# RIO-TYPE INEQUALITY FOR THE EXPECTATION OF PRODUCTS OF RANDOM VARIABLES

```RIO-TYPE INEQUALITY FOR THE EXPECTATION OF
PRODUCTS OF RANDOM VARIABLES
B. L. S. PRAKASA RAO
We develop an inequality for the expectation of a product of n random variables generalizing the recent work of Dedecker and Doukhan (2003) and the earlier results of Rio
(1993).
1. Introduction
Let (Ω,Ᏺ,P) be a probability space and let (X,Y ) be a bivariate random vector defined
on it. Suppose that E(X 2 ) &lt; ∞ and E(Y 2 ) &lt; ∞. Hoeﬀding proved that
Cov(X,Y ) =
R
2
P(X ≤ x, Y ≤ y) − P(X ≤ x)P(Y ≤ y) dx d y.
(1.1)
In [5], Lehmann gave a simple proof of this identity and used it in his study of some
concepts of dependence. This identity was generalized to functions h(X) and g(Y ) with
E[h2 (X)] &lt; ∞ and E[g 2 (Y )] &lt; ∞ and with finite derivatives h (&middot;) and g (&middot;) by Newman
[6]. Multidimensional versions of these results were proved by Block and Fang [1], Yu
[13], and more recently by Prakasa Rao [7]. Related covariance identities for exponential
and other distributions are given by Prakasa Rao in [9, 10].
Suppose that ᏹ is a sub-σ-algebra of Ᏺ and Y is measurable with respect to ᏹ. Let
σ(X) be the sub-σ-algebra generated by the random variable X. Define
α(ᏹ,X) = sup P(A ∩ B) − P(A)P(B), A ∈ ᏹ, B ∈ σ(X) .
(1.2)
Define
z
QX (u) = inf x : P |X | &gt; x ≤ u ,
GX (s) = inf z :
0
QX (t)dt ≥ s ,
HX,Y (s) = inf t : E |X |I[|Y |&gt;t] ≤ s .
Copyright &copy; 2005 Hindawi Publishing Corporation
Journal of Inequalities and Applications 2005:1 (2005) 7–14
DOI: 10.1155/JIA.2005.7
(1.3)
8
Rio-type inequality
Rio [11] proved that
Cov(X,Y ) ≤ 2
α(ᏹ,X)/2
QY (u)QX (u)du.
0
(1.4)
Related results are given in [12, page 9]. These results were generalized by Bradley [2]
for a strong-mixing process and by Prakasa Rao [8] for rth-order joint cumulant under
rth-order strong mixing. In a recent work, Dedecker and Doukhan [3] proved that
E(XY ) ≤
E(X |ᏹ)1
0
HX,Y (t)dt ≤
E(X |ᏹ)1
QY oGX (t)dt
0
(1.5)
and obtained an improved version of the above inequality. If Xi , 1 ≤ i ≤ n, are positivevalued random variables, it is easy to see that
E X1 X2 &middot; &middot; &middot; Xn ≤
1
0
QX1 (u)QX2 (u) &middot; &middot; &middot; QXn (u)du.
(1.6)
For a proof, see [12, Lemma 2.1, page 35].
We now obtain an improved version of the above inequality following the techniques
of Dedecker and Doukhan [3] and Block and Fang [1].
2. Main result
Let {Xi , 1 ≤ i ≤ n} be a sequence of nonnegative random variables defined on a probability space (Ω,Ᏺ,P). Then the random variable Xi can be represented in the form
Xi =
∞
0
I(xi ,∞) Xi dxi ,
(2.1)
where

1
if Xi &gt; xi ,
I(xi ,∞) Xi = 
0 if Xi ≤ xi .
