Universality of Persistence of Random Polynomials
Abstract
We investigate the probability that a random polynomial with independent, mean-zero and finite variance coefficients has no real zeros. Specifically, we consider a random polynomial of degree with coefficients given by an i.i.d. sequence of mean-zero, variance-1 random variables, multiplied by an -regularly varying sequence for . We show that the probability of no real zeros is asymptotically , where is the persistence exponents of a mean-zero, one-dimensional stationary Gaussian processes with covariance function . Our work generalizes the previous results of Dembo et al. [DPSZ02] and Dembo & Mukherjee [DM15] by removing the requirement of finite moments of all order or Gaussianity. In particular, in the special case , our findings confirm a conjecture by Poonen and Stoll [PS99, Section 9.1] concerning random polynomials with i.i.d. coefficients.
keywords:
[class=MSC]keywords:
m1Department of Statistics, University of Chicago. PG’s research is Partially supported by NSF Grant DMS-2346685. m2Department of Statistics, University of Columbia University. SM’s research is partially supported by NSF Grant DMS-1712037.
and
1 Introduction
In this paper, we consider the random polynomials
where , with i.i.d. random variables and . Hhere is a regularly varying function of order , with . The regularly varying assumption implies we can write for some , and is slowly varying at infinity, i.e.,
We seek to study the decay of the persistence probability, i.e., the probability of having no real zero. To this goals, we introduce the following:
(1.1) |
Since an odd degree polynomial necessarily has a real root, we have for odd. For even, the quantity is anticipated to converge to a constant as . This constant is known as the persistence exponent of the random polynomial . In the scenario when , i.e. are i.i.d. random variables,—Poonen and Stoll [PS99, Section 9.1] conjectured that this constant remains invariant across different distributions of , provided that satisfy the following conditions: (1) mean zero, (2) , and (3) belongs to the domain of attraction of the normal distribution. Essentially, this conjecture suggests that the persistence exponent of is universal within the class of i.i.d. random variables with zero mean and finite variance.
Our main result, outlined below, resolves this conjecture and extends it by demonstrating the universality of persistence exponents for across a significantly broader range of random polynomial classes.
Theorem 1.1.
Fix , and assume with i.i.d. random variables with and . Then we have
where is defined via no-zero crossing probability of a stationary Gaussian process , i.e.,
(1.2) |
where is a mean zero stationary Gaussian process with the following covariance function .
Remark 1.1.
The limit in (1.2) exists by Slepian’s Lemma, on using stationarity and non-negativity of the covariance function (see [DM15, Lem 1.1]). Theorem 1.1 can be extended to the case where are independent but not identically distributed random variables. To establish this generalization, an appropriate uniform integrability condition on is required. However, our proof techniques remain largely unchanged in this scenario. To avoid unnecessary complications, we present and prove the theorem for the case where are identically and independently distributed.
Significant strides have been made in the investigation of persistence in random polynomials, starting with [DPSZ02], where it was shown that when the coefficients of are i.i.d. and possess finite moments of all orders, the persistence exponents of are independent of the distribution of and are expressed as . This seminal result laid the foundation for subsequent research in this field. In [DM15], the persistence probability was examined under the assumption that the coefficients of random polynomials are independent Gaussian variables with mean zero and variances that vary regularly. It was demonstrated therein that the persistence exponent in this context is given by , as stated in Theorem 1.1.
The persistence probabilities of random polynomials and other stochastic systems have also been closely scrutinized in the physics literature; see, for instance, [Maj99, SM07, SM08] for further exploration. In a recent contribution [PS18], the authors computed precisely by evaluating the probability of non-real eigenvalues within the random orthogonal matrix ensemble [KPT+16]. In fact, [PS18] found also an alternative way to arrive at the same exponent by matching the persistence probabilities between a 2-dimensional heat equation started from random initial data (studied in [DM15]) and the Glauber dynamics of a semi-infinite Ising spin chain (studied in [DHP95, DHP96]).
The exploration of the zero-count of random polynomials has a rich historical backdrop. A particularly well-studied scenario involves the coefficients being independent and identically distributed as standard Gaussian random variables. Early investigations, dating back to the work of Littlewood and Offord [LO38, LO39, LO43], yielded both upper and lower bounds for the expected total number of real roots, denoted as . Their contributions also provided insights into the tails of this distribution, offering an initial estimate that scales as when the coefficients follow the distribution .
Kac [Kac43] made a significant stride by deriving the exact formula for the expected value of when ’s are independent and identically distributed as standard normal random variables. Subsequently, over several decades, researchers honed in on the precise asymptotic behavior of , refining and extending Kac’s findings. In a more recent development, as highlighted by [NNV16], it was demonstrated that the asymptotic expectation of equals for the case where follows a distribution with and .
The challenge of determining the probability of a random polynomial having no real roots remained a formidable task until the seminal work of [DPSZ02]. Their work established that for even values of , exhibits polynomial decay, specifically as . They also showed that the probability of to have exact (simple) zeros also behave in the same way. The study of non-identically distributed, but independent coefficients has been considered in [DM15], where the authors verify Theorem 1.1, under the assumption that all coefficients are Gaussian. Theorem 1.1 represents a significant generalization of the findings in [DM15] and [DPSZ02]. It achieves this by removing the requirement that possess finite moments of all orders or follow Gaussian distributions. In other word, it shows universality of the persistence of random polynomials. This pivotal development resolves Poonen and Stole’s conjecture on persistence of random polynomials.
