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Probability
└── Homework
    └── W8
        └── Q4.tex

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\fancyhead[l]{Li Yifeng}
\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}