This section is dedicated to the proof of Theorems 5.1 and 5.2 and we need the following lemmas.
Lemma 7.1. Let(X , K, π)be an irreducible birth and death chain on{0, 1, ..., n}and τi,eτibe the first hitting times to state iof the discrete time chain and the associated continuous time chain. Letλi be the smallest eigenvalue of the submatrix of I − K indexed by0, ..., i − 1. Then, fori < j,
π([0, i])
2π([0, j − 1])(Eiτej)2≤Vari(τej) ≤ 2 λjEiτej
and
δπ([0, i])
2π([0, j − 1])(Eiτj)2≤Vari(τj) ≤ 2 λjEiτj, whereδ = miniK(i, i). In particular,
Eiτj= Eieτj ≤4π([0, j − 1]) π([0, i])λj .
Lemma 7.2. LetK be the transition matrix of an irreducible birth and death chain on{0, 1, ..., n} andeτi be the first hitting time to state ifor the continuous time chain associated withK. For0 < i ≤ nanda ∈ (0, 1),
P0(eτi> aE0eτi) ≥ min
e−
√a, (1 − a)2
√a + (1 − a)2
.
Lemma 7.3. LetKbe the transition matrix of an irreducible birth and death chain on X = {0, 1, ..., n}with transition ratespi, qi, riand stationary distributionπ. Letτi,τeibe as in Lemma 7.1. Then, fori < j < k,
Ejmin{τi, τk} = Ejmin{τei,τek} = A/B, where
A = X
i+1≤`1≤j j≤`2≤k−1
π([`1, `2]) π(`1)q`1π(`2)p`2
, B =
k−1
X
`=i
1 π(`)p`
.
Lemma 7.4. Let(X , K, π)be an irreducible birth and death chain on{0, 1, ..., n}and Ht= e−t(I−K). Then,
(1) Ht(0, i)/π(i) ≥ Ht(0, i + 1)/π(i + 1)for0 ≤ i < nandt ≥ 0,
(2) Assume that miniK(i, i) ≥ 1/2. Then, Km(0, i)/π(i) ≥ Km(0, i + 1)/π(i + 1) for 0 ≤ i < nandm ≥ 0.
We relegate the proofs of Lemmas 7.1, 7.2 and 7.3 to the appendix and refer the reader to Lemma 4.1 in [14] for a proof of Lemma 7.4.
Proof of Theorem 5.1. We first prove the equivalence for cutoffs. Note thatπn(0) → 0is necessary for the total variation cutoff since
lim inf
n→∞ d(c)n,TV(0, t) ≤ lim inf
n→∞ d(c)n,TV(0, 0) = 1 − lim sup
n→∞
πn(0).
Under the assumption thatπn(0) → 0, it is easy to see that, for anya ∈ (0, 1),Mn(a) ≥ 1 ifnis large enough. Fora ∈ (0, 1)andn ≥ 1such thatMn(a) ≥ 1, we let
λn,1(a) < · · · < λn,Mn(a)(a)
be the eigenvalues of the submatrix ofI − Kn indexed by0, 1, ..., Mn(a) − 1. Clearly, λn(a) = λn,1(a)and, by Lemma 2.1,
un(a) =
Mn(a)
X
i=1
1
λn,i(a), vn2(a) =
Mn(a)
X
i=1
1 λ2n,i(a). As in the proof of (2.4), we have
pun(a)λn(a) ≤ un(a)
vn(a) ≤ un(a)λn(a).
This implies the equivalence of (2) and (3).
To prove the remaining equivalences, we letd(c)n,TVbe the total variation distance of thenth chains. By Lemma 3.1, one has
d(c)n,TV(0, t)
(≤ P0(eτi(n)> t) + πn([i + 1, n]),
≥ P0(eτi(n)> t) − πn([0, i − 1]). (7.1) As a result of the one-sided Chebyshev inequality, this implies
Tn,(c)TV(0, )
≤ E0eτi(n)+ q
(1−δδ )Var0(eτi(n)) for = δ + πn([i + 1, n]),
≥ E0eτi(n)− q
(1−δδ )Var0(eτi(n)) for = δ − πn([0, i − 1]),
(7.2)
whereδ ∈ (0, 1).
