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5108 K:IProj/Acc&Pros/ClRC/Dateom.indd

7 _ . _fi ..—..____ ,__ —_—_—

THREE ESSAYS ON ECONOMETRICS
BY

WEI SIANG WANG

A DISSERTATION

Submitted to
Michigan State University
in partial fulfillment of the requirements
for the degree of

DOCTOR OF PHILOSOPHY

ECONOMICS

2009

ABSTRACT
THREE ESSAYS ON ECONOMETRICS
BY

WEI SIANG WANG

The first essay, “On the Distribution of Estimated Technical Efficiency in Stochastic
Frontier Models,” considers a stochastic frontier model with error 8 = v — u , where v is
normal and u is half normal. We derive the distribution of the usual estimate of u ,

E(u l 8). We show that as the variance of v approaches zero, E (u | a) — u converges to
zero, while as the variance of v approaches infinity, E(u | 8) converges to E(u). We
graph the density of E (u | a) for intermediate cases. To show that E (u [ 6‘) is a shrinkage
of u towards its mean, we derive and graph the distribution of E (u l a) conditional on u.

We also consider the distribution of estimated inefficiency in the fixed-effects panel data
setting.

The second essay, “Goodness of Fit Tests in Stochastic Frontier Models,” discusses
goodness of fit tests for the distribution of technical inefficiency in stochastic frontier
models. If we maintain the hypothesis that the assumed normal distribution for statistical
noise is correct, the assumed distribution for technical inefficiency is testable. We show
that a goodness of fit test can be based on the distribution of estimated technical
efficiency, or equivalently on the distribution of the composed error term. We consider
both the Pearson chi-squared test and the Kolmogorov-Smirnov test. The bootstrap can
be used to account for the effects of parameter estimation. Alternatively, for the Pearson

test, we use existing results in the literature to account for the fact that estimated

parameters are used to construct the actual and/or the expected cell counts. Finally, we
provide simulation results to show the extent to which the tests are reliable in finite
samples.

The third essay, “Testing Equality of Distribution for Two Correlated Variables,”
discusses how to test the null hypothesis that y] , y2 ,..., y,7 and x1 ,x2 ,...,x,, from a

’5

correlated paired sample of size n: (y,-,x,-) , i = l, 2, .3, ...,n , have the same distribution.

We implement the Pearson chi-squared test, based on differences of frequencies in non-
overlapping intervals (cells) that span the support of the variables, in a GMM setting.
This procedure makes no assumption about the correlation between the two variables.
We also suggest a novel bootstrapping procedure that enables us to generate
asymptotically valid critical values for the Kolmogorov-Smirnov and Baumgartner-

Weiss-Schindler tests.

T 0 my parents

ACKNOWLEDGEMENTS

This dissertation would not have been possible without the help of so many
people in so many ways. First, I express my most heartfelt thanks to Professor Peter
Schmidt. He is an exemplary mentor through his continued patience, understanding, and
guidance. He went beyond normal means in helping me and giving his support for my
endeavors. His guidance is priceless, and for this I will be forever grateful. I wish to
express my sincere gratitude to Professor Christine Amsler for her contribution on my
dissertation. In addition, she is always enthusiastic and willing to share her wealth of
knowledge and life experience with me since the first day I met her. She has broadened
my outlook and mind. Her encouragement has made East Lansing a more welcome place.

In addition, I would like. to thank Professor Emma Iglesias for listening and
giving me sage comments as I struggled to solve problems. I wanted to thank Professor
Scott Swinton for his suggestions on my dissertation and research career. I also owe a
debt of gratitude to Jennifer and Stephanie for their help whenever I needed it.

I wish to thank my fellow graduate students. My special thanks go to Pei-Lin
Simon, Kampon, Gabriel, Brian, Dooyeon, Do Won, Sung-Guam, Meng-Chi, and Tum.

Finally and most importantly, I want to thank my parents, Hong Kiew and Swee
Yong, and my Uncle Jerry and Aunt Janice for their unconditional love, advice, and

support over the years.

TABLE OF CONTENTS

TABLE OF CONTENTS .......................................................................... vi
LIST OF TABLES ................................................................................. vii
LIST OF FIGURES ................................................................................ ix
1. ON THE DISTRIBUTION OF ESTIMATED TECHNICAL EFFICIENCY IN
STOCHASTIC FRONTIER MODELS l
1.1 Introduction ....................................................................................... 2
1.2 The Distribution of ii ........................................................................... 5
1.3 The Distribution of a? Conditional on u ....................................................... 8
1.4 Panel Data ......................................................................................... 11
1.5 Concluding Remarks ........................................................................... 14
Appendix 1 .......................................................................................... 16
2. GOODNESS OF FIT TESTS IN STOCHASTIC FRONTIER MODELS 31
2.1 Introductron31
2.2 Tests Based on the Distribution of s ......................................................... 33
2.3 Simple Hypotheses .............................................................................. 35
2.4 Composite Hypotheses ......................................................................... 37
2.5 An Introductory Example: Testing Normality .............................................. 44
2.6 The Stochastic Frontier Model ................................................................ 47
2.6.1 Size of the Test ....................................................................... 48
2.6.2 Power of the Test .................................................................... 51
2.7 Concluding Remarks ...................... 54
Appendix 2 ......................... , ................................................................. 57
3. TESTING EQUALITY OF DISTRIBUTION FOR TWO CORRELATED
VARIABLES 81
3.1 Introduction ...................................................................................... 81
3.2 Tests of Equality of Two Distributions ...................................................... 82
3.2.1 Chi-squared Test ..................................................................... 82
3.2.2 Kolmogorov-Smimov Test ......................................................... 84
3.2.3 The Baumgartner-Weiss-Schindler Test .......................................... 85
3.2.4 The Bootstrap Resampling Method ............................................... 86
3.2.5 Pairwise T-Test ...................................................................... 87
3.3 Monte Carlo Simulations ...................................................................... 88
3.3.1 Size ofthe Test ....................................................................... 88
3.3.2 Power of the Test .................................................................... 90
3.4 Concluding Remarks ........................................................................... 91
BIBLIOGRAPHY ............... . ................................................................. 111

vi

LIST OF TABLES

Table 2.1 Size of the test of the hypothesis that the data are N(O, 1). ...................... 62
Table 2.2 Size of the test of the hypothesis that the data are normal ........................ 63
Table 2.3 Quantiles of the distribution of the normal / half normal composed error. .....64

Table 2.4 Size of test of the hypothesis that the data are normal / half-normal
(A = 0.1) ................................................................................. 67

Table 2.5 Size of test of the hypothesis that the data are normal / half-normal
(k = 0.5) ................................................................................. 68

Table 2.6 Size of test of the hypothesis that the data are normal / half-normal
(k = 1) ................................................................................... 69

Table 2.7 Size of test of the hypothesis that the data are normal / half-normal
(7» = 2) ................................................................................... 70

Table 2.8 Size of test of the hypothesis that the data are normal / half-normal
(7t = 10) ................................................................................. 71

Table 2.9 Power of the test of the hypothesis that the data are normal / half-normal
Alternative: the data are normal / exponential(6’)
Number of cells: k = 3 ................................................................ 72

Table 2.10 Power of the test of the hypothesis that the data are normal / half-normal
Alternative: the data are normal / exponential(0)
Number of cells: k = 5 ................................................................. 73

Table 2.11 Power of the test of the hypothesis that the data are normal / half-normal
Alternative: the data are normal / exponential(6)
Number of cells: k = 10 ............................................................... 74

Table 2.12 Power of the test of the hypothesis that the data are normal / half-normal
Alternative: the data are normal / gamma (u is c times gamma(m))
”1:01 .................................................................. 75

Table 2.13 Power of the test of the hypothesis that the data are normal / half-normal
Alternative: the data are normal / gamma (u is c times gamma(m))
m = 0.5 .................................................................................. 76

vii

Table 2.14 Power of the test of the hypothesis that the data are normal / half-normal
Alternative: the data are normal / gamma (u is c times gamma(m))
m = 2 .................................................................................... 77

Table 2.15 Power of the test of the hypothesis that the data are normal / half-normal
Alternative: the data are normal / gamma (u is c times gamma(m))

m=10 .................................................................................... 78

Table 2.16 Size of the test of the hypothesis that the data are normal / half normal
with 03 = 03 =1 (simple hypothesis) ............................................. 79

Table 2.17 Size of the test of the hypothesis that the data are normal / half normal

Results conditional on negative skew (Tauchen version of Pearson test) ...... 80

Table 3.1 Size of tests of the hypothesis that the data are bivariate- normal .............. 93
Table 3.2 Power of the tests when p = 0.00 ................................................... 96
Table 3.3 Power of the tests when p = 0.20 ................................................... 97
Table 3.4 Power of the tests when p = 0.40 ................................................... 98
Table 3.5 Power of the tests when p = 0.60 ................................................... 99
Table 3.6 Power of the tests when p = 0.80 ................................................ 100
Table 3.7 Power of the tests when p = 0.90 ................................................. 101
Table 3.8 Power of the tests when p = 0.95 .................................................. 102
Table 3.9 Power of the tests when p = -0.20 ................................................ 103
Table 3.10 Power of the tests when p = -0.40 ................................................ 104
Table 3.1 1 Power of the tests when p = -0.60 ................................................ 105
Table 3.12 Power of the tests when p = -0.80 ................................................ 106
Table 3.13 Power of the tests when p = -0.90 .................... 7 ............................ 107
Table 3.14 Power of the tests when p = -0.95 ................................................ 108

viii

FIGURE 1.1

FIGURE 1.2

FIGURE 1.3

FIGURE 1.4

FIGURE 1.5

FIGURE 1.6

FIGURE 1.7

FIGURE 1.8

FIGURE 1.9

LIST OF FIGURES

The relationship between sand 1} with 03 = 03' = 1 ........................ 20
Density of z? with 03‘ =1 and 03:.1, 1, 10, 100 ........................... 21
The combined graph for Figure 1.2 ............................................. 22

Density ofiilu = .1 with of, =1 and 03:001. .01,.1,1,10,100 ........ 23
Density 0le | u = 2 with 03 =1 and 0'3 :00]. .01, .1, 1, 10, 100 ......25
Density off: | u for u =.l, .5. l and 2 with 03 =1 and 03 = 100 ............ 27

Density oft} and 17 with 03:1,03/T=1andN =10,100,1000
Density 0le and it with 03 =1,03/T= 0.1 and N =10,100,1000
Density oft: and {1 with 03 =1,03/T= 0.01 and N = 10. 100, 1000 ...... 28

Density ofti|u =0.1 and tilu :01 with 03 =1,o—3/T= land N =10,100,
1000

Density oft; l u =01 and fill: :01 with 03 =1,03/T= 0.1and N =10.
100.1000

Density 0le I u =0] and 27121 =0.1 with 03 =1,0'\2,/T= 0.01 and N =10,
100,1000 ........................................................................... 29

Density ofzi|u=2 and 27|u=2 with c73=1,03/T=1 and N =10,100,
1000

Density ofti|u = 2 and ii|u =2 with afi- =1,03/T= 0.1andN =10,100,
1000

Density ofz'ilu =2 and 17|u=2 with 03 =1,0'3'/T= 0.01 and N = 10,100,
1000 ................... l ............................................................. 30

Essay 1

Wang, W.S. and P. Schmidt (2009), “On the Distribution of Estimated Technical
Efficiency in Stochastic Frontier Models,” Journal of Econometrics 148, 36-45.

Essay 1
ON THE DISTRIBUTION OF ESTIMATED
TECHNICAL EFFICIENCY IN STOCHASTIC

FRONTIER MODELS

1.1 INTRODUCTION

In this paper we consider the stochastic frontier model introduced by Aigner,
Lovell and Schmidt (1977) and Meeusen and van den Broeck (1977). We write the model

as

(1.1) yI'T-Xifl'l'gi .. EIZVi-lli , ul-ZO.

Here typically y,- is log output. X ,- is a vector of input measures (e.g., log inputs in the

Cobb-Douglas case), v,- is a norinal error with mean zero and variance 0'3 , and u,- 2 0
represents technical inefficiency. Technical efficiency is defined as TEi = exp(—u,-) , and
the point of the model is to estimate 11,- or TEi.

A specific distributional assumption on ”i is required. The papers cited above

considered the case that u,- is half normal (that is, it is the absolute value of a normal with

mean zero and variance 0;) and also the case that it IS exponential. Other dlstnbutlons

proposed in the literature include general truncated normal (Stevenson (1980)) and gamma

(Greene (1980a, 1980b, 1990) and Stevenson (1980)). In this paper we will consider only

the half normal case, but similar results would apply to the other cases.

Define ,3 to be the MLE of ,6, and 5,- = yi — X130 . Then the usual estimate of u,-,

suggested by Jondrow et a1. (1982), is E (ui lei) , evaluated at 8,- = éi. We can estimate TE,-
by fE: = exp(—1i,-) but a preferred estimate is TE,- = E {exp(—u,~)|8,~} evaluated at 8,- = 53;.

See Battese and Coelli (1988), who also show how to define 12,- and TE in the case of
panel data.

In this paper we derive the distribution of 12,-. (The same method of derivation
would also apply to TE) , though we do not give the details.) It is important to realize that
this is not, and should not be expected to be, the same as the distribution of u,-. In other

words, if one assumes that the u,- are half normal, it is tempting to look at the 1?,- and see if

their distribution looks half normal. It should not, unless 0'3 is very small. We show that

the distribution of 13,- becomes the same as the distribution of ul- as 03 —> 0 (with 0'3

fixed), and that the distribution of 13,- collapses on the point E (u) as 0'3 —> 00. We also

graph the distribution for intermediate values of 05'.
One way to understand the difference between the distributions of z},- and ui is to
realize that z},- is a shrinkage of ui toward its mean. This reflects the familiar principle that

an optimal (conditional expectation) forecast is less variable than the thing being forecast.

The usual breakdown of variance into explained and unexplained parts says:

(1.2) var(u,-) = var[E(u,~ |s,)] + E[var(ui |s,)]

b.)

so that var(11,-) is greater than var(1l,) by the amount E[var(11,- |1€,-)].l An implication of
shrinkage is that on average we will overestimate 11,- when it is small, and underestimate u,-

when it is large. To see the exact sense in which this is true, we also derive the distribution

of 11,- conditional on 11,. We show that as 03 —> 0 (with 03 fixed), the distribution of 1?,-

conditional on 11,- collapses on 11,-, while as 03 —> oo , the distribution of 11,- conditional on

u,- does not depend on u,- (it collapses on the point E(u)). Once again we graph the

distribution for intermediate values of 0'3 , for various values of 11,-.

The relevance of these results is not for inference about u,. To construct
confidence intervals for 11,-, we simply need the distribution of 11,- conditional on 5,, as in
Horrace and Schmidt (1996). Rather, the practical usefulness ofour results is for testing
the adequacy (goodness of fit) of the assumed distribution of u,. Partly our message is
negative: as noted above, it is not legitimate to test the model’s distributional assumptions
by comparing the observed distribution of 1?,- to the assumed distribution of 11,-. However,
there is also a positive message: it is legitimate to test the model’s distributional
assumptions by comparing the observed distribution of 1?,- to the distribution that it should
have if the model’s distributional assumptions are correct. That distribution is what is
given in this paper. The mechanics of such a test are the subject of a subsequent paper.

Although the exposition so far is for the cross-sectional case, our analysis also

applies to the case of panel data, as in Pitt and Lee (1981) and Battese and Coelli (1988).

However, in the case of panel data, one can also consider the alternative of a fixed-effects

 

The expectation is over the distribution of the conditioning variable, 8".

treatment as in Schmidt and Sickles (1984). We analyze the distribution of the

fixed-effects estimate of u,- via simulations, and compare it to the distribution of the

random-effects estimate that assumes a distribution for 11,-. The fixed-effect estimates
show serious bias, as expected, unless 0; IS qu1te small and/or the time-series sample Size,

is quite large. However, when 03 /T is small and the number of firms is not too large, the
fixed-effect estimates are a reasonable alternative to 11,-.

The plan of the paper is as follows. Section 1.2 considers the distribution of 11,-.
Section 1.3 considers the distribution of 17,- conditional on u,-. Section 1.4 discusses the

case of panel data. Section 1.5 gives our concluding remarks. There is also an Appendix

which contains some of the derivations.

1.2 THE DISTRIBUTION OF 11

In this section we derive and discuss the distribution of 1?,- = E (u,- '13,). This is a
random variable because it is a function of a,- , which is a random variable, and its
distribution follows from the distribution of 19,-.

Our discussion will ignore estimation error in ,6 . That is, we consider
1?,- : E(u, 15,-) , whereas in practice 1?,- = E01, 18,-) evaluated at 8,- = 53,-. The difference
between 8,- and 5,- is that 8,- = y,- —X,-fl whereas 5,- = y,- -X,,E ; that is, the difference is
just the contribution of estimation error in ,6 . The justification for ignoring this is that, in
any application we can envision, the intrinsic randomness in E (1.1,- '19,) due to its being a

function of 8,- will dwarf the randomness due to estimation error in ,6 . More formally, the

former is Op(1) while the latter is Op(1/ x/N ). Also, for notational simplicity, we will

66'99
1

henceforth omit subscript from 11,11,v and 8.

Since 11 = E (11 [8) it is a function of 8 , and we can write 11 = 11(8). The function h

was given by Jondrow et a1. (1982):

(1.3) 11=h(s)=—,9—L2—[
0'; +03,

-8+00°4(3/00)l , where 011 =(011+03)'0\1/021’

2(3) = ¢(s) / [1 — (D(s)] , and where ¢ and (I) are the standard normal density and cdf,

respectively.
The function h is a monotonic (strictly decreasing) function, so it can be inverted.

That is, we can formally write

(1.4) s=h"(1i)=g(ti) .

We cannot express the function g analytically, but it is well defined and we can calculate it.
For example, Figure 1.1 shows the function g for the case that 03‘ = 0'3 =1.

Let f8 and f,; represent the densities of 8 and 11. Then making the simple

change of variables in (1.4), we have

(1.5) 112(11)=.fs(g(l1))°

 

8'01)! -

The density of 8 is given by Aigner, Lovell and Schmidt:

(1.6) fg(8)=(2/a)-¢(8/a)-CD(—8b/a) , a=,/05+03 , b=0',,/0',,.2

 

This notation is slightly different from Aigner, Lovell and Schmidt. Our 0 is their 0' and our b is their /1 .
But we have already used xi. for the inverse Mill’s ratio, and there are enough different 0' ’5 already without
introducing another one.

 

Also, we can calculate the Jacobian term g'(11)|. We show in Appendix l-A that

2
a

2 , where x1'(s) = —S/1(s) + 12(8).
011 '1-1 + ”(801) / 0‘0 )1

(1-7) g'(11)=

 

So, substituting (1.6) and (1.7) into (1.5), we obtain

20-18801) / a)-<D(-g(fi)b / a) .
03 l-1+ ,i'(g(ti) / 0'0 )|

 

(13) f1; (12) =

Clearly this is not the same as f,, , the half normal density.

The following result shows what happens in the limit as 0'3 approaches zero and

infinity, respectively. The proof is given in Appendix l-B.

THEOREM 1.2.1:
(1) As 0,? =0, (L1—ll)—)p 0.

(2) As 0'3 —> 0, f,; —> f,, (pointwise).

(3) As 0'3 —>00, 11 —>,, E(11).

(4) As 03 —> oo, [72' / (7t — 2)]-(0',, 103 )-(12 — E(u)) —>d N(O, 1) .
These results make sense if we realize that we are treating 8 = v — u as our

observable quantity. If 0'3 = 0 , so that v E 0 , we effectively observe 11, and so in the limit

11 = 11 and the distribution of 11 equals the distribution of 11. Conversely, when 03 = 00 , 8

contains no useful information about u, and the best estimate of 11 is simply 11 = E (11) . Part
(4) says that, for large 0'3 , 11 is approximately normally distributed around E (u) , with
variance [(7r — 2) / 71']2 -(0',‘,1 /0',2.).

For values of 0'3, between zero and infinity, the density of 11 represents the

shrinkage of u towards its mean, which is ,/(2 / 71)-0',, , or about 0.80 -0',, .3 Figure 1.2

gives the density of 11 for 0'3 = 0.1, 1, 10 and 100. None of these densities looks much

like the half normal. Comparing the densities in the different figures requires some care,

since the axes are scaled differently. However, it is clearly the case that, as 0'3 increases,

the density of 11 becomes more peaked and concentrated more tightly about the mean of

0.80. As 03 becomes large, the distribution of 11 collapses onto the point E(u) , as

indicated in part (3) of Theorem 1.2.1. The approximate normality of the distribution of 11

A-

for large 0', is evident in the last panel (corresponding to 03 = 100).

