Abstract

We study the array of partial sums, PX, of a given array X in terms of its h-type indices. Concretely, we show that h(PX) can be described in terms of the Lorenz curve of the array X and obtain a relation between the sum of the components of PX and the Gini index of X. Moreover, we obtain sharp lower and upper bounds for h-type indices of PX.

1. INTRODUCTION

h-type indices such as the h-index itself and the g-index have interesting mathematical properties as shown, for example, in Egghe and Rousseau (2019a), although they are only probably approximately correct (PAC) in research evaluation exercises (Bouyssou & Marchant, 2011; Waltman & van Eck, 2012; Rousseau, 2016). In this investigation we continue our theoretical investigation of the mechanism leading to h-type indices. Concretely, we study properties related to h-type indices of the array of partial sums of a given array X. We recall that these partial sums form the basis of the Lorenz curve and the related Gini index. As a consequence, we also obtain relations with the Gini index and the Lorenz curve of the original array X. We will further derive sharp lower and upper bounds for h-type indices of PX.

In the following section we recall the definitions we will use in this investigation.

2. DEFINITIONS

Let (R+)N be the set of all arrays of length N with nonnegative real values. An array X = (xr)r=1,2,…,N in (R+)N is said to be decreasing if, for all r = 1, 2, …, N, xrxr+1. The set (R+)N has a natural partial order defined by XY if, for all r = 1, 2, …, N, xryr. Equality between X and Y only occurs if xr = yr for all r. We denote the set of all decreasing arrays in (R+)N with at least one component larger than or equal to 1 by ΦN.

Next we recall the definition of some h-type indices for arrays in (R+)N.

2.1. The h-index (Hirsch, 2005)

Let X = (xr)r=1,2,…,N ∈ ΦN. The h-index of X, denoted h(X), is the largest natural number such that the first h coordinates have each at least a value h. If all components of a decreasing array X are strictly smaller than 1, then h(X) = 0. Such arrays are not considered further on because we will work with arrays in ΦN. If xN, the last element in X, is larger than or equal to N, then h(X) = N.

Recall that an h-index (and similarly for the other h-type indices defined further on) can only be defined for decreasing arrays. Moreover, for rh(X), xrr; conversely, if for all rn, xrn, then nh(X). Further, if r > h(X), then xr < h(X) + 1.

2.2. The g-index (Egghe, 2006a, b)

Let X = (xr)r=1,2,…,N ∈ ΦN. The g-index X, denoted g(X), is defined as the highest natural number g such that the sum of the first g coordinates is at least equal to g2. If the sum of all coordinates of X is strictly larger than N2, then we extend the array X with coordinates equal to zero, making it into an array in ΦM, M > N, until it is possible to apply the definition.

2.3. The R-index (Jin et al., 2007)

Let X = (xr)r=1,2,…,N ∈ ΦN. The R-index of X is defined as the square root of the sum of all coordinates up to and including the one with index h(X). Omitting the square root yields the R2-index. As it is easier to work with R2 than with R if their properties are for our purposes the same, all concrete examples will be given for R2.

2.4. Kosmulski’s h(2) Index (Kosmulski, 2006)

Let X = (xr)r=1,2,…,N ∈ ΦN. The h(2) or Kosmulski index of X, denoted h(2)(X), is the largest natural number h(2) such that the first h(2) coordinates have each at least a value (h(2))2.

2.5. The Majorization Order (Hardy et al., 1934)

Let X, Y ∈ ΦN, where N is any finite number in N = {1, 2, 3, …}. The array X is majorized by Y, or X is smaller than or equal to Y in the majorization order, denoted as XY if for all i = 1, …, N
i=1Nxi=i=1Nyiandj=1ixjj=1iyj;i=1,,N.
We note that this definition is also valid for arrays in which all values are between zero (included) and 1 (not included).

3. AN INEQUALITY RELATED TO THE g-INDEX AND THE MAJORIZATION ORDER

Theorem 1. ∀ NN: X, Y ∈ ΦN: X − < Yg(X) ≤ g(Y) (*)

Proof. Although this theorem is implied in Egghe (2009, p. 487) we present here two short proofs.

First proof: For each ig(X), j=1iyjj=1ixji2. This implies that g(Y) ≥ g(X).

