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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.

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    $\begingroup$ 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
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    $\begingroup$ 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
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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 $\Psi(u) = \lambda \left(\mathrm{e}^{i u x} - 1 \right)$.

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