2
$\begingroup$

The Levy-Khintchine formula tells us what the characteristic function of a Levy process looks like. Given a process $Y_t$, the characteristic function of $Y_1$ is given by \begin{equation} \phi_1(u) = e^{\Psi(u)}, \end{equation} where \begin{equation} \Psi(u) =i \mu u - \frac{1}{2}\sigma^2 u^2 +\int_{\mathbb{R}} \left(e^{i u x} -1 - i u x \mathbf{1}_{(-1,1)} \right) \nu(\mathtt{d} x ). \end{equation} Is $e^{-\Psi(u)}$ also the characteristic function of a Levy process? Thanks in advance.

$\endgroup$
  • 1
    $\begingroup$ I don't know much about this stuff (Levy processes, etc) but wouldn't the minus sign ruin the boundedness of $\exp(\Psi(u))$? $\endgroup$ – guy Aug 10 '12 at 15:43
  • $\begingroup$ No, $\Psi$ is complex valued: $\Psi: \mathbb{R}\to \mathbb{C}$.And $|\phi_1(u)|\leq 1$ for all $u$ (characteristic function of a probability measure). $\endgroup$ – angry_pacifist Aug 10 '12 at 16:46
  • 1
    $\begingroup$ Take the Wiener process; $-\Psi(u) = \frac 1 2 \sigma^2 u^2$ so $\exp(-\Psi(u))$ is unbounded for real valued $u$. $\endgroup$ – guy Aug 10 '12 at 17:04
2
$\begingroup$

HINT: Consider simple examples of Levy processes: Wiener process, Poisson process and check if $\exp\left(-\Psi(u)\right)$ is a characteristic function of a random variable. (see criteria, and use examples: first, second, third, fourth)

For Wiener process, $\Psi(u) = -\frac{1}{2} \sigma^2 u^2$, for Poisson process $\Psi(u) = \lambda \left(\mathrm{e}^{i u x} - 1 \right)$.

$\endgroup$

Your Answer

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

Not the answer you're looking for? Browse other questions tagged or ask your own question.