Proof of Theorem 6.3.16:

We will prove (a) of the Theorem. The remaining results follow via straightforward modifications of

the arguments for (a). For (a) recall that we first grow the tree using the uniform attachment scheme with

f

0

≡1 till it is of sizen

γ

and then use the preferential attachment scheme. We will assume thatT

nθ

has

been constructed as follows:

(a) Generate the genealogical tree according to a rate one Yule process©T

Yule

_{(t) :t}

≥0ªas in Definition

6.5.3 run for ever.

(b) To obtainT

nθ

, letT

nγ

=T

Yule

(T

nγ

). Now every vertex inT

nγ

switches to offspring dynamics giving

birth to children at rate corresponding to the number of children+1+α(thus modulated by the

function

f

1

). Write BP

n

(·) for the combined process and stop this process at timeT

n

and letT

nθ

=

BP

n

(T

n

).

The following describes asymptotics for the above continuous time construction.

Proposition 6.8.14.

For the processBP

n

(·)as constructed above:

(a)

The stopping time T

nγ

satisfies,

T

nγ

−γlogn−→

a.e.

W˜,

whereW˜

= −logW and W

=exp(1).

(b)

Letω

n

→ ∞arbitrarily slowly. Then there exists a constant C>0independent ofω

n

such that

P

µ

sup

t≥0

¯

¯

¯

¯

e

−(2+α)t

|BP

n

(t+T

nγ

)|

n

γ

−1

¯

¯

¯

¯>

ω

n

n

γ/2

¶

≤

C

ω

2 n

.

In particular whp as n→ ∞,

¯

¯

¯

¯T

n

−

1−γ

2+αlogn

¯

¯

¯

¯≤

ω

n

n

γ/2

.

Proof.

Part(a) follows from Lemma 6.5.4. To prove (b), recall that fort

>T

nγ

, all individuals switch to

offspring dynamics modulated byf

1

. For the rest of the proof, we proceed conditional on the history of

the process till timeT

nγ

. Using Proposition 6.5.7,

M

1

(t) :=¡e

−(2+α)t

|B P

n

(t+T

nγ

)| −n

γ

¢+

1−e

−(2+α)t

(2+α)

,

t≥0,

and

M

2

(t) :=e

−2(2+α)t

|B P

n

(t+T

nγ

)|

2

−

Z

t 0

α

e

−2(2+α)s

|B P

n

(s+T

nγ

)|d s−

e

−2(2+α)t

2(2+α),

t≥0,

are martingales. Using these expressions, it can be deduced that

sup

t≥0

E

¡

M

₁2

(t)¢

≤C n

γ

for some constantC>0. An appeal to Doob’sL

2

-maximal inequality then proves the first assertion of

Proposition 6.8.14(b) which then results in the second assertion.

■

Fix constantBand a sequenceω

n

=o(n

γ/2

)↑ ∞and consider the following construction ˜T

n+

(B,ω

n

)

related to the above continuous time construction ofT

nθ

:

(a) Run a rate one Yule process for timeγlogn+B.

(b) Now every vertex in the Yule process switches dynamics so that it reproduces at rate equal to the

number of children+1+α. Grow this process foran additionaltimet

_n+

:=

1−_2+αγ

logn+

ωn

nγ/2

.

Analogously define ˜T

n−

(B,ω

n

) where in the above construction we wait till time logn−Bbefore switching

dynamics and run the new dynamics for timet

−

n

:=

₂₊1−_αγ

logn−

_nωγn/2

. By Proposition 6.8.14 given anyε>0

we can choose a constantB=B(ε) such that for anyω

n

↑ ∞, we can produce a coupling betweenT

nθ

and ˜T

_n+

(B,ω

n

) such that for all largen, with probability at least 1−εT

nθ

⊆T˜

n+

(B,ω

n

) where we see the

object on the left as a subtree of the object on the right with the same root. A similar assertion holds

with ˜T

_n−

(B,ω

n

)⊆T

nθ

. Using these couplings, the following Proposition completes the proof of Theorem

6.3.16 with part(a) of the Proposition proving the lower bound while part(b) proving the upper bound.

Proposition 6.8.15.

Fix B>0andω

n

=o(logn)↑ ∞.

(a)

Consider the degree of the root D

−

(b)

Consider the maximal degree M

_n+

(1)inT˜

_n+

(B,ω

n

). Then∃A>0such that whp as n→ ∞, M

n+

(1)¿

An

(1−γ)/(2+α)

(logn)

2

.

Proof:We start with (a). Note that each individual in the original Yule process reproduces according to

a rate one Poisson process. In particular standard bounds for a Poisson random variable implies that

the degree of the root in ˜T

_n−

(B,ω

n

) by timeγlogn−B

when the dynamics is switched to preferential

attachment dynamics satisfies

|deg

_n

(ρ,γlogn−B)−γlogn| =O

P

(

q

logn).

