SOME MEASURES OF SKEWNESS
By
Clarence Edwards Davis
and
Dana Quade
Department of Biostatistics
University of North Carolina at Chapel Hill
Institute of Statistics Mimeo Series No. 663
January 1970
ABSTRACT
DAVIS, CLARENCE EDWARDS.
Some
Measures
of
Skewness.
(Under the
direction of DANA QUADE).
Two new statistics, ql and qz, for describing the symmetry or lack
of symmetry of a distribution are introduced.
Since these statistics are
U statistics and thus are asymptotically normally distributed, either
can be used to test the hypothesis of population symmetry for large
samples.
The variances of the qk depend on the underlying distribution;
hence these tests are not distribution free.
However, consistent
estimators of the variances are obtained and whence asymptotically
distribution free tests.
for most parent
Using Monte Carlo methods it is shown that
distributions a sample of size 50 is adequate to assume
that the distribution of qk; k
= 1,2
is approximately normal.
For tests
of the simple hypothesis of population normality against five simple
222
alternatives; viz. the Xl , X2 ' X ' log normal and Poisson ( A
4
= 4)
distributions, empirical sampling methods are used to compare the small
sample power of the tests based on the qk with that based on the moment
statistic b .
1
For a sample of 10, b l is found to be more powerful than
either q1 or Q2'
However, for samples of size 15 there is little
difference among the powers of the three tests.
Letting the population
probability density function be
f(x)
=
(1+0) g(x) I[x<O]+(l-o)g(x) I[x>O]' where g(x)
is the density of a distribution symmetric about the origin, the
asymptotic relative efficiencies of ql and q2 to b l are computed.
When g(x) is the uniform, normal, Laplace, or Cauchy distributions,
these asymptotic relative efficiencies are all greater than one.
ABSTRACT (Continued)
ii
A U,statistic, h, for describing kurtosis is also defined.
Combining the appropriate test based on h and either of the above tests
of symmetry, a test of the normality of a distribution is obtained.
As
the Monte Carolo work indicates that under the hypothesis of normality,
the distributions of the qk and h are approximately normal for samples
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of 30, the test for normality can be applied with as few as 30 observations.
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BIOGRAPHY
The author was born January 14, 1941 in Houston, Texas.
He was
reared on a farm near Waco, Texas and was graduated from LaVega High
School in 1959.
In 1963 he received the Bachelor of Arts degree with a major in
mathematics from Howard Payne College.
He received a grant from the
National Institutes of Health which enabled him to begin studies in
biostatistics at Tulane University where he received the Master of
Science degree in 1965.
From 1965 through 1969, the author was a
doctoral student in Experimental Statistics at North Carolina State
University.
During that time he also was a research fellow in the
Department of Biostatistics at the University of North Carolina.
The
requirements for the Ph. D. in Experimental Statistics were completed
in 1970.
He then accepted a position as Assistant Professor of
Statistics at the University of Florida.
The author is married to the former Sherri1yn Elizabeth Crawford
of Abilene, Texas.
They have one child, Terisa Elizabeth.
iv
ACKNOWLEDGMENTS
The author expresses his appreciation to his advisor, Dr. Dana
Appreciation-is also expressed to
the other members·of-the-advisory-committee, -Professors H. A. David,
B. B. Bhattacharyya, P. K. Sen and H. A. Tyroler.
Appreciation-is also expressed to Professor Tom Dowling for his
assistance with one -of the proofs -and·- to Professor Ron Helms and
Mr. Charles Holzer-for help and-advice-regarding"computer usage.
The manuscript-was - typed -by -the
authorls~w1fe,
She1:'rilyn.
Her
assistance -and patience -throughout -his -studies ,- are'-mueh appreciated.
Finally,-the author-expresses-his-thanks-ta-his
pare~ts,
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who suggested the topic of this dissertation; and who was always
available for counsel-and-guidance,
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Mr. and Mrs. H. -J. Turk, "for their encouragement-at,ld-understanding
throughout his studies.
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TABLE OF CONTENTS
Page
LIST OF TABLES
vi
LIST OF FIGURES.
1.
INTRODUCTION AND REVIEW OF LITERATURE .
1.1
1.2
2.
1
TWO NEW MEASURES OF SKEWNESS . .
14
2.3
3.3
3.4
3.5
.
On Defining Skewness
Definitions of the New Statistics for
Describing Non-symmetry.
Some Examples of Q and Q2 for Non-symmetrical
1
Distributions.
·....
.... ..·.
...·
.........
The Tests of Symmetry. .
The Rate of Convergence of the
Distributions to Normality •
The Small Sample Power of the Tests. .
Some Asymptotic Relative Efficiency Considerations
Example. . . • .
. . . • •
. . . . .
16
19
35
43
48
56
80
82
82
87
A Heasure of Kurtosis.
Example • . . .
SUMMARY AND SUGGESTIONS FOR FURTHER RESEARCH.
5 .1
Sutnmary. . .
5.2
Suggestions for Further Research.
It
14
35
A MEASURE OF KURTOSIS .
4.1
4.2
5.
1
TESTS OF SYMMETRY BASED ON ql AND q2'
3.1
3.2
4.
1
Introduction . . . . .
Review of Literature.
2.1
2.2
3.
. vii
•
•
•
•
•
0
•
•
•
88
88
89
6.
LIST OF REFERENCES.
90
7.
APPENDICES . . . . .
92
7.1
7.2
Computer Program for Calculating q1 and q2 .
Computer Program for Calculating h . • • • .
92
93
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LIST OF TABLES
Page
....
3.1
Ok 2 ; k • 1,2 for Four Symmetric Distributions • • • • .
3.2
Number of Observed Izkl; k • 1,2 Which Exceeded the 5
and 1% Points of the Normal Distribution in 200
Samples of n • 10, 20, 30 from the Normal Distribution ••
44
Number of Observed Izkl; k .. 1,2 Which Exceeded the 5
and 1% Points of the Normal Distribution in 200
Samples of n = 30, 40, 50 from the ~iform
Distribution • • • •
• • • • •
46
3.3
3.4
3.5
3.6
Number of Observed IZkl; k • 1,2 Which Exceeded the 5
and 1% Points of the Normal Distribution in 200
Samples of n .. 30, 40, 50, 60 from the Cauchy
Distribution
• • • •
• • . • . • • •
47
Number of Observed I'zkl; k .. 1,2 Which Exceeded the 5
and 1% Points of the Normal Distribution in 200
Samples of n = 30 from the Laplace Distribution
47
Sample Size Needed to Assume Approximate Normality
.for zk;.k • 1,2 .
.
= 0.1
3.7
Empirical Powers far ex
3.8
Comparison of the Estimates of the Functionals with Their
True Values for the Uniform Distribution; N = 50,
n - 50
3.9
4.1
41
.
•
• • • • • • • ••
48
50
•••••.
Comparison of the Estimates of the Functiona1s and the
Values Obtained by Numerical Integration for the
Narmal Distribution; N .. 500, n • 50 • • • •
74
75
Number of Observed Izi Which Exceeded the 5 and 1% Points
of the Normal Distribution in 200 Samples of
n
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20. 30. . . . . . . . . . . . . . . . . . . . . .
86
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LIST OF FIGURES
Page
3.1
3.2
3.3
Merit Curve for the Chi Squared Distribution with pne
Degree. af .Freedam;.. n = 10
51
Merit Curve for
Degrees of
51
. · · ·.. · · . . . . . · . · .
the Chi Squared Distribution with Two
Freedom; n = 10
···.··.• ...·.·.
Merit Curve for the Chi Squared Distribution with Four
Degrees of Freedom; n = 10
•
·· .··
·.
3.4
Merit Curve for the Log Normal Distribution; n = 10
3.5
Merit Curve for the Poisson Distribution; A
3.6
Merit Curve for the Chi Squared Distribution with One
Degree of Freedam; n = 15 • • • • • • • • • • • •
3.7
3.$
3.9
= 4,
n
52
= 10.
52
53
· .. . .
53
Merit Curve for the Chi Squared Distribution with Two
Degrees af Freedom; n = 15 • • • • • • • • • . •
54
Merit Curve for the Chi Squared Distribution with Four
Degrees of Freedom; n = 15 ••• • • • •
54
Merit Curve for the Log Normal Distribution; n
= 15
3.10 Merit Curve for the Poisson Distribution; A a 4, n
55
= 15.
.
55
CHAPTER 1.
1.1
INTRODUCTION AND REVIEW OF LITERATURE
Introduction
Let Xl' X2, •..• Xn be n independently and identically distributed
random variables, with cumulative distribution function (c .d. f.) F.
Various functions of the observed values of the X's are used to
describe the properties of F.
The characteristics of the distribution
most commonly considered are location, scale, and skewness or deviation
from symmetry.
We shall be primarily concerned with statistics for
describing the skewness of a distribution.
izing the skewness of F will be introduced.
Two statistics for characterBased on these statistics
tests of the hypothesis that the distribution F is symmetric versus
either the one-sided alternative of directed skewness of the twosided alternative of skewness are obtained.
As an extension of these
statistics, a statistic for describing the kurtosis of a distribution
and a corresponding test are developed.
1.2
Review of the Literature
The most commonly used measure of the skewness of a distribution
is the moment measure given by
13 1 ::: 113/ 11123
where llk; k ::: 2,3, is the k th central moment of the distribution.
Note that we have departed from the conventional notation in which
13 1
The estimate of 61 is
=
ll/ / 112 3
2
n
- k
th
,~(xl' - X) ; k = 2, 3; !.~. m is the k
sample moment.
k
1=1
Assuming population moments through order 6 exist, under the hypothesis
~ =
where
n
-1
of symmetry, E (b )
1
= 0,
02(b )
1
and b
1
=
(1.2.1)
is asymptotically normally distributed.
Thus for large samples,
bl/O(b l ) has the standard normal distribution and this statistic can
be used to test the symmetry of the parent distribution.
Gupta (1967)
notes that since 0 2 (b l ) depends on the underlying distribution, this test
It can be made so by estimating 02(b )
l
is not distribution free.
by the corresponding function of the sample moments and studentizing.
However, Pearson (1932) notes that even if the underlying distribution
is the normal distribution, a sample of at least 100 is required before
the distribution of b 1 is sufficiently close to the normal distribution
to use the above test.
For samples of less than 100, the 5% and 1%
points of the distribution are tabulated, under the assumption of a
normal parent (see Pearson and Hartley (1962)).
If F is symmetric and its mean exists, then the mean, median,
and mode coincide.
This has led to two other measures of skewness:
Yl
=
(~
Y2
=
(~
-
xO)/O , where
Xo
is the mode of F
and
- 8)/0
,where 8 is the median of F.
These measures are estimated by replacing the population values with
their corresponding sample values.
Yl has the disadvantage of the
difficulty in determining the sample mode.
Also, little is known
concerning the sampling distributions of either of these statistics.
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Assume that F is strictly increasing, so that F (a) < F (a+b)
for b > O.
o<
X < 1.
Let 1;x
be the Xth quantile of F;
i.e. F(1;X)
= X
The skewness of F can then be described by
X3 (X) • [(1;1_~- 1;1/2) - (1;1/2 - 1;X)]/(1;l_X -~X); 0 < X < l/Z.
If F is not strictly increasing, Y3 (X) may not be unique.
To estimate
Y (X) we replace the population quantiles by the corresponding sample
3
quanti1es.
viz.
This leads to a second problem in the use of Y3 (X),
the determination of the sample quanti1es.
There are several
ways of defining sample quanti1es, which may lead to different estimates
of Y3 (X).
One advantage of Y3 (X) over the previously mentioned
measures of skewness is the lack of assumptions concerning the existence
of any of the moments of F.
Thus,
Y (X) is applicable to a wider
3
class of distributions than these measures.
However, as in the case
of Y and Y , the sampling distribution of Y (X) is not known, in
1
3
2
general.
David and Johnson (1956) suggested using functions of the order
statistics to describe the skewness of a distribution.
be the ordered
XIS.
K
r
Their statistics are, assuming n
• Yr + Yn - r +1 - ZYm+1
Yr - Yn- r +1
Let Y1 yZ'"
= 2m + 1,
r > m
and
where g ; k and g, k > m.
Using the Pitman definition of asymptotic
relative efficiency and comparing the statistics with b 1 , they
determined the
g
= 0.9775
~ptimum
(n+1, k
r, g, k, and ag/ak to be r = 0.9875 (n+1),
= 0.097
(n+1), and ag/ak
= 2.515.
Since they
are primarily interested in tests for censored data, they
Yn
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conclude that these values are too near the extremes to provide useful
tests for their purposes.
A = 1/4 and n
example suppose
~1/2
and
= Ym;
Note that Kr is similar to Y3 (A).
i·~·
= 4r-1,
= Yr'
then ~1/4
~3/4
For
= Yn-r+1'
the sample values of Kr and Y3 (1/4) are equivalent.
In a recent paper, Gupta (1967) introduces two new statistics
for describing departures from symmetry in the positive direction.
If the median,
a,
of F is known, he defines
where
if min (Xi' Xj )<8< max
X. + X < 2 a
1
j
otherwise.
~(a,
If the median is not known, he uses the statistic
unknown value of
A
J (J)
e
3,
is replaced by its sample value.
in which the
The statistic
is based on the number of positive deviations from the (sample)
median that exceed the absolute values of the negative deviation from
the (sample) median.
If F is symmetric, E (J)
and J has an asymptotic normal distribution.
= 1/4,
0 2 (J)
= 1/12n,
If F is positively skewed,
E (J) will be greater than 1/4; hence the test of symmetry versus the
one sided alternative
2 13n (J - 1/4).
o~
positive skewness is based on the statistic
The limiting distribution of
I1n (J - 1/4)/20(J) is
the standard normal distribution, where
Thus the test based on J is asymptotically distribution free, while
A
the test based on J is noto
However, Gupta obtains a consistent
estimator of 0 2 (3), and hence obtains an asymptotically distribution
free test.
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A fourth group of descriptive statistics is made up of those
which describe the kurtosis of the distribution.
kurtosis corresponding to 81 is 82 • ~4/~22.
tion, 8 • 3.
2
The measure of
For the normal distribu-
When this measure was first introduced, it was thought
that distributions for which 82 > 3 were more sharply peaked than the
normal distribution, while those for which 82 < 3 were more f1attopped.
However. as Kendall and Stuart (1958) point out, this is not
necessarily true.
Hence this idea of kurtosis is not widely used as
a descriptive statistic, due to the difficulty in understanding what
deviations from the value 3 really mean concerning the distribution.
82 is estimated by b 2 - m4/m22.
Assuming moments through order 8
exist, the variance of b 2 is
where
For the normal distribution cr 2 (b )
2
normally distributed.
1/2/1[
= 24/n
and b 'is asymptotically
2
This has 1ed,tothe use of the statistic
(b 2 - 3) in combinationwiththe~testbased on'b 1 as a test
for normality.
Pearson (1932) points,out that'the convergence of the
distribution of bi to normality is extremely slow and a sample of
size lOOO'is'not really'adequate for the distribution to be considered
normal, even for a normal parent'distribution.
