# Two-stage linear regression

If I do regression in two stages:

Stage 1: $y\sim x_1 + 1$

Stage 2: resid_1st_stage $\sim x_2 + 1$

Will the resid_2nd_stage be orthogonal to $x_1$?

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## 1 Answer

Suppose that the two models involved are:

$\newcommand{\Cov}{\mathrm{Cov}}$ $y=\alpha_1 + \beta_1 x_1 + \epsilon_1$

and

$\hat\epsilon_1=\alpha_2 + \beta_2 x_2 + \epsilon_2$

The process goes as follows:

1: estimates of $\hat \beta_1$ and $\hat\alpha_1$ (assuming OLS validity)

2: calculate the residuals $\hat\epsilon_1=y-\hat\alpha_1-\hat\beta_1x_1$

3: repeat the above with the second model, thus obtaining $\hat\epsilon_2=\hat\epsilon_1-\hat\alpha_2-\hat\beta_2x_2$

4: check that $\Cov(\hat\epsilon_2,x_1)=0$:

$\Cov(\hat\epsilon_2,x_1)=\Cov(\hat\epsilon_1-\hat\alpha_2-\hat\beta_2x_2,x_1)= \Cov(y-\hat\alpha_1-\hat\beta_1x_1 -\hat\alpha_2-\hat\beta_2x_2,x_1)= \Cov(y-\hat\beta_1x_1 -\hat\beta_2x_2,x_1)$

Which is clearly not null, given the presence of $x_1$ itself in the result, by which

$\Cov(\hat\beta_1x_1,x_1)=\hat\beta_1\sigma_{x_1}^2$

not to mention the interaction between $y$ and $x_1$ or $x_1$ and $x_2$.

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