# Comparison of Some Tests of Fit for the Inverse Gaussian Distribution

Authors: D. J. Best; J. C. W. Rayner; O. Thas
Journal: Advances in Decision Sciences
 (2012)

---

## Abstract

This paper gives an empirical investigation of some tests of goodness of fit for the inverse Gaussian distribution.


---

## Body

## 1. Introduction

The inverse Gaussian(IG) distribution is an important statistical model for the analysis of positive data. See, for example, Seshadri [1]. In its standard form the distribution, denoted IG(λ,μ), depends on the shape parameter λ>0 and the mean μ>0. Its probability density function is
(1.1)λ2πx3exp{-λ(x-μ)22μ2x},forx>0,andzerootherwise.LetX1,X2,…,Xn be a sequence of independent observations. We wish to test H0: X is distributed as IG(λ,μ) for λ>0 and μ>0 against HA and not H0.The maximum likelihood estimators are(1.2)λ^={∑i=1n(1/Xi-1/X-)n}-1,μ^=X-.
Put φ^=λ^/μ^.We consider the following test of fit statistics:(i)
the smooth test statistic of Ducharme [2], denoted by R^3,(ii)
the Laplace transform test statistic of Henze and Klar [3], denoted by V0,(iii)
the empirical likelihood ratio test statistic of Vexler et al. [4], denoted by TKn,(iv)
the traditional Anderson-Darling test statistic, denoted byA2,(v)
conventional smooth tests following Rayner et al. [5].We strongly suggest thatP values for all tests be found using the parametric bootstrap. A link for smooth test statistics for many distributions and bootstrap P values for their tests of fit is http://www.biomath.ugent.be/~othas/smooth2.


## 2. Definitions

### 2.1. Ducharme’sR^3

The usual approach to constructing a smooth test as outlined, for example, in Rayner et al. [5], produces inconsistent and less powerful tests; see Ducharme [2, page 279] and Henze and Klar [3, Tables 1 and 2]. Ducharme [2] suggests testing H0 above by first defining Zi=μ/Xi,i=1,2,…,n; he then states that Z has a random walk distribution, RW(φ) say. The Ducharme [2] statistic is R^3=V^22+V^32, where V^2 and V^3 are, under H0, asymptotically independent and asymptotically standard normal variables, and hence R^3 is asymptotically χ22 distributed. Thus R^3 does not depend on φ, as do V0 (see Section 2.2) and the Anderson-Darling test statistic A2. Ducharme [2, page 279] notes this implies A2 loses power for some values φ can take. The components V^22 and V^32 are defined by
(2.1)V^22=nφ^424+6φ^{Z^-2-(1+3φ^+3φ^2)}2,V^32=nφ^624(4+φ^)+(120+75φ^+15φ^2+φ^3)×{Z^-3(4+φ^)-Z^-2(60φ^+30+4φ)+120φ^3+195φ^2+123φ^+32+3φ^3}2
in which Z^-r=∑i=1n(μ^/Xi)r/n.A positive feature of smooth tests is that their components can often shed light on how the data differ from the hypothesised distribution. This is somewhat less evident with Ducharme’s test given that he transforms the data. Another positive feature of smooth tests is that their components often give highly focused tests with good power. Ducharme’s test has components that are likely to fulfil this role.


### 2.2. Henze and Klar’sV0

This is defined using the exponentially scaled complementary error functionerfce(x)=ex2erfc(x) where erfc(x)=2∫x∞e-t2dt/π. Note that in erfc(x) we divide by π and not π as in Henze and Klar [3, page 428], as we believe a typographical error was made in their paper. Now let Zj=Xj/μ^ and Zjk=Zj+Zk. Then
(2.2)V0=1n∑j,k=1nZjk-1-2∑j=1nZj-1{1+πφ^2Zjerfce((Zj+1)φ^2Zj)}+n(1+2φ^)4φ^.
Tests based on the empirical Laplace transform, as is V0, have produced powerful tests for other distributions, and so it is useful to compare V0 with other recently suggested tests.


