2. Preliminary calculations

Throughout Sections 2 and 3, we restrict our attention to $p = p(n)$ , where $p =n^{-2/3 + \epsilon } < p < 1/ \log (n)^2$ for some $\epsilon>0$ . This is valid as the classical results of Bollobás and Erdős and Matula treat the case $ p> 1/ \log (n)^2$ . It is a classical result of Frieze that for all such p we have

(2) $$ \begin{align} \alpha( G_{n,p}) = \frac{2 \left[ \log(np) - \log(\log(np)) + \log(e/2) \pm o(1) \right] }{p} \end{align} $$

with high probability. As we are interested in refinements of this result, we will need some estimates regarding numbers in this range.

Recall that $ X_k$ is the number of independent sets of size k in $ G_{n,p}$ . The upper bound in Frieze’s result is achieved by simply identifying a value of k such that $\mathbb {E} [X_k] = o(1)$ . In order to make comparisons with this bound, we define $ k_x$ to be the largest value of k for which

$$\begin{align*}\mathbb{E}[X_k] = \binom{n}{k} (1-p)^{\binom{k}{2}}> n^{2\epsilon}. \end{align*}$$

We note in passing that the exact value in the exponent of this cut-off function is not crucial; we simply need it to be small (in particular, we need $ \mathbb {E}[ X_{k_x+2}] = o(1)$ ) and greater than the exponent in the polynomial function bound that we use to define the variable $k_z$ below.

Now, suppose $ k = k_x \pm o(1/p)$ . In other words, consider

(3) $$ \begin{align} k = \frac{2 \left[ \log(np) - \log(\log(np)) + \log( e/2) \pm o(1) \right] }{p}. \end{align} $$

(Note that impact of the $ 2 \epsilon $ in the definition of $ k_x$ is absorbed in the $ o(1)$ term.) Our first observation is that, for such a k, we have

$$ \begin{align*} \frac{ne}k (1-p)^{k/2} = (1+o(1)) \frac{ne}k \exp\{ - \log(np) + \log( \log(np)) - \log( e/2) + o(1)\} \\ = ( 1 +o(1)) \frac{ \log(np) }{ \log(np) - \log\log(np) + \log(e/2) - o(1)} = 1+o(1). \end{align*} $$

So, we have

(4) $$ \begin{align} (1-p)^{k/2} = (1+o(1)) \frac{k}{e n}. \end{align} $$

It follows that we have

(5) $$ \begin{align} \left( 1 - ( 1 -p)^k\right)^{n-k} = \left( 1 - (1+o(1)) \left( \frac{k}{en}\right)^2 \right)^{n-k} = e^{ - (1 +o(1)) \frac{k^2}{ e^2 n}}. \end{align} $$

Similarly,

(6) $$ \begin{align} \left[ 1 - (1-p)^k - pk(1-p)^{k-1} \right] ^{n-k} = e^{ - (1 +o(1)) \frac{ p k^3}{ e^{\vphantom{A^A}2} n} }. \end{align} $$

3. Two-point concentration

In this section, we show that two point concentration of $\alpha (G_{n,p})$ persists for p as low as $n^{-2/3+\epsilon }$ but at a value that does not equal $k_x$ for small p. Our main technical result is as follows. We write $ f(n) \sim g(n)$ if $ \lim _{n \to \infty } f/g =1 $ .

Theorem 3. If $ n^{-2/3+\epsilon } < p < 1 / \log (n)^2 $ for some $\epsilon>0$ , then there exists an integer $k_{z} = k_{z}(n)$ such that $\alpha (G_{n,p}) \in \{k_{z},k_{z}+1\}$ whp. Furthermore, we have

$$\begin{align*}k_x - k_{z} \begin{cases} = 0 & \text{ if } p = \omega\left( \log(n) n^{-1/2} \right)\\ = \xi_n & \text{ if } p = C \log(n) n^{-1/2} \\ \sim \frac{k_x^2}{e^2 n} \sim \frac{ 4 \log(np)^2}{ e^2 p^2 n} & \text{ if } p = o\left( \log(n) n^{-1/2} \right), \end{cases} \end{align*}$$

where $ \xi _n \in \left ( \frac {1}{ e^2C^2} -\frac {5}{2}, \frac {1}{ e^2C^2} +\frac {3}{2} \right ) $ .

We emphasize that the sequence $ \xi _n$ for the case $ p = C \log (n) n^{-1/2}$ does not converge to a particular value; rather, it ranges over the specified interval. Indeed, $ \xi _n$ is more precisely given by the following expression

$$\begin{align*}\xi_n = \left\lceil \frac{1}{e^2C^2} - \frac{2\log(\mathbb{E}[X_{k_x}])}{\log(n)} + 2\epsilon \pm o(1) \right\rceil. \end{align*}$$

As $ \mathbb {E}[ X_{k_x}]$ can take a wide range of values as n varies, we have persistent variation of $ \xi _n$ over a small list of values. The remainder of this section is a proof of Theorem 3.

Recall that an augmented independent set of order k in a graph G is defined to be a set of vertices S for which there exists some $r \geq 0$ such that $|S| = k+r$ and S contains exactly r edges that form a matching (i.e., these edges are pairwise disjoint) and all vertices outside of S have at least two neighbors in S. To motivate this definition, first note that such a set S contains $2^r$ independent sets of size k, so this structure is well suited for isolating large variations in the number of independent sets of size k. The fact that we are interested in studying the number of independent sets of size k where k is close to the independence number of G motivates the condition regarding vertices outside of S. Indeed, suppose S is a set such that the induced graph on S is a (not necessarily perfect) matching and v is a vertex outside of S that has one neighbor in S. If v is adjacent to a vertex u that is not in the matching then adding v to S creates an augmented independent set of order k with an additional edge in its matching. So in this situation we would want to include v in S in order to isolate as much variation in the number of independent sets of k as possible. On the other hand, if v is adjacent to a vertex u that is part of the matching, then $ S \setminus \{u\} \cup \{v\}$ contains an independent set of size $k+1$ .

Proof. Let $ \hat {\alpha }(G)$ be the largest k for which there is an augmented independent set of order k. As an augmented independent set of order k contains an independent set on k vertices, we have

$$\begin{align*}\hat{\alpha}(G) \le \alpha(G). \end{align*}$$

Now, suppose that G has a maximum independent set S of size k. Let T be a maximum set of vertices that contains S and has the additional property that the graph induced on T is a matching. We claim that T is an augmented independent set; that is, every vertex outside of T has two neighbors in T. Clearly, every vertex outside T has a neighbor in T (since S is a maximum independent set). If v is a vertex outside of T with only one neighbor in T, then this neighbor u is in one of the matching edges (since T is chosen to be a maximum set with the given property), but in this situation the set $ (T \setminus \{u\}) \cup \{v\}$ contains an independent set that is larger than S, which is a contradiction. So, T is an augmented independent set of order at least k. Therefore,

$$\begin{align*}\alpha(G) \le \hat{\alpha}(G).\\[-36pt] \end{align*}$$

We prove Theorem 3 by applying the second moment method to the random variable Z, which counts the number of augmented independent sets of order $k_z$ with a prescribed number of edges $r_z$ . This requires some relatively delicate estimates to determine the optimal values of the parameters $k_z$ and $r_z$ . We begin with those calculations. We then move on to the first moment and second moment calculations for Z that suffice to prove Theorem 3.

