3 ms·
If you think of the covariance matrix, entry i,j for i ≠ j will be floor(n / (p[i]*p[j])) / n - floor(n / p[i]) * floor(n/p[j]) / n^2 and the ith diagonal
by murbard2 8y ago
If you think of the covariance matrix, entry i,j for i ≠ j will be
floor(n / (p[i]*p[j])) / n - floor(n / p[i]) * floor(n/p[j]) / n^2
and the ith diagonal entry will be
floor(n / p[i]) / n - ( floor(n / p[i]) / n )^2
for n large, you approximately get a diagonal matrix with diagonal entries / eigenvalues 1/p[i] - 1/p[i]^2.
- mturmon 8y agoSmart observation. Another way to say it is that, for distinct primes p1 and p2, the events “p1 divides n”, and “p2 divides n”, are approximately statistically independent. So you get a near-diagonal covariance with entries as you wrote.