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Non universality for the variance of the number of real roots of random trigonometric polynomials

Abstract

In this article, we consider the following family of random trigonometric polynomials \(p_n(t,Y)=\sum _{k=1}^n Y_{k}^1 \cos (kt)+Y_{k}^2\sin (kt)\) for a given sequence of i.i.d. random variables \(Y^i_{k}\), \(i\in \{1,2\}\), \(k\ge 1\), which are centered and standardized. We set \({\mathcal {N}}([0,\pi ],Y)\) the number of real roots over \([0,\pi ]\) and \({\mathcal {N}}([0,\pi ],G)\) the corresponding quantity when the coefficients follow a standard Gaussian distribution. We prove under a Doeblin’s condition on the distribution of the coefficients that

$$\begin{aligned} \lim _{n\rightarrow \infty }\frac{\text {Var}\left( {\mathcal {N}}_n([0,\pi ],Y)\right) }{n} =\lim _{n\rightarrow \infty }\frac{\text {Var}\left( {\mathcal {N}}_n([0,\pi ],G)\right) }{n} +\frac{1}{30}\left( {\mathbb {E}}\left( \left( Y_{1}^1\right) ^4\right) -3\right) . \end{aligned}$$

The latter establishes that the behavior of the variance is not universal and depends on the distribution of the underlying coefficients through their kurtosis. Actually, a more general result is proven in this article, which does not require that the coefficients are identically distributed. The proof mixes a recent result regarding Edgeworth expansions for distribution norms established in Bally et al. (Electron J Probab 23(45):1–51, 2018) with the celebrated Kac–Rice formula.

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Correspondence to Guillaume Poly.

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Appendices

A Ergodic lemma

The following lemmas are based on the ergodic action of irrational rotations on the Torus.

Lemma A.1

Set \(\alpha \) a positive number such that \(\frac{\alpha }{\pi }\in \mathbb {R}/ \mathbb {Q}\), f a \(2\pi \)–periodic function and \(q\ge 1\) a positive integer. One gets

$$\begin{aligned} \lim _{n\rightarrow \infty }\frac{1}{n}\sum _{k=1}^{n}f(k\alpha )= & {} \frac{1}{ 2\pi }\int _{0}^{2\pi }f(t)dt, \end{aligned}$$
(A.1)
$$\begin{aligned} \lim _{n\rightarrow \infty }\frac{1}{n}\sum _{k=1}^{n}\frac{k^{q}}{n^{q}} f(k\alpha )= & {} \frac{1}{(q+1)2\pi }\int _{0}^{2\pi }f(t)dt \end{aligned}$$
(A.2)

Proof

We denote by \(\mathcal {C}_{2\pi }^{0}(\mathbb {R})\) the space of continuous \(2\pi \) periodic functions. We set

$$\begin{aligned} \mathcal {H}_{0}=\left\{ \phi \in \mathcal {C}_{2\pi }^{0}(\mathbb {R})\,\, \Big |\,\,\int _{0}^{2\pi }\phi (t)dt=0\right\} , \end{aligned}$$

and

$$\begin{aligned} \mathcal {E}=\left\{ f(x)=\phi (x+\alpha )-\phi (x)\,\,\Big |\,\,\phi \in \mathcal {C}_{2\pi }^{0}(\mathbb {R})\right\} . \end{aligned}$$

Let us first prove that \(\mathcal {E}\) is dense in \(\mathcal {H}_{0}\). We take T a continuous linear form on \(\mathcal {H}_{0}\) and we extend it to \( \mathcal {C}_{2\pi }^{0}(\mathbb {R})\) by taking \(T(\phi )=T(\phi -m(\phi ))\) with \(m(\phi )=\int _{0}^{2\pi }\phi (t)dt\). We have to prove that if T vanishes on \(\mathcal {E}\) then \(T=0\) (in virtue of the Hahn-Banach Theorem, this implies that \(\mathcal {E}\) is dense in \(\mathcal {H}_{0})\). The Riesz Theorem ensures us that there exists a finite measure \(\mu \) on \( \mathbb {R}/2\pi \mathbb {Z}\) such that

$$\begin{aligned} \forall \phi \in \mathcal {C}^0_{2\pi }(\mathbb {R}),\,\,T(\phi )=\int _0^{2\pi } \phi (x) d\mu (x). \end{aligned}$$

