<p>In standard linear regression model</p>
<p><span class="math-container">$$y = x^\top \beta + \epsilon$$</span></p>
<p>with exogeneity <span class="math-container">$\mathbb E[x\epsilon] = \mathbf 0$</span> you have <span class="math-container">$K$</span> parameters because <span class="math-container">$\beta$</span> is <span class="math-container">$K \times 1$</span> and you have <span class="math-container">$K$</span> equations</p>
<p><span class="math-container">$$\mathbb E[x\epsilon] = \mathbb E[x(y - x^\top \beta)] = \mathbf 0 \Leftrightarrow \mathbb E[xy] = \mathbb E[xx^\top]\beta,$$</span></p>
<p>where <span class="math-container">$\mathbb E[xy] = \mathbb E[xx^\top]\beta$</span> is a system of linear equations. Only if the matrix <span class="math-container">$\mathbb E[xx^\top]$</span> is invertible it follows that <span class="math-container">$\beta := \mathbb E[xx^\top]^{-1}\mathbb E[xy]$</span>.</p>
<p>For each endogenous variable in <span class="math-container">$x_k$</span> you lose one identifying equation because for the endogenous variable <span class="math-container">$x_k$</span> it does not hold that <span class="math-container">$\mathbb E[x_k\epsilon] = 0$</span>. So for example with 2 endogenous variables you would be trying to identify <span class="math-container">$K$</span> parameters using only <span class="math-container">$K-2$</span> equations which amounts to trying to solve a system of <span class="math-container">$K-2$</span> linear equations for <span class="math-container">$K$</span> unknowns. So assuming that you know how systems of linear eqautions work it should come as no surprise that you need to find at least two instruments. This would allow using the true equation <span class="math-container">$\mathbb E[z_k (y - x^\top \beta)] =0$</span> instead of the untrue <span class="math-container">$\mathbb E[x_k (y - x^\top \beta)] =0$</span> for each <span class="math-container">$k$</span> (which amounts to assuming <span class="math-container">$z_k$</span> is exogenous while <span class="math-container">$x_k$</span> is not).</p>

CC BY-SA 4.0
2021-01-25T18:40:20.420Z
2021-01-25T18:40:20.420Z
Jesper Hybel
<p>As the title says. Why do we need at least as many instrumental variables as endogenous regressors to identify parameters?</p>

<econometrics><statistics><instrumental-variable>
Why do we need at least as many instrumental variables as endogenous regressors to identify parameters in 2SLS?
Elias
# Why do we need at least as many instrumental variables as endogenous regressors to identify parameters in 2SLS?

(Asked by: username_0 on 2021-01-25)

As the title says. Why do we need at least as many instrumental variables as endogenous regressors to identify parameters?

**Tags:** econometrics, statistics, instrumental-variable

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## 1 answers

### Answer by username_1 on 2021-01-25. Score: 5

In standard linear regression model

$$y = x^\top \beta + \epsilon$$

with exogeneity $\mathbb E[x\epsilon] = \mathbf 0$ you have $K$ parameters because $\beta$ is $K \times 1$ and you have $K$ equations

$$\mathbb E[x\epsilon] = \mathbb E[x(y - x^\top \beta)] = \mathbf 0 \Leftrightarrow \mathbb E[xy] = \mathbb E[xx^\top]\beta,$$

where $\mathbb E[xy] = \mathbb E[xx^\top]\beta$ is a system of linear equations. Only if the matrix $\mathbb E[xx^\top]$ is invertible it follows that $\beta := \mathbb E[xx^\top]^{-1}\mathbb E[xy]$.

For each endogenous variable in $x\_k$ you lose one identifying equation because for the endogenous variable $x\_k$ it does not hold that $\mathbb E[x\_k\epsilon] = 0$. So for example with 2 endogenous variables you would be trying to identify $K$ parameters using only $K-2$ equations which amounts to trying to solve a system of $K-2$ linear equations for $K$ unknowns. So assuming that you know how systems of linear eqautions work it should come as no surprise that you need to find at least two instruments. This would allow using the true equation $\mathbb E[z\_k (y - x^\top \beta)] =0$ instead of the untrue $\mathbb E[x\_k (y - x^\top \beta)] =0$ for each $k$ (which amounts to assuming $z\_k$ is exogenous while $x\_k$ is not).


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