Second proof: For each i > g(Y), j=1ixjj=1iyj < i2. Also this inequality implies that g(X) ≤ g(Y).

Comments

  • A. 

    This theorem proves that g is an order-preserving mapping from (R+)N with the majorization order to the positive real numbers with their natural order.

  • B. 

    The converse of inequality (*) does not hold. Consider for instance X = (5, 2, 2) and Y = (4, 4, 1). Then g(X) = g(Y) = 3 but neither XY nor YX holds.

  • C. 

    The inequality (*) can be strict. Indeed, take X = (2, 1, 1) and Y = (2, 2, 0). Then XY, but g(X) = 1 and g(Y) = 2.

  • D. 

    Yet, inequality (*) cannot be strict for N = 2. Indeed, consider X = (x1, x2) and Y = (y1, y2), with XY. Then 1 ≤ x1y1 (hence g(X), g(Y) ≥ 1) and x1 + x2 = y1 + y2. As N = 2, this sum completely determines the value of the g-index. Hence this value must be equal for X and Y. We note that even here there is no upper bound to the value of g(X) = g(Y).

  • E. 

    If it were allowed that x1 < 1 then the previous Comment D is not valid. Indeed, take X = (½, ½) and Y = (1, 0) then XY, and g(X) = 0 and g(Y) = 1.

  • F. 

    Inequality (*) does not hold for the h- or the R2-index. Consider X = (3, 3, 3) and Y = (5, 2, 2). Then XY but h(X) = 3 > h(Y) = 2. Moreover, R2(X) = 9 > R2(Y) = 7.

For small N we even have the opposite relation for the h-index. This is shown in the next proposition.

Proposition 1.

  • (a) 

    For X, Y ∈ Φ2 and XY, h(X) ≥ h(Y)

  • (b) 

    For X, Y ∈ Φ3, XY and if the components of X and Y are strictly positive natural numbers, then h(X) ≥ h(Y).

Proof.

  • (a). 

    N = 2, XY then x1y1 and x1 + x2 = y1 + y2. Hence x2y2. If now h(X) = 1, then 1 ≤ y1 and 2 > x2y2. This implies that h(Y) = 1 = h(X). The case h(X) = 2 is trivial: As N = 2, it follows that h(Y) ≤ 2 = h(X).

  • (b). 

    N = 3 and XY, then x1y1; x1 + x2y1 + y2 and x1 + x2 + x3 = y1 + y2 + y3. This already implies that x3y3. We now consider three cases: h(Y) = 3, h(Y) = 2 and h(Y) = 1.

Assume first that h(Y) = 3. Then y3 ≥ 3. Hence we see that x1x2x3y3 ≥ 3, from which we derive that h(X) = 3 = h(Y).

Assume next that h(Y) = 2. Then y3 < 3, hence y3 = 2 or y3 = 1, and y1y2 ≥ 2. We first consider the case y3 = 1. We know already that x3 is at least equal to 1. So, if x3 is equal to 1, then x1 + x2 = y1 + y2. As x1y1 the previous equality implies that y2x2. Now h(Y) = 2 leads to 2 ≤ y2x2, or h(X) ≥ 2 = h(Y). Still with y3 = 1 we now consider the case that x3 > 1. Then x1x2x3 ≥ 2, leading to h(X) ≥ 2 = h(Y).

Next we consider the case y3 = 2. Then x1x2x3y3 = 2, which implies that h(X) ≥ 2 = h(Y).

Finally, as components are assumed to be strictly positive natural numbers, h(X) and h(Y) are at least equal to 1. Hence h(Y) = 1 implies h(X) ≥ h(Y).

Comments

  • A. 

    Proposition 1(b) is not valid if some of the components are zero. This is illustrated as follows. Let X = (2, 1, 1) and Y = (2, 2, 0). Then XY, but h(X) = 1 and h(Y) = 2.

  • B. 

    Proposition 1(b) is also not valid if some of the components are not natural numbers. Indeed, let X = (2, 1.5, 1.5) and Y = (2, 2, 1). Then XY, but h(X) = 1 and h(Y) = 2.

  • C. 

    Propositions 1(a) and 1(b) are not valid for R2.

