At time $0$ there is one red ball and one green ball in the urn. At time $n$, we draw out a ball chosen at random. We return it to the urn and add one more ball of the color chosen.
- Compute the probability that the first $k$ draws are red and the next $n-k$ are green.
- Generalize from this to conclude that if $X_n$ is a fraction of red balls at time $n$ then $P(X_n=k/(n+2))=1/n+1$ for $1\leq k\leq n+1$.
- Now suppose we start with $5$ green and $1$ red. Use the stopping theorem to find an upper bound on the probability of $A=\{X_n>1/2\text{ for some } n\geq0\}$
- Let $R_n=${the $n$-th ball is red} and $G_n=${the $n$-th ball is blue}. Then we know $P(R_n)=\frac{R}{R+G}=\frac{1}{2}$ and $P(G_n)=\frac{G}{G+R}=\frac{1}{2}$ (where $R$ and $G$ represents the initial number of red and green balls respectively). Then the probability that the first $k$ draws are red would be $\frac{1}{2}^k$ and the probability that the rest are green is $\frac{1}{2}^{n-k}$. Putting this together, we have $\frac{1}{2}^{k}\frac{1}{2}^{n-k}=\frac{1}{2}^n$.
- The Optional Stopping Theorem: Suppose $X_n$ is a submartingale and $N$ is a stopping time. Assume that $X_n$ is bounded or $N$ is bounded. That is, for some $M<\infty$, $|X_n|\leq M$ with probability $1$, or $N<M$ with probability $1$. Then $E[X_1]=E[X_N]$. If $N\leq n$ with probability $1$ then $E[X_1]\leq E[X_N]\leq E[X_n]$. I'm not sure where to proceed with this.
I am having trouble with this problem. I'd appreciate it if I could get some pointers on this. Thank you.
Maybe I misunderstand the question but I get the answer to the first question to be like this instead.
$\left( \frac{1}{2}\right) \left( \frac{2}{3}\right) ...\left( \frac{k}{1+k}% \right) \left( \frac{1}{2+k}\right) \left( \frac{2}{3+k}\right) ...\left( \frac{n-k}{n+1}\right) =\frac{k!(n-k)!}{(n+1)!}$