Extremal degrees at a fixed logarithmic mean

Fix $a>0$, let $k=\lfloor a\mu \rfloor$, and let $N=n-1$ or $n-2$. For sufficiently large $n$, $1\le k<N$. By [F2] and the logarithm laws in [F6], the logarithm of the mass at $k$ is ${\sum }_{j=0}^{k-1}\log (N-j)+k\log p-\log (k!)+(N-k)\log (1-p)$. Since $k=O(\log n)$, [F6] gives ${\sum }_{j=0}^{k-1}\log (N-j)=k\log N+O(k^2\mathrm{/}N)$ and $(N-k)\log (1-p)=-Np+O(kp+N{p}^2)$. By [F3], $\log (k!)=k\log k-k+O(\log k)$. Thus the log mass is $-\mu I(a)+o(\log n)$, because $Np=\mu +o(1)$ and $k\mathrm{/}\mu \to a$. Consequently this mass is $n^{-\gamma I(a)+o(1)}$. The same conclusion holds if $k$ is changed by a bounded integer amount.

For $X\sim \mathrm{Bin}(n-1,p)$, [F2] gives $\mathbb{P}(X=j-1)\mathrm{/}\mathbb{P}(X=j)=j(1-p)\mathrm{/}((n-j)p)$ for $1\le j\le n-1$. If $0<a<1$, this ratio is at most a fixed $\rho <1$ for all $j\le \lfloor a\mu \rfloor$ and all sufficiently large $n$. If $a>1$, the reciprocal ratio $\mathbb{P}(X=j+1)\mathrm{/}\mathbb{P}(X=j)$ is at most a fixed $\rho <1$ for all $j\ge \lfloor a\mu \rfloor$ with $j<n-1$. Summing the corresponding finite geometric bounds and using step 1.1 shows that each respective lower or upper tail is at most $n^{-\gamma I(a)+o(1)}$. If $a<a_{-}$ or $a>a_{+}$, [F4] makes $\gamma I(a)>1$; the union bound [F7] over the $n$ vertices then makes the probability of a degree in that respective tail tend to zero.

Fix $a>0$ with $\gamma I(a)<1$, and set $k=\lfloor a\mu \rfloor$. Let $Z$ count vertices of degree exactly $k$, and write $q=\mathbb{P}(d(v)=k)$. By step 1.1, $\mathbb{E}Z=nq=n^{1-\gamma I(a)+o(1)}\to \mathrm{\infty }$. For distinct $u,v$, let $b_j=\mathbb{P}(R_u=j)$ and put $r=b_{k-1}\mathrm{/}{b}_k$. The mass formula gives $r=k(1-p)\mathrm{/}((n-1-k)p)\to a$. By the independent shared-edge decomposition in [F1], $q=b_k((1-p)+pr)$ and $\mathbb{P}(d(u)=k,d(v)=k)=b_k^2((1-p)+p{r}^2)$. Hence the ratio of this joint probability to $q^2$ is $1+p(1-p)(r-1{)}^2\mathrm{/}((1-p)+pr{)}^2=1+O(p)=1+o(1)$. Expanding the variance using [F5] yields $\mathrm{Var}(Z)\mathrm{/}(\mathbb{E}Z{)}^2\le 1\mathrm{/}(nq)+o(1)\to 0$. Chebyshev's inequality therefore gives $\mathbb{P}(Z=0)\to 0$.

Given $\epsilon >0$, choose $b_{-},c_{-},b_{+},c_{+}$ with $0<b_{-}<a_{-}<c_{-}<1<c_{+}<a_{+}<b_{+}$, all within $\epsilon$ of their respective roots. By [F4], the outer choices have $\gamma I>1$ and the inner choices have $\gamma I<1$. Step 2.1 excludes degrees at or below $\lfloor b_{-}\mu \rfloor$ and at or above $\lfloor b_{+}\mu \rfloor$ with probability tending to one. Step 2.2 supplies vertices of degrees $\lfloor c_{-}\mu \rfloor$ and $\lfloor c_{+}\mu \rfloor$ with probability tending to one. Since $\mu \to \mathrm{\infty }$, the floor errors divided by $\mu$ vanish. By [F7], these four bounds hold simultaneously with probability tending to one and bracket $d_{\min }\mathrm{/}\mu$ and $d_{\max }\mathrm{/}\mu$ within $\epsilon$ of their respective roots with probability tending to one, proving both asserted convergences.