# Random Trigonometric Polynomials with Nonidentically Distributed Coefficients

Authors: K. Farahmand; T. Li
Journal: International Journal of Stochastic Analysis
 (2010)

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## Abstract

This paper provides asymptotic estimates for the expected number of real zeros of two different forms of random trigonometric polynomials, where the coefficients of polynomials are normally distributed random variables with different means and variances. For the polynomials in the form ofa0+a1cosθ+a2cos2θ+⋯+ancosnθ and a0+a1cosθ+b1sinθ+a2cos2θ+b2sin2θ+⋯+ancosnθ+bnsinnθ, we give a closed form for the above expected value. With some mild assumptions on the coefficients we allow the
means and variances of the coefficients to differ from each others. A case of
reciprocal random polynomials for both above cases is studied.


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## Body

## 1. Introduction

There are mainly two different forms of random trigonometric polynomial previously studied. They are(1.1)T(θ)=∑j=0najcosjθ,D(θ)=∑j=0n(ajcosjθ+bjsinjθ).
Dunnage [1] first studied the classical random trigonometric polynomial T(θ). He showed that in the case of identically and normally distributed coefficients {aj}j=0n with μj≡0 and σj2≡1, j=0,1,2,…,n, the number of real zeros in the interval (0,2π), outside of an exceptional set of measure zero, is 2n/3+O{n11/13(logn)3/13}, when n is large. Subsequent papers mostly assumed an identical distribution for the coefficients and obtained 2n/3 as the asymptotic formula for the expected number of real zeros. In [2–4] it is shown that this asymptotic formula remains valid when the expected number of real zeros of the equation T(θ)=K, known as K-level crossing, is considered. The work of Sambandham and Renganathan [5] and Farahmand [6] among others obtained this result for different assumptions on the distribution of the coefficients. Earlier works on random polynomials have been reviewed in Bharucha-Reid and Sambandham [7], which includes a comprehensive reference.Later Farahmand and Sambandham [8] study a case of coefficients with different means and variances, which shows an interesting result for the expected number of level crossings in the interval (0,2π). Based on this work, we study the following two cases in order to better understand how the behavior of random trigonometric polynomials is affected by the different assumptions of the distribution on the coefficients for both T(θ) and D(θ), defined above.To this end we allow all the coefficients to have different means and variances. Also, motivated by the recent developments on random reciprocal polynomials, we assume the coefficientsaj and an-j have the same distribution. In [9] for the case of random algebraic polynomial aj≡an-j is assumed. Further in order to overcome the analysis we have to make the following assumptions on the means and variances. Let max{σj2}=σn*2=O(n2/3) and also |σj2-σj+12|≤σn*2/n. For the means, we assume max{|μj|}=μn*=O(n) and |μj-μj+2|≤μn*/n. We also need σn*2,σ*n2, where σ*n2=min{σj2} and μn* is chosen such that for any positive constant δ, (σn*3nδ-1/σ*n3)→0 and (μn*nδ-1/2/σ*n)→0 as n→∞. Then for σ*n2 finite, we have the following theorem.Theorem 1.1.
If the coefficientsaj, j=1,…,n of T(θ)=∑j=0najcosjθ are normally distributed with mean μj and variance σj2, where σj2=σn-j2, then the mathematical expectation of the number of real zeros of the T(θ) satisfies
(1.2)ENT(0,2π)=2∑j=0(n-1)/2σj2(n2/2-nj+j2)∑j=0(n-1)/2σj2.We study the case ofD(θ) in Theorem 3.1 later. We first give some necessary identities.


