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009 Random Marked SetsFelix Ballani, Zakhar Kablu hko, Martin S hlatherInstitute for Mathemati al Sto hasti s, Georgia Augusta University,Golds hmidtstr. 7, D-37077 Göttingen, Germany,ballanimath.uni-goettingen.de, kablu hmath.uni-goettingen.de,s hlathermath.uni-goettingen.deMar h 13, 2009Abstra t. We introdu e a new lass of sto hasti pro esses whi h aredened on a random set in R

d. These pro esses an be seen as a link be-tween random elds and marked point pro esses. Unlike for random elds,the mark ovarian e fun tion need in general not be positive denite. Thisimplies that in many situations the use of simple geostatisti al methods ap-pears to be questionable. Surprisingly, for a spe ial lass of pro esses basedon Gaussian random elds, we do have positive deniteness for the orre-sponding mark ovarian e fun tion and mark orrelation fun tion.Classi ation. Primary: 60G60, 60G55; se ondary: 60G15, 60D05Keywords. random eld, random set, marked point pro ess, mark or-relation fun tion, mark ovarian e fun tion1 Introdu tionQuantities measured in Rd are mostly modelled as so- alled regionalizedvariables, i. e., one usually assumes that these quantities an, in prin iple, bemeasured everywhere in Rd and that the hoi e of sampling points does notdepend on the values of these quantities. Based on this assumption, severalgeostatisti al methods like variogram analysis or kriging an be applied, see[6. However, there are two types of situations where this assumption doesnot hold [23 and hen e, un riti al use of geostatisti al methods might ausein orre t or meaningless results. 1

The rst type of problems is aused by the investigators themselves bysome kind of preferential sampling [9. For instan e, this happens when dataare sampled only at pla es where high values of the variable of interest areexpe ted. The se ond type of problems is intrinsi to the investigated obje titself. An obvious situation is the investigation of individuals, e. g. trees in aforest, where intera tions among individuals are present. In this parti ularsituation the theory of marked point pro esses provides a formal frameworkfor data analysis [10, 12.Here, we like to draw the attention to some further, de eptive situationswhere impli it onditioning has been mostly ignored in literature [13, 16, 29.For instan e, the investigation of pesti ides in soil is restri ted to roplandand the height of forest litter is restri ted to silvi ultural areas. In both ases,a presele tion annot be ex luded sin e environmental onditions dire tlyinuen e the kind of land use. A further, simple example has motivated thiswork and appears when the altitude is predi ted by geostatisti al methodsbased on measurements that are taken above sealevel only.Su h kind of onditioning might be onsidered as minor, but an ausemajor ee ts, nonetheless. We advise aution be ause of the following fa ts:1. Any hara teristi , su h as the ovarian e fun tion or the variogram,has to be understood as a onditional quantity given measurements an betaken at ertain lo ations.2. In general, neither the ovarian e fun tion is positive denite nor thevariogram is onditionally negative denite.Sin e Gaussian random elds are rather popular, a bigger part of thispaper deals with the following model: the sealevel is at 0 and the altitude isgiven by some (smooth) stationary Gaussian random eld Z with mean −tand varian e 1. Then we fa e the following oddities when inferen e is basedon measurements above sealevel only:1. The theoreti al variogram is not onditionally negative denite, ingeneral.2. A naive denition of the ovarian e fun tion C(x, y) byC(x, y) = E[Z(x)Z(y) | Z(x) ≥ 0, Z(y) ≥ 0] − m2leads in general to a fun tion whi h is not positive denite, for any m ∈ R.3. A more suitable denition of the ovarian e fun tion for the altitudeabove sealevel as the onditional ovarian e given that Z(x) ≥ 0 and Z(y) ≥

0 leads to a fun tion whi h is never dierentiable.4. If t = 0, the onditional ovarian e fun tion is positive denite.Though, no random eld exists that is independent of the sampling lo a-tions and that an model the altiditude above sealevel.2

Before dis ussing the above set up in detail, we will introdu e a theo-reti al framework so that both a meaningful denition of se ond-order har-a teristi s is possible and usual random elds as well as marked point pro- esses are in luded as parti ular ases. For this reason we extend the no-tion of a random upper semi- ontinuous (u. s. .) fun tion (taking values inR = [−∞,∞]) on R

d su h that the domain is a random subset of Rd. Tothis end we make use of Matheron's [17 idea and onsider the hypograph

Af = (x, t) ∈ X × R : t ≤ f(x), X ⊆ Rdof a fun tion f : X → R. In fa t, Af is losed if and only if f is u. s. . on losed X, and the mapping f 7→ Af is a bije tion.The paper is organized as follows. In Se tion 2 we formally introdu ethe notion of a random marked losed set and dis uss some examples. InSe tion 3 we generalise the denition of several hara teristi s for randomelds to random marked sets. We show that, in general, they do not sharethe same deniteness properties as their random eld analogues. In Se tion4 we study Gaussian random elds Z(x) given that Z(x) ex eeds a ertainthreshold t ∈ R. In Se tion 5 some results on the dierentiablity of the mark ovarian e fun tion of random marked sets are given. In Se tion 6, we olle tthe proofs of the statements of the pre eding se tions.2 Random marked losed setsDenote by R = R ∪ −∞,+∞ the extended real line. Let

Φusc = (X, f) : X ⊆ Rd is losed, f : X → R is u. s. ..

Φusc is isomorphi to the system Ucl of all losed sets A ⊆ Rd × R whi hsatisfy

∀x ∈ Rd ∀t ∈ R : (x, t) ∈ A ⇒ x × [−∞, t] ⊆ A (1)by the bije tion

τ : Φusc → Ucl

(X, f) 7→ (x, t) ∈ X × R : t ≤ f(x), (X, f) ∈ Φusc.The subsequent proposition follows immediately from the fa t that the spa eF(Rd × R) of losed subsets of R

d × R is ompa t [17, 19 and that Ucl is losed in F(Rd × R). 3

Proposition 1. Φusc is ompa t in the topology indu ed by Ucl.Denition 1. Let be (Ω,A, P) a omplete probability spa e and let (Ξ, Z) :Ω → Φusc be a mapping with

ω ∈ Ω : τ(Ξ, Z) ∩ B 6= ∅ ∈ Afor every ompa t set B in Rd × R. Then (Ξ, Z) is alled a random marked losed set.The distribution law of a random losed set is hara terized by the prob-abilities of hitting ompa t sets [18, 19 whereas, by [18, Prop. 2.3.1, itsu es to restri t to a suitable base. When hoosing the same base of allnite unions of half ylinders Bi × [ti,∞] as in [24, Thm. XII-6 for randomu. s. . fun tions on R

d, we obtain the following hara terization of randommarked losed sets.Theorem 1. The distribution of a random marked losed set (Ξ, Z) (as aprobability measure on Φusc) is ompletely determined by the joint probabili-tiesP(

supx∈Bi∩Ξ

Z(x) < ti, Bi ∩ Ξ 6= ∅, i ∈ I; Bj ∩ Ξ = ∅, j ∈ 1, . . . , n \ I),where B1, . . . , Bn are ompa t subsets of R

d, t1, . . . , tn ∈ R, and I is a subsetof 1, . . . , n, n ∈ N.Denition 2. A random marked losed set (Ξ, Z) is alled stationary ifP(τ(Ξ, Z) + (x, 0) ∈ · ) = P(τ(Ξ, Z) ∈ · )for all x ∈ R

d, and it is alled isotropi ifP(θτ(Ξ, Z) ∈ · ) = P(τ(Ξ, Z) ∈ · )for all rotations θ ∈ SOd+1 with θ(Rd × 0) = R

d × 0.Example 1. A parti ular model of a random marked losed set that de-s ribes an unbiased sampling of a random eld [27 is given when Z is arandom u. s. . fun tion on Rd that is independent of the random losed set

Ξ. We all (Ξ, Z) a random-eld model.If the data are onsistent with a random-eld model, any analysis sim-plies onsiderably sin e the domain and the marks an be investigated sep-arately (see also Remark 4) by using standard te hniques for random sets[26 and for geostatisti al data [6, 10. For the parti ular ase of markedpoint pro esses, several tests for the random-eld model hypothesis havebeen developed [11, 23. 4