2 Proof Ideas
The study of random polynomials has been a dynamic field, continually yielding fresh insights and relevant methodologies. A significant portion of this research has been devoted to investigating the presence of real roots in random polynomials. In particular, the problem of a random polynomial having no real zeros has attracted widespread attention within the mathematics and physics communities.
Many prior approaches to establish the persistence of such random polynomials have leaned heavily on comparisons with Gaussian distributions, often employing techniques like Komlos-Major-Tusnady embedding and tools such as Slepian’s inequality [DPSZ02, DM15]. However, a notable challenge in establishing the universality of this persistence lies in the absence of KMT-type results for random variables with only finite second moments. Moreover, the challenge becomes even more complex when dealing with non-identically distributed random variables. We seek to resolve this by innovating a new approach for showing the persistence. We divide the proofs into two parts which are given as follows:
(2.1) |
We show and in Sections 3 and 4 respectively. The proof ideas for showing these two claims are different and in what follows, we discuss those proof ideas in details. We divide the proofs of and into two steps. We begin with the steps of the proof of and then proceed to the steps in the proof of . From this point forward, we assume that is an even integer. However, it turns out that the proofs concerning and do not make use of this assumption, with the exception of Propositions 4.5 and 4.4.
Step I of : To establish , we initially bound by considering the probability of having no real zeros for within four small intervals. These intervals are located around two points: around which is associated to the intervals and , and associated to intervals and .
Now, we need to analyze the probability of no real roots of in each of these four intervals. However, it is important to note that the events of having no zeros in these intervals are not independent. To address this challenge, we demonstrate that the probability of having no real zeros of in disjoint intervals are dictated by the disjoint components of . For instance, the probability of no zero of in the interval is mainly governed by and this holds roughly because the variance of is greater than any other portions of when (see Theorem ). Consequently, we can express the probability of having no real zeros in any of these four intervals as approximately the product of the probabilities of having no real zeros in each of these intervals.
(2.2) |
Step II of : Now we need to upper bound the probability of no real zero in the each of the intervals and . We show that probability of no zeros of in the interval behaves as and the probability of no real zeros in the interval behaves as (see (1.2) for the definition of for any ). By symmetry, the probabilities of no real zero in the interval and are respectively and . More concretely, we show the following:
and,
Sum of these yields the desired upper bound following (2.2).
We now explain why the probability of no zeros for the intervals and are different and how we bound those probabilities. Recall that the probability of no real zero of in the interval depends on whereas the probability of no real zero of in is governed by . We further divide the polynomial into almost number of sub-polynomials (see (3.1) for ) of varying sizes. Each of these sub-polynomials (for instance, ) when restricted to a particular sub-interval ( of (3.3) in the case of ) of , converges to a Gaussian process which is directly linked to . These sub-polynomials are constructed in Section 3.1 (see (3.4)). Such construction also ensures that those sub-polynomials when restricted to some other sub-intervals of , the covariance function decays (see Lemma 3.5). As a result, the probability of no real zeros of in any of those sub-intervals of can be bounded by the probability of the respective sub-polynomials not exceeding an arbitrary small negative number (Lemma 3.2). Since all these sub-polynomials are independent of each other and they weakly converges (when restricted to the respective sub-intervals) to the same Gaussian process, we do have
where for any fixed . The constant of proportionality in the above equation is . Taking logarithm in both sides in the above display, dividing by and letting shows that probability of no real zeros in is bounded by . On the other hand, to bound the similar probability for , we divide into almost many sub-polynomials corresponding to index sets (see (3.2)). However, unlike the case, each of these sub-polynomials (for instance, ) when restricted to a particular sub-interval ( of (3.3) in the case of ) of , converges to a Gaussian process which is directly linked to . Now, by a similar argument, we show
where the right hand side of the above display decays as . Combining these probabilities for and yields the proof of . Although the steps to find the exact asymptotics are straightforward, the assumption of only finite second moments enforces careful accounting of probability bounds which is executed mainly in the proof of Lemma 3.2 through employing a combinatorial selection criterion ‘move, flush and repeat’ and the required probability bounds are assimilated from several important lemmas stated and proved in Section 3.2.2.
Step I of : The proof of is carried out by investigating the lower bound of the probability of for all . This latter event can be rewritten as where the intervals (for ) are shown in (4.1) (see Figure 2).
Like as in the proof of , we notice that is dictated by the sub-polynomials (see (4.2) for the definitions of the sets ). From the definition of , it follows that . Such decomposition of implies
where is the standard deviation of the random variable . Since the sub-polynomials are independent, the probability of the event of the right hand side is equal to the product of the the following probabilities;
(2.3) |
The main thrust of the proof of lies in the lower bound of the above probabilities. Indeed, we derive these lower bounds in Proposition 4.1, 4.2, 4.3, 4.5 and 4.4 respectively. We show in Proposition 4.1 and 4.2, that the above probabilities for and are bounded below by the probabilities of no-zero crossing of two Gaussian processes which later is bounded by the non-zero crossing probabilities of the Gaussian processes and respectively (through Lemma 4.22). Thus the probabilities of the above display for and roughly contributes and in the lower bound to the probability of for all . Furthermore, Propositions 4.3, 4.5 and 4.4 shows that the probabilities of the above display corresponding to are bound below by . Assembling these bounds together yields the proof of . In the next step, we discuss the ideas that we use in lower bounding the probabilities in the above display for and . It is important to note that, while the probabilities in (2.3) for do not directly contribute to the exact value of the persistence exponent, the proof of their lower bound (which is of the order ) in Propositions 4.5 and 4.4 relies heavily on the fact that is a polynomial of even degree.
