Columbia University MA in Economics
GR 5411 Econometrics I Seyhan Erden
SOLUTIONS TO Problem Set 5 due on Dec. 7th at 10am through Gradescope
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1. (Practice question, not graded) Consider the regression model 𝑌𝑌 = 𝑋𝑋𝑋𝑋 + 𝑈𝑈. Partition 𝑋𝑋 as [𝑋𝑋1 𝑋𝑋2] and 𝑋𝑋 as [𝑋𝑋1′ 𝑋𝑋2′ ], where 𝑋𝑋1 has 𝑘𝑘1 columns and 𝑋𝑋2 has 𝑘𝑘2 columns. Suppose that
𝑋𝑋2′𝑌𝑌=0𝑘𝑘 ×1.Let𝑅𝑅=�𝐼𝐼𝑘𝑘 0𝑘𝑘 ×𝑘𝑘 �. 1 112
′′ �′′−1−1 (a) Show that 𝑋𝑋̂ (𝑋𝑋 𝑋𝑋)𝑋𝑋̂ = �𝑅𝑅𝑋𝑋̂� [𝑅𝑅(𝑋𝑋 𝑋𝑋) 𝑅𝑅′]
�𝑅𝑅𝑋𝑋̂�
(b) Consider the regression 𝑈𝑈𝑖𝑖𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇 = 𝛿𝛿0 + 𝛿𝛿1𝑍𝑍1𝑖𝑖 + ⋯ + 𝛿𝛿𝑚𝑚𝑍𝑍𝑚𝑚𝑖𝑖 + 𝛿𝛿𝑚𝑚+1𝑊𝑊1𝑖𝑖 + ⋯ +
𝛿𝛿 𝑊𝑊 +𝑒𝑒where𝑒𝑒istheregressionerrorterm.Let𝑊𝑊=[1𝑊𝑊 𝑊𝑊…𝑊𝑊]where1is 𝑚𝑚+𝑟𝑟 𝑟𝑟𝑖𝑖 𝑖𝑖 𝑖𝑖 � 1 2 𝑟𝑟
the 𝑛𝑛 × 1 vector with 𝑖𝑖𝑡𝑡h element 𝑊𝑊 , and so forth. Let 𝑈𝑈𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇 denote the vector of two
1𝑖𝑖 �
stage least squares residuals. (i) Show that 𝑊𝑊′𝑈𝑈𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇 = 0 (ii) Show that the method for
computing the 𝐽𝐽 statistic, using homoskedasticity-only F statistic and that using the
formula
𝐽𝐽 = � � 𝑈𝑈� ′ 𝑃𝑃 𝑍𝑍 𝑈𝑈� 𝑈𝑈′𝑀𝑀𝑍𝑍𝑈𝑈⁄(𝑛𝑛 − 𝑚𝑚 − 𝑟𝑟)
produce the same value for the 𝐽𝐽-statistic. (hint: use the results in part (a) and (b.i))
Solutions:
(a) 𝑋𝑋̂′(𝑋𝑋′𝑋𝑋)𝑋𝑋̂ = 𝑌𝑌′𝑋𝑋(𝑋𝑋′𝑋𝑋)−1𝑋𝑋′𝑌𝑌 = 𝑌𝑌′𝑋𝑋1𝐻𝐻𝑋𝑋1′𝑌𝑌, where 𝐻𝐻 is the upper 𝑘𝑘1 × 𝑘𝑘1 block of
(𝑋𝑋′𝑋𝑋)−1. Also 𝑅𝑅𝑋𝑋̂ = 𝐻𝐻𝑋𝑋1′𝑌𝑌 and 𝑅𝑅(𝑋𝑋′𝑋𝑋)−1𝑅𝑅 = 𝐻𝐻. Thus �𝑅𝑅𝑋𝑋̂�′[𝑅𝑅(𝑋𝑋′𝑋𝑋)−1]−1�𝑅𝑅𝑋𝑋̂� =
𝑌𝑌 ′ 𝑋𝑋 1 𝐻𝐻 𝑋𝑋 1′ 𝑌𝑌 . � �
(b) (i) Write the 2 stage regression as 𝑌𝑌 = 𝑋𝑋𝑋𝑋 + 𝑈𝑈, where 𝑋𝑋 are the fitted values from
the 1st stage regression. Note that 𝑈𝑈�′𝑋𝑋� = 0, where 𝑈𝑈� = 𝑌𝑌 − 𝑋𝑋�𝑋𝑋̂ because OLS
nd � ̂� �̂
residuals are orthogonal to the regressors. Now, 𝑈𝑈𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇 = 𝑌𝑌 − 𝑋𝑋𝑋𝑋 = 𝑈𝑈 − �𝑋𝑋 − 𝑋𝑋�𝑋𝑋 = 𝑈𝑈� − 𝑉𝑉�𝑋𝑋̂, where 𝑉𝑉� is the residual from the 1 stage regression. But since 𝑊𝑊is a
st ′ st 𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇 ′ ′′ regressor in the 1�stage regression, 𝑉𝑉� 𝑊𝑊 = 0. Thus 𝑈𝑈� ′𝑊𝑊 = 𝑈𝑈� 𝑊𝑊 − 𝑋𝑋̂ 𝑉𝑉� 𝑊𝑊 = 0.
