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Suppose we have two variables $X_2$ and $X_3$ and two regressions are run:

$$X_2=a+bX_3+v_2$$

and

$$X_3=c+dX_2+v_3$$

$v_2$ and $v_3$ are the observed residuals.

Can we say if we regress $v_3$ on $v_2$, the coefficient of $v_2$ will be $d$? Is there a way this can be shown?

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  • $\begingroup$ What are $v2$ and $v3$ in your models? They look like error terms, which you cannot observe. $\endgroup$
    – Ale
    Commented Sep 26, 2020 at 8:24
  • $\begingroup$ They are the observed residuals. Sorry forgot to mention. I've added it now. $\endgroup$
    – user584534
    Commented Sep 26, 2020 at 9:08
  • $\begingroup$ Have you tried some analytical solution? $\endgroup$
    – Ale
    Commented Sep 26, 2020 at 14:56
  • $\begingroup$ Please check if my edit solve your problem. $\endgroup$
    – Ale
    Commented Sep 26, 2020 at 22:18
  • $\begingroup$ Yes it does! Thanks! $\endgroup$
    – user584534
    Commented Sep 27, 2020 at 7:27

2 Answers 2

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As a hint, consider this simulation in R:

rm(list=ls())
set.seed(42)

n=1000
x3= rnorm(n)
x2 = 1 + 2 * x3 + rnorm(n)

lm(x2 ~ x3)
Coefficients:
(Intercept)    x3  
  0.9949       2.0098

# let's store the residual v2: 
v2 = lm(x2~x3)$res

# now let's consider the second model:
lm(x3~x2)
Coefficients:
(Intercept)   x2  
-0.4044       0.4014

# and store the residual v3:
v3 = lm(x3~x2)$res

# let's regress v3 on v2:
lm(v3~v2)
Coefficients:
(Intercept)    v2  
 -2.255e-17   -4.014e-01

They're not equal but they look related.

Edit:

Plug the equation of $X_2$ in the second model and isolate $v_3$. You will see that the parameter on $v_2$ is $-d$.

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From the first regression, we have $$X_3 = -\frac{a}{b} + \frac{1}{b} X_2 - \frac{v_2}{b}$$ Comparing this with $$ X_3 = c + dX_2 + v_3$$

and assuming the two regressions are performed similarly, we have $$c = -\frac{a}{b}$$ $$d = \frac{1}{b} $$ and $$v_3 = \frac{-1}{b}v2$$

Therefore we have $$v3 = -d v2$$

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