arXiv:1608.05994v1 [] 21 Aug 2016

Inefficient Best Invariant Tests
Richard A. Lockhart∗
Department of Mathematics and Statistics
Simon Fraser University
Burnaby, BC
Canada V5A 1S6
November 10, 2017

Abstract
Test statistics which are invariant under various subgroups of the
orthogonal group are shown to provide tests whose powers are asymptotically equal to their level against the usual type of contiguous alternative in models where the number of parameters is allowed to grow
as the sample size increases. The result is applied to the usual analysis of variance test in the Neyman-Scott many means problem and
to an analogous problem in exponential families. Proofs are based
on a method used by Čibisov(1961) to study spacings statistics in a
goodness-of-fit problem. We review the scope of the technique in this
context.

Keywords. Asymptotic relative efficiency, Neyman-Scott many means problem, goodness-of-fit, spacings statistics, many parameter problems, permutation central limit theorem, bootstrap.
AMS 1980 Classification. Primary 62G20, Secondary 62G30.
Running title: Invariant tests
∗
This manuscript was written in 1992 but the reviewers were so caustic I never tried
again to publish it. It seems relevant today. Supported by the Natural Sciences and
Engineering Research Council of Canada.

1

1

Introduction

Consider the problem of data X coming from a model indexed by a parameter
space M. Suppose there is a group G which acts both on the data and on
the parameter space so that gX has the same distribution under m ∈ M as
X has under g −1 m. The problem of testing Ho : m ∈ M0 ⊂ M is invariant
under G if m ∈ M0 iff gm ∈ M0 for all g ∈ G. In what follows we shall
impose the stronger condition that gm = m for all m ∈ M0 and all g ∈ G.
In this note we use the observation that for an invariant test the power
at alternative m minus the level of the test is the covariance between the
test function and the likelihood ratio averaged over the orbit of m under
G to study the power of invariant tests under contiguous alternatives. The
technique was used in Čibisov(1961) to study the asymptotic behaviour of
tests of uniformity based on sample spacings.
Our results may be summarized as follows. When for a given point m
in the null hypothesis, the number of possible directions of departure from
m into the alternative hypothesis grows with the sample size the power of
invariant tests may be expected to be low. We will exhibit a variety of
examples in which the power minus the level of invariant tests converges to
0 uniformly in the class of invariant tests.
We begin in section 2 with a simple normal example to illustrate the
technique and set down the basic identities and inequalities. In section 3 we
extend the normal example to general exponential families using a version of
the permutation central limit theorem. In section 4 we examine the Neyman
Scott many means problem and extend our results to more general models.
In section 5 we revisit Čibisov’s example to show that a variety of goodnessof-fit tests have ARE 0 over a large class of alternatives. The results of the
first 5 sections suggest that either invariance is not always desirable or that
contiguity calculations are not always the right way to compare powers of
tests in such models. Section 6 is a discussion of the relevance of contiguity
calculations in this context, together with some open problems and suggestions for further work. An appendix contains some technical steps in the
proofs.

2

2

Basic Results: A Normal Example

Suppose X ∈ Rn has a multivariate normal distribution with mean vector
m ∈ Rn and identity covariance matrix. The problem of testing the null
hypothesis Ho : m = 0 is invariant under the group of orthogonal transformations. Under Ho , X and PX have the same distribution for any orthogonal
matrix P. Consider a simple alternative m 6= 0. The likelihood ratio of m to
0 is L(X) = exp{mT X − kmk2 /2}. Thus the Neyman Pearson test rejects
when mT X/kmk is too large. Under Ho this statistic has a standard normal
distribution while under the alternative m the mean is shifted to kmk. Thus
as n → ∞ non-trivial limiting power (i.e. a limiting power larger than the
level and less than 1) results when kmk → δ 6= 0. (Throughout this paper
objects named by Roman letters depend on n; wherever possible the dependence is suppressed in the notation. Objects named by Greek letters do not
depend on n.)
We may analyse the efficiency of an invariant test relative to the Neyman
Pearson test for a given sequence of alternatives as follows. Let T be the
class of all test functions T (X) for testing Ho which are invariant under the
group of orthogonal transformations, that is, for which T (PX) = T (X) for
each orthogonal transformation P. We have the following theorem.
Theorem 1 As n → ∞,
sup{|Em (T ) − E0 (T )| : T ∈ T , kmk ≤ δ} → 0.
Since the parameter space depends on the sample size the usual definition
of relative efficiency does not make sense. Instead, given an alternative m
we will define the efficiency of a (level α where α is fixed) test relative to the
Neyman Pearson test to be 1/c2 where c is chosen so that the test under consideration has power against cm equal to the Neyman Pearson power against
the alternative m. In traditional finite dimensional parametric models this
notion agrees with the usual notion of Pitman relative efficiency (asymptotically). Thus ARE will be the limit of 1/c2 ; the alternative sequence m must
be contiguous so that the asymptotic power of the Neyman Pearson test is
not 1. For a discussion of the relevance of the Neyman-Pearson test as a
standard for the relative efficiency see section 6.
3

Corollary 1 The best invariant test of Ho has ARE 0.
Proof of Theorem 1
Let L(X) be the likelihood ratio for m to 0. Then
Em (T (X)) = E0 (T (X)L(X)) .
Since PX and X have the same distribution under Ho we have
Em (T (X)) = E0 (T (PX)L(PX))
= E0 (T (X)L(PX))

(1)

for all orthogonal P. Since P appears only on the right hand side of (1) we
may average over orthogonal P to obtain
Em (T (X)) =

Z

E0 (T (X)L(PX)) F (dP)


= E0 T (X)

Z

L(PX)F (dP)



(2)

where F is any probability measure on the compact group of orthogonal
matrices. Let
Z
L(X) = L(PX)F (dP).
Then

|Em (T ) − E0 (T )| = |E0 (T (L − 1))|
≤ E0 (|L − 1|).

(3)

Since the last quantity is free of T we need only show that
sup{E0 (|L − 1|) : kmk ≤ δ} → 0.
Since E(L) = 1 the dominated convergence theorem shows that it suffices to
prove for an arbitrary sequence of alternatives m with kmk ≤ δ and for a
suitably chosen sequence of measures F that L → 1 in probability. We take
F to be Haar measure on the compact group of orthogonal transformations;
that is, we give P a uniform distribution.
4

For each fixed X, when P has the distribution for which F is Haar measure
on the orthogonal group, the vector PX has the uniform distribution on the
sphere of radius kXk. Using the fact that a standard multivariate normal
vector divided by its length is also uniform on a sphere we find that
L = H(kmk · kXk)/H(0)
where
H(t) =

Z

π
0

exp(t cos θ) sinn−2 θ dθ.

