<p>Yes, you are correct about the conclusion that <span class="math-container">$J_{n1}=0$</span> and hence <span class="math-container">$C_{n}=J_{n}-J_{n1}=J_{n}$</span>. However, despite the test statistics are the same, the asymptotical distributions and  critical values are the same, they are derived from different procedure under different null hypothesis, I prefer to keep using the name C-test instead of Sargan-Hansen test.</p>

CC BY-SA 4.0
2021-08-22T07:32:04.943Z
2021-08-22T07:32:04.943Z
Q9y5
<p>My question is motivated by <a href="https://www.sciencedirect.com/science/article/pii/S0304387818302050" rel="noreferrer">Do, Joshi, and Stopler's paper &quot;Can Environmental Policy Reduce Infant Mortality? Evidence from the Ganga Pollution Cases&quot;</a>.</p>
<p>A simplified form of their models that describes the question is:</p>
<p><span class="math-container">$$y_i =\beta_0 +\beta_1 X_i +\varepsilon_i $$</span></p>
<p>Which would be estimated (1) using only <span class="math-container">$Z_{1i}$</span> as an instrument and (2) using <span class="math-container">$Z_{1i}$</span> and <span class="math-container">$Z_{2i}$</span> as instruments. The goal is to determine if <span class="math-container">$Z_{2i}$</span> is a valid instrument (uncorrelated with <span class="math-container">$\varepsilon_i$</span>). They &quot;know&quot; <span class="math-container">$Z_{1i}$</span> is valid.</p>
<p>Their method is to use a C-statistic, which as far as I can tell is the difference in Sargan-Hansen test statistics between the two models, and would be distributed as <span class="math-container">$\chi^2_1$</span> because the number of instruments differs by 1.</p>
<p>My question is, wouldn't the Sargan-Hansen test from (1) just give 0? The equation is exactly identified. Thus, doesn't the C-statistic just reduce to a Sargan-Hansen test of (2)? Is there something I am missing? Are they, perhaps, just trying to show they are aware of complicated techniques to signal intelligence? Or is it more appropriate to call it a C-statistic due to the comparative nature?</p>
<p>TIA!</p>

<econometrics><instrumental-variable>
C-statistic when one estimation is just identified
Michael Gmeiner
# C-statistic when one estimation is just identified

(Asked by: username_0 on 2021-08-21)

My question is motivated by [Do, Joshi, and Stopler's paper "Can Environmental Policy Reduce Infant Mortality? Evidence from the Ganga Pollution Cases"](https://www.sciencedirect.com/science/article/pii/S0304387818302050).

A simplified form of their models that describes the question is:

$$y\_i =\beta\_0 +\beta\_1 X\_i +\varepsilon\_i $$

Which would be estimated (1) using only $Z\_{1i}$ as an instrument and (2) using $Z\_{1i}$ and $Z\_{2i}$ as instruments. The goal is to determine if $Z\_{2i}$ is a valid instrument (uncorrelated with $\varepsilon\_i$). They "know" $Z\_{1i}$ is valid.

Their method is to use a C-statistic, which as far as I can tell is the difference in Sargan-Hansen test statistics between the two models, and would be distributed as $\chi^2\_1$ because the number of instruments differs by 1.

My question is, wouldn't the Sargan-Hansen test from (1) just give 0? The equation is exactly identified. Thus, doesn't the C-statistic just reduce to a Sargan-Hansen test of (2)? Is there something I am missing? Are they, perhaps, just trying to show they are aware of complicated techniques to signal intelligence? Or is it more appropriate to call it a C-statistic due to the comparative nature?

TIA!

**Tags:** econometrics, instrumental-variable

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## 1 answers

### Answer by username_1 on 2021-08-22. Score: 4 | Accepted answer

Yes, you are correct about the conclusion that $J\_{n1}=0$ and hence $C\_{n}=J\_{n}-J\_{n1}=J\_{n}$. However, despite the test statistics are the same, the asymptotical distributions and critical values are the same, they are derived from different procedure under different null hypothesis, I prefer to keep using the name C-test instead of Sargan-Hansen test.


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