Persistence of iterated partial sums

0.15 MB·

by  Amir Dembo

Persistence of iterated partial sums
Amir Dembo∗ Fuchang Gao†
February 1, 2011
1
1
0
2
Abstract

n
a Let S(2) denote the iterated partial sums. That is, S(2) = S +S + +S , where
n n 1 2 n
J ···
S =X +X + +X . AssumingX ,X ,...,X areintegrable,zero-mean,i.i.d. random
0 vairiable1s, we2sh·o·w· thait the persisten1ce p2robabilitnies
3

PR] p(n2) :=P(cid:18)1m≤ia≤xnSi(2) <0(cid:19)≤cs(nE+|S1n)E+1|X| 1|,
.
h withc 6√30(andc=2wheneverX1issymmetric). Furthermore,theconverseinequality
t holds w≤henever P( X > t) e−αt for some α > 0 or P( X > t)1/t 0 as t .
a − 1 ≍ − 1 → → ∞
m Consequently,for these randomvariableswe havethatp(2) n−1/4 ifX has finite second
n 1
≍
[ moment. In contrast, we show that for any 0 <γ <1/4 there exist integrable, zero-mean
1 random variables for which the rate of decay of p(n2) is n−γ.
v
3
1 Introduction
4
7
5 The estimation of probabilities of rare events is one of the central themes of research in the
.
1 theory of probability. Of particular note are persistence probabilities, formulated as
0
1
1 qn = P max Yk < y , (1.1)
: 1≤k≤n
v (cid:16) (cid:17)
i where Y n isasequenceofzero-meanrandomvariables. ForindependentY thepersistence
X { k}k=1 i
probabilityiseasilydeterminedtobetheproductofP(Y < y)andtoalargeextentthisextends
r k
a to the case of sufficiently weakly dependent and similarly distributed Y , where typically q
i n
decays exponentially in n. In contrast, in the classical case of partial sums, namely Y = S =
k k
k X with X i.i.d. zero-mean random variables, it is well known that q = O(n−1/2)
i=1 i { j} n
decays as a power law. This seems to be one of the very few cases in which a power law decay
P
for q can be proved and its exponent is explicitly known. Indeed, within the large class of
n
similar problems where dependencebetween Y is strong enough to rule out exponential decay,
i
thebehaviorof q is very sensitive totheprecisestructureof dependencebetween thevariables
n
Y and even merely determining its asymptotic rate can be very challenging (for example, see
i
[3]forrecentresultsincaseY = n X (1 c )i arethevaluesofarandomKacpolynomials
k i=1 i − k,n
evaluated at certain non-random c ).
k,n
P{ }
∗The research was supported in part by NSFgrant DMS-0806211.
†Thisresearch waspartially donewhenthesecondnamedauthorvisitedStanforduniversityinMarch2010.
The authors also thank NSFgrant DMS-0806211 for thesupport.
1
We focus here on iterated sums of i.i.d. squared integrable, zero-mean, random variables
X . That is, with S = n X and
{ i} n k=1 k
P n n
S(2) = S = (n i+1)X , (1.2)
n k − i
k=1 i=1
X X
we are interested in the asymptotics as n of the persistence probabilities
→ ∞
p(2)(y) := P max S(2) < y , p(2)(y) := P max S(2) y , (1.3)
n 1≤k≤n k n 1≤k≤n k ≤
(cid:18) (cid:19) (cid:18) (cid:19)
where y 0 is independent of n. With y n it immediately follows from Lindeberg’s clt
≥ ≪
(2)
that p (y) 0 as n and our goal is thus to find a sharp rate for this decay to zero.
