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Question Number 19 by user1 last updated on 25/Jan/15

Let θ be the angle between the regression line of y on x and the regression line of x on y. Then prove that tan θ={(((1−r^2 ))/r)×((σ_x ×σ_y )/((σ_x ^2 +σ_y ^2 )))}

$$\mathrm{Let}\:\theta\:\mathrm{be}\:\mathrm{the}\:\mathrm{angle}\:\mathrm{between}\:\mathrm{the}\:\mathrm{regression} \\ $$$$\mathrm{line}\:\mathrm{of}\:{y}\:\mathrm{on}\:{x}\:\mathrm{and}\:\mathrm{the}\:\mathrm{regression}\:\mathrm{line}\:\mathrm{of} \\ $$$${x}\:\mathrm{on}\:{y}.\:\mathrm{Then}\:\mathrm{prove}\:\mathrm{that}\: \\ $$$$\mathrm{tan}\:\theta=\left\{\frac{\left(\mathrm{1}−{r}^{\mathrm{2}} \right)}{{r}}×\frac{\sigma_{{x}} ×\sigma_{{y}} }{\left(\sigma_{{x}} ^{\mathrm{2}} +\sigma_{{y}} ^{\mathrm{2}} \right)}\right\} \\ $$

Answered by user1 last updated on 30/Oct/14

The equation of the line of regression of x on y is (x−x^− )= r×(σ_x /σ_y )(y−y^− ) ...(i) And the equation of the line of regression of y on x is (y−y^− )= r×(σ_y /σ_x )(x−x^− ) ...(ii) Let m_1 and m_2 be the slopes of (i) and (ii) res. Then, m_1 =(σ_y /(r∙σ_x )) and m_2 =((r∙σ_y )/σ_x ) ∴ tan θ = (((m_1 −m_2 ))/((1+m_1 m_2 ))) = (((σ_y /(r∙σ_x ))−((r∙σ_y )/σ_x ))/(1+(((σ_y )^2 )/((σ_x )^2 ))))={(((1−r^2 ))/r)×((σ_x ∙σ_y )/((σ_x ^2 +σ_y ^2 )))}

$$\mathrm{The}\:\mathrm{equation}\:\mathrm{of}\:\mathrm{the}\:\mathrm{line}\:\mathrm{of}\:\mathrm{regression}\: \\ $$$$\mathrm{of}\:{x}\:\mathrm{on}\:{y}\:\mathrm{is} \\ $$$$\:\:\:\left({x}−\overset{−} {{x}}\right)=\:{r}×\frac{\sigma_{{x}} }{\sigma_{{y}} }\left({y}−\overset{−} {{y}}\right)\:\:\:\:\:\:\:\:\:\:...\left(\mathrm{i}\right) \\ $$$$\mathrm{And}\:\mathrm{the}\:\mathrm{equation}\:\mathrm{of}\:\mathrm{the}\:\mathrm{line}\:\mathrm{of}\:\mathrm{regression}\: \\ $$$$\mathrm{of}\:{y}\:\mathrm{on}\:{x}\:\mathrm{is} \\ $$$$\left({y}−\overset{−} {{y}}\right)=\:{r}×\frac{\sigma_{{y}} }{\sigma_{{x}} }\left({x}−\overset{−} {{x}}\right)\:\:\:\:\:\:\:\:\:\:\:\:\:...\left(\mathrm{ii}\right) \\ $$$$\mathrm{Let}\:{m}_{\mathrm{1}} \:\mathrm{and}\:{m}_{\mathrm{2}} \mathrm{be}\:\mathrm{the}\:\mathrm{slopes}\:\mathrm{of}\:\left(\mathrm{i}\right)\:\mathrm{and}\:\left(\mathrm{ii}\right)\:\mathrm{res}. \\ $$$$\mathrm{Then},\:{m}_{\mathrm{1}} =\frac{\sigma_{{y}} }{{r}\centerdot\sigma_{{x}} }\:\mathrm{and}\:{m}_{\mathrm{2}} =\frac{{r}\centerdot\sigma_{{y}} }{\sigma_{{x}} } \\ $$$$\therefore\:\:\:\mathrm{tan}\:\theta\:=\:\frac{\left({m}_{\mathrm{1}} −{m}_{\mathrm{2}} \right)}{\left(\mathrm{1}+{m}_{\mathrm{1}} {m}_{\mathrm{2}} \right)} \\ $$$$\:\:\:\:=\:\frac{\frac{\sigma_{{y}} }{{r}\centerdot\sigma_{{x}} }−\frac{{r}\centerdot\sigma_{{y}} }{\sigma_{{x}} }}{\mathrm{1}+\frac{\left(\sigma_{{y}} \right)^{\mathrm{2}} }{\left(\sigma_{{x}} \right)^{\mathrm{2}} }}=\left\{\frac{\left(\mathrm{1}−{r}^{\mathrm{2}} \right)}{{r}}×\frac{\sigma_{{x}} \centerdot\sigma_{{y}} }{\left(\sigma_{{x}} ^{\mathrm{2}} +\sigma_{{y}} ^{\mathrm{2}} \right)}\right\} \\ $$$$ \\ $$

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