Random polynomials: central limit theorems for the real roots
Oanh Nguyen, Van Vu

TL;DR
This paper establishes a new, more general central limit theorem for the number of real roots in a broad class of random polynomials, extending Maslova's classical result from 1974.
Contribution
It introduces a novel approach that generalizes and strengthens the existing CLT for real roots of random polynomials with polynomially growing coefficients.
Findings
Derives a general CLT for real roots of random polynomials
Extends Maslova's theorem to broader classes of polynomials
Provides a new methodology for analyzing real roots
Abstract
The number of real roots has been a central subject in the theory of random polynomials and random functions since the fundamental papers of Littlewood-Offord and Kac in the 1940s. The main task here is to determine the limiting distribution of this random variable. In 1974, Maslova famously proved a central limit theorem (CLT) for the number of real roots of Kac polynomials. It has remained the only limiting theorem available for the number of real roots for more than four decades. In this paper, using a new approach, we derive a general CLT for the number of real roots of a large class of random polynomials with coefficients growing polynomially. Our result both generalizes and strengthens Maslova's theorem.
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Random polynomials: central limit theorems for the real roots
Oanh Nguyen
Department of Mathematics, Princeton University, Princeton, NJ 08544, USA
Department of Mathematics, University of Illinois at Urbana–Champaign, Urbana, IL 61801, USA
[email protected], [email protected]
and
Van Vu
Department of Mathematics, Yale University, New Haven, CT 06520, USA
Abstract.
The number of real roots has been a central subject in the theory of random polynomials and random functions since the fundamental papers of Littlewood-Offord and Kac in the 1940s. The main task here is to determine the limiting distribution of this random variable.
In 1974, Maslova famously proved a central limit theorem (CLT) for the number of real roots of Kac polynomials. It has remained the only limiting theorem available for the number of real roots for more than four decades.
In this paper, using a new approach, we derive a general CLT for the number of real roots of a large class of random polynomials with coefficients growing polynomially. Our result both generalizes and strengthens Maslova’s theorem.
This work is partially supported by VIASM (Vietnam); O. Nguyen is supported by NSF DMS 1954174; V. Vu is partially supported by NSF AWD0000777 and AWDD02154
1. Introduction
Random polynomials, so simple to define but difficult to understand, have attracted generations of mathematicians. Typically, a random (algebraic) polynomial has the form
[TABLE]
where are iid copies of an (atom) random variable with zero mean and unit variance, and are deterministic coefficients which may depend on both and . Different definitions of give rise to different classes of random polynomials, which have distinct behaviors.
When for all , the polynomial is often referred to as the Kac polynomial. Even this special class provides great challenges, which have led to rich literature (see, for example, the books [3, 13] and the references therein).
Let denote the number of real roots of . A key problem in the theory of random polynomials is to understand the behavior of the random variable , with tending to infinity. As a matter of fact, this is the problem that started the whole field, with fundamental works of Littlewood-Offord [24, 25, 26] and Kac [22] from the 1940s.
The first natural question is to determine the expectation of . It took more than 20 years and the works of Kac [22], Erdős-Offord [12] and Ibragimov-Maslova [19, 20] to settle this problem for the Kac polynomial (the case ). By now, the problem has been solved for many classes of random polynomials, with various choices for and under very general assumptions for (see the introduction of [30]; also [17, 11, 38, 35, 39, 42, 15, 33, 34, 9, 40, 41] and the references therein).
The next, and more important, problem is to determine the variance and limiting distribution of . This problem is much harder and our understanding is far from complete. In the 1970s, Maslova [28] proved the Central Limit Theorem (CLT) for the Kac polynomial. Here and later, means convergence in distribution; denotes the standard normal distribution, , .
Theorem 1.1**.**
[27, 28]** Let be a positive constant. Consider the Kac polynomial with the random variables being iid with mean zero, variance one, bounded moment, and . We have, as tends to infinity,
[TABLE]
Furthermore, , where .
The proof of Maslova relied heavily on explicit computation that requires all the to be equal. Only very recently, Central Limit Theorems have been established for other classes of polynomials, via new methods. In 2015, Dalmao [5] established the CLT for binomial polynomials (the case when ), and in 2018, Do and the second author [10] handled Weyl polynomials ()). However, in both papers, the authors need to assume that the random variables are standard Gaussian and their arguments rely strongly on special properties of Gaussian processes. It remains a major challenge to extend these results to other random variables (Rademacher, for example). For related results concerning random trigonometric polynomials, see [16, 2, 1].
The goal of this paper is to establish CLT for a large class of random polynomials where the deterministic coefficients grow polynomially. We will only need a mild assumption on the , which is satisfied by most random variables used in practice (in particular, this assumption is weaker than Maslova’s). In fact, we can also handle the more general setting when the are not iid.
We consider
[TABLE]
where are independent random variables and are deterministic coefficients satisfying the following conditions for some positive constants and some constant .
- (A1)
The random variables are independent (but not necessarily identically distributed) real-valued random variables with unit variance and bounded moments, namely , 2. (A2)
for all 3. (A3)
The coefficients are deterministic real numbers that grow polynomially, namely
[TABLE]
and
[TABLE]
This class contains many interesting ensembles of polynomials including
- •
the Kac polynomial (all )
- •
semi-Kac polynomials with (for some constant ) and all other taking arbitrary values from a fixed set of non-zero constants. (For example, we can have and all are either 2 or 3 in arbitrary fashion.)
- •
all derivatives of the Kac polynomial (the zeroes of these polynomials are thus the critical points of the Kac polynomial),
- •
hyperbolic polynomials where is a positive constant (see [17, 9, 14] and the references therein).
- •
has the form where is a polynomial in of a fixed degree and is any function satisfying .
Our main result establishes the CLT for these random polynomials.
Theorem 1.2**.**
Assume that the polynomial satisfies Conditions (A1)-(A3) and that for some constant . Then where , .
The condition is guaranteed by the following lemma.
Lemma 1.3**.**
Assume that the polynomial satisfies Conditions (A1)-(A3) and there exist constants such that for all ,
[TABLE]
Then for some constant .
The condition in this lemma is satisfied by all classes listed above. We obtain
Corollary 1.4**.**
The CLT holds for the Kac polynomial and its derivatives. It also holds for hyperbolic polynomials.
Remark 1.5**.**
When restricted to the Kac polynomial with being iid copies of an atom variable , our result strengthens Maslova’s, as the condition in Theorem 1.1 is removed.
Remark 1.6**.**
The real roots of are the real solutions of the equation . The flexibility in condition (A2) enables to extend Theorem 1.2 to the equation , where is any fixed polynomial with bounded degree. In particular, taking for a constant , we conclude that CLT holds for any level set of .
Related literature. Random polynomials with coefficients growing polynomially, also known as generalized Kac polynomials, have attracted researches in different areas including Probability and Mathematical Physics. For example, we refer to Das [6], Schehr–Majumdar [37, 38], Do and the authors [9]. It has been established in [37, 38] that the real roots of these polynomials are closely connected to zero crossing properties of the diffusion equation with random initial conditions. The connection has been applied in a paper by Dembo and Mukherjee [7] to study the probability that these random polynomials have no real roots, which is known as persistence probability as it is related to persistence properties of physically systems. See also [8] and the references therein. In [7], the random variables are Gaussian and the persistence probability is . It is shown that the power exponent is universal roughly in the sense that it depends on (in Condition (A3)) rather than the specific values of . We also refer to Poplavskyi and Schehr [32] for a recent development in finding the power exponent for the Kac polynomial. It would be interesting to see if the power exponent is universal in the sense presented in this paper that is if we replace the Gaussian distribution by other distributions.
Notations. We use standard asymptotic notations under the assumption that tends to infinity. For two positive sequences and , we say that or if there exists a constant such that . If for some sequence , we also write .
If , we say that . If , we say that . If , we sometimes employ the notations and to make the idea intuitively clearer or the writing less cumbersome; for example, if is the quantity of interest, we may write instead of , and instead of .
2. The Universality Method
The key ingredient of our proof is the universality method. The general idea of this method is to show that limiting laws do not depend too much on the distribution of the atom variable (or the variables in general, if they are not iid). Once universality has been established, then it suffices to prove the desired law for the case in which the are Gaussian, and here one can bring extra powerful tools such as properties of Gaussian processes; see [17, 11, 38, 35, 39, 42, 15, 39, 18, 21, 33, 34].
The universality method has been powerful in studying local statistics such as the density or correlation functions concerning the number of roots in a small region (where the expectation is of order ) (see, for example, [42, 29, 9, 30]). However, universality arguments are tailored to the local settings and in order to use them to prove the global law in this paper, we need to perform a number of considerably technical steps, linking local statistics to the global one. The proof for the Gaussian case itself also requires new ideas.
To study the real roots of , we divide the real line into two regions: a core region that contains most of the real roots and the remaining one that contains an insignificant number of real roots. Consider small numbers that depend on and satisfy the following property for all constants :
[TABLE]
For example, . We define
[TABLE]
where for any given set , we define , , and . For appropriate choices of and , this will be our core region.
For a subset , let be the number of roots of in . Let be iid standard Gaussian random variables and set
[TABLE]
We denote by the number of zeros of in .
Our main result on global universality of the real roots states that on the core , the distributions of the roots of and are approximately the same.
Theorem 2.1**.**
Assume that the polynomial satisfies Conditions (A1)-(A3). There exist positive constants and such that for every satisfying (2), for sufficiently large and every function whose derivatives up to order are bounded by , we have
[TABLE]
Since is always an integer, for every real number ,
[TABLE]
where is any smooth function that takes values in and . Therefore, Theorem 2.1 implies
[TABLE]
Using Theorem 2.1 (not in the straightforward way), we deduce the following corollary
Corollary 2.2**.**
Assume that the polynomial satisfies Conditions (A1)-(A3). Let be an integer. There exist positive constants and such that for every satisfying (2) and for sufficiently large , we have
[TABLE]
In particular,
[TABLE]
Next, we show that the contribution outside of the core is negligible.
Proposition 2.3**.**
Assume that the polynomial satisfies Conditions (A1)-(A3). Let be an integer. There exists a positive constant such that for every satisfying (2) and for sufficiently large , we have
[TABLE]
To prove Theorem 1.2 and Lemma 1.3, we use the universality results stated in Theorem 2.1, Corollary 2.2 and Proposition 2.3 to reduce to the Gaussian case (i.e., the case in which the are iid standard Gaussian) with roots restricted to the core . In particular, we prove
Lemma 2.4**.**
Assume that the polynomial satisfies Conditions (A1)-(A3). Let be any positive constant, then for any satisfying
[TABLE]
we have
[TABLE]
And we also prove the following special case of Lemma 1.3 for Gaussian.
Lemma 2.5**.**
Assume that the polynomial satisfies Conditions (A1)-(A3) and there exist constants such that for all ,
[TABLE]
Then
[TABLE]
To illustrate the method of universality, we include here the short proof of Theorem 1.2 and Lemma 1.3 assuming the Gaussian case (Lemma 2.4 and Lemma 2.5) together with the universality results (Corollary 2.2 and Proposition 2.3).
Proof of Lemma 1.3.
We first choose and that satisfy all the conditions in Corollary 2.2 and make the right-hand side of (5) as small as when . In particular, we let
[TABLE]
and
[TABLE]
By the triangle inequality on the 2-norm, we obtain
[TABLE]
where in the last equation, we used Proposition 2.3. Since is just a special case of (where the random variables are iid Gaussian), we also have
[TABLE]
Combining this with Lemma 2.5, we obtain
[TABLE]
Applying Corollary 2.2 and (8) yields
[TABLE]
From this and (7),
[TABLE]
This completes the proof. ∎
Proof of Theorem 1.2.
Let and be as in the proof of Lemma 1.3. By the assumption that and by (7), we have
[TABLE]
By this and Corollary 2.2, we also have
[TABLE]
Thus, (6) holds and so we can apply Lemma 2.4 to get
[TABLE]
Hence,
[TABLE]
because by (4), for any fixed ,
[TABLE]
By Corollary 2.2,
[TABLE]
Combining these with (9), we get
[TABLE]
From Proposition 2.3, we have
[TABLE]
By Markov’s inequality, for any fixed , we have
[TABLE]
Thus,
[TABLE]
Adding (10) and (11) completes the proof. ∎
In Section 7, we use universality again to prove Lemma 2.5. But in this case, we will reduce general coefficients to the case when . In other words, we could swap random variables with different means or variances. This deviates significantly from standard swapping arguments that swap random variables with the same mean and variance.
The rest of the paper is organized as follows. Section 3 is devoted to the proof of Theorem 2.1, Section 4 for Corollary 2.2, Section 5 for Proposition 2.3, Section 6 for Lemma 2.4, and Section 7 for Lemma 2.5.
3. Proof of Theorem 2.1
Under the hypothesis of Theorem 2.1, we need to show that
[TABLE]
We first restrict to the interval and prove that
[TABLE]
The proof of (12) follows from the same arguments with some (merely technical) modifications explained in Section 3.7. We choose to start by presenting the proof of (13) as it already captures all of the ideas without having to deal with the tedious technical and notational complications detailed in Section 3.7. This way, it would make the proofs clearer and easier to follow.
3.1. Partition into dyadic intervals and preliminary results
Recall that
[TABLE]
We start by partitioning the interval into dyadic intervals: To be more specific, let for where is the smallest number such that . Let . Note that . For each , let be the number of real roots of in the interval . Let be the number of real roots of in the interval . We have, . 111 If , we set and to be the number of real roots of in the interval .
Generally, there is no difference in our proof if an interval of interest includes one of its endpoints or not. So, for example, if one cares about instead of , one can use the exact same analysis.
