Large deviations for renewal processes
aa r X i v : . [ m a t h . P R ] S e p LARGE DEVIATIONS FOR RENEWAL PROCESSES
RAPHA¨EL LEFEVERE, MAURO MARIANI, AND LORENZO ZAMBOTTI
Abstract.
We investigate large deviations for the empirical measure of the for-ward and backward recurrence time processes associated with a classical renewalprocess with arbitrary waiting-time distribution. The Donsker-Varadhan theorycannot be applied in this case, and indeed it turns out that the large deviationsrate functional differs from the one suggested by such a theory. In particular, anon-strictly convex and non-analytic rate functional is obtained. Introduction
Motivations from Statistical Physics.
In large deviations theory, the ap-pearance of rate functionals with singular points (that is, points of non-differentiabi-lity or non-analiticity) is a feature marking the existence of critical phenomena inthe underlying stochastic processes. Existence of such singularities is of particularinterest in a number of situations, for instance whenever these functionals are asso-ciated with deviations of physical quantities in Statistical Mechanics models, as theyidentify phase transitions. Moreover, values of the parameters in which deviationsfunctionals are convex, or affine, or non-convex are related to different behaviors ofthe system.In this respect, this work has been initially motivated by the appearance of affinestretches in large deviations rate functionals of Statistical Mechanics models, whosedynamics depends on renewal processes. In [11] a heat conduction model is in-troduced, and it is shown that the rate functional of the energy current is convexbut not strictly convex, with an affine behavior over two distinct intervals, fromwhich the appearance of critical points. In these conditions, the classical G¨artner-Ellis Theorem does not yield the full large deviations principle and a more detailedunderstanding of the random dynamics is necessary.In this paper we do not pursue this Statistical Mechanics interpretation, but rathershow how affine stretches in large deviations rate functionals of renewal processesarise when the inter-arrival times have heavy tails. We argue that in such situa-tions the Donsker-Varadhan approach [5] does not yield a good rate functional andtherefore the classical framework must be modified.Before detailing the main result, we recall an example concerning large deviationsof the renewal cumulative process, with the aim to underline that our Theorem 1.4below may have interesting consequences not related to Statistical Mechanics.1.2.
A motivating example.
Suppose a sequence of tasks i = 1 , . . . is given,where the task i takes a service time τ i to be accomplished. If the reward paid for Mathematics Subject Classification.
Key words and phrases.
Large Deviations; Renewal Process; Cumulative Process; Heavy Tails. executing such a task i is function F ( τ i ) of the time elapsed to accomplish it, thenthe total amount C t gained at time t > C t := N t − X i =1 F ( τ i ) , t > , (1.1)where N t := inf ( n ≥ n X i =1 τ i > t ) , and C t = 0 if N t = 1. When the service times τ i are random, the study of the cumulative process ( C t ) t ≥ can be of interest in many applications, for instancequeueing and risk theory.We assume throughout the paper that the sequence ( τ i ) i ≥ is an i.i.d. sequence ofpositive random variables and that F : ]0 , + ∞ [ [0 , + ∞ [ is bounded and continu-ous. The law of τ i is an arbitrary probability measure ψ on ]0 , + ∞ [, without anymoment assumption. Then N t is a so called renewal counting process and it is easilyseen that a.s.lim t → + ∞ C t t = lim t → + ∞ N t − t N t − N t − X i =1 F ( τ i ) = E ( F ( τ )) E ( τ ) ∈ [0 , + ∞ [ . This is therefore the total cost per unit of time on a large time interval. A natu-ral question, especially in the interpretations provided above, is the study of largedeviations for the mean payoff C t /t as t → + ∞ .Define Λ ∗ : [0 , + ∞ [ [0 , + ∞ ], the Legendre transform of the map Λ( x, y ) :=log ψ ( e xτ + yF ), namelyΛ ∗ ( a, b ) := sup x,y ∈ R (cid:8) ax + by − log ψ (cid:0) e xτ + yF (cid:1) (cid:9) , a, b ≥ . (1.2)A first result obtained as a consequence of the theory developed below is Theorem 1.1.
The law of the random variable C t /t defined by (1.1) satisfies a largedeviations principle with good rate functional J F : [0 , + ∞ [ → [0 , + ∞ ] defined as J F ( m ) := inf { β Λ ∗ (1 /β, m/β ) : β > } (1.3) i.e. for each closed set C ⊂ R lim t → + ∞ t log P ( C t /t ∈ C ) ≤ − inf C J F and for each open set O ⊂ R lim t → + ∞ t log P ( C t /t ∈ O ) ≥ − inf O J F . This result is known for a broader class of cumulative processes, but in the contestof a bounded sequence ( τ i ) i ≥ , see [6, 13], or in more generality for F ≡
1, corre-sponding to the large deviations of N t /t , see [7]. Here we address the case where τ i has an arbitrary distribution, and indeed large deviations display a more interestingbehavior in the case of heavy tailed distribution of τ i , as explained in section 1.7below. ARGE DEVIATIONS FOR RENEWAL PROCESSES 3
Empirical measures.
We refer to [4] for general large deviations theory and[1] for renewal processes. We denote the classical renewal process associated with( τ i ) i ≥ by S := 0 , S n := τ + · · · + τ n , n ≥ , so that the number of renewals before time t > N t := ∞ X n =0 ( S n ≤ t ) = inf { n ≥ S n > t } . Recall that the backward recurrence time process ( A t ) t ≥ and the forward recurrencetime process ( B t ) t ≥ are defined by A t := t − S N t − , B t := S N t − t, t ≥ . It is well known and easy to prove that the process ( A t , B t ) t ≥ is Markov. Onecan consider its empirical measure µ t := 1 t Z [0 ,t [ δ ( A s ,B s ) ds ∈ P (]0 , + ∞ [ ) , (1.4)i.e. for all bounded continuous f : ]0 , + ∞ [ → R µ t ( f ) := 1 t Z [0 ,t [ f ( A s , B s ) ds. The Donsker-Varadhan (DV) theory [5], [4, Chap. 6], provides a general result forthe large deviations of the empirical measure of Markov processes on metric spaces.However, the standard assumptions of classical DV theorems do not hold here. Infact, even formally, the DV rate functional does not provide the right large deviationsfunctional, see Section 1.6 below for a discussion.The main result of this paper, in Theorem 1.4 below, is a large deviations prin-ciple for the law P t of µ t as t → + ∞ with an explicit rate functional I defined in(1.11). This allows to deduce Theorem 1.1 with a contraction principle and obtaina relationship between I and J F , see (1.14) below.1.4. The large deviations rate functional.
In order to properly define the ratefunctional I for the large deviations of the law P t of µ t , some preliminary notationis needed.For a Polish space X , C b ( X ) denotes the space of real bounded continuous func-tions on X , and P ( X ) denotes the Polish space of Borel probability measures on X , equipped with its narrow (weak) topology. For µ ∈ P ( X ) is a Borel probabilitymeasure on a metric space X and f : X → [0 , + ∞ ] a Borel function, the notation µ ( f ) := Z X f dµ, is used throughout the paper. We also adopt the conventions0 · ∞ = 0 , ∞ = 0 . R. LEFEVERE, M. MARIANI, AND L. ZAMBOTTI
The space ]0 , + ∞ ] will be endowed throughout the paper with a metric which makesit isometric to [0 , + ∞ [, for instance by setting t : ]0 , + ∞ ] → [0 , + ∞ [, t ( p ) := 1 p , d ( p, p ′ ) := | t ( p ) − t ( p ′ ) | , p, p ′ ∈ ]0 , + ∞ ] . Thus (]0 , + ∞ ] , d ) is a Polish space. Let τ : ]0 , + ∞ ] × ]0 , + ∞ ] → ]0 , + ∞ ] , τ ( a, b ) := a + b, while we understand τ : ]0 , + ∞ ] → ]0 , + ∞ ] to be the identity map. Thus for µ ∈P (]0 , + ∞ ] ) and π ∈ P (]0 , + ∞ ]) µ (1 /τ ) = µ (1 / ( a + b )) = Z ]0 , + ∞ ] a + b µ ( da, db ) ,π ( τ ) = Z ]0 , + ∞ ] τ π ( dτ ) , π (1 /τ ) = Z ]0 , + ∞ ] τ π ( dτ ) . (1.5)Let us define ∆ ⊂ P (]0 , + ∞ ] ) as∆ := n µ ∈ P (]0 , + ∞ ] ) : µ ( da, db ) = Z [0 , × ]0 , + ∞ [ δ ( uτ, (1 − u ) τ ) ( da, db ) du ⊗ π ( dτ ) ,π ∈ P (]0 , + ∞ [) , π (1 /τ ) < + ∞ o . (1.6)In other words, µ is the law of ( U P, (1 − U ) P ), where U and P are independent, U is uniform on [0 ,
1] and P has law π ∈ P (]0 , + ∞ [). We also set ∆ ⊂ P (]0 , + ∞ ] )∆ := n µ = αµ + (1 − α ) δ (+ ∞ , + ∞ ) : µ ∈ ∆ , α ∈ [0 , o . (1.7)If µ ∈ ∆ then the writing (1.7) is unique up to the trivial arbitrary choice of µ when α = 0.If ν, µ ∈ P ( X ) then H( ν | µ ) denotes the relative entropy of ν with respect to µ ;this notation is used regardless of the space X . Finally, we set ξ := sup n c ∈ R : ψ ( e cτ ) < + ∞ o ∈ [0 , + ∞ ] , (1.8)where we recall that ψ denotes the law of τ i . Definition 1.2.
