Journal of Inequalities in Pure and
Applied Mathematics
A NOTE ON THE MARTINGALE INEQUALITY
YU MIAO
College of Mathematics and Information Science
Henan Normal University
453007 Henan, China.
volume 7, issue 5, article 187,
2006.
Received 26 April, 2006;
accepted 29 September, 2006.
Communicated by: F. Qi
EMail: yumiao728@yahoo.com.cn
Abstract
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2000
Victoria University
ISSN (electronic): 1443-5756
123-06
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Abstract
In this paper, we will establish a martingale inequality, which extends the classic
Hoeffding inequality in some sense. In addition, our inequality improves the
results of Lee and Su [7] (2002) in some cases.
2000 Mathematics Subject Classification: 60G42, 60E15.
Key words: Bounded Martingale; Deviation bound; Hoeffding inequality; Martingale
inequality.
I wish to thank Y.Q. Chen and K. Zheng of Henan Normal University for many suggestions and helpful discussions during our writing of this paper. I am also grateful
to the conscientious anonymous referee for his very serious and valuable report. His
suggestions and comments have largely contributed to Section 3.
A Note on the Martingale
Inequality
Yu Miao
Title Page
Contents
Contents
1
Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3
2
Main Results . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5
3
The Longest Increasing Subsequence . . . . . . . . . . . . . . . . . . . . 12
References
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1.
Introduction
Given a probability space (Ω, F, P) and a filtration F0 = {φ, Ω} ⊂ F1 ⊂ · · · ⊂
Fn = F, an integrable random variable X ∈ L1 (Ω, F, P) can be written as
n
n
X
X
X − EX =
E(X|Fk ) − E(X|Fk−1 ) :=
dk ,
k=1
k=1
where dk is a martingale difference. An early inequality result is the following.
If for any k, there exist constants ak and bk , such that P(dk ∈ [ak , bk ]) = 1, then
for any t > 0, we have the following classic Hoeffding inequality (cf. [5])
2t2
P(|X − EX| ≥ t) ≤ 2 exp − Pn
.
2
k=1 (bk − ak )
De la Peña [2, 3] discussed a general class of exponential inequalities for
bounded martingale difference and ratios by the decoupling theory. Andreas [9]
gave exponential deviation inequalities for one-sided bounded martingale difference sequences. In the case of the length of longest increasing subsequences
and the independence number of sparse random graphs, Lee and Su [7] have
utilised the symmetry argument in the martingale inequality.
For these phenomena of measure concentration, the usual procedure in analysis is via martingale methods, information-theoretic methods and Talagrand’s
induction method (see [6, 8, 10]). In most applications, X is a function of
n independent (possibly vector valued) random variables ξ1 , ξ2 , . . . , ξn and the
0
0
0
filtration is Fk = σ(ξ1 , ξ2 , . . . , ξn ). In this case we let {ξ1 , ξ2 , . . . , ξn } be an
independent copy of {ξ1 , ξ2 , . . . , ξn } and define
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Inequality
Yu Miao
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0
0
0
0
0
∆k = X(ξ1 , ξ2 , . . . , ξk−1 , ξk , ξk+1 , . . . , ξn )−X(ξ1 , ξ2 , . . . , ξk−1 , ξk , ξk+1 , . . . , ξn ).
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Let dk = E(∆k |Fk ). By definition, ∆k is the change in the value of X
resulting from a change only in one coordinate. So, if there exists a constant ck ,
such that |∆k | ≤ ck a.s., then |dk | ≤ ck a.s. and we can apply the Hoeffding
inequality to obtain a tail bound for X. However, in many cases, ck grows too
rapidly and so the Hoeffding inequality does not provide any reasonable tail
bound. For improving the Hoeffding inequality, Lee and Su [7] obtained the
following reasonable tail bound for X.
Theorem 1.1 (See Theorem 1 in Lee and Su [7]). Assume that there exists a
positive and finite constant c such that for all k ≤ n, |∆k | ≤ c a.s. and there
exist 0 < pk < 1 such that for each k ≤ n, P(0 < |∆k | ≤ c|Fk−1 ) ≤ pk a.s.
Then, for every t > 0,
t2
(1.1)
P(|X − EX| ≥ t) ≤ 2 exp − 2 Pn
.
2c
k=1 pk + 2ct/3
In this paper, we will demonstrate that if c Pnt pk is larger, especially if
k=1
2.83
Pnt
≥
2.83e
,
we
can
obtain
a
more
precise
inequality than (1.1). In
c k=1 pk
Section 2, we will give the main results and show our inequalities are more
precise than (1.1) in some cases. In Section 3, we apply our results to the
longest increasing subsequence.
