2017.3.12 Question 12

1. First, note that

   $$\begin{array}{llll}\hfill 1& =\sum _{x,y=1}^{x=n}\text{P}<!--\; FUNCTION\; APPLICATION\; -->(X=x,Y=y)& \hfill & \\ \hfill & =\sum _{x=1}^n\sum _{y=1}^{n}k(x+y)& \hfill & \\ \hfill & =\sum _{x=1}^n\sum _{y=1}^n(\mathit{kx}+\mathit{ky})& \hfill & \\ \hfill & =\sum _{x=1}^n\left(n\cdot \mathit{kx}+k\sum _{y=1}^{n}y\right)& \hfill & \\ \hfill & =\mathit{nk}\sum _{x=1}^{n}x+\mathit{nk}\sum _{y=1}^{n}y& \hfill & \\ \hfill & =n^2(n+1)k& \hfill & \end{array}$$

   Therefore, $k=\frac{1}{n^2(n+1)}$

   $$\begin{array}{llll}\hfill \text{P}<!--\; FUNCTION\; APPLICATION\; -->(X=x)& =\sum _{y=1}^n\text{P}<!--\; FUNCTION\; APPLICATION\; -->(X=x,Y=y)& \hfill & \\ \hfill & =\sum _{y=1}^{n}k(x+y)& \hfill & \\ \hfill & =\mathit{nkx}+k\sum _{y=1}^{n}y& \hfill & \\ \hfill & =\mathit{nkx}+\frac{\mathit{kn}(n+1)}{2}& \hfill & \\ \hfill & =\frac{x}{n(n+1)}+\frac{1}{2n}& \hfill & \\ \hfill & =\frac{2x+n+1}{2n(n+1)},& \hfill & \end{array}$$ as desired.

   By symmetry, $\text{P}<!--\; FUNCTION\; APPLICATION\; -->(Y=y)=\frac{2y+n+1}{2n(n+1)}$.

   We have

   $$\text{P}<!--\; FUNCTION\; APPLICATION\; -->(X=x)\cdot \text{P}<!--\; FUNCTION\; APPLICATION\; -->(Y=y)=\frac{(2x+n+1)(2y+n+1)}{4n^2{(n+1)}^2}.$$

   But $\text{P}<!--\; FUNCTION\; APPLICATION\; -->(X=x,Y=y)=\frac{x+y}{n^2(n+1)}$ is not equal to this. So $X$ and $Y$ are not independent.

2. By definition,

   $$\text{Cov}<!--\; FUNCTION\; APPLICATION\; -->(X,Y)=\text{E}<!--\; FUNCTION\; APPLICATION\; -->(\mathit{XY})-\text{E}<!--\; FUNCTION\; APPLICATION\; -->(X)\text{E}<!--\; FUNCTION\; APPLICATION\; -->(Y).$$

   We have

   $$\begin{array}{llll}\hfill \text{E}<!--\; FUNCTION\; APPLICATION\; -->(X)=\text{E}<!--\; FUNCTION\; APPLICATION\; -->(Y)& =\sum _{t=1}^{n}t\cdot \text{P}<!--\; FUNCTION\; APPLICATION\; -->(X=t)& \hfill & \\ \hfill & =\sum _{t=1}^n\frac{t\cdot (2t+n+1)}{2n(n+1)}& \hfill & \\ \hfill & =\frac{1}{n(n+1)}\sum _{t=1}^n{t}^2+\frac{1}{2n}\sum _{t=1}^{n}t& \hfill & \\ \hfill & =\frac{n(n+1)(2n+1)}{6n(n+1)}+\frac{n(n+1)}{4n}& \hfill & \\ \hfill & =\frac{2n+1}{6}+\frac{n+1}{4}& \hfill & \\ \hfill & =\frac{4n+2+3n+3}{12}& \hfill & \\ \hfill & =\frac{7n+5}{12},& \hfill & \end{array}$$

   and

   $$\begin{array}{llll}\hfill \text{E}<!--\; FUNCTION\; APPLICATION\; -->(\mathit{XY})& =\sum _{x,y=1}^n\mathit{xy}\cdot \text{P}<!--\; FUNCTION\; APPLICATION\; -->(X=x,Y=y)& \hfill & \\ \hfill & =\sum _{x=1}^n\sum _{y=1}^n\frac{\mathit{xy}(x+y)}{n^2(n+1)}& \hfill & \\ \hfill & =\frac{1}{n^2(n+1)}\sum _{x=1}^n\sum _{y=1}^n\mathit{xy}(x+y)& \hfill & \\ \hfill & =\frac{1}{n^2(n+1)}\sum _{x=1}^n\sum _{y=1}^n(x^{2}y+x{y}^2)& \hfill & \\ \hfill & =\frac{1}{n^2(n+1)}\left[\sum _{x=1}^n{x}^2\sum _{y=1}^{n}y+\sum _{x=1}^{n}x\sum _{y=1}^n{y}^2\right]& \hfill & \\ \hfill & =\frac{1}{n^2(n+1)}\cdot 2\cdot \frac{n(n+1)(2n+1)}{6}\cdot \frac{n(n+1)}{2}& \hfill & \\ \hfill & =\frac{(2n+1)(n+1)}{6}.& \hfill & \end{array}$$

   Therefore,

   $$\begin{array}{llll}\hfill \text{Cov}<!--\; FUNCTION\; APPLICATION\; -->(X,Y)& =\text{E}<!--\; FUNCTION\; APPLICATION\; -->(\mathit{XY})-\text{E}<!--\; FUNCTION\; APPLICATION\; -->(X)\text{E}<!--\; FUNCTION\; APPLICATION\; -->(Y)& \hfill & \\ \hfill & =\frac{(2n+1)(n+1)}{6}-\frac{{(7n+5)}^2}{144}& \hfill & \\ \hfill & =\frac{48n^2+72n+24}{144}-\frac{49n^2+70n+25}{144}& \hfill & \\ \hfill & =\frac{-n^2+2n-1}{144}& \hfill & \\ \hfill & =-\frac{{(n-1)}^2}{144}& \hfill & \\ \hfill & <0,& \hfill & \end{array}$$

   as desired.