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**Instrumental Variables Estimation and Two Stage Least Square**

Research Method Lecture 11-3 (Ch15) Instrumental Variables Estimation and Two Stage Least Square

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**IV solution to Errors in variable problems: Example 1**

Consider the following model Y=β0+β1x1*+β2x2+u Where x1* is the correctly measured variable. Suppose, however, that you only have error ridden variable x1=x1*+e1. Thus, the actual estimation model becomes Y=β0+β1x1+β2x2+(u- β1e1) Thus, the OLS estimate of β1 is biased. This is the error-in-variable bias.

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The error-in-variable bias cannot be corrected with the panel data method. But IV method can solve the problem. Suppose that you have another measure for x1*. Call this z1. For example consider x1* is the husband’s annual salary, and x1 is the annual salary reported by the husband, which is reported with errors. Sometimes, the data also asks the wife to report her husband’s annual salary. Then z1 is the husband’s annual salary reported by the wife.

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**In this case, z1=x1*+a1 where a1 is the measurement error.**

Although z1 is measured with errors, it can serve as the instrument for x1. Why? First, x1 and z1 should be correlated. Second, since e1 and a1 are just measurement errors, they are unlikely to be correlated, which means that z1 is uncorrelated with the error term (u- β1e1). Y=β0+β1x1+β2x2+(u- β1e1) So, 2SLS with z1 as an instrument can eliminate this bias.

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**IV solution to Errors in variable problems: Example 2**

This is a more complicated example. Consider the following model. where we have the unobserved ability problem. Suppose that you have two test scores that are the indicators of the ability. test1 = γabil+e1 test2 = δabil+e2

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**If you use test1 as the proxy variable for ability, you have the following model.**

where Thus, test1 is correlated with the error term: It has the error-in-variable problem. In this case, a simple plug-in-solution does not work. However, since you have test2, another measure of abil, you can use test2 as an instrument for test1 in the 2SLS procedure to eliminate the bias.

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Exercise Using WAGE2.dta, consider a log-wage regression with explanatory variables educ exper tenure married south urban and black. Using IQ and KWW (knowledge of the world of work) as two measures of the unobserved ability, estimate the model that correct for the bias in educ.

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OLS

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**Simple plug in solution**

Plug in + IV using KWW as the instrument for IQ

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**2SLS with heteroskedasticity**

When heteroskedasticity is present, we have to modify the standard error formula. The derivation of the formula is not the scope of this class. However, STATA automatically compute this. Just use robust option.

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**Testing overidentifying restrictions**

Usually, the instrument exogeneity cannot be tested. However, when you have extra instruments, you can effectively test this. This is the test of overidentifying restrictions.

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**The basic idea behind the test of overidentifying restrictions**

Before presenting the procedure, I will provide you with the basic idea of the test. Consider the following model. y1=β0+β1y2+β2z1+β3z2+u1 Suppose you have two instruments for y2: z3 z4. If both instruments are valid instruments, using either z3 or z4 as an instrument will produce consistent estimates. Let be the IV estimator when z3 is used as an instrument. Let be the IV estimate when z4 is used as an instrument

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**The idea is to check if and are similar. That is, you test H0: .**

If you reject this null, it means that either z3 or z4, or both of them are not exogenous. We do not know which one is not exogenous. So the rejection of the null typically means that your choice of instruments is invalid.

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On the other hand, if you fail to reject the null hypothesis, we can have some confidence in the overall set of instruments used. However, caution is necessarily. Even if you fail to reject the null, this does not always mean that the set of instruments are valid. For example, consider wage regression with education being the endogenous variable. And you have mother and father’s education as instruments.

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Even if mother and father’s education do not satisfy the instrument exogeneity, may be very close to zero since the direction of the biases are the same. In this case, even if they are invalid instruments, we may fail to reject the null (i.e., erraneously judge that they satisfy the instrument exogeneity).

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**The procedure of the test of overidentifying restrictions**

Estimate the structural equation by 2SLS and obtain the 2SLS residuals, . Regress on al exogenous variables. Obtain R-squared. Say R12. Under the null that all IVs are uncorrelated with the structural error u1, where q is the number of extra instruments. If you fail to reject the null (i.e., if nR12 is small), then you have some confidence about the instrument exogeneity. If you reject it, at least some of the instruments are not exogenous.

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**The NR12 statistic is valid when homoskedasticity assumption holds**

The NR12 statistic is valid when homoskedasticity assumption holds. NR12 statistic is also calld the Sargan’s statistic. When we assume heteroskedasticity, we have to use another statistic called the Hansen’s J statistic. Both tests can be done automatically using STATA.

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**Exercise Consider the following model.**

Log(wage)=β0+β1(educ)+β2Exper+β3Exper2+u Using Mroz.dta, estimate the above equation using motheduc & fathereduc as instruments for (educ). Test the overidentifying restrictions.

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Answers 1 OLS 2SLS

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**Answer: 2 First, conduct the test “manually”. 1. First, estimate 2SLS.**

2. Second, generate the 2sls residual. Call this uhat.

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**3. Third, regress uhat on all the exogenous variables**

3. Third, regress uhat on all the exogenous variables. Don’t forget to include exogenous variables in the structural equation: exper and expersq 4. Fourth, get the R-squared from this regression. You can use this, but this is rounded. To compute more precisely, type this. 5. Finally, compute NR2. This is NR2 stat. This is also called the Sargan’s statistic

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The NR2 stat follows χ2(1).The degree of freedom is equal to the number of extra instruments. In our case it is 1. (In our mode, there is only one endogenous variable. Thus, you need only one instrument. But we have two instruments. Therefore the number of extra instrumetn is 1. ) Since the 5% cutoff point for χ2(1) is 3.84, we failed to reject the null hypothesis that exogenous variables are not correlated with the structural error. Thus, we have some confidence in the choice of instruments. In other word, our instruments have ‘passed’ the test of overidentifying restrictions.

