A Discontinuity Test for Identification in Nonparametric Models with Endogeneity
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1 A Discontinuity Test for Identification in Nonparametric Models with Endogeneity Carolina Caetano 1 Christoph Rothe 2 Nese Yildiz 1 1 Department of Economics 2 Department of Economics University of Rochester Columbia University April 9, 2013 Caetano, Rothe and Yildiz (Rochester and Columbia) Test for Identification with Endogeneity 04/09/ / 25
2 Outline 1 Introduction 2 Intuition 3 Control Function 4 Test Parameter 5 The Test 6 Conclusion Caetano, Rothe and Yildiz (Rochester and Columbia) Test for Identification with Endogeneity 04/09/ / 25
3 Introduction Control function approach is often used for identifying structural quantities in models with endogeneity. (See Newey, Powell, and Vella (1999), Blundell and Powell (2003, 2004), Imbens (2007), Imbens and Newey (2009), Rothe (2009) and Kasy (2011) for examples of nonparametric identification using control function approach). Caetano, Rothe and Yildiz (Rochester and Columbia) Test for Identification with Endogeneity 04/09/ / 25
4 Introduction Control function approach is often used for identifying structural quantities in models with endogeneity. (See Newey, Powell, and Vella (1999), Blundell and Powell (2003, 2004), Imbens (2007), Imbens and Newey (2009), Rothe (2009) and Kasy (2011) for examples of nonparametric identification using control function approach). In a general triangular model, the conditions that are necessary to justify a control variable approach have no testable implications. Caetano, Rothe and Yildiz (Rochester and Columbia) Test for Identification with Endogeneity 04/09/ / 25
5 Introduction Control function approach is often used for identifying structural quantities in models with endogeneity. (See Newey, Powell, and Vella (1999), Blundell and Powell (2003, 2004), Imbens (2007), Imbens and Newey (2009), Rothe (2009) and Kasy (2011) for examples of nonparametric identification using control function approach). In a general triangular model, the conditions that are necessary to justify a control variable approach have no testable implications. In this paper, we argue that it is sometimes possible to test the validity of the control variable approach. Caetano, Rothe and Yildiz (Rochester and Columbia) Test for Identification with Endogeneity 04/09/ / 25
6 Introduction Control function approach is often used for identifying structural quantities in models with endogeneity. (See Newey, Powell, and Vella (1999), Blundell and Powell (2003, 2004), Imbens (2007), Imbens and Newey (2009), Rothe (2009) and Kasy (2011) for examples of nonparametric identification using control function approach). In a general triangular model, the conditions that are necessary to justify a control variable approach have no testable implications. In this paper, we argue that it is sometimes possible to test the validity of the control variable approach. To the best of our knowledge, our paper is the first to propose a test for this type of hypothesis. Caetano, Rothe and Yildiz (Rochester and Columbia) Test for Identification with Endogeneity 04/09/ / 25
7 Control function approach (Imbens and Newey (2009)) Y = m(x,u) X = h(z,v) Caetano, Rothe and Yildiz (Rochester and Columbia) Test for Identification with Endogeneity 04/09/ / 25
8 Control function approach (Imbens and Newey (2009)) Y = m(x,u) X = h(z,v) Although X and U may be dependent, it is assumed that (i) Z U, V, (ii) V is scalar and continuously distributed, (iii) h is strictly increasing in V, and (iv) the support of the distribution of V is large. Caetano, Rothe and Yildiz (Rochester and Columbia) Test for Identification with Endogeneity 04/09/ / 25
