This lemma is extremely useful in the probabilistic method: if you can prove that the set of objects satisfying a certain number of constraints has positive probability (or density), then this set cannot be empty. In particular, an object satisfying those constraints do exist. However, this method does not explicitly build the desired object. The constructive version of the Lovász local lemma was given by Moser and Tardos, as explained in this previous post.

**1. Lovász Local Lemma**

Let $A_1,\dots,A_n$ be a set of bad events. These events are not necessarily mutually independent. For each index $1\le i \le n$, we denote by $D_i \subseteq \{1,\dots,n\}$ the set of indices such that the events $A_i, A_j$ with $j \not\in D_i$ are mutually independent. In other words, $D_i$ is the dependency set of the event $A_i$.

The lemma states:

The proof goes as follows. We first show that for any index $1\le i \le n$ and any strict subset $S \subset \{1,\dots,n\}$, we haveIf there exists real values $x_1,\dots,x_n \in [0,1)$ such that $$ \mathbb{P}(A_i) \le x_i \cdot \prod_{j \in D_i} (1-x_j) ~~~~~ (\star) $$ then $$\mathbb{P}\left(\bigwedge\limits_{i=1}^n \overline{A}_i\right) \ge \prod_{i=1}^n (1-x_i).$$

$$

\mathbb{P}\left( A_i ~\bigg|~ \bigwedge\limits_{j \in S} \overline{A_j} \right) \le x_i.

$$

We prove this claim by induction on the size $m = |S|$. The case $m = 0$, i.e., $S = \emptyset$, follows from $(\star)$. Now, assume the result holds for all subset of size less than $m$. We split the set $S$ in two disjoint parts $S_D = S \cap D_i$ and $S_I = S - S_D$. The indices in $S_D$ (resp. $S_I$) are exactly the indices in $S$ of events that are not independent (resp. that are independent) from $A_i$. We have, by the chain rule of conditional probabilities:

$$

\begin{align*}

\mathbb{P}\left(A_i ~\bigg|~ \bigwedge\limits_{j \in S} \overline{A_j} \right)

&= \mathbb{P}\left(A_i ~\bigg|~ \bigwedge\limits_{k \in S_D} \overline{A_k} \land \bigwedge\limits_{l \in S_I} \overline{A_l} \right) \\

&= \frac{\mathbb{P}\left(A_i \land \bigwedge\limits_{k \in S_D} \overline{A_k} ~\bigg|~ \bigwedge\limits_{l \in S_I} \overline{A_l} \right)}{\mathbb{P}\left(\bigwedge\limits_{k \in S_D} \overline{A_k} ~\bigg|~ \bigwedge\limits_{l \in S_I} \overline{A_l} \right)}\\

&\le \frac{\mathbb{P}\left(A_i ~\bigg|~ \bigwedge\limits_{l \in S_I} \overline{A_l} \right)}{\mathbb{P}\left(\bigwedge\limits_{k \in S_D} \overline{A_k} ~\bigg|~ \bigwedge\limits_{l \in S_I} \overline{A_l} \right)}

\end{align*}

$$

Since the events $A_i, A_l$ , $l \in S_I$, are mutually independent, we have:

$$

\begin{align*}

\mathbb{P}\left(A_i ~\bigg|~ \bigwedge\limits_{j \in S} \overline{A_j} \right)

&\le \frac{\mathbb{P}\left(A_i \right)}{\mathbb{P}\left(\bigwedge\limits_{k \in S_D} \overline{A_k} ~\bigg|~ \bigwedge\limits_{l \in S_I} \overline{A_l} \right)}\\

&\le \frac{x_i \cdot \prod_{j\in D_i} (1-x_j)}{\mathbb{P}\left(\bigwedge\limits_{k \in S_D} \overline{A_k} ~\bigg|~ \bigwedge\limits_{l \in S_I} \overline{A_l} \right)}

\end{align*}

$$

Regarding the denominator, denoting $S_D = \{k_1,\dots,k_s\}$, and $S_D^r = \{k_1,\dots,k_r\}$ for any $1 \le r \le s$, we have

$$

\begin{align*}

\mathbb{P}\left(\bigwedge\limits_{k \in S_D} \overline{A_k} ~\bigg|~ \bigwedge\limits_{l \in S_I} \overline{A_l} \right)

&= \prod_{r = 1}^s \mathbb{P}\left( \overline{A}_{k_r} ~\bigg|~ \bigwedge\limits_{l \in S_I \sqcup S_D^r} \overline{A_l} \right)\\

&= \prod_{r = 1}^s \left(1 - \mathbb{P}\left( A_{k_r} ~\bigg|~ \bigwedge\limits_{l \in S_I \sqcup S_D^r} \overline{A_l} \right)\right) \\

&\ge \prod_{r = 1}^s (1 - x_{k_r}) = \prod_{j \in S_I} (1 - x_j).

\end{align*}

$$

The last inequality follows from the induction hypothesis. We conclude that

$$

\begin{align*}

\mathbb{P}\left(A_i ~\bigg|~ \bigwedge\limits_{j \in S} \overline{A_j} \right)

&\le x_i \cdot \frac{\prod_{j \in D_i} (1 - x_j)}{\prod_{j \in S_I} (1 - x_j)} \\

&\le x_i ~~~ \text{ since } S_I \subseteq D_i.

\end{align*}

$$

This concludes the induction proof.

We apply this inequality to prove the Lovász lemma. Indeed, we have

$$

\begin{align*}

\mathbb{P}\left(\bigwedge\limits_{i=1}^n \overline{A}_i\right)

&= \prod_{i=1}^n \left( 1 - \mathbb{P}\left( A_i ~\bigg|~ \bigwedge\limits_{1 \le j < i} \overline{A}_j \right)\right) \\

&\ge \prod_{i=1}^n (1-x_i).

\end{align*}

$$

**2. Other formulations**

The previous formulation is quite general. A restricted formulation is given by the following:

Assume $\mathbb{P}(A_i) \le p < 1$ for all $1 \le i \le n$. Let $d$ be the maximum size of the dependency sets $D_i$'s. If $e\cdot p\cdot (d+1) \le 1$, thenIndeed,let's take $x_i = 1/(d+1)$. It suffices to prove that$$\mathbb{P}\left(\bigwedge\limits_{i=1}^n \overline{A}_i\right) \ge (1- \frac{1}{d+1})^n > 0.$$

$$

p \le \frac{1}{d+1} \cdot \left(1-\frac{1}{d+1}\right)^d

$$

But we have

$$

\begin{align*}

\left(1-\frac{1}{d+1}\right)^d \ge \left(1-\frac{1}{d}\right)^d

\ge \frac{1}{e}

\end{align*}

$$

Another restricted formulation is given as follows:

If $\sum_{j \in D_i} \mathbb{P}(A_j) \le 1/4$ for all $i$, thenIndeed, let's take $x_i = 2\mathbb{P}(A_i)$. The condition imposes that each $x_i \in [0,1]$. One can easily shows by induction on the size of $D_i$ that$$\mathbb{P}\left(\bigwedge\limits_{i=1}^n \overline{A}_i\right) \ge \prod_{i=1}^n \left(1- 2\mathbb{P}(A_i)\right) > 0.$$

$$

\begin{align*}

\prod_{j \in D_i} \left(1- 2\mathbb{P}(A_j)\right) &\ge 1 - \sum_{j\in D_i} 2\mathbb{P}(A_j) \\

&\ge \frac{1}{2}

\end{align*}

$$

Therefore, $\mathbb{P}(A_i) \le x_i \prod_{D_i} (1-x_j)$ is satisfied for all $i$.

pb