\Big{[}1+2\beta(t-l)\mathrm{e}^{-\beta
ml}+\frac{(v-m)(m^{2}-2m+2)}{m(m-1)^{2}}\mathrm{e}^{-\beta(ml-(m-1)t)}\Big{]}.$
(4.2)

###### Proof.

We follow the same steps as in the proof of Proposition 3.1. Let us fix $l$
and let $t>l$. From (3) we get that

$\big{(}\hat{N}_{t}^{l}\big{)}^{2}=\mathbf{1}_{\\{l<T\leq
t\\}}+2\cdot\mathbf{1}_{\\{l\leq T\leq
t\\}}\sum_{k=1}^{\xi}\hat{N}_{t-T}^{l}(k)+\mathbf{1}_{\\{0\leq T\leq
t\\}}\Big{(}\sum_{k=1}^{\xi}\hat{N}_{t-T}^{l}(k)\Big{)}^{2}$

Taking the expectation and using the Markov property we get the integral
equation for $\hat{v}_{t}^{l}$:

$\displaystyle\hat{v}_{t}^{l}=$ $\displaystyle(\mathrm{e}^{-\beta
l}-\mathrm{e}^{-\beta
t})+2\mathbb{E}(\xi)\int_{l}^{t}\hat{m}_{t-s}^{l}\beta\mathrm{e}^{-\beta
s}\mathrm{d}s+\int_{0}^{t}\big{(}\mathbb{E}(\xi)\cdot\hat{v}_{t-s}^{l}+\mathbb{E}(\xi^{2}-\xi)\cdot(\hat{m}_{t-s}^{l})^{2}\big{)}\beta\mathrm{e}^{-\beta
s}\mathrm{d}s$ $\displaystyle=$ $\displaystyle(\mathrm{e}^{-\beta
l}-\mathrm{e}^{-\beta t})+2m\beta\mathrm{e}^{-\beta
t}\int_{0}^{t}\hat{m}_{u}^{l}\mathrm{e}^{\beta
u}\mathrm{d}u+\beta\mathrm{e}^{-\beta
t}\int_{0}^{t}\big{(}m\hat{v}_{u}^{l}+(v-m)(\hat{m}_{u}^{l})^{2}\big{)}\mathrm{e}^{\beta
u}\mathrm{d}u$

Differentiating it with respect to $t$ yields the differential equation

$\left\\{\begin{array}[]{rl}(\hat{v}_{t}^{l})^{\prime}&=\beta(m-1)\hat{v}_{t}^{l}+\beta\mathrm{e}^{-\beta
l}+2\beta m\mathrm{e}^{-\beta
l}\hat{m}_{t-l}^{l}+\beta(v-m)(\hat{m}_{t}^{l})^{2}\quad,\ t>l\\\
\hat{v}_{l+}^{l}&=0\end{array}\right.$

Multiplying by the integrating factor $\mathrm{e}^{-\beta(m-1)t}$ gives

$\displaystyle(\hat{v}_{t}^{l})^{\prime}\mathrm{e}^{-\beta(m-1)t}-\beta(m-1)\mathrm{e}^{-\beta(m-1)t}\hat{v}_{t}^{l}=$
$\displaystyle\beta\mathrm{e}^{-\beta l}\mathrm{e}^{-\beta(m-1)t}+2\beta
m\mathrm{e}^{-\beta l}\mathrm{e}^{-\beta(m-1)t}\hat{m}_{t-l}^{l}$
$\displaystyle+\beta(v-m)\mathrm{e}^{-\beta(m-1)t}(\hat{m}_{t}^{l})^{2}$

Integrating this equation between $l$ and $t$ and then multiplying by
$\mathrm{e}^{\beta(m-1)t}$ gives