### 2.3. Thelog(TKn) Statistic of Vexler et al.

Order the observations so thatX(1)≤X(2)≤⋯≤X(n), and let Y(i)=1/X(n-i+1). Then
(2.3)log(TKn)=min1≤m≤nδ[n2log(πe2λ^)+∑i=1nlog{2mn(Y(i+m)-Y(i-m))}],
where δ can be taken to be 0.5. Observe that, for small m, such as m=1, the statistic can take an infinite value when there are tied data. Vexler et al. [4] do not appear to note this. For the Poisson alternative in Table 1 and δ=0.5 the log(TKn) statistic is often infinite. Choi and Lim [6] show that an entropy-type statistic-like log(TKn) has good power for the Laplace distribution and so it is of interest to see how the entropy method works with a skewed distribution.Table 1
Powers (in percentages) of goodness of fit tests for the inverse Gaussian distribution forn=20 and (a) α=0.10, (b) α=0.05.
(a)
Alternative
R
^
3
V
^
2
2
V
^
3
2
log(TKn)
V
0
A
2
U
^
3
2
U
^
4
2
U
~
3
2
U
~
4
2
IG(0.25)
10
10
11
9
9
10
10
10
3
6
IG(0.5)
10
10
9
11
10
10
10
10
4
8
IG(1)
10
10
10
11
10
10
10
8
6
9
IG(3)
10
10
10
10
10
10
9
8
8
10
IG(5)
11
11
10
12
10
11
8
9
8
10
IG(10)
9
9
10
10
10
10
8
10
9
11
IG(20)
10
8
11
9
10
9
9
11
9
9
IG(100)
10
9
9
8
9
10
9
10
11
10
W(0.5)
95
92
83
93
96
95
6
17
0
8
W(0.8)
84
79
72
68
83
82
52
61
6
21
W(1)
74
70
61
55
75
73
65
63
10
30
W(1.2)
67
64
55
49
67
63
61
50
18
36
W(1.5)
64
60
53
42
58
55
52
34
29
47
W(2)
58
54
44
37
54
48
43
12
40
54
W(3)
50
47
37
30
46
39
16
15
50
56
LN(0.6)
15
15
13
9
13
14
13
8
9
12
LN(1)
23
23
19
10
22
22
20
12
4
10
LN(1.4)
42
39
31
18
39
36
23
26
2
7
LN(2)
70
66
57
47
70
68
9
20
0
4
LN(3)
94
90
80
87
95
95
0
1
0
1
G(0.6)
90
87
77
83
91
91
41
51
8
21
G(2)
49
46
37
26
47
41
39
20
21
34
χ
1
2
94
89
86
89
95
95
25
36
4
19
χ
10
2
28
26
21
15
24
21
8
5
19
25
HN(0, 1)
79
74
65
65
80
78
74
64
36
59
HC(0, 1)
56
56
41
30
55
49
18
25
2
4
U(0, 1)
90
84
77
93
91
91
88
68
94
98
POI(3)
52
39
49
49
50
92
39
1
40
57
(b)
Alternative
R
^
3
V
^
2
2
V
^
3
2
log(TKn)
V
0
A
2
U
^
3
2
U
^
4
2
U
~
3
2
U
~
4
2
IG(0.25)
5
5
5
5
4
5
4
5
2
5
IG(0.5)
5
5
6
5
4
5
5
4
2
5
IG(1)
5
6
5
5
4
6
6
4
2
5
IG(3)
6
5
6
4
4
4
6
6
1
3
IG(5)
5
5
5
4
5
6
4
5
1
2
IG(10)
5
5
5
5
4
6
3
4
0
1
IG(20)
5
4
5
5
5
5
2
4
0
1
IG(100)
5
5
5
5
5
6
2
2
0
0
W(0.5)
93
91
83
87
87
94
1
3
0
2
W(0.8)
76
72
60
61
77
77
26
42
1
10
W(1)
67
65
54
47
70
63
45
49
4
14
W(1.2)
62
59
47
38
63
57
47
42
8
21
W(1.5)
54
53
37
31
50
48
45
28
15
32
W(2)
45
43
31
22
41
42
37
12
23
39
W(3)
41
39
29
20
34
29
13
3
35
47
LN(0.6)
8
8
7
4
5
8
6
2
3
6
LN(1)
16
18
10
5
15
13
9
2
1
4
LN(1.4)
35
34
24
11
27
27
8
2
1
2
LN(2)
64
61
47
38
55
63
2
1
0
1
LN(3)
90
86
76
81
92
94
0
1
0
0
G(0.6)
87
85
75
78
88
88
20
35
1
11
G(2)
37
37
26
18
36
33
28
15
8
21
χ
1
2
89
87
83
85
93
92
7
17
1
8
χ
10
2
17
19
10
6
12
13
6
2
10
18
HN(0, 1)
74
70
60
55
70
72
63
55
19
39
HC(0, 1)
51
51
36
23
46
46
7
14
0
2
U(0, 1)
85
81
67
87
83
88
81
61
80
94
POI(3)
37
30
35
39
34
77
8
0
24
43


