A technical report on hitting times, mixing and cutoff

0.58 MB·English

by  Jonathan Hermon

A technical report on hitting times, mixing and cutoff
Jonathan Hermon
∗
5
1
0 Abstract
2
Consider a sequence of continuous-time irreducible reversible Markov chains and a
r
a sequence of initial distributions, µ . Instead of performing a worst case analysis, one
n
M
can study the rate of convergence to the stationary distribution starting from these
initial distributions. The sequence is said to exhibit (total variation) µ -cutoff if the
2 n
1 convergence to stationarity in total variation distance is abrupt, w.r.t. (with respect

to) this sequence of initial distributions.
]
R In this work we give a characterization of µ -cutoff (and more generally, total-
n
P variation mixing) for an arbitrary sequence of initial distributions µ (in the above
n
h. setup). Our characterization is expressed in terms of hitting times of sets which are
t “worst” (in some sense) w.r.t. µ .
a n
m Consider a Markov chain on Ω whose stationary distribution in π. Let t (α) :=
H
[ maxx∈Ω,A⊂Ω:π(A)≥αEx[TA], be the expected hitting time of the set of stationary prob-

ability at least α which is “worst in expectation” (starting from the worst starting
3
v state). The connection between tH( ) and the mixing time of the chain was previ-
·
9 ously studied by Aldous and later by Lov´asz and Winkler, and was recently refined
6
by Peres and Sousi and independently by Oliveira. In this work we further refine this
8
1 connection and show that µ -cutoff can be characterized in terms of concentration of
n
0
hitting times (starting from µ ) of sets which are worst in expectation w.r.t. µ . Con-
n n
.
1 versely, we construct a surprising counter-example which demonstrates that in general
0
cutoff (as opposed to cutoff w.r.t. a certain sequence of initial distributions) cannot be
5
1 characterized in this manner.
: In addition, we prove a decomposition theorem which asserts that for reversible
v
i Markov chains on a finite state space, mixing with respect to some relaxations of L∞-
X
mixing and separation mixing is in fact equivalent to total variation-mixing.
r
a Finally, we also prove that there exists an absolute constant C such that for any
reversible chain (cid:15)(t ((cid:15)) t (1 (cid:15))) Ct log(cid:15) , for all 0 < (cid:15) < 1/2, where t is the
H H rel rel
− − ≤ | |
inverse of the spectral gap of the chain.
Keywords: Mixing-time, hittingtimes, cutoff, finitereversibleMarkovchains, maximalinequal-
ity, counter-example.
∗Department of Statistics, UC Berkeley, USA. E-mail: [email protected].
1
1 Introduction
This work is a continuation of [3], in which Starr’s maximal inequality was used to character-
ize the cutoff phenomenon for reversible Markov chains in terms of concentration of hitting
times. Here using the same technique we present several new related results.
Generically, we shall denote the state space of a Markov chain by Ω and its stationary
distribution by π (or Ω and π , respectively, for the n-th chain in a sequence of chains). We
n n
say that the chain is finite, whenever Ω is finite. Let (Y )∞ be an irreducible Markov chain
t t=0
on a finite state space Ω with transition matrix P and stationary distribution π. We denote
suchachainby(Ω,P,π). Achain(Ω,P,π)iscalledreversible ifπ(x)P(x,y) = π(y)P(y,x),
for all x,y Ω.
∈
Periodicity issues can be avoided by considering the continuous-time version of the chain,
(X ) . This is a continuous time Markov chain whose heat kernel is defined by H (x,y) :=
t t≥0 t
(cid:80)∞ e−ttkPt(x,y). It is a classic result of probability theory that for any initial condition
k=o k!
the distribution of X converges to π when t goes to infinity. The object of the theory of
t
Mixing time for Markov chain is to study the characteristic of this convergence (see [6] for a
self-contained introduction to the subject). Throughout, we shall consider only continuous
timechains, althoughall ourresults canbe statedalso indiscrete time, assumingP(x,x) δ
≥
for some δ > 0 for all x Ω.
∈
We denote by Ht (H ) the distribution of X ((X ) ), given that the initial distribution
µ µ t t t≥0
is µ. When µ = δ (where δ (y) = 1 ), for some x Ω, we simply write Ht (H ).
We denote thex set of prxobabilityx=dyistributions o∈n a (finite) set B by Px(B)x. For any
µ,ν P(B), their total-variation distance is defined to be
∈
1 (cid:88) (cid:88)
µ ν := µ(x) ν(x) = µ(x) ν(x).
