2
$\begingroup$

Consider a stationary sequence $X_1,X_2,...X_n$ such that $X_i \sim N(0,1)$ and their correlation sequence is given by $r_n=E(X_iX_{i+n})$. It is well known that if $r_n \log n \rightarrow 0$ then the maxima $M_n=\max_{1 \leq i \leq n} X_i$ converges in distribution to a Gumbel distribution and if $r_n \log n \rightarrow \lambda>0$ then $M_n$ converges to a mixture of Gumbel and standard normal distribution.

My question is what would happen when the dependence is large, i.e when $r_n \log n \rightarrow \infty$? I can see some related results in literature which show that $M_n \xrightarrow{d} N(0,1)$ in this case, under some extra assumptions about the smoothness of $r_n$. Would it be true if we assume nothing other than $r_n \log n \rightarrow \infty$?

$\endgroup$
2
  • 1
    $\begingroup$ No, for example, the exchangeable case which is much like i.i.d. gaussians. $\endgroup$ – mike Mar 16 at 8:13
  • $\begingroup$ @mike if the sequence is exchangeable, wouldn't that imply the correlation is constant? It's well known that in the constant correlation case (triangular array settings), $r_n \log n \rightarrow \infty$ implies the normal limiting distribution for $M_n$ (no extra assumptions needed). $\endgroup$ – random Mar 16 at 22:17
2
$\begingroup$

The answer to your second question is apparently no. In Leadbetter, Lindgren and Rootzén (Extremes and related properties of random sequences and processes), they state below Theorem 6.6.4 that there exists a class of possible limiting distributions that can occur when the covariance function decreases irregularly as a function of $n$.

$\endgroup$
3
  • $\begingroup$ Thank you for the reference. I did look into their books/papers and they didn't provide an actual counter example, it was an example in which $r_n=0$ on a subsequence and the limit is the maxima of 2 independent standard normal. $\endgroup$ – random Apr 11 at 0:26
  • $\begingroup$ Sorry that it didn't help. From the way it was written, it looked like there was an example where $r_n \to 0$ and $r_n \log(n) \to \infty$ such that the limit is not normal as they explicitly state theorem 6.6.4. My apologies if my answer turned out incorrect. $\endgroup$ – Gilles Mordant Apr 11 at 8:53
  • $\begingroup$ It was very helpful, indeed. The condition $r_n \log n \rightarrow \infty$ is violated in their example as it only holds along a subsequence (on a different subsequence you have $r_n \log n=0$). $\endgroup$ – random Apr 11 at 21:26

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.