Example 2. Let Ξ be a random losed set and Z(x) = d(∂Ξ, x) the Euk-lidean distan e of x ∈ Rd to the boundary of Ξ. Then Z is even ontinuouson Ξ. Sin e lo al maxima of Z are only attained at lo ations in the interiorof Ξ, the random marked set (Ξ, Z) is a random-eld model if and only if

Ξ = ∂Ξ almost surely, in whi h ase Z is trivial.Example 3. Cressie et al. [7 onsider the spatial predi tion on a rivernetwork. Here, Ξ is the ow of the river (as a one-dimensional line or atwo-dimensional stripe) and Z models the dissolved oxygen.Example 4. Let Ξ be a random losed set represented as a lo ally niteunion of losed C2-smooth hypersurfa es in Rd su h that any two hypersur-fa es interse t at most in a set of measure zero with respe t to the (d − 1)-dimensional Hausdor measure. For any x ∈ Ξ, the mark Z(x) is the maxi-mum of the mean urvatures of the hypersurfa es at x. The mean urvaturehas its importan e for example in the analysis of foams [15.3 Chara teristi s for random marked losed setsFor the des ription of random elds a set of se ond-order hara teristi slike the variogram, the ovarian e fun tion and the orrelation fun tion areused [6. In analogy to these summary fun tions, several se ond-order har-a teristi s for marked point pro esses have been introdu ed as onditionalquantities given the existen e of points of the respe tive unmarked point pro- ess [22, 26. Sin e point pro esses an be des ribed as random ( ounting)measures, these quantities have been derived as Radon-Nikodym derivativesof ertain se ond-order moment measures [3, Se tion 2.7. Nevertheless, ran-dom measures are not always appropriate for the denition of se ond-order hara teristi s as the following example illustrates.Example 5. Let the stationary random losed set Ξ in R

1 be given byΞ = ξ +

⋃

z∈Z

[2z − p, 2z + p] ∪ 2z + 1,where p ∈ (0, 13) and ξ is uniformly distributed in [0, 1]. Obviously, interpointdistan es r ∈ (0, 2p] are only possible if both points belong to the samesegment ξ + [2z − p, 2z + p], and interpoint distan es r ∈ (1 − p, 1 + p]are only possible if one point belongs to a segment ξ + [2z − p, 2z + p] andthe other is from one of the singletons, ξ + 2z − 1 or ξ + 2z + 1. Sin e

P(o, r ∈ Ξ) = 0 for all r ∈ (1−p, 1+p], the approa h of dening se ond-order5

hara teristi s using a random measure, whi h is here based on the Lebesguemeasure on R1, annot a ount for segment-singleton point pairs, and hen e,these hara teristi s are undened for r ∈ (1− p, 1+ p]. Nonetheless, it doesmake sense also to onsider the orrelation of two marks given that the orresponding points are a distan e r, r ∈ (1 − p, 1 + p], apart.In what follows, Bε(x) denotes the Eu lidean ball in R

d with entrex ∈ R

d and radius ε ≥ 0, ⊕ denotes Minkowski addition, and we writeshortly Ξ⊕ε for Ξ ⊕ Bε(o).Let (Ξ, Z) be a stationary random marked losed set in Rd with marksin R. For any ε ≥ 0 dene the (stationary) random eld Zε by

Zε(x) =

maxy∈Ξ∩Bε(x)

Z(y), x ∈ Ξ⊕ε,

0, otherwise.Let f : R2 → R be a right- ontinuous fun tion. For all h ∈ R

d deneκf (h) = lim

ε→0+E

[f(Zε(o), Zε(h)

)| o, h ∈ Ξ⊕ε

] (2)whenever κ|f |(h) < ∞ and P(o, h ∈ Ξ⊕ε) > 0 for all ε > 0, otherwise κf (h)is undened.In parti ular, for the following hoi es of f ,e(m1,m2) = m1, c(m1,m2) = m1m2, v(m1,m2) = m2

1, (3)deneE(h) = κe(h) (4)γ(h) =

1

2(κv(h) + κv(−h)) − κc(h) (5)

cov(h) = κc(h) − κe(h)κe(−h) (6)cor(h) =

κc(h) − κe(h)κe(−h)

(κv(h) − κe(h)2)1/2(κv(−h) − κe(−h)2)1/2(7)

kmm(h) = (m)−2 κc(h), (m 6= 0), (8)wherem = E[Z(o) | o ∈ Ξ]is the mean mark. 6

We all γ the mark variogram, cov the mark ovarian e fun tion, cor themark orrelation fun tion and kmm Stoyan's kmm-fun tion of (Ξ, Z) [22.Note that, if Ξ ≡ Rd, these denitions are ompatible with the lassi aldenitions for random elds (see Remark 4).Whenever (Ξ, Z) is assumed to be both stationary and isotropi the har-a teristi s given by (4)(8) are rotation invariant. By slight abuse of notationwe will write E(r), r ∈ [0,∞), instead of E(h), h ∈ R

d. The same appliesfor the fun tions dened in Eq. (5)(8).Remark 1. Let Ψε = νd( ·∩Ξ⊕ε) be the random volume measure asso iatedwith the random losed set Ξ⊕ε. Here, νd is the d-dimensional Lebesguemeasure. If µ(2)ε denotes the se ond-order moment measure of Ψε then, for

B1, B2 ∈ B(Rd), we have∫

B2

∫

B1

E

[f(Zε(x), Zε(y)

)1Ξ⊕ε

(x)1Ξ⊕ε)(y)]

dxdy

= E

[∫

B2

∫

B1

f(Zε(x), Zε(y)

)Ψε(dx)Ψε(dy)

]

=

∫

B1×B2

∫

R2

f(m1,m2)Qε;x,y(d(m1,m2))µ(2)ε (d(x, y))

=

∫

B2

∫

B1

∫

R2

f(m1,m2)Qε;x,y(d(m1,m2)) P(x, y ∈ Ξ⊕ε) dxdy,where Qε;x,y is the two-point mark distribution of the weighted random mea-sure (Ψε, Zε) [3. Hen e, for almost all (x, y) with P(x, y ∈ Ξ⊕ε) > 0, wehaveE

[f(Zε(x), Zε(y)

)| x, y ∈ Ξ⊕ε

]=

∫

R2

f(m1,m2)Qε;x,y(d(m1,m2)).Remark 2. In ase P(o, h ∈ Ξ) > 0, h ∈ Rd, the above denition takes thesimpler form

κf (h) = E[f(Z(o), Z(h)) | o, h ∈ Ξ]if we impose the integrability onditionsE[|Z(o)|1Ξ(o)1Ξ(h)] < ∞, κ|e|(h) < ∞,for f = e, and,E[|Z(o)|21Ξ(o)1Ξ(h)] < ∞, κ|v|(h) < ∞,7

for f = c or f = v. This an be seen as follows. Denoting by a+ and a− thepositive and the negative part of a ∈ R, respe tively, we always haveZε(o)+1Ξ⊕ε

(o)1Ξ⊕ε(h) ≤ Zε(o)+1Ξ⊕ε

(o)1Ξ⊕ε(h) ≤ |Zε(o)|1Ξ⊕ε

(o)1Ξ⊕ε(h)for all ε ≤ ε, where the right-hand side is integrable due to κ|e|(h) < ∞.Similarly,

Zε(o)−1Ξ⊕ε(o)1Ξ⊕ε

(h) ≤ Z(o)−1Ξ(o)1Ξ(h) ≤ |Z(o)|1Ξ(o)1Ξ(h).In the same way we obtain|Zε(o)|21Ξ⊕ε

(o)1Ξ⊕ε(h) + |Z(o)|21Ξ(o)1Ξ(h)as an integrable upper bound of |Zε(o)|21Ξ⊕ε

(o)1Ξ⊕ε(h)| and

(|Z(o)|2 + |Z(h)|2

)1Ξ(o)1Ξ(h) +

(|Zε(o)|2 + |Zε(h)|2

)1Ξ⊕ε

(o)1Ξ⊕ε(h)as an integrable upper bound of |Zε(o)Zε(h)|1Ξ⊕ε

(o)1Ξ⊕ε(h). Sin e Z isu. s. . on Ξ, a value ε > 0 exists for every x ∈ Ξ and for every δ > 0 su hthat Z(y) ≤ Z(x) + δ for all y ∈ Bε(x) ∩ Ξ. Hen e, we have Zε(x) → Z(x)from above as ε → 0+. Further, x /∈ Ξ implies x /∈ Ξ⊕ε for all su ientlysmall ε. We then have

f(Zε(o), Zε(h))1Ξ⊕ε(o)1Ξ⊕ε

(h) → f(Z(o), Z(h))1Ξ(o)1Ξ(h) a. s.as ε → 0+. Hen e, by the dominated onvergen e theorem, we haveκf (h) = lim

