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I am trying to prove that, when given a bipartite graph $G=(X \cup Y, E)$ with $|X|=|Y|=n$ and edge probability $\frac{1}{2}$, as $n\rightarrow \infty$ the probability of the graph having a perfect matching approaches 1.

After looking at this question, my inclination is to use Hall's theorem, although I can't quite figure out how to prove this.

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My answer to the linked question also gives you the proof here, but in the $p = \frac12$ case we can simplify the calculations required.

Our goal is to show that the expected number of sets $S \subseteq X$ or $S \subseteq Y$ with $|N(S)| < |S| \le \frac n2$ goes to $0$. (Here's why it's enough to consider $|S|\le \frac n2$. If $|S| > \frac n2$, say $S \subseteq X$, let $T$ be a subset of $Y \setminus N(S)$ with $n-|S|+1$ elements. Then $N(T) \subseteq X \setminus S$, so $|N(T)| \le n - |S| < |T|$, and we can use $T$ in place of $S$.)

We can bound this expected number by $$ 2\sum_{k=1}^{n/2} \binom nk \binom n{k-1} 2^{-k(n-k+1)} $$ which comes from the following:

  • We $\sum_{k=1}^{n/2}$ to look at the different possible values of $k = |S|$.
  • There are $\binom nk$ ways to pick $S$ and $\binom{n}{k-1}$ ways to pick some set of $k-1$ vertices containing $N(S)$.
  • There is a $2^{-k(n-k+1)}$ probability that there are no edges from $S$ to the other $n-k+1$ vertices on the other side.
  • We multiply by $2$ because $S$ could be in $X$ or in $Y$.

To show that this goes to $0$, bound $\binom nk \le n^k$, $\binom n{k-1} \le n^k$, and $2^{-k(n-k+1)} \le 2^{-kn/2}$. Then $$ 2\sum_{k=1}^{n/2} \binom nk \binom n{k-1} 2^{-k(n-k+1)} \le 2 \sum_{k=1}^{n/2} n^{2k} 2^{-kn/2} \le 2 \sum_{k=1}^\infty (n^2 2^{-n/2})^k = 2 \cdot \frac{n^2 2^{-n/2}}{1 - n^2 2^{-n/2}}. $$ As $n \to \infty$, $n^2 2^{-n/2} \to 0$, and therefore this expected value goes to $0$ as well.

Misha Lavrov
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  • It's true that you only need to consider $|S|\le\frac n2$. However, a) you don't need this, since the summand is symmetric under $k\to n-k+1$, so it would just be doubled, and b) your argument for it isn't quite right – there need not be an $\frac n2$-element subset of $Y\setminus N(S)$. But you can take all of $Y\setminus N(S)$, with at least $n-|S|+1$ elements, and its neighbourhood is in $X\setminus S$, with $n-|S|$ elements (which is the same duality as $k\to n-k+1$ in the sum). – joriki Mar 25 '20 at 14:39
  • We can take an at-most-$\frac n2$-element subset. All of $Y \setminus N(S)$ requires us to first argue that we can assume $|N(S)| = |S|-1$, since otherwise the new set could also be big. But I'll think about your point (a), maybe that can simplify my argument. – Misha Lavrov Mar 25 '20 at 14:48
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    The problem still persists after your edit, since you use $\frac n2=|T|$ later on. But your argument is right – I think the right mix of both is to use an $(n-|S|+1)$-element subset? – joriki Mar 25 '20 at 14:50
  • Thanks, that works. (And we can just take it to be exactly size $n-|S|+1$, since $N(S)$ excludes at most $|S|-1$ elements.) I think it takes more work to get approach (a) to work, since the inside sum is not exactly symmetric, and we want to avoid the endpoint where $k=n$ since that turns into a term $k=0$ we don't really like. – Misha Lavrov Mar 25 '20 at 14:53
  • No, it is exactly symmetric. $k=n$ is transformed to $k=1$, not $k=0$. It really is exactly the same duality that you're using to restrict to $\le\frac n2$. – joriki Mar 25 '20 at 14:56