4
$\begingroup$

For a nondecreasing Levy process $\mathbf{X}$ with values in $[0,\infty)$ (i.e. a subordinator) Jean Bertoin defines the potential measure of $\mathbf{X}$ in his book "Levy processes" as follows (p. 74): \begin{align*} U=U(0,A):=\mathbb{E}\Big[\int_{0}^{\infty}\mathbb{1}_{\{X_t\in A\}}dt\Big]. \end{align*} I already figured out that for an $\alpha$-stable subordinator one can find the density of $U$ wrt Lebesgue: \begin{align*} U(dy)=\frac{1}{\Gamma(\alpha)}y^{\alpha-1}dy. \end{align*}

Now I wonder if it is possible to find the density of the potential measure of $\mathbf{X}:=(\mathbf{X}_1,\mathbf{X}_2)$, where the $\mathbf{X}_i$ are independent $\alpha$-stable subordinators, as well?

EDIT: The following calculation may indicate that it is NOT the product measure:

\begin{align*} U(0,A\times B)=\mathbb{E}\Big[\int_{0}^{\infty}\mathbb{1}_{X_1(t)\in A}\mathbb{1}_{X_2(t)\in B}dt\Big]=\int_{0}^{\infty}\mathbb{P}[X_1(t)\in A]\mathbb{P}[X_2(t)\in B] dt \end{align*}

by Fubini and the fact that $\mathbf{X}_1$ and $\mathbf{X}_2$ are independent. The RHS is in general different from $\int_{0}^{\infty}\mathbb{P}[X_1(s)\in A] ds\int_{0}^{\infty}\mathbb{P}[X_2(t)\in B] dt$ I would guess.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.