A sharp estimate for cover times on binary trees

Transcription

1 A sharp estimate for cover times on binary trees Jian Ding U. C. Berkeley Ofer Zeitouni University of Minnesota & Weizmann institute March 7, 0 Abstract We compute the second order correction for the cover time of the binary tree of depth n by continuous-time) random walk, and show that with probability approaching as n increases, τcov = E [ log n log n/ log + Olog log n) 8 ], thus showing that the second order correction differs from the corresponding one for the maximum of the Gaussian free field on the tree. Introduction The cover time of a random walk on a graph, which is the time it takes the walk to visit every vertex in the graph, is a basic parameter and has been researched intensively over the last several decades see [3,, 3] for background). One often studied aspect concerns precise estimates for cover times on specific graphs including D lattices and regular trees. For the D discrete torus, the asymptotics of the cover time were established by Dembo, Peres, Rosen and Zeitouni [9]. For regular trees, the asymtotics of the cover time were evaluated by Aldous [4] and a tightness result for the cover time after suitable normalization was demonstrated by Bramson and Zeitouni [7]. It was conjectured in [7] that the cover time of D discrete torii exhibits a similar tightness behavior. Meanwhile, the supremum of the Gaussian free field GFF) was also heavily studied. For squares in the D lattice, the first order asymptotics were evaluated by Bolthausen, Deuschel and Giacomin [5]. Interestingly, both [5] and [9] are based on the study of similar tree structures for the D lattice; in fact, the square of the GFF has the same first order asymptotics as the cover time after proper normalization. Recently, Ding, Lee, and Peres [0] demonstrated a useful connection between cover times and GFFs, by showing that, for any graph, the cover time is equivalent, up to a universal multiplicative constant, to the product of the number of edges and the supremum of the GFF. An important ingredient in [0] is a version of the so-called Dynkin Isomorphism theorem, which completely characterizes the distribution of local times closely related to the cover time) using GFFs. All these connections seem to suggest that a detailed study of fluctuations for one model should carry over to the other with moderate work. A particular motivating example in this direction is the case of squares in the D discrete lattice. Recently, Bramson and Zeitouni Partially supported by Microsoft Research. Much of the work was carried out during a visit of J.D. to Weizmann institute. Partially supported by NSF grant DMS and a grant from the Israel Science Foundation.

2 [6] established a tightness result for the supremum of GFF there with proper centering, but no other normalization), and further computed the centering up to an additive constant. One could hope that transfering this result to the cover time problem is now purely technical ; an essential part of such a program would be to verify that the supremum of the GFF correctly predicts the second order correction for the rescaled) cover time. The present paper is a cautionary note in that direction. We study the cover time on binary trees and obtain the sharp second order term. Interestingly, we demonstrate that the latter is larger than the corresponding one for the binary tree GFF. Our result improves the estimate in [4], and complements the result of [7]. We focus here on binary trees, but it should be clear from the proof that the method applies to more general Galton Watson trees. Let T = V,E) be a binary tree rooted at ρ of height n, and consider a continuous-time random walk X t ) started at ρ. Let τ cov be the first time when the random walk visited every single vertex in the tree. Our main result is the following. Theorem.. Consider a random walk on a binary tree T = V,E) of height n, started at the root ρ. Then, with high probability, τcov E = log n log n log + Olog log n) 8 ). ) At the cost of a more refined analysis, we believe that the error term Olog log n) 8 ) can be improved to O). To relate Theorem. to the GFF {η v } v V on the tree, recall that the latter can be defined as follows. Let {X e } e E be i.i.d. standard Gaussian variables and set η v = e:e ρ v where the sum is over all the edges that belong to the path from ρ to v. By adapting Bramson s arguments on branching Brownian motion [8] to the discrete setup, as in Addario-Berry and Reed [], one can show that E supη v = log n v X e, 3log n + O). ) log The lower bound in ) follows directly from [, Theorem 3]. The upper bound, that involves also the internal nodes of the tree, requires the use of [, Lemma 3] and a union bound over the levels.) Comparing ) and ), we do observe agreement in the first order and a discrepancy in the second order terms. Our proof uses ideas from [8] and is based on the study of the local times associated with the random walk. For any v V, we define the local time L v t to be the time that the random walk spends at v up to t, with a normalization by the degree of v. More precisely, L v t = t d {Xs=v}ds. v 0 Define the inverse local time τt) to be the first time when the local time at the root achieves t, by τt) = inf{s 0 : L ρ s t}.

