Directory UMM :Data Elmu:jurnal:O:Operations Research Letters:Vol26.Issue5.2000:

Operations Research Letters 26 (2000) 237–245
www.elsevier.com/locate/dsw

Comparison of probability measures: Dominance of the
third degree
Lars Thorlund-Petersen ∗
Department of Operations Managements, Copenhagen Business School, Solbjerg Pl. 3, DK-2000 Frederiksberg, Denmark
Received 1 August 1997; received in revised form 1 March 2000

Abstract
This note provides necessary and sucient conditions for third-degree stochastic dominance between the probability
measures on a nite grid of n points. Verifying these conditions involve n linear and n quadratic inequalities; the arithmetic
operation of division is not required. The results are compared with those of Fishburn and Lavalle, Math. Operat. Res. 20
(1995) 513–525 and Milne and Neave, Management Sci. 40 (1994) 1343–1352. Finally, the solution to an open problem
c 2000 Elsevier Science B.V. All rights
in Fishburn and Lavalle, Math. Operat. Res. 20 (1995) 513–525 is presented.
reserved.
Keywords: Decision theory; Stochastic dominance; Third-degree dominance

1. Introduction
Stochastic-dominance rules are widely used in analysis of decision making under risk or uncertainty, see,

e.g., Ref. [3]. Such rules are dened by partial specication of the decision maker’s preferences. In the usual
framework of the expected-utility model and unidimensional random incomes or payos, dominance of the
rst degree corresponds to the assumption that more income is preferred to less; dominance of the second
degree corresponds to the stronger assumption that the decision maker in addition is risk averse. In many
applications further restrictions on preferences are assumed, such as “aversion to downside risk”, see Ref. [4].
This assumption corresponds to the dominance of the third degree dened in Ref. [9], which is necessary
for the dominance of any higher degree. For example, by considering the dominance of the third degree and
Also at: Bodo= Graduate School of Business, N-8002 Bodo=, Norway.
Correspondence address: Copenhagen Business School, Department of Operations Management, Solbjerg Pl. 3, DK-2000 Frederiksberg,
Denmark.
E-mail address: ltp.om@cbs.dk (L. Thorlund-Petersen).
∗

c 2000 Elsevier Science B.V. All rights reserved.
0167-6377/00/$ - see front matter
PII: S 0 1 6 7 - 6 3 7 7 ( 0 0 ) 0 0 0 3 1 - 6

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L. Thorlund-Petersen / Operations Research Letters 26 (2000) 237–245

beyond, Ritchken and Kuo in Ref. [6] have obtained stronger option-pricing bounds than those of (simple)
risk aversion.
In the following, suppose that N = {1; 2; : : : ; n} is a given nite set of states of the world. Correspondincome and p = (p1 ; : : : ; pn ); q =
ingly, let x = (x1 ; : : : ; x n ); y = (y1 ; : : : ; yn ) be n-vectors ofPrandom P
(q1 ; : : : ; qn ) be probability measures on 2N , thus pi ; qi ¿0;
p
=
i
N
N qi = 1. Furthermore, let U be a
given set of functions on the real line. Then (y; q) (stochastically) dominates (x; p) with respect to U if and
only if
n
X

u(xi )pi 6

n
X

u(i)ti 60

k
X

ti ¿0

(3a)

(k + 1 − i)ti ¿0;

(3b)

i=1

n
X

u(yi )qi ;

for all u ∈ U:

(1)

i=1

Predominantly, three classes of utility functions commonly studied in the literature are considered here. Let
U 1 be the class of increasing functions, U 2 ⊂ U 1 the subclass of concave functions, and U 3 ⊂ U 2 the further subclass of functions having a convex derivative. If U equals U 1 ; U 2 , or U 3 , then dominance with
respect to U is called dominance of the rst, second, or third degree, respectively. The third degree generally is more dicult to analyse than the rst and second degree and it is of interest to consider special cases such as random incomes being conned to a nite set of equally spaced points as in Refs.
[2,5].
Therefore, consider in (1) the case x = y and x2 − x1 = · · · = x n − x n−1 ¿ 0. Without loss of generality, it is
furthermore assumed throughout that x = (1; 2; : : : ; n), thus all possible incomes xi lie in the grid {1; 2; : : : ; n}.
This distributional assumption may be interpreted as an approximation of a more general framework, for
example, such an approximation argument is implicit in Ref. [6]. On the other hand, in some cases only
integer values are relevant, hence the grid assumption follows naturally as for example in Ref. [8].
Given the above assumptions, (1) denes a dominance relation on the given set of probability measures.
Thus dominance of p by q with respect to U means that
for all u ∈ U;

