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As the title mentioned, I want to calculte \begin{equation} \sum_{k=0}^{m} \binom{n}{k}x^k(1-x)^{n-k}. \end{equation} Note that if $m=n$, the result is simply 1. However, when $m<n$, this seems to become a open problem of the tail of beta distribution. Therefore, I only expect to obtain an upper bound w.r.t. $x$. By the way, in my exact problem, $m\approx \frac n3$. Asymptotic is also welcomed, but strict upper bound is favourable.

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1 Answer 1

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Here is an approximation approach.

Instead of refering to a beta distribution, consider a binomial $Bin(n,p)$ random variable $X$. In this framework :

$$\sum_{k=0}^{m} \binom{n}{k}x^k(1-x)^{n-k}$$

can be interpreted as $P(X \le m)$.

Now work with the normal approximation $N(np,\sqrt{np(1-p)})$ of a $Bin(n,p)$.

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  • $\begingroup$ Thank you for your reply. This can generate a fine asymptotic. Is there any way to obtain a strict upper bound with normal approximation (can we analyze the remainer terms)? $\endgroup$
    – Jobs Adam
    13 hours ago
  • $\begingroup$ Maybe by using a Gram-Charlier/Edgeworth series but I am not sure... $\endgroup$
    – Jean Marie
    8 hours ago

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