Robust winner determination in positional scoring rules with uncertain weights

Abstract

Scoring rules constitute a particularly popular technique for aggregating a set of rankings. However, setting the weights associated with rank positions is a crucial task, as different instantiations of the weights can often lead to different winners. In this work we adopt minimax regret as a robust criterion for determining the winner in the presence of uncertainty over the weights. Focusing on two general settings (non-increasing weights and convex sequences of non-increasing weights) we provide a characterization of the minimax regret rule in terms of cumulative ranks, allowing a quick computation of the winner. We then analyze the properties of using minimax regret as a social choice function. Finally we provide some test cases of rank aggregation using the proposed method.

A Appendix: Proofs

We omit the proof of Proposition 3 that is trivial.

Proposition 4

Let $x, y \in A$. The following statements connect max regret and dominance relations:

• 1. For an arbitrary set W of scoring vectors:

  If x weakly dominates y, then ${{\,\mathrm{MR}\,}}(x; W) \le {{\,\mathrm{MR}\,}}(y; W)$;

  if x strongly dominates y, then ${{\,\mathrm{MR}\,}}(x; W) < {{\,\mathrm{MR}\,}}(y; W)$.

• 2. Moreover, when considering decreasing weights, i.e. the scoring vector belongs to $W^{D}$:

  If $V^{x} \succeq V^{y}$ then ${{\,\mathrm{MR}\,}}(x; W^{D}) \le {{\,\mathrm{MR}\,}}(y; W^{D})$;

  if $V^{x} \succ V^{y}$ then ${{\,\mathrm{MR}\,}}(x; W^{D}) < {{\,\mathrm{MR}\,}}(y; W^{D})$.

• 3. Finally, when considering convex weights, i.e. the scoring vector belongs to $W^{C}$:

  If ${\mathcal {V}}^{x} \succeq {\mathcal {V}}^{y}$ then ${{\,\mathrm{MR}\,}}(x; W^{C}) \le {{\,\mathrm{MR}\,}}(y; W^{C})$;

  if ${\mathcal {V}}^{x} \succ {\mathcal {V}}^{y}$ then ${{\,\mathrm{MR}\,}}(x; W^{C}) < {{\,\mathrm{MR}\,}}(y; W^{C})$.

Proof

1) If x weakly dominates y, by definition $\forall w \in W \;\; s(x; w) \ge s(y; w)$ (Definition 3). For any $z \in A$, we have that:

$$\begin{aligned} {{\,\mathrm{PMR}\,}}(x, z; W) =&\max _{w \in W} \{ s(z; w) - s(x; w) \} \le \max _{w \in W} \{ s(z; w)\nonumber \\&- s(y; w) \} = {{\,\mathrm{PMR}\,}}(y, z;W). \end{aligned}$$

(31)

Therefore ${{\,\mathrm{MR}\,}}(x; W) = \max _{z \in A} {{\,\mathrm{PMR}\,}}(x, z; W) \le \max _{z \in A} {{\,\mathrm{PMR}\,}}(y, z;W) = {{\,\mathrm{MR}\,}}(y;W)$.

2) The result follows directly from part 1 of this proposition and Proposition 1

3) The result follows directly from part 1 of this proposition and Proposition 2. $\square$

Proposition 5

The followings statements hold:

• 1. For any $x \in S^{*}(W)$, x is weakly undominated, i.e. there is no $y \in A$ such that y strongly dominates x.

• 2. For all $x \in S^{*}(W)$, there is no $y \in A\setminus S^{*}(W)$ such that y weakly dominates x.

Proof

• 1) Let $x \in S^{*}(W)$, by definition $x \in \arg \min _{z \in A} {{\,\mathrm{MR}\,}}(z; W)$. Now assume y strongly dominates x. Then, by Proposition 4, we have ${{\,\mathrm{MR}\,}}(y;W) < {{\,\mathrm{MR}\,}}(x;W)$, but this is absurd.

• 2) Let $x \in S^{*}(W)$. Now assume $y \in A\setminus S^{*}(W)$ and that y weakly dominates x. Then, by Proposition 4, we have ${{\,\mathrm{MR}\,}}(y;W) \le {{\,\mathrm{MR}\,}}(x;W)$ and, therefore, ${{\,\mathrm{MR}\,}}(y;W)={{\,\mathrm{MMR}\,}}(W)$, meaning $y \in S^{*}(W)$ . We obtain a contradiction. $\square$

Proposition 6

Let $x \in A$. The following statements hold:

• 1. Alternative x is a necessary co-winner if and only if ${{\,\mathrm{MR}\,}}(x;W)=0$.

