Download ON PROBABILITY AND MOMENT INEQUALITIES FOR

THEORY PROBAB. APPL.
Vol. 45, No. 1
Translated from Russian Journal
ON PROBABILITY AND MOMENT INEQUALITIES FOR
DEPENDENT RANDOM VARIABLES∗
S. V. NAGAEV†
(Translated by M. V. Khatuntseva)
Abstract. The paper obtains the upper estimate for the probability that a norm of a sum of
dependent random variables with values in the Banach space exceeds a given level. This estimate is
principally different from the probability inequalities for sums of dependent random variables known
up to now both by form and method of proof. It contains only one of the countable number of
mixing coefficients. Due to the introduction of a quantile the estimate does not contain moments.
The constants in the estimate are calculated explicitly. As in the case of independent summands,
the moment inequalities are derived with the help of the estimate obtained.
Key words. Banach space, Gaussian random vector, Hilbert space, quantile, uniform mixing
coefficient, Hoffman–Jørgensen inequality, Marcinkiewicz–Zygmund inequality, Euler function
PII. S0040585X97978142
1. Introduction. Statement and discussion of the results. Let Xk , k =
1, . . . , n, be a sequence
a separable Banach space with a
kof random variables with values in n
norm | · |. Set Sk = 1 Xi , Mn = max1kn |Sk |, Skn = k+1 Xi .
Let
P(AB)
k
n
φ(m) = sup − P(B); 1 k n − m, A ∈ F1 , B ∈ Fk+m , P(A) = 0
P(A)
be a uniform mixing coefficient. Here Fjk denotes a σ-algebra generated by a random variable Xl , l = j, . . . , k.
Define
φ+ (m) = sup
P(AB)
P(A)
n
, P(A) = 0 .
− P(B); 1 k n − m, A ∈ F1k , B ∈ Fk+m
Remark that φ+ (n) < φ(n). Let φ+ (1) < 1 and let δ > 0 satisfy the condition δ + φ+ (1) < 1.
Set ρ = δ + φ+ (1). Let α be a number such that P{2Mn > α} < δ. Define
n
Q(r) =
P{|Xj | > r},
1
At =
n
E|Xj |t .
1
Let us formulate an estimate from the above for P{Mn > r} obtained in the paper.
Theorem 1. For any r > α and 0 < ε < 16 ,
(1)
P{Mn > r} <
2
αρ
r
Q
0
rαε2
2u
du
εr
+ ρ−1 1 +
α
(1 + εu/α)s(ε)+1
−s(ε)
,
where s(ε) = − log ρ/ log(1 + ε).
∗ Received by the editors March 10, 1998. This work was supported by Russian Foundation for
Basic Research grant 96-01-01529.
http://www.siam.org/journals/tvp/45-1/97814.html
† Sobolev Mathematical Institute SO RAN, Akad. Koptyug Pr., 4, 630090 Novosibirsk, Russia
([email protected]).
152
153
ON PROBABILITY AND MOMENT INEQUALITIES
Note that the integral of the right-hand side of (1) is a convolution of two functions
given on a multiplicative group of positive real numbers.
The proof of estimate (1) is based on a modification of the known Hoffman–Jørgensen
inequality [6] (see also [4]). Inequality (1) is close to inequality (3 ) of [2] (see also [1,
Theorem 4] and [3, Corollaries 1.5 and 1.6]).
If φ+ (1) = 1 but φ+ (2) < 1, then we can consider subsequences X1 , X3 , . . . , X2k+1 , . . .
and X2 , X4 , . . . , X2k , . . . . For each of them φ+ (1) < 1. If Mn and Mn are defined for the
first and second subsequences, respectively, then
P{Mn > r} < P Mn >
r
2
+ P Mn >
r
.
2
Obviously, this approach can be applied if φ+ (k) = 0, 1 k < m, φ+ (m) < 1.
Corollary 1. For any t > 0 and 0 < ε < 16 such that s(ε) > t,
EMnt < c1 (t) At + c2 (t) αt ,
(2)
where
(3)
c1 (t)
c2 (t)
2t+1
B t + 1, s(ε) − t + 1 ,
ε3t+1 ρ ρ−1 ε−t B t + 1, s(ε) − t t + 1,
B(·, ·) is the Euler function.
Inequality (1) is principally different from the known probability inequalities for sums
of dependent random variables both by form and method of proof. It contains only one
coefficient of the countable number of mixing coefficients and it is one-sided. This inequality
does not contain any moments because of the introduction of a quantile. The constants of
the right-hand side of the inequality are calculated explicitly. All this also concerns moment
inequality (2).
Inequality (2) is universal in the sense that it permits us to include the case 0 < t < 2
and also EXj = 0.
Naturally, the problem of calculating the quantile α arises. We propose one of the
possible approaches to this problem, assuming for simplicity that the Xj take real values.
Let τx be the first exit time of a walk Sk over the level x, x > 0. Define
φ− (k) = sup P(B) −
P(AB)
n
, P(A) = 0 .
; 1 k n − m, A ∈ F1k , B ∈ Fk+m
P(A)
Obviously, {Sn > x} ⊃ ∪n−1
{Skn > −x, τ2x = k} ∪ {τ2x = n}. Let us use the inequality
1
P{Skn > −x, τ2x = k} > P{Skn > −x} − φ− (1) P{τ2x = k}.
Further, P{Skn > −x} > 1 − B 2 (k, n)/x2 , where B 2 (k, n) = E|Skn |2 . As a result we have
(4)
B 2 (0, n)
B 2 (k, n)
>
P
|S
|
>
x
>
P{M
>
2x}
1
−
sup
− φ− (1) .
n
n
x2
x2
0kn−1
So the problem is reduced to estimating the second moments of B 2 (k, n), k = 0, . . . ,
n − 1. ∞ 1/2
If
φ (k) < ∞ and EXj = 0, j = 1, . . . , n, then
1
(5)
2
B (k, n) <
1+2
∞
1
2
1/2
φ
n
(k)
EXi2 .
1
We can also estimate the variances B (k, n) using the results of [10]. We see that we only
need all the mixing coefficients to estimate the quantile α.
154
S. V. NAGAEV
Corollary 2. Let the random variables Xj take real values and EXj = 0. Then for
any t > 2 and 0 < ε < 16 such that s(ε) > t,
(6)
n
EMnt < c1 (t) At + c2 (t) Bnt ,
EXj2 , c2 (t) = c2 (t) (32c(φ)/((1 − φ− (1))(1 − φ+ (1))))t/2 , c(φ) = 1 +
where Bn2 =
1
∞ 1/2
2 1 φ (k), c1 (t) and c2 (t) satisfy conditions (3).
In the case of the arbitrary Banach space we can use the formula presented above to
estimate
the quantile α and as a result arrive at the inequality of type (6) with Bn2 =
n
E|Xj |2 (naturally, we need the restriction EXj = 0). We do not do it now, and will
1
apply a new approach based on reducing the infinite-dimensional case to the one-dimensional
case. This approach was proposed in [5]. However, it is appropriate only for a Hilbert space.
∞
Let φ = {φ(k)}∞
1 and ψ = {ψ(k)}1 be two sequences of uniform mixing coefficients.
We shall assume that φ < ψ if φ(k) ψ(k), k = 1, . . . , ∞, and φ(k) < ψ(k) for at least
one k.
Theorem 2. Let constants c1 (t, φ) and c2 (t, φ) monotonically depend on the sequence
φ = {φ(k)}∞
1 of uniform mixing coefficients, i.e., ci (t, φ ) < ci (t, φ ), if φ < φ , and
for t > 2,
(7)
EMnt < c1 (t, φ) At + c2 (t, φ) Bnt
for any sequence of one-dimensional random variables with EXj = 0.
Then for any sequence of random variables with values in the separable Hilbert space H,
zero means, and the sequence of uniform mixing coefficients ψ = {ψ(k)}∞
1 φ,
EMnt < c1 (t, φ) At + c2 (t, φ) E|Y |t βt−1 ,
t > 2,
where Y is a Gaussian random variable in H with the same covariance operator as Sn
and βt is the absolute moment of order t of the one-dimensional standard Gaussian law.
By the Marcinkiewicz–Zygmund inequality for t 1,
E|Y |t < at E|Y |2
t/2
,
where at is a constant dependent only on t (see, for example, [7, p. 78]). However, the more
exact result is valid.
Theorem 3. Let Y be a Gaussian random vector in H and EY = 0. Then for any
t > 2,
t/2
E|Y |t E|Y |2
βt .
The last inequality can be considered isoperimetric. It shows that a maximum of absolute
moment of order t 1 of the norm of the Gaussian vector with fixed second moment is
achieved on a one-dimensional distribution. In what follows (see section 7) we compare
Theorem 3 with one inequality from the monograph of Bogachev [14].
Corollary 3 follows from Theorems 2 and 3.
Corollary 3. If the conditions of Theorem 2 are satisfied, then
EMnt c1 (t, φ) At + c2 (t, φ) E|Y |2
t/2
,
t > 2.
