Statistical inference versus mean field limit for Hawkes processes

Transcription

1 Eleconic Jounal of Saisics Vol ISS: DOI: /16-EJS1142 Saisical infeence vesus mean field limi fo Hawkes pocesses Sylvain Delae Laboaoie de Pobabiliés e Modèles Aléaoies, UMR 7599, Univesié Pais Dideo, Case couie 712, avenue de Fance, 7525 Pais Cedex 13, Fance sylvain.delae@univ-pais-dideo.f and icolas Founie Laboaoie de Pobabiliés e Modèles Aléaoies, UMR 7599, Univesié Piee-e-Maie Cuie, Case 188, 4 place Jussieu, F Pais Cedex 5, Fance nicolas.founie@upmc.f Absac: We conside a populaion of individuals, of which we obseve he numbe of acions unil ime. Fo each couple of individuals i, j, j may o no influence i, which we model by i.i.d. Benoullip-andom vaiables, fo some unknown paamee p, 1]. Each individual acs auonomously a some unknown ae μ>andacsby mimeism a some ae popoional o he sum of some funcion ϕ of he ages of he acions of he individuals which influence him. The funcion ϕ is unknown bu assumed, oughly, o be deceasing and wih fas decay. The goal of his pape is o esimae p, which is he main chaaceisic of he gaph of ineacions, in he asympoic,. The main issue is ha he mean field limi as of his model is unidenifiable, in ha i only depends on he paamees μ and pϕ. Founaely, his mean field limi is no valid fo lage imes. We disinguish he subciical case, whee, oughly, he mean numbe m of acions pe individual inceases linealy and he supeciical case, whee m inceases exponenially. Alhough he nuisance paamee ϕ is non-paameic, we ae able, in boh cases, o esimae p wihou esimaing ϕ in a nonpaameic way, wih a pecision of ode 1/2 + 1/2 m 1, up o some abiaily small loss. We explain, using a Gaussian oy model, he eason why his ae of convegence migh be almos opimal. MSC 21 subjec classificaions: 62M9, 6J75, 6K35. Keywods and phases: Mulivaiae Hawkes pocesses, poin pocesses, saisical infeence, ineacion gaph, sochasic ineacing paicle sysems, popagaion of chaos, mean field limi. Received Sepembe Inoducion and main esuls 1.1. Seing We conside some unknown paamees p, 1], μ>andϕ :[, [,. Fo 1, we conside an i.i.d. family π i d, dz,..., of Poisson measues 1223

2 1224 S. Delae and. Founie on [, [, wih inensiy measue ddz, independen of an i.i.d. family θ ij i,j=1,..., of Benoullip-disibued andom vaiables. We also conside he sysem of equaions, fo i =1,...,, whee Z i, = λ i, = μ {z λ i, s } πi ds, dz, θ ij j=1 ϕ sdz j, s. 1 Hee and in he whole pape, means [,] and means. The soluion [, Z i,,..., is a family of couning pocesses ha is, a.s. inegevalued, càdlàg and non-deceasing. The following well-posedness esul is moe o less well-known, see e.g. Bémaud-Massoulié [9] and[13] we will apply diecly he lae efeence. Poposiion 1. Assume ha ϕ is locally inegable and fix 1. The sysem 1.1 hasauniquecàdlàg F -adaped soluion Z i,,..., such ha E[Zi, ] < fo all, wheef = σπ i A :A B[,] [,,i= 1,..., σθ ij : i, j =1,...,. Le us povide a bief heuisic descipion of his pocess. We have individuals and Z i, sands fo he numbe of acions of he i-h individual unil. We say ha j influences i if and only if θ ij = 1 wih possibly i = j. Each individual i acs, a ime, wih ae λ i,. In ohe wods, each individual has an auonomous ae of acion μ as well as a subodinae ae of acion 1 j=1 θ j, ij ϕ sdz s, which depends on he numbe of acions of he individuals ha influence him, wih a weigh 1 and aking ino accoun he age of hese acions hough ϕ. If fo example ϕ = a1 [,K], hen he subodinae ae of acion of i is simply a/ imes he oal numbe of acions, duing [ K, ], of all he individuals ha influence him. As is well-known, a phase-ansiion occus fo such a model, see Hawkes- Oakes [18] o [13] fo such consideaions on lage newoks: seing Λ = ϕd, in he subciical case whee Λp <1, we will see ha Z 1, inceases linealy wih ime, a leas on he even whee he family θ ij i,j=1,..., behaves easonably; in he supeciical case whee Λp >1, we will see ha Z 1, inceases exponenially fas wih ime, a leas on he even whee he family θ ij i,j=1,..., behaves easonably. The limi heoems, and hus he saisical infeence, compleely diffe in boh cases, so ha he pesen pape conains essenially wo independen pas. We will no sudy he ciical case whee Λp = 1 because i is a vey paicula case. Howeve, i would be vey ineesing o undesand wha happens nea he ciical case. Ou esuls say nohing abou his poblem.

3 Hawkes pocesses Assumpions Recalling ha Λ = ϕsds, we will wok unde one of he wo following condiions: eihe fo some q 1, o μ,, Λp, 1 and s q ϕsds < Hq μ,, Λp 1, and dϕs inceases a mos polynomially. A In many applicaions, ϕ is smooh and has a fas decay, so ha, excep in he ciical case, eihe Hq is saisfied fo all q 1oA is saisfied Refeences and fields of applicaion Hawkes pocesses have been inoduced by Hawkes [17] and Oakes-Hawkes [18] have found a noiceable epesenaion of such pocesses in ems of Galon- Wason ees. Since hen, hee has been a huge lieaue on Hawkes pocesses, see e.g. Daley and Vee-Jones [12] fo an inoducion, Massoulié [24], Bémaud- Massoulié [9] and[13] fo sabiliy esuls, Bémaud-appo-Toisi [1], Zhu [35, 36] and[3] fo limi heoems, ec. Hawkes pocesses ae used in vaious fields of applicaions: eahquake eplicas in seismology, see Helmsee-Sonee [19], Kagan [23], Ogaa [26], Bacy-Muzy [5], spike ains fo bain aciviy in neuoscience, see Gün e al. [15], Okaan e al. [27], Pillow e al. [28], Reynaud e al. [31, 32], genome analysis, see Reynaud-Schbah [3], vaious fields of mahemaical finance, see Aï-Sahalia e al. [1], Bauwens- Hausch [6], Hewle [2], Bacy e al. [2], Bacy-Muzy [4, 5], social newoks ineacions, see Blundell e al. [8] and Zhou e al. [34]. Concening he saisical infeence fo Hawkes pocesses, only he case of fixed finie dimension has been sudied, o ou knowledge, in he asympoic fo possibly moe geneal shapes of ineacion. Some paameic and nonpaameic esimaion pocedues fo μ and ϕ have been poposed, wih o wihou igoous poofs. Le us menion Ogaa [25], Bacy-Muzzy [5], [2], he vaious ecen esuls of Hansen e al. [16] andreynaudeal.[3, 31, 32], as well as he Bayesian sudy of Rasmussen [29] Goals and moivaion In many applicaions, he numbe of individuals is vey lage hink of neuons, financial agens o of social newoks. Then we need some esimaos in he asympoic whee and end simulaneously o infiniy. This poblem seems o be compleely open.

4 1226 S. Delae and. Founie We assume ha we obseve Z i, s,...,,s [,] o Zs i,,...,,s [,2] fo convenience, ha is all he acions of he individuals on some lage ime ineval. In ou poin of view, we only obseve he aciviy of he individuals, we do no know he gaph of ineacions. A vey simila poblem was sudied in [32], alhough in fixed finie dimension. Ou goal is o esimae p, which can be seen as he main chaaceisic of he gaph of ineacions, since i epesens he popoion of open edges. We conside μ and ϕ as nuisance paamees, alhough his is debaable. In he supeciical case, we will be able o esimae p wihou esimaing μ no ϕ. In he subciical case, we will be able o ecove p esimaing only μ and he inegal Λ of ϕ. In any case, we will no need o povide a nonpaameic esimaion of ϕ, and we believe i is a vey good poin: i would equie egulaiy assumpions and would complicae a lo he sudy. The main goal of his pape is o povide he basic ools fo he saisical esimaion of Hawkes pocesses when boh he gaph size and he obsevaion ime incease. Of couse, his is only a oy model and we have no pecise idea of eal wold applicaions, alhough we can hink e.g. of neuons spiking: hey ae clealy numeous so is lage, we can only obseve hei aciviies each ime hey spike, and we would like o have an idea of he gaph of ineacions. See again [32] fo a moe convincing biological backgound. Think also of financial agens: hey ae also numeous, we can obseve hei acions each ime hey buy o sell a poduc, and we would like o ecove he ineacion gaph Mean field limi We quickly descibe he expeced chaoic behavio of Z i,,..., as. We efe o Szniman [33] fo an inoducion o popagaion of chaos. Exending he mehod of [13, Theoem 8], i is no had o check, assuming ha ϕ 2 sds <, ha fo each given k 1andT >, he sample [,T ],...,k goes in law, as, o a family Y i [,T ],...,k of i.i.d. inhomogeneous Poisson pocesses wih inensiy λ, unique locally Z i, bounded nonnegaive soluion o λ = μ + pϕ sλ sds. On he one hand, appoximae independence is of couse a good poin fo saisical infeence. On he ohe hand, he mean-field limi i.e. he Y i s depends on p and ϕ only hough λ and hus hough pϕ, whichisa negaive poin: he mean-field limi is unidenifiable. The siuaion is howeve no hopeless because oughly, he mean-field limi does no hold ue fo he whole sample Z i,,..., and is less and less ue as ime becomes lage and lage Main esul in he subciical case Fo 1andfoZ i,,..., he soluion o 1.1, we inoduce Z = 1. We menion in he following emak, ha we will pove lae, Zi,

