I am interested in finding sharp enough bound for tail probability $\Pr(X\gt t)$ given $X\sim \operatorname{Beta}(a,b)$ when $a$ is very close to 0 while $b$ is fixed value greater than 1. Numeric results show that when $b$ is fixed, as $a$ goes to 0, the tail probability goes to zero too. Is there any sharp enough bound involving $a$ to characterize this trend? Thank you very much!


The following analysis obtains bounds that hold for sufficiently small $\alpha$ and are expressed in terms of elementary functions.

The tail probability, written as a function of $\alpha\gt 0,$ is

$$p_{t,\beta}(\alpha) = {\Pr}_{\alpha,\beta}(X\gt t) = \frac{\Gamma(\alpha+\beta)}{\Gamma(\alpha)\Gamma(\beta)}\int_t^1 s^{\alpha-1}(1-s)^{\beta-1}\mathrm{d}s.$$

For an upper bound, observe that over the interval of integration $(t,1),$

$$s^{\alpha-1}(1-s)^{\beta-1} \gt t^{\alpha-1}(1-s)^{\beta-1}.$$

The relative error is no greater than $t^{\alpha-1},$ which will be satisfactory for $t\approx 1.$

Integrate that upper bound and use the fundamental relationship $z\Gamma(z)=\Gamma(z+1)$ to produce

$$p_{t,\beta}(\alpha) \le \frac{\Gamma(\alpha+\beta)}{\Gamma(\alpha)\Gamma(\beta)}\frac{t^{\alpha-1}(1-t)^\beta}{\beta} = \frac{\Gamma(\alpha+\beta)}{\Gamma(\alpha+1)\Gamma(\beta+1)}\,\alpha\, t^{\alpha-1}(1-t)^\beta.$$

This is a multiple of $\alpha.$ The multiplier, although depending on $\alpha,$ is differentiable at $\alpha=0$ where its value is

$$\frac{\Gamma(0+\beta)}{\Gamma(0+1)\Gamma(\beta+1)} t^{0-1}(1-t)^\beta = \frac{(1-t)^\beta}{t\beta}.$$

Thus, at least for sufficiently small $\alpha,$

$$p_{t,\beta}(\alpha) \lt \alpha \left(\frac{(1-t)^\beta}{t\beta}\right).\tag{1}$$

Taking $0\lt t\le 1,$ the Binomial expansion of

$$\eqalign{s^{\alpha-1}&=(1+(s-1))^{\alpha-1} \\ &= 1 + ({\alpha-1})(s-1) + \cdots + \frac{(\alpha-1)(\alpha-2)\cdots(\alpha-i)}{i!}(s-1)^i + \cdots\\ &= \sum_{i=0}^\infty \binom{\alpha}{i}(s-1)^i}$$

converges absolutely, permitting the integral of $s^{\alpha-1}(1-s)^{\beta-1}$ to be performed term by term as

$$\eqalign{ p_{t,\beta}(\alpha) &= \frac{\Gamma(\alpha+\beta)}{\Gamma(\alpha)\Gamma(\beta)}\sum_{i=0}^\infty \int_t^1 \binom{\alpha}{i}(s-1)^i(1-s)^{\beta-1}\mathrm{d}s\\ &= \frac{\Gamma(\alpha+\beta)}{\Gamma(\alpha)\Gamma(\beta)}\sum_{i=0}^\infty \binom{\alpha}{i}(-1)^i\frac{1}{i+\beta}(1-t)^{i+\beta}. }$$

The assumption $0\lt \alpha\lt 1$ implies $\alpha-1$ is negative, in which case the Binomial coefficients may be expressed as

$$\binom{\alpha}{i}(-1)^i = \frac{(1-\alpha)(2-\alpha)\cdots(i-\alpha)}{i!}=\frac{\Gamma(i+1-\alpha)}{\Gamma(1-\alpha)i!}.$$

Using the formula

$$\Gamma(\alpha)\Gamma(1-\alpha) = \frac{\pi}{\sin(\pi\alpha)},$$ re-express $p$ as

$$\eqalign{ p_{t,\beta}(\alpha) &= \frac{\sin(\pi\alpha)\Gamma(\alpha+\beta)}{\pi\Gamma(\beta)}(1-t)^\beta\left(\frac{\Gamma(1-\alpha)}{\beta}+\sum_{i=1}^\infty \frac{\Gamma(i+1-\alpha)}{i!}\frac{(1-t)^{i}}{i+\beta}\right). }$$

Since each term is positive, lower bounds can be obtained by truncating the sum. By stopping at the zeroth term (which is explicitly written out) we obtain

$$p_{t,\beta}(\alpha) \ge \frac{\sin(\pi\alpha)\Gamma(\alpha+\beta)}{\pi\Gamma(\beta)}(1-t)^\beta\frac{\Gamma(1-\alpha)}{\beta}\approx \alpha \left(\frac{(1-t)^\beta}{\beta}\right).\tag{2}$$

The relative error is on the order of $1-t,$ which will be excellent for $t\approx 1.$

Together, the bounds $(1)$ and $(2)$ give

$$\alpha \left(\frac{(1-t)^\beta}{\beta}\right) \lt p_{t,\beta}(\alpha) \lt \alpha \left(\frac{(1-t)^\beta}{t\beta}\right)$$

for $\alpha \approx 0.$

For $t\approx 1$ and $\alpha \approx 0$ these inequalities work well. Here, to illustrate, are plots (colored curves) of $p_{t,\beta}(\alpha)$ for the range $10^{-6}\lt \alpha\lt 10^{-1}.$ The first row is for $\beta=1$ and the second for $\beta=20.$ Both scales are logarithmic.


The bounds $(1)$ and $(2)$ are plotted with dotted lines. Evidently they are correct and, for larger $t,$ are very accurate.

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Graphical comment: CDF plots of $\mathsf{Beta}(a, 2)$ for $a=.01, .05, .1, .15, .2, .25, .3$ (respective colors red through purple).

enter image description here

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