The difference between manipulability indices in the IC and IANC models

Abstract

We consider the calculation of Nitzan–Kelly’s manipulability index in the impartial anonymous and neutral culture (IANC) model. We provide a new theoretical study of this model and an estimation for the maximal difference between manipulability indices in the IANC model and a basic model, the impartial culture (IC). The asymptotic behavior of this difference is studied with the help of the impartial anonymous culture (IAC) model. It is shown that the difference between the IAC and IANC models tends to zero as the number of alternatives or the number of voters grows. These results hold for any other probabilistic measure that is anonymous and neutral. Finally, we calculate Nitzan–Kelly’s index in the IANC model for four social choice rules and compare it with the IC model.

Appendices

Appendix 1

Theorem 1

(ANECs with the minimal and maximal number of elements)

• (a) The minimal number of elements in an anonymous and neutral equivalence class is m!. This class is unique for the case of $n\ge 3$.

• (b) The maximal number of elements in an anonymous and neutral equivalence class is m!n! if and only if $m!>n$.

Proof

• (a) If profile $\vec {P}$ consists of equal columns, then $G_{\vec {P}} =\{(\tau _0 ,\sigma ):\sigma \in S_n\}$ and $|G_{\vec {P}} |=n!$. Thus, $|\theta _{\min } |\le m!n!/n!=m!$. For any $\sigma \in S_n$ and $\tau _1 \ne \tau _2$, it is impossible that both $(\tau _1,\sigma )\in G_{\vec {P}}$ and $(\tau _2 ,\sigma )\in G_{\vec {P}}$ because $(\vec {P}^{\sigma })^{\tau _1 }\ne (\vec {P}^{\sigma })^{\tau _2}$. Then, for any $\vec {P}$, $|G_{\vec {P}} |\le n!$. Hence, $|\theta _{\min } |\ge m!n!/n!=m!$. Therefore, $|\theta _{\min } |=m!$.

  Let us consider a preference profile consisting of two columns, $P_1 =(a_1 ,a_2 ,\ldots ,a_m )$ and $P_2 =(a_2 ,a_1 ,a_3 \ldots ,a_m)$, where $\tau$ is of order 2. Take $\tau =(12)(3)\ldots (m)$ and $\sigma =(12)$; then, $|G_{\vec {P}} |=2=n!$. Thus, the number of ANECs with a minimal number of elements is greater than 1 when $n=2$.

  Now consider a profile $\vec {P}$ with $n\ge 3$ and at least two distinct columns, $P_i$ and $P_j$. Let $\sigma \in S_n$ permute columns i and j, and fix column l, $l\ne i$, $l\ne j$. However, there is no permutation $\tau$ s.t. $(\tau ,\sigma )\in G_{\vec {P}}$. Thus, $|G_{\vec {P}} |<n!$.

  We conclude that $|G_{\vec {P}} |=n!$ with $n\ge 3$ if and only if $P_1 =P_2 =\cdots =P_n$. Since all such profiles belong to the same ANEC, it is unique.

• (b) First, we show that $|\theta _{\max } |=m!n!$ implies $m!>n$. Suppose, that $|\theta _{\max } |=m!n!$, but $m!\le n$. If $m!<n$ for some $\vec {P}$, then there exist i, j such that $P_i =P_j$. Let $\sigma _1$ permute only columns i and j. Then, $g_0 =(\tau _0 ,\sigma _0 )\in G_{\vec {P}}$ and $g_1 =(\tau _0 ,\sigma _1 )\in G_{\vec {P}}$, $|G_{\vec {P}} |>1$ and $|\theta _{\max } |<m!n!$. On the other hand, if $m!=n$, consider the preference profile that consists of all m! different columns. Consider any permutation of order m, for example, ($1 2 {\ldots } m$), which splits up a set of m! columns into $m!-1$ non-intersecting orbits. Again, $|G_{\vec {P}} |>1$. Consequently, if $|\theta _{\max } |=m!n!$ then $m!>n$. Second, we prove that if $m!>n$, then $|\theta _{\max } |=m!n!$. Let L be the set of all linear orders. Suppose that there exist such m and n such that for each $\vec {P}$ consisting of distinct columns, $|G_{\vec {P}} |\ge 2$. Take any such $\vec {P}$ and construct a set of preference profiles replacing its n-th column, $P_n$, by any ${P}'\in L\backslash \{P_1 ,\ldots ,P_{n-1} \}$. We get

  $$\begin{aligned} M=\{{\vec {P}}'=(P_1 ,\ldots ,P_{n-1} ,{P}'):{P}'\in L\backslash \{P_1 ,\ldots ,P_{n-1} \}\}. \end{aligned}$$