ε→0+

E

[f(Zε(o), Zε(h)

)1Ξ⊕ε

(o)1Ξ⊕ε(h)]

P(o, h ∈ Ξ⊕ε)

=E [f (Z(o), Z(h)) 1Ξ(o)1Ξ(h)]

P(o, h ∈ Ξ).Remark 3. There exists an alternative on ept of random marked sets whi his inspired by the notion of random elds and where se ond-order hara ter-isti s in the sense of the pre eding remark an be dened.Let R∅ = R ∪ ζ∅ be the extension of R by some ζ∅. We denote by

B(R∅) the respe tive Borel σ-eld whi h is generated by all sets B1 ∪B2 forB1 ∈ B(R) and B2 ⊂ −∞,∞, ζ∅.A family of random variables Z(·, x) : Ω → R∅, x ∈ R

d, on the probabil-ity spa e (Ω,A, P) is alled a random eld with random domain Ξ, ifΞ = x ∈ R

d : Z(·, x) 6= ζ∅.8

Clearly, when Z takes only values dierent from ζ∅ or −∞,∞, ζ∅ thisnotion of a random marked set in ludes usual R- or R-valued random eldson Rd.Note that Ξ is a random set in a very general sense [18, entirely deter-mined by its indi ator 1Ξ(x) = 1

R(Z(x)). If Z is jointly measurable, i. e.,

Z is (A ⊗ B(Rd),B(R∅))-measurable, then the realizations of Ξ are almostsurely Borel measurable. If we have even almost surely losed (open) realiza-tions of Ξ then Z is alled a random eld with random losed (open) domain,see also [19.If P(o ∈ Ξ) > 0 holds for a stationary random eld Z with randomdomain Ξ we an dene se ond-order hara teristi s without any furtherassumption on path regularity. Let Z be the (stationary) random eld givenby Z(x) = Z(x) for x ∈ Ξ, and Z(x) = 0 otherwise. Let f : R2 → R be ameasurable fun tion. For all h ∈ R

d deneκf (h) = E[f(Z(o), Z(h)) | o, h ∈ Ξ]whenever P(o, h ∈ Ξ) > 0 and E[|f(Z(o), Z(h))|1Ξ(o)1Ξ(h)] < ∞.Remark 4. Let (Ξ, Z) be a stationary real-valued random-eld model and

Zε(x) =

maxy∈Ξ∩Bε(x)

Z(y), x ∈ Ξ⊕ε,

Z(x), otherwise.Sin e Z is u. s. . on Ξ we have Zε(x) → Z(x) from above for x ∈ Ξ, andhen e, by the denition of Zε, for all x ∈ Rd as ε → 0+. Then, using theindependen e of Z and Ξ, we obtain

κf (h) = limε→0+

E [f (Zε(o), Zε(h)) | o, h ∈ Ξ⊕ε]

= limε→0+

E [f (Zε(o), Zε(h))]

= E [f (Z(o), Z(h))]for all h ∈ Rd whi h satisfy P(o, h ∈ Ξ⊕ε) > 0 for all ε > 0 and, dependingon the hoi e of f a ording to (3), one of the integrability onditions inRemark 2 with Ξ repla ed by R

d.Remark 5. The denition of κf a ording to (2) is, in important situtations, onsistent with the lassi al denition of the se ond-order hara teristi s ofstationary marked point pro esses [22: Let Φ be a stationary simple markedpoint pro ess on Rd × R. Then, Ξ is the support of the unmarked point9

pro ess Φ = Φ( · × R). We assume that the se ond-order moment measureµ(2) of Φ is lo ally nite. For ‖h‖ > 0 we have

E

[f(Zε(o), Zε(h)

)1Ξ⊕Bε(o)(o)1Ξ⊕Bε(o)(h)

]

= E

[f(Zε(o), Zε(h)

)1Φ(Bε(o))=11Φ(Bε(h))=1

]

+ E

[f(Zε(o), Zε(h)

)1Φ(Bε(o))>11Φ(Bε(h))≥1

]

+ E

[f(Zε(o), Zε(h)

)1Φ(Bε(o))=11Φ(Bε(h))>1

].For any 0 < ε < ‖h‖/2 the rst summand equals

E

∑

(x1,m1),(x2,m2)∈eΦ

f(m1,m2)1Bε(o)(x1)1Bε(h)(x2)

=: µ

(2)f (Bε(o) × Bε(h)).We an extend the argumentation in [8, Prop. 9.3.XV in order to on ludethat

P(o, h ∈ Ξ⊕ε)

µ(2)(Bε(o) × Bε(h))=

P(Φ(Bε(o)) ≥ 1,Φ(Bε(h)) ≥ 1)

µ(2)(Bε(o) × Bε(h))→ 1as ε → 0+. If we additionally impose the ondition that for some ε > 0,

supε∈(0,ε)

E

[∣∣∣f(Zε(o), Zε(h)

)∣∣∣1Φ(Bε(o))>11Φ(Bε(h))≥11|f( eZε(o), eZε(h))|>M

]

P(Φ(Bε(o)) ≥ 1, Φ(Bε(h)) ≥ 1)→ 0as M → ∞, we obtain

κf (h) = limε→0+

µ(2)f (Bε(o) × Bε(h))

µ(2)(Bε(o) × Bε(h))whi h equals µ(2)-a. e. the Radon-Nikodym derivativedµ

(2)f (x, x + h)

dµ(2)(x, x + h).For instan e, the above ondition is satised if E

∣∣∣f(Zε(o), Zε(h)

)∣∣∣α is uni-formly bounded on (0, ε) for some α > 1.10

A fun tion f : Rd → R is alled positive denite if

n∑

i=1

n∑

j=1

aiajf(xi − xj) ≥ 0for any n ∈ N, x1, . . . , xn ∈ Rd, and a1, . . . , an ∈ R, and f is alled ondi-tionally negative denite if

n∑

i=1

n∑

j=1

aiajf(xi − xj) ≤ 0for any n ∈ N, x1, . . . , xn ∈ Rd, and for all a1, . . . , an ∈ R with ∑n

i=1 ai = 0.For a random-eld model all se ond-order hara teristi s oin ide withthat of a random eld with u. s. . paths, see Remark 4, and thus, share thesame deniteness properties. On the other hand, for marked point pro essesit has been shown by examples [28 and systemati ally [22 that the mark o-varian e fun tion, the mark orrelation fun tion and the kmm-fun tion neednot be positive denite, and the mark variogram need not be onditionallynegative denite in ontrast to random elds. Some of the onstru tions usedin [22 are based on the fa t that for a marked point pro ess, Ξ is a lo allynite subset of Rd and has therefore Lebesgue measure zero. However, thenext example shows that in general we annot expe t that the mark ovari-an e fun tion is positive denite (and the mark orrelation fun tion and the

kmm-fun tion either) even when we have E[νd(Ξ ∩ [0, 1]d)] = P (o ∈ Ξ) > 0.Example 6 (Continuation of Example 2). Let be p ∈ (23 , 1], ξ a randomvariable uniformly distributed on [0, 1], Ξ = Z ⊕ [ξ, p + ξ], and Z(ξ, · ) a1-periodi fun tion dened by

Z(ξ, x) =

x − ξ, x ∈ Z ⊕ [ξ, p2 + ξ),

p − (x − ξ), x ∈ Z ⊕ [p2 + ξ, p + ξ),

0, x ∈ Z ⊕ [p + ξ, 1 + ξ).Then Z and Ξ are jointly stationary and ea h of the hara teristi s given by(4)(8) is 1-periodi . In parti ular, on [0, 1/2) we havecov(r) =

p4−4p3r−12p2r2+48pr3−36r4

48(p−r)2, r ∈ [0, 1 − p),

−3p4−24p3r+12p3−48p2r2+48p2r−12p2

48(2p−1)2

+64pr3+48pr2−48pr+8p−32r3−24r2+24r−448(2p−1)2

, r ∈ [1 − p, p2 ),

−4p4−8p3+6p2−2p+12r4−24r3+18r2−6r+112(2p−1)2

, r ∈ [p2 , 12 ],11

and, by symmetry, cov(r) = cov(1− r) for r ∈ (12 , 1). Sin e cov is 1-periodi the 0th oe ient of the Fourier series of cov is proportional to

