The Mann-Whitney U-statistic for α-dependent sequences

arXiv:1611.06828v1 [] 21 Nov 2016

Jérôme Dedeckera and Guillaume Saulièreb
November 22, 2016
a

Université Paris Descartes, Sorbonne Paris Cité, Laboratoire MAP5 (UMR 8145).
Email: jerome.dedecker@parisdescartes.fr
b

Université Paris Sud, Institut de Recherche bioMédicale et d’Epidémiologie du Sport (IRMES),

Institut National du Sport de l’Expertise et de la Performance (INSEP).
Email: Guillaume.SAULIERE@insep.fr
Abstract
We give the asymptotic behavior of the Mann-Whitney U-statistic for two independent stationary
sequences. The result applies to a large class of short-range dependent sequences, including many
non-mixing processes in the sense of Rosenblatt [15]. We also give some partial results in the longrange dependent case, and we investigate other related questions. Based on the theoretical results, we
propose some simple corrections of the usual tests for stochastic domination; next we simulate different
(non-mixing) stationary processes to see that the corrected tests perform well.

1

Introduction and main result

Let (Xi )i∈Z and (Yi )i∈Z be two independent stationary sequences of real-valued random variables.
Let m = mn be a sequence of positive integers such that mn → ∞ as n → ∞. The MannWhitney U-statistic may be defined as
n

Un = Un,m =

m

1 XX
1X <Y .
nm i=1 j=1 i j

We shall also use the notations π = P(X1 < Y1 ), HY (t) = P(Y1 > t) and GX (t) = P(X1 < t).
It is well known that, if (Xi )i∈Z and (Yi )i∈Z are two independent sequences of independent
and identically (iid) random variables, then Un may be used to test π = 1/2 against π 6= 1/2
(or π > 1/2, π < 1/2). When π > 1/2 we shall say that Y stochastically dominates X in a
1

weak sense. This property of weak domination is true for instance if Y stochastically dominates
X, that is if HY (t) ≥ HX (t) for any real t, with a strict inequality for some t0 . But it also
holds in many other situations, for instance if X and Y are two Gaussian random variables with
E(X) < E(Y ), whatever the variances of X and Y .
In this paper, we study the asymptotic behavior of Un under some conditions on the αdependence coefficients of the sequences (Xi )i∈Z and (Yi )i∈Z . Let us recall the definition of these
coefficients.
Definition 1.1. Let (Xi )i∈Z be a strictly stationary sequence, and let F0 = σ(Xk , k ≤ 0). For
any non-negative integer n, let
α1,X (n) = sup kE (1Xn ≤x |F0 ) − F (x)k1 ,
x∈R

where F is the distribution function of X0 .
We shall also always require that the sequences are 2-ergodic in the following sense.
Definition 1.2. A strictly stationary sequence (Xi )i∈Z is 2-ergodic if, for any bounded mesurable
function f from R2 to R, and any non-negative integer k
n

1X
f (Xi , Xi+k ) = E(f (X0 , Xk )) almost surely.
lim
n→∞ n
i=1

(1.1)

Note that the almost sure limit in (1.1) always exists, by the ergodic theorem. The property of
2-ergodicity means only that this limit is constant. This property is weaker than usual ergodicity,
which would implies the same property for any bounded function f from R` to R, ` ∈ N − {0}.
Our main result is the following theorem.
Theorem 1.1. Let (Xi )i∈Z and (Yi )i∈Z be two independent stationary 2-ergodic sequences of
real-valued random variables. Assume that
∞
∞
X
X
α1,X (k) < ∞ and
α1,Y (k) < ∞ .
(1.2)
k=1

k=1

Let (mn ) be a sequence of positive integers such that
n
lim
= c , for some c ∈ [0, ∞).
n→∞ mn
√
Then the random variables n(Un − π) converge in distribution to N (0, V ), with
V = Var(HY (X0 )) + 2

∞
X

(1.3)

Cov(HY (X0 ), HY (Xk ))

k=1

+ c Var(GX (Y0 )) + 2

∞
X
k=1

2

!
Cov(GX (Y0 ), GX (Yk ))

. (1.4)

Note that the result of Theorem 1.1 has been obtained by Serfling [16], provided that the
more restrictive condition


X
1
α(k) < ∞ and α(k) = O
(1.5)
k
log
k
k≥0
is satisfied by the two sequences (Xi )i∈Z and (Yi )i∈Z , where the α(k)’s are the usual α-mixing
coefficients of Rosenblatt [15].
The notion of α-dependence that we consider in the present paper is much weaker than αmixing in the sense of Rosenblatt [15]. Some properties of these α-dependence coefficients are
stated in the book by Rio [14], and many examples are given in the two papers [7], [8]. Let us
recall one of these examples. Let
X
Xi =
ak εi−k
(1.6)
k≥0

where (ak )k≥0 ∈ `2 , and the sequence (εi )i∈Z is a sequence of independent and identically distributed real-valued random variables, with mean zero and finite variance. If the cumulative
distribution function F of X0 is γ-Hölder for some γ ∈ (0, 1], then
γ
! γ+2

X

α1,X (n) ≤ C

a2k

k≥n

for some positive constant C (this follows from the computations in Section 6.1 of [8]). In
particular, if ak is geometrically decreasing, then so is α1,X (n) (whatever the index γ).
Note that, without extra assumptions on the distribution of ε0 , such linear processes have no
reasons to be mixing in the sense of Rosenblatt. For instance, it is well known that the linear
process
X εi−k
Xi =
, where P(ε1 = −1/2) = P(ε1 = 1/2) = 1/2,
2k+1
k≥0
is not α-mixing (see for instance [1]). By contrast, for this particular example, α1,X (n) is
geometrically decreasing.
To conclude, let us also mention the paper [5] where it is proved that the coefficients α1,X (n)
can be computed for a large class of dynamical systems on the unit interval (see Subsection 6.2
of the present paper for more details).
Remark 1.1. Other types of dependence are considered in the papers by Dewan and Prakasa
Rao [11] and Dehling and Fried [9]. The paper [11] deals with the U-statistics Un for two
independent samples of associated random variables. The second paper [9] deals with general
U-statistics of functions of β-mixing sequences, for either two independent samples, or for two
3

adjacent samples from the same time series (see Subsection 5.2 for an extension of our results
to this more difficult context). The class of unctions of β-mixing sequences contains also many
examples, with a large intersection with the class of α-dependent sequences. The advantage of
our approach is that it leads to conditions that are optimal in a precise sense (see Section 3). On
another hand, it seems more difficult to deal with general U-statistics in our context (some kind
of regularity is needed, for instance we could certainly deal with U-statistics based on functions
which have bounded variation in each coordinates).