TV
(cid:107) − (cid:107) 2 | − | −
x x∈B:µ(x)>ν(x)
The worst-case total variation distance at time t is defined as d(t) := max d (t), where
x∈Ω x
for any µ P(Ω),
∈
1 (cid:88) (cid:88)
d (t) := H (X ) π = Ht π = µ(x)H (x,y) π(y) .
µ (cid:107) µ t ∈ · − (cid:107)TV (cid:107) µ − (cid:107)TV 2 | t − |
y∈Ω x∈Ω
The (cid:15)-mixing-time is defined as
t ((cid:15)) := inf t : d(t) (cid:15) .
mix
{ ≤ }
We also define the (cid:15)-mixing-time w.r.t. a fixed initial distribution µ to be t ((cid:15)) :=
mix,µ
inf t : d (t) (cid:15) . When (cid:15) = 1/4 we simply write t and t .
µ mix mix,µ
{ ≤ }
Recall that if (Ω,P,π) is a finite reversible irreducible chain, then P is self-adjoint
w.r.t. the inner product induced by π (see Definition 2.1) and hence has Ω real eigen-
| |
values. Throughout we shall denote them by 1 = λ > λ ... λ 1 (where λ < 1,
1 2 |Ω| 2
≥ ≥ ≥ −
by irreducibility). Define the relaxation-time of P as t := (1 λ )−1. The following
rel 2
−
general relation holds for reversible chains,
(cid:18) (cid:19) (cid:18) (cid:19)
1 1
t log t ((cid:15)) log t , (1.1)
rel mix rel
2(cid:15) ≤ ≤ (cid:15)min π(x)
x
2
(see [6] Lemmas 20.5 and 20.11).
The following mixing parameter, introduced in [3], shall play a key role in this work.
Definition 1.1. Let (Ω,P,π) be an irreducible chain. For any µ P(Ω), δ,(cid:15) (0,1) and
∈ ∈
t 0, define p (δ,t) := max H [T > t], where T := inf t : X B is the
µ B⊂Ω:π(A)≥δ µ B B t
≥ { ∈ }
hitting time of the set B. Set p(δ,t) := max p (δ,t). We define
x∈Ω x
hit ((cid:15)) := min t : p (δ,t) (cid:15) and hit ((cid:15)) := min t : p(δ,t) (cid:15) ,
δ,µ µ δ
{ ≤ } { ≤ }
Next, consider a sequence of such chains, ((Ω ,P ,π ) : n N), each with its correspond-
n n n
∈
ing worst-distance from stationarity d (t), its mixing-time t(n), etc.. Loosely speaking, the
n mix
(total variation) cutoff phenomenon occurs when over a negligible period of time, known
as the cutoff window, the (worst-case) total variation distance (of a certain finite Markov
chain from its stationary distribution) drops abruptly from a value close to 1 to near 0. In
other words, one should run the n-th chain until time (1 o(1))t(n) for it to even slightly
− mix
mix in total variation, whereas running it any further after time (1+o(1))t(n) is essentially
mix
redundant. Formally, we say that a sequence of chains exhibits a cutoff if the following
sharp transition in its convergence to stationarity occurs:
t(n)((cid:15))
lim mix = 1, for every 0 < (cid:15) < 1.
n→∞ t(n)(1 (cid:15))
mix −
Similarly, for a sequence of initial distributions µ P(Ω ), we say that a sequence of
n n
∈
chains exhibits a µ -cutoff if
n
t(n) ((cid:15))
lim mix,µn = 1, for every 0 < (cid:15) < 1.
n→∞ t(n) (1 (cid:15))
mix,µn −
We say that the sequence has a cutoff window w , if w = o(t(n)) and for any (cid:15) (0,1)
n n mix ∈
there exists c > 0 such that for all n
(cid:15)
t(n)((cid:15)) t(n)(1 (cid:15)) c w
mix − mix − ≤ (cid:15) n (1.2)
(respectively, t(n) ((cid:15)) t(n) (1 (cid:15)) c w ).
mix,µn − mix,µn − ≤ (cid:15) n
We say that a family of chains satisfies the product condition if (1 λ(n))t(n) as
− 2 mix → ∞
n (or equivalently, t(n) = o(t(n))). The following well-known fact follows easily from
→ ∞ rel mix
the first inequality in (1.1) (c.f. [6], Proposition 18.4).
