Problem set 2 for the course on. Markov chains and mixing times

Transcription

1 J. Seif T. Hirscher Soluions o Proble se for he course on Markov chains and ixing ies February 7, 04 Exercise 7 (Reversible chains). (i) Assue ha we have a Markov chain wih ransiion arix P, such ha here exis a posiive funcion f and a non-negaive funcion g on S wih p ij = f(i) g(j) for all i,j S. Show ha such a Markov chain is reversible and derive he corresponding saionary probabiliy vecor. Why is he saionary disribuion unique in his case? How any seps will i ake o reach equilibriu? (ii) Can -sae Markov chains be non-reversible? Exhibi a 3-sae Markov chain wih p ij > 0 for all i,j S which is no reversible. Soluion. (i) Le E := {i S, g(i) > 0}. Since p ij > 0 for all i S, j E, all saes in E are essenial, saes ouside E us be inessenial due o p ij = 0 for all i S, j / E. In conclusion, E is he subse of all essenial saes. Tha P is sochasic yields = j S p ij = f(i) j S g(j). Hence, for all saes i, we ge f(i) =, where c := c j S g(j). If we undersand g as a vecor in R n, every uliple of i will saisfy he deailed balance equaions, due o f being consan. Obviously, π = g is he righ scaling o c ge a probabiliy disribuion. Since p ij > 0 for all i,j E, we have only one essenial counicaing class. By Prop..6 which iediaely follows fro Ex. 5 on he firs assignen he saionary disribuion π is unique. Since p ij does no depend on i, due o f being consan, he disribuion afer he firs sep will be π regardless of he iniial disribuion. So afer sep, equilibriu is reached. page

2 (ii) In Ex. 3 on he firs assignen i was shown ha a general -sae MC wih ransiion probabiliies p = p, p = q has he saionary disribuion π = ( q, p ), if i is no he case ha p = q = 0. Since π saisfies he p+q p+q deailed balance equaion (here is jus one having only saes) and in he case p = q = 0 boh sides equal 0, every -sae MC is reversible. As an exaple for a non-reversible chain on hree saes wih p ij > 0, we can ake he ransiion arix P =. 4 Since P is doubly sochasic, he unifor disribuion is saionary. The corresponding Markov chain is obviously irreducible, which ells us ha π = ( 3, 3, 3 ) is he unique saionary disribuion. This allows us o conclude fro π() p = 3 = 6 π() p = 3 4 = ha he chain is non-reversible, since every disribuion saisfying he deailed balance equaions would be saionary. For he nex exercise you will need he following corollary o he Convergence Theore (Th. 4.9): A finie Markov chain (X ) N0 which is irreducible and aperiodic forges is iniial sae and is disribuion converges o equilibriu in he sense ha for all saes i,j. li p() ij = li P[X = j X 0 = i] = π(j) Exercise 8. Le (X ) N0 be a finie irreducible Markov chain having saionary disribuion π and furher N(i,) denoe he (rando) nuber of visis of sae i aong X,...,X. Wihou using Proposiion.4 or Theore 4.6, show ha E[N(i,)] π(i) as and ha N(i,) π(i) in probabiliy. Noe ha Theore 4.6 acually iplies ha his laer convergence holds alos surely. Hin: Wrie N(i,) as a su of indicaor variables and bound is variance. Soluion. Le us firs consider only aperiodic chains, wrie q s := P(X s = i) and N(i,) = s= {Xs=i}. page

