LOCAL LARGE DEVIATIONS
AND THE STRONG RENEWAL THEOREM
FRANCESCO CARAVENNA AND RON DONEY
Abstract. We establish two different, but related results for random walks in the domain
of attraction of a stable law of index α. The first result is a local large deviation upper
bound, valid for α ∈ (0, 1) ∪ (1, 2), which improves on the classical Gnedenko and Stone
local limit theorems. The second result, valid for α ∈ (0, 1), is the derivation of necessary
and sufficient conditions for the random walk to satisfy the strong renewal theorem (SRT).
This solves a long standing problem, which dates back to the 1963 paper of Garsia and
Lamperti [GL63] for renewal processes (i.e. random walks with non-negative increments),
and to the 1968 paper of Williamson [Wil68] for general random walks.
1. Introduction and results
This paper contains new results about asymptotically stable random walks. We first
present a local large deviation estimate which improves the error term in the classical local
limit theorems, without making any further assumptions (cf. Theorem 1.1). Then we exploit
this bound to solve a long-standing problem, namely we establish necessary and sufficient
conditions for the validity of the strong renewal theorem (SRT), both for renewal processes
(cf. Theorem 1.4) and for general random walks (cf. Theorem 1.12). The corresponding
result for Lévy processes is also presented (cf. Theorem 1.18).
This paper supersedes the individual preprints [Car15] and [Don15].
Notation. We set N = {1, 2, 3, . . .} and N0 = N ∪ {0}. We denote by RV (γ) the class of
regularly varying functions with index γ, namely f ∈ RV (γ) if and only if f (x) = xγ `(x)
for some slowly varying function ` ∈ RV (0), cf. [BGT89]. Given f, g : [0, ∞) → (0, ∞) we
write f ∼ g as a shorthand for lims→∞ f (s)/g(s) = 1.
1.1. Local large deviations. Let (Xi )i∈N be i.i.d. real-valued random variables, with law
F . Let S0 := 0, Sn := X1 + . . . + Xn be the associated random walk and
Mn := max{X1 , X2 , . . . , Xn } .
(1.1)
We assume that the law F is in the domain of attraction of a strictly stable law with index
α ∈ (0, 1) ∪ (1, 2). More explicitly, setting F (x) := F ((x, ∞)) and F (x) := F ((−∞, x]),
p
q
F (x) ∼
and
F (−x) ∼
for some A ∈ RV (α) ,
(1.2)
x→∞ A(x)
x→∞ A(x)
with p > 0 and q ≥ 0. For α > 1 we further assume that E[X] = 0.
We may assume that A ∈ RV (α) is strictly increasing. Introducing the norming sequence
an := A−1 (n) ,
n ∈ N,
Date: December 22, 2016.
2010 Mathematics Subject Classification. 60K05; 60G50.
Key words and phrases. Renewal Theorem, Local Limit Theorem, Regular Variation.
1
(1.3)
2
FRANCESCO CARAVENNA AND RON DONEY
Sn /an converges in law to a random variable Y with a stable law of index α and positivity
1
πα
parameter ρ = P(Y > 0) = 21 + πα
arctan( p−q
p+q tan 2 ) > 0 (since we assume that p > 0).
Our first result is a local large deviation estimate for Sn , constrained on Mn . This extends
a result for α = 2 due to Negaev [Nag79].
Theorem 1.1 (Local Large Deviations). Let F satisfy (1.2) with α ∈ (0, 1) ∪ (1, 2) and
p > 0, and E[X] = 0 if α > 1. Fix a bounded measurable J ⊆ R. Given γ ∈ (0, ∞), there
is C0 = C0 (γ, J) < ∞ such that, for all n ∈ N and x ≥ 0, the following relations hold:
d1/γe
n
1
P(Sn ∈ x + J, Mn ≤ γx) ≤ C0
,
(1.4)
an A(x)
where dxe := min{n ∈ N : n ≥ x} is the upper integer part of x. More explicitly:
k
1
n
1
1
∀k ∈ N, ∀γ ∈ [ k , k−1 ) :
P(Sn ∈ x + J, Mn ≤ γx) ≤ C0
.
an A(x)
(1.5)
Moreover, for some C00 = C00 (J) < ∞,
P(Sn ∈ x + J) ≤ C00
1 n
.
an A(x)
(1.6)
Remark 1.2. The classical Gnedenko and Stone local limit theorems (that we recall in
Subsection 2.3) yield the bound P(Sn ∈ x + J) = O( a1n ). Relation (1.6) improves on this
bound as soon as A(x)/n → ∞, that is when x/an → ∞.
A heuristic explanation of (1.6) goes as follows: for large x, if Sn ∈ x + J, it is likely
that a single step Xi takes values comparable to x. Since P(Xi > cx) ≈ 1/A(x) by (1.2),
and since there are n available steps, we get the factor n/A(x) in (1.6).
A similar argument sheds light on (1.5). Under the constraint Mn ≤ γx, with γ ∈
1
), the most likely way to have Sn ∈ x + J is that exactly k steps Xi1 , . . . , Xik take
[ k1 , k−1
values comparable to x/k, and this yields the factor (n/A(x))k in (1.5).
1.2. The strong renewal theorem. Henceforth we assume that α ∈ (0, 1). We say that
F is arithmetic if it is supported by hZ for some h > 0, in which case the maximal value
of h > 0 with this property is called the arithmetic span of F . It is convenient to set
(
arithmetic span of F
(if F is arithmetic)
I = (−h, 0]
where
h :=
(1.7)
any fixed number > 0 (if F is non-arithmetic) .
The renewal measure U (·) associated to F is defined by
X
X
U (dx) :=
F ∗n (dx) =
P(Sn ∈ dx) .
n≥0
(1.8)
n≥0
It is well known [BGT89, Theorem 8.6.3] that (1.2) implies the following renewal theorem:
C
A(x) ,
with
C = C(α, ρ) = α E[Y −α 1{Y >0} ]
(1.9)
U ([0, x]) ∼
x→∞ α
(recall that Y denotes a random variable with the limiting stable law). In the special case
when p = 1 and q = 0 in (1.2) (so that ρ = 1) one has C = π1 α sin(πα).
It is natural to wonder whether the local version of (1.9) holds, namely
U (x + I) = U ((x − h, x])
∼
x→∞
Ch
A(x)
.
x
(SRT)
THE STRONG RENEWAL THEOREM
3
This relation, called strong renewal theorem (SRT), follows from (1.2) for α > 12 , cf. [GL63,
Wil68, Eri70, Eri71], while for α ≤ 21 there are F satisfying (1.2) but not (SRT).
Our next results are necessary and sufficient conditions for the SRT. Let us set for k ≥ 0
and x ∈ R
A(|x|)k
bk (x) :=
.
(1.10)
|x| ∨ 1
Definition 1.3 (Asymptotic Negligibility). A function J(δ; x) is asymptotically negligible
(a.n.) if and only if
lim lim sup
δ→0 x→+∞
J(δ; x)
x
= lim lim sup
J(δ; x) = 0 .
δ→0
b1 (x)
x→+∞ A(x)
(1.11)
The SRT turns out to be equivalent to the a.n. of suitable integral quantities defined in
terms of F and bk . We start with the case of renewal processes, which is simpler.
1.3. The renewal process case. Assume that F is a law on [0, ∞) such that
1
F (x) ∼
for some A ∈ RV (α) .
x→∞ A(x)
For δ > 0 and x ≥ 0 we set
Z
I1+ (δ; x) :=
0≤z≤δx
Z
F (x − dz) b2 (z) =
F (x − dz)
0≤z≤δx
A(z)2
.
z∨1
(1.12)
(1.13)
The following is our main result for renewal processes.
Theorem 1.4 (SRT for Renewal Processes). Let F be a probability on [0, ∞) satisfying
(1.12) with α ∈ (0, 1). Define I = (−h, 0] with h > 0 as in (1.7).
• If α > 12 , the SRT holds with no extra assumption on F .
• If α ≤ 12 , the SRT holds if and only if I1+ (δ; x) is a.n. (cf. Definition 1.3).
In the “boundary” case α = 21 , we can say precisely under which conditions on A(x) the
SRT holds with no further assumption on F besides (1.12) (like it happens for α > 12 ).
Theorem 1.5 (SRT for Renewal Processes with α = 12 ). Let F be a probability on [0, ∞)
√
satisfying (1.12) with α = 12 (so that A(x)/ x is a slowly varying function). If
A(s)
A(x)
sup √
= O √
,
(1.14)
s x→∞
x
1≤s≤x
√
then the SRT holds with no extra assumption on F . (This includes the case A(x) ∼ c x.)
If condition (1.14) fails, there are examples of F for which the SRT fails.
1.4. Sufficient conditions for renewal processes. Given a probability F on [0, ∞)
which satisfies (1.12), a classical sufficient condition for the SRT is
1
F (x + I) = O
,
(1.15)
x→∞
xA(x)
as proved by Doney [Don97] in the arithmetic case (extending previous results of Williamson
[Wil68]), and by Vatutin and Topchii [VT13] in the non-arithmetic case.
Interestingly, if one only looks at the growth of the “local” probabilities F (x + I), no
sharper condition than (1.15) can ensure that the SRT holds, as the following result shows.
4
FRANCESCO CARAVENNA AND RON DONEY
Proposition 1.6. Fix A ∈ RV (α) with α ∈ (0, 12 ), and let ζ : (0, ∞) → (0, ∞) be an
arbitrary non-decreasing function with limx→∞ ζ(x) = ∞. Then there exists a probability
ζ(x)
F on [0, ∞) which satisfies (1.12), such that F (x + I) = O( xA(x)
), for which the SRT fails.
Intuitively, when condition (1.15) is not satisfied, in order for the SRT to hold, the
1
must not be “too cluttered”. We can give a precise
points x for which F (x + I) xA(x)
formulation of this loose statement by looking at the probability of intervals F ((x − y, x]).
Proposition 1.7. Let F be a probability on [0, ∞) satisfying (1.12) with α ∈ (0, 12 ].
• A sufficient condition for the SRT is that there exists γ > 1 − 2α such that
1 y γ
,
for any y = yx ≥ 1 with yx = o(x) .
F ((x − y, x]) = o
x→∞
A(x) x
(1.16)
• A necessary condition for the SRT is that (1.16) holds for any γ < 1 − 2α.
Remark 1.8. In the sufficient condition (1.16), one can equivalently replace o(·) by O(·).
In fact, if (1.16) holds with O(·) for some γ > 1 − 2α, it holds with o(·) for any γ 0 < γ.
Remark 1.9. The sufficient condition (1.16) is a generalization of (1.15). In fact, if (1.15)
Pdy/he
1
holds, then (1.16) holds with γ = 1, since F ((x−y, x]) ≤ j=0 F (x−j +I) ≤ 2y
h O( xA(x) ).
Remark 1.10. Other sufficient conditions for the SRT, which generalize (1.15), were given
by Chi in [Chi15, Chi13]. These can be deduced from Theorem 1.4.
Conditions similar to (1.16), in a different context, appear in [CSZ16].
1
We point out that if F satisfies (1.12), we can easily deduce that F ((x − y, x]) = o( A(x)
)
for any y = o(x), that is (1.16) holds with γ = 0. However, with no extra assumption, one
cannot hope to improve this estimate, as Lemma 9.2 below shows.
To see how condition (1.16) appears, let us introduce the following variant of (1.13):
Z δx
Z δx
F ((x − z, x]) A(z)2
F ((x − z, x])
+
˜
b2 (z) dz =
dz .
(1.17)
I1 (δ; x) :=
z
z
z
1
1
Our next result shows that one can look at I˜+ (δ; x) instead of I + (δ; x).
1
1
Proposition 1.11. Let F be a probability on [0, ∞) satisfying (1.12) for some α ∈ (0, 21 ].
• If I˜1+ (δ; x) is a.n., then also I1+ (δ; x) is a.n., hence the SRT holds.
• When α < 21 , the converse is also true: I˜1+ (δ; x) is a.n. if and only if I1+ (δ; x) is a.n..
1.5. The general random walk case. We now turn to the general random walk case,
that is we assume that F is a probability on R which satisfies (1.2) for some α ∈ (0, 1),
p > 0 and q ≥ 0. Let us generalize (1.13) as follows: for δ > 0 and x ≥ 0 we set:
Z
I1 (δ; x) :=
F (x + dy) b2 (y) .
(1.18)
|y|≤δx
For k ∈ N with k ≥ 2, we introduce a further parameter η ∈ (0, 1) and we set
Z
Z
Z
Ik (δ, η; x) :=
F (x + dy1 )
···
Py1 (dy2 , . . . , dyk ) bk+1 (yk ) ,
|y1 |≤δx
where
|yj |≤η|yj−1 | for 2≤j≤k
Py1 (dy2 , . . . , dyk ) := F (−y1 + dy2 )F (−y2 + dy3 ) · · · F (−yk−1 + dyk ) .
(1.19)
THE STRONG RENEWAL THEOREM
5
Note that Py1 (dy2 , . . . , dyk ) is the law of (S2 , . . . , Sk ) conditionally on S1 = y1 , hence
Ik (δ, η; x) = E bk+1 (Sk ) 1{|Sj |≤η|Sj−1 | for 2≤j≤k} 1{|S1 |≤δx} S0 = −x .
(1.20)
The same formula holds also for k = 1 (where
Let us define

0






1
1
κα :=
−1= 2

α
..



.



m
the first indicator function equals 1).
if α ∈ ( 21 , 1)
if α ∈ ( 13 , 12 ]
if α ∈ ( 14 , 13 ]
,
(1.21)
1
1
if α ∈ ( m+2
, m+1
]
We are going to see that, when 1/α 6∈ N, necessary and sufficient conditions for the SRT
involve the a.n. of Ik (δ; x) for k = κα . The case 1/α ∈ N is slightly more involved. We need
to introduce a suitable modification of (1.10), namely
Z |z|
bk (t)
dt ,
(1.22)
b̃k (z, x) := b̃k (|z|, |x|) :=
|x| t ∨ 1
where the integral vanishes if |x| > |z|. We then define I˜1 (δ; x) and I˜k (δ, η; x) in analogy
with (1.18) and (1.19), replacing b2 (y) by b̃2 (δx, y) and bk+1 (yk ) by b̃k+1 (yk−1 , yk ):
Z
˜
F (x + dy) b̃2 (δx, y) ,
(1.23)
I1 (δ; x) :=
|y|≤δx
and for k ≥ 2:
I˜k (δ, η; x) :=
Z
Z
F (x + dy1 )
|y1 |≤δx
Z
···
Py1 (dy2 , . . . , dyk ) b̃k+1 (yk−1 , yk ) .
(1.24)
|yj |≤η|yj−1 | for 2≤j≤k
Note that, by Fubini’s theorem, we can equivalently rewrite (1.23) as
Z δx
F ((x − t, x + t])
I˜1 (δ; x) =
b2 (t) dt ,
t∨1
0
(1.25)
which is a natural random walk generalization of (1.17).
We can now state our main result for random walks.
Theorem 1.12 (SRT for Random Walks). Let F be a probability on R satisfying (1.12)
with α ∈ (0, 1) and with p, q > 0. Define I = (−h, 0] with h > 0 as in (1.7).
• If α > 12 , the SRT holds with no extra assumption on F .
• If α ≤
1
2
1
α 6∈ N, we distinguish two cases:
( 13 , 12 ), i.e. κα = 1, the SRT holds if
1
1
( k+2
, k+1
) for some k = κα ≥ 2, the
and
- if α ∈
- if α ∈
is a.n., for every fixed η ∈ (0, 1).
• If α ≤
1
2
and
1
α
and only if I1 (δ; x) is a.n..
SRT holds if and only if Iκα (δ, η; x)
∈ N, the same statement holds if we replace Ik by I˜k , namely:
- if α = 12 , i.e. κα = 1, the SRT holds if and only if I˜1 (δ; x) is a.n..
1
- if α = k+1
, for some k = κα ≥ 2, the SRT holds if and only if I˜κα (δ, η; x) is
a.n., for every fixed η ∈ (0, 1).
6
FRANCESCO CARAVENNA AND RON DONEY
In Appendix A we show some relations between the quantities Ik and I˜k . These lead to
the following clarifying remarks.
Remark 1.13. When α1 6∈ N, I˜κα is a.n. if and only if Iκα is a.n. (cf. Corollary A.5 below).
As a consequence, we could rephrase Theorem 1.12 in a more compact way as follows:


if α > 12

with no extra assumption
The SRT holds: iff I˜1 (δ; x) is a.n.
(1.26)
if 13 < α ≤ 12


iff I˜ (δ, η; x) is a.n. for every η ∈ (0, 1) if α ≤ 1
κα
3
α
When α ≤
our proof actually shows that if I˜κα (δ, η; x) is a.n. for some η > 1 − 1−α
,
then (the SRT holds and consequently) it is a.n. for every η ∈ (0, 1). It is not clear whether
the a.n. of I˜κα (δ, η; x) for some η ≤ 1 − α also implies the a.n. for any η ∈ (0, 1).
1
3,
1−α
Remark 1.14. If α1 6∈ N, the condition that Iκα (δ, η; x) is a.n. is equivalent to the seemingly
stronger condition that Ik (δ, η; x) is a.n. for all k ∈ N (cf. Corollary A.2 below). Similarly,
I˜κα (δ, η; x) is a.n. if and only if I˜k (δ, η; x) is a.n. for all k ∈ N (cf. Corollary A.4 below).
Remark 1.15. In Theorem 1.12 we require q > 0 (that is the positivity index ρ is strictly
less than one), but a large part of it actually extends to q = 0. More precisely, when q = 0,
our proof shows that if α > 12 the SRT holds with no extra assumption on F , while if α ≤ 12
the a.n. of Iκα (if α1 6∈ N) or I˜κα (if α1 ∈ N) are sufficient conditions for the SRT. However,
when q = 0, we do not expect the a.n. of Iκα or I˜κα to be necessary, in general.
1.6. Sufficient conditions for random walks. Necessary and sufficient conditions for
the SRT in the random walk case involve the a.n. of I˜k for a suitable k = κα ∈ N. Unlike
the renewal process case, this cannot be reduced to the a.n. of just I˜1 .
Proposition 1.16. For any α ∈ (0, 31 ), there is a probability F on R which satisfies (1.12),
such that I˜1 (δ; x) is a.n. but I˜2 (δ, η; x) is not a.n., for any η ∈ (0, 1) (hence the SRT fails).
Let us now give simpler sufficient conditions which ensure the a.n. of I˜k . Note that the
condition that I˜1 (δ; x) is a.n. only involves the right tail of F (cf. Definition 1.3). To express
conditions on the left tail of F , we define
Z δx
F ((−x − t, −x + t])
∗
˜
I1 (δ; x) :=
b2 (t) dt ,
(1.27)
t∨1
0
which is nothing but I˜1 (δ; x) in (1.25) applied to the reflected probability F ∗ (A) := F (−A).
Proposition 1.17. Let F be a probability on R satisfying (1.12) with α ∈ (0, 21 ] and p > 0,
q ≥ 0. If both I˜1 (δ; x) and I˜1∗ (δ; x) are a.n., then the SRT holds.
In particular, a sufficient condition for the SRT is that there exists γ > 1 − 2α such that
relation (1.16) holds both for F and for F ∗ (i.e., both as x → +∞ and as x → −∞). As a
special case, the SRT holds when the classical condition (1.15) holds both for F and F ∗
1.7. Lévy processes. Let X = (Xt )t≥0 be a Lévy process with Lévy measure Π, Brownian
coefficient σ 2 and linear term µ in its Lévy-Khintchine representation, that is
Z
σ2 2
iθX1
log E[e
] = iµθ − θ +
(eiθx − 1 − iθx1{|x|≤1} ) Π(dx) .
(1.28)
2
R\{0}
THE STRONG RENEWAL THEOREM
7
Whenever X is transient, we can define its potential or renewal measure by
Z ∞
P(Xt ∈ dx) dt .
G(dx) :=
0
We assume that X is asymptotically stable: more precisely, there is a norming function
a(t) such that Xt /a(t) converges in law as t → ∞ to a random variable Y with a stable
law of index α ∈ (0, 1) and positivity parameter ρ > 0. In this case
1
1
A(x) :=
=
∈ RV (α)
as x → +∞ ,
Π((x, ∞))
Π(x)
and we can take a(·) = A−1 (·). Under these assumptions, the renewal theorem (1.9) holds,
just replacing U ([0, x]) by G([0, x]). It is natural to wonder whether the corresponding local
version (SRT) holds as well, in which case we say that X satisfies the SRT.
Our next result shows that this question can be reduced to the validity of the SRT for
a random walk whose step distribution F only depends on the Lévy measure Π, namely:

