Range of Reverse Regression Slope Given Forward Slope
Consider two regressions on centered data (zero-intercept):
- Regress $y$ on $x$: $y \sim bx$, with OLS slope $b = \text{Cov}(x,y) / \text{Var}(x)$
- Regress $x$ on $y$: $x \sim cy$, with OLS slope $c = \text{Cov}(x,y) / \text{Var}(y)$
Given that $b = 1$, determine the range of possible values for $c$. Include detailed steps and reasoning.
Hints
- Write out the formulas for both OLS slopes in terms of $\text{Cov}(x,y)$, $\text{Var}(x)$, and $\text{Var}(y)$. What happens when you multiply them?
- The product of the forward and reverse regression slopes equals $r^2$, the squared correlation coefficient: $bc = r^2$.
- Since $r^2 \in [0, 1]$ by Cauchy-Schwarz and $b = 1$, you get $c = r^2 \in [0, 1]$.
Worked Solution
How to Think About It: This is the classic forward-vs-reverse regression slope problem. The bridge between the two slopes is the identity $bc = r^2$, the squared correlation. The subtle part the careless version misses: the constraint $b=1$ is not just "$r^2 \le 1$"; it actually forces $\mathrm{Cov}(x,y)$ to be strictly positive (for a nondegenerate regressor), which rules out $c=0$. So the range is the half-open interval $(0,1]$, not the closed $[0,1]$.
Quick Estimate: From $bc=r^2$ and $b=1$ we get $c=r^2$. Since $0 \le r^2 \le 1$, naively $c\in[0,1]$. But $b=1$ means $\mathrm{Cov}(x,y)=\mathrm{Var}(x)$, and for a non-constant $x$ that is strictly positive, so $r^2>0$ -- the value $0$ is only a limit, never attained. The ceiling $c=1$ is reached at perfect correlation. So $c\in(0,1]$.
Approach: Derive $bc=r^2$ from the OLS formulas, then use the extra information in $b=1$ to pin the sign of the covariance.
Formal Solution:
The zero-intercept OLS slopes on centered data are
$$b = \frac{\mathrm{Cov}(x,y)}{\mathrm{Var}(x)}, \qquad c = \frac{\mathrm{Cov}(x,y)}{\mathrm{Var}(y)}.$$
Multiplying,
$$bc = \frac{\mathrm{Cov}(x,y)^2}{\mathrm{Var}(x)\,\mathrm{Var}(y)} = r^2,$$
where $r$ is the Pearson correlation. By Cauchy-Schwarz, $r^2 \in [0,1]$, with $r^2=1$ iff $x$ and $y$ are perfectly linearly related.
Now use $b=1$. From $b = \mathrm{Cov}(x,y)/\mathrm{Var}(x) = 1$ we get
$$\mathrm{Cov}(x,y) = \mathrm{Var}(x).$$
For a nondegenerate regressor ($\mathrm{Var}(x) > 0$, otherwise the slope $b$ is undefined), this makes $\mathrm{Cov}(x,y) = \mathrm{Var}(x) > 0$ -- strictly positive. Therefore
$$c = \frac{r^2}{b} = r^2 = \frac{\mathrm{Cov}(x,y)^2}{\mathrm{Var}(x)\,\mathrm{Var}(y)} > 0.$$
Boundary behaviour: - $c \to 0^{+}$ is a limit, never reached. Since $\mathrm{Cov}(x,y)=\mathrm{Var}(x)$ is fixed and positive, $c = \mathrm{Cov}(x,y)/\mathrm{Var}(y) = \mathrm{Var}(x)/\mathrm{Var}(y)$ shrinks toward $0$ only as $\mathrm{Var}(y)\to\infty$; it is strictly positive for every finite configuration. So $c=0$ is excluded. - $c = 1$ is attained at perfect correlation $r^2=1$: then $\mathrm{Var}(y)=\mathrm{Var}(x)$, both regressions recover the same line $y=x$ (up to centering), and $c=1$. - $c>1$ is impossible, since $c=r^2 \le 1$.
Practical note: The asymmetry between $b$ and $c$ trips people up -- regressing $y$ on $x$ minimizes vertical residuals while regressing $x$ on $y$ minimizes horizontal ones, so the slopes differ unless the fit is perfect. The identity $bc=r^2$ ties them together, and the constraint $b=1$ is what forbids the degenerate $c=0$ endpoint.
Answer: $c \in (0,\,1]$ -- strictly positive (because $b=1$ forces $\mathrm{Cov}(x,y)=\mathrm{Var}(x)>0$, hence $r^2>0$), reaching $1$ only at perfect correlation.
Intuition
The identity $bc = r^2$ is one of those elegant facts that connects regression to correlation in a fundamental way. It tells you that the two regression slopes are not independent -- they are linked through the strength of the linear relationship. When the correlation is weak, the two slopes diverge (one can be large while the other is small). When the correlation is perfect, they converge to the same value.
In practice this matters when people naively invert a regression. If you estimate $y = bx$ and then claim $x = y/b$, you are implicitly assuming $r^2 = 1$. The correct reverse regression slope is $c = r^2/b$, which is always closer to zero than $1/b$. This is the "regression to the mean" effect -- and it is the reason you cannot just flip a regression equation to go the other direction.