There are several versions of Chernoff bounds. I was wodering which versions are applied to computing the probabilities of a Binomial distribution in the following two examples, but couldn't. I have tried to find them out but not succeeded.
From Wikipedia
The cumulative distribution function can be expressed as: $ F(x;n,p) = \Pr(X \le x) = \sum_{i=0}^{\lfloor x \rfloor} {n\choose i}p^i(1-p)^{n-i} $ For $k ≤ np$, Chernoff's inequality can be used to derive the bound $ F(k;n,p) \leq \exp\left(-\frac{1}{2\,p} \frac{(np-k)^2}{n}\right). \! $
From Wikipedia
Let $X_1, \dots, X_n$ be independent Bernoulli random variables, each having probability $p > 1/2$. Then the probability of simultaneous occurrence of more than n/2 of the events $\{X_k = 1\}$ has an exact value $S$, where $ S=\sum\limits_{i = \lfloor \frac{n}{2} \rfloor + 1}^n \binom{n}{i}p^i (1 - p)^{n - i} . $ The Chernoff bound shows that $S$ has the following lower bound: $ S \ge 1 - \mathrm{e}^{- 2n \left( {p - \frac{1}{2}} \right)^2} . $
The second one, if I understand correctly, is equivalent to $ F(n/2;n,p) \leq \mathrm{e}^{- 2n \left( {p - \frac{1}{2}} \right)^2}. $
I don't think part 2 can be derived from part 1. Granted that part 1 is correct based on some unknown result, then since $n/2 < np$, $ F(n/2;n,p) \leq \mathrm{e}^{- \left( {np - \frac{n}{2}} \right)^2 / (2np)} = \mathrm{e}^{- n \left( {p - \frac{1}{2}} \right)^2 / (2p)} $ The RHS is $\leq \mathrm{e}^{- 2n \left( {p - \frac{1}{2}} \right)^2}$ if and only if $p \leq 1/4$ which is contrary to $p >1/2$.
Thanks!