Now, we prove (2)⇒(1) and assume that (2) holds. By the last inequality of Lemma 7.1, we have, for0 < δ < < 1,
0 ≤ un() − un(δ) ≤ 4
δλn()≤ 4vn()
δ = o(un()). (7.3)
Fix ∈ (0, 1) and let0 < 1 < < 2 < 1. By (7.2), the replacement of i = Mn(2), δ = 2− in the first inequality and the replacement ofi = Mn(1),δ = 1 − + 1in the second inequality yield
Tn,(c)TV(0, 1 − ) ≤ un(2) +q (1
2− − 1)vn(2) = (1 + o(1))un(2), Tn,(c)TV(0, 1 − ) ≥ un(1) −q
(−1
1 − 1)vn(1) = (1 + o(1))un(1).
As a result of (7.3), we obtain thatTn,(c)TV(0, ) = (1 + o(1))un(η)for any, η ∈ (0, 1), which proves (1).
Next, we prove (4)⇒(3). Assume that (tn)∞n=0 is a positive sequence satisfying tn = O(un(c))for allc ∈ (0, 1)anda ∈ (0, 1)is a constant such that
n→∞lim P0 eτM(n)
n(b)> (1 − )tn
= 1, ∀b ∈ (a, 1), (7.4)
and, for anyb ∈ (a, 1), there corresponds a constantαb∈ (0, 1)such that lim sup
n→∞ P0
τeM(n)
n(b)> (1 + )tn
≤ αb, (7.5)
for all ∈ (0, 1). Note thatλn(a2) ≤ λn(a1)for0 < a1< a2< 1. To prove (3), it suffices to show thattnλn(b) → ∞for all b ∈ (a, 1). Now, we fix b ∈ (a, 1). Sinceπn(0) → 0, it is clear that Mn(b) ≥ 1 for n large enough. By [5], if Mn(b) ≥ 1, we may write eτM(n)
n(b)= Tn(b) + Sn(b), whereTn(b)andSn(b)are independent,Tn(b)is an exponential random variable with parameterλn(b)andSn(b)is a sum of independent exponential random variables with parametersλn,2(b), ..., λn,Mn(b)(b). Note that
P0 eτM(n)
n(b)> (1 − )tn
= Z ∞
0
λn(b)e−λn(b)sP0(Sn(b) > (1 − )tn− s)ds
≤ (1 − e−λn(b)t)P0(Sn(b) > (1 − )tn− t) + e−λn(b)t, where the inequality is obtained by separating the region of integration into(0, t)and [t, ∞), and
P0
eτM(n)
n(b)> (1 + )tn
= Z ∞
0
λn(b)e−λn(b)sP0(Sn(b) > (1 + )tn− s)ds
≥ P0(Sn(b) > (1 + )tn− r)e−λn(b)r.
By (7.4) and (7.5), the replacement oft = C/λn(b)andr = 2C/λn(b)withC =14logα1
b in the above inequalities yields that, for all ∈ (0, 1),
n→∞lim P0(Sn(b) > (1 − )tn− C/λn(b)) = 1 and
lim sup
n→∞ P0(Sn(b) > (1 + )tn− 2C/λn(b)) ≤√ αb< 1.
As a consequence, for ∈ (0, 1), ifnis large enough, one has (1 + )tn− 2C/λn(b) ≥ (1 − )tn− C/λn(b), which impliestnλn(b) ≥ C/(2). This provestnλn(b) → ∞.
To finish the proof of those equivalences, it remains to show (1)⇒(4). Assume thatFc
has a cutoff with cutoff timetn. The replacement ofi = Mn(a)in (7.1) implies that, for all ∈ (0, 1),
lim inf
n→∞ P0 τeM(n)
n(a)> (1 − )tn
≥ a (7.6)
and
lim sup
n→∞ P0 eτM(n)
n(a)> (1 + )tn
≤ a. (7.7)
By the Markov inequality, (7.6) implies thattn= O(un(a))for alla ∈ (0, 1). As a result of Lemma 7.2, (7.7) implies thatun(a) = O(tn)for alla ∈ (0, 1), which leads totn un(a) for alla ∈ (0, 1).
