1
$\begingroup$

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.

$\endgroup$

1 Answer 1

1
$\begingroup$

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.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .