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\begin{fcdefn}[]
  Let $f : \Rbb^n \to \Rbb$ and let $\rho \in [-1, 1]$.
  Then $U_\rho f(x) = \Ebb_{y \sim N_\rho(x)} f(y)$.
\end{fcdefn}

\begin{remark*}
  If $x \sim N(0, 1)$ and $y \sim N_\rho(x)$, then $y \sim N(0, 1)$ and
  $x \sim N_\rho(y)$, from which it follows that $U_\rho$ is self-adjoint.
  If $y \sim N_\rho(x)$ and $z \sim N_\sigma(y)$, then there are independent
  Gaussians $g_1$, $g_2$ with $y \sim \rho x + \sqrt{1 - \rho^2} g_1$, $z
  \sim \sigma(\rho x + \sqrt{1 - \rho^2} g_1) + \sqrt{1 - \sigma^2} g_2$.
  But $\sigma^2 (1 - \rho^2) + 1 - \sigma^2 = 1 - \rho^2 \sigma^2$, so $z
  \sim N_{\rho \sigma}(x)$.
  From this, it follows easily that $U_\rho U_\sigma = U_{\rho \sigma}$, i.e.
  $\{U_\rho : \rho \in [-1, 1]\}$ forms a semigroup, called the
  \emph{Orstein--Uhlenbeck semigroup}.
\end{remark*}

We define, if $f : \Rbb^n \to \Rbb$, $\Stab_\rho f$ to be
\[
  \langle f, U_\rho f \rangle
  = \Ebb_{x \sim_\rho y} f(x) f(y)
.\]

\begin{fcthm}[Sheppard's formula]
  % Theorem 8.4
  \label{thm:8.4}
  Let $A \subset \Rbb^n$ be a half space, i.e. a set of the form $\{x : \langle
  x, n \rangle \ge 0\}$ for some non-zero $u$.
  Then $\Stab_\rho \indicator{A} = \half - \frac{\cos^{-1} \rho}{2\pi}$.
\end{fcthm}

\begin{proof}
  We are interested in
  \[
    \Ebb_{x \sim_\rho y} \indicator{A}(x) \indicator{A}(y)
    = \Pbb_{x \sim_\rho y} [\text{$\langle x, u \rangle \ge 0$ and $\langle y, u
    \rangle \ge 0$}]
  .\]
  Without loss of generality $u$ is a unit vector.
  Then $\langle x, u \rangle$ and $\langle y, u \rangle$ are $\rho$-correlated
  $1$-dimensional Gaussians (by rotational invariance, we can think of $u$ as
  just being $u = e_1$).

  So pick unit vectors $u, v \in \Rbb^2$, with $\langle v, w \rangle = \rho$,
  and consider $\langle v, g \rangle$, $\langle w, g \rangle$, $g \sim N(0,
  1)^2$.
  Then draw a picture:
  \begin{center}
    \includegraphics[width=0.3\linewidth]{images/67569673e4b8438c.png}
  \end{center}
  From this we get
  \[
    \Pbb[\text{$\langle v, g \rangle \ge 0$ and $\langle w, g \rangle \ge 0$}]
    = \frac{\pi - \cos^{-1} \rho}{2\pi}
    = \half - \frac{\cos^{-1} \rho}{2\pi}
    .
    \qedhere
  \]
\end{proof}

\begin{fcdefn}[Rotation sensitivity]
  Let $A \subset \Rbb^n$.
  The \emph{rotation sensitivity} $\RS_\delta(A)$ is
  \[
    \Pbb_{x \sim_{\cos \delta} y} [\indicator{A}(x)
    \neq \indicator{A}(Y)]
  .\]
\end{fcdefn}

If $A$ is \emph{balanced} (i.e. has Gaussian measure $\half$), then $\Pbb[x \in
A, y \in A] = \Pbb[x \notin A, y \notin A]$, so
\[
  \Pbb_{x \sim_{\cos \delta} y} [\indicator{A}(x) \neq \indicator{A}(y)]
  = 1 - 2 \Pbb_{x \sim_{\cos \delta} y} [\indicator{A}(x) = \indicator{A}(y)]
  = 1 - 2\Stab_{\cos \delta} \indicator{A}(x)
.\]
The statement $\Stab_{\cos \delta} \indicator{A}(x) \le \half -
\frac{\delta}{2\pi}$ is equivalent to $\RS_\delta(A) \ge \frac{\delta}{\pi}$.

\begin{fclemma}[Subadditivity of $RS$]
  Let $A$ be a balanced set in $\Rbb^n$.
  Then for any $\delta_1, \ldots, \delta_k \ge 0$ we have
  \[
    \RS_{\delta_1 + \cdots + \delta_k}(A)
    \le \RS_{\delta_1}(A) + \cdots + \RS_{\delta_k}(A)
  .\]
\end{fclemma}

\begin{proof}
  For $i = 0, \ldots, k$, let $\theta_i = \delta_1 + \cdots + \delta_i$.
  Let $g$ and $g'$ be independent $n$-dimensional Gaussians and let $x_i = \cos
  \theta_i g + \sin \theta_i g'$ for each $i$.
  Then $x_0$ and $x_k$ are $\cos \theta_k$-correlated, so
  \[
    \RS_{\delta_1 + \cdots + \delta_k}(A)
    = \Pbb[\indicator{A}(x_0) \neq \indicator{A}(x_k)]
  .\]
  Also, $x_{i - 1}$ and $x_i$ are ($\cos \theta_{i - 1} \cos \theta_i + \sin
  \theta_{i - 1} \sin \theta_i = \cos(\theta_i - \theta_{i - 1}) = \cos
  \delta_i$)-correlated.
  So the RHS equals $\sum_{i = 1}^{k} \Pbb[\indicator{A}(x_{i - 1}) \neq
  \indicator{A}(x_i)]$.
  The result now follows from a union bound.
\end{proof}

\begin{fccoro}[Special case of Borell's isoperimetric inequality]
  % Corollary 8.6
  \label{coro:8.6}
  Let $A$ be balanced and let $k \in \Nbb$.
  Then $\RS_k(A) \ge \frac{1}{2k}$.
\end{fccoro}

\begin{proof}
  $\RS_{\frac{\pi}{2}}(A) = \half$ because $\cos \frac{\pi}{2}$-correlated
  variables are independent.
  Setting $\delta_1 = \cdots = \delta_k = \frac{\pi}{2k}$, we deduce that $k
  \RS_{\frac{\pi}{2k}} \ge \half$, hence $\RS_{\frac{\pi}{2k}} \ge
  \frac{1}{2k}$.
\end{proof}
