9
$\begingroup$

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')

enter image description here

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?

$\endgroup$
  • 1
    $\begingroup$ 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? $\endgroup$ – Mark Fischler May 9 '19 at 20:19
  • $\begingroup$ @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 $-1/(n-1) < c < 1$. $\endgroup$ – Robert Israel May 9 '19 at 20:32
  • $\begingroup$ @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). $\endgroup$ – kimchi lover May 9 '19 at 20:35
  • $\begingroup$ 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. $\endgroup$ – Robert Israel May 9 '19 at 20:37
  • $\begingroup$ 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. $\endgroup$ – Robert Israel May 9 '19 at 20:43
0
$\begingroup$

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.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ Thanks for the answer. It is not immediately obvious to me why the eigenvalues of the covariance matrix are (1+(n-1)c, 1-c, 1-c, ...., 1-c). Actually, I also don't follow the reasoning about the eigenvector with eigenvalue (1-c) either, but I'm sure that's because I'm missing something in the first part. $\endgroup$ – The_Anomaly May 12 '19 at 1:20

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.