A Note on the Martingale
Inequality
Yu Miao
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2.
Main Results
In this section, we will continue to use the notions of Section 1.
Theorem 2.1. Let X be an integrable random variable defined on a probability
space (Ω, F, P) which is in fact a function of n independent random variables
ξ1 , ξ2 , . . . , ξn . We define Fk , ∆k , dk as in Section 1. Assume that there exist
positive and finite constants ck such that for all k ≤ n,
|∆k | ≤ ck
(2.1)
a.s.
A Note on the Martingale
Inequality
and there exist 0 < pk < 1 such that for each k ≤ n,
P(0 < |∆k | ≤ ck |Fk−1 ) ≤ pk
(2.2)
Yu Miao
a.s.
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Then, for every t > 0,
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2
(2.3)
t
P(|X − EX| ≥ t) ≤ 2 exp − Pn
2 k=1 esck c2k pk
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where s satisfies the equation s = Pn etsck c2 pk . In addition, if there exists a
k=1
k
constant b, such that s ≥ b, we will obtain
(2.4)
−bt/2
P(|X − EX| ≥ t) ≤ 2e
.
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Proof. In fact, we only prove the form P(X − EX ≥ t), and the other form
P(X − EX ≤ −t) is similar. By Jensen’s inequality, for any s > 0, we have
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E(esdk |Fk−1 ) = E(esE(∆k |Fk ) |Fk−1 ) ≤ E(es∆k |Fk−1 ),
a.e.
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From (2.1), (2.2) and the following elementary inequality,
ex ≤ 1 + x +
∀x ∈ R,
|x|2 |x|
e ,
2
we can obtain
s∆k
E(e
|s∆k |2 |s∆k |
e
|Fk−1
|Fk−1 ) ≤ E 1 + s∆k +
2
s2 sck
≤ 1 + e E(∆2k |Fk−1 )
2
s2 sck 2
≤ 1 + e ck pk
2 2
s sck 2
≤ exp
e ck pk
a.e.
2
It is easy to check that
X − EX =
n
X
dk .
k=1
Thus, by Markov’s inequality, for any s > 0,
A Note on the Martingale
Inequality
Yu Miao
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−st
P(X − EX ≥ t) ≤ e
−st
s(X−EX)
Ee
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Pn
Ees k=1 dk
h Pn−1
i
≤ e−st E es k=1 dk E esdn |Fn−1
≤e
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Pn−1
s2 scn 2
≤ exp −st + e cn pn Ees k=1 dk
2
≤ ···
(
)
n
s2 X sck 2
≤ exp −st +
e ck pk .
2 k=1
If we could take
t
,
sck c2 p
k k
k=1 e
P
(2.3) can be shown. In fact, putting fn (s) = nk=1 esck sc2k pk , it is easy to see
that for any n, fn (s) is a continuous function in s, and is nondecreasing on
[0, ∞) with fn (0) = 0. Thus, for any t > 0, there exists only one solution that
satisfies equation s = Pn etsck c2 pk . The remainder of the proof is straightfork=1
k
ward.
(2.5)
s = Pn
Remark 1. It is easy to see that the solution of the equation s = Pn etsck c2 pk
k=1
k
could not be given concretely. However, we can use the formula (2.4), by obtaining a low bound of s in many cases.
Corollary 2.2. Under the conditions of Theorem 1.1, we assume that for all
1 ≤ k ≤ n, ck = c. Then, for every t > 0,
t2
(2.6)
P(|X − EX| ≥ t) ≤ 2 exp − sc 2 Pn
,
2e c
k=1 pk
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where s satisfies the equation s = esc c2 Pt n pk . In addition, if there exists a
k=1
constant b, such that s ≥ b, we obtain
P(|X − EX| ≥ t) ≤ 2e−bt/2 .
(2.7)
Next, we will show that, in some cases, the condition s ≥ b in Corollary 2.2
could be obtained and our results are better than inequality (1.1).
Proposition 2.3. Under the conditions of Corollary 2.2,
A Note on the Martingale
Inequality
(R1 ) Assuming that for any given t > 0,
Yu Miao
(2.8)
t
c
Pn
k=1 pk
≥ 2.83e2.83 ,
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then we have the following inequality
(2.9)
Contents
P(|X − EX| ≥ t) ≤ 2e−2.83t/(2c) ,
and in this case, our bound e−2.83t/(2c) is better than (1.1).
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(R2 ) Conversely, if for any given t > 0,
Close
(2.10)
t
c
Pn
k=1
2.82
pk
≤ 2.82e
,
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then (1.1) is better than our result.