An alternative statistic which may be used for detecting changes
in kurtosis was introduced by Geary (1935);' This statistic is
a -
n
n
_ 2 1/2
I x - xl )/[n E (Xi - x)]
.
i-1
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i-1
(E
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For the normal distributionE(a) =.f2Ti.
give
~ables
Pearson and Hartley (1962)
of the 10, 5, and 1% points of the distribution of a,
under the hypothesis of normality.
David and Johnson (1956) also propose a statistic based on the
order statistics, for describing kurtosis:
Yg - Yn-g+l
Hg,k • - - - - - Yk - Y'n - k+l
The optimum values of g and k based on asymptotic efficiency comparison
with
B2
1.3
Definitions and Known Results
are g • 0.78 (0+1) and k· 0.998 (n+l).
In this section various definitions and results which will be
used in the remainder of the paper are presented.
In lieu of giving
proofs for the-theorems; references where proofs may be found are
Definition 1.3.1:
The distribution F(x) is said"to be symmetric about
e)
+ F (
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cited.
F (x +
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) • 1, for all x.
without loss "of generality that
a·
If
e is
a,
if
known, we may assume
OJ which"we shall do in what
follows.
Definition 1.3.2:
A real valued parameter
~
(F) is called estimable if it has an
unbiased estimator.
Definition 1.3.3:
The degree, m, of an-estimable parameter is.defined to be the
smallest sample size' for" which the parameter has an .unbiased estimator.
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Let
¢(X l ,
~
(F) be an estimable parameter of degree
X2""~)
be an unbiased estimator of
~
m(~n)
and let
which is symmetric
in its arguments.
Definition 1.3.4 Hoeffding (1948):
The U statistic for estimating
where the summation (c) extends over 1
is
~
c
< C < ••• <c ~
12m
n.
The assumption that (Xl' X2,""Xm) is symmetric in its
arguments is not serious.
estimator of
~
which is not symmetric, a symmetric unbiased estimator
can be constructed by taking
1
=m.;"l
where the summation (p) is over the m! permutations of (1, 2, .•. ,m).
U statistics have a number of desirable properties, several of
which are summarized in the following theorems.
The proof for the first
is in Berk (l966); the remainder can be found in Fraser (1957).
Theorem 1.3.1:
If E[¢(X l , X2 , .•. ,X )]
m
exists, then the corresponding U statistic
for a sample of size n is a consistent estimator of
~.
Theorem 1.3.2:
If E[¢ 2 (X 1 ,X 2 , .•. ,X )] exists, then as n
m
+
00
the limiting
distribution of In(U - ~ ) is the normal distribution with mean 0
2
and variance m sl' where
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Sen (1960) defines the components of a U statistic by
-.
Definition 1.3.5:
The components of the U statistic corresponding to
••• , Xc
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m-l
where the summation (ci) extends over 1
S cl
< c2 <••• <cm_ls n,
except that c j - i must not occur for any j.
Sen also proves the following results:
Theorem 1. 3. 3 :
Let s
to
2
1
E(V _U)2;
n-r
i-l i
-
then s2 converges in
prob~bility
r,;l'
Thus s2 is proportional to the sum of
the squared deviations of a set of scores from their average.
In
view of theorems 1.3.2 and 1.3.3 and a theorem of Cramer (1946) we
have
I
Theorem 1. 3.4:
If E[~2(Xl' X2 ••••• Xm)] exists, then as n ~~. the limiting
distribution of
Suppose
k - 1, 2.
~e
Let
In
(U - ~ )/ms is the standard normal distribution.
have two estimable parameters
~k
~k(F)
of degree
~;
be their symmetric unbiased estimators. Uk the
corresponding U statistic with componentsVki,asymptotic variances
mk 2 r,;k' and estimated variances mk2sk2; k - 1, 2.
and
_I
Define
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Z
=
9
Z Z
Z Z
ml ~Z ~l - ZmlmZ~l~Z~lZ + mZ ~l ~Z
~ 4
Z
Quade (1967.) proves the following result:
Theorem 1. 3.5:
If Ul and U2 are two U statistics as defined above, and if
~2
> 0
and 0 < 1"2 <
00,
then the random variable
converges in distribution to the standard normal distribution, where
+ m Zu 28 2
212
U4
2
The following definitions and theorem will be useful in the sequel.
Definition 1.3.6:
~(Xl'
A function
X2 , ..• X ) is bounded
m
for all (Xl' ••• , Xm) and some finite M.
Definition 1.3.7:
A function
~(Xl,X2'·'·' X )
m
~(Xl,X2, ••.
= -~
,Xm) is antisymmetric if
(-Xl' -X 2 ,,··, -Xm)·
Theorem 1.3.6:
Let f(x) be a probability density function, symmetric about 0
and let
~(Xl'X2"o.,
Xm) be antisymmetric and bounded.
If Xl ,X 2 , •.• ,Xm
are independent and identically distributed random variables with
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Proof:
-.
Note that E exists since ~ is bounded.
In fact, lEI SM.
In the
integral expression of E, make the transformation Yi--xi; i-l,2, •.• ,m.
The Jacobian is (_l)m.
Hence
Reversing the limits of integration, using the antisymmetry property
of
~,
and the symmetry property of f, we obtain
Hence E • E[~(Xl'."'Xm)] •
o.
Q.E.D.
We shall also make use of the following well-known result which
is proved in Sarhan and Greenberg (1962).
Theorem 1. 3 • 7 :
Let Yl S Y2 S Y3 s.o.SYn denote the order statistics in a sample
of n from the standard exponential distribution, for which
f(x) - e- x
oS
x.
Moreover, let Zl - nY l , and Zi - (n-i+l)(Yi-Y i _ l ); i - 2,3, ••• ,n.
Then the Zi are independent and identically distributed as the
standard exponential distribution.
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The remaining four theorems of this chapter summarize the results
needed to compare the asymptotic efficiency of two tests in the
Pitman sense.
A complete discussion can be found in Fraser (1957).
Suppose we wish to test a hypothesis which-is characterized by a
parameter, 0, in such a way that 0=0 corresponds to the null hypothesis
and
corresponds to the alternative hypothesis.
O~O
tests {R (x)}, and {R
n
'Ie
and {t n (x)}
°02 (Tn)
'Ie
'Ie
n
(x)}
respectively.
Suppose that the
are based on the test statistics {tn(x)}
Let E (T n ) ,
5
~
°02 (Tn)'
Eo(Tn ) and
'Ie
be the mean and variance of tn(X) and-t n (X), respectively.
First we have two theorems concerning the Hlimiting-power of a sequence
of tests {R (x)} •
n
Theorem 1. 3 . 8 :
> 0 ,
If
lim•.
> 0
n~-
and if, for the
~equence
of alternatives
lim
t;I
n~
lim
n~
... "" 1 ,
1 ,
12
k
and if,corresponding to the parameter value on • -- ,T
rn
n
is
asymptotically normal with mean E~ (T ) and variance 002(T ), then the
un n
n n
limiting power of the size a test Ru(x) is 1 - ~ (ua - kc).
Theorem 1. 3 • 9 :
If for
o· 0
• c fot some y > 0 ,
lim
n-+c>o
and if for the sequence
~
un
• .!L.
n'Y
dm
E (Tn)
dam on
lim
n-+c>o
lim
nt+oO
elm
-- E
dom 0
• 1
(Tn)
• 1
.~. for k ~ 0
n nY'
,
T is asymptotically normally distributed with mean E~ (Tn) and
and if corresponding to the parametricv.alueo
n
~u
variance 0 0 2 (Tn) , then-the limiting-power of the size a test Rn(x) is
n
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Theorem 1.3.10:
'Ie
If {Tn} and {Tn} satisfy the conditions of Theorem 1.3.8, then
the asymptotic relative
.
effi~iency
'Ie
of {R } with respect to {R } is
n
n
2
0'02(Tn 'le)
lim
n-+oo
0'02(Tn)
Theorem 1.3.11:
If {Tn} and {Tn *} satisfy the conditions of Theorem 1.3.9
*
*
and if Y • Y , m • m , then the asymptotic relative efficiency of
{Rn}
re {Rn'le}
is
lImY
lim
n-+oo
..........- --
dm EJ.:(T
dam
u
'Ie)
n
~.: (T )
"'0 n
CHAPTER 2.
2.1
TWO NEW MEASURES OF SKEWNESS
On Defining Skewness
In the preceding discussion we have used the term skewness as
though it were a well defined property of a distribution.
Such is
not the case, for no satisfactory definition of skewness exists.
Thus before defining the new measures of skewness, we discuss the
problem of defining the skewness of a distribution.
There are several statistics which describe the location of a
distribution;
~'A'
mean, median, and mode.
Yet for only special types
of distributions do these measures of location coincide.
A distri-
bution may be located in one place by the mean measure, but in a
different place by the median measure.
Thus when speaking of the
location of a distribution, we must indicate which measure of location
is being used.
Exactly the same situation exists when describing the
skewness or non-symmetry of a distribution.
Most measures of skewness take the value zero if the distribution
is symmetric.
If the distribution is not symmetric, the measures 9f
skewness can be quite different and indeed may have opposite signs.
For example, consider the following distribution:
l O s x s 1/2
f
(x)
s
~
2/3
1/2 < x
2
7/8 < x s 1
E(x) • 0.53125, E(x2 ) • 0.38281, E(x3 ) • 0.30981.
7/8
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-.
• -0000043,
• 0.10059,
and 13 1 • 113/ v'"?
2 • -0.01339.
and
[,3/4 • 7/8.
On the other hand, F,1/4 • 1/4, [,1/2 • 1/2,
Hence
(7/8 - 1/2) - (1/2 - 1/4)
• 1/5
7/8 - 1/4
Thus for this distribution, 13
and y 3 (1/4) have opposite signs. The
1
distribution is therefore positively skewed in the Y3(1/4) sense,
while it is negatively skewed in the 13 1 sense.
From this example we
see that it is meaningless to say that a distribution is positively
skewed or negatively skewed 1f the criterion of measuring skewness is
not indicated.
Also a distribution which has zero skewness by one criterion
may have non-zero skewness by another.
f(x) •
Then F,1/4 • 1/4, F,1/2
D
[
For example, suppose
1
o~
2
3/4 < x
x
~
3/4
~
7/8 •
1/2, F,3/4 • 3/4 and hence Y3(1/4)
= O.
However, E(x) • 31/64 and thus Y2 < O.
The conclusion is that no one definition of skewness can be
formulated.
Instead one must be specific about how the skewness of
the distribution is measured.
Yl - skewness, etc.
We_shall then speak of 13 1 - skewness,
When we are describing skewness of a distribution.
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2.2
Definitions of the New Statistics for Describing Non-srmmetry
As before, let Xl' X2 ,
"0,
Xn be n independently and identically
distributed random variables with cod.f.
mean,
~,
and median,
e,
coincide.
Fo
If F is symmetric, the
Based on this observation, we define
a U statistic for describing the deviation from symmetry of the sample.
Definition 2.2.1:
Let
( 1
~1(Xl,X2,X3) •
l-:
i f mean (X l ,X2 ,X3 ) > median (X l ,X 2 ,X 3 )
i f mean (Xl,X2'X3 ) • median (X l ,X 2 ,X3)
i f mean (X l ,X 2 ,X3 ) < median (X l ,X2 ,X3 )·
The corresponding U statistic is
Let Yl
~
Y2
~
Y3 be the order statistics corresponding to
If the distribution is· symmetric we would expect that,
on the average, Y3 - Y2 • Y - Yl •
2
This leads to a second U statistic
for describing non-symmetry.
Definition 2.2.2:
Let
(Y 3 - Y2) - (Y2 - Yl )
Y3 - Yl
o
otherwise
The corresponding U statistic is
q2 •
(~)-l il<i2<i3~2(Xil,Xi2,Xi3)'
i f Yl
;t
Y3
17
If F is discrete, a modified version of ql and q2 might be proposed.
Definition 2.2.3:
Let N be the number of sample triplets for which Yl
~
Y3 , then
~ot
statistics, and in fact are the ratio of two U statistics.
[
1
if Yl
o
otherwise
~
U
Let
Y
3
then the corresponding U statistic is
• N(3)-1
Hence qk* • qk/U*;k • 1,2, the ratio of two U statistics.
This, as
we shall see in Chapter 3, complicates the distribution theory.
AI
a result, we shall focus primary attention for the remainder of our
discussion on ql and q2'
Note that
~r'
while
~2(Xl,X2,X3)
~1(Xl ,~
for a sample of 3.
is David and Johnson's (1956) statistic,
'X3 ) is proportional to the samplevalue of Y2
Hence, in this sense, ql and q2 are U statistics
based on Y2 and Kr , respectively.
Since mean
(Xl,~,X3)·
1/3
3
i~lYi
Also, ql and q2 are related.
and median (Xl,X2,X3) • Y2 ,
3
~1(Xl,X2,X3) •
rl
1
if 1/3 i~IYi > Y2
0
if 1/3 rY i • Y2
-1
if 1/3 EY i < Y2
-.
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define as the measures of skewness
While these statistics may be nicer intuitively, they are
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1 i f Y3 - Y2 > Y2 - Y1
•
0
if
Y3 - Y2
-1
if
Y3 - Y2 < Y2 - Yl
sgn
=
Y2 - Yl
[CP2 (Xl ,X 2 ,X 3)] ,
(2.2.1)
where
sgn
(u) •
1 if
u > 0
0
u - 0
-1
if
if u <
o.
Alternately,
CP2(Xl,X2,X3)·
[(Y 3 - Y2) - (Y 2 - Yl )]!(Y3 - Y1 )
•
(Y 3 + Yl - 2Y 2)!(Y 3 - Y1)
•
3(X3 -8 3)!w3'
'"
x3 '
8'" 3 , and w3 are the sample mean, median, and range,
respectively, in a sample of 3.
where
Intuitively, we have argued that if F is symmetric,
E(q1) • E(q2) • O.
The following theorem formalizes the argument.
Theorem 2.2.1:
Let E(qk) • Qk; k • 1,2.
If F is symmetric, then Qk • 0;
k • 1,2.
Proof:
Since Qk· E[CPk(X 1 ,X2 ,X3)]; the proof follows from Theorem 1.3.6
upon noting that ICPk(Xl,X2,X3)
Is 1
and epk(Xl,X2,X3)' is antisymmetric;
k • 1, 2.
Q and Q2 are then measures of the skewness of a distribution, in
l
the traditional sense. If Qk < 0(>0) the distribution will be said to
be negatively (positively) Qk - skewed; k • 1,2.
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2.3
Some Examples of Q1 and Q2 for Non-symmetrical.Distributions
Before presenting several examples of Q1 and Q2' we prove the
following results which will be of use in the computations.
-.
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Theorem 2.3.1:
If F is a continuous distribution with density f, then
(i)
Q1
a
12_~X F[(x + y)/2]f(y)f(x)dy dx - 3
CO
Q2. 6.-.00-00
I IX-00fY
(ii)
[(x + z - 2y)/(x - z)]f(z)f(y)f(x)dz dy dx .