  The cardinality of this set is $m!-n+1$. According to our assumption, for each $\vec {P}$ in M, $|G_{\vec {P}} |\ge 2$. This means that there exist $(\tau ,\sigma )\ne (\tau _0 ,\sigma _0)$, and $\sigma$ does not have cycles of length 1. Then, for every ${P}'=(a_1 ,\ldots ,a_m)$, $(a_{\tau ^{i}(1)} ,\ldots ,a_{\tau ^{i}(m)} )\in \{P_1 ,\ldots ,P_{n-1} \}$, where $i=1..t$ and t is the order of $\tau$; however, this is impossible. Consequently, for any m and n, there exists a preference profile with only one stabilizing permutation, $(\tau _0 ,\sigma _0)$. The cardinality of the equivalence class $\theta _{\max }$ consisting of such profiles is

  $$\begin{aligned} | {\theta _{\max } }|=\frac{m!n!}{1}=m!n!. \end{aligned}$$

$\square$

Lemma 1

The number of fixed-points from $\tilde{\Omega }_{m,n}$ for a permutation $g=(\tau ,\sigma )$ is equal to

$$\begin{aligned} | {\tilde{F}_g } |=\left\{ {\begin{array}{l@{\quad }l} \mathop {\prod }\limits _{j=0}^\alpha {(m!-j\cdot d} ),&{} if\quad \lambda _1 =\lambda _2 =\cdots =\lambda _\alpha =d, \\ 0, &{}\quad otherwise. \\ \end{array}} \right. \end{aligned}$$

where $d=LCM(\mu )$.

Proof

For some $\vec {P}\in \tilde{\Omega }_{m,n}$ and some $g\in G$, suppose that $\vec {P}^{g}=\vec {P}$. Let us consider the permutation of alternatives $\tau$. The order of this permutation, d, is equal to the least common multiple of cycle lengths, $d=LCM(\mu )$. If a column is $(a_1 ,a_2 ,\ldots ,a_m)$, then after the action of permutation $\tau$, it becomes $(a_{\tau (1)} ,a_{\tau (2)} ,\ldots ,a_{\tau (m)})$; after the second action of $\tau$, it becomes $(a_{\tau ^{2}(1)} ,a_{\tau ^{2}(2)} ,\ldots ,a_{\tau ^{2}(m)})$; and so on. Finally, $(a_{\tau ^{d-1}(1)} ,a_{\tau ^{d-1}(2)} ,\ldots ,a_{\tau ^{d-1}(m)})$ again becomes $(a_1 ,a_2 ,\ldots ,a_m)$.

Next, for a given permutation $\sigma$ of voters, we take any cycle $(i_1 \quad i_2 \ldots \quad i_k)$. Let the $i_1$-th column be $(a_1 ,a_2 ,\ldots ,a_m)$. Then, after the action of permutation $\sigma$, it becomes the column number $i_2$. At the same time, $\tau$ permutes alternatives and the column number $i_2$ should be equal to $(a_{\tau (1)} ,a_{\tau (2)} ,\ldots ,a_{\tau (m)})$ because the preference profile after permuting both voters and alternatives should be the same. Thus, if the $i_2$-th column $(a_{\tau (1)} ,a_{\tau (2)} ,\ldots ,a_{\tau (m)})$ after permuting columns becomes the $i_3$-th column and $(a_{\tau (1)} ,a_{\tau (2)} ,\ldots ,a_{\tau (m)})$ is mapped to $(a_{\tau ^{2}(1)} ,a_{\tau ^{2}(2)} ,\ldots ,a_{\tau ^{2}(m)})$ by $\tau$, then the $i_3$-th column should be $(a_{\tau ^{2}(1)} ,a_{\tau ^{2}(2)} ,\ldots ,a_{\tau ^{2}(m)})$, etc. Finally, the $i_k$-th column becomes the $i_1$-th column, and we conclude that it should be equal to $(a_{\tau ^{d-1}(1)} ,a_{\tau ^{d-1}(2)} ,\ldots ,a_{\tau ^{d-1}(m)})$ in order to be mapped to $(a_1 ,a_2 ,\ldots ,a_m)$ by permutation $\tau$. Since repeated columns are not permitted, the length of the cycle $(i_1 \quad i_2 \ldots \quad i_k)$, k, should be equal to $d=LCM(\mu )$. The same result holds for any other cycle of permutation $\sigma$; in other words, the length of all cycles in $\sigma$ is the same,$\lambda _1 =\lambda _2 =\ldots =\lambda _\alpha$, or $GCD(\lambda )=LCM(\lambda )$, and equals d.