3.1. Defining and estimating $k_z$ and $r_z$

In this section, we define $ k_z$ and $r_z$ , and establish a number of estimates of these quantities.

To start, let $E(n,k,r)$ denote the expected number of augmented independent sets of order k with exactly r edges in $G_{n,p}$ . There are $\binom {n}{k+r}$ ways to choose the location of our set, and given this set there are $\frac {(k+r)!}{(k-r)! 2^r r!}$ ways to choose the locations of the r disjoint edges. The probability of choosing internal edges accordingly is

$$\begin{align*}p^r(1-p)^{\binom{k+r}{2}-r},\end{align*}$$

and finally, the probability that every vertex outside our set has at least two neighbors within the set is $ F(n,k,r)^{n-k-r}$ where we set

(7) $$ \begin{align} F(n,k,r) := 1 - (1-p)^{k+r} - (k+r)p(1-p)^{k+r-1}. \end{align} $$

Thus, we have

(8) $$ \begin{align} E(n,k,r) =\binom{n}{k+r}\frac{(k+r)!}{(k-r)! 2^r r!}p^r(1-p)^{\binom{k+r}{2}-r} F(n,k,r)^{n-k-r}. \end{align} $$

Now, we define $k_z = k_z(n)$ to be the largest $k \le k_x$ for which there exists an r such that $ E(n,k,r)> n^{\epsilon }$ .Footnote ¹

Our first observation is that, while $k_z$ is not equal to $k_x$ when p is sufficiently small, the difference between the two numbers is relatively small.

Lemma 5. If $ p = \omega ( \log (n)/ \sqrt {n} )$ , then $ k_z = k_x$ . Otherwise, we have

$$ \begin{align*} k_z \ge k_x - O \left( \frac{ k_x^2}{n} \right) \ge k_x - o( k_x^{1/2}). \end{align*} $$

Proof. This observation follows from considering $r=0$ and $ k = k_x - \frac { C k_x^2}{n}$ , where C is a sufficiently large constant and applying Equations (4) and (6). We first note that if $ p = \omega ( \log (n)/ \sqrt {n})$ , then the expression in Equation (6) is subpolynomial. In this case, we have $ E(n,k_x,0)> n^{\epsilon }$ and $ k_z = k_x$ .

Restricting our attention to $ p = O ( \log (n)/ \sqrt {n})$ , note that for $ k = k_x - O( k_x^2/n)$ we have

(9) $$ \begin{align} \frac{ \binom{n}{k} (1-p)^{\binom{k}{2}} }{ \binom{n}{k+1} (1-p)^{\binom{k+1}{2}} } = \frac{ k+1}{n-k} (1-p)^{-k} = (1+o(1)) \frac{ ne^2}{k}, \end{align} $$

and thus

$$ \begin{align*} \begin{split} E(n,k,0) &=\binom{n}{k} (1-p)^{ \binom{k}{2} } \exp \left\{- (1+o(1)) \frac{p k^3}{ e^2 n} \right\} \nonumber \\& \ge \left( \frac{ ne^2}{ 2k} \right)^{ \frac{ C k^2}{n}} \exp \left\{- (1+o(1)) \frac{p k^3}{ e^2 n} \right\}. \end{split} \end{align*} $$

Therefore, $E(n,k,0)> n^{\epsilon }$ if C is a sufficiently large constant.

Note that we may henceforth assume that the estimates given in Equations 3–6 hold for $k_z \le k \le k_x$ .

Next, for $ k_z \le k \le k_x$ define

$$\begin{align*}r_M(k) = r_M(n,k) := \arg \displaystyle\max_{r} \{E(n,k,r)\}. \end{align*}$$

We are now ready to define the random variable Z. Set $ r_z = r_M(k_z)$ , and let the variable Z count the number of augmented independent sets of order $k_z$ with exactly $r_z$ internal edges. Note that such sets have $k_{z}+r_z$ vertices and $\mathbb {E}(Z) = E(n,k_z,r_z)> n^\epsilon $ .

We now move on to more accurate estimates for $k_z$ and $r_z$ , starting with an estimate for $r_z$ that we obtain by estimating the ratio of consecutive terms of Equation (8). Note that we always have $r \leq k$ , which we will use a couple of times below. First, we show that $F(n,k,r)^{n-k-r}$ will have an insignificant effect on the ratio. One can easily verify, using Equation (4), that $F(n,k,r) = 1-o(1)$ . Furthermore,

$$\begin{align*}F(n,k,r+1) - F(n,k,r) =(k+r)p^2(1 -p)^{k+r-1}\end{align*}$$

exactly. Therefore, we have

$$ \begin{align*} \frac{F(n,k,r+1)^{n-k-r-1}}{F(n,k,r)^{n-k-r}} & = \left(\frac{F(n,k,r+1)}{F(n,k,r)}\right)^{n-k-r} F(n,k,r+1)^{-1} \\ & = (1 +o(1)) \left( 1 + \frac{ (k+r)p^2(1 -p)^{k+r-1}}{ F(n,k,r)} \right)^{n-k-r} \\ & = (1 +o(1)) \left( 1 + (1+o(1)) (k+r)p^2(1 -p)^{k+r-1} \right)^{n-k-r}. \end{align*} $$

Since

$$ \begin{align*}(k+r)p^2(1-p)^{k+r-1}(n-k-r) \leq 2 k p^2 n(1-p)^{k-1} < \frac{k^3 p^2}{n} = o(1),\end{align*} $$

it follows that

(10) $$ \begin{align} \frac{F(n,k,r+1)^{n-k-r-1}}{F(n,k,r)^{n-k-r}} = 1+o(1), \end{align} $$

as desired.

With this bound on the ratio $ F(n,k,r+1)^{n-k-r-1}/ F(n,k,r)^{n-k-r}$ in hand and applying Equation (4) as in Equation (9), we conclude that if $r = o(1/p)$ , then we have

(11) $$ \begin{align}\nonumber \frac{E(n,k,r+1)}{E(n,k,r)} &=(1+o(1)) \frac{ \binom{n}{k+r+1} (1-p)^{\binom{k+r+1}{2}} }{ \binom{n}{k+r} (1-p)^{\binom{k+r}{2}} } \left(\frac{(k+r+1)(k-r)p}{2(r+1)}\right)\\& = (1+o(1)) \left(\frac{k}{e^2 n}\right) \left(\frac{(k+r+1)(k-r)p}{2(r+1)}\right) \end{align} $$

(12) $$ \begin{align} &= (1+o(1))\frac{k^3p}{2e^2n(r+1)}(1-r^2/k^2).\qquad \end{align} $$

This calculation provides the desired estimate for $r_z$ .

Lemma 6.

$$\begin{align*}r_z = \left\lceil (1+o(1)) \frac{4 \log(np)^3 }{ e^2 n p^2} -1 \right\rceil. \end{align*}$$

Proof. First, note that the form of Equation (12) suggests that $ r_M(k)$ should be roughly $ k^3p/n$ as this is the regime in which the ratio is roughly one. Since $k^3p/n = o(1/p)$ , the estimate given by Equation (12) holds in the vicinity of $ r = k^3p/n$ . Furthermore, for $ r= \Omega (1/p)$ the arguments above can be easily adapted to show that $ E(n,k,r+1)/E(n,k,r)$ is bounded above by the expression in Equation (12). It follows that we have

(13) $$ \begin{align} r \ge k^3p/(e^2n) \ \ \ \Rightarrow \ \ \ \frac{E(n,k,r+1)}{E(n,k,r)} < 1. \end{align} $$

Hence, we can conclude that

(14) $$ \begin{align} r_M(k) \le \frac{k^3p}{e^2n} = O\left( \frac{\log(np) k^2}{n} \right). \end{align} $$

Note that if $r = O( k^3p/n )$ then the $(1-r^2/k^2) $ term in Equation (12) is equal to $ 1+o(1)$ . We conclude that we have

(15) $$ \begin{align} r_M(k) = \left\lceil \frac{k^3p}{ 2 e^2n} (1+o(1)) -1 \right\rceil. \end{align} $$

In light of Lemma 5, the proof is complete.

