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:

Theorem 8.1

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 $\Delta X_t:=X_t-X_{t-}$.


1 Answer 1


The second equality is a consequence of the strong Markov property.

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.

  • $\begingroup$ Thank you for your answer. Could you explain how exactly the second equality follows from the strong Markov property? (And could you include your definition of the strong Markov property?) $\endgroup$
    – 0xbadf00d
    Nov 3, 2021 at 15:28
  • $\begingroup$ I'd still be interested in this. Would be great if you could add an explanation. $\endgroup$
    – 0xbadf00d
    Nov 16, 2021 at 16:11

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