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Let $(X_n)_n$ be an i.i.d sequence of random variables defined on the same probability space and denote $S_n=\frac{1}{n}\sum_{k=1}^n X_k.$

Assume that $\mathbb{E}(X_1)\ne0$ and given $\alpha>0$ that $$\sqrt{n}(X_n-\alpha\mathbb{1}_{\{S_n\ne 0\}})\overset{d}{\to} \frac{X_1}{\mathbb{E}(X_1)}.$$

I would like to deduce that $\sqrt{n}(X_n-\alpha)$ converge in distribution. My idea is to use the strong long of large numbers that $\mathbb{1}_{\{S_n\ne 0\}}\overset{\text{a.e.}}\to 1$ but not sure how can I conclude.

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  • $\begingroup$ Maybe there is a typo $\endgroup$
    – anonymus
    May 23, 2018 at 17:56
  • $\begingroup$ @anonymus where ? $\endgroup$
    – user331066
    May 23, 2018 at 18:19
  • $\begingroup$ The indicator function is on the whole term : ? $\endgroup$
    – anonymus
    May 23, 2018 at 19:41
  • $\begingroup$ @anonymus ah perhaps, the result is true if the indicator is on the whole term ? $\endgroup$
    – user331066
    May 23, 2018 at 20:51
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    $\begingroup$ Please do not just delete question that have an answer. If there is a particular reason why you want to do this, please at least explain it. $\endgroup$
    – quid
    May 24, 2018 at 13:03

1 Answer 1

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Write $$\sqrt{n}(X_n-\alpha)=\sqrt{n}(X_n-\alpha \mathbb{1}_{\{S_n\ne 0\}})+\sqrt{n}\alpha(\mathbb{1}_{\{S_n\ne 0\}}-1).$$ Since $\mathbb{1}_{\{S_n\ne 0\}}$ can take values $0\vee 1$ only, convergence $\mathbb{1}_{\{S_n\ne 0\}}\overset{\text{a.e.}}\to 1$ means that for every elementary event $\omega\in A$, where $\mathbb P(A)=1$, there exists $N=N(\omega)$ s.t. for all $n\geq N$ $$ \mathbb{1}_{\{S_n\ne 0\}}(\omega):=\mathbb{1}_{\{S_n(\omega)\ne 0\}} = 1. $$ So, for every $\omega\in A$ and for each $n\geq N$ the additional term vanishes: $$\sqrt{n}\alpha(\mathbb{1}_{\{S_n\ne 0\}}-1)(\omega)=0.$$ We proved that $$ \sqrt{n}\alpha(\mathbb{1}_{\{S_n\ne 0\}}-1)\overset{\text{a.e.}}\to 0. $$ Next use Slutsky theorem as proposed by anonymus.

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