We note in passing that Lemma 6 implies that $ r_z $ becomes relevant at $ p = \Theta ( \log (n)^{3/2}/ n^{1/2}) $ , as $ r_z=0$ for p larger than this regime. Furthermore, for all p that we contemplate in this section (i.e., for $p> n^{-2/3 + \epsilon }$ ) we have

(16) $$ \begin{align} r_z = o( k_x^{1/2} ). \end{align} $$

Finally, we turn to estimating $ k_z $ . Note that we may assume $ p = O( \log (n) n^{-1/2})$ as we have $ k_z= k_x $ for larger p by Lemma 5. Note further that Equation (15) implies that $ r_M(k) = \Omega ( \log (n))$ for $ k_z \le k \le k_x$ for p in this regime. Recall that, appealing to Equation (6), we have

$$\begin{align*}E(n,k_x,0) = \mathbb{E}[ X_{k_x}] e^{- (1+o(1)) \frac{p k_x^3}{ e^2 n} }. \end{align*}$$

Next, observe that, since $F(n,k+1,r) = F(n,k,r+1)$ , we can easily adapt the argument that proves Equation (10) to establish

$$\begin{align*}\frac{F(n,k+1,r)^{n-k-r-1}}{F(n,k,r)^{n-k-r}} = 1+o(1) \end{align*}$$

for $ k_z \le k \le k_x$ and $r = o(1/p)$ . So we can calculate the ratio over a change in k in the same way that we established Equation (12). Thus, for $ k_z \le k \le k_x$ and $r = o(1/p)$ we have

(17) $$ \begin{align} \frac{E(n,k+1,r)}{E(n,k,r)} = \frac{k}{e^2 n}(1+o(1)). \end{align} $$

Using Equations (6), (17), (12) and (15), we write

$$ \begin{align*} \begin{split} E(n,k,r_M(k)) & = E(n,k_x,0) \prod_{\ell=k}^{k_x-1} \frac{E (n,\ell,0)}{ E(n, \ell+1,0) } \prod_{r=0}^{r_M(k)-1} \frac{E(n,k,r+1)}{ E(n,k,r)} \\ & = \mathbb{E}[ X_{k_x}] e^{- (1+o(1)) \frac{p k_x^3}{ e^2 n} } \left( \frac{e^2n}{k} (1+o(1)) \right)^{k_x-k} \left((1+o(1))\frac{k^3p}{2e^2n} \right)^{r_M(k)} \frac{1}{ r_M(k)!} \\ & = \mathbb{E}[ X_{k_x}] e^{- (1+o(1)) \frac{p k_x^3}{ 2 e^2 n} } \left( \frac{e^2n}{k} (1+o(1)) \right)^{k_x-k}. \end{split} \end{align*} $$

This estimate suffices to establish the following.

Lemma 7.

$$ \begin{align*} n^{-2/3 + \epsilon} < p \ll \log(n) n^{-1/2} & \hskip5mm \Rightarrow & k_x - k_z = (1+o(1)) \frac{ k_x^2 }{ e^2 n } = (1+o(1)) \frac{ 4 \log(np)^2 }{ e^2p^2n} \\ p = C \log(n) n^{-1/2} & \hskip5mm \Rightarrow & k_x - k_z = \left\lceil \frac{1}{e^2C^2} - \frac{2\log(\mathbb{E}[X_{k_x}])}{\log(n)} + 2 \epsilon \pm o(1) \right\rceil. \end{align*} $$

Proof. First, note that for p in these ranges we have

$$\begin{align*}\log( n/k) = (1+o(1)) \log(np) = (1+o(1))\frac{pk}{2}.\end{align*}$$

It follows that $ k_x - k_z$ is the smallest integer $ \kappa $ such that

$$\begin{align*}\log \mathbb{E}[ X_{k}] - ( 1+o(1)) \frac{ k_x^3p}{2e^2n} + \kappa (1+o(1))\frac{pk_x}{2}> \epsilon \log(n). \end{align*}$$

If $ p \ll \log (n) n^{-1/2}$ , then $ \log \mathbb {E}[X_k] = O( \log (n)) = o( k_x^3p/n)$ and the first part of the Lemma follows. On the other hand, if $ p = C \log (n) n^{-1/2}$ , then $ k_x^3 p /n = (1+o(1))\log (n)/C^2$ and $ p k_x = \log (n) (1+o(1))$ and we recover the second part of the lemma.

3.2. First moment

Here, we show that, with high probability, no augmented independent set of order k appears for any $ k_{z}+2 \le k \le k_x+1$ . This is sufficient because we can simply consider the expected number of independent sets of size $k_x + 2$ to rule out the appearance of any larger independent set. We emphasize that we are restricting our attention to k that satisfy Equation (3). We begin with a simple observation.

Lemma 8. For all k that satisfy Equation (3), we have

$$ \begin{align*}\displaystyle\sum_{r=0}^{k} E(n,k,r) \leq 4 \sum_{r=0}^{2r_M(k)} E(n,k,r).\end{align*} $$

Proof. We begin by noting that Equation (12) can be adapted to show

$$\begin{align*}\frac{ E(n,k,r+1) }{ E(n,k,r)} \le (1+o(1))\frac{k^3p}{2e^2n(r+1)}. \end{align*}$$

Now, we consider cases depending on the value of $ r_M = r_M(k)$ . First, note that if $ r_M = 0$ then we have $ k^3p/ (2e^2n) < 1 +o(1) $ and

$$\begin{align*}\sum_{r=0}^k E(n,k,r) \le (1+o(1)) E(n,k,0) \sum_{r=0}^\infty \frac{1}{r!} \le 4 E(n,k,0). \end{align*}$$

Now, suppose $ r_M \ge 1$ . Recalling Equation (15), note that if $r> 2r_M$ , then we have

(18) $$ \begin{align} E(n,k,r+1)/E(n,k,r) \le 2/3. \end{align} $$

So we have

$$ \begin{align*} \displaystyle\sum_{r=0}^{k} E(n,k,r) &= \sum_{r=0}^{2r_M} E(n,k,r) + \sum_{r =2r_M+1}^k E(n,k,r) \\ &\le \left(\sum_{r=0}^{2r_M} E(n,k,r)\right) + 3 E(n,k,2r_M+1) \le \left(\sum_{r=0}^{2r_M} E(n,k,r)\right) + 3 E(n,k,2r_M).\quad\qquad\\[-52pt] \end{align*} $$

Note that it follows from the proof of Lemma 8 that we have

$$\begin{align*}E(n, k+1,k+1) \leq \frac{2}{3} E(n,k+1,k). \end{align*}$$

Applying this observation together with Lemma 8 and Equation (17), we have

$$ \begin{align*} \sum_{r=0}^{k+1}E(n,k+1,r) &< \frac{5}{3} \sum_{r=0}^k E(n,k+1,r) \\&< \frac{20}{3} \sum_{r=0}^{2 r_M(k+1)} E(n,k+1,r)\\&< \frac{k}{n} \sum_{r=0}^{2 r_M(k+1)} E(n,k,r)\\&< \frac{k}{n} \sum_{r=0}^{k}E(n,k,r). \end{align*} $$

