Prove the normal approximation of beta distribution How should I prove the normal approximation of beta distribution as follows:
Let $\mathrm B_{r_1, r_2}\sim \mathrm{Beta}(r_1, r_2)$, then prove that
$\sqrt{r_1+r_2} (\mathrm B_{r_1, r_2}- \dfrac{r_1}{r_1+r_2}) \to \mathrm N(0, \gamma(1-\gamma))$
where $r_1, r_2 \to \infty$ and $\frac{r_1}{r_1+r_2}\to \gamma$  $(0<\gamma<1)$.
My attempt: By some calculation, I figured out that $\mathbb E(\mathrm B_{r_1, r_2})=\dfrac{r_1}{r_1+r_2}$ and $\operatorname{Var}(\mathrm B_{r_1, r_2})=\dfrac{r_1 r_2}{(r_1 + r_2)^2 (r_1 +r_2 +1)}$.
Therefore, it remains to prove that $\sqrt{r_1 + r_2}\dfrac{\mathrm B_{r_1, r_2}-\mathbb E(\mathrm B_{r_1, r_2})}{\sqrt{\operatorname{Var}(\mathrm B_{r_1, r_2})}}$ converges to $Z \sim \mathrm N(0, 1)$.
Here I think I have to apply CLT, but I don't know how to because the given quantity does not contain sample mean. Does anyone have ideas?
Thanks for your help!
 A: Represent your $B_{r_1,r_2}$ as $\Gamma_{r_1}/(\Gamma_{r_1}+\Gamma_{r_2}).$ The $\Gamma_r$ distributions are asymptotically normal as $r\to\infty$ by the CLT.  Now get your result by applying the delta method.
A: Want
$\sqrt{a+b}(B(a,b)-\frac{a}{a+b})
\to N(0, c(1-c))
$
where
$\frac{a}{a+b}
\to c$.
If
$\frac{a}{a+b}
\to c$,
then
$a \approx cr$
and
$b \approx (1-c)r$
as
$r \to \infty$.
However,
I get,
for large $r$
and
$0 < c < 1$,
that
$B(cr, (1-c)r)
\approx \sqrt{2\pi}\dfrac{(c^c(1-c)^{1-c})^r}{\sqrt{c(1-c)r}}
$
so
$\sqrt{r}B(cr, (1-c)r)
\approx \sqrt{2\pi}\dfrac{(c^c(1-c)^{1-c})^r}{\sqrt{c(1-c)}}
$
which does not 
seem to agree.
Here is my derivation.
$\begin{array}\\
B(a, b)
&=\int_0^1 t^{a-1}(1-t)^{b-1}dt\\
&=\dfrac{\Gamma(a)\Gamma(b)}{\Gamma(a+b)}\\
&\approx\dfrac{\sqrt{2\pi /a}(a/e)^a\sqrt{2\pi /b}(b/e)^b}{\sqrt{2\pi/ (a+b)}((a+b)/e)^{a+b}}
\qquad\text{for large } a, b\\
&=\sqrt{2\pi}\dfrac{a^{a-1/2}b^{b-1/2}}{(a+b)^{a+b-1/2}}\\
\end{array}
$
For
$a=cr, b=(1-c)r$
we have
$r = a+b$
and
$\dfrac{a}{a+b}
=c
$
so that
$\begin{array}\\
B(a, b)
&=B(cr, (1-c)r)\\
&\sim \sqrt{2\pi}\dfrac{(cr)^{cr-1/2}((1-c)r)^{(1-c)r-1/2}}{r^{r-1/2}}\\
&= \sqrt{2\pi}\dfrac{c^{cr-1/2}(1-c)^{(1-c)r-1/2}r^{cr-1/2}r^{(1-c)r-1/2}}{r^{r-1/2}}\\
&= \sqrt{2\pi}\dfrac{c^{cr-1/2}(1-c)^{(1-c)r-1/2}r^{r-1}}{r^{r-1/2}}\\
&= \sqrt{2\pi}\dfrac{c^{cr}(1-c)^{(1-c)r}}{\sqrt{c(1-c)r}}\\
&= \sqrt{2\pi}\dfrac{(c^c(1-c)^{1-c})^r}{\sqrt{c(1-c)r}}\\
\end{array}
$
A: I finally figured out the answer:
Let $\mathrm B_{r_1, r_2} = \dfrac{V_1}{V_1+V_2}$ where $V_1 \sim \chi^2(2r_1 ) = \mathrm{Gamma}(r_1, 2)$ and $V_2 \sim \chi^2(2r_2 ) = \mathrm{Gamma}(r_2, 2)$ and $V_1, V_2$ are independent.
Now we know that $\dfrac{V_1-2r_1}{\sqrt{4r_1}} \to \mathrm N(0, 1)$ as $r_1\to\infty$ and $\dfrac{V_2-2r_2}{\sqrt{4r_2}} \to \mathrm N(0, 1)$ as $r_2\to\infty$. Therefore, $\sqrt{r_1+r_2}(\dfrac{V_1-2r_1}{\sqrt{4r_1(r_1+r_2)}}, \dfrac{V_2-2r_2}{\sqrt{4r_2(r_1+r_2)}})^t \to \mathrm N_2(0, I)$ where $I$ is the identity matrix.
Also, $\sqrt{r_1+r_2}\left(\dfrac{V_1-2r_1}{\sqrt{4r_1(r_1+r_2)}}, \dfrac{V_2-2r_2}{\sqrt{4r_2(r_1+r_2)}}\right)^t\sim \sqrt{n}\left(\dfrac {V_1}{\sqrt{4\gamma}n}-\sqrt{\gamma}, \dfrac {V_2}{\sqrt{4(1-\gamma)}n}-\sqrt{1-\gamma} \right)^t$
Where $r_1+r_2=n, r_1 \sim \gamma n, r_2 \sim (1-\gamma)n$.
Now let $g(x, y)=\dfrac{\sqrt{\gamma}x}{\sqrt{\gamma}x+\sqrt{1-\gamma}y}$, then $g'(x, y)=\left(\dfrac{\sqrt{\gamma (1-\gamma)}y}{\left(\sqrt{\gamma}x+\sqrt{1-\gamma}y\right)^2}, -\dfrac{\sqrt{\gamma (1-\gamma)}x}{\left(\sqrt{\gamma}x+\sqrt{1-\gamma}y\right)^2} \right)^t$.
Therefore, $\sqrt{n}\left( g\left(\dfrac {V_1}{\sqrt{4\gamma}n}, \dfrac {V_2}{\sqrt{4(1-\gamma)}n} \right)-g\left(\sqrt{\gamma}, \sqrt{1-\gamma}\right)  \right)=\sqrt{n}(\mathrm B_{r_1, r_2}-\gamma)\to g'\left(\sqrt{\gamma}, \sqrt{1-\gamma}\right)^t\mathrm N(0, I) = \mathrm N(0, \gamma(1-\gamma))$
Am I right?
A: A straightforward approach is to compute the Taylor series expansion of $f(\mu + \sigma t)$ where $f$ is the pdf of Beta$(r_1, r_2)$ and $\mu$ and $\sigma$ are its mean and variance. An advantage of this approach is that you get convergence rates, but it may be overkill for what you've asked.
I have done this computation in Lemma A.1 of the paper https://projecteuclid.org/euclid.ejs/1472829397
