Stationary distribution and cover time of random walks on random digraphs

Stationary distribution and co v er time of random walks on random digraphs. Coli n Co o p er ∗ Alan F rieze † Octob er 13 , 20 18 Abstract W e study prop erties of a simp le random w alk on the random digraph D n,p when np = d log n, d > 1. W e pro v e that whp th e stationary probabilit y π v of a v ertex v is asymptotic to deg − ( v ) /m where deg − ( v ) is the in-degree of v and m = n ( n − 1) p is the e xp ected n umber of e dges of D n,p . If d = d ( n ) → ∞ w ith n , the stationary distribution is asymptotically uniform whp . Using th is r esult we prov e that, for d > 1, whp the co v er time of D n,p is asymptotic to d log( d/ ( d − 1)) n log n . I f d = d ( n ) → ∞ with n , then the co v er time is asymp totic to n log n . 1 In tro duction Let D = ( V , E ) b e a strongly connected digraph w ith | V | = n , and | E | = m . F or the simple random walk W v = ( W v ( t ) , t = 0 , 1 , . . . ) on D starting at v ∈ V , let C v b e t he exp ected time tak en to visit ev ery vertex of D . The c over time C D of D is defined as C D = max v ∈ V C v . F or connected undirected graphs, the co v er time is w ell understo o d, and has b een extensiv ely studied. It is an old result of Aleliunas, Karp, Lipton, Lov´ asz and Rac k off [2] tha t C G ≤ 2 m ( n − 1). I t was sho wn b y F eige [10], [11], that for an y connected graph G , the co v er time satisfies (1 − o ( 1 )) n lo g n ≤ C G ≤ (1 + o (1) ) 4 27 n 3 , where log n is the natural logarithm. An example of a graph achiev ing the low e r b ound is the complete graph K n whic h has co v er ∗ Department of Infor matics, King’s Co llege, Universit y of London, Lo ndon WC2R 2LS, UK † Department of Ma thematical Sciences, Car negie Mellon Univ ersity , Pittsbur gh P A15213 , USA. Supported in part by NSF grant CCF050 2793. 1 time determined by the Coupo n Collector problem. T he lol lip op graph consisting of a pat h of length n/ 3 joined to a clique of size 2 n/ 3 has co v er time asymptotic to the upp er b ound of (4 / 27) n 3 . F or directed graphs co ve r time is less w ell understo o d, a nd t here are strongly connected digraphs with cov er time exp onen tial in n . An example of this is the digraph consisting of a directed cycle (1 , 2 , ..., n, 1) , and edges ( j, 1), fr o m v ertices j = 2 , ..., n − 1. Starting from v ertex 1 , the expected time for a random walk to reac h v ertex n is Ω(2 n ). In earlier pap ers, we inv estigated the co v er time of v arious classes of (undirected) random graphs, a nd derive d precise r esults for their co v er times. The main results can b e summarized as follo ws: • [4] If p = d log n/n and d > 1 then whp C G n,p ∼ d log  d d − 1  n log n . • [7, 8] Let d > 1 and let x denote the solution in ( 0 , 1 ) o f x = 1 − e − dx . Let X g b e the gian t comp o nen t of G n,p , p = d/n . Then whp C X g ∼ dx (2 − x ) 4( dx − log d ) n (log n ) 2 . • [5] If r ≥ 3 is a constan t and G n,r denotes a random r -regular graph on v ertex set [ n ] with r ≥ 3 then whp C G n,r ∼ r − 1 r − 2 n log n . • [6] If m ≥ 2 is constan t and G m denotes a p r efer ential attachment gr aph of a v erage degree 2 m then whp C G m ∼ 2 m m − 1 n log n . • [9] If k ≥ 3 and G r,k is a random geometric graph in ℜ k of ball size r suc h that the exp ected degree of a v ertex is asymptotic to d log n , then whp C G r,k ∼ d log  d d − 1  n log n . A few remarks on notation: W e use the notation a ( n ) ∼ b ( n ) to mean that a ( n ) /b ( n ) → 1 as n → ∞ . Some inequalities in this pa p er only hold for la rge n . W e assume henceforth that n is sufficie n tly large for a ll claimed inequalities to hold. All whp statemen ts in this pap er are relativ e to the class of random digraphs D n,p under discus sion, and not the random w a lk. In this pap er we turn our attention to the co v er time o f ra ndom directed graphs. Let D n,p b e the random digraph with vertex set V = [ n ] where eac h p o ssible directed edge ( i, j ) , i 6 = j is indep enden tly inc luded with probabilit y p . It is known that if np = d log n = log n + γ where γ = ( d − 1) log n → ∞ then D n,p is strongly connected whp . If γ as defined t ends to −∞ then whp D n,p is not strongly connected. As we do not hav e a direct reference to this r esult, w e next giv e a brief pro of of this. It is easy to show that if np = log n − γ where γ → ∞ , there are v ertices of in-degree zero whp . On the o ther hand, if np = log n + γ where γ → ∞ then [12] sho ws that the random digraph is Hamiltonian and hence strongly connected. Strong connectivit y for np = log n + γ where γ → ∞ also follows directly fr om the pro of o f (60). W e determine the co v er time of D n,p for v a lues of p at or a b ov e t he thr eshold for strong connectivit y . 2 Theorem 1. L et np = d log n wher e d = d ( n ) is such that γ = np − log n → ∞ . Th e n whp C D n,p ∼ d log  d d − 1  n log n. Note that if d = d ( n ) → ∞ with n , then w e hav e C D n,p ∼ n lo g n . The metho d w e use to find the cov er time of D n,p requires us t o know the stationary dis- tribution of the random walk. F or an undirected graph G , the stat io nary distribution is π v = deg( v ) / 2 m , where deg( v ) denotes the degree of ve rtex v , and m is the num b er of edges in G . F or a digraph D , let deg − ( v ) denote the in-degree of v ertex v , deg + ( v ) denote the out- degree, and let m b e the nu m b er of edges in D . F or strongly connected digraphs in which each v ertex v has in-degree equal to out-degree (deg − ( v ) = deg + ( v )) , then π v = deg − ( v ) / m . F o r general digra phs, how ev er, there is no simple form ula for the stationary distribution. Indeed, there ma y not b e a unique stationary measure. T he main tec hnical task of this pap er is t o find go o d estimates for π v in the case of D n,p . Along the w ay , this implies uniqueness of the stationary measure whp . W e summarize our result conc erning the stationary distribution in The orem 2 b elo w. F or a giv en v ertex v , define a quan tit y ς ∗ ( v ), whic h in essence depends on the in-neighbour w of v with minim um out- degree: ς ∗ ( v ) = max w ∈ N − ( v )  deg − ( w ) deg + ( w )  . (1) Theorem 2. L et np = d log n wher e d = d ( n ) is such that γ = np − log n → ∞ . L et m = n ( n − 1) p . Then whp , for al l v ∈ V , π v ∼ de g − ( v ) + ς ∗ ( v ) m . W e note the following sp ecial cases. Remark 1. We pr ove in L emma 14 that whp ς ∗ ( v ) = o ( de g − ( v )) for almost al l v e rtic es v . F or these vertic e s , the ς ∗ ( v ) term c an b e absorb e d into the err or term of π v . Remark 2. If γ = ω (log log n ) then whp ς ∗ ( v ) = o ( de g − ( v )) for a l l ve rtic es v . In p a rticular when d = 1 + δ, δ > 0 and c onstant then the minimum out-de gr e e is Ω(log n ) , in which c as e , π v ∼ de g − ( v ) / m . Remark 3. If d = d ( n ) → ∞ with n , whp the stationary distribution of D n,p is π v ∼ 1 /n . 2 Outline of the pap er A t the heart of our appro a c h to the co v er time is the following claim: Supp o se that T is a ”mixing time” for a simple random w alk and A v ( t ) is the ev en t that W u do es no t visit v in 3 steps T , T + 1 , . . . , t . Then, essen tially , Pr ( A v ( t )) ∼ e − tπ v /R v . (2) Here R v ≥ 1 , v ∈ V is the exp ected n um b er of v isits/returns to v b y the walk W v within T time steps. This is the conten t of Lemma 3 and w e ha v e used it to prov e our previous r esults on this topic. Giv en (2) w e can estimate the cov er time from ab ov e via C u ≤ t + 1 + X v X s ≥ t Pr ( A v ( s )) . This is ( 9 3) and we ha v e used this inequalit y previously . Here C u is the exp ected time for W u to visit ev ery v ertex. It is v alid for arbitrary t and w e get our upp er b ound for C D b y c ho osing t large enough so that the double sum is o ( t ). W e estimate the co v er time from b elo w b y using the Che b yshev inequalit y . W e c ho ose a set of v ertices V ∗∗ that are candidates for taking a long time to visit and estimate the exp ected size of the set V † of v ertices in V ∗∗ that hav e not b een visite d within our estimate of the co v er time. W e sho w that E | V † | → ∞ . W e then t ak e pairs of vertice s v , w ∈ V ∗∗ and con tract them to a single v ertex γ and then use (2) to sho w that Pr ( A γ ( t )) ∼ P r ( A v ( t )) Pr ( A w ( t )). The main problem here is that we do not kno w π v and muc h of the pap er is devoted to pro ving that, esse n tially , whp , π v ∼ deg − ( v ) m for all v ∈ V . (3) Our pro of of this leads easily to a claim that whp T = O (lo g 2 n ) and w e will find then find that it is easy to pro v e that R v = 1 + o ( 1) f or all v ∈ V . W e approx imate the statio nary distribution π using the express ion π = π P k , where P is the transition matrix. F or suitable c hoices of k w e find w e can b ound P ( k ) x ( y ) = Pr ( W x ( k ) = y ) from a b ov e and b elo w b y v a lues indep enden t of x and obtain, essen tially , P ( k ) x ( y ) ∼ deg − ( y ) m an expression indep endent of x . (3) f ollo ws easily from this. T o estimate P ( k ) x ( y ) from b elow we pro ceed a s follows : W e let k = 2 ℓ = 2 3 log np n . W e consider t w o Breadth First Searc h trees of depth ℓ . T low x branc hes out from x to depth ℓ and T low y branc hes in to y from depth ℓ . Almost a ll of the w alk measure asso ciat ed with w alks of length 2 ℓ + 1 from x to y will go from x lev el b y lev el to the b oundary of T low x , jump across to the b oundary of T low y and then go lev el by lev el to y . W e ana lyse suc h w alks and pr o duce a lo w er b ound. 4 T o estimate P ( k ) x ( y ) from ab ov e we change the depths of the out- tree from x and the in-tree to y . This eliminates some complexities. In computing the lo wer b ound, w e ignored some paths that take more circuitous ro utes from x to y and w e hav e to sho w that these do not add m uc h in w alk measure. The structure o f the pap er is no w as follow s: Section 3 describ es Lemma 3 that w e ha v e often used b efore in the analysis of the co v er time. Section 4 establishes many structural prop erties of D n,p . In Section 5 we prov e the low er and upp er b ounds giv en in Theorem 2. These b ounds hold for any digr aph with the high probability structures elicited in Section 4. Sections 4 and 5, whic h form the main b o dy of this paper, are first pro v ed under the assumption that 2 ≤ d ≤ n δ , for some sm all δ > 0, an assumption w e refer to as A ssumption 1 . In Section 6, we extend the pro of of Theorem 2 by remo ving Assumption 1. Section 7 is short and establishes tha t the conditions of Lemma 3 hold. T o do this, w e use a b o und on the mixing time, based on results obtained in Sections 5, 6. F inally , in Section 8 w e establish the whp co v er time, as g iv en in Theorem 1. 