Therefore, by Lemma 8 and Equation (14), we have

$$ \begin{align*} \displaystyle\sum_{k=k_z+2}^{k_x+1} \left(\sum_{r=0}^{k}E(n,k,r)\right) &< \frac{2k_z}{n} \sum_{r=0}^{k_z+1} E(n,k_z+1,r) \\[4pt]&\leq \frac{8k_z}{n} \sum_{r=0}^{2r_M(k_z+1)} E(n,k_z+1,r) \\[4pt]&\leq \left(\frac{8k_z}{n}\right)\left(\frac{3k_z^3 p}{e^2 n}\right) E(n,k_z+1,r_M( k_z+1)) \\[4pt]&\leq \left(\frac{8k_z}{n}\right)\left(\frac{3k_z^3 p}{e^2 n}\right) n^\epsilon \\[4pt]&= O(\log(n)^4n^{-2\epsilon}). \end{align*} $$

Hence, whp no augmented independent set of order k for any $k \geq k_z+2$ appears and by Lemma 4, we have $\alpha (G) \leq k_z+1$ whp.

3.3. Second moment method applied to Z

Here, we prove by the second moment method that an augmented independent set of order $k_z$ appears with high probability; more specifically, we prove that such a set with $k_z+r_z$ vertices and $r_z = r_M( k_z)$ internal edges appears whp. For convenience we let

throughout the rest of this subsection. Recall that the random variable Z counts the number of augmented independent sets of order k with r edges, and by the definition of $r_M$ and Z we have

$$\begin{align*}\mathbb{E}(Z)> n^\epsilon,\end{align*}$$

and our goal is to show that $ Z>0$ whp.

We now break up Z into a sum of indicator random variables. Let $ \mathbb {S}$ be the collection of all pairs $ (S, m_S)$ , where S is a set of vertices and $ m_S$ is a matching consisting of r edges all of which are contained in S. Note that

For each pair $ (S, m_S) \in \mathbb {S}$ , let $ Z_{S,m_S}$ be the indicator random variable for the event that S is an augmented independent set with matching $m_S$ . We have

$$ \begin{align*} &\text{Var}(Z) \leq \sum_{(S,m_S) \in \mathbb{S}} \text{Var}(Z_{S,m_S})+ \sum_{S \neq T} \text{Cov}(Z_{S,m_S},Z_{T,m_T})\qquad\qquad\qquad\qquad\qquad\qquad \\[4pt] &\qquad\qquad\qquad\qquad\qquad\qquad\qquad\leq \mathbb{E}(Z) + \sum_{S \neq T} \mathbb{E}(Z_{S,m_S} Z_{T,m_T}) - \mathbb{E}(Z_{S,m_S})\mathbb{E}(Z_{T,m_T}), \end{align*} $$

where here and throughout this section such summations are over all $ (S,m_S), (T,m_T) \in \mathbb {S}$ that have the specified S and/or T. Note that we are making use of the full covariance; we will see below that this is necessary to handle sets $S,T$ such that $ |S\cap T|$ is roughly $ k^2/n$ . Next, let

(19) $$ \begin{align} h_i &:= \frac{1}{\mathbb{E}(Z)^2}\sum_{|S \cap T| = i} \mathbb{E}(Z_{S,m_S} Z_{T,m_T}) - \mathbb{E}(Z_{S,m_S}) \mathbb{E}(Z_{T,m_T}) \end{align} $$

(20) $$ \begin{align}&= \left(\frac{1 }{ |\mathbb{S}| }\right)^2\sum_{|S \cap T| = i} \left(\frac{\mathbb{E}(Z_{T,m_T} | Z_{S, m_S} = 1)}{\mathbb{E}(Z_{S,m_S})} - 1\right).\qquad\quad\ \end{align} $$

Applying the second moment method, we have $ \mathbb {P}( Z = 0) \le \text {Var}[Z]/ \mathbb {E}[Z]^2 $ . Thus, our goal is to show

We emphasize that $ \mathbb {E}(Z)$ can be significantly different from $ \mathbb {E}(X)$ here. We consider three ranges of i to establish the desired bounds on $ \sum h_i$ .

3.3.1. Case 1: $i < \frac {1}{ \log (n) p}$ .

Fix a pair $(S, m_S) \in \mathbb {S}$ , and define $ \mathcal {E} $ to be the event that $ Z_{S,m_S} =1 $ . In other words, $ \mathcal {E} $ is the event that S gives an augmented independent set with the prescribed matching $m_S$ . Let $ \mathcal {E}_T$ be the event $ Z_{T,m_T} =1 $ . Recalling Equation (20), we seek to bound

(21) $$ \begin{align} h_i = \frac{1}{|\mathbb{S}|} \sum_{T: |S \cap T|=i } \left( \frac{ \mathbb{P}( {\mathcal{E}}_T \mid \mathcal{E})}{ \mathbb{P} ( \mathcal{E}) } - 1 \right). \end{align} $$

Now, for $ \ell = 0, \dots , r $ let $ \mathbb {S}_\ell $ be the collection of pairs $(T, m_T) \in \mathbb {S}$ with the property that $ m_S$ and $m_T$ have exactly $\ell $ edges in common. Define

$$\begin{align*}B = \left\lceil \frac{1}{\sqrt{p\ln(n)}}\right\rceil,\end{align*}$$

and, for $ B < i < B^2$ , let $\mathbb {S}_{i,\ell }$ be the collection of pairs $(T, m_T) \in \mathbb {S}_\ell $ such that $|T \cap S| = i$ . Finally, let $\mathbb {S}^{\prime }_\ell $ be the collection of pairs $(T, m_T) \in \mathbb {S}_\ell $ with the additional property that $ |T \cap S| \le B$ . We make two claims.

Lemma 9. If $(T, m_T) \in \mathbb {S}_{i,\ell }$ and $ i < \frac {1}{\log (n)p }$ then

$$\begin{align*}\mathbb{P}( \mathcal{E}_T \mid \mathcal{E} ) \le e^{O(k^3p^2i/n) +o(1)} \frac{ \mathbb{P}( \mathcal{E}) }{ p^\ell (1-p)^{\binom{i}{2}}}. \end{align*}$$

Proof. In order to discuss the conditioning on the event $ \mathcal {E}$ , we need to define an additional parameter. Let w be the number of edges in $m_T$ that have exactly one vertex in $ S\cap T$ and let W be the vertices in $T \setminus S$ that are included in such edges. To be precise, we define

$$\begin{align*}W =\{ v \in T \setminus S: \exists \ u \in S \cap T \text{ such that } uv \in m_T \}. \end{align*}$$

For each vertex $ v \in W$ , let $ vv'$ be the edge of $m_T$ that contains v.

