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I'm interested in the asymptotics of the ${\rm Beta}(\alpha,\beta)$ density $$ f(x) = {\rm B}(\alpha,\beta) x^{\alpha - 1}(1-x)^{\beta - 1} $$ as $\alpha, \beta \to \infty$ with $\lim \alpha/\beta \in (0, 1)$. (The case $\alpha$ fixed, $\beta \to \infty$ is already known to converge to a rescaled $\Gamma(\alpha, 1)$.) For moderately large $\beta$, numerical evidence suggests that $f(x)/n$ (where $n=\alpha+\beta$) is well approximated on $[0,1]$ by a Gaussian density $$ \varphi(x) = \exp \left\{\log f(p) -\frac{(x-p)^2}{2p(1-p)/n}\right\}. $$ for $p=\alpha/(\alpha+\beta)$. Obviously this phenomenon is related to the CLT and limiting distribution of a binomial random variable. However, from staring at lots of plots, the convergence appears to be a) uniform in $x$ and (possibly) b) exponentially fast in $\beta$, which I don't think are implied by the standard CLT.

I can prove by a simple argument (expand $\log f$ about $x=p$ and bound the sum of the order 3 and higher terms) that $|f/n-\varphi|_\infty \in O(\log^{1/2}\beta / \beta^{1/4})$, but doubt this is sharp. Is the correct rate $\beta^{-1/2}$, and/or does it fall easily out of some more general theorem?

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    Should there be a minus sign in front of $(x-p)^2\text{ ?} \qquad$ – 2017-02-12
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    Of course, thanks. – 2017-02-12
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    Here's a bit of preliminary scratchwork. The expected value is of course $p=\alpha/(\alpha + \beta).$ $$ p+q=1. $$ $$ p = \frac \alpha {\alpha+\beta}, \qquad q = \frac \beta {\alpha+\beta}. $$ $$ \frac{pq} n = \text{variance} = \frac{\left(\dfrac \alpha {\alpha+\beta}\right) \left( \dfrac \beta {\alpha+\beta}\right) }{ \alpha+\beta+ 1}. $$ $$ \text{Therefore } n = \alpha+\beta+1. $$ – 2017-02-12
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    Sorry, I was sloppy with my $\propto$'s and omitted a scaling factor on $f$. Should be correct now. – 2017-02-12
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    It seems simpler to use $n=\alpha+\beta+1$ than $n=\alpha+\beta. \qquad$ – 2017-02-12

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In case anyone is interested, I figured out the way to get what I think is the correct rate. It's easier to write out if I switch notation slightly and let $\alpha,\beta$ be fixed constants and consider $X\sim\text{Beta}(n\alpha,n\beta)$ as $n\to\infty$. $X$ has mean $\mu=\alpha/(\alpha + \beta)$ and variance $\sigma^2_n = [c_n/(\alpha+\beta)]^2$ for $c_n^2 := \alpha \beta/(n\alpha+n\beta+1)$. The standardized variable $Y = (X-\mu)/\sigma_n$ has density $$ f_Y(y) \propto (\alpha+c_n y)^{n\alpha - 1}(\beta-c_n y)^{n\beta - 1},\,-\alpha/c_n \le y \le \beta/c_n. $$ The function $g_n(y) = n \log (1 + y / \sqrt{n})$ has the asymptotic expansion $$ g_n(y) = y\sqrt{n} - y^2/2 + O(n^{-1/2}), $$ where the $O(n^{-1/2})$ term holds uniformly over $y$ in a compact subset. Applying this to the above equation we obtain after some manipulations $$ \log f_Y(y) = \text{const} -y^2/2 + O(n^{-1/2}). $$