Q.If $b_{yx} > 1$ then $b_{xy}$ is ______.

Since the product of the regression coefficients equals the square of the correlation coefficient, $b_{yx}\, b_{xy} = r^2 \le 1$; hence $b_{yx} > 1$ forces $b_{xy} < 1$.

A key property of the regression coefficients is

$b_{yx}\times b_{xy} = r^2$

where $r$ is the correlation coefficient and $-1 \le r \le 1$, so $r^2 \le 1$. Therefore

$b_{yx}\times b_{xy} \le 1$

If $b_{yx} > 1$, then to keep the product at most $1$ we must have

$b_{xy} \le \frac{1}{b_{yx}} < 1$

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Since $|r|=\sqrt{b_{yx}\cdot b_{xy}}$ (GM) and AM $\ge$ GM, we must have $\dfrac{b_{yx}+b_{xy}}{2}\ge |r|$. Testing: $\dfrac{1.3}{2}=0.65$, but $|r|=0.75$, and $0.65<0.75$ violates the property, so the data is inconsistent and the statement is True.

Property used. The correlation coefficient is the geometric mean of the two regression coefficients:

$r = \pm\sqrt{b_{yx}\cdot b_{xy}}, \qquad \text{so}\qquad |r| = \sqrt{b_{yx}\cdot b_{xy}}.$

By the AM–GM inequality, the arithmetic mean of the two coefficients is at least their geometric mean:

$\frac{b_{yx} + b_{xy}}{2} \;\ge\; \sqrt{b_{yx}\cdot b_{xy}} = |r|.$

Testing the given data. Here $b_{yx}+b_{xy}=1.3$ and $r=0.75$, so …

A standard property of regression coefficients is that $b_{xy}$ and $b_{yx}$ are independent of a change of origin but not of a change of scale — option (b).

Suppose we transform the variables by $u=\dfrac{x-a}{h}$ and $v=\dfrac{y-b}{k}$, where $a,b$ shift the origin and $h,k$ scale the data. Then the regression coefficients transform as:

$b_{yx}=\frac{k}{h}\,b_{vu}, \qquad b_{xy}=\frac{h}{k}\,b_{uv}$ …

$b_{YX}$ is the slope of the line of regression of $Y$ on $X$ ($Y$ dependent, $X$ independent), given by $b_{YX} = r\dfrac{\sigma_Y}{\sigma_X}$. Hence option (A).

The subscript convention is: in $b_{YX}$ the first letter ($Y$) is the variable being estimated (dependent) and the second ($X$) is the variable used to estimate it (independent). Thus $b_{YX}$ is the regression coefficient of $Y$ on $X$, appearing in the line

$Y - \bar{Y} = b_{YX}(X - \bar{X}), \qquad b_{YX} = r\,\dfrac{\sigma_Y}{\sigma_X} = \dfrac{\mathrm{Cov}(X,Y)}{\sigma_X^{2}}$

…

Since $r = \pm\sqrt{b_{XY}\cdot b_{YX}}$, the product of the regression coefficients equals $r^2$.

The two regression coefficients are

$b_{YX} = r\dfrac{\sigma_y}{\sigma_x}$ and $b_{XY} = r\dfrac{\sigma_x}{\sigma_y}$.

Multiplying them:

$b_{XY}\cdot b_{YX} = \left(r\dfrac{\sigma_x}{\sigma_y}\right)\left(r\dfrac{\sigma_y}{\sigma_x}\right) = r^2$.

…

$b_{xy}=\dfrac{n\sum xy-\sum x\sum y}{n\sum y^2-(\sum y)^2}=\dfrac{5(76)-20(20)}{5(90)-20^2}=\dfrac{-20}{50}=-0.4$.

The regression coefficient of $x$ on $y$ is

$b_{xy} = \dfrac{n\sum xy - \sum x \sum y}{n\sum y^2 - (\sum y)^2}$.

With $n = 5$, $\sum x = 20$, $\sum y = 20$, $\sum y^2 = 90$ and $\sum xy = 76$:

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