To fulfill the requirement in (4), one has to prove that there isa ∈ (0, 1)such that
n→∞lim P0 eτM(n)
n(a)> (1 − )tn
= 1, ∀ ∈ (0, 1). (7.8)
To see the above limit, we fix ∈ (0, 1)and show that, for any subsequence of positive integers, there is a further subsequence satisfying (7.8). Letkn be a subsequence of positive integers and set
R(a) := lim
b→1lim inf
n→∞
EMkn(a)eτM(kn)
kn(b)
tkn .
Clearly,R(a)is nonnegative and non-increasing ina.
We consider the following two cases ofR(a). First, assume thatR(a) = 0for some a ∈ (0, 1)and letbn be a sequence in(a, 1)that converges to1. SinceR(b1) = 0, we may choose`1∈ {k1, k2, ...}such thatEM`1(a)eτM(`1)
`1(b1)< t`1/2. Inductively, forn ≥ 1, we may select, according to the factR(bn+1) = 0, a constant`n+1 ∈ {k1, k2, ...}satisfying
`n+1> `n and
EM`n+1(a)τeM(`n+1)
`n+1(bn+1)< t`n+1/2n+1. This implies
EM`n(a)τeM(`n)
`n(b)= o(t`n), ∀b ∈ (a, 1).
By Lemma 2.1, un(a) tn implies 1/λn(a) = O(tn) and, by Lemma 7.1, this yields VarM`n(a)eτM(`n)
`n(b)= o(t2`
n)for allb ∈ (a, 1). As a consequence of the one-sided Chebyshev inequality, we obtain
n→∞lim PM`n(a)
eτM(`n)
`n(b)≤ ηt`n
= 1, ∀b ∈ (a, 1), η > 0.
This leads to
lim inf
n→∞ P0 eτM(`n)
`n(a)> (1 − )t`n
≥ lim inf
n→∞ P0
eτM(`n)
`n(b)> (1 − /2)t`n,eτM(`n)
`n(b)−eτM(`n)
`n(a)≤ t`n/2
= lim inf
n→∞ P0 eτM(`n)
`n(b)> (1 − /2)t`n
≥ b,
for allb ∈ (a, 1), where the last inequality uses (7.6). Lettingbtend to1gives the desired limit.
Next, we assume that R(a) > 0for all a ∈ (0, 1). Along with this fact un(a) tn for alla ∈ (0, 1), it is easy to see that, for any a ∈ (0, 1), there isb ∈ (a, 1)such that EMkn(a)eτM(kn)
kn(b) tkn. To prove (7.8) for the subsequencekn, we need the following discussion. Forn ≥ 1, set Hn,t = e−t(I−Kn) and let(Xn,t)t≥0 be a realization of the semigroupHn,tand, forη ∈ (0, 1), let
Nn(η) = max{0 ≤ i ≤ n|Hn,(1−η)tn(0, i) > πn(i)}.
By Lemma 7.4, we have
d(c)n,TV(0, (1 − η)tn) = Hn,(1−η)tn(0, [0, Nn(η)]) − πn([0, Nn(η)]).
SinceFchas a cutoff with cutoff time(tn)∞n=1, this implies
n→∞lim Hn,(1−η)tn(0, [0, Nn(η)]) = 1, lim
n→∞πn([0, Nn(η)]) = 0.
Obviously, this yields
n→∞lim P0 Xn,(1−η)tn≤ Mn(a) = 1, ∀a, η ∈ (0, 1). (7.9) Back to the case thatR(a) > 0for alla ∈ (0, 1), one may choose0 < b < a− < a <
a+< c < 1such that
EMkn(b)eτM(kn)
kn(a−) tkn EMkn(a+)eτM(kn)
kn(c). (7.10)
This implies thatMkn(b) < Mkn(a−)andMkn(a+) < Mkn(c)fornlarge enough. Next, let Lbe a positive integer and set
∆n= ∆n(L) :=(1 − )tn
L .
