# Show that $Z_t = Z_0 \exp\left( \mu t + X_t \right)$ is well defined where $X_t$ is a Jump process (Lévy process)

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?

• (1.) $\delta_n$? (2.) Surely you have (been given) some hints?
– Did
Jan 30 '13 at 8:46
• @did: (1) it's $\delta_{-n}$ actually. (2) I'm really not familiar with lévy process so I don't see the definition problem of $Z_t$. Could you please enlighten me?
– Paul
Jan 30 '13 at 18:14
• @Did : Can you help me ? This question had no too much attention. I'll appretiate your enlightement
– Paul
Jan 31 '13 at 18:40

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 $\nu((x,+\infty))=\mathrm e^{-\lambda x+o(x)}$.

• It's $\delta_{-n}$ not $\delta_n$. So at the end you should have $\mathbb E(\mathrm e^{\theta X_t})=\exp\left(t\sum_{n\geqslant1}p^n(\mathrm e^{-\theta n}-1)\right).$. Of course $p \in (0,1)$, but I forgot to say. Also, $A,B, \lambda, \mu \geq$
– Paul
Jan 31 '13 at 23:45
• Most important, I didn't get why and how have you past from the expected value of Fourier trasformation for Laplace's one. Could you please detail it formally? Thank you in advance!
– Paul
Jan 31 '13 at 23:54
• See Edit.  
– Did
Feb 1 '13 at 15:23
• Thank you very much for your answer. It really helps me to understand it all.
– Paul
Feb 1 '13 at 21:32