Figure 1.3 contains the four graphs that were in Figure '1 .2, plus the half-normal
density, on a common set of axes. The use of a common set of axes makes it hard to see the

detail in any one of the graphs, but seeing them all together does make clear what happens

2
as 0',, changes.

1.3 THE DISTRIBUTION OF 11 CONDITONAL ON 11

 

3 . . . .
Note that, by the law of iterated expectations. the mean of 11 is the same as the mean of 11.

In the previous section, we saw that the distribution of 11 is a shrinkage toward the
mean of the distribution of u. Intuitively, this means that we should expect that on average
we will overestimate small realizations of 11 and underestimate large ones. To see the
precise sense in which this is true, in this section we derive and graph the density of 11
conditional on 11.

The density of 11 conditional on u is given by the following equation.

az-expi—(I12a§><g00+u>31
,/(27r)-0'3 .03 . —1 + k'(g(11) / 0'0)l .

 

(1.9) f(z1|11) =

 

The derivation is given in Appendix 1-C.
Theorem 1.2.1 above gives some guidance as to what we should expect this density

to look like. As 0'3 —> 0 , the distribution of 11 conditional on 11 should collapse onto the

. 7 . . . . . .
pomt u. Conversely, as 0,“; —> oo , the distribution of 11 conditional on 11 no longer depends

on u; it collapses onto the point E(u).

The following result shows that, approximately normalized, 11 conditional on 11 is

asymptotically normal both as 0'3? ——) 0 , and as 0'3 —) 00. (The normalization obviously

must differ in the two cases.) The proof is given in Appendix 1-D.

THEOREM 1.3.1:

11 —11

 

(1) As afi- —> 0. —>d N(O,1).

V

” >-( 0: >121 — E(u)) =1 N(0,1>.

(2) As 03 ——>oo,(
71"‘2 0'17

 

 

Results (1) and (2) hold treating 11 as fixed. That is, they deal with the distribution

of 11 conditional on u. Result (2) is, however, the same as the unconditional result given in

result (4) of Theorem 1.2.1.
Figure 1.4 gives the density of 11 conditional on u, for 11 = 0.1, 0'3 =1, and 0'3 =

0.001, 0.01, 0.1, l, 10 and 100. The value 11 = 0.1 is a small value (in the left tail of the

distribution) and so we expect to overestimate it, on average. This does occur except

perhaps for the very smallest value of 0'3. We do not have a strict shrinkage to the mean,

in the sense that there is probability mass for 11 to the left of the true value of u, but except

when 0'3 is very small the vast majority of the probability mass is to the right of u. For the

larger values of 0'3 most of the probability mass is near the mean, E(u). The approximate
normality of the distribution of 11 conditional on 11 for small 0'3 and for large 0'3 can be

seen in the first and last panels of Figure 1.4, respectively. For intermediate values of 0'3

the distribution does not look normal.
Figure 1.5 gives the same results, but now for the case that 11 = 2. The value u = 2
is a large value (in the right tail of the distribution) and so we expect to underestimate it, on

average. This does occur, and again the amount of shrinkage to the mean is small when 0'3

is small and large when 03: is big.
Figure 1.6 illustrates the point that, when 0'3 is large enough, the density of 11

conditional on u no longer depends on 11. In Figure 1.6 we have 0'3 = 1 and 0'3 = 100, and

we display the density of 11 conditional on u for 11 = 0.1, 0.5, 1 and 2. These densities are

10

not much different. With enough noise, the data are no longer very relevant in estimating
u, or equivalently the estimate is not very different depending on the true value of u that
generated the data.

We emphasize that the fact that the conditional expectations estimate 11
underestimates large realizations of u and overestimates small realizations does not mean
that there is anything “wrong” with this estimator. It is, after all, the minimum mean square

error estimate of 11, and it is unbiased in the unconditional sense [ E (11 - 11) = 0] even though
it is not unbiased in the conditional sense [E (11l11) = u ]. Waldman (1984) considers two
alternatives: (i) the “best linear predictor” 17 = a + b8 , where b = — var(u) / [var(u) + var(v)]
and a = (1 + ,6)E(11); and (ii) the “linear unbiased estimator” 11' = ——8. The best linear
predictor is also a shrinkage estimator and so it also underestimates large 11 and
overestimates small 11. The linear unbiased estimator has the conditional unbiasedness
property and is the only estimator that does, so far as we are aware. However, we do not
find it very appealing, because it makes no attempt to remove noise, and indeed Waldman’s

calculations show that it perfonris very poorly, in terms of mean square error or in terms of

correlation with 11, if there is substantial noise in the model.

1.4 PANEL DATA

Although the exposition so far is for the cross-sectional case, our analysis also
applies to the case of panel data, as in Pitt and Lee (1981) and Battese and Coelli (1988).

Now the model is

(1.10) y” :Xilfl+£il , 8i, :1)” —ll,' , i=I,...,]V ,1=I,...,T.

11

Note that the 11,- are time-invariant. For the moment we assume that the v,-, are i.i.d.
normal and the 11,- are i.i.d. half normal. We will call this the random effects case.

Ignoring the effect of parameter estimation and suppressing the subscript i, as

above, we observe 812'"ng and the estimate of u is 11 = E (Zl!8l ,...,€T) , as suggested by

Battese and Coelli (1988). However, for the case that the v ’s are normal, this is the same

as 11 = E (11 IE ) where obviously 8 = V — 11 and V is normal with mean zero and variance

0'3 / T . Therefore the results of Sections 1.2 and 1.3 apply also to the random effects panel

. . . 7
data case, if we Simply reinterpret 03 as 0'; /T.

We now consider the fixed effects case, as in Schmidt and Sickles (1984). Here we

would obtain an estimated intercept, say 11,- , for each firm, and then 11,: (max ,- 61,-) -01,-.

Ignoring estimation error in ,6 , this is the same as 11,: (max ,- 8—,) —8,-. For small (fixed)

. . . * .
N, this 18 best regarded as an estimate of i1,- = 11,- —(minj11j)< 11,. As
. . . . ‘1' .
N —) oo, (minyzl u j) ——> Oand the distinction between u,- and 11,- disappears. The
relevance of this distinction is that 11,- of the previous section is an estimate of u,. While

~ . . . * . . .
u,- is biased upward as an estimate of 11,- , because of the “max” operation, it may or may

not be biased upward as an estimate of 11,- , unless N is large.
The 8,- are the difference between a variable distributed as N (0, 0'3 / T) and a

variable distributed as N (0, 031+ . So the distribution of 11,- depends on 03‘ / T , 0'3 and N.

12

We cannot find an analytical (closed form) expression for the density of i7 ,4 so we resort
to simulation to generate it. Our simulations are based on 100,000 replications. We can
then compare the density of i7 (for various values of N) to the density of i? (which does not
depend on N). For 27 we will consider N = 10, 100 and 1000. We also did simulations for
N = 2000 but they were not very different from those for N = 1000, and in any case such
large values of N do not seem empirically relevant for stochastic frontier models.

Figure 1.7 compares the density of 22 , and the densities of 17 for N = 10, 100 and

1000, for 03:1 and for 0,2. /T= 1, 0.1 and 0.01. When 0'3“ /T= 1, it is easy to see the
upward bias of i7 , in the sense that the distributions are centered well to the right of E(u)
= 0.8. As expected, this is so especially for the larger values of N. This bias diminishes as
0'3 / T decreases. When 03 / T = 0.01, there does not appear to be much bias and the

density of 17 for N = 10 bears a fairly close resemblance to half normal.

Perhaps a more interesting issue is the behavior of the densities conditional on u.
Here we know that ii is not conditionally unbiased because it is a shrinkage toward the
mean, and we expect that {i will also fail to be conditionally unbiased because it is biased
upward due to the max operation. Figure 1.8 gives the results for the distribution

conditional on u = 0.1, a very small value of u. In the first panel, with lots of noise

(0,? / T =1), all of the point estimates are biased upward, and L7 is worse than i? . As

03: / T decreases all of the estimates improve, but ii is still pretty bad for N = 100 and

1 000.

We can derive thejoint distribution of (max / 51-) and E), but the density of the difference between these

two quantities requires an integral that we cannot calculate.

13

Figure 1.9 gives the same kinds of results but for the distribution conditional on u

= 2, a very large value of u. When there is lots of noise (0'3 / T =1), the downward bias of

z? and the upward bias of ii are apparent. As the amount of noise decreases, all of the

estimates improve, but once again 27 is still pretty bad for N = 100 and 1000. For N = 10,
ii is nearly conditionally unbiased for the two smaller values of 03 / T .

To interpret this last result, one should remember that for small N the difference
between 14,- and 21:: u,- —(minj uj) becomes relevant. With 03 =1, the expected value of
(minj uj) is about 0.13 for N = 10. This explains why, in the last panel of Figure 1.9, the
density of L7 is centered to the left of the true value of u = 2, and also why its bias

performance in the panel corresponding to 0'3 / T = 0.1 is as favorable as it is. However,

having said that, it remains the case that when N is small and there is not much noise, ii is

a reasonably good estimator.

1.5 CONCLUDING REMARKS

This paper derived the distribution of the technical efficiency estimate 1? = E (ule),

and also the distribution of z? conditional on u. We used these distributions to make two
main points. The first point is that the distribution of z? is not, and should not be expected
to be, the same as that of u. So, for example, if we assume a half normal distribution for u,
and we plot the distribution of z? , we should not be disturbed when it does not look half
normal. A goodness of fit test, whether formal or informal, should compare the distribution
of i? to the distribution it should have when u is half normal, which is what this paper

provides. The second point is that i? is (in a probabilistic sense) a shrinkage of u toward

14

the mean. On average, we will overestimate the smaller realizations of u and underestimate

the larger realizations. The amount of shrinkage depends on the amount of noise in the

model; it is large when there is lots of noise (0'3 is large) and it is small when there is little

noise.

We also consider the distribution of the estimate 17 from the fixed effects panel data
model. This suffers from a well-known upward bias due to the maximum involved in the
estimation of the efficient frontier. On average we overestimate both small realizations and
large realizations of u. This bias can be severe, but it is not large when N is relatively small
and when there is not much noise.

Our summary of these results is straightforward. If we have the distributional
assumptions correct, it is hard to argue with z? , which after all is the optimal (rational,

minimum mean square error, ...) forecast of u. However, if we have panel data where N is

not too large and where 03 /T is small relative to 0'3 , the estimate 17 based on the fixed

effects model is a plausible alternative, and it has the advantage of not depending on a

distributional assumption.

15

APPENDIX 1

Appendix l-A Derivation of the Jacobian in equation (1.7)

From equation (1.3), we have 1? '2 h(6‘) = k°[—6‘ + Goo/1(8/0'0)], where k 2 0'3 /(0'3 + 03 ).

 

 

 

So
dz? , .
(1.11) —=ko[—l+00-}t (8/00)o(l/0'0)]=k-[—1+/i(8/00)]. Then
a
A —l 2 2
(1.12) g'(zi)=df=[fl‘i] = 1' = 7 ”NOVA
dll d8 k'[—I '1' ll. (8/00)] 0;.[-1+ A'(g(u)/o'0)]

and the Jacobian is just the absolute value of this expression.
Appendix l-B Proof of Theorem 1.2.1

First we give some facts about the inverse Mill’s ratio 1(3) 2 (15(3) / [1 — (D(s)]. As

5 —> -oo , (i) Ms) —> 0 , (ii) 52(3) —> 0, (iii) x1'(s) = —S/i(s) + 22(5) —> 0. (Note that (i) and
(iii) follow from (ii), and (ii) follows from the existence of the integral defining the mean
of the standard normal.)

Now we start with the expression for 12 , as given above. As 0'3 ——) 0 , k —> 1,

—£ —>p u (since as 03. —> 0,v ——>,, 0 ), 0'0 —+ 0 , and 002(8/0'0) —> 0ozi(—oo) = 0.
Therefore 23 —>p u (in the sense that the difference between 13 and u goes to zero). This

proves part (1) of Theorem 1.2.1.

To prove part (2), consider the density of L? as given in equation (1.8) of the text.

2 2 .
As 0,, —>0,wehave a—>0'u, a/0'u —>1/0'u, g(u)—>—u,

l6

¢(g(1i)/a) —* ¢(—U / 01,) = (15(1)! / (71,) . and ¢(—g(fl)b / a) —> <D(oo) = 1. Also the Jacobian
term —> 1 because A'(—oo) = 0. Therefore f;,(zi) ——> 2o(1/0'u)o¢(zi/0'u), which is the half

normal density.

To prove part (3), of the Theorem, we return to the expression for Li given above,

. . . 7 7
which we write as u = —k8 + k0'O/i(8 / 0'0). As 0; —> oo , k —> 0,06 —> oo,k-0'0 —> 0'“

and 2(8/0'0) —> 1(0) 2 ,/(2/7r). Therefore z? —>,, Ulla/(Z/fl' = E(u).

To prove part (4), we write

 

 

 

 

 

(1.13) 0,2, -(1? — E(u)) = — 0': .k.g + 0': ol:k-00 -/i(—8—) — 0'”. 3] .
Cu 0:7 017 ‘70 7’
The first term on the r.h.s. of(l.13) equals
(1.14) 70v 7:),— 70v 7. V bi,
0;;+0'; 0;;+0'; 0v 0v

where “ A z B ” means that A — B —> 0 with probability one as 03‘ —> 00. Note that
—v/0'v isN(0,1).

The second term on the r.h.s. of (l .14) is

(1.15) 0:; {boo-2(1) — a, \F—J
0'“ 0'0 72'

ll

 

 

 

l7

Now use the mean value theoren'i (delta method) to write

8

 

(1.16) Mi) = 1(0)+/1'(0)°( 1
0'

0 00

2 2 8
: —+——o
72' 71' 0'0

and so the term in (1.15) becomes

 

 

 

 

 

 

 

 

 

 

 

0' 2 6‘
(1.17) ‘._.
u 7! 0'0
Also
2 0' 8 2 0' —u v
(1.18) —0 v0 =—o v.( +_)
71' 0"“ 0'0 71' 0'“ 0'0 0'0
NZ 0", v _2_0',,- 0'" _v
7! 0' 0' 7r 0' l 2 2 0'
u 0 u 0'u+0'v V
~- v
0V

Combining (1.18) with (1.14), we have

0' V

2
0v

(1.19)

 

 

3-(27 — E(u)) s: (—l + 3)-
. it

all

which is distributed as N(0,[(7r-2) / 7:12).
Appendix l-C Derivation of f (zilu) in equation (1.9)
We begin by noting that the joint density of (we) is f (u,8) = f” (u)ofv(£ + u). Now

transform to (11,2?) where as before 8 = g(zi). The Jacobian of this transformation is

 

g'(1i)| as given in equation (1.7) of the text. Therefore the joint density of (11.1?) is

18

(1.20) f(u,1?) = ,f;,(14)-fv(g(ii + u)- Jacobian

and the conditional density of 1.? given u is

(1.21) f(zi|u) = f(u,zi)/ f, (u) = f,(g(zi) + u)-Jacobian .

Substituting the normal density for fv and the Jacobian expression in (1 .7), we arrive at the

expression in equation (1.9) of the text.
Appendix l-D Proof of Theorem 1.3.1
To prove part (1) of the Theorem, we write

fi—u v 21 0'0 v—u

=—ko +(k—1)o +k- oxi(

0'v 0". 0", UV 0'0

 

 

 

 

 

(1.22) ) .

As 0'3 —> 0 ,k —> 1 , so the first term on the r.h.s. z -v/0'v . The second term is:

k—l 0,,
cu:— 7011—)

O'v 0'" + 0", 0'

 

 

 

(1.23) 2-u=0.

11
(Remember u is fixed in this calculation.) The third term is

00.2(V—ll

0'v 0'0

(1.24) k-

 

 

)=A<—oo)=0

where we have used the facts that, as 0'3 —9 0 , k —> 1 and 0'0 /0',, ——) 1. Therefore

(1? —u)/0'v —~— —v/0'v which is N(0.l).

The proof of part (2) is essentially the same as the proof of part (4) of Theorem 1.2.1,

and is therefore omitted.

19

Figure 1.1

The relationship between a and z? with 0'3 =

0'3:

1

 

 

 

 

20

 

)

3.0

Figure 1.2
Density ofa with (:3 =1 and 03 =.1, 1,10,100

 

 

 

 

 

 

 

 

 

 

 

 

 

 

03:.1 1 03:
f(u)6l l f(u)T
" . 121 .
l
0.8'L J
0.4!»
o '14 " 0:8 12 1:6 i ‘21 1.1 32 5‘ (14 0511 1:6 2 21 2'8 9.1

 

 

0.8L

 

 

 

 

 

 

 

 

 

115 1.1 0.l5 111 iii 11 it)

u

 

21

Density

Figure 1.3 The combined graph for Figure 1.2

 

 

 

 

 

 

 

 

 

 

 

Density of a half normal and 12
N
‘— l l l l l j l I I
i
,_ 1
II
CD ' H
II
II
,0. ll
I l
l l
v . l l
1'21
5' l — aigrnoV2 : Half normal
N ~ El l'o - - siqmoV2 = 100
° '2 0000000 simgoVQ = 10
..m... "2'1": ---° sigmoV2 = 1
° ' simgoV2 = 0.1
0‘ "o . l L
0.0 0.4 0.8 1.2 1.5 2.0 2.4 2.8 3.2 3.5 4.0
u and z?

22

fining mniq

 

 

 

 

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26

f0? 1 u'i

10

Density of

Figure 1.6
zilufor u =.1,.5, i and2with 03=1and a3=1oo

 

 

 

 

 

 

 

 

 

I f l l l l
. I
:I ‘1
_ . .\ ..
0’ \
.I 1
0’ '\
3’ l
0’ ..‘
- J A -1
'1 l
.’ .1
:1 1
.l .1
0' .\
_. .I J ..
'I '\
0' .\
'I '\
.1 :1
- " ‘1 .
.l :1
a, 1‘
.l .1
0’ .\
I .\
_ 'I \ _ U =.1 _
’ 0‘ - - U =.5
I" ' .\ can.“ U :1
" ‘\\ I--. U :2
0.65 0.70 0.75 0.80 0.85 0.90 0.95 1.00
12 | u

27

1.6

 

Density
1 .2

0.8

0.4

 

16v

Density

1.2~

0.8

0.4

1 . 6
Density

 

Figure 1.7
2
Density oft: and a with 03 =1,“7v =1 and N=10, 100, 1000

_ .______T—_

 

 

 

 

 

 

 

2

Density oft“. and a with 0'3=1,%¥-=.1andN=10,100,1000

 

F—.‘ '—T_ —' """T I" ' r

 

 

 

 

Density Offi and ii with 03 =1,3Tv— = .01 and N=10, 100, 1000

r—‘-—--— - .Afi

N=10

 

 

 

2

 

-..._ _ _._]

 

 

28

Figure 1.8
2
Density offilu =.1 and a 1 u =.1 with (:3 =1,? =1 and N=10,100,1000

,—______—_., __. .v. ==_._ —

2.4.. 1 " 1 "fl__‘__——ffl T ' _ l 'A _J
Density 2_

1.6”
1.2
O.8~
0.4

 

 

 

 

 

 

Densityoffi|u=.1 andti|u=.1 with 03 =1,

4 2 .. .. - . _—
Density

3.2

 

 

 

2.4

 

1.6

0.8‘

 

 

 

2
Density oftilu =.1 and {Zlu =.1 with 03:1,37¥—=.01andN=10,100,1000

 

 

a“- . . .,. 2 . _.. ,2, -. s 2-- -. ..-_._-_._ ., 2
Density 7r ,
6 ~

5 4

4 ~

3 - s

2 s

1 .

l

 

 

 

Figure 1.9
2
Density offiIu = 2 and tiiu =2 with 03=1,"7V=1andN=10,1oo,1000

Density 1 f

0.8~
0.6
Q4.
0.2

 

 

 

2
Density offilu = 2 and fllu =2 with 0,3 =1,i'§1’—=.1 and N=10, 100, 1000

1.6=—-— , ,
Density
1.2

 

 

0.8

0.41

 

 

 

 

2
Density oft: | u = 2 and ti 1 u = 2 with (:3 =1,? =.01 and N=10,100,1000

 

5 fi—W— ' ' I "'—_ ""T_" ' ' ' __ _"m'T'T' _' ' i 1 '_'—'_

Density
4

3-

 

 

 

 

Essay 2
GOODNESS OF FIT TESTS IN STOCHASTIC

FRONTIER MODELS

2.1 INTRODUCTION

In this paper we consider the stochastic frontier model introduced by Aigner,
Lovell and Schmidt (1977) and Meeusen and van den Broeck (1977). We write the model

as

(2.1) inXl'fl‘l'El' , 8":Vi—ll° , 1.120 , i=l,...n.

