An Analysis of Winsorized Weighted Means

Abstract

The Winsorized mean is a well-known robust estimator of the population mean. It can also be seen as a symmetric aggregation function (in fact, it is an ordered weighted averaging operator), which means that the information sources (for instance, criteria or experts’ opinions) have the same importance. However, in many practical applications (for instance, in many multiattribute decision making problems) it is necessary to consider that the information sources have different importance. For this reason, in this paper we propose a natural generalization of the Winsorized means so that the sources of information can be weighted differently. The new functions, which we will call Winsorized weighted means, are a specific case of the Choquet integral and they are analyzed through several indices for which we give closed-form expressions: the orness degree, k-conjunctiveness and k-disjunctiveness indices, veto and favor indices, Shapley values and interaction indices. We also provide a closed-form expression for the Möbius transform and we show how we can aggregate data so that each information source has the desired weighting and outliers have no influence in the aggregated value.

Proofs

We first recall the definition and some properties of binomial coefficients (see, for instance, Riordan (1968, pp. 1–3) and Grabisch (2016, p. 3).

Remark 2

Let $m\in \mathbb {N}$ and $k \in \mathbb {Z}$. Then:

• 1. $\displaystyle \left( {\begin{array}{c}m\\ k\end{array}}\right) ={\left\{ \begin{array}{ll} \displaystyle \frac{m!}{k!(m-k)!} &{} \text {if } 0\le k \le m,\\ 0 &{} \text {otherwise}. \end{array}\right. }$

• 2. $\displaystyle \left( {\begin{array}{c}m\\ k\end{array}}\right) = \left( {\begin{array}{c}m\\ m-k\end{array}}\right)$.

• 3. If $0 \le k \le m$, then $\displaystyle \sum _{j=0}^{k} (-1)^{j} \left( {\begin{array}{c}m\\ j\end{array}}\right) = (-1)^{k} \left( {\begin{array}{c}m-1\\ k\end{array}}\right)$.

The following simple remarks on summation will be useful in some of the proofs.

Remark 3

Let $p,q\in \mathbb {N}$, with $p\le q+1$.⁵ Then:

$$\begin{aligned} \sum _{t=p}^{q} t = \sum _{t=1}^{q} t - \sum _{t=1}^{p-1} t = \frac{q(q+1) - p(p-1)}{2}. \end{aligned}$$

Remark 4

Let $\varvec{p}$ be a weighting vector. If $\varnothing \subsetneq A \subseteq N$ and $t\ge 1$, then

$$\begin{aligned} \sum _{\begin{array}{c} T\subseteq A \\ |T|=t \end{array}} \sum _{i\in T} p_i = \left( {\begin{array}{c}|A|-1\\ t-1\end{array}}\right) \sum _{i\in A} p_i. \end{aligned}$$

In particular, when $A = N$ we have

$$\begin{aligned} \sum _{\begin{array}{c} T\subseteq N \\ |T|=t \end{array}} \sum _{i\in T} p_i = \left( {\begin{array}{c}n-1\\ t-1\end{array}}\right) \sum _{i=1}^n p_i = \left( {\begin{array}{c}n-1\\ t-1\end{array}}\right) =\left( {\begin{array}{c}n\\ t\end{array}}\right) \frac{t}{n}. \end{aligned}$$

Remark 5

Let $\varvec{p}$ be a weighting vector and $j\in N$. If $t\ge 1$, then

$$\begin{aligned} \sum _{\begin{array}{c} T\subseteq N{\setminus } \{j\} \\ |T|=t \end{array}} \sum _{i\in T} p_i = \left( {\begin{array}{c}n-2\\ t-1\end{array}}\right) \sum _{\begin{array}{c} i=1 \\ i\ne j \end{array}}^{n} p_i = \left( {\begin{array}{c}n-2\\ t-1\end{array}}\right) (1-p_{j}) = \left( {\begin{array}{c}n-1\\ t\end{array}}\right) \frac{t(1-p_{j})}{n-1}. \end{aligned}$$

Remark 6

Let $\varvec{p}$ be a weighting vector and $j,k\in N$. If $t\ge 1$, then

$$\begin{aligned} \sum _{\begin{array}{c} T\subseteq N{\setminus } \{j,k\} \\ |T|=t \end{array}} \sum _{i\in T} p_i&= \left( {\begin{array}{c}n-3\\ t-1\end{array}}\right) \sum _{\begin{array}{c} i=1 \\ i\ne j,k \end{array}}^{n} p_i = \left( {\begin{array}{c}n-3\\ t-1\end{array}}\right) (1-p_{j}-p_{k})\\&= \left( {\begin{array}{c}n-2\\ t\end{array}}\right) \frac{t(1-p_{j}-p_{k})}{n-2}. \end{aligned}$$

