# Confusion with linear regression coefficient when variables are reversed

Given $$Y$$ and $$X$$ in a typical linear regression model, where

$$Y = \beta_1 X + \epsilon_1$$

We know that $$\hat{\beta_1} = (X^TX)^{-1}X^TY$$.

Assuming that the sample mean of $$X$$ and $$Y$$ is zero, and denoting the sample s.d. as $$\sigma_x$$ and $$\sigma_y$$, and the sample correlation as $$\rho_{xy}$$, we have:

$$\hat{\beta_1} = \frac{\sigma_y}{\sigma_x} \rho_{xy}$$

If $$\sigma_y = \sigma_x$$, this simplifies to

$$\hat{\beta_1} = \rho_{xy}$$

Now, if we carry out the same analysis on the following regression problem:

$$X = \beta_2 Y + \epsilon_2$$

We will also get $$\hat{\beta_2} = \rho_{xy} = \hat{\beta_1}$$

My confusion is the following: Why is $$\hat{\beta_2} = \hat{\beta_1}$$ and not $$\hat{\beta_2} = \hat{\beta_1^{-1}}$$? I'm looking at this from a geometric viewpoint (i.e. $$X = \frac{1}{m}Y$$ if $$Y = mX$$) and I can't seem to understand why this is (not) the case.

Note that the OLS of $$y=\beta x + \epsilon$$ is $$\hat{\beta} = \frac{ \sum x_i y_i }{ \sum x_i^2 } = \frac{<\mathrm{x},\mathrm{y} >}{ \| \mathrm{x} \|^2},$$ now, take the variance to be $$1$$ (it can be any other constant), thus $$\hat{\beta} = <\mathrm{x},\mathrm{y} > = \| \mathrm{x} \| \| \mathrm{x} \| \cos \theta = \cos \theta,$$ if you reverse the order $$x = \beta y+\epsilon$$ the OLS of $$\beta$$ remains $$\cos \theta$$, i.e., the cosine between the two random vectors. Therefore, you have the same slope.