- Email: [email protected]
A note on Slepian’s inequality based on majorization
Statistics and Probability Letters 124 (2017) 132–139
Contents lists available at ScienceDirect
Statistics and Probability Letters journal homepage: www.elsevier.com/locate/stapro
A note on Slepian’s inequality based on majorization Ying Ding a,∗ , Xinsheng Zhang b a
Department of Applied Mathematics, Zhejiang University of Technology, Zhejiang, 310023, PR China
b
Department of Statistics, Fudan University, Shanghai, 200433, PR China
article
info
Article history: Received 22 September 2016 Received in revised form 10 January 2017 Accepted 15 January 2017 Available online 25 January 2017
abstract In this paper, we generalize Slepian’s inequality based on majorization. We obtain some sufficient conditions for ordering results of multivariate normal distributions with different variance vectors and different correlation coefficients. It is shown these conditions are also necessary in some cases. © 2017 Elsevier B.V. All rights reserved.
MSC: 60E15 60E05 Keywords: Slepian’s inequality Majorization Multivariate normal distribution
1. Introduction Among the discussions of stochastic comparisons on multivariate normal distributions, order statistics play prominent roles because of their important behaviors both in mathematics and statistics. For example, for centered normal distributions with the same variance vectors, the well-known Slepian’s inequality (1962) provides the comparisons of the maximum and minimum order statistics based on correlation coefficient comparisons. It is a useful tool in Gaussian process (see Maurer, 2009 and Adler and Taylor, 2007). For independent normal distributions, Huang and Zhang (2009) give some comparisons of the maximum and minimum order statistics based on variance and mean vector majorization, respectively. Then Fang and Zhang (2011) generalize it to a special dependent case in which all the correlation coefficients are the same. Now the question is, for those normal distributions with different variance (mean) vectors and different correlation coefficients, what are their order statistic comparisons? Let X = (X1 , . . . , Xn ) be a Gaussian random vector with density function f (t ) = (2π )−n/2 |6|−1/2 exp − 21 (t − µ)T 6−1
(t − µ) , t ∈ Rn . Here µ is the mean vector, and 6 is the positive definite covariance matrix. We denote it as X ∼ N (µ, 6). Let X1:n = min{X1 , . . . , Xn } and Xn:n = max{X1 , . . . , Xn }. In this paper, we study the stochastic comparisons of X1:n and Xn:n from dependent multivariate normal distributions based on variance (mean) vector majorization and correlation coefficient comparisons. Our results generalize Slepian’s inequality. Explicit definitions of stochastic and majorization orders are given in Section 2. Let D = {(a1 , . . . , an ) : a1 ≥ · · · ≥ an }.
∗
Corresponding author. E-mail addresses: [email protected] (Y. Ding), [email protected] (X. Zhang).
http://dx.doi.org/10.1016/j.spl.2017.01.010 0167-7152/© 2017 Elsevier B.V. All rights reserved.
Y. Ding, X. Zhang / Statistics and Probability Letters 124 (2017) 132–139
133
Theorem 1.1. Suppose X ∼ N (µ, 6) and Y ∼ N (µ, 6∗ ), where µ = (µ, . . . , µ)T ,
σ12 ρ12 σ1 σ2 6= ρ1n σ1 σn
ρ12 σ1 σ2 σ22 ··· ρ2n σ2 σn
ρ1n σ1 σn ρ2n σ2 σn , 2 σn
··· ··· ···
σ1∗2 ∗ ∗ ∗ ρ12 σ1 σ2 6∗ = ∗ ∗ ∗ ρ1n σ1 σn
∗ ∗ ∗ ρ12 σ1 σ2 σ2∗2 ··· ∗ ∗ ∗ ρ2n σ2 σn
··· ··· ···
∗ ∗ ∗ ρ1n σ1 σn ∗ ∗ ∗ ρ2n σ2 σn ∗2 σn
with (1/σ1 , . . . , 1/σn ), (1/σ1 , . . . , 1/σn ) ∈ D. If ∗
∗
(1/σ1 , . . . , 1/σn ) ≥m (1/σ1∗ , . . . , 1/σn∗ ), and
ρjl ≤ ρjl∗ ,
∀j, l = 1, . . . , n,
then X1:n ≤st Y1:n and Xn:n ≥st Yn:n . Theorem 1.2. Suppose X ∼ N (µ, 6) and Y ∼ N (µ∗ , 6), where µ = (µ1 , . . . , µn )T , µ∗ = (µ∗1 , . . . , µ∗n )T , and
σ2 ρ12 σ 2 6= ρ1n σ 2
ρ12 σ 2 σ2 ··· ρ2n σ 2
··· ··· ···
ρ1n σ 2 ρ2n σ 2 σ2
with (µ1 , . . . , µn ), (µ∗1 , . . . , µ∗n ) ∈ D. If
(µ1 , . . . , µn ) ≥m (µ∗1 , . . . , µ∗n ), then X1:n ≤st Y1:n and Xn:n ≥st Yn:n . The rest of this paper is organized as follows. Some notations and terminologies are recalled in Section 2. Section 3 carries out the proofs of Theorems 1.1 and 1.2. Two examples are also given in this section. 2. Preliminary Let {x(1) ≤ x(2) ≤ · · · ≤ x(n) } and {x[1] ≥ x[2] ≥ · · · ≥ x[n] } be the increasing and decreasing arrangement of components from the vector x = (x1 , · · ·, xn ), respectively. Definition 2.1. Given two vectors x = (x1 , · · ·, xn ) and y = (y1 , · · ·, yn ).
j
j n n i=1 y(i) for j = 1, . . . , n − 1, and i=1 x(i) = i=1 y(i) . j j (ii) x is said to weakly supermajorize y (written as x ≥w y) if i=1 x(i) ≤ y for j = 1 , . . . , n. (i) j i =1 j y for j = 1, . . . , n. (iii) x is said to weakly submajorize y (written as x ≥w y) if i=1 x[i] ≥ [ i ] i=1 (i) x is said to majorize y (written as x ≥m y) if
i =1
x(i) ≤
Let X and Y be two random variables with F X and F Y as their survival functions, respectively. Definition 2.2. X is said to be smaller than Y in the usual stochastic order (written as X ≤st Y ) if F X (x) ≤ F Y (x), ∀x. For more details on stochastic orders, one may refer to Marshall et al. (2011) and Müller and Stoyan (2002). 3. Main results Before we prove the main results, we need some notations and lemmas. Denote
ρ11 ρ12 6i = (ρjl ) =
ρ1i
ρ12 ρ22 ··· ρ2i
··· ··· ···
ρ1i ρ2i ρii
as the i × i positive definite covariance matrix with ρjj = 1 and ρjl ∈ (−1, 1) when j ̸= l, Ai = |6i | as the determinant of 6i , and ai,jl as the (j, l) cofactor of ρjl in 6i , j, l = 1, . . . , i, i = 2, . . . , n. We have Ai > 0 and ai,ii = Ai−1 > 0, i = 1, . . . , n.
134
Y. Ding, X. Zhang / Statistics and Probability Letters 124 (2017) 132–139
Also denote 1
0 a2,22
a2,12 A2 a2,22 a 3,13 Pn = A3 a3,33 an,1n
A2 a2,22
A3 a3,33
a3,23
0
···
0
0
0
···
0
0
···
0
,
a3,33
A3 a3,33
··· an,2n
An an,nn a2,22
0
an,3n
An an,nn a2,12 A2 a2,22 1
a3,23
A3 a3,33
An an,nn
An an,nn
0
0
···
0
0
0
···
0
0
···
0
a3,33
an,nn
···
An an,nn
A2 a2,22 a0 3,13 Qn = A3 a3,33 an,1n
an,4n
A3 a3,33
,
··· an,2n
An an,nn
an,3n
An an,nn
an,4n
An an,nn
an,nn
···
An an,nn
An an,nn
and En as the n × n identity matrix. Note the differences between Pn and Qn are only in the first two lines. Lemma 3.1. Given an n × n positive definite covariance matrix 6n . Then Pn 6n PnT = Qn 6n QnT = En . Proof. We prove the conclusions by induction. For clarity, we only prove Pn 6n PnT = En . The case of Qn is similar. When n = 2,
1
P2 62 P2T
= −ρ12 2 1 − ρ12
0
1
1 ρ 12
2 1 − ρ12
−ρ12
1
ρ12 1 0
2 1 − ρ12
1
2 1 − ρ12
= E2 .
When n = k, we assume Pk 6k PkT = Ek . When n = k + 1, we denote
Pk+1 =
where
0 B22
Pk B21
BT21
,
6k+1 =
ak+1,1(k+1)
√
=
Ak+1 ak+1,(k+1)(k+1)
6k
T C12
C12 C22
,
ak+1,k(k+1)
,..., √
T
ak+1,(k+1)(k+1)
be a k × 1 vector, B22 = √
Ak+1 ak+1,(k+1)(k+1)
Ak+1 ak+1,(k+1)(k+1)
be a number,
0 = (0, . . . , 0) be a k × 1 vector, C12 = (ρ1(k+1) , . . . , ρk(k+1) ) be a k × 1 vector, and C22 = 1 be a number. Then T
T
Pk+1 6k+1 PkT+1
=
Pk 6k PkT
Pk (6k BT21 + C12 B22 )
T (B21 6k + B22 C12 )PkT
T (B21 6k + B22 C12 )BT21 + (B21 C12 + B22 C22 )B22
.
