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\begin{fcdefn}[]
  \glssymboldefn{Nrho}%
  \glsadjdefn{rhocorrel}{$\rho$-correlated}{boolean variables}%
  Let $\rho \in [-1, 1]$.
  If $x \in \{-1, 1\}^n$, we say that $y \sim N_\rho(x)$ if for each $i$,
  \[
    y_i
    = \begin{cases}
      x_i
      & \text{probability $\frac{1 + \rho}{2}$} \\
      -x_i
      & \text{probability $\frac{1 - \rho}{2}$}
    \end{cases}
  \]
  (Equivalently: if $\rho \ge 0$, then with probability $\rho$ take $y_i = x_i$
  and with probability $1 - \rho$ take $y_i$ random.)

  All choices independent.

  If $x$ is uniform and $y \sim N_\rho(x)$, then $y$ is uniform and $x \sim
  N_\rho(y)$.
  In this case, we write $x \sim_\rho y$ and say that $x$ and $y$ are
  \emph{$\rho$-correlated}.
\end{fcdefn}

\begin{fcdefn}[]
  \glssymboldefn{Trho}%
  If $\rho \in [-1, 1]$, then the \emph{noise operator} $T_\rho$ takes a
  function $f : \{-1, 1\}^n \to \Rbb$ to $T_\rho f$, defined by $T_\rho f =
  \Ebb_{y \sim \N_\rho(x)} f(y)$.
\end{fcdefn}

\begin{remark*}
  This is a convolution: $T_\rho f$ is a convolution of $f$ with a particular
  probability measure.
  So we should expect a nice formula about its Fourier coefficients.
\end{remark*}

Reminder: $x_A = \prod_{i \in A} x_i$.

\begin{align*}
  \noise_\rho
  &= \Ebb_{y \in \N_\rho(x)} y_A \\
  &= \Ebb_{y \sim \N_\rho(x)} \prod_{i \in A} y_i \\
  &= \prod_{i \in A} \Ebb_{y \sim \N_\rho(x)} y_i \\
  &= \prod_{i \in A} \left( \frac{1 + \rho}{2} x_i + \frac{1 - \rho}{2}(-x_i)
  \right) \\
  &= \prod_{i \in A} (\rho x_i) \\
  &= \prod^{|A|} x_A
\end{align*}
Therefore, by the inversion formula,
\[
  \noise_\rho f
  = \noise_\rho \left( \sum_A \ft{f}(A) x_A \right)
  = \sum_A \rho^{|A|} \ft{f}(A) x_A
,\]
so
\[
  \ft{\noise_\rho f}(A)
  = \rho^{|A|} \ft{f}(A)
.\]

\begin{fcdefn}[]
  \glssymboldefn{Stab}%
  Let $\rho \in [-1, 1]$ and let $f : \{-1, 1\}^n \to \Rbb$.
  Then
  \begin{align*}
    \Stabinternal_\rho(f)
    &= \Ebb_{x \sim_\rho y} f(x) f(y) \\
    &= \Ebb_x f(x) \Ebb_{y \sim N_{\rho(x)}} f(y) \\
    &= \Ebb_x f(x) \noise_\rho f(x) \\
    &= \ip\langle f, \noise_\rho f \rangle \\
    &= \dualip\langle \ft{f}, \ft{\noise_\rho f} \rangle \\
    &= \sum_A \rho^{|A|} \ft{f}(A)^2
  \end{align*}
\end{fcdefn}

\newpage
\section{Kalai's proof of Arrow's theorem}

\begin{fcdefn}[Ranking, vote, voting rule, restriction]
  \glsnoundefn{ranking}{ranking}{rankings}%
  \glsnoundefn{vote}{vote}{votes}%
  \glsnoundefn{vrule}{voting rule}{voting rules}%
  \glsnoundefn{restriction}{restriction}{restrictions}%
  Let $A = \{A_1, \ldots, A_k\}$ be a set of candidates.
  A \emph{ranking} of $A$ is a total order on $A$.

  Let $R_A$ be the set of all rankings of $A$.
  A \emph{vote} with $n$ voters and candidate set $A$ is an element of $R_A^n$.

  A \emph{voting rule} is a function $f : R_A^n \to R_A$.

  If $B \subset A$, then the \emph{restriction} of a vote on $A$ to $B$ is the
  element $v$ of $R_B^n$ where each $v_i$ is the restriction of the $i$-th
  voter's ranking of $A$.
\end{fcdefn}

\begin{fcthm}[Arrow's Theorem]
  % Theorem 3.1
  \label{thm:3.1}
  If $|A| \ge 3$, then there is no \gls{vrule} with the following three
  properties:
  \begin{enumerate}[(1)]
    \item \emph{Weak monotonicity:} If every voter prefers $A_i$ to $A_j$, then
      $A_i$ beats $A_j$.
    \item \emph{No dictators:} There is no $i$ such that the outcome is
      determined by the $i$-th \gls{vote}.
    \item \glsnoundefn{irrelalt}{Irrelevant alternatives have no effect}{}%
      \emph{Irrelevant alternatives have no effect:} The \gls{restriction}
      of the result to a subset $B$ only on the \gls{restriction} of the
      \gls{vote} to $B$.
  \end{enumerate}
\end{fcthm}

Note that in particular, property (3) implies that for any two candidates $A_i,
A_j$, their eventual relative ranking depends only on their relative rankings by
each voter.
So if we encode the relative rankings by $x \in \{-1, 1\}^n$ where
\[
  x_r
  = \begin{cases}
    1
    & \text{$r$ prefers $A_i$ to $A_j$} \\
    0
    & \text{$r$ prefers $A_j$ to $A_i$}
  \end{cases}
\]
Then the relative ranking in the outcome is given by a Boolean function $u_{ij}
: \{-1, 1\}^n \to \{-1, 1\}$.

If we calculate all these pairwise rankings, then they must lead to a total
order (defining $A_i > A_j$ if $A_i$ is ranked above $A_j$), i.e. we don't have
a \emph{Condorcet paradox}, where voters prefer $A_i$ to $A_j$, $A_j$ to $A_k$
and $A_k$ to $A_i$.
So to prove \nameref{thm:3.1}, it will be enough to show that such an
incompatibility must happen.

\begin{fclemma}[]
  % Lemma 3.2
  \label{lemma:3.2}
  Let $f : \{-1, 1\}^n \to \{-1, 1\}$ be a linear function (i.e. of \gls{degree}
  $1$), with $\ft{f}(\emptyset) = 0$.
  Then there exists $i$ such that $\forall x, f(x) = x$ or $\forall x, f(x) =
  -x$.
\end{fclemma}

\begin{proof}
  If $f$ takes values in $\{-1, 1\}$, then changing $x_i$ changes $f$ by $-2$,
  $0$ or $2$.
  Since $f$ is linear, it has a formula of the form $f(x) = \sum_i a_i x_i$.
  Therefore, each $a_i$ belongs to $\{-1, 0, 1\}$.
  At least one $a_i$ is non-zero.
  If $a_i$ and $a_j$ are non-zero, $i \neq j$, then $a_i x_i + a_j x_j$ takes at
  least three distinct values, contradiction.
\end{proof}
