# Recurrent Lévy process implies $\limsup_{t\rightarrow\infty}X_{t}=\infty$ and $\liminf_{t\rightarrow\infty}X_{t}=-\infty$

Let $\{X_{t}\}_{t\geq 0}$ be a non-zero recurrent Lévy process on $\mathbb{R}.$ Then $$\space\displaystyle\limsup_{t\rightarrow\infty}X_{t}=\infty\space\text{a.s.}\space\text{and}\space\displaystyle\liminf_{t\rightarrow\infty}X_{t}=-\infty\space a.s.$$

If $\{X_{t}\}_{t\geq 0}$ is recurrent then $\displaystyle\liminf_{t\rightarrow\infty}|X_{t}|=0.$ But I can't see why this implies the above.

Intuitively, a process is recurrent at $x\in\mathbb{R}$ if the process visits every neighbourhood of $x$ an infinite number of time. So this implies that $\limsup$ and $\liminf$ take infinite value, however, I don't get why such limits take infinite positive and negative values. I'd like to have a formal reason for this but I don't get how $\displaystyle\liminf_{t\rightarrow\infty}|X_{t}|=0$ implies the desired.

Any kind of help is thanked in advanced.

So, you want to show $\liminf |X_t| = 0$ implies $X$ is recurrent and then show that $\limsup X_t=\infty$ & $\limsup X_t=- \infty$?

For a Lévy process, let $R$ be the set of recurrent states: $$R = \big\{ x \in \mathbb{R} \, | \, \forall t>0 \text{ and }\forall \, \epsilon>0, \, \ \mathbb{P}( \exists s>t, \text{ such that } X_s \in B_\epsilon(x))=1 \big\}$$ For non-pathologic Lévy processes (taking values in $\mathbb{R}$), $R= \emptyset$ or $R=\mathbb{R}$.

1-Let's first show that $R$ is nonempty.

If $\liminf |X_t| =0$ a.s. then outisde of a null set $N$:

$$0=\liminf |X_t| = \lim_{t \rightarrow \infty} \inf_{s >t} |X_s|$$ For any arbitrary $\epsilon$, let $B_\epsilon(0)$ be a ball centered in 0. If $\omega \in N^c$, $\forall t>0 \ \exists s_\omega>t$ such that $X_{s_\omega} \in B_\epsilon(0)$. (Because if it didn't happened, then $|X|$ stays away from 0 after t and contradicts $0=\liminf |X_t|$ ).

You can deduce from this that $0 \in R$. Since $R$ is nonempty, $R=\mathbb{R}$ and $X$ is recurrent. Now let's show the second claim:

2-Let's show that $\limsup X_t = \infty$

Since $R = \mathbb{R}$, for any $t>0$ and $n \in \mathbb{N}$, $$\mathbb{P}(\exists s>t \ \text{ such that } X_s \in B_1(n)) = 1$$

Let: $$M^t := \bigcap _{n \in \mathbb{N}} \big\{ \exists s >t \, \text{ such that } X_s \in B_1(n) \big\}$$ Then for any $t$, $\mathbb{P}(M^t) =1$. This means that, after time $t$, $X$ becomes arbitrarly large a.s. since $X$ enters all the balls $B_1(n)$. However,it's not enought: we need to show this stil happens for arbitrarelly large $t$.

Now take an arbitrary sequence $\{t_k \}_{k \in \mathbb{N}}$ of times such that $t_k \rightarrow \infty$. Since $$\mathbb{P}(M^{t_k})= \mathbb{P} \left( \bigcap _n \big\{ \exists s >t_k \, \text{ such that } X_s \in B_1(n) \big\}\right) = 1$$ for all $k$, then $$1=\mathbb{P} \left(\bigcap_{k}M^{t_k}\right) = \mathbb{P}\left( \bigcap_{k}\bigcap _{n} \big\{ \exists s >t_k \, \text{ such that } X_s \in B_1(n) \big\}\right)$$ Let $t>0.$For any $w \in \bigcap_{k}\bigcap _{n} \big\{ \exists s >t_k \, \text{ such that } X_s \in B_1(n) \big\}$, pick a $t_k >t$. By construction of this set, for all $n$, $X_s \in B_1(n)$ for a large enought $s>t_k$. Since $n$ is arbitrary, $\sup_{s>t}X_s = \infty$. Now since $t$ is arbitrary, then a.s. for all $t>0$, $\, \sup_{s>t}X_s = \infty$ and then$$\limsup X_s = \lim_{t \rightarrow \infty} \sup_{s>t} X_s = \infty$$

The proof for $\liminf X_t = - \infty$ is similar.

• Very helpul @Skorohod. Thanks for the help! – Squird37 May 8 '18 at 6:09