3 Main Lemma In this section D denotes a fixed strongly connected digra ph with n v ertices. A random w a lk W u is started from a v ertex u . Let W u ( t ) b e the v ertex reac hed at step t , let P b e the matrix of transition probabilities of the walk and let P ( t ) u ( v ) = Pr ( W u ( t ) = v ). W e a ssume that the random walk W u on D is ergo dic with stationary distribution π . Let d ( t ) = max u,x ∈ V | P ( t ) u ( x ) − π x | , and let T be a p ositiv e in teger suc h that for t ≥ T max u,x ∈ V | P ( t ) u ( x ) − π x | ≤ n − 3 . (4) Consider the walk W v , starting at vertex v . Let r t = r t ( v ) = P r ( W v ( t ) = v ) b e t he probabilit y that this walk returns to v at step t = 0 , 1 , ... . Let R T ( z ) = T − 1 X j =0 r j z j (5) and let R v = R T (1) . Lemma 3. F ix a vertex u ∈ V and for v ∈ V and t ≥ T le t A v ( t ) b e the event that W u do es not vis i t v in steps T , T + 1 , . . . , t . Supp ose that 5 (a) F or some c onstant θ > 0 , we have min | z |≤ 1+ λ | R T ( z ) | ≥ θ . (b) T 2 π v = o (1) and T π v = Ω( n − 2 ) fo r al l v ∈ V . L et K > 0 b e a sufficiently lar ge absolute c ons tant and let λ = 1 K T . (6) Then, with p v = π v R v (1 + O ( T π v )) , (7) we have that for al l v ∈ V an d t ≥ T , Pr ( A v ( t )) = (1 + O ( T π v )) (1 + p v ) t + O ( T 2 π v e − λt/ 2 ) . (8) 4 Structur al Prop ert i e s of D n,p In this section w e gather together some whp prop erties of D n,p needed for the pro of o f Theorem 2. Some are quite elab orate and so we will try to motiv at e them where w e can. W e stress that t hroughout this section, the probabilit y space is the space of D n,p and not the space of walks on an instance. Once w e complete this section how ev er, w e can concen trate on estimating the co v er t ime of a digraph with the giv en pro p erties, as long as Assumption 1 of (20) holds. F or large p outside Assumption 1, the pro of is quite simple. 4.1 Degree Sequence etc Chernoff Bounds The fo llowing inequalities are used extensiv ely throughout this pap er. Let Z = Z 1 + Z 2 + · · · Z N b e t he sum of the indep enden t ra ndom v a riables 0 ≤ Z i ≤ 1 , i = 1 , 2 , . . . , N with E ( Z 1 + Z 2 + · · · + Z N ) = N µ . Then Pr ( | Z − N µ | ≥ ǫN µ ) ≤ 2 e − ǫ 2 N µ/ 3 . (9) Pr ( Z ≥ αN µ ) ≤ ( e/α ) αN µ . (10) F or pr o ofs see for example Alon and Sp encer [3]. 6 The next lemma giv es some prop erties of the degree sequence of D n,p . The lemma can b e pro v ed b y the use of the first and second momen t methods (see [4] for v ery similar calculations). Let np = d log n and let ∆ 0 = C 0 np w her e C 0 = 30 . (11) Lemma 4. (i) First assume that np = d log n wher e 1 < d = O (1) and ( d − 1) log n → ∞ . L et D ( k ) = n  n − 1 k  p k (1 − p ) n − 1 − k denote the exp e cte d numb er of vertic es v with d e g − ( v ) = k ≤ ∆ 0 . Note that D ( k ) ≤ 2 n d − 1  nep k  k . (12) L et D ( k ) denote the actual numb er o f ve rtic es of in-de gr e e k , and let K 0 = { k ∈ [1 , ∆ 0 ] : D ( k ) ≤ (log n ) − 2 } . K 1 = { 1 ≤ k ≤ 15 : (log n ) − 2 ≤ D ( k ) ≤ log log n } . K 2 = { k ∈ [16 , ∆ 0 ] : (log n ) − 2 ≤ D ( k ) ≤ (log n ) 2 } . K 3 = [1 , ∆ 0 ] \ ( K 0 ∪ K 1 ∪ K 2 ) . The de gr e e s e quenc e h as the fol lowing pr o p erties. (a) If d − 1 ≥ (log n ) − 1 / 3 then K 1 = ∅ , min { k ∈ K 2 } ≥ (log n ) 1 / 2 , | K 2 | = O (log log n ) . (b) The fol lowing holds whp : F or al l de gr e es k ∈ [1 , ∆ 0 ] . k ∈ K 0 , D ( k ) = 0 , k ∈ K 1 , D ( k ) ≤ (log log n ) 2 , (13) k ∈ K 2 , D ( k ) ≤ (log n ) 4 , (14) k ∈ K 3 , D ( k ) 2 ≤ D ( k ) ≤ 2 D ( k ) . (15) (ii) Supp ose that 1 < d ≤ n δ wher e δ is a smal l p ositive c onstant. L et k ∗ = ⌈ ( d − 1) log n ⌉ . L et V ∗ =  v ∈ V : de g − ( v ) = k ∗ and de g + ( v ) = k † = ⌈ d log n ⌉  and let γ d = ( d − 1) log  d d − 1  . Then whp | V ∗ | ≥ n γ d 10 d log n . 7 (iii) L et D b e the event  ∃ v ∈ V : de g + ( v ) ≥ ∆ 0 or de g − ( v ) ≥ ∆ 0  , (16) then Pr ( D ) ≤ n − 10 e − 10 np . (17) (iv) The numb er of e dges | E ( D n,p | ∼ m = n ( n − 1) p whp . (v) de g ± ( v ) ∼ np for al l v ∈ V whp if d → ∞ . Pro of W e will only giv e an outline pro of of (ii) as the other claims hav e (essen tially) b een pro v ed in [4]. W e hav e E ( | V ∗ | ) = n  n − 1 k ∗  n − 1 k †  p k ∗ + k † (1 − p ) 2 n − 2 − k ∗ − k † Using,  n k  ≥ 1 3 k 1 / 2  ne k  k , whic h is obt a inable fro m Stirling’s approximation, w e see that E ( | V ∗ | ) ≥ (1 − o (1 ) ) 1 9( d − 1) 1 / 2 d 1 / 2 log n  d d − 1  ( d − 1) log n . (18) The Cheb yshev inequalit y w ill sho w t ha t | V ∗ | is concen trated around its mean, pro vided w e v erif y that the mean tends to ∞ . If d > 2 then  d d − 1  ( d − 1) log n ≥ exp  ( d − 1) log n  1 d − 1 − 1 2( d − 1) 2  ≥ n 1 / 2 . No w d ≤ n δ and so E ( | V ∗ | ) → ∞ follo ws fro m (18). If d = 1 + ǫ ≤ 2 and ǫ is b ounded a w a y from zero then so is  d d − 1  d − 1 and so E ( | V ∗ | ) = n Ω(1) . So now supp ose that ǫ = ω log n where ω = ω ( n ) → ∞ and ω = o (log n ). Then,  d d − 1  d − 1 ≥  log n ω  ω . If ω ≥ log 1 / 2 n then  d d − 1  d − 1 ≥ e log 1 / 2 n and if ω ≤ log 1 / 2 n then  d d − 1  d − 1 ≥ log ω / 2 n . In either case E ( | V ∗ | ) ≥ log θ n where θ = θ ( n ) → ∞ (19) ✷ 4.1.1 Assumption 1: a con v enien t restr iction W e will first carry out the main b o dy of the pro of under the followin g assumption: Assumption 1 : 2 ≤ d ≤ n δ . (20) 8 Here 0 < δ ≪ 1 is some small fixe d po sitive constant. W e note that o ur ch oice of the v alue d ≥ 2 is somewhat arbitrar y , and any constan t larger than 1 w ould suffice. W e w a it un til Section 6 to remo v e Assumption 1. The pro of for d > n δ is m uc h simpler and is giv en separately in Section 6.1 . The pro of for 1 < d ≤ 2 is giv en in Section 6.2. Under Assum ption 1 a nd d = O (1) there is a constan t c > 0 and an in t erv a l I = [ c 0 np, ∆ 0 ] , (21) suc h that if ν ∈ [3 n/ 4 , n ] then there exists γ = γ ( c ) > 0 suc h that Pr ( B in ( ν, p ) ∈ I ) = 1 − o ( n − 1 − γ ) . (22) When d → ∞ w e can take c = 0 . 9 99 and C 0 = 1 . 001 . Let E + S (resp. E − S ) b e the ev en t that the in-degree (resp. out-degree) of all v ertices in S ⊆ V are in the in terv al I . Th us e.g. E + S =  D n,p : ∀ v ∈ S ⊆ V , deg + ( v ) ∈ [ c 0 np, ∆ 0 ]  . (23) Let E S = E + S ∩ E − S . Then for an y S ⊆ V w e ha v e Pr ( E S ) = 1 − O ( n − γ ) . (24) 4.2 Prop erties needed for a lo w er b ound on the stati onary distri- bution The calculations in this section are made under Assumption 1. Fix v ertices x, y where x = y is allow ed. Most short random w alks f r o m v ertex x to ve rtex y tak e the form of a simple directed path, or cycle if x = y . W e can coun t suc h paths (or cycles) with the help of a breadth first o ut-tree T low x ro oted at x , and a breadth first in-tree T low y ro oted at y . W e build these trees to depth ℓ , where ℓ =  2 3 log np n  . (25) F or a v ertex v let N − ( v ) be the set of in-neighbours of v and for a set S , let N − ( S ) = S v ∈ S N − ( v ). Define N + ( v ) , N + ( S ) similarly with respect to out-neighbours. Construction of in-tree T low y . F or fixed y ∈ V , w e build a tree T y = T low y ro oted at y , in breadth-first fashion. Define Y 0 = { y } . T he tree T low y has lev el sets Y i , i = 0 , 1 , ..., ℓ , a nd v ertex set Y = ∪ ℓ i =0 Y i . Let Y ≤ i = ∪ j ≤ i Y j . 9 Let T y ( i ) b e the tree consisting of the first i lev els of the breadth first tree T y ( ℓ ) = T low y . Giv en T y ( i ) we construct T y ( i + 1 ) by adding the in-neigh b ours of Y i in V \ Y ≤ i . T o remo v e am biguit y , the v ertices of Y i are pro cessed in increasing o rder of v ertex lab el. L et this order b e ( v 1 , v 2 , ..., v | Y i | ). F or v ∈ Y i , the set N − T ( v ) is the subset of Y i +1 ∩ N − ( v ) whose edges in the tree T y ( i + 1) p oint to v . F ormally these sets are defi ned as follows : N − T ( v 1 ) = N − ( v 1 ) \ Y ≤ i , and in general N − T ( v k ) = N − ( v k ) \ ( Y ≤ i ∪ N − ( v 1 , ..., v k − 1 )). Th us if v ∈ Y i and w ∈ Y i +1 and ( w , v ) is an edge of T y ( i + 1), then v = v k is the first out-neigh b our of w in Y i in the order ( v 1 , ..., v k ). As w ∈ Y i +1 , there are no edges from w to Y ≤ i − 1 , and thus no edges b et w een w and Y ≤ i − 1 ∪ { v 1 , ..., v k − 1 } . Let deg − T ( v ) denote the in- degree o f v ∈ Y in T low y . If deg − T ( v ) > 0 for all v ∈ Y ≤ ℓ − 1 , w e say the c onstruction of T low y suc c e e ds . Asso ciated with this construction of T low y is a set of parameters and random v aria bles. • F o r v j ∈ Y i , let σ ( v j ) = | V \ [ Y ≤ i ∪ N − T ( v 1 , ..., v j − 1 )] | . Th us σ ( v j ) is the num b er of v ertices not in T y after a ll in-neigh b ours of v 1 , ..., v j − 1 ha v e b een added to T y . • Let B ( v j ) ∼ B in ( σ ( v j ) , p ). Th us B ( v j ) = | N − T ( v j ) | = deg − T ( v j ), t he in-degree of v j in T y . • Let σ ′ ( v j ) = | V \ [ Y ≤ i − 1 ∪ { v 1 , ..., v j − 1 }| . • Let D ∼ 1 + B in ( σ ′ ( v j ) , p ), and let D ( j, k ) , k = 1 , ..., B ( v j ) b e independen t copies of D . The in terpretation of the random v ariable D ( j, k ) is as follow s. If w k ∈ N − T ( v j ) then D ( j, k ) is the out-degree of w k in D n,p . The one ar ises from ( w k , v j ) b eing the first edge from w k to Y i . Construction of out-tree T low x . Give n the set of v ertices Y o f T low y , w e define X 0 = { x } , X 1 , . . . , X ℓ where X i +1 = N + ( X i ) \ ( Y ∪ X 0 ∪ · · · ∪ X i ) for 0 ≤ i < ℓ . If w ∈ X i +1 is the out-neighbour o f more than one v ertex of X i , we only k eep the edge ( z , w ) with z as small as p ossible as in the construction of T low y . Let X = S ℓ i =0 X i and let T low x denote the BFS tree constructed in this manner. Let deg + T ( v ) = | N + T ( v ) | denote the out-degree of v ∈ X in T low x . Similarly to the construction of T low y , the v alue of deg + T ( v ) is give n by a random v ariable B ( v ) ∼ B in ( σ ( v ) , p ). If deg + X ( v ) > 0 for all v ∈ X ≤ ℓ − 1 , we sa y the construction of T low x succeeds . W e ga ther tog ether a few facts ab out T low x , T low y that w e need f or t he pro ofs o f t his section. Lemma 5. (i) With pr ob ab i l i ty 1 − O ( n − γ ) , the c on s truction of T low x , T low y suc c e e ds for al l x, y ∈ V . 