Let $X$ be a càdlàg Lévy process, $\pi$ denote the unique random measure with $$\pi_\omega([0,t]\times B)=\sum_{s\in[0,\:t]}1_B\left({\Delta X_s(\omega)}\right),$$ where $\Delta x(t):=x(t)-\lim_{s\to t-}x(s)$, $$\zeta_t(\omega,B):=\pi_\omega([0,t]\times B)$$ and $$\tau:=\inf\{t>0:\Delta X_t\in B\}$$ for some Borel measurable $B$ with $0\not\in\overline B$.

We can show that $$\operatorname P[\tau>s+t]=\operatorname P[\tau>s]\operatorname P[\tau>t]\tag1.$$ Since $$\phi(t):=\operatorname P[\tau>t]$$ is right-continuous, this implies that $$\phi(t)=e^{t\ln\phi(1)}\tag2.$$

Question: By the result above, $\tau$ is exponentially distributed, but how do we see that $\tau\sim\operatorname{Exp}(\lambda)$, where $\lambda:=\operatorname E\left[\zeta_1(\;\cdot\;,B)\right]$?

Clearly, $$\operatorname{Exp}(\lambda)((t,\infty))=e^{-\lambda t}\tag3.$$ So, it seems like we need to show $$\ln\phi(1)=-\lambda(B).\tag4$$ But how do we do this?

**BOUNTY EDIT**

I still don't understand how we obtain $(4)$.

## Best Answer

$A \to \pi_{\omega}(A \times B)$ is the random measure corresponding to a Poisson process on $[0,\infty)$. The rate of this Poisson process, is also the rate of the exponential waiting time for the first point of the process. See Jumps of Lévy process. So what is being used here is a simple but powerful property of the Poisson process on a halfline: The rate $\lambda$ of the exponential waiting time for the first point of the process, equals the expected number of points of the process that land in the unit interval. https://en.wikipedia.org/wiki/Poisson_point_process#Interpreted_as_a_point_process_on_the_real_line