Question

If the acute angle between the two regression lines is $\theta$, then

• A. $\sin \theta \ge 1–r^2$
• B. $\tan \theta \ge 1–r^2$
• C. $Sin\theta \le 1–r^2$
• D. $\tan \theta \le 1–r^2$

Answer 1

The correct option is C

$Sin\theta \le 1–r^2$

Explanation for Correct Option

Step 1: Finding the angle of regression

If $\theta$ is the angle between the two lines of regression

$$\begin{gathered} \Rightarrow \tan \theta =\left[\frac{r\times {\displaystyle \frac{{\sigma }_y}{{\sigma }_x}}-r\times \frac{{\sigma }_x}{{\sigma }_y}}{(1+r\times r\times \frac{{\sigma }_y}{{\sigma }_x}\times \frac{{\sigma }_x}{{\sigma }_y})}\right] \\ \Rightarrow \tan \theta =\left[\frac{\left(r^2-1\right)\times {\displaystyle {\sigma }_x{\sigma }_y}}{\left(r\times {{\sigma }_x}^2{{\sigma }_y}^2\right)}\right] \\ \Rightarrow \tan \theta =\left[\frac{\left(1-r^2\right)\times {\displaystyle {\sigma }_x{\sigma }_y}}{\left(r\times {{\sigma }_x}^2{{\sigma }_y}^2\right)}\right] \\ \Rightarrow \theta ={\tan }^{-1}\left\{\frac{\left(1-r^2\right)\times {\displaystyle {\sigma }_x{\sigma }_y}}{\left(r\times {{\sigma }_x}^2{{\sigma }_y}^2\right)}\right\} \end{gathered}$$

The angle between two lines of regression is $\tan \theta =\left[\frac{r\times {\displaystyle \frac{{\sigma }_y}{{\sigma }_x}}-r\times \frac{{\sigma }_x}{{\sigma }_y}}{(1+r\times r\times \frac{{\sigma }_y}{{\sigma }_x}\times \frac{{\sigma }_x}{{\sigma }_y})}\right]$

Step 2:Since AM of two numbers is greater than GM of the same quantities

$$\begin{gathered} \frac{\left({{\sigma }_x}^2+{{\sigma }_y}^2\right)}{2}\ge {\sigma }_x{\sigma }_y \\ \frac{{\sigma }_x{\sigma }_y}{\left({{\sigma }_x}^2+{{\sigma }_y}^2\right)}\le \frac{1}{2} \\ \frac{{\sigma }_x{\sigma }_y}{\left({{\sigma }_x}^2+{{\sigma }_y}^2\right)}\le \frac{1}{2} \\ {\tan }^2\theta \le {\left\{\frac{[1–r^2]}{2r}\right\}}^2 \end{gathered}$$

Since, ${\sin }^2\theta =\frac{{\tan }^2\theta }{(1+{\tan }^2\theta )}...\left(\mathrm{i}\right)$

Step 3: Put the value of ${\tan }^2\theta$ in the equation (i)

$$\begin{gathered} {\sin }^2\theta =\frac{{\left\{\frac{[1–r^2]}{2r}\right\}}^2}{\left(1+{\left\{\frac{[1–r^2]}{2r}\right\}}^2\right)} \\ {\sin }^2\theta ={\left\{\frac{[1–r^2]}{[1+r^2]}\right\}}^2 \\ \mathrm{Since}1–{\mathrm{r}}^2<1+{\mathrm{r}}^2\mathrm{and}\mathrm{sin\theta }\le 1–{\mathrm{r}}^2 \end{gathered}$$

Hence Option (3) is Correct