I have recently started to learn about Lévy processes, and have learnt about the Lévy-Khintchine theorem. My question is about the Poisson process, it stated to have Lévy triplet (0,0,$\lambda\delta(1)$), where $\delta(1)$ is the dirac measure with unity at 1, and zero elsewhere. But I can't seem to find an example showing this, is the Lévy triplet derived more from intuition about the process than through the Lévy-Khintchine theorem?
How to find Lévy Triplets
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1 Answers
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The Lévy - Kinchine theorem tells you that the characteristic function of a Lévy process has a certain form. Not how to compute this. For the Poisson process it is an easy exercise to compute the characteristic triplet:
$$\mathbb{E} e^{i \xi N_1} = \sum_{k \le 0} e^{i \xi k} \frac{ \lambda^k}{k!} e^{- \lambda} = exp( \lambda( e^{i \xi} - 1))$$
Taking the logarithm we get our representation:
$$\lambda( e^{i \xi} - 1) = \int \lbrace e^{i \xi u} - 1 \rbrace \ \lambda \delta_1(du)$$
Eventually I think it is important to understand that the Lévy - Kinchine theorem tells you that a general Lévy process actually behaves quite like a Poisson process, up to a drift and a Brownian motion. In fact the term involving the Lévy measure is always a compound Poisson Process unless the measure is infinite.
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0Are the triplets unique? Or can there be a characteristic be describe by the Lévy-Khintchine form with different triplet values? – 2017-02-14
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0Yes the triplets are unique. For example suppose that you have a triplet such that the associated $\psi$ is zero. Now the second order term prevails on the others at $\xi \to \infty$ so it has to vanish. Finally the term under the integral is not linear, thus both the remaining terms have to vanish as well. – 2017-02-14