352 Electronic Communications in Probability
2 The Mixer Chain on Z
We now consider the mixer chain on Z, with {1, −1} as the symmetric generating set. We denote
by ω
t
= S
t
, σ
t t
≥0
the mixer chain on Z. For
ω ∈ Z ⋉ Σ we denote by Dω the distance of ω from 0, id with respect to the generating
set Υ, see Definition 1.2. Denote by D
t
= Dω
t
the distance of the chain at time t from the origin.
As stated above, we show that the mixer chain on Z has degree of escape 3 4. In fact, we prove
slightly stronger bounds on the distance to the origin at time t.
Theorem 2.1. Let D
t
be the distance to the origin of the mixer chain on Z. Then, there exist constants c, C
0 such that for all t ≥ 0, ct
3 4
≤ E[D
t
] ≤ C t
3 4
. The proof of Theorem 2.1 is in Section 3.
For z ∈ Z, denote by X
t
z = |σ
t
z − z|, the distance of the tile marked z to its origin at time t. Define
X
t
= X
z ∈Z
X
t
z, which is a finite sum for any given t. As shown in Propositions 1.3 and 1.4, X
t
approximates D
t
up to certain factors. For z
∈ Z define V
t
z =
t
X
j=0
1
S
t
= σ
t
z .
V
t
z is the number of times that the mixer visits the tile marked z, up to time t.
2.1 Distribution of X
t
z
The following proposition states that the “mirror image” of the mixer chain has the same distribu- tion as the original chain. We omit a formal proof, as the proposition follows from the symmetry
of the walk.
Proposition 2.2. For σ ∈ Σ define σ
′
∈ Σ by σ
′
z = −σ−z for all z ∈ Z. Then, for any t ≥ 1, S
1
, σ
1
, . . . , S
t
, σ
t
and −S
1
, σ
′ 1
, . . . , −S
t
, σ
′ t
have the same distribution. By a lazy random walk on Z, we refer to the integer valued process W
t
, such that W
t+1
− W
t
are i.i.d. random variables with the distribution P[W
t+1
− W
t
= 1] = P[W
t+1
− W
t
= 1] = 14 and P
[W
t+1
− W
t
= 0] = 12.
Lemma 2.3. Let t ≥ 0 and z ∈ Z. Let k ≥ 1 be such that P[V
t
z = k] 0. Then, conditioned on V
t
z = k, the distribution of σ
t
z − z is the same as W
k −1
+ B, where W
k
is a lazy random walk on Z and B is a random variable independent of
W
k
such that |B| ≤ 2. Essentially the proof is as follows. We consider the successive time at which the mixer visits the
tile marked z. The movement of the tile at these times is a lazy random walk with the number of steps equal to the number of visits. The difference between the position of the tile at time t and its
position at the last visit is at most 1, and the difference between the tile at time 0 and its position at the first visit is at most 1. B is the random variable that measures these two differences.
Proof. Define inductively the following random times: T z = 0, and for j ≥ 1,
T
j
z = inf ¦
t ≥ T
j −1
z + 1 : S
t
= σ
t
z ©
.
Rate of Escape of the Mixer Chain 353
Claim 2.4. Let T = T
1
0. For all ℓ such that P[T = ℓ] 0, P
σ
T
0 = 1 T =
ℓ
= P
σ
T
0 = −1 T =
ℓ
= 14, and
P
σ
T
0 = 0 T =
ℓ
= 12, Proof. Note that
|S
1
−σ
1
0| = 1 and that for all 1 ≤ t T , σ
t
0 = σ
1
0. Thus, σ
T −1
0 = σ
1
and S
T −1
= S
1
. So we have the equality of events {T = ℓ} =
ℓ−1
\
t=1
S
t
6= σ
t
\ S
ℓ−1
= S
1
, σ
ℓ−1
0 = σ
1
\ S
ℓ
= σ
1
0 or σ
ℓ
0 = S
1
. Hence, if we denote
E = T
ℓ−1 t=1
S
t
6= σ
t
T S
ℓ−1
= S
1
, σ
ℓ−1
0 = σ
1
, then P
[T = ℓ] = P[E ] · P
S
ℓ
= σ
1
0 or σ
ℓ
0 = S
1
E
= P[E ] · 1
2 .
