On the rate of growth of Lévy processes with no positive jumps conditioned to stay positive

: In this article, we study the asymptotic behaviour of L´evy processes with no positive jumps conditioned to stay positive. We establish integral tests for the lower envelope at 0 and at + ∞ and an analogue of Khintchin’s law of the iterated logarithm at 0 and + ∞ , for the upper envelope.

1 Introduction and main results.
Lévy processes conditioned to stay positive have been introduced by Bertoin at the beginning of the nineties (see [1], [2], [3]).These first studies are devoted to the special case where the Lévy processes are spectrally one-sided.In a recent work Chaumont and Doney [9] studied more general cases.In this article, we are interested in the case when Lévy processes have no positive jumps (or spectrally negative Lévy processes).This case has been deeply studied by Bertoin in Chapter VII of [5].This will be our basic reference.The aim of this note is to describe the lower and upper envelope at 0 and at +∞ of Lévy processes with no positive jumps conditioned to stay positive throughout integral tests and laws of the iterated logarithm.For our purpose, we will first introduce some important properties of Lévy processes with no positive jumps and then we will define the "conditioned" version.Let D denote the Skorokhod's space of càdlàg paths with real values and defined on the positive real half-line [0, ∞) and P a probability measure defined on D under which ξ will be a real-valued Lévy process with no positive jumps, that is its Lévy measure has support in the negative real-half line.From the general theory of Lévy processes (see Bertoin [5] or Sato [19] for background), we know that ξ has finite exponential moments of arbitrary positive order.In particular ξ satisfies E exp λξ t = exp tψ(λ) , λ, t ≥ 0, where e λx − 1 − λx1I {x>−1} Π(dx), λ ≥ 0, a ∈ IR, σ ≥ 0 and Π is a measure that satisfies The measure Π is well-known as the Lévy measure of the process ξ.According to Bertoin [5], the mapping ψ : [0, ∞) → (−∞, ∞) is convex and ultimately increasing.We denote its right-inverse on [0, ∞) by Φ.Let us introduce the first passage time of ξ by From Theorem VII.1 in [5], we know that the process T = (T x , x ≥ 0) is a subordinator, killed at an independent exponential time if ξ drifts towards −∞.The Laplace exponent of T is given by Φ, E exp − λT x = exp − xΦ(λ) , λ, t ≥ 0.
According to Bertoin [5] Chapter III, we know that where k is the killing rate, d is the drift coefficient and ν is the Lévy measure of the subordinator T which fulfils the following condition, (0,∞) It is sometimes convenient to perform an integration by parts and rewrite Φ as Note that the killing rate and the drift coefficient are given by k = Φ(0) and d = lim λ→+∞ Φ(λ) λ .
In particular, the life time ζ has an exponential distribution with parameter k ≥ 0 (ζ = +∞ for k = 0).Hence, it is not difficult to deduce that ξ drifts towards −∞ if and only if Φ(0) > 0.
In order to study the case when ξ drifts towards −∞, we will define the following probability measure, where F t is the P-complete sigma-field generated by (ξ s , s ≤ t).Note that under P ♮ , the process ξ is still a Lévy process with no positive jumps which drifts towards +∞ and its Laplace exponent is defined by ψ ♮ (λ) = ψ(Φ(0) + λ), λ ≥ 0. Moreover the first passage process T is still a subordinator with Laplace exponent Φ ♮ (λ) = Φ(λ) − Φ(0).Since ξ has no positive jumps, we can solve the so-called two-sided-exit problem in explicit form.This problem consists in determining the probability that ξ makes its first exit from an interval [−x, y] (x, y > 0) at the upper boundary point.More precisely, where The function W is well-known as the scale function and is necessary for the definition of Lévy processes conditioned to stay positive.
Using the Doob's theory of h-transforms, we construct a new Markov process by an h-transform of the law of the Lévy process killed at time R = inf{t ≥ 0 : ξ t < 0} with the harmonic function W (see for instance Chapter VII in Bertoin [5], Chaumont [8] or Chaumont and Doney [9]), and its semigroup is given by where P x denotes the law of ξ starting from x > 0. Under P ↑ x , ξ is a process taking values in [0, ∞).It will be referred to as the Lévy process started at x and conditioned to stay positive.It is important to note that when ξ drifts towards −∞, we have that P ↑ x = P ♮↑ x , for all x ≥ 0. Hence the study of this case is reduced to the study of the processes which drift towards +∞.Bertoin proved in [5] the existence of a measure P ↑ 0 under which the process starts at 0 and stays positive.In fact, the author in [5] proved that the probability measures P ↑ x converge as x goes to 0+ in the sense of finite-dimensional distributions to P ↑ 0 := P ↑ and noted that this convergence also holds in the sense of Skorokhod.Several authors have studied this convergence, the most recent work is due to Chaumont and Doney [9].In [9], the authors proved that this convergence holds in the sense of Skorokhod under general hypothesis.Chaumont and Doney [9] also give a path decomposition of the process (ξ, P ↑ x ) at the time of its minimum.In particular, if m is the time at which ξ, under P ↑ x , attains its minimum, we have that under P ↑ x the pre-infimum process, (ξ t , 0 ≤ t < m), and the post-infimum process, (ξ t+m − ξ m , t ≥ 0), are independent and the later has by law P ↑ .Moreover, the process (ξ, P ↑ x ) reaches its absolute minimum once only and its law is given by Recently, Chaumont and