Finding the parameters a and b for the Beta prior

Recall that a Beta distribution has variance

$$\sigma^2 = {{ab} \over {(a+b+1) (a+b)^2}}$$ (7)

If we make the change of variable from a and b to n and pusing equation 4 we get.

$$\sigma^2 = {{\left( 1 + n\,\left( 1 - p \right) \right) \,\... ...) }^2}\, \left( 3 + n\,\left( 1 - p \right) + n\,p \right) }}$$ (8)

which simplifies to

$$\sigma^2 = {{\left( 1 + n - n\,p \right) \,\left( 1 + n\,p \r... ... }\over {{{\left( 2 + n \right) }^2}\,\left( 3 + n \right) }}$$ (9)

We know $\sigma^2$ and p and would like to solve for n. This cannot be done directly, so it is necessary to use Newton's Method. Of course, Newton's Method requires the derivative of equation 8 which is given by

$$\matrix{ {{{\rm d}\sigma^2} \over {{\rm d}n}} = -{{\left( 1... ... }^2}\, \left( 3 + n\,\left( 1 - p \right) + n\,p \right) }}}$$ (10)

This equation simplifies to

$${{{\rm d}\sigma^2} \over {{\rm d}n}} = {{-2 - 6\,n - 2\,{n^2}... ...r {{{\left( 2 + n \right) }^3}\,{{\left( 3 + n \right) }^2}}}$$ (11)

Once the parameter n has been solved, it is possible to calculate a and b, giving us the exact distribution of the parameter p. From here it is possible to generate a confidence interval for p.