$\displaystyle\hat{v}_{t}^{l}=$
$\displaystyle\mathrm{e}^{\beta(m-1)t}\int_{l}^{t}\mathrm{e}^{-\beta(m-1)s}\Big{(}\beta\mathrm{e}^{-\beta
l}+2\beta m\mathrm{e}^{-\beta
l}\hat{m}_{s-l}^{l}+\beta(v-m)(\hat{m}_{s}^{l})^{2}\Big{)}\mathrm{d}s$
$\displaystyle=$
$\displaystyle\mathrm{e}^{\beta(m-1)t}\Big{[}\beta\mathrm{e}^{-\beta
ml}\int_{l}^{t}\mathrm{e}^{-\beta(m-1)(s-l)}\mathrm{d}s$
$\displaystyle\qquad\qquad+\frac{2\beta m}{m-1}\mathrm{e}^{-2\beta
ml}\int_{2l}^{t}\mathrm{e}^{-\beta(m-1)(s-2l)}\big{(}\mathrm{e}^{\beta(m-1)(s-2l)}-1\big{)}\mathrm{d}s\mathbf{1}_{\\{t\geq
2l\\}}$
$\displaystyle\qquad\qquad+\frac{\beta(v-m)}{(m-1)^{2}}\mathrm{e}^{-\beta(m+1)l}\int_{l}^{t}\mathrm{e}^{-\beta(m-1)(s-l)}\big{(}\mathrm{e}^{\beta(m-1)(s-l)}-1\big{)}^{2}\mathrm{d}s\Big{]}$
$\displaystyle=$
$\displaystyle\frac{1}{m-1}\mathrm{e}^{-\beta(ml-(m-1)t)}\Big{[}\beta(m-1)\int_{l}^{t}\mathrm{e}^{-\beta(m-1)(s-l)}\mathrm{d}s$
$\displaystyle\qquad\qquad\qquad\qquad\qquad+2\beta m\mathrm{e}^{-\beta
ml}\int_{2l}^{t}1-\mathrm{e}^{-\beta(m-1)(s-2l)}\mathrm{d}s\mathbf{1}_{\\{t\geq
2l\\}}$
$\displaystyle\qquad\qquad\qquad\qquad\qquad+\frac{\beta(v-m)}{m-1}\mathrm{e}^{-\beta
l}\int_{l}^{t}\mathrm{e}^{-\beta(m-1)(s-l)}\big{(}\mathrm{e}^{\beta(m-1)(s-l)}-1\big{)}^{2}\mathrm{d}s\Big{]}.$
(4.3)

Let us now give an upper bound on each of the three terms inside the square
brackets, which we refer to as (I), (II) and (III):

$(I)=\int_{l}^{t}\mathrm{e}^{-\beta(m-1)(s-l)}\beta(m-1)\mathrm{d}s\leq\int_{0}^{\infty}\mathrm{e}^{-\beta(m-1)\tau}\beta(m-1)\mathrm{d}\tau=1,$
$(II)=2\beta m\mathrm{e}^{-\beta
ml}\int_{2l}^{t}1-\mathrm{e}^{-\beta(m-1)(s-2l)}\mathrm{d}s\mathbf{1}_{\\{t\geq
2l\\}}\leq 2\beta m\mathrm{e}^{-\beta ml}(t-2l)\mathbf{1}_{\\{t\geq 2l\\}}\leq
2\beta m\mathrm{e}^{-\beta ml}(t-l)$

and

$\displaystyle(III)\leq$
$\displaystyle\frac{\beta(v-m)}{m-1}\mathrm{e}^{-\beta
l}\int_{l}^{t}\mathrm{e}^{-\beta(m-1)(s-l)}\big{(}\mathrm{e}^{\beta(m-1)(s-l)}-1\big{)}^{2}\mathrm{d}s$
$\displaystyle=$ $\displaystyle\frac{v-m}{(m-1)^{2}}\mathrm{e}^{-\beta
l}\int_{0}^{t-l}\mathrm{e}^{-\beta(m-1)\tau}\big{(}\mathrm{e}^{\beta(m-1)\tau}-1\big{)}^{2}\beta(m-1)\mathrm{d}\tau$
$\displaystyle\leq$ $\displaystyle\frac{v-m}{(m-1)^{2}}\mathrm{e}^{-\beta
l}\int_{0}^{t-l}\mathrm{e}^{\beta(m-1)\tau}\beta(m-1)\mathrm{d}\tau$
$\displaystyle\leq$ $\displaystyle\frac{v-m}{(m-1)^{2}}\mathrm{e}^{-\beta
l}\mathrm{e}^{\beta(m-1)(t-l)}$ $\displaystyle=$
$\displaystyle\frac{v-m}{(m-1)^{2}}\mathrm{e}^{-\beta(ml-(m-1)t)},$

which establishes (4.1). Inequality (4.2) is then easily derived from (4.1)
and (2.5) using the fact that
$\tilde{v}_{t}^{l}\leq\hat{v}_{t}^{l}+v_{t}^{l}$. ∎

