I'm curious about how well the Chernoff bound approximates the value of the upper tail of a binomial distribution.
It is well known that, for $X\sim B(n,p),\ \delta >0$:
$$P(X \geq (1+\delta)np) \leq \exp\left(-\frac{\delta^2np}{2+\delta}\right)$$
What, if anything, can be said about how closely this upper bound approximates the probability on the left hand side? That is, how does the following function vary with $n,p$:
$$f(n,p)=\lvert{P(X \geq (1+\delta)np) - \exp\left(-\frac{\delta^2np}{2+\delta}\right)}\rvert$$
For concreteness, let's take $\delta = \frac{n+1}{2np} - 1$ with $p<0.5$, so that this is really a question about:
$$f(n,p)=\lvert{P(X \geq (n+1)/2) - \exp\left(-\frac{((1-2p)n+1)^2}{2((1+2p)n+1)}\right)}\rvert$$
Plugging a few test cases in, for small $n$, I found that these two quantities can differ by orders of magnitude. But a more complete picture eludes me, as well as the regime in terms of $n,p$ where the Chernoff bound is actually a good approximation to the Binomial tail. Any suggestions very welcome!
I've reposted this here as I suspect it will get a better response in math.se. I hope that is within the rules.
EDIT: From plotting, it seems that for small (odd) values of $n$ and restricting $0.05 <p < 0.4$, we can construct a modified approximation that bounds the below function $|g|<0.1$:
$$g(n,p)={{P(X \geq (n+1)/2) - \frac{1}{\sqrt n}\exp\left(-\frac{((1-2p)n+1)^2}{2((1+2p)n+1)}\right)}}$$
Curious now whether this is provable? Any light on this would be of great interest.