T )PkT , it is equal to For the item Pk 6k PkT , we know it is equal to Ek according to the assumption. For the item (B21 6k + B22 C12
B21
B22
6k T C12
PkT .
That is
a(k+1),1(k+1)
Ak+1 a(k+1),(k+1)(k+1)
,
···
,
a(k+1),(k+1)(k+1)
Ak+1 a(k+1),(k+1)(k+1)
1
ρ12
ρ1(k+1)
ρ12 1
··· ···
··· ρ2(k+1)
···
ρ1k ρ2k T T T T Pk = 0 Pk = 0
ρk(k+1)
Y. Ding, X. Zhang / Statistics and Probability Letters 124 (2017) 132–139
135
T T
T since a(k+1),j(k+1) is the cofactor of ρj(k+1) in 6k+1 , j = 1, . . . , k+1. Similarly, Pk (6k BT21 +C12 B22 ) = (B21 6k + B22 C12 )Pk T T T 0 and (B21 6k + B22 C12 )B21 = 0 . For the item (B21 C12 + B22 C22 )B22 , it is equal to
B21
B22
C12 C22
=
B22 .
That is a(k+1),1(k+1)
Ak+1 a(k+1),(k+1)(k+1)
,
a(k+1),(k+1)(k+1)
···,
Ak+1 a(k+1),(k+1)(k+1)
ρ1(k+1) .. . a(k+1),(k+1)(k+1) =1 A a ρk(k+1) k+1 (k+1),(k+1)(k+1) 1
also since a(k+1),j(k+1) is the cofactor of ρj(k+1) in 6k+1 , j = 1, . . . , k + 1. Thus we obtain Pk+1 6k+1 PkT+1 = Ek+1 and the conclusion. 2 Lemma 3.2. (i) Let g (x) = ex /2
+∞ x
2 e−t /2 dt , x ∈ R. Then g (x) is nonincreasing.
2 2 x (ii) Let l(x) = ex /2 −∞ e−t /2 dt , x ∈ R. Then l(x) is nondecreasing.
The above lemma is easy to prove, so we omit the proof here. The following two lemmas are from Marshall et al. (2011) and Slepian (1962), respectively. Lemma 3.3. Let φ be a real-valued function defined and continuous on D = {(a1 , . . . , an ) : a1 ≥ · · · ≥ an } and continuously differentiable on the interior of D. Then x ≥m y on D implies φ(x) ≥ φ(y) if and only if ∇φ(z) = (∂φ(z)/∂ z1 , . . . , ∂φ(z)/∂ zn ) ∈ D for all z in the interior of D. Lemma 3.4. Suppose X ∼ N (µ, 6) and Y ∼ N (µ, 6′ ), where µ = (µ, . . . , µ)T ,
σ12 ρ12 σ1 σ2 6= ρ1n σ1 σn
ρ12 σ1 σ2 σ22 ··· ρ2n σ2 σn
ρ1n σ1 σn ρ2n σ2 σn , σn2
··· ··· ···
σ12 ∗ ρ 12 σ1 σ2 6′ = ∗ ρ1n σ1 σn
∗ ρ12 σ1 σ2 σ22 ··· ∗ ρ2n σ2 σn
··· ··· ···
∗ ρ1n σ1 σn ∗ ρ2n σ2 σn . σn2
If
ρjl ≤ ρjl∗ ,
∀j, l = 1, . . . , n,
then X1:n ≤st Y1:n and Xn:n ≥st Yn:n . Theorem 3.5. Suppose X ∼ N (µ, 6) and Y ∼ N (µ, 6′′ ), where µ = (µ, . . . , µ)T ,
σ12 ρ12 σ1 σ2 6= ρ1n σ1 σn
ρ12 σ1 σ2 σ22 ··· ρ2n σ2 σn
ρ1n σ1 σn ρ2n σ2 σn , σn2
··· ··· ···
σ1∗2 ∗ ∗ ρ 12 σ1 σ2 6′′ = ρ1n σ1∗ σn∗
ρ12 σ1∗ σ2∗ σ2∗2 ··· ρ2n σ2∗ σn∗
··· ··· ···
ρ1n σ1∗ σn∗ ρ2n σ2∗ σn∗ σn∗2
with (1/σ1 , . . . , 1/σn ), (1/σ1∗ , . . . , 1/σn∗ ) ∈ D. If
(1/σ1 , . . . , 1/σn ) ≥m (1/σ1∗ , . . . , 1/σn∗ ), then X1:n ≤st Y1:n and Xn:n ≥st Yn:n . Proof. Without losing generality, we assume µ = (0, . . . , 0)T . We denote λi = 1/σi and λ∗i = 1/σi∗ , i = 1, . . . , n. Then we have (λ1 , . . . , λn ) ≥m (λ∗1 , . . . , λ∗n ) with λ1 ≥ · · · ≥ λn and λ∗1 ≥ · · · ≥ λ∗n due to the assumptions. We prove X1:n ≤st Y1:n . According to Lemma 3.3, we need to prove ∂ F 1:n /∂λ1 ≤ · · · ≤ ∂ F 1:n /∂λn , where F 1:n (x) =
+∞
+∞
···
x
x
+∞
=
+∞
··· λ1 x
λn x
1
1 exp − tT 6−1 t dt1 · · · dtn n / 2 1 / 2 (2π ) |6| 2
1 1/2
(2π )n/2 An
1
exp − t 6n t dt1 · · · dtn , 2 T
−1
x∈R
136
Y. Ding, X. Zhang / Statistics and Probability Letters 124 (2017) 132–139
is the survival function of X1:n and t = (t1 , . . . , tn )T . For clarity, we only prove ∂ F 1:n /∂λ1 ≤ ∂ F 1:n /∂λ2 . The other cases are similar. Set (λ1 x, s2 , . . . , sn )T = Pn (λ1 x, t2 , . . . , tn )T . Then using Lemma 3.1, we have
∂ F 1:n = −x ∂λ1
+∞
λ2 x +∞
+∞
···
λn x +∞
1/2
(2π )n/2 An +∞
···
= −x
1
exp − (λ1 x, t2 , . . . , tn )6n (λ1 x, t2 , . . . , tn ) 2 −1
T
dt2 · · · dtn
1
−1 1 exp − (λ1 x, s2 , . . . , sn )T Pn 6n PnT 2
n/2 A1/2 n
(2π )
mn
m3
m2
1
× (λ1 x, s2 , . . . , sn )T A1n/2 ds2 · · · dsn = −xe
− 21 λ21 x2 − 12 m22
e
g (m2 )
+∞
+∞
···
(2π )n/2
mn
m3
1 2 2 exp − (s3 + · · · + sn ) ds3 · · · dsn .
1
2
Here (λ1 x, m2 , . . . , mn )T = Pn (λ1 x, λ2 x, . . . , λn x)T and g (x) = e
∂ F 1:n 1 2 1 2 2 = −xe− 2 λ2 x e− 2 m1 g (m1 ) ∂λ2
+∞
+∞
x
2 e−t /2 dt. Similarly,
1
1 exp − (s23 + · · · + s2n ) ds3 · · · dsn , (2π )n/2 2
··· mn
m3
x2 /2 +∞
where (m1 , λ2 x, m3 , . . . , mn ) = Qn (λ1 x, λ2 x, . . . , λn x)T . Since λ21 x2 + m22 = λ22 x2 + m21 , to prove ∂ F 1:n /∂λ1 ≤ ∂ F 1:n /∂λ2 equals to prove −xg (m2 ) ≤ −xg (m1 ). Since g is nonincreasing due to the (i) Lemma 3.2, it equals to prove T
a2,12
A2 a2,22
λ1 +
a2,22 A2 a2,22
λ2 ≤
a2,22 A2 a2,22
λ1 +
a2,12
A2 a2,22
λ2 .
That is −ρ12 λ1 + λ2 ≤ λ1 − ρ12 λ2 , and this is satisfied since λ1 ≥ λ2 . Using Lemmas 3.1, (ii) in 3.2 and 3.3 similarly, we obtain Xn:n ≥st Yn:n . We omit the proof here.
Remark 3.6. When ρjl = ρ, j, l = 1, . . . , n, Theorem 3.5 recovers the results in Fang and Zhang (2011). Remark 3.7. Theorem 3.5 tells us, given X = (X1 , . . . , Xn ) and Y = (Y1 , . . . , Yn ) with EX = EY , if the places of VarXi and VarXj in the increasing arrangement of (VarX1 , . . . , VarXn ) are the same as the places of VarYi and VarYj in the increasing arrangement of (VarY1 , . . . , VarYn ), and the correlation coefficients Corr (Xi , Xj ) = Corr (Yi , Yj ), ∀i, j = 1, . . . , n, then
(1/ VarX1 , . . . , 1/ VarXn ) ≥m (1/ VarY1 , . . . , 1/ VarYn ) H⇒ X1:n ≤st Y1:n , Xn:n ≥st Yn:n . When n = 2, Theorem 3.5 can be strengthened as follows. Theorem 3.8. Suppose X ∼ N (µ, 6) and Y ∼ N (µ, 6′′ ), where µ = (µ, µ)T ,
σ12 6= ρσ1 σ2
ρσ1 σ2 , σ22
6 = ′′
σ1∗2
ρσ1∗ σ2∗
ρσ1∗ σ2∗
σ2∗2
.
