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Overcrowding Estimates for zero count and nodal length of stationary Gaussian processes

Lakshmi Priya Department of Mathematics, Indian Institute of Science, Bangalore 560012, India [email protected]
Abstract.

Assuming certain conditions on the spectral measures of centered stationary Gaussian processes on {\mathbb{R}} (or 2{\mathbb{R}}^{2}), we show that the probability of the event that their zero count in an interval (resp., nodal length in a square domain) is larger than nn, where nn is much larger than the expected value of the zero count in that interval (resp., nodal length in that square domain), is exponentially small in nn.

This work is supported by CSIR-SPM fellowship (File No. SPM-07/079(0260)/2017-EMR-I)), CSIR, Government of India and by a UGC CAS-II grant (Grant No. F.510/25/CAS-II/2018(SAP-I))

1. Introduction

1.1. Main Results

In this paper we consider centered stationary Gaussian processes on {\mathbb{R}}/2{\mathbb{R}}^{2}, study the unlikely event that there is an excessive zero count/nodal length in a region and obtain probability estimates for them. Let g:2g:{\mathbb{R}}^{2}\rightarrow{\mathbb{R}} be a C1C^{1} function such that for every z2z\in{\mathbb{R}}^{2} we have |g(z)|+|g(z)|0|g(z)|+|\nabla g(z)|\neq 0. Then its zero/nodal set 𝒵(g):=g1{0}\mathcal{Z}(g):=g^{-1}\{0\} is a smooth one-dimensional submanifold. Let f:f:{\mathbb{R}}\rightarrow{\mathbb{R}} be any function and T>0T>0, we define 𝒩T(f)\mathscr{N}_{T}(f) and T(g)\mathscr{L}_{T}(g) to be the zero count of ff in [0,T][0,T] and the nodal length of gg in [0,T]2[0,T]^{2} respectively, that is

𝒩T(f)\displaystyle\mathscr{N}_{T}(f) :=#{x[0,T]:f(x)=0},\displaystyle:=\#\{x\in[0,T]:f(x)=0\},
T(g)\displaystyle\mathscr{L}_{T}(g) :=Length{z[0,T]2:g(z)=0}.\displaystyle:=\text{Length}\{z\in[0,T]^{2}:g(z)=0\}.

When there is no ambiguity of the function under consideration, we simply use 𝒩T\mathscr{N}_{T} and T\mathscr{L}_{T}.

Let XX be a centered stationary Gaussian process on {\mathbb{R}}/2{\mathbb{R}}^{2} whose spectral measure μ\mu is a symmetric Borel probability measure. Stationarity of XX implies that 𝔼[𝒩T]T\mathbb{E}[\mathscr{N}_{T}]\propto T and 𝔼[T]T2\mathbb{E}[\mathscr{L}_{T}]\propto T^{2}, under the assumption that μ\mu has a finite second moment these expectations are also finite. Under some assumptions on μ\mu, we get probability estimates for the unlikely events 𝒩T𝔼[𝒩T]\mathscr{N}_{T}\gg\mathbb{E}[\mathscr{N}_{T}] and T𝔼[T]\mathscr{L}_{T}\gg\mathbb{E}[\mathscr{L}_{T}]. The following are the assumptions we make on μ\mu and their significance.

  • All moments CnC_{n}(1)/Cm,nC_{m,n}(2) of μ\mu are finite: The higher order moments CnC_{n} of μ\mu control the higher order derivatives of XX, hence a control on CnC_{n} gives a control on the oscillations of XX, which in turn controls the number of 0-crossings of XX.

  • (A1A1)/(A2A2): The event 𝒩T𝔼[𝒩T]\mathscr{N}_{T}\gg\mathbb{E}[\mathscr{N}_{T}]/T𝔼[T]\mathscr{L}_{T}\gg\mathbb{E}[\mathscr{L}_{T}] is intimately related to the event that XX takes small values in a region (small ball event). Under assumption (A1A1)/(A2A2), the small ball event becomes highly unlikely and it is possible to estimate their probabilities.

Lemmas 2.1 and 2.7 are the quantitative versions of the above statements.

Notations.

Before stating the main results, we introduce some useful notations.

  1. 1.

    Let μ\mu be a positive Borel measure on {\mathbb{R}}. For nn\in\mathbb{N}, we define CnC_{n} and DnD_{n} by

    Cn:=|x|n𝑑μ(x) and Dn:=max{1,C2n,C2n+2}.\begin{gathered}C_{n}:=\int_{{\mathbb{R}}}|x|^{n}d\mu(x)\text{ and }D_{n}:=\max\{1,\sqrt{C_{2n}},\sqrt{C_{2n+2}}\}.\end{gathered} (1)
  2. 2.

    For μ\mu a positive Borel measure on 2{\mathbb{R}}^{2}, we denote by μr\mu_{r} the push forward of μ\mu by the map ϕ\phi, where ϕ:2[0,)\phi:{\mathbb{R}}^{2}\rightarrow[0,\infty) is defined by ϕ(x,y):=x2+y2\phi(x,y):=\sqrt{x^{2}+y^{2}}. For m,n{0}m,n\in\mathbb{N}\cup\{0\}, we define the following quantities

    Cm,n:=2|x|m|y|n𝑑μ(x,y) and R~n:=max{Ck,2nk:0k2n},Rn:=max{1,R~1,R~n,R~n+1}.L~n:=(0t2n𝑑μr(t))1/2 and Ln:=max{1,L~1,L~n,L~n+1}.\begin{gathered}C_{m,n}:=\int_{{\mathbb{R}}^{2}}|x|^{m}|y|^{n}d\mu(x,y)\text{ and }\widetilde{R}_{n}:=\max\{\sqrt{C_{k,2n-k}}:0\leq k\leq 2n\},\\ R_{n}:=\max\{1,\widetilde{R}_{1},\widetilde{R}_{n},\widetilde{R}_{n+1}\}.\\ \widetilde{L}_{n}:=\left(\int_{0}^{\infty}t^{2n}d\mu_{r}(t)\right)^{1/2}\text{ and }L_{n}:=\max\{1,\widetilde{L}_{1},\widetilde{L}_{n},\widetilde{L}_{n+1}\}.\end{gathered} (2)

    The relation between the moments LnL_{n} and RnR_{n} is discussed in Remark 3.10.

  3. 3.

    Let μ\mu be a positive Borel measure on {\mathbb{R}}/2{\mathbb{R}}^{2}. Then μ\mu is said to satisfy assumption (A1)(A1)/ (A2)(A2) if:

    μ\mu is a symmetric Borel probability measure on {\mathbb{R}} which has a nontrivial absolutely continuous part w.r.t. the Lebesgue measure on {\mathbb{R}}. (A1A1)
    μ\mu is a symmetric Borel probability measure on 2{\mathbb{R}}^{2} such that there exist v1v_{1}, v2𝕊1v_{2}\in\mathbb{S}^{1} satisfying v1v2v_{1}\perp v_{2} and the marginals of μ\mu on v1{\mathbb{R}}v_{1} and v2{\mathbb{R}}v_{2} satisfy (A1A1). (A2A2)
Theorem 1.1.

Let XX be a centered stationary Gaussian process on {\mathbb{R}} whose spectral measure μ\mu is such that it satisfies (A1A1) and all its moments CnC_{n} are finite. Then there exist constants c>0c>0, b,B(0,1)b,B\in(0,1) such that for every ϵ(0,1/2)\epsilon\in(0,1/2), nn\in\mathbb{N} such that n1/ϵ2n\geq 1/\epsilon^{2} and T(0,bϵn)T\in(0,b\lfloor\epsilon n\rfloor) satisfying (B/Dn1/n)(n/T)12ϵe(B/D_{n}^{1/n})\cdot(n/T)^{1-2\epsilon}\geq e, we have

exp(n2log(cn/T))(𝒩Tn)2exp{ϵn22log(BDn1/n(nT)12ϵ)}.\displaystyle\exp(-n^{2}\log(cn/T))\leq\mathbb{P}(\mathscr{N}_{T}\geq n)\leq 2\exp\left\{-\frac{\epsilon n^{2}}{2}\log\left(\frac{B}{D_{n}^{1/n}}\left(\frac{n}{T}\right)^{1-2\epsilon}\right)\right\}. (3)
Theorem 1.2.

Let XX be a centered stationary Gaussian process on 2{\mathbb{R}}^{2} whose spectral measure μ\mu is such that it satisfies (A2A2) and all its moments Cm,nC_{m,n} (or equivalently L~n\widetilde{L}_{n}) are finite. Then there are constants b,B(0,1)b,B\in(0,1) such that for every ϵ(0,1/4)\epsilon\in(0,1/4), n1/ϵ2n\geq 1/\epsilon^{2} and T(0,bϵn)T\in(0,b\lfloor\epsilon n\rfloor) such that B/Ln1/n(n/T)14ϵeB/L_{n}^{1/n}\cdot(n/T)^{1-4\epsilon}\geq e, we have

(T>4nT)6exp{ϵn22log(BLn1/n(nT)14ϵ)}.\displaystyle\mathbb{P}(\mathscr{L}_{T}>4nT)\leq 6\exp\left\{-\frac{\epsilon n^{2}}{2}\log\left(\frac{B}{L_{n}^{1/n}}\left(\frac{n}{T}\right)^{1-4\epsilon}\right)\right\}. (4)
Remark 1.3.

If we fix n16n\geq 16 and take ϵ=1/4\epsilon=1/4 in (3), we can conclude that there are constants bn,bn>0b_{n},b^{\prime}_{n}>0 such that for every δ(0,bn)\delta\in(0,b_{n}^{\prime}) we have

(𝒩δn)bnδn2/16,\displaystyle\mathbb{P}(\mathscr{N}_{\delta}\geq n)\leq b_{n}~{}\delta^{n^{2}/16},

and this indicates short-range repulsion of the zeros of XX.

In Section 6, we apply Theorems 1.1 and 1.2 to specific classes of spectral measures μ\mu and get tail estimates for 𝒩T\mathscr{N}_{T} and T\mathscr{L}_{T}; these results are summarized in the following tables.

Growth of CnC_{n} Example of μ\mu Constraints on TT and nn log(𝒩Tn)\log\mathbb{P}(\mathscr{N}_{T}\geq n) (𝔼[𝒩Tm])1m(\mathbb{E}[\mathscr{N}_{T}^{m}])^{\frac{1}{m}}\lesssim
CnqnC_{n}\leq q^{n}, for q>0q>0 Any μ\mu with supp(μ)[q,q](\mu)\subseteq[-q,q] T1,nCTT\geq 1,~{}n\geq CT, for C1C\gg 1 n2log(nT)\asymp-n^{2}\log(\frac{n}{T}) TmT\vee\sqrt{m}
CnnαnC_{n}\leq n^{\alpha n}, for α(0,1)\alpha\in(0,1) μ𝒩(0,1)\mu\sim\mathcal{N}(0,1), with α=1/2\alpha=1/2 T1,nT1/κT\geq 1,~{}n\geq T^{1/\kappa}, for κ(0,1α)\kappa\in(0,1-\alpha) n2logn\asymp-n^{2}\log n T1/κmT^{1/\kappa}\vee\sqrt{m}
CnnαnC_{n}\leq n^{\alpha n}, for α1\alpha\geq 1 dμ(x)e|x|1αdxd\mu(x)\propto e^{-|x|^{\frac{1}{\alpha}}}dx T=1,nT=1,n\in\mathbb{N} n2α+κ\lesssim-n^{\frac{2}{\alpha+\kappa}}, for any κ>0\kappa>0 m(α+κ)/2m^{(\alpha+\kappa)/2}
logCnnγ+1γ\log C_{n}\lesssim n^{\frac{\gamma+1}{\gamma}}, for γ>1/2\gamma>1/2 dμ(x)e(log|x|)1+γ𝟙|x|1dxd\mu(x)\propto e^{-(\log|x|)^{1+\gamma}}\mathbbm{1}_{|x|\geq 1}dx T=1,nT=1,n\in\mathbb{N} (logn)2γ\lesssim-(\log n)^{2\gamma} exp(cm12γ1)\exp({c}m^{\frac{1}{2\gamma-1}})

