\section{Mallows}

We consider the following single-agent preference model. Let $\mathcal{M}=(\Theta=\{(R_0, \varphi)\}, \mathcal{L}(\mathcal{A}), \Pi=\{\pi_{(R_0,\varphi)}\})$, which has been defined in Example \ref{ex:mallows}. We suppose $R_0=[a\succ c_1 \succ c_2\cdots\succ  c_m]$, without loss of generality.
Given a voting rule $r$, we are interested in the likelihood of 
\[
\widetilde{\mallows}(r, R_0, \varphi)(n):=\mathrm{Pr}_{P\sim\pi_{(R_0, \varphi)}^n}(r(P)\neq a).
\]

\subsection{Likelihood of outcome of Mallows' model different from ground truth}

\begin{theorem} \label{thm:Mallows_psr_gt}
    Given $m\geq 3$ and any score vector $\vec s$, 
    \[
    \widetilde{\mallows}(r_{\vec s}, R_0, \varphi) (n)=
    \begin{cases}
        \Theta(1) & \varphi = 1 \\
        \exp(-\Theta(n)) & \varphi <1.
    \end{cases}
    \]
\end{theorem}

\begin{proof}
    When $\varphi =1$, we have $\pi_{(R_0,\varphi)}=\pi_{uni}$, and hence $\widetilde{\mallows}(r_{\vec s}, R_0, 1)(n)=\Theta(1)$ can be derived from Theorem \ref{thm:diffposi}. When $\varphi<1$, we consider $m-1$ events $\{Y_j\}_{j\in[m-1]}$, where $Y_j$ is ``the score of $c_j$ is higher than that of $a$'', we have 
    \[
    \begin{aligned}
        \widetilde{\mallows}(r_{\vec s}, R_0, \varphi) & =\mathrm{Pr}_{P\sim\pi_{(R_0, \varphi)}^n}(r(P)\neq a)  \\
        & \leq \sum_{j=1}^{m-1}\mathrm{Pr}_{P\sim\pi_{(R_0, \varphi)}^n}(Y_j)
    \end{aligned}
    \]
    %We then characterize each $Y_j$ by polyhedron $\mathcal{H}^j$, and verify $\pi_{(R_0, \varphi)}\notin (\mathcal{H}^j)_{\leq 0}$. Finally, we derive the conclusion by Theorem \ref{thm:pmv1}.
    
    For each $j$, we characterize $\mathcal{H}^j$ by $Score_{1,j}^{\vec s}\cdot\vec x\leq -\vec 1$, and it is easy to verify the equivalence between $\Hist(P)\in\mathcal{H}^j$ and $Y_j$. Let $\vec s(R, a, c_j)$ be the score difference of $a$ and $c_j$ under $\vec s$, we then verify that $\pi_{(R_0,\varphi)}\notin(\mathcal{H}^j)_{\leq 0}$. Partition $\mathcal{L}(\mathcal{A})$ by $\mathcal{L}_{a\succ c_j}$ and $\mathcal{L}_{c_j\succ a}$, where $\mathcal{L}_{i\succ j}$ consists of linear orders with $i\succ j$. Given $R_1\in\mathcal{L}_{a\succ c_j}$, we consider pair $(R_1, R_2)$ where $R_2\in \mathcal{L}_{c_j\succ a}$ is generated by reversing the ranking of $a$ and $c_j$ in $R_1$. It can be verified that 
    $KT(R_2,R_0)\geq KT(R_1,R_0)+1$.
    Hence we have 
    \[
    \begin{aligned}
        Score_{a,c_j}^{\vec{s}}\cdot \pi_{(R_0,\varphi)} &=\frac{1}{Z_\varphi}\sum_{R\in\mathcal{L}(\mathcal{A})} \varphi^{KT(R,R_0)}\cdot \vec s(R,a,c_j)\\
        & = \frac{1}{Z_\varphi}\left(\sum_{R_1\in \mathcal{L}_{a\succ c_j}}\varphi^{KT(R_1,R_0)}\vec s(R_1, a, c_j) -\sum_{R_2\in \mathcal{L}_{c_j\succ a}} \varphi^{KT(R_2,R_0)} \vec s(R_2, c_j, a)\right) \\
        & = \frac{1}{Z_\varphi}\sum_{R_1\in \mathcal{L}_{a\succ c_j}}\left(\varphi^{KT(R_1,R_0)}-\varphi^{KT(R_2,R_0)}\right)\vec s(R_1, a, c_j) \\
        & \geq \frac{1}{Z_\varphi}\sum_{R_1\in \mathcal{L}_{a\succ c_j}}(1-\varphi)\varphi^{KT(R_1,R_0)}\cdot\vec s(R_1, a, c_j)>0,
    \end{aligned}
    \]
    and $\pi_{(R_0,\varphi)}\notin(\mathcal{H}^j)_{\leq 0}$ as a result. Finally, it is clear that $(\mathcal{H}^j)^{\mathbb{Z}}_n\neq \emptyset$ and also $(\mathcal{H}^j)_{\leq 0}\cap CH(\Pi)=\emptyset$, we have 
    \[
    \mathrm{Pr}_{P\sim\pi_{(R_0, \varphi)}^n}(Y_j)=\exp(-\Theta(n))
    \]
    by Theorem \ref{thm:pmv1}, and hence 
    \[
    \widetilde{\mallows}(r_{\vec s}, R_0, \varphi) (n)\leq \sum_{j=1}^{m-1}\mathrm{Pr}_{P\sim\pi_{(R_0, \varphi)}^n}(Y_j) =\exp(-\Theta(n)).
    \]
    Also, there exists $j_0$ whose score is higher than that of $a$ when $a$ is not the positional rule winner, and hence, 
    \[
    \begin{aligned}
        \widetilde{\mallows}(r_{\vec s}, R_0, \varphi) (n)& =\mathrm{Pr}_{P\sim\pi_{(R_0, \varphi)}^n}(r(P)\neq a) \\
        & \geq \mathrm{Pr}_{P\sim\pi_{(R_0, \varphi)}^n}(Y_{j_0})=\exp(-\Theta(n)),
    \end{aligned}
    \]
    which completes the proof.
\end{proof}