(2)

i=1

P
convex cone of directions t for which N ti = 0
where t = p − q. For U = U  ;  = 1; 2; 3; let C  denote the
P
and (2) holds. Clearly, t ∈ C  entails that the mean Et = N iti satises Et60.
Necessary and sucient conditions for rst- and second-degree dominance follows from Refs. [2,5]: q
dominates p of the rst or second degree if and only if t = p − q for k = 1; : : : ; n − 1 satises

i=1

or

k
X
i=1

P
1
2
respectively. Accordingly, if
N ti = 0, then t ∈ C or t ∈ C if and only if t satises (3a) or (3b). In
1
2
particular, C and C are nite cones for any n.
In Ref. [2] the authors study dominance with respect to a set of utility functions U which is dened with
regard to higher dierences calculated on the grid under consideration, hence U depends on this grid. Such
dominance is called partial-sums dominance in Ref. [2] and is, in eect, also studied in Ref. [5]. While
partial-sums dominance of the rst and second degree coincide with (true) dominance, this is not the case
for the third degree. In this note the emphasis is on dominance and partial-sums dominance of the third
degree, except for Section 5. As pointed out in Ref. [2], partial-sums dominance is sucient for dominance
and computationally simple to check. In spite of this, the exact relationship between such sucient conditions

L. Thorlund-Petersen / Operations Research Letters 26 (2000) 237–245

239

and stochastic dominance has so far not been fully explored in the literature. This note contains a complete
characterization for the third degree, in terms of necessary and sucient conditions for a given dierence
t = p − q determining a direction of the cone C 3 .
The main result on third-degree dominance is given in Section 2, followed by numerical examples in Section
3. In Section 4 the extreme directions of the cone C 3 are discussed and in Section 5 the results of Section 2
are used to solve an open problem in Ref. [2].
The present characterization of third-degree dominance is possible basically due to the following simple
geometric property of the graph of a quadratic function. Consider real numbers A; B; C where A 6= 0, and let f
be the quadratic function f()=A2 +B+C. For given , let L ()=f′ () (−)+f()=(2A+B)−A2 +C
denote the ane approximation (or tangent) of f at . It is a simple exercise to show that for any two points
 ¡ , the equality L () = L () holds if and only if  bisects the interval [; ];  = ( + )=2. In summary,
a quadratic function has the tangential bisection property.

2. Third-degree dominance
In studying third-degree stochastic dominance, one encounters the diculty that the cone C 3 fails to be
nite for n¿4. In spite of this, there exists a rather simple set of conditions for t ∈ C 3 which is in nature
quadratic. First, for k = 1; : : : ; n − 1 dene the linear form in t1 ; : : : ; tk−1 by
Vk (t) =

k−1

X

(k − i + 1)(k − i)ti ;

(4)

i=1

where V1 (t) = 0. Secondly, for k = 1; : : : ; n let Dk (t) denote the quadratic form in t1 ; : : : ; tk determined by the
k × k matrix −2[(i − j)2 ]ki; j=1 ,
Dk (t) = −2

k
k X
X

(i − j)2 ti tj ;

(5)

i=1 j=1

thus, D1 (t) = 0; D2 (t) = −4t1 t2 .
The main result can now easily be stated. If Et60, t ∈ C 3 if and only if the largest of the two numbers
Vk (t); −Dk (t) is nonnegative for k = 2; : : : ; n − 1 (see Theorem 1). However, some further concepts must be
introduced before Theorem 1 can be proved.
It suces to check (2) against functions u which generate an extreme direction in U 3 . These directions
are well known to be determined by the functions of the following form. For every real number , dene
u () = −( − )2 if 6 and u () = 0 otherwise; for  = ∞, dene u () = . Thus, in order for t ∈ C 3 to
hold, (2) must be satised for every such function u .
P
For given n-vector t, dene the function ft on the real line by ft () = − N u (i)ti . It is readily seen that
for any k ∈ {1; : : : ; n − 1}, one has

0;
61;



(6)

ft () = ( − 1)2 t1 + ( − 2)2 t2 + · · · + ( − k)2 tk ; k ¡ 6k + 16n;