Step II of :
We discuss here strategies to bound (2.3) for . As in the proof of , we divide the intervals and into many intervals. These are now denoted as and respectively. On the other hand, the index sets and are also divided into and (see (4.3) and (4.4)). We show that the probability in (2.3) could be lower bounded by product of certain probabilities of the polynomials when and when . We show below how this bound follows for case:
where and are defined in (4.40), (4.41), (4.42) and (4.43) respectively. In words, signifies the event that (when scaled by ) to be less then a small negative constant for all inside and around a neighborhood of whereas and signifies to be less than positive numbers which decays as moves away . We show that and can be lower bounded by probability of the Gaussian process being smaller a small negative number everywhere inside an interval whose length grows as . This last probability is indeed same as the no-zeros crossing probability of the Gaussian process when the small constant is taken to . We show the above mentioned lower bound by combining Lemma 4.8, Lemma 4.9, Lemma 4.10 with Lemma 4.22. The first three lemma required careful derivation of the decay of probabilities of exceeding a positive constants as moves away from . Since we have assumed only second moments of ’s finite, the derivation of these probabilities poses several challenges which are overcome in Lemma 4.6. Taking the product of the aforementioned lower bound over all shows that the probability in (2.3) for can be bounded from below by the product of many no-zero crossing probability of . Similarly, the probability in (2.3) for can be bounded from below by the product of many no-zero crossing probability of . These explain the bounds obtained in Proposition 4.1 and 4.2 which are the most import components in the proof of .
3 Upper bound
3.1 Fixation of notations
Fix , and let
Define the positive weight function as
Also, define
Fixing , let . Define the sets , by setting
(3.1) |
and use the bound to note that
Similarly, define the sets , where
(3.2) |
and again use to note that
We will now partition a proper subset , for . For , define
(3.3) |
Then we have , for
For set
(3.4) |
We will create a back-log of objects. For , define
(3.5) |
We further define the index set
Let us also define , and set
3.2 Proof of
We claim that it suffices to show that
(3.6) |
Indeed, using the above display for the polynomial , we get
The upper bound follows by combining the above two bounds. We now proceed now to show (3.6). We assume is an even integer. To this effect, first note the trivial upper bound
(3.7) |
Recalling that for (see (3.3)), a union bound gives, for any ,
(3.8) |
where is chosen such that , and the last line follows on noting that for we can write
The r.h.s. of (3.2) can be further bounded above by
(3.9) | ||||
(3.10) |
We now state a lemma to bound the second term in the RHS of (3.9), deferring the proof of the Lemma to the end of the section.
Lemma 3.1.
There exist such that
(3.11) |
Lemma 3.1 is proved in Section 3.3. As a result of Lemma 3.1, the second term in the RHS of (3.9) is bounded above by . Now we proceed to bound the first term. Notice that the events and are jointly independent of and . As a result, we can write the first term in the RHS of (3.9)
where . To bound the RHS of the above display, by union bound, we write
(3.12) |
where . Notice that is upper bounded by . So the number of terms in the above sums are finite. We now proceed to bound the two terms in the right hand side of (3.12).
Lemma 3.2.
(i) For any , there exists a positive integer (depending on ) such that for all we have
(3.13) |
Here are i.i.d. centered Gaussian process with covariance Definition 3.3.
(ii) For any , there exists a positive integer (depending on ) and an absolute constant such that for all and we have
(3.14) |
where if , and if . Here denotes where and are defined as follows
and they satisfy the relation
(3.15) |
Before proceeding to the proof of Lemma 3.2 we first complete the proof of the upper bound. Owing to (3.2), (3.9), (3.12), (3.13) and (3.2) and Lemma 3.1, we obtain
where
Taking logarithm on both sides and using the inequality to bound the logarithm of the right hand sides of the above display yields,
(3.16) |
Now we divide both sides of (3.16) by . Notice that
(3.17) | ||||
(3.18) | ||||
Since is arbitrarily small when gets large and furthermore, , we have
The second to the last inequality follows since and are independent Gaussian processes, and, furthermore, as we show below, the covariance functions of and are bounded respectively by the covariance functions of and . Here, and are the centered Gaussian processes same as in Theorem 1.1. To see this, recall from Definition 3.3 that
where . As , increases up to which is equal to . Similarly, increases up to which is equal to the covariance function . Due to this, the second to the last inequality follows by the Slepian’s inequality.
Since is arbitrary, letting to yields the right hand side of the above display to be when and when . On the other hand, we have
(3.19) | ||||
By using the same argument as in above (using Slepian’s lemma), we can bound the probability by when and by when .
Following [PS99, Lemma 2.5] and [DM15], we know that
for large and very small. On the other hand, both and is upper bounded by for all large and small. Hence we get (first taking ),
where . The last inequality follows by combining (3.15) with also letting . Combining (3.16), (3.17) and (3.19) yields that
3.2.1 Proof of Lemma 3.2
We first state the following definition and a lemma, which we will use to verify Lemma 3.2. We defer the proof of the lemma to Section 3.2.2.
Definition 3.3.
Set for ,
Also, for any integer and set
Fixing , define a process on by setting
Also let be i.i.d. centered Gaussian processes, with
where
Lemma 3.4.
For any and with , we have
where the convergence is in the weak topology of .
Proof of Lemma 3.2.
-
(a)
Proof of (3.13)
-
(b)
Proof of (3.2)
We prove the lemma for , noting that the proof of follows by a similar argument.