3 We will let τt) be defined as above throughout the paper. We also set t + = log n log n log + 00log log n) and t = log n log n log 00log log n)8). 3) The following is the key to the proof of Theorem.. Theorem.. Consider a random walk on a binary tree T of height n, started at the root ρ. Then, Pτt ) τ cov τt + )) = + o), as n. In the next two sections, we prove the upper and lower bounds for the preceding theorem respectively; we conclude the paper by deriving Theorem. from Theorem.. Notation and convention: Throughout, C, c denote generic constants that may change from line to line, but are independent of n. Further, the phrase with high probability should be understood as the statement with probability approaching as n. Upper bound We establish an upper bound on the cover time in this section, as formulated in the next theorem. Theorem.. With notation as in Theorem., we have Pτ cov τt + )) = + o), as n. Theorem. is equivalent to the statement that at time τt + ), all the leaf-nodes have positive local times, with high probability. To this end, we consider a leaf-node of local time 0 with typical and non-typical profiles, respectively. For the latter, we show its unlikeliness directly; for the former, we prove it is a rare event by comparing to the same type of event for Gaussian free field.. Unlikeliness for non-typical profile As preparation, we prove a large deviation result which will be used to control the pairwise concentration of local times. Definition.. For r,λ > 0, let N be a Poisson variable with mean r and Y i be i.i.d. exponential variables with mean λ. Then, the random variable Z = N i= Y i is said to follow the distribution PoiGammar, λ), and we write Z PoiGammar, λ). Lemma.3. For α,r > 0, let Z PoiGammar,λ). Then for α < λr, Furthermore, for all α > 0, PZ λr α) exp rr α/λ) + α/λ r ). 4) PZ λr + α) exp rr + α/λ) r α/λ ). 5) 3

4 Proof. As in the definition of the PoiGammar, λ) distribution, let N be Poisson variable with mean r and let Y be an independent exponential variable with mean λ. For θ > 0, we have Ee θz/λ = EEe θy/λ ) N = E/ + θ)) N = e θr +θ. Combined with Markov s inequality, it follows that PZ λr α) = Pe θz/λ e θλr α)/λ ) e θr +θ e θr α/λ) = exp θ r +θ θα ) λ. For α < λr, optimizing the exponent at θ = r leads to inequality 4). To prove 5), consider 0 < θ <. We have r α/λ Ee θz/λ = EEe θy/λ ) N = E/ θ)) N = e θr θ. Another application of Markov s inequality gives that PZ λr + α) Pe θz/λ e θλr+α)/λ ) = exp θ r θ θα ) λ. Optimizing the exponent at θ = r, we deduce the inequality 5). r+α/λ Remark. The right side of 4) can be bounded by e α /4λ r. In this form, it is closely related to the discrete time bound in [, Lemma 5.]. We have the following immediate and useful corollary. Corollary.4. With notation as in Lemma.3, we have for any β > 0, P Z β) λr) e rβ, 6) and P Z + β) λr) e rβ. 7) For k N, we denote by V k V the set of vertices in k-th level of the tree. We next show that it is unlikely to have a too small local time for a vertex in intermediate levels. Lemma.5. With notation as in Theorem., define Then, PA) = o) as n. A = n log n k= u Vk {L u τt + ) k/n) t + 3log n) }. 8) Proof. Throughout the proof, we write t for t +. Consider u V k such that k n log n. It is clear that L u τt) has the distribution PoiGammat/k,k). Applying 4), we obtain that P Z k/n) t 3log n) ) exp k tk/n + 3log n) ) k n. Now a simple union bound gives that PA) n log n k= k k n /n = o). 4

