Assuming that for each fixed $k$, $(X_{n,1},\ldots,X_{n,k})\Longrightarrow(X_1,\ldots,X_k)$ where $X_1,\ldots,X_k$ are i.i.d. with mean zero and variance $\sigma^2$, will the array inherit the CLT from its limit? i.e. do I have (if $r_n\to\infty$): $\frac{1}{\sqrt{r_n}}(X_{n,1}+\cdots+X_{n,r_{n}})\Longrightarrow N(0,\sigma^2)$?
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$\begingroup$ You haven't said that each of $X_1,\ldots,X_k$ is normally distributed. Did you intend that to be understood? If so, it should be mentioned. $\endgroup$– Michael HardyCommented Jul 9, 2021 at 17:45
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$\begingroup$ It is not generally true that if $U,V$ are normally distributed then so is $U+V.$ For example, let $V=\pm U,$ where $\text{“$\pm$”}$ it is $\text{“$+$”}$ with probability $1/2$ and otherwise $\text{“$-$”},$ independently of $U.$ Then $\Pr(U+V=0)>0$ and $\Pr(U+V\ne0)>0,$ so this sum is not normally distributed. $\qquad$ $\endgroup$– Michael HardyCommented Jul 9, 2021 at 17:48
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$\begingroup$ None of $X_1,\ldots,X_k$ has actually to be normally distributed let alone jointly. But since they are i.i.d. they will satisfy CLT. $\endgroup$– mdouCommented Jul 10, 2021 at 1:14
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No. Take any sequence $r_n \to \infty$ and let $X_{n,k}$ be iid $N(0,1)$ for $k \ne r_n$, and $X_{n,r_n} = r_n$. The hypothesis is satisfied because $(X_{n,1}, \dots, X_{n,k})$ are iid $N(0,1)$ as soon as $n$ is so large that $r_n > k$. But $$\frac{1}{\sqrt{r_n}}(X_{n,1} + \dots + X_{n,r_n}) \sim N(\sqrt{r_n}, \frac{r_n-1}{r_n})$$ which is not even tight.
answered Jul 9, 2021 at 7:03
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$\begingroup$ sorry. I forgot to say that $r_n=o(n)$. $\endgroup$– mdouCommented Jul 9, 2021 at 7:10
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$\begingroup$ @mdou: The same idea still works, see edit. $\endgroup$ Commented Jul 9, 2021 at 7:15
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$\begingroup$ I see. Excellent. Thank you very much for this answer. $\endgroup$– mdouCommented Jul 9, 2021 at 7:16