We condition as follows. First, we simply condition on all pairs within S appearing as edges and nonedges as required for the event $ \mathcal {E}$ . But we proceed more carefully for vertices v that are not in S. For such vertices, we must reveal information about the neighbors of v in S. We do this by observing whether or not the potential edges of the form $vu$ , where $u \in S$ actually are edges one at a time, starting with potential neighbors $ u \in S \setminus T$ , and stopping as soon as we observe at least two such neighbors. For vertices $v\in W$ , we add the restriction that the edge $ vv' \in m_T $ is the last edge observed in this process. So, for a vertex $v \in T \setminus S$ , this process either observes two edges between v and $ S \setminus T$ – and therefore observes no potential edges between v and $S \cap T$ – or the process observes at most one edge between v and $ S \setminus T$ and therefore observes edges between v and $S \cap T$ and reveals that at least one such edge appears. In this latter case, the conditional probability of $ \mathcal {E}_T$ is zero unless this vertex v is in W and the observed edge is $vv'$ .

For each set $ U \subset T \setminus S$ , we define $ \mathcal {F}_{U}$ to be the event that $\mathcal {E}$ holds and U is the set of vertices $ v \in T \setminus S$ with the property that our process reveals edges between $ v$ and $ S \cap T$ . Note that we have

$$\begin{align*}\mathcal{E} = \bigvee_{ U \subseteq T \setminus S } \mathcal{F}_{U} \end{align*}$$

and

$$\begin{align*}U \not\subseteq W \ \ \ \Rightarrow \ \ \ \mathbb{P} ( \mathcal{E}_T \wedge \mathcal{F}_{U}) = 0.\end{align*}$$

Next, let $I = S \cap T$ , and define $ \mathcal {I}$ to be the event that the process detailed above resorts to revealing potential edges between v and I for more than $ \log ^2(n) k $ vertices $ v \not \in S$ . Note that the probability that a particular vertex $ v \not \in S $ resorts to observing potential edges to I is

Thus, the expected number of vertices that observe edges to $ I$ is $ O(\log (n)^2 k^2/n) = o(k)$ . As these events are independent, the Chernoff bound (see Corollary 21.9 in [Reference Frieze and Karońksi10]) then implies

(22) $$ \begin{align} \mathbb{P}( \mathcal{I}) \le \exp \{ - \log^2(n) k \}. \end{align} $$

Now, we are ready to put everything together. Applying the law of total probability, we have

$$ \begin{align*} \begin{split} \mathbb{P}( \mathcal{E}_T \mid \mathcal{E} ) & = \sum_{ U \subseteq T \setminus S} \left(\mathbb{P}( \mathcal{E}_T \mid \mathcal{F}_{U} \wedge \mathcal{I}) \cdot \frac{ \mathbb{P}( \mathcal{F}_{U} \wedge \mathcal{I})}{ \mathbb{P}( \mathcal{E})} + \mathbb{P}( \mathcal{E}_T \mid \mathcal{F}_{U} \wedge \overline{\mathcal{I}}) \cdot \frac{ \mathbb{P}( \mathcal{F}_{U} \wedge \overline{ \mathcal{I}})}{ \mathbb{P}( \mathcal{E})} \right) \\ & \le \frac{ \mathbb{P}( \mathcal{I})}{ \mathbb{P}( \mathcal{E} ) } + \sum_{ U \subseteq T \setminus S } \mathbb{P}( \mathcal{E}_T \mid \mathcal{F}_{U} \wedge \overline{\mathcal{I}}) \cdot \frac{ \mathbb{P}( \mathcal{F}_{U} )}{ \mathbb{P}( \mathcal{E})} \\ & = \frac{ \mathbb{P}( \mathcal{I})}{ \mathbb{P}( \mathcal{E} ) } + \sum_{ U \subseteq W } \mathbb{P}( \mathcal{E}_T \mid \mathcal{F}_{U} \wedge \overline{\mathcal{I}}) \cdot \frac{ \mathbb{P}( \mathcal{F}_{U} )}{ \mathbb{P}( \mathcal{E})}. \end{split} \end{align*} $$

Now, observe that we have

and if $ U \subseteq W $ , then we have

where we use $ |U| \le \min \{i,r\} $ and $ k^4 p < n^{2 - 2\epsilon } $ . Thus,

$$ \begin{align*} &\sum_{ U \subseteq W } \mathbb{P}( \mathcal{E}_T \mid \mathcal{F}_{U} \wedge \overline{\mathcal{I}}) \cdot \frac{ \mathbb{P}( \mathcal{F}_{U} )}{ \mathbb{P}( \mathcal{E})} \\ &\quad\le \frac{ (1+o(1)) e^{O(k^3p^2i/n)} \mathbb{P}( \mathcal{E} )}{ p^\ell (1-p)^{i^2/2}} \sum_{u=0}^{w} \binom{w}{u} \left[ (1+o(1))\frac{ k^3}{ e^2 i n^2} \right] ^{u} = \frac{ (1+o(1)) e^{O(k^3p^2i/n)} \mathbb{P}( \mathcal{E} )}{ p^\ell (1-p)^{i^2/2}}, \end{align*} $$

where we use $ w \le i$ to bound the sum. We finally note that we have

$$\begin{align*}\frac{ \mathbb{P}( \mathcal{I})}{ \mathbb{P}( \mathcal{E})^2 } \le \exp \{ - \log^2(n) k + O(\log(n) k) \} = \exp \{ - \Omega\left( \log^2(n)k\right) \} = o(1). \end{align*}$$

And the proof of the lemma is complete.

Lemma 10. For $\ell =1, \dots , r$ , we have

$$\begin{align*}| \mathbb{S}_\ell| \le | \mathbb{S}| \left( \frac{ 3 r^2}{ n^2} \right)^\ell. \end{align*}$$

Furthermore, we have

Proof. We have

Thus,

The calculation for the second part of the lemma is similar.

Thus,

We are now ready to show that $\sum _{i=0}^{1/(p \log (n))} h_i = o(1)$ using Lemmas 9 and 10. We will split this into two subranges.

Subcase 1.1: $i \leq B$ .

In this case, by Lemma 9 we have $\mathbb {P}(\mathcal {E}_T \mid \mathcal {E}) \leq (1+o(1))\frac {\mathbb {P}(\mathcal {E})}{p^{\ell }}$ , hence

$$ \begin{align*} \begin{split} \sum_{i=0}^{B}h_i & = \frac{1}{|\mathbb{S}|} \sum_{\ell = 0}^r \sum_{ (T, m_T) \in \mathbb{S}_\ell' } \left( \frac{ \mathbb{P}( \mathcal{E}_T \mid \mathcal{E})}{ \mathbb{P} ( \mathcal{E}) } - 1 \right) \\ & = \frac{ o( | \mathbb{S}_0'| )}{ |\mathbb{S}|} + \sum_{\ell=1}^r \left( \frac{ 4 r^2}{ p n^2} \right)^\ell \\ & \le o(1) + \sum_{\ell=1}^r \left( \frac{1}{ n^{2/3}} \right)^\ell \\ & = o(1). \end{split} \end{align*} $$

Subcase 1.2: $B < i < B^2$ .

Here, we have

Since and , we have

$$ \begin{align*} \sum_{i=B}^{B^2} h_i = o(1). \end{align*} $$

3.3.2. Case 2: .

In this case, we can afford to be lenient with our bounds; in particular, we do not need to make use of the condition that vertices outside S have at least two neighbors in S. We begin by bounding $ \frac {1}{ \mathbb {E}( Z_{S,m_S})} \sum _{m_S, m_T} \mathbb {E}( Z_{S, m_S} Z_{T, m_T}) $ for a fixed pair of sets $S,T$ such that $ |S \cap T| =i$ . If there are $\ell $ edges in the intersection, a bound for the number of ways to choose $m_S$ and $m_T$ is (we have $2r-\ell $ edges total; each has less than choices.) Thus, we have

Recalling Equation (6), we have

It follows that

As $ r =o(k^{1/2}) = o(i) $ , we have

3.3.3. Case 3:

We calculate $h_i$ as defined in Equation (19). For simplicity, let . For some fixed i (hence j), there are ways to choose S and T. Now, fix sets S and T.

We now consider the number of ways to choose the matchings $m_S$ and $m_T$ that are compatible with the event $ \{ Z_{S,m_S} Z_{T, m_T} =1 \}$ . Let $R_I$ be the set of matching edges contained in $S \cap T$ , and let $r_I:=|R_I|$ (so $r_I$ is the same as $\ell $ from Cases 1 and 2). Secondly, let $R_O$ be the set of edges in $m_S$ that are not in $R_I$ , and let $R^{\prime }_O$ be the set of edges of $m_T$ that are not in $R_I$ , and define $r_O := |R_O| = |R^{\prime }_O|$ (‘I’ stands for ‘inner’, and ‘O’ stands for ‘outer’). Hence, $r = r_O + r_I$ . Finally, let $R_S$ be the matching edges in $R_O$ which are contained completely in $S \setminus T$ , and let $r_S = |R_S|$ . See Figure 1.

Figure 1 A depiction of the edges appearing in intersecting augmented independent sets.

We now count the number of ways to choose edges in $R_O$ if $r_O$ is fixed. To do this, we first note that if $r_S$ is also fixed, then the number of ways to choose the edges that comprise $R_S$ is

$$ \begin{align*}\frac{j!}{(j-2r_S)! r_S! 2^{r_S}}.\end{align*} $$

Then the number of ways to choose the rest of the rest of the edges in $R_O$ is

therefore, the total number of ways to choose $R_O$ is

(23)

By a symmetric argument, Equation (23) is an upper bound for the number of ways to choose the edges in $R^{\prime }_O$ as well.

Next, note that the number of ways to choose edges in $R_I$ is bounded by . Putting these together, the number of ways to choose $m_S$ and $m_T$ is at most

(24)

(assuming that $r^{r_O} = 1$ if $r = r_O = 0$ ).

Now, we estimate the probability of the event $ \{ Z_{S,{m_S}} Z_{T,m_T} = 1 \}$ . The probability the edges within S and within T are chosen accordingly is

Now, consider pairs of vertices from $S \backslash T$ to $T \backslash S$ . Recall that a vertex $ v \in S \setminus T$ has at least two neighbors in T. So, if $ v $ is already incident with an edge that goes into T among the edges already specified, then there is at least one edge from v to $T \backslash S$ (since each from $S \backslash T$ can send at most one edge to $S \cap T$ ). Such an edge appears with probability $1 - (1-p)^{j} \le jp$ . If a vertex $ v \in S \backslash T$ is not in an edge of $m_S$ that intersects $ S \cap T$ , then v is incident with at least two edges going to $T \backslash S$ . The probability of these two edges appearing is at most $j^2 p^2$ by the union bound. Since the number of vertices in the first category is at most $r_O$ , the total probability of all edges between $S \backslash T$ and $T \backslash S$ being chosen accordingly is at most $(jp)^{2j-r_O}$ . Hence, the probability that the edges within $ S \cup T$ appear in accordance with the event $ \{ Z_{S,{m_S}} Z_{T,m_T} = 1 \}$ is at most

Finally, the probability that all vertices outside $S \cup T$ have at least two neighbors in S is at most . Note that, since

we have

Multiplying this probability estimate by the estimate for the number of choices for $ m_S$ and $m_T$ given by Equation (24) and summing over $r_O$ , we see that for a fixed S and T we have

$$\begin{align*}\frac{ 1 }{\mathbb{P}( Z_{S, m_S} =1 ) } \sum_{ m_S, m_T} \mathbb{P}\left( Z_{S,m_S} Z_{T,m_T} = 1 \right) \end{align*}$$

is at most

where we use $ r = o( k^{1/2}) $ in the last step (and assume $r^{r_O} = 1$ if $r = r_O = 0$ ). Recalling that and that the number of choices for the pair of sets $S,T$ is , we have

As $ r = o( k^{1/2})$ and $ k = O ( \log (n) n^{2/3 - \epsilon })$ , we have , as desired.

4. Weak anti-concentration of $\alpha (G_{n,p})$ for $ n^{-1} < p < n^{-2/3}$

In this section, we present the argument of Sah and Sawhney that shows that two-point concentration of the independence number does not extend to $p = n^{\gamma }$ if $\gamma \leq -2/3$ ; that is, we prove Theorem 2. We note in passing that this proof is similar to an argument that appears in [Reference Alon and Krivelevich2] (see the last item in the ‘concluding remarks and open problems’ section of [Reference Alon and Krivelevich2]). Throughout this section, we assume $ \omega ( 1/n ) < p< ( \log (n)/n)^{2/3} $ . Recall Theorem 2:

Theorem 2 (Sah and Sawhney [Reference Sah and Sawhney18]).

Let $p = p(n)$ satisfy $ \omega ( 1/n ) < p< ( \log (n)/(12n))^{2/3} $ , and set

$$ \begin{align*}\ell = n^{-1}p^{-3/2} \log(np) / 2.\end{align*} $$

Then there exists $ q = q(n)$ such that $p \leq q \leq 2p$ such that $ \alpha ( G_{n,q})$ is not concentrated on $ \ell $ values.

For p in the specified range, we define

$$\begin{align*}p' = p + n^{-1} \sqrt{p}.\end{align*}$$

We first observe that this choice of $p'$ is close enough to p to ensure that there is no ‘separation’ of the intervals over which $ G_{n,p}$ and $ G_{n,p'}$ are respectively concentrated. To be precise, we establish the following:

Lemma 11. If $ \omega (n^{-1}) < p < o(1)$ , then there is a sequence $ k = k(n)$ such that

$$ \begin{align*} \mathbb{P}[\alpha(G_{n,p}) \leq k]> \frac{1}{20} \ \ \ \text{ and } \ \ \ \mathbb{P}[ \alpha(G_{n,p'}) \geq k] > \frac{1}{20} \end{align*} $$

for n sufficiently large.

Proof. First, observe that, since the distributions of $e(G_{n,p})$ and $e\left (G_{n,p'}\right )$ are approximately Gaussian with equal variances and with means that are $1/\sqrt {2}$ standard deviations apart, there is some value $ m$ such that

$$\begin{align*}\mathbb{P}[ e( G_{n,p} )> m ] > \frac{1}{10} \ \ \ \text{ and } \ \ \ \mathbb{P}[ e( G_{n,p'} ) < m ]> \frac{1}{10}. \end{align*}$$

Now let k be the median value of $ \alpha ( G_{n,m} )$ . As the addition of edges does not increase the independence number, the lemma follows.

Now, we prove Theorem 2. Assume for the sake of contradiction that if n is sufficiently large and $ p \le q \le 2p$ , then there is an interval I such that $ |I| \le \ell $ and

$$\begin{align*}\mathbb{P}[ \alpha( G_{n,q}) \in I ]> \frac{49}{50}. \end{align*}$$

Consider the sequence $p_0, p_1, p_2, \dots $ , where $p_0 = p$ and $p_{i+1} = p_i'$ for $i \geq 0$ , and let z be the lowest integer such that $p_z \geq 2p$ ; it is easy to see that $z \leq n\sqrt {p}$ . Let $ I_0 = [a_0,b_0], I_1 = [a_1, b_1], \dots , I_{z-1} = [a_{z-1},b_{z-1}]$ be intervals of $\ell $ values such that $ \alpha ( G_{n,p_i} )$ lies in $ I_i$ with probability at least $ 49/50$ for each i. By Lemma 11, we have $a_{i} \leq b_{i+1}$ for all $i < z$ . This implies $ b_{i+1} \ge b_i - \ell $ . Iterating this observation gives

$$\begin{align*}b_z \ge b_0 - z \ell \geq b_0 - \frac{ \log{ np}}{ 2p}. \end{align*}$$

On the other hand, by Equation (1), we have

$$\begin{align*}b_0 - b_z = (1+o(1)) \frac{2 \log(np) }{p} - (1+o(1)) \frac{ 2 \log(2np)}{ 2p} = (1+o(1)) \frac{ \log(np)}{ p}. \end{align*}$$

This is a contradiction.

5. Anti-concentration of $\alpha (G_{n,p})$ for $p = c/n$

In this section, we establish anti-concentration of the independence number of $ G_{n,p}$ for $ p = c/n$ , where c is a constant. We note in passing that for very small $p(n)$ it is easy to show that $\alpha (G)$ is not narrowly concentrated. Indeed, if $p = n^{\gamma }$ for some $\gamma \in (-2,-3/2)$ , then whp no two edges intersect, and hence the independence number is determined by the number of edges that appear. Since the distribution of the number of edges is roughly Poisson with mean $\mu \approx n^{2+\gamma }/2$ , then we will not have concentration of $\alpha (G)$ on fewer than $\Theta (n^{1+\gamma /2})$ values.

We now state our anti-concentration result.

Theorem 12 implies that $ \alpha ( G_{n, c/n})$ is not concentrated on fewer than $ \sqrt {n}/(2L)$ values. Note that this is the best possible anti-concentration result in this regime (up to the constant) as standard martingale methods (such as Hoeffding–Azuma or Talagrand’s inequality) show that, for any function ${k(n) = \omega (\sqrt {n})}$ , the independence number is concentrated on some interval of length $k(n)$ (this assertion holds for any value of p).

For a fixed graph G, let $T (G)$ be the subgraph of G given by all connected components that are trees on $4$ vertices. Let $t(G)$ be the number of components in $T(G)$ . Finally, let $\bar {T}(G)$ be the remainder of the graph G (so $ G = T(G) + \overline {T}(G)$ ). Let $p =c/n$ and let $ G = G_{n,p}$ . We set $ \mu = \mu (c)=\frac {2 c^3 e^{-4c}}{ 3}$ . Note that we have

$$\begin{align*}E[t(G)] = \binom{n}{4} \cdot 16 \cdot p^3 (1-p)^{3+4(n-4)} =(1+o(1)) \frac{16 c^3 e^{-4c}n}{ 24} = (1+o(1)) \mu n. \end{align*}$$

We note that $t(G)$ is a well concentrated random variable.

Lemma 13.

$$\begin{align*}\mathbb{P}( t(G) < \mu n/2) = O \left( \frac{1}{n} \right). \end{align*}$$

Proof. Here, we use standard second moment method once again. For vertex set S with $|S| = 4$ , let $W_S$ be the random variable that indicates if the graph induced on S is an isolated tree. Set $ W =t(G)$ , and note that $ t(G)$ is the sum of all $\binom {n}{4}$ indicators. By Chebychev, we have

$$ \begin{align*} \mathbb{P}\left(W < \frac{\mu n}{2}\right) \leq \mathbb{P}\left(|W - \mathbb{E}(W)| \geq \frac{\mathbb{E}(W)}{3}\right) \leq \frac{9\text{Var}(W)}{\mathbb{E}(W)^2}. \end{align*} $$

Hence

$$ \begin{align*} \mathbb{P}\left(W < \frac{ \mu n}{2}\right) \leq \frac{9\text{Var}(W)}{\mathbb{E}(W)^2} < 9\left(\frac{1}{\mathbb{E}(W)} + \max_{S,T}\left\{\frac{\mathbb{E}(W_S W_{T}) - \mathbb{E}(W_S)\mathbb{E}(W_{T})}{\mathbb{E}(W_S)^2}\right\}\right). \end{align*} $$

Note that the only case where we get a positive covariance is where $S \cap T = \emptyset $ , and in this case, we have

$$ \begin{align*} \frac{\mathbb{E}(W_S W_{T}) - \mathbb{E}(W_S)\mathbb{E}(W_{T})}{\mathbb{E}(W_S)^2} = (1-p)^{-16}-1 = \frac{16c+o(1)}{n}. \end{align*} $$

Therefore,

$$ \begin{align*} \mathbb{P}\left( t(G) < \frac{ \mu n}{2}\right) < \frac{9 +o(1)}{ \mu n} + \frac{144 c+o(1)}{n} =O\left(\frac{1}{n} \right).\\[-46pt] \end{align*} $$

With this observation in hand, we are ready to proceed to the proof of Theorem 12. Let k be a fixed integer. Let $ \mathcal {T}$ be the collection of all graphs H on $v_H \le n$ vertices such that $ 4 \mid n - v_H$ and H has no connected component that is a tree on four vertices. Let $ \mathcal {T}'$ be the collection of graphs in $H \in \mathcal {T}$ such that $ v_H \le n - 2 \mu n$ vertices. Applying the law of total probability, we have

$$ \begin{align*} \mathbb{P}( \alpha( G) = k) &= \sum_{H \in \mathcal{T}} \mathbb{P}( \alpha( G) = k \mid \overline{T}( G) = H) \mathbb{P}( \overline{T}(G) = H) \\ &\le \mathbb{P} (t(G) \le \mu n/2) + \sum_{H \in \mathcal{T}'} \mathbb{P}( \alpha( G) = k \mid \overline{T}( G) = H) \mathbb{P}( \overline{T}(G) = H). \end{align*} $$

So, it suffices to bound the conditional probability

$$\begin{align*}\mathbb{P}( \alpha( G) = k \mid \overline{T}( G) = H), \end{align*}$$

where H is a graph in $ \mathcal {T}'$ . Under this conditioning, the graph G is H together with a forest of $ (n - v_H)/4$ trees on four vertices. Furthermore, the number of trees in this forest which are stars on four vertices is a binomial random variable with $ (n - v_H)/4$ trials with probability of success $ 1/4$ on each trial. Let B be this random variable, and note that if G is drawn from the conditional probability space on the collection of graphs with $\overline {T}( G) = H $ , then we have

$$\begin{align*}\alpha(G) = \alpha(H) + (n - v_H)/2 + B. \end{align*}$$

Therefore, we have

$$\begin{align*}\mathbb{P}( \alpha( G) = k \mid \overline{T}( G) = H) = \mathbb{P}( B = k -\alpha(H) - (n - v_H)/2 \mid \overline{T}( G) = H). \end{align*}$$

But B is a binomial random variable with a linear number of trials, and therefore the probability that B is any particular value is at most $ L/\sqrt {n}$ for some constant L.

7. Appendix: two-point concentration of $ \alpha (G_{n,p})$ for $n^{-1/2+ \epsilon } < p< 1/ \log (n)^2$

Here, we show that the second moment method applied to the random variable $Y_k$ , which is the number of maximal independent sets of size k, suffices to establish two point concentration of $ \alpha (G_{n,p}) $ for $n^{-1/2+ \epsilon } < p< 1/ \log (n)^2$ .