For a dyadic interval , we can control the moments of the number of roots. More generally, the following result works not just for dyadic intervals but also for balls on the complex plane.
Lemma 3.1** (Bounded number of roots).**
For any positive constants and , there exists a constant such that for every , every and with , we have
[TABLE]
and
[TABLE]
where is the disk with center and radius in the complex plane.
As a consequence, for and for the dyadic interval , applying Lemma 3.1 for , we obtain
[TABLE]
and
[TABLE]
Proof of Lemma 3.1.
We shall prove that for a large constant and for every ,
[TABLE]
where the implicit constant only depends on and . Setting , we obtain (14). Setting , letting run from to and using the fact that with probability 1, we obtain
[TABLE]
where in the first inequality, the first term bounds on the event
[TABLE]
the second term comes from the events for each , and the third term comes from the event . This proves (15), completing the proof.
It remains to prove (16). To that end, we use the following version of Jensen’s inequality which asserts that for every entire function , every and ,
[TABLE]
where and . This is a consequence of the classical Jensen’s formula (see, for example, [36]). We add a proof of this inequality in Appendix 8.1 for completeness.
Applying Jensen’s inequality to the polynomial gives
[TABLE]
where and .
From (18), to prove (16), it suffices to show that
[TABLE]
and
[TABLE]
Since
[TABLE]
it follows that by Conditions (A1) and (A3). The bound (19) then follows from Markov’s inequality.
For (20), writing and observing that the set is a subset of , we have
[TABLE]
By taking the supremum outside, the right-hand side is at most
[TABLE]
and hence, by projecting onto the real line and conditioning on the random variables , it is bounded by
[TABLE]
We use the following anti-concentration lemma from [30].
Lemma 3.2**.**
[30, Lemma 9.2]** Let be an index set of size , and let be independent random variables satisfying Condition (A1). Let be deterministic (real or complex) coefficients with for all and for some number . Then for any constant , any interval of length at least , there exists such that
[TABLE]
where the implicit constant depends only on and the constants in Condition (A1).
Applying Lemma 3.2 with , , and (where we use Condition (A3) and the assumption that to get ), we obtain such that for a sufficiently large constant ,
[TABLE]
which gives (20) and completes the proof of Lemma 3.1. ∎
3.2. A generalization
Theorem 2.1 is deduced from the following more general result that can be of independent interest.
Proposition 3.3**.**
Let : be any function whose every partial derivative up to order is bounded by . We have
[TABLE]
To deduce (13) (which is essentially Theorem 2.1 as mentioned at the beginning of this section), let be the function defined by . It is easy to check that where being the supremum of all partial derivatives up to order 3 of . By applying Proposition 3.3 to this , (13) follows. ∎
The rest of this section is devoted to the proof of Proposition 3.3.
3.3. Approximate the indicator function by smooth functions
To apply analytical tools, we first approximate the indicator function in counting the number of real roots by smooth functions.
Let be a sufficiently small positive constant. Let be a smooth function taking values in , supported on and equal at [math] with for all . For example, we can take the classical bump function
[TABLE]
For , let be a smooth function taking values in , supported on and equal on with for all . An example of can be obtained by translating and scaling as follows.
[TABLE]
The indicator of the dyadic interval shall be approximated by the following function defined on the complex plane
[TABLE]
In other words, the number of roots in , which is just , is approximated by
[TABLE]
where be the roots of .
To control the error terms in this approximation, note that for all ,
[TABLE]
The following lemma estimates the error term in approximating by .
Lemma 3.4**.**
We have
[TABLE]
Proof of Lemma 3.4.
By the derivative assumption on , we have
[TABLE]
For each is bounded by the number of roots of in the union of the sets , where is the set of all complex numbers whose real part lies in and imaginary part in , , and
Thus, Lemma 3.4 follows by proving that for
[TABLE]
To this end, we use the following lemma from [9].
Lemma 3.5**.**
[9, Lemma 5.1]** There exists a constant such that for all and all with ,
[TABLE]
To show that , we note that is contained in a union of small balls of radius . By Lemma 3.5, the union bound and the fact that the complex roots come in conjugate pairs, the probability that is nonzero is in fact negligible
[TABLE]
Thus, except on an event, named , of probability at most
The expectation on the tail event is controlled as the higher moments of are bounded by Lemma 3.1. More specifically, since , applying (15) to this ball and Hölder’s inequality, we obtain
[TABLE]
For , [9, Theorem 2.4] implies that for such intervals as and , the expected number of real roots is universal in the sense that
[TABLE]
where is the number of real roots of in a set . It thus remains to show that . To this end, we use Kac-Rice formula (see [22, 11]; here we use [13, Formula 3.12]),
[TABLE]
Algebraic manipulations show that
[TABLE]
We add the verification of this estimate in Appendix 8.2 for completeness. Putting the bounds together gives .
By combining this with (23), it follows that the left-hand side of (22) is bounded by , proving (22) and Lemma 3.4. ∎
3.4. Reducing to an explicit function of the random polynomial: log
In light of Lemma 3.4, to show Proposition 3.3, it remains to show that
[TABLE]
where are the roots of
In this section, we reduce the sums to an explicit function of . The starting point for this reduction is to apply the Green’s second identity to the compactly supported function
[TABLE]
where we note that is supported on .
We show that the integral on the right-most side is well approximated by its Riemann sum. In particular, we prove that for
[TABLE]
with probability at least , where are chosen independently, uniformly at random from the ball and are independent of all previous random variables.
Proof of equation (28).
This proof is based on [9, Equation (4.20)].
For notational convenience, we skip the subscript and write , , and . Let , the center of the ball.
Since is compactly supported in , by the Green’s second identity, it holds that
[TABLE]
We shall think about the integral on the right-most side of (29) as an expectation with respect to , up to some rescale; so in approximating an expectation by a sample mean (which is the second summation in (28)), it is sufficient to control the variance. To this end, we would need to require that is bounded above and below. For that purpose, we introduce the good event on which the following hold for :
- (T1)
for all . 2. (T2)
for some .
By Jensen’s inequality (17), these conditions imply
[TABLE]
We will show later that
[TABLE]
Assuming (31), it suffices to show that (28), conditioned on , holds with probability . The following lemma provides the required variance bound, conditioned on .
Lemma 3.6**.**
On the event , we have
[TABLE]
Assuming Lemma 3.6 and the fact that by the definition of , we conclude that on the event ,
[TABLE]
where \mathchoice{{\vbox{\hbox{\textstyle- }}\kern-7.83337pt}}{{\vbox{\hbox{\scriptstyle- }}\kern-5.90005pt}}{{\vbox{\hbox{\scriptscriptstyle- }}\kern-4.75003pt}}{{\vbox{\hbox{\scriptscriptstyle- }}\kern-4.25003pt}}\!\int_{B(x_{0},2\delta/3)}f(z)dz:=\frac{1}{|{B(x_{0},2\delta/3)}|}\int_{B(x_{0},2\delta/3)}f(z)dz is the average of on the domain of integration.