Proof:
Consider first
Q1
= E[~1(X1'X2,X3)]
= P(Y 1 + Y - 2Y>
.3
. 2
0) - P(Y 1
+
Y - 2Y < 0),
2
3
where Y1 < Y2 < Y3 are the ordered X's.
a 2P(Y
1 + Y3 - 2Y 2 >·0) -1.
Since the joint density of
Y1'Y2~Y3
is
8(Y1'Y2'Y3) • 6f(Yl)f(Y2)f(Y3)~ Y1 < Y2< Y3
Q1 • 12[Yl+/~{_ 2Y2 > 0] f(Y1)f(Y2)f(Y3)dY1dY2dY3 - 1
1
Y i(Y + Y )
12_~_~3/Yl 1 3 f(Y1)f(Y2)f(Y3)dY2dY1dY3 - 1
Cl5l
a
But,
12~_!3F(Y1)f(Y1)f(Y3)dY1dY3•
6_lF2(Y3)f(YJ)dY3 • 2
Hence
coX
Q • 12-£~F[~ (x + y)]f(y)f(x)dy dx -3.
1
(2.3.1)
el
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(ii) follows immediately from the joint density of Yl'Y2'Y3 and the
definition of expectation.
Theorem 2.3.2:
If X is a random variable with probability function
P(X" j) .. Pj ; j • 0, 1, 2, ••• ,
(i)
!Pj
= 1,
then
Q1· 6i <I<kPi Pj Pk sgn(k-2j
and
(ii)
Proof:
00
00
00
Qm • i~O j~O k~O PiPjPk~m(i,j,k); m· 1,2
•
6i<f<kPiPjPk~m(i~j~k) + 3i~jPi2pj~m(i,i,j)
+
But
00
3
i~oPi ~m(i,i,i).
~
m(i,i,i) • 0 and ~m(i,i,j) • sgn(j - i); m • 1,2 so that
Qm • 6i<f<kPiPjPk~m(i,j,k) + 3i~jPiPj(Pi-Pj) m • 1,2.
Q.E.D.
Consider the exponential distribution for which f(x) .. e~x;
x > O.
From equation (2.3.1) and the transformation of Theorem 1.3.7
we have
Q1 .. 2P(Z3 - Z2/2 > 0) - 1,
where the joint density of Zl and Z2 is
f(z3'Z2) .. e
-ZS-Z2
Let u1 .. z3 - z2/ 2 and u
IJI· •
1.
2
; z3' z2 > O.
• z3' then z2 • 2(u2 - u 3), z3 .. u 2 and
Thus
and
Q1 • 2P(u1 > 0) - 1.
(2.3.2)
21
P(u
1
> 0)
-.
Thus
In (2.3.2) let u 1
Zz -
u 2 (l-u).
z3 -
c
(z3-z2/2)/(z3+z2/2) and u 2 - z3+z2/2.
IJI - Uz
u 2 (1+u1 )/Z.
Then
and
f(u1. u 2) - u2 exp[-u2(3- u1)/2]; -1 <u1 < 1. u 2 >
o.
100
Thus Q2
-_{£
u1u 2 exp[-u2(3-u1)/2]duZdu1
2
1
- 4 _{[u l /(3-u l ) ]du l
- 3 + 4 log 2--410g 4
- 0.2274
Thus for the exponential distribution. Ql • ~ and Q • 0.2274.
2
For comparison
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- 2, Y2 • 0.3070, and y (1/4) • 0.2623.
3
As an illustration of the use of Theorem 2.3.2 consider the
geometric distribution: P(X-j)-Pj-pqj; i-0,1,2, •••• 0<p<1, q=1-p.
Then
3 E p p (p -p ) - 3 E p3 q i +j(qi_ qj)
i<j i j j i
i<j
•
-
3i~0 j_~+lP3qi+j(qi_qj)
3
r
Z 3i+1
i-O P q
00
-
3 ~
3 3i+2
p q
iIiO~.-1-';"q-2"--
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=
=
3P2q
l-q j
3P3q 2
-(l---q-2)-(-1-_q-3-)
-
3p 2q(1_q2) _ 3p 3q 2
(1-q3) (1_q2)
3pq
=
(l+q) (1+q+q2)
(2.3.3)
6i<1<kPiPjPks~n(k-2j+i) •
(2.3.4)
Consider now
Le t t • k - j and s • j - i, then j
k ... 2j + i ..
t
=i
+ s, 'k = i + 8 + t,
8 and (2.3.4) becomes
-
(2.3.5)
Now
00
(xy)k (y-x)
00
8~k t~kx8ytsgn(t-s)
=
(l~x)(l-y)(l-xy).
Substituting this into (2.3.5) yields
6i<1<kPiPjPksgn(k-2j+i)
(1-q2) (l-q) (1_q3)
..
6q 4
(l+q) (l+q+q2) 2
Thus
Ql ...
..
69 4
(l+q) (l+q+q2)Z
3q(l-q+q2)
(l+q+q2)2
+
3pg
(l+q) (l+q+q2)
23
Note that lim
q-+()
dQ1
Q1 • 0 and ~1! Q1
= ~.
Now
-.
3(1-q) (l+q) (1-3q+q2)
•
(1+q+q2)3
dq
and
dQ1
-dq
± 1 or
• 0 if q •
3±15
2
For 0 < q < 1, this is :~1 • 0 if q • 1 or
When q •
3-
2
rs
' Q1 •
3
8'
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Thus the maximum value of Q1 for the
geometric distribution is ~ •
Although the result for Q2 in Theorem 2.3.2 is .imi1ar to that
for Q1' the application of it is difficult except for distributions
which have positive probability on only a few of· the integers.
I
have been unable to evaluate Q2 for any of the standard discrete
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distributions.
We now consider an application of·Theorem 2.3.1.
f(x) • kx k- l ; 0 < x < 1, k· 1,2,....
We shall compute Q1 and Q 2
as functions of k.
From Theorem 2.3.1
1 x
Q1(k) • 12 { {IF[~(X+Y)]f(Y)f(X) dy dx - 3
• 3k
2
2'1t-!
}1(x+y)kxk-1 yk-1 dy dx - 3
00
Let
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k
= 2k-2
k k 1
~ (~)j=O J k+j
-
(2.3.6)
3.
We now prove the following result:
~ma
2.3,1:
~ (k) k then
j=aO j k+j ,
Let 2k+1A- =
-K
Proof:
Let am(k)
k
·k
m
= jE o0j )m+j
; m
= 1,
Note that
2,
2k+l Ak = ak(k) and
a (k) - a (k-1)
m
m
= ~
j=O
(k) _ (k-1)]-ffij
j
m+j
¥ (k-l)
m
j-1 m+j
= jt!!tJ.
m
= m+1 am+1(k-1)
Thus ~_(k)
-m
a
~(a
m-l
m-
l(k+l) - a
m-
•
l(k»).
(2.3.7)
The following identity can be verified by noting that it holds for mcl
and satisfies the recursion (2.3.7).
Putting m = k, and ignoring the term (_l)k-l(~k)-l, yields
25
Let bk(j) • (~k)-1(2~=J)
= k(k-1) ••• (k-j+1)
s k, and
j
-.
2k(2k-1) ••• (2k-j+l)
co
bk(j) ~ 0 if j > k.
Then bk
j-1
bk(j) •
(~)j t~O
(1 -
lim A
k~
k
=
=
•
=
= jE 1 (-1)j- 1bk (j).
2~-t)'
Now
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Thus for any fixed j
lim b k
k-+oo
E <-1)j-1 bk (j)
lim
k-+oo j =1
j~l (_1)j-1 ~.;:
co
y
j"l
b j (k)
j 1 1
(-1) - (-)j
2
= -13
Q.E.D.
Applying this result to (2.3.6) yields
• _1
3
Thus, as k gets large, F becomes progressively more negatively
Ql-skewed, but the limiting value of Q1 is not -1, as one might
expect.
Before evaluating Q for this same distribution we rewrite the
2
integral expression for Q2.
in a sample of 3 is
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Letting vl
= Yl
+ Y3 - 2Y2 , v 2 = Y3 - Yl' and v3 = Y3' we obtain
Y3 - Yl
Thus
Q2
1
j co
= E(Vl) =-{-co
£ZV1V2f(V3-V2) f(V3) f{~[2v3 - V2(1+vl)]}dv2dv3dvl .
= kx k - l
For the distribution with density f(x)
; O<x<l; k
=
1,2, ... ,
this is upon interchanging the order of integration
3k3 1 Y3 1
k-l
k-l
k-l
[2v 3 - V2 (l+v l )]
Q2(k) = ~l£ {_{V1V2V3
(V 3-V2)
dv l dv 2dv 3
Consider the integral
1
k 1
I =_f1Vl[Zv3 -V2(1+v 1)] - dV 1
where a
= 2V3
- v2 and b = -VZ'
1 a+b k
k 1
I = ~ f (x - ax - ) dx
b a-b
•
bZ(k+l)
=
Vl(a+bvl)k-l dVl'
Let x = a + bv 1 , then
[(a+b)k+l _ (a_b)k+l]
1
1
=-1f
a [(a+b)k - (a_b)k]
kb 2
[(2V3-2V2)k+l_(2V3)k+l]ZV3- V2[(2V3-ZvZ)k_(Zv3)k].
kV2Z
vZZ(k+l)
1
\
After considerable algebraic manipulation, we obtain:
I = 2k (1-k) (v 3-v Z)k _ Zkv 3k _ 2k+1
~ (-1)j(~)VZj-Zv3k-j+l
k(k+l)v Z
kvZ
k(k+l) j=l
= A1
- AZ - A (say) •
3
Thus Q2(k)
= 3k
3
Zk-l
1~
fJ
00
k 1
k-l
(Al -AZ-A3)v2v3 - (v3-v2)
dvZ dV3
27
-.
I
_ 6k 2 (1-k) r V3k-l~v 2k dv
k+l
0
2k 3
3
-
3k(1-k) I 3k-l
k+l
£V3
dV3
-l-k
k+l
.
Finally,
3 IVa
- ~fj V3 k - 1 (V3-V2)2k-l A3 dV2 dV 3
3
2k-10 0
B
12k2 k
j k I V3 j-l 2k-l
k-1
- ~j~1(-1) (j)££ V2
V3
(V 3-V2)
dV2 dV 3
k
IV a k-1
_ 12k2 I: (-i)j (k)f f I: (-1) i(k-l)v i+j-~ 3k-i-j-1dv dv
k+l J-1
j 00 i-O
i
2
3
2 3
12k2 k
k-l
IVS
- k+l j~l(-l)j(~)i~O(-l)i(kil)££V2i+j-lv33k-i-j-ldv2dV3
2 k
k-l
I
_ ~ I: (_l)j(k) ~ (_l)i(k-l)-!-J v 3k-1 dv
j i 0
i i +j 0 3
3
k+1 j -1
_ 4k
~ (_1)j(k)k 1(_1)i(k-l)
k+1 j-l
j i-O
i
i+j
-!-
E
Hence
Lemma 2.3.2:
k-l (_1)i(k-l) __i
i~O
i
i+j
I
I
(k+j-l) !
j
j
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Proof:
Note that
k;l ( k-I
. ) x ,i-I( -X), i
i=O
1
{"
Integrating both sides from 0 to 1 with respect to x yields the
desired result.
Applying Lemma 2.3.2 we obtain
1-k
k+1
= ---
~im
k~co
Q2(k) has not been evaluated algebraically, but a computer program
was written to compute Q2(k) for k
= 1,2, ••• ,50.
From this program
it was found that Q2(k) is a monotonically decreasing function of k,
with Q2(1) • 0, Q2(49) • -0.2228 and QZ(sO)
= -0.2229.
Thus we see
that Q2(k) is apparently converging as k gets large with the limit
being approximately -00223.
The
~act
that neither Q1(k) or Q2(k) attain the value -1 even
though as k gets large F becomes progressively more non-symmetric,
leads one to the conjecture that sharper bounds than IQkl
k
= 1,2 can be found.
$
1;
That this is not the case is shown by the
following counterexamples.
Consider first Q and let Xl be a random variable with probability
1
function P (Xl
a
-1)
=p
and P (Xl
= 1) = l-par.
. If tho:-ee observat ions
are made on Xl' there are four possible outcomes:
Outcome
Probability
-1,-1,-1
p3
0
-1,-1,1
3p2r
1
-1,1,1
3p'r 2
-1
1,1,1
r3
h
0
29
Thus the expected value of ql' for the random variable Xl' is
Ql(l) • 3p2r - 3pr 2
~
3pr(p-r)
• 3pr(2p-l)
The conditional expected value of ql given no triple ties is
Q (e) • 3pr(2p-l~
1
2
3p r+3pr
• 2p-l
Now let X2 be the random variable with probability function
2
P(X2 • -4) • p , P(X 2 • -2) - pr, P(X2 • 2) • pr, and
P(X 2 • 4) • r 2 ; !.~. we divide P(X l < 0) • p into two parts by the
formula p(p+r) • p2+pr and similarly for P(Xl > 0) and let X2 take
the values ±3 ± 1.
Since the two negative values X2 can assume
are two units apart and the associated probabilities are
P(X2 • -4) • pP(Xl • -1) and P(X2 • ~2) • pP(Xl • 1), the expected
value of ql (2) given that all three observations are negative is
E(ql(2)1 all three observations are <0) • p30,(1) • Q (1) •
p3
1
Similarly,
E(ql(2)1 all three observations are> 0) • Ql(l).
If two observations are negative and one is positive, then
Y3-Y2 > Y2- Y1' where the y's are the ordered observations on X2 .
Hence in this case ~1(2) • 1.
E(ql (2)
I
Thus
two observations < 0 and one observation > 0)
= 1.
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Similarly
E(ql (2)1 two observations> 0 and 1 observation> 0) = -1.
Thus
E(ql (2)12 observations of one sign and 1 of the other)
= 3p 2 r
- 3pr 2
3p 2r + 3pr 2
Hence
Continuing
in the same manner, define the probability function
P(X3 - -13)
= p3;
P(X
3
= -11) = p2r ;
p(X3 • -7) • p2r ;
P(X3 • -5) - pr 2 ; P(X3 • 5) = p2 r ; P(X3
P(X3 - 11) • pr 2 ; P(X3 = 13) = r 3 •
= 7) = pr 2 ;
The same argument as that used above shows that
In general, if the probability functions are constructed in the
same manner, we have for the k th random variable
(2.3.8)
Applying this formula k-l times yields
Ql(k)
=
(p3 + rJ)k-1Q1(1)
31
Now p3 + r 3 = 1 - 3pr < 1,
-,
Hence
• 2p - 1.
Thus by choosing k large enough and p sufficiently close to
~.
Q1 (k) can be made as close to unity as desired.
For Q2 • E(q2)' let Xl be distributed exactly as Xl above;
1,e.
P(X1 • -1) - P and P(X I • 1) = r.