Let π ∈ P (]0 , + ∞ [) satisfy π (1 /τ ) ∈ ]0 , + ∞ [ , and set ˜ π ( dτ ) := 1 π (1 /τ ) 1 τ π ( dτ ) ∈ P (]0 , + ∞ [) . (1.9) Then the functionals I , I : P (]0 , + ∞ ] ) → [0 , + ∞ ] are defined by I ( µ ) := ( π (1 /τ ) H (cid:0) ˜ π (cid:12)(cid:12) ψ (cid:1) if µ ∈ ∆ is given by (1.6)+ ∞ if µ / ∈ ∆ , (1.10)I( µ ) := ( α π (1 /τ ) H (cid:0) ˜ π (cid:12)(cid:12) ψ (cid:1) + (1 − α ) ξ if µ ∈ ∆ is given by (1.6) - (1.7)+ ∞ if µ / ∈ ∆ . (1.11) ARGE DEVIATIONS FOR RENEWAL PROCESSES 5
Any µ ∈ ∆ can be written in the form (1.7), with the only caveat that π is notuniquely defined if α = 0. Notice that for π and ˜ π as in (1.9) the following relationshold ˜ π ( τ ) = 1 π (1 /τ ) , π ( dτ ) := 1˜ π ( τ ) τ ˜ π ( dτ ) . (1.12) Proposition 1.3.
The functional I is good , namely its sublevel sets are compact.Moreover I is the lower-semicontinuous envelope of I .For all bounded and continuous F : ]0 , + ∞ [ → [0 , + ∞ [ the functional J F definedin (1.3) is related to I and I by the formulae J F ( m ) = min (cid:26) I ( µ ) : µ ∈ P (]0 , + ∞ ] ) , Z ]0 , + ∞ ] F ( a + b ) a + b µ ( da, db ) = m (cid:27) , (1.13) J F ( m ) = min (cid:26) I( µ ) : µ ∈ P (]0 , + ∞ ] ) , Z ]0 , + ∞ ] F ( a + b ) a + b µ ( da, db ) = m (cid:27) . (1.14)1.5. The large deviations principle for the empirical measure.
We give herethe main result of this paper.
Theorem 1.4.
The family ( P t ) t> satisfies a large deviations principle with goodrate I defined by (1.11) as t ↑ + ∞ with speed t , i.e. for each closed set C ⊂P (]0 , + ∞ ] ) lim t → + ∞ t log P t ( C ) ≤ − inf u ∈C I( u ) (1.15) and for each open set O ⊂ P (]0 , + ∞ ] )lim t → + ∞ t log P t ( O ) ≥ − inf u ∈O I( u ) . (1.16) Some comments on the rate functional I . We stress again that the probability dis-tribution ψ on ]0 , + ∞ [ is completely arbitrary. However the fine properties ofthe associated renewal process depend on ψ , and the same is true for I. Define¯ µ ∈ P (]0 , + ∞ ] ) as¯ µ ( da, db ) := (R [0 , × ]0 , + ∞ [ τψ ( τ ) δ ( uτ, (1 − u ) τ ) ( da, db ) du ⊗ ψ ( dτ ) if ψ ( τ ) < + ∞ δ (+ ∞ , + ∞ ) if ψ ( τ ) = + ∞ It follow from our results that µ t ⇀ ¯ µ as t → + ∞ . Then Remark 1.5. (1) If ξ = + ∞ , i.e. if ψ has all exponential moments, then I ≡ I and I( µ ) = 0iff µ = ¯ µ ∈ P (]0 , + ∞ [ ).(2) If ξ < + ∞ and ψ ( τ ) = + ∞ , then I = I , and I( µ ) = 0 iff µ = δ (+ ∞ , + ∞ ) = ¯ µ .(3) If ξ < + ∞ and ψ ( τ ) < + ∞ . Then I(¯ µ ) = 0 and thusI( α ¯ µ + (1 − α ) δ (+ ∞ , + ∞ ) ) = (1 − α ) ξ Therefore in this case the functional I is not strictly convex . Still, if ξ > µ ) = 0 iff µ = ¯ µ . On the other hand, if ξ = 0, then I vanishes identicallyon the segment { α ¯ µ + (1 − α ) δ (+ ∞ , + ∞ ) , α ∈ [0 , } . Therefore the largedeviations at speed t do not yield the full large deviations behavior if ξ = 0,and we shall study large deviations at a slower speed in a future work. R. LEFEVERE, M. MARIANI, AND L. ZAMBOTTI
For the reader interested in the relation with the Statistical Mechanics models[12, 11] already cited above, we point out that such Gaussian models are related tothe case (3) with ξ = 0, so that the non-exponential decay of slow currents thereobserved is a consequence of the fact that I − ( { } ) is a whole segment in this case.We refer to [11, section 3] for further details.1.6. Relation with Donsker-Varadhan approach.
In the case of heavy-taileddistribution of τ i , the DV theory would yield I , defined in (1.10), as rate functional,while Theorem 1.4 shows that I is the correct functional. In fact, if ξ < + ∞ , longinter-arrival times τ i of length comparable with t may occur with a probability whichis not super-exponentially small in t . Thus I( µ ) is finite at µ = δ (+ ∞ , + ∞ ) , while theDV functional I is finite only on probability measures supported by ]0 , + ∞ [ .However I is in general not a good rate functional on P (]0 , + ∞ [ ) by proposition1.3 and it is good if and only if all exponential moments of τ are finite, i.e. ξ = + ∞ .As long as one exponential moment of τ is infinite, then the sublevels of I in P (]0 , + ∞ [ ) are not compact, and the law P t of µ t as t → + ∞ does not satisfy afull large deviations principle on P (]0 , + ∞ [ ).There are various extensions of DV theory, dealing with the lack of regularityproperties of the Markov process, e.g. [8], or ergodicity [14, 9]. However, even suchextensions do not take into account the model studied in this paper, and at the sametime do not provide the right large deviations rate functional in this case.We finally remark that this criticality is not a special feature of ( A t , B t ), but alsoother processes feature singular behavior. In the same setting, one may consider forinstance the Markov process σ t := ( τ N t , t − S Nt − τ Nt ). If the tail of ψ has an oscillatingbehavior, then the empirical measure of ( σ t ) t does not even satisfy a large deviationsprinciple, but it satisfies optimal upper and a lower large deviations bounds withfunctionals which may be different. This issue is not addressed here and will be thesubject of a forthcoming work.1.7. Affine stretches.
In this section we detail how the structure of the rate func-tional I explains the appearance of flat stretches in large deviations rate function-als J F . Let us consider the case of F ≡
1, i.e. the large deviations of N t /t as t ↑ + ∞ , where N t is the counting process. Recall that the rate functional is J ( m ) = m Λ ∗ (1 /m ), where Λ ∗ ( a ) := sup x ( ax − log ψ ( e xτ )). Here we suppose that ξ < + ∞ , i.e. that ψ has some infinite exponential moment, and that T := sup c<ξ E ( τ e cτ ) E ( e cτ ) < + ∞ . It is then easily seen that J ( · ) is strictly convex on [1 /T, + ∞ [, while J ( m ) = m Λ ∗ ( T ) + (1 − mT ) ξ, m ∈ [0 , /T ] . If ξ = 0 and T < + ∞ (which is the case if for instance ψ has polynomial tails andfinite mean) , J vanishes on [0 , /T ].Therefore, there is a transition between a strictly-convex regime and an affineregime. However, if we go back to the formula (1.13) above, which becomes for ARGE DEVIATIONS FOR RENEWAL PROCESSES 7 F ≡ J ( m ) = inf (cid:8) π (1 /τ ) H (cid:0) ˜ π (cid:12)(cid:12) ψ (cid:1) : π ∈ P (]0 , + ∞ [) , π (1 /τ ) = m (cid:9) = m inf (cid:8) H (cid:0) ζ (cid:12)(cid:12) ψ (cid:1) : ζ ∈ P (]0 , + ∞ [) , ζ ( τ ) = 1 /m (cid:9) , (1.17)then it is hard to understand what makes this inf strictly convex for m > /T andaffine for m ≤ /T . This apparent paradox is solved if we take into account formula(1.14) above, which becomes in this case J ( m ) = inf n m H (cid:0) ζ (cid:12)(cid:12) ψ (cid:1) + (1 − α ) ξ : ζ ∈ P (]0 , + ∞ [) , α ∈ [0 , , ζ ( τ ) = α/m o , (1.18)In (1.18) the appearance of the two regimes is clear. • For m ≥ /T , there exists a measure ζ m ∈ P (]0 , + ∞ [) which minimizesthe relative entropy H (cid:0) ζ (cid:12)(cid:12) ψ (cid:1) under the constraint ζ ( τ ) = 1 /m , and thisminimizer is an exponential tilt of ψ , i.e. ζ m ( dτ ) = 1 ψ ( e c ( m ) τ ) e c ( m ) τ ψ ( dτ ) , where c ( m ) is fixed by ψ ( τ e c ( m ) τ ) ψ ( e c ( m ) τ ) = 1 m and H (cid:0) ζ m (cid:12)(cid:12) ψ (cid:1) = Λ ∗ (1 /m ). Then the minimizer of (1.17) is ζ m and theminimizer of (1.18) is ζ m and α = 1. • For m < /T , on the other hand, no minimizer of (1.17) exists and theadditional parameter α in (1.18) starts to play a role; it turns out that theminimizer of (1.18) is given by α m = T m and ζ /T , and therefore we obtainthe correct value of J ( m ).The same picture is correct for more general functions F . Although J F can beexpressed as an inf in terms of I , in general this inf is not attained and it is noteasy to guess a minimizing sequence; on the other hand this problem is easily solvedif one expresses J F as a min in terms of I over a larger set of probability measures.This phenomenon is discussed in detail in [11] with applications to a heat con-duction model. Although the results of this paper are not explicitly applied there,the intuition behind the proof of [11, Theorem 3.4] comes from the understandingof the structure of the functional I defined above.For more on minimization of entropy functionals, see [2].2. The functional
IIn this section we analyze the properties of the functional I and prove in particularProposition 1.3. We also prove the following stability result which will come usefulin the following. Recall the definitions (1.6), (1.7), (1.8) and (1.11). Then
Proposition 2.1.
