By Tormod Naes
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Proof. On account of the strict convexity ψ it follows that Φ(0) > 0 if and only if ψ (0+ ) < 0 and hence lim q↓0 q = Φ(q) 0 ψ (0+ ) if ψ (0+ ) 0 if ψ (0+ ) > 0. By taking q to zero in the identity (10) we now have that E eαX ∞ = 0 ψ (0+ )α/ψ(α) if ψ (0+ ) 0 if ψ (0+ ) > 0. Next, recall from (12) that for α > 0 E e−αX eq = Φ(q) Φ(q) + α and hence by taking the limit of both sides as q tends to zero, E e−αX ∞ = (α/Φ (0) + 1)−1 0 if ψ (0+ ) < 0 if ψ (0+ ) 0. Parts (i)–(iii) follow immediately from the previous two identities by considering their limits as α ↓ 0.
The second author would like to thank EURANDOM for their support. In addition, both authors gratefully acknowledge grant nr. 310 from Nederlandse Organisatie voor Wetenschappelijk Onderzoek. References 1. E. and Pistorius, M. (2004) Exit problems for spectrally negative L´evy processes and applications to (Canadized) Russian options, Ann. Appl. Probab. 14, 215-238. A martingale review of some ﬂuctuation theory 29 2. Bertoin, J. (1996a) L´evy processes, Cambridge University Press. 3. Bertoin, J.
The diﬀerence-diﬀerential equation (12) yields 1 ˆ ˆ −e−λ h(1) + λh(λ) − β h(λ) + e−λ ˆ ˆ e−λx h(0)eαx dx + e−λ h(λ) − h(λ) =0 0 which can be rewritten, taking into account the previous results, ˆ (λ − β + e−λ − 1)h(λ) − e−λ h(0) eα − eα−λ − 1 α−λ =0 or, since β = ψ(−α), h(0) ˆ ψ(−α) − ψ(λ) ψ(−λ) − ψ(−α) h(λ) = eα−λ α−λ This gives: eα−λ ˆ = h(0) h(λ) = h(0) λ−α Hence, h(x) = h(0)eαx for all x Wald martingale. ∞ e−λx eαx dx. 1 0, and we have found again the classical We now turn, until the end of this section, to the special case when X is spectrally negative.