n
→ → ∞
(2) (2) (2)
Note that for any fixed y > 0 we have that p (y) p (y) p (0) up to a constant
n n n
∼ ∼
depending only on y. Indeed, because EX− > 0, clearly P(X < ε) > 0 for ε = y/k and
1
−
some integer k 1. Now, for any n 1 and z 0,
≥ ≥ ≥
p(2)(z) p(2)(z) P(X < ε)p(2) (z+ε) P(X < ε)p(2)(z+ε)
n ≥ n ≥ n − n−1 ≥ 1 − n
and applying this inequality for z = iε, i= 0,1,...,k 1 we conclude that
−
p(2)(0) [P(X < ε)]kp(2)(y). (1.4)
n ≥ 1 − n
(2) (2) (2)
Of course, we also have the complementary trivial relations p (0) p (0) p (y)
n n n
≤ ≤ ≤
(2) (2) (2) (2)
p (y), so it suffices to consider only p (0) and p (0) which we denote hereafter by p and
n n n n
(2) (2) (2) (2) (2)
p , respectively. Obviously, p and p have the same order (with p = p whenever X
n n n n n 1
has a density), and we consider both only in order to draw the reader’s attention to potential
(2) (2)
identities connecting the two sequences p and p .
n n
{ } { }
(2)
Persistence probabilities such as p appearin many applications. For example, theprecise
n
problemweconsiderherearisesinthestudyoftheso-called sticky particlesystems(c.f. [9]and
the references therein). In case of standard normal X it is also related to entropic repulsion
i
for 2-Gaussian fields (c.f. [2] and the references therein), though here we consider the easiest
∇
version, namely a one dimensional 2-Gaussian field. In his 1992 seminal paper, Sinai [8]
∇
proved that if P(X = 1) = P(X = 1) = 1/2, then p(2) n−1/4. However, his method
1 1 n
− ≍
(2)
relies on the fact that for Bernoulli X all local minima of k S correspond to values
{ k} 7→ k
of k where S = 0, and as such form a sequence of regeneration times. For this reason,
k
Sinai’s method can not be readily extended to most other distributions. Using a different
approach, more recently Vysotsky [10] managed to extend Sinai’s result that p(2) n−1/4 to
n
≍
X which are double-sided exponential, and a few other special types of random walks. At
i
about the same time, Aurzada and Dereich [1] used strong approximation techniques to prove
the bounds n−1/4(logn)−4 . p(2) . n−1/4(logn)4 for zero-mean random variables X such
n i
{ }
that E[eβ|X1|] < for some β > 0. However, even for X which are standard normal variables
i
∞
it was not known before whether these logarithmic terms are needed, and if not, how to get
rid of them. Our main result, stated below, resolves this question, at least when the lower tail
of X− = min(X ,0) decays exponentially:
1 − 1
Decay Assumption: We assume that there exist constants r > 0, θ > 1/r and finite K, L,
such that for all t >0,s > 0,
P( X > t+s) KP( X > t)P( X > s)+L[P( X > r)]θ(t+s). (1.5)
1 1 1 1
− ≤ − − −
2
Remark 1.1. With β = limsup 1 logP( X > t), it is easy to check that our condition
− t→∞ t − 1
holds (with K = 0), whenever P( X > r) > e−βr for some r > 0. Obviously this applies
1
−
whenever X decays super-exponentially, that is, for β = . Conversely, considering K =
1
− ∞
L > 0 and t = s= 2ℓr, ℓ = 0,1,... it is not hard to show that some exponential decay, namely
β > 0, is necessary for our decay assumption. In the borderline case of an exponential decay,
that is, when β > 0 is finite, our condition holds also whenever P( X > t) e−βt (taking
1
− ≍
now L = 0 and K > 0, with K = 1 corresponding to X− having a New Better than Used
1
distribution).
