I am having trouble understanding the different regularity conditions for different versions of the finite population central limit theorem. I would greatly appreciate any help or insight anyone has.

Consider the following infinite triangular matrix: \begin{align} \begin{bmatrix} v_{11} \\ v_{21} & v_{22} \\ . & . & . \\ . & . & . & . \\ v_{N1} & v_{N2} & . & . & v_{NN} \\ . & . & . & . & . & . \end{bmatrix}, \end{align} whose entries are all real elements. Each row in the infinite triangular matrix is a finite population, which we can denote by $\Pi_1, \Pi_2, \ldots$, where $\Pi_1$ consists of a single element with the value $v_{11}$, $\Pi_2$ consists of two elements with the values $v_{21}, v_{22}$ and so on and so forth. The population $\Pi_N$ refers to row $N = 1, 2, \ldots$ of the infinite triangular matrix that consists of elements with values $v_{N1}, v_{N2}, \ldots, v_{NN}$.

Focusing on the sample sum (call it $S_N$), Theorem 6 of Lehmann (1975, Appendix 4) shows that the standardized sample sum converges in distribution to $\mathcal{N}(0, 1)$ as $N \to \infty$ when the following condition holds: \begin{eqnarray} \label{eq: regularity conds} n, N - n \to \infty \text{ and } \dfrac{\max \limits_{1 \leq i \leq N} \left(v_{Ni} - \bar{v}_N\right)^2}{\sum \limits_{i = 1}^N\left(v_{Ni} - \bar{v}_N\right)^2} \max\left(\dfrac{N - n}{n}, \dfrac{n}{N - n}\right) & \to & 0 \text{ as } N \to \infty \end{eqnarray} or, equivalently, $\dfrac{n}{N} \to p \in \left(0, 1\right)$ and $\dfrac{\max \limits_{1 \leq i \leq N} \left(v_{Ni} - \bar{v}_N\right)^2}{\sum \limits_{i = 1}^N\left(v_{Ni} - \bar{v}_N\right)^2}$ is bounded as $N \to \infty$, where $n$ is the sample size and $\bar{v}_N$ is the population mean.

However, in David Freedman's 2008 article in The Annals of Applied Statistics, he shows that $\sqrt{N}(S_N - \mathbb{E}[S_N])$ converges in distribution to $\mathcal{N}(0, \nu)$, where $\mathbb{E}[\cdot]$ is the expected value operator, $\nu$ is $\lim \limits_{N \to \infty} \mathbb{V}[\sqrt{N}(S_N - \mathbb{E}[S_N])]$ and $\mathbb{V}[\cdot]$ is the variance operator, when the following three conditions hold:

(1) For all $N = 1, 2, \ldots$, $\mathbf{v}_N$ has a bounded fourth central moment, i.e., $\dfrac{1}{N} \sum \limits_{i = 1}^N \left\lvert v_{Ni} \right\rvert^4 < L < \infty$.

(2) As $N \to \infty$, $\dfrac{n}{N} \to p \in \left(0, 1\right)$.

(3) The population quantities $\dfrac{1}{N} \sum \limits_{i = 1}^N v_{Ni}$ and $\dfrac{1}{N} \sum \limits_{i = 1}^N v_{Ni}^2$ tend to finite limits as $N \to \infty$.

(Technically Freedman proves this result for the difference-in-means estimator in the context of finite population causal inference with two finite populations of fixed potential outcomes, but I am just drawing on the fact that the difference-in-means estimator can be represented as a sample sum via scale and shift factors that do not depend on treatment assignment.)

I think I see why Freedman needs condition (3). Is it to ensure that the variance of the sample sum converges to a positive, finite value? Presumably Lehmann doesn't need this assumption because standardizing the sample sum ensures that the variance is 1 for all $N = 1, 2, \ldots$. I can also see that Freedman's condition (2) is equivalent to Lehmann's assumption that $\dfrac{n}{N} \to p \in \left(0, 1\right)$.

However, I am having trouble understanding the role of Freedman's assumption (1) and whether it implies Lehmann's assumption that $\dfrac{\max \limits_{1 \leq i \leq N} \left(v_{Ni} - \bar{v}_N\right)^2}{\sum \limits_{i = 1}^N\left(v_{Ni} - \bar{v}_N\right)^2}$ is bounded as $N \to \infty$.

Does Freedman's condition (1) of a bounded fourth central moment imply Lehmann's condition that $\dfrac{\max \limits_{1 \leq i \leq N} \left(v_{Ni} - \bar{v}_N\right)^2}{\sum \limits_{i = 1}^N\left(v_{Ni} - \bar{v}_N\right)^2}$ is bounded as $N \to \infty$? And if so, does anyone have a proof of this? Or am I missing something altogether about the relationship between the regularity conditions in these two version of the finite population CLT?


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