Factorial moments of isolated vertices at the connectivity window

For each ordered tuple $t$ of $r$ distinct vertices, let $A_t$ be the event that every vertex in $t$ is isolated. In each graph, the number of such tuples is $(X{)}_r$, so $(X{)}_r={\sum }_t{\mathbf{1}}_{A_t}$. By the finite expectation formula, $\mathbb{E}{\mathbf{1}}_{A_t}={\sum }_{\omega \in A_t}w(\omega )=\mathbb{P}(A_t)$, and the two finite sums can be interchanged. The edges required absent consist of $r(n-r)$ edges to the complement and $r(r-1)\mathrm{/}2$ internal pairs, for a total of $m_{n,r}=rn-r(r+1)\mathrm{/}2$. Independence gives $\mathbb{P}(A_t)=(1-p{)}^{m_{n,r}}$. There are $(n{)}_r$ tuples, proving the exact formula. When $m_{n,r}=0$ the absent-edge product is $1$, even at $p=1$; for $r=0$ there is one empty tuple and its event is the whole probability space.

Fix $0<u<1$. The rational integrands below are continuous by [F12], since $x\ge 1-u>0$ on $[1-u,1]$ and $1-t\ge 1-u>0$ on $[0,u]$; hence they are bounded and integrable by [F13]. The decreasing affine surjection $\varphi (t)=1-t$ from $[0,u]$ onto $[1-u,1]$ has derivative $-1$ at every point, including the one-sided endpoints (its difference quotient is identically $-1$), so its derivative is integrable. Applying [F15], then [F5], [F13] and [F14], gives $\log (1-u)=-{\int }_{1-u}^{1}dx\mathrm{/}x=-{\int }_0^{u}dt\mathrm{/}(1-t)$ and $\log (1-u)+u=-{\int }_0^{u}t\mathrm{/}(1-t)dt$. To evaluate $I:={\int }_0^{u}tdt$, use the affine surjection $\psi (t)=u-t$ from $[0,u]$ onto itself, with the same derivative $-1$ and the continuous bounded integrand $f(x)=x$. By [F15] and [F14], $I={\int }_0^u(u-t)dt=u^2-I$, using the constant-integral rule in [F13]; hence $I=u^2\mathrm{/}2$. On $[0,u]$ we have $0\le t\mathrm{/}(1-t)\le t\mathrm{/}(1-u)$, so [F13] and [F14] give $\mathrm{\mid }\log (1-u)+u\mathrm{\mid }={\int }_0^{u}t\mathrm{/}(1-t)dt\le u^2\mathrm{/}[2(1-u)]$. At $u=0$ the identity and bound follow from the zero-length integral rule and $\log 1=0$.

Since $c_n\to c$, eventually $\mathrm{\mid }{c}_n\mathrm{\mid }\le \mathrm{\mid }c\mathrm{\mid }+1$. The logarithm is unbounded above by [F11], so $p=O(\log n\mathrm{/}n)\to 0$, and $n{p}^2=O((\log n{)}^2\mathrm{/}n)\to 0$ by [F6] with $a=1\mathrm{/}2$. For fixed $r$, therefore $m_{n,r}\log (1-p)=-m_{n,r}p+o(1)=-r(\log n+c_n)+o(1)$. Also $q_n:=(n{)}_r\mathrm{/}{n}^r={\prod }_{j=0}^{r-1}(1-j\mathrm{/}n)\to 1$ by the finite product limit rule, and $q_n>0$ eventually.

Fix $r\ge 1$. Eventually $n\ge r$ and $0\le p<1$, so the exact moment in step 1.1 is positive. The product and quotient laws for logarithms, applied repeatedly to its integer powers, give $\log (\mathbb{E}(X{)}_r)=\log q_n+r\log n+m_{n,r}\log (1-p)=-r{c}_n+o(1)\to -rc$, since $\log q_n\to \log 1=0$. Apply the inverse relation and continuity of the exponential to conclude $\mathbb{E}(X{)}_r\to \exp (-rc)=e^{-cr}$. For $r=0$ the moment is identically $1=e^0$.