Random trigonometric polynomials

Jordan Bell
April 23, 2016

Borwein and Lockhart [1].

1 L2 norm

Let 𝕋=ℝ/ℤ, and for f:𝕋→ℂ let

∥f∥Lpp=∫01|f⁢(θ)|p⁢𝑑θ.

Let X1,X2,…:(Ω,ℱ,P)→(ℝ,ℬℝ) be independent identically distributed random variables with mean 0 and variance 1, and define

qn⁢(θ)=∑j=0n-1Xj⁢e2⁢π⁢i⁢j⁢θ,θ∈𝕋.

By Plancherel’s theorem,

∥qn∥L22=∑j=0n-1Xj2.

Let Yj=Xj2-1, which are independent and identically distributed. Then

∥qn∥L22-n=∑j=0n-1Yj.

We have

E⁢(Yj)=E⁢(Xj2)-1=0.

Write

σ2=E⁢(Yj2)=E⁢(Xj4-2⁢Xj2+1)=E⁢(Xj4)-2⁢E⁢(Xj2)+1=E⁢(Xj4)-1,

and let

Zn=∑j=0n-1Yjσ⁢n=∥qn∥L22-nσ⁢n,

which has mean 0 and variance 1. Because Y1,Y2,… are independent and identically distributed with mean 0 and variance σ2, by the central limit theorem, Zn→γ1 in distribution, where γt2 is the Gaussian measure on ℝ with variance t2.

Theorem 1.
E⁢(∥qn∥L2)=n-18⁢σ2n+O⁢(n-1),

where σ2=E⁢(Xj4)-1.

Proof.

Because

∥qn∥L2=n+σ⁢n⁢Zn,

we have

∥qn∥L2-n=n⁢(1+σ⁢Znn-1).

Using the binomial series,

1+σ⁢Znn=1+σ⁢Zn2⁢n-18⁢σ2⁢Zn2n+O⁢(Zn3n3/2),

so

∥qn∥L2-n=σ⁢Zn2-18⁢σ2⁢Zn2n+O⁢(Zn3n). (1)

We expand Zn4: it is

σ-4⁢n-2⁢(∑jYj4+∑j≠kYj3⁢Yk+∑j≠kYj2⁢Yk2+∑j≠k≠pYk2⁢Yj⁢Yp+∑j≠k≠p≠qYj⁢Yk⁢Yp⁢Yq).

Then, because E⁢(Yj)=0 and E⁢(Yj2)=σ2, we get

E⁢(Zn4)=σ-4⁢n-2⁢(n⁢E⁢(Y14)+n⁢(n-1)⁢σ4).

Now define

τ=E⁢(Yj4),

so

E⁢(Zn4)=σ-4⁢n-2⁢(n⁢τ+n⁢(n-1)⁢σ4)=1+1n⁢(τσ4-1).

But E⁢(|Zn|3)1/3≤E⁢(|Zn|4)1/4, so

E⁢(|Zn|3)≤(1+1n⁢(τσ4-1))3/4≤1.

Taking the expectation of (1), because E⁢(Zn)=0 and E⁢(Zn2)=1,

E⁢(∥qn∥L2)=n-18⁢σ2n+O⁢(n-1).

∎

2 Berry-Esseen

Theorem 2.
∥qn∥L2-n→σ2⁢Z

in distribution.

Proof.

Write

ρ=E⁢(|Yj|3)

and

Sn=∑j=0n-1Yj,

and let

Fn(x)=P(Sn≤σn1/2x)

and

Φ(x)=P(Z≤x).

The Berry-Esseen theorem [2, p. 262, Theorem 5.6.1] states that there is some C, not depending on the random variables Yj, such that for all n and for all x∈ℝ,

|Fn⁢(x)-Φ⁢(x)|≤C⁢ρσ3⁢n.

Now,

Zn=1σ⁢n1/2⁢∑j=0n-1Yj=Snσ⁢n1/2,

so

Fn(x)=P(Snσ⁢n1/2≤x)=P(Zn≤x).

For A>0,

P(|Zn|≥A)=P(Zn≥A)+P(Zn≤-A)=1-P(Zn<A)+P(Zn≤-A).
P(|Zn|≥A)-P(|Z|≥A)=P(Z<A)-P(Zn<A)+P(Zn≤A)-P(Z≤-A).

Then

|P(|Zn|≥A)-P(|Z|≥A)|≤|Φ(A)-Fn(A)|+P(Zn=A)+|Fn(-A)-Φ(-A)|,

so by the Berry-Esseen theorem,

|P(|Zn|≥A)-P(|Z|≥A)|≤P(Zn=A)+2C⁢ρσ3⁢n1/2.

Markov’s inequality tells us

P(|Z|≥A)≤2πe-A2/2A.