Note that, for0 ≤ j ≤ L − 1,
By (7.9), summing up the above inequalities overj and then passingnto the infinity yields
Observe that if there isL > 0such that lim sup
as desired. To get the limit in (7.11), it suffices to show that there isL > 0such that lim sup
It is easy to see from the first identity in Lemma A.1 that An≥ (a+− a−)EMn(b)eτM(n)
where the first inequality holds fornlarge enough. SinceTn≤eτM(n)
n(c), one also has As a result of the one-sided Chebyshev inequality, this implies
lim sup
cutoff with cutoff time(un(a))∞n=1for anya ∈ (0, 1). In the assumption of (4), we lettnbe a sequence and0 < a1< a2< 1be constants such that
n→∞lim P0
eτM(n)
n(a1)> (1 − )tn
= 1, ∀ ∈ (0, 1), (7.12)
and
lim sup
n→∞ P0 eτM(n)
n(a2)> (1 + )tn
< 1, ∀ > 0. (7.13) To show thattn is a cutoff time forFcL, it suffices to prove, based on the equivalence of assumptions (2) and (4), thattn ∼ un(a)for somea ∈ (0, 1). First, by the Markov inequality, (7.12) implies
lim inf
n→∞
un(a1) tn
≥ 1.
On the other hand, by the Chebyshev inequality, one has
lim sup
n→∞ P0 eτM(n)
n(a2)− un(a2)
> un(a2)
≤ lim sup
n→∞
vn(a2)2
2un(a2)2 = 0, ∀ > 0, where the last equality uses assumption (2). Clearly, this implies that
n→∞lim P0 eτM(n)
n(a2)> (1 − )un(a2)
= 1, ∀ ∈ (0, 1).
A comparison of the tail probabilities in (7.13) and in the above limit says that, for
∈ (0, 1),
(1 − )un(a2) ≤ (1 + )tn, fornlarge enough. This yields
lim sup
n→∞
un(a2) tn
≤ 1.
As a result of the factun(a1) ∼ un(a2), we obtainun(a1) ∼ tn, as desired.
Proof of Theorem 5.2. Set δ = inf
i,nKn(i, i), Kn(δ) = (Kn− δI)/(1 − δ), Hn,t(δ)= et(K(δ)n −I).
It is easy to see that(Xn, Kn, πn)and(Xn, Hn,t(δ), πn)are respectively theδ-lazy walk and the continuous time chain associated with(Xn, Kn(δ), πn). Letdn,TV, d(c,δ)n,TV andTn,TV, Tn,(c,δ)TV andτi(n), τi(n,δ)be respectively the total variation distances, the total variation mixing times and the first hitting times to stateiof chains(Xn, Kn, πn)and(Xn, Hn,t(δ), πn). As a result of the following observation
Hn,t(δ)= et(Kn(δ)−I)= et(Kn−I)/(1−δ), (7.14) it is easy to see that the ratio of the spectral gaps of(Xn, Kn, πn)and(Xn, Hn,t(δ), πn)is constant innand, further,
E0eτM(n,δ)
n(a)= (1 − δ)un(a), Var0τeM(n,δ)
n(a) wn(a), (7.15) where the latter also uses Remark 5.5. This is consistent with (3.3).
SetFc(δ)= (Xn, Hn,t(δ), πn)∞n=1and letFc(δ,L)denote the family of chains inFc(δ)started at the left boundary points. The remaining proof for the equivalence of (1), (2) and (3) is very similar to the proof of the discrete time case in Theorem 1.4 if (3.1) and (3.2) hold under the replacement ofF , Fc(δ)byFL, Fc(δ,L). These two equivalences are given
by Theorem 3.4 in [8] but the prerequisite of this theorem asks the existence of some
∈ (0, 1)such thatTn,TV(0, ) → ∞andTn,(c,δ)TV (0, ) → ∞. (The authors of [8] point out the observation that such a requirement is missed in their article.) First, consider the requirementTn,(c,δ)TV (0, ) → ∞. Recall the second inequality in Lemma 3.1 in the following
d(c,δ)n,TV(0, t) ≥ P0 eτM(n,δ)
n(a)> t
− a.
By Lemma 7.2, (7.15) and the fact Var0eτM(n,δ)
n(a)≤ (E0τeM(n,δ)
n(a))2, the above inequality implies d(c,δ)n,TV(0, α(1 − δ)un(a)) ≥ min
e−
√α, (1 − α)2
√α + (1 − α)2
− a, ∀α ∈ (0, 1).
