Talagrand's equivalence conjecture for integral and fractional expectation-thresholds

Let XX be a finite set and let F2X{\mathcal F}\subseteq2^X be increasing. Let q(F)q({\mathcal F}) be the expectation-threshold and qf(F)q_f({\mathcal F}) the fractional expectation-threshold. Talagrand's equivalence conjecture. There is a universal constant KK such that

q(F)qf(F)K.q({\mathcal F})\geq\frac{q_f({\mathcal F})}{K}.

The source describes this as a problem proposed by Talagrand that would imply the equivalence of the integral and fractional forms of the threshold conjectures. Its resolution status is not given in the supplied text.

Sources & referencesView supporting material

Primary source

Keith Frankston, Jeff Kahn, Bhargav Narayanan and Jinyoung Park, “Thresholds versus fractional expectation-thresholds”, arXiv:1910.13433 (2019).

Progress summary

Refreshed
Partially solved

The conjecture remains open, but several restricted cases and related logarithmic bounds have been proved.

Talagrand proposed in 2010 that the integral and fractional expectation-thresholds should differ by at most a universal constant, a statement that would connect the corresponding threshold conjectures.

Known results

  • 2021: the fractional expectation-threshold conjecture was proved with a logarithmic loss, pc(F)Kqf(F)logXp_c({\mathcal F})\leq Kq_f({\mathcal F})\log |X|; this does not prove Talagrand’s constant-factor comparison.
  • Talagrand proved the singleton-support case; support on pairs was identified as a harder test case.
  • Park and Pham proved the relevant selector-process conjecture, with a quantitative strengthening by Bednorz, Martynek, and Meller, yielding restricted threshold results.

2024–2025 restricted progress

Results in 2024 established rounding and the conjecture for fractional solutions supported on sets of bounded size, with losses involving logt\log t. A 2025 note settled additional special cases via a method of DeMarco and Kahn. Neither work claims the unrestricted universal-constant conjecture.

Current status (as of August 2026): Restricted cases and logarithmic-factor theorems are settled, but the full universal-constant comparison q(F)qf(F)/Kq({\mathcal F})\geq q_f({\mathcal F})/K remains open.

Sources

Solutions 0

No solutions have been posted yet.