5 Hawkes pocesses 1227 ha he numbe of acions pe individual inceases linealy in he subciical case. Remak 2. Assume H1. Then fo all ε>, lim P Z,, We nex inoduce E = Z 2 Z, V = μ ε =. 1 Λp Z i, 2 W Δ, =2Z 2Δ, Z Δ,, whee Z Δ, = Z i, 2/Δ k=/δ+1 E 2 E, Z kδ Z k 1Δ ΔE In he las expession, Δ, is equied o be such ha /2Δ. Theoem 3. We assume Hq fo some q > 3. Fo 1, we pu Δ = /2 1 4/q+1 : i holds ha /2Δ and ha Δ 4/q+1 /2 as. Thee is a consan C depending only on p, μ, ϕ and q such ha fo all ε, 1, all 1, all 1, E P μ ε C 1 1 Λp ε , q V P μ2 Λ 2 p1 p ε 1 Λp 2 C + 1, ε W μ P Δ, ε 1 Λp 3 C 1 ε /q+1 We will easily deduce he following coollay. Coollay 4. We assume Hq fo some q > 3. Fo 1, wepuδ = /2 1 4/q+1. Thee is a consan C depending only on p, μ, ϕ and q such ha fo all ε, 1, all 1, all 1, Ψ P E, V, WΔ, μ, Λ,p ε C ε /q+1 2C 1 +, ε 1 4/q+1 whee Ψ=1 D Φ wih D = {u, v, w R 3 : w>u>and v } and Φ:D R 3 defined by u Φ 1 u, v, w =u w, Φ 2u, v, w = v +[u Φ 1u, v, w] 2, u[u Φ 1 u, v, w] Φ 3 u, v, w = 1 u 1 Φ 1 u, v, w. Φ 2 u, v, w We did no opimize he dependence in q: in many applicaions, Hq holds fo all q 1.

6 1228 S. Delae and. Founie 1.7. Main esul in he supeciical case Fo 1andfoZ i,,..., he soluion o 1.1, we se Z = 1 Zi,. We will check lae he following emak, which saes ha he mean numbe of acions pe individual inceases exponenially in he supeciical case. Remak 5. Assume A and conside α > defined by p e α ϕd =1. Then fo all η>, lim lim P Z [e α η,e α+η ] =1. We nex inoduce [ Z i, U = Z 2 ] Z Z 1 { Z >} and P 1 = U +1 1 {U }. Theoem 6. Assume A and conside α > defined in Remak 5. Fo all η>, hee is a consan C η > depending only on p, μ, ϕ, η such ha fo all 1, all 1, allε, 1, P P p ε C ηe η ε e α Deecing subciicaliy and supeciicaliy In pacise, we may of couse no know if we ae in he subciical o supeciical case. Poposiion 7. i Unde H1, hee ae some consans <c<cdepending only on p, μ, ϕ such ha fo all 1, all 1, Plog Z log 2 Ce c + 1/2 e log 2. ii Unde A, fo all η>, hee is a consan C η depending only on p, μ, ϕ, η such ha fo all 1, all 1, Plog Z log 2 C η e η 1/2 + e α. I is hen no had o check ha, wih he noaion of Coollay 4 and Theoem 6, unde Hq fosomeq>3 o A, he esimao ˆp = 1 {log Z <log }Ψ 2 3 E/2, V /2, W Δ /2,/2 +1 {log Z >log }P 2, which is based on he obsevaion of Zs i, s [,],,...,, conveges in pobabiliy o p, wih he same speed of convegence as in Coollay 4 unde Hq fo some q>3 o as in Theoem 6 unde A Abou opimaliy In Subecion 2.3, we will see on a oy model ha hee is no eal hope o find an esimao of p wih a bee pecision han 1/2 + 1/2 m 1,whee m is somehing like he mean numbe of jumps pe individual duing [,].

7 Hawkes pocesses 1229 Consequenly, we believe ha he pecision we found in Coollay 4 is almos opimal, since hen m by Remak 2 and since we each he pecision 1/2 + 1/2 α 1 fo any α>ifϕ has a fas decay, so ha he loss is abiaily small. Similaly, he pecision found in Theoem 6 is ahe saisfying, since hen m e α by Remak 5 and since we each he pecision e η 1/2 + 1/2 e α fo any η>, so ha he loss is, hee also, abiaily small. The main defaul of he pesen pape is ha he consans in Coollay 4 and in Theoem 6 songly depend on he paamees μ, Λ,p. They also depend on q in he subciical case. In paicula, i would be quie delicae o undesand how hey behave when appoaching, fom below o fom above, he ciical case Abou he modeling Thee ae wo main limiaions in ou seing. Assuming ha he θ ij s ae i.i.d. is of couse a song assumpion. Wha we eally need is ha he family θ ij i,j=1,..., saisfies simila popeies as hose shown in Subsecion 4.1 in he subciical case and in Subsecion 5.1 in he supeciical case. This clealy equies ha he family θ ij i,j=1,..., is no oo fa fom being i.i.d., and i does no suffice ha lim 2 i,j=1 θ ij = p. Howeve, we believe ha all he conclusions of he pesen pape ae sill ue if one assumes ha θ ij 1 i j is i.i.d. and ha θ ji = θ ij fo all 1 i< j, which migh be he case in some applicaions whee he ineacions ae symmeic. A igoous poof would equie some wok bu should no be oo had. We will sudy his poblem numeically a he end of he pape. Assuming ha we obseve all he populaion is also ahe singen. I would be ineesing o sudy wha happens if one obseves only Zs i,,...,k,s [,], fo some K lage bu smalle han. I is no difficul o guess how o adap he esimaos o such a siuaion see Secion 7 fo pecise fomulae. The heoeical analysis would equie a caeful and edious sudy. Again, we will discuss his numeically oaion We denoe by P θ he condiional pobabiliy knowing θ ij i,j=1,...,.weinoduce E θ, Va θ and Cov θ accodingly. Fo wo funcions f,g :[, R, we inoduce if i exiss f g = f sgsds. The funcions ϕ n will play an impoan ole in he pape. Obseve ha, since ϕsds =Λ, ϕ n sds =Λ n. We adop he convenions ϕ sds = δ ds andϕ sds = δ ds. We also adop he convenion ha ϕ n s =fos<. All he finie consans used in he uppebounds ae denoed by C, heposiive consans used in he lowebounds ae denoed by c and hei values change fom line o line. They ae allowed o depend only on μ, p and ϕ and on q unde Hq, bu neve on no on. Any ohe dependence will be indicaed

8 123 S. Delae and. Founie in subscip. Fo example, C η is a finie consan depending only on μ, p, ϕ and η and on q unde Hq Plan of he pape In he nex secion, we y o give he main easons why ou esimaos should be convegen, which should help he eade o undesand he saegies of he poofs. We also biefly and fomally inoduce a Gaussian oy model in Secion 2.3 o show ha he aes of convegence we obain ae no fa fom being he bes we can hope fo. In Secion 3, we pove Poposiion 1 song exisence and uniqueness of he pocess and check a few moe o less explici,...,, of consan use. Secion 4 is devoed o he poof of Theoem 3 and Coollay 4 main esuls in he subciical case. Theoem 6 main esul in he supeciical case is poved in Secion 5. We check Poposiion 7 in Secion 6. Finally, we illusae numeically he esuls of he pape and some possible exensions in he las secion. fomulae concening Z i, 2. Heuisics This secion is compleely infomal and he symbol means nohing pecise. Fo example, Z i, E θ [Z i, ]folage should be undesood as we hope ha Z i, /E θ [Z i, ] ends o 1 as in pobabiliy o in anohe sense The subciical case We assume ha Λp [, 1 and y o explain oughly he asympoics of Z i,,...,, and whee he hee esimaos E, V and WΔ, come fom. We inoduce he maices A i, j = 1 θ ij and Q =I ΛA 1,which exiss wih high pobabiliy because Λp <1. We also se l i = j=1 Q i, j and c i = j=1 Q j, i. Fixing and knowing θ ij i,j=1,...,, we expec ha Z i, E θ [Z i, ]fo lage by a law of lage numbes. ex, i is no had o check ha E θ [Z i, ]=μ+ 1 j=1 θ ij ϕ se θ[zs j, ]ds. Assume ha γ i = lim 1 E θ [Z i, ] exiss fo each i =1,...,. Then, using ha ϕ ssds Λ fo lage, we find ha he veco γ mus solve γ = μ1 +ΛA γ,whee1 is he -dimensional veco wih all coodinaes equal o 1. This implies ha γ = μi ΛA 1 1 = μl. We hus expec ha Z i, E θ [Z i, ] μl i. Based on his and seing l = 1 l i, we expec ha Z μ l fo lage values of, whence Ẽ := 1 Z μ l. Knowing θ ij i,j=1,...,, Z 1, should esemble, oughly, a Poisson pocess, so ha i should appoximaely hold ue ha Va θ Z 1, E θ [Z 1, ]. Consequenly, 1 Zi, Z 2 should esemble Va Z 1, =Va E θ [Z 1, ]+