Suppose a permutation $g=(\tau ,\sigma )$ is such that $\lambda _1 =\lambda _2 =\cdots =\lambda _\alpha =d$ and $\sigma$ consists of $\alpha =n/d$ cycles. The first column of the first cycle in $\sigma$ is chosen from m! different columns. The rest of the columns in this cycle are determined by permutation $\tau$. Then, the first column of the second cycle in $\sigma$ can be represented by any of $m!-d$ variants, as d columns are already used in the first cycle. The first column of the third cycle can be defined $m!-2d$ different ways and so on. Inside any cycle, there cannot appear any column that is already used because a permutation $\tau$ splits up the set of all different columns into non-intersecting orbits of columns that can be produced one from another by this permutation. Columns in every cycle form one of such orbits.

Finally, we obtain an exact formula for the number of fixed-points from $\tilde{\Omega }_{m,n}$ for some permutation $g=(\tau ,\sigma )$

$$\begin{aligned} | {\tilde{F}_g } |=\left\{ {\begin{array}{l@{\quad }l} \mathop {\prod }\limits _{j=0}^\alpha {(m!-j\cdot d} ),&{} if\quad \lambda _1 =\lambda _2 =\cdots =\lambda _\alpha =d, \\ 0,&{}\quad otherwise. \\ \end{array}} \right. \end{aligned}$$

$\square$

Theorem 2

For any m and n such that $m!>n$, the number of equivalence classes on $\tilde{\Omega }_{m,n}$ is equal to

$$\begin{aligned} \tilde{R}(m,n)=\sum _\lambda {\sum _\mu {z_\lambda ^{-1} z_\mu ^{-1} \chi (S(\mu ,\lambda ))\mathop {\prod }\limits _{j=0}^{\alpha -1} {(m!-j\cdot d)} } } , \end{aligned}$$

where $S(\mu ,\lambda )$ is the statement “$\lambda _1 =\lambda _2 =\cdots =\lambda _\alpha =d$”.

Proof

Let $\tilde{R}$ be the number of ANECs on $\tilde{\Omega }_{m,n}$, and $\vec {P}^{1},\vec {P}^{2},\ldots ,\vec {P}^{\tilde{R}}$ be the representatives of these classes. For any preference profile $\vec {P} \quad | {\theta _{\vec {P}} } |\cdot | {G_{\vec {P}} } |=| G |$ holds. Then, take the sum of these equalities over representatives

$$\begin{aligned}&\sum _{i=1}^{\tilde{R}} {| {\theta _{\vec {P}^{i}} } |\cdot | {G_{\vec {P}^{i}} } |} =\tilde{R}\cdot | G |,\\&\tilde{R}=\frac{1}{| G |}\sum _{i=1}^{\tilde{R}} {| {\theta _{\vec {P}^{i}} } |\cdot | {G_{\vec {P}^{i}} }|} . \end{aligned}$$

Since $| {G_{\vec {P}^{k}} } |=| {G_{\vec {P}^{l}} }|$ if $\vec {P}^{k}$ and $\vec {P}^{l}$ belong to the same ANEC, we can rewrite

$$\begin{aligned}&| {\theta _{\vec {P}^{i}} } |\cdot | {G_{\vec {P}^{i}} } |=\sum _{\vec {P}\in \theta _{\vec {P}^{i}} } {| {G_{\vec {P}^{i}} } |} .\\&\tilde{R}=\frac{1}{| G |}\sum _{i=1}^{\tilde{R}} {| {\theta _{\vec {P}^{i}} } |\cdot | {G_{\vec {P}^{i}} } |} =\frac{1}{| G |}\sum _{i=1}^{\tilde{R}} {\sum _{\vec {P}\in \theta _{\vec {P}^{i}} } {| {G_{\vec {P}^{i}} } |} =} \frac{1}{| G |}\sum _{\vec {P}\in \tilde{\Omega }_{m,n} } {| {G_{\vec {P}} }|} \end{aligned}$$

The sum of stabilizing permutations over all preference profiles from $\tilde{\Omega }_{m,n}$ is equal to the sum of all fixed-points from $\tilde{\Omega }_{m,n}$ for all permutations

$$\begin{aligned}&\sum _{\vec {P}\in \tilde{\Omega }} {| {G_{\vec {P}} } |} =\sum _{\vec {P}\in \tilde{\Omega }} {\sum _{g\in G_{\vec {P}} } 1 =} \sum _{g\in G} {\sum _{\vec {P}\in \tilde{F}_g } 1 =} \sum _{g\in G} {| {\tilde{F}_g } |} .\\&\tilde{R}=\frac{1}{| G |}\sum _{g\in G} {| {\tilde{F}_g } |}. \end{aligned}$$

