Small-time Behaviour of L¶evy Processes

. In this paper a necesary and su–cient condition is established for the probability that a L¶evy process is positive at time t to tend to 1 as t tends to 0. This condition is expressed in terms of the characteristics of the process, and is also shown to be equivalent to two probabilistic statements about the behaviour of the process for small t .


Introduction
The quantity ρ(t) = P (X t > 0) where X = (X t , t ≥ 0) is a Lévy process is of fundamental importance in fluctuation theory.For example, combining results in [1] and [2] shows that, both as t → ∞ and as t ↓ 0, where A ρ denotes a random variable with an arc-sine law of parameter ρ if 0 < ρ < 1, and a random variable degenerate at ρ if ρ = 0, 1.It would therefore be useful to find a necessary and sufficient condition for (1) to hold, ideally expressed in terms of the characteristics of X, that is its Lévy measure Π, its Brownian coefficient σ 2 , and γ, the coefficient of the linear term in the Lévy-Itô decomposition (2) below.This problem is obviously difficult, and has so far only been solved for large t in the special case ρ = 0, 1.This result is in Theorem 3.3 of [5], and in an extended form in Theorem 1.3 in [4].In both cases the results are deduced from the corresponding results for random walks due to Kesten and Maller in [7] and [8].Here we consider the corresponding question for small t, where apparently the large t results will have no relevance, but in fact it turns out that there is a striking formal similarity, both in the statement and proof.
Our Lévy process will be written as where B is a standard BM, Y (1) is a pure jump martingale formed from the jumps whose absolute values are less than or equal to 1, Y (2) is a compound Poisson process formed from the jumps whose absolute values exceed 1, and B, Y (1) , and Y (2) are independent.We denote the Lévy measure of X by Π, and introduce the tail functions and the tail sum and difference The rôles of truncated first and second moments are played by A(x) = γ + D(1) + x 1 D(y)dy, U (x) = σ 2 + 2 x 0 yT (y)dy, x > 0. (5) We will denote the jump process of X by ∆ = (∆ s , s ≥ 0), and put for the magnitude of the largest negative jump which occurs by time t.
For the case t ↓ 0, a sufficient condition for (1) with ρ = 1 was given in Theorem 2.3 of [5]; the following, together with Lemma 5 shows that the condition given there is also necessary: Theorem 1 Suppose that the Lévy process X has σ = 0, Π(R) = ∞, and M (0+) > 0; then the following are equivalent.
for some deterministic d which decreases to 0 and is regularly varying of index 1 at 0, Remark 2 None of the above assumptions are really restrictive.First, if σ = 0, it was shown in [5] that P (X t > 0) → 1/2.It was also shown there that when M (0+) = 0, i.e.X is spectrally positive, and N (0+) = ∞, then (7) occurs iff X is a subordinator iff A(x) ≥ 0 for all small x, and it is easy to see that these are equivalent to (9) in this case.(Of course (8) is not relevant here, as ∆ (1) t ≡ 0.) Finally the case when Π(R) < ∞ is of no real interest; then X is a compound Poisson process plus linear drift, and the behaviour of ρ t as t ↓ 0 is determined by whether the drift is positive or not.

Remark 3
Comparing the above results with the large-time results we see that each of ( 7)-(10) has a formally similar counterpart at ∞. (Actually the counterpart of (8) was omitted in [4], but it is easy to establish.)One difference is that (7) as t → ∞ implies that X t P → ∞, and of course this can't happen as t ↓ 0. At first sight the appearance of A in both (10) and its counterpart at ∞ is surprising.However this can be understood by realising that A acts both as a generalised mean at ∞ and a generalised drift at 0. To be precise, t −1 X t P → c, as t → ∞ or as t ↓ 0 is equivalent to xT (x) → 0 and A(x) → c as x → ∞ or as x ↓ 0; in the first case if X has finite mean µ then c = µ, and in the second if X has bounded variation and drift δ then c = δ.(See Theorems 2.1 and 3.1 of [5].)Remark 4 The structure of the following proof also shows a strong similarity to the proof of the random walk results in [7] and [8].There are of course differences in detail, and some simplifications due to the advantages of working in continuous time and the ability to decompose X into independent components in various ways.There are also some extra difficulties; for example we need to establish results related to the Cental Limit Theorem which are standard for random walks but apparently not previously written down for Lévy processs at zero.Also the case where Π has atoms presents technical difficulties which are absent in the random walk situation; compare the argument on page 1499 of [7] to the upcoming Lemma 9.