Let ( ψ n ) be a sequence in P (]0 , + ∞ [) . Let ξ n and I n be definedas in (1.8) and (1.11) respectively, with ψ replaced by ψ n . Assume that ψ n ⇀ ψ and ξ n → ξ as n → + ∞ . Then (1) Any sequence ( µ n ) in P (]0 , + ∞ ] ) such that lim n I n ( µ n ) < + ∞ is tight, andthus precompact in P (]0 , + ∞ ] ) . (2) For any µ and any sequence ( µ n ) in P (]0 , + ∞ ] ) such that µ n ⇀ µ , we have lim n I n ( µ n ) ≥ I( µ ) . R. LEFEVERE, M. MARIANI, AND L. ZAMBOTTI (3)
For any µ in P (]0 , + ∞ ] ) with I( µ ) < + ∞ , there exists a sequence ( µ n ) suchthat µ n ⇀ µ , µ n ∈ ∆ for all n , and lim n I n ( µ n ) ≤ I( µ ) . In the setting of [3], Proposition 2.1 states that I n Γ-converges to I, and that∆ is I-dense in P (]0 , + ∞ ] ). Before proving Proposition 2.1, let us show howProposition 1.3 follows immediately from it. Proof of Proposition 1.3.
In Proposition 2.1 take ψ n = ψ . Then the statement (1)implies that I has precompact sublevel set (namely it is coercive), statement (2)implies that I has closed sublevel sets (namely it is lower semicontinuous), and thus(1) and (2) imply that I is good. Since I ≤ I , statement (2) implies that I is smalleror equal than the lower semicontinuous envelope of I , while (3) states that I isgreater or equal to it. (cid:3) Lemma 2.2.
For all π ∈ P (]0 , + ∞ [) such that π (1 /τ ) < + ∞ and a > π (1 /τ ) H (cid:0) ˜ π (cid:12)(cid:12) ψ (cid:1) = sup ϕ ( π ( ϕ/τ ) − π (1 /τ ) log ψ ( e ϕ ))= sup ϕ : ψ ( e ϕ )= a ( π ( ϕ/τ ) − π (1 /τ ) log ψ ( e ϕ ))= sup f (cid:0) π ( f ) − π (1 /τ ) log ψ ( e τf ) (cid:1) where the suprema are taken over ϕ ∈ C b (]0 , + ∞ [) , and over and f ∈ C (]0 , + ∞ [) bounded from below and such that π ( f ) < + ∞ .In particular π π ( τ ) H( π | ψ ) is convex on { π ∈ P (]0 , + ∞ [) : π (1 /τ ) < + ∞} and thus I is convex.Proof. It is well known thatH (cid:0) ˜ π (cid:12)(cid:12) ψ (cid:1) = sup ϕ ∈ C b (]0 , + ∞ [) (˜ π ( ϕ ) − log ψ ( e ϕ )) . Now, suppose that ψ ( e ϕ ) = a > ϕ a := ϕ − log a . Then π ( ϕ/τ ) − π (1 /τ ) log ψ ( e ϕ ) = π ( ϕ a /τ ) − π (1 /τ ) log ψ ( e ϕ a )and ψ ( e ϕ a ) = 1. Therefore the quantitysup ϕ : ψ ( e ϕ )= a ( π ( ϕ/τ ) − π (1 /τ ) log ψ ( e ϕ )) = sup ϕ : ψ ( e ϕ )=1 ( π ( ϕ/τ ) − π (1 /τ ) log ψ ( e ϕ ))does not depend on a > a sup ϕ : ψ ( e ϕ )= a ( π ( ϕ/τ ) − π (1 /τ ) log ψ ( e ϕ )) = sup ϕ ( π ( ϕ/τ ) − π (1 /τ ) log ψ ( e ϕ ))= π (1 /τ ) H (cid:0) ˜ π (cid:12)(cid:12) ψ (cid:1) , where all suprema are taken over ϕ ∈ C b (]0 , + ∞ [).A standard approximation argument proves that one can take ϕ = τ f in thesupremum, provided the conditions on f in the statement of the lemma hold. (cid:3) Proof of Proposition 2.1-(1).
Since lim n I n ( µ n ) < + ∞ , µ n ∈ ∆ for n large enough,and thus µ n admits the writing (1.6)- (1.7), for some α n ∈ [0 ,
1] and π n ∈ P (]0 , + ∞ [)with π n (1 /τ ) < + ∞ . We first show thatlim n α n π n (1 /τ ) < + ∞ . (2.1) ARGE DEVIATIONS FOR RENEWAL PROCESSES 9
Notice that I n ( µ n ) ≥ α n π n (1 /τ ) H(˜ π n | ψ n ) , π n (1 /τ ) = 1˜ π n ( τ ) , so that lim n α n π n (1 /τ ) ≤ lim n I n ( µ n )H( π n | ψ n ) ∧ α n ˜ π n ( τ ) ≤ C lim n H(˜ π n | ψ n ) ∨ ˜ π n ( τ )for some C >
0. The denominator in the right hand side above is uniformly boundedaway from 0. Indeed, if lim k H(˜ π n k | ψ n k ) vanishes on some subsequence n k , thenlim k ˜ π n k = lim k ψ n k = ψ , and therefore lim k ˜ π n k ( τ ) ≥ ψ ( τ ) > µ n ) is precompact.It is easy to see that for each M > M := { µ ∈ ∆ : µ (1 / ( a + b ) ≤ M } is compact in P (]0 , + ∞ ] ). Now by (2.1)lim n µ n (1 / ( a + b )) = lim n α n π n (1 /τ ) < + ∞ namely µ n ∈ ∆ M for n and M large enough. (cid:3) Lemma 2.3.
Let π n ∈ P (]0 , + ∞ [) be such that π n (1 /τ ) < + ∞ and lim n π n = βπ + (1 − β ) δ + ∞ (2.2) for some β ∈ [0 , and π ∈ P (]0 , + ∞ [) such that π (1 /τ ) < + ∞ . Then lim n π n (1 /τ ) H(˜ π n | ψ n ) ≥ βπ (1 /τ ) H(˜ π | ψ ) + (1 − β ) ξ. (2.3) Proof.
By Lemma 2.2 π n (1 /τ ) H(˜ π n | ψ n ) = sup f (cid:0) π n ( f ) − π n (1 /τ ) log ψ n ( e τf ) (cid:1) (2.4)where the supremum is carried over continuous functions f bounded from below andsuch that π n ( f ) < + ∞ .Fix a ϕ ∈ C b (]0 , + ∞ [) such that ψ ( e ϕ ) <
1. Fix also c ∈ [0 , ξ [ if ξ > c = 0 if ξ = 0. For an arbitrary M >
0, let χ M be a smooth function on ]0 , + ∞ [such that χ M ( τ ) = 1 for τ ≤ / ( M + 1) or τ ≥ M + 1 ,χ M ( τ ) = 0 for 1 /M ≤ τ ≤ M .
Since ψ ( e ϕ ) <
1, there exists M ′ ≡ M ′ ( ϕ, c ) such that ψ ( e c τ χ M + ϕ (1 − χ M ) ) < M ≥ M ′ ( ϕ, c )and since ψ n → ψ and ξ n → ξ > c (the case ξ = 0 is easily taken care), for n largeenough depending on M , ϕ and cψ n ( e c τ χ M + ϕ (1 − χ M ) ) < M ≥ M ′ ( ϕ, c ) and n large enough. (2.5)Now, in (2.4) consider a f of the form f ( τ ) = c χ M ( τ ) + ϕ ( τ ) (1 − χ M ( τ )) /τ , whichis allowed for n large enough such that (2.5) holds. Then the logarithm in the righthand side of (2.4) is negative, and therefore recalling (2.2)lim n π n (1 /τ ) H(˜ π n | ψ n ) ≥ lim n π n ( ϕ (1 − χ M ) /τ ) + π n ( cχ M )= βπ ( ϕ (1 − χ M ) /τ ) + β c π ( χ M ) + (1 − β ) c. Taking the limit M → ∞ , since π (1 /τ ) < + ∞ and π ( { + ∞} ) = 0, by dominatedconvergence lim n π n (1 /τ ) H(˜ π n | ψ n ) ≥ βπ ( ϕ/τ ) + (1 − β ) c. Optimizing over c < ξ and ϕ such that ψ ( e ϕ ) < n π n (1 /τ ) H(˜ π n | ψ n ) ≥ sup c<ξ sup ϕ βπ ( ϕ/τ ) + (1 − β ) c = β sup ϕ π ( ϕ/τ ) + (1 − β ) ξ. (2.6)Still by Lemma (2.2)sup ψ ( e ϕ ) < π ( ϕ/τ ) = sup a< sup ψ ( e ϕ )= a π ( ϕ/τ )= sup a< (cid:2) log a + sup ψ ( e ϕ )= a π ( ϕ/τ ) − log ψ ( e ϕ ) (cid:3) = sup a< log a + π (1 /τ ) H(˜ π | ψ ) = π (1 /τ ) H(˜ π | ψ )which concludes the proof in view of (2.6). (cid:3) Proof of Proposition 2.1-(2).