Standard asymptotic expansions of Bessel functions (see, e.g. Abramowitz
and Stegun, 1965, p 376ff) then make it easy to show that L → 1 in probability, finishing the proof.•
A test invariant under the group of orthogonal transformations is a function of kXk2 and the analysis above can be made directly and easily using
the fact that this statistic has a chi-squared distribution with non-centrality
parameter kmk2 . Our interest centres on the technique of proof. Equations
(1-3) and the argument following (3) use only the group structure of the
problem and the absolute continuity of the alternative with respect the null.
The remainder of the argument depends on an asymptotic approximation to
the likelihood ratio averaged over alternatives. Whenever such an approximation is available we can expect to obtain efficiency results for the family of
invariant tests. In the next section we apply the technique to a more general
model by replacing the explicit Bessel function calculation with a version of
the permutation central limit theorem.

3

Exponential Families

Suppose now that X = (X1 , . . . , Xn )T with the Xi independent and Xi having the exponential family density exp(mi xi − β(mi )) relative to some fixed
measure. We assume the mi ’s take values in Θ an open subset of R. The
parameter space is then M = Θn . Let Θ0 be a fixed compact subset of Θ.
Consider the null hypothesis Ho : m1 = · · · = mn . This problem is invariant
under the subgroup of the orthogonal group consisting of all permutation
5

matrices P. Let m = mi /n; we also use m to denote the vector of length
n all of whose entries are m. The calculations leading to (1-3) establish that
for any test T in T , the family of all tests invariant under permutations of
the entries of X, we have
P

|Em (T ) − Em (T )| ≤ Em (|L − 1|).
where now
n

L = exp −

X

(β(mi ) − β(m))

oX





exp (m − m)T PX /n!

P

is the likelihood ratio averaged over all permutations of the alternative vector
m.

3.1

Heuristic Computations

Think now of X as fixed and P as a randomly chosen permutation matrix.
Then (m − m)T PX = (m − m)T P(X − X) has moment generating function
g(s) =

X

exp{s(m − m)T P(X − X)}/n!.

P

If max{|mi − m| : 1 ≤ i ≤ n} → 0 the permutation central limit theorem
suggests that (m − m)T PX has approximately a normal distribution with
P
P
mean 0 and variance (mi − m)2 (Xi − X)2 /n. Thus heuristically
g(s) ≈ exp{s2

X

(mi − m)2

X

(Xi − X)2 /(2n)}.

Under the same condition, max{|mi − m| : 1 ≤ i ≤ n} → 0, we may expand
X

(β(mi ) − β(m)) ≈

X

(mi − m)2 β ′′ (m)/2.

We are led to the heuristic calculation
L(X) ≈ exp{

X

(mi − m)2 (S 2 − β ′′ (m))/2}

where S 2 = (Xi − X)2 /n is the sample variance. Since Varm (X) = β ′′ (m)
P
we see that L should converge to 1 provided (mi − m)2 = o(n1/2 ) or
P

6

km − mk = o(n1/4 ). In the simple normal example of the previous section
we are actually able to prove (again using asymptotic expansions of Bessel
functions) this strengthened version of Theorem 1, namely, if a = o(n1/4 )
then
sup{|Em (T ) − E0 (T )| : T ∈ T , kmk ≤ a} → 0.
In the more general exponential family problem we do not know a permutation central limit theorem which extends in a useful way to give convergence
of moment generating functions. For contiguous alternative sequences we are
able to replace the moment generating function by a characteristic function
calculation. We can then prove that invariant tests have power converging
to their level uniformly on compact subsets of Θ.
Theorem 2 As n → ∞,
sup{|Em (T ) − Em (T )| : T ∈ T , km − mk ≤ δ, mi ∈ Θ0 } → 0.

3.2

Proof of Theorem 2

Our proof uses contiguity techniques to replace the moment generating function used in the heuristics above by a corresponding characteristic function
calculation. A standard compactness argument reduces the problem to showing that
Em (T ) − Em (T ) → 0
for an arbitrary sequence of alternatives m satisfying (mi − m)2 ≤ δ 2 ,
mi ∈ Θ0 , and an arbitrary sequence T ∈ T . Our proof is now in three stages.
First we prove that any such sequence is contiguous to the null sequence
indexed by m. The second step is to eliminate the particular statistics, T , as
at (3), reducing the problem to showing that the permutation characteristic
function of the log-likelihood ratio is nearly non-random. The final step is
to establish the latter fact appealing to Theorem 3.
P

Step 1: Contiguity of m to m.
The log-likelihood ratio of m to m is
ℓ(X) =

X

(mi − m)(Xi − β(m)) −
7

X

(β(mi ) − β(m)).

Under m the mean of ℓ(X) is
−

X

(β(mi ) − β(m)) = −

X

(mi − m)2 β ′′ (ti )

for some ti between mi and m. In view of the conditions on m and the
compactness of Θ0 this sequence of means is bounded. Also under m the
variance of ℓ(x) is
X
(mi − m)2 β ′′ (m).

Again this is bounded. Thus the sequence of log-likelihood ratios is tight under the null sequence m and so the sequence of alternatives, m, is contiguous
to the null sequence m.

Lemma 1 Suppose Q is a sequence of measures contiguous to a sequence of
measures P . If T is a bounded sequence of statistics such that
EP (T exp(iτ log dQ/dP )) − EP (T )EP (exp(iτ log dQ/dP )) → 0
for each real τ then
EQ (T ) − EP (T ) → 0.
Remark: In the Lemma the random variable log dQ/dP can be replaced by
any random variable S such that log dQ/dP −S tends to 0 in probability
under P .
Remark: The Lemma is very closely connected with LeCam’s third lemma
(see Hájek and Šidák, 1967, page 209, their formula 4) which could also
be applied here to the Q characteristic function of T .
Before proving the lemma we finish the theorem.
Step 2: Elimination of the sequence T .
Let H(t, X) = exp(itℓ(X)). According to the lemma we must show
Em (T (X)H(τ, X)) − Em (T )Em (H(τ, X)) → 0 .
Arguing as in (1-3) the quantity in (4) may be seen to be
Em (T (X)H(τ, X)) − Em (T (X))Em (H(τ, X))
8

(4)

where H(τ, X) = P H(τ, PX)/n! = E(H(τ, PX)|X). Since both T and
H(τ, X) are bounded it suffices to prove that
P