J. Ineq. Pure and Appl. Math. 7(5) Art. 187, 2006
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Proof. By s =
esc c2
t
P
n
k=1
pk
and
c
Pnt
k=1
pk
scesc ≥ 2.83e2.83
≥ 2.83e2.83 , it is easy to see that
and
sc ≥ 2.83.
From Corollary 2.2, (2.9) can be obtained.
Next we will show that our bound e−2.83t/(2c) is better than (1.1). For c Pnt pk
k=1
≥ 3e3 , we know
(2.11)
c
t
Pn
⇔
(1/c − s/3) < s,
k=1 pk
c
t
P
n
2
k=1
⇔ sesc <
⇔ esc <
pk
3c
3c
<
ts
Pn
k=1 pk
k=1
n
X
pk
pk
k=1
ts
Pn
t
Pn
⇔ c(esc − 1)
3c
s=
+ s,
esc c2
pk < t/3,
sesc =
⇔ 2c2 esc
n
X
k=1
pk < 2c2
n
X
k=1
c
sesc =
k=1
pk + 2ct/3,
k=1 pk
sesc =
+ s,
sesc =
+ 1,
t
Pn
c
;
t
P
n
2
k=1
t
c2
Pn
k=1
t
Pn
2
k=1
c2
t
Pn
pk
pk
k=1
A Note on the Martingale
Inequality
pk
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;
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;
pk
sesc =
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c2
t
Pn
k=1 pk
.
Thus, by comparing (2.6) and (1.1), the proof of (R1 ) is given under the condition c Pnt pk ≥ 3e3 . To proving remainders, by (2.11), we only show the
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following relations

 c Pnt pk (1/c − s/3) ≥ s, if 2.83e2.83 ≤ c Pnt pk < 3e3 ;
k=1
k=1
(2.12)
 Pt
(1/c − s/3) ≤ s, if c Pnt pk < 2.82e2.82 .
c n pk
k=1
Since s =
(2.13)
esc c2
t
P
n
k=1
k=1
pk
, (2.12) is equivalent to the following relations

cesc (1/c − s/3) ≥ 1, if 2.83e2.83 ≤
cesc (1/c − s/3) ≤ 1, if
c
Pnt
k=1
c
Pnt
k=1
2.82
pk
< 2.82e
< 3e3 ;
pk
.
Letting f (s) = cesc (1/c−s/3)−1 and sc = x, we have f (x) = ex (1−x/3)−1.
It is not difficult to check that f (x) is an increasing function in [0, 2.82] and a
decreasing function in [2.83, ∞) (or see Figure 1). And f (0) = 0, f (x0 ) = 0,
where x0 ∈ [2.82, 2.83]. The rest is obvious.
Remark 2. In the above proposition, though the bounds 2.82e2.82 and 2.83e2.83
are coarser, we can easily determine which inequalities are a little sharper by
using these bounds.
Remark 3. The above results show that for given n (resp. t), our inequality is
more precise in the case of sufficiently large t (resp. small n). However, in many
cases, we need computer power to use our inequality. For example, assuming
P t
= B, where B is given, then we often need to control the solution of
c n
k=1 pk
the equation xex = B.
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5
0
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Yu Miao
−5
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−10
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−20
−1
−0.5
0
0.5
1
1.5
Figure 1
2
2.5
3
3.5
4
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3.
The Longest Increasing Subsequence
In this section, we discuss the longest increasing subsequence as in Lee and
Su [7] (2002) and show our results are little sharper. Consider the symmetric
group Sn of permutations π on the number 1, 2, . . . , n, equipped with the uniform probability measure. Given a permutation π = (π(1), π(2), . . . , π(n)), an
increasing subsequence i1 , i2 , . . . , ik is a subsequence of 1, 2, . . . , n such that
i1 < i2 < · · · < ik , π(i1 ) < π(i2 ) < · · · < π(ik ).
We write Ln (π) for the length of longest increasing subsequences of π.
Let Ui = (Xi , Yi ), i = 1, 2, . . . , n, be a sequence of i.i.d. uniform sample on
the unit square [0, 1]2 . Ui1 , Ui2 , . . . , Uik is called a monotone increasing chain
of height k if
Xij < Xij +1 , Yij < Yij +1 for j = 1, 2, . . . , k − 1.
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Yu Miao
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Define Ln (U ) to be the maximum height of the chains in the sample U1 , U2 , . . . , Un .
By Hammersley [4] (1972) and Aldous and Diaconis [1] (1999), the following facts are known:
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(F1 ) Ln (π) has the same distribution as Ln (U ).