Let us now apply upper bounds (4.1) and (4.2) on $\hat{v}_{t}^{l}$ and
$\tilde{v}_{t}^{l}$ to prove Proposition 3.2. Namely, that if $l$ depends on
$t$ in any such way that $t-l\to\infty$ and $ml-(m-1)t\to\infty$ as
$t\to\infty$ (conditions (3.4) and (3.5)) then

$\mathbb{P}\big{(}\hat{N}_{t}^{l}>0\big{)}\sim\hat{m}_{t}^{l}\sim\frac{1}{m-1}\mathrm{e}^{-\beta(ml-(m-1)t)}$

and

$\mathbb{P}\big{(}\tilde{N}_{t}^{l}>0\big{)}\sim\tilde{m}_{t}^{l}\sim\frac{m}{m-1}\mathrm{e}^{-\beta(ml-(m-1)t)}.$

###### Proof of Proposition 3.2.

Trivially,
$\mathbb{P}\big{(}\hat{N}_{t}^{l}>0\big{)}\leq\mathbb{E}\hat{N}_{t}^{l}=\hat{m}_{t}^{l}$
and
$\mathbb{P}\big{(}\tilde{N}_{t}^{l}>0\big{)}\leq\mathbb{E}\tilde{N}_{t}^{l}=\tilde{m}_{t}^{l}$.
Therefore,

$\limsup_{t\to\infty}\mathrm{e}^{\beta(ml-(m-1)t)}\mathbb{P}\big{(}\hat{N}_{t}^{l}>0\big{)}\leq\frac{1}{m-1}$
(4.4)

and

$\limsup_{t\to\infty}\mathrm{e}^{\beta(ml-(m-1)t)}\mathbb{P}\big{(}\tilde{N}_{t}^{l}>0\big{)}\leq\frac{m}{m-1}.$
(4.5)

To establish corresponding lower bounds let us consider two cases: $v<\infty$
and $v=\infty$.

Case (I): $v<\infty$: Note that under conditions (3.4) and (3.5) inequalities
(4.1) and (4.2) say that

$\hat{v}_{t}^{l}\leq\frac{1}{m-1}\mathrm{e}^{-\beta(ml-(m-1)t)}\Big{[}1+o(1)\Big{]}\sim\hat{m}^{l}_{t}$

and

$\tilde{v}_{t}^{l}\leq\frac{m}{m-1}\mathrm{e}^{-\beta(ml-(m-1)t)}\Big{[}1+o(1)\Big{]}\sim\tilde{m}^{l}_{t}$

as $t\to\infty$. Then, since also $\hat{v}_{t}^{l}\geq\hat{m}_{t}^{l}$ and
$\tilde{v}_{t}^{l}\geq\tilde{m}_{t}^{l}$, it follows that
$\hat{v}_{t}^{l}\sim\hat{m}_{t}^{l}$ and
$\tilde{v}_{t}^{l}\sim\tilde{m}_{t}^{l}$. Hence by Paley-Zygmund inequality

$\mathbb{P}\big{(}\hat{N}_{t}^{l}>0\big{)}\geq\frac{\big{(}\mathbb{E}\hat{N}_{t}^{l}\big{)}^{2}}{\mathbb{E}\big{[}\big{(}\hat{N}_{t}^{l}\big{)}^{2}\big{]}}=\frac{(\hat{m}_{t}^{l})^{2}}{\hat{v}_{t}^{l}}\sim\hat{m}^{l}_{t}$

as $t\to\infty$ and likewise for
$\mathbb{P}\big{(}\tilde{N}_{t}^{l}>0\big{)}$.

