Let $p\in (0,1)$ be fixed and let $X$ be a binomial random variable with parameters n and p. Consider a related normal random variable $N$ with mean $np$ and variance $np(1-p)$. Is it true that for some $x=x(n)$ we have $P(X>np+x)=(1+o(1))P(N>np+x)$? That is, if true, I would like to know how large $x$ can be. Is it true that the natural borderline is $x=O(\sqrt{n})$?
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2$\begingroup$ If you follow the proof of the Bahadur-Rao theorem (theorem 3.7.4 in Dembo-Zeitouni's book on large deviations), you will note it works with $q=q_n$; Indeed, it will work as long as you are in the moderate deviations regime, i.e. $x=o(n)$. $\endgroup$– ofer zeitouniCommented Aug 31, 2013 at 8:29
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$\begingroup$ @Ofer: If my answer is correct, the last part of your comment is not correct. $\endgroup$– Brendan McKayCommented Sep 1, 2013 at 11:18
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2$\begingroup$ @Brendan: In my original comment, I just checked that the B-R proof works, but didn't look at what it gives in comparison to the Normal {\it with fixed parameters}. In particular, the Gaussian you get is not $e^{-cx^2/n}$ but rather $e^{-n \Lambda^*(x/n)}$ where $\Lambda^*$ is the rate function for the Bernoulli. This gives in the range $x>>\sqrt{n}$ a correction which is I think the same as yours (and a Gaussian, but with slightly modified mean and variance); this seems to be valid not just for Bernoulli but rather in greater generality. $\endgroup$– ofer zeitouniCommented Sep 1, 2013 at 12:10
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$\begingroup$ Thank you both very much, the references are very useful! $\endgroup$– TOMCommented Sep 2, 2013 at 8:12
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1 Answer
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There are quite a lot of approximations more accurate than the pure normal distribution, with a large literature difficult to sort out. You can see several in this paper (Adv. Appl. Prob., 21 (1989) 475-478). If I am expanding Theorem 1 correctly, for fixed $p$ that is away from $\frac 12$, the relative error is $o(1)$ iff $x=o(n^{2/3})$. For $p=\frac12$, the symmetry makes the leading error term vanish so the error is still $o(1)$ further out, maybe for $x=o(n^{4/5})$.
All of this is for $x>0$. For $x<0$ the accuracy is fine all the way to the end since it approaches 1.
answered Aug 31, 2013 at 10:31