9 Hawkes pocesses 1231 E[Va θ Z 1, ] Va E θ [Z 1, ] + E[Z 1, ], which iself should be appoximaely equal o 1 E θ[z i, ] E θ [ Z ] 2 + Z 1 μ 2 2 l i l 2 + Z. All in all, we expec ha Ṽ := 2 [ Zi, Z 2 Z ] μ 2 l i l 2 fo lage. The empoal empiical vaiance Δ 1 /Δ k=1 [ Z kδ Z k 1Δ Δ 1 Z ] 2 should esemble Va θ [ Z Δ ]if1 Δ. Thus W Δ, := 1 /Δ k=1 [ Z kδ Z k 1Δ Δ 1 Z ] 2 Δ 1 Va θ [ Z Δ i, ]. Inoducing he maingales M = Z i, C i, whee C i, is he compensao of Z i,, he ceneed pocesses U i, = Z i, E θ [Z i, ], and he -dimensional vecos U and M wih coodinaes U i, and M i,, we will see in Secion 3 ha U = M + A ϕ su s ds, so ha fo lage imes, U M +ΛA U and hus U Q M. μl j, so ha Va θ Q M μ 2 j=1 Q i, j 2 l j, which is nohing bu μ 2 j=1 c j 2 l j. Finally, Va θ [ Z ] = Va θ [Ū ] hus esembles μ 2 j=1 c j 2 l j and we hope ha W Δ, Δ 1 Va θ [ Z Δ ] μ 1 j=1 c j 2 l j if1 Δ. We need o find he limis as of l, Consequenly, we hope ha Ū Q M, whee Ū = 1 U i, and Q M = 1 Q M i. A lile sudy shows ha he maingales M j, ae ohogonal and ha [M j,,m j, ] = Z j, l i l 2 and 1 l ic i 2. I is no easy o make igoous, bu i holds ue ha l i 1+Λ1 Λp 1 L i, whee L i = j=1 A i, j. This comes fom j=1 A2 i, j = j=1 A i, j k=1 A j, k p j=1 A i, j =pl i, j=1 A3 i, j p2 L i fo simila easons, ec. I is vey ough, bu i will imply ha l i = n Λn j=1 An i, j 1+ n 1 Λn p n 1 L i =1+Λ1 Λp 1 L i. Once his is seen as well as a simila fac fo he columns, we ge convinced, L being a veco of i.i.d. Binomial,p-disibued andom vaiables, ha l 1/1 Λp, ha l i l 2 Λ 2 p1 p/1 Λp 2 and ha 1 l ic i 2 1/1 Λp 3. A he end, i should be moe o less ue ha, fo, Δand lage enough and in a suiable egime, Ẽ μ/1 Λp, Ṽ μ 2 Λ 2 p1 p/1 Λp 2,and W Δ, μ/1 Λp3. Of couse, all his is compleely infomal and many poins have o be claified. Obseve ha concening Ṽ, we use ha Z 1, while concening W Δ,, we use ha Z esembles a Poisson pocess, does no esemble a Poisson pocess. The hee esimaos E, V, WΔ, we sudy in he pape esemble much Ẽ, Ṽ, W Δ, and should convege o he same limis. Le us explain why we have modified he expessions. We saed his subsecion by he obsevaion ha E θ [Z i, ] μl i, on which he consucion of he esimaos elies. A deailed sudy shows ha, unde Hq, E θ [Z i, ]=μl i + χ i ± 1 q,fo some finie andom vaiable χ i. As a consequence, 1 E θ [Z i, 2 Z i, ] con-

10 1232 S. Delae and. Founie veges o μl i consideably fase wih an eo in q han 1 E θ [Z i, ] fo which he eo is of ode 1. This explains ou modificaions and why hese modificaions ae cucial. Le us conclude his subsecion wih a echnical issue. If Λ > 1whichisno fobidden even in he subciical case, hee is a posiive pobabiliy ha an anomalously high popoion of he θ ij s equal 1, so ha I ΛA is no inveible and ou mulivaiae Hawkes pocess is supeciical on his even wih small pobabiliy. We will hus wok on an even Ω 1 on which such poblems do no occu and show ha his even has a high pobabiliy The supeciical case We now assume ha Λp >1 and explain he asympoics of Z i,,...,, and whee he esimao U comes fom. We inoduce A i, j = 1 θ ij. Fixing and knowing θ ij i,j=1,...,, we expec ha Z i, H E θ [Z i, ], fo some andom H > no depending on i and wih H almos consan fo lage. This is ypically a supeciical phenomenon, ha can aleady be obseved on Galon-Wason pocesses. Founaely, we will no eally need o check i no o sudy H, essenially because we will use he aios Z i, / Z, which makes disappea H. ex, we believe ha E θ [Z i, ] γ ie α fo lage, fo some veco γ wih posiive enies and some exponen α >. Inseing his ino E θ [Z i, ]= μ + 1 j=1 θ ij ϕ se θ[zs j, ]ds, we find γ = A γ e α s ϕsds. The veco γ being posiive, i is necessaily a Peon-Fobenius eigenveco of A,sohaρ = e α s ϕsds 1 is is Peon-Fobenius eigenvalue i.e. is specal adius. We now conside he nomalized Peon-Fobenius eigenveco V such ha V i 2 = and conclude ha Z i, fo all i =1,...,,wheeK =[ 1 γ i 2 ] 1/2 H. K V ie α As in he subciical case, he empiical vaiance 1 Zi, Z 2 esembles 1 E θ[z i, ] E θ [ Z ] 2 + Z 1 K 2 e2α V i V 2 + Z.Wealsoguessha Z K V e α,whee V = 1 V i. Hence we expec, ha fo lage, U = Z 2 [ Zi, Z 2 Z ] V 2 V i V 2. We now seach fo he limi of V 2 V i V 2 as. Roughly, A 2 i, j p2 /, whence, saing fom A 2 V = ρ 2 V, we see ha ρ 2 V p 2 V 1,whee1 is he -dimensional veco wih all coodinaes equal o 1. Consequenly, V =A V /ρ κ A 1,wheeκ = p 2 /ρ 3 V. In ohe wods, V is almos colinea o L := A 1,andL is a veco of i.i.d. Binomial,p-disibued andom vaiables. I is hus easonable o expec ha V 2 V i V 2 L 2 L i L 2 p 2 p1 p =1/p 1. All in all, we hope ha fo and lage and in a suiable egime, U 1/p 1.

