Finding the Levy triplet of a Levy process

Question

I know the levy triplet of a Poisson process

$N_t$- $(0,0,\lambda\delta_{1}(y))$ and its characteristic function is

$\phi_N=exp[-t\Bigl(\intop_{0}^{\infty}(1-e^{iuy}+iuy1_{\{\mathbf{|}\mathbf{y}|<1\}})\delta_{1}(y)\Big)]$

and also that of the standard $\alpha$ stable subordinator

$D_t$ - $(\frac{iu\alpha}{\Gamma(1-\alpha)},0,\frac{\alpha}{\Gamma(1-\alpha)}y^{-\alpha-1}dy)$ and its characteristic function is

$\phi_{D}(u)=exp[-t\bigl(-\frac{iu\alpha}{\Gamma(1-\alpha)}+\intop_{0}^{\infty}(1-e^{iuy}+iuy1_{\{\mathbf{|}\mathbf{y}|<1\}})\frac{\alpha}{\Gamma(1-\alpha)}y^{-\alpha-1}dy\bigr)]$

My question is how do I find the triplet of $(N_t,D_t)$ where $N_t$ and $D_t$ are independent?

Since in my case the two processes are independent the Levy measure of $(N_t,D_t)$ should be $v_{x}(dy_{1},dy_{2})=\delta_{1}(y_{1})\frac{\alpha}{\Gamma(1-\alpha)}y_{2}^{-\alpha-1}dy_{2}$ (or am I wrong?), in that case the characteristic function should look like this

$exp[-t\Bigl(-iu_{1}-\frac{iu_{2}\alpha}{\Gamma(1-\alpha)}+\intop_{\mathbb{R}^{2}/\{0\}}\bigl(1-e^{iu_{1}y_{1}+iu_{2}y_{2}}+iu_{1}y_{1}1_{\{\mathbf{|}\mathbf{y}|<1\}}+iu_{2}y_{2}1_{\{\mathbf{|}\mathbf{y}|<1\}}\bigr)\delta_{1}(y_{1})\frac{\alpha}{\Gamma(1-\alpha)}y_{2}^{-\alpha-1}dy_{2}\Bigr)]$

so the Levy triplet should be of the form $\bigl((b_1,b_2),0,\delta_{1}(y_{1})\frac{\alpha}{\Gamma(1-\alpha)}y_{2}^{-\alpha-1}dy_{2}\bigr) $

so my problem boils down to what is the vector $(b_1,b_2)$?

---

I'm not sure I can take the cut-off to be $1_{\{\mathbf{|}\mathbf{y}|<1\}}$ since in that case the integral $\intop_{0}^{\infty}\bigl(1-e^{iu_{1}y_{1}+iu_{2}y_{2}}+iu_{1}y_{1}1_{\{\mathbf{|}\mathbf{y}|<1\}}+iu_{2}y_{2}1_{\{\mathbf{|}\mathbf{y}|<1\}}\bigr)\delta_{1}(y_{1})\frac{\alpha}{\Gamma(1-\alpha)}y_{2}^{-\alpha-1}dy_{2}\Bigr)]$ diverges, or can I?

Answer 1

You are looking for the Lévy symbol of a 2-dimensional process $(N,D)$, so your symbol will be a function on $\mathbb R^2$. It will be $\phi_{N,D}(u_1, u_2) = \phi_N(u_1) + \phi_D(u_2)$, by the following trivial argument.

By independence of $N$ and $D$, the characteristic function of $(N_t, D_t)$ is $\mathbb E\left[ \exp\left(i (u_1 N_t + u_2 D_t) \right) \right] = \mathbb E \left[ \exp \left( i u_1 N_t \right) \right] \mathbb E \left[ \exp \left( i u_2 D_t \right) \right] = \exp \left( t \phi_N(u_1) \right) \exp \left( t \phi_D(u_2) \right)$.