I

Here typically y,- is log output, X ,- is a vector of input measures (e.g., log inputs in the

Cobb-Douglas case), v,- is a normal error with mean zero and variance 0'3 , and ul- 2 0
represents technical inefficiency: Technical efficiency is defined as TE ,- = exp(—u,-) , and
the point of the model is to estimate it, or TE).

A specific distributional assumption on u,- is required. The papers cited above

considered the case that ui is half normal (that is. it is the absolute value of a normal with

mean zero and variance 0'3) and also the case that it is exponential. Other distributions

proposed in the literature include general truncated normal (Stevenson (1980)) and gamma
(Greene (1980a, 1980b, 1990) and. Stevenson (1980)). Our exposition is for the

cross-sectional case, but we could also consider panel data as in Pitt and Lee (1981).

31

Our interest is in testing the distributional assumption on ui. We will do this while

maintaining the other assumptions that underlie the model, such as the functional form of

the regression, the exogeneity of the X ,- . and the normality of vi. This viewpoint is
motivated by the fact that in this literature the specification of the distribution of ul- is often
regarded as being subject to the most doubt.

The problem then arises that u,- is not observable, and in fact cannot be consistently
estimated. To be more precise, define 0 to be the MLE of ,6 and 8',- = yl- —— X [,0 . Then the

usual estimate of u,- , suggested by Jondrow et al. (1982), is 12,- = E(ui lei) , evaluated at

8,- = 5,. The distribution of 1?,- h'as been derived by Wang and Schmidt (2009). It is not the
same as the distribution of u,- . even for large n. Therefore it is not legitimate to test

goodness of fit by comparing the observed distribution of ii to the assumed distribution of
u. It is legitimate to test goodness of fit by comparing the observed distribution of it to the
distribution derived by Wang and Schmidt. However, it is easier to base the tests instead

on the distribution of a,- that is implied by normality of v,- and the assumed distribution of
ui. This is reasonable because, given that normality of v,- is maintained, a rejection of the

implied distribution of g,- is a rejection of the assumed distribution of ui.

We consider the usual 12 goodness of fit test based on expected and actual

numbers of observations in cells, and also the Kolmogorov-Smirnov test based on the
maximal difference between the empirical and theoretical cdf. For these tests the only

technical problem of note is how to handle the issue of parameter estimation. This is

relevant because both the “observations” 5,- = y,- — X [,0 and the expected numbers of

32

observations in various cells depend on estimated parameters. For the chi-squared test, the
relevant asymptotic theory was developed by Heckman (1984), Tauchen (1985) and Newey
(1985), and we explain how this theory allows asymptotically valid tests in the stochastic
frontier setting. For the Kolmogorov-Smimov test, the comparable asymptotic theory does
not exist. However, the bootstrap can be used to construct asymptotically valid tests (either
for the chi-squared test or for the Kolmogorov-Smirnov test).

The plan of the paper is as follows. In Section 2.2 we discuss further the basics of
goodness of fit testing in the stochastic frontier model. Sections 2.3 and 2.4 contain a
general exposition of goodness of fit tests for simple and composite hypotheses,
respectively. Section 2.5 gives a brief discussion of a prototypical problem, testing for
normality, and presents some simulations. In Section 2.6 we discuss the problem of main
interest, testing the error distribution in the stochastic frontier model, and we present
detailed simulation evidence on the accuracy (size) and the power of various tests. Finally,

Section 2.7 gives our concluding remarks.

2.2 TESTS BASED ON THE DISTRIBUTION OF 8

As noted above, the usual. estimate of u,- is ii,- = E (u,- lay). (This is evaluated at

8,- = 53,- , a point that we ignore in the rest of this section but address subsequently, when we

discuss the relevance of allowing for the effects of parameter estimation.) The distribution

of ii,- is given by Wang and Schmidt (2009). It depends on the assumed distributions for
both v,- and u,- , and it is not the same as the distribution of u,-. Therefore it is not

legitimate to test goodness of fit by comparing the observed distribution of z? to the

33

assumed distribution of u. So, for example, if u is assumed to be half-normal, this does not
imply that 1? should be half-normal, and it is not correct to test the half-normal assumption
by seeing whether the distribution of 22 appears to be half-normal.

This does not mean that the observed distribution of 1? is uninformative. It is
perfectly legitimate to test goodness of fit by comparing the observed distribution of i) to
the distribution that it should have under the distributional assumptions being made, as
derived by Wang and Schmidt. Because this distribution depends on the distribution of
both v and u, we have to maintain the correctness of the assumed (normal) distribution of
v to test the correctness of the distributional assumption on it. This issue is inevitable in
this context.

Such a comparison is complicated because the distribution of L? is complicated. It
is much easier to base a goodness of fit test on the distribution of .9. The distribution of 8
also follows from the assumed distributions of v and u, and so if we maintain the
correctness of the assumed distribution for v, we can test the correctness of the assumed
distribution for u a goodness of fit test based on the distribution of a. This is
computationally easier than a test based on the distribution of 1?. The following simple
point is therefore relevant: ii is a monotonic function of a. This implies that most
goodness of fit tests based on the distribution of 2? will be equivalent to the same goodness
of fit tests based on the distribution of a . For example, the Kolmogorov-Smirrnov test will

be exactly the same whether it is based on the distribution of 1.? or the distribution of 8 . For
the Pearson ,1/2 tests based on the observed versus actual numbers of observations in cells,

again the test is exactly the same whether it is based on the distribution of 1? or the

distribution. of g , provided that the cells are defined conformably.

34

Therefore, for reasons of computational simplicity, we will consider tests based on
the distribution of 8 that is implied by the assumed distributions for v and u. We maintain
the correctness of the assumed (normal) distribution of v, and therefore interpret the tests

as tests of the correctness of the assumed distribution of u.

2.3 SIMPLE HYPOTHESES
Suppose that we have a random sample y1,y2,...,yn and we wish to test the
hypothesis that the population distribution is characterized by the pdf f ( y, 190). The

subscript “zero” on 6 indicates the true value of the parameter 6 , which we assume to be
the same as the value specified by the hypothesis being tested. That is, in this section we

take 60 as given. Thus, for example, we could be testing the simple hypothesis that y is

distributed as N(0,1), as opposed to the composite hypothesis that y is normal with y and

0'2 unspecified.

To define the Kolmogorov-Smirnov statistic, let F ( y, 60) be the cdf corresponding
to the pdf f ( y, (90). Also let F,,( y) be the empirical cdf of the sample: F" (y) = (number
of y,- S y )/n. Then the Kolmogorov-Smimov statistic is
(2.2) KS = sup... lFty.60>— F..(y)|.

The asymptotic distribution of KS is known and widely tabulated. It does not depend on

the form of the distribution (f, or F).

Now consider the Pearson 12 statistic. Let the possible range of y be split into k

“cells” (intervals) A1,..., Ak , such that any value of y is in one and only one cell. Let

35

1( y e A j) be the “indicator function” that equals one if y is in cell A j , and equals zero
otherwise. Let p j = pJ-(QO) = P( y 6 A1): E[1( y e A j )]. With 11 observations as above,

we define the observed (0) and expected (E) numbers of observations in each cell:
(2.3) 0j=Zf=li(y,e'/ij) , 15].:an , j=1,...,k.
Then the Pearson 12 statistic is given by:

2_ k 2
(2.4) x —Zj:l(()j—Ej) HEj

Asymptotically (as n —-> 00) its distribution is chi-squared with (k-l) degrees of freedom.
It is interesting (and later it is useful) to put these results into a generalized method

of moments (GMM) framework. We begin with the set of moment conditions

(25) E[g(y~ 6’0 )1 = 0

where g(y,t9) is a vector of dimension (k-l) whose j’h element equals

[1( y e A j ) — .01-(6)]. The subscript “zero” on 19 reinforces the point that the expectation in

(2.5) equals zero only at 60 , the true value of 6. Also, note that we have omitted one cell

so as to avoid a subsequent singularity. We have omitted cell Ak but the choice of which

cell to omit does not matter. Now define

_ 1 n
(2.6) gm) = 3; 2,2, got-.6)

and note that the jth element of §(0) is equal to 1(0j — E j (6)). We also need to define
it

the variance matrix of the moment conditions g(y,60) . This variance matrix is the matrix

V ((90) , of dimension (k-l) by (k-l), whose j’h diagonal element equals ( p j — p} ). and

36

whose i’h , j’h offdiagonal element (1' :2 j) equals (— p,- p j ), with all probabilities
evaluated at 60.

A central limit theorem implies that the asymptotic distribution of J;§(60) is

N(O, V (60) ). From this fact it follows that

_ , _ _ 2
(2.71 "8(90) V00) 'g(60)—>d 21-1-
To link this to the distributional result given above for the test of the simple hypothesis that

y has density f ( y,60) , we simply observe that
_ . . —1 — k 2
(2.8) news) v (60) 54090) = 210(0)- —E,,-) /E).

the Pearson 12 statistic. The equality in (2.8) is proved in Appendix 2-A. So this

establishes the distributional result given in the sentence following equation (2.4).

2.4 COMPOSITE HYPOTHESES

Now suppose that we wish to test the composite hypothesis that the population
distribution is characterized by the pdf f ( y, 6) for some (unspecified and unknown) value
of 6. This is the empirically relevant case.

For the Kolmogorov-Smimov test, we can estimate 6 by MLE. Denote this
estimate by 6. Now we can use 6 in place of 60 in equation (2.2) to calculate the statistic.

The problem is that the distribution of the statistic is changed, even asymptotically, and
furthermore there is no general asymptotic theory to show how to alter the asymptotic
distribution to reflect the effects of parameter estimation. For some cases (e. g. the case that

the distribution being tested is exponential), it is known that the distribution of the KS

37

statistic using 6 does not depend on the value of 60 , so that critical values can be

calculated by simulation. However, there is no such result that would apply to the
stochastic frontier model.

An asymptotically valid Kolmogorov-Smirnov test for a composite hypothesis can

be constructed using bootstrapping. Let f(y,6) be the hypothesized density, evaluated at
the estimate 6. Now we use a “parametric bootstrap”: for b = 1, 2, . . ., B, where B (the

number of bootstrap draws) is large, draw yl(b), y2(b),..., ynlb) from f ( y, 6) . Based on

this data, calculate the estimate 6(1)) and the KS statistic in (2.2) based on 6(b). Then use
the critical values derived from the appropriate quantiles of the empirical distribution of
these B values of the statistic. The asymptotic validity of this procedure has been
established by Giné and Zinn (1990) and Stute, Gonzales and Presedo (1993).

Next we will consider the Pearson 12 test. As for the Kolmogorov-Smirnov test. it
is not legitimate to ignore parameter estimation. Also as for the Kolmogorov-Smimov test,
an asymptotically valid test can be obtained using critical values from the parametric
bootstrap. However, for the 12 test the necessary asymptotic theory to correct for

parameter estimation is known, and tests based on this theory are an alternative to tests
using the bootstrap.

To discuss this asymptotic theory, recall that the number of cells was k, and let the

dimension of 6 be m, with m S k —- 1. We have pj = _pj (6) and Ej (6) = npj (6); that is,
the expected numbers of observations in the cells depend on 6. Thus the value of the

statistic in (2.4) or (2.8), say 12(6) , depends on 6. As above, let 6 be the MLE of 6, and

38

let 6 be the value of 6 that minimizes 12(6). A famous result is that 12(6) is

asymptotically distributed as chi-squared with (k-l-m) degrees of freedom. That is, we still
have a chi-squared distribution but the number of degrees of freedom is reduced by one for

every estimated parameter. This is a nice result but it is not altogether satisfying, since we

have reduced the number of degrees of freedom, and because 6 is in general an inefficient

estimator. It is much more natural to consider the statistic 352 (6) which uses the MLE.
However, this is not asymptotically distributed as chi-squared, and using the chi-squared
distribution with (k-l-m) degrees of freedom results in a test which is conservative, and
therefore presumably less powerful than is possible. (For this result, and the result referred
to above as famous, see, e.g., Tallis (1983, p. 457).)

To understand these results, and the way in which parameter estimation by MLE is
successfully accommodated, we return to the GMM interpretation of the 12 test given at
the end of Section 2.3. The value of 6 is unknown, but we can estimate 6 by GMM based
on the moment conditions (2.5). In this case we will minimize the GMM criterion function
(2.9) n§(0)'V"§(9) .
where I? is either V(6) , in the case of the “continuous updating” GMM estimator, or is any

consistent estimate of V(60) , in the case of the “two step” GMM estimator. The first

possibility corresponds to the minimization of 12(6) with respect to 6 , and yields the

estimator 6 discussed in the previous paragraph. In either the continuous updating case or
the two step case, standard GMM results indicate that the minimized value of the criterion

function (2.9) is asymptotically distributed as chi-squared with degrees of freedom equal to

39

the number of moment conditions minus the number of parameters estimated, that is,
(k-l-m) degrees of freedom. (This is generally referred to in the GMM literature as the “test
of overidentifying restrictions”) This argument establishes the “famous result” referred to
above.

The estimator 6 is not generally efficient, and so we ought to be able to do better

than this. As above, let 6 be the MLE, which is (asymptotically) efficient. Unfortunately
12(6) does not generally have a chi-squared distribution. This raises the question of

whether we can construct a goodness of fit statistic based on 6 that does have a
chi-squared distribution. The answer is yes, as was shown by Heckman (1984), Tauchen
(1985) and Newey (1985). Our discussion will follow Tauchen. We wish to test the

composite hypothesis that the density of y is f (y, 6). Define the “score function”

0'11 f(y.9)

(2.10) s( y. 6) = 66

The MLE satisfies the first order condition 2;] s(y,-,6) = 0 and is the GMM estimator

based on the (exactly identified) set of moment conditions: Es( y, 60) = 0 . Now the

technical trick that leads to the test is to combine these moment conditions based on the
score function with the moment conditions that we want to test, based on numbers of

observations falling into various cells. Formally we write the full set of moment conditions

as Eh(y,60) = 0 , where

17102.6?) S(y.6)
( ) (y ) lh2(y.9)l lgty,9)l

40

Here h] = s is the score function and h2 = g is the vector of (k-l) functions given in

equation (2.5) above. We wish to maintain the correctness of hi (to obtain 6) and test the
correctness of 112.

The test statistic is of the form
(2.12) CMT = 1752(6)sz gm“) .

22

where C will be defined in the next paragraph. The relevant distributional result is that

CMT is asymptotically distributed as chi-squared with (k-l) degrees of freedom. That is,
we do obtain a chi-squared limiting distribution and there is no loss in degrees of freedom

due to estimation of 6.

The difference between this statistic and [2(6) is that the conditional moment test

(CMT) uses C22 where 12(6) uses V(6)—l. The matrix C22 is defined as follows. Let

C be the variance matrix of the vector h(y.6). Its dimension is (m+k-l) by (m+k—l). Let

C ’1 be its inverse. We partition C and C" correspondingly to the partitioning of h into

11' and 112 :

7 " ll tl2
(213) elf" ("2] (7": C C
('21 C22 C21 C22

So C22 is the lower right submatrix, ofdimension(k-1) by (k-l), of C—l . Then C22 is any
consistent estimate of C22. (A specific estimate will be discussed below.) We can note

that C22 = V (60) is the variance matrix of g(y,6) and so basically 12(6) uses (an

41

estimate of) CE; whereas CMT uses (an estimate of) (722. A standard matrix equality

says that

7

(2'14) ('2' = (C22 — '21C1—1'C12Y'
which is bigger than C23}. That is the sense in which the CMT adjusts for the fact that
12(6) is too conservative.

From equation (2.14), an estimate of C 22 requires an estimate of all of C. The

most commonly used estimate is the “CFO” (for “outer product of the gradient”) estimate:
A A1 A 1 A A I
(2.15) (7:09) =—Z;’=,h<yt,6)h(ys9).
n

The CMT using this estimate can be calculated as the uncentered R2 in a regression of a
constant (one) on h(y,-.6). See Newey (1985, p. 1052) for this expression.

Alternatively, for some problems we may be able to obtain an analytical expression
for C , and then evaluate it at 6. The submatrix CI] is the information matrix for the
estimation of 6 by MLE, and may be evaluated using expectations of second derivatives
or cross products of first derivatives. The submatrix C22 can be evaluated as V(6) where

the matrix V is defined in the discussion following equation (2.16). This leaves

C12 = Ehlhz' for which there may be an analytical expression in ceitain simple cases (e.g.

testing for normality), but not in general.

An alternative to the CMT test is to estimate 6 by GMM using the full set of
moment conditions h, and then perform the usual GMM overidentification test. Let 6 be

this estimate. It is different from the MLE 6, but it has the same asymptotic distribution

42

since the second set of moment conditions (g) is statistically redundant for estimation given
the score (5). The GMM overidentification test statistic is nl7(6)'C—ll_1(6). This can be
compared to the CMT statistic 1132(6)'C22 33(6) = n17(6)'C_lk_z(6) , where the last equality
follows from the fact that §(6) = 0. So the difference between the CMT and the

overidentification test is just due to the difference between 6 and 6 , which is
asymptotically negligible. The reason the CMT is preferable is a matter of simplicity — if
one has estimated the model by MLE, which would typically be the case, then there is no
need to reestimate the model.

There is a further theoretical point worth making. The discussion above takes the

cells (A j) as given, so that the probabilities of observations falling into the cells are what

depend on 6. This fits well into the GMM setting because the usual asymptotic distribution
theory for GMM depends on the moment conditions being differentiable with respect to 6.
In practice, however, the cell definitions will naturally depend on 6. For example, if we
test normality based on five equi-probable cells, the first cell will be (—oo, ,u — 0840']. So
then the probabilities of being in the various cells are given, but the observed numbers in

the cells depend on ,u and 0'. Now the moment conditions depend on the indicators
1( y 6 AI (6)) which are not differentiable with respect to 6. However, Tauchen shows that
the distributional theory for the CMT still holds in this case.

An interesting theoretical detail that we have not found in the literature is the
following. Instead of comparing observed and expected numbers of observations in cells,
we could compare the sample and population quantiles. For example, in the normal

example mentioned above, in the first case we fix the cell boundary at ([1 — 0.840") and see

43

how close the number of observations less than this is to 0.2. Alternatively, we could

calculate the 20th percentile in the sample and see how close it is to (,0 — 0.840') . Intuition

would suggest that the difference between these two tests ought to be asymptotically
negligible. The sense in which that is true is discussed in Appendix 2-B.

A final point is that it is also possible to apply the bootstrap to the asymptotically

valid (Tauchen) form of the 12 statistic. It is well known that the bootstrap can provide

higher-order refinements to the asymptotically valid distributions of estimators or test
statistics. That is, using asymptotic theory plus the bootstrap may give better (more
accurate in finite samples) critical values than using asymptotic theory alone or the

bootstrap alone. We will investigate that issue in our simulations.

2.5 AN INTRODUCTORY EXAMPLE: TESTING NORMALITY

Our interest in this paper is testing distributional assumptions in stochastic frontier
models. However, first we will present a few simulations for a simpler problem, testing
normality. The point is to see how the tests work in a very simple setting, and in particular
in one where we can do some things analytically that we cannot do in the more complicated
model.

All of our tests will be based on standard normal data. The number of replications
in the simulations is 10,000, except that for the bootstrap tests we use 1000 replications and
999 bootstrap samples per replication. We will consider sample sizes n = 50, 100, 250 and
500. We use tests with nominal size of 0.05. (We also calculated results for sizes of 0.01

and 0.10 but they would not lead to different conclusions than the results for size of 0.05.)

44

For the 12 tests we will consider three different numbers of cells: k = 3, 5 and 10. In all
cases we use equiprobable cells.

We first consider the test of the simple hypothesis that the data are N(0,l). That is,
the values #0 = 0 and 00 = 1 are specified by the null hypothesis (i.e. assumed known).
The 12 statistic is n§(60)'V(60)_l§(60) as given in equation (2.8) above, while the KS
statistic is as given in equation (2.2). The results for these tests are given in Table 2.1,
using the asymptotic critical values and the bootstrapped critical values. They are very easy
to summarize. All of the tests are quite accurate, in the sense that actual size is quite close

to nominal size. The KS test is very slightly undersized for the smaller sample sizes (n =

50 and 100) but this is not a serious discrepancy. Bootstrapping fixes this problem.