Π(dx)


for |x| ≥ 1
Π(R
\ (−1, 1))
F (dx) :=
.
(1.29)

0
for |x| < 1
Theorem 1.18 (SRT for Lévy Processes). Let X be any Lévy process that is in the domain
of attraction of a stable law of index α ∈ (0, 1) and positivity parameter ρ > 0 as t → ∞.
Suppose also that its Lévy measure is non-arithmetic. Then X satisfies the SRT, i.e.
lim x Π(x) G((x − h, x]) = h α E[Y −α 1{Y >0} ] ,
x→∞
∀h > 0 ,
(1.30)
if and only if the random walk with step distribution F defined in (1.29) satisfies the SRT.
As a consequence, Theorems 1.4 and 1.12 can be applied to X.
The proof of Theorem 1.18, given in Section 9, is obtained comparing the Lévy process
X with a compound Poisson process with step distribution F .
Remark 1.19. It is known, cf. [Ber96, Proof of Theorem 21 on page 38], that the potential
measure G(dx) of any Lévy process X coincides for x 6= 0R with the renewal measure of a
∞
random walk (Sn )n≥0 with step distribution P(S1 ∈ dx) := 0 e−t P(Xt ∈ dx) dt. It is also
easy to see that X is in the domain of attraction of a stable law of index α ∈ (0, 1) and
positivity parameter ρ > 0, with norming function a(t), if and only if the random walk S
is in the domain of attraction of a the same stable law with norming function a(n).
So, if we write down necessary and sufficient conditions for S to verify the SRT, these
will be necessary and sufficient conditions for X to verify the SRT. However this approach
is unsatisfactory, because one would like conditions expressed in terms of the characteristics
of X, i.e. the quantities Π, σ 2 , µ appearing in the Lévy-Khintchine representation (1.28),
and the technical problem of expressing our necessary and sufficient conditions for S to
satisfy the SRT in terms of these characteristics seems quite challenging.
1.8. Structure of the paper. The paper is organized as follows.
• In Section 2 we recall some standard results.
• In Section 3 we prove Theorem 1.1.
• Sections 4–8 are devoted to the proofs of Theorems 1.4 and 1.12.
- In Section 4 we reformulate the SRT and we give two key bounds.
8
FRANCESCO CARAVENNA AND RON DONEY
- In Section 5 we prove the necessity part for both Theorems 1.4 and 1.12.
- In Section 6 we prove the sufficiency part of Theorem 1.4.
- The sufficiency part of Theorem 1.12 is proved in Section 7 for the case α > 13 .
The case α ≤ 13 is treated in Section 8 and is much more technical.
• In Section 9 we prove some “soft” results (such as Theorem 1.5, Propositions 1.6, 1.7,
1.11, 1.16, 1.17, and Theorem 1.18) which are corollaries of our main results.
• In Appendix A we prove some technical results.
2. Setup
2.1. Notation. We recall that f (s) . g(s) or f . g means f (s) = O(g(s)), i.e. for a
suitable constant C < ∞ one has f (s) ≤ C g(s) for all s in the range under consideration.
The constant C may depend on the probability F (in particular, on α) and on h. When
some extra parameter enters the constant C = C , we write f (s) . g(s). If both f . g
and g . f , we write f ≈ g. We recall that f (s) ∼ g(s) means lims→∞ f (s)/g(s) = 1.
2.2. Regular variation. Without loss of generality [BGT89, §1.3.2], we can assume that
A : [0, ∞) → (0, ∞) is differentiable, strictly increasing and such that
A0 (s) ∼ α
A(s)
,
s
as s → ∞ .
(2.1)
We fix A(0) := 12 and A(1) := 1, so that both A and A−1 map [1, ∞) onto itself. We also
write au = A−1 (u) for all u ∈ [ 12 , ∞), in agreement with (1.3).
We observe that, by Potter’s bounds, for every > 0 one has
ρα+ .
A(ρx)
. ρα− ,
A(x)
∀ρ ∈ (0, 1], x ∈ (0, ∞) such that ρx ≥ 1 .
(2.2)
More precisely, part (i) of [BGT89, Theorem 1.5.6] shows that relation (2.2) holds for
ρx ≥ x̄ , for a suitable x̄ < ∞; the extension to 1 ≤ ρx ≤ x̄ follows as in part (ii) of the
same theorem, because A(y) is bounded away from zero and infinity for y ∈ [1, x̄ ].
We also recall Karamata’s Theorem [BGT89, Propositions 1.5.8 and 1.5.10]:
X
1
if f (n) ∈ RV (ζ) with ζ > −1 :
f (n) ∼
t f (t) ,
(2.3)
t→∞ ζ + 1
n≤t
if f (n) ∈ RV (ζ) with ζ < −1 :
X
n>t
f (n) ∼
t→∞
−1
t f (t) .
ζ +1
(2.4)
2.3. Local limit theorems. We call a probability F on R lattice if it is supported by
vZ + a for some v > 0 and 0 ≤ a < v, and the maximal value of v > 0 with this property
is called the lattice span of F . If F is arithmetic (i.e. supported by hZ), then it is also
lattice, but the spans might differ (for instance, F ({−1}) = F ({+1}) = 12 has arithmetic
span h = 1 and lattice span v = 2). A lattice distribution is not necessarily arithmetic.†
†
If F is lattice, say supported by vZ + a where v is the lattice span and a ∈ [0, v), then F is arithmetic
if and only if a/v ∈ Q, in which case its arithmetic span equals h = v/m for some m ∈ N.
THE STRONG RENEWAL THEOREM
9
Recall that, under (1.2), Sn /an converges in distribution as n → ∞ toward a stable law,
whose density we denote by φ (the norming sequence an is defined in (1.3)). If we set
(
lattice span of F
(if F is lattice)
J = (−v, 0]
with
v=
,
(2.5)
any fixed number > 0 (if F is non-lattice)
Gnedenko’s and Stone’s local limit theorems [BGT89, Theorems 8.4.1 and 8.4.2] yield
x lim sup an P(Sn ∈ x + J) − v φ
= 0.
(2.6)
n→∞ x∈R
an Since supz∈R φ(z) < ∞, we obtain the useful estimate
sup P(Sn ∈ (x − w, x]) .w
x∈R
1
,
an
(2.7)
which, plainly, holds for any fixed w > 0 (not necessarily the lattice span of F ).
3. Proof of Theorem 1.1
We prove (1.4), equivalently (1.5), by steps. Without loss of generality, we assume that
J ⊆ [0, ∞) (it suffices to redefine x 7→ x0 := x + min J and J 7→ J 0 := J − min J).
Step 1. We first prove an integrated version of (1.4), namely:
1/γ
n
∀γ ∈ (0, ∞) :
P(Sn ≥ x, Mn ≤ γx) .γ
.
A(x)
(3.1)
By Markov’s inequality, for any µ ≥ 0
P(Sn ≥ x, Mn ≤ γx) ≤ e−µx E eµSn 1{Mn ≤γx} = e−µx (m0 )n ,
where m0 := E[eµX1 1{X1 ≤γx} . We are going to fix
µ :=
1
log Λ ,
γx
where
Λ :=
A(x)
.
n
Since e−µx = Λ−1/γ , it remains to show that m0 ≤ 1 + nc for some c < ∞.
For x = O(an ) we have n/A(x) ≥ (const.) > 0 and (3.1) holds trivially. Then we take
x ≥ C an , with C large so that Λ = A(x)/n ≥ e2 and consequently µ2 ≤ γx. If we split
Z
Z
Z
µy
µy
m0 =
e F (dy) +
e F (dy) +
eµy F (dy) ,
2
(−∞,− µ
]
2 2
(− µ
,µ]
2
(µ
,γx]
by 1 ≥ F ( µ2 ) we obtain
Z
m0 − 1 ≤
(e
2 2
,µ]
(− µ
µy
Z
− 1) F (dy) +
eµy F (dy) .
(3.2)
2
,γx]
(µ
Assume first that α ∈ (0, 1). Writing eµy = y eµy−log y and observing that µy − log y is
increasing for y ≥ µ1 , we get
Z
Z
eµγx
m0 − 1 . µ
|y| F (dy) +
y F (dy) .
γx (0,γx]
(− 2 , 2 ]
µ µ
10
FRANCESCO CARAVENNA AND RON DONEY
R
Rw
Note that (0,w] y F (dy) = 0 F (z) dz − wF (w) . w F (w) for w ≥ 0 by (2.3), because
α < 1, hence
1
Λ
1 A(an )
m0 − 1 . F (− µ2 ) + F ( µ2 ) + eµγx F (γx) .γ
+
=
+
1
.
(3.3)
n A( µ1 )
A( µ1 ) A(x)
It remains to show that A(an ) . A( µ1 ). Since A ∈ RV (α), it is enough to show that an . µ1 .
First we note that, by Potter’s bounds (2.2), for x ≥ Can
α/2
3α/2
A(an )
an
an
−1
.Λ =
,
(3.4)
.
x
A(x)
x
and we conclude that, always for x ≥ Can ,
an
1
µ
=
1 log Λ
log(x/an )3α/2
.γ
. 1.
γ x/an
x/an
Next we assume that α ∈ (1, 2). Subtracting µ E[X1 ] = 0 from (3.2), we can write
Z
Z
Z
µy
m0 − 1 ≤ −µ
y F (dy) +
(e − 1 − µy) F (dy) +
eµy F (dy)
2
(−∞,− µ
]
Z
.µ
2 2
(− µ
,µ]
|y| F (dy) + µ2
2
(−∞,− µ
]
Z
2
(µ
,γx]
y 2 F (dy) +
2 2
(− µ
,µ]
eµγx
(γx)2
Z
(3.5)
y 2 F (dy) ,
(0,γx]
where the last inequality holds by eµy = y 2 eµy−2 log y , because µy − 2 log y is increasing for
y ≥ µ2 . Since α > 1, it follows by (2.4) that for w ≥ 0
Z ∞Z ∞
Z ∞
Z ∞
y F (dy) =
1{z<y} dz F (dy) = wF (w) +
F (z) dz . w F (w) ,
w
w
0
w
while using also α < 2 and (2.3), always for w ≥ 0,
Z w
Z w Z ∞
Z
y 2 F (dy) = 2
z 1{z<y} dz F (dy) .
0
0
0
w
z F (z) dz . w2 F (w) .
0
Similar bounds hold for the integrals over y ≤ 0. Then from (3.5) we get
m0 − 1 . F (− µ2 ) + F ( µ2 ) + eµγx F (γx) .γ
1
1
Λ
1 + A(x) . n ,
A( µ )
where the last inequality holds as we showed in (3.3). This completes the proof of (3.1).
Step 2. Next we deduce from (3.1) a rougher version of (1.4), namely
1/γ
1
n
1
P(Sn ∈ x + J, Mn ≤ 2 γx) .
.
an A(x)
(3.6)
Define X̂i := Xn+1−i , for 1 ≤ i ≤ n, and let (Ŝk := X̂1 + . . . + X̂k = Sn − Sn−k )1≤k≤n be
the corresponding random walk, which has the same law as (Sk )1≤k≤n . Then
P(Sn ∈ x + J, Sbn/2c < x2 , Mn ≤ 12 γx) = P(Ŝn ∈ x + J, Ŝbn/2c < x2 , M̂n ≤ 12 γx)
= P(Sn ∈ x + J, Sn − Sdn/2e < x2 , Mn ≤ 12 γx)
≤ P(Sn ∈ x + J, Sdn/2e > x2 , Mn ≤ 12 γx) ,
THE STRONG RENEWAL THEOREM
11
where the second equality holds because Ŝn = Sn and M̂n = Mn , while for the inequality
note that Sn ≥ x (by J ⊆ [0, ∞)). To lighten notation, henceforth we assume that n is
even (the odd case is analogous). It follows from the previous inequality that
P(Sn ∈ x + J, Mn ≤ 21 γx) ≤ 2 P(Sn ∈ x + J, Sn/2 ≥ x2 , Mn ≤ 21 γx)
Z
P(Sn/2 ∈ dz, Mn/2 ≤ 12 γx) P(Sn/2 ∈ x − z + J)
≤2
z≥ x2
.
1
an/2
P(Sn/2 ≥
1
2 x,
Mn/2 ≤
1
2 γx)
1
.
an
n
A(x)
1/γ
,
where we have used (2.7) and (3.1).
Step 3. Next we show that
P(Sn ∈ x + J) .
1 n
.
an A(x)
(3.7)
This is easy: if we fix = 21 , by (2.7) we can write
P(Sn ∈ x + J, Mn > x) ≤ n P(Sn ∈ x + J, X1 > x)
Z
F (dy) P(Sn−1 ∈ x − y + J) .
=n
y> x
1 n
n
F ( x) .
.
an−1
an A(x)
(3.8)
Applying (3.6) with γ = 1, we see that (3.7) holds.
Step 4. Finally we prove (1.5). The case k = 1, that is γ ∈ [1, ∞), follows by (3.7).
1
, k1 ), assuming that it
Inductively, we fix k ∈ N and we prove that (1.5) holds for γ ∈ [ k+1
1
1
1
holds for γ ∈ [ k , k−1 ). Let us fix := 2(k+1) . By (3.6) (where we choose γ = 2) we get
1
P(Sn ∈ x + J, Mn ≤ x) .
an
n
A(x)
1/(2)
1
=
an
n
A(x)
k+1
.
It remains to consider
P(Sn ∈ x + J, x < Mn ≤ γx) ≤ n P(Sn ∈ x + J, X1 > x , Mn ≤ γx)
Z
≤n
F (dy) P(Sn−1 ∈ x − y + J, Mn−1 ≤ γx)
y∈( x,γx]
≤ n F (x)
sup
P(Sn−1 ∈ z + J, Mn−1 ≤ γx) .
z≥(1−γ)x
Observe that, for z ≥ (1 − γ)x, we can bound
P(Sn−1 ∈ z + J, Mn−1 ≤ γx) ≤ P(Sn−1 ∈ z + J, Mn−1 ≤ γ 0 z) ,
with
γ 0 :=
γ
.
1−γ
1
1
The key observation is that γ 0 ∈ [ k1 , k−1
), since γ ∈ [ k+1
, k1 ). By our inductive assumption,
n k
relation (1.5) holds for γ 0 , so P(Sn−1 ∈ z + J, Mn−1 ≤ γ 0 z) . a1n ( A(z)
) and we get
1
P(Sn ∈ x + J, x < Mn ≤ γx) .γ n F (x)
an
which completes the proof.
n
A(x)
k
1
.
an
n
A(x)
k+1
,
12
FRANCESCO CARAVENNA AND RON DONEY
4. Strategy and key bounds for Theorems 1.4 and 1.12
4.1. Reformulation of the SRT. It turns out that proving the SRT amounts to showing
that small values of n give a negligible contribution to the renewal measure. More precisely,
if F is a probability on R satisfying (1.2), it is known that (SRT) holds if and only if
X
P(Sn ∈ x + I)
is a.n. ,
(4.1)
T (δ; x) :=
1≤n≤A(δx)
see [Chi15, Appendix] or Remark 4.1 below.
Applying Theorem 1.1, it is easy to show that (4.1) always holds for α > 12 . Since n/an
is regularly varying with index 1 − 1/α > −1, by (1.6) and (2.3)
X
X
n
1 A(δx)2
A(x)
1
.
∼ δ 2α−1
,
P(Sn ∈ x + I) .
x→∞
A(x)
an
A(x) δx
x
1≤n≤A(δx)
1≤n≤A(δx)
from which (4.1) follows, since 2α − 1 > 0. We have just proved Theorems 1.4 and 1.12 for
α > 12 . In the next sections, we will focus on the case α ≤ 12 .
Remark 4.1. It is easy to see how (4.1) arises. For fixed δ > 0, by (1.8) we can write
X
P(Sn ∈ x + I) .
(4.2)
U (x + I) ≥
A(δx)<n≤A( 1δ x)
Since P(Sn ∈ x + I) ∼ ahn φ( axn ) by (2.6) (where we take h = v for simplicity), a Riemann
sum approximation yields (see [Chi15, Lemma 3.4])
Z 1
X
δ
A(x)
z α−2 φ( z1 ) dz .
P(Sn ∈ x + I) ∼ h
C(δ) ,
with
C(δ) = α
x
δ
1
A(δx)<n≤A( δ x)
Since limδ→0 C(δ) = C, proving (SRT) amounts to controlling the ranges excluded from
(4.2), i.e. {n ≤ A(δx)} and {n > A( 1δ x)}. The latter gives a negligible contribution by
P(Sn ∈ x + I) ≤ C/an (recall (2.7)), while the former is controlled precisely by (4.1).
4.2. Some key bounds. First we note that, for any bounded J ⊆ R,
I1 (δ; x) is a.n.
=⇒
F (x + J) = o b1 (x) ,
(4.3)
x→∞
as it follows by (1.18), because I1 (δ; x) ≥ F (x + J)(inf z∈J b2 (z)) &J F (x + J) for large x.
Relation (4.3) implies that
I1 (δ; x) is a.n.
=⇒
for every fixed ` ∈ N : P(S` ∈ x + J) = o b1 (x) , (4.4)
x→∞
just because (assuming for simplicity that J ⊆ (−∞, 0])
P(S` ∈ x + J) ≤ ` P(S` ∈ x + J, X1 = max{X1 , . . . , X` }) ≤ `
sup
F (z + J) .
(4.5)
z∈[ 1` x,∞)
Next we give two useful lemmas. We recall that κα was defined in (1.21).
Lemma 4.2 (Big jumps). Let F satisfy (1.2) for some A ∈ RV (α), with α ∈ (0, 1). There
is η = ηα > 0 such that for all δ ∈ (0, 1], γ ∈ (0, 1) and x ∈ [0, ∞) the following holds:
X
∀` ≥ κα :
n` sup P (Sn ∈ z + I, Mn > γx) .γ,` δ η b`+1 (x) .
(4.6)
1≤n≤A(δx)
z∈R
THE STRONG RENEWAL THEOREM
13
Proof. For x < 1 one has δx < 1, hence the left hand side of (4.6) vanishes (since A(1) = 1).
Henceforth we assume that x ≥ 1. Recalling (2.7), we can write
P Sn ∈ z + I, Mn > γx ≤ n P (Sn ∈ z + I, X1 > γx)
Z
P (X ∈ dw) P(Sn−1 ∈ z − w + I)
=n
w>γx
(
)
(4.7)
≤ n P (X > γx) sup P(Sn−1 ∈ y + I)
y∈R
.
n
n 1
1
.γ
,
A(γx) an
A(x) an
therefore
n sup P (Sn ∈ z + I, Mn > γx) .γ
X
`
z∈R
1≤n≤A(δx)
1
A(x)
X
1≤n≤A(δx)
n`+1
an
(4.8)
1 A(δx)`+2
.`
,
A(x)
δx
because n`+1 /an is regularly varying with index (` + 1) − α1 ≥ (κα + 1) − α1 =
by1 (2.3),
1
α − α > −1. Let us introduce a parameter b = bα ∈ (0, 1), depending only on α, that will
be fixed in a moment. Since (4.6) holds trivially for δx < 1 (the left hand side vanishes, by
A(0) < 1), we may assume that δx ≥ 1. We can then apply the upper bound in (2.2) with
= (1 − b)α and ρ = δ, that is A(δx) . δ bα A(x), which shows that (4.8) is
A(x)`+1
A(x)`+1
. δ bα(κα +2)−1
,
x
x
because δ ≤ 1 and ` ≥ κα by assumption. Since α(κα + 2) > 1 (because κα + 2 =
b α1 c + 1 > α1 ), we can choose b = bα < 1 so that the exponent of δ is strictly positive (e.g.
bα = {α(κα + 2)}−1/2 ). This completes the proof.
. δ bα(`+2)−1
Our second lemma analyzes the case of “no big jump”. The proof exploits in an essential
way the large deviation estimate provided by Theorem 1.1.
α
Lemma 4.3 (No big jump). Let F satisfy (1.2) with α ∈ (0, 1). For any γ ∈ (0, 1−α
) there
is θ = θα,γ > 0 such that for all δ ∈ (0, 1] and x ∈ [0, ∞) the following holds:
X
∀` ≥ 0 :
n` P (Sn ∈ x + I, Mn ≤ γx) .γ,` δθ b`+1 (x) .
(4.9)
1≤n≤A(δx)
Proof. We may assume that x ≥ 1, since for x < 1 there is nothing to prove (the left hand
side of (4.9) vanishes). By (1.4)
`+ 1
`+ γ1 +1
n γ
1 A(δx)
n P (Sn ∈ x + I, Mn ≤ γx) .
.
1
1
δx
A(x) γ 1≤n≤A(δx) an
A(x) γ
1≤n≤A(δx)
X
1
`
X
`+ 1
,
where we applied (2.3), because the sequence n γ /an is regularly varying with index
` + γ1 − γ1 ≥ 0 > −1. Since the left hand side of (4.6) vanishes for δx < 1, by A(0) < 1, we
may assume that δx ≥ 1. By the upper bound in (2.2), since ` ≥ 0 and δ ≤ 1 we can write
`+ γ1 +1
A(δx)
. δ
(α−)(`+ γ1 +1)
`+ γ1 +1
A(x)
.δ
(α−)( γ1 +1)
`+ γ1 +1
A(x)
.
14
FRANCESCO CARAVENNA AND RON DONEY
Since γ <
α
1−α ,
we can choose = α,γ > 0 small enough so that (α − )( γ1 + 1) > 1.
5. Proof of Theorems 1.4 and 1.12: necessity
5.1. Necessity for Theorem 1.4. Let F be a probability on [0, ∞) satisfying (1.12) with
α ∈ (0, 1). We assume (4.1), which is equivalent to the strong renewal theorem (SRT), and
we deduce that I1+ (δ; x) is a.n. (recall (1.13)).
If F is lattice (cf. section 2.3), we actually assume (4.1) with I = (−h, 0] replaced by