11 Hawkes pocesses 1233 Le us menion ha α α ecall Remak 5 because e α s ϕsds = 1/ρ, because e αs ϕsds =1/p and because ρ p. This las asseion follows fom he fac ha A 2 i, j p2 /, so ha he lages eigenvalue of A 2 should esemble p 2, whence ha of A should esemble p. Of couse, all his is no clea and has o be made igoous. Le us menion ha we will use a quanified vesion of he Peon-Fobenius of G. Bikhoff [7]. As we will see, he pojecion ono he eigenveco V will be vey fas almos immediae fo vey lage. As in he subciical case, we will have o wok on an even Ω 2,ofhigh pobabiliy, on which he θ ij s behave easonably. Fo example, o apply he Peon-Fobenius heoem, we have o be sue ha he maix A is ieducible, which is no a.s. ue Abou opimaliy: A elaed oy model Conside α and wo unknown paamees Γ > andp, 1]. Fo 1, conside an i.i.d. family θ ij i,j=1,..., of Benoullip-disibued andom vaiables, pu λ i, = 1 Γe α j=1 θ ij and, condiionally on θ ij i,j=1,...,,afamily Z 1,,...,Z, of independen inhomogeneous Poisson pocesses wih inensiies λ 1,,...,λ,.WeobseveZs i, s [,],,..., and we wan o esimae p in he asympoic,,. This poblem can be seen as a songly simplified vesion of he one sudied in he pesen pape, wih α = in he subciical case and α > inhe supeciical case. Roughly, he mean numbe of jumps pe individual esembles m = eαs ds, which is of ode when α =ande α else. Thee is classically no loss of infomaion, since α is known, if we only obseve,..., : afe a deeminisic and known change of ime, he pocesses Z i,,..., become homogeneous Poisson pocesses wih unkown paamees condiionally on θ ij i,j=1,...,, and he condiional law of a Poisson pocess on [,] knowing is value a ime does no depend on is paamee. We nex poceed o a Gaussian appoximaion: we have λ i, Γe α [p + Z i, 1 p1 p]g i and Z i, λi, s ds + λi, s dsh i, fo wo independen i.i.d. families G i,...,,h i,..., of, 1-disibued andom vaiables. Using finally ha m 1 1/2 m 1 in ou asympoic, we conclude ha m 1 Z i, Γp +Γ 1 p1 pg i + m 1 ΓpH i, of which he law is Γp, 1 Γ 2 p1 p+m 1 Γp. Ou oy poblem is hus he following: esimae p when obseving a -sample X i,,..., of he Γp, 1 Γ 2 p1 p+m 1 Γp-disibuion. We assume ha Γp is known, which can only make easie he esimaion of p. As is wellknown he saisic S = 1 Xi, Γp 2 is hen sufficien and is he bes esimao in all he usual senses, fo 1and 1fixed,of 1 Γ 2 p1 p+m 1 Γp, sohat = Γp 2 S m 1 Γp is moe o less he bes esimao of 1/p 1. Bu Va S = Γ 2 p1 p +m 1 Γp 2, whence Va T = 2Γp 4 1/2 Γ 2 p1 p+ 1/2 m 1 Γp 2.Iishusno

12 1234 S. Delae and. Founie possible o esimae 1/p 1 wih a bee pecision han 1/2 + 1/2 m 1. This of couse implies ha we canno esimae p wih a bee pecision han 1/2 + 1/2 m Well-posedness and explici fomulae We fis give he Poof of Poposiion 1. Condiionally on θ ij i,j=1,...,, we can apply diecly [13, Theoem 6], of which he assumpion is saisfied hee, see [13, Remak 5- i]: condiionally on θ ij i,j=1,...,, hee is a unique soluion Z i,,,..., o 1.1 such ha E θ[z i, ] < fo all. Since now θ ij i,j=1,..., can only ake a finie numbe of values, we immediaely deduce ha indeed E[Zi, ] < fo all. We cay on wih a classical lemma. Recall ha ϕ sds = δ ds by convenion. Lemma 8. Conside d 1, A M d d R, m, g :[, R d locally bounded and assume ha ϕ :[, [, is locally inegable. If m = g + ϕ sam s ds fo all, henm = n ϕ n sa n g s ds. Poof. The equaion m = g + ϕ sam sds wih unkown m has a mos one locally bounded soluion. Indeed, conside wo such soluions m, m, obseve ha u = m m saisfies u A ϕ su sds, and conclude ha u = by he genealized Gonwall lemma, see e.g. [13, Lemma 23-i]. We hus jus have o pove ha m := n ϕ n sa n g s ds is locally bounded and solves m = g + Aϕ m. We inoduce k n = A n ϕ n sds, which is locally bounded because ϕ is locally inegable and which saisfies k n+1 A kn s ϕ sds. Weuse[13, Lemma 23-ii] o conclude ha n kn is locally bounded. Consequenly, m sup [,] g s n kn is locally bounded. Finally, we wie m = g+ n 1 An ϕ n g = g+aϕ n An ϕ n g = g+aϕ m as desied. We nex inoduce a few pocesses. oaion 9. Assume only ha ϕ is locally inegable, fix 1 and conside he soluion Z i,,,..., o 1.1. Foeachi =1,...,,weinoducehe maingale ecall ha λ i, was defined in 1.1 M i, = 1 {z λ i, } πi s ds, dz, whee π i ds, dz =π i ds, dz dsdz is he compensaed Poisson measue associaed o π i.wealsoinoducem i,, =sup [,] Ms i,, as well as he condiionally ceneed pocess U i, = Z i, E θ [Z i, ].

13 Fo each, wedenoebyz veco wih coodinaes Z i, Hawkes pocesses 1235 esp. M, M, esp. M i,, M i,,, U he-dimensional, U i,. We also se Z = 1 Zi,, M = 1 M i, and Ū = 1 U i, We efe o Jacod-Shiyaev [22, Chape 1, Secion 4e] fo definiions and popeies of pue jump maingales and of hei quadaic vaiaions. Remak 1. Since he Poisson measues π i ae independen, he maingales M i, ae ohogonal. Moe pecisely, we have [M i,,m j, ] =if i j, while [M i,,m i, ] = Z i, because Z i, couns he jumps of M i,, which ae all of size 1. Consequenly, E θ [Ms i, M j, ]=1 {i=j} E θ [Z i, s ]. We now give some moe o less explici fomulas. We denoe by 1 he - dimensional veco wih all enies equal o 1 and we se A i, j = 1 θ ij fo i, j =1,...,. Lemma 11. Assume only ha ϕ is locally inegable. We have ecall ha ϕ sds = δ ds: Z =M + μ1 +. ϕ sa Z s ds, 2 E θ [Z ]=μ [ ] sϕ n sds A n 1, 3 n U = ϕ n sa n M s ds. 4 n Poof. The fis expession is no difficul: saing fom 1.1, Z i, = M i, + λ i, s ds = M i, + μ + Using [13, Lemma 22], we see ha whence indeed, Z i, = M i, + μ + s A i, j j=1 ϕs udz j, u ϕ s j=1 s ds = A i, jzs j, ds, ϕs udzu j, ds. ϕ szj, s ds, which is nohing bu 2. Taking condiional expecaions in 2, we find ha E θ [Z ]=μ1 + ϕ sa E θ [Z s ]ds and hus also U = M + ϕ sa U s ds. Sincenowϕ is a.s. locally inegable, since μ1 and M ae a.s. locally bounded, as well as E θ [Z ]andu,3 and4 diecly follow fom Lemma The subciical case Hee we conside he subciical case. We fis sudy he lage -asympoic of he maix Q =I ΛA 1, which plays a cenal ole in he es of he

14 1236 S. Delae and. Founie secion. In Subsecion 4.2, we finely sudy he behavio of ϕ n. In Subsecion 4.3, we handle a few compuaions o be used seveal imes lae. Subsecions 4.4, 4.5 and 4.6 ae devoed o he sudies of he hee esimaos E, V and WΔ,. We conclude he poofs of Theoem 3 and Coollay 4 in Subsecion Sudy of a andom maix We use he following sandad noaion: fo x =x 1,...,x R and [1,, we se x = x i 1/ and x =max,..., x i.fo [1, ], we denoe by he opeao nom on M R associaed o. We ecall ha M 1 = sup j=1,..., and ha fo all 1,, M ij, M = sup,..., M ij j=1 M M 1/ 1 M 1 1/. 5 oaion 12. We assume ha Λp <1. Foeach 1, weinoducehe andom maix A defined by A i, j = 1 θ ij, as well as he even Ω 1 = { } Λ A a fo all [1, ], whee a = 1+Λp Λp, On Ω 1,he maix Q = n Λn A n =I ΛA 1 is well-defined and we inoduce, fo each i = 1,...,, l i = j=1 Q i, j, c i = j=1 Q j, i, aswellas l = 1 l i and c = 1 c i. We of couse have l = c. Le us emak once fo all ha, wih C =1/1 a <, { } { Ω 1 Q C fo all [1, ] sup Ω 1 1 i } max{l i,c i} C, 7 { } 1 {i=j} Q i, j 1 {i=j} +ΛC 1 fo all i, j =1,...,. 8 Indeed, 7 is saighfowad since Q = n Λn A n.tocheck8, we fis obseve ha Q i, j Λ A i, j =1 {i=j}. ex, we use ha A i, j 1 while, if n 2, A n i, j = k=1 A i, ka n 1 k, j 1 k=1 An 1 k, j 1 A n A n 1 1.ThusA n i, j 1 A n 1 1 fo all n 1. Hence on Ω 1, i holds ha Q i, j 1 {i=j} + 1 n 1 Λn A n {i=j} + 1 Λ/1 a as desied. Lemma 13. Assume ha Λp <1. I holds ha PΩ 1 1 C exp c.