Recall that $X_k$ is the random variable which counts the number of independent sets of size k in $ G_{n,p}$ , and we defined $k_x$ to be the largest integer such that

(25) $$ \begin{align} \mathbb{E}(X_k)> n^{2 \epsilon}. \end{align} $$

Set $ X = X_{k_x}$ and $ Y = Y_{x_k}$ .

Proof. Set $k = k_x$ . First, we show by the standard first moment method that no independent set of size $k+2$ appear whp:

$$ \begin{align*} \mathbb{P}( \alpha( G_{n,p}) \ge k +2) &\leq \binom{n}{k +2}(1-p)^{\binom{k+2}{2}} \\&= \binom{n}{k+1}(1-p)^{\binom{k+1}{2}}\Bigg(\frac{n-k-1}{k+2}(1-p)^{k+1} \Bigg) \\&= (1+o(1))\binom{n}{k+1}(1-p)^{\binom{k+1}{2}} \frac{n}{k} \left( \frac{k}{ne} \right)^2 \\& = o \left( \frac{k}{n^{1/2}} \right) \\&= o(1), \end{align*} $$

where we use the fact that $ k = o(n^{1/2}) $ when $p> n^{-1/2+\epsilon }$ in the last line.

Next, we show that an independent set of size k exists whp. Here, we use the second moment method, but the variance of X itself is too large. So we work with the random variable Y which counts the number of maximal independent sets of size k. The first moment calculation is straightforward, using Equation (5):

(26) $$ \begin{align} \mathbb{E}(Y) &= \binom{n}{k}(1-p)^{\binom{k}{2}} (1-(1-p)^{k})^{n-k} \nonumber \\&= \binom{n}{k}(1-p)^{\binom{k}{2}}e^{- O( k^2/n)} \end{align} $$

(27) $$ \begin{align}&= \mathbb{E}(X)(1-o(1))\qquad\qquad\ \ \,\\&= \omega(1).\qquad\qquad\ \ \, \nonumber \end{align} $$

By Chebyshev’s inequality, all that is left to show is that $\frac {Var(Y)}{\mathbb {E}(Y)^2} = o(1)$ . We write Y as the sum of indicator variables $Y_S$ , over all sets S such that $|K| = k$ , of the random variable $Y_S$ which is the indicator for the event that S is a maximal independent vertex set. We define indicator variables $X_S$ for not necessarily maximal independent sets analogously. Then we have

$$ \begin{align*} \text{Var}(Y) = \sum_S \text{Var}(Y_S)+ \sum_{S \neq T} \text{Cov}(Y_S,Y_{T}) \leq \mathbb{E}(Y)+ \sum_{S \neq T} \mathbb{E}(Y_S Y_{T}). \end{align*} $$

Next, define, for all relevant i:

$$ \begin{align*} &f_i = \frac{1}{\mathbb{E}(X)^2}\sum_{|S \cap T| = i} \mathbb{E}(X_S X_{T}) = \frac{\binom{k}{i}\binom{n-k}{k-i}}{\binom{n}{k}} (1-p)^{-\binom{i}{2}}, \\& g_i = \frac{1}{\mathbb{E}(Y)^2}\sum_{|S \cap T| = i} \mathbb{E}(Y_S Y_{T}), \text{ and} \\& \kappa_i = \frac{f_{i+1}}{f_i}. \end{align*} $$

Since $\mathbb {E}(Y) = \omega (1)$ , our goal is to show that

$$ \begin{align*} \sum_{i=0}^{k-1} g_i = o(1). \end{align*} $$

We consider three ranges of i: $i < \frac {\epsilon }{2} k_x$ , $\frac {\epsilon }{2} k_x \leq i \leq (1-\epsilon )k_x$ , and $i> (1-\epsilon ) k_x$ (where $\epsilon $ is the constant in the statement of the Theorem). For each range, we show that the sum of variables $g_i$ within that range is $o(1)$ . For the first two ranges, we will actually work with variables $f_i$ instead, noting that, since $Y_S \leq X_S$ and by similar reasoning to Equation (27), $g_i \leq f_i(1+o(1))$ .

First, consider $i < \frac {\epsilon }{2} k$ . Using the explicit formula for $f_i$ we have

$$ \begin{align*} \kappa_i = \frac{(k-i)^2(1-p)^{-i}}{(i+1)(n-2k+i+1)} < \frac{ k^2 (1-p)^{-\epsilon k/2}}{n} = (1+o(1)) \frac{ k^2}{n} \left( \frac{ ne}{k} \right)^{\epsilon} = O \left( \frac{ \log(n)^2}{ n^{\epsilon}} \right) = o(1). \end{align*} $$

As $ f_2 = O( k^4/n^2) = o(1)$ then $\sum _{i=2}^{\epsilon k/2} f_i$ is a geometric sum with leading term and ratio in $o(1)$ .

Next, consider the range $\frac {\epsilon }{2} k \leq i \leq (1-\epsilon ) k$ . We have

(28) $$ \begin{align} f_i & \le 2^{k} \left( \frac{k}{n} \right)^i (1-p)^{-\binom{i}{2}} \nonumber\\ & \le \left[ 2^{k/i} \frac{k}{n} \left( (1-p)^{-k/2} \right)^{i/k} \right]^i \nonumber\\ & \le \left[ (1+o(1)) 2^{ 2/ \epsilon } \frac{k}{n} \left( \frac{ne}{k} \right)^{1-\epsilon} \right]^i \\ & \le \left[ (1+o(1)) 2^{ 2/ \epsilon } e^{1-\epsilon} \left( \frac{k}{n} \right)^\epsilon \right]^i. \nonumber \end{align} $$

Hence, the sum of variables $f_i$ over $\frac {\epsilon }{2} k_x \leq i \leq (1-\epsilon )k_x$ is $o(1)$ .

The range where $i> (1-\epsilon ) k$ is where we use the variables $g_i$ instead of $f_i$ . Given two vertex sets S and T with intersection size i, and given that they are both independent, consider the probability that both are maximal independent sets. This probability is bounded above by the probability that each vertex in $S \backslash T$ is adjacent to at least one vertex in $T \backslash S$ , which is equal to $(1-(1-p)^{k-i})^{k-i} \leq ((k-i)p)^{k-i}$ . Therefore, we have

$$ \begin{align*} g_i &\leq (1+o(1)) \frac{\binom{k}{i}\binom{n-k}{k-i}}{\binom{n}{k}} (1-p)^{-\binom{i}{2}} ((k-i)p)^{k-i} \\&= \frac{1+o(1)}{\mathbb{E}[X]} \Bigg(\binom{k}{k-i} (k-i)^{k-i}\Bigg) \Bigg(\binom{n-k}{k-i}(1-p)^{\binom{k}{2}-\binom{i}{2}}p^{k-i} \Bigg) \\&< \frac{1+o(1)}{\mathbb{E}[X]} (ke)^{k-i} n^{k-i} (1-p)^{(1-\epsilon)k(k-i)}p^{k-i} \\&< \frac{1}{\mathbb{E}[X]} \Bigg((1+o(1)) e k n p \left( \frac{ k}{ ne} \right)^{2(1 -\epsilon)} \Bigg)^{k-i} \\&< \frac{1}{\mathbb{E}[X]} \Bigg( 22 \log(n)^3 n^{-\epsilon + 2\epsilon^2} \Bigg)^{k-i}, \end{align*} $$

so $\sum _{i>(1-\epsilon ) k} g_i = o(1)$ , as desired.