Then the following statements are equivalent. (i) (1/σ1 , 1/σ2 ) ≥m (1/σ1∗ , 1/σ2∗ ); (ii) X1:2 ≤st Y1:2 ; (iii) X2:2 ≥st Y2:2 . Proof. We denote λi = 1/σi and λ∗i = 1/σi∗ , i = 1, 2. Without generality, we assume µ = (0, 0)T , λ1 ≥ λ2 and λ∗1 ≥ λ∗2 . The parts (i) ⇒ (ii), (iii) have been proved in Theorem 3.5. ∗ ∗ (ii) ⇒ (i) If X1:2 ≤st Y1:2 , then for any x ∈ R, F 1:2 (x) ≤ F 1:2 (x), where F 1:2 (x) and F 1:2 (x) are the survival functions of X1:2 and Y1:2 . Using Taylor’s expansion, we have for any x ∈ R, ∗
∗
F 1:2 (0) + dF 1:2 (0)/dx · x + ◦(x) ≤ F 1:2 (0) + dF 1:2 (0)/dx · x + ◦(x). ∗
√
∗
By some computation, we know F 1:2 (0) = F 1:2 (0) = 1/4, dF 1:2 (0)/dx = −(λ1 + λ2 )/2 2π , and dF 1:2 (0)/dx =
√
−(λ∗1 + λ2 )∗ /2 2π . Then we have λ1 + λ2 = λ∗1 + λ∗2
since x ∈ R is arbitrary. If λ1 ≥ λ∗1 , then we have (i) immediately. If λ1 < λ∗1 , we then have λ∗2 < λ2 ≤ λ1 < λ∗1 which implies (λ1 , λ2 ) ≤m (λ∗1 , λ∗2 ). This leads to X1:2 =st Y1:2 and X2:2 ≤st Y2:2 due to parts (i) ⇒ (ii), (i) ⇒ (iii), and the assumption in (ii). Then for any x ∈ R, P (X1 ∈ (λ1 x, +∞), X2 ∈ (λ2 x, +∞)) = P Y1 ∈ (λ∗1 x, +∞), Y2 ∈ (λ∗2 x, +∞) ,
Y. Ding, X. Zhang / Statistics and Probability Letters 124 (2017) 132–139
137
and
P (X1 ∈ (−∞, λ1 x), X2 ∈ (−∞, λ2 x)) ≥ P Y1 ∈ (−∞, λ∗1 x), Y2 ∈ (−∞, λ∗2 x) ,
where P is the probability function of two dimensional centered normal distribution with variances σ12 = σ22 = 1 and correlation coefficients ρ ∈ (−1, 1). Since
P (X1 ∈ (λ1 x, +∞), X2 ∈ (λ2 x, +∞)) = 1 − P (X1 ∈ (−∞, λ1 x), X2 ∈ R) − P (X1 ∈ R, X2 ∈ (−∞, λ2 x))
+ P (X1 ∈ (−∞, λ1 x), X2 ∈ (−∞, λ2 x)) , it follows for any x ∈ R,
P (X1 ∈ (−∞, λ1 x), X2 ∈ R) + P (X1 ∈ R, X2 ∈ (−∞, λ2 x))
≥ P Y1 ∈ (−∞, λ∗1 x), Y2 ∈ R + P Y1 ∈ R, Y2 ∈ (−∞, λ∗2 x) . That is for any x ∈ R,
Φ (λ1 x) + Φ (λ2 x) ≥ Φ (λ∗1 x) + Φ (λ∗2 x), where Φ (x) is the cumulated distribution function of N (0, 1) on R. Making use of the property Φ (−x) = 1 − Φ (x) and taking x = ±1, we have
Φ (λ1 ) + Φ (λ2 ) = Φ (λ∗1 ) + Φ (λ∗2 ). According to the mean value theorem, we know there exist some ξ and η satisfying 0 < λ∗2 < ξ < λ2 ≤ λ1 < η < λ∗1 such that
φ(ξ )(λ1 − λ∗1 ) = φ(η)(λ∗2 − λ2 ), where φ is the density function of N (0, 1) on R. Since λ1 − λ∗1 = λ∗2 − λ2 , we then have ξ = η. But this is in contradiction to the fact ξ < η. Hence λ1 ≥ λ∗1 and (λ1 , λ2 ) ≥m (λ∗1 , λ∗2 ). The (iii) ⇒ (i) part is similar to the (ii) ⇒ (i) part. We omit it here. Now we prove Theorem 1.1. Proof. Let the random vector Z ∼ N (µ, 6′′ ), where µ = (µ, . . . , µ)T and
σ1∗2
ρ12 σ1 σ2 6′′ = ∗
∗
ρ1n σ1∗ σn∗
ρ12 σ1∗ σ2∗
···
ρ1n σ1∗ σn∗
σ2∗2 ··· ρ2n σ2∗ σn∗
···
ρ2n σ2∗ σn∗ .
σn∗2
···
Then X1:n ≤st Z1:n and Xn:n ≥st Zn:n due to Theorem 3.5. Meanwhile, Lemma 3.4 tells us Z1:n ≤st Y1:n and Zn:n ≥st Yn:n since ρjl ≤ ρjl∗ , ∀j, l = 1, . . . , n. Combining these two facts, we obtain the conclusion. Remark 3.9. When σj = σj∗ and ρjl ≤ ρjl∗ , j, l = 1, . . . , n, Theorem 1.1 reduces to Slepian’s inequality. Theorem 3.10. Suppose X ∼ N (µ, 6) and Y ∼ N (µ∗ , 6), where µ = (µ1 , . . . , µn )T , µ∗ = (µ∗1 , . . . , µ∗n )T , and
σ2 ρ12 σ 2 6= ρ1n σ 2
ρ12 σ 2 σ2 ··· ρ2n σ 2
··· ··· ···
ρ1n σ 2 ρ2n σ 2 σ2
with (µ1 , . . . , µn ), (µ1 , . . . , µ∗n ) ∈ D. Let f : R → R be a twice differentiable function satisfying ∗
(f (µ1 ), . . . , f (µn )) ≥m f (µ∗1 ), . . . , f (µ∗n ) . (i) If f ′ (x)f ′′ (x) ≥ 0, then X1:n ≤st Y1:n ; (ii) If f ′ (x)f ′′ (x) ≤ 0, then Xn:n ≥st Yn:n . Proof. For clarity, we assume σ = 1. (i) We prove X1:n ≤st Y1:n when f ′ ≥ 0 and f ′′ ≥ 0. Since (µ1 , . . . , µn ), (µ∗1 , . . . , µ∗n ) ∈ D, then f (µ1 ) ≥ · · · ≥ f (µn ), f (µ∗1 ) ≥ · · · ≥ f (µ∗n ), and f ′ (µ1 ) ≥ · · · ≥ f ′ (µn ) ≥ 0. According to Lemma 3.3, we need to prove ∂ F 1:n /∂ f (µ1 ) ≤ · · · ≤ ∂ F 1:n /∂ f (µn ), where F 1:n (x) =
+∞
+∞
··· x−µ1
x−µn
1 1/2
(2π )n/2 An
1 T −1 exp − t 6n t dt1 · · · dtn , 2
x∈R
138
Y. Ding, X. Zhang / Statistics and Probability Letters 124 (2017) 132–139
is the survival function of X1:n and t = (t1 , . . . , tn )T . For clarity, we only prove ∂ F 1:n /∂ f (µ1 ) ≤ ∂ F 1:n /∂ f (µ2 ). The other cases are similar. Set (x − µ1 , s2 , . . . , sn )T = Pn (x − µ1 , t2 , . . . , tn )T . Then using Lemma 3.1, we have
∂ F 1:n = ∂µ1
+∞
··· x−µn +∞
x−µ2
+∞
=
+∞
1 1/2
(2π )n/2 An +∞
···
1 −1 T exp − (x − µ1 , t2 , . . . , tn ) 6n (x − µ1 , t2 , . . . , tn ) dt2 · · · dtn 2
1 n/2 A1/2 n
exp −
(2π ) T × (x − µ1 , s2 , . . . , sn ) A1n/2 ds2 · · · dsn
=e
−
pn
p3
p2
(x−µ1 )2 2
e
−
p2 2 2
g (p2 )
+∞
+∞
··· pn
p3
1
−1 (x − µ1 , s2 , . . . , sn )T Pn 6n PnT
2
1
(2π )n/2
1
exp − ( 2
s23
2 where (x − µ1 , p2 , . . . , pn )T = Pn (x − µ1 , x − µ2 , . . . , x − µn )T and g (x) = ex /2 p2 (x−µ2 )2 ∂ F 1:n 1 = e− 2 e− 2 g (p1 ) ∂µ2
+∞
+∞
··· p3
pn
+∞ x
1
+ · · · + ) ds3 · · · dsn , s2n
2 e−t /2 dt. Similarly,
1 exp − (s23 + · · · + s2n ) ds3 · · · dsn , n / 2 (2π ) 2
where (p1 , x − µ2 , p3 · · · , pn ) = Qn (x − µ1 , x − µ2 , . . . , x − µn )T . Since (x − µ1 )2 + p22 = (x − µ2 )2 + p21 , then to 1 1 prove ∂ F 1:n /∂ f (µ1 ) ≤ ∂ F 1:n /∂ f (µ2 ) equals to prove f ′ (µ g (p2 ) ≤ f ′ (µ g (p1 ). Since f ′ (µ1 ) ≥ f ′ (µ2 ) ≥ 0 and g is ) ) T
1
2
a2,22
nonincreasing, it is enough to prove p1 ≤ p2 . This is satisfied since p1 = √
A2 a2,22
√a2,12 (x − µ1 ) + √a2,22 (x − µ2 ), and µ2 ≤ µ1 . A2 a2,22
a (x − µ1 ) + √ 2,12 (x − µ2 ), p2 = A2 a2,22
A2 a2,22
The case of f ′ ≤ 0 is similar and we omit it here. (ii) Using Lemmas 3.1, (ii) in 3.2 and 3.3, we obtain Xn:n ≥st Yn:n similarly. We omit it for clarity.