For any set finite set , we denote the number of elements in by . If the event occurs, for every there exists such that occurs. Also, since , we either have or . Without loss of generality assume that we are in the first case. In this case we construct a subset of the set , using the following algorithm which we call as ‘move, flush and repeat’:
-
•
Move: Pick the smallest element in the set , say , and put it in the subset . By definition of we have .
-
•
Flush: Remove all elements of in the interval .
-
•
Repeat: If is empty, stop and output the subset . If not, go back to step 1.
With this construction, we have . Let , and let denote the elements of . Also, set and note that the set necessarily consists of distinct elements in . Thus the number of choices for the set is at most . Set
(3.20) where , and . Setting and we have
The right hand side of the last inequality could be further written as
The fits inequality in the above display follows from the union bound. The following equality holds true and are mutually independent by the construction of set . The second inequality follows by noticing that
as . Here there exists depending on such that is less than for all . Furthermore, are bounded by for by Lemma 3.8 and bounded by for by Lemma 3.7. To bound the exponent in the RHS of ((b)), using , and we note
(3.21) As a result, we get
Combining the above two displays, the exponent of in the second term in the RHS of ((b)) equals
Along with ((b)), this gives
which completes the proof of part (ii).
-
•
∎
3.2.2 Proof of Lemma 3.4
Before proceeding to the proof Lemma 3.4, we derive few facts about the process (see Definition 3.3) in Lemma 3.5. We use these facts in our proof of Lemma 3.4. We first state Lemma 3.5 and use it prove Lemma 3.4 and thereafter, complete the proof of Lemma 3.5. In the proof of Lemma 3.4, one needs tightness of the processes which we prove in Lemma 3.6.
Lemma 3.5.
Set for . Recall , and from Definition 3.3.
-
(1)
Consider the case .
(i) For all large enough (depending on and ) we have
(ii) For any positive integer we have
(3.22) (iii) If , for any positive integer and we have
where for .
-
(2)
Consider the case .
(i) For all large enough (depending on and ) we have
(ii) For any , and positive integer we have
(iii) If , for any positive integer and we have
-
(3)
If then for any and positive integer we have
Proof of Lemma 3.4.
We show the desired convergence of the stochastic process by checking convergence of finite dimensional distributions and tightness.
Step 1 - Convergence of finite dimensional distributions: For showing convergence of finite dimensional distributions, we will show that for any real vector we have
For this, fixing it suffices to show that
To this effect, note that
is a sum of independent components with mean . To verify the convergence in distribution of we will use the Lindeberg Feller CLT. Since
on taking and using parts (a), (b), (c) of Lemma 3.5 we get
(3.23) |
To complete the proof, it suffices to verify the Lindeberg Feller condition, which in this case is equivalent to verifying that for every we have
where has mean and variance . We now claim that
(3.24) |
Before proceeding to the proof of (3.24), let us show how this this claim proves the Lindeberg Feller condition. Notice that
By our assumption, are uniformly integrable which implies that
as . To show the Lindeberg Feller condition holds, it suffices now to prove that has a finite limit as . Notice that (a), (b), (c) of Lemma 3.5 implies converges to a finite limit. This proves the Lindeberg-Feller condition. It remains to show (3.24).
For verifying (3.24), we split the proof into two cases, depending on the value of . If , using the regular variation of along with part (b) of Lemma 3.5, for any we have
where we use the fact that . Similarly, if , using the regular variation of along with part (c) of Lemma 3.5, for any we have
Since , on combining the above two displays (3.24) follows.
Step 2 - Tightness: Proceeding to show tightness in we will invoke the Kolmogorov-Chentsov criterion, for which it suffices to show that for any in this interval we have
But this follows from Lemma 3.6 for any .
∎
Proof of Lemma 3.5.
To begin, for any , we have
(3.25) |
Now we proceed to prove the claim made in the parts (1), (2), (3) of Lemma 3.5.
-
(1)
We start by recalling that
Writing with slowly varying, fixing for all large (depending on , and the function ) we have
(3.26) for some which is close to as approaches to . Also, noting that
(3.27) shows us
(3.28) Finally, a change of variable gives
(3.29) Combining (3.26), (3.28) and (3.29), we get that for all large (depending only on and ) , any we have
The conclusion of part (i) now follows from the above inequalities since approaches to as goes to .
The conclusion of part (ii) follows on noting that stays bounded, and stays fixed. The conclusion of part (iii) follows on noting that if the first term in the RHS of (3.2.2) equals 1, and taking ratios using part (ii).
- (2)
-
(3)
If , the first term in the RHS of (3.2.2) converges to , on using the fact that
The desired conclusion follows, on noting that the second term converges to a finite number as , on invoking parts (a) and (b).
∎
The next two lemmas aims to show the tightness (Lemma 3.6) and the tail probabilities (Lemma 3.7) of the processes for . The latter is shown using Proposition 3.9 which derives a maximal inequality of a stochastic process based on the pointwise bound on the second moments of the stochastic process. Finally the tail probabilities of are used to derive bounds (recall the definitions of from (3.1)) in Lemma 3.8, a crucial input needed to complete of the proof of the upper bound in Theorem 1.1.
Lemma 3.6.
Setting for , for all large enough (depending only on ) we have the following bounds:
Proof of Lemma 3.6.
Recalling that , a direct computation gives
(3.32) |
where in the second equality we have used the mean value theorem. Note that . We now analyze the right hand side of (3.2.2) by splitting the argument depending on whether or .