Pardo [10] studied the lower envelope of positive self-similar Markov processes (or pssMp for short).In particular, the authors obtained integral tests at 0 and at +∞, for the lower envelope of stable Lévy processes with no positive jumps conditioned to stay positive and with index α ∈ (1, 2]; such processes are wellknown examples of pssMp.More precisely, when X (x) is the stable Lévy process with no positive jumps conditioned to stay positive and with starting point x ≥ 0, we have the following integral test for the lower envelope at 0 and at +∞: let f be an increasing function, then Let f be an increasing function, then for all x ≥ 0, Later, Pardo [17] studied the upper envelope of pssMp at 0 and at +∞.In particular, the author noted that X (x) , the stable Lévy process with no positive jumps conditioned to stay positive, satisfies the following law of the iterated logarithm where c(α) is a positive constant which depends on α.Moreover, Pardo [17] also established that where J (x) is the future infimum process of X (x) , i.e.J s .It is important to note that the above laws of the iterated logarithm have been also obtained at +∞, for any starting point x ≥ 0. Bertoin [4] studied the local rate of growth of Lévy processes with no positive jumps.In [4], the author noted that the sample path behaviour of a Lévy process with no positive jumps ξ, immediately after a local minimum is the same as that of its conditioned version (ξ, P ↑ ) at the origin.The main result in [4] gives us a remarkable law of the iterated logarithm at an instant when the path attains a local minimum on the interval [0, 1].We will recall this result for Lévy processes conditioned to stay positive.With a misuse of notation, we will denote by Φ and ψ for the functions Φ ♮ and ψ ♮ , respectively; when we are in the case where the process ξ drifts towards −∞.
Theorem 1 (Bertoin, [4]) There is a finite positive constant k such that lim sup It is important to note that the constant found by Bertoin satisfies k ∈ [c, c + γ], where c is the same constant found below in Theorem 3 and γ ≥ 6. Bertoin presented in [4] a "geometric" proof using some path transformations that connect ξ and its conditioned version (ξ, P ↑ ).Here, we will present standard arguments involving probability tail estimates.Our main results requires the following hypothesis, for all β > 1 Theorem 2 Let us suppose that (H2) is satisfaied, then there is a positive constant k such that, Note that with our arguments, we found that k ∈ [c, cη] where of course η ≥ 1 and cη > 3. We also remark that in particular (H1) and (H2) are satisfied under the assumption that ψ is regularly varying at 0 and at ∞ with index α > 1.Under this The next result gives us a law of the iterated logarithm at 0 and at +∞ of the future infimum of (ξ, P ↑ ).This result extends the result of Pardo [17] for the stable case.
Theorem 3 Let J denote the future infimum process of ξ, defined by J t = inf s≥t ξ s , then there is a positive constant c such that, If we assume that (H2) is satisfied, then i.e. that c = k.
We now turn our attention to the Lévy process conditioned to stay positive reflected at its future infimum.The following theorem extends the law of the iterated logarithm of Pardo [17] for the stable case.
Let us suppose that for all β < 1 Theorem 4 Under the hypothesis (H2) and (H3), we have that lim sup Moreover if (H1) and (H4) are satisfied, then lim sup Again conditions (H3) and (H4) are satisfied when ψ is regularly varying at 0 and at ∞ with index α > 1.
The lower envelope of (ξ, P ↑ ) at 0 and at ∞ is determined as follows, The lower envelope at +∞ is determined as follows: if ξ oscillates or drifts towards −∞ and the function f The following result describes the lower envelope of the future infimum of (ξ, P ↑ ).In fact, we will deduce that (ξ, P ↑ ) and its future infimum have the same lower functions.
Theorem 6 (i) If ξ has bounded variation, one has Moreover, we have the following integral test for the lower functions in terms of Φ.
The rest of this note consists of two sections, which are devoted to the following topics: Section 2 provides asymptotic results for the first and the last passage times of the process (ξ, P ↑ ).In Section 3, we will prove the results presented above.
2 First and last passage times.
Let us recall the definition of the first and last passage time of (ξ, P ↑ ) or ξ ↑ to simplify our notation, From Theorem VII.18 in [5], we know that (ξ t , 0 ≤ t ≤ T x ), the Lévy process killed at its first passage time above x, under P ♮ , has the same law as the Lévy process conditioned to stay positive time-reversed at its last passage time below x, (x−ξ ↑ In particular, we deduce that U ↑ x has the same law as T x and that U ↑ = (U ↑ x , x ≥ 0) is a subordinator with Laplace exponent Φ(λ) and therefore we obtain that the process ξ ↑ drifts towards +∞.There exist a huge variety of results on the upper envelope of subordinators.Fristedt and Pruitt [12] proved a general law of the iterated logarithm which is valid for a wide class of subordinators.The sharper result on the lower envelope for subordinators is due to Pruitt [18].In his main result, he gave an important integral test.Bertoin [5] presents a more precise law of the iterated logarithm of subordinators than the result obtained by Fristedt and Pruitt but for a more restrictive class of subordinators.In his result, Bertoin supposes that ψ is regularly varying at +∞ with index α > 1 (see Theorems III.11 and III.14 in [5]).In particular, we have the following lemma.

