A permuted random wal exts faster Rchard Pymar Perla Sous Aprl 4, 013 Abstract Let σ be a permutaton of {0,..., n}. We consder the Marov chan X whch jumps from 0, n to σ( + 1) or σ( 1), equally lely. When X s at 0 t jumps to ether σ(0) or σ(1) equally lely, and when X s at n t jumps to ether σ(n) or σ(n 1), equally lely. We show that the dentty permutaton maxmzes the expected httng tme of n, when the wal starts at 0. More generally, we prove that the httng tme of a random wal on a strongly connected d-drected graph s maxmzed when the graph s the lne 0, n Z wth d self-loops at every vertex and d 1 self-loops at 0 and n. Keywords and phrases. Marov chan, drected graph, httng tme. MSC 010 subject classfcatons. Prmary 60J10. 1 Introducton Let σ be a permutaton of {0,..., n} and (ξ ) be..d. unform random varables n { 1, 1}. We defne the process X σ by settng X σ 0 = 0 and Xσ t+1 = σ(xσ t + ξ t+1 ) f X σ t 0, n. Otherwse, f X σ t = 0, then X σ t+1 s unformly random n the set {σ(0), σ(1)} and f Xσ t = n, then t s unformly random n the set {σ(n), σ(n 1)}. In ths paper we address the queston of maxmzng the httng tme of n startng from 0 by the process X σ. In partcular we show that the dentty permutaton gves the slowest httng tme of n startng from 0,.e. n ths case X σ s a smple random wal on {0,..., n} wth a self-loop at 0 and at n. Theorem 1.1. Let σ be a permutaton of {0,..., n} and X σ the Marov chan defned above. If τ n = nf{t 0 : Xt σ = n}, then E 0 τ n n + n. Equalty s acheved f and only f σ s the dentty permutaton. As we explan n Secton 3, the process X σ can be vewed as a random wal on a strongly connected graph where every vertex has outdegree and ndegree equal to. In Secton we prove a more general result (Theorem 1.4) concernng drected graphs n whch every vertex has ndegree equal to the outdegree equal to d. Then n Secton 3 we gve the proof of Theorem 1.1 by applyng Theorem 1.4 for d =. Unversty College London, London, UK; r.pymar@ucl.ac.u Unversty of Cambrdge, Cambrdge, UK; p.sous@statslab.cam.ac.u 1
Defnton 1.. Let d and n N. We defne L(d, n) to be the graph on 0, n Z wth the followng propertes: 1) 0 and n have d 1 self loops each and all other vertces have d self loops ) for every 0 and l n there s a drected edge from to 1 and from l to l + 1. d 1 d d d d d d 1 0 1 n 1 n Fgure 1: The graph L(d, n) Defnton 1.3. A drected graph G s connected f the graph G obtaned by removng the drectons from the edges s connected. A drected graph G s strongly connected f for every par of vertces x, y there s a drected path from x to y. We denote ths by x y. We denote by V (G) the vertex set of a graph G and we wrte outdeg G (x) and ndeg G (x) for the outdegree and ndegree of the vertex x n the graph G. Theorem 1.4. Let G be a drected graph on n + 1 vertces whch s strongly connected and such that the ndegree of every vertex s equal to ts outdegree and equal to d (allowng (multple) self-loops and multple edges). Then f τ x s the frst httng tme of x by a smple random wal on G, then for all x and y we have E x τ y d n(n + 1). Equalty s acheved f and only f G s somorphc to L(d, n). To date, much of the wor on Marov chans has focused on random wals on undrected graphs. Random wals on drected graphs have receved relatvely less attenton and there are many nterestng unexplored questons n ths area. In partcular, the frst nown bounds for mxng tme parameters of a smple random wal on a drected graph have been studed by Fll n 4 and for the Euleran case by Montenegro n 5. Although the methods and deas of the proofs are completely dfferent, at a phlosophcal level the problem of maxmzng the httng tme of {0, n} by X σ over all permutatons σ s related to applcatons of rearrangement nequaltes as n 3 and 6. We state a related result that was proved by Azenman and Smon n 1: among all open sets of equal area, the ball maxmzes the ext tme by a Brownan moton. An analogous statement for a dscrete lazy random wal s proved n 7. Drected graphs In ths secton we gve the proof of Theorem 1.4. We start by statng a standard result about Euleran graphs whose proof can be found n the dscusson followng Theorem 1 n. We then state and prove some prelmnary results about drected graphs that wll be used n the proof.
