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Martingale & Probability

  1. Nov 29, 2008 #1
    1. The problem statement, all variables and given/known data

    [​IMG]

    3. The attempt at a solution
    If I define [tex]Z_n = \frac{X_n}{n+2} [/tex] with Xn the number of white balls in stage n then how can I prove that it's martingale?
     
  2. jcsd
  3. Nov 29, 2008 #2
    Re: martingale

    To start with, what is the definition of a martingale?
     
  4. Nov 30, 2008 #3
    Re: martingale

    Do you know the answer to the question or are you teaching me?

    [tex] E[ Z_{n+1}| Z_1 ,Z_2,...,Z_n] =Z_n [/tex]
     
  5. Nov 30, 2008 #4
    Re: martingale

    1) After the nth replication, there are [tex]Z_n(n+2)[/tex] white balls and [tex](1-Z_n)(n+2)[/tex] black balls in the urn.

    2) In the (n+1)th replication, there will either be a white ball or a black ball added into the urn. For each of these cases, what will the new proportion of white balls in the urn be and what are the associated probabilities? Use the definition of expectation to calculate [tex]E \ [\ Z_{n+1}\ | \ Z_{1} , \ Z_{2} \, ..., \ Z_{n} \ ][/tex]
     
    Last edited: Nov 30, 2008
  6. Nov 30, 2008 #5
    Re: martingale

    I thought that at each replication two white balls are added or removed. But suppose you're right then I get:

    [tex] Z_{n+1}= \frac{Z_n(n+2)+1}{n+3} [/tex]

    [tex] Z_{n+1} = \frac{Z_n(n+2)-1}{n+3} [/tex]

    with probabilities 1/2 for each?
     
  7. Dec 1, 2008 #6
    Re: martingale

    At each replication, a ball is drawn from the urn. The colour of the ball is noted, then that ball, together with a new ball of the same colour, are placed into the urn. Note then that after each replication, the number of white balls in the urn either stays the same or increases by one.

    Now, what is the probability of drawing a white ball (similarly, a black ball) from the urn during the (n+1)th replication? (Not necessarily half!)

    If a white ball is drawn during the (n+1)th replication, then that white ball will be put back into the urn and a new white ball added. So, what will the proportion of white balls be after the (n+1)th replication? Similarly, if a black ball is drawn during the (n+1)th replication, what will the new proportion of white balls in the urn be?
     
    Last edited: Dec 1, 2008
  8. Dec 1, 2008 #7
    Re: martingale

    So

    [tex]E[Z_{n+1}|Z_1,...,Z_n] = Z_n \cdot \frac{Z_n (n+2) +1}{n+3} +(1-Z_n) \cdot \frac{Z_n (n+2) +1}{n+3} =Z_n[/tex]

    Ok thanks. The next question is:

    if T is the first stage at which a black ball is drawn prove that then:[tex] E \left[ \frac{1}{T+2} \right] - \frac{1}{4} [/tex] via the martingale stopping theorem.

    [​IMG]


    If X_N is the number of white balls in stage n then:

    [tex] E[Z_T] = E[Z_1] = \frac{1}{2} = E \left[ \frac{X_T}{T+2} \right] = E[X_T] E \left[ \frac{1}{T+2} \right] [/tex]

    Is this correct?
     
    Last edited: Dec 1, 2008
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