<p>What you are describing is the so called &quot;<em>forbidden regression</em>&quot;, which (in general) does not give consistent estimates. This is a summary of the <a href="https://benjamindwilliams.weebly.com/uploads/6/8/5/7/68575765/notesonforbiddenregressions.pdf" rel="nofollow noreferrer">notes of Ben Williams</a></p>
<p>Consider a (nonlinear) first stage regression of <span class="math-container">$X$</span> on the instruments <span class="math-container">$Z$</span> giving fitted values (e.g. using a log-log specification):
<span class="math-container">$$
\hat X = \hat \mu(Z)
$$</span>
Consider the structural (causal) equation:
<span class="math-container">$$
Y = X'\beta + u
$$</span>
What you propose is to use <span class="math-container">$\hat X : \hat \mu(Z)$</span> instead of <span class="math-container">$X$</span> in the second stage. This gives:
<span class="math-container">$$
\begin{align*}
\hat \beta &amp;= (\hat X' \hat X)^{-1} \hat X Y,\\
&amp;= (\hat X' \hat X)^{-1} \hat X (X' \beta + u),\\
&amp;= (\hat X' \hat X)^{-1} \hat X (\hat X' \beta) + (\hat X' \hat X)' \hat X'(X - \hat X)'\beta + (\hat X' \hat X)^{-1}\hat X u,\\
&amp;= \beta + \underbrace{(\hat X' \hat X)' \hat X'(X - \hat X)'\beta}_A + \underbrace{(\hat X' \hat X)^{-1}\hat X u}_B,
\end{align*}
$$</span>
If <span class="math-container">$Z$</span> is a valid instrument, on can expect that the <span class="math-container">$B$</span> vanishes as <span class="math-container">$\hat X = \hat \mu(Z)$</span> and by assumption <span class="math-container">$\mathbb{E}(u|Z) = 0$</span>.</p>
<p>Now the <span class="math-container">$A$</span> terms is the real problem. Notice that we can always write:
<span class="math-container">$$
X = \mathbb{E}(X|Z) + \eta,\\
\text{ with } \mathbb{E}(\eta|Z) = 0
$$</span>
(here <span class="math-container">$\eta$</span> is simply <span class="math-container">$X - \mathbb{E}(X|Z)$</span>).</p>
<p>Then taking the middle part of the <span class="math-container">$A$</span> term gives:
<span class="math-container">$$
\hat X'(X - \hat X) = \hat X'(\mathbb{E}(X|Z) - \hat X) + \hat X'\eta,\\
$$</span>
The last term should vanish as <span class="math-container">$\mathbb{E}(\eta|Z) = 0$</span>. The first term however, will (in general) only vanish if <span class="math-container">$\hat X = \hat \mu(Z)$</span> is consistent for <span class="math-container">$\mathbb{E}(X|Z)$</span>, which will be the case if <span class="math-container">$\mu(Z)$</span> is a correct specification of <span class="math-container">$\mathbb{E}(X|Z)$</span>.</p>
<p>The usual 2SLS however is consistent as in this case:
<span class="math-container">$$
\hat X = Z(Z'Z)^{-1}Z'X.
$$</span>
Then:
<span class="math-container">$$
\begin{align*}
\hat X'(X - \hat X) &amp;= X'Z(Z'Z)^{-1}Z'(X - Z(Z'Z)^{-1}Z'X),\\
&amp;= X'Z(Z'Z)^{-1}Z'X - X'Z(Z'Z)^{-1}Z'Z(Z'Z)^{-1}Z'X,\\
&amp;= X'Z(Z'Z)^{-1}Z'X - X'Z(Z'Z)^{-1}Z'X = 0
\end{align*}
$$</span>
So either you do normal 2SLS, which will be consistent if <span class="math-container">$Z$</span> is uncorrelated with <span class="math-container">$u$</span> or you can use what is called indirect least squares.</p>
<ol>
<li><p>Regress <span class="math-container">$X$</span> on <span class="math-container">$Z$</span> using a nonlinear regression (e.g. loglinear regression).</p>
</li>
<li><p>Use the fitted values <span class="math-container">$\hat X = \hat \mu(Z)$</span> as instruments themselves in a 2SLS of <span class="math-container">$Y$</span> on <span class="math-container">$X$</span>. So run 2SLS with instruments <span class="math-container">$\hat X = \hat \mu(Z)$</span> instead of <span class="math-container">$Z$</span>.  As <span class="math-container">$\mu(Z)$</span> is a function of <span class="math-container">$Z$</span>, we also have that <span class="math-container">$\mathbb{E}(u|\mu(Z)) = 0$</span>, so these should be valid instruments.</p>
</li>
</ol>

CC BY-SA 4.0
2021-10-03T08:06:21.987Z
The second equation you wrote is the causal model or structural equation, clearly it is not the first stage equation
Giorgetto
CC BY-SA 4.0
2021-10-03T08:08:22.630Z
@Giorgetto: thanks, I corrected it. Feel free to edit any mistakes.
tdm
CC BY-SA 4.0
2021-10-02T11:18:50.507Z
2021-10-03T08:07:56.527Z
2021-10-03T08:07:56.527Z
tdm
<p>I have a regression in levels, derived from theory.</p>
<p>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.</p>
<p>Is it possible to somehow instrument with the first stage in logs, and the second stage in levels?</p>

<econometrics><instrumental-variable>
IV regression: first stage in logs, second stage in levels?
Econ_User
# IV regression: first stage in logs, second stage in levels?

(Asked by: username_0 on 2021-10-01)

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?

**Tags:** econometrics, instrumental-variable

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## 1 answers

### Answer by username_1 on 2021-10-02. Score: 5 | Accepted answer

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](https://benjamindwilliams.weebly.com/uploads/6/8/5/7/68575765/notesonforbiddenregressions.pdf)

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.

> **Comments:**
> **username_2**: The second equation you wrote is the causal model or structural equation, clearly it is not the first stage equation  
> **username_1**: @Giorgetto: thanks, I corrected it. Feel free to edit any mistakes.  

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