Having bounded the -norm, we now use the following sampling lemma which is a direct application of Chebyshev’s inequality.
Lemma 3.7** (Monte Carlo sampling Lemma).**
([42, Lemma 38]) Let be a probability space, and be a square integrable function. Let , let be drawn independently at random from with distribution , and let be the empirical average
[TABLE]
Then has mean and variance . In particular, by Chebyshev’s inequality, we have
[TABLE]
Conditioning on and applying this sampling lemma with together with (33), we obtain
[TABLE]
with probability at least where we recall that .
Combining this with (29) gives (28), conditioned on as claimed. ∎
It is left to verify (31) and Lemma 3.6.
Proof of (31).
Since
[TABLE]
has mean at most , applying Markov’s inequality to the random variable
[TABLE]
we conclude that the event (T1) happens with probability at least for any constant .
For (T2), writing and observing that the set is a subset of , we have
[TABLE]
By taking the supremum outside, the right-hand side is at most
[TABLE]
and hence is bounded by
[TABLE]
by projecting onto the real line and conditioning on the random variables .
Applying the anti-concentration Lemma 3.2 with , , and (where we use Condition (A3) and the assumption that to get ), we obtain such that for all ,
[TABLE]
This proves that (T2) holds with probability at least , concluding the proof of (31). ∎
Proof of Lemma 3.6.
This proof is based on [9, Lemma 4.8]. Since , it suffices to show that
[TABLE]
By (30), there exists an such that does not have zeros in the (closed) annulus with center at and radii where .
It is now sufficient to show that
[TABLE]
Let be all zeros of in , then and where is a polynomial having no zeros on the closed ball . By triangle inequality,
[TABLE]
where in the last inequality, we used
[TABLE]
Next, we will bound by finding a uniform upper bound and lower bound for . Since is harmonic in , it attains its extrema on the boundary. Thus,
[TABLE]
Notice that is also harmonic on the ball . For the upper bound of , we claim that for all in ,
[TABLE]
Indeed, since a harmonic function attains its extrema on the boundary, we can assume that . By Condition (T1), . Additionally, by noticing that for all , we get
[TABLE]
as claimed.
As for the lower bound, let , then is a non-negative harmonic function on the ball . By Harnack’s inequality (see [36, Chapter 11]) for the subset of the above ball, we have that for every ,
[TABLE]
where . Hence,
[TABLE]
And so,
[TABLE]
Thus, we reduce to bounding . From Lemma 38 and condition (T2), we have
[TABLE]
And so, , which together with (40) give
[TABLE]
From (36), (37), and (41), we obtain (35) and hence Lemma 3.6. ∎
3.5. Universality of log
Since , by applying (28), the left-hand side of (26) equals
[TABLE]
and the right-hand side of (26) equals
[TABLE]
where
[TABLE]
In this section, we show that the difference between these two identities is small (Lemma 3.8). Before stating the result, note that by (21) and the assumption on the derivatives of , it holds that
[TABLE]
for all .
Lemma 3.8** (Universality of log ).**
There exists a constant such that for every constant , every function that satisfies (43) and every in , we have
[TABLE]
In order to prove Lemma 3.8, we first prove the following smooth version where the log function is replaced by a smooth function. The proof of Lemma 3.8 follows by a routine smoothening argument that we defer to Appendix 8.3. Both proofs are based on [42].
Lemma 3.9**.**
There exists a constant such that for every , every smooth 222By “smooth”, we mean that has continuous derivatives up to order 3. function that satisfies (43) and every in , we have
[TABLE]
where and is the constant in Conditions (A2) and (A3).
Proof of Lemma 3.9..
We use the Lindeberg swapping argument. Let for . Then and and is obtained from by replacing the random variable by . Let
[TABLE]
The left-hand side of (44) is bounded by . Fix (where is the constant in Conditions (A2) and (A3)) and let
[TABLE]
for . We have
[TABLE]
Conditioned on the and for all , the are fixed. To bound , we reduce to bounding
[TABLE]
where . Note that this function also satisfies (43) because does.
Let . By Condition (A3), we have
[TABLE]
and
[TABLE]
Since , we have from (46) and (47) that
[TABLE]
for some constant . Taylor expanding around the origin, we obtain
[TABLE]
where
[TABLE]
Since satisfies (43), we have
[TABLE]
Expanding to the next derivative, we have, in a similar manner,
[TABLE]
where and
[TABLE]
By definition, . Using interpolation, Hölder’s inequality and , we get
[TABLE]
All of these estimates also hold for in place of . Since and have the same first and second moments and they both have bounded moments, we get
[TABLE]
Taking expectation with respect to the remaining variables shows that the same upper bound holds for for all . By (48), choosing to be sufficiently small compared to , we have . Hence,
[TABLE]
where we used , and (2).
For , instead of (49) and (50), we use mean value theorem to get a rough bound
[TABLE]
which by the same arguments as above gives
[TABLE]
Taking all these bounds together, we get . This completes the proof of Lemma 3.9. ∎
3.6. Finishing the proof of Proposition 3.3
In Lemma 3.4, we approximated the number of real roots in dyadic intervals, , by the sums and estimated the error term to be
[TABLE]
as in (22).
Applying this bound for the specific Gaussian case yields
[TABLE]
It remains to show that
[TABLE]
where are the roots of
By (28), the sums can be well approximated by the sample sum
[TABLE]
with probability at least , where are chosen independently, uniformly at random from the ball and are independent of all previous random variables.
Since , by applying (54), the left-hand side of (26) equals
[TABLE]
and the right-hand side of (26) equals
[TABLE]
where is the function defined in (42).
For any fixed , Lemma 3.8 asserts that the difference between these two identities is small
[TABLE]
This gives Proposition 3.3.∎
3.7. From to
In this section, we detail the modifications needed to prove Theorem 2.1 from the proof of (13). For Theorem 2.1, we need to show that
[TABLE]
Inequality (13) restricts to the interval and says that
[TABLE]
We first decompose and into the sum of the numbers of real roots in the intervals , , , and and denote by and with the corresponding number of real roots. For example,
[TABLE]
[TABLE]
and so on.
Note that
[TABLE]
It has been shown in proving (13) how to deal with . To deal with , note that there is a one-to-one correspondence between the real roots of in and the real roots in of the polynomial . Denote this new polynomial by and the original polynomial by . All arguments that have been used for to handle can be applied to to handle .
For , there is a one-to-one correspondence between the roots of in and the roots in of the polynomial . The coefficients of satisfy Condition A3 with , except for a negligible number of , and hence the same arguments as for also apply for .
Similarly, for , there is a one-to-one correspondence between the roots of in and the roots in of the polynomial . All arguments that work for also works for .