Next we consider the test ofthe composite hypothesis that the data are N( ,u, 0'2 ) for

unknown ,u and 02. In Table 2.2 we give the empirical size of the four asymptotically

valid tests discussed above. The first (under the heading “Pearson (Tauchen)”) is the

Tauchen version of the 12 statistic, equal to r1§(6)'C22 §(6) , as given in equation (2.12).
. . ~ . . . _ . l _ .
The MLE of 6 is 6: (21,02) where y = y and 0'2 =—Z'.1_l(y,- —y)2. Then C is
n _

obtained from the inverse of C , the CFO estimate as in equation (2.15), with h( y, 6)

evaluated at the MLE 6. This expression requires the score function for the normal
density, which is given in Appendix 2-C. The second test (“Bootstrap Pearson (Tauchen)”)

uses the same test statistic but uses bootstrapped critical values. The third test (“Bootstrap

Pearson”) is the usual Pearson 12 test given in (2.8), using 6 in place of 60 , with

45

bootstrapped critical values. The fourth test (“Bootstrap KS”) is the Kolmogorov-Smimov
test given in (2.2), but using 6 in place of 60 , with bootstrapped critical values.

The results for these four tests in Table 2.2 are easy to summarize. All of the tests
that use critical values from the bootstrap are quite accurate, in the sense that empirical size
is close to nominal size. The results for the Tauchen version of the Pearson test, which
relies on asymptotic theory instead of the bootstrap, are less favorable. For this test there
are moderate to large size distortions in small samples, especially when a large number of
cells is used. For example, with k = 3 the actual size of the nominal 0.05 level test is 0.067
for n = 50 and 0.055 for n = 100, which is not too bad. For k = 5, we have actual sizes of
0.086, 0.070 and 0.057 for n = 50, 100 and 250, respectively, so the size distortions
disappear more slowly. For k = 10, we have actual sizes of 0.185, 0.117, 0.079 and 0.060
for n = 50, 100, 250 and 500, respectively, so that a rather large sample size is required to
have an accurate test. An obvious implication is not to use too many cells unless the
sample size is large (or, to use the bootstrap).

It is interesting to ask why it is that this test is less accurate than the Pearson test for

the simple hypothesis (Table 2.1). The current test differs from the Pearson test of the
simple hypothesis in a number of ways. First, it uses cells defined on the basis of 6 rather
than 60. Second, it obtains the weighting matrix as a submatrix of the variance matrix of
the moment conditions after they have been augmented with the score. Third, it evaluates
the variance matrix of the (augmented) moment conditions using the OPG estimate, as
opposed to using an analytical expression.

The question is which of these differences is the one that matters. To provide

evidence on this question, we consider two other tests. One (“Simple Hypothesis OPG” in

46

Table 2.2) is based on the test statistic for the simple hypothesis (equation (2.8) above)

except that we replace the variance matrix V(60) with the OPG estimate

~ 1 . . .
C 22 (60) = — 2:11 g(y,-,60 )g(y,~.60 )' . So we are usmg the OPG estimate unnecessarily,
n _

but we do not have estimation error in 6. The second test (“Composite Hypothesis C

Known” in Table 2.2) is based on the test stastistic ng—r(6)'C22 (60)§(6) , where ('22 is the

relevant submatrix of the true (not estimated) variance matrix C of the augmented moment
conditions. This is possible because for this problem we can calculate C analytically. This
expression is given in Appendix 2-C. This test statistic uses the cell definitions based on 6
but uses the analytical expression for C and evaluates it at 60. This would be infeasible in
actual practice but it is feasible in the Monte Carlo setting because we know the value of
60.

The results for “Composite Hypothesis C Known” are quite good. The results for
“Simple Hypothesis OPG” are much less accurate. The size distortions follow the same
pattern as for “Pearson (Tauchen)” though they are a little smaller. Thus it appears that the

finite sample inaccuracy of the non-bootstrapped Tauchen version of the Pearson test is due

primarily to the use of the OPG estimate of C. The true value of C accounts properly for the

effects of parameter estimation but the OPG estimated C does not.

2.6 THE STOCHASTIC FRONTIER MODEL

We now return to the stochastic frontier model. The model is as given in equation

(2.1) above. We will consider the case in which v is normal and u is half-normal. The

47

7 7 . .
parameters of the model are ,6, 0'; and 0'; . We follow the usual convention that 0'3 18 the

variance of the normal random variable of which it is the absolute value, so

7r - 2 .
that var(u) = — 0'3 . We also 'adopt the standard notation xi = 0'" /0,, and
7r

0'2 = 03 + 03 .

We will refer to 8 = v — u as the “composed error”. Its density is known (e.g.
Aigner, Lovell and Schmidt (1977, p. 26)) but its cdf does not have any known closed-form
expression. We need the cdf, or a tabulation of it, to calculate the cell probabilities or the
cell boundaries. We have therefore tabulated the cdf via a simulation, with 10,000,000

replications for each simulation. In Table 3 we present the quantiles 0.1, 0.2, ..., 0.9 for
values of xi between zero and 10,000, and for 0'2 =1. For a given value of xi and for a
different value of 02 , one just needs to multiply the quantile by 0'. (For example, for

xi = l and 02 =1, the 0.40 quantile is -0.754. So for xi = 1 and 0'2 = 2, the 0.40 quantile

is (-0.754)(72 ) = -1.066.)
2.6.1 Size of the Test
Primarily to check these tabulations, we first briefly consider the case in which the

composed error a is observed (equivalently, ,6 is known or specified by the null

hypothesis), and the null hypothesis specifies the value of 0'3 and 03 . Thus we are testing

a simple null hypothesis under which the distribution of 6‘ is completely specified. We
note that, apart from the randomness of the simulation, the results should be exactly the
same as the results in Table 2.1, where we were testing the null that the data are standard

normal. (If the null completely and correctly specifies the distribution of the data. the test

48

is the same as if we compared the percentiles corresponding to the observations to the

uniform distribution, and the nature of the parent distribution is irrelevant.) We also give

results for the composed error with 03 = 0'3 =1 in (Supplemental ) Table 2.16. This table

corresponds to Table 2.1 of this paper for the standard normal case.
Now we turn to the case of main interest, in which we wish to test that the
composed error has the normal fhalf—normal distribution with unspecified (unknown)

values of the parameters. In this case our model for the simulations will be:
(2.16) yi=a+8i , 8,-=V,-—u,- ,
7 7
and the unknown parameters are (1.0;; and 03‘}. There are no regressors other than

intercept, and no slope coefficients, because estimation error in these is not likely to be
important. However, the presence of intercept is important. It is obviously empirically

relevant, and it affects the results because, by demeaning the data, it prevents the level of

the series from containing information about 0'3. In performing our experiments, we
. . 7 .
parameterize in terms of a, xi = 0,, / 0,, and 0',‘,‘. The results depend only on xi (for a given

value of /i , the chi-squared statistics are not affected by changes in a or 0'3) but in

estimation we estimate three parameters.
A technical complication worth mentioning is the “wrong skew” problem pointed
out by Waldman (1982). The distribution of a has a negative skew (negative third central

moment). However, in the data we may encounter a positive skew of the residuals. In our
. . - 1 _
case, this would correspond to m3 > 0 where m3 = —Z;7_l (y,- — y)3 , an occurrence of the
n _

“wrong skew.” When we have the wrong skew, the MLE’s are as follows:

49

A _ .7 .7 l _7
(2.17) a=y,0';=0,0; =;Z;l(yi—y)‘ .

This happens with a positive probability that depends on 21 and. n. For example, when A
is near zero and n is small, the wrong skew problem occurs nearly half of the time. It is
widely argued (e.g. Simar and Wilson (2009)) that the wrong skew problem causes
considerable difficulties in inference in stochastic frontier models. One of the points of our
experiments will be to see whether this is true for goodness of fit testing.

Our results for the cases where the null is true are given in Tables 2.4-2.8, which
correspond to A = 0.1, 0.5, l, 2 and 10, respectively. These tables have essentially the same
format as Table 2.2 (minus its last two columns), except that they also report the frequency
of occurrence of the wrong skew problem.

One striking result in these tables is that the frequency of rejection (size of the test)
does not depend. very strongly on A . That is, for a given value of n and for a given test, the
size of the test is approximately the same in all five tables. In fact, the results in these tables
are very similar to the results in Table 2.2, which was for the case of testing normality with
unknown mean and variance. The parameter estimation problem is much simpler in the
normal case than in the normal (half-normal case, so we might expect larger size
distortions in Tables 2.4-2.8 than in Table 2.2. However, we don’t actually find that; any
differences are very slight. Correspondingly, the main conclusions are the same as in
Section 2.4. All of the tests that use bootstrapped critical values are quite accurate (size
close to nominal size). The Tauchen version of the Pearson test, which relies on asymptotic
theory instead of the bootstrap, is less reliable. There are noticeable size distortions unless

the sample size is very large or the number of cells used is small. Based on these results

50

we would recommend using critical values from the bootstrap. The choice of which test to
use logically would depend on considerations of power, which we will discuss in the next
subsection.

The frequencies of occurrence of the wrong skew problem are in line with previous
evaluations, such as in Simar and Wilson (2009). The fact that the frequency of occurrence
of the wrong skew problem varies strongly with /l. , but the size of the test does not, would
seem to imply that any size distortions we encounter are not primarily a reflection of this
problem. As a matter of curiosity, we also calculated the frequency of rejection for those
samples where the wrong skew problem did and did not occur. We did this for the Tauchen
version of the Pearson test only, since that was the only test with significant size
distortions. These results are given in (Supplemental) Table 2.17. The frequencies of
rejection are different but not too different for the samples with the wrong skew than they
are for the samples with the correct skew. For example, with n = 50 and xi =1 , we have

6399 replications with the correct (negative) skew and 3601 with the wrong (positive)

skew. For the Tauchen [2 test with k = 3, we have rejection frequencies of 0.094

conditional on correct skew and 0.055 conditional on wrong skew; for k = 5 we have 0.104
and 0.074. These numbers are clearly different, but it is not the case that the rejections are
coming overwhelmingly from one case or the other.
2.6.2 Power of the Test

Now we turn to the question of the power of the test. This requires specification of
the alternative hypothesis. The null is exactly as in the previous section: the model is as
given in equation (2.16), and the null is that the composed error e = v — u has the

distribution implied by v being normal and u being half-normal. The alternatives that we

51

consider will be based on the same model, except that u will follow some other one-sided
distribution. Specifically, we will consider exponential and gamma distributions for u.

For the simulations in this subsection, we still use 10,000 replications for the
Tauchen version of the Pearson test, and 1000 replications with 999 bootstrap samples for
the tests based on the bootstrap, except that for the bootstrapped KS test, we use 1000
replications with 399 bootstrap samples.

Tables 2.9, 2.10 and 2.1 1. give the power of the test when v is N(O,l) and u is
exponential with mean equal to 6 (and, correspondingly, variance equal to 6 2 ). We
consider 6 = 0.1, 0.5, 1, 2, 5 and 10. Varying 6 changes the relative importance of noise and
one-sided error. Since the results of the tests are invariant to linear transformation of the

data, we could equally have kept 6 fixed and changed the variance of v. (For example. the

results with 6 = 5 and 0'3? =1 are the same as with 6 = l and 0'3 = 1/25.) Larger values of

6 correspond to less noise relative to one-sided error, and presumably should lead to higher
power, since it is easier to distinguish half-normal and exponential data if they are
contaminated with less noise. As a result, as we move down in each section of these tables,
power should increase as 6 increases. However, it is not the case that the power goes to one
as 6 —> oo . Rather, as 6 —> oo , power should approach the power that we would have if
there were no noise and we were testing the null that the data are half-normal against the
alternative that they are exponential.

The KS test using bootstrapped critical values is generally the most powerful test.
It clearly dominates the other two tests that use bootstrapped critical values. Its comparison
to the non-bootstrapped Tauchen version of the Pearson test is somewhat ambiguous,

because the Tauchen test sometimes appears to be more powerful, but this occurs in cases

52

(small n and/or large k) in which the Tauchen test had non-trivial size distortions. Even in
those cases the bootstrapped KSitest is more powerful if 6 is large enough that power is
non-trivial. Basically whenever power is over 0.2, the bootstrapped KS test is best.

Comparing results for the various Pearson tests across the three tables, we see that
power is generally higher when less cells are used. That is, power is higher with three cells
than with five, and higher with five cells than with ten. (There are a few exceptions for the
non-bootstrapped Tauchen test when power is low and n and k are such that size distortions
were found under the null.) Since size distortions are smaller and power is higher when a
small number of cells is used, it is obvious to recommend using a small number of cells.
Precisely how small is a question that could be investigated further.

Unfortunately, we can also see that power is rather low unless the sample size is
quite large and/or the variance of u is much larger than the variance of v. For example,
when 6 = 1, which corresponds to equal variance for v and u, power for the bootstrapped
KS test is only 0.054 for n = 50, 0.089 for n = 100, and 0.150 for n = 250. When 6 is bigger
the situation is more favorable. For example, when 6 = 5, power is 0.530 for n = 100 and
0.930 for n = 250. However, 6 = 5 corresponds to var(u) = 25var(v), which is generally not
common in empirical applications.

Another way to summarize these results is that we can expect to distinguish
exponential data from half-normal, but that this becomes difficult if the data are
contaminated by normal noise.

In Tables 2.12-2.15 we consider the case that the one-sided error has a gamma

distribution. Now u = cu* where u* follows the standard gamma distribution with density

__ _ 1|!
(u*yn 16 u

P(m)

 

(2-18) f(u*) =

The parameter m governs the shape of u*. When m = 1 we have the exponential
distribution which we have just considered. Values of m less than one lead to densities
with a mode at zero and very steep decline as u * increases. Values of m larger than one lead
to a positive mode, and the distribution approaches normality as m —) oo . We consider m

= 0.1, 0.5, 2 and 10. The mean and variance of the standard gamma distribution in (2.16)

both equal m, so the mean of u = cu* equals cm and the variance equals czm. Thus for a
given value of m, we expect power to increase when 0 increases.

In Tables 212-215, the results for the Pearson tests are for k = 3 only.

The general pattern of results is similar to what was found for the exponential case.

Power increases as n increases and as 0 increases. The Kolmogorov—Smimov test is
generally the most powerful. And, again, the power of the tests is low over the part of the
parameter space that would seem to be empirically most common.

An interesting feature of these results is that the power is quite low for the values
of m greater than one, even for large values of c. This is so despite the fact that the gamma
distribution with m greater than one does not at all resemble the half-normal distribution.
The reason for this low power is presumably that the gamma distribution with large m

resembles the normal distribution, and therefore is mistaken for part of the noise.

2.7 CONCLUDING REMARKS

In this paper we have considered goodness of fit tests for the stochastic frontier

model. We are interested in testing the distributional assumption for the one—sided error

54

(inefficiency term). The essential difficulty is that we can only observe the composed error,
which is the sum of the one-sided error and normal random noise. So in the end we test the
hypothesis that the composed error has the distribution that is implied by normality of the

noise and the assumed distribution for the one-sided error.

We considered Pearson 12 goodness of fit tests based on expected and actual

numbers of observations in cells defined by values of the composed error, and also the
Kolmogorov-Smimov test. We discussed the asymptotic theory that corrects the Pearson
test for the effects of parameter estimation. We also noted that asymptotically correct
critical values can be found by boostrapping, for either the Pearson test or the
Kolmogorov-Smimov test.

We performed simulations to investigate the size and power properties of the tests.

In terms of size, bootstrapping works better than asymptotic theory. In terms of power, the

Kolmogorov-Smirnov test dominates the Pearson tests, so that the best test overall appears
to be the Kolmogorov-Smimov test using critical values from the bootstrap.

The remaining problem is that the power of these tests against plausible alternative
distributions is disappointingly low. Reasonable power seems to require sample sizes
and/or signal to noise ratios that are not commonly found empirical applications. Making
the same point somewhat differently, it is easy to distinguish an exponential distribution
from a half-normal. However, it is hard to distinguish the sum of a normal and an
exponential from the sum of a normal and a half-normal, unless the variance of the normal
component is very small or the sample size is very large.

Further research is needed to understand the empirical significance of these

findings. Philosophically, it does not matter if different models yield different results if we

55

can distinguish statistically between the models; conversely, it does not matter if we cannot
distinguish statistically between models, if the models give more or less the same results.
It is only a problem if we cannot distinguish statistically between models and the models
give substantively different results. Intuitively, it seems reasonable to conjecture that data
sets for which it is hard to distinguish between different distributions of inefficiency are
also data sets for which the different distributions lead to similar empirical results.
(Presumably these are cases in which different distributions for inefficiency lead to
essentially the same distribution of the composed error 8 .) Therefore the relationship
between robustness of results and the power of goodness of fit tests (or, more generally, the
ability of any model selection method to distinguish between different distributions of

inefficiency) is obviously an important issue to investigate.

56

APPENDIX 2
Appendix 2-A

A

In this Appendix we establish equation (8) of the text. We write §(60) = P — P ,

where P is the (k-l)-dimensional vector with j’h element p j = p j(60) and P is the
(k-1)-dimensional vector with j’h element 131- = 0 j / n. Also we write V(60) = TI — PP'

where II is the diagonal matrix with j’h diagonal element equal to p j' Now we use the

fact (e.g. Abadir and Magnus (2005), p. 87) that

l

——l—H_'PP'FI"
1—PH"P

(2.19) [rI—PP'1"=n"+

Therefore

n§(90)'V(90)—l§(90)=n(13-P)TI_I(13-P)
(2.20) n . , _l .. —1 .
+—-——T(P—P)H PPH (P—P)
l—P'II P

The first term on the right hand side of(2.20) equals ’12:: (13/ _ pj)2 / p]. =

k—l

2:: (Oj — Ej )2 / Ej. For the second term, note that 1— P'II’IP 21— 2.1.2]

Pj =Pk and

that (P — P)'II—IP = (P — P)'ek_l (where e/H is a vector of dimension (k—l) with each

element equal to one) = [(1— 13k)—(1 — pk )] = (pk — 13k ). Therefore
_ , _ _ k—l _ .
ng(90) H60) 'gwo): Zj=1<0,—E,-)’/Ej + n(pk —pk>2 /pk=

k 2
Zj=l(0]—E./) ”51°

57

Appendix 2-B
In this Appendix we discuss the goodness of fit test based on quantiles and its
relationship to the Pearson test based on actual and expected cell counts. Suppose that we

pick (k-l) probabilities 0 < Pl < p2 < Pk—l <1. Let the corresponding population
quantiles be ml (6) < m2 (6) - -- < mk—l (6), so that P( y S m j (6)) = p j , and let the sample
quantiles be ”“21 S "“12 S [fl/(_l . So now the test will depend on (61 — m), the vector whose
j”7 element equals ( 61 j — m j (6) ), and the test statistic equals 21(0) — m(6))'W(rfz — m(6))
with an appropriate choice of W.

To see how this compares to the CMT test, we note that @051]- —- m j (6)) is

asymptotically normal, and so it must be expressable as an average (plus an asymptotically

negligible term). This is the “influence function representation,” which is given by:
. 1
(2.21) Jam].—mj(9))=—\/:Zf:lry(a)+op(1),

where 0,,(1) is an asymptotically negligible term (i.e., it converges in probability to zero),

and where

l

2.22 -- 6 =——————
( ) r“ ) ./‘(m,-<6»

[17)- -1(y,- S m,,-(6))].

where f is the pdf of y. See, for example, Ruppert and Carroll (1980), p. 832. Therefore
the test based on (62 — m) is equivalent in large samples to the CMT test based on the

moment conditions E[1(y S m 1(6)) — p j], j = 1, 2, ..., k-l. This is an overlapping set of

cells. However, it is also equivalent to consider the non-overlapping cells:

58

AI = {yly S ml (6)} , A2 = {y|m1(6) < y S m2 (6)} , etc. The resulting test is the CMT test

based on observed versus actual'cell counts, as discussed in the text.
Appendix 2-C

In this Appendix we derive analytically the variance matrix C used in the
conditional moment test, for the case of a normal distribution. We wish to evaluate

(2.23) C,” = E(.S'.S") , C12 = E(ssg') _, C22 = E(gg') .

Here s = s( y, 6) is the score function for the normal distribution, given by

_ 1 _
70/ - #)
(2.24) s( y, a) = ‘7

" 1 2
+ —— (y — u)
_20'2 '20'4

 

 

 

-

and g = g(y,6) is the vector whose j’h element equals [1(y e Aj)— pj ].

It is well known that C I I is the information matrix for the normal distribution,

given by

(2.25) ‘7
0 _...—

 

 

Also C 22 equals the matrix V (6) as defined in the discussion following equation (2.4) of

the text.

