# Convergence in distribution using strong law of large numbers

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.

• Maybe there is a typo – anonymus May 23 '18 at 17:56
• @anonymus where ? – user331066 May 23 '18 at 18:19
• The indicator function is on the whole term : ? – anonymus May 23 '18 at 19:41
• @anonymus ah perhaps, the result is true if the indicator is on the whole term ? – user331066 May 23 '18 at 20:51
• 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. – quid May 24 '18 at 13:03

## 1 Answer

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.