Fact 1.2. For a sequence of irreducible reversible Markov chains, if the sequence exhibits a
cutoff, then t(n) = o(t(n)).
rel mix
Definition 1.3. Let (Ω ,P ,π ) be a sequence of irreducible chains and let α (0,1). We
n n n
∈
say that the sequence exhibits a hit -cutoff (resp. hit -cutoff), if for every 0 < (cid:15) < 1/4,
α α,µn
(cid:16) (cid:17)
hit(n)((cid:15)) hit(n)(1 (cid:15)) = o hit(n)(1/4)
α − α − α
(cid:16) (cid:17)
(respectively, hit(n) ((cid:15)) hit(n) (1 (cid:15)) = o hit(n) (1/4) ).
α,µn − α,µn − α,µn
3
The main abstract result in [3] is the following theorem.
Theorem 1. Let (Ω ,P ,π ) be a sequence of reversible irreducible finite Markov chains.
n n n
The following are equivalent:
(1) The sequence exhibits a cutoff.
(2) The sequence exhibits a hit -cutoff for some α (0,1) and t(n) = o(t(n)).
α ∈ rel mix
Definition 1.4. Let µ P(Ω) and 0 < α < 1. Define
∈
t (α) := max E [T ] and t (α) := max E [T ].
H,µ µ A H x A
A⊂Ω:π(A)≥α x∈Ω,A⊂Ω:π(A)≥α
This work was greatly motivated by the results of Peres and Sousi in [9]. Similar results
were obtained independently by Oliviera [8]. Both papers refine previous results of Aldous
[1] and of Lov´asz and Winkler [7]. Their results share the general theme of describing
mixing-times in terms of hitting-times. Their approach relied on the theory of random
times to stationarity combined with a certain “de-randomization” argument which shows
that for any reversible irreducible finite chain and any stopping time T such that X π,
T
t = O(max E [T]). As a consequence, they showed that for any 0 < α < 1/2 ∼(this
mix x∈Ω x
was extended to α = 1/2 in [5]), there exist some constants c ,c(cid:48) > 0 such that for any
α α
reversible irreducible finite chain
c(cid:48) t (α) t c t (α). (1.3)
α H ≤ mix ≤ α H
It is natural to ask whether the more studied mixing parameter t (α) could be used in
H
Theorem 1 instead of the mixing parameter hit ( ).
α
ThefollowingtheoremextendsTheorem1toar·bitrarystartingdistributionsµ P(Ω ),
n n
∈
such that t(n) = o(t(n) ). In addition, it asserts that “cutoff” w.r.t. these initial distribu-
rel mix,µn
tions (i.e. µ -cutoff), is in fact equivalent to concentration of hitting times of sets which are
n
“worst in expectation” w.r.t. these initial distributions (in the sense of Definition 1.4).
Theorem 2. Let (Ω ,P ,π ) be a sequence of finite irreducible reversible chains. Let µ
n n n n
∈
P(Ω ) be such that t(n) = o(t(n) (δ)), for some 0 < δ < 1. Then the following are
n rel mix,µn
equivalent:
i) The sequence exhibits a µ -cutoff.
n
ii) There exists some α (0,1) such that the sequence exhibits a hit -cutoff.
∈ α,µn
iii) There exist some α (0,1) and a sequence of sets A Ω with π (A ) α satisfying
n n n n
∈ ⊂ ≥
that E [T ] = t(n) (α), such that
µn An H,µn
lim H [ T t(n) < (cid:15)E [T ]] = 1, for every (cid:15) > 0.
n→∞ µn | An − mix,µn| µn An
Corollary 1.5. Let (Ω ,P ,π ) be a sequence of finite irreducible reversible transitive chains.
n n n
Then the following are equivalent
4
i) The sequence exhibits a cutoff.
ii) The sequence satisfy the product condition, and for some sequence x Ω and some
n n
∈
0 < α < 1, there exists a sequence of sets A Ω with π (A ) α satisfying that
n n n n
⊂ ≥
E [T ] = t(n) (α), such that
xn An H,xn
lim H [ T t(n) < (cid:15)E [T ]] = 1, for every (cid:15) > 0.
n→∞ xn | An − mix| xn An
The following proposition asserts that in general cutoff (as opposed to cutoff from a se-
quenceoffixedinitialdistributions)cannotbecharacterizedintermsofthemixingparameter
t (α).
H
Proposition 1.6. There exists a sequence (Ω ,P ,π ) of finite irreducible reversible chains
n n n
satisfying the product condition such that the following holds:
There exist A Ω and x Ω such that π (A ) 1/2 and E [T ] = t(n)(1/2).
