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We have the standard linear multivariate regression model $y=\beta_0+\beta_1x_1+\beta_2x_2+u$ under the Gauss-Markov assumptions. Suppose we estimate $\gamma_0, \gamma_1, \gamma_2$ from $x_2=\gamma_0+\gamma_1x_1+\gamma_2y+v$ and get $\hat{\gamma_0}, \hat{\gamma_1}, \hat{\gamma_2}$. Is $\frac{1}{\hat{\gamma_2}}$ an unbiased estimate of $\beta_2$? I don't know how to answer the question.

First of all it seems that to run the regression of $x_2$ on $y$ and $x_1$ we need $\beta_2$ be nonzero. And is the expression $x_2=\frac{-\beta_0}{\beta_2}+\frac{-\beta_1}{\beta_2}x_1+\frac{1}{\beta_2}y+\frac{-1}{\beta_2}u$ somehow related to the regression model $x_2=\gamma_0+\gamma_1x_1+\gamma_2y+v$?

Another question is whether the Gauss-Markov assumptions hold in the case of this new regression, seems like some of them can fail, for instance $E(v|x_1,y)=0$ may not be true(is it important here?).

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  • $\begingroup$ Not a stupid question. I'm not sure though whether you have either: a system of two equations (because y is somehow caused by x1 and x2, and x2 is also caused by x1 and y); or just one equation,about which you don't mind the causality, that you have switched around and want to estimate the second for some reason (what?). If the former, you have a problem, as you won't be able to fit both equations simultaneously (as they both have the same sets of variables - not possible to separate out the two way causality). If the latter, it's not clear why you don't just fit your first equation. $\endgroup$
    – Peter Ellis
    Jan 25, 2012 at 9:09
  • $\begingroup$ I understand the task as follows: we are given a linear regression model ($y$ on $x_1$, $x_2$) under G-M assumptions. We can estimate it's parameters through ols. After that we consider the regression of $x_2$ on $y$, $x_1$. It is asked if the reciprocal of $\hat{\beta_2}$ (which is an estimator of $\beta_2$) is an unbiased estimate of the parameter $\gamma_2$ in the new regression model. By "why you don't just fit your first equation" you mean to plug $x_2$ from second equation into the first one? $\endgroup$
    – marcin63
    Jan 25, 2012 at 17:43
  • $\begingroup$ Is this homework? It's hard to see why this would be a necessary thing to do in an applied situation. $\endgroup$
    – Peter Ellis
    Jan 25, 2012 at 18:02
  • $\begingroup$ yes, it's my HA $\endgroup$
    – marcin63
    Jan 25, 2012 at 18:04
  • $\begingroup$ By "fit your first equation" I just meant that if you have values of y, x1 and x2 why do you need to muck around with reciprocals of anything, why not just use OLS to fit whichever of the two equations you want estimates for. But the exercise looks like one to see the relationship between different estimators, not just a practical one of trying to fit a model. $\endgroup$
    – Peter Ellis
    Jan 25, 2012 at 18:36

1 Answer 1

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Rewrite the model in the following way:

$x_{2} =\gamma _{0}+\gamma _{1}x_{1}+\gamma _{2}y+v$ $=\left( \frac{\beta _{0}}{-\beta _{2}}\right) +\left( \frac{1}{\beta _{2}}\right) y+\left( \frac{\beta _{1}}{-\beta _{2}}\right) x_{1}+\left( \frac{u}{-\beta _{2}}\right) $ $ =\left( \frac{\beta _{0}}{-\beta _{2}}\right) +\left( \frac{1}{\beta _{2}}\right) \left( \beta _{0}+\beta _{1}x_{1}+\beta _{2}x_{2}+u\right)+\left( \frac{\beta _{1}}{-\beta _{2}}\right) x_{1}+\left( \frac{u}{-\beta_{2}}\right).$

You can see that y regressor is correlated with the error term since they both include u. This causes $\gamma_{2}$ to be downward-biased, which means your estimate of $\beta_{2}$ will be upward-biased. This is because the E[v|y]!=0.

Another general approach when you get stuck with the math is to simulate the problem. This will give you intuition for the proof. Here's some Stata code:

#delimit;
clear;
matrix C = (1,0,0\0,1,0\0,0,1);
drawnorm x1 x2 u, n(100000000) corr(C);
gen y=2+x1+5*x2+u;
reg y x1 x2;
reg x2 y x1;
di 1/_b[y];
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  • $\begingroup$ Sorry, I can't seem to get the LaTeX to work. $\endgroup$
    – dimitriy
    Jan 25, 2012 at 20:07
  • $\begingroup$ Thanks. So, to make things clear: in the regression of $x_2$ on $x_1,y$ the regressor($y$) is correlated with the regressand $x_2$, therefore one of the G-M assumptions fails and we can't expect the estimate of $\gamma_2$ to be unbiased? Another point is about the case $\beta_2=0$(so, $x_2$ has no ceteris paribus effect on $y$). If we run the regression of $y$ on $x_1,x_2$ we will get $\hat{\beta_2}=0$ (or will it be only $E(\hat{\beta_2})=0$?), so in this case estimate $\hat{\gamma_2}$ is not defined. $\endgroup$
    – marcin63
    Jan 25, 2012 at 20:32
  • $\begingroup$ I would say you don't want regressors ($y$ in this case) to be correlated with the error term $v$ since both contain the $u$ term: $v=\frac{u}{-\beta_{2}}$. You certainly want your regressand and regressor to be correlated. $\endgroup$
    – dimitriy
    Jan 25, 2012 at 20:38
  • $\begingroup$ Seems like $\beta_2$ will almost never be equal to zero in the case if true value $\beta_2$ equals zero, but we will always have $E(\hat{\beta_2})=0$. $\endgroup$
    – marcin63
    Jan 25, 2012 at 20:40
  • $\begingroup$ If $\beta_{2}$ is zero, then $y$ and $x_{2}$ have no relationship, but your estimate of $\beta_{2}$ using $\frac{1}{\hat\gamma_{2}}$ will be enormous. The bias will go to infinity... and beyond. Well, at least in theory. $\endgroup$
    – dimitriy
    Jan 25, 2012 at 20:48

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