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What is the argument to prove asymptotic normality of logit transformed of binomial proportion which follows Beta distribution?

$\theta$ has beta prior and model $y|\theta$ follows $Binom(\theta,n)$. Then posterior $p(\theta|y)$ follows beta distribution.

The following is asserted in Gelman's Bayesian Data Analysis 3rd Ed, Sec 24, page 35 right before "conjugate prior distributions".

"For the binomial parameter θ, the normal distribution is a more accurate approximation in practice if we transform θ to the logit scale; that is, performing inference for log(θ/(1 − θ)) instead of θ itself, thus expanding the probability space from [0, 1] to (−∞,∞), which is more fitting for a normal approximation."

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    $\begingroup$ The ostensible reasons are embedded right there in the quotation. They're not convincing, though. The issue really is about the asymmetry of Beta distributions with means close to $0$ or $1$ and the ability of the logit transformation to make them approximately symmetric. $\endgroup$
    – whuber ♦
    Commented Dec 31, 2021 at 16:25
  • $\begingroup$ @whuber I am not sure that symmetrization argument is even correct. If I consider any reasonable (canonical/non-canonical) link function associated to some specific family of distributions, I do not see why link function transformed variable necessarily symmetric. I agree that link function is used to spread the interval out to $(-\infty,\infty)$ but I do not believe such transformation will always give symmetry or asymptotic symmetry here. $\endgroup$
    – user45765
    Commented Dec 31, 2021 at 17:51
  • $\begingroup$ I do not claim exact symmetry. But check it out--it really does render these Beta distributions much more symmetrical. There is a theoretical reason: when the variance of a positive distribution family is related to its mean by a power law, then there is a Box-Cox transformation that approximately symmetrizes all members of the family. For Betas with a mean near $0,$ that is a log--which is closely approximated by the logit. For Betas with a mean near $1,$ the same thing occurs (by symmetry). $\endgroup$
    – whuber ♦
    Commented Dec 31, 2021 at 20:46

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