Consider the equation $$p^\alpha(1-p)^\beta = cB(\alpha+1,\beta+1)$$ where $c,\alpha,\beta>0$ and $B(\alpha,\beta)$ is the Beta function.

Question: What is the probability $p$ as a function of $c,\alpha,\beta$?

  • $\begingroup$ The Beta function in this equation is irrelevant. You simply want the inverse of the function $p^\alpha(1-p)^\beta$ which I believe can only be found for specific values of $\alpha,\beta$. $\endgroup$ – Peter Foreman Apr 25 at 22:35

As @Peter Foreman commented, you only need to find the zero of function $$f(p)=p^\alpha(1-p)^\beta-k$$ and, for the most general case, you will need some numerical method.

From a practical point of view, I think that it would be better to consider $$g(p)=\alpha \log(p)+\beta \log(1-p)-\log(k)$$ Its derivatives $$g'(p)=\frac{\alpha }{p}-\frac{\beta }{1-p}\qquad \text{and} \qquad g''(p)=-\frac{\alpha }{p^2}-\frac{\beta }{(1-p)^2} <0 \,\,\, \forall p$$ The first derivative cancels at $$p_*=\frac{\alpha }{\alpha +\beta }$$ which corresponds to a maximum (by the second derivative test. So, if $$g\left(\frac{\alpha }{\alpha +\beta }\right)=\alpha \log \left(\frac{\alpha }{\alpha +\beta }\right)+\beta \log \left(\frac{\beta }{\alpha +\beta }\right)-\log (k)$$ is positive, there will be roots to the equation. One root will be $0 < p_1 <\frac{\alpha }{\alpha +\beta }$ and the second one $\frac{\alpha }{\alpha +\beta }< p_2 <1$.

Start the search at the midpoint of the interval and use Newton method.

For example, let us use $\alpha=10$, $\beta=17$ and $c=\frac 1 {15}$. For the first root, the iterates would be $$\left( \begin{array}{cc} n & p_n \\ 0 & 0.185185 \\ 1 & 0.238691 \\ 2 & 0.258600 \\ 3 & 0.261070 \\ 4 & 0.261105 \end{array} \right)$$ and, for the second root, $$\left( \begin{array}{cc} n & p_n \\ 0 & 0.685185 \\ 1 & 0.561947 \\ 2 & 0.503743 \\ 3 & 0.490163 \\ 4 & 0.489394 \\ 5 & 0.489392 \end{array} \right)$$


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