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Consider a Lévy process $\left( X_t\right)_{t\geq 0} $ whose Lévy measure is $\nu$.

Why have we for all $u \in \mathbb R$ the following result?

$\mathbb E \left[ \exp (u X_t )\right] < \infty$ if and only if $\int_{|z|>1} e^{uz} ~\nu(dz) < \infty$

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up vote 1 down vote accepted

Call (1) the condition that $\mathbb E(\mathrm e^{uX_t})$ is finite for every nonnegative $t$ and $u$, and (2) the condition that $\displaystyle\int_{x\geqslant1}\mathrm e^{ux}\nu(\mathrm dx)$ is finite for every nonnegative $u$.

The equivalence of (1) and (2) is Theorem 25.3 in K.-I. Sato's book Lévy processes and infinitely divisible distributions.

Likewise, $\mathbb E(\mathrm e^{-uX_t})$ is finite for every nonnegative $t$ and $u$ if and only if $\displaystyle\int_{x\leqslant-1}\mathrm e^{-ux}\nu(\mathrm dx)$ is finite for every nonnegative $u$.

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Thank you one more time. – Paul Feb 2 '13 at 15:55

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