Lecture 8. Here we show the Asymptotics for Green s funcion and application for exiting annuli in dimension d 3 Reminders: P(S n = x) 2 ( d

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1 Random Walks and Brownian Motion Tel Aviv University Spring 0 Lecture date: Apr 9, 0 Lecture 8 Instructor: Ron Peled Scribe: Uri Grupel In this lecture we compute asumptotics estimates for the Green s function and apply it to the exiting annuli problem. Also we define Capacity, Polar set Prove the B-P-P theorem of Martin s Capacity for Markov chains[3] and use apply it on the intersection of RW problem. Tags for today s lecture: Green s function, exiting annuli, Capacity, Polar set, Martin Capacity, intersection of RW. Asymptotics for Green s funcion Here we show the Asymptotics for Green s funcion and application for exiting annuli in dimension d 3 Reminders: Local Centeral Limit Theorem (LCLT): sup P(S n = x) x Z d x and n of same parity ( )d d πn e d x n = O(n d+ as n We ll denote p(n,x) = ( )d d πn e d x n and E(n,x) = P(S n = x) ( d Large Deviation: C d,c d > 0 such that r,n P( S n r n) C d e c dr Green s Function: In dimension d 3 the Green s functions is defined by G(x,y) = G(x y) = E x (number of vists to y) = P x (S n = y) n= ) )d πn e d x n Since in d =, this sum is always In dimesion d =, the ptential kernel plays a similar role a(x) = P(S n = 0) P(S n = x) = G(0) G(x) n= The use of quation marks is due to the fact that in dimension d =, the sum in the Green s function definition would be 8-

2 Theorem For d 3, G(x) a d x d as x where a d = d Γ( d ) π d = (d )w d and w d = vol(b d ) the volume of the unit ball in R d. More precisley α < d, G(x) = a d x d +o( x α ) as x ProofFix x 0 of even parity, G(x) = E(number of visits to x) = n= P(S n = x). Note first p(n,x) = n= n= ( )d d e d x πn 4n = = d Γ ( d )π d x d +O( x d ) 0 ( )d d e d x 4t +O( x d ) = πt Where O( x d ) is with respect to x. Note that the estimate in is according to the order of the first term for x n Now G(x) = n= p(n,x) + E(n,x) we need only to show that for any α < d n= E(n,x) = o( x α ), to see this we pick a threshold n 0 close to x E.g, n 0 = c x, log x then write E(n,x) = n= n 0 E(n, x) n= By large deviation and choice of c =O( x d ) For x of odd parity note G(x) is harmonic for x 0, so G(x) = e i are the unit vectors) giving the result. + n 0 + E(n, x) By LCLT =O(n d 0 )=o( x α ) for any α<d d i= G(x+e i) (where Remark. In the continuem, for Brownien Motion in R d (d 3) the Green function is exactly a d x d, this function is harmonic except at x = 0 and has laplacian δ 0 at 0.. The error in the previus theorem is even O( x d ) Application (exiting annuli or quantitative transience) In d 3 denote A n,m = {x Z d ; n < x < m}, τ n,m = min{j; S j / A n,m }, then for x A n,m P x ( Sτ n) = x d m d +O(n d ) n d m d In particular taking m, we get P x ( j; S j n) = x d n d +o(n ) ProofSince G(x) is harmonic except in x = 0, M j := G(S j τ) is a bounded martingale so G(x) = E x M j = optional sampeling x d d 8-