9 Control function approach (Imbens and Newey (2009)) Y = m(x,u) X = h(z,v) Although X and U may be dependent, it is assumed that (i) Z U, V, (ii) V is scalar and continuously distributed, (iii) h is strictly increasing in V, and (iv) the support of the distribution of V is large. These assumptions imply that V is identifiable, and that U X V. Caetano, Rothe and Yildiz (Rochester and Columbia) Test for Identification with Endogeneity 04/09/ / 25
10 Control function approach (Imbens and Newey (2009)) Y = m(x,u) X = h(z,v) Although X and U may be dependent, it is assumed that (i) Z U, V, (ii) V is scalar and continuously distributed, (iii) h is strictly increasing in V, and (iv) the support of the distribution of V is large. These assumptions imply that V is identifiable, and that U X V. Identification of functions of m are obtained through integration over V. For example, ASF(x)=E[m(x,U)]= E[Y X = x,v = v]df V (v) Caetano, Rothe and Yildiz (Rochester and Columbia) Test for Identification with Endogeneity 04/09/ / 25
11 Summary We devise a direct test of the primitive conditions which yield the key identifying condition U X V The requirements are that X is continuously distributed. m(x,u) is continuous in in x with probability one. The test has power if F U X,V ( X = x,v) is discontinuous in x at zero. Caetano, Rothe and Yildiz (Rochester and Columbia) Test for Identification with Endogeneity 04/09/ / 25
12 Outline 1 Introduction 2 Intuition 3 Control Function 4 Test Parameter 5 The Test 6 Conclusion Caetano, Rothe and Yildiz (Rochester and Columbia) Test for Identification with Endogeneity 04/09/ / 25
13 Identification condition Let Y = m(x,u) The test idea is similar to the exogeneity test developed by Caetano (2012). That paper implements a test of the condition U X W, where W is a vector of observable controls. Caetano, Rothe and Yildiz (Rochester and Columbia) Test for Identification with Endogeneity 04/09/ / 25
14 Identification condition Let Y = m(x,u) The test idea is similar to the exogeneity test developed by Caetano (2012). That paper implements a test of the condition U X W, where W is a vector of observable controls. The key identification condition in the control function approach is analogous U X V, with the added difficulty that the control function V is unobservable, and must be estimated. Caetano, Rothe and Yildiz (Rochester and Columbia) Test for Identification with Endogeneity 04/09/ / 25
15 Intuition Assumption 1: continuous birth weight cigarettes unobservables Assumption 2: discontinuous Caetano, Rothe and Yildiz (Rochester and Columbia) Test for Identification with Endogeneity 04/09/ / 25
16 Intuition Assumption 1: continuous birth weight expected birth weight among the non-smoking mothers expected birth weight cigarettes unobservables 0 cigarettes Assumption 2: discontinuous Caetano, Rothe and Yildiz (Rochester and Columbia) Test for Identification with Endogeneity 04/09/ / 25
17 Intuition Assumption 1: continuous birth weight expected birth weight among the non-smoking mothers V expected birth weight cigarettes? unobservables 0 cigarettes Caetano, Rothe and Yildiz (Rochester and Columbia) Test for Identification with Endogeneity 04/09/ / 25
18 Intuition Assumption 1: continuous birth weight expected birth weight among the non-smoking mothers for a fixed value of V cigarettes? V unobservables 0 expected birth weight for a fixed value of V cigarettes Caetano, Rothe and Yildiz (Rochester and Columbia) Test for Identification with Endogeneity 04/09/ / 25
19 Expected birthweight per amount smoked Caetano, Rothe and Yildiz (Rochester and Columbia) Test for Identification with Endogeneity 04/09/ / 25
20 Caetano (2012) separable case, no controls Y = m(x)+u Caetano, Rothe and Yildiz (Rochester and Columbia) Test for Identification with Endogeneity 04/09/ / 25
