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 of a 0 a 1 cos θ a 2 cos 2θ · · · a n cos nθ and a 0 a 1 cos θ b 1 sin θ a 2 cos 2θ b 2 sin 2θ · · · a n cos nθ b n sin nθ, 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

K Farahmand

T Li

CiteSeerX

Hindawi Publishing Corporation
International Journal of Stochastic Analysis
Volume 2010, Article ID 931565, 10 pages
doi:10.1155/2010/931565
Research Article
Random Trigonometric Polynomials with
Nonidentically Distributed Coefficients
K. Farahmand and T. Li
Department of Mathematics, University of Ulster at Jordanstown, Co. Antrim, BT37 0QB, UK
Correspondence should be addressed to K. Farahmand, k.farahmand@ulster.ac.uk
Received 17 December 2009; Accepted 9 February 2010
Academic Editor: Bradford Allen
Copyright q 2010 K. Farahmand and T. Li. This is an open access article distributed under the
Creative Commons Attribution License, which permits unrestricted use, distribution, and
reproduction in any medium, provided the original work is properly cited.
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
of a0+a1 cos θ+a2 cos 2θ+· · ·+an cosnθ and a0+a1 cos θ+b1 sin θ+a2 cos 2θ+b2 sin 2θ+· · ·+an cosnθ+
bn sinnθ,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.
1. Introduction
There are mainly two different forms of random trigonometric polynomial previously
studied. They are
T(θ) =
n∑
j=0
aj cos jθ,
D(θ) =
n∑
j=0
(
aj cos jθ + bj sin jθ
)
.
(1.1)
Dunnage [1] first studied the classical random trigonometric polynomial T(θ). He showed
that in the case of identically and normally distributed coefficients {aj}nj=0 with μj ≡ 0
and σ2j ≡ 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
2 International Journal of Stochastic Analysis
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
coefficients aj 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 themeans and variances. Letmax{σ2j } = σ∗n2 = O(n2/3) and also
|σ2j −σ2j+1| ≤ σ∗n2/n. For the means, we assume max{|μj |} = μ∗n = O(
√
n) and |μj −μj+2| ≤ μ∗n/n.
We also need σ∗2n , σ
2
∗n, where σ∗n
2 = min{σ2j } and μ∗n is chosen such that for any positive
constant δ, (σ∗3n n
δ−1/σ3∗n) → 0 and (μ∗nnδ−1/2/σ∗n) → 0 as n → ∞. Then for σ2∗n finite, we
have the following theorem.
Theorem 1.1. If the coefficients aj , j = 1, . . . , n of T(θ) =
∑n
j=0 aj cos jθ are normally distributed
with mean μj and variance σ2j , where σ
2
j = σ
2
n−j , then the mathematical expectation of the number of
real zeros of the T(θ) satisfies
ENT (0, 2π) = 2
√√√√√
∑(n−1)/2
j=0 σ
2
j
(
n2/2 − nj + j2)
∑(n−1)/2
j=0 σ
2
j
. (1.2)
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
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)}.
(2.1)
Then from Farahmand [10, page 43], we have the extension of the Kac-Rice formula for our
case as
EN(a, b) = I1(a, b) + I2(a, b), (2.2)
International Journal of Stochastic Analysis 3
where
I1(a, b) =
∫b
a
Δ
πA2
exp
(
−α
2B2 + β2A2 − 2αβC
2Δ2
)
dθ,
I2(a, b) =
∫b
a
√
2
∣∣βA2 − Cα∣∣
πA3
exp
(
− α
2
2A2
)
erf
(∣∣βA2 − Cα∣∣√
2AΔ
)
dθ.
(2.3)
As usual, erf (|βA2 − Cα|/√2AΔ) is the function defined as
erf(x) =
∫x
0
exp
(
−t2
)
dt =
√
πΦ
(
x
√
2
)
−
√
π
2
. (2.4)
Now we are going to define the following functions to make the estimations. At first, we
define S(θ) = sin(2n + 1)θ/ sin θ 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 ∈ (ε, π − ε), (π +
ε, 2π − ε), we have |S(θ)| < 1/ sin ε, we can obtain
S(θ) = O(1/ε). (2.5)
Furthermore,
S′(θ) =
(2n + 1) cos(2n + 1)θ
sin θ
− cot θS(θ) = O
(
n
ε
)
,
S′′(θ) =
−(2n + 1)2 sin(2n + 1)θ
sin θ
− (2n + 1) cos θ cos(2n + 1)θ
sin 2θ
− cot θS′(θ) + csc 2θS(θ)
= O
(
n2
ε
)
.
