investigate further this issue, we consider the following five DGPs: where X$_{1}$ , X$_{2}$ and ε are defined as before, { X$_{i}$ } 10 k =3 are independent standard normal random variables, D = 1 { D ∗ > 0 } , and ε ⊥ X , with X = (1 ,X$_{1}$,X$_{2}$,... ,X$_{10}$ ) $^{′}$. For each of these DGPs we consider the potential outcomes where m$_{2}$ ( X ) = 1 + ∑ 10 j $_{=1}$X$_{j}$ , u (1) and u (0) are independent normal random variables with mean zero and variance 0 . 1. The observed outcome is Y = DY (1)+(1 − D ) Y (0), and the true ATE is 1. For DGP 6 − DGP 10, the H$_{0}$ considered is We estimate θ$_{0}$ by ML . Note that DGP 6 falls under H$_{0}$, whereas DGP 7- DGP 10 fall under H$_{1}$ . The simulation results for DGP 6- DGP 10 are presented in Table 2. As before, we first discuss the size properties of the tests. From the results of DGP 6, we find that KS$_{n}$ and CvM$_{n}$ tests are oversized when n = 100, but as sample size n increases, the empirical size gets closer to its nominal value. The same holds true for ICM tests based on the unprojected process (2.10). Shaikh et al. (2009)’s test tends to be conservative (with the exception when c = 0 . 01) , and sensitive to the bandwidth choice. The traditional balancing tests, and the ICM tests based on (4.1) are conservative, reflecting the “curse of dimensionality”. Finally, note that when the propensity score is correctly specified, the finite sample properties of the ATE estimator (4.3) are good: the bias and the length of the 95% confidence interval get smaller when sample size increases, and the coverage probability is relatively close to its 23 H$_{0}$ : ∃ θ$_{0}$ = ( β$_{0}$,β$_{1}$,β$_{2}$,... ,β$_{10}$ ) ′ ∈ Θ : E [ D | Φ ( X $^{′}$θ$_{0}$ )] = Φ ( X $^{′}$θ$_{0}$ ) a.s.. (4.4) Y (1) = 2 m$_{2}$ ( X ) + u (1) and Y (0) = m$_{2}$ ( X ) + u (0) , DGP 6 . D ∗ = − ∑ 10 j $_{=1}$X$_{j}$ 6 − ε, ; DGP 7 . D ∗ = − 1 − ∑ 10 j $_{=1}$X$_{j}$ 10 + X$_{1}$X$_{2}$ 2 − ε, ; DGP 8 . D ∗ = − 1 − ∑ 10 j $_{=1}$X$_{j}$ 10 + X$_{1}$ ∑ 5 k $_{=2}$X$_{k}$ 4 − ε, ; DGP 9 . D ∗ = − 1 . 5 − ∑ 10 j $_{=1}$X$_{j}$ 6 + ∑ 10 k $_{=1}$X 2 k 10 − ε ; DGP 10 . D ∗ = − 0 . 1 + 0 . 1 ∑ 5 j $_{=1}$X$_{j}$ exp ( − 0 . 2 10 k $_{=1}$X$_{j}$ ) − ε,,