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Let $X$, $Y$ be discrete random variables taking
values in an abelian group. What is $X + Y$ when
$X$ and $Y$ are independent?

For each $z$, $\Pbb(X + Y = z) = \sum_{x + y = z}
\Pbb(X = x)\Pbb(Y = y)$. Writing $p_x$ and $q_y$
for $\Pbb(X = x)$ and $\Pbb(Y = y)$ respectively,
this givesim $\sum_{x + y = z} p_x q_y = p * q(z)$
where $p(x) = p_x$ and $q(y) = z_y$.

So, sums of independent random variables
$\leftrightarrow$ convolutions.

\begin{fcdefnstar}[Entropic Ruzsa distance]
  \glssymboldefn{entd}%
  Let $G$ be an abelian group and let $X$, $Y$ be
  $G$-valued random variables. The \emph{entropic
  Ruzsa distance} $d[X; Y]$ is
  \[
    \ent{X' - Y'} - \half \ent X - \half \ent Y
  \]
  where $X'$, $Y'$ are independent copies of $X$
  and $Y$.
\end{fcdefnstar}

\begin{fclemma}[]
  \label{lemma:6.3}
  Assuming:
  - $A$, $B$ are finite subsets of $G$
  - $X$, $Y$ are uniformly distributed on $A$, $B$
  respectively
  Then:
  \[
    \entd XY
    \le \log \ruzd AB
  .\]
\end{fclemma}

\begin{proof}
  Without loss of generality $X$, $Y$ are
  indepent. Then
  \begin{align*}
    \entd X Y
    &= \ent{X - Y} - \half \ent X - \half \ent Y
    \\
    &\le \log |A - B| - \half \log A - \half \log
    B \\
    &= \log \ruzd AB
    \qedhere
  \end{align*}
\end{proof}

\begin{fclemma}[]
  \label{lemma:6.4}
  Assuming:
  - $X$, $Y$ are $G$-valued random variables
  Then:
  \[
    \ent{X + Y}
    \ge \max \{\ent X, \ent Y\}
    - \muti XY
  .\]
\end{fclemma}

\begin{proof}
  \begin{align*}
    \ent{X + Y}
    &\ge \cent{X + Y}{Y}
    &&\text{(by \nameref{subadd})} \\
    &= \ent{X + Y, Y} - \ent Y \\
    &= \ent{X, Y} - \ent Y \\
    &= \ent X + \ent Y - \ent Y - \muti XY \\
    &= \ent X - \muti XY
  \end{align*}
  By symmetry we also have
  \[
    \ent{X + Y}
    \ge \ent Y - \muti XY
    .
    \qedhere
  \]
\end{proof}

\begin{corollary*}[]
  Assuming that:
  \begin{itemize}
    \item $X$, $Y$ are $G$-valued random variables
  \end{itemize}
  Then:
  \[
    \ent{X - Y}
    \ge \max \{\ent X, \ent Y\}
    - \muti XY
  .\]
\end{corollary*}

\begin{fccoro}[]
  \label{coro:6.5}
  Assuming:
  - $X$, $Y$ are $G$-valued random variables
  Then:
  \[
    \entd XY \ge 0
  .\]
\end{fccoro}

\begin{proof}
  Without loss of generality $X$, $Y$ are
  independent. Then $\muti XY = 0$, so
  \begin{align*}
    \ent{X - Y}
    &\ge \max \{\ent X, \ent Y\} \\
    &\ge \half (\ent X + \ent Y)
    \qedhere
  \end{align*}
\end{proof}

\begin{fclemma}[]
  \label{lemma:6.6}
  Assuming:
  - $X$, $Y$ are $G$-valued random variables
  Then:
  \begin{iffc}
    \lhs
    $\entd XY = 0$
    \rhs
    there is some (finite) subgroup $H$ of $G$
    such that $X$ and $Y$ are uniform on cosets of
    $H$.
  \end{iffc}
\end{fclemma}

\begin{iffproof}
  \leftimpl
  If $X$, $Y$ are uniform on $x + H$, $y + H$,
  then $X' - Y'$ is uniform on $x - y + H$, so
  \[
    \ent{X' - Y'} = \ent X = \ent Y
  .\]
  So $\entd XY = 0$.

