# Asymptotic behavior of Lévy processes

I'm trying to prove the next proposition:

Let $\{X_{t}\}_{t\geq 0}$ be a non-zero Lévy process on $\mathbb{R}.$ Then it satisfies one of the following three conditions: $$i)\space\displaystyle\lim_{t\rightarrow\infty}X_{t}=\infty\space a.s.;$$ $$ii)\displaystyle\lim_{t\rightarrow\infty}X_{t}=-\infty\space a.s.;$$ $$iii)\space\displaystyle\limsup_{t\rightarrow\infty}X_{t}=-\infty\space a.s.\space\text{and}\space\displaystyle\liminf_{t\rightarrow\infty}X_{t}=-\infty\space a.s.$$ To prove this consider the next:

We can suppose that $\{X_t\}_{t\geq 0}$ is recurrent or transient because of the dicotomy of Lévy processes. Then, if $\{X_{t}\}_{t\geq 0}$ is recurrent, we have that $\displaystyle\liminf_{t\rightarrow\infty}|X_{t}|=0\space a.s.$ Then $iii)$ holds.

Now, if $\{X_t\}_{t\geq 0}$ is transient $\displaystyle\lim_{t\rightarrow\infty}|X_{t}|=\infty.$ Let $$M=\displaystyle\limsup_{t\rightarrow\infty}X_{t}\space\text{and}\space N=\displaystyle\liminf_{t\rightarrow\infty}X_{t}.$$

Since $|X_{t}(\omega)|\rightarrow\infty,$ no finite points is a limit point of $X_{t}(\omega)$ as $t\rightarrow\infty.$ Hence $$P(M=\infty\space\text{or}\space M=-\infty)=1.$$

Kolmogorov's $0-1$ law implies that $P(M=\infty)=1$ or $0.$ It follows that either $P(M=\infty)=1$ or $P(M=-\infty)=1.$ Similarly, either $P(N=\infty)=1$ or $P(N=-\infty)=1.$ Hence one of $i),ii)$ and $iii)$ holds.

$\textbf{My first doubt is:}$ How $\{X_t\}_{t\geq 0}$ recurrent implies $iii)$ holds? I don't get how such limitis have extreme values.

$\textbf{Second doubt:}$ How Kolmogorov's law works to get $P(M=\infty)=1$ or $0?$ I don't get how to prove that $\{M=\infty\}$ belongs to the tail $\sigma-$algebra and the random variables $X_{t}$ are independent.

Any kind of help is thanked in advanced.

$\textbf{Edit:}$ Now I know how to use Kolmogorov's $0-1$ law in this case. The only thing that I don't get yet is my first doubt. Any kind of help is thanked in advanced.

Set

$$M := \limsup_{t \to \infty} X_t.$$

Since $(X_t)_{t \geq 0}$ is a non-trivial Lévy process, we can find some $t_0>0$ such that $X_{t_0}$ is non-trivial. If we consider the restarted process

$$\tilde{X}_t := X_{t+t_0}-X_{t_0}$$

then $(\tilde{X}_t)_{t \geq 0}$ is a Lévy process which equals in law $(X_t)_{t \geq 0}$. In particular,

$$M' := \limsup_{t \to \infty} \tilde{X}_t$$

equals in law $M$. On the other hand, we clearly have by the very definition of $\tilde{X}_t$ that

$$M' = M-X_{t_0}.$$

The independence of the increments yields that $M'$ is independent from $X_{t_0}$, and so

$$\mathbb{E}(\exp(i \xi M') 1_{\{|M'|<\infty\}}) = \mathbb{E}(\exp(i \xi M) 1_{\{|M|<\infty\}}) \mathbb{E}\exp(-i \xi X_{t_0})\tag{1}$$

for any $\xi \in \mathbb{R}$ As $X_{t_0}$ is non-trivial there exists a sequence $(\xi_n)_{n \in \mathbb{N}}$ such that $\xi_n \to 0$ and $0<|\mathbb{E}\exp(i \xi X_{t_0})|<1$ for all $n \in \mathbb{N}$. Equation $(1)$ gives

$$\mathbb{E}(\exp(i \xi_n M') 1_{\{|M'|<\infty\}}) = \mathbb{E}(\exp(i \xi_n M) 1_{\{|M|<\infty\}}) =0$$ and letting $n \to \infty$ we get

$$\mathbb{P}(|M|< \infty) =0.$$

On the other hand, Kolmogorov's zero one law shows that $\mathbb{P}(M = \infty) \in \{0,1\}$ and $\mathbb{P}(M=-\infty) \in \{0,1\}$. Combining these considerations we find that

$$M = \limsup_{t \to \infty} X_t = \infty \, \, \text{a.s.} \quad \text{or} \quad M = \limsup_{t \to \infty} X_t= - \infty \, \, \text{a.s.}$$

A very similiar reasoning applies for $\liminf$ and noting that $\liminf \leq \limsup$ this yields that always one of the three conditions (i)-(iii) holds true.

• Thanks @saz. Very clear! – Squird37 May 8 '18 at 6:07
• @Squird37 You are welcome. If you like the answer (... it seems like that...), you can upvote it by clicking on the up arrow next to it. – saz May 8 '18 at 6:21
• Of course @saz. I've done it. I think I have a problem with my computer because I always upvote when the answer is really good, but some people say me that I haven't done it when actually already I've done! – Squird37 May 8 '18 at 6:39
• @Squird37 I see; that's odd. Well, never mind... Glad that I could help you. – saz May 8 '18 at 7:20
• Sorry for asking you now; I thought that I understand every step but I have a couple doubts: How the independent increments yields that $M^{'}$ is independent of $X_{t_{0}}?$ And the equation $(1)$ becomes zero because $M$ and $M^{'}$ has the same distribution, right? Thanks for the help @zas. – Squird37 May 9 '18 at 6:16