{ "id": "2008.07215", "version": "v1", "published": "2020-08-17T10:50:17.000Z", "updated": "2020-08-17T10:50:17.000Z", "title": "Clustering of consecutive numbers in permutations under a Mallows distribution", "authors": [ "Ross G. Pinsky" ], "categories": [ "math.PR" ], "abstract": "Let $A^{(n)}_{l;k}\\subset S_n$ denote the set of permutations of $[n]$ for which the set of $l$ consecutive numbers $\\{k, k+1,\\cdots, k+l-1\\}$ appears in a set of consecutive positions. Under the uniformly random measure $P_n$ on $S_n$, one has $P_n(A^{(n)}_{l;k})\\sim\\frac{l!}{n^{l-1}}$ as $n\\to\\infty$. In this paper we consider the probability of the clustering of consecutive numbers under Mallows distributions $P_n^q$, $q>0$. Because of a duality, it suffices to consider $q\\in(0,1)$. In particular, we show that for fixed $q$, $\\lim_{l\\to\\infty}\\lim_{n\\to\\infty}P_n^q(A^{(n)}_{l;k_n})=\\big(\\prod_{i=1}^\\infty(1-q^i)\\big)^2,\\ \\text{if}\\ \\lim_{n\\to\\infty}\\min(k_n,n-k_n)=\\infty$, and that for $q_n=1-\\frac c{n^\\alpha}$, with $c>0$ and $\\alpha\\in(0,1)$, $P_n^q(A^{(n)}_{l;k_n})$ is on the order $\\frac1{n^{\\alpha(l-1)}}$, uniformly over all sequences $\\{k_n\\}_{n=1}^\\infty$. Thus, letting $N^{(n)}_l=\\sum_{k=1}^{n-l+1}1_{A^{(n)}_{l;k}}$ denote the number of sets of $l$ consecutive numbers appearing in sets of consecutive positions, we have \\begin{equation*} \\lim_{n\\to\\infty}E_n^{q_n}N^{(n)}_l=\\begin{cases}\\infty,\\ \\text{if}\\ l<\\frac{1+\\alpha}\\alpha;\\\\ 0,\\ \\text{if} \\ l>\\frac{1+\\alpha}\\alpha. \\end{cases}. \\end{equation*}", "revisions": [ { "version": "v1", "updated": "2020-08-17T10:50:17.000Z" } ], "analyses": { "subjects": [ "60C05", "05A05" ], "keywords": [ "consecutive numbers", "mallows distribution", "permutations", "clustering", "consecutive positions" ], "note": { "typesetting": "TeX", "pages": 0, "language": "en", "license": "arXiv", "status": "editable" } } }