Proof of Proposition 1

Let $\varvec{p}$ be a weighting vector and $(r,s)\in \mathcal {R}$. Since

$$\begin{aligned} {{\,\mathrm{orness}\,}}\!\big (M_{\varvec{p}}^{(r,s)}\big ) = \frac{1}{n-1}\sum _{t=1}^{n-1}\frac{1}{\left( {\begin{array}{c}n\\ t\end{array}}\right) } \sum _{\begin{array}{c} T\subseteq N \\ |T|=t \end{array}} \mu _{\varvec{p}}^{(r,s)}(T), \end{aligned}$$

and $\mu _{\varvec{p}}^{(r,s)}$ is given by expression (1) (or expression (2) when $r+s=n-1$), we distinguish two cases:

• 1. If $r+s=n-1$, then

  $$\begin{aligned} {{\,\mathrm{orness}\,}}\big (M_{\varvec{p}}^{(r,s)}\big ) = \frac{1}{n-1}\sum _{t=s+1}^{n-1} 1 = \frac{n-s-1}{n-1} = \frac{r}{n-1}. \end{aligned}$$

• 2. If $r+s<n-1$, then, by Remarks 4 and 3, we have

  $$\begin{aligned}&{{\,\mathrm{orness}\,}}\big (M_{\varvec{p}}^{(r,s)}\big )\\&\quad = \frac{1}{n-1} \left( \sum _{t=s+1}^{n-r-1}\frac{1}{\left( {\begin{array}{c}n\\ t\end{array}}\right) } \sum _{\begin{array}{c} T\subseteq N \\ |T|=t \end{array}} \sum _{i\in T} p_i + r \right) \\&\quad = \frac{1}{n-1}\left( \frac{1}{n}\sum _{t=s+1}^{n-r-1} t + r \right) \\&\quad = \frac{1}{n-1}\left( \frac{(n-r)(n-r-1) - s(s+1)}{2n} + r \right) \\&\quad = \frac{1}{n-1} \frac{n(n-1) + r(r+1) - s(s+1)}{2n}\\&\quad = \frac{1}{2} + \frac{r(r+1) - s(s+1)}{2n(n-1)}. \end{aligned}$$

  Notice that, when $r+s=n-1$, the previous expression returns $r/(n-1)$. So, it is also valid in the case $r+s=n-1$.

$\square$

Proof of Proposition 2

Let $\varvec{p}$ be a weighting vector, $(r,s)\in \mathcal {R}$, and $k\in N{\setminus } \{n\}$. Since

$$\begin{aligned} {{\,\mathrm{conj}\,}}_{k}\!\big (M_{\varvec{p}}^{(r,s)}\big ) = 1 - \frac{1}{n-k}\sum _{t=1}^{n-k}\frac{1}{\left( {\begin{array}{c}n\\ t\end{array}}\right) } \sum _{\begin{array}{c} T\subseteq N \\ |T|=t \end{array}}\mu _{\varvec{p}}^{(r,s)}(T), \end{aligned}$$

and $\mu _{\varvec{p}}^{(r,s)}$ is given by expression (1) (or expression (2) when $r+s=n-1$), we distinguish the following cases:

• 1. If $n-k \le s$ (or, equivalently, $k\ge n-s$), then

  $$\begin{aligned} {{\,\mathrm{conj}\,}}_{k}\!\big (M_{\varvec{p}}^{(r,s)}\big ) = 1. \end{aligned}$$

• 2. If $s< n-k < n-r$ (or, equivalently, $r< k < n-s$), by Remarks 4 and 3 we have

  $$\begin{aligned} {{\,\mathrm{conj}\,}}_{k}\!\big (M_{\varvec{p}}^{(r,s)}\big )&= 1 - \frac{1}{n-k} \sum _{t=s+1}^{n-k}\frac{1}{\left( {\begin{array}{c}n\\ t\end{array}}\right) } \sum _{\begin{array}{c} T\subseteq N \\ |T|=t \end{array}} \sum _{i\in T} p_{i} = 1 - \frac{1}{n-k} \sum _{t=s+1}^{n-k} \frac{t}{n} \\&= 1 - \frac{1}{n-k} \frac{(n-k)(n-k+1) - s(s+1)}{2n}\\&= 1 - \frac{n-k+1}{2n} + \frac{s(s+1)}{2n(n-k)} \\&= \frac{n+k-1}{2n} + \frac{s(s+1)}{2n(n-k)}\\&= \frac{n(n-1) + s(s+1) - k(k-1)}{2n(n-k)}. \end{aligned}$$

• 3. If $n-k \ge n-r$ (or, equivalently, $k \le r$), we distinguish two cases:

  • (a) If $r+s=n-1$, then

    $$\begin{aligned} {{\,\mathrm{conj}\,}}_{k}\!\big (M_{\varvec{p}}^{(r,s)}\big ) = 1 - \frac{1}{n-k} \sum _{t=n-r}^{n-k} 1 = 1 - \frac{r-k+1}{n-k} = \frac{n-(r+1)}{n-k}. \end{aligned}$$

  • (b) If $r+s<n-1$, then

    $$\begin{aligned}&{{\,\mathrm{conj}\,}}_{k}\!\big (M_{\varvec{p}}^{(r,s)}\big )\\&= 1 - \frac{1}{n-k} \left( \sum _{t=s+1}^{n-r-1}\frac{1}{\left( {\begin{array}{c}n\\ t\end{array}}\right) } \sum _{\begin{array}{c} T\subseteq N \\ |T|=t \end{array}} \sum _{i\in T} p_{i} + r-k+1 \right) \\&= 1 - \frac{1}{n-k} \left( \sum _{t=s+1}^{n-r-1} \frac{t}{n} + r-k+1 \right) \\&= 1 - \frac{1}{n-k} \left( \frac{(n-r)(n-r-1)-s(s+1)}{2n} + r-k+1 \right) \\&= 1 - \frac{1}{n-k} \left( \frac{n+1-2k}{2} + \frac{r(r+1)-s(s+1)}{2n} \right) \\&= \frac{n(n-1) + s(s+1) - r(r+1)}{2n(n-k)} . \end{aligned}$$

    Notice also that, when $r+s=n-1$, the previous expression returns $(n-r-1)/(n-k)$. So, it is also valid in the case $r+s=n-1$.

$\square$

Proof of Proposition 3

Let $\varvec{p}$ be a weighting vector, $(r,s)\in \mathcal {R}$, and $k\in N{\setminus } \{n\}$. Since

$$\begin{aligned} {{\,\mathrm{disj}\,}}_{k}\!\big (M_{\varvec{p}}^{(r,s)}\big ) = \frac{1}{n-k}\sum _{t=k}^{n-1}\frac{1}{\left( {\begin{array}{c}n\\ t\end{array}}\right) } \sum _{\begin{array}{c} T\subseteq N \\ |T|=t \end{array}} \mu _{\varvec{p}}^{(r,s)}(T), \end{aligned}$$

and $\mu _{\varvec{p}}^{(r,s)}$ is given by expression (1) (or expression (2) when $r+s=n-1$), we distinguish the following cases:

• 1. If $k \ge n-r$, then

  $$\begin{aligned} {{\,\mathrm{disj}\,}}_{k}\!\big (M_{\varvec{p}}^{(r,s)}\big ) = \frac{1}{n-k}\sum _{t=k}^{n-1} 1 = 1. \end{aligned}$$

• 2. If $s< k < n-r$, by Remarks 4 and 3 we have

  $$\begin{aligned} {{\,\mathrm{disj}\,}}_{k}\!\big (M_{\varvec{p}}^{(r,s)}\big )&= \frac{1}{n-k} \left( \sum _{t=k}^{n-r-1}\frac{1}{\left( {\begin{array}{c}n\\ t\end{array}}\right) } \sum _{\begin{array}{c} T\subseteq N \\ |T|=t \end{array}} \sum _{i\in T} p_{i} + r \right) \\&= \frac{1}{n-k} \left( \sum _{t=k}^{n-r-1} \frac{t}{n} + r \right) \\&= \frac{1}{n-k} \left( \frac{(n-r)(n-r-1) - k(k-1)}{2n} + r \right) \\&= \frac{n(n-1)+r(r+1)-k(k-1)}{2n(n-k)}. \end{aligned}$$

• 3. If $k\le s$, we distinguish two cases:

  • (a) If $r+s=n-1$, then

    $$\begin{aligned} {{\,\mathrm{disj}\,}}_{k}\!\big (M_{\varvec{p}}^{(r,s)}\big ) = \frac{1}{n-k}\sum _{t=n-r}^{n-1} 1 = \frac{r}{n-k}. \end{aligned}$$