\begin{theorem}
    Given $m\geq 3$ and $\varphi<1$. For any voting rule $r$, where $r(P)=a$ whenever $a$ is the Condorcet winner under profile $P$, 
    we have 
    \[
    \widetilde{\mallows}(r, R_0, \varphi)(n) = \exp(-\Theta(n)).
    \]
\end{theorem}

\begin{proof}
    %If $a$ is a Condorcet winner, then $r(P)=a$ for any WMG-based rules. 
    A fact is that $r(P)\neq a$ implies $a$ is not a Condorcet winner. Let $Y$ denote the event where $a$ is not the Condorcet winner and let $Y_j$ be the event where $c_j\in\{c_1,\cdots,c_{m-1}\}$ is an alternative win the pairwise comparison to $a$. We have 
    \[
    \begin{aligned}
        \widetilde{\mallows}(r, R_0, \varphi)(n)& = \mathrm{Pr}_{P\sim\pi_{(R_0, \varphi)}^n}(r(P)\neq a) \\
        & \leq \mathrm{Pr}_{P\sim\pi_{(R_0, \varphi)}^n}(Y) \\
        & \leq \sum_{j\in[m-1]} \mathrm{Pr}_{P\sim\pi_{(R_0, \varphi)}^n}(Y_j).
    \end{aligned}
    \]
    Event $Y_j$ can be characterized by $Pair_{a,c_j}\cdot \Hist(P)\leq -\vec 1$ for each $j$. Let $\mathcal{H}^j$ denote $Pair_{a,c_j}$, we verify $\pi_{(R_0,\varphi)}\notin (\mathcal{H}^j)_{\leq 0}$. 
    \[
        \begin{aligned}
            Pair_{a,c_j}\cdot & = \frac{1}{Z_\varphi}\left(\sum_{R_1\in \mathcal{L}_{a\succ c_j}}\varphi^{KT(R_1,R_0)} -\sum_{R_2\in \mathcal{L}_{c_j\succ a}} \varphi^{KT(R_2,R_0)}\right) \\
            & = \frac{1}{Z_\varphi}\sum_{R_1\in \mathcal{L}_{a\succ c_j}}\left(\varphi^{KT(R_1,R_0)}-\varphi^{KT(R_2,R_0)}\right)\\ 
            & \geq \frac{1}{Z_\varphi}\sum_{R_1\in \mathcal{L}_{a\succ c_j}}(1-\varphi)\varphi^{KT(R_1,R_0)} >0.
        \end{aligned}
    \]
    Hence, $\pi_{(R_0,\varphi)}\notin(\mathcal{H}^j)_{\leq 0}$. We have 
    \[
    \mathrm{Pr}_{P\sim\pi_{(R_0, \varphi)}^n}(Y_j)=\exp(-\Theta(n))
    \]
    by Theorem \ref{thm:pmv1}, and hence 
    \[
    \widetilde{\mallows}(r, R_0, \varphi)(n)\leq \sum_{j=1}^{m-1}\mathrm{Pr}_{P\sim\pi_{(R_0, \varphi)}^n}(Y_j) =\exp(-\Theta(n)).
    \]


