Ability Bias, Errors in Variables and Sibling Methods James J. Heckman University of Chicago Econ 312 This draft, May 26, 2006 1
1 Ability Bias Consider the model: log = 0 + 1 + where =income, = schooling, and 0 and 1 are parameters of interest. What we have omitted from the above specification is unobserved ability, which is captured in the residual term.wethusre-writetheaboveas: log = 0 + 1 + + where is ability, ( 0 ) ( 0), andwebelievethat ( ) 6= 0.Thus,( ) 6= 0,sothatOLS on our original specification gives biased and inconsistent estimates. 2
1.1 Strategies for Estimation 1. Use proxies for ability: Find proxies for ability and include them as regressors. Examples may include: height, weight, etc. The problem with this approach is that proxies may measure ability with error and thus introduce additional bias (see Section 1.3). 3
2. Fixed Eect Method: Find a paired comparison. Examples may include a genetic twin or sibling with similar or identical ability. Consider two individuals and 0 : log log 0 = ( 0 + 1 + ) ( 0 + 1 0 + 0 ) = 1 ( 0)+( 0)+( 0 ) Note: if = 0,thenOLS performedonourfixedeect 4
estimator is unbiased and consistent. If 6= 0,thenwe just get a dierent bias (see Section 1.2). Further, if is measured with error, we may exacerbate the bias in our fixed eect estimator (see Section 1.3). 1.2 OLS vs. Fixed Eect (FE) In the OLS case with ability bias, we have: plim ( 1 )= 1 + ( ) () (See derivation of Equation (2.2) for more background on the above derivation). 5
We also impose: () = ( 0 ) ( ) = ( 0 0 ) ( 0 ) = ( 0 ) With these assumptions, our fixed eect estimator is given by: plim 1 = 1 + ( 0 ( 0 )+( 0 )) ( 0 ) = 1 + ( ) (0 ) () ( 0 ). Note that if ( 0 )=0 and ability is positively correlated with schooling, then the fixed eect estimator is upward biased. 6
From the preceding, we see that the fixed eect estimator has more asymptotic bias if: ( ) ( 0 ) () ( 0 ) ( ) () ( ) () (0 ) ( 0 ). 7
1.3 Measurement Error Say = + where is observed schooling. Our model now becomes: log = 0 + 1 + = 0 + 1 +( + 1 ) and the fixed eect estimator gives: log log 0 = ( 0 + 1 + ) ( 0 + 1 0 + 0 ) = 1 ( 0 )+( 0 )+ 1 ( 0 ) Now we wish to examine which estimator (OLS or fixed eect), has more asymptotic bias given our measurement error problem. For the remaining arguments of this section, we assume: ( ) = ( 0 ) = ( 0 )=0 so that the OLS estimator gives: 8
plim 1 = 1 + ( + 1 ) ( ) = 1 + ( ) 1(). ()+() The fixed eect estimator gives: plim 1 = 1 + ³ 0 ( 0 )+ 1 ( 0 ) ( 0 ) = 1 + (( 0 ) ( 0 )) 1 ( 0 ) ( 0 )+( 0 ) = 1 + ( ) ( 0 ) 1 () ()+() ( 0 ). 9
Under what conditions will the fixed eectbiasbegreater? From the above, we know that this will be true if and only if: ( ) ( 0 ) 1 () ( ) 1() ()+() ( 0 ) ()+() ( 0 )(()+()) ( 1 () ( )) ( 0 ) ( ) 1() ()+() ( 0 ) ( 0 ). If this inequality holds, taking dierences can actually worsen the fit over OLS alone. Intuitively, we see that we have dierenced out the true component,, and compounded our measurement error problem with the fixed eect estimator. 10
In the special case = 0, the condition is 1 () ()+() ( 0 ) ( ) 1() ()+() 11
2 Errors in Variables 2.1 The Model Suppose that the equation for earnings is given by: = 1 1 + 2 2 + where ( 1 2 )=0 0.Alsodefine: 1 = 1 + 1 and 2 = 2 + 2 12
Here, 1 and 2 are observed and measure 1 and 2 with error. We also impose that. So, our initial model can be equivalently re-written as: = 1 1 + 2 2 +( 1 1 2 2 ). Finally, by assumed independence of and, wewrite: = +. 13
2.2 McCallum s Problem Question: Is it better for estimation of 1 to include other variables measured with error? Suppose that 1 is not measured with error, in the sense that 1 =0while 2 is measured with error. In 2.2.1 and 2.2.2 below, we consider both excluding and including 2 and investigate the asymptotic properties of both cases. 2.2.1 Excluded 2 The equation for earnings with omitted 2 is: = 1 1 +( + 2 2 ) 14
Therefore, by arguments similar to those in the appendix, we know: plim 1 = 1 + 12 2. (2.1) 11 Here, 12 is the covariance between the regressors, and 11 is thevarianceof 1 Before moving on to a more general model for the inclusion of 2 let us first consider the classical case for including both variables. Suppose = 11 0 11 0 0 =. 22 0 22 We know that: plim ˆ = ( ) 1 ( ) (2.2) 15
where the coecient and regressor vectors have been stacked appropriately (see Appendix for derivation). Note that represents the variance-covariance matrix of the measurement errors, and is the variance-covariance matrix of the regressors. Straightforward computations thus give: plim ˆ = = " 11 + 11 0 0 22 + 22 11 0 11 + 11 22 0 22 + 22 1 11 0 1 2 0 22 # 1 2 16
2.2.2 Included 2 In McCallum s problem we suppose that 12 =0 Further, as 1 is not measured with error, 11 =0 Substituting this into equation 2.2 yields: 1 plim ˆ 11 = 12 0 0 12 22 + 22 0 22 With a little algebra, the above gives: μ plim ˆ 12 1 = 1 + 2 22 11 22 + 22 2 12 μ μ 11 12 22 = 1 + 2 11 22 (1 2 12)+ 22 17
where 2 12 is simply the correlation coecient, we know that: 0 2 12 1 2 12 11 22 Further, so including 2 results in less asymptotic bias (inconsistency). (We get this result by comparing the above with the bias from excluding 2 in section 2.2.1, the result captured in equation (2.1)). So, we have justified the kitchen sink approach. This result generalizes to the multiple regressor case - 1 badly measured variable with good ones (Econometrica, 1972). 18
2.3 General Case In the most general case, we have: plim ˆ = ( ) 1 11 + = 11 12 + 1 12 11 12 + 12 22 + 12 22 12 22 With a little algebra we find: 1 2. det( )= 11 22 + 11 22+ 11 22 + 11 22 2 122 12 12 2 12 19
Therefore: plim ˆ = 1 det( ) 11 12 12 22 Supposing 12 =0 we get: 22 + 22 ( 12 + 12) ( 12 + 12) 11 + 11 1 2 det( )=det( ) 12 =0 = 11 22 + 11 22 + 11 22 + 11 22 2 12 20
and thus: plim ˆ = ( 22 + 22) 11 det( ) 11 12 det( ) 12 22 det( ) ( 11 + 11) 22 det( ) 1 2 Note that if 2 12 0 OLS may not be downward biased for 1.If 2 =0 we get: plim ˆ 2 = 1 12 11 det( ) so, if 2 were a race variable and blacks get lower quality schooling, (where schooling is measured by 1 ) then 12 0 and hence ˆ 2 0 This would be a finding in support of labor market discrimination. 21
2.4 The Kitchen Sink Revisited McCallum s analysis suggests that one should toss in a variable measured with error if there is no measurement error in 1 But suppose that there is measurement error in 1 Is it still better to include the additional variable measured with error as a regressor? We proceed by imposing 2 =0. (i) Excluded X 2. The equation for earnings with measurement error in 1 and excluded 2 is: = (1 + 1 ) 1 +( + 2 2 ) = 1 1 +( + 2 2 + 1 1 ) 22
Therefore: μ plim 1 = 1 11 1 11 + 11 = 1 1 1+ 11 11 μ 11 = 1 11 + 11 (2.3) (ii) Included X 2. From our analysis in the General Case (Section 2.3), we know that: μ plim ˆ (22 + 1 = 22) 11 2 12 1. (2.4) det( ) 23
If 22 =0 so that 2 is not measured with error: μ plim ˆ 11 22 2 12 1 = 1 11 22 2 12 + Ã! 11 22 1 2 12 = 1. 1 2 12 + 11 11 (2.5) Comparing eqn (2.4) and eqn (2.5), we see that adding the variable measured without error always exacerbates the bias. 24
