Beta distribution, as $\epsilon \to 0$, $(-\epsilon\,\log(B_\epsilon), -\epsilon\,\log(1 - B_\epsilon)) \implies (\xi E_a, (1 - \xi)E_b)$. Fix $a$, $b > 0$. For $\epsilon > 0$, let $B_\epsilon$ be distributed according to a Beta distribution with parameters $\epsilon a$ and $\epsilon b$. Now, I wish to show that as $\epsilon \to 0$,$$(-\epsilon\,\log(B_\epsilon), -\epsilon\,\log(1 - B_\epsilon)) \implies (\xi E_a, (1 - \xi)E_b)$$where $\xi$ is a Bernoulli $(0, 1)$-valued random variable with $\mathbb{P}(\xi = 1) = b/(a + b)$, and $E_a$, $E_b$ are independent (of each other and of $\xi$) exponential random variables with rates $a$ and $b$.
But I'm not sure on where to start. Is it possible somebody could give me a hint, get me started in the right direction?
 A: To show convergence in distribution, we must show for $x, y \in \mathbb{R}$:
$$\lim_{\epsilon \to 0}\mathbb{P}(-\epsilon \text{log}(B_{\epsilon}) \leq x, -\epsilon \text{log}(1 - B_{\epsilon}) \leq y) = \mathbb{P}(\zeta E_{a} \leq x, (1 - \zeta)E_{b} \leq y)$$
PART I (Left-hand side)
We start be rewriting some expressions on the left-hand side:
$$ -\epsilon \text{log}(B_{\epsilon}) \leq x \Rightarrow B_{\epsilon} \geq e^{-x/ \epsilon}$$
$$ -\epsilon \text{log}(1 - B_{\epsilon}) \leq y \Rightarrow B_{\epsilon} \leq 1 - e^{-y / \epsilon}$$
We use these back in the left-hand side of our original expression:
$$ \mathbb{P}(-\epsilon \text{log}(B_{\epsilon}) \leq x, -\epsilon \text{log}(1 - B_{\epsilon}) \leq y) = \mathbb{P}(e^{-x / \epsilon} \leq B_{\epsilon} \leq 1 - e^{-y / \epsilon})$$
Without loss of generality, assume $e^{-x / \epsilon} \leq 1 - e^{-y / \epsilon}$ for $x,y \geq 0$. The above becomes:
$$ \frac{1}{B(\epsilon a, \epsilon b)} \int_{e^{-x/ \epsilon}}^{1 - e^{-y / \epsilon}} t^{a \epsilon - 1}(1 - t)^{b \epsilon - 1} dt $$
$$ = \frac{1}{B(\epsilon a, \epsilon b)}[\int_{0}^{1} t^{a \epsilon - 1}(1 - t)^{b \epsilon - 1} dt - \int_{e^{1 - e^{-y / \epsilon}}}^{1} t^{a \epsilon - 1}(1 - t)^{b \epsilon - 1} dt - \int_{0}^{e^{-x / \epsilon}} t^{a \epsilon - 1}(1 - t)^{b \epsilon - 1} dt] $$
$$ = 1 - \frac{1}{B(\epsilon a, \epsilon b)}(\int_{0}^{e^{-y / \epsilon}} t^{b \epsilon - 1}(1 - t)^{a \epsilon - 1} dt - \int_{0}^{e^{-x / \epsilon}} t^{a \epsilon - 1}(1 - t)^{b \epsilon - 1} dt)$$
by integration by substitution; here, $B$ is the beta function. Now, let us look at the quantity $\int_{0}^{z} t^{\alpha - 1}(1 - t)^{\beta - 1}dt$, where $0 < z \leq 1$ and $0  < \alpha, \beta, < 1$. We expand $(1 - t)^{\beta - 1}$ by the binomial theorem:
$$\int_{0}^{z} t^{\alpha - 1}(1 - t)^{\beta - 1}dt$$
$$ = \int_{0}^{z} (t^{\alpha - 1} + \sum_{n = 1}^{\infty}(-1)^{n}\frac{(\beta - 1)(\beta - 2)\dots (\beta - 1 - (n - 1))}{n!}t^{n + \alpha - 1} )dt$$
$$ = \int_{0}^{z} t^{\alpha - 1}dt + \sum_{n = 1}^{\infty} (-1)^{n}\frac{(\beta - 1)(\beta - 2)\dots (\beta - 1 - (n - 1))}{n!} \int_{0}^{z}t^{n + \alpha - 1}dt, \text{ by interchanging integral and sum}$$
$$ = z^{\alpha}(\frac{1}{\alpha} + \sum_{n = 1}^{\infty} (-1)^{n}\frac{(\beta - 1)(\beta - 2)\dots (\beta - 1 - (n - 1))}{n!(n + \alpha)} z^{n}) \text{   (*)}$$
When $z = 1, \alpha = \epsilon a, \beta = \epsilon b$, the above is $B(\epsilon a, \epsilon b)$, and from how $B(\alpha, \beta) = B(\beta, \alpha)$, $(*)$ can be written as either:
$$ \frac{1}{\epsilon a} + \sum_{n = 1}^{\infty} (-1)^{n}\frac{(\epsilon b - 1)(\epsilon b - 2)\dots (\epsilon b - 1 - (n - 1))}{n!(n + \epsilon a)}$$
or
$$ \frac{1}{\epsilon b} + \sum_{n = 1}^{\infty} (-1)^{n}\frac{(\epsilon a - 1)(\epsilon a - 2)\dots (\epsilon a - 1 - (n - 1))}{n!(n + \epsilon b)}$$
From these two expressions, another way to write $B(\epsilon a, \epsilon b)$ is:
