The Integrated

7/30/2019 The Integrated

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The IntegratedConditional Moment Test

Herman J. Bierens

AbstractIn this lecture I will review the integrated con-

ditional moment (ICM) test for functional form

of a conditional expectation model. This is a

consistent test: the ICM test has asymptotic

power 1 against all deviations from the null

hypothesis. Moreover, this test has non-trivialn local power.I will focus on the mathematical foundations

of the ICM approach, in particular the consis-

tency proof and the derivation of the asymp-

totic null distribution.

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1 The Fourier transform of

a Borel measurable function

Let g(x) be a Borel measurable real functionon Rk. The Fourier transform of g(x) relativeto a probability measure X(.) on the Borelsets in Rk is defined by

() = Zexp(i.0x) g(x)dX(x), i = 1,provided that

R|g(x)|dX(x) < .

LEMMA 1: Let g1(x) and g2(x) be Borel mea-surable functions on Rk satisfying R|g1(x)|dX(x)< , R|g2(x)|dX(x) < , with Fouriertransforms 1 () , 2 () , respectively, rela-tive to a probability measure (.) on the Borelsets in Rk. Then g1(x) = g2(x) a.s. X(.),i.e.,

B0 = x Rk : g1(x) g2(x) = 0X(B0) = 1, if and only if 1 () 2 () .2


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Proof: Suppose1 ()

2 () and X(B0) 0. Then we can

define the probability measures

m (B) =1

c ZBrm(x)dX(x), m = 1, 2,

with corresponding characteristic functions

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m () = Zexp(i.0y) dm (y)=

1

c

Zexp(i.0x) rm(x)dX(x)

for m = 1, 2. But I have just established that

Rexp(i.0x) r1(x)dX(x)

Rexp(i.0x) r2(x)dX(x)

hence 1 ()

2 () , which by the unique-

ness of characteristic functions implies that1 (B)= 2 (B) for all Borel sets B Rk. It is nowan easy (ECON 501) exercise to verify that the

latter implies r1(x) = r2(x) a.s. X(.), henceg1(x) = g2(x) a.s. X(.).

Corollary:

LEMMA 2: Let Ube a random variable satis-fying E[|U|] < , and let X Rk be a ran-dom vector. If P [E(U|X) = 0] < 1 then thereexists a Rk such that E[Uexp(i.0X)] 6=0.

Question: Where to look for such a ?

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LEMMA 3: If X is bounded then under the

conditions of Lemma 2, for each > 0 thereexists a satisfying kk < such thatE[Uexp(i.0X)] 6= 0.

Proof: Let X R. Then

E[Uexp(i.X)] = E"U Xm=0

immXm

m! #=

Xm=0

imm

m!E[U.Xm]

Since E[Uexp(i.X)] 6= 0 for some wemust have that E[U.Xm] 6= 0 for some integer

m 0. Let m0 be the smallest m for whichE[U.Xm] 6= 0. Thendm0E[Uexp(i.X)]

(d)m0

=0

= im0E[U.Xm0] 6= 0

which implies that E[Uexp(i.X)] 6= 0 for6= 0 arbitrarily close to zero.

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LEMMA 4: Under the conditions of Lemma 3,

the set S0 = Rk : E[U. exp(i.0X)] = 0has Lebesgue measure zero and is nowhere dense.

Proof: Let k = 1 and 0 S0. Define U0 =Uexp(i.0X) . Then P(E[U0|X] = 0) < 1,hence for an arbitrarily small > 0 there exists

a (, 0) (0, ) such thatE[Uexp(i.0X)exp(i.X)] 6= 0.

By continuity it follows now that for each 0 S0 there exists an > 0 such that

/ S0 for all (0 , 0) (0, 0 + ) .Consequently, in the case k = 1, the set S0 iscountable and is nowhere dense. In the general

case k 1, S0 has Lebesgue measure zero andis nowhere dense.

More generally, we have:

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LEMMA 5: Let w(u) be a real or complexvalued function of the type

w(u) =X

s=0

(s/s!) us

where |s| < and at most a finite numberof ss are zero. Then under the conditions of

Lemma 3, the setS0 = Rk : E[U.w (0X)] = 0

has Lebesgue measure zero and is nowhere dense.

For example, let w(u) = cos(u) + sin(u), orw(u) = exp(u).

The condition that the random vector X isbounded can be get rid of by replacing X with(X), where is a Borel measurable boundedone-to-one mapping, because the-algebra gen-

erated by X is then the same as the -algebragenerated by(X), hence conditioning on(X)

is equivalent to conditioning on X.