    • (a) 

      N = 2. Consider X = (1, 1) and Y = (2, 0). Then XY, h(X) = h(Y) = 1; R2(X) = 1 and R2(Y) = 2.

    • (b) 

      N = 3. Consider X = (2, 2, 2) and Y = (3, 2, 1). Then XY, h(X) = 2 = h(Y) and R2(X) = 4 < R2(Y) = 5.

  • D. 

    D. 6 with N ≥ 4 is not valid for the h-index h.

Consider X = (6, 5, 2, 2) and Y = (6, 5, 3, 1). Then XY, h(X) = 2 < h(Y) = 3. We further remark that R2(X) = 11 < R2(Y) = 14.

4. INTRODUCING THE ARRAY OF PARTIAL SUMS

Now we come to the main part of this article. First we introduce some notation. Let X = (xr)r=1,2,…,N ∈ ΦN and consider the partial sums:
j=1ixj,i=1,,N.
Ranking these partial sums again in decreasing order leads to the array PX. The ith component of PX, denoted as yi, is equal to j=1Ni+1xj. An example: Let X = (4, 3, 2, 1). Then PX = (10, 9, 7, 4) and h(PX) = 4.

Remarks

  • 1. 

    (PX)1 = A = j=1Nxj; (PX)N = x1.

  • 2. 

    If X ends with p zeros, then PX starts with p + 1 As.

  • 3. 

    Clearly XPX, as (PX)N = x1. Hence h(X) ≤ h(PX), g(X) ≤ g(PX) and R(X) ≤ R(PX). (Egghe & Rousseau, 2019a; Proposition 2).

If X denotes the number of received citations of an author’s publications, then the indicator value h(PX) shows how many of the less cited publications can be removed so that the total number of the remaining items of X is higher than the rank of this total in the array PX. This is another way of describing the impact of the most cited publications. Contrary to the case of h(X), h(PX) may increase if a publication in X’s h-core, not necessarily the most cited one, increases its number of citations. We provide an example: let X = (4, 2, 0, 0, 0, 0, 0, 0). Then h(X) = 2; PX = (6, 6, 6, 6, 6, 6, 6, 4) and h(PX) = 6. Consider now X′ = (4, 3, 0, 0, 0, 0, 0, 0). Then h(X′) = 2; PX′ = (7, 7, 7, 7, 7, 7, 7, 4) and h(PX′) = 7.

5. A RELATION WITH THE GINI INDEX

We recall (Rousseau et al., 2018, formula (4.19)) that the Gini concentration index of a decreasing array X of nonnegative real numbers, (xj)j=1,…,N is obtained as
GX=1NN+12Aj=1Njxj,
(1)
where A = j=1Nxj. Consider now PX. The sum of all components of PX, denoted as S(X), is
j=1Nxj+j=1N1xj++j=11xj=j=1NNj+1xj=N+1Aj=1Njxj.
From this result we obtain a relation between G(X) and S(X):
GX=1NN+12AN+1ASX=1NN+12N+1+2SXA=1NA2SXAN+1.
(2)
Conversely
SX=A2N1+GX+1.
(3)
An example: If X = (a, a, a, a), a > 0, then G(X) = 0 (by definition), N = 4, A = 4a and S(X) = 10a. Now we check formula 3 and find that, indeed, 10a = 4a2(4 + 1).

6. A GEOMETRIC INTERPRETATION OF h(PX) IN TERMS OF THE LORENZ CURVE LX

For the decreasing array X of nonnegative real numbers, (xj)j=1,…,N and for aj = xjk=1Nxk = xjNx¯, the Lorenz curve of X, denoted as LX, connects points with coordinates s=iNj=1iaj. The average of array X is denoted as x¯=1Ni=1Nxi.

Now h(PX) is equal to the largest natural number i such that yi = j=1Ni+1xji, which is also equal to (N + 1) minus the smallest natural number i such that yNi+1 = j=1ixjNi + 1. Dividing by the sum of all elements in X this yields
j=1iajNi+1N*x¯=1x¯1s+1N.
Consequently, h(PX) is equal to N (1 − the smallest s such that LX(s) ≥ 1x¯1s+1N) + 1.

