Problem Set - Instrumental Variables
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1 Problem Set - Instrumental Variables 1. Consider a simple model to estimate the effect of personal computer (PC) ownership on college grade point average for graduating seniors at a large public university: GP A = β 0 + β 1 P C + u where PC is a binary variable indicating PC ownership. (a) Why might PC ownership be correlated with u? (b) Explain why PC is likely to be related to parents annual income. Does this mean parental income is a good IV for PC? Why or why not? (c) Suppose that, four years ago, the university gave grants to buy computers to roughly one-half of the incoming students, and the students who received grants were randomly chosen. Carefully explain how you would use this information to construct an instrumental variable for PC. (a) More able, more motivated, or more wealthy students may have PC s, all of which may be associated with their GPA. (b) Richer parents can better afford PC s. But they may also (i) better afford other inputs, (ii) be a select group who perform differently, (iii) install vices or virtues that influence GPA. (c) Denote by D = 1 and D = 0 students given grants and not given grants, respectively. Since grants are random, E (ɛ D) = 0. Therefore E (GP A D = 1) = β 0 + β 1 E (P C D = 1) E (GP A D = 0) = β 0 + β 1 E (P C D = 0) E (GP A D = 1) E (GP A D = 0) = β 1 [E (P C D = 1) E (P C D = 0)] β 1 = E (GP A D = 1) E (GP A D = 0) E (P C D = 1) E (P C D = 0) A natural consistent estimator of β 1 is then the Wald estimator, using the sample analogues of the means β 1 = GP A (D=1) GP A (D=0) P C (D=1) P C (D=0) 1
2 2. Suppose that you wish to estimate the effect of class attendance on student performance. A basic model is examscore = β 0 + β 1 attendance + β 2 priorgp A + u where examscore is students score on the exam (from 1 to 6), attendance is the number of seminar meetings attended (from 1 to 12), and priorgp A is the average exam grade last year. (a) Let dist be the distance from the students living quarters to the lecture hall. Do you think dist is uncorrelated with u? (b) Assuming that dist and u are uncorrelated, what other assumption must dist satisfy in order to be a valid IV for attendance? (c) Suppose, we add the interaction term priorgp A attendance. If attendance is correlated with u, then, in general, so is priorgp A attendance. What might be a good IV for priorgp A attendance? (a) If living far away implies lower socioeconomic status, then probably not. If it is mixed, or if we observe most of those relevant dimensions, then perhaps. (b) Relevance (i.e. cov(attendance, dist) 0: Living far away from the lecture hall will likely affect attendance, so it should be a relevant instrument. (c) Use as an instrument priorgp A dist. 3. Consider the model y 1 = z 1 δ 1 + α 1 y 2 + u 1 (1) y 2 = zπ 2 + v 2 (2) where z 1 z. (a) How would you estimate α 1 using 2SLS? What is your instrument? i. Regress y 2 on z and form the linear predictions ŷ 2. ii. Regress y 1 on z 1 and ŷ 2. The instruments are the set of variables in z but not in z 1. 2
3 (b) Consider an alternative estimator of (δ 1, α 1 ): (a) estimate equation (2) by OLS and save the residuals ˆv 2. (b) estimate the following equation by OLS y 1 = z 1 δ 1 + α 1 y 2 + ρˆv 2 + error. Show that the OLS estimates of δ 1 and α 1 from this regression are identical to the 2SLS estimators. (Hint: Use the partitioned regression algebra of OLS. In particular, if ŷ = x 1 ˆβ1 +x 2 ˆβ2 is an OLS regression, ˆβ 1 can be obtained by first regressing x 1 on x 2, getting the residuals, say ẍ 1 and then regressing y on ẍ 1. You must also use the fact that z 1 and ˆv 2 are orthogonal in the sample.) We want to estimate the parameters δ 1 and α 1. Using the hint, the OLSestimates of these parameters can also be obtained by partitioned regression: i. For every variable in x 1 (z 1, y 2 ), regress it onto ˆv 2 and save the residual. Collect all the residuals in ẍ 1. ii. Regress y 1 onto the residuals ẍ 1. By definition, ˆv 2 is orthogonal in sample to z. Therefore, the residuals from regressing z 1 onto ˆv 2 are just z 1. (More precisely, n i=1 z 1iˆv 2i = 0.) Also by definition, y 2 = ŷ 2 + ˆv 2 where ŷ 2 and ˆv 2 are orthogonal in sample, which implies that the residuals from regressing y 2 onto ˆv 2 are simply the first stage fitted values, ŷ 2. In other words, ẍ 1 = (z 1, ŷ 2 ). The 2SLS estimator of β 1 is obtained exactly from the OLS regression of y 1 on (z 1, ŷ 2 ). The intuition is the following: ˆv 2 contains all the endogenous variation in y 2. Since it is also orthogonal (by definition) to all the other included regressors, controlling for ˆv 2 is sufficient to control for the endogeneity bias. This illustrates partitioned regression. There is a much simpler way: y 1 = z 1 δ + αy 2 + ρˆv 2 + error = z 1 δ + αŷ 2 + ρˆv 2 + error where ρ = ρ + α. Since cov (ˆv 2, ŷ 2 ) = cov (ˆv 2, z 1 ) ˆπ 2,1 = 0 in sample, omitting ˆv 2 from the regression does not influence the estimates of δ and α. But then this is just the 2SLS-regression. 3
