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In the Ehrenfest Chain model:

There are M balls which are divided between urn A and urn B. At each stage, if a ball is chosen, then it would be moved into a different urn.

Let $X_n$ be the # of balls in urn A, then the transistion is

$P_{i,i+1}=1-i/m$,

$P_{i,i-1}=i/m$.

It turns out that the stationary distribution is $\pi_j=$$M\choose j$$2^{-M}$.

The textbook claims this is quit intuitive. I don't understand what is "by symmetry, it is equally likely to be in either urn".

This, however, is quite intuitive, for if we focus on any one ball, it becomes quite clear that its position will be independent of the positions of the other balls (since no matter where the other M − 1 balls are, the ball under consideration at each stage will be moved with probability 1/M) and by symmetry, it is equally likely to be in either urn.

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    $\begingroup$ Have you heard of "projections" or "lumpings" of Markov chains? $\endgroup$ – Omnomnomnom Feb 8 '16 at 20:20
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This is how I would view the symmetry associated with the stationary distribution:

  • Put each ball in either urn with equal probability. Pick that ball up a random number of times unrelated to which urn it is in, and move it to the other urn.

  • The probability that that ball started in urn A was $\frac12$ and in urn B also $\frac12$. Similarly with its finishing position. The same is true for each of the other balls.

  • In addition the starting position of each ball was independent of the starting position of the other balls and the finishing position of each ball is independent of the finishing position of the other balls.

  • This symmetry and independence between the balls means that the probability that urn A finishes with $j$ out of $m$ balls stays an invariant ${m \choose j}2^{-m}$

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