This paper summarizes research conducted by Josh Young during a 10-week internship supervised by Dr. Matthew Roberts at the University of Bath. The paper applies existing results on Galton-Watson trees and branching random walks to more specific cases to make them understandable to undergraduate mathematics students. It begins by looking at properties of size-biased Galton-Watson trees, which have random infinite spines, and uses these to prove the Kesten-Stigum theorem. It then applies spine techniques to analyze the asymptotic maximal growth rate of a binary branching random walk.

1. The asymptotic maxima of a branching random walk via
spine techniques
Josh Young
Supervised by Dr Matthew Roberts
August 19, 2016
Summary
This paper is the product of a 10 week research internship granted by the Bath Institute
for Mathematical Innovation, under the supervision of Dr Matthew Roberts of the University
of Bath. Most of the results within have been well studied; the aim of this paper is to apply
these results to more specific cases, in a manner understandable to the average mathematics
undergraduate. We begin by looking at some elementary properties of Galton-Watson trees
with random infinite spines, also called size-biased Galton-Watson trees, and using these to
prove the Kesten-Stigum theorem. We then apply these spine techniques to the case of a
binary branching random walk to derive the asymptotic maximal growth rate.
Contents
1 Size-biased Galton-Watson trees 2
1.1 The canonical Galton-Watson Process . . . . . . . . . . . . . . . . . . . . . . . . . . 2
1.2 Size-biasing . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2
1.3 Spine Decomposition and the Kesten-Stigum theorem . . . . . . . . . . . . . . . . . 5
2 A discrete-time branching process on the unit square 8
2.1 Preliminaries and heuristics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8
2.2 Change of measure and Spine Decomposition . . . . . . . . . . . . . . . . . . . . . . 10
2.3 Asymptotic growth of the maximal particle . . . . . . . . . . . . . . . . . . . . . . . 13
1

2. 1 Size-biased Galton-Watson trees
1.1 The canonical Galton-Watson Process
We will define a Galton-Watson tree (henceforth abbreviated to GWT) in the standard manner.
Let L be a random variable with P(L = k) = pk for k ∈ N ∪ {0}. Let (L
(n)
i ; n, i ∈ N ∪ {0}) be
independent copies of L. We define a sequence (Zn, n ≥ 0) inductively by
Zn+1 :=
Zn
i=1
L
(n)
i
with the convention that Z0 = 1. This can be visualised as a breeding process from a single ancestor,
where Zn is the number of descendants in the nth generation, and L
(n)
i is the number of children
produced by the ith descendant in generation n. We will use the notation |u| = n to indicate that
particle u belongs to generation n. We also denote the mean of the process by m := E[L]
1.2 Size-biasing
Galton-Watson trees can also be endowed with a spine, which we will construct as follows. Label
the root of the tree ξ0. For each i ∈ N ∪ {0}, uniformly select one of the children of ξi, and label
this ξi+1. The sequence Ξ := (ξn, n ∈ N ∪ {0}) is called the spine. We will denote the number of
children of ξi by Lξi . In the next few sections, we will define the filtrations and martingales used
in the study of GW trees with spines.
Definition 1.1. For all n ∈ N, define:
1. Wn := Zn
mn
2. Mn := 1
mn
n−1
i=0
Lξn
Definition 1.2. For all n ∈ N, let:
1. Fn be the σ-algebra generated by the first n generations of the process
2. Gn be the σ-algebra generated by the first n spinal particles, and the children of the first n−1.
3. Fn := σ(Fn ∪ Gn)
Proposition 1.3. 1. The process W := (Wn, n ∈ N) is a non-negative Fn-martingale
2. The process M := (Mn, n ∈ N) is a non-negative Fn-martingale
Proof of (1). We have:
2

