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I am interested in central limit theorems for the following rather simple setup of finite uniform distributions:

Let $X_{ni}$ for $i \leq n$ and $n \in \{1,2,3,\ldots\}$ be independent discrete random variables such that $X_{ni}$ is uniformly distributed on the interval $[-a_{ni},a_{ni}] \subset \mathbb{Z}$. Let $s_n^2 = \mathbb{V}(X_n) = \sum_i \mathbb{V}(X_{ni})$ be the variance of $X_n = \sum_i X_{ni}$.

Are there necessary and sufficient conditions on the array $(a_{ni})$ implying a central limit $\frac{X_n}{s_n} \rightarrow N(0,1)$ ?

It is well-known that under the assumption $\frac{1}{s_n^2}\max_i \mathbb{V}(X_{ni}) \rightarrow 0$ for $n \rightarrow \infty$, the Lindeberg condition is both necessary and sufficient. My question thus can be given in two parts:

Assuming $\frac{1}{s_n^2}\max_i \mathbb{V}(X_{ni}) \rightarrow 0$.

Is it known how the Lindeberg condition translate to a (hopefully simple) condition that can be directly written in terms of the parameters $(a_{ni})$ ?

Assuming $\frac{1}{s_n^2}\max_i \mathbb{V}(X_{ni}) \not\rightarrow 0$.

Is it anyways possible for $X_{ni}$ to satisfy a central limit?

I finally add that I have very little background on probability theory. References to books/papers answering these or related questions are highly appreciated.

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This is an attempt to solve the first part of my question assuming $\frac{max_i \mathbb{V}(X_{ni})}{s_n^2} \rightarrow 0$.

Since resorting doesn't change $X_n$, we also use w.l.o.g. that $a_{n1} \leq \dots \leq a_{nn}$ for any $n$.

Claim: The Lindeberg condition holds. This is, for any $\epsilon > 0$, $$\frac{1}{s_n^2}\sum_{i=1}^n\mathbb{E}\big(X_{ni}^2\cdot I\big\{|X_{ni}| \geq \epsilon s_n\big\}\big) \rightarrow 0.$$

Proof: The support of $X_{ni}$ is bounded by $a_{ni}$. By this, I mean $$|x| > a_{ni} \Rightarrow Prob(X_{ni} = x) = 0.$$ The variance is $$\mathbb{V}(X_{ni}) = \tfrac{1}{3} a_{ni}(a_{ni}+1), \quad s_n^2 = \frac{1}{3} \sum_{i=1}^n a_{ni}(a_{ni}+1)$$ For any $k$ consider the sequence in $n$ given by $a_{n,n-k}$ for $n > k$. Since the $a_{ni}$ are sorted in $i$, the sequence $a_{nn}$ grows at least as fast as any of the sequences $a_{n,n-k}$. This is, $$ a_{n,n-k} \in \mathcal{O}(a_{nn})$$ for any $k$. The assumed condition $\frac{\mathbb{V}(X_{nn})}{s_n^2} \rightarrow 0$ says that $\mathbb{V}(X_{nn})$ grows strictly slower than $s_n^2$. In symbols, $$\mathbb{V}(X_{nn}) \in \mathcal{o}(s_n^2).$$ Since $\mathbb{V}(X_{nn}) \sim a_{nn}^2$ this implies that $$a_{nn} \sim \sqrt{\mathbb{V}(X_{nn})} \in \mathcal{o}(s_n).$$ Therefore, it exists a global integer $N$ such that $\epsilon s_n > a_{nn}$ for all $n \geq N$.

I finally want to conclude that $$\mathbb{E}\big(X_{ni}^2\cdot I\big\{|X_{ni}| \geq \epsilon s_n\big\}\big) = 0$$ for $n \geq N$ because the support of $X_{ni}$ is completely contained in the excluded interval. Thus, the sum equals zero for large enough $n$ implying the Lindeberg condition.

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  • $\begingroup$ Looks good, that's exactly, what I meant. $\endgroup$ – zhoraster Jan 10 '18 at 13:09
  • $\begingroup$ Great, thanks for checking! $\endgroup$ – Christian Jan 10 '18 at 13:11
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Is it known how the Lindeberg condition translate to a (hopefully simple) condition...

Yes. In your situation the assumption $$ \frac{1}{s_n^2}\max_i \mathbb{V}(X_{ni}) \rightarrow 0, n\to\infty, \tag{1} $$ (equivalently, $\frac{\max_i a_{ni}^2}{\sum_{i} a_{ni}^2}\to 0,n\to\infty$) implies the Lindeberg condition. Try to prove this (hint: for any $\varepsilon>0$, the expression in question is zero for $n$ large enough).

Assuming $\frac{1}{s_n^2}\max_i \mathbb{V}(X_{ni}) \not\rightarrow 0$, is it anyways possible for $X_{ni}$ to satisfy a central limit?

No. (Provided the meaning of "central limit" is standard, which is normal distribution.) You can write characteristic to check that the uniform smallness assumption $(1)$ is necessary in your case.

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  • $\begingroup$ For the second, I have to think more about your answer given that I maybe have never written down such a characteristic function. Thanks again! $\endgroup$ – Christian Jan 10 '18 at 12:56
  • $\begingroup$ @Christian, then I suspect that using cumulants will be much simpler approach for you. Try to prove that the fourth cumulant of your sum does not converge to $0$ without the uniform smallness assumption. $\endgroup$ – zhoraster Jan 10 '18 at 13:16
  • $\begingroup$ Will do. But I'll also try to write down the characteristic function. $\endgroup$ – Christian Jan 10 '18 at 13:19

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