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This is a question about central limit theorems when the dimension is increasing. Suppose now I have a random vector $X_N = (X_{N1}, \cdots, X_{Np})\in\mathbb{R}^p$. For all $c_p\in\mathbb{R}^p$ with $\|c_p\|_2 = 1$, suppose we have $c_p^\top X_N \xrightarrow{d} N(0,1)$ as $N\to\infty, p\to\infty$. What can we say about the asymptotic distribution of $\sum_{i=1}^p X_{Ni}^2$ (after normalization if needed)? Is it normal or there is a counterexample?

Several thoughts:

  1. If $X_N$ are i.i.d. $N(0,1)$, we have CLT.
  2. If $p$ is fixed, by Cramer-Wood theorem we have asymptotic normality for the $p$-dimensional random vector; then by continuous mapping theorem we have chi-square distribution. But sum of chi-square(1) converge to a normal eventually!

So what I need here is probably a combination of CLT and CMT in increasing dimensions. But I have limited knowledge in this direction.

Background information: why would we care about this type of CLT? Suppose $X_N$ is an aggregation of $p$ summary statistics. Typically we can show any finite linear combination of them are asymptotically normal. If we care about testing a global null to see whether the $p$ true estimands are zero, it is natural to consider a procedure involving a sum-of-square statistics.

More concretely, think about a multiple testing problem where we have $p$ different null hypothesis $H_{0j},j=1,\cdots,p$. For each null hypothesis, we have some testing statistics, say $X_{Nj} = N^{-1}\sum_{i=1}^Nx_{ij}$. Each of these testing statistics are asymptotically normal. If we consider whether they are jointly normal, it's likely to be true since we do linear combination of the $p$ coordinates we have a linear combination of all the samples involved in the $p$ studies, for which we could apply the general CLT. That's why we have a condition such as "for all $c_p$...". If we test the global null(that is, all null hypotheses are true) we might consider aggregating the testing statistics by taking a summation of squares. But now, how do we rigorously justify this squared summation also has certain stable distibution?

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In such generality, virtually nothing can be said about the asymptotic distribution of $V_{Np}:=\sum_{i=1}^p X_{Ni}^2$ or even about the existence of such an asymptotic distribution. In particular, $V_{Np}$ may have a non-normal asymptotic distribution or no asymptotic distribution at all.

Indeed, consider the following three simple settings.

Setting 1: All the $X_{Ni}$'s are iid standard normal and $c_p^\top c_p=1$. Then $Y_{Np}:=c_p^\top X_N\sim N(0,1)$ and $V_{Np}\sim\chi^2_p\approx N(p,2p)$ (as $p\to\infty$), so that $V_{Np}$ is asymptotically normal.

Setting 2: $X_{N1}\sim N(0,1)$, $X_{N2}=\cdots=X_{Np}=0$, and $c_p=[1,0,\dots,0]^\top$. Then $Y_{Np}=X_{N1}\sim N(0,1)$ and $V_{Np}=X_{N1}^2\sim\chi^2_1$, so that the asymptotic distribution of $V_{Np}$ is not normal.

Setting 3: This is a combination of Settings 1 and 2: for odd $N$ we use Setting 1, and for even $N$ we use Setting 2. Then there is no asymptotic distribution at all.

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  • $\begingroup$ Thanks for the response! I agree with setting 1 but setting 2 actually violates my assumption: for ANY $c_p$, $c_p^\top X_N$ is asymptotically normal. But you are only considering a specific $c_p$... $\endgroup$ Commented Mar 8, 2022 at 1:16
  • $\begingroup$ Actually it might be more intuitive to consider the coordinate $X_{Np}$ as a summation statistic by itself to see why we have the $c_p$ condition; i.e., $X_{Np} = \sum_{i=1}^N Y_{ip}$. Now they are asymptotically normal by themselves; how about summarizing all parts as a squared summation? $\endgroup$ Commented Mar 8, 2022 at 1:30
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    $\begingroup$ You placed the quantifier "any" not where you say you wanted it: instead of "For any $c_p\in\mathbb{R}^p$ with $\|c_p\|_2 = 1$, suppose we have $c_p^\top X_N \xrightarrow{d} N(0,1)$ as $N\to\infty, p\to\infty$", you should have said "Suppose that for all [rather than any] $c_p\in\mathbb{R}^p$ with $\|c_p\|_2 = 1$ we have $c_p^\top X_N \xrightarrow{d} N(0,1)$ as $N\to\infty, p\to\infty$". $\endgroup$ Commented Mar 8, 2022 at 1:58
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    $\begingroup$ Last comment continued: However, the latter condition is way too strong to be of any use -- because then, for any fixed natural $j$, you can choose $c_p$ to have $1$ as its $j$th coordinate (with all other coordinates of $c_p$ equal $0$), which would imply that each $X_{Nj}$ can only be standard normal (and nothing else). $\endgroup$ Commented Mar 8, 2022 at 2:08
  • $\begingroup$ Thanks for the suggestion. I've changed any to all to avoid ambiguity. For you second comment, I have to say this is actually not a strong condition. We only know each $X_{Nj}$ are asymptotically normal but not exactly normal. $\endgroup$ Commented Mar 8, 2022 at 22:42
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As Iosif Pinellis pointed out, virtually nothing can be said about the limiting distribution.

Still, along with $\sum_{i=1}^p X_{Ni}^2$ consider $\sum_{i=1}^p \tilde{X}_{Ni}^2$ where the $\tilde{X}_i$ is a Gaussian random variable with $E\tilde{X}_i=E{X}_i$ and $\operatorname{Var}\tilde{X}_i=\operatorname{Var}{X}_i$, for all $i$. Then, for a sequence of sets $\mathcal{A}_p$ that is not too rich (in a suitable sense), you might still have $$ \lim_{p \to \infty} \sup_{A \in \mathcal{A}_p }\left\lvert \rm{P}\left(\sum_{i=1}^p X_{Ni}^2 \in A\right) - \rm{P}\left(\sum_{i=1}^p \tilde{X}_{Ni}^2 \in A\right) \right\rvert =0, $$ which is more than enough for some statistical applications. If I do recall correctly, hypercubes and balls should constitue such not-too-rich classes.

An example thereof is the paper https://arxiv.org/pdf/1212.6906.pdf. Other works in the same direction have been done by Kengo Kato. I believe that it constitutes a correct entry-point to the literature you might be interested in.

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  • $\begingroup$ Thanks! I am working on proving such type of a Berry-Esseen bound for $\ell_2$ balls. Many results are for convex sets and they give rates polynomial in dimension and sample size which I feel have been quite enough for my use. $\endgroup$ Commented Mar 9, 2022 at 18:45

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