CSE 525: Randomized Algorithms Spring 2025 Lecture 3: Strong Concentration Bounds Lecturer: Shayan Oveis Gharan 04/02/2026

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

We have seen how knowledge of the variance of a random variable $X$ can be used to control deviation of $X$ from its mean. This is the heart of the second moment method. But often we can control even higher moments, and this allows us to obtain much stronger concentration properties. A prototypical example is when $X_1,X_2,\mathrm{\dots },X_n$ is a family of independent (but not necessarily identically distributed) $\{0,1\}$ random variables and $X=X_1+X_2+\mathrm{\cdots }+X_n$. Let $p_i=𝔼\left[X_i\right]$ and define $\mu =𝔼\left[X\right]={\sum }_{i=1}^n{p}_1+p_2+\mathrm{\cdots }+p_n$. In that case, we have the following multiplicative form of the ”Chernoff bound”.

Theorem 3.1 (Multiplicative Chernoff bound).

. For every $\delta \ge 0$, it holds that

$$\mathbb{P}\left[X\ge (1+\delta )\mu \right]\le {\left(\frac{e^{\delta }}{{(1+\delta )}^{1+\delta }}\right)}^{\mu }.$$

and

Consequently,

$$\mathbb{P}\left[X\ge (1+\delta )\mu \right]\le e^{-{\delta }^2\mu /(2+\delta },\mathbb{P}\left[X\le (1-\delta )\mu \right]\le e^{-{\delta }^2\mu /2}$$

Proof.

Let $t$ be a parameter that we choose later.

$$\mathbb{P}\left[X\ge (1+\delta )\mu \right]=\mathbb{P}\left[e^{tX}\ge e^{t(1+\delta )\mu }\right]\underset{\text{Markov’s Inequality}}{\le }\frac{𝔼\left[e^{tX}\right]}{e^{t(1+\delta )\mu }}.$$ (3.1)

The first inequality uses that the exponential function is a monotone function.

Now, we can write

$$𝔼\left[e^{tX}\right]=𝔼\left[e^{t{\sum }_i{X}_i}\right]=𝔼\left[\prod _{i=1}^n{e}^{t{X}_i}\right]\underset{\text{independence}}{=}\prod _{i=1}^n𝔼\left[e^{t{X}_i}\right].$$

Now, observe that

$$𝔼\left[e^{tX}\right]=p_i{e}^t+(1-p_i)=1+p_i(e^t-1)\underset{1+x\le e^x}{\le }{e}^{p_i(e^t-1)}$$

Plugging this back we obtain

$$𝔼\left[e^{tX}\right]\le \prod _{i=1}^n{e}^{p_i(e^t-1)}=e^{\mu (e^t-1)}$$

Putting back in (3.1), we obtain

$$\mathbb{P}\left[X\ge (1+\delta )\mu \right]\le \frac{e^{\mu (e^t-1)}}{e^{t(1+\delta )\mu }}=e^{\mu (e^t-1-(1+\delta )t)}\underset{\text{set }t=\ln (1+\delta )}{=}{\left(\frac{e^{\delta }}{{(1+\delta )}^{1+\delta }}\right)}^{\mu }$$

The other case can be proven similarly. ∎

3.1 Giant Connected Components in Erdös-Réyni Graphs

In this section we prove the following theorem.

Theorem 3.2.

Theorem 1 Let $ϵ>0$ be a small enough constant. Let $G$ be an Erdös-Réyni random graph with parameter $p$.

1. Let $p=\frac{1-ϵ}{n}$. Then whp all connected components of $G$ are of size at most $\frac{7}{{ϵ}^2}\ln n$.

2. Let $p=\frac{1+ϵ}{n}$. Then whp $G$ contains a path of length at least $\frac{{ϵ}^{2}n}{5}$.

We run the DFS algorithm to prove the theorem. First, let us recall this algorithm: Fix a natural order on the vertices of $G$ we assume that algorithm prioritizes vertices according to this natural order. DFS maintains three sets of vertices, letting $X$ be the set of vertices whose exploration is complete, i.e., explored, $U$ be the set of unvisited vertices, and $T=[n]\setminus X\setminus U$ be the set of active vertices in the stack.