### 2.4. The Anderson-Darling Statistic

Again order the data from smallest to largest to obtainX(1)≤X(2)≤⋯≤X(n) and take Z(i)=F(X(i);λ^,μ^) where F() is the distribution function for the IG distribution. Then the Anderson-Darling statistic is
(2.4)A2=-n-1n∑i=1n(2i-1){logZ(i)+log(1-Z(n-i+1))}.
The Anderson-Darling has stood the test of time as a useful general option for tests of fit for many distributions. Have newer tests improved on its power performance?


### 2.5. Conventional Smooth Test Third and Fourth Components

Henze and Klar [3] consider the test based on the conventional second-order component U^22 and show it has poor power for some alternatives. Ducharme [2] notes that these conventional smooth tests discussed, for example, in Rayner et al. [5], can be inconsistent. However we decided to include tests based on U^32 and U^42 in our comparisons.In general therth-order component is Ur=∑j=1nπr(xj)/n, in which πr() is an orthonormal polynomial of degree r on the inverse Gaussian distribution. We find
(2.5)π3(z)=(z3-a3z2-b3z-c3)√(ne3)
in which z=x-μ, w=λ+3μ, y=λ3+12λ2μ+48λμ2+60μ3, a3=3μ2(3λ+10μ)/(wλ), b3=3μ3(λ2-λμ+15μ2)/(wλ2), c3=-3μ5(2λ+7μ)/(wλ2), and e3=6μ9y/(wλ5).Moreover(2.6)π4(z)=(z4+a4z3+b4z2+c4z+d4)√(ne4)
in which z, w and y are as for π3(z), a4=-6μ2(3λ3+38λ2μ+160λμ2+210μ3)/(yλ), b4=-3μ3(3λ5+22λ4μ-135λ3μ2-1800λ2μ3-5940μ4λ-6300μ5)/(wyλ2), c4=3μ5(17λ5+253λ4μ+1284λ3μ2+1980λ2μ3-2280λμ4-6300μ5)/(wyλ3), d4=6μ6(λ5+10λ4μ-6λ3μ2-390λ2μ3-1500λμ4-1710μ5)/(wyλ3), and e4=6μ12(7λ7+198λ6μ+2439λ5μ2+16920λ4μ3+71100λ3μ4+180180λ2μ5-253440λμ6+151200μ7)/(wyλ7).The parametersλ and μ can be estimated by maximum likelihood (ML) estimation using the previous formula for λ^ and μ^, thereby giving U^3 and U^4. We also looked at U~3 and U~4 where the parameters are estimated using method of moments (MOM) estimators  λ^=nμ~3/∑j=1n(Xj-μ~)2 and μ~=μ^=X-.As indicated previously, smooth tests can indicate in terms of moments how data and the hypothesised distribution differ. This feature, and good power in previous studies for testing for other distributions, prompted us to include conventional smooth tests in our comparisons.


## 2.1. Ducharme’sR^3

The usual approach to constructing a smooth test as outlined, for example, in Rayner et al. [5], produces inconsistent and less powerful tests; see Ducharme [2, page 279] and Henze and Klar [3, Tables 1 and 2]. Ducharme [2] suggests testing H0 above by first defining Zi=μ/Xi,i=1,2,…,n; he then states that Z has a random walk distribution, RW(φ) say. The Ducharme [2] statistic is R^3=V^22+V^32, where V^2 and V^3 are, under H0, asymptotically independent and asymptotically standard normal variables, and hence R^3 is asymptotically χ22 distributed. Thus R^3 does not depend on φ, as do V0 (see Section 2.2) and the Anderson-Darling test statistic A2. Ducharme [2, page 279] notes this implies A2 loses power for some values φ can take. The components V^22 and V^32 are defined by
(2.1)V^22=nφ^424+6φ^{Z^-2-(1+3φ^+3φ^2)}2,V^32=nφ^624(4+φ^)+(120+75φ^+15φ^2+φ^3)×{Z^-3(4+φ^)-Z^-2(60φ^+30+4φ)+120φ^3+195φ^2+123φ^+32+3φ^3}2
in which Z^-r=∑i=1n(μ^/Xi)r/n.A positive feature of smooth tests is that their components can often shed light on how the data differ from the hypothesised distribution. This is somewhat less evident with Ducharme’s test given that he transforms the data. Another positive feature of smooth tests is that their components often give highly focused tests with good power. Ducharme’s test has components that are likely to fulfil this role.


