convergence in distribution of a sum of random variables

Question

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

At what rate do $a_i$ converge to 0? – Learner Dec 13 '12 at 01:39 — Learner, Dec 13 '12 at 01:39

score 0 · Answer 1 · answered Dec 13 '12 at 01:38

0

If $X_i$ is a positive random variable and $a_i = 1/\sqrt{i}$ then you will not get this convergence to $0$.

answered Dec 13 '12 at 01:38

Henry

157,058

Hi @henry, We need to show that it converges #in distribution#...does that make any difference? – Dec 13 '12 at 01:53
No - You certainly have a problem if $\sum_1^n a_i \ge \sqrt{n}$ and $E[X_i]\not = 0$ – Henry Dec 13 '12 at 07:40
I see. Now let's consider $E[X_i] = 0$. I guess that should fix it...I'll add this to the question. – Dec 13 '12 at 16:13

score 0 · Accepted Answer · answered Dec 13 '12 at 23:20

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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.

answered Dec 13 '12 at 23:20

Yury

7,025

@Eli Do you know why $\sum\limits_{i\leqslant n}a_i^2/n\to0$? – Did Sep 20 '13 at 17:56

convergence in distribution of a sum of random variables

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