TITLE: Improvement of Chernoff bound in Binomial case
QUESTION [6 upvotes]: We know from Chernoff bound
$P\bigg(X \leq (\frac{1}{2}-\epsilon)N\bigg)\leq e^{-2\epsilon^2 N}$ where 
$X$ follows Binomial($N, \frac{1}{2}$). 
If I take $N=1000, \epsilon=0.01$, the upper bound is 0.82. 
However, the actual value is 0.27. Can we improve this Chernoff bound?

REPLY [7 votes]: An explicit high-accuracy bound on the Kolmogorov distance between a properly standardized binomial distribution $B(N,p)$ and the standard normal distribution is provided by the theorem on page 129 of Uspensky's book "Introduction to Mathematical Probability" (1937). In your case, of $p=1/2$, this theorem implies that for any integer $x$ and any natural $N\ge100$
\begin{equation}
 |P(X\le x)-\Phi(z)|<\delta:=\frac{0.52}N+\exp(-\tfrac34\,\sqrt N),
\end{equation}
where $\Phi$ is the standard normal cumulative distribution function and $z:=(x-Np+1/2)/\sqrt{N/4}$. In particular, for your $N=1000$, we have $\delta=0.000520000050\ldots<\frac1{1923}$. So, for $N=1000, \epsilon=0.01$, Uspensky's result implies that the true probability (which is $0.27398\ldots$) is in the interval $[0.27397\ldots-\frac1{1923},0.27397\ldots+\frac1{1923}]\subset[0.27345, 0.27450]$, which is much more informative than 
the Chernoff and entropy upper bounds on $0.27398\ldots$, which both are $\approx0.82$. 
As was commented by user RaphaelB4, the Chernoff and entropy upper bounds are bounds on probabilities of large deviations. Those probabilities are small, in contrast with your probability $0.27398\ldots$, which makes large deviation bounds unsuitable in your example. Also, there are large deviation bounds much better than Chernoff's. E.g., in a more general situation one has an upper bound on large deviation probabilities that is asymptotic to the corresponding Gaussian tail.