(ii) 𝑋𝑋̂′(𝑋𝑋′𝑋𝑋)𝑋𝑋̂ = �𝑅𝑅𝑋𝑋̂�′[𝑅𝑅(𝑋𝑋′𝑋𝑋)−1𝑅𝑅′]−1�𝑅𝑅𝑋𝑋̂� = 𝑆𝑆𝑆𝑆𝑅𝑅𝑟𝑟𝑟𝑟𝑟𝑟𝑡𝑡𝑟𝑟𝑖𝑖𝑟𝑟𝑡𝑡𝑟𝑟𝑟𝑟 − 𝑆𝑆𝑆𝑆𝑅𝑅𝑢𝑢𝑢𝑢𝑟𝑟𝑟𝑟𝑟𝑟𝑡𝑡𝑟𝑟𝑖𝑖𝑟𝑟𝑡𝑡𝑟𝑟𝑟𝑟 for the regression in 𝑈𝑈𝑖𝑖𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇 = 𝛿𝛿0 + 𝛿𝛿1𝑍𝑍1𝑖𝑖 + ⋯ + 𝛿𝛿𝑚𝑚𝑍𝑍𝑚𝑚𝑖𝑖 + 𝛿𝛿𝑚𝑚+1𝑊𝑊1𝑖𝑖 + ⋯ + 𝛿𝛿𝑚𝑚+𝑟𝑟𝑊𝑊𝑟𝑟𝑖𝑖 + 𝑒𝑒𝑖𝑖
result follows directly.
2. (10p) Consider a simple linear model:
𝑦𝑦𝑖𝑖 = 𝑋𝑋1 + 𝑋𝑋2𝑋𝑋𝑖𝑖 + 𝜀𝜀𝑖𝑖
1
with 𝐶𝐶𝐶𝐶𝐶𝐶(𝑋𝑋𝑖𝑖 , 𝜀𝜀𝑖𝑖 ) ≠ 0. Let � be an exogenous, relevant instrument for this model, and assume that � is binary – taking on values of 0 or 1. Show the algebraic formula for the
OLS estimators and IV estimators for both 𝑋𝑋 and 𝑋𝑋
Solution:
The two ortogonality conditions for the IV estimator:
̂𝑖𝑖=1 ̂
The first one implies that 𝑋𝑋1𝑖𝑖𝑖𝑖 = 𝑦𝑦 − 𝑥𝑥̅′𝑋𝑋2𝑖𝑖𝑖𝑖. Substituting this into second eq., we get
1
2
𝑢𝑢̂̂
� � 𝑦𝑦 𝑖𝑖 − 𝑥𝑥 𝑖𝑖′ 𝑋𝑋 2𝑖𝑖 𝑖𝑖 − 𝑋𝑋 1𝑖𝑖 𝑖𝑖 � = 0
𝑖𝑖=1
𝑢𝑢̂̂
� 𝑧𝑧 𝑖𝑖 � 𝑦𝑦 𝑖𝑖 − 𝑥𝑥 𝑖𝑖′ 𝑋𝑋 2𝑖𝑖 𝑖𝑖 − 𝑋𝑋 1𝑖𝑖 𝑖𝑖 � = 0
̂ 𝑖𝑖=1
𝑢𝑢̂
� 𝑧𝑧 𝑖𝑖 � ( 𝑦𝑦 𝑖𝑖 − 𝑦𝑦 ) − ( 𝑥𝑥 𝑖𝑖 − 𝑥𝑥 ) ′ 𝑋𝑋 2𝑖𝑖 𝑖𝑖 � = 0
And solving for 𝑋𝑋2𝑖𝑖𝑖𝑖, we can summarize the results as: ̂𝑢𝑢 −1𝑢𝑢
𝑋𝑋2𝑖𝑖𝑖𝑖 = ��(𝑧𝑧𝑖𝑖 − 𝑧𝑧)(𝑥𝑥𝑖𝑖 − 𝑥𝑥)� ��(𝑧𝑧𝑖𝑖 − 𝑧𝑧)(𝑦𝑦𝑖𝑖 − 𝑦𝑦)�
1𝑖𝑖𝑖𝑖 ′ 2𝑖𝑖𝑖𝑖 𝑖𝑖=1 𝑋𝑋̂ =𝑦𝑦−𝑥𝑥̅𝑋𝑋̂𝑖𝑖=1
Using a similar ap̂ proach one can find the OLS coefficients 𝑢𝑢 −1𝑢𝑢
