0 0 11 views

**Unformatted text preview:**

Non symmetric Simple Random Walks along Orbits of Ergodic Automorphisms V Yu Kaloshin Ya G Sinai Consider an ergodic measure preserving automorphism T of probability space M M Having a measurable function p 0 p x 1 a e we can construct Markov chain whose phase space is M and a moving point x M jumps from x to T x with probability p x and to T 1 x with probability 1 p x Such Markov chains are called simple random walks along orbits of T If x0 is an initial position then the position xn at any moment of time n can be written as xn T bn x0 For the sequence bn we have b0 0 bn 1 bn 1 i e bn is a simple random walk on Z1 for which P bn 1 k 1 bn k p T k x P bn 1 k 1 bn k 1 p T k x The first question which arises in this situation is whether this Markov chain has a stationary measure absolutely continuous wrt to The density r of this measure must satisfy the equation r x r T 1 x p T 1 x r T x 1 p T x 1 Assume that 0 c1 const p x c2 const 1 a e and put q x r x p x Then q x 1 p T x q T 1 x q T x p x p T x or q x q T x 1 p T x 1 p x q T 1 x q x p T x p x x This shows that q x q T x 1 p T is invariant under T and must be a constant a e p T x since T is ergodic q x q T x 1 p T x C p T x Mathematics Department of Princeton University Landau Institute of Theoretical Physics Moscow 1 or q x q T x 1 p T x C p T x C const 2 R Definition 1 Simple random walks is called symmetric if ln p x d x R ln 1 p x d x Otherwise it is called non symmetric Let us show that in the non symmetric case 2 and therefore 1 always have a solution Assume for the definiteness that Z Z ln 1 p x d x ln p x d x 3 It is easy to see that the series below converges for a e x 1 p T x 1 p T x 1 p T 2 x q x C C C p T x p T x p T 2 x 4 Indeed the product of k terms can be rewritten as k k Y X 1 p T i x 1 p T i x exp ln i x i x p T p T i 1 i 1 The last sum behaves a e as ck R R as k in a view R of ergodic theorem where c p x ln 1 p x d x 0 The case ln 1 p x d ln p x d x can be considered in the same way If the random walk is symmetric then C 0 Indeed we must have Z Z ln q x d x 1 p T x ln q T x C d x p T x 5 The right hand side is a monotone function of C Since 5 holds for C 0 we get the result In this case 2 is reduced to q x 1 p T x q T x p T x or ln q x ln q T x ln 1 p T x ln p T x 2 6 The last equation has a solution for generic p only if T is a shift on a compact abelian group i e it is an automorphism with pure point spectrum S2 Lect 4 This follows easily from the spectral theory of dynamical systems This case with M T ord was considered in S1 For other transformations having a continuous component in the spectrum the equation 6 has no solution for generic p Fix x M and consider the probabilities p2n x of random walks b m 0 m 2n see above for which b m 0 for 0 m 2n b 2n 0 Following P Levy we call such walks positive excursions One can consider also negative excursions and corresponding probabilities p2n x It is easy to see that p2 x p x 1 p T x 7 X p2n x p x s 1 X p2n1 T x p2ns T x 1 p T x 8 n1 n2 ns n 1 ni 0 1 i s P For the generating x n 1 p2n x 2n 1 we derive from 7 and 8 the equation see also S1 x p x 1 p T x 2 1 T x x Lemma 1 If 3 is valid then x 1 p x x 1 where x The last series converges a e Proof For x 1 we have from 9 x 1 p x 1 p T x 1 T x 1 9 Q k P k 1 i 1 1 p T i x p T i x 10 Then for x 1 x 1 p x 1 we have x 1 1 p T x 1 p T x p T x T x 1 3 11 Since y 1 0 one can write T k x 1 x x 1 T k x T k x 1 Then 11 gives 1 p T x T x 1 p T x 1 p T x 1 T x p T x p T x or x 1 p T x 1 p T x T x p T x p T x 12 This shows that x Y k X 1 p T i x k 1 i 1 p T i x The last series converges a e in view of 3 and gives the result QED Notice that x C 1 q x where q x is defined by formula 4 If we calculate in the same way the distribution of negative Levy excursions we shall easily get for the corresponding generating functions that x 1 1 p x Thus in the non symmetric case 3 the probability of the negative excursion is x 1 p x the probability of the positive excursion is p x x 1 and the probability that 1 the moving point jumps from x to T x and does not return back to x is p x x 1 Let us calculate the expectation of the length of the positive excursion m1 x P 2np2n x We have by differentiating 9 n 1 m1 x The ratio x 1 2p x 1 p T x p x 1 p T x m T x 1 T x 1 1 T x 1 2 1 p x 1 p T x 1 T x 1 2 13 equals p x 1 p T x T x 1 2 p x 1 p T …