Since \(Tf=0\) for every \(f\in \mathcal {E}\), for any integer \(n\ge 1\) one has

$$\begin{aligned} \int _{0}^{2\pi } \phi (x +n \alpha ) d\mu (x) =\int _0^{2\pi } \phi (x) d\mu (x), \end{aligned}$$

and since the sequence \(n\alpha \) is dense modulo \(2\pi \) one deduces that for any \(y\in \mathbb {R}{\setminus } \mathbb {Q}\):

$$\begin{aligned} \int _{0}^{2\pi } \phi (x +y) d\mu (x) =\int _0^{2\pi } \phi (x) d\mu (x). \end{aligned}$$

By the continuity of \(\phi \), this is true for each \(y\in \mathbb {R}\). As a result, \(\mu \) is invariant under translations and necessarily it is the Lebesgue measure up to a multiplicative constant. Hence, we get that \(T=0\) over \(\mathcal {H}_0\) and that \(\mathcal {E}\) is dense for the uniform topology. Finally, this preliminary consideration enables us to consider that \(f(x)=\phi (x+\alpha )-\phi (x)\) in the statements (A.1) and (A.2). Then, the conclusion is immediate since an Abel transforms gives us

$$\begin{aligned} \Big |\frac{1}{n}\sum _{k=1} \frac{k^q}{n^q} \left( \varphi ((k+1)\alpha )- \varphi (k\alpha )\right) \Big |\le 2 \Vert \phi \Vert _\infty \frac{1}{n} \sum _{k=1}^n \left( \frac{(k+1)^q-k^q}{n^q}\right) \xrightarrow [n\rightarrow \infty ]~0. \end{aligned}$$

\(\square \)

In the following \(C_{n}(k,t)\) is the matrix introduced in (3.25).

Lemma A.2

For every \(i,j,l,l^{\prime }\in \{1,2\}\) and every ts such that \( t,s,t+s,t-s\) are irrational one has

$$\begin{aligned} \lim _{n}\frac{1}{n}\sum _{k=1}^{n}C_{n}^{i,l}(k,nt)^{2}C_{n}^{j,l^{\prime }}(k,ns)^{2}=\frac{1}{4(1+2(i+j-2))} \end{aligned}$$
(A.3)

Proof

We treat just two examples: take \(i=1,j=2,l=1,l^{\prime }=2\). Then

$$\begin{aligned}&C_{n}^{i,l}(k,nt)^{2}C_{n}^{j,l^{\prime }}(k,ns)^{2} =\left( \cos kt\times \frac{ k}{n}\cos ks\right) ^{2}\\&\quad =\frac{1}{4}\times \frac{k^{2}}{n^{2}}(\cos (k(t+s))+\cos (k(t-s)))^{2} \\&\quad =\frac{1}{4}\times \frac{k^{2}}{n^{2}}(\cos ^{2}(k(t+s))+\cos ^{2}(k(t-s))+2\cos (k(t+s))\cos (k(t-s))) \\&\quad =\frac{1}{4}\times \frac{k^{2}}{n^{2}}(\cos ^{2}(k(t+s))+\cos ^{2}(k(t-s))+\cos (2kt)-\cos (2ks))). \end{aligned}$$

Then, the ergodic lemma (with \(q=2)\) gives

$$\begin{aligned} \lim _{n}\frac{1}{n}\sum _{k=1}^{n}C_{n}^{i,l}(k,nt)^{2}C_{n}^{j,l^{\prime }}(k,ns)^{2}=2\times \frac{1}{4}\times \frac{1}{2\pi \times 3}\int _{0}^{2\pi }(\cos ^{2}(u)+\cos (u))du=\frac{1}{12}. \end{aligned}$$

Take now \(i=2,j=2,l=1,l^{\prime }=2\). Then

$$\begin{aligned} C_{n}^{i,l}(k,nt)^{2}C_{n}^{j,l^{\prime }}(k,ns)^{2}= & {} \Big ( \frac{k}{n}\sin kt\times \frac{k}{n}\cos ks\Big )^{2}\\= & {} \frac{1}{4}\times \frac{k^{4}}{n^{4}}(\sin (k(t+s))+\sin (k(t-s))^{2} \\= & {} \frac{1}{4}\times \frac{k^{4}}{n^{4}} ( \sin ^{2}(k(t+s))+\sin ^{2}(k(t-s))+\cos (2kt)\\&\quad +\cos (2ks)) \end{aligned}$$

