Why is the number of vertices of degree $d$ in $G(n,p)$ equal to $0$ with high probability if $np-\log(n)-d \log\log(n)$ tends to infinity?

Question

Let $Y_d$ be the random variable corresponding to the number of vertices of degree $d$ (not depending on $n$) in $G(n,p)$. The degree of a vertex is distributed as $\operatorname{Bin}(n-1,p)$. So the expected number of vertices of degree $d$ is given by: \begin{align} \mathbb{E}(Y_d) &= n \binom{n-1}{d} p^d (1-p)^{n-1-d} \\ &\sim \exp\bigl((d+1) \log(n) + d \log(p) + (n-1-d) \log(1-p) \bigr) \end{align} How do we prove that $\left((d+1) \log(n) + d \log(p) + (n-1-d) \log(1-p) \right)$ is tending to $-\infty$ if $np -\log(n)-d\log\log(n)$ tends to $\infty$? I tried some approximation of $p$ which did not work.

Answer 1

The assumption that $np−\log(n)−d\log\log(n)$ tends to $\infty$ with $n$ means that there exists a function $\omega(n)$ such that $\omega(n) \to \infty$ as $n \to \infty$ where $$p = \frac{\log(n)+d\log\log(n)+\omega(n)}{n}.$$ You can use this to show that $(d+1)\log(n)+d\log(p) \sim \log(n) + d\log\log(n)$. That and the fact that $\log(1 - p) \sim -p$ should give you what you need.