The standard approach to the derivation of probability inequalities is based on an application of inequalities of the Rosenthal type. In order to derive these inequalities, a straightforward approach is used based on a representation of the even moment of the sum or the
norm
sum of random variables as a sum of mixing moments. In addition, as a rule,
n of the
(E|Xj |t+ε )t/(t+ε) is used instead of At . An exception is the work by Utev [8]. He applies
1
the straightforward approach in the case of the Hilbert space, but he does not replace At
n
by
(E|Xj |t+ε )t/(t+ε) . Since the combinatorial reasonings of Utev are very complicated
1
155
ON PROBABILITY AND MOMENT INEQUALITIES
and intricate, they generate doubts (which are quite valid from our point of view) in the
correctness of his proof (see [9, reference on p. 27]).
The inequalities obtained in this paper permit us, in particular, to improve many results
in estimating the rate of convergence in laws of large numbers for sequences of random
variables with uniform mixing.
A review of probability and moment inequalities for random processes and fields with
mixing and their applications can be found, for example, in [9]. We also mention the paper
by Rio [11], where the Bennet–Hoeffding and the Nagaev–Fuk inequalities are transferred to
a sum of random variables satisfying the Rosenblatt strong mixing condition.
2. Proof of Theorem 1. We need the following lemma.
Lemma. For any nonnegative s, t, and u,
P{Mn > t + s + u} P{Mn > t} P{M n > s} + φ+ (1) +
n
P |Xj | > u ,
1
1
where M n = max0k<jn |Skj |.
Proof. Without loss of generality, we can assume that P{Mn t} > 0. It is not difficult
to see that
(8)
{Mn > t + s + u} ⊂
Mn > t + s + u, max |Xj | u
1jn
max |Xj | > u .
1jn
Furthermore,
(9)
Mn > t + s + u, max |Xj | u
1jn
⊂
n τ = k, max |Skj | s ,
k<jn
1
where τ = inf{k: |Sk | > t}. By the definition of φ+ (·),
(10)
P τ = k, max |Skj | > s
k<jn
P{τ = k} P
+
max |Skj | s + φ (1) .
k<jn
Combining (8)–(10) and taking into account that
P
max |Skj | > s
k<jn
P{M n > s},
n
{τ = k} = {Mn > t},
1
we arrive at the statement of the lemma. This lemma is a generalization of Lemma 4.4 of [6]
(see also [4, Proposition 3]).
By virtue of the lemma, for any t > 0, s > 0, and u > 0,
(11)
P{Mn > t + s + u} P{Mn > t} P Mn >
s
2
+ φ+ (1)
+
n
P |Xj | > u ,
1
since M n 2Mn .
Let us consider a recurrent sequence ym = (1 + ε) ym−1 + α, m 1, y0 = 0. Obviously,
(12)
ym =
(1 + ε)m − 1
α.
ε
Assuming in (11) t = ym−1 , s = α, u = εym−1 , for m > 1, we have P{Mn > ym } < P{Mn >
ym−1 } ρ + Q(εym−1 ). Hence
m−1
(13)
P{Mn > ym } <
Q(εym−k ) ρk−1 + ρm .
1
1 For
real random variables the close result was obtained by Peligrad [12, Lemma 3.1].
156
S. V. NAGAEV
By (12), k = [log(1 + εyk /α)]/[log(1 + ε)]. Consequently,
ρk = 1 +
log ρ/ log(1+ε)
εyk
α
.
Substituting this expression in (13), we obtain the estimate
m−1
P{Mn > ym } <
(14)
1
Q(εym−k )
εym
+ 1+
α
(1 + εyk−1 /α)s(ε)
where s(ε) = − log ρ/ log(1 + ε).
Furthermore,
1
yk − yk−1
Q(εym−k ) <
ρ
(1 + εyk−1 /α)s(ε)
(15)
yk
Q
yk−1
ym+1 ε2 α
2u
−s(ε)
,
du
,
(1 + εu/α)s(ε)
since (1 + ε)k−1 α < yk < (1 + ε)k α/ε and, so, for 0 < ε < 16 ,
αε
αε
yk−1 ym−k
>
>
.
ym+1
(1 + ε)4
2
We note that
1 + εyk /α
ε(1 + ε) yk−1 /α + 2
=
< 2.