Then, the ergodic lemma (with \(q=4)\) gives

$$\begin{aligned} \lim _{n}\frac{1}{n}\sum _{k=1}^{n}C_{n}^{i,l}(k,nt)^{2}C_{n}^{j,l^{\prime }}(k,ns)^{2}=\frac{1}{4}\times \frac{1}{2\pi \times 5}\times 2\int _{0}^{2\pi }(\sin ^{2}(u)+\cos (u))du=\frac{1}{20}. \end{aligned}$$

\(\square \)

Lemma A.3

For every \(j,i,l\in \{1,2\}\) and every ts such that \(t,s,t+s,t-s\) are irrational one has

$$\begin{aligned}&\lim _{n}\frac{1}{n} \sum _{k=1}^{n}C_{n}^{i,1}(k,nt)C_{n}^{i,2}(k,nt)C_{n}^{j,l}(k,ns)^{2} \nonumber \\&\quad =\lim _{n}\frac{1}{n} \sum _{k=1}^{n}C_{n}^{i,1}(k,nt)C_{n}^{i,2}(k,nt)C_{n}^{j,1}(k,ns)C_{n}^{j,2}(k,ns)=0. \end{aligned}$$
(A.4)

Proof

All the computations are analogous so we treat just an example: \(l=i=j=1\). So we have

$$\begin{aligned} C_{n}^{1,1}(k,nt)C_{n}^{1,2}(k,nt)C_{n}^{1,1}(k,ns)^{2}= & {} \cos (kt)\sin (kt)\cos ^{2}(ks) \\= & {} \frac{1}{2}\sin (2kt)\cos ^{2}(ks) \\= & {} \frac{1}{4}(\sin (k(2t+s))+\sin (k(2t-s)))\cos (ks) \\= & {} \frac{1}{8}(\sin (k(2t+2s))+2\sin (2kt)+\sin (k(2t-2s))) \end{aligned}$$

and using the ergodic lemma with \(q=0\) we get

$$\begin{aligned} \lim _{n}\frac{1}{n} \sum _{k=1}^{n}C_{n}^{1,1}(k,nt)C_{n}^{1,2}(k,nt)C_{n}^{1,1}(k,ns)^{2}=\frac{1 }{8}\times \frac{1}{2\pi }\times 4\int _{0}^{2\pi }\sin (u)du=0. \end{aligned}$$

\(\square \)

Lemma A.4

For every \(i_{1},i_{2},i_{3},l_{1},l_{2},l_{3}\in \{1,2\}\) and every ts such that \(t,s,t+s,t-s\) are irrational one has

$$\begin{aligned} \lim _{n}\frac{1}{n} \sum _{k=1}^{n}C_{n}^{i_{1},l_{1}}(k,nt)C_{n}^{i_{2},l_{2}}(k,nt)C_{n}^{i_{3},l_{3}}(k,ns)=0. \end{aligned}$$
(A.5)

Proof

The poof is similar in all cases so we treat just an example: \(i_{1}=1,i_{2}=2,i_{3}=2,l_{1}=l_{2}=l_{3}=1\). Then we deal with

$$\begin{aligned} \cos (kt)\times \frac{k}{n}\sin (kt)\times \frac{k}{n}\sin (ks)= & {} \frac{ k^{2}}{n^{2}}\times \frac{1}{2}\sin (2kt)\sin (ks) \\= & {} \frac{k^{2}}{n^{2}}\times \frac{1}{4}(\cos (k(2t+s))-\cos (k(2t-s))) \end{aligned}$$

And using the ergodic lemma with \(q=2\) we get

$$\begin{aligned}&\lim _{n}\frac{1}{n}\sum _{k=1}^{n}\frac{k^{2}}{n^{2}}\times \frac{1}{4} (\cos (k(2t+s))-\cos (k(2t-s))) \\&\quad =\frac{1}{24\pi }\left( \int _{0}^{2\pi }\cos (u)du-\int _{0}^{2\pi }\cos (u)du\right) =0. \end{aligned}$$

