Bounds on the incomplete beta function (beta CDF)

The incomplete beta function is defined by

$B(x, y, z) = \int_0^z t^{x-1} (1-t)^{y-1} \, dx$

It is called incomplete because the limit of integration doesn’t necessarily run all the way up to 1. When z = 1 the incomplete beta function is simply the beta function

discussed in the previous post [1].

The incomplete beta function is proportional to the CDF of a beta random variable, and the proportionality constant is 1/B(x, y). That is, if X is a beta(a, b) random variable, then

$\text{Prob}(X \leq x) = \frac{B(a, b, x)}{B(a, b)}$

The right-hand side of the equation above is often called the regularized incomplete beta function. This is what an analyst would call it. A statistician would call it the beta distribution function.

The previous post gave simple bounds on the (complete) beta function. This post gives bounds on the incomplete beta function. These bounds are more complicated but also more useful. The incomplete beta function is more difficult to work with than the beta function, and so upper and lower bounds are more appreciated.

One application of these bounds would be to speed up software that evaluates beta probabilities. If you only need to know whether a probability is above or below some threshold, you could first compute an upper or lower bound. Maybe computing the exact probability is unnecessary because you can decide from your bounds whether the probability is above or below your threshold. If not, then you compute the exact probability. This is less work if your bounds are usually sufficient, and more work if they’re usually not. In the former case this could result in a significant speed up because the incomplete beta function is relatively expensive to evaluate.

Cerone, cited in the previous post, gives two lower bounds, L₁ and L₂, and two upper bounds, U₁ and U₂, for B(x, y, z). Therefore the maximum of L₁ and L₂ is a lower bound and the minimum of U₁ and U₂ is an upper bound. The bounds are complicated, but easier to work with than the incomplete beta function.

$\begin{align*} L_1 &= \frac{z^{x-1}}{y} \left[ \left(1 - z + \frac{z}{x} \right)^y - (1-z)^y \right] \\ U_1 &= \frac{z^{x-1}}{y} \left[ 1 - \left(1 - \frac{z}{x}\right)^y\right] \\ L_2 &= \frac{\lambda(z)^x}{x} + (1-z)^{y-1}\, \frac{z^x - \lambda(z)^x}{x} \\ U_2 &= (1-z)^{y-1} \frac{\left( x - \lambda(z) \right)^x}{x} + \frac{z^x - \left(z - \lambda(z)\right)^x}{x} \\ \end{align*}$

where

$\lambda(z) = \frac{1 - (1-z)[1 - z(1-y)]}{y\left[1-(1-z)^{y-1}\right]}$

[1] Incidentally, Mathematica overloads the Beta function just as we overload B. Mathematica’s Beta[x, y] corresponds to our B(x, y), but the three-argument version of Beta, the incomplete beta function, puts the limit of integration z first. That is, our B(x, y, z) corresponds to Beta[z, x, y] in Mathematica.

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