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Problem. Let $W_1, W_2,...$ be independent and identically distributed random variables such that $E(W_1)=0$ and $\sigma^2 := V(W_1) \in (0,\infty)$. Let $T_n = \frac{1}{\sqrt{n}} \sum_{j=1}^n a_j W_j$ where $a_j\neq 0$ for all $j\in \Bbb{N}$. If $$\lim_{n\to \infty} \frac{\max_{j=1,...,n}|a_j|}{\sqrt{\sum_{j=1}^na_j^2}}=0,$$ then $$\frac{T_n}{\sqrt{V(T_n)}} \longrightarrow_d N(0,1) \quad\text{(convergence in distribution}).$$

Here is my attempt: If we define $X_{nj}= \frac{a_j}{\sqrt{n}}W_j$ for $j=1,...,n$, then $T_n=\sum_{j=1}^n X_{nj}$. So if we can check that the Lindeberg condition holds for this triangular array, then the central limit theorem of Lindeberg-Feller implies the claim that $\frac{T_n}{\sqrt{V(T_n)}} \longrightarrow_d N(0,1)$. To this end, we need to show that for any $\varepsilon>0$ $$\lim_{n\to \infty} \frac{1}{\sigma_n^2} \sum_{j=1}^n E( X_{nj}^2 \cdot \mathbf 1\{ |X_{nj}|\gt \varepsilon \sigma_n \})=0$$ where $\sigma_n^2=\sum_{j=1}^n V(X_{nj}) = \frac{\sigma^2}{n}\sum_{j=1}^n a_j^2$.

I have not been able to prove this and would be very thankful for any help.

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    $\begingroup$ Did you try to compute $$\frac{1}{\sigma_n^2} \sum_{j=1}^n E( X_{nj}^2 \cdot \mathbf 1\{ |X_{nj}|\gt \varepsilon \sigma_n \})$$ in terms of expectations involving $\varepsilon$, $W$ and the coefficients $a_j^2$, $m_n^2=\max\limits_{j=1}^na_j^2$ and $s_n^2=\sum\limits_{j=1}^na_j^2$? $\endgroup$ – Did Jun 25 '15 at 14:28
  • $\begingroup$ Well, I don't know how to compute $E( X_{nj}^2 \cdot \mathbf 1\{ |X_{nj}|\gt \varepsilon \sigma_n \})$. I'm having trouble dealing with the indicator function. All the upper bounds I've put on it turned out to be useless. $\endgroup$ – iwriteonbananas Jun 25 '15 at 14:47
  • $\begingroup$ This might be so but, since you showed nothing of "All the upper bounds (you) (ha)ve put on it turned out to be useless", how can we help? To begin with, every $X_{nj}$ is distributed like a multiple of a single random variable $W$, this allows some simplications, no? $\endgroup$ – Did Jun 25 '15 at 15:11
  • $\begingroup$ @Did Right, I posted an answer since it didn't fit in the comment box. Is that what you had in mind? And is it correct? Thanks $\endgroup$ – iwriteonbananas Jun 26 '15 at 6:29
  • $\begingroup$ The answer is correct. $\endgroup$ – Did Jun 26 '15 at 6:38
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We consider $m_n=\max\limits_{j=1}^n|a_j|$, $s_n^2=\sum\limits_{j=1}^na_j^2$, $W$ some random variable distributed like every $W_j$, and we simplify the quantities involved in Lindeberg's condition as follows; $$\begin{align} \sum_{j=1}^nE( X_{nj}^2 \cdot \mathbf 1\{ |X_{nj}|\gt \varepsilon \sigma_n \}) &= \sum_{j=1}^n \frac{a_j^2}{n} E\left( W^2 \cdot \mathbf 1\{ |W|\gt \frac{\varepsilon \sigma_n \sqrt n}{|a_j|} \}\right) \\&\leq\frac{s_n^2}{n}E\left( W^2 \cdot \mathbf 1\{ |W|\gt \frac{\varepsilon \sigma_n \sqrt n}{m_n} \}\right) \\&=\frac{s_n^2}{n}E\left( W^2 \cdot \mathbf 1\{ \frac{|W|}{\sigma}\gt \frac{\varepsilon s_n}{m_n} \}\right) \end{align}$$ Since $1/\sigma_n^2=n/(\sigma^2 s_n^2)$, this yields $$\begin{align} \frac{1}{\sigma_n^2} \sum_{j=1}^n E( X_{nj}^2 \cdot \mathbf 1\{ |X_{nj}|\gt \varepsilon \sigma_n \}) &\leq \frac{1}{\sigma^2} E\left( W^2 \cdot \mathbf 1\{ \frac{|W|}{\sigma}\gt \frac{\varepsilon s_n}{m_n} \}\right) \\&= E\left( \frac{W^2}{\sigma^2} \cdot \mathbf 1\{ \frac{|W|}{\sigma}\gt \frac{\varepsilon s_n}{m_n} \}\right) \quad(*) \end{align}$$ Since $m_n/s_n\to0$ when $n\to\infty$, one knows that $\mathbf 1\{ |W|/\sigma\gt \varepsilon s_n/m_n \}\to0$ almost surely, when $n\to\infty$. Since $W$ is square integrable, by Lebesgue dominated convergence theorem, the expression $(*)$ converges to $0$ as $n\to \infty$, QED.

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