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:

  1. 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).
  2. Each $X_{n,k}$ has zero mean, variance $0<\sigma^2<\infty$, and third absolute moment $0<\rho<\infty$
  3. 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.

  • $\begingroup$ How do you quantify convergence for distributions of vectors? $\:$ $\endgroup$
    – user5810
    Feb 25, 2013 at 19:44
  • $\begingroup$ I am looking for results of the form "the supremum of |p1(A) - p2(A)| over all sets A tends to 0" (preferably at a given rate, say n^-0.5), where p1 is the probability measure of interest and p2 is the corresponding Gaussian probability. This definition should suit both the univariate and multivariate case. $\endgroup$
    – jmscarlett
    Feb 25, 2013 at 20:42
  • $\begingroup$ In addition to the obstacles indicated by Douglas's answer, the kind of result you say you're looking for is far to strong to hope for, without much stronger assumptions on the distributions of $X_{n,k}$, if you really insist on letting A be an arbitrary set. What you described in your comment would be convergence in total variation, which fails to be true if, for example, p1 is a discrete measure. In Berry-Esseen-type results you typically only let A range over all intervals. $\endgroup$ Feb 26, 2013 at 14:48

2 Answers 2


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.


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 $(X_{n,k})$.


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .