I'm trying to order sequences by their regression coefficient $y=\beta x$. I thought we should have a total order on $\mathbb{R}^n$ given by by $x\lesssim y$ if $$\beta_{xy}=\frac{x \cdot y}{x \cdot x}\geq 1$$
However, I'm can't prove it satisfies the transitivity, although geometrically it seems to hold up.
$$x \lesssim y,\quad y \lesssim z \implies x \lesssim z$$ $$ x \cdot x \leq x \cdot y,\quad y \cdot y \leq z \cdot y \implies x\cdot x \leq z \cdot x$$
What am I missing to prove this?
Here's a simple counter example in $\mathbb{R}^2$: take $x=(1,0)$, $y=(1,1)$ and $z=(0,2)$, then $\beta_{xy}=1$, $\beta_{yz}=1$ but $\beta_{xz}=0$.
Note that with help from the Cauchy-Schwarz inequality you can show that:
$$\beta_{xy}^2\beta_{yz}^2\beta_{zx}^2\leq 1$$
and hence that $\beta_{xy} \geq 1$ and $\beta_{yz} \geq 1$ imply $\beta_{zx}^2 \leq 1$. But it does not automatically follow that $\beta_{xz}\geq 1$.