· n ⊂ n n ∈ n n n ≥ xn An H
The distribution of the hitting times of A are concentrated w.r.t. the initial states x .
n n
·
The sequence does not exhibit a cutoff.
·
In Example 5.1 we construct an sequence of chains which exhibits the behavior described
in Proposition 1.6.
Remark 1.7. It was shown in [4] that a sequence of finite continuous-time Markov chains
exhibits a cutoff iff t(n)((cid:15)) t(n)(1 (cid:15)) = o(t(n)(1/4)), where t(n)((cid:15)) is the (cid:15)-mixing-time of the
L − L − L L
associated lazy chain. They also showed that the same holds for a sequence of fixed initial
distributions. Hence in part (i) of Theorem 2 and of Corollary 1.5 we could have considered
the lazy version of the chain, rather than its continuous-time version.
The main ingredient in the proof of Theorem 2 is the following proposition
Proposition 1.8. Let (Ω,P,π) be a finite irreducible reversible Markov chain. Let µ
P(Ω). Let 0 < (cid:15) 1/2. Then ∈
≤
hit (3(cid:15)/2) 2t log(cid:15) t ((cid:15)) hit (3(cid:15)/4)+t log(16/(cid:15)),
1/2,µ rel mix,µ 1/2,µ rel
− | | ≤ ≤
1
hit (1 (cid:15)/2) 2t log(cid:15) t (1 (cid:15)) hit (1 2(cid:15))+ t log8.
1/2,µ rel mix,µ 1/2,µ rel
− − | | ≤ − ≤ − 2
The separation of y Ω w.r.t. µ P(Ω) is defined as 1 µ(y)/π(y). An important
∈ ∈ −
notion of distance from stationarity, intimately related to total variation distance, is the
separation distance of µ (from π) defined as s := max 1 µ(y)/π(y). Generally,
µ y∈Ω
µ π s , for any µ P(Ω) (see e.g. Lemma 6.13 in [6]). −
TV µ
(cid:107) − (cid:107) ≤ ∈
Another notions of distance from stationarity are the L∞ and Lp distances (1 p < ),
(cid:104) ≤ ∞(cid:105)1/p
defined,respectively,as µ π := max µ(y) 1 and µ π := (cid:80) π(y)(µ(y) 1)p .
(cid:107) − (cid:107)∞,π y|π(y)− | (cid:107) − (cid:107)p,π y π(y) −
By Jensen’s inequality,
2 µ π = µ π µ π µ π , for any 1 < p < .
TV 1,π p,π ∞,π
(cid:107) − (cid:107) (cid:107) − (cid:107) ≤ (cid:107) − (cid:107) ≤ (cid:107) − (cid:107) ∞
5
The (cid:15) separation and Lp mixing times are defined as t ((cid:15)) = inf t : max s (cid:15), and
sep x Ht
{ x ≤ }
τ ((cid:15)) = inf t : max Ht π (cid:15) , for 1 p . Then
p { x(cid:107) x − (cid:107)p,π ≤ } ≤ ≤ ∞
t ((cid:15)) t ((cid:15)) and t ((cid:15)/2) τ ((cid:15)) τ ((cid:15)), for any 0 < (cid:15) < 1 and 1 p < p(cid:48) .
mix sep mix p p(cid:48)
≤ ≤ ≤ ≤ ≤ ∞
In general, one always has that t ((cid:15)) 2t ((cid:15)/4) (c.f. [6] Lemma 19.3). Conversely,
sep mix
≤
in many cases t(n)(1 o(1)) (2 o(1))t(n) (e.g. lazy simple random walk on the n-
sep − ≥ − mix
dimensional hypercube, see Theorem 18.8 in [6]). The L2 mixing-time in many cases satisfies
(cid:16) (cid:17)
t(n) = o τ(n)(1/2) . The following theorem gives a surprising quantitative converse, when
mix 2
one replaces the separation and Lp distances by weaker notions of distance from stationarity
which have a similar from.
