Beta Function

The integral with $ℜ (z_{1}), ℜ (z_{2}) > 0$

B (z_{1}, z_{2}) = \int_{0}^{1} t^{z_{1} - 1} (1 - t)^{z_{2} - 1} d t

We have that $B (z_{1}, z_{2}) = B (z_{2}, z_{1})$ , we also have the equivalente integral definition

B (z_{1}, z_{2}) = 2 \int_{0}^{π / 2} (\sin θ)^{2 z_{1} - 1} (\cos θ)^{2 z_{2} - 1} d θ

it has a relationship with the Gamma Function as

B (z_{1}, z_{2}) = \frac{Γ (z_{1}) Γ (z_{2})}{Γ (z_{1} + z_{2})}

We have a simplification

\int_{0}^{1} x^{p} (1 - x^{m})^{q} d x = \frac{1}{m} B (\frac{p + 1}{m}, q + 1)

Also, the integrals

\int_{0}^{\frac{π}{2}} (\sin θ)^{n} d θ = \int_{0}^{\frac{π}{2}} (\cos θ)^{n} d θ = \frac{1}{2} B (\frac{n + 1}{2}, \frac{1}{2})

We can look at special values of $B (a, b)$ :

$B (a, 1) = 1 / a$
$B (a + 1, b) = \frac{a}{b} B (a, b + 1)$
$B (a + 1, b) = \frac{a}{a + b} B (a, b)$

Incomplete Beta Function

We can define the incomplete beta function, as the integral

B (x; a, b) = \int_{0}^{x} t^{a - 1} (1 - t)^{b - 1} d t

we have that $B (1; a, b) = B (a, b)$ .

We can define, the regularized incomplete beta function, or regularized beta function as

I_{x} (a, b) = \frac{B (x; a, b)}{B (a, b)}

This has several important properties. It is used as the cdf of a random variable with a Beta Distribution, and is related the cdf $F (x; n, p)$ of a random variable $X$ following a binomial distribution with a probability of a single success $p$ and number of Bernulliy trials

F (k; n, p) = P (X \leq k) = I_{1 - p} (n - k, k + 1) = 1 - I_{p} (k + 1, n - k)

Properties

$I_{0} (a, b) = 0$
$I_{1} (a, b) = 1$
$I_{x} (a, 1) = x^{a}$
$I_{x} (1, b) = 1 - (1 - x)^{b}$
$I_{x} (a, b) = 1 - I_{1 - x} (b, a)$