\(\square \)

B Estimates of some trigonometric sums

For \(n\in \mathbb {N},\)\(i=0,1,2\) and \(b\in \mathbb {R}_{+}\backprime \{2\pi p;p\in \mathbb {N}\}\) we put

$$\begin{aligned} S_{b,i}(c)=\frac{1}{n}\sum _{k=1}^{n}\frac{k^{i}}{n^{i}}\cos (bk). \end{aligned}$$

We also denote

$$\begin{aligned} \overline{b}=\inf _{p\in \mathbb {N}}\frac{\left| 2\pi p-b\right| }{p\vee 1}. \end{aligned}$$
(B.1)

The aim of this section is to prove the following lemma:

Lemma B.1

There exists a universal constant \(C\ge 1\) such that for every \(n\in \mathbb {N}\)\( i=0,1,2\) and \(\in \mathbb {R}_{+}\backprime \{2\pi p;p\in \mathbb {N}\}\)

$$\begin{aligned} \left| S_{b,i}(c)\right| \le \frac{C}{n\overline{b}}. \end{aligned}$$
(B.2)

The first step is the following abstract estimate:

Lemma B.2

A. Let \(f\in L^{2}(0,1)\) and let \(\phi (x)=\sum _{k=0}^{\infty }f(x-k)1_{[k,k+1)}(x)\). There exists a universal constant such that for every \(k<n\)

$$\begin{aligned} \left| \int _{k}^{n}\phi (x)\cos (bx)\right| \le \frac{C}{\overline{b }}\left\| f\right\| _{2}. \end{aligned}$$
(B.3)

B. Moreover there exists a universal constant C such that, for \( i=0,1,2\)

$$\begin{aligned} \left| \int _{0}^{n}\frac{x^{i}}{n^{i}}\phi (x)\cos (bx)dx\right| \le \frac{C}{\overline{b}}\left\| f\right\| _{2} \end{aligned}$$
(B.4)

and in particular, taking \(f=1,\)

$$\begin{aligned} \int _{0}^{n}\frac{x^{i}}{n^{i}}\cos (bx)dx\le \frac{C}{\overline{b}} \end{aligned}$$
(B.5)

The same estimates hold if we replace \(\cos \) by \(\sin \).

Proof of A

We denote \(\alpha _{0}=\int _{0}^{1}f(x)dx\) and

$$\begin{aligned} \alpha _{p}=\int _{0}^{1}f(x)\cos (2\pi px)dx,\quad \beta _{p}=\int _{0}^{1}f(x)\sin (2\pi px)dx. \end{aligned}$$

Then using the development in Fourier series of \(\phi \) we obtain

$$\begin{aligned} \int _{k}^{n}\phi (x)\cos (bx)dx= & {} \alpha _{0}\int _{k}^{n}\cos (bx)dx +\sum _{p=1}^{\infty }\alpha _{p}\int _{k}^{n}\cos (2\pi px)\cos (bx)dx\\&+\beta _{p}\int _{k}^{n}\sin (2\pi px)\cos (bx)dx. \end{aligned}$$

We write

$$\begin{aligned} \cos (2\pi px)\cos (bx)=\frac{1}{2}(\cos ((2\pi p+b)x)+\cos ((2\pi p-b)x). \end{aligned}$$

and we use a similar decomposition for \(\sin (2\pi px)\cos (bx)\).

Notice that for every \(\theta >0\) one has

$$\begin{aligned} \left| \int _{k}^{n}\cos (\theta x)dx\right| \le \frac{2\pi }{\theta }\quad and\quad \left| \int _{k}^{n}\sin (\theta x)dx\right| \le \frac{2\pi }{\theta }. \end{aligned}$$

Using these inequalities we obtain

$$\begin{aligned} \left| \int _{k}^{n}\phi (x)\cos (bx)\right|= & {} \frac{2\pi }{b} \left| \alpha _{0}\right| +\sum _{p=1}^{\infty }\left( \left| \alpha _{p}\right| +\left| \beta _{p}\right| \left( \frac{2\pi }{2\pi p+b}+ \frac{2\pi }{\left| 2\pi p-b\right| }\right) \right) \\\le & {} \frac{2\pi }{b}\left| \alpha _{0}\right| +\frac{4\pi }{ \overline{b}}\sum _{p=1}^{\infty }\left( \left| \alpha _{p}\right| +\left| \beta _{p}\right| \right) \frac{1}{p} \\\le & {} \frac{2\pi }{b}\left| \alpha _{0}\right| +\frac{C}{\overline{b }}\left( \sum _{p=1}^{\infty }\left( \left| \alpha _{p}\right| +\left| \beta _{p}\right| \right) ^{2}\right) ^{1/2}\le \frac{C}{\overline{b}}\left\| f\right\| _{2}. \end{aligned}$$

