Law of large numbers for dependent random variables with fixed covariance

Given identically distributed random variables $$X_1,...,X_n$$, where $$\mathbb E X_i = 0$$, $$\mathbb E [X_i^2] = 1$$, and $$\mathbb E [X_i X_j] = c<1$$, define $$S_n = X_1 + ... + X_n$$. Will

$$\frac{S_n}{n} \rightarrow 0$$ in probability as $$n\rightarrow \infty$$?

The closest question I could find is this one, where an additional constraint is placed on the covariances, such that Chebyshev inequality can be used to bound $$\left|\frac{S_n}{n} \right|$$ by $$\text{Var}\left(\frac{S_n}{n}\right)$$. However, in my case $$\text{Var}\left(\frac{S_n}{n}\right)$$ is constant, so this approach will not work.

Indeed, when I simulate using MATLAB on an example where I generate Gaussian random variables which all have covariance $$c=0.1$$, I do not see it approaching zero.

rng(1)
NN =20000;
C = ones(NN,NN)*0.1 + 0.9*diag(ones(NN,1));
rndnums = mvnrnd(zeros(NN,1), C, 1);
NNs = 1:20:NN
means = -99*ones(1,length(NNs));
for ti=1:length(NNs)
means(ti) = mean(rndnums(1,1:NNs(ti)));
end
scatter(NNs ,means, '.'); xlabel('n'); ylabel('mean')


Intuitively, since all the variables are all "locked" to eachother in correlation, I can't imagine how their mean will "eventually" reach zero.

Is there a law of large numbers in this context? If so, what is the rate of convergence? Or if not, is there any way to tell what it will converge to?

• How can you show that $c \leq \frac1{n-1}$? What is wrong, then, with the OP's example, where it looks like $c$ can be 0.1 regardless of how large $n$ is? – Mark Fischler May 9 '19 at 20:19
• @kimchilover Not true. The $n \times n$ matrix with diagonal entries $1$ and other entries $c$ has eigenvalues $(n-1) c + 1$ and $1-c$ (with multiplicity $n-1$), and thus is positive definite if $-1/(n-1) < c < 1$. – Robert Israel May 9 '19 at 20:32
• @RobertIsrael, and others: I'm sorry; I mixed up signs. What I should have said was $1+(n-1)c\ge0$, that is, $c\ge -1/(n-1)$ (which is implicit in RobertIsrael's construction). – kimchi lover May 9 '19 at 20:35
• If $X_1, \ldots, X_n$ are jointly normal with means $0$ and covariance matrix having diagonal entries $1$ and all other entries $c$, you can define $X_{n+1}$ as a suitable linear combination of $X_1 + \ldots + X_n$ and an independent normal random variable so that $X_1, \ldots, X_{n+1}$ are iid and satisfy the condition. – Robert Israel May 9 '19 at 20:37
• Of course, in the jointly normal case $S_n/n$ is normal with mean $0$ and variance $c + (1-c)/n$, and so it doesn't go to $0$ in probability. I don't know if this is true in general. – Robert Israel May 9 '19 at 20:43

Letting $$A$$ be the ($$n\times n$$) matrix with ones in all entries, and $$I$$ be the $$n\times n$$ identity matrix, the covariance matrix is $$M = (1-c)I + cA = P D P^{-1}$$ where $$D$$ is diagonal and $$P$$ is orthogonal, and (as @RobertIsrael observed) we are free to order the eigenvalues such that the upper left element of $$D$$ is $$1+(n-1)c$$ and the remaining diagonal elements are $$1-c$$.
When we do this, we find that $$(1,1,1,\ldots,1) P = (\sqrt{n},0,0,0...)$$ which tells us that the linear combinations of the variates that do not constitute the eigenvector with eigenvalue $$(1-c)$$ do not contribute to the sum $$S_n$$.
So the sum $$S_n$$ is distributed as $$\sqrt{n}$$ times a variate with mean zero (of course) and variance $$1+(n-1)c$$. This need not be a Gaussian but if the tails are not thick (so that the law of large numbers applies) then for large $$n$$ the standard deviation of $$S_n/n$$ goes like $$\frac{\sqrt{n}\sqrt{1+(n-1)c}}{n} \sim \sqrt{c}$$ So $$S_n/n$$ will not approach zero with probability one.
I think even if the individual variates have thick tails, that only makes matters worse for saying that $$S_n/n$$ will approach zero with probability one, so the above statement still holds.