  • (b) If $r+s<n-1$, then

    $$\begin{aligned} {{\,\mathrm{disj}\,}}_{k}\!\big (M_{\varvec{p}}^{(r,s)}\big ) = \frac{1}{n-k} \left( \sum _{t=s+1}^{n-r-1}\frac{1}{\left( {\begin{array}{c}n\\ t\end{array}}\right) } \sum _{\begin{array}{c} T\subseteq N \\ |T|=t \end{array}} \sum _{i\in T} p_{i} + r \right) . \end{aligned}$$

    Notice that the above expression coincides with that of the second item when $k=s+1$. Therefore,

    $$\begin{aligned} {{\,\mathrm{disj}\,}}_{k}\!\big (M_{\varvec{p}}^{(r,s)}\big ) = \frac{n(n-1) + r(r+1) - s(s+1)}{2n(n-k)}. \end{aligned}$$

    Notice also that, when $r+s=n-1$, the previous expression returns $r/(n-k)$. So, it is also valid in the case $r+s=n-1$.

$\square$

Proof of Proposition 4

Let $\varvec{p}$ be a weighting vector, $(r,s)\in \mathcal {R}$ and $j\in N$. Since

$$\begin{aligned} {{\,\mathrm{veto}\,}}\!\big (M_{\varvec{p}}^{(r,s)},j\big ) = 1 - \frac{1}{n-1}\sum _{t=1}^{n-1}\frac{1}{\left( {\begin{array}{c}n-1\\ t\end{array}}\right) } \sum _{\begin{array}{c} T\subseteq N{\setminus } \{j\} \\ |T|=t \end{array}} \mu _{\varvec{p}}^{(r,s)}(T), \end{aligned}$$

and $\mu _{\varvec{p}}^{(r,s)}$ is given by expression (1) (or expression (2) when $r+s=n-1$), we distinguish two cases:

• 1. If $r+s=n-1$, then

  $$\begin{aligned} {{\,\mathrm{veto}\,}}\!\big (M_{\varvec{p}}^{(r,s)},j\big ) = 1 - \frac{1}{n-1}\sum _{t=s+1}^{n-1} 1 = 1 - \frac{r}{n-1} = \frac{s}{n-1}. \end{aligned}$$

• 2. If $r+s<n-1$, then, by Remarks 5 and 3, we get

  $$\begin{aligned} {{\,\mathrm{veto}\,}}\!\big (M_{\varvec{p}}^{(r,s)},j\big )&= 1 - \frac{1}{n-1}\left( \sum _{t=s+1}^{n-r-1} \frac{1}{\left( {\begin{array}{c}n-1\\ t\end{array}}\right) } \sum _{\begin{array}{c} T\subseteq N{\setminus } \{j\} \\ |T|=t \end{array}} \sum _{i\in T} p_{i} + r \right) \\&= 1 - \frac{1}{n-1}\left( \frac{1-p_{j}}{n-1} \sum _{t=s+1}^{n-r-1} t + r \right) \\&= 1 - \frac{r}{n-1} - \frac{(1-p_{j})\big ((n-r)(n-r-1)-s(s+1)\big )}{2(n-1)^2}. \end{aligned}$$

  Notice that, when $r+s=n-1$, the previous expression returns $s/(n-1)$. So, it is also valid in the case $r+s=n-1$.

$\square$

Proof of Proposition 5

Let $\varvec{p}$ be a weighting vector, $(r,s)\in \mathcal {R}$ and $j\in N$. Since

$$\begin{aligned} {{\,\mathrm{favor}\,}}\!\big (M_{\varvec{p}}^{(r,s)},j\big ) = \frac{1}{n-1}\sum _{t=0}^{n-1} \frac{1}{\left( {\begin{array}{c}n-1\\ t\end{array}}\right) } \sum _{\begin{array}{c} T\subseteq N{\setminus } \{j\} \\ |T|=t \end{array}} \mu _{\varvec{p}}^{(r,s)}(T\cup \{j\}) - \frac{1}{n-1}, \end{aligned}$$

and $\mu _{\varvec{p}}^{(r,s)}$ is given by expression (1) (or expression (2) when $r+s=n-1$), we distinguish two cases:

• 1. If $r+s=n-1$, then

  $$\begin{aligned} {{\,\mathrm{favor}\,}}\!\big (M_{\varvec{p}}^{(r,s)},j\big ) = \frac{1}{n-1}\sum _{t=s}^{n-1} 1 - \frac{1}{n-1} = \frac{n-s-1}{n-1}= \frac{r}{n-1}. \end{aligned}$$