Using Lemma 1 and denoting the statement “$\lambda _1 =\lambda _2 =\cdots =\lambda _\alpha =d$” by $S(\mu ,\lambda )$ we get

$$\begin{aligned} \tilde{R}(m,n)=\frac{1}{| G |}\sum _{g\in G} {\chi (S(\lambda ,\mu ))\mathop {\prod }\limits _{k=0}^\alpha {(m!-k\cdot LCM(\mu )} )} . \end{aligned}$$

Since $| {\tilde{F}_g }|$ depends only on the cycle type of permutation $g=(\tau ,\sigma )$, we can take the sum over all partitions $\lambda$ and $\mu$ and multiply every component by the number of permutations of n with partition $\lambda$ and the number of permutations of m with partition $\mu , z_\lambda ^{-1} n!$ and $z_\mu ^{-1} m!$, respectively.

$$\begin{aligned} \tilde{R}(m,n)= & {} \frac{1}{m!n!}\sum _\lambda {\sum _\mu {z_\lambda ^{-1} n!\cdot z_\mu ^{-1} m!\cdot \chi (S(\lambda ,\mu ))\mathop {\prod }\limits _{j=0}^{\alpha -1} {(m!-j\cdot d)} } } \\= & {} \sum _\lambda {\sum _\mu {z_\lambda ^{-1} z_\mu ^{-1} \chi (S(\lambda ,\mu ))\mathop {\prod }\limits _{j=0}^{\alpha -1} {(m!-j\cdot d)} } } . \end{aligned}$$

$\square$

Corollary 1

For any m and n such that $m!>n$:

• (a) The number of maximal ANEC satisfies the following inequality

  $$\begin{aligned} \frac{2(m!-1)!}{(m!-n)!n!}-\tilde{R}(m,n)\le R_{\max } (m,n)\le \tilde{R}(m,n). \end{aligned}$$

• (b) If for m and n we have $n>m$ and n is a prime number, then the number of maximal ANECs is equal to $\tilde{R}(m,n)$.

Proof

• (a) The second inequality is obvious, because preference profiles from maximal equivalence class always consist of different columns for $m!>n$. The total number of preference profiles consisting of pairwise different columns is

  $$\begin{aligned} m!\cdot (m!-1)\cdot \cdots \cdot (m!+1-n)=\frac{(m!)!}{(m!-n)!}. \end{aligned}$$

  At the same time, it is the number of fixed points from $\tilde{\Omega }_{m,n}$ for the identity permutation. Since for the identity permutation the statement $S(\lambda , \mu )$ is true, $(m!)!/(m!-n)!$ is included in the sum $m!n!\tilde{R}(m,n)$. The rest of the components of the sum, i.e.

  $$\begin{aligned} m!n!\sum _\lambda {\sum _\mu {z_\lambda ^{-1} z_\mu ^{-1} \chi (S(\lambda ,\mu ))\mathop {\prod }\limits _{j=0}^{\alpha -1} {(m!-j\cdot d)} } } -\frac{(m!)!}{(m!-n)!} \end{aligned}$$

  form the sum of fixed points from $\tilde{\Omega }_{m,n}$ for all permutations except the identity permutation. The problem is that the sets $\tilde{F}_g$ intersect and we cannot find an exact number of preference profiles that have more than one stabilizing permutation. However, we can be sure that this number is not more than this sum. Consequently, the number of profiles having only one stabilizing permutation is not less than

  $$\begin{aligned}&\frac{(m!)!}{(m!-n)!}-\left( {m!n!\sum _\lambda {\sum _\mu {z_\lambda ^{-1} z_\mu ^{-1} \chi (S(\lambda ,\mu ))\mathop {\prod }\limits _{j=0}^{\alpha -1} {(m!-j\cdot d)} } } -\frac{(m!)!}{(m!-n)!}} \right) \\&\quad = \frac{2(m!)!}{(m!-n)!}-m!n!\sum _\lambda {\sum _\mu {z_\lambda ^{-1} z_\mu ^{-1} \chi (S(\lambda ,\mu ))\mathop {\prod }\limits _{j=0}^{\alpha -1} {(m!-j\cdot t)} } } . \end{aligned}$$

