# How to find Lévy Triplets

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?

$$\mathbb{E} e^{i \xi N_1} = \sum_{k \le 0} e^{i \xi k} \frac{ \lambda^k}{k!} e^{- \lambda} = exp( \lambda( e^{i \xi} - 1))$$
$$\lambda( e^{i \xi} - 1) = \int \lbrace e^{i \xi u} - 1 \rbrace \ \lambda \delta_1(du)$$
• Yes 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. – Kore-N Feb 14 '17 at 16:20