B. We just treat the case \(i=1\) (the other ones are similar). We write

$$\begin{aligned} \frac{1}{n}\int _{0}^{n}x\phi (x)\cos (bx)dx=\frac{1}{n}\int _{0}^{n}\psi (x)\cos (bx)dx+\frac{1}{n}\sum _{k=1}^{n}k\int _{k}^{k+1}\phi (x)\cos (bx)dx \end{aligned}$$

with \(\psi \) associated to g( \(x)=xf(x)\). Using (B.3) (notice that \( \left\| g\right\| _{2}\le \left\| f\right\| _{2})\)

$$\begin{aligned} \frac{1}{n}\left| \int _{0}^{n}\psi (x)\cos (bx)dx\right| \le \frac{C }{n}\times \frac{1}{\overline{b}}\left\| f\right\| _{2}. \end{aligned}$$

Moreover

$$\begin{aligned} \frac{1}{n}\sum _{k=1}^{n}k\int _{k}^{k+1}\phi (x)\cos (bx)dx=\frac{1}{n} \sum _{k=1}^{n}\int _{k}^{n}\phi (x)\cos (bx)dx \end{aligned}$$

so that, by (B.3) we upper bound the above term by

$$\begin{aligned} \frac{1}{n}\sum _{k=1}^{n}\left| \int _{k}^{n}\phi (x)\cos (bx)dx\right| \le \frac{C}{n}\times n\times \frac{1}{\overline{b}} \left\| f\right\| _{2}=\frac{C}{\overline{b}}\left\| f\right\| _{2}. \end{aligned}$$

And (B.5) is a particular case of (B.4) with \(f=1\) (so that \(\phi =1)\). \(\square \)

We recall that

$$\begin{aligned} S_{b,i}(c)=\frac{1}{n}\sum _{k=1}^{n}\frac{k^{i}}{n^{i}}\cos \left( \frac{ak}{n} \right) ,\quad \text{ with }\quad a=nb. \end{aligned}$$

We also denote

$$\begin{aligned} \phi (x)= & {} \sum _{k=0}^{\infty }(k+1-x)^{2}1_{\{k\le x<k+1\}}\quad and\quad J_{b,i}(c)\nonumber \\&\quad =\frac{1}{n^{1+i}}\int _{0}^{n}x^{i}\phi (x)\cos (bx)dx \end{aligned}$$
(B.6)
$$\begin{aligned} I_{b,i}(c)= & {} \frac{1}{n^{1+i}}\int _{0}^{n}x^{i}\cos (bx)dx=\int _{0}^{1}x^{i}\cos (ax)dx. \end{aligned}$$
(B.7)

We also define \(S_{b,i}(s)\) and \(J_{b,i}(s)\) by replacing the \(\cos \) by \( \sin \).

We will prove the following estimates:

Lemma B.3

Let \(a=bn\) with \(0<b\). There exists a universal constant C such that, for \(i=0,1,2\)

$$\begin{aligned} \left| S_{b,i}(c)\right| =\frac{1}{1+b^{2}/4}\Big (\frac{b^{3}}{4} J_{b,i}(s)+\frac{b^{2}}{2}J_{b,i}(c)+I_{b,i}(c)-\frac{b}{2} I_{b,i}(s)\Big )+\varepsilon _{n} \end{aligned}$$
(B.8)

with \(\left| \varepsilon _{n}\right| \le C/n\).