𝑋𝑋2𝑜𝑜𝑜𝑜𝑟𝑟 = ��(𝑥𝑥𝑖𝑖 − 𝑥𝑥)(𝑥𝑥𝑖𝑖 − 𝑥𝑥)� ��(𝑥𝑥𝑖𝑖 − 𝑥𝑥)(𝑦𝑦𝑖𝑖 − 𝑦𝑦)�
1𝑜𝑜𝑜𝑜𝑟𝑟 ′ 2𝑜𝑜𝑜𝑜𝑟𝑟 𝑖𝑖 = 1 𝑋𝑋̂ = 𝑦𝑦 − 𝑥𝑥 ̅ 𝑋𝑋̂ 𝑖𝑖 = 1
2
3. (10p) Consider the linear model:
𝑦𝑦 = 𝑋𝑋𝑋𝑋 + 𝜀𝜀 where � is a 𝑘𝑘 × 1 parameter vector to be estimated.
Assume there exists instruments �, with � > � columns, that satisfy: 𝐸𝐸[𝜀𝜀|𝑍𝑍] = 0 and 𝐸𝐸[𝜀𝜀𝜀𝜀′|𝑍𝑍] = 𝜎𝜎2𝐼𝐼.
Premultiplication of the linear model above by �′ yields the transformed model 𝑍𝑍′𝑦𝑦 = 𝑍𝑍′𝑋𝑋𝑋𝑋 + 𝑍𝑍′𝜀𝜀
Show that the efficient GLS estimator for � in this transformed model is the same as the 2SLS estimator discussed in class.
Solution:
We know that the GLS estimator for the model 𝑦𝑦 = 𝑋𝑋𝑋𝑋 + 𝑒𝑒 can be expressed as:
� 𝑋𝑋̂𝐺𝐺𝑇𝑇𝑇𝑇 = �𝑋𝑋′Ω�−1𝑋𝑋�−1𝑋𝑋′Ω�−1𝑦𝑦
Where 𝛺𝛺 is the estimated covariance matrix 𝐸𝐸[𝑒𝑒𝑒𝑒′] = 𝛺𝛺 (up to scale).
Hence, for the transformed model we regress Z’Y on Z’X, and the error term is 𝑍𝑍′𝜀𝜀. The covariance matrix is therefore
𝛺𝛺 = 𝐸𝐸[𝑍𝑍′𝜀𝜀(𝑍𝑍′𝜀𝜀)′] = 𝐸𝐸[𝑍𝑍′𝐸𝐸[𝜀𝜀𝜀𝜀′|𝑍𝑍]𝑍𝑍] = 𝜎𝜎2𝐸𝐸[𝑍𝑍′𝑍𝑍]
And so we can use 𝛺𝛺̂ = 𝑍𝑍′𝑍𝑍 since the matrix matters only up to scale. By combining these facts we obtain a formula for our GLS estimator:
3
𝐸𝐸𝑥𝑥𝐸𝐸 𝑊𝑊𝑘𝑘𝑊𝑊 𝑂𝑂𝑂𝑂𝑂𝑂 𝐼𝐼𝑛𝑛𝐼𝐼 𝑆𝑆𝐶𝐶𝑆𝑆𝑆𝑆h 𝑆𝑆𝑀𝑀𝑆𝑆𝑆𝑆 𝑀𝑀𝑆𝑆 𝑈𝑈𝑛𝑛𝑖𝑖𝐶𝐶𝑛𝑛 𝐸𝐸𝐼𝐼 𝐹𝐹𝑒𝑒𝑚𝑚 𝐵𝐵𝐵𝐵𝑘𝑘