    On the other hand, let $Y_1'$ be the event where $c_1$ is the Condorcet winner. Clearly $Y_1'$ implies $r(P)\neq a$.
    Also, this event can be characterized by $Ax\leq \vec b$, where $A$ is the matrix with row vectors $\{Pair_{k ,a}: k\in\{a, c_2, \cdots, c_m\} \}$ and $\vec b= -\vec 1$. Let $\mathcal{H}'$ be the corresponding polyhedron, then clearly $\pi_{(R_0,\varphi)}\notin\mathcal{H}'_{\leq 0}$ and $\mathcal{H}_n^{\mathbb{Z}}\neq \emptyset$, and hence 
    
    \[
    \begin{aligned}
        \widetilde{\mallows}(r, R_0, \varphi)(n) & =\mathrm{Pr}_{P\sim\pi_{(R_0, \varphi)}^n}(r(P)\neq a) \\
        & \geq \mathrm{Pr}_{P\sim\pi_{(R_0, \varphi)}^n}(Y_1')=\exp(-\Theta(n)),
    \end{aligned}
    \]
    which completes the proof.

    
\end{proof}

\subsection{Likelihood of Two voting rules different under Mallows' model}

In this section, we are interested in the likelihood of 
\[
\widetilde{\mallows}(r_1, r_2, R_0, \varphi)(n):=\mathrm{Pr}_{P\sim \pi^n_{(R_0, \varphi)}}(r_1(P)\neq r_2(P)),
\]
where $r_1, r_2$ are two voting rules.

\begin{theorem}
    Given two different positional scoring rule $r_{\Vsi}$ and $r_{\Vsii}$,
    \[
        \widetilde{\mallows}(r_{\Vsi}, r_{\Vsii}, R_0, \varphi)(n)=\exp(-\Theta(n)).
    \]
\end{theorem}
\begin{proof}
    We draw this conclusion by giving a sub-event and a super-event, both probabilities of whom are $\exp(-\Theta(n))$, and hence the likelihood of a different outcome is also $\exp(-\Theta(n))$.

    On the one hand, given $r_{\Vsi}(P)\neq r_{\Vsii}(P)$, at least one of the voting rules $r_{\Vsi}$ and $r_{\Vsii}$ has outcome other than $a$, which is the first place of ground truth $R_0$. Hence, $\widetilde{\mallows}(r_1, r_2, R_0, \varphi)(n)\leq \exp(-\Theta(n))$ by Theorem~\ref{thm:Mallows_psr_gt}.

    On the other hand, $r_{\Vsi}(P)=c_1, r_{\Vsii}(P)=c_2$ implies $r_{\Vsi}\neq r_{\Vsii}$, hence it suffices to show 
    \[
    \mathrm{Pr}_{P\sim \pi^n_{(R_0, \varphi)}}(r_{\Vsi}(P)=c_1, r_{\Vsii}(P)=c_2)=\exp(-\Theta(n)).
    \]
    This event can be characterized by $\mathcal{H}^{c_1, c_2}:=\{\vec x: A^{\Vsi, \Vsii}\vec x\leq \vec b\}$, where $A^{\Vsi, \Vsii}=\begin{pmatrix}
        A^{\Vsi, c_1} \\ A^{\Vsii, c_2}
    \end{pmatrix}$ and $A^{c_i}$ is consists of $\{Score^{\vec s^i}_{c_i, k}: k\neq c_i\}$, and $\vec b=-\vec 1$. Clearly $\pi_{(R_0, \varphi)}\notin (\mathcal{H}^{C_1, c_2})_{\leq 0}$ and $(\mathcal{H}^{c_1, c_2})_{\leq 0}^{\mathbf{Z}}\neq \emptyset$. Hence, $\widetilde{\mallows}(r_1, r_2, R_0, \varphi)(n)\geq \mathrm{Pr}_{P\sim \pi^n_{(R_0, \varphi)}}(r_{\Vsi}(P)=c_1, r_{\Vsii}(P)=c_2)=\exp(-\Theta(n))$, which derives 

    \[
        \widetilde{\mallows}(r_{\Vsi}, r_{\Vsii}, R_0, \varphi)(n)=\exp(-\Theta(n)).
    \]
\end{proof}