Q2(1)
= 3pr(2p
Then
- 1) and Q2(c) • 2p - 1. as before.
For the second stage. let P(X 2 - _~1)=p2; P(X2 • -m+1) = prj
P(X 2 • m-l) = pr and P(X2 • m+l) = r 2 j m = 3.4,5, •••• As with
Ql(2), we have
E(q2(2)lall observations <0) = Q2(1)= E(q2(2)lall observations >0).
If 2 observations are negative and one is positive, then the
minimu~
value of ~2(2) occurs when the observations are (-m-l. -m+1, m-l. m+l),
for which
~2(2). [(m-l)-(-m+l)] - [(-m+l) - (-m-l)]
[(m-l) - (-m+l)] + [(-m+l) - (-m-l)]
1 - ~
m
Thus E(q2 (2)1 2 observations < 0 and 1 observation> 0) ~ 1 - ~ •
If two observations are positive and one is negative, the minimum
value ~2(2)can attain is -1 and E(q2(2)1 2 observations> 0 and
1 observation < 0)
~
-1,
I
I
I
I
I
I
I
I
_I
I
I
I
I
I
I
I
-.
I
32
Hence
E(q2(2)lz observations are of one sign and one is of the other)
3p2r (l-~) - 3pr Z
3pr 2 + 3p2r
jE.]
3pr( (p-r) -
=------3pr
=
Q2 (cL ~
.
Thus
QZ(2)~ (p3+r 3) QZ(l)+ 3pr(Qz(c);o,1I!.); m-3,4,S...
m
(2.3.9)
Note that the right hand side of (2.3.9) is a monotonically increasing
function of m and
Thus for large m,
Continuing in the same manner, for the k th random variable,
E(q2 (k)lare 3 observations <O)=E(Q2(k)lal1 3 observations >O)=Q2(k-l).
If two observations are negative and one is positive, then
2(mk- 1 _ 2mk- Z Qz (k) ~
2mk - 1
e ••
-
2
1 +-r+."+~
1
1 )
- 1 - 2( m
m"
m~-.L
- 1 - Z l(
~
(1 -
'ml-k ).]
1 - 1
m
- 1 - 2 [
33
Thus
E(q2(k) 12 observations <0, 1 observation >0)
~
1 k
1- 2(1- m - )
m-1
Also
E(qz(k)lz observations >0, 1 observation< 0) ~ - 1.
Hence
E(qZ(k)lz observations with one sign and one with the other)
=
Q (c) - ...
2p...(o..::l:--_....:.;m;...l_-k_.~)
2
m- 1
Thus
I
I
-.
I
I
I
I
I
I
eI
QZ(k) is a monotonically increasing function of m, and
1 k
lim 1 - m m+oo
m-l
=
Hence
lim Q (k). (p3 + r 3 ) QZ(k-1) + 3pr Qz(C)
m-+oo
and for large m
Z
I
I
I
I
I
I
I
-.
I
34
This is the same relation for Q2(k) as (2.3.8).
Hence for large m,
Zp-1; Le,
Thus by choosing k and m large enough and p sufficiently close to 1,
QZ can be made as close to unity as desired.
CHAPTER 3.
301
TESTS OF SYMMETRY BASED ON ql AND q2
The Tests of Symmetry
Suppose it is desired to test the hypothesis that F(x) is symmetric
against the alternative that it is qk-skewed; k • 1,2.
Either of these
statistics can be used as a basis for a test of this hypothesis.
The tests can be either two-sided or one-sided;
i.~.
the alternative
hypothesis can be either general qk-skewness or directed qk-skewness.
Explicitly, ql and q2 are U statistics for estimating Ql and Q2
and we have proved in Theorem 2.2.1 that Qk ,. 0; k ,. 1,2 if F is
symmetric.
Thus from Theorem 1.3.4, under the hypothesis of symmetry,
tbe large sample distribution of
k ,. 1, 2,
is the standard normal distribution, where
(3.1.1)
Thus if 0k2 were known, by comparing zk* with the tabulated values
of the standard normal distribution, inferences could be made concerning
th~
symmetry of F, for sufficiently large n.
Theorem 3.1.1:
If F is symmetric with density f, non-zero on the interval
(-A, A), where A may • +
00
then
A
2 (X)
°k 2 = 2JQk
o
f(x) dx; k ,. 1,2,
I
I
36
where
(i) Q1 (x)
A
A
-A
x
-.
= 4 !F[~(X+Y)]f(Y)dY - 4!F(2x-y)f(y)dy - 2F 2 (x) - 1,
and
Ay
Q2(x) • 2!! y+X-2z f (y) fez) dz dy
xx y-x
Ax
+ 2! f y +z- 2x fez) fey) dz dy
x-A y-z
(it)
I
I
xx
+ 2 !!x+y-2z fez) fey) dz dye
-Az x-y
Proof:
Let Qk(x) • E[ ~k(x1,x2,x3) IX1 = x]; k = 1,2, then since F
is symmetric, ak 2 = E[Qk 2 (x)]. Note that Qk(x); k • 1,2 is an antisymmetric function of x.
I
I
Hence
A
E [Qk 2 (x)] • 2~ Qk 2 (x) f(x) dx; k • 1,2.
Now let
Qki(x) • E[~k (x1,x2'x 3)\ Yi
= x];
Y1 < Y2 < Y3 are the ordered XIS.
Then
k· 1,2; i
I
I
= 1,2,3,
_I
where
Qk(x) • [1-F(x)]2 Qkl (X) +2F(x) [1-F(x)]Qk2(X)+F 2 (x)Qk3(x); k=l,2.
(3.1.2)
From the joint distribution of two order statistics conddtional on
the third in a sample of 3, and definitions 2.2.1 and 2.2.2, we have
AI3
Qk(x) • 2!! ~k(Y2' Y3' x) f(Y2) f(Y3) dY2 dY3
xx
Ax
+ 2!x-A
! ~k(Yl' Y3' x) f(Yl) f(Y3) dYl dY3
xx
+ 2 !! ~k(Yl' Y2' x) f(Yl) f(Y2)dY2dYl; k • 1,2
-AYI
(3.1.3)
I
I
I
I
I
I
--I
I
37
From equation (2,201) we have
AY3
Ql(x) = 2ff sgn(Y3 + x - 2Y2) f(Y2) f(Y3) dY2 dY3
xx
+ zZ_ZSgn(Y3 + Y1 - Zx) f(Y1) f(Y3) dYl dY3
xx
+ Z ff sgn(Yl + x - ZYZ) f(Yl) f(yZ) dyz dYl'
-AYl
0.1.4)
Consider
AY3
Z~~
sgn(Y3 + x - ZY Z) f(yZ) f(Y3) dyZ dY 3
1
A Z(x+Y3)
= Zff
AY3
. f(Yz)f(Y3)dYzdYr2f1
x ~(X+Y3)
xx
f(yz)f(Y3)dy zdY
3
A
• ZI {F[~(X+Y) - F(x) - F(y) + F[~(X+Y)]} fey) dy
x
A
A
A
x
x
x
• 4f F[~(X+Y)] f(y)dy - ZF(x)f f(y)dy - Zf F(y)f(y)dy
A
• 4~ F[~(X+Y)] f(y)dy - ZF(x) [l-F(x)] - 1 + FZ(x)
A
1
Z
• 4/ F[2(x+y)] f(y)dy - 2F(x) + 3F (x) - 10
x
(3.1.5)
For the second integral of (3.1.4), for x > 0, we have
x x
2 1 J sgn(Y3 + Yl - 2x) f(Yl) f(Y3) dYl dY3
-A Yl
Ax
A ZX-Y3
- Zff
f(Yl)f(Y3)dy l dY3 - 2fJ
f(Yl)f(Y3)dYldY3
x2x-Y3
xA
= zt {F(x) - F(Zx-y) - F(Zx-y)}f(y)dy
x
A
• ZF(x) [l-F(x)] - 4 fF(Zx-y) fey) dYe
x
(3.1.6)
38
Finally,
xx
Z If sgn(y 1+x-ZyZ) f(Y1) f(yz) dyZ dY1
-AYI
1
x ~(Y1+X)
xx
• Z_if
f(Y1)f(YZ)dYZ dY 1- Z ff
f(Y1)f(YZ) dY1dy Z
y1
-A lZ(Y1~
x
~
x
• 4 JF[~(x+Y)]f(Y)dy-2 fF(y)f(y)dy-2F(x) ff(y)dy
-A
-A
-A
x 1
Z
• 4 fF[-Z(x+y)] fey) dy - 3F (x).
-A
(3.1.7)
Substituting (3.1.5), (3.1.6), and (3.1.7) into (3.1.4) we obtain
A
Q1(x) • 4 fF[~(X+Y)]f(y)dy ~ 2F(x) + 3F 2 (x) - 1
x
A
2
+ ZF(x) [1-F(x)]-4fF(2x-y)f(y)dy-2F(x)-1+3F (x)
x
2
+ 4 fF[-zl(x+Y)]f(y) dy - 3F (x).
-A
A
A
2
• 4 fF[!2(x+Y)]f(y)dy-4fF(2x-y)f(y)dy-2F (x)-l.
-A
x
Thus' (i) is proved.
of
~2
(ii) follows immediately from the definition
(x 1 ,x 2 ,x 3) and equation (3.1.3).
As an example, consider the uniform distribution on the interval
(-1, 1), for which
I
I
-,
I
I
I
I
I
I
_I
I
I
I
I
I
I
--I
I
39
and
F(x) = !Cx+1); -1 ~ x ~ 1.
2
Thus from Theorem 3.1.1,
a
X
+ 2 _
+ ~x2 _ 1
2
2
X
1 2
-x
2
3
X - -
2
- x(x - 1).
Hence
I
0 12
Q2(x) •
2fJ
xx
Q21 ( x),. 1
2
1
2 {x 2 (x-1)2 dx
a
1 y+x-2z dxdy +
4
irr .
2f j ! y+z-2~
y-z
y-x
x-I 4
xx
+ 2 II l x+y-2z dy dz
-lz 4 X Y
ly
fl
XX
dz dy
2
y+x- z d d
y-x
.z Y
1.M.v
• - f[ ( L.:..=. )z -
2 x
y-x
2
!
- ]y. dy
y-x X
1
• 1 Iy +
2
=
X
1 X
- 11 I
Q22 ()
X
2x-1
X -
Y-
y+z-2x
y-z
X
dy
= O.
dz dy
1
- I[(y-x)log(y+1)-(y-x)10g(y-x)- l (l+x)] dy
2
X
=~
Q23(x) • 1
2
(1+x)2 log(l+x)- ~(1_x)2 10g(1-x) - 2x log 2
Xx
If
_1?/
= 1, j [(
2
1
=2
-1
X
x+y-2z dz dy
X
Y
x+y ) z _ ~ ] X dy
x-y
x-y Y
I(x + y) -
_1
(X
+ y) dy
= O.
40
Thus
1
2
121
2
2 • f [-(1+X) log (l+x) - -(1-x)2 log(1-x)-2x~10g2] dx
O
O2
2
0 2 2
was evaluated using a 32 point
'
Gauss-~drature method.
program used was the I.B.M. Scientific Subprogram DQG32.
(1968)~.
TI1e
(See I.B.M.
The resulting value was
02 2 • 0.012724 •
Unfortunately, since 01 2 and-0 22 depend on the form of the
parent distribution, e~'en-under the -~ypothesis- of symmetry, the tests
based onzl* and-z 2l!1 are not distribution free. In addition, the
evaluation of 0k2 ; k . 1,2 is not easy,exceptin'special cases.
Thus the tests -outlines-above will-notbevery--useful- in practice.
The value of
Ok 2 ; k ·-1,2 has-beennullerieally determined for
three other- symmetrie-<distributions: - the normal, Cauchy, and
Laplace distributions, fbr'which- the -qensities 'are:
I
I
-.
I
I
I
I
I
I
_I
Normalfelt)
1
.--'
I2iT
Cauchy
-OO<X<ClO
Laplace
f(x) • 1:,
2
e-Ixl
The DQG32 program was again-used for this purpose; -Since-the numerical.
integration assumes finite-l±mits-on-theintegrationithe range of
xwal truncated-for-each-distribut!on.-Thus-0it2;-k- 1,2 was evaluated
I
I
I
I
I
I
_.
I
I
41
for the function
f(x)
--m
f(x) _
1
-6
~
x
~
6
1
7T(1+x2)
-10
~
x
~
10
£(x) - J: e-Ixl
-10
~
x
~
10.
2
In this sense, the-resulting values of the O'k2-are-less "than the actual
values.
However; since we are using a 32potnt quadrature method,
as the range -of -the -integration-increases; -the --approximation of the
function in the -middle- range, -where - the probability --density is greater,
is not as accurate;u!hos-theva1oes-of-O'k2 ;k--Cl;2wi11 depend on
the limits we-choose;- -The-results-of the-evaluat1on'of the O'k2 are
presented in Table 3.1.
Table 3. L - 0'k,2; -k -1; 2 "for Four -Symmetric Dtstrtbutions
Distribution
O't 2
°' 2
.2..
Uniform
0.033333
0.012724
Normal
0.034994
0.009493
Caochy
0.033080
0.015056
Laplace
0.. 032664
0.013824
Sen's (1960) results can
be"osed~to"obtain-consistent
el:ltimators
of O'k 2 ; k ."1,2~ and-thus:provideasymptoticallydistribution free
tests.
Following Sen, define-the components"of the qk by
Vki • (n21)-1
E
Cl<C2
cl~i;l:C2
~k(Xi'Xc ,X -);1-1.2,; •• ,n; k=l,2.
1
c
2
42
From Theorem 1;a.3, it follows that-a consistent-estimator of
Ok2 ; k - 1.2 is
sk2
n
L (V - - qk) 2 ; k - 1. 2.
n-J.i_1 k i
-..lr
(3.1.8)
Hence by Theorem 1.3.4.
wlllhave the;standard--nermal-distr1bl1t1en';unc!er:;the"hypothesis of
symmetry-,-if n-is"s1;lffieientlyp large.' Thoa'CzLo21"-zr-ean be used to
test 'the hypothes1s'that--""!s symmetric;" -These--t••t8-are-a.ymptClltically
distribution free.
If-the-statistics' af qk tllt ; k'· 1.2,a8 g1ven-by"deftnition (2.2.3)
are preferred, Theorem -' 1.3 ';.Smay: be ueed-:to - obtain'" eatimates of the
respective ,variancas. -·l.et'-VkH k ·-1,2; "i-·'-1;.2 ••• tn be tbe
components :of'qk"and'let-V:i~"be the -c~.ponellta::(Jf'-U~.
fnqk tilt
u
.
. tk
k
; k - 1.2
is asymptetieallydistr1buted -as' thecstandard':nermal'-distribution.
where
and
s
*2
1
n
tilt
* 2
- I: (V
- U)
•
n-1 i-1
i
-.
I
I
I
I
I
I
_I
Then from
Theorem: L 3; 5, -it -.follows that
- .-, ~ " .•
I
I
I
I
I
I
I
I
_.