Proof

Let us prove (B.8) for \(i=1\) (the proof is analogous for \(i=0\) and \(i=2)\). We write

$$\begin{aligned} S_{b,1}(c)=I_{b,1}(c)-\sum _{k=1}^{n}\int _{k/n}^{(k+1)/n}\Big (x\cos (ax)-\frac{k}{ n}\cos \Big (\frac{ak}{n}\Big )\Big )dx. \end{aligned}$$

Moreover

$$\begin{aligned} \int _{k/n}^{(k+1)/n}\Big (x\cos (ax)-\frac{k}{n}\cos \left( \frac{ak}{n}\right) \Big )dx=\frac{k}{n} \int _{k/n}^{(k+1)/n}\Big (\cos (ax)-\cos \Big (\frac{ak}{n}\Big )\Big )dx+\delta _{n,k} \end{aligned}$$

with \(\left| \delta _{n,k}\right| \le 1/n^{2}\) so that \( \sum _{k=1}^{n}\delta _{n,k}=\varepsilon _{n},\) with \(\left| \varepsilon _{n}\right| \le C/n\). We write now (recall that \(a=nb)\)

$$\begin{aligned}&\frac{k}{n}\int _{k/n}^{(k+1)/n}\Big (\cos (ax)-\cos \Big (\frac{ak}{n}\Big )\Big )dx\\&\quad =-\frac{ak }{n}\int _{k/n}^{(k+1)/n}dx\int _{k/n}^{x}dy\sin ay \\&\qquad -\frac{ak}{2n^{3}}\sin \frac{ak}{n}-\frac{ak}{n}\int _{k/n}^{(k+1)/n}dx \int _{k/n}^{x}dy\Big (\sin ay-\sin \frac{ak}{n}\Big ) \\&\quad =-\frac{bk}{2n^{2}}\sin \frac{ak}{n}-\frac{a^{2}k}{n} \int _{k/n}^{(k+1)/n}dx\int _{k/n}^{x}dy\int _{k/n}^{y}dz\cos az \\&\quad =-\frac{bk}{2n^{2}}\sin \frac{ak}{n}-\frac{a^{2}k}{2n} \int _{k/n}^{(k+1)/n}(k+1-z)^{2}\cos (az)dz \end{aligned}$$

Summing over k this gives (with \(\varepsilon _{n}\) of order \(\frac{1}{n}\) and which changes from a line to another)

$$\begin{aligned} S_{b,1}(c)= & {} I_{b,1}(c)+\varepsilon _{n}+\frac{b}{2}S_{b,1}(s)+\frac{a^{2}}{ 2}\sum _{k=1}^{n}\frac{k}{n}\int _{k/n}^{(k+1)/n}\cos (ax)\Big (\frac{k+1}{n}-x\Big )^{2} \\= & {} I_{b,1}(c)+\varepsilon _{n}+\frac{b}{2}S_{b,1}(s)+\frac{a^{2}}{2} \sum _{k=1}^{n}\int _{k/n}^{(k+1)/n}x\cos (ax)\Big (\frac{k+1}{n}-x\Big )^{2} \\= & {} I_{b,1}(c)+\varepsilon _{n}+\frac{b}{2}S_{b,1}(s)+\frac{b^{2}}{2}\times \frac{1}{n^{2}}\int _{0}^{n}x\phi (x)\cos (bx)dx \\= & {} I_{b,1}(c)+\varepsilon _{n}+\frac{b}{2}S_{b,1}(s)+\frac{b^{2}}{2} J_{b,1}(c). \end{aligned}$$

The same computations give

$$\begin{aligned} S_{n,1}(s)=I_{b,1}(s)+\varepsilon _{n}-\frac{b}{2}S_{n,1}(c)+\frac{b^{2}}{2} J_{b,1}(s). \end{aligned}$$

We insert this in the previous estimate and we get

$$\begin{aligned} S_{b,1}(c)=\varepsilon _{n}+I_{1}(c)-\frac{b}{2}I_{b,1}(s)-\frac{b^{2}}{4} S_{b,1}(c)+\frac{b^{3}}{4}J_{b,1}(s)+\frac{b^{2}}{2}J_{b,1}(c) \end{aligned}$$

and we are done. \(\square \)

Proof of (B.2)

By (B.4) and (B.5)

$$\begin{aligned} \left| J_{b,i}(s)\right| +\left| J_{b,i}(c)\right| +\left| I_{b,i}(s)\right| +\left| I_{b,i}(c)\right| \le \frac{C}{n\overline{b}} \end{aligned}$$

so (B.2) follows. \(\square \)

