Directory UMM :Data Elmu:jurnal:S:Stochastic Processes And Their Applications:Vol89.Issue2.2000:
Stochastic Processes and their Applications 89 (2000) 175–191
www.elsevier.com/locate/spa
Empirical law of the iterated logarithm for Markov chains
with a countable state space
Tsung-Hsi Tsai ∗
Institute of Statistical Science, Academia Sinica, Taipei 115, Taiwan
Received 3 November 1999; received in revised form 17 February 2000; accepted 23 February 2000
Abstract
We nd conditions which are sucient and nearly necessary for the compact and bounded
c 2000 Elsevier
law of the iterated logarithm for Markov chains with a countable state space.
Science B.V. All rights reserved.
Keywords: Markov chain; Empirical processes; Law of the iterated logarithm; Absolute
regularity
1. Introduction
Let (S; G; P) be a probability space and let F be a set of measurable functions on S
with an envelope function F nite everywhere. Let X1 ; X2 ; : : : be a strictly stationary
sequence of random variables with distribution P.
We say F satis es the compact LIL with respect to {Xi } if there exists a compact
set K in l∞ (F), where l∞ (F) is the space of all bounded function on F with
sup norm, such that, with probability 1,
)∞
(
n
X
1
p
(f(Xi ) − Ef(X1 )) : f ∈ F
2n log log n i=1
n=1
is relatively compact and its limit set is K, and F satis es the bounded LIL with
respect to {Xi } if, with probability 1,
n
X
1
(f(Xi ) − Ef(X1 )) ¡ ∞:
sup sup p
2n log log n
n f∈F
i=1
We say the uniform CLT holds for F if
)∞
(
n
1 X
√
(f(Xi ) − Ef(X1 )) : f ∈ F
n i=1
n=1
converges weakly, in the space l∞ (F) to a Gaussian process. If F is a class of
indicator functions on S then the uniform CLT holding for F implies F satis es the
compact LIL for i.i.d. sequence of random variables (Kuelbs and Dudley, 1980).
∗
Corresponding author.
c 2000 Elsevier Science B.V. All rights reserved.
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T.-H. Tsai / Stochastic Processes and their Applications 89 (2000) 175–191
In this paper, {Xi }i¿0 is a positive recurrent irreducible Markov chain taking values
in countable state space S ={1; 2; 3; : : :} with the unique invariant probability measure .
Let Ni be the ith hitting time of state 1, and mi; j be the expected minimal number of
steps from state i to state j. Levental (1990) proved that
∞
X
k=1
√
(k) m1; k ¡ ∞
(1.1)
is a necessary and sucient condition for the uniform CLT for F = {1A : A ⊂ S} for
Markov chains satisfying E(N2 −N1 )2 ¡ ∞. However, the family of indicator functions
can be generalized to unbounded classes of functions, F = {f: |f|6F}, where F is
a non-negative function on S, corresponding to the condition (Tsai, 1997)
∞
X
k=1
√
F(k)(k) m1; k ¡ ∞:
(1.2)
It should also be observed that Levental’s result generalizes the Durst and Dudley
condition (Durst and Dudley, 1981)
∞
X
k=1
1=2 (k) ¡ ∞
(1.3)
for the uniform CLT for i.i.d. data. That is, since m1; k = mk; k = ((k))−1 in the i.i.d.
case, (1.1) coincides with (1.3).
We prove the compact LIL and bounded LIL for Markov chains under a weaker
condition than condition (1.2) of the uniform CLT. Suppose E(N2 − N1 )2 ¡ ∞ and let
F be a compact subset of L2 (S; ) with envelope function F satisfying
2
X
F(Xj ) ¡ ∞:
E
N1 ¡j6N2
If
n
X
1
√
F(k)(k) m1; k → 0
log log n k=1
p
(1.4)
as n → ∞ for all orderings of S, (the convergence is relative to the ordering of S)
then F satis es the compact LIL with respect to {Xj }:
On the other hand, suppose that mk; 1 and F(k) are both of polynomial rate, i.e. there
are c ¿ 0 and ¿ 0 such that
max{mk; 1 ; F(k)}6ck
for all k ∈ S:
(1.5)
If {F1A : A ⊆ S} satis es the compact LIL with respect to {Xi }, then (1.4) holds for
all ordering satisfying (1.5).
We also have the bounded LIL result if (1.4) is replaced by
n
sup p
n
X
1
√
F(k)(k) m1; k ¡ ∞:
log log n k=1
(1.6)
T.-H. Tsai / Stochastic Processes and their Applications 89 (2000) 175–191
177
In particular, if {Xi }i¿0 is i.i.d., then mk; 1 is a constant and the condition
E(N2 − N1 )2 ¡ ∞ holds. Thus we have that {1A : A ⊆ S} satis es the compact LIL
(bounded LIL) if and only if
!
n
n
X
X
p
p
−1=2
−1=2
(k) → 0
sup (log log n)
(k) ¡ ∞
(log log n)
n
k=1
k=1
for all ordering of S.
Arcones (1995) proved the compact LIL for stationary sequences satisfying absolutely regular mixing conditions under the following two conditions:
∞
X
k k 2=(p−2) log log k ¡ ∞
k=1
and the envelope function F satis es
∞
X
F(k)(P(X1 = k))1=p ¡ ∞;
k=1
where p ¿ 2 and k is absolutely regular mixing coecients. Since a positive recurrent irreducible Markov chain has convergent absolutely regular mixing coecients
(Davydov, 1973), one can also obtain LIL results by using empirical process LIL for
stationary sequences satisfying absolutely regular mixing conditions. We have an example to show that our conditions are less restrictive than those required for a mixing
process application to these problems.
2. Statement of the results
Let {Xj }j¿0 be a positive recurrent irreducible Markov chain taking values in S =
{1; 2; 3; : : :} with an invariant probability measure . Let Ni be the ith hitting time of
state 1, i.e.
N1 = min{n: n¿1; Xn = 1}
and for i ¿ 1
Ni = min{n: n ¿ Ni−1 ; Xn = 1}
and mi; j be the expected minimal number of steps from state i to state j, i.e.
mi; j = E(min{n: n¿1; Xn = j} | X0 = i):
We have sucient and nearly necessary conditions for the compact LIL and bounded
LIL.
Theorem 1. Suppose
E(N2 − N1 )2 ¡ ∞
(2.1)
2
and let F be a compact subset of L (S; ) with envelope function F satisfying
2
X
F(Xj ) ¡ ∞:
(2.2)
E
N1 ¡j6N2
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T.-H. Tsai / Stochastic Processes and their Applications 89 (2000) 175–191
If
n
X
1
√
F(k)(k) m1; k → 0
log log n k=1
(2.3)
p
as n → ∞ for all orderings of S (the convergence is relative to the ordering of S)
then F satis es the compact LIL with respect to {Xj }.
On the other hand; suppose that mk; 1 and F(k) are of polynomial rate; i.e. there
are c ¿ 0 and ¿ 0 such that
max{mk; 1 ; F(k)}6ck
for all k ∈ S:
(2.4)
If {F1A : A ⊆ S} satis es the compact LIL with respect to {Xi }; then (2:3) holds for
all orderings satisfying (2:4).
Theorem 2. Suppose (2:1) and let F be a compact subset of L2 (S; ) with envelope
function F satisfying (2:2). If
n
sup p
n
X
1
√
F(k)(k) m1; k ¡ ∞
log log n k=1
(2.5)
for all orderings of S (this is relative to the ordering of S) then F satis es the
bounded LIL with respect to {Xi }:
On the other hand; suppose that F(k) and mk; 1 are of polynomial rate; i.e. there
are c ¿ 0 and ¿ 0 such that (2:4) holds. If {F1A : A ⊆ S} satis es the bounded LIL
with respect to {Xi }; then (2:5) holds for all orderings satisfying (2:4).
