Random walks conditioned to avoid certain sets

By Lemma 3.2, for any w ∈ ∂ B 2l , P w ξ ξ K ∧ σ n = X y ∈∂ B l P w S ξ l = y; ξ l ξ K ∧ σ n P y ξ ξ K ∧ σ n ≤ P w ξ l ξ K ∧ σ n P z ξ ξ K ∧ σ n ≤ 1 − cP z ξ ξ K ∧ σ n . Thus, P z ξ ξ K ∧ σ n ≤ P z ξ σ 2l + 1 − cP z ξ ξ K ∧ σ n P z σ 2l ξ ≤ P z ξ σ 2l + 1 − cP z ξ ξ K ∧ σ n , and therefore P z ξ ξ K ∧ σ n ≤ 1 c P z ξ σ 2l , which completes the proof.

3.2 Random walks conditioned to avoid certain sets

Proposition 3.5. There exist constants N and c 0 such that for all n ≥ N the following holds. Suppose that K ⊂ Λ \ B n n, 0 where B n n, 0 denotes the ball of radius n centered at n, 0 see Figure 2. Then, P § argS σ n ∈ [− π 4 , π 4 ] σ n ξ K ª ≥ c. − π 4 π 4 K Figure 2: The setup for Proposition 3.5 Proof. For z ∈ D, let hz = P z § arg[W τ D ] ∈ [− π 4 , π 4 ] ª 1029 where W denotes standard two-dimensional Brownian motion. Then h is the solution to the Dirichlet problem with boundary value 1 [−π4,π4] . Therefore, we can express h as hz = 1 2 π Z π 4 − π 4 H D z, e i θ dθ where H D z, e i θ = 1 − |z| 2 e i θ − z 2 is the Poisson kernel for the unit disk. One can compute h it is easier to consider the problem on H and then map back via a conformal transformation: hz = 1 π – arctan ‚ p 2 − 1 |1 + z| 2 + 2 Imz 1 − |z| 2 Œ + arctan ‚ p 2 − 1 |1 + z| 2 − 2 Imz 1 − |z| 2 Œ™ . We now establish three basic facts about h that we will use below. 1. Let D 1 1 be the disk of radius 1 centered at the point 1. We claim that for all z ∈ D \ D 1 1, hz ≤ h0. By the maximal principle for harmonic functions, h restricted to D \ D 1 1 takes its maximal value on ∂ D 1 1 ∩ D since h vanishes on ∂ D \ D 1 1. Thus, to prove the claim, it suffices to show that f t = h1 + cos t, sin t takes its maximal value at t = π for 2π3 ≤ t ≤ 4π3. Since one has an explicit formula for h, this is left as an exercise for the reader or the reader’s Calculus students. 2. Next, fixing w = e i θ , it is a basic calculation to show that ∂ H D ., w ∂ x 0 = 2 cos θ , ∂ H D ., w ∂ y 0 = 2 sin θ . Therefore, ∂ h ∂ x 0 = 1 2 π Z π 4 − π 4 2 cos θ dθ = p 2 π , and ∂ h ∂ y 0 = 1 2 π Z π 4 − π 4 2 sin θ dθ = 0. These results can also be obtained from the explicit formula for h. 1030 3. Finally, there exists 0 r 1 such that for all r |z| 1, and argz π3, hz 18. This follows from the fact that if |z| r and | argz| π3 then for w ∈ ∂ D with | argw| π4, H D z, w = 1 − |z| 2 |w − z| 2 ≤ c1 − r 2 , which can be made to be arbitrarily close to 0. Assume that n is large enough so that B r n ⊂ nD where r is as in the previous paragraph. We let h n z = hzn which is harmonic in nD. Then for z ∈ B r n , define eh n z = E z h n Sσ r n . Then e h n is discrete harmonic in B r n and agrees with h n on ∂ B r n . A natural question to ask is how close does the discrete harmonic solution e h n approximate the continuous harmonic solution h n ? By [16, Corollary 6.2.4], for all z ∈ B r n , h n z = e h n z − E z    σ r n −1 X j=0 L h n S j    , where L f x = − f x+ P y ∈Λ p y f x + y is the generator