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I have another problem from Koralov-Sinai that I don't know how to do and I would appreciate help please. It goes like this:

5.16. Consider a Markov chain whose state space is the unit circle. Let the density of the transition function $P(x,dy)$ be given by $p(x,y) = \frac{1}{2 \epsilon}$ if the angle $(y,x) < \epsilon$ and $0$ otherwise. Find the stationary distribution.

Thank you all!

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The definite article is a giveaway -- if there's only one stationary distribution and the transition function exhibits rotational symmetry, what would you expect regarding the symmetry properties of the stationary distribution? –  joriki Nov 6 '12 at 2:56
    
can you post a complete answer please? I'm struggling with the concepts... –  Dquik Nov 6 '12 at 3:33
    
I'm assuming you mean we get a reversible distribution... but I don't see how to prove it –  Dquik Nov 6 '12 at 16:55
    
I'm still lost with this; can anyone help? –  Dquik Nov 7 '12 at 5:32

1 Answer 1

The rotationally invariant density $p(\phi)=(2\pi)^{-1}$ for the angle $\phi$ is stationary because the transition density exhibits rotational symmetry,

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I'm following Koralov and Sinai and there they don't talk about general Markov chains very clearly, so that's why I'm finding difficuties following. Again, would you be kind and write me a more detailed answer so that I could understand what you're saying? It seems like everything should be pretty easy - so I just want to understand the theory –  Dquik Nov 7 '12 at 5:51
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@Dquik: Sorry, I thought I'd spelled it out by writing out the density -- could you be more specific about what aspect of this you don't understand? –  joriki Nov 7 '12 at 5:55
    
why $2/\pi$ and not $1 / 2\pi$? –  Dquik Nov 7 '12 at 9:33
    
@Dquik: Sorry about that; you're right of course. Fixed. I hope that implies you've gained some understanding of the answer? :-) –  joriki Nov 7 '12 at 12:19

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