# Distribution and Martingale in Polya's urn

This is exercise 10.1 from Probability with Martingale by David Williams:

At time $$0,$$ an urn contains $$1$$ black ball and $$1$$ white ball. At each time $$1,2,3,...,$$ a ball is chosen at random from the urn and is replaced together with a new ball of the same color. Just after time $$n$$, there are therefore $$n+2$$ balls in the urn, of which $$B_n+1$$ are black, where $$B_n$$ is the number of black balls chosen by time $$n.$$

Let $$M_n=\frac{(B_n+1)}{n+2},$$ the proportion of black balls in the urn just after time $$n.$$ Prove that (relative to a natural filtration) $$M$$ is martingale.

Prove that $$P(B_n=k)=(n+1)^{-1}\text{ for } 0\le k\le n.$$ What is the distribution of $$\theta \text{ where } \theta =\lim M_n.$$

Prove that for $$0<\theta<1,$$ $$N^{\theta}_n=\frac{(n+1)!}{B_n!(n-B_n)!}\theta^{B_n}(1-\theta)^{n-B_n}$$ defines a martingale $$N^{\theta}$$

Now I think I have proven the first two statements:

First, I chose the natural filtration to be $${B_t+1},$$ then $$M$$ is adapted. Then for $$s=t+1$$ $$E(M_s|B_t+1)=\frac{E(B_s+1|B_t)}{(s+2)}\\ E(B_s+1|B_t+1)=\frac{B_t+1}{t+2}(B_t+2)+\frac{(t+1-B_t)}{t+2}(B_t+1)\\ =\frac{3(B_t+1)+t(B_t+1)}{t+2}\\ =\frac{(t+3)(B_t+1)}{(t+2)}$$ so we have: $$E(M_s|B_t)=M_t\text{ for } s\ge t$$ by induction.

To prove $$P(B_n=k)$$ for $$0\le k\le n.$$ my approach is using conditional probability, then I found there are $$\binom{n}{k}$$ ways of choosing $$k$$ black balls in $$n$$ drawings. And the probability for each of them to occur is:$$\frac{k! (n-k)!}{(n+1)!}$$

so we have $$P(B_n=k)=(n+1)^{-1}.$$

This is so far I got, just need some idea of figuring out the distribution of $$\theta$$ and some hints on the filtration of $$N^{\theta}$$.

The weak limit of the uniform distribution over $$\{\frac1{n+1},\frac2{n+1}\ldots\frac{n+1}{n+2} \}$$ is the uniform distribution over $$[0,1]$$, so the latter is the distribution of $$\theta$$.