for S. Consider the associated operator f L f x = 1 2 X y ∈Λ κ y ∂ 2 f ∂ y 2 x. By Taylor’s theorem, for any C 4 function f and z ∈ Λ, L f z − f L f z ≤ CR 4 M 4 f z, where R is the range of the walk S and M 4 f is the L ∞ norm of the sum of the fourth derivatives of f in the disk D R z. Since the random walk S has covariance matrix the identity we have been assuming that S has this property but this is the first place we use it, one can show that f L is actually a multiple of the continuous Laplacian. Thus, f L h n = 0. Furthermore, since the fourth derivatives of h are bounded on rD, M 4 h n is bounded by C n −4 in B r n . Therefore, combining all the previous remarks and letting CR 4 = C since R depends only on the random walk S which we’ve fixed, we obtain that for z ∈ B r n , h n z − eh n z ≤ E z    σ r n −1 X j=0 CR 4 M 4 h n z    ≤ C n −4 E z σ r n − 1 = C n −4 X x ∈B n G n z, x ≤ C n −4 X x ∈B 2n G 2n 0, x ≤ C n −4 X x ∈B 2n log 2n − log |x| + C ′ ≤ C n −2 . 1031 We now have all the pieces we need to prove the proposition. Let z be any point in B r n \ B n n, 0, and fix x ∈ Λ such that Rex 0. Then by Taylor’s theorem and our previous observations about h, if n is large enough so that x is in B r n , eh n x − eh n z = [e h n x − h n x] + [h n x − h n 0] + [h n 0 − h n z] + [h n z − eh n z] ≥ −C n −2 + n −1 ∂ h ∂ x 0 Rex − n −2 M 2 h |x| 2 + 0 − C n −2 where M 2 h is the L ∞ norm of the sum of the second order derivatives of h in rD. Since ∂ h ∂ x 0 0, it is clear that for n sufficiently large, eh n x ≥ eh n z, for all z ∈ B r n \ B n n, 0. Since K ⊂ Λ \ B n n, 0, E x h n Sσ r n ξ K σ r n ≤ max z ∈K∩B r n E z h n Sσ r n = max z ∈K∩B r n eh n z ≤ eh n x = E x h n Sσ r n . Thus, E x h n Sσ r n = E x h n Sσ r n ξ K σ r n P x ξ K σ r n + E x h n Sσ r n σ r n ξ K P x σ r n ξ K ≤ E x h n Sσ r n P x ξ K σ r n + E x h n Sσ r n σ r n ξ K P x σ r n ξ K . This implies that E x h n Sσ r n σ r n ξ K ≥ E x h n Sσ r n = e h n x ≥ eh n ≥ h0 − C n −2 ≥ 1 5 for n sufficiently large, since h0 = 1 4. Recall that r was defined so that for all z such that r |z| 1, and argz π3, hz 18. 1032 Therefore, 1 5 ≤ E x h n Sσ r n σ r n ξ K = X | argz|π3 h n zP x S σ r n = z σ r n ξ K + X | argz|≤π3 h n zP x S σ r n = z σ r n ξ K ≤ 1 8  1 − P x § arg Sσ r n ≤ π 3 σ r n ξ K ª‹ + P x § arg Sσ r n ≤ π 3 σ r n ξ K ª . Thus, P x § arg Sσ r n ≤ π 3 σ r n ξ K ª ≥ c 0. Since x is independent of K and n, P ξ x ξ K ∧ σ r n ≥ c, and hence, P § arg Sσ r n ≤ π 3 σ r n ξ K ª ≥ c 0. Finally, P § arg Sσ n ≤ π 4 σ n ξ K ª ≥ P § arg Sσ n ≤ π 4 ; σ n ξ K σ r n σ K ª ≥ cP § arg Sσ r n ≤ π 3 σ r n ξ K ª ≥ c. Lemma 3.6. For 0 θ π, there exist cθ , N θ and αθ such that the following holds. For n N , and z ∈ Λ with N |z| n, let W be the wedge W = {x ∈ Λ : 0 ≤ |x| ≤ n, argx − argz ≤ θ}. Then, P z σ W = σ n ≥ c |z| n α . Remark By comparison with Brownian motion, one expects that αθ = πθ would be the optimal constant. However, in this paper we will only need the existence of α and not its exact value. Proof. It is clear that we can make αθ non-increasing in θ , therefore, without loss of generality, take θ π2. Also, without loss of generality, assume argz = 0. Let f W be the cone f W = {x ∈ Λ : argx ≤ θ}. 