Lemma 1
The last passage time process U ↑ under the assumption that ψ is regularly varying at +∞ with index α > 1, satisfies , almost surely, and for large times, if we suppose that ψ is regularly varying at 0 with index α > 1, then , almost surely.Now, we turn our attention to the first passage time process.Note that due to the absence of positive jumps, for all x ≥ 0, ξ ↑ T ↑ x = x, a.s.Hence from the strong Markov property, we have that T ↑ = (T ↑ x , x ≥ 0) is an increasing process with independent increments but not stationary.Here we will use the results presented in Bertoin [5] and Lemma 1 to obtain the following law of the iterated logarithm for the first and last passage time of ξ ↑ .Proposition 2 Suppose that ψ is regularly varying at +∞ with index α > 1.Then the first passage time process satisfies the following law of the iterated logarithm, , almost surely, and for large times, if we suppose that ψ is regularly varying at 0 with index α > 1, we have Proof of Proposition 2: We will only prove the result for small times since the proof for large times is very similar.For all x ≥ 0, we see that T ↑ x ≤ U ↑ x , then from Lemma 1 we obtain the upper bound Next, we prove the lower bound.With this purpose we establish the following lemma.
Lemma 2 Assume that ψ is regularly varying at +∞ with index α > 1.Then for every constant c 1 > 0, we have

Proof of Lemma 2:
We know that U ↑ is a subordinator and that ψ is the inverse of the function Φ, then from Lemma III.12 in Bertoin [5], we see that Then the upper bound is clear since for all x > 0 we have that T ↑ x ≤ U ↑ x .For the lower bound, let us first define the supremum process S = (S t , t ≥ 0) by S t = sup 0≤s≤t ξ s .Next, we fix ǫ > 0, then by the Markov property (2.3) From the definition of the future infimum process, it is clear that J 0 is the absolute minimum of (ξ, P ↑ x ) then by (1.2) On the other hand, from (1.1) and applying the Tauberian and Monotone density theorems (see for instance Bertoin [5] or Bingham et al [7]) we deduce that Now, since the last passage time process is the right inverse of the future infimum process, we have that and applying Chebyshev's inequality, we have that for every λ > 0 and thus Since ψ is regularly varying at ∞ with index α, we see that

This implies
We now choose K in such way that , and In conclusion, we have established that Hence from the inequality (2.3) and (2.4) and (2.5), we deduce and since ǫ can be chosen arbitrarily small, the lemma is proved.
Now we can prove the lower bound of the law of the iterated logarithm for T ↑ .Let (x n ) be a decreasing sequence of positive real numbers which converges to 0 and let us define the event for all large n.Since the function g and the process T ↑ are increasing, we have Then, it is enough to prove that n P(A n ) < ∞.In this direction, we take Since ψ is regularly varying and we can chose r close enough to 1, we see that for n 0 sufficiently large and from Lemma 2 this last integral is finite since with this we finish the proof.
There also exist a huge variety of results for the upper envelope of subordinators, see for instance Chapter III of Bertoin [5].Here, we will state with out proofs the main results for the upper envelope of U ↑ .The proofs of the following results can be found in Chapter III of Bertoin [5].
3 Proofs of the main results.
For simplicity, we introduce the notation We start with the proof of the first part of Theorem 3, since a key result on subordinators due to Fristed and Pruitt [12] easily yields the result.The second part will be proved after the proof of Theorem 2, since the latter is necessary for its proof.
Proof of Theorem 3 (first part): First we will observe that ψ(λ) = O(λ 2 ), as λ goes to +∞, then λ 1/2 = O(Φ(λ)).Since the last passage time process U ↑ is a subordinator with Laplace exponent Φ and the future infimum process is the right-inverse of the last passage times U ↑ , then according to Theorem 2 and Remark on p. 176 in Fristed and Pruitt [12], there exists a positive constant c such that lim sup then the first part of Theorem 3 is proved.

