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\subsubsection*{Noise and stability}

Fix $p$, so that we don't have to write subscripts everywhere.

\glssymboldefn{pN}%
Let $x \in \{-1, 1\}$ and let $\rho \in [0, 1]$.
We say that $y \sim N_\rho(x)$ if with probability $\rho$ $y_i = x_i$ and with
probability $1 - \rho$ $y_i$ is $\mup$-random, and all $y_i$s independent.

If $x \sim \mup$, then
\[
  \Pbb[y_i = 1]
  = pq + (1 - p)q
  = q
.\]
So $y \sim \mup$.
Then we say that $x$ and $y$ are $\rho$-correlated, and write $x \sim_\rho y$.

Note: $\sim_\rho$ depends on $p$, but we don't write $\sum_{\rho, p}$ (i.e. we
omit the $p$ for convenience).

\begin{fcdefn}[]
  \glssymboldefn{pnoise}%
  The \emph{$p$-biased noise operator} $T_\rho$ is given by the formula
  \[
    T_\rho f(x)
    = \Ebb_{y \sim \pN_\rho(x)} f(y)
  .\]
\end{fcdefn}

\begin{fclemma}[]
  % Lemma 6.2
  \label{lemma:6.2}
  For every $A \subset [n]$, $\pnoise_\rho \pphi_A = \rho^{|A|} \pphi_A$.
\end{fclemma}

Reminders:
\begin{align*}
  \pphi(t)
  &= \frac{t - \mu}{\sigma} \\
  \mu
  &= q - p = 2q - 1 = 1 - 2p \\
  \sigma
  &= 2\sqrt{pq} \\
  \pphi(1)
  &= \sqrt{\frac{p}{q}} \\
  \pphi(-1)
  &= -\sqrt{\frac{q}{p}} \\
  \pphi_A(x)
  &= \prod_{i \in A} \pphi(x_i)
\end{align*}

\begin{proof}
  \begin{align*}
    \pnoise_\rho \pphi_A(x)
    &= \Ebb_{y \sim \pnoise_\rho(x)} \prod_{i \in A} \pphi(y_i) \\
    &=\prod_{i \in A} \Ebb_{y \sim \pnoise_\rho(x)} \pphi(y_i) \\
    &= \prod_{i \in A} (\rho \pphi(x_i) + (1 - \rho) \Ebb \pphi) \\
    &= \rho^{|A|} \pphi_A(x)
    \qedhere
  \end{align*}
\end{proof}

\begin{fccoro}[]
  % Corollary 6.3
  \label{coro:6.3}
  $\pft{\pnoise_\rho f}(A) = \rho^{|A|} \pft{f}(A)$.
\end{fccoro}

\begin{proof}
  $\pnoise_\rho f = \sum_A \pft{f}(A) \pnoise_\rho \pphi_A = \sum_A \rho^{|A|}
  \pft{f}(A) \pphi_A$.
\end{proof}

\begin{fcdefn}[]
  \glssymboldefn{pstab}%
  Let $\rho \in [0, 1]$, $f : \{-1, 1\}^n \to \Rbb$.
  Then
  \[
    \Stabinternal_\rho f
    = \pip \langle f, \pnoise_\rho f \rangle
    = \Ebb_{x \sim_\rho y} f(x) f(y)
  .\]
\end{fcdefn}

\begin{remark*}
  By \cref{coro:6.3}, $\pstab_\rho f = \sum_A \rho^{|A|} \pft{f}(A)^2$, as in
  unbiased case.
\end{remark*}

\subsubsection*{The Margulis--Russo formula}

We shall be considering more than one value of $p$.

\textbf{Convention:} Let $f : \{-1, 1\}^n \to \Rbb$.
If we write $f^{(p)}$, then all definitions should be understood to be
$p$-biased.

\begin{fclemma}[]
  % Lemma 6.4
  \label{lemma:6.4}
  Let $f : \Rbb^n \to \Rbb$ be a multilinear function, and write $f$ also for
  its restriction to $\{-1, 1\}^n$.
  Then $\Ebb_{x \sim \mup} f^{(p)}(x) = f(\ol{\mu}, \mu, \ldots, \mu)$.
\end{fclemma}

Note: whenever we write $f^{(p)}$, it is implicit that we are restricting to
$\{-1, 1\}^n$, since $\mup$ is only defined there.

We will give 3 proofs!

\begin{proof}[Proof 1]
  Write $f = \sum_A \theta_A x_A$.
  Then $\Ebb_{x \sim \mup} x_A = \prod_{i \in A} \Ebb_{x \sim \mup} x_i =
  \prod_{i \in A} (q - p) = \mu^{|A|} = x_A(\mu, \mu, \ldots, \mu)$.
  Then by linearity, we're done.
\end{proof}

You might think the above proof is a bit odd, since it uses the $x_A$ even
though we're in the $p$-biased case.
I would agree with you!