5 For v V and k < n, let v k V k be the ancestor of v in the k-th level. Define γk) = min{ k log k, n k logn k)} +. 9) Lemma.6. With notation as in Theorem., define { B = v V n, k < n log n : L v τt + ) = 0, L v k+ τt + ) L v k τt + ) q } L v k τt + ) γk) A c. 0) Then, PB) = o) as n. Proof. We continue to write t = t +. Consider v V n and k < n log n. Note that conditioned on L v k τt), the collection of random variables {Lv k+j τt) } j 0 possess the same law as {L v k+j τ v kl v } k τt) ) j 0 this is an instance of the second Ray-Knight theorem in this context). Abusing notation, this implies in particular that conditioned on {L v k τt) = x}, Lv k+ τt) has distribution PoiGammax,). We will employ such an abuse of notation repeatedly throughout the paper.) Fixing x k/n) t 3log n, an application of Corollary.4 gives for j, P j x γk) L v k+ τt) x j + ) x L γk),lv τt) = 0 v k τt) = x) e j x /γk)) e x j+)/γk)) n k. Note that the right hand side in the above decays geometrically with j. Thus, summing over j, we obtain that P L v k+ τt) x x L γk),lv τt) = 0 v k τt) = x) 4e x 4x n k e n k)γk)e x γk)) 4e x n k e x γk)). Noting that PL v τt) = 0 Lv k τt) = x ) = e x n k, we obtain that P L v k+ τt) x x γk) L v τt) = 0,Lv k τt) = x) e x γk)) e log3/ n, where the last nequality follows from the fact that x k/n) t 3log n) and k n log n. Therefore, q ) P A c,l v τt) = 0, L v k+ τt) L v k L v k τt) τt) γk) PL v τt) = 0) e log3/ n = e t/n e log3/ n n n. At this point, a simple union bound completes the proof.. Unlikeliness for typical profile We next compare the density of local times and Gaussian variables. This comparison of density is of significance for the proof of both upper and lower bounds. Lemma.7. For l > 0, let Z PoiGammal,) and let f ) denote the density function of Z on R +, with f0) = PZ = 0). Denote by W a standard Gaussian variable, and denote by g ) the density function of W/. Then, for any w such that w l/, we have fl + w) = w l + O w + l )) gw). 5

6 Proof. Write y = l + w, and let h ) be the density function of Z. Then for z > 0, we have hz) = k= Applying a change of variables, we obtain that fy) = y e k= l lk k! l lk z k e k! e z k )!. e y yk ) k )! = le l +y ) where I x) is a modified Bessel function defined by I x) = k=0 k=0 x/) k+ k!k + )!. yl) k+ k!k + )! = le l +y ) I yl), For the modified Bessel function I x), the following expansion is known when x is large see []): I x) = Plugging into the preceding expansion, we get that e yl ex πx 3 8x + O x )). fy) = le l +y ) 3 πyl 8 yl + O ) ) y l = e w w π l + O )) w + l. Combined with the fact that gw) = π e w, the desired estimate follows immediately. We single out the next calculation, which will be used repeatedly. Claim.8. Consider z i,l i R with l i+ = l i + z i for i = 0,...,m such that z i l i / for all i. Assume that i z i + l i Proof. On one hand, note that l m l 0 = m i= = O). Then, m i= z i l i + O zi + )) l0 l = Θ). i lm l i = m i= + z i ) l l i = exp m i= z i l i + O m i On the other hand, we have zi i= l i m i= z i l i + O zi + )) l = exp m i= z i l i i + O m i= Combining these estimates completes the proof. = exp m i= z i l i + O) ) = )) = exp m i= z i l i + O) ). zi + l i )) l 0 l m expo)). We next demonstrate that it is unlikely to have a leaf-node of local time 0, even with a typical profile for local times along the path from ρ to the leaf. 6