First note that it is enough to prove the statementfor a subsequence of ( µ n ), and subsequences will be often indexed by the same n in this proof. Therefore one can assume sup n I n ( µ n ) < + ∞ , the statement beingtrivial otherwise. Thus, up to passing to a subsequence, µ n ∈ ∆ and according to(1.6)-(1.7) one can write µ n = α n µ ,n + (1 − α n ) δ (+ ∞ , + ∞ ) ,µ ,n := Z [0 , × ]0 , + ∞ [ δ ( uτ, (1 − u ) τ ) du ⊗ π n ( dτ ) , (2.7)for some α n ∈ [0 ,
1] and π n ∈ P (]0 , + ∞ [) with π n (1 /τ ) < + ∞ . If lim n α n = 0, then µ = lim n µ n = δ (+ ∞ , + ∞ ) and thereforelim n I n ( µ n ) ≥ lim n (1 − α n ) ξ n = ξ = I( µ ) . Let us turn to the case lim n α n =: ¯ α >
0. Up to passing to a subsequence, one canassume lim n α n = ¯ α >
0. Since sup n I n ( µ n ) < + ∞ , the bound on (2.1) holds, andsince ¯ α > n π n (1 /τ ) < + ∞ . In particular π n is tight in P (]0 , + ∞ ]) (note that + ∞ is and should be includedhere). Thus, up to passing to a further subsequencelim n π n = βπ + (1 − β ) δ + ∞ (2.8)for some β ∈ [0 , β > π (1 /τ ) ≤ β lim n π n (1 /τ ) < + ∞ while one can choose an arbitrary π satisfying π (1 /τ ) < + ∞ if β = 0. In particularthe conditions of Lemma 2.3 are fulfilled, and therefore (2.3) holds. ARGE DEVIATIONS FOR RENEWAL PROCESSES 11
Patching (2.7) and (2.1) together µ = lim n µ n = ¯ αβµ + (1 − ¯ αβ ) δ (+ ∞ , + ∞ ) ,µ := Z [0 , × ]0 , + ∞ [ δ ( uτ, (1 − u ) τ ) du ⊗ π ( dτ ) . In particular µ ∈ ∆ with α = ¯ αβ . And recalling α n → ¯ α and ξ n → ξ I( µ ) = ¯ αβπ (1 /τ ) H(˜ π | ψ ) + (1 − ¯ αβ ) ξ = ¯ α (cid:2) βπ (1 /τ ) H(˜ π | ψ ) + (1 − β ) ξ (cid:3) + (1 − ¯ α ) ξ = ¯ απ n (1 /τ ) H(˜ π n | ψ n ) + (1 − ¯ α ) ξ + ¯ α (cid:2) βπ (1 /τ ) H(˜ π | ψ ) + (1 − β ) ξ − π n (1 /τ ) H(˜ π n | ψ n ) (cid:3) ≤ lim n I n ( µ n ) + ¯ α lim n (cid:2) βπ (1 /τ ) H(˜ π | ψ ) + (1 − β ) ξ − π n (1 /τ ) H(˜ π n | ψ n ) (cid:3) . The limit in square brackets in the last line is negative, by (2.8) and Lemma 2.3.The wanted inequality follows. (cid:3)
Proof of Proposition 2.1-(3).
Since I( µ ) < + ∞ , µ ∈ ∆ and let α and π be as in(1.6)-(1.7) (again, the choice of π is not relevant if α = 0).Fix δ > , L > M > ψ ( { /M } ) = ψ ( { M } ) = 0 = ψ ( { L } ) = 0. Thenthere exist N ∈ N and 1 /M = T < T < . . . < T N = M such that T i +1 − T i ≤ δ and ψ ( { T i } ) = 0 for all i = 1 , . . . , N . Here of course N ≡ N ( M, δ ) and T i ≡ T i ( M, δ );we also use the shorthand notation A i = [ T i , T i +1 [ and A = ∪ Ni =1 A i in this proof.Then for L > M define π δ,M,Ln ( dτ ) ∈ P (]0 , + ∞ [) as π δ,M,Ln ( dτ ) = τ ˜ π δ,M,Ln ( dτ )˜ π δ,M,Ln ( τ ) , ˜ π δ,M,Ln ( dτ ) = α N X i =1 ˜ π ( A i )˜ π ( A ) ψ n ( dτ | A i ) + (1 − α ) ψ n ( dτ | [ M, L [) . The above definition is well posed if
L > M is large enough, and n is large enoughdepending on L and M ( n will be sent to + ∞ before L , and L before M ). Indeed,since I( µ ) < + ∞ and ψ ( ∂A i ) = 0, if ψ n ( A i ) = 0 for n large, then π ( A i ) = 0, andsimilarly if ψ n ([ M, L [) = 0 then α = 1.We want to provelim M → + ∞ lim L → + ∞ lim δ ↓ lim n π δ,M,Ln ( dτ ) = απ + (1 − α ) δ + ∞ , (2.9)lim M lim L lim δ lim n π δ,M,Ln (1 /τ ) H(˜ π δ,M,Ln | ψ n ) ≤ απ (1 /τ ) H(˜ π | ψ )+(1 − α ) ξ = I( µ ) , (2.10)where the limits in M and L are understood to run over M and L satisfying theabove conditions.Indeed, once (2.9)-(2.10) are proved, one can extract subsequences δ n → L n , M n → + ∞ such that, defining π n := π δ n ,M n ,L n n , one has π n ⇀ π and alsolim n π n (1 /τ ) H(˜ π n | ψ n ) ≤ I( µ ). It is then easy to verify that the sequence ( µ n ) de-fined by µ n := Z [0 , × ]0 , + ∞ [ δ ( uτ, (1 − u ) τ ) ( da, db ) du ⊗ π n ( dτ ) fullfills the wanted requirements.Note that the convergence ˜ π δ,M,Ln ⇀ ˜ π is immediate, so that (2.9) readily follows.In order to prove (2.10) define˜ π ,n ( dτ ) ≡ ˜ π δ,M ,n ( dτ ) = N X i =1 ˜ π ( A i )˜ π ( A ) ψ n ( dτ | A i ) , ˜ π δ,M ( dτ ) ≡ ˜ π δ,M ( dτ ) = N X i =1 ˜ π ( A i )˜ π ( A ) ψ ( dτ | A i ) . By the convexity statement in Lemma 2.2 π δ,M,Ln (1 /τ ) H(˜ π δ,M,Ln | ψ n ) = 1˜ π δ,M,Ln ( τ ) H(˜ π δ,M,Ln | ψ n ) ≤ α π ,n ( τ ) H(˜ π ,n | ψ n ) + (1 − α ) 1 ψ n ( τ | [ M, L [) H( ψ n ( ·| [ M, L [) | ψ n ) . (2.11)All the terms above can be explicitly calculated. In particular, since ψ ( { M } ) = ψ ( { L } ) = 0,lim n ψ n ( τ | [ M, L [) H( ψ n ( ·| [ M, L [) | ψ n ) = 1 ψ ( τ | [ M, L [) H( ψ ( ·| [ M, L [) | ψ ) , (2.12)and it is easy to check thatlim M lim L ψ ( τ | [ M, L [) H( ψ ( ·| [ M, L [) | ψ ) ≤ − lim M M log ψ ([ M, + ∞ [) = ξ. (2.13)On the other hand, since ψ ( ∂A i ) = 0, one haslim n π ,n ( τ ) → ˜ π ( τ ) , lim δ ↓ ˜ π ( τ ) = ˜ π ( τ | [1 /M, M [) , lim M → + ∞ ˜ π ( τ | [1 /M, M [) = ˜ π ( τ ) , namely lim M → + ∞ lim δ ↓ lim n ˜ π ,n ( τ ) = ˜ π ( τ ) , (2.14)and lim n H(˜ π ,n | ψ n ) = H(˜ π | ψ ) = N X i =1 ˜ π ( A i )˜ π ( A ) log ˜ π ( A i )˜ π ( A ) ψ ( A i )= 1˜ π ( A ) h ˜ π ( A c ) log ˜ π ( A c ) ψ ( A c ) + N X i =1 ˜ π ( A i ) log ˜ π ( A i ) ψ ( A i ) i − (cid:2) log ˜ π ( A ) + ˜ π ( A c ) log ˜ π ( A c ) ψ ( A c ) ] ≤ H(˜ π | ψ ) − (cid:2) log ˜ π ( A ) + ˜ π ( A c ) log π ( A c )] . The term in square brackets in the last line above vanishes as M → + ∞ , so thatlim M sup δ< lim n H(˜ π ,n | ψ n ) ≤ H(˜ π | ψ ) . (2.15)The inequality (2.10) finally follows from (2.11), (2.12), (2.13), (2.14), (2.15). (cid:3) ARGE DEVIATIONS FOR RENEWAL PROCESSES 13
This concludes the proof of Proposition 2.1. We end this section with someadditional results concerning the functional I which will come useful in the following.
Lemma 2.4.