Case (II): $v=\infty$:

The issue here is that now $\hat{v}_{t}^{l}=\tilde{v}^{l}_{t}=\infty$ and so
Paley-Zygmund inequality is of no help to us. To overcome this problem we
consider the truncated process $\mathcal{T}_{t}^{K}$, $t\geq 0$ derived from
$\mathcal{T}_{t}$, $t\geq 0$ by having all the offspring killed at each birth
event which produces more than $K$ particles (where $K\geq 1$ is some given
number). Then every edge in $\mathcal{T}_{t}^{K}$ is an edge in
$\mathcal{T}_{t}$ and hence
$\mathbb{P}\big{(}\hat{N}_{t}^{l}>0\big{)}\geq\mathbb{P}\big{(}\hat{N}_{t}^{K,l}>0\big{)}$
and
$\mathbb{P}\big{(}\tilde{N}_{t}^{l}>0\big{)}\geq\mathbb{P}\big{(}\tilde{N}_{t}^{K,l}>0\big{)}$,
where $\hat{N}_{t}^{K,l}$ and $\tilde{N}_{t}^{K,l}$ are numbers of interior
and all (interior and pendant) edges of length $\geq l$ in
$\mathcal{T}_{t}^{K}$.

Moreover, the truncated process $\mathcal{T}_{t}^{K}$, $t\geq 0$ is a Galton-
Watson process with branching rate $\beta$ and offspring distribution
$p^{K}_{k}$, $k\geq 0$, where

$p_{0}^{K}=\mathbb{P}(\xi=0)+\mathbb{P}(\xi>K)=p_{0}+p_{K+1}+p_{K+2}+\cdots,$
$p_{1}^{K}=\mathbb{P}(\xi=1)=p_{1},\ p_{2}^{K}=\mathbb{P}(\xi=2)=p_{2},\
\cdots,\ p_{K}^{K}=\mathbb{P}(\xi=K)=p_{K}$

and

$p_{K+1}^{K}=p_{K+2}^{K}=\cdots=0.$

If we let $m^{K}=\sum_{k\geq 0}kp_{k}^{K}=\sum_{k=0}^{K}kp_{k}$ and
$v^{K}=\sum_{k\geq 0}k^{2}p_{k}^{K}=\sum_{k=0}^{K}k^{2}p_{k}$ denote the first
and second moments of the offspring distribution in the truncated Galton-
Watson process then we see that $v^{K}\leq K^{2}$ and

$m-m^{K}=\mathbb{E}\big{[}\xi\mathbf{1}_{\\{\xi>K\\}}\big{]}\leq\mathbb{E}\Big{[}\xi\frac{\log^{+}\xi}{\log
K}\mathbf{1}_{\\{\xi>K\\}}\Big{]}\leq\frac{F(K)}{\log K},$

where $F(K)=\mathbb{E}\big{[}\xi\log^{+}\xi\mathbf{1}_{\\{\xi>K\\}}\big{]}$ is
a function such that $F(K)\to 0$ as $K\to\infty$. Hence

$\displaystyle m-\frac{F(K)}{\log K}\leq m^{K}\leq m.$ (4.6)

We then have that

$\mathbb{P}\big{(}\hat{N}_{t}^{l}>0\big{)}\geq\mathbb{P}\big{(}\hat{N}_{t}^{K,l}>0\big{)}\geq\frac{\big{(}\mathbb{E}\hat{N}_{t}^{K,l}\big{)}^{2}}{\mathbb{E}\big{[}\big{(}\hat{N}_{t}^{K,l}\big{)}^{2}\big{]}}$
(4.7)

for any $K\geq 1$. Let us now take
$K=K(t)=\lfloor\mathrm{e}^{c(ml-(m-1)t)}\rfloor$ for any
$c\in(0,\frac{\beta}{2})$. Then from (3.1) we have that

$\mathbb{E}\hat{N}_{t}^{K,l}=\frac{1}{m^{K}-1}\mathrm{e}^{-\beta
l}\big{(}\mathrm{e}^{\beta(m^{K}-1)(t-l)}-1\big{)}.$