An alternative definition of s is that it is the smallest number a such that one can
Ht
x
write Ht = aν + (1 a)π. A natural relaxation of this can be looking for a small a such
x −
that one can write Ht = aν +(1 a)µ for µ satisfying that π( y : µ(y) < (1 (cid:15))π(y) ) < (cid:15).
x − (cid:80) { − }
Similarly, one can require µ to be a mixture of the form c π , for some collection of sets
(cid:80) i∈I i Ai
(A ) such that π(A ) 1 (cid:15) for all i, where c = 1 and π denotes π conditioned on A
i i∈I i ≥ − i i A
(i.e. π (y) = 1 π(y)/π(A)). In the same spirit, a relaxation of the Lp distance from π is
A y∈A
to look at a small a such that one can write Ht = aν+(1 a)µ for µ such that µ π < (cid:15).
x − (cid:107) − (cid:107)p
This following theorem, which we call the Decomposition Theorem, asserts that for
t which is slightly larger than hit (p), one can indeed write Ht as such a mixture, with a
1−(cid:15),µ µ
slightly larger than p. The statement of the Decomposition Theorem may seem cumbersome
at first sight. The reader may find it easier to first think of p and (cid:15) as constants and of w as
tending to infinity.
Theorem 1.9. Let (Ω,P,π) be a finite irreducible reversible Markov chain. Let σ P(Ω).
(cid:16) (cid:17) ∈
Let w R and 0 < (cid:15),p < 1. Denote C := 1 log 16 . Let
∈ (cid:15),p 2 (cid:15)p2
τ = τ := t [w+C ] and ρ = ρ := pe−w.
w,(cid:15),p rel (cid:15),p p,w
Then for any w C , there exist some 0 < c a := p+ρ(1 p) and two distributions
(cid:15),p τ τ
≥ − ≤ −
ν = ν and µ = µ such that the following hold
τ τ
Ht+τ = c ν +(1 c )µ, where t := hit (p),
σ τ − τ 1−(cid:15),σ
1
µ(x)/π(x) = 1+Θ(pe−w), for any x Ω, (1.4)
≤ 1 ρ ∈
−
and for any b [0,1],
∈
1
(cid:0) (cid:1)
π x : µ(x) (1 b)π(x) = O p2b2e−2w . (1.5)
{ ≤ − } ≤ 1+(2b/ρ)2
Moreover,
ρ2
(cid:2) (cid:3) (cid:0) (cid:1)
µ π 2 (ρ/2)2log 1+(2/ρ)2 + = Θ wp2e−2w . (1.6)
(cid:107) − (cid:107)2,π ≤ (1 ρ)2
−
6
In addition, let d := ρ +pe−w/2/2 = Θ(cid:0)pe−w/2(cid:1). Then there exists κ d such that
τ 1−ρ τ ≤ τ
(cid:88)
µ := κ µ +(1 κ )µ , where µ ,µ P(Ω) and µ := µ¯(A)π ,
τ 1 τ 2 1 2 2 A
− ∈
A⊂Ω
(1.7)
(cid:26) ew (cid:27)
where µ¯ is some distribution on A Ω : π(A) .
⊂ ≥ 1+ew
Recall the definition of t ((cid:15)) from Definition 1.4. In [5] the following general inequality
H
was proved (without a reversibility assumption).
Theorem 1.10. Fix 0 < (cid:15) < 1/2. For any irreducible finite Markov chain (cid:15)t ((cid:15)) t (1/2).
H H
≤
We prove a specialized version of the above result for the reversible setup. In many
cases the bounds obtained from Proposition 1.11 are considerably better than the bound in
Theorem 1.10 (in particular, this is the case when the product condition holds).
Proposition 1.11. There exists an absolute constant C > 0 such that for any finite reversible
irreducible chain (Ω,P,π),
t ((cid:15)) t (1 (cid:15)) C(cid:15)−1t , for every 0 < (cid:15) < 1/2 and µ P(Ω).
H,µ H,µ rel
− − ≤ ∈
1.1 A remark about our approach
The approach taken here for relating t ( ) and hit ( ) follows that taken in [3]. Namely,
mix,µ ·,µ
· ·
we define for any B Ω, the set G = G (B,m), which we call the good set for B from
s
⊂
time s within m standard-deviations. This set is defined formally in (2.3). The reason for
the name would become clear once this set is defined formally. Fix some 0 < (cid:15) < 1 and set
m = m := 2/√(cid:15). Fix some initial distribution µ. From the definition of G, it would be
(cid:15)
easy to see that for any t 0, hitting G by time t serves as a “certificate” that the chain is
≥
”(cid:15)-mixed w.r.t. B“ at time t+s.
