Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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!

share|improve this question
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,

share|improve this answer
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
@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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.