Take two dependent Levy processes $L_1(t)$ and $L_2(t)$ with law $\mathcal{L}(L_1(1),L_2(1)$. If we stop the first process at a general time $t=s_1$ and stop the second process at another general time $t=s_2$ can we say anything about the distribution of the random variables $(L_1(s_1),L_2(s_2))$?
Any reference will be highly appreciated. Thank you.
Edit : The bivariate process $(L_1,L_2)$ is assumed Levy as well.
Edit : The time the process is stopped is a deterministic time.
This is what I worked out. The characteristic function of $(L_1(s_1),L_2(s_2))$ can be written as:
\begin{equation} E[exp(i(u_1L_1(s_1)+u_2L_2(s_2))]=E[exp(i(u_1L_1(s_1)+u_2L_2(s_1)+u_2I(s_2-s_1))] \end{equation} for $s_1<s_2$ and I(t)representing the increment from $s_1$ to $s_2$ in the component $L_2$.
Next, we condition on $I$ (or better on the filtration generated by it)(this might be a problem) : \begin{equation} E[exp(i(u_1L_1(s_1)+u_2L_2(s_1)+u_2I(s_2-s_1))]=E[e^{iu_2I(s_2-s_1)}E[e^{i(u_1L_1(s_1)+u_2L_2(s_1)}|I]] \end{equation} Now, by the properties of Levy processes we get: \begin{equation} E[e^{iu_2I(s_2-s_1)}E[e^{i(u_1L_1(s_1)+u_2L_2(s_1)}|I]]=E[e^{iu_2I(s_2-s_1)}e^{s_1\Psi_{L_1,L_2}(u_1,u_2)}]=e^{s_1\Psi_{L_1,L_2}(u_1,u_2)}E[e^{u_2I(s_2-s_1)}]=e^{s_1\Psi_{L_1,L_2}(u_1,u_2)}e^{(s_2-s_1)\Psi_I(u_2)} \end{equation} where $\Psi_{L_1,L_2}$ and $\Psi_I$ are the characteristic exponents of the respective processes.
Can you spot any mistakes or flaws in the above reasoning? Thank you.