• 2. If x is a necessary winner then ${{\,\mathrm{MR}\,}}(x;W)=0$ and $S^{*}(W)=\{x\}$.

• 3. Alternative x is a necessary winner if and only if ${{\,\mathrm{PMR}\,}}(x,z;W)<0$ for all $z \in A {\setminus } \{x\}$.

Proof

• 1. x is a necessary co-winner $\iff$$s(x;w) \ge s(z;w) \;\; \forall z \in A, \forall w \in W$$\iff$${{\,\mathrm{PMR}\,}}(x, z; W) \le 0$ for all $z \in A$$\iff$${{\,\mathrm{MR}\,}}(x; W) = 0$.

• 2. If x is a necessary winner then it is also a necessary co-winner, and part 1 of this proposition implies ${{\,\mathrm{MR}\,}}(x;W)=0$, and $x \in S^{*}(W)$. Let $y \in A {\setminus }\{x\}$. Since x is a necessary winner, $s(x;w)>s(y;w)$ for all $w \in W$; therefore, ${{\,\mathrm{PMR}\,}}(y,x;W) > 0$ and ${{\,\mathrm{MR}\,}}(y; W) \ge {{\,\mathrm{PMR}\,}}(y,x;W) > 0$. Hence $S^{*}(W)=\{x\}$.

• 3. x is a necessary winner $\iff$$s(x;w) > s(z;w) \;\; \forall z \in A {\setminus } \{x\}, \forall w \in W$$\iff$${{\,\mathrm{PMR}\,}}(x, z; W) < 0$ for all $z \in A {\setminus } \{x\}$

$\square$

Lemma 1

In the case of decreasing weights, for any pairs $x, y \in A$, it holds

$$\begin{aligned} {{\,\mathrm{PMR}\,}}(x,y; W^{D}) = \max _{j \in [\![m-1 ]\!]} \{ V^y_j - V^x_j \}. \end{aligned}$$

(21)

Proof

By noticing that the optimal solution of a bounded linear program is attained in one of the vertices, we derive

$$\begin{aligned} {{\,\mathrm{PMR}\,}}(x,y; W^D)&= \max \left\{ \sum _{j=1}^{m} [w_j v^y_j - w_j v^x_j] \Big | 1 = w_1 \ge w_{2} \ge \cdots \ge w_{m-1} \ge w_m = 0 \right\} \nonumber \\&= \max \left\{ \sum _{j=1}^{m-1} \delta _j ( V^y_j - V^x_j) \Big | \delta _1 \ge 0, \ldots , \delta _{m-1} \ge 0 \wedge \sum _{j=1}^{m-1} \delta _j = 1 \right\} \nonumber \\&=\max \left\{ \sum _{j=1}^{m-1} \delta _j ( V^y_j - V^x_j) \Big | \delta \in \{ \text {oneat(1)},\ldots ,\text {oneat(m-1)} \} \right\} \nonumber \\&= \max _{j \in [\![m-1 ]\!]} \{ V^y_j - V^x_j \} \end{aligned}$$

(32)

where $\text {oneat(j)}$ is the vector with 0 everywhere except in position j where the value is 1. $\square$

Theorem 1

For any alternative $x \in A$ it holds

$$\begin{aligned} {{\,\mathrm{MR}\,}}(x; W^D) = \max _{j \in [\![m-1 ]\!]} \{ V^*_j - V^x_j \} \end{aligned}$$

(22)

where $V^*_k = \max _{x \in A} V^x_k$.