3 = P( S τ n)e x (G(S τ ) S τ n) +( P( S τ n))e x (G(S τ ) S τ m) n d m d Noting that G(x) = x d +O( x d ) (weaker than the error in the theorem). By isolation P( S τ n) we get the result. Similarly one can prove Proposition 3 In d 3 letting C n = {x Z d ; x < n} and τ = τ n = min{j; S j / C n ors j = 0} for x C n as x P(S τ = 0) = a d G(0) ( x d n d )+O( x d ) Reminder: (Second type of Green s function) For any A Z d, G A (x,y) = E x (number of visits to y before exiting A) Question 4 How do we estimate G a (x,y)? Definition 5 For A Z d we define the boundry of A as A = {x Z d ; x / A, x y for some y A} Proposition 6 For finite A Z d x,z A G A (x,z) = G(x,z) y A where H A (x,y) = P x (y is first exit point of A) ProofLet τ = min{j; S j / A} then τ G A (x,z) = E x j=0 Sj =z H A (x,y)g(y,z) = E x Sj =z j=0 j=τ Sj =z The first term is G(x,z), and the second is the same as in the proposition. Combining proposition 3 and proposition 6 we have the following proposiotion Proposition 7 G Cn (x,0) = a d ( x d n d )+O( x d ) ProofExercise. Analogous results for dimension d =, [][] 8-3

4 In d = : a(x) = x. In d = : k such that α < a(x) = π log x +k +o( x α ) With k = γ π log where γ = lim n n j= j logn the Eulrt constant. G A (x,z) = y A H A(x,y)a(y x) a(z x) for some finite A and x,y A For d = In the annulus A n,m, x A n,m (Quantitative recurrence) For d = x C n, α < P x ( S τ n) = logm log x +O(n ) logm logn G Cn (x,0) = π (logn log x )+o( x α )+O(n ) Capacity In this section we define Capacity and Polar sets, intoduce Kakutanis theorem charactrizing polar sets, and prove the Benjamini-Pemantle-Peres theorem which gives a quantitive connection between the Capacity of a set and a Markov Chain on the set. In the end of the section we see some application to the intersection of random walks. This section follow [3] Definition 8 Given a measure space (,F), a measurable function F : [0, ) (kernel) and a finite measure µ of, the underlinef - energy of µ with respect to F is I F (µ) = F(x, y)dµ(x)dµ(y) Remark 9 It is useful to think of R 3, µ charge density of a certain material and F(x,y) = x y Definition 0 The capacity of with respect to F is cap F () = ( ) inf I F(µ) µ Where µ is a probability measure on, and =

5 Remark Capacity is monotone increasing in the set. It is a measure of the size of λ with respect to F. Remark For is will be a countable with F = {all subsets of } Sometimes we ll also mention R d with F Borel σ-field. Kakutani s theorem (944): Definition 3 A Borel A R d is called Polar if P x ( t > 0,B(t) A) = 0 x R d for a Brownian Motion B(t) Theorem 4 (Kakutani) A Borel A R d is Polar if and only if cap F (A) = 0 for F(x,y) = { log ( x y ) d = x y d d 3 Now let {X n } be a Markov chain on a countable set Y, with transition probability p(x,y). Let G(x,y) = E x (# visits to y) = p (n) (x,y) n=0 Theorem 5 (Benjamini, Pemantle, Peres) If {X n } is a transient chain (any state is visited onlt finitly many times a.s) then for all starting point ρ Y and any Y cap K() P ρ ( n; X n ) cap K () where K is the Martin kernel K(x,y) = G(x,y) G(ρ,y). Furthermore, letting cap( ) K := inf 0 finitecap K (\ 0 ) cap( ) K () P(X n i.o) cap ( ) K () ProofTo get upper bound it is enough to find one µ. Let τ = min{n; S n } (τ = if is not hit), for x ν(x) = P ρ (τ <, S τ = x) ν() = P ρ (τ < ) if ν() = 0, nothing to prove. Assume () > 0. Note, y, g(x,y)dν(x) = x ν(x) P x (X n = y) = G(ρ,y) n=0 8-5