21 Caetano (2012) separable case, no controls Y = m(x)+u potentially endogenous unobservables Caetano, Rothe and Yildiz (Rochester and Columbia) Test for Identification with Endogeneity 04/09/ / 25
22 Caetano (2012) separable case, no controls birth weight Y = m(x)+u cigarettes? Caetano, Rothe and Yildiz (Rochester and Columbia) Test for Identification with Endogeneity 04/09/ / 25
23 Caetano (2012) separable case, no controls Y = m(x)+u Suppose that X is a continuous r.v. and m is continuous, then Caetano, Rothe and Yildiz (Rochester and Columbia) Test for Identification with Endogeneity 04/09/ / 25
24 Caetano (2012) separable case, no controls Y = m(x)+u Suppose that X is a continuous r.v. and m is continuous, then if X is exogenous,e[y X = x]=m(x)+e[u X = x] must be continuous in x. Caetano, Rothe and Yildiz (Rochester and Columbia) Test for Identification with Endogeneity 04/09/ / 25
25 Caetano (2012) separable case, no controls Y = m(x)+u Suppose that X is a continuous r.v. and m is continuous, then if X is exogenous,e[y X = x]=m(x)+e[u X = x] must be continuous in x. Additionally, suppose that when X is endogenous, E[U X = x] is discontinuous in x at zero, then Caetano, Rothe and Yildiz (Rochester and Columbia) Test for Identification with Endogeneity 04/09/ / 25
26 Caetano (2012) separable case, no controls Y = m(x)+u Suppose that X is a continuous r.v. and m is continuous, then if X is exogenous,e[y X = x]=m(x)+e[u X = x] must be continuous in x. Additionally, suppose that when X is endogenous, E[U X = x] is discontinuous in x at zero, then if X is endogenous,e[y X = x]=m(x)+e[u X=x] must be discontinuous in x at zero. Caetano, Rothe and Yildiz (Rochester and Columbia) Test for Identification with Endogeneity 04/09/ / 25
27 Caetano (2012) separable case, no controls Y = m(x)+u Suppose that X is a continuous r.v. and m is continuous, then if X is exogenous,e[y X = x]=m(x)+e[u X = x] must be continuous in x. Additionally, suppose that when X is endogenous, E[U X = x] is discontinuous in x at zero, then if X is endogenous,e[y X = x]=m(x)+e[u X=x] must be discontinuous in x at zero. Caetano (2012) s test of exogeneity: ise[y X = x] discontinuous at zero? Caetano, Rothe and Yildiz (Rochester and Columbia) Test for Identification with Endogeneity 04/09/ / 25
28 This paper separable case, no controls Y = m(x)+u probably endogenous Caetano, Rothe and Yildiz (Rochester and Columbia) Test for Identification with Endogeneity 04/09/ / 25
29 This paper separable case, no controls Y = m(x)+u Suppose that X is a continuous r.v. and m is continuous, then Caetano, Rothe and Yildiz (Rochester and Columbia) Test for Identification with Endogeneity 04/09/ / 25
30 This paper separable case, no controls Y = m(x)+u Suppose that X is a continuous r.v. and m is continuous, then if E[U X, V] = E[U V], E[Y X = x,v]=m(x)+e[u X = x,v]=m(x)+e[u V] must be continuous in x. Caetano, Rothe and Yildiz (Rochester and Columbia) Test for Identification with Endogeneity 04/09/ / 25
31 This paper separable case, no controls Y = m(x)+u Suppose that X is a continuous r.v. and m is continuous, then if E[U X, V] = E[U V], E[Y X = x,v]=m(x)+e[u X = x,v]=m(x)+e[u V] must be continuous in x. Additionally, suppose that when X is endogenous, E[U X = x, V] is discontinuous in x at zero, then Caetano, Rothe and Yildiz (Rochester and Columbia) Test for Identification with Endogeneity 04/09/ / 25
32 This paper separable case, no controls Y = m(x)+u Suppose that X is a continuous r.v. and m is continuous, then if E[U X, V] = E[U V], E[Y X = x,v]=m(x)+e[u X = x,v]=m(x)+e[u V] must be continuous in x. Additionally, suppose that when X is endogenous, E[U X = x, V] is discontinuous in x at zero, then if X is endogenous,e[y X = x,v]=m(x)+e[u X = x,v] must be discontinuous in x at zero. Caetano, Rothe and Yildiz (Rochester and Columbia) Test for Identification with Endogeneity 04/09/ / 25