The algorithm starts with $X=T=\mathrm{\varnothing }$ and $U=V$, and runs till $T\cup U=\mathrm{\varnothing }$. At each round of the algorithm, if the set $T$ is non-empty, the algorithm queries $U$ for neighbors of the last vertex $v$ that has been added to $T$, scanning $U$ according to the natural order. If $v$ has a neighbor $u\in U$, the algorithm deletes $u$ from $U$ and inserts it into $T$ . If $v$ does not have a neighbor in $U$, then $v$ is popped out of $T$ and is moved to $X$. If $T$ is empty, the algorithm chooses the first vertex of $U$ according to the natural order, deletes it from $U$ and pushes it into $T$. In order to complete the exploration of the graph, whenever the sets $T$ and $U$ have both become empty (at this stage all connected components of $G$ have been revealed), we make the algorithm query all remaining pairs of vertices in $S$, not queried before.

The following properties of DFS are immediate:

• At each round of the algorithm one vertex moves, either from $U$ to $T$ , or from $T$ to $X$;

• At any time during the algorithm, it has been revealed already that the graph $G$ has no edges between the current set $X$ and the current set of unvisited vertices $U$;

• The set $T$ always spans a path (indeed, when a vertex $u$ is added to $T$ , it happens because $u$ is a neighbor of the last vertex $v$ in $T$ ; thus, $u$ augments the path spanned by $T$, of which $v$ is the last vertex).

Let $N=\left(\genfrac{}{}{0pt}{}{n}{2}\right)$ To prove the theorem we run DFS on a random input $G(n,p)$. Thus we feed DFS algorithm with a sequence of i.i.d. Bernoulli(p) random variables $Y_1,\mathrm{\dots },Y_N$ so that is gets its i-th query answered positively if $Y_i=1$ and answered negatively otherwise, the so obtained graph is clearly distributed according to G(n, p). Thus, studying the component structure of $G$ can be reduced to studying the properties of the random sequence $X$. In particular, observe crucially that as long as $U\ne \mathrm{\varnothing }$, every positive answer to a query results in a vertex being moved from $U$ to $T$ , and thus after $t$ queries and assuming $T\ne \mathrm{\varnothing }$ still, we have $|X\cup T|\ge {\sum }_{i=1}^t{Y}_i$. (The last inequality is strict in fact as the first vertex of each connected component is moved from $T$ to $U$ ”for free”, i.e., without need to get a positive answer to a query.) On the other hand, since the addition of every vertex, but the first one in a connected component, to U is caused by a positive answer to a query, we have at time t: $|T|\le 1+{\sum }_{i=1}^t{Y}_i$.

The following lemma gives us the tool that we need to prove the theorem.

Lemma 3.3.

Let $ϵ>0$ be a small enough constant. Consider the sequence of iid Bernoulli random variables with parameter $p$. $Y_1,\mathrm{\dots },Y_N$.

1. Let $p=\frac{1-ϵ}{n}$ and $k=\frac{7}{{ϵ}^2}\ln n$. Then, with probability $\gtrsim 1-1/\sqrt{n}$, there is no interval of length $kn$ where at least $k$ of the Bernoullis are 1.

2. Let $p=\frac{1+ϵ}{n}$ and $N_0=\frac{ϵn^2}{2}$. Then,

Proof.

Consider an interval $I$ of length $kn$ in $[N]$. Let $Y={\sum }_{i\in I}{Y}_i$. Notice $𝔼\left[Y\right]=knp$. By the multiplicative Chernoff bound,

$$\mathbb{P}\left[Y\ge k\right]=\mathbb{P}\left[Y\ge \frac{𝔼\left[Y\right]}{np}\right]=\mathbb{P}\left[Y\ge \frac{𝔼\left[Y\right]}{1-ϵ}\right]\le \exp (-\frac{{ϵ}^2𝔼\left[Y\right]}{2+ϵ})\le n^{-\frac{7}{2+ϵ}}$$

where the last inequality follows by $k=\frac{7}{{ϵ}^2}\ln n$. By a union bound the probability, since there are only $O(n^2)$ many such intervals the claim follows.

