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?


1 Answer 1


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, 2021 at 8:06
  • $\begingroup$ @Giorgetto: thanks, I corrected it. Feel free to edit any mistakes. $\endgroup$
    – tdm
    Oct 3, 2021 at 8:08

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