%! TEX root = ABF.tex
% vim: tw=80 ft=tex
% 12/03/2026 12PM

\begin{fcdefn}[]
  Let $f : \Rbb^n \to \Rbb$ be a multilinear function $f = \sum_A \ft{f}(A)
  x_A$.
  Alternatively, think of $f$ as a formal multilinear polynomial.
  Define
  \begin{itemize}
    \item $\Ebb f = \ft{f}(\emptyset)$,
    \item $\Ebb f^2 = \sum_A \ft{f}(A)^2$,
    \item $\Var f = \sum_{A \neq \emptyset} \ft{f}(A)^2$,
    \item $\ip \langle f, g \rangle = \sum_A \ft{f}(A) \ft{g}(A)$,
    \item $E_i f = \sum_{A \not\ni i} \ft{f}(A) x_A$,
    \item $\D_i f = \sum_{A \ni i} \ft{f}(A) x_{A \setminus \{i\}}$.
  \end{itemize}
  Note that $E_i f$ and $\D_i f$ do not depend on $x_i$ and also that $f = E_i f
  + x_i \D_i f$, and $\langle E_i f, x_i \D_i f \rangle = 0$.
  Define
  \begin{itemize}
    \item $\Inf_i f = \sum_{A \ni i} \ft{f}(A)^2$,
    \item $\totinf(f) = \sum_i \Inf_i f = \sum_A |A| \ft{f}(A)^2$.
  \end{itemize}
  One could also define $\noise_\rho f$, $\Stab_\rho f$.
\end{fcdefn}

Now let $X = (X_1, \ldots, X_n)$ be independent random variables with $\Ebb X_i
= 0$, $\Ebb X_i^2 = 1$.
Then define $F(X)$ to be $f$ evaluated at $(X_1, \ldots, X_n)$, i.e. $\sum_A
\ft{f}(A) \prod_{i \in A} X_i$.
Then it is easy to check that $\langle F, G \rangle = \langle f, g \rangle$,
$\|F\|_2^2 = \|f\|_2^2$.
For example,
\begin{align*}
  \langle F, G \rangle
  &= \Ebb_X \left(\sum_A \ft{f}(A) \prod_{i \in A} X_i\right) \left( \sum_B
  \ft{g}(B) \prod_{i \in B} X_i \right) \\
  &= \sum_{A, B} \ft{f}(A) \ft{g}(B) \prod_{i \in A \symdiff B} X_i \prod_{i \in
  A \cap B} X_i^2 \\
  &= \sum_A \ft{f}(A) \ft{g}(A)
\end{align*}
Also, defining $E_i F(X)$ to be $\Ebb[f(X_1, \ldots, X_n) \mid X_1, \ldots, X_{i
- 1}, X_{i + 1}, \ldots, X_n]$ we have that $E_i F(X) = E_i f(X)$.

Suppose that $(X_1, \ldots, X_n)$ are independent and
\[
  \Ebb X_i = 0, \qquad
  \Ebb X_i^2 = 1, \qquad
  \Ebb X_i^3 = 0, \qquad
  \Ebb X_i^4 \le 9
  \tag{$*$}
  \label{lec15:eqstar}
.\]
Then the proof of \nameref{lemma:4.1} straightforwardly gives that if $f$ has
degree at most $k$, then $\Ebb f(X_1, \ldots, X_n)^4 \le 9^k (\Ebb f(X_1,
\ldots, X_n)^2)^2$.

\begin{fcthm}[Invariance principle]
  % Theorem 8.2
  \label{thm:8.2}
  Let $(X_1, \ldots, X_n)$ and $(Y_1, \ldots, Y_n)$ be sequences of random
  variables satisfying condition \eqref{lec15:eqstar}.
  Let $f$ be a multilinear polynomial of degree at most $k$ and let $\psi : \Rbb
  \to \Rbb$ satisfy that $\|\psi''''\|_\infty \le C$ (bounded fourth
  derivative).
  Then
  \[
    |\Ebb \psi(f(X_1, \ldots, X_n)) - \Ebb \psi(f(Y_1, \ldots, Y_n))|
    \le \frac{C}{12} \cdot 9^k \sum_{i = 1}^{n} (\Inf_i f)^2
  .\]
\end{fcthm}

\begin{remark*}
  It is possible to get a stronger result than this, but we prove this version
  because it can be proved with the version of Bonami's Lemma mentioned above.
\end{remark*}

\begin{example*}
  Examples of $f$ where the LHS of the above Theorem is large if we set $X_i
  \sim \Unif(\{-1, 1\})$, $Y_i \sim N(0, 1)$:
  \begin{itemize}
    \item $f(Z) = Z_1$.
      Then the distribution of $f(X)$ is then uniform on $\{-1, 1\}$, so LHS is
      large.
      The RHS is large because $\Inf_1 f$ is large.
    \item $f(Z) = Z_1 \cdots Z_n$.
      Again, distribution of $f(X)$ is uniform on $\{-1, 1\}$.
      The RHS is large because $k$ is large, and also because all $\Inf$ terms
      are large.
  \end{itemize}
\end{example*}