Table 1. Consequences of Theorem 1.1
Growth of LnL_{n} Example of μ\mu or XX Constraints on TT and \ell log(T)\log\mathbb{P}(\mathscr{L}_{T}\geq\ell) (𝔼[Tm])1m(\mathbb{E}[\mathscr{L}_{T}^{m}])^{\frac{1}{m}}\lesssim
LnqnL_{n}\leq q^{n}, for q>0q>0 Random plane wave (dμ(θ)(d\mu(\theta)\sim dθ/2πd\theta/2\pi on 𝕊1)\mathbb{S}^{1}) T1,CT2T\geq 1,~{}\ell\geq CT^{2}, for C1C\gg 1 (2T2logT2)\lesssim-(\frac{\ell^{2}}{T^{2}}\log\frac{\ell}{T^{2}}) T(Tm)T(T\vee\sqrt{m})
LnnαnL_{n}\leq n^{\alpha n}, for α(0,1)\alpha\in(0,1) μ𝒩(0,I)\mu\sim\mathcal{N}(0,I), with α=1/2\alpha=1/2 T1,Tκ+1κT\gg 1,~{}\ell\geq T^{\frac{\kappa+1}{\kappa}}, for κ(0,1α)\kappa\in(0,1-\alpha) (2T2log)\lesssim-(\frac{\ell^{2}}{T^{2}}\log\ell) T(T1κm)T(T^{\frac{1}{\kappa}}\vee\sqrt{m})
LnnαnL_{n}\leq n^{\alpha n}, for α1\alpha\geq 1 dμr(x)ex1α𝟙x>0dxd\mu_{r}(x)\propto e^{-x^{\frac{1}{\alpha}}}~{}\mathbbm{1}_{x>0}~{}dx T=1,1T=1,\ell\gg 1 2α+κ\lesssim-\ell^{\frac{2}{\alpha+\kappa}}, for any κ>0\kappa>0 m(α+κ)/2m^{(\alpha+\kappa)/2}
logLnnγ+1γ\log L_{n}\lesssim n^{\frac{\gamma+1}{\gamma}}, for γ>1/2\gamma>1/2 dμr(x)e(logx)1+γ𝟙x1dxd\mu_{r}(x)\propto e^{-(\log x)^{1+\gamma}}~{}\mathbbm{1}_{x\geq 1}~{}dx T=1,1T=1,\ell\gg 1 (log)2γ\lesssim-(\log\ell)^{2\gamma} exp(cm12γ1)\exp({c}m^{\frac{1}{2\gamma-1}})

Table 2. Consequences of Theorem 1.2

1.2. Prior work &\& comparison with our results

Before we comment about how our results compare with what is already known, let us briefly review the results known about zero count and nodal length of stationary Gaussian processes. A more detailed and comprehensive account of results known so far can be found in [1, 5, 15].

As before, let XX be a centered stationary Gaussian processes on {\mathbb{R}}/2{\mathbb{R}}^{2} whose spectral measure μ\mu is a symmetric Borel probability measure. We denote its covariance kernel by k=μ^k=\widehat{\mu}. The exact value of the expectations of 𝒩T\mathscr{N}_{T} and T\mathscr{L}_{T} are given by the Kac-Rice formulas. Under the assumption that the tails of μ\mu are light enough and some integrablility assumptions on kk and k′′k^{\prime\prime}, it was shown in [6] that Var𝒩TT\mathscr{N}_{T}\asymp T and a central limit theorem was also established for 𝒩T\mathscr{N}_{T}. Finiteness of moments of 𝒩T\mathscr{N}_{T} and bounds for these were established in [3, 4, 14] under the assumption that the moments of μ\mu are finite and more recently the asymptotics of the central moments of 𝒩T\mathscr{N}_{T} were obtained in [1] assuming certain decay of kk. Under the assumption that μ\mu has very light tails and kk is integrable, it was shown in [5] that 𝒩T\mathscr{N}_{T} concentrates exponentially around its mean. Central limit theorems and variance asymptotics for T\mathscr{L}_{T} in specific examples of stationary Gaussian processes were obtained in [8, 13]. Under the assumption that moments of μ\mu are finite, finiteness of moments of T\mathscr{L}_{T} were established in [2].

These results can be broadly classified into two categories based on the assumptions made on their spectral measure or covariance.

  • Decay of covariance kk (and light tails of μ\mu): The philosophy behind such an assumption is that decay of kk implies quasi-independence of events in well separated regions and hence for a large TT, 𝒩T\mathscr{N}_{T} is approximately a sum of identically distributed random variables which are MM-dependent. Although it is quite easy to state this idea, building on it and making it work is far from trivial. Hence the concentration results in [5, 6] can be viewed as stemming from some underlying independence.

  • Finiteness of moments of μ\mu: In the other results like [3, 4, 14] which established finiteness of moments of 𝒩T\mathscr{N}_{T}, the only assumption made on μ\mu is finiteness of its moments. The significance of this, as discussed earlier, is that a control on the moments imply a control on the oscillations and hence on the zero count also.

The nature of the results from these two categories also differ: in the former, the assumptions made are strong and so are the conclusions (concentration) and in the latter, the assumptions are much weaker and the conclusions (non-explicit moment bounds) are also weak. In terms of the assumptions made on the spectral measure/covariance, our result falls in the second category, but surprisingly we get quite strong conclusions. To illustrate this claim, we present some consequences of Theorem 1.1; the following are two instances where we get the exact asymptotics for the deviation probabilities (𝒩Tn)\mathbb{P}(\mathscr{N}_{T}\geq n):

  • If μ\mu is compactly supported, there is C>1C>1 such that if T>1T>1 and CTnCT\leq n, then

    log(𝒩Tn)n2log(n/T).\displaystyle-\log\mathbb{P}(\mathscr{N}_{T}\geq n)\asymp n^{2}\log(n/T).
  • If μ\mu is such that CnnαnC_{n}\leq n^{\alpha n} for some α(0,1)\alpha\in(0,1), then for every κ(0,1α)\kappa\in(0,1-\alpha) and n,Tn,T such that 1Tnκ1\leq T\leq n^{\kappa}, we have

    log(𝒩Tn)n2logn.\displaystyle-\log\mathbb{P}(\mathscr{N}_{T}\geq n)\asymp n^{2}\log n.

It is also interesting to compare the exponential concentration result ([5], Theorem 1.1) with Theorem 1.1; both the results give estimates for (|𝒩T𝔼[𝒩T]|F(T))\mathbb{P}(|\mathscr{N}_{T}-\mathbb{E}[\mathscr{N}_{T}]|\geq F(T)). The former gives estimates when the fluctuation F(T)F(T) is of order TT, while our result gives estimates when F(T)/T1F(T)/T\gg 1 and F(T)F(T) is large (how large depends on the growth of moments CnC_{n}) enough. Hence only in certain situations can we actually compare these two results and one such instance is when μ\mu is compactly supported and has a density which belongs to W1,2W^{1,2}. There exists C>1C>1 (depending on μ\mu) such that for η>0\eta>0 and F(T)=ηTF(T)=\eta T, the estimates for the deviation probability from [5] and Theorem 1.1 are exp(cηT)\exp(-c_{\eta}T), η>0\forall\eta>0 and exp(cη2T2)\exp(-c\eta^{2}T^{2}), ηC\forall\eta\geq C respectively. So this naturally raises a question about the true deviation probabilities: either the true deviations for η(0,C)\eta\in(0,C) are much smaller than exp(cηT)\exp(-c_{\eta}T) or there is some c0(0,C)c_{0}\in(0,C) such that the behaviour of the deviation probabilities are different for η(0,c0)\eta\in(0,c_{0}) and η(c0,C)\eta\in(c_{0},C) which hints at a possible JLM law.

1.3. Plan of the paper

In Section 2, we present the main ideas in the proofs of Theorems 1.1 and 1.2 and also give a sketch of their proofs. The relevant details and calculations required to complete the proofs are presented in Sections 4 and 5. In Section 3, we recall some known results and present other preliminary results required in proving our main theorems. In Section 6, we deduce the tail bounds for 𝒩T\mathscr{N}_{T} and T\mathscr{L}_{T} presented in Tables 1 and 2 from Theorems 1.1 and 1.2 respectively.

2. Idea of the proof

In this section, we present the key ideas in the proofs of Theorems 1.1 and 1.2; we also give a brief sketch of their proofs here. The calculations and detailed arguments will be presented in Sections 4 and 5.

2.1. Proof idea of Theorem 1.1

We now present a simple yet very useful idea regarding the zero count of a smooth function and this serves as the starting point for proving the upper bound in Theorem 1.1. This idea has previously been employed in [3, 4, 14] to study moments of the zero count of random functions. The following lemma is inspired by these earlier versions and is a slight modification of those.

Lemma 2.1.

Let T,M>0T,M>0 and nn\in\mathbb{N}. Let f:[0,2T]f:[0,2T]\rightarrow{\mathbb{R}} be a smooth function which has at least nn distinct roots in [0,T][0,T] and f(n)L[0,2T]M\|f^{(n)}\|_{L^{\infty}[0,2T]}~{}\leq M. Then for every 0kn0\leq k\leq n, we have

f(nk)L[T,2T]M(2T)kk!.\displaystyle\|f^{(n-k)}\|_{L^{\infty}[T,2T]}\leq M\frac{(2T)^{k}}{k!}.
Proof.

Let g:[0,T]g:[0,T]\rightarrow{\mathbb{R}} be a smooth function and let x1<x2x_{1}<x_{2} be two zeros of gg, then there exists a zero of gg^{\prime} in [x1,x2][x_{1},x_{2}]. This along with the fact that ff has at least nn distinct roots in [0,T][0,T] implies that there are points 0αn1αn2α1α0T0\leq\alpha_{n-1}\leq\alpha_{n-2}\leq\cdots\leq\alpha_{1}\leq\alpha_{0}\leq T such that for every 0kn10\leq k\leq n-1, we have f(k)(αk)=0f^{(k)}(\alpha_{k})=0. For t[αn1,2T]t\in[\alpha_{n-1},2T], we have

|f(n1)(t)|=|αn1tf(n)(x)𝑑x|M(tαn1)Mt.\begin{gathered}|f^{(n-1)}(t)|=|\int_{\alpha_{n-1}}^{t}f^{(n)}(x)~{}dx|\leq M(t-\alpha_{n-1})\leq Mt.\end{gathered} (5)

Using the estimate for |f(n1)||f^{(n-1)}| in (5), we have for every t[αn2,2T]t\in[\alpha_{n-2},2T]

|f(n2)(t)|=|αn2tf(n1)(x)𝑑x|M(t2αn222)Mt22.\displaystyle|f^{(n-2)}(t)|=|\int_{\alpha_{n-2}}^{t}f^{(n-1)}(x)~{}dx|\leq M\left(\frac{t^{2}-\alpha_{n-2}^{2}}{2}\right)\leq M\frac{t^{2}}{2}.

Thus inductively we can establish that for every kn1k\leq n-1 and every t[αnk,2T]t\in[\alpha_{n-k},2T]

|f(nk)(t)|Mtkk!M(2T)kk!,\displaystyle|f^{(n-k)}(t)|\leq M\frac{t^{k}}{k!}\leq M\frac{(2T)^{k}}{k!},

and this establishes our claim. ∎

We get the following result as an immediate consequence of Lemma 2.1.

Lemma 2.2.

Let F:F:{\mathbb{R}}\rightarrow{\mathbb{R}} be a smooth random function, then for every T,M>0T,M>0 and nn\in\mathbb{N} we have

(𝒩Tn)(FL[T,2T]M(2T)nn!)+(F(n)L[0,2T]>M).\mathbb{P}(\mathscr{N}_{T}\geq n)\leq\mathbb{P}\left(\lVert F\rVert_{L^{\infty}[T,2T]}\leq M\frac{(2T)^{n}}{n!}\right)+\mathbb{P}(\lVert F^{(n)}\rVert_{L^{\infty}[0,2T]}>M). (6)

Upper bound in Theorem 1.1

We use Lemma 2.2 to prove the upper bound in Theorem 1.1 and hence we only need to estimate the two terms on the r.h.s. of (6). The second term is estimated using well known tail bounds for the supremum of Gaussian processes; we recall these standard results in Section 3.2. Since we are working with stationary Gaussian processes, calculating the metric entropy and the corresponding Dudley integral are quite straightforward.

