Let $E$ be a normed $\mathbb R$-vector space and $(X_t)_{t\ge0}$ be an $E$-valued Lévy process on a filtered probability space $(\Omega,\mathcal A,(\mathcal F_t)_{t\ge0},\operatorname P)$, i.e.

  1. $X_0=0$;
  2. $X$ is $\mathcal F$-adapted and continuous in probability;
  3. $X_{s+t}-X_s$ is independent of $\sigma(X_r,r\le s)$ for all $s,t\ge0$;
  4. $X_{s+t}-X_s\sim X_t$ for all $s,t\ge0$.

We can show that if $X$ is almost surely right-continuous, $\tau$ is a finite $\mathcal F$-stopping time, $$Y_t:=X_{\tau+t}-X_\tau$$ and $$\mathcal G_t:=\mathcal F_{\tau+t}$$ for $t\ge0$, then $Y$ is $\mathcal G$-Lévy and $X\sim Y$.

Now assume $X$ is càdlàg and $$c:=\sup_{t\ge0}\left\|\Delta X_t\right\|_E<\infty$$ (as usual, $\Delta X_t:=X_t-\lim_{s\to t-}X_s$). Let $\tau_0:=0$ and $$\tau_n:=\inf\left\{t\ge\tau_{n-1}:\left\|X_t-X_{\tau_{n-1}}\right\|_E\ge c\right\}$$ for $n\in\mathbb N$.

Question: How can we show that $(\tau_n)_{n\in\mathbb N}$ is an independent identically distributed process?

The claim must obviously follow from (3.) and (4.), but how do we need to argue precisely? My idea is that the mentioned fact about the "restarted" Lévy process $Y$ might be useful ...

In fact, if we define $\Omega_n:=\{\tau_n<\infty\}$, $\mathcal A_n:=\left.\mathcal A\right|_{\Omega_n}$, $\operatorname P_n:=\frac{\left.\operatorname P\right|_{\Omega_n}}{\operatorname P[\Omega_n]}$ and $X^{(n)}_t:=\left.X_t\right|_{\Omega_n}$, $\mathcal F^{(n)}_t:=\left.\mathcal F_t\right|_{\Omega_n}$ for $t\ge0$ and $n\in\mathbb N$ with $\operatorname P[\Omega_n]>0$, then we obtain, for fixed $t\ge0$, by (3.) and (4.) that $X^{(n)}_{\left.\tau_n\right|_{\Omega_n}+t}-X^{(n)}_{\left.\tau_n\right|_{\Omega_n}}$ is independent of $\mathcal F^{(n)}_{\left.\tau_n\right|_{\Omega_n}}$ and distributed as $X^{(n)}_{\left.\tau_n\right|_{\Omega_n}}$ ...

But is this useful?


2 Answers 2


Formulated this way the answer is no. The reason is that with your definition you will have $\tau_0\leq\tau_1\leq\dotsc$.

Imagine the situation where $X$ is a Poisson process in $\mathbb{R}$ and $c=1/2$. Then the $\tau_n$ are the jump times of $X$ which are not independent. However, in this case the differences $\tau_n-\tau_{n-1}$ are i.i.d.

I suggest you have a look a the differences instead. You should be able to write something like $$ \tau_n-\tau_{n-1}=\inf\{t\geq0:\Vert X_{\tau_{n-1}+t}-X_{\tau_{n-1}}\Vert_E\geq c\}. $$ Now the answer by user1118 applies.


First let's show independence. Since $\tau_n$ is a stopping time, it is $\mathcal F_{\tau_n}$-measurable. In particular, if some random variable is independent of $\mathcal F_{\tau_n}$ then it is also independent of $\tau_n$.

Let $Y^{(n)}$ be the stochastic process defined by $Y^{(n)}_t = X_{\tau_n + t} - X_{\tau_n}$, $t\ge0$. As you said, this is Lévy with respect to the filtration $(\mathcal F_{\tau_n + t}, t\ge0)$. Therefore for all $t\ge 0$, $$ X_{\tau_n + t} - X_{\tau_n} = Y^{(n)}_t = Y^{(n)}_t -Y^{(n)}_0 \text{ is independent of }\mathcal F_{\tau_n + 0} = \mathcal F_{\tau_n}. $$ Since $\tau_{n+1}$ is by definition $\sigma(Y^{(n)}_t,t\ge0)$-measurable, this means that $\tau_{n+1}$ is independent of $\mathcal F_{\tau_n}$, and therefore of $\tau_n$. Induction finishes this part of the proof.

Now we show identical distribution. Since $Y^{(n)}$ has the same law as $X$, $$ ||X_{\tau_n + t} - X_{\tau_n}||_E = ||Y^{(n)}_t||_E \sim ||X_t||_E \qquad \text{for all }n\in \mathbb N_0\text{, all }t\ge0. $$ In particular, each $\tau_n$ has the same law as $$ \inf\{t\ge0 : ||X_t||_E \ge c\}. $$


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .