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Let us have an Erdos-Renyi graph $\mathcal{G} = \mathcal{G}(n,p)$. For a subset of vertices $S$ we define the cut-size $c(S)$ as the number of edges $(u,v)$ such that $u \notin S$ and $v \in S$. Let us assume that $p$ is a constant probability and finite.

For the maximum cut it is known that, with high-probability, ${\rm max} \ c(S) \leq \frac{1}{4}n^{2}p + \mathcal{O}(n^{3/2})$ - a bound which becomes sufficiently tight as $n \rightarrow \infty$.

I am wondering if I can make a more broad statement about the cut sizes. Specifically let |S| equal the number of vertices in the cut, and assume that |S| is proprtional to $n$, i.e. $|S| = \lambda n$.

\begin{equation} \lim_{n \rightarrow \infty} \frac{c(S)}{n^{2}} = \lim_{n \rightarrow \infty}\frac{E[c(S)]}{n^{2}} = \lambda (1-\lambda)p, \end{equation}

where $E[c(S)] = |S|(n-|S|)p$ is the expected value of the cut.

More specifically I am asking if the cut cannot differ in any significant (quadratic) way from it's expected value, so you have some inequality like

\begin{equation} |c(S) - E[c(S)]|\leq \mathcal{O}(n^{\alpha}) \ \qquad \alpha < 2. \end{equation}

Can this be proven? Or does anyone know of any references which show something like this?

EducationalFerret
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  • The statement $$\lim_{n \rightarrow \infty} \frac{c(S)}{n^{2}} = \frac{E[c(S)]}{n^{2}}$$ is not true because it does not make sense in a number of ways. First of all, if we take the limit as $n \to \infty$, the result will not have an $n$ in it. Second, $c(S)$ is not a function of $n$, because $S$ is a subset of the vertices of $\mathcal G(n,p)$ for one particular $n$. Third, if we arrange for $\mathcal G(n,p)$ to use consistent vertices for different values of $n$, we'll have $c(S) \le |S|(n-|S|) = \mathcal O(n)$, so $\lim_{n \to \infty} \frac{c(S)}{n^2}$ will be $0$. – Misha Lavrov Dec 08 '21 at 20:33
  • I think that you mean to ask something other than what you've actually written, but could you clarify what that is? – Misha Lavrov Dec 08 '21 at 20:33
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    I get the feeling the correct question would be answered by Cor 2.3 in https://people.math.ethz.ch/~sudakovb/pseudo-random-survey.pdf – dbal Dec 08 '21 at 21:21
  • @MishaLavrov Yes I agree that the RHS won't have an $n$ in it. But this is not untrue of my equation because if |S| is independent of $n$ the expression clearly vanishes (as you said) and for $|S|$ proportional to $n$, i.e. $|S| = \lambda n$ then the RHS is $\lambda (1-\lambda)p$, independent of $n$. So I guess I mean to assume that |S| is proportional to $n$ (because any other case is trivially $0$ like you said). Will edit. – EducationalFerret Dec 09 '21 at 09:34
  • @dbal Thank you for the reference, that looks to be exactly what I am looking for! Although rather strange they state $p \leq 0.99$, does the proof not just extend to $p < 1$? – EducationalFerret Dec 09 '21 at 09:41
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    I think the distinction is that if you write p <1, then p may be a function tending to 1. They want to indicate that p is bounded away from 1 – dbal Dec 09 '21 at 14:47

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