7 Lemma.9. With notation as in Theorem. and A,B as in 8) and 0), define Then, PD v ) = o n ). D v = {L v τt + ) = 0} \ A B), for v V n. ) Proof. Again, we write t = t +. Write n = n log n. Let Ω R n be such that for z,...,z n Ω, we have { } n k= L v k τt) L v k τt) = z k D v. Let α ) and β ) be density functions for L v k τt) L v k τt) ) k n and η v k / η vk / ) k n, respectively. Denote by l k = t + k i= z i. Note that for z,...,z n ) Ω, we have n +zi i= l i = O) n i= Applying Lemma.7 and Claim.8, we obtain that Therefore, we obtain that PD v ) = Ω n i) + γi)) ) = O). αz,...,z n ) βz,...,z n ) = n i= z i l i + O zi + )) l = Θ) n log n. i αz,...,z n )PL v τt) = 0 L v n n τt) = l n )dz O) log n Ω βz,...,z n )e l n n n dz. Write s = n /n) t 3log n. Let βx) = {l n =x} βz,...,z n )dz for x s. Note that βx) = x e n Pη vk / k/n) t 3log n for k n η vn πn / = x). 3) ) Conditioning on η vn / = x, we have {η vk / ) k n η vn / = x} law = {W k / + k/n )x) k n }, where W r ) 0 r n, is a Brownian Bridge of length n, i.e., a Brownian motion conditioned on hitting 0 at both time 0 and n. It is well-known that the maximum of a Brownian bridge W r ) on [0,q] follows the Rayleigh distribution see, e.g., [4]), i.e., Therefore, we obtain that Pmax 0 r q W r λ) = e λ q, for all λ 0. 4) Pη vk / k/n) t 3log n for k n η vn / = x) Pmin r n W r / 3log n x s)) = Pmax r n W r 3log n + x s))) Plugging the above estimate into 3), we obtain that 43log n+x s)) n. βx) 43log n + x s)) n ) 3/ e x n. 7

8 Together with ), we obtain that PD v ) O) s log n + x s)) n log n Using the change of variables y = x + PD v ) O) e t n n log n e x t+x) n e n n dx O) tn n, we obtain that log n + x s)) n log n 4log n + y) e n + n n )y dy = n olog 6 n), where we used the fact that n = n log n, completing the proof. e x t+x) n e n n dx. Proof of Theorem.. The proof now follows trivially. Since v V n PD v ) = n n o) = o) as well as PA) = o) and PB) = o), we see that with high probability, every leaf-node has positive local time by τt), implying the desired upper bound on cover time. 3 Lower bound This section is devoted to the proof of the following lower bound on the cover time for a binary tree T. Theorem 3.. With notation as in Theorem., Pτ cov τt )) = + o), as n. The proof consists of an analysis for exceptionally large values in the Gaussian free field and a comparison argument based on Lemma Exceptional points for Gaussian free field We first study the Gaussian free field {η v } v V on the tree T of height n, with η ρ = 0. For k < n, let ψk) = logk n k)). Denote by log a k = k/n) log n log n ) log ψk) +, for k < n, and an = log n log n log. 5) Consider = a n + log 4 n. Recall the definition of γk) in 9). For v V n, define E v = {η vk / a k, for all k < n,a n η v / a n + }, 6) F v = {E v, k n : η vk η vk η vk / /γk)}. 7) We start with a lower bound on the probability for event E v. Lemma 3.. There exists a constant c > 0 such that for all v V n, we have PE v ) c n n. 8

9 Proof. It is clear that PE v ) Pa n η v / a n +) min PE v η v = x) a n x a n+ n 5 n min a n x a n+ PE v η v = x), where the second inequality follows from a bound on the Gaussian density. Denote by W t ) 0 t n a Brownian bridge. We note that {η vl : 0 l n} η v = ) law } x = {W l + l n x : 0 l n. This implies that, for x a n, PE v η v = x) PW l ψl) for 0 l n). By [8, Proposition ], we have that PW l ψl) for 0 l n) c/n for a constant c > 0. Altogether, we obtain that PE v ) c 5 n n. We now show that the event F v is extremely rare. Lemma 3.3. For any v V n, we have Proof. Take v V n. It is clear that PF v ) = n o/n), as n. PF v ) Pa n η v / a n + )J n nj, where J = max an x an+ y a k n P η vk η vk η vk / /γk) η v = x,η vk = y). k= Conditioning on η v = x,η vk = y, we have η vk η vk distributed as a Gaussian variable with mean n k n k+ x y) and variance n k+. For x,y that is under consideration, we have n k+ x y) = ). Therefore, we obtain that o y γk) P η vk η vk η vk / /γk) η v = x,η vk = y) e y) 4γk)) e log n, for large enough n. This implies that J ne log n, and thus PF v ) = n o/n). We next study the correlation for events E u and E v. For u,v V, denote by u v the least common ancestor of u and v. Lemma 3.4. Consider u,v V n and assume that u v V k. Then, 0log n PE u E v ) PE u ) n k). n k n k) k) 9

