Testing exclusion restrictions and additive separability in sample selection models

Abstract

Standard sample selection models with non-randomly censored outcomes assume (i) an exclusion restriction (i.e., a variable affecting selection, but not the outcome) and (ii) additive separability of the errors in the selection process. This paper proposes tests for the joint satisfaction of these assumptions by applying the approach of Huber and Mellace (Testing instrument validity for LATE identification based on inequality moment constraints, 2011) (for testing instrument validity under treatment endogeneity) to the sample selection framework. We show that the exclusion restriction and additive separability imply two testable inequality constraints that come from both point identifying and bounding the outcome distribution of the subpopulation that is always selected/observed. We apply the tests to two variables for which the exclusion restriction is frequently invoked in female wage regressions: non-wife/husband’s income and the number of (young) children. Considering eight empirical applications, our results suggest that the identifying assumptions are likely violated for the former variable, but cannot be refuted for the latter on statistical grounds.

Appendix

1.1 Link to Kitagawa (2010)

The subsequent discussion links the testable implications of Sect. 3 to Kitagawa (2010), who derives a testable implication based on comparable model assumptions. Considering only positive monotonicity, Kitagawa (2010) shows in his Proposition 2.3 that under Assumptions 1 and  2,

$$\begin{aligned} f(y,S=1|Z=0) \le f(y,S=1|Z=1) \hbox { for all }y \hbox { in the support of }Y, \end{aligned}$$

(13)

i.e., the joint density of $Y$ and $S=1$ given $Z=1$ must nest the joint density of $Y$ and $S=1$ given $Z=0$ for any value of $Y.$ Rearranging terms such that $f(y,S=1|Z=1)-f(y,S=1|Z=0) \ge 0$ gives the intuitive interpretation that the pdf of the compliers’ outcome cannot be smaller than zero, as densities must not be negative.

Note that (7) in Sect. 3 is equivalent to

$$\begin{aligned} \frac{\Pr (Y\in A,S=1|Z=1)}{P_{1|0}}-\frac{P_{1|1}-P_{1|0}}{P_{1|0}}&\le \frac{\Pr (Y\in A,S=1|Z=0)}{P_{1|0}} \nonumber \\&\le \frac{\Pr (Y\in A,S=1|Z=1)}{P_{1|0}} \end{aligned}$$

(14)

for all $A$ in the support of $Y,$ because

$$\begin{aligned} \frac{\Pr (Y\in A|Z=1,S=1)-(1-q)}{q}&= \frac{\Pr (Y\in A,S=1|Z=1)}{q\cdot \Pr (S=1|Z=1)}-\frac{(1-q)}{q} \\&= \frac{\Pr (Y\in A,S=1|Z=1)}{P_{1|0}}-\frac{P_{1|1}-P_{1|0}}{P_{1|0}},\\ \frac{\Pr (Y\in A|Z=1,S=1)}{q}&= \frac{\Pr (Y\in A,S=1|Z=1)}{q\cdot \Pr (S=1|Z=1)}\\&= \frac{\Pr (Y\in V,D=1|Z=1)}{P_{1|0}},\\ \Pr (Y\in A|Z=0,S=1)&= \frac{\Pr (Y\in A,S=1|Z=0)}{P_{1|0}}, \end{aligned}$$

by using basic probability theory. (14) in turn implies that $\hbox { for all }A \hbox { in the support of }Y,$

$$\begin{aligned} \Pr (Y\in A,S=1|Z=1)-(P_{1|1}-P_{1|0})&\le \Pr (Y\in A,S=1|Z=0)\nonumber \\&\le \Pr (Y\in A,S=1|Z=1), \end{aligned}$$

(15)

and when applied to the pdf, that $\hbox {for all }y \hbox { in the support of }Y$

$$\begin{aligned} f(y,S=1|Z=1)-(P_{1|1}-P_{1|0})&\le f(y,S=1|Z=0)\nonumber \\&\le f(y,S=1|Z=1), \end{aligned}$$