Preliminary Results
We start by showing that (10) can be replaced by the simpler Lemma 5 (i) If (10) holds then (11) holds.
Proof For i) just note that For (ii) we first show that where we write so that U (x) = U − (x) + U + (x).Given ε > 0 take x 0 > 0 such that εA(x) ≥ xM (x) for x ∈ (0, x 0 ], and hence Note also that for 0 < x < 1 we can write M (y)dy, and and similarly Since ε is arbitrary, (14) follows.Also for 0 and (12) follows.Since The main part of the proof consists of showing that (11) holds whenever ρ t → 1.We first dispose of one situation where the argument is straightforward.
Proof In this case we can write is a compound Poisson process which is independent of the spectrally positive process X (0) .Clearly

P {X
(1) t = 0} = e −tM (0) → 1 as t ↓ 0, so we have But, as previously mentioned, it was shown in [5] that this happens iff X (0) is a subordinator, i.e. it has bounded variation, so that ) → 0 as x ↓ 0, and we can write where the drift δ (0) is non-negative.Comparing this to the representation (2) of X we see that , where Π * denotes the Lévy measure of −X.If δ (0) > 0 the alternative expression which results from ( 5) by integration by parts shows that The next result allows us to make some additional assumptions about X in the remaining case.
Lemma 7 Let X # be any Lévy process with no Brownian component which has M # (0+) = ∞ and ρ # t = P (X # t > 0) → 1 as t ↓ 0. Then there is a Lévy process X with no Brownian component such that ρ t = P (X t > 0) → 1 as t ↓ 0 whose Lévy measure can be chosen so that (i) (ii) each of N and M is continuous and strictly decreasing on (0, c] for some c > 0. Moreover, (iii) (11) holds for X # if and only if it holds for X.
Proof Note that if X (1) has the same characteristics as X # except that Π # is replaced by where λ denotes Lebesgue measure, we have X (1) + Y (1) = X # + Y (2) , where Y (1) is a compound Poisson process independent of X (1) and Y (2) is a compound Poisson process independent of X # .Since P (Y are strictly decreasing, and (16) shows that A (1) as x ↓ 0 and we see that (11) holds for X (1) if and only if it holds for X # .This establishes (i), and allows us to assume in the remainder of the proof that N # (1) = M # (1) = 0, and both N # and M # are strictly decreasing on (0, 1].For (ii) it remains only to show that we can take N and M to be continuous.
So suppose that Π # has atoms of size a n and b n located at has only finitely many atoms there is nothing to prove, and the case when the restriction of Π # to (0, 1] or [−1, 0) has only finitely many atoms can be dealt with in a similar way to what follows.Note that from Now let Π (c) denote the continuous part of Π # , so that where δ(x) denotes a unit mass at x.With U [a, b] denoting a uniform probability distribution on [a, b] we introduce the measure We choose α n > 0, β n > 0 to satisfy the following conditions; for n = 1, 2, • • • , and Note that (17) and (19) imply that and hence (21) Now let X be a Lévy process with Lévy measure Π, no Brownian component, and having γ = γ # + c + c * .Since T (1) = T # (1) = 0 and we have got Π by 'moving some of the mass of Π # to the right', we have, for each fixed t > 0, Thus P (X t > 0) → 1, and to conclude we only need to show that (11) holds for X # if and only if it holds for X.Again using D(1) = D # (1) = 0 we have, for x ∈ (0, 1), where we have used (20) to see that and similarly for c * .Since (19) gives and the same argument applies to the second integral in (22), we see that The next piece of information we need is reminiscent of the Berry-Esseen Theorem; Lemma 8 Let µ be any Lévy measure, and write µ t for the restriction of µ to the interval [−b t , b t ], where b t ↓ 0 as t ↓ 0. Suppose that for each t > 0 Z t has an infinitely divisible distribution determined by and write Φ for the standard Normal distribution function.Then for any ε > 0 there is a positive constant W ε such that for all x Proof Note first that EZ t = 0 and EZ 3 t = t R x 3 µ t (dx) := tζ t , and write We will apply the inequality (3.13), p 512 of [6], with t fixed, where φ is the standard Normal density function.From it we deduce that for any T > 0 the LHS of (25) is bounded above by Here where M is an absolute constant, for all t satisfying |ν t | ≤ 1.But if (26) holds we have so this will hold provided 6W ε ≥ 1.Now fix T = 72M πε , so that the third term in ( 27) is no greater than ε/3.The same argument shows that the first term in ( 27) is also no greater than ε/3 provided (26) holds and 3W ε ≥ 1/ε.Finally, to deal with the middle term we write θ = θ √ tσt , and note that where for some positive constant c ε It follows from this and (28) that, increasing the value of W ε if necessary, we can make and clearly we can also arrange that whenever (26) holds.Putting these bounds into (27) finishes the proof.
Next we record a variant of the Lévy-Khintchine decomposition (2) which is important for us: Lemma 9 If X is any Lévy process with no Brownian component and b, b * ∈ (0, 1) and t > 0 are fixed we can write where are independent, and Proof This is proved in the same way as (2), except we compensate over the interval (−b * , b) rather than (−1, 1).
Finally we are in a position to establish the main technical estimate we need in the proof of Theorem 1; Proposition 10 Suppose that X is a Lévy process with no Brownian component whose Lévy measure satisfies N (0+) = M (0+) = ∞, and suppose d(t) and d * (t) satisfy for all small enough t > 0. Then there is a finite constant K such that, for any λ > 0, ρ > 0, L ≥ 0 there exists C = C(X, λ, ρ, L) > 0 with for all small enough t.
Proof We start by noting that if we use decomposition (31) for each fixed t with b and b * replaced by d(λt) and d * (ρλt), (32) gives (1,+) t has mean zero and since d(t) ↓ 0, because N (0+) = ∞, we can apply Lemma 8 to Y x 2 Π(dx).Choosing x = 0 and writing W for the W ε of Lemma 8 with ε = 1/4 we conclude from (25) that P {Y Thus in all cases we can fix K large enough that An exactly similar argument shows that we can fix a finite K * with P {Y Finally we note that if Z is a random variable with a Poisson(1/ρλ) distribution Combining ( 34)-(37) gives the required conclusion.

