Conditions for the finiteness of the moments of the volume of level sets

Let X(t) be a Gaussian random field R d $\rightarrow$ R. Using the notion of (d -- 1)-integral geometric measures, we establish a relation between (a) the volume of the level set (b) the number of crossings of the restriction of the random field to a line. Using this relation we prove the equivalence between the finiteness of the expectation and the finiteness of the second spectral moment matrix. Sufficient conditions for finiteness of higher moments are also established.


Introduction
Let X(t) be a centered, stationary, Gaussian random field X : Ω × R d → R, with continuous sample paths. By a scaling argument, and without loss of generality, we may assume that X(t) is centered with variance 1. On the other hand, for a given u ∈ R, let us consider the level set restricted to some compact set K ⊂ R d C u,K := {t ∈ K : X(t) = u}. (1.1) already have a sufficient condition of finiteness (but as we will see, the latter is not necessary). For higher moments the KRF provides a multiple integral, the integrand of which is degenerate on the diagonal so the study of finiteness is not straightforward.
When d = 1, C u,K is a.s. a set of points and its measure is just the number of points. We have several result on finiteness of moments, see Sections 3 and 4. They all use at some stage the Intermediate Value Theorem. Unfortunately these methods are completely inoperative in higher dimensions. Here, we appeal to integral geometry in order to establish dimension-independent necessary and sufficient conditions of almost sure finiteness of level set volumes that boil down to one-dimensional results.
In Section 2 we recall the definition of the (d − 1)-dimensional integral-geometric measure, which is defined as the integral of the number of points over a family of lines.
Our three main results follow • In Section 3 we establish the equivalence between (a) the finiteness of the expectation of the (d − 1)-dimensional integral-geometric measure of the level set and (b) the finiteness of the second spectral moment matrix. This result gives a simpler presentation and shorter proof of the results of Wschebor [7] which uses De Gorgi perimeters.
• In Section 4 we give sufficient conditions for finiteness of the second moment (Theorem 4.1) using the Geman condition (see [5]).
• In the same section, we prove finiteness of all moments (Theorem 4.4), under some conditions, when the sample paths are smooth.

Integral geometric measure, Crofton formula
Let B be a Borel set in R d . Following Morgan [6] (and also Federer [4]) we define the (d − 1)-integral geometric measure of B by where S d−1 is the unit sphere in R d with its induced Riemannian measure, and v,y is the affine linear space {y + tv : t ∈ R}. The constant can be easily computed, using the Crofton formula below and considering the particular case of the sphere, yielding, The integrand in (2.1) is measurable (see for example Morgan[6, page 13]), and since it is non-negative, the integral is always well defined, finite or infinite.
In particular, if B is (d − 1)-rectifiable, then Crofton's formula [6] p. 31 yields 3 Characterisation for the finiteness of the expected volume of the level set.
The spectral measure F of X(·) is a symmetric measure with mass one: it is a probability measure.
Let Λ 2 be the second spectral moment matrix defined by This matrix may be finite or infinite, infinite meaning by convention that at least one entry is infinite.
When Λ 2 is finite, i.e. all its entries are finite, it is easy to prove that X(·) is differentiable in quadratic mean. If in addition the sample paths are almost surely differentiable (which is a little stronger) and if a.s. there exist no point t such that X(t) = u, ∇X(t) = 0, we have: • the level set C u,K is almost surely a submanifold of codimension 1, and its Riemannian volume can be defined and coincides with its (d − 1)-Hausdorff measure, namely, H d−1 (C u,K ); • the KRF (see Adler-Taylor [1] or Azaïs-Wschebor [2]) and (2.2) implies that The second equality in (3.1) is the true Kac-Rice formula, the third is due to classical integration.
When Λ 2 is infinite, i.e. at least one of its entries is infinite, we need to extend the definition of F(Λ 2 ) by setting it to +∞. So we consider the following relation: Note that its terms on both hand sides are now always well defined, finite or infinite.
The goal of this section is to prove that in a broad sense this formula is always true: • Whenever Λ 2 is finite, of course the RHS of (3.1) is finite, but so is the LHS as well and equality holds true.
• If Λ 2 is infinite then both sides of (3.1) are infinite.
Such a kind of property is known since the work of Cramér-Leadbetter [3] for d = 1 and from the work of Wschebor [7] for d > 1. Our proof uses Cramer-Leadbetter's result and generalised Crofton's formula. We first recall a result due to Cramér-Leadbetter, main result of Section 10.3 of [3]. The In case λ 2 is finite we have furthermore This result is based on polygonal approximation and intermediate values theorem, so it heavily relies on one-dimensional settings.
We now turn to our first main result.
Theorem 3.1. Let X(t) be a centered, stationary random field X : R d → R, with continuous sample paths. Then, for u ∈ R, K compact set of R d and C u,K defined by (1.1) we have equivalence between: ECP 24 (2019), paper 17.
In such a case we have Proof. Since X is almost surely continuous, then C u,K is a Borel set on R d a.s., and therefore its integral geometric measure is well defined. By Fubini theorem we get that As a matter of fact, because of stationarity of the process and by Cramér-Leadbetter applied to the process t → X(y + tv), we get and by Fubini, we obtain that Integrating in polar coordinates the expression F(Λ 2 ), given in (3.2), we obtain Furthermore, making the change of variable u = ρ 2 /2, it is straighforward to conclude and therefore from (3.4) yields We consider the two following cases.
• When Λ 2 is finite the integral on the RHS of (3.6) is finite and therefore we get the desired result in this case.
• When Λ 2 is infinite, we define the linear subspace We prove that G(Λ 2 ) is of dimension strictly smaller than d. Let v 1 , . . . , v d0 be a maximal set of linearly independent vectors of G(Λ 2 ). Then by standard linear algebra: this implies that G(Λ 2 ) = span(v 1 , . . . , v d0 ).
In conclusion the integrand in (3.6) is almost everywhere infinite so the integral is infinite and by consequence the expectation of the integral geometric measure is infinite.