J = (−v, 0], where v is the lattice span of F . This is no problem, since
bv/hc
P(Sn ∈ x + J) ≤
X
P(Sn ∈ x` + I) ,
where
x` := x − `h .
(5.1)
`=0
We also observe that, by (4.1), it follows that
F (x + J) = P(S1 ∈ x + J) = o(b1 (x)) ,
x→∞
(5.2)
and, by subadditivity, this relation actually holds for any fixed J ⊆ R.
We need some preparation. Let us define the compact interval
K := [ 12 , 1] .
By (2.6), since inf z∈K φ(z) > 0, there are n1 ∈ N and c1 , c2 ∈ (0, ∞) such that
c1
∀n ≥ n1 :
inf
P(Sn ∈ z + J) ≥
,
an
z∈R: z/an ∈K
c2
.
∀n ∈ N :
sup P(Sn ∈ z + J) ≤
a
n
z∈R
(5.3)
(5.4)
(5.5)
Then, since F ((−∞, −x] ∪ [x, ∞)) . 1/A(x), we can fix C ∈ (0, ∞) such that
c1 1
∀n ∈ N :
F ((−∞, −Can ] ∪ [Can , ∞)) ≤
.
(5.6)
2c2 n
(Of course, we could just take F ([Can , ∞)), since F ((−∞, 0)) = 0, but this estimate will
be useful later for random walks.) We also claim that
c1 1
∀n ≥ n1 :
inf
P (Sn ∈ z + J, max{|X1 |, . . . , |Xn |} < Can ) ≥
. (5.7)
2 an
z∈R: z/an ∈K
This follows because P(Sn ∈ z + J) ≥ c1 /an , by (5.4), and applying (5.5), (5.6) we get
P(Sn ∈ z + J , ∃1 ≤ j ≤ n with |Xj | ≥ Can )
Z
c1
n F ((−∞, −Can ] ∪ [Can , ∞)) c2
≤
.
≤n
F (dy) P(Sn−1 ∈ z − y + J) ≤
an
2 an
|y|≥Can
We can now start the proof. The events Bi := {Xi ≥ Can , maxj∈{1,...,n+1}\{i} Xj < Can }
are disjoint for i = 1, . . . , n, hence for n ≥ n1 we can write
P(Sn+1 ∈ x + J)
≥ (n + 1) P (Sn+1 ∈ x + J, max{X1 , . . . , Xn } < Can , Xn+1 ≥ Can )
Z
≥n
P (Xn+1 ∈ x − dz) P (Sn ∈ z + J, max{X1 , . . . , Xn } < Can )
{z≤x−Can }
Z
c1 n
≥
F (x − dz)
1
,
2 an {z/an ∈K}
{z≤x−Can }
(5.8)
THE STRONG RENEWAL THEOREM
15
where the last inequality holds by (5.7). We are going to choose n ≤ A(δx), in particular
x − Can ≥ x − Cδx ≥ 2δ x for δ > 0 small enough. Restricting the integral, we get
!
Z
A(δx)−1
A(δx)
X n
X
F (x − dz)
1{z/an ∈K} .
P(Sn+1 ∈ x + J) &
δ
a
n
{a
≤z≤
x}
n
n=n
n=n
1
2
1
1
Note that z/an ∈ K means 12 an ≤ z ≤ an , that
integration, we have A(z) ≥ n1 and A(2z) ≤ A(δ),
is A(z) ≤ n ≤ A(2z). In the range of
hence
A(2z)
X n
X n
A(z)2
A(z)
1{z/an ∈K} ≥
&
A(2z) − A(z) &
= b2 (z) ,
an
an
2z
z∨1
n=n
A(δx)
n=A(z)
1
where the last inequality holds since z ≥ an1 ≥ 1 (if we take n1 large enough). Then
Z
X
P(Sn+1 ∈ x + J) &
F (x − dz) b2 (z)
z∈[an1 , 2δ x]
n1 ≤n≤A(δx)
≥ I1+ ( 2δ ; x) − Ĉ F ([x − an1 , x]) ,
where Ĉ := sup|z|≤an1 b2 (z) < ∞. The left hand side is a.n. by (4.1), hence the right hand
side is a.n. too. Since F ([x − an1 , x]) is a.n. by (5.2), it follows that I1+ (δ; x) is a.n..
5.2. Necessity for Theorem 1.12. Let F be a probability on R satisfying (1.2) with
α ∈ (0, 1) and p, q > 0. We assume (4.1) and we deduce that I˜1 (δ; x) is a.n. and, moreover,
I˜k (δ, η; x) is a.n., for any k ≥ 2 and for all fixed η ∈ (0, 1). This completes the proof of the
necessity part in Theorem 1.12, see the discussion in Remark 1.14.
Remark 5.1. For |x| ≥ 1 we can rewrite (1.22) as
Z |z|
Z A(|z|)
Z A(|z|) k−1
bk (t)
bk (A−1 (s))
1
s
b̃k (z, x) =
dt =
ds '
ds ,
−1
0
−1
−1
t
A (s) A (A (s))
|x|
A(|x|)
A(|x|) A (s)
where we have used (2.1) and (1.10). Recalling also (1.3), we can write
A(|z|)
b̃k (z, x) '
X
n=A(|x|)
nk−1
.
an
(5.9)
Since we assume that p, q > 0 in (1.2), the density φ(·) of the limiting Lévy process is
strictly positive on the whole real line. In particular, instead of (5.3), we can define
K := [−1, 1] ,
(5.10)
and relations (5.4), (5.5), (5.6), (5.7) still hold.
Let us show that I˜1 (δ; x) is a.n.. This is similar to the case of renewal processes in
subsection 5.1. In fact, relation (5.8) with Xi replaced by |Xi | and z replaced by −y gives
Z
n
P(Sn+1 ∈ x + J) &
F (x + dy)
1
,
(5.11)
an {|y|≤an }
|x+y|≥Can
because K = [−1, 1]. Note that for n ≤ A(δx) we have an ≤ δx ≤ x − Can for δ > 0 small,
hence we can ignore the restriction |x + y| ≥ Can . Then we can write
A(δx)
X n
1{|y|≤an } = 1{|y|≤δx}
a
n
n=n
1
A(δx)
X
n=A(y)∨n1
n
& 1{|y|≤δx} b̃2 (δx, |y| ∨ an1 ) .
an
16
FRANCESCO CARAVENNA AND RON DONEY
For |y| < an1 , b̃2 (δx, |y| ∨ an1 ) = b̃2 (δx, an1 ) differs from b̃2 (δx, |y|) by at most the constant
P
n
n≤n1 an . It follows that b̃2 (δx, |y| ∨ an1 ) & b̃2 (δx, |y|) for large x, hence, by (1.23),
A(δx)
X
P(Sn+1 ∈ x + J) & I˜1 (δ; x) .
n=n1
Since we assume that (4.1) holds, it follows that I˜1 (δ; x) is a.n..
Next we fix k ≥ 2 and η ∈ (0, 1) and we generalize the previous arguments in order to
show that I˜k (δ, η; x) is a.n., cf. (1.24). (Inductively, we assume that we already know that
I˜1 (δ; x), I˜2 (δ, η; x), . . . , I˜k−1 (δ, η; x) are a.n..) Suppose that z1 , · · · zk ∈ R satisfy
min |zj | ≥ Can ,
1≤j≤k
|(z1 + . . . + zk ) − x| ≤ an ,
and set yk := x − (z1 + . . . + zk ). Then, for n ≥ n1 , we can write
P (∃1 ≤ j1 < j2 < · · · < jk ≤ n with Xj1 ∈ dz1 , . . . , Xjk ∈ dzk , and Sn+k ∈ x + I)
n+k
≥
P(Xr ∈ dzr , ∀1 ≤ r ≤ k , Xj ∈
/ {dz1 , · · · dzk }, ∀k < j ≤ n + k , Sn+k ∈ x + I)
k
& nk P(Xr ∈ dzr , ∀1 ≤ r ≤ k) P(|Xj | ≤ Can , ∀1 ≤ j ≤ n , Sn ∈ yk + I)
&
nk
P(Xr ∈ dzr , ∀1 ≤ r ≤ k) ,
an
It follows that for n ≥ n1 we have the the bound
nk
P min |Xr | ≥ Can , |(X1 + . . . + Xk ) − x| ≤ an
P(Sn+k ∈ x + I) &
1≤r≤k
an
k
n
P−x min |Sr − Sr−1 | ≥ Can , |Sk | ≤ an ,
=
1≤r≤k
an
where P−x denotes the law of the random walk Sr := −x + (X1 + . . . + Xr ), r ≥ 1, which
starts from S0 := −x.
If we fix η ∈ (0, 1), and define η := 1 − η, we can write
n
o
min |Sr − Sr−1 | ≥ Can ⊇ |Sr − Sr−1 | ≥ η|Sr−1 | and |Sr−1 | ≥ Cη an , ∀1 ≤ r ≤ k .
1≤r≤k
For r = 1, |Sr−1 | ≥ Cη an reduces to x ≥ Cη an , which holds automatically, since we take
n ≤ A(δx) with δ > 0 small, while |Sr − Sr−1 | ≥ η|Sr−1 | becomes |S1 + x| ≥ ηx, which
is implied by |S1 | ≤ Cη δx, for δ > 0 small. For r ≥ 2, |Sr − Sr−1 | ≥ η|Sr−1 | is implied by
|Sr | ≤ η|Sr−1 |, since η = 1 − η. Thus
n
o
min |Sr − Sr−1 | ≥ Can ⊇ |S1 | ≤ Cη δx , |Sr | ≤ η|Sr−1 | , ∀2 ≤ r ≤ k , |Sk−1 | ≥ Cη an ,
1≤r≤k
where the last term is justified because |Sk−1 | = min2≤r≤k−1 |Sr−1 | on the event. Thus
"
k
#
n
P(Sn+k ∈ x + I) & E−x 1{|S1 |≤ C δx, |Sr |≤η|Sr−1 | , ∀2≤r≤k}
1
.
η
η
an {A(|Sk |)≤n≤A( C |Sk−1 |)}
THE STRONG RENEWAL THEOREM
17
Let us now sum over n1 ≤ n ≤ A(δx). Note that A( Cη |Sk−1 |) ≤ A( Cη |S1 |) ≤ A(δx), hence
X
P(Sn+k ∈ x + I)
n1 ≤n≤A(δx)
(5.12)
h
& E−x 1{|S1 |≤ C δx,
η
η
|Sr |≤η|Sr−1 | , ∀2≤r≤k} b̃k+1 C |Sk−1 |,
|Sk | ∨ an1
i
,
where we recall that b̃k+1 is given by (5.9). The right hand side can be rewritten as
Z
F (x + dy1 ) Py1 (dy2 , . . . , dyk ) b̃k+1 |yk−1 |, |yk | ∨ c ,
|y1 |≤δ 0 x
|yr |≤η|yr−1 | for all 2≤r≤k
(5.13)
where
δ0 =
C
ηδ,
=
η
C
,
c = an1 .
˜ 0
This looks very
similar to Ik (δ ,η; x), cf. (1.24), the only difference being that
the term
bk+1 (yk−1 , yk = bk+1 (|yk−1 |, |yk | in (1.24) is replaced by bk+1 (|yk−1 |, |yk | ∨ c . It remains
to show that this difference is immaterial. More precisely, since we assume (4.1), it follows
by (5.12) that (5.13) is a.n.. From this, we now deduce that I˜k (δ, η; x) is a.n..
For any B ∈ (0, ∞), the contribution of |yk−1 | ≤ B to I˜k (δ, η; x) in (1.24) is at most
b̃k+1 (B, 0) P(| − x + Sk−1 | ≤ B) .B P(Sk−1 ∈ [x − B, x + B])
and this is a.n. as x → ∞, in analogy with (5.2), because we assume that (4.1) holds.
We are left with the contribution
of {|yk−1 | > B} to I˜k (δ, η; x) in (1.24). We claim that
we can bound b̃k+1 |yk−1 |, |yk | . b̃k+1 |yk−1 |, |yk | ∨ c , in this regime. In fact, the two
terms differ only when |yk | < c, in which case their ratio is bounded, for |yk−1 | > B, by
b̃k+1 |yk−1 |, |yk |
b̃k+1 |yk−1 |, c + b̃k+1 c, |yk |
b̃k+1 c, 0
≤1+
≤
.
b̃k+1 |yk−1 |, |yk | ∨ c
b̃k+1 |yk−1 |, c
b̃k+1 B, c
If we fix B > c so that b̃k+1 B, c > 0, we have b̃k+1 |yk−1 |, |yk | . b̃k+1 |yk−1 |, |yk | ∨ c .
Finally, we write bk+1 |yk−1 |, |yk | ∨ c ≤ bk+1 |yk−1 |, |yk | ∨ c + bk+1 |yk−1 |, |yk−1 |
and we note that the contribution of the first term to I˜k (δ, η; x) in (1.24) is a.n., because
we know that (5.13) is a.n.. For the second term, just note that
A(|yk−1 |)
bk+1 |yk−1 |, |yk−1 | =
X
n=A(|yk−1 |)
nk
A(|yk−1 |)k+1
≤
. bk+1 (yk−1 ) ,
an
|yk−1 |
hence
Z
F (−yk−1 + dyk ) b̃k+1 (|yk−1 |, |yk−1 |) . bk+1 (yk−1 ) F (−(1 − η)|yk−1 |)
|yk |≤η|yk−1 |
.η bk (yk−1 ) ,
and the corresponding contribution to I˜k (δ, η; x) is .,η Ik−1 (δ, η; x). Since I˜k−1 is a.n. by
our inductive assumption, also Ik−1 is a.n., cf. (A.7), hence we are done.
18
FRANCESCO CARAVENNA AND RON DONEY
6. Proof of Theorem 1.4: sufficiency
In this section we prove the sufficiency part of Theorem 1.4: we assume that I1+ (δ; x) is
a.n. and we deduce (4.1), which is equivalent to the SRT. Let us set
X
n` P(Sn ∈ x + I) .
(6.1)
T` (δ; x) :=
1≤n≤A(δx)
Theorem 6.1. Let F be a probability on [0, ∞) satisfying (1.12) with α ∈ (0, 1). Assume
that I1+ (δ; x) is a.n.. Then for every ` ∈ N0 :
lim lim sup
δ→0
x→∞
T` (δ; x)
= 0.
b`+1 (x)
(6.2)
In particular, setting ` = 0, relation (4.1) holds.
The proof exploits the general bounds provided by Lemmas 4.2 and 4.3, together with
the next Lemma, which is specialized to renewal processes.
Lemma 6.2. If F is a probability on [0, ∞) which satisfies (1.12) with α ∈ (0, 1), there
are C, c ∈ (0, ∞) such that for all n ∈ N0 and z ∈ [0, ∞)
P(Sn ∈ z + I) ≤
n
C −c A(z)
e
.
an
(6.3)
Remark 6.3. We observe that T` is uniformly bounded, by Lemma 6.2: for some C < ∞
T` (δ; x) ≤ C .
(6.4)
Proof of Lemma 6.2. Assume that n is even (the odd case is analogous). By (2.7), we get
Z
1
P(S n2 ≤ z)
P (Sn ∈ z + I) =
P(S n2 ∈ dy) P(S n2 ∈ z − y + I) .
n
a
y∈[0,z]
2
!
n
n
−c n
2
1
(1 − P(X > z))
e− 2 P(X>z)
e A(z)
.
P maxn Xi ≤ z =
≤
≤
,
an
an
an
an
1≤i≤ 2
provided c > 0 is chosen such that P(X > z) ≥ 2c/A(z) for all z ≥ 0. This is possible by
(1.12) and because z 7→ A(z) is increasing and continuous, with A(0) > 0 (see §2.2).
α
Proof of Theorem 6.1. We fix, once and for all, γ ∈ (0, 1−α
), and we decompose
X
T` (δ; x) =
n` P(Sn ∈ x + I, Mn > γx)
1≤n≤A(δx)
+
X
n` P(Sn ∈ x + I, Mn ≤ γx) .
1≤n≤A(δx)
Then it follows by Lemma 4.2 and Lemma 4.3 that (6.2) holds for every ` ≥ κα .
It remains to prove that (6.2) holds for ` < κα . We proceed by backward induction: we
fix ` ∈ {0, 1, . . . , κα − 1} and, assuming that
lim lim sup
δ→0
x→∞
T`+1 (δ; x)
= 0,
b`+2 (x)
we deduce (6.2). We need to estimate T` (δ; x) and we split it in some pieces.
We start by writing
P (Sn ∈ x + I) = P (Sn ∈ x + I, Mn ≤ γx) + P (Sn ∈ x + I, Mn > γx) ,
(6.5)
THE STRONG RENEWAL THEOREM
19
and note that the contribution of the first term in the right hand side is negligible for (6.2),
by Lemma 4.3. Next we bound
P (Sn ∈ x + I, Mn > γx) ≤ n P (Sn ∈ x + I, X1 > γx)
Z
F (x − dz) P (Sn−1 ∈ z + I) .
=n
(6.6)
0≤z<(1−γ)x
Looking back at (6.1), we may restrict the sum to n ≥ 2, because the contribution of the
term n = 1 is negligible for (6.2), since F (x + I) = o(b1 (x)) = o(b`+1 (x)) by (4.3). As a
consequence, it remains to prove that (6.2) holds with T` (δ; x) replaced by
Z
X
`+1
e
F (x − dz) P (Sn−1 ∈ z + I) .
n
T` (δ; x) :=
2≤n≤A(δx)
0≤z<(1−γ)x
We can bound n`+1 . (n − 1)`+1 , since n ≥ 2, and rename n − 1 as n, to get
(
)
Z
X
`+1
Te` (δ; x) ≤
F (x − dz)
n
P (Sn ∈ z + I) + o(b`+1 (x)) , (6.7)
1≤z<(1−γ)x
1≤n≤A(δx)−1
where we have restricted the integral to z ≥ 1 and estimated the contribution of z ∈ [0, 1)
as o(b1 (x)) = o(b`+1 (x)), thanks to (4.3) and (6.4).
Let us fix ∈ (0, 1) and consider the contribution to the sum in (6.7) given by n > A(z).
Applying Lemma 6.2, since an ≥ z & z, we get
X
X
n`+1 P (Sn ∈ z + I) .
A(z)<n≤A(δx)
A(z)<n≤A(δx)
A(z)`+2
1{z< δ x}
.
z
(
X
n∈N
n
n`+1 −c A(z)
e
an
1
A(z)
n
A(z)
`+1
n
−c A(z)
e
(6.8)
)
.
R∞
The bracket is a Riemann sum which converges to 0 t`+1 e−ct dt < ∞ as z → ∞, hence
it is uniformly bounded for z ∈ [0, ∞). The contribution to the integral in (6.7) is then
Z
A(z)`+2
.
≤ A( δ x)` I1+ ( δ ; x) . A(x)` I1+ ( δ ; x) .
F (x − dz)
δ
z
1≤z< x
This is negligible for (6.2), for any fixed > 0, by the assumption that I1+ is a.n..
Finally, the contribution to the integral in (6.7) given by n ≤ A(z) is, by (6.1),
Z
F (x − dz) T`+1 (; z) .
(6.9)
1≤z<(1−γ)x
By the inductive assumption (6.5), for every η > 0 we can choose > 0 small enough and
x̄ < ∞ large enough so that T`+1 (; x) ≤ ηb`+2 (x) for every x ≥ x̄ . Recalling (6.4), we
see that (6.9) is bounded by
C F ((x − x̄ , x − 1]) + η b`+2 (x) F ([γx, ∞)) .,γ o(b`+1 (x)) + η b`+1 (x) .
If we let x → ∞ and then η → 0, we have proved (6.2).
20
FRANCESCO CARAVENNA AND RON DONEY
7. Proof of Theorem 1.12: sufficiency in case α ∈ ( 13 , 12 ]
Let F be a probability on R that satisfies (1.2) with p, q ≥ 0 and α ∈ ( 13 , 12 ]. In this
section, we assume that I˜1 (δ, x) is a.n., cf. (1.23), and we deduce (4.1), which is equivalent
to the SRT. By Remark 1.14, this proves the sufficiency part in Theorem 1.12 when κα = 1.
Let us set
Z1 := Mn = max{X1 , . . . , Xn } .
(7.1)
α
) and define the events
We fix γ ∈ (0, 1−α
(1)
(2)
E1 := {Z1 ≤ γx} ,
E1 := {|Z1 − x| ≤ an } ,
(7.2)
(3)
E1 := {Z1 > γx, |Z1 − x| > an }
By Lemma 4.3 with ` = 0, with no extra assumptions on F , we already know that
X
(1)
P(Sn ∈ x + I, E1 ) is always a.n. .
(7.3)
1≤n≤A(δx)
(2)
Next we look at E1 . Note that by (2.7)
(2)
P(Sn ∈ x + I, E1 ) ≤
n
X
P(Sn ∈ x + I, |Xi − x| ≤ an )
i=1
Z
=n
F (x + dy) P(Sn−1
|y|≤an
n
∈ I − y) .
an
(7.4)
Z
F (x + dy) .
|y|≤an
Using the fact that n/an is regularly varying and recalling (5.9), we obtain
Z
X
X
n
(2)
P(Sn ∈ x + I, E1 ) .
F (x + dy)
an
|y|≤δx
1≤n≤A(δx)
A(|y|)≤n≤A(δx)
Z
.
F (x + dy) b̃2 (δx, y) = I˜1 (δ; x) .
(7.5)
|y|≤δx
Recalling (1.23), we have shown that
I˜1 (δ; x) is a.n.
(2)
X
=⇒
P(Sn ∈ x + I, E1 )
is a.n. .
(7.6)
1≤n≤A(δx)
(Actually, the reverse implication also holds, as shown in Section 5.)
(3)
We finally turn to E1 . Arguing as in (7.4) and setting γ := 1 − γ, we have by (1.6)
Z
(3)
P(Sn ∈ x + I, E1 ) . n
F (x + dy) P(Sn−1 ∈ I − y)
|y|>an , y>−γx
.
n2
an
Z
F (x + dy)
|y|>an , y>−γx
hence, recalling (1.10), we get
Z
X
(3)
P(Sn ∈ x + I, E1 ) .
F (x + dy)
y>−γx
1≤n≤A(δx)
Z
F (x + dy)
.
y>−γx
1
,
A(y)
1
A(y)
X
1≤n≤A(δx∧|y|)
1
b3 (δx ∧ |y|) ,
A(y)
n2
an
THE STRONG RENEWAL THEOREM
21
where the last inequality holds for α > 13 , thanks to (2.3), because n2 /an is regularly varying
with index 2 − 1/α > −1. The right hand side can be estimated by
Z
Z
1
F (x + dy) b2 (y) + b3 (δx)
F (x + dy)
A(y)
y>−γx, |y|>δx
|y|≤δx
. I1 (δ; x) +
b3 (δx)
F ((γx, ∞)) . I1 (δ; x) + b1 (δx) .
A(δx)
Therefore
X
(3)
P(Sn ∈ x + I, E1 )
1≤n≤A(δx)
is a.n.
if α >
1
and I1 (δ; x) is a.n. .
3
(7.7)
Relations (7.3), (7.6), (7.7) prove Theorem 6.1 in case κα = 1 (recall Lemma A.3).
8. Proof of Theorem 1.12: sufficiency in case α ≤
1
3
Let F be a probability on R that satisfies (1.2) with p, q ≥ 0 and α ∈ (0, 13 ]. In this
section, we assume that I˜κα (δ, η; x) is a.n. and we deduce (4.1), which is equivalent to the
SRT. By Remark 1.14, this proves the sufficiency part in Theorem 1.12, in case κα ≥ 2.
By Remark 1.14, I˜1 (δ; x) is a.n., hence also I1 (δ; x) is a.n. (cf. Lemma A.3). Throughout
α
this section we fix γ ∈ (0, 1−α
), we set η = γ := 1 − γ and we drop η from notations.
8.1. Preparation. Let us rewrite Ik (δ; x) as follows (recall (1.18), (1.19)):
Z
Ik (δ; x) :=
F (x + dy1 ) gk (y1 ) ,
(8.1)
|y1 |≤δx
where we set


