Q1
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└── Homework
└── W5
└── Q1.tex
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\fancyhead[l]{Li Yifeng}
\fancyhead[c]{Homework \#5}
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\begin{document}
\section*{Question 1}
\noindent Consider a binary communication channel, with input $X$ having a Bernoulli distribution with parameter $p = 0.9$. The common error probability is $\epsilon = 0.05$ (i.e., the probability that the received character differs from the input character is $\epsilon$). Let $Y$ denote the output character.
\bigskip
\begin{enumerate}[start=1,label={\bfseries Part \arabic*:},leftmargin=0in]
\bigskip\item Show that $Y$ is a Bernoulli distribution with parameter $q$.
\subsection*{Solution}
We know that $X$ has a Bernoulli distribution in sample space $\Omega = \{0,1\}$ (let 0 denotes $not\enspace received$ and 1 denotes $received$) with parameter $p = 0.9$, then easy to see that
\[
\begin{aligned}
\bP(Y=0) &= \bP(X=0)\bP_{X=0}(Y=0) + \bP(X=1)\bP_{X=1}(Y=0)\\
&= (1-p)(1-\epsilon) + p\epsilon\\
&= 2p\epsilon - p - \epsilon + 1
\end{aligned}
\]
Similiarly, we have
\[
\begin{aligned}
\bP(Y=1) &= \bP(X=0)\bP_{X=0}(Y=1) + \bP(X=1)\bP_{X=1}(Y=1)\\
&= (1-p)\epsilon + p(1-\epsilon)\\
&= -2p\epsilon + p + \epsilon
\end{aligned}
\]
Then we have
\[
0 < \bP(Y=0), \bP(Y=1) < 1,\quad\text{and}\quad \bP(Y=0) + \bP(Y=1) = 1
\]
Then we can say that $Y$ is a Bernouli distribution in sample space $\Omega = \{0,1\}$ with parameter
\[
\begin{aligned}
q &= \bP(Y = 1)\\
&= -2p\epsilon + p + \epsilon
\end{aligned}
\]
\subsection*{Answer}
\[\boxed{\text{See above.}}\]
\bigskip\item Determine $q$.
\subsection*{Solution}
With the explanation and formula above, easy to see that
\[q = -2p\epsilon + p + \epsilon = 0.86\]
\subsection*{Answer}
\[\boxed{q = 0.86}\]
\bigskip\item Compute the joint probability distribution function of $(X,Y)$.
\subsection*{Solution}
Easy to compute the joint distribution function of $(X,Y)$ as
\[
f(x,y) =
\begin{cases}
\begin{aligned}
\bP(X=0,Y=0) &= \bP(X=0)\bP_{X = 0}(Y = 0) = &(1-p)1 - \epsilon &= 0.095,&\quad x = 0,\quad y = 0\\
\bP(X=0,Y=1) &= \bP(X=0)\bP_{X = 0}(Y = 1) = &(1-p)\epsilon &= 0.005,&\quad x = 0,\quad y = 1\\
\bP(X=1,Y=0) &= \bP(X=1)\bP_{X = 1}(Y = 0) = &p\epsilon &= 0.045,&\quad x = 1,\quad y = 0\\
\bP(X=1,Y=1) &= \bP(X=1)\bP_{X = 1}(Y = 1) = &p(1 - \epsilon) &= 0.855,&\quad x = 1,\quad y = 1
\end{aligned}
\end{cases}
\]
\subsection*{Answer}
\[\boxed{f(x,y) =
\begin{cases}
\begin{aligned}
0.095,&\quad x = 0,\quad y = 0\\
0.005,&\quad x = 0,\quad y = 1\\
0.045,&\quad x = 1,\quad y = 0\\
0.855,&\quad x = 1,\quad y = 1
\end{aligned}
\end{cases}}\]
\end{enumerate}
\end{document}