  \rightimpl
  Suppose that $X$, $Y$ are independent and
  $\ent{X - Y} = \half (\ent X + \ent Y)$.

  From the first line of the proof of
  \cref{lemma:6.4}, it follows that $\cent{X -
  Y}{Y} = \ent{X - Y}$. Therefore, $X - Y$ and $Y$
  are independent. So for every $z \in A - B$ and
  every $y_1, y_2 \in B$,
  \[
    \Pbb(X - Y = z \mid Y = y_1)
    = \Pbb(X - Y = z \mid Y = y_2)
  \]
  where $A = \{x : p_x \neq 0\}$, $B = \{y : q_y
  \neq 0\}$, i.e. for all $y_1, y_2 \in B$,
  \[
    \Pbb(X = y_1 + z)
    = \Pbb(X = y_2 + z)
  .\]
  So $p_x$ is constant on $z + B$.

  In particular, $A \supset z + B$.

  By symmetry, $B \supset A - z$.

  So $A = B + z$ for any $z \in A - B$. So for
  every $x \in A$, $y \in B$, $A = B + x - y$, so
  $A - x = B - y$. So $A - x$ is the same for
  every $x \in A$. Therefore, $A - x = A - A$ for
  every $x \in A$.

  It follows that
  \[
    A - A + A - A
    = (A - x) - (A - x)
    = A - A
  .\]
  So $A - A$ is a subgroup. Also, $A = A - A + c$,
  so $A$ is a coset of $A - A$. $B = A + z$, so
  $B$ is also a coset of $A - A$.
\end{iffproof}

Recall \cref{lemma:1.16}: If $Z = f(X) = g(Y)$,
then:
\[
  \ent{X, Y} + \ent Z
  \le \ent X + \ent Y
.\]

\begin{fclemma}[The entropic Ruzsa triangle inequality]
  \label{lemma:6.7}
  Assuming:
  - $X$, $Y$, $Z$ are $G$-valued random variables
  Then:
  \[
    \entd XZ
    \le \entd XY + \entd YZ
  .\]
\end{fclemma}

\begin{proof}
  We must show that (assuming without loss of
  generality that $X$, $Y$ and $Z$ are
  independent) that
  \[
    \ent{X - Z} - \half \ent X - \half \ent Z
    \le \ent{X - Y} - \half \ent X - \half \ent Y
    + \ent{Y - Z} - \half \ent Y - \half \ent Z
  ,\]
  i.e. that
  \[
    \ent{X - Z} + \ent Y
    \le \ent{X - Y} + \ent{Y - Z}
    \tag{$*$}
    \label{lec11eq1}
  .\]
  Since $X - Z$ is a function of $(X - Y, Y - Z)$
  and is also a function of $(X, Z)$, we get using
  \cref{lemma:1.16} that
  \[
    \ent{X - Y, Y - Z, X, Z}
    + \ent{X - Z}
    \le \ent{X - Y, Y - Z} + \ent{X, Z}
  .\]
  This is the same as
  \[
    \ent{X, Y, Z} + \ent{X - Z}
    \le \ent{X, Z} + \ent{X - Y, Y - Z}
  .\]
  By independence, cancelling common terms and
  \nameref{subadd}, we get \eqref{lec11eq1}.
\end{proof}

\begin{fclemma}[Submodularity for sums]
  \label{lemma:6.8}
  Assuming:
  - $X$, $Y$, $Z$ are independent $G$-valued
  random variables
  Then:
  \[
    \ent{X + Y + Z} + \ent Z
    \le \ent{X + Z} + \ent{Y + Z}
  .\]
\end{fclemma}

\begin{proof}
  $X + Y + Z$ is a function of $(X + Z, Y)$ and
  also a function of $(X, Y + Z)$. Therefore
  (using \cref{lemma:1.16}),
  \[
    \ent{X + Z, Y X, Y + Z} + \ent{X + Y + Z}
    \le \ent{X + Z, Y} + \ent{X, Y + Z}
  .\]
  Hence
  \[
    \ent{X, Y, Z} + \ent{X + Y + Z}
    \le \ent{X + Z} + \ent Y + \ent X + \ent{Y + Z}
  .\]
  By independence and cancellation, we get the
  desired inequality.
\end{proof}