b2 (y1 )
gk (y1 ) := Z

Py1 (dy2 , . . . , dyk ) bk+1 (yk )

if k = 1
if k ≥ 2
,
(8.2)
Ωk (y1 )
Ωk (y1 ) := |yj | ≤ γ|yj−1 | for 2 ≤ j ≤ k ,
(8.3)
and we recall that Py1 (dy2 , . . . , dyk ) := F (−y1 + dy2 )F (−y2 + dy3 ) · · · F (−yk−1 + dyk ).
If Ir (δ, x) is a.n., then for δ > 0 small we have the bound Ir (δ, x) . b1 (x) for all x ≥ 0.
The following crucial Proposition shows that the same bound holds also when the integral
in (8.1) is enlarged to {y1 > −κx}, for any fixed κ < 1.
Proposition 8.1. If Ir (δ, x) is a.n. for some r ≥ 1, then for any 0 < κ < 1
Z
F (x + dy) gr (y) .κ,γ b1 (x)
(∀x ≥ 0) .
(8.4)
y>−κx
Proof. The case r = 1 is easy: since b2 ∈ RV (2α − 1) and 2α − 1 < 0, for any fixed δ0 > 0
Z
b2 (x)
F (x + dy) b2 (y) ≤
sup b2 (y) F ((1 − κ)x) .δ0 ,κ
= b1 (x) . (8.5)
A(x)
|y|>δ0 x, y>−κx
|y|>δ0 x
On the other hand, the contribution to the integral of |y| ≤ δ0 x gives I1 (δ0 ; x) which is
. b1 (x) for δ0 > 0 small enough, as we already observed, because I1 (δ; x) is a.n..
Next we fix k ≥ 1, we assume that Ik+1 (δ; x) is a.n.. By induction, we can assume that
(8.4) holds for 1 ≤ r ≤ k and our goal is to prove it for r = k + 1.
22
FRANCESCO CARAVENNA AND RON DONEY
We start estimating gk+1 (−z) for z ≥ 0. In case k = 1, we can write
Z
Z
F (z + dy2 ) b3 (y2 ) ≤ A(z)
g2 (−z) =
|y2 |≤γz
F (z + dy2 ) b2 (y2 ) . b2 (z) ,
|y2 |≤γz
where in the last inequality we applied the inductive hypothesis (8.4) for r = 1 (using the
fact that z ≥ 0). Similarly, for k ≥ 2,
Z
Z
gk+1 (−z) =
Py2 (dy3 , · · · , dyk+1 ) bk+2 (yk+1 )
F (z + dy2 )
|y2 |≤γz
Ωk (y2 )
Z
Z
≤ A(z)
Py2 (dy3 , · · · , dyk+1 ) bk+1 (yk+1 )
F (z + dy2 )
|y2 |≤γz
Ωk (y2 )
Z
= A(z)
F (z + dy2 ) gk (y2 ) . b2 (z) ,
|y2 |≤γz
by (8.4) for r = k. As a consequence, the contribution of y ≤ 0 to (8.4) for r = k + 1 is
Z
Z
F (x − dz) gk+1 (−z) .
F (x − dz) b2 (z) . b1 (x) ,
0≤z<κx
0≤z<κx
where the last inequality holds again by (8.4) for r = 1.
It remains to look at the contribution of y > 0 in (8.4) for r = k + 1. By (8.1), the
contribution of {0 < y ≤ δ1 x} is bounded by Ik+1 (δ1 , x) which is a.n. by assumption, hence
it is . b1 (x) provided δ1 > 0 is small enough. It remains to focus on {y > δ1 x}.
We need a simple observation; let I = (a1 , a2 ) where 0 ≤ a1 < a2 ≤ ∞ and put
I 0 = (γa1 , (2 − γ)a2 ). Then for all non-negative f, g
Z
Z
F (x + dy) f (y)
F (−y + dz) g(z)
|z|≤γy
y∈xI
Z
Z
F (x + dy) f (y)
F (−dw) g(y − w)
γy≤w≤(2−γ)y
Z
Z
≤
F (−dw)
F (x + dy) f (y) g(y − w)
w∈xI 0
(2−γ)−1 w≤y≤γ −1 w
Z
Z
F (−dw)
F (x + w + dv) f (w + v) g(v),
=
=
y∈xI
w∈xI 0
(8.6)
−γ1 w≤v≤γ2 w
where γ1 = 1 − (2 − γ)−1 and γ2 = γ −1 − 1. Using this, we can write the contribution of
{y > δ1 x} to (8.4) for r = k + 1 as follows:
Z
∞
Z
∞
Z
γy
F (x + dy) gk+1 (y) ≤
δ1 x
F (x + dy)
δ1 x
Z ∞
Z
=
F (−dw)
γδ1 x
Z ∞
F (−y + dz) A(z) gk (z)
−γy
γ2 w
F (x + w + dv) A(v) gk (v)
−γ1 w
Z
γ2 (x+w)
F (−dw) A(x + w)
.γ
γδ1 x
F (x + w + dv) gk (v) .
−γ1 (x+w)
THE STRONG RENEWAL THEOREM
23
Applying (8.4) for r = k, and the fact that b2 (·) is asymptotically decreasing, we obtain
Z ∞
Z ∞
Z ∞
F (−dw) b2 (x + w)
F (−dw) A(x + w) b1 (x + w) =
F (x + dy) gk+1 (y) .
γδ1 x
γδ1 x
δ1 x
Z ∞
. b2 (x)
F (−dw) .δ1 ,γ b1 (x) .
γδ1 x
We need to generalize 8.2: for any non-negative, even function f : R → R we denote by
gk (y1 , f ) what we get by replacing bk+1 (yk ) by bk (yk )f (yk ) in 8.2, that is

if k = 1

b1 (y1 ) f (y1 )
Z
gk (y1 , f ) :=
.
(8.7)