## 2. Preliminary Analysis

In order to be able to prove the theorem, we need to define some auxiliary results. Let(2.1)A2=var{T(θ)},B2=var{T′(θ)},C=cov{T(θ),T′(θ)},Δ=A2B2-C2,α=E{T(θ)},β=E{T′(θ)}.
Then from Farahmand [10, page 43], we have the extension of the Kac-Rice formula for our case as(2.2)EN(a,b)=I1(a,b)+I2(a,b),
where(2.3)I1(a,b)=∫abΔπA2exp(-α2B2+β2A2-2αβC2Δ2)dθ,I2(a,b)=∫ab2|βA2-Cα|πA3exp(-α22A2)erf(|βA2-Cα|2AΔ)dθ.
As usual, erf(|βA2-Cα|/2AΔ) is the function defined as(2.4)erf(x)=∫0xexp(-t2)dt=πΦ(x2)-π2.
Now we are going to define the following functions to make the estimations. At first, we define S(θ)=sin(2n+1)θ/sinθ and to be continuous at θ=jπ see also [10, page 74]. Let ϵ be any positive value arbitrary at this point, to be defined later. Since for θ∈(ɛ,π-ɛ),(π+ɛ,2π-ɛ), we have |S(θ)|<1/sinɛ, we can obtain(2.5)S(θ)=O(1/ɛ).
Furthermore,(2.6)S′(θ)=(2n+1)cos(2n+1)θsinθ-cotθS(θ)=O(nɛ),S′′(θ)=-(2n+1)2sin(2n+1)θsinθ-(2n+1)cosθcos(2n+1)θsin2θ-cotθS′(θ)+csc2θS(θ)=O(n2ɛ).Now using the above identities and by expandingsinθ(1+2∑j=1ncos2jθ), we can show(2.7)∑j=1ncos2jθ=sin(2n+1)θ2sinθ-12=S(θ)-12=O(1ɛ),∑j=1njsin2jθ=-12{∑j=1ncos2jθ}′=-14S′(θ)=O(nɛ),∑j=1nj2cos2jθ=-14{∑j=1ncos2jθ}′′=-18S′′(θ)=O(n2ɛ).
In a similar way to [10], we define Q(θ)=cosθ-cos(2n+1)θ/2sinθ, then for θ∈(ɛ,π-ɛ),(π+ɛ,2π-ɛ), since cosθ-cos(2n+1)θ=2sin(n+1)θsinnθ, we have |Q(θ)|<1/sinɛ. Hence, we can obtain(2.8)Q(θ)=O(1ɛ).
Furthermore, we have
(2.9)Q′(θ)=-sinθ+(2n+1)sin(2n+1)θ2sinθ-cotθQ(θ)=O(nɛ),Q′′(θ)=-cosθ+(2n+1)2cos(2n+1)θ2sinθ+cosθ[sinθ-(2n+1)sin(2n+1)θ]2sin2θ-cotθQ′(θ)+csc2θQ(θ)=O(n2ɛ).
Now using these identities for Q(θ), Q′(θ), and Q′′(θ) and by expanding 2sinθ∑j=1nsin2jθ, we can get a series of the following results:(2.10)∑j=1nsin2jθ=cosθ-cos(2n+1)θ2sinθ=Q(θ)=O(1ɛ),∑j=1njcos2jθ=12{∑j=1nsin2jθ}′=12Q′(θ)=O(nɛ),∑j=1nj2sin2jθ=-14{∑j=1nsin2jθ}′′=-14Q′′(θ)=O(n2ɛ).
Now we are in position to give a proof for Theorem 1.1 for T(θ) in the intervals (ϵ,π-ϵ) and (π+ϵ,2π-ϵ). In order to avoid duplication the remaining intervals for both cases of T(θ) and D(θ) are discussed together later.