Case: In this case Lemma 3.5 part (1)(i) shows that for all large enough (depending on and ),
which on taking ratio gives
where the last inequality uses the estimate
(3.33) |
for , for any , for the particular choices . Along with (3.2.2), this gives
(3.34) |
We obtained the second inequality by using Using Lemma 3.5 part (a)(i), the third inequality by using (3.26), and and the last inequality by (3.33).
We now consider cases, depending on the relative values of .
Sub-case: A change of variable gives that for any we have
(3.35) |
Combining (3.33) and (3.35) with , and using the bound the RHS of (3.2.2) can be bounded by
and so the conclusion of the lemma holds in this case.
Sub-case: To this effect, for any we have
(3.36) |
where the last inequality holds for all large enough (depending only on ). Combining (3.33) and (3.36) with , the RHS of (3.2.2) can be bounded by
and so again the conclusion of the lemma holds in this case.
Case: As before, we need to bound the RHS of (3.2.2). Note that
which gives the following bound to the RHS of (3.2.2) (without the factor ):
(3.37) |
Using Lemma 3.5 part (b)(i) along with (3.33) gives
(3.38) |
On the other hand, invoking (3.28) and (3.29) gives
(3.39) | |||
(3.40) |
where the last estimate of the second inequality uses (3.35) and (3.36). Using (3.38) and (3.39) with the RHS of (3.2.2) can be bounded by
This completes the proof of the lemma. ∎
Lemma 3.7.
For any and we have the following bounds:
-
(i)
If , then
-
(ii)
If , then
Proof.
To prove this lemma we will use Proposition 3.9 for the choices
,and
We split the proof into two cases, depending on the value of .
-
(i)
In this case we have
where the inequality in the first line uses Lemma 3.5 part (a)(i) and in the second line uses (3.33) for the denominator, and (3.35) and (3.36) for the numerator. Thus we can take , where is a universal constant. Proceeding to estimate of Proposition 3.9, using Lemma 3.6 we can take . Invoking Proposition 3.9 the desired conclusions follow.
-
(ii)
∎
Lemma 3.8.
For any and , for all and we have
-
1.
when for some fixed integer ,
-
2.
for such that ,
Proof of Lemma 3.8.
Using Lemma 3.4, we get that whenever , we have
(3.41) |
We show how the conclusions in (1) follows from the above convergence. Lemma 3.7 shows that are uniformly tight. Furthermore, for any and , we have . Recall that for and if . In the case when , we get the following bound
Without loss of generality, one can bound the right hand side of the above equation by . By the maximal inequality of Gaussian processes, the probability could now be bounded by . The conclusion in (1) now follows from this bound and the weak convergence in the display (3.41).
To establish conclusion (2), it suffices to show that for any there exists , such that for all we have
(3.42) |
To this effect, we use Lemma 3.7 to get the appropriate (depending on ), such that
The desired conclusion then follows by an application of Borel-TIS inequality ([AT09, Chapter 2]) for all large enough, depending on . ∎
Proposition 3.9.
Suppose be a continuous time stochastic process with mean and continuous sample paths, such that for some constants positive and we have
Then there exists a constant depending only on such that the following hold:
Proof.
Since the second conclusion follows from the first one, it suffices to verify the first conclusion. To this effect, for every , partition the interval into dyadic rationals of the form for . Then continuity of sample paths gives
and so
where is chosen such that . The desired conclusion then follows by an application of Chebyshev’s inequality, on noting that
∎
3.3 Proof of Lemma 3.1
Recall that
where . To bound uniformly for , notice that behaves very similarly as for the case when . By computing the variance, we get
Plugging and using the variance bound to first bound the probability of and then, to bound using Borel-TIS [AT09] proves the result.
4 Lower Bound
4.1 Fixation of Notations
Before proceeding to the proof, we introduce few notations. To begin, fix positive reals . Define the sets
(4.1) | ||||
(4.2) |
and note that are disjoint partitions, where . We intend to partition the interval and into sub-intervals (blocks) of size . The choice of are made in a way such that For any , define
We now divide and into many sub-intervals. Define the sets by setting
(4.3) |
and note that
Similarly, define the sets where
(4.4) |
and note that
For , define
(4.5) |
Define the positive weight function as
(4.6) |
4.2 Proof of
For any , we have
(4.7) | ||||
where the last line uses the fact that are independent. For any , define
Below we find lower bound to for each .
Proposition 4.1.
Consider any . There exists such that for all and large,
(4.8) |
where the Gaussian process is defined in Definition 4.18.
Proposition 4.2.
Consider any . There exists such that for all and large,
(4.9) |
The Gaussian process is defined in Definition 4.18.
Proposition 4.3.
There exists and such that for all and large and ,
(4.10) |
Proposition 4.4.
Assume that there exists and a small such that for ,
Then for all large ,
(4.11) |
where is an arbitrarily small number.
Proposition 4.5.
Assume that there exists and , such that for ,
Then for all large ,
(4.12) |
for some arbitrarily small number .
Final Stage of the proof: Continuing (4.7), we can write
Taking logarithm on both sides and dividing both sides by shows
(4.13) |
As and , we know
and
Combining these limits and plugging back into (4.2) yields the results
It remains to show that the conditions of Proposition 4.4 and 4.5 are satisfied when are i.i.d. random variables with and . Since , there must exists such that . If any of the probabilities or is positive, then the conditions of the propositions are satisfied. Otherwise there must exists such that since . Therefore we arrive at the condition and for some small constant. Under these revised conditions, using the same arguments as in Proposition 4.4 and 4.5, we obtain
for . By flipping the sign of the coefficients for and using Proposition 4.1, 4.3 and 4.2 respectively, we obtain for ,
and,
These bounds together shows that
4.3 Proof of Proposition 4.1: Case
We consider the following decomposition , and so
(4.14) |
where for a non negative sequence satisfying and a large but fixed integer , we define
(4.15) | ||||
(4.16) | ||||
(4.17) | ||||
(4.18) |
For simplicity, we will choose to work with for some small constant .
