# C-statistic when one estimation is just identified

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!

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.