  Dividing by the cardinality m!n! of the maximal ANEC, we get

  $$\begin{aligned} R_{\max } (m,n)\ge \frac{2(m!-1)!}{(m!-n)!n!}-\tilde{R}(m,n). \end{aligned}$$

• (b) If n is a prime number, then $\sigma$ either contains only one cycle of length n, or n cycles of length one. The first case means that the least common divisor of the $\tau$ cycle lengths should also be n, but we assumed $n>m$. Thus, this is impossible. The second case means that we only have an identity permutation in a stabilizer group of any profile from $\tilde{\Omega }_{m,n}$, i.e. $\tilde{\Omega }_{m,n}$ consists of preference profiles from maximal equivalence classes, and we have only one component in the sum $\tilde{R}(m,n)$

  $$\begin{aligned} \tilde{R}(m,n)=R_{\max } (m,n)=\frac{(m!-1)!}{(m!-n)!n!}. \end{aligned}$$

$\square$

Lemma 2

For differences between any probabilistic models $M_1, M_2$, $M_3$, the triangular inequality holds

$$\begin{aligned} \Delta _{M_1 -M_2 } (m,n)+\Delta _{M_2 -M_3 } (m,n)\ge \Delta _{M_1 -M_3 } (m,n). \end{aligned}$$

Proof

Let $E_1$, $E_2$ and $E_3$ denote events that maximize the difference of probabilities between $M_1$ and $M_2$, $M_2$ and $M_3$, $M_1$ and $M_3$ respectively.

$$\begin{aligned} E_1= & {} \arg \mathop {\max }\limits _E (|\Pr (E,M_1 ,m,n)-\Pr (E,M_2 ,m,n)|),\\ E_2= & {} \arg \mathop {\max }\limits _E (|\Pr (E,M_2 ,m,n)-\Pr (E,M_3 ,m,n)|),\\ E_3= & {} \arg \mathop {\max }\limits _E (|\Pr (E,M_1 ,m,n)-\Pr (E,M_3 ,m,n)|). \end{aligned}$$

Since the absolute difference s maximal both for $E_1$ and the complementary event $E_1 ^{c}$, without loss of generality, suppose that all differences are positive for $E_1$, $E_2$ and $E_3$. Then,

$$\begin{aligned}&\Pr (E_3 ,M_1 ,m,n)-\Pr (E_3 ,M_3 ,m,n)=\Pr (E_3 ,M_1 ,m,n)-\Pr (E_3 ,M_2 ,m,n)\\&\quad +(\Pr (E_3 ,M_2 ,m,n)-\Pr (E_3 ,M_3 ,m,n)).\\&\Pr (E_3 ,M_1 ,m,n)-\Pr (E_3 ,M_2 ,m,n)\\&\quad \le \Pr (E_1 ,M_1 ,m,n)-\Pr (E_1 ,M_2 ,m,n)=\Delta _{M_1 -M_2 } (m,n)\\&\Pr (E_3 ,M_2 ,m,n)-\Pr (E_3 ,M_3 ,m,n)\\&\quad \le \Pr (E_2 ,M_2 ,m,n)-\Pr (E_2 ,M_3 ,m,n)=\Delta _{M_2 -M_3 } (m,n). \end{aligned}$$

Therefore,

$$\begin{aligned} \Delta _{M_1 -M_3 } (m,n)= & {} \Pr (E_3 ,M_1 ,m,n)-\Pr (E_3 ,M_3 ,m,n)\\\le & {} \Delta _{M_1 -M_2 } (m,n)+\Delta _{M_2 -M_3 } (m,n). \end{aligned}$$

$\square$

Theorem 3

The share of maximal N-AECs and the number of elements in them tends to 1 as $n\rightarrow \infty$ or $m\rightarrow \infty$.

• (1) $\mathop {\lim }\nolimits _{n\rightarrow \infty }\frac{m!R_r (m,n)}{K(m,n)}=1$ and $\mathop {\lim }\nolimits _{m\rightarrow \infty }\frac{m!R_r (m,n)}{K(m,n)}=1$;

• (2) $\mathop {\lim }\nolimits _{n\rightarrow \infty }\frac{R_r (m,n)}{R(m,n)}=1$ and $\mathop {\lim }\nolimits _{m\rightarrow \infty }\frac{R_r (m,n)}{R(m,n)}=1$.

