# Do I need Sargan test with equal numbers of instruments and endogenous variables when one instrument can affect more than one endogenous variables?

I have a instrumental variable logistic regression I run with three instruments (z1, z2, z3) and three endogenous variables (k1, k2, k3).

Therefore, since the number of instruments "3" equals the number of endogenous variables "3", the statistics says I do not need to run the Sargan test to check over-identification.

My model also shows strong instruments (passing Instrument relevance test) and endogeneity tests (by passing the tests for exogeneity) with these three instruments and endogenous variables.

However, my economic interpretation says z1 may not only affect k1 but also k2.

Does this mean I am still having over-identification problem? But even though I do so, I will still have zero degrees of freedom that won't make me run the Sargan test at all.

Shall I assume I still do not need over-identification test in this case?

May I also have some literature support for the arguments as well please?

Also, if I interpret that z1 may not only affect k1 but also k2, is this acceptable in general as part of the instrumental variable regression result interpretation? If not, is there a way to make this interpretation acceptable?

I need a clear answer without disregarding the last sentence above in bold texts please.

• I'd rather say you cannot, rather than do not need to, perform a test for identification, because a Sargan type statistic will be identically zero regardless of whether your instruments are valid or not. – Christoph Hanck Jul 15 '19 at 11:57
• – Christoph Hanck Jul 15 '19 at 11:58
• @ChristophHanck: Then if I interpret that z1 may not only affect k1 but also k2, is this acceptable in general as part of the instrumental variable regression result interpretation? – Eric Jul 15 '19 at 20:47
• You can of course compute partial correlations (first stage regressions) to inform yourself about which instruments correlate with which regressors, but in general, all instruments jointly identify the coefficients of interest. – Christoph Hanck Jul 16 '19 at 3:54

I should start this answer by warning that it is quite difficult in general to give a causal interpretation of IV when there are multiple instruments and multiple endogenous regressors (chapter 4 of Angrist and Pischke should show how hard it is even with a single endogenous regressor) when we allow for heterogeneous treatment effect. Even when the exogeneity conditions for instruments are satisfied, a Sargan test of an over-identified system can still reject the null because different instruments will measure different local average treatment effects.

With that caveat in mind (maybe you're estimating some structural model, or maybe you are willing to make the strong homogeneity assumptions necessary to simply the analysis of IVs), we can discuss your problem. In what follows, I will replace your endogenous regressors $$K$$ with $$X$$, since this is more typical notation (the tl;dr is that when we look at the IV assumptions, none of it depends on the relationship between your instruments and the endogenous regressors, just that your instrument induces enough variation in the endogenous regressors, so your concerns about how $$z_1$$ relates to $$k_1$$,$$k_2$$ is simply irrelevant to the question of whether the IV model is valid).

With 3 instruments and 3 endogenous regressors, our population model is:

$$$$Y = \beta_0 + \beta_1 X_1 + \beta_2 X_2 + \beta_3 X_3 + \varepsilon,\quad \mathbb E[\varepsilon Z_k] = 0, k = 1,2,3\tag{1}$$$$ the 2SLS estimating procedure is really just trying to find $$\beta_0, \beta_1, \beta_2, \beta_3$$ to satisfy the following moment conditions: $$\mathbb E[Y - \beta_0 - \beta_1 X_1 - \beta_2 X_2 - \beta_3 X_3] = 0$$ $$\mathbb E[(Y - \beta_0 - \beta_1 X_1 - \beta_2 X_2 - \beta_3 X_3) Z_1] = 0$$ $$\mathbb E[(Y - \beta_0 - \beta_1 X_1 - \beta_2 X_2 - \beta_3 X_3) Z_2] = 0$$ $$\mathbb E[(Y - \beta_0 - \beta_1 X_1 - \beta_2 X_2 - \beta_3 X_3) Z_3] = 0$$

(the first equation comes from the fact that we can always choose a $$\beta_0$$ to make the error term 0, while the last 3 are from the exogeneity assumption, which in particular implies that $$\mathbb E[\varepsilon Z_k] = 0$$. Under the instrument relevance assumption that $$\mathbb E[\mathbf Z \mathbf X^T]$$ is full rank where $$\mathbf Z = (1, Z_1, Z_2, Z_3)^T$$ and $$\mathbf X = (1, X_1, X_2, X_3)^T$$ (note that this assumption is testable in the data), the above moments uniquely map into values of $$\beta_0,\beta_1,\beta_2,\beta_3$$. In other words, give me any dataset where $$\mathbb Z^T \mathbb X$$ (using the matrix notation typically found in econometrics texts) is invertible, and I can find a set of coefficients consistent that make the data consistent with the model. This is why, as the commenters above have said, you cannot do a Sargan type test on this data. The linear model itself has no empirical content, because you can force it to be true with any population by choosing the right parameters; we can only test models when there is some population that is not consistent with the model (note that clearly, there is a theoretical difference between cases when the $$\beta$$'s that end up being estimated represent some sort of causal parameter or just some sort of way of summarizing correlations; the above discussion, however, shows that there is no fully data driven way to distinguish between these cases, so some sort of assumption will always be necessary).

The Sargan test comes into play when you have more instruments than endogenous regressors. The reason why is because in that case, we will have that $$\mathbb E[Z X^T]$$ will be an $$m \times 4$$ matrix, which will not be fully invertible even if it is full rank. This means that some populations cannot possibly be represented by model (1). In this case, with enough data, we will be able to distinguish between datasets that are consistent with (1) and datasets that are not. Notice that even among the data that are consistent with (1), there are some data generating processes where the $$\beta$$s have a causal interpretation and others where a causal interpretation is inappropriate, so even when the Sargan test fails to reject the null, it is still possible that the underlying model isn't identifying what you're really interested in.

Finally, going back to your bolded question, notice that none of the above analysis of what IV tries to do depends directly on the specific relationship between $$Z$$ and $$X$$ except that this relationship is "full rank". A data generating process where $$Z_1$$ affects $$X_1$$ only or where $$Z_1$$ affects both $$X_1$$ and $$X_2$$ are equally valid IV models as long as $$Z_2, Z_3$$ induce enough variation in $$X_2,X_3$$. Note that it is possible to test whatever restriction you want to place on your first stage relationship (i.e. the relationship between $$X$$ and $$Z$$), but any such restrictions will not tell us anything meaningful of the structural equation (i.e. the relationship between $$Y$$ and $$X$$).

To summarize more conceptually, the Sargan test for linear IV is about testing the hypothesis: "If different combinations of instruments induce different variation in my endogenous regressors, then does this variation relate to the outcome in the same way?". Meanwhile, your question seems to be more "how many of/which of the regressors do my instruments affect". If you had more than 3 instruments, your question to some extent addresses the "if" part of the Sargan test, but the "then" part is what really determines whether the Sargan test returns a null result or not. When you only have 3 instruments, the if part of the Sargan test is vacuously false. There are no different combinations of instruments because all 3 instruments are needed to identify the model in the first place.

• Thank you. If I confirm only about the bold text, in short shall I assume your answer is "yes" in this case then? – Eric Jul 16 '19 at 7:09
• Yes, but with the caveat that multivariate IV like this is really hard to interpret in general. – stats_model Jul 16 '19 at 15:33