𝑋𝑋̂𝐺𝐺𝑇𝑇𝑇𝑇 = ((𝑍𝑍′𝑋𝑋)′𝛺𝛺�−1(𝑍𝑍′𝑋𝑋))−1(𝑍𝑍′𝑋𝑋)′𝛺𝛺�−1𝑍𝑍′𝑦𝑦
= (𝑋𝑋′𝑍𝑍(𝑍𝑍′𝑍𝑍)−1𝑍𝑍′𝑋𝑋)−1𝑋𝑋′𝑍𝑍(𝑍𝑍′𝑍𝑍)−1𝑍𝑍′𝑦𝑦 = (𝑋𝑋′𝑃𝑃𝑍𝑍𝑋𝑋)−1𝑋𝑋′𝑃𝑃𝑍𝑍𝑦𝑦
= 𝑋𝑋̂2𝑇𝑇𝑇𝑇𝑇𝑇
4. (Practice question, not graded) Instrumental variable estimation of a labor supply equation: Cornwell and Rupert (1988)1 analyzed return to schooling in panel data set of 595 observations on heads of households. The estimating equation is
𝐿𝐿𝑛𝑛(𝑊𝑊𝑊𝑊𝑊𝑊𝑒𝑒) =𝛼𝛼 +𝛼𝛼 𝐸𝐸𝑥𝑥𝐸𝐸 +𝛼𝛼 𝐸𝐸𝑥𝑥𝐸𝐸2 +𝛼𝛼 𝑊𝑊𝑘𝑘𝑊𝑊 +𝛼𝛼 𝑂𝑂𝑂𝑂𝑂𝑂 +𝛼𝛼 𝐼𝐼𝑛𝑛𝐼𝐼 +𝛼𝛼 𝑆𝑆𝐶𝐶𝑆𝑆𝑆𝑆h 𝑖𝑖𝑡𝑡 1 2 𝑖𝑖𝑡𝑡 3 𝑖𝑖𝑡𝑡 4 𝑖𝑖𝑡𝑡 5 𝑖𝑖𝑡𝑡 6 𝑖𝑖𝑡𝑡 7 𝑖𝑖𝑡𝑡
+𝛼𝛼8𝑆𝑆𝑀𝑀𝑆𝑆𝑆𝑆𝑖𝑖𝑡𝑡 + 𝛼𝛼9𝑀𝑀𝑆𝑆𝑖𝑖𝑡𝑡 + 𝛼𝛼10𝑈𝑈𝑛𝑛𝑖𝑖𝐶𝐶𝑛𝑛𝑖𝑖𝑡𝑡 + 𝛼𝛼11𝐸𝐸𝐼𝐼𝑖𝑖𝑡𝑡 + 𝛼𝛼12𝐹𝐹𝑒𝑒𝑚𝑚𝑖𝑖𝑡𝑡 + 𝛼𝛼13𝐵𝐵𝐵𝐵𝑘𝑘𝑖𝑖𝑡𝑡 + 𝜀𝜀𝑖𝑖𝑡𝑡
Description of Variables
Years of full time work experience.
Weeks worked.
1 if blue-collar occupation, 0 otherwise.
1 if the individual works in a manufacturing industry, 0 otherwise.
1 if the individual resides in the south, 0 otherwise.
1 if the individual resides in an SMSA, 0 otherwise.
1 if the individual is married, 0 otherwise.
1 if the individual’s wage is set by a union contract, 0 otherwise.
Years of education as of 1981
1 if the individual is female, 0 otherwise.
1 if the individual is black, 0 otherwise.