1∫

0

cov(r) dr = 76p3 ln

(p

2p−1

)+ 409p5−790p4+565p3−280p2+120p−24

120(2p−1)2,whi h is negative for 2

3 ≤ p < 1 (and vanishes for p = 1, whi h is the randomeld ase). Sin e cov is ontinuous, Bo hner's theorem [21 implies thatcov(r) annot be a positive denite fun tion.Our major example is analysed within an own se tion sin e some resultsmight be of interest not only to the eld of random marked sets but also tothe theory of positive denite fun tions.4 Gaussian random elds ex eeding t ∈ RLet Z be a stationary and isotropi entered unit varian e Gaussian randomeld in R

d. Then, for t ∈ R, we deneΞt = x ∈ R

d : Z(x) ≥ t.If, in parti ular, Z is almost surely ontinuous [1, 2 then Ξt is almost surely losed, i. e., (Ξt, Z) is a random marked losed set. Note that Ξt is a so- alledex ursion set whi h has been extensively studied in the literature, see [1, 2and the referen es therein.Sin e Z is assumed to be both stationary and isotropi , its ovarian efun tion, Cov(x, y) = E[Z(x)Z(y)], x, y ∈ Rd, is translation and rotationinvariant, i. e., there exists a fun tion R : [0,∞) → R su h that

Cov(x, y) = R(‖x − y‖), x, y ∈ Rd.First, we onsider the ase t = 0.Theorem 2. Let Z be a stationary and isotropi entered unit varian eGaussian random eld in R

d with ovarian e fun tion given by R : [0,∞) →R. Then, for r ∈ [0,∞), the se ond-order hara teristi s of (Ξ0, Z) are given

12

byE(r) =

√π

2

1 + R(r)

arcsin(R(r)) + π2

, (9)cov(r) = R(r) +

√1 − R(r)2

arcsin(R(r)) + π2

− π

2

(1 + R(r))2

(arcsin(R(r)) + π

2

)2 , (10)γ(r) = (1 − R(r))

(1 −

√1 − R(r)2

arcsin(R(r)) + π2

), (11)

kmm(r) =π

2

(R(r) +

√1 − R(r)2

arcsin(R(r)) + π2

), (12)

cor(r) =R(r)(arcsin(R(r))+ π

2)2+

√1−R(r)2(arcsin(R(r))+ π

2)−π

2(1+R(r))2

(arcsin(R(r))+ π2)2+R(r)

√1−R(r)2(arcsin(R(r))+ π

2)−π

2(1+R(r))2

. (13)Obviously, ea h of the se ond-order hara teristi s of (Ξ0, Z) is a on-tinuous transform of R. In parti ular, this means that ontinuity of R ispreserved. Vi e versa, due to the monotoni ity of the transform for cov (seeTheorem 3 below) already from cov it an be dedu ed whether or not R ontains a nugget ee t [6.Sin e, for every stationary random-eld model, E(h), h ∈ Rd, is onstant,equality (9) implies that (Ξ0, Z) is not a random-eld model, i. e., there doesnot exist a random eld in R

d whose se ond-order hara teristi s oin idewith that of (Ξ0, Z). It is therefore quite surprising to see that we are not ableto falsify that (Ξ0, Z) is a random-eld model by using the mark ovarian efun tion or the mark orrelation fun tion of (Ξ0, Z).Theorem 3. The fun tions f0 : [−1, 1] → R,f0(ρ) = ρ +

√1 − ρ2

arcsin ρ + π2

− π2

(1 + ρ)2

(arcsin ρ + π2 )2

,and g0 : [−1, 1] → R,g0(ρ) =

ρ(arcsin ρ + π2 )2 +

√1 − ρ2(arcsin ρ + π

2 ) − π2 (1 + ρ)2

(arcsin ρ + π2 )2 + ρ

√1 − ρ2(arcsin ρ + π

2 ) − π2 (1 + ρ)2

,are absolutely monotone on [0, 1], i. e., they have only nonnegative derivativesthere.Corollary 1. cov(r) and cor(r) are positive denite fun tions.13

However, the following proposition shows that the kmm-fun tion of (Ξ0, Z)is not positive denite, in general.Proposition 2. Let Z be a stationary and isotropi entered unit varian eGaussian random eld in Rd with ovarian e fun tion given by a ontinuousfun tion R : [0,∞) → R. Then the kmm-fun tion of (Ξ0, Z) is positivedenite if and only if R ≡ 1.Proof. Let

q(ρ) = ρ +

√1 − ρ2

arcsin(ρ) + π2

.Then q(1) = 1 andq(ρ) ≥ ρ + π−1

√1 − ρ2 > 1, ρ ∈ ((π2 − 1)/(π2 + 1), 1).Hen e, kmm(r) is not a positive denite fun tion if R 6≡ 1 [21, Theorem1.4.1.The mark variogram of (Ξ0, Z) is in general not onditionally negativedenite, whi h an be seen as follows. Consider (Ξ0, Z) for dimension d = 1and R(r) = cos(r). Sin e γ is onditionally negative denite if and only if

e−sγ is positive denite for all s > 0 [21, Theorem 6.1.9 it su es to showthat e−γ is not positive denite. e−γ(r) inherits 2π-periodi ity from cos(r),and hen e, it is positive denite if and only if its Fourier oe ients arenonnegative. Numeri al al ulations yield that the rst Fourier oe ient isnearby -0.03364.Now we swit h over to the more general ase t ∈ R. Unfortunately,unlike the ase t = 0, we annot express all the se ond-order hara teristi sof (Ξt, Z) in losed form. In parti ular, for a stationary and isotropi enteredunit varian e Gaussian random eld Z in Rd with ovarian e fun tion givenby R : [0,∞) → R, we have

P(o, h ∈ Ξt) =

R(‖h‖)∫

0

ϕ(t, t, s) ds + Ψ(t)2,see [5, Eqn. (10.8.3). Here,Ψ(t) =

∞∫

t

ϕ(s) ds, ϕ(t) =1√2π

e−t2

2 , t ∈ R,14

denotes the tail probability fun tion of the standard Gaussian distribution.By ϕ(s, t, ρ) we denote the density of the bivariate Gaussian distributionwith unit varian e and orrelation ρ. In the following we on entrate on themark ovarian e fun tion of (Ξt, Z).Lemma 1. Let Z be a stationary and isotropi entered unit varian e Gaus-sian random eld in Rd with ovarian e fun tion given by R : [0,∞) → R.Then, for t ∈ R and h ∈ R

d, we haveE[Z(o)1[t,∞)(o)1[t,∞)(h)] = Et(R(‖h‖)), Et(ρ) = ϕ(t)

(ρ + 1

)Ψ

(t

√1 − ρ

1 + ρ

),and

E[Z(o)Z(h)1[t,∞)(o)1[t,∞)(h)] = Ct(R(‖h‖))whereCt(ρ) =

∫ ρ

0ϕ(t, t, s)(ρ − s − t2) ds + ρΨ2(t) + ϕ2(t)

+ 2tϕ(t)[Ψ(t) + (ρ + 1)Ψ

(t√

(1 − ρ)/(1 + ρ))]

.If we writePt(ρ) =

∫ ρ

0ϕ(t, t, s) ds + Ψ(t)2, ρ ∈ [−1, 1],then the mark ovarian e fun tion of (Ξt, Z) is given by

cov(r) = ft(R(r)),whereft(ρ) =

Ct(ρ)

Pt(ρ)− Et(ρ)2

Pt(ρ)2.There is strong eviden e that also the mark ovarian e fun tion of (Ξt, Z),

t 6= 0, is positive denite for a ertain lass of Gaussian random elds Z.Figure 1 shows ft(ρ) and f ′t(ρ) for several t, indi ating that for these t thefun tions ft(ρ) are both in reasing and onvex for ρ ∈ [0, 1]. Hen e, if thisis really true, for instan e Pólya's riterion [20 would imply that, for any ontinuous and onvex fun tion R : [0,∞) → R satisfying R(0) = 1 and

limr→∞ R(r) = 0, the fun tion ft(R(| · |)) is positive denite on R.15

Figure 1: ft (left) and f ′t (right) for t = −1 (dashed), t = 0 (dotted) and

t = 1 (solid).5 Dierentiability at 0In this se tion we ontinue the example of the pre eding se tion and showthat the orresponding mark ovarian e fun tion has a right-hand derivativeat 0 whi h does not vanish.Lemma 2. Let Z be a stationary and isotropi entered unit varian e Gaus-sian random eld in Rd with ontinuous ovarian e fun tion given by R :

