Q4
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Probability
└── Homework
└── W6
└── Q4.tex
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\begin{document}
\section*{Question 4}
\noindent Let $X$ be a discrete random variable with the following probability distribution function (PDF):
\[
p_X(k) =
\begin{cases}
\begin{aligned}
\frac{1}{4}, &\quad k = -2\\
\frac{1}{8}, &\quad k = -1\\
\frac{1}{8}, &\quad k = 0\\
\frac{1}{4}, &\quad k = 1\\
\frac{1}{4}, &\quad k = 2\\
0, &\quad \text{otherwise}
\end{aligned}
\end{cases}
\]
\noindent We define a new random variable $Y$ as $Y = (X + 1)^2$. Find the probability distribution function (PDF) of $Y$.
\bigskip
\begin{enumerate}[label={},leftmargin=0in]\item
\subsection*{Solution}
Easy to see that
\[
\begin{tabular}{|c|c|c|}
\hline
$X$ & $(X+1)^2=Y$ & $\bP(X)$ \\ \hline
$-2$ & $1$ & $\frac{1}{4}$ \\ \hline
$-1$ & $0$ & $\frac{1}{8}$ \\ \hline
$0$ & $1$ & $\frac{1}{8}$ \\ \hline
$1$ & $4$ & $\frac{1}{4}$ \\ \hline
$2$ & $9$ & $\frac{1}{4}$ \\ \hline
\end{tabular}
\]
Then we have the $pdf$ of $Y$
\[
p_Y(k) =
\begin{cases}
\begin{aligned}
\frac{1}{8}, &\quad k = 0\\
\frac{1}{4} + \frac{1}{8} = \frac{3}{8}, &\quad k = 1\\
\frac{1}{4}, &\quad k = 4\\
\frac{1}{4}, &\quad k = 9\\
0, &\quad \text{otherwise}
\end{aligned}
\end{cases}
\]
\subsection*{Answer}
\[\boxed{p_Y(k) =
\begin{cases}
\begin{aligned}
\frac{1}{8}, &\quad k = 0\\
\frac{3}{8}, &\quad k = 1\\
\frac{1}{4}, &\quad k = 4\\
\frac{1}{4}, &\quad k = 9\\
0, &\quad \text{otherwise}
\end{aligned}
\end{cases}}\]
\end{enumerate}
\end{document}