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Math Help - Using a series to model an infinite set of consecutive dependent probabilities

  1. #1
    Oct 2012

    Using a series to model an infinite set of consecutive dependent probabilities

    Hello everyone.

    I am currently trying to model a simple system in programming using a geometric series. The problem at hand is:

    For every one unit of "x" I receive, I have a 30% chance of obtaining a second "x". This second "x" has in turn a 30% chance to spawn a third x, which can further propagate to 4, 5, and 6 to infinity, each with a 30% chance of initiating the subsequent term. If at any point the probability condition is not satisfied, the propagation stops with that term.

    In order to determine how many "x" I end up with, I wanted to model the system as the geometric series:

    \sum\limits_{n = 0}^{\infty} {(0.3)^{n}x }

    Where x is however many units I start with.

    This converges to 1.428

    My questions are:

    1. I'm having trouble interpreting the meaning of the converging value. Is it accurate to say the "true" probability of spawning an additional x is 42.8%, rather than 30%, because the series converges to that? If not, what is the significance of this converging value? My qualm here is that the n = 1 term is still dependent on a 30% chance to occur, and not a 42.8% chance, which leads me to my next question.

    2. Since each additional term in the series is not arbitrarily present but dependent on a previous term occurring, I don't know if it is correct to model it as a geometric series. Factoring in this probability has given me trouble. In light of this, I have also considered the following:

    Assuming I have ten units, I can expand the series into the following terms:

    1st term: 10 units
    2nd term: 0.3 * 10 units
    3rd term: 0.3 * (0.3 *10 units) =  {(0.3)^{2}}*10

    Since the 3rd term is dependent on the probability of the 2nd term occurring, I could rewrite the 3rd term as 3rd'' term:

    3rd'' term = 3rd term * probability of 2nd term occurring = [(0.3)^{2}*10]*0.3

    The corresponding series would be:

    x + \sum\limits_{n = 1}^{\infty} {(0.3)^{n}x\,(0.3)^{n-1} } = x + \sum\limits_{n = 1}^{\infty} {(0.3)^{2n-1}x } = \sum\limits_{n = 0}^{\infty} {(0.3)^{2n}x }

    Using the same derivation for convergance of a geometric series, I got that the above series converges to \frac{a}{1 - r^{2}} = \frac{1}{1 - 0.3^{2}} = 1.099

    I have no idea what the significance of this convergance would be.

    Is this going in the right direction? Thank you for any advice/help!
    Last edited by blaisem; February 18th 2013 at 06:13 AM.
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