• 2. If $r+s<n-1$, then, by Remarks 5 and 3, we get

  $$\begin{aligned}&{{\,\mathrm{favor}\,}}\!\big (M_{\varvec{p}}^{(r,s)},j\big )\\&\quad = \frac{1}{n-1} \left( \sum _{t=s}^{n-r-2} \frac{1}{\left( {\begin{array}{c}n-1\\ t\end{array}}\right) } \sum _{\begin{array}{c} T\subseteq N{\setminus } \{j\} \\ |T|=t \end{array}} \left( \sum _{i\in T} p_{i} + p_{j}\right) + r + 1 \right) - \frac{1}{n-1}\\&\quad = \frac{1}{n-1} \left( \frac{1-p_{j}}{n-1} \sum _{t=s}^{n-r-2} t + p_{j}(n-r-s-1) + r \right) \\&\quad = \frac{(1-p_{j})\big ((n-r-2)(n-r-1)-s(s-1) + 2r(n-1) \big )}{2(n-1)^2}\\&\qquad + \frac{p_{j}(n-s-1)}{n-1}\\&\quad = (1-p_{j}) \frac{(n-1)(n-2) + r(r+1) - s(s-1)}{2(n-1)^2} + p_{j}\left( 1 - \frac{s}{n-1}\right) . \end{aligned}$$

  Notice that, when $r+s=n-1$, the previous expression returns $r/(n-1)$. So, it is also valid in the case $r+s=n-1$.

$\square$

Proof of Proposition 6

Let $\varvec{p}$ be a weighting vector, $(r,s)\in \mathcal {R}$ and $j\in N$. Since

$$\begin{aligned} \phi _{j}\big (\mu _{\varvec{p}}^{(r,s)}\big )&= \frac{1}{n}\sum _{t=0}^{n-1}\frac{1}{\left( {\begin{array}{c}n-1\\ t\end{array}}\right) } \sum _{\begin{array}{c} T\subseteq N{\setminus } \{j\} \\ |T|=t \end{array}} \mu _{\varvec{p}}^{(r,s)}(T\cup \{j\})\\&\quad - \frac{1}{n}\sum _{t=0}^{n-1}\frac{1}{\left( {\begin{array}{c}n-1\\ t\end{array}}\right) } \sum _{\begin{array}{c} T\subseteq N{\setminus } \{j\} \\ |T|=t \end{array}} \mu _{\varvec{p}}^{(r,s)}(T). \end{aligned}$$

and $\mu _{\varvec{p}}^{(r,s)}$ is given by expression (1) (or expression (2) when $r+s=n-1$), we distinguish two cases:

• 1. If $r+s=n-1$, then

  $$\begin{aligned} \phi _{j}\big (\mu _{\varvec{p}}^{(r,s)}\big ) = \frac{1}{n}\sum _{t=s}^{n-1} 1 - \frac{1}{n}\sum _{t=s+1}^{n-1} 1 = \frac{1}{n}. \end{aligned}$$

• 2. If $r+s<n-1$, then, by Remark 5, we have

  $$\begin{aligned} \phi _{j}\big (\mu _{\varvec{p}}^{(r,s)}\big )&= \frac{1}{n} \left( \sum _{t=s}^{n-r-2}\frac{1}{\left( {\begin{array}{c}n-1\\ t\end{array}}\right) } \sum _{\begin{array}{c} T\subseteq N{\setminus } \{j\} \\ |T|=t \end{array}} \left( p_{j} + \sum _{i\in T} p_i\right) + r + 1 \right) \\&\quad - \frac{1}{n} \left( \sum _{t=s+1}^{n-r-1}\frac{1}{\left( {\begin{array}{c}n-1\\ t\end{array}}\right) } \sum _{\begin{array}{c} T\subseteq N{\setminus } \{j\} \\ |T|=t \end{array}} \sum _{i\in T} p_i + r \right) \\&= \frac{1}{n} \left( (n-r-s-1)p_{j} + \frac{1-p_{j}}{n-1} (s-(n-r-1)) + 1 \right) \\&= \frac{1}{n}\left( \frac{s+r + n(n-r-s-1)p_j}{n-1}\right) \\&= \frac{r+s}{n-1}\,\frac{1}{n} + \frac{n-1-r-s}{n-1} p_{j}\\&= \frac{r+s}{n-1}\,\frac{1}{n} + \left( 1 - \frac{r+s}{n-1}\right) p_{j}. \end{aligned}$$

  Notice that, when $r+s=n-1$, the previous expression returns 1 / n. So, it is also valid in the case $r+s=n-1$.