10 (ii) With pr ob abi l i ty 1 − O ( n − γ ) , F or a l l x , and for al l v ∈ X ≤ ℓ − 1 , de g + T ( v ) ∈ [ c 0 np (1 − o (1)) , C 0 np ] , F or a l l y and for al l v ∈ Y ≤ ℓ − 1 , de g − T ( v ) ∈ [ c 0 np (1 − o (1)) , C 0 np ] . (iii) Given E + x , E − y , then for i ≤ ℓ , | X i | ∼ d e g + ( x )( np ) i − 1 , | Y i | ∼ de g − ( y )( np ) i − 1 qs 1 Pro of W e give pro o fs for T low x , the pro ofs for T low y are similar. P art (i), ( ii). Let X = { x 0 = x, x 1 , . . . , x N } where x i is the i -th vertex added to T low x . F or x j ∈ X , let f ( x j ) = | N + ( x j ) ∩ ( Y ∪ { x 0 , x 1 , . . . , x j − 1 } ) | . Th us deg + ( v ) = deg + T ( v ) + f ( v ). W e can b ound f ( v ) by the binomial B in ( N X , p ) where N X = | Y | + | X | . This is true eve n after constructing T low x , T low y , b ecause the out-edges of v counted b y f ( v ) hav e not b een exp osed. Assuming ¬D , see (17 ) , w e hav e N X ≤ 2 ℓ X i =1 ∆ i 0 = n 2 / 3+ o (1) . Using the Chernoff b ound (10 ) , w e hav e with ω = log 1 / 2 n that Pr  f ( v ) ≥ np ω  ≤ Pr  B in ( n 2 / 3+ o (1) , p ) ≥ np ω  = O ( n − 10 ) . (26) The eve n t S x ∈ v E + X ≤ ℓ − 1 ⊆ E + V and the lat t er holds with probabilit y 1 − O ( n − γ ). Th us giv en (26), and E + V w e ha v e deg + T ( v ) > 0 for v ∈ X ≤ ℓ − 1 for all x ∈ V . In summary , whp t he construction of T low x succeeds for a ll x ∈ V , and deg + T ( v ) ∈ [ c 0 np (1 − o (1)) , C 0 np ] for all v ∈ X ≤ ℓ − 1 in all trees T low x , x ∈ V . P art (iii). By construction T low x w a s made after T low y , so | X i | dep ends on T low x and T low y . Assume E + x =  deg + ( x ) ∈ I = [ c 0 np, C 0 np ]  , and that | Y | ≤ n 2 / 3+ o (1) . (Strictly speaking w e should v erify t hat | Y | ≤ n 2 / 3+ o (1) b efore considering T low x . On the other hand, the pro of w e presen t no w can b e applied to T low y ). F o r i ≥ 1, | X i +1 | is distributed as B in ( n − o ( n ) , 1 − (1 − p ) | X i | ). The num b er of trials n − o ( n ) is based on the inductiv e assumption that | X j +1 | = (1 + o (1 ) ) n | X j | p and that | X ≤ j | p = o (1). That these ass umptions hold qs . follows from the Chernoff b ounds. W e th us ha v e that | X ℓ | ∼ deg + ( x )( np ) ℓ − 1 , qs . (2 7) ✷ F or u ∈ X i let P u denote the path of length i from x to u in T low x and α i,u = Y w ∈ P u w 6 = u 1 deg + ( w ) . 1 A sequence o f even ts A n , n ≥ 0 ho ld ‘quite surely ’( qs ) if Pr ( A n ) = 1 − O ( n − K ) for any constant K > 0. 11 In the eve n t that the construction of T low x fails to complete to depth ℓ , let P u ∈ X ℓ α ℓ,u = 0. Similarly , f or v ∈ Y i let Q v denote the path fro m v to y in T low y and β i,v = Y w ∈ Q v w 6 = y 1 deg + ( w ) . (28) In the eve n t that the construction of T low y fails to complete to depth ℓ , let P v ∈ Y ℓ β ℓ,v = 0. Let Z ( x, y ) = Z low ( x, y ) = X u ∈ X ℓ v ∈ Y ℓ α ℓ,u β ℓ,v 1 uv deg + ( u ) (29) where 1 uv is the indicator for the existence of the edge ( u , v ) and w e take 1 uv deg + ( u ) = 0 if deg + ( u ) = 0 . Note that Z ( x, y ) = 0 if w e fail to construct T low x or T low y . Remark 4. The imp ortanc e o f the quantity Z ( x, y ) lies in the fact that it is a lower b ound on the pr ob ability that W x (2 ℓ + 1 ) = y . The aim of the next few lemm as is to pr ove (se e L emma 9), under Assumption 1, that with m = n ( n − 1) p , Pr  ∃ x,y ∈ V such that Z ( x, y ) 6 = (1 + o ( 1)) de g − ( y ) m  = O ( n − γ ) . (30) The first two lemmas give whp b ounds for P u ∈ X ℓ α ℓ,u , P v ∈ Y ℓ β ℓ,v r esp e c tively, as use d in the thir d lemma and its c or ol lary. Lemma 6. L et A 1 ( x, y ) = ( 1 − ǫ X ≤ X u ∈ X ℓ α ℓ,u ≤ 1 ) . (31) F or ǫ X = O (1 / √ log n ) , Pr  ¬A 1 ( x, y ) ∩ E + x  = o ( n − 10 ) . (32) Pro of F or u ∈ X ℓ let xP u = ( u 0 = x, u 1 , . . . , u ℓ = u ) denote the path from x to u in T low x . F or the random w alk on the digraph T low x , starting at x ; X ℓ is reac hed with proba bility Φ = 1 in exactly ℓ steps, after whic h the w alk halts. Th us 1 = Φ = X u ∈ X ℓ Y v ∈ P u v 6 = u 1 deg + T ( v ) ≥ X u ∈ X ℓ α ℓ,u . (33) 12 W e assume that the construction of T low x succeeds , and that deg + T ( v ) > 0 for v ∈ X ≤ ℓ − 1 , as established in Lemma 5. In the nota t ion of that lemma, deg + ( v ) = deg + T ( v ) + f ( v ). No w Φ = X u ∈ X ℓ Y v ∈ P u 1 deg + ( v ) − f ( v ) = X u ∈ X ℓ Y v ∈ P u 1 deg + ( v ) ! Y v ∈ P u 1 1 − f ( v ) / deg + ( v ) ! = X u ∈ X ℓ α ℓ,u Y v ∈ P u 1 1 − f ( v ) / deg + ( v ) ! . No w if Y v ∈ P u 1 1 − f ( v ) / deg + ( v ) ≤ 1 + h ∀ u ∈ X ℓ , (34) then P u ∈ X ℓ α ℓ,u = 1 − o (1) provided h = o (1 ). W e next prov e we can c ho ose h = O (1 / √ log n ), whic h dete rmines our v alue of ǫ X . Similar to the pro of of (26 ) o f Lemma 5 w e hav e, with ω = √ log n that Pr X v ∈ xP u f ( v ) ≥ np ω ! ≤ Pr  B in ( n 2 / 3+ o (1) , p ) ≥ np ω  = O ( n − 10 ) . (35) Using (26), and (35) it f ollo ws that X v ∈ P u f ( v ) deg + ( v ) − f ( v ) ≤ 1 cω − 1 . F or 0 < x < 1, (1 − x ) − 1 ≤ e x/ (1 − x ) , and so Y v ∈ P u 1 1 − f ( v ) / deg + d ( v ) ≤ exp X v ∈ P u  f ( v ) deg + ( v ) − f ( v )  ! ≤ exp  1 cω − 1  = 1 + O  1 ω  . (36) There are at most n trees and n paths p er tree and so (34), with ǫ X = h = O (1 /ω ) = O (1 / √ log n ), follows from (36). This completes the pro o f of (32). ✷ The next step is to o bta in an estimate o f P v ∈ Y ℓ β ℓ,v . The pro of is inductiv e, moving down the tree T y lev el b y lev el. F o r brevity we write d + ( u ) = deg + ( u ) , d − T ( u ) = deg − T ( u ) etc. Let the r a ndom v ar ia ble W ( y , i ) b e defined b y 13 W ( y , i ) = X u ∈ N − ( y ) X v ∈ Y i Y z ∈ v P u 1 d + ( z ) , where for v ∈ Y i the notation means tha t the the unique path v P uy from v to y in T y passes through u , and that v P u is written a s v = z i , ..., z j , ..., z 1 = u in the pro duct term. Note that W ( y , ℓ ) = X v ∈ Y ℓ β ℓ,v . Define W ∗ ( y , i ) by W ∗ ( y , i ) = X u ∈ N − ( y ) X v ∈ Y i d − T ( v ) Y z ∈ v P u 1 d + ( z ) , where for v ∈ Y ℓ w e define d − T ( v ) = 1 so that W ( y , ℓ ) = W ∗ ( y , ℓ ). Note that W ∗ ( y , 1) = X u ∈ N − ( y ) d − T ( u ) d + ( u ) . W e prov e the following lemma for a more general v alue of ℓ , as it is also used in our pro of of the upp er b ound. Lemma 7. L et A 2 ( y ) =    D n,p : X v ∈ Y ℓ β ℓ,v = (1 + O (1 / p log n )) 1 np X u ∈ N − ( y ) d − T ( u ) d + ( u )    . (37) L et ℓ = η log np n whe r e 0 < η ≤ 2 / 3 . Then unde r Assumption 1, Pr ( ∃ y ∈ V such that ¬A 2 ( y )) = O ( n − γ ) . Pro of The lemma is prov ed inductiv ely assuming E − y and E + Y \{ y } . W e pro v e the induction for 2 ≤ i ≤ ℓ , where b y assumption ( np ) ℓ = O ( n 0 . 67 ). Let E [ d + ( i )] W ( y , i ) b e t he exp ectation of W ( y , i ) ov er ( d + ( v ) , v ∈ Y i ), conditional on all other degrees d + ( u ) > 0 , d − T ( u ) , u ∈ Y ≤ i − 1 b eing fixed suc h that | Y ≤ i − 1 | ∼ d − ( y )( np ) i − 1 ≤ n 0 . 67 whic h is true qs from Lemma 5. F or v ∈ Y i , d + ( v ) is distributed as D ( v ) ∼ 1 + B in ( σ ′ ( v ) , p ), for some σ ′ ( v ) ∈ I 0 = [ n − O ( n 0 . 67 ) , n ]. G iv en the v a lues σ ′ ( v ) for v ∈ Y i , the D ( v ) are indep enden t random v a riables. 14 F or v ∈ Y i , let v P u b e written v w P u , where ( v , w ) ∈ T y . Then E [ d + ( i )] Y z ∈ v P u 1 d + ( z ) ! = E  1 d + ( v )  Y z ∈ w P u 1 d + ( z ) ! , where giv en E + Y \{ y } , and δ = max( n − 0 . 33 , n − γ ), E  1 D ( v )  = (1 + O ( δ )) 1 np . This follo ws from the iden tit y N X j =0 1 j + 1  N j  p j q N − j x j +1 = 1 ( N + 1) p ( q + px ) N +1 , obtained b y integrating ( q + px ) N ; and fro m Pr ( ¬E + v ) = O ( n − 1 − γ ). Th us E [ d + ( i )] W ( y , i ) = (1 + O ( δ )) 1 np X w ∈ Y i − 1 X v ∈ N − T ( w ) Y z ∈ w P u 1 d + ( z ) ! = (1 + O ( δ )) 1 np X w ∈ Y i − 1 d − T ( w ) Y z ∈ w P u 1 d + ( z ) ! = (1 + O ( δ )) 1 np W ∗ ( y , i − 1) . T o obtain a concen tration result, let U ( i ) = W ( y , i ) · ((1 − o (1)) c 0 np ) i , w e can w rite U ( i ) = P v ∈ Y i U v , where U v are independen t random v ariables. Assuming E + Y \{ y } and that Lemma 5(ii) holds we hav e ( c (1 − o (1 )) /C 0 ) i ≤ U v ≤ 1. Let ǫ i = p 3 K lo g n/ ( E U ) fo r some large constan t K . Then Pr ( | U ( i ) − E U | ≥ ǫ E U ) ≤ 2 e − ǫ 2 3 E U = O ( n − K ) , and so Pr ( | W ( y , i ) − E W | ≥ ǫ E W ) = O ( n − K ) . Note tha t E U ≥ | Y i | ( c (1 − o (1)) /C 0 ) i ≥ ( c/ 2 )( cnp/ 2 C 0 ) i . Th us ǫ i ≤ 1 / p ( A log n ) i − 1 for some A > 0 constan t. F o r i ≥ 2 , ǫ i = O (1 / √ log n ), and thus ǫ i = o (1). In summary , with probability 1 − O ( n − K ), W ( y , i ) = (1 + O ( δ ) + O ( ǫ i )) 1 np W ∗ ( y , i − 1) . 15 Con tin uing in this v ein, let E [ d − T ( i − 1)] W ∗ ( y , i − 1) b e the exp ectation of W ∗ ( y , i − 1) ov er ( d − T ( v ) , v ∈ Y i − 1 ), conditional on all other degrees ( d + ( u ) , d − T ( u ) , u ∈ Y ≤ i − 2 ) being fixed. F or v ∈ Y i − 1 , d − T ( v ) is distributed as B ( v ) ∼ B in ( σ ( v ) , p ) conditional on E − Y \ Y ℓ . Let 1 X denote the indicator for an ev en t X , then E B ( v ) = E ( B ( v ) · 1 E − Y \ Y ℓ ) + E ( B ( v ) · 1 ¬E − Y \ Y ℓ ) , and, splitting the second ev en t on D giv es E ( B ( v ) · 1 ¬E − Y \ Y ℓ ) = O (∆ 0 n − γ ) + O ( nn − 10 ) . Th us, giv en E − Y \ Y ℓ w e hav e E d − T ( v ) = ( 1 + O ( δ )) np . Th us E [ d − T ( i − 1)] d − T ( v ) Y z ∈ v P u 1 d + ( z ) ! = ( E d − T ( v )) Y z ∈ v P u 1 d + ( z ) ! , and E [ d − T ( i − 1)] W ∗ ( y , i − 1) = (1 + O ( δ )) np W ( y , i − 1) . Using Lemma 5 (ii) and argumen ts similar to ab o v e, for i ≥ 3 with probabilit y 1 − O ( n − K ) W ∗ ( y , i − 1) = (1 + O ( δ ) + O ( ǫ i − 1 )) np W ( y , i − 1 ) completing the induction for i ≥ 3 . The final step is to use W ( y , 2) = (1 + O ( δ ) + O ( ǫ 2 )) 1 np W ∗ ( y , 1) , and thus whp W ( y , ℓ ) = ℓ Y i =2 (1 + O ( δ ) + O ( ǫ i )) 2 1 np W ∗ ( y , 1) =  1 + O  1 √ log n  1 np X u ∈ N − ( y ) d − T ( u ) d + ( u ) . Th us from (22) Pr ( ∃ y ∈ V suc h tha t ¬A 2 ( y )) = O ( P r ( ∃ v ∈ V : deg ± ( v ) 6∈ I )) = O ( n − γ ) . ✷ 16 Corollary 8. Pr ovid e d Assumption 1 holds, let A 2 ( y ) = ( X v ∈ Y ℓ β ℓ,v = (1 + O (1 / p log n )) de g − ( y ) np ) , (38) then Pr ( ∃ y ∈ V such that ¬A 2 ( y )) = O ( n − γ ) . (39) Pro of Referring t o (37), under As sumption 1 and ¬D , then d − T ( u ) = deg − ( u )(1 − o (1)) sim ulta neously fo r a ll u ∈ N − ( y ) with probabilit y 1 − O ( n − 1 − γ ). Let ζ = 1 / log log log n . A v ertex is normal if at most ζ 0 = ⌈ 4 / ( ζ 3 d ) ⌉ of its in- neighbours ha ve out-degrees whic h are not in the range [(1 − ζ ) np, (1 + ζ ) np ], and similarly for in-degrees. Let N ( y ) b e the ev en t y is normal. W e observ e that Pr ( ¬N ( y ) | E − y ) ≤ 2 C 0 np X s = c 0 np  s ζ 0  (2 e − ζ 2 np/ 3 ) ζ 0 = O ( n − Ω(log log log n ) ) , where E − y is giv en b y (23), and th us (see (22)) Pr ( ¬ ( N ( y ) ∩E − y )) = O ( n − 1 − γ ) . (40) No w if y is normal then deg − ( y ) 1 − ζ 1 + ζ − O ( ζ 0 ) ≤ X u ∈ N − ( y ) deg − ( u ) deg + ( u ) ≤ deg − ( y ) 1 + ζ 1 − ζ + O ( ζ 0 ) . ✷ Recall the definition of Z ( x, y ), Z ( x, y ) = X u ∈ X ℓ v ∈ Y ℓ α ℓ,u β ℓ,v 1 uv deg + ( u ) , (4 1) where 1 uv is the indicator for the existence of the edge ( u , v ) and w e take 1 uv deg + ( u ) = 0 if deg + ( u ) = 0 . The next lemma gives a high probabilit y b ound for Z ( x, y ). Lemma 9. L et A 3 ( x, y ) =  Z ( x, y ) = (1 + O ( ǫ Z )) de g − ( y ) m  , wher e ǫ Z = 1 / ( √ log n ) . Then given Assumption 1, Pr ( ∃ x, y : ¬A 3 ( x, y )) = O ( n − γ ) . (42) 17 Pro of Let B = B ( x, y ) = ( E + X \ X ℓ ∩ E − Y \ Y ℓ ∩ A 1 ( x, y ) ∩ A 2 ( y ) ∩ L ) , where E is giv en b y (23 ), A 1 , A 2 b y (31) , (38), and L is the ev en t tha t Lemma 5 holds. Let u ∈ X ℓ and let w ∈ Y \ Y ℓ . As X ℓ ∩ Y = ∅ , w e know that u is not an in-neighbour of w . Other out-edges of u are unconditioned by the construction of T low x , T low y . Giv en Y \ Y ℓ ≤ n 2 / 3+ o (1) , the distribution of deg + ( u ) is B in ( ν , p ) f o r some n − n 0 . 67 ≤ ν ≤ n − 1. Th us E  1 uv deg + ( u )     B  = ν X k =1  ν k  p k (1 − p ) ν − k k ν 1 k = 1 n  1 + O ( n − 0 . 33 )  . ( 4 3) Here k / ν is the conditional probabilit y that edge ( u , v ) is presen t , giv en that u has k o ut- neigh b ours. W e use the notation Pr C ( · ) = P r ( · | C ) etc, fo r any ev en t C . F rom (31), (38), ( 43), E B ( Z ) = (1 + O ( ǫ Z )) deg − ( y ) m . (44) Conditional o n B , | Y ℓ | ≤ n 2 / 3+ o (1) b y construction, a nd as t he edges fr om u to Y ℓ are unexp osed, Pr B  | N + ( u ) ∩ Y ℓ | ≥ 1 000  ≤ Pr ( B in ( n 2 / 3+ o (1) , p ) ≥ 1000) ≤ n − 10 . (45) Let F = F ( x, y ) =  | N + ( u ) ∩ Y ℓ | < 10 0 0 , ∀ u ∈ X ℓ  , and let G ( x, y ) = B ( x, y ) ∩ F ( x, y ) ∩ E + X ℓ . The quan tity of intere st to us is the v alue of Z ( x, y ) conditional on G ( x, y ). W e first obtain E G ( Z ) f rom E B ( Z ) using E B ( Z ) = E B ( Z · 1 F ( x, y ) ∩E + X ℓ ) + E B ( Z · 1 ¬ [ F ( x,y ) ∩E + X ℓ ] ) . (46) The eve n t ¬ [ F ( x, y ) ∩ E + X ℓ ] ⊆  F ( x, y ) ∩ ¬E + X ℓ  ∪ [ ¬F ( x, y )]. Using (41), w e obtain E B ( Z · 1 ¬ [ F ( x,y ) ∩E + X ℓ ] ) = O ( E B ( Z ) n − γ ) + O  1000 ( c 0 np ) 2 ℓ  | X ℓ |  O ( n − (1+ γ ) ) + O ( n − 10 )  + O ( | X ℓ || Y ℓ | ) O ( n − 10 ) (47) = E B ( Z ) O ( n − γ ) . T o see this, partition the v ertices of X ℓ in to sets R, S , where v ertices in R ha v e out-degree in [ c 0 np, C 0 np ], and vertice s of S do not. The first term in (47) is the con tribution to the first term in the RHS o f (46) from the vertice s in R , m ultiplied by the probabilit y of ¬E X ℓ . Assuming F ( x, y ) holds, the second term in the RHS o f (47) is the contribution to the first term in the 18 RHS of (46 ) f r om the v ertices in S . The last term in the RHS of (47) is the con tribution is the con tributio n to the first term in the RHS o f (4 6) in the case where ¬F ( x, y ) holds. Th us E G ( Z ) = E ( Z · 1 B · 1 F ( x, y ) ∩E + X ℓ ) Pr ( G ) = E B ( Z . 1 F ( x, y ) ∩E + X ℓ ) Pr ( B ) Pr ( G ) , and so E G ( Z ) = E B ( Z )(1 + O ( n − γ )) = (1 + O ( ǫ Z )) deg − ( y ) np 1 n . (48) W e no w examine the concen tration of ( Z | G ). Let A = 1000 / ((1 − o (1)) c 0 np ) 2 ℓ +1 . It follows from Lemma 5(ii) that giv en G we ha v e Z u ≤ A . Let b Z u = Z u / A , then for u ∈ X ℓ , the b Z u are indep enden t random v ariables, and 0 ≤ b Z u ≤ 1. Let b Z = P u ∈ X ℓ b Z u and let b µ = E G ( b Z ). Th us b µ = n 1 / 3+ o (1) . (49) It fo llows f rom (9 ) tha t if 0 ≤ θ ≤ 1, Pr G ( | b Z − b µ | ≥ θ b µ ) ≤ 2 e − θ 2 b µ/ 3 . With θ = 4( np/ b µ ) 1 / 2 w e find that, Pr G ( | b Z − b µ | ≥ 4( np b µ ) 1 / 2 ) = o ( n − 4 ) , and hence that Pr G ( | Z − E G Z | ≥ 4 A ( np b µ ) 1 / 2 ) = o ( n − 4 ) . Using (49) w e ha v e 4 A ( np b µ ) 1 / 2 = O ( n − 7 / 6+ o (1) ), and so Pr G  | Z − E G ( Z ) | = O  1 n 7 / 6+ o (1)  = 1 − o ( n − 4 ) . W e see from (48) that E G ( Z ) = (1 + O ( ǫ Z )) deg − ( y ) m . Th us Pr G  Z ( x, y ) 6 = (1 + O ( ǫ Z )) deg − ( y ) m  = o ( n − 4 ) . (50) Using (24), (32), (39) a nd (45), Pr [ x,y ¬G ( x, y ) ! ≤ Pr ( ¬E V ) + P r ( ¬L ) + P r [ x,y ¬F ( x, y ) ! + Pr [ x,y ¬A 1 ( x, y ) ! + Pr [ y ¬A 2 ( y ) ! = O ( n − γ ) . (51) Th us finally , from (50) and (51 ) Pr ( ∃ x, y : ¬A 3 ( x, y )) = O ( n − γ ) . (52) ✷ 19 4.3 Prop erties needed for an upp er b ound on the stationary dis- tribution W e r emind the reader t ha t np ≤ n δ . W e first sho w that small sets of ve rtices ar e sparse whp . Lemma 10. L et ζ b e a n a rb i tr a ry p o sitive c onstant. F or al l S ⊆ V , | S | ≤ s 0 = (1 − 2 ζ )Λ , S c ontains a t most | S | e dges whp . Pro of The exp ected n um b er o f sets S with more t ha n | S | edges can b e b o unded b y s 0 X s =3  n s  s 2 s + 1  p s +1 ≤ s 0 X s =3 ( e 2 np ) s sep ≤ exp ( − ζ (1 + o (1)) log n ) = o ( n − ζ / 2 ) . ✷ Let Λ = log np n. W e will use the follow ing v a lues: ℓ 0 = (1 + η )Λ , ℓ 1 = (1 − 10 η )Λ , ℓ 2 = 11 η Λ . F or the upper b ound we need to sligh tly alter our definition of breadth-first trees and call them T up x , T up y . This time w e grow T up x to a depth ℓ 1 and T up y to a relativ ely small depth ℓ 2 . With this c hoice, Lemma 1 0 implies that Y will con tain no more than | Y | edges whp . This reduces the complexit y o f the argumen t. W e fix x, y and gro w T up x from x to a depth ℓ 1 , and T up y in to y to a depth ℓ 2 . The definition of T up x is sligh tly differen t from T low , but we retain some o f the notatio n. Construction of T up x . W e build a tree T up x , m uc h as in Section 4.2 , by gro wing a breadth-first out-tree from x t o depth ℓ . The difference is that we construct T up x b efore T up y , so that T up x is not disjoin t from Y . As b efore, let X 0 = { x } , and X i , i ≥ 1 b e the i -th lev el set of the tree. Let T up x ( i ) denote the BFS tree up to and including lev el i , and let T up x = T up x ( ℓ 1 ). Let X ≤ i = ∪ j ≤ i X j , and let X = X ≤ ℓ 1 . In Section 5.2 b elow w e will need to consider a la r g er set X ≤ ℓ 3 where ℓ 3 = (1 − η / 10)Λ. Construction of T up y . Our upp er b ound construction of T up y is the same as for the low er b ound, except that we only grow it to depth ℓ 2 . Our aim is to prov e an upp er b ound similar to the lo w er b ound pro v ed in Lemma 9. F or u ∈ X ℓ 1 w e let α ℓ 1 ,u = Pr ( W x ( ℓ 1 ) = u ) 20 where X u ∈ X α ℓ 1 ,u ≤ 1 . (53) The RHS of (53 ) is one, except when w e fail to construct T up y to lev el ℓ 2 . This is the only place where w e w rite a structural prop ert y of D n,p in terms of a walk proba- bilit y . This is of course v a lid, since α ℓ 1 ,u is the sum o v er w alks o f length ℓ 1 from x to u of the pro duct of recipro cals of out- degrees. F ortunat ely , all w e need is (53 ) . W e also define the β i,v as w e did in (28) and no w we let Z ( x, y ) = Z up ( x, y ) = X u ∈ X v ∈ Y ℓ 2 \ X α ℓ 1 ,u β ℓ 2 ,v 1 uv deg + ( u ) . (54) The fo llo wing lemma follow s from Coro llary 8. Lemma 11. L et ℓ b e as in (25) . If 2 ≤ k ≤ ℓ then for some ǫ Y = o (1) we have Pr X v ∈ Y k β k ,v ≥ (1 + ǫ Y ) de g − ( y ) np ! = o ( n − 1 − γ / 2 ) wher e γ is as in (2 2) . It fo llows by an argument similar to that for L emma 9 that Lemma 12. Pr  ∃ x, y : Z ( x, y ) ≥ (1 + o (1)) de g − ( y ) m  = O ( n − γ / 2 ) . (55) In computing the exp ectation of Z , some of the v ertices in X o f T up x ma y b e insp ected in our construction of T up y , or o f T up x up to lev el ℓ 1 . Th us E (1 uv / deg + ( u )) ≤ (1 /n )( 1 + o (1)), (see (43)). Remark 5. The upp er b ound for Z ( x, y ) obtaine d ab ove is p ar am eterize d by ℓ 0 = (1 + η )Λ . Pr ovide d η > 0 c onstant, so that L emma 12 holds, we c an apply this ar gument simultane ously for n γ / 3 differ ent va lues of η . W e next prov e a lemma ab out non-tree edges inside X , and edges from X to Y \ Y ℓ 2 . Lemma 13. (a) L et ℓ 3 = (1 − η / 10)Λ and L a ( ℓ 3 ) = {∀ z ∈ X ≤ ℓ 3 : z ha s ≤ 10 0 /η in-nei g hb ours in X ≤ ℓ 3 } . Then Pr ( ¬L a ( ℓ 3 )) = O ( n − 9 ) . 