I try to solve this property of Regularized Incomplete Beta Function. How can you solve a statement :
$I_x (a,a) = 1 - \frac{1}{2} I_{4x(1-x)}(a,\frac{1}{2}) $ where $1/2 \leq x \leq 1$.
Some definitions :
$I_x (a,b) = \frac{B(x;a,b)}{B(a,b)}$ with B(x;a,b) is incomplete Beta function and B(a,b) is Beta function.
I tried to use substitution rule with z = 1 - x. But it doesn't go any further.
Could someone give me idea more please ?
Using the definitions you wrote $$\text{lhs}=\frac{B_x(a,a)}{B(a,a)}$$ $$\text{rhs}=1-\frac{B_{4 (1-x) x}\left(a,\frac{1}{2}\right)}{2 B\left(a,\frac{1}{2}\right)}$$ Now, for $ \frac 12 \leq x \leq 1$ $$\frac d {dx}\text{[lhs-rhs]}=\frac{(1-x)^{a-1} x^{a-1}}{B(a,a)}-\frac{2^{2 a-3} (4 (1-x)-4 x) ((1-x) x)^{a-1}}{\sqrt{1-4 (1-x) x} B\left(a,\frac{1}{2}\right)}$$ Since $ \frac 12 \leq x \leq 1$, the last term simplify and write $$\frac{2^{2 a-1} ((1-x) x)^{a-1}}{B\left(a,\frac{1}{2}\right)}$$ $$\frac d {dx}\text{[lhs-rhs]}=\frac{(1-x)^{a-1} x^{a-1}}{B(a,a)}-\frac{2^{2 a-1} ((1-x) x)^{a-1}}{B\left(a,\frac{1}{2}\right)}$$ Divide by $(1-x)^{a-1} x^{a-1}$ to get $$\frac{\frac d {dx}\text{[lhs-rhs]} } {(1-x)^{a-1} x^{a-1} }=\frac{1}{B(a,a)}-\frac{2^{2 a-1}}{B\left(a,\frac{1}{2}\right)}=0$$ by the properties of the beta function.
So, for $ \frac 12 \leq x \leq 1$, $\text{[lhs-rhs]}$ is a constant. For $x=1$ for example $$\text{lhs}=\frac{B_1(a,a)}{B(a,a)}=1 \qquad \text{and} \qquad \text{rhs}=1-\frac{\Gamma \left(a+\frac{1}{2}\right) B_0\left(a,\frac{1}{2}\right)}{2 \sqrt{\pi } \Gamma (a)}=1$$ Then, the identity.