15 Hawkes pocesses 1237 Poof. By 5, i suffices o pove ha PΛ A 1 >a C exp c and PΛ A >a C exp c. Since A = A 1 and since A he anspose of A has he same law as A, i acually suffices o veify he fis inequaliy. Fis, A 1 =max{x 1,...,X },wheex i = j=1 θ ij is Binomial,p-disibued fo each i. Consequenly, PΛ A 1 > a PX 1 a/λ P X 1 p a/λ p. Since a/λ > p, we can use he Hoeffding inequaliy [21] o obain PΛ A 1 > a 2 exp 2a/Λ p 2 C exp a/λ p 2 as desied. The nex esul is much hade bu cucial. Poposiion 14. Assume ha Λp <1. I holds ha [ E 1 Ω 1 l 1 2 ] C 1 Λp 2, [ 1 E 1 Ω 1 l ic i 2 1 2] 1 Λp 3 C 2, E [1 Ω l 1 i l 2 Λ2 p1 p ] 1 Λp 2 C. Poof. Recall ha 1 is he -dimensional veco of which all he coodinaes equal 1. Le l esp. c be he veco wih coodinaes l 1,...,l esp. c 1,...,c. We also inoduce, fo all i = 1,...,, L i = j=1 A i, j andc i = j=1 A j, i, as well as he coesponding vecos L and C. Le us obseve ha, wih obvious noaion, l = c and L = C. Finally, we inoduce he vecos x = l l 1, y = c c 1, X = L L 1, Y = C C 1. We ecall ha a =1+Λp/2, 1 and we inoduce b =2+Λp/3 a, 1. Sep 1. We inoduce he even { A = L p1 2 + C p1 2 1/4} { X 2 + Y 2 1/4}. The inclusion comes fom he fac ha a.s., X 2 = L L 1 2 L x1 2 fo any x R. SinceL =Z1,...,Z wihz i i.i.d. and Binomial,p-disibued, i is vey classical ha fo any α>, E[ L p1 α 2 ] C α unifomly in, we have similaly E[ C p1 α 2 ] C α,so ha PA 1 C α α/4. Sep 2. We now check he following poins: i E[ L p 2 ] C 2, ii E[ X 4 2] C, iii E[ X 2 2 p1 p 2 ] C 1 and iv E[ A X 2 2] C 1. Poin i is clea, because L = 2 i,j=1 θ ij is nohing bu he empiical mean of 2 independen Benoullip-andom vaiables. Poins ii and iii

16 1238 S. Delae and. Founie ae vey classical, since X 2 2 is he empiical vaiance of independen Binomial,p-andom vaiables. We now pove iv: E[ A X 2 2]= [ E j=1 = 1 E [ j=1 θ ij L j L 2 ] θ 1j L j L 2 ] by symmey. We now wie E[ A X 2 2] 4 1 I + J + K, whee I = E [ L p 2 2 [ 2 ] θ 1j ], J = E θ 11 L 1 p, j=1 [ 2 ] K = E θ 1j L j p. j=2 Fis, I 2 E[ L p 2 ] C by i. ex, i is obvious ha J 1 because θ 11 {, 1} and L 1 [, 1]. Finally, he andom vaiables θ 1j L j p being i.i.d. and ceneed fo j =2,...,, we may wie [ 2 ] [ K = 1E θ 12 L 2 p 1E L 2 p 2] C, since L 2 follows a Binomial,p-disibuion. This complees he sep. Sep 3. We nex pove ha i x =ΛA x Λ 1 +Λ l X on Ω 1, whee = 2 i,j=1 θ ij px j and ha ii 3/4 x 2 on Ω 1 A. We sa fom l = Q 1 =I ΛA 1 1, whence l = 1 +ΛA l. Since l = 1 l, 1, we see ha l =1+Λ 1 A l, 1 hee, is he usual scala poduc on R and hus x =ΛA l Λ 1 A l, 1 1 =ΛA x Λ 1 A x, l ΛA 1 l Λ 1 A 1, 1 1. I only emains o veify ha 1 A x, 1 =, which follows fom he facs ha 1 A x, 1 = 2 i,j=1 θ ijx j, ha j=1 x j = ; and ha A 1 1 A 1, 1 1 = X, which is clea since A 1 = L. To veify ii, we obseve ha = 1 j=1 C j px j, whence, by he Cauchy-Schwaz inequaliy, 1 x 2 C p1 2 3/4 x 2 on Ω 1 A. Sep 4. Le be he smalles inege such ha a +Λ 1/4 b. Wecheck ha fo all, 1 Ω 1 A x 2 C X 2. Using Sep 3 and ha 1 2 = 1/2,wewie x 2 Λ A 2 x 2 + Λ 1/4 x 2 +Λ l X 2. Bu on Ω 1, Λ A 2 a and l C,

17 Hawkes pocesses 1239 see 6 and7. Hence, fo,onω 1 A,wehave x 2 a + Λ 1/4 x 2 + C X 2 b x 2 + C X 2.Sinceb<1, he conclusion follows. Sep 5. We now pove ha fo, [ E 1 Ω 1 A l 1 2 ] C 1 Λp 2. Using Sep 3, we know ha on Ω 1 A, l = 1 +ΛA l, whence l =1+ Λ A i, jl j =1+ Λ i,j=1 C jl j =1+Λp l + S, j=1 whee S =Λ 1 j=1 C j pl j. Consequenly, l =1 Λp S, and we only have o pove ha E[1 Ω 1 A S 2 ] C 2. To his end, we wie S = Λ 1 a + b, whee a = j=1 C j px j and b = l j=1 C j p. Fis, since l C on Ω 1 by 7, we can wie E[1 Ω 1 b 2 ] CE[ j=1 C j p 2 ]=C 2 E[ C p 2 ] C, he las inequaliy coming fom Sep 2-i since C = L. ex, we use he Cauchy-Schwaz inequaliy: a 2 C p1 2 x 2 C C p1 2 X 2 on Ω 1 A by Sep 4. Consequenly, E[1 Ω 1 A a 2 ] CE[ X 2 2] 1/2 E[ C p1 2 2] 1/2. Bu E[ X 2 2] C by Sep 2-ii and we have seen a he end of Sep 1 ha E[ C p1 2 2] C. Sep 6. Hee we veify ha, sill fo, [ 1 E 1 Ω 1 A l ic i 2 1 2] 1 Λp 3 C 2. We wie, using ha c = l, 1 l ic i 2 = 1 l ic i c 2 + l l l ic i c. Fis, since l C on Ω 1,wehave l 3 1 Λp 3 C l 1 Λp 1, whence E[1 Ω 1 A l 3 1 Λp 3 2 ] C 2 by Sep 5. I hus suffices o veify ha E[1 Ω 1 A a 2 +b 2 ] C,wheea = l ic i c 2 and b = l ic i c. Fis, b = l iy i = x iy i because y i =. Hence b x 2 y 2.BuonΩ 1 A,weknowfomSep4ha x 2 C X 2, and i obviously also holds ue ha y 2 C Y 2.Wehus conclude ha E[1 Ω 1 A b 2 ] CE[ X 4 2] 1/2 E[ Y 4 2] 1/2 = E[ X 4 2]by

18 124 S. Delae and. Founie symmey. Using finally Sep 2-ii, we deduce ha indeed, E[1 Ω 1 A b 2 ] C. ex, since l i C on Ω 1 by 7, we can wie a C c c = C y 2 2. We conclude as peviously ha E[1 Ω 1 A a 2 ] C. Sep 7. The goal of his sep is o esablish ha, fo all, [ x E 1 Ω 1 A 2 2 Λ2 p1 p ] 1 Λp 2 C. Saing fom Sep 3, we wie x Λ l X =ΛA x Λ 1 =ΛA x Λ l X +Λ 2 l A X Λ 1. Thus x Λ l X 2 Λ A 2 x Λ l X 2 +Λ 2 l A X 2 +Λ 1/2 C p1 2 x 2, wheeweusedha 1 2 = 1/2 and ha 1 C p1 2 x 2 on Ω 1 A, as checked a he end of Sep 3. Using now ha Λ A 2 a<1 and l C on Ω 1 and ha x 2 C X 2 on Ω 1 A by Sep 4, we conclude ha, sill on Ω 1 A, x Λ l X 2 2 C A X C 1 C p1 2 2 X 2 2. Since now E[ A X 2 2] C 1 by Sep 2-iv, since E[ X 4 2] C by Sep 2-ii and since E[ C p1 4 2] C see he end of Sep 1, we deduce ha [ ] E 1 Ω 1 A x Λ l X 2 2 C. ex, we obseve ha x 2 2 Λ l 2 X 2 2 x Λ l X 2 x 2 + Λ l X 2 C x Λ l X 2 X 2 on Ω 1 A by Sep 4 and since l is bounded on Ω 1. Hence [ ] x E 1 Ω 1 A 2 2 Λ l 2 X 2 2 C E[ X 2 2] 1/2 C by Sep 2-ii. To complee he sep, i only emains o veify ha ] d = E [1 Ω 1 A l 2 X 2 2 p1 p1 Λp 2 C. We naually wie d a + b,whee [ ] a =E 1 Ω 1 A l 2 1 Λp 2 X 2 2, [ ] b =1 Λp 2 X E 1 Ω 1 A 2 2 p1 p.