If for some s = s we had that π(G (B,m )) 1 (cid:15) for all B Ω, then if by time t any
(cid:15) s (cid:15)
≥ − ⊂
set of stationary probability at least 1 (cid:15) is hit with probability at least 1 (cid:15) given that
− −
X µ (i.e. t hit ((cid:15))), then by using the certificate T t and then maximizing
0 ∼ ≥ 1−(cid:15),µ Gs(B,m(cid:15)) ≤
over all B we get that t (2(cid:15)) t+s.
mix,µ
≤
As in [3], we shall use Starr’s maximal inequality (Theorem 2.3) in conjunction with the
L2 contraction Lemma (Lemma 2.2) to show that there exists an absolute constant C > 0
such that π(G (B,m )) 1 (cid:15) for all B, where s := Ct log(cid:15) .
s(cid:15) (cid:15) ≥ − (cid:15) rel| |
In order to relate t ( ) and t ( ), we relate t (α) and hit ( ) by showing that
mix,µ H,µ H,µ α+(cid:15),µ
· · ·
for any (cid:15),α,p > 0 such that (cid:15)+α < 1 and (cid:15)+p < 1, if π(A) α is “worst in expectation”
(t (α) = E [T ]) then hit (p+(cid:15)) min t : H [T > t] p≥+C t (see Lemma 4.7).
H,µ µ A α,µ µ A (cid:15),α rel
≤ { ≤ }
2 Starr’s Maximal inequality
In this section we present the machinery that will be utilized in the proof of the main results.
The most important tool we shall utilize is Starr’s L2 maximal inequality (Theorem 2.3).
We start with a few basic definitions and facts. We denote Z := n N : n 0 and
+
R := t R : t 0 . { ∈ ≥ }
+
{ ∈ ≥ }
7
Definition 2.1. Let (Ω,P,π) be a finite reversible chain. For any f RΩ, let E [f] :=
π
(cid:80) π(x)f(x) and Var f := E [(f E f)2]. The inner-product , ∈and L2 norm are
x∈Ω π π − π (cid:104)· ·(cid:105)π
f,g := E [fg] and f := (cid:0)E [ f 2](cid:1)1/2.
π π 2 π
(cid:104) (cid:105) (cid:107) (cid:107) | |
We identify H with the operator H : L2(RΩ,π) L2(RΩ,π), defined by
t t
→
(cid:88)
H f(x) := H (x,y)f(y) = E [f(X )],
t t x t
y∈Ω
By reversibility H is self-adjoint (w.r.t. , ).
t π
(cid:104)· ·(cid:105)
The following lemma is standard and follows from elementary linear algebra using the
the spectral decomposition of a function f RΩ (see e.g. Lemma 20.5 in [6]).
∈
Lemma 2.2 (L2-contraction Lemma). Let (Ω,P,π) be a finite reversible irreducible Markov
chain. Let f RΩ. Then
∈
Var H f e−2t/trelVar f, for any t 0, and (2.1)
π t π
≤ ≥
We now state a particular case of Starr’s maximal inequality ([10] Proposition 3). The
proof in the discrete time setup could be found in [3]. Let f RΩ. Define its maximal
∈
function by f∗(x) := sup H f(x) .
t≥0 t
| |
Theorem 2.3 (Maximal inequality [10]). Let (Ω,P,π) be a finite reversible irreducible
Markov chain. Then for any f RΩ
∈
f∗ 2 f . (2.2)
2 2
(cid:107) (cid:107) ≤ (cid:107) (cid:107)
(cid:112)
For any B Ω and s R , set ρ(B) := π(B)(1 π(B)) = √Var 1 , σ :=
+ π B s
⊂ ∈ −
ρ(B)e−s/trel. Note that by Lemma 2.2, σ √Var H 1 . We define the good set for B
s π s B
≥
from time s within m standard-deviations to be
(cid:8) (cid:9)
G (B,m) := y : Ht(B) π(B) < mσ , for all t s . (2.3)
s | y − | s ≥
The motivation behind the definitions in (2.3) was previously explained in 1.1. The fol-
§
lowing corollary follows by combining Lemma 2.2 with Theorem 2.3.
Corollary 2.4. Let (Ω,P,π) be a finite reversible irreducible chain. Then
4
π(G (B,m)) 1 , for all B Ω, s 0 and m > 0. (2.4)
s ≥ − m2 ⊂ ≥
Proof. For any s 0, let f (x) := H (1 (x) π(B)) = Hs(B) π(B). Then by Lemma 2.2
≥ s s B − x −
and Theorem 2.3
f∗ 2 f 2e−s/trel 1 (x) π(B) = 2σ . (2.5)
(cid:107) s(cid:107)2 ≤ (cid:107) s(cid:107)2 ≤ (cid:107) B − (cid:107)2 s
For any t 0, H f (x) = f (x) = Ht+s(B) π(B). Then, in the notation of Theorem 2.3,
≥ t s t+s x −
f∗(x) := sup H f (x) = sup Ht(B) π(B) .
s | t s | | x − |
t≥0 t≥s
Hence D := x Ω : f∗(x) mσ is the complements of G (B,m). Thus by Markov
{ ∈ s ≥ s} s
inequality and (2.5)
1 π(G (B,m)) = π(D) = π(cid:8)(f∗)2 (mσ )2(cid:9) 4m−2.