The set ∆ defined in (1.7) is closed in P (]0 , + ∞ ] ) .Proof. Let µ n ∈ ∆ such that µ n ⇀ µ ∈ P (]0 , + ∞ ] ) in ]0 , + ∞ ] , with µ n given by(1.6)-(1.7) with α n ∈ [0 ,
1] and π n ∈ P (]0 , + ∞ [). We can assume that α n convergesto some ¯ α and π n ⇀ π ∈ P ([0 , + ∞ ]). By Skorohod’s representation theorem, thereexists a sequence ( P n ) n of random variables such that P n has law π n , P n ∈ ]0 , + ∞ [converges a.s. to P ∈ [0 , + ∞ ] and P has law π . If U is uniform on [0 ,
1] andindependent of ( P n ) n then for any f ∈ C b ([0 , + ∞ ] ) we obtain that µ n ( f ) = α n E ( f ( U P n , (1 − U ) P n )) + (1 − α n ) f (+ ∞ , + ∞ ) → α E ( f ( U P, (1 − U ) P )) + (1 − α ) f (+ ∞ , + ∞ )and this limit must be equal to µ ( f ). Since µ ∈ P (]0 , + ∞ ] ), then P ( P = 0) = 0. If P ( P < + ∞ ) ∈ { , } then µ ∈ ∆. If β := P ( P < + ∞ ) ∈ ]0 ,
1[ then µ ( f ) = αβ E ( f ( U P, (1 − U ) P ) | P < + ∞ ) + (1 − αβ ) f (+ ∞ , + ∞ )and therefore µ ∈ ∆. (cid:3) For a bounded measurable f : ]0 , + ∞ [ × ]0 , + ∞ [ → R set f ( r, τ ) := Z r f ( uτ, (1 − u ) τ ) du, r ∈ [0 , , τ > . (2.16)Let Γ be the set of all bounded lower semicontinuous f : ]0 , + ∞ ] × ]0 , + ∞ ] → R such that C f := Z ]0 , + ∞ [ ψ ( dτ ) e τf (1 ,τ ) < . (2.17) D f := sup s> Z ] s, + ∞ [ ψ ( dτ ) e τf ( s/τ,τ ) < + ∞ . (2.18) Lemma 2.5.
For all µ ∈ ∆ I( µ ) ≤ sup f ∈ Γ µ ( f ) (2.19) Proof.
Let ϕ ∈ C c (]0 , + ∞ ]), c < ξ if ξ > c := 0 if ξ = 0 and M >
0. Let f c,ϕ,M ( a, b ) := ϕ ( a + b ) a + b + c ] M, + ∞ ] ( a + b ) , ( a, b ) ∈ ]0 , + ∞ ] . Then f c,ϕ,M is lower semicontinuous on ]0 , + ∞ ] and f c,ϕ,M ( r, τ ) := r (cid:18) ϕ ( τ ) τ + c ] M, + ∞ ] ( τ ) (cid:19) , τ > , r ∈ [0 , . Then Z [ s, + ∞ [ ψ ( dτ ) e τf c,ϕ,M ( s/τ,τ ) = Z [ s, + ∞ [ ψ ( dτ ) exp (cid:16) sτ (cid:0) ϕ ( τ ) + cτ ] M, + ∞ ] ( τ ) (cid:1)(cid:17) ≤ e k ϕ k ∞ ψ ( e cτ )which is bounded uniformly in s , so that (2.18) holds for f = f c,ϕ,M . Let now a < ψ ( e ϕ ) = a < then there exists M = M ( c, ϕ ) such that for all M > M C f c,ϕ,M = ψ (cid:0) e ϕ + cτ ] M, + ∞ ] (cid:1) < f c,ϕ,M ∈ Γ. Now, if µ is given by (1.6)-(1.7) then µ ( f c,ϕ,M ) = α π ( ϕ/τ ) + c (1 − α ) . Since π ( ϕ/τ ) = π (1 /τ ) ˜ π ( ϕ ) then (with the usual convention 0 · ∞ = 0)sup f ∈ Γ µ ( f ) ≥ sup ϕ sup c,m sup M µ ( f c,ϕ,M )= α π (1 /τ ) sup ϕ { ˜ π ( ϕ ) − log ψ ( e ϕ ) } + π (1 /τ ) log a + (1 − α ) ξ where in the right hand side, the supremum on M is performed over ] M ( c, ϕ ) , + ∞ [,the supremum on c over [0 , ξ [ and the supremum on ϕ over ϕ ∈ C c (]0 , + ∞ ]) satisfy-ing (2.20). By Lemma 2.2 the supremum over ϕ satisfying (2.20) does not dependon a and the first term equals α π (1 /τ ) H(˜ π | ψ ), so that optimizing over a sup f ∈ Γ µ ( f ) ≥ sup a< { α π (1 /τ ) H(˜ π | ψ ) + (1 − α ) ξ + α π (1 /τ ) log a } = I( µ ) . (cid:3) Upper bound
In this section we prove the upper bound (1.15) in Theorem 1.4.3.1.
Exponential tightness.Lemma 3.1. lim M → + ∞ lim t → + ∞ t log P ( µ t (1 / ( a + b )) > M ) = −∞ . (3.1) In particular the sequence ( P t ) t> is exponentially tight with speed t , namely inf K⊂⊂P (]0 , + ∞ ] ) lim t → + ∞ t log P t ( K ) = −∞ . Proof.
We recall that { S n ≤ t } = { N t > n } . Note that if ⌊ M t ⌋ ≥ { µ t (1 /τ ) > M } = (cid:26) N t − t + t − S N t t τ N t > M (cid:27) ⊂ (cid:8) N t > ⌊ M t ⌋ (cid:9) = (cid:8) S ⌊ Mt ⌋ ≤ t (cid:9) . Therefore by the Markov inequality P ( µ t (1 /τ ) ≥ M ) ≤ P (cid:0) S ⌊ Mt ⌋ ≤ t (cid:1) ≤ e t E (cid:0) e − S ⌊ Mt ⌋ (cid:1) = e t + ⌊ Mt ⌋ log c where c := E ( e − τ ) <
1, and inequality (3.1) follows easily. Since for any
M > { µ ∈ P (]0 , + ∞ ] ) : µ (1 / ( a + b )) ≤ M } is tight in ]0 , + ∞ ] , exponentialtightness follows. (cid:3) ARGE DEVIATIONS FOR RENEWAL PROCESSES 15
The empirical measure is asymptotically close to ∆ . We give here themain argument to show that the rate functional at speed t of µ t must be equal to+ ∞ outside ∆. It will follow from the Lemma 2.4, and the following Lemma statingthat µ t belongs to an arbitrary neighborhood of ∆ in P (]0 , + ∞ ] ) for t large enough. Lemma 3.2.
For f ∈ C b (]0 , + ∞ ] ) , set ν t ( f ) := 1 t N t − X i =1 τ i Z f ( uτ i , (1 − u ) τ i ) du + t − S N t − t f (+ ∞ , + ∞ ) (3.2) then ν t ∈ ∆ . For all f ∈ C b (]0 , + ∞ ] ) and δ > , there exists t large enough suchthat the event {| µ t ( f ) − ν t ( f ) | > δ } is empty.Proof. It is easy to see that ν t ∈ ∆, and that it is given as in (1.6)-(1.7) with α = S N t − t , π = 1 S N t − N t − X i =1 τ i δ τ i . Recall the definition (2.16). Then for all f ∈ C b (]0 , + ∞ ] ) µ t ( f ) = 1 t N t − X i =1 τ i Z f ( uτ i , (1 − u ) τ i ) du + τ N t t Z t − SNt − τNt f ( uτ N t , (1 − u ) τ N t ) du = 1 t N t − X i =1 τ i f (1 , τ i ) + τ N t t f (cid:18) t − S N t − τ N t , τ N t (cid:19) . (3.3)We can rewrite τ N t t f (cid:18) t − S N t − τ N t , τ N t (cid:19) = t − S N t − t Z f ( u ( t − S N t − ) , τ N t − u ( t − S N t − )) du. Then | µ t ( f ) − ν t ( f ) | = t − S N t − t (cid:12)(cid:12)(cid:12)(cid:12)Z [ f ( u ( t − S N t − ) , τ N t − u ( t − S N t − )) − f (+ ∞ , + ∞ )] du (cid:12)(cid:12)(cid:12)(cid:12) . Since f ( a, b ) → f (+ ∞ , + ∞ ) as ( a, b ) → (+ ∞ , + ∞ ) and f is bounded, then thefunction ζ ( s ) := Z sup τ ≥ s | f ( us, τ − us )) − f (+ ∞ , + ∞ ) | du is bounded, monotone non-increasing and tends to 0 as s → + ∞ . Then {| µ t ( f ) − ν t ( f ) | > δ } ⊂ (cid:26) t − S N t − t ζ ( t − S N t − ) > δ (cid:27) = { x t ζ ( tx t ) > δ } where x t = t − S Nt − t ∈ [0 , x ∈ [0 ,
1] satisfies x ζ ( tx ) > δ , then δ < ζ ( tx ) and x ≤ ζ − ( δ ) /t , so that δ < C δ /t and this is impossible as soon as t ≥ C δ /δ . Therefore,for t large enough the event {| µ t ( f ) − ν t ( f ) | > δ } is empty. (cid:3) Free energy.
Recall the definition of Γ and (2.17)-(2.18).
Proposition 3.3.
For all f ∈ Γsup t> E e tµ t ( f ) = sup t> E exp (cid:18)Z t f ( A s , B s ) ds (cid:19) ≤ D f − C f < + ∞ . (3.4) Proof.
