Non-symmetric Simple Random Walks along Orbitsof Ergodic Automorphisms.V. Yu. Kaloshin∗, Ya. G. Sinai∗†Consider an ergodic measure-preserving automorphism T of probability space(M, M, µ). Having a measurable function p, 0 < p(x) < 1 a.e., we can constructMarkov chain whose phase space is M and a moving point x ∈ M jumps from x toT x with probability p(x) and to T−1x with probability 1 −p(x). Such Markov chainsare called simple random walks along orbits of T . If x0is an initial position then theposition xnat any moment of time n can be written as xn= Tbnx0. For the sequencebnwe have b0= 0, bn+1− bn= ±1, i.e. bnis a simple random walk on Z1, for whichP {bn+1= k + 1|bn= k} = p(Tkx), P {bn+1= k − 1|bn= k} = 1 − p(Tkx). The firstquestion which arises in this situation is whether this Markov chain has a stationarymeasure absolutely continuous wrt toµ. The densityrof this measure must satisfythe equationr(x) = r(T−1x)p(T−1x) + 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). Thenq(x)p(x)= q(T−1x) + q(T x )1 − p(Tx)p(T x)orq(x) −q(T x)1 − p(T x)p(T x)= q(T−1x) − q(x)1 − p(x)p(x).This shows that q(x) −q(T x)1−p(T x)p(T x)is invariant under T and must be a constant a.e.since T is ergodic:q(x) − q(T x)1 − p(T x)p(T x)= C∗Mathematics Department of Princeton University.†Landau Institute of Theoretical Physics, Moscow1orq(x) = q(Tx)1 − p(T x)p(T x)+ C, C = const (2)Definition 1. Simple random walks is called symmetric ifRln p(x)dµ(x) =Rln(1 − p(x))dµ(x). Otherwise it is called non-symmetric.Let us show that in the non-symmetric case (2) and therefore (1) always have asolution. Assume for the definiteness thatZln(1 − p(x))dµ(x) <Zln p(x)dµ(x). (3)It is easy to see that the series below converges for a.e. xq(x) = C + C1 − p(T x)p(T x)+ C1 − p(T x)p(T x)·1 − p(T2x)p(T2x)+ ··· (4)Indeed, the product of k terms can be rewritten askYi=1(1 − p(Tix))p(Tix)= exp"kXi=1ln(1 − p(Tix))p(Tix)#.The last sum behaves a.e. as −ck as k → ∞ in a view of ergodic theorem where c =Rlnp(x)1−p(x)dµ(x) > 0. The caseRln(1 − p(x))dµ >Rln p(x)dµ(x) can be consideredin the same way.If the random walk is symmetric then C = 0. Indeed, we must haveZln q(x)dµ(x) =Zlnµq(T x) ·1 − p(T x)p(T x)+ C¶dµ(x) (5)The right-hand side is a monotone function of C. Since (5) holds for C = 0 weget the result. In this case (2) is reduced toq(x)q(T x)=1 − p(T x)p(T x)orln q(x) − ln q(Tx) = ln(1 − p(T x)) − ln p(T x) (6)2The last equation has a solution for generic p only if T is a shift on a com-pact 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 withM = T ordwas considered in [S1]. For other transformations having a continuouscomponent in the spectrum the equation (6) has no solution for generic p.Fix x ∈ M and consider the probabilities p(+)2n(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 wecall such walks positive excursions. One can consider also negative excursions andcorresponding probabilities p(−)2n(x). It is easy to see thatp(+)2(x) = p(x)(1 − p(T