3. E
Zn
mn
Fn−1 = E
1
mn
Zn−1
i=1
L
(n−1)
i Fn−1
=
1
mn
Zn−1
i=1
E L
(n−1)
i Fn−1
=
1
mn
Zn−1
i=1
m =
Zn−1
mn−1
Hence Wn is a martingale. Non-negativity is trivial.
Proof of (2) We have:
E[Mn|Fn−1] =
1
mn
E
n−1
i=0
Lξi Fn−1
=
1
mn
×
n−2
i=0
Lξi × E[Lξn−1 |Fn−1]
=
1
mn−1
n−2
i=0
Lξi = Mn−1
Once again, non-negativity is trivial.
We will now use the martingale Mn to define a new probability measure, Q by setting:
dQ
dP Fn
= Mn
The following lemma and subsequent proposition will allow us to visualise how GW trees behave
under this new measure.
Lemma 1.4. E[Mn|Fn] = Wn
Proof. From the definitions, we immediately have that E[Mn|Fn] = 1
mn E[
n−1
i=0 Lξn
|Fn]. We now
sum over the indicator of j ∈ Ξ:
1
mn
E
Zn
j=1
n−1
i=0
L
(i)
j 1{j∈Ξ} Fn
Now, both Zn and L
(i)
j are Fn-measurable, and the indicator of j ∈ Ξ is independent of Fn. Hence
this reduces to
3

4. 1
mn
Zn
j=1
n−1
i=0
L
(i)
j P(j ∈ Ξ) =
The probability of any given j with |j| = n being a spinal particle is 1
n−1
i=0 Lξi
. The two products
now cancel out, leaving us with 1
mn
Zn
j=1 1 = Zn
mn = Wn
Proposition 1.5. Let |u| = n. Then
Q(L(n)
u = k) =
kpk
m u ∈ ξ
pk u ∈ ξ
Proof. Consider u ∈ ξ. That is, u = ξn. Hence Q(L
(n)
u = k) = Q(Lξn
= k) = E[Mn+11{Lξn+1
=k}].
The law of total expectation tells us that
E[Mn1{Lξn=k}] = E[Mn+1|Lξn
= k] × P(Lξn
= k)
= E
k
mn+1
n−1
i=0
Lξi
× pk
=
kpk
mn+1
n−1
i=0
E[Lξi
]
=
kpk
mn+1
× mn
=
kpk
m
We now consider u ∈ ξ. This gives us Q(L
(n)
u = k) = E[Mn+11{L
(n)
u =k}
]. Since u ∈ ξ, we have
that Mn and the indicator function are independent. We can use this and reduce the expression to
E[Mn+1]P(L
(n)
u = k) = E[Mn+1]pk. We have as a corollary to Lemma 1.4 that E[Mn] = E[Wn] = 1,
completing the proof.
This proposition tells us that the offspring of particles in the spine follow a size-biased distribution,
whereas the other particles behave in the usual manner. In particular, the probability of any spinal
particle having no children is 0. Since the root of the tree is in the spine, this tells us that under Q,
the event of extinction almost surely does not occur. Lyons and Peres [2] call GWTs under the law
Q size-biased Galton-Watson trees. An example of a size-biased GWT is given in figure 1, where
each non-spine particle forms the root of an independent GWT.
4

5. ξ0
ξ1
ξ2
ξ3
GW GW
GW
GW
Figure 1: An example tree after 4 generations
1.3 Spine Decomposition and the Kesten-Stigum theorem
We will now use the properties of sized-biased trees to prove Kesten and Stigum’s classic limit
theorem, stated below.
Theorem 1.6. The Kesten-Stigum Theorem [1]
Let L be the offspring random variable of a Galton-Watson process with mean m ∈ (1, ∞) and
martingale limit W. Then the following are equivalent:
a. P[W = 0] = q
b. E[W] = 1
c. E[L log+
L] < ∞
Proposition 1.7. Spine Decomposition
For all n ≥ 1
EQ[Wn|G∞] =
n−1
i=0
(Lξi
− 1)m−(i+1)
+ 1
Proof. We will prove this result inductively. Since the root is also the first particle in the spine,
we have that Z1 = Lξ0
. Therefore EQ[Wn|G∞] = 1
m Lξ0
. This settles the base case. Now, we will
consider EQ[Wk+1|G∞] for some k > 1, which we will denote Ek+1. From the definitions of Wn and
Zn, we have that
Ek+1 =
1
mk+1
EQ
Zk
i=0
L
(k)
i G∞
5

