Say, you have a model $$ y_{it} = \beta_1 a_{it} + \beta_2 b_{it} $$ And you find that $a_{it}$ is endogenous, therefore, you need to find an instrument for it.

Say you find an instrument $z_{it} = \Delta y_{it-1}$ that is relevant and exogenous,

Would you be able to estimate $\beta_1$ and $\beta_2$ by estimating

$$ \Delta y_{it} = \beta_1 \Delta y_{it-1} + \beta_2 \Delta b_{it} $$

And if so, how would this be done? I am running into a problem where the matrix of $\Delta y_{it-1}$ has fewer rows than $\Delta b_{it}$.

Would you just reshape $\Delta b_{it}$? Is this allowed?


1 Answer 1


It is allowed to use lags. In fact using using lags as instruments is actually quite common in some fields like macroeconomics (See Romer Advanced Macroeconomics pp 376).

However, note:

$$ \Delta y_{it} = \beta_1 \Delta y_{it-1} + \beta_2 \Delta b_{it} +\epsilon_{it} $$

Is the reduced form of IV so it will give you estimate of: $\beta_1 \pi$, instead of just $\beta_1$ but you are getting rid of endogeneity. If you want to get just $\beta_1$ you could run 2SLS.

This is because you are basically substituting $a_{it} = \pi_0 + \pi y_{it-1} +e_{it}$ so:

$$ \Delta y_{it} = \beta_1 \pi_0 + \beta_1 \pi y_{it-1} + \beta_2 \Delta b_{it} + \beta_1 e_{it} +\epsilon_{it} $$

You should not get an error doing that using common packages/programs.

If you are trying to program the IV estimator yourself in something like R you are likely getting the error because when you create matrix containing $\Delta y_{t-1}$ it won't match the matrix of $\Delta y_t$ as there is no way of having $\Delta y_{t-1}$ for the first row. A simple solution to this would be just to exclude the first row of your data matrix after you create $\Delta y_{t-1}$ variable from regression (or you could even outright delete it but I presume that data can still be useful for some summary statistics/visualizations).


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