3 I is no hard o show ha any sequence (a s ) s N converging o soe lii a is Cesàro suable and he Cesàro su equals a, i.e. li a s = a. Using he convergence of p () ij q = P(X = i) = j S Cobining boh facs yields E[N(i,)] = s= as saed above, we can conclude p () ji P(X 0 = j) π(i) as. q s π(i) as. s= As o he second saeen, Chebyshev s inequaliy gives for all ε > 0: P ( N(i,) π(i) > ε ) P ( N(i,) E[N(i,)] > ε ) 4 var ( N(i,) ), ε if is large enough s.. E[N(i,)] π(i) < ε. In order o prove ha his ends o 0, which iplies N(i,) π(i) in probabiliy, i is lef o show ha var ( N(i,) ) 0 as. Using lineariy of expecaion, we can conclude var ( N(i,) ) = E[N(i,) ] ( E[N(i,)] ) = E [ = {Xr=i,X s=i} ] ( E [ {Xs=i} + s= r s s= q s qs + r = i,x s = i) q r q s ) r s(p(x s= s= q s + r<s s= q r (p (s r) ii q s ). ]) {Xs=i} The firs su divided by converges o π(i), so o 0 if divided by. As o he second su, we know fro above ha for δ > 0, here exiss T N such ha boh p (s) ii π(i) and q s π(i) are a os δ for s T. For s T, using he riangle inequaliy, we ge s q r (p (s r) ii q s ) s T q r δ + s q r δ q r + T δ π(i), as. r= This iplies for T : r<s r= r=s T + q r (p (s r) ii q s ) = s= r= r= s q r (p (s r) ii q s ) ( T + ( δ s=t q r + T ) ). r= page 3

4 Since his las upper bound converges o δ π(i) as well and δ > 0 was arbirary, we have shown ha var ( N(i,) ) = o( ) which concludes he proof for he aperiodic case. If he considered Markov chain is no aperiodic, due o irreducibiliy all saes have he sae period d. We won ge he convergence of probabiliies as saed before he exercise, bu if we consider P d, we can order he saes such ha he ransiion arix has block-diagonal for: Le define an equivalence relaion on S by i j p (sd) ij > 0 for soe s > 0. Le A,...,A d denoe he corresponding equivalence classes, ordered such ha P[X A k+ X 0 A k ] =. If we look a he d-sep MC, i has he irreducible coponens A,...,A d. If π denoes he saionary disribuion, πp d = π and for X 0 π: π(a k+ ) = P(X A k+ ) = P(X 0 A k ) = π(a k ). Hence π(a k ) = for all k. The d-sep MC resriced o A d k, for fixed k, is irreducible, aperiodic and has saionary disribuion π π(a k ) Ak = d π Ak. So by he above we ge for i A k : E[N(i,s)] d π(i) and N(i,s) d π(i) in probabiliy, s s where s is he ie in he d-sep MC. If denoes he ie in he original periodic chain, we ge = sd, hence jus as claied. E[N(i,)] π(i) and N(i,) π(i) in probabiliy, Exercise 9. You are given wo probabiliy easures µ and ν on a finie se S. Bob is required o flip a fair coin (which you canno see he resul of) and if he coin is heads, he us give you an eleen of S which has disribuion µ and if he coin is ails, he us give you an eleen of S which has disribuion ν (independenly chosen of he coin oss). Based upon wha you end up receiving, your job is o ry o guess if he coin was heads or ails and o axiize he probabiliy ha you are correc. Of course you can be correc wih probabiliy by jus always guessing heads bu you wan o do beer han his. (i) Show ha if µ ν TV δ, hen here is a sraegy which gives a probabiliy of being correc which is a leas + δ. (ii) Show ha if µ ν TV δ, hen he axial probabiliy of being correc is a os + δ. In conclusion, oal variaion easures he degree o which you can saisically ell apar wo disribuions. page 4

5 Soluion. By Prop. 4. we know ha µ ν TV = µ(i) ν(i) = µ(i) ν(i), i B where B := {i S, µ(i) ν(i)}. Le X denoe he rando eleen of S which we are given. Then for all i S: P(heads, X = i) = µ(i) and P(ails, X = i) = ν(i). (i) If µ ν TV δ, le us adop he sraegy o guess µ whenever we receive an eleen of B and ν oherwise. The probabiliy p of guessing correc becoes p = P(heads, X = i) + P(ails, X = i) i B i/ B = ( µ(i) + ) ν(i) i B i/ B = ( + ) µ(i) ν(i) = ( + µ ν TV) i B + δ. (ii) No aer which sraegy we adop, if we are given eleen i we are incorrec wih our guess wih probabiliy a leas in{p[heads X = i], P[ails X = i]}, even for a randoized decision. Hence, if µ ν TV δ: p P(X = i) in{p[heads X = i], P[ails X = i]} = in { µ(i), ν(i) } = ( µ(i) + i/ B i B = ( µ(i) + i B i/ B δ, which iplies p + δ. ) ν(i) ) ν(i) = µ ν TV Exercise 0. Le P be he ransiion arix of a finie Markov chain (X ) N0 wih saionary disribuion π and saring disribuion µ = L(X 0 ). Show ha he oal variaion disance of he disribuion of X, i.e. µ P, o π is non-increasing wih, i.e. µ P π TV µ P + π TV for all N 0. Explain how his iplies ha d() = ax P (i, ) π TV is non-increasing. page 5