This yields that
lim inf
n→∞
Tn,(c,δ)TV (0, )
un(a) > 0 forsmall enough. (7.16) Sinceun(a) → ∞, we haveTn,(c,δ)TV (0, ) → ∞forsmall enough.
Next, we prove Tn,TV(0, ) → ∞. Note that one may use (7.14) and the triangle inequality to derive
d(c,δ)n,TV(0, t) ≤ P(Nt≤ m) + P(Nt> m)dn,TV(0, m), (7.17) where(Nt)t≥0is a Poisson process with parameter1/(1 − δ). A simple application of the weak law of large numbers says thatNt/tconverges to1/(1 − δ)in probability asttends to infinity. By (7.16) and the assumptionun(a) → ∞, the replacement oft = βun(a)and m = dβun(a)ein (7.17) with smallβimplies that
lim inf
n→∞
Tn,TV(0, )
un(a) > 0 forsmall enough. (7.18) This yields thatTn,TV(0, ) → ∞forsmall enough.
To show (1)⇔(4), let(Nt)t≥0be the Poisson process as before. It is easy to see from (7.14) that if(Xm(n))∞m=0is a realization of(Xn, Kn, πn), then(XN(n)
t )t≥0is a realization of (Xn, Hn,t(δ), πn). This implies
P0
τei(n,δ)> s
= P0
XN(n)
r < i, ∀0 ≤ r ≤ s
= P0(Xm(n)< i, ∀m ≤ Ns) = P0
τi(n)> Ns .
(7.19)
Sinceun(a) → ∞for somea ∈ (0, 1), we obtain
FLhas a cutoff ⇔ Fc(δ,L)has a cutoff.
By Theorem 5.1, the latter is equivalent to the existence of a sequencetn > 0and a constanta ∈ (0, 1)satisfying
tn = O
E0τeM(n,δ)
n(c)
, ∀c ∈ (0, 1) (7.20)
and
n→∞lim P0 eτM(n,δ)
n(a)> (1 − )tn
= 1, ∀ ∈ (0, 1) (7.21)
and, for anyb ∈ (a, 1), there isαb∈ (0, 1)such that lim sup
n→∞ P0 eτM(n,δ)
n(b)> (1 + )tn
≤ αb, ∈ (0, 1). (7.22)
As a result of (7.15), one can see that (7.20) is equivalent totn = O(un(a))and further, by (7.19), (7.21) implies
lim inf
n→∞ P0
τM(n)
n(a)>(1 − )tn
1 − δ
≥ lim inf
n→∞ P0 τM(n)
n(a)> N(1−/2)tn
= lim inf
n→∞ P0
eτM(n,δ)
n(a)> (1 − /2)tn
= 1.
and (7.22) implies lim sup
n→∞ P0
τM(n)
n(b)>(1 + )tn
1 − δ
≤ lim sup
n→∞ P0 τM(n)
n(b)> N(1+/2)tn
= lim sup
n→∞ P0 τeM(n,δ)
n(b)> (1 + /2)tn
≤ αb,
for all ∈ (0, 1). This gives the desired properties in (4). Conversely, one may use a similar statement to prove (7.21) and (7.22) based on the observation of (4) and this part is omitted.
For a choice of the cutoff time, if (2) or (3) holds, the proof for the selected cutoff time is given by (7.15) and Theorem 3.4 in [8]. If (4) holds, the proof is exactly the same as that of Theorem 5.1 and we skip it here.
Proof of Corollary 5.4. The(un(a), bn)cutoff ofFcLis immediately from (7.2) and Lemma 7.1. For the(un(a), bn)cutoff ofFL, the assumptioninfnbn > 0andbn= o(un(a))implies thatun(a) → ∞for alla ∈ (0, 1), which means that the cutoff time tends to infinity. The remaining proof also uses Theorem 3.4 in [8] and is similar to the proof of the discrete time case in Theorem 1.4. We refer the reader to Section 3 for details.
