1. (a) The joint probability mass function of $X$ and $Y$ $p_{X,Y}: \mathbb{R}^{2} \to [0,1]$ is given by $p_{X,Y}(x,y)= \begin{cases} \frac{1}{8} & (x,y) \in \{ (1,0), (3,0), (1,2), (3,2) \} \\ \frac{1}{2} & (x,y) = (2,1) \\ 0 & \text{otherwise} \end{cases}$Its support is given by $D_{X,Y}= \{ (1,0), (3,0), (1,2), (3,2), (2,1) \}.$ (b) Note that the supports of $X$ and $Y$ are given by $D_{X} = \{ 1,2,3 \} \text{ and } D_{Y} = \{ 0,1,2 \}$So using $p_{X}(x)=\sum_{y \in D_{Y}} p_{X,Y}(x,y)$, we obtain $p_{X}(x) = \begin{cases}\frac{1}{4} & x=1 \\ \frac{1}{2} & x= 2 \\ \frac{1}{4} & x= 3 \\ 0 & \text{otherwise} \end{cases}$ Similarly $p_{Y}(y) = \begin{cases} \frac{1}{4} & y= 0 \\ \frac{1}{2} & y = 1 \\ \frac{1}{4} & y=2 \\ 0 & \text{otherwise} \end{cases}$ (c) Note that $\mathbb{E}[X]=\sum_{x\in D_{X}} x \cdot p_{X}(x) = 1 \times \frac{1}{4} + 2 \times \frac{1}{2} + 3 \times \frac{1}{4}= 2$. Similarly $\mathbb{E}[Y] = 0 \times \frac{1}{4} + 1 \times \frac{1}{2} + 2 \times \frac{1}{4} = 1$ and $\mathbb{E}[XY]= 0 \times \frac{1}{4} +2 \times \left( \frac{1}{2} + \frac{1}{8} \right) +6 \times \frac{1}{8} = 2$. Now $\text{Cov(X,Y)}= \mathbb{E}[XY] - \mathbb{E}[X] \cdot \mathbb{E}[Y]= 2 - 2 \times 1 = 0$. So $\rho(X,Y)=0$. (d) Note that $p_{X,Y}(1,1) = 0$. However $p_{X}(1) \times p_{Y}(1) = \frac{1}{4} \times \frac{1}{2} = \frac{1}{8} \neq 0$ which shows that $X$ and $Y$ are not independent. 2. Note that $\mathbb{E}\left[ \frac{X_{1}+\dots+X_{n}}{n} \right] = \mu$ by linearity of expectation since each $X_{i}$ has the same mean $\mu$. Also $\text{Var}(X_{i}) = \text{Cov}(X_{i},X_{i}) = \sigma^{2}$ for $i=1,2,\dots,n$. Hence $\begin{align}\text{Var}\left( \frac{\sum_{i=1}^{n} X_{i}}{n } \right) & = \frac{1}{n^{2}} \text{Var} \left( \sum_{i=1}^{n} X_{i} \right) \\ &= \frac{1}{n^{2}} \left[ \sum_{i=1}^{n} \text{Var}(X_{i }) + 2\sum_{1\leq j<k \leq n} \text{Cov} (X_{j}, X_{k}) \right] \\ &= \frac{1}{n^{2}} \left[ \sum_{i=1}^{n} \sigma^{2} + 2 \sum_{\substack{j=k-1 \\ 2 \leq k \leq n}} \text{Cov} (X_{j}, X_{k}) \right] \quad \text{(since Cov} (X_{j},X_{k}) = 0 \text{ for } j<k-1 \text{)} \\ &= \frac{1}{n^{2}} (n \sigma^{2} + 2(n-1) r) \quad \text{(since Cov} (X_{j},X_{k}) = r \text{ for } j=k-1 \text{)} \\ &= \frac{\sigma^{2} + 2r}{n} - \frac{2r}{n^{2}} \end{align}$So using Chebyshev's inequality, $\begin{align}\mathbb{P}\left( \mu- a < \frac{X_{1}+\dots+X_{n}}{n} <\mu+a \right) &= 1- \mathbb{P}\left( \left| \frac{(X_{1}+\dots+X_{n})}{n} - \mu \right | \geq a \right) \\ &\geq 1- \frac{\text{Var}\left( \frac{X_{1}+\dots+X_{n}}{n} \right)}{a^{2}} \\ &= 1 - \frac{\sigma^{2} + 2r}{a^{2}n} + \frac{2r}{a^{2} n^{2}} > 1 - \frac{\sigma^{2} + 2r}{a^{2}n} \end{align}$since $\frac{2r}{a^{2} n^{2}} > 0$, which concludes the proof. 3. The mean and standard deviation of each random variable are given by $\mu= \frac{1}{2}$ and $\sigma= \frac{1}{2}$. Now $\begin{align} &\mathbb{P}\left( \left\lvert \frac{X_{1}+\dots+X_{n}}{n} - \frac{1}{2} \right\rvert \geq 0.0196 \right) \leq 0.05 \\ \iff &1- \mathbb{P}\left( -0.0196 < \frac{ 2(X_{1}+\dots+X_{n}) -n }{2n} < 0.0196 \right) \leq 0.05 \\ \iff & \mathbb{P} \left( - 0.0196 \times 2 \sqrt{ n} < \frac{X_{1}+\dots+X_{n} - \frac{n}{2}}{\frac{1}{2} \cdot \sqrt{ n }} <0.0196 \times 2 \sqrt{ n } \right) \geq 0.95 \end{align}$Using CLT and the datum, this inequality is satisfied by $2\sqrt{ n } = 100$ which gives $n = 2500$. Using linearity of expectation, $\mathbb{E}\left[ \frac{X_{1}+\dots+X_{n}}{n} \right] = \frac{1}{2}$ and $\text{Var}\left( \frac{X_{1}+\dots+X_{n}}{n} \right) = \frac{\sigma^{2}}{n} = \frac{1}{4n}$ since the random variables are independent. Using Chebyshev's inequality $\begin{align}\mathbb{P}\left( \left\lvert \frac{X_{1}+\dots+X_{n}}{n} - \frac{1}{2} \right\rvert \geq 0.0196 \right) \leq \frac{\text{Var}\left( \frac{X_{1}+\dots+X_{n}}{n} \right)}{0.0196^{2}} = \frac{1}{4 \times 0.0196^{2} \times n} \end{align}$ so we want $\frac{1}{4 \times 0.0196^{2} \times n} \leq 0.05$ which gives $n \geq 13015.41$.