Remark. (i) Note that (2.1) and (2.2) are also necessary conditions of the compact
LIL or bounded LIL.
(ii) Eq. (2.2) implies F ∈ L2 (S; ). Thus {F1A : A ⊆ S} and {f: |f|6F} are both
compact subsets in L2 (S; ).
(iii) De ne for all f ∈ F
X
(f(Xi ) − (f));
Z1 (f) =
N1 ¡i6N2
then Z1 (·) is almost surely continuous on compact set F ⊂ L2 (S; ). We consider Z1
taking value in C(F), where C(F) is the class of continuous functions on F. Since
F is compact, C(F) is a separable Banach space. We de ne HL(Z1 ) in C(F) by
canonical construction (Kuelbs, 1976), and let K be the unit ball of HL(Z1 ) . The limit
set of
)∞
(
n
X
1
p
(f(Xi ) − (f)) : f ∈ F
2n log log n i=1
n=1
√
is K= m1; 1 .
T.-H. Tsai / Stochastic Processes and their Applications 89 (2000) 175–191
179
3. Equivalent condition of the compact LIL and bounded LIL
Denote an =
Sn (f) =
p
2n log log n. We de ne for every f ∈ L1 (S; ) centered sum
n
X
i=1
(f(Xi ) − (f))
and centered sum of blocks
X
(f(Xi ) − (f))
Zj (f) =
Nj ¡i6Nj+1
for all j¿1: Then the Zj (f) are i.i.d. Let
l(n) = max{i : Ni 6n};
then
Sn (f) =
X
16i6N1
l(n)−1
(f(Xi ) − (f)) +
X
j=1
Zj (f) +
X
Nl(n) ¡i6n
(f(Xi ) − (f)):
We have
X
1
(f(Xi ) − (f))
lim sup √
n 16i6N or N ¡i6n
n→∞f∈F
1
l(n)
X
1
(F(Xi ) + (F))
6 lim √
n→∞
n 16i6N or N ¡i6n
1
l(n)
= 0 a:s:
from Chung’s proof of Theorem 5 in Chung (1967, p. 106) (use assumption (2:2)).
Thus,
l(n)−1
X
1
(3.1)
Zj (f) → 0 a:s:
sup Sn (f) −
an f∈F
j=1
as n → ∞. We have the following lemma.
Lemma 1. Suppose E(N2 −N1 )2 ¡ ∞ and let F be a compact subset of L2 (S; ) with
envelope function F satisfying
2
X
F(Xj ) ¡ ∞:
E
N1 ¡j6N2
Then F satis es the compact LIL with respect to {Xi } is equivalent to
X
n
1
Zj (f) → 0 in probability
sup
an f∈F
j=1
(3.2)
180
T.-H. Tsai / Stochastic Processes and their Applications 89 (2000) 175–191
and F satis es the bounded LIL with respect to {Xj } is equivalent to
X
n
1
Zj (f) is bounded in probability:
sup
an f∈F
(3.3)
j=1
Proof. F satis es the compact LIL with respect to {Xi } is equivalent to that there
exist a compact subset K in l∞ (F) such that
K
1
→→ √
Sn (f)
a:s:
(3.4)
an
m
1; 1
f∈F
(the notation {x n } →→ A means both limn d(x n ; A) = 0 and the cluster set of {x n }
is A, for a metric space (X; d) and a sequence {x n } of points in X ).
Using (3.1), (3.4) is equivalent to
1 l(n)−1
X
K
a:s:
Zj (f)
→→ √
an
m1; 1
j=1
f∈F
√
Since an =al(n)−1 → m1; 1 ; where al(n)−1 =
expression is equivalent to
1 l(n)−1
X
Zj (f)
→→ K a:s:
al(n)−1
j=1
f∈F
and then
n
1 X
Zj (f)
an
j=1
p
2(l(n) − 1)log log(l(n) − 1); the last
→→ K
a:s:;
(3.5)
f∈F
since {l(n): n = 1; 2; : : :} = {1; 2; : : :}:
Since Zj (·) are almost surely continuous on compact set F ⊂ L2 (S; ), we can consider Zj taking value in C(F), where C(F) is the class of continuous functions on
F with sup norm. Since F is compact, C(F) is a separable Banach space. Thus we
can apply those limit theorems in separable Banach spaces.
Note that
2
X
F(Xi ) + 2 (F) E (N2 − N1 ) 2 ¡ ∞
EkZ1 k2 62 E
N1 ¡j6N2
and EZ1 = 0. By applying Theorem 4:1 in Kuelbs (1977) we have that (3.5) is equivalent to
X
n
1
Zj
→ 0 in probability:
an
j=1
Thus F satis es the compact LIL with respect to {Xi } is equivalent to (3.2).
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T.-H. Tsai / Stochastic Processes and their Applications 89 (2000) 175–191
Using similar arguments we can obtain that F satis es the bounded LIL with respect
to {Xi } is equivalent to
X
n
1
(3.6)
Zj
sup
¡ ∞ a:s:
n an
j=1
and (3.6) is equivalent to
n
1 X
Zj
an
is bounded in probability
j=1
by Theorem 4:2 in Kuelbs (1977).
4. Proof of sucient part of the compact LIL
Suppose
E(N2 − N1 )2 ¡ ∞;
and
E
X
N1 ¡j6N2
2
F(Xj ) ¡ ∞
n
X
1
√
F(k)(k) m1; k → 0
log log n k=1
(4.1)
p
for all ordering of k in S. By Lemma 1 we only have to show
n
X
1
sup
Zj (f) → 0 in probability:
an f∈F
j=1
First, we have
n
n
n
∞
∞
X
X
X
X
X
Zj (f) 6 sup
sup
|f(k)|
Zj (1{k} ) 6
F(k)
Zj (1{k} ) :
f∈F j=1
f∈F k=1
j=1
k=1
j=1
(4.2)
Thus,
n
X
n2
n
X
X
1
1
F(k)
Zj (1{k} ) ¿
Zj (f) ¿ 6 P
P sup
an f∈F
an
2
j=1
j=1
k=1
|
{z
}
I
n
∞
X
X
1
+P
F(k)
Zj (1{k} ) ¿ :
an
2
j=1
k=n2 +1
{z
}
|
II
(4.3)
182
T.-H. Tsai / Stochastic Processes and their Applications 89 (2000) 175–191
By Markov’s inequality
X
n
n2
X
2
−1=2
−1=2
(log log n)
Zj (1{k} ) :
F(k)E n
I6
√
j=1
k=1
Since Zi (·) are i.i.d. and centered,
2 1=2
n
!
n
X
X
Zi (1{k} )
Zi (1{k} ) 6 n−1 E
n−1=2
E
i=1
i=1
= (E(Z12 (1{k} )))1=2 :
(4.4)
Thus, we need to show that
2
(log log n)
−1=2
n
X
k=1
F(k)(E(Z12 (1{k} )))1=2 → 0;
and this is equivalent to
(log log n)−1=2
n
X
k=1
F(k)(E(Z12 (1{k} )))1=2 → 0:
(4.5)
From Chung (1967, p. 88)
E
X
N1 ¡j6N2
2
1{k} (Xj ) = 2m1; 1 2 (k)(m1; k + mk; 1 ) − m1; 1 (k) for k¿1:
(4.6)
Thus,
(E(Z12 (1{k} )))1=2 6 E
6
X
N1 ¡j6N2
2 1=2
1{k} (Xj )
p
√
√
2m1; 1 ((k) m1; k + (k) mk; 1 ):
(4.7)
We can obtain the convergence of (4.5) from (4.1) and
n
X
1
√
F(k)(k) mk; 1 → 0
log log n k=1
p
(4.8)
for all ordering of k. To show (4.8), from Chung (1967, p. 88),
E(N2 − N1 )2 = 2m1; 1
∞
X
k=1
(k)mk; 1 − m1; 1 :
(4.9)
T.-H. Tsai / Stochastic Processes and their Applications 89 (2000) 175–191
183
Since E(N2 − N1 )2 ¡ ∞, we thus have
∞
X
k=1
(k)mk; 1 ¡ ∞:
Hence,
∞
X
√
F(k)(k) mk; 1 6
∞
X
!
