# Does the Levy process stay in any open ball at any fixed time with positive probability?

Question: Let $X$ be a $d$-dimensional Levy process. Then for every $t>0$ and $a>0$, $$\tag{1} \mathbf P\{|X_t|<a\}>0\ ?$$

The question comes from the proof of Lemma 35.5 in Sato's book. I think that it is true, and I try to prove it.

Firstly, it is easy to see that the proposition holds true when $X$ is a Brownian motion or a compound Poisson process. Indeed, if $X$ is a Brownian motion, then the random variable $X_t$ is Gaussian and stays with positive probability in the ball $B_a(0)$ centering at $0$ with radius $a$. If $X$ is a compound Poisson process, then the first waiting time $T_1:=\inf\{t\ge0\mid X_t\neq 0\}$ has exponential distribution. This implies that $\mathbf P\{|X_t|<a\} \ge \mathbf P\{X_t= 0\}\ge\mathbf P\{T_1\ge t\}>0$.

Now if $X$ is a general Levy process with triplet $(b,A,\nu)$, then we can use Levy-Ito decomposition to write $$\tag{2} X_t=bt+B^A_t+C_t^{(0)}+\sum_{n=1}^\infty C_t^{(i)},$$ where $B^A$ is a Brownian motion with covariance $A$, $C^{(0)}$ is a compound Poisson process, $\{C^{(i)}\mid i=1,2,\cdots\}$ is a family of compensated compound Poisson processes, and all the processes on the right hand side of (2) are mutually independent.

If there are only finite terms in the summation in (2), then we can show (1) by the inequality $$\mathbf P\left\{\left|\sum_{i=1}^n \xi_i \right|<a\right\}\ge \prod_{i=1}^n\mathbf P\left\{|\xi_i|<\frac{a}{n}\right\},\quad \text{for independent sequence } \{\xi_i\mid i=1,2,\cdots,n\}.$$

But how to deal with the general case that it is a infinite sum in (2)?

I have no idea then... Could anyone give some hints or comments? TIA!

Edit: We would like to assume in addition that there is no drift $bt$ in the decomposition (2) of the Levy process $X$, due to the counterexample provided by @Skorohod (see below).

If $X$ is a Levy process in $\mathbb{R}$ with only postive jumps ($C_t$ a standard poisson process for example) and positive drift:$$X_t = t + C_t$$ if $t >a$, $\ \mathbb{P}(|X_t|< a )=0$ since at time $a$, $X_a\geq a$.
Basically, pick any subordinator $\sigma_t$ add some drift $t$, and it will not satisfy your claim after time $a$.