Proof

• (1) First prove that

$$\begin{aligned}&\mathop {\lim }\limits _{n\rightarrow \infty }\frac{K(m,n)}{m!R(m,n)}=1\hbox { and }\mathop {\lim }\limits _{m\rightarrow \infty } \frac{K(m,n)}{m!R(m,n)}=1.\\&\quad \frac{K(m,n)}{m!R(m,n)} = {K(m,n)}\Big /\left( m!\cdot \frac{1}{m!}\left( {{\begin{array}{l} {m!+n-1} \\ {m!-1} \\ \end{array} }} \right) \right. \\&\qquad \left. +m!\sum _{\mu \in \Pi _m \backslash \{(1,1,\ldots ,1)\}} {z_\mu ^{-1} \left( {{\begin{array}{l} {n/d+m!/d-1} \\ {m!/d-1} \\ \end{array} }} \right) } \right) \\&\quad ={K(m,n)}\Big /{\left( {K(m,n)+m!\sum _{\mu \in \Pi _m \backslash \{(1,1,\ldots ,1)\}} {z_\mu ^{-1} \left( {{\begin{array}{l} {n/d+m!/d-1} \\ {m!/d-1} \\ \end{array} }} \right) } } \right) } \end{aligned}$$

Let us denote the sum in the denominator by $\mathop {\sum }\nolimits _{\mu \in \Pi _m \backslash \{(1,1,\ldots ,1)\}} {A_\mu }$, where

$$\begin{aligned} A_\mu =z_\mu ^{-1} \left( {{\begin{array}{l} {n/d+m!/d-1} \\ {m!/d-1} \\ \end{array} }} \right) \quad \hbox { for }\mu \in \Pi _m \backslash \{(1,1,\ldots ,1)\}. \end{aligned}$$

Then

$$\begin{aligned} \frac{K(m,n)}{K(m,n)+m!\sum _{\begin{array}{l} \mu \in \Pi _m \backslash \\ \{(1,1,\ldots ,1)\} \\ \end{array}} {A_\mu } }=\frac{1}{1+\mathop {\sum }\limits _{\begin{array}{l} \mu \in \Pi _m \backslash \\ \{(1,1,\ldots ,1)\} \\ \end{array}} {\frac{m!A_\mu }{K(m,n)}} }. \end{aligned}$$

Suppose, that n / d is integer, and the binomial coefficient is non-zero

$$\begin{aligned} \frac{m!A_\mu }{K(m,n)}=\frac{m!}{z_\mu }\cdot \frac{\left( {{\begin{array}{l} {n/d+m!/d-1} \\ {m!/d-1} \\ \end{array} }} \right) }{\left( {{\begin{array}{l} {m!+n-1} \\ {m!-1} \\ \end{array} }} \right) }=\frac{m!}{z_\mu }\cdot \frac{(n/d+m!/d-1)!(m!-1)!n!}{(m!/d-1)!(n/d)!(m!+n-1)!}. \end{aligned}$$

• (2) Fix the value of m and let $n\rightarrow \infty$

$$\begin{aligned}&\mathop {\lim }\limits _{n\rightarrow \infty }\frac{m!}{z_\mu }\cdot \frac{(n/d+m!/d-1)!(m!-1)!n!}{(m!/d-1)!(n/d)!(m!+n-1)!}\\&\quad =\frac{m!(m!-1)!}{z_\mu (m!/d-1)!}\mathop {\lim }\limits _{n\rightarrow \infty }\frac{(n/d+m!/d-1)!n!}{(n/d)!(m!+n-1)!}\\&\quad = \frac{m!(m!-1)!}{z_\mu (m!/d-1)!}\mathop {\lim } \limits _{n\rightarrow \infty }\frac{(n/d+m!/d-1)!}{(n/d)!}\cdot \frac{n!}{(m!+n-1)!}\\&\quad = \frac{m!(m!-1)!}{z_\mu (m!/d-1)!}\mathop {\lim }\limits _{n\rightarrow \infty }\frac{(n/d+1)\cdot \cdots \cdot (n/d+m!/d-1)}{(n+1)\cdot \cdots \cdot (m!+n-1)}=0. \end{aligned}$$

