Random walks
EPFL - Spring Semester 2014-2015
Solutions 10
1. a) We have
P(X 6= Y ) = 1 − P(X = Y ) = 1 −
X
ξi = 1 −
i∈S
X
=
(µi − νi ) =
i∈S : µi >νi
X
X
min(µi , νi )
i∈S
(νi − µi )
i∈S : νi >µi
where the last two equalities hold because both µ and ν are distributions. Summing these last two
equalities, we obtain
X
2 P(X 6= Y ) =
|µi − νi | = 2 kµ − νkTV
i∈S
b) Observe first that if
obtain for i ∈ S:
P
i∈S ξi
= 1, then X = Y with probability one. When
P
i∈S ξi
< 1, we
X (µi − ξi ) (νj − ξj )
P
1 − k∈S ξk
j∈S\i
j∈S\i


X
X
µi − ξi
µi − ξi
1 − νi −
P
P
= ξi +
(νj − ξj ) = ξi +
ξj + ξi 
1 − k∈S ξk
1 − k∈S ξk
P(X = i) = P(X = Y = i) +
X
P(X = i, Y = j) = ξi +
j∈S
j∈S\i
= ξi + (µi − ξi ) −
(µi − ξi ) (νi − ξi )
P
= µi
1 − k∈S ξk
as (µi − ξi ) (νi − ξi ) = 0 necessarily. A similar reasoning shows that P(Y = j) = νj for all j ∈ S.
c) Fix i ∈ S, let X0 = i and Y0 ∼ π the stationary distribution, and fix also a time n. By parts a)
and b), we can find a coupling of Xn and Yn such that di (n) = kPin − πkTV = P(Xn 6= Yn ). We
can now define a new coupling for Xn+1 and Yn+1 in the following way:
• If Xn = Yn , then Xn+1 = Yn+1 ;
• Else, let X and Y evolve independently according to P .
Then
di (n + 1) = kPin+1 − πkTV ≤ P(Xn+1 6= Yn+1 ) ≤ P(Xn 6= Yn ) = di (n)
The first inequality holds by the coupling lemma, and the second inequality is by construction.
Observe finally that d(n) = maxi∈S di (n) is also non-increasing in n (being the maximum of nonincreasing functions).
1
2. Because of the assumptions made, aij > 0 if ψij > 0, so the chain with transition probabilities
pij is also irreducible and aperiodic, therefore ergodic, as the state space S is finite. Let us check
the detailed balance equation:
πi pij = πi ψij aij =
πi ψij πj ψji
πj ψji + πi ψij
which is clearly symmetric in i and j, and therefore equal to πj pji .
3. First note that Z =
1−θN
1−θ
'
1
1−θ
for large N .
a) The weights defined in class are given in this case by wi =
πi
ψi
=
(
1
wj
= min 1, θj−i =
aij = min 1,
wi
θj−i
which leads to
pij =

1

N
1
N

1
N
b) From the course, we know that
where
λ∗ = 1 −
θj−i
+
1
N
PN
k=i+1
1 − θk−i
N
Z
θi−1 , so that for j 6= i:
if j < i
if j > i
if j < i
if j > i
if j = i
λn
kPin − πkTV ≤ √∗
2 πi
1
w∗
and w∗ = max wi = w1 =
i∈S
N
Z
We conclude therefore that
kPin − πkTV
√
Z
Z n
≤ √
1−
N
2 θi−1
For i = 1 and large N , this bound leads to:
1
n
kP1n − πkTV ≤ √
exp −
N (1 − θ)
2 1−θ
while for i = N and large N , this bound leads to:
n
1
1
N −1
n
n
exp −
= √
exp
log(1/θ) −
kPN − πkTV ≤ p
N (1 − θ)
2
N (1 − θ)
2 1−θ
2 (1 − θ) θN −1
c) Because of the last estimate, in order for maxi∈S kPin − πkTV to be smaller than ε, we need that
n N 2 , which gives the desired upper bound on the mixing time. What can actually be shown in
this case (but this was not asked) is the following: using the more precise estimate
v
uN −1
X
1u
(k) 2
n
kPi − πkTV ≤ t
λ2n
φ
i
k
2
k=1
we find that this quantity is small (uniformly in i) for n N already.
2
4. a) There is no stationary distribution. By computing the stationary distribution, one would
find an infinite normalization constant. The walk is irreducible but since there is no stationary
distribution it cannot be positive-recurrent.
b) The acceptance probability for the move i → i + 1 is
(
1, i ≤ −1,
−a((i+1)2 −i2 )
ai,i+1 = min(1, e
)=
e−a(2i+1) , i ≥ 0.
and for the move i → i − 1 is
−a((i−1)2 −i2 )
ai,i−1 = min(1, e
(
1, i ≥ 1,
)=
ea(2i−1) , i ≤ 0.
In words: a move towards the origin i = 0 is always accepted and a move away from the origin is
accepted with probability e−a(1±2i) .
The transition probabilities are
pi,i+1
1
2
2
= min(1, e−a((i+1) −i ) ) =
2
(
1
2 , i ≤ −1,
1 −a(2i+1)
,
2e
i ≥ 0.
and
pi,i−1
1
2
2
= min(1, e−a((i−1) −i ) ) =
2
(
1
2 , i ≥ 1,
1 a(2i−1)
,
2e
i ≤ 0.
and for the self-loops
pii = 1 − pi,i−1 − pi,i+1

1 a(2i−1)

− 12 = 12 (1 − ea(2i−1) ), i ≤ −1,
1 − 2 e
= 1 − 12 e−a − 12 e−a = 1 − e−a , i = 0,


1 − 21 − 12 e−a(2i+1) = 12 (1 − e−a(2i+1) ), i ≥ 1.
c) The walk is obviously irreducible. A stationary distribution πi∗ exists by construction. Thus by
the first fundamental theorem in class the walk is positive recurrent. Moreover because of the self
loops it is aperiodic. An irreducible, positive recurrent, aperiodic walk satisfies the ergodic theorem
and therefore
lim (P n )ij = πj∗
n→+∞
where P is the transition matrix elements (P n )ij = P(Xn = j | X0 = i).
d) We start on the right of the origin because z > 0. The initial distance between the two walkers is
d. When we propose a move towards the left ξ1 = −1 both walkers accept the move with probability
1 so the distance stays equal to d. When we propose a move towards the right ξ1 = +1 the walkers
may accept-accept the move, accept-reject the move, reject-accept the move, or reject-reject the
move. In the first case and in the last case the distance stays equal to d, in the second case it
decreases to d − 1 and the third case it increases to d + 1. Reasoning like this, we see that the
minimum coalescence time corresponds to the events where we always propose a right move and
the walkers accept-reject.
3
This minimum coalescence time is thus equal to d. The condition z > d ensures that the origin is
cannot be reached in d steps. We have
1
P(T = d) = d
2
z+d−1
Y
e
−a(2i+1)
i=z
4
(1 − e−a(2(z+d)+1) )d