This leaves the submatrix C12. It is of dimension 2 by (k-l ). We will evaluate in

turn the (l , j) and (2, j) elements of this matrix. To do so we make the reasonable

assumption that the cells are intervals, so that A j = (0,1)] , where for notational simplicity

we do not express the subscript “j” that should appear on a and b.

59

Then element (1, j) of (7'2 equals

1 1
jEU-IUIIU’E Aj)-le = —2Ey[l(ye Aj)—le
0 0
_ l 1
‘ 7Eyl(yeAJ-)——2-pj,u
0' 0
p.
= +41£<ylcz<ySb)—#1
0,-

=.‘zlwl";”l-wtb;"ll

where “ ¢ ” is the standard normal density function. Here we have evaluated the conditional

 

 

 

expectation E(yla < y S b) =,u + —l—[(o(fllj —(o[b—-—'ti]] as in Johnson and Kotz

(1970), equation (79), p. 81.

Similarly element (2, j) of (712 equals

 

-l l 7
E +—(y-#)“][1(yeA')-p-1
[202 204 J J

l 2
= E[ngflu) ]ll(y€ Aj)_pj]

1 p-
= —;E(y-m21(a<y5b)———J

 

20' 20'4
1 7 1 1 7 p)-
=——Ey“1(yeA‘)+——(—2#)Ey1(yeA-)+—/fp--——
20'4 J 20'4 j 20'4 J 20'4
1 p #2 p
_ 2 x! i /
—-—--Ey1(yeAh)-—phE(yy€A-)+ ‘ — - .
20'4 ‘l 0'4 j j 20'4 20'4

60

2

where Ey21(y E Aj) = P} var(yly e Aj)+ pf [E(y’y E Aj):l

F urtherrnore,

 

 

 

Eyzuye A»: pjaz {1_b¢<b)—a¢(a)_[ ¢<h)—¢<a) H+ ”PM, ¢<b)—¢(a) ]

(13(1)) — d)(a) <D(b) — (13(0) 0(1)) - Cl>(a)

61

Size of the test of the hypothesis that the data are N(0,1)

Table 2.1

Nominal size = 0.05

 

Pearson

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

k n Bootstrap KS Bootstrap
Pearson KS
3 50 0.049 0.053 0.040 0.045
100 0.054 0.047 0.040 0.048
250 0.053 0.039 0.046 0.048
500 0.047 0.045 0.050 0.056
5 50 0.043 0.050 * *
100 0.049 0.044 * *
250 0.049 0.041 * *
500 0.051 0.058 * *
10 50 0.048 0.052 * *
100 0.049 0.043 * *
250 0.052 0.056 * *
500 0.051 0.058 * *

 

* The number of cells (k) is not relevant for the Kolmogorov-Smirnov test.

62

 

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63

Quantiles of the distribution of the normal / half normal composed error

Table 2.3

2 2 2 -
0' 20'“ +0'v =1, varlous/i

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Quantile
.10 .20 .30 .40 .50 .60 .70 .80 .90
-1.281 —0.841 -0.524 -0.253 0.000 0.253 0.524 0.841 1.281
-1.357 -0.918 -0.602 -0.332 -0.080 0.173 0.444 0.759 1.198
-1 .423 -0.987 -0.675 -0.407 -0.156 0.094 0.362 0.675 1.109
-1.477 -1.048 -0.739 -0.475 -0.228 0.018 0.282 0.590 1.017
-1.522 -1.099 -0.796 -0.537 -0.294 -0.053 0.206 0.508 0.926
-1 .556 -1.141 -0.843 -0.589 -0.353 -0.117 0.135 0.430 0.837
-1.582 -1 . 176 -0.884 -0.635 -0.405 -0.174 0.071 0.358 0.754
-1.602 -1.202 -0.91 7 -0.674 -0.449 -0.224 0.014 0.292 0.675
-1.616 -1.223 -0.943 -0.706 -0.486 -0.269 -0.037 0.233 0.603
-1.626 -1.238 -0.964 -0.733 -0.518 -0.306 -0.081 0.180 0.537
-1 .632 —l .250 -0.981 -0.754 -0.545 -0.339 -0.120 0.133 0.478
-1.637 -1.259 -0.994 -0.772 -0.567 -0.366 -0.153 0.091 0.425
-1.640 -1.266 -1.004 -0.786 -0.586 -0.389 -0.183 0.054 0.377
-1.642 —1.271 -1.012 -0.797 -0.601 -0.409 -0.209 0.022 0.334
-1.643 -1.274 -1.018 -O.806 -0.614 -0.426 -0.230 -0.006 0.295
-1.644 -1.276 -1 .023 -0.814 -0.625 -0.440 -0.249 -0.032 0.261
-1.644 -1.278 -1.026 -0.819 -0.633 -0.453 -0.266 -0.054 0.230
-1.644 -1.279 -1 .029 -0.824 -0.640 -0.463 -0.280 -0.074 0.202
-1.645 -1 .280 -l .031 -0.828 -0.646 -0.472 -0.293 -0.091 0.177
-1.645 -1.280 -1.033 -0.831 -0.651 -0.480 -0.304 -0.107 0.154
-1.645 -1.281 -l.034 -0.833 -0.656 -0.486 -0.313 -0.120 0.134
-1.645 -1.281 -1 .034 -0.835 -0.659 -0.491 -0.322 -0.132 0.115
-1.645 -1.281 -1.035 -0.837 -0.662 -0.496 -0.329 -0. 144 0.098
-1.645 -1.282 -1 .035 -0.838 -0.664 -0.500 -0.335 -0.154 0.083
-1.645 -1.282 -1.036 -0.838 -0.666 -0.504 -0.341 -0. 162 0.069
-1.645 -1 .282 -l .036 -0.839 -0.667 -0.507 -0.346 -0.170 0.056
-1.645 -1.282 -1.036 -0.840 -0.669 -0.509 -0.350 -0. l 77 0.044
-1.645 -1.282 -1.036 -0.840 -0.670 -0.512 -0.3 54 -0.184 0.034
-1.645 -1.282 -1.036 -0.841 -0.671 -0.513 -0.358 -0.190 0.024
-1 .645 -l .282 -l .036 -0.841 -0.672 -0.515 -0.361 -0.195 0.014
-1.645 -1 .282 -1.036 -0.841 -0.672 -0.516 -0.364 -0.200 0.006

 

64

 

Table 2.3 (cont’d.)

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

k Quantile
.10 .20 .30 .40 .50 .60 .70 .80 .90
3.1 -1.645 -1.282 -1.036 0841 -0.673 -0.518 -0.366 -0.204 -0.002
3.2 -1.645 -1.282 -1.036 -0.841 -0.673 -0.519 —0.368 -0.208 -0.009
3.3 -1.645 -1.282 -1.036 -0.841 -0.673 -0.519 -0.370 -0.212 -0.016
3.4 -1.645 -1.282 -1.036 -0.841 -0.674 -0.520 -0.372 -0.215 -0.022
3.5 -1.645 -1.282 -1.036 -0.841 -0.674 -0.521 -0.373 -0.218 -0.027
3.6 -l.645 -1.282 -1.036 -0.841 -0.674 -0.521 -0.374 -0.221 -0.033
3.7 -l .645 -1.282 -1.036 -0.841 -0.674 -0.522 -0.376 -0.224 -0.038
3.8 -1.645 -1.282 -1.036 —0.841 -0.674 -0.522 -0.377 -0.226 -0.043
3.9 -l .645 -1.282 -1.036 -0.841 -0.674 -0.523 -0.378 -0.228 -0.047
4.0 -l .645 -1.282 -1 .036 -0.842 -0.674 -0.523 -0.379 -0.230 -0.051
4.1 -1.645 -1 .282 -1.036 -0.842 —0.674 —0.523 -0.379 -0.232 -0.055
4.2 -1.645 -1.282 -1.036 -0.842 -0.674 -0.523 -0.380 -0.233 -0.059
4.3 -1.645 -1.282 -1.036 -0.842 -0.674 -0.523 -0.381' -0.235 -0.062
4.4 -1.645 -1.282 -1.036 -0.842 -O.674 -0.524 -0.381- -0.236 -0.065
4.5 -1.645 -1.282 -1.036 -0.842 -0.674 -0.524 -0.382 -0.238 -0.068
4.6 -1.645 -1.282 -1.036 -0.842 -0.674 -0.524 -0.382 -0.239 -0.071
4.7 -l .645 -1.282 -1.036 -0.842 -0.674 -0.524 -0.383 -0.240 -0.073
4.8 -1.645 -1.282 -1 .036 -0.842 -0.674 -0.524 -0.383 -0.241 -0.076
4.9 -1.645 -1.282 -1.036 -0.842 -0.674 -0.524 -0.383 -0.242 -0.078
5.0 -1.645 -1.282 -1.036 -0.842 -0.674 -0.524 -0.383 -0.243 -0.081
5.1 -1.645 -1 .282 —1.036 -0.842 -0.675 -0.524 -0.3 83 -0.244 -0.083
5.2 -1.645 -1.282 -1.036 -0.842 -0.675 —0.524 -0.384 -0.244 -0.085
5.3 -1.645 -1.282 -1.036 -0.842 -0.675 -0.524 -0.3 84 -0.245 -0.086
5.4 -1.645 -1.282 -1 .036 -0.842 -0.675 -0.524 -0.384 -0.246 -0.088
5.5 -1.645 -1 .282 -1.036 -0.842 -0.675 -0.524 -0.3 84 -0.246 -0.090
5.6 -1.645 -1.282 -1 .036 -0.842 -0.675 -0.524 -0.384 -0.247 -0.092
5.7 -1.645 -1.282 -1.036 -0.842 -0.675 -0.524 -0.384 -0.247 -0.093
5.8 -1.645 -1.282 —1.036 -0.842 -0.675 -0.524 -0.385 -0.248 -0.094
5.9 -l .645 -1 .282 -1.036 -0.842 -0.675 -0.524 -0.385 -0.248 -0.096
6.0 -1.645 -1.282 -1.036 -0.842 -0.675 -0.524 -0.3 85 -0.248 -0.097
6.1 -1.645 -1.282 -l.036 -0.842 -0.675 -0.524 -0.385 -0.249 -0.098
6.2 -1.645 -1.282 -1.036 -0.842 -0.675 -0.525 -0.385 -0.249 -0. 100
6.3 -l .645 -1.282 -1.036 -0.842 -0.675 -0.524 -0.385 -0.250 -0.101
6.4 -1.645 -1.282 -1.036 -0.842 -0.675 -0.524 -0.385 -0.250 -0.102
6.5 -1.645 -1.282 -1.036 -0.842 -0.675 -0.524 -0.385 -0.250 0103
6.6 -1 .645 -1.282 -1.036 -0.842 -0.675 -0.524 -0.385 -0.250 -0.104
6.7 -1.645 -1 .282 -1.036 -0.842 -0.675 -0.524 -0.385 -0.251 -0.105
6.8 -l.645 -1.282 -1.036 -0.842 -0.675 -0.524 -0.385 -0.251 -0.105
6.9 -1.645 -1.282 -1.036 -0.842 -0.675 -0.524 -0.385 -0.251 -0.106
7.0 -1.645 —1.282 -1.036 -0.842 -0.675 -0.524 -0.385 -0.251 -0.107

 

65

 

Table 2.3 (cont’d.)

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Quantile
.20 .30 .40 .50 .60 .70 .80 .90
-1.282 -1 036 -0.842 -0.675 -0.524 -0.385 -0.251 -0.108
-1.282 -1 036 -0.842 -0.675 -0.524 -0.385 -0.251 -0.109
-1.282 -1 036 -0.842 -0.675 -0.524 -O.385 -0.252 -0.109
-l.282 -l 036 -0.842 -0.675 -0.524 -0.385 -0.252 -0.110
-1.282 -1 036 -0.842 —0.675 -0.524 -0.385 -0.252 -0.111
-1.282 -1 036 -0.842 -0.675 -0.524 -0.385 -O.252 -0.111
-1.282 -1 036 -0.842 -0.675 -0 524 -0.385 -0.252 -0.112
-1.282 -1 036 -0.842 -0.675 -0.524 -0.385 -0.252 -0.112
-1.282 -1 036 -0.842 -0.675 -0.524 -0.385 -0.252 -0.113
-1.282 —1 036 -0.842 -0.675 —0.524 -0.385 -0.252 -0.113
-1.282 -1 .036 -0.842 -0.675 -0.524 -0.385 -0.252 -0.114
-1.282 -1.036 -0.842 -0.675 -0.524 -0.385 -0.253 -0.114
-l.282 -1.036 -0.842 -0.675 -0 524 -0.385 -0.253 -0.115
-1.282 -1.036 -0.842 -0.675 -0.524 -0.385 -0.253 -0.115
-1.282 -1 036 -0.842 -0.675 -0.525 -0.385 -O.253 -0.116
-1.282 -1 .036 -0.842 -0.675 -0.524 -0.385 -0.253 -0.116
-1.282 -1.036 -0.842 -0.675 -0.524 —O.385 -0.253 -0.116
-1.282 -1.036 -0.842 -0.675 -0.524 -0.385 -0.253 -0.117
-1 .282 -1.036 -0.842 -0.675 -0.524 -0.385 -0.253 -0.117
-1.282 —1 .036 -0.842 -0.675 -0.524 -0.385 -0.253 -0.117
-1.282 -1.036 -0.842 -0.675 -0.524 -0.385 -0.253 -0.118
-1.282 -1.036 -0.842 -0.675 -0.524 -0.385 -0.253 -0.118
-1.282 -1 .036 -0.842 -0.675 -0.524 -0.385 -0.253 -0.118
-1.282 -1 036 -0.842 -0.675 -0.524 -0.385 -0.253 -0.119
-1.282 -1 036 -0.842 -0.675 -0.524 -0.385 -0.253 0.119
-1.282 -1.036 -0.842 -0.675 -0.524 -0.385 -0.253 -0.119
-1.282 -1 .036 -0.842 -0.675 -0.524 -0.385 -0.253 -0.1 19
-1.282 -1.036 -0.842 -0.675 -0 524 -0.385 -0.253 -0 120
-1.282 -1 036 -0.842 -0.675 -0 524 -0.385 -0.253 —0 120
-1.282 -1 036 -0.842 -0.675 -0.524 -0.385 -0.253 -0 120
-1.282 -1 036 -0.842 -0.675 -0.524 -0.385 -0.253 -0 122
-l.282 —1 036 -0.842 -0.675 -0 524 -0.385 -0.253 -0 123
-l.282 -1 036 -0.842 -0.675 -0.524 -0.385 —0.253 -0 124
-1.282 -1 036 -0.842 -0.675 -0.524 -0.385 -0.253 -0 124
-1.282 -1 036 -0.842 -0.675 -0.524 —0.385 -0.253 -0 125
-l.282 —l 036 -0.842 -0.675 -0.524 -0.385 -0.253 -0 125
-1.282 -1 036 -0.842 -0.675 -0.524 —0.385 -0.253 -O 125
-1.282 -1 036 -0.842 -O.675 -0 524 -0.385 -0.253 -0.125
-1.282 -1 036 -0.842 -0.675 -O 524 -0.385 -0.253 -0.126
-l.282 -1 036 -0.842 -0.675 -0 524 -0.385 -0.253 -0.126

 

 

 

 

66

 

Table 2.4

Size of test of the hypothesis that the data are normal / half-normal

7t =0.l

Nominal size = 0.05

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

k n Wrong Pearson Bootstrap Bootstrap Bootstrap
Skew (%) (Tauchen) Pearson Pearson KS
(Tauchen)

3 50 50.2 0.077 0.051 0.057 0.048
100 50.9 0.062 0.053 0.055 0.062
250 50.2 0.056 0.049 0.040 0.041

5 50 50.2 0.092 0.043 0.043 *
100 50.9 0.068 0.046 0.046 *
250 50.2 0.060 0.049 0.049 *

10 50 50.2 0.200 0.045 0.047 *
100 50.9 0.119 0.049 0.054 *
250 50.2 0.076 0.051 0.044 *

 

 

 

 

 

*The number of cells (k) is not relevant for the Kolmogorov-Smimov test.

67

 

Table 2.5

Size of the test of the hypothesis that the data are normal / half-normal

)t. = 0.5

Nominal size = 0.05

 

 

 

 

 

 

 

 

 

 

 

 

 

k n Wrong Pearson Bootstrap Bootstrap Bootstrap
Skew (%) (Tauchen) Pearson Pearson KS
(Tauchen)

3 50 47.8 0.079 0.048 0.031 0.047
100 47.4 0.055 0.037 0.041 0.050
250 44.3 0.051 0.050 0.046 0.045

5 50 47.8 0.089 0.044 0.044 *
100 47.4 0.065 0.056 0.055
250 44.3 0.055 0.044 0.057

10 50 47.8 0.191 0.053 0.053 *
100 47.4 0.113 0.056 0.055 *
250 44.3 0.073 0.055 0.049 *

 

 

 

 

 

 

 

* The number of cells (k) is not relevant for the Kolmogorov-Smimov test.

68

 

 

Table 2.6

Size of the test of the hypothesis that the data are normal / half-normal

k=l

Nominal size = 0.05

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

k n Wrong Pearson Bootstrap Bootstrap Bootstrap
Skew (%) (Tauchen) Pearson Pearson KS
Tauchen)

3 50 36.0 0.080 0.051 0.041 0.042
100 30.8 0.059 0.053 0.037 0.041
250 19.5 0.049 0.048 0.053 0.040

5 50 36.0 0.093 0.042 0.039 *
100 30.8 0.065 0.056 0.042 *
250 19.5 0.053 0.043 0.043 *

10 50 36.0 0.190 0.039 0.050 *
100 30.8 0.111 0.041 0.044 *
250 19.5 0.068 0.061 0.055 *

 

 

 

 

 

* The number of cells (k) is not relevant for the Kolmogorov-Smirnov test.

69

 

Table 2.7

Size of the test of the hypothesis that the data are normal / half-normal

3:2

Nominal size = 0.05

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

k 11 Wrong Pearson Bootstrap Bootstrap Bootstrap
Skew (%) (Tauchen) Pearson Pearson KS
(Tauchen)

3 50 11.5 0.074 0.053 0.034 0.037
100 4.1 0.067 0.046 0.039 0.042
250 0.2 0.050 0.056 0.044 0.045

5 50 l 1.5 0.107 0.043 0.042 *
100 4.1 0.072 0.064 0.048 *
250 0.2 0.055 0.060 0.053 *

10 50 11.5 0.233 0.040 0.043 *
100 4.1 0.122 0.048 0.046 *
250 0.2 0.069 0.058 0.052 *

 

* The number of cells (k) is not relevant for the Kolmogorov-Smirnov test.

70

 

Table 2.8

Size of the test of the hypothesis that the data are normal / half-normal

1:10

Nominal size = 0.05

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

k 11 Wrong Pearson Bootstrap Bootstrap Bootstrap
Skew (%) (Tauchen) Pearson Pearson KS
(Tauchen)

3 50 0.1 0.060 0.049 0.050 0.044
100 0 0.053 0.057 0.051 0.038
250 0 0.051 0.048 0.048 0.045

5 50 0.1 0.090 0.041 0.051
100 0 0.064 0.049 0.048
250 0 0.059 0.053 0.048

10 50 0.1 0.238 0.038 0.045 *
100 0 0.116 0.054 0.057 *
250 0 0.073 0.052 0.043 *

 

"‘ The number of cells (k) is not relevant for the Kolmogorov-Smirnov test.