2
2
2
( − 1) t1 + ( − 2) t2 + · · · + ( − n) tn ; n ¡ :
The piecewise quadratic function ft consists of at most n+1 pieces determined by the subdivision −∞; 1; 2; : : : ;
′′
n; ∞ of the real line. Furthermore, ft has a piecewise ane derivative ft′ at every
P  and a second derivative ft
at  6= 1; 2; : : : ; n. Moreover, ft is constant on 61 and ane on  ¿ n since N ti =0. The derivatives satisfy

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L. Thorlund-Petersen / Operations Research Letters 26 (2000) 237–245

for  ¿ 1:
(

( − 1)t1 + ( − 2)t2 + · · · + ( − k)tk ; k ¡ 6k + 16n;

1
2

ft′ ()

1
2

ft′′ () = t1 + t2 + · · · + tk ;

=

−Et;

n ¡ ;

k ¡  ¡ k + 16n:

(7)

(8)

It is useful to consider some examples. Firstly, if t = (1; −1; 0; : : : ; 0) ∈ C 1 , then ft is convex for 1 ¡ 62
and ane for  ¿ 2. Secondly, if t = (1; −2; 1; 0; : : : ; 0) ∈ C 2 , then ft is convex for 1 ¡ 62, concave for
2 ¡ 63, and constant for  ¿ 3. Thirdly, for t = (1; −3; 3; −1; 0; : : : ; 0) ∈ C 3 ; ft is convex–concave–convex
on 1 ¡ 62; 2 ¡ 63; 3 ¡ 64 and zero for  ¿ 4. Furthermore, ft is symmetric at  = 2:5 and ft () ¿ 0
if and only if 1 ¡  ¡ 4.
Dominance of the rst, second, and third degree can
in terms of conditions on
P conveniently be expressed
2
t
=
0,
then
t
∈
C
(or
t
∈ C 1 ) if and only if the
ft . It follows from (3), (7), and (8) that whenever
N i
function ft is nonnegative and increasing (or nonnegative, increasing, and convex). Since it suces to check
(2) against the functions u , then t ∈ C 3 holds if and only if ft is nonnegative.
The remaining part of this section is organized as follows. Firstly, for every function ft , an auxiliary
function f̂t is dened and analysed. This function has the important property that it being nonnegative is
equivalent to third-degree partial-sums dominance. Secondly, it turns out that in order to test nonnegativity of
ft , it suces to consider  for which f̂t () ¡ 0. In this manner the exact relationship between the dominance
and partial-sums dominance is established.
2.1. Third-degree partial-sums dominance and the function f̂t
For any ft in (6), the function f̂t is dened as follows. The ane approximation of ft at any integer j is
given by Lj (; t)=ft′ (j)(−j)+f(j). If  ¡ 1 or n ¡ , then dene f̂t ()=ft () and for 16k ¡ 6k +16n,
dene
(
max{Lk (; t); Lk+1 (; t)}; t1 + · · · + tk ¿0;
(9)
f̂t () =
min{Lk (; t); Lk+1 (; t)}; t1 + · · · + tk ¡ 0:
By (9), f̂t is piecewise ane and f̂t (k) = ft (k) for k ∈ {1; : : : ; n}. The number ˆ is called an in
ection point
of f̂t if for any  ¿ 0; f̂t is nonane on [ˆ − ; ˆ + ]. Recall from Section 1 that any quadratic function has
the tangential bisection property. Therefore, the function f̂t is piecewise ane with respect to a particularly
neat subdivision.
Lemma 1. The set of in
ection points of f̂t is contained in {k + 1=2|k = 1; : : : ; n − 1} and the value of f̂t at
any possible in
ection point k + 1=2 equals Vk (t) dened in (4).
Proof. First, assume for given k ∈ {1; : : : ; n − 1} that 2(t1 + · · · + tk ) = ft′ (k + 1) − ft′ (k) 6= 0, hence, ft is
quadratic and nonane on the interval [k; k + 1]. Then there exists ˆk ∈ ]k; k + 1[, uniquely determined by
Lk (ˆk ; t) = Lk+1 (ˆk ; t), hence ˆk − k = (ft′ (k + 1) − ft (k + 1) + ft (k))=2(t1 + · · · + tk ) = 21 , the latter equality by
(6) and (7). Secondly, if t1 +· · ·+tk =0, then f̂t ()=ft () is ane on  ∈ [k; k +1]. This proves that f̂t has at
most n − 1 in
ection points k + 12 ; k = 1; : : : ; n − 1. Finally, f̂t (k + 21 ) = Lk (k + 21 ; t) = ( 21 )ft′ (k) + ft (k) = Vk (t).