The equation suggested is a reduced form equation; it contains all the variables in the model but does not specify the underlying structural relationship. In contrast, the following three equation model is a structural equation system
(𝐷𝐷𝑒𝑒𝑚𝑚𝑊𝑊𝑛𝑛𝐼𝐼) 𝑄𝑄𝑆𝑆𝑊𝑊𝑛𝑛𝑆𝑆𝑖𝑖𝑆𝑆𝑦𝑦𝐷𝐷 = 𝛼𝛼0 + 𝛼𝛼1𝑃𝑃𝑟𝑟𝑖𝑖𝑂𝑂𝑒𝑒 + 𝛼𝛼2𝐼𝐼𝑛𝑛𝑂𝑂𝐶𝐶𝑚𝑚𝑒𝑒 + 𝜀𝜀𝐷𝐷
(𝑆𝑆𝑆𝑆𝐸𝐸𝐸𝐸𝐵𝐵𝑦𝑦) 𝑄𝑄𝑆𝑆𝑊𝑊𝑛𝑛𝑆𝑆𝑖𝑖𝑆𝑆𝑦𝑦𝑇𝑇 = 𝑋𝑋0 + 𝑋𝑋1𝑃𝑃𝑟𝑟𝑖𝑖𝑂𝑂𝑒𝑒 + 𝑋𝑋2𝐼𝐼𝑛𝑛𝐸𝐸𝑆𝑆𝑆𝑆 𝑃𝑃𝑟𝑟𝑖𝑖𝑂𝑂𝑒𝑒 + 𝑋𝑋3𝑅𝑅𝑊𝑊𝑖𝑖𝑛𝑛 𝑓𝑓𝑊𝑊𝐵𝐵𝐵𝐵 + 𝜀𝜀𝑇𝑇
1 Cornwell, C. and P. Rupert. “Efficient Estimation with Panel Data: An Empirical Comparison of Instrumental Variable Estimators” Journal of Applied Econometrics, 3, 1988, pp. 149-155
4
(𝐸𝐸𝐸𝐸𝑆𝑆𝑖𝑖𝐵𝐵𝑖𝑖𝐸𝐸𝑟𝑟𝑖𝑖𝑆𝑆𝑚𝑚) 𝑄𝑄𝑆𝑆𝑊𝑊𝑛𝑛𝑆𝑆𝑖𝑖𝑆𝑆𝑦𝑦𝐷𝐷 = 𝑄𝑄𝑆𝑆𝑊𝑊𝑛𝑛𝑆𝑆𝑖𝑖𝑆𝑆𝑦𝑦𝑇𝑇
Arguably, the supply side of this market might consist of a household labor supply equation such as
𝑊𝑊𝑘𝑘𝑊𝑊𝑖𝑖𝑡𝑡 = 𝑋𝑋1 + 𝑋𝑋2𝐿𝐿𝑛𝑛(𝑊𝑊𝑊𝑊𝑊𝑊𝑒𝑒)𝑖𝑖𝑡𝑡 + 𝑋𝑋3𝐸𝐸𝐼𝐼𝑖𝑖𝑡𝑡 + 𝑋𝑋4𝑈𝑈𝑛𝑛𝑖𝑖𝐶𝐶𝑛𝑛𝑖𝑖𝑡𝑡 + 𝑋𝑋5𝐹𝐹𝑒𝑒𝑚𝑚𝑖𝑖𝑡𝑡 + 𝜀𝜀𝑖𝑖𝑡𝑡
𝐿𝐿𝑛𝑛(𝑊𝑊𝑊𝑊𝑊𝑊𝑒𝑒)
estimator sets for 𝐿𝐿𝑛𝑛(𝑊𝑊𝑊𝑊𝑊𝑊𝑒𝑒)𝑖𝑖𝑡𝑡 based on:
If the number of weeks worked and the accepted wage offer are determined jointly, then
𝑖𝑖𝑡𝑡
and errors are correlated due to simultaneous causality. We consider two IV and 𝑍𝑍1 = [𝐼𝐼𝑛𝑛𝐼𝐼𝑖𝑖𝑡𝑡,𝐸𝐸𝐼𝐼𝑖𝑖𝑡𝑡,𝑈𝑈𝑛𝑛𝑖𝑖𝐶𝐶𝑛𝑛𝑖𝑖𝑡𝑡,𝐹𝐹𝑒𝑒𝑚𝑚𝑖𝑖𝑡𝑡]
𝑍𝑍2 = [𝐼𝐼𝑛𝑛𝐼𝐼𝑖𝑖𝑡𝑡,𝐸𝐸𝐼𝐼𝑖𝑖𝑡𝑡,𝑈𝑈𝑛𝑛𝑖𝑖𝐶𝐶𝑛𝑛𝑖𝑖𝑡𝑡,𝐹𝐹𝑒𝑒𝑚𝑚𝑖𝑖𝑡𝑡,𝑆𝑆𝑀𝑀𝑆𝑆𝑆𝑆𝑖𝑖𝑡𝑡]
(a) Endogenous variables are 𝐿𝐿𝑛𝑛(𝑊𝑊𝑊𝑊𝑊𝑊𝑒𝑒)𝑖𝑖𝑡𝑡 and 𝑊𝑊𝑘𝑘𝑊𝑊𝑖𝑖𝑡𝑡, and all other variables are exogenous. Assume that 𝐿𝐿𝑛𝑛(𝑊𝑊𝑊𝑊𝑊𝑊𝑒𝑒)𝑖𝑖𝑡𝑡 is determined by 𝑊𝑊𝑘𝑘𝑊𝑊𝑖𝑖𝑡𝑡 and other “appropriate” exogenous variables. From the discussion above, deduce what variables would appear in a labor “demand” equation for 𝐿𝐿𝑛𝑛(𝑊𝑊𝑊𝑊𝑊𝑊𝑒𝑒) and what variables
would serve as IV?