Lemma.1. Let G be a drected graph wth the property that the outdegree of every vertex equals ts ndegree. If G s connected, then t s strongly connected. Lemma.. Let G be a fnte strongly connected graph. Suppose there exsts a vertex such that ndeg G () outdeg G () and for all x we have ndeg G (x) outdeg G (x). If τ + s the frst return tme to by a smple random wal on G, then E τ + x outdeg G(x) + outdeg G () ndeg G (). outdeg G () Proof. Snce n every drected graph the sum of the outdegrees must equal the sum of the ndegrees, we get ndeg G () outdeg G () = x (outdeg G (x) ndeg G (x)). (.1) Consder the set A = {x : ndeg G (x) < outdeg G (x)}. We start addng fcttous edges from to all vertces j A untl the total number of edges that come nto j equals outdeg G (j). We call the new graph G as shown n Fgure. A Fgure : The graph G, where the red dashed lnes are the new fcttous edges and the blac lnes are edges n the orgnal graph. In vew of (.1) n the new graph ndeg G () = ndeg G () = outdeg G (). Also ndeg G (x) = outdeg G (x) = outdeg G (x) for all x V (G) \ {}, and hence f π denotes the statonary dstrbuton of a smple random wal on the drected graph G, we obtan π() = ndeg G () x outdeg G(x) + ndeg G (). Snce the drected graph G s strongly connected, the smple random wal on G s rreducble, and hence the return tme to n the graph G s 1/π(),.e., x outdeg G(x) + ndeg G () = outdeg G() ndeg G () outdeg G () E τ + + j A (ndeg G (j) ndeg G (j))s (1 + E j τ ), outdeg G () 3
where τ s the httng tme of vertex n G. However, snce outdeg G () = ndeg G () and E j τ 1 for all j by rearrangng we get E τ + x outdeg G(x) + outdeg G () ndeg G () outdeg G () and ths concludes the proof of the lemma. In the next results, we wll usually need to construct new graphs that come from a drected graph G wth a dstngushed vertex u. In order to avod repettons of the same constructon n many of the statements and proofs we now gve the defnton of the new graph. Defnton.3. Let G be a drected graph and u V (G) a dstngushed vertex. We wrte I u for the set of vertces havng a drected edge to u. For each I u, we construct the graph G as follows: frst we remove u from the graph G together wth all the edges ncdent to t and then we connect every j I u \ {} to usng multple edges f there are multple edges n the orgnal graph between j and u so that outdeg G (j) = outdeg G (j). We defne A = {x : x n G } and the graph (A, E ) to be the subset of G nduced by A. We wrte ndeg A (x) and outdeg A (x) for the ndegree and outdegree of x A n the graph (A, E ). I u \ {} A Fgure 3: The graph (A, E ), where the red dashed lnes represent the new edges and the blac lnes are edges n the orgnal graph. Lemma.4. Let G be a strongly connected graph such that ndeg G (x) = outdeg G (x) for all x. Let u be a dstngushed vertex of G. Fx and suppose that A. The graph (A, E ) s strongly connected and contans. Furthermore f r s the number of drected edges from to u n the graph G, then outdeg A (x) = outdeg G (x) and ndeg A (x) ndeg G (x), outdeg A () = outdeg G () r and ndeg A () outdeg A (). for all x A \ {} 4
Proof. Frst we establsh that f x A, then there s a drected path from x to usng only vertces of A. Indeed, n the orgnal graph G, there s a path from x to, snce G was assumed to be strongly connected. If ths path does not use the vertex u, then we have nothng to show. If t does, then f t uses the edge (, u), then we are done agan. If not, then t uses an edge of the form ( l, u), n whch case snce l I u s connected to by at least one edge n G, t follows that x. Clearly by the defnton of the set A all the vertces n the path from x to are n A. Furthermore, A, snce ts neghbours are n A by defnton. Snce all vertces n A are connected to n both drectons, t follows that the graph defned by A s strongly connected. Agan by defnton t follows that all the neghbours of x A \ {} are n A. Hence f x A \ {} we have outdeg A (x) = outdeg G (x) and outdeg A () = outdeg G () r. th we deduce that ndeg A (x) ndeg G (x) for all x A \ {}. Usng these nequaltes together wth the fact that (outdeg A (x) ndeg A (x)) = ndeg A () outdeg A () x A \{} we deduce that ndeg A () outdeg A () and ths fnshes the proof of the lemma. Lemma.5. Let G be a strongly connected graph on n vertces such that for all vertces x t satsfes ndeg G (x) = outdeg G (x) = d. We fx a vertex u and wrte I u for the set of vertces that have an edge leadng to u. If µ s a probablty measure supported on I u and τ u s the frst httng tme of u by a smple random wal on G, then E µ τ u nd d. Proof. Let X be a smple random wal on G wth X 0 µ and I u = { 1,..., } wth d, snce there could be multple edges. For any I u we wrte r for the number of drected edges from to u. Every tme the random wal s at a vertex I u t has probablty r /d of jumpng drectly to u. If t does not jump, then t starts walng n the remanng graph untl the frst tme that t hts I u agan. Defne (ξ () ) to be the lengths of..d. excursons startng from I u untl the frst tme that they come bac to the set I u wthout httng u ndependently for dfferent. It s clear that addng drected edges from every l I u \ {} to cannot affect ξ () 1. Hence we can upper bound ξ () 1 by the return tme to n the graph (A, E ) constructed n Defnton.3. In ths graph we have outdeg A () = d r. Lemma.4 gves that A satsfes the assumptons of Lemma.. Therefore snce A < n and r 1 for all I u we deduce E ξ () 1 dn d 1. (.) d r We now defne ndependent collectons of random varables B ( 1) 1, B ( 1),.....d. B(r 1 /d) B ( ) 1, B ( ),.....d. B(r /d). B ( ) 1, B ( ),.....d. B(r /d), where B(p) stands for the Bernoull dstrbuton of parameter p. 5
We can realze the random wal X untl the frst tme that t hts u n the followng way: at tme 0 f B (X 0) 1 = 1, then t jumps drectly to u. Otherwse t maes an excurson of length ξ (X 0) 1 untl the frst tme that t comes bac to I u. We defne l(1) = X 0 and ζ 1 = ξ (X 0) 1. Inductvely we defne +1 l( + 1) = X ζ and ζ +1 = ξ s (l(s)). In words, ζ s the tme whch has passed untl the end of the -th excurson and l() s the poston of the random wal at the end of the ( 1)-th excurson. At the end of the ( 1)-th excurson X hts u drectly wth probablty r l() /d. If t does not, thene we attach another excurson of length ξ (l()) and we contnue n the same way untl the frst tme that X hts u. We fnally defne s=1 T = mn{s : B (l(s)) s = 1},.e. T s the number of used Bernoull random varables untl the frst tme that a Bernoull s equal to 1. Hence we can now wrte By the defnton of ζ we get T 1 Eζ T 1 = E = = =1 =1 ξ (l()) (d rl(1) ) E d =1 E µ τ u = Eζ T 1 + 1. (.3) E ξ (l()) 1(T > ) =... (d r l()) E ξ (l()) d Usng (.) we now mmedately get that =1 E E ξ (l()) 1(T > ) ξ (l()), (l(j)) j (l(j)) j. E ξ (l()) (l(j)) j dn d 1, d r l() snce ξ (l()) s the length of an excurson started from the vertex l(). Hence pluggng that nto (.4) and usng that r 1 for all we get Eζ T 1 Ths together wth (.3) gves (d 1) 1 (dn d 1) d = dn d 1. =1 and ths concludes the proof of the lemma. E µ τ u dn d We are now ready to gve the proof of Theorem 1.4. Proof of Theorem 1.4. Note that f G s somorphc to L(d, n), then max E xτ y = d n(n + 1). x,y 6 (.4)
We prove the strct nequalty by nducton on n. For n = 1 t s trvally true. Suppose that for any strongly connected graph on n + 1 vertces wth ndeg(x) = outdeg(x) = d for all x and whch s not somorphc to L(d, n) we have max E xτ y < d n(n + 1). x,y Let G be a strongly connected graph on n + vertces wth ndeg G (x) = outdeg G (x) = d for all x whch s not somorphc to L(d, n + 1). We wll show that for all x and y E x τ y < d (n + 1)(n + ). (.5) Let I = { 1,..., } be the vertces such that there s a drected edge from every l to y. Note that d, snce we are allowng multple edges and self-loops. Clearly we can wrte E x τ y = E x τ I + E µ τ y, (.6) where τ I s the frst httng tme of the set I by the smple random wal on G and µ s a measure on I wth µ( l ) = P x (X τi = l ). By Lemma.5 we mmedately obtan E µ τ y nd + d. (.7) If x I, then (.7) fnshes the proof. So from now on we assume that x / I. Snce n order to ht y we must frst ht the set I, we are gong to loo at the graph not contanng the vertex y and the edges ncdent to t. Clearly addng edges comng out of ponts of I s not gong to change the frst httng tme of the set I startng from x. We now explan how we add extra edges comng out of the set I n order to apply the nducton hypothess to a graph of smaller sze. Let J = {j 1,..., j m } be the vertces such that there s a drected edge from y to every j l. Note that as above m d. Removng the vertex y and ts ncdent edges removes the edges from y to vertces n J as well as edges from vertces n I to y. Therefore n order to eep the n and out degrees of all the vertces n the new graph (obtaned by removng y) equal to d we shall add edges gong from I to J. We descrbe how to acheve ths whlst eepng the graph connected. Let J 1 be the subset of J contanng only those vertces j r whch n graph G have an undrected path from j r to 1 that does not vst vertex y. Suppose we have defned the sets J 1,..., J l. We next defne J l+1 to be the set of j r J \ s l J s whch n G have an undrected path from j r to l+1 that does not vst vertex y. Note that some of the sets J l could be empty, nevertheless snce the orgnal graph s connected we have J r = J. r We now consder only the non-empty subsets J r, whch we ndex by s 1,..., s l so that f j J s1, then t s connected to s1. We then connect s1 to an element of J s (chosen arbtrarly) and s to an element of J s3 and so on. Fnally we connect sm to an element of J s1. At the end of ths procedure, we add edges between I and J so that n the resultng graph every vertex has ndegree equal to outdegree equal to d. Ths s possble, snce the ndegree of y s equal to ts outdegree. We call the resultng graph G as shown n Fgure 4. 7
s1 s sm J s1 J s J sm Fgure 4: The graph G, where the red dashed lnes are the new edges and the blac lnes are edges n the orgnal graph G. We now clam that ths new graph G s connected. Indeed, all the vertces n J r are connected to each other by the defnton of the set J r. Let j 1 J s and j J s+1. Then snce we connect j 1 to s+1 t follows that j 1 s connected to j. Furthermore, as each I s connected to at least one j J, the graph s connected. By Lemma.1 t follows that t s strongly connected (snce we ept the n and out degrees at every vertex equal to d) on n + 1 vertces. If the graph G s not somorphc to L(d, n), then by the nducton hypothess we get E x τ I < d n(n + 1) and ths together wth (.6) and (.7) fnshes the proof of (.5) n ths case. If G s somorphc to L(d, n) (n whch case we dentfy these two graphs), then we shall consder two separate cases: I = 1 and I. We start wth the case I = 1. If I = { 1 } and from 1 the only vertex we can reach n one step s y n G, then E 1 τ y d. (.8) Snce the graph G s a strongly connected graph on n + 1 vertces wth n and out degree of every vertex equal to d, from the nducton hypothess t follows that E x τ I d n(n + 1). Hence ths together wth (.6) and (.8) gves that n ths case E x τ y < d (n + 1)(n + ). If 1 has another out-neghbour h y n G, then 1 cannot be an endpont of the lne, so the httng tme of 1 wll be bounded by the maxmum httng tme on L(d, m) for m n 1 and thus we get E x τ 1 < d n(n + 1). Ths fnshes the proof of (.5) n the case I = 1. It remans to show that f I, then E x τ I < d n(n + 1). 8