6. Now, under Q we know that there is exactly 1 spine particle in generation k, specifically ξk.
Removing ξk’s children from the sum gives
1
mk+1
EQ
Zk−1
i=0
L
(k)
i + Lξk
G∞
We can now recall that Lξn
is G∞-measurable, and use the iid nature of the L
(k)
i s to rewrite this as
1
mk+1
(EQ[Zk|G∞] − 1)m + Lξk
=
1
mk+1
(mk
Ek − 1)m + Lξk
.
The remaining steps are simply an exercise in algebraic manipulation. This proves the result.
Lemma 1.8. Let X1, X2, ... be a sequence of non-negative iid random variables. Then
lim sup
n→∞
1
n
Xn =
0 if E[X] < ∞
∞ if E[X] = ∞
almost surely.
Proof. We aim to use the Borel-Cantelli lemma to show that Xn/n is positive only finitely often. To
do this, we consider the event {Xn
n ≥ } Since the Xn’s are iid, this event has the same probability
as {X
n ≥ }. We have:
∞
k=1
P
X
k
≥ =
∞
k=1
P
X
≥ k =
∞
k=1
E[1{X/ ≥k}] = E
∞
k=1
1{X/ ≥k} = E X/ .
Now by the Borel-Cantelli lemma, if E[X] < ∞ then the event {Xn/n ≥ } happens only finitely
often for any > 0 (no matter how small), so Xn/n → 0. On the other hand, by the second
Borel-Cantelli lemma, if E[X] = ∞, then the event {Xn/n ≥ } happens infinitely often for any
> 0 (no matter how large), so lim supn→∞ Xn/n = ∞.
Lemma 1.9. Let X1, X2, ... be a sequence of non-negative iid random variables. Then for any
c ∈ (0, 1),
∞
k=1
eXk
ck < ∞ if E[X] < ∞
= ∞ if E[X] = ∞
almost surely.
Proof. Suppose first that E[X] < ∞. By Lemma 1.8, we have that lim sup eXk/k
= e0
= 1. Using
the definition of limsup, we have that for all > 0, there exists an M such that eXk/k
≤ 1 + , for
each k ≥ M. To prove our result, fix c ∈ (0, 1) and choose > 0 such that c(1 + ) < 1. Then select
an M as above. Now:
∞
k=M
eXk
ck
≤
∞
k=M
(1 + )k
ck
.
Since (1 + )c < 1, we have that the right hand side is finite. Finally, we write
∞
k=1
eXk
ck
=
M−1
k=1
eXk
ck
+
∞
k=M
eXk
ck
.
6

7. Since both of these sums are almost surely finite, we have proven the result in the case E[X] < ∞.
Now suppose E[X] = ∞. By Lemma 1.8, lim sup Xn/n = ∞, so for any K, we can find n1, n2, . . . →
∞ such that Xni
/ni ≥ K for all i. Fix c ∈ (0, 1), and choose ni as above with K = − log c. Then
∞
n=1
eXn
cn
≥
∞
i=1
eXni cni
≥
∞
i=1
e−ni log c
cni
=
∞
i=1
1 = ∞
as required.
We will now use these two lemmas to prove that (c) implies (b) in the Kesten-Stigum theorem. We
need one more tool, which is proved in [2].
Lemma 1.10. Suppose that µ and ν are probability measures and dµ
dν |Fn = Xn. Let X∞ =
lim supn→∞ Xn. Then
X∞ < ∞ ν-almost surely ⇔ Eµ[X∞] = 1
and
X∞ = ∞ ν-almost surely ⇔ Eµ[X∞] = 0.
We can now prove the Kesten-Stigum thoerem.
Proposition 1.11. Let W and L be as in Theorem 1.6. Then:
E[L log+
L] < ∞ ⇔ E[W] = 1
Proof. We first show that 1/Wn is a supermartingale under Q. Recall from the definition that
EQ[1/Wn|Fn−1] is the (almost surely unique) Fn−1-measurable random variable Y such that EQ[1/Wn1A] =
EQ[Y 1A] for all A ∈ Fn−1. Now,
EQ[
1
Wn
1A] = EP[1A1{Wn>0}] = EP[1A1{Wn−1>0}P(Wn > 0|Fn−1)] = EQ[
1
Wn−1
1AP(Wn > 0|Fn−1)].
Therefore
EQ[
1
Wn
|Fn−1] =
1
Wn−1
P(Wn > 0|Fn−1) ≤
1
Wn−1
.
So 1/Wn is a non-negative supermartingale under Q, so it converges almost surely to a limit 1/W∞
(which may be 0). In particular, lim supn→∞ Wn = lim infn→∞ Wn, Q-almost surely.
Now we demonstrate that lim supn→∞ EQ[Wn|G∞] is almost surely finite if E[L log+
L] < ∞. From
Lemma 1.7, we have that
EQ[Wn|G∞] = 1 +
n−1
i=0
(Lξi
− 1)m−(i+1)
≤ 1 +
n
i=1
elog+
(Lξi−1
−1)
m−i
Clearly, by Lemma 1.9 this converges when EQ[log+
(Lξi − 1)] < EQ[log+
(Lξi )] < ∞. Now,
7

