In other words, given $x_1$, $x_2$, $p_1$ and $p_2$ in the interval $(0,1)$, is there any algorithm to find $\alpha$ and $\beta$ such that $F(x_i)=p_i$ for $i = 1,2$? Here $F$ is the (cumulative) distribution function associated to the Beta probability distribution $B(\alpha,\beta)$.
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When the system has a solution then it can be found using any root-finding method. I tried this a naive implementation of Newtons method and it works well unless we start too far away from the root for which it breaks down. One can probably fix this by being smarter in the choice of root-function or one can try other methods. – Winther Sep 08 '15 at 23:35
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It is interesting how the beta.select function from the LearnBayes module solves this problem. Instead of finding $\alpha, \beta$ that solves the original problem, the function finds $K$ and $\delta$ that approximately solves $\alpha = \delta K$ and $\beta = (1-\delta)K$.
Every $K$ is taken from a logarithmic list, and for each $K_j$, $\delta_j$ is the $\delta$ that has the best result solving the problem for the first percentile using the bisect method.
Finally, to choose among the different $K_j$, it uses the second percentile
jburroni
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