2

Testing the stochastic domination from two independent time series

√
The statistic n(Un − 1/2) cannot be used to test H0 : π = 1/2 versus one of the possible
√
alternatives, because under H0 the asymptotic distribution of n(Un − 1/2) depends of the
unknow quantity V defined in (1.4). In the next Proposition, we propose an estimator of V
under some slightly stronger conditions on the dependence coefficients.
Definition 2.1. Let (Xi )i∈Z be a strictly stationary sequence of real-valued random variables,
and let F0 = σ(Xk , k ≤ 0). Let fx (z) = 1z≤x − F (x), where F is the cumulative distribution
function of X0 . For any non-negative integer n, let


α2,X (n) = max α1,X (n),
sup
kE (fx (Xi )fy (Xj )|F0 ) − E (fx (Xi )fy (Xj ))k1 .
x,y∈R,j≥i≥n

Proposition 2.1. Let (Xi )i∈Z and (Yi )i∈Z be two independent stationary sequences, and let (mn )
be a sequence of positive inetegers satisfying (1.3). Assume that
∞
X

α2,X (k) < ∞

and

k=1

Let

∞
X

α2,Y (k) < ∞ .

(2.1)

k=1

m

1 X
Hm (t) =
1Y >t
m j=1 j

n

and

1X
Gn (t) =
1X <t ,
n i=1 i

and
n−k

1X
(Hm (Xi )− H̄m )(Hm (Xi+k )− H̄m ) ,
γ̂X (k) =
n i=1

m−`

1 X
γ̂Y (`) =
(Gn (Yj )− Ḡn )(Gn (Yj+` )− Ḡn ) ,
m j=1

P
P
where nH̄m = ni=1 Hm (Xi ) and mḠn = m
j=1 Gn (Yj ). Let also (an ) and (bn ) be two sequences
√
of positive integers tending to infinity as n tends to infinity, such that an = o( n) and bn =
4

√
o( mn ). Then
an
X

n
Vn = γ̂X (0) + 2
γ̂X (k) +
mn
k=1

γ̂Y (0) + 2

bn
X

!
γ̂Y (`)

`=1

converges in L2 to the quantity V defined in (1.4).
Combining Theorem 1.1 and Proposition 2.1, we obtain that, under H0 : π = 1/2, if V > 0,
the random variables
√
n(Un − 1/2)
(2.2)
Tn = p
max{Vn , 0}
converge in distribution to N (0, 1).
Remark 2.1. The condition (2.1) is not restrictive: in all the natural examples we know, the
coefficients α2,X (k) behave as α1,X (k) (this is the case for instance for the linear process (1.6)).
Note that, if we assume (2.1) instead of (1.2), the ergodicity is no longer required in Theorem
1.1. Finally, following the proof of Theorem 4.1 page 89 in [2], one can also build a consistent
estimator of V under the condition
r
r
∞
∞
X
X
α1,X (k)
α1,Y (k)
< ∞ and
< ∞,
(2.3)
k
k
k=1
k=1
which is not directly comparable to (2.1) (for instance, the first part of (2.1) is satisfied if
α2,X (k) = O(k −1 (log k)−a ) for some a > 1, while the first part of (2.3) requires α1,X (k) =
O(k −1 (log k)−a ) for some a > 2).
Remark 2.2. The choice of the two sequences (an ) and (bn ) is a delicate matter. If the coefficients α2,X (k) decrease very quickly, then an should increase very slowly (it suffices to take
an ≡ 0 in the iid setting). On the contrary, if α2,X (k) = O(k −1 (log k)−a ) for some a > 1, then
the terms in the covariance series have no reason to be small, and one should take an close to
√
n to estimate many of these covariance terms. A data-driven criterion for choosing an and bn
is an interesting (but probably difficult) question, which is far beyond the scope of the present
paper.
However, from a practical point of view, there is an easy way to proceed: one can plot the
estimated covariances γ̂X (k)’s and choose an (not too large) in such a way that
γ̂X (0) + 2

an
X

γ̂X (k)

k=1

should represent an important part of the covariance series
Var(HY (X0 )) + 2

∞
X

Cov(HY (X0 ), HY (Xk )) .

k=1

5

Similarly, one may choose bn from the graph of the γ̂Y (`)’s. As we shall see in the simulations
(Section 6), if the decays of the true covariances γX (k) and γY (`) are not too slow, this provides
an easy and reasonable choice for an and bn .

3

Long range dependence

In this section, we shall see that the condition (1.2) cannot be weakened for the validity of
Theorem 1.1. More precisely, we shall give some examples where (1.2) fails, and the asymptotic
distribution of an (Un − π) (if any) can be non Gaussian.
Let us begin with the boundary case, where
α2,X (k) = O((k + 1)−1 ) and α2,Y (k) = O((k + 1)−1 ) .

(3.1)

In that case, one can prove the following result.
Proposition 3.1. Let (Xi )i∈Z and (Yi )i∈Z be two independent stationary sequences. Assume
that (3.1) is satisfied, and let (mn ) be a sequence a positive integers satisfying (1.3). Then:
1. For any r ∈ (2, 4), there exists a positive constant Cr such that, for any x > 0,
 C
p
r
n/ log n(Un − π) > x ≤ r .
lim sup P
x
n→∞
2. There are some examples of stationary Markov chains (Xi )i∈Z and (Yi )i∈Z for which the
p
sequence n/ log n(Un − π) converges in distribution to a non-degenerate normal distribution.
We consider now the case where
∞
X

α1,X (k) < ∞ and α1,Y (k) = O((k + 1)−(p−1) ) , for some p ∈ (1, 2).

(3.2)

k=1

This means that (Xi )i∈Z is short-range dependent, while (Yi )i∈Z is possibly long-range dependent.
In that case, we shall prove the following result.
Proposition 3.2. Let (Xi )i∈Z and (Yi )i∈Z be two independent stationary sequences. Assume
that (3.2) is satisfied for some p ∈ (1, 2), and that (Xi )i∈Z is 2-ergodic. Let (mn ) be a sequence
a positive integers such that
lim

n

n→∞ m2(p−1)/p
n

= c,

Then:
6

for some c ∈ [0, ∞).

(3.3)

1. There exists a positive constant Cp such that, for any x > 0,
lim sup P
n→∞

 Cp
√
n(Un − π) > x ≤ p .
x

2. There are some examples of stationary Markov chains (Xi )i∈Z and (Yi )i∈Z for which the
√
√
sequence n(Un − π) converges in distribution to Z + cS, where Z is independent of S,
Z is a Gaussian random variable, and S is a stable random variable of order p.
Remark 3.1. Note that we do not deal with the case where the two samples are long-range
dependent; in that case the degenerate U-statistic from the Hoeffding decomposition (see (4.2))
is no longer negligible, which gives rise to other possible limits. See the paper [10] for a treatment
of this difficult question in the case of functions of Gaussian processes.