Py1 (dy2 , . . . , dyk ) bk (yk ) f (yk ) if k ≥ 2

Ωk (y1 )
In particular, gk (y) is gk (y, A). The following result is in the same spirit as Proposition 8.1.
Proposition 8.2. Assume that I`−1 (δ, x) is a.n. for some ` ≥ 1 (if ` = 1, no assumption
is made). For any f ∈ RV (β) with 0 < β < 1 − α, and for all 0 < δ0 < κ < 1,
Z
f (x)
(∀x ≥ 0) .
(8.8)
F (x + dy) g` (y, f ) .δ0 ,κ,γ
x∨1
|y|>δ0 x, y>−κx
Proof. Since g1 (·) = b1 (·)f (·) ∈ RV (α + β − 1) is asymptotically decreasing,
Z
Z
F (x + dy) g1 (y, f ) .δ0 b1 (x)f (x)
F (x + dy)
|y|>δ0 x, y>−κx
y>−κx
.κ
b1 (x)f (x)
f (x)
=
,
A(x)
x∨1
which proves (8.8) if ` = 1. Henceforth we assume that ` ≥ 2 and proceed by induction.
For ` = 2 and z ≥ 0, since f (·) is asymptotically increasing, we can bound
Z
Z
g2 (−z, f ) =
F (z + dy2 ) b2 (y2 ) f (y2 ) . f (z)
F (z + dy2 ) b2 (y2 )
|y2 |≤γz
|y2 |≤γz
Z
. f (z)
F (z + dy2 ) b2 (y2 ) . f (z) b1 (z) ,
y2 ≥−γz
where for the last inequality we apply Proposition 8.1 for r = ` − 1 = 1 (since z ≥ 0).
Similarly, for ` ≥ 3 and z ≥ 0,
Z
Z
F (z + dy2 )
Py2 (dy3 , · · · , dy` ) b` (y` ) f (y` )
g` (−z, f ) =
|y2 |≤γz
Ω`−1 (y2 )
Z
Z
. f (z)
F (z + dy2 )
Py2 (dy3 , · · · , dy` ) b` (y` )
|y2 |≤γz
Ω`−1 (y2 )
Z
= f (z)
F (z + dy2 ) g`−1 (y2 ) .γ f (z) b1 (z) ,
|y2 |≤γz
again by Proposition 8.1 for r = ` − 1, because z ≥ 0. Then the contribution of y ≤ 0 to
(8.8) is easily estimated (recall that f (·)b1 (·) is asymptotically increasing):
Z
Z
f (x) b1 (x)
f (x)
=
.
F (x − dz) g` (−z, f ) .δ0 f (x) b1 (x)
F (x − dz) .κ
A(x)
x∨1
δ0 x≤z≤κx
δ0 x≤z≤κx
24
FRANCESCO CARAVENNA AND RON DONEY
It remains to control the contribution to (8.8) of y > 0. Let us consider the case ` ≥ 3:
by (8.7), since f (·) is asymptotically increasing,
Z
Py (dy2 , . . . , dy` ) b` (y` ) f (y` )
g` (y, f ) =
Ω` (y)
Z
Z
=
F (−y + dz)
Pz (dy3 , . . . , dy` ) b` (y` ) f (y` )
|z|≤γy
Ω`−1 (z)
Z
Z
.
F (−y + dz) f (z)
Pz (dy3 , . . . , dy` ) b` (y` )
|z|≤γy
Ω`−1 (z)
Z
=
F (−y + dz) f (z) g`−1 (z) .
|z|≤γy
and the same bound holds also for ` = 2. Then, applying (8.6),
Z γy
Z ∞
Z ∞
F (x + dy)
F (x + dy) g` (y, f ) ≤
F (−y + dz) f (z) g`−1 (z)
δ0 x
δ0 x
−γy
Z ∞
Z γ2 w
.
F (−dw)
F (x + w + dz) f (z) g`−1 (z)
γδ x
−γ1 w
Z 0∞
Z ∞
.γ
F (−dw) f (w)
F (x + w + dz) g`−1 (z) .
(8.9)
−γ1 (x+w)
γδ0 x
Applying again Proposition 8.1 for r = ` − 1 we get
Z ∞
Z ∞
F (x + dy) g` (y, f ) .
F (−dw) f (x + w) b1 (x + w)
δ0 x
γδ0 x
Z
∞
. f (x) b1 (x)
F (−dw) .γ,δ0
γδ0 x
f (x)
.
x∨1
(8.10)
We need to look at a generalization of the quantities that we have studied so far. Let us
enlarge the set Ωk (y1 ) in (8.3), defining for k ≥ 2 a new set Θk (y1 ) ⊆ Rk−1 by
Θk (y1 ) := |yj − yj−1 | ≥ γ|yj−1 | for 2 ≤ j ≤ k ,
:= yj ∈ θ(yj−1 ) for 2 ≤ j ≤ k ,
(
(8.11)
(−∞, γy] if y ≥ 0
where
θ(y) :=
.
[γy, +∞) if y < 0
Then we define hk (y1 ) in analogy with gk (y1 ), cf. (8.2), just replacing Ωk (y1 ) by Θk (y1 ):

if k = 1

b2 (y1 )
Z
hk (y1 ) :=
.
(8.12)

Py1 (dy2 , . . . , dyk ) bk+1 (yk ) if k ≥ 2

Θk (y1 )
Similarly to (8.7), for any non-negative, even function f : R → R we also define