1=2
2
F (k)(k)
k=1
k=1
∞
X
(k)mk; 1
k=1
!
1=2
¡ ∞;
(4.10)
and thus obtain (4.8).
For the other part of (4.3) we have
√
n
1=2 X
∞
∞
X
X
n
8m1; 1
2
Zj (1{k} ) 6
F(k)(k);
F(k)E
II6
an
log log n
j=1
k=n2 +1
k=n2 +1
since
X
n
Zj (1{k} ) 6nE|Z1 (1{k} )|62m1; 1 (k)n:
E
j=1
Now we have to show that
1=2 X
∞
n
F(k)(k) → 0:
log log n
2
k=n +1
Note
∞
X
F(k)(k) 6
∞
X
2
6
∞
X
∞
X
F (k)(k)
k=n2 +1
k=n2 +1
!1=2
2
!1=2
F (k)(k)
k=n2 +1
k=n2 +1
!1=2
(k)
;
we will show
∞
X
n
F 2 (k)(k) → 0:
log log n
2
k=n +1
Since
p
√
√
√
(k) = (k) mk; k 6(k) m1; k + (k) mk; 1 ;
(4.1) and (4.8) imply
n
X
p
1
F(k) (k) → 0
log log n k=1
p
p
for all ordering of k. Thus we can assume F(k) (k) is decreasing, then for n large
enough
n
p
p
1X
1p
log log n:
F(k) (k)6
F(n) (n)6
n
n
k=1
184
T.-H. Tsai / Stochastic Processes and their Applications 89 (2000) 175–191
Thus for n large enough
∞
∞
X
X
n
log log k
n
F 2 (k)(k)6
;
log log n
log
log
n
k2
2
2
k=n +1
k=n +1
and the right-hand side converges to zero as n → ∞.
5. Proof of necessary part of the compact LIL
In this section F = {F1A : A ⊆ S} and k · k is the sup norm in C(F). Suppose
E(N2 − N1 )2 ¡ ∞;
and
E
X
N1 ¡j6N2
(5.1)
2
F(Xj ) ¡ ∞
X
n
1
Z
j
→ 0
an
j=1
in probability:
We need to show that
n
X
1
√
p
F(k)(k) m1; k → 0
log log n k=1
(5.2)
(5.3)
(5.4)
for all orderings such that mk; 1 and F(k) are of a polynomial rate, i.e. there are c ¿ 0
and ¿ 0 such that
max{mk; 1 ; F(k)}6ck
for all k ∈ S:
Recall
Zj (f) =
X
(f(Xi ) − (f))
X
f(Xi ) − m1; 1 (f)
Nj ¡i6Nj+1
and de ne
Yj (f) =
Nj ¡i6Nj+1
in this section. To prove (5.4), rst we show the following lemma.
Lemma 2. Suppose (5:1); (5:2) and (5:3) hold; then
n
∞
X
1 X
F(k)E
Yj (1{k} ) → 0 as n → ∞:
an
j=1
k=1
Proof. Let
Uj (f) = ((Nj+1 − Nj ) − m1; 1 )(f);
(5.5)
T.-H. Tsai / Stochastic Processes and their Applications 89 (2000) 175–191
185
then
Zj (f) = Yj (f) + Uj (f):
Kolmogorov’s LIL holds for the i.i.d. sequence {Nj+1 −Nj }j¿1 since E(N2 −N1 )2 ¡ ∞,
and that is equivalent to
n
1 X
((Nj+1 − Nj ) − m1; 1 ) → 0 in probability:
an
j=1
We have
1
an
X
n
Uj
→0
j=1
in probability;
since supf∈F (f) = (F) ¡ ∞. Thus
X
n
1
Yj
→ 0 in probability:
an
j=1
(5.6)
We then claim
n
X
1
Yj
E
→ 0:
an
j=1
(5.7)
Then use Corollary 6:12 in Ledoux and Talagrad (1991)
n
X
1
Y
E sup
j
¡ ∞
n an
(5.8)
We have the bounded LIL for {Yj } since the bounded LIL holds for {Zj } and {Uj },
that is
X
n
1
Yj
sup
¡ ∞ a:s:
n an
j=1
j=1
is equivalent to
1
E sup kYn k ¡ ∞:
n an
Since (5.2) holds we have
X
1
F(Xi ) ¡ ∞:
E sup
n an
N ¡i6N
n
(5.9)
n+1
(Here we use that if {Xi } is i.i.d. and real valued, then E X12 ¡ ∞ implies
E[supn n−1=2 |Xn |] ¡ ∞:) Since
X
F(Xi ) + m1; 1 (F);
kYn k6
Nn ¡i6Nn+1
we have (5.9) and thus (5.8) holds. Combine (5.6) and (5.8), we obtain (5.7).
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T.-H. Tsai / Stochastic Processes and their Applications 89 (2000) 175–191
Pointwise, for all !, we have
X
X
n
n
Yj
= sup
Yj (f)
j=1
f∈F j=1
X
n
X
∞
f(k)
Yj (1{k} )
= sup
f∈F k=1
j=1
n
∞
X
X
P
Yj (1{k} ) :
F(k) 1
¿
n
Yj (1{k} )¿0
j=1
j=1
k=1
Thus,
n
X
∞
1X
X
n
Y
¿
F(k)E
Y
(1
)
E
j
j {k} :
j=1
j=1
2 k=1
From (5.7) we have
n
∞
X
1 X
Yj (1{k} ) → 0:
F(k)E
an
j=1
k=1
Then we need the following lemma.