## 2.2. Henze and Klar’sV0

This is defined using the exponentially scaled complementary error functionerfce(x)=ex2erfc(x) where erfc(x)=2∫x∞e-t2dt/π. Note that in erfc(x) we divide by π and not π as in Henze and Klar [3, page 428], as we believe a typographical error was made in their paper. Now let Zj=Xj/μ^ and Zjk=Zj+Zk. Then
(2.2)V0=1n∑j,k=1nZjk-1-2∑j=1nZj-1{1+πφ^2Zjerfce((Zj+1)φ^2Zj)}+n(1+2φ^)4φ^.
Tests based on the empirical Laplace transform, as is V0, have produced powerful tests for other distributions, and so it is useful to compare V0 with other recently suggested tests.


## 2.3. Thelog(TKn) Statistic of Vexler et al.

Order the observations so thatX(1)≤X(2)≤⋯≤X(n), and let Y(i)=1/X(n-i+1). Then
(2.3)log(TKn)=min1≤m≤nδ[n2log(πe2λ^)+∑i=1nlog{2mn(Y(i+m)-Y(i-m))}],
where δ can be taken to be 0.5. Observe that, for small m, such as m=1, the statistic can take an infinite value when there are tied data. Vexler et al. [4] do not appear to note this. For the Poisson alternative in Table 1 and δ=0.5 the log(TKn) statistic is often infinite. Choi and Lim [6] show that an entropy-type statistic-like log(TKn) has good power for the Laplace distribution and so it is of interest to see how the entropy method works with a skewed distribution.Table 1
Powers (in percentages) of goodness of fit tests for the inverse Gaussian distribution forn=20 and (a) α=0.10, (b) α=0.05.
(a)
Alternative
R
^
3
V
^
2
2
V
^
3
2
log(TKn)
V
0
A
2
U
^
3
2
U
^
4
2
U
~
3
2
U
~
4
2
IG(0.25)
10
10
11
9
9
10
10
10
3
6
IG(0.5)
10
10
9
11
10
10
10
10
4
8
IG(1)
10
10
10
11
10
10
10
8
6
9
IG(3)
10
10
10
10
10
10
9
8
8
10
IG(5)
11
11
10
12
10
11
8
9
8
10
IG(10)
9
9
10
10
10
10
8
10
9
11
IG(20)
10
8
11
9
10
9
9
11
9
9
IG(100)
10
9
9
8
9
10
9
10
11
10
W(0.5)
95
92
83
93
96
95
6
17
0
8
W(0.8)
84
79
72
68
83
82
52
61
6
21
W(1)
74
70
61
55
75
73
65
63
10
30
W(1.2)
67
64
55
49
67
63
61
50
18
36
W(1.5)
64
60
53
42
58
55
52
34
29
47
W(2)
58
54
44
37
54
48
43
12
40
54
W(3)
50
47
37
30
46
39
16
15
50
56
LN(0.6)
15
15
13
9
13
14
13
8
9
12
LN(1)
23
23
19
10
22
22
20
12
4
10
LN(1.4)
42
39
31
18
39
36
23
26
2
7
LN(2)
70
66
57
47
70
68
9
20
0
4
LN(3)
94
90
80
87
95
95
0
1
0
1
G(0.6)
90
87
77
83
91
91
41
51
8
21
G(2)
49
46
37
26
47
41
39
20
21
34
χ
1
2
94
89
86
89
95
95
25
36
4
19
χ
10
2
28
26
21
15
24
21
8
5
19
25
HN(0, 1)
79
74
65
65
80
78
74
64
36
59
HC(0, 1)
56
56
41
30
55
49
18
25
2
4
U(0, 1)
90
84
77
93
91
91
88
68
94
98
POI(3)
52
39
49
49
50
92
39
1
40
57
(b)
Alternative
R
^
3
V
^
2
2
V
^
3
2
log(TKn)
V
0
A
2
U
^
3
2
U
^
4
2
U
~
3
2
U
~
4
2
IG(0.25)
5
5
5
5
4
5
4
5
2
5
IG(0.5)
5
5
6
5
4
5
5
4
2
5
IG(1)
5
6
5
5
4
6
6
4
2
5
IG(3)
6
5
6
4
4
4
6
6
1
3
IG(5)
5
5
5
4
5
6
4
5
1
2
IG(10)
5
5
5
5
4
6
3
4
0
1
IG(20)
5
4
5
5
5
5
2
4
0
1
IG(100)
5
5
5
5
5
6
2
2
0
0
W(0.5)
93
91
83
87
87
94
1
3
0
2
W(0.8)
76
72
60
61
77
77
26
42
1
10
W(1)
67
65
54
47
70
63
45
49
4
14
W(1.2)
62
59
47
38
63
57
47
42
8
21
W(1.5)
54
53
37
31
50
48
45
28
15
32
W(2)
45
43
31
22
41
42
37
12
23
39
W(3)
41
39
29
20
34
29
13
3
35
47
LN(0.6)
8
8
7
4
5
8
6
2
3
6
LN(1)
16
18
10
5
15
13
9
2
1
4
LN(1.4)
35
34
24
11
27
27
8
2
1
2
LN(2)
64
61
47
38
55
63
2
1
0
1
LN(3)
90
86
76
81
92
94
0
1
0
0
G(0.6)
87
85
75
78
88
88
20
35
1
11
G(2)
37
37
26
18
36
33
28
15
8
21
χ
1
2
89
87
83
85
93
92
7
17
1
8
χ
10
2
17
19
10
6
12
13
6
2
10
18
HN(0, 1)
74
70
60
55
70
72
63
55
19
39
HC(0, 1)
51
51
36
23
46
46
7
14
0
2
U(0, 1)
85
81
67
87
83
88
81
61
80
94
POI(3)
37
30
35
39
34
77
8
0
24
43