(F2 )
Ln (π)
√
n
0
Close
→ 2, in probability and in mean.
0
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0
Let {U1 , U2 , . . . , Un } be an independent copy of {U1 , U2 , . . . , Un }. It is easy
to see that, letting
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0
0
0
0
0
∆k = Ln (U1 , . . . , Uk−1 , Uk , Uk+1 , . . . , Un )−Ln (U1 , . . . , Uk−1 , Uk , Uk+1 , . . . , Un )
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∆k takes values only +1, 0, −1. Moreover, since E(∆k |Fk−1 ), where Fk−1 =
σ(U1 , U2 , . . . , Uk−1 ), we have
P(∆k = +1|Fk−1 ) = P(∆k = −1|Fk−1 ).
Letting pk = 2ELn−k+1 (Uk , Uk+1 , . . . , Un )/(n − k + 1), from Lee and Su [7]
(2002), there is the following fact:
(F3 ) P(∆k = +1|Fk−1 ) ≤ pk /2.
For the longest increasing subsequence, we have the following result.
Theorem 3.1. There exists a constant δ < 1/2, such that for all sufficiently
large n and any r > 0,
δrn log n
(3.1)
P(|Ln (U ) − ELn (U )| > rn) ≤ 2 exp −
.
2
Proof. For any r > 0 and sufficiently large n, s in Corollary 2.2 needs to satisfy
the equation s = es Prn
. Since
n
pk
k=1
1
√ ELn (U ) → 2 as
n
n → ∞,
n
n → ∞,
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we have
1 X ELk (U )
√
→ 4 as
k
n k=1
A Note on the Martingale
Inequality
i.e.,
n
−1/2
n
X
Close
pk → 4.
k=1
By the equation s = es Prn
, we know that for sufficiently large n, ses =
n
k=1 pk
√
O( n). Thus there exists a constant δ < 1/2, such that ses > eδ log n δ log n,
i.e., s ≥ δ log n. By Corollary 2.2, we have the result.
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Remark 4. By Proposition 2.3, we know our results are sharper than the ones
in Lee and Su [7] to a certainty. Lee and Su [7] gave the following result by an
application of inequality (1.1).
Theorem LS. Given any ε > 0, for all sufficiently large n and any t > 0,
t2
√
(3.2)
P (|Ln (π) − ELn (π)| ≥ t) ≤ 2 −
.
(16 + ε) n + 2t/3
A Note on the Martingale
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Yu Miao
Here if taking t = rn, then P (|Ln (π) − ELn (π)| ≥ rn) ≤ O(e−n ), which is
coarser than (3.1)
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References
[1] D.J. ALDOUS AND P. DIACONIS, Longest increasing subsequences:
From patience sorting to the Baik-Deift-Johansson theorem, Bull. Amer.
Math. Soc., 36 (1999), 413–432.
[2] V.H. DE LA PEÑA, A bound on the moment generating function of a
sum of dependent variables with an application to simple random sampling
without replacement, Ann. Inst. H. Poincaré Probab. Staticst., 30 (1994),
197–211.
A Note on the Martingale
Inequality
[3] V.H. DE LA PEÑA, A general class of exponential inequalities for martingales and ratios, Ann. Probab., 27 (1999), 537–564.
[4] J.M. HAMMERSLEY, A few seedlings of research, Proceedings of the
Sixth Berkeley Symposium on Mathematical and Statistical Probability,
University of California Press, Berkeley, CA, (1972), 345–394.
[5] W. HOEFFDING, Probability inequalities for sums of bounded random
variables, J. Amer. Statist. Assoc., 58 (1963), 12–20.
[6] M. LEDOUX, Isoperimetry and gaussian analysis, Lectures on Probability Theory and Statistics, Ecole d’Et’e de Probabilités de St-Flour XXIV1994, P. Bernard (Editor), LNM 1648, Springer-Verlag, Berlin, 1996, 165–
294.
[7] S. LEE AND Z.G. SU, The symmetry in the martingale inequality, Stat.
Probab. Letters, 56 (2002), 83–91.
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[8] P. MASSART, About the constants in Talagrand’s concentration inequalities for empirical processes, Ann. Probab., 28(2) (2000), 863–884.
[9] A. MAURER, A bound on the deviation probability for sums of nonnegative random variables, J. Inequal. Pure. Appl. Math., 4(1) (2003), Art.
15. [ONLINE: http://jipam.vu.edu.au/article.php?sid=
251].
[10] M. TALAGRAND, Concentration of measure and isoperimetric inequalities in product spaces, Publ. Math. del’IHES, 81 (1995), 73–205.
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