Finiteness of a converging random number series

  1. 1. Imagine a positive point x not equal to zero.

    2. Consider a randomly chosen point y with distance to zero less than x.

    3. Let y=x. Repeat #2.

    4. Is the sum of the y-values finite as y approaches zero?
     
  2. jcsd
  3. For step three, it's supposed to be the other way around, right? x is supposed to equal y? Otherwise there's no reason for y to approach zero (or any other number).

    I don't know if it always converges, but on average it converges to x (by "average" I mean that for any given random y value, the average of all choices is x/2, so y, on average, equals x/2).
     
  4. Hi, I understand this as follows: denote by ran(x) a random number between 0 and x. Let x1=ran(1), and let xi=ran(xi-1) for x>1.

    Let S be the sum Ʃxi.

    As noted above, the expected value of S is 1 (does require a very minor argument). The chance of the series not converging is 0. For example the chance of S>N must be less than 1/N, for the average sum to be 1, so the chance of divergence is less than 1/N for any positive N. (A small simulation shows that the chance of the sum exceeding 7 is about 1 in 10 million)
     
    Last edited: Feb 29, 2012
  5. Thanks kindly both of you for your information, which I am attempting to cogitate.
     
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