## 2.4. The Anderson-Darling Statistic

Again order the data from smallest to largest to obtainX(1)≤X(2)≤⋯≤X(n) and take Z(i)=F(X(i);λ^,μ^) where F() is the distribution function for the IG distribution. Then the Anderson-Darling statistic is
(2.4)A2=-n-1n∑i=1n(2i-1){logZ(i)+log(1-Z(n-i+1))}.
The Anderson-Darling has stood the test of time as a useful general option for tests of fit for many distributions. Have newer tests improved on its power performance?


## 2.5. Conventional Smooth Test Third and Fourth Components

Henze and Klar [3] consider the test based on the conventional second-order component U^22 and show it has poor power for some alternatives. Ducharme [2] notes that these conventional smooth tests discussed, for example, in Rayner et al. [5], can be inconsistent. However we decided to include tests based on U^32 and U^42 in our comparisons.In general therth-order component is Ur=∑j=1nπr(xj)/n, in which πr() is an orthonormal polynomial of degree r on the inverse Gaussian distribution. We find
(2.5)π3(z)=(z3-a3z2-b3z-c3)√(ne3)
in which z=x-μ, w=λ+3μ, y=λ3+12λ2μ+48λμ2+60μ3, a3=3μ2(3λ+10μ)/(wλ), b3=3μ3(λ2-λμ+15μ2)/(wλ2), c3=-3μ5(2λ+7μ)/(wλ2), and e3=6μ9y/(wλ5).Moreover(2.6)π4(z)=(z4+a4z3+b4z2+c4z+d4)√(ne4)
in which z, w and y are as for π3(z), a4=-6μ2(3λ3+38λ2μ+160λμ2+210μ3)/(yλ), b4=-3μ3(3λ5+22λ4μ-135λ3μ2-1800λ2μ3-5940μ4λ-6300μ5)/(wyλ2), c4=3μ5(17λ5+253λ4μ+1284λ3μ2+1980λ2μ3-2280λμ4-6300μ5)/(wyλ3), d4=6μ6(λ5+10λ4μ-6λ3μ2-390λ2μ3-1500λμ4-1710μ5)/(wyλ3), and e4=6μ12(7λ7+198λ6μ+2439λ5μ2+16920λ4μ3+71100λ3μ4+180180λ2μ5-253440λμ6+151200μ7)/(wyλ7).The parametersλ and μ can be estimated by maximum likelihood (ML) estimation using the previous formula for λ^ and μ^, thereby giving U^3 and U^4. We also looked at U~3 and U~4 where the parameters are estimated using method of moments (MOM) estimators  λ^=nμ~3/∑j=1n(Xj-μ~)2 and μ~=μ^=X-.As indicated previously, smooth tests can indicate in terms of moments how data and the hypothesised distribution differ. This feature, and good power in previous studies for testing for other distributions, prompted us to include conventional smooth tests in our comparisons.