Lemma 3. There are c′ ; M ¿ 0 such that
n
X
−1=2
Yj (1{k} ) ¿c′ [EY12 (1{k} )]1=2
E
n
j=1
for all n¿M=k ; where k = P( k ¡ 1 | X0 = 1) and i = min{n¿1: Xn = i}:
Proof. By the Marcinkiewicz–Zygmund inequality
1=2
X
n
n
X
Yj2 (1{k} ) ;
Yj (1{k} ) ¿c1 E n−1
n−1=2 E
j=1
j=1
where c1 ¿ 0 is a constant which is independent of the random variables. Note that
1=2
n
n
X
X
1
1
E
Yj2 (1{k} ) ¿[EY12 (1{k} )]1=2 P
Yj2 (1{k} )¿EY12 (1{k} ) :
n
n
j=1
j=1
By the Berry–Esseen theorem
n
X
1 3n−1=2 EY16 (1{k} )
1
:
Yj2 (1{k} )¿EY12 (1{k} ) ¿ −
P
n
2
(EY14 (1{k} ))3=2
j=1
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T.-H. Tsai / Stochastic Processes and their Applications 89 (2000) 175–191
Let i = min{n¿1: Xn = i}, k = P( k ¡ 1 | X0 = 1) and k = P( 1 ¡ k | X0 = k). Thus
it is enough to show that there is a M ¿ 0 such that
3n−1=2 EY16 (1{k} ) 1
M
for all n¿ :
6
4
k
(EY14 (1{k} ))3=2
P
Denote W (k) = N1 ¡i6N2 1{k} (Xi ) then
(5.10)
Y1 (1{k} ) = W (k) − EW (k)
and
l
EW (k) =
∞
X
ml P(W (k) = m)
m=1
=
∞
X
m=1
ml [k (1 − k )m−1 k ]
= k (1 − k )−1 k
∞
X
m=1
ml (1 − k )m :
(5.11)
Thus by computation EW (k) = k =k ,
EY16 (1{k} ) 6 EW 6 (k)
=
k
(1 + 57(1 − k ) + 302(1 − k )2 + 302(1 − k )3
k6
+ 57(1 − k )4 + (1 − k )5 )
6
720k
k6
and
EY14 (1{k} )
4
k
= E W (k) −
k
k
= 4
k
1 + 11(1 − k ) + 11(1 − k )2 + (1 − k )3 − 4k (1 − k )2
−16k (1 − k ) − 4k + 62k (1 − k ) + 62k − 33k
Since
∞
X
k=1
∞
∞
∞
X
X
k X
k 6
=
EW (k) =
m1; 1 (k) = m1; 1 ¡ ∞;
k
k=1
k=1
k=1
most k are small. Thus
k
EY14 (1{k} )¿ 4
2k
except for nitely many k. Hence there is a M1 ¿ 0 such that
EY16 (1{k} )
6M1 k−1=2 :
(EY14 (1{k} ))3=2
!
:
188
T.-H. Tsai / Stochastic Processes and their Applications 89 (2000) 175–191
Thus
3n−1=2 EY16 (1{k} ) 1
6
4
(EY14 (1{k} ))3=2
for all n¿
144M1
:
k
Proof of (5.4). Since
(EY12 (1{k} ))1=2 ¿ E
¿
√
X
!2 1=2
1{k} (Xi )
N1 ¡i6N2
− m1; 1 (k)
1=2
m1; 1 (k)m1;
k − 2m1; 1 (k);
from (4:6), it is enough to show
(log log n)−1=2
n
X
k=1
F(k)(EY12 (1{k} ))1=2 → 0:
Note that by using (5.11)
2 − k − k
2k
6 2
EY12 (1{k} ) = k
2
k
k
(5.12)
(5.13)
and
mk; 1 ¿
∞
X
j(P( 1 ¿ k | X0 = k)) j−1 P( 1 ¡ k | X0 = k)
∞
X
j(1 − k ) j−1 k
j=1
=
j=1
= k−1 :
(5.14)
Let
U = {k: F(k)(EY12 (1{k} ))1=2 ¿k −2 }:
Use (5.13), (5.14) and assumption (5.5) for all k ∈ U
k −(4 +4)
1 1 k −4
1
¿
:
k ¿ k2 EY12 (1{k} )¿
2
2 m2k; 1 F 2 (k)
2c4
From Lemmas 2 and 3 we have
X
F(k)[EY12 (1{k} )]1=2 → 0:
(log log n)−1=2
k: k ¿M=n
Note
k6
implies
n 1=(4 +4)
2Mc4
k −(4 +4) M
¿ ;
2c4
n
(5.15)
189
T.-H. Tsai / Stochastic Processes and their Applications 89 (2000) 175–191
thus (5.15) implies
(log log n)−1=2
k∈U and
Since
P
k6∈U
X
k6(n=2Mc4 )1=(4 +4)
F(k)(EY12 (1{k} ))1=2 6
[(n=2Mc4 )1=(4 +4) ]
(log log n)
1=2
X
P
k6∈U
F(k)[EY12 (1{k} )]1=2 → 0:
k −2 ¡ ∞,
F(k)(EY12 (1{k} ))1=2 → 0:
k=1
and this is equivalent to (5.12).
6. Proof of the bounded LIL
In view of (3.3) in Lemma 1, the proof of the bounded LIL now follows as indicated
in Sections 4 and 5 for the compact LIL. Here, of course, to show (3.3), one uses
(2.5) rather than (2.3) at various places. The details are straightforward.
7. Comparison with mixing results
Arcones (1995) proved the compact LIL for stationary sequences satisfying absolutely regular mixing conditions under the following two conditions:
∞
X
k=1
k k 2=(p−2) log log k ¡ ∞
(7.1)
and the envelope function F satis es
∞
X
k=1
F(k)(P(X1 = k))1=p ¡ ∞;
(7.2)
where p ¿ 2 and k is absolutely regular mixing coecients.
To show our conditions are less restrictive than Arcones’ conditions for Markov
chains, we present an example such that the conditions E(N2 − N1 )2 ¡ ∞;
2
X
(7.3)
F(Xj ) ¡ ∞
E
N1 ¡j6N2
and
∞
X
k=1
v
u
u
X
u
F(k)tE
N1 ¡j6N2
2
1{k} (Xj ) ¡ ∞
(7.4)
hold (condition (7.4) implies (2.3)), but Arcones’ conditions (7.1) and (7.2) fail.
190
T.-H. Tsai / Stochastic Processes and their Applications 89 (2000) 175–191
Example. Let {Xi } be a stationary Markov chain with transition probability
s
n
; Pn; 1 = 1 − Pn; n+1 for all n¿1 and some s ¿ 1:
Pn; n+1 =
n+1
Let F(k) = k t for some t¿0. We choose s and t such that (7.3) and (7.4) hold, but
(7.1) and (7.2) cannot both hold.
From example 3.2 in Tsai (1997) (k) ≈ k −s ; P(N2 − N1 = n) ≈ n−s−1 ,
2
X
1{k} (Xj ) ≈ k −s
E
N1 ¡j6N2
and k /k 1−s . Thus
∞
X
k=1
and
F(k)((k))1=p ≈
X
∞
X
2
F(Xj ) =
E
N1 ¡j6N2
∞
X
v
u
u
X
u
F(k)tE
k=1
k=1
k t−(s=p) ¡ ∞ only if
∞
X
n=1
N1 ¡j6N2
n
X
kt
k=1
2
!2
1{k} (Xj ) ≈
p6
s
;
t+1
P(N2 − N1 = n) ≈
∞
X
∞
X
n2t−s+1
n=1
k t−(s=2) :
k=1
Hence (7.3) and (7.4) hold if s ¿ 2t + 2. We take s = 4:2 and t = 1. If (7.2) holds,
then p6 21
10 . But
∞
X
k k 2=(p−2) log log k/
k=1
thus (7.1) fails.
∞
X
k=1
k −3:2 k 20 log log k = ∞;
Acknowledgements
The author wishes to express his gratitude to Professor Jim Kuelbs for advice and
guidance.
References
Arcones, M.A., 1995. The law of the iterated logarithm for empirical processes under absolute regularity.
J. Theoret. Probab. 8 (2), 433–451.
Chung, K.L., 1967. Markov Chain with Stationary Probability. Springer, Berlin.
Davydov, Y.A., 1973. Mixing conditions for Markov chains. Theory Probab. Appl. 28, 313–318.
Durst, M., Dudley, R.M., 1981. Empirical processes. Vapnik–Cervonenkis classes and Poisson processes.
Probab. Math. Statist. 1, 109–115.
Kuelbs, J., 1976. A strong convergence theorem for Banach space values random variables. Ann. Probab. 4,
744–771.