8. EQ[log+
(Lξi )] =
∞
k=0
Q(Lξi = k) log+
k
=
∞
k=0
kpk
m
log+
k
=
1
m
E[L log+
L]
Hence, E[L log+
L] < ∞ =⇒ lim sup EQ[Wn|G∞] < ∞ almost surely. Now Fatou’s lemma,
combined with the fact (proven above) that lim supn→∞ Wn = lim infn→∞ Wn, Q-almost surely,
gives
EQ[lim sup Wn|G∞] = EQ[lim inf Wn|G∞] ≤ lim inf EQ[lim inf Wn|G∞] ≤ lim sup EQ[lim inf Wn|G∞] < ∞.
Therefore lim sup Wn < ∞ Q-almost surely, so by Lemma 1.10 we have EP[W] = 1.
2 A discrete-time branching process on the unit square
2.1 Preliminaries and heuristics
Our process begins with the unit square. It splits into two rectangles of area U and 1 − U respec-
tively, where U is a random variable uniform on [0, 1]. In each subsequent generation, each of the
rectangles splits in a manner similar to the original square. The orientation of these splits is not
relevant to the following results, so we assume that each split occurs either vertically or horizontally
with probability 1
2 . Figure 2 shows a simulated outcome of this process after ten generations.
A natural question to ask is: what size would we expect the smallest and largest rectangles to be
after n generations? There are many ways one could interpret the notion of size in this context;
for our purposes, we will be considering the area of each rectangle to be its size. After n ≥ 1
generations, the area of any given rectangle in the nth generation, here denoted An, can be written:
An =
n
k=1
Uk
where (Uk : k ∈ N) is a sequence of iid unif(0, 1) random variables. We will denote the set of
rectangles in generation n by Nn.
We will now use our simulation to plot the areas of the rectangles in each generation, and hopefully
provide some heuristic justification for the main result of this paper.
8

9. Figure 2: The process after 10 generations
9

10. Here we see what our intuition tells us: the area of the rectangles decreases exponentially quickly.
This plot however is not particularly useful due to the nature of exponential growth. To rectify
this, we will plot the logarithm of the area of each rectangle.
Clearly, the growth of the log-area in this plot is linear, somewhat confirming our suspicion that
the growth is exponential. It also appears that there are upper and lower boundary lines between
which all of the rectangles fall. It is the upper boundary line which, over the course of this sec-
tion, we shall not only demonstrate the existence of, but also give an explicit expression for its slope.
Before we do this, we shall construct a spine for this process in a manner very similar to that in
Section 1.3. We will henceforth refer to the rectangles as particles. As before, we shall define the
spine recursively. Define ξ0 as the root particle. For each ξk, uniformly select one of its children,
and set that particle to be ξk+1. Finally, define Ξ = {ξk|k ∈ N0}. This forms our spine.
2.2 Change of measure and Spine Decomposition
Definition 2.1. Important filtrations
1. Fn is the filtration defined by the first n generations of the process
2. Gn is the filtration defined by the first n generations of the spine
3. Fn = σ(Fn ∪ Gn)
10