[0,∞) → R ex luding R ≡ 1. Let be CΞt(‖h‖) = P(o, h ∈ Ξt) the set ovari-an e of the ex ursion set Ξt. If Z is mean-square dierentiable thenC ′

Ξt(0+) = −ϕ(t)√

2π

√−R′′(0+) < 0,otherwise C ′

Ξt(0+) = −∞.Theorem 4. Let Z be a stationary and isotropi entered unit varian eGaussian random eld in R

d with ontinuous ovarian e fun tion given byR : [0,∞) → R ex luding R ≡ 1. Then the mark ovarian e fun tioncov(r) = ft(R(r)) of (Ξt, Z) has a negative right-hand derivative at r = 0,in parti ular, we have

cov′(0+) = −(t2 − 1)Ψ(t)2 − 3tϕ(t)Ψ(t) + 2ϕ(t)2

Ψ(t)3· ϕ(t)

√−R′′(0+)√2π

< 0in ase Z is mean-square dierentiable, and cov′(0+) = −∞ otherwise.16

6 ProofsFirst, we prove Lemma 1 of Se tion 4 whi h is needed in the proof of Theorem2.6.1 Proof of Lemma 1The proof is based on te hniques used in [14, p. 351. We haveEt(ρ) =

∫ ∞

t

∫ ∞

tx1ϕ(x1, x2, ρ) dx1 dx2

=

∫ ∞

t

∫ ∞

tx1

(1

4π2

∫ ∞

−∞

∫ ∞

−∞e−iyT xe−

12yT Σy dy1 dy2

)dx1 dx2

=1

4π2

∫ ∞

−∞

∫ ∞

−∞e−

12yT Σye−it(y1+y2) 1

iy2

(t

iy1− 1

y21

)dy1 dy2after reversing the order of integration and integrating out x1 and x2. Forthe part

Et;1(ρ) = t1

4π2

∫ ∞

−∞

∫ ∞

−∞e−

12yT Σy e−it(y1+y2)

−y1y2dy1 dy2we have

Et;1(ρ) = tPt(ρ) = t

∫ ρ

0ϕ(t, t, s) ds + tΨ(t)2.For the part

Et;2(ρ) =1

4π2

∫ ∞

−∞

∫ ∞

−∞e−

12yT Σy e−it(y1+y2)

−iy21y2

dy1 dy2we obtain∂2Et;2(ρ)

∂ρ2=

i

4π2

∫ ∞

−∞

∫ ∞

−∞y2e

− 12yT Σye−it(y1+y2) dy1 dy2

= − ∂ϕ(x1, x2, ρ)

∂x2

∣∣∣∣(x1,x2)=(t,t)

= ϕ(t, t, ρ)t

1 + ρ.By Et(0) = ϕ(t)Ψ(t) and

∂Et(ρ)

∂ρ

∣∣∣∣ρ=0

=

∫ ∞

t

∫ ∞

tx1

∂ϕ(x1, x2, ρ)

∂ρ

∣∣∣∣ρ=0

dx1 dx2 = (tϕ(t) + Ψ(t)) ϕ(t),17

we getEt(ρ) = t

∫ ρ

0

∫eρ

0

ϕ(t, t, s)

1 + sds dρ + t

∫ ρ

0ϕ(t, t, s) ds + (ρ + 1)ϕ(t)Ψ(t)

= (ρ + 1)

[t

∫ ρ

0

ϕ(t, t, s)

1 + sds + ϕ(t)Ψ(t)

].Further,

t

ρ∫

0

ϕ(t, t, s)

1 + sds =

t

2π

1∫

11+ρ

e−t2u

√2u − 1

du =te−t2/2

4π

1∫

1−ρ

1+ρ

e−t2v/2

√v

du (14)=

e−t2/2

2π

t∫

t√

(1−ρ)/(1+ρ)

e−w2/2 du

= ϕ(t)[Ψ(t√

(1 − ρ)/(1 + ρ))− Ψ(t)

]whi h gives the rst equality in the lemma.The se ond equality is shown similarily:Ct(ρ) = E[Z(o)Z(r)1[t,∞)(o)1[t,∞)(r)]

=

∫ ∞

t

∫ ∞

tx1x2ϕ(x1, x2, ρ) dx1 dx2

=

∫ ∞

t

∫ ∞

tx1x2

(1

4π2

∫ ∞

−∞

∫ ∞

−∞e−iyT xe−

12yT Σy dy1 dy2

)dx1 dx2

=1

4π2

∫ ∞

−∞

∫ ∞

−∞e−

12yT Σye−it(y1+y2)

(t

iy1− 1

y21

)(t

iy2− 1

y22

)dy1 dy2.Thus, we an write Ct(ρ) = Ct;1(ρ) + Ct;2(ρ) + Ct;3(ρ), where

Ct;1(ρ) = t21

4π2

∫ ∞

−∞

∫ ∞

−∞e−

12yT Σy e−it(y1+y2)

−y1y2dy1 dy2

= t2Pt(ρ),

Ct;2(ρ) = t1

4π2

∫ ∞

−∞

∫ ∞

−∞e−

12yT Σy e−it(y1+y2)

−iy21y2

dy1 dy2

+ t1

4π2

∫ ∞

−∞

∫ ∞

−∞e−

12yT Σy e−it(y1+y2)

−iy1y22

dy1 dy2

= 2tEt;2(ρ), 18

andCt;3(ρ) =

1

4π2

∫ ∞

−∞

∫ ∞

−∞e−

12yT Σy e−it(y1+y2)

y21y

22

dy1 dy2For the latter expression we have∂Ct;3(ρ)

∂ρ=

1

4π2

∫ ∞

−∞

∫ ∞

−∞e−

12yT Σy e−it(y1+y2)

−y1y2dy1 dy2

= Pt(ρ).Sin e Ct(0) = ϕ(t)/√

2π we then haveCt(ρ) = 2t2

∫ ρ

0

∫eρ

0

ϕ(t, t, s)

1 + sds dρ +

∫ ρ

0

∫eρ

0ϕ(t, t, s) ds dρ

+ t2∫ ρ

0ϕ(t, t, s) ds + ρΨ(t)2 + 2ρtϕ(t)Ψ(t) + ϕ2(t).

=

∫ ρ

0ϕ(t, t, s)(ρ − s − t2) ds + ρΨ2(t) + ϕ2(t)

+ 2tϕ(t)[Ψ(t) + (ρ + 1)Ψ

(t√

(1 − ρ)/(1 + ρ))]6.2 Proof of Theorem 2As

P(Z(o) ≥ 0, Z(r) ≥ 0) =1

2π

(arcsin(R(r)) +

π

2

)the formulae for E and cov follow immediately from Lemma 1. Now,m = E[Z(o) | Z(o) ≥ 0] =

√2

π,and it remains to show that

E[(Z(o))21[0,∞)(Z(o))1[0,∞)(Z(r))

]

=R(r)

√1 − R(r)2 + arcsin(R(r)) + π

2

2π=: V0(R(r))We have

V0(ρ) =

∫ ∞

0

∫ ∞

0x2

1 ϕ(x1, x2, ρ) dx1 dx2

=

∫ ∞

0

∫ ∞

0

(1

4π2

∫ ∞

−∞

∫ ∞

−∞x2

1 e−iyT xe−12yT Σy dy1 dy2

)dx1 dx2

=2

4π2

∫ ∞

−∞

∫ ∞

−∞e−

12yT Σy 1

y31y2

dy1 dy2.19

The oe ient of ρ in the exponent of the integrand is (−y1y2). Hen e,dierentiating the last equation three times with respe t to ρ yields∂3V0(ρ)

∂ρ3=

−2

4π2

∫ ∞

−∞

∫ ∞

−∞y22 e−

12yT Σy dy1 dy2

=−2√

det(Σ−1)

2π

1

2π√

det(Σ−1)

∫ ∞

−∞

∫ ∞

−∞y22 e−

12yT Σy dy1 dy2

=−2√

det(Σ−1)

2π

(Σ−1

)22

=−2

2π

1√

1 − ρ23 .Be ause of

∂2V0(ρ)