## 3. The Proof

Case 1.
Here we study the random trigonometric polynomial in the classical form ofT(θ)=∑j=0najcosjθ as assumed in Theorem 1.1 and prove the theorem in this section. To this end, we have to get all the terms in the Kac-Rice formula, such as A2,B2,C,α,andβ. Since the property σj2=σn-j2, using the results obtained in Section 2 of (2.7) and (2.10), we can have all the terms needed to calculate formula (2.2).At first, we get the variance of the polynomial, that is,
(3.1)AT2=∑j=0(n-1)/2var{aj}{cos2jθ+cos2(n-j)θ}=∑j=0(n-1)/2σj2+cosnθ∑j=0(n-1)/2σj2{cos(n-2j)θ}=∑j=0(n-1)/2σj2+O(σn*2ɛ).
Next, we calculate the variance of its derivative T′(θ) with respect to θ:
(3.2)BT2=∑j=0(n-1)/2var{aj}{j2sin2jθ+(n-j)2sin2(n-j)θ}=∑j=0(n-1)/2σj2(n22-nj+j2)-∑j=0(n-1)/2σj2j22cos2jθ-∑j=0(n-1)/2σj2(n22-nj+j2)cos2(n-j)θ=∑j=0(n-1)/2σj2(n22-nj+j2)+O(n2σn*2ɛ).
At last, it turns to the covariance between the polynomial and its derivative
(3.3)CT=∑j=0(n-1)/2-jσj2sinjθcosjθ=-∑j=0(n-1)/2j2σj2{sin2jθ-sin2(n-j)θ}-n2∑j=0(n-1)/2σj2sin2(n-j)θ=O(nσn*2ɛ).
Then, from (3.1), (3.2), and (3.3), we can get
(3.4)ΔT2=AT2BT2-CT2=∑j=0(n-1)/2σj2∑j=0(n-1)/2σj2(n22-nj+j2)+O(n2σn*2ɛ).
It is also easy to obtain the means of T(θ) and its derivative as
(3.5)αT=∑j=0nμjcosjθ=O(μn*ɛ),βT=-∑j=0njμjsinjθ=O(nμn*ɛ).
From (2.3) and the results of (3.1)–(3.5), we therefore have
(3.6)ENT(ɛ,π-ɛ)~∫ɛπ-ɛΔπA2dθ~∫ɛπ-ɛ∑j=0(n-1)/2σj2(n2/2-nj+j2)π2∑j=0(n-1)/2σj2dθ=∑j=0(n-1)/2σj2(n2/2-nj+j2)∑j=0(n-1)/2σj2.
Now before considering the zeros in the small interval of length ϵ we consider the polynomial D(θ).Case 2.
We have to make the assumptions a little different. In this case letmax|{σaj2}-{σbj2}|=σn*2 and |σaj2+σbj2-σaj+12-σbj+12|≤σn*2/n. For the means, we assume max{μaj}=μan* and |μaj-μaj+2|≤μan*/n,max{μbj}=μbn* and |μbj-μbj+2|≤μbn*/n, and max|{μaj}-{μbj}|=μn*.Theorem 3.1.
Consider the polynomialD(θ)=∑j=0najcosjθ+bjsinjθ, where aj,bj are independent, normally distributed random variables, divided into n groups each with its own mean μaj,μbj and variance σaj2,σbj2, j=1,2,…,n,. The expected number of real zeros of D(θ) satisfies
(3.7)END(0,2π)=2∑j=0nj2{σaj2+σbj2}∑j=0n{σaj2+σbj2}.Similarly, using the same results obtained from Section2, we can get the following terms. At first, we get the means of the polynomial and its derivative separately
(3.8)αD=∑j=0nμajcosjθ+∑j=0nμbjsinjθ=O(μan*+μbn*ɛ),βD=-∑j=0njμajsinjθ+∑j=0njμbjcosjθ=O(nμn*ɛ).
Then we obtain the variance of the polynomial(3.9)AD2=var{∑j=0najcosjθ+bjsinjθ}=12∑j=0n{σaj2+σbj2}+12∑j=0n{σaj2-σbj2}cos2jθ=12∑j=0n{σaj2+σbj2}+O(σn*2ɛ).
Next, we calculate the variance of its derivative with respect to θ:(3.10)BD2=var{∑j=0n-jajsinjθ+bjcosjθ}=12∑j=0nj2{σaj2+σbj2}+12∑j=0nj2{σbj2-σaj2}cos2jθ=12∑j=0nj2{σaj2+σbj2}+O(n2σn*2ɛ).
At last, it turns to the covariance between the polynomial and its derivative:(3.11)CD=-E{aj2}jsinjθcosjθ+E{bj2}jsinjθcosjθ+E2{aj}jsinjθcosjθ-E2{bj}jsinjθcosjθ=∑j=0n-j{σbj2-σaj2}sinjθcosjθ=O(nσn*2ɛ).
Then, from (3.9), (3.10), and (3.11), we can get(3.12)ΔD2=14∑j=0n{σaj2+σbj2}∑j=0nj2{σaj2+σbj2}+O(nɛ).
From (2.3) and (3.8)–(3.12), we therefore have(3.13)END(ϵ,π-ϵ)~∫ɛπ-ɛΔπA2dθ~∫ɛπ-ɛ∑j=0nj2{σaj2+σbj2}π2∑j=0n{σaj2+σbj2}dθ=∑j=0nj2{σaj2+σbj2}∑j=0n{σaj2+σbj2}.This is the main contribution to the number of real zeros. In the following we show there is a negligible number of zeros in the remaining intervals of lengthϵ. For the number of real roots in the interval (0,ɛ), (2π-ϵ,2π) or (π-ɛ,π+ɛ), we use Jensen's theorem [11, page 300]. The method we used here is applicable to both of the cases we discussed above. Here we take the first case as the example to prove that the roots of these intervals are negligible. Let m=∑j=0nμj and s2=∑j=0nσj2. As T(0) is normally distributed with mean m and variance s2, for any constant k(3.14)Pr(-n-k≤D(0)≤n-k)=(2πs2)-1/2∫-n-kn-kexp{-(t-m)22s2}dt<2nkπs2.
Also since |cos(2ɛeiθ)|≤2e2nɛ we have(3.15)D(2ɛeiθ)≤2e2nɛ∑j=1n|aj|≤2(n+1)e2nɛmaxj=1,…,n|aj|.
Now by Chebyshev's inequality, for any k>2, we can find a positive constant d such that for 0≤j≤n,(3.16)Pr(|aj|≥n-k)≤E(aj2)n2k≤dnk.
since μn*2+σn*2≤nk. Therefore, except for sample functions in an ω-set of measures not exceeding dn-k,(3.17)T(2ɛeiθ)≤3n1+2kexp(2nɛ).
So we obtain(3.18)|T(2ɛeiθ,ω)T(0,ω)|≤3n1+2kexp(2nɛ),
except for the sample functions in an ω-set of measure not exceeding dn-k+2/nkπs2.This implies that we can find an absolute constantd' such that(3.19)Pr{N(ɛ)>2nɛ+(1+2k)logn+log3}≤n-k(d+2s2π)≤d′n-k.
Let [3nɛ] be the greatest integer less than or equal to 3nɛ. Then since the number of real zeros of D(θ) is at most 2n we have(3.20)EN(ɛ)=∑j>02nPr{N(ɛ)≥j}≤3nɛ+∑j=[3nɛ]+12nPr{N(ɛ)>j}≤3nɛ+d′n1-k=O(nɛ),
where we choose(3.21)ϵ=max{μn*σ*nn1/2-δ,σn*3σ*n3n1-δ},
and δ is any positive number, so the error terms become small and we can prove the theorem.



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