$\square$

Proof of Proposition 7

Let $\varvec{p}$ be a weighting vector, $(r,s)\in \mathcal {R}$ and $j,k\in N$. Notice that $I_{jk}\big (\mu _{\varvec{p}}^{(r,s)}\big )$ can be written as

$$\begin{aligned} I_{jk}\big (\mu _{\varvec{p}}^{(r,s)}\big ) = \frac{1}{n-1}\left( I_{\{j,k\}} - I_{\{j\}} - I_{\{k\}} + I_{\varnothing }\right) , \end{aligned}$$

where $I_{K}$, $K \subseteq N$, is defined by

$$\begin{aligned} I_{K} = \sum _{t=0}^{n-2}\frac{1}{\left( {\begin{array}{c}n-2\\ t\end{array}}\right) } \sum _{\begin{array}{c} T\subseteq N{\setminus } \{j,k\} \\ |T|=t \end{array}} \mu _{\varvec{p}}^{(r,s)}(T\cup K). \end{aligned}$$

Since $\mu _{\varvec{p}}^{(r,s)}$ is given by expression (1) (or expression (2) when $r+s=n-1$), we distinguish two cases:

• 1. If $r+s=n-1$, then it is easy to check that $I_{\{j,k\}}, I_{\{j\}}, I_{\{k\}}$, and $I_{\varnothing }$ take the following values:

  $$\begin{aligned} I_{\{j,k\}}= & {} {\left\{ \begin{array}{ll} \sum \nolimits _{t=0}^{n-2} 1 = n-1 &{} \text {if } s=0, \\ \sum \nolimits _{t=s-1}^{n-2} 1 = n - s&{} \text {otherwise}, \end{array}\right. } \\ I_{\{j\}}= & {} I_{\{k\}} = \sum \limits _{t=s}^{n-2} 1 = n - 1 - s, \\ I_{\varnothing }= & {} {\left\{ \begin{array}{ll} \sum \nolimits _{t=s+1}^{n-2} 1 = n-2-s &{} \text {if } s< n-1, \\ 0 &{} \text {otherwise}. \end{array}\right. } \end{aligned}$$

  We now calculate $I_{jk}\big (\mu _{\varvec{p}}^{(r,s)}\big )$ taking into account the different values of $I_{\{j,k\}}$, $I_{\{j\}}$, $I_{\{k\}}$, and $I_{\varnothing }$. We distinguish three cases:

  • (a) If $s=0$, which is equivalent to $r=n-1$, we get

    $$\begin{aligned} I_{jk}\big (\mu _{\varvec{p}}^{(r,s)}\big ) = \frac{1}{n-1} (n-1 -2(n-1)+n-2) = - \frac{1}{n-1}. \end{aligned}$$

  • (b) If $0< s < n-1$ we have

    $$\begin{aligned} I_{jk}\big (\mu _{\varvec{p}}^{(r,s)}\big ) = \frac{1}{n-1} (n-s -2(n-1-s) + n-2-s) = 0. \end{aligned}$$

  • (c) If $s=n-1$, which is equivalent to $r=0$, we obtain

    $$\begin{aligned} I_{jk}\big (\mu _{\varvec{p}}^{(r,s)}\big ) = \frac{1}{n-1}. \end{aligned}$$

    Therefore,

    $$\begin{aligned} I_{jk}\big (\mu _{\varvec{p}}^{(r,s)}\big ) = {\left\{ \begin{array}{ll} \frac{1}{n-1} &{} \text {if } r=0, \\ -\frac{1}{n-1} &{} \text {if } r=n-1, \\ 0 &{} \text {otherwise}. \end{array}\right. } \end{aligned}$$

    (4)