\begin{proof}
  By the triangle inequality, the quantity we wish to bound is at most
  \[
    \sum_{i = 1}^{n} |\Ebb \psi(f(Y_1, \ldots, Y_{i - 1}, X_i, X_{i + 1},
    \ldots, X_n)) - \Ebb \psi(f(Y_1, \ldots, Y_{i - 1}, Y_i, X_{i + 1}, \ldots,
    X_n))|
  .\]
  Write $U_i = (Y_1, \ldots, Y_{i - 1}, X_{i + 1}, \ldots, X_n)$.
  Then we can rewrite each summand as
  \[
    |\Ebb \psi(E_i f(U_i) + X_i \D_i f(U_i)) - \Ebb \psi(E_i f(U_i) + Y_i \D_i
    f(U_i))|
  .\]
  Let $u_i = E_i f(U_i)$, $v_i = \D_i f(U_i)$.
  So we can rewrite as
  \[
    |\Ebb \psi(u_i + X_i v_i) - \Ebb \psi(u_i + Y_i v_i)|
  .\]
  But
  \[
    \psi(u_i + X_i v_i)
    = \psi(u_i)
    + X_i v_i \psi'(u_i)
    + \half X_i^2 v_i^2 \psi''(u_i)
    + \frac{1}{6} X_i^3 v_i^3 \psi'''(u_i)
    + \frac{1}{24} X_i^4 v_i^4 \psi''''(w_i)
  \]
  and
  \[
    \psi(u_i + Y_i v_i)
    = \psi(u_i)
    + Y_i v_i \psi'(u_i)
    + \half Y_i^2 v_i^2 \psi''(u_i)
    + \frac{1}{6} Y_i^3 v_i^3 \psi'''(u_i)
    + \frac{1}{24} Y_i^4 v_i^4 \psi''''(z_i)
  .\]
  Taking expectations and subtracting, noting condition \eqref{lec15:eqstar} and
  that $X_i$ and $Y_i$ are independent of $u_i$ and $v_i$, we see that
  everything cancels apart from the error terms, so we get
  \[
    \frac{1}{24} |
      \Ebb X_i^4 v_i^4 \psi''''(w_i)
      - \Ebb Y_i^4 v_i^4 \psi''''(z_i)
    |
    \le \frac{C}{24} |\Ebb X_i^4 v_i^4 + \Ebb Y_i^4 v_i^4|
  .\]
  But
  \[
    \Ebb X_i^4 v_i^4
    = \Ebb (X_i \D_i f(U_i))^4
    = \Ebb (x_i \D_i f)(Y_1, \ldots, Y_{i - 1} X_i X_{i + 1}, \ldots, X_n)^4
  .\]
  But $(Y_1, \ldots, Y_{n - 1}, X_i, X_{i + 1}, \ldots, X_n)$ satisfies
  \eqref{lec15:eqstar} and $x_i \D_i f$ has degree at most $k$.
  So Bonami's Lemma applies, and we get an upper bound of $9^k (\Ebb X_i^2
  v_i^2)^2 = 9^k (\Ebb (\D_i f)^2)^2 = 9^k (\Inf_i f)^2$.
  Same for $Y_i$, so summing over $i$ gives the result.
\end{proof}

\subsubsection*{Gaussian Space}

Let $x \in \Rbb$.
We say that $y \sim N_\rho(x)$, \emph{$y$ is $\rho$-correlated with $x$} if $y
\sim \rho x + \sqrt{1 - \rho^2} g$, where $g \sim N(0, 1)$.
If $x \sim N(0, 1)$ and $y \sim N_\rho(x)$, thene there are independent
Gaussians $g_1$, $g_2$ with $x = g_1$, $y = \rho g_1 + \sqrt{1 - \rho^2} g_2$,
so $y \sim N(0, 1)$ and $\Ebb xy = \rho \Ebb g_1^2 + \sqrt{1 - \rho^2} \Ebb g_1
g_2 = \rho$.

A nice way to construct a pair $(x, y)$ of $\rho$-correlated Gaussians is to
take unit vectors $u, v \in \Rbb^2$, $g \in N(0, 1)^2$ and set $x = \langle u, g
\rangle$, $y = \langle v, g \rangle$, choosing $u, v$ so that $\langle u, v
\rangle = \rho$.
Writing $g = (g_1, g_2)$, we have
\[
  \Ebb xy
  = \Ebb u_1 v_1 g_1^2 + \Ebb u_2 v_2 g_2^2
  = u_1 v_1 + u_2 v_2
  = \langle u, v \rangle
  = \rho
.\]