4

Proofs

4.1

Proof of Theorem 1.1

The main ingredient of the proof is the following proposition:
Proposition 4.1. Let (Xi )i∈Z and (Yi )i∈Z be two independent stationary sequences. Let also
f (x, y) = 1x<y − HY (x) − GX (y) + π .
Then



n
m
1 XX

f (Xi , Yj )
E
nm i=1 j=1

!2 

(4.1)

n−1 m−1

XX
 ≤ 16
min{α1,X (i), α1,Y (j)} .
mn i=0 j=0

Let us admit this proposition for a while, and see how it can be used to prove the theorem.
Starting from (4.1), we easily see that
√
n(Un − π) =

√ X
√ X
n X
m
n
m
n
1 X
n
f (Xi , Yj ) + √
(HY (Xi ) − π) +
(GX (Yj ) − π) .
nm i=1 j=1
m j=1
n i=1

(4.2)

This is the well known Hoeffding decomposition of U-statistics; the first term on the right-hand
side of (4.2) is called a degenerate U-statistic, and is handled via Proposition 4.1. Indeed, it
follows easily from this proposition that

!2 
n X
m
n−1
m−1
X
1
16 X
16 X


nE
≤
f (Xi , Yj )
(i + 1)α1,X (i) +
(j + 1)α1,Y (j) .
(4.3)
nm i=1 j=1
m i=0
m j=0
7

Since (1.2) holds, then α1,X (n) = o(n−1 ) and α1,Y (n) = o(m−1 ) (because these coefficients
decrease) and by Cesaro’s lemma
n

1X
lim
(i + 1)α1,X (i) = 0 and
n→∞ n
i=0

m

1 X
lim
(j + 1)α1,Y (j) = 0 .
m→∞ m
j=0

From (4.3) and Condition (1.3), we infer that
√ X
n X
m
n
f (Xi , Yj ) converges to 0 in L2 as n → ∞.
nm i=1 j=1

(4.4)

It remains to control the two last terms on the right-hand side of (4.2). We first note that these
two terms are independent, so it is enough to prove the convergence in distribution for both
terms. The functions HY and GX being monotonic, the central limit theorem for stationary and
2-ergodic α-dependent sequences (see for instance [5]) implies that
n

1 X
√
(HY (Xi ) − π) converges in distribution to N (0, v1 ), where
n i=1
v1 = Var(HY (X0 )) + 2

∞
X

Cov(HY (X0 ), HY (Xk )) , (4.5)

k=1

and
m

1 X
√
(GX (Yj ) − π) converges in distribution to N (0, v2 ), where
m j=1
v2 = Var(GX (Y0 )) + 2

∞
X

Cov(GX (Y0 ), GX (Yk )) . (4.6)

k=1

Combining (4.2), (4.4), (4.5) and (4.6), the proof of Theorem 1.1 is complete.
It remains to prove Proposition 4.1. We start from the elementary inequality

!2 
n X
m
n
m
n
m
X
1
4 XXXX
E
f (Xi , Yj )  ≤ 2 2
E(f (Xi , Yj )f (Xk , Y` )) .
nm i=1 j=1
n m i=1 j=1 k=i `=j

(4.7)

To control the term E(f (Xi , Yj )f (Xk , Y` )), we first work conditionally on Y = (Yi )i∈Z . Note first
that, since Y is independent of X, E(f (Xi , Yj )|Y) = 0. Since |f (x, y)| ≤ 2,
|E(f (Xi , Yj )f (Xk , Yl )|Y = y)| ≤ 2kE(f (Xk , y` )|Fi )k1 ,

(4.8)

where Fi = σ(Xk , k ≤ i). Now, by definition of f ,
kE(f (Xk , y` )|Fi )k1 ≤ kE(1Xk <y` |Fi ) − GX (y` )k1 + kE(HY (Xk )|Fi ) − πk1 ≤ 2α1,X (k − i) . (4.9)
8

From (4.8) and (4.9), it follows that
|E(f (Xi , Yj )f (Xk , Y` ))| ≤ kE(f (Xi , Yj )f (Xk , Y` )|Y)k1 ≤ 4α1,X (k − i) .

(4.10)

In the same way
|E(f (Xi , Yj )f (Xk , Y` ))| ≤ 4α1,Y (` − j) .

(4.11)

Proposition 4.1 follows from (4.7), (4.10) and (4.11).

4.2

Proof of Proposition 2.1

We proceed in two steps.
Step 1. Let
n−k

∗
γX
(k)

m−`

1X
(HY (Xi ) − π)(HY (Xi+k ) − π) ,
=
n i=1

γY∗ (`)

1 X
=
(GX (Yj ) − π)(GX (Yi+k ) − π) ,
m j=1

and note that
n−k
n−k
Cov(HY (X0 ), HY (Xk )) and E(γY∗ (`)) =
Cov(GX (Y0 ), GX (Y` )) .
n
n
(4.12)
In this first step, we shall prove that
!
an
bn
X
X
n
∗
∗
γX
(k) +
(0) + 2
Vn∗ = γX
γY∗ (`) ,
γY∗ (0) + 2
m
n
k=1
`=1

∗
E(γX
(k)) =

converges in L2 to V . Clearly, by (4.12), it suffices to show that Var(Vn∗ ) converges to 0 as
n → ∞. We only deal with the second term in the expression of Vn∗ , the other ones may be
handled in the same way. Letting Zi = HY (Xi ) − π, and γk = Cov(HY (X0 ), HY (Xk )) we have
!
an
an X
an X
n−k X
n−`
X
1 X
∗
Var
γX
(k) = 2
E((Zi Zi+k − γk )(Zj Zj+` − γ` )) .
(4.13)
n
k=1
k=1 `=1 i=1 j=1
We must examine several cases.
On Γ1 = {j ≥ i + k},
|E((Zi Zi+k − γk )(Zj Zj+` − γ` ))| ≤ 2α2,X (j − i − k).
Consequently
!
an X
an X
n−k X
n−`
an X
an X
n
X
1 X
2 X
Ca2n
|E((Z
Z
−
γ
)(Z
Z
−
γ
))|1
≤
≤
α
(j)
i i+k
k
j j+`
`
Γ1
2,X
n2 k=1 `=1 i=1 j=1
n2 k=1 `=1 i=1 j≥0
n
(4.14)
9

for some positive constant C.
On Γ2 = {i ≤ j < i + k} ∩ {i + k ≤ j + `},
|E((Zi Zi+k − γk )(Zj Zj+` − γ` ))| = |E((Zi Zi+k − γk )Zj Zj+` )| ≤ 2α1,X (j + ` − i − k).
Consequently
!
an X
an X
n−k X
an X
an X
n
n−`
X
1 X
Ca2n
2 X
.
α
(j)
≤
|E((Z
Z
−
γ
)(Z
Z
−
γ
))|1
≤
1,X
i i+k
k
j j+`
`
Γ2
n2 k=1 `=1 i=1 j=1
n2 k=1 `=1 i=1 j≥0
n
(4.15)
On Γ3 = {i ≤ j} ∩ {i + k > j + `},
|E((Zi Zi+k − γk )(Zj Zj+` − γ` ))| = |E((Zj Zj+` − γ` )Zi Zi+k )| ≤ 2α1,X (i + k − j − `).
Consequently
n−k X
n−`
n
an X
an X
an X
an X
X
2 X
1 X
|E((Z
Z
−
γ
)(Z
Z
−
γ
))|1
α1,X (i)
≤
i i+k
k
j j+`
`
Γ3
n2 k=1 `=1 i=1 j=1
n2 k=1 `=1 j=1 i≥0

!
≤

Ca2n
.
n
(4.16)

From (4.14), (4.15) and (4.16), setting Γ = {i ≤ j} = Γ1 ∪ Γ2 ∪ Γ3 , one gets
n−k X
n−`
an X
an X
3Ca2n
1 X
|E((Zi Zi+k − γk )(Zj Zj+` − γ` ))|1Γ ≤
n2 k=1 `=1 i=1 j=1
n