(2.2)
Hence
E X1 X2 &middot; &middot; &middot; Xn = E X1 Πni=2
=
=
Rn+−1
Rn+−1
∞
0
I(xi ,∞) Xi dxi
E X1 Πni=2 I(xi ,∞) Xi dx2 &middot; &middot; &middot; dxn
(2.3)
E X1 I[Xi &gt;xi , 2≤i≤n] X2 ,...,Xn dx2 &middot; &middot; &middot; dxn
by the Fubini’s theorem, where R+n−1 = {(x2 ,...,xn ) : xi ≥ 0,2 ≤ i ≤ n}. Observe that
E X1 I[Xi &gt;xi , 2≤i≤n] X2 ,...,Xn
≤ min E[X1 ],E X1 I[Xi &gt;xi , 2≤i≤n] X2 ,...,Xn
(2.4)
B. L. S. Prakasa Rao 9
and hence
E X1 X2 &middot; &middot; &middot; Xn ≤
Rn+−1
EX1
0
χ(E[X1 I[Xi &gt;xi , 2≤i≤n] (X2 ,...,Xn )]&gt;u) (u)du dx2 &middot; &middot; &middot; dxn .
(2.5)
Here χA (&middot;) denotes the indicator function of the set A. Let
gX1 x2 ,...,xn = E X1 I[Xi &gt;xi , 2≤i≤n] X2 ,...,Xn ].
(2.6)
Then
E X1 X2 &middot; &middot; &middot; Xn ≤
=
Rn+−1
EX1
0
χ[gX1 (x2 ,...,xn )&gt;u] (u)du dx2 &middot; &middot; &middot; dxn
E(X1 ) [(x2 ,...,xn ):gX1 (x2 ,...,xn )&gt;u]
0
(2.7)
1dx2 &middot; &middot; &middot; dxn du.
Let
HX1 ,X2 ,...,Xn (u) = λ x2 ,...,xn : gX1 x2 ,...,xn &gt; u ,
(2.8)
where λ is the Lebesgue measure on the space R+n−1 . Hence
E X1 X2 &middot; &middot; &middot; Xn ≤
E(X1 )
HX1 ,X2 ,...,Xn (u)du.
0
(2.9)
Observe that
gX1 x2 ,...,xn = E X1 I[Xi &gt;xi , 2≤i≤n] X2 ,...,Xn
≤
E[I[X &gt;x , 2≤i≤n] (X2 ,...,Xn )]
i
i
QX1 (u)du
0
(2.10)
from the Fréchet’s inequality [4]. Here QX1 (&middot;) is the generalized inverse of the function
TX1 (x) = P(X1 &gt; x) as defined earlier. Let
MX1 (y) =
y
0
QX1 (t)dt.
(2.11)
Observe that MX1 (&middot;) is nondecreasing in y. Let GX1 (u) = inf {z : MX1 (z) ≥ u} as defined
earlier. Let
TX2 ,...,Xn x2 ,...,xn = P Xi &gt; xi , 2 ≤ i ≤ n .
(2.12)
Note that
gX1 x2 ,...,xn ≤ MX1 E I[Xi &gt;xi , 2≤i≤n] X2 ,...,Xn
gX1 x2 ,...,xn &gt; u =⇒ MX1 E I[Xi &gt;xi , 2≤i≤n] X2 ,...,Xn
,
&gt;u
=⇒ E I[Xi &gt;xi , 2≤i≤n] X2 ,...,Xn &gt; GX1 (u)
=⇒ P Xi &gt; xi , 2 ≤ i ≤ n &gt; GX1 (u).
(2.13)
10
Rio-type inequality
Hence the set
x2 ,...,xn ∈ R+n−1 : gX1 x2 ,...,xn &gt; u
(2.14)
is contained in the set
x2 ,...,xn ∈ R+n−1 : P Xi &gt; xi , 2 ≤ i ≤ n &gt; GX1 (u) .
(2.15)
In particular, it follows that the Lebesgue measure of the former set is less than or equal
to that of the latter. Let
QX∗2 ,...,Xn GX1 (u)
(2.16)
denote the Lebesgue measure of the set (2.15).
Then
HX1 ,X2 ,...,Xn (u) ≤ QX∗2 ,...,Xn GX1 (u)
(2.17)
for all 0 ≤ u ≤ 1. Hence
E X1 X2 &middot; &middot; &middot; Xn ≤
E(X1 )
0
QX∗2 ,...,Xn GX1 (u) du.