  • 2. If $r+s<n-1$, then, by Remark 6, we can see that $I_{\{j,k\}}$, $I_{\{j\}}$, $I_{\{k\}}$, and $I_{\varnothing }$ take the following values:

    $$\begin{aligned} I_{\{j,k\}}= & {} {\left\{ \begin{array}{ll} \sum \nolimits _{t=0}^{n-2} 1 = n-1 &{} \text {if } s=0 \text { and } r=n-2,\\ \sum \nolimits _{t=0}^{n-r-3} \left( p_{j} + p_{k} + \frac{t(1-p_{j}-p_{k})}{n-2}\right) + r + 1 &{} \text {if } s=0 \text { and } r< n-2,\\ \sum \nolimits _{t=s-1}^{n-r-3} \left( p_{j} + p_{k} + \frac{t(1-p_{j}-p_{k})}{n-2}\right) + r + 1&{} \text {otherwise}, \end{array}\right. } \\ I_{\{j\}}= & {} \sum _{t=s}^{n-r-2} \left( p_{j} + \frac{t(1-p_{j}-p_{k})}{n-2}\right) + r, \\ I_{\{k\}}= & {} \sum _{t=s}^{n-r-2} \left( p_{k} + \frac{t(1-p_{j}-p_{k})}{n-2}\right) + r, \\ I_{\varnothing }= & {} {\left\{ \begin{array}{ll} 0 &{} \text {if } r=0 \text { and } s=n-2,\\ \sum \nolimits _{t=s+1}^{n-2} \frac{t(1-p_{j}-p_{k})}{n-2} &{} \text {if } r=0 \text { and } s < n-2,\\ \sum \nolimits _{t=s+1}^{n-r-1} \frac{t(1-p_{j}-p_{k})}{n-2} + r - 1&{} \text {otherwise}, \end{array}\right. } \end{aligned}$$

    We now calculate $I_{jk}\big (\mu _{\varvec{p}}^{(r,s)}\big )$ taking into account the different values of $I_{\{j,k\}}$, $I_{\{j\}}$, $I_{\{k\}}$, and $I_{\varnothing }$. For instance, when $s=0$ and $r=n-2$ we get

    $$\begin{aligned}&I_{jk}\big (\mu _{\varvec{p}}^{(r,s)}\big )\\&\quad = \frac{1}{n-1}\left( n-1 - p_{j} - (n-2) - p_{k} - (n-2) + \frac{1-p_{j}-p_{k}}{n-2} + n-3\right) \\&\quad = \frac{1}{n-1}\frac{1-(n-1)(p_{j}+p_{k})}{n-2} = -\frac{1}{n-2}\left( p_{j}+p_{k} -\frac{1}{n-1}\right) . \end{aligned}$$

    Once all the cases have been analyzed (to avoid tedious calculations, the remaining cases are left to the reader), we have

    $$\begin{aligned} I_{jk}\big (\mu _{\varvec{p}}^{(r,s)}\big ) = {\left\{ \begin{array}{ll} \frac{1}{n-2}\left( p_{j}+p_{k} -\frac{1}{n-1}\right) &{} \text {if } r = 0 \text { and } 0< s< n-1,\\ -\frac{1}{n-2}\left( p_{j}+p_{k} -\frac{1}{n-1}\right) &{} \text {if } s = 0 \text { and } 0< r < n-1,\\ 0 &{} \text {otherwise}. \end{array}\right. } \end{aligned}$$

    (5)

Expressions (4) and (5) together establish the truth of Proposition 7. $\square$

Proof of Proposition 8

Let $\varvec{p}$ be a weighting vector and $(r,s)\in \mathcal {R}$. Given $A\subseteq N$, since

$$\begin{aligned} m^{\mu _{\varvec{p}}^{(r,s)}}(A) = \sum _{t=1}^{|A|} (-1)^{|A|-t} \sum _{\begin{array}{c} T\subseteq A \\ |T|=t \end{array}} \mu _{\varvec{p}}^{(r,s)}(T), \end{aligned}$$

and $\mu _{\varvec{p}}^{(r,s)}$ is given by expression (1) (or expression (2) when $r+s=n-1$), we distinguish the following cases:

