Proof of Theorem

We can make the condition of Theorem 5.3.5 a little easier to check for our case by observing that our thinning process narrows down the types of sequences ¯s we need to consider.

In the context of our thinned branching process, the requirement thatpn↓0 implies that the sequence ¯s must satisfy sn→1. It also should not be surprising that any finite number of leading terms of the sequence (sn) don’t matter so that it is enough for us to check this condition as n→ ∞. Let us summarize these arguments in a lemma:

Lemma 5.3.6. For a thinned branching process driven by thinned distributions with pgfs {fn}n>0, e < 1 if and only if there exists a sequence {sn}n>0, sn ∈ [0,1) and sn → 1 such that for some N

fn(sn+1)6sn, ∀n>N (5.4) Proof:

We begin by arguing that there is no loss of generality in considering only {sn} with sn →1. If there does in fact exist ¯s such that ¯f(¯s)6¯s, then it is necessary thatsn →1. To see why, note that fn(s)> fn(0) for all s ∈[0,1]. Therefore if fn(sn+1) 6sn for alln, then sn >fn(0) for alln. Sincepn →0, then fn(0)→1 so sn→1 also.

The fact that s → 1 is necessary also implies that if there does not exist ¯s satisfying ¯

f(¯s) ₆ ¯s with sn → 1, then there cannot exist any other such sequence ˜s ∈ I∞ satisfying ¯

f(˜s)₆s˜.

To see how the tail condition 5.4 implies the full condition on ¯s in Theorem 5.3.5, note that we can start from sN and then construct the initial elements of the sequence s1, . . . , sN−1 satisfying condition 5.3 by setting sN−1 =fN−1(sN)∈(0,1).

Now letting g(s, α) be the pgf of the offspring distribution, the thinned distribution f(s, p, α) for a fixed thinning probability p may be written g(1− p+ps, α) so that the survival criteria in Lemma 5.3.6 is that there exists{s } ,s ∈[0,1) for all nwith s →1

such that

g(1−pn+pnsn+1, α)6sn, for all n

It will be convenient to work on a simpler approximating function for g(·). For future reference we introduce the notation

h(s) := (−log(s))α (5.5)

The approximation we will work with is

g(1−p+ps)∼1−C·h(1−p+ps) = 1−C(−log(1−p+ps))α

holding for allC > 0 ass→1. Clearly, any function converging to 0 can be used in place of h for this approximation, but the reason for our particular choice will become clear shortly.

The survival criteria is now

1−Ch(1−pn(1−sn+1))6sn, some sequence sn→1

To establish the possibility of (sn)n>0 satisfying this criteria, we will make use of the following

deterministic lemma:

Lemma 5.3.7. For brevity, say that (pn)n>0 satisfies (?) if there exists b ∈(0,1) such that:

− n

X

k=1

γ−klogpk→ −logb

Fix someγ >1and suppose that(pn)n>0 is a sequence∈(0,1)which converges to 0. Note: in

what follows below the sequences(an)depend on the constantC but we suppress this notation for simplicity.

1. If (pn)n>1 satisfies (?), then for every C > 0 there exists a sequence (an)n>0 ∈ (0,1]

also converging to 0 such that pn>C a γ n

2. If for some C > 0 there exists a sequence (an)n>0 ∈ (0,1] also converging to 0 such

that pn>C a γ n

an+1 for all n >1, then (pn)n>1 satisfies (?).

Proof: For the direct portion we show that a sequence (an)n>1 satisfying the theorem can

be constructed. Observe that for the (n+ 1)th terman+1 to satisfy the stated condition, we

must have:

an+1 >C aγ_n pn Iterating this n times we see that

an+1 >Cγ+γ 2_+...+γn aγ n 1 Qn k=1p γ(n−k) k

Now assume without loss of generality that C <1 so that the term involving C is bounded above by 1 (ifC _>1 then it can be absorbed into the pk’s by pk 7→pk/C and the rest of the argument goes through). If we show that aγ₁n/Qn

k=1p

γ(n−k)

k → 0, then this guarantees that we will be able to pick (an) satisfying the stated inequality with an ∈ (0,1) for all n and an →0.

Take logs and set the initial valuea1 =b− where∈(0, b). Then

log a γn 1 Qn k=1p γ(n−k) k =γnlogb− n X k=1 γ(n−k)logpk =γn logb− n X k=1 γ−klogpk ! Since Pn k=1γ −k_logp

k→logb then the quantity inside the parenthesis eventually stays neg- ative and the entire RHS → −∞, which implies that the original ratio →0.