Of course, interchanging i and j, the same is true on Γc = {i > j}, and finally
n−k X
n−`
an X
an X
6Ca2n
1 X
|E((Zi Zi+k − γk )(Zj Zj+` − γ` ))| ≤
n2 k=1 `=1 i=1 j=1
n

(4.17)

Pn ∗
√
From (4.13), (4.17) and the fact that an = o( n), we infer that ak=1
γX (k) converges in L2
P
to k>0 Cov(HY (X0 ), HY (Xk )). Since the other terms in the definition of Vn∗ can be handled in
the same way, it follows that Vn∗ converges in L2 to V .
∗
Step 2. In this second step, in the expression of γX
(k), we replace HY (Xi ) by Hm (Xi ),
HY (Xi+k ) by Hm (Xi+k ), and π by H̄m . Similarly, in the expression of γY∗ (`), we replace GX (Yj )
by Gn (Yj ), GX (Yj+` ) by Gn (Yj+` ), and π by Ḡn . If we can prove that these replacements in the
expression of Vn∗ are negligible in L2 , the conclusion of Proposition 2.1 will follow.
Let
n−k
1X
γ1,X (k) =
(Hm (Xi ) − π)(HY (Xi+k ) − π) .
n i=1

10

Then
∗
kγX
(k)

1/2

Z
− γ1,X (k)k2 ≤ kHm (Xi ) − HY (Xi )k2 =

kHm (x) −

HY (x)k22 PX (dx)

,

(4.18)

the last equality being true because (Xi )i∈Z and (Yi )i∈Z are independent. Now,
!
Pn−1
n−1
X
2 k=0
α1,Y (k)
1
2
kHm (x) − HY (x)k2 ≤
Var(1Y0 >x ) + 2
. (4.19)
|Cov(1Y0 >x , 1Yk >x )| ≤
m
m
k=1
From (4.18) and (4.19), we infer that there exist a positive constant C1 such that
∗
γX
(0) − γ1,X (0) + 2

an
X
∗
(γX
(k) − γ1,X (k))
k=1

2

C 1 an
≤ √ .
m

√
By condition (1.3) and the fact that an = o( n), this last quantity tends to 0 as n → ∞.
Let now
n−k
1X
γ2,X (k) =
(Hm (Xi ) − H̄m )(HY (Xi+k ) − π) .
n i=1
Then
n

kγ1,X (k) − γ2,X (k)k2 ≤ H̄m −

1X
HY (Xi )
n i=1

n

1X
HY (Xi ) − π
n i=1

+
2

.

(4.20)

2

Proceeding as in (4.19) for the first term on the right-hand side of (4.20), and noting that
n

1X
HY (Xi ) − π
n i=1

2

≤

2

Pn−1

2

α1,X (k)
,
n

k=0

we get that there exist a positive constant C2 such that
an
X
γ1,X (0) − γ2,X (0) + 2
(γ1,X (k) − γ2,X (k))
k=1

2

C 2 an C 2 an
≤ √ + √ .
m
n

Again, this last quantity tends to 0 as n → ∞.
Finally, all the other replacements may be done in the same way, and lead to negligible
contributions in L2 . Hence
lim kVn − Vn∗ k2 = 0
n→∞

and the result follows from Step 1.

11

4.3

Proof of Proposition 3.1

We start from (4.2) again, but with the normalization
holds,

p
√
n/ log n instead of n. Since (3.1)

n

lim sup
n→∞

1X
(i + 1)α1,X (i) < ∞ and
n i=0

m

lim sup
m→∞

1 X
(j + 1)α1,Y (j) < ∞ .
m j=0

From Inequality (4.3) and Condition (1.3), we infer that
√
n X
m
X
n
√
f (Xi , Yj ) converges to 0 in L2 as n → ∞.
nm log n i=1 j=1

(4.21)

Let us prove Item 1. From (4.2) and (4.21),
!
n

p
X
p
n/ log n(Un − π) > x ≤ lim sup P
(HY (Xi ) − π) > x n log n/2
lim sup P
n→∞

n→∞

+ lim sup P
n→∞

n
m

i=1
m
X

!
p
(GX (Yi ) − π) > x n log n/2 . (4.22)

j=1

Since (3.1) holds, it follows from Proposition 5.1 in [6] that: for any r ∈ (2, 4), there exists a
positive constant c1,r such that, for any x > 0
!
n
X
p
c1,r
lim sup P
(HY (Xi ) − π) > x n log n/2 ≤ r .
x
n→∞
i=1
Since (1.3) holds, the same is true for the second term on the right-hand side of (4.22) (for a
positive constant c2,r ), and Item 1 follows.
Let us prove Item 2. We use a result by Gouëzel [12] about intermittent maps. Let θ1/2
be the map described in Subsection 6.2 of the present paper. As explained in Subsection 6.2,
there exists an unique absolutely continuous θ1/2 -invariant measure ν1/2 on [0, 1], and (see the
comments on the case α = 1/2 in Section 1.3 of Gouëzel’s paper), on the probability space
([0, 1], ν1/2 ),
n

1 X
k
√
HY (θ1/2
) − ν1/2 (HY )
n k=1
converges in distribution to a non-degenerate normal distribution provided HY (0) 6= ν(HY )
(which is true for instance if Y has distribution ν1/2 , since in that case HY (0) = 1 and ν(HY ) =
1/2). As fully explained in Subsection 6.2, there exists a stationary Markov chain (Xi )i∈Z such

12

n
that the distribution of (X1 , . . . , Xn ) is the same as (θ1/2
, . . . , θ1/2 ) on the probability space
([0, 1], ν1/2 ). Take (Yi )i∈Z an independent copy of the chain (Xi )i∈Z . For such chains, both
n

1 X
√
(HY (Xi ) − π) and
n i=1

m

1 X
√
(GX (Yj ) − π)
m j=1

converge to the same non-degenerate normal distribution (note that π = 1/2 in that case).
Moreover, as recalled in Subsection 6.2, there exist two positive constants A, B such that
B
A
≤ α2,X (k) = α2,Y (k) ≤
.
k+1
k+1
This completes the proof of Item 2.

4.4

Proof of Proposition 3.2

We start from (4.2) again. At the end of the proof, we shall prove that if (3.2) and (3.3) are
satisfied, then (4.4) holds.
Let us now prove Item 1. From (4.2) and (4.4),
lim sup P
n→∞

√

!
n
X

√
n(Un − π) > x ≤ lim sup P
(HY (Xi ) − π) > x n/2
n→∞

i=1

+ lim sup P
n→∞

!
m
√
n X
(GX (Yi ) − π) > x n/2 . (4.23)
m j=1

Since the first part of (3.2) is satisfied, it follows from the central limit theorem for stationary
and 2-ergodic α-dependent sequences that
!
n
X
√
lim P
(HY (Xi ) − π) > x n/2 = P(Z > x/2) ,
(4.24)
n→∞

i=1

where Z is a Gaussian random variable. From Condition (3.2) and Remark 3.3 in [6], we infer
that if (3.3) holds, then there exists a positive constant κp such that, for any x > 0,
!
m
√
n X
κp
lim sup P
(GX (Yi ) − π) > x n/2 ≤ p .
(4.25)
m j=1
x
n→∞
Item 1 follows from (4.23), (4.24) and (4.25).
Let us prove Item 2. For the sequence (Xi )i∈Z , take a sequence of iid random variables
uniformly distributed over [0, 1]. To find the sequence (Yi )i∈Z we use again a result by Gouëzel
[12] about intermittent maps. Let θ1/p be the map described in Subsection 6.2 of the present
13

paper. As explained is Subsection 6.2, there exists an unique absolutely continuous θ1/p -invariant
measure ν1/p on [0, 1], and (see Theorem 1.3 in [12]), on the probability space ([0, 1], ν1/p ),
m