(16)

i.e., (16) yields one additional testable implication compared to (13). If we rearrange the first part in (15) $\Pr (Y\in A,S=1|Z=1)-(P_{1|1}-P_{1|0})\le \Pr (Y\in A,S=1|Z=0)$ to be $\Pr (Y\in A,S=1|Z=1)-\Pr (Y\in A,S=1|Z=0)\le (P_{1|1}-P_{1|0}),$ our additional implication gets an intuitive interpretation: The joint probability of being a complier and having a particular value of the outcome (and any sum of joint probabilities defined by non-overlapping subsets $A$) must not be larger than the unconditional probability of being a complier, because

$$\begin{aligned} \int [f(y,S=1|Z=1)-f(y,S=1|Z=0)] dy = P_{1|1}-P_{1|0}. \end{aligned}$$

(17)

It is worth noting that if testing is based on subsets $A$ that are non-overlapping and jointly cover the entire support of $Y,$ then our additional testable implication in (16) is are already taken into account by (13) and thus redundant. The prevalence of some $\Pr (Y\in A,S=1|Z=1)-\Pr (Y\in A,S=1|Z=0)>(P_{1|1}-P_{1|0})$ then necessarily implies the existence of at least one distinct $A'$ for which $\Pr (Y\in A',S=1|Z=1)-\Pr (Y\in A',S=1|Z=0)<0$ so that (13) is violated, too. Therefore, power gains from the additional testable implication might possibly only be realized when using subsets $A$ that overlap (so that violations may be averaged out) and/or do not cover the entire support of $Y,$ see also the discussion in Huber and Mellace (2011).

1.2 Chen and Szroeter’s test algorithm

This section provides the algorithm of the Chen and Szroeter (2012) test when testing the constraints on the mean outcome given in (12), but testing the probability constraints in (8) is analogous. Let $\hat{\theta }$ denote the sample analog of $\theta =(\theta ^m_{1},\theta ^m_{2})'.$ The algorithm can be sketched as follows:

• 1. Estimate the vector of parameters $\hat{\theta }$ and the asymptotic variance $\hat{J}$ of $\sqrt{n}\cdot (\hat{\theta }-\theta ).$

• 2. Let $\hat{\eta }_i=1/\sqrt{\hat{J}_{i}}, \ i=1,2,$ where $\hat{J}_{i}$ is the ith element of the main diagonal of $\hat{J},$ and compute the smoothing function $\hat{\Psi }_i(\delta _n^{-1}\cdot \hat{\eta }_i\cdot \hat{\theta }_i)=\Phi (\delta _n^{-1}\cdot \hat{\eta }_i\cdot \theta _i),$ where $\Phi$ is the standard normal cdf and the tuning parameter $\delta _n$ is a sequence satisfying $\delta _n\rightarrow 0$ and $\sqrt{n}\cdot \delta _n\rightarrow \infty$ as $n\rightarrow \infty .$ In the applications, we choose $\delta _n=\sqrt{\frac{2\cdot \ln (\ln (n))}{n}}\cdot \hat{\sigma }_{\theta _i},$ where $\hat{\sigma }_{\theta _i}$ is the estimated standard deviation of the ith inequality constraint.

• 3. Compute the approximation term $\hat{\Lambda }_i=\phi (\delta _n^{-1}\cdot \hat{\eta }_i\cdot \hat{\theta }_i)\cdot \frac{1}{\delta _n\cdot \sqrt{n}}, \quad i=1,2,$ with $\phi$ being the standard normal pdf.

• 4. Define the vectors $\hat{\Psi }=\left( \hat{\Psi }_1(\delta _n^{-1}\cdot \hat{\eta }_1\cdot \hat{\theta }_1), \hat{\Psi }_2(\delta _n^{-1}\cdot \hat{\eta }_2\cdot \hat{\theta }_2)\right) ^T,\,\hat{\Lambda }=\left( \hat{\Lambda }_1, \hat{\Lambda }_2\right) ^T,\iota _2=(1,1)^T,\,\hat{\Delta }=diag(\hat{J}_1, \hat{J}_2).$

• 5. Let $\hat{Q}_1=\sqrt{(}n)\cdot \hat{\Psi }^T\hat{\Delta }\hat{\theta }-\iota _2^T\hat{\Lambda }$ and $\hat{Q}_2=\sqrt{\hat{\Psi }^T\hat{\Delta }\hat{J}\hat{\Delta }\hat{\Psi }}.$

• 6. Compute the p-value as $\hat{p}=\left\{ \begin{array}{cc} 1-\Phi \left( \frac{\hat{Q}_1}{\hat{Q}_2}\right) &{} \hbox { if }\hat{Q}_2>0\\ 1 &{}\hbox { if }\hat{Q}_2=0. \end{array}\right.$