In this paper, we will prove:
Theorem 1.2. For i.i.d. X of zero mean and E X < , let S(2) = S +S + +S ,
k 1 n 1 2 n
{ } | | ∞ ···
where S = X +X + +X . Then,
i 1 2 i
···
n E S
p(2)p(2) c2 | n+1|, (1.6)
k n−k ≤ 1 E X
1
k=0 | |
X
where c 6√30, and c = 2 if X is symmetric. Furthermore, if X− satisfies the decay
1 ≤ 1 1 1
assumption (1.5), the converse inequality
n 1 E S
(2) (2) n+1
p p | | (1.7)
k n−k ≥ c E X
2 1
k=0 | |
X
holds for some finite c =c (K,L,θ,r). Taken together, these bounds imply that
2 2
1 E S E S
| n+1| p(2) c | n+1| . (1.8)
4c1c2s(n+1)E X1 ≤ n ≤ 1s(n+1)E X1
| | | |
Remark 1.3. If X has finite second moment, then by the clt, there exist finite constants
1
C > C > 0 such that C √n E S C √n for all n 1. Consequently, we then have
2 1 2 n 1
≥ | | ≥ ≥
p(2) n−1/4 under the decay assumption of Theorem 1.2. In contrast, for any 0 < γ < 1/4
n
≍
there exists integrable, zero-mean variable X for which p(2) n−γ. Indeed, considering P(Y >
1 n 1
≍
y)= y−α1 with 1< α < 2, our decay assumption applies for the bounded below, zero-mean,
y≥1
integrable random variable X = Y EY . Setting a = n1/α, clearly nP(X > a x) x−α
1 1 1 n 1 n
− | | →
as n , hence a−1S b converges in distribution to a zero-mean, one-sided Stable
→ ∞ n n − n α
variable Z , and it is further easy to check that b = a−1nE[X 1 ] b = EY . In
α n n 1 |X1|≤an → ∞ − 1
fact, it is not hard to verify that a−1S is a uniformly integrable sequence and consequently
{ n n}
n−1/αE S E Z EY finite and positive. From Theorem 1.2 we then deduce that p(2)
n α 1 n
| | → | − | ≍
n−γ for γ = (1 1/α)/2.
−
(2)
The sequences S and S are special cases of the class of auto-regressive processes
{ k} { k }
Y = L a Y + X with zero initial conditions, i.e. Y 0 when k 0 (where S
k ℓ=1 ℓ k−ℓ k k ≡ ≤ k
(2)
corresponds to L = a =1 and S corresponds to L = a =2, a = 1). While each of these
P 1 k 1 2 −
stochastic processes is a time-homogeneous Markov chain of state space R and q = P(τ > n)
n
is merely the upper tail of the first hitting time τ of [y, ) by the relevant chain, the general
∞
theory of Markov chains does not provide the precise decay of q , which even in case L = 1
n
ranges from exponential decay for a > 0 small enough, via the O(n−1/2) decay for a = 1 to
1
a constant n q in the limit a . While we do not pursue this here, we believe that
n 1
7→ ↑ ∞
3
the approach we develop for proving Theorem 1.2 can potentially determine the asymptotic
behavior of q for a large collection of auto-regressive processes. This is of much interest, since
n
for example, as shown in [5], the asymptotic tail probability that random Kac polynomials
have no (or few) real roots is determined in terms of the limit as r of the power law tail
→ ∞
decay exponents for the iterates S(r) = k S(r−1), r 3.
k i=1 i ≥
Ourapproachfurthersuggeststhattheremightbesomeidentitiesconnectingthesequences
P
(2) (2)
p and p . Note that, if we denote
n n
{ } { }
p(1) = P max S < 0 , p(1) = P max S 0 ,
n 1≤k≤n k n 1≤k≤n k ≤
(cid:18) (cid:19) (cid:18) (cid:19)
(1)
then as we show in Section 2 there are indeed identities connecting the sequences p and
n
{ }
p(1) . As we mentioned earlier, it is well-known that p(1) is of the order n−1/2. The next
n n
{ }
(1)
proposition provides the exact value of p for symmetric random variables with a density,
n
and the elegant argument of its proof serves as the starting point of our approach to the study
(2)
of p .
n
Proposition 1.4. If X are mean zero i.i.d. symmetric random variables then for all n 1,
i
≥
(2n 1)!!
p(1) − p(1). (1.9)
n ≤ (2n)!! ≤ n
In particular, if X also has a density, then
1
(2n 1)!!
p(1) = − . (1.10)
n (2n)!!
