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Suppose $X_1, X_2,...$ is a sequence of independent, identically distributed random variables with $\mathbb{E}[X_n] = 0$ and $\mathbb{E}[X_n^2] = 1$. Find sequences of constants $a_n$ and $b_n$ such that,

$$ \frac{\left(\sum\limits_{1 \leq i < j \leq n}X_iX_j - a_n\right)}{b_n} $$

converges in distribution and find the limit distribution.

My approach was to try and use the Lindeberg condition. Let $S_n = \sum\limits_{1 \leq i < j \leq n}X_iX_j = \sum\limits_{j=1}^n\sum\limits_{i=1}^{j-1}X_iX_j$. and define, $Y_j = \sum\limits_{i=1}^{j-1}X_iX_j$. Then $\mathbb{E}[Y_j] = 0$ and $\mathbb{E}[Y_j^2] = j-1$. Let $b_n^2 = \sum\limits_{j=1}^n\mathbb{E}[Y_j^2] = \frac{n(n+1)}{2} - n$. To verify the Lindeberg condition, we need to investigate the asymptotic behaviour of,

$$ \frac{1}{b_n^2}\sum\limits_{j=1}^n\mathbb{E}[Y_j^2\mathbb{1}_{\{|Y_j|>\epsilon b_n\}}] \;\;\;\;\; \forall \epsilon > 0 $$

If the above quantity goes to $0$ as $n\rightarrow\infty$, then $S_n/b_n$ will converge in distribution to $N(0,1)$. Since we're not given any other information about $X_1, X_2,...$, can we possibly use this approach? It makes no assumption of boundedness.

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Hint: $$\sum_{1 \leq i < j \leq n} X_i X_j = \sum_{1 \leq i \leq j \leq n} X_i X_j - \sum_{i=1}^n X_i^2 = \left( \sum_{i=1}^n X_i \right)^2 - \sum_{i=1}^n X_i^2$$

and so

$$\frac{1}{n} \sum_{1 \leq i < j \leq n} X_i X_j = \left( \frac{1}{\sqrt{n}} \sum_{i=1}^n X_i \right)^2 - \frac{1}{n} \sum_{i=1}^n X_i^2.$$

Use the CLT for the first term and the SLLN for the second.

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  • $\begingroup$ Wow this is beautifully concise. Thanks! $\endgroup$ – Flowsnake Jan 21 '18 at 16:52
  • $\begingroup$ @Flowsnake You are welcome. $\endgroup$ – saz Jan 21 '18 at 16:53

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