4 4. (Difficult) Consider the following structural equation y i = β 0 + β 1 x 1i + β 2 x 2i + u i where cov(x 1i, u i ) 0 and cov(x 2i, u i ) = 0. You have an instrumental variable z i for x 1i : cov(z i, x 1i ) 0 and cov(z i, u i ) = 0. Which you would use to estimate the first-stage x 1i = π 0 + π 1 z i + π 2 x 2i + v i (a) Show that plim ˆβ 1 = β 1 but plim ˆβ 2 β 2 when x 2 is included in the first-stage estimation, but not used to construct the instrument ˆx 1 (i.e., ˆx 1 = ˆπ 0 + ˆπ 1 z 1 ) (b) Show that both plim β 1 β 1 and plim β 2 β 2 when x 2 is not included in the first-stage estimation (Hint: Remember that x 1 = ˆx 1 +(x 1 ˆx 1 ) which means that the second stage becomes y i = β 0 + β 1ˆx 1i + β 2 x 2i + u i + β 1 (x 1i ˆx 1i ) }{{} ũ i Hint: The 2SLS estimate of β 1 is ˆβ 1 = n i=1 ˆr 1iy i / n i=1 ˆr2 1i where ˆr 1i = ˆx 1i ˆα 0 ˆα 1 x 2i, the residual from the following regression: ˆx 1i = α 0 + α 1 x 2i + r 1i. The 2SLS estimate of β 2 has a similar expression.) The hint gives us y = β 0 + β 1 x 1 + β 2 x 2 + u = β 0 + β 1 x 1 + β 2 x 2 + [β 1 (x 1 ˆx 1 ) + u] = β 0 + β 1ˆx 1 + β 2 + ũ It also uses partitioned regression (Frisch Waugh Lovell theorem), as in the previous exercise. Define ˆr 1 = ˆx 1 ˆα 0 ˆα 1 x 2 as the residual from a regression of ˆx 1 onto x 2. The theorem then says that the OLSestimate of β 1 from the original regression is equivalent to the estimator from the 4
5 bivariate regression of y on ˆr 1, ie. n ˆβ 1 = Mˆr 1,y i = i=1 ˆr 1iy i n. i=1 ˆr2 1i n i=1 = ˆr 1i (β 0 + β 1ˆr 1i + ũ) n i=1 ˆr2 1i = β 1 + Mˆr 1,ũ so that ˆβ 1 is consistent if and only if Mˆr1,ũ p 0. Inserting from ũ above, ˆβ 1 = β 1 + Mˆr 1,β 1 (x 1 ˆx 1 )+u = β 1 + β 1 Mˆr1,x 1 ˆx 1 + Mˆr 1,u (3) Since Mˆr1,u = ˆπ 1 M z,u p 0 in both (a) and (b), I will disregard this term below. Similarly, we can show that ˆβ 2 = β 2 + β 1 Mˆr2,x 1 ˆx 1 Mˆr2,ˆr 2 (a) In (a), we have that ˆx 1 = ˆπ a 0 + ˆπ a 1z, while the correct first stage is x 1 = π 0 + π 1 z + π 2 x 2 + v. Then, x 1 ˆx 1 = (π 0 ˆπ a 0) + (π 1 ˆπ a 1) z + π 2 x 2 + v. Since we estimate the correct first stage, plim ˆπ a s = π s for s = 0, 1, 2. Then plim (x 1 ˆx 1 ) = π 2 x 2 + v. Therefore, plim ˆβ a 1 = β 1 + β 1 cov (ˆr 1, π 2 x 2 + v) varˆr 1 = β 1 plim ˆβ a 2 = β 2 + β 1 cov (ˆr 2, π 2 x 2 + v) varˆr 2 = β 2 + β 1 cov (ˆr 2, π 2 (ˆx 2 + ˆr 2 )) varˆr 2 = β 2 + β 1 π 2 where the third equality uses that x 2 can be decomposed into the orthogonal parts ˆx 2 + ˆr 2, and that v is independent of ˆr 2. 5
6 (b) To solve (b) exactly is a bit more complicated, but the path is very similar. We first note that plim ˆπ b 1 = π 1 + π 2 cov (z, x 2 ) var (x 2 ) π 1 + B 1 π 1 which implies that plim Mˆr1,x 1 ˆx 1 = B 1 cov (ˆr 1, z) which is obviously nonzero from the definition of ˆr 1, any time B 1 0. This is enough to show that the estimator is inconsistent, and shows that we get a consistent estimate of β 1 only if the instrument z is uncorrelated with x 2 (or, obviously, if x 2 is uncorrelated with x 1, i.e. π 2 = 0, in which case we don t need x 2 in the first OR the second stage). Therefore, if the exclusion restriction holds conditional on x 2, we need to control for x 2 in the first stage! To get the exact bias, we derive the following { plim ˆπ 0 b = π 0 + π 2 E [x 2 ] E [z] cov (z, x } 2) var (x 2 ) plim (x 1 ˆx 1 ) = B 0 B 1 z + π 2 x 2 + v. plim = plim Mˆr1,ˆπ b 0 +ˆπb 1 z ˆα 0 ˆα 1 x 2 = plim (ˆπ b 1) plim Mˆr1,z = (π 1 + B 1 ) cov (ˆr 1, z) which gives the probability limit of the estimator, plim ˆβ 1 b B 1 cov (ˆr 1, z) = β 1 β 1 (π 1 + B 1 ) cov (ˆr 1, z) ( ) π1 = β 1 π 1 + B ( 1 ) π 1 = β 1 π 1 + π 2 cov (z, x 2 ) /var (x 2 ) π 0 + B 0 π 0 where the final equality just replaces B 1 with its definition from above. For ˆβ 2 we proceed similarly. 6
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