Remark 1.5. Let B(s) denote a Brownian motion starting at B(0) = 0 and consider the inte-
t
grated Brownian motion Y(t) = B(s)ds. Sinai [8] proved the existence of positive constants
0
A and A such that for any T > 0,
1 2
R
A T−1/4 P sup Y(t) 1 A T−1/4. (1.11)
1 2
≤ ≤ ≤
t∈[0,T]
(cid:16) (cid:17)
Upon setting ε = T−3/2 and t = uT, by Brownian motion scaling this is equivalent up to a
constant to the following result that can be derived from an implicit formula of McKean [7]
(c.f. [4]):
3Γ(5/4)
lim ε−1/6P sup Y(u) ε = .
ε→0+ u∈[0,1] ≤ 4π 2√2π
(cid:16) (cid:17)
Since the iterated partial sums S(2) corresponding to i.i.d stpandard normal random variables
n
X , forms a discretization of Y(t), the right-most inequality in (1.11) readily follows from
i
{ }
Theorem 1.2. Indeed, with E[Y(k)Y(m)] = k2(3m k)/6 and E[S(2)S(2)] = k(k+1)(3m k+
− k m −
1)/6 for m k, setting Z(k) = (1+1/k)(1+1/(2k)Y(k), results with E[(S(2))2]= E[Z(k)2]
≥ k
and it is further not hard to show that f(m,k) := E[S(2)S(2)]/E[Z(m)Z(k)] 1 for all m = k
p m k ≥ 6
(as f(k+1,k) 1 and df(m,k)/dm > 0 for any m k+1). Thus, by Slepian’s lemma, we
≥ ≥
have that for any y
P max Z(k)< y p(2)(y),
1≤k≤n ≤ n
(cid:16) (cid:17)
4
and setting n as the integer part of T 1 it follows that
≥
P sup Y(t) 1 P max Y(k) 1 P max Z(k)< 2 p(2)(2).
t∈[0,T] ≤ ≤ 1≤k≤n ≤ ≤ 1≤k≤n ≤ n
(cid:16) (cid:17) (cid:16) (cid:17) (cid:16) (cid:17)
(2) (2)
Since p (2) cp for some finite constant c and all n, we conclude from Theorem 1.2 that
n n
≤
P sup Y(t) 1 2c(n+1)−1/4 2cT−1/4.
≤ ≤ ≤
t∈[0,T]
(cid:16) (cid:17)
2 Proof of Proposition 1.4
Setting S = 0 let M = max S and consider the 0,1,2,...,n -valued random variable
0 n 0≤j≤n j
{ }
= min l 0: S = M .
l n
N { ≥ }
For each k = 1,2,...,n 1 we have that
−
= k = X > 0,X +X > 0,...,X +X + +X > 0;
k k k−1 k k−1 1
{N } { ···
X 0,X +X 0,...,X +X + +X 0 .
k+1 k+1 k+2 k+1 k+2 n
≤ ≤ ··· ≤ }
By the independence of X , the latter identity implies that
i
{ }
P( = k)= P(X >0,X +X > 0,...,X +X + +X > 0)
k k k−1 k k−1 1
N ···
P(X 0,X +X 0,...,X +X + +X 0)
k+1 k+1 k+2 k+1 k+2 n
× ≤ ≤ ··· ≤
(1) (1)
= p p ,
k n−k
where the last equality follows from our assumptions that X are i.i.d symmetric random
i
variables. Also note that P( = 0) = p(1) and
n
N
P( = n) = P(X > 0,X +X > 0,...,X +X + +X > 0) = p(1).
N n n n−1 n n−1 ··· 1 n
(1) (1)
Thus, setting p = p = 1 we arrive at the identity
0 0
n n
p(1)p(1) = P( = k) = 1, (2.1)
k n−k N
k=0 k=0
X X
holding for all n 0.
