Beta CDFs and Incomplete-Beta Probabilities

From the Beta Density to Probabilities

A Beta density describes a continuous distribution on the bounded interval from 0 to 1. To find a probability, we integrate that density over the relevant interval. The incomplete beta function packages this integral in a standard form; its regularized version is exactly the Beta distribution function. This connection is useful both for expressing probabilities compactly and for evaluating them with numerical software.

Throughout, let $a>0$ and $b>0$. The complete beta function is $B(a,b)=\int_0^1 t^{a-1}(1-t)^{b-1}\,dt$, which is finite and positive for these parameters. A random variable $X\sim\operatorname{Beta}(a,b)$ has density $f(x)=x^{a-1}(1-x)^{b-1}/B(a,b)$ for $0<x<1$, and density zero outside that interval.

Definition: For $a,b>0$ and $0\leq x\leq 1$, the lower incomplete beta function is $$ B_x(a,b)=\int_0^x t^{a-1}(1-t)^{b-1}\,dt. $$ The regularized incomplete beta function is $$ I_x(a,b)=\frac{B_x(a,b)}{B(a,b)}. $$ For $x\leq 0$, set $I_x(a,b)=0$; for $x\geq 1$, set $I_x(a,b)=1$. These extensions agree with the distribution function outside the support.

The word regularized means that the integral has been divided by the complete beta integral. That division normalizes the accumulated density to a probability. In particular, $I_0(a,b)=0$ and $I_1(a,b)=1$.

The Beta CDF and Interval Probabilities

For a continuous random variable, the distribution function at $x$ is the probability of falling at or below $x$. Integrating the Beta density gives the following result.

Theorem: If $X\sim\operatorname{Beta}(a,b)$, where $a,b>0$, then for every real $x$, $$ F(x)=P(X\leq x)=I_x(a,b), $$ using the extension of $I_x$ outside $[0,1]$. For real numbers $r\leq s$, $$ P(r<X\leq s)=I_s(a,b)-I_r(a,b). $$ The same probability is obtained if either endpoint is included or excluded, because $X$ has a density and hence assigns probability zero to every single point.

Proof. If $0<x<1$, the density formula gives $$ P(X\leq x)=\int_0^x \frac{t^{a-1}(1-t)^{b-1}}{B(a,b)}\,dt =\frac{B_x(a,b)}{B(a,b)} =I_x(a,b). $$ For $x\leq 0$, the probability is zero because $X$ lies in $(0,1)$ with probability one; for $x\geq1$, it is one. This verifies the CDF formula on the whole real line. Subtracting the two CDF values yields $P(r<X\leq s)=F(s)-F(r)$. Finally, for any fixed $c$, $P(X=c)=\int_{\{c\}}f(t)\,dt=0$, since a singleton has length zero. Thus endpoint inclusion does not change the interval probability. $\square$

The extension outside the support makes the interval formula convenient without a separate case for every endpoint. For instance, if $r<0<s<1$, then $I_r(a,b)=0$, so the probability is just $I_s(a,b)$. If $r<0$ and $s>1$, the difference is $1-0=1$, as it should be.

Worked Example: An Interval Probability for a Uniform Distribution

Let $X\sim\operatorname{Beta}(1,1)$. Since $B(1,1)=\int_0^1 1\,dt=1$, its density is 1 on $(0,1)$. Find $P(0.2<X\leq0.65)$.

For these parameters, $$ B_x(1,1)=\int_0^x 1\,dt=x,\qquad I_x(1,1)=x,\quad 0\leq x\leq1. $$ Therefore, $$ P(0.2<X\leq0.65)=I_{0.65}(1,1)-I_{0.2}(1,1) =0.65-0.20=0.45. $$ This is the uniform interval length $0.65-0.20$, since the density is 1 across a support of length 1.

Integer Parameters and Finite Binomial Sums

When $a$ and $b$ are positive integers, the incomplete-beta CDF also has a finite-sum expression. This can be useful for hand calculations and reveals a connection to binomial probabilities. Set $N=a+b-1$, and write $\binom{N}{k}$ for the binomial coefficient.

Theorem: If $a,b$ are positive integers and $0\leq x\leq1$, then $$ I_x(a,b)=\sum_{k=a}^{a+b-1}\binom{a+b-1}{k}x^k(1-x)^{a+b-1-k}. $$ Equivalently, if $Y$ has a binomial distribution with $a+b-1$ trials and success probability $x$, then $I_x(a,b)=P(Y\geq a)$.

Proof. Let $N=a+b-1$ and denote the sum on the right by $S(x)$. For $0<x<1$, differentiating each term and using the binomial-coefficient identities gives $$ \frac{d}{dx}\left[\binom Nk x^k(1-x)^{N-k}\right] =N\binom{N-1}{k-1}x^{k-1}(1-x)^{N-k} -N\binom{N-1}{k}x^k(1-x)^{N-k-1}. $$ On summing from $k=a$ to $N$, the negative term at each index cancels the positive term at the next index. The final negative term is zero because $\binom{N-1}{N}=0$. Consequently, $$ S'(x)=N\binom{N-1}{a-1}x^{a-1}(1-x)^{N-a} =\frac{N!}{(a-1)!(b-1)!}x^{a-1}(1-x)^{b-1}. $$ The beta-gamma identity, or the beta integral for positive integers, gives $B(a,b)=(a-1)!(b-1)!/(a+b-1)!$. Since $N=a+b-1$, the derivative just found is the Beta$(a,b)$ density. Also, $S(0)=0=I_0(a,b)$ because every term has $k\geq a\geq1$. The functions $S(x)$ and $I_x(a,b)$ have the same derivative on $(0,1)$ and agree at 0; hence they agree throughout $[0,1]$ by continuity. The binomial probability statement follows from the binomial probability mass function. $\square$

Worked Example: A Beta CDF From a Binomial Tail

Let $X\sim\operatorname{Beta}(3,2)$. Find $P(0.25<X\leq0.75)$ using the finite-sum identity.