8 Examples
In this section, we consider some classical examples and use the developed theory to examine the existence of cutoff and, in particular, compute the cutoff time. First, we writeF = (Xn, Kn, πn)∞n=1for a family of irreducible birth and death chains with Xn = {0, 1, ..., n}and writeFL, FRfor families of chains inF started at the left and right boundary states. For the continuous time case, those families are written asFc, FcL, FcR instead. Forn ≥ 1, let pn,i, qn,i, rn,i be the birth, death and holding rates inKn and τi(n),τei(n) be the first hitting times to state iof thenth chains inF , Fc. Fora ∈ (0, 1), Mn(a)denotes a state inXn satisfyingπn([0, Mn(a)]) ≥ aandπn([Mn(a), n]) ≥ 1 − a.
(1) Biased random walk. Forn ∈ N, let
pn,i= rn,n= p, qn,i+1= rn,0= q, ∀0 ≤ i < n, n ≥ 1, withq = 1 − p ∈ (0, 1/2). Note that the stationary distribution satisfies
πn(i) = p/q − 1 (p/q)n+1− 1
p q
i
, ∀0 ≤ i ≤ n.
This implies
πn([0, i])
πn(i) = p/q − (p/q)−i
p/q − 1 , ∀0 ≤ i ≤ n. (8.1)
By Lemma A.1, one has
E0eτn(n)=
n−1
X
i=0
πn([0, i])
pπn(i) , ζn,i≤Variτei+1(n)≤ 2ζn,i,
where
ζn,i= 1 p2πn(i)
i
X
`=0
πn([0, `]) πn(`)
2 πn(`).
Applying (8.1) to the computation ofE0eτn(n)andζn,iyields
E0eτn(n)= n
p − q − p2 q(p − q)2
1 − q
p
n , 1
p2 ≤ ζn,i≤ p
(p − q)3, ∀i,
where the bound ofζn,ileads to Var0τen(n) n. Observe thatπn([0, n]) = 1andπn(n) → 1−
q/p. As a consequence of Theorems 1.3, 1.4 and 6.1 withMn= n, the familiesFc, FcLhave a(p−qn ,√
n)cutoff in total variation and separation. To examine the existence of cutoff forFcR, we fixa ∈ (q/p2− 1, q/p). Based on the observation thatπn(n − 1) → (p − q)/p2, one hasMn(a) = n − 1fornlarge enough and this implies VarneτM(n)
n(a)= (EneτM(n)
n(a))2. By Theorem 5.1,FcRhas no cutoff in total variation.
(2) Metropolis chains for exponential distributions Consider an increasing positive functionf on(0, ∞). Forn ≥ 1, letπn(i) = πn(0)f (i)and
pn,i= rn,0= 1/2, qn,i+1= f (i)
2f (i + 1), ∀0 ≤ i < n, (8.2) and
rn,i+1 =1
2 − f (i)
2f (i + 1), ∀0 ≤ i < n − 1, rn,n= 1 −f (n − 1)
f (n) . (8.3) One can check that thenth chain is the Metropolis chain forπn with base chain the simple random walk onXn with holding probability 1/2 at boundaries. We refer the reader to [10] for details of Metropolis chains.
It is worthwhile to note thatKn is monotonic, i.e. pn,i+ qn,i+1 ≤ 1for all0 ≤ i < n. By Corollary 4.2 in [14], separation of thenth chain inF , FL, FR(and respectively in Fc, FcL, FcR) is the same. As a result of Theorem 1.1, the existence of separation cutoff of Fis equivalent to that ofFc and the cutoff time and window forFc given by Theorem 1.1 is applicable toF. For the total variation distance, ifinfn,irn,i> 0is assumed, then Theorems 1.4, 5.1 and 5.2 and Remarks 1.6 and 5.5 imply that the existence of cutoff of F(respectivelyFL, FR) is equivalent to that ofFc (respectivelyFcL, FcR). Furthermore, the cutoff times and windows forF , Fc given by Theorem 1.4 (respectively forFL, FcL and forFR, FcR given by Theorems 5.1 and 5.2) are consistent in the way that the cutoff times are equal and the cutoff windows are of the same order.
In this example,f (x) = exp{αxβ} withα > 0andβ > 0. Note thatinfn,irn,i > 0 if β ≥ 1andinfn,irn,i= 0ifβ ∈ (0, 1). In what follows, the cutoff phenomenon is discussed case by case according toβ.