In Section 3.1, is partitioned into a sum of , which is the number of roots of in a dyadic interval . Denote these by . Denote by , and the number of roots in the same interval of and , respectively. We have
[TABLE]
All other steps in the proof of (13) can now be written for the proof of Theorem 2.1 by placing these in place of . For example, Proposition 3.3 becomes
[TABLE]
where : is any function whose every partial derivative up to order is bounded by .
4. Proof of Corollary 2.2
We define as in the beginning of the proof of Theorem 2.1. Note that and .
To prove the first part of Corollary 2.2, we first reduce to the interval as explained in Section 3.7, namely, it suffices to show that
[TABLE]
We write , . Let be the event on which (here, can be replaced by any large constant). Let be a smooth function that is supported on the interval and for all . Since is always an integer, it holds that . The function can be chosen such that all of its derivatives up to order are bounded by . Applying Theorem 2.1 to the rescaled function , we obtain
[TABLE]
for some small constant where is the corresponding event on which .
To finish the proof of the first part, we show that the contribution from the complement of is negligible, i.e.,
[TABLE]
Since by (2) and since , it suffices to show that for all , . Let be the event on which . Note that . Let be a large constant. By (14) of Lemma 3.1, . Thus,
[TABLE]
This together with (15) of Lemma 3.1 give
[TABLE]
Since , the right most side is at most by (2) and by choosing .
The second part of Corollary 2.2 follows from the first part by observing that
[TABLE]
where in the first inequality, we used the first part of Corollary 2.2 for , in the second inequality, we used (15) to get that
[TABLE]
and in the last inequality, we used (2). follows immediately by observing that
[TABLE]
where in the first inequality, we used the first part of Corollary 2.2 for , in the second inequality, we used (15) to get that
[TABLE]
and in the last inequality, we used (2). This completes the proof of Corollary 2.2.
5. Proof of Proposition 2.3
5.1. Probability of multiple roots
We start by proving a useful tool that control the probability that the polynomial has many roots in a small interval. For any , let
[TABLE]
and
[TABLE]
Lemma 5.1**.**
Assume that the random variables are iid standard Gaussian. There exists a constant such that for any , any , and satisfying
[TABLE]
for some , we have
[TABLE]
and
[TABLE]
where the implicit constants depend only on , not on .
Proof of Lemma 5.1.
We start by proving (59). By Rolle’s theorem and the fundamental theorem of calculus, if has at least zeros in the interval then
[TABLE]
Therefore,
[TABLE]
where , and
[TABLE]
By (A3), we have the following estimate whose proof is deferred to Appendix 8.4 as it is merely algebraic:
[TABLE]
Since is a Gaussian vector with mean 0 and covariance matrix
\left[\begin{array}[]{cc}1&r(y,t)\\ r(y,t)&1\end{array}\right], we have
[TABLE]
It remains to show that
[TABLE]
Since , there exists such that . By Markov’s inequality, we have
[TABLE]
By Hölder’s inequality, the right-hand side is at most
[TABLE]
and so
[TABLE]
For each , since is a Gaussian random variable, using the hypercontractivity inequality for the Gaussian distribution (see, for example, [4, Corollary 5.21]), we have for some constant ,
[TABLE]
where in the last inequality, we used an estimate similar to (61).
Plugging this and (61) into (63), we obtain
[TABLE]
which gives
[TABLE]
Using , for and , we get
[TABLE]
by choosing . This proves (59).
The inequality (58) is obtained by the same reasoning:
[TABLE]
Thus
[TABLE]
This completes the proof of Lemma 5.1. ∎
5.2. Partition into pieces
Let be the right-hand side of (5):
[TABLE]
Writing as a union of four sets , , and and using triangle inequality, we reduce Proposition 2.3 to showing that for each ,
[TABLE]
We only prove (64) for ; the proofs for the remaining are similar. Since , by triangle inequality, (64) follows from showing that for some large constant ,
[TABLE]
[TABLE]
[TABLE]
and
[TABLE]
where we note that if then the interval is empty and (68) is vacuously true.
The bound (65) is precisely the content of [9, Lemma 2.5]. In the following sections, we show (66), (67), and (68).
5.3. Proof of (66)
Dividing the interval into dyadic intervals , , , (where ) and applying triangle inequality together with (15), we obtain
[TABLE]
Thus,
[TABLE]
proving (66).
5.4. Reducing to Gaussian
To prove (67) and (68), applying (55) to the intervals
and , we get
[TABLE]
and
[TABLE]
Thus, it remains to prove (67) and (68) when the random variables are iid standard Gaussian. So for the rest of this proof, we assume that it is the case.
5.5. Proof of (67)
For (67), we use Hölder’s inequality and (15) to conclude that
[TABLE]
Using the Kac-Rice formula (24), we get
[TABLE]
where we used and . Plugging this into (69) and using (2), we obtain
[TABLE]
5.6. Proof of (68)
By (58), for every interval with
[TABLE]
where is a sufficiently large constant, we have
[TABLE]
Dividing the interval into intervals that satisfy (72), we obtain
[TABLE]
So, (68) follows from (73) and the following
[TABLE]
which can be deduced from
[TABLE]
and
[TABLE]
To prove (75), let . We have for all thanks to assumption A3.
Using Kac-Rice formula (24), we have
[TABLE]
We use [9, Lemma 10.3] with which estimates the above integrand uniformly over the interval and asserts that
[TABLE]
which is by the assumption .
That gives (75) because
[TABLE]
For (76), we use the same bound as in (70) to obtain
[TABLE]
This proves (76) and completes the proof of (68).
6. Proof of Lemma 2.4
Since the lemma only involves Gaussian random variables , we simplify the notation and write for and for (this helps us to avoid multiple superscripts later on). Thus, for this section, for all .
We will adapt the argument in Maslova [28], which is to approximate the number of roots by a sum of independent random variables. Since the random variables are now standard Gaussian, numerous technical steps in [28], which may be impossible to reproduce without having , can be greatly simplified and applied to our general setting thanks to special properties of Gaussian variables.
6.1. Approximate the number of real roots by the number of sign changes.
Let and be defined as in (56) and (57). Lemma 5.1 asserts that in a small interval, it is unlikely that the polynomial has more than root. If has at most root in an interval and does not vanish at and then if and have different signs and otherwise. Hence, on a small interval , it is reasonable to approximate by the number of sign changes:
[TABLE]
where
[TABLE]
The following lemma estimates the accuracy of this approximation for a long interval.
Lemma 6.1** (Approximate by sign changes).**
Assume that the are iid standard Gaussian. For any positive constant , there exist constants such that the following holds. Let and be such that and . Let and where is any number with
[TABLE]
Assume (without loss of generality) that and are integers and let for all Let
[TABLE]
Then
[TABLE]
Proof of Lemma 6.1.
Note that , and
We have
[TABLE]
where , ) and ).
By Lemma 5.1, we have
[TABLE]
For each , we have
[TABLE]
Let . We split the right-hand side into three sums: for the first sum, and for the second sum, and and for the third sum, and denote the corresponding sums by , respectively.