From (4.6) we know that

$m(t-l)-\frac{F(K)(t-l)}{ct-cm(t-l)}\leq m^{K}(t-l)\leq m(t-l)$

from which it follows that $m^{K}(t-l)=m(t-l)+o(1)$ and hence

$\mathrm{e}^{\beta(m^{K}-1)(t-l)}\sim\mathrm{e}^{\beta(m-1)(t-l)}$

as $t\to\infty$. Thus

$\mathbb{E}\hat{N}_{t}^{K,l}\sim\frac{1}{m-1}\mathrm{e}^{-\beta
l}\mathrm{e}^{\beta(m-1)(t-l)}=\frac{1}{m-1}\mathrm{e}^{-\beta(ml-(m-1)t)}$
(4.8)

as $t\to\infty$. Also, from (4.1) we have that

$\displaystyle\mathbb{E}\hat{N}_{t}^{K,l}\leq\mathbb{E}\Big{[}\big{(}\hat{N}_{t}^{K,l}\big{)}^{2}\Big{]}\leq\frac{1}{m^{K}-1}\mathrm{e}^{-\beta(m^{K}l-(m^{K}-1)t)}\Big{[}1+$
$\displaystyle 2\beta m^{K}(t-l)\mathrm{e}^{-\beta m^{K}l}$ $\displaystyle+$
$\displaystyle\frac{v^{K}-m^{K}}{(m^{K}-1)^{2}}\mathrm{e}^{-\beta(m^{K}l-(m^{K}-1)t)}\Big{]}.$
(4.9)

For the same reason as above,

$\mathrm{e}^{-\beta(m^{K}l-(m^{K}-1)t)}\sim\mathrm{e}^{-\beta(ml-(m-1)t)}$

as $t\to\infty$. We see that the second term inside the square brackets of (4)
converges to $0$ as $t\to\infty$. So does the third term since

$\displaystyle v^{K}\mathrm{e}^{-\beta(m^{K}l-(m^{K}-1)t)}\leq
K^{2}\mathrm{e}^{-\beta(m^{K}l-(m^{K}-1)t)}\leq$
$\displaystyle\mathrm{e}^{2c(ml-(m-1)t)}\mathrm{e}^{-\beta(m^{K}l-(m^{K}-1)t)}$
$\displaystyle\leq$ $\displaystyle\mathrm{e}^{(2c-\beta)(ml-(m-1)t)}\to 0$

as $t\to\infty$. It follows that

$\displaystyle\mathbb{E}\Big{[}\big{(}\hat{N}_{t}^{K,l}\big{)}^{2}\Big{]}\sim\frac{1}{m-1}\mathrm{e}^{-\beta(ml-(m-1)t)}.$
(4.10)

Combining (4.7), (4.8) and (4.10) yields

$\liminf_{t\to\infty}\mathrm{e}^{\beta(ml-(m-1)t)}\mathbb{P}\big{(}\hat{N}_{t}^{l}>0\big{)}\geq\frac{1}{m-1}.$
(4.11)

Similarly it can be shown that

$\liminf_{t\to\infty}\mathrm{e}^{\beta(ml-(m-1)t)}\mathbb{P}\big{(}\tilde{N}_{t}^{l}>0\big{)}\geq\frac{m}{m-1}.$
(4.12)

Inequalities (4.4), (4.11), (4.5) and (4.12) establish asymptotic relations
(3.7) and (3.8). ∎

##  5 Proof of Theorem 1.6

In this last section we prove that $\frac{L_{t}^{(k)}}{t}$,
$\frac{\hat{L}_{t}^{(k)}}{t}$ and $\frac{\tilde{L}_{t}^{(k)}}{t}$ for any
choice of $k=1,2,3,$… all converge to $\alpha^{\ast}$ as $t\to\infty$ a.s. on
survival.

We do this by first establishing convergence along integers and then expanding
it to convergence along real numbers.