(b) Estimate the parameters of labor demand equation you suggested in part (a) by OLS
and by 2SLS and compare the results (ignore the panel data nature of the data set. Just
pool the data)
(c) Are the instruments used relevant? Explain.
(d) Are the instruments used exogeneous? Explain.
Solution:
(a) The statement of the problem is actually a bit optimistic. Given the way it is stated, it would imply that the exogenous variables in the “demand” equation would be, in principle, (Ed, Union, Fem), which are also in the supply equation, plus the remainder, (Exp, Exp2, Occ, Ind, South, SMSA, Blk). The problem is that the model as stated would not be identified—the supply equation would, but the demand equation would not be. The way out would be to assume that at least one of (Ed, Union, Fem) does not appear in the demand equation. Since surely education would, that leaves one or both of Union and Fem. We will assume both of them are omitted. So, our equation is
𝐿𝐿𝑛𝑛(𝑊𝑊𝑊𝑊𝑊𝑊𝑒𝑒) =𝛼𝛼 +𝛼𝛼 𝐸𝐸𝐼𝐼 +𝛼𝛼 𝐸𝐸𝑥𝑥𝐸𝐸 +𝛼𝛼 𝐸𝐸𝑥𝑥𝐸𝐸 +𝛼𝛼 𝑂𝑂𝑂𝑂𝑂𝑂 +𝛼𝛼 𝐼𝐼𝑛𝑛𝐼𝐼 +𝛼𝛼 𝑆𝑆𝐶𝐶𝑆𝑆𝑆𝑆h 𝑖𝑖𝑡𝑡 1 2 𝑖𝑖𝑡𝑡 3 𝑖𝑖𝑡𝑡 4 𝑖𝑖𝑡𝑡 5 𝑖𝑖𝑡𝑡 6 𝑖𝑖𝑡𝑡 7 𝑖𝑖𝑡𝑡
2
+𝛼𝛼8𝑆𝑆𝑀𝑀𝑆𝑆𝑆𝑆𝑖𝑖𝑡𝑡 + 𝛼𝛼9𝐵𝐵𝐵𝐵𝑘𝑘𝑖𝑖𝑡𝑡 + 𝛾𝛾𝑊𝑊𝑘𝑘𝑊𝑊𝑖𝑖𝑡𝑡 + 𝑆𝑆𝑖𝑖𝑡𝑡
𝑖𝑖𝑡𝑡
5
(b) through (d) see pr4.log
5. (Practice Question, not graded) Suggest an estimator that we discussed in class under
endogeneity, show the asymptotic properties of the estimator you suggested.
Assume that there is a set of additional variables, 𝑍𝑍 that have two essential properties: 1. Relevance: 𝑍𝑍 are correlated with endogenous regressors (covariate) 𝑋𝑋
Solution:
(𝑂𝑂𝐶𝐶𝑟𝑟𝑟𝑟(𝑋𝑋, 𝑍𝑍) ≠ 0)