≥
Fixing x [0,1), upon multiplying (2.1) by xn and summing over n 0, we arrive at
∈ ≥
P(x)P(x) = 1 , where P(x) = ∞ p(1)xk and P(x) = ∞ p(1)xk. Now, if X also has
1−x k=0 k k=0 k 1
a density then p(1) = p(1) for all k and so by the preceding P(x) = P(x) = (1 x)−1/2.
k k P P −
Consequently, p(1) is merely the coefficient of xn in the Taylor expansion at x = 0 of the
n
function (1 x)−1/2, from which we immediately deduce the identity (1.10).
−
If X does not have a density, let Y be i.i.d. standard normal random variables, inde-
1 i
pendent of the sequence X and den{ote} by S and S˜ the partial sums of X and Y ,
i k k i i
{ } { } { }
respectively. Note that for any ε > 0, each of the i.i.d. variables X +εY is symmetric and
i i
5
has a density, with the corresponding partial sums being S +εS˜ . Hence, for any δ > 0 we
k k
have that
P max S < δ P max (S +εS˜ ) 0 +P max εS˜ δ
k k k k
1≤k≤n − ≤ 1≤k≤n ≤ 1≤k≤n ≥
(cid:16) (cid:17) (cid:16) (cid:17) (cid:16) (cid:17)
(2n 1)!!
= − +P max εS˜ δ .
k
(2n)!! 1≤k≤n ≥
(cid:16) (cid:17)
Taking first ε 0 followed by δ 0, we conclude that
↓ ↓
(2n 1)!!
P max S < 0 − ,
k
1≤k≤n ≤ (2n)!!
(cid:16) (cid:17)
and a similar argument works for the remaining inequality in (1.9).
Remark 2.1. The argument of the next section allows us to modify this proof and deduce order
of (n+1)−1/2 upper and lower bounds for p(1) even in the non-symmetric case. However, since
n
this result is already known, we shall not do so here.
3 Proof of Theorem 1.2
By otherwise considering X /E X , we assume without loss of generality that E X = 1. To
i i 1
| | | |
(2)
adapt the method of Section 2 for dealing with the iterated partial sums S , we introduce the
n
parameter t R and consider the iterates S(2)(t) = S (t)+ +S (t), j 0, of the translated
∈ j 0 ··· j ≥
(1) (2) (2)
partial sums S (t) = t+S , k 0. That is, S (t) = (j +1)t+S for each j 0.
k k ≥ j j ≥
Having fixed the value of t, we define the following 0,1,2,...,n -valued random variable
{ }
(2) (2)
= min l 0 :S (t)= max S (t) .
Kt ≥ l 0≤j≤n j
n o
Then, for each k = 2,3,...,n 2, we have the identity
−
= k = S (t)> 0,S (t)+S (t) > 0,...,S (t)+S (t)+ +S (t)> 0;
t k k k−1 k k−1 1
{K } { ···
S (t) 0,S (t)+S (t) 0,...,S (t)+S (t)+ +S (t) 0
k+1 k+1 k+2 k+1 k+2 n
≤ ≤ ··· ≤ }
= S (t)> 0;X < 2S (t),...,(k 1)X + +X < kS (t) S (t) 0
k k k k 2 k k+1
{ − ··· }∩{ ≤ }
X 2S (t),...,(n k 1)X + +X (n k)S (t) .
k+2 k+1 k+2 n k+1
∩{ ≤ − − − ··· ≤ − − }
Next, for 2 k n we define Y σ(X ,...,X ) and Y σ(X ,...,X ) such that
k,2 2 k k,n k n
≤ ≤ ∈ ∈
X 2X +X (k 1)X + +X
k k k−1 k 2
Y = max , ,..., − ··· ,
k,2
2 3 k
(cid:26) (cid:27)
X 2X +X (n k+1)X + +X
k k k+1 k n
Y = max , ,..., − ··· .