The first term corresponds to a small ball event and the discussion in Section 3.1 is devoted to obtaining probability estimates for such small ball events. We briefly explain how this is done: say η>0\eta>0 and we want to estimate (XL[0,T]η)\mathbb{P}(\|X\|_{L^{\infty}[0,T]}\leq\eta). For mm\in\mathbb{N} and 0km0\leq k\leq m, we define tk:=kT/mt_{k}:=kT/m which are (m+1)(m+1) equispaced points in [0,T][0,T]. Consider the Gaussian vector VmV_{m} defined by Vm:=(Xt0,,Xtm).V_{m}:=(X_{t_{0}},\ldots,X_{t_{m}}). Let Σm\Sigma_{m} and ϕm\phi_{m} be the covariance matrix and the density of VmV_{m} respectively, then

ϕm(x)=1(2π)m+1|Σm|1/2exp(Σm1x,x2),\displaystyle\phi_{m}(x)=\frac{1}{(\sqrt{2\pi})^{m+1}~{}|\Sigma_{m}|^{1/2}}\exp\left(-\frac{\langle\Sigma_{m}^{-1}x,x\rangle}{2}\right),
hence ϕmL(m+1)1(2π)m+1|Σm|1/21(2π)m+1λmm+1/2,\displaystyle\text{hence }\|\phi_{m}\|_{L^{\infty}({\mathbb{R}}^{m+1})}\leq\frac{1}{(\sqrt{2\pi})^{m+1}~{}|\Sigma_{m}|^{1/2}}\leq\frac{1}{(\sqrt{2\pi})^{m+1}~{}\lambda_{m}^{m+1/2}},

where λm\lambda_{m} is the smallest eigenvalue of Σm\Sigma_{m} and hence the following is an estimate for the small ball probability

(XL[0,T]η)[η,η]m+1ϕm(x)𝑑x(2η2πλm1/2)m+1.\displaystyle\mathbb{P}(\|X\|_{L^{\infty}[0,T]}\leq\eta)\leq\int_{[-\eta,\eta]^{m+1}}\phi_{m}(x)dx\leq\left(\frac{2\eta}{\sqrt{2\pi}~{}\lambda_{m}^{1/2}}\right)^{m+1}. (7)

Result 3.3 gives a lower bound for λm\lambda_{m} and hence an upper bound for the small ball probability in (7). We then use (7) with η=M(2T)n/n!\eta=M(2T)^{n}/n! and an optimal choice of mm to get an esimate for the first term in (6).

For given values of nn and TT, we need to make an optimal choice of MM (it should be small enough so that the first event is indeed a small ball event and large enough so that the second event is unlikely) so that the r.h.s. of (6) is as small as possible.

Lower bound in Theorem 1.1

We use the same notations as above and take m=nm=n. If the sign of a continuous function defined on [0,T][0,T] alternates at the points tkt_{k}, then there is necessarily a 0-crossing of the function between every two successive points tkt_{k} and hence the zero count in [0,T][0,T] is at least nn. Thus we have

(Xt0<0,Xt1>0,,(1)n+1Xtn>0)(𝒩Tn),\displaystyle\mathbb{P}(X_{t_{0}}<0,X_{t_{1}}>0,\ldots,(-1)^{n+1}X_{t_{n}}>0)\leq\mathbb{P}(\mathscr{N}_{T}\geq n), (8)

and hence to get a lower bound for the above probability, we need a lower bound on the density ϕn\phi_{n} and we get this below. Σn\Sigma_{n} being a Gram-matrix with all diagonal entries equal to 1, we have |Σn|1|\Sigma_{n}|\leq 1 ([9], Lemma 3) and hence

ϕn(x)1(2π)n+1exp(Σn1x,x2)1(2π)n+1exp(x22λn).\displaystyle\phi_{n}(x)\geq\frac{1}{(\sqrt{2\pi})^{n+1}}\exp\left(-\frac{\langle\Sigma_{n}^{-1}x,x\rangle}{2}\right)\geq\frac{1}{(\sqrt{2\pi})^{n+1}}\exp\left(-\frac{\|x\|^{2}}{2\lambda_{n}}\right). (9)

As before, we use the lower bound for λn\lambda_{n} from Result 3.3 to get a lower bound on the density in (9) and then use this to get a lower bound for the l.h.s. of (8).

Remark 2.3.

Result 3.3 gives a lower bound for λn\lambda_{n} and the only assumption on μ\mu required to establish this result is (A1A1). Hence it is the same assumption, namely (A1A1), which gives both the following bounds:

  • an upper bound for the small ball estimate in (7),

  • a lower bound for the probability of the event {Xt0<0,Xt1>0,}\{X_{t_{0}}<0,X_{t_{1}}>0,\ldots\} in (8).

This is not very surprising and we can perceive this as: the propensity of XX to oscillate makes it difficult for it to be confined to a small ball.

2.2. Proof idea of Theorem 1.2

We now present some deterministic results about nodal length of a smooth function which will be useful in proving Theorem 1.2. Let us first introduce some notations. For f:2f:{\mathbb{R}}^{2}\rightarrow{\mathbb{R}} a smooth function, T>0T>0 and t[0,T]t\in[0,T], we define N1,TN_{1,T} and N2,TN_{2,T} as follows

N1,T(t)\displaystyle N_{1,T}(t) :=#{y[0,T]:f(t,y)=0},\displaystyle:=\#\{y\in[0,T]:f(t,y)=0\},
N2,T(t)\displaystyle N_{2,T}(t) :=#{x[0,T]:f(x,t)=0}.\displaystyle:=\#\{x\in[0,T]:f(x,t)=0\}.

The following result in an easy consequence of Lemma 2.1.

Corollary 2.4.

Let T,M>0T,M>0 and nn\in\mathbb{N} be such that 2T<n2T<n. Suppose f:[0,n]f:[0,n]\rightarrow{\mathbb{R}} is a smooth function such that f(n)L[0,n]M\|f^{(n)}\|_{L^{\infty}[0,n]}\leq M and fL[T,2T]>M(2T)n/n!\|f\|_{L^{\infty}[T,2T]}>M(2T)^{n}/n!, then ff has no more than (n1)(n-1) zeros in the interval [0,T][0,T].

The following integral geometric result gives a bound for the nodal length of a smooth function and it appears in the proof of Lemma 5.11 from [7].

Result 2.5.

Let f:2f:{\mathbb{R}}^{2}\rightarrow{\mathbb{R}} be a smooth function and let T>0T>0, then we have the following bound for the nodal length of ff

length{z[0,T]2:f(z)=0\displaystyle\mbox{length}\{z\in[0,T]^{2}:f(z)=0~{} andf(z)0}\displaystyle\mbox{and}~{}\nabla f(z)\neq 0\}
2(0TN1,T(x)𝑑x+0TN2,T(y)𝑑y).\displaystyle\leq\sqrt{2}\left(\int_{0}^{T}N_{1,T}(x)dx+\int_{0}^{T}N_{2,T}(y)dy\right).

The following lemma is an immediate consequence of Corollary 2.4 and is inspired by Lemma 5.11 in [7].

Lemma 2.6.

Let T,M>0T,M>0, nn\in\mathbb{N} be such that 2Tn2T\leq n. Let f:2f:{\mathbb{R}}^{2}\rightarrow{\mathbb{R}} be a smooth function satisfying the following

max{2fL[0,n]2,1nfL[0,n]2}M/2 and f(,0)L[T,2T]>M(2T)n/n!.\displaystyle\max\left\{\left\|\partial_{2}f\right\|_{L^{\infty}[0,n]^{2}},\|\partial_{1}^{n}f\|_{L^{\infty}[0,n]^{2}}\right\}\leq M/2\text{ and }\|f(\cdot,0)\|_{L^{\infty}[T,2T]}>M(2T)^{n}/n!.

Then for every t[0,(2T)n/n!]t\in[0,(2T)^{n}/n!], we have N2,T(t)<nN_{2,T}(t)<n.

Proof.

Let x0[T,2T]x_{0}\in[T,2T] be such that |f(x0,0)|>M(2T)n/n!|f(x_{0},0)|>M(2T)^{n}/n!, then for every t[0,(2T)n/n!]t\in[0,(2T)^{n}/n!] we have

f(x0,t)f(x0,0)=0t2f(x0,s)ds,hence |f(x0,t)f(x0,0)|<M2(2T)nn!,hence |f(x0,t)|>M(2T)n2n!.\begin{gathered}f(x_{0},t)-f(x_{0},0)=\int_{0}^{t}\partial_{2}f(x_{0},s)~{}ds,\\ \text{hence }|f(x_{0},t)-f(x_{0},0)|<\frac{M}{2}\cdot\frac{(2T)^{n}}{n!},\\ \text{hence }|f(x_{0},t)|>\frac{M(2T)^{n}}{2n!}.\end{gathered}

Thus we have f(,t)L[T,2T]M(2T)n/2n!\|f(\cdot,t)\|_{L^{\infty}[T,2T]}\geq M(2T)^{n}/2n! and 1nf(,t)L[0,n]2M/2\|\partial_{1}^{n}f(\cdot,t)\|_{L^{\infty}[0,n]^{2}}\leq M/2. Corollary 2.4 now gives the desired result. ∎

The following lemma which is a direct and easy consequence of Lemma 2.6 is the main ingredient in the proof of Theorem 1.2.

Lemma 2.7.

Let T,M>0T,M>0, nn\in\mathbb{N} be such that 2Tn2T\leq n and define δ:=(2T)n/n!\delta:=(2T)^{n}/n!. Suppose that f:2f:{\mathbb{R}}^{2}\rightarrow{\mathbb{R}} is a smooth function, then we have the following.

  • If for every r{0,1,2,,δ1T}r\in\{0,1,2,\ldots,\lfloor\delta^{-1}T\rfloor\}, the following bounds hold for ff

    max{2fL[0,n]2,1nfL[0,n]2}M/2 and f(,rδ)L[T,2T]>Mδ,\displaystyle\max\left\{\left\|\partial_{2}f\right\|_{L^{\infty}[0,n]^{2}},\|\partial_{1}^{n}f\|_{L^{\infty}[0,n]^{2}}\right\}\leq M/2\text{ and }\|f(\cdot,r\delta)\|_{L^{\infty}[T,2T]}>M\delta, (10)

    then for every t[0,T]t\in[0,T] we have N2,T(t)<nN_{2,T}(t)<n.

  • If for every r{0,1,2,,δ1T}r\in\{0,1,2,\ldots,\lfloor\delta^{-1}T\rfloor\}, the following bounds hold for ff

    max{1fL[0,n]2,2nfL[0,n]2}M/2 and f(rδ,)L[T,2T]>Mδ,\displaystyle\max\left\{\left\|\partial_{1}f\right\|_{L^{\infty}[0,n]^{2}},\|\partial_{2}^{n}f\|_{L^{\infty}[0,n]^{2}}\right\}\leq M/2\text{ and }\|f(r\delta,\cdot)\|_{L^{\infty}[T,2T]}>M\delta, (11)

    then for every t[0,T]t\in[0,T] we have N1,T(t)<nN_{1,T}(t)<n.

Thus if both (10) and (11) hold for ff, we can conclude from Result 2.5 that

length{z[0,T]2:f(z)=0andf(z)0}4nT.\displaystyle\mbox{length}\{z\in[0,T]^{2}:f(z)=0~{}\mbox{and}~{}\nabla f(z)\neq 0\}\leq 4nT.
Refer to caption0TT2T2TnnTT2T2Tnn
Figure 1. An illustration of Lemma 2.7. Derivative bounds for ff on [0,n]2[0,n]^{2} and lower bound for |f||f| on the δ\delta-separated horizontal and vertical lines in [T,2T]×[0,T][T,2T]\times[0,T] and [0,T]×[T,2T][0,T]\times[T,2T] respectively give an upper bound for the nodal length of ff in [0,T]2[0,T]^{2}.

The following result is an immediate consequence of Lemma 2.7.

Lemma 2.8.