2. Exogeneity: they are uncorrelated with the disturbance (𝐸𝐸(𝜀𝜀𝑖𝑖|𝑍𝑍𝑖𝑖) = 0).
We have
𝐸𝐸𝐵𝐵𝑖𝑖𝑚𝑚 𝑍𝑍 𝑍𝑍 = 𝑄𝑄 a finite p.d. matrix (well behaved data)
𝑢𝑢1′ 𝑍𝑍𝑍𝑍
𝐸𝐸𝐵𝐵𝑖𝑖𝑚𝑚 𝑢𝑢1 𝑍𝑍′𝑋𝑋 = 𝑄𝑄𝑍𝑍𝑍𝑍 a finite 𝐵𝐵 × 𝑘𝑘 matrix with rank k (relevance)
𝐸𝐸𝐵𝐵𝑖𝑖𝑚𝑚 𝑢𝑢1 𝑍𝑍′𝜀𝜀 = 0 (exogeneity)
Let 𝐵𝐵 = 𝑘𝑘 for now, that is the number of 𝑍𝑍′𝑊𝑊 and the number of 𝑋𝑋′𝑊𝑊 are the same
Due to exogeneity 𝐸𝐸(𝑧𝑧𝑖𝑖𝜀𝜀𝑖𝑖) = 0, however left hand side can be written as follows (which in turn must be equal to zero as well,
𝐸𝐸(𝑧𝑧 (𝑦𝑦 − 𝑥𝑥 𝑋𝑋)) = 0 𝑖𝑖 𝑖𝑖 𝑖𝑖′
𝐸𝐸(𝑧𝑧𝑖𝑖𝑦𝑦𝑖𝑖) − 𝐸𝐸(𝑧𝑧𝑖𝑖𝑥𝑥𝑖𝑖′)𝑋𝑋 = 0
Then 𝑖𝑖 𝑖𝑖′ −1 𝑖𝑖 𝑖𝑖 𝑋𝑋=[𝐸𝐸(𝑧𝑧𝑥𝑥)] 𝐸𝐸(𝑧𝑧𝑦𝑦)
𝑋𝑋 = [𝐸𝐸(𝑧𝑧𝑖𝑖𝑥𝑥𝑖𝑖′)]−1𝐸𝐸(𝑧𝑧𝑖𝑖𝑦𝑦𝑖𝑖)
̂ = �𝐸𝐸𝐵𝐵𝑖𝑖𝑚𝑚 𝑛𝑛1 𝑍𝑍′𝑋𝑋�−1 𝐸𝐸𝐵𝐵𝑖𝑖𝑚𝑚 �𝑛𝑛1 𝑍𝑍′𝑦𝑦�
′ −1 ′ 𝑋𝑋 =(𝑍𝑍𝑋𝑋) 𝑍𝑍𝑦𝑦
This leads to the instrumental variable estimator,
̂𝐼𝐼𝐼𝐼
For a model with a constant term and a single 𝑥𝑥 and an instrumental variable 𝑧𝑧, we have
𝑋𝑋 =∑̂(𝑧𝑧𝑖𝑖−𝑧𝑧̅)(𝑦𝑦𝑖𝑖−𝑦𝑦�)=𝑂𝑂𝐶𝐶𝐶𝐶(𝑧𝑧,𝑦𝑦) →𝑝𝑝 𝑋𝑋
∑(𝑧𝑧𝑖𝑖 − 𝑧𝑧̅)(𝑥𝑥𝑖𝑖 − 𝑥𝑥̅) 𝑂𝑂𝐶𝐶𝐶𝐶(𝑥𝑥, 𝑦𝑦)
𝐼𝐼𝐼𝐼
Since, 𝑋𝑋𝐼𝐼𝐼𝐼 = (𝑍𝑍′𝑋𝑋)−1𝑍𝑍′𝑦𝑦
6
𝑋𝑋̂ =(𝑍𝑍𝑋𝑋) 𝑍𝑍(𝑋𝑋𝑋𝑋+𝜀𝜀)
We can write
′ −1 ′ ′ −1 ′ 𝐼𝐼𝐼𝐼 = (𝑍𝑍′𝑋𝑋)−1𝑍𝑍′𝑋𝑋𝑋𝑋 + (𝑍𝑍 𝑋𝑋) 𝑍𝑍 𝜀𝜀
̂ = 𝑋𝑋 + (𝑍𝑍′𝑋𝑋)−1𝑍𝑍′𝜀𝜀
𝐼𝐼𝐼𝐼 ′ −1 ′ 𝑋𝑋 −𝑋𝑋=(𝑍𝑍𝑋𝑋) 𝑍𝑍𝜀𝜀
Then, 𝑍𝑍′𝑋𝑋 −1 1 √𝑛𝑛�𝑋𝑋̂𝐼𝐼𝐼𝐼−𝑋𝑋�=�𝑛𝑛� √𝑛𝑛𝑍𝑍′𝜀𝜀
Then, we can express
√𝑛𝑛�𝑋𝑋̂ −𝑋𝑋�=�𝑍𝑍′𝑋𝑋�−1 1 𝑍𝑍′𝜀𝜀
𝐼𝐼𝐼𝐼 𝑛𝑛√𝑛𝑛
1
This has the same limiting distribution as
𝑄𝑄−1 ��√𝑛𝑛� 𝑍𝑍′𝜀𝜀� 𝑧𝑧𝑧𝑧
1𝑟𝑟
�√𝑛𝑛𝑍𝑍′𝜀𝜀� → 𝑁𝑁[0,𝜎𝜎2𝑄𝑄𝑧𝑧𝑧𝑧]
We know that
and −1 𝑍𝑍𝑋𝑋1𝑟𝑟
�′�� 𝑍𝑍′𝜀𝜀�→𝑁𝑁�0,𝜎𝜎2𝑄𝑄−1𝑄𝑄𝑄𝑄−1� 𝑛𝑛 √𝑛𝑛 𝑧𝑧𝑧𝑧𝑧𝑧𝑧𝑧𝑧𝑧𝑧𝑧
This step completes the derivation for the next theorem:
Theorem: Asymptotic Distribution of the Instrumental Variables Estimator…
