I am trying to find an uninformative prior for a Beta distribution $Beta(a, b)$. This page suggests using a Logistic distribution for the sum $a + b$ but I am having difficulty understanding how this distribution is being derived as a prior for a Beta distribution. I've included everythign here for completeness.
A natural parameterization of the prior distribution $p(a,b)$ is $p(\mu,\eta)$, where $\mu=a/a+b$ is the hyperprior mean and $\eta=a+b$ is the hyperprior sample size. We can rewrite the hyperprior distribution in terms of the new parameters $\mu$ and $\eta$ as follows:
$$ a = \mu \eta $$ $$ b = (1-\mu)\eta $$
We assume that $\mu$ and $\eta$ are independent which means that our beliefs about the prior mean are independent of the beliefs about the prior sample size. We assign $\mu$ assigns this parameter a Uniform prior (which is equivalent to a $Beta(1,1)$) as we have no beliefs about the hyperprior mean.
The prior for the hyperparameter sample size, $\eta$, is more complicated.
Imagine the case where the hyperparameter values are known. If $k^*$ and $n^*$ are respectively the number of conversion and number of users for one arm, then the posterior mean of the conversions rate $p^*$ is given by
$$ E\left(p \mid k\right) = \frac{a+k}{a+b+n} = \left(1-\lambda_n\right)\frac{a}{a+b} + \lambda_n \frac{k}{n} $$
Where
$$ \lambda_n = \frac{n}{n+1} $$
$\lambda_n$ falls in (0, 1) and represents the degree of shrinkage of the posterior mean away from the sample proportion $k^*/n^*$ towards $\mu$. So
$$ \lambda = \frac{\eta}{\bar{n} + \eta} = \frac{\bar{n}}{\bar{n} + 1} \sim \mathrm{Unif}(0, 1) $$
It is useful to remember that the inverse logit function is:
$$ \mathrm{logit}^{-1}(x) = \frac{\exp{x}}{\exp{x} + 1} $$
Putting this all together $\theta = \log\eta$, has a Logistic distribution with location $\log\bar{n}$ and scale 1. We represent this distribution as $\mathrm{Logistic}(\log\bar{n}, 1)$, with pdf:
$$ p(\theta)= \frac{ e^{−(\theta−\bar{n})} }{(1+e^{−(\theta−\bar{n})})^2} $$
I can sort of see how they arrive at it, but also it's not 100% clear and I'd like to clear it up.