Let T,M>0T,M>0, nn\in\mathbb{N} be such that 2Tn2T\leq n and define δ:=(2T)n/n!\delta:=(2T)^{n}/n!. Suppose that F:2F:{\mathbb{R}}^{2}\rightarrow{\mathbb{R}} is a smooth random function such that almost surely FF has no singular zeros, then we have

(T4nT)\displaystyle\mathbb{P}(\mathscr{L}_{T}\geq 4nT) j=1,2[(j)+(j)]+rδ1T[(𝒜r)+(r)],\displaystyle\leq\sum_{j=1,2}[\mathbb{P}(\mathscr{E}_{j})+\mathbb{P}(\mathscr{F}_{j})]+\sum_{r\leq\lfloor\delta^{-1}T\rfloor}[\mathbb{P}(\mathscr{A}_{r})+\mathbb{P}(\mathscr{B}_{r})], (12)

where the events j\mathscr{E}_{j}, j\mathscr{F}_{j}, 𝒜r\mathscr{A}_{r} and r\mathscr{B}_{r} are defined as follows

j:={jFL[0,n]2M/2},j:={jnFL[0,n]2M/2},𝒜r:={F(,rδ)L[T,2T]Mδ},r:={F(rδ,)L[T,2T]Mδ}.\begin{gathered}\mathscr{E}_{j}:=\{\|\partial_{j}F\|_{L^{\infty}[0,n]^{2}}\geq M/2\},~{}\mathscr{F}_{j}:=\{\|\partial_{j}^{n}F\|_{L^{\infty}[0,n]^{2}}\geq M/2\},\\ \mathscr{A}_{r}:=\{\|F(\cdot,r\delta)\|_{L^{\infty}[T,2T]}\leq M\delta\},~{}\mathscr{B}_{r}:=\{\|F(r\delta,\cdot)\|_{L^{\infty}[T,2T]}\leq M\delta\}.\end{gathered}

Proof sketch of Theorem 1.2

The spectral measure μ\mu of XX satisfies (A2A2), we assume without loss of generality that v1=(1,0)v_{1}=(1,0) and v2=(0,1)v_{2}=(0,1) in (A2A2). Hence both the marginals μ1\mu_{1} and μ2\mu_{2} are symmetric probability measures on {\mathbb{R}} which satisfy (A1A1). Let X1X_{1} and X2X_{2} be centered stationary Gaussian processes on {\mathbb{R}} whose spectral measures are μ1\mu_{1} and μ2\mu_{2} respectively.

We further assume that μ\mu is not supported on any line v{\mathbb{R}}v, the degenerate case supp(μ)v(\mu)\subseteq{\mathbb{R}}v (for some v2v\in{\mathbb{R}}^{2}) is much easier and will be analysed in Section 5. In this case, it follows from Bulinskaya’s lemma (Result 3.9) that almost surely XX does not have singular zeros. Hence we can use Lemma 2.8 and the stationarity of XX to conclude that

(T4nT)j=1,2[(j)+(j)]+δ1T[(𝒜0)+(0)],\displaystyle\mathbb{P}(\mathscr{L}_{T}\geq 4nT)\leq\sum_{j=1,2}[\mathbb{P}(\mathscr{E}_{j})+\mathbb{P}(\mathscr{F}_{j})]+\lceil\delta^{-1}T\rceil~{}[\mathbb{P}(\mathscr{A}_{0})+\mathbb{P}(\mathscr{B}_{0})], (13)

and we now see how to get bounds for the terms on the r.h.s. of (13). Since XX is a stationary Gaussian process, so are all its derivatives. Like in the one dimensional case, estimating (j)\mathbb{P}(\mathscr{E}_{j}) and (j)\mathbb{P}(\mathscr{F}_{j}) using the well known tail bounds for supremum of Gaussian processes is quite straightforward and we do this in Section 5.

We now observe that X(,0)=𝑑X1X(\cdot,0)\overset{d}{=}X_{1} and X(0,)=𝑑X2X(0,\cdot)\overset{d}{=}X_{2}. Hence 𝒜0\mathscr{A}_{0} and 0\mathscr{B}_{0} correspond to small ball events for X1X_{1} and X2X_{2} respectively. By our assumption, both their spectral measures satisfy (A1A1) and a way to estimate these small ball probabilities was already discussed in the few lines leading up to (7).

Similar to the one dimensional case, for fixed n,Tn,T we need to make an optimal choice of MM so that the r.h.s. of (13) is minimized.

3. Preliminaries

In this section we present some preliminary results and recall other known results which will be used in establishing Theorems 1.1 and 1.2.

3.1. Small Ball Probability

We now get small ball probability estimates for stationary Gaussian processes on {\mathbb{R}} whose spectral measures satisfy (A1A1).

Let XX be a Gaussian process on an interval II\subseteq{\mathbb{R}} and let {tj}j=1mI\{t_{j}\}_{j=1}^{m}\subset I be distinct points in II such that the Gaussian vector (Xt1,Xt2,,Xtm)(X_{t_{1}},X_{t_{2}},\ldots,X_{t_{m}}) is non-degenerate and let ϕ\phi be its density. Then for every η>0\eta>0, we have

(XL(I)η)\displaystyle\mathbb{P}(\|X\|_{L^{\infty}(I)}\leq\eta) (|Xt1|η,,|Xtm|η),\displaystyle\leq\mathbb{P}(|X_{t_{1}}|\leq\eta,\ldots,|X_{t_{m}}|\leq\eta),
=[η,η]mϕ(x)𝑑x2mϕL(m)ηm.\displaystyle=\int_{[-\eta,\eta]^{m}}\phi(x)dx\leq 2^{m}\|\phi\|_{L^{\infty}({\mathbb{R}}^{m})}~{}\eta^{m}. (14)

For a stationary Gaussian process whose spectral measure satisfies (A1A1), we can get an estimate for ϕL(m)\|\phi\|_{L^{\infty}({\mathbb{R}}^{m})} which appears in (14) and this is what is done below.

Result 3.1 ([12], Turan’s lemma).

Let p(t)=k=1nckeiλktp(t)=\sum_{k=1}^{n}c_{k}e^{i\lambda_{k}t}, where ckc_{k}\in\mathbb{C} and λk\lambda_{k}\in\mathbb{R}. Then there is a constant A>0A>0 such that for every interval II\subseteq{\mathbb{R}}, every measurable set EIE\subseteq I and every q[0,]q\in[0,\infty], we have

pLq(I)(A|I||E|)n1pLq(E),\displaystyle\|p\|_{L^{q}(I)}\leq\left(\frac{A|I|}{|E|}\right)^{n-1}\|p\|_{L^{q}(E)},

where |||\cdot| denotes the Lebesgue measure.

Fact 3.2.

Let XX be a centered stationary Gaussian process on {\mathbb{R}} whose spectral measure μ\mu satisfies (A1A1). Hence μ\mu admits the following decomposition dμ(x)=f(x)dx+dμs(x)d\mu(x)=f(x)dx+d\mu_{s}(x), where f0f\not\equiv 0. Then there is a constant δ0>0\delta_{0}>0 such that |{fδ0}|δ0|\{f\geq\delta_{0}\}|\geq\delta_{0} and hence there is a large enough M0>πM_{0}>\pi such that |S|δ0/2|S|\geq\delta_{0}/2, where S:={fδ0}(M0,M0)S:=\{f\geq\delta_{0}\}\cap(-M_{0},M_{0}). For mm\in\mathbb{N} and T>0T>0 consider the stationary Gaussian process YY on \mathbb{Z} defined by

Y:=XT/m,for .\displaystyle Y_{\ell}:=X_{\ell T/m},~{}\text{for }\ell\in\mathbb{Z}. (15)

Then the spectral measure μm,T\mu_{m,T} of YY is a symmetric probability measure on [π,π][-\pi,\pi] which is the push forward of μ\mu by the map ψ\psi given by

ψ:\displaystyle\psi:{\mathbb{R}}\longrightarrow 𝕊1[π,π)\displaystyle~{}\mathbb{S}^{1}\simeq[-\pi,\pi)
x\displaystyle x\longmapsto Txm(mod2π).\displaystyle~{}\frac{Tx}{m}~{}(\text{mod}~{}2\pi).

Hence μm,T\mu_{m,T} considered as a measure on [π,π][-\pi,\pi] has a nontrivial absolutely continuous part w.r.t. the Lebesgue measure on [π,π][-\pi,\pi] given by fm,T(x)dxf_{m,T}(x)dx, where

fm,T(x):=nmTf(mT(x+2πn))mTf(mxT), for x[π,π].\displaystyle f_{m,T}(x):=\sum_{n\in\mathbb{Z}}\frac{m}{T}~{}f\left(\frac{m}{T}(x+2\pi n)\right)\geq\frac{m}{T}~{}f\left(\frac{mx}{T}\right),\text{ for $x\in[-\pi,\pi]$}.

Define b:=π/M0b:=\pi/M_{0}, if TT and mm are such that TbmT\leq bm, then (T/m)S[π,π](T/m)S\subset[-\pi,\pi] and hence

|{fm,Tmδ0/T}||(T/m)S|Tδ0/2m.\displaystyle|\{f_{m,T}\geq m\delta_{0}/T\}|\geq|(T/m)S|\geq T\delta_{0}/2m. (16)

The following result taken from the proofs of Lemma 3 and Theorem 2 of [9] is the main tool in obtaining the small ball probability estimates.

Result 3.3.

Let XX be a centered stationary Gaussian process on {\mathbb{R}} whose spectral measure μ\mu satisfies (A1A1). Let T>0T>0, mm\in\mathbb{N} and YY be the Gaussian process on \mathbb{Z} defined in (15). Let Σ\Sigma and ϕ\phi denote the covariance matrix and density of the Gaussian vector (Y1,,Ym)(Y_{1},\ldots,Y_{m}) respectively. Let λ\lambda be the smallest eigenvalue of Σ\Sigma. Then there exists b,c(0,1)b,c\in(0,1) and C>1C>1 such that whenever TbmT\leq bm, we have

λ(cTm)2(m1) and ϕL(m)(CmT)m2.\displaystyle\lambda\geq\left(\frac{cT}{m}\right)^{2(m-1)}\text{ and }~{}\|\phi\|_{L^{\infty}({\mathbb{R}}^{m})}\leq\left(\frac{Cm}{T}\right)^{m^{2}}.
Proof.

Let ff, M0M_{0}, bb, δ0\delta_{0} and S{S} be as in Fact 3.2. The density ϕ\phi is given by

ϕ(x)\displaystyle\phi(x) =1(2π)m/2|Σ|1/2exp(Σ1x,x2),\displaystyle=\frac{1}{(2\pi)^{m/2}|\Sigma|^{1/2}}\exp\left(-\frac{\langle\Sigma^{-1}x,x\rangle}{2}\right),
hence ϕL(m)\displaystyle\text{hence }\|\phi\|_{L^{\infty}({\mathbb{R}}^{m})} 1(2π)m/2|Σ|1/21(2π)m/21λm/2.\displaystyle\leq\frac{1}{(2\pi)^{m/2}|\Sigma|^{1/2}}\leq\frac{1}{(2\pi)^{m/2}}\frac{1}{\lambda^{m/2}}. (17)

We now get a lower bound on λ\lambda and use this to get an upper bound on ϕ\phi. Let u=(u1,u2,,um)mu=(u_{1},u_{2},\ldots,u_{m})\in{\mathbb{R}}^{m} and define U(x):=k=1mukeikxU(x):=\sum_{k=1}^{m}u_{k}e^{ikx}. Then we have

ππ|U(x)|2𝑑x=2πu2 and ππ|U(x)|2𝑑μm,T(x)=Σu,u.\displaystyle\int_{-\pi}^{\pi}|U(x)|^{2}dx=2\pi\|u\|^{2}\text{ and }\int_{-\pi}^{\pi}|U(x)|^{2}d\mu_{m,T}(x)=\langle\Sigma u,u\rangle.

Hence it follows from (16) that

Σu,u\displaystyle\langle\Sigma u,u\rangle =ππ|U(x)|2𝑑μm,T(x)ππfm,T(x)|U(x)|2𝑑x,\displaystyle=\int_{-\pi}^{\pi}|U(x)|^{2}d\mu_{m,T}(x)\geq\int_{-\pi}^{\pi}f_{m,T}(x)|U(x)|^{2}dx, (18)
mTδ0(T/m)S|U(t)|2𝑑t.\displaystyle\geq\frac{m}{T}\delta_{0}\int_{(T/m)S}|U(t)|^{2}dt.

We now use Turan’s lemma (Result 3.1) to get

(T/m)S|U(t)|2𝑑t{2πA|(T/m)S|}2(m1)ππ|U(t)|2𝑑t,\displaystyle\int_{(T/m)S}|U(t)|^{2}dt\geq\left\{\frac{2\pi A}{|(T/m)S|}\right\}^{-2(m-1)}\int_{-\pi}^{\pi}|U(t)|^{2}dt, (19)

and hence we conclude the following from (16), (18) and (19)

Σu,umδ0T(Tδ04πmA)2(m1)2πu2,\displaystyle\langle\Sigma u,u\rangle\geq\frac{m\delta_{0}}{T}\left(\frac{T\delta_{0}}{4\pi mA}\right)^{2(m-1)}2\pi\|u\|^{2},
hence, λ2πmδ0T(Tδ04πmA)2(m1)(cTm)2(m1),\displaystyle\text{hence, }\lambda\geq 2\pi\cdot\frac{m\delta_{0}}{T}\left(\frac{T\delta_{0}}{4\pi mA}\right)^{2(m-1)}\geq\left(\frac{cT}{m}\right)^{2(m-1)}, (20)

for some c(0,1)c\in(0,1). Using the lower bound we get for λ\lambda from (20) in (17), we get the desired result

ϕL(m)\displaystyle\|\phi\|_{L^{\infty}({\mathbb{R}}^{m})} (CmT)m2.\displaystyle\leq\left(\frac{Cm}{T}\right)^{m^{2}}.