Proceeding to estimate the probability of the sets above, we claim the following Lemma, whose proof we defer to the end of the section.
Lemma 4.6.
Fix any . Fix an interval such that is far away from , i.e., either or for some . Denote .
-
(a)
If , for any we have
(4.19) -
(b)
If , for any we have
(4.20)
Proof of Lemma 4.6.
We prove the lemma for . Proof in the other case follows from same argument.
-
(a)
To begin, for any with , we have
We obtain the last inequality by using Lemma 4.7. Invoking Lemma 4.21 with gives
Finally we use Chebyshev’s inequality to get
which along with the bound in the previous display gives the desired estimates of (4.19).
- (b)
∎
Lemma 4.7.
Consider the following function
(4.21) |
Fix some positive number . There exists such that for all , and ,
(4.22) |
Proof.
Note that which we define as is increasing on the interval and decreasing on the interval . Bounding the sum by its integral approximation shows
(4.23) |
Since and , we know
This completes the proof. ∎
4.3.1 Lower Bound on
Lemma 4.8.
Consider the Gaussian process defined in 4.18. For all large , we have
(4.24) |
Proof.
Recall the definition of from (4.5). By Lemma 4.19, we have
(4.25) |
as where is a centered Gaussian process as defined in 4.18.
Now we proceed to prove (4.24) using (4.25). By the weak convergence, for all large
(4.26) | ||||
(4.27) | ||||
(4.28) | ||||
(4.29) |
where the last inequality follows by applying the Slepian’s inequality for the Gaussian processes.
Recall that and are independent centered Gaussian processes. As a result, we have
Plugging this bound in the last line of (4.47) yields
r.h.s. of (4.47) | |||
for all large . The equality in the last display follows since and are independent Gaussian process and their marginal laws are same. This completes the proof.
∎
4.3.2 Upper Bound on
Lemma 4.9.
Recall from (4.41). For all large , we have
(4.30) |
Proof.
Recall that
Notice that
where . For fixed with , applying part (a) of Lemma 4.6 with gives
(4.31) |
where the second inequality follows since for all and . On the other hand, for such that , using part (b) of Lemma 4.6 with we have
(4.32) | ||||
(4.33) |
Using union bound and combining the inequalities in (4.38) and (4.32) as varies over the set yields
(4.34) |
Note that the above inequality follows since . This completes the proof. ∎
4.3.3 Upper Bound on &
Lemma 4.10.
We have
(4.35) |
Proof.
Bound on : We write where for . By the union bound, we have
(4.36) | ||||
(4.37) |
Now we first bound and then, bound .
Fix any . For any ,
A similar calculation gives
Applying part (b) of Lemma 4.21 with gives
By summing the above bounds, we get
Now we turn to bound . For , we have
A similar calculation gives
Using Lemma 4.21 with where gives
which implies
Combining the bound on and shows the upper bound on
Bound on : We now write . Let us denote . For any and , we note that . Applying part (a) of Lemma 4.6 with gives
(4.38) |
Therefore by the union bound, we get
This proves the desired claim.
∎
4.3.4 Proof of Proposition 4.1
4.4 Proof of Proposition 4.2: Case
We divide into many sub-intervals. Recall the set of intervals where
and note that
As in (4.5), we define
(4.39) |
As a result, we have and in the same spirit as in (4.14), we get
where
(4.40) | ||||
(4.41) | ||||
(4.42) | ||||
(4.43) |
Recall that for any .
Proposition 4.2 will be proved by showing lower bound to and upper bounds to , and . We show these as follows by using Lemma 3.5, 4.20 and 4.21.
4.4.1 Lower Bound on
Lemma 4.11.
Consider the Gaussian process defined in 4.18. For all large , we have
(4.44) |
Proof.
Recall the process from Definition 4.18. By Lemma 4.19, we know
(4.45) |
as where is a centered Gaussian process as defined in 4.18. We now show (4.44) using (4.45). This proof is very similar to the proof of Lemma 4.8.
Now we proceed to prove (4.24) using (4.25). By the weak convergence, for all large
(4.46) | ||||
(4.47) |
where the last inequality follows by applying the Slepian’s inequality for the Gaussian processes. Since is centered Gaussian process, the product in the last line of the above display is lower bounded by . Substituting this into the right hand side of the above inequality completes the proof.
∎
4.4.2 Upper bound on
Lemma 4.12.
Recall the event . For all large ,
(4.48) |
Proof.
Recall that
By the union bound, can be bounded above by . Throughout the rest of the proof, we bound . We claim that for all large and uniformly for any such that ,
(4.49) |
From (4.49), the inequality in (4.48) follows by the union bound. We divide the proof of (4.49) into two stage. In Stage 1, we consider the case when (where ) and in Stage 2, we consider the case when .
Stage 1: For and we have
A similar calculation gives
Using Lemma 4.21 and Chebyshev’s inequality, we have
Stage 2: For and , we have
By a similar computation, we get
4.4.3 Upper bound on &
Lemma 4.13.
For all large , we have
(4.50) |
Proof.