11. Proposition 2.2. Let α > −1. Then:
1. The process W(α)
= (W
(α)
n )n∈N defined by W
(α)
n = (α+1
2 )n
u∈Nn
Aα
u is a Fn-martingale
2. The process M = (Mn)n∈N defined by Mn = Aα
ξn
(α + 1)n
is a Fn-martingale
We now define a new measure Q by setting dQ
dP Fn
= Mn. In the following propositions, we will
show that the spine decomposition converges Q almost-surely to a finite limit. First, we need to
consider how our process changes under this new measure.
Lemma 2.3. Let u ∈ Nn. Then for all k ∈ [0, 1] and β ∈ N:
(i) Q(Uu ≤ x) =
xα+1
u ∈ Ξ
x u ∈ Ξ
(ii) EQ[Uβ
u ] =
α+1
α+β+1 u ∈ Ξ
1
β+1 u ∈ Ξ
(iii) EQ[log Uξn ] = − 1
α+1
Proof. Consider the case that u ∈ Ξ. Then u = ξn, and:
Q(Uξn ≤ x) = E[Mn|Uξn ≤ x] × P(Uξn ≤ x)
= x(α + 1)n
× E[Aα
ξn−1
Uα
ξn
|Uξn ≤ x]
= x(α + 1)n
× E[Aα
ξn−1
] × E[Uα
ξn
|Uξn ≤ x]
= x(α + 1)n
×
1
(a + 1)n−1
×
x
0
1
x
xα
dx
= x(α + 1) ×
1
x
×
xα+1
α + 1
= xα+1
Define gu(x) to be the pdf of Uu under Q. Let u = ξn. We now know that
gξn (x) :=
dQ(Uξn
≤ x)
dx
= (α + 1)xα
Hence,
EQ[Uβ
u ] = (α + 1)
1
0
xα+β
dx
=
α + 1
α + β + 1
As required. The results for u ∈ Ξ are trivial. Finally, we have:
EQ[log Uξn ] = (α + 1)
1
0
log(x)xα
dx
= −
α + 1
(α + 1)2
= −
1
α + 1
11

12. Proposition 2.4. The Spine Decomposition
Let α ∈ [0, 1]. Then:
EQ W(α)
n G∞ = Aα
ξn
α + 1
2
n
+
α + 1
2
n−1
k=0
α + 1
2
k
Aξk
− Aξk+1
α
Proof. Let us consider what particles exist at time n. A trivial examination of this tree structure
reveals that for any k < n, there will be 2n−k−1
particles alive at time n whose last spinal ancestor
was ξk. We will denote the set of non-spine descendants of ξk alive at time n by Cn(ξk). Each
u ∈ Cn(ξk) has size distribution Aξk
×
n−k
Ui, where the Ui’s are uniform (0, 1) random variables.
Now, we can once again use the binary structure of the process to remove a degree of randomness
from this expression. ξk has two children. One of these children is in the spine, and hence contributes
no descendants to Cn(ξk). However, the other child has size distribution Aξk+1
− Aξk
, and its
descendants at time n form precisely the set Cn(ξk). As such, each u ∈ Cn(ξk) has distribution
(Aξk+1
− Aξk
) ×
n−k−1
Ui. The keen observer will notice however that | k Cn(ξk)| = 2n
− 1; we
are missing ξn, which trivially has size distribution Aξn . This completes our characterisation of the
particles in Nn. Hence:
EQ W(α)
n |G∞ =
α + 1
2
n
× EQ Aα
ξn
|G∞ +
n−1
k=0
2n−k−1
EQ Aξk+1
− Aξk
×
n−k−1
i=1
Ui
α
G∞
Now, the Aξk
’s are G∞-measurable, so we can take them out of the expectations. The Ui’s are also
independent of G∞, so we simply take their Q-expectations. By Proposition 2.3.(ii), we have that
EQ [Uα
i ] = 1
α+1 . Hence our full expression for the spine decomposition is
EQ W(α)
n |G∞ =
α + 1
2
n
× Aα
ξn
+
n−1
k=0
2
α + 1
n−k−1
Aξk+1
− Aξk
α
some simple algebraic manipulation gives the required result.
Proposition 2.5. Set α such that e− α
α+1 < (α+1
2 ). Then:
lim sup EQ[W(α)
n |G∞] < ∞
almost surely.
Proof. First we will consider the convergence of
∞
k=0
α+1
2
k
Aξk
− Aξk+1
α
. We will first rewrite
this as
∞
k=0
α+1
2 e
α
k log(Aξk
−Aξk+1 )
k
. We have that:
1
k
log Aξk
− Aξk+1
=
1
k
log 1 − Uξk+1
k
i=0
Uξi
=
log(1 − Uξk+1
)
k
+
1
k
k
i=1
log(Uξi )
12