71

 

Table 2.9

Power of the test of the hypothesis that the data are normal / half-normal

Nominal size = 0.05

- Number of cells: k = 3

Alternative: the data are normal / exponentia1(6)

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

n 6 Pearson Bootstrap Bootstrap Bootstrap
(Tauchen) Pearson Pearson KS
(Tauchen)
50 0.1 0.069 0.059 0.059 0.046
0.5 0.071 0.055 0.053 0.041
1 0.079 0.056 0.059 0.054
2 0.142 0.110 0.098 0.147
5 0.269 0.217 0.195 0.340
10 0.367 0.285 0.294 0.494
100 0.1 0.064 0.049 0.054 0.056
0.5 0.060 0.055 0.050 0.048
1 0.080 0.084 0.069 0.089
2 0.196 0.188 0.207 0.239
5 0.468 0.440 0.384 0.530
10 0.633 0.563 0.537 0.700
250 0.1 0.064 0.049 0.051 0.057
0.5 0.052 0.051 0.041 0.048
1 0.101 0.084 0.108 0.150
2 0.416 0.407 0.476 0.507
5 0.879 0.842 0.789 0.930
10 0.959 0.945 0.922 0.966

 

 

 

 

 

 

 

Table 2.10

Power of the test of the hypothesis that the data are normal / half-normal

Nominal size = 0.05

Number of cells: k = 5

Alternative: the data are normal / exponentia1(6)

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

n 6 Pearson Bootstrap Bootstrap Bootstrap
(Tauchen) Pearson Pearson KS
- (Tauchen)
50 0.1 0.086 0.058 0.052 0.046
0.5 0.087 0.042 0.057 0.041
1 0.097 0.065 0.064 0.054
2 0.150 0.097 0.079 0.147
5 0.265 0.137 0.123 0.340
10 0.343 0.167 0.185 0.494
100 0.1 0.062 0.054 0.055 0.056
0.5 0.087 0.045 0.044 0.048
1 0.107 0.076 0.071 0.089
2 0.202 0.167 0.135 0.239
5 0.399 0.354 0.282 0.530
10 0.537 0.435 0.383 0.700
250 0.1 0.061 0.056 0.051 0.057
0.5 0.059 0.061 0.056 0.048
1 0.110 0.101 0.091 0.150
2 0.383 0.382 0.302 0.507
5 0.851 0.799 0.695 0.930
10 0.944 0.917 0.833 0.966

 

 

 

 

 

 

 

73

 

Table 2.11

Power of the test of thehypothesis that the data are normal / half-normal

Alternative: the data are normal / exponential(6)

Nominal size = 0.05

Number ofcells: k = 10

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

n 6 Pearson Bootstrap Bootstrap Bootstrap
(Tauchen) Pearson Pearson KS
(Tauchen)

50 0.1 0.191 0.044 0.047 0.046
0.5 0.167 0.049 0.045 0.041

1 0.203 0.050 0.037 0.054

2 0.245 0.065 0.063 0.147

5 0.370 0.079 0.102 0.340

10 0.485 0.093 0.110 0.494

100 0.1 0.119 0.050 0.042 0.056
0.5 0.104 0.049 0.047 0.048

1 0.140 0.061 0.061 0.089

2 0.225 0.108 0.101 0.239

5 0.364 0.209 0.210 0.530

10 0.499 0.295 0.260 0.700

250 0.1 0.082 0.052 0.045 0.057
0.5 0.077 0.045 0.052 0.048

1 0.124 0.075 0.059 0.150

2 0.338 0.267 0.217 0.507

5 0.716 0.671 0.494 0.930

10 0.879 0.815 0.789 0.966

 

 

 

 

 

 

 

Table 2.12

Power of the test of the hypothesis that the data are normal / half-normal

Nominal size = 5%

Alternative: the data are normal / gamma (u is c times gamma(m))

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

m = 0.1
n 6 Pearson Bootstrap Bootstrap Bootstrap
(Tauchen) Pearson Pearson KS
(Tauchen)
50 0.1 0.069 0.051 0.046 0.042
0.5 0.060 0.061 0.061 0.051
1 0.073 0.065 0.066 0.058
2 0.081 0.135 0.120 0.117
5 0.407 0.365 0.403 0.467
10 0.785 0.695 0.730 0.814
100 0.1 0.059 0.055 0.046 0.055
0.5 0.061 0.052 0.049 0.068
1 ' 0.075 0.046 0.046 0.056
2 0.099 0.095 0.127 0.177
5 0.614 0.602 0.607 0.978
10 0.962 0.960 0.965 0.984
250 0.1 0.062 0.058 0.060 0.060
0.5 0.052 0.055 0.051 0.052
1 0.063 0.060 0.062 0.059
2 0129 0.134 0.175 0.308
5 0.888 0.921 0.954 1.000
10 1.000 1.000 1.000 1.000

 

 

 

 

 

 

75

 

Table 2.13

Power of the test of the hypothesis that the data are normal / half-normal

Nominal size = 5%

Alternative: the data are normal / gamma (u is 6 times gamma(m))

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

m 0.5
n c Pearson Bootstrap Bootstrap Bootstrap
(Tauchen) Pearson Pearson KS
' (Tauchen)

50 0.1 0.068 0.054 0.049 0.061
0.5 0.049 0.046 0.046 0.046

1 0.073 0.066 0.066 0.056

2 0.153 0.132 0.132 0.154

5 0.511 0.416 0.416 0.621

10 0.791 0.740 0.740 0.887

100 0.1 0.074 0.061 0.061 0.053
0.5 0.055 0.061 0.061 0.050

1 0.061 0.078 0.078 0.078

2 0.260 0.271 0.271 0.328

5 0.784 0.732 0.732 0.869

10 0.974 0.948 0.948 0.945

250 0.1 0.067 0.053 0.053 0.060
0.5 0.061 0.069 0.069 0.067

1 0.080 0.082 0.082 0.120

2 0.516 0.583 0.583 0.685

5 0.995 0.973 0.973 1.000

10 1.000 1.000 1.000 1.000

 

 

 

 

 

 

 

 

Table 2.14

Power of the test of the'hypothesis that the data are normal / half-normal

Nominal size = 5%

Alternative: the data are normal / gamma (u is c times gamma(m))

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

m = 2
n 6 Pearson Bootstrap Bootstrap Bootstrap
(Tauchen) Pearson Pearson KS
(Tauchen)
50 0.1 0.058 0.054 0.061 0.043
0.5 0.050 0.038 0.043 0.029
1 0.073 0.057 0.053 0.050
2 0.072 0.075 0.067 0.049
5 0.095 0.048 0.064 0.073
10 0.1 10 0.076 0.070 0.082
100 0.1 0.071 0.053 0.050 0.060
0.5 0.044 0.063 0.049 0.046
1 0.066 0.062 0.053 0.068
2 0.091 0.075 0.092 0.107
5 0.106 0.096 0.094 0.127
10 0.125 0.110 0.107 0.132
250 0.1 0.040 0.061 0.057 0.057
0.5 0.049 0.052 0.054 0.063
1 0.080 0.088 0.079 0.100
2 0.164 0.141 0.158 0.198
5 0.210 0.196 0.188 0.220
10 0.219 0.210 0.217 0.233

 

 

 

 

 

 

 

 

 

Table 2.15

Power of the test of the hypothesis that the data are normal / half-normal

Nominal size = 5%

Alternative: the data are normal / gamma (u is 0 times gamma(m))

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

m = 10
n 0 Pearson Bootstrap Bootstrap Bootstrap
(Tauchen) Pearson Pearson KS
(Tauchen)
50 0.1 0.061 0.050 0.052 0.060
0.5 0.062 0.054 0.043 0.055
1 0.065 0.042 0.052 0.072
2 0.065 0.054 0.044 0.068
5 0.095 0.074 0.076 0.098
10 0.110 0.098 0.099 0.115
100 0.1 0.062 0.051 0.049 0.055
0.5 , 0.064 0.060 0.055 0.053
1 0.065 0.048 0.052 0.068
2 0.068 0.062 0.063 0.075
5 0.093 0.070 0.065 0.099
10 0.100 0.082 0.085 0.121
250 0.1 0.055 0.055 0.054 0.039
0.5 0.059 0.063 0.059 0.055
1 0.055 0.066 0.066 0.043
2 0.071 0.069 0.072 0.058
5 0.082 0.079 0.090 0.087
10 0.104 0.095 0.088 0.119

 

 

 

 

 

 

78

 

 

(Supplemental) Table 2.16

Size of the test of the hypothesis that the data are normal / half normal

Nominal size = 0.05

with 0'3 = 0'3. =1 (simple hypothesis)

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

K N . Pearson Bootstrap KS Bootstrap
Pearson KS
3 50 0.050 0.044 0.042 0.048
100 0.049 0.047 0.040 0.053
250 0.050 0.051 0.047 0.048
5 50 0.043 0.045 * *
100 0.049 0.044 * *
250 0.047 0.050 * *
10 50 0.045 0.054 * *
100 0.048 0.046 * *
250 0.040 0.052 * *

 

* The number of cells (k) is not relevant for the Kolmogorov-Smirnov test.

79

 

 

 

(Supplemental) Table 2.17
Size of the test of the hypothesis that the data are normal / half normal
Results conditional on negative skew

Tauchen version of Pearson test

Nominal size = 0.05

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

k n k=0.1 k=0.5 k= 1:2 k=10
3 50 0.089 0.092 0.094 0.124 0.060
100 0.066 0.055 0.065 0.068 0.053
250 0.058 0.051 0.052 0.050 0.052
500 0.057 0.058 0.052 0.052 0.052
5 50 0.097 0.092 0.104 0.111 0.090
100 0.068 0.065 0.067 0.072 0.064
250 0.064 0.059 0.057 0.056 0.059
500 0.057 0.060 0.055 0.053 0.054
10 50 0.209 0.207 0.206 0.241 0.238
100 0.119 0.111 0.111 0.122 0.116
250 0.080 0.078 0.079 0.070 0.073
500 0.069 0.072 0.065 0.061 0.064

 

 

 

 

 

 

 

 

80

 

Essay 3
TESTING EQUALITY OF DISTRIBUTION FOR

TWO CORRELATED VARIABLES

3.1 INTRODUCTION

In this paper, we consider the problem of testing equality of distribution for two
variables. One is often confronted with the question whether two samples have the same
distribution, for instance, whether a government policy program changes the earnings
distribution, or whether students from two high schools in the same county have the same
SAT score distribution, or whether the temperature distribution changes before and after
the Kyoto Protocol.

Many nonparametric and parametric tests are available for testing equality of two
distribution functions. Examples include the Kolmogorov-Smimov (KS) test, the
Cramer-von Mises (CM) test, and the Baumgartner-Weiss-Schindler (BWS) test of
Baumgartner et a1. (1998). These tests assume the two data series are independent.
However, this assumption is restrictive. As an example, if we have a cross-section of
cities, and for each we have an average temperature before and after the Kyoto Protocol,
obviously these two observations are correlated for each city. In this paper we wish to
allow pairwise correlation across the two data sets, to accommodate cases like this.

Li et a1. (2009) is the only paper of which we are aware that allows this type of

pairwise correlation. Their test is somewhat complicated because they consider the case

81

 

that each observation is a vector random variable, and they allow mixed continuous and
categorical variables. Our tests are much simpler, since we simply ask how we can

modify the simple tests of the previous paragraph to allow for pairwise correlation.

In this paper we show how to implement a Pearson [2 test, based on differences

of frequencies in non-overlapping intervals (cells) that span the support of the variables,
in a GMM setting. This procedure makes no assumption about the correlation between
the two variables. We also suggest a novel bootstrapping procedure that enables us to
generate asymptotically valid critical values for the KS and BWS tests.

The plan of the paper is as follows. In section 3.2 we present the tests used to test
equality of distribution. Section 3.3 gives simulation results to assess the size and power

of the different relevant tests. Section 3.4 gives our concluding remarks.

3.2 TESTS OF EQUALITY OF TWO DISTRIBUTIONS

Suppose that we have a paired sample of sizen: (yi,x,-) , i = l, 2, 3, ...,n. We
have independence over i, but for a given 1', y,- and x,- may be correlated. We wish to
test the null hypothesis that y, , yz ,..., y" and xl ,x2 ,...,x,, are from the same distribution.

That is, our interest is in testing the hypothesis that the dependent random variables Y
and X have the same distribution.
3.2.1 Chi-squared Test

We follow Wang, Amsler and Schmidt (2009) in considering a Pearson 12

statistic using the GMM framework. They shOwed how to test whether a single sample is

 

from a hypothesized distribution. Here we extend their methods to the case of testing for

equality of distribution when the two samples are dependent.

Consider the GMM structure for the Pearson 12 statistic, as follows. Let the
possible range of the random variables Y and X be split intok cells (intervals) Al": Ak ,
such that any value of y,- and x,-, respectively, are in one and only one cell. Furthermore,

let 1(y e A j) (orl(x e A j)) be the indicator function that equals one if yl- (or x,) is in cell
A j , and equals zero otherwise. Under the null, the events 1( y e A j)and 1(x e A j) have

the same probability, and thus we have the set of moment conditions:
(3-1) E[g(y,X)1= 0

where g(y,x) is a vector of dimension (k-l) whose j’h element equals
[1( y e A j)“ 1(x e A j )]. We have omitted one cell so as to avoid a subsequent singularity.
We have omitted cell Ak but the choice of which cell to omit does not matter. If the
dependent sample y and x are not from the same distribution, the observed frequencies

of y would not be similar to the corresponding observed frequencies of x for some or

perhaps all cells. As a result, the sample analog of equation (3.1) would not be close to
zero.

Now define

_ 1
(3.2) g =;Z,’.’=,g(y,-,x,-)

and note that the j’h element of g is equal to “1"[27—110’1' e Aj) —Z:.1_ll(x i6 241)].
n _ _

We also need to define the variance matrix of the moment conditions g( y. x). If the two

83

 

variables are independent, we can use the variance matrix of dimension (k—l) by (k-l),

.th .th
I

whose j’ih diagonal element equals 2*( p j — p3) , and whose z ,‘ off diagonal

element (i = j) equals —2* pipj, with p,- = p j equal to the probability of each cell. But

this does not apply here as it does not take into account the correlation structure that we
allow.
However, we can use the outer product of the gradient (OPG) estimate of the

variance matrix of the moment conditions, which is:
(3.3) V“ = £272, g<y,-.x,->g(y,-.x,-)'.
This is a consistent estimate of V , the variance matrix of the moment conditions; that is,
V = Eg(y,'.X,-)g(y,-.x,-)'.

A central limit theorem then implies that the asymptotic distribution of fig is
N (0, V) . From this fact it follows that
(3.4) new? —>.. xii] .

To implement the test. for a given value of k , we use equiprobable cells, where
“equiprobable” is defined using the sample quantiles of x1,x2,...,x,,. We could

alternatively use the sample quantiles of y] , y2,..., y” or of the combined sample but we

do not pursue these alternatives here. Tauchen (1985) showed that using sample
quantiles to define the cells does not invalidate the distributional result (3.4).
3.2.2 Kolmogorov-Smirnov Test

The nonparametric KS two sample test is widely applied to test whether two

independent samples are from the same distribution. The test is sensitive to any kind of

84

 

 

distributional differences, for example, in the mean, variance, and kurtosis; however, the
test assumes independent samples.
The KS test compares the vertical difference of the empirical cdf between the two

random samples yl, yz...., ym and xl,x2,....x,,1 and is defined by the largest value of this

difference. If the two random samples are from the same distribution, then both

 

 

empirical cdfs should be expected to be very similar to one another. The statistic is l
1 l

(3.5) KS =sup —Z'.7' 1(y,- Sx)——Z’.72 1(x,- Sx)
I11 1 n2 1

If KS o\/n1 onz / (”I + 11;) > K a where K0, is the critical value1 at the a significance E

level, then we reject the null hypothesis that two random variables have the same
distribution at the a level. In our case, we assume paired data and so 11] = n2 = n.
3.2.3 The Baumgartner-Weiss-Schindler Test

Baumgartner et al. (1998) introduced a new nonparametric rank test---the
Baumgartner-Weiss-Schindler (BWS) test. It is also used to test the null hypothesis that
two independently samples have the same distribution. The test only requires that the
underlying distribution function is continuous.

They suggested that the test should be based on the square of the difference

between the empirical cdfs of two samples and be weighted by its variance. The statistic
is B = (”I 0 n2 )/(nl + 11; ) - E(PnI (z) — P7,, (z))/(z o (1 — 2))dz . To calculate this statistic,

suppose that we have the two samples yl , y2,..., y"l and x1, x2 ,...,xnj . Then we mix the

 

' Pearson et a1. (1972) tabulated the critical points for finite samples.

85

two samples together and define H j and G,- as the rank numbers of y j and x; in the

pooled sample. We can calculate the BWS test statistic B , as follows:

(3), + 31")

 

 

(3.6) B = , where
11+}?
8 :_l_ nI n]
Y I7] I=IJ l --.j ).£Z(nl+n7)
711+] I’ll-1'1 n1
, n+n 2
(or—Ll )
g zL n. "2
X m i=1 1 z n](n1+m)
__ -.(1_ )
”7+1 112+] n7

 

The critical values are tabulated in their paper.

Baumgartner et al. showed that the power of the test is at least as good as
previously proposed nonparametric tests such as the KS, CM, and Wilcoxon tests. In
addition, they showed that the BWS test tends to be a more powerful test for samples
from a distribution with heavy tails because it takes into consideration the weighting
variance, 2 o (1 — z) .

3.2.4 The Bootstrap Resampling Method

Although the regular KS and BWS tests are only valid for independent samples,
we can use the percentile bootstrap approach to find the critical values of the KS and
BWS tests for dependent samples. The bootstrap can be used to construct asymptotically
valid tests provided the resampling method is correct.

A prOper bootstrapping procedure to obtain critical values must ensure that (i) the
bootstrap samples satisfy the null, and (ii) the correlation structure of the assumed DGP is

maintained. Li et a1. (2009) propose a bootstrapping procedure that satisfies (i) but not

86

(ii). They merged the two data sets X and Y into one combined series Z , and then they
drew the bootstrap samples of X (or Y ) for the data set Z . So half of the bootstrap X
observations are original X observations and half are original Y observations, and
similarly for the bootstrap Y observations. Obviously, the bootstrap X and bootstrap Y
series are from the same (Z ) distribution, so they satisfy the null, but the pairwise
correlation pattern has been destroyed.

Our bootstrap procedure is as follows. We construct a bootstrap sample of pairs l.

(yi,x,-) from the original sample. Because we draw pairs, the pairwise correlation

structure is maintained. Then with probability 0.5, we accept the pair as it is, but with

 

probability 0.5, we rename yl- as x,- and x,- as yi. So now, as in Li et a1. (2009), half of

the bootstrap X observations are original X observations and half are original Y
observations, and similarly for the bootstrap Y observations. So the null is true (the
bootstrap X and bootstrap Y series are from the same distribution) and the correlation
structure is maintained.
3.2.5 Pairwise T-Test

The pairwise t-test is used to test the null hypothesis that two samples

y]. y2,..., y,, and x1 ,x2 ,...,xn from a normal distribution have the same mean. It is
simply a test of whether the differences D,- = y,- — x,- have zero mean. First, we sum the

differences across n pairs and then calculate the sample variance of the differences.

Second, we use the classical t-test statistic

 

(3.7) t =1/./;Zf:l Di/J2721wi _ 5]. )21/,,.-"(n_1) .

87

If the statistic value is larger than ’n—La , we reject the null that both samples have the

equal mean. Therefore. two distributions are not equal.

This test is asymptotically (large n) valid without normality. Our interest in it is
that its power should serve as a standard of comparison for our two-sample tests if the
two series are normal but with different means, a case that we will consider in our

simulations. E

3.3 MONTE CARLO SIMULATIONS

Our interest is to test equality of distribution for correlated variables. We will

 

consider a paired sample (yi.x,-)f7=l that is drawn from a bivariate normal distribution

with varying degree of correlation between Y and X . Therefore we simulate the samples

based on the DGP

(3.8) J’;=5+p'xi+\fl-_/?3°zi .. |p|<1

where p is the correlation coefficient between y,- and xi, and x,- and z,- are independent
and identically distributed as N(0,1). So X ~ N(0,1) while Y ~ N(6,1). The null
hypothesis is 6 = 0 and the alternative is (5 = 0.

3.3.1 Size of the Test

In this section, we show the sizes of the various tests at the 0.05 nominal level.

We will consider sample sizes of 50, 100, 250, 500, and 1000 with the correlation

coefficient ,0 varying from -0.99 to 0.95. In the 12 test, we will consider three different

numbers of non-overlapping equiprobable cells: k = 4, 7, and 10.

88

In Table 3.], columns 3,- 4 and 5 show the 12 GMM test results based on 20,000
replications. The test is relatively accurate in all cases when the sample size is greater or
equal to 250, in the sense that the actual size level is close to the nominal level. The test
is slightly undersized for the smaller sample sizes (n = 50 and 100), but this is not a

serious discrepancy for the 4 cell case.

For the smaller sample size cases, the simulations show that the 12 test with

fewer cells has smaller size distortions. However, the size differences between tests with
larger and smaller numbers of cells used in the test are not statistically different from zero

when the sample size is large enough, e.g., 500 or 1000.

The size of the 12 test does not depend much on p. This agrees with the

theoretical result that the test is asymptotically valid even if X and Y are correlated.

In Table 3.], column 6 shows the size for the KS test. When p = 0, so that X
and Y are independent, the size of the test is more or less correct so long as the sample
size is not too small. However, the KS test is undersized when p is positive and
oversized when ,0 is negative. These size distortions are fairly severe when l pl is large.
For the BWS test, column 8 of Table 3.1 shows a similar pattern of size distortions, but
the degree of oversize for negative p is larger than for the KS test.

