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\begin{fclemma}[]
  \label{lemma:1.15}
  Assuming:
  - $X, Y, Z$ random variables
  - $Z = f(Y)$
  Then:
  \[
    \cent XY \le \cent XZ
  .\]
\end{fclemma}

\begin{proof}
  \begin{align*}
    \cent XY
    &= \ent{X, Y} - \ent Y \\
    &= \ent{X, Y, Z} - \ent{Y, Z} \\
    &\le \ent{X, Z} - \ent Z 
    &&\text{(\nameref{submod})} \\
    &= \cent XZ
    &&
    \qedhere
  \end{align*}
\end{proof}

\begin{fclemma}[]
  \label{lemma:1.16}
  Assuming:
  - $X, Y, Z$ random variables
  - $Z = f(X) = g(Y)$
  Then:
  \[
    \ent{X, Y} + \ent Z \le \ent X + \ent Y
  .\]
\end{fclemma}

\begin{proof}
  \nameref{submod} says:
  \[
    \ent{X, Y, Z} = \ent Z
    \le \ent{X, Z} + \ent{Y, Z}
  \]
  which implies the result since $Z$ depends on
  $X$ and $Y$.
\end{proof}

\begin{fclemma}[]
  \label{lemma:1.17}
  Assuming:
  - $X$ takes values in a finite set $A$
  - $Y$ is uniform on $A$
  - $\ent X = \ent Y$
  Then:
  $X$ is uniform.
\end{fclemma}

\begin{proof}
  Let $p_a = \Pbb[X = a]$. Then
  \begin{align*}
    \ent X
    &= \sum_{a \in A} p_a \log \left(
    \frac{1}{p_a} \right) \\
    &= |A| \sum_{a \in A} p_a \log \left(
      \frac{1}{p_a} \right)
  \end{align*}
  The function $x \mapsto x \log \frac{1}{x}$ is
  concave on $[0, 1]$. So, by Jensen's inequality
  this is at most
  \[
    |A| (\Ebb_a p_a) \log \left( \frac{1}{\Ebb_a
    p_a} \right) = \log (|A|) = \ent Y
  .\]
  Equality holds if and only if $a \mapsto p_a$ is
  constant -- i.e. $X$ is uniform.
\end{proof}

\begin{fccoro}[]
  \label{coro:1.18}
  Assuming:
  - $X, Y$ random variables
  - $\ent{X, Y} = \ent X + \ent Y$
  Then:
  $X$ and $Y$ are independent.
\end{fccoro}

\begin{proof}
  We go through the proof of \nameref{subadd} and
  check when equality holds.

  Suppose that $X$ is uniform on $A$. Then
  \begin{align*}
    \cent XY
    &= \sum_y \Pbb[Y = y] \cent X{Y = y} \\
    &\le \ent X
  \end{align*}
  with equality if and only if $\cent X{Y = y}$ is
  uniform on $A$ for all $y$ (by
  \cref{lemma:1.17}), which implies that $X$ and
  $Y$ are independent.

  At the last stage of the proof we used
  \[
    \cent XY = \cent X{Y, W} = \cent XW \le \ent X
  \]
  where $W$ was uniform. So equality holds only
  if $X$ and $W$ are independent, which implies
  (since $Y$ depends on $W$) that $X$ and $Y$ are
  indpendent.
\end{proof}

\begin{fcdefnstar}[Mutual information]
  \glsnoundefn{muti}{mutual information}{NA}%
  Let $X$ and $Y$ be random variables. The
  \emph{mutual information} $\mathbf{I}[X : Y]$ is
  \begin{align*}
    \mathcloze{\ent X + \ent Y - \ent{X, Y}}
    &= \mathcloze{\ent X - \cent XY} \\
    &= \mathcloze{\ent Y - \cent YX}
  \end{align*}
\end{fcdefnstar}

\nameref{subadd} is equivalent to the statement
that $\muti XY \ge 0$ and \cref{coro:1.18} implies
that $\muti XY = 0$ if and only if $X$ and $Y$ are
independent.

Note that
\[
  \ent{X, Y} = \ent X + \ent Y - \muti XY
.\]

\begin{fcdefnstar}[Conditional mutual information]
  \glsnoundefn{cmuti}{conditional mutual
  information}{NA}%
  Let $X$, $Y$ and $Z$ be random variables. The
  \emph{conditional mutual information} of $X$ and
  $Y$ given $Z$, denoted by $\mathbf{I}[X : Y | Z]$ is
  \begin{align*}
    &\mathcloze{\sum_z \Pbb[Z = z] \muti{X \mid Z = z}{Y \mid
    Z = z}} \\
    &= \mathcloze{\sum_z \Pbb[Z = z] (\cent X{Z = z} + \cent
    Y{Z = z} - \cent{X, Y}{Z = z})} \\
    &= \mathcloze{\cent XZ + \cent YZ - \cent{X, Y}Z} \\
    &= \mathcloze{\ent{X, Z} + \ent{Y, Z} - \ent{X, Y, Z} -
    \ent Z}
  \end{align*}
\end{fcdefnstar}

\nameref{submod} is equivalent to the statement
that $\cmuti XYZ \ge 0$.

\newpage
\section{A special case of Sidorenko's conjecture}

Let $G$ be a bipartite graph with vertex sets $X$
and $Y$ (finite) and density $\alpha$ (defined to
be $\frac{|E(G)|}{|X||Y|}$). Let $H$ be another
(think of it as `small') bipartite graph with
vertex sets $U$ and $V$ and $m$ edges.

Now let $\phi : U \to X$ and $\psi : V \to Y$ be
random functions. Say that $(\phi, \psi)$ is a
\emph{homomorphism} if $\phi(x)\phi(y) \in E(G)$
for every $xy \in E(H)$.

Sidorenko conjectured that: for every $G, H$, we
have
\[
  \Pbb[\text{$(\phi, \psi)$ is a homomorphism}]
  \ge \alpha^m
.\]
Not hard to prove when $H$ is $K_{r, s}$. Also not
hard to prove when $H$ is $K_{2, 2}$ (use Cauchy
Schwarz).

\begin{fcthm}[]
  \label{thm:2.1}
  Sidorenko's conjecture is true if $H$ is
  a path of length $3$.
\end{fcthm}

\begin{proof}
  We want to show that if $G$ is a bipartite
  graph of density $\alpha$ with vertex sets $X,
  Y$ of size $m$ and $n$ and we choose $x_1, x_2
  \in X$, $y_1, y_2 \in Y$ independently at
  random, then
  \[
    \Pbb[x_1 y_1, x_2 y_2, x_3 y_3 \in E(G)] \ge
    \alpha^3
  .\]
  It would be enough to let $P$ be a P3 chosen
  uniformly at random and show that $\ent P \ge
  \log(\alpha^3 m^2 n^2)$.

  Instead we shall define a \emph{different}
  random variable taking values in the set of all
  P3s (and then apply \gls{maximality}).

  To do this, let $(X_1, Y_1)$ be a random edge of
  $G$ (with $X_1, \in X$, $Y_1 \in Y$). Now let
  $X_2$ be a random neighbour of $Y_1$ and let
  $Y_2$ be a random neighbour of $X_2$.

  It will be enough to prove that
  \[
    \ent{X_1, Y_1, X_2, Y_2} \ge \log(\alpha^3
    m^2 n^2)
  .\]