To prove 2, let $Y={\sum }_{i=1}^{N_0}{Y}_i$. Then,

$$𝔼\left[Y\right]=N_0\cdot p=\frac{(1+ϵ)ϵn}{2}$$

Now, again by multiplicative Chernoff bound, for $\delta =\frac{2n^{-1/3}}{ϵ(1+ϵ)}$

$$\mathbb{P}\left[\left|\sum _{i=1}^{N_0}{Y}_i-\frac{ϵ(1+ϵ)n}{2}\right|>n^{2/3}\right]\le \exp (-{\delta }^2\mu /3)\le \exp (-n^{1/3})$$

∎

We are now ready to prove the theorem.

Part 1.

Assume to the contrary that $G$ contains a connected component $C$ with more than $k=\frac{7}{{ϵ}^2}\ln n$ vertices. Let us look at the epoch of the DFS when $C$ was created (an epoch is a period during which the stack gets empty again). Consider the moment inside this epoch when the algorithm has found the $(k+1)$-st vertex of $C$ and is about to move it to $T$. Denote $X_C=X\cap C$ at that moment. Then $|X_C\cup T|=k$, and thus the algorithm got exactly $k$ positive answers to its queries to random variables $Y_i$ during the epoch, with each positive answer being responsible for revealing a new vertex of $C$, after the first vertex of $C$ was put into $T$ in the beginning of the epoch. During the epoch only pairs of edges touching $X_C\cup T$ have been queried, and the number of such pairs is therefore at most $\left(\genfrac{}{}{0pt}{}{k}{2}\right)+k(n-k)\le kn$. It thus follows that the sequence $Y$ contains an interval of length at most $kn$ with at least $k$ 1’s which is a contradiction.

Part 2.

Now, assume that the sequence $Y$ satisfies Property 2 of 3.3. We claim that after the first $N_0=\frac{ϵn^2}{2}$ queries of the DFS algorithm, the set $T$ contains at least $\frac{{ϵ}^{2}n}{5}$ vertices (with the contents of $T$ forming a path of desired length at that moment).

First observe that at time $N_0$. Indeed, if $|X|\ge \frac{n}{3}$, then let us look at a moment $t$ where $|X|=\frac{n}{3}$. At that moment by Property 2 of the Lemma. Then $|U|=n-|X|-|T|\ge \frac{n}{3}$, and the algorithm has examined all $|X|\cdot |U|\ge \frac{n^2}{9}>N_0$ pairs between $X$ and $U$ (and found them to be non-edges) – a contradiction.

Getting back to time $N_0$; now assume and then, we have $U\ne \mathrm{\varnothing }$. This means in particular that the algorithm is still revealing connected components of $G$, and each positive answer it got resulted in moving a vertex from $U$ to $T$ (some of these vertices may have already moved further from $T$ to $X$). By Property 2 of 3.3 the number of positive answers at that point is at least $\frac{ϵ(1+ϵ)n}{2}-n^{2/3}$. Hence, we have $|X\cup T|\ge \frac{ϵ(1+ϵ)n}{2}-n^{2/3}$. If $|T|\le \frac{{ϵ}^{2}n}{5}$, then $|X|\ge \frac{ϵn}{2}+\frac{3{ϵ}^{2}n}{10}-n^{2/3}$. Therefore, all pairs of vertices between $X,U$ are queried already (and received a negative answer), i.e., $|X|\cdot |U|$ many pairs. It follows that

${\displaystyle \frac{{ϵ}^{2}n}{2}}=N_0$ $\ge |X|\cdot |U|\ge |X|\cdot \left(n-|X|-{\displaystyle \frac{{ϵ}^{2}n}{5}}\right)$

$\ge \left({\displaystyle \frac{ϵn}{2}}+{\displaystyle \frac{3{ϵ}^{2}n}{10}}-n^{2/3}\right)\cdot \left(n-{\displaystyle \frac{ϵn}{2}}-{\displaystyle \frac{{ϵ}^{2}n}{2}}+n^{2/3}\right)$

$\ge {\displaystyle \frac{ϵn^2}{2}}+{\displaystyle \frac{{ϵ}^2{n}^2}{20}}-O({ϵ}^3)n^2>{\displaystyle \frac{ϵn^2}{2}}$

as desired.