Since C f ∈ ]0 , + ∞ [, we can introduce the probability measure ψ f ( dτ ) := 1 C f ψ ( dτ ) e τf (1 ,τ ) and denote by ζ n the law of S n if ( τ i ) i ∈ N ∗ is i.i.d. with common law ψ f . Recalling(2.16) and (3.3) E exp (cid:18)Z t f ( A s , B s ) ds (cid:19) = E (cid:0) ( N t =1) exp (cid:0) τ f ( t/τ , τ ) (cid:1)(cid:1) + ∞ X n =1 E ( N t = n +1) exp n X i =1 τ i f (1 , τ i ) + τ n +1 f (cid:18) t − S n τ n +1 , τ n +1 (cid:19)!! = Z ] t, + ∞ [ ψ ( dτ ) e τ f ( t/τ,τ ) + ∞ X n =1 Z [0 ,t ] C nf ζ n ( ds ) Z ] t − s, + ∞ [ ψ ( dτ ) e τ f (( t − s ) /τ,τ ) ≤ D f ∞ X n =0 C nf = D f − C f . (cid:3) Proof of Theorem 1.4, upper bound.
For
M > g ∈ C b (]0 , + ∞ ] ) and δ >
0, let∆
M,g,δ = (cid:8) µ ∈ P (]0 , + ∞ ] ) : ∃ ν ∈ ∆ , | µ ( g ) − ν ( g ) | ≤ δ, µ (1 / ( a + b )) ≤ M (cid:9) and R M,g,δ := − lim t → + ∞ t log P t (∆ cM,g,δ ) . For A measurable subset of P (]0 , + ∞ ] ) and for f ∈ Γ, by (3.4),1 t log P t ( A ) = 1 t log E (cid:0) e tµ t ( f ) e − tµ t ( f ) A ( µ t ) (cid:1) ≤ t log (cid:2) e − t inf µ ∈A µ ( f ) E (cid:0) e tµ t ( f ) (cid:1)(cid:3) ≤ − inf µ ∈A µ ( f ) + 1 t log D f − C f and therefore lim t → + ∞ t log P t ( A ) ≤ − inf µ ∈A µ ( f ) . (3.5)Let now O be an open subset of P (]0 , + ∞ ] ). Then applying (3.5) for A = O∩ ∆ M,g,δ lim t → + ∞ t log P t ( O ) ≤ lim t → + ∞ t log (cid:2) P t ( O ∩ ∆ M,g,δ ) , P t (∆ cM,g,δ )) (cid:3) ≤ max (cid:18) − inf µ ∈O∩ ∆ M,g,δ µ ( f ) , − R M,g,δ (cid:19) = − inf µ ∈O∩ ∆ M,g,δ µ ( f ) ∧ R M,g,δ which can be restated aslim t → + ∞ t log P t ( O ) ≤ − inf µ ∈O I f,M,g,δ ( µ ) (3.6) ARGE DEVIATIONS FOR RENEWAL PROCESSES 17 for any open set O , f ∈ Γ and
M >
0, where the functional I f,M,g,δ is defined asI f,M,g,δ ( µ ) := ( µ ( f ) ∧ R M,g,δ if µ ∈ ∆ M,g,δ + ∞ otherwise . Since f is lower semicontinuous and ∆ M,g,δ is compact by Lemma 3.1, then I f,M,g,δ is lower semicontinuous. By minimizing (3.6) over { f, M, g, δ } we obtainlim t → + ∞ t log P t ( O ) ≤ − sup f,M,g,δ inf µ ∈O I f,M,g,δ ( µ )and by applying the minimax lemma [10, Appendix 2.3, Lemma 3.3], we get thatfor all compact set K lim t → + ∞ t log P t ( K ) ≤ − inf µ ∈K sup f,M,g,δ I f,M,g,δ ( µ )i.e. ( P t ) t ≥ satisfies a large deviations upper bound on compact sets with speed t and rate ˜I( µ ) for µ ∈ P (]0 , + ∞ ] )˜I( µ ) := sup { I f,M,g,δ ( µ ) : f ∈ Γ , M > , g ∈ C b (]0 , + ∞ ] ) , δ > } . By Lemma 2.4 we have ∩ g,δ ∆ M,g,δ ⊂ ∆, so that ˜I( µ ) = + ∞ if µ / ∈ ∆. By Lemma 3.1and Lemma 3.2, lim M → + ∞ R M,g,δ = + ∞ , ∀ g ∈ C b (]0 , + ∞ ] ) , δ > . Therefore for all µ ∈ P (]0 , + ∞ ] )˜I( µ ) ≥ sup { I f ( µ ) , f ∈ Γ } where I f ( µ ) := µ ( f ) if µ ∈ ∆+ ∞ otherwiseThus ˜I( µ ) ≥ I( µ ) by Lemma 2.5. Therefore ( P t ) t ≥ satisfies a large deviations upperbound with rate I on compact sets. By Lemma 3.1 and [4, Lemma 1.2.18], ( P t ) t ≥ satisfies the full large deviations upper bound on closed sets. (cid:3) Lower bound
In this section we prove the lower bound (1.16) in Theorem 1.4.4.1.
Law of large numbers for µ t . For any π ∈ P (]0 , + ∞ [) with π (1 /τ ) ∈ ]0 , + ∞ [we recall that ˜ π ( dτ ) := 1 π (1 /τ ) 1 τ π ( dτ ) . and we denote by P ˜ π the law of an i.i.d. sequence ( τ i ) i ≥ with marginal distribution˜ π , i.e. P ˜ π := ⊗ i ∈ N ∗ ˜ π ( dτ i ) . (4.1) Proposition 4.1.
Let π ∈ P (]0 , + ∞ [) with π (1 /τ ) ∈ ]0 , + ∞ [ . Under P ˜ π , a.s. µ t ⇀ Z [0 , × ]0 , + ∞ [ δ ( uτ, (1 − u ) τ ) du ⊗ π ( dτ ) on ]0 , + ∞ ] , t → + ∞ . Proof.
For all f ∈ C (]0 , + ∞ ] ) we recall the notation (2.16) f ( r, τ ) := Z r f ( uτ, (1 − u ) τ ) du, r ∈ [0 , , τ > , and, by (3.3) µ t ( f ) = 1 t N t − X i =1 τ i f (1 , τ i ) + τ N t t f (cid:18) t − S N t − τ N t , τ N t (cid:19) . By the strong law of large numbers a.s.lim n → + ∞ n n X i =1 τ i f (1 , τ i ) = ˜ π ( τ f (1 , τ )) = 1 π (1 /τ ) π ( f (1 , τ )) . By the renewal Theorem, a.s.lim t → + ∞ N t − t = 1 E ˜ π ( τ ) = π (1 /τ ) R τ τ π ( dτ ) = π (1 /τ ) ∈ ]0 , + ∞ [ . Therefore a.s. lim t → + ∞ N t − t N t − N t − X i =1 τ i f (1 , τ i ) = π (cid:0) f (1 , τ ) (cid:1) . On the other hand, by the law of large numbers a.s.lim n → + ∞ S n n = ˜ π ( τ ) = 1 π (1 /τ ) , so that a.s.lim t → + ∞ S N t − t = lim t → + ∞ S N t − N t − N t − t = 1 , lim t → + ∞ t − S N t − t = 0 . It follows that a.s.lim t → + ∞ (cid:12)(cid:12)(cid:12)(cid:12) τ N t t f (cid:18) t − S N t − τ N t , τ N t (cid:19)(cid:12)(cid:12)(cid:12)(cid:12) ≤ lim t → + ∞ t − S N t − t k f k ∞ = 0 . (cid:3) Proof of the lower bound.
For the proof of the lower bound, it is well knownthat it is enough to show the following
Proposition 4.2.
For every µ ∈ ∆ there exists a family Q t of probability measureson P (]0 , + ∞ ] ) such that Q t ⇀ δ µ and lim t → + ∞ t H( Q t | P t ) ≤ I( µ ) . Indeed, if Proposition 4.2 is proved, then we reason as follows. Let µ ∈ ∆ and let V be an open neighborhood of µ in the weak topology. Thenlog P t ( V ) = log Z V d P t d Q t d Q t = log (cid:18) Q t ( V ) Z V d P t d Q t d Q t (cid:19) + log Q t ( V ) ≥ Q t ( V ) Z V log (cid:18) d P t d Q t (cid:19) d Q t + log Q t ( V ) ARGE DEVIATIONS FOR RENEWAL PROCESSES 19 by using Jensen’s inequality. Now, since x log x ≥ − e − for all x ≥
0, we obtainlog P t ( V ) ≥ Q t ( V ) (cid:18) − H( Q t | P t ) + Z V c log (cid:18) d Q t d P t (cid:19) d Q t d P t d P t (cid:19) + log Q t ( V ) ≥ Q t ( V ) (cid:0) − H( Q t | P t ) − e − (cid:1) + log Q t ( V ) . Since µ ∈ V , Q t ⇀ δ µ and V is open, then Q t ( V ) → t → + ∞ . We obtainlim t → + ∞ t log P t ( V ) ≥ − lim t → + ∞ t H( Q t | P t ) ≥ − I( µ ) . Therefore, for any open set O and for any µ ∈ O lim t → + ∞ t log P t ( O ) ≥ − I( µ ) , and by optimizing over µ ∈ O we have the lower bound. Proof of Proposition 4.2.