For the converse portion, suppose that the contrapositive is not true. That is, suppose −Pn

k=1γ

−k_logp

k → ∞but that there exists (an)n>0 such that pk >C·

aγ_k ak+1

Rearranging this we have the relation

ak 6

p_ka_k+1

C

1/γ

Now starting at a1 and expanding the right-hand side recursively n times we obtain

a1 6 (pγ₁−1pγ₂−2· · ·pγ−n n )a γ−n n+1 Cγ−1_+γ−2_+···+γ−n

SinceC < 1 thenγ−1+· · ·+γ−n₆γ/(γ−1) for allnso the denominator is bounded above by C0 = Cγ/(γ−1). If C > 1 then this quantity is bounded by C0 = C1/γ. Combining this

with the fact that an <1 for all n, the inequality simplifies to

a1 6 Qn k=1p γ−k k C0 However, −Pn k=1γ −k_logp k → ∞ implies Qn_k=1pγ −k

k → 0, thus sending n → ∞ above

implies a1 = 0 and we have a contradiction.

From there it is only a short hop to the result for our approximation sequence:

Lemma 5.3.8. Again for brevity, say that (pn)n>0 satisfies (?) if there exists b∈(0,1)such

that: − n X k=1 γ−klogpk→ −logb

1. If (pn)n>1 satisfies (?), then for every C > 0 there exists a sequence (sn)n>0 ∈ [0,1)

converging to 1 such that 1−C·h(1−pn(1−sn+1))6sn for all n>1.

2. If for some C > 0 there exists a sequence (sn)n>0 ∈ [0,1) converging to 1 such that

1−C·h(1−pn(1−sn+1))6sn for all n >1, then (pn)n>1 satisfies (?).

It is sufficient to show the relevant properties for sequences (sn)n>0 satisfying the easier

inequality

1−C·hpn(1−sn+1)

iα

6sn for all n >1 (5.6) since, by the identity 1−x₆−log(x), x > 0 and the facts thatpn, sn∈[0,1) the following inequality holds for all n:

1−C·h(1−pn(1−sn+1))61−C·

h

pn(1−sn+1)

iα

Inequality (5.6) implies that we want to study sequences (sn)n>0, sn ∈ [0,1) for all n such that:

pn>C·

(1−sn)1/α 1−sn+1

, for all n_>1 (5.7)

Now applying the previous lemma with the map 1−sn=an gives the result.

Finally we need to tighten the result above concerning the approximating functionh(·) to the pgfg(·). The key tool is a Tauberian theorem giving the behavior of the Laplace-Stieltjes transform of a heavy-tailed random variable.

Theorem 5.3.9. ([16], Theorem 8.1.6) Let F(·)be the CDF of some probability distribution and φ(·) its Laplace-Stieltjes transform. Then for 0 ₆α <1 and a slowly-varying function `, the following are equivalent

(1) 1−φ(s)∼sα`(1/s), s↓0 (2) 1−F(x)∼ `(x)

xα_Γ(1−α), x→ ∞

The relationship between the Laplace-Stieltjes transform φ(·) and the probability gen- erating function g(·) is φ(−logs) = f(s). Also, for our offspring distributions the function `(x) is a constantCα >0 for allα∈(0,1). So making the maps7→ −(logs), we can rewrite

Corollary 5.3.10. Letg(·)be the pgf of an offspring distribution satisfying 5.2 from Theorem 5.3.3. Then there exists C >0 such that

1−g(s)∼C·h(s), s→1

Where h(·) is defined in 5.5.

This implies the following simple Lemma:

Lemma 5.3.11. Given an offspring distribution satisfying 5.2, then there exist constants C0, C1 >0 such that

1. There exists s0 such that 1−g(s)6C0·h(s) for all s > s0

2. There exists s1 such that 1−g(s)>C1·h(s) for all s > s1

Proof: Solving (1) for C0 gives

1−g(s)

h(s) 6C0 for all s > s0

Now by Corollary 5.3.10, there exists C > 0 such that (1−g(s))/h(s) → C. Thus setting C0 > C implies the result. The proof for (2) is exactly analogous.

Combining Lemma 5.3.11 and Lemma 5.3.8, extracting the {pn}n>0 criterion from the

CHAPTER 6 Future directions

We discuss some potential extensions of the two works presented above, and some ideas for new work in unrelated areas.