An illustration: If X = (3, 2, 1, 0), N = 4, x¯ = 32, PX = (6, 6, 5, 3) and h(PX) = 3. Now LX14 = 36 < 23114+14 = 23; but LX24 = 5623124+14 = 24. Hence, the smallest s is equal to 2/4 and h(PX) = 4(1 − 2/4) + 1 = 2 + 1 = 3.

In Egghe and Rousseau (2019b) we studied h(PX) and its relation with the Lorenz curve in a continuous context. This led to a new geometric interpretation of the h-index.

7. BOUNDS ON h-TYPE INDICES

In the next sections we derive bounds for h-type indices of PX. This is of importance for the following reason. A function relates an input to a unique output. In this way the standard h-index is a function which maps an array to a natural number. Yet it is not an explicit function, such as the function that maps the real number x to x2 + 4x + 7 or the function which maps a finite array to its sum. Finding an h-index needs a procedure and hence it is not possible to study properties in an analytical way (e.g., using integrals). The bounds obtained in this article are explicit functions which can be studied using analytical methods.

We denote by ⌊a, the floor function of a (i.e., the largest integer smaller than or equal to a). We note that a ≥ ⌊a⌋ > a − 1. Using the notation just introduced we come to the following interesting theorem.

Theorem 2. Let X = (xr)r=1,2,…,N ∈ ΦN, then
minNA=minNNx¯hPXN+1x¯x¯+1,
(4)

Before proving 10 we make three remarks:

  • 1. 

    The first inequality, namely min(N, A) ≥ h(PX) is easy to see because, on the one hand, an h-index can never be larger than the length of the array and on the other 1 ≤ h(PX) ≤ j=1NhPX+1xjA.

  • 2. 

    h(PX) = N if and only if x1N.

  • 3. 

    h(PX) = 0 can never occur in our context. Indeed, this may only occur if all components are strictly smaller than 1, which is excluded. Yet, in Egghe and Rousseau (2019c) we showed that formula 4 is also correct in cases for which h(PX) = 0.

Proof of Theorem 2. We only have to show the second inequality. By definition we know that h(PX) is equal to the largest index i such that yi = j=1Ni+1xji. We know that yi = j=1Ni+1xj = (Ni + 1).(the average of (x1, x2, …, xNi+1)) ≥ (Ni + 1). x¯ (as the array X is ranked in decreasing order).

Now, if (Ni + 1). x¯i then certainly yii. Solving this inequality for i leads to
iN+1x¯x¯+1.
As the index i is a natural number it follows that h(PX) ≥ N+1x¯x¯+1. This proves this theorem.

In order to make these bounds more concrete we provide a table (Table 1) for some values of N and x¯ (or A), showing how sharp these bounds often are. Largest differences occur when the average number of items is one.

Table 1. 
Some specific bounds for h(PX) according to Eq. 4
Nx¯
0.10.512510
10 1 ≥ h(PX) ≥ 1 5 ≥ h(PX) ≥ 3 10 ≥ h(PX) ≥ 5 10 ≥ h(PX) ≥ 7 10 ≥ h(PX) ≥ 9 10 ≥ h(PX) ≥ 10 
30 3 ≥ h(PX) ≥ 2 15 ≥ h(PX) ≥ 10 30 ≥ h(PX) ≥ 15 30 ≥ h(PX) ≥ 20 30 ≥ h(PX) ≥ 25 30 ≥ h(PX) ≥ 28 
100 10 ≥ h(PX) ≥9 50 ≥ h(PX) ≥ 33 100 ≥ h(PX) ≥ 50 100 ≥ h(PX) ≥ 67 100 ≥ h(PX) ≥ 84 100 ≥ h(PX) ≥ 91 
200 20 ≥ h(PX) ≥ 18 100 ≥ h(PX) ≥ 67 200 ≥ h(PX) ≥ 100 200 ≥ h(PX) ≥ 134 200 ≥ h(PX) ≥ 167 200 ≥ h(PX) ≥ 182 
Nx¯
0.10.512510
10 1 ≥ h(PX) ≥ 1 5 ≥ h(PX) ≥ 3 10 ≥ h(PX) ≥ 5 10 ≥ h(PX) ≥ 7 10 ≥ h(PX) ≥ 9 10 ≥ h(PX) ≥ 10 
30 3 ≥ h(PX) ≥ 2 15 ≥ h(PX) ≥ 10 30 ≥ h(PX) ≥ 15 30 ≥ h(PX) ≥ 20 30 ≥ h(PX) ≥ 25 30 ≥ h(PX) ≥ 28 
100 10 ≥ h(PX) ≥9 50 ≥ h(PX) ≥ 33 100 ≥ h(PX) ≥ 50 100 ≥ h(PX) ≥ 67 100 ≥ h(PX) ≥ 84 100 ≥ h(PX) ≥ 91 
200 20 ≥ h(PX) ≥ 18 100 ≥ h(PX) ≥ 67 200 ≥ h(PX) ≥ 100 200 ≥ h(PX) ≥ 134 200 ≥ h(PX) ≥ 167 200 ≥ h(PX) ≥ 182 

