Q4
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Probability
└── Homework
└── W8
└── Q4.tex
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\fancyhead[c]{Homework \#8}
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\begin{document}
\section*{Question 4}
\noindent Given the joint PMF:
\bigskip
\[
\begin{tabular}{|c|c|c|c|}
\hline
$ $ & $X = -2$ & $X = 0$ & $X = 2$ \\ \hline
$Y = 1$ & $0$ & $2a $ & $a $ \\ \hline
$Y = 2$ & $2a$ & $0 $ & $2a $ \\ \hline
$Y = 4$ & $a$ & $2a $ & $0 $ \\ \hline
\end{tabular}
\]
\bigskip
\noindent Let $S = X + Y$, $Z = X - Y$.
\bigskip
\begin{enumerate}[start=1,label={\bfseries Part \arabic*:},leftmargin=0in]
\bigskip\item Find the value of $a$ and the marginal probability density function of $X$.
\subsection*{Solution}
Summing all entries: $(3a)+(4a)+(3a)=10a=1$, so $a=\tfrac{1}{10}$. Marginally,
\[
\begin{aligned}
P(X=-2)&=3a&=0.3\\
P(X=0)&=4a&=0.4\\
P(X=2)&=3a&=0.3
\end{aligned}
\]
\subsection*{Answer}
\[\boxed{a=0.1,\quad P_X(-2)=0.3,\;P_X(0)=0.4,\;P_X(2)=0.3}\]
\bigskip\item Are $X$ and $Y$ independent?
\subsection*{Solution}
For example, $P(X=-2,Y=1)=0\ne P(X=-2)P(Y=1)=0.3\cdot0.3=0.09$, so not independent.
\subsection*{Answer}
\[\boxed{No.}\]
\bigskip\item Compute the covariance of $S$ and $Z$.
\subsection*{Solution}
Note $S=X+Y$, $Z=X-Y$, so
\[
\begin{aligned}
\mathrm{Cov}(S,Z)&=\mathrm{Cov}(X+Y,X-Y)\\
&=\mathrm{Var}(X)-\mathrm{Var}(Y)
\end{aligned}
\]
We have
\[
\begin{aligned}
E[X]&=0\\
\mathrm{Var}(X)&=4(0.3+0.3)=2.4\\
E[Y]&=2.3\\
\mathrm{Var}(Y)&=6.7-(2.3)^2=1.41
\end{aligned}
\]
Thus
\[
\mathrm{Cov}(S,Z)=2.4-1.41=0.99
\]
\subsection*{Answer}
\[\boxed{\mathrm{Cov}(S,Z)=0.99}\]
\bigskip\item Are $S$ and $Z$ independent?
\subsection*{Solution}
Since $\mathrm{Cov}(S,Z)=0.99\ne0$, they cannot be independent.
\subsection*{Answer}
\[\boxed{No.}\]
\end{enumerate}
\end{document}