Recursive relationship for incomplete beta function

Question

Consider the incomplete Beta function $I_x(a, b)$ $$ I_x(a, b) = \dfrac{B(x; a, b)}{B(a, b)} = \dfrac{\int_0^x t^{a-1} \left( 1-t \right)^{b-1} dt}{\int_0^1 t^{a-1} \left( 1-t \right)^{b-1} dt}. $$ How can I prove the recursive property given in Wikipedia $$ I_x(a+1, b) = I_x(a, b) - \dfrac{x^a \left( 1-x \right)^b}{a \, B\left( a, b \right)} $$

Answer 1

Using the Beta function property $B(a+1,b)=aB(a.b)/(a+b)$ compute the derivative $$I_x'(a+1, b) = \frac{x^a (1-x)^{b-1}}{B(a+1,b)} = \frac{x^a (1-x)^{b-1}}{B(a,b)a/(b+a)}= (a+b)\frac{x^a (1-x)^{b-1}}{B(a,b)a}$$

Now let $$ f(a,b,x) = I_x(a+1, b) - I_x(a, b) + \dfrac{x^a \left( 1-x \right)^b}{a \, B\left( a, b \right)} $$ Then $f(a,b,0)=0$ and $$f'(a,b,x) = I_x'(a+1, b) - I_x'(a, b) + \frac{d}{dx}\Big(\frac{x^a \left( 1-x \right)^b}{a \, B\left( a, b \right)}\Big)$$

$$f'(a,b,x)= (a+b)\frac{x^a (1-x)^{b-1}}{B(a,b)a} - \frac{x^{a-1} (1-x)^{b-1}}{B(a,b)} + \frac{x^{a-1}(1-x)^b}{B(a,b)} - \frac{b x^a(1-x)^{b-1}}{aB(a,b)}$$

$$f'(a,b,x)= \frac{x^{a-1}(1-x)^{b-1}}{B(a,b)} \Big((a+b)x/a - 1 + 1 -x -bx/a\Big) = 0 $$

therefore $f \equiv 0$ and the recursive property is proven.