Counting records in a random, non-uniform, permutation
Boris Pittel
TL;DR
This paper analyzes the number of records (left-to-right maxima) in a non-uniform random permutation of $[n]$, where an initial permutation is thinned by independent marking with probability $p$ and the marked elements are randomly permuted. The authors decompose the total number of records into unmarked and marked contributions and derive tight asymptotic bounds, showing that $\max_{\bar{X}} E[\lambda(X)]=\Theta\bigl((1-p)\sqrt{n/p}\bigr)$ whenever $p\gg 1/n$ and $1-p\ge \mathrm{const}\, n^{-1/2}\log n$. This refines prior results (BMB) and establishes matching upper and lower bounds in the stated regime, enabling precise characterization of average-case behavior under this smoothed/randomized input model. The methods combine probabilistic decomposition, integral representations, and binomial tail bounds to control unmarked and marked records, with implications for understanding algorithmic performance on restricted random instances.
Abstract
Counting permutations of $[n]$ by the number of records, i.e. left-to-right maxima, is a classic problem in combinatorial enumeration. In the first volume of ``The Art of Computer Programming", Donald Knuth demonstrated its relevance for analysis of average case complexity of a basic algorithm for determining a maximum in a linear list of numbers. It is well known that the expected, and likely, number of those records in a {\it uniformly\/} random permutation is asymptotic to $\log n$. Cyril Banderier, Rene Beier, and Kurt Mehlhorn studied the case of a non-uniform random permutation, which is obtained from a generic permutation of $[n]$ by selecting its elements one after another independently with probability $p$, and permuting the selected elements uniformly at random. They proved that $E_n(p)$, the largest expected number of the maxima, is between $\text{const}\sqrt{n/p}$ and $O\bigl(\sqrt{(n/p)\log n}\bigr)$ if $p$ is fixed. For $p\gg 1/n$ and simultaneously $1-p\ge \text{const }n^{-1/2}\log n$, we prove that $E_n(p)$ is exactly of order $(1-p)\sqrt{n/p}$.