T.-H. Tsai / Stochastic Processes and their Applications 89 (2000) 175–191
191
Kuelbs, J., 1977. Kolmogorov’s law of iterated logarithm for Banach space valued random variables. Illinois
J. Math. 21, 784–800.
Kuelbs, J., Dudley, R.M., 1980. Log log laws for empirical measures. Ann. Probab. 8, 405–418.
Ledoux, M., Talagrad, M., 1991. Probability in Banach Spaces. Springer, Berlin.
Levental, S., 1990. Uniform CLT for Markov chain with a countable state space. Stochastic Process. Appl.
34, 245–253.
Tsai, T., 1997. Uniform CLT for Markov chains with a countable state space. Taiwanese J. Math. 1 (4),
481–498.
www.elsevier.com/locate/spa
Empirical law of the iterated logarithm for Markov chains
with a countable state space
Tsung-Hsi Tsai ∗
Institute of Statistical Science, Academia Sinica, Taipei 115, Taiwan
Received 3 November 1999; received in revised form 17 February 2000; accepted 23 February 2000
Abstract
We nd conditions which are sucient and nearly necessary for the compact and bounded
c 2000 Elsevier
law of the iterated logarithm for Markov chains with a countable state space.
Science B.V. All rights reserved.
Keywords: Markov chain; Empirical processes; Law of the iterated logarithm; Absolute
regularity
1. Introduction
Let (S; G; P) be a probability space and let F be a set of measurable functions on S
with an envelope function F nite everywhere. Let X1 ; X2 ; : : : be a strictly stationary
sequence of random variables with distribution P.
We say F satis es the compact LIL with respect to {Xi } if there exists a compact
set K in l∞ (F), where l∞ (F) is the space of all bounded function on F with
sup norm, such that, with probability 1,
)∞
(
n
X
1
p
(f(Xi ) − Ef(X1 )) : f ∈ F
2n log log n i=1
n=1
is relatively compact and its limit set is K, and F satis es the bounded LIL with
respect to {Xi } if, with probability 1,
n
X
1
(f(Xi ) − Ef(X1 )) ¡ ∞:
sup sup p
2n log log n
n f∈F
i=1
We say the uniform CLT holds for F if
)∞
(
n
1 X
√
(f(Xi ) − Ef(X1 )) : f ∈ F
n i=1
n=1
converges weakly, in the space l∞ (F) to a Gaussian process. If F is a class of
indicator functions on S then the uniform CLT holding for F implies F satis es the
compact LIL for i.i.d. sequence of random variables (Kuelbs and Dudley, 1980).
∗
Corresponding author.
c 2000 Elsevier Science B.V. All rights reserved.
0304-4149/00/$ - see front matter
PII: S 0 3 0 4 - 4 1 4 9 ( 0 0 ) 0 0 0 1 9 - 3
176
T.-H. Tsai / Stochastic Processes and their Applications 89 (2000) 175–191
In this paper, {Xi }i¿0 is a positive recurrent irreducible Markov chain taking values
in countable state space S ={1; 2; 3; : : :} with the unique invariant probability measure .
Let Ni be the ith hitting time of state 1, and mi; j be the expected minimal number of
steps from state i to state j. Levental (1990) proved that
∞
X
k=1
√
(k) m1; k ¡ ∞
(1.1)
is a necessary and sucient condition for the uniform CLT for F = {1A : A ⊂ S} for
Markov chains satisfying E(N2 −N1 )2 ¡ ∞. However, the family of indicator functions
can be generalized to unbounded classes of functions, F = {f: |f|6F}, where F is
a non-negative function on S, corresponding to the condition (Tsai, 1997)
∞
X
k=1
√
F(k)(k) m1; k ¡ ∞:
(1.2)
It should also be observed that Levental’s result generalizes the Durst and Dudley
condition (Durst and Dudley, 1981)
∞
X
k=1
1=2 (k) ¡ ∞
(1.3)
for the uniform CLT for i.i.d. data. That is, since m1; k = mk; k = ((k))−1 in the i.i.d.
case, (1.1) coincides with (1.3).
We prove the compact LIL and bounded LIL for Markov chains under a weaker
condition than condition (1.2) of the uniform CLT. Suppose E(N2 − N1 )2 ¡ ∞ and let
F be a compact subset of L2 (S; ) with envelope function F satisfying
2
X
F(Xj ) ¡ ∞:
E
N1 ¡j6N2
If
n
X
1
√
F(k)(k) m1; k → 0
log log n k=1
p
(1.4)
as n → ∞ for all orderings of S, (the convergence is relative to the ordering of S)
then F satis es the compact LIL with respect to {Xj }:
On the other hand, suppose that mk; 1 and F(k) are both of polynomial rate, i.e. there
are c ¿ 0 and ¿ 0 such that
max{mk; 1 ; F(k)}6ck
for all k ∈ S:
(1.5)
If {F1A : A ⊆ S} satis es the compact LIL with respect to {Xi }, then (1.4) holds for
all ordering satisfying (1.5).
We also have the bounded LIL result if (1.4) is replaced by
n
sup p
n
X
1
√
F(k)(k) m1; k ¡ ∞:
log log n k=1
(1.6)
T.-H. Tsai / Stochastic Processes and their Applications 89 (2000) 175–191
177
In particular, if {Xi }i¿0 is i.i.d., then mk; 1 is a constant and the condition
E(N2 − N1 )2 ¡ ∞ holds. Thus we have that {1A : A ⊆ S} satis es the compact LIL
(bounded LIL) if and only if
!
n
n
X
X
p
p
−1=2
−1=2
(k) → 0
sup (log log n)
(k) ¡ ∞
(log log n)
n
k=1
k=1
for all ordering of S.
Arcones (1995) proved the compact LIL for stationary sequences satisfying absolutely regular mixing conditions under the following two conditions:
∞
X
k k 2=(p−2) log log k ¡ ∞
k=1
and the envelope function F satis es
∞
X
F(k)(P(X1 = k))1=p ¡ ∞;
k=1
where p ¿ 2 and k is absolutely regular mixing coecients. Since a positive recurrent irreducible Markov chain has convergent absolutely regular mixing coecients
(Davydov, 1973), one can also obtain LIL results by using empirical process LIL for
stationary sequences satisfying absolutely regular mixing conditions. We have an example to show that our conditions are less restrictive than those required for a mixing
process application to these problems.
2. Statement of the results
Let {Xj }j¿0 be a positive recurrent irreducible Markov chain taking values in S =
{1; 2; 3; : : :} with an invariant probability measure . Let Ni be the ith hitting time of
state 1, i.e.
N1 = min{n: n¿1; Xn = 1}
and for i ¿ 1
Ni = min{n: n ¿ Ni−1 ; Xn = 1}
and mi; j be the expected minimal number of steps from state i to state j, i.e.
mi; j = E(min{n: n¿1; Xn = j} | X0 = i):
We have sucient and nearly necessary conditions for the compact LIL and bounded
LIL.
Theorem 1. Suppose
E(N2 − N1 )2 ¡ ∞
(2.1)
2
and let F be a compact subset of L (S; ) with envelope function F satisfying
2
X
F(Xj ) ¡ ∞:
(2.2)
E
N1 ¡j6N2
178
T.-H. Tsai / Stochastic Processes and their Applications 89 (2000) 175–191
If
n
X
1
√
F(k)(k) m1; k → 0
log log n k=1
(2.3)
p
as n → ∞ for all orderings of S (the convergence is relative to the ordering of S)
then F satis es the compact LIL with respect to {Xj }.
