Central Limit Theorem (and Berry-Esseen theorem) for non-independent variables Consider the triangular array $X_{n,k}$ such that, for each $n>0$, the variables $(X_{n,1},\cdots,X_{n,n})$ have the following properties:


*

*For any given $1 \le L \le n$, all
subsets of
$(X_{n,1},\cdots,X_{n,n})$ of size
$L$ have the same joint distribution
(even after applying an arbitrary permutation). 

*Each $X_{n,k}$ has
zero mean, variance
$0<\sigma^2<\infty$, and third
absolute moment $0<\rho<\infty$

*Each $(X_{n,k_1},X_{n,k_2})$ pair
($k_1 \ne k_2$) has covariance
$\frac{-C}{n}$ for some $0 < C < \infty$ (hence the variables are all negatively correlated, and the correlation tends to zero)


Let $S_n = \sum_{k=1}^{n} X_{n,k}$.  Is $\frac{S_n}{\sqrt{\mathrm{Var}[S_n]}}$ asymptotically distributed as $N(0,1)$?  If so, is a convergence rate of $O(\frac{1}{\sqrt{n}})$ achieved (cf. Berry-Esseen theorem)?  What about in the multivariate setting where each $X_{n,k}$ is a random vector (and the relevant quantities above are replaced by vectors/matrices)?
If it helps, we can also assume that the $X_{n,k}$ are uniformly bounded in $n$ and $k$ with probability one.  Answers with further assumptions than the ones listed will also be appreciated.
 A: These types of conditions (exchangeability and $O(1/n)$ covariance) do not even ensure that $S_n$ is approximately normal.
Ignore odd $n$. (I believe similar examples can be constructed for odd $n$.) Let $n=2m$. Let each $X_{n,k} \in \lbrace -1,1 \rbrace$. Let each set of $m$ positive signs have probability ${2m \choose m}^{-1}$. Then the sum is always $0$. This satisfies conditions 1 and 2. $P(X_{n,1} = X_{n,2}) = {2m-2 \choose m}/{2m-1 \choose m} = \frac{m-1}{2m-1}$ so the covariance is $\frac{m-1}{2m-1} - \frac{m}{2m-1} = \frac{-1}{2m-1}$.
The point mass at $0$ is still normal, but degenerate. However, this example can be modified slightly so that the sum has positive standard deviation. With probability $\frac{1}{4m^2}$ let all $X_{n,i}$ be equal ($\frac{1}{8m^2}$ chance of all $+1$, $\frac{1}{8m^2}$ all $-1$), and with probability $\frac{4m^2-1}{4m^2}$ let the positive indices be a uniformly random subset of size $m$. Then the covariance $\text{Cov}(X_{n,1},X_{n,2})$ is $\frac{-1}{n}$, and $P(S_n =0) = \frac{n^2-1}{n^2}, P(S_n = \pm n) = \frac{1}{2n^2}$ so $\text{Var}(S_n) = 1$, and $\frac{S_n}{\sqrt{\text{Var}(S_n)}}$ is far from normal.  
A: No.  You can even suppose the variables are all identically distributed, bounded and the covariances are all zero, and you still won't have convergence to a normal distribution.  To see this, take my main example $(X_n)$ from here (for any fixed $N>1$), arrange it into a triangular array by letting $Y_{n,k} = X_k$, and then let $X_{n,k} = Y_{n,\pi_n(k)}$ where $\pi_n$ is a random permutation (with all permutations equally likely) of $1,...,n$.  Since $Y_{n,1},...,Y_{n,n}$ is $N$-tuplewise independent, the variables in the randomly permuted row will still have zero covariance.  And since the $(X_n)$ don't satisfy the CLT, neither will the $(X_{n,k})$.
