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Poisson Martingales and Gambler's Ruin

  1. Feb 7, 2010 #1
    I posted this in the HW help section, but I had no responses. I figure that this place may be better to answer this question. If this is against the rules or anything, mods please remove it!

    1. The problem statement, all variables and given/known data
    Find the probability of an outcome of Gambler's Ruin using a Poisson martingale.

    Let m be the parameter of a Poisson Process (ie the lambda)
    Let N(t) be a continuous Poisson process at time t>=0
    Let M(t) = N(t) - mt

    2. Relevant equations
    Now, define s = inf{t | M(t) <= -a or M(t) >= b} (ie the stopping time)
    Let r = P(M(s)>=b)

    3. The attempt at a solution
    I have shown that M(t) is a martingale with E[M(t)] = 0 for all t
    Using the martingale principle, I have:
    E[M(t)] = 0 = a*(1-r) + b*r
    r = a/(a+b).

    This is the solution for discrete gambler's ruin. However, my professor says that this value should be an upper bound for r, and the lower bound should be a/(a+b+1) <= r. I'm not sure how to get this.

    My guess should be that it has something to do with the Poisson distribution, and the only difference is that b is bigger by 1. But, if there is a Poisson jump at some episilon of time, its just as if we have a+1 dollars, and our opponent has b-1 dollars; hence I should see a change in the numerator of the bound (and not the denominator). Is there any intuition for this? I'm not necessarily just looking for the answer (or equation), but more of understanding why there should be two bounds to my probability.

  2. jcsd
  3. Feb 8, 2010 #2


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    Can you describe parameters a and b?
  4. Feb 9, 2010 #3

    I think of it like this:

    You start out with a dollars, and your opponent starts out with b dollars. The variable M(t) is the sum of all your winnings and losses. You play until one person beats out the other, so you play until M(t) is either -a (where you are bust) or +b (where you bust your opponent). You want to find the probability that you win.
  5. Feb 9, 2010 #4


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    If I'm starting out with $a, and M is a martingale, isn't E[M(t)] = E[M(0)] = a for any t > 0 ?
  6. Feb 9, 2010 #5
    M(t) represents the net change from your starting amount, so E[(M(t)] = 0.
  7. Feb 9, 2010 #6


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    You're making an arithmetical error: 0 = a*(1 - r) + b*r implies r = a/(a - b), which raises a further issue: a = b implies r = a/0.

    [I am guessing you had a typo in the equation, you meant 0 = -a*(1-r) + b*r. That makes sense. See this, where h = a in your problem and N = a + b in your problem. With [itex]\alpha[/itex] = 1/2, you do obtain the symmetrical solution h/N = a/(a+b). However, you'd get a different probability if [itex]\alpha\ne[/itex] 1/2.]
    Last edited: Feb 9, 2010
  8. Feb 9, 2010 #7
    Yeah, sorry, that was a typo; what you have written above is correct.
  9. Feb 9, 2010 #8


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    See my edits.
  10. Feb 10, 2010 #9
    First you'll need the optional stopping theorem to get E[M(s)]=M(0)=0.

    Also you know the process first crosses the upper barrier via a jump, or the lower barrier via a non-jump, i.e. M(s)<=-a -> M(s)=a, and M(s)>=b -> M(s)<b+1.

    Apply those to E[M(s)] to get two inequalities involving r.

  11. Feb 10, 2010 #10
    I figured it had to do with a Poisson jump. Thanks!
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