Is the multiplicative Chernoff bound stronger than additive one? The multiplicative Chernoff Bound says for $X_i \in \{0,1\}$ that satisfies $\mathbb{E}[X_i] = p$,
$$
\mathbb P\left(\sum\limits_i^n{X_i} \geq np(1+\delta)\right) \leq e^{-\frac{1}{3} np\delta^2} \>.
$$ 
The additive version says that 
$$
\mathbb P\left(\sum\limits_{i}^nX_i \geq np+n\epsilon \right) \leq e^{-2n\epsilon^2} \>.
$$ 
I wonder if the multiplicative version could be stronger. Let $\epsilon = p \delta$, then the additive Chernoff bound is reduced to 
$$
\mathbb P\left(\sum\limits_{i}^{n}{X_i} \geq np(1+\delta)\right) \leq e^{-2np^2\delta^2} \>.
$$ This is a much weaker bound when $p \ll 1$. 
How can these two versions of Chernoff bounds have such a difference? I mean, which step in deriving these two bounds diverges, causing the fact I have illustrated?
 A: No version is stronger than the other. Look at the proofs of both statements and you'll notice (e.g. http://en.wikipedia.org/wiki/Chernoff_bound) that one case uses $1+x\lt e^x$ while the other does not. This explains the divergence you mention.
A: The addictive one is stronger than the multiplicative one, since the multiplicative one can be derived from the addictive one.You should see the derivation in this material www.cs.berkeley.edu/~sinclair/cs271/n13.pdf‎ . The multiplicative one emerge as a corollary of the addictive one.
A: Here is one proof of the additive Chernoff bound you mentioned. Let $t$ be any positive real number. We have
$$
\begin{align*}
\Pr\left[ \bar{X} > p + \epsilon \right] & = \Pr\left[ \exp\left( t \sum_{i = 1}^n (X_i - p) \right) > \exp( t n \epsilon ) \right] \\ 
& \leq \frac{ \prod_{i = 1}^n \mathrm{E}[ \exp(t(X_i - p)) ] }{\exp(tn\epsilon) }  \\
& \leq \frac{ \prod_{i=1}^n \exp( t^2/8 ) }{\exp(tn\epsilon) } \\
& = \exp( n(t^2/8 - t\epsilon) ),
\end{align*}
$$
where second inequality is due to Markov's inequality, the last but one inequality is due to Hoeffding's lemma by considering $(X_i - p)$ as a random variable. By setting $t = 4\epsilon$, your additive version of Chernoff bound is achieved. 
Since Hoeffding's lemma holds for any random variable with domain $[a, b]$ such that $b - a \leq 1$, it does not make most use of domain $[0, 1]$. Hence it is weaker. And this explains why it is much weaker when $p \ll 1$.
The multiplicative Chernoff bound you mentioned is derived by the multiplicative one in Chernoff Bound, Wikipedia which does not use Hoeffding's lemma in its proof. And the proof is subject to the domain $[0, 1]$. 
Considering the multiplicative and additive ones in Wikipedia.  I think the additive one is technically stronger than the multiplicative one since the multiplicative one uses one more relaxation like @ivan said.