The proof is divided in two stages: in Stage 1, we prove the bound on and Stage 2 will contain the bound on .
Stage 1: From the definition of ,
(4.51) |
In what follows, we seek to bound the right hand side of the above inequality. We write where . For any fixed and , we have
A similar calculation gives
Using Lemma 4.21, this gives
A union bound then gives
from which the desired conclusion follows on noting that . This proves the bound on of (4.50).
Stage 2: By the union bound, we write
In what follows, we seek to bound and separately. We claim and prove that for all large ,
Bound on : For for we have
A similar calculation gives
Noting that gives
The desired bound on follows by taking large, or equivalently .
Bound on : Note that . For we have
A similar calculation gives
Noting that gives
Summing over gives the bound .
∎
4.5 Proof of Proposition 4.3: Case
Proposition 4.3 follows from the following theorem.
Theorem 4.14.
Fix . Consider the following event:
(4.52) |
where and if for any . Then, there exists , and such that for all and ,
(4.53) |
Proof.
We use the following shorthand notations:
It is straightforward to see that . Thus, is bounded below by . In what follows, we show there exist and such that for all and ,
(4.54) |
where is the complement of the event . Combining the bounds on and and substituting those into the inequality proves (4.53).
Proof of : Lemma 4.16 shows that weakly converges to as . Thus, can be bounded below by (see (4.17) for the definition of ). The lower bound of now follows from Lemma 4.17.
Proof of : Since , we get that for any ,
Similarly, for any ,
As a consequence, we get
This upper bound in conjunction with Kolmogorov-Centsov’s type argument shows there exists such that for all large integer and ,
(4.55) |
This completes the proof of (4.54) and hence, completes the proof the result. ∎
Lemma 4.15.
Consider the stochastic process . Then, there exists such that for all ,
(4.56) |
Proof.
For any ,
(4.57) |
By the Kolmogorov’s maxmimal inequality, there exists such that
(4.58) |
Note that there exists such that . Furthermore, we have . Combining this with the inequality in the above display shows that there exists such that
(4.59) |
This completes the proof. ∎
Lemma 4.16.
Then the sequence of stochastic processes is uniformly equi-continuous and converges to the Gaussian process as where has the following covariance structure:
Fix any interval . Then, there exists such that for any ,
(4.60) |
Similarly, for any
(4.61) |
Proof.
By the mean value theorem, for all .
(4.62) |
Combining this with Lemma 4.15, there exists such that for all ,
(4.63) |
This shows that is uniformly equi-continuous as goes to . Moreover, for any ,
(4.64) |
By Lindeberg-Feller’s theorem, for any ,
(4.65) |
where is the same Gaussian process as stated in the lemma. The above display shows the finite dimensional convergence of process . Allying this with the tightness in (4.63) yields the weak convergence.
It remains to show (4.60) and (4.61). We only show (4.60). The proof of (4.61) follows from similar argument. By the mean-value theorem, for any
(4.66) | ||||
(4.67) | ||||
(4.68) |
where the last inequality follows by expanding . Approximating the sum by the integral and substituting into the right hand side of the above display yields
r.h.s. of (4.68) | (4.69) | |||
(4.70) |
Note that there exists constant such that can be bounded above by . The maximum value of this lower bound as varies in is . By using this upper bound for and Doobs’s maximal inequality for the martingale to control the tail probability of shows
(4.71) |
Now (4.60) follows from the above inequality by letting on both sides. ∎
Lemma 4.17.
Fix . Consider the Gaussian process as in Lemma 4.16. Define
(4.72) |
where
for . Then, there exists and such that for all
(4.73) |
Proof.
Fix some large number . We use the following shorthand notations:
Notice that is equal to . Since is a Gaussian process with positive correlation, we may apply Slepian’s inequality to write
(4.74) |
Fix . In what follows, we claim and show the following: there exists such that for all
(4.75) |
Substituting these lower bounds on to the right hand side of (4.74) proves (4.73). In the rest of the proof, we focus on showing (4.75). We only show the lower bound for , and . The bound for and follow from similar argument for and respectively.
Proof : Note that there exists such that is less than for all . Furthermore, it is straightforward to check that for any
For any , define . By Dudley’s entropy theorem [Dud10, Theorem 7.1], we have
for some . Let us define
Note that . As a result, we get
where the last inequality follows by Borel-Tis inequality. This shows
For large, this lower bound is bounded below by . This proves the lower bound for .
Proof : By Slepian’s inequality which we can apply since is a Gaussian process,
(4.76) |
By symmetry of between , it suffices to show
(4.77) |
for some constant . We show this follows.
We divide the interval into many intervals of equal length and denote them as where . By Slepian’s inequality,
(4.78) |
We first lower bound each term of the product of the right hand side of the above display. Applying (4.60) of Lemma 4.16, we notice
Fix . Note that is less than . Letting , we see that the right hand side of the above inequality is bounded by . This implies
Suppose for some . Since for all large , we write
where is the cumulative distribution function of a standard normal distribution. The last line of the above display can be bounded below by for some constant which does not depend on . By taking small, we can lower bound by for some . This shows a lower bound to -th term of the product in (4.78). Substituting all these bounds into the right hand side of (4.78) shows (4.77). This completes showing the lower bound on .