Remark 3.11. If we take f (x) = x in Theorem 3.10, we have Theorem 1.2 which is the generalization of Fang and Zhang (2011). Remark 3.12. Theorem 1.2 tells us, given X = (X1 , . . . , Xn ) and Y = (Y1 , . . . , Yn ) with the same variances of each components, if the places of EXi and EXj in the increasing arrangement of (EX1 , . . . , EXn ) are the same as the places of EYi and EYj in the increasing arrangement of (EY1 , . . . , EYn ), and the correlation coefficients Corr (Xi , Xj ) = Corr (Yi , Yj ), ∀i, j = 1, . . . , n, then
(EX1 , . . . , EXn ) ≥m (EY1 , . . . , EYn ) H⇒ X1:n ≤st Y1:n , Xn:n ≥st Yn:n . Corollary 3.13. Suppose X ∼ N (µ, 6) and Y ∼ N (µ, 6′ ), where µ = (µ1 , . . . , µn )T , µ = (µ, . . . , µ)T , µ =
σ ρ12 σ 2 6= ρ1n σ 2
2
ρ12 σ σ2 ··· ρ2n σ 2
2
··· ··· ···
ρ1n σ ρ2n σ 2 , 2 σ 2
σ ∗ 2 ρ 12 σ 6′ = ∗ 2 ρ1n σ
2
ρ12 σ σ2 ··· ∗ 2 ρ2n σ ∗
2
··· ··· ···
n
i =1
µi /n,
ρ1n σ ∗ 2 ρ2n σ 2 σ ∗
2
with (µ1 , . . . , µn ) ∈ D. If
ρjl ≤ ρjl∗ ,
∀j, l = 1, . . . , n,
then X1:n ≤st Y1:n and Xn:n ≥st Yn:n . Proof. Let the random vector Z ∼ N (µ, 6). Since (µ1 , . . . , µn ) ≥m (µ, . . . , µ), then we have X1:n ≤st Z1:n and Xn:n ≥st Zn:n according to Theorem 1.2. Meanwhile Lemma 3.4 tells us Z1:n ≤st Y1:n and Zn:n ≥st Yn:n since ρjl ≤ ρjl∗ , ∀j, l = 1, . . . , n. Combining these two facts, we have the conclusion. Corollary 3.14. Suppose X ∼ N (µ, 6) and Y ∼ N (µ∗ , 6), where µ = (µ1 , . . . , µn )T , µ∗ = (µ∗1 , . . . , µ∗n )T , and
σ2 ρ12 σ 2 6= ρ1n σ 2
ρ12 σ 2 σ2 ··· ρ2n σ 2
··· ··· ···
ρ1n σ 2 ρ2n σ 2 σ2
with (µ1 , . . . , µn ), (µ∗1 , . . . , µ∗n ) ∈ D. Let f : R → R be a twice differentiable function.
Y. Ding, X. Zhang / Statistics and Probability Letters 124 (2017) 132–139
(i) When f ′ (x) > 0, f ′′ (x) ≥ 0, if (ii) When f ′ (x) < 0, f ′′ (x) ≤ 0, if (iii) When f ′ (x) > 0, f ′′ (x) ≤ 0, if (iv) When f ′ (x) < 0, f ′′ (x) ≥ 0, if
139
(f (µ1 ), . . . , f (µn )) ≥w f (µ∗1 ), . . . , f (µ∗n ) , then X1:n ≤st Y1:n ; ∗ (f (µ1 ), . . . , f (µn )) ≥w f (µ1 ), . . . , f (µ∗n ) , then X1:n ≤st Y1:n . ∗ (f (µ1 ), . . . , f (µn )) ≥w f (µ1 ), . . . , f (µ∗n ) , then Xn:n ≥st Yn:n ; (f (µ1 ), . . . , f (µn )) ≥w f (µ∗1 ), . . . , f (µ∗n ) , then Xn:n ≥st Yn:n .
Proof. We only prove (i) since the other cases are similar. Because f ′ > 0 and (f (µ1 ), . . . , f (µn )) ≥w (f (µ∗1 ), . . . , f (µ∗n )), there exists a vector µ′ = (µ′1 , µ2 , . . . , µn )T such that (f (µ′1 ), . . . , f (µn )) ≥m (f (µ∗1 ), . . . , f (µ∗n )) and µ′1 ≥ µ1 . Let the random vector Z ∼ N (µ′ , 6). According to Theorem 3.10, we have Z1:n ≤st Y1:n . Meanwhile, we have X1:n ≤st Z1:n since µ′1 ≥ µ1 . Combining these two facts, we have the conclusion. We give two examples which cannot be solved either by Slepian (1962) or by Fang and Zhang (2011). Example 1. Let X ∼ N (µ, 6) and Y ∼ N (µ, 6∗ ), where µ = (1, 1, 1)T ,
6=
1/25 −1/30 −1/15
−1/30 1/9 1/12
−1/15 1/12 ,
6 = ∗
1
1/16 −1/36 −1/16
−1/36 1 /9 1 /8
1/16 1/8 1/4
∗ ∗ ∗ with (σ1 , σ2 , σ3 ) = (1/5, 1/3, 1), (σ1∗ , σ2∗ , σ3∗ ) = (1/4, 1/3, 1/2), (ρ12 , ρ13 , ρ23 ) = (−1/2, −1/3, 1/4), (ρ12 , ρ13 , ρ23 )= ∗ ∗ ∗ ∗ ∗ (−1/3, 1/2, 3/4). It is obvious (1/σ1 , 1/σ2 , 1/σ3 ), (1/σ1 , 1/σ2 , 1/σ3 ) ∈ D, (1/σ1 , 1/σ2 , 1/σ3 ) ≥m (1/σ1 , 1/σ2 , 1/σ3∗ ), and ρjl ≤ ρjl∗ , ∀j, l = 1, 2, 3. Then according to Theorem 1.1, we have X1:3 ≤st Y1:3 and X3:3 ≥st Y3:3 .
Example 2. Let X ∼ N (µ, 6) and Y ∼ N (µ, 6∗ ), where µ = (3, 2, 1)T , µ = (2, 2, 2)T ,
6=
1 −1/3 2/3
−1/3 1 1/2
2/3 1/2 , 1
6 = ∗
1 −1/4 4/5
−1/4 1 2/3
4/5 2/3 1
∗ ∗ ∗ with σ = σ ∗ = 1, (ρ12 , ρ13 , ρ23 ) = (−1/3, 2/3, 1/2), (ρ12 , ρ13 , ρ23 ) = (−1/4, 4/5, 2/3). It is obvious that µT ∈ D, T ∗ T µ ≥m µ , and ρjl ≤ ρjl , ∀j, l = 1, 2, 3. Then according to Corollary 3.13, we have X1:3 ≤st Y1:3 and X3:3 ≥st Y3:3 .
Acknowledgments We are grateful to the editors and referees for careful reading of the manuscript. Their comments are all valuable and useful for improving our paper. First author is supported by National Natural Science Foundation of China (Grant No. 11601483) and Zhejiang Provincial Natural Science Foundation of China (Grant No. LQ16A010008). Second author is supported by National Natural Science Foundation of China (Grant No. 11571080). References Adler, R., Taylor, J., 2007. Random Fields and Geometry. Springer, New York. Fang, L., Zhang, X., 2011. Slepian’s inequality with respect to majorization. Linear Algebra Appl. 434 (4), 1107–1118. Huang, Y., Zhang, X., 2009. On stochastic orders for order statistics from normal distributions. Chin. J. Appl. Probab. Statist. 25 (04), 381–388. Marshall, A., Olkin, I., Arnold, B., 2011. Inequalities: Theory of Majorization and its Applications. Springer-Verlag, New York. Maurer, A., 2009. Transfer bounds for linear feature learning. Mach. Learn. 75 (3), 327–350. Müller, D., Stoyan, D., 2002. Comparison Methods for Stochastic Models and Risks. John Wiley and Sons, New York. Slepian, D., 1962. The one-sided barrier problem for Gaussian processes. Bell Syst. Tech. J. 41, 463–501.
