 Solution

Write $N={\sum }_i{I}_i$ as in part (ii). For $i\ne j$, the two vertices are both isolated exactly when their combined $2n-3$ incident edges are absent, soThereforeLet $A=\mathbb{E}N=n(1-p{)}^{n-1}$. The displayed formula gives

Now take $p=c\log n/n$ with $c<1$, and choose $\alpha >1$ such that $\alpha c<1$. For all sufficiently large $n$, the supplied inequality gives $1-p\ge e^{-\alpha p}$, whenceAlso $p\to 0$, so both terms in the variance ratio tend to zero. Part (iv) now applies and proves the lower side of the isolated-vertex threshold: