I am trying to see the connection between the Lévy-Khintchine and the integrability conditions of a Lévy measure. The literature seems to always connect both, but I cannot make sense of this relation nor found a proof. Perhaps somebody can give me some insight into the matter:

The Lévy-Khintchine formula for Lévy processes is given by $$ \varphi(u) := i\alpha u - \frac{1}{2}\sigma^2u^2 + \int_{|z|<1}(e^{iuz}-1-iuz)\,\nu(dz) + \int_{|z|\geq 1}(e^{iuz}-1)\,\nu(dz) $$ where the parameters $\alpha \in \mathbb R$ and $\sigma^2>0$ are constants. From this formula, it seems like we need to impose that and $\nu$ is a finite measure satisfying $$ \int_{\mathbb R \backslash \,0} \min(1,z^2)\,\nu(dz) < \infty $$

Why does it follow from the expressions above described that $\nu$ is a valid Lévy measure of some Lévy process? In particular, why the minimum expression?

Thanks in advance

  • $\begingroup$ Try a Taylor expansion of the function $z\mapsto \mathrm{e}^{\mathrm{i}uz}-1-\mathrm{i}uz$ for $|z|<1$. $\endgroup$ – Stefan Hansen Aug 6 '14 at 11:06
  • 1
    $\begingroup$ @StefanHansen I guess that's what Adam already figured out - that this condition is necessary.. but he wants to know why it is natural for a Lévy process to satisfy these conditions. Am I right, Adam? $\endgroup$ – saz Aug 6 '14 at 19:23
  • $\begingroup$ @saz yes you are correct, that would help very much. $\endgroup$ – Adam Aug 6 '14 at 19:50

The condition

$$\int_{\mathbb{R} \backslash \{0\}} \min\{1,z^2\} \nu(dz)<\infty$$

is equivalent to

$$\int_{|z| \leq 1} z^2 \, \nu(dz) < \infty \quad \text{and} \quad \int_{|z| \geq 1} \nu(dz) < \infty.$$

Let's discuss them separately; for simplicity of notation we consider the $1$-dimensional case.

  1. Any Lévy process $(X_t)_{t \geq 0}$ has càdlàg sample paths. This implies that almost surely the sample paths $t \mapsto X_t(\omega)$ have only finitely many jumps with jump height $>\varepsilon$ on compact ($t$-)sets for any $\varepsilon>0$. And that's exactly why the condition $$\int_{|z| \geq 1} \nu(dz)<\infty \tag{1}$$ holds. In order to make this more precise, we define for $U \subseteq \mathbb{R}$ $$N^U_t(\omega) := \sharp \{s \leq t; \Delta X_s(\omega) \in U\}$$ the corresponding counting measure. So, basically, we count the jumps up to time $t$ with jump height in a set $U$. Then one can show that $(N_t^U)_{t \geq 0}$ is a Poisson process with intensity $\nu(U)$ whenever $U$ satisfies $$U \cap B(0,\varepsilon) = \emptyset \tag{2}$$ for some $\varepsilon>0$. In $(1)$, we are interested in the set $U := \{z; |z| \geq 1\}$ which obviously satisfies $(2)$.
  2. The condition $$\int_{|z| \leq 1} z^2 \, \nu(dz) \tag{3}$$ is basically due to the fact that the jump heights of a Lévy process $(X_t)_{t \geq 0}$ are square summable; in fact, $$\mathbb{E} \left( \sum_{s \leq t} |\Delta X_s|^2 \right)<\infty.$$ (Note that this implies in particular $\sum_{s \leq t} |\Delta X_s|^2 < \infty$ almost surely.) This result is due to some (more general) results on stochastic integrals with respect to random measures. One might still wonder why it is natural for a Lévy process to have square-summable jump heights. Unfortunately, I don't have an intuitive explanation for this fact; I just want to provide a short proof why $(3)$ holds (and which does not rely on the Lévy-Khinchine formula). To this end, we define for fixed $\varepsilon>0$ $$X_t^{\varepsilon} := \sum_{s \leq t} \Delta X_s \cdot 1_{\{1>|\Delta X_s|>\varepsilon\}}.$$ Then this process is again a Lévy process. The infinitely divisibility entails that $$0 \neq |\mathbb{E}e^{\imath \xi X_t}| \leq |\mathbb{E}e^{\imath \, \xi X_t^{\varepsilon}}| \tag{4}$$ for all $\varepsilon>0$. Actually, $(X_t^{\varepsilon})_{t \geq 0}$ is even a compound Poisson process and this means that we can calculate the characteristic function explicitly. This yields $$|\mathbb{E}e^{\imath \, \xi X_t^{\varepsilon}}| = \exp \left(- \int_{\delta<|z|<1} (1-\cos(z \xi)) \, \nu(dz) \right).$$ Now using $(1-\cos(x)) \approx \frac{1}{4} x^2$ for $|x| \leq 1$, we see that $$0 \neq |\mathbb{E}e^{\imath \xi X_t}| \leq \exp \left(- \int_{\delta<|z|<1} (\xi z)^2 \, \nu(dz) \right)$$ for $|\xi| \leq 1$. Now monotone convergence proves $$\int_{|z| \leq 1} z^2 \, \nu(dz)<\infty.$$

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