Problem
Let $X_1, X_2, \cdots, X_n$ be iid random variables. The cdf and empirical cdf are $F(t)=P[X\leq t]$ and $\hat{F}_n(t)=\frac{1}{n} \sum_{i=1}^n 1(X_i\leq t)$. The Glivenko-Cantelli theorem says,
$$ \sup_{t\in\mathbb{R}}\vert \hat{F}_n(t)-F(t)\vert \rightarrow 0\ a.s. $$
Then prove that with probability at least $(1-\delta)$ $$ \sup_{t\in\mathbb{R}}\vert \hat{F}_n(t)-F(t)\vert\leq 2\sqrt{\frac{2\log(n+1)}{n}} +\sqrt{\frac{1}{2n}\log \frac{1}{\delta}} $$
What I Have Done
Denote $\Delta s(t) := \hat{F}_n(t)-F(t)$, then by union bound, we have $$ P[\sup_{t\in\mathbb{R}}\vert \Delta s(t)\vert - \mathbb{E}[\vert \Delta s(t)\vert]\geq \alpha]\leq P[\sup_{t\in\mathbb{R}}\Delta s(t)-\mathbb{E}[\Delta s(t)]\geq \alpha]+ P[\sup_{t\in\mathbb{R}}\mathbb{E}[\Delta s(t)]-\Delta s(t)\geq \alpha] $$
The two terms on LHS could be easily bounded using MacDiarmid inequality with $\exp(-2n\alpha^2)$, then we have $$ P[\sup_{t\in\mathbb{R}}\vert \Delta s(t)\vert - \mathbb{E}[\vert \Delta s(t)\vert]\geq \alpha] \leq 2\exp(-2n\alpha^2) $$
However, if we set RHS with $\delta$, then we will end up getting $\sqrt{\frac{1}{2n}\log \frac{2}{\delta}}$. I am not sure how to get rid of this and get the required form?
Could anyone help me, thank you in advance!
