Polya's urn model - limit distribution Let an urn contain $w$ white and $b$ black balls. Draw a ball randomly from the urn and return it together with another ball of the same color. Let $b_n$ be the number of black balls and $w_n$ the number of white balls after the $n$-th draw-and-replacement. Let $X_n$ be the relative proportion of white balls after the $n$-th draw-and-replacement.
I start with $b=w=1$, so the total number of balls after the $n$-th draw-and-replacement is $n+2$.
Now I want to find the limit distribution of $X_n$; I already showed that $X_n$ is a martingale and that it converges a.s.
It is
$$X_n = \dfrac{w_n}{n+2} \quad\text{for}\quad n \in \mathbb{N}_0. $$
I've read that the limit distribution is a beta distribution, but I don't know how to get there.
I could write $w_n$ as the sum of $Y_i$ where $Y_i$ is $0$, if the $i$-th ball is black and $1$, if the $i$-th ball is black. Then I'd have
$$ w_n = 1+\sum_{i=1}^{n} Y_i. $$
Does this help? How can I proceed?
Thanks! :)

Refer to this?

Assuming $B_n$ is uniform on $\{0,1,...,n\}$ (proven by induction):
$$M_{\Theta}(t) = E[\exp(t\Theta)]$$
$$= E[\exp(t\lim \frac{B_n + 1}{n+2})]$$
$$= E[\lim\exp(t \frac{B_n + 1}{n+2})]$$
$$= \lim E[\exp(t \frac{B_n + 1}{n+2})]$$
$$= \lim \frac{1}{n+1}[\exp(t \frac{1}{n+2}) + \exp(t \frac{2}{n+2}) + ... + \exp(t \frac{n+1}{n+2})]$$
Case 1: $t \ne 0$
$$= \lim \frac{a(n)}{(n+1)(1-a(n))} (1-a(n)^{n+1}), \ \text{where} \ a(n) := e^{\frac{t}{n+2}}$$
$$= \lim \frac{a(n)}{(n+1)(1-a(n))} \lim (1-a(n)^{n+1})$$
$$= \lim \frac{a(n)}{(n+1)(1-a(n))} (1-e^t)$$
$$= \frac{1-e^t}{-t}$$
$$= \frac{e^t-1}{t}$$
Case 2: $t = 0$
$$= \lim \frac{1}{n+1}[\exp((0) \frac{1}{n+2}) + \exp((0) \frac{2}{n+2}) + ... + \exp((0) \frac{n+1}{n+2})]$$
$$= \lim \frac{1}{n+1} (1)(n+1) = 1$$
This is the mgf of $Unif(0,1)$