1033 We define a random sequence of points {z k } ⊂ f W as follows. We let z = z. Then, given z k , we let B k be the largest ball centered at z k such that B k ⊂ f W , r k be the radius of B k and let z k+1 = Sσ B k where S is a random walk starting at z k . We note that z j = z k for all j ≥ k if and only if z k ∈ ∂ i f W . We make N θ large enough to ensure that if |z| N then z ∈ ∂ i f W . In this case, there exists c ′ θ 0 such that r ≥ c ′ θ |z|. Let E k denote the event that z k+1 6= z k and that argz k+1 − z k ≤ θ 2 . On the event E k , r k+1 ≥ 1 + 2 sinθ 4r k = ecθ r k , and z k+1 2 ≥ z k 2 + r 2 k we use the fact that θ π2 for the second assertion. Therefore, if E , . . . , E j all hold, then r k ≥ ec k r ≥ ec k c ′ |z| for k = 1, . . . , j. Therefore, z j+1 2 ≥ |z| 2 + j X k=0 r 2 k ≥ |z| 2 + j X k=0 c ′ 2 ec 2k |z| 2 ≥ c ′ 2 ec 2 j |z| 2 . Since ec 1, it follows that if we let j be the smallest integer such that j ≥ logn c ′ |z| log ec , then j \ k=0 E j ⊂ {σ W = σ n }. Finally, by the invariance principle, there exists a constant c ′′ θ and N such that for n ≥ N, P E k ≥ c ′′ for all k. Therefore, P z σ W = σ n ≥ P    j \ k=0 E k    = j Y k=0 P E k ≥ c ′′ j+1 ≥ cθ |z| n αθ , where α = − logc ′′ log ec and c = c ′′ c ′ α . Corollary 3.7. Fix θ 1 , θ 2 ∈ 0, π2. There exist N, α and c 0 depending only on θ 1 + θ 2 such that the following holds. Let N ≤ l m n, and z ∈ ∂ B m . Let W be the half-wedge W = {x ∈ Λ : l ≤ |x| ≤ n, −θ 1 ≤ argx − argz ≤ θ 2 }. 1034 1. Let r = min {m sin θ 1 , m sin θ 2 , m − l}. Then for any K ⊂ B m , P z S[0, σ n ] ⊂ W σ n ξ K ≥ c  r n ‹ α . 2. Let r ′ = min{m sin θ 1 , m sin θ 2 , n − m}. There exists β = βθ 1 + θ 2 , l m such that for any K ⊂ Λ \ B m , P z S[0, ξ l ] ⊂ W ξ l ξ K ≥ c ‚ r ′ m Œ β . Notice that in both cases, the right hand side depends only on θ 1 , θ 2 , and the ratios l n and mn. n m l z K W Figure 3: The setup for part 1 of Corollary 3.7 Proof. Both parts of the corollary are proved similarly. We prove 1 in detail, and indicate the modi- fications needed to prove 2. Without loss of generality, assume that argz = 0. The quantity r defined in the statement of the corollary is the radius of the largest ball with center z whose closure is contained in the half-infinite wedge c W = {x ∈ Λ : |x| ≥ l, −θ 1 ≤ argx ≤ θ 2 }. We can apply Proposition 3.5 to the ball B = Bz, r, to obtain that there exists a constant c 0 such that P z § argSσ B − z ≤ π 4 σ B ξ K ª ≥ c. Let y be any point on ∂ B such that argy − z ≤ π4, and let eB = By, r2. Note that eB ⊂ c W \B m . There exists a point w ∈ ∂ e B such that y is on the bisector of the angle formed from the lines joining w to the two outermost corners of W , x 1 = ne −iθ 1 and x 2 = ne i θ 2 . Let s = max { x 1 − w , x 2 − w } ≤ 2n, 1035 and let f W be the wedge with vertex w, radius s and such that x 1 and x 