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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.

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1 Answer 1

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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).

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  • $\begingroup$ Thank you, every thing is clear! $\endgroup$
    – mathex
    Apr 7, 2020 at 21:40

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