组合数学 (Fall 2016)/Problem Set 1

Given a sequence of events [math]\displaystyle{ A_1,A_2,\ldots,A_n }[/math], we use the dependency graph to describe the dependencies between these events.

Definition (dependency graph)

Let [math]\displaystyle{ A_1,A_2,\ldots,A_n }[/math] be a sequence of events. A graph [math]\displaystyle{ D=(V,E) }[/math] on the set of vertices [math]\displaystyle{ V=\{1,2,\ldots,n\} }[/math] is called a dependency graph for the events [math]\displaystyle{ A_1,\ldots,A_n }[/math] if for each [math]\displaystyle{ i }[/math], [math]\displaystyle{ 1\le i\le n }[/math], the event [math]\displaystyle{ A_i }[/math] is mutually independent of all the events [math]\displaystyle{ \{A_j\mid (i,j)\not\in E\} }[/math].

The notion of mutual independence between an event and a set of events is formally defined as follows.

Definition (mutual independence)

An event [math]\displaystyle{ A }[/math] is said to be mutually independent of events [math]\displaystyle{ B_1,B_2,\ldots, B_k }[/math], if for any disjoint [math]\displaystyle{ I^+,I^-\subseteq\{1,2,\ldots,k\} }[/math], it holds that [math]\displaystyle{ \Pr\left[A \mid \left(\bigwedge_{i\in I^+}B_i\right) \wedge \left(\bigwedge_{i\in I^-}\overline{B_i}\right)\right]=\Pr[A] }[/math].

Example

Let [math]\displaystyle{ X_1,X_2,\ldots,X_m }[/math] be a set of mutually independent random variables. Each event [math]\displaystyle{ A_i }[/math] is a predicate defined on a number of variables among [math]\displaystyle{ X_1,X_2,\ldots,X_m }[/math]. Let [math]\displaystyle{ v(A_i) }[/math] be the unique smallest set of variables which determine [math]\displaystyle{ A_i }[/math]. The dependency graph [math]\displaystyle{ D=(V,E) }[/math] is defined by

[math]\displaystyle{ (i,j)\in E }[/math] iff [math]\displaystyle{ v(A_i)\cap v(A_j)\neq \emptyset }[/math].

The following lemma, known as the Lovász local lemma, first proved by Erdős and Lovász in 1975, is an extremely powerful tool, as it supplies a way for dealing with rare events.

Lovász Local Lemma (symmetric case)

Let [math]\displaystyle{ A_1,A_2,\ldots,A_n }[/math] be a set of events, and assume that there is a [math]\displaystyle{ p\in[0,1) }[/math] such that the followings are satisfied: for all [math]\displaystyle{ 1\le i\le n }[/math], [math]\displaystyle{ \Pr[A_i]\le p }[/math]; the maximum degree of the dependency graph for the events [math]\displaystyle{ A_1,A_2,\ldots,A_n }[/math] is [math]\displaystyle{ d }[/math], and [math]\displaystyle{ ep\cdot (d+1)\le 1 }[/math]. Then [math]\displaystyle{ \Pr\left[\bigwedge_{i=1}^n\overline{A_i}\right]\gt 0 }[/math].

We will prove a general version of the local lemma, where the events [math]\displaystyle{ A_i }[/math] are not symmetric. This generalization is due to Spencer.

Lovász Local Lemma (general case)

Let [math]\displaystyle{ D=(V,E) }[/math] be the dependency graph of events [math]\displaystyle{ A_1,A_2,\ldots,A_n }[/math]. Suppose there exist real numbers [math]\displaystyle{ x_1,x_2,\ldots, x_n }[/math] such that [math]\displaystyle{ 0\le x_i\lt 1 }[/math] and for all [math]\displaystyle{ 1\le i\le n }[/math], [math]\displaystyle{ \Pr[A_i]\le x_i\prod_{(i,j)\in E}(1-x_j) }[/math]. Then [math]\displaystyle{ \Pr\left[\bigwedge_{i=1}^n\overline{A_i}\right]\ge\prod_{i=1}^n(1-x_i) }[/math].

To see that the general LLL implies symmetric LLL, we set [math]\displaystyle{ x_i=\frac{1}{d+1} }[/math] for all [math]\displaystyle{ i=1,2,\ldots,n }[/math]. Then we have [math]\displaystyle{ \left(1-\frac{1}{d+1}\right)^d\gt \frac{1}{\mathrm{e}} }[/math].

Assume the condition in the symmetric LLL:

1. for all [math]\displaystyle{ 1\le i\le n }[/math], [math]\displaystyle{ \Pr[A_i]\le p }[/math];
2. [math]\displaystyle{ ep(d+1)\le 1 }[/math];

then it is easy to verify that for all [math]\displaystyle{ 1\le i\le n }[/math],

[math]\displaystyle{ \Pr[A_i]\le p\le\frac{1}{e(d+1)}\lt \frac{1}{d+1}\left(1-\frac{1}{d+1}\right)^d\le x_i\prod_{(i,j)\in E}(1-x_j) }[/math].