Proof of : We divide the interval into many sub-intervals of equal length for where
Applying the Slepian’s inequality, we may write
(4.79) |
In what follows, we find a lower bound to the to each term of the product in the right hand side of the above display. Fix and suppose for some . Let us denote
where . Notice that . Thus, can be bounded below by where denotes the complement of the event . Since is a Gaussian r.v. with mean zero, is bounded below by . To obtain a lower bound to , it suffices to bound from above. Due to (4.60),
Note that is less than when
This shows is bounded above by the left hand side of the above display when is equal to . Furthermore, by setting the following , the right hand side is bounded by . As a consequence, is bounded above by . Combining
(4.80) |
For large , the right side of the last inequality is bounded below by . This provides a lower bound to the -th term of the product (4.79). Substituting these lower bounds into the right hand side of (4.79) yields the lower bound of in (4.75). ∎
4.6 Proof of Proposition 4.4: Case
Recall that and .
Proof of Proposition 4.4.
Recall that . We divide the proof into two cases: when and for some and when and .
Case : Consider the following event:
for some and . Since ’s are independent, we write
(4.81) |
where the last inequality follows since by our assumption and . We now claim and prove that for all large and such that ,
(4.82) |
On the event , we have
where and . Fix such that . Recall that . Thus for all large , we have
For large , on the event for , we have
Note that is greater than for all since the discriminant of is less than because . Therefore, for all where is some positive constant which only depends on and . We start by recalling that for and furthermore, for ,
Recall that for all on the event . Hence for all and for all . This shows the claim in (4.82).
Case : Consider the event
Note that is bounded above since ’s are independent, and . In the same spirit of (4.82), we now show that
(4.83) |
On the event , we have
(4.84) |
Recall that and let be such that . For all large , we have
Note that the discriminant of the quadratic polynomial is since the choice of is made in such a way to satisfy this inequality. This shows the right hand side of (4.84) is bounded above for some positive constant . This implies (4.83) and hence, completes the proof. ∎
4.7 Proof of Proposition 4.5: Case
The dominant interval here is and the corresponding interval for the coefficient indices is .
4.8 Supporting Lemmas
We first state the following two lemmas, which we will use to lower bound the terms in the product (4.2). We begin by introducing the following notations.
Definition 4.18.
Fixing , define a process on by setting
Note that
(4.85) |
where is defined as
(4.86) |
For any integer and set
Also let be i.i.d. centered Gaussian processes defined on the interval , with
(4.87) |
Lemma 4.19.
For any and with , we have
Here the convergence is in the topology of .
Lemma 4.20.
Recall the functions and .
-
(1)
Suppose .
-
(i)
For all large enough (depending on and ) we have
-
(ii)
For any positive integer , we have
(4.88) -
(iii)
If , for any positive integer and we have
where for .
-
(i)
-
(2)
Suppose .
-
(i)
For all large enough (depending on and ) we have
-
(ii)
For any , and positive integer we have
-
(iii)
If , for any positive integer and we have
-
(i)
-
(3)
If then for any and positive integer we have
The proof of Lemma 4.19 and 4.20 are very similar to that of Lemma 3.4 and Lemma 4.20. We will skip the details.
Lemma 4.21.
Suppose is a continuous time stochastic process with continuous sample paths, such that and . Then the following holds:
(4.89) | ||||
(4.90) |
Proof of Lemma 4.21.
For a non negative integer and , setting we can write . For and let if . Then, using continuity of sample paths and noting that we can write
(4.91) |
To prove (4.89), use (4.91) along with a union bound to get
where the last line uses Chebyshev’s inequality. The desired conclusion is immediate from this. ∎
Lemma 4.22.
Consider the centered Gaussian processes of Definition 4.18. For any given , there exists such that for any one gets satisfying
(4.92) | ||||
(4.93) |
for all .
Proof.
Here we show how to prove (4.92). Proof of (4.93) follows from similar arguments. Fix small. To this end, we divide into three sub-intervals; and . Recall that the covariance function of from (4.87). Since the covariance is non-negative, by Slepian’s inequality of the Gaussian process, we get
For any fixed such that , we have and . This shows
as approaches . Recall that is the correlation function of the stationary Gaussian process . We now intend to apply Theorem 1.6 of [DM15] which will imply that
(4.94) |
However, according to [DM15, Theorem 1.6], the above limit holds when the covariance function of the Gaussian processes satisfy condition (1.15) of [DM15, Theorem 1.6]. For this it suffices to check Condition (1.23) of [DM15, Theorem 1.6] which is given as
(4.95) |
for some . To show the above condition, we first claim and prove that there exists such that
(4.96) |
for all . Recall that for any
(4.97) |
Suppose that . In that case, we get the following following bound
(4.98) |
where the second inequality follows since strictly decreases as increases to and for all . Now we claim and prove that there exists some such that
for all . To show this, we consider the function defined as . Note that
This shows that if is chosen such that
will be strictly increasing on the interval for all . Hence, for all satisfying the above inequality, we get for all
Taking and shows (4.96). From (4.96), it follows that
uniformly for all . The above inequality shows that (LABEL:eq:DMCondition) holds for any since as .
To complete the proof, it suffices to show that there exists such that for all and large, one has
(4.99) |
We only show the first inequality. The second follows from similar arguments. Note that transforms to under the change of variable . Furthermore, by the stationarity of , we have
This shows why takes values on the interval in the right hand of (4.8). Now we we lower bound . By Slepian’s inequality, we know
Consider a mean zero Gaussian process such that for . Due to (4.96) and Slepian’s inequality, we get
(4.100) |
Note that is a stationary Gaussian process. By [DM15, Theorem 1.6],
Combining the above fact with (4.100) shows that there exists such that for all and large large, first inequality of (4.8) holds. The proof of the second inequality is similar. ∎
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