Let $S_n=\dfrac{B_n – np}{\sqrt{n\cdot p\cdot (1-p)}}$ be a random variable which has the standardized binomial distribution. From Chebyshev's inequality I know that $$P(|S_n| \ge x) \le \frac{1}{x^2}$$ Is there also a good approximation of the form $P(|S_n| \ge x) \ge \ldots\,{}$?
[Math] Approximation of the binomial distribution
inequalityprobability distributionsprobability theoryreference-request
Best Answer
Start by expressing the binomial distribution as $$ P(B_n = k) = \binom{n}{k}p^k(1-p)^{n-k} $$ so that $$ P(|S_n| \leq x) = \sum_{i=np-x\sqrt{np(1-p)}}^{np-x\sqrt{np(1-p)}}\binom{n}{k}p^k(1-p)^{n-k} $$ and express $$ \binom{n}{k} = \frac{n!}{k!(n-k)!} $$ Then use the Stirling approximation with the first 2 non-trivial terms: $$ \sqrt{2\pi n} \left(\frac{n}{e}\right)^n e^{\frac{1}{12n}-\frac{1}{360n^3}} < m! < \sqrt{2\pi n} \left(\frac{n}{e}\right)^n e^{\frac{1}{12n}} $$ This yields really tight bounds on $P(|S_n| \leq x)$.