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**Now, let us conduct the test of overidentifying restriction automatically.**

This is NR2 stat. It is also called the Sargan’s statistic.

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**The heteroskedasticity version can also be done automatically.**

Use robust option when estimating 2SLS. Then type the same command. Heteroskedasticity robust version is called the Hansen’s J statistic

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Note Even if you fail to reject the null hypothesis in the test, there is a possibility that your instruments are still invalid. Thus, even if your instruments “pass the test”, in general, you should try to provide a plausible “story” why your instruments satisfy the instrument exogeneity. (Quarter of birth is a good example).

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**Testing the endogeneity**

Consider again the following model. y1=β0+β1y2+β2z1+β3z2+u1 Where y2 is the suspected endogenous variable and you have instruments z3 and z4. If y2 is actually exogenous, OLS is better. If you have valid instruments, you can test if y2 is exogenous or not.

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**then, test if δ is zero or not.**

Before laying out the procedure, let us understand the basic idea behind the test. Structural eq:y1=β0+β1y2+β2z1+β3z2+u1 Reduced eq :y2=π0+π1z1+ π2z2+ π3z3+ π4z4+v2 You can check that y2 is correlated with u1 only if v2 is correlated with u1. Further, let u1=δv2+e1. Then u1 and v2 are correlated only if δ =0. Thus, consider y1=β0+β1y2+β2z1+β3z2+ δv2+e1 then, test if δ is zero or not.

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**The test of endogeneity: procedure**

Estimate the reduced form equation using OLS. y2=π0+π1z1+ π2z2+ π3z3+ π4z4+v2 Then obtain the residual . (ii) Add to the structural equation and estimate using OLS y1=β0+β1y2+β2z1+β3z2+α +e1 Then, test H0: α=0. If we reject H0, then we conclude that y2 is endogenous because u1 and v2 are correlated.

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**Exercise Consider the following model.**

Log(wage)=β0+β1(educ)+β2Exper+β3Exper2+u Suppose that father and mother’ education satisfy the instrument exogeneity. Conduct the Hausman test of endogeneity to check if (educ) is exogenous or not.

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**Answer First, conduct the test “manually”.**

To use the same observations as 2SLS, run 2SLS once and generate this variable

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**Then check if this coefficient is different from zero.**

Now run the reduced for regression, then get the residual. Then check if this coefficient is different from zero.

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**The coefficient on uhat is significant at 10% level**

The coefficient on uhat is significant at 10% level. Thus, you reject the null hypothesis that educ is exogenous (not correlated with the structural error) at 10% level. This is a moderate evidence that educ is endogenous and thus 2SLS should be reported (along with OLS).

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**Stata conduct the test of endogeneity automatically**

Stata conduct the test of endogeneity automatically. Stata uses a different version of the test.

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Note that the test of endogeneity is valid only if that the instruments satisfy the instrument exogeneity. Thus, test the overidentifying restrictions first to check if the instruments satisfy the instrument exogeneity. If instruments “pass” the overidentifying test, then conduct the test of endogeneity.

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**Applying 2SLS to pooled cross sections**

When you simply apply 2SLS to the pooled cross section data, there is no new difficulty. You can just apply 2SLS.

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**Combining panel data method and IV method**

Suppose you have two period panel data. The period is 1987 and Consider the following model. Log(scrap)it=β0+δ0d88t+β1(hrsemp)it+ai+uit Where (scrap) is the scrap rate. (Hrsemp) is the hours of employee training. You have data

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**Correlation between ai and (hrsemp)it causes a bias in β1**

Correlation between ai and (hrsemp)it causes a bias in β1. In the first differened model, we difference to remove ai: that is, we estimate ∆Log(scrap)it=δ0+β1∆ (hrsemp)it+∆uit…(1) In some case, ∆ (hrsemp)it and ∆uit can still be correlated. For example, when a firm hires more skilled workers, they may reduce the job training.

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In this case, the quality of the worker is time varying, so it is not contained in ai, but it is contained in uit. In this case ∆(hrsemp)it and∆uit may be negatively correlated. This would cause OLS estimate of β1 to be biased upward (bias towards not finding the productivity enhancing effect of training). To eliminate the bias, we can apply IV method to equation (1).

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**One possible instrument for ∆(hrsemp)it is the ∆(Grant)it**

One possible instrument for ∆(hrsemp)it is the ∆(Grant)it. (Grant)it is a variable indicating if the company received job training grant. Since the grant designation is given at the beginning of 1988, ∆(Grant)it may be uncorrelated with ∆uit. At the same time, it would be correlated with ∆(hrsemp)it. Thus, we can use ∆(Grant)it as an IV for ∆(hrsemp)it.

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**Exercise Using JTRAIN.dta, estimate the following model.**

∆Log(scrap)it=δ0+β1∆ (hrsemp)it+∆uit…(1) Use ∆(Grant)it as an instrument for ∆(hrsemp)it. Use the data between 1987 and 1988 only.

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**Answer First estimate it manually.**

This is the simple first differenced model

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**This is the first-differenced model + IV method**

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**Now, estimate the model automatically.**

First differenced model + IV method.

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