1 X
k
GX (θ1/p
) − ν1/p (GX )
1/p
m k=1

converges in distribution to a non-degenerate stable distribution of order p provided GX (0) 6=
ν(GX ) (this condition is satisfied here, because GX (0) = 0 and ν1/p (GX ) > 0). As fully explained
in Subsection 6.2, there exists a stationary Markov chain (Yi )i∈Z such that the distribution of
n
, . . . , θ1/p ) on the probability space ([0, 1], ν1/p ). For such a chain,
(Y1 , . . . , Yn ) is the same as (θ1/p
m
1 X
(GX (Yj ) − π)
m1/p j=1

converges in distribution to a non-degenerate stable distribution of order p. Moreover, as recalled
in Subsection 6.2, there exist two positive constants A, B such that
A
B
≤
α
(k)
≤
.
1,Y
(k + 1)p−1
(k + 1)p−1
This completes the proof of Item 2.
It remains to prove (4.4). From Proposition 4.1 we get that

!2 
!
n
m
∞
m−1
XX
X
X
1
16
nE 
f (Xi , Yj )  ≤
min{α1,X (i), α1,Y (j)} .
nm i=1 j=1
m
i=0
j=0
Note that

m−1
16 X
min{α1,X (i), α1,Y (j)} ≤ 16α1,X (i) ,
m j=0

(4.26)

(4.27)

and that, for any positive integer i, since α1,Y (j) → 0 as j → ∞,
m−1
16 X
lim
min{α1,X (i), α1,Y (j)} = 0 ,
m→∞ m
j=0

(4.28)

P
by Cesaro’s lemma. Since i>0 α1,X (i) < ∞, it follows from (4.26), (4.27), (4.28), and the
dominated convergence theorem that

!2 
n X
m
X
1
lim nE 
f (Xi , Yj )  = 0 ,
n→∞
nm i=1 j=1
proving (4.4).
14

5
5.1

Other related results
Testing the stochastic domination by a known distribution

We briefly describes here a procedure for testing the (weak) domination by a known distribution
µ. Let H(t) = µ((t, ∞)), and let (Xi )i∈Z be a stationary 2-ergodic sequence of real-valued
random variables. Let π = E(H(X0 )), and define
n

1X
H̄n =
H(Xi ) .
n i=1
If

∞
X

α1,X (k) < ∞ ,

(5.1)

k=1

√
then the random variables n(H̄n − π) converge in distribution to N (0, V1 ), where
V1 = Var(H(X0 )) + 2

∞
X

Cov(H(X0 ), H(Xk )) .

(5.2)

k=1

Now, we want to test H0 : π = 1/2 (no relation of weak domination between µ and PX )
against one of the possible alternatives: H1 : π 6= 1/2 (one of the two distributions weakly
dominates the other one), H1 : π > 1/2 (µ weakly dominates PX ), H1 : π < 1/2 (PX weakly
dominates µ).
√
Again, one cannot use directly the statistics n(H̄n − 1/2) to test H0 : π = 1/2 versus one
√
of the possibles alternatives, because under H0 the asymptotic distribution of n(H̄n − 1/2)
depends of the unknown quantity V1 defined in (5.2).
To estimate V1 , we take
an
X
γ̂(k)
V1,n = γ̂(0) + 2
k=1

√
for some sequence an = o( n), where
n−k

γ̂(k) =

1X
(H(Xi ) − H̄n )(H(Xi+k ) − H̄n ) .
n i=1

Now, as in Proposition 2.1, if (5.1) holds for α2,X (k) instead of α1,X (k), then vn converges in L2
to v. It follows that, under H0 : π = 1/2, if v > 0, the random variables
√
n(H̄n − 1/2)
0
Tn = p
max{V1,n , 0}
converges in distribution to N (0, 1).
15

5.2

The case of two adjacent samples from the same time series

Let (Xi )i∈Z be a stationary sequence of real-valued random variables. Let m = mn be a sequence
of positive integers such that mn → ∞ as n → ∞. We consider here the U-statistic
Un = Un,m =

n n+m
1 X X
1X <Y ,
nm i=1 j=n+1 i j

where Yj = f (Xj ), f being a monotonic function. This statistic can be used to test if there is a
change in the time series after the time n, and more precisely if Yn+1 weakly dominates X1 . Let
π = P(X1 < f (X1∗ )), where X1∗ is an independent copy of X1 . Let also HY (t) = P(f (X1 ) > t)
and GX (t) = P(X1 < t).
Finding the asymptotic behavior of Un is a slightly more difficult problem than for two
independent samples, because we cannot use the independence to control the degenerate Ustatistic as in Proposition 4.1. However, it can be done by using a more restrictive coefficient
than α2,X .
Definition 5.1. Let (Xi )i∈Z be a strictly stationary sequence of real-valued random variables,
and let F0 = σ(Xk , k ≤ 0). Let P be the law of X0 and P(Xi ,Xj ) be the law of (Xi , Xj ). Let PXk |F0
be the conditional distribution of Xk given F0 , and let P(Xi ,Xj )|F0 be the conditional distribution
of (Xi , Xj ) given F0 . Define the random variables
b(k) = sup PXk |F0 (fx ) ,
x∈R

b(i, j) =

sup
(x,y)∈R2

P(Xi ,Xj )|F0 (fx ⊗ fy ) − P(Xi ,Xj ) (fx ⊗ fy ) ,

where the function fx has been defined in Definition 2.1. Define now the coefficients


β1,X (k) = E(b(k)) and β2,X (k) = max β̃1,X (k), sup E((b(i, j))) .
i>j≥k

Of course, by definition of the coefficients, we always have that α2,X (k) ≤ β2,X (k). The
coefficients β2,X are weaker than the usual β-mixing coefficients of (Xi )i∈Z . Many examples of
non-mixing process for which β2,X can be computed are given in [8]. For all the examples that
we mention in the present paper, the coefficient β2,X (k) is of the same order than the coefficient
α2,X (k) (see the paper [4] for the example of the intermittent maps described in Subsection 6.2).
Theorem 5.1. Let (Xi )i∈Z be a stationary sequence of real-valued random variables. Assume
that
∞
X
β2,X (k) < ∞ .
(5.3)
k=1

16

Let (mn ) be a sequence of positive integers satisfying (1.3). Then
distribution to N (0, V ), with
V = Var(HY (X0 )) + 2

∞
X

√
n(Un − π) converge in

Cov(HY (X0 ), HY (Xk ))

k=1

+ c Var(GX (f (X0 ))) + 2

∞
X

!
Cov(GX (f (X0 )), GX (f (Xk )))