− s s ≥ s ≤
8
3 Inequalities relating t ( ) and hit ( )
mix
· · ·
Our aim in this section is to obtain inequalities relating t ((cid:15)) and hit (δ) for suitable
mix β
values of β, (cid:15) and δ using Corollary 2.4. As was shown in [3], these two notions of mixing are
intimately connected to each other. In this section we refine the analysis from [3]. Corollary
3.3 below contains the more difficult half of Proposition 1.8. We end the section with a proof
of the Decomposition Theorem.
Lemma 3.1. Let (Ω,P,π) be a finite irreducible reversible chain. Let µ P(Ω), (cid:15),p (0,1),
∈ ∈
s 0 and B Ω. Denote ρ(B) := Var 1 = π(B)(1 π(B). Denote τ := hit (p) and
π B (cid:15),p 1−(cid:15),µ
≥ ⊂ (cid:112) −
(cid:96) := 2(1 p)e−s/trel 1/(cid:15). Then
s,p,(cid:15)
−
(cid:112) (cid:112)
π(B) Hτ(cid:15),p+s(B) pπ(B)+(cid:96) ρ(B) = pπ(B)+2(1 p)e−s/trel ρ(B)/(cid:15). (3.1)
− µ ≤ s,p,(cid:15) −
Proof. Consider the set
(cid:110) (cid:112) (cid:111)
H := y : Ht(B) π(B) < (cid:96) ρ(B) for all t s .
| y − | s,p,(cid:15) ≥
Then by Corollary 2.4, π(H) 1 (cid:15). By the Markov property and the definition of H,
≥ −
(cid:112)
H [X B T τ ] π(B) (cid:96) ρ(B).
µ τ(cid:15),p+s ∈ | H ≤ (cid:15),p ≥ − s,p,(cid:15)
By the definition of τ and the fact that π(H) 1 (cid:15), we get that H [T τ ] 1 p.
(cid:15),p µ H (cid:15),p
≥ − ≤ ≥ −
Thus
π(B) Hτ(cid:15),p+s(B) π(B) H [X B,T τ ]
− µ ≤ − µ τ(cid:15),p+s ∈ H ≤ (cid:15),p
(cid:0) (cid:1)
H [T > τ ]π(B)+H [T τ ] π(B) H [X B T τ ]
≤ µ H (cid:15),p µ H ≤ (cid:15),p − µ τ(cid:15),p+s ∈ | H ≤ (cid:15),p
(cid:112) (cid:112)
H [T > τ ]π(B)+H [T τ ](cid:96) ρ(B) pπ(B)+(cid:96) ρ(B).
µ H (cid:15),p µ H (cid:15),p s,p,(cid:15) s,p,(cid:15)
≤ ≤ ≤
This concludes the proof of (3.1).
Definition 3.2. Let 0 r,p 1. Denote D = D := x : Ht+s(x) (1 p r)π(x) .
≤ ≤ r t,s,p,r,µ { µ ≤ − − }
Corollary 3.3. Let (Ω,P,π) be a reversible irreducible finite chain. Let µ P(Ω), w R
∈ ∈
and p,(cid:15) (0,1).
∈
(1) Let
(cid:0) (cid:1)
s = s(w,(cid:15),p) := t [w+k ], where k := log 2(cid:15)−1/2(1 p) . (3.2)
rel (cid:15),p (cid:15),p
−
Then
t (p+2e−w) hit (p)+s, for all w k . (3.3)
mix,µ 1−(cid:15),µ (cid:15),p
≤ ≥ −
(2) For every 0 < (cid:15) 1/2,
≤
t ((cid:15)) hit (3(cid:15)/4)+t log(16/(cid:15)),
mix,µ 1/2,µ rel
≤
1 (3.4)
t (1 (cid:15)/2) hit (1 (cid:15))+ t log(64/(cid:15)) and
mix,µ 1−(cid:15),µ rel
− ≤ − 2
3
hit (5(cid:15)/4) t ((cid:15)) hit (3(cid:15)/4)+ t log(16/(cid:15)). (3.5)
1−(cid:15)/4,µ mix,µ 1−(cid:15)/4,µ rel
≤ ≤ 2
9
(3) (1)-(2) remain valid when µ is omitted.