###### Proof of convergence in Theorem 1.6 along integers.

We choose $k=1,2,3,$… and observe that for any $\alpha\in(\alpha^{\ast},1)$
and $n\in\mathbb{N}$

$\mathbb{P}\big{(}\frac{L^{(k)}_{n}}{n}\geq\alpha\big{)}=\mathbb{P}\big{(}L^{(k)}_{n}\geq\alpha
n\big{)}=\mathbb{P}\big{(}N_{n}^{\alpha n}\geq
k\big{)}\leq\frac{1}{k}\mathbb{E}N_{n}^{\alpha
n}=\frac{1}{k}\mathrm{e}^{-\beta m(\alpha-\alpha^{\ast})n},$

which is summable over $n$ and hence by Borel-Cantelli Lemma
$\mathbb{P}\big{(}\frac{L^{(k)}_{n}}{n}\geq\alpha\text{ i.o.}\big{)}=0$ and
also $\mathbb{P}\big{(}\frac{L^{(k)}_{n}}{n}\geq\alpha\text{
i.o.}|$survival$\big{)}=0$. Therefore for any $\alpha\in(\alpha^{\ast},1)$

$\limsup_{n\to\infty}\frac{L^{(k)}_{n}}{n}\leq\alpha$

a. s. on survival and hence

$\limsup_{n\to\infty}\frac{L^{(k)}_{n}}{n}\leq\alpha^{\ast}$ (5.1)

a. s. on survival. The same argument applies to $\hat{L}^{(k)}_{n}$ and
$\tilde{L}^{(k)}_{n}$.

On the other hand, we may take any $\alpha\in(0,\alpha^{\ast})$ and let
$\delta=1-\frac{\alpha}{\alpha^{\ast}}$ (so that
$\alpha=\alpha^{\ast}(1-\delta)$). Then

$\big{\\{}N_{n}^{\alpha n}<k\big{\\}}\subseteq\bigcap_{j=1}^{N_{\delta
n}}\big{\\{}N_{(1-\delta)n}^{\alpha n}(j)<k\big{\\}},$

where $N_{(1-\delta)n}^{\alpha
n}(j)=N_{(1-\delta)n}^{\alpha^{\ast}(1-\delta)n}(j)$ are the numbers of
pendant edges of length $\geq\alpha n$ in subtrees
$\mathcal{T}_{(1-\delta)n}(j)$ initiated by particles $j=1$, $2$, …,
$N_{\delta n}$ at time $\delta n$ and evolved up to time $n$. Then we check
that

$\displaystyle\mathbb{P}\big{(}\frac{L_{n}^{(k)}}{n}<\alpha,\ N_{\delta
n}>n\big{)}=$ $\displaystyle\mathbb{P}\big{(}N_{n}^{\alpha n}<k,\ N_{\delta
n}>n\big{)}$ $\displaystyle\leq$
$\displaystyle\mathbb{P}\Big{(}\bigcap_{j=1}^{N_{\delta
n}}\big{\\{}N_{(1-\delta)n}^{\alpha n}(j)<k\big{\\}},\ N_{\delta n}>n\Big{)}$
$\displaystyle=$
$\displaystyle\mathbb{E}\Big{[}\Big{(}\mathbb{P}\big{(}N_{(1-\delta)n}^{\alpha
n}<k\big{)}\Big{)}^{N_{\delta n}}\mathbf{1}_{\\{N_{\delta n}>n\\}}\Big{]}$
$\displaystyle\leq$
$\displaystyle\mathbb{P}\big{(}N_{(1-\delta)n}^{\alpha^{\ast}(1-\delta)n}<k\big{)}^{n}\mathbb{P}\big{(}N_{\delta
n}>n\big{)}.$

Since
$\mathbb{P}\big{(}N_{(1-\delta)n}^{\alpha^{\ast}(1-\delta)n}<k\big{)}\to\mathbb{P}(V_{0}<k)<1$
as $n\to\infty$ it follows that $\mathbb{P}(\frac{L_{n}^{(k)}}{n}<\alpha,\
N_{\delta n}>n)$ is summable over $n$. Therefore

$\mathbb{P}\big{(}\frac{L_{n}^{(k)}}{n}<\alpha,\ N_{\delta n}>n\text{
i.o.}\big{)}=0$

and so also

$\mathbb{P}\big{(}\frac{L_{n}^{(k)}}{n}<\alpha,\ N_{\delta n}>n\text{
i.o.}\big{|}\text{survival}\big{)}=0.$

