I am given a graph $G(n,n^{-\varepsilon})$, so a random graph with each edge drawn independenly with the probability $n^{-\varepsilon}$ and I want to somehow bound the maximum degree $d_\max$, such that $\Pr(d_\max = x)$ goes to $1$ as $n$ goes to infinity.
My idea: Let $X$ be the degree of a node $v$. Then $E[X] = np=n^{1-\varepsilon}$. Then I would bound the probability with chernoff, that the degree is bigger than something:
$$\Pr[X> (1+d)] \leq \exp\left(-\frac{n^{1-\varepsilon} d^2}{3}\right)$$
If I choose $d=\sqrt{ n^{o(-1+\varepsilon)}}$, the right term should go to $0$, so I guess the choice for $d=O(\sqrt{n^{-1+\varepsilon}})$. Is my reasoning correct or can one find a better bound?