Measuring centrality by a generalization of degree

Abstract

Network analysis has emerged as a key technique in communication studies, economics, geography, history and sociology, among others. A fundamental issue is how to identify key nodes, for which purpose a number of centrality measures have been developed. This paper proposes a new parametric family of centrality measures called generalized degree. It is based on the idea that a relationship to a more interconnected node contributes to centrality in a greater extent than a connection to a less central one. Generalized degree improves on degree by redistributing its sum over the network with the consideration of the global structure. Application of the measure is supported by a set of basic properties. A sufficient condition is given for generalized degree to be rank monotonic, excluding counter-intuitive changes in the centrality ranking after certain modifications of the network. The measure has a graph interpretation and can be calculated iteratively. Generalized degree is recommended to apply besides degree since it preserves most favourable attributes of degree, but better reflects the role of the nodes in the network and has an increased ability to distinguish among their importance.

Appendix

Proof of Theorem 1

Let $x_\ell = x_\ell (\varepsilon )(N,A)$ and $x_\ell ^{\prime } = x_\ell (\varepsilon )(N,A^{\prime })$ for all $\ell \in N$. It can be assumed without loss of generality that $x_i^{\prime }-x_i \ge x_j^{\prime }-x_j$. Let $s = x_i^{\prime }-x_i$, $t = x_j^{\prime }-x_j$, $u = \max \lbrace x_\ell ^{\prime }-x_\ell : \ell \in N {\setminus } \{ i,j \} \rbrace = x_k^{\prime } - x_k$ and $v = \min \{ x_\ell ^{\prime }-x_\ell : \ell \in N {\setminus } \{ i,j \} = x_m^{\prime } - x_m$. It will be verified that $t \ge u$.

Assume to the contrary that $t < u$. The difference of equations from (1) for node k gives:

$$\begin{aligned}&\left( x_k^{\prime } - x_k \right) + \varepsilon \sum _{\ell \in N {\setminus } \{ i,j,k \} } a_{k \ell } \left[ \left( x_k^{\prime } - x_k \right) - \left( x_\ell ^{\prime } - x_\ell \right) \right] \\&\quad + \varepsilon a_{k i} \left[ \left( x_k^{\prime } - x_k \right) - \left( x_i^{\prime } - x_i \right) \right] + \varepsilon a_{k j} \left[ \left( x_k^{\prime } - x_k \right) - \left( x_j^{\prime } - x_j \right) \right] = d_k^{\prime }-d_k = 0. \end{aligned}$$

Since $u = x_k^{\prime } - x_k \ge x_\ell ^{\prime } - x_\ell$ for all $\ell \in N {\setminus } \{ i,j \}$ and $u \ge t$, we get

$$\begin{aligned} u \le \varepsilon a_{k i} (s-u) \Leftrightarrow u \le \frac{\varepsilon a_{k i}}{1+\varepsilon a_{k i}} s. \end{aligned}$$

(2)

Now take the difference of equations from (1) for node i:

$$\begin{aligned} s + \varepsilon \sum _{\ell \in N {\setminus } \{ i,j \} } a_{i \ell } \left[ s - \left( x_\ell ^{\prime } - x_\ell \right) \right] + \varepsilon \left( x_i^{\prime } - x_j^{\prime } \right) = d_i^{\prime }-d_i = 1. \end{aligned}$$

Since $u \ge x_\ell ^{\prime } - x_\ell$ for all $\ell \in N {\setminus } \{ i,j \}$, we get

$$\begin{aligned} s + \varepsilon \sum _{\ell \in N {\setminus } \{ i,j \} } a_{i \ell } (s-u) + \varepsilon \left( x_i^{\prime } - x_j^{\prime } \right) = s + \varepsilon d_i (s-u) + \varepsilon \left( x_i^{\prime } - x_j^{\prime } \right) \le 1. \end{aligned}$$

An upper bound for u is known from (2), thus

$$\begin{aligned} 1 + \varepsilon \left( x_j^{\prime } - x_i^{\prime } \right) \ge s + \varepsilon d_i (s-u) \ge s + \varepsilon d_i \left( 1- \frac{\varepsilon a_{k i}}{1 + \varepsilon a_{k i}} \right) s = \frac{1 + \varepsilon a_{k i} + \varepsilon d_i}{1 + \varepsilon a_{k i}} s. \end{aligned}$$

(3)

By combining (2) and (3):

$$\begin{aligned} u \!\le \!\frac{\varepsilon a_{k i}}{1\!+\!\varepsilon a_{k i}} \frac{1 \!+\! \varepsilon a_{k i}}{1 \!+\! \varepsilon a_{k i} + \varepsilon d_i} \left[ 1 \!+\! \varepsilon \left( x_i^{\prime } - x_j^{\prime } \right) \right] \!= \!\frac{\varepsilon a_{k i}}{1 + \varepsilon a_{k i} + \varepsilon d_i} \left[ 1 + \varepsilon \left( x_j^{\prime } - x_i^{\prime } \right) \right] . \end{aligned}$$

(4)

Take the difference of equations from (1) for node m:

$$\begin{aligned}&\left( x_m^{\prime } - x_m \right) + \varepsilon \sum _{\ell \in N {\setminus } \{ i,j \} } a_{m \ell } \left[ \left( x_m^{\prime } - x_m \right) - \left( x_\ell ^{\prime } - x_\ell \right) \right] \\&\quad + \varepsilon a_{m i} \left[ \left( x_m^{\prime } \!-\! x_m \right) \!-\! \left( x_i^{\prime } \!-\! x_i \right) \right] \!+\! \varepsilon a_{m j} \left[ \left( x_m^{\prime } \!-\! x_m \right) \!-\! \left( x_j^{\prime } - x_j \right) \right] \! =\! d_m^{\prime }-d_m = 0. \end{aligned}$$

