Random trigonometric polynomials

Borwein and Lockhart [1].

1 $L^2$ norm

Let $𝕋=ℝ/\mathbb{Z}$, and for $f:𝕋\to ℂ$ let

$${\parallel f\parallel }_{L^p}^p={\int }_0^1{|f(\theta )|}^p𝑑\theta .$$

Let $X_1,X_2,\mathrm{\dots }:(\mathrm{\Omega },ℱ,P)\to (ℝ,{ℬ}_{ℝ})$ be independent identically distributed random variables with mean $0$ and variance $1$, and define

$$q_n(\theta )=\sum _{j=0}^{n-1}{X}_j{e}^{2\pi ij\theta },\theta \in 𝕋.$$

By Plancherel’s theorem,

$${\parallel q_n\parallel }_{L^2}^2=\sum _{j=0}^{n-1}{X}_j^2.$$

Let $Y_j=X_j^2-1$, which are independent and identically distributed. Then

$${\parallel q_n\parallel }_{L^2}^2-n=\sum _{j=0}^{n-1}{Y}_j.$$

We have

$$E(Y_j)=E(X_j^2)-1=0.$$

Write

$${\sigma }^2=E(Y_j^2)=E(X_j^4-2X_j^2+1)=E(X_j^4)-2E(X_j^2)+1=E(X_j^4)-1,$$

and let

$$Z_n=\frac{{\sum }_{j=0}^{n-1}{Y}_j}{\sigma \sqrt{n}}=\frac{{\parallel q_n\parallel }_{L^2}^2-n}{\sigma \sqrt{n}},$$

which has mean $0$ and variance $1$. Because $Y_1,Y_2,\mathrm{\dots }$ are independent and identically distributed with mean $0$ and variance ${\sigma }^2$, by the central limit theorem, $Z_n\to {\gamma }_1$ in distribution, where ${\gamma }_{t^2}$ is the Gaussian measure on $ℝ$ with variance $t^2$.

Theorem 1.

$$E({\parallel q_n\parallel }_{L^2})=\sqrt{n}-\frac{1}{8}\frac{{\sigma }^2}{\sqrt{n}}+O(n^{-1}),$$

where ${\sigma }^{\mathrm{2}}\mathrm{=}E\mathit{}\mathrm{(}{X}_j^{\mathrm{4}}\mathrm{)}\mathrm{-}\mathrm{1}$.

Proof.

Because

$${\parallel q_n\parallel }_{L^2}=\sqrt{n+\sigma \sqrt{n}{Z}_n},$$

we have

$${\parallel q_n\parallel }_{L^2}-\sqrt{n}=\sqrt{n}\left(\sqrt{1+\frac{\sigma Z_n}{\sqrt{n}}}-1\right).$$

Using the binomial series,

$$\sqrt{1+\frac{\sigma Z_n}{\sqrt{n}}}=1+\frac{\sigma Z_n}{2\sqrt{n}}-\frac{1}{8}\frac{{\sigma }^2{Z}_n^2}{n}+O\left(\frac{Z_n^3}{n^{3/2}}\right),$$

so

$${\parallel q_n\parallel }_{L^2}-\sqrt{n}=\frac{\sigma Z_n}{2}-\frac{1}{8}\frac{{\sigma }^2{Z}_n^2}{\sqrt{n}}+O\left(\frac{Z_n^3}{n}\right).$$ (1)

We expand $Z_n^4$: it is

$${\sigma }^{-4}{n}^{-2}\left(\sum _j{Y}_j^4+\sum _{j\ne k}{Y}_j^3{Y}_k+\sum _{j\ne k}{Y}_j^2{Y}_k^2+\sum _{j\ne k\ne p}{Y}_k^2{Y}_j{Y}_p+\sum _{j\ne k\ne p\ne q}{Y}_j{Y}_k{Y}_p{Y}_q\right).$$

Then, because $E(Y_j)=0$ and $E(Y_j^2)={\sigma }^2$, we get

$$E(Z_n^4)={\sigma }^{-4}{n}^{-2}(nE(Y_1^4)+n(n-1){\sigma }^4).$$

Now define

$$\tau =E(Y_j^4),$$

so

$$E(Z_n^4)={\sigma }^{-4}{n}^{-2}(n\tau +n(n-1){\sigma }^4)=1+\frac{1}{n}\left(\frac{\tau }{{\sigma }^4}-1\right).$$