Case 1: β > 1. We first make some computations. Note that d
dxf (x) = αβxβ−1f (x) ≥ αβf (x) ∀x ≥ 1.
This implies
i
X
j=0
f (j) ≤ 1 + f (i) + Z i
1
f (x)dx ≤ 1 + f (i) +f (i) − f (1)
αβ ≤
2 + 1
αβ
f (i).
Whenitends to infinity, one has
1 − 1
i
β
= 1 − β
i + O 1 i2
.
This leads to As a result, we obtain
1 ≤ πn([0, i])
The estimation of the variance implies Var0eτn(n) n. By Theorem 1.3, 1.4 and Theorem 6.1, bothFc andFcL have a(2n,√
n)cutoff in total variation and separation. For the familyFcR, the observation,πn(n) → 1, implies that the total variation mixing time of the nth chain is equal to0whennis large enough.
Case 2: β = 1. Setδ = (1 − e−α)/2. Note that(Kn− δI)/(1 − δ)is the biased random walk onXn with p = 1/(1 + e−α). The result for biased random walks implies thatFc
andFcLhave a(1−e2n−α,√
n)cutoff in total variation and separation butFcRhas no total variation cutoff.
In Cases 1 and 2, one has infn,irn,i > 0. This implies that, in the total variation distance, the conclusion on the existence of cutoff, the cutoff time and the cutoff window also applies toFc, FcL, FcR.
Replacingi, jwithn, bn/2cin (8.5) gives 1
Next, we fixc > 0and letcnbe a sequence converging tocsuch thatcnn1−β ∈ Xn. SetMn= n − cnn1−β. Replacingi, jwithMn, bMn/2cin (8.5) yields
n→∞lim πn([0, Mn]) = lim
n→∞πn(0)
Mn
X
`=0
f (`) = e−cαβ∈ (0, 1).
By Lemma A.1, one has, when2jn≤ in ≤ Mnandjn→ ∞,
E0eτM(n)
n= 2
Mn−1
X
`=0
f (0) + · · · + f (`)
f (`) = 2n2−β
αβ(2 − β) + O n2−βjn−β+ i2−βn + n , where the second equality is given by separatingP
`<Mn intoP
`<inandP
in≤`<Mnand then applying (8.4) and (8.5) respectively, and
Var0τeM(n)
n X
0≤`≤i<Mn
(f (0) + · · · + f (`))2 f (i)f (`)
Mn−1
X
i=0
1 f (i)
i
X
`=0
`2−2βf (`),
where the computation uses (8.4). Observe that4 − 3β > 2 − β. Settingjn= bn1/2cand in= bn4−3β4−2βc. Clearly,in≥ 2jnfornlarge enough and, in the computation of expectation, this leads to
E0τeM(n)
n= 2n2−β
αβ(2 − β)+ O
n2−32β+ n .
Applying the following fact d
dx(x3−3βf (x)) = [(3 − 3β)x2−3β+ αβx2−2β]f (x) x2−2βf (x), ∀x ≥ 1, to the computation of the variance yields
Var0eτM(n)
n
Mn−1
X
i=0
i3−3β n4−3β.
Similarly, one may use the observation thatf (Mf (n)n)=→ e−cαβ to derive qn,i 1 uniformly forMn≤ i ≤ n.
By Lemma A.1, this implies
EneτM(n)
n
n
X
i=Mn+1
f (i) + · · · + f (n)
f (i) n2−2β and
VarneτM(n)
n X
Mn<i≤`≤n
(f (`) + · · · + f (n))2
f (i)f (`) n4−4β.
As a consequence of Theorem 1.3, 1.4 and 6.1,Fc andFcLhave a(αβ(2−β)2n2−β , n2−32β+ n) cutoff in total variation and separation but, by Theorem 5.1,FcR has no total variation cutoff. Note that, whenβ ∈ (2/3, 1), a better choice of the cutoff window isn2−32β. To have this cutoff window, a more subtle estimation of the cutoff time is required.
We summarize the above results in the following theorem.
