# Relationship between correlation coefficient and the angle between two regression lines

Geometric interpretation Regression lines for $$y = g_X(x)$$ [red] and $$x = g_Y(y)$$ [blue] Regression lines for $$y = g_X(x)$$ [red] and $$x = g_Y(y)$$ [blue]

For uncentered data, there is a relation between the correlation coefficient and the angle $$\varphi$$ between the two regression lines, $$y = g_X(x)$$ and $$x = gY(y)$$, obtained by regressing $$y$$ on $$x$$ and $$x$$ on $$y$$ respectively. (Here, $$\varphi$$ is measured counterclockwise within the first quadrant formed around the lines' intersection point if $$r > 0$$, or counterclockwise from the fourth to the second quadrant if $$r < 0$$.) One can show[18] that if the standard deviations are equal, then $$r = \sec \varphi$$ − tan $$\varphi$$, where sec and tan are trigonometric functions.

How can we prove this relationship mentioned above: $$r = \sec \varphi − \tan \varphi$$?

Working based on the listed reference ([18]).

The equations of the two regression lines are

$$x= a_{xy} + b_{xy} y \\ y= a_{yx} + b_{yx} x$$

If we graph these, then they will have gradients of $$b_{yx}$$ and $$\frac{1}{b_{xy}}$$, respectively.

The relationship between these regression coefficients, the standard deviations $${s_x}, {s_y}$$ on $$x$$ and $$y$$, and the correlation $$r$$ is:

$$b_{xy} = r\frac{s_x}{s_y},\quad \text{and} \quad b_{yx} = r\frac{s_y}{s_x}$$

So, in the case that $$s_x=s_y$$, the two regression lines have gradients of $$r$$ and $$1/r$$.

In the following, I will assume that $$0. In the case that $$r=1$$, the two lines are the same line, and so the angle is zero; and in the case that $$r=0$$, the two lines are orthogonal. You can show that each of these cases satisfy the identity that you care about with substitution. For the case that $$r<0$$, the same idea of the proof works, you just need to flip the entire picture upside down in your mind.

Lemma 1: the angle between two lines with gradients $$m_1$$ and $$m_2$$ is given by the equation

$$\tan (\phi) = \frac{m_1 - m_2}{1+m_1m_2}$$

Proof: The line with gradient $$m_1$$ makes an angle of $$\theta_1 = \tan^{-1}(m_1)$$ with the $$x$$-axis. The line with gradient $$m_2$$ makes an angle of $$\theta_2 = \tan^{-1}(m_2)$$ with the $$x$$-axis. So the angle between the lines is:

$$\phi = \theta_1 - \theta_2 = \tan^{-1}(m_1) - \tan^{-1}(m_2)$$

Then we can use the compound angle formula for tan:

$$\tan(\phi) = \tan\left( \tan^{-1}(m_1) - \tan^{-1}(m_2) \right) \\ =\frac{m_1-m_2}{1+m_1m_2} \qquad \square$$

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So, applying this to our equation, we learn that

$$\tan(\phi) = \frac{1}{2}\left(\frac{1}{r} - r\right)$$

We can show that this equation is satisfied by $$r=\sec(\phi)-\tan(\phi)$$ by substituting this equation into the RHS, and showing that it reduces to the LHS.

Let $$r=\frac{1-\sin(\phi)}{\cos{\phi}}$$. Then:

$$RHS = \frac{1}{2}\left(\frac{1}{r} - r\right) = \frac{1}{2}\left(\frac{\cos{\phi}}{1-\sin(\phi)} - \frac{1-\sin(\phi)}{\cos{\phi}}\right) \\ = \frac{1}{2}\left(\frac{ \cos (\phi)^2 - (1-\sin(\phi))^2}{\cos(\phi)(1-\sin(\phi))}\right) \\ = \frac{1}{2}\left(\frac{ 1 - \sin(\phi)^2 - \left(1-2\sin(\phi)+\sin(\phi)^2\right)}{\cos(\phi)(1-\sin(\phi))}\right) \\ = \frac{1}{2}\left(\frac{ 2\sin(\phi) - 2\sin(\phi)^2}{\cos(\phi)(1-\sin(\phi))}\right) \\ = \frac{\sin(\phi)}{\cos(\phi)} = LHS \quad \square$$