Proof

We use Lemma 1 and substitute Eq. (21) into the formula of ${{\,\mathrm{MR}\,}}(x)$ given by Eq. (17):

$$\begin{aligned} {{\,\mathrm{MR}\,}}(x; W^D)= & {} \max _{y \in A} \max _{j \in [\![m-1 ]\!]} \{ V^y_j - V^x_j \} = \max _{j \in [\![m-1 ]\!]} \max _{y \in A} \{V^y_j - V^x_j\} \\= & {} \max _{j \in [\![m-1 ]\!]} \{ \max _{y \in A} \{V^y_j \}- V^x_j\} = \max _{j \in [\![m-1 ]\!]} \{V^*_j - V^x_j\}. \end{aligned}$$

$\square$

Lemma 2

Assuming convex weights, the pairwise max regret of an alternative x against y can be computed as follows:

$$\begin{aligned} {{\,\mathrm{PMR}\,}}(x,y; W^C) = \max _{j \in [\![m-1 ]\!]} \left\{ \frac{{\mathcal {V}}^y_j - {\mathcal {V}}^x_j}{j} \right\} . \end{aligned}$$

(23)

Proof

The proof is similar to that of Lemma 1. The pairwise max regret can be computed using the following linear program:

$$\begin{aligned} {{\,\mathrm{PMR}\,}}(x,y; W^C) = \max \;\;\;&\sum _{j=1}^{m-1} \phi _j ({\mathcal {V}}^y_j - {\mathcal {V}}^x_j) \end{aligned}$$

(33)

$$\begin{aligned} \text {s.t. } \;\;\;&\sum _{j=1}^{m-1} j \, \phi _{j} = 1 \end{aligned}$$

(34)

$$\begin{aligned}&\phi _1 \ge 0, \ldots , \phi _{m-1} \ge 0 \end{aligned}$$

(35)

Let $\text {oneat(j)}$ be the vector with 0 everywhere except in position j where the value is 1.

We know from the theory of linear programming that the optimum is attained in one of the vertices. Therefore, the optimal $\phi$ must be of the type $j \cdot \text {oneat}(j)$, that is $(1,0,\ldots ,0)$, $(0,0.5,0,\ldots ,0)$, ..., $(0,\ldots ,0, \frac{1}{m-1})$. The corresponding optimal $\delta$ is among $(1,0,\ldots ,0)$, $(\frac{1}{2}, \frac{1}{2}, 0 , \ldots , 0)$, $(\frac{1}{3}, \frac{1}{3}, \frac{1}{3}, 0, \ldots , 0)$, etc.

$$\begin{aligned} {{\,\mathrm{PMR}\,}}(x,y; W^C)&= \max \left\{ \sum _{j=1}^{m-1} \phi _{j} ( {\mathcal {V}}^y_j - {\mathcal {V}}^x_j) \, \Big | \, \phi \in \left\{ \frac{ \text {oneat(j)} }{j} \right\} _{j=1}^{m-1} \right\} \nonumber \\&= \max _{j \in [\![m-1 ]\!]} \frac{{\mathcal {V}}^y_j - {\mathcal {V}}^x_j}{j}. \end{aligned}$$

(36)

$\square$

Theorem 2

Assuming convex weights, the max regret of alternative x can be computed as follows:

$$\begin{aligned} {{\,\mathrm{MR}\,}}(x; W^C) = \max _{j \in [\![m-1 ]\!]} \Big \{ \frac{{\mathcal {V}}^*_j - {\mathcal {V}}^x_j}{j} \Big \} \end{aligned}$$

(24)

where ${\mathcal {V}}^*_j = \max _{x \in A} {\mathcal {V}}^x_j$.

Proof

The result follows by using Lemma 2 in order to substitute Eqs. (23) in (17); we then exchange the order of the two $\max$.

$$\begin{aligned} {{\,\mathrm{MR}\,}}(x; W^C)= & {} \max _{y \in A} \max _{j \in [\![m-1 ]\!]} \left\{ \frac{{\mathcal {V}}^y_j - {\mathcal {V}}^x_j}{j} \right\} = \max _{j \in [\![m-1 ]\!]} \max _{y \in A} \left\{ \frac{{\mathcal {V}}^y_j - {\mathcal {V}}^x_j}{j} \right\} \\= & {} \max _{j \in [\![m-1 ]\!]} \left\{ \frac{{\mathcal {V}}^*_j - {\mathcal {V}}^x_j}{j} \right\} . \end{aligned}$$

$\square$

Proposition 7

For any pair of alternatives $x, y \in A$, it holds

$$\begin{aligned} {{\,\mathrm{PMR}\,}}(x,y; W^{D,t}) = \left[ \left( 1 - \sum _{j=1}^{m-1} t_{j} \right) \max _{j\in [\![m-1 ]\!]} \{ V^y_j - V^x_j \} \right] + \sum _{j=1}^{m-1} t_{j} (V^y_j - V^x_j ) \end{aligned}$$

(26)

or, equivalently,

$$\begin{aligned} {{\,\mathrm{PMR}\,}}(x,y; W^{D,t}) = \left[ \left( 1 - \sum _{j=1}^{m-1} t_{j} \right) {{\,\mathrm{PMR}\,}}(x,y; W^{D}) \right] + \sum _{j=1}^{m-1} t_{j} (V^y_j - V^x_j) \end{aligned}$$

where ${{\,\mathrm{PMR}\,}}(x,y; W^{D})$ is pairwise max regret computed with no discriminative thresholds.