L. Thorlund-Petersen / Operations Research Letters 26 (2000) 237–245

241

As an example, if n = 5, then f̂t is nonnegative if Vk (t)¿0 for k = 2; 3; 4 and −Et¿0, thus,
 
  

 t1

0
V2 (t)
1 0 0 0 0  
t


2
 3 1 0 0 0     V (t)   0 
   3   

2
¿ ;
 t3  = 
 6 3 1 0 0
   V4 (t)   0 
t
 4
0
−2Et
−1 −2 −3 −4 −5
t5

hence, partial-sums dominance holds by Refs. [2,5]. In general, one has

Proposition 1. Let p; q be given probability measures and t = p − q. Then q third-degree partial-sums
dominates p if and only if the auxiliary function f̂t is nonnegative; a property which again is equivalent to
(10):
V2 (t); : : : ; Vn−1 (t); −Et¿0:

(10)

Proof. By Lemma 1, the piecewise ane function f̂t is nonnegative if and only if (10) holds. Furthermore,
(10) is equivalent to conditions for third-degree partial-sums dominance in Refs. [2,5].
2.2. Third-degree dominance and the functions ft , f̂t
If 16k ¡ 6k + 16n, then by (6)
ft () = (t1 + t2 + · · · + tk )2 − 2(t1 + 2t2 + · · · + ktk ) + (t1 + 4t2 + · · · + k 2 tk ):
The discriminant of this second-degree polynomial in  is by denition:
4(t1 + 2t2 + · · · + ktk )2 − 4(t1 + t2 + · · · + tk )(t1 + 4t2 + · · · + k 2 tk ):

(11)

Lemma 2. For every t; the discriminant (11) equals Dk (t) dened in (5). Furthermore; for k = 1; : : : ; n − 1;
one has the iterative formula
Dk+1 (t) = Dk (t) − 4ft (k + 1)tk+1 :

(12)

Proof. In (11) the product corresponding to i; j equals −4(i − j)2 ti tj . Furthermore, ft (k + 1) = k 2 t1 + (k − 1)2 t2
+ · · · + tk , hence, (12) follows from (5).
The characterization of third-degree dominance can now be proved.
Theorem 1. Let p; q be given probability measures and t=p−q. Then q third-degree stochastically dominates
p if and only if the function ft is nonnegative; a property which again is equivalent to
max{Vk (t); −Dk (t)}¿0;
− Et¿0:

k = 2; : : : ; n − 1;

(13a)
(13b)

Proof. Equivalence of third-degree dominance and nonnegativity of ft has already been established. Thus,
suppose that ft is nonnegative. Clearly (13b) holds. If Vk (t) ¡ 0 for some k = 2; : : : ; n − 1, then as ft is
quadratic on the interval [k; k + 1] and ft (k); ft (k + 1)¿0, it follows that ft′ (k)606ft′ (k + 1), see (9).
Therefore, ft is quadratic, convex, and nonnegative on this interval. Thus, by Lemma 2, it must be the case
that Dk (t)60, proving (13a).

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L. Thorlund-Petersen / Operations Research Letters 26 (2000) 237–245
Table 1
35tk

14

−27

0

21

0

0

−8

Vk (35t)
Dk (35t)

0
0

28
1512

30
1512

6
0

−2
0

3
0

∗
576

Table 2
tk

0.5

−1:75

3.5

−2:25

Vk (t)
Dk (t)

0
0

1
3.5

−0:5
0

∗
9

On the other hand, suppose that (13) holds. Then Et60 and ft is nonnegative outside the interval [1; n].
If ft attains its global minimum at ∗ ∈ [k; k + 1] ⊂ [1; n] and ft (∗ ) ¡ 0, then k¿2 and ft is quadratic and
convex on [k; k + 1], hence, by Lemma 2, Vk (t) ¡ 0 ¡ Dk (t); this contradicts (13a).

3. Examples
The analysis in Ref. [2] of partial-sums dominance is in part motivated by considerations of computational
simplicity. By Proposition 1 and Theorem 1 it follows that third-degree stochastic dominance is almost as
simple in this respect. In testing for third-degree dominance, one must in addition to the linear form Vk (t)
calculate the quadratic form Dk (t) and these calculations do not require the arithmetic operation of
P division
involving t. Note that Vn (t) is not dened (hence, the ∗ in Tables 1–5) and Dn (t) = 4(Et)2 as
N ti = 0.
Theorem 1 is illustrated by two examples.
21
27
8
Example 1. Consider as in Ref. [2, p. 522], n = 7; p = ( 14
35 ; 0; 0; 35 ; 0; 0; 0); q = (0; 35 ; 0; 0; 0; 0; 35 ). Then
35Et = −12, and for k = 1; : : : ; 7, see Table 1.