Contents lists available at ScienceDirect
Statistics and Probability Letters journal homepage: www.elsevier.com/locate/stapro
A note on Slepian’s inequality based on majorization Ying Ding a,∗ , Xinsheng Zhang b a
Department of Applied Mathematics, Zhejiang University of Technology, Zhejiang, 310023, PR China
b
Department of Statistics, Fudan University, Shanghai, 200433, PR China
article
info
Article history: Received 22 September 2016 Received in revised form 10 January 2017 Accepted 15 January 2017 Available online 25 January 2017
abstract In this paper, we generalize Slepian’s inequality based on majorization. We obtain some sufficient conditions for ordering results of multivariate normal distributions with different variance vectors and different correlation coefficients. It is shown these conditions are also necessary in some cases. © 2017 Elsevier B.V. All rights reserved.
MSC: 60E15 60E05 Keywords: Slepian’s inequality Majorization Multivariate normal distribution
1. Introduction Among the discussions of stochastic comparisons on multivariate normal distributions, order statistics play prominent roles because of their important behaviors both in mathematics and statistics. For example, for centered normal distributions with the same variance vectors, the well-known Slepian’s inequality (1962) provides the comparisons of the maximum and minimum order statistics based on correlation coefficient comparisons. It is a useful tool in Gaussian process (see Maurer, 2009 and Adler and Taylor, 2007). For independent normal distributions, Huang and Zhang (2009) give some comparisons of the maximum and minimum order statistics based on variance and mean vector majorization, respectively. Then Fang and Zhang (2011) generalize it to a special dependent case in which all the correlation coefficients are the same. Now the question is, for those normal distributions with different variance (mean) vectors and different correlation coefficients, what are their order statistic comparisons? Let X = (X1 , . . . , Xn ) be a Gaussian random vector with density function f (t ) = (2π )−n/2 |6|−1/2 exp − 21 (t − µ)T 6−1
(t − µ) , t ∈ Rn . Here µ is the mean vector, and 6 is the positive definite covariance matrix. We denote it as X ∼ N (µ, 6). Let X1:n = min{X1 , . . . , Xn } and Xn:n = max{X1 , . . . , Xn }. In this paper, we study the stochastic comparisons of X1:n and Xn:n from dependent multivariate normal distributions based on variance (mean) vector majorization and correlation coefficient comparisons. Our results generalize Slepian’s inequality. Explicit definitions of stochastic and majorization orders are given in Section 2. Let D = {(a1 , . . . , an ) : a1 ≥ · · · ≥ an }.
∗
Corresponding author. E-mail addresses: [email protected] (Y. Ding), [email protected] (X. Zhang).
http://dx.doi.org/10.1016/j.spl.2017.01.010 0167-7152/© 2017 Elsevier B.V. All rights reserved.
Y. Ding, X. Zhang / Statistics and Probability Letters 124 (2017) 132–139
133
Theorem 1.1. Suppose X ∼ N (µ, 6) and Y ∼ N (µ, 6∗ ), where µ = (µ, . . . , µ)T ,
σ12 ρ12 σ1 σ2 6= ρ1n σ1 σn
ρ12 σ1 σ2 σ22 ··· ρ2n σ2 σn
ρ1n σ1 σn ρ2n σ2 σn , 2 σn
··· ··· ···
σ1∗2 ∗ ∗ ∗ ρ12 σ1 σ2 6∗ = ∗ ∗ ∗ ρ1n σ1 σn
∗ ∗ ∗ ρ12 σ1 σ2 σ2∗2 ··· ∗ ∗ ∗ ρ2n σ2 σn
··· ··· ···
∗ ∗ ∗ ρ1n σ1 σn ∗ ∗ ∗ ρ2n σ2 σn ∗2 σn
with (1/σ1 , . . . , 1/σn ), (1/σ1 , . . . , 1/σn ) ∈ D. If ∗
∗
(1/σ1 , . . . , 1/σn ) ≥m (1/σ1∗ , . . . , 1/σn∗ ), and
ρjl ≤ ρjl∗ ,
∀j, l = 1, . . . , n,
then X1:n ≤st Y1:n and Xn:n ≥st Yn:n . Theorem 1.2. Suppose X ∼ N (µ, 6) and Y ∼ N (µ∗ , 6), where µ = (µ1 , . . . , µn )T , µ∗ = (µ∗1 , . . . , µ∗n )T , and
σ2 ρ12 σ 2 6= ρ1n σ 2
ρ12 σ 2 σ2 ··· ρ2n σ 2
··· ··· ···
ρ1n σ 2 ρ2n σ 2 σ2
with (µ1 , . . . , µn ), (µ∗1 , . . . , µ∗n ) ∈ D. If
(µ1 , . . . , µn ) ≥m (µ∗1 , . . . , µ∗n ), then X1:n ≤st Y1:n and Xn:n ≥st Yn:n . The rest of this paper is organized as follows. Some notations and terminologies are recalled in Section 2. Section 3 carries out the proofs of Theorems 1.1 and 1.2. Two examples are also given in this section. 2. Preliminary Let {x(1) ≤ x(2) ≤ · · · ≤ x(n) } and {x[1] ≥ x[2] ≥ · · · ≥ x[n] } be the increasing and decreasing arrangement of components from the vector x = (x1 , · · ·, xn ), respectively. Definition 2.1. Given two vectors x = (x1 , · · ·, xn ) and y = (y1 , · · ·, yn ).
j
j n n i=1 y(i) for j = 1, . . . , n − 1, and i=1 x(i) = i=1 y(i) . j j (ii) x is said to weakly supermajorize y (written as x ≥w y) if i=1 x(i) ≤ y for j = 1 , . . . , n. (i) j i =1 j y for j = 1, . . . , n. (iii) x is said to weakly submajorize y (written as x ≥w y) if i=1 x[i] ≥ [ i ] i=1 (i) x is said to majorize y (written as x ≥m y) if
i =1
x(i) ≤
Let X and Y be two random variables with F X and F Y as their survival functions, respectively. Definition 2.2. X is said to be smaller than Y in the usual stochastic order (written as X ≤st Y ) if F X (x) ≤ F Y (x), ∀x. For more details on stochastic orders, one may refer to Marshall et al. (2011) and Müller and Stoyan (2002). 3. Main results Before we prove the main results, we need some notations and lemmas. Denote
ρ11 ρ12 6i = (ρjl ) =
ρ1i
ρ12 ρ22 ··· ρ2i
··· ··· ···
ρ1i ρ2i ρii
as the i × i positive definite covariance matrix with ρjj = 1 and ρjl ∈ (−1, 1) when j ̸= l, Ai = |6i | as the determinant of 6i , and ai,jl as the (j, l) cofactor of ρjl in 6i , j, l = 1, . . . , i, i = 2, . . . , n. We have Ai > 0 and ai,ii = Ai−1 > 0, i = 1, . . . , n.
134
Y. Ding, X. Zhang / Statistics and Probability Letters 124 (2017) 132–139
Also denote 1
0 a2,22
a2,12 A2 a2,22 a 3,13 Pn = A3 a3,33 an,1n
A2 a2,22
A3 a3,33
a3,23
0
···
0
0
0
···
0
0
···
0
,
a3,33
A3 a3,33
··· an,2n
An an,nn a2,22
0
an,3n
An an,nn a2,12 A2 a2,22 1
a3,23
A3 a3,33
An an,nn
An an,nn
0
0
···
0
0
0
···
0
0
···
0
a3,33
an,nn
···
An an,nn
A2 a2,22 a0 3,13 Qn = A3 a3,33 an,1n
an,4n
A3 a3,33
,
··· an,2n
An an,nn
an,3n
An an,nn
an,4n
An an,nn
an,nn
···
An an,nn
An an,nn
and En as the n × n identity matrix. Note the differences between Pn and Qn are only in the first two lines. Lemma 3.1. Given an n × n positive definite covariance matrix 6n . Then Pn 6n PnT = Qn 6n QnT = En . Proof. We prove the conclusions by induction. For clarity, we only prove Pn 6n PnT = En . The case of Qn is similar. When n = 2,
1
P2 62 P2T
= −ρ12 2 1 − ρ12
0
1
1 ρ 12
2 1 − ρ12
−ρ12
1
ρ12 1 0
2 1 − ρ12
1
2 1 − ρ12
= E2 .
When n = k, we assume Pk 6k PkT = Ek . When n = k + 1, we denote
Pk+1 =
where
0 B22
Pk B21
BT21
,
6k+1 =
ak+1,1(k+1)
√
=
Ak+1 ak+1,(k+1)(k+1)
6k
T C12
C12 C22
,
ak+1,k(k+1)
,..., √
T
ak+1,(k+1)(k+1)
be a k × 1 vector, B22 = √
Ak+1 ak+1,(k+1)(k+1)
Ak+1 ak+1,(k+1)(k+1)
be a number,
0 = (0, . . . , 0) be a k × 1 vector, C12 = (ρ1(k+1) , . . . , ρk(k+1) ) be a k × 1 vector, and C22 = 1 be a number. Then T
T
Pk+1 6k+1 PkT+1
=
Pk 6k PkT
Pk (6k BT21 + C12 B22 )
T (B21 6k + B22 C12 )PkT
T (B21 6k + B22 C12 )BT21 + (B21 C12 + B22 C22 )B22
.