Proof

We let ${\hat{\delta }}_{j}$ to be the slack between the value of $\delta _{j}$ and its discriminative value $t_{j}$:

$$\begin{aligned} \hat{\delta _j} = w_j - w_{j+1} - t_{j} = \delta _j - t_{j} \end{aligned}$$

(37)

for $j \in [\![m-1 ]\!]$. Equation (25) ensures that $\hat{\delta _j} \ge 0$. Consider the difference of score between two alternatives y and x:

$$\begin{aligned} s(y; w)-s(x; w)&= \sum _{j=1}^{m-1} {\hat{\delta }}_{j} (V^{y}_{j} - V^{x}_{j}) + \sum _{j=1}^{m-1} t_{j} (V^{y}_{j} - V^{x}_{j}) \end{aligned}$$

(38)

We now consider the maximum $s(y; w)-s(x; w)$ when picking a scoring rule whose scoring vector w belongs to $W^{D,t}$ (weakly decreasing and with discriminative values). Note that, the second addendum on the right-hand side of Eq. (38) can be viewed as constant.

$$\begin{aligned}&\max _{w \in W^{D,t}} s(y; w)-s(x; w) \nonumber \\&\quad = \max \left\{ \sum _{j=1}^{m-1} \delta _j (V^y_j - V^x_j) \mid 0 \le \delta _j \le 1 \wedge \sum _1^{m-1} \delta _j = 1 \wedge \delta _j \ge t_j \,\cdot \, \forall j \in [\![m-1 ]\!]\right\} \end{aligned}$$

(39)

$$\begin{aligned}&\quad = \sum _{j=1}^{m-1} t_{j} (V^y_j - V^x_j ) + \max \left\{ \sum _{j=1}^{m-1} \hat{\delta _j} (V^y_j - V^x_j) \mid 0 \le {\hat{\delta }}_j \le 1 \wedge \sum _1^{m-1} \hat{\delta _j} \right. \nonumber \\&\left. \quad = 1-\sum _{j=1}^{m-1} t_{j} \,\cdot \, \forall j \in [\![m-1 ]\!]\right\} \end{aligned}$$

(40)

$$\begin{aligned}&\quad = \sum _{j=1}^{m-1} t_{j} (V^y_j - V^x_j ) + \left( 1-\sum _{j=1}^{m-1} t_{j} \right) \max _{j \in [\![m-1 ]\!]} (V^y_j - V^x_j). \end{aligned}$$

(41)

$\square$

Proposition 8

Let p be the profile.

• When an alternative $x \in A$ is ranked first at least $\frac{2}{3} n$ times, then ${{\,\mathrm{MMR\hbox {-}Dec}\,}}(p)=\{x\}$.

  In other words, if x is such that $v_{1}^{x} > \frac{2}{3} n$ then $S ^{*}(W^{D})=\{x\}$

• When an alternative $x \in A$ is ranked first at least $\frac{2m-2}{3m-2} n$ times, then ${{\,\mathrm{MMR\hbox {-}Con}\,}}(p)=\{x\}$.

  In other words, if x is such that $v_{1}^{x} > \frac{2m-2}{3m-2} n$, then $S^{*}(W^{C})=\{x\}$.

Proof

• 1. Decreasing weights$W^{D}$: Consider a profile p in which x is such that $v^{x}=(a,0,\ldots ,0,n-a)$ and $v^{y}=(n-a,a,0,\ldots ,0)$ and $a > \frac{n}{2}$ (x is ranked first at least a times and is last all other times; y is first $n-a$ times and second a times). Then $V^{x}=(a,\ldots ,a)$ and $V^{y}=(n-a,n,\ldots ,n)$. According to Lemma 1, ${{\,\mathrm{PMR}\,}}(x,y; W^{D})= V^{y}_{2} - V^{x}_{2} = n-a$ and ${{\,\mathrm{PMR}\,}}(y,x; W^{D})= V^{x}_{1} - V^{y}_{1}=2a-n$. It can be shown that in this profile ${{\,\mathrm{MR}\,}}(x; W^{D})={{\,\mathrm{PMR}\,}}(x,y; W^{D})$ and ${{\,\mathrm{MR}\,}}(y; W^{D})={{\,\mathrm{PMR}\,}}(y, x; W^{D})$ since every alternative other than x and y is (weakly) dominated. Therefore, x is a winner if