19 Hawkes pocesses 1241 Sep 2-iii diecly implies ha b C 1/2. Using ha l is bounded on Ω 1, we deduce ha l 2 1 Λp 2 C l 1 Λp 1.Thus [ a CE 1 Ω 1 A l 1 Λp 1 2] 1/2E[ X 4 2] 1/2. Sep 2-ii and Sep 5 imply ha a C 1 C 1/2 as desied. Sep 8. I emains o conclude. I clealy suffices o ea he case whee, because l i andc i ae unifomly bounded on Ω 1 by 7, so ha he inequaliies of he saemen ae ivial when if he consan C is lage enough. Since l is unifomly bounded on Ω 1,wehave [ E 1 Ω 1 l 1 2 ] [ E 1 1 Λp Ω 1 A l 1 2 ] + C PA c. 1 Λp The fis em is bounded by C 2 by Sep 5, as well as he second one use he las inequaliy of Sep 1 wih α = 8. Similaly, using Sep 6 and ha l i andc i ae unifomly bounded on, we see ha Ω 1 [ 1 E 1 Ω 1 l ic i 2 1 2] 1 Λp 3 C 2 + C PA c C 2. Finally, obseve ha l i l 2 = x 2 2 is bounded by C on Ω 1, so ha by Sep 7, E [1 Ω l 1 i l 2 Λ2 p1 p ] 1 Λp 2 We used he las inequaliy of Sep 1 wih α = Peliminay analyic esimaes C + C PA c C. In view of 3and4, i will be necessay fo ou pupose o sudy vey pecisely he behavio of ϕ n, which we now do. The following saemens may seem ahe edious, bu hey ae exacly he ones we need. Recall ha ϕ sds = δ ds and ha ϕ n s =fos< by convenion. Lemma 15. Recall ha ϕ :[, [, and ha Λ= ϕsds. Assume ha hee is q 1 such ha s q ϕsds < and se κ =Λ 1 sϕsds. i Fo n and, we have sϕ n sds =Λ n nλ n κ + ε n, whee ε n Cn q Λ n 1 q and ε n nλ n κ. ii Fo n, fo z and s [,z], weseβ n, z, s =ϕ n z s ϕ n s. Then z β n, z, s ds 2Λ n and fo all Δ and all z [, +Δ], z β n, z, sds Cn q Λ n q,

20 1242 S. Delae and. Founie Δ z β n, z, s ds + β n, z, sds Cn q Λ n Δ q. Δ iii Fo m, n, fo z, wepu γ m,n, z = z z s uβ m, z, sβ n, z, ududs. I holds ha γ m,n, +Δ Λ m+n Δ, fo all, allδ. Fuhemoe, hee is a family κ m,n saisfying κ m,n m + nκ such ha, fo all Δ, γ m,n, +Δ=ΔΛ m+n κ m,n Λ m+n + ε m,n, +Δ, wih ε m,n, +Δ Cm + n q Λ m+n Δ q. Poof. We inoduce some i.i.d. andom vaiables X 1,X 2,... wih densiy Λ 1 ϕ and se S = as well as S n = X X n fo all n 1. We obseve ha, by he Minkowski inequaliy, E[Sn] q n q E[X q 1 ] Cnq,sinceE[X q 1 ]= Λ 1 s q ϕsds < by assumpion. To check i, we use ha S n has fo densiy Λ n ϕ n,sohawecanwie sϕ n sds = sϕ n sds =Λ n E[ S n + ]=Λ n Λ n E[S n ]+ε n, whee ε n =Λ n E[S n 1 {Sn }]. We clealy have ha E[S n ]=nκ, ha ε n and ha ε n Λ n E[S n ]=nλ n κ. Finally, ε n Λ n E[S n 1 {Sn }] Λ n 1 q E[Sn] q Cn q Λ n 1 q. To check ii, we obseve ha z β n, z, s ds 2Λ n is obvious because ϕ n sds =Λ n and ha, since E[S q n] Cn q, ϕ n udu =Λ n PS n Cn q Λ n q. We wie z β n, z, sds = z ϕ n z sds ϕ n sds = z ϕ n udu, which implies ha z β n, z, sds ϕ n udu Cn q Λ n q. ex, we see ha Δ β n, z, s ds Δ ϕ n z udu + Δ ϕ n udu, which is bounded by 2 Δ ϕ n udu Cn q Λ n Δ q. Finally, using he wo pevious bounds, we find z Δ β n, z, sds z β n, z, sds + Δ β n, z, sds Cn q Λ n q + Cn q Λ n Δ q Cn q Λ n Δ q because Δ [,] by assumpion. We finally pove iii and hus conside Δ and m, n. We sa fom +Δ +Δ [ γ m,n, +Δ= s u ϕ m +Δ sϕ n +Δ u + ϕ m sϕ n u ϕ m +Δ sϕ n u ] ϕ m sϕ n +Δ u duds.

21 Hawkes pocesses 1243 Using anohe independen i.i.d. family Y 1,Y 2,... of andom vaiables wih densiy Λ 1 ϕ and seing T m = Y Y m o T m =ifm =, we may wie γ m,n, +Δ [ =Λ m+n E +Δ T m + +Δ S n + + T m + S n + +Δ T m + S n + T m + +Δ S n + ]. This pecisely ewies γ m,n, +Δ = Λ m+n E[ +Δ T m S n + T m S n + + ], which implies ha γ m,n, +Δ Λ m+n Δ. We nex inoduce δ m,n, +Δ=Λ m+n E[ +Δ T m S n T m S n ], which equals δ m,n, +Δ = Λ m+n Δ κ m,n, whee κ m,n = E[ T m S n ] saisfies κ m,n κm + n. Thus γ m,n, +Δ = Λ m+n Δ κ m,n + ε m,n, + Δ, whee ε m,n, +Δ=γ m,n, +Δ δ m,n, + Δ. Finally, i is clea ha, since Δ, ε m,n, +Δ Λ m+n +ΔPT m S n +ΔoT m S n o T m S n Δ 2Λ m+n PT m ΔoS n Δ. This is, as usual, bounded by CΛ m+n m q + n q Δ q Peliminay sochasic analysis We handle once fo all a numbe of useful compuaions concening he pocesses inoduced in oaion 9. Lemma 16. We assume Hq fo some q 1. Recall ha Ω 1 and l wee defined in oaion 12 and ha all he pocesses below have been inoduced in oaion 9. i Fo any [1, ], fo all, 1 Ω 1 E θ [Z ] C 1. ii Fo any [1, ], fo all s, ] Eθ 1 Ω 1 [Z Z s μ sl C1 s 1 q 1. iii Fo all s +1 1, [ 1 Ω 1 sup E θ Z i, Zs i, 2 +sup M i, Ms i, 4] [ + 1 Ω 1 E θ Z Z s 2],..., [s,] C s 2.

22 1244 S. Delae and. Founie Poof. Recall 3, which asses ha E θ [Z ]=μ n [ sϕ n sds]a n 1. Since sϕ n sds Λ n, we ge E θ [Z ] μ n Λn A n 1. This is clealy bounded, on Ω 1,byC 1, which poves i. Using nex Lemma 15-i, E θ [Z ] = μ n [Λn nλ n κ + ε n ]A n 1, whee ε n Cn q Λ n 1 q 1. Hence E θ [Z ] E θ [Z s ]=μ s Λ n A n 1 + μ n ε n s]a n n [ε n 1. Bu n Λn A n 1 = Q 1 = l on Ω 1. Thus, sill on Ω1,sinces and q 1, E θ [Z Z s Since [M i,,m i, ] = Z i, E θ [sup [s,] M i, ] μ sl C1 s 1 q n n q Λ n A n 1 ] Ms i, 4 inequaliy ells us ha E θ [ Z sup,..., E θ [Z i, Z i, s 2 ]. C1 s 1 q 1. by Remak 1, he Doob inequaliy implies ha CE θ [Z i, Zs i, 2 ]. Also, he Cauchy-Schwaz Z s 2 ] 1 E θ[z i, Hence we jus have o pove ha sup,..., E θ [Z i, E θ [Z i, Zs i, 2 ] Zs i, 2 ] C s 2. Recalling ha Z i, = U i, + E θ [Z i, ], we have o show ha, on Ω 1, a ] E θ [Zs i, ] 2 C s 2 and b E θ [U i, Us i, 2 ] C s 2. To pove a, we use ii wih = and find ha, on Ω 1, E θ[z i, ] E θ [Zs i, ] μ s l + C 1 C s, since l is bounded on Ω 1 and since s 1 by assumpion. To pove b, we use 4 owie U i, Us i, = β n s,, A n i, jm j, d, n j=1 wheewehaveseβ n s,, =ϕ n ϕ n s as in Lemma 15. We deduce ha E[U i, Us i, 2 ]= β m s,, uβ n s,, v m,n By Remak 1, E θ [M j, u =, we see ha x j, j,k=1 Mv k, ]=1 {j=k} E θ [Z j, := E θ [Z j, C on Ω 1.WehuswieE θ[u i, I =μ m,n β m s,, uβ n s,, v A m i, ja n i, ke θ [M j, u Mv k, ]dvdu. u v]. Using now ii wih s =and ] μl j saisfies sup,j=1,..., x j, Us i, 2 ]=I + J, whee j=1 A m i, ja n i, ju vl jdudv,

