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\begin{fclemma}[]
  \label{lemma:6.9}
  Assuming:
  - $G$ an abelian group
  - $X$ a $G$-valued random variable
  Then:
  \[
    \entd{X}{-X}
    \le 2\entd{X}{X}
  .\]
\end{fclemma}

\begin{proof}
  Let $X_1$, $X_2$, $X_3$ be independent copies of
  $X$. Then
  \begin{align*}
    \entd{X}{-X}
    &= \ent{X_1 + X_2} - \half \ent{X_1} - \half
    \ent{X_2} \\
    &\le \ent{X_1 + X_2 - X_3} - \ent X \\
    &\le \ent{X_1 - X_3} + \ent{X_2 - X_3} -
    \ent{X_3} - \ent X \\
    &= 2\entd{X}{X}
  \end{align*}
  (as $X_1, X_2, X_3$ are all copies of $X$).
\end{proof}

\begin{fccoro}[]
  \label{lemma:6.10}
  Assuming:
  - $X$ and $Y$ are $G$-valued random variables
  Then:
  \[
    \entd{X}{-Y}
    \le 5 \entd{X}{Y}
  .\]
\end{fccoro}

\begin{proof}
  \begin{align*}
    \entd{X}{-Y}
    &\le \entd XY + \entd{Y}{-Y} \\
    &\le \entd XY + 2\entd YY \\
    &\le \entd XY + 2(\entd YX + \entd XY) \\
    &= 5 \entd XY
    \qedhere
  \end{align*}
\end{proof}

\subsubsection*{Conditional Distances}

\begin{fcdefnstar}[Conditional distance]
  \glssymboldefn{centdist}%
  Let $X, Y, U, V$ be $G$-valued random variables
  (in fact, $U$ and $V$ don't have to be
  $G$-valued for the definition to make sense).
  Then the \emph{conditional distance} is
  \[
    d[X \mid U ; Y \mid V]
    = \sum_{u, v} \Pbb[U = u] \Pbb[V = v] \entd{X
    \mid U = u}{Y \mid V = v}
  .\]
\end{fcdefnstar}

The next definition is not completely standard.

\begin{fcdefnstar}[Simultaneous conditional distance]
  \glssymboldefn{scentd}%
  \glsnoundefn{condindept}{conditionally
  independent trials}{NA}%
  Let $X, Y, U$ be $G$-valued random variables.
  The \emph{simultaneous conditional distance of
  $X$ to $Y$ given $U$} is
  \[
    d[X ; Y \| U]
    = \sum_u \Pbb[U = u] \entd{X \mid U = u}{Y
    \mid U = u}
  .\]
  We say that $X'$, $Y'$ are \emph{conditionally
  independent trials} of $X$, $Y$ given $U$ if:
  \begin{itemize}
    \item $X'$ is distributed like $X$.
    \item $Y'$ is distributed like $Y$.
    \item For each $u \in U$, $X' \mid U = u$ is
      distributed like $X \mid U = u$,
    \item For each $u \in U$, $Y' \mid U = u$
      is distributed like $Y \mid U = u$.
    \item $X' \mid U = u$ and $Y' \mid U = u$ are
      independent.
  \end{itemize}
  Then
  \[
    d[X; Y \| U]
    = \cent{X' - Y'}U - \half \cent{X'}U - \half
    \cent{Y'}U
  \]
  (as can be seen directly from the formula).
\end{fcdefnstar}

\begin{fclemma}[The entropic BSG theorem]
  \label{lemma:6.11}
  Assuming:
  - $A$ and $B$ are $G$-valued random variables
  Then:
  \[
    \scentd{A}{B}{A+B}
    \le 3 \muti AB + 2\ent{A + B} - \ent A - \ent
    B
  .\]
\end{fclemma}

\begin{remark*}
  The last few terms look like $2\entd{A}{-B}$.
  But they aren't equal to it, because $A$ and $B$
  aren't (necessarily) independent!
\end{remark*}

\begin{proof}
  \[
    \scentd{A}{B}{A + B}
    = \cent{A' - B'}{A + B} - \half \cent{A'}{A +
    B} - \half \cent{B'}{A + B}
  \]
  where $A'$, $B'$ are \gls{condindept} of $A$,
  $B$ given $A + B$. Now calculate
  \begin{align*}
    \cent{A'}{A + B}
    &= \cent{A}{A + B} \\
    &= \ent{A, A + B} - \ent{A + B} \\
    &= \ent{A, B} - \ent{A + B} \\
    &= \ent A + \ent B - \muti AB - \ent{A + B}
  \end{align*}
  Similarly, $\cent{B'}{A + B}$ is the same, so
  $\half \cent{A'}{A + B} + \half \cent{B'}{A +
  B}$ is also the same.
  \[
    \cent{A' - B'}{A + B}
    \le \ent{A' - B'}
  .\]
  Let $(A_1, B_1)$ and $(A_2, B_2)$ be
  conditionally independent trials of $(A, B)$
  given $A + B$. Then $\ent{A' - B'} = \ent{A_1 -
  B_2}$. By \nameref{submod},
  \begin{align*}
    \ent{A_1 - B_2}
    &\le \ent{A_1 - B_2, A} + \ent{A_1 - B_2, B_1}
    - \ent{A_1 - B_2, A_1, B_1} \\
    \ent{A_1 - B_2, A_1}
    &= \ent{A_1, B_2} \\
    &\le \ent{A_1} + \ent{B_2} \\
    &= \ent A + \ent B \\
    \ent{A_1 - B_2, B_1}
    &= \ent{A_2 - B_1, B_1}
    &&\text{(since $A_1 + B_1 = A_2 + B_2$)} \\
    &= \ent{A_2, B_1} \\
    &\le \ent A + \ent B
  \end{align*}
  Finally,
  \begin{align*}
    \ent{A_1 - B_2, A_1, B_1}
    &= \ent{A_1, B_1, A_2, B_2} \\
    &= \cent{A_1, B_1, A_2, B_2}{A + B}
    + \ent{A + B} \\
    &= 2\ent{A, B}{A + B} + \ent{A + B}
    &&\text{(by conditional independence of $(A_1,
    B_1)$ and $(A_2, B_2)$)} \\
    &= 2\ent{A, B} - \ent{A + B} \\
    &= 2\ent A + 2\ent B - 2\muti AB - \ent{A + B}
  \end{align*}
  Adding or subtracting as appropriate all these
  terms gives the required inequality.
\end{proof}