∂ρ2=

2

4π2

∫ ∞

−∞

∫ ∞

−∞

y2

y1e−

12yT Σy dy1 dy2we have ∂2V0

∂ρ2 (0) = 0. Hen e, we obtain∂2V0(ρ)

∂ρ2=

−2

2π

ρ√1 − ρ2after integration with respe t to ρ. Be ause of ∂V0

∂ρ (0) = 1/π and V0(0) = 1/4we then obtain after further integrations with respe t to ρ

V0(ρ) =1

2π

(ρ√

1 − ρ2 + arcsin ρ +π

2

).6.3 Proof of Theorem 3The idea of the proof is to show that the oe ients of the Taylor expansionof f0 and g0 are all nonnegative. We give the proof for f0 only; for g0, thesame te hniques an be applied.Sin e arcsin an be extended to the omplex plain C by

arcsin z = −i ln(iz +√

1 − z2), z ∈ C,with bran h uts (−∞,−1) and (1,∞), also f0 an be extended to C. Werestri t our attention tof(z) =

√1 − z2

arcsin z + π2

− π2

(1 + z)2

(arcsin z + π2 )2

.Sin e arcsin z = −π/2 is satised only for z = −1 and the denumerators ofall derivatives of f onsist of a produ t of integral powers of arcsin z + π/220

and √1 − z2, f is not only holomorphi in the open unit dis D = z ∈

C : |z| < 1 but also on C \ ((−∞,−1] ∪ [1,∞)). This holds be ause √z isholomorphi everywhere ex ept on (−∞, 0] where √z is not even ontinuous.The expansion of arcsin z around z = −1

arcsin z = −π2 +

√2√

1 + z(1 + O(1 + z)

), |z| < 1,shows that the derivatives of any order of f an be ontinuously extendedin z = −1 as z → −1, |z| < 1. Hen e, the only bran hing point of f in D is

z = 1. From the expansion of f around z = 1

f(z) = − 2π + c1(1 − z)

12 +

(2π + 2

π2 − 12π3

)(1 − z) + c2(1 − z)

32

+ O((1 − z)2), |z| < 1withc1 = −

√2(

4π2 − 1

π

), c2 =

√2(−14π + 11

3π2 + 2π3 − 16

π4

),we see that the asymptoti behaviour of f around z = 1 is essentially de-s ribed by

g(z) = c1(1 − z)12 + c2(1 − z)

32 .At rst, we derive an upper bound for the oe ients bn of the Taylor ex-pansion of

h(z) = f(z) − g(z)at z = 0. The fun tion h is holomorphi in D and by onstru tion twi e ontinuously dierentiable in D sin e h(2)(z) behaves around z = 1 likec + O((1 − z)

12 ), |z| < 1. As a onsequen e of Cau hy's integral formula wehave

bn =h(n)(0)

n!=

1

2πi

∫

∂D1−ε

h(ζ)

ζn+1dζ, n = 0, 1, 2, . . . ,for all ε ∈ (0, 1) and D1−ε = z ∈ C : |z| < 1 − ε. After twi e partiallyintegrating we have

bn =1

2πi

1

n(n − 1)

∫

∂D1−ε

h(2)(ζ)

ζn−1dζ, n = 2, 3, 4, . . . .21

Hen e, for n ≥ 2, we obtain|bn| ≤

1

2π

1

n(n − 1)(1 − ε)1−n max

ϕ∈[0,2π]

∣∣∣h(2)((1 − ε)eiϕ)∣∣∣ · 2π(1 − ε).Sin e h(2)(z) is ontinuous in D we an on lude that for n ≥ 2

|bn| ≤1

n(n − 1)max

ϕ∈[0,2π]

∣∣∣h(2)(eiϕ)∣∣∣ <

1

n(n − 1)0.182by Lemma 3 below.In the following we derive a lower bound for the oe ients an of theTaylor expansion of g(z) at z = 0. Note that we have the series expansions

(1 − z)12 =

∞∑

n=0

(−1)n(1

2

n

)zn = 1 −

∞∑

n=1

(2n − 2)!

n!(n − 1)!22n−1zn, |z| < 1,and

(1 − z)32 = 1 − 3

2z + 3∞∑

n=2

(2n − 4)!

n!(n − 2)!22n−2zn, |z| < 1.Hen e, for n ≥ 2, we have

an = −c1 ·(2n − 2)!

n!(n − 1)!22n−1+ c2 · 3

(2n − 4)!

n!(n − 2)!22n−2.Sin e c1 < 0 and c2 > 0, we obtain

an > −c1 ·(2n − 2)!

n!(n − 1)!22n−1≥ −c1

e

2√

π

(n − 1

n

)n−1

n− 32 e

− 112n

− 112(n−1)by Stirling's approximation. At least for n ≥ 6 we have

e(

n−1n

)n ·(1 + 1

n−1

)e− 1

12n− 1

12(n−1) ≥(1 + 1

n−1

)e− 1

6(n−1) ≥ 1.Thus,an >

4 − π

π2√

2πn− 3

2 , n ≥ 6.Combining this with the upper bound for |bn| it is easily seen that the nth oe ient of the Taylor expansion of f0 at x = 0 is nonnegative for all n ≥ 30.The remaining oe ients of order n = 0, 1, . . . , 29 an be determined froma series expansion and are easily evaluated to be nonnegative.22

Lemma 3. We havemax

ϕ∈[0,2π]

∣∣∣h(2)(eiϕ)∣∣∣ < 0.182.Proof. Figure 2 indi ates that this maximum is a hieved for ϕ = 0. Fromthe expansion of f around z = 1 we obtain the value of this maximum: It istwi e the absolute value of the oe ent of (1 − z)2, i. e., we have

maxϕ∈[0,2π]

∣∣∣h(2)(eiϕ)∣∣∣ = |h(2)(1)| =

(−1π − 2

3π2 + 20π3 + 8

π4 − 80π5

)< 0.08.Formally, we an a tually prove that an upper bound is 0.182. Sin e the onjugate- omplex of h(2)(eiϕ) equals h(2)(e−iϕ) we an restri t to the interval

[0, π].First we make an estimate for [π3 , π] then for [0, π3 ]. From the seriesexpansion

(1 − z)12 = 1 −

∞∑

n=1

(2n − 2)!

n!(n − 1)!22n−1zn, |z| < 1, (15)we obtain the two series expansions at z = −1

(1 − z)12 =

√2

√1 − 1 + z

2=

√2

(1 −

∞∑

n=1

(2n − 2)!

n!(n − 1)!22n−1

(1 + z

2

)n)

=

∞∑

n=0

vn(1 + z)nandarcsin z + π

2 =√

2∞∑

n=0

(2n)!

n!n!23n+1

1

n + 12

(1 + z)n+ 12

= (1 + z)12

∞∑

n=0

u(1)n (1 + z)n,whi h are onvergent for |1+z| < 2. From the latter we also obtain a ordingseries expansions

(arcsin z + π2 )k = (1 + z)

k2

∞∑

n=0

u(k)n (1 + z)n, k = 2, 3, 4.23

LetP

(k)N (z) = (1 + z)

k2

N∑

n=0

u(k)n (1 + z)n, k = 1, 2, 3, 4

p(k)N (z) =

arcsin z + π2 − P

(k)N (z)

(1 + z)72

, k = 1, 2, 3, 4,andQN (z) =

N∑

n=0

vn(1 + z)n, qN (z) =(1 − z)

12 − QN (z)

(1 + z)72

.We write h(2)(z) as a fra tion with numeratorhnum(z) = 2(arcsin z + π

2 )(1 + z)(1 − z) − (arcsin z + π2 )3(1 + z)(1 − z)

− π(arcsin z + π2 )2(1 + z)

32 (1 − z)

32

+ z(arcsin z + π2 )2(1 + z)

12 (1 − z)

12 − z2(arcsin z + π

2 )3

+c1

4(arcsin z + π

2 )4(1 + z)32 − 3π(1 + z)

52 (1 − z)

12

+ 4π(arcsin z + π2 )(1 + z)2(1 − z)

+ πz(arcsin z + π2 )(1 + z)2 − 3 c2

4(arcsin z + π

2 )4(1 + z)32 (1 − z)and denumerator

hdenum(z) = (arcsin z + π2 )4(1 + z)

32 (1 − z)

