Past exam of the mathematics course of the University of Cambridge / 2022 / ii / Paper 4 / 17F / Solution

In the Erdős-Rényi model $G(n,p)$, the vertex set is $[n]$ and every one of the $\left(\genfrac{}{}{0ex}{}{n}{2}\right)$ possible edges is present independently with probability $p$.

Let $X$ count isolated vertices. If $p\ge (1+\epsilon )\log n/n$, the union bound givesThus $\mathbb{P}(E_n)\to 1$.

Chebyshev inequality states thatNow suppose $p\le (1-\epsilon )\log n/n$. Monotonicity lets us use the largest allowed $p$. Withthe supplied lower exponential estimate showsFor distinct vertices $i,j$,HenceandChebyshev with $t=\mathbb{E}X$ gives $\mathbb{P}(X=0)\to 0$, so $\mathbb{P}(E_n)\to 0$.

For connectivity, the lower-threshold result is immediate because a connected graph has no isolated vertex:For the upper threshold, if a graph is disconnected it has a component with vertex set $S$ of some size $1\le k\le n/2$. In particular, every one of the $k(n-k)$ edges from $S$ to its complement is absent. ThereforeChooseFor $1\le k\le \eta n$, the standard bound $\left(\genfrac{}{}{0ex}{}{n}{k}\right)\le (en/k{)}^k$ givesThe sum of these terms tends to zero. For $\eta n\le k\le n/2$, use $\left(\genfrac{}{}{0ex}{}{n}{k}\right)\le 2^n$ and $k(n-k)\ge \eta n^2/2$:Thus $\mathbb{P}(F_n)\to 1$ above the threshold. Together,This is the connectivity threshold in the Erdős-Rényi model.

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