I am playing with the chernoff bounds and derived a bound from its additive version. The method I used seems a little suspicious and circular in logic, but I cant figure out what the mistake is.
Additive form of Chernoff bound
The following theorem is due to Wassily Hoeffding and hence is called the Chernoff-Hoeffding theorem.
Chernoff-Hoeffding theorem
Suppose $X_1, \ldots, X_n$ are i.i.d. random variables, taking values in $\{0,1\}$. Let $p=\mathrm{E}\left[X_i\right]$ and $\delta>0$. $$ \begin{aligned} & \operatorname{Pr}\left(\frac{1}{n} \sum X_i \geq p+\delta\right) \leq\left(\left(\frac{p}{p+\delta}\right)^{p+\delta}\left(\frac{1-p}{1-p-\delta}\right)^{1-p-\delta}\right)^n=e^{-D(p+\delta \| p) n} \\ & \operatorname{Pr}\left(\frac{1}{n} \sum X_i \leq p-\delta\right) \leq\left(\left(\frac{p}{p-\delta}\right)^{p-\delta}\left(\frac{1-p}{1-p+\delta}\right)^{1-p+\delta}\right)^n=e^{-D(p-\delta \| p) n} \end{aligned} $$ where $$ D(x \| y)=x \ln \frac{x}{y}+(1-x) \ln \left(\frac{1-x}{1-y}\right) $$ is the Kullback-Leibler divergence between Bernoulli distributed random variables with parameters $x$ and $y$ respectively.
What if we dont know $\mathrm{E}\left[X_i\right]$?
Let $$\mu=n p$$ $$X=\sum X_i$$ Then, \begin{equation} \operatorname{Pr}\left(X \geq \mu+n\delta\right) \leq e^{-D(\frac{\mu}{n}+\delta \| \frac{\mu}{n}) n}\label{ub} \\ \end{equation}
Suppose we fix the value of $\delta=\delta^{lb}$,then,
\begin{equation} \operatorname{Pr}\left(X \geq \mu+n\delta^{lb}\right) \leq f\left( \mu,\delta^{lb},n\right)=\epsilon^{lb}, \label{ubmod} \end{equation} where $f\left( \mu,\delta,n\right)=e^{-D(\frac{\mu}{n}+\delta \| \frac{\mu}{n}) n}$.
Then the above equation implies $\operatorname{Pr}\left(X-n\delta^{lb} \geq \mu\right) \leq \epsilon^{lb} \implies $
\begin{equation}\operatorname{Pr}\left(X-n\delta^{lb} < \mu\right) > 1-\epsilon^{lb}.\label{ubfunc} \end{equation}
If $f\left( \mu,\delta^{lb},n\right)=\epsilon^{lb}\leq f\left( \left(X-n \delta^{lb}\right),\delta^{lb},n\right)=\epsilon'^{lb}$ when $X-n\delta^{lb} < \mu$, then we have, \begin{eqnarray} \operatorname{Pr}\left(X \geq \mu+n\delta^{lb}\right) &\leq& f\left( \mu,\delta^{lb},n\right),\nonumber\\ &\leq& f\left( \left(X-n \delta^{lb}\right),\delta^{lb},n\right)\nonumber\\ &\leq&\epsilon'^{lb}. \end{eqnarray} which holds with probability $>1-\epsilon^{lb}\geq1-\epsilon'^{lb}$.
Is this correct?
If not, is there a way to convert this into a legit one? I need to avoid using the expectation value of the random variable to calculate the bound.
EDIT
One could also restatwe the proof the following way,
If $f\left( \mu,\delta^{lb},n\right)=\epsilon^{lb}\leq f\left( \left(X-n \delta^{lb}\right),\delta^{lb},n\right)=\epsilon'^{lb}$ when $X-n\delta^{lb} < \mu$, then we have, \begin{eqnarray} \operatorname{Pr}\left(X-n\delta^{lb} < \mu\right) &>& 1- f\left( \mu,\delta^{lb},n\right),\nonumber\\ &>& 1- f\left( \left(X-n \delta^{lb}\right),\delta^{lb},n\right)\nonumber\\ &>& 1 - \epsilon'^{lb}. \end{eqnarray} which holds with probability $>1-\epsilon^{lb}\geq1-\epsilon'^{lb}$.