k,n
2 3 n k+2
(cid:26) − (cid:27)
It is then not hard to verify that the preceding identities translate into
= k = S (t) > 0 S (t) Y < S (t) Y S (t) (3.1)
t k k+1 k,2 k k+2,n k+1
{K } { ≥ }∩{ }∩{ ≤ − }
= S +(Y )+ < t X S (Y )+ (3.2)
k k,2 k+1 k k+2,n
{− ≤ − − − }
6
holding for each k = 2,...,n 2. Further, for k = 1 and k = n 1 we have that
− −
= 1 = S (t)> 0 S (t) 0 Y S (t) ,
t 1 2 3,n 2
{K } { }∩{ ≤ }∩{ ≤ − }
= n 1 = S (t) > 0 Y < S (t) S (t) 0 ,
t n−1 n−1,2 n−1 n
{K − } { }∩{ }∩{ ≤ }
so upon setting Y = Y = , the identities (3.1) and (3.2) extend to all 1 k n 1.
1,2 n+1,n
−∞ ≤ ≤ −
For the remaining cases, that is, for k = 0 and k = n, we have instead that
= 0 = t X (Y )+ , (3.3)
t 1 2,n
{K } { ≤ − − }
= n = S +(Y )+ < t . (3.4)
t n n,2
{K } {− }
In contrast with the proof of Proposition 1.4, here we have events (Y )+ < S (t) and
k,2 k
{ }
(Y )+ S (t) that are linked through S (t) and consequently not independent of
k+2,n k+1 k
{ ≤ − }
each other. Our goal is to unhook this relation and in fact the parameter t was introduced
precisely for this purpose.
3.1 Upper bound
For any integer n> 1, let
A = max S , B = max S .
n k+1 n k
1≤k≤n{− } −1≤k≤n{ }
By definition A B . Further, for any 1 k n 1, from (3.1) we have that the
n n
≥ ≤ ≤ −
event = k implies that S (t) > 0 S (t) = S < t S and hence that
t k k+1 k k+1
{K } { ≥ } {− ≤ − }
B < t A . From (3.2) we also see that for any 1 k n 1,
n−1 n−1
{ ≤ } ≤ ≤ −
1 dt (X )−1 1
R {Kt=k} ≥ k+1 {Yk,2<0} {Yk+2,n≤0}
Z
and consequently,
n−1
A B = 1 dt 1 dt (3.5)
n−1− n−1 R {Bn−1<t≤An−1} ≥ R {Kt=k}
Z k=1Z
X
n−1
(X )−1 1 .
≥ k+1 {Yk,2<0} {Yk+2,n≤0}
k=1
X
Taking the expectation of both sides we deduce from the mutual independence of Y , X
k,2 k+1
and Y that
k+2,n
n−1
E[A B ] E[(X )−]P(Y < 0)P(Y 0).
n−1 n−1 k+1 k,2 k+2,n
− ≥ ≤
k=1
X
Next, observe that since the sequence X has an exchangeable law,
i
{ }
P(Y < 0) = P(X < 0,2X +X <0,...,(k 1)X + +X < 0)
k,2 k k k−1 k 2
− ···
= P(X < 0,2X +X < 0,...,(k 1)X + +X < 0) = p(2) . (3.6)
1 1 2 − 1 ··· k−1 k−1
7
Similarly, P(Y 0) = p(2) . With X having zero mean, we have that E[(X )−]=
k+2,n ≤ n−1−k k+1 k+1
E[(X )+] = 1/2 (by our assumption that E X = E X = 1). Consequently, for any
k+1 k+1 1
| | | |
n > 2,
n−1 n−2
1 1
E[A B ] p(2) p(2) = p(2)p(2) .
n−1− n−1 ≥ 2 k−1 n−1−k 2 k n−2−k
k=1 k=0
X X
With E[S ]= 0 and X exchangeable, we clearly have that
n+1 k
{ }
E[A B ] = E[ max S ]+E[ max S S ] = 2E[ max S ]. (3.7)
n n k n+1 k+1 k
− 1≤k≤n 1≤k≤n{ − } 1≤k≤n
Recall Ottaviani’s maximal inequality that for a symmetric random walk P(maxn S t)
k=1 k ≥ ≤
2P(S t) for any n,t 0, hence in this case
n
≥ ≥
∞
E[ max S ] 2 P(S t)dt = E S .