\begin{proof}[Proof 2]
  Write $f^{(p)} = \sum_A \pft{f}(A) \pphi_A$.
  Then
  \[
    \Ebb_{x \sim \mup} \pphi_A
    = \prod_{i \in A} \Ebb_{x \sim \mup} \pphi_i
    = \begin{cases}
      0
      & A \neq \emptyset \\
      1
      & A = \emptyset
    \end{cases}
    = \pphi_A(\mu, \mu, \ldots, \mu)
    \qedhere
  \]
\end{proof}

\begin{proof}[Proof 3]
  Induction on $n$.
  \[
    \Ebb f^{(p)}(x)
    = \Ebb (q f^{(p)} (x_{n \subst 1}) + p f^{(p)} (x_{n \subst -1}))
    = \Ebb f^{(p)}(x_{n \subst \mu})
    = f^{(p)}(\mu, \ldots, \mu)
  .\]
  Where the second equality used linearity in the last coordinate, and the last
  equality is by induction hypothesis.
\end{proof}

\begin{fcthm}[The Margulis--Russo formula]
  % Theorem 6.5
  \label{thm:6.5}
  Let $f$ be as above.
  Then
  \[
    \frac{\dd}{\dd \mu} \Ebb f^{(p)}
    = \frac{1}{\sigma} \sum_{i = 1}^n \pft{f^{(p)}}(i)
  .\]
\end{fcthm}

\begin{remark*}
  Later we will care instead about $\frac{\dd}{\dd p}$.
  But for now it is easier to work with $\frac{\dd}{\dd \mu}$.
\end{remark*}

\begin{proof}
  By \cref{lemma:6.4},
  \begin{align*}
    \frac{\dd}{\dd \mu} \Ebb f^{(p)}
    &= \frac{\dd}{\dd \mu} f(\mu, \mu, \ldots, \mu) \\
    &= \sum_{i = 1}^{n} \frac{\partial}{\partial x_i} f(\mu, \mu, \ldots, \mu)
    \\
    &= \sum_{i = 1}^{n} \half (f(\bm{\mu}_{i \subst 1}) - f(\bm{\mu}_{i \subst
    -1}))
    &&\text{(by multilinearity)} \\
    &= \frac{1}{\sigma} \sum_{i = 1}^{n}  \pD_i f(\bf{\mu}) \\
    &= \frac{1}{\sigma} \sum_{i = 1}^{n}  \Ebb_{x \sim \mup} \pD_i f^{(p)}(x)
  \end{align*}
  But $\pD_i f = \sum_{A \ni i} \ft{f}(A) \pphi_{A \setminus \{i\}}$, so $\Ebb
  \pD_i f = \pft{f}(i)$.
  The result follows.
\end{proof}

\begin{fccoro}[]
  % Corollary 6.6
  \label{coro:6.6}
  Let $f : \{-1, 1\}^n \to \{-1, 1\}$ be a monotone Boolean function.
  Then
  \[
    \frac{\dd}{\dd p} \Pbb[f^{(p)}(x) = -1]
    = \frac{1}{\sigma^2} \totinf(f^{(p)})
  .\]
\end{fccoro}

\begin{proof}
  $\Ebb f^{(p)} = 1 - 2\Pbb[f^{(p)}(x) = -1]$, so
  \[
    \Pbb[f^{(p)}(x) = -1]
    = \frac{1 - \Ebb f^{(p)}}{2}
  .\]
  Therefore,
  \begin{align*}
    \frac{\dd}{\dd p} \Pbb[f^{(p)}(x) = -1]
    &= -\half \frac{\dd}{\dd \mu} \Ebb f^{(p)} \frac{\dd \mu}{\dd p} \\
    &= \frac{1}{\sigma} \sum_{i = 1}^{n} \pft{f^{(p)}}(i)
    &&\text{(by \cref{thm:6.5})}
  \end{align*}
  From the proof of \cref{thm:6.5}, this is $\frac{1}{\sigma} \sum_{i = 1}^{n}
  \Ebb \pD_i f^{(p)}$.
  Since $f$ is monotone, $\pD_i f^{(p)}(x) \in \{0, \sigma\}$, so this equals
  \[
    \frac{1}{\sigma^2} \sum_{i = 1}^{n} \Ebb(\pD_i f^{(p)})^2
    = \frac{1}{\sigma^2} \sum_i \|\pD_i f^{(p)}\|_2^2
    = \totinf(f^{(p)})
    .
    \qedhere
  \]
\end{proof}

\begin{remark*}
  Suppose that $p_1 < p_2$ are such that $\Pbb[f^{(p_1)}(x) = -1] = \eps$,
  $\Pbb[f^{(p_2)} = -1] = 1 - \eps$.
  Then by MVT there exists $p \in (p_1, p_2)$ such that
  \[
    \frac{\dd}{\dd p} \Pbb[f^{(p)}(x) = -1]
    = \frac{1 - 2\eps}{p_2 - p_1}
  .\]
  So by \nameref{thm:6.5}, there exists $p \in (p_1, p_2)$ such that
  $\totinf(f^{(p)}) = \sigma^2 \left( \frac{1 - 2\eps}{p_2 - p_1} \right)$.

  So if $p_2 - p_1$ isn't small, then there exists $p$ such that $\eps \le
  \Pbb[f^{(p)} = -1] \le 1 - \eps$ and $\totinf(f^{(p)})$ isn't too large.
\end{remark*}