Now fix n and let $m\rightarrow \infty$

$$\begin{aligned}&\mathop {\lim }\limits _{m\rightarrow \infty }\frac{m!}{z_\mu }\cdot \frac{(n/d+m!/d-1)!(m!-1)!n!}{(m!/d-1)!(n/d)!(m!+n-1)!}\\&\quad =\frac{n!}{(n/d)!}\mathop {\lim }\limits _{n\rightarrow \infty }\frac{m!}{z_\mu }\cdot \frac{(n/d+m!/d-1)!(m!-1)!}{(m!/d-1)!(m!+n-1)!} \frac{m!}{z_\mu }\cdot \frac{(n/d+m!/d-1)!(m!-1)!}{(m!/d-1)!(m!+n-1)!}\\&\quad < \frac{m!(n/d+m!/d-1)!(m!-1)!}{(m!/d-1)!(m!+n-1)!}, \quad \mathop {\lim }\limits _{m\rightarrow \infty }\frac{m!(n/d+m!/d-1)!(m!-1)!}{(m!/d-1)!(m!+n-1)!}\\&\quad =\mathop {\lim } \limits _{m\rightarrow \infty }\frac{m!(n/d+m!/d-1)\cdot \cdots \cdot (m!/d)}{(m!+n-1)\cdot \cdots \cdot (m!+1)m!}\\&\quad =\mathop {\lim }\limits _{m\rightarrow \infty }\frac{(n/d+m!/d-1)}{(m!+n-1)}\cdot \cdots \cdot \frac{(m!/d)}{(m!+n-n/d)}\\&\qquad \quad \cdot \frac{1}{(m!+n-n/d-1)\cdot \cdots \cdot (m!+1)}=0. \end{aligned}$$

Consequently,

$$\begin{aligned} \mathop {\lim }\limits _{m\rightarrow \infty }\frac{m!}{z_\mu }\cdot \frac{(n/d+m!/d-1)!(m!-1)!n!}{(m!/d-1)!(n/d)!(m!+n-1)!}=0. \end{aligned}$$

• (3) Now sum the terms $m!A_\mu /K(m,n)$ over all partitions $\mu$ from $\Pi _m \backslash \{(1,1,\ldots ,1)\}$ and s.t. n / d is integer. Let us denote this set of partitions by ${\Pi }^{\prime }_{m}$. The number of elements in ${\Pi }^{\prime }_{m}$ is less or equal than in $\Pi _m \backslash \{(1,1,\ldots ,1)\}$. Further, $| {\Pi _m \backslash \{(1,1,\ldots ,1)\}} |<(m-1)!$ for $m\ge 4$ and for $m=3 | {\Pi _m \backslash \{(1,1,\ldots ,1)\}} |=(m-1)!=2$.

$$\begin{aligned}&\frac{(n/d+m!/d-1)}{(m!+n-1)}\cdot \cdots \cdot \frac{(m!/d)}{(m!+n-n/d)}\cdot \frac{1}{(m!+n-n/d-1)\cdot \cdots \cdot (m!+1)}\\&\quad \le \frac{(n/d+m!/d-1)}{(m!+n-1)}\cdot \cdots \cdot \frac{(m!/d)}{(m!+n-n/d)}\cdot \frac{1}{(m!+1)}\le \frac{1}{(m!+1)}<\frac{1}{m!} \end{aligned}$$

Therefore,

$$\begin{aligned} \mathop {\sum }\limits _{\mu \in {\Pi }'_m } {\frac{m!A_\mu }{K(m,n)}} \le \frac{(m-1)!}{m!}. \end{aligned}$$

Finally,

Since

$$\begin{aligned}&\mathop {\lim }\limits _{m\rightarrow \infty }\frac{(m-1)!}{m!}=0,\\&\mathop {\lim }\limits _{m\rightarrow \infty }\mathop {\sum }\limits _{\mu \in {\Pi }'_m } {\frac{m!A_\mu }{K(m,n)}} =0\quad \hbox { and }\quad \mathop {\lim }\limits _{n\rightarrow \infty }\mathop {\sum }\limits _{\mu \in {\Pi }'_m } {\frac{m!A_\mu }{K(m,n)}=0},\\&\mathop {\lim }\limits _{n\rightarrow \infty }\frac{K(m,n)}{K(m,n)+m!\mathop {\sum }\limits _{\mu \in {\Pi }'_m } {A_\mu } }=1\quad \hbox { and }\quad \mathop {\lim }\limits _{m\rightarrow \infty }\frac{K(m,n)}{K(m,n)+m!\mathop {\sum }\limits _{\mu \in {\Pi }'_m } {A_\mu } }=1,\\&\mathop {\lim }\limits _{n\rightarrow \infty }\frac{K(m,n)}{m!R(m,n)}=1\quad \hbox { and }\quad \mathop {\lim }\limits _{m\rightarrow \infty } \frac{K(m,n)}{m!R(m,n)}=1. \end{aligned}$$

• (4) In this part the proof is done for $n\rightarrow \infty$, for $m\rightarrow \infty$ the proof is analogous.