13. We now have two summands to consider. By the SLLN, 1
k
k
i=1 log Uξi → EQ[log Uξi ] Q-almost
surely. By Lemma 2.3 we know that this is − 1
α+1 . We will now consider the convergence of
1
k log(1 − Uξk+1
). By Lemmas 1.8 and 2.3, we have that −1
k log(1 − Uξk+1
) converges to 0. Hence
we also have 1
k log(1 − Uξk+1
) → 0, Q-almost-surely. This gives us a precise limit:
lim
k→∞
1
k
log Aξk
− Aξk+1
= −
1
α + 1
Hence, we have that
lim
k→∞
α + 1
2
exp
α
k
log Aξk
− Aξk+1
=
α + 1
2
e− α
α+1
We have now that the sum
∞
k=0
α+1
2
k
Aξk
− Aξk+1
α
converges Q-almost surely if (α+1
2 )e− α
α+1 <
1. It remains now to find which values of α ensure the convergence of Aα
ξn
α+1
2
n
. Now:
lim
n→∞
1
n
log(Aξn ) = lim
n→∞
1
n
log
n
i=1
Uξi
= lim
n→∞
1
n
n
i=1
log(Uξn )
= EQ[log Uξn ] = −
1
α + 1
Hence, similar to before,
lim sup
n→∞
α + 1
2
n
Aα
ξn
= lim sup
n→∞
α + 1
2
e
α
n log(Aξn )
n
= lim sup
n→∞
α + 1
2
e− α
α+1
n
Which is finite iff (α+1
2 )e− α
α+1 < 1. This fact, combined with the convergence of the aforementioned
sum on the same interval gives us our required condition for convergence.
2.3 Asymptotic growth of the maximal particle
The following three lemmas are essential in proving our main result.
Lemma 2.6. Set α such that e− α
α+1 < (α+1
2 ). Then:
E lim
n→∞
W(α)
n = 1
Q(α)
-almost surely.
Proof. We first use the proof of Proposition 1.11 to deduce that lim supn→∞ W
(α)
n = lim infn→∞ W
(α)
n .
We now have
EQ[lim sup W(α)
n |G∞] = EQ[lim inf W(α)
n |G∞] ≤ lim inf EQ[W(α)
n |G∞] ≤ lim sup EQ[W(α)
n |G∞] < ∞
From a combination of Fatou’s Lemma and Proposition 2.5. Therefore, lim W
(α)
n < ∞ Q-almost
surely, and by Lemma 1.10 we have that E limn→∞ W
(α)
n = 1 as required.
13

14. Lemma 2.7. Let A ∈ Fn. Then:
Q(α)
(A) = EP W(α)
∞ 1E + P(A ∩ {W(α)
∞ = ∞})
Proof is a standard result of measure theory
Lemma 2.8. Let Eδ be the event {supu∈Nn
1
n log Au > δ i.o.}. Then P(Eδ) is either 0 or 1.
Proof. We shall consider the probability of the complement of Eδ given the filtration Fk.
We have that:
P(Ec
δ|Fk) = P ∀v ∈ Nk, sup
u∈Nn, v≤u
log Au
n
≤ δ e.v. Fk
= P
v∈Nk
sup
u∈Nn, v≤u
log Au
n
≤ δ e.v. Fk
=
v∈Nk
P sup
u∈Nn, v≤u
log Au
n
≤ δ e.v. Fk
=
v∈Nk
P sup
u∈Nn−k
log Au
n
≤ δ e.v.
=
v∈Nk
P sup
u∈Nn
log Au
n
·
n + k
n
≤ δ e.v.
≤
v∈Nk
P sup
u∈Nn
log Au
n
≤ δ + e.v.
= P Ec
δ+
|Nk|
To summarise, we now have that P(Ec
δ|Fk) ≤ P(Ec
δ+ )|Nk|
. We can now take the limit as → 0 and
take the expectation of both sides to obtain P(Ec
δ) ≤ P(Ec
δ)2k
. This proves the result.
The following definition and lemma are purely technical, serving only to simplify the statement of
the main theorem.
Definition 2.9. The Lambert W Function
Let z be any complex number. Then W is the unique function satisfying the equation
z = W(z)eW (z)
Lemma 2.10.
min
x>0
1
x
log
2
x + 1
= W −
1
2e
Proof. Let f(x) = 1
x log 2
x+1 . Simple calculus shows that
f (x) = −
1
x2
log
2
x + 1
−
1
x(x + 1)
14

