Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Let $X_t$ be the compound Poisson process $$ X_t = t - \sum_{i=1}^{N_t} \xi_i, \tag{1} $$ where $N$ is a Poisson process with rate $\lambda$ and the $\xi_i$ are i.i.d., positive, with common distribution $F$.
The characteristic exponent of $X_t$ is $$ \Psi(\theta) = \theta - \lambda \int_{(0, \infty)} \left( 1 - e^{-\theta x}\right)F(dx). \tag{2} $$ Assume $\Psi$ has a root $\theta^* \ne 0$. Define the stopping time $$ \tau = \inf_{t > 0} \left\{ X_t > x \right\} , x > 0. \tag{3} $$
Show $$ E\left[\exp\left(\theta^*X_\tau -\Psi(\theta^*)\tau\right)1_{\{\tau < \infty\}}\right] = e^{\theta^*x}P\left( \tau < \infty \right). \tag{4} $$

This is exercise 1.9 in Kyprianou, Fluctuation of Levy Process.
The solution is given, but I don't see why (4) holds.
As $\theta^*$ is a root, the left of (4) reduce to $$ E\left[\exp\left(\theta^*X_\tau \right)1_{\{\tau < \infty\}}\right]. $$ But, by (2), $X_t$ is a jump process. So, I cannot infer that $$ X_\tau = x, \text{a.s.} $$

share|cite|improve this question
up vote 2 down vote accepted

In (4), first replace $\Psi(\theta^*)$ by its definition, $1$. Then note that $t\mapsto X_t$ has only negative jumps and $X_0=0\lt x$ hence $X_\tau=x$ on the event [$\tau$ finite] and $\exp(\theta^*X_\tau)\mathbf 1_{\tau\lt\infty}=\exp(\theta^*x)\mathbf 1_{\tau\lt\infty}$ almost surely.

share|cite|improve this answer
Very clear, thank you. – Nicolas Essis-Breton Jan 19 '13 at 0:45

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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