What are the most elegant approximations by using Stirling's approximation for the sum of binomial terms as below?

$Sum(r) = \sum\limits_{k=r}^{n} \binom{n}{k} p^k (1-p)^{n-k}$

I tried to plug in the approximation of factorial $x!$ into the formula, but the result looks crazily clumsy. Is there any elegant way to do this?

Many thanks!

  • $\begingroup$ There are a lot of different approximations depending on the precise range you want. For your purposes how large is $r$? If $r$ is close to $n$ then this is easier to approximate. $\endgroup$ – JoshuaZ Sep 9 '18 at 21:15
  • $\begingroup$ Many thanks for your reply. $r$ would be half of $n$. A typical one would be $\frac {n+1}{2}$ when $n$ is odd. $\endgroup$ – H42 Sep 9 '18 at 21:41

From 'Decoding Generalized Hyperoctohedral Groups and Asymptotic Analysis of Correctible Error Patterns,' R. Bailey and T. Prellberg, Contribut. Discrete Math. vol. 7 # 1, p 1-14 (2012). The following asymptotic expression is derived:

Define $$ F_{m,r}(x)=\sum_{k=0}^r{\binom{m}{k}x^k} $$ $$ \beta=r/m \text{ with } 0<\beta<1$$ $$ \rho(\beta,x)=\sqrt{\log{(1+x)} - \beta \, \log{x} + \beta\log{\beta} + (1-\beta)\log{(1-\beta}) } $$ Then $$ F_{m,r}(x) \sim (1+x)^m \frac{1}{2} \text{erfc}\big( \sqrt{m}\, \rho(\beta, x) \big) $$ The paper gives the next term in the expansion as well. The sign associated with $\rho$ because of the square root is tricky. I did the math once and I believe it is equivalent to $\text{sign}(x-\beta/(1-\beta))$.

Your problem can be reduced to this one by using the full binomial theorem and subtracting the 'head' of the series. Of course, to use this formula $x=p/(1-p)$ and there will be an external factor of $(1-p)^n.$

Comment: 9/26/2018

When I wrote this I tested the approximation with $\beta$ > 0.6 and for several $0<x<1.$ In this region the formula appears to work well. For $\beta$ < 1/2 and $x$ in the upper range, this formula seems to be off by a factor that can get disturbingly high. For example, let m=400 and r=50. Then for x=1/10 $F_{400,50}(1/10)/'approx' = 1.0029$, acceptable to me for most situations. However for x=9/10 and the same m,n, $F_{400,50}(9/10)/'approx' = 2.685.$ I would not find this discrepancy satisfactory. There must be some implicit assumption in the paper's setup that precludes this formula from being universally applicable.


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