1 + εyk−1 /α
εyk−1 /α + 1
yk − yk−1 = εyk−1 + α,
Thus, for yk−1 u yk ,
1
2
.
yk − yk−1
α(1 + εu/α)
(16)
From (14)–(16) it follows that
2
P{Mn > ym } <
αρ
ym−1
Q
0
ym+1 ε2 α
2u
du
εym+1
+ ρ−1 1 +
α
(1 + εu/α)s(ε)+1
−s(ε)
For any y, we choose m so that ym y < ym+1 . As a result we obtain that, for y > α,
2
P{Mn > y} <
αρ
y
Q
0
yε2 α
2u
du
εy
+ ρ−1 1 +
α
(1 + εu/α)s(ε)+1
−s(ε)
3. Proof of Corollary 1. First we note that
∞
(17)
rt−1 P{Mn > r} dr = EMnt .
t
0
Multiplying by rt−1 both parts of (1) and integrating with respect to r, we have
(18)
∞
ρ
rt−1 P{Mn > r} dr α
where
∞
I1 =
rt−1 Q
1
Here we used the relation
0
∞
rt−1 dr
r
g
0
2
I1 I2 +
α
0
αε2
r dr,
2
∞
∞
I2 =
0
r
f (u) du =
u
rt−1 dr
,
(1 + εr/α)s(ε)
rt dr
.
(1 + εr/α)s(ε)+1
∞
∞
f (u)du
0 ∞
=
u
t
rt−1 g
u f (u) du
0
1
∞
r
dr
u
v t−1 g(v) dv.
.
.
ON PROBABILITY AND MOMENT INEQUALITIES
Obviously,
I1 (19)
2
αε2
t
157
At
.
t
Further,
(20)
t+1 α
I2 B t + 1, s(ε) − t + 1 ,
ε
where B(·, ·) is the Euler function (see [13, formula 11]). Similarly,
∞
(21)
0
Finally,
rt−1 dr
(1 + εr/α)s(ε)
α
(22)
t α
B t, s(ε) − t .
ε
rt−1 P{Mn > r} dr 0
αt
.
t
Combining equality (17) and estimates (18)–(22), we obtain the needed result.
4. Proof of Corollary 2. Let x0 be a positive root of the equation
1 − φ− (1)
c(φ) Bn2
=
.
x2
2
If x > x0 , then 1 − c(φ) Bn2 /x2 − φ− (1) > (1 − φ− (1))/2. Hence, by (4) and (5), for x > x0 ,
P{Mn > 2x} <
If
2c(φ)Bn2
.
− φ− (1))
x2 (1
32c(φ)Bn2
< 1 − φ+ (1),
x2 (1 − φ− (1))
(23)
then P{2Mn > x} < 1 − φ+ (1). Assuming in (2) α equal to the minimal of x satisfying
inequality (23), we obtain the needed result.
5. Proof of Theorem 2. Without loss of generality, we can assume that H = l2 .
Let γN be a Gaussian random vector in l2 whose first N coordinates form a standard Gaussian
vector in RN and the others are equal to zero. In [5] it is proved that for any random vector X
with values in H,
t
(24)
E|X|t = lim βt−1 EγN EX (X, γN )
N →∞
if γN and X are independent. Here (·, ·) is a scalar product. We write Eξ f (ξ, η) when the
mathematical expectation is taken with respect to ξ for fixed η.
By our proposition (7), for fixed γN
t/2
n
n
t
t
2
(25) ESn (Sn , γN ) < c1 (t, φ)
EXj (Xj , γN ) + c2 (t, φ)
EXj (Xj , γN )
.
1
By virtue of (24),
(26)
Further,
(27)
1
t
lim EγN EXj (Xj , γN ) = E|Xj |t βt .
N →∞
n
EXj (Xj , γN )2 = ESn (Sn , γN )2 = (Tn γN , γN ),
1
where Tn is a covariance operator Sn .
Without loss of generality, we can assume that the operator Tn is diagonal. On the
other hand,
158
S. V. NAGAEV
(28)
(Tn γN , γN ) = Tn1/2 γN , Tn1/2 γN = |YN |2 ,
1/2
where YN = Tn γN .
Obviously, we can consider YN as a Gaussian variable in RN with the covariance operator Tn (N ) which is a limitation of Tn on RN . It is not difficult to see that
lim E|YN |t = E|Y |t ,
(29)
N →∞
where Y is a Gaussian random variable in l2 with the covariance operator Tn . It follows
from (27)–(29) that
n
lim EγN
(30)
N →∞
2
t/2
EXj (Xj , γN )
= E|Y |t .
1
Comparing (24)–(26) and (30), we obtain that
E|Sn |t = βt−1 lim EγN ESn |(Sn , γN )|t < c1 (t, φ) At + c2 (t, φ) E|Y |t βt−1 .
