By Tormod Naes

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**Sample text**

Since µ± (0+) are automatically ﬁnite and µ+ (0+) is ﬁnite because H+ is a compound Poisson process, we see immediately that 1 |x|Π(dx) = Π + (0+) + Π − (0+) < ∞. 4 A First Passage Quintuple Identity 49 This, together with σ = 0, means that X has bounded variation, and this contradiction establishes the claim. 12) is immediate, because it is equivalent to the fact that for any a, Xt +at visits both half-lines immediately, and of course Xt + at is also an inﬁnite variation L´evy process. 4). for the process killed at time Tx , which played a rˆ The corresponding result for random walks is easily established, but again the proof for L´evy process is more complicated.

As t → ∞ if ˜ − = ∞, and this process satisﬁes t−1 X ˜ + = EX EX 1 1 and only if m+ = ∞. 4 Creeping Let us ﬁrst dispose of some easy cases. As we have seen (Corollary 4, Chapter 4) σ 2 = 2δ + δ − , so we will take σ 2 = 0; then at least one of these drifts has to be 0. e. t−1 Xt → γ˜ as t ↓ 0), and this is similar to the subordinator case: δ + > 0 if and only if γ˜ > 0. 1), the compound Poisson term Y (2) has no eﬀect on whether X creeps, since it is zero until the time at which the ﬁrst ‘large jump’ occurs.

For the process killed at time Tx , which played a rˆ The corresponding result for random walks is easily established, but again the proof for L´evy process is more complicated. Recall the notation Gt = sup {s ≤ t : Xs = Ss } , put γ x = G(Tx −) for the time at which the last maximum prior to ﬁrst passage over x occurs, and denote the overshoot and undershoot of X and undershoot of H+ by Ox = X(Tx ) − x, Dx = x − X(Tx −), and Dx(H) = x − S(Tx −). Theorem 18. Suppose that X is not a compound Poisson process.