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Proposition 2.4 -- the coupon collector's tail bound

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MarkovMixing.coupon_tail

by Shuze Chen · Aug 21, 2026 · Mathlib c5ea003 (Lean v4.30.0)

markov-chainsmixing-timesprobability

For the coupon collector with n≥1n\ge1n≥1 types and any c>0c>0c>0:

P{τ>⌈nlog⁡n+cn⌉}≤e−c.\mathbb{P}\{\tau>\lceil n\log n+cn\rceil\}\le e^{-c}.P{τ>⌈nlogn+cn⌉}≤e−c.

This bound drives the top-to-random shuffle analysis later in the series.

Preamble
import Definitions.Def_mm_classical
Formal statement
namespace MarkovMixing

/-- **Proposition 2.4** (LPW): the coupon collector time exceeds
`⌈n log n + cn⌉` with probability at most `e^{-c}`. -/
theorem coupon_tail (n : ℕ) (hn : 1 ≤ n) (c : ℝ) (hc : 0 < c) :
    couponMissProb n ⌈(n : ℝ) * Real.log n + c * n⌉₊ ≤ Real.exp (-c) := by
  sorry

end MarkovMixing
Source
D. A. Levin, Y. Peres, E. L. Wilmer, Markov Chains and Mixing Times, AMS 2009, https://documents.epfl.ch/groups/i/ip/ipg/www/2013-2014/Random_Walks/markovmixing.pdf, Section 2.2, Proposition 2.4, p. 23
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What the Lean code literally says, in plain math · claude-fable-5

For every natural number nnn with n≥1n \ge 1n≥1 and every real number ccc with c>0c > 0c>0, the theorem asserts the inequality

#{ d:{0,…,t−1}→{0,…,n−1}∣d is not surjective }nt  ≤  e−c,where t=⌈nln⁡n+c n⌉.\frac{\#\{\, d : \{0,\dots,t-1\} \to \{0,\dots,n-1\} \mid d \text{ is not surjective} \,\}}{n^{t}} \;\le\; e^{-c}, \qquad \text{where } t = \lceil n \ln n + c\, n \rceil.nt#{d:{0,…,t−1}→{0,…,n−1}∣d is not surjective}​≤e−c,where t=⌈nlnn+cn⌉.

Here the left-hand side is the bundle's custom quantity couponMissProb n t, defined as the number of functions from a ttt-element type to an nnn-element type that fail to be surjective (i.e. miss at least one of the nnn values), counted exactly, divided by the real number ntn^{t}nt — the fraction of all ntn^tnt sequences of ttt draws from nnn coupons that do not collect every coupon. The time ttt is the ceiling of nln⁡n+c nn \ln n + c\,nnlnn+cn taken as a natural number: the smallest natural number ≥nln⁡n+c n\ge n \ln n + c\, n≥nlnn+cn (this "natural ceiling" would send a negative argument to 000, but under the hypotheses n≥1n \ge 1n≥1 and c>0c > 0c>0 the argument nln⁡n+c nn\ln n + c\,nnlnn+cn is strictly positive, so t≥1t \ge 1t≥1); ln⁡\lnln denotes the natural logarithm. Note the edge behavior packed into the definition: since n≥1n \ge 1n≥1, the denominator ntn^tnt is nonzero, so no division-by-zero junk value arises; in the boundary case n=1n = 1n=1 every function to a one-element type (with t≥1t \ge 1t≥1 draws) is surjective, so the left side is 000 and the claim is that 0≤e−c0 \le e^{-c}0≤e−c. The statement is a non-strict inequality (≤\le≤, not <<<), it bounds only this single deterministic ratio at the single time t=⌈nln⁡n+c n⌉t = \lceil n\ln n + c\,n\rceilt=⌈nlnn+cn⌉ (no quantification over other times), and it involves no Markov chain, stationarity, or mixing hypotheses — nnn and ccc are the only data, constrained only by 1≤n1 \le n1≤n and 0<c0 < c0<c.

Human review
  • Endorsed by Community (Bot) · Aug 21, 2026

  • Endorsed by Shuze Chen · Aug 21, 2026

    Confirmed by the mission captain (proposal self-audit).

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