As V5 (t) ¡ 0 and D5 (t) = 0, partial-sums dominance fails by (10) whereas stochastic dominance holds by
(13). The smallest n for which this can happen is n = 4; if t = (0:5; −1:75; 3:5; −2:25), then Et = −1:5, and
for k = 1; : : : ; 4, see Table 2.
n
)i−1 (1 −
Example 2. For  ∈ [0; 1] let p be given by binomial probabilities on the grid {1; : : : ; 7}; pi = ( i−1

)7−i ; i = 1; : : : ; 7, and let q = (0; 16 ; : : : ; 61 ) on {1; : : : ; 7}. The choice between p and q has the following simple
economic interpretation. A decision maker who initially has riskless income 1 dollar faces the choice between
the outcome in dollars equal to the number of heads from six successive tosses of a (possibly nonregular)
coin, or receiving the dollar amount equal to the points resulting from a single throw of a die.
Clearly, if  = 0, then q rst-degree dominates p = (1; 0; : : : ; 0). Using conditions (3a), (3b), or (13),
one nds that q rst-, second-, or third-degree dominates p if 6 0:4475; 6 0:4668, or 60:4846, respectively. For  = 0:4846, binomial probabilities are p = (0:0187; 0:1057; 0:2486; 0:3116; 0:2197; 0:0826; 0:0130);
Et = −0:5924, and for k = 1; 2; : : : ; 7, see Table 3.
Condition (10) is violated at k = 3, whereas (13) holds.

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L. Thorlund-Petersen / Operations Research Letters 26 (2000) 237–245
Table 3
tk

0.0187

−0:0609

0.0819

0.1450

0.0531

−0:0840

−0:1537

Vk (t)
Dk (t)

0
0

0.0375
0.0046

−0:0094
0.0000

0.0232
−0:0040

0.4251
−0:0516

1.3025
0.2387

∗
1.4038

4. Extreme directions
For n = 4, the cones C 1 and C 2 have a nite number of extreme directions determined by the columns of
the matrices


1
0
0


 −1
1
0
;

(14a)
 0 −1
1


0

0

−1

1

0

0


 −2

 1

0

1


0
;
1

−1



−2
1



(14b)

see Ref. [5]. The rst two columns of (14b) correspond to a mean-preserving spread as dened in Ref. [7]
and the last column is a pure income decrease. While (14) can be generalized in a straightforward fashion to
any n, the structure of C 3 is more complicated. A complete characterization of extreme directions for any n
is beyond the scope of this note and only the cases n = 4; 5; 7 are discussed.
Any direction t has support {i | ti 6= 0} and length max{j − i | ti ; tj 6= 0}. For C 1 or C 2 the largest length
of any extreme direction is 1 or 2, independently of n. Apparently, there is no similar upper bound for the
length of extreme directions of C 3 . In order to determine extreme directions, the following lemma is needed.
Lemma 3. If s∗ = (0; : : : ; 0; 1; −3; 3; −1; 0; : : : ; 0) has support j; : : : ; j + 36n; then fs∗ () equals zero outside
the interval ]j; j + 3[. If s is an extreme direction of C 3 of length at least 4; then on every interval
]j; j + 3[ ⊂ [1; n]; fs (0 ) = 0 holds for some 0 in this interval.
Proof. The rst property follows from direct calculation based on (6). Thus, let s be an extreme direction of
length at least 4. If fs () ¿ 0 for all  ∈ ]j; j + 3[, then one would have s − s∗ ∈ C 3 for some suciently
small  ¿ 0. Consequently, since direction s∗ is extreme, then s = s∗ for some  ¿ 0, a contradiction of s
having length at least 4.
As a consequence of Lemma 3, any extreme direction s with s1 ¿ 0, must satisfy D3 (s) = 0; D5 (s) = 0; : : : :
Thus, using the condition D3 (s) = 0, one can determine all extreme directions of C 3 for n = 4:


1−
0


 −3 + 2 + 2
1


(15)
 ; 0661;


3+
−2 


−1 − 2 − 2

1

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L. Thorlund-Petersen / Operations Research Letters 26 (2000) 237–245
Table 4
sk

1

−3:5

7

−12

48

−76:5

36

Vk (s)
Dk (s)