T )PkT , it is equal to For the item Pk 6k PkT , we know it is equal to Ek according to the assumption. For the item (B21 6k + B22 C12
B21
B22
6k T C12
PkT .
That is
a(k+1),1(k+1)
Ak+1 a(k+1),(k+1)(k+1)
,
···
,
a(k+1),(k+1)(k+1)
Ak+1 a(k+1),(k+1)(k+1)
1
ρ12
ρ1(k+1)
ρ12 1
··· ···
··· ρ2(k+1)
···
ρ1k ρ2k T T T T Pk = 0 Pk = 0
ρk(k+1)
Y. Ding, X. Zhang / Statistics and Probability Letters 124 (2017) 132–139
135
T T
T since a(k+1),j(k+1) is the cofactor of ρj(k+1) in 6k+1 , j = 1, . . . , k+1. Similarly, Pk (6k BT21 +C12 B22 ) = (B21 6k + B22 C12 )Pk T T T 0 and (B21 6k + B22 C12 )B21 = 0 . For the item (B21 C12 + B22 C22 )B22 , it is equal to
B21
B22
C12 C22
=
B22 .
That is a(k+1),1(k+1)
Ak+1 a(k+1),(k+1)(k+1)
,
a(k+1),(k+1)(k+1)
···,
Ak+1 a(k+1),(k+1)(k+1)
ρ1(k+1) .. . a(k+1),(k+1)(k+1) =1 A a ρk(k+1) k+1 (k+1),(k+1)(k+1) 1
also since a(k+1),j(k+1) is the cofactor of ρj(k+1) in 6k+1 , j = 1, . . . , k + 1. Thus we obtain Pk+1 6k+1 PkT+1 = Ek+1 and the conclusion. 2 Lemma 3.2. (i) Let g (x) = ex /2
+∞ x
2 e−t /2 dt , x ∈ R. Then g (x) is nonincreasing.
2 2 x (ii) Let l(x) = ex /2 −∞ e−t /2 dt , x ∈ R. Then l(x) is nondecreasing.
The above lemma is easy to prove, so we omit the proof here. The following two lemmas are from Marshall et al. (2011) and Slepian (1962), respectively. Lemma 3.3. Let φ be a real-valued function defined and continuous on D = {(a1 , . . . , an ) : a1 ≥ · · · ≥ an } and continuously differentiable on the interior of D. Then x ≥m y on D implies φ(x) ≥ φ(y) if and only if ∇φ(z) = (∂φ(z)/∂ z1 , . . . , ∂φ(z)/∂ zn ) ∈ D for all z in the interior of D. Lemma 3.4. Suppose X ∼ N (µ, 6) and Y ∼ N (µ, 6′ ), where µ = (µ, . . . , µ)T ,
σ12 ρ12 σ1 σ2 6= ρ1n σ1 σn
ρ12 σ1 σ2 σ22 ··· ρ2n σ2 σn
ρ1n σ1 σn ρ2n σ2 σn , σn2
··· ··· ···
σ12 ∗ ρ 12 σ1 σ2 6′ = ∗ ρ1n σ1 σn
∗ ρ12 σ1 σ2 σ22 ··· ∗ ρ2n σ2 σn
··· ··· ···
∗ ρ1n σ1 σn ∗ ρ2n σ2 σn . σn2
If
ρjl ≤ ρjl∗ ,
∀j, l = 1, . . . , n,
then X1:n ≤st Y1:n and Xn:n ≥st Yn:n . Theorem 3.5. Suppose X ∼ N (µ, 6) and Y ∼ N (µ, 6′′ ), where µ = (µ, . . . , µ)T ,
σ12 ρ12 σ1 σ2 6= ρ1n σ1 σn
ρ12 σ1 σ2 σ22 ··· ρ2n σ2 σn
ρ1n σ1 σn ρ2n σ2 σn , σn2
··· ··· ···
σ1∗2 ∗ ∗ ρ 12 σ1 σ2 6′′ = ρ1n σ1∗ σn∗
ρ12 σ1∗ σ2∗ σ2∗2 ··· ρ2n σ2∗ σn∗
··· ··· ···
ρ1n σ1∗ σn∗ ρ2n σ2∗ σn∗ σn∗2
with (1/σ1 , . . . , 1/σn ), (1/σ1∗ , . . . , 1/σn∗ ) ∈ D. If
(1/σ1 , . . . , 1/σn ) ≥m (1/σ1∗ , . . . , 1/σn∗ ), then X1:n ≤st Y1:n and Xn:n ≥st Yn:n . Proof. Without losing generality, we assume µ = (0, . . . , 0)T . We denote λi = 1/σi and λ∗i = 1/σi∗ , i = 1, . . . , n. Then we have (λ1 , . . . , λn ) ≥m (λ∗1 , . . . , λ∗n ) with λ1 ≥ · · · ≥ λn and λ∗1 ≥ · · · ≥ λ∗n due to the assumptions. We prove X1:n ≤st Y1:n . According to Lemma 3.3, we need to prove ∂ F 1:n /∂λ1 ≤ · · · ≤ ∂ F 1:n /∂λn , where F 1:n (x) =
+∞
+∞
···
x
x
+∞
=
+∞
··· λ1 x
λn x
1
1 exp − tT 6−1 t dt1 · · · dtn n / 2 1 / 2 (2π ) |6| 2
1 1/2
(2π )n/2 An
1
exp − t 6n t dt1 · · · dtn , 2 T
−1
x∈R
136
Y. Ding, X. Zhang / Statistics and Probability Letters 124 (2017) 132–139
is the survival function of X1:n and t = (t1 , . . . , tn )T . For clarity, we only prove ∂ F 1:n /∂λ1 ≤ ∂ F 1:n /∂λ2 . The other cases are similar. Set (λ1 x, s2 , . . . , sn )T = Pn (λ1 x, t2 , . . . , tn )T . Then using Lemma 3.1, we have
∂ F 1:n = −x ∂λ1
+∞
λ2 x +∞
+∞
···
λn x +∞
1/2
(2π )n/2 An +∞
···
= −x
1
exp − (λ1 x, t2 , . . . , tn )6n (λ1 x, t2 , . . . , tn ) 2 −1
T
dt2 · · · dtn
1
−1 1 exp − (λ1 x, s2 , . . . , sn )T Pn 6n PnT 2
n/2 A1/2 n
(2π )
mn
m3
m2
1
× (λ1 x, s2 , . . . , sn )T A1n/2 ds2 · · · dsn = −xe
− 21 λ21 x2 − 12 m22
e
g (m2 )
+∞
+∞
···
(2π )n/2
mn
m3
1 2 2 exp − (s3 + · · · + sn ) ds3 · · · dsn .
1
2
Here (λ1 x, m2 , . . . , mn )T = Pn (λ1 x, λ2 x, . . . , λn x)T and g (x) = e
∂ F 1:n 1 2 1 2 2 = −xe− 2 λ2 x e− 2 m1 g (m1 ) ∂λ2
+∞
+∞
x
2 e−t /2 dt. Similarly,
1
1 exp − (s23 + · · · + s2n ) ds3 · · · dsn , (2π )n/2 2
··· mn
m3
x2 /2 +∞
where (m1 , λ2 x, m3 , . . . , mn ) = Qn (λ1 x, λ2 x, . . . , λn x)T . Since λ21 x2 + m22 = λ22 x2 + m21 , to prove ∂ F 1:n /∂λ1 ≤ ∂ F 1:n /∂λ2 equals to prove −xg (m2 ) ≤ −xg (m1 ). Since g is nonincreasing due to the (i) Lemma 3.2, it equals to prove T
a2,12
A2 a2,22
λ1 +
a2,22 A2 a2,22
λ2 ≤
a2,22 A2 a2,22
λ1 +
a2,12
A2 a2,22
λ2 .
That is −ρ12 λ1 + λ2 ≤ λ1 − ρ12 λ2 , and this is satisfied since λ1 ≥ λ2 . Using Lemmas 3.1, (ii) in 3.2 and 3.3 similarly, we obtain Xn:n ≥st Yn:n . We omit the proof here.
Remark 3.6. When ρjl = ρ, j, l = 1, . . . , n, Theorem 3.5 recovers the results in Fang and Zhang (2011). Remark 3.7. Theorem 3.5 tells us, given X = (X1 , . . . , Xn ) and Y = (Y1 , . . . , Yn ) with EX = EY , if the places of VarXi and VarXj in the increasing arrangement of (VarX1 , . . . , VarXn ) are the same as the places of VarYi and VarYj in the increasing arrangement of (VarY1 , . . . , VarYn ), and the correlation coefficients Corr (Xi , Xj ) = Corr (Yi , Yj ), ∀i, j = 1, . . . , n, then
(1/ VarX1 , . . . , 1/ VarXn ) ≥m (1/ VarY1 , . . . , 1/ VarYn ) H⇒ X1:n ≤st Y1:n , Xn:n ≥st Yn:n . When n = 2, Theorem 3.5 can be strengthened as follows. Theorem 3.8. Suppose X ∼ N (µ, 6) and Y ∼ N (µ, 6′′ ), where µ = (µ, µ)T ,
σ12 6= ρσ1 σ2
ρσ1 σ2 , σ22
6 = ′′
σ1∗2
ρσ1∗ σ2∗
ρσ1∗ σ2∗
σ2∗2
.