(2.18)
We have proved the following inequality.
Theorem 2.1. Let Xi , 1 ≤ i ≤ n, be nonnegative random variables defined on a probability
space (Ω,Ᏺ,P). Then
E X1 X2 &middot; &middot; &middot; Xn ≤
E(X1 )
0
HX1 ,X2 ,...,Xn (u)du ≤
E(X1 )
0
QX∗2 ,...,Xn oGX1 (u)du,
(2.19)
where the functions H,Q∗ , and G are as defined earlier.
3. Applications
We now suppose that the random variables {Xi , 1 ≤ i ≤ n} are arbitrary but with
EX1 X2 &middot; &middot; &middot; Xn &lt; ∞.
(3.1)
Define
gX1 x2 ,...,xn = E X1 I[|Xi |&gt;xi , 2≤i≤n] X2 ,...,Xn ,
HX1 ,X2 ,...,Xn (u) = λ x2 ,...,xn : gX1 x2 ,...,xn ≤ u ,
TX2 ,...,Xn x2 ,...,xn
= P Xi &gt; xi , 2 ≤ i ≤ n ,
(3.2)
and define MX1 (&middot;),QX1 (&middot;),QX∗2 ,...,Xn , and GX1 accordingly. The following theorem follows
by arguments analogous to those given in Section 2.
B. L. S. Prakasa Rao 11
Theorem 3.1. Let Xi , 1 ≤ i ≤ n, be arbitrary random variables defined on a probability
space (Ω,Ᏺ,P). Then
E X1 X2 &middot; &middot; &middot; Xn ≤
E(|X1 |)
0
HX1 ,X2 ,...,Xn (u)du ≤
E(|X1 |)
0
QX∗2 ,...,Xn oGX1 (u)du,
(3.3)
QX2 oGX1 (u)du
(3.4)
where the functions H,Q∗ , and G are as defined above.
In particular, for n = 2, we have
E X1 X2 ≤
E(|X1 |)
0
HX1 ,X2 (u)du ≤
E(|X1 |)
0
since QX∗ = QX for any univariate random variable X. Furthermore,
GX1 −E(X1 ) (u) ≥ GX1
u
,
2
0≤u≤1
(3.5)
QX2 (u)QX1 (u)du.
(3.6)
(cf. [3]). Hence
E X1 X2 ≤
GX−1 (E(|X1 |)/2)
1
0
Therefore, for any two functions fi (&middot;), i = 1,2, with fi (0) = 0 such that E| f1 (X1) f2 (X2)| &lt;
∞, we obtain that
E f1 X1 f2 X2 ≤
G−f 1(X
1
1)
(E(| f1 (X1 )|)/2)
Q f2 (X2 ) (u)Q f1 (X1 ) (u)du.
0
(3.7)
Applying Theorem 3.1 for the random variables X1 − E(X1 ),X2 ,...,Xn , we get that
E X1 − E X1 X2 &middot; &middot; &middot; Xn ≤
E(|X1 −E(X1 )|)
0
QX∗2 ,...,Xn oGX1 −E(X1 ) (u)du.
(3.8)
But
GX1 −E(X1 ) (u) ≥ GX1
u
,
2
u≥0
(3.9)
(cf. [3]). Hence
E X1 − E X1 X2 &middot; &middot; &middot; Xn ≤
E(|X1 −E(X1 )|)/2
0
QX∗2 ,...,Xn oGX1 (u)du.
Observing that GX1 (&middot;) is the inverse of the function MX1 (y) =
E X1 − E X1 X2 &middot; &middot; &middot; Xn ≤
GX−1 (E(|X1 −E(X1 )|)/2)
Hence we have the following result.
1
0
y
0
(3.10)
QX1 (t)dt, it follows that
QX∗2 ,...,Xn (u)QX1 (u)du.
(3.11)
12
Rio-type inequality
Theorem 3.2. Let Xi , 1 ≤ i ≤ n, be arbitrary random variables defined on a probability
space (Ω,Ᏺ,P) with E|X1 | &lt; ∞ and E|X1 X2 &middot; &middot; &middot; Xn | &lt; ∞. Then (3.11) holds.
