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Beta Function

Let h(x)=xα1-1(1-x)α2-1 for 0<x<1. The Beta function B(α1,α2) determines how the integral of this function over the range (0,1) varies with α1>0 and α2>0

B(α1,α2)=01xα1-1(1-x)α2-1dx.

Properties:

  1. 1.

    B(1,1)=1.

  2. 2.

    B(α1,α2)=B(α2,α1)

  3. 3.

    B(α1,α2)=Γ(α1)Γ(α2)/Γ(α1+α2)

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

> gamma(4)*gamma(0.5)/gamma(4+0.5)        # Calc Beta(4,0.5)
[1] 0.9142857

We now show that

B(α1,α2)=01xα1-1(1-x)α2-1dx=Γ(α1)Γ(α2)Γ(α1+α2).

Firstly, consider the product:

Γ(α1)Γ(α2) =0sα1-1e-sds0tα2-1e-tdt
=00sα1-1tα2-1e-(s+t)dsdt
=00(ss+t)α1-1(ts+t)α2-1(s+t)α1+α2-2e-(s+t)dsdt.

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

|(x,y)(s,t)|=1s+t=1y,

and the range for x is now from 0 to 1, with y from 0 to . Hence

Γ(α1)Γ(α2) =010xα1-1(1-x)α2-1yα1+α2-2e-yydxdy
=01xα1-1(1-x)α2-1dx0yα1+α2-1e-ydy
=01xα1-1(1-x)α2-1dxΓ(α1+α2).