By Lemma 5.1, letting gives
[TABLE]
For , we use Hölder’s inequality to get
[TABLE]
where is a sufficiently large constant and in the next to last inequality, we used Lemma 5.1 in a similar way as in (81). Similarly, . Hence,
[TABLE]
and so
[TABLE]
To complete the proof of the lemma, it remains to bound from below. For each , let
[TABLE]
By Condition (A3), for all and thus, for all ,
[TABLE]
and
[TABLE]
Therefore, in order to bound from below, it suffices to assume that for all for the rest of the proof of Lemma 6.1.
For , we have for every ,
[TABLE]
where . We defer the simple verification of (84) and (85) to Appendix 8.5.
Letting and yields
[TABLE]
Let . To estimate , let us first estimate . We have
[TABLE]
Thus, if then and if then . Combining this with the assumption that , we have for all . This implies
[TABLE]
and
[TABLE]
Plugging this into (83), we obtain
[TABLE]
completing the proof of Lemma 6.1. ∎
6.2. Truncate the polynomial to get independence.
We now show that and (in some rough sense) are independent, whenever the intervals and are relatively far apart. This allows us to approximate by a sum of independent random variables, from which we can derive a Central Limit Theorem.
For any , let
[TABLE]
where is a large constant to be chosen.
Define a truncated version of by
[TABLE]
We get from by a truncation in which the truncation points and depend on the value of . Let
[TABLE]
The following lemma asserts that is a good approximation of and that and are independent when and are far apart.
Lemma 6.2**.**
For every , it holds that
[TABLE]
Moreover, if and if then and are independent because
[TABLE]
Proof of Lemma 6.2.
Since , for all , . We write .
By (61), on the right-most side of (87), we have
[TABLE]
On the other side, we have
[TABLE]
By the same argument as in (118), the right-most sum is at most
[TABLE]
where we used and by the definition of and . Thus,
[TABLE]
where we used by the definition of . This proves (87).
As for the second part of Lemma 6.2, writing and , we have and , so
[TABLE]
This proves Lemma 6.2. ∎
6.3. Approximating sign changes of by those of : short intervals
Let
[TABLE]
be the sign change of on the interval . In the next lemma, we show that is a good approximation of the corresponding sign change of defined in (80).
Lemma 6.3** (Approximation by truncation I).**
Assume that the are iid standard Gaussian. Let be any positive constant. Let with . Then
[TABLE]
Proof of Lemma 6.3.
Using the formula
[TABLE]
we have
[TABLE]
where
[TABLE]
Decompose the plane of into two regions: the square
and its complement. We denote the corresponding integrals on these regions by and , respectively.
First, we show that the contribution from is negligible. Indeed, using the estimates
[TABLE]
we obtain
[TABLE]
From this and the Gaussianity of and , we have
[TABLE]
where .
For , we need to make use of the cancellation between and . We rewrite as
[TABLE]
Using for all , for all , and (87), we get
[TABLE]
where we used Lemma 6.2 (recalling that the random variables are iid standard Gaussian and hence have mean 0) to get
[TABLE]
This completes the proof of Lemma 6.3. ∎
6.4. Approximating sign changes of by those of : long intervals
Lemma 6.4** (Approximation by truncation II).**
Assume that the are iid standard Gaussian. There exist constants such that the following holds. Let and be such that and . Let and where is any number in . Assume (without loss of generality) that and are integers and let for all Let
[TABLE]
Then
[TABLE]
Proof of Lemma 6.4.
By Lemma 6.3, we have
[TABLE]
By the definition of and , we get
[TABLE]
proving Lemma 6.4. ∎
6.5. Control of the fourth moment
The following lemma controls the fourth moment of .
Lemma 6.5** (Bounded forth moment).**
Under the setting of Lemma 6.4 and an additional assumption that , we have
[TABLE]
Proof of Lemma 6.5.
Let be the constant in Lemma 5.1.
Case 1. . Since , it suffices to show that
[TABLE]
For simplicity, we write for . Let as in the setting of Lemma 6.4. By the definition of sign changes, we have with probability ,
[TABLE]
Hence, by Lemma 6.4, Hölder’s inequality, and the assumption that , we have
[TABLE]
Thus, it suffices to show that . Since for any interval ,
[TABLE]
Partition the interval into smaller intervals such that . Since , the number of such sub-intervals is . By (58), for each of these intervals , we have
[TABLE]
Using this and the assumption that of Case 1, we have as desired.
Case 2. . We decompose the sum in into blocks of size of the form
[TABLE]
for each , where is the number of blocks. Notice that .
We have
[TABLE]
Note that each is of the form for some that satisfy . Thus, (91) implies that for all . By Hölder’s inequality, each term in the summation of is of order and so,
[TABLE]
To bound and , we use the independence in Lemma 6.2 to conclude that if then and are independent. Together with the fact that for all , we observe that most terms in the sums are zero. Ignoring these zero terms, we have
[TABLE]
and
[TABLE]
Putting the above bounds together, we obtain Lemma 6.5. ∎
6.6. Proof of Lemma 2.4
Using the results in Sections 6.1 and 6.2, we shall approximate by a sum of independent random variables to prove that it satisfies the CLT. We again recall that in this proof, the are iid standard Gaussian as mentioned at the beginning of this section. Recall the hypothesis (6) that
[TABLE]
In particular, satisfies Condition (2). Let be any constants satisfying
[TABLE]
Let
[TABLE]
[TABLE]
We have . Let
[TABLE]
Observe that and grows with . For simplicity, we will assume that and are integers. In the case that they are not, we only need to replace them by their integer part. As before, let for
Let be defined as in (88). By Lemmas 6.1 and 6.4, we can approximate by
[TABLE]
and get an error term
[TABLE]
where in the last inequality, we used (92) and (93).
Combining this with the assumption that , we get
[TABLE]
Similarly, for the interval , we approximate the number of real roots by
[TABLE]
And for the intervals and , we respectively use
[TABLE]
where . Let . We note that all of the lemmas proven earlier in this section hold for in place of (with the value of being changed to [math] as in Section 3.7). From (95) and its analog for , we have
[TABLE]
Making use of Lemma 6.2, we now approximate by a sum of independent random variables as follows. Let
[TABLE]
and
[TABLE]
where
[TABLE]
By Lemma 6.2, the random variables are mutually independent because . Similarly for the random variables . Moreover, all random variables are mutually independent because the only involve the random variables where (by the definition (86) and the left-most inequality in (92)) while the only involve the random variables where, again, .
To evaluate the accuracy of the approximation of by , consider
[TABLE]
where
[TABLE]
and are defined similarly with respect to .