Finitness of k-moments of the volume of the level set
Proof. Without loss of generality we assume that δ < 2. Let r be the covariance function of X, and let us first consider the field restricted to the line y,v : y + tv, t ∈ R:X y,v (t) = X(y + tv). Its covariance function is given by Note that because of stationarity it does not depend on y.
It is sufficient to prove the assertion of the theorem for a set K being a centred ball B a with sufficiently small diameter a. In that case note that the integral in the right-hand side of (2.1), for B = C u,K , is finite since the regularity of the sample path. Since the second spectral moment matrix is finite and non degenerate is bounded below and above. On the other hand, using a monotone convergence argument, as b tends to infinity So it is easy to conclude that for b sufficiently large, for any v ∈ S d−1 In the rest of the paper C will denote some unimportant constant, its value may change from an occurence to another.
Applying the Jensen inequality (with respect to the integral) yields As already remarked, the integral is over a bounded domain and it is sufficient to prove that the integrand is uniformly bounded.
Remark also that B a ∩ y,v is always a centred interval with length 2c less that 2a. Consequently, It remains to prove that is uniformly bounded.
In fact, because of the Rolle theorem, if U u is the number of up-crossings of the level u on the line 0,v , then #{C u,K ∩ 0,v } ≤ 2U u + 1.
So it is sufficient to bound the second moment of U u and even, because we have proved in the previous section that the first moment is uniformly bounded, it is sufficient to bound the second factorial moment. Since the variance has been assumed to be 1 and Λ 2 is non-degenerate, the Kac-Rice formula applies and yields By a standard regression formula, see for example [2] page 99, . Set θ v (τ ) := r v (τ ) − 1 + λ 2,v τ 2 /2, using the inequality z + t + ≤ (z + t) 2 /4 and the fact that This implies in turn that for τ < 1/b where b has been defined in (4.2) ECP 24 (2019), paper 17.
where F v is the spectral measure along the line 0,v (for convenience it is on (0, +∞)). The penultimate inequality uses (4.2), the last inequality is due to the fact that Λ 2 is non-degenerate.
On the other hand it is direct to prove that 1 − r v (τ ) ≤ λ 2,v τ 2 and, by compactness, the quantity λ 2,v is bounded as a function of v giving that 1 + r v (τ ) ≥ 1 as soon as the radius a of the ball is sufficiently small. This yields As a consequence The integrand in (4.3) can be bounded because We have Define R(u) := (u − sin(u)).
Its behaviour at zero and at infinity implies that for every δ, 0 < δ < 2, there exist a constant C δ such that 0 ≤ R(u) ≤ C δ u 1+δ .
This implies that resulting in the convergence of the integral in (4.3), uniformly in v.
Next, we consider a Gaussian field having C ∞ sample paths. This is for instance the case of Gaussian random trigonometric polynomials in several variables or the random plane wave model [8]. A result of Nualart-Wschebor, quoted as Theorem 3.6 in the book [2], can be used for obtaining that all the moments of the random variable I d−1 (C u,K ) are finite. The background result is the following: Proposition 4.3. Consider a real Gaussian process Y over R satisfying Var(Y (t)) > κ for all t ∈ I, a compact interval of R, and some κ > 0. Then for all u ∈ R, and m, p ∈ N such that p > 2m, it holds where N u is the number of points t ∈ I such that χ(t) = u, C p,m is a constant depending only on p, m and the length of the interval I, and C is a bound for the density of Y (t).
Let us assume now that the field X has C ∞ sample paths and we assume that the variance is bounded below. As in the proof on Theorem 4.1 the processX y,v (t) = X(y + tv) is a real process but now with C ∞ trajectories. Chose p = 2m + 1 then from Proposition 4.3, for every m, It is an easy consequence of the Borel-Sudakov-Tsirelson inequality that E X (2m+2) y,v ∞ is finite. An argument of continuity shows that it is uniformly bounded. A further application of Jensen's inequality gives our third main result Theorem 4.4. Let X(t) be a real valued Gaussian random field on R d with C ∞ sample paths and with variance bounded below. Then for every integer m and every compact set K, E(I d−1 (C u,K )) m < ∞.