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Say I want to regress a variable $y_t$ on $X_t$ using ordinary least squares, with the inclusion of a constant, when $X_t$ exhibits a deterministic trend.

Let's say I detrend $X_t$ by running a regression on a time index variable and obtain the residuals $(\hat{u}_t)$.

When it is time to regress $y_t$ on $X_t$, do I use $\hat{u}_t$ instead of $X_t$?

For example:

$y_t=\beta_0+\beta_1\hat{u}_t$

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The most obvious thing to do would seem to include a time trend in the regression of $y_t$ on $X_t$, i.e. $$ y_t=\beta_0+\beta_1t+\beta_2X_t+error $$ Now, by the Frisch-Waugh-Lovell theorem, the estimate for $\beta_2$ of that regression will be exactly the same if you

  • first regress $y_t$ on a constant and the trend, save the residuals, call them $\hat{u}_{yt}$,
  • then regress $X_t$ on a constant and the trend, save the residuals, call them $\hat{u}_{Xt}$,
  • and finally regress $\hat{u}_{yt}$ on $\hat{u}_{Xt}$ and take the resulting coefficient.
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    $\begingroup$ Actually, you don't need the first step. The estimate of $\beta_2$ will be the same even if the original dep var is used. So the answer to the question is yes, even without the addition of step one. $\endgroup$
    – hejseb
    Commented Oct 21, 2015 at 19:40

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