H(τ, X) − Em (H(τ, X)) → 0

(5)

in probability.
Remark: If ℓ(X) = S(X)+oP (1) under m then in view of the remark following
the lemma the random variable ℓ(X) can be replaced in the definition of
H by the random variable S(X). Moreover, if S ∗ (X) = S(X) + a then
(5) will hold for S ∗ replacing ℓ if and only if it holds for S replacing ℓ
because the sequence exp(iτ a) is bounded in modulus.
Step 3: Application of the Permutation Central Limit Theorem
Using the last remark take S ∗ (X) = (m − m)T Px. The variable H
P
then becomes P exp(iτ (m − m)T Px)/n! whereas Em (H(τ, X)) becomes
Em (exp(iτ (m − m)T X)).
Now let F̂ be the empirical distribution of the X1 , . . . , Xn . Let X1∗ , . . . , Xn∗
be independent and identically distributed according to F̂ . We will show in
the next section that
X



exp(iτ mT Px)/n! − EF̂ exp(iτ

P

and



EF̂ exp(iτ

X





X

mj Xj∗ ) − Em exp(iτ



mj Xj∗ ) → 0

X



mj Xj ) → 0

(6)

(7)

in probability for each fixed τ . This will finish the proof of Theorem 2 except
for establishing the lemma.
To prove the (undoubtedly well-known) lemma argue as follows. Letting
ℓ = log dQ/dP , the condition shows that EP (T φ(ℓ)) − EP (T )EP (φ(ℓ)) → 0
for each bounded continuous function φ. There is then a sequence a tending
to infinity so slowly that EP (T f (ℓ)) − EP (T )EP (f (ℓ)) → 0 where f (x) =
min(ex , a). In view of contiguity EQ (T ) − EP (T eℓ ) → 0 and for any sequence
a tending to infinity |EP (T (eℓ − f (ℓ)))| ≤ EP (|eℓ − f (ℓ)|) → 0. The lemma
follows.•

9

3.3

The Permutation Limit Theorem

Suppose m ∈ Rn and x ∈ Rn are two (non-random) vectors with m = 0 for
convenience. Suppose P is a random n × n permutation matrix. The random
P
P
variable mT Px has mean nm x and variance s2 = (mi −m)2 (xi −x)2 /(n−
1). An alternative description of the random variable mT Px is as follows.
Let J1 , . . . , Jn be a simple random sample drawn without replacement from
the set {1, . . . , n}. Then J1 , . . . , Jn is a random permutation of {1, . . . , n}
P
and mT Px has the same distribution as mi x(Ji ); we now use functional
rather than subscript notation for legibility. Hájek(1961, see his formula
3.11) shows that it is possible to construct, on a single probability space,
J1 , . . . , Jn together with J1∗ , . . . , Jn∗ , a random sample drawn with replacement
from {1, . . . , n} in such a way that
E[(

X

mi x(Ji ) −

X

mi x(Ji∗ ))2 ] ≤ 3s max |xi − x|/(n − 1)1/2 .

(8)

Since the Xi have an exponential family distribution and Θ is compact it is
straightforward to check that
S max |Xi − X|/(n − 1)1/2 = OP (log n/n1/2 ) = oP (1)
under m where S denotes the sample standard deviation. In view of the
elementary inequality
|E(exp(itW )) − E(exp(itW ′ ))| ≤ t2 E((W − W ′ )2 ) + |t|E1/2 ((W − W ′ )2 ) (9)
this establishes (6).
It seems easiest to deal with (7) after recalling some facts about weak
convergence in the presence of moment conditions. Consider the set, ∆,
of distribution functions F on Rπ which have finite variance. Throughout
2
this paper we write F ⇒Φ if EF (γ(X)) → EΦ (γ(X)) for each fixed continuous function
γ such that γ(x)/(1 + kxk2 ) is bounded. (The notation is
R
EF (γ(X)) = γ(x)F (dx).) This notion of convergence defines a topology
on ∆ which can be metrized by a metric ρ2 in such a way that ∆ becomes a
complete separable metric space. In fact the metric ρ2 may be taken to be
the Wasserstein metric; see Shorack and Wellner (1986, pp 62-65).
Suppose that ∆0 is a subset of ∆. The following are equivalent:
10

i. ∆0 has compact closure.
ii. for each ǫ > 0 there is a fixed compact subset Ψ of Rπ such that
EF (kXk2 1(X 6∈ Ψ)) ≤ ǫ for all F ∈ ∆0 .




iii. there is a fixed function Ψ such that EF kXk2 1(kXk ≥ t) ≤ Ψ(t) for
all F ∈ ∆0 and all t where Ψ has the property that limt→∞ Ψ(t) = 0.
iv. the family ∆0 makes kXk2 uniformly integrable.
Notice that for each
limt→∞ Ψ(t) = 0 the family of distribu fixed Ψ such that

2
tions F with EF kXk 1(|X| ≥ t) ≤ Ψ(t) for all t is a compact metric space.
2

Finally note that F ⇒Φ if and only if F converges in distribution to Φ and
EF (kXk2 ) → EΦ (kXk2 ). All the foregoing results are trivial modifications
of the usual results for convergence in distribution; see Billingsley (1968) pp
31-41. It will be useful to let ρ0 denote a metric analogous to ρ2 for which
the space of all distributions on Rπ becomes a complete separable metric
space with the topology of convergence in distribution. If ∆0 is a compact
subset of ∆ for the metric ρ2 then it is a compact subset of the space of all
distributions for the metric ρ0 .
To state the result let F̂ be the empirical distribution of the numbers
X1 , . . . , Xn . Let X1∗ , . . . , Xn∗ be independent and identically distributed acP
cording to F̂ . Let LF̂ be the (conditional given X1 , . . . , Xn ) law of mi Xi∗
P
and LF be the law of mi Xi .

Theorem 3 Let ∆0 be a fixed compact subset of ∆. Suppose F is any sequence in ∆0 and that for each n the vector X = (X1 , . . . , Xn )T has independent entries distributed according to F . Assume m is an arbitrary sequence
P
satisfying m = 0 and m2i ≤ δ 2 . If F̂ is the empirical distribution function
of X1 , . . . , Xn then ρ2 (LF , LF̂ ) → 0 in probability.
Corollary 2 Under the conditions of Theorem 3


EF̂ exp(iτ

X





mj Xj ) − EF exp(iτ

in probability for each fixed τ .