Proof of Theorem 2:
We only prove the result for small times since the proof for large times is very similar.The lower bound is easy to deduce from Theorem 3 and since J ↑ t ≤ ξ ↑ t , where J ↑ t denotes the future infimum of ξ ↑ .Hence Now, we prove the upper bound.Let (x n ) be a decreasing sequence of positive real numbers which converges to 0, in particular we choose x n = r n , for r < 1.
From the first Borel-Cantelli's Lemma, for all large n.Since the function h and the process S are increasing in a neighbourhood of 0, we have S t ≤ ηch(t) for r n+1 ≤ t ≤ r n , under P ↑ .
Then, it is enough to prove that n P ↑ (A n ) < ∞.In this direction, we will prove the following lemma, Lemma 3 Let 0 < ǫ < 1 and r < 1.If we assume that condition (H2) is satisfied then there exists a positive constant C(ǫ) such that Proof of Lemma 3: From the inequality (2.3), we have that and since J 0 is the absolute minimum of (ξ, P ↑ ηch(r n+1 ) ) then by (1.2) ) .
On the other hand, an application of Proposition III.1 in Bertoin [5] gives that there exist a positive real number K 1 such that then it is clear that ) .
From this inequality and condition (H2), there exist a positive constant C(ǫ) such that for n sufficiently large which proves our result.Now, we prove the upper bound for the law of the iterated logarithm of (ξ, P ↑ ).Fix 0 < ǫ < 1/(2 + r −1 ).Since J can be seen as the right inverse of U, it is straightforward that and this probability is bounded from above by for every λ ≥ 0. We choose λ = r −(n+1) log | log r n+1 |, then hence from the above inequality and Lemma 3, we have that From the above definition, it is clear that R n ≥ n and that R n diverge a.s. as n goes to +∞.From Theorem 2, we deduce that R n is finite a.s.Now, by (1.2) and since (ξ, P ↑ ) is a strong Markov process with no positive jumps, we have that Now applying (3.7), we have that and since the hypothesis (H2) is satisfied, an application of the Fatou-Lebesgue Theorem shows that lim inf n→+∞ Therefore, for all ǫ ∈ (0, 1/2) The event on the left hand side is in the upper-tail sigma-field ∩ t σ{ξ ↑ s : s ≥ t} which is trivial from Bertoin's construction of (ξ, P ↑ ) (see Theorem VII.20 in [5]).Hence lim sup and since ǫ can be chosen arbitrarily small, the result for large times is proved.In order to prove the law of the iterated logarithm for small times, we now define the following stopping time Following same argument as above and assuming that (H1) is satisfied, we get that for a fixed ǫ ∈ (0, 1/2) and n sufficiently large Next, we note that ≥ (1 − 2ǫ), for some p ≥ n ≥ P ↑ J Rn kh(R n ) ≥ (1 − 2ǫ) .
Since R n converge a.s. to 0 as n goes to ∞, the conclusion follows taking the limit when n goes towards to +∞.

Proof of Theorem 4:
The proof of this theorem is very similar to the proof of the previous result.Following same arguments, we first prove the law of the iterated logarithm for large times.Assume that the hypothesis (H2) and (H3) are satisfied.
Since the hypothesis (H3) is satisfied, an application of the Fatou-Lebesgue Theorem shows that lim sup Since f is increasing the following inclusion holds On the other hand where f −1 is the right-inverse of f .Now, we take x n = cf (r n ), for r < 1.Since f is increasing and from the above equality, we get that P ↑ x n+1 < cf U xn ≤ P ↑ r n+1 < U cf (r n ) The obvious inequality applied for t = cf (r n ) and a = r n+1 entails that Since the mapping t → tΦ(1/t) increases, it is not difficult to deduce that the function Φ satisfies that Φ(r −(n+1) ) ≤ r −2 Φ(r (n−1) ).x −1 f (x)Φ(1/x)dx.
In order to prove that the future infimum satisfies the same result, we note first that we can replace ξ ↑ by its future infimum in the sets A n , and then the same arguments will give us the desired result.