# stopping time almost surely finite

Let $$(X_n)_n$$ be a sequence of independent random variables and identically distributed such that $$P_{X_1}=p\delta_1+q\delta_{-1}+r\delta_0$$ where $$0 \leq p,q,r<1$$ and $$p+q+r=1.$$ Let $$\alpha, \beta \in \mathbb{Z}$$ such that $$\alpha<0<\beta.$$ Let $$Y_n=\sum_{k=1}^nX_k$$ for $$n \in \mathbb{N^*}$$ and $$T=\inf(n \in \mathbb{N^*};Y_n \notin ]\alpha,\beta[).$$ $$T$$ is a stopping time for $$(\mathcal{F_n})_n$$ where $$\mathcal{F_n}=\sigma(X_1,...,X_n)$$.

I need to prove that $$T$$ is a.s finite ($$T<+\infty$$ a.s.) by:

1) using the central limit theorem,

2) proving that there exists $$\epsilon>0$$ and $$n_0 \in \mathbb{N^*}$$ such that $$\forall n \in \mathbb{N^*}, P(T \leq n+n_0|\mathcal{F_n})>\epsilon.$$

I know that 2) is a characterisation which can prove that $$T$$ is finite a.s.
So how can we prove that $$T=\inf(n \in \mathbb{N^*};Y_n \notin ]\alpha,\beta[)$$ by applying the central limit theorem and by verifying the property 2).

I am thankful for any idea.

I will just consider the case where $$\mathrm E[X_1]=0$$, as otherwise thanks to the strong law of large numbers, $$|Y_n|\to \infty$$, $$n\to\infty$$, supplying the finiteness of $$T$$.
The central limit theorem implies the convergence in distribution $$Z_n:=\frac{Y_n}{\sigma \sqrt{n}}\overset{d}{\longrightarrow} Z\simeq N(0,1), n\to\infty,$$ where $$\sigma^2=\mathrm{Var}(X_1$$). Choose some $$\epsilon\in(0,1)$$. Let $$n_0$$ be such that $$\beta-\alpha<\epsilon\sigma\sqrt{n_0}$$ and $$\mathrm{P}(|Z_{n_0}|\le \epsilon)< \mathrm{P}(|Z|\le \epsilon) + \frac{\epsilon}{5}.$$ Noting that $$\mathrm{P}(|Z|\le \epsilon) \le \frac{2 \epsilon}{\sqrt{2\pi}}< \frac{4\epsilon}{5},$$ we get that $$\mathrm{P}(|Z_{n_0}|\le \epsilon)<\epsilon$$. Thanks to the choice of $$n_0$$, $$\mathrm{P}(|Y_{n_0}|\le \beta-\alpha) <\epsilon.$$ However, $$\mathrm{P} (T > n+n_0\mid \mathcal{F_n})\le \mathrm{P} (|Y_{n+n_0}- Y_n| \le \beta-\alpha\mid \mathcal{F_n})\\ = \mathrm{P} (|Y_{n+n_0}- Y_n| \le \beta-\alpha) = \mathrm{P}(|Y_{n_0}|\le \beta-\alpha) <\epsilon,$$ as required (though with $$\epsilon$$ replaced by $$1-\epsilon$$, but this does not matter).