(2.6)
Now using the above identities and by expanding sin θ(1 + 2
∑n
j=1 cos 2jθ), we can
show
n∑
j=1
cos 2jθ =
sin(2n + 1)θ
2 sin θ
− 1
2
=
S(θ) − 1
2
= O
(
1
ε
)
,
n∑
j=1
j sin 2jθ = −1
2
⎧
⎨
⎩
n∑
j=1
cos 2jθ
⎫
⎬
⎭
′
= −1
4
S′(θ) = O
(
n
ε
)
,
n∑
j=1
j2 cos 2jθ = −1
4
⎧
⎨
⎩
n∑
j=1
cos 2jθ
⎫
⎬
⎭
′′
= −1
8
S′′(θ) = O
(
n2
ε
)
.
(2.7)
4 International Journal of Stochastic Analysis
In a similar way to [10], we define Q(θ) = cos θ − cos(2n + 1)θ/2 sin θ, then for θ ∈ (ε, π −
ε), (π + ε, 2π − ε), since cos θ − cos(2n + 1)θ = 2 sin(n + 1)θ sinnθ, we have |Q(θ)| < 1/ sin ε.
Hence, we can obtain
Q(θ) = O
(
1
ε
)
. (2.8)
Furthermore, we have
Q′(θ) =
− sin θ + (2n + 1) sin(2n + 1)θ
2 sin θ
− cot θQ(θ)
= O
(
n
ε
)
,
Q′′(θ) =
− cos θ + (2n + 1)2 cos(2n + 1)θ
2 sin θ
+
cos θ[sin θ − (2n + 1) sin(2n + 1)θ]
2 sin 2θ
− cot θQ′(θ) + csc 2θQ(θ)
= O
(
n2
ε
)
.
(2.9)
Now using these identities for Q(θ), Q′(θ), and Q′′(θ) and by expanding 2 sin θ
∑n
j=1 sin 2jθ,
we can get a series of the following results:
n∑
j=1
sin 2jθ =
cos θ − cos(2n + 1)θ
2 sin θ
= Q(θ) = O
(
1
ε
)
,
n∑
j=1
j cos 2jθ =
1
2
⎧
⎨
⎩
n∑
j=1
sin 2jθ
⎫
⎬
⎭
′
=
1
2
Q′(θ) = O
(
n
ε
)
,
n∑
j=1
j2 sin 2jθ = −1
4
⎧
⎨
⎩
n∑
j=1
sin 2jθ
⎫
⎬
⎭
′′
= −1
4
Q′′(θ) = O
(
n2
ε
)
.
(2.10)
Nowwe 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 of T(θ) =∑n
j=0 aj cos jθ 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, and β. Since the
International Journal of Stochastic Analysis 5
property σ2j = σ
2
n−j , 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,
A2T =
(n−1)/2∑
j=0
var
{
aj
}{
cos2jθ + cos2
(
n − j)θ
}
=
(n−1)/2∑
j=0
σ2j + cosnθ
(n−1)/2∑
j=0
σ2j
{
cos
(
n − 2j)θ}
=
(n−1)/2∑
j=0
σ2j +O
(
σ∗2n
ε
)
.
(3.1)
Next, we calculate the variance of its derivative T ′(θ) with respect to θ:
B2T =
(n−1)/2∑
j=0
var
{
aj
}{
j2sin2jθ +
(
n − j)2sin2(n − j)θ
}
=
(n−1)/2∑
j=0
σ2j
(
n2
2
− nj + j2
)
−
(n−1)/2∑
j=0
σ2j
j2
2
cos 2jθ
−
(n−1)/2∑
j=0
σ2j
(
n2
2
− nj + j2
)
cos 2
(
n − j)θ
=
(n−1)/2∑
j=0
σ2j
(
n2
2
− nj + j2
)
+O
(
n2σ∗2n
ε
)
.
