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Suppose I have some triangular array $\{X_{n,j}\}$ of random variables, which need not be independent or identically distributed. Suppose I further know that $$Var\left(\sum_{j=1}^n X_{n,j}\right)\to \sigma^2 \text{ and } \sum_{j=1}^n \mathbb{E}|X_{n,j}|^{2+\epsilon}\to 0,$$ for some $\epsilon>0$. Then this triangular array satisfies the Lyapunov condition. However, for Lyapunov's CLT we further require independence for each $n$. Suppose instead that we have asymptotic independence in the sense that for any $t\in\mathbb{R},$ $$\left|\mathbb{E}\left[\prod_{j=1}^n\exp\left(itX_{n,j}\right)\right] - \prod_{j=1}^n\mathbb{E}\left[\exp\left(itX_{n,j}\right)\right]\right|\to 0,$$ where $i$ denotes the imaginary unit. Can we then still conclude that $$\sum_{j=1}^n X_{n,j}\to N(0, \sigma^2)?$$

It is easy to see (by a Levy continuity theorem argument) that this is true if the covariances also fade fast enough, so that $$\sum_{j=1}^nVar\left( X_{n,j}\right)\to \sigma^2,$$ but do we need this or can we also conclude asymptotic normality without this additional condition?

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The answer is no. E.g., suppose that \begin{equation*} (X_{n,j})=(X_{n,j})_{j=1}^n\sim(1-u^2/n)N_n(0,I_n/n)+(u^2/n) N_n(0,J_n/n), \end{equation*} where $u\in(0,\infty)$, $I_n$ is the $n\times n$ identity matrix, and $J_n$ is the $n\times n$ matrix with all entries equal $1$, so that $N_n(0,J_n/n)$ is the distribution of the random vector $(Y,\dots,Y)/\sqrt n$ with $Y\sim N(0,1)$. Thus, the distribution of $(X_{n,j})$ is the mixture of the $n$-variate normal distributions $N_n(0,I_n/n)$ and $N_n(0,J_n/n)$ with respective weights $1-u^2/n$ and $u^2/n$.

The idea here is to make the variance of the sum $\sum_j X_{n,j}$ greater (than it would be with $u=0$) without significantly affecting the distribution of the sum, which will still be $\approx N(0,1)$ even if $u\ne0$.

Then \begin{equation*} Cov(X_{n,j},X_{n,k})=(1-u^2/n)1_{j=k}/n+(u^2/n)/n \end{equation*} and hence \begin{equation*} Var\sum_jX_{n,j}=\sum_{j,k}Cov(X_{n,j},X_{n,k}) \\ =(1-u^2/n)+n(n-1)(u^2/n)/n\to1+u^2=:\sigma^2. \tag{1} \end{equation*} Next, for any real $\epsilon>0$ \begin{equation*} E|X_{n,j}|^{2+\epsilon}=(1-u^2/n)/n^{1+\epsilon/2}+(u^2/n)/n^{1+\epsilon/2} =1/n^{1+\epsilon/2} \end{equation*} and hence \begin{equation*} \sum_jE|X_{n,j}|^{2+\epsilon}=n/n^{1+\epsilon/2}\to0. \end{equation*} Similarly, $Ee^{itX_{n,j}}=e^{-t^2/(2n)}$ and hence \begin{equation*} \prod_1^n Ee^{itX_{n,j}}=e^{-t^2/2}. \end{equation*} Further, \begin{equation*} E\prod_1^n e^{itX_{n,j}}=(1-u^2/n)e^{-t^2/2}+O(u^2/n)\to e^{-t^2/2}, \end{equation*} so that the condition \begin{equation*} E\prod_1^n e^{itX_{n,j}}-\prod_1^n Ee^{itX_{n,j}}\to0 \end{equation*} holds as well. However, \begin{equation*} \sum_j X_{n,j}\sim(1-u^2/n)N(0,1)+(u^2/n) N(0,n)\to N(0,1)\ne N(0,\sigma^2), \end{equation*} in view of (1).

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  • $\begingroup$ Very nice example, I very much liked the intuitive explanation you added. Thank you! $\endgroup$
    – Dasherman
    May 9 '20 at 0:09

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