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 $N^{\theta}$.


1 Answer 1


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 $\theta$.


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