Due to the general LLL, we have

[math]\displaystyle{ \Pr\left[\bigwedge_{i=1}^n\overline{A_i}\right]\ge\prod_{i=1}^n(1-x_i)=\left(1-\frac{1}{d+1}\right)^n\gt 0 }[/math].

This proves the symmetric LLL.

Now we prove the general LLL by the original induction proof.

Proof. First, apply the chain rule. We have [math]\displaystyle{ \Pr\left[\bigwedge_{i=1}^n\overline{A_i}\right]=\prod_{i=1}^n\Pr\left[\overline{A_i}\mid \bigwedge_{j=1}^{i-1}\overline{A_{j}}\right]=\prod_{i=1}^n\left(1-\Pr\left[{A_i}\mid \bigwedge_{j=1}^{i-1}\overline{A_{j}}\right]\right) }[/math]. Next we prove by induction on [math]\displaystyle{ m }[/math] that for any set of [math]\displaystyle{ m }[/math] events [math]\displaystyle{ i_1,\ldots,i_m }[/math], [math]\displaystyle{ \Pr\left[A_{i_1}\mid \bigwedge_{j=2}^m\overline{A_{i_j}}\right]\le x_{i_1} }[/math]. The local lemma follows immediately by the above chain rule. For [math]\displaystyle{ m=1 }[/math], this is obvious because [math]\displaystyle{ \Pr[A_{i_1}]\le x_{i_1}\prod_{(i_1,j)\in E}(1-x_j)\le x_{i_1} }[/math]. For general [math]\displaystyle{ m }[/math], let [math]\displaystyle{ i_2,\ldots,i_k }[/math] be the set of vertices adjacent to [math]\displaystyle{ i_1 }[/math] in the dependency graph, i.e. event [math]\displaystyle{ A_{i_1} }[/math] is mutually independent of [math]\displaystyle{ A_{i_{k+1}},A_{i_{k+2}},\ldots, A_{i_{m}} }[/math]. By conditional probability, we have [math]\displaystyle{ \Pr\left[A_{i_1}\mid \bigwedge_{j=2}^m\overline{A_{i_j}}\right] =\frac{\Pr\left[ A_i\wedge \bigwedge_{j=2}^k\overline{A_{i_j}}\mid \bigwedge_{j=k+1}^m\overline{A_{i_j}}\right]} {\Pr\left[\bigwedge_{j=2}^k\overline{A_{i_j}}\mid \bigwedge_{j=k+1}^m\overline{A_{i_j}}\right]} }[/math]. First, we bound the numerator. Due to that [math]\displaystyle{ A_{i_1} }[/math] is mutually independent of [math]\displaystyle{ A_{i_{k+1}},A_{i_{k+2}},\ldots, A_{i_{m}} }[/math], we have [math]\displaystyle{ \begin{align} \Pr\left[ A_{i_1}\wedge \bigwedge_{j=2}^k\overline{A_{i_j}}\mid \bigwedge_{j=k+1}^m\overline{A_{i_j}}\right] &\le\Pr\left[ A_{i_1}\mid \bigwedge_{j=k+1}^m\overline{A_{i_j}}\right]\\ &=\Pr[A_{i_1}]\\ &\le x_{i_1}\prod_{(i_1,j)\in E}(1-x_j). \end{align} }[/math] Next, we bound the denominator. Applying the chain rule, we have [math]\displaystyle{ \Pr\left[\bigwedge_{j=2}^k\overline{A_{i_j}}\mid \bigwedge_{j=k+1}^m\overline{A_{i_j}}\right] =\prod_{j=2}^k\Pr\left[\overline{A_{i_j}}\mid \bigwedge_{\ell=j+1}^m\overline{A_{i_\ell}}\right] }[/math] which by the induction hypothesis, is at least [math]\displaystyle{ \prod_{j=2}^k(1-x_{i_j})=\prod_{\{i_1,i_j\}\in E}(1-x_j) }[/math] where [math]\displaystyle{ E }[/math] is the set of edges in the dependency graph. Altogether, we prove the induction hypothesis [math]\displaystyle{ \Pr\left[A_{i_1}\mid \bigwedge_{j=2}^m\overline{A_{i_j}}\right] \le\frac{x_{i_1}\prod_{(i_1,j)\in E}(1-x_j)}{\prod_{\{i_1,i_j\}\in E}(1-x_j)}\le x_{i_1}. }[/math] Due to the chain rule, it holds that [math]\displaystyle{ \begin{align} \Pr\left[\bigwedge_{i=1}^n\overline{A_i}\right] &=\prod_{i=1}^n\Pr\left[\overline{A_i}\mid \bigwedge_{j=1}^{i-1}\overline{A_{j}}\right]\\ &=\prod_{i=1}^n\left(1-\Pr\left[A_i\mid \bigwedge_{j=1}^{i-1}\overline{A_{j}}\right]\right)\\ &\ge\prod_{i=1}^n\left(1-x_i\right). \end{align} }[/math] [math]\displaystyle{ \square }[/math]