Question

If the regression line of $y$ on $x$ is $ax+by+c=0$ and the regression line of $x$ on $y$ is $a_{1}x+b_{1}y+c_1=0$, then prove that: $a{b}_1\le a_{1}b$.

Answer 1

Regression line of $y$ on $x$ is.
$ax+by+c=0$
$\Rightarrow by=-ax-c$
$\Rightarrow y=-\frac{a}{b}x-\frac{c}{b}$
$\therefore b_{yx}=-\frac{a}{b}$
Regression line of $x$ on $y$ is:
$a_{1}x+b_{1}y+c_1=0$
$\Rightarrow a_{1}x=-b_{1}y-c_1$
$\Rightarrow x=-\frac{b_1}{a_1}-y\frac{c_1}{a_1}$
$\therefore b_{xy}=-\frac{b_1}{a_1}$
We know that $r^2\le 1$
$\Rightarrow b_{yx}\cdot b_{xy}\le 1$
$\Rightarrow (-\frac{a}{b})\times (-\frac{b_1}{a_1})\le 1$
$\Rightarrow \frac{a{b}_1}{a_{1}b}\le 1$
$a{b}_1\le a_{1}b$.