2.1 Since the events
S
1
= 0 and σ
1
0 = 0 are disjoint and their union is the whole space, we get
that P
[σ
T
0 = 0, T = ℓ] = P[E , S
ℓ
= σ
1
0 = 0] + P[E , σ
ℓ
0 = S
1
= 0] = P
E, σ
1
0 = 0 · P
S
ℓ
= σ
1
S
ℓ−1
= S
1
, σ
ℓ−1
0 = σ
1
0 = 0
+ P E, S
1
0 = 0 · P
σ
ℓ
0 = S
1
S
ℓ−1
= S
1
= 0, σ
ℓ−1
0 = σ
1
= P[E ] ·
1 4
. 2.2
Combining 2.1 and 2.2 we get that P
σ
T
0 = 0 T =
ℓ
= 1
2 .
Finally, by Proposition 2.2, P
σ
T
0 = 1, T = ℓ = P E , S
ℓ
= σ
ℓ
0 = 1 = P
σ
T
0 = −1, T = ℓ .
Since the possible values for σ
T
0 are −1, 0, 1, the claim follows. ⊓
⊔ We continue with the proof of Lemma 2.3.
We have the equality of events V
t
z = k = T
k
z ≤ t T
k+1
z .
Let t
1
, t
2
, . . . , t
k
, t
k+1
be such that P
[T
1
z = t
1
, . . . , T
k+1
z = t
k+1
] 0, and condition on the event
E = T
1
z = t
1
, . . . , T
k+1
z = t
k+1
. Assume further that t
k
≤ t t
k+1
, so that V
t
z = k. Write σ
t
z − z = σ
t
z − σ
T
k
z
z +
k
X
j=2
σ
T
j
z
z − σ
T
j −1
z
z + σ
T
1
z
z − z. 2.3
354 Electronic Communications in Probability
For 1 ≤ j ≤ k − 1 denote Y
j
= σ
T
j+1
z
z − σ
T
j
z
z. By Claim 2.4 and the Markov property, conditioned on
E , ¦
Y
j
© are independent with the distribution P[Y
j
= 1|E ] = P[Y
j
= −1|E ] = 14 and P[Y
j
= 0|E ] = 12. So conditioned on E , P
k −1
j=1
Y
j
has the same distribution of W
k −1
. Finally,
|σ
t
z − σ
T
k
z
z| ≤ 1, and |σ
T
1
z
z − z| ≤ 1. Since conditioned on E , σ
t
z − σ
T
k
z
z, and
σ
T
1
z
z − z are independent of ¦
Y
j
© , this completes the proof of the lemma.
⊓ ⊔
Corollary 2.5. There exist constants c, C
0 such that for all t ≥ 0 and all z ∈ Z, c E[
p V
t
z] − 2 P[V
t
z ≥ 1] ≤ E[X
t
z] ≤ C E[ p
V
t
z] + 2 P[V
t
z ≥ 1]. Proof. Let
W
t
be a lazy random walk on Z. Note that 2W
t
has the same distribution as ¦
S
′ 2t
© where
¦ S
′ t
© is a simple random walk on Z. It is well known see e.g. [5], that there exist universal
constants c
1
, C
1
0 such that for all t ≥ 0, c
1
p t
≤ E[|S
′ 2t
|] = 2 E[|W
t
|] ≤ C
1
p t.
By Lemma 2.3, we know that for any k ≥ 0,
E [|W
k
|] − 2 ≤ E
X
t
z V
t
z = k + 1
≤ E[|W
k
|] + 2. Thus, summing over all k, there exists constants c
2
, C
2
0 such that c
2
E [
p V
t
z] − 2 P[V
t
z ≥ 1] ≤ E[X
t
z] ≤ C
2
E [
p V
t
z] + 2 P[V
t
z ≥ 1]. ⊓
⊔
Lemma 2.6. Let
¦ S
′ t
© be a simple random walk on Z started at S
′
= 0, and let L
t
z =
t
X
j=0
1
n S
′ j
= z o
. Then, for any z
∈ Z, and any k ∈ N, P
[L
2t
2z ≥ k] ≤ P[V
t
z ≥ k]. Specifically, E[
p L
2t
2z] ≤ E[ p
V
t
z]. Proof. Fix z
∈ Z. For t ≥ 0 define M
t
= S
t
− σ
t
z + z. Note that V
t
z =
t
X
j=0
1
¦ M
j
= z ©
, so V
t
z is the number of times M
t
visits z up to time t. M
t
is a Markov chain on Z with the following step distribution.