The following small ball probability estimate is a consequence of (14) and Result 3.3.

Lemma 3.4 (Small ball probability).

Let XX be a centered stationary Gaussian process on {\mathbb{R}} whose spectral measure satisfies (A1A1). Then there are constants b(0,1)b\in(0,1) and C>1C>1 (both depending only on XX) such that for every η>0\eta>0 and every T>0T>0, mm\in\mathbb{N} satisfying TbmT\leq bm, we have

(XL[0,T]η)(CmT)m2ηm.\displaystyle\mathbb{P}(\|X\|_{L^{\infty}[0,T]}\leq\eta)\leq\left(\frac{Cm}{T}\right)^{m^{2}}\eta^{m}. (21)

In the following Lemma, we get small ball probability estimates by using Lemma 3.4 with specific values of η\eta, mm and TT. This will be used in the proofs of Theorems 1.1 and 1.2.

Lemma 3.5.

Let XX, bb, CC, TT and mm be as in Lemma 3.4. Let ϵ(0,1/2)\epsilon\in(0,1/2), nn\in\mathbb{N} be such that m=ϵnm=\lfloor\epsilon n\rfloor, n1/ϵ2n\geq 1/\epsilon^{2}. Let A>1A>1, B=(4eAC)1B=(4eAC)^{-1}, M=2A𝒟nnM=2A\mathscr{D}_{n}\mathscr{H}_{n} where 𝒟n1\mathscr{D}_{n}\geq 1 and n\mathscr{H}_{n} is defined by

n:=n2hn, where hn:=log(B𝒟n1/n(nT)12ϵ).\displaystyle\mathscr{H}_{n}:=\sqrt{n^{2}h_{n}},\text{ where }h_{n}:=\log\left(\frac{B}{\mathscr{D}_{n}^{1/n}}\cdot\left(\frac{n}{T}\right)^{1-2\epsilon}\right).

Suppose hn1h_{n}\geq 1, then with η=M(2T)n/n!\eta=M(2T)^{n}/n! the estimate in Lemma 3.4 becomes

(XL[0,T]M(2T)n/n!)exp{ϵn22log(B𝒟n1/n(nT)12ϵ)}.\displaystyle\mathbb{P}(\|X\|_{L^{\infty}[0,T]}\leq M(2T)^{n}/n!)\leq\exp\left\{\frac{-\epsilon n^{2}}{2}\log\left(\frac{B}{\mathscr{D}_{n}^{1/n}}\cdot\left(\frac{n}{T}\right)^{1-2\epsilon}\right)\right\}.
Proof.

Let WW denote the expression in the r.h.s. of (21) for η=M(2T)n/n!\eta=M(2T)^{n}/n!, then we have

W\displaystyle W =(CmT)m2(Mn!(2T)n)m,\displaystyle=\left(\frac{Cm}{T}\right)^{m^{2}}\left(\frac{M}{n!}(2T)^{n}\right)^{m},
Cmnnm2Tm2(2e)mnTmn(2An𝒟n)mnnm,\displaystyle\leq\frac{C^{mn}n^{m^{2}}}{T^{m^{2}}}\cdot\frac{(2e)^{mn}~{}T^{mn}~{}(2A\mathscr{H}_{n}\mathscr{D}_{n})^{m}}{n^{nm}},
(nT)m2nm(B1𝒟n1/n)mnnm.\displaystyle\leq\left(\frac{n}{T}\right)^{m^{2}-nm}\cdot(B^{-1}\mathscr{D}_{n}^{1/n})^{mn}\cdot\mathscr{H}_{n}^{m}. (22)

Since m=ϵnm=\lfloor\epsilon n\rfloor and n1/ϵ2n\geq 1/\epsilon^{2}, we have

m2mnϵ2n2(ϵn1)n=ϵn2(1ϵ(1/ϵn))ϵ(12ϵ)n2.\displaystyle m^{2}-mn\leq\epsilon^{2}n^{2}-(\epsilon n-1)n=-\epsilon n^{2}\left(1-\epsilon-(1/\epsilon n)\right)\leq-\epsilon(1-2\epsilon)n^{2}.

Also because n/Tn/T, B1𝒟n1/nB^{-1}\mathscr{D}_{n}^{1/n} and n\mathscr{H}_{n} are all greater than 11, we have the following upper bound for the expression on the r.h.s. of (22)

W\displaystyle W exp(ϵ(12ϵ)n2log(n/T)+ϵn2log(B1𝒟n1/n)+ϵnlogn)),\displaystyle\leq\exp(-\epsilon(1-2\epsilon)n^{2}\log(n/T)+\epsilon n^{2}\log(B^{-1}\mathscr{D}_{n}^{1/n})+\epsilon n\log\mathscr{H}_{n})),
exp{ϵn2log(B𝒟n1/n(nT)12ϵ)+ϵnlogn},\displaystyle\leq\exp\left\{-\epsilon n^{2}\log\left(\frac{B}{\mathscr{D}_{n}^{1/n}}\cdot\left(\frac{n}{T}\right)^{1-2\epsilon}\right)+\epsilon n\log\mathscr{H}_{n}\right\},
=exp(ϵn2hn+ϵnlogn2hn),\displaystyle=\exp(-\epsilon n^{2}h_{n}+\epsilon n\log\sqrt{n^{2}h_{n}}),
exp(ϵn(nhnlog(nhn))exp(ϵn2hn/2),\displaystyle\leq\exp(-\epsilon n(nh_{n}-\log(nh_{n}))\leq\exp(-\epsilon n^{2}h_{n}/2), (23)

where the inequality in (23) follows from our assumption that hn1h_{n}\geq 1 and this establishes our claim. ∎

3.2. Metric Entropy and Supremum of Gaussian Processes

We now recall well known results which give tail bounds for the supremum of Gaussian processes and we shall use them to get tail bounds for higher derivatives of stationary Gaussian processes. The definitions and results in this section are taken from Section 3.4 of [10].

Let 𝒯\mathscr{T} be an index set and let (Xt)t𝒯(X_{t})_{t\in\mathscr{T}} be a centered Gaussian process on 𝒯\mathscr{T}. Then XX induces a pseudo-metric dd on 𝒯\mathscr{T} given by

d(s,t):=(𝔼[XsXt]2)1/2.d(s,t):=(\mathbb{E}[X_{s}-X_{t}]^{2})^{1/2}.

Suppose (𝒯,d)(\mathscr{T},d) is totally bounded and let ϵ>0\epsilon>0. The ϵ\epsilon-covering number denoted by N(ϵ,𝒯)N^{\prime}(\epsilon,\mathscr{T}) is defined to be the minimal number of ϵ\epsilon-balls required to cover (𝒯,d)(\mathscr{T},d). The ϵ\epsilon-packing number denoted by N(ϵ,𝒯)N(\epsilon,\mathscr{T}) is defined to be the maximal possible cardinality of subset A𝒯A\subset\mathscr{T} which is such that d(s,t)>ϵd(s,t)>\epsilon, for distinct s,tAs,t\in A. The quantities NN and NN^{\prime} are related by

N(ϵ,𝒯)N(ϵ,𝒯)N(ϵ/2,𝒯).N^{\prime}(\epsilon,\mathscr{T})\leq N(\epsilon,\mathscr{T})\leq N^{\prime}\left(\epsilon/2,\mathscr{T}\right). (24)

The ϵ\epsilon-entropy number H(ϵ,𝒯)H(\epsilon,\mathscr{T}) is defined as H(ϵ,𝒯):=logN(ϵ,𝒯)H(\epsilon,\mathscr{T}):=\log N(\epsilon,\mathscr{T}). The following results are stated in terms of NN, but because of (24) similar results also hold with NN replaced by NN^{\prime}.

Result 3.6 ([10], Theorem 3.18).

Let (Xt)t𝒯(X_{t})_{t\in\mathscr{T}} be a centered Gaussian process and assume that (𝒯,d)(\mathscr{T},d) is totally bounded. If H(,𝒯)\sqrt{H(\cdot,\mathscr{T})} is integrable at 0, then XtX_{t} admits a version which is almost surely uniformly continuous on (𝒯,d)(\mathscr{T},d) and for such a version we have

𝔼[supt𝒯Xt]120σH(x,𝒯)𝑑x,\mathbb{E}[\sup_{t\in\mathscr{T}}X_{t}]\leq 12\int_{0}^{\sigma}\sqrt{H(x,\mathscr{T})}~{}dx,

where σ=(supt𝒯𝔼[Xt2])1/2\sigma=(\sup_{t\in\mathscr{T}}\mathbb{E}[X_{t}^{2}])^{1/2}.

Fact 3.7.

For (Xt)t𝒯(X_{t})_{t\in\mathscr{T}} a centered Gaussian process on an index set 𝒯\mathscr{T}, we have

𝔼[supt𝒯|Xt|]𝔼[|X0|]+2𝔼[supt𝒯Xt].\mathbb{E}[\sup_{t\in\mathscr{T}}|X_{t}|]\leq\mathbb{E}[|X_{0}|]+2\mathbb{E}[\sup_{t\in\mathscr{T}}X_{t}]. (25)

The following result gives tail bounds for the supremum of a Gaussian process.

Result 3.8 ([10], Proposition 3.19).

Let (Xt)t𝒯(X_{t})_{t\in\mathscr{T}} be an almost surely continuous centered Gaussian process on the totally bounded set (𝒯,d)(\mathscr{T},d). Let ZZ denote either supt𝒯Xt\sup_{t\in\mathscr{T}}X_{t} or supt𝒯|Xt|\sup_{t\in\mathscr{T}}|X_{t}| and let σ=(supt𝒯𝔼[Xt2])1/2\sigma=(\sup_{t\in\mathscr{T}}\mathbb{E}[X_{t}^{2}])^{1/2}. Then for every x>0x>0, we have

(Z𝔼[Z]σ2x)exp(x),\displaystyle\mathbb{P}(Z-\mathbb{E}[Z]\geq\sigma\sqrt{2x})\leq\exp(-x),
(𝔼[Z]Zσ2x)exp(x).\displaystyle\mathbb{P}(\mathbb{E}[Z]-Z\geq\sigma\sqrt{2x})\leq\exp(-x).

3.3. Derivatives of a stationary Gaussian process

We now present some basic facts about stationary Gaussian processes on {\mathbb{R}}/2{\mathbb{R}}^{2}. Let XX be a centered stationary Gaussian process on {\mathbb{R}}/2{\mathbb{R}}^{2} with spectral measure μ\mu and covariance function 𝔼[XtX0]=k(t)=μ^(t)\mathbb{E}[X_{t}X_{0}]=k(t)=\hat{\mu}(t). Further assume that μ\mu is a probability measure all of whose moments CnC_{n}/Cm,nC_{m,n} are finite. Then we have the following.

  1. 1.

    For XX on {\mathbb{R}} and tt\in{\mathbb{R}}, we have

    • XX is smooth and for every nn\in\mathbb{N}, X(n)X^{(n)} is a centered stationary Gaussian process with covariance function given by

      𝔼[Xt(n)X0(n)]=(1)nk(2n)(t).\displaystyle\mathbb{E}[X^{(n)}_{t}X^{(n)}_{0}]=(-1)^{n}k^{(2n)}(t). (26)
    • The derivatives of kk are given by

      k(n)(t)\displaystyle k^{(n)}(t) =(ix)neitx𝑑μ(x),\displaystyle=\int_{{\mathbb{R}}}(ix)^{n}e^{-itx}d\mu(x),
      k(n)L()\displaystyle\|k^{(n)}\|_{L^{\infty}({\mathbb{R}})} |x|n𝑑μ(x)=Cn.\displaystyle\leq\int_{{\mathbb{R}}}|x|^{n}d\mu(x)=C_{n}. (27)
  2. 2.