6 so ) k( K(x,y)dν(x) =. Hence I ν ν() = ν() and cap K (λ) ν() For the lower bound, use the second moment method. Given a probability measure µ on, let z = G(ρ,y) dµ(y) By Fubini, E rho z = E rho z = E ρ E ρ n=0 Xn=y E ρ ()=G(ρ,y) G(ρ,z) G(ρ,z) G(ρ,y) G(ρ,z) m,n>0 0 m n Xm=z, X n=ydµ(y)dµ(z) Xm=z, X n=ydµ(y)dµ(z) For each m, E rho n m X m=z, X n=y = P rho (X m = z)g(z,y) by the markov property. Taking the sum over m E ρ z G(z, y) G(ρ,y) dµ(z)dµ(y) = I K(µ) By Cauchy-Schwartz: P rho ( n,x n ) P rho (z > 0) (Eρ z) E ρ z Proving the first part of the theorem. For the second part, note that by transience I K (µ) P( n; X n ) cap K() {X n i.o} = stackrelmod 0= 0 finite {visits\ 0 } Hence if n is a sequence of finite sets increasing to, the by monotonicity and P(visits i.o) = lim n Pρ (visits\ 0 ) cap ( ) K () = lim cap K(\ 0 ) n Corollary 6 In Z d, d 3, for any A Z d, we deduce cap( ) K (A) P(A is hit i.o) cap ( ) K (A) {0,} by Hewitt Savage 0 law So A is hit i.o a.s if and only if cap ( ) K (A) > 0, now note that this remains true if we replace K by F(x,y) = y d since by the asyptotics of G(x) we have c F(x,y) + x y d K(x,y) c for some c,c >

7 Remark 7 In a different application, the B-P-P theorem show a theorem of Lyons that the chance that a path from root to leaves survive under a general Perculation (keep each eage with probability p e independet) Process is given up to by a certain capacity on leaves, by looking at the Markov chain of surviving leaves from left to right. Remark 8 x,y. The theorem still applies if the chainis not transient but G(x,y) <. You can replace K by its symmetrized version (K(x,y) + K(y,x)) in the theorem, since it doesn t affect capacity. Intersections of Random Walks Define S[n,m] := {S j ; j [n,m]} Theorem 9 c,c and ϕ such that n c ϕ(n) P(S[0,n] S[n,3n] ) P(S[0,n] S[n, ) ) c ϕ(n) d 3 ϕ(n) = (logn) d = 4 n d 4 d 5 ProofTrivial uper bound for d 3 and the lower bound for d followes from lower bound for d = 3, therefore we can assume d 3. Proof is by second moment method, with additional trick for d = 4. Denote J n = n 3n j=0 k=n Sj =S k, K n = n j=0 k=n Sj =S k By the Markov property and LCLT we get, (when k j is even) So P(S j = S k ) = P(S k j = 0) c n d/ c n 4 d EJ n c n 4 d we can even get (with a bit more calculation) cn d = 3 EJn clogn d = 4 d 5 And we get the lower bound for all d since cn d P(S[0,n] S[n,3n] ) P(J n > 0) (E(J n)) EJ n 8-7

8 We get the upper bpund for all d 4 since For the upper bound in d = 4 need to show P(S[0,n] S[n, ) ) = P(K n ) EK n E(K n K n ) clogn this is the expected number of pairs of times of intesec for s SRW starting in origin, this is like what happens after first intersection of walks. Note c = EK n = P(K n )E(K n K n ) Remark 0 By taking n we get that for BM in R d P(B[0,] B[,3] ) From the other side, it is intresting to consider { > 0 d =,,3 = 0 d 4 q(n) = P(S [0,n] S (0,n] ) for d =,3,4 where S,S are independet SRW. Defining Y n = P(S [0,n] S (0,n] S [0,n]) We have EY n = q n, and it is possible to calculate EY n = P(S [0,n] (S (0,n] S 3 (0,n]) ) up to a constant { logn d = 4 n d 4,,3 Since 0 Y n we have EYn EY n EYn Finally one can show that in d = 4 the upper inequality us sharp q(n) logn in d = 4 Known: q(n) n f d. where f =, f = 5 const 8, for d = 3 it is an open question, by simulation it is known that f References [] G. Lawler. Intersection of Random Walks Birkha user Boston; edition (August 8, 996) [] G. Lawler, V. Limic. Random Walk: a Modern Introduction Cambridge Studies in Advanced Mathematics (No. 3) June 00 [3] I. Benjamini, R. Pemantle, Y. Peres. Martin Capacity for Markov Chains Ann. Probab. Volume 3, Number 3 (995),