Snce the subgraph G s somorphc to L(d, n), then the vertex of I closest to x satsfes E x τ < d n(n + 1). Ths together wth (.7) fnshes the proof of the theorem. 3 Permutaton wal Lemma 3.1. Let σ be a permutaton of {0,..., n}. Then the Marov chan X σ s rreducble. Proof. We frst observe that the Marov chan can be represented as a random wal on a drected graph such that the outdegree of every vertex s equal to ts ndegree (note that we also count self-loops). Hence f we establsh that gnorng orentatons, the underlyng graph s connected, then we can apply Lemma.1 and fnsh the proof. It s easy to see that the undrected graph s connected. Indeed, from the descrpton of the process, all the odd ponts are connected to each other and all the even ponts are connected to each other, snce 1 and + 1 both lead to σ(). Snce 0 and 1 both lead to σ(0), t follows that the two sets (odd and even ponts) are connected, and hence ths concludes the proof. Proof of Theorem 1.1. As we already noted n the proof of Lemma 3.1 above, the Marov chan X σ can be vewed as a random wal on a drected graph such that the n and out degree of every vertex s equal to. Furthermore, from Lemma 3.1 we now that ths graph s strongly connected. Hence applyng Theorem 1.4 shows that E 0 τ n n + n. From Theorem 1.4 we get that equalty s acheved only f the resultng graph s somorphc to L(, n) and σ(0) = 0 and σ(n) = n. It thus follows that σ has to be the dentty permutaton and ths fnshes the proof. Remar 3.. We note that the statement of Theorem 1.1 remans true f we change the Marov chan as follows: whenever at the next step s ether σ() + 1 or σ() 1 equally lely. Indeed, t s easy to see that ths Marov chan s agan a smple random wal on a drected graph whch s strongly connected, and hence Theorem 1.4 apples. 4 Open problem The followng problem was communcated to us by Yuval Peres, but we could not trace ts orgns. We state t here: Open problem: Let σ be a permutaton of { n,..., n}. Let (ξ ) be..d. tang values n { 1, 1} equally lely and set X0 σ = 0 and Xσ t+1 = σ(xσ t + ξ t+1 ) f Xt σ n, n, otherwse Xt+1 σ taes values n {σ(n), σ(n 1)} or {σ( n), σ( n + 1)} respectvely equally lely. Show that the dentty permutaton maxmzes E 0 τ{ n,n}, where τ{ n,n} s the frst httng tme of the set { n, n} by X σ. 9
Remar 4.1. In contrast to Theorem 1.1 where equalty s acheved only when σ s the dentty permutaton, we note that for ths problem ths s no longer the case. In other words, the dentty s not the unque permutaton that maxmzes E 0 τ{ n,n}. Indeed, f σ(x) = x for all x n, n Z, then E 0 τ{ n,n} = n. Also, t s easy to chec that f σ s the permutaton that transposes 0 and 1, then t also acheves the same upper bound. Nevertheless, f σ only transposes wth + 1, then for all 0 < n 1, n. E 0 τ{ n,n} = n n 3 n 1 + n(n ) By argung n a smlar way as n the proof of Lemma 3.1, t s easy to see that the process X σ can be vewed as a random wal on a drected graph whch s strongly connected. Hence by Theorem 1.4 we mmedately get that for any permutaton σ E 0 τ{ n,n} 4n + 6n +. Acnowledgements We are grateful to Yuval Peres for tellng us about the open problem stated n Secton 4. We than Jason Mller and James Norrs for helpful dscussons. Part of ths wor was completed whle the frst author was a postdoc at the Unversty of Angers, France. References 1 M. Azenman and B. Smon. Brownan moton and Harnac nequalty for Schrödnger operators. Comm. Pure Appl. Math., 35():09 73, 198. Béla Bollobás. Modern graph theory, volume 184 of Graduate Texts n Mathematcs. Sprnger- Verlag, New Yor, 1998. 3 A. Burchard and M. Schmucenschläger. Comparson theorems for ext tmes. Geom. Funct. Anal., 11(4):651 69, 001. 4 James Allen Fll. Egenvalue bounds on convergence to statonarty for nonreversble Marov chans, wth an applcaton to the excluson process. Ann. Appl. Probab., 1(1):6 87, 1991. 5 Rav Montenegro. The smple random wal and max-degree wal on a drected graph. Random Structures Algorthms, 34(3):395 407, 009. 6 Y. Peres and P. Sous. An sopermetrc nequalty for the Wener sausage. Geom. Funct. Anal., (4):1000 1014, 01. 7 P. Sous and P. Wnler. Mxng tmes and movng targets, 01. Avalable at arxv:110.536. 10