By Lemma 6.2, the random variables are also mutually independent. Note that each is of the form defined in Lemma 6.4 for some and with By (93) and the definition of in (94), ; this allows us to use Lemma 6.5 to get
[TABLE]
One can obtain a similar estimate for . Thus, the error term of the approximation of by has variance
[TABLE]
Combining this with (95), we get
[TABLE]
The sum is a sum of independent random variables satisfying forth moment bound
[TABLE]
where in the first inequality, we used Lemma 6.5. By the Lyapunov Central Limit Theorem (see for example, [31]), the sum satisfies the Central Limit Theorem.
This and (96) imply that also satisfies the Central Limit Theorem, completing the proof of Lemma 2.4. ∎
7. Proof of Lemma 2.5
Since in this section we only deal with Gaussian random variables, we again use to denote iid standard Gaussian variables (instead of ). This would help avoid complicated notation (such as double superscripts) later on. By symmetry of the Gaussian distribution, we can assume that for all .
Let
[TABLE]
and
[TABLE]
Note that this satisfies Condition (2).
By Proposition 2.3,
[TABLE]
Thus, to prove Lemma 2.5, it suffices to show that
[TABLE]
We have
[TABLE]
where .
Since for any two real random variables and , it suffices to show that
[TABLE]
and
[TABLE]
7.1. Universality for
In order to verify (98), we use the universality method in a novel way. Instead of swapping the random variables , we swap the deterministic coefficients . This allow us to couple with the Kac polynomial and the desired bound follows by known results concerning the variance of the Kac polynomial. This swapping is possible thanks to the fact that the “important” coefficients are which are close to 1 by (1).
Let
[TABLE]
be the corresponding Kac polynomial. We prove the following analogs of Theorem 2.1 and Corollary 2.2 for and .
Proposition 7.1**.**
Assume that the are iid standard Gaussian. Let be any constant. There exists a constant such that for every function whose derivative up to order 3 are bounded by 1 and for every , we have
[TABLE]
Proposition 7.2**.**
Assume that the are iid standard Gaussian. Let be any constant. There exists a constant such that for every , we have
[TABLE]
for . In particular,
[TABLE]
Proposition 7.1 implies Proposition 7.2, using the same arguments as in the proof of Corollary 2.2.
Proof of Proposition 7.1.
We use the same arguments as in the proof of Theorem 2.1 with the following modifications. First, is replaced by and is replaced by , and all of the in the former for a small constant with be replaced by for a large constant . For example, Lemma 3.5 is replaced by the following variant that can be proved using the same argument.
Lemma 7.3**.**
Assume that the are iid standard Gaussian. Let . For any constant and with , we have
[TABLE]
The only remaining difference compared to the proof of Theorem 2.1 is in the proof of the analog of Lemma 3.9, namely for and ( is the largest integer such that ), and for ,
Lemma 7.4**.**
Assume that the are iid standard Gaussian. Let be any positive constant. Let be a smooth function with all derivatives up to order 3 being bounded by . Then for every in , we have
[TABLE]
where .
Assuming this lemma, the rest of the proof of Theorem 2.1 can be adapted in a straightforward manner to complete the proof of Proposition 7.1. ∎
Proof of Lemma 7.4.
While for Lemma 3.9, going from to , we need to swap the general random variables to the Gaussian ones , here, going from to , we need to swap the coefficients to and keep the Gaussian random variables intact. Keeping that in mind, we set for each ,
[TABLE]
We have , and is obtained from by replacing the coefficient by 1.
The difference in (45) for now becomes
[TABLE]
where is obtained from by translation and thus has all derivatives up to order 3 bounded by . The task is to show that
[TABLE]
By the Taylor expansion of order 2, we get
[TABLE]
where
[TABLE]
and
[TABLE]
where we used (1) to get that
[TABLE]
Similarly, we get the expansion for . Subtracting the two expansions and taking expectation both sides (noting again that all of the are iid standard Gaussian and in particular, have mean 0 and variance 1), we obtain
[TABLE]
where in the last inequality, we used , Cauchy-Schwartz inequality and the fact that . Note that for each ,
[TABLE]
where in the second to last inequality, we used Condition (A3) and in the last inequality, we used by the choice of in (97).
Thus, plugging this into (104) and using for any constant ,
[TABLE]
Let
[TABLE]
Splitting the double sum
[TABLE]
into , , and and denoting the corresponding sums by , we obtain
[TABLE]
By assumption (1), we have for every ,
[TABLE]
Hence,
[TABLE]
For , we observe that for all by Condition (A3) and so
[TABLE]
For , we observe that for all by Condition (A3) and that for all ,
[TABLE]
And so,
[TABLE]
Combining these bounds, we obtain
[TABLE]
proving (102) and completing the proof of Lemma 7.4. ∎
7.2. Proof of (98)
As shown in [27], for the Kac polynomial (recall that the random variables are iid standard Gaussian), .
By Proposition 2.3 for the Kac polynomial and the choice of in (97),
[TABLE]
So, by the triangle inequality,
[TABLE]
This together with Proposition 7.2 imply (98). ∎
7.3. Proof of (99)
By a classical formula [23, Theorem 1], we have that for every and for every nonzero polynomial ,
[TABLE]
We will apply this formula for both and . To avoid the improper integrals, we need to cut off the domain of integration. Let , and approximate by
[TABLE]
and
[TABLE]
We first show that is a good approximation of . We claim that for any ,
[TABLE]
To show this, let be all the roots of in the interval and let . We have . Since keeps the same sign on each interval , it holds that
[TABLE]
where we used (89). Thus,
[TABLE]
Divide the interval into equal intervals by the points .
Let (or any small constant). For each , assume that for some . If and then
[TABLE]
and so
[TABLE]
This happens with small probability
[TABLE]
We defer the proof of (107) to Appendix 8.6 as it is similar to the proof of Lemma 5.1. Using this and the union bound over all possible values of , we get
[TABLE]
where we used the fact that is a Gaussian random variable with variance . Plugging this into (106) and using yield
[TABLE]
This proves (105) which means that is a good approximation of .
Next, we show that for all , is also a good approximation of , namely,
[TABLE]
To start, using the fact that for every real number , we have
[TABLE]
Taking the second moment of both sides, we get
[TABLE]
where we again used the fact that satisfies (2). This proves (108).
Combining this with (105), we conclude that for any ,
[TABLE]
We can obtain a similar estimate for . Therefore, in order to prove (99), it suffices to show
[TABLE]
To prove this bound, we need to make a critical use of a property of Gaussian variable. For a standard Gaussian random variable and any real number , . Since are Gaussian for any value of , we have for ,
[TABLE]
where
[TABLE]
in which the sums are taken over all possible assignments of and signs in place of the and
[TABLE]
These formulas follow directly from the definition of ; we provide the tedious derivation in Appendix 8.7 for the reader’s convenience.