I
I
43
Hence, for sufficiently large n, uk can be used to,test the hypothesis
.
of symmetry, against the alternative of qk -skewness; k
K
1,2.
We see
then that the computations are somewhat more complicated for uk'
This will not be a serious drawback for computer calculations,
.
Whether to use qk or qk ; k • 1,2 will in the end be a matter
of personal preference.
The situation is similar to that discussed
by Quade (1961) with respect to measures of order association.
As
he says, Quade, 1967 p, 27.
" ••• 1 prefer'to'retain'theties, in general, on the
grounds that the variables X and Y, [in our case the
single variable X] even if not continuous as recorded,
usually represent underlying continua,with'Ues occuring
only because of imprecision in measurement-Cor grouping
later), In that case'the descriptive measure we adopt
should involve a'certain'penalty (in the"sense of giving
values closer to zero) 'for such'imprecision."
Of course,
if
X does not representan-underlying'continum, -this argument
would not apply.
'It
My reasons for preferring qk to qk ; k • 1,2 are
the above statement by Quade and:the'slightsiip.pl1fication in theory
and computations afforded by the qk'
In the remaining sections-of this chapter some 'of-the properties
of the tests based,on ql and'q2will be'investigated.
3,2
The ' Rate' of- Convergence,- of' the' Distributions ~ to Normality
While Theorem'L3;4 proves that 'zland-zi converge in'distribution
to the'standard'normal'distribuUon; it-gives-no'indication of the
rate of' convergence; - -If the- tests are - to~be" usefu1i .some-idea concerning
the sample size-necessary-for'the-distribution-:to-be:approximately
normal is needed;·' - In' order- to investigate' this -problem, computer
simulation-methods were'used;
We examine the-rate-of convergence to
normality of the-distribution of-z k ;k=1,2; -£or£our parent distributions:
the normal, uniform';' Cauchy; "and -Laplace' distributions,
44
For the normaldist:ributioft, 200 samples of 'size 10 were generated
using Bell's (1968) modification of the Box-Muller teehnique.
From
each of these samples zk;k-=l,2was computed,and the sample distribution was then compared'with the standard normal distribution.
zk' are
the data generated, .the
From
too spread -out "imp1ying . that sk2 is an
underestimate'of the·true'variance'of-qk •. Thus 'the-tails of the
sample distributions' contained-more observations than'-would be expected
if the true distribution"were the normal distribution (See Table 3.2).
Table 3~2.Number of, observed Izkl' ';1t '··--l,2.-which-exceeded the
5 and-!%'pointsof thenormal-distr1bution in 200 samples
of n • 10, 20, '30'from the normal distribution
Statistic
Observations greater than z(a) 'in'ablolute values
ex
= 0.05
n-10
12
24
9
n-20
16
14
8
7
n-30
Zl
z2
Expected
frequency
12
11
3
3
10
2
Thus we lee that if either test is used with a sample of 10, the null
hypothesis will be rejected too often.
The same method was used to
generate 200 samples of 20 and subsequently 200 samples of 30.
-.
I
I
I
I
I
I
_I
ex • 0.01
28
I
I
From
I
I
I
I
I
I
_.
I
I
Table 3.2, we see that for n • 20, there are-againmore'observations
in the tails than' would be expected.
n
However; -for n" - ' 30, the tails
of the sample dis tributien c10sely approximate' the-' tails of the
theore tical -dis tri'butien.. :- As ',a 'fu1l'ther eOlilpar is on -with the normal
distribution; .the "Kelmogorov-Smirnov- statistte "'was'-:used .to compare the
distribution of
I Zk,h'
k --1,2 with thehalf-ncsrmal distribution:
!
2
f(x) =(i~2
inc~~eory
Since
1 2
~,":~x.
o < x.
the distribution of zk; k • 1,2 is symmetric, the
comparison with the half-normal 'will 'be more' likely-to 'detect differences
in the shapes of
the'distributions~Usingthi8-procedure,
the
t
hypothesis -of 'normality-was "not rejected ( :,a'.,O;l) for either ~~l
or z2.
We conclude"that a sample of size 30'is:sufficiently large
for the distribution"of
zkY"k. -,?-,2 to be considered approximately
normally distributed.
Far the' remaining"' three"" dis~ributions,~the'" soec-methods were
used except· that instead" of 'beginning" at'n -·'10,; -the'starting point
was chosen as n - 30.
"The "I;B;M;' Seielltific'SubprograIlF,RAN!)Uwas;:used:ta generate the
unifarm:variat.s. 'Far
tha~J"aplace 'distribut:ten:the
following
algorithm'wasused;" Two"unifCDrm -(0,1) ,variates' X;i 'and,)l;2 were
generated.
If~i~<-~'
~l > ~ , y - - log(x 2);
the Laplace variate-in y-.':leg(Jt ) , while if
2
TheCauchyvar1atl~~weregenerated by obtaining
a uniform'(O,l) 'variate"alld-transforming it to 'a-Cauchy variate as
indicated in'thefollQwing..Let'~be a-random'vadable with the
unifarm (0,1) distribution;
46
-.
.!..!..
f (x) • 1 ; 0 < x
Let y •
Let
Z -
1
2)
,
< 1.
then· the density of y is
1
1T
fey)
-1T , - -2 < y < :IT2
tan(y), •then-the density of zis
7T (x .-
-
fez) _
.
1
. - ~ < z < ~
1T(1+z2) ,
1.!..
the density of theeauchy distribution.
Tables 3;3 .-
3.~
are the tables for tha·uniform; Cauchy, and
Laplace distributions·corresponding to Table 3.1.
Table 3.3.
Statistic
Number of observedzk ; k - 1,2 whioh.exceeded the
5 and·1% points af thenarmal disttibution in 200
samples·of n- 30, 40, 50'fram the·'unifom·distribution
a. • 0.01
n-30
17
16
7
7
n-40
15
15
5
4
n-50
Expected
frequency
I
I
I
I
I
I
Observatbns -greater· than·ls (a.)1 in absolute values
a. • 0.05
zl
z2
I
I
12
3
1
10
2
13
el
I
I
I
I
I
I
_.
I
I
47
Table 3.4.
Statistic
Number of observed Izkl;k .. 1,2 which exceeded the
5 and 1% points of the normal distribution in 200
samples of n = 30, 40, 50, 60 from the Cauchy
distribution
Observations greater than z(a) in absolute value
a
= 0.05
a .. 0.01
n=30
19
18
n-40
19
19
n=50
18
18
n-60
11
11
10
Table 3.5.
Statistic
Number of observed Izkl ; k '" 1,2 which exceeded the
5 and 1% points of the normal distribution in 200
samples of n = 30 from the Laplace distribution
Observations greater than z(a) in absolute value
CL '"
zl
z2
Expected
frequency
0.05
a
= 0.01
12
11
4
4
10
2
48
If the tails of the sample distributions were considered to be
relatively close to the expected frequencies. the Ko1mogorov-Smirnov
test was applied comparing the absolute values of the zk's with the
half-normal distribution.
was not rejected (a
Table 3.6.
= 0.1).
In each case the hypothesis of normality
The results are summarized in Table 3.6.
Sample size needed to assume approximate normality for
zk; k = 1.2
Parent distribution
Sample size
Normal
Uniform
Cauchy
Laplace
30
50
60
30
Recall that the moment statistic. b 1 • is approximately normally
distributed if n
~
100. when the underlying distribution is the normal
distribution •. The above results indicate that the convergence of the
distributions of ql and q2 is much faster.
However, different criteria
have been used to determine the rate of convergence.
Whereas Monte
Carlo methods have been used to obtain the sample sizes necessary
for q1 and q2 to be approximately normally distributed. Pearson (1932)
used the method of moments to fit the distribution of bl to a Type IV
Pearson curve; whence he obtained the rule of thumb
n~
100.
Unfortu-
nate1y, b 1 was not computed in the simulation. so that we do not have
a direct comparison of its distribution with that of qk; k
3.3
= 1,2.
The Small Sample Power of the Tests
Shapiro,
~.
a1. (1968) empirically compared the sensitivities
of nine procedures for testing normality.
In this section, we use
I
I
-.
I
I
I
I
I
I
_I
I
I
I
I
I
I
JI
I
49
their methods to compare three tests of symmetry, those based on
b 1 , q1 and q2·
The null hypothesis is that F is the normal distribution.
2
Five alternative distributions are employed: X ; v
v
normal (x
= eZ ,
where
Z
= 1,2,4,
the log
is a standard normal random variable), and
the Poisson distribution with A = 4.
Two sample sizes, n
= 10
and
n • 15 are used.
The method is an empirical sampling method.
In order to
approximate the null distribution of the three statistics, M • 500
samples of each sample size were generated from the normal distribution,
and the empirical c.d.f.'s tabulated.
For the non-null distributions,
the empirical c.d.f.'s were based on M = 200 samples.
samples were submitted to each of the statistics.
The same
Given these
c.d.f.'s, the empirical power of the tests for significance level a
can be evaluated.
For example, suppose ca / and c l - a / are the
2
2
empirical critical points of a size a test based on b l •
The power
of the test based on b l against a particular alternative distribution
can then be estimated by determining the proportion of sample values
in the non-null distribution which lie outside the interval
It might well be argued that a one-sided test wou1• •e more
appropriate here, since all five alternative distributions are
positively Sl-skewed.
The two sided test was used in following
Shapiro et. a1. (1968).
As it turns out, for all of the alternate
distributions except the Poisson (4) the empirical c.d.f.'s are such
that the power of a two sided test of size a
~essential1y
the same
as the power of a one sided (positively directed) test of size a/2;
i.~.
for these four distributions there are no points in the empirical
c.d.f. lying below ca /2.
50
As in the previous section, the modified Box-Muller technique
The X1 2 deviates were
was used to generate the normal deviates.
generated by obtaining normal deviates and squaring them.
For the
X2 2 distribution, a uniform (0,1) deviate, u, was generated and
transformed by x • -log u.
summing two X2
Z
2
The X4 2 deviates were obtained by
deviates and the log normal deviates by x
is a standard normal deviate.
= eZ ,
where
Finally, to generate a sample from
the Poisson (4) distribution, an algorithm of Knuth (1969) was used.
..
Table 3.7 summarizes the results for the tests with a • 0.1.
This corresponds to Table 3 of Shapiro,
~.
a1. (1968).
Figures
3.1 - 3.10 give the merit curves, which are plots of the power against
the significance level.
These correspond to Figure 1.01 - 1.36 of
Shapiro.
Table 3.7.
b1
q1
q2
10
15
10
15
10
15
.70
.88
.52
.88
.64
.90
X2
2
.55
.65
.31
.66
.41
.69
2
.42
.47
.22
.42
.27
.45
Log Normal
.72
.81
.42
.80
.55
.80
Poisson (4)
.15
.14
.10
.16
.07
.16
Sample Size
Alternate Distribution
2
X1
XIt
••I
I
I
I
I
I
el
Empirical powers for a • 0.1
Test Statistic
I
I
I
I
I
I
I
I
•••I
I
1.0
'0
--b,
.......... q,
-~b,
··....·...·ql
---q2
---q2
B
, .•..
,/
/
,/
A_---:.:::::·::::::·:::·::::::I
",,.-' •....•..
...•.
.6
a:
'"o
~
.•...
.
Q.
If,:!/,:-;;f>'-''
•
I ..···
/'
,f
05
'0
'5
05
20
SIGNIFICANCE LEVEL;.
Figure 3.1.
,...•
2
Merit Curve for the Chi Squared
Distribution with One Degree of
Freedom; n=10
'0
.5
20
SIGNIFICANCE LEVEL;.
Figure 3.2.
Merit Curve for the Chi Squared
Distribution with Two Degrees of
Freedom; n=10
....
VI
- "
10
10
-~'I
.....······ql
-_···_ql
---q2
---q2
~. J
~.
ffi
a.
4
ffi
~../'''
£~
/~
~
•.. . ;:".
:::.,.~
///:.=.~,-:.~;
-
,;~:.::.:.~.~.."._..._...
.....
?/~::.:~:r-
2~' 1,11l/
ri"
05
10
15
a
20
_e_
10
'"
20
SIGNIFICANCE LEVEL; a
SIGNIFICANCE LEVEL; a
Figure 3.3.
00
Figure 3.4.
Merit Curve for the Chi Squared
Distribution with Four Degrees
of Freedom; n=lO
Merit Curve for the Log Normal
Distribution; n=lO
V1
N
e
e
10
--0,
'0
-------.-.-
..·········111
---112
.8
II::
W
~
•
6
II::
W
~
o
0
"-
"- .4
.4
--0,
- ·...-·111
---liZ
.2
05
.to
'5
0
20
SIGNIFICANCE LEVEL;.
Figure 3.5.
Merit Curve for the Poisson
Distribution; A=4, n=10
05
'0
'5
20
SIGNIFICANCE LEVEL;.
Figure 3.6.
Merit Curve for the Chi Squared
Distribution with One Degree of
Freedom; n=15
VI
W
'0
--0,
........_(1,
--0,
····..·····ql
'0
---(12
---q2
.B
a::
a::
~
~
6
r---;.~;:::::::~:;:::.::::~
w
w
o
"-
"-4
.4
.i'
2
.11'
,_/
r/:
.05
'0
15
05
20
_e_
15
20
SIGNIFICANCE LEVEL; a
SIGNIFICANCE LEVEL; a
Figure 3.7.
'0
Figure 3.8.
Merit Curve for the Chi Squared
Distribution with Two Degrees
of Freedom; n=15
e
Merit Curve for the Chi Squared
Distribution with Four Degrees
of Freedom; n=15
VI
::-
e
LO
LO
---(12
..... "
-.~.~.:-;
~iii.,"·=
0:
6
--OL
·····_··.. ql
0:
W
~
OJ
~
o
o
Q.
"- 4
--OL
·......····ql
---q2
o
05
LO
L5
20
05
Merit Curve for the Log Normal
Distribution; n=15
L5
20
SIGNIFICANCE LEVEL;"
SIGNIFICANCE LEVEL;"
Figure 3.9.
LO
Figure 3.10.
Merit Curve for the Poisson
Distribution; A=4, n=15
V1
V1
56
It is evident from both the table and the merit curves that for
n = 10, b i is superior to both ql and q2.
However, for samples of
15, there is very little apparent difference in power of the three
tests.
It is noteworthy that the power for b l for samples of 10 in our
study are considerably higher than those found by Shapiro, et.
For n
3.4
~.
= 15, these results agree closely with theirs.
Some Asymptotic Relative Efficiency Considerations
In this section we consider the asymptotic relative efficiencies
(ARE)
of the tests based on the qk to that of b l , using the Pitman
definition of ARE, against alternatives of the following special type
(1
+ 0) g(x) if x <·0
o
f(x)
(1
where g(x)
If
= g(-x);
0 = 0, f(x)
!.~.
-0 ) g(x) if x
g(x) is the
~
0
~
> 0
d~nsity
of a symmetric distribution.
= g(x) and f(x) is symmetric, while if 0
is not symmetric.