But $E{({|Z_n|}^3)}^{1/3}\le E{({|Z_n|}^4)}^{1/4}$, so

$$E({|Z_n|}^3)\le {\left(1+\frac{1}{n}\left(\frac{\tau }{{\sigma }^4}-1\right)\right)}^{3/4}\le 1.$$

Taking the expectation of (1), because $E(Z_n)=0$ and $E(Z_n^2)=1$,

$$E({\parallel q_n\parallel }_{L^2})=\sqrt{n}-\frac{1}{8}\frac{{\sigma }^2}{\sqrt{n}}+O(n^{-1}).$$

∎

2 Berry-Esseen

Theorem 2.

$${\parallel q_n\parallel }_{L^2}-\sqrt{n}\to \frac{\sigma }{2}Z$$

in distribution.

Proof.

Write

$$\rho =E({|Y_j|}^3)$$

and

$$S_n=\sum _{j=0}^{n-1}{Y}_j,$$

and let

$$F_n(x)=P(S_n\le \sigma n^{1/2}x)$$

and

$$\mathrm{\Phi }(x)=P(Z\le x).$$

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

$$|F_n(x)-\mathrm{\Phi }(x)|\le \frac{C\rho }{{\sigma }^3\sqrt{n}}.$$

Now,

$$Z_n=\frac{1}{\sigma n^{1/2}}\sum _{j=0}^{n-1}{Y}_j=\frac{S_n}{\sigma n^{1/2}},$$

so

$$F_n(x)=P(\frac{S_n}{\sigma n^{1/2}}\le x)=P(Z_n\le x).$$

For $A>0$,

Then

$$|P(|Z_n|\ge A)-P(|Z|\ge A)|\le |\mathrm{\Phi }(A)-F_n(A)|+P(Z_n=A)+|F_n(-A)-\mathrm{\Phi }(-A)|,$$

so by the Berry-Esseen theorem,

$$|P(|Z_n|\ge A)-P(|Z|\ge A)|\le P(Z_n=A)+2\frac{C\rho }{{\sigma }^3{n}^{1/2}}.$$

Markov’s inequality tells us

$$P(|Z|\ge A)\le \sqrt{\frac{2}{\pi }}\frac{e^{-A^2/2}}{A}.$$

For $ϵ>0$ and for $A=n^{1/4}{e}^{1/2}$,

$$P(|Z|\ge A)=P(|Z{|}^2\ge n^{1/2}ϵ)\le \frac{\sqrt{\frac{2}{\pi }}\exp \left(-\frac{n^{1/2}ϵ}{2}\right)}{n^{1/4}{ϵ}^{1/2}}.$$

Therefore $\frac{Z_n^2}{n^{1/2}}\to 0$ in probability and $\frac{{|Z_n|}^3}{n}\to 0$ in probability, and because $\frac{\sigma Z_n}{2}\to \frac{\sigma }{2}Z$ in distribution, it follows that

$${\parallel q_n\parallel }_{L^2}-\sqrt{n}\to \frac{\sigma }{2}Z$$

in distribution. ∎

3 $L^4$ norm

Theorem 3.

$$E({\parallel q_n\parallel }_{L^4}^4)=2n^2+n(E(X_j^4)-2).$$

Proof.

$${\parallel q_n\parallel }_{L^4}^4={\int }_0^1{q}_n{(\theta )}^2\overline{q_n{(\theta )}^2}𝑑\theta .$$

Write $e(\theta )=e^{2\pi i\theta }$.

$$q_n^2=\sum _j{X}_j^{2}e(2j\theta )+\sum _{j\ne k}{X}_j{X}_{k}e(j\theta +k\theta ).$$

$q_n^2\overline{q_n^2}$ $={\displaystyle \sum _{j,p}}{X}_j^{2}e(2k\theta )X_p^{2}e(-2p\theta )+{\displaystyle \sum _p}{X}_p^{2}e(-2p\theta ){\displaystyle \sum _{j\ne k}}{X}_j{X}_{k}e(k\theta +j\theta )$

$+{\displaystyle \sum _{p\ne q}}{X}_p{X}_{q}e(-p\theta -q\theta ){\displaystyle \sum _j}{X}_j^{2}e(2j\theta )$