For, the bias in the excluded case will be smaller if: 1 1 1+ 11 11 1 μ 1 2 12 + 11 11 0 2 12 1 2 12 1 2 12 + 11 μ 11 1+ 11 11 11 11 1 2 12 whichisalwaysthecase,provided 2 12 0 (Notethatthe coecients on 1 for both the excluded and included case are less than one. So, the larger coecient is the one with less bias, as stated above.) 25
Now suppose that 22 0 so that both variables are measured with error. Then: μ plim ˆ (22 + 1 = 22) 11 2 12 1 det( ) 1+ 22 2 12 22 = 1 1+ 11 11 + 11 11 22 22 + 22 22 2 12 Intuitively, adding measurement error in 2 can only worsen the bias, and thus exclusion should again be preferred to inclusion. Formally, including 2 givesmorebiasifandonly 26
if: 1 1+ 11 11 + 11 1+ 22 22 2 12 22 + 22 2 12 11 22 μ 22 1+ 11 11 1 μ 1+ 22 22 2 12 μ 1+ 11 11 + 11 11 22 22 + 22 22 2 12 1 1+ 11 11 2 12 11 11 0. 27
Thus, provided 2 12 0 including 2 resultsinmorebias than excluding it. If 2 12 =0 the bias from including 2 is obviously seen to be: 1 1+ 22 22 1+ 11 11 + 11 11 22 22 + 22 22 = 1 μ 1+ 22 = 1 1+ 22 22 22 μ 1+ 11 1 1+ 11 11 so that including and excluding 2 yields the same result. 11 28
Finally, from the General Case section, we have: plim ˆ 1 = 1 ( 22 + 22) 11 2 12 + 2 ( 12 22). 11 22 2 12 + 11 22 + 11 22 + 11 22 L Hôpital s rule on the above shows that: 11 lim ³plim 1 ˆ ³ lim plim 1 ˆ = 1 11 + 2 12 22 11 + 11 =0 and = 1 11 11 + 11 + 2 12. 11 + 11 29
Appendix Derivation of Equation (2.2) We can write = +( 1 1 2 2 ) where: = 1 2 and = 1 2 and 1 2 are 1. 30
So: ³ ˆ = 0 1 ( 0 ) = + ³ 0 1 ³ 0 ( 1 1 2 2 ) Ã 0! 1 = + μμ μ μ 0 0 1 1 0 2 2 ³ + ³ 0 1 ³ ³ 0 ³ 0 1 1 ³ 0 2 2 31
0 = 0 μ = 0 1 1 2 1 0 1 1 0 1 1 0 0 2 1 1 2 2 2 1 + 0 0 1 2 0 2 1 2 2 0 1 1 0 0 2 1 0 1 2 2 2 1 0 1 2 0 2 2 1 1 2 2 32
= 0 μ ( ) 1 1 + 0 0 1 1 1 + 0 1 2 0 2 + 0 0 2 1 2 + 0 2 2 1 2 = ( ) 1 ( ) where the second-to-last step follows from the independence of and This type of argument is also used to derive the probability limit of the s in section 1. 33
3 Sibling Models: Components of Variance Scheme Suppose that data on two brothers, say and is at our disposal Without loss of generality, we will consider how to estimate parameters of interest for person in what follows. We will begin by introducing a general model and then focus on the two-person case mentioned above. Consider the following triangular system: 1 = 1 2 = 12 1 + 2 3 = 13 1 + 23 2 + 3 34
Here, indexes the person in the group. We assume that and 0 are uncorrelated (i.e., uncorrelated across groups). Further, we suppose: = + = +, for =1 2 3. We assume is uncorrelated across equations and across within the group, is i.i.d. across groups, and is i.i.d. within groups and uncorrelated with. 35
3.1 Estimation We specialize the above model into a two person framework and propose a similar three equation system. Let 1 = early (preschool) test score, 2 = schooling (years), and 3 = earnings. It seems plausible to write the equation system 1 = + 1 2 = 2 + 2 3 = 23 2 + 3 + 3 where = ability. Regressing 3 on 2 clearly gives biased estimates of 23 as ( 2 ) 6= 0 If 3 0 then OLS estimates of 23 are upward biased. One estimation approach is to use 1 as a proxy for ability: 3 = 23 2 + 3 ( 1 1 )+ 3 36
However, this results in a similar problem regressing 3 on 1 and 2 will give biased estimates as 1 is correlated with our residual. (i.e., 1 is an imperfect proxy). Solutions: Onesolutionistouse 1 as an instrument for 1 Why is this a valid IV? From our construction of the model, we know that the are uncorrelated across equations and groups. Further, test scores are correlated across siblings. That is, ( 1 1 ) 6= 0by our group structure. Another solution is possible if there exists an additional early reading on the same person: 0 = 0 + 0 Then if 0 6=0 0 is a valid proxy for 1 and we can perform 2SLS. 37