Proof. The proof of (3) is parallel to that of (1)-(2) and thus omitted. We first prove (3.3).
Denote t := hit (p). First observe that from the definition of D := D we have that
1−(cid:15),µ r t,s,p,r,µ
(1 p r)π(D ) Ht+s[D ] 0.
− − r − µ r ≥
(cid:0) (cid:1)
Fix some w k = log 2(cid:15)−1/2(1 p) . Then s = s(w,(cid:15),p) = t [w+k ] 0. By the
(cid:15),p rel (cid:15),p
≥ − − − (cid:112) ≥
definition of k we get that 2e−k(cid:15),p(1 p) 1/(cid:15) = 1. By the definition of s we have that
(cid:15),p
(cid:112) −
(cid:96) := 2e−s/trel(1 p) 1/(cid:15) = e−w. By (3.1) we have that
s,p,(cid:15)
−
(cid:112)
0 (1 p r)π(D ) Ht+s[D ] rπ(D )+e−w π(D )(1 π(D )).
≤ − − r − µ r ≤ − r r − r
(cid:112)
Let c > 0. Consider the function h : [0,1] R defined as h(x) = cx+ x(1 x). Then
→ − −
f 0 iff x (1+c2)−1. Taking c = rew and x = π(D ) we get that
r
≥ ≤
1
π(D ) . (3.6)
r ≤ 1+r2e2w
(cid:16) (cid:17)
Define f : Ω R by f(x) := (1 p) Htµ+s(x) 1 . Then 0 f 1 p and
→ − − π(x) x∈D0 ≤ ≤ −
f r = D , for any 0 r 1 p. Note that
r
{ ≥ } ≤ ≤ −
(cid:90) 1−p (cid:90) 1−p dr
(1 p)π(D ) Ht+s[D ] = E f = π f r dr
− 0 − µ 0 π { ≥ } ≤ 1+r2e2w
r=0 r=0 (3.7)
(cid:90) (1−p)ew d(cid:96) (cid:90) (1−p)ew 2d(cid:96)
= e−w e−w 2e−w.
1+(cid:96)2 ≤ (1+(cid:96))2 ≤
r=0 r=0
Let B := x Ω : (1 p)π(x) < Ht+s(x) < π(x) . Then by definition we have
{ ∈ − µ }
that π(B) Ht+s(B) < pπ(B). Denote C = B D . Then by the definition of the total
− µ ∪ 0
variation distance (first equality) together with (3.7) (second inequality) and the fact that
π(B) Ht+s(B) < pπ(B) (second inequality) we get that
− µ
Ht+s π = π(C) Ht+s(C) = π(B) Ht+s(B)+pπ(D )+(1 p)π(D ) Ht+s[D ]
(cid:107) µ − (cid:107)TV − µ − µ 0 − 0 − µ 0
pπ(B)+pπ(D )+2e−w < pπ(C)+2e−w p+2e−w.
0
≤ ≤
This concludes the proof of (3.3).
The first inequality in (3.5) follows directly from the definition of the total variation
distance. To see this, let E Ω be an arbitrary set with π(E) 1 (cid:15)/4. Let t := t ((cid:15)).
1 mix,µ
⊂ ≥ −
Then H [T t ] H [X E] π(E) Ht1 π 1 5(cid:15)/4. In particular, we get
µ A ≤ 1 ≥ µ t1 ∈ ≥ −(cid:107) µ − (cid:107)TV ≥ −
directly from Definition 1.3 that hit (5(cid:15)/4) t = t ((cid:15)).
1−(cid:15)/4,µ 1 mix,µ
≤
The remaining inequalities are obtained by applying the previously established ones with
particular choices of (w,(cid:15),p). We now specify the required choice of parameters for each
inequality, leaving the necessary calculations to the reader. For the second inequality in
(3.5), use (3.3) with (w,(cid:15),p) being (log(8/(cid:15)),(cid:15)/4,3(cid:15)/4).
For the first inequality in (3.4), apply (3.3) with (w,(cid:15),p) = (log(8/(cid:15)),1/2,3(cid:15)/4). For the
second inequality in (3.4), apply (3.3) with (w,(cid:15),p) being (log(4/(cid:15)),(cid:15),1 (cid:15)).
−
10