11

X



mj Xj ) → 0

It is straightforward to rephrase a (weakened) version of Hájek’s permutation central limit theorem in the notation of Theorem 3. Let LP be the
law of mT Px.
Theorem 4 Suppose that m ∈ Rn and x ∈ Rn are two sequences of vectors
such that
nm x → 0,
(10)
X

(mi − m)2 ≤ δ 2

(11)

for a fixed δ and such that x satisfies
X

(xi − x)2 1(|xi − x| ≥ t)/n ≤ Ψ(t)

(12)

where Ψ is a fixed function such that limt→∞ Ψ(t) = 0. If F̂ is the empirical
distribution of the numbers {1, . . . , xn } then ρ2 (LP , LF̂ ) → 0.
The proof of Theorem 3 is in the Appendix.
Remark: Theorem 3 asserts the validity of the bootstrap approximation to
LF . Theorem 4 says that the bootstrapping carried out in Theorem 3 by
sampling from the list X1 , . . . , Xn with replacement can also be carried
out without replacement. Notice that the result applies only to contrasts
in the Xi ; the condition m = 0 is crucial to this bootstrap interpretation
of Theorem 4.

4
4.1

Extensions
Non-exponential families

The exponential family model gives the log-likelihood ratio a rather special
form. For general models, however, a Taylor expansion can be used to show
that the log-likelihood ratio has almost that form. Rather than try to discover
the weakest possible conditions on a general model permitting the conclusion
we try to illustrate the idea with assumptions which are far from best possible.
12

Suppose X = (X1 , . . . , Xn )T with the individual Xi independent and Xi
having density exp(φ(·; mi )) where mi ∈ Θ an open subset of R. Again
consider the null hypothesis Ho : m1 = · · · = mm . Let Θ0 denote some
fixed compact subset of Θ. The log-likelihood ratio for m to m is ℓ(X) =
P
{φ(Xi ; mi ) − φ(Xi ; m)}. Assume that φ is twice differentiable with respect
to the parameter. Let φi denote the i-th derivative of φ with respect to the
parameter. Assume that the usual identities of large sample theory of likelihood hold, namely, that φ1 (X, m) has mean 0 and finite variance ι(m) under
m and that φ2 (X, m) has mean −ι(m). Then we may write
ℓ(X) =

(mi − m)φ1 (Xi , m) − (mi − m)2 ι(m)/2
X
+ (mi − m)2 {φ2 (Xi , m) + ι(m)}/2
X
+ (mi − m)2 {φ2 (Xi , ti ) − φ2 (Xi , m)}

X

X

(13)

where ti is between mi and m.

We see that, under the usual sort of regularity conditions, there will exist
a constant α such that
|Em (φ(Xi ; m∗ ) − φ(Xi ; m))| ≤ α(m∗ − m)2

(14)

Varm (φ(Xi ; m∗ ) − φ(Xi ; m)) ≤ α(m∗ − m)2

(15)

and such that

for all m and m∗ in Θ0 . Under these two conditions we see that |E(ℓ(X))| ≤
P
P
α (mi − m)2 and Varm (ℓ(X)) ≤ α (mi − m)2 . Thus any sequence m with
P
(mi − m)2 ≤ δ 2 and all mi ∈ Θ0 is contiguous to the null sequence m.
The first two steps in the proof of Theorem 3 may therefore be seen to apply to general one-parameter models under regularity conditions, specifically
whenever (14) and (15) hold.
Further regularity assumptions are necessary in order for Step 3 of Theorem 2 to go through in the present context. To get some insight consider
the situation where max{|mi − m|; 1 ≤ i ≤ n} → 0. Under further regularity
conditions we will have
X

(mi − m)2 {φ2 (Xi ; ti ) − φ2 (Xi ; m)} → 0
13

(16)

and
X

(mi − m)2 {φ2 (Xi ; m) + ι(m)} → 0

(17)

in probability. Assuming that (16) and (17) hold we see
ℓ(X) =

X

(mi − m)φ1 (Xi ; m) −

X

(mi − m)2 ι(m)/2 + oP (1).

(18)

Define S ∗ (X) = (mi − m)φ1 (Xi ; m). Step 3 of the proof of Theorem 2
may now be carried through with Xi replaced by Ui = φ1 (Xi ; m) provided
that the map from Θ0 to ∆ which associates m with the m distribution of
φ1 (Xi ; m) is continuous.
P

The assumption that max{|mi − m|; 1 ≤ i ≤ n} → 0 can be avoided; we
now make this assertion precise. We will need three more assumptions:
sup{Em (|φ2 (X, m) + ι(m)|1(|φ2 (X, m) + ι(m)| ≥ t)); m ∈ Θ0 } → 0

(19)

as t → ∞. Define
W (X, ǫ, m) = sup{|φ2 (X, m′ ) − φ2 (X, m)|; |m′ − m| ≤ ǫ, m′ ∈ Θ0 }.
The second assumption will then be
lim sup{Em (W (X, ǫ, m)); m ∈ Θ0 } = 0.
ǫ→0

(20)

Finally we will assume that the map
(m, m) 7→



L((φ(X, m) − φ(X, m))/(m − m)|m) m 6= m
L(φ1 (X, m))
m=m

(21)

is continuous from Θ0 × Θ0 to ∆ where L(X|m) denotes the law of X when
m is true.
Theorem 5 Assume conditions (14, 15, 19, 20, 21). Then as n → ∞,
sup{|Em (T ) − Em (T )| : T ∈ T , km − mk ≤ δ, mi ∈ Θ0 } → 0.
Proof
Assume, without loss of generality that the entries in m have been sorted
so that |m1 − m| ≥ · · · ≥ |mn − m|. Let k = k(n) be any sequence tending to
14

infinity. Then (mi − m)2 ≤ δ 2 implies that max{|mi − m|; k ≤ i ≤ n} → 0.
The assumptions now imply that
P

ℓ(X) =

X

{φ(Xi , mi ) − φ(Xi , m)} +

(mi − m)φ1 (Xi , mi )

i>k

i≤k

−

X

X

(mi − m)2 ι(m)/2 + oP (1).

i>k

Define
S ∗ (X) =

X

{φ(Xi , mi ) − φ(Xi , m) − Em (φ(Xi , mi ) − φ(Xi , m))}

i≤k

+

X

(mi − m)φ1 (Xi , mi ).

i>k

Let H(τ, X) =

P

P

exp(iτ S ∗ (PX))/n!. As before it suffices to prove that
H(τ, X) − Em (H(τ, X)) → 0

(22)

in probability. The proof may be found in the Appendix; its length is due to
our failure to impose the condition that max{|mi − m|; k ≤ i ≤ n} → 0 .