23 J = m,n β m s,, uβ n s,, v Hawkes pocesses 1245 j=1 A m i, ja n i, jx j, u vdudv. Fis, using ha x j, is bounded on Ω 1 and ha β ms,, u du 2Λ m, we find J C m,n Λm+n j=1 Am i, jan i, j =C j=1 Q i, j 2 on Ω 1, whence J C j=1 1 {i=j} by 8. We conclude ha J C C s 2. ex, we ealize ha, wih he noaion of Lemma 15-iii, I = μ m,n γ m,n s, A m i, ja n i, jl j. j=1 Bu we know ha γ m,n s, Λ m+n s, so ha I μ s Q i, j 2 l j C s, j=1 since l is bounded on Ω 1 and since, as aleady seen, j=1 Q i, j 2 is also bounded on Ω 1. We conclude ha E θ[u i, U i, 2 ] C s C s 2 on Ω 1, as desied Fis esimao We ecall ha E = Z 2 Z /, ha he maices A and Q and he even Ω 1 wee defined in oaion 12, aswellasl i = j=1 Q i, j and l = 1 l i. The goal of his subsecion is o esablish he following esimae. Poposiion 17. Assume Hq fo some q 1. Then fo 1, [ E 2] 1 1 Ω 1 E θ μ l C 2q + 1. We sa wih he following lemma ecall ha Ū was defined in oaion 9. Lemma 18. Assume Hq fo some q 1. ThenonΩ 1,fo 1, E θ [E ] μ l C q and E θ [ Ū 2 ] C 1. Poof. Applying Lemma 16-ii wih = 1, we immediaely find, on Ω 1, E θ [E ex, Ū [ ] μ l 1 Z E 2 Z θ = 1 n ϕ n s i,j=1 An ] μl 1 C 1 q 1 1 = C q. j, i, jms ds by 4. Hence

24 1246 S. Delae and. Founie E θ [ Ū 2 ] 1/2 1 n ϕ n se θ [ i,j=1 2 ] 1/2ds A n i, jms j, by he Minkowski inequaliy. Bu ecalling Remak 1, i.e. E θ [Ms j, Ms l, ]= 1 {j=l} E θ [Zs j, ], E θ [ i,j=1 2 ] A n i, jms j, = j=1 A 2n 1 2Eθ A n i, j [Zs j, ] j=1 E θ [Z j, s ]. We know fom Lemma 16-i wih = 1 ha j=1 E θ[zs j, ] Cs on Ω 1. Hence, sill on Ω 1, E θ [ Ū 2 ] 1/2 C A n 1 sϕ n sds C1/2 Λ n A 1/2 n 1, n which is smalle han C 1/2 1/2 as desied. We can now give he Poof of Poposiion 17. I suffices o wie [ E 2] [ E E θ μ l 2E θ E θ [E ] 2 ] +2 E θ [E 2 ] μ l and o obseve ha E E θ [E ] = Ū 2 Ū / Ū 2 / + Ū /, whence finally [ E 2] E θ μ l 4 2 E θ[ Ū 2 2 ]+E θ [ Ū 2 ] + 2 E θ [E 2 ] μ l. Then he poposiion immediaely follows fom Lemma 18. n 4.5. Second esimao We ecall ha V = [Zi, 2 Z i, / E ] 2 E / whee E = Z 2 Z /, ha he maices A and Q and he even Ω 1 wee defined in oaion 12, aswellasl i = j=1 Q i, j and l = 1 l i. We also inoduce V = μ 2 [l i l ] 2. Poposiion 19. Assume Hq fo some q 1. Then fo 1, a.s., [ ] V 1 Ω 1 E θ V [ C 1+ l i l ] 2 1/2 q Obseve ha he em [l i l ] 2 will no cause any poblem, since is expecaion esiced o Ω 1 is unifomly bounded, see Poposiion 14.

25 We wie V V Δ,1 +Δ,2 Δ,1 = Δ,2 = Δ,3 =2 [Z i, 2 [Z i, 2 [Z i, 2 Hawkes pocesses Δ,3 Z i, / E ] 2 We nex wie Δ,2 Δ,21 +Δ,22 Δ,21 = Δ,22 = Δ,23 =2 [ Z i, 2,whee [Z i, 2 Z i, / μl i] 2 E /, Z i, / μl i][μl i μ l ]. +Δ,23,whee Z i, / E θ [Z i, 2 Z i, [ 2, E θ [Z i, 2 Z i, /] μl i] [ Z i, 2 Z i, / E θ [Z i, 2 We will also need o wie, ecalling ha U i, Δ,21 = whee Δ,211 = Δ,212 = Δ,213 = Z i, / μ l ] 2, 2 /]] E /, ] Z i, /] [ E θ [Z i, 2 Z i,. /] μl i] = Z i, [ 2 U i, 2 U i, /] E / Δ,211 { U i, 2 E θ [ U i, 2 E / E θ [E /]. Finally, we will use ha Δ,3 Δ,31 =2 Δ,31 =2 Δ,32 [ Z i, 2 [ E θ [Z i, 2 E θ [Z i, ], +Δ,212 +Δ,213, 2 [ 2 ]} U i, / Eθ U i, 2 U i,, / 2 ] U i, / E θ [E +Δ,32 Z i, / E θ [Z i, 2,whee /], ][ ] Z i,, /] μl i μ l ][ ] Z i,. /] μl i μl i μ l

26 1248 S. Delae and. Founie To summaize, we have o bound Δ,1,Δ,211,Δ,212,Δ,213,Δ,22,Δ,23, Δ,31 and Δ,32. Only he em Δ,211 is eally difficul. In he following lemma, we ea he easy ems. We do no y o be opimal when no useful: fo example in iv below, some shape esimae could pobably be obained wih moe wok, bu since we aleady have a em in 1/2 1 see Lemma 24, his would be useless. We also ecall ha we do no eally y o opimize he dependence in q: i is likely ha q could be eplaced by 2q hee and hee. Lemma 2. Assume Hq fo some q 1. Thena.s.onΩ 1,fo 1, i E θ [Δ,1 ] C 2q + 1, ii E θ [Δ,22 ] C 2q, iii E θ [Δ,23 ] C q, iv E θ [Δ,213 ] C 1/2 3/2, v E θ [Δ,32 ] C q. Poof. We wok on Ω 1 duing he whole poof. Using ha E = 1 [Zi, 2 Z i, /], one easily checks ha Δ,1 = E μ l 2. Thus poin i follows fom Poposiion 17. ex, we obseve ha Δ,22 = E θ [Z 2 Z /] μl 2 2. Applying Lemma 16-ii wih = 2, we conclude ha indeed, Δ,22 C 2q = C 2q. We wie Δ,23 2 Z 2 Z / E θ [Z 2 Z Eθ /] [Z 2 Z /] μl. 1 Applying Lemma 16-ii wih =, we deduce ha E θ [Z 2 Z /] μl C q. Lemma 16-i wih =1givesushaE θ [ Z 2 Z / E θ [Z 2 Z /] 1 ] 2 1 E θ [Z 2 + Z ] 1 C. We hus find ha indeed, E θ [Δ,23 ] C q. Since Δ,213 =/ E E θ [E ] = 2 Ū 2 Ū 2 Ū 2 + Ū, we deduce fom Lemma 18 ha E θ [Δ,213 ] C 2 / = C 1/2 3/2. Finally, saing fom Δ,32 2μ E θ [Z 2 Z /] μl l l 1 1 and using ha, as aleady seen when sudying Δ,23, E θ [Z 2 Z /] μl C q, we conclude ha Δ,32 C q l l 1 1 C q,since l is bounded see 7 on Ω 1. ex, we ea he em Δ,212. Lemma 21. Assume Hq fo some q 1. Then a.s. on Ω 1, fo 1, E θ [Δ,212 ] C 1. Poof. We wok on Ω 1. Recalling ha E = 1 Zi, 2 Z i,, we may wie E θ [Δ,212 ] 2 a i,wheea i = E θ [U i, 2 U i, 2 Z i, 2 Z i, ]. We infe fom 4 hau i, = M i, + n 1 ϕ n s j=1 An j, i, jms ds,