Theorem 8.1. Let f (x) = exp{αxβ} with α > 0, β > 0. Consider the family F = (Xn, Kn, πn)∞n=1, whereXn = {0, 1, ..., n},πn(i) = π(0)f (i) andKn is a birth and death
chain with transition rates
pn,i= rn,0= 1/2, qn,i+1 = f (i)
2f (i + 1), rn,i+1=1
2 − f (i)
2f (i + 1), ∀0 ≤ i < n.
Then,Fc andFcL have a(tn, bn)cutoff in total variation and separation butFcR has no total variation cutoff, where
tn=
2n forβ > 1
2n
1−e−α forβ = 1
2n2−β
αβ(2−β) for0 < β < 1
, bn=
(√n forβ ≥ 1
n2−32β+ n for0 < β < 1.
(3) Metropolis chains for polynomial distributions In this example, we consider the family of Metropolis chains given by (8.2) and (8.3) with the replacement of f (x)by g(x) = exp{α(log(x + 1))β}, whereα, βare positive. It has been shown in [9] thatFchas a cutoff in total variation and separation whenβ > 1but has no cutoff when0 < β ≤ 1. The following theorem provides a cutoff time and a cutoff window whenβ > 1.
Theorem 8.2. Letg(x) = exp{α(log(x + 1))β}withα > 0andβ > 1. Consider the family F = (Xn, Kn, πn)∞n=1, whereXn = {0, 1, ..., n}, πn(i) = π(0)g(i)andKn is a birth and death chain with transition rates
pn,i= rn,0= 1/2, qn,i+1= g(i)
2g(i + 1), rn,i+1=1
2 − g(i)
2g(i + 1), ∀0 ≤ i < n.
Then,Fc andFcL have a(tn, bn)cutoff in total variation and separation butFcR has no total variation cutoff, where
tn =
N
X
`=0
n2
αβB`(log n)β+`−1, bn = n2 (log n)32(β−1) andB0= 1,B`= 2`(β − 1)β · · · (β + ` − 2),N = dβ−32 e ≥ 0.
Remark 8.1. Note that, in Theorem 8.2,β + N − 1 < 32(β − 1) ≤ β + N.
The proof of Theorem 8.2 is similar to the proof of the caseβ ∈ (0, 1)in Theorem 8.1 and is placed in the appendix.
(4) Metropolis chains for binomial distributions Forn ≥ 1, letπn(i) = 2−n ni and pn,i= qn,n−i=1
2, qn,i+1 = pn,n−i−1= i + 1
2(n − i), ∀0 ≤ i < n/2,
andrn,i= 1 − pn,i− qn,ifor0 ≤ i ≤ n. It is easy to check thatKnis the Metropolis chain forπn with base chain the simple random walk onXn with holding probability1/2at the boundary states. The separation cutoff of this family is proved in [12] and we will discuss the cutoff time and the cutoff window in this example. First, one may use Lemma A.1 and (5.3) to derive
Mn−1
X
i=0
πn([0, i])
πn(i)(1 −ni) =n log n
4 + O(n), X
0≤`≤i<Mn
πn([0, `])2
πn(`)πn(i) n2, (8.6) for any sequenceMn∈ Xn satisfying|Mn−n2| = O(√
n). Note that
πn(i)
1 − i
n
=πn−1(i)
2 , πn([0, i]) = πn−1([0, i]) −πn−1(i)
2 .
This implies
πn([0, i])
πn(i)(1 −ni)= 2πn−1([0, i])
πn−1(i) − 1. (8.7)
SetMn= bn/2c. By Lemma A.1, (8.6) and (8.7), we obtain
E0eτM(n)
n=
Mn−1
X
i=0
2πn([0, i]) πn(i) =
Mn−1
X
i=0
πn+1([0, i]) πn+1(i)(1 −n+1i )+ 1
!
=n log n
4 + O(n) and
Var0τeM(n)
n X
0≤`≤i<Mn
πn([0, `])2 πn(`)πn(i) n2. In a similar way, one has
EnτeM(n)
n =n log n
4 + O(n), VarneτM(n)
n n2.
As a consequence of Theorems 1.3, 1.4 and 6.1,Fchas a(12n log n, n)separation cutoff andFc, FcLhave a(14n log n, n)total variation cutoff.