Next we consider the bootstrapped version of the KS and BWS tests. We use
1000 replications in the simulations, and 399 bootstrap replications to calculate the
critical values used in each replication of the simulations. We use the “novel” bootstrap

method described in section 3.2.4.

89

 

In Table 3.1, columns 7 and 8 give these results. They are easy to summarize.
The bootstrap-based tests are quite accurate (size is close to the nominal significance
level.) except when p is positive and large, and n is small. In those cases, the bootstrap
KS test is a little better (in terms of size) than the bootstrap BWS test.

As expected, the pairwise t-test (column 10) has correct size.
3.3.2 Power of the Test

Now we turn to the question of the power of the test. The alternative is (3 ¢ 0 in
(3.8) above. Tables 3.2-3.8 give the results for different non negative value of p. Also
Tables 3.9-3.14 give similar results, which we will display but not discuss, for negative

values of p .

Columns 3, 4, and 5 in Tables 3.2-3.8 show the power of the 12 test for the cases
of k = 4, 7, and 10 cells. As expected, the power of the tests increases as the sample size
is larger and as 6 is larger. The power increases also when p is larger (and, from Table

3.1, this is not a reflection of size distortions). Finally, the power is larger when we use
less cells. From Table 3.1, this last result may be a reflection of larger size distortion
under the null when more cells are used.

When ,0 = 0 , results not reported here show that the KS and BWS tests (using the

usual tabulated critical values) are more powerful than the 12 test. However, we do not
recommend these tests because we have seen that they have large size distortion when
p ¢ 0.

Columns 6 and 7 in Tables 3.2-3.8 show the powers of the bootstrap KS and the

bootstrap BWS tests. The results show both bootstrap tests are much more powerful than

90

 

the 12 test. Moreover, the bootstrap BWS test is superior to the bootstrap KS test. The
possible reason that the bootstrap BWS is better is that the weighting emphasizes the tails
of the distribution functions, which increases the power of the test. (Baumgartner et al.
(1998)). As for the 12 test, the power of the bootstrapped KS and BWS tests increases as
n increases, as 6 increases, and as p increases.

Finally, the last column of Tables 3.2-3.8 gives the power of the t-test. We expect
this test to be more powerful than the other tests because the DGP satisfies exactly the

assumption that underlies the test: normal distributions with equal variance but unequal

 

'I‘. '41.”! 1

means. And, indeed, the power of the t-test is greater than the power of the other tests
(except when p = 0 ). However, the difference in power between the t-test and the

bootstrapped BWS test is not very large. We interpret this as evidence favorable to the

good power properties of the bootstrapped BWS test.

3.4 CONCLUDING REMARKS

In this paper, we have considered tests for equality of distributions. More
specifically, we are interested in testing whether two correlated variables have the same
distribution. This is different from many classical tests, e.g., the KS, the CM, and the
Wilcoxon tests, which are widely used to test whether two independent samples belong to

the same distribution.

We first apply the Pearson 12 GMM test based on equiprobable cells over the

support to perform the test. This test compares the difference of the samples’ observation

counts in each cell. Cell boundaries are based on sample quantiles. Simulations show

91

that the size of the tests is acceptably accurate in finite samples, provided not too many
cells are used.

Next we consider the KS and BWS tests. These tests assume independence and
have substantial size distortions when X and Y are correlated. We therefore propose a
novel bootstrapping resampling scheme to obtain valid critical values. The bootstrapped

KS and BWS tests have more or less correct size, and they are more powerful than the

12 tests. The bootstrapped BWS test appears to be the most powerful. In fact, its
power is almost as large as that of the t-test for equality means, despite the fact that the
DGP for the simulations favors the t-test.

In further research we plan to consider different types of alternatives. The results
here were based on an alternative which is different from the null only by a constant
mean. We plan to use copulas to simulate more flexible underlying distributions,

including correlated distributions with non-normal marginals. In such cases, we can see

whether the BWS test still outperforms the 12 test.

92

 

Table 3.1

Size of tests of the hypothesis that the data are bivariate- normal

Nominal size = 0.05

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

P n 13 (3 Z 2 (6) Z 3 (9) KS Bt.strp BWS Bt.strp
KS BWS T-test
-0.99 50 0.041 0.021 0.012 0.108 0.050 0.154 0.058 0.051
100 0.043 0.040 0.028 0.131 0.044 0.154 0.057 0.053
250 0.049 0.045 0.040 0.113 0.049 0.150 0.041 0.049
500 0.047 0.047 0.045 0.125 0.042 0.157 0.046 0.050
1000 0.050 0.046 0.047 0.124 0.045 0.151 0.049 0.051
-0.95 50 0.044 0.029 0.016 0.105 0.050 0.150 0.056 0.050
100 0.044 0.043 0.034 0.131 0.054 0.149 0.049 0.053
250 0.047 0.046 0.043 0.115 0.064 0.143 0.042 0.047
500 0.051 0.048 0.049 0.124 0.055 0.151 0.044 0.050
1000 0.051 0.050 0.049 0.126 0.044 0.146 0.054 0.051
-0.90 50 0.044 0.030 0.017 0.103 0.049 0.146 0.052 0.050
100 0.044 0.046 0.033 0.131 0.054 0.146 0.054 0.053
250 0.045 0.046 0.045 0.109 0.051 0.138 0.048 0.048
500 0.049 0.048 0.047 0.124 0.053 0.146 0.052 0.051
1000 0.051 0.049 0.049 0.123 0.049 0.141 0.051 0.051
-0.80 50 0.044 0.033 0.018 0.097 0.046 0.138 0.056 0.050
100 0.047 0.046 0.035 0.124 0.048 0.135 0.056 0.053
250 0.049 0.046 0.042 0.102 0.049 0.128 0.043 0.048
500 0.047 0.047 0.046 0.1 16 0.049 0.135 0.048 0.051
1000 0.050 0.050 0.052 0.114 0.051 0.134 0.053 0.051
-0.60 50 0.045 0.031 0.017 0.082 0.046 0.1 15 0.055 0.050
100 0.046 0.045 0.034 0.106 0.056 0.1 16 0.059 0.053
250 0.050 0.046 0.044 0.091 0.061 0.109 0.046 0.047
500 0.046 0.050 0.046 0.098 0.048 0.1 15 0.054 0.050
1000 0.051 0.049 0.048 0.101 0.050 0.116 0.051 0.051
-0.40 50 0.046 0.034 0.018 0.067 0.040 0.095 0.048 0.050
100 0.045 0.042 0.032 0.089 0.056 0.096 0.053 0.053
250 0.049 0.043 0.044 0.072 0.054 0.900 0.041 0.048
500 0.048 0.047 0.044 0.082 0.051 0.093 0.057 0.050
1000 0.051 0.049 0.050 0.087 0.052 0.096 0.048 0.051

 

 

 

 

 

 

 

 

 

 

93

 

 

Table 3.1 (cont’d.)

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

94

,0 n I 3 (3) Z 3 (6) Z 3 (9) KS Bt.strp BWS Bt.strp T-test

KS BWS
-0.20 50 0.046 0.031 0.017 0.050 0.041 0.072 0.043 0.051
100 0.045 0.043 0.034 0.070 0.052 0.076 0.055 0.053
250 0.048 0.046 0.044 0.057 0.051 0.069 0.046 0.049
500 0.048 0.047 0.043 0.064 0.061 0.072 0.046 0.050
1000 0.052 0.049 0.050 0.072 0.054 0.076 0.050 0.052
0.00 50 0.043 0.033 0.017 0.037 0.048 0.052 0.036 0.050
100 0.045 0.041 0.034 0.054 0.053 0.054 0.053 0.053
250 0.050 0.045 0.043 0.042 0.048 0.052 0.045 0.049
500 0.049 0.05] 0.043 0.049 0.066 0.052 0.049 0.050
1000 0.053 0.049 0.052 0.053 0.049 0.054 0.051 0.052
0.20 50 0.042 0.032 0.016 0.024 0.041 0.030 0.036 0.049
100 0.044 0.042 0.035 0.036 0.045 0.033 0.046 0.054
250 0.048 0.048 0.046 0.029 0.05] 0.028 0.044 0.048
500 0.051 0.048 0.044 0.034 0.039 0.029 0.050 0.051
1000 0.053 0.050 0.051 0.035 0.047 0.033 0.050 0.052
0.40 50 0.044 0.030 0.015 0.012 0.044 0.015 0.033 0.049
100 0.044 0.044 0.034 0.018 0.050 0.013 0.051 0.052
250 0.049 0.045 0.045 0.015 0.045 0.013 0.044 0.049
500 0.048 0.049 0.049 0.018 0.041 0.012 0.050 0.051
1000 0.054 0.050 0.050 0.030 0.046 0.016 0.052 0.053
160 50 0.043 0.029 0.015 0.004 0.053 0.003 0.0300 0.050
_ 100 0.046 0.041 0.033 0.006 0.050 0.030 0.045 0.052
x 250 0.048 0.047 0.045 0.006 0.045 0.002 0.041 0.048
x 500 0.048 0.048 0.048 0.006 0.040 0.002 0.051 0.052
K 1000 0.052 0.051 0.049 0.007 0.045 0.003 0.049 0.053
£0 50 0.039 0.026 0.013 0.000 0.019 0.000 0.022 0.050
x 100 0.045 0.041 0.033 0.000 0.039 0.000 0.038 0.051
X 250 0.048 0.048 0.043 0.000 0.045 0.000 0.034 0.050
x 500 0.048 0.048 0.045 0.000 0.042 0.000 0.053 0.052
L\ 1000 0.049 0.049 0.050 0.000 0.048 0.000 0.048 0.051

 

 

 

Table 3.1 (cont’d.)

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

[:2 n 12(3) 12(6) 12(9) KS Bt.strp BWS Bt.strp T-test
KS BWS
0.90 50 0.034 0.023 0.010 0.000 0.025 0.000 0.007 0.051
100 0.043 0.040 0.031 0.000 0.036 0.000 0.025 0.051
250 0.046 0.044 0.044 0.000 0.037 0.000 0.030 0.050
500 0.047 0.044 0.044 0.000 0.054 0.000 0.051 0.053
1000 0.049 0.049 0.049 0.000 0.051 0.000 0.050 0.050
0.95 50 0.027 0.014 0.007 0.000 0.012 0.000 0.004 0.050
100 0.034 0.033 0.023 0.000 0.024 0.000 0.020 0.051
250 0.046 0.042 0.040 0.000 0.053 0.000 0.028 0.049
500 0.047 0.047 0.044 0.000 0.055 0.000 0.050 0.053
1000 0.049 0.050 0.047 0.000 0.049 0.000 0.053 0.051

 

95

 

 

 

Table 3.2
Power of the tests when p = 0.00

Nominal size = 0.05

Null: the data are bivariate-normal with the equal means and equal variances
Alternative: the data are from bivariate-normal with the mean difference 6

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

5 n 2’ 3 (3) 2’2 (6) 2’ 3 (9) Bt.strp Bt.strp T-test
KS BWS

0.10 50 0.061 0.036 0.020 0.066 0.060 0.056

100 0.068 0.057 0.045 0.066 0.098 0.095

250 0.122 0.087 0.078 0.150 0.175 0.141

500 0.198 0.161 0.133 0.280 0.328 0.269

1000 0.388 0.305 0.270 0.505 0.585 0.483

0.20 50 0.105 0.060 0.033 0.143 0.147 0.113

100 0.156 0.116 0.088 0.211 0.271 0.235

250 0.392 0.292 0.244 0.500 0.553 0.462

500 0.681 0.594 0.423 0.774 0.863 0.777

1000 0.951 , 0.926 0.890 0.979 0.990 0.972

0.30 50 0.182 0.103 0.054 0.238 0.273 0.216

100 0.318 0.244 0.181 0.459 0.519 0.455

250 0.753 0.649 0.577 0.883 0.905 0.810

500 0.969 0.949 0.913 0.984 0.994 0.985

1000 0.999 0.999 0.999 1.000 1.000 1.000

0.40 50 0.300 0.175 0.103 0.374 0.445 0.356

100 0.545 0.440 0.351 0.684 0.759 0.698

250 0.954 0.908 0.8693 0.973 0.993 0.970

500 0.996 0.991 0.998 1.000 1.000 1.000

1000 1.000 1 .000 1.000 1 .000 1.000 1.000

0.50 50 0.447 0.283 0.171 0.509 0.627 0.528

100 0.763 0.669 0.563 0.856 0.915 0.865

250 0.997 0.990 0.982 0.998 1.000 0.999

500 1.000 1.000 1.000 1.000 1.000 1.000

1000 1 .000 1 .000 1.000 1 .000 1.000 1.000

1.00 50 0.975 0.925 0.827 1.000 1.000 0.998

100 1.000 1.000 0.993 1.000 1.000 1.000

250 1.000 1.000 1.000 1.000 1.000 1.000

500 1.000 1.000 1.000 1.000 1.000 1.000

1000 1.000 1.000 1.000 1.000 1.000 1.000

 

 

 

 

 

 

 

 

96

 

 

 

Table 3.3
Power of the tests when p = 0.20

Nominal size = 0.05

Null: the data are bivariate-normal with the equal means and equal variances
Alternative: the data are from bivariate-normal with the mean difference 6

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

6 n I 3 ( 3 ) 13(6) 25 3 (9) Bt.strp Bt.strp T-test
- KS BWS
0.10 50 0.064 0.038 0.020 0.059 0.068 0.084
100 0.076 0.061 0.049 0.089 0.107 0.120
250 0.145 0.103 0.092 0.182 0.205 0.237
500 0.232 0.186 0.158 0.310 0.377 0.420
1000 0.459 0.379 0.329 0.578 0.630 0.711
0.20 50 0.115 0.063 0.034 0.134 0.159 0.192
100 0.178 0.141 0.103 0.242 0.319 0.345
250 0.468 0.364 0.307 0.564 0.651 0.706
500 0.773 0.699 0.623 0.829 0.918 0.939
1000 0.981 0.969 0.953 0.993 1.000 0.999
0.30 50 0.211 0.120 0.062 0.262 0.314 0.376
100 0.380 0.298 0.224 0.489 0.606 0.645
250 0.835 . 0.758 0.689 0.878 0.953 0.963
500 0.990 0.982 0.967 0.995 1.000 1.000
1000 1.000 1.000 0.999 1.000 1.000 1.000
0.40 50 0.356 0.217 0.125 0.447 0.507 0.591
100 0.637 0.536 0.439 0.739 0.842 0.876
250 0.981 0.964 0.942 0.990 1.000 0.991
500 1.000 1.000 0.999 1.000 1.000 1.000
1000 1.000 1.000 1.000 1.000 1.000 1.000
0.50 50 0.523 0.357 0.221 0.635 0.714 0.784
100 0.843 0.771 0.676 0.904 0.961 0.974
250 0.999 0.999 0.996 0.999 1.000 1.000
500 1.000 1.000 1.000 1.000 1.000 1.000
1000 1.000 1.000 1.000 1.000 1.000 1.000
1.00 50 0.991 0.972 0.918 0.997 0.999 0.999
100 1.000 1.000 1.000 1.000 1.000 1.000
250 1.000 1.000 1.000 1.000 1.000 1.000
500 1.000 1.000 1.000 1.000 1.000 1.000
1000 1.000 1.000 1.000 1.000 1.000 1.000

 

 

 

 

 

 

 

 

97

 

 

Table 3.4
Power of the tests when p = 0.40

Nominal size = 0.05

Null: the data are bivariate-normal with the equal means and equal variances
Alternative: the data are from bivariate-normal with the mean difference 6

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

6 n I 3 (3) Z 3 (6) Z 3(9) Bt.strp Bt.strp T-test
KS BWS

0.10 50 0.067 0.041 0.021 0.070 0.065 0.096

100 0.079 0.069 0.052 0.105 0.188 0.146

250 0.172 0.131 0.109 0.197 0.247 0.303

500 0.290 0.243 0.251 0.361 0.457 0.526

1000 0.564 0.494 0.433 0.648 0.745 0.828

0.20 50 0.136 0.075 0.038 0.165 0.181 0.245

100 0.221 0.169 0.131 0.297 0.384 0.436

250 0.576 0.475 0.409 0.640 0.763 0.826

500 0.871 0.831 0.776 0.904 0.969 0.984

1000 0.995 0.999 0.990 1.000 1.000 0.999

0.30 50 0.262 0.150 0.088 0.295 0.381 0.473

100 0.477 0.385 0.310 0.558 0.702 0.768

250 0.918 0.872 0.831 0.936 0.985 0.990

500 0.999 0.998 0.959 1.000 1.000 1.000

1000 1 .000 1 .000 1.000 1 .000 1 .000 1.000

0.40 50 0.439 0.283 0177 0.506 0.610 0.717

100 0.752 0.670 0.580 0.847 0.915 0.949

250 0.996 - 0.672 0.985 1.000 1.000 1.000

500 1 .000 0.999 1 .000 1.000 1 .000 1.000

1000 1 .000 1 .000 1.000 1.000 1 .000 1.000

0.50 50 0.522 0.466 0.316 0.696 0.836 0.887

100 0.843 0.884 0.825 0.951 0.987 0.994

250 0.999 1.000 0.999 1.000 1.000 1.000

500 1.000 1.000 1.000 1.000 1.000 1.000

1000 1 .000 l .000 l .000 1 .000 1.000 1.000

1.00 50 0.991 0.994 0.980 1.000 1.000 1.000

100 1.000 1.000 1.000 1.000 1.000 1.000

250 1.000 1.000 1.000 1.000 1.000 1.000

500 1.000 1.000 1.000 1.000 1.000 1.000

1000 1.000 1.000 1.000 1.000 1.000 1.000

 

 

 

 

 

 

 

 

98

 

 

Table 3.5
Power of the tests when p = 0.60

Null: the data are bivariate-normal with the equal means and equal variances
Alternative: the data are from bivariate-normal with the mean difference 6

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

6 n 2’2 (3) 2’ 3 (6) I 3 (9) Bt.strp Bt.strp T-test
KS BWS

0.10 50 0.075 0.041 0.023 0.056 0.079 0.121

100 0.094 0.083 0.055 0.112 0.139 0.197

250 0.224 0.174 0.149 0.270 0.334 0.423

500 0.394 0.349 0.302 0.430 0.602 0.702

1000 0.721 0.683 0.626 0.758 0.899 0.946

0.20 50 0.172 0.097 0.052 0.189 0.236 0.350

100 0.295 0.246 0.184 0.346 0.482 0.595

250 0.731 0.659 0.599 0.770 0.891 0.944

500 0.959 0.949 0.932 0.969 0.996 0.999

1000 0.999 ‘ 1.000 1.000 1.000 1.000 1.000

0.30 50 0.353 0.222 0.132 0.352 0.484 0.642

100 0.625 0.558 0.473 0.676 0.837 0.909

250 0.981 0.970 0.957 0.978 1.000 0.997

500 1.000 1.000 1 .000 l .000 1 .000 1.000

1000 1.000 1.000 1.000 1.000 1.000 1.000

0.40 50 0.438 0.420 0.279 0.604 0.752 0.875

100 0.752 0.851 0.782 0.906 0.976 0.993

250 0.996 0.999 0.999 1.000 1.000 1.000

500 1.000 1.000 1.000 1.000 1.000 1.000

1000 1.000 1.000 1.000 1.000 1.000 1.000

0.50 50 0.634 0.649 0.496 0.696 0.946 0.973

100 0.925 ' 0.974 0.956 0.951 1.000 1.000

250 0.999 1.000 1.000 1.000 1.000 1.000

500 1.000 1.000 1.000 1.000 1.000 1.000

1000 l .000 1 .000 1 .000 1 .000 l .000 1 .000

1.00 50 1.000 0.999 0.999 1.000 1.000 1.000

100 l .000 1 .000 1 .000 1 .000 1 .000 1.000

250 1.000 1.000 1.000 1.000 1.000 1.000

500 1.000 1.000 1.000 1.000 1.000 1.000

1000 1.000 1.000 1.000 1.000 1.000 1.000

 

 

 

 

 

 

 

 

99

 

 

 

Table 3.6
Power of the tests when p = 0.80

Null: the data are bivariate-normal with the equal means and equal variances
Alternative: the data are from bivariate-normal with the mean difference 6