27 Hawkes pocesses 1249 so ha U i, 2 U i, R i, = M i, 2 = n 1 M i, 2 β n, 2, s + R i,,whee j=1 A n i, jms j, ds. We have se β n, 2, s =ϕ n 2 s ϕ n s as in Lemma 15 and he only hing we will use is ha 2 β n, 2, s ds 2Λ n. Recalling ha M i, is a maingale wih quadaic vaiaion [M i,,m i, ] = Z i,,seeremak1, we deduce ha E θ [M i, 2 M i, 2 ]=E θ [Z i, 2 Z i, ]. Hence a i = E θ [R i, 2 ]+2E θ [M i, 2 M i, R i, ]=b i + d i, he las equaliy sanding fo a definiion. We fis wie b i = 2 2 β m, 2, sβ n, 2, u m,n 1 Bu we know ha E θ [M j, s E θ [Z j, s u] C on Ω 1 b i C m,n 1 Λ m+n j,k=1 A m i, ja n i, ke θ [M j, s Mu k, ]=1 {j=k} E θ [Z j, by Lemma 16-i wih =. Hence j=1 A m i, ja n i, j =C Mu k, ]duds. s u] by Remak 1 and ha 2. Λ n A n i, j j=1 n 1 Bu n 1 Λn A n i, j = Q i, j 1 {i=j} C 1 on Ω 1 b i C 1. ex, we sa fom by 8, so ha d i =2 n 1 2 β n, 2, s j=1 As peviously, we see ha E θ [M i, 2 E θ [M i, 2 whence M i, M i, s A n i, je θ [M i, 2 M i, M j, s M i, ]=E θ [Z i, 2 s Zi, s ] C on Ω 1 d i C Λ n A n i, i =CQ i, i 1 C 1 n 1 Ms j, ]ds. ] = if i j and ha by Lemma 16-i, on Ω 1 by 8 again. Finally, a i C 1,sohaE θ [Δ,212 ] 2 a i C 1 on Ω 1. We nex compue some covaiances in he following edious lemma. Lemma 22. Assume Hq fo some q 1. Then a.s., on Ω 1, fo all 1, all k, l, a, b {1,...,}, all, s, u, v [,],

28 125 S. Delae and. Founie i Cov θ Z k, ii Cov θ Z k,,z l, s,m l, s iii Cov θ Z k,, s M l, han C 3/ {k=l}, iv E θ [M k, v Cov θ M k, M k, s vi Cov θ M k, vii Cov θ M k, viii Cov θ M k, M l, s = Cov θ U k, = Cov θ U k, τ dmτ l, M l, M k, s M k, s Poof. We wok on Ω 1 Z k, E θ [Z k, fom 4 ha Cov θ U k, M l, s,u l,,m l, s C {k=l}, s C {k=l},, s is smalle = Cov θ U k, M l, τ dmτ l, u ] C 1 if #{k, l} =2,,Mu a, Mv b, =if #{k, l, a, b} =4,,Mu a, Mv b, C 2 if #{k, a, b} =3,,Mu a, Mv a, C 1 3/2 if #{k, a} =2, C 2 wihou condiion.,m a, u M b, v and sa wih poin i. Fis, i is clea, since U k, ], ha Cov θ Z k,,us l, = m,n s i,j=1,z l, s =Cov θ U k, ϕ m xϕ n s y,u l, s A m k, ia n l, jcov θ M i, x =. Then we infe,my j, dydx. BuweknowseeRemak1 hacov θ Mx i,,my j, =1 {i=j} E θ [Zx y] i, C1 {i=j} by Lemma 16-i wih =. Thus Cov θ U k,,us l, C Λ m+n m,n =C Q k, iq l, i. A m k, ia n l, i Recalling 8, Q k, iq l, i C 1 +1 {k=i} 1 +1 {l=i} C {k=l}. Poin i is checked. Fo poin ii, we again have Cov θ Z k,,ms l, =Cov θ U k,,ms l, and, using again 4, Cov θ U k, Since Cov θ M i, x,ms l, = n ϕ n x A n k, icov θ M i, x,ms l, C1 {i=l} as in i, we conclude ha,ms l, dx. Cov θ U k,,ms l, C Λ n A n k, l =CQ k, l C {k=l}. n Poin iii is checked similaly as poin ii: we jus have o veify ha Cov θ Mx i,, s M l, τ dmτ l, C1 {i=l} 3/2. This is obvious if i l because

29 he maingales M i, and if i = l, ha Cov θ M i, x, s Hawkes pocesses 1251 M l, τ dmτ l, [ s M i, τ dmτ i, Eθ [ Mx i, 2 ] 1/2 E θ C 3/2. The las inequaliy uses ha E θ [ M i, x and Lemma 16-i and ha E θ [ s M i, [ M i, τ dm i, τ, M i, E θ [ s M i, τ dmτ i, τ dmτ i, 2 ] ] s = s E θ [M i,, s ae ohogonal, and elies on he fac, M i, τ dmτ i, 2 ] 1/2 2 ] = E θ [Zx i, ] C by Remak 1 τ 2 ] C 2. Indeed, we have τ 2 dz i, τ Ms i,, 2 Zs i,, whence τ dm i, i, M 2 Z i, s ] E θ [Ms i,, 4 ] 1/2 E θ [Zs i, 2 ] 1/2, which is bounded by C 2 by Lemma 16-iii. Fo poin iv, we assume e.g. ha s and fis noe ha E θ [M k, M k, s M l, u ]=E θ [M k, E θ [M k, s M l, u F ]] = E θ [M k, 2 Mu ] l, because he maingales M k, and M l, ae ohogonal. Since [M k,,m k, ] = Z k,,wehavem k, 2 =2 M k, τ dmτ k, + Z k,.since M k, τ dmτ k, and M l, ae ohogonal, we conclude ha E θ [M k, 2 Mu ] l, =E θ [Z k, Mu ] l, = Cov θ Z k,,mu. l, Since k l, we conclude using poin ii. Poin v is obvious, since when k, l, a, b ae paiwise diffeen, he maingales M k,, M l,, M a, and M b, ae ohogonal. Poin vi is hade. Recall ha #{k, a, b} = 3, so ha clealy, Cov θ M k, M k, s,m a, u M b, v =E θ [M k, We assume e.g. ha s and we obseve ha E θ [M k, M k, s M a, u M b, v ]=E θ [M k, M k, s E θ [M k, s =E θ [M k, 2 Mu a, Mv ] b, Mu a, Mv b, ]. Mu a, Mv b, F ]] because M k,, M a, and M b, ae ohogonal. We hus have o pove ha fo all, u, v [,]wihu, v, E θ [M k, 2 Mu a, Mv b, ] C 2.We wie M k, 2 =2 M k, τ dmτ k, + Z k, as in he poof of iv. The hee maingales M k, τ dmτ k,, M a, and M b, being ohogonal, we find ha E θ [M k, 2 Mu a, Mv b, ]=E θ [Z k, Mu a, Mv b, ]=E θ [U k, Mu a, Mv b, ]. We now wie, saing again fom 4, E θ [U k, M a, u Mv b, ]= n ϕ n x j=1 A n k, je θ [M j, x M a, u Mv b, ]dx. Bu E θ [Mx j, Mu a, Mv b, ] is zeo if j / {a, b} because he maingales M j,, M a, and M b, ae ohogonal, and is bounded by C 1 else by poin iv.

30 1252 S. Delae and. Founie As a consequence, E θ [U k, Mu a, Mv b, ] C 1 Λ n A n k, a+a n k, b n =C 1 Q k, a+q k, b. Since k a and k b, his is bounded by C 2 by 8. Fo vii, we assume e.g. ha s and u v and we ecall ha k a. We have Cov θ M k, M k, s =Cov θ M k, 2, Mu a, + Cov θ M k, + Cov θ M k, Ms k, =I + J + K + L.,Mu a, Mv a, 2 +Cov θ M k, Ms k, 2,Mu a, Mv a, Mu a, M k,,mu a, Mv a, Mu a, Fis, L =. Indeed, assuming e.g. ha u, wehave L =E θ [M k, Ms k, =E θ [M k, + E θ [M k, M k, Mu a, E θ [Ms k, Mu a, M a, M k,, Mu a, 2 Mu a, Mv a, M a, + M a, Mu a, ] M k, Mv a, M a, F ]] Mu a, E θ [Ms k, M k, F ]] and in boh ems, he condiional expecaion vanishes. ex, we wie as usual M k, 2 =2 M k, τ dmτ k, + Z k, and Mu a, 2 =2 u M a, τ dmτ a, + Zu a,. By ohogonaliy of he maingales M k, τ dmτ k, and M a, τ dmτ a,,we find u I =Cov θ Z k,,zu a, +2Cov θ Z k,, M a, τ dmτ a, +2Cov θ M k, τ,zu a,. τ dm k, We deduce fom poins i and iii, since k a,ha I C /2 C 1 3/2. We now ea K.Ivanishesifu, because E θ [Mv a, Mu a, F u ]=. We hus assume ha u <. We wie as usual M k, 2 = Mu k, u M k, τ dmτ k, + Z k, Zu k, and K =E θ [Mu k, 2 Mu a, Mv a, Mu a, ] [ ] +2E M k, τ dmτ k, Mu a, Mv a, Mu a, u + E θ [Z k, Z k, u The fis em vanishes because E θ [Mv a, one because E θ [ u M k, τ dmτ k, Mu a, Mv a, Mu a, ]. M a, v he involved maingales. Consequenly, K =E θ [Z k, Z k, u M a, u M a, u F u ] =, as well as he second F u ] = by ohogonaliy of Mu a, Mv a, Mu a, ]