Linear Regression • Topic 2 of 2

Regression Coefficients & Correlation

The slopes of the two regression lines are the regression coefficients $b_{yx}$ and $b_{xy}$.

Formulas

$$b_{yx}=r\,\frac{\sigma_y}{\sigma_x},\qquad b_{xy}=r\,\frac{\sigma_x}{\sigma_y},$$

where $r$ is the coefficient of correlation and $\sigma_x,\sigma_y$ are the standard deviations.

Link to correlation

Multiplying the two coefficients gives

$$b_{yx}\cdot b_{xy}=r^{2}\quad\Longrightarrow\quad r=\pm\sqrt{b_{yx}\,b_{xy}}.$$

  • $r$ takes the same sign as the two coefficients (which always share a sign).
  • Since $|r|\le1$, we must have $|b_{yx}\,b_{xy}|\le1$ — a useful consistency check.
  • If one coefficient exceeds $1$ in magnitude, the other must be below $1$.
1
Worked Example
If $b_{yx}=0.8$ and $b_{xy}=0.2$, find $r$.
Solution

$r=\sqrt{0.8\times0.2}=\sqrt{0.16}=0.4$ (positive, since both coefficients are positive).

2
Worked Example
If $b_{yx}=-0.6$ and $b_{xy}=-0.6$, find $r$.
Solution

$r=-\sqrt{(-0.6)(-0.6)}=-\sqrt{0.36}=-0.6$ (negative sign matches the coefficients).

3
Worked Example
Can $b_{yx}=2$ and $b_{xy}=3$ occur together?
Solution

$b_{yx}b_{xy}=6>1$ would give $r^2=6$, impossible since $|r|\le1$. So no — these cannot both be regression coefficients.

4
Worked Example
Given $r=0.5$ and $\sigma_y=4,\ \sigma_x=2$, find $b_{yx}$.
Solution

$b_{yx}=r\dfrac{\sigma_y}{\sigma_x}=0.5\times\dfrac{4}{2}=0.5\times2=1.$

Key Points

  • $b_{yx}=r\dfrac{\sigma_y}{\sigma_x}$, $\ b_{xy}=r\dfrac{\sigma_x}{\sigma_y}$.
  • $b_{yx}\,b_{xy}=r^{2}$, so $r=\pm\sqrt{b_{yx}b_{xy}}$.
  • $r$ shares the (common) sign of the two coefficients.
  • $|r|\le1 \Rightarrow |b_{yx}b_{xy}|\le1$ (consistency check).
Tap an option to check your answer0 / 4
Q1.$b_{yx}\cdot b_{xy}$ equals:
Explanation: Product of regression coefficients is $r^2$.
Q2.If $b_{yx}=0.8,b_{xy}=0.2$, then $r=$
Explanation: $\sqrt{0.16}=0.4$.
Q3.The sign of $r$ is:
Explanation: $r$ shares their common sign.
Q4.Which pair is impossible as regression coefficients?
Explanation: $2\times3=6>1$ gives $r^2>1$, impossible.