Remark 2.1.

CSE 525: Randomized Algorithms Spring 2026 Lecture 2: Second Moment Method Lecturer: Shayan Oveis Gharan 04/02/2026 Scribe:

Disclaimer: These notes have not been subjected to the usual scrutiny reserved for formal publications.

Consider a positive integer $n$ and $p\in[0,1]$ . Perhaps the simplest model of random (undirected) graphs is $G_{n,p}$ . To sample a graph from $G_{n,p}$ , we add every edge $\{u,v\}$ (for $u\neq v$ and $u,v\in\{1,\dots,n\}$ ) independently with probability $p$ .

For example, if $X$ denotes the number of edges in a $G_{n,p}$ random graph, then we have

\mathbb{E}\left[X\right]={n\choose 2}\cdot p.

A 4-clique in a graph is a set of four nodes such that all ${4\choose 2}=6$ possible edges between the nodes are present. Let $G$ be a random graph sampled according to $G_{n,p}$ , and let ${\cal C}_{4}$ denote the event that $G$ contains a 4-clique. It will turn out that if $p\gg n^{-2/3}$ , then $G$ contains a 4-clique with probability close to 1, while if $p\ll n^{-2/3}$ , then $\mathbb{P}\left[{\cal C}_{4}\right]$ will be close to 0. Thus $p=n^{-2/3}$ is a “threshold” for the appearance of a 4-clique.

Remark 2.1.

Here we use the asymptotic notation $f(n)\gg g(n)$ to denote that $\lim_{n\to\infty}f(n)/g(n)\to\infty$ . Similarly, we write $f(n)\ll g(n)$ to denote that $\lim_{n\to\infty}f(n)/g(n)\to 0$ .

We can use a simple first moment calculation for one side of our desired threshold behavior.

Lemma 2.2.

If $p\ll n^{-2/3}$ then $\mathbb{P}\left[{\cal C}_{4}\right]\to 0$ as $n\to\infty$ .

Proof.

Let $X$ denote the number of 4-cliques in $G\sim G_{n,p}$ . We can write $X=\sum_{S}X_{S}$ where the set S runs over all ${n\choose 4}$ subsets of four vertices in $G$ , and $X_{S}$ be the indicator random variable that there is a 4-clique on S. We have $\mathbb{P}\left[X_{S}=1\right]=p^{6}$ since all 6 edges must be present and are independent, thus by linearity of expectation $\mathbb{E}\left[X\right]=p^{6}\cdot{n\choose 4}$ . So if $p\ll n^{-2/3}$ , then $\mathbb{E}\left[X\right]\to 0$ as $n\to\infty$ . But now Markov’s inequality implies that

\mathbb{P}\left[{\cal C}_{4}\right]=\mathbb{P}\left[X\geq 1\right]\leq\mathbb{% E}\left[X\right]\to 0.

∎

On the other hand, proving that $p\gg n^{-2/3}\Rightarrow\mathbb{P}\left[{\cal C}_{4}\right]\to 1$ is more delicate. Even though a first moment calculation implies that, in this case, $\mathbb{E}\left[X\right]\to\infty$ , this is not enough to conclude that $\mathbb{P}\left[{\cal C}_{4}\right]\to 1$ . For instance, it could be the case that with probability $1-\frac{1}{n^{2}}$ , we have no 4-cliques, but we see all ${n\choose 4}$ many 4-cliques otherwise. In that case, $\mathbb{E}\left[X\right]=\Theta(n^{2})$ , but still the probability of seeing a 4-clique would be $\frac{1}{n^{2}}$ In other words, if the only thing we know about the random variable $X$ is its expectation we cannot say it is non-zero with high probability. We need to know higher order moments of $X$ .

2.1 Chebyshev’s Inequality

Definition 2.3 (Variance).

The variance of a random variable $X$ is defined as

\textup{Var}(X)=\mathbb{E}\left[(X-\mathbb{E}X)^{2}\right]=\mathbb{E}\left[X^{% 2}\right]-\mathbb{E}\left[X\right]^{2}

Theorem 2.4 (Chebyshev’s Inequality).

For any random variable $X$ ,

\mathbb{P}\left[|X-\mathbb{E}X|>\epsilon\right]<\frac{\operatorname{Var}(X)}{% \epsilon^{2}}

In the probabilistic method, the following statement is very handy.

Corollary 2.5.

For any random variable $X$ ,

\mathbb{P}\left[X=0\right]\leq\frac{\operatorname{Var}(X)}{(\mathbb{E}X)^{2}}

Proof.

Let $\epsilon=\mathbb{E}X$ in the Chebyshev’s inequality. Then,

\mathbb{P}\left[X=0\right]\leq\mathbb{P}\left[|X-\mathbb{E}X|\geq\mathbb{E}X% \right]\leq\frac{\operatorname{Var}(X)}{(\mathbb{E}X)^{2}}.

∎

Lemma 2.6.

If $X$ is a non-negative random variable, then

\mathbb{P}\left[X>0\right]\geq\frac{(\mathbb{E}\left[X\right])^{2}}{\mathbb{E}% \left[X^{2}\right]}.

Proof.

