How do I find the Levy triplet of a Levy process

Question

I know the levy triplet of a Poisson process

$N_t$- $(0,0,λδ_1(y))$ and its characteristic function is

$exp[-t\Bigl(\intop_{0}^{\infty}(1-e^{iuy}+iuy1_{\{\mathbf{|}\mathbf{y}|<1\}})\delta_{1}(y)\Big)]$ and also that of the standard α stable subordinator

$D_t - (αΓ(1−α),0,αΓ(1−α)y^{−α−1}dy)$ 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 $\delta_{1}(y_{1})\frac{\alpha}{\Gamma(1-\alpha)}y_{2}^{-\alpha-1}dy_{2}$ (or am I wrong?), in that case the characteristic function should look like this

$exp[-t\Bigl(-ib_1u_{1}-ib_2u_{2}+\intop_{\mathbb{R}^{2}/\{0\}}\bigl(1-e^{iu_{1}y_{1}+iu_{2}y_{2}}+iu_{1}y_{1}1_{\{\mathbf{|}\mathbf{y}|<1\}}+iu_{2}y_{2}1_{\{\mathbf{|}\mathbf{y}|<1\}}\bigr)\delta_{1}(y_{1})\frac{\alpha}{\Gamma(1-\alpha)}y_{2}^{-\alpha-1}dy_{2}\Bigr)]$

so the Levy triplet should be of the form $((b_1,b_2),0,δ_1(y_1)αΓ(1−α)y^{−α−1}_2dy_2)$ 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 $iuy1_{\{\mathbf{|}\mathbf{y}|<1\}}$ since in that case the integral $\intop_{\mathbb{R}^{2}/\{0\}}\bigl(1-e^{iu_{1}y_{1}+iu_{2}y_{2}}+iu_{1}y_{1}1_{\{\mathbf{|}\mathbf{y}|<1\}}+iu_{2}y_{2}1_{\{\mathbf{|}\mathbf{y}|<1\}}\bigr)\delta_{1}(y_{1})\frac{\alpha}{\Gamma(1-\alpha)}y_{2}^{-\alpha-1}dy_{2}\Bigr)]$ diverges, or can I?

Answer 1

Let $(N_t)_{t \geq 0}$, $(D_t)_{t \geq 0}$ (arbritrary) one-dimensional Lévy processes with Lévy triplet $(\ell_1,q_1,\nu_1)$ and $(\ell_2,q_2,\nu_2)$, respectively. Note that we can determine the Lévy triplet of $(D_t,N_t)$ (uniquely) from the characteristic function of $(D_1,N_1)$, i.e.

$$\mathbb{E}\exp \left( \imath \, (\xi,\eta) \cdot (D_1,N_1) \right)$$

By the independence of the processes, we have

$$\mathbb{E}\exp \left( \imath \, (\xi,\eta) \cdot (D_1,N_1) \right) = \mathbb{E}\exp(\imath \, \xi \cdot D_1) \cdot \mathbb{E}\exp(\imath \, \xi \cdot N_1)$$

Using Lévy-Khinchine's formula we conclude

$$\begin{align*} & \mathbb{E}\exp \left( \imath \, (\xi,\eta) \cdot (D_1,N_1) \right) \\ &= \exp \left( - \imath \, \ell_1 \xi - \frac{1}{2} q_1 \xi^2 - \int (e^{\imath \, y \xi}-1-\imath \, y \xi 1_{|y|<1}) \, d\nu_1(y) \right) \\ & \quad \cdot \left( - \imath \, \ell_2 \eta - \frac{1}{2} q_2 \eta^2 - \int (e^{\imath \, y \eta}-1-\imath \, y \eta 1_{|y|<1}) \, d\nu_2(y) \right) \\ &= \exp \left(- \imath \, \ell \cdot (\xi,\eta) - \frac{1}{2} (\xi,\eta) Q (\xi,\eta) - \int (e^{\imath \, (y_1,y_2) \cdot (\xi,\eta)}-1-\imath \, (\xi,\eta) \cdot (y_1,y_2) \cdot 1_{|y|<1}) \, d\nu(y_1,y_2) \right) \end{align*}$$

where

$$\begin{align*} \ell &:= \begin{pmatrix} \ell_1 \\ \ell_2 \end{pmatrix} \\ Q &:= \begin{pmatrix} q_1 & 0 & \\ 0 & q_2 \end{pmatrix} \\ \nu &:= \nu_1 \otimes \delta_{0} + \delta_0 \otimes \nu_2 \end{align*}$$

Consequently, $(\ell,Q,\nu)$ equals the Lévy triplet of $(D_t,N_t)$.

In your case, we have

$$\ell_1 = 0 \quad q_1 = 0 \quad \nu_1 = \lambda \delta_1 \\ \ell_2 = \alpha \Gamma(1-\alpha) \quad q_2 = 0 \quad d\nu_2(y) = \alpha \Gamma(1-\alpha) y^{-\alpha-1} \, dy$$

Thus,

$$\begin{align*} \ell&=\begin{pmatrix} 0 \\ \alpha \Gamma(1-\alpha) \end{pmatrix} \\ Q &= 0 \\ d\nu(y_1,y_2) &= \lambda d\delta_1(y_1) d\delta_0(y_2) + \alpha \Gamma(1-\alpha) {y_2}^{-\alpha-1} \, dy_2 d\delta_0(y_1) \end{align*}$$