Let us first suppose that µ ∈ ∆ as in (1.6). Notice that µ (1 /τ ) = π (1 /τ ) ∈ ]0 , + ∞ [. Fix δ > T t := ⌊ π (1 /τ ) (1 + δ ) t ⌋ . For t > /π (1 /τ ), let us denote by P t,δ the law on ]0 , + ∞ ] N ∗ such that under P t,δ thesequence ( τ i ) τ ≥ is independent and(1) for all i ≤ T t , τ i has law ˜ π (2) for all i ≥ T t + 1, τ i has law ψ .Let us set Q t,δ := P t,δ ◦ µ − t . Let us prove now thatlim δ ↓ lim t ↑ + ∞ Q t,δ = δ µ . (4.2)By the law of large numbers of Proposition 4.1, under P ˜ π we have a.s.lim t → + ∞ S T t t = lim t → + ∞ S T t T t T t t = 1 π (1 /τ ) π (1 /τ ) (1 + δ ) = 1 + δ. However S T t has the same law under P ˜ π and under P t,δ , so we obtain for any δ > t → + ∞ P t,δ ( S T t ≤ t ) = lim t → + ∞ P ˜ π (cid:18) S T t t ≤ (cid:19) = 0 . (4.3)Therefore, if we set D t,δ := { S T t > t } then, by (4.3) we obtain that for all δ > t → + ∞ P t,δ ( D t,δ ) = 1 . (4.4)We recall that { S n > t } = { N t ≤ n } . Therefore on D t,δ we have N t ≤ T t andtherefore by (3.3) for any f ∈ C b (]0 , + ∞ ] ) P t,δ ( | µ t ( f ) − µ ( f ) | > ε ) ≤ P ˜ π ( {| µ t ( f ) − µ ( f ) | > ε } ∩ D t,δ ) + P t,δ ( D ct,δ )By Proposition 4.1 lim δ ↓ lim t ↑ + ∞ P ˜ π ( {| µ t ( f ) − µ ( f ) | > ε } ∩ D t,δ ) = 0 , which, in view of (4.4), implies (4.2). Now we estimate the entropyH( Q t,δ | P t ) ≤ H (cid:0) P t,δ | P ψ (cid:1) = T t H(˜ π | ψ ) , (4.5)so that lim δ ↓ lim t ↑ + ∞ t H( Q t,δ | P t ) ≤ π (1 /τ ) H(˜ π | ψ ) . Then there exists a map t δ ( t ) > t ↑ + ∞ such that Q t := Q t,δ ( t ) → δ µ and lim t t − H( Q t | P t ) ≤ I( µ ).Let now µ ∈ ∆ \ ∆ . Then, by Proposition 2.1-(3) (applied with ψ n = ψ ) we canfind a sequence ( µ n ) n in ∆ such that µ n ⇀ µ and lim n I( µ n ) ≤ I( µ ). Moreover,we now know that there exists for all n a family Q nt of probability measures on P (]0 , + ∞ ] ) such that Q nt ⇀ δ µ n andlim t → + ∞ t H( Q nt | P t ) ≤ I( µ n ) . With a standard diagonal procedure we can find a family Q t such that Q t ⇀ δ µ andlim t → + ∞ t H( Q t | P t ) ≤ I( µ ) . (cid:3) Large deviations of C t /t In this section we prove Theorem 1.1, with F : ]0 , + ∞ [ → [0 , + ∞ [ continuous andbounded, and we set ˜ F ( τ ) := F ( τ ) /τ , τ ∈ ]0 , + ∞ ]. We remark that A t + B t = τ N t and we define the empirical measure ν t of ( τ N s ) s ≥ ν t ( O ) := 1 t Z [0 ,t [ O ( τ N s ) ds = Z O ( a + b ) µ t ( da, db ) , O ⊂ ]0 , + ∞ [ . Notice that by (3.3) ν t ( O ) = 1 t N t − X i =1 τ i O ( τ i ) + t − S N t − t O ( τ N t ) . (5.1)Then ν t ( ˜ F ) = ν t ( F/τ ) = 1 t N t − X i =1 F ( τ i ) + 1 t t − S N t − τ N t F ( τ N t ) , t > , by the representation (3.3). So that a.s. (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) ν t ( ˜ F ) − t N t − X i =1 F ( τ i ) (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) ≤ k F k ∞ t . (5.2)In particular, t P N t − i =1 F ( τ i ) and ν t ( ˜ F ) are exponentially equivalent , i.e.lim t → + ∞ t log P (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) ν t ( ˜ F ) − t N t − X i =1 F ( τ i ) (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) > δ ! = −∞ , ∀ δ > . ARGE DEVIATIONS FOR RENEWAL PROCESSES 21
By [4, Theorem 4.2.13], if the law of ν t ( ˜ F ) satisfies a large deviations principle, thesame large deviation principle holds for the law of t P N t − i =1 F ( τ i ). Moreover we have ν t ( ˜ F ) = µ t ( G ) where G : ]0 , + ∞ ] → [0 , + ∞ [ , G ( a, b ) := ˜ F ( a + b ) . This suggests to derive large deviations for t P N t − i =1 F ( τ i ) by using the classicalcontraction principle [4, Theorem 4.2.1] over the map P (]0 , + ∞ ] ) ∋ µ µ ( G ) ∈ [0 , + ∞ [. We shall start off by computing the candidate rate functional, then weconsider the case of ˜ F bounded, and finally we show how to remove this assumption.5.1. The case of a bounded ˜ F . In the above setting, set J ( m ) := inf (cid:8) I( µ ) : µ ∈ P (]0 , + ∞ ] ) , µ ( G ) = m (cid:9) , m ∈ [0 , + ∞ ] . (5.3)We compute now this rate functional. Lemma 5.1.
Recall (1.2) and (1.3) . For each m ∈ [0 , + ∞ ] we have J ( m ) = J F ( m ) .Moreover the inf in (5.3) is attained for all finite m .Proof. By (1.3)inf (cid:8) I( µ ) : µ ∈ ∆ , µ ( G ) = m (cid:9) = inf (cid:8) I( µ ) : µ as in (1.6)-(1.7) , απ ( ˜ F ) = m (cid:9) = inf (cid:8) I( µ ) : µ as in (1.6)-(1.7) , ˜ π ( τ ) = β, ˜ π ( τ ˜ F ) = βm/α, β > (cid:9) = inf (cid:8) ( α/β ) H(˜ π | ψ ) + (1 − α ) ξ : ˜ π ( τ ) = β, ˜ π ( F ) = βm/α, α ∈ [0 , , β > (cid:9) , where we have used that by (1.12) π ( ˜ F ) = ˜ π ( τ ˜ F )˜ π ( τ ) = ˜ π ( F )˜ π ( τ ) , π (1 /τ ) = 1˜ π ( τ ) . Now, setting p ( a, b ) := inf { H( ζ | ψ ) : ζ ( τ ) = a, ζ ( F ) = b (cid:9) , then p ≡ Λ ∗ by [2, Theorem 3], where, in the notation (1.2), Λ( x, y ) = log ψ ( e xτ + yF )and Λ ∗ is the Legendre transform of Λ. Another way to check that p ≡ Λ ∗ is thefollowing: p and Λ ∗ are easily seen to be lower semicontinuous convex functions of( a, b ) and moreover the Legendre transform of p is p ∗ ( x, y ) = sup a,b ( ax + by − p ( a, b )) = sup a,b,ζ { ax + by − H( ζ | ψ ) : ζ ( τ ) = a, ζ ( F ) = b } = sup ζ { ζ ( xτ + yF ) − H( ζ | ψ ) } = log ψ ( e xτ + yF ) = Λ( x, y ) , so that p = Λ ∗ . Therefore J ( m ) = inf (cid:8) I( µ ) , µ ∈ ∆ , : µ ( ˜ F ) = m (cid:9) = inf { ( α/β )Λ ∗ ( β, βm/α ) + (1 − α ) ξ, α ∈ [0 , , β > } = inf { β Λ ∗ ( α/β, m/β ) + (1 − α ) ξ, α ∈ [0 , , β > } . We want now to prove that J ( m ) = J F ( m ), recall (1.3). In particular we show thatfor all β > α ∈ [0 , { β Λ ∗ ( α/β, m/β ) + (1 − α ) ξ } = β Λ ∗ (1 /β, m/β ) . (5.4) First notice that the left hand side of (5.4) is clearly less or equal to the right handside by choosing α = 1. We now prove the converse inequality. For all α ∈ [0 , β Λ ∗ ( α/β, m/β ) + (1 − α ) ξ = sup x,y ( αx + (1 − α ) ξ + my − β Λ( x, y )) ≥ sup x,y ( x ∧ ξ + my − β Λ( x, y )) . Now, since F is bounded, then Λ( x, y ) = + ∞ for all x > ξ , so that the supremumover x can be restricted to a supremum over { x ≤ ξ } . Therefore we obtain β Λ ∗ ( α/β, m/β ) + (1 − α ) ξ ≥ sup x,y ( x + my − β Λ( x, y )) = β Λ ∗ (1 /β, m/β )and (5.4) is proven.Finally, in order to prove that the inf in (5.3) is attained, let us use the formulaobtained at the beginning of the proofinf (cid:8) I( µ ) , µ ∈ ∆ : µ ( G ) = m (cid:9) == inf (cid:8) ( α/β ) H(˜ π | ψ ) + (1 − α ) ξ, ˜ π ( τ ) = β, ˜ π ( F ) = βm/α, α ∈ [0 , , β > (cid:9) . We consider a minimizing sequence ( α n , ˜ π n , β n ) and the associated µ n ∈ ∆. usecoercivity and lower semi- continuity of the relative entropy and the bound | β | ≤k F k ∞ /m , and extract a sequence converging to ( α, ζ , β ). Now we have to prove thatthe limit still satisfies the required constraint, in particular that ζ ( τ ) = β , since therest follows easily. Let us notice that for all δ > Z ]0 ,δ ] τ π n ( dτ ) = ˜ π n (]0 , δ ])˜ π n ( τ ) = ˜ π n (]0 , δ ]) β and since (˜ π n ) is tight in ]0 , + ∞ [ we obtainlim δ → sup n Z ]0 ,δ ] τ π n ( dτ ) = 0 . It follows that ( π n ) n is tight in ]0 , + ∞ ]; if π n k ⇀ π in P (]0 , + ∞ ]), by a uniformintegrability argument, we obtain that π n (1 /τ ) → π (1 /τ ) and that ζ = ˜ π , i.e. inparticular π n ⇀ π . Since π n (1 /τ ) = 1 / ˜ π n ( τ ) = 1 /β , we obtain that ζ ( τ ) = β andthe inf above is attained, so that we can reconstruct µ ∈ ∆ attaining the minimumin (5.3). (cid:3) If ˜ F is bounded, then G is bounded too and we have the following Remark 5.2.