33 This paper separable case, no controls Y = m(x)+u Suppose that X is a continuous r.v. and m is continuous, then if E[U X, V] = E[U V], E[Y X = x,v]=m(x)+e[u X = x,v]=m(x)+e[u V] must be continuous in x. Additionally, suppose that when X is endogenous, E[U X = x, V] is discontinuous in x at zero, then if X is endogenous,e[y X = x,v]=m(x)+e[u X = x,v] must be discontinuous in x at zero. Our test: is E[Y X = x, V] discontinuous at zero? Caetano, Rothe and Yildiz (Rochester and Columbia) Test for Identification with Endogeneity 04/09/ / 25
34 When does the test have power? Y = m(x,u) When is F(U X = x,v) discontinuous in x at a known point? joint cumulative distribution function Caetano, Rothe and Yildiz (Rochester and Columbia) Test for Identification with Endogeneity 04/09/ / 25
35 When does the test have power? Y = m(x,u) When is F(U X = x,v) discontinuous in x at a known point? E(U X = x, V) x x Caetano, Rothe and Yildiz (Rochester and Columbia) Test for Identification with Endogeneity 04/09/ / 25
36 When does the test have power? Y = m(x,u) When is F(U X = x,v) discontinuous in x at a known point? 1. Censoring: X = max{0,x } E(U X = x,v) E(U X = x, V) x x Caetano, Rothe and Yildiz (Rochester and Columbia) Test for Identification with Endogeneity 04/09/ / 25
37 When does the test have power? Y = m(x,u) When is F(U X = x,v) discontinuous in x at a known point? 1. Censoring: X = max{0,x }, X = daily number of cigarettes U includes mother s type expected type of non-smoking mothers expected type of mothers per number of cigarettes smoked cigarettes Caetano, Rothe and Yildiz (Rochester and Columbia) Test for Identification with Endogeneity 04/09/ / 25
38 Censoring example: maternal smoking Mother s education Mother s marital status years of education years of education Father s education cigarettes per day cigarettes per day likelihood of drinking during pregnancy likelihood of not being married Mother s drinking cigarettes per day cigarettes per day Caetano, Rothe and Yildiz (Rochester and Columbia) Test for Identification with Endogeneity 04/09/ / 25
39 Other examples generated by censoring: Observations that chose the threshold value may be constrained, and are therefore discontinuously different form those observations that chose values just above the threshold. Natural constraints: zero Consumption goods and services Other: hours of work, days breastfeeding, number of crimes in a neighborhood Caetano and Maheshri (2012). Law-imposed constraints: Years of education (due to minimum attendance laws and minimum age for working) Caetano and Caetano (2012). Caetano, Rothe and Yildiz (Rochester and Columbia) Test for Identification with Endogeneity 04/09/ / 25
40 Outline 1 Introduction 2 Intuition 3 Control Function 4 Test Parameter 5 The Test 6 Conclusion Caetano, Rothe and Yildiz (Rochester and Columbia) Test for Identification with Endogeneity 04/09/ / 25
41 Problem We can test the identification condition U X V in models of the type Y = m(x,u) X = max{0,h(z,v)} All the previously cited papers use the control function approach for models in which h is strictly monotone in V. What variable is suitable as a possible control function in our model? How can we identify a valid control function? Caetano, Rothe and Yildiz (Rochester and Columbia) Test for Identification with Endogeneity 04/09/ / 25
42 Control Function Assumption 1 1 Z and (U, V) are stochastically independent. 2 V is scalar and continuously distributed, and without loss of generality its distribution is normalized such that V Unif[0, 1]. 3 The function v h(z, v) is strictly increasing with probability 1. Assumption 1 is analogous to the conditions of Theorem 1 of Imbens and Newey (2009), and is central for the validity of control variable arguments. It implies that U X V. V is not identified in this model, but V := V 1(X > 0) is identified through the relationship V = F X Z (X;Z)1(X > 0). Taken together, these two statements imply that U X (V,X > 0). Caetano, Rothe and Yildiz (Rochester and Columbia) Test for Identification with Endogeneity 04/09/ / 25