2 are on ∂ f W . The wedge f W will have aperture θ ≥ θ 1 + θ 2 2 and y will be on the axis of symmetry of f W . Therefore, by Lemma 3.6, P y S[0, σ n ] ⊂ W ; σ n ξ K ≥ P y ¦ σ f W = σ Bw,s © ≥ cθ  r s ‹ αθ ≥ cθ 1 + θ 2  r n ‹ αθ 1 +θ 2 . To finish the proof of 1, let c ∗ = c  r n ‹ α . Then, P z S[0, σ n ] ⊂ W ; σ n ξ K ≥ X y ∈∂ Bz,R | arg y −z | ≤π4 P y S[0, σ n ] ⊂ W ; σ n ξ K P z σ B ξ K ; S σ B = y ≥ c ∗ P z § argSσ B − z ≤ π 4 ; σ B ξ K ª ≥ c ∗ P z σ B ξ K ≥ c ∗ P z σ n ξ K . The proof of 2 is similar. In this case, the angle θ of the wedge f W will be such that θ ≥ tan −1 2l sin θ 1 + θ 2 2 m . This is why β will also depend on lm. Besides this observation, the proof of 2 is identical to the proof of 1. The following corollary is similar to the previous one, except that here we are conditioning to avoid sets that are on either side of the half-wedge. Corollary 3.8. Let θ ∈ 0, 2π] and 0 a 1 b ∞. Then there exists N and c 0 depending on a, b and θ such that for n N the following holds. Let W be the half-wedge W = {x : an ≤ |x| ≤ 4bn, argx ≤ θ}. Suppose that K 1 ⊂ B n contains a path connecting ∂ B an to ∂ i B n , and K 2 ⊂ Λ \ B 4n contains a path connecting ∂ B 4n to ∂ B 4bn . Let K = K 1 ∪ K 2 . Then for any z ∈ ∂ B n , y ∈ ∂ B 4n with argz θ2, argy θ2, P z ¦ S[0, ξ y ] ⊂ W ξ y ξ K © ≥ c. 1036 1 00 11 z y an n 4n 4bn K1 K2 Figure 4: The setup for Corollary 3.8 Proof. Applying part 1 of Corollary 3.7 to the half-wedge W ∗ = {x : an ≤ |x| ≤ 2n, argx − argz ≤ θ 4 }, one obtains that there exists a constant c = ca, θ such that P z S[0, σ 2n ] ⊂ W ∗ σ 2n ξ K ≥ c. Now suppose that w ∈ ∂ B 2n ∩ W ∗ . Then by Lemma 3.1, P w ¦ S[0, ξ y ] ⊂ W ξ y ξ K © = P w ¦ ξ y σ W ∧ ξ K © P w ¦ ξ y ξ K © = Gw; W \ { y} ∪ KP y ¦ ξ w σ W ∧ ξ K ∧ ξ y © Gw; Λ \ { y} ∪ KP y ¦ ξ w ξ K ∧ ξ y © . However, for w ∈ ∂ B 2n ∩ W ∗ , Gw; W \ { y} ∪ K ≥ Gw; B c θ n w, and therefore by Lemma 3.3, Gw; W \ { y} ∪ K Gw; Λ \ { y} ∪ K ≥ cθ , b. Thus, P w ¦ S[0, ξ y ] ⊂ W ξ y σ K © ≥ cθ , b P y ¦ ξ w σ W ∧ ξ K ∧ ξ y © P y ¦ ξ w ξ K ∧ ξ y © . 1037 By the strong Markov property, P y ¦ ξ w σ W ∧ ξ K ∧ ξ y © ≥ X x ∈∂ i B 3n ∩W ∗ P x ¦ ξ w σ W ∧ ξ K ∧ ξ y © P y ¦ S ξ 3n = x; ξ 3n σ W ∗ ∧ ξ K ∧ ξ y © . However, by Lemma 3.4, there exists c θ , b such that for x ∈ ∂ i B 3n ∩ W ∗ , P x ¦ ξ w σ W ∧ ξ K ∧ ξ y © ≥ cθ , bP x ¦ ξ w ξ K ∧ ξ y © . Furthermore, by the discrete Harnack inequality, there exists c 0 such that for all x, x ′ ∈ ∂ i B 3n , P x ¦ ξ w ξ K ∧ ξ y © ≥ cP x ′ ¦ ξ w ξ K ∧ ξ y © . Thus, fixing x ′ ∈ ∂ i B 3n , P y ¦ ξ w σ W ∧ ξ K ∧ ξ y © ≥ cθ , bP x ′ ¦ ξ w ξ K ∧ ξ y © P y ¦ ξ 3n σ W ∗ ∧ ξ K ∧ ξ y © . Similarly, P y ¦ ξ w ξ K ∧ ξ y © ≤ cP x ′ ¦ ξ w ξ K ∧ ξ y © P y ¦ ξ 3n ξ K ∧ ξ y © . Therefore using part 2 of Corollary 3.7, P w ¦ S[0, ξ y ] ⊂ W ξ y σ K © ≥ cθ , b P y ¦ ξ w σ W ∧ ξ K ∧ ξ y © P y ¦ ξ w ξ K ∧ ξ y © ≥ cθ , b P y ¦ ξ 3n σ W ∗ ∧ ξ K ∧ ξ y © P y ¦ ξ 3n ξ K ∧ ξ y © = c θ , bP y ¦ ξ 3n σ W ∗ ξ 3n ξ K ∧ ξ y © ≥ cθ , b.

3.3 Random walk approximations to hitting probabilities of curves by Brownian mo-