On the other hand; suppose that mk; 1 and F(k) are of polynomial rate; i.e. there
are c ¿ 0 and ¿ 0 such that
max{mk; 1 ; F(k)}6ck
for all k ∈ S:
(2.4)
If {F1A : A ⊆ S} satis es the compact LIL with respect to {Xi }; then (2:3) holds for
all orderings satisfying (2:4).
Theorem 2. Suppose (2:1) and let F be a compact subset of L2 (S; ) with envelope
function F satisfying (2:2). If
n
sup p
n
X
1
√
F(k)(k) m1; k ¡ ∞
log log n k=1
(2.5)
for all orderings of S (this is relative to the ordering of S) then F satis es the
bounded LIL with respect to {Xi }:
On the other hand; suppose that F(k) and mk; 1 are of polynomial rate; i.e. there
are c ¿ 0 and ¿ 0 such that (2:4) holds. If {F1A : A ⊆ S} satis es the bounded LIL
with respect to {Xi }; then (2:5) holds for all orderings satisfying (2:4).
Remark. (i) Note that (2.1) and (2.2) are also necessary conditions of the compact
LIL or bounded LIL.
(ii) Eq. (2.2) implies F ∈ L2 (S; ). Thus {F1A : A ⊆ S} and {f: |f|6F} are both
compact subsets in L2 (S; ).
(iii) De ne for all f ∈ F
X
(f(Xi ) − (f));
Z1 (f) =
N1 ¡i6N2
then Z1 (·) is almost surely continuous on compact set F ⊂ L2 (S; ). We consider Z1
taking value in C(F), where C(F) is the class of continuous functions on F. Since
F is compact, C(F) is a separable Banach space. We de ne HL(Z1 ) in C(F) by
canonical construction (Kuelbs, 1976), and let K be the unit ball of HL(Z1 ) . The limit
set of
)∞
(
n
X
1
p
(f(Xi ) − (f)) : f ∈ F
2n log log n i=1
n=1
√
is K= m1; 1 .
T.-H. Tsai / Stochastic Processes and their Applications 89 (2000) 175–191
179
3. Equivalent condition of the compact LIL and bounded LIL
Denote an =
Sn (f) =
p
2n log log n. We de ne for every f ∈ L1 (S; ) centered sum
n
X
i=1
(f(Xi ) − (f))
and centered sum of blocks
X
(f(Xi ) − (f))
Zj (f) =
Nj ¡i6Nj+1
for all j¿1: Then the Zj (f) are i.i.d. Let
l(n) = max{i : Ni 6n};
then
Sn (f) =
X
16i6N1
l(n)−1
(f(Xi ) − (f)) +
X
j=1
Zj (f) +
X
Nl(n) ¡i6n
(f(Xi ) − (f)):
We have
X
1
(f(Xi ) − (f))
lim sup √
n 16i6N or N ¡i6n
n→∞f∈F
1
l(n)
X
1
(F(Xi ) + (F))
6 lim √
n→∞
n 16i6N or N ¡i6n
1
l(n)
= 0 a:s:
from Chung’s proof of Theorem 5 in Chung (1967, p. 106) (use assumption (2:2)).
Thus,
l(n)−1
X
1
(3.1)
Zj (f) → 0 a:s:
sup Sn (f) −
an f∈F
j=1
as n → ∞. We have the following lemma.
Lemma 1. Suppose E(N2 −N1 )2 ¡ ∞ and let F be a compact subset of L2 (S; ) with
envelope function F satisfying
2
X
F(Xj ) ¡ ∞:
E
N1 ¡j6N2
Then F satis es the compact LIL with respect to {Xi } is equivalent to
X
n
1
Zj (f) → 0 in probability
sup
an f∈F
j=1
(3.2)
180
T.-H. Tsai / Stochastic Processes and their Applications 89 (2000) 175–191
and F satis es the bounded LIL with respect to {Xj } is equivalent to
X
n
1
Zj (f) is bounded in probability:
sup
an f∈F
(3.3)
j=1
Proof. F satis es the compact LIL with respect to {Xi } is equivalent to that there
exist a compact subset K in l∞ (F) such that
K
1
→→ √
Sn (f)
a:s:
(3.4)
an
m
1; 1
f∈F
(the notation {x n } →→ A means both limn d(x n ; A) = 0 and the cluster set of {x n }
is A, for a metric space (X; d) and a sequence {x n } of points in X ).
Using (3.1), (3.4) is equivalent to
1 l(n)−1
X
K
a:s:
Zj (f)
→→ √
an
m1; 1
j=1
f∈F
√
Since an =al(n)−1 → m1; 1 ; where al(n)−1 =
expression is equivalent to
1 l(n)−1
X
Zj (f)
→→ K a:s:
al(n)−1
j=1
f∈F
and then
n
1 X
Zj (f)
an
j=1
p
2(l(n) − 1)log log(l(n) − 1); the last
→→ K
a:s:;
(3.5)
f∈F
since {l(n): n = 1; 2; : : :} = {1; 2; : : :}:
Since Zj (·) are almost surely continuous on compact set F ⊂ L2 (S; ), we can consider Zj taking value in C(F), where C(F) is the class of continuous functions on
F with sup norm. Since F is compact, C(F) is a separable Banach space. Thus we
can apply those limit theorems in separable Banach spaces.
Note that
2
X
F(Xi ) + 2 (F) E (N2 − N1 ) 2 ¡ ∞
EkZ1 k2 62 E
N1 ¡j6N2
and EZ1 = 0. By applying Theorem 4:1 in Kuelbs (1977) we have that (3.5) is equivalent to
X
n
1
Zj
→ 0 in probability:
an
j=1
Thus F satis es the compact LIL with respect to {Xi } is equivalent to (3.2).
181
T.-H. Tsai / Stochastic Processes and their Applications 89 (2000) 175–191
Using similar arguments we can obtain that F satis es the bounded LIL with respect
to {Xi } is equivalent to
X
n
1
(3.6)
Zj
sup
¡ ∞ a:s:
n an
j=1
and (3.6) is equivalent to
n
1 X
Zj
an
is bounded in probability
j=1
by Theorem 4:2 in Kuelbs (1977).
4. Proof of sucient part of the compact LIL
Suppose
E(N2 − N1 )2 ¡ ∞;
and
E
X
N1 ¡j6N2
2
F(Xj ) ¡ ∞
n
X
1
√
F(k)(k) m1; k → 0
log log n k=1
(4.1)
p
for all ordering of k in S. By Lemma 1 we only have to show
n
X
1
sup
Zj (f) → 0 in probability:
an f∈F
j=1
First, we have
n
n
n
∞
∞
X
X
X
X
X
Zj (f) 6 sup
sup
|f(k)|
Zj (1{k} ) 6
F(k)
Zj (1{k} ) :
f∈F j=1
f∈F k=1
j=1
k=1
j=1
(4.2)
Thus,
n
X
n2
n
X
X
1
1
F(k)
Zj (1{k} ) ¿
Zj (f) ¿ 6 P
P sup
an f∈F
an
2
j=1
j=1
k=1
|
{z
}
I
n
∞
X
X
1
+P
F(k)
Zj (1{k} ) ¿ :
an
2
j=1
k=n2 +1
{z
}
|
II
(4.3)
182
T.-H. Tsai / Stochastic Processes and their Applications 89 (2000) 175–191
By Markov’s inequality
X
n
n2
X
2
−1=2
−1=2
(log log n)
Zj (1{k} ) :
F(k)E n
I6
√
j=1
k=1
Since Zi (·) are i.i.d. and centered,
2 1=2
n
!
