Consider $X= \left( X_t \right)_{t\geq 0}$ is a Lévy process whose characteristic triplet is $\left( \gamma, \sigma ^2, \nu \right)$ and where its Lévy measure is $$ \nu \left( dx\right) = A \sum_{n=1} ^{\infty} p^n \delta_{-n} \left( dx \right) + Bx^{\beta-1}\left( 1+x \right)^{-\alpha -\beta}e^{-\lambda x } \mathbf{1}_{\left ]0,+\infty \right[}\left( x\right)dx.$$
I'd like to know how to show that $Z_t = Z_0 \exp\left( \mu t + X_t \right)$ is well defined and admits first and second order moments.
I'm kind of lost here. I don't see what is the problem with this definition. Could someone please enlighten me ?
Must I show that $Z_t < \infty \ a.s.$ ?
Or maybe aplly Itô-Lévy lemma for derive the SDE $Z_t$ satisfies and so conclude that it's well defined as the unique strong solution of this SDE?
Or maybe another thing I've not even think about?
Let us look at the simple case where $\gamma=\sigma^2=B=\mu=0$ and $A=Z_0=1$. By definition of the Lévy measure $\nu$, for every real number $\theta$, $$ \mathbb E(\mathrm e^{\mathrm i\theta X_t})=\exp\left(t\int_{\mathbb R\setminus\{0\}}(\mathrm e^{\mathrm i\theta x}-1-\mathrm i\theta x\mathbf 1_{|x|\lt1})\,\nu(\mathrm dx)\right), $$ thus, $$ \mathbb E(\mathrm e^{\mathrm i\theta X_t})=A(\mathrm i\theta),\qquad A(z)=\exp\left(t\sum_{n\geqslant1}p^n(\mathrm e^{-nz}-1)\right). $$ The function $A$ is analytical on the disk $D=\{z\in\mathbb C\mid |z|\lt-\log p\}$ hence there exists some complex sequence $(A_n)_n$ such that, for every $z$ in $D$, $$ A(z)=\sum_{n\geqslant0}A_n\frac{z^n}{n!}. $$ Differentiating $2n$ times the identity $\mathbb E(\mathrm e^{\mathrm i\theta X_t})=A(\mathrm i\theta)$ with respect to $\theta$ at $\theta=0$ yields $\mathbb E(X_t^{2n})=A^{(2n)}(0)=A_{2n}$. This proves that, for every $z$ in $D$, $\mathbb E(\cosh(zX_t))$ converges, hence $z\mapsto\mathbb E(\mathrm e^{zX_t})$ is analytic on $D$, and equal to $A$ there. Thus, for every real number $\theta$ such that $|\theta|\lt-\log p$, $$ \mathbb E(\mathrm e^{\theta X_t})=\exp\left(t\sum_{n\geqslant1}p^n(\mathrm e^{-\theta n}-1)\right). $$ For example $\mathrm e^{X_t}$ and $\mathrm e^{-X_t}$ are integrable if $p\lt\mathrm e^{-1}$. What happens if $p\geqslant\mathrm e^{-1}$ is that the jumps of length $-n$ for $n\geqslant1$ generated by the discrete part of $\nu$ are too numerous hence $X_t=-\infty$ almost surely, for every $t\gt0$.
The proof when one includes the continuous part of $\nu$ on $(0,+\infty)$ is similar but the final result might depend on a balance between the parameters $p$ and $\lambda$ which describe the behaviour of $\nu((-\infty,-x))$ and $\nu((x,+\infty))$ when $x\to+\infty$ since $\nu((-\infty,-x))=p^{x+o(x)}$ and $\nu((x,+\infty))=\mathrm e^{-\lambda x+o(x)}$.