6 Soluion. Le us wrie µ := µ P. Since πp = π, applying Prop. 4. and he riangle inequaliy yields Choosing µ = δ i shows µ P + π TV = µ P π P TV = µ (j)p (j,i) π(j)p (j,i) j S j S P (j,i) µ (j) π(j) j S = µ (j) π(j) P (j,i) j S = µ (j) π(j) = µ π TV. j S P + (i, ) π TV P (i, ) π TV, axiizing his over i gives d( + ) d(). Exercise. (i) Consider an aperiodic irreducible finie Markov chain having saionary disribuion π and he propery ha here exiss a sae i and a se A S wih π(a) = i A π(i) >, and d(i,a), where d(i,a) is he shores pah 4 disance fro i o any node in A in he (direced) Markov chain graph. Show ha ix. (ii) Use his o obain lower bounds for he ixing ie for a lazy rando walk on he hypercube Z d and on he orus Z, d >. How sharp a bound can you ge for he hypercube arguing his way? For wha cobinaions of and d can you prove, using (i), ha he chain is no rapidly ixing? Soluion. (i) If <, we find P[X A X 0 = i] = 0. This enails d() P (i, ) π TV π(a) j A P (i,j) = π(a) > 4. Hence ix >, which gives ix. (ii) In he case of he hypercube Z d, ake i = (0,...,0) and A o be he se of saes having a leas d coordinaes wih value, giving d(i,a) d. As he graph is regular, π is unifor and we ge π(a), which iplies ix d by he above. Fro Chebyshev s inequaliy, we know ha for a Bin(d, )-disribued rando variable Z P( Z EZ c) var(z) c = d 4c. page 6

7 As he disribuion of Z is syeric around is ean, his iediaely iplies P(Z EZ c) d. Hence, already he se B of vecors having a leas 8c d + d ones is of insufficien size, since π(b) = d B = P(Z EZ d) 8 < 4. So using he arguen above, when i coes o he asypoically leading er we won ge anyhing beer han d as a lower bound. In he case of he orus Z, d ake again i = (0,...,0) and his ie A o be he se of vecors which have a leas d coordinaes in { 4,..., 3 4 }. Then d(i,a) d 4 and since π is again unifor, π(a). Consequenly, ix d 4. Noe ha he size of he sae space is Z d = d. Using he lower bound on ix jus derived, we can conclude ha a sequence of lazy rando walks on ori Z d is no rapidly ixing, if is no polynoial in d, i.e. he sequence of (,d) is such ha for all k N. dk Exercise. Show ha here exiss a finie sae Markov chain so ha for wo of is saes i and j, li P (i, ) P (j, ) TV > 0 bu here exiss a coupling of he Markov chain saring respecively a i and a j, (X,Y ) N0 so ha T := inf{ : X = Y } is finie wih probabiliy. Noe ha his ells you ha condiion (5.) is very essenial in he saeen of Theore 5.. Soluion. By he convergence heore, such a chain can no be irreducible: In he aperiodic case boh P (i, ) and P (j, ) converge o π; in he periodic case define and he corresponding equivalence classes as in he soluion o Exercise 8, hen eiher i j, which will give convergence of boh P d (i, ) and P d (j, ) o π([i]) π [i] by he sae reasoning as above and hus li P (i, ) P (j, ) TV = 0, or i j which will ake P(T < ) = ipossible. A concree exaple can be found as Prop. 4. in he paper A Noe on Disagreeen Percolaion by Olle Häggsrö. Noe ha he coupling described here is no Markovian. If we had a Markovian coupling wih P(T < ) =, we could odify i for > T o enforce (5.) which hen would give li P (i, ) P (j, ) TV = 0, by Th. 5.. page 7