Îf̂assumptions listed above all hold for �𝑦𝑦 , 𝑥𝑥 ,𝑧𝑧 ,𝜀𝜀 �, where 𝑧𝑧 is a valid set of 𝐵𝐵 = 𝑘𝑘
𝐼𝐼𝐼𝐼 ′−1′ 𝑋𝑋 =(𝑍𝑍𝑋𝑋) 𝑍𝑍𝑛𝑛𝑦𝑦
𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖
instrumental variables estimator
𝑋𝑋 →𝑟𝑟 𝑁𝑁�𝑋𝑋,𝜎𝜎2𝑄𝑄−1𝑄𝑄 𝑄𝑄−1� 𝐼𝐼𝐼𝐼 𝑧𝑧𝑧𝑧 𝑧𝑧𝑧𝑧 𝑧𝑧𝑧𝑧
6. (10p) Given the following regression 𝑦𝑦𝑖𝑖 = 𝑋𝑋0 + 𝑋𝑋1𝑋𝑋1𝑖𝑖 + 𝑆𝑆𝑖𝑖
7
Assume that X is endogenous and Z is an instrumental variable (IV). By studying the probability limit (plim) of the IV estimator we can see that when Z and u are possibly correlated, we can write
interesting part of this equation involves the correlation terms. It shows that, even if
𝐸𝐸𝐵𝐵𝑖𝑖𝑚𝑚 𝑋𝑋̂1,𝐼𝐼𝐼𝐼 = 𝑋𝑋1+ 𝐶𝐶𝑜𝑜𝑟𝑟𝑟𝑟(𝑍𝑍,𝑢𝑢) 𝜎𝜎𝑢𝑢 (1) 𝐶𝐶𝑜𝑜𝑟𝑟𝑟𝑟(𝑍𝑍,𝑍𝑍) 𝜎𝜎𝑥𝑥
where 𝜎𝜎𝑢𝑢 and 𝜎𝜎𝑧𝑧 are the standard deviation of u and X in the population, respectively. The
Corr(Z,u) is small, the inconsistency in the IV estimator can be very large if Corr(Z,X) is
also small. Thus, even if we focus only on consistency, it is not necessarily better to use
Using the fact that 𝐶𝐶𝐶𝐶𝑟𝑟𝑟𝑟(𝑋𝑋, 𝑆𝑆) = 𝐶𝐶𝐶𝐶𝐶𝐶(𝑋𝑋, 𝑆𝑆)⁄(𝜎𝜎 . 𝜎𝜎 ) along with the fact that 𝐸𝐸𝐵𝐵𝑖𝑖𝑚𝑚 𝑋𝑋̂ =
𝜎𝜎𝑥𝑥
IV than OLS if the correlation between Z and u are smaller than that between X and u.
𝑋𝑋1 + 𝐶𝐶𝑜𝑜𝑖𝑖(𝑍𝑍,𝑢𝑢) = 𝑋𝑋 when 𝐶𝐶𝐶𝐶𝐶𝐶(𝑋𝑋, 𝑆𝑆) = 0, we can write the plim of OLS estimator – call 𝐼𝐼𝑉𝑉𝑟𝑟(𝑍𝑍)
it 𝐸𝐸𝐵𝐵𝑖𝑖𝑚𝑚 𝑋𝑋̂1,𝑂𝑂𝑇𝑇𝑇𝑇 – as
𝐸𝐸𝐵𝐵𝑖𝑖𝑚𝑚 𝑋𝑋̂1,𝑂𝑂𝑇𝑇𝑇𝑇 = 𝑋𝑋1+ 𝐶𝐶𝐶𝐶𝑟𝑟𝑟𝑟(𝑋𝑋, 𝑆𝑆) 𝜎𝜎𝑢𝑢 (2)
𝑢𝑢𝑧𝑧1
Assume that 𝜎𝜎𝑢𝑢 = 𝜎𝜎𝑧𝑧, so that the population variance in the error term is the same as it is in X. Suppose the instrumental variable, Z, is slightly correlated with u: 𝐶𝐶𝐶𝐶𝐶𝐶(𝑍𝑍, 𝑆𝑆) = 0.1. Suppose also that Z and X have somewhat stronger correlation: 𝐶𝐶𝐶𝐶𝐶𝐶(𝑍𝑍, 𝑋𝑋) = 0.2.
(