  $$\begin{aligned} {{\,\mathrm{MR}\,}}(x; W^{D})< {{\,\mathrm{MR}\,}}(y; W^{D}) \iff n-a < 2a-n \iff a > \frac{2}{3}n. \end{aligned}$$

  The last step is to show that the profile p is the most challenging situation to alternative x: in any profile $p'$ in which x is ranked first at least a times, x will be at least as well ranked as in p, and any challenger will be at most ranked as well as y in p. Indeed monotonicity guarantees us that the max regret of x will be minimum.

• 2. Convex weights$W^{C}$: Consider a profile p in which x is such that $v^{x}=(a,0,\ldots ,0,n-a)$ and $v^{y}=(n-a,a,0,\ldots ,0)$ and $a > \frac{n}{2}$ (x is ranked first at least a times and is last all other times; y is first $n-a$ times and second a times). The cumulative ranks of x and y are $V^{x}=(a,\ldots ,a)$ and $V^{y}=(n-a,n,\ldots ,n)$; the double cumulative ranks ${\mathcal {V}}^{x}$ and ${\mathcal {V}}^{y}$ are such that that ${\mathcal {V}}_{j}^{x} = j a$ and ${\mathcal {V}}_{j}^{y}=jn-a$, with $j \in [\![m-1 ]\!]$. We impose that the max regret of x is lower than that of y, and then use Lemma 1:

  $$\begin{aligned}&{{\,\mathrm{MR}\,}}(x; W^{C})< {{\,\mathrm{MR}\,}}(y; W^{C}) \\&\quad \iff {{\,\mathrm{PMR}\,}}(x,y; W^{C})< {{\,\mathrm{PMR}\,}}(y,x; W^{C}) \\&\quad \iff \max _{j \in [\![m-1 ]\!]} \left\{ \frac{{\mathcal {V}}^{y}_{j} - {\mathcal {V}}^{x}_{j}}{j} \right\}< \max _{j \in [\![m-1 ]\!]} \left\{ \frac{{\mathcal {V}}^{x}_{j} - {\mathcal {V}}^{y}_{j}}{j} \right\} \\&\quad \iff \underbrace{n- \frac{a}{m-1} }_{\frac{{\mathcal {V}}^{y}_{m-1}}{m-1}} - \underbrace{a}_{\frac{{\mathcal {V}}^{x}_{m-1}}{m-1}} < \underbrace{a}_{\frac{{\mathcal {V}}^{x}_{1}}{1}} - \underbrace{(n-a)}_{\frac{{\mathcal {V}}^{y}_{1}}{1}} \iff a > \frac{2 (m-1)}{3m-2} n. \end{aligned}$$

  Notice that alternative x is the “worst ranked” among all possible rank distribution satisfying the above condition and y is ranked as well as possible (since we filled up the profile in the most disadvantageous way to x). Now to conclude the proof, as in the previous case, we make use of monotonicity to prove the argument that in any other profile where alternative x satisfies the above condition on the number of first positions, x will achieve minimax regret.

$\square$

Theorem 3

The social choice functions ${{\,\mathrm{MMR\hbox {-}Dec}\,}}$ and ${{\,\mathrm{MMR\hbox {-}Con}\,}}$ satisfy anonymity, neutrality, unanimity, monotonicity, IRDA, homogeneity and independence from symmetric profiles.

Proof

Neutrality and anonymity are trivial to check. Proof sketches for homogeneity and independence from symmetric profiles were given in the main text.

For unanimity, we need to show that if all voters place an alternative x in first position, then x will have a max regret value of zero and, therefore, x will be declared the winner. x has rank distribution $v^x=(n,0,\ldots ,0)$ and cumulative standings $V^x=(n,\ldots ,n)$. All other alternatives $y \ne x$ have 0 in the first component of their cumulative ranks, $V^{y}_{1}=0$, and a value less than n in the other components. Using Theorem 1, ${{\,\mathrm{MR}\,}}(y; W^{D})=n$, for any y, while ${{\,\mathrm{MR}\,}}(x; W^{D})=0$; therefore, x is the only winner. The proof is similar for $W^{C}$.

