Consider a random walk on the real time, starting from $0$. But this time assume that we can decide, for each step $i$, a step size $t\_i>0$ to the left or the right with equal probabilities. 


To formalize this, we have $(X\_n)\_{n\geq 0}$ such that $X\_0=0$, $Pr[X\_n=t\_n]=0.5$ and $Pr[X\_n=-t\_n]=0.5$ (hence still $E(X\_i)=0$). Let $I=[-1,1]$ and $$p=Pr[\sum\_{i=1}^n X\_i \in I. \forall n\geq 0]$$ 


The question is: does there exist a sequence $(t\_i)$ such that $t\_i>0$ and $\sum\_{i=0}^\infty=\infty$ (i.e. $(t\_i)$ diverges), to ensure $p>0$. 


Clearly such a sequence does not exist to let $p=1$, and intuitively it seems to me that neither exists $(t\_i)$ to force $p>0$, but I am looking for a rigorous proof or disproof. Moreover, I am not exactly sure whether this kind of questions has been studied, so references would be appreciated. 



Random walk to stay in an interval forever
Yes. Indeed, if $s = \sum\_{i \geq 1} t\_i^2 <1$, then 
$$
\mathbb{P}[ \ \ \forall n, \sum\_{i=1}^n X\_i \in [-1,1] \ \ ] \geq 1-s > 0.
$$
To see this, note that $M\_n = |\sum\_{i=1}^n X\_i|$ is a nonnegative submartingale, so that Doob's martingale inequality yields
$$
\mathbb{P}[ \max\_{1 \leq j \leq n} M\_j > 1 ] \leq \mathbb{E}[M\_n^2] = \sum\_{i=1}^n t\_i^2 \leq s.
$$
Letting $n$ tends to infinity, one gets
$$
\mathbb{P}[ \sup\_{j \geq 1} M\_j > 1 ] \leq s,
$$
hence the result by taking the complement.


Remark : In the other direction, one can show that if
$$
\mathbb{P}[ \ \ \forall n, \sum\_{i=1}^n X\_i \in [-1,1] \ \ ] \geq c > 0,
$$
then 
$$
s = \sum\_{i \geq 1} t\_i^2 \leq \frac{14}{c^2}.
$$
One first note that one must have $|t\_i| \leq 2$ for all $i$. One has $|\cos(2 \pi \xi)| \leq e^{-2 \pi^2 \xi^2 }$ whenever $|\xi| \leq 2 \delta$, with $\delta = 0.14$, hence
$$
|\mathbb{E}[e^{-2i \pi \xi \sum\_{j=1}^n X\_j} ]| = \prod\_{j=1}^n |\cos(2 \pi \xi t\_j)| \leq e^{-2 \pi^2 \xi^2 \sum\_{i=1}^n t\_i^2}
$$
whenever $|\xi| \leq \delta$. Let $\chi$ be the Beurling-Selberg majorant of $\mathbb{1}\_{[-1,1]}$ with parameter $\delta$ : one has $\chi \geq \mathbb{1}\_{[-1,1]}$, $||\chi||\_{L^1}= 2 + \delta^{-1}$, and its Fourier transform $\hat{\chi}$ is supported on $[-\delta,\delta]$. In particular :
$$
c \leq \mathbb{E}[\chi(\sum\_{i=1}^n X\_i)] = \int\_{\mathbb{R}} \hat{\chi}(\xi) \mathbb{E}[e^{-2i \pi \xi \sum\_{j=1}^n X\_j} ] d \xi \leq (2 + \delta^{-1}) \int\_{\mathbb{R}} e^{-2 \pi^2 \xi^2 \sum\_{i=1}^n t\_i^2} d \xi,
$$
and thus
$$
c \leq \frac{2 + \delta^{-1}}{\sqrt{2 \pi \sum\_{i=1}^n t\_i^2}}.
$$
This implies
$$
\sum\_{i=1}^n t\_i^2 \leq \frac{(2 + \delta^{-1})^2}{2 \pi c^2} \leq \frac{14}{c^2},
$$
hence the result.