Since

$\mathbb{P}\big{(}N_{\delta n}>n\text{ ev.}\big{|}\text{survival}\big{)}=1.$

it must be that

$\mathbb{P}\big{(}\frac{L_{n}^{(k)}}{n}<\alpha\text{
i.o.}\big{|}\text{survival}\big{)}=0.$

and thus

$\liminf_{n\to\infty}\frac{L_{n}^{(k)}}{n}\geq\alpha$

a.s. on survival for all $\alpha\in(\alpha^{\ast},1)$. Therefore

$\liminf_{n\to\infty}\frac{L_{n}^{(k)}}{n}\geq\alpha^{\ast}$ (5.2)

a.s. on survival and the same argument applies to $\hat{L}_{n}^{(k)}$ and
$\tilde{L}_{n}^{(k)}$. Putting together (5.1) and (5.2) we see that

$\lim_{n\to\infty}\frac{L_{n}^{(k)}}{n}=\lim_{n\to\infty}\frac{\hat{L}_{n}^{(k)}}{n}=\lim_{n\to\infty}\frac{\tilde{L}_{n}^{(k)}}{n}=\alpha^{\ast}$

a.s. on survival. ∎

###### Proof of convergence in Theorem 1.6 along integers.

The key observations are that for any times $s\leq t$ it is true that

$L_{t}^{(k)}\leq L_{s}^{(k)}+(t-s)$

(since over the time interval $[s,t]$ the $k$th longest pendant edge cannot
increase by more than $t-s$),

$\hat{L}_{s}^{(k)}\leq\hat{L}_{t}^{(k)}$

and

$\tilde{L}_{s}^{(k)}\leq\tilde{L}_{t}^{(k)}$

(since $k$th longest edge and $k$th longest interior edge can not get shorter
over time). Thus for any $t\in[0,\infty)$ it is true that

$\frac{L_{\lceil t\rceil}^{(k)}}{\lceil t\rceil}\cdot\frac{\lceil
t\rceil}{t}-\frac{1}{t}=\frac{L_{\lceil
t\rceil}^{(k)}-1}{t}\leq\frac{L_{t}^{(k)}}{t}\leq\frac{L_{\lfloor
t\rfloor}^{(k)}+1}{t}=\frac{L_{\lfloor t\rfloor}^{(k)}}{\lfloor
t\rfloor}\cdot\frac{\lfloor t\rfloor}{t}+\frac{1}{t},$ $\frac{\hat{L}_{\lfloor
t\rfloor}^{(k)}}{\lfloor t\rfloor}\cdot\frac{\lfloor
t\rfloor}{t}=\frac{\hat{L}_{\lfloor
t\rfloor}^{(k)}}{t}\leq\frac{\hat{L}_{t}^{(k)}}{t}\leq\frac{\hat{L}_{\lceil
t\rceil}^{(k)}}{t}=\frac{\hat{L}_{\lceil t\rceil}^{(k)}}{\lceil
t\rceil}\cdot\frac{\lceil t\rceil}{t}$

and

$\frac{\tilde{L}_{\lfloor t\rfloor}^{(k)}}{\lfloor t\rfloor}\cdot\frac{\lfloor
t\rfloor}{t}=\frac{\tilde{L}_{\lfloor
t\rfloor}^{(k)}}{t}\leq\frac{\tilde{L}_{t}^{(k)}}{t}\leq\frac{\tilde{L}_{\lceil
t\rceil}^{(k)}}{t}=\frac{\tilde{L}_{\lceil t\rceil}^{(k)}}{\lceil
t\rceil}\cdot\frac{\lceil t\rceil}{t}.$

Taking $t\to\infty$ yields the results. ∎

## References

  * [1] S.Bocharov, S.Harris, E.Kominek, A.Mooers and M.Steel “Predicting long pendant edges in model phylogenies with applications to biodiversity and tree inference,” Systematic Biology. 72(3), 575–589 (2023)
  * [2] T.E. Harris “The Theory of Branching Processes”, Springer, Berlin (1964)
  * [3] P.Olofsson “The $x\log x$ Condition for General Branching Processes,” Journal of Applied Probability. 35(3), 537–544 (1998)