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The symmetric Lovász Local Lemma under $e p (d + 1) \leq 1$

Statement

Let $d \in N$ and $p \geq 0$ . Let $(A_{i})_{i \in I}$ be a finite family of events with a dependency digraph of maximum out-degree at most $d$ . If $P (A_{i}) \leq p$ for every $i$ and $e p (d + 1) \leq 1,$ then $P (⋂_{i} A_{i}^{c}) > 0$ .

Facts & Assumptions

Given: A finite event family, its dependency digraph, and $p, d$ satisfying the Statement.

[L1]

The asymmetric Local Lemma applies when $P (A_{i}) \leq x_{i} \prod_{i \to j} (1 - x_{j})$ with $0 \leq x_{i} < 1$ (The asymmetric Lovász Local Lemma for finitely many events).

[L2]

$1 + y \leq \exp (y)$ for every real $y$ ( $1 + x \leq \exp (x)$ for every real $x$ , hence $(1 - p)^{m} \leq \exp (- m p)$ ).

[L3]

$\exp (u + v) = \exp (u) \exp (v)$ ; natural powers preserve order on nonnegative bases; and positive inequalities may be multiplied and inverted using the ordered-field laws (The exponential addition formula $\exp (x + y) = \exp (x) \exp (y)$ , Integer powers $a^{m}$ , Laws of integer exponents, Monotonicity of $x \mapsto x^{n}$ and of $n \mapsto a^{n}$ , The reals form a totally ordered field).

Proof

technique · cases

1.1

Suppose $d = 0$ and set every $x_{i} = 1 / e$ . Applying [L2] at $y = 1$ gives $2 \leq e$ , so $0 < x_{i} < 1$ . The hypothesis gives $p \leq 1 / e = x_{i}$ , and the empty neighbour product is $1$ , so [L1] applies.

assume-case zeroL1L2algebra

1.2

Suppose $d \geq 1$ and set every $x_{i} = 1 / (d + 1)$ . From [L2] at $y = 1 / d$ and [L3], $(1 + 1 / d)^{d} \leq e$ , hence $(1 - 1 / (d + 1))^{d} = (d / (d + 1))^{d} \geq 1 / e$ .

assume-case positiveL2L3algebra

2.1

Each vertex has at most $d$ out-neighbours, so $x_{i} \prod_{i \to j} (1 - x_{j}) \geq 1 / (e (d + 1)) \geq p$ by the hypothesis. Thus [L1] applies.

step 1.2L1L3algebra

3.1

The cases $d = 0$ and $d \geq 1$ are exhaustive and both give positive probability that no bad event occurs.

step 1.1step 2.1cases-exhaustive∎

Depends on

Used by

A k-uniform hypergraph is 2-colourable when every edge meets at most d other edges and e(d+1)≤2ᵏ⁻¹ Theorem

Dependency tree · two levels

46 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

J. Matousek and J. Vondrak, The Probabilistic Method, Corollary 5.1.2 (standard reference, not scraped)
Y. Zhao, MIT 18.218 Probabilistic Method in Combinatorics, Corollary 5.3 (standard reference, not scraped)

The symmetric Lovász Local Lemma under ep(d+1)≤1ep(d+1)≤1