15. setting f (x) = 0, we obtain
1
x
log
2
x + 1
= −
1
x + 1
(1)
Let x∗
denote the solution to this equation, and let f∗
denote the minimum of f. Then clearly f∗
is given by
f∗
= f(x∗
) = −
1
x∗ + 1
Rearranging (1), we can obtain
−
1
2e
= −
1
x + 1
e− 1
x+1
Hence the f∗
is the solution to the equation
−
1
2e
= f∗
ef∗
The definition of the Lambert W Function says that for any real number z, we have
z = W(z)eW (z)
Therefore, f∗
= W(− 1
2e )
Theorem 2.11. Let both Au and Nn be defined as in Section 2.1. Then,
lim sup
n→∞
max
u∈Nn
log Au
n
= W −
1
2e
Almost surely.
Proof. This proof is split into two parts, in which we will derive both upper and lower bounds for
the limit. We will begin with the former. Fix some γ ∈ R, and suppose that there exists some
particle u ∈ Nn such that log Au
n > γ. Then,
W(α)
n =
α + 1
2
n
v∈Nn
Aα
u ≥
α + 1
2
n
Aα
u >
α + 1
2
n
eαγn
Now, since W
(α)
n is a non-negative martingale, it converges almost surely to a finite limit. Hence
to ensure convergence, we must have that α+1
2 eαγ
> 1 only finitely often. Rewriting this, we
see that this implies maxu∈Nn
log Au
n > γ > 1
α log 2
α+1 happens only finitely often for all α > 0.
This inequality relies on varying α, so we will optimise over α via Lemma 2.10, giving us that
minα>0
1
α log 2
α+1 = W(− 1
2e ) Therefore, we have
lim sup
n→∞
max
u∈Nn
log Au
n
≤ W −
1
2e
almost surely.
Now, we will use our spine decomposition to provide the lower bound. Let W
(α)
∞ := lim sup W
(α)
n .
By Lemma 2.6 we have that E[W
(α)
∞ ] = 1. Hence P(W
(α)
∞ = ∞) = 0. Now, combining this with
Lemma 2.7, we have
15

16. Q(α)
(A) = EP W(α)
∞ 1A
Let E be the event {supu∈Nn
1
n log Au ≥ − 1
α+1 i.o.}. In Proposition 2.5, we showed that limn→∞
1
n log Aξn
=
− 1
α+1 , i.e. Q(α)
{ 1
n log Aξn
= − 1
α+1 i.o.} = 1. The following sequence of set inclusions will make it
obvious that Q(α)
(E) = 1:
1
n
log Aξn
= −
1
α + 1
i.o. ⊆
1
n
log Aξn
≥ −
1
α + 1
i.o.
⊆ sup
u∈Nn
1
n
log Au ≥ −
1
α + 1
i.o.
= E
We also have that
Q(α)
(E) = EP W(α)
∞ 1E
Hence P(E) > 0. Finally, by Lemma 2.8 we have that P(E) = 0 or 1. We have just shown that
P(E) > 0, therefore we necessarily have that P(E) = 0, i.e.
lim sup
n→∞
sup
u∈Nn
log Au
n
≥ −
1
α + 1
= W −
1
2e
Proving our result.
References
[1] H. Kesten and B. P. Stigum. A limit theorem for multidimensional galton-watson processes.
Ann. Math. Statist., 37(5):1211–1223, 10 1966.
[2] R. Lyons and Y. Peres. Probability on Trees and Networks. Cambridge University Press, 2016.
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