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

6 n 13 (3) 12(6) 2’ 3 (9) Bt.strp Bt.strp T-test
KS BWS

0.10 50 0.095 . 0.046 0.025 0.083 0.081 0.195

100 0.138 0.114 0.086 0.158 0.205 0.342

250 0.369 0.309 0.274 0.372 0.524 0.710

500 0.633 0.629 0.577 0.646 0.856 0.941

1000 0.933 0.941 0.931 0.945 0.993 0.998

0.20 50 0.269 0.159 0.092 0.256 0.340 0.592

100 0.492 0.454 0.376 0.481 0.710 0.876

250 0.934 0.928 0.901 0.926 0.999 0.999

500 0.999 1.000 0.999 1 .000 l .000 1.000

1000 1.000 1 .000 1.000 1.000 1.000 1.000

0.30 50 0.570 0.420 0.287 0.537 0.724 0.908

100 0.874 0.864 0.815 0.881 0.974 0.996

250 0.999 1.000 1.000 1.000 1.000 1.000

500 1.000 , 1.000 1.000 1.000 1.000 1.000

1000 1.000 1.000 1.000 1.000 1.000 1.000

0.40 50 0.832 0.733 0.592 0.789 0.955 0.992

100 0.989 0.991 0.983 0.881 1.000 1.000

250 1.000 1.000 1.000 1.000 1.000 1.000

500 1.000 1.000 1.000 1.000 1.000 1.000

1000 1 .000 1 .000 1 .000 1 .000 1 .000 1.000

0.50 50 0.963 0.930 0.861 0.931 0.997 1.000

100 0.999 1 .000 1.000 0.994 0.999 1 .000

250 1.000 1.000 1.000 1.000 1.000 1.000

500 1.000 1.000 1.000 1.000 1.000 1.000

1000 1.000 1.000 1 .000 1.000 1.000 1.000

1.00 50 1.000 - 1.000 1.000 1.000 1.000 1.000

100 l .000 1 .000 1 .000 1 .000 1 .000 1.000

250 1.000 1.000 1.000 1.000 1.000 1.000

500 1.000 1.000 1.000 1.000 1.000 1.000

1000 1.000 1.000 1.000 1.000 1.000 1.000

 

 

 

 

 

 

 

 

100

 

 

Table 3.7
Power of the tests when p = 0.90

Nominal size = 0.05

Null: the data are bivariate-normal with the equal means and equal variances
Alternative: the data are from bivariate-normal with the mean difference 6

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

6 n 13(3) 12(6) )5 3 (9) Bt.strp Bt.strp T-test
KS BWS

0.10 50 0.121 0.064 0.030 0.086 0.099 0.340

100 0.202 0.184 0.147 0.207 0.324 0.592

250 0.574 0.562 0.529 0.531 0.797 0.941

500 0.865 0.901 0.890 0.877 0.988 0.999

1000 0.995 0.999 0.998 0.999 1.000 1.000

0.20 50 0.426 0.293 0.188 0.308 0.515 0.875

100 0.730 0.752 0.701 0.674 0.926 0.993

250 0.996 0.998 0.998 0.995 1.000 1.000

500 1.000 1.000 1.000 1.000 1.000 1.000

1000 1.000 1.000 1.000 1.000 1.000 1.000

0.30 50 0.799 0.719 0.600 0.653 0.932 0.996

100 0.984 0.990 0.990 0.977 0.999 1.000

250 1.000 1.000 1.000 1.000 1.000 1.000

500 1.000 1.000 1.000 1.000 1.000 1.000

1000 1.000 1.000 1.000 1.000 1.000 1.000

0.40 50 0.971 0.958 0.922 0.919 0.998 1.000

100 1.000 1.000 1.000 1.000 1.000 1.000

250 1.000 1.000 1.000 1.000 1.000 1.000

500 1.000 I 1.000 1.000 1.000 1.000 1.000

1000 1 .000 1 .000 1 .000 l .000 1 .000 1.000

0.50 50 0.998 0.998 1.000 0.980 1.000 1.000

100 1.000 1.000 1.000 1.000 1.000 1.000

250 1.000 1.000 1.000 1.000 1.000 1.000

500 1.000 1.000 1.000 1.000 1.000 1.000

1000 1.000 1.000 1.000 1.000 1.000 1.000

1.00 50 1.000 1.000 1.000 1.000 1.000 1.000

100 1.000 1.000 1.000 1.000 1.000 1.000

250 1.000 1.000 1.000 1.000 1.000 1.000

500 1 .000 1 .000 1.000 1.000 1 .000 1.000

1000 1.000 1.000 1.000 1.000 1.000 1.000

 

 

 

 

 

 

 

 

101

 

 

Table 3.8
Power of the tests when p = 0.95

Nominal size = 0.05

Null: the data are bivariate-normal with the equal means and equal variances
Alternative: the data are from bivariate-normal with the mean difference 6

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

6 n 2’ 3 (3) 2’ 3 (6) 2’ 3 (9) Bt.strp Bt.strp T-test
KS BWS

0.10 50 0.160 0.084 0.042 0.102 0.112 0.595

100 0.300 0.313 0.279 0.270 0.484 0.876

250 0.780 0.834 0.836 0.736 0.966 0.999

500 0.974 0.995 0.997 0.975 0.999 1.000

1000 0.999 1 .000 1.000 1 .000 1.000 1 .000

0.20 50 0.612 0.528 0.412 0.431 0.758 0.993

100 0.960 0.958 0.951 0.853 0.995 1.000

250 1.000 1.000 1.000 1.000 1.000 1.000

500 1.000 1.000 1 .000 1.000 1 .000 1.000

1000 1.000 1.000 1.000 1.000 1.000 1.000

0.30 50 0.940 0.943 0.915 0.803 1.000 1.000

100 0.999 1.000 1.000 1.000 1.000 1.000

250 1.000 1.000 1.000 1.000 1.000 1.000

500 1.000 1.000 1.000 1.000 1.000 1.000

1000 1.000 1.000 1.000 1.000 1.000 1.000

0.40 50 0.998 0.999 1.000 0.973 1.000 1.000

100 1.000 1.000 1.000 1.000 1.000 1.000

250 1.000 1.000 1.000 1.000 1.000 1.000

500 1.000 1.000 1 .000 1.000 1.000 1.000

1000 1.000 1.000 1.000 1.000 1.000 1.000

0.50 50 1.000 1.000 1.000 1.000 1.000 1.000

100 1.000 1.000 1.000 1.000 1.000 1.000

250 1.000 1.000 1.000 1.000 1.000 1.000

500 1.000 1.000 1.000 1.000 1.000 1.000

1000 1 .000 1.000 1.000 1.000 1.000 1.000

1.00 50 1.000 1.000 1.000 1.000 1.000 1.000

100 1.000 1.000 1.000 1.000 1.000 1.000

250 1.000 1.000 1.000 1.000 1.000 1.000

500 1.000 1.000 1.000 1.000 1.000 1.000

1000 1.000 1.000 1.000 1.000 1.000 1.000

 

 

 

 

 

 

102

 

Table 3.9
Power of the tests when p = - 0.20

Nominal size = 0.05

Null: the data are bivariate-normal with the equal means and equal variances
Alternative: the data are from bivariate-normal with the mean difference 6

 

12(3)

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

6 n I 3 (6) 2’2 (9) Bt.strp Bt.strp T-test
. KS BWS
0.10 50 0.055 0.045 0.024 0.056 0.060 0.063
100 0.055 0.054 0.047 0.099 0.095 0.094
250 0.110 0.088 0.078 0.132 0.151 0.167
500 0.187 0.255 0.102 0.253 0.279 0.298
1000 0.317 0.255 0.218 0.432 0.482 0.508
0.20 50 0.090 0.046 0.033 0.121 0.137 0.141
100 0.141 0.108 0.008 0.192 0.225 0.242
250 0.319 0.240 0.209 0.415 0.495 0.524
500 0.609 0.514 0.417 0.722 0.804 0.818
1000 0.914 0.854 0.808 0.958 0.980 0.985
0.30 50 0.163 0.082 0.043 0.215 0.241 0.265
100 0.286 0.216 0.158 0.398 0.473 0.496
250 0.661 g 0.562 0.467 0.763 0.846 0.863
500 0.930 0.903 0.864 0.967 0.981 0.987
1000 0.999 0.998 0.994 1.000 1.000 1.000
0.40 50 0.245 0.137 0.074 0.314 0.394 0.425
100 0.480 0.376 0.286 0.590 0.695 0.719
250 0.921 0.855 0.798 0.954 0.983 0.981
500 0.998 0.994 0.988 1.000 1.000 1.000
1000 1.000 1.000 1.000 1.000 1.000 1.000
0.50 50 0.369 0.219 0.136 0.490 0.570 0.611
100 0.674 0.573 0.464 0.789 0.864 0.878
250 0.992 0.976 0.962 1.000 1.000 1.000
500 1.000 1.000 1.000 1.000 1.000 1.000
1000 1 .000 l .000 1 .000 1 .000 1 .000 l .000
1.00 50 0.954 0.881 0.714 0.975 0.990 0.993
100 0.999 0.999 0.996 1.000 1.000 1.000
250 1.000 1.000 1.000 1.000 1.000 1.000
500 1.000 1.000 1.000 1.000 1.000 1.000
1000 1.000 1.000 1.000 1.000 1.000 1.000

 

 

 

 

 

103

 

Table 3.10
Power of the tests when p = - 0.40

Nominal size = 0.05

Null: the data are bivariate-normal with the equal means and equal variances
Alternative: the data are from bivariate-normal with the mean difference 6

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

6 n 2’ 3 (3) Z 3 (6) Z 3 (9) Bt.strp Bt.strp T-test
KS BWS

0.10 50 0.060 0.032 0.028 0.061 0.067 0.065
100 0.050 0.048 0.050 0.090 0.090 0.090

250 0.097 0.077 0.068 0.119 0.143 0.153

500 0.171 0.128 0.100 0.238 0.253 0.262

1000 0.278 - 0.209 0.197 0.394 0.432 0.439

0.20 50 0.077 0.059 0.027 0.110 0.126 0.117
100 0.129 0.081 0.066 0.163 0.224 0.220

250 0.289 0.206 0.192 0.405 0.434 0.466

500 0.539 0.438 0.379 0.651 0.763 0.758

1000 0.865 0.787 0.732 0.926 0.970 0.974

0.30 50 0.126 0.077 0.042 0.203 0.225 0.233
100 0.256 0.191 0.135 0.359 0.426 0.432

250 0.598 0.481 0.412 0.729 0.795 0.812

500 0.896 0.845 0.777 0.949 0.973 0.975

1000 0.997 0.993 0.988 1.000 1.000 1.000

0.40 50 0.224 0.1 17 0.061 0.317 0.351 0.372
100 0.439 . 0.315 0.251 0.576 0.633 0.657

250 0.884 0.772 0.701 0.919 0.960 0.968

500 0.993 0.982 0.975 1.000 1.000 1.000

1000 1.000 1.00 1.000 1.000 1.000 1.000

0.50 50 0.339 0.210 0.177 0.437 0.512 0.544
100 0.622 0.506 0.380 0.722 0.823 0.840

250 0.979 0.950 0.927 0.986 0.990 0.999

500 1.000 1.000 0.999 1.000 1.000 1.000

1000 1 .000 1 .000 1.000 1 .000 l .000 1.000

1.00 50 0.927 0.806 0.636 0.958 0.979 0.989
100 0.999 0.994 0.989 1.000 1.000 1.000

250 1.000 1.000 1.000 1.000 1.000 1.000

500 1.000 1.000 1.000 1.000 1.000 1.000

1000 1.000 , 1.000 1.000 1.000 1.000 1.000

 

 

 

 

 

 

 

104

 

Table 3.11
Power of the tests when p = - 0.60

Nominal size = 0.05

Null: the data are bivariate-normal with the equal means and equal variances
Alternative: the data are from bivariate-normal with the mean difference 6

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

6 n 2’ 3 (3) Z 3 (6) Z— 3 (9) Bt.strp Bt.strp T-test
KS BWS

0.10 50 0.064 0.035 0.021 0.059 0.070 0.065

100 0.077 0.061 0.048 0.080 0.084 0.086

250 0.096 0.078 0.077 0.118 0.126 0.142

500 0.141 0.120 0.109 0.220 0.232 0.232

1000 0.242 0.175 0.177 0.349 0.389 0.394

0.20 50 0.086 0.044 0.018 0.111 0.115 0.116

100 0.120 0.105 0.066 0.152 0.205 0.194

250 0.267 0181 0.160 0.350 0.410 0.419

500 0.481 - 0.400 0.332 0.613 0.673 0.683

1000 0.804 0.718 0.656 0.896 0.947 0.951

0.30 50 0.132 0.070 0.035 0.184 0.200 0.205

100 0.217 0.166 0.118 0.333 0.386 0.387

250 0.528 0.424 0.366 0.671 0.744 0.754

500 0.862 0.776 0.712 0.915 0.955 0.960

1000 0.991 0.984 0.974 1.000 1.000 1.000

0.40 50 0.195 0.115 0.070 0.279 0.321 0.332

100 0.379 0.281 0.209 0.509 0.574 0.582

250 0.828 0.716 0.632 0.885 0.930 1.000

500 0.982 0.965 0.947 1.000 1.000 1.000

1000 1 .000 1 .000 1 .000 1 .000 1 .000 1 .000

0.50 50 0.295 - 0.185 0.108 0.394 0.464 0.478

100 0.554 0.460 0.365 0.699 0.788 0.787

250 0.959 0.915 0.881 0.988 0.997 0.996

500 1.000 1.000 0.993 1.000 1.000 1.000

1000 1.000 1.000 1.000 1.000 1.000 1.000

1.00 50 0.877 0.737 0.574 0.935 0.964 0.971

100 0.997 0.985 0.965 1.000 1.000 1.000

250 1.000 1.000 1.000 1.000 1.000 1.000

500 1.000 1.000 1.000 1 .000 1.000 1.000

1000 1.000 1.000 1.000 1.000 1.000 1.000

 

 

 

 

 

 

 

 

105

 

 

 

 

Table 3.12
Power of the tests when p = - 0.80

Nominal size = 0.05

Null: the data are bivariate-normal with the equal means and equal variances
Alternative: the data are from bivariate-normal with the mean difference 6

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

6 n I 3(3) 2’2 (6) Z 3(9) Bt.strp Bt.strp T-test
' KS BWS
0.10 50 0.051 0.036 0.023 0.061 0.073 0.066
100 0.057 0.055 0.054 0.082 0.083 0.088
250 0.083 0.076 0.067 0.101 0.121 0.132
500 0.134 0.105 0.108 0.205 0.205 0.208
1000 0.225 0.166 0.157 0.312 0.357 0.363
0.20 50 0.082 0.043 0.024 0.102 0.110 0.106
100 0.102 0.076 0.068 0.153 0.187 0.173
250 0.226 0.165 0.150 0.317 0.364 0.379
500 0.435 0.356 0.301 0.553 0.625 0.632
1000 0.747 0.661 0.603 0.855 0.918 0.919
0.30 50 0.129 0.065 0.036 0.1650 0.178 0.188
100 0.199 0.145 0.100 0.286 0.336 0.338
250 0.477 ' 0.347 0.309 0.608 0.681 0.705
500 0.800 0.718 0.663 0.886 0.921 0.927
1000 0.989 0.973 0.956 0.993 0.995 0.999
0.40 50 0.197 0.095 0.057 0.271 0.300 0.298
100 0.333 0.259 0.183 0.464 0.522 0.535
250 0.764 0.636 0.566 0.885 0.904 0.907
500 0.964 0.940 0.918 0.987 0.990 0.992
1000 1.000 1.000 0.998 1.000 1.000 1.000
0.50 50 0.280 0.149 0.084 0.362 0.419 0.445
100 0.504 0.402 0.323 0.643 0.725 0.731
250 0.930 0.862 0.806 0.965 0.988 0.988
500 0.999 0.999 0.990 1.000 1.000 1.000
1000 1.000 1.000 1.000 1.000 1.000 1.000
1.00 50 0.843 0.675 0.512 0.905 0.942 0.950
100 0.992 0.969 0.942 0.907 0.989 0.999
250 1.000 1.000 1.000 1.000 1.000 1.000
500 1.000 1.000 1.000 1.000 1.000 1.000
1000 1.000 1.000 1.000 1 .000 1.000 1.000

 

 

 

 

 

 

 

 

106

 

Power of the tests when p = - 0.90
Nominal size = 0.05

Table 3.13

Null: the data are bivariate-normal with the equal means and equal variances
Alternative: the data are from bivariate-normal with the mean difference 6

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

6 n 2’ 3(3) 12(6) Z 3(9) Bt.strp Bt.strp T-test
KS BWS

0.10 50 0.054 0.034 0.020 0.065 0.079 0.071

100 0.051 0.047 0.044 0.081 0.090 0.093

250 0.083 0.062 0.058 0.104 0.123 0.128

500 0.126 0.119 0.093 0.184 0.193 0.197

1000 0.207 0.159 0.126 0.297 0.338 0.345

0.20 50 0.069 0.033 0.022 0.105 0.109 0.105

100 0.109 0.085 0.061 0.160 0.179 0.172

250 0.215 0.147 0.132 0.293 0.346 0.360

500 0.397 0.328 0.282 0.515 0.588 0.613

1000 0.712 0.621 0.568 0.821 0.895 0.904

0.30 50 0.116 0.072 0.042 0.159 0.171 0.177

100 0.189 0.138 0.107 0.263 0.317 0.326

250 0.451 0.346 0.298 0.578 0.655 0.680

500 0.771 0.682 0.644 0.860 0.903 0.916

1000 0.986 0.963 0.940 0.995 0.999 0.999

0.40 50 0.180 0.093 0.050 0.224 0.276 0.284

100 0.307 0.236 0.168 0.435 0.506 0.518

250 0.746 ' 0.623 0.522 0.824 0.890 0.889

500 0.952 0.937 0.911 0.980 0.990 0.990

1000 1.000 0.999 0.998 1.000 1.000 1.000

0.50 50 0.268 0.142 0.079 0.351 0.414 0.418

100 0.477 0.384 0.288 0.605 0.703 0.708

250 0.920 0.854 0.792 0.953 0.979 0.983

500 0.998 0.994 0.986 1.000 1.000 1.000

1000 1.000 1.000 1.000 1.000 1.000 1.000

1.00 50 0.830 0.646 0.478 0.881 0.929 0.940

100 0.991 0.968 0.936 0.991 0.995 0.990

250 1.000 1.000 1.000 1.000 1.000 1.000

500 1.000 1.000 1.000 1.000 1.000 1.000

1000 1.000 1.000 1.000 1.000 1.000

 

 

 

 

 

 

 

 

1 .000

107

 

Table 3.14
Power of the tests when p = - 0.95

Nominal size = 0.05

Null: the data are bivariate-normal with the equal means and equal variances
Alternative: the data are from bivariate-normal with the mean difference 6

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

6
6 n 2’ 3(3) 2’ 3(6) Z 3(9) Bt.strp Bt.strp T-test
KS BWS
0.10 50 0.056 0.035 0.023 0.063 0.078 0.072
100 0.048 0.053 0.042 0.085 0.088 0.095
250 0.085 0.660 0.059 0.097 0.116 0.126
500 0.132 0.108 0.088 0.191 0.188 0.190
1000 0.210 0.162 0.131 0.287 0.298 0.338
0.20 50 0.073 0.038 0.019 0.091 0.105 0.110
100 0.090 0.072 0.069 0.144 0.174 0.172
250 0.211 0.155 0.135 0.294 0.342 0.352
500 0.388 I 0.323 0.260 0.503 0.503 0.597
1000 0.699 0.617 0.562 0.819 0.884 0.897
0.30 50 0.106 0.057 0.040 0.167 0.172 0.172
100 0.190 0.137 0.101 0.243 0.308 0.317
250 0.438 0.342 0.295 0.592 0.655 0.667
500 0.754 0.668 0.602 0.855 0.909 0.910
1000 0.985 0.953 0.934 1.000 1.000 1.000
0.40 50 0.170 0.089 0.050 0.239 0.284 0.281
100 0.304 0.230 0.166 0.411 0.493 0.495
250 0.727 0.617 0.520 0.811 0.874 0.882
500 0.994 0.925 0.892 1.000 1.000 1.000
1000 1.000 0.999 0.999 1.000 1.000 1.000
0.50 50 0.259 ' 0.153 0.083 0.344 0.404 0.416
100 0.452 0.368 0.277 0.616 0.679 0.697
250 0.913 0.837 0.781 0.943 0.981 0.978
500 0.996 0.990 0.982 1.000 1.000 1.000
1000 1.000 1.000 1.000 1.000 1.000 1.000
1.00 50 0.814 0.639 0.460 0.879 0.924 0.936
100 0.986 0.962 0.914 0.993 0.990 0.990
250 1.000 1.000 1.000 1.000 1.000 1.000
500 1.000 1.000 1.000 1.000 1.000 1.000
1000 1.000 1.000 1.000 1.000 1.000 1.000

 

 

 

 

 

 

 

 

108

 

 

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