$\require{mathtools}
%
%%% GENERIC MATH %%%
%
% Environments
\newcommand{\al}[1]{\begin{align}#1\end{align}} % need this for \tag{} to work
\renewcommand{\r}{\mathrm}
%
% Greek
\newcommand{\eps}{\epsilon}
\newcommand{\veps}{\varepsilon}
\let\fi\phi % because it looks like an f
\let\phi\varphi % because it looks like a p
%
% Miscellaneous shortcuts
% .. over and under
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% .. relations
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% .. miscer
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%
% Delimiters
% (I needed to create my own because the MathJax version of \DeclarePairedDelimiter doesn't have \mathopen{} and that messes up the spacing)
% .. one-part
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% .... (use phantom to force at least the standard height of double bars)
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%% .. two-part
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% ..... (use phantom to force at least the standard height of double bar)
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% .. vanilla versions (is it within a constant?)
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% .. dotted versions (is it equal in the limit?)
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% .. log versions (is it equal up to log?)
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%
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% .. operators
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\DeclareMathOperator{\negl}{negl}
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% .. functions
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% .. analysis
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% .. sets
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%
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% Logic
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% Linear algebra
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% ... named tensors
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%
% Probability
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% ... information theory
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%
%%% SPECIALIZED COMPUTER SCIENCE %%%
%
% Complexity classes
% .. classical
\newcommand{\Poly}{\mathsf{P}}
\newcommand{\NP}{\mathsf{NP}}
\newcommand{\PH}{\mathsf{PH}}
\newcommand{\PSPACE}{\mathsf{PSPACE}}
\renewcommand{\L}{\mathsf{L}}
% .. probabilistic
\newcommand{\formost}{\mathsf{Я}}
\newcommand{\RP}{\mathsf{RP}}
\newcommand{\BPP}{\mathsf{BPP}}
\newcommand{\MA}{\mathsf{MA}}
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% .. circuits
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% .. resources
\newcommand{\TIME}{\mathsf{TIME}}
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% .. keywords
\newcommand{\co}{\mathsf{co}}
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%
% Boolean analysis
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% .. functions (small caps sadly doesn't work)
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%
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% In text
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% .. bold
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\newcommand{\TZ}{\mathtt{Z}}$
Adapted from Chapter 9.1 of “Analysis of boolean functions” by Ryan O’Donnell.
Reasonableness
Don’t you hate it when a random variable:
- is never close to its expectation?
- is almost always bigger than its mean, or almost never?
- has small first, second, third moment but its fourth moment blows up?
Wouldn’t it be nice if random variables were reasonable people? As a first try, let’s say $\BX$ is $B$-reasonable if $\norm{\BX}_4 \leq B \norm{\BX}_2$: its fourth moment isn’t too much bigger than its second moment. Note that we always have $\norm{\BX}_2 \leq \norm{\BX}_4$, so this says that two moments are within a factor $B$ of each other.
For example:
- unbiased $\pm 1$ bits are $1$-reasonable (the best!);
- standard Gaussians are $3$-reasonable;
- if $\BX$ is a Bernoulli of mean $2^{-n}$, then $\norm{\BX}_2 = 2^{-n/2}$ while $\norm{\BX}_4 = 2^{-n/4}$, so $\BX$ is $2^{n/4}$-(un)reasonable.
Low-degree functions are reasonable
If we think of the outputs of boolean functions $f:\pmo^n \to \R$ as random variables, then $f$ is $B$-reasonable if $\norm{f}_4 \le B\norm{f}_2$. Taking the analogous examples:
- dictators are $1$-reasonable;
- the sum $\frac{x_1+\cdots+x_n}{\sqrt{n}}$ is basically a standard Gaussian and is $3$-reasonable;
- the AND function (with $0,1$ outputs) is $2^{n/4}$-(un)reasonable.
To be unreasonable, a function $f$ needs to have some rare but extreme values, and that intuitively, that only happens when $f$ has high degree. This is the Bonami lemma: if $\deg(f)\leq k$, then
\[
\norm{f}_4 \leq \sqrt{3}^k\norm{f}_2 \Leftrightarrow \E[f^4] \leq 9^k\E[f^2]^2
\]
Intuition for the proof
Say you were trying to prove $\E[f^4] \leq \E[f^2]^2$ instead, without the $9^k$. This is true if $f$ is constant. But in general, this will be impossible because of convexity: since $f^4$ is “more convex” than $f^2$, it averages to higher values when there is variance.
For example, let $f(x) = e + dx_1$ for some constants $e,d$. Then this is what $\E[f^4]$ and $\E[f^2]^2$ look like:
TODO: insert #figure from Notability
We have
\[
\E[f^4] = \frac{(e-d)^4+(e+d)^4}{2} = e^4 + 6e^2d^2 + d^4
\]
and
\[
\E[f^2]^2 = \p{\frac{(e-d)^2 + (e+d)^2}{2}} = (e^2+d^2)^2 = e^4 + 2e^2d^2 + d^4,
\]
so there is a gap of
\[
\E[\Bf^4] - \E[\Bf^2]^2 = 6e^2d^2 - 2e^2d^2 = 4e^2d^2
\]
between them, and we fail.
In the actual proof, we will need to “pay for the factor $3$” between $6e^2d^2$ and $2e^2d^2$ by using the fact that the “$d$ part” has lower degree than $f$, and is therefore more reasonable.
Proof
Let’s try to do induction over $n$. We can decompose the function as $f(x) = e(x) + x_1d(x)$ into the monomials that include the first variables and those that don’t. Then, using the shortcuts
\[
\Bf, \Be, \Bd \ce f(\BX), e(\BX), d(\BX)
\]
for a random input $\BX$, the left-hand side is
\[
%\al{
\E[\Bf^4] = \E[(\Be+\BX_1\Bd)^4] = \E[\Be^4] + 6\E[\Be^2\Bd^2] + \E[\Bd^4].
%}
\]
To deal with $\E[\Be^2\Bd^2]$, we can use Cauchy-Schwarz:
\[
\E[\Bf^4] \leq \E[\Be^4] + 6\sqrt{\E[\Be^4]\E[\Bd^4]} + \E[\Bd^4].
\]
Comparing this to
\[
9^k\E[\Bf^2]^2 = 9^k\p{\E[\Be^2] + \E[\Bd^2]}^2 = 9^k\p{\E[\Be^2]^2 + 2\E[\Be^2]\E[\Bd^2] + \E[\Bd^2]^2},
\]
we’re in good shape by induction on $d$ and $e$, except for the middle term, where we need
\[
6\sqrt{\E[\Be^4]\E[\Bd^4]} \stackrel{?}{\le} 9^k \cdot 2\E[\Be^2]\E[\Bd^2],
\]
even though $6>2$.
This is where the decrease in degree saves us: since $d$ has degree $\leq k-1$, we get $\E[\Bd^4] \leq 9^{k-1}\E[\Bd^2]^2$ instead of just $\le 9^k\E[\Bd^2]^2$, so we gain a factor $\frac{1}{\sqrt{9}} = \frac{1}{3}$:
\[
6\sqrt{\E[\Be^4]\E[\Bd^4]} \le 6\sqrt{\p{9^k\E[\Be^2]^2}\cdot \p{9^{k-1}\E[\Bd^2]^2}} =\frac{6}{3}\sqrt{\p{9^k\E[\Be^2]^2}\cdot \p{9^k\E[\Bd^2]^2}} = 9^k \cdot 2\E[\Be^2]\E[\Bd^2].
\]