$$ B(\epsilon a, \epsilon b) = \frac{1}{2}(B(\epsilon a, \epsilon b) + B(\epsilon b, \epsilon a))$$
$$ = \frac{a + b}{\epsilon ab} + \frac{1}{2} \sum_{n = 1}^{\infty} \frac{(-1)^{n}}{n!} (\frac{(\epsilon b - 1)(\epsilon b - 2)\dots (\epsilon b - 1 - (n - 1))}{n + \epsilon a} + \frac{(\epsilon a - 1)(\epsilon a - 2)\dots (\epsilon a - 1 - (n - 1))}{n + \epsilon b} )$$
When $z = e^{-y / \epsilon}, \alpha = \epsilon b, \beta = \epsilon a$, $(*)$ becomes
$$ e^{-by}(\frac{1}{\epsilon b} + \sum_{n = 1}^{\infty} (-1)^{n} \frac{(\epsilon a - 1) \dots (\epsilon a - n)}{n!(n + \epsilon b)}e^{-yn / \epsilon})$$
and when $z = e^{-x / \epsilon}, \alpha = \epsilon a, \beta = \epsilon b$, $(*)$ becomes
$$ e^{-ay}(\frac{1}{\epsilon a} + \sum_{n = 1}^{\infty} (-1)^{n} \frac{(\epsilon b - 1) \dots (\epsilon b - n)}{n!(n + \epsilon a)}e^{-xn / \epsilon})$$
We now compute the following limits:
$$ \lim_{\epsilon \to 0} \frac{\int_{0}^{e^{-y/ \epsilon}}t^{b \epsilon - 1}(1 - t)^{a \epsilon - 1} dt}{B(\epsilon a, \epsilon b)} $$
$$ = e^{-by}\lim_{\epsilon \to 0}\frac{\frac{1}{b} + \epsilon \sum_{n = 1}^{\infty} (-1)^{n} \frac{(\epsilon a - 1) \dots (\epsilon a - n)}{n!(n + \epsilon b)}e^{-yn / \epsilon}}{\frac{a + b}{ab} + \frac{1}{2} \epsilon \sum_{n = 1}^{\infty} \frac{(-1)^{n}}{n!} (\frac{(\epsilon b - 1)(\epsilon b - 2)\dots (\epsilon b - 1 - (n - 1))}{n + \epsilon a} + \frac{(\epsilon a - 1)(\epsilon a - 2)\dots (\epsilon a - 1 - (n - 1))}{n + \epsilon b} )}$$
Note the following:
1) For fixed $n$, $e^{-yn/ \epsilon} \to 0$ as $\epsilon \to 0$ and as $n \to \infty$, $e^{-yn/ \epsilon} \to 0$ as $\epsilon \to 0$, which implies $\epsilon \sum_{n = 1}^{\infty} (-1)^{n} \frac{(\epsilon a - 1) \dots (\epsilon a - n)}{n!(n + \epsilon b)}e^{-yn / \epsilon} \to 0$ as $\epsilon \to 0$.
2) $\frac{1}{2} \epsilon \sum_{n = 1}^{\infty} \frac{(-1)^{n}}{n!} (\frac{(\epsilon b - 1)(\epsilon b - 2)\dots (\epsilon b - 1 - (n - 1))}{n + \epsilon a} + \frac{(\epsilon a - 1)(\epsilon a - 2)\dots (\epsilon a - 1 - (n - 1))}{n + \epsilon b} ) \to 0$ as $\epsilon \to 0$; this should be able to be proved by writing $\epsilon$ as $\frac{1}{1/ \epsilon}$ and using l'Hopital's rule; one will have to differentiate under the series.
From these observations, the above is $e^{-by}\frac{1/b}{(a + b)/ab} = \frac{a}{a + b}e^{-by}$. A similar computation shows $\lim_{\epsilon \to 0}\frac{\int_{0}^{e^{-x/ \epsilon}}t^{a \epsilon - 1}(1 - t)^{b \epsilon - 1}dt}{B(\epsilon b, \epsilon a)} = \frac{b}{a + b}e^{-ay}$. Plugging all of this back into our original computation shows:
$$ \lim_{\epsilon \to 0}\mathbb{P}(-\epsilon \text{log}(B_{\epsilon}) \leq x, -\epsilon \text{log}(1 - B_{\epsilon}) \leq y) = 1 - \frac{1}{a + b}(ae^{-by} + be^{-ax})$$
PART II (Right-hand side)
Now, let us look at the right-hand side of our original expression. Using that $\zeta$, $E_{a}$, and $E_{b}$ are independent and how $\zeta$ is a Bernoulli $(0,1)$-random variable, and assuming that $x, y \geq 0$:
$$\mathbb{P}(\zeta E_{a} \leq x, (1 - \zeta)E_{b} \leq y) = \mathbb{P}(E_{a} \leq x)\mathbb{P}(\zeta = 1) + \mathbb{P}(E_{b} \leq y)\mathbb{P}(\zeta = 0)$$
$$ = \frac{b}{a + b}(1 - e^{-ax}) + \frac{a}{a + b}(1 - e^{-by})$$
$$ = 1 - \frac{1}{a + b}(be^{-ay} + ae^{-by})$$
$$ = \lim_{\epsilon \to 0}\mathbb{P}(-\epsilon \text{log}(B_{\epsilon}) \leq x, -\epsilon \text{log}(1 - B_{\epsilon}) \leq y), \text{what we have shown in part I}$$
Thus, $(-\epsilon \text{log}(B_{\epsilon}), -\epsilon \text{log}(1 - B_{\epsilon}))$ converges to $(\zeta E_{a}, (1 - \zeta)E_{b})$ in distribution as $\epsilon \to 0$.
(Note to Ivan Corwin, who posed this problem: this question was asked and answered on Math Stack Exchange on 12/30, well after the final exam.)