21 (b) L et X ◦ ℓ = { v ∈ X ℓ : N + ( v ) ∩ X ≤ ℓ 6 = ∅} and L b ( ℓ ) =  | X ◦ ℓ | ≤ 1 8∆ 0 2 ℓ p + log 2 n  . Then Pr ( ¬L b ( ℓ )) = O ( n − 10 ) fo r ℓ ≤ ℓ 3 . (c) L et t 0 =  K Λ log np  wher e K = 2 log(100 C 0 /η c 0 ) . (56) Fix t ≤ t 0 and i > 2 η Λ and let S ◦ i,t = { z ∈ X : z i s r e achable fr om X ◦ i in a t most t steps } . Then let A ◦ = A ◦ ( x, y , t ) b e the numb er of e dges fr om S ◦ i,t ∩ X to Y ℓ 2 − t . Then, Pr ( ∃ x, y , t : A ◦ ≥ log n ) = O ( n − 10 ) . (d) L et A = A ( x, y , t ) b e the numb er of e dges b etwe en X ℓ 1 and Y ℓ 2 − t \ X , wher e t 0 < t ≤ ℓ 2 − 1 . Pr ( ∃ x, y , t : A ≥ 9 | X ℓ 1 || Y ℓ 2 − t | p + log 2 n ) = o ( n − 10 ) . Pro of (a) Let z ∈ X ( ℓ 3 ). Let ζ b e the n umber of in-neigh b ours o f z in X ≤ ℓ 3 . In the construction of T up x ( ℓ 3 ), w e only exposed one in-neigh b our of z . Th us ζ is distributed as 1 + B in ( | X ≤ ℓ 3 | , p ) ≤ 1 + B in (∆ ℓ 3 0 , p ) + n Pr ( D ). W e apply (10 ) and (17) to deal with the binomial. Hence if r + 1 = 100 /η , Pr ( ζ ≥ r + 1) ≤ ∆ r ℓ 3 0 p r + n − 10 e − 10 np ≤ 2 n r ( δ − η / 10) + n − 10 e − 10 np = O ( n − 9 ) . P art (a) of the lemma follow s. (b) F o r v ∈ X ℓ the out edges of v are unconditioned during the construction of T up x ( ℓ ). The n um b er of out edges o f v to X ≤ ℓ is B in ( | X ≤ ℓ | , p ) . Unless D o ccurs, | X ≤ ℓ | ≤ 2 ∆ 0 ℓ and Pr ( | N + ( v ) ∩ X ≤ ℓ | > 0 | ¬D ) ≤ 1 − (1 − p ) 2∆ 0 ℓ ≤ 2∆ 0 ℓ p, and E ( | X ◦ ℓ | | ¬D ) ≤ 2∆ 0 2 ℓ p. By (10) Pr ( | X ◦ ℓ | ≥ 18∆ 0 2 ℓ p + log 2 n ) = O ( n − 10 ) + P r ( D ) = O ( n − 10 ) . (c) Let S ( u, t ′ ) b e the set of v ertices in X that a w alk starting from u ∈ X i can reac h in ℓ 1 − i + t − t ′ steps. Th us unless D o ccurs, | S ( u, t ′ ) | ≤ ∆ 0 ℓ 1 − i + t − t ′ . So, giv en ¬D , | S ◦ i,t | ≤ 2 | X ◦ i | ∆ 0 ℓ 1 − i + t . (57) 22 W e can assume that, after constructing T up x w e construct T up y to lev el Y ℓ 2 − t , and then insp ect the edges from S ◦ i,t to Y ℓ 2 − t \ X . These edges are unconditioned at this p o in t and their n um b er A is sto c hastically dominated by B in ( | S ◦ i,t | | Y ℓ 2 − t | , p ) . Giv en L b ( i ) of part ( b) of this lemma, | X ◦ i | ≤ 1 8∆ 0 2 i p + log 2 n. (58) Let i = a Λ, where 2 η ≤ a ≤ 1 − 1 0 η . Case 2 η ≤ a ≤ ( 1 + ǫ ) / 2 for some small ǫ > 0 constant. Using (57), (58) and | Y ℓ 2 − t | ≤ ∆ 0 ℓ 2 − t giv es E A ◦ ≤ (18∆ 0 2 i p + log 2 n )2∆ 0 ℓ 1 − i + t ∆ 0 ℓ 2 − t p + n 2 ( Pr ( D ) + Pr ( L b ( i ))) ≤ 36 C ℓ 0 + ℓ 1 0 p 2 ( np ) ℓ 0 + i + 2 C ℓ 0 0 (log 2 n ) p ( np ) ℓ 0 − i + O ( n − 9 ) ≤ 36 C ℓ 0 + ℓ 1 0 ( np ) 2 n − 1 2 + η + ǫ + 2 C ℓ 0 0 log 2 n ( np ) n − η + O ( n − 9 ) = O ( n − η/ 2 ) . Case (1 + ǫ ) / 2 ≤ a ≤ 1 − 10 η . F or i ≥ (1 + ǫ ) / 2 Λ, | X ◦ i | ≤ 20∆ 0 2 i p . Th us E A ◦ ≤ 20∆ 0 2 i p · 2∆ 0 ℓ 1 − i + t ∆ 0 ℓ 2 − t p + n 2 ( Pr ( D ) + Pr ( L b ( i ))) ≤ 40 C ℓ 0 + ℓ 1 0 p 2 ( np ) ℓ 0 + i + O ( n − 9 ) ≤ 40 C ℓ 0 + ℓ 1 0 ( np ) 2 n − 9 η + O ( n − 9 ) = O ( n − η/ 2 ) . In either case, with probability 1 − o ( n − 10 ), A ≤ log n . (d) After gro wing T up x to lev el ℓ 1 , w e gro w T up y to lev el ℓ 2 − t . T hen A ( t ) has a binomial distribution and E A ( t ) ≤ | X ℓ 1 || Y ℓ 2 − t | p . The result follo ws from the Chernoff inequalit y . ✷ 4.4 Small a v erage degree: 1 + o (1) ≤ d ≤ 2 This section contains f ur t her lemmas needed for the case 1 + o (1) ≤ d ≤ 2 . W e will assume no w that 1 + o (1 ) ≤ d ≤ 2 . Let a v ertex b e smal l if it has in-degree or out-degree at most np/ 20 and lar ge otherwise. Let we ak d i s tanc e refer t o distance in the underlying undirected gra ph of D n,p . Lemma 14. (a) Whp ther e ar e fewer than n 1 / 5 smal l ve rtic es. 23 (b) If np ≥ 2 log n then whp ther e a r e n o smal l vertic es. (c) Whp every p air of smal l vertic es ar e at we ak distanc e at le ast ℓ 10 = log n 10 log log n ap art. (d) Whp ther e do es not exist a vertex v with max  de g + ( v ) , de g − ( v )  ≤ log n/ 20 . (e) L et ς ∗ ( v ) b e given by (1) . Whp for al l vertic es y , X u ∈ N − ( y ) de g − ( u ) de g + ( u ) = (1 + o (1))( de g − ( y ) + ς ∗ ( y )) . Pro of (a) The exp ected num b er o f small ve rtices is at most n log n/ 20 X k =0  n − 1 k  p k q n − 1 − k = O ( n . 1998 ) . ( 5 9) P art (a) no w follows from the Mark o v inequalit y . (b) F o r np ≥ 2 log n t he RHS o f (59) is o (1 ) . (c) The exp ected n um b er of pairs of small vertice s at distance ℓ 10 or less is at most n 2 ℓ 10 X k =0 2 k n k p k +1   2 log n / 20 X l =0  n − 1 l  p l q n − 1 − l   2 = O ( nℓ 10 (2 d log n ) ℓ 10 +1 (20 ed ) log n/ 10 n − 2 d ) = O ( n · n 1 / 10+ o (1) · n 1 / 2 · n − 2 ) = o (1) . (d) The exp ected num b er of v ertices with small out- and in-degree is O ( n 1 − 2 × . 8002 ) = o (1). (e) F or 1 ≤ k ≤ ∆ 0 let λ k = ( 1 1 ≤ k ≤ log n (log log n ) 4 (log log n ) 4 log n (log log n ) 4 ≤ k ≤ ∆ 0 . Let ǫ = 1 log l og n . The probability that there exists a ve rtex o f in-degree k ∈ [1 , ∆ 0 ] with λ k in-neigh b ours of in or out- degree o utside (1 ± ǫ ) np , is b ounded b y ∆ 0 X k =1 n  n − 1 k  p k q n − 1 − k  k λ k  (4 e − ǫ 2 np/ 3 ) λ k ≤ ∆ 0 X k =1 2 n 1 − d  nep k · 2 · n − ǫ 2 dλ k / (4 k )  k = o (1) . 24 No w assume that t here are few er than λ k neigh b ours of v of in or out-degree outside (1 ± ǫ ) np . Assuming at most one neigh b our w of y is small, X u ∈ N − ( y ) \{ w } deg − ( u ) deg + ( u ) = ( (1 + O ( ǫ )) k 1 ≤ k ≤ log n (log log n ) 4 (1 + O ( ǫ ))( k − λ k ) + O ( λ k ) log n (log log n ) 4 ≤ k ≤ ∆ 0 . This completes t he pro of of the lemma. ✷ Let we ak distance refer to distance in the underlying graph of D n,p , and let a cycle in the underlying graph b e called a we ak cycle. Lemma 15. Whp ther e do es not exist a sm al l vertex that is within we ak distanc e ℓ 10 of a we ak c ycle C of length at most ℓ 10 . Pro of Let v , C b e suc h a pair. Let | C | = i a nd j b e t he w eak distance of v from C . The probabilit y that suc h a pair exists is a t most ℓ 10 X i =3 (2 np ) i i ℓ 10 X j =0 (2 np ) j log n / 20 X l =0 2  n − 1 l  p l q n − 1 − l = O ( n 1 / 10+ o (1) · n 1 / 10+ o (1) · n − 4 / 5+ o (1) ) = o (1) . ✷ 5 Analysis o f the random w alk: Esti mating t he statio n- ary dist ributio n In this section w e k eep Assumption 1 and assume that we are dealing with a digraph whic h has all of the high probabilit y prop erties of the previous section. 5.1 Lo w er Bound on the stationary distribution W e use t he prop erties describ ed in Section 4.2. W e deriv e a low er b ound on P 2 ℓ +1 x ( y ). F or this lo w er b ound w e only consider ( x, y )-paths of length 2 ℓ + 1 consisting of a T low x path from x to X ℓ follo w ed by a n edge fro m X ℓ to Y ℓ and then a T low y path to y . The probability of follo wing suc h a path is Z ( x, y ), see (29 ). Lemma 42 implies tha t P (2 ℓ +1) x ( y ) ≥ (1 − o ( 1 )) deg − ( y ) m for all v ∈ V . (60) 25 Lemma 16. F or al l y ∈ V , π y ≥ (1 − o ( 1 )) de g − ( y ) m . Pro of It fo llows f rom (6 0) that for an y y ∈ V , π y = X x ∈ V π x P (2 ℓ +1) x ( y ) ≥ (1 − o ( 1 )) deg − ( y ) m X x ∈ V π x = (1 − o (1)) deg − ( y ) m . (61) ✷ 5.2 Upp er Bound on the stationary distribution Lemma 16 ab o v e prov es that the expression in Theorem 2 is a lo w er b ound on the stationary distribution. As P π y = 1, this can b e used to deriv e a n upp er b o und of π y ≤ (1 + o (1)) deg − ( y ) m whic h holds f o r all but o ( n ) v ertices y . In this section w e extend this upp er b ound to al l y ∈ V . W e use the prop erties describ ed in Section 4.3. W e now conside r the probability of v arious t yp es of w alks o f length ℓ 0 + 1 from x to y . Some of these w alks are simple directed paths in BFS trees constructed in a similar w a y to the low er b ound, and some use back edges of these BFS tr ees, or con tain cycles etc. W e will upp er b ound P ℓ 0 +1 x ( y ) as a sum P ℓ 0 +1 x ( y ) ≤ Z ℓ 0 +1 x ( y ) + S ℓ 0 +1 x ( y ) + Q ℓ 0 +1 x ( y ) + R ℓ 0 +1 x ( y ) , ( 62) where the definitions o f the probabilities on the righ t hand side a re described b elo w. Z ℓ 0 +1 x ( y ). This is the probability that W x ( ℓ 0 + 1) = y and the ( ℓ 1 + 1)th edge ( u, v ) is suc h that u ∈ X a nd v ∈ Y ℓ 2 \ X , and the last ℓ 2 steps of the walk use edges of the tree T up y . These are the simplest w alks to describ e. T hey go t hrough T up x for ℓ 1 steps and then lev el b y lev el through T up y . They mak e up almost all of the w alk probabilit y . S ℓ 0 +1 x ( y ). This is the probability that W x ( ℓ 0 + 1) = y g o es from x to y without lea ving X . This includes an y special cases such as, for example, a w alk xy xy ...xy based on the existence of a cycle ( x, y ) , ( y , x ) in the digraph. Q ℓ 0 +1 x ( y ). This is the probability that W x ( ℓ 0 + 1 ) = y and the ( ℓ 1 + 1 ) t h edge ( u , v ) is such that v ∈ Y ℓ 2 ∩ X and the last ℓ 2 steps of the w a lk use edges of the tree T up y . W e exclude w a lks within X that are counted in S ℓ 0 +1 x ( y ) . R ℓ 0 +1 x ( y ). This is the probabilit y tha t W x ( ℓ 0 + 1) = y and during the last ℓ 2 steps, the w a lk uses some edge whic h is a bac k or cross edge with resp ect to the tree T up y . 26 Upp er b ound for Z ℓ 0 +1 x ( y ) . It fo llows f rom (5 5) that Z ℓ 0 +1 x ( y ) ≤ (1 + o (1)) deg − y m . (63) Upp er b ound for S ℓ 0 +1 x ( y ) . Let W x b e a walk of length t in X , and let W x ( t ) = v . Let d − max = max w ∈ X | N − ( w ) ∩ X | . T r a cing bac k from v for t steps, the n umber of w alks length t in X terminating at v is at most ( d − max ) t ; so this serv es as an upp er b ound on the n um b er of w alks from x to v of this length. By Lemma 13 ( a), we may assume that d − max ≤ 100 /η . Applying this des cription, there can b e at most (1 00 /η ) ℓ 0 +1 w a lks of length ℓ 0 + 1 from x to y , whic h do not exit from X . W e conclude that S ℓ 0 +1 x ( y ) ≤  (100 /η ) c 0 np  ℓ 0 +1 = o  1 n 1+ η/ 2  . (64) Upp er b ound for Q ℓ 0 +1 x ( y ) . W e say t ha t a walk W x delays for t steps , if W x exits X for the first time at step ℓ 1 + t . A w a lk delays at lev el i , if the w alk tak es a cross edge (to the same lev el i ) or a bac k edge (to a lev el j < i ) i.e. a non-tr e e e dge e = ( u, v ) con t a ined in X that is no t par t of T up x . Lemma 17. L et t 0 = l K Λ log n p m wher e K = 2 log(100 C 0 /η c 0 ) , then Pr ( W x ( ℓ 0 + 1) = y and W x delays for t 0 or m o r e steps ) = o (1 /n ) . Pro of The only w a y for a w alk to exit from X is via X ℓ 1 (recall that edges oriented o ut from X i end in X i +1 ). Let W x b e an ( x, y )-walk whic h delays for t steps, and then tak es edge e = ( u , v ) b et w een X ℓ 1 and Y ℓ 2 − t \ X . There are at most (1 0 0 /η ) ℓ 1 + t w a lks of length ℓ 1 + t from x to u within X . After reachin g vertex v , W x follo ws the unique pa t h from v to y in T Y . Applying Lemma 13(d) w e se e that the t o tal probability P † ( t ) o f suc h ( x, y )-w alks of length ℓ 0 + 1 and dela y t is P † ( t ) ≤ (100 /η ) ℓ 1 + t  9 | X ℓ 1 || Y ℓ 2 − t | p + log 2 n  ( c 0 np ) ℓ 0 +1 ≤ (100 /η ) ℓ 1 + t  9 C ℓ 0 0 ( np ) ℓ 0 − t p + log 2 n  ( c 0 np ) ℓ 0 +1 = O 1 n  (100 /η ) C 0 c 0  ℓ 0  1 ( np ) t + log 2 n n η  ! . 