if k = 1

b1 (y1 ) f (y1 )
Z
hk (y1 , f ) :=
,

Py1 (dy2 , . . . , dyk ) bk (yk ) f (yk ) if k ≥ 2

Ωk (y1 )
(8.13)
THE STRONG RENEWAL THEOREM
and note that hk (y1 , A) = hk (y1 ). Finally, recalling (8.1), we set
Z
F (x + dy) hk (y) .
Jk (δ; x) :=
25
(8.14)
|y|≤δx
Note that h1 = g1 , hence J1 (δ; x) = I1 (δ; x). For k ≥ 2, since Θk ⊇ Ωk , we have hk ≥ gk ,
hence Jk (δ; x) ≥ Ik (δ; x).
Out next result shows that we can often work with Jk rather than Ik .
Proposition 8.3. Fix r ∈ N and assume α <
1
r+1 .
If Ir (δ, x) is a.n., then
Jr (δ, x) is a.n..
(8.15)
Also if f ∈ RV (β) with 0 < β < 1 − rα then for any 0 < δ0 < κ < 1
Z
f (x)
F (x + dy) hr (y, f ) .δ0 ,κ,γ
(∀x ≥ 0) .
x
∨1
|y|>δ0 x, y>−κx
(8.16)
Proof. We claim that for all f satisfying the assumptions the following bound holds:
hr (y, f ) .γ
r
X
gj (y, f )
(∀y ∈ R) .
(8.17)
j=1
Applying this bound, relation (8.16) follows immediately by Proposition 8.2 (compare (8.7)
and (8.13)). If we choose f = A in (8.17) (this isPpossible because β = α satisfies β < 1−rα,
1
) we obtain hr (y) .γ rj=1 gj (y), which plugged into (8.14) shows
since we assume α < r+1
Pr
that Jr (δ; x) .γ j=1 Ir (δ; x), from which (8.15) follows.
Note that (8.17) holds trivially for r = 1, since h1 (y, f ) = g1 (y, f ). Henceforth we fix
r ≥ 2 and, inductively, we assume that (8.15), (8.16) and (8.17) have been proved when r
is replaced by r − 1, r − 2, etc.. Our goal is to prove (8.17).
Let us first consider the case y < 0. By (8.13) we can write
Z
r−1 Z
X
F (−y + dy2 ) hr−1 (y2 , Af ) .γ
F (−y + dy2 ) gj (y2 , Af ) ,
hr (y, f ) =
y2 ≥−γ|y|
j=1
y2 ≥−γ|y|
by the induction hypothesis, applying (8.17) with r −1 instead of r (this is possible because
Af ∈ RV (β 0 ) with β 0 = α + β which satisfies 0 < β 0 < 1 − (r − 1)α). We now split the
domain of integration in the two subsets [−γ|y|, γ|y|] ∪ (γ|y|, ∞). The first subset gives
Z
F (−y + dy2 ) gj (y2 , Af ) = gj+1 (y, f ) ,
|y2 |≤γ|y|
by (8.7). For the second subset we can apply Proposition 8.2 (since −y ≥ 0), getting
Z
A(y)f (y)
F (−y + dy2 ) gj (y2 , Af ) .γ
= b1 (y) f (y) = g1 (y, f ) ,
y∨1
y2 >γ|y|
hence (8.17) is proved.
Next we consider the case y ≥ 0. If we restrict the domain of integration Θr (y) in (8.13)
to y2 ≥ 0, y3 ≥ 0, . . . , yr ≥ 0, then the domain becomes {0 ≤ yj ≤ γyj−1 for 2 ≤ j ≤ r}
which is included in Ωr (y), cf. (8.3). This contribution to hr (y, f ) is then bounded from
above by gr (y, f ), cf. (8.7), in accordance with (8.17).
It remains to estimate hr (y, f ) for y ≥ 0, when some of the coordinates y2 , y3 , . . . , yr in
the integral in (8.13) are negative. Let us define H := min{j ∈ {2, . . . , r} : yj ≤ 0}.
26
FRANCESCO CARAVENNA AND RON DONEY
In the extreme case H = r, the corresponding contribution to hr (y, f ) is
Z γy
Z γyr−2 Z 0
Py (dy2 , · · · , dyr ) br (yr )f (yr )
···
y2 =0
γy
yr−1 =0 yr =−γyr−1
γyr−2 Z −γyr−1
Z
···
+
(8.18)
Z
y2 =0
Py (dy2 , · · · , dyr ) br (yr )f (yr ) .
yr−1 =0
yr =−∞
If r = 2, one should ignore the first integrals, that is we have
Z 0
Z γy
F (−y + dy2 ) b2 (y2 ) f (y) .
F (−y + dy2 ) b2 (y2 ) f (y) +
y2 =−γy
(8.19)
y2 =−∞
The first integral in (8.18)-(8.19) is bounded by gr (y, f ) (recall (8.7)). For the second
integral, we use the fact that br (·)f (·) ∈ Rrα−1+β is asymptotically decreasing (since
rα − 1 + β < 0 by assumption), hence we can bound br (yr )f (yr ) . br (yr−1 )f (yr−1 ).
Since Py (dy2 , · · · , dyr ) = Py (dy2 , · · · , dyr−1 )F (−yr−1 + dyr ), when we integrate over yr ∈
(−∞, −γyr−1 ] we get a factor .γ 1/A(yr−1 ). Overall, we can bound (8.18) by
Z γy
Z γyr−2
1
Py (dy2 , · · · , dyr−1 )
···
≤ gr (y, f ) +
br (yr−1 )f (yr−1 )
A(y
r−1 )
yr−1 =0
y2 =0
≤ gr (y, f ) + gr−1 (y, f ) ,
and the same bound holds also for r = 2. This proves (8.17) when H = r.
Finally, if H = j ∈ {2, . . . , r − 1}, we can estimate
Z
Z γy
Z γyj−2 Z 0
Py (dy2 , · · · , dyj )
···
F (|yj | + dyj+1 ) hr−j (yj+1 , Aj f )
y2 =0
Z
γy
···
.γ
y2 =0
yj+1 >−γ|yj |
yj−1 =0 yj =−∞
Z γyj−2 Z 0
Py (dy2 , · · · , dyj ) bj (yj ) f (yj ) ,
yj−1 =0
yj =−∞
thanks to the induction hypothesis (8.16). Splitting the integral over yj in the domains
(−∞, −γyj−1 ) ∪ [−γyj−1 , 0], we obtain precisely (8.18) with j in place of r. With the same
arguments, we see that this contribution is . gj (y, f ) + gj−1 (y, f ).
1
We finally need to consider, in the case α = k+1
for some k ≥ 2, the following quantity:
Z
J˜k (δ, x) =
F (x + dy1 ) h̃k (y1 ) ,
(8.20)
|y1 |≤δx
with
Z
Py (dy2 , · · · , dyk ) b̃k+1 (yk−1 , yk ),
h̃k (y1 ) =
(8.21)
Θk−1 (y1 )∩{|yk |≤γ|yk−1 |}
where Θk (y1 ) and b̃(x, z) are defined in (8.11) and (1.22) (or equivalently (5.9)). In case
k = 2, the constraint Θk−1 (y1 ) in (8.21) should be ignored (since Θr (y1 ) is only defined for
r ≥ 2) and J˜2 (δ, x) coincides with I˜2 (δ, x), defined in (1.24) (recall that γ = 1 − γ = η).
1
From (5.9), (2.4) and (1.10), for any λ ∈ (0, k), with α = k+1
, we have the bound
b̃k+1 (y, z) ≤ A(y)λ b̃k+1−λ (y, z) ≤ A(y)λ
∞
X
m(k−λ)
. A(y)λ bk+1−λ (z).
am
(8.22)
m=A(z)
Proposition 8.4. Fix k ∈ N with k ≥ 2 and assume that α =
J˜k (δ, x) is a.n..
1
k+1 .
If I˜k (δ, x) is a.n., then
THE STRONG RENEWAL THEOREM
27
Proof. In case k = 2 there is nothing to prove, since J˜2 (δ, x) = I˜2 (δ, x).
We now fix k ≥ 3. If we consider the contribution to the integrals in (8.20)-(8.21) of
y1 ≥ 0, y2 ≥ 0, . . . , yk−1 ≥ 0, the domain of integration, cf. (8.11), reduces to
{0 ≤ y1 ≤ δx} ∩ {0 ≤ yi ≤ γyi−1 for 2 ≤ i ≤ k − 1} ∩ {|yk | ≤ γyk−1 } .
This contribution is bounded from above by I˜k (δ, x), cf. (1.24), which is a.n. by assumption.
Next we consider the contribution to (8.20)-(8.21) coming from y1 , . . . , yk−1 such that
yi < 0 for some 1 ≤ i ≤ k − 1. Let us define H̃ = max{j ∈ {1, . . . , k − 1} : yj < 0}. If
H̃ = k − 1, the bound (8.22) with λ = k − 1 and the fact that yk−1 < 0 show that
Z
F (−yk−1 + dyk ) b̃k+1 (yk−1, yk )
|yk |≤γ|yk−1 |
Z
k−1
F (−yk−1 + dyk ) b2 (yk ) . A(yk−1 )k−1 b1 (yk−1 ) = bk (yk−1 ),
. A(yk−1 )
|yk |≤γ|yk−1 |
where the second inequality comes from Proposition 8.1 with r = 1 (we recall that Ik (δ, x) is
a.n. follows from I˜k (δ, x) is a.n., thanks to Lemma A.3). Plugging this bound into (8.21), we
see that the contribution to h̃k (y) is at most hk−1 (y) (recall (8.12)), hence the contribution
to H̃k (δ, x) is at most Jk−1 (δ, x) (recall (8.14)), which is a.n. by Proposition 8.3.
We finally consider the contribution of H̃ = k − j with j ≥ 2. This means that yk−j < 0,
while yk−j+1 ≥ 0, . . . , yk−1 ≥ 0, and the range of integration in (8.21) is a subset of
Θk−j−1 (y1 ) ∩ {yk−j < 0} ∩ {yk−j+1 ≥ 0} ∩ {|yk−j+1+r | ≤ γ|yk−j+r |, r = 1, · · · j − 1}.
We split this into the two subsets {0 ≤ yk−j+1 ≤ γ|yk−j |} and {yk−j+1 > γ|yk−j |}.
On the first subset {0 ≤ yk−j+1 ≤ γ|yk−j |}, we bound b̃k+1 (yk−1 , yk ) . A(yk−1 )k−j bj+1 (yk ),
by (8.22) with λ = k − j, and then A(yk−1 ) . A(yk−j ). Recalling the definition (8.2) of
gj (·), we see that this part of the integral with respect to yk−j+1 , · · · , yk is
Z
. A(yk−j )k−j
F (|yk−j | + dyk−j+1 ) gj (yk−j+1 )
0≤yk−j+1 ≤γ|yk−j |
Z
≤ A(yk−j )k−j
F (|yk−j | + dz) gj (z)
z≥−γ|yk−j |
k−j
. A(yk−j )
b1 (yk−j ) = bk−j+1 (yk−j ) ,
where the last inequality follows by Proposition 8.1. The contribution to (8.21) is
Z
.
Py (dy2 , . . . , dyk−j ) bk−j+1 (yk−j ) ≤ hk−j (y1 ) ,
(8.23)
Θk−j (y1 )∩{yk−j <0}
hence the contribution to J˜k (δ, x) is . Jk−j (δ, x), which is a.n. by Proposition 8.3.
On the second subset {yk−j+1 > γ|yk−j |}, we we use the bound b̃k+1 (yk−1 , yk ) .
A(yk−1 )k−j+1 bj (yk ), by (8.22) with λ = k − j + 1, and then A(yk−1 ) . A(yk−j+1 ), getting
Z
Z
.
F (|yk−j | + dyk−j+1 ) A(yk−j+1 )k−j+1
F (−yk−j+1 + dyk−j+2 ) gj−1 (yk−j+2 )
yk−j+1 >γ|yk−j |
Z
=
y>γ|yk−j |
F (|yk−j | + dy) A(y)k−j+1
|yk−j+2 |≤γyk−j+1
Z
F (−y + dz) gj−1 (z) ,
|z|≤γy
28
FRANCESCO CARAVENNA AND RON DONEY
where we have set y = yk−j+1 and z = yk−j+2 for short. Applying (8.6), where γ1 =
1 − (2 − γ)−1 and γ2 = γ −1 − 1, we get
Z
Z
.
F (|yk−j | + w + dv) A(w + v)k−j+1 gj−1 (v)
F (−dw)
−γ1 w≤v≤γ2 w
w≥γγ|yk−j |
Z
k−j+1
Z
F (|yk−j | + w + dv) gj−1 (v)
F (−dw) A(w)
.γ
v≥−γ1 (|yk−j |+w)
w≥γγ|yk−j |
Z
F (−dw) A(w)k−j+1 b1 (|yk−j | + w) ,
.γ
w≥γγ|yk−j |
by Proposition 8.1 with r = j − 1. Finally, this is easily bounded by
Z
F (−dw) bk−j+2 (|yk−j | + w) . F (−γγ|yk−j |) bk−j+2 (|yk−j |) . bk−j+1 (|yk−j |) ,
w≥γγ|yk−j |
because bk−j+2 (·) ∈ RV (α(k − j + 2) − 1) is asymptotically decreasing, since j ≥ 2 and
1
α = k+1
. Arguing as in (8.23), the contribution to J˜k (δ, x) is . Jk−j+1 (δ, x), which is a.n.
by Proposition 8.3. This completes the proof.
8.2. Proof of Sufficiency. Throughout the proof we fix k = κα = b1/αc − 1 (cf. (1.21)).
We generalize (7.1), defining two sequences Z1 , Z2 , . . . Zk and Y1 , Y2 , . . . , Yk as follows:
Z1 := max{X1 , . . . , Xn } ,
Y1 := Z1 − x ,
and for r ∈ {2, . . . , k}
(
max {Xj , 1 ≤ j ≤ n} \ {Zj , 1 ≤ j ≤ r − 1}
Zr :=
min {Xj , 1 ≤ j ≤ n} \ {Zj , 1 ≤ j ≤ r − 1}
Yr :=
r
X
if Yr−1 ≤ 0,
(8.24)
if Yr−1 > 0.
Zi − x
(8.25)
i=1
Intuitively, Zr is the largest available step towards x from Z1 + . . . + Zr−1 . In fact, we may
assume that the following holds:
Zr > 0
if Yr−1 ≤ 0 ,
while
Zr < 0
if Yr−1 > 0 ,
(8.26)
because the event that (8.26) fails to be true can be ignored. In fact, this event occurs if,
for some r ≤ k, either Yr−1 ≤ 0 and {Xj , 1 ≤ j ≤ n} \ {Zj , 1 ≤ j ≤ r − 1} contains no
positive terms or Yr−1 > 0 and this set contains no negative terms. Call En,r such event.
Clearly P(En,r , Sn ∈ x + I, Yr−1 ∈
/ I) = 0 for all n ≥ r (we recall that I = (−h, 0]), while
n
P(En,r , Yr−1 ∈ I) ≤
P(Sr−1 ∈ x + I, Xr , Xr+1 , . . . , Xn ≤ 0)
r−1
≤ nr−1 cn−(r−1) P(Sr−1 ∈ x + I) ,
with c = P(X1 ≤ 0) < 1, hence by (4.4)
∞
X
n=r
P(En,r , Sn ∈ x + I) ≤
∞
X
n=r
P(En,r , Yr−1 ∈ I) .r P(Sr−1 ∈ x + I) = o b1 (x) .
x→∞
THE STRONG RENEWAL THEOREM
29
We recall from (7.2) that
(3)
E1 = {Z1 > γx, |Y1 | > an } .
(1)
(2)
(3)
Next we define events Er , Er , Er
for r ≥ 2 inductively by
(3)
(3)
Er(1) = Er−1 ∩ {|Zr | ≤ γ|Yr−1 |} ,
Er(2) = Er−1 ∩ {|Yr | ≤ an } ,
(8.27)
(3)
Er(3) = Er−1 ∩ {|Zr | > γ|Yr−1 |, |Yr | > an }.
We can partition the probability space Ω as follows:
Ω=
k
[
r=1
Er(1) ∪
k
[
(3)
Er(2) ∪ Ek .
r=1
(1)
(2)
The argument to show that P(Sn ∈ x + I, E1 ∪ E1 ) is a.n. presented in Section 7 is still
(1)
(2)
valid, so it suffices to show that P(Sn ∈ x + I, Er ) and P(Sn ∈ x + I, Er ) are a.n. for
(3)
2 ≤ r ≤ k, and furthermore P(Sn ∈ x + I, Ek ) is a.n..
(3)
Remark 8.5. We can rewrite E`
(3)
E`
more explicitly as follows:
= {Z1 > γx , |Zi | > γ|Yi−1 | for 2 ≤ i ≤ `} ∩ {|Yi | > an , for 1 ≤ i ≤ `} .
(3)
Recalling the definition (8.11) of Θk (·), we claim that E`
(3)
E`
can also be rewritten as
= {Y1 > −γx, (Y2 , . . . , Y` ) ∈ Θ` (Y1 )} ∩ {|Yi | > an , for 1 ≤ i ≤ `} .
(8.28)
To prove the claim, we show that |Zi | > γ|Yi−1 | is equivalent to Yi ∈ θ(Yi−1 ), for 2 ≤ i ≤ k,
with θ(·) defined in (8.11). We recall that Zi = Yi − Yi−1 , cf. (8.25). If Yi−1 > 0, then
Zi ≤ 0 by (8.26), hence |Zi | > γ|Yi−1 | becomes Yi−1 − Yi > γYi−1 , which is precisely
Yi ∈ (−∞, −γYi−1 ) = θ(Yi−1 ). Similar arguments apply if Yi−1 ≤ 0, in which case Zi ≥ 0.
(1)
Estimate of Er . We fix r ∈ {2, . . . , k}. By exchangeability,
P(Sn ∈ x + I, Er(1) ) ≤ nr−1 P (Z1 , . . . Zr−1 ) = (X1 , . . . , Xr−1 ), Sn ∈ x + I, Er(1) . (8.29)
0
Conditionally on (X1 , . . . , Xr−1 ) = (z1 , . . . , zr−1 ), we have Sn = (z1 +. . .+zr−1 )+Sn−(r−1)
,
0
0
0
0
where we set Sk := X1 + . . . + Xk with Xi := X(r−1)+i . Motivated by (8.25), if we set
yr−1 := (z1 + . . . + zr−1 ) − x ,
0
then {Sn ∈ x + I} = {Sn−(r−1)
∈ −yr−1 + I}. Assume first that yr−1 ≤ 0. Then Zr =
0
0
Mn−(r−1) := max{Xi , 1 ≤ i ≤ n − (r − 1)}. Recalling (8.27) and (8.24), we need to evaluate
0
0
P Sn−(r−1)
∈ −yr−1 + I, |Mn−(r−1)
| ≤ γ |yr−1 | .
(8.30)
Since this probability is increasing in γ, applying (1.4), we get the bound
d
1
n
1
.
,
for all d ≤ .
an A(|yr−1 |)
γ
(8.31)
30
FRANCESCO CARAVENNA AND RON DONEY
In case yr−1 > 0, relation (8.30) holds with Mk0 replaced by (Mk0 )∗ := min1≤i≤k Xi0 . Applying
(1.4) to the reflected walk (S 0 )∗ = −S 0 , we see that the bound (8.31) still holds. Looking
back at (8.29), and recalling (8.27) and (8.28), we have the key bound
Z
nr+d−1
(1)
. (8.32)
P(Sn ∈ x+I, Er ) .
F (x+dy1 ) Py1 (dy2 , · · · , dyr−1 )
an A(|yr−1 |)d
y1 >−γx,
(y2 ,...,yr−1 )∈Θr−1 (y1 ),
min{|y1 |,...,|yr−1 |}≥an
α
Henceforth we fix d ∈ ( α1 − r, α1 − r + 1). Since γ < 1−α
by assumption, and r ≥ 2, we
have γ1 > α1 − 1 ≥ α1 − (r − 1) > d, hence the constraint d ≤ γ1 is satisfied. The sequence
nd+r−1 /an is regularly varying with exponent (d + r − 1) − α1 > −1, hence by (2.3)
X
1≤n≤A(w)
A(w)r+d
nr+d−1
.
= br+d (w) ,
an
w∧1
∀w ≥ 0 ,
where we recall that bk (·) was defined in (1.10). It is convenient to set
ỹr−1 := min{|y1 |, . . . , |yr−1 |} ,
(8.33)
so that the integral in (8.32) is restricted to n ≤ A(ỹr−1 ). It follows that
X
P(Sn ∈ x + I, Er(1) )
1≤n≤A(δx)
Z
F (x + dy1 ) Py1 (dy2 , · · · , dyr−1 )
.
br+d (ỹr−1 ∧ δx)
.
A(|yr−1 |)d
(8.34)
y1 >−γx
(y2 ,...,yr−1 )∈Θr−1 (y1 )
To show that this integral is a.n., we separate the contributions of |y1 | ≤ δx and |y1 | > δx.
• Since ỹr−1 ≤ |yr−1 |, and br+d (·) is increasing (because r + d > α1 ), we have
br+d (ỹr−1 ∧ δx)
br+d (|yr−1 |)
≤
= br (|yr−1 |) ,
A(|yr−1 |)d
A(|yr−1 |)d
hence the contribution of |y1 | ≤ δx to the integral in (8.34) is bounded by
Z
F (x + dy1 ) Py1 (dy2 , · · · , dyr−1 ) br (|yr−1 |) =: Jr−1 (δ, x) ,
|y1 |≤δx
(y2 ,...,yr−1 )∈Θr−1 (y1 )
where Jk (δ, x) was defined in (8.14). By Proposition 8.3, this is a.n..
• If we define f1 (y) := 1/br+d−1 (y), then recalling (8.13) we can write, for fixed y1 ,
Z
1
= hr−1 (y1 , f1 ) .
Py1 (dy2 , · · · , dyr−1 )
A(|yr−1 |)d
(y2 ,...,yr−1 )∈Θr−1 (y1 )
As a consequence, the contribution of |y1 | > δx to the integral in (8.34) is at most
Z
br+d (δx)
F (x + dy1 ) hr−1 (y1 , f1 ) .
|y1 |>δx, y1 >−γx
THE STRONG RENEWAL THEOREM
31
Note that f1 (·) ∈ RV (β) with β = 1 − (r + d − 1)α. Our choice of d implies that
1
1
1
α < r + d < α + 1, hence 0 < β < α. Since we are in the regime α < 2 , we also have
0 < β < 1 − α, hence Proposition 8.2 with f = f1 , shows that this term is a.n.
(2)
Estimate of Er . Always for r ∈ {2, . . . , k}, in analogy with (8.29), we have
P(Sn ∈ x + I, Er(2) ) ≤ nr P (Z1 , . . . Zr ) = (X1 , . . . , Xr ), Sn ∈ x + I, Er(2)
r
(2)
sup P(Sn−r ∈ z + I)
≤ n P (Z1 , . . . Zr ) = (X1 , . . . , Xr ), Er
z∈R
.
nr
(8.35)
P (Z1 , . . . Zr ) = (X1 , . . . , Xr ), Er(2) ,
an
where the last inequality follows by (2.7). By (8.27) and (8.28), we obtain
Z
nr
F (x + dy1 ) Py1 (dy2 , · · · , dyr )
P(Sn ∈ x + I, Er(2) ) .