n
X
X
Zi (1{k} )
Zi (1{k} ) 6 n−1 E
n−1=2
E
i=1
i=1
= (E(Z12 (1{k} )))1=2 :
(4.4)
Thus, we need to show that
2
(log log n)
−1=2
n
X
k=1
F(k)(E(Z12 (1{k} )))1=2 → 0;
and this is equivalent to
(log log n)−1=2
n
X
k=1
F(k)(E(Z12 (1{k} )))1=2 → 0:
(4.5)
From Chung (1967, p. 88)
E
X
N1 ¡j6N2
2
1{k} (Xj ) = 2m1; 1 2 (k)(m1; k + mk; 1 ) − m1; 1 (k) for k¿1:
(4.6)
Thus,
(E(Z12 (1{k} )))1=2 6 E
6
X
N1 ¡j6N2
2 1=2
1{k} (Xj )
p
√
√
2m1; 1 ((k) m1; k + (k) mk; 1 ):
(4.7)
We can obtain the convergence of (4.5) from (4.1) and
n
X
1
√
F(k)(k) mk; 1 → 0
log log n k=1
p
(4.8)
for all ordering of k. To show (4.8), from Chung (1967, p. 88),
E(N2 − N1 )2 = 2m1; 1
∞
X
k=1
(k)mk; 1 − m1; 1 :
(4.9)
T.-H. Tsai / Stochastic Processes and their Applications 89 (2000) 175–191
183
Since E(N2 − N1 )2 ¡ ∞, we thus have
∞
X
k=1
(k)mk; 1 ¡ ∞:
Hence,
∞
X
√
F(k)(k) mk; 1 6
∞
X
!
1=2
2
F (k)(k)
k=1
k=1
∞
X
(k)mk; 1
k=1
!
1=2
¡ ∞;
(4.10)
and thus obtain (4.8).
For the other part of (4.3) we have
√
n
1=2 X
∞
∞
X
X
n
8m1; 1
2
Zj (1{k} ) 6
F(k)(k);
F(k)E
II6
an
log log n
j=1
k=n2 +1
k=n2 +1
since
X
n
Zj (1{k} ) 6nE|Z1 (1{k} )|62m1; 1 (k)n:
E
j=1
Now we have to show that
1=2 X
∞
n
F(k)(k) → 0:
log log n
2
k=n +1
Note
∞
X
F(k)(k) 6
∞
X
2
6
∞
X
∞
X
F (k)(k)
k=n2 +1
k=n2 +1
!1=2
2
!1=2
F (k)(k)
k=n2 +1
k=n2 +1
!1=2
(k)
;
we will show
∞
X
n
F 2 (k)(k) → 0:
log log n
2
k=n +1
Since
p
√
√
√
(k) = (k) mk; k 6(k) m1; k + (k) mk; 1 ;
(4.1) and (4.8) imply
n
X
p
1
F(k) (k) → 0
log log n k=1
p
p
for all ordering of k. Thus we can assume F(k) (k) is decreasing, then for n large
enough
n
p
p
1X
1p
log log n:
F(k) (k)6
F(n) (n)6
n
n
k=1
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T.-H. Tsai / Stochastic Processes and their Applications 89 (2000) 175–191
Thus for n large enough
∞
∞
X
X
n
log log k
n
F 2 (k)(k)6
;
log log n
log
log
n
k2
2
2
k=n +1
k=n +1
and the right-hand side converges to zero as n → ∞.
5. Proof of necessary part of the compact LIL
In this section F = {F1A : A ⊆ S} and k · k is the sup norm in C(F). Suppose
E(N2 − N1 )2 ¡ ∞;
and
E
X
N1 ¡j6N2
(5.1)
2
F(Xj ) ¡ ∞
X
n
1
Z
j
→ 0
an
j=1
in probability:
We need to show that
n
X
1
√
p
F(k)(k) m1; k → 0
log log n k=1
(5.2)
(5.3)
(5.4)
for all orderings such that mk; 1 and F(k) are of a polynomial rate, i.e. there are c ¿ 0
and ¿ 0 such that
max{mk; 1 ; F(k)}6ck
for all k ∈ S:
Recall
Zj (f) =
X
(f(Xi ) − (f))
X
f(Xi ) − m1; 1 (f)
Nj ¡i6Nj+1
and de ne
Yj (f) =
Nj ¡i6Nj+1
in this section. To prove (5.4), rst we show the following lemma.
Lemma 2. Suppose (5:1); (5:2) and (5:3) hold; then
n
∞
X
1 X
F(k)E
Yj (1{k} ) → 0 as n → ∞:
an
j=1
k=1
Proof. Let
Uj (f) = ((Nj+1 − Nj ) − m1; 1 )(f);
(5.5)
T.-H. Tsai / Stochastic Processes and their Applications 89 (2000) 175–191
185
then
Zj (f) = Yj (f) + Uj (f):
Kolmogorov’s LIL holds for the i.i.d. sequence {Nj+1 −Nj }j¿1 since E(N2 −N1 )2 ¡ ∞,
and that is equivalent to
n
1 X
((Nj+1 − Nj ) − m1; 1 ) → 0 in probability:
an
j=1
We have
1
an
X
n
Uj
→0
j=1
in probability;
since supf∈F (f) = (F) ¡ ∞. Thus
X
n
1
Yj
→ 0 in probability:
an
j=1
(5.6)
We then claim
n
X
1
Yj
E
→ 0:
an
j=1
(5.7)
Then use Corollary 6:12 in Ledoux and Talagrad (1991)
n
X
1
Y
E sup
j
¡ ∞
n an
(5.8)
We have the bounded LIL for {Yj } since the bounded LIL holds for {Zj } and {Uj },
that is
X
n
1
Yj
sup
¡ ∞ a:s:
n an
j=1
j=1
is equivalent to
1
E sup kYn k ¡ ∞:
n an
Since (5.2) holds we have
X
1
F(Xi ) ¡ ∞:
E sup
n an
N ¡i6N
n
(5.9)
n+1
(Here we use that if {Xi } is i.i.d. and real valued, then E X12 ¡ ∞ implies
E[supn n−1=2 |Xn |] ¡ ∞:) Since
X
F(Xi ) + m1; 1 (F);
kYn k6
Nn ¡i6Nn+1
we have (5.9) and thus (5.8) holds. Combine (5.6) and (5.8), we obtain (5.7).
186
T.-H. Tsai / Stochastic Processes and their Applications 89 (2000) 175–191
Pointwise, for all !, we have
X
X
n
n
Yj
= sup
Yj (f)
j=1
f∈F j=1
X
n
X
∞
f(k)
Yj (1{k} )
= sup
f∈F k=1
j=1
n
∞
X
X
P
Yj (1{k} ) :
F(k) 1
¿
n
Yj (1{k} )¿0
j=1
j=1
k=1
Thus,
n
X
∞
1X
X
n
Y
¿
F(k)E
Y
(1
)
E
j
j {k} :
j=1
j=1
2 k=1
From (5.7) we have
n
∞
X
1 X
Yj (1{k} ) → 0:
F(k)E
an
j=1
k=1
Then we need the following lemma.