We use the Cauchy-Schwartz inequality: For any two random variables $X, Y$ we can write

\mathbb{E}\left[X\cdot Y\right]\leq\sqrt{\mathbb{E}\left[X^{2}\right]}\cdot% \sqrt{\mathbb{E}\left[Y^{2}\right]}.

Having this we write,

\mathbb{E}\left[X\right]=\mathbb{E}\left[X{\bf 1}_{X>0}\right]\leq\sqrt{% \mathbb{E}\left[X^{2}\right]}\sqrt{\mathbb{E}\left[{\bf 1}_{X>0}\right]}=\sqrt% {\mathbb{E}\left[X^{2}\right]}\sqrt{\mathbb{P}\left[X>0\right]}.

∎

For random variables $X, Y$ let

\operatorname{Cov}(X,Y)=\mathbb{E}\left[XY\right]-\mathbb{E}\left[X\right]% \mathbb{E}\left[Y\right].

In particular, if $X, Y$ is independent, then $\operatorname{Cov}(X,Y)=\mathbb{E}\left[XY\right]$ .

Fact 2.7.

If $X=X_{1}+\dots+X_{n}$ , then

\operatorname{Var}(X)=\sum_{i}\operatorname{Var}(X_{i})+\sum_{i\neq j}% \operatorname{Cov}(X_{i},X_{j}).

In particular, if all $X_{i}$ ’s are independent then $\operatorname{Var}(X)=\sum_{i}\operatorname{Var}(X_{i})$ .

Proof.

First, observe

\operatorname{Var}(X)=\mathbb{E}(\sum_{i}X_{i})^{2}-\left(\mathbb{E}\sum_{i}X_% {i}\right)^{2}

Expanding the terms and combining the terms corresponding to $X_{i},X_{j}$ gives the desired identity. ∎

Lemma 2.8.

If $p\gg n^{-2/3}$ , then $\mathbb{P}\left[{\cal C}_{4}\right]\to 1$ as $n\to\infty$ .

Proof.

Let $X_{S}$ be the indicator random variable of having a clique on $S$ and $X=\sum_{S}X_{S}$ as before. Using 2.5,

\mathbb{P}\left[{\cal C}_{4}\right]=\mathbb{P}\left[X>0\right]\geq 1-\frac{% \operatorname{Var}(X)}{(\mathbb{E}X)^{2}}

our goal is to show that $\operatorname{Var}(X)\ll(\mathbb{E}X)^{2}$ .

First, notice that for any $S$ ,

\operatorname{Var}(X_{S})=\mathbb{E}\left[X_{S}\right]-(\mathbb{E}\left[X_{S}% \right])^{2}\leq\mathbb{E}\left[X_{S}\right]=p^{6}.

So, $\sum_{S}\operatorname{Var}(X_{S})\leq{n\choose 4}p^{6}$ .

Now, fix two sets $S,T\in{n\choose 4}$ . Obviously if $|S\cap T|\leq 1$ , then $S, T$ do not share any ”potential” edges. So, by independence of edges $\mathbb{P}\left[X_{S}X_{T}\right]=\mathbb{P}\left[X_{S}\right]\mathbb{P}\left[% X_{T}\right]=p^{12}$ .

On the other hand, if $|S\cap T|=2$ . Then,

\mathbb{P}\left[X_{S}X_{T}\right]=\mathbb{P}\left[X_{S}\right]\mathbb{P}\left[% X_{T}|X_{S}\right]=p^{6}\mathbb{P}\left[X_{T}|X_{S}\right]=p^{11}.

The last identity is because since $X_{S}$ occurs we know that there is an edge in the common pair. So, we only need 5 more edges to get $X_{T}$ . Similarly, if $|S\cap T|=3$ , then $\mathbb{P}\left[X_{S}X_{T}\right]=p^{9}$ . In summary,

\mathbb{P}\left[X_{S}X_{T}\right]=\begin{cases}\mathbb{P}\left[X_{S}\right]% \mathbb{P}\left[X_{T}\right]&\text{if }|S\cap T|\leq 1\\ p^{11}&\text{if }|S\cap T|=2\\ p^{9}&\text{if }|S\cap T|=3.\end{cases}

It follows that

	$\displaystyle\sum_{S\neq T}\operatorname{Cov}(X_{S},X_{T})$	$\displaystyle=\sum_{S}\left(\sum_{T:\|T\cap S\|=2}\operatorname{Cov}(X_{S},X_{T}% )+\sum_{T:\|T\cap S\|=3}\operatorname{Cov}(X_{S},X_{T})\right)$
		$\displaystyle=\sum_{S}\left(6{n-4\choose 2}(p^{11}-p^{12})+4{n-4\choose 1}(p^{% 9}-p^{12})\right)$
		$\displaystyle\leq{n\choose 4}(3n^{2}p^{11}+4np^{9})$

Lastly,

\displaystyle\mathbb{P}\left[X=0\right]\leq\frac{\operatorname{Var}(X)}{(% \mathbb{E}X)^{2}}\leq\frac{{n\choose 4}p^{6}+{n\choose 4}(3n^{2}p^{11}+3np^{9}% ))}{({n\choose 4}p^{6})^{2}}\leq\frac{1+3n^{2}p^{5}+4np^{3}}{{n\choose 4}p^{6}}

Observe that for $p\gg n^{-2/3}$ the ratio goes to infinity as $n\to\infty$ . ∎