0
0

2
14

−1
0

5
96

−4
0

68
9792

∗
0

compare with (14). For n = 5, the following directions are extreme:


1−
0



 −3 + 2 +  2
1−




2

3+
−3 + 2 +   ; 06; 61;




 −1 − 4 − 3 2
3+


2
2
−1 − 2 − 
2 + 2

(16)

The mean of the rst column of (15) and the second one of (16) equals −2(1 + ). The rst column of
(16) has zero mean and length 4 if 0 ¡  ¡ 1. Furthermore, if 0 ¡ ;  ¡ 1, then the rst column of (15)
(for  = 12 , see Table 2) and both columns of (16) dene directions in C 3 for which partial-sums dominance
does not hold.
Finally, if n = 7, then an extreme direction of length 6 and zero mean can be constructed as follows. For
0 ¡  ¡ 1 choose s1 ; s2 ; s3 by setting s1 = 1; s2 = (−3 + 2 +  2 )(1 − )−1 ; s3 = (3 + )(1 − )−1 . Then s1 ¿ 0,
D3 (s) = 0, fs (3) = 4s1 + s2 ¿ 0, fs (4) = 9s1 + 4s2 + s3 ¿ 0 (see (16)). Choose s4 such that s1 + s2 + s3 + s4 60,
fs (5) = 16s1 + 9s2 + 4s3 + s4 ¿ 0, thus, s4 must satisfy −(1 + 3)2 (1 − )−1 ¡ s4 6 − (1 + )2 (1 − )−1 . Then
s4 fs (4) + s5 fs (5) = 0.
choose s5 such that D5 (s) = 0; by the iterative formula (12), this entails that s5 satises P
7
The construction is completed by determining s6 ; s7 such that s = (s1 ; : : : ; s7 ) satises 1 si = 0 and Es = 0.
For  = 12 and s4 = −12, the extreme direction s is shown in Table 4 for k = 1; : : : ; 7.
Thus, under the assumption of a xed grid, there exist extreme directions of length larger than that of
s∗ = (: : : ; 1; −3; 3; −1; : : :), in contradistinction to a related but dierent result in Ref. [1].
Generally, it is conjectured that for every odd n¿5 there exist extreme directions in C 3 of zero mean and
length n − 1.
5. The solution to an open problem in Ref. [2]
In Ref. [2, p. 523] the authors ask whether it is possible to have third-degree stochastic dominance and not
partial-sums dominance of some higher degree. Due to the tangential bisection property of quadratic functions,
inspection of the graphs of ft and f̂t provides the answer.
First, it follows from Ref. [2, Theorem 2] and (8) that for given measures p; q; t = p − q, and n¿5, q
fourth-degree partial-sums dominates p if and only if (17) holds:
k
X

Vi (t)¿0;

k = 2; : : : ; n − 2;

(17a)

i=1

Vn−1 (t)¿0;

(17b)

− Et¿0:

(17c)

For example, direction s of Table 4 satises Et = 0 and (17); this is no coincidence.

245

L. Thorlund-Petersen / Operations Research Letters 26 (2000) 237–245
Table 5
tk

0

0.5

−1:75

3.5

−2:25

Vk (t)
Dk (t)

0
0

0
0

1
3.5

−0:5
0

∗
9

To see this, consider t such that the function ft is nonnegative, t ∈ C 3 . Then by Lemma 1 the function f̂t
is ane on every interval [k − 1=2; k + 1=2] ⊂ [1; n] and nonnegative at  = k, f̂t (k) = ft (k)¿0. Accordingly,
the values of f̂t at the endpoints on such intervals must satisfy f̂t (k − 21 ) + f̂t (k + 12 ) = 2f̂t (k)¿0. Therefore,
it follows that if t ∈ C 3 , then
Vk−1 (t) + Vk (t)¿0;

k = 2; : : : ; n − 1:

(18)

If furthermore Et = 0, then f̂t must be nonnegative on [n − 1; n] hence Vn−1 (t)¿0. Thus, (17) follows from
(18).
In summary, third-degree dominance of p by q and E(p − q) = 0 implies fourth-degree partial-sums
dominance as well. The provision of zero mean cannot in general be vaiwed; consider for example the second
column of (16) with  = 21 which gives the simple variant of Table 2, for k = 1; : : : ; 5, see Table 5.
As V4 (t) ¡ 0, (17b) is violated and hence fourth-degree partial-sums dominance does not hold.
Acknowledgements
The author has beneted from comments by E.H. Neave and critical remarks from an associate editor have
resulted in several improvements.
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