3.2 Griliches and Chamberlain model Here we have a modified triangular system as follows: 1 = 1 + 1 2 = 12 1 + 2 + 2 3 = 13 1 + 23 2 + 3 + 3 where 1 = years schooling, 2 = late test score (SAT), and 3 = earnings. Notethattherearealternativemodelswith other dependent variables. For example, { 1 = schooling, 2 = early earnings, and 3 = late earnings}, and { 1 = schooling, 2 = consumption, and 3 = earnings}. Getting the equation system into reduced form and expressing as matrix notation, we write = +, 38
where: and: = = 1 2 3 1 2 3 = = 1 2 + 12 1 3 + 13 1 + 23 ( 2 + 12 1 ) 1 2 + 12 1 3 + 13 1 + 23 ( 2 + 12 1 ) Estimation. For estimation, we impose that 23 =0 In our second example of section 3.2, this would be equivalent to stating that there is no correlation between transient income and consumption (permanent income hypothesis). In general, with one factor, we need one more exclusion than that implied by triangularity. 39
(i) 1 proxies. so that = 1 1 1 2 = 2 1 1 2 1 1 + 2. We can then estimate 2 1 consistently by using 1 as an instrument for 1 in the equation above. (ii) Get residuals from (i): = 2 2 1 1. (iii). Use the residuals as an instrument for 1 in the 3 equation. is valid since it is both uncorrelated with and 3 and it is correlated with 1 : 40
μ ( 1 )= 1 2 2 1 μ 1 = 1 + 1 2 + 12 1 2 + 12 1 1 μ 1 = 1 + 1 2 2 1 6= 0 1 if 1 6=0 and, 2 6=0 Thus we can estimate 13. (iv). Interchange the role of 2 and 3 to estimate 12. (v). Form the residual (and recall that 13 is known and 23 =0) = 3 13 1 = 3 + 3. 41
(vi) Use 1 as a proxy for ability. Substituting this into gives: = 3 1 1 + 3 1 3 3 1 1. (vii) Now use 1 as an instrument for 1 in the above to get an estimate of 3 1. (viii) Interchange the role of 2 and 3 to estimate 2 1. 42
3.3 Triangular systems more generally Without loss of generality, suppose that 2 is excluded from the equation of our system. (We are supposing the existence of an extra exclusion than that implied by triangularity). We seek to estimate the parameters of the system in equation as well as equations before and after Equation t. i. Use 1 as a proxy for ability. Solving for and substituting into the equation: = + 43
We get: = 1 1 1 1 + and we are considering =2 1 The ratio 1 can then be identified using 1 as an instrument for 1 ii. Form the residuals: = 1 1 =2 1 Now we have 2 IV s ( 2 3 1 ) for the 2 independent variables in the equation ( 1 3 1 ) so we can consistently estimate the coecients in the equation. 44
Equations before t. iii. Form: = 1 1 1 1 We can use 1 1 to form 1 purged IV s and is used as a proxy for unobserved ability,. Inthisway,wecanestimatealloftheparameters in equations (Note the sequential order implicit in this triangular system. We must first estimate before this step can be made.) Example. Suppose 3 and 3 = 13 1 + 23 2 + 3 + 3. 45
Use = + as a proxy for. Substituting this into our 3 equation yields: 3 = 13 1 + 23 2 + μ 3 + 3 3. Observe that 1 and 2, are independent of our residual, but is not. We can use as an instrument for to estimate the parameters above. This obviously generalizes for all equations less than. 46
Equations after t. iv. Assume identification for all equations through via an exclusion restriction in equation. Example. As an example, consider the following: Define: 4 = 14 1 + 24 2 + 34 3 + 4 + 4 2 2 12 1 3 3 13 1 23 2 Solving for 1 and 2 and substituting into the equation for 4 we find: 47