We now show that for and for all ,
[TABLE]
We will show it for . The cases are completely similar. We have
[TABLE]
where
[TABLE]
Since and for any constant , we have
[TABLE]
Thus, bounding the first two exponents in (112) by 1, using and , we get that on the domain of integration in (111),
[TABLE]
Finally, using , we have
[TABLE]
proving (111) and completing the proof of (99). ∎
Acknowledgements. Oanh would like to thank François Baccelli and Terence Tao for valuable conversations concerning random polynomials. We thank the anonymous referees for their helpful suggestions.
8. Appendix
8.1. Proof of the Jensen’s inequality (17)
By setting and prove the corresponding inequality for , it suffices to assume that and . Let be the zeros of in . For each inside the unit disk , consider the map
[TABLE]
For and , one can show by algebraic manipulation that
[TABLE]
Moreover, for all and , we have
[TABLE]
Let . Then is an analytic function on . By maximum principle, we have for every ,
[TABLE]
Thus, for all , completing the proof.∎
8.2. Proof of (25)
We first reduce to the hyperbolic polynomials for which the Kac-Rice formula (24) is easier to handle. Consider the hyperbolic polynomial with coefficients , .
By condition (1), for all . Using the Kac-Rice formula (24), we have
[TABLE]
We use [9, Lemma 10.3] with which estimates the above integrand uniformly over the interval for some sufficiently large constant and asserts that
[TABLE]
This together with (113) give (25) for as in this case, .
If , since , for all , we have
[TABLE]
Plugging this into (113) and using the fact that give
[TABLE]
and hence (25) for , completing the proof of (25) for all values of . ∎
8.3. Proof of Lemma 3.8
In this section, we deduce Lemma 3.8 from Lemma 3.9.
The constant in this proof will be a small fraction of the in Lemma 3.9. Let . Then, still satisfies (43) and we can reduce the problem to showing that
[TABLE]
Ideally, we would like to set and apply Lemma 3.9 for this function . However, the singularity of the log function at 0 prevents from satisfying (43). To handle this difficulty, we split the space of into two regions and where is the image of the log function around 0 and show that the contribution from is insignificant. On , the log function is well-behaved and we can then apply Lemma 3.9 there.
More specifically, for , let
[TABLE]
and
[TABLE]
Let be a smooth function taking values in such that is supported in , on the complement of and for all . Put , and . We have and both satisfy (43) with ,
We now show that the contribution from is negligible. Set and
[TABLE]
Since , we observe that satisfies
- •
,
- •
,
- •
is constant on ,
- •
satisfies (43) (with the power being replaced by but that doesn’t affect the argument).
Choose to be small enough such that is at most the constant in Lemma 3.9 where is some sufficiently large absolute constant. Applying Lemma 3.9, we get
[TABLE]
Since the variables are Gaussian, we have
[TABLE]
Thus, . Finally, we will show that
[TABLE]
Define by . Since ,
[TABLE]
Thus, is well-defined and satisfies (43) (with the power being replaced by ). Applying Lemma 3.9 gives
[TABLE]
This completes the proof of Lemma 3.8. ∎
8.4. Proof of (61)
In this section, we prove (61), namely, for a sufficiently large constant , we have
[TABLE]
To this end, we will repeatedly use (A3) and the assumption that .
If , we have
[TABLE]
For the lower bound, we have and so
[TABLE]
These bounds prove (115) for .
If , letting , we have and
[TABLE]
As for the upper bound, we have for any constant ,
[TABLE]
Since , the right-most sum is at most
[TABLE]
By choosing sufficiently large (depending only on ) such that the function is decreasing on , we have
[TABLE]
Plugging this into (116) and (117), we obtain which is the desired upper bound. ∎
8.5. Proof of (84) and (85)
Proof of (84).
We need to show that
[TABLE]
By Condition (A3), for all , and so the left-hand side of (119) is at most the order of the right-hand side. To prove the reverse, by Condition (A3), for all , so
[TABLE]
Thus, it remains to show that the remaining terms on the right-hand side of (119) are of smaller order, namely, for all ,
[TABLE]
Since is a constant, for all and since , we have for ,
[TABLE]
Assume without loss of generality that . Using the simple observation that
[TABLE]
we have
[TABLE]
And so, the right-most side of (121) is of order at most the right-most side of (120), proving (119). ∎
Proof of (85).
We want to show that for every ,
[TABLE]
where . By Taylor’s expansion, we have
[TABLE]
Thus, it suffices to show that
[TABLE]
We have
[TABLE]
and so, it is left to verify that for all ,
[TABLE]
Indeed, we have
[TABLE]
Using by the assumption (6), we obtain
[TABLE]
∎
8.6. Proof of (107)
Let . We want to show that for any interval with ,
[TABLE]
Let denote the above double integral. By Markov’s inequality and Hőlder’s inequality, for a large constant to be chosen, we have
[TABLE]
and so,
[TABLE]
Since is a Gaussian random variable, by the hypercontractivity of Gaussian distribution, we have
[TABLE]
Thus, by choosing ,
[TABLE]
where we used for any constant as satisfies Condition (2). ∎
8.7. Proof of (7.3)
We have
[TABLE]
Thus, we have by definition of that
[TABLE]
where
[TABLE]
Note that we can use Fubini’s theorem in the above calculation because the integrands are absolutely integrable.
We have
[TABLE]
and so
[TABLE]
We recall that the random variables are iid standard Gaussian and for a standard Gaussian random variable and any real number , . Thus,
[TABLE]
Similarly,
[TABLE]
and
[TABLE]
The reference list from the paper itself. Each links out to its DOI / PubMed record.
- 1[1] Jean-Marc Azaïs, Federico Dalmao, and José R León. CLT for the zeros of classical random trigonometric polynomials. In Annales de l’Institut Henri Poincaré, Probabilités et Statistiques , volume 52, pages 804–820. Institut Henri Poincaré, 2016.
- 2[2] Jean-Marc Azaïs and José León. CLT for crossings of random trigonometric polynomials. Electronic Journal of Probability , 18, 2013.
- 3[3] Albert T Bharucha-Reid and M Sambandham. Random Polynomials: Probability and Mathematical Statistics: a Series of Monographs and Textbooks . Academic Press, 1986.
- 4[4] Stéphane Boucheron, Gábor Lugosi, and Pascal Massart. Concentration inequalities: A nonasymptotic theory of independence . Oxford University Press, 2013.
- 5[5] Federico Dalmao. Asymptotic variance and CLT for the number of zeros of Kostlan Shub Smale random polynomials. Comptes Rendus Mathematique , 353(12):1141–1145, 2015.
- 6[6] Minaketan Das. Real zeros of a class of random algebraic polynomials. J. Indian Math. Soc , 36:53–63, 1972.
- 7[7] Amir Dembo, Sumit Mukherjee, et al. No zero-crossings for random polynomials and the heat equation. Annals of Probability , 43(1):85–118, 2015.
- 8[8] Amir Dembo, Bjorn Poonen, Qi-Man Shao, and Ofer Zeitouni. Random polynomials having few or no real zeros. Journal of the American Mathematical Society , 15(4):857–892, 2002.
