<p>Usually, $\hat{\beta_1^{IV}} = \beta_1 + \frac{cov(z,u)}{cov(z,x)}$.  The denominator will go to zero. </p>

<p>That is true unless there is some correlation between the instrument and the error term, and the nominator is the strength of the relationship between the instrument and the endogenous variable. The smaller the denominator gets, the greater the bias $\left[\frac{cov(z,u)}{cov(z,x)}\right]$.</p>

<p>In addition,  weak instrument will have no precision, so that the variance will have a big upward bias. 
\begin{eqnarray} var(\hat{\beta_1}) &amp;\to_p&amp; \frac{\sigma^2}{n \sigma^2_x} \nonumber \\
\hat{\beta_1^{IV}} &amp;=&amp; \frac{\sum (z_i - \bar{z})y_i}{\sum(z_i - \bar{z})x_i} = \beta_1 + \frac{\sum (z_i - \bar{z})u_i}{\sum(z_i - \bar{z})x_i} \nonumber \\
var(\hat{\beta_1^{IV}} &amp;=&amp; var \left( \frac{\sum (z_i - \bar{z})u_i}{\sum(z_i - \bar{z})x_i} \right) \nonumber\\
var(u | z) &amp;=&amp; \sigma^2 \nonumber\\
var(\hat{\beta_1^{IV}}) &amp;=&amp; \frac{\sigma^2 \frac{1}{n} \sum (z_i - \bar{z})}{n[ \frac{1}{n} \sum (z_i - \bar{z})(x_i - \bar{x})]^2} \nonumber
\end{eqnarray}</p>

<p>As $n \to \inf$
\begin{eqnarray} var(\hat{\beta_1^{IV}}) &amp;\to_p&amp; \frac{\sigma^2 \sigma_z^2 }{\sigma^2_{zx}} \nonumber\\
var(\hat{\beta_1^{IV}}) &amp;\to_p&amp;  \sigma^2 \frac{1}{n \sigma^2_x} \frac{1}{\rho^2_{xz}}\nonumber \\
\rho^2_{xz} &amp;=&amp; \frac{[\sigma^2 _{xz}]^2}{\sigma_x^2 \sigma_z^2} \textit{for} \rho \in [0,1]\nonumber
\end{eqnarray}</p>

<p>That is why if your instrument is weak, then you  may be better off running an OLS regression.    </p>

CC BY-SA 3.0
2016-11-20T16:12:29.820Z
The line following "$var(u|z)=\sigma^2$" is not exactly the variance (also the numerator misses the square notation, just a typo). The denominator is random (because $x_i$ are endogenous) and the variance is much more complicated.
chan1142
CC BY-SA 3.0
2015-05-26T19:56:49.193Z
In the equation for the first variance of the IV estimator, I do believe that the variance of the unbiased beta one is missing -- right? You only assign the variance to the portion related with the bias of the IV estimator. If I am wrong, please explain to me what I am missing.
John Doe
CC BY-SA 3.0
2015-05-22T21:31:09.707Z
2015-05-22T21:41:12.367Z
2015-05-22T21:41:12.367Z
Nox
<p>Weak instruments combined with slight <em>instrumental endogeneity</em> can lead to a larger bias than OLS. As Nox's answer shows, the probability limit of the IV estimator is $\beta_1 + cov(z,u)/cov(z,x)$. When $cov(z,u) \ne 0$ though small, if $cov(z,x)$ is small, then the bias can be large. See Bound, Jaeger and Baker's (1995, JASA) remark following equation (7) on page 444. </p>

<p><a href="http://www.djaeger.org/research/pubs/jasav90n430.pdf" rel="noreferrer">http://www.djaeger.org/research/pubs/jasav90n430.pdf</a></p>

<p>"It is clear from Equation (7) that a weak correlation between the potentially endogenous variable, $x$, and the instruments, $z_1$, will exacerbate any problems associated with the correlation between the instrument and the error, $\varepsilon$. If the correlation between the instrument and the endogenous explanatory variable is weak, then even a small correlation between the instrument and the error can produce a larger inconsistency in the IV estimate of $\beta$ than in the OLS estimate."</p>

<p>Without instrumental endogeneity, I don't think the IV estimator's bias (of the limit distribution, there may be no probability limit) is larger than OLS's inconsistency.</p>

<p>Another thing to consider is that the variance of the IV estimator using very weak instruments can be large even with a very large $n$, and thus you may have an IV estimate more nonsense than OLS for a data set just by chance.</p>

CC BY-SA 3.0
2016-11-20T15:41:28.400Z
2016-11-20T15:41:28.400Z
chan1142
<p>Some papers argue that OLS can produce less bias than IV estimation depending on the quality of your instruments. Suppose we consider a demand estimation equation. </p>

<p>Suppose the demand elasticity is negative in OLS. By my intuition weak instruments should produce biased estimates towards OLS, but no less negative. Can you guys produce an example? I cannot really grasp how it could lead to a more biased estimation with IV estimation.</p>

CC BY-SA 3.0
2015-07-28T14:50:09.337Z
Which are "some papers" (preferably well known ones, or lit review-type) that you refer to in your first sentence? I'm interested in looking at them. Thanks.
Kim Jong Un
CC BY-SA 3.0
2015-05-22T21:17:52.653Z
IV is biased but it consistent, so I imagine your statement is true. but I suppose it all depends on your objectives. prediction vs inference.
user157623
<econometrics>
OLS bias in demand estimation: the bias always underestimate the demand's elasticity?
John Doe
# OLS bias in demand estimation: the bias always underestimate the demand's elasticity?