(3.2)
At last, it turns to the covariance between the polynomial and its derivative
CT =
(n−1)/2∑
j=0
− jσ2j sin jθ cos jθ
= −
(n−1)/2∑
j=0
j
2
σ2j
{
sin 2jθ − sin 2(n − j)θ} − n
2
(n−1)/2∑
j=0
σ2j sin 2
(
n − j)θ
= O
(
nσ∗2n
ε
)
.
(3.3)
Then, from (3.1), (3.2), and (3.3), we can get
Δ2T = A
2
TB
2
T − C2T =
(n−1)/2∑
j=0
σ2j
(n−1)/2∑
j=0
σ2j
(
n2
2
− nj + j2
)
+O
(
n2σ∗2n
ε
)
. (3.4)
6 International Journal of Stochastic Analysis
It is also easy to obtain the means of T(θ) and its derivative as
αT =
n∑
j=0
μj cos jθ = O
(
μ∗n
ε
)
,
βT = −
n∑
j=0
jμj sin jθ = O
(
nμ∗n
ε
)
.
(3.5)
From (2.3) and the results of (3.1)–(3.5), we therefore have
ENT (ε, π − ε) ∼
∫π−ε
ε
Δ
πA2
dθ
∼
∫π−ε
ε
√√√√√
∑(n−1)/2
j=0 σ
2
j
(
n2/2 − nj + j2)
π2
∑(n−1)/2
j=0 σ
2
j
dθ
=
√√√√√
∑(n−1)/2
j=0 σ
2
j
(
n2/2 − nj + j2)
∑(n−1)/2
j=0 σ
2
j
.
(3.6)
Now before considering the zeros in the small interval of length εwe consider the polynomial
D(θ).
Case 2. We have to make the assumptions a little different. In this case let max |{σ2aj}−{σ2bj}| =
σ∗n
2 and |σ2aj + σ2bj − σ2aj+1 − σ2bj+1| ≤ σ∗n2/n. For the means, we assume max{μaj} = μ∗an and
|μaj − μaj+2| ≤ μ∗an/n,max{μbj} = μ∗bn and |μbj − μbj+2| ≤ μ∗bn/n, and max |{μaj} − {μbj}| = μ∗n.
Theorem 3.1. Consider the polynomial D(θ) =
∑n
j=0 aj cos jθ + bj sin jθ, where aj , bj are
independent, normally distributed random variables, divided into n groups each with its own mean
μaj , μbj and variance σ2aj , σ
2
bj
, j = 1, 2, . . . , n,. The expected number of real zeros of D(θ) satisfies
END(0, 2π) = 2
√√√√√√
∑n
j=0 j
2
{
σ2aj + σ
2
bj
}
∑n
j=0
{
σ2aj + σ
2
bj
} . (3.7)
Similarly, using the same results obtained from Section 2, we can get the following
terms. At first, we get the means of the polynomial and its derivative separately
αD =
n∑
j=0
μaj cos jθ +
n∑
j=0
μbj sin jθ = O
(
μ∗an + μ
∗
bn
ε
)
,
βD = −
n∑
j=0
jμaj sin jθ +
n∑
j=0
jμbj cos jθ = O
(
nμ∗n
ε
)
.
(3.8)
International Journal of Stochastic Analysis 7
Then we obtain the variance of the polynomial
A2D = var
⎧
⎨
⎩
n∑
j=0
aj cos jθ + bj sin jθ
⎫
⎬
⎭
=
1
2
n∑
j=0
{
σ2aj + σ
2
bj
}
+
1
2
n∑
j=0
{
σ2aj − σ2bj
}
cos 2jθ
=
1
2
n∑
j=0
{
σ2aj + σ
2
bj
}
+O
(
σ∗n
2
ε
)
.
(3.9)
Next, we calculate the variance of its derivative with respect to θ:
B2D = var
⎧
⎨
⎩
n∑
j=0
− jaj sin jθ + bj cos jθ
⎫
⎬
⎭
=
1
2
n∑
j=0
j2
{
σ2aj + σ
2
bj
}
+
1
2
n∑
j=0
j2
{
σ2bj − σ2aj
}
cos 2jθ
=
1
2
n∑
j=0
j2
{
σ2aj + σ
2
bj
}
+O
(
n2σ∗n
2
ε
)
.
