# How do we see that the jump time of a Lévy process is exponentially distributed using the Markov property?

Let $$E$$ be a normed $$\mathbb R$$-vector space, $$X$$ be a càdlàg Lévy process on a probability space $$(\Omega,\mathcal A,\operatorname P)$$, $$B\in\mathcal B(E)$$ with $$0\not\in\overline B$$ and$$^1$$ $$\tau:=\inf\{t>0:\Delta X_t\in B\}.$$

How do we show that $$\tau$$ has exponential distribution?

As usual, we "only" need to show that $$\operatorname P[\tau>s+t]=\operatorname P[\tau>s]\operatorname P[\tau>t]\tag1$$ for all $$s,t\ge0$$.

We know that $$X$$ is a time-homogeneous Markov process with transition semigroup $$\kappa_t(x,B):=\operatorname P\left[x+X_t\in B\right]\;\;\;\text{for }(x,B)\in E\times\mathcal B(E)\text{ and }t\ge0.$$

We can show that $$\kappa(x,B):=\operatorname P[x+X\in B]\;\;\;\text{for }(x,B)\in E\times\mathcal B(E)^{\otimes[0,\:\infty)}$$ with source $$(E,\mathcal B(E))$$ and target $$\left(E^{[0,\:\infty)},\mathcal B(E)^{\otimes[0,\:\infty)}\right)$$. We can now define $$\operatorname P^{\mu}:=\mu\kappa$$ as the composition of a probability measure $$\mu$$ on $$(E,\mathcal B(E))$$ and $$\kappa$$. In particular, we define $$\operatorname P^x:=\operatorname P^{\delta_x}$$, where $$\delta_x$$ is the Dirac measure at $$x\in E$$.

In the special case $$E=\mathbb R$$ and $$B=\{1\}$$, we can find the following proof in Theorem 8.1 of From Lévy-Type Processes to Parabolic SPDEs:

Above, $$\tau_1=\tau$$ and $$\operatorname P^{X_s}=\kappa(X_s,\;\cdot\;)$$.

How do the authors obtain the second and third equality above? And how do we generalize this to the setting of my question?

I don't know, but might it be easier to proof this by considering the coordinate process $$(\pi_t)_{t\ge0}$$ on $$E^{[0,\:\infty)}$$? For every probability measure $$\mu$$ on $$(E,\mathcal B(E))$$, this process is clearly a time-homogeneous Markov process with respect to $$\operatorname P^\mu$$ with transition kernel $$(\kappa_t)_{t\ge0}$$.

Moreover, if we define $$\theta_s:E^{[0,\:\infty)}\to E^{[0,\:\infty)}\;,\;\;\;x\mapsto(x_{s+t})_{t\ge0}$$ for $$s\ge0$$, we've got $$\operatorname E\left[f\circ\theta_s\circ X\mid\mathcal F^X_s\right]=\int\kappa(X_s,{\rm d}y)f(y)\tag2$$ and, analogously, $$\operatorname E^{\mu}\left[f\circ\theta_s\mid\mathcal F^\pi_s\right]=\int\kappa(\pi_s,{\rm d}y)f(y)\;\;\;\text{for every probability measure }\mu$$ for all $$s\ge0$$ and bounded $$\mathcal B(E)^{\otimes[0,\:\infty)}$$-measurable $$f:E^{[0,\:\infty)}\to\mathbb R$$, where $$\mathcal F^Y_t:=\sigma(Y_s,s\le t)$$ for $$t\ge0$$ is the filtration generated by a process $$Y$$.

$$^1$$ As usual, $$X_{t-}:=\lim_{s\to t-}X_s$$ and $$\Delta X_t:=X_t-X_{t-}$$.

The third is because of the spatial homogeneity of a Levy process: $$X=(X_t)_{t\ge 0}$$ under $$\kappa(x,\cdot)$$ has the same distribution as $$x+X=(x+X_t)_{t\ge 0}$$ under $$\kappa(0,\cdot)$$. Moreover, $$X$$ and $$x+X$$ have identical jumps.