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Bose-Einstein physics question

  1. Jan 22, 2008 #1
    Suppose that [tex] M [/tex] molecules are distributed among two urns; and at each time point one of the molecules it chosen at random, removed from its urn, and placed in the other one. So this is a time-reversible Markov process right?

    So [tex] P_{i,i+1} = \frac{M-i}{M} [/tex]. What do the limiting probabilities mean in words?

    Like [tex] \pi_0 = \left[ 1 + \sum_{j=1}^{M} \frac{(M-j+1) \cdots (M-1)M}{j(j-1) \cdots 1} \right ]^{-1} [/tex]

    [tex] = \left [\sum_{j=0}^{M} \binom{M}{j} \right]^{-1} = \left(\frac{1}{2} \right)^{M} [/tex]


    and [tex] \pi_i = \binom{M}{i} \left(\frac{1}{2} \right)^{M}, \ i = 0,1, \ldots, M [/tex].

    What do these really signify?

    Source: Introduction to Probability Models by Sheldon Ross

    Thanks
     
  2. jcsd
  3. Jan 22, 2008 #2

    EnumaElish

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    Limiting probability of a state-transition probability is a probability value that the state-transition probability converges to, as the number of steps approaches infinity.

    See http://en.wikipedia.org/wiki/Markov_chain
     
  4. Jan 23, 2008 #3

    ssd

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    I am not sure though, but the significance appears as-
    The stated scheme in the long run is equivalent to distribute M distinguishable particles
    in two urns where each particle has probability 1/2 to go into an urn. Pai(i) is the probability that one specified urn will contain i particles. That is, in the stated scheme, whatever be the initial distribution of particles in the urns, in a long run they will be distributed as in case of a Binomial distribution.
     
  5. Jan 23, 2008 #4

    EnumaElish

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    Usually a transition probability is expressed as p(i,j) where i and j are the two states. p(i,j) = Prob{X(n+1) = i given X(n) = j}.
     
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