.
an
y1 >−γx, (y2 ,...,yr−1 )∈Θr−1 (y1 ),
min{|y1 |,...,|yr−1 |}≥an , |yr |<an
Recalling (8.33) and (5.9), we can write
Z
X
(2)
P(Sn ∈ x + I, Er ) .
1≤n≤A(δx)
(8.36)
A(δx∧ỹr−1 )
F (x + dy1 ) Py1 (dy2 , · · · , dyr )
y1 >−γx,
(y2 ,...,yr−1 )∈Θr−1 (y1 )
X
n=A(|yr |)
nr
an
Z
F (x + dy1 ) Py1 (dy2 , · · · , dyr ) b̃r+1 (δx ∧ ỹr−1 , |yr |) .
=
y1 >−γx,
(y2 ,...,yr−1 )∈Θr−1 (y1 )
• Note that b̃r+1 (δx ∧ ỹr−1 , |yr |) ≤ b̃r+1 (|yr−1 |, |yr |) 1{|yr |≤δx} , hence the contribution of
|y1 | ≤ δx is bounded by J˜r (δ, x), cf. (8.20), which is a.n. by Proposition 8.4.
• Next we deal with |y1 | > δx. Note that r ∈ {2, . . . , k}, hence α(r + 1) ≤ α(k + 1) ≤ 1
1
1
(since k = κα means α ∈ ( k+2
, k+1
]). If α(r+1) < 1 we set ψ = 0, while if α(r+1) = 1
we fix arbitrarily ψ ∈ (0, 1), so that α(r + 1 − ψ) < 1 and f2 (y) := A(y)1−ψ ∈ RV (β),
with β = α(1 − ψ), satisfies the conditions of Proposition 8.3. Then in both cases
b̃r+1 (δx ∧ ỹr−1 , |yr |) . A(δx)ψ b̃r+1−ψ (ỹr−1 , |yr |) . A(δx)ψ br+1−ψ (yr ),
and we conclude that the contribution of |y1 | > δx is
Z
ψ
F (x + dy1 ) hr (y1 , f2 ),
. A(δx)
y1 >−γx,|y1 |>δx
which is a.n. by Proposition 8.3.
(3)
Estimate of Ek . Finally we write
(3)
Z
P (Sn ∈ x + I, Ek ) .
F (x + dy1 ) Py1 (dy2 , · · · , dyk )
y1 >−γx,
(y2 ,...,yk )∈Θk (y1 ),
min{|y1 |,...,|yk |}≥an
nk+1
.
an A(yk )
32
FRANCESCO CARAVENNA AND RON DONEY
Note that nk+1 /an ∈ RV (ζ) with ζ = k + 1 − α1 > −1, by k = κα . Therefore, by (2.3),
Z
X
bk+2 (δx ∧ ỹk )
(3)
P (Sn ∈ x + I, Ek ) .
F (x + dy1 ) Py1 (dy2 , · · · , dyk )
.
A(yk )
1≤n≤A(δx)
y1 >−γx,
(y2 ,...,yk )∈Θk (y1 )
• Note that bk+2 (δx ∧ ỹk )/A(yk ) ≤ bk+1 (yk ), hence the contribution of {|y1 | ≤ δx} is
bounded by Jk (δ; x), cf. (8.14), which is a.n. by Proposition 8.3.
• To deal with {|y1 | > δx} we put ν = 1 if α(k + 1) < 1, while if α(k + 1) = 1 we
choose ν ∈ (0, 1) with α(k + ν) < 1. We also put f3 (y) = 1/bk+ν (y) ∈ RV (β) with
β = 1 − α(k + ν) and note that f3 satisfies the assumptions of Proposition 8.3 with
r = k. Then in both cases we have the bound
bk+1+ν (δx)A(yk )1−ν
bk+1+ν (δx)
bk+2 (δx ∧ ỹk )
.
=
= bk+1+ν (δx) bk (yk ) f3 (yk ) ,
A(yk )
A(yk )
A(yk )ν
and we conclude that the contribution of {|y1 | > δx} is
Z
. br+1+ν (δx)
F (x + dy1 ) hk (y1 , f3 ) ,
y1 >−γx,|y1 |>δ0 x
which is also a.n. by Proposition 8.3 with r = k.
9. Soft results
In this section we prove several corollaries of our main results. Let us first describe a
practical way to build counter-examples.
Remark 9.1. Let us fix A ∈ RV (α). Let F1 be a probability on (0, ∞) which satisfies
2
1
F1 (x) ∼
,
F1 ((x − h, x]) = O
, ∀h > 0 .
(9.1)
x→∞ A(x)
x→∞
xA(x)
P
2α
(For instance, fix n0 ∈ N such that c1 := n>n0 n A(n)
< 1 and define F1 ({n0 }) := 1 − c1 ,
2α
F1 ({n}) := n A(n) for n ∈ N with n > n0 .) Let F2 be a probability on (0, ∞) such that
1
F2 (x) = o
.
(9.2)
x→∞
A(x)
If we define F := 12 (F1 + F2 ), we obtain a new probability on (0, ∞) which satisfies
F (x) ∼
x→∞
1
,
A(x)
1
F (x + I) ≥ F2 (x + I) .
2
(9.3)
1
Next we state a useful result. If F satisfies (1.12), then necessarily F (x + I) = o( A(x)
)
1
as x → ∞ (because F (x − h) ∼ F (x) ∼ A(x)
). Interestingly, this bound can be approached
as close as one wishes, in the following sense.
Lemma 9.2. Fix two arbitrary positive positive sequences xn → ∞ and n → 0. For any
A ∈ RV (α), with α ∈ (0, 1), there is a probability F on (0, ∞) satisfying (1.12) such that
n
F ({xn }) ≥
for infinitely many n ∈ N .
(9.4)
A(xn )
THE STRONG RENEWAL THEOREM
33
Proof. Let us fix A ∈ RV (α). By Remark 9.1, it is enough to build a probability F2 on
(0, ∞), supported on the sequence {xn }n∈N , which satisfies (9.2) and
n
F2 ({xn }) ≥ 2
for infinitely many n ∈ N .
(9.5)
A(xn )
Then, if we define F := 21 (F1 + F2 ), the proof is completed (recall (9.3)).
By assumption xn → ∞ and n → 0, hence we can fix a subsequence (nk )k∈N such that
nk+1
1 nk
≤
,
A(xnk+1 )
2 A(xnk )
∀k ∈ N .
(9.6)
P
n
This ensures that k∈N A(xnk ) < ∞ (the series converges geometrically) and we fix k0 ∈ N
k
P
n
so that k≥k0 A(xnk ) ≤ 12 . We now define F2 , supported on the set {xnk : k ≥ k0 }, by
k
nk
F2 ({xnk }) := c2
A(xnk )
for k ≥ k0 ,
where
X
c2 :=
k≥k0
nk
A(xnk )
!−1
≥ 2.
In this way, (9.5) is satisfied. It remains to check that (9.2) holds. Given x ∈ (0, ∞), if we
set k̄ := min{k ≥ k0 : xnk > x}, we can write
X
nk̄
nk̄ X
nk
1
F2 ((x, ∞)) =
≤ 2 c2
c2
,
≤ c2
k−
k̄(x)
A(xnk )
A(xnk̄ )
A(x)
2
k≥k̄
k≥k̄
where we used (9.6), and the last inequality holds because xnk̄ > x, by definition of k̄.
Since n → 0 by assumption, and k̄ → ∞ as x → ∞, the proof is completed.
9.1. Proof of Theorem 1.5. Assume that condition (1.14) holds. By (1.10) we can write
A(z)2
A(x)2
.
.
z
x
1≤z≤x
sup b2 (z) = sup
1≤z≤x
2
For 0 ≤ z ≤ 1 we can also write b2 (z) ≤ A(1)2 = b2 (1) . A(x)
hence by (1.13)
x
A(x)2
A(x)2
1
1
+
I1 (δ; x) .
F ([x − δx, x]) ∼
−
x→∞
x
x
A((1 − δ)x) A(x)
A(x)
1
A(x)
∼
−1 =
O(δ) .
α
x→∞
δ→0
x
(1 − δ)
x
This shows that I1+ (δ; x) is a.n., hence the SRT holds by Theorem 1.4.
Next we prove the second part of Theorem 1.5: we assume that condition (1.14) is not
satisfied, and we build a probability F for which the SRT fails. Since A ∈ RV ( 12 ), we
√
can write A(x) = L(x) x where L is slowly varying. By assumption, cf. (1.14), there is a
subsequence xn → ∞ such that sup1≤s≤xn L(s) L(xn ), hence we can find 1 ≤ sn ≤ xn for
which L(sn ) L(xn ). We have necessarily sn = o(xn ), because L(s)/L(xn ) → 1 uniformly
for s ∈ [xn , xn ], for any fixed > 0, by the uniform convergence theorem of slowly varying
functions [BGT89, Theorem 1.2.1]. Summarizing:
xn → ∞ ,
sn = o(xn ) ,
n :=
L(xn )
→ 0.
L(sn )
(9.7)
34
FRANCESCO CARAVENNA AND RON DONEY
By Lemma 9.2, there is√a probability F on (0, ∞), which satisfies (1.12), such that (9.4)
holds. Since A(x) = L(x) x, recalling (1.13), for infinitely many n ∈ N we can write
I1+ (δ; xn + sn ) ≥
n
L(sn )
1 A(xn )
A(xn + sn )
A(sn )2
F ({xn }) ≥ L(sn )2
= √
=
,
sn
A(xn )
xn
n xn
xn + s n
where the last inequality holds because n → 0 and xn + sn ∼ xn , cf. (9.7). This shows that
I1+ (δ; x) is not a.n., hence the SRT fails, by Theorem 1.4.
9.2. Proof of Proposition 1.6. Let us fix A ∈ RV (α) with α ∈ (0, 12 ). By Remark 9.1,
it is enough to build a probability F2 on (0, ∞) which satisfies (9.2) and moreover
ζ(x)
F2 (x + I) = O
,
I1+ (δ; x; F2 ) is not a.n. ,
(9.8)
xA(x)
where I1+ (δ; x; F2 ) denotes the quantity I1+ (δ; x) in (1.13) with F replaced by F2 . Once this
is done, we can set F := 21 (F1 + F2 ) and the proof is completed (recall (9.3)).
We recall that ζ(·) is non-decreasing with limx→∞ ζ(x) = ∞. Let us define xn := 2n ,
and fix n0 ∈ N large enough so that ζ(xn0 −1 ) ≥ 1. Let us define
1
xn
zn :=
,
where θ > 0 will be fixed later .
(9.9)
2 ζ(xn−1 )1+θ
Note that zn ≤ 21 xn for n ≥ n0 (because ζ(xn−1 ) ≥ 1), hence the intervals (xn − zn , xn ] are
disjoint. We may also assume that zn ≥ 1 (if we decrease ζ(·) we get a stronger statement,
so we can replace ζ(x) by min{ζ(x), log x}, which ensures
S that zn → ∞).
We define a probability F2 supported on the set n≥n0 (xn − zn , xn ], with a constant
density on each interval, as follows:
ζ(xn−1 )
F2 (xn − ds) := c
1
(s) ds ,
∀n ≥ n0 ,
(9.10)
xn A(xn ) [0,zn ]
where c ∈ (0, ∞) is the normalizing constant that makes F2 a probability. Note that
F2 ((xn − zn , xn ]) = c
ζ(xn−1 )
c
zn =
.
xn A(xn )
2 A(xn ) ζ(xn−1 )θ
Since A(xn ) = A(2xn−1 ) ∼ 2α A(xn−1 ) as n → ∞, we may assume that A(xn ) ≥ 2α/2 A(xn−1 )
for all n ≥ n0 + 1 (possibly enlarging n0 ). Since ζ(·) is non-decreasing, we obtain
F2 ((xn − zn , xn ]) ≤ 2−α/2 F2 ((xn−1 − zn−1 , xn−1 ]) ,
∀n ≥ n0 + 1 .
It follows that, for every n ≥ n0 ,
X
X
F2 ((xm − zm , xm ]) ≤ F2 ((xn − zn , xn ])
(2−α/2 )m−n = C F2 ((xn − zn , xn ]) ,
m≥n
m≥n
where C := (1 − 2−α/2 )−1 < ∞. This relation, for n = n0 , shows that F2 is indeed a finite
measure, so we can make it a probability by suitably choosing c in (9.10).
For every x ∈ (0, ∞) large enough, we have xn−1 < x ≤ xn for a unique n ≥ n0 and
X
cC
1
= o
F2 (x) ≤
F2 ((xm − zm , xm ]) ≤ C F2 ((xn − zn , xn ]) ≤
,
A(x)
2 A(x) ζ(xn−1 )θ x→∞
m≥n
so that (9.2) holds. Similarly, for xn−1 < x ≤ xn we can write
F2 (x + I) = F2 ((x − h, x]) ≤ c h
ζ(xn−1 )
ζ(x)
≤ ch
,
xn A(xn )
x A(x)
THE STRONG RENEWAL THEOREM
35
because ζ(·) and A(·) are non-decreasing, hence the first relation in (9.8) holds. Finally, for
x = xn and δ < 12 , since zn ≤ δxn for n large enough, we have by (2.3)
Z
A(z)2
ζ(xn−1 )
ζ(xn−1 )
+
I1 (δ; xn ; F2 ) = c
∼ c
A(zn )2 .
xn A(xn ) 0≤z≤zn z ∨ 1 n→∞ xn A(xn )
Recalling (9.9), we can apply Potter’s bounds (2.2), since zn ≥ 1, to get, for any > 0,
ζ(xn−1 )
A(xn )2
A(xn )
A(xn )
= ζ(xn−1 )1−2(1+θ)(α+)
,
2(1+θ)(α+)
xn A(xn ) ζ(xn−1 )
xn
xn
where the last inequality holds provided we choose θ > 0 and > 0 small enough, depending
only on α, so that 1−2(1+θ)(α+) > 0 (we recall that α < 12 ). This shows that I1+ (δ; xn ; F2 )
is not a.n. and completes the proof.
I1+ (δ; xn ; F2 ) &
9.3. Proof of Proposition 1.7. We first prove that relation (1.16) for γ < 1 − 2α is a
necessary condition for the SRT. We can assume that α < 21 , because for α = 12 relation
(1.16) follows directly from (1.12). If we restrict the integral (1.13) to z ∈ [0, y), where
y = yx = o(x), for arbitrary δ > 0 and large x we obtain I1+ (δ; x) & b2 (y) F ((x − y, x]),
because b2 (z) ∈ RV (2α − 1) is asymptotically decreasing. If the SRT holds, I1+ (δ; x) is a.n.
by Theorem 1.4. It follows that b2 (y) F ((x − y, x]) = o(b1 (x)), or equivalently
1 b2 (x)
,
for any y = yx ≥ 1 with yx = o(x) .
F ((x − y, x]) = o
x→∞
A(x) b2 (y)
Since b2 ∈ RV (2α − 1), it follows by Potter’s bounds (2.2) that, for any given γ < 1 − 2α,
y γ
we have bb22 (x)
(y) .γ ( x ) , hence (1.16) holds as claimed.
We now turn to the sufficiency part. Let F be a probability on [0, ∞) which satisfies
(1.12), for some α ∈ (0, 21 ], such that relation (1.16) holds for some γ > 1 − 2α. We prove
that the SRT holds by showing that I˜1+ (δ; x) defined in (1.17) is a.n., by Proposition 1.11.
By Remark 1.8, we can assume that (1.16) holds with o(·) replaced by O(·). We claim
that this is equivalent to the following relation:
1 y γ
∃x0 , C < ∞ :
F ((x − y, x]) ≤ C
(9.11)
,
∀x ≥ x0 , ∀y ∈ [1, 12 x] .
A(x) x
It is clear that (9.11) implies (1.16) with o(·) replaced by O(·). On the other hand, if (9.11)
fails, there are sequences xn → ∞, Cn → ∞ and yn ∈ [1, 12 xn ] such that
1 y n γ
F ((xn − yn , xn ]) > Cn
.
(9.12)
A(xn ) xn
By extracting subsequences, we may assume that xynn → ρ ∈ [0, 12 ]. If ρ = 0, then yn = o(xn )
and (9.12) contradicts (1.16). If ρ > 0, then (9.12) contradicts (1.12), because it yields
1
1
F ( 12 xn ) ≥ F ((xn − yn , xn ]) > Cn
ργ + o(1) .
1
A(xn )
A( 2 xn )
Applying (9.11) and recalling (1.17), for δ < 12 we get
Z δx
C
A(z)2
C0
A(δx)2
+
˜
I1 (δ; x) ≤
dz
∼
x→∞ A(x) xγ (δx)1−γ
A(x) xγ 1 z 2−γ
∼
x→∞
C 0 δ 2α−1+γ
A(x)
,
x
where C 0 := C/(2α − 1 − γ) and the first asymptotic equivalence holds by (2.3), because
z 7→ A(z)2 /z 2−γ is regularly varying with index 2α − (2 − γ) > −1 (since γ > 1 − 2α). This
shows that I˜1+ (δ; x) is a.n. and completes the proof of Proposition 1.7.
36
FRANCESCO CARAVENNA AND RON DONEY
9.4. Proof of Proposition 1.11. By (1.17) we can write
Z δx Z
b2 (z)
+
I˜1 (δ; x) =
1{y∈[0,z)} F (x − dy)
dz
z
1
R
Z δx
Z
b2 (z)
F (x − dy)
=
dz .
z
y∈[0,δx)
1∨y
(9.13)
We recall that bk is defined in (1.10). Assume that α < 12 . Then the function z 7→ b2 (z)/z
is regularly varying with index 2α − 2 < −1, hence by (2.4), for y ≥ 0 we can write
Z δx
Z ∞
b2 (z)
b2 (z)
dz ≤
dz . b2 (1 ∨ y) . b2 (y) ,
z
z
1∨y
1∨y
because for 0 ≤ y < 1 we have b2 (y) ≥ A(0)2 > 0, cf. §2.2. Recalling (1.13), we have shown
that I˜1+ (δ; x) . I1+ (δ; x) when α < 21 . Then, if I1+ (δ; x) is a.n., also I˜1+ (δ; x) is a.n..
We now work for α ≤ 21 . Let us restrict the outer integral in (9.13) to y ∈ [0, 2δ x), and
the inner integral to z ∈ [1 ∨ y, 2 ∨ 2y). For y ≥ 1 we have
Z 2y
Z 2∨2y
b2 (z)
b2 (y)
1
b2 (z)
dz =
dz ≥
(2y − y) = b2 (y) ,
z
z
2y
2
y
1∨y
R 2∨2y
R2
while for 0 ≤ y < 1 we can write 1∨y b2z(z) dz = 1 b2z(z) dz = C & b2 (y). Overall, it
follows from (9.13) that I˜+ (δ; x) & I + ( δ ; x). This completes the proof.
1
1
2
9.5. Proof of Proposition 1.16. We fix α ∈ (0, 13 ) and choose for simplicity A(x) := xα .
We are going to build a probability F on R which satisfies (1.2) with p = q = 1, such that
I˜1 (δ; x) is a.n. but I˜2 (δ, η; x) is not a.n., for any η ∈ (0, 1). It suffices to show that I1 (δ; x)
is a.n. but I2 (δ, η; x) is not a.n., thanks to (A.5) and (A.6).
In analogy with Remark 9.1, we fix a probability F1 , this time on the whole real line R,
1
which satisfies (1.2) with p = q = 3 and such that F1 ((x − h, x]) = O( |x|A(x)
) as x → ±∞.
Then we define two probabilities F2 , F3 on (0, ∞) which both satisfy (9.2), and we set
1
F := (F1 + F2 + F3∗ ) ,
(9.14)
3
where F3∗ (A) := F3 (−A) is the reflection of F3 (so that it is a probability on (−∞, 0)).
Clearly, (1.2) holds for F with p = q = 1. It remains to build F2 and F3 .
We are going to define F2 so that
I1 (δ; x; F2 ) is a.n. ,
(9.15)
(where I1 (δ; x; F2 ) denotes the quantity in (1.18) with F replaced by F2 ). This implies that
I1 (δ; x) = I1 (δ; x; F ) is a.n., because I1 (δ; x; F1 ) is clearly a.n., while F3∗ is supported on
(−∞, 0) and gives no contribution.
2k
1
). We set En,k := [2n + 2k , 2n + 2k + 2k
We fix a parameter p ∈ (1, 3α
p ) for n ∈ N with
n
k
n
k+1
n ≥ 2 and for 1 ≤ k ≤ n − 1. Note that En,k ⊆ [2 + 2 , 2 + 2 ) are disjoint intervals,
S
n n+1 ). We define F by stipulating that it has a constant
and moreover n−1
2
k=1 En,k ⊆ [2 , 2
density in each interval En,k (for n ≥ 2 and 1 ≤ k ≤ n − 1) and zero otherwise, given by
F2 (2n + 2k + dw) :=
c
1
1 2k ) (w) dw ,
n
1−α
k
p
`(n) (2 )
(2 )2α [0, 2k
(9.16)
THE STRONG RENEWAL THEOREM
37
where c ∈ (0, ∞) is a suitable normalizing constant and we set for short
`(n) := log(1 + n) .
(9.17)
Note that
F2 (En,k ) =
c
1
2k
c
(2k )1−2α
=
,
`(n) (2n )1−α (2k )2α 2k p
`(n) (2n )1−α
2k p
(9.18)
hence
n−1
X
c (2n )1−2α
c
k 1−2α
(2
)
≤
`(n) (2n )1−α
`(n) (2n )1−α
k=1
k=1
c
1
=
.
= o
`(n) (2n )α n→∞
(2n )α
F2 ([2n , 2n+1 )) =
n−1
X
F2 (En,k ) ≤
Note that F2 ([2n , 2n+1 )) decreases exponentially fast in n, hence for x ∈ [2n , 2n+1 ) we have
F2 (x) ≤ F2 (2n ) . F2 ([2n , 2n+1 )) = o(1/A(x)), which shows that (9.2) is fulfilled. It remains
to check (9.15). We do this by showing that, for any δ < 41 ,
1
(9.19)
I1 (δ; x; F2 ) = o b1 (x) = o 1−α .
x→∞
x
This is elementary but slightly technical, and it is shown below.
k
Finally, we define F3 , We introduce the disjoint intervals Gk := [2k , 2k + k2p ) for k ≥ 2.
