Is there a way of solving $I_x^{-1}(a_1,b_1)w_1+I_x^{-1}(a_2,b_2)w_2=k$?

Question

Is there any way to rearrange for $x$ in the following formula: $$I_x^{-1}(a_1,b_1)w_1+I_x^{-1}(a_2,b_2)w_2=k$$

$I_x^{-1}(a,b)$ is the inverse incomplete beta function and $1\leq a_1,b_1,a_2,b_2$; $0\leq x\leq 1$; $k\leq w_1+w_2$

Answer 1

There's probably no exact formula that can be found, but one can prove there is a solution for the case $k\geq 0$ and you can therefore use some numerical method to find it. Define the function

$$f(x) = w_1 I^{-1}_x(a_1,b_1) + w_2 I^{-1}_x(a_2,b_2) - k$$

Then, $f(1)=w_1+w_2-k \geq 0$ and $f(0) = -k$. This means that if $k>0$, by the intermediate value theorem, there exists an $x^*$ in $(0,1)$ such that $f(x^*)=0$. Moreover, I think it's possible to argue even further that this solution is unique by using the properties of $I^{-1}$ which I think is injective.

(Note: I assumed $I^{-1}_x(a,b)$ is the inverse of the regularized incomplete beta function as it seemed to coincide with the notation of wikipedia. But I think the argument can easily be tweaked for the non-regularized case.)