A word over the alphabet $\{0,1\}$ of length $n$ may contain squares, cubes, and generally $k$th powers, where $2\le k\le n$. Let $O_k(w)$ denote the number of $k$th power occurrences in the word $w$.
For instance, $w:=000101$ contains three square occurrences (two occurrences of 00 and one occurrence of 0101), and one cube (000). So $O_2(w)=3$ and $O_3(w)=1$.
If $w$ is selected uniformly randomly from among all words of length $n$, we may ask about the covariance matrix $\Sigma_n$ of the random variables $O_k$, $2\le k\le n$.
For $n\le 4$ I got the following (symmetric) matrices: $$ \Sigma_2 = \frac14\left(\begin{matrix} 1\end{matrix}\right) $$ $$ \Sigma_3 = \frac1{16}\left(\begin{matrix} 8 & 4 \\ . & 3\end{matrix}\right) $$ $$ \Sigma_4 = \frac1{64}\left(\begin{matrix} 60&40&18\\ . & 32 & 12 \\ .& .& 7\end{matrix}\right) $$ At this point I am wondering: can we assert that the entries of $\Sigma_n$ are rather small compared to the means $\mathbb E(O_k)$ (which I calculated in an answer to another question)?
Intuitively it seems the answer must be yes: most of the contribution to $O_k$ comes from short subwords, and the number of such powers is almost a sum of independent random variables. Perhaps the multivariate Chebyshev applies?