P
M
t+1
= M
t
+ ǫ M
t
=
1
2 M
t
= z, ǫ ∈ {−1, 1} ,
1 2
|M
t
− z| = 1, ǫ = −M
t
+ z, 1
4 |M
t
− z| = 1, ǫ = 0, 1
4 |M
t
− z| = 1, ǫ = M
t
− z, 1
4 |M
t
− z| 1, ǫ ∈ {−1, 1} , 1
2 |M
t
− z| 1, ǫ = 0.
Rate of Escape of the Mixer Chain 355
Specifically, M
t
is simple symmetric when at z, lazy symmetric when not adjacent to z, and has a drift towards z when adjacent to z.
Define N
t
to be the following Markov chain on Z: N = 0, and for all t ≥ 0,
P
N
t+1
= N
t
+ ǫ N
t
=
1
2 N
t
= z, ǫ ∈ {−1, 1} ,
1 2
N
t
6= z, ǫ = 0, 1
4 N
t
6= z, ǫ ∈ {−1, 1} . So
N
t
is simple symmetric at z, and lazy symmetric when not at z. Let V
′ t
z =
t
X
j=0
1
¦ N
j
= z ©
, be the number of times
N
t
visits z up to time t. Define inductively
ρ = ρ
′
= 0 and for j ≥ 0, ρ
j+1
= min n
t ≥ 1 : M
ρ
j
+t
= z o
, ρ
′ j+1
= min n
t ≥ 1 : N
ρ
′ j
+t
= z o
. If N
t
≥ M
t
z then P
M
t+1
= M
t
+ 1 M
t
= P
N
t+1
= N
t
+ 1 N
t
,
and P
M
t+1
= M
t
− 1 M
t
≥ P
N
t+1
= N
t
− 1 N
t
.
Thus, we can couple M
t+1
and N
t+1
so that M
t+1
≤ N
t+1
. Similarly, if N
t
≤ M
t
z then M
t+1
moves towards z with higher probability than N
t+1
, and they both move away from z with proba- bility 1
4. So we can couple M
t+1
and N
t+1
so that M
t+1
≥ N
t+1
. If N
t
= M
t
= z then M
t+1
and N
t+1
have the same distribution, so they can be coupled so that N
t+1
= M
t+1
. Thus, we can couple
M
t
and N
t
so that for all j ≥ 0, ρ
j
≤ ρ
′ j
a.s. Let
¦ S
′ t
© be a simple random walk on Z. For x
∈ Z, let τ
x
= min ¦
2t ≥ 2 : S
′ 2t
= 2z , S
′
= 2x ©
. That is,
τ
x
is the first time a simple random walk started at 2x hits 2z this is necessarily an even number. In [5, Chapter 9] it is shown that
τ
x
has the same distribution as τ
2z
− 2|z − x|. Note that if N
t
6= z then S
′ 2t+2
− S
′ 2t
has the same distribution as 2N
t+1
− N
t
. Since |N
ρ
′ j
−1
+1
− z| = 1, we get that for all j
≥ 2, ρ
′ j
has the same distribution as
1 2
τ
2z
− 2 + 1. Also, ρ
′ 1
has the same distribution as
1 2
τ if z
6= 0, and the same distribution as
1 2
τ
2z
− 2 + 1 if z = 0. Hence, we conclude that for any k
≥ 1, P
k j=1
ρ
′ j
has the same distribution as
1 2
P
k j=1
˜ ρ
j
, where ¦
˜ ρ
j
©
j ≥1
are defined by
˜ ρ
j+1
= min n
2t ≥ 2 : S
′ ˜
ρ
j
+2t
= 2z o
. Finally note that V
t
z ≥ k if and only if P
k j=1
ρ
j
≤ t, V
′ t
z ≥ k if and only if P
k j=1
ρ
′ j
≤ t, and L
t
2z ≥ k if and only if P
k j=1
˜ ρ
j
≤ t. Thus, under the above coupling, for all t ≥ 0, V
t
z ≥ V
′ t
z a.s. Also, V
′ t
z has the same distribution as L
2t
2z. The lemma follows. ⊓
⊔
356 Electronic Communications in Probability
2.2 The Expectation of X