The next theorem shows that the second inequality in 10 becomes an equality for the array X¯ = (x¯, x¯, …, x¯) ∈ (R+)N.

Theorem 3. Let X = (xr)r=1,2,…,N ∈ ΦN, then
minNA=minNNx¯hPX¯=N+1x¯x¯+1,
(5)

Proof. We see that PX¯ = (Nx¯, (N − 1) x¯, …, 2x¯, x¯). Then h(PX¯) is the largest natural number i such that (Ni + 1) x¯i. We observe that then h(PX¯) is equal to the largest natural number i such that i ≤ (N + 1) x¯x¯+1 and hence h(PX¯) = N+1x¯x¯+1. This proves 12.

We next present some examples, illustrating different aspects of the previous results.

Example 1. Returning to the example introduced before, we have X = (4, 3, 2, 1), with x¯ = 2.5 and PX = (10, 9, 7, 4). Now N = h(PX) = 4 > N+1x¯x¯+1 = 5×2.52.5+1 = ⌊3.571⌋ = 3. This illustrates that the second inequality in 10 can be strict. Continuing now with X¯ we see that N = 4 > h(PX¯) = h(10, 7.5, 5, 2.5) = 3 = 5×2.53.5.

Example 2. Consider X = (4, 2, 1, 1) with x¯ = 2 and PX = (8, 7, 6, 4). Now N = h(PX) = 4 > N+1x¯x¯+1 = 5×22+1 = 103 = 3. Continuing with X¯ we see that N = 4 > h(PX¯) = h(8, 6, 4, 2) = 3 = 5×23 = 103. This example illustrates that the floor function is really needed, because 3 < 10/3.

Example 3. Let X = (4, 0, 0, 0) with x¯ = 1 and PX = (4, 4, 4, 4). Now N = h(PX) = 4 > 5×11+1 = 5×12 = ⌊2.5⌋ = 2. This is another example that the second inequality in 10 can be strict. Continuing with X¯ we see that N = A = 4 > h(PX¯) = h(4, 3, 2, 1) = 2 = 5×12 = ⌊2.5⌋. This is not only another example that the floor function is really needed, but it also illustrates that the first inequality in 12, and hence also in 10, can be strict.

Example 4. In the previous examples h(PX) = N. Next we present an example where h(PX) < N. Let X = (3, 2, 1, 0). Then x¯ = 3/2 and PX = (6, 6, 5, 3). Now N = 4 > h(PX) = 3 ≥ 5×3/23/2+1 = 5×35 = ⌊3⌋ = 3. Continuing with X¯ we see that N = 4 > h(PX¯) = h(6, 4.5, 3, 1.5) = 3 = 5×35 = 3.

Example 5. Finally, we present an example where min(N, A) = A < N. Let X = (2, 0, 0, 0). Then x¯ = ½ and PX = (2, 2, 2, 2). Now A = 2 = min(N, A) = h(PX) = 2 ≥ 5×1/21/2+1 = 5×13 = 1. Continuing with X¯ we see that A = 2 = min(N, A) > h(PX¯) = h(2, 3/2, 1, 1/2) = 1 = 5×13 = 1.

Corollaries

  • A. 

    If x¯N, then h(PX) = N.

Proof. As limt⇒∞tt+1 = 1, there exist a number t0 such that for all t > t0.
NN+1tt+1<N+1.
This double inequality clearly holds if we take t0 = N. With x¯ in the role of t we see that in these circumstances N+1x¯x¯+1 = N and thus by 10, Corollary A is proved.

