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as a follow up from this question :

Suppose that we have a simpler problem, where the probability $p$ is fixed. Of course we could use the above result to proove that almost every graph in the model $\mathcal{G}_{n,n,p}$ contains a perfect matching, saying for instance that $$ p > \frac{\sqrt{n} + \log{n}}{n}$$ for big enough $n$.

However, couldn't we find a more direct result? I've been looking at the probability of findind a set $S$ (size $k$) giving a contradiction for Hall's mariage theorem. With $q=1-p$, we need each element of $S$ not connected to the $n-k-1$ elements not in $N(S)$, therefore : $$ Pr[ |N(S)|<|S| ] \leq q^{nk-k(k-1)}$$

And $$Pr[\nexists \text{ a perfect matching}] \leq \sum_{k=1}^n \binom{n}{k}q^{nk-k(k-1)}$$

But from there, trying to proove that this sum tends to 0, simple binomial approximation $\sim n^k/k!$ gives too large a bound. Would you recommand any other calculation? Thanks

Thomas Lesgourgues
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1 Answers1

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Many approximations to the binomial $\binom nk$ are not symmetric in $k$ and $n-k$, so they perform badly when $k$ is close to $n$. (For example, $\frac{n^k}{k!}$ is somewhere around $e^n$ when $k=n$, even though actually $\binom nn = 1$.)

Here, the sum is almost but not quite symmetric: we can write the power of $q$ as $q^{k(n+1-k)} < q^{k(n-k)}$, and the latter is symmetric. So we can bound the sum by $$ 2 \sum_{k=1}^{n/2} \binom nk q^{k(n-k)} $$ and avoid the problematic values of $k$, getting a sum that's easy to bound using your approximation. It's even enough to write $\binom nk \le n^k$ and get $\binom nk q^{k(n-k)} \le (n \cdot q^{n-k})^k \le (n \cdot q^{n/2})^k$.

Misha Lavrov
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