43 Control Function Assumption 2 1 The support Supp(V X = x) of V conditional on X = x is equal to [0,1] for all x Supp(X X > 0). 2 The function x m(x, U) is continuous at x = 0 with probability 1. Assumption 2(1) is analogous Assumption 2 of Imbens and Newey (2009). This assumption is testable. Assumption 2(2) is new, and allows us to identify the ASF at zero in spite of the cesoring. Caetano, Rothe and Yildiz (Rochester and Columbia) Test for Identification with Endogeneity 04/09/ / 25
44 Identification Result Proposition 1 1 Suppose that Assumption 1 and Assumption 2(1) hold. Then the ASF(x) = E[m(x, U)] is identified for all x Supp(X X > 0). 2 Suppose that Assumptions 1 and 2 hold. Then the ASF(x)=E[m(x,U)] is identified for all x Supp(X). A sketch of the identification argument is that E[m(x,U)]=E[E[m(x,U) V]]= m(x,u)df U V (u v)df V (v) = m(x,u)df U X,V (u x,v)df V (v)= E[Y X = x,v = v]df V (v) and since V is identified for X > 0, so is the ASF(x) for x>0. The identification of the ASF(0) follows from continuity. Caetano, Rothe and Yildiz (Rochester and Columbia) Test for Identification with Endogeneity 04/09/ / 25
45 Outline 1 Introduction 2 Intuition 3 Control Function 4 Test Parameter 5 The Test 6 Conclusion Caetano, Rothe and Yildiz (Rochester and Columbia) Test for Identification with Endogeneity 04/09/ / 25
46 The test H 0 : Assumption 1 is satisfied H 1 : Assumption 1 is violated Assumption 2 is maintained. As discussed before, we base the test on the discontinuity. (V)=lim x 0 E[Y X = x,v] E[Y X = 0,V] If Assumption 1 holds, then X U V, and thus (V)=0. Caetano, Rothe and Yildiz (Rochester and Columbia) Test for Identification with Endogeneity 04/09/ / 25
47 Power of the test X = max{0,h(z,v)} { E[m(x,U) h(z,v)=x,v = v] if x>0 E[Y X = x,v = v]= E[m(x,U) h(z,v) x,v = v] if x=0. Since the conditioning sets in the two conditional expectations on the right-hand side of the previous equation differ, we would in general expect the function x E(Y X = x, V = v) to be discontinuous at x = 0 for at least some (and potentially all) v [0, 1]. Caetano, Rothe and Yildiz (Rochester and Columbia) Test for Identification with Endogeneity 04/09/ / 25
48 Problem The test could be based on (V)=lim x 0 E[Y X = x,v] E[Y X = 0,V] Unfortunately, under H 0 we are only able to identify the censored variable V1(X > 0), but not V itself, and thus this approach is not feasible: while identification of V1(X > 0) is sufficient for identifying E[Y X = x, V = v] for x > 0, it does not suffice for learning E[Y X = 0,V = v]. Caetano, Rothe and Yildiz (Rochester and Columbia) Test for Identification with Endogeneity 04/09/ / 25
49 Aggregation trick (V)=lim x 0 E[Y X = x,v] E[Y X = 0,V] We therefore consider testing H 0 by checking a necessary condition for (V) = 0, namely that = 0, where := = (v)df V X (v 0) lime[y X = x,v = v]df V X (v 0) E[Y X = 0]. x 0 Although this aggregation may lose some power, it is identifiable. E[Y X = x,v = v] is identifiable for x>0, because V = F X Z (X Z) is identified when X > 0. Caetano, Rothe and Yildiz (Rochester and Columbia) Test for Identification with Endogeneity 04/09/ / 25
50 Identifying the measure The conditional CDF F V X ( ;0) of V given X = 0 is identified. To see this, recall that by Assumption 1(2) the marginal distribution of V is normalized to be the uniform distribution on the interval [0,1], and thus its unconditional CDF F V is known. V = V1(X > 0)=F X Z (X Z)1(X > 0) is identified. It then follows from the Law of Total Probabiliy that F V X (v 0)= v P(V v,x > 0). P(X = 0) Caetano, Rothe and Yildiz (Rochester and Columbia) Test for Identification with Endogeneity 04/09/ / 25