. (5.4)

k=1

For the estimation of V , one can prove the following analogue of Proposition 2.1.
Proposition 5.2. Let (Xi )i∈Z be an independent stationary sequence, and let Yj = f (Xj ) where
f is a monotonic function. Let (mn ) be a sequence of positive integers satisfying (1.3). Assume
that (5.3) is satisfied. Let
n+m
1 X
1Y >t
Hm (t) =
m j=n+1 j

n

and

1X
Gn (t) =
1X <t ,
n i=1 i

and
n−k
n+m−`
1X
1 X
γ̂X (k) =
(Hm (Xi )−H̄m )(Hm (Xi+k )−H̄m ) , γ̂Y (`) =
(Gn (Yj )−Ḡn )(Gn (Yj+` )−Ḡn ) ,
n i=1
m j=n+1

Pn+m
P
Gn (Yj ). Let also (an ) and (bn ) be two sequences
where nH̄m = ni=1 Hm (Xi ) and mḠn = j=n+1
√
of positive integers tending to infinity as n tends to infinity, such that an = o( n/ log(n)) and
√
bn = o( mn / log(mn )). Then
!
an
bn
X
X
n
γ̂X (k) +
γ̂Y (`)
Vn = γ̂X (0) + 2
γ̂Y (0) + 2
m
n
k=1
`=1
converges in L2 to the quantity V defined in (5.4).
Combining Theorem 5.1 and Proposition 5.2, we obtain that, under H0 : π = 1/2, if V > 0,
the random variables
√
n(Un − 1/2)
Tn = p
(5.5)
max{Vn , 0}
converge in distribution to N (0, 1).
The proof of Proposition 5.2 follows the same lines as that of Proposition 2.1, so we do not
give the details here. The only main change concerns Step 2, Inequalities (4.18) and (4.19),
where we used the independence between the samples. Here instead, one can use the fact that
Pm−1
2
2
k=0 β1,X (k)
sup |Hn (x) − H(x)| ≤ C (log(m))
.
m
x∈R
2
17

This can be easily proved by following the proof of Proposition 7.1 in Rio [14] up to (7.8) and
by using the computations in the proof of Theorem 2.1 in [3].
The proof of Theorem 5.1 follows also the line of that of Theorem 1.1; the only non-trivial
change concerns Proposition 4.1. Here, we have to deal with
n n+m
1 X X
f (Xi , Yj )
Ln =
nm i=1 j=n+1

where f is defined in (4.1). We want to prove that nE(L2n ) tends to 0 as n tends to infinity. As
in the proof of Proposition 4.1, we start from the elementary inequality
E(L2n ) ≤

n n+m n n+m
4 X X XX
E(f (Xi , Yj )f (Xk , Y` )) .
n2 m2 i=1 j=n+1 k=i `=j

(5.6)

Let (Xi∗ )i∈Z be an independent copy of (Xi∗ )i∈Z and Yi∗ = f (Xi∗ ), and write
E(f (Xi , Yj )f (Xk , Y` )) = E(f (Xi , Yj )f (Xk , Y` )) − E(f (Xi , Yj∗ )f (Xk , Y`∗ ))
+ E(f (Xi , Yj∗ )f (Xk , Y`∗ )) . (5.7)
Now, using the independence, the second term on the right-hand side of (5.7) can be handled
exactly as in Proposition 4.1. It remains to deal with the first term in the right-hand side of
(5.7). Note that
|E(f (Xi , Yj )f (Xk , Y` )) − E(f (Xi , Yj∗ )f (Xk , Y`∗ ))|
Z
≤ |E(f (xi , Yj )f (xk , Y` )|Xi = xi , Xk = xk ) − E(f (xi , Yj )f (xk , Y` ))|P(Xi ,Xk ) (dxi , dxk ) (5.8)
For any fixed x, the function y → hx (y) = f (x, y) is a bounded variation function whose variation
is bounded by 2, and E(hx (Xi )) = 0. Proceeding as in Lemma 1 of [7], one can then prove that
Z
|E(f (xi , Yj )f (xk , Y` )|Xi = xi , Xk = xk ) − E(f (xi , Yj )f (xk , Y` ))|P(Xi ,Xk ) (dxi , dxk )
≤ 4β2,X (j − k) . (5.9)
Considering separately the two cases j − k > n1/4 or j − k ≤ n1/4 , we get the upper bound
n n+m n n+m
1 X X XX
|E(f (Xi , Yj )f (Xk , Y` ) − E(f (Xi , Yj∗ )f (Xk , Y`∗ ))|
n2 m2 i=1 j=n+1 k=i `=j

4
4 X
≤√
+
β2,X (k) (5.10)
nm m
1/4
k>n

from which we easily derive that nE(L2n ) tends to 0 as n tends to infinity (since (1.3) and (5.3)
are satisfied).
18

6

Simulations

6.1

Example 1: functions of an auto-regressive process

In this section, we first simulate (Z1 , . . . , Zn ), according to the simple AR(1) equation
Zk+1 =

1
(Zk + εk+1 ) ,
2

where Z1 is uniformly distributed over [0, 1], and (εi )i≥2 is a sequence of iid random variables
with distribution B(1/2), independent of Z1 .
One can check that the transition Kernel of the chain (Zi )i≥1 is
  


x
x+1
1
f
+f
,
K(f )(x) =
2
2
2
and that the uniform distribution on [0, 1] is the unique invariant distribution by K. Hence, the
chain (Zi )i≥1 is strictly stationary.
It is well known that the chain (Zi )i≥1 is not α-mixing in the sense of Rosenblatt [15] (see
for instance [1]). However, one can prove that the coefficients α2,X of the chain (Zi )i≥1 are such
that
α2,Z (k) ≤ 2−k
(and the same upper bound is true for β2,Z (k), see for instance Section 6.1 in [8]).
Let now Qµ,σ2 be the inverse of the cumulative distribution function of the law N (µ, σ 2 ). Let
then
Xi = Qµ1 ,σ12 (Zi ) .
The sequence (Xi )i≤1 is also a stationary Markov chain (as an invertible function of a stationary
Markov chain), and one can easily check that α2,X (k) = α2,Z (k). By construction, Xi is N (µ, σ 2 )distributed, but the sequence (Xi )i≥1 is not a Gaussian process (otherwise it would be mixing
in the sense of Rosenblatt).
Next, we simulate iid random variables (Y1 , . . . , Ym ) with common distribution N (µ2 , σ22 ).
Note that, if µ1 = µ2 , the hypothesis H0 : π = 1/2 is satisfied and the random variables Tn
defined in (2.2) converge in distribution to N (0, 1).
We shall study the behavior of Tn for σ12 = 4, σ22 = 1 and different choices of µ1 , µ2 , an and
bn . As explained in Section 2, the quantities an and bn may be chosen by analyzing the graph
of the estimated auto-covariances γ̂X (k) and γ̂Y (`). For this examples, these graphs suggest a
choice of an = 3 or 4 and bn = 0 (see Figure 1).