10 Proof. Denote by w = u v, and let f ) be the density function of η w /. For i < j, write = {η vl / a l, for all i l < j}. Then, E i,j v PE u E v ) = PE u )PE v E u ) PE u ) max an+ PE u ) max x a k a n x a k PE k,n e y x) n k n k v,a n η v a n + η w / = x) PE k,n v η w / = x,η v / = y)dy. 8) For x a k and a n ya n +, we first analyze the probability PEv k,n η w / = x,η v / = y). Let W s ) 0 s n k be a Brownian bridge. It is clear that { η vl / } : k l n η w / = x,η v / ) law = y = {W l k / } + l k n k y + n l n k x : k l n. Combined with 4), it follows that PE k,n v η w / = x,η v / = y) Pmax W s log n + a k x))) 4log n + a k x)). 9) s n k By a straightforward calculation, we have that e y x) n k e an a k ) n k e an a k )a k x) n k n k) e n k n log n logn k) k) e log x a k ) n k) n k log n k) k e x ak ). Combined with 9), it follows that an+ a n e y x) n k n k PE k,n v η w = x,η v = y)dy n k) 4log n + a k x)) n k n k) k) e log x a k ) n k) 0log n. n k n k) k) Together with 8), we deduce that 0log n PE u E v ) PE u ) n k). n k n k) k) 3. Lower bound for cover times We now turn to study the cover time. The key estimate lies in the following proposition. Proposition 3.5. With notation as in Theorem., let s = log n log n log + log4 n). Then there exists a constant c > 0 such that Pmin v Vn L v τs) log8 n) c log 5 n. 0

11 Proof. Let Z v = L v τs) for v V. Let a k be defined as in 5). For v V n, define Ẽ v = { s Z vk a k, for all k < n,a n s Z v a n + }, 0) F v = {Ẽv, k n : Z vk Z vk Z vk /γk)}. ) Define Ω v R n such that for any z,...,z n ) Ω v It is clear from 6) and 7) that {Z vk Z vk = z k for all k n} Ẽv \ F v. {η vk / η vk / = z k for all k n} E v \ F v. Let α v ),β v ) be density functions over Ω for Z vk Z vk ) k n and η vk / η vk / ) k n, respectively. Consider z,...,z n ) Ω v. By Lemma.7, we have α v z,...,z n ) = β v z,...,z n ) n k= z k l k + O z k + l k ) ), ) where l k = s k i= z k. Since z,...,z n ) Ω v, we have that n k= z k + l k n k= γk)) + 4 n k) +log 4 n ) = O). Applying Claim.8, we obtain that n α v z,...,z n ) = Θ) log n β vz,...,z n ). Integrating over both sides and recalling Lemmas 3.3 and 3., we obtain that PẼv \ F n v ) = Θ) log n PE v \ F v ) = Θ) c n log n PE v) Θ) log n n. 3) We next analyze the correlation of Ẽv \ F v and Ẽu \ F u. Consider u,v V n and assume that u v V k. We write Z = Z v Z v0,...,z vn Z vn,z uk+ Z uk,...,z un Z un ), η = η v0 η v,...,η vn η vn,η uk η uk+,...,η un η un ). Define Ω u,v R n k such that for all z = z v,,...,z v,n,z u,k+,...,z u,n ) Ω u,v, {Z = z} Ẽv \ F v ) Ẽu \ F u ). It is then clear that {η = z} E v \F v ) E u \F u ). Let α u,v ) and β u,v ) be density functions for Z and η, respectively. Let z u,i = z v,i for all i k. For w {u,v}, write l w,j = t j i= z w,j. By Lemma.7, we get that that α u,v z) = β u,v z) n j= z v,j l v,j + O zv,j + )) l v,j n j=k z u,j l u,j + O zu,j + )) l. u,j