Proofs
Proof of Theorem 1.1 Since we have demonstrated in Lemma 5 the equivalence of ( 10) and (11), and Theorem 2.3 of [5] shows that (11) implies (7) under our assumptions, we will first show that ( 7) implies (11), and later their equivalence to (9) and ( 8) So assume (7), and also that M (0+) = ∞, since Lemma 6 deals with the contrary case.Then X satisfies the assumptions we made about X # in Lemma 7, so that result allows us to save extra notation by assuming that the conclusions (i) and (ii) of that lemma apply to X.For the moment assume also that N (0+) = ∞, so that we can define d and d * as the unique solutions of (32) on (0, t 0 ] for some fixed t 0 > 0. Our first aim is to show that lim inf To see this we use Proposition 10 with L = 0, which since P (X t ≤ 0) → 0 implies that for all sufficiently small t tγ(d(λt), d * (λρt)) + Kd(λt) ≥ 0. (39) Writing ν(x) = 1 x yΠ(dy) and ν * (x) = But in this we may choose λ arbitrarily small and ρ arbitrarily large, so (38) follows.
Assume next that (11) fails, and recall that Then for some sequence x k ↓ 0 and some , or equivalently then from (38) we have γ(x k , x k ) ≥ − 1 2 x k N (x k ) for all large enough k, and hence, using (40), and hence We now invoke Proposition 10 again, this time choosing t = t k , λ = K −1 , and ρ = 2D, to get ) whenever t = t k and k is large enough.However, in view of (41) the term on the right of the inequality is bounded above by If now we choose L = 2K we see that (42) contradicts (7); this contradiction implies that (11) is in fact correct.
We reached this conclusion making the additional assumption that N (0+) = ∞, but it is easy to see that it also holds if N (0+) < ∞.In this case by an argument we have used previously there is no loss of generality in taking N (0+) = 0, so that X is spectrally negative.We can then repeat the proof of Proposition 10 with b(t) ≡ 0, the conclusion being that for any L ≥ 0 there exists C = C(X, L) > 0 with for all sufficiently small t.When (7) holds this clearly implies that γ +ν * (x) ≥ 0 for all sufficiently small x.If (11) were false, we would have we again get a contradiction by choosing L sufficiently large.This completes the proof of the equivalence of ( 7) and (10).
Next, given 0 where ).If ln denotes the base 2 logarithm, we have k(x) ≥ ln 1/x, so (46) holds with Clearly f δ (x) increases as x ↓ 0, and if j = j(x) is the unique integer with We are now in a position to prove (9).We use Lemma 9 with b = b * = b δ (t).Replacing M {b δ (t)} by 0, observing that Y