3

Consider a simple formulation for the Polya urn model. An urn contains two balls at time 0, one is white and the other is black. At time $n\in\mathbb{N}$, one of the balls is chosen uniformly at random. Another ball of the same color is put into the urn. Let $X_n$ be the number of black balls at time $n$. My task is to:

  1. Give a formal description of the process.
  2. Show that $\lim_{n\to\infty}\frac{X_n}{n+2}$ exists almost surely and is uniformly random in $(0,1)$.
  3. Show that $A:=\left\{\lim_{n\to\infty}\frac{X_n}{n+2}>\frac12\right\}\in\mathcal{T}$, i.e. $A$ is a tail event. Moreover, show that $P(A)=\frac12$.

Point 3 is easy to show in my opinion and the last part follows simply by the uniformity of the random variable. I am unsure how to prove 2 and also what is meant exactly by "formal description of the process". Is it enough to give the conditional probabilities of the form $P(X_{n+1}=k|X_n=l)$? I understand the proof of almost sure convergence follows easily from the martingale convergence theorem. Is there any other way to prove it without using martingales?

Irene
  • 208
  • https://math.stackexchange.com/questions/1125320/polyas-urn-model-limit-distribution –  Nov 22 '21 at 20:56
  • Thank you for the comment and also for the answer. I am specifically interested if there is a proof of the a.s. convergence without using martingale theory. – Irene Nov 23 '21 at 09:51

1 Answers1

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, $$ \frac{B_n}{n+2}=\frac{n}{n+2}\times \frac{\hat{B}_n}{n}+\frac{1}{n+2}\xrightarrow{d} \text{U}[0,1]. $$