C Non degeneracy

In this section we discuss the non degeneracy of the matrix \(\Sigma _{n}(t,s) \) which is the covariance matrix of \(S_{n}(t,s,Y)\). Direct computations show that:

$$\begin{aligned} \Sigma _{n}^{1,1}(t,s)= & {} \Sigma _{n}^{3,,3}(t,s)=1,\quad \Sigma _{n}^{2,2}(t,s)=\Sigma _{n}^{4,4}(t,s)=\frac{1}{n}\sum _{k=1}^{n}\frac{k^{2}}{ n^{2}},\quad \\ \Sigma _{n}^{1,3}(t,s)= & {} \Sigma _{n}^{3,1}(t,s)=\frac{1}{n} \sum _{k=1}^{n}\cos \frac{k(t-s)}{n}\\ \Sigma _{n}^{2,4}(t,s)= & {} \Sigma _{n}^{4,2}(t,s)=\frac{1}{n}\sum _{k=1}^{n}\frac{k^{2}}{n^{2}}\cos \frac{k(t-s) }{n}\\ \Sigma _{n}^{1,4}(t,s)= & {} \Sigma _{n}^{4,1}(t,s)=-\Sigma _{n}^{2,3}(t,s)=-\Sigma _{n}^{3,2}(t,s)=\frac{1}{n}\sum _{k=1}^{n}\frac{k}{n} \cos \frac{k(t-s)}{n} \\ \Sigma _{n}^{1,2}(t,s)= & {} \Sigma _{n}^{2,1}(t,s)=\Sigma _{n}^{3,4}(t,s)=\Sigma _{n}^{4,3}(t,s)=0. \end{aligned}$$

We define \(\Sigma (t,s)\) just by passing to the limit (for fixed t and s):

$$\begin{aligned} \Sigma ^{1,1}(t,s)= & {} \Sigma ^{3,3}(t,s)=1,\quad \Sigma ^{2,2}(t,s)=\Sigma ^{4,4}(t,s)=\frac{1}{3} \\ \Sigma ^{1,3}(t,s)= & {} \Sigma ^{3,1}(t,s) =\frac{\sin \tau }{\tau }\Big |_{\tau =t-s},\\ \Sigma ^{2,4}(t,s)= & {} \Sigma ^{4,2}(t,s) =\frac{2\tau \cos \tau -2\sin \tau +\tau ^2\sin \tau }{\tau ^3}\Big |_{\tau =t-s},\\ \Sigma ^{1,4}(t,s)= & {} \Sigma ^{4,1}(t,s)=-\Sigma ^{2,3}(t,s)=-\Sigma ^{3,2}(t,s) =\frac{\sin \tau -\tau \cos \tau }{\tau ^2}\Big |_{\tau =t-s},\\ \Sigma ^{1,2}(t,s)= & {} \Sigma ^{2,1}(t,s)=\Sigma ^{3,4}(t,s)=\Sigma ^{4,3}(t,s)=0. \end{aligned}$$

Then it is easy to check that there exists a universal constant \(C\ge 1\) such that for every \(i,j=1,\ldots ,4\) and every \(0<s<t\)

$$\begin{aligned} \sup _{\left| t-s\right| \le n^{\rho }}\left| \Sigma _{n}^{i,j}(t,s)-\Sigma ^{i,j} (t,s)\right| \le \frac{C(t-s)}{n}. \end{aligned}$$
(C.1)

Notice however that, if \(t-s\approx n\) the above inequality says nothing. So our strategy will be the following: we consider a first case, when \(t-s\le \sqrt{n}\) and then we use the non degeneracy of \(\Sigma (t,s)\) (which we prove in the following lemma) in order to obtain the non degeneracy of \( \Sigma _{n}(t,s)\). And in the case \(\sqrt{n}\le t-s\le n\pi \) we use the estimates from the previous section in order to obtain directly the non degeneracy of \(\Sigma _{n}(t,s)\).