We now prove monotonicity. Indeed, assume that, starting from a profile p, we modify the ranking associated with one of the voters so that an alternative x moves from some position $i_2$ to some position $i_1<i_2$ and call the resulting profile $p'$. Then the rank distribution of the new profile is such that $v^{x}_{i_1}[p']=v^{x}_{i_1}[p]+1$ and $v^{x}_{i_2}[p']=v^{x}_{i_2}[p]-1$. It follows that the cumulative rank distribution $V^x[p']$ of alternative x in the new profile $p'$ is such that

$$\begin{aligned} V^{x}_j[p'] = {\left\{ \begin{array}{ll} V^x_j[p] &{} j=1,\ldots ,i_1-1 \\ V^x_j[p]+1 &{} j=i_{1},\ldots ,i_2-1\\ V^x_j[p] &{} j=i_2,\ldots ,m \\ \end{array}\right. } \end{aligned}$$

We, therefore, have $V^{x}[p'] \succeq V^{x}[p]$. We also have $V^{y}[p'] \preceq V^{y}[p]$ for any $y \ne x$ since y can either have the same ranks as in $p'$ or may have lose a position.

Now, let $J = \{ j \in [\![m-1 ]\!] | x \in \arg \max _{z} V^{z}_{j}[p'] \}$, that is the set of positions for which x has maximum cumulative rank in $p'$. We have (using Theorem 1)

$$\begin{aligned} {{\,\mathrm{MR}\,}}_{p'}(x; W^{D})&= \max _{j=1}^{m-1} \{ V^{*}_{j}[p'] - V_{j}^{x}[p'] \} = \max _{j \not \in J} \{ \max _{y \in A - \{x\}} \{ V^{y}_{j}[p'] \} - V_{j}^{x}[p'] \} \le \\&\le \max _{j \not \in J} \{ \max _{y \in A - \{x\}} \{ V^{y}_{j}[p] \} - V_{j}^{x}[p] \} = \max _{j=1}^{m-1} \{ V^{*}_{j}[p] - V_{j}^{x}[p] \} \\&= {{\,\mathrm{MR}\,}}_{p}(x; W^{D}). \end{aligned}$$

and this means regret of x in $p'$ cannot be higher than the regret of x in p. Assume y is any alternative that is not a winner in p, meaning that ${{\,\mathrm{MR}\,}}_{p}(x; W^{D}) < {{\,\mathrm{MR}\,}}_{p}(y;W^{D})$. It follows

$$\begin{aligned} {{\,\mathrm{MR}\,}}_{p'}(x;W^{D}) \le {{\,\mathrm{MR}\,}}_{p}(x; W) < {{\,\mathrm{MR}\,}}_{p}(y;W^{D}) \le {{\,\mathrm{MR}\,}}_{p'}(y;W^{D}). \end{aligned}$$

Therefore, if x is a winner in p, it has to continue to have minimax regret, and, therefore, to be a winner in $p'$; we conclude that the rule is monotone in $W^{D}$. Analogous reasoning can be made for $W^{C}$.

To prove IRDA note that, according to Eq. (22), the max regret of an alternative depends only on its rank distribution, and on the values $V^{*}_{1},\ldots ,V^{*}_{m-1}$. But, for any j, $V^{*}_{j} = \max _{y} V^{y}_{j}$ and the associated alternatives in $\arg \max _{y} V^{y}_{j}$ have maximal cumulative rank in the j-th position, meaning that are either undominated or dominated by another alternative that is also in $\arg \max _{y} V^{y}_{j}$ (said in another way, for each j, there is at least 1 undominated alternative in $\arg \max _{y} V^{y}_{j}$).

Thus any change in the rank of dominated alternatives does not change the values $V^{*}_{1},\ldots ,V^{*}_{m-1}$, when moving from p to $p'$. This means that by changing the ranks of dominated alternatives will not change ${{\,\mathrm{MR}\,}}(x; W^{D})$, and, therefore, the winners according to ${{\,\mathrm{MMR}\,}}$ will be the same, and, therefore, ${{\,\mathrm{MMR}\,}}$ is IRDA. $\square$