4 = 14 1 + 24 2 + 34 ( 3 + 13 1 + 23 2 )+ 4 + 4 where: = ( 14 + 34 13 ) 1 +( 24 + 34 23 ) 2 + 34 3 + 4 + 4 = ( 14 + 34 13 ) 1 +( 24 + 34 23 )(2 + 12 1 ) + 34 3 + 4 + 4 = 14 1 + 24 2 + 34 3 + 4 + 4 14 = 14 + 24 12 + 34 13 24 = 24 + 23 34 Using 1 as a proxy for and substituting we get: μ 4 = 1 1 + 242 + 34 3 + 4 4 1 1 48
where 1 = 14 + 4 1 We can then use 1 2 and 3 as instruments to get an estimate of 34 Define: 4 = 4 34 3 = 14 1 + 24 2 + 4 + 4 (Excluding 3 allows us to estimate the remaining parameters). Using 3 as a proxy for yields: 4 = 14 1 + 24 2 + μ 4 3 + 4 4 3. 3 3 We can then estimate 14 and 24 by using 1 2 and 3 as an IV. We can continue estimating. For example, consider the 5 equation: (i) Rewrite in terms of 1 2 3 and 4 49
(ii) Use 1 to proxy. (iii) Use a cross-member IV for 1 in addition to =2 3 4 which gives our estimate of 45 (iv) Now form 5 = 5 45 4 (v) With 4 excluded, we can use purged IV s on 5 as before. 50
3.4 Comments 1. One needs to check the rank order conditions for identification (requires imposing an exclusion restriction). 2. Griliches and Chamberlain (IER, 1976) find a small ability bias - 3 decimal point dierence in schooling coecient. 51
4 Twin Methods Basic Principle: Monozygotic or MZ (identical) twins are more similar than Dizygotic or DZ (fraternal) twins. The key assumption is that if environmental factors are the same for both types of twins, then we can estimate genetic components to outcomes. 52
4.1 Univariate Twin Model Let = observed phenotypic variable, = unobserved genotype, and = environment. Further, suppose that we can write our model additively: = + and assume independence of and so that 2 = 2 + 2. Now suppose that we have data on another individual: = + Then our phenotypic covariance is: ( )= ( )+ ( ) 53
where we are imposing the assumption: ( )=( )=0 Defining standardized forms and some simplifying notation, let 2 2 2 2 2 2 Thus, = + which implies = + We can also derive the identity: 2 + 2 = 2 2 + 2 2 =1 where the last step follows from our assumption of independence. Now we wish to consider the correlation between ob- 54
served phenotypes of our two individuals: = ( ) = ( + + ) = 2 ( ) + 2 ( ) ( ) ( ) = 2 + 2 say, with and defined as above. We assume that =1 and that 1 Thatis,thegenotypicvariableisperfectly correlated among identical twins, but less than perfectly correlated among fraternal twins. Replacing this result into the above produces: = 2 + 2 = 2 + 2 55
Therefore: = (1 ) 2 +( ) 2 = (1 ) 2 +( )(1 2 ) where the last equality follows from our established identity. Solving for 2, we find: 2 = ( ) ( ) (1 ) ( ) The only known in the right hand side of the above equality is the expression ( ), which is simply the correlation coecient of the observed phenotypic variable. The remaining two expressions, (1 ) and ( ) can not be computed as they represent statistics on variables we don t observe. 56
One could impose = so that: 2 = 1 The expression is a measure of how closely the genetic variable is correlated across our two observations. One could then guess or estimate a value for this parameter to derive corresponding estimates of 2 the ratio of how much variance in the phenotypic variable is explained by variance in the genetic component. Other studies have attempted to include ( ) 6= 0but this presents an identification problem. A typical value of the estimable portion of the above,, is commonly reported in the literature to be 02. 57