(Asked by: username_0 on 2015-05-22)

Some papers argue that OLS can produce less bias than IV estimation depending on the quality of your instruments. Suppose we consider a demand estimation equation.

Suppose the demand elasticity is negative in OLS. By my intuition weak instruments should produce biased estimates towards OLS, but no less negative. Can you guys produce an example? I cannot really grasp how it could lead to a more biased estimation with IV estimation.

**Tags:** econometrics

> **Comments:**
> **username_2**: IV is biased but it consistent, so I imagine your statement is true. but I suppose it all depends on your objectives. prediction vs inference.  
> **username_1**: Which are "some papers" (preferably well known ones, or lit review-type) that you refer to in your first sentence? I'm interested in looking at them. Thanks.  

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## 2 answers

### Answer by username_3 on 2015-05-22. Score: 8

Usually, $\hat{\beta\_1^{IV}} = \beta\_1 + \frac{cov(z,u)}{cov(z,x)}$. The denominator will go to zero.

That is true unless there is some correlation between the instrument and the error term, and the nominator is the strength of the relationship between the instrument and the endogenous variable. The smaller the denominator gets, the greater the bias $\left[\frac{cov(z,u)}{cov(z,x)}\right]$.

In addition, weak instrument will have no precision, so that the variance will have a big upward bias.
\begin{eqnarray} var(\hat{\beta\_1}) &\to\_p& \frac{\sigma^2}{n \sigma^2\_x} \nonumber \\
\hat{\beta\_1^{IV}} &=& \frac{\sum (z\_i - \bar{z})y\_i}{\sum(z\_i - \bar{z})x\_i} = \beta\_1 + \frac{\sum (z\_i - \bar{z})u\_i}{\sum(z\_i - \bar{z})x\_i} \nonumber \\
var(\hat{\beta\_1^{IV}} &=& var \left( \frac{\sum (z\_i - \bar{z})u\_i}{\sum(z\_i - \bar{z})x\_i} \right) \nonumber\\
var(u | z) &=& \sigma^2 \nonumber\\
var(\hat{\beta\_1^{IV}}) &=& \frac{\sigma^2 \frac{1}{n} \sum (z\_i - \bar{z})}{n[ \frac{1}{n} \sum (z\_i - \bar{z})(x\_i - \bar{x})]^2} \nonumber
\end{eqnarray}

As $n \to \inf$
\begin{eqnarray} var(\hat{\beta\_1^{IV}}) &\to\_p& \frac{\sigma^2 \sigma\_z^2 }{\sigma^2\_{zx}} \nonumber\\
var(\hat{\beta\_1^{IV}}) &\to\_p& \sigma^2 \frac{1}{n \sigma^2\_x} \frac{1}{\rho^2\_{xz}}\nonumber \\
\rho^2\_{xz} &=& \frac{[\sigma^2 \_{xz}]^2}{\sigma\_x^2 \sigma\_z^2} \textit{for} \rho \in [0,1]\nonumber
\end{eqnarray}

That is why if your instrument is weak, then you may be better off running an OLS regression.

> **Comments:**
> **username_0**: In the equation for the first variance of the IV estimator, I do believe that the variance of the unbiased beta one is missing -- right? You only assign the variance to the portion related with the bias of the IV estimator. If I am wrong, please explain to me what I am missing.  
> **username_4**: The line following "$var(u|z)=\sigma^2$" is not exactly the variance (also the numerator misses the square notation, just a typo). The denominator is random (because $x_i$ are endogenous) and the variance is much more complicated.  

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### Answer by username_4 on 2016-11-20. Score: 6

Weak instruments combined with slight *instrumental endogeneity* can lead to a larger bias than OLS. As Nox's answer shows, the probability limit of the IV estimator is $\beta\_1 + cov(z,u)/cov(z,x)$. When $cov(z,u) \ne 0$ though small, if $cov(z,x)$ is small, then the bias can be large. See Bound, Jaeger and Baker's (1995, JASA) remark following equation (7) on page 444.

<http://www.djaeger.org/research/pubs/jasav90n430.pdf>

"It is clear from Equation (7) that a weak correlation between the potentially endogenous variable, $x$, and the instruments, $z\_1$, will exacerbate any problems associated with the correlation between the instrument and the error, $\varepsilon$. If the correlation between the instrument and the endogenous explanatory variable is weak, then even a small correlation between the instrument and the error can produce a larger inconsistency in the IV estimate of $\beta$ than in the OLS estimate."

Without instrumental endogeneity, I don't think the IV estimator's bias (of the limit distribution, there may be no probability limit) is larger than OLS's inconsistency.

Another thing to consider is that the variance of the IV estimator using very weak instruments can be large even with a very large $n$, and thus you may have an IV estimate more nonsense than OLS for a data set just by chance.


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