(3.10)
At last, it turns to the covariance between the polynomial and its derivative:
CD = −E
{
a2j
}
j sin jθ cos jθ + E
{
b2j
}
j sin jθ cos jθ
+ E2
{
aj
}
j sin jθ cos jθ − E2{bj
}
j sin jθ cos jθ
=
n∑
j=0
− j
{
σ2bj − σ2aj
}
sin jθ cos jθ
= O
(
nσ∗n
2
ε
)
.
(3.11)
Then, from (3.9), (3.10), and (3.11), we can get
Δ2D =
1
4
n∑
j=0
{
σ2aj + σ
2
bj
} n∑
j=0
j2
{
σ2aj + σ
2
bj
}
+O
(
n
ε
)
. (3.12)
8 International Journal of Stochastic Analysis
From (2.3) and (3.8)–(3.12), we therefore have
END(ε, π − ε) ∼
∫π−ε
ε
Δ
πA2
dθ
∼
∫π−ε
ε
√√√√√√
∑n
j=0 j
2
{
σ2aj + σ
2
bj
}
π2
∑n
j=0
{
σ2aj + σ
2
bj
} dθ
=
√√√√√√
∑n
j=0 j
2
{
σ2aj + σ
2
bj
}
∑n
j=0
{
σ2aj + σ
2
bj
} .
(3.13)
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 length ε. For the number of
real roots in the interval (0, ε), (2π − ε, 2π) or (π − ε, π + ε), 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 =
∑n
j=0 μj and s
2 =
∑n
j=0 σ
2
j . As T(0) is normally distributed with mean m and variance
s2, for any constant k
Pr
(
−n−k ≤ D(0) ≤ n−k
)
=
(
2πs2
)−1/2∫n−k
−n−k
exp
{
− (t −m)
2
2s2
}
dt
<
2
nk
√
πs2
.
(3.14)
Also since | cos(2εeiθ)| ≤ 2e2nε we have
D
(
2εeiθ
)
≤ 2e2nε
n∑
j=1
∣∣aj
∣∣ ≤ 2(n + 1)e2nε max
j=1,...,n
∣∣aj
∣∣. (3.15)
Now by Chebyshev’s inequality, for any k > 2, we can find a positive constant d such that for
0 ≤ j ≤ n,
Pr
(∣∣aj
∣∣ ≥ n−k
)
≤
E
(
a2j
)
n2k
≤ d
nk
. (3.16)
since μ∗2n + σ
∗2
n ≤ nk. Therefore, except for sample functions in an ω-set of measures not
exceeding dn−k,
T
(
2εeiθ
)
≤ 3n1+2k exp(2nε). (3.17)
International Journal of Stochastic Analysis 9
So we obtain
∣∣∣∣∣
T
(
2εeiθ, ω
)
T(0, ω)
∣∣∣∣∣ ≤ 3n
1+2k exp(2nε), (3.18)
except for the sample functions in an ω-set of measure not exceeding dn−k + 2/nk
√
πs2.
This implies that we can find an absolute constant d′ such that
Pr
{
N(ε) > 2nε + (1 + 2k) logn + log 3
} ≤ n−k
(
d +
√
2
s2
√
π
)
≤ d′n−k. (3.19)
Let [3nε] be the greatest integer less than or equal to 3nε. Then since the number of real zeros
of D(θ) is at most 2n we have
EN(ε) =
2n∑
j>0
Pr
{
N(ε) ≥ j}
≤ 3nε +
2n∑
j=[3nε]+1
Pr
{
N(ε) > j
}
≤ 3nε + d′n1−k = O(nε),
(3.20)
where we choose
ε = max
{
μ∗n
σ∗nn1/2−δ
,
σ∗3n
σ3∗nn1−δ
}
, (3.21)
and δ is any positive number, so the error terms become small andwe can prove the theorem.
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10 International Journal of Stochastic Analysis
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Random trigonometric polynomials with nonidentically distributed coefficients

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Random trigonometric polynomials with nonidentically distributed coefficients

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