12 Applying Claim.8 again, we obtain that nn k) αu,v)z) = O) log β u,v z). n Integrating over both sides and applying Lemma 3.4, we get that PẼv \ F v ) Ẽu \ F n u ) = O) n k) k PE v) n k). This implies that for a constant C > 0 E w V n Ẽw\ F w ) C n n PE v ) n j= w:w v V j n j) n j) j C n n PE v ) 4log n. At this point, an application of the second moment method together with 3) gives that P w V n : Ẽw \ F w ) E w V n Ẽw\ F w ) E ) n PẼv \ F v )) w V Ẽw\ F C n n PE v ) n w C log 5 n, for a constant C > 0. Recalling the definition of Ẽv, we complete the proof of the proposition. Next, we bootstrap the above estimate and prove the main result in this section. Proof of Theorem 3.. Throughout the proof, we write t = t. Let n = 30log log n, n 3 = log 4 n log, n 4 = 0log log n) 8, and n = n n n 3 n 4. For k N, write b k = log k log k log. Note that t + 50n b n + b n3. Our proof is divided into 4 steps. Step. Write t = t + n ). Since for all v V n we have L v τt) PoiGammat/n,n ), an application of 7) yields that P v V n : L v τt) t ) n PPoiGammat/n,n ) t ) n e 4n = o). 4) Step. For v V n, let T v be the subtree rooted at v of height n. Write t = t b n ). By 4), we assume in what follows that L v τt) t. Applying Proposition 3.5 to the subtree T v, we deduce that for a constant c > 0, P min u T v V n +n L u τt) t ) c log 5 n. Let S = {v V n : min u Tv V n L u τt) t }. By independence of the random walk on different subtrees, we obtain that with high probability S n /log 6 n log log n. We assume this in what follows. Define S = {u V n +n : L u τt) t }. We see that S S log log n. Step 3. For u S, consider the subtree T u rooted at u of height n 3. Since t b n3 ) log log n) 8. We can apply Proposition 3.5 again to the subtree T u and obtain that for a constant c > 0 P min w T u V n +n +n 3 L w τt) log log n)8 ) c log log n) 5.

13 Let S 3 = {w T u V n +n +n 3 : L w τt) log log n)8 }. We can then obtain that with high probability S 3 log log n, and we assume this in what follows. Step 4. For w S 3, let T w be the subtree rooted at w that contains all its descendants. We trivially have that P min w T w L w τt) = 0). Since S 3 log log n, we see that with high probability there exists a vertex w w S3 T w V with L w τt) = 0, completing the proof. 3.3 Concentration of inverse local time We have been measuring the cover time via the inverse local time so far. In this subsection, we prove that the inverse local is well-concentrated around the mean and thus it indeed yields a good estimate on the cover time. Lemma 3.6. Consider a random walk on a rooted binary tree T = V,E) of height n. Then, Varτt)) = O) n t. Proof. Note that τt) = v V d vl v τt). Consider u,v V and write w = u v. Assume that w V k. Then, EL v τt) Lu τt) ) = EELv τt) Lu τt) ) Lw τt) ) = ELw τt) ) ) = t + VarL w τt) ) t + 6tk, where the last transition follows from the fact that L w τt) PoiGammat/k,k) and a simple application of the total variance formula VarX = EVarX Y )) + VarEX Y )). We then have CovL v τt),lu τt) ) 6tk. Therefore, Varτt)) = u,v V where we used the fact that d v 3. n d v d u CovL v τt),lu τt) ) k n k) 3 6tk = O) n t, Now it is obvious that Theorems., 3. imply Theorem.. Together with Lemma 3.6, we complete the proof of Theorem., by noting that Eτt)) = t E = n+ 4)t. Acknowledgement k= We thank Amir Dembo and Yuval Peres for helpful discussions. References [] Handbook of mathematical functions, with formulas, graphs, and mathematical tables, volume 55 of Edited by Milton Abramowitz and Irene A. Stegun. Third printing, with corrections. National Bureau of Standards Applied Mathematics Series. Superintendent of Documents, U.S. Government Printing Office, Washington, D.C.,

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