19

We shall compute Tn for different choices of (n, m) (from (150, 100) to (750, 500) with n/m =
1.5). We estimate the three quantities Var(Tn ), P(Tn > 1.645) and P(|Tn | > 1.96) via a classical
Monte-Carlo procedure, by averaging over N = 2000 independent trials.
If µ1 = µ2 , we say that P(Tn > 1.645) is level 1 (the level of the one-sided test) and
P(|Tn | > 1.96) is level 2 (the level of the two-sided test). If an , bn are well chosen, the estimated
variance should be close to 1 and the estimated levels 1 and 2 should be close to 0.05.
If µ1 6= µ2 , we say that P(Tn > 1.645) is power 1 (the power of the one-sided test) and
P(|Tn | > 1.96) is power 2 (the power of the two-sided test).
• Case µ1 = µ2 = 0 and an = 0, bn = 0 (no correction):
n; m 150 ; 100 300 ; 200 450 ; 300 600 ; 400 750 ; 500
Estimated variance
2.13
2.199
2.118
2.112
2.273
Estimated level 1
0.129
0.145
0.121
0.132
0.14
Estimated level 2
0.177
0.194
0.18
0.18
0.191
Here, since an = 0, we do not estimate any of the covariance terms γX (`); hence, we compute
Tn as if both series (Xi )i≥1 and (Yj )j≥1 were uncorrelated. The result is that the estimated
variance is too large (around 2.1 whatever (n, m)) and so are the estimated levels 1 and 2
(around 0.13 for level 1 and 0.18 for level 2, whatever (n, m)). This means that the-one sided or
two-sided tests build with Tn will reject the null hypothesis too often.
• Case µ1 = µ2 = 0 and an = 3, bn = 0:
n; m 150 ; 100 300 ; 200 450 ; 300 600 ; 400 750 ; 500
Estimated variance
1.227
1.148
1.172
1.119
1.018
Estimated level 1
0.071
0.061
0.065
0.06
0.046
Estimated level 2
0.074
0.069
0.071
0.066
0.054
As suggested by the graphs of the estimated auto-covariances (see Figure 1), the choice
an = 3, bn = 0 should give a reasonable estimation of V . This is indeed the case: the estimated
variance is now around 1.1, the estimated level 1 is around 0.06 and the estimated level 2 around
0.07.
• Case µ1 = µ2 = 0 and an = 4, bn = 0:

20

0.15
0.10
0.05
0.00

5

10

15

20

0.00

0.01

0.02

0.03

0

0

5

10

15

20

Figure 1: Estimation of the γX (k)’s (top) and γY (`)’s (bottom) for Example 1, with µ1 = µ2 = 0,
n = 300 and m = 200.
n; m 150 ; 100 300 ; 200 450 ; 300 600 ; 400 750 ; 500
Estimated variance
1.151
1.11
1.121
1.03
0.987
Estimated level 1
0.06
0.05
0.058
0.049
0.042
Estimated level 2
0.062
0.067
0.06
0.057
0.046
Here, we see that the choice an = 4, bn = 0 works well also, and seems even slightly better
than the choice an = 3, bn = 0.
• Case µ1 = 0, µ2 = 0.2 and an = 4, bn = 0 :
In this example, H0 is not satisfied, and π > 1/2. However, µ2 is close to µ1 , and since
σ1 = 2 and σ2 = 1, Y1 does not stochastically dominate X1 (there is only stochastic domination

21

in a weak sense). We shall see how the two tests are able to see this weak domination, by giving
some estimation of the powers.
n; m 150 ; 100 300 ; 200 450 ; 300 600 ; 400 750 ; 500
Estimated variance
1.229
1.136
1.107
1.066
1.059
Estimated power 1
0.185
0.276
0.352
0.4
0.476
Estimated power 2
0.13
0.187
0.25
0.291
0.344
As one can see, the powers of the one sided and two-sided tests are always greater than
0.05, as expected. Still as expected, these powers increase with the size of the samples. For
n = 750, m = 500, the power of the one-sided test is around 0.47 and that of the two-sided test
is around 0.34.

6.2

Example 2: Intermittent maps

For γ in ]0, 1[, we consider the intermittent map θγ from [0, 1] to [0, 1], introduced by Liverani,
Saussol and Vaienti [13]:

x(1 + 2γ xγ ) if x ∈ [0, 1/2[
θγ (x) =
(6.1)
2x − 1
if x ∈ [1/2, 1].
It follows from [13] that there exists a unique absolutely continuous θγ -invariant probability
measure νγ , with density hγ .
Let us briefly describe the Markov chain associated with θγ , and its properties. Let first Kγ
be the Perron-Frobenius operator of θγ with respect to νγ , defined as follows: for any functions
u, v in L2 ([0, 1], νγ )
νγ (u · v ◦ θγ ) = νγ (Kγ (u) · v) .
(6.2)
The relation (6.2) states that Kγ is the adjoint operator of the isometry U : u 7→ u ◦ θγ acting
on L2 ([0, 1], νγ ). It is easy to see that the operator Kγ is a transition kernel, and that νγ is
invariant by Kγ . Let now (ξi )i≥1 be a stationary Markov chain with invariant measure νγ and
transition kernel Kγ . It is well known that on the probability space ([0, 1], νγ ), the random vector
(θγ , θγ2 , . . . , θγn ) is distributed as (ξn , ξn−1 , . . . , ξ1 ). Now it is proved in [5] that there exist two
positive constants A, B such that
B
A
≤ α2,ξ (n) ≤
(1−γ)/γ
(n + 1)
(n + 1)(1−γ)/γ
(this is also true for the coefficient β2,ξ (n), see [4]). Moreover, the chain (ξi )i≥1 is not α-mixing
in the sense of Rosenblatt [15].
22

Let Xi (x) = θγi 1 (x) and Yj (y) = θγj 2 (y). On the probability space ([0, 1] × [0, 1], νγ1 ⊗ νγ2 ) the
two sequences (Xi )i≥1 and (Yj )j≥1 are independent.
Note that, if γ1 = γ2 , the hypothesis H0 : π = 1/2 is satisfied and the random variables Tn
defined in (2.2) converge in distribution to N (0, 1).
We shall study the behavior of Tn for different choices of γ1 , γ2 , an and bn . As explained
in Section 2, the quantities an and bn may be chosen by analyzing the graph of the estimated
auto-covariances γ̂X (k) and γ̂Y (`).
As in Subsection 6.1, we shall compute Tn for different choices of (n, m) (from (150, 100)
to (750, 500) with n/m = 1.5). We estimate the three quantities Var(Tn ), P(Tn > 1.645) and
P(|Tn | > 1.96) via a classical Monte-Carlo procedure, by averaging over N = 2000 independent
trials.
If γ1 = γ2 , we say that P(Tn > 1.645) is level 1 (the level of the one-sided test) and P(|Tn | >
1.96) is level 2 (the level of the two-sided test). If an , bn are well chosen, the estimated variance
should be close to 1 and the estimated levels 1 and 2 should be close to 0.05.
If γ1 6= γ2 , we say that P(Tn > 1.645) is power 1 (the power of the one-sided test) and
P(|Tn | > 1.96) is power 2 (the power of the two-sided test).
Note that we cannot simulate exactly (Xi )i≥1 and (Yj )j≥1 , since the distribution ν1/4 is
unknown. We start by picking X1 and Y1 independently over [0, 0.05] (in order to start near the
neutral fixed point 0) and then generate Xi = θγi−1
(X1 ) and Yj = θγj−1
(Y1 ) for i > 1, j > 1. We
1
2
shall see that the tests are robust to this (slight) lack of stationarity.
• Case γ1 = γ2 = 1/4 and an = 0, bn = 0 (no correction):
n; m 150 ; 100 300 ; 200 450 ; 300 600 ; 400 750 ; 500
Estimated variance
5.172
4.658
4.858
4.764
4.751
Estimated level 1
0.204
0.196
0.19
0.206
0.218
Estimated level 2
0.358
0.359
0.361
0.361
0.37
Here again, since an = bn = 0, we compute Tn as if both series (Xi )i≥1 and (Yj )j≥1 were
uncorrelated. It leads to a disastrous result, with an estimated variance around 4.8, an estimated
level 1 around 0.2, and an estimated level 2 around 0.36. This means that the-one sided or twosided tests build with Tn will reject the null hypothesis much too often.
• Case γ1 = γ2 = 1/4 and an = 4, bn = 4:

