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Distribution of Number of Pieces

  1. Jul 18, 2010 #1
    Let [tex]\{A_i\}[/tex] be independent random variables, real numbers selected uniformly from the interval (0,1-v) for some constant v, 0<v<1.
    Let [tex]B_i = \cup^i_{j=1} (A_j,A_j+v)[/tex]
    Let [tex]C_i[/tex] be the number of disconnected pieces of [tex]B_i[/tex].

    Problem: What is the distribution of [tex]C_i[/tex]? I doubt that a closed form expression is possible but it's tough to even find a computer program to calculate it except via monte carlo.
     
  2. jcsd
  3. Jul 19, 2010 #2
    I haven't worked out the details, but I think I see a way to approach this. Even if it doesn't result in a closed form formula, you might be able to work out an algorithm to compute it. Basically, the hard part is finding the volume left of an i-cube of side 1-v after several pieces have been sliced off. It should be easier for v >= 1/3, but tricky for v < 1/3. The idea is to work with the order statistics of the Aj, and find the probability that there are exactly k-1 "gaps" of the form A(m+1)-A(m) > v. In this case Ci = k. Is that making sense?

    Mind if I ask what is the application of this?
     
  4. Jul 21, 2010 #3
    Conjecture:

    [tex]P\{C_i = k\} = \frac{1}{(1-v)^i}\binom{i-1}{k-1}\sum_{j=0}^{k-1}(-1)^j\binom{k-1}{j}G(i-k+j, i; v)[/tex]

    where

    [tex]G(m,i;v) = \sum_{n=1}^{r}(-1)^{n+1}\binom{m}{n-1}(1-nv)^i[/tex]

    for

    [tex]\frac{1}{r+1} \leq v < \frac{1}{r}[/tex]
     
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