$+{\displaystyle \sum _{p\ne q}}{X}_p{X}_{q}e(-p\theta -q\theta ){\displaystyle \sum _{j\ne k}}{X}_j{X}_{k}e(k\theta +j\theta ).$

Then

${\displaystyle {\int }_0^1}{q}_n^2\overline{q_n^2}𝑑\theta$ $={\displaystyle \sum _j}{\displaystyle \sum _p}{X}_j^2{X}_p^2{\delta }_{j-p,0}$

$+{\displaystyle \sum _p}{\displaystyle \sum _{j\ne k}}{X}_p^2{X}_j{X}_k{\delta }_{j+k-2p,0}$

$+{\displaystyle \sum _j}{\displaystyle \sum _{p\ne q}}{X}_j^2{X}_p{X}_q{\delta }_{2j-p-q,0}$

$+{\displaystyle \sum _{j\ne k}}{\displaystyle \sum _{p\ne q}}{X}_j{X}_k{X}_p{X}_q{\delta }_{j+k-p-q,0}.$

That is

${\parallel q_n\parallel }_{L^4}^4$ $={\displaystyle \sum _j}{\displaystyle \sum _p}{X}_j^2{X}_p^2{\delta }_{j,p}$

$+{\displaystyle \sum _p}{\displaystyle \sum _{j\ne k}}{X}_p^2{X}_j{X}_k{\delta }_{j+k,2p}$

$+{\displaystyle \sum _j}{\displaystyle \sum _{p\ne q}}{X}_j^2{X}_p{X}_q{\delta }_{p+q,2j}$

$+{\displaystyle \sum _{j\ne k}}{\displaystyle \sum _{p\ne q}}{X}_j{X}_k{X}_p{X}_q{\delta }_{j+k,p+q}$

$={\displaystyle \sum _j}{X}_j^4+2{\displaystyle \sum _p}{\displaystyle \sum _{j\ne p}}{\displaystyle \sum _{k\ne p,k\ne j}}{X}_p^2{X}_j{X}_k{\delta }_{j+k,2p}$

$+{\displaystyle \sum _j}{\displaystyle \sum _{k\ne j}}{\displaystyle \sum _{p\ne j,p\ne k}}{\displaystyle \sum _{q\ne p,q\ne j,q\ne k}}{X}_j{X}_k{X}_p{X}_q{\delta }_{j+k,p+q}$

$+{\displaystyle \sum _j}{\displaystyle \sum _{k\ne j}}{X}_j^2{X}_k^2+{\displaystyle \sum _j}{\displaystyle \sum _{k\ne j}}{X}_j^2{X}_k^2.$

Then

$E({\parallel q_n\parallel }_{L^4}^4)$ $={\displaystyle \sum _j}E(X_j^4)+2{\displaystyle \sum _j}{\displaystyle \sum _{k\ne j}}E(X_j^2)E(X_k^2)$

$=nE(X_j^4)+2n^2-2n.$

∎

4 Gaussian random variables

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

$$S_{n,\theta }=\sum _{j=0}^{n-1}{X}_j{e}^{2\pi ij\theta }=\sum _{j=0}^{n-1}{X}_j\cos 2\pi j\theta +i\sum _{j=0}^{n-1}{X}_j\sin 2\pi j\theta ,n\ge 1,\theta \in 𝕋.$$

Then for each $\theta \in 𝕋$, there are $Z_{\theta }$ and $W_{\theta }$, each random variables with the standard Gaussian distribution, such that

$$S_{n,\theta }={(n/2)}^{1/2}{Z}_{\theta }+i{(n/2)}^{1/2}{W}_{\theta }.$$

Now, $|Z_{\theta }+i{W}_{\theta }|$ has density $t\mapsto t{e}^{-t^2/2}$, and then

$$E({|Z_{\theta }+i{W}_{\theta }|}^p)=2^{p/2}\mathrm{\Gamma }\left(1+\frac{p}{2}\right).$$

Then

$$E({|S_{n,\theta }|}^p)={\left(\frac{n}{2}\right)}^{p/2}{2}^{p/2}\mathrm{\Gamma }\left(1+\frac{p}{2}\right)=n^{p/2}\mathrm{\Gamma }\left(1+\frac{p}{2}\right).$$