Correlation Coefficient between $x$ and $y$ and between $x$ and $\begin{Bmatrix} y+x \end{Bmatrix}$ 
Let  Let Correlation Coefficient between $x$ and $y$  be $R$ and between
  $x$ and $\begin{Bmatrix} y+x \end{Bmatrix}$ be $S$ , then 
a) $R<S$
b) $R>S$
c)$R=S$
d) None of the above

The correct answer is none of the above, can anyone explain why? , then 
 A: If you take $x = (1,2,3)$ and $y = (4,5,6)$, there are linearly dependent and you can see that $R=S=1$ (then a/ and b/ cannot be true).
If you take $x = (1,2,3)$ and $y = (-2,-4,-3)$, then $0.5 = S \geq R = 1$ (then c/ cannot be true).
Generally, we have $R \leq S$ (assuming the correlation exists).
You need to use the triangular inequality for the standard deviation:
$\sigma_{X+Y} \leq \sigma_{X} + \sigma_{Y}.$
Then, you can write: 
$$R = \frac{Cov(X,Y)}{\sigma_{X}\sigma_{Y}},$$ whereas:
$$S = \frac{Cov(X,X+Y)}{\sigma_{X}\sigma_{X+Y}}$$
For $S$, using bilinearity of $Cov$,
$$S = \frac{\sigma_{X}^2 + Cov(X,Y)}{\sigma_{X}\sigma_{X+Y}}.$$
Using the triangular inequality:
$$S \geq \frac{\sigma_{X}^2 + Cov(X,Y)}{\sigma_{X}(\sigma_{X}+\sigma_{Y})} = \frac{\sigma_{X}^2 + Cov(X,Y)}{\sigma_{X}^2+\sigma_{X}\sigma_{Y}}.$$
You can then compare $A:= \frac{\sigma_{X}^2 + Cov(X,Y)}{\sigma_{X}^2+\sigma_{X}\sigma_{Y}}$ with $R$ and conclude that $R \leq S$.
A: Here are the examples of random variables $X$ and $Y$ to demonstrate that none of the first three options holds.


*

*When $Y = -X$, then correlation between $X$ and $Y$ is $R = -1$ and between $X$ and $X+Y = 0$ is $S = 0$. This rules out options (b) and (c).

*When $Y = X$, then correlation between $X$ and $Y$ is $R = 1$ and between $X$ and $X+Y = 2X$ is also $S = 1$. This rules out options (a) and (b).