Then the following statements are equivalent. (i) (1/σ1 , 1/σ2 ) ≥m (1/σ1∗ , 1/σ2∗ ); (ii) X1:2 ≤st Y1:2 ; (iii) X2:2 ≥st Y2:2 . Proof. We denote λi = 1/σi and λ∗i = 1/σi∗ , i = 1, 2. Without generality, we assume µ = (0, 0)T , λ1 ≥ λ2 and λ∗1 ≥ λ∗2 . The parts (i) ⇒ (ii), (iii) have been proved in Theorem 3.5. ∗ ∗ (ii) ⇒ (i) If X1:2 ≤st Y1:2 , then for any x ∈ R, F 1:2 (x) ≤ F 1:2 (x), where F 1:2 (x) and F 1:2 (x) are the survival functions of X1:2 and Y1:2 . Using Taylor’s expansion, we have for any x ∈ R, ∗
∗
F 1:2 (0) + dF 1:2 (0)/dx · x + ◦(x) ≤ F 1:2 (0) + dF 1:2 (0)/dx · x + ◦(x). ∗
√
∗
By some computation, we know F 1:2 (0) = F 1:2 (0) = 1/4, dF 1:2 (0)/dx = −(λ1 + λ2 )/2 2π , and dF 1:2 (0)/dx =
√
−(λ∗1 + λ2 )∗ /2 2π . Then we have λ1 + λ2 = λ∗1 + λ∗2
since x ∈ R is arbitrary. If λ1 ≥ λ∗1 , then we have (i) immediately. If λ1 < λ∗1 , we then have λ∗2 < λ2 ≤ λ1 < λ∗1 which implies (λ1 , λ2 ) ≤m (λ∗1 , λ∗2 ). This leads to X1:2 =st Y1:2 and X2:2 ≤st Y2:2 due to parts (i) ⇒ (ii), (i) ⇒ (iii), and the assumption in (ii). Then for any x ∈ R, P (X1 ∈ (λ1 x, +∞), X2 ∈ (λ2 x, +∞)) = P Y1 ∈ (λ∗1 x, +∞), Y2 ∈ (λ∗2 x, +∞) ,
Y. Ding, X. Zhang / Statistics and Probability Letters 124 (2017) 132–139
137
and
P (X1 ∈ (−∞, λ1 x), X2 ∈ (−∞, λ2 x)) ≥ P Y1 ∈ (−∞, λ∗1 x), Y2 ∈ (−∞, λ∗2 x) ,
where P is the probability function of two dimensional centered normal distribution with variances σ12 = σ22 = 1 and correlation coefficients ρ ∈ (−1, 1). Since
P (X1 ∈ (λ1 x, +∞), X2 ∈ (λ2 x, +∞)) = 1 − P (X1 ∈ (−∞, λ1 x), X2 ∈ R) − P (X1 ∈ R, X2 ∈ (−∞, λ2 x))
+ P (X1 ∈ (−∞, λ1 x), X2 ∈ (−∞, λ2 x)) , it follows for any x ∈ R,
P (X1 ∈ (−∞, λ1 x), X2 ∈ R) + P (X1 ∈ R, X2 ∈ (−∞, λ2 x))
≥ P Y1 ∈ (−∞, λ∗1 x), Y2 ∈ R + P Y1 ∈ R, Y2 ∈ (−∞, λ∗2 x) . That is for any x ∈ R,
Φ (λ1 x) + Φ (λ2 x) ≥ Φ (λ∗1 x) + Φ (λ∗2 x), where Φ (x) is the cumulated distribution function of N (0, 1) on R. Making use of the property Φ (−x) = 1 − Φ (x) and taking x = ±1, we have
Φ (λ1 ) + Φ (λ2 ) = Φ (λ∗1 ) + Φ (λ∗2 ). According to the mean value theorem, we know there exist some ξ and η satisfying 0 < λ∗2 < ξ < λ2 ≤ λ1 < η < λ∗1 such that
φ(ξ )(λ1 − λ∗1 ) = φ(η)(λ∗2 − λ2 ), where φ is the density function of N (0, 1) on R. Since λ1 − λ∗1 = λ∗2 − λ2 , we then have ξ = η. But this is in contradiction to the fact ξ < η. Hence λ1 ≥ λ∗1 and (λ1 , λ2 ) ≥m (λ∗1 , λ∗2 ). The (iii) ⇒ (i) part is similar to the (ii) ⇒ (i) part. We omit it here. Now we prove Theorem 1.1. Proof. Let the random vector Z ∼ N (µ, 6′′ ), where µ = (µ, . . . , µ)T and
σ1∗2
ρ12 σ1 σ2 6′′ = ∗
∗
ρ1n σ1∗ σn∗
ρ12 σ1∗ σ2∗
···
ρ1n σ1∗ σn∗
σ2∗2 ··· ρ2n σ2∗ σn∗
···
ρ2n σ2∗ σn∗ .
σn∗2
···
Then X1:n ≤st Z1:n and Xn:n ≥st Zn:n due to Theorem 3.5. Meanwhile, Lemma 3.4 tells us Z1:n ≤st Y1:n and Zn:n ≥st Yn:n since ρjl ≤ ρjl∗ , ∀j, l = 1, . . . , n. Combining these two facts, we obtain the conclusion. Remark 3.9. When σj = σj∗ and ρjl ≤ ρjl∗ , j, l = 1, . . . , n, Theorem 1.1 reduces to Slepian’s inequality. Theorem 3.10. Suppose X ∼ N (µ, 6) and Y ∼ N (µ∗ , 6), where µ = (µ1 , . . . , µn )T , µ∗ = (µ∗1 , . . . , µ∗n )T , and
σ2 ρ12 σ 2 6= ρ1n σ 2
ρ12 σ 2 σ2 ··· ρ2n σ 2
··· ··· ···
ρ1n σ 2 ρ2n σ 2 σ2
with (µ1 , . . . , µn ), (µ1 , . . . , µ∗n ) ∈ D. Let f : R → R be a twice differentiable function satisfying ∗
(f (µ1 ), . . . , f (µn )) ≥m f (µ∗1 ), . . . , f (µ∗n ) . (i) If f ′ (x)f ′′ (x) ≥ 0, then X1:n ≤st Y1:n ; (ii) If f ′ (x)f ′′ (x) ≤ 0, then Xn:n ≥st Yn:n . Proof. For clarity, we assume σ = 1. (i) We prove X1:n ≤st Y1:n when f ′ ≥ 0 and f ′′ ≥ 0. Since (µ1 , . . . , µn ), (µ∗1 , . . . , µ∗n ) ∈ D, then f (µ1 ) ≥ · · · ≥ f (µn ), f (µ∗1 ) ≥ · · · ≥ f (µ∗n ), and f ′ (µ1 ) ≥ · · · ≥ f ′ (µn ) ≥ 0. According to Lemma 3.3, we need to prove ∂ F 1:n /∂ f (µ1 ) ≤ · · · ≤ ∂ F 1:n /∂ f (µn ), where F 1:n (x) =
+∞
+∞
··· x−µ1
x−µn
1 1/2
(2π )n/2 An
1 T −1 exp − t 6n t dt1 · · · dtn , 2
x∈R
138
Y. Ding, X. Zhang / Statistics and Probability Letters 124 (2017) 132–139
is the survival function of X1:n and t = (t1 , . . . , tn )T . For clarity, we only prove ∂ F 1:n /∂ f (µ1 ) ≤ ∂ F 1:n /∂ f (µ2 ). The other cases are similar. Set (x − µ1 , s2 , . . . , sn )T = Pn (x − µ1 , t2 , . . . , tn )T . Then using Lemma 3.1, we have
∂ F 1:n = ∂µ1
+∞
··· x−µn +∞
x−µ2
+∞
=
+∞
1 1/2
(2π )n/2 An +∞
···
1 −1 T exp − (x − µ1 , t2 , . . . , tn ) 6n (x − µ1 , t2 , . . . , tn ) dt2 · · · dtn 2
1 n/2 A1/2 n
exp −
(2π ) T × (x − µ1 , s2 , . . . , sn ) A1n/2 ds2 · · · dsn
=e
−
pn
p3
p2
(x−µ1 )2 2
e
−
p2 2 2
g (p2 )
+∞
+∞
··· pn
p3
1
−1 (x − µ1 , s2 , . . . , sn )T Pn 6n PnT
2
1
(2π )n/2
1
exp − ( 2
s23
2 where (x − µ1 , p2 , . . . , pn )T = Pn (x − µ1 , x − µ2 , . . . , x − µn )T and g (x) = ex /2 p2 (x−µ2 )2 ∂ F 1:n 1 = e− 2 e− 2 g (p1 ) ∂µ2
+∞
+∞
··· p3
pn
+∞ x
1
+ · · · + ) ds3 · · · dsn , s2n
2 e−t /2 dt. Similarly,
1 exp − (s23 + · · · + s2n ) ds3 · · · dsn , n / 2 (2π ) 2
where (p1 , x − µ2 , p3 · · · , pn ) = Qn (x − µ1 , x − µ2 , . . . , x − µn )T . Since (x − µ1 )2 + p22 = (x − µ2 )2 + p21 , then to 1 1 prove ∂ F 1:n /∂ f (µ1 ) ≤ ∂ F 1:n /∂ f (µ2 ) equals to prove f ′ (µ g (p2 ) ≤ f ′ (µ g (p1 ). Since f ′ (µ1 ) ≥ f ′ (µ2 ) ≥ 0 and g is ) ) T
1
2
a2,22
nonincreasing, it is enough to prove p1 ≤ p2 . This is satisfied since p1 = √
A2 a2,22
√a2,12 (x − µ1 ) + √a2,22 (x − µ2 ), and µ2 ≤ µ1 . A2 a2,22
a (x − µ1 ) + √ 2,12 (x − µ2 ), p2 = A2 a2,22
A2 a2,22
The case of f ′ ≤ 0 is similar and we omit it here. (ii) Using Lemmas 3.1, (ii) in 3.2 and 3.3, we obtain Xn:n ≥st Yn:n similarly. We omit it for clarity.