  • B. 

    limx¯h(PX) = N

This follows immediately from Corollary A.

Remark

When applied to publications, corollaries A and B show that for large x¯ we only need those publications in X with the highest citations to determine h(PX). This is in accordance with the principle and meaning of an h-index.

8. PARTIAL SUMS AND THE G-INDEX

Using the same notations as before, we next prove the analogue of 10 for the g-index. We recall that the g-index has no upper limit.

Theorem 4. Let X = (xr)r=1,2,…,N ∈ ΦN. Then
gPX2N+1x¯x¯+2ifj=1Nyj<N26aNN+1x¯2ifj=1NyjN26b
Proof.
gPX=maxiNj=1iyji2ifj=1Nyj<N2maxiNj=1Nyji2ifj=1NyjN2
Now, j=1iyj=j=1ik=1Nj+1xk=s=1Nxs+s=1N1xs++s=1Ni+1xsNx¯ + (N − 1) x¯ + … + (Ni + 1) x¯ (because X is ordered decreasingly) = x¯ × NN+12NiNi+12 = x¯2i(2Ni + 1).

Now we require that this expression is larger than or equal to i2. This leads to:
i2Ni+1x¯2.
Solving for i yields: i ≤ (2N + 1) x¯x¯+2. Taking into account that g(PX) is an integer, we obtain that if j=1Nyj < N2 then g(PX) ≥ 2N+1x¯x¯+2.

If j=1NyjN2 then we have to study j=1Nyji2. In the same way as above we find that j=1Nyjx¯. NN+12i2 (is all we need). Hence ix¯2NN+1 or imax = x¯2NN+1, where imax denotes the maximal value the index i can take here.

Consequently, if j=1NyjN2 then g(PX) ≥ x¯2NN+1.

Similar to the theory for the h-index, the next theorem shows that inequality in 23 becomes an equality for the array X¯.

Theorem 5. Let X = (xr)r=1,2,…,N ∈ ΦN, then
gPX¯=2N+1x¯x¯+2ifx¯<2NN+17aNN+1x¯2ifx¯2NN+17b

Proof. Now: j=1Nyj=j=1Nk=1Nj+1x¯=s=1Nx¯+s=1N1x¯++s=11x¯ = Nx¯ + (N − 1) x¯ + … + 1. x¯ = x¯. NN+12.

Hence, j=1NNj+1x¯<N2x¯<2NN+1.

Similarly, j=1NNj+1x¯N2x¯2NN+1.

Comment. Also here we can make the remark that lower bounds for g(PX) and g(PX¯) depend only on N and x¯.

Examples

Example 1. Take X = (4, 4, 4, 4), x¯ = 4 and PX = (16, 12, 8, 4). Then g(PX) = 6 (as 40 > 62 and 40 < 72). As this is a case where 40 > 42 we have to check formula 6b. This formula states that 6 = g(PX) ≥ 4×4×52 = 40 = 6. Thanks to the use of the floor function we obtain an equality.

Example 2. Take X = (4, 3, 2, 1), x¯ = 2.5 and PX = (10, 9, 7, 4). Then g(PX) = 5 (as 30 > 52 and 30 < 62). Also here we have to check formula 6b. We see that 5 = g(PX) ≥ 2.5×4×52 = 25 = 5. This is an example where the floor function is not necessary.

Example 3. For X = (4, 0, 0, 0), x¯ = 1 and PX = (4, 4, 4, 4). Here the sum, namely 16, is larger than or equal to N2 = 42; hence we have to check formula 6b. This leads to 4 = g(PX) ≥ 12×12 = 6 = 2. This is another case where we have strict inequality.

Example 4. For X = (2, 0, 0, 0), x¯ = 0.5 and PX = (2, 2, 2, 2). Here the sum, namely 8 < 42, hence we have to check formula (6a). This leads to 2 = g(PX) ≥ 9*0.52.5 = 1. Also here we have strict inequality.

e) Finally we consider a case for which N ≠ 4. Let X = (5, 4, 3, 2, 1), x¯ = 3 and PX = (15, 14, 12, 9, 5). Here the sum namely 55 > 52; hence we check formula 6b. We first note that g(PX) = 7 (55 > 72 and 55 < 82). Now 7 = g(PX) ≥ 3×5×62 = 45 = 6. This is again a case with a strict inequality.