    For XX on 2{\mathbb{R}}^{2} and z=(x,y)2z=(x,y)\in{\mathbb{R}}^{2} and s2s\in{\mathbb{R}}^{2}, we have

    • XX is smooth and for every m,n{0}m,n\in\mathbb{N}\cup\{0\}, 1m2nX\partial_{1}^{m}\partial_{2}^{n}X is a centered stationary Gaussian process with covariance function given by

      𝔼[1m2nX(z)1m2nX(0)]=(1)m+n12m12nk(z).\displaystyle\mathbb{E}[\partial_{1}^{m}\partial_{2}^{n}X(z)\cdot\partial_{1}^{m}\partial_{2}^{n}X(0)]=(-1)^{m+n}\partial_{1}^{2m}\partial_{1}^{2n}k(z).
    • The derivatives of kk are given by

      1m2nk(s)\displaystyle\partial_{1}^{m}\partial_{2}^{n}k(s) =2(ix)m(iy)neis,z𝑑μ(z),\displaystyle=\int_{{\mathbb{R}}^{2}}(ix)^{m}(iy)^{n}e^{i\langle s,z\rangle}d\mu(z),
      1m2nkL(2)\displaystyle\|\partial_{1}^{m}\partial_{2}^{n}k\|_{L^{\infty}({\mathbb{R}}^{2})} 2|x|m|y|n𝑑μ(z)=Cm,n.\displaystyle\leq\int_{{\mathbb{R}}^{2}}|x|^{m}|y|^{n}~{}d\mu(z)=C_{m,n}.

Bulinskaya’s lemma ([11], Lemma 6) for a stationary Gaussian process reads as follows.

Result 3.9.

Let XX be a stationary Gaussian process on 2{\mathbb{R}}^{2} with spectral measure μ\mu which is a symmetric probability measure on 2{\mathbb{R}}^{2}. Then almost surely XX does not have any singular zero (a singular zero is a point z2z\in{\mathbb{R}}^{2} such that X(z)=|X(z)|=0X(z)=|\nabla X(z)|=0) if the following conditions holds.

  • max{C4,0,C0,4}<\max\{C_{4,0},C_{0,4}\}<\infty.

  • The Gaussian vector X(0)=(1X(0),2X(0))\nabla X(0)=(\partial_{1}X(0),\partial_{2}X(0)) is non-degenerate, which holds iff there is no v2v\in{\mathbb{R}}^{2} for which supp(μ)v(\mu)\subseteq{\mathbb{R}}v.

Remark 3.10.

We now see how the moments LnL_{n} and RnR_{n} defined in (2) are related. We first note that

Cm,n2(x2+y2)m+n2𝑑μ(x,y)=0tm+n𝑑μr(t),\displaystyle C_{m,n}\leq\int_{{\mathbb{R}}^{2}}(x^{2}+y^{2})^{\frac{m+n}{2}}d\mu(x,y)=\int_{0}^{\infty}t^{m+n}d\mu_{r}(t),

and hence R~nL~n\widetilde{R}_{n}\leq\widetilde{L}_{n}. We also have

L~n2\displaystyle\widetilde{L}_{n}^{2} =2(x2+y2)n𝑑μ(x,y)22n(x2n+y2n2)𝑑μ(x,y)2nR~n2\displaystyle=\int_{{\mathbb{R}}^{2}}(x^{2}+y^{2})^{n}d\mu(x,y)\leq\int_{{\mathbb{R}}^{2}}2^{n}\left(\frac{x^{2n}+y^{2n}}{2}\right)d\mu(x,y)\leq 2^{n}\widetilde{R}_{n}^{2}

and hence we conclude that

Rn1/nLn1/n2Rn1/n.\displaystyle R_{n}^{1/n}\leq L_{n}^{1/n}\leq\sqrt{2}R_{n}^{1/n}. (28)

4. Proof of Theorem 1.1

Upper bound

As discussed in Section 2.1, we use Lemma 2.2 to prove the upper bound in Theorem 1.1 and hence we only need to estimate the two terms on the r.h.s. of (6). Let bb, CC be as in Lemma 3.4 and define B:=(4eAC)1B:=(4eAC)^{-1}, where A=192πA=192\sqrt{\pi}. Let ϵ(0,1/2)\epsilon\in(0,1/2), n,mn,m\in\mathbb{N} and T>0T>0 be such that n1/ϵ2,m=ϵn and Tbϵnn\geq 1/\epsilon^{2},~{}m=\lfloor\epsilon n\rfloor\text{ and }{T\leq b\lfloor\epsilon n\rfloor}. We also assume that hn1h_{n}\geq 1, where hnh_{n} is defined by

hn:=log(BDn1/n(nT)12ϵ).\displaystyle h_{n}:=\log\left(\frac{B}{D_{n}^{1/n}}\cdot\left(\frac{n}{T}\right)^{1-2\epsilon}\right).

With DnD_{n} as in (1), we choose M=2AHnDnM=2AH_{n}D_{n}, where Hn:=n2hnH_{n}:=\sqrt{n^{2}h_{n}}.

For this choice of MM, we get the following estimate for the small ball probability in (6) from Lemma 3.5

(XL[T,2T]Mn!(2T)n)exp{ϵn22log(BDn1/n(nT)12ϵ)}.\displaystyle\mathbb{P}\left(\lVert X\rVert_{L^{\infty}[T,2T]}\leq\frac{M}{n!}(2T)^{n}\right)\leq\exp\left\{\frac{-\epsilon n^{2}}{2}\log\left(\frac{B}{{D}_{n}^{1/n}}\cdot\left(\frac{n}{T}\right)^{1-2\epsilon}\right)\right\}. (29)

We now estimate (X(n)L[0,2T]>M)\mathbb{P}(\lVert X^{(n)}\rVert_{L^{\infty}[0,2T]}>M), which is the other term in (6), using Results 3.6 and 3.8. Let k:=μ^k:=\widehat{\mu} be the covariance function of XX. The following calculation is to estimate the entropy numbers for the process (Xt(n))t[0,2T](X^{(n)}_{t})_{t\in[0,2T]}. For this we first get an estimate of the ϵ\epsilon-covering number. We denote by dnd_{n} the pseudo-metric on {\mathbb{R}} induced by X(n)X^{(n)}.

dn(t,0)2\displaystyle d_{n}(t,0)^{2} =𝔼[(Xt(n)X0(n))2]=2(1)n[k(2n)(0)k(2n)(t)],\displaystyle=\mathbb{E}[(X^{(n)}_{t}-X^{(n)}_{0})^{2}]=2(-1)^{n}[k^{(2n)}(0)-k^{(2n)}(t)],
=2(1)n+10t[k(2n+1)(s)k(2n+1)(0)]𝑑s,\displaystyle=2(-1)^{n+1}\int_{0}^{t}[k^{(2n+1)}(s)-k^{(2n+1)}(0)]ds,
20tk(2n+2)L()s𝑑sC2n+2t2,\displaystyle\leq 2\int_{0}^{t}\|k^{(2n+2)}\|_{L^{\infty}({\mathbb{R}})}~{}sds\leq C_{2n+2}~{}t^{2}, (30)

where the last inequality follows from mean value theorem and (27). Let n(x,r)\mathcal{B}_{n}(x,r) denote the ball in (,dn)({\mathbb{R}},d_{n}) with center xx and radius rr. Then it follows from (30) that

(xϵ,x+ϵ)n(x,C2n+2ϵ),\displaystyle(x-\epsilon,x+\epsilon)\subseteq\mathcal{B}_{n}\left(x,\sqrt{C_{2n+2}}~{}\epsilon\right),

and hence we get the following bound for the ϵ\epsilon-covering number N(ϵ)N^{\prime}(\epsilon) of ([0,2T],dn)([0,2T],d_{n})

N(ϵ)\displaystyle N^{\prime}(\epsilon) C2n+2ϵ2T.\displaystyle\leq\frac{\sqrt{C_{2n+2}}}{\epsilon}~{}2T. (31)

We now get a bound on the ϵ\epsilon-packing number N(ϵ)N(\epsilon) using (24) and (31)

N(ϵ)\displaystyle N(\epsilon) N(ϵ2)4C2n+2ϵT=βϵ,where β=4C2n+2T.\displaystyle\leq N^{\prime}\left(\frac{\epsilon}{2}\right)\leq\frac{4\sqrt{C_{2n+2}}}{\epsilon}~{}T=\frac{\beta}{\epsilon},~{}\mbox{where $\beta=4\sqrt{C_{2n+2}}~{}T$}.

The above inequality holds for ϵ<β\epsilon<\beta and for ϵβ\epsilon\geq\beta, we have N(ϵ)=1N(\epsilon)=1. Hence we have

H(ϵ){=0,if ϵβ,log(β/ϵ),if ϵ<β.H(\epsilon)\begin{cases}=0,&\text{if }\epsilon\geq\beta,\\ \leq\log\left(\beta/\epsilon\right),&\text{if }\epsilon<\beta.\end{cases}
0H(x)𝑑x=0βH(x)𝑑x\displaystyle\int_{0}^{\infty}\sqrt{H(x)}dx=\int_{0}^{\beta}\sqrt{H(x)}dx 0βlog(βx)𝑑x=2β0y2ey2𝑑y=πβ.\displaystyle\leq\int_{0}^{\beta}\sqrt{\log\left(\frac{\beta}{x}\right)}dx=2\beta\int_{0}^{\infty}y^{2}e^{-y^{2}}dy=\sqrt{\pi}\beta. (32)

We now conclude from Result 3.6 that

𝔼[supt[0,2T]Xt(n)]12πβ=48πC2n+2T.\mathbb{E}[\sup_{t\in[0,2T]}X^{(n)}_{t}]\leq 12\sqrt{\pi}\beta=48\sqrt{\pi}\sqrt{C_{2n+2}}~{}T. (33)

It now follows from (25) and (33) that with A=192πA=192\sqrt{\pi}, we have

𝔼[supt[0,2T]|Xt(n)|]\displaystyle\mathbb{E}[\sup_{t\in[0,2T]}|X^{(n)}_{t}|] 𝔼[|X0(n)|]+96πC2n+2T,\displaystyle\leq\mathbb{E}[|X^{(n)}_{0}|]+96\sqrt{\pi}\sqrt{C_{2n+2}}~{}T, (34)
(26)|k(2n)(0)|+96πC2n+2T,\displaystyle\leq_{\eqref{covkerder}}\sqrt{|k^{(2n)}(0)|}+96\sqrt{\pi}\sqrt{C_{2n+2}}~{}T,
96π(C2n+C2n+2T),\displaystyle\leq 96\sqrt{\pi}~{}(\sqrt{C_{2n}}+\sqrt{C_{2n+2}}~{}T),
Amax{C2n,C2n+2T}AnDn.\displaystyle\leq A\max\{\sqrt{C_{2n}},\sqrt{C_{2n+2}}~{}T\}\leq AnD_{n}.

It follows from Result 3.8 that for every nn\in\mathbb{N} and every x>0x>0, we have

(supt[0,2T]|Xt(n)|>AnDn+C2nx)exp(x2/2),\mathbb{P}(\sup_{t\in[0,2T]}|X^{(n)}_{t}|>AnD_{n}+\sqrt{C_{2n}}~{}x)\leq\exp(-x^{2}/2),

and since M=2AHnDnAnDn+DnHnAnDn+C2nHnM=2AH_{n}D_{n}\geq AnD_{n}+D_{n}H_{n}\geq AnD_{n}+\sqrt{C_{2n}}H_{n}, we conclude that

(supt[0,2T]|Xt(n)|M)exp(Hn2/2)=exp(n2hn/2).\displaystyle\mathbb{P}(\sup_{t\in[0,2T]}|X^{(n)}_{t}|\geq M)\leq\exp(-H_{n}^{2}/2)=\exp(-n^{2}h_{n}/2). (35)

Hence we conclude from (6), (29) and (35) that

(𝒩Tn)2exp{ϵn22log(BDn1/n(nT)12ϵ)}.\displaystyle\mathbb{P}(\mathscr{N}_{T}\geq n)\leq 2\exp\left\{-\frac{\epsilon n^{2}}{2}\log\left(\frac{B}{D_{n}^{1/n}}\cdot\left(\frac{n}{T}\right)^{1-2\epsilon}\right)\right\}.

