Is it always true that $\operatorname{Corr}(ax,by) = \operatorname{Corr}(x,y)$ for any constants $a$ and $b$?

Question

We know that $$\operatorname{Corr}(x,y) = \frac{\operatorname{Cov}(x,y)}{\sqrt{\operatorname{Var}(x) \operatorname{Var}(y))}}.$$

So $$\operatorname{Corr}(ax,by) = \frac{\operatorname{Cov}(ax,by)}{\sqrt{\operatorname{Var}(ax) \operatorname{Var}(by))}}$$ $$\Rightarrow \operatorname{Corr}(ax,by) = \frac{a b\operatorname{Cov}(x,y)}{\sqrt{(ab)^{2}}\sqrt{\operatorname{Var}(x) \operatorname{Var}(y))}}$$ $$\Rightarrow \operatorname{Corr}(ax,by) = \frac{a b\operatorname{Cov}(x,y)}{|ab|\sqrt{\operatorname{Var}(x) \operatorname{Var}(y))}}$$ $$\Rightarrow \operatorname{Corr}(ax,by) = \frac{ab}{|ab|} \operatorname{Corr}(x,y)$$

If $ab>0$, $\operatorname{Corr}(ax,by) = \operatorname{Corr}(x,y)$ but if $ab<0$ then $\operatorname{Corr}(ax,by) = -\operatorname{Corr}(x,y)$

So it is not true for all values of $a$ and $b$. Is this correct?

Answer 1

That's right, it comes down to the sign of $ab$. As a very simple example consider $y=x,\,a=1,\,b=-1$ viz. $\operatorname{corr}(x,\,x)=1,\,\operatorname{corr}(x,\,-x)=-1$.