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THEOREM 1: Let U be a random variable

satisfying E[|U|] < and let X Rk be arandom vector. Denote

S = Rk : E[U.w (0(X))] = 0

,

where w(.) is defined in Lemma 5, and (.) isa Borel measurable bounded one-to-one map-

ping. If P [E(U|X) = 0] < 1 then S has

Lebesgue measure zero and is nowhere dense,

whereas if P [E(U|X) = 0] = 1 then S =R

k.

2 The ICM test

Given a random sample (Yj, Xj), j = 1, . . ,n,Xj Rk, and a conditional expectation model

E(Yj|Xj) = g(Xj, 0), 0 ,where Rm is the parameter space, The-orem 1 suggests to test the correctness of the

functional specification of this model on thebasis of following ICM statistic:

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Z|bz()|2 d ()In this expression, () is an absolutely con-tinuous (w.r.t. Lebesgue measure) probability

measure with compact support Rk, and

bz() =

1n

n

Xj=1bUjw (

0(Xj)) .

where bUj = Yjg(Xj,b), withb the NLLS es-timator of0, is a bounded one-to-one map-ping, and w(.) is a weight function satisfyingthe conditions of Theorem 1.

More formally, the null hypothesis to be tested

is thatH0: There exists a 0 such thatP [E(yj|xj) = g(xj, 0)] = 1,

and the alternative hypothesis is that H0 is false:H1: For all ,

P [E(yj|xj) = g(xj, )] < 1,

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Under the null hypothesis and standard reg-

ularity conditions,

nb 0 = A1 1

n

nXj=1

g(Xj, )

0

=0

Uj

+op(1)

where

A = p limn 1n

nXj=1

g(Xj, )0

g(Xj, )0

0=0

Hence, by the uniform law of large numbers,

bz() =

1n

nXj=1

bUjw (

0(Xj))

= 1n

nXj=1

Ujw (0(Xj))

1n

nXj=1

g(Xj,b) g(Xj, 0)w (0(Xj))

=

1

n

n

Xj=1 Ujj () + op(1),10


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say, where

j () = w (0(Xj))

b ()0 A1 g(Xj, )0

=0

with

b () = p limn

1

n

n

Xj=1g(Xj, )

0 =0 w (0(Xj)) ,

and op(1) is uniform in .

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THEOREM 2: Under the null hypothesis and

some regularity conditions (one of these con-ditions is that is compact),

bz() = 1n

nXj=1

bUjw (0(Xj)) z() on ,where z() is a zero-mean Gaussian process

on

, with covariance function(1, 2) = E[z(1)z(2)] .

Hence by the continuous mapping theorem,Z|bz()|2 d () d Z|z()|2 d () .

Under the alternative that the null is false,

bz()/n p () uniformly on , where () 6=0 except on a set with zero Lebesgue measure,so that

(1/n)

Z|bz()|2 d () p Z|()|2 d () > 0,

provided that () is absolutely continuous w.r.t.Lebesgue measure and its support has pos-

itive Lebesgue measure.

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3 The null distribution of

the ICM test

If we choose the weight function w real-valued,for example w(u) = cos(u)+sin(u), then z()is a real-valued zero-mean Gaussian process

on , with real-valued covariance function

(1, 2) = E[z(1)z(2)]

= limn

1

n

nXj=1

E

U2j j (1)j (2)

.

This covariance function is symmetric and pos-

itive semidefinite, in the following sense:

Z Z(1)(1, 2)(2)d (1) d (2) 0for all Lebesgue integrable functions () on. Such functions have non-negative eigenval-ues and corresponding orthonormal eigenfunc-

tions:

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THEOREM 3: The functional eigenvalue prob-

lem R(1, 2)(2)d (2) = .(1) a.e. on has a countable number of solutionsZ

(1, 2)j(2)d (2) = j.j(1)a.e. on ,

j = 1, 2, .....

where

j 0, Xj=1

j < ,Z

j1()j2()d ()

= 0 if j1 6= j2= 1 if j1 = j2

Moreover,

THEOREM 4 (Mercers Theorem): The co-

variance function (1, 2) can be written as(1, 2) =

Pj=1 jj(1)j(2).

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The sequence {t()}t=1 is an orthonormal

basis for a Hilbert space H () of Lebesgue in-tegrable functions on , with inner product

hf, gi =

Zf() g () d () .

so that every function finH () can be writtenas

f() =X

t=1

tt(),X

t=1

2t < , where

t = hf,ti , t = 1, 2, 3, .....

It can be shown that z() is a random elementofH () , hence

z() =

Xt=1

ztt(), where

zt =

Z

z()t()d () , t = 1, 2, 3,....