N →∞
6. Proof of Theorem 3. Without loss of generality, we can assume that H = l2 and
Y = (σ1 ξ1 , σ2 ξ2 , . . . , σn ξn , . . . ), where ξj are independent one-dimensional standard normal
random variables.
Let us consider a finite-dimensional Gaussian random variable Y with values in RN . It
is easy to see that in this case
E|Y |t =
1
(2π)N/2
∞
∞
···
−∞
N
−∞
σj2 x2j
t/2
N
x2j
exp −
1
1
2
dx1 dx2 · · · dxN .
Show that
E|Y |t E|Y |t ,
(31)
where Y = (σ1 ξ1 , σ2 ξ2 , . . . , σN −2 ξN −2 , η), η does not depend on ξj , j = 1, . . . , N − 2, and
2
2
has a normal distribution, Eη 2 = σN
−1 + σN , Eη = 0.
To this end we consider the function
f (u) =
∞
∞
···
−∞
N
−2
−∞
σj2 x2j
ux2N −1
+
2
+ (b −
u) x2N
t/2
N
x2j
exp −
1
1
2
2
2
where b2 = σN
−1 + σN , 0 u b .
Obviously, f (b2 ) = f (0) = (2π)N/2 E|Y |t .
It is easy to show that, for t > 2,
t(t − 2)
d2
f (u) =
du2
4
∞
∞
···
−∞
−∞
N
−2
σj2 x2j
1
+
ux2N −1
2
+ (b −
N
x2j
× exp −
1
2
u) x2N
2
dx1 dx2 · · · dxN ,
t/2−2
(x2N −1 − x2N )2
dx1 dx2 · · · dxN 0
(the integral is finite for 0 u b2 ). Therefore, for t > 2, the function f (u) is convex in
the interval (0, b2 ) and, therefore, achieves its maximum on the ends of the interval. So we
proved inequality (31). Sequentially applying (31), we arrive at
E|Y |t N
σj2
t/2
βt
1
if Y is of the dimension N .
In the general case, it is sufficient to note that E|Y |t = limN →∞ E|YN |t , where YN =
(σ1 ξ1 , σ2 ξ2 , . . . , σN ξN , 0, . . .).
ON PROBABILITY AND MOMENT INEQUALITIES
159
7. Remarks. In [14, p. 207, Corollary 5.63] the inequality
In (f )p (p − 1)n/2 f 2 ,
(32)
p 2,
was obtained. Here f ∈ L2 (γ), γ is the Gaussian measure on a locally convex space, · is
a norm in Lp (γ), and In (·) is a projection on χn , where χn are orthogonal spaces L2 (γ),
described in terms of Hermite polynomials.
It is natural to try to obtain from (32) an inequality of the same type as in Theorem 3,
choosing the respective function f . If γ is a one-dimensional Gaussian distribution, then the
space χn is generated by the Hermite polynomial Hn (x). Assuming f (x) = x and n = 1, we
arrive at
p/2
.
E|Y |p (p − 1)p/2 E|Y |2
This estimate is slightly overstated.
Now let γ be a standard Gaussian distribution in RN . Set
f (x) =
N
σj2 x2j ,
1
and calculate I2 f . It is easy to see that
other hand,
∞
−∞
x = (x1 , x2 , . . . , xN ),
∞
−∞
√
f (x) H2 (xj ) dγ = σj2 (β4 − β2 )/ 2. On the
f (x) H1 (xj ) H1 (xk ) dγ = 0, j = k. Thus
2
1
I2 f = √ (β4 − β2 )
σj H2 (xj ).
2
1
N
Let Y = (σ1 ξ1 , σ2 ξ2 , . . . , σN ξN ), where ξj are independent and normal N (0, 1). In terms
of Y , we write I2 f as
β 4 − β2 2
(33)
|Y | − E|Y |2 .
I2 f = √
2
By (32) and (33),
(34)
p
β4 − β2 1/p 2
√
E
|Y | − E|Y |2 < (p − 1)E1/2 |Y |4 .
2
This inequality is extended by the limit pass on the finite-dimensional case. This is not our
aim, but inequality (34) is itself interesting.
If in order to replace the fourth moment in the right-hand side of (34) with the second
moment, we set
f (x) =
N
σj2 x2j
1/2
,
1
then difficulties in the calculation of I2 f arise.
Acknowledgments. The author is grateful to the referee, who called the author’s
attention to [14] and made remarks improving the original text.
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