23

0.08
0.06
0.04
0.02
0.00

5

10

15

20

0.00

0.02

0.04

0.06

0.08

0

0

5

10

15

20

Figure 2: Estimation of the γX (k)’s (top) and γY (`)’s (bottom) for Example 2, with γ1 = γ2 =
1/4, n = 300 and m = 200.
n; m 150 ; 100 300 ; 200 450 ; 300 600 ; 400 750 ; 500
Estimated variance
1.352
1.308
1.329
1.324
1.28
Estimated level 1
0.057
0.0645
0.062
0.069
0.067
Estimated level 2
0.089
0.082
0.087
0.087
0.075
As suggested by the graphs of the estimated auto-covariances (see Figure 2), the choice
an = 4, bn = 4 should give a reasonable estimation of V . We see that the procedure is much
better than if no correction is made: the estimated variance is now around 1.3, the estimated
level 1 is around 0.065 and the estimated level 2 is around 0.085.
• Case γ1 = γ2 = 1/4 and an = 5, bn = 5:
24

n; m 150 ; 100 300 ; 200 450 ; 300 600 ; 400 750 ; 500
Estimated variance
1.349
1.249
1.164
1.181
1.185
Estimated level 1
0.057
0.062
0.055
0.055
0.063
Estimated level 2
0.084
0.077
0.068
0.071
0.07
For an = 5, bn = 5, the results seem slightly better than for an = 4, bn = 4. Even for large
samples, the estimated level 2 is still larger than the expected level: around 0.07 instead of 0.05
(but 0.07 has to be compared to 0.36 which is the estimated level 2 without correction). Recall
that the coefficients α2,X (n) and α2,Y (n) are exactly of order n−1/3 , which is quite slow. The
procedure can certainly be improved for larger n, m by estimating more covariance terms (in
accordance with Proposition 2.1).
• Case γ1 = 0.25, γ2 = 0.1 and an = 5, bn = 4.
We first estimate the quantity π = P(X1 < Y1 ) via a realization of
n

m

1 XX
1X <Y .
nm i=1 j=1 i j
For n = m = 30000, this gives an estimation around 0.529 for π, meaning that Y weakly
dominates X. Let us see how the tests based on Tn can see this departure from H0 : π = 1/2.
Based on the estimated covariances, a reasonable choice is an = 5, bn = 4. This choice leads
to the following estimated powers (based on N = 2000 independent trials).
n; m 150 ; 100 300 ; 200 450 ; 300 600 ; 400 750 ; 500
Estimated variance
1.24
1.2
1.2
1.23
1.117
Estimated power 1
0.11
0.165
0.185
0.226
0.241
Estimated power 2
0.091
0.12
0.13
0.158
0.17
As one can see, the powers of the one sided and two-sided tests are always greater than
0.05, as expected. Still as expected, these powers increase with the size of the samples. For
n = 750, m = 500, the power of the one-sided test is around 0.24 and that of the two-sided test
is around 0.17. This has to be compared with the estimated levels when Tn is centered on 0.529
instead of 0.5 (recall that 0.529 is the previous estimation of π on very large samples): in that
case, for n = 750, m = 500 and an = 5, bn = 4, one gets an estimated level 1 = 0.059, and an
estimated level 2 = 0.068, which are both close to 0.05 as expected.

25

6.3

Intermittents maps: the case of two adjacent samples

Here, we shall illustrate the result of Subsection 5.2. As in Subsection 6.2, we first simulate X1
according to the uniform distribution over [0, 0.05], and then generate Xi = θγi−1 (Z1 ), but now
for the indexes i ∈ {2, . . . , n + m}. For j ∈ {n + 1, . . . , n + m}, we take Yj = f (Xj ) for a given
monotonic function f .
In our first experiment, we simply take f =Id, so that Yj = Xj . Hence the assumption
H0 : π = 1/2 is satisfied.
For the second experiment, we take f (x) = x4/5 , so that f (X1 ) > X1 almost surely. Hence
π > 1/2.
Let Tn be defined in (5.5). We follow the same scheme as in Subsection 6.2, and we estimate
the three quantities Var(Tn ), P(Tn > 1.645) and P(|Tn | > 1.96) via a classical Monte-Carlo
procedure, by averaging over N = 2000 independent trials.
• Case γ = 1/4, f =Id and an = 0, bn = 0 (no correction):
n; m 150 ; 100 300 ; 200 450 ; 300 600 ; 400 750 ; 500
Estimated variance
4.851
4.908
4.551
4.926
4.85
Estimated level 1
0.294
0.276
0.252
0.256
0.251
Estimated level 2
0.368
0.373
0.354
0.366
0.362
Here, as in the corresponding two-sample case (see the table for γ1 = γ2 = 1/4), the uncorrected tests lead to a disastrous result, with an estimated level 1 around 0.25 and an estimated
level 2 around 0.36.
• Case γ = 1/4, f =Id and an = 5, bn = 5:
n; m 150 ; 100 300 ; 200 450 ; 300 600 ; 400 750 ; 500
Estimated variance
1.369
1.302
1.272
1.205
1.171
Estimated level 1
0.114
0.097
0.086
0.081
0.075
Estimated level 2
0.109
0.096
0.079
0.072
0.066
The corrected tests with the choice an = bn = 5 give a much better result than if no
correction is made. One can observe, however, that the estimated levels are still too high
(≥ 8%) for moderately large samples (up to n = 450, m = 300). The results seem less good than
in the corresponding two-sample case (see the table for γ1 = γ2 = 1/4), which is of course not
surprising.
• Case γ = 1/4, f (x) = x4/5 and an = 5, bn = 5:
26

n; m 150 ; 100 300 ; 200 450 ; 300 600 ; 400 750 ; 500
Estimated variance
1.596
1.448
1.334
1.311
1.451
Estimated power 1
0.321
0.385
0.451
0.49
0.548
Estimated power 2
0.242
0.293
0.34
0.383
0.442
As one can see, the powers of the one sided and two-sided tests are always greater than
0.05, as expected. Still as expected, these powers increase with the size of the samples. For
n = 750, m = 500, the power of the one-sided test is around 0.55 and that of the two-sided test
is around 0.44.

Acknowledgments. We thank Herold Dehling for helpful comments on a preliminary
version of the paper.

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28

