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I have a question about . Let us denote the Lévy measure $\nu$ defined on $\mathbb{R}^d\setminus\{0\}$ and with $N$ the Poisson random measure such that $E[N(t,S)]=t\,\nu(A)$ for all $t>0$ and $S\subset\mathbb{R}^d\setminus\{0\}$.
If $\nu(\mathbb{R}^d\setminus\{0\})<\infty$, then the expected number of jumps per unit time is finite and we have that the process is of finite activity, i.e. the number of jumps for any compact interval of time is finite. For this reason, it was natural to me thinking that $\nu(\mathbb{R}^d\setminus\{0\})<\infty$ implied also $\int_{\mathbb{R}^d} |x|^n\,\nu(dx)<\infty$ for any $n\in\mathbb{N}$ (since I only have a finite number of jumps). But, actually, this is not true and it is linked to the behavior of small jumps. Has someone an intuitive explanation for this?

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Take any random variable $X$ such that $\mathbb{E}(|X|^n) = \infty$ and set $\nu := \mathbb{P}_{X}$. Then $\nu( \mathbb{R}^d \backslash \{0\}) \leq 1$, but $\int |x|^n \, \nu(dx) = \infty$.


The reason why your intuition fails is the following: For each fixed $\omega \in \Omega$, the number of jumps of $t \mapsto X_t(\omega)$ is finite for compact $t$-intervals. However, the number of jumps as well as the jump heights depend on $\omega$. If we take the expectation of $X_t$, then we have to take the average over all $\omega \in \Omega$ and this may be infinite.

Have a look at the following example (this is not a Lévy process, but it shows where your argumentation fails): Consider $(0,1)$ endowed with the Lebesgue measure and define

$$X_t(\omega) := \frac{1}{\omega} \cdot 1_{\big[\frac{1}{2},\infty \big)}(t), \qquad t \geq 0, \omega \in (0,1).$$

Then, by definition, $X_t(\omega)=0$ for all $t \in [0,1/2)$ and $X_t(\omega) = 1/\omega$ for $t \geq 1/2$; in particular, we have a finite number of jumps on compact time intervals. For $t=1$, we have $X_1(\omega) = \frac{1}{\omega}$ and therefore $\mathbb{E}X_1 = \infty$.

Similarly, one can construct an example where not the jump height, but the frequency of the jumps does depend on $\omega$.

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  • $\begingroup$ Thanks for the answer. One little question: why $X_t(\omega) =1$ for $t\geq1$? $\endgroup$ – NH2 Jun 12 '15 at 8:02
  • $\begingroup$ @NH2 sorry, this was a typo. $\endgroup$ – saz Jun 12 '15 at 8:29

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