(6.8.32)

Now let {Y

i

(·) :i≥1} be a collection of independent rate one Yule processes. Comparing rates, the degree

of the root afterγlogn−Bwe get that

deg

n

(γlogn−B+ ·)ºst

degn(ρ,γlogn−B)

X

i=1

Y

i

(·),

(6.8.33)

Using (6.8.32), Lemma 6.5.4 and standard tail bounds for the Geometric distribution now completes the

proof.

Let us now prove (b). Recall that after the change point, dynamics are modulated byf

1

(·) := · +1+α.

LetAdenote the smallest integer≥α+1. Letξ

f1

be the corresponding continuous time offspring point

process. Comparing rates we see that

ξ

f1

(·)≤

st

A+2

X

i=1

Y

i

(·),

(6.8.34)

where as before {Y

i

(·) :i≥1} is a collection of independent rate one Yule processes. For every vertexv

write deg

_n

(v) for the degree of the vertex at time logn+B+t

_n+

when we have finished constructing the

process ˜T

n+

(B,ω

n

). Abusing notation, writeT

v

for the time of birth of vertexv. We will break up the

proof of (b) into two cases:

(b1) Maximal degree for vertices born afterlogn+B:Define

A

n

=

n

v∈T˜

n+

(B,ω

n

) :T

v

∈[logn+B, logn+B+t

n+

], deg

n

(v)>C n

1−γ

2+α

_(logn)

2

_.

o

whereCis an appropriate large constant that will be chosen later. The aim is to show that we can choose

Csuch thatE(|A

n

|)→0, asn→ ∞. This would then imply

P(∃v∈T˜

n+

(B,ω

n

),T

v

≥logn+B

deg

n

(v)>C n

1−γ

2+α

(logn)

2

)→0.

(6.8.35)

Letk

n

:=C n

1−γ

2+α

(logn)

2

) and let ˜T

_n+

(t) denote the tree at timet. Since the offspring distribution of

each new vertex born att>logn+Bis a Yule process then, by Lemma 6.5.4 the probably a new vertex

has degree greater thank

n

by timet

n+

is given by

P(Geom(e

t−tn+

_)≥k

_n

_)≤e

knet−tn+

Note that new vertices are produced at rate (2+α)|T˜

n+

(t)|−1. As in the proof of Proposition 6.8.14M(t) :=

e

−(2+α)t

|T˜

n+

(t)| +

₍₂₊1_α)

e

−(2+α)

t

_,t

≥logn+Bis a martingale. NotingE|T˜

n+

(logn+B)| =e

B

n

γ

we get that

E|T˜

n+

(t)| =C

0

n

γ

e

(2+α)t

fort≥logn+B

whereC

0

is a constant depending only onB,α. Thus

E(|A

n

|)≤C

00

n

γ

Z

t_n+ 0

e

knet−t+n

_e

(2+α)t

_{d t}

whereC

00

depends only onB,αand it is sufficient to check the following lemma.

Lemma 6.8.16.

Let

I

n

:=n

γ

Z

t+_n 0

e

−C(logn)2n 1−γ 2+αet−tn+

e

(2+α)t

d t

(6.8.36)

For sufficiently large C , I

n

→0as n→ ∞.

Proof.

Writinga:=

₂1₊−_αγ

andb:=2+α, algebraic manipulations result in the form:

I

n

≤n

γ

(logn)

−2b

e

b wn nγ/2

Γ

³

b,C(logn)

2

e

−nwnγ/2

´

:=E

n

.

(6.8.37)

whereΓ(b,z)=R

z∞

e

−t

t

b−1

d t

is the upper incomplete Gamma function. Known asymptotics for the

incomplete Gamma functionΓ(b,z)=Ω(z

b−1

e

−z

) asz→ ∞imply

E

n

∼n

γ−Clogne −wn

nγ/2

(logn)

−2

e

−nwnγ/2

→0.

■

(b2) Maximal degree for vertices born beforelogn+B:

We prove that vertices born beforeγlogn+B

cannot have too large of a maximal degree in ˜T

+

n

(B,ω

n

). To simplify notation, write the following for the

two times:

∆

n

:=γlogn+B,

Υ

n

:=γlogn+B+t

n+

.

(6.8.38)

Further write deg(v,t) for the degree of a vertexv

at timet

with the convention that deg(v,t) :=0 for

t<T

v

. Write deg

n

(v) :=deg(v,Υ

n

) for the final degree ofvin ˜T

n+

(B,ω

n

). Finally in the construction of

the tree ˜T

n+

(B,ω

n

), for any 0≤t≤Υ

n

, write ˜T

n+

(t) for the tree at timet.

FixC>0 and letB

n

be the set of vertices born before logn+Bwhose final degree is too large i.e.

B

n

:={v∈BP

n

:T

v

≤logn+B, deg

n

(v)>C n

1−γ

2+α

_(logn)

2

_.}

where deg

_n

(v) is the degree of vertexvin the final tree ˜T

_n+

(B,ω

n

).

Proposition 6.8.17.

We can choose C< ∞such thatP(B

n

≥1)→0as n→ ∞.

The plan is as follows: we control the maximal degree of vertices born in the early (pre∆

n

) tree then

show that none of these early vertices have time to accumulate too many edges in the remainingΥ

n

−∆

n

time period.

