Q4
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└── Homework
└── W10
└── Q4.tex
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\begin{document}
\section*{Question 4}
\noindent Consider a binary communication channel with input $X\sim\mathrm{Bern}(p)$, output $Y\sim\mathrm{Bern}(q)$, and crossover (error) probability $\varepsilon$. It is known that
\[
q \;=\; p + \varepsilon \;-\; 2\varepsilon p.
\]
The channel capacity is
\[
C \;=\;\max_{0\le p\le1} I(X;Y).
\]
Find the value of $p$ that maximises $C$ (i.e.\ maximises $I(X;Y)$).
\bigskip
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\subsection*{Solution}
The mutual information can be written
\[
I(X;Y)
=H(Y)-H(Y\mid X)
=H\bigl(q\bigr)-H\bigl(\varepsilon\bigr),
\]
where $H(\varepsilon)=-\varepsilon\log_2\varepsilon-(1-\varepsilon)\log_2(1-\varepsilon)$ is constant in $p$, and
\[
q(p)=p + \varepsilon -2\varepsilon p
=(1-2\varepsilon)p + \varepsilon.
\]
Differentiate with respect to $p$:
\[
\frac{dI}{dp}
=\frac{d}{dp}H(q)
=(1-2\varepsilon)\,\bigl[-\log_2 q + \log_2(1-q)\bigr]
\stackrel{!}{=}0.
\]
Hence
\[
\log_2\frac{1-q}{q}=0
\;\Longrightarrow\;q=\tfrac12.
\]
Solving $q=(1-2\varepsilon)p+\varepsilon=\tfrac12$ gives
\[
(1-2\varepsilon)p=\tfrac12-\varepsilon
\;\Longrightarrow\;
p=\frac{\tfrac12-\varepsilon}{1-2\varepsilon}
=\tfrac12,
\]
for all $\varepsilon\neq\tfrac12$.
\subsection*{Answer}
\[
\boxed{p=\frac12.}
\]
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