1,
~
0, f(x)
Thus the hypothesis of symmetry and its alternative
can be written as HO : 0
=0
vs. HI : 0
~
O.
Under this set up, the efficacy of a test Tn is
(3.4.1)
where 002(Tn ) is the variance of Tn under HO.
The ARE of Tn to
another test Tn * is
ARE(T
n'
T *)
n
= lim
ef(Tn)
n
n~ ef(T *)
(3.4.2)
I
I
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I
I
I
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I
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el
I
I
I
I
I
I
JI
I
57
Consider the efficacy of the test based on b l •
~3/~2 ~
E(b l ) ;
For large n
•
Hence
0-0
3
•
~20
1
3
- d~21
~ 0-0-2 ~30~202 ~ 0-0
2 d~31
-------------~203
where ~kO is the k th central moment under the null hypothesis;
i.e. the k th central moment of g.
~kO· 0
Since g is symmetric,
if k is odd, hence
origift of f.
Thus
d~ 3 ,
d0
I
d~l'
0=0
-3~20' ~I
,
,since
6=0
~lO = O.
(3.4.3)
58
,
Consider
d~3
--I
do
d
=0=0
de
3
I x f (x) dx
00
-00
= £-[(1+0)
de
=
o
o
I
0=0
3
00
3
I x g(x)dx + (l-o)/x g(x)dx)l~
3
u-o
0
-00
00
3
I x g(x)dx - I x g(x)dx
0
-00
00
3
= -2 Ix g(x)dx.
(3.4.4)
o
Similarly,
,
00
_d~_l_1
de
0=0
= i- I xf(x)dx
do -00
I
0=0
o
00
= i-[(1+0) I xg(x)dx + (1-0) Ixg(x)dx]I
do
-00
0
0-0
I
I
-.
I
I
I
I
I
I
00
= -2 I xy(x)dx •
(3.4.5)
o
Substituting (3.4.4) and (3.4.5) into (3.4.3) yields
(3.4.6)
Thus from (1.2.1), (3.4.6), and (3.4.1), we have
(3.4.7)
Now from Theorem 2.3.1
co x
1
E(q1) = Q1 = 12 f IF[i(x+y)] fey) f(x) dy dx - 3
-eo-co
•
el
I
I
I
I
I
I
--I
I
59
Hence
X
0:>
= 12_~_~
{F[~(X+Y)]f(Y)f(x)}IO=Odydx
(3.4.8)
Now
:0
{F[~(X+Y)]f(y)f(x)}lo=o
= {f(y)f(x)~ F[!(X+y)] + f(X)F[!2(X+y)]d~ f(y)
do
2
du
+ f(Y)F[~(X+Y)]%of(x)}lo=O
,
.
= f(y)g(x)S- F[!(X+y)]I
+ g(X)G[-21(X+y)]d~f(y)1
do
2
0=0·
du
0=0
,,.. ..
+ g(y)G[!(x+y)]d~f(x)1
2
du
0=0
Now
.
(3.4.9)
1
Z(x+y)
f(t) dt
F[~(X+Y)] = J
-0:>
=
~(x+Y)
1,
{(1+<S)g(t)l[t<0] +(1-0)g(t)I[t>O]} dt
Hence
= I 1
[2(X+Y)<0]
1 (x+y)
JZ
g(t)dt
..JJO
o
+1 1
[2(x+y»0]
Jg(t)dt
_0:>
1
-(x+y)
2
g(t)dt
I
1
J
- [2(x+y) >0 ] 0
=
G[~(X+Y)]I[~(X+Y)<O]+ {1-G[~(X+Y)]}I[~(X+Y»0]
(3.4.10)
60
Similarly
(3.4.11)
df(u)1
- g(u)I
g(u)I
do
0=0 [u<O] [u>O]
Substituting (3.4.10) and (3.4.11) into (3.4.9) yields
1
1
= g(y)g(x){ G[-2(x+y )]I[x+y<0] + [l-G[-(x+y)]I
}
2
[x+y>O]
+
G[~(X+y)]g(X) {g(y)I[y<O] - g(y) I[y>O]}
+
G[~(X+Y)] g(y) {g(x)I[x<O] - ,g(x)I[X>O]}
and hence from (3.4.8).letting H(x.y)= G[~(x+y)]g(y)g(x). we have
dQ
~I
do
0
0=0
x
= 12 1 1 H(x.y)dydx + 12
_00-00
~x
x
1o-~IH(x.y)
~
+ 12 1 1 H(x.y) dy dx + 12 1 }H(x,y) dy dx
0-00
-~-oo
~
x
0
x
- 121o !H(x.y)
dy dx + 12 1 1 H(x.y) dy dx
0
-00-00
~x
- 1210-001 H(x.y) dy dx
= 36
f j H(x.y)
dy dx - 36j
_00-00
~
~(x.y)
0 0
~
dy dx
0
+ 12 1-1H(x.y) dy dx - 12 1 IH(x.y) dy dx
0-00
a-x
~
x
+ 12 1 Ig(y)g(x) dy dx
o-x
I
I
I
I
I
I
I
I
I
I
I
I
~x,
~
-.
_I
dy dx
+ 12 1 Ig(y)g(x) dy dx - 12 1 IH(x.y) dy dx
o-x
o-x
o x
I
I
(3.4.12)
_.
I
I
61
Now
= -x
Letting u
and v
= -y,
o x
we have
o:>V
1
00
2
36 f !H(x,y)dy dx = 36ff{1-G[-(u+v)]}g(u) g(v) du dv
-00_00
o:>V
o:>V
= 36ffg(u)g(v)dudv
00
=
=
1
- 36ffG[-2(u+v)] g(u) g(v) du dv
00
36 i{G(v) - G(O)} g(v)dv _0 36nG[!(u+v) ]g(u)g(v)dudv
o
2
00
[18G 2 (v) - 36G(0)g(v)] 0:> - 36fjj1
G[-(u+v)g(u)g(v)dudv
o
= -92 -
00
36jj G[!(x+y)]g(y)g(x) dy dx
00
2
12f fH(x,y)dydx
= -2
2
•
(3.4.13)
Similarly,
o:>_x
0-00
3
00
0
1
- 12f fG[-2(x+y)]g(y)g(x)dy dx ,
o-x
(3.4.14)
and
0:>
x
12f !g(y)g(x) dy dx
o-x
= 12f[G(x)
0
- G(-x)]g(x) dx
= 3.
(3.4.15)
Hence, substituting (3.4.13), (3.4.14), and (3.4.15) into
(3.4.12), yields
d Ql
dO
1
0=0
=9
«>x 1
- 72ffG[Z(x+y)]g(y) g(x) dy dx
00
0
- 24i fG[!2(x+y)]g(y)g(x) dy dx
o-x
(3.4.16)
62
Let Sex) be the c.d.f. of lxi, where x has c.d.f. G(x).
G is symmetric about 0, for u > 0,
S
(u)
==
2g(u) and S(u)
==
Since
2G(u) - '1.
Now consider
Q3(G)
!£G[~(x+y)]
==
g(y) g(x) dy dx
1 co~ 1
fJS[-(x+y)] + 1 s(y) sex) dy dx
8 00 2
== -
==
1
1
2
coX
-8 ffS[-(x+y)]s(y) sex) dy dx
00
+ -81
ifs(y) sex) dy dx
00
Using Theorem 2.3.1, this is
=9
Q (G)
3
+ Q1(S)
96
(3.4.17)
Similarly,
==
ff G[-2(x-u)]g(u) g(x) du dx
00
==
==
1
1
cox
8 ££{S[2(x-u)]
+ l}s(u) sex) du dx
1 + 2 Q4*(S)
16
where
Q4*(S)
cox
==
ffS[12(x-u)] s(u) sex) du dx.
(3.4.18)
00
Substituting (3.4.17) and (3.4.18) into (3.4.16) yields
d Q1
do
1
==
0=0
3
4
-.
I
I
I
I
I
I
el
1
coX
I
I
(3.4.19)
I
I
I
I
I
I
_.
I
I
63
For Q2' from Theorem 2.3.1,
ooxy-
d
'" ~6 1 1 Ix+z-2y f(z)f(y)f(x)dzdydx}1
de -00-00-00 x-z
6-0
'"
00
I
x Yx+z-2y d
6_~_~r x-y
d6 {f(z)f(y)f(x)} 6=odzdydx.
(3.4.20)
Now
:6 {f(Z)f(y)f(x)}lo",o ...
g(Y)f(x)~(z)lo=o + g(Z)f(X)~f(y)16=0
+ g(z)g(y) dd~ f(x)1
u
0=0
•
From (3.4.11), this is
d
do {f(z)f(y)f(x)}lo=o '" g(y)g(x) [g(z)I[z<O] - g(z)1[z>O]]
+ g(z)g(x) [g(y)I[y<O] - g(y)1[y>O]]
+ g(z)g(y) [g(x)1[x<O] - g(x)1[x>O]] •
Thus from (3.4.20) ,
~l
do
•
0.0
6
00
x y
+ 2
1 1 1 xx-z
z-
Y g(y)g(x) [g(z)I[z<Ol- g(z)I[z>O]] dzdydx
-00-00-00
Letting h(x,y,z) ... x~~;2y g(z)g(y)g(x), this is
O~y
d
~I
do
0=0
.
ooxy
= IS 1 J Ih(x,y,z)dzdydx - ISlllh(x,y,z) dzdydx
-<Xl-<Xl-<Xl
000Y.
+ 61 1
j
o-00-00
00
cox
°
0
h(x,y,z)dzdydx - 611 [h(x,y,z) dzdydx.
0 0 -00
(3.4.21)
64
-.
In
o x Y
18 f f fh(x,y,z) dzdydx, let u
::0
-x, a • -y, and w • -z,
_ClO-<lO-CO
then we have
oowv
18 fff h(-u,-v,-w) dudvdw
I
I
I
I
I
I
000
::0
wv 2
181ff
v-u-w g(u) g(v) g(w) du dv dw
w-u
000
ooxy. + 2
• _ 18ffT
x z- Y g(z) g(y) g(x) dz dy dx.
x-z
000
Similarly,
6fooo'j
f Jh(x,y,z) dzdydx
::0
0_00-00
coX °
-6ff
fh(x,y,z) dz dy dx.
00-00
Thus
d Q2
---d~
~_O::o
u 1 u-
coxy. +z 2
-36ffT
x - Y g(z)g(y)g(x) dz dy dx
000
_ 12
x-z
ooX 0
ff /x+z-2y g(z)g(y)g(x) dz dy
00-00
x-z
_I
d~
(3.4.22)
::0
iflh(x y,z) g(z)g(y)g(x) dz dy dx
j
000
::0
J.8 ~fy.
)) r h(x,y,z) s(z) s(y) sex) dz dy dx
000
From Theorem 2.3.1 this is
(3.4.23)
Similarly,
coX 0
Q6(G) • ff fh(x,y,z) g(z) g(y) g(x) dz dy dx
00_00
ooXoo
• fff h(x,y,-z) g(z) g(y) g(x) dz dy dx
000
I
I
I
I
I
I
I
,I
_'I
I
I
65
= -81
COxco
! ! !h(x,y,-z) s(z)s(y)s(x) dz dy dx
o
0
0
j f
1
jx+z s(z)s(y)s(x) dz dy dx
8 0 0 ox+Z
coxco
- 1 ! ! fY+z s(z)s(y)s(x) dz dy dx
4
0
oX+z
0
coX co
= -l - 1 ! ! f Y+z s(z)s(y)s(x) dz dy dx
16
4
= 161 - 2;1
0
0
oy+x
Q6
*
(S)
(3.4.24)
Substituting (3.4.23)and (3.4.24) into (3.4.22) gives
(3.4.25)
As a first example, suppose
g(x) = ~ ; - 1 < x < 1.
Then
1
! 1 x 2 dx
-1
2
1
1
! - x dx
o 2
= -31
= -14
'
'
1
! ! x 3 dx ... 1
o 2
8 '
and
~kO
1
=
! ! xk dx
-1
2
=
-l,
k+1
if k is even.
Hence from (3.4.7)
Also, since sex) = 1; 0 < x < 1, Q1(S)
=0
and
66
While from (3.4.18),
1
x 1
o
0
f f -(x-u) du dx
=
2
=.L
12
Thus Q4(G)
= ~6
and from (3.4.16), Theorem
3.1.1
and (3.4.1):
Sn
=-
6
Hence ARE (ql' h )
l
For Q2, Q2(S)
=
(Sn) / (10Sn) - 64
6
128
- 63
= 0, and
1 X
1
Q/(S) = f f f
o
0
1
1 X
0
y+z dz dy dx
y+x
1 1
1 f f x(x+2z) dz dx
2 0 0
x+z
= -1 + -1 f1 x 2 log( x ) dx
Integrating hy parts, with u
Q6*(S)
= 12
+
x:FT
21 +
= log (x~r) and dv = x
!2 { _13 log
1 1
2( 6
= 2- - log
12
- 32
2.
2 -
log 2)
13
.,.
I
I
I
I
I
I
_I
dy dz dx
= fo f0 f0 y+z
y+x
220
I
I
2
2
j x+l
x dx
0
, gives
}
I
I
I
I
I
:1
.-I
I
67
Therefore
(log 2 - 1) /12
=
= -0.02557.
Hence
[(-12)(-0.02557)]2n
9(0.012724)
= 0.8225n,
and
= 0.8225n (i~~n)
ARE(q2' b 1)
1.0027
=
We conclude that for the uniform distribution and the alternative
considered here that the tests based on q1 and q2 are more
efficient than that based on b1 , for large samples.
1 - lx2
Now suppose g(x) = ~(x) = I2TI e 2 ;- 00 < x <
00
We shall
use the notation
=
~(x)
00
Now f
o
00
x~(x)
~(t)
f
dt.
1
dx· ---
I
3
f x ~(x) dx
o
~20 =
x
_00
1, ~40
2TI
= 127IT,
= 3, and
n [ -
~60 =
12
6
I2i
15.
-2 --- ]
liT
Therefore from (3.4.7 ),
2
=n
3TI
6
From (3.4.16),
d~l
du
I0=0 = 9
-
72Q3(~) - 24 Q4 (~).
I
I
68
-.
Also from (3.4.22)
dQ2
I
do
The
Qj(~);
Q5(~)
... - 36
0=0
- 12 Q
6
(~).
j ... 3,4,5,6 are not readily evaluated.
There are two
alternatives to explicit evaluation: numerical evaluation on the
computer or estimation using Monte Carlo methods.
Consider first numerical evaluation.
We first write Q3 and
Q4 as single integrals.
co X
Let s
= X-Y
z
... ! ! H~(x+y)'] <!>(y)<!>(x) dy dx.
o 0
,{.
and t = x+y. Then
z
co t
Q3(~)
... 2 ! !
o 0
~(t)<!>(t+s)<!>(t-s)
ds dt
But
] <!>(t+s)<!>(t-s)ds ...
o
f ( 1-- )2 e-~[(t+s)2+(t-s)2]dS
~
0
1
0=
-
e
_t2
Y"2'i
1
... e
21TI
_t 2
t
I
! e
o Y"2'i
2
Similarly,
co co
... 2 ! !¢(t)<!>(t+s)<!>(t-s) ds dt
o
1
co
_t 2
ds
I
dt.
