Beta Function

Let $h(x)=x^{{\alpha }_1-1}{(1-x)}^{{\alpha }_2-1}$ for . The Beta function $B({\alpha }_1,{\alpha }_2)$ determines how the integral of this function over the range $(0,1)$ varies with ${\alpha }_1>0$ and ${\alpha }_2>0$

$B({\alpha }_1,{\alpha }_2)={\displaystyle {\int }_0^1}{x}^{{\alpha }_1-1}{(1-x)}^{{\alpha }_2-1}dx.$

Properties:

1. 1.

   $B(1,1)=1$.

2. 2.

   $B({\alpha }_1,{\alpha }_2)=B({\alpha }_2,{\alpha }_1)$

3. 3.

   $B({\alpha }_1,{\alpha }_2)=\mathrm{\Gamma }({\alpha }_1)\mathrm{\Gamma }({\alpha }_2)/\mathrm{\Gamma }({\alpha }_1+{\alpha }_2)$

Thus the Beta function can easily be evaluated using the Gamma function. In R:

We now show that

$B({\alpha }_1,{\alpha }_2)={\displaystyle {\int }_0^1}{x}^{{\alpha }_1-1}{(1-x)}^{{\alpha }_2-1}dx={\displaystyle \frac{\mathrm{\Gamma }({\alpha }_1)\mathrm{\Gamma }({\alpha }_2)}{\mathrm{\Gamma }({\alpha }_1+{\alpha }_2)}}.$

Firstly, consider the product:

	$\mathrm{\Gamma }({\alpha }_1)\mathrm{\Gamma }({\alpha }_2)$	$={\displaystyle {\int }_0^{\mathrm{\infty }}}{s}^{{\alpha }_1-1}{e}^{-s}ds{\displaystyle {\int }_0^{\mathrm{\infty }}}{t}^{{\alpha }_2-1}{e}^{-t}dt$
		$={\displaystyle {\int }_0^{\mathrm{\infty }}}{\displaystyle {\int }_0^{\mathrm{\infty }}}{s}^{{\alpha }_1-1}{t}^{{\alpha }_2-1}{e}^{-(s+t)}dsdt$
		$={\displaystyle {\int }_0^{\mathrm{\infty }}}{\displaystyle {\int }_0^{\mathrm{\infty }}}{\left({\displaystyle \frac{s}{s+t}}\right)}^{{\alpha }_1-1}{\left({\displaystyle \frac{t}{s+t}}\right)}^{{\alpha }_2-1}{(s+t)}^{{\alpha }_1+{\alpha }_2-2}{e}^{-(s+t)}dsdt.$

Now apply the transformation $x=s/(s+t)$ and $y=s+t$. The Jacobian of this is

$\left|{\displaystyle \frac{\partial (x,y)}{\partial (s,t)}}\right|={\displaystyle \frac{1}{s+t}}={\displaystyle \frac{1}{y}},$

and the range for $x$ is now from $0$ to $1$, with $y$ from $0$ to $\mathrm{\infty }$. Hence

	$\mathrm{\Gamma }({\alpha }_1)\mathrm{\Gamma }({\alpha }_2)$	$={\displaystyle {\int }_0^1}{\displaystyle {\int }_0^{\mathrm{\infty }}}{x}^{{\alpha }_1-1}{(1-x)}^{{\alpha }_2-1}{y}^{{\alpha }_1+{\alpha }_2-2}{e}^{-y}ydxdy$
		$={\displaystyle {\int }_0^1}{x}^{{\alpha }_1-1}{(1-x)}^{{\alpha }_2-1}dx{\displaystyle {\int }_0^{\mathrm{\infty }}}{y}^{{\alpha }_1+{\alpha }_2-1}{e}^{-y}dy$
		$={\displaystyle {\int }_0^1}{x}^{{\alpha }_1-1}{(1-x)}^{{\alpha }_2-1}dx\mathrm{\Gamma }({\alpha }_1+{\alpha }_2).$