4.2

The Neyman-Scott Problem

Consider now the Neyman-Scott many means problem in the following form.
Let {Xij ; 1 ≤ j ≤ ν, 1 ≤ i ≤ n} be independent normals with mean mi and
standard deviation σ. The usual Analysis of Variance F -test of the hypothesis
that m1 = · · · = mn is invariant under permutations of the indices i. Any
level α test of this null hypothesis for this model with σ unknown is a level
α test in any submodel with a known value of σ. When σ is known the
argument of section 3 can be applied to the vector X = (X 1· , . . . , X n· ) of
cell means to conclude that the ARE of ANOVA is 0 along any contiguous
sequence of alternatives.
This Analysis of Variance problem may be extended to the following
multiparameter exponential family setting. Suppose that for i = 1, . . . , n
the Rπ -valued random variable Xi has density exp{mT
i xi − β(mi )} relative
to some fixed measure on Rπ . The natural parameter space for a single
15

observation Xi is some Θ ⊂ Rπ . Let a(m) be some parameter of interest
and consider the problem of testing Ho : a(m1 ) = · · · = a(mn ). The problem
is again permutation invariant.
Let M be the n × π matrix with ith row mi and M be the n × π matrix
with ith row m. Let X be the n × π matrix with ith row XiT . The log
likelihood ratio of M to M is
ℓ(X) = tr((M − M )T X) −

X

(β(mi ) − β(m))

Denote kMk2 = tr(M T M). Let Θ0 be some fixed compact subset of Θ. Let
T be the family of all permutation invariant test functions, T (X).
Theorem 6 As n → ∞,
sup{|EM (T ) − EM (T )| : T ∈ T , kM − M k ≤ δ, mi ∈ Θ0 } → 0.
The proof of this theorem is entirely analogous to that of Theorem 2
needing only a multivariate extension of Theorems 3 and 4. Suppose M and
x are sequences of n × π matrices. Let M1 , . . . , Mπ and x1 , . . . , xπ be the
columns of M and x respectively. Let P be a random n × n permutation
matrix and let F̂ be the empirical distribution (measure on Rπ ) of the n
rows of x. Let X be an n × π matrix whose rows are iid according to F̂ . Let
LP denote the joint law of (M1T Px1 , . . . , MπT Pxπ ). Let LF̂ denote the law of
MiT Xi where Xi is the ith column of X.
Theorem 7 Suppose that each Mi satisfies (11) and has Mi = 0. Suppose
that each xi satisfies (12). Then ρ2 (LP , LF̂ ) → 0.
The obvious analogue of Theorem 3 also holds.
Theorem 8 Let ∆0 be a fixed compact subset of ∆. Suppose F is any sequence in ∆0 and that for each n the n × π matrix X has independent rows
distributed according to F . Assume M is an arbitrary sequence of n × π
matrices whose columns Mi each satisfy (11) and have Mi = 0. If F̂ is the
empirical distribution function of the rows of X then ρ2 (LF , LF̂ ) → 0 in
probability.
16

It should be noted that the actual null hypothesis plays no role in these
theorems. If the theorems are to be used to deduce that any particular
sequence of permutation invariant tests has poor power properties it is necessary that m1 = . . . = mn imply the assertion that the null hypothesis is
true and that there be some alternative sequence satisfying the conditions of
the preceding theorems.

5

Spacings Statistics

Suppose U1 ≤ · · · ≤ Un are the order statistics for a sample of size n from
a distribution on the unit interval. To test the null hypothesis that this distribution is uniform many authors have suggested tests based on the sample
spacings Di = Ui − Ui−1 where we take U0 = 0 and Un+1 = 1. Examples
P
of statistics include Moran’s statistic log(Di ) and Greenwood’s statistic
P 2
Di . See Guttorp and Lockhart (1989) and the references therein for a
detailed discussion. Notice that these statistics are invariant under permutations of the Di . Also note that the joint distribution of the Di is permutation
invariant.
Consider a sequence of alternative densities 1 + h(x)/n1/2 . Čibisov(1961)
showed (though his proof seems to rely on convergence of the permutation
moment generating function which does not seem to me to follow from the
form of the permutation central limit theorem which he cites) under differentiability conditions on h that the power of any spacings statistic invariant
under permutations of the Di is asymptotically equal to its level using essentially the method of proof used above. We can relax the conditions on h
somewhat to achieve the following.
Theorem 9 Let ∆0 be a compact subset of L2 , the Hilbert space of square
integrable functions on the unit interval. Let T be the family of permutation
invariant test functions T (D). As n → ∞,
sup{|Eh (T ) − E0 (T )| : T ∈ T , h ∈ ∆0 } → 0
.
17

In Guttorp and Lockhart(1988) it is established that, under the condiP
tions of the theorem, ℓ(D), the log-likelihood ratio, is equal to hi (Di −
R
1/(n + 1)) − h2 (x) dx/2 + OP (1) for any sequence of alternatives h converging in L2 where the hi are suitable constants derived from h. Using the
remark following Lemma 1, the fact that the joint distribution of the Di is
permutation invariant under the null hypothesis, and the characterization of
the spacings as n + 1 independent exponentials divided by their total, the
theorem may be proved by following the argument leading to Theorem 2.

6
6.1

Discussion
Relevance of Contiguity

All the theorems establish that permutation invariant tests are much less
powerful than the Neyman-Pearson likelihood ratio test for alternatives which
are sufficiently different from the null that the Neyman-Pearson test has nontrivial power. Thus if, in practice, it is suspected which parameters mi are
the ones most likely to be different from all the others there will be scope for
much more sensitive tests than the invariant tests.
On the other hand, some readers will argue that the analysis of variance
is often used in situations where no such prior information is available. Such
readers, I suspect, will be inclined to argue that this sort of contiguity calculation is irrelevant to practical people. Some readers may feel that a user
would need unreasonable amounts of prior knowledge to derive a better test
than the F -test. Consider the situation of the normal example in the first
section. Suppose that the mi can be sorted so that adjacent entries are rather
similar. Define h(s) = m[ns] . If the sequence of functions are reasonably close
to some square integrable limit η then, without knowing η, we can construct
a test whose power stays larger than its level if the Neyman-Pearson test has
the same property. Specifically consider the exponential family example. Let
γi ; i = 1, 2, . . . be an orthogonal basis of L2 [0, 1] with each γi continuous and
let λi be any sequence of summable positive constants. Define a test statistic
P
P
P
of the form T (X) = i λi ( γi (j/n)Xj )/( j γi2 (j/n))1/2 . If the sequence h
18