32 .By happrox

num (z) we denote the approximation of hnum(z) if we repla e (arcsin z+π2 )k by P

(k)N (z) and √

1 − z by QN (z) within hnum(z). On the set eiϕ : ϕ ∈[π3 , π] we an make the following estimate for the error

η(ϕ) =

∣∣∣∣∣hnum

(eiϕ)− happrox

num

(eiϕ)

hdenum (eiϕ)

∣∣∣∣∣ .Sin e we have|(arcsin z + π

2 )4| ≥ 4|1 + z|2

24

we obtainη(ϕ) ≤ 1

4|1 − eiϕ|− 3

2

(2|p(1)

N (eiϕ)| · |1 + eiϕ| · |1 − eiϕ|

+ |p(3)N (eiϕ)| · |1 + eiϕ| · |1 − eiϕ|

+ π|p(2)N (eiϕ)| · |1 + eiϕ| 32 · |1 − eiϕ| · |QN (eiϕ)|

+ π|P (2)N (eiϕ)||1 + eiϕ| 32 |1 − eiϕ| · |qN (eiϕ)|

+ π|p(2)N (eiϕ)| · |1 + eiϕ|5 · |1 − eiϕ| · |qN (eiϕ)|

+ |p(2)N (eiϕ)| · |1 + eiϕ| 12 · |QN (eiϕ)|

+ |P (2)N (eiϕ)| · |1 + eiϕ| 12 · |qN (eiϕ)| + |p(2)

N (eiϕ)| · |1 + eiϕ|4 · |qN (eiϕ)|

+ |p(3)N (eiϕ)| + |c1|

4|p(4)

N (eiϕ)| · |1 + eiϕ| 32 + 3π|1 + eiϕ| 52 · |qN (eiϕ)|

+ 4π|1 + eiϕ|2 · |1 − eiϕ| · |p(1)N (eiϕ)| + π|1 + eiϕ|2 · |p(1)

N (eiϕ)|

+3|c2|

4|p(4)

N (eiϕ)| · |1 + eiϕ| 32 · |1 − eiϕ|)

.Sin e |1 + eiϕ| is de reasing on [0, π] we have|1 + eiϕ| ≤ |1 + ei π

3 | =√

3on [π3 , π]. The oe ients u(k)n , k = 1, 2, 3, 4, and vn are non-alternating for

n ≥ 1. Hen e, on [π3 , π], we have|p(k)

N (eiϕ)| ≤ |p(k)N (ei π

3 )| =: δ(k)N , k = 1, 2, 3, 4,

|qN (eiϕ)| ≤ |qN (ei π3 )| =: εNfor all N ≥ 1. Furthermore, on [π3 , π], we have

|1 − eiϕ|−1 ≤ 1and, for N ≥ 20, we an estimate|P (2)

N (eiϕ)| < 2.122

|Q(2)N (eiϕ)| < 1.001 · |1 − eiϕ|

12 .

25

Hen e we obtainη <

1

4

(2√

3δ(1)N +

√3δ

(3)N + π3

34 δ

(2)N · 1.001 + 2.122π 3

34 εN

+ π 352 δ

(2)N εN + 3

14 δ

(2)N · 1.001 + 2.122 · 3 1

4 δ(2)N εN + δ

(3)N

+|c1|4

334 δ

(4)N + 3π3

54 εN + 12πδ

(1)N + 3πδ

(1)N +

3|c2|4

334 δ

(4)N

).Thus, for N ≥ 20 we have η < 0.005041. Furthermore, we an estimate

∣∣∣∣∣happrox

num

(eiϕ)

hdenum (eiϕ)

∣∣∣∣∣ ≤

∣∣∣∣∣∣happrox

num

(eiϕ)

4|1 + eiϕ|72

∣∣∣∣∣∣.By onstru tion, the right-hand side of this inequality is the absolute valueof a polynomial in 1 + eiϕ whi h takes its maximum on [π3 , π] at ϕ = π. For

N = 20 this maximum value is less than 0.1764. Combining both estimateswe end up with maxϕ∈[π3 ,π]

∣∣h(2)(eiϕ)∣∣ < 0.182.Similarly, from (15) we obtain at z = 1 the series expansions

(1 + z)12 =

√2

√1 − 1 − z

2=

√2

(1 −

∞∑

n=1

(2n − 2)!

n!(n − 1)!22n−1

(1 − z

2

)n)

=∞∑

n=0

vn(1 − z)n,

(1 + z)−12 =

√2

∞∑

n=0

(2n)!

n!n!23n+1(1 − z)n,

(1 + z)−32 =

∞∑

n=0

wn(1 − z)n,andarcsin z + π

2 = π −√

2

∞∑

n=0

(2n)!

n!n!23n+1

1

n + 12

(1 − z)n+ 12

= π +

∞∑

n=0

u(1)n (1 − z)n+ 1

2 ,whi h are onvergent for |1−z| < 2. From the latter we also obtain a ordingseries expansions(arcsin z + π

2 )k = πk +∞∑

m=1

u(k)m (1 − z)

m2 , k = 2, 3, 4.26

LetP

(k)N (z) = πk +

2N∑

m=1

u(k)m (1 − z)

m2 , k = 1, 2, 3, 4,

QN (z) =N∑

n=0

vn(1 − z)n,

RN (z) =

N∑

n=0

wn(1 − z)nandp(k)N (z) =

arcsin z + π2 − P

(k)N (z)

(1 − z)32

, k = 1, 2, 3, 4,

qN (z) =(1 + z)

12 − QN (z)

(1 − z)32

,

rN (z) =(1 + z)−

32 − RN (z)

(1 − z)32

.We write h(2)(z) as a fra tion with numeratorhnum(z) = (1 + z)−

32(2(arcsin z + π

2 )(1 + z)(1 − z)

− (arcsin z + π2 )3(1 + z)(1 − z)

− π(arcsin z + π2 )2(1 + z)

32 (1 − z)

32

+ z(arcsin z + π2 )2(1 + z)

12 (1 − z)

12 − z2(arcsin z + π

2 )3

+c1

4(arcsin z + π

2 )4(1 + z)32 − 3π(1 + z)

52 (1 − z)

12

+ 4π(arcsin z + π2 )(1 + z)2(1 − z)

+ πz(arcsin z + π2 )(1 + z)2 − 3 c2

4(arcsin z + π

2 )4(1 + z)32 (1 − z)

)and denumeratorhdenum(z) = (arcsin z + π

2 )4(1 − z)32 .By happrox

num (z) we denote the approximation of hnum(z) if we repla e (arcsin z+

π2 )k by P

(k)N (z), √1 + z by QN (z) and (1 + z)−

12 by RN (z) within hnum(z).27

On the set eiϕ : ϕ ∈ [0, π3 ] we an make the following estimate for the error

η(ϕ) =

∣∣∣∣∣hnum

(eiϕ)− happrox

num

(eiϕ)

hdenum (eiϕ)

∣∣∣∣∣ .Sin e on [0, π3 ] we have

| arcsin(eiϕ) +π

2| ≥ | arcsin(ei π

3 ) +π

2| > 2.11we obtain

2.114η(ϕ) <∣∣hnum(eiϕ)

∣∣ rN (eiϕ)

+

[2|p(1)

N (eiϕ))||1 + eiϕ||1 − eiϕ| + |p(3)N (eiϕ)||1 + eiϕ||1 − eiϕ|

+ π|P (2)N (eiϕ)||qN (eiϕ)||1 + eiϕ||1 − eiϕ|

32

+ π|p(2)N (eiϕ)||QN (eiϕ)||1 + eiϕ||1 − eiϕ|

32

+ π|p(2)N (eiϕ)||qN (eiϕ)||1 + eiϕ||1 − eiϕ|3

+ |P (2)N (eiϕ)||qN (eiϕ)||1 − eiϕ|

12

+ |p(2)N (eiϕ)||QN (eiϕ)||1 − eiϕ|

12

+ |p(2)N (eiϕ)||qN (eiϕ)||1 − eiϕ|2

+ |p(3)N (eiϕ)| + |c1|

4|P (4)

N (eiϕ)||qN (eiϕ)||1 + eiϕ|

+|c1|4

|p(4)N (eiϕ)||QN (eiϕ)||1 + eiϕ|

+|c1|4

|p(4)N (eiϕ)||qN (eiϕ)||1 − eiϕ|

32 |1 + eiϕ|

+ 3π|qN (eiϕ)||1 − eiϕ|12 |1 + eiϕ|2

+ 4π|1 − eiϕ||1 + eiϕ|2|p(1)N (eiϕ)| + π|1 + eiϕ|2|p(1)