Remark 3.11. If we take f (x) = x in Theorem 3.10, we have Theorem 1.2 which is the generalization of Fang and Zhang (2011). Remark 3.12. Theorem 1.2 tells us, given X = (X1 , . . . , Xn ) and Y = (Y1 , . . . , Yn ) with the same variances of each components, if the places of EXi and EXj in the increasing arrangement of (EX1 , . . . , EXn ) are the same as the places of EYi and EYj in the increasing arrangement of (EY1 , . . . , EYn ), and the correlation coefficients Corr (Xi , Xj ) = Corr (Yi , Yj ), ∀i, j = 1, . . . , n, then
(EX1 , . . . , EXn ) ≥m (EY1 , . . . , EYn ) H⇒ X1:n ≤st Y1:n , Xn:n ≥st Yn:n . Corollary 3.13. Suppose X ∼ N (µ, 6) and Y ∼ N (µ, 6′ ), where µ = (µ1 , . . . , µn )T , µ = (µ, . . . , µ)T , µ =
σ ρ12 σ 2 6= ρ1n σ 2
2
ρ12 σ σ2 ··· ρ2n σ 2
2
··· ··· ···
ρ1n σ ρ2n σ 2 , 2 σ 2
σ ∗ 2 ρ 12 σ 6′ = ∗ 2 ρ1n σ
2
ρ12 σ σ2 ··· ∗ 2 ρ2n σ ∗
2
··· ··· ···
n
i =1
µi /n,
ρ1n σ ∗ 2 ρ2n σ 2 σ ∗
2
with (µ1 , . . . , µn ) ∈ D. If
ρjl ≤ ρjl∗ ,
∀j, l = 1, . . . , n,
then X1:n ≤st Y1:n and Xn:n ≥st Yn:n . Proof. Let the random vector Z ∼ N (µ, 6). Since (µ1 , . . . , µn ) ≥m (µ, . . . , µ), then we have X1:n ≤st Z1:n and Xn:n ≥st Zn:n according to Theorem 1.2. Meanwhile Lemma 3.4 tells us Z1:n ≤st Y1:n and Zn:n ≥st Yn:n since ρjl ≤ ρjl∗ , ∀j, l = 1, . . . , n. Combining these two facts, we have the conclusion. Corollary 3.14. Suppose X ∼ N (µ, 6) and Y ∼ N (µ∗ , 6), where µ = (µ1 , . . . , µn )T , µ∗ = (µ∗1 , . . . , µ∗n )T , and
σ2 ρ12 σ 2 6= ρ1n σ 2
ρ12 σ 2 σ2 ··· ρ2n σ 2
··· ··· ···
ρ1n σ 2 ρ2n σ 2 σ2
with (µ1 , . . . , µn ), (µ∗1 , . . . , µ∗n ) ∈ D. Let f : R → R be a twice differentiable function.
Y. Ding, X. Zhang / Statistics and Probability Letters 124 (2017) 132–139
(i) When f ′ (x) > 0, f ′′ (x) ≥ 0, if (ii) When f ′ (x) < 0, f ′′ (x) ≤ 0, if (iii) When f ′ (x) > 0, f ′′ (x) ≤ 0, if (iv) When f ′ (x) < 0, f ′′ (x) ≥ 0, if
139
(f (µ1 ), . . . , f (µn )) ≥w f (µ∗1 ), . . . , f (µ∗n ) , then X1:n ≤st Y1:n ; ∗ (f (µ1 ), . . . , f (µn )) ≥w f (µ1 ), . . . , f (µ∗n ) , then X1:n ≤st Y1:n . ∗ (f (µ1 ), . . . , f (µn )) ≥w f (µ1 ), . . . , f (µ∗n ) , then Xn:n ≥st Yn:n ; (f (µ1 ), . . . , f (µn )) ≥w f (µ∗1 ), . . . , f (µ∗n ) , then Xn:n ≥st Yn:n .
Proof. We only prove (i) since the other cases are similar. Because f ′ > 0 and (f (µ1 ), . . . , f (µn )) ≥w (f (µ∗1 ), . . . , f (µ∗n )), there exists a vector µ′ = (µ′1 , µ2 , . . . , µn )T such that (f (µ′1 ), . . . , f (µn )) ≥m (f (µ∗1 ), . . . , f (µ∗n )) and µ′1 ≥ µ1 . Let the random vector Z ∼ N (µ′ , 6). According to Theorem 3.10, we have Z1:n ≤st Y1:n . Meanwhile, we have X1:n ≤st Z1:n since µ′1 ≥ µ1 . Combining these two facts, we have the conclusion. We give two examples which cannot be solved either by Slepian (1962) or by Fang and Zhang (2011). Example 1. Let X ∼ N (µ, 6) and Y ∼ N (µ, 6∗ ), where µ = (1, 1, 1)T ,
6=
1/25 −1/30 −1/15
−1/30 1/9 1/12
−1/15 1/12 ,
6 = ∗
1
1/16 −1/36 −1/16
−1/36 1 /9 1 /8
1/16 1/8 1/4
∗ ∗ ∗ with (σ1 , σ2 , σ3 ) = (1/5, 1/3, 1), (σ1∗ , σ2∗ , σ3∗ ) = (1/4, 1/3, 1/2), (ρ12 , ρ13 , ρ23 ) = (−1/2, −1/3, 1/4), (ρ12 , ρ13 , ρ23 )= ∗ ∗ ∗ ∗ ∗ (−1/3, 1/2, 3/4). It is obvious (1/σ1 , 1/σ2 , 1/σ3 ), (1/σ1 , 1/σ2 , 1/σ3 ) ∈ D, (1/σ1 , 1/σ2 , 1/σ3 ) ≥m (1/σ1 , 1/σ2 , 1/σ3∗ ), and ρjl ≤ ρjl∗ , ∀j, l = 1, 2, 3. Then according to Theorem 1.1, we have X1:3 ≤st Y1:3 and X3:3 ≥st Y3:3 .
Example 2. Let X ∼ N (µ, 6) and Y ∼ N (µ, 6∗ ), where µ = (3, 2, 1)T , µ = (2, 2, 2)T ,
6=
1 −1/3 2/3
−1/3 1 1/2
2/3 1/2 , 1
6 = ∗
1 −1/4 4/5
−1/4 1 2/3
4/5 2/3 1
∗ ∗ ∗ with σ = σ ∗ = 1, (ρ12 , ρ13 , ρ23 ) = (−1/3, 2/3, 1/2), (ρ12 , ρ13 , ρ23 ) = (−1/4, 4/5, 2/3). It is obvious that µT ∈ D, T ∗ T µ ≥m µ , and ρjl ≤ ρjl , ∀j, l = 1, 2, 3. Then according to Corollary 3.13, we have X1:3 ≤st Y1:3 and X3:3 ≥st Y3:3 .
Acknowledgments We are grateful to the editors and referees for careful reading of the manuscript. Their comments are all valuable and useful for improving our paper. First author is supported by National Natural Science Foundation of China (Grant No. 11601483) and Zhejiang Provincial Natural Science Foundation of China (Grant No. LQ16A010008). Second author is supported by National Natural Science Foundation of China (Grant No. 11571080). References Adler, R., Taylor, J., 2007. Random Fields and Geometry. Springer, New York. Fang, L., Zhang, X., 2011. Slepian’s inequality with respect to majorization. Linear Algebra Appl. 434 (4), 1107–1118. Huang, Y., Zhang, X., 2009. On stochastic orders for order statistics from normal distributions. Chin. J. Appl. Probab. Statist. 25 (04), 381–388. Marshall, A., Olkin, I., Arnold, B., 2011. Inequalities: Theory of Majorization and its Applications. Springer-Verlag, New York. Maurer, A., 2009. Transfer bounds for linear feature learning. Mach. Learn. 75 (3), 327–350. Müller, D., Stoyan, D., 2002. Comparison Methods for Stochastic Models and Risks. John Wiley and Sons, New York. Slepian, D., 1962. The one-sided barrier problem for Gaussian processes. Bell Syst. Tech. J. 41, 463–501.