## 3. Sizes and Powers

In this section wherever possible we have used IMSL routines to generate random deviates. Calculations were done using double precision arithmetic and FORTRAN code. For the inverse Gaussian random deviates were found as in Michaels et al. [7].We examine a similar range of alternatives to that given by Henze and Klar [3] so that comparisons can be made with the other statistics in their Table 1(a). We note that(i)
for the lognormal alternativeLN(φ) the probability density function should be exp{-log2(x)/2φ2}/(xφ2π),(ii)
theW(1) alternative is a standard exponential alternative. Note that in Henze and Klar [3, Table 1] W(1), G(1), and χ22 are equivalent.It appears that the tests based onR^3 and V0 generally do well while that based on log(TKn) is only competitive for the symmetric uniform alternative. The smooth tests based on U^32 and U^42, like that based on U^22 in Henze and Klar [3], were not competitive. The tests based on U~32 and U~42 were generally even less competitive. This is unfortunate as these components help describe the data and this facility is not available with the other tests. All powers were calculated using parametric bootstrap.The alternatives in Table1 are defined in Henze and Klar [3]. However note that IG(λ) in Henze and Klar [3] is IG(λ,1) here. There is, however, one exception, and that is the Poisson-type alternative POI(3) which has probability function e-θθx/x! in which θ=3 here and if a random x value is zero we take this to be x=0.5. This alternative was suggested by the comment in Henze and Klar [3] that for this shelf life data they examine A2 and the other EDF statistics have a much smaller P value than V0 and the other empirical Laplace transform statistics. A feature of the shelf life data was that there were tied observations — not something one would expect for an inverse Gaussian distribution. The POI(3) alternative gives parametric bootstrap simulated samples with many ties and, as can be seen in Table 1, the power of the test based on A2 is much greater than those for R^3 or V0. The test based on log(TKn) classifies infinite values as rejections of the null hypothesis. We have no explanation as to why the Anderson-Darling test is quite powerful for tied data when the null hypothesis specifies an inverse Gaussian distribution.In Table1(a) our powers for the tests based on V0 and A2 are very similar to those obtained by Henze and Klar [3]. Table 1(b) gives powers for the same alternatives as Table 1(a) but with α=0.05 as this choice of α is commonly used in practice. The relative performance of the tests in Tables 1(a) and 1(b) is similar. The traditional smooth tests, based on the U statistics, are sometimes particularly poor in Table 1(b).


## 4. The Approach toχ2

An advantage of the smooth test statistics and their components is that under the null hypothesis they have asymptoticχ2 distributions. Thus for larger sample sizes approximate P values can be found using the χ2 distribution. However for V^22 and R^3 Table 2 indicates that, to give actual test sizes close to the nominal 5%, for λ=2 and μ=1 a sample size of 200 might be needed, while an even greater sample size might be needed for V^32. This ties in with the suggestion, made in Section 1, to use the parametric bootstrap.Table 2
Empirical null 95% points forV^22, V^32, R^3, U~32, and T2/v(T) for λ=2 and μ=1.
n
V
^
2
2
V
^
3
2
R
^
3
U
~
3
2
T
2/v(T)
20
2.20
1.77
3.89
0.97
0.70
50
2.89
1.96
4.64
1.20
0.83
100
3.26
2.24
4.89
1.58
1.08
200
3.51
2.54
5.44
2.12
1.42
500
3.72
3.07
5.62
2.98
1.99
1000
3.75
3.20
5.76
3.97
2.62
2000
3.82
3.34
5.82
4.34
2.89
5000
3.82
3.54
5.84
5.39
3.58
∞
3.84
3.84
5.99
—
3.84We did not expect the conventional smooth test statistics to be asymptoticallyχ2 distributed; see, for example, Rayner et al. [5, Section 9.3]. As an illustration of this Table 2 shows that 95% points of U~32 do not converge to 3.84. If m3=∑i(Xi-X-)3/n, we observe that because of the MOM estimators used in U~32 we can write the numerator of U~3 as n(m3-3μ~5/λ~2)=T say. Then T2/v(T), where v(T) is the variance of T, should be asymptotically χ2. Here v(T)=6μ9(λ2+12λμ+36μ2)/(nλ5). This formula can be found by the delta method. No powers for T2/v(T) are shown in Table 1 as they are similar to those for U~32.