converges to some η 6= 0 in L2 [0, 1] then the asymptotic power of T will be
larger than its level. The test is the analogue of the usual sort of quadratic
goodness-of-fit test of the Cramer-von Mises type.
It is worth noting that the calculations compute the power function by
averaging over alternative vectors which are a permutation of a basic vector
m. Another approach to problems with large numbers of different populations (labelled here by the index i) is to model the mi themselves as an iid
sequence chosen from some measure G. In this case the null hypothesis is
that G is point mass at some unknown value m. I note that alternative
measures G which make the resulting model contiguous to the null make
VarG (mi ) = O(n−1/2 ) which means that a typical mi deviates from m by
the n−1/4 discrepancy which arises in our first example and in analysis of
spacings tests. In other words when any ordering of the mi is equally likely
vectors m differing from m by the amount we have used here are indistinguishable from the null hypothesis according to this empirical Bayes model.
It is important to note, however, that for this empirical Bayes model the
hypothesis of permutation invariance of the statistic T is unimportant: if
VarG (mi ) = o(n−1/2 ) then every test statistic, permutation invariant or not,
has power approaching its level.
The proofs hinge rather critically on the exact invariance properties of the
statistics considered. In the Neyman-Scott problem for instance if a single
sample size were to differ from all the others the whole argument would
come apart. As long as the sample sizes are bounded the ARE of ANOVA
is 0 nevertheless, as may be seen by direct calculation with the alternative
non-central F -distribution. In the spacings problem of section 5 the sample
2-spacings defined by Di = Ui+2 −Ui still provide tests with non-trivial power
only at alternatives at the n−1/4 distance; the joint distribution of these 2spacings is not permutation invariant and our ideas do not help. Our ideas
do apply, however, to the non-overlapping statistics of Del Pino(1971).
The definition of ARE offered here may well be challenged since the comparison is relative to the Neyman-Pearson test which would not be used for
a composite null versus composite alternative situation. Nevertheless there
seems to us to be a sharp distinction between procedures for which our definition yields an ARE of 0 and the quadratic tests mentioned above whose

19

ARE is then positive.

6.2

Open Problems and Conjectures

The results presented here lead to some open problems and obvious areas for
further work. Hájek’s proof of the permutation central limit theorem guarantees convergence of the characteristic function and moments up to order 2
of the variables mT Px. Our heuristic calculations suggest that a good deal
more information could be extracted if the characteristic function could be
replaced by the moment generating function and if convergence of the moment generating function could be established not only for fixed arguments
but for arguments growing at rates slower than n1/2 . Such an extension would
eliminate, as in the normal example, the need for considering only contiguous alternatives. Large sample theory for spacings statistics suggests that
the results presented here hold out to alternatives at such distances. If, in
addition, approximations were available to the moment generating function
for arguments actually growing at the rate n1/2 the technique might extend
to providing power calculations for alternatives so distant that permutation
invariant tests have non-trivial limiting power. Another possible extension
would use Edgeworth type expansions in the permutation central limit theorem to get approximations for the difference between the power and the level
in the situation, covered by our theorems, where this difference tends to 0.
Consider the exponential family model of section 3 for the special case of
the normal distribution. The problem of testing m1 = · · · = mn is invariant under the permutation group and under the sub-group of the orthogonal
group composed of all orthogonal transformations fixing the vector 1 all of
whose entries are 1. It is instructive to compare our results for the two different groups. The example illustrates the trade-off. For statistics invariant
under larger groups it may be easier to prove the required convergence of
the average likelihood ratio; the easier proof is balanced against applying the
conclusion to a smaller family of statistics.
For statistics invariant under the larger group of orthogonal transformations fixing 1 we can modify the argument of section 2 and extend the
conclusion described in the heuristic problem of section three to relatively
20

large values for km − mk. Rather than describe the details we follow a suggestion made to us by Peter Hooper. Suppose Y = Xb + se where e has a
multivariate normal distribution with mean vector m and variance covariance
matrix the identity and where s is an unknown constant, X is an n×p matrix
of regression covariate values of rank p and b is an unknown p dimensional
vector. Let H = X(X T X)−1 X T be the hat matrix. Suppose we wish to test
the hypothesis (I − H)m = 0. (If X = 1 this is equivalent to the problem
mentioned above of testing m1 = · · · = mn . The problem is identifiable only
if Hm = 0 or equivalently if X T m = 0.) The problem is invariant under the
group OX of orthogonal matrices P for which PX = X.
Suppose T (Y ) is a family of statistics such that T (PY ) = T (Y ) for any
P in OX . Consider the likelihood ratio of m to Hm (the latter is a point in
the null). Following equations (1-3) we are lead to study
L(Y ) =

Z

L(PY )F (dP).

where now F is Haar measure on OX and L(Y ) = exp(mT (I − H)(Y −
Xb) − k(I − H)mk2 /2). Since PX = x we see that L(PY ) = exp(mT (I −
H)P(Y − Xb) − k(I − H)mk2 /2). If P is distributed according to F and
Z is standard multivariate normal then PT (I − H)m/k(I − H)mk and (I −
H)Z/k(I − H)Zk have the same distribution. This fact and expansions of
Bessel functions show that L → 1 in probability provided k(I − H)mk =
o((n − p)1/4 ).
The family of statistics invariant under the group of permutations of the
entries of Y will be different than the family invariant under OX . When
X is simply a vector which is a non-zero multiple of 1 the family of statistics invariant under the permutation group is much larger than the family invariant under O1 . For this case we are led to study the variable
P
L = P exp((m − m)T Pe − km − mk2 /2)/n!. We find that Em (L) = 1
P
and Varm (L) = P exp((m − m)T P(m − m))/n! − 1. Just when this variance
goes to 0 depends on extending the permutation central limit theorem to
give convergence of moment generating functions. Since the random variable
(m − m)T P(m − m) has mean 0 and variance km − mk4 /n we are again led
to the heuristic rate km − mk = o(n1/4 ). However, by taking m to have
exactly one non-zero entry it is not too hard to check that this heuristic calculation cannot be made rigorous without further conditions on m to control
21

the largest entries.
Finally, if X is not a scale multiple of 1 the problem is not invariant under
the permutation group. Is there some natural extension of our techniques to
this context for a group smaller than OX ?

Appendix
Proof of Theorem 3
We prove below (cf Shorack and Wellner, p 63 their formula 5, except that
there the distribution F does not depend on n) that
ρ2 (F, F̂ ) → 0

(23)

in probability. If Theorem 3 were false then from any counterexample sequence we could extract a subsequence which is a counterexample and along
which the convergence in (23) is almost sure. The theorem then follows from
the assertion that
ρ2 (F, G) → 0

implies

ρ2 (LF , LG ) → 0

(24)

whenever F is any sequence of distributions with compact closure in ∆.
Assertion (24) is a consequence of Lemma 1 of Guttorp and Lockhart (1988).
To prove (23) we may assume without loss, in view of the compactness of
2
∆0 that F ⇒Φ for some Φ. Elementary moment calculations assure that F̂ (τ )
converges in probability to Φ(τ ) for each τ which is a continuity point of Φ.
This guarantees that ρ0 (F̂ , Φ) → 0 in probability. We need only show that
P
EF̂ (X 2 ) → EΦ (X 2 ). But EF̂ (X 2 ) = Xi2 /n. The triangular array version of
the law of large numbers given in Lemma 2 of Guttorp and Lockhart(1988)
P
2
shows that Xi2 /n − EF (X 2 ) → 0 in probability. Since F ⇒Φ implies that
EF (X 2 ) → EΦ (X 2 ) we are done.