Consequently,

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Z|z()|2 d () = Z Xt=1

ztt()!2 d ()=

Xt1=1

Xt2=1

zt1zt2

Zt1()t2()d ()

=

Xt1=1

Xt2=1 zt1zt2I(t1 = t2) =

Xt=1 z2tThe sequence zt is a zero-mean Gaussian process,with variance function

E

z2t

= E

"Zz()t()d ()

2#

= Z ZE[z(1)z(2)]t(1)t(2)d (1) d (2)=

Z Z Xj=1

jj(1)j(2)

t(1)t(2)

d (1) d (2)

where the latter equality follows from Mercers

theorem.

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Thus by the orthonormality of the eigenfunc-

tions,

E

z2t

=X

j=1

j

Zj()t()d ()

2=

Xj=1jI(j = t) = t.

Moreover, by a similar argument it follows that

E[zt1zt2] =

Xj=1

jI(j = t1) I(j = t2)

= 0 ift1 6= t2.

Hence, denoting t = zt/t ift > 0, we

have:

THEOREM 5:R|z()|2 d () =

Pt=1 t

2

t ,where the ts are i.i.d. N(0, 1) and the ts arethe eigenvalues of the covariance function .

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4 Critical values

The problem is that the eigenvaluest are case-

dependent: They depend on the distribution of

the regressors, the functional form of the NLLS

model, and the conditional variance of the er-

rors. Therefore, the distribution of

R|z()|2 d ()

cannot be tabulated. A possible way to get aroundthis problem is to bootstrap this distribution.

However, a convenient way to get around this

problem is to derive upper bounds of the criti-

cal values, as follows.

Without loss of generality we may assume

that the ts are positive and arranged in de-creasing order. Moreover, it follows from Mer-

cers theorem thatZ(, )d () =

Xj=1

j

Zj()

2d ()

=X

j=1

j

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THEOREM 6: Denoting pt = t/Pj=1 j,we haveR|z()|2 d ()R(, )d ()

=X

t=1

pt2

t

supp1p2.....,

Pt=1

pt=1

Xt=1pt

2

t

= supm1

1m

mXi=1

2i = T ,

say,

so that asymptotic critical values can be de-

rived from the latter distribution. The actual

test statistic of the ICM test is thereforebTICM = R|bz()|2 d()Rb(, )d() ,where

b(1, 2) is a consistent estimator of(2, 2),

uniformly on .

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5 Local power of the ICM

test

Consider the local alternative hypothesis

HL1

: E[Yj|Xj] = g(Xj, 0) +h (Xj)

na.s.,

where h (Xj) is not constant:

P [h (Xj) = E(h (Xj))] < 1.Then under HL

1,bz() z() + () on ,

where z() is the same zero-mean Gaussian processon as before, and() is a deterministic meanfunction satisfying 0 < R()2d () < .Similar to the case under the null hypothesis,we can write

z() + () =X

t=1

ztt()

where now

zt =

tpt + twith t i.i.d. N(0, 1) andt = R()t()d () .20


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Hence,Z|bz()|2 d () d Z|z() + ()|2 d ()=

Xt=1

tpt + t

2

THEOREM 7: The ICM test has nontrivialn-local power, in the sense that for every K >

0,

P

" Xt=1

tpt + t

2 K

#

< P" Xt=1

t2t K#

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Proof:

Let

Cn =X

t=1

tpt + t

2npn + n

2and suppose that n 6= 0. Then

P"

Xt=1 tpt + t2

K#

= P

npn + n

2 K Cn and Cn K

= Php

K Cn npn + n

pK Cn

and Cn K

< PhpK Cn npn pK Cnand Cn K

= P

2nn + Cn K and Cn K

= P

2nn + Cn K

where the inequality is due to the symmetry

and unimodality of the N(0, n) distribution.The result of Theorem 7 follows now by in-duction.

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6 Bibliography

Bierens, H. J. (1982): Consistent Model Spec-

ification Tests, Journal of Econometrics 20,

105-134.

Bierens, H. J. (1984): Model Specification

Testing of Time Series Regressions, Journal

of Econometrics 26, 323-353.Bierens, H. J. (1990): A Consistent Condi-

tional Moment Test of Functional Form,Econo-

metrica 58, 1443-1458.

Bierens, H. J. and W. Ploberger (1997): As-

ymptotic Theory of Integrated Conditional Mo-

ment Tests, Econometrica 65, 1129-1151.De Jong, R. M. (1996): On the Bierens Test

Under Data Dependence, Journal of Econo-

metrics 72, 1-32.

Stinchcombe, M. B., and H. White (1998):

Consistent Specification Testing with Nuisance

Parameters Present Only Under the Alternative,

Econometric Theory 14, 295-325.

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