Find the correlation coefficient In studying the relation between the two variables $x$ and $y$ , if the equation of the regression line of $y$ on $x$ was 
$$y=0.421x+0.67$$ and the equation of the regression line of $x$ on $y$ was 
$$x=1.58y+3.9$$ \Find\
\ The linear correlation coefficient between $x$ and $y$ 
My solution is 
$$r= \pm\sqrt{0.421\times 1.58}= \pm0.8155$$
Does my solution correct or i would not take the negative value into account ? 
 A: You are right, but you can find the sign too.
The correlation coefficient of $x$ and $y$ is $\dfrac{\operatorname{cov}(x,y)}{\sqrt{\operatorname{var}(x)\operatorname{var}(y)}}$.
The slope in the regression $y=ax+b$ is given by $a=\dfrac{\operatorname{cov}(x,y)}{\operatorname{var}(x)}$.
Likewise, the slope in the regression $x=a'y+b'$ is given by $a'=\dfrac{\operatorname{cov}(x,y)}{\operatorname{var}(y)}$.
Hence the correlation coefficient is $\pm\sqrt{aa'}$. But it's also of the same sign as $a$ (or $a'$), hence positive here.
A: The two least-squares line are:
$$y=0.421x+0.67 \\ x=1.58y+3.9$$
You should have
$$
\frac{y-\nu}{\tau} = \rho\left( \frac{x-\mu} \sigma  \right) \\
\frac{x-\mu}\sigma = \rho\left( \frac{y-\nu} \tau \right)
$$
where


*

*$\mu$ is the average $x$-value,

*$\nu$ is the average $y$-value,

*$\sigma$ is the standard deviation of the $x$-values,

*$\tau$ is the standard deviation of the $y$-values,

*$\rho$ is the correlation.


Thus
\begin{align}
& \frac{\rho\tau} \sigma = 0.421, \\[10pt]
& \frac{\rho\sigma} \tau = 1.58.
\end{align}
Multiplying left sides and right sides, you get $\rho^2 = 0.421\times 1.58.$
But notice also that $\rho$ must be positive since the slopes are positive.