22

Proof of Theorem 5
It remains to choose a sequence k = k(n) in such a way that we can
check (22). In view of permutation invariance we may assume without
loss that |m1 − m| ≥ · · · ≥ |mn − m|. Define matrices C and D by
setting C(i, j) = φ(Xj , mi ) − φ(Xj , m) − Em (φ(Xj , mi ) − φ(Xj , m)) and
D(i, j) = (mi − m)φ1 (Xj ; m). We will eventually choose a sequence k and
put B(i, j) = C(i, j) for i ≤ k and B(i, j) = D(i, j) for i > k. Note that
P
∗
i B(i, i) is simply S (X).
If P is a random permutation matrix then in row i there is precisely 1 nonzero entry; let Ji be the column where this entry occurs. Then S ∗ (PX) =
P
i B(i, Ji ). The variables J1 , . . . , Jn are a random permutation of the set
{1, . . . , n}. As in the proof of Theorem 2 let J1∗ , . . . , Jn∗ be a set of independent
random variables uniformly distributed on {1, . . . , n}. We will show that for
each fixed κ


ρ2 L(

and



ρ2 L(

X

C(i, Ji∗ )|X), L(

i≤κ

X

X
i≤κ

C(i, Ji∗ )|X), L(

X
i≤κ

i≤κ

C(i, i)) → 0

(25)



C(i, Ji )|X) → 0

(26)

in probability. We will also show that for any sequence k tending to ∞ with
k 2 = o(n) we have


ρ2 L(

X

(mi − m)φ1 (X(Ji∗ ), m)|X), L(

i>k

X
i>k



D(i, i)) → 0

(27)

in probability. There is then a single sequence k tending to infinity so slowly
that (25) and (26) hold with κ replaced by k and so that (27) holds. We use
this sequence k to define B.
We will then show that
ρ2 L(

X

B(i, Ji∗ )|X), L(

i

X
i

23

!

B(i, i)) → 0

(28)

in probability and that
ρ2 L(

X

B(i, Ji )|X), L(

X

B(i, Ji∗ )|X)

i

i

!

→0

(29)

in probability. These two are enough to imply (22) as in Corollary 2 and the
obvious (but unstated) corresponding corollary to Theorem 4.
Proof of (25)
For each fixed i we may apply (23) with the vector (X1 , . . . , Xn ) replaced
by (C(i, 1), . . . , C(i, n)) to conclude that
ρ2 (L(C(i, Ji∗ )|X), L(C(i, i))) → 0
in probability; the condition imposed on F leading to (23) is implied by (21).
Use the independence properties to conclude
ρ2 (L(C(1, J1∗), . . . , C(κ, Jκ∗ )|X), L(C(1, 1), . . . , C(κ, κ))) → 0
for each fixed κ. Assertion (25) follows.
Proof of (26)
For each fixed i we have L(C(i, Ji∗ )|X) = L(C(i, Ji )|X). Furthermore it
is possible to construct J and J ∗ in such a way that for each fixed κ we have
P (Ji = Ji∗ ; 1 ≤ i ≤ κ) → 1. This establishes (26).
Proof of (27)
Let m−k = i>k mi /(n − k) and let U be the vector with ith entry
φ1 (Xi , m). Arguing as in the proof of Theorem 3 (see 24 above) we see that
P



ρ2 L(

X

(mi − m−k )U(Ji∗ )|X), L(

i>k

X
i>k



(mi − m−k )Ui ) → 0

(30)

in probability. We need to replace m−k by m in order to verify (27). Elementary algebra shows m−k − m = O(k/(n − k)). Temporarily let
T1 =

X

(mi − m−k )U(Ji∗ ) −

X
i>k

i>k

24

(mi − m)U(Ji∗ ),

and
T2 =

X

(mi − m−k )Ui −

i>k

X

(mi − m)Ui

i>k

In view of (21) we see that Var(U1 ) = O(1). Hence
Var(T2 ) = (m−k − m)2 (n − k)Var(U1 ) → 0.
Since the central limit theorem shows that the sequence L(
compact closure in ∆ we may use (31) to show that


ρ2 L(

X

(mi − m−k )Ui ), L(

i>k

X
i>k

(31)

P

i>k

D(i, i)) has



D(i, i)) → 0.

(32)

Next
Var(T1 |X) = (m−k − m)2 (n − k)Var(U(J1∗ )|X).

(33)

Since
Var(U(J1∗ )|X) =

X

Ui2 /n − (

X

Ui /n)2

we may apply the triangular array law of large numbers given in Guttorp and
Lockhart(1988, Lemma 2) to conclude that Var(U(J1∗ )|X) = OP (1). Since
k 2 = o(n) we see that the right hand side of (33) tends to 0 in probability.
Hence


ρ2 L(

X

(mi − m−k )U(Ji∗ )|X), L(

i>k

i>k



X

(mi − m)U(Ji∗ )|X) → 0

(34)

in probability. Assembling (30), (32) and (34) we have established (27).
Proof of (28)
Given X, the variables i≤k C(i, Ji∗ ) and i>k (mi − m)φ1 (X(Ji∗ ), m) are
P
P
independent. Similarly i≤k C(i, i) and i>k D(i, i) are independent. Statement (28) then follows from (25) and (27).
P

P

Proof of (29)
To deal with (29) we must cope with the lack of independence among
the Ji . A random permutation of {1, . . . , n} can be generated as follows.
25

Pick J1 , . . . , Jk a simple random sample of size k from {1, . . . , n}. Let P−k ,
independent of J1 , . . . , Jk , be a random permutation of {1, . . . , n − k}. Then
P
B(i, Ji ) has the same distribution (given X) as
k
X

C(i, Ji ) + (m−k − m−k )T P−k U−k

i=1

where the subscript −k on m denotes deletion of m1 , . . . , mk while that on
U denotes deletion of the entries U(J1 ), . . . , U(Jk ).
Let Z0 denote a random variable, independent of J and J ∗ whose condiP
tional distribution given X is normal with mean 0 and variance i>k (mi −
m)2 S where S is the sample variance of the Uj ’s. Statement (29) is a consequence of the following 3 assertions:


ρ2 L(

X

B(i, Ji∗ )|X), L(

X
i≤k

in probability,


ρ2 (L(

X

C(i, Ji ) + Z0 |X), L(

i≤k



C(i, Ji∗ ) + Z0 |X) → 0
X
i≤k

in probability, and


ρ2 L(

X

C(i, Ji ) + Z0 |X), L(



C(i, Ji∗ ) + Z0 |X) → 0

X
i

i≤k

in probability.