We let F3 have a constant density on every Gk (for k ≥ 2) and zero otherwise, given by
F3 (2k + dz) :=
c0
kp
1 k (z) dz ,
k
`(k) (2 )1+α [0, k2 p )
(9.20)
where c0 ∈ (0, ∞) is a normalizing constant. Then
k
k+1
F3 ([2 , 2
kp
2k
c0
1
1
c0
=
=o
.
)) = F3 (Gk ) =
`(k) (2k )1+α k p
`(k) (2k )α
(2k )α
Then for x ∈ [2k , 2k+1 ) we have F3 (x) ≤ F3 (2k ) . F3 ([2k , 2k+1 )) = o(1/A(x)) as x → ∞,
hence (9.2) holds. Given η ∈ (0, 1), fix k0 = k0 (η) large enough so that 2k1p < η for k ≥ k0 .
Then, recalling (1.19) and (9.14), we can write
Z
Z
I2 (δ,η; 2n ; F ) ≥
F (2n + dy)
F (−y + dz) b3 (z)
0≤y≤δ2n
1
≥
9
blog2
(δ2n )c
X
|z|≤ηy
Z
n
2k
0≤w≤ 2k
p
k=k0
Z
k
F2 (2 + 2 + dw)
2k
0≤z≤ 2k
p
F3 2k + w + dz (1 ∨ z)3α−1 .
k
2
3α−1 (we recall that α < 1 ) and by (9.20)
Note that (1 ∨ z)3α−1 ≥ ( 2k
p)
3
∀0 ≤ w ≤
2k
2kp
:
2k
c0
kp
1
1
2k
F3 2k + w + [0, 2k
)
=
&
.
p
p
k
1+α
k
`(k) (2 )
2k
`(k) (2 )α
Since log2 (δ2n ) = n + log2 δ, recalling (9.18), we can write for large n
n
I2 (δ, η; 2 ; F ) &
n/2
X
k=k0
1
1
F2 (En,k )
k
`(k) (2 )α
2k
2k p
3α−1
n/2
X
1
1
&
.
n
1−α
`(n) (2 )
`(k) k 3αp
k=k0
38
FRANCESCO CARAVENNA AND RON DONEY
Since we have fixed p <
1
3α ,
applying (2.3) and recalling (9.17) we finally obtain
I2 (δ, η; 2n ; F ) &
1
1
n1−3αp
n 1−α = b1 (2n ) .
2
n
1−α
`(n) (2 )
(2 )
This shows that I2 (δ, η; x; F ) is not a.n..
k
2
Proof of (9.19). We recall that F2 is supported on the intervals En,k := [2n +2k , 2n +2k + 2k
p)
Sn−1
n
n+1
).
with n ≥ 2 and 1 ≤ k ≤ n − 1. Let us set En := k=1 En,k ⊆ [2 , 2
For large x ≥ 0, we define n ≥ 2 such that 2n ≤ x < 2n+1 . For small δ ∈ (0, 14 ) and large
x, the interval (x − δx, x + δx) can intersect at most En and En+1 (because the rightmost
2n−2
3 n
point in En−1 is 2n−1 + 2n−2 + 2(n−2)
as n → ∞). Consequently we can write
p ∼ 4 2
Z
Z
F2 (dz) b2 (z − x) .
(9.21)
F2 (x + dy) b2 (y) ≤
I1 (δ; x; F2 ) =
z∈En ∪En+1
|y|≤δx
For z ∈ En+1 we have z ∈ En+1,k for some 1 ≤ k ≤ n, in which case z ≥ 2n+1 + 2k .
Since x < 2n+1 , we have |z − x| = z − x > 2k which yields b2 (z − x) ≤ b2 (2k ) = (2k )2α−1 .
Recalling (9.18), we see that the contribution of En+1 to (9.21) is at most
n
∞
X
X
c
1
(2k )1−2α k 2α−1
1
1
(2
)
.
=
o
,
`(n) (2n+1 )1−α
2k p
`(n) (2n+1 )1−α
kp
(2n+1 )1−α
k=1
k=1
1
which is o x1−α , so it is negligible for (9.19).
Then we look at the contribution of En to (9.21). Assume first that 2n + 2 ≤ x < 2n+1 .
Then we can write 2n + 2k̃ ≤ x < 2n + 2k̃+1 for a unique k̃ ∈ {1, . . . , n − 1}. For z ∈ En we
have z ∈ En,k for some 1 ≤ k ≤ n − 1. We distinguish three cases.
• If k ≤ k̃ − 1, then
|z − x| = x − z & (2n + 2k̃ ) − (2n + 2k̃−1 +
2k̃−1
)
2(k̃−1)p
& 2k̃ ,
hence b2 (z − x) . b2 (2k̃ ) = (2k̃ )2α−1 . By (9.18), the contribution to (9.21) is at most
k̃−1
X
k̃−1
k̃ 2α−1
F2 (En,k ) (2 )
X
c
c
k̃ 2α−1
≤
(2
)
(2k )1−2α .
,
n
1−α
`(n) (2 )
`(n) (2n )1−α
k=1
k=1
which is negligible for (9.19).
• If k ≥ k̃ + 2, then |z − x| = z − x & (2n + 2k ) − (2n + 2k̃+1 ) ≥ 2k − 2k−1 & 2k , hence
b2 (z − x) . b2 (2k ) = (2k )2α−1 and we get
n−1
X
F2 (En,k ) (2k )2α−1 ≤
k=k̃+1
∞
X
c
1
c
.
,
`(n) (2n )1−α
2k p
`(n) (2n )1−α
k=k̃+1
because p > 1, hence this contribution is also negligible for (9.19).
• If k ∈ {k̃, k̃ + 1}, then |z − x| ≤ 2k̃+2 − 2k̃ = 3 · 2k̃ . By (9.16), since the density of F2
is larger in En,k̃ then in En,k̃+1 , we see the contribution to (9.21) is at most
Z
c
1
1
(|w| ∨ 1)2α−1 dw .
,
n
1−α
`(n) (2 )
`(n) (2n )1−α
(2k̃ )2α |w|≤3·2k̃
which is negligible for (9.19).
THE STRONG RENEWAL THEOREM
39
Finally, the regime 2n ≤ x < 2n + 2 is treated similarly. For z ∈ En,k , we distinguish the
cases k ≥ 2 and with k = 1. If we set k̃ := 0, the estimates in the two cases k ≥ k̃ + 2 and
k ∈ {k̃, k̃ + 1} treated above apply with no change.
9.6. Proof of Proposition 1.17. Assume that both I˜1 (δ; x) and I˜1∗ (δ; x) are a.n., cf.
(1.25) and (1.27). We claim that, for any η ∈ (0, 1),
Z
∀z ∈ R, ∀` ∈ N :
F (−z + dy) b̃`+1 (z, y) .η b` (z) .
(9.22)
|y|≤η|z|
Since b̃`+1 (z, y) ≤ A(z)`−1 b̃2 (z, y) and b` (z) = A(z)`−1 b1 (z), cf. (1.22) and (1.10), it is
enough to prove (9.22) for ` = 1. Let
R us fix 0 < δ0 < η. For |y| > δ0 |z| we can bound
b̃2 (z, y) . b̃2 (z, δ0 z) .δ0 b2 (z) and δ0 |z|<|y|≤η|z| F (−z + dy) .δ0 ,η 1/A(z). It remains to
prove (9.22) for ` = 1 and with η replaced by an arbitrary δ0 > 0. The left hand side
of (9.22) equals I˜1 (η; z) for z ≥ 0 and I˜1∗ (η; −z) for z ≤ 0 (recall (1.23)), which are a.n.
by assumption, hence we can fix η = δ0 > 0 small enough so that the inequality (9.22)
holds for |z| > x0 , for a suitable x0 ∈ (0, ∞). Finally, for |z| ≤ x0 both sides of (9.22) are
uniformly bounded away from 0 and ∞, hence the inequality (9.22) still holds.
Observe that, for |z| ≤ η|w|, we can bound b` (z) .η b̃` ( η1 z, z) ≤ b̃` (w, z), so (9.22) yields
Z
∀z, w ∈ R with |z| ≤ η|w|, ∀` ∈ N :
F (−z +dy) b̃`+1 (z, y) .η b̃` (w, z) . (9.23)
|y|≤η|z|
If we plug this inequality into (1.24), we see that I˜2 (δ, η; x) .η I˜1 (δ; x) and, similarly,
I˜k (δ, η; x) .η I˜k−1 (δ, η; x) for any k ≥ 3. Since I˜1 (δ; x) is a.n. by assumption, it follows that
I˜k (δ, η; x) is a.n. for any k ≥ 2, hence the SRT holds by Theorem 1.12.
Finally, if relation (1.16) holds both for F and for F ∗ , the same arguments as in the
proof of Proposition 1.7, cf. §9.3, show that both I˜1 (δ; x) and I˜1∗ (δ; x) are a.n..
9.7. Proof of Theorem 1.18. Since Stone’s local limit theorem applies equally to Lévy
processes, cf. [BD97, Proposition 2], an argument similar to the random walk case (see
Subsection 4.1) shows that the SRT (1.30) holds if and only if Tb(δ; x) is a.n., where
Z δA(x)
b
T (δ; x) :=
P(Xt ∈ (x − h, x]) dt .
(9.24)
0
Let Js := Xs − Xs− be the jump of X at time s > 0. If we write
X
(1)
(2)
(1)
Xt = Xt + Xt ,
where
Xt :=
Js 1{|Js |≥1} ,
s≤t
then X (1) and X (2) are independent Lévy processes.
(1)
• The process X (1) is compound Poisson: we can write Xt = SNλt , where N = (Nt )t≥0
is a standard Poisson process, S = (Sn )n∈N0 is a random walk with step distribution
P(S1 ∈ dx) = F (dx) given in (1.29), and λ = Π(R \ (−1, 1)) ∈ (0, ∞).
(2)
(2)
• The process Xt can be written as Xt = σBt + µt + Mt , where M is the martingale
formed from the compensated sum of jumps with modulus less than 1.
To complete the proof, we show that the SRT holds for the random walk S if and only if
it holds for X (1) (step 1) if and only if it holds for X (step 2).
40
FRANCESCO CARAVENNA AND RON DONEY
(1)
Step 1. Since Xt
= SNλt , we have
(1)
P(Xt
∈ (x − h, x]) =
X
e−λt
n∈N0
Rz
(λt)n
P(Sn ∈ (x − h, x]) .
n!
(λt)n
Note that 0 e−λt n! dt = λ1 P(Zn,λ ≤ z), where Zn,λ denotes a random variable with a
Gamma(n, λ) distribution. Then the quantity Tb(δ; x) = TbX (1) (δ; x) for X (1) equals
1 X
P(Zn,λ ≤ δA(x)) P(Sn ∈ (x − h, x]) .
(9.25)
TbX (1) (δ; x) =
λ
n∈N0
For n ≤ λδA(x) we have P(Zn,λ ≤ δA(x)) ≥ P(Zn,λ ≤ nλ ) → 12 as n → ∞, by the
central limit theorem (recall that Zn,λ ∼ λ1 (Y1 + . . . + Yn ), where Yi are i.i.d. Exp(1)
random variables). Denoting by TS (δ; x) the quantity in (4.1) for the random walk S, and
restricting the sum in (9.25) to n ≤ λδA(x), we get
TbX (1) (δ; x) & TS (λδ; x) .
To prove a reverse inequality, we observe that for all z ≤ 21 nλ we can write, for > 0,
1 n
eλz
e2
λz
−λZn,λ
P(Zn,λ ≤ z) ≤ e E[e
]=
≤
≤ e−cn ,
n
(1 + )
1+
where the last inequality holds with c = c > 0, provided we fix > 0 small. Then, splitting
the sum in (9.25) according to n ≤ 2λδA(x) and n > 2λδA(x), we get
X
X
1
TbX (1) (δ; x) ≤
P(Sn ∈ (x − h, x]) +
e−cn . TS (2λδ; x) + e−2cλδA(x) .
λ
n≤2λδA(x)
n>2λδA(x)
These inequalities show that TbX (1) (δ; x) is a.n. if and only if TS (δ; x) is a.n., that is, the
SRT holds for X (1) if and only if it holds for S.
Step 2. Assume that X = X (1) + X (2) and the SRT holds for X (1) , that is TbX (1) (δ; x) is
a.n.. Then, given ε > 0, there are δ0 , x0 such that, for all 0 < δ < δ0 ,
Z δA(y)
A(y)
(1)
.
(9.26)
∀y > x0 :
TbX (1) (δ; y) =
P(Xt ∈ (y − h, y]) dt ≤ ε
y
0
Let us now write
Z δA(x)
(2)
P(Xt ∈ (x − h, x], Xt ≤ x/2) dt
0
Z
x/2
=
−∞
(2)
P(Xt
δA(x)
Z
∈ dz)
0
(1)
P(Xt
∈ (x − z − h, x − z]) dt .
For z ≤ x/2 we can write A(x) ≤ cA(x/2), for any c > 2α and for large x. Then the inner
A(x)
integral is bounded by TbX (1) (cδ; x − z) ≤ ε A(x−z)
x−z . ε x , by (9.26). This shows that
Z δA(x)
A(x)
(2)
P(Xt ∈ (x − h, x], Xt ≤ x/2) dt . ε
.
(9.27)
x
0
Note that X (2) has finite exponential moments, because its Lévy measure Π( · ∩(−1, 1)) is
(2)
(2)
(2)
compactly supported, hence E[e|Xt | ] ≤ E[eXt ] + E[e−Xt ] ≤ ect for a suitable c ∈ (0, ∞).
THE STRONG RENEWAL THEOREM
41
(2)
This yields the exponential bound P(|Xt | > a) ≤ e−a ect , for all a ≥ 0, hence
Z δA(x)
(2)
P(Xt ∈ (x − h, x], Xt > x/2) dt
0
δA(x)
Z
≤
0
(2)
P(Xt
> x/2) dt . e
−x/2 cδA(x)
e
=
x→∞
o
A(x)
x
.
Together with (9.27), this shows that TbX (δ; x) is a.n., that is the SRT holds for X.
If the SRT holds for X, to show that it holds for X (1) we can repeat the previous
arguments switching X and X (1) (no special feature of X (1) was used in this step).
Appendix A. Some technical results
The next Lemmas show some relations between the quantities Ik and I˜k defined in (1.18),
(1.19) and in (1.23), (1.24), respectively.
Lemma A.1. Let F be a probability on R satisfying (1.12) with α ∈ (0, 1) and with p, q > 0.
Fix k ∈ N with k ≥ 2 and η ∈ (0, 1). For x ≥ 0, the following relations hold:
if k <
if k >
Corollary A.2. If
1
α
1
α
1
α
−1:
Ik−1 (δ, η; x) .η Ik (δ, η; x) ,
(A.1)
−1:
Ik (δ, η; x) .η Ik−1 (δ, η; x) .
(A.2)
6∈ N, Ikα (δ, η; x) is a.n. implies Ik (δ, η; x) is a.n. for all k ∈ N.
It follows by (A.1) and (A.2) that
max Ik−1 (δ, η; x), Ik (δ, η; x) ≈η
(
Ik (δ, η; x)
Ik−1 (δ, η; x)
if k <
if k >
1
α
1
α
−1
−1
.
(A.3)
Lemma A.3. Let F be a probability on R satisfying (1.12) with α ∈ (0, 1) and with p, q > 0.
For x ≥ 0, the following relations hold:
I1 ( 2δ ; x) . I˜1 (δ; x) ,
if α < 21 :
I˜1 (δ; x) . I1 (δ; x) .
Fix now k ∈ N with k ≥ 2 and η ∈ (0, 1). For x ≥ 0, the following relations hold:
max Ik−1 (δ, η; x), Ik (δ, η; x) .η I˜k (δ, η; x) ,
if k 6= 1 − 1 :
I˜k (δ, η; x) .η max Ik−1 (δ, η; x), Ik (δ, η; x) .
α
(A.4)
(A.5)
(A.6)
(A.7)
It follows that
if k ≤
if k >
where we stress that k =
1
α
1
α
1
α
−1:
−1:
I˜k−1 (δ, η; x) .η I˜k (δ, η; x) ,
I˜k ( δ , η; x) .η I˜k−1 (δ, η; x) ,
2
(A.8)
(A.9)
− 1 is included in (A.8).
Corollary A.4. I˜κα (δ, η; x) is a.n. implies I˜k (δ, η; x) is a.n. for all k ∈ N.
Corollary A.5. If α1 6∈ N, I˜κα (δ, η; x) is a.n. if and only if Iκα (δ, η; x) is a.n..
If α1 ∈ N, I˜κα (δ, η; x) is a.n. implies Iκα (δ, η; x) is a.n. (but not the other way round).
42
FRANCESCO CARAVENNA AND RON DONEY
A.1. Proof of Lemma A.1. For k < α1 − 1, the function bk+1 (y), cf. (1.10), is regularly
varying with index (k + 1)α − 1 < 0, hence it is asymptotically decreasing: bounding
bk+1 (yk ) & bk+1 (ηyk−1 ) for |yk | ≤ η|yk−1 | gives
Z
F (−yk−1 + dyk ) bk+1 (yk ) & F (−(1 − η)|yk−1 |) bk+1 (ηyk−1 ) &η bk (yk−1 ) ,
|yk |≤η|yk−1 |
which plugged into (1.19) shows that Ik (δ, η; x) &η Ik−1 (δ, η; x), proving (A.1). When
1
, the function bk+1 (y) is asymptotically increasing, and we use a similar argument
α > k+1
to get Ik (δ, η; x) .η Ik−1 (δ, η; x), that is (A.2).
A.2. Proof of Corollary A.2. If α1 6∈ N, then κα < α1 − 1, hence if Iκα is a.n. also
Iκα −1 , Iκα −2 , . . . are a.n., by (A.1), while Iκα +1 , Iκα +2 , . . . are a.n., by (A.2).
A.3. Proof of Lemma A.3. By (5.9), for |yk | ≤ η|yk−1 | with η ∈ (0, 1) we can write
A(|yk |/η)
b̃k+1 (yk−1 , yk ) ≥
X
m=A(|yk |)
A(|yk |)k
mk
&
1 + A(|yk |/η) − A(|yk |) &η bk+1 (yk ) ,
am
(|yk |/η) ∨ 1
The same arguments show that, for |y| ≤ 2δ x, we have b̃2 (δx, y) ≥ b̃2 (2y, y) & b2 (y).
Plugging these bounds into (1.23) and (1.24) proves (A.4) and also I˜k (δ, η; x) &η Ik (δ, η; x),
which is half of (A.6). For the other half, note that for |yk | ≤ η|yk−1 |, always by (5.9),
A(|yk−1 |)
b̃k+1 (yk−1 , yk ) &
X
m=A(η|yk−1 |)
A(η|yk−1 |)k
mk
&
1+A(|yk−1 |)−A(η|yk−1 |) &η bk+1 (yk−1 ) ,
am
|yk−1 | ∨ 1
hence
Z
F (−yk−1 + dyk ) b̃k+1 (yk−1 , yk ) &η bk+1 (yk−1 )F (−(1 − η)|yk−1 |) &η bk (yk−1 ) .
|yk |≤η|yk−1 |
(A.10)
From (1.23) we get I˜k (δ, η; x) &η Ik−1 (δ, η; x), which completes the proof of (A.6).
Next we prove (A.5) and (A.7). Recalling (A.3), we distinguish two cases.
• If k < α1 − 1, the sequence mk /am is regularly varying with index k −
(2.4), we can write
b̃k+1 (yk−1 , yk ) ≤
∞
X
m=A(|yk |)
1
α
< −1. By
mk
A(|yk |)k+1
.
= bk+1 (yk ) ,
am
|yk | ∨ 1
which yields I˜k (δ, η; x) .η Ik (δ, η; x). For k = 1, we have proved (A.5), since k <
means precisely α < 12 , while for k ≥ 2 we have proved half of (A.7).
1
α −1
• If k > α1 − 1, with k ≥ 2, the sequence mk /am is regularly varying with index
k − α1 > −1 and by (2.3) we get
A(|yk−1 |)
b̃k+1 (yk−1 , yk ) ≤
X
m=1
mk
A(|yk−1 |)k+1
.
= bk+1 (yk−1 ) ,
am
|yk−1 | ∨ 1
and in analogy with (A.10) we get I˜k (δ, η; x) .η Ik−1 (δ, η; x). Relation (A.7) is proved.
THE STRONG RENEWAL THEOREM
43
We finally prove (A.8) and (A.9). Fix k ≤ α1 − 1 and assume first that k ≥ 3. By (A.7)
and (A.3) (with k − 1 in place of k; note that k − 1 < α1 − 1), and then by (A.6), we have
I˜k−1 .η max{Ik−2 , Ik−1 } ≈η Ik−1 ≤ max{Ik−1 , Ik } .η I˜k .
If k = 2, the assumption k ≤ α1 − 1 means α ≤ 13 , hence we can apply (A.5) followed by
(A.6) to get I˜1 . I1 ≤ max{I1 , I2 } .η I˜2 . This completes the proof of (A.8).
Fix now k > α1 − 1 with k ≥ 2. By (A.7) and (A.3), we can write
I˜k ( 2δ ) .η max{Ik−1 ( 2δ ), Ik ( 2δ )} .η Ik−1 ( 2δ ) .
If k ≥ 3, we apply (A.6) with k − 1 in place of k, to get
Ik−1 ( 2δ ) ≤ max{Ik−2 ( 2δ ), Ik−1 ( 2δ )} .η I˜k−1 ( 2δ ) ≤ I˜k−1 (δ) .
This yields I˜k ( 2δ ) .η I˜k−1 (δ), which is precisely (A.9). If k = 2, we apply (A.4) to see that
Ik−1 ( 2δ ) = I1 ( 2δ ) ≤ I˜1 (δ). This completes the proof of (A.9).
A.4. Proof of Corollary A.4. Assume that I˜κα is a.n.. Since κα ≤ α1 − 1, we can apply
(A.8) iteratively to see that I˜κα −1 , I˜κα −2 , . . . are a.n.. Similarly, since κα + 1 > α1 − 1,
relation (A.9) shows that I˜κα +1 , I˜κα +2 , . . . are a.n..
A.5. Proof of Corollary A.5. With no restriction on α, if I˜κα is a.n. then Iκα is also
a.n., by (A.4) and (A.6). The reverse implication holds when α1 6∈ N, because we can apply
(A.5) if κα = 1 (in which case α < 21 , since α1 6∈ N) or (A.7) if κα > 1.
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44
FRANCESCO CARAVENNA AND RON DONEY
Dipartimento di Matematica e Applicazioni, Università degli Studi di Milano-Bicocca, via
Cozzi 55, 20125 Milano, Italy
E-mail address: [email protected]
School of Mathematics, University of Manchester, Oxford Road, Manchester M13 9PL,
UK
E-mail address: [email protected]