$$\begin{aligned}&\mathop {\lim }\limits _{n\rightarrow \infty }\frac{m!R(m,n)-K(m,n)}{m!R(m,n)}=0,\\&\frac{m!R(m,n)-K(m,n)}{m!R(m,n)}\ge \frac{(R(m,n)-R_r (m,n))m!/2}{m!R(m,n)},\\&\mathop {\lim }\limits _{n\rightarrow \infty }\frac{(R(m,n)-R_r (m,n))m!/2}{m!R(m,n)}=0\quad \hbox { and }\quad \mathop {\lim }\limits _{n\rightarrow \infty } \frac{R(m,n)-R_r (m,n)}{R(m,n)}=0,\\&\mathop {\lim }\limits _{n\rightarrow \infty }\frac{R_r (m,n)}{R(m,n)}=1. \end{aligned}$$

Finally,

$$\begin{aligned}&\frac{m!R_r (m,n)}{m!R(m,n)}\le \frac{m!R_r (m,n)}{K(m,n)},\\&\mathop {\lim }\limits _{n\rightarrow \infty }\frac{m!R_r (m,n)}{K(m,n)}=1. \end{aligned}$$

$\square$

Corollary 2

The difference between the IAC and IANC models $\Delta _{\textit{IAC}-\textit{IANC}}$ tends to zero as $n\rightarrow \infty$ or $m\rightarrow \infty$.

Proof

$\Delta _{\textit{IAC}-\textit{IANC}}$ is calculated as the difference of two components: (1) the share of N-AECs $\vartheta$ whose cardinality is grater than average cardinality, $av(|\vartheta |)$; (2) the share of AECs $\rho$ which belong to such $\vartheta , |\vartheta |>av(|\vartheta |)$. Suppose, the set $\{\vartheta :\vartheta >av(|\vartheta |)\}$ consists of $R_1$ N-AECs of cardinality m! (containing $K_1$ AECs), $R_2$ N-AECs of cardinality m! / 2 (containing $K_2$ AECs), $R_3$ N-AECs of cardinality m! / 3(containing $K_3$ AECs) and so on. Then,

$$\begin{aligned} \Delta _{\textit{IAC}-\textit{IANC}} =\frac{R_1 }{R(m,n)}-\frac{K_1 }{K(m,n)}+\frac{R_2 }{R(m,n)}-\frac{K_2 }{K(m,n)}+\cdots \end{aligned}$$

The next part of the proof is done for $n\rightarrow \infty$, for $m\rightarrow \infty$ the proof is analogous. By Theorem 3,

$$\begin{aligned}&\mathop {\lim }\limits _{n\rightarrow \infty }\frac{R_1 }{R(m,n)}=1\quad \hbox { and }\quad \mathop {\lim }\limits _{n\rightarrow \infty }\frac{K_1 }{K(m,n)}=\mathop {\lim }\limits _{n\rightarrow \infty }\frac{m!R_r (m,n)}{K(m,n)}=1.\\&\forall i\ge 2 \quad \frac{R_i }{R(m,n)}\le \frac{R(m,n)-R_1 }{R(m,n)}, \quad \frac{K_i }{K(m,n)}\le \frac{K(m,n)-K_1 }{K(m,n)}.\\&\mathop {\lim }\limits _{n\rightarrow \infty }\frac{R(m,n)-R_1 }{R(m,n)}=0\quad \hbox { and }\quad \mathop {\lim }\limits _{n\rightarrow \infty }\frac{K(m,n)-K_1 }{K(m,n)}=0,\\&\forall i\ge 2 \quad \mathop {\lim }\limits _{n\rightarrow \infty }\frac{R_i }{R(m,n)}=0\quad \hbox { and }\quad \mathop {\lim }\limits _{n\rightarrow \infty } \frac{K_i }{K(m,n)}=0. \end{aligned}$$

Summing up,

$$\begin{aligned}&\mathop {\lim }\limits _{n\rightarrow \infty }\Delta _{\textit{IAC}-\textit{IANC}}\\&\quad =\mathop {\lim }\limits _{n\rightarrow \infty }\left( {\frac{R_1 }{R(m,n)}-\frac{K_1 }{K(m,n)}+\frac{R_2 }{R(m,n)}-\frac{K_2 }{K(m,n)}+\cdots } \right) =0. \end{aligned}$$

$\square$

Appendix 2

See Figs. 8, 9 and 10

Fig. 8

The Nitzan–Kelly’s index for the Leximax method and 3 alternatives in the IC model

Fig. 9

The Nitzan–Kelly’s index for the Leximax method and 3 alternatives in the IANC model

Fig. 10

The difference of the Nitzan–Kelly’s index for 3 alternatives in IC and IANC, Leximax