Corollary
limx¯gPX=

Proof. If x¯ is large, then we have to consider formula 6b. Then the right-hand side of formula 6b becomes unlimited large and hence this also holds for g(PX). This result confirms the fact that the g-index has no upper limit.

9. PARTIAL SUMS, THE R(R2)-INDEX AND KOSMULSKI’S INDEX H(2)(X)

In the previous sections we studied the h-index and the g-index. As a final case we mention the R2-index and Kosmulski’s h(2)-index. For proofs of the results we refer the reader to Egghe and Rousseau (2019c).

Theorem 6. Let X = (xr)r=1,2,…,N ∈ ΦN. Then
R2PX>x¯×N+12x¯+12×N×x¯2+2N×x¯x¯2.
(8)

Theorem 7. If (N + 1) x¯x¯+1 is a natural number and X = (xr)r=1,2,…,N ∈ ΦN, then
R2PX¯=x¯2N+1Nx¯+2N+12x¯+12.
(9)
Finally, we extend our results to the case of Kosmulski’s index, denoted as h(2), referring to Egghe and Rousseau (2019c) for proofs.

Theorem 8. Let X = (xr)r=1,2,…,N ∈ ΦN. Then
minNAh2PX12x¯2+4N+1x¯x¯
(10)

Similarly to 12 we have

Theorem 9. Let X = (xr)r=1,2,…,N ∈ ΦN. Then
minNAh2PX¯=12x¯2+4N+1x¯x¯
(11)

10. DISCUSSION AND CONCLUSION

In this article we studied arrays of partial sums, PX, of a given array X in terms of their h-type indices. We showed that h(PX) can be described in terms of the Lorenz curve of the array X. Moreover, we obtained sharp lower and upper bounds for these h-type indices. We found bounds that only depend on N, the length of the array, and the average of array X, or equivalently, on the length of the array and the total sum of all items in the array.

As h(PX) is an h-index it is not surprising that it is not strictly independent in the sense of Bouyssou and Marchant (2011). This means that if h(PX) < h(PY) and if one adds to X and Y the same items (X becomes X′, and Y becomes Y′) then it is possible that h(PX′) > h(PY). An example: Let X = (2, 0, 0, 0, 0) and Y = (1, 1, 1, 1, 1). Then PX = (2, 2, 2, 2, 2) with h(PX) = 2, and PY = (5, 4, 3, 2, 1) with h(PY) = 3, hence h(PX) < h(PY). Adding 5 times 1 to each of them yields X′ = (2, 1, 1, 1, 1, 1, 0, 0, 0, 0), PX′ = (7, 7, 7, 7, 7, 6, 5, 4, 3, 2) with h(PX) = 6, and Y′ = (1, 1, 1, 1, 1, 1, 1, 1, 1, 1), PY = (10, 9, 8, 7, 6, 5, 4, 3, 2, 1) with h(PY) = 5, hence h(PX) > h(PY).

A reviewer asked if h(PX) can be described in terms of Vannucci’s (2010) dominance dimension. It can: Using Vannucci’s notation we see that h(PX) = domprojnnj=1N+1nxjnn1n, where X = (xn)n=1,…,N is an array of length N.

Our investigation illustrated the rich mathematical structure hidden in the mechanism leading to h-type indices (see also Egghe & Rousseau, 2019d). In this article we considered the discrete case, requiring the floor function in order to get the correct results. In further research we intend to study the continuous case, where by definition no floor function will be needed. Then, bounds will be differentiable and integrable functions.

AUTHOR CONTRIBUTIONS

Leo Egghe: conceptualization; formal analysis; investigation; methodology; writing—original draft; writing—review and editing. Ronald Rousseau: validation; writing—review and editing.

COMPETING INTERESTS

The authors have no competing interests.

FUNDING INFORMATION

No funding has been received.

DATA AVAILABILITY

Not applicable.

ACKNOWLEDGMENTS

The authors thank anonymous reviewers for useful suggestions to improve the presentation of this article.

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Author notes

Handling Editor: Vincent Larivière

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