Proof.

Consider the tree ˜T

_n+

(∆

n

). LetM

n

(∆

n

) :=max

v∈T˜+

n(∆n)

d eg(v,∆

n

) be the maximal degree of ver-

tices in ˜T

_n+

(∆

n

) at time∆

n

. Let`

n

:=10elognand fix a sequenceω

n

↑ ∞. By the union bound,

P(B

n

≥1)≤P(B

n

≥1,|T˜

n+

(∆

n

)| <ω

n

n

γ

,M

n

≤`

n

)

Lemmas 6.8.18 and 6.8.19 which bound the three terms on the right complete the proof of the Proposi-

tion.

■

Lemma 6.8.18.

For C large enoughP(B

n

≥1,|T˜

n+

(∆

n

)| <ω

n

n

γ

,M

n

≤`

n

)→0as n→ ∞.

Proof.

LetG

n

={|T˜

n+

(∆

n

)| <ω

n

n

γ

,M

n

≤`

n

}. It is sufficient to showP(B

n

≥1|G

n

)→0. Conditional on

G

n

, we will construct a random variable that stochastically bounds the growth of degrees in the process

˜

T

+ n

(t) fort≥∆

n

. Let

©

X

i

(·) : 1≤i≤n

γ

ω

n

ª

be a collection of independent rate one Yule processes each

starting with`

n

+ dαeindividuals at time 0 and run each for timet

n+

=

₂₊1−_αγ

logn+

_nωγn/2

. ConsiderM

n

=

max

_1≤i≤ωnnγ

X

i

(t

n+

).

On the eventG

n

, the degree evolution of ˜T

n+

after time∆

n

is as follows: Sample ˜T

n+

(∆

n

) conditional

onG

n

i.e. the event that there are fewer than

ω

n

n

γ

vertices and the maximal degree is less than`

n

.

For each vertex,v, in ˜T

_n+

(∆

n

) we run an independent, rate 1 Yule process starting with deg(v,∆

n

)+α

individuals for time

t

_n+

. Our new process starts each Yule process as if each individual has maximal

degree at timeγlogn+B. In particular on the eventG

n

, the maximal degreeM

n

(Υ

n

) at timeΥ

n

satisfies

M

n

(Υ

n

)¹stM

n

. The rest of the proof analyzesM

n

. Using the union bound gives,

P(B

n

≥1|G

n

)≤P

³

M

n

≥C n

1−γ 2+α

_(logn)

2

´_≤ω

_n

_n

γ

P³_X

_i

_(t

_n+

_{)≥C n}

1−γ 2+α

_(logn)

2

´_.

Now for a rate one Yule process started withmindividuals at time zero sayY

m

(·) for fixedt,Y

m

(t) is

distributed as the sum ofmiid geometric random variables withp=e

−t

_{. Thus}

P¡

Y

m

(t)>λ¢≤mP

µ

geom(e

−t

)>

λ

m

¶

≤mexp

·

−λ

me

−t

¸_.

Plugging inm=`

n

+ dαe,t=t

n+

,λ=C n

1−γ 2+α

_(logn)

2

we get,

ω

n

n

γ

P

³

X

i

(t

n+

)≥C n

1−γ 2+α

_(logn)

2

´

≤Kω

n

n

γ

lognn

−C

which goes to zero for sufficiently largeC.

Lemma 6.8.19.

For C large enough as n→ ∞,

P(|T˜

n+

(∆

n

)| ≥ω

n

n

γ

)→0,

P(M

n

(∆

n

)>`

n

)→0.

Proof.

We first prove the assertion on|T˜

n

(∆

n

)|. Note the size of the tree grows according to a rate one

Yule process. Thus by Lemma 6.5.4,|T˜

n

(∆

n

)| ∼Geom¡e

−γl og n−B

¢. Thus

P¡

|T˜

n+

(∆

n

)| ≥ω

n

n

γ

¢

≤exph−ω

n

n

γ

e

−γlogn−B

i

→0,

asn→ ∞.

For the second assertion, note that for any 0≤t≤∆

n

, the rate at which a new vertex is born is|T˜

n+

(t)|.

Since the offspring distribution of each new vertex (before time∆

n

) is a Poisson process, the probability

that this new vertex has degree greater than`

n

conditional on ˜T

n+

(t) is

P(Poisson(∆

n

−t)≥`

n

)≤P(Poisson(∆

n

)≥`

n

).

Thus writingN

n

(∆

n

) for the number of vertices with degree at least`

n

by time∆

n

and recalling that for

t≤∆

n

,E( ˜T

n+

(t))=e

t

we have,

E(N

n

(∆

n

))=

Z

∆n 0

P

(Poisson(∆

n

−t)≥`

n

)e

t

d t≤

e

B

n

γ

P(Poisson(∆

n

)≥`

n

).

Since∆

n

=γlogn+B

withγ<1, exponential tail bounds for the Poisson distribution completes the

proof.

■

In document Carmichael_unc_0153D_18458.pdf (Page 175-181)