Lower bound

Let b>0b>0 be as in Result 3.3. For T>0T>0 and nn\in\mathbb{N} such that TbnT\leq bn, we define the (n+1)(n+1) dimensional Gaussian vector VnV_{n} by

Vn:=(Y0,Y1,Y2,,Yn),\displaystyle V_{n}:=(Y_{0},Y_{1},Y_{2},\ldots,Y_{n}),

where Yk:=XkT/nY_{k}:=X_{kT/n}. Let Σ\Sigma denote the covariance matrix of VnV_{n}. The density ϕn\phi_{n} of VnV_{n} is

ϕn(x)=1(2π)n+1|Σ|1/2exp(Σ1x,x2).\displaystyle\phi_{n}(x)=\frac{1}{(\sqrt{2\pi})^{n+1}~{}|\Sigma|^{1/2}}\exp\left(-\frac{\langle\Sigma^{-1}x,x\rangle}{2}\right).

Let λ\lambda be the smallest eigenvalue of Σ\Sigma, then we get the following lower bound for λ\lambda from Result 3.3. There is c>0c>0 such that λ(cT/n)2n\lambda\geq(cT/n)^{2n}, hence we have

Σ1x,x1λx2(ncT)2nx2.\displaystyle\langle\Sigma^{-1}x,x\rangle\leq\frac{1}{\lambda}\|x\|^{2}\leq\left(\frac{n}{cT}\right)^{2n}\|x\|^{2}.

We have from Lemma 3 in [9] that |Σ|1|\Sigma|\leq 1 and hence we have the following lower bound for the density ϕn\phi_{n}

ϕn(x)1(2π)n+1exp((ncT)2nx22).\displaystyle\phi_{n}(x)\geq\frac{1}{(\sqrt{2\pi})^{n+1}}\exp\left(-\left(\frac{n}{cT}\right)^{2n}\frac{\|x\|^{2}}{2}\right).

It follows from (8) that

(𝒩Tn)\displaystyle\mathbb{P}(\mathscr{N}_{T}\geq n) 1(2π)n+100exp((ncT)2nx22)𝑑x,\displaystyle\geq\frac{1}{(\sqrt{2\pi})^{n+1}}\int\cdots\int_{-\infty}^{0}\int_{0}^{\infty}\exp\left(-\left(\frac{n}{cT}\right)^{2n}\frac{\|x\|^{2}}{2}\right)dx,
1(2π)n+1(2π2(cTn)n)n+1(Tcn)n2,\displaystyle\geq\frac{1}{(\sqrt{2\pi})^{n+1}}\cdot\left(\frac{\sqrt{2\pi}}{2}\left(\frac{cT}{n}\right)^{n}\right)^{n+1}\geq\left(\frac{T}{c^{\prime}n}\right)^{n^{2}},
exp(n2log(cn/T)).\displaystyle\geq\exp\left(-n^{2}\log\left(c^{\prime}n/T\right)\right).

5. Proof of Theorem 1.2

We continue to work with the assumptions and the notations of Section 2.2. The starting point of our analysis here is (13), which is

(T4nT)j=1,2[(j)+(j)]+δ1T[(𝒜0)+(0)],\displaystyle\mathbb{P}(\mathscr{L}_{T}\geq 4nT)\leq\sum_{j=1,2}[\mathbb{P}(\mathscr{E}_{j})+\mathbb{P}(\mathscr{F}_{j})]+\lceil\delta^{-1}T\rceil~{}[\mathbb{P}(\mathscr{A}_{0})+\mathbb{P}(\mathscr{B}_{0})],

By our assumption, both X1X_{1}, X2X_{2} satisfy the assumptions of Lemma 3.4 and let b(0,1)b\in(0,1), C>1C>1 be such that the conclusion of Lemma 3.4 holds for both X1X_{1}, X2X_{2} with these constants. Define B:=(4eAC)1{B}:=(4eAC)^{-1}, where A=384πA=384\sqrt{\pi}. Let ϵ(0,1/4)\epsilon\in(0,1/4), n,mn,m\in\mathbb{N} and T>0T>0 be such that n1/ϵ2,m=ϵn, Tbϵnn\geq 1/\epsilon^{2},~{}m=\lfloor\epsilon n\rfloor,\text{ }{T\leq b\lfloor\epsilon n\rfloor} and δ:=(2T)n/n!\delta:=(2T)^{n}/n!. We further assume that (B/Ln1/n)(n/T)14ϵ>e(B/L_{n}^{1/n})\cdot(n/T)^{1-4\epsilon}>e and since we have from (28) that LnRnL_{n}\geq R_{n}, it follows that gn1g_{n}\geq 1, where gng_{n} is defined by

gn:=log(BRn1/n(nT)12ϵ).\displaystyle g_{n}:=\log\left(\frac{{B}}{R_{n}^{1/n}}\cdot\left(\frac{n}{T}\right)^{1-2\epsilon}\right).

We choose M=2ARnGnM=2AR_{n}G_{n}, where Gn:=n2gnG_{n}:=\sqrt{n^{2}g_{n}} and estimate the terms on the r.h.s. of (13). We first estimate the small ball probabilities (𝒜0),(0)\mathbb{P}(\mathscr{A}_{0}),~{}\mathbb{P}(\mathscr{B}_{0}) using Lemma 3.5 with 𝒟n=Rn\mathscr{D}_{n}=R_{n} and n=Gn\mathscr{H}_{n}=G_{n}

(𝒜0),(0)exp{ϵn22log(BRn1/n(nT)12ϵ)},\displaystyle\mathbb{P}(\mathscr{A}_{0}),~{}\mathbb{P}(\mathscr{B}_{0})\leq\exp\left\{\frac{-\epsilon n^{2}}{2}\log\left(\frac{B}{R_{n}^{1/n}}\cdot\left(\frac{n}{T}\right)^{1-2\epsilon}\right)\right\},

and since δ1T=(n!T)/(2T)n(n/T)n(T/2n)(n/T)n\delta^{-1}T=(n!T)/(2T)^{n}\leq(n/T)^{n}\cdot(T/2^{n})\ll(n/T)^{n}, we have

δ1T[(𝒜0)+(0)]\displaystyle\lceil\delta^{-1}T\rceil~{}[\mathbb{P}(\mathscr{A}_{0})+\mathbb{P}(\mathscr{B}_{0})] 2exp{ϵn22log(BRn1/n(nT)12ϵ2ϵn)},\displaystyle\leq 2\exp\left\{\frac{-\epsilon n^{2}}{2}\log\left(\frac{B}{R_{n}^{1/n}}\cdot\left(\frac{n}{T}\right)^{1-2\epsilon-\frac{2}{\epsilon n}}\right)\right\},
2exp{ϵn22log(BRn1/n(nT)14ϵ)},\displaystyle\leq 2\exp\left\{\frac{-\epsilon n^{2}}{2}\log\left(\frac{B}{R_{n}^{1/n}}\cdot\left(\frac{n}{T}\right)^{1-4\epsilon}\right)\right\}, (36)

where the last inequality follows from our assumption that n1/ϵ2n\geq 1/\epsilon^{2}.

Like in the one dimensional case, Results 3.6 and 3.8 will be used to estimate the probability of the events j\mathscr{E}_{j} and j\mathscr{F}_{j}. We now calculate the expected supremum of the process 1nX\partial_{1}^{n}X on [0,n]2[0,n]^{2}. Let dnd_{n} be the pseudo-metric on 2{\mathbb{R}}^{2} induced by the Gaussian process 1nX\partial_{1}^{n}X, then for z2z\in{\mathbb{R}}^{2} we have

dn(z,0)2\displaystyle d_{n}(z,0)^{2} =𝔼[1nX(z)1nX(0)]2=2(1)n[12nk(0)12nk(z)].\displaystyle=\mathbb{E}\left[\partial_{1}^{n}X(z)-\partial_{1}^{n}X(0)\right]^{2}=2(-1)^{n}[\partial_{1}^{2n}k(0)-\partial_{1}^{2n}k(z)].

For s=(s1,s2)2s=(s_{1},s_{2})\in{\mathbb{R}}^{2}, we have

|12nk(0)12nk(s)|\displaystyle|\partial_{1}^{2n}k(0)-\partial_{1}^{2n}k(s)| =|0112nk(ts)sdt|,\displaystyle=|\int_{0}^{1}\nabla\partial_{1}^{2n}k(ts)\cdot s~{}dt|,
=|s101[12n+1k(ts)12n+1k(0)]dt\displaystyle=|s_{1}\int_{0}^{1}[\partial_{1}^{2n+1}k(ts)-\partial_{1}^{2n+1}k(0)]~{}dt
+s201[12n2k(ts)12n2k(0)]dt|,\displaystyle\hskip 34.14322pt+s_{2}\int_{0}^{1}[\partial_{1}^{2n}\partial_{2}k(ts)-\partial_{1}^{2n}\partial_{2}k(0)]dt|,
|s1|01(|s1|C2n+2,0+|s2|C2n+1,1)t𝑑t\displaystyle\leq|s_{1}|\int_{0}^{1}(|s_{1}|C_{2n+2,0}+|s_{2}|C_{2n+1,1})~{}tdt
+|s2|01(|s1|C2n+1,1+|s2|C2n,2)t𝑑t,\displaystyle\hskip 34.14322pt+|s_{2}|\int_{0}^{1}(|s_{1}|C_{2n+1,1}+|s_{2}|C_{2n,2})~{}tdt,
max{C2n+2,0,C2n,2,C2n+1,1}s2.\displaystyle\leq\max\{C_{2n+2,0},C_{2n,2},C_{2n+1,1}\}\|s\|^{2}.

Hence we have dn(s,0)R~n+1sRnsd_{n}(s,0)\leq\widetilde{R}_{n+1}\|s\|\leq{R}_{n}\|s\|, where R~n\widetilde{R}_{n} and RnR_{n} are defined in (2). We let D(x,)D(x,\ell) and n(x,)\mathcal{B}_{n}(x,\ell) denote the balls centered at xx with radius \ell in the Euclidean metric and dnd_{n} respectively, then D(x,ϵ/Rn)n(x,ϵ)D(x,\epsilon/R_{n})\subseteq\mathcal{B}_{n}(x,\epsilon) and hence N(ϵ)N^{\prime}(\epsilon), the ϵ\epsilon-covering number of ([0,n]2,dn)([0,n]^{2},d_{n}), has the following upper bound N(ϵ)4Rn2n2/ϵ2N^{\prime}(\epsilon)\leq{4R_{n}^{2}}n^{2}/\epsilon^{2} and hence the ϵ\epsilon-packing number N(ϵ)N(\epsilon) satisfies N(ϵ)8Rn2n2/ϵ2N(\epsilon)\leq{8R_{n}^{2}}n^{2}/\epsilon^{2}. Similar to (32), we can get the following estimate for the Dudley integral in this case also

0logN(x)𝑑x4πnRn.\displaystyle\int_{0}^{\infty}\sqrt{\log N(x)}~{}dx\leq 4\sqrt{\pi}nR_{n}.

We thus conclude from Result 3.6 that

𝔼[sup[0,n]21nX]48πnRn,\mathbb{E}[\sup_{[0,n]^{2}}\partial_{1}^{n}X]\leq 48\sqrt{\pi}nR_{n},

and by a calculation similar to (34), we conclude that with A=384πA=384\sqrt{\pi} we have

𝔼[1nXL[0,n]2]AnRn/2,\displaystyle\mathbb{E}[~{}\|\partial_{1}^{n}X\|_{L^{\infty}[0,n]^{2}}]\leq AnR_{n}/2,

thus we get the following tail bound from Result 3.8 and the fact that RnC2n,0R_{n}\geq\sqrt{C_{2n,0}}

(1nXL[0,n]2>(AnRn/2)+xRn)exp(x2/2).\displaystyle\mathbb{P}(\|\partial_{1}^{n}X\|_{L^{\infty}[0,n]^{2}}>(AnR_{n}/2)+xR_{n})\leq\exp(-x^{2}/2).