Here $N=3+2-1=4$, so $$ I_x(3,2)=\sum_{k=3}^{4}\binom{4}{k}x^k(1-x)^{4-k} =4x^3(1-x)+x^4 =4x^3-3x^4. $$ At $x=0.75$, $$ I_{0.75}(3,2)=4(0.75)^3-3(0.75)^4 =1.6875-0.94921875 =0.73828125. $$ At $x=0.25$, $$ I_{0.25}(3,2)=4(0.25)^3-3(0.25)^4 =0.0625-0.01171875 =0.05078125. $$ Subtracting, $$ P(0.25<X\leq0.75)=0.73828125-0.05078125=0.6875. $$ The same calculation can be read as subtracting two probabilities of at least three successes in four binomial trials, with success probabilities $0.75$ and $0.25$, respectively.

Worked Example: A Quantile With a Closed-Form Answer

Let $X\sim\operatorname{Beta}(2,1)$. Find its 90th percentile, meaning the value $q$ for which $P(X\leq q)=0.90$.

For $0\leq x\leq1$, $B_x(2,1)=\int_0^x t\,dt=x^2/2$, while $B(2,1)=1/2$. Thus $I_x(2,1)=x^2$. Solving the percentile equation gives $$ I_q(2,1)=q^2=0.90,\qquad q=\sqrt{0.90}\approx0.9487. $$ The solution is in the support, and $I_{0.9487}(2,1)$ is approximately $0.90$, as required.

Quantiles and Numerical Evaluation

For a general Beta distribution, the equation $I_q(a,b)=p$ may not have a convenient algebraic solution. A quantile is still defined precisely, and numerical methods can find it by inverting the CDF.

Definition: For $0<p<1$, the $p$-quantile of $X\sim\operatorname{Beta}(a,b)$ is the value $q_p$ satisfying $I_{q_p}(a,b)=p$. It is unique and lies in $(0,1)$. The endpoint quantiles are $q_0=0$ and $q_1=1$.

To see why the interior solution is unique, the fundamental theorem of calculus gives $$ \frac{d}{dx}I_x(a,b)=\frac{x^{a-1}(1-x)^{b-1}}{B(a,b)}>0,\qquad 0<x<1. $$ The derivative is positive throughout the interior for every $a,b>0$. Therefore $I_x(a,b)$ is strictly increasing there; it is continuous, with endpoint values 0 and 1. The intermediate value theorem gives a solution to $I_q(a,b)=p$ for every $0<p<1$, and strict increase makes that solution unique. $\square$

In practice, a numerical library may provide the regularized incomplete beta function and its inverse. The parameters must be supplied in the same order as the density: $a$ is the exponent parameter on $x$, and $b$ is the exponent parameter on $1-x$. Swapping them generally changes the distribution and its CDF.

Common Pitfall: Using the Unregularized Integral as a Probability

The incomplete beta function $B_x(a,b)$ is an integral, but it is not generally a probability. The density includes the normalizing factor $1/B(a,b)$, so the CDF is $B_x(a,b)/B(a,b)$, not $B_x(a,b)$ alone. For example, for $X\sim\operatorname{Beta}(2,3)$, $B(2,3)=1/12$; omitting division by $1/12$ would make the purported CDF twelve times too small.

Worked Example: Normalizing a Beta CDF Before Subtracting

For $X\sim\operatorname{Beta}(2,3)$, find $P(0.2<X\leq0.7)$. The complete beta integral is $B(2,3)=1/12$, so the density is $12x(1-x)^2$. Integrating from 0 to $x$ gives $$ I_x(2,3)=12\int_0^x t(1-t)^2\,dt =6x^2-8x^3+3x^4,\qquad 0\leq x\leq1. $$ At $x=0.7$, this is $$ 6(0.7)^2-8(0.7)^3+3(0.7)^4 =2.94-2.744+0.7203 =0.9163. $$ At $x=0.2$, it is $$ 6(0.2)^2-8(0.2)^3+3(0.2)^4 =0.24-0.064+0.0048 =0.1808. $$ Therefore, $$ P(0.2<X\leq0.7)=0.9163-0.1808=0.7355. $$ The normalization is essential: integrating $t(1-t)^2$ without multiplying by 12 would give an accumulated density that reaches only $1/12$, not 1, at the right endpoint.

The main practical rule is to distinguish the integral from its normalized probability. Use $I_x(a,b)$ for a Beta CDF value, subtract two such values for an interval probability, and solve $I_q(a,b)=p$ for a quantile. When $a$ and $b$ are positive integers, the finite binomial sum provides an additional way to evaluate the CDF.

Check Your Understanding

Try these questions using the definitions and results above.

Write $P(r<X\leq s)$ in terms of the regularized incomplete beta function when $X\sim\operatorname{Beta}(a,b)$.
Why is $B_x(a,b)$ alone not generally equal to a Beta CDF?
For positive integers $a,b$, how many binomial trials appear in the finite-sum expression for $I_x(a,b)$?
Explain why the interior $p$-quantile of a Beta distribution is unique.
For $X\sim\operatorname{Beta}(2,1)$, use $I_x(2,1)=x^2$ to find the median.