k n n
1≤k≤n ≤ Z0 ≥ | |
To deal with the general case, we replace Ottaviani’s maximal inequality by Montgomery-
Smith’s inequality
P( max S t) 3 max P(S t/3) 9P(S t/30)
k k n
1≤k≤n| |≥ ≤ 1≤k≤n | | ≥ ≤ | | ≥
(see [6]), from which we deduce that
∞
E[ max S ] 9 P(S t/30)dt = 270E S (3.8)
k n n
1≤k≤n ≤ Z0 | |≥ | |
(2) (2) (2)
and thereby get (1.6). Finally, since n p is non-increasing and p p , the upper
n n n
7→ ≤
bound of (1.8) is an immediate consequence of (1.6).
3.2 Lower bound
Turning to obtain the lower bound, let
m := X (Y )+, M := S +(Y )+.
n 1 2,n n n n,2
− − −
Note that for any n 2,
≥
1 1
Y +Y [(n 1)X + +X ]+ [(n 1)X + +X ] = S X ,
n,2 2,n n 2 2 n n 1
≥ n − ··· n − ··· −
and consequently,
M m X S +(Y +Y )+ (X S )+ = (X + +X )−. (3.9)
n n 1 n 2,n n,2 1 n 2 n
− ≥ − ≥ − ···
In particular, M m . From (3.3) and (3.4) we know that if m < t M then necessarily
n n n n
≥ ≤
1 n 1. Therefore,
t
≤ K ≤ −
n−1
M m = 1 dt 1 dt. (3.10)
n − n R {mn<t≤Mn} ≤ R {Kt=k}
Z k=1Z
X
8
In view of (3.2) we have that for any 1 k n 1,
≤ ≤ −
b := E[ 1 dt] = E X +(Y )+ +(Y )+ − .
k {Kt=k} k+1 k,2 k+2,n
R
Z h i
(cid:0) (cid:1)
By the mutual independence of the three variables on the right side, denoting by F and
k,2
F the distribution functions of (Y )+ and (Y )+, respectively, we thus find that
k+2,n k,2 k+2,n
∞ ∞
b = E[(X +x+y)−]dF (x)dF (y)
k k+1 k,2 k+2,n
Z0 Z0
∞ ∞
= g (x+y)dF (x)dF (y),
k+1 k,2 k+2,n
Z0 Z0
where g (t) = ∞P( X > t+u)du. Since X have identical distributions, g (t) = g (t)
k 0 − k { k} k 1
does not depend on k and we have already seen that g (0) = E[(X )−] = 1/2. Thus, setting
1 1
R
α = logP( X > r), applying our decay assumption (1.5) first for t = x+y, s = u and then
1
− −
for t = x, s = y, we find that
∞
g (x+y) KP( X > x+y)g (0)+L e−θα(x+y+u)du
1 1 1
≤ −
Z0
K2
P( X > x)P( X > y)+L e−θα(x+y)
1 1 1
≤ 2 − −
where L = L[K/2+(θα)−1]. Consequently, we arrive at the bound
1
K2 ∞ ∞
b P( X > x)dF (x) P( X > y)dF (y)
k 1 k,2 1 k+2,n
≤ 2 − −
Z0 Z0
∞
+L e−θαxdF (x)dF (y)
1 k,2 k+2,n
Z0
K2
= P(X +(Y )+ < 0)P(X +(Y )+ < 0)+L E[e−θα(Yk,2)+]E[e−θα(Yk+2,n)+].
k+1 k,2 k+1 k+2,n 1
2
Next, observe that just as we did in deriving the identity (3.6),
P(X +(Y )+ < 0) = P(X < 0,2X +X < 0,...,kX + +X < 0)
k+1 k,2 k+1 k+1 k k+1 2
···
= P(X < 0,2X +X < 0,...,kX + +X < 0) =p(2).
