Polya's urn as a counterexample for the Kolmogorov 0-1 law

Solution 1:

You need to use a martingale structure to show the almost sure convergence. Specifically, $$ \mathsf{E}\!\left[\frac{B_n}{n+2}\mid B_{n-1}\right]=\frac{B_{n-1}}{n+2}\left(1-\frac{B_{n-1}}{n+1}\right)+\frac{B_{n-1}+1}{n+2}\frac{B_{n-1}}{n+1}=\frac{B_{n-1}}{n+1}, $$ where $B_n$ is the number of black balls after $n$ steps ($X_n$ in your notation). That is, $B_n/(n+2)$ is a bounded martingale (w.r.t. the natural filtration), and thus, it converges almost surely.

As for the asymptotic distribution note that if $\hat{B}_n$ denote the number of draws of black balls in $n$ trials, then $B_n=\hat{B}_n+1$, and $$ \mathsf{P}(\hat{B}_n=k)=\frac{k!(n-k)!}{(n+1)!}\times \binom{n}{k}=\frac{1}{n}. $$ Thus, for $x\in [0,1]$, $$ \mathsf{P}(\hat{B}_n\le nx)=\sum_{k=0}^{\lfloor nx\rfloor}\frac{1}{n}=\frac{\lfloor nx\rfloor}{n}\to x $$ as $n\to\infty$. Finally, $$ \frac{B_n}{n+2}=\frac{n}{n+2}\times \frac{\hat{B}_n}{n}+\frac{1}{n+2}\xrightarrow{d} \text{U}[0,1]. $$