Lemma C.1

For every \(\varepsilon >0\) there exists \(\lambda (\varepsilon )>0\) such that for every t and s such that \(\left| t-s\right| >\varepsilon \) one has

$$\begin{aligned} \det \Sigma (t,s)\ge \lambda (\varepsilon ). \end{aligned}$$
(C.2)

Proof

We first prove that \(\det \Sigma (t,s)>0\) for every \(t\ne s\). Let \((X_t)_{t\ge 0}\) denote a centred stationary Gaussian process whose covariance function is \(r(t,s)=\Gamma (t-s)\), with \(\Gamma (\tau )=\sin \tau /\tau \). Recall that X has smooth paths (see e.g. [4], Ch. 1, Sect. 4.3). It is known (see Ex. 3.5 in [4]) that if the spectral measure of X has at least an accumulation point (in our case the spectral measure is \(1_{|x|<1}dx\)) then, for \(t\ne s\), the law of \(\xi =(X_t,X'_t, X_s, X'_s)\) is non degenerated—consequently the covariance matrix is non degenerated. Straightforward computations give that \(\Sigma (t,s)\) is the covariance matrix of \(\xi \), so \(\det \Sigma (t,s)>0\) when \(t\ne s\).

Now, if \(i\ne j\) one has \(\lim _{t-s\rightarrow \infty } \Sigma ^{i,j}(t,s)=0\), so that \(\lim _{t-s\rightarrow \infty } {\mathrm {det}\,}\Sigma (t,s)= 1/9\). As a consequence, for some \(M>0\) one gets \({\mathrm {det}\,}\Sigma ^{i,j}(t,s)\ge 1/18\) for \(|t-s|>M\). If \(|t-s|\in [\varepsilon ,M]\) the function \((t,s)\mapsto {\mathrm {det}\,}\Sigma (t,s)\) is continuous, so it achieves a strictly positive minimum. The statement now follows. \(\square \)

Corollary C.2

Let \(b_{*}<2\pi \). For \(\varepsilon >0\) let \(\lambda (\varepsilon )\) be given as in Lemma C.1. Then there exists \( n(\varepsilon )\) such that for \(n\ge n(\varepsilon )\) one has

$$\begin{aligned} \inf _{\varepsilon <\left| t-s\right| \le b_{*}n}\det \Sigma _{n}(t,s)\ge \frac{1}{2}\lambda (\varepsilon ). \end{aligned}$$
(C.3)

Proof

Suppose first that \(\varepsilon <t-s\le n^{1/2}\). Then

$$\begin{aligned} \det \Sigma _{n}(t,s)\ge \det \Sigma (t,s)-\left| \det \Sigma _{n}(t,s)-\det \Sigma (t,s)\right| \ge \lambda (\varepsilon )-\frac{C}{ n^{1/2}}\ge \frac{1}{2}\lambda (\varepsilon ) \end{aligned}$$

for sufficiently large n.

We consider now the case \(t-s>n^{1/2}\). We will use (B.2) with \(b= \frac{t-s}{n}\) in order to prove that all the terms out of the diagonal are very small, so the determinant will be close to the product of the terms of the diagonal which is (almost) \(\frac{1}{9}\). We look to

$$\begin{aligned} \Sigma _{n}^{4,2}(t,s)=\frac{1}{n}\sum _{k=1}^{n}\frac{k^{2}}{n^{2}}\cos \frac{k(t-s)}{n}=S_{b,2}(c) \end{aligned}$$

Since \(t-s\le b_{*}n\) it follows that \(b=\frac{t-s}{n}\le b_{*}<2\pi \) and this guarantees that \(\overline{b}=\min \{b,2\pi -b_{*}\}\). Since \(nb=t-s\ge \sqrt{n},\) for sufficiently large n we have \(\overline{b} n\ge \sqrt{n}\) and so, by (B.2)

$$\begin{aligned} \left| \Sigma _{n}^{4,2}(t,s)\right| \le \frac{C}{\sqrt{n}} \rightarrow 0. \end{aligned}$$

The same is true for the other terms out of the diagonal. \(\square \)

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Bally, V., Caramellino, L. & Poly, G. Non universality for the variance of the number of real roots of random trigonometric polynomials. Probab. Theory Relat. Fields 174, 887–927 (2019). https://doi.org/10.1007/s00440-018-0869-2

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  • DOI: https://doi.org/10.1007/s00440-018-0869-2

Keywords

  • Random trigonometric polynomials
  • Edgeworth expansion for non smooth functions
  • Kac–Rice formula
  • Small balls estimates

Mathematics Subject Classification

  • 60G50
  • 60F05