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 $(b\_1,b\_2)$?




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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?



Finding the Levy triplet of a Levy process
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
$\mathbb E\left[ \exp\left(i (u\_1 N\_t + u\_2 D\_t) \right) \right] = \mathbb E \left[ \exp \left( i u\_1 N\_t \right) \right] \mathbb E \left[ \exp \left( i u\_2 D\_t \right) \right] = \exp \left( t \phi\_N(u\_1) \right) \exp \left( t \phi\_D(u\_2) \right)$.