... /iT {e
_s2
{~(tv'2) - -
Hence
¢(t){ l-<!>(tv'2)} dt.
I
I
I
I
I
}
I
_I
I
I
I
I
I
I
I
-,
I
69
Thus
dQ11
do
=
9
0=0
24
72
00
liT
0
00
f ¢(t) e
- -liT f0 e
48
12
+ -=
In
oof
1
{¢ (tl:2 - - } dt
2
_t2p;...
¢(t){ 1 - ¢(tY2J }dt
00
r.= f
y'IT 0
9 -
-t 2
e -t
e
2
_t2
¢(t) ¢(tl:2) dt +
¢ ()
t dt.
(3.4.26)
0
Using the program DQG32 once again, theee integrals were evaluated:
1o
e
_t 2
¢(t) ¢(t!1)dt
= 0.478462
and
00
f e- t
2
¢(t) dt = 0.616737.
o
Substitution into (3.4.25) yields
0.218203.
Using this result and the value of 01 2 for the normal distribution
given in Table 3.1.1 in (3.4.17) gives
e£(q1) ~ n(0.218203)2
- 9 ( .034994)
0.151178n,
and thus
:. 0.151178n
= 1.425.
(n/3n)
Q5(¢) and Q6(¢) were also evaluated numerically using DQG32:
70
and
do
~ 0.132432
I
0=0
and fronl (3.4.22) And the value of 02 2 for the nonn;1.1 distribution
n('132432)2
9( .012724)
• 0.153151n,
.
0.153),.51n
=
= 1.443
(n/3~)
In view of the qut:\st·ion cOncerning the preci ~j nn of the
J
numerical ~valuation of 0k 2 ; k ... 1,2 in section 3.1.1, the
precision of these ARE's is also not known.
A method of avoiding
this is to use Honte-Carlo-methods. to obtain .estimates 0'£ Qj (4));
j = 3,4,5,6 and Ok 2 ; ,k • 1,2.
(3.1.8) gives a fonnula for obtaining consistent-estimates
2
of the ok'
Sen (1966) notes that a functional of a distribution,
say A(f), can be estimated using U statistics, under certain
regularity conditions.
All that need be done is to find the degree
of A , find a symmetric unbiased estimator of it, and compute the
cQrresponding U Itatistic from a sample.
k - 3,4,5,6 and hence
since the estimates
• stimat...
Pk.
a~e
.
d
dQkI
dO 6-0
Using this method Qk(G);
1
can b~ estimated.
I
I
I
I
I
given in Table 3.1.1,
and then
-.
I
lience
dQ2
I
I
Moreover,
U statistics, they will be consistent
If a sample of n is generated from G, we can find estimates
_I
I
I
I
I
I
I
I
-•
I
71
of the Qk(G) as close to the true value as desired, by taking n
sufficiently large.
It would have been better if the simulation
methods used below had been applied to Q1(S), Q4 * (s), Q2(S), and
Q6 * (S), where S is the c.d.f. of the half normal distribution.
However, the simplifications (3.4.19) and (3.4.25) were not
pointed out to me until after the Monte Carlo work had been
completed.
Consider first the estimation of
00 x
1
Q3(G) = f f G(- (x+y)] g(y) g(x) dy dx.
o 0
2
Let Xl' X2 , X3 be independent and identically distributed random
variables with common c.d.f.
G, and define
o otherwise
¢3 * (X ,X ,X ) is an unbiased estimator of Q3(G), but it is not
1 2 3
symmetric in its arguments.
A symmetric unbiased estimator of
where the summation (A) is over the 3! permutation of (Xl. ,X2 ,X3 ).
Thus the U statistic for estimating Q2(G) is
l:
(p)
¢3 * (Xi ,X. ,Xi ),
1
where the summation (P) is over the (n)3
of the Xi's.
1. 2
3
= n(n-1) (n-2) permutations
72
Similarly, estimates of Q4, Qs and Q6 are obtained by defining
</>4 * (X1'X 2 'X 3 ) ...
</>S * (X1 ,X 2 ,X 3 ) -
3 <~(X1 +X 2 ) and 0 < - Xl < X2 •
1
[
if X
X1+X3- 2X 2
6 (XrX1)
l
if
I
0< Xl < X2 < X3
0
otherwise
X1+X 3-2X 2
6 (X3-X1)
i f Xl < 0
and
r
</>6* (Xl ,X 2 ,X3 ) =
I
l
,
X3 >
Xl < X2 < X3
o and
0
_I
't"
q4 = (n) -31 (p)
u
</>4*'\Xi 1 ,Xi 2 ,Xi 3 ) '
and
(n) -1
3
>::
i 1 <12<i 3 </>6
I
I
I
I
The corresponding U statistics are
q6 =
-.
I
0 otherwise
r
I
I
(
)
Xil'Xi2'Xi3 •
As the qk; k=3,4,S,6 are consistent estimations of the Qk(G);
k=3,4,S,6, respectively, consistent estimators of the Dk ; k-1,2 are
and
I
I
I
I
I,
These estimators can be used to obtain estimates of ef(qk); k=1,2:
k
= 1,2.
(3.4.27)
I
I
-.
I
73
In order to estimate ef(qk); k=1,2, we generate a sample of n,
compute the qj; j
=
3,4,5,6 and sk 2 ;k = 1,Z and use these estimates
in (3.4.27).
It requires a great deal of computer time to generate large
samples from a particular distribution, and compute the estimates
of the functiona1s and variances.
However, as the estimates of
the functiona1s are U statistics, they are asymptotically normally
distributed.
We have seen in section 3.2 that for most distributions
the statistics q1 and q2, which are similar to qj; j
are approximately normally distributed for n = 50.
that this is also true for qj ; j
a
3,4,5,6,
If we assume
= 3,4,5,6 we can generate N
samples of size 50, compute the estimates of the functions for each
of the N samples and then average these estimates to obtain estimates
of the Qj(G).
If the asymptotic variance of qj is Oj2, the variance
of the average of N estimates is OJ
2
IN.
Thus by taking N large
enough, the desired precision can be attained.
Returning to the uniform distribution, if the above method is
used, an idea of the precision of the estimates can be obtained, since
the values of the functionals and the variances are known.
samples of n=50 observations were generated.
N-SO
The results are
summarized in Table 3.8.
We note that for all 4 functiona1s, the estimates agree with
the true values quite well.
estimated.
However, the variances are both over-
For q1' Dl and 01 2 are both overestimated and the net
effect is that the estimate of ARE (ql,b l ) is not far off. However,
for qz, DZ is slightly underestimated while 02 2 is overestimated and
74
Table 3.8
Comparison of the estimates of the functiona1s with
their true values for the uniform distribution;
N = SO, n = SO
Parameter
True Value
0/
0.033333
0.035172
Q3 (G)
0.093750
0.093614
Q4 (G)
0.072917
0.072153
0.500000
0.528120
1.0159
1.0741
0.012724
0.014133
0
0.000105
D1
ARE(ql,b 1 )
02
2
Q5(G)
Q6(G)
Estimate
-0.025617
-0.025389
0.307404
0.300893
1.0063
0.8677
D2
AR.E(Q2,b 1 )
hence ARE(Q2,b 1 ) is grossly underestimated.
Presumably, if more
their true values.
= sao
samples of SO were generated
and the estimates of the functions and variances computed.
These
results are summarized in Table 3.9.
From the table it is apparent that the values of the ARE's
from the numerical integration and those from the estimation method
do not agree.
are correct.
Unfortunately, we cannot say which values, if either,
The conclusion we draw is that the ARE's of Q1 and
Q2 to b1 are both at least 1.
Thus for the normal distribution and
the alternative considered here, the tests based on q1 and Q2 are at
least as efficient as that based on b 1 , for large samples.
-.
I
I
I
I
I
I
samples were generated, the estimates would agree more closely with
For the normal distribution N
I
I
_I
I
I
I
I
I
I
I
-,
I
75
Table 3.9
Comparison of the estimates of the functionals and
the values obtained by numerical integration for the
normal distribution; n = 50, N = 500
Value from Numerical
Integration
Estimate
0.034994
0.029167
Q3
0.095965
0.096629
Q4
0.078013
0.077812
D
l
0.218203
0.175224
1.425
1.102
ol
0.009493
0.011011
Q5
0.002912
0.002964
Q6
-0.019772
-0.019802
D2
0.132432
0.131004
1.443
1.632
Parameter
01
2
ARE(ql,b l )
ARE(Q2,b l )
As our next example we use the Laplace distribution:
g(x) =
and s (x) = e -x
x > O.
00
_~
l eX ; x < 0
2
[
1
e-x ,. x > 0
2
Now
2
x g(x)dx = 2
00
J0 xg(x)dx
=
1
2
and
j0
x 3 g(x)dx = 3.
Thus
~6
(31'6=0
0=
(6-6)/ 1\120 3
=
o.
76
For q2' we have seen in Chapter 2 that QI(S) = ~ , hence
Q3(G) = ~.
72
Now
Q4 * (S)
=
IX>
X
o
0
I I S[~(x-u)] s(u) s(x) du dx
IX>X
=
II {l - e
-~~¥u
2
} e
-x-u
du dx
00
IX>
= Ie-x (I_e-x ) dx o
lX>
2
Ie
3
--x
2
0
_!x
(I-e 2 ) dx
= I6
QI
= 9 _ 72 (..2) - 24 (.:1.) '" 0
1
do 0=0
72
12
d
Since the tests based on qi and b 1 do not meet the conditions
of Theorem 1.3.10, we apply Theorem 1.3.11.
theorem we must determine m.
In order to apply this
To this end note that
IX>
Ix
3
f(x) dx
-00
'" - 60
1
0 2 x
1
IX> 2 -x
Ix e dx + 2(1-0) Ix e dx
112 ' '" -(1+0)
2
0
-<Xl
= 2,
and
o x
1
IX>
-x
dx
111 ' = !(l+o)_£xe dx + 2(1-0) Ixe
0
= - 0
I
I
-,
I
I
I
I
I
I
_I
I
I
I
I
I
I
I
-,
I
77
Hence )..13
':
-60 + 68 - 26-
=
,3
-2(
~
a'td
3
6 = _28 3 (2 _ 8 2 )-2
1
Therefore
and
~
I
= O.'
W1 0=0
=-
as before.
3
2 -2
120(2-0)
+
Now
2
terms involving 0 ,
3
2
= - 12(2 - 0 )-2 + terms involving
0,
and
d3
-8
3
d0
I
6
1 0=0
/T
Note that
F(u)
=
(1+0)G(U)I[u<0] + [o+(l-o)G(u)]I[u>O]
Thus for the Laplace distribution and y < x
F[~ (x+y)] fey) f(x)
~(x+y)
1(1+0)3 e k
I
8
[x+y<O ~ x <0]
1
2
~(3y-x)
+ 8(1+0)(1-0 ) e
I [x+y<O, x>O, y<O]
78
Hence
x
_!-c!F[~(X+Y)] fey) f(x) dy dx
o x 3(x+y)
= 1(1+0)3
f feZ
8
dy dx
-oo-<lO
+
~(l+O)(1-02) 1-1 e~(3Y-X)
o
dy dx
0-00
00 0
_!(x+y)
+ l(1-0 2)f f[0+(1-0)(1-].2e 2
)]e Y- x dy dx
o-x
4
ooX
1 (x+y)
)e-x - y dy dx
+ l(1_0)2 ff[0+(1-0)(1-!e- Z
4
2
00
.
+ !-(1_0 2) (1-0) + ]. 0(1-0) +2- (1-0)3
12
8
72
= ]. - l:..
4
Hence Q1 • - !03.
3
36
03 •
Therefore
~
d03
Q1. -2.
The variance of q1 under the null hypothesis is, as we have
seen in section 3.1, 0.293976/n. From 1.2.1, the variance of b 1 is
For the Laplace distribution this is
=
178.1909/n.
I
I
-,
I
I
I
I
I
I
_I
I
I
I
I
I
I
I
-,
I
79
Thus from Theorem 1.3.11, with m = 3 and y = 3
2
(
l
ARE(Q1,b 1 )
=
yields
-2
13 .34882
(-6/12)
1.293976
r
134.696
We now show that the conditions of Theorem 1.3.11 with m
=3
are not met for Q2 and hence Q2 and b1 are not comparable using
our alternative and Pitman efficiency.
00
Q2
x X
= 6_~~-Lh(x,y,z)
f(x) f(y) f(z) dz dy dx,
where h(x,y,z)
= x+z-2y
x-z.
In this case,
3
3 0x y
Q2 • 4(1+0) ~-£-r h(x,y,z) e X+Y+z dz dy dx
+ 4d(l- 0 2) (1+0) j
J 1 h(x,y,z)
e-X+Y+Z dz dy dx
0-00-00
3
2)(1+0) 00f ~J f0 h(x,y,z) e-x - y+z dz dy dx
+ 4-(1-0
o 0..00
+
4~(1-C)3 j
o
1 1 h(x,y,z)
0
e-x-y-z dz dy dx
0
Now in
let u = -x, v
= -y,
w = -z.
ooWv
Then
• f f f 2v-w-u e-u - v- w du dv dw
II
0 0 0
w-u
=
80
-,
= -1 3 , so
Similarly, 12
Hence
2
9
3
d Q2
2 1 4 - -2 I 3' --I
do 2 6=0
and
3
d Q2
~loaO
-
-91 4 ,
In order to meet thel conditt~s of Theorem 1.3.11 with m.~ we must
have simultaneously 314 • 13' 13 • 0, and 14 ~ 0 which is obvipus1y
not possible.
Thus q2 the conditions of Theorem 1.3.11 are not
met by q2 and b 1 with the same value of m.
Finally we note that
a1
is not defined for the Cauchy distribution,
thus ARE(qk' b l ) ; k • 1,2 is infinite.
3.5
As an example we U$e some data of Pearson as quoted in Pearson
and Hartley (1962):
29
-6
5
-14
19
-5
-15
4
-9
-7
6
0
-8
-11
2
-12
-1
-16
-2
9
29
-17
0
11
15
0
-10
-3
14
1
-8
9
20
-2
-13
-1~
-14
-11
-7
-4
11
-11
-8
-13
-11
0
-15
-4
-2
Computing ql' q2 ' sl 2 ' and 82
2
by hand are quite tedious.
Hotvever,
in the appendix a computer program written in Fortran.' is provided.
Using this program we obtain:
q1 • 0.2633
3s1
--.
• 0.0562
q2 • 0.1662
-3s2
15'0"
• 0.0392
150
I
I
I
I
I
I
_I
Example
,
I
I
I
I
I
I
I
I
I
-,
I
81
Thus
0.2633
0.0562
". 4.69
and
0.1662
z2
= 0.0392
=
4.24
Hence with either ql or q2 we reject the hypothesis that the
distribution from which these observations came is symmetric.
of comparison, b l
= 1.086,
By way
which is "well beyond the upper 1% point"
of the tabulated distribution of b l , according to Pearson and Hartley.
We conclude that the distribution from which this sample came is
positively skewed in the ql' q2 and b l senses.
CHAPTER 4.
4.1
A MEASURE OF KURTOSIS
A Measure of Kurtosis
In section 2.1 the problem of defining skewness was discussed.
~.~.
Much of that discussion would also apply to kurtosis;
no one set definition of kurtosis.
there is
Kurtosis is, in general, thought
of as a measure of how much a symmetric distribution is either more
peaked or less peaked than the normal distribution.
Thus measures
of kurtosis have been used with tests of symmetry to test the normality
of a sample distribution.
In this connection, we define a U statistic
for describing the peakedness or kurtosis of a distribution.
Definition 4.1.1:
Let Xl' X2 , •.. , Xn be i.i.d. random variables with common c.d.f. F.
Y3- Y2
Y4- Y1
Define
if Y4 ~ Y1
;: 0 otherwise
where Y1
~
Y2
~
Y3
S;
Y4 are the ordered X' s.
The corresponding
U statistic is
where the summation (c) is over 1
~
i1 < i2 < i3 < i4
S;
n.
Since kurtosis is used to compare the peakedness of a symmetric
distribution with that of
E(h)
= H,
t~e
normal distribution, we evaluate
for the normal distribution.
83
-•
.. E(~+ Y3- Y4 + Y1- YZ )
Y4 - Y1
Since the normal distribution is symmetric,
Therefore
(4.1.1)
Dixon (1951) investigates ratios of this type as tests for outliers
In the course of his work, he shows that the density of
r ..
I
I
I
I
I
I
el
in a sample from the normal distribution is
Thus H - 1-2E(r), where r has the above density.
I
I
Using the program
DQG32, we find that
E(r) - 0.350627
and hence
H .. 0.298746.
Thus to compare the kurtosis, in the H-sense, of a distribution with
that of the normal distribution, we compare the sample h with 0.2987.
I
I
I
I
I
I
.-I
I
84
Theorem 1.3.2 could be applied to obtain the asymptotic variance
of h; however,
it is much easier to use Theorem 1.3.3 and obtain an
an estimate of the variance.
Defining the components of h in the
usual way:
where the summation (ci) is over 1
jk
=
~
jl < j2 < j3
~
n, except that
i must not occur, we obtain the corresponding estimate of the
asymptotic variance of h:
s 2
h
=
2
n
1
n-l
L (Wi - h) .
i=l
Thus under the hypothesis of normality,
Z
= m(h-O.2987)
4s
h
is asymptotically distributed as the standard normal distribution.
For large samples, this variate can be used to test the equality
of the H-kurtosis of F with that of the normal distribution.
Consider the evaluation of H for the uniform distribution.
The joint density of the order statistics in a sample of 4 from the
uniform distribution is
Let zl
= Y4
f(Zl' z2' z3' z4)
= 24;
O<z4+z 2-
z3<zl<z4 < 1
O<z2<z3<z4 < 1
Thus
1
f(zl,z2)
=J
Zl
= 24(zl
- z2)(1 - zl);
Zl+
J
24dz 3dz 4
o<
z2 < zl < 1.
ZI+- Z l+ z 2
85
Letting u 1
Zz
= Z1
and
U
z = zl'
we have
g(ul'u Z)= Z4u Z(uZ - u1uZ) (l-u Z); 0 <
Thus H
= E(u
1
U
z
< 1,0 < u
1
< 1-
)
1
3
We now evaluate H for the Laplace distribution.
In the joint
density of the order statistics in a sample of 4 let
Then
the joint density of the z's is
I
I
-I
I
I
I
I
I
I
_I
Thus
100 00 Z4
H-Z4{{_L-L zl z Zf (z4- z Z) f(z3- z 1 z Z) f(z3) f(z4) dz 3dz 4 dz Zdz 1 ,
For the Laplace distribution,
Z4
~
f(z3- z 1 z Z) f(z3) dZ 3
I
I
I
I
I
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_I
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86
Continuing with the integration we obtain
00
Z4
~~
f(z4 -z2)f(z4)f(z3- z l z 2)f(z3) dZ3dz4
~d
H
= ~ - li log 2' - 3~ log 3
= 0.3786.
It is easily verified that 8 2 (U) < 82'(N) < 82 (L), where
82(U), 82(N), and 82(L) are the population values of 82 for the
uniform, normal and Laplace distribution, respectively.
We have
just shown that this inequality does not hold for H and in fact
H(N) < H(ll) < H(L).
The same method as used for ql and q2 was used to determine
the sample size necessary for distribution of z to be approximately
normal.
Corresponding to Table 3.2
Table 4.1.
, we have
Number of observed Izi which exceeded the 5 and 1% points
of the normal distribution in 200 samples of n • 20,30
Sample Size
Observations Greater than z(a)
in Absolute Value
a
20
30
Expected frequency
= 0.05
a • 0.01
18
8
1
10
2
4
87
Using the Kolmogorov-Smirnov statistic to compare the sample with the
normal distribution for n
rejected (a
= 0.1).
= 30,
the hypothesis of normality was not
Pearson (1932) found that the distribution
of b
is approximately the normal distribution for n ~ 1000. Again,
2
however, we are using different criteria for determining approximate
normality.
4.2
Example
Consider the data used in the example of section 3.5.
Using
the computer program in the appendix, we obtain h • 0.4333, with
standard error 0.0667.
For the test of equal H-kurtosis with the
normal distribution
0.4333 - 0.2987
0.0667
Thus the hypothesis of equality of H-kurtosis for the sample and
Actually, this test has very
little meaning since the hypothesis of symmetry for this distribution
was previously rejected.
By comparison
b2
= 3.661
a
= 0.778.
and
Using the test based on Geary's statistic, the hypothesis of
equality of a-kurtosis is not rejected (a
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= 2.02
normal distributions is rejected.
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CHAPTER 5.
5.1
SUMMARY AND SUGGESTIONS FOR FURTHER RESEARCH
Summary
Two new descriptive statistics s ql and q2 s for describing the
symmetry or lack of symmetry of a distribution have been introduced.
Since these statistics are U statistics and
thus are asymptotically
normally distributed s either can be used to test the hypothesis
of population symmetry for large samples.
Using Monte Carlo methods s
it was found that the sample size necessary for the distributions of
ql and q2 to approximately normal depends on the parent distribution.
It appears s however s that a sample of size 50 is adequate for most
cases.
Using computer simulation methods the small sample powers
of blS qls and q2 were compared.
Based on this work bl is more
powerful than either ql or q2 in small samples.
The ARE of both
ql and q2 to the moment statistic for skewness s bls was investigated.
For the alternative considered s both ARE(qls bl) and ARE(q2s b l )
are greater than one even when the parent distribution is the normal
distribution.
A U statistics h s for describing kurtosis has also been
introduced.
Combining the appropriate test based on h and either
of the above tests of symmetrys a test of the normality of a distribution is obtained.
for n
~
Since h is approximately normally distributed
30 s under the hypothesis of normalitys the test for the
normality of a distribution can be applied with as few as 30 observations.
89
5.2
Suggestions for Further· Research
i)
The small sample distributions of q1' Q2' and h could be
investigated more thoroughly, and possibly the three distributions
tabulated.
ii)
Since the empirical study of the powers of Q1, Q2 and b 1
showed that b 1 was clearly superior for n • 10 but not for n • 15,
it would be interesting to see the work carried out to n • 20 to
see if Ql and/or Q2 surpass b •
l
iii)
The ARE of the three tests could be determined for
different types of alternatives than those used here.
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CHAPTER 6.
LIST OF REFERENCES
Bell, J. R. 1968. Algorithm No. 334. Communications of the
Association of Computing Machinery 11:498.
Berk, R. H. 1966. Limiting behavior of posterior distributions
when the model is incorrect. Annals of }1athematical Statistics
37:51.
Cramer, H. 1946. Mathematical Methods of Statistics.
University Press, Princeton, New Jersey.
Princeton
David, F. N., and Johnson, N. L. 1956. Some tests of significance
with ordered variables. Journal of the Royal Statistical
Society, Series B, 19:1.
Dixon, W. J. 1951. Ratios involving extremedvalues.
Mathematical Statistics 22:68.
Annals of
Fraser, D. A. S. 1957. Nonparametric Methods in Statistics.
Wiley and Sons, Inc., New York.
John
Geary, R. C. 1935. The ratio of the mean deviation to the standard
deviation as a test for normality. Biometrika 27:310.
General Electric. 1966. Time-Sharing Fortran Reference Manual.
Schenectady, New York.
Gupta, M. K. 1967. An asymptotically nonparametric test of symmetry.
Annals of Mathematical Statistics 38:849.
Hoeffding, W. 1948. A class of statistics with asymptotic normal
distribution. Annals of Mathematical Statistics 19:293.
I.B.M. Applications Programs. 1968. H20-0205-3 System/360,
Scientific Subroutine Package, Version III, Fourth Edition.
White Plains, New York.
Kendall, M. G., and Stuart, A.
Volume 1, Second Edition.
1963. The Advanced Theory of Statistics,
Hafner Publishing Company, New York.
Knuth, D. E. 1969. The Art of Computer Programming, Volume 2.
Addison and Wesley, Reading, Massachusetts.
Pearson, E. S. 1932. A further development of tests of normality.
Biometrika 22:239.
Pearson, E. 5., and Hartley, H. O. 1962. Biometrika Tables for
Statisticians, Volume 1, Second Edition. Cambridge University
Press, Cambridge, England.
91
Quade, Dana. 1967. Nonparametric Partial Correlation. Institute
of Statistics Mimeo Series No. 526. University of North Carolina,
Chapel Hill, North Carolina.
Sen, P. K. 1960. On some convergence properties of U statistics.
Calcutta Statistical Association Bulletin 10:1.
Sen, P. K. 1966. On a distribution free method of estimating
asymptotic efficiency of a class of nonparametric tests.
Annals of Mathematical Statistics 37:1759.
Shapiro, S. S., Wi1k, M. B., and Chen, H. J. 1968. A comparative
study of various tests for normality. Journal of the American
Statistical Association 63:1343.
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CHAPTER 7.
7.1
APPENDICES
Computer Program for Calculating q1 and q2
The following computer program can be used to calculate
qk: k .. 1,2.
It is written in Time-Sharing Fortran; see General
Electric (1966).
Note that the subroutine SORTR is a routine which
orders the observations from the largest to the smallest.
10
20
30
40
50
60
70
80
90
100
110
120
130
140
150
160
170
180
190
200
210
220
230
240
250
260
270
280
290
300
310
320
330
340
350
360
370
DIMENSION A(100),B(100),C(100)
INPUT,N
INPUT,(A(I),I-1,N)
DO 3 I -l,N
3 S-S+A(I)
paN
S=S/P
DO 4 I .. 1,N
T"T+(A(I)-S) t2
4 R=R+(A(I) -S) t3
SK.. (R/T)*SQRT(P/T)
SD=SQRT(T/(P-1.»
PRINT 5,N,S,SD,SK
.
5 FORMAT(//"N=",I3,//"MEAN=",F8.4,//"STANDARD DEVIATION-"F8.4
+,//"Bl..",F8.4)
CALL SORTR(A,N,1,1)
Nl=N-l
N2-N-2
DO 6 I a 1,N2
Z=A(I)
IP=I+l
DO 6 J-IP,Nl
Y=A(J)
JP=J+l
DO 6 K=JP,N
X=A(K)
IF(Z-X)6,6,7
7 F=(Z+X-2.*Y)/(Z-X)
B(I)-B(I)+F
B(J)-B(J)+F
B(K)=B(K)+F
IF(F)8,6,9
8 (CI)=C(I)-l.
C(J)=C(J)-l.
C(K)=C(K)-l •.
GO TO 6
9 C(I)=C(I)+1.
93
380
390
400
410
420
430
440
450
460
470
480
490
500
510
520
530
540
550
560
570
580
590
600
610
620
-630
];.2
10
20
30
40
50
60
70
80
90
100
110
120
130
140
150
160
170
180
190
200
210
220
230
C(J)=C(J)+l.
C(K)-C(K)+l.
6 CONTINUE
Pl-Nl
P2-N2
CON-2./(Pl*P2)
DO 10 I-l,N
B(I)-B(I)*CON
C(I)-C(I)*CON
V-V+(C(I)*C(I»)
U-U+(B(I)*B(I»
Ql-Ql+C(I)
10 Q2-Q2+B(I)
Ql-Ql/P
Q2-Q2/P
V-9.*(V-(P*Ql*Ql»/(P*Pl)
U-9.*(U-(P*Q2*Q2»/(P*Pl)
SE1-SQRT(V)
SE2-SQRT(U)
PRINT 11,Ql,SEl
11 FORMAT(/ /"Ql_" ,F8.4, II
PRINT l2,Q2,SE2
12 FORMAT (/ /"Q2_" , F8. 4, II
$USE SORTRM***
RETURN
END
-STANDARD ERROR-" ,F8. 4)
STANDARD ERROR-",F8.4)
Computer Program-for Calculating h
DIMENSION A(50),B(50)
INPUT,N
INPUT,(A(I),I-l,N)
DO 3 I-l,N
3 S-S+A(I)
paN
S-S/P
DO 4 I-l,N
T-T+(A(I)-S)t2
4 R-R+(A(I)-S)t4
XK-(P*R)/(T*T)·
SD-SQRT(T/(~-l.»
PRINT 5,N,S,SD,XK
5 FORMAT(/ /"N-" ,13, / /"MEAN- U ,Fa. 4, / r'S'l'Al(DARDDEVIATION=",
+FB.4,//"B2-",F8.4)
CALL SORTR(A,N,l,l)
NI-N-1
N2-N-2
N3-N-3
DO 6 I-1,N3
Z-A(I)
IP-I+1
nQ 6· J-IP ,N2
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240
250
260
270
280
290
300
310
320
330
340
350
360
370
380
390
400
410
420
430
440
450
455
460
470
480
490
500
510
520
W-A(J)
JP-J+l
DO 6 K-JP,Nl
Y"'A(K)
KP-K+l
DO 6 KK=KP,N
X-A(KK)
IF(Z-X)6,6,7
7F-(W-Y)/(Z-X)
B(l) ..B(I)+F
B(J)-B(J)+F
(BK)aB(K)+F
B(KK)-B(KK)+F
6 CONTINUE
PI-Nl
P2-N2
P3..N3
COM-6./(Pl*P2*P3)
DO 10 I ..l,N
B(I)-B(I)*COM
V-V+(B(I)*B(I»
10 H..H+B(I)
H-H/P
V-16.*(V-P*H*H»/(P*Pl)
SE-SQRT(V)
PRINT 11,H,SE
11 FORMAT(//"H- 1 ,F8.4,"
$USE SORTRM***
RETURN
USE
STANDARD ERROR-",F8.4)