x)), (7)p(+)2n(x) = p(x)Xs≥1Xn1+n2+···+ns=n−1ni>0,1≤i≤sp(+)2n1(T x) · ··· · p(+)2ns(T x)(1 − p(T x)) (8)For the generating ϕ(+)(x, Θ) =Pn≥1p(+)2n(x)Θ2n, |Θ| ≤ 1, we derive from (7) and(8) the equation (see also [S1])ϕ(+)(x, Θ) =p(x)(1 − p(T x))Θ21 − ϕ(+)(T x, Θ). (9)Lemma 1. If (3) is valid then ϕ(+)(x, 1) = p(x)Λ(x)Λ(x)+1where Λ(x) =∞Pk=1kQi=11−p(Tix)p(Tix).The last series converges a.e.Proof. For ϕ(+)(x, 1) we have from (9)ϕ(+)(x, 1) =p(x)(1 − p(T x))1 − ϕ(+)(T x, 1). (10)Then for ψ(+)(x, 1) = ϕ(+)(x, 1)/p(x) ≤ 1 we haveψ(+)(x, 1) =1−p(T x)p(T x)1p(T x)− ψ(+)(T x, 1)(11)3Since ψ(+)(y, 1) > 0 one can write ψ(+)¡Tkx, 1¢=Λ(Tkx)Λ(Tkx)+1. Then (11) givesΛ(x)Λ(x) + 1=1−p(T x)p(T x)(Λ(T x) + 1)1−p(T x)p(T x)Λ(T x) +1p(T x),orΛ(x) =1 − p(T x)p(T x)+1 − p(T x)p(T x)Λ(T x) (12)This shows thatΛ(x) =∞Xk=1kYi=11 − p(Tix)p(Tix)The last series converges a.e. in view of (3) and gives the result, QED. Noticethat Λ(x) = C−1q(x), where q(x) is defined by formula (4).If we calculate in the same way the distribution of negative Levy excursions weshall easily get for the corresponding generating functions that ϕ(−)(x, 1) = 1 −p(x).Thus in the non-symmetric case (3) the probability of the negative excursion is1−p(x), the probability of the positive excursion is p(x)Λ(x)Λ(x)+1and the probability thatthe moving point jumps from x to T x and does not return back to x is p(x)1Λ(x)+1.Let us calculate the expectation of the length of the positive excursion m(+)1(x) =Pn≥12np(+)2n(x). We have by differentiating (9)m(+)1(x) =∂ϕ(+)(x, 1)∂Θ=2p(x)(1 − p(T x))1 − ϕ(+)(T x, 1)+p(x)(1 − p(T x))(1 − ϕ(+)(T x, 1))2m(+)1(T x) (13)The ratiop(x)(1−p(T x))(1−ϕ(+)(T x,1))2equalsp(x)(1 − p(T x))(Λ(T x) + 1)2((1 − p(T x))Λ(T x) + 1)2=p(x)p(T x)·1 − p(T x)p(T x)·(Λ(T x) + 1)2³1−p(T x)p(T x)· Λ(T x) +1−p(T x)p(T x)+ 1´2==p(x)p(T x)·(Λ(T x) + 1)2(Λ(x) + 1)2·1 − p(T x)p(T x).4The last step was based on (12). If we put h(x) =2p(x)(1−p(T x))1−ϕ(+)(T x,1)then we can writem(+)1(x) in the formm(+)1(x) = h(x) +p(x)(Λ(x) + 1)2·∞Xk=1h(Tkx)kYi=11 − p(Tix)p(Tix)·(Λ(Tkx) + 1)2p(Tk+1x)(14)In view of our assumption concerning p the function h is bounded from above andbelow. We needAssumption 1. The function Λ ∈ L1(M, M, µ).Under this assumption the density r(x) ∈ L1(M, M, µ) (see (1) and (4)).In a similar way one can write down the expressions for the derivatives m(+)r(x) =drϕ(+)(x,1)dΘrwhich are finite a.e. if the Assumption 1 holds.The same analysis can be applied to ϕ(−)(x, 1) and its derivatives. One canshow that if (3) holds then the probabilities p(−)2n(x) decay exponentially. The sumϕ(x, Θ) = ϕ(+)(x, Θ) + ϕ(−)(x, Θ) is generating function for excursions, both negativeand positive.Consider the probability distribution P(n)xconcentrated on the setS|k|≤nTkx whereP(n)x(k) is the probability of random walks starting from x = x0for which xn= Tkx0or b(n) = k. For k > 0 such random walks have the following structure. First itmakes some numbers ν0of positive or negative excursions coming back to x0. Thenit jumps to T