Observe that QX∗ = QX for any univariate random variable X. Let n = 2 in Theorem 3.2.
Then QX∗2 = QX2 and the above result reduces to
GX−1 (E(|X1 −E(X1 )|)/2)
E X1 − E X1 X2 ≤
1
0
QX2 (u)QX1 (u)du.
(3.12)
QX2 −E(X2 ) (u)QX1 (u)du.
(3.13)
As a further consequence, we get that
E X1 − E X1 X2 − E X2 ≤
GX−1 (E(|X1 −E(X1 )|)/2)
1
0
Since
QX2 −E(X2 ) ≤ QX2 + EX2 ,
(3.14)
we obtain that
E X1 − E X1 X2 − E X2 ≤
GX−1 (E(|X1 −E(X1 )|)/2)
1
0
QX2 (u)QX1 (u)du + EX2 GX−1 (E(|X1 −E(X1 )|)/2)
1
0
QX1 (u)du.
(3.15)
Let
−1
α X1 ,X2 = max GX1
E X1 − E X1 2
−1
,GX2
E X2 − E X2 2
.
(3.16)
Then it follows that
E X1 − E X1 X2 − E X2 ≤
α(X1 ,X2 )
0
1 QX1 (u)QX2 (u)du +
EX1 2
α(X1 ,X2 )
0
QX1 (u)du + EX2 α(X1 ,X2 )
0
QX2 (u)du .
(3.17)
This inequality is diﬀerent from the inequality in [12, page 9].
Let f1 and f2 be diﬀerentiable functions on R+ with fi (0) = 0. Let Xi , i = 1,2, be nonnegative random variables. Suppose that E[ fi2 (Xi )] &lt; ∞, i = 1,2. It is easy to see that
fi Xi =
Then
E f1 X1 f2 X2
∞
0
fi Xi I(xi ,∞) Xi dxi .
∞ f2 X2 I(x2 ,∞) X2 dx2
= E f1 X1
0
E f1 X1 f2 X2 I(x2 ,∞) (X2 ) dx2
=
R+
(3.18)
(3.19)
B. L. S. Prakasa Rao 13
by the Fubini’s theorem. Observe that
E f1 X1 f2 X2 I[X2 &gt;x2 ] X2
≤ min E f1 X1 f2 X2 ,E f1 X1 f2 X2 I[X2 &gt;x2 ] X2
(3.20)
and hence
E f1 X1 f2 X2 ≤
E[| f1 (X1 ) f2 (X2 )|]
R+
χ(E[| f1 (X1 ) f2 (X2 )|I[X2 &gt;x2 ] (X2 )]&gt;u) (u)du dx2 .
0
(3.21)
Here χA (&middot;) denotes the indicator function of the set A. Let
g f1 (X1 ), f2 (X2 ) x2 = E f1 X1 f2 X2 I[X2 &gt;x2 ], X2 .
(3.22)
Then
E f1 X1 f2 X2 ≤
≤
R+
E[| f1 (X1 ) f2 (X2 )|]
χ([g f1 (X1 ), f (X2 ) (x2 )]&gt;u) (u)du dx2
2
0
E[| f1 (X1 ) f2 (X2 )|]
0
[x2 :g f1 (X1 ), f (X2 ) (x2 )&gt;u]
(3.23)
1dx2 du.
2
Let
H f1 (X1 ), f2 (X2 ) (u) = inf x2 : g f1 (X1 ), f2 (X2 ) (x2 ) ≤ u .
(3.24)
Then it follows that
E f1 X1 f2 X2 ≤
E[| f1 (X1 ) f2 (X2 )|]
0
H f1 (X1 ), f2 (X2 ) (u)du.
(3.25)
An analogous inequality holds by interchanging f1 (X1 ) and f2 (X2 ):
E f1 X1 f2 X2 ≤
E[| f1 (X1 ) f2 (X2 )|]
0
H f1 (X1 ), f2 (X2 ) (u)du.
(3.26)
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