Lemma 3. There are c′ ; M ¿ 0 such that
n
X
−1=2
Yj (1{k} ) ¿c′ [EY12 (1{k} )]1=2
E
n
j=1
for all n¿M=k ; where k = P( k ¡ 1 | X0 = 1) and i = min{n¿1: Xn = i}:
Proof. By the Marcinkiewicz–Zygmund inequality
1=2
X
n
n
X
Yj2 (1{k} ) ;
Yj (1{k} ) ¿c1 E n−1
n−1=2 E
j=1
j=1
where c1 ¿ 0 is a constant which is independent of the random variables. Note that
1=2
n
n
X
X
1
1
E
Yj2 (1{k} ) ¿[EY12 (1{k} )]1=2 P
Yj2 (1{k} )¿EY12 (1{k} ) :
n
n
j=1
j=1
By the Berry–Esseen theorem
n
X
1 3n−1=2 EY16 (1{k} )
1
:
Yj2 (1{k} )¿EY12 (1{k} ) ¿ −
P
n
2
(EY14 (1{k} ))3=2
j=1
187
T.-H. Tsai / Stochastic Processes and their Applications 89 (2000) 175–191
Let i = min{n¿1: Xn = i}, k = P( k ¡ 1 | X0 = 1) and k = P( 1 ¡ k | X0 = k). Thus
it is enough to show that there is a M ¿ 0 such that
3n−1=2 EY16 (1{k} ) 1
M
for all n¿ :
6
4
k
(EY14 (1{k} ))3=2
P
Denote W (k) = N1 ¡i6N2 1{k} (Xi ) then
(5.10)
Y1 (1{k} ) = W (k) − EW (k)
and
l
EW (k) =
∞
X
ml P(W (k) = m)
m=1
=
∞
X
m=1
ml [k (1 − k )m−1 k ]
= k (1 − k )−1 k
∞
X
m=1
ml (1 − k )m :
(5.11)
Thus by computation EW (k) = k =k ,
EY16 (1{k} ) 6 EW 6 (k)
=
k
(1 + 57(1 − k ) + 302(1 − k )2 + 302(1 − k )3
k6
+ 57(1 − k )4 + (1 − k )5 )
6
720k
k6
and
EY14 (1{k} )
4
k
= E W (k) −
k
k
= 4
k
1 + 11(1 − k ) + 11(1 − k )2 + (1 − k )3 − 4k (1 − k )2
−16k (1 − k ) − 4k + 62k (1 − k ) + 62k − 33k
Since
∞
X
k=1
∞
∞
∞
X
X
k X
k 6
=
EW (k) =
m1; 1 (k) = m1; 1 ¡ ∞;
k
k=1
k=1
k=1
most k are small. Thus
k
EY14 (1{k} )¿ 4
2k
except for nitely many k. Hence there is a M1 ¿ 0 such that
EY16 (1{k} )
6M1 k−1=2 :
(EY14 (1{k} ))3=2
!
:
188
T.-H. Tsai / Stochastic Processes and their Applications 89 (2000) 175–191
Thus
3n−1=2 EY16 (1{k} ) 1
6
4
(EY14 (1{k} ))3=2
for all n¿
144M1
:
k
Proof of (5.4). Since
(EY12 (1{k} ))1=2 ¿ E
¿
√
X
!2 1=2
1{k} (Xi )
N1 ¡i6N2
− m1; 1 (k)
1=2
m1; 1 (k)m1;
k − 2m1; 1 (k);
from (4:6), it is enough to show
(log log n)−1=2
n
X
k=1
F(k)(EY12 (1{k} ))1=2 → 0:
Note that by using (5.11)
2 − k − k
2k
6 2
EY12 (1{k} ) = k
2
k
k
(5.12)
(5.13)
and
mk; 1 ¿
∞
X
j(P( 1 ¿ k | X0 = k)) j−1 P( 1 ¡ k | X0 = k)
∞
X
j(1 − k ) j−1 k
j=1
=
j=1
= k−1 :
(5.14)
Let
U = {k: F(k)(EY12 (1{k} ))1=2 ¿k −2 }:
Use (5.13), (5.14) and assumption (5.5) for all k ∈ U
k −(4 +4)
1 1 k −4
1
¿
:
k ¿ k2 EY12 (1{k} )¿
2
2 m2k; 1 F 2 (k)
2c4
From Lemmas 2 and 3 we have
X
F(k)[EY12 (1{k} )]1=2 → 0:
(log log n)−1=2
k: k ¿M=n
Note
k6
implies
n 1=(4 +4)
2Mc4
k −(4 +4) M
¿ ;
2c4
n
(5.15)
189
T.-H. Tsai / Stochastic Processes and their Applications 89 (2000) 175–191
thus (5.15) implies
(log log n)−1=2
k∈U and
Since
P
k6∈U
X
k6(n=2Mc4 )1=(4 +4)
F(k)(EY12 (1{k} ))1=2 6
[(n=2Mc4 )1=(4 +4) ]
(log log n)
1=2
X
P
k6∈U
F(k)[EY12 (1{k} )]1=2 → 0:
k −2 ¡ ∞,
F(k)(EY12 (1{k} ))1=2 → 0:
k=1
and this is equivalent to (5.12).
6. Proof of the bounded LIL
In view of (3.3) in Lemma 1, the proof of the bounded LIL now follows as indicated
in Sections 4 and 5 for the compact LIL. Here, of course, to show (3.3), one uses
(2.5) rather than (2.3) at various places. The details are straightforward.
7. Comparison with mixing results
Arcones (1995) proved the compact LIL for stationary sequences satisfying absolutely regular mixing conditions under the following two conditions:
∞
X
k=1
k k 2=(p−2) log log k ¡ ∞
(7.1)
and the envelope function F satis es
∞
X
k=1
F(k)(P(X1 = k))1=p ¡ ∞;
(7.2)
where p ¿ 2 and k is absolutely regular mixing coecients.
To show our conditions are less restrictive than Arcones’ conditions for Markov
chains, we present an example such that the conditions E(N2 − N1 )2 ¡ ∞;
2
X
(7.3)
F(Xj ) ¡ ∞
E
N1 ¡j6N2
and
∞
X
k=1
v
u
u
X
u
F(k)tE
N1 ¡j6N2
2
1{k} (Xj ) ¡ ∞
(7.4)
hold (condition (7.4) implies (2.3)), but Arcones’ conditions (7.1) and (7.2) fail.
190
T.-H. Tsai / Stochastic Processes and their Applications 89 (2000) 175–191
Example. Let {Xi } be a stationary Markov chain with transition probability
s
n
; Pn; 1 = 1 − Pn; n+1 for all n¿1 and some s ¿ 1:
Pn; n+1 =
n+1
Let F(k) = k t for some t¿0. We choose s and t such that (7.3) and (7.4) hold, but
(7.1) and (7.2) cannot both hold.
From example 3.2 in Tsai (1997) (k) ≈ k −s ; P(N2 − N1 = n) ≈ n−s−1 ,
2
X
1{k} (Xj ) ≈ k −s
E
N1 ¡j6N2
and k /k 1−s . Thus
∞
X
k=1
and
F(k)((k))1=p ≈
X
∞
X
2
F(Xj ) =
E
N1 ¡j6N2
∞
X
v
u
u
X
u
F(k)tE
k=1
k=1
k t−(s=p) ¡ ∞ only if
∞
X
n=1
N1 ¡j6N2
n
X
kt
k=1
2
!2
1{k} (Xj ) ≈
p6
s
;
t+1
P(N2 − N1 = n) ≈
∞
X
∞
X
n2t−s+1
n=1
k t−(s=2) :
k=1
Hence (7.3) and (7.4) hold if s ¿ 2t + 2. We take s = 4:2 and t = 1. If (7.2) holds,
then p6 21
10 . But
∞
X
k k 2=(p−2) log log k/
k=1
thus (7.1) fails.
∞
X
k=1
k −3:2 k 20 log log k = ∞;
Acknowledgements
The author wishes to express his gratitude to Professor Jim Kuelbs for advice and
guidance.
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