B(i, Ji )|X) → 0

(35)

(36)

(37)

Condition (36) follows from (26), the conditional independence of Z0 and
J, J ∗ and the fact that the conditional variance of Z0 is bounded. Condition
(35) is implicit in the proof of (28) after noting that the variance S of the
entries in U is negligibly different from ι(m).
It remains to establish (37). We will condition on (J1 , . . . , Jk ) as well as
X and apply the Permutation Central Limit Theorem. The application of
the conditions of that theorem is a bit delicate since the conditions will only
be shown to hold in probability. We present the argument in the form of a
technical lemma.
26

Lemma 2 Suppose (W1 , W2 , W3 ) is a sequence of random variables. Suppose
g and h are sequences of measurable functions defined on the range spaces of
(W1 , W2 ) and (W1 , W2 , W3 ). Let ζ1 , ζ2 be two independent real valued random
variables. Suppose that there are functions fi for i = 0, 1, . . . (also indexed
as usual by the hidden index n) such that
f0 (w1 ) → 0

implies

2

g(w1 , W2 ) ⇒ζ1

(38)

and
f1 (w1 , w2 ) → 0

i = 1, 2, . . .

implies

2

h(w1 , w2 , W3 ) ⇒ζ2 .

(39)

If f0 (W1 ) → 0 in probability and fi (W1 , W2 ) → 0 in probability then
L(g(W1, W2 ) + h(W1 , W2 , W3 )|W1 ) ⇒ ζ1 + ζ2
in probability. If in addition
E(g(W1 , W2 ) + h(W1 , W2 , W3 )|W1 ) → E(ζ1 + ζ2 )

(40)

Var(g(W1 , W2 )|W1 ) → Var(ζ1 )

(41)

Var(h(W1 , W2 , W3 )|W1 ) → Var(ζ2 )

(42)

in probability,
in probability and

in probability then
2

L(g(W1 , W2 ) + h(W1 , W2 , W3 )|W1 ) ⇒ζ1 + ζ2
in probability.
The lemma is to be applied with W1 = X, with W2 = (J1 , . . . , Jk ) and
with W3 = P−k . Conclusion (37) can be reduced to the form given by a comP
pactness argument. The sequence of laws of i≤k C(i, i) has compact closure
in ∆ in view of (21). Then apply (25) and (26) to conclude that the sequence
P
of laws of i≤k C(i, Ji ) also has compact closure. Let the distribution of ζ1
be any limit point of this sequence of laws. The random variable ζ2 will be
the normal limit in distribution of (m−k − m)T P−k U−k .
27

In order to apply the lemma we must give the conditions of the Permutation Central Limit Theorem in a form in which we have only a countable
family of convergences, as required in (39), to check. Note that (12) in Theorem 4 can be replaced by the assertion that there is a sequence τ1 , . . . of real
numbers increasing to ∞ such that
max(n−1
for each j ≥ 1.

(xi − x)2 1(|xi − x| > τj ), Ψ(τj )) − Ψ(τj ) → 0

X

(43)

We now apply Theorem 4 with x replaced by U−k , with m replaced by
m−k −m and with P replaced by P−k . It is easy to check by conditioning on J
that E(U−k ) = 0 and Var(U−k ) = ι(m)/(n−k). Hence(n−k)(m−k −m)U−k →
0 in probability. Set
T3 =
Then

X

(U−k,i − U−k )2 1(|U−k,i − U−k | > t)/(n − k)

T3 ≤ 2

X

2
U−k,i
1(|U−k,i − U−k | > t)/(n − k)

+2U−k

2

X

1(|U−k,i − U−k | > t)/(n − k).

The second term on the right is OP (1/(n − k)) = oP (1). Since
1(|U−k,i − U−k | > t) ≤ 1(|U−k,i | > t/2) + 1(|U−k | > t/2)
we see that
T3 ≤ 2(n − k)−1

X

Ui2 1(|Ui | > t/2) + oP (1) .

i

Take Ψ(t) = 2 sup{Em (φ1 (X1 , m)1(|φ1 (X1 , m)| > t/2); m ∈ Θ0 } and apply Lemma 2 of Guttorp and Lockhart together with (21) to check that (43)
holds. To finish the proof of (37) we need to check convergence of second moments as in (40), (41) and (42). This can be done using (25), (26) and direct
P
calculation of the conditional mean and variance given X of i>k D(i, Ji ).
Theorem (5) follows.
The technical lemma itself may be proved as follows. From any counterexample sequence we may extract by a diagonalization argument a subsequence
which is still a counterexample and for which f0 (W1 ) → 0 almost surely and
28

fj (W1 , W2 ) → 0 almost surely for each j ≥ 1. For any sample sequence for
which all these convergences occur we have L(h(W1 , W2 , W3 )|W1 , W2 ) ⇒ ζ2
and L(g(W1, W2 )|W1 ) ⇒ ζ1 . Evaluation of the conditional characteristic
function of g(W1 , W2 ) + h(W1 , W2 , W3 ) given W1 by further conditioning on
W2 yields the convergence in distribution asserted in the lemma. The remaining conclusions concerning moments are more elementary analogues of
the same idea.

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Abramowitz, M. and Stegun, I. A. (1965). Handbook of Mathematical Functions. New York: Dover.
Billingsley, Patrick (1968). Convergence of Probability Measures. New York:
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Čibisov, D.M. (1961). On the tests of fit based on sample spacings. Teor.
Verojatnos. i Primenen. 6, 354–8.
Del Pino, G.E. (1979). On the asymptotic distribution of k-spacings with
applications to goodness-of-fit tests. Ann. Statist., 7, 1058-1065.
Guttorp, P. and Lockhart, R. A. (1988). On the asymptotic distribution of
quadratic forms in uniform order statistics. Ann. Statist., 16, 433-449.
Hájek, J. (1961). Some estensions of the Wald-Wolfowitz-Noether Theorem.
Ann. Math. Statist., 32, 506-523.
Hájek, J. and Šidák, Z. (1967). Theory of Rank Tests. Academic Press: New
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Shorack, G. R. and Wellner, J. A. (1986). Empirical Processes with Applications to Statistics. New York: Wiley.

29