51 More on power Recall = lime[y X = x,v = v]df V X (v 0) E[Y X = 0]. x 0 When the test has no power: Pathological cases: the conditional distribution of Y given (X,V ) is such that the integral in is incidentally equal toe[y X = 0] even though V is not a control function. More meaningful cases: if U X V, but U X (V,0 X < δ) for some small δ > 0. Our test is thus generally unable to detect violations of the null hypothesis where V is only able to capture the dependence between X and U local to X = 0. We would argue, however, that one is unlikely to encounter such alternatives in empirical applications. Caetano, Rothe and Yildiz (Rochester and Columbia) Test for Identification with Endogeneity 04/09/ / 25
52 Outline 1 Introduction 2 Intuition 3 Control Function 4 Test Parameter 5 The Test 6 Conclusion Caetano, Rothe and Yildiz (Rochester and Columbia) Test for Identification with Endogeneity 04/09/ / 25
53 Test Statistic Recall that = Define lime[y X = x,v = v]df V X (v 0) E[Y X = 0]. x 0 ˆm + Y X,V (0,V)=ÊLL h [Y X = 0 +, ˆV = v] 1 ˆm Y X (0)= n ˆP(X = 0) n i=1 Y i 1(X i = 0), Caetano, Rothe and Yildiz (Rochester and Columbia) Test for Identification with Endogeneity 04/09/ / 25
54 Test Statistic Recall that = Define lime[y X = x,v = v]df V X (v 0) E[Y X = 0]. x 0 ˆm + Y X,V (0,V)=ÊLL h [Y X = 0 +, ˆV = v] 1 ˆm Y X (0)= n ˆP(X = 0) n i=1 Y i 1(X i = 0), ˆ = ˆm + Y X,V (0,v)d ˆF V X (v 0) ˆm Y X (0) Caetano, Rothe and Yildiz (Rochester and Columbia) Test for Identification with Endogeneity 04/09/ / 25
55 Test Statistic Recall that = Define lime[y X = x,v = v]df V X (v 0) E[Y X = 0]. x 0 ˆm + Y X,V (0,V)=ÊLL h [Y X = 0 +, ˆV = v] 1 ˆm Y X (0)= n ˆP(X = 0) n i=1 Y i 1(X i = 0), ˆ = ˆm + Y X,V (0,v)d ˆF V X (v 0) ˆm Y X (0) Recall the measure identification: F V X (v 0)= v P(V v,x > 0) P(X = 0) Caetano, Rothe and Yildiz (Rochester and Columbia) Test for Identification with Endogeneity 04/09/ / 25
56 Test Statistic Recall that = Define lime[y X = x,v = v]df V X (v 0) E[Y X = 0]. x 0 ˆm + Y X,V (0,V)=ÊLL h [Y X = 0+, ˆV = v] 1 ˆm Y X (0)= n ˆP(X = 0) n i=1 Y i 1(X i = 0), ˆ = 1 1 ˆP(X = 0) 0 ˆm + Y X,V (0,v)dv 1 n n i=1 ˆm + Y X,V (0, ˆV i )1(X i > 0) ˆm Y X (0) Recall the measure identification: F V X (v 0)= v P(V v,x > 0) P(X = 0) Caetano, Rothe and Yildiz (Rochester and Columbia) Test for Identification with Endogeneity 04/09/ / 25
57 Asymptotic distribution Theorem 1 Under some standard regularity conditions, the following statements hold: 1 Under the null hypothesis, i.e. = 0, ( nh ˆ d N 0,C σ + 2 (0) ) f X + (0) 2 Under any fixed alternative that implies 0, ( lim P ) nh ˆ >c = 1 n Caetano, Rothe and Yildiz (Rochester and Columbia) Test for Identification with Endogeneity 04/09/ / 25
58 Asymptotic variance The asymptotic variance of ˆ is C σ + 2 (0) f X +, where (0) with, for j {0,1,2}, σ 2 +(0)= lim x 0 E(η 2 X = x), η =(Y E(Y X = x,v = v)f V X (V,0)/f + V X (V,0), C= κ2 2 λ 0 2κ 1 κ 2 λ 1 + κ 2 1 λ 2 (κ 2 κ 0 κ 2 1 )2, κ j = x j K (x)dx and λ j = x j K (x) 2 dx, 0 0 Caetano, Rothe and Yildiz (Rochester and Columbia) Test for Identification with Endogeneity 04/09/ / 25
59 Outline 1 Introduction 2 Intuition 3 Control Function 4 Test Parameter 5 The Test 6 Conclusion Caetano, Rothe and Yildiz (Rochester and Columbia) Test for Identification with Endogeneity 04/09/ / 25
60 Conclusion Proposed a set up which causes unobservables in outcome equation to change discontinuously. Gave conditions for identification of a control function as well as ASF (could also do QSF). Proposed novel tests of the identifying assumption. Thank you! Caetano, Rothe and Yildiz (Rochester and Columbia) Test for Identification with Endogeneity 04/09/ / 25
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