Say $X\sim\mathrm{Beta}(\alpha,\beta)$. Are there any "nice" closed form upper bounds for the tail probability $P(X\geq\epsilon)$, that are reasonably tight when $\beta$ is large? By "nice" I mean involving only elementary functions, and not for instance the incomplete beta function.

In my specific setting I have $\alpha=1/2$ and $\beta=d/2$ for some (typically large) integer $d$, and so one way of bounding the tail is to write $X=\frac{Y}{Y+Z}$ where $Y\sim\chi^2_1$ and $Z\sim\chi^2_d$, and using an upper bound on $Y$ and lower bound on $Z$ (e.g. from here: What are the sharpest known tail bounds for $\chi_k^2$ distributed variables?). This gives an answer, but it's somewhat messy (requiring the small $d$ case to be handled separately, etc). Is there a cleaner way?


  • $\begingroup$ Do you intend $\epsilon$ to be close to $1$? Would it have any definite relationship to the parameters, such as exceeding the mean, median, or mode (to justify calling this a "tail probability")? $\endgroup$ – whuber Nov 12 '14 at 19:18
  • $\begingroup$ In my specific case when $\beta$ is big and $\alpha$ is a constant, I'd want $\epsilon$ to be just slightly larger than the mean. A clearer way to state the question may be to say that for instance I want a bound like $X\leq f(\log(1/\delta))$ with probability at least $1-\delta$ for some monotonic function $f$ (e.g. $\log(1/\delta)/d$, or something of the sort). $\endgroup$ – martin Nov 12 '14 at 20:10

We write $I_{\epsilon}(A,B,\epsilon)$ the incomplete beta function. Thus $P(X>\epsilon) = 1-I_{\epsilon}(A,B,\epsilon)$

A simple bound is

$$\frac{2\sqrt{\epsilon}}{2\sqrt{\epsilon}+B\left(\frac{1}{2},\frac{d}{2}\right)}< I_\epsilon\left(\frac{1}{2},\frac{d}{2},\epsilon\right)<\frac{2 \sqrt{\epsilon}}{B\left(\frac{1}{2},\frac{d}{2}\right)}$$

A tighter lower bound (so upper bound for the tail) would be

$$\frac{2 \left(\sqrt{\epsilon} B\left(\frac{1}{2},\frac{d}{2}\right)+2 \epsilon\right)}{2 \sqrt{\epsilon} B\left(\frac{1}{2},\frac{d}{2}\right)+B\left(\frac{1}{2},\frac{d}{2}\right)^2+4 \epsilon}$$

Mathematica will spit them out for you with

1 - 1 / Normal[Series[1/(1 - BetaRegularized[\epsilon, 1/2, d/2]), {\epsilon, 0, n}]]
  • $\begingroup$ How is $x$ related to the $\epsilon$ of the question? $\endgroup$ – whuber Nov 12 '14 at 19:16
  • $\begingroup$ $x = \epsilon$ ... $\endgroup$ – Arthur B. Nov 12 '14 at 20:35
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    $\begingroup$ To make your answer readable, why don't you adopt a consistent notation and--wherever it may differ from what is in the question--explain the differences. That will help readers understand how your post answers the question. $\endgroup$ – whuber Nov 12 '14 at 21:20
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    $\begingroup$ Thanks Arthur. This is an asymptotic bound right, not exact? $\endgroup$ – martin Nov 13 '14 at 10:39
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    $\begingroup$ No, it's always true $\endgroup$ – Arthur B. Nov 13 '14 at 12:06

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