Solutions - 5240-01

(a)

$$\begin{array}{c}E[X+Y]=\sum _x\sum _y(x+y)p(x,y)\\ \\ \gg \gg 1\left(\frac{1}{15}\right)+2(\frac{1}{10}+\frac{1}{15})+3(\frac{1}{30}+\frac{2}{15}+\frac{1}{10})\\ \\ \gg \gg +4(\frac{1}{15}+\frac{1}{5}+\frac{1}{30})+5\left(\frac{1}{10}\right)+6\left(\frac{1}{10}\right)\gg \gg \frac{7}{2}\end{array}$$

---

(b)

$$\begin{array}{cc}E\left[{(X-Y)}^2\right]& =\sum _x\sum _y{(x-y)}^{2}p(x,y)\\ & =1(\frac{1}{15}+\frac{1}{10}\frac{2}{15}+\frac{1}{10})+2(\frac{1}{10}+\frac{1}{30}+\frac{1}{15})+9\left(\frac{1}{30}\right)\\ & =1.5\end{array}$$

---

(c)

(1) Recall the formula for $\text{Cov}[X,Y]$:

$$\text{Cov}[X,Y]=E[XY]-E[X]E[Y]$$

---

(2) Compute $E[X]$ and $E[Y]$:

$$\begin{array}{cc}E[X]& =1(\frac{1}{15}+\frac{1}{10}+\frac{1}{30})+2(\frac{2}{15}+\frac{1}{5})+3(\frac{1}{15}+\frac{1}{10}+\frac{1}{10})\\ & =\frac{5}{3}\\ E[Y]& =1(\frac{1}{15}+\frac{1}{15}+\frac{2}{15}+\frac{1}{15})+2(\frac{1}{10}+\frac{1}{10}+\frac{1}{5}+\frac{1}{10})+3(\frac{1}{30}+\frac{1}{30}+\frac{1}{10})\\ & =\frac{11}{6}\end{array}$$

---

(3) Compute $E[XY]$:

$$E[XY]=1\left(\frac{1}{15}\right)+2(\frac{1}{10}+\frac{2}{15})+3(\frac{1}{30}+\frac{1}{15})+4\left(\frac{1}{5}\right)+6\left(\frac{1}{10}\right)+9\left(\frac{1}{10}\right)=\frac{47}{15}$$

---

(4) Compute $\text{Cov}[X,Y]$:

$$\begin{array}{c}\text{Cov}[X,Y]=\frac{47}{15}-\frac{5}{3}\cdot \frac{11}{6}\gg \gg \frac{7}{90}\end{array}$$

---

(d)

(1) Recall the formula for $\rho [X,Y]$:

$$\rho [X,Y]=\frac{\text{Cov}[X,Y]}{\sqrt{\text{Var}[X]\text{Var}[Y]}}$$

---

(2) Compute $E\left[X^2\right]$ and $E\left[Y^2\right]$:

$$\begin{array}{cc}E\left[X^2\right]& =1(\frac{1}{15}+\frac{1}{10}+\frac{1}{30})+4(\frac{2}{15}+\frac{1}{5})+9(\frac{1}{15}+\frac{1}{10}+\frac{1}{10})\\ & =\frac{59}{15}\\ E\left[Y^2\right]& =1(\frac{1}{15}+\frac{1}{15}+\frac{2}{15}+\frac{1}{15})+4(\frac{1}{10}+\frac{1}{10}+\frac{1}{5}+\frac{1}{10})+9(\frac{1}{30}+\frac{1}{30}+\frac{1}{10})\\ & =\frac{23}{6}\end{array}$$

---

(3) Compute $\text{Var}[X]$ and $\text{Var}[Y]$:

$$\begin{array}{cc}\text{Var}[X]& =E\left[X^2\right]-{\left(E[X]\right)}^2=\frac{59}{15}-{\left(\frac{5}{3}\right)}^2=\frac{52}{45}\\ \text{Var}[Y]& =E\left[Y^2\right]-{\left(E[Y]\right)}^2=\frac{23}{6}-{\left(\frac{11}{6}\right)}^2=\frac{17}{36}\end{array}$$

---

(4) Compute $\rho [X,Y]$:

$$\begin{array}{c}\rho [X,Y]=\frac{\frac{7}{90}}{\sqrt{\frac{52}{45}\cdot \frac{17}{36}}}\gg \gg \approx 0.1053\end{array}$$