I have a regression in levels, derived from theory.

I want to instrument one of the variables, but the best instrument I find has a weak correlation to the endogenous variable in levels, and a strong correlation in logs. Both are very heteroskedastic.

Is it possible to somehow instrument with the first stage in logs, and the second stage in levels?


What you are describing is the so called "forbidden regression", which (in general) does not give consistent estimates. This is a summary of the notes of Ben Williams

Consider a (nonlinear) first stage regression of $X$ on the instruments $Z$ giving fitted values (e.g. using a log-log specification): $$ \hat X = \hat \mu(Z) $$ Consider the structural (causal) equation: $$ Y = X'\beta + u $$ What you propose is to use $\hat X : \hat \mu(Z)$ instead of $X$ in the second stage. This gives: $$ \begin{align*} \hat \beta &= (\hat X' \hat X)^{-1} \hat X Y,\\ &= (\hat X' \hat X)^{-1} \hat X (X' \beta + u),\\ &= (\hat X' \hat X)^{-1} \hat X (\hat X' \beta) + (\hat X' \hat X)' \hat X'(X - \hat X)'\beta + (\hat X' \hat X)^{-1}\hat X u,\\ &= \beta + \underbrace{(\hat X' \hat X)' \hat X'(X - \hat X)'\beta}_A + \underbrace{(\hat X' \hat X)^{-1}\hat X u}_B, \end{align*} $$ If $Z$ is a valid instrument, on can expect that the $B$ vanishes as $\hat X = \hat \mu(Z)$ and by assumption $\mathbb{E}(u|Z) = 0$.

Now the $A$ terms is the real problem. Notice that we can always write: $$ X = \mathbb{E}(X|Z) + \eta,\\ \text{ with } \mathbb{E}(\eta|Z) = 0 $$ (here $\eta$ is simply $X - \mathbb{E}(X|Z)$).

Then taking the middle part of the $A$ term gives: $$ \hat X'(X - \hat X) = \hat X'(\mathbb{E}(X|Z) - \hat X) + \hat X'\eta,\\ $$ The last term should vanish as $\mathbb{E}(\eta|Z) = 0$. The first term however, will (in general) only vanish if $\hat X = \hat \mu(Z)$ is consistent for $\mathbb{E}(X|Z)$, which will be the case if $\mu(Z)$ is a correct specification of $\mathbb{E}(X|Z)$.

The usual 2SLS however is consistent as in this case: $$ \hat X = Z(Z'Z)^{-1}Z'X. $$ Then: $$ \begin{align*} \hat X'(X - \hat X) &= X'Z(Z'Z)^{-1}Z'(X - Z(Z'Z)^{-1}Z'X),\\ &= X'Z(Z'Z)^{-1}Z'X - X'Z(Z'Z)^{-1}Z'Z(Z'Z)^{-1}Z'X,\\ &= X'Z(Z'Z)^{-1}Z'X - X'Z(Z'Z)^{-1}Z'X = 0 \end{align*} $$ So either you do normal 2SLS, which will be consistent if $Z$ is uncorrelated with $u$ or you can use what is called indirect least squares.

  1. Regress $X$ on $Z$ using a nonlinear regression (e.g. loglinear regression).

  2. Use the fitted values $\hat X = \hat \mu(Z)$ as instruments themselves in a 2SLS of $Y$ on $X$. So run 2SLS with instruments $\hat X = \hat \mu(Z)$ instead of $Z$. As $\mu(Z)$ is a function of $Z$, we also have that $\mathbb{E}(u|\mu(Z)) = 0$, so these should be valid instruments.

  • $\begingroup$ The second equation you wrote is the causal model or structural equation, clearly it is not the first stage equation $\endgroup$
    – Giorgetto
    Oct 3 at 8:06
  • $\begingroup$ @Giorgetto: thanks, I corrected it. Feel free to edit any mistakes. $\endgroup$
    – tdm
    Oct 3 at 8:08

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