N (eiϕ)|

+3|c2|

4|P (4)

N (eiϕ)||qN (eiϕ)||1 + eiϕ||1 − eiϕ|

+3|c2|

4|p(4)

N (eiϕ)||QN (eiϕ)||1 + eiϕ||1 − eiϕ|

+3|c2|

4|p(4)

N (eiϕ)||qN (eiϕ)||1 + eiϕ||1 − eiϕ|52

]· |RN (eiϕ)|.28

Sin e |1 − eiϕ| is in reasing on [0, π] we have|1 − eiϕ|−1 ≤ |1 − ei π

3 |−1 = 1on [0, π3 ]. The oe ients u

(1)n , vn, and wn are non-alternating for n ≥ 1.Furthermore, for m ≥ 1, we have u

(k)2m > 0 and |u(k)

2m| > |u(k)2m+1|, k = 2, 3, 4.Sin e we have

|1 − eiϕ| ≤ |1 − ei π3 | = 1on [0, π

3 ] we an on lude that there|p(k)

N (eiϕ)| ≤ |p(k)N (ei π

3 )| =: δ(k)N , k = 1, 2, 3, 4,

|qN (eiϕ)| ≤ |qN (ei π3 )| =: εN ,

|rN (eiϕ)| ≤ |rN (ei π3 )| =: κNis satised. Finally, we have

|P (k)N (eiϕ)| ≤ πk, k = 1, 2, 3, 4,

|1 + eiϕ| ≤ 2, |QN (eiϕ) ≤ 2.Hen e we obtainη ≤

(4π + 2π3 + 2

√2π3 +

√2π2 + π3 + |c1|

2

√2π4

+ 12√

2π + 16π2 + 4π2 + 3|c2|2

√2π4)κN

+ 2δ(1)N 3−

14 + δ

(3)N 3−

14 + π3εN3−

14 + πδ

(2)N + πδ

(2)N εN3−

14

+ π2εN + δ23− 1

2 + δ(2)N εN

+ δ(3)N + |c1|

4 π4εN3−14 + |c1|

4 δ(4)N + |c1|

4 δ(4)N εN3−

14 + 3πεN

√2 + 4π

√2δ

(1)N

+ π√

2δ(1)N + 3

4 |c2|π4εN3−14 + 3

4 |c2|δ(4)N + 3

4 |c2|δ(4)N εN3−

14 .For N ≥ 12 we have η < 0.01052.Both the fun tions | arcsin(eiϕ)+ π

2 | and |happroxnum (eiϕ)(1−eiϕ)−3/2| for N = 12are de reasing on [0, π

3 ]. In order to estimate the quotient of both we divide[0, π

3 ] into small subintervals of length π300 . For ea h subinterval it is easy toevaluate that the quotient is less than 0.1. Hen e we have

∣∣∣∣∣happrox

num

(eiϕ)

hdenum (eiϕ)

∣∣∣∣∣ < 0.129

Figure 2: ∣∣h(2)(eiϕ)∣∣ for ϕ ∈ [0, 2π].whi h implies together with the estimate for η that also on [0, π

3 ]∣∣h(2)(eiϕ)

∣∣is less than 0.182.6.4 Proof of Lemma 2We have Pt(1) = P(Z(o) ≥ t) = Ψ(t), and a series expansion of Pt(ρ) aroundρ = 1 yields

Pt(ρ) = Ψ(t) +ϕ(t)

2√

π

(−2√

1 − ρ + O((1 − ρ)32 ))

.From this it is easy to see that in aseR(0+) = lim

ε→0+

R(ε) − 1

ε< 0(in luding R(0+) = −∞) we always have

C ′Ξt

(0+) = limε→0+

Pt(R(ε)) − Pt(1)

R(ε) − 1

R(ε) − 1

ε

=ϕ(t)

2√

πlim

ε→0+

2√1 − R(ε)

R(ε) − 1

ε= −∞.In ase R(0+) = 0, whi h means that the ovarian e fun tion of Z is dier-entiable at the origin, dene

R′′(0+) = 2 limε→0+

R(ε) − 1

ε2.30

Then we haveC ′

Ξt(0+) = −ϕ(t)√

2πlim

ε→0+

√1 − R(ε)

ε= −ϕ(t)√

2π

√−R′′(0+).Hen e, if R′′(0+) = −∞, whi h means that the ovarian e fun tion of Z isnot twi e dierentiable at the origin, then also C ′

Ξt(0+) = −∞. Otherwisewe are in the ase where R′′(0+) is nite and Z by denition mean-squaredierentiable [1.6.5 Proof of Theorem 4The de ompositions Et(ρ) = tPt(ρ)+Et;2(ρ) and Ct(ρ) = t2Pt(ρ)+2tEt;2(ρ)+

Ct;3(ρ) (see subse tion 6.1) imply that ft simplies toft(ρ) =

Ct;3(ρ)

Pt(ρ)− Et;2(ρ)2

Pt(ρ)2.Sin e C ′

t;3(ρ) = Pt(ρ) we obtaincov′(0+) =

(1 −

2Et;2(1)E′t;2(1)

Pt(1)2

)R′(0+) +

2Et;2(1)2 − Ct;3(1)Pt(1)

Pt(1)3C ′

Ξt(0+).We have Pt(1) = Ψ(t), Et(1) = ϕ(t), Ct(1) = tϕ(t) + Ψ(t), and

E′t,2(1) = t

1∫

0

ϕ(t, t, s)

1 + sds + ϕ(t)Ψ(t).Hen e, Et;2(1) = ϕ(t) − tΨ(t) and Ct;3(1) = (1 + t2)Ψ(t) − tϕ(t). By meansof eq. (14) we get

Ψ2(t) − 2Et;2(1)E′t;2(1)

= Ψ2(t) − (ϕ(t) − tΨ(t))ϕ(t)

≥ ϕ2(t)[(t−1 − t−3 + 3t−5 − 15t−7)2 − t−2 + 3t−4 − 15t−6

]

= ϕ2(t)

[1

t4− 8

t6− 36

t8+

39

t10− 90

t12+

225

t14

]> 0, t ≥ 4,sin e

t−1 − t−3 + 3t−5 − 15t−7 ≤ Ψ(t)/ϕ(t), t ≥ 0.31

Numeri al inspe tion for t ∈ [0, 4] yields that 1−2[Pt(1)]−2Et;2(1)E

′t;2(1)is positive for all t ≥ 0. Similar, but simpler argumentation holds for t < 0.With

−945t−11 ≤ Ψ(t)/ϕ(t) − (t−1 − t−3 + 3t−5 − 15t−7 + 105t−9) ≤ 0, t ≥ 0,we have2Et;2(1)

2 − Ct;3(1)Pt(1)

= (t2 − 1)Ψ(t)2 − 3tϕ(t)Ψ(t) + 2ϕ(t)2

≥ ϕ2(t)[(t2 − 1)(t−1 − t−3 + 3t−5 − 15t−7 + 105t−9 − 945t−11)2

− 3t(t−1 − t−3 + 3t−5 − 15t−7 + 105t−9) + 2]

= ϕ2(t)

[2

t6− 30

t8− 2439

t10+

4935

t12− 11565

t14+

48195

t16

− 237825

t18+

1091475

t20− 893025

t22

]

> 0, t ≥ 8.Again, numeri al inspe tion and similar onsiderations for t < 0 yield that2Et;2(1)

2 − Ct;3(1)Pt(1) > 0, t ∈ R.Sin e R′′(0+) = 0 implies that R is onstant [21, Lemma,1.10.16, Lemma2 yields that cov′(0+) ∈ [−∞, 0). Further, cov′(0+) = −∞ if and only if Zis not mean-square dierentiable.7 A knowledgementZakhar Kablu hko has been supported by DFG-SNF FOR 916 Statisti alRegularisation and Qualitative Constraints.Referen es[1 Adler, R. J. (1981). The Geometry of Random Fields. John Wiley &Sons, London.[2 Adler, R. J., and Taylor, J. E. (2007). Random Fields and Geometry.Springer, New York.[3 Bene², V., and Rataj, J. (2004). Sto hasti Geometry: Sele ted Topi s.Kluwer, Boston. 32

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