Since $v = x_m^{\prime } - x_m \le x_\ell ^{\prime } - x_\ell$ for all $\ell \in N {\setminus } \{ i,j \}$ and $s \ge t$, we get

$$\begin{aligned} v \ge \varepsilon a_{m i} (s-v) + \varepsilon a_{m j} (t-v) \Rightarrow v \ge \frac{\varepsilon \left( a_{m i} + a_{m j} \right) }{1+\varepsilon \left( a_{m i} + a_{m j} \right) } t. \end{aligned}$$

(5)

The difference of equations from (1) for node j results in:

$$\begin{aligned} t + \varepsilon \sum _{\ell \in N {\setminus } \{ i,j \} } a_{j \ell } \left[ t - \left( x_\ell ^{\prime } - x_\ell \right) \right] + \varepsilon \left( x_j^{\prime } - x_i^{\prime } \right) = d_j^{\prime }-d_j = 1. \end{aligned}$$

Since $v \le x_\ell ^{\prime } - x_\ell$ for all $\ell \in N {\setminus } \{ i,j \}$, we get

$$\begin{aligned} t + \varepsilon \sum _{\ell \in N {\setminus } \{ i,j \} } a_{j \ell } (t-v) + \varepsilon \left( x_j^{\prime } - x_i^{\prime } \right) = t + \varepsilon d_j (t-v) + \varepsilon \left( x_j^{\prime } - x_i^{\prime } \right) \ge 1. \end{aligned}$$

A lower bound for v is known from (5), thus

$$\begin{aligned} 1 + \varepsilon \left( x_i^{\prime } - x_j^{\prime } \right) \le t + \varepsilon d_j \left[ 1- \frac{\varepsilon \left( a_{m i} + a_{m j} \right) }{1 + \varepsilon \left( a_{m i} + a_{m j} \right) } \right] t = \frac{1 + \varepsilon \left( a_{m i} + a_{m j} \right) + \varepsilon d_j}{1 + \varepsilon \left( a_{m i} + a_{m j} \right) } t. \end{aligned}$$

(6)

According to our assumption $t < u$, therefore from (4) and (6):

$$\begin{aligned} \frac{1 + \varepsilon \left( a_{m i} + a_{m j} \right) }{1 + \varepsilon \left( a_{m i} + a_{m j} \right) + \varepsilon d_j} \left[ 1 + \varepsilon \left( x_i^{\prime } - x_j^{\prime } \right) \right] < \frac{\varepsilon a_{k i}}{1 + \varepsilon a_{k i} + \varepsilon d_i} \left[ 1 + \varepsilon \left( x_j^{\prime } - x_i^{\prime } \right) \right] . \end{aligned}$$

Obviously, it does not hold if $\varepsilon \rightarrow 0$. Now an upper bound is determined for the parameter $\varepsilon$. After some calculations we get:

$$\begin{aligned} \frac{1 \!+\! \varepsilon \left( a_{m i} \!+\! a_{m j} + d_i \right) \!+\! \varepsilon ^2 \left[ d_i \left( a_{m i} + a_{m j} \right) - a_{k i} d_j \right] }{1 + \varepsilon \left( a_{m i} + a_{m j} + 2 a_{k i} \!+\! d_i \right) \!+\! \varepsilon ^2 \left[ \left( 2 a_{k i} \!+\! d_i \right) \left( a_{m i} \!+ \!a_{m j} \right) + a_{k i} d_j \right] } \!<\! \varepsilon \left( x_j^{\prime } - x_i^{\prime } \right) . \end{aligned}$$

Let introduce the notations $\alpha = \varepsilon \left( a_{m i} + a_{m j} + d_i \right) + \varepsilon ^2 \left[ d_i \left( a_{m i} + a_{m j} \right) - a_{k i} d_j \right]$ and $y = 1 + 2 \varepsilon a_{ki} + \varepsilon ^2 \left[ 2 a_{k i} \left( a_{m i} + a_{m j} + d_j \right) \right] > 1$. Then the fraction on the left-hand side can be written as $(1 + \alpha ) / (y + \alpha )$. However, $(1 + \alpha ) / (y + \alpha ) > 1 / y$ because $y + \alpha > 0$ and $y > 1$, hence

$$\begin{aligned} \frac{1}{1 + 2 \varepsilon a_{k i} + \varepsilon ^2 \left[ 2 a_{k i} \left( a_{m i} + a_{m j} + d_j \right) \right] } < \varepsilon \left( x_j^{\prime } - x_i^{\prime } \right) . \end{aligned}$$

Here $a_{k i},a_{m i},a_{m j} \le 1$ and $d_j \le \mathfrak {d}$. As $x_j^{\prime } - x_i^{\prime } \le x_j - x_i \le \mathfrak {d} - \mathsf {d}$ from boundedness (Proposition 1):

$$\begin{aligned} 1 < (\mathfrak {d} - \mathsf {d}) \left[ (2\mathfrak {d} + 4) \varepsilon ^3 + 2 \varepsilon ^2 + \varepsilon \right] , \end{aligned}$$

which contradicts to the condition of Theorem 1. $\square$