27 So, P † ( ≥ t 0 ) = X t ≥ t 0 P † ( t ) = O  A Λ n  1 ( np ) t 0 + log 2 n n η  , (65) where A = (100 C 0 /η c 0 ) 1+ η . No w A Λ = n o (1) . Also A Λ /np = o (1) if log 2 np ≥ 2(log A )(log n ) in whic h case the RHS of (65) is o (1 /n ), whic h is what we need to sho w. So assume no w that log 2 np ≤ 2(lo g A )(log n ). This means that Λ → ∞ and then A Λ ( np ) t 0 ≤  A e K  Λ → 0 . Th us in b oth cases P † ( ≥ t 0 ) = o (1 / n ) . (66) ✷ W e can now fo cus on w alks with dela y t , where 1 ≤ t < t 0 . A non-tr e e edge of X is an edge induced b y X whic h is not an edge of T up x . F o r 4 i ≤ (1 − ǫ )Λ, Lemma 10 implies that whp the set U = X ≤ i con tains at most | S | edges. F or, if U contained more than | U | + 1 edges then it w ould contain t w o distinct cycles C 1 , C 2 . In whic h case, C 1 , C 2 and the shortest undirected path in U joining them would form a set S whic h satisfies the conditions of Lemma 10 . Th us there is at most one non-tr e e e d g e e = ( u, v ) contained in X ≤ (1 − ǫ )Λ / 4 . Let θ = 2 η Λ . W e classify walks into tw o t yp es. T yp e 1 W alks. These hav e a dela y caused b y using a non-tree edge o f X ≤ θ , but no dela y arising at an y lev el i > θ . Th us, o nce the w alk finally exits X θ to X θ + 1 it mo ve s forw ard at each step tow ards X ℓ 1 , and then exits to Y ℓ 2 − t \ X . T yp e 2 W alks. The se ha ve a dela y arising at some lev el X i , i > θ . W e do not exclude previous dela ys o ccurring in X ≤ θ , or subsequen t dela ys at an y lev el. T yp e 1 W alks . W e can assume that X ≤ θ induces exactly one non- tree edge e = ( u, v ). Let u ∈ X i then v ∈ X j , j ≤ i . There are tw o cases. (a) e is a cross edge, or back edge not inducing a directed cycle . Here t he delay is t = i + 1 − j and this is less than t 0 b y assumption. Then, as w e will see, Pr (T yp e 1(a) w a lk ) ≤ 1 ( c 0 np ) 2 1 deg + ( w ) X w ∈ N + ( v ) Z ( ℓ 0 − i + j +1) w ( y ) = O  1 n ( np ) 2  . (67) The term 1 / ( c 0 np ) 2 arises from the w alk ha ving to take the out-neigh b our of x that leads to u in T up x and then ha ving to tak e the edge ( u, v ). The next step of the w alk is to c ho ose 28 w ∈ N + ( v ) a nd it must then fo llo w a path to y lev el b y leve l through the t wo trees. The v alue of Z ℓ 0 − i + j +1 w ( y ) can b e obtained as fo llo ws. Let ℓ ′ 0 = ℓ 0 − ( i − j ) − 1 , then as t < t 0 = o (Λ) we ha v e that ℓ ′ 0 ∼ ℓ 0 . F o r w ∈ N + ( v ) replace ℓ 0 b y ℓ ′ 0 in (55) ab o v e, to obtain Z ( w , y ) = O (1 /n ), see Remark 5 . This v erifies (67). (b) e is a bac k edge inducing a directed cycle . Let xP u b e the path from x to u in T up x ( θ ). As v is a ve rtex of xP u , w e can write xP u = xP v , v P u and cycle C = v C v = v P u, ( u, v ). Let σ ≥ 2 be the length of C . F or some w in v P u the w alk is of the form xP v , v P w , ( w C w ) k , w P z , where w C w is C started at w , t he w alk go es round w C w , k times and exits at w to u ′ ∈ N + ( w ) \ C and then mo ves forw ard along w P z to z ∈ X ℓ 1 and then o n to y . The dela y is t = k σ and t his is less than t 0 b y assumption W e claim that Pr (T yp e 1(b) w alk) ≤ X w ∈ C X k ≥ 1 ( c 0 np ) − k σ 1 deg + ( w ) − 1 X u ′ ∈ N + ( w ) \ C Z ℓ 0 − k σ +1 u ′ ( y ) = O  1 n ( np ) 2  . (68) The term ( c 0 np ) − k σ accoun ts for having to go round C k times and w e can a rgue that Z ℓ 0 − k σ +1 u ′ ( y ) = O (1 /n ) as w e did for T yp e 1(a) w alks. So fr o m (67) and (68) w e ha v e that Pr (T yp e 1 w alk) = O  1 n ( np ) 2  . (69) T yp e 2 W alks . Suppo se W x is a w alk whic h exits X at step ℓ 1 + t and is delay ed at some lev el i > θ b y using an edge ( u, v ). The walk arriv es at v ertex u ∈ X i for the first time at some step i + t ′ and trav erses a cross or bac k edge to v ∈ X j , j ≤ i . A con tr ibuting w alk will hav e to use one of the A ◦ ( x, y , t ) ≤ lo g n edges describ ed in Lemma 13(c). By Lemma 13(a) there are at most (1 00 /η ) ℓ 1 + t log n from x to u ∈ X ◦ i . Once the w a lk reac hes w ∈ Y ℓ 2 − t there is (b y assum ption) a unique path in T up y from w to y . Let P ( i, t ) b e the probabilit y of these T yp e 2 walks , then P ( i, t ) ≤ (100 /η ) ℓ 1 + t log n ( c 0 np ) ℓ 0 +1 = O  1 n 1+ η/ 2  . (70) Th us finally from (66), (69), (70) Q ℓ 0 +1 x ( y ) = P † ( ≥ t 0 ) + P r (Ty p e 1 walk) + X 1 ≤ t ≤ t 0 X θ ≤ i ≤ ℓ i P ( i, t ) = O  1 n 1 ( np ) 2  . (71) Upp er b ound for R ℓ 0 +1 x ( y ) . 29 Let Y = Y ≤ ℓ 2 b e the vertex set o f T up y ( ℓ 2 ). W e assume that Y induces a unique edge e = ( u, v ) whic h is not in T up y . Note that the condition that | Y | induces at most | Y | edges holds, ev en if w e replace ℓ 2 with 2 ℓ 2 based on the construction of T Y (2 ℓ 2 ) to depth 2 ℓ 2 , by branc hing bac kw ards from y . W e consider t wo cases. (i) e is a cr oss or forw ard edge, or bac k edge not inducing a directed cycle. W e ha v e u ∈ Y i , v ∈ Y j for some i ≤ j ≤ ℓ 2 . W e suppo se the ( x, y )- w alk is of the form xW u, ( u, v ) , v W y where u 6∈ v W y , so that v W y is a unique path in T up y . Case 1: i > (4 η / 5)Λ. Let ℓ 3 = (1 − η / 10) Λ . The length of the path ( u, v ) , v W y is j , so the length of the walk xW u is ℓ 0 − j + 1. Let h b e the distance from u t o X ≤ ℓ 3 in T up y . Then h = max { 0 , ℓ 0 − ℓ 3 − j + 1 } ≤ max { 0 , ℓ 0 − ℓ 3 − i + 1 } . Let w ∈ X ≤ ℓ 3 . By Lemma 13, the n um b er of ( x, w )-w alks of length ℓ ≤ ℓ 3 in X ≤ ℓ 3 passing through w at step ℓ is b ounded b y (100 /η ) ℓ 3 . The the num b er of w alks length h from u to X ≤ ℓ 3 is at most ∆ 0 h . Th us, t he n um b er of ( x, y )- w alks passing thr o ugh e = ( u , v ) is b ounded b y (100 /η ) ℓ 3 ∆ 0 h . Th us R ℓ 0 +1 x ( y ) = O (100 /η ) ℓ 3 ∆ 0 9 η Λ / 10 ( c 0 np ) ℓ 0 +1 ! = O ( n − 1 − η/ 20 ) . (72) Case 2: 0 < i ≤ (4 η / 5 )Λ. Let i = a Λ. Let η ′ = η (1 − a ) , ℓ ′ 1 = (1 − 10 η ′ )Λ , ℓ ′ 2 = 11 η ′ Λ, and let ℓ ′ 0 = ℓ ′ 1 + ℓ ′ 2 . As observ ed ab ov e, the v ertex set U of the tree T U of heigh t ℓ ′ 2 ab ov e u induces no extra edges, so we can apply the upp er b ound result for walks of length ℓ ′ 0 + 1 from x to u ba sed on the assumption R ℓ ′ 0 +1 x ( u ) = 0. Thu s P ℓ ′ 0 +1 x ( u ) ≤ (1 + o (1)) deg − ( u ) m . The pro ba bility the walk then follows the path ( u, v ) , v P y is O (1 / ( np ) 2 ). Thu s R ℓ 0 +1 x ( y ) = O  deg − ( u ) m ( np ) 2  . (73) (ii) e is a back edge inducing a directed cycle. In this case, there is an edge e = ( u, v ) where u ∈ Y i , v ∈ Y j and j > i . Let v P u denote the path from v to u in T up y , and C the cycle v P u, ( u, v ). There is some k ≥ 1 suc h that the walk is P 0 = xP u, ( uC u ) k , u P y . Let σ b e the length of C , let τ b e the distance from u to y in T up y , 30 and let s = τ + k σ . Let ℓ = ℓ 0 − s . Then ℓ + 1 is the length of the w a lk xP u fro m x to u prior to the final s steps. Either ℓ < (1 + 4 η / 5 )Λ and the argumen t in Case 1 ( i ≥ 4 η / 5 )Λ) ab ov e can b e applied, giving us the b ound R ℓ 0 +1 x ( y ) = O (100 /η ) ℓ 3 ∆ 0 9 η Λ / 10 ( c 0 np ) ℓ 0 +1 ! = O ( n − 1 − η/ 20 ) . (74) Or ℓ ≥ (1 + 4 η / 5)Λ and w e adapt Case 2. Let w b e t he predeces sor of u on P 0 . W e can use Remark 5 as ab ov e to o bt a in P ℓ x ( w ) ≤ (1 + o (1))deg − ( w ) /m . As k σ ≥ 2 , τ ≥ 0, (the w o r st case is u = y , w ∈ N − ( y )), w e obtain R ℓ 0 +1 x ( y ) = O  deg − ( w ) m ( np ) 2  . (75) Th us, us ing (72), (73 ), (74), (75) w e hav e R ℓ 0 +1 x ( y ) = O  1 n · 1 ( np ) 2  . (76) W e hav e therefore shown that S ℓ 0 +1 x ( y ) + Q ℓ 0 +1 x ( y ) + R ℓ 0 +1 x ( y ) = o (1 /n ) completing the pro of that P ℓ 0 +1 x ( y ) ≤ (1 + o (1)) deg − ( y ) m (77) Lemma 18. F or al l y ∈ V , π y = (1 + o (1)) de g − ( y ) m . Pro of It fo llows f rom (6 0) that for an y y ∈ V , π y = X x ∈ V π x P ( ℓ 0 +1) x ( y ) ≤ (1 + o (1)) deg − ( y ) m X x ∈ V π x = (1 + o (1)) deg − ( y ) m . ( 7 8) The lemma now follows from Lemma 16. ✷ 31 6 Stationary di s tributi o n: Remo ving Assumpt ion 1 6.1 Large a v erage degree case 6.1.1 np ≥ n δ . W e can deal with this case by using a concen tration inequalit y (79) from Kim and V u [13]: Let Υ = ( W, E ) b e a hypergraph where e ∈ E implies that | e | ≤ s . Let Z = X e ∈ E w e Y i ∈ e z i where the w e , e ∈ E are p ositiv e reals and the z i , i ∈ W are indep enden t random v ariables taking v alues in [0 , 1]. F or A ⊆ W , | A | ≤ s let Z A = X e ∈ E e ⊇ A w e Y i ∈ e \ A z i . Let M A = E ( Z A ) and M j ( Z ) = max A, | A |≥ j M A for j ≥ 0. There exist po sitiv e constan ts a and b suc h that for an y λ > 0 , Pr ( | Z − E ( Z ) | ≥ aλ s p M 0 M 1 ) ≤ b | W | s − 1 e − λ . (79) F or us, W will b e the set of edges of ~ K n the complete digraph on n v ertices. z i will b e the indicator v ariable f or t he presenc e of the i th edge of ~ K n . E will b e the set of sets of edges in w a lks of length s = ⌈ 2 / δ ⌉ b et w een t w o fixed v ertices x and y in ~ K n , and w e = 1. Z will b e the n umber o f walks of length s that are in D n,p . In whic h case w e hav e E ( Z ) = ( 1 + o (1)) n s − 1 p s M j ≤ (1 + o (1)) n s − j − 1 p s − j ≤ (1 + o (1)) E ( Z ) /np f or j ≥ 1 . So M 0 = E ( Z ) and applying (79) with λ = (log n ) 2 w e se e that for an y x, y w e ha v e Pr ( | Z − E ( Z ) | = O ( E ( Z ) n − δ/ 2 log O (1) n )) = 1 − O ( n − 3 ) . Th us whp P s x ( y ) = (1 + o (1)) n s − 1 p s ((1 − ǫ 1 ) np ) s ∼ 1 n , ∀ x, y ∈ V . W e now finish with the argumen ts of Lemmas 16 and 18. ✷ 32 6.2 Small a v erage degree case 6.2.1 Lo w er b ound on stationary distribution A v ertex is smal l if it has in-degree or out-degree at most np/ 20 and lar ge otherwise . In the pro ofs of Section 4.2 w e assumed x, y were large. W e pro ceed as in Section 5.1 but initially restrict our analysis to large x, y . Also, with the exce ption of Y 1 w e do not include small v ertices when creating the X i , Y i . Av oiding the ≤ n 1 / 5 small ve rtices (see Lemma 14(a)) is easily incorp or a ted b ecause in t he pro of w e ha v e a llow ed for the a v oidance of n 2 / 3+ o (1) v ertices from S i X i etc. Provide d there are no small v ertices in N − ( y ), our pr evious lo wer b ound analysis holds. In this w ay , we sho w for all large x, y that, P (2 ℓ +1) x ( y ) ≥ (1 − o ( 1 )) deg − ( y ) m . (80) If x is small, then it will only hav e large out- neighbours (see Lemma 14( c)) and so if y is la rge then P (2 ℓ +2) x ( y ) = 1 deg + ( x ) X z ∈ N + ( x ) P (2 ℓ +1) z ( y ) ≥ (1 − o (1)) deg − ( y ) m . (81) A similar a rgumen t deals with s mall y and x arbitra r y i.e. P (2 ℓ +2) x ( y ) = X z ∈ N − ( y ) P (2 ℓ +1) x ( z ) deg + ( z ) ≥ (1 − o (1)) X z ∈ N − ( y ) deg − ( z ) m 1 deg + ( z ) ≥ (1 − o (1)) deg − ( y ) m . (82) W e hav e used Lemma 14(e) to justify the last inequalit y . In the case that some u ∈ N − ( y ) ha s small out-degree, then b y Lemma 14(c) there is at most one suc h u whp . F or z ∈ N − ( y ), w e rep eat the argumen t a b ov e fo r eac h factor P 2 ℓ +1 x ( z ). The extra term ς ∗ ( y ) no w arises from deg − ( u ) / deg + ( u ) and P (2 ℓ +2) x ( y ) = X z ∈ N − ( y ) P (2 ℓ +1) x ( z ) deg + ( z ) ≥ (1 − o (1)) 1 m X z ∈ N − ( y ) deg − ( z ) deg + ( z ) ≥ (1 − o (1)) deg − ( y ) + ς ∗ ( y ) m . W e can now pro ceed as in (61). 6.2.2 Upp er b ound on stationary distr ibution W e first explain how the upp er b ound pro of in Section 5 .2 alters if Assumption 1 is r emov ed. The assumption that the minim um degree w as at least c 0 np w as used in the follo wing places: 1. W e assumed in Section 4.2 that deg + ( x ) , deg − ( y ) ≥ c 0 np . These assumptions can b e circum ve n ted b y using Lemma 14(c) with the metho ds used in the low er b ound case. 33 2. In (64), (70 ), (72). I n these case s w e use d ( c 0 np ) ℓ 0 as a lo we r bo und on the pro duct of out-degrees o n a path of length λ for some λ ≥ ℓ 1 . Using Lemmas 14 and 15, w e see that small v ertices are at w eak distance at least ℓ 10 and therefore there can b e at most 1 1 suc h v ertices on any w alk length ℓ 0 + 1. Th us, after dropping Assum ption 1, w e replace this lo wer b ound by ( c 0 np ) λ − 11 , and the pro o f contin ues es sen tially unchanged. 3. In the pro of of Lemma 9 w e made a re-scaling B = 100 0 / ( c 0 np ) 2 ℓ +1 . The exp onent 2 ℓ + 1 was replaced b y ℓ 0 + 1 in the pro o f of (55) in L emma 12. W e no w replace ℓ 0 + 1 b y ℓ 0 − 10. 4. In the pro of of Lemma 7 w e made a re-scaling U ( i ) = W ( y , i ) · ( c 0 np ) i at each lev el 3 ≤ i ≤ ℓ . Assum e that 2 ℓ 2 < ℓ 10 i.e. η ≤ 1 / 2 50 so tha t there is at most one small v ertex u in Y . If w e replace ( c 0 np ) i b y ( c 0 np ) i − 1 do es not a ffect our concen tration results, pro vided i ≥ 3. The b ounds on U v are no w ( c 0 /np )( c 0 /C 0 ) i ≤ U v ≤ 1, and ǫ i = 1 / p ( A log n ) i − 2 . If the small v ertex u ∈ N − ( y ) then the direct calculations used in the lo w er b ound hold. If the small v ertex u is in lev els i = 2 , 3 this adds an extra term of O (deg − ( u ) / ( m ( np ) i − 1 )) to o ur estimate of Z ℓ 0 +1 x ( y ) in Section 5.2. 5. In (68), (73) , (75). It follows from Lemma 15, that if e.g. T up y con tains a non-tree edge, then no v ertex of T up y is small, and the calculations in the pro of are unaltered. Th us the pro of as is w orks p erfectly w ell if we assume that y is large and if it has no small in-neigh b ours and there is no small v ertex in Y . W e call suc h a v ertex y or dinary . If y is small then from Lemmas 14 and 15 w e can assume that all o f its in-neighbours are ordinary . This is under the assu mption that 2 ℓ 2 < ℓ 10 e.g. if η ≤ 1 / 250. So in this case we can use Lemma 14(e) and obtain P ( ℓ 0 +2) x ( y ) = X ξ ∈ N − ( y ) P ( ℓ 0 +1) x ( ξ ) deg + ( ξ ) ≤ 1 + o (1 ) m X ξ ∈ N − ( y ) deg − ( ξ ) deg + ( ξ ) = (1 + o (1)) deg − ( y ) m . Supp ose no w that y is lar g e and that there is a small ve rtex u ∈ Y . W e can assume from Lemma 15 that Y do es not con ta in a ny edge not in T up y . Either u ∈ N − ( y ) or , if not, from p oin t 4. of the discussion ab o v e, an extra O (deg − ( u ) / ( m ( np ))) is added to Z ℓ 0 +1 x ( y ) fo r the probabilit y of the ( x, y )- w alk going via u . In the case where u ∈ N − ( y ) then as in the low er b ound P ( ℓ 0 +1) x ( y ) ≤ 1 + o (1 ) m   deg − ( u ) deg + ( u ) + X u ∈ N − ( y ) \ w deg − ( u ) deg + ( u )   = (1 + o ( 1)) m  deg − ( y ) + ς ∗ ( y )  . W e hav e now completed the pro of of the a symptotic steady state without Assumption 1. 34 7 Mixing time and the c onditio ns o f Lemma 3 7.1 Upp er Bound on M ixing time Let T b e a mixing time as defin ed in (4) and let ℓ = O (log np n ) b e giv en b y (25). W e pro ve that ( whp ) T satisfies T = o ( ℓ log n ) = o ((log n ) 2 ) . (83) Define ¯ d ( t ) = max x,x ′ ∈ V | P ( t ) x − P ( t ) x ′ | (84) to b e the maximum ov er x, x ′ of the v aria tion distance b et w een P ( t ) x and P ( t ) x ′ . It is pro v ed in Lemma 20 of Chapter 2 of Aldous and Fill [1] that ¯ d ( s + t ) ≤ ¯ d ( s ) ¯ d ( t ) and max x | P ( t ) x − π x | ≤ ¯ d ( t ) . (85) Equation (42) implies that whp ¯ d (2 ℓ + 1) = o (1) , (86) and so (83) follows immediately from (85) and (86). 7.2 Conditions of L emma 3 W e see immediately from (83) that Condition (b) of Lemma 3 is satisfied. W e show b elow that whp for all v ∈ V R T (1) = 1 + o (1) . (87) Using (87), the pr o of that Condition (a) of Lemma 3 is satisfied, is as follow s. Let λ = 1 / K T as in (6). F o r | z | ≤ 1 + λ , we ha v e R T ( z ) ≥ 1 − T X t =1 r t | z | t ≥ 1 − (1 + λ ) T T X t =1 r t = 1 − o (1) . Th us for v ∈ V , the v alue of p v in (7) is give n by p v = (1 + o (1)) deg − ( v ) m . (88) 35 Pro of of (87) : If d ≥ (log n ) 2 , then t he minim um out- degree of D n,p is Ω( d log n ). In whic h case we hav e for an y x, y Pr ( W v ( t ) = y | W v ( t − 1) = x ) = O  1 d lo g n  . (89) The expected n um b er of returns to v ∈ V by W v during T step s, is t herefore O ( T /d log n )= o (1) . No w assume tha t d ≤ (log n ) 2 . (i) Lemma 10 implies that if H is the subgraph of D n,p induced b y v ertices a t w eak distance at mo st Λ / 2 0 f rom v then H contains at most | V ( H ) | edges. (ii) Lemma 14 implies that there is at most one small v ertex in H . (iii) Lemma 15 implies that there is no small vertex within w eak distance 10 of a w eak cycle of length ≤ 10. Assume that conditions (i), (ii), (iii) hold. Let A 4 denote the set of v ertices u 6 = v suc h that D n,p has a path of length at most 4 fr o m u to v . W e sho w next that: With probabilit y 1 − O (1 / ( np ) 2 ) , W v ( i ) / ∈ A 4 , 1 ≤ i ≤ 4 . (90) F or this to happ en, there has t o b e a cycle C of length at most 8 con taining v . If suc h a cycle exists then all v ertices within we ak distance 10 of v ha ve degree at least np/ 20. F urthermore, the only wa y that the walk can reach A 4 in 4 or less steps is via this cycle. This v erifies (90). Assume t hen that W v ( i ) / ∈ A 4 , 1 ≤ i ≤ 4. Supp ose next that there is a time T 1 ≤ T suc h that W v ( T 1 ) = v . Let T 2 = min { τ ≤ T 1 : W v ( t ) ∈ A 4 , τ ≤ t ≤ T 1 } . It m ust b e the case that d ( T 2 ) = 4 where d ( t ) is the distance from W v ( t ) to v . If A 4 do es not con tain a small w eak cycle then the w alk mus t pro ceed dir ectly to v in 4 steps. The probability of t his is O (1 / ( np ) 3 ), since at most one v ertex on t he path of length 4 from x = W ( T 2 ) to v will b e of degree at most np/ 20. If there is a small w eak cycle C then there is an edge e of C whose remo v al leav es an in- branc hing of depth 4 into v . There are no w 2 pat hs that W can f o llo w from x t o v . One uses e and o ne do es not. Eac h path has a probability of O (1 / ( np ) 4 ) of b eing follo w ed. Putting t his altogether w e see that the exp ected n um b er of returns to v is O (1 / ( np ) 2 + T / ( np ) 3 ) = o (1). This completes t he pro of of (87 ). 36 8 The Co v er Time of D n,p 8.1 Upp er Bound on the Co v er Time F or np = d log n , d constan t, let t 0 = (1 + ǫ )  d lo g  d d − 1  n log n . F or np = d log n d = d ( n ) → ∞ let t 0 = (1 + ǫ ) n log n . In b ot h cases w e assume ǫ → 0 sufficien tly slowly to ensure that all inequalities b elo w are v alid. Let T D ( u ) b e the time tak en b y the random walk W u to visit ev ery v ertex of D . Let U t b e the n um b er of v ertices of D whic h ha v e not b een visited b y W u at step t . W e note the follo wing: C u = E ( T D ( u )) = X t> 0 Pr ( T D ( u ) ≥ t ) , (91) Pr ( T D ( u ) ≥ t ) = Pr ( T D ( u ) > t − 1) = Pr ( U t − 1 > 0) ≤ min { 1 , E ( U t − 1 ) } . (92) Recall that A v ( t ) denotes the ev ent that W u ( t ) did not visit v ertex v in the inte rv al [ T , t ]. It follo ws from (9 1), ( 92) that for a ny t ≥ T , C u ≤ t + 1 + X s ≥ t E ( U s ) ≤ t + 1 + X v X s ≥ t Pr ( A v ( s )) . (93) Assume first that d ( n ) → ∞ . If s/T → ∞ then (8) of Lemma 3 together with the v alue of p v giv en b y (88), and concen tration of in-degrees implies that Pr ( A v ( s )) ≤ ( 1 + o (1)) exp  − (1 − o (1)) s n  + O ( e − Ω( s/T ) ) . (94) Plugging (94) in to (93) w e get C u ≤ t 0 + 1 + 2 n X s ≥ t 0  exp  − (1 − o (1)) s n  + O ( e − Ω( s/T ) )  (95) ≤ t 0 + 1 + 3 n 2 exp  − (1 − o (1)) t 0 n  + O ( nT e − Ω( t 0 /T ) ) = (1 + o ( 1 )) t 0 . W e no w assume that d is b ounded as n → ∞ , and the conditions of Lemma 4 hold. F or v ∈ V w e hav e Pr ( A v ( s )) = (1 + o (1)) exp {− (1 + o (1 / log n )) π v s } + O ( e − Ω( s/T ) ) where, b y Lemma 16, π v ≥ (1 − o ( 1 )) deg − ( v ) m . 