The map P (]0 , + ∞ ] ) ∋ µ µ ( G ) ∈ [0 , + ∞ [ is continuous in theweak topology.Proof. Notice that ˜ F is bounded and continuous on ]0 , + ∞ [ and ˜ F ( τ ) = F ( τ ) /τ → τ → + ∞ , so that it has a unique continuous extension to ]0 , + ∞ ]. Then the map G defined above is bounded and continuous and thus µ µ ( G ) is continuous. (cid:3) By the contraction principle [4, Theorem 4.2.1], we obtain that the law of µ t ( G )satisfies a large deviations principle with speed t and rate functional J given by(5.3), which is equal to J F by Lemma 5.1. ARGE DEVIATIONS FOR RENEWAL PROCESSES 23
The case of general ˜ F . Now we remove the assumption that ˜ F be bounded,always assuming F to be bounded and continuous. In this case, the map ν ν ( ˜ F )is no more necessarily continuous as in Remark 5.2.We introduce now the approximation that will allow us to justify the use of theclassical contraction principle. We fix ε > S εn := n X i =1 τ i ∨ ε, n ≥ , N εt := { n ≥ S εn ≤ t } = inf { n ≥ S εn > t } , and for all t ≥ A εt := t − S εN εt − , B εt := S εN εt − t. Define the empirical measure µ εt := 1 t Z [0 ,t [ δ ( A εs ,B εs ) ds ∈ P (]0 , + ∞ ] )and denote by P εt the law of µ εt . Notice that ( S εn , N εt , A εt , B εt , µ εt ) under P have thesame law as ( S n , N t , A t , B t , µ t ) under P ψ ε (recall (4.1)), where ψ ε ( dτ ) := ψ (]0 , ε ]) δ ε ( dτ ) + ( τ>ε ) ψ ( dτ ) . We denote by Λ ε , ξ ε and J εF the quantities defined by (1.2), (1.8) and (1.3) replacing ψ by ψ ε and remark that in fact ξ = ξ ε . Then we have the following Lemma 5.3.
The law of the random variable t P N εt − i =1 F ( τ εi ) satisfies a large devi-ations principle with rate J εF .Proof. By Theorem 1.4, P εt satisfies a large deviations principle with good ratefunctionalI ε = ( α π (1 /τ ) H (cid:0) ˜ π (cid:12)(cid:12) ψ ε (cid:1) + (1 − α ) ξ if µ ∈ ∆ is given by (1.6)-(1.7)+ ∞ if µ / ∈ ∆ . (5.5)For each ε >
0, the map P (]0 , + ∞ ] ) ∋ ν ν ( F/ ( τ ∨ ε )) ∈ [0 , ε − ] is continuous.Since µ t ( F/τ ) = µ t ( F/ ( τ ∨ ε )) almost surely under P ψ ε , Lemma 5.1 and the classi-cal contraction principle imply that the law of µ εt ( F/τ ) satisfies a large deviationsprinciple with speed ( t ) and rate J εF . By (5.2), µ εt ( F/τ ) and t P N εt − i =1 F ( τ εi ) areexponentially close, so that by [4, Theorem 4.2.13] we obtain the desired result. (cid:3) The following lemma states that ( N εt /t ) t> is an exponentially good approximationof ( N t /t ) t> . Lemma 5.4.
For all δ > ε ↓ lim t → + ∞ t log P ( | N t − N εt | > tδ ) = −∞ . Proof.
Notice that N t ≥ N εt . For δ > M > P ( N t − N εt > tδ ) ≤ ⌊ Mt ⌋ X n =0 P ( N t − N εt > tδ, N t = n ) + P ( N t > M t )= ⌊ Mt ⌋ X n =0 n X k =0 P N t − N εt > tδ, N t = n, n X i =1 ( τ i <ε ) = k ! + P ( S ⌊ Mt ⌋ ≤ t ) . Now, for k ≤ n ≤ M t and m ≤ n , on the event (cid:8) N t = n, N εt = m, P ni =1 ( τ i <ε ) = k (cid:9) we have that t < S εm ≤ S m + kε = ⇒ S m > t − kε = ⇒ N t − kε ≤ m = N εt and finally N t − Mtε ≤ N εt . Therefore P ( N t − N εt > tδ ) ≤ ( M t ) P ( N t − N t − Mtε > tδ ) + P ( S ⌊ Mt ⌋ ≤ t ) . Now, we can write for s < t and k ∈ N { N t − N s > k } = { S N s + k ≤ t } = { S N s + ˆ S k ≤ t } ⊂ { ˆ S k ≤ t − s } where ( ˆ S k := S N s + k − S N s , k ≥ S k , k ≥ P ( N t − N εt > tδ ) ≤ ( M t ) P (cid:0) S ⌊ tδ ⌋ ≤ M tε (cid:1) + P ( S ⌊ Mt ⌋ ≤ t ) . Thereforelim ε ↓ lim t → + ∞ t log P ( N t − N εt > tδ ) ≤ lim M → + ∞ lim ε ↓ lim t → + ∞ t log (cid:20) (cid:26) ( M t ) P (cid:0) S ⌊ tδ ⌋ ≤ M tε (cid:1) , P ( S ⌊ Mt ⌋ ≤ t ) (cid:27)(cid:21) ≤ max (cid:26) lim ε ↓ lim t → + ∞ t log P (cid:0) S ⌊ tδ ⌋ ≤ tε (cid:1) , lim M → + ∞ lim t → + ∞ t log P ( S ⌊ Mt ⌋ ≤ t ) (cid:27) . Arguing as in the proof of Lemma 3.1, by the Markov inequality P (cid:0) S ⌊ tδ ⌋ ≤ tε (cid:1) = P (cid:0) e − S ⌊ tδ ⌋ /ε ≥ e − t (cid:1) ≤ e t + ⌊ tδ ⌋ log E ( e − τ /ε ) so that lim ε ↓ lim t → + ∞ t log P (cid:0) S ⌊ tδ ⌋ ≤ tε (cid:1) ≤ lim ε ↓ (cid:0) δ log E ( e − τ /ε ) (cid:1) = −∞ , and analogouslylim M → + ∞ lim t → + ∞ t log P ( S ⌊ Mt ⌋ ≤ t ) ≤ lim M → + ∞ lim t → + ∞ t log e t +( ⌊ Mt ⌋ ) log E ( e − τ ) = −∞ . (cid:3) Let us define C εt := N εt − X i =1 F ( τ εi ) , t > . We deduce from Lemma 5.4 that
Lemma 5.5.
The process ( C εt /t ) t> is an exponentially good approximation of theprocess ( C t /t ) t> , i.e. for all δ > ε ↓ lim t → + ∞ t log P (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) t N εt − X i =1 F ( τ εi ) − t N t − X i =1 F ( τ i ) (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) > δ ! = −∞ . (5.6) ARGE DEVIATIONS FOR RENEWAL PROCESSES 25
Proof.
Let ω F ( ε ) := sup {| F ( ε ) − F ( η ) | , η ∈ [0 , ε ] } . Since N t ≥ N εt , (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) t N εt − X i =1 F ( τ εi ) − t N t − X i =1 F ( τ i ) (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) ≤ t N εt − X i =1 | F ( τ εi ) − F ( τ i ) | + 1 t N t − X i = N εt − | F ( τ i ) |≤ N t − t ω F ( ε ) + N t − N εt t k F k ∞ . Since F is continuous in 0, ω F ( ε ) → ε ↓ (cid:3) Since ( C εt /t ) t> is an exponentially good approximation of the process ( C t /t ) t> by Lemma 5.5, then by Lemma 5.3 and [4, Theorem 4.2.16] we have that ( C t /t )satisfies a large deviations principle with rate˜ J ( m ) := sup δ> lim ε ↓ inf z : | z − m |≤ δ J εF ( z ) . Proof of Theorem 1.1.
By Lemma 5.5, ( C εt /t ) t> is an exponentially good approx-imation of the process ( C t /t ) t> , then by Lemma 5.3 and [4, Theorem 4.2.16] wehave that ( C t /t ) satisfies a large deviations principle with ratesup δ> lim ε ↓ inf z : | z − m |≤ δ J εF ( z ) . which equals J F as a straightforward consequence of Proposition 2.1. Remark thatwe have also proved (1.14) and, still by Proposition 2.1-(3), (1.13). (cid:3) References [1] S. Asmussen,
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