For ϵ>0 and for A=n1/4⁢e1/2,

P(|Z|≥A)=P(|Z|2≥n1/2ϵ)≤2π⁢exp⁡(-n1/2⁢ϵ2)n1/4⁢ϵ1/2.

Therefore Zn2n1/2→0 in probability and |Zn|3n→0 in probability, and because σ⁢Zn2→σ2⁢Z in distribution, it follows that

∥qn∥L2-n→σ2⁢Z

in distribution. ∎

3 L4 norm

Theorem 3.
E⁢(∥qn∥L44)=2⁢n2+n⁢(E⁢(Xj4)-2).
Proof.
∥qn∥L44=∫01qn⁢(θ)2⁢qn⁢(θ)2¯⁢𝑑θ.

Write e⁢(θ)=e2⁢π⁢i⁢θ.

qn2=∑jXj2⁢e⁢(2⁢j⁢θ)+∑j≠kXj⁢Xk⁢e⁢(j⁢θ+k⁢θ).
qn2⁢qn2¯ =∑j,pXj2⁢e⁢(2⁢k⁢θ)⁢Xp2⁢e⁢(-2⁢p⁢θ)+∑pXp2⁢e⁢(-2⁢p⁢θ)⁢∑j≠kXj⁢Xk⁢e⁢(k⁢θ+j⁢θ)
+∑p≠qXp⁢Xq⁢e⁢(-p⁢θ-q⁢θ)⁢∑jXj2⁢e⁢(2⁢j⁢θ)
+∑p≠qXp⁢Xq⁢e⁢(-p⁢θ-q⁢θ)⁢∑j≠kXj⁢Xk⁢e⁢(k⁢θ+j⁢θ).

Then

∫01qn2⁢qn2¯⁢𝑑θ =∑j∑pXj2⁢Xp2⁢δj-p,0
+∑p∑j≠kXp2⁢Xj⁢Xk⁢δj+k-2⁢p,0
+∑j∑p≠qXj2⁢Xp⁢Xq⁢δ2⁢j-p-q,0
+∑j≠k∑p≠qXj⁢Xk⁢Xp⁢Xq⁢δj+k-p-q,0.

That is

∥qn∥L44 =∑j∑pXj2⁢Xp2⁢δj,p
+∑p∑j≠kXp2⁢Xj⁢Xk⁢δj+k,2⁢p
+∑j∑p≠qXj2⁢Xp⁢Xq⁢δp+q,2⁢j
+∑j≠k∑p≠qXj⁢Xk⁢Xp⁢Xq⁢δj+k,p+q
=∑jXj4+2⁢∑p∑j≠p∑k≠p,k≠jXp2⁢Xj⁢Xk⁢δj+k,2⁢p
+∑j∑k≠j∑p≠j,p≠k∑q≠p,q≠j,q≠kXj⁢Xk⁢Xp⁢Xq⁢δj+k,p+q
+∑j∑k≠jXj2⁢Xk2+∑j∑k≠jXj2⁢Xk2.

Then

E⁢(∥qn∥L44) =∑jE⁢(Xj4)+2⁢∑j∑k≠jE⁢(Xj2)⁢E⁢(Xk2)
=n⁢E⁢(Xj4)+2⁢n2-2⁢n.

∎

4 Gaussian random variables

Suppose that the distribution of each Xj is the standard Gaussian measure on ℝ, and write

Sn,θ=∑j=0n-1Xj⁢e2⁢π⁢i⁢j⁢θ=∑j=0n-1Xj⁢cos⁡2⁢π⁢j⁢θ+i⁢∑j=0n-1Xj⁢sin⁡2⁢π⁢j⁢θ,n≥1,θ∈𝕋.

Then for each θ∈𝕋, there are Zθ and Wθ, each random variables with the standard Gaussian distribution, such that

Sn,θ=(n/2)1/2⁢Zθ+i⁢(n/2)1/2⁢Wθ.

Now, |Zθ+i⁢Wθ| has density t↦t⁢e-t2/2, and then

E⁢(|Zθ+i⁢Wθ|p)=2p/2⁢Γ⁢(1+p2).

Then

E⁢(|Sn,θ|p)=(n2)p/2⁢2p/2⁢Γ⁢(1+p2)=np/2⁢Γ⁢(1+p2).

References

  • [1] P. Borwein and R. Lockhart (2001) The expected Lp norm of random polynomials. Proc. Amer. Math. Soc. 129 (5), pp. 1463–1472. Cited by: Random trigonometric polynomials.
  • [2] M. A. Pinsky (2009) Introduction to Fourier analysis and wavelets. Graduate Studies in Mathematics, Vol. 102, American Mathematical Society, Providence, RI. Cited by: §2.