37 In place of (95) w e us e the b ounds on the num b er of v ertices of degree k giv en in Lemm a 4, in terms of the sets K i , i = 0 , 1 , 2 , 3. Th us C u ≤ t 0 + 1 + o (1) + 3 X i =0 S i (96) where S i = X k ∈ K i D ( k ) X s ≥ t 0 exp  − (1 − o (1)) k s m  ≤ 2 m X k ∈ K i D ( k ) k e − (1 − o (1)) kt 0 /m ≤ 2 m X k ∈ K i D ( k ) k  d − 1 d  (1+ ǫ/ 2) k . The main term o ccurs at i = 3. Using (1 2), (15), the fact that ( nep ( d − 1)) / ( k d )) k is maximize d at k = np ( d − 1) /d , a nd m = dn log n (1 + o (1) ) whp , w e see that S 3 ≤ 8 m n d − 1 ∆ 0 X k = c 0 np  nep k  k  d − 1 d  (1+ ǫ/ 2) k ≤ 8 m ∆ 0 e − ǫc 0 np/ 2 d = o ( t 0 ) . (97) Note that K 0 = 0 . W e next consider the cases i = 1 , 2. F or i = 1, w e refer first to Lemma 4(i-a). If d − 1 ≥ (log n ) − 1 / 3 then K 1 = ∅ . If d − 1 < (log n ) − 1 / 3 , then D ( k ) ≤ (log log n ) 2 , from ( 13). In this case t 0 = O ((1 / ( d − 1))) d n log n . Th us S 1 ≤ m X k ∈ K 1 D ( k ) k  d − 1 d  (1+ ǫ/ 2) k ≤ m 15 X k =1 (log log n ) 2 k  d − 1 d  (1+ ǫ/ 2) k = O ( t 0 )(log log n ) 2 ( d − 1) − ǫ/ 2 = o ( t 0 ) (98) F or i = 2, b y Lemma 4 if d − 1 < ( log n ) − 1 / 3 and k ≥ 16, and using (14 ) w e hav e D ( k ) ≤ 38 (log n ) 4 . Th us S 2 ≤ m X k ∈ K 2 D ( k ) k  d − 1 d  (1+ ǫ/ 2) k ≤ O ( t 0 ) X k ∈ K 2 log 4 n k ( d − 1)  d − 1 d  (1+ ǫ/ 2) k = O ( t 0 ) log 4 n (log n ) − (19 / 3+ ǫ/ 8) = o ( t 0 ) . (99) If d − 1 ≥ (log n ) − 1 / 3 then b y Lemma 4(i-a ) min { k ∈ K 2 } ≥ (log n ) 1 / 2 , and | K 2 | = O (log log n ). Th us, as d is b ounded S 2 = O ( t 0 ) X k ≥ (log n ) 1 / 2 log log n k ( d − 1)  d − 1 d  (1+ ǫ/ 2) k = o ( t 0 ) (100) The upp er b ound on co ver t ime of C u ≤ t 0 + o ( t 0 ) now follo ws f r o m (96 )–(100). 8.2 Lo w er Bound on the Co v er T ime F or np = d log n , let t 1 = (1 − ǫ ) d log  d d − 1  n log n . Here ǫ → 0 sufficien tly slo wly so that all inequalities claimed b elo w are v alid. Case 1: np ≤ n δ where 0 < δ ≪ η is a p ositiv e constant. Let k ∗ = ( d − 1) log n , and let V ∗ =  v : deg − ( v ) = k ∗ and deg + ( v ) = d log n  . Whp the size | V ∗ | ≥ n ∗ = n γ d 4 π log n ( d ( d − 1)) 1 / 2 (see Lemma 4(ii)). Let us first work assuming d ≥ 1 . 05. In this case γ d = ( d − 1) ln( d/ ( d − 1)) ≥ . 15 and w e write n ∗ = n γ d − o (1) . The maxim um degree in D is at most ∆ 0 = O ( np ) and so V ∗ con tains a sub-set V ∗ 1 of size n γ d / 2 suc h that v , w ∈ V ∗ 1 and x ∈ V implies dist ( x, v ) + dist ( x, w ) > Λ / 100 . (101) dist ( y , x ) + dist ( x, y ) > Λ / 50 , f or y = v , w . (102) Here ” dist” refers to directed distance in D n,p and recall that Λ = lo g np n . Eac h v ∈ V ∗ 1 has π v ∼ d − 1 dn and so w e can choose a subse t V ∗∗ of size ≥ n γ d / 3 suc h that if v 1 , v 2 ∈ V ∗∗ then | π v 1 − π v 2 | ≤ 1 n log 10 n . (103) 39 Indeed, supp ose that π v ∈ h d − 1 2 dn , 2( d − 1) dn i for v ∈ V ∗ 1 . Divide this in terv al in to log 10 n equal sized sub-in terv als and then use the pigeon- hole principle. No w c ho o se u / ∈ V ∗∗ and let V † denote the set of v ertices in V ∗∗ that ha v e not b een visited b y W u b y time t 1 . Then E ( | V † | ) → ∞ , as the followin g calculation sho ws; E ( | V † | ) ≥ n γ d / 3  exp  − (1 + o ( 1)) k ∗ t 1 m  − o ( e − Ω( t 1 /T ) )  − T , where the la st term accoun ts for p ossible visits b efore time T . No w assume that 1 + o (1) ≤ d ≤ 1 . 0 5. In these circumstances w e hav e n ∗ = log ω n where ω → ∞ , see (19). Equations (1 0 1), (102) no w hold fo r all v , w ∈ V ∗ . This follo ws from Lemma 14 b ecause the v ertices of V ∗ are small. The size of V ∗∗ is at least n ∗ / (log n ) 10 and w e can again write E ( | V † | ) ≥ n ∗ (log n ) 10  exp  − (1 + o ( 1)) k ∗ t 1 m  − o ( e − Ω( t 1 /T ) )  − T → ∞ . As in previous pa p ers, see for example [5], we will finish our pro of b y using, the Che b yshev inequalit y to sho w that V † 6 = ∅ whp , th us completing the pro of of Theorem 1. This will follo w if we can pro v e that V ar ( | V † | ) = o ( E ( | V † | 2 ) + O ( | V ∗∗ | 2 n − 2 ) = o ( E ( | V † | ) 2 ) . T o establish this inequalit y , w e will sho w that if v , w ∈ V ∗∗ then Pr ( A v ( t 1 ) ∩ A w ( t 1 )) ≤ (1 + o (1)) Pr ( A v ( t 1 )) Pr ( A w ( t 1 )) . (104) T o prov e this, w e iden tify ve rtices v , w in to a “sup erno de” σ to obtain a digraph D σ with n − 1 v ertices. In this digraph σ has in-degree deg − ( v ) + deg − ( w ) = 2 k ∗ and out- degree 2 d log n . The stationary distribution of D σ . Let π ∗ denote the v ector of steady states in D σ . The argumen ts w e used in Sections 4 a nd 5 remain v alid in D σ , and th us π ∗ σ ∼ (1 − o (1)) 2 k ∗ m . Ho w ev er, w e need to b e more precise. F or a ve rtex x of D σ let ˆ π x = ( π x x 6 = σ π v + π w x = σ . W e will prov e for all x ∈ V ( D σ ), tha t | π ∗ x − ˆ π x | = O  1 n (log n ) 8  . (105) 40 Pro of of (105) . Let ξ = ˆ π − π ∗ b e the difference b et w een ˆ π and π ∗ . Let P ∗ b e the transition matrix of the w a lk on D σ , then P ∗ ( x, y ) =      P ( x, y ) x , y 6 = σ ( P ( v , y ) + P ( w , y )) / 2 x = σ P ( x, v ) + P ( x, w ) y = σ . Let ξ ′ b e the transp ose of ξ . It follo ws from the steady state equations that ( ξ ′ P ∗ ) x =      ˆ π x − π ∗ x x / ∈ N + ( { v , w } ) ˆ π x − π ∗ x + π w − π v 2 P ( v , x ) x ∈ N + ( v ) ˆ π x − π ∗ x + π v − π w 2 P ( w , x ) x ∈ N + ( w ) . W e rewrite this as ξ ′ ( I − P ∗ ) = η ′ (106) where η x = 0 for x / ∈ N + ( { v , w } ) and | η x | ≤ | π v − π w | / 2 otherwise. Multiplying (106) on the righ t b y M = P T − 1 t =0 ( P ∗ ) t w e ha v e ξ ′ ( I − P ∗ ) M = ξ ′ ( I − ( P ∗ ) T ) = η ′ M . (107) Let ( P ∗ ) T = Π + E (108) where Π is the ( n − 1) × ( n − 1) matrix with each row equal to ( π ∗ ) ′ . The definition of T implies that each entry of E ha s absolute v alue b ounded by n − 3 . No w write ξ = α π ∗ + ζ where ζ ⊥ π ∗ . It follo ws from ( π ∗ ) ′ P ∗ = ( π ∗ ) ′ and (107) that ( απ ∗ + ζ ) ′ ( I − ( P ∗ ) T ) = ζ ′ ( I − ( P ∗ ) T ) = ζ ′ ( I − Π − E ) = η ′ M . No w ζ ′ ( I − E ) = ζ ′ ( I − ( P ∗ ) T + Π) = η ′ M + ζ ′ Π . As ζ ⊥ π ∗ this implies tha t ζ ′ ( I − E ) ζ = η ′ M ζ . (1 09) Note that | η ′ M ζ | ≤ T − 1 X t =0 | η ′ ( P ∗ ) t ζ | ≤ T | η | | ζ | , (110) where | z | denotes the ℓ 2 norm of z . No w | ζ ′ ( I − E ) ζ | ≥ | ζ | 2 − | ζ ′ E ζ | ≥ | ζ | 2 − n − 3 n − 1 X i =1 | ζ i | ! 2 ≥ | ζ | 2 (1 − n − 2 ) . (111) 41 It fo llows f rom (1 09), (110) and (111 ) that | ζ | 2 (1 − n − 2 ) ≤ T | η || ζ | and so using (103) we find that | ζ | = O  1 n (log n ) 8  . ( 1 12) No w let 1 denote the ( n − 1)-v ector of 1’s. Then 0 = 1 − 1 = ( ˆ π − π ∗ ) ′ 1 = ξ ′ 1 = α + ζ ′ 1 . Using (112) this giv es | α | ≤ | 1 | | ζ | = O  1 n 1 / 2 (log n ) 8  . No w ξ x = α π ∗ x + ζ x for all x and so ξ 2 x ≤ 2 α 2 ( π ∗ x ) 2 + 2 ζ 2 x = O  1 n (log n ) 16 · 1 n 2 + 1 n 2 (log n ) 16  = O  1 n 2 (log n ) 16  . This completes t he pro of of (10 5). ✷ Pro of of (104) . F or v ∈ V ∗∗ , w e first tighte n (87) to R v = 1 + o (1 / (log n ) 2 ) . (113) Assume first that np ≤ log 10 n . Then (101) and (1 0 2) imply that for 1 ≤ t ≤ (log n ) 2 / 3 , v ertex v will b e at distance ≥ 2 log 2 / 3 n − t from W v ( t ). Then once t he w alk is a t a ve rtex w within distance lo g 2 / 3 n of v its ch ance o f getting closer is only O (1 / log n ). This b eing true with at most one exception for a v ertex of lo w o ut-degree. The pro ba bilit y that there is a time t suc h that W v is within log 2 / 3 n of v and it mak es 10 ste ps closer to v in the next 100 steps is O ( T / log 9 n ) = O (1 / log 7 n ). This implies (1 1 3). If np ≥ log 10 n then w e use R v ≤ 1 + (1 + o (1)) T /np . Similarly , R σ = 1 + o (1 / (log n ) 2 ) . (114) The mixing time T in wh at follo ws is the max im um of the mixing t imes for D and the maxim um ov er v , w for D σ . Using the suffix Pr σ to denote pro ba bilities related to random w a lks in D σ and using (105), it follo ws that Pr σ ( A σ ( t 1 )) ≤ exp  − (1 + O ( T π ∗ σ )) π ∗ σ t 1 m  − o ( e − Ω( t 1 /T ) ) ≤ exp  − (1 + o ( 1 / log n ))( π v + π w ) t 1 m  − o ( e − Ω( t 1 /T ) ) = (1 + o ( 1 )) Pr ( A v ( t 1 )) Pr ( A w ( t 1 )) . (115) 42 But, using r a pid mixing in D σ , Pr σ ( A σ ( t 1 )) = X x 6 = σ P T σ ,u ( x ) Pr σ ( W x ( t ) 6 = σ , 1 ≤ t ≤ t 1 − T ) = X x 6 = σ ( π ∗ x + O ( n − 3 )) Pr σ ( W x ( t ) 6 = σ, 1 ≤ t ≤ t 1 − T ) On the other hand, Pr ( A v ( t 1 ) ∩ A w ( t 1 )) = X x 6 = v, w P T u ( x ) Pr ( W x ( t ) 6 = v , w , T ≤ t ≤ t 1 ) = X x 6 = v, w ( π x + O ( n − 3 )) Pr ( W x ( t ) 6 = v , w , 1 ≤ t ≤ t 1 − T ) But, Pr σ ( W x ( t ) 6 = σ, T ≤ 1 ≤ t 1 − T ) = Pr ( W x ( t ) 6 = v , w, 1 ≤ t ≤ t 1 − T ) b ecause random w alks from x that do not meet v , w or σ hav e the same measure in b oth digraphs. It fo llows t ha t Pr ( A v ( t 1 ) ∩ A w ( t 1 )) − Pr σ ( A σ ( t 1 )) = X x 6 = v, w ( π x − π ∗ x + O ( n − 3 )) Pr ( W x ( t ) 6 = v , w , 1 ≤ t ≤ t 1 − T ) ≤ O  1 n log 8 n  X x 6 = v, w Pr ( W x ( t ) 6 = v , w , 1 ≤ t ≤ t 1 − T ) ≤ O  1 n log 8 n  X x 6 = v, w P T u ( x ) P T u ( x ) Pr ( W x ( t ) 6 = v , w, 1 ≤ t ≤ t 1 − T ) ≤ O  1 n log 8 n  O ( n log n ) X x 6 = v, w P T u ( x ) Pr ( W x ( t ) 6 = v , w , 1 ≤ t ≤ t 1 − T ) since P T u ( x ) = Ω(1 / n log n ) ≤ O  1 log 7 n  Pr ( A v ( t 1 ) ∩ A w ( t 1 )) . (116) Equations (115) and (11 6) together imply (104). ✷ Case 2: np ≥ n δ . In this range w e tak e t 1 = (1 − ǫ ) n log n and let V ∗ b e the set of v ertices of degree ⌊ np ⌋ . A simple second momen t calculation sho ws that whp we hav e | V ∗ | = Ω(( np ) 1 / 2 − o (1) ). W e then 43 c ho ose ǫ so that E ( | V † | ) ≥ ( np ) 1 / 4 . It is then only a matter of verifying (104 ). The details are as in the previous case. 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