• 1. If $|A| \le s$, then

  $$\begin{aligned} m^{\mu _{\varvec{p}}^{(r,s)}}(A) = 0. \end{aligned}$$

• 2. If $s< |A| < n-r$, by Remark 4 and the third item of Remark 2, we get

  $$\begin{aligned}&m^{\mu _{\varvec{p}}^{(r,s)}}(A) \\&\quad = \sum _{t=s+1}^{|A|} (-1)^{|A|-t} \sum _{\begin{array}{c} T\subseteq A \\ |T|=t \end{array}} \sum _{i\in T} p_{i} = \sum _{t=s+1}^{|A|} (-1)^{|A|-t} \left( {\begin{array}{c}|A|-1\\ t-1\end{array}}\right) \sum _{i\in A} p_{i}\\&\quad =\! \left( \sum _{i\in A} p_{i} \!\!\right) \!\sum _{t=s+1}^{|A|} \!(-1)^{|A|-t} \left( {\begin{array}{c}|A|-1\\ |A|-t\end{array}}\right) = \left( \sum _{i\in A} p_{i} \!\right) \!\sum _{j=0}^{|A|-s-1} \!(-1)^{j} \left( {\begin{array}{c}|A|-1\\ j\end{array}}\right) \\&\quad =\! \left( \sum _{i\in A} p_{i}\!\! \right) \! (-1)^{|A|-s-1} \!\left( {\begin{array}{c}|A|-2\\ |A|-s-1\end{array}}\right) = (-1)^{|A|-s-1} \!\left( {\begin{array}{c}|A|-2\\ s-1\end{array}}\right) \!\! \left( \sum _{i\in A} p_{i} \!\!\right) \!\!. \end{aligned}$$

  Notice that when $s=0$ we have $m^{\mu _{\varvec{p}}^{(r,s)}}(A) = 0$.

• 3 If $|A| \ge n-r$, we distinguish two cases:

  • (a) If $r+s=n-1$, then, by the third item of Remark 2 we have

    $$\begin{aligned} m^{\mu _{\varvec{p}}^{(r,s)}}(A)&= \sum _{t=s+1}^{|A|} (-1)^{|A|-t} \left( {\begin{array}{c}|A|\\ t\end{array}}\right) = \sum _{t=s+1}^{|A|} (-1)^{|A|-t} \left( {\begin{array}{c}|A|\\ |A|-t\end{array}}\right) \\&= \sum _{j=0}^{|A|-s-1} (-1)^{j} \left( {\begin{array}{c}|A|\\ j\end{array}}\right) = (-1)^{|A|-s-1} \left( {\begin{array}{c}|A|-1\\ |A|-s-1\end{array}}\right) \\&= (-1)^{|A|-s-1} \left( {\begin{array}{c}|A|-1\\ s\end{array}}\right) . \end{aligned}$$

  • (b) If $r+s<n-1$, by Remark 4 and the third item of Remark 2, we have

    $$\begin{aligned}&m^{\mu _{\varvec{p}}^{(r,s)}}(A)\\&\quad = \sum _{t=s+1}^{n-r-1} (-1)^{|A|-t} \left( {\begin{array}{c}|A|-1\\ t-1\end{array}}\right) \sum _{i\in A} p_{i} + \sum _{t=n-r}^{|A|} (-1)^{|A|-t} \left( {\begin{array}{c}|A|\\ t\end{array}}\right) \\&\quad = \left( \sum _{i\in A} p_{i} \right) \sum _{t=s+1}^{n-r-1} (-1)^{|A|-t} \left( {\begin{array}{c}|A|-1\\ |A|-t\end{array}}\right) + \sum _{t=n-r}^{|A|} (-1)^{|A|-t} \left( {\begin{array}{c}|A|\\ |A|-t\end{array}}\right) \\&\quad = \left( \sum _{i\in A} p_{i} \right) \sum _{j=|A|-n+r+1}^{|A|-s-1} (-1)^{j} \left( {\begin{array}{c}|A|-1\\ j\end{array}}\right) + \sum _{j=0}^{|A|-n+r} (-1)^{j} \left( {\begin{array}{c}|A|\\ j\end{array}}\right) \\&\quad = \left( \sum _{i\in A} p_{i} \right) \left( (-1)^{|A|-s-1} \left( {\begin{array}{c}|A|-2\\ |A|-s-1\end{array}}\right) \right. \\&\qquad \left. - (-1)^{|A|-n+r} \left( {\begin{array}{c}|A|-2\\ |A|-n+r\end{array}}\right) \right) + (-1)^{|A|-n+r} \left( {\begin{array}{c}|A|-1\\ |A|-n+r\end{array}}\right) \\&\quad = \left( \sum _{i\in A} p_{i} \right) \left( (-1)^{|A|-s-1} \left( {\begin{array}{c}|A|-2\\ s-1\end{array}}\right) - (-1)^{|A|-n+r} \left( {\begin{array}{c}|A|-2\\ n-r-2\end{array}}\right) \right) \\&\qquad + (-1)^{|A|-n+r} \left( {\begin{array}{c}|A|-1\\ n-r-1\end{array}}\right) . \end{aligned}$$

    Notice also that, when $r+s=n-1$, the previous expression returns $(-1)^{|A|-s-1} \left( {\begin{array}{c}|A|-1\\ s\end{array}}\right)$. So, it is also valid in the case $r+s=n-1$.

$\square$