1 1 2 1 ··· k k
By a similar reasoning one verifies that p(2) = P(X +(Y )+ < 0). Now, by exchange-
n−k k+1 k+2,n
ability of X , the sequence k P(Y < z) is non-increasing for any fixed value of z.
k k,2
{ } 7→
Furthermore, if z r > 0 then
≥
P(Y < z r)= P(X < 2(z r),...,(k 1)X + +X < k(z r))
k,2 k k 2
− − − ··· −
P(X < r)P X < 2z,...,(k 2)X + +X < (k 1)z
k k−1 k−1 2
≥ − − ··· −
= P( X > r)P(Y < z) P(X < r)P(Y < z).
1 (cid:0) k−1,2 1 k,2 (cid:1)
− ≥ −
Iterating this inequality for z = jr, j = 1,...,k := x/r we deduce that for any x > 0,
⌈ ⌉
P(Y < x) P(Y < kr) [P( X > r)]−kP(Y < 0) eα(x/r+1)p(2)
k,2 ≤ k,2 ≤ − 1 k,2 ≤ k−1
9
(relying on the identity (3.6) for the right-most inequality). Consequently,
∞
E[e−θα(Yk,2)+]= θα P(0 Y < x)e−θαxdx κp(2) ,
≤ k,2 ≤ k−1
Z0
with κ = eαθ/(θ 1/r) finite. Similarly, we find that for any y > 0,
−
P(Y <y) = p(2) (y) eα(y/r+1)p(2) ,
k+2,n n−k−1 ≤ n−k−1
hence
E[e−θα(Yk+2,n)+] κp(2) .
≤ n−k−1
(2)
Combining all these bounds we have by the monotonicite of k p that
7→ k
K2 c
b p(2)p(2) +L κ2p(2) p(2) 2p(2) p(2) ,
k ≤ 2 k n−k 1 k−1 n−k−1 ≤ 2 k−1 n−k−1
where c = K2+2L κ2 is a finite constant. Thus, considering the expectation of both sides of
2 1
(3.10) we deduce that for any n > 2,
n−1
c
E(M m ) 2 p(2) p(2) .
n− n ≤ 2 k−1 n−k−1
k=1
X
In view of (3.9) we also have that E(M m ) E[(S )−] = 1E S , from which we
n − n ≥ n−1 2 | n−1|
conclude that (1.7) holds for all n 1.
≥
(2) (2)
Turning to lower bound p as in (1.8), recall that n p is non-increasing. Hence,
n n
7→
applying (1.7) for n = 2m+1 and utilizing the previously derived upper bound of (1.8) we
have that
1 m m E S
E S 2 p(2)p(2) 2c p(2) | k+1|
c | 2(m+1)|≤ k m ≤ 1 m k+1
2 r
k=0 k=0
X X
4c p(2) (m+1)E S , (3.11)
≤ 1 m | m+1|
p
whereinthelastinequality weusethefactthatforindependent,zero-mean X ,thesequence
k
{ }
S is a sub-martingle, hence k E S is non-decreasing. This proves the lower bound of
k k
| | 7→ | |
(1.8).
References
[1] Aurzada, F.andDereich, S.,Universality of theasymptotics of theone-sided exit problem
for integrated processes. Preprint (2009).
[2] Caravenna, F. and Deuschel, J.-D., Pinning and wetting transition for (1+1)-dimensional
fields with Laplacian interaction, Ann. Probab. 36 (2008), 2388-2433.
[3] Dembo, A., Poonen, B., Shao, Q. and Zeitouni, O., Random polynomials having few or
no real zeros. J. Amer. Math. Soc. 15 (2002), 857-892.
[4] Devulder, A., Shi, Z. and Simon, T., The lower tail problem for the area of a symmetric
stable process. Unpublished manuscript (2007).
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