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I am interested in estimating a regression that looks like this:

$(x_{1,i} - y_{i} )_{i} = x’_{i}*\beta + \epsilon_{i}$ (1)

However, I am not sure if doing this—in this form—is appropriate. The reason for this is that $y_{i}$ and $x_{1,i}$ are correlated. Put differently, $x_{1,i}$ is an significant explanatory variable of $y_{i}$, which is commonly estimated using the following linear (or sometimes using a log-linear) regression:

$y_{i} = \beta_{0} + \beta_{1}*x_{1,i} + \beta_{2}*x_{2,i} + \beta_{3}*x_{3,i} + \epsilon_{i}$ (2)

My thinking (and please correct me if I am wrong) is that in order to interpret the regression results in (1) correctly I would have to rewrite those in (2) as follows:

$(y_{i} - x_{1,i}) = \beta_{0} + (\beta_{1} - 1)*x_{1,i} + \beta_{2}*x_{2,i} + \beta_{3}*x_{3,i} + \epsilon_{i}$

where $u_{i} = ((\beta_{1} - 1)*x_{1,i} + \epsilon_{i})$

IFF

$(x_{1,i} - y_{i}) = (-1)*(\beta_{0} + \beta_{2}*x_{2,i} + \beta_{3}*x_{3,i} + u_{i})$

As such, in running (1) I need to worry about: (i) possible endogeneity due to the possibility that the $(\beta_{1} - 1)*x_{1,i}$ term in $u_{i}$ is correlated with the other regressors; and (ii) I need to multiply the coefficients from (1) by $-1$ in order to get the appropriate sign?

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1 Answer 1

up vote 5 down vote accepted

No problem here. If we accept that the specification

$$y_{i} = \beta_{0} + \beta_{1}x_{1,i} + \beta_{2}x_{2,i} + \beta_{3}x_{3,i} + \epsilon_{i} \tag{1}$$

is correct, then the transformation by subtracting one regressor from the dependent variable,

$$(y_{i} - x_{1,i}) = \beta_{0} + (\beta_{1} - 1)x_{1,i} + \beta_{2}x_{2,i} + \beta_{3}x_{3,i} + \epsilon_{i} \tag{2}$$

is equivalent to the specification

$$z_i = \beta_{0} + \gamma_1 x_{1,i} + \beta_{2}x_{2,i} + \beta_{3}x_{3,i} + \epsilon_{i} \tag{3}$$

with $z_i \equiv y_{i} - x_{1,i}$ and $\gamma_1 \equiv \beta_{1} - 1$ which simply means that if we estimate $(3)$ the obtained estimate $\hat \gamma_1$ will be estimating $\beta_1-1$ and not $\beta$, while all others coefficient estimates will be numerically the same with those that would be obtained from specification $(1)$.

No endogeneity issue arises, and no other transformation becomes necessary.

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Thanks Alecos--very helpful. But what if I want $x_i - y_i$ rather than $y_i - x_i$ on the LHS? If I run such a regression in e.g. Stat--will I need to premultiply the estimates by -1 in order to correctly interpret them? –  Seb Aug 18 at 18:44
    
Yes. What you do on the LHS, you must do on the RHS. And you obtain $x-y$ by multiplying the LHS by $-1$. –  Alecos Papadopoulos Aug 18 at 18:47

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