## 5. Examples

(i) Failure Times
Proschan [8] has given failure times for air conditioning in Boeing 720 jets. For jet number  7912 the 30 times were
(5.1)23,261,87,7,120,14,62,47,225,71,246,21,42,20,5,12,120,11,3,14,71,11,14,11,16,90,1,16,52,95.
Does the inverse Gaussian provide a good model for these times? We find λ^=13.76, μ^=μ~=59.60, and λ~=42.38. With nδ=5, log(TKn)=6.936 with P value 0.152. Further, R^3=22.07  with P value 0.0007 and V0=1.158 with P value 0.012. If α=0.05, say, it makes a difference which test is used: those based on R^3 and V0 are significant but that based on log(TKn) is not. If α=0.01 only the test based on R^3 is significant. Further results with P values in parentheses are V^22=19.42(0.001), V^32=2.65(0.02), U~32=0.32(0.27), U~42=1.02(0.002), U^32=0.77(0.06), and U^42=0.71(0.06). We see the tests based on V^22 and U~42 are significant at the 5% level; the latter suggests the lack of fit is due to kurtosis differences between the model and the data. See Figure 1. Observe that in Figures 1 and 2 the height of the histogram bars is class frequency/number of observations/class width and that this height is labelled “density” so as to be on the same scale as the probability “density” curve. Figure 1 uses MOM estimators for this curve.
Aside from that, we note that in Henze and Klar [3, Table 3] the value 3.7 should be 3.0. This does not affect the conclusions of Henze and Klar for this data set.Figure 1
Air conditioner failure times in Boeing jets.Figure 2
Storm precipitation at Jug bridge, MD, USA.(ii) Precipitation at Jug Bridge
Ang and Tang [9, page 266] consider precipitation from storms in inches at the Jug bridge in MD, USA. Their data were
(5.2)1.01.1.11,1.13,1.15,1.16,1.17,1.17,1.2,1.52,1.54,1.54,1.57,1.64,1.73,1.79,2.09,2.09,2.57,2.75,2.93,3.19,3.54,3.57,5.11,5.62.
Figure 2 indicates the inverse Gaussian fit is marginal. We find λ^=8.08, μ^=μ~=2.16, and λ~=6.72  with V0=0.33, giving a P value of 0.09. Further test statistics with P values in parentheses are R^3=2.86(0.09), log(TKn)=9.60(0.01), V^22=1.90(0.08), V^32=0.97(0.16), U~32=0.05(0.78), U~42=0.01(0.91), U^32=0.36(0.07), and U^42=0.09(0.72). As the test based on log(TKn) is most critical of the IG hypothesis, the data may be more symmetric than the IG. The inverse Gaussian curve in Figure 2 uses ML estimators.
In passing we note that the exponential distribution with parameter 0.463 does not provide a good fit to the data. When testing for an exponential distributionA2=3.226 with an approximate P value of 0.03. Visual inspection of Figure 2 may have indicated the exponential model may have been appropriate.


## 6. Conclusion

The tests based onV0 and R^3 do well in the power comparisons while that based on U~42 indicates possible kurtosis differences from the IG distribution in the failure time example. For the precipitation data the test based on log(TKn) is most critical of the fit of the IG model. In fact apart from the tests based on U^32 and U^42, all of the tests studied here had something to recommend them: reasonable power or interpretability. No test was uniformly superior to the others.



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*Source: 150303-2012-11-11.xml*