I remember there is a result from the early 1980s, which states that the tail probability of a binomial distribution is always at least as large as the tail probability of the normal distribution (at least when $p \leq 1/4$ or something like this for the binomial distribution is assumed). However, I cannot rememeber the name of the theorem and the author, resp. Can someone help me?

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    $\begingroup$ The comparison is rather vice versa: the normal tail is heavier than the binomial, at least far enough from the mean, where the binomial tail is just 0. $\endgroup$ – Iosif Pinelis Feb 19 '18 at 12:05
  • $\begingroup$ Oh well, probably there is a requirement to stay away from the extreme tail. Still, there is a theorem which says that (within some range, probably) the tail of the binomial distribution is heavier than that of the normal distribution. I am sure it exists, I just cannot find it anymore. $\endgroup$ – Kurisuto Asutora Feb 19 '18 at 12:43
  • $\begingroup$ you should find the answer in this earlier MO question $\endgroup$ – Carlo Beenakker Feb 19 '18 at 18:32
  • $\begingroup$ No, that's not what I mean. As Iosif pointed out the desired result cannot hold for the whole tail, but only up to some reasonable threshold. However, the range for the Berry-Essen is too small as a large deviations inequality, for the purpose that I have in mind. $\endgroup$ – Kurisuto Asutora Feb 20 '18 at 6:57

You might be thinking of Eric V. Slud (1977), Distribution inequalities for the binomial law, https://projecteuclid.org/download/pdf_1/euclid.aop/1176995801, and you can find further related results proved or referenced in the thesis of Jona Schulz (2016), The optimal Berry–Esseen constant in the binomial case, http://ubt.opus.hbz-nrw.de/volltexte/2016/1007/pdf/Dissertation_Schulz.pdf

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    $\begingroup$ It is amazing to me that the condition $p\le1/4$ is enough for the binomial tail $P(B_{n,p}\ge k)$ to be heavier than the corresponding normal tail for all integers $k$ between $np$ and $n$. Of course, here one cannot replace $\ge k$ by $>k$. $\endgroup$ – Iosif Pinelis Feb 21 '18 at 3:33
  • $\begingroup$ Yes, thank you Lutz, this is exactly what I had in mind. Thanks! (It's not early 80s but late 70s, I realize now.) $\endgroup$ – Kurisuto Asutora Feb 21 '18 at 10:28

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