For j=1,2j=1,2 and Y=jXL[0,n]2Y=\|\partial_{j}X\|_{L^{\infty}[0,n]^{2}}, jnXL[0,n]2\|\partial_{j}^{n}X\|_{L^{\infty}[0,n]^{2}}, similar calculations as above yield the following tail bounds. For every x>0x>0, we have

(Y>(AnRn/2)+xRn)exp(x2/2).\displaystyle\mathbb{P}(Y>(AnR_{n}/2)+xR_{n})\leq\exp(-x^{2}/2). (37)

Since M/2=AGnRn(AnRn/2)+GnRnM/2=AG_{n}R_{n}\geq(AnR_{n}/2)+G_{n}R_{n}, it follows from (37) that

(Y>M/2)exp(Gn2/2)exp(n2gn/2).\displaystyle\mathbb{P}(Y>M/2)\leq\exp(-G_{n}^{2}/2)\leq\exp(-n^{2}g_{n}/2). (38)

It now follows from (13), (5) and (38) that

(T4nT)6exp{ϵn22log(BRn1/n(nT)14ϵ)}.\displaystyle\mathbb{P}(\mathscr{L}_{T}\geq 4nT)\leq 6\exp\left\{\frac{-\epsilon n^{2}}{2}\log\left(\frac{B}{R_{n}^{1/n}}\cdot\left(\frac{n}{T}\right)^{1-4\epsilon}\right)\right\}. (39)

We have from (28) that Ln1/nRn1/nL_{n}^{1/n}\geq R_{n}^{1/n} and now the desired result follows from (39).

Now suppose that there exists v2v\in{\mathbb{R}}^{2} such that supp(μ)v(\mu)\subseteq{\mathbb{R}}v, without loss of generality we may assume that v=(1,0)v=(1,0). In this case μ\mu considered as a measure on {\mathbb{R}} satisfies (A1A1) and almost surely for every (x,y)2(x,y)\in{\mathbb{R}}^{2}, we have X(x,y)=X(x,0)X(x,y)=X(x,0) and hence the zero set of XX is t:X(t,0)=0{(t,y):y}\cup_{t:X(t,0)=0}\{(t,y):y\in{\mathbb{R}}\}. If we let 𝒩T\mathscr{N}_{T} denote the zero count of X(,0)X(\cdot,0) in [0,T][0,T], we have TnT\mathscr{L}_{T}\geq nT iff 𝒩Tn\mathscr{N}_{T}\geq n. The result now follows by using the estimates for (𝒩Tn)\mathbb{P}(\mathscr{N}_{T}\geq n) obtained in Theorem 1.1.

6. Consequences of Theorems 1.1 &\& 1.2

6.1. One dimension

In this section, we deduce from Theorem 1.1 the tail estimates for the zero count given in Table 1.

  1. 1.

    Let μ\mu be compactly supported, say supp(μ)[q,q](\mu)\subset[-q,q] for some q>1q>1. Then Dnq2nD_{n}\leq q^{2n} and hence Dn1/nq2D_{n}^{1/n}\leq q^{2}. Taking ϵ=1/4\epsilon=1/4 in (3), we conclude that there exists c,C>0c,C>0 such that whenever T1T\geq 1 and nCTn\geq CT we have

    (𝒩Tn)\displaystyle\mathbb{P}(\mathscr{N}_{T}\geq n) exp(cn2log(n/T)).\displaystyle\lesssim\exp\left(-cn^{2}\log\left(n/T\right)\right). (40)

    Moment bounds. We use (40) to get bounds for the moments of 𝒩T\mathscr{N}_{T}, for mm\in\mathbb{N} we have

    𝔼[𝒩Tm]\displaystyle\mathbb{E}[\mathscr{N}_{T}^{m}] (CT)m+nCTnmecn2,\displaystyle\lesssim(CT)^{m}+\sum_{n\geq CT}n^{m}e^{-cn^{2}}, (41)
    (CT)m+(supx0xmecx2/2)n1ecn2/2,\displaystyle\leq(CT)^{m}+(\sup_{x\geq 0}x^{m}e^{-cx^{2}/2})\sum_{n\geq 1}e^{-cn^{2}/2},
    (CT)m+(m/c)m/2(c~(Tm))m.\displaystyle\lesssim(CT)^{m}+(m/c)^{m/2}\leq(\tilde{c}(T\vee\sqrt{m}))^{m}.
  2. 2.

    Let μ\mu be such that CnnαnC_{n}\lesssim n^{\alpha n} for some α>0\alpha>0, then Dn1/nnαD_{n}^{1/n}\lesssim n^{\alpha}. We consider the cases α<1\alpha<1 and α1\alpha\geq 1 separately and make the following conclusions from (3).

    • Let α<1\alpha<1, then for every κ(0,1α)\kappa\in(0,1-\alpha) there is cκ>0c_{\kappa}>0 such that whenever TnκT\leq n^{\kappa}, we have

      (𝒩Tn)exp(cκn2logn).\displaystyle\mathbb{P}(\mathscr{N}_{T}\geq n)\lesssim\exp({-c_{\kappa}n^{2}\log n}). (42)

      Justification for (42). Let κ,κ′′\kappa^{\prime},\kappa^{\prime\prime} be such κ<κ<κ′′<(1α)\kappa<\kappa^{\prime}<\kappa^{\prime\prime}<(1-\alpha), then we can choose ϵ(0,(1α)/2)\epsilon\in(0,(1-\alpha)/2) such that (1α2ϵ)/(12ϵ)=κ′′(1-\alpha-2\epsilon)/(1-2\epsilon)=\kappa^{\prime\prime} and hence for this choice of ϵ\epsilon and large enough nn, we have

      (Bnα(nT)12ϵ)112ϵ=B112ϵn1α2ϵ12ϵTnκTnκnκ=nκκ,\displaystyle\left(\frac{B}{n^{\alpha}}\left(\frac{n}{T}\right)^{1-2\epsilon}\right)^{\frac{1}{1-2\epsilon}}=B^{\frac{1}{1-2\epsilon}}\frac{n^{\frac{1-\alpha-2\epsilon}{1-2\epsilon}}}{T}\geq\frac{n^{\kappa^{\prime}}}{T}\geq\frac{n^{\kappa^{\prime}}}{n^{\kappa}}=n^{\kappa^{\prime}-\kappa},

      and hence

      Bnα(nT)12ϵn(κκ)(12ϵ)e.\displaystyle\frac{B}{n^{\alpha}}\left(\frac{n}{T}\right)^{1-2\epsilon}\geq n^{(\kappa^{\prime}-\kappa)(1-2\epsilon)}\geq e.

      And if we choose nn large enough, the fact that TnκT\leq n^{\kappa} will imply that TbϵnT\leq b\lfloor\epsilon n\rfloor and hence all the conditions of Theorem 1.1 are satisfied. Using ϵ\epsilon chosen above in (3) and letting cκ=ϵ(12ϵ)(κκ)/2c_{\kappa}=\epsilon(1-2\epsilon)(\kappa^{\prime}-\kappa)/2, we get (42). Similar arguments can also be used to justify the tail bounds we get for 𝒩T\mathscr{N}_{T} in the other cases, namely (43) and (46). Moment bounds. By a calculation similar to (41), we conclude using (42) that

      𝔼[𝒩Tm](cκ(T1/κm))m.\displaystyle\mathbb{E}[\mathscr{N}_{T}^{m}]\leq(c^{\prime}_{\kappa}(T^{1/\kappa}\vee\sqrt{m}))^{m}.
    • Let α1\alpha\geq 1, then for every κ>0\kappa>0 there exists cκ>0c_{\kappa}>0 such that whenever nn is large enough and T1/nα1+κT\leq 1/n^{\alpha-1+\kappa}, we have

      (𝒩Tn)\displaystyle\mathbb{P}(\mathscr{N}_{T}\geq n) exp(cκn2log(eTnα1+κ)).\displaystyle\lesssim\exp\left(-c_{\kappa}n^{2}\log\left(\frac{e}{Tn^{\alpha-1+\kappa}}\right)\right). (43)

      We write [0,1][0,1] as a union of nα1+κ\lceil n^{\alpha-1+\kappa}\rceil many subintervals {I}\{I_{\ell}\}, each of length n(α1+κ)n^{-(\alpha-1+\kappa)}. Then if 𝒩1nα+κ\mathscr{N}_{1}\geq n^{\alpha+\kappa}, at least one of the intervals II_{\ell} must contain more than nn zeros and hence a union bound gives

      (𝒩1nnα1+κ)(𝒩n(α1+κ)n)nα1+κexp(cκn2),\displaystyle\mathbb{P}(\mathscr{N}_{1}\geq n\cdot n^{\alpha-1+\kappa})\leq\sum_{\ell}\mathbb{P}(\mathscr{N}_{n^{-(\alpha-1+\kappa)}}\geq n)\lesssim n^{\alpha-1+\kappa}\exp(-c_{\kappa}n^{2}), (44)

      and hence there exists c~κ>0\tilde{c}_{\kappa}>0 such that

      (𝒩1n)\displaystyle\mathbb{P}(\mathscr{N}_{1}\geq n) exp(c~κn2α+κ).\displaystyle\lesssim\exp({-\tilde{c}_{\kappa}~{}n^{\frac{2}{\alpha+\kappa}}}). (45)

      Moment bounds. Using (45) we get the following moment bounds for 𝒩1\mathscr{N}_{1}. There exists cκ>0c_{\kappa}>0 such that for every mm\in\mathbb{N}, we have

      𝔼[𝒩1m](cκmα+κ2)m.\displaystyle\mathbb{E}[\mathscr{N}_{1}^{m}]\leq(c_{\kappa}m^{\frac{\alpha+\kappa}{2}})^{m}.
  3. 3.

    Let μ\mu be such that logCnn1+1γ\log C_{n}\lesssim n^{1+\frac{1}{\gamma}} for some γ>1/2\gamma>1/2. We conclude from (3) that there exist constants c,c>0c,c^{\prime}>0 such that whenever nn is large enough and Tecn1/γT\leq e^{-cn^{1/\gamma}}, we have

    (𝒩Tn)exp(cn2),\displaystyle\mathbb{P}(\mathscr{N}_{T}\geq n)\lesssim\exp(-c^{\prime}n^{2}), (46)

    and hence by partitioning [0,1][0,1] into subintervals of length ecn1/γe^{-cn^{1/\gamma}}, using the above estimate in each of them and a union bound as in (44) implies there is c~>0\tilde{c}>0 such that

    (𝒩1n)exp(c~(logn)2γ).\displaystyle\mathbb{P}(\mathscr{N}_{1}\geq n)\lesssim\exp(-\tilde{c}(\log n)^{2\gamma}). (47)

    Moment bounds. We use (47) to conclude that there is c>0c>0 such that for every mm\in\mathbb{N}, we have

    𝔼[𝒩1m]exp(cm1+12γ1).\displaystyle\mathbb{E}[\mathscr{N}_{1}^{m}]\leq\exp\left(cm^{1+\frac{1}{2\gamma-1}}\right).

6.2. Two dimensions

We now illustrate how the overcrowding estimates for nodal length given in Table 2 are deduced from Theorem 1.2. Since this analysis is similar to the one dimensional case considered above, we do this only for the first example in Table 2.

Let μ\mu be compactly supported, say supp(μ)q𝔻(\mu)\subset q\mathbb{D} for some q>1q>1. Then Ln1/nqL_{n}^{1/n}\lesssim q and hence by taking ϵ=1/8\epsilon=1/8 in (4) and by letting =4nT\ell=4nT, we conclude that there are constants c,C>0c,C>0 such that whenever T1T\geq 1 and CT2\ell\geq CT^{2} is large enough, we have

(T)\displaystyle\mathbb{P}(\mathscr{L}_{T}\geq\ell) exp(c2T2log(/T2)).\displaystyle\lesssim\exp\left(-\frac{c\ell^{2}}{T^{2}}\log\left(\ell/T^{2}\right)\right).

Moment bounds. We get moment bounds using the above tail estimates. For mm\in\mathbb{N},

𝔼[Tm]\displaystyle\mathbb{E}[\mathscr{L}_{T}^{m}] =0mxm1(Tx)𝑑x(CT2)m+mCT2xm1ecx2/T2𝑑x,\displaystyle=\int_{0}^{\infty}mx^{m-1}\mathbb{P}(\mathscr{L}_{T}\geq x)dx\lesssim(CT^{2})^{m}+m\int_{CT^{2}}^{\infty}x^{m-1}e^{-cx^{2}/T^{2}}dx,
(CT2)m+m(T/c)m0ym1ey2𝑑y(cT)m(Tm)m.\displaystyle\lesssim(CT^{2})^{m}+m(T/\sqrt{c})^{m}\int_{0}^{\infty}y^{m-1}e^{-y^{2}}dy\leq(c^{\prime}T)^{m}(T\vee\sqrt{m})^{m}.

Acknowledgements

This work was carried out during my Ph.D. under the guidance of Manjunath Krishnapur. I thank him for suggesting me the questions considered in this paper, his patience and encouragement throughout this study and for being very generous with his time and insights on the subject of this paper. I also thank Riddhipratim Basu for helpful discussions.

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