7
$\begingroup$

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

$\endgroup$
  • $\begingroup$ 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
10
$\begingroup$

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 $$\int_{|z| \leq 1} z^2 \, \nu(dz)<\infty.$$
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.