# Levy process : $E\left[\exp\left(\theta^*X_\tau -\Psi(\theta^*)\tau\right)1_{\{\tau < \infty\}}\right] = e^{\theta^*x}P\left( \tau < \infty \right)$

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 $$X_\tau = x, \text{a.s.}$$

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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.