After my last post went horribly wrong I want to do things right this time. I'm researching Lévy process in my free time as part of my interest in stochastic processes. Unlike different processes I've seen that the moments of Lévy process are not easy to find.
My first step in this journey was reading the paper: First exit times of SDEs driven by stable Lévy processes, there in eq 4.46 , they calculated the expectation of a Lévy $\xi$ process time $\epsilon$ squared, i.e $\mathop{\mathbb{E}}[\epsilon\xi]^2$, where $\xi$ follows this CF:
$E[e^{i\lambda\xi}]=exp\{-d\frac{\lambda^2}{2}+\int_{ \mathbb{R} \mathbin{/}\{0\}}(e^{i\lambda y}-1-i\lambda y\mathbb{I}\{|y|<1\}\mathbb{I}\{|y|\leq \frac{1}{\sqrt{\epsilon}}\frac{dy}{|y|^{1+\alpha}}\}) \}$
As written in the paper in Eq.(2.5). Two note,first $0<\epsilon<1$, second, I don't know why they didn't write the drift term. The calculation of $\mathop{\mathbb{E}}[\epsilon\xi]^2$ was as follows:
$\mathop{\mathbb{E}}[\epsilon\xi]^2 = t\epsilon^2(\frac{2}{2-\alpha}\epsilon^{\frac{\alpha}{2}-1}+d)$
After a long searching on the web, I found this paper , that claims (page 10):
$E[\xi] = t(\int_{\{|y|>1}y\frac{dy}{|y|^{1+\alpha}\}}+b)$
Where b is the drift. So I tried to use it, but I got:
$E[\epsilon\xi] = \epsilon t(\int_{\{1<|y|<\epsilon^{-\frac{1}{2}}} y\frac{dy}{|y|^{1+\alpha}\}}+b) =\frac{\epsilon t}{1-\alpha}[\epsilon^{-\frac{1}{2}(1-\alpha)}-1]$
I'm just not able to get the same term, So I kept on looking, and I found this
paper, there in equation 2.6 they claimed:
$E[\xi] \leq t(\int_{\{|y|>1}y\frac{dy}{|y|^{1+\alpha}\}}+b)$
i.e that there is an inequality and not equality, where of course $t>0,\epsilon>0$. I'm really confused about how to calculate the expectation of a Lévy process, I would happy for some advice. Thanks!!