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Consider $X_i$'s as iid random variables with mean 0 that are not point mass (they are non-degenerate) and they have finite variance. $a_i$'s are constants that are finite which converge to $0$. How can I show that that $\sum_{i = 1}^{n} a_iX_i /\sqrt{n} \rightarrow 0$ in distribution?

Thank you

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    @Learner Irrelevant.2013-09-20

2 Answers 2

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Denote $Y_n = \sum_{i = 1}^{n} a_iX_i /\sqrt{n}$. We have $\mathrm{Var}[Y_n] = \mathrm{Var} [X] \times \frac{\sum_{i = 1}^{n} a_i^2 }{n} \to 0$ as $n\to \infty$. Therefore, by Chebyshev's inequality for every $\varepsilon > 0$ $\Pr[|Y_n| > \varepsilon] \leq \mathrm{Var}[Y_n]/\varepsilon^2 \to 0 \quad \text{ as } n\to\infty.$ Therefore, $Y_n \to 0$ in probability and thus $Y_n \to 0$ in distribution.

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    @Eli Do you know why $\sum\limits_{i\leqslant n}a_i^2/n\to0$?2013-09-20
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If $X_i$ is a positive random variable and $a_i = 1/\sqrt{i}$ then you will not get this convergence to $0$.

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    I see. Now let's consider $E[X_i] = 0$. I guess that should fix it...I'll add this to the question.2012-12-13