Consider the two simple linear regression models:

$Y = \alpha_0 + \alpha_1X + \epsilon(0, \sigma^2)$ and $X = \beta_0 + \beta_1Y + \epsilon(0, \sigma^2)$.

Let $(\hat{\alpha_0}, \hat{\alpha_1})$ and $(\hat{\beta_0}, \hat{\beta_1})$ be the LSE of the regression coefficients in these models from the same set of observations $\{(x_i, y_i) : 1 \leq i \leq n\}$. Do the equations $y = \hat{\alpha_0} + \hat{\alpha_1}x$ and $x = \hat{\beta_0} + \hat{\beta_1}y$ define the same straight line in the plane ?

Here are my thoughts so far:

I was able to show that $\hat{\alpha_1} = \frac{\sum_{i = 1}^n x_iy_i - n\bar{x}\bar{y}}{\sum_{i = 1}^n x_i^2 - n\bar{x}^2}$ and $\hat{\beta_1} = \frac{\sum_{i = 1}^n x_iy_i - n\bar{x}\bar{y}}{\sum_{i = 1}^n y_i^2 - n\bar{y}^2}$. From here, I was also able to show that $\hat{\alpha_1} \cdot \hat{\beta_1} \leq 1$. Is there some reason that equality must hold here ? That is, $\hat{\alpha_1} \cdot \hat{\beta_1} = 1$ ? If so, how can I proceed from there to show that the two given equations define the same straight line in the plane ? Is the only route to take to show that $\hat{\alpha_0} = \hat{\beta_0}$ and $\hat{\alpha_1} = \hat{\beta_1}$ ?

Thanks for your help. I really appreciate it. I'm relatively new to this subject.


No, they don't. Just run one, and invert the equation, to see that this is so. The reason is that if you run the $y=mx+b$ equation, you're minimizing the sums of squares of $y$ distances, whereas if you run the $x=my+b$, you're minimizing the sums of squares of $x$ distances. Now if you minimized the right-angle distances from points to the best-fit line, you would get the same equation either way.

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