If X and Y are perfectly correlated, what is the correlation of X+Y and X-Y? I've started with thinking that if X and Y are perfectly correlated, then it is the same as looking at the correlation of Y and -Y (since the X provides no new information) and thus the correlation is -1.
Is this correct?
 A: Hint: In general, \begin{align}
\rho_{A,B} &= \frac{\operatorname{cov}(A,B)}{\sqrt{\operatorname{var}(A)\operatorname{var}(B)}},\\
\operatorname{var}(X\pm Y)&= \operatorname{var}(X)+\operatorname{var}(Y)
\pm 2\operatorname{cov}(X,Y),\\
\text{and}\qquad\operatorname{cov}(X+Y,X-Y)&=\operatorname{var}(X)-\operatorname{var}(Y)\end{align} So, work out what $\rho_{X+Y,X-Y}$ is in general, and in the special case when $Y = aX+b$. You might be surprised at the result.
A: I'll treat this as self-study, and I'd encourage you to read its wiki and add the tag.
Your argument is already very good. Here are a few pointers. Feel free to write a comment so we can discuss and work towards a good answer.


*

*I assume you are looking at Pearson's correlation, right? (Does your argument work for other measures of correlation?)

*What does a perfect correlation mean graphically?

*What will $X+Y$ and $X-Y$ look like graphically if $X$ and $Y$ are perfectly Pearson-correlated?

A: $X,Y$ perfectly correlate $\implies Cov(X,Y)=\sqrt{Var(X)Var(Y)}$
$Cov(X+Y,X-Y)=Cov(X,X)-Cov(X,Y)+Cov(Y,X)-Cov(Y,Y)=Var(X)-Var(Y)$
$Var(X+Y)Var(X-Y)=[Var(X)+Var(Y)+2Cov(X,Y)][Var(X)+Var(Y)-2Cov(X,Y)]=[Var(X)+Var(Y)]^2-4Cov(X,Y)Cov(X,Y)=_{\rho_{X,Y}=1}=[Var(X)+Var(Y)]^2-4Var(X)Var(Y)=[Var(X)-Var(Y)]^2$
Hence, $\rho_{X+Y,X-Y}=Cov(X+Y,X-Y)/\sqrt{Var(X+Y)Var(X-Y)}=\pm1$.
As @whuber pointed out, when $Var(X)=Var(Y)$, $\rho_{X+Y,X-Y}$ is undefined.
@Dilip Sarwate has provided the answer in the comment to his post. I added some details.
Actually, if $Var(X)=Var(Y)$, given they are perfectly correlated, $Y=\pm X+b$, hence either $X+Y\equiv0$ or $X-Y\equiv0$. Therefore, $\rho=0$.
A: If $X$ is a linear function of $Y$ (definition of perfect correlation), then both $X-Y$ and $X+Y$ will be linear functions of $Y$, and therefore are linear functions of each other. 
So, $X-Y$ and $X+Y$ are perfectly correlated.  
