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For a standardized binomial distributed random variable $\tilde B_n$ we have

$$P(\tilde B_n\le x) = \Phi(x) + \frac {q-p}{6\sqrt{npq}} (1-x^2) \phi(x) + \frac{R_1\left(np+x\sqrt{npq}\right)}{\sqrt{npq}}\phi(x) + O\left(\frac 1n\right)$$

with $R_1(x)=\lfloor x \rfloor -x +\frac 12$ [1].

Question: What is the Edgeworth Expension for $P( a \le \tilde B_n \le b)$? Is it true, that it is

$$\begin{align} P( a \le \tilde B_n \le b) &= \Phi(b)-\Phi(a) + \frac {q-p}{6\sqrt{npq}} (1-b^2) \phi(b) - \frac {q-p}{6\sqrt{npq}} (1-a^2) \phi(a) \\ & + \frac{R_1\left(np+b\sqrt{npq}\right)}{\sqrt{npq}}\phi(b)-\frac{R_2\left(np+a\sqrt{npq}\right)}{\sqrt{npq}}\phi(a) + O\left(\frac 1n\right) \end{align}$$

with $R_2(x)=\lceil x \rceil -x -\frac 12$ ?

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Stephan Kulla
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  • Shouldn't $R_2 = \lfloor a \rfloor -a +\frac12$? Other that that, your calculation looks correct to me. Since $O(\frac1n)-O(\frac1n)=O(\frac1n)$. – Thomas Ahle May 23 '16 at 13:08
  • I guess I chose $R_2$ because $P(a \le \tilde B_n \le b)= P(\tilde B_n \le b)- P(\tilde B_n < a)$ and for $P(\tilde B_n < a)$ the Edgeworth Expension need to be adjusted... But it's too long ago since I asked the question. So I am not quite sure... – Stephan Kulla Jun 12 '16 at 19:25

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