I have a problem with the following exercise:

We have an urn with one black and one white ball. At time 1 you take one of the balls from the urn randomly. Then you take this ball and replace it by two balls of the same colour. For example you take one white ball, then you replace it by two white balls and you put two white balls back in the urn. Then you continue doing this procedure.

$W_n$ denotes the number of white balls at time $n$, then define $X_n=\frac{W_n}{n+1}$. This $X_n$ should converge to something , lets call it $S$. I want to compute the expectation and the variance from $S$.

My first thought was to get the expectation of $W_n$. Therefore I need some formula for $W_n$. $W_2=1$ with prob. $1-p$ and $W_2=2$ with prob $p$. Probability $p$ is the probability that you take a white ball. Continuing leads me to $W_n=1$ with prob. $1-p$, $W_n=2$ with prob $p$....$W_n=n$ with prob $p^{n-1}$, therefore $E[W_n]=1-p+2p+3p^2+...+np^{n-1}$ How can I proceed?


1 Answer 1


Since the problem is symmetric with respect to white balls and black balls, the $E[W_n]= \frac{n+1}{2}$

  • $\begingroup$ May you expand the argument a little bit more, I do not see that quaility at the moment? $\endgroup$ Commented Nov 7, 2013 at 16:12
  • $\begingroup$ I do see the equality when the probability is 1/2 for getting a black or white ball but what happens if the probability is different? $\endgroup$ Commented Nov 7, 2013 at 16:23
  • $\begingroup$ Ulrich, I think you should give Chaturi credit for her reply! $\endgroup$
    – DannyDan
    Commented Nov 7, 2013 at 16:25
  • $\begingroup$ Thanks @DannyDan! @Ulrich I think that when the problem says "randomly" it's assumed that each ball has an equal probability of being chosen. $\endgroup$ Commented Nov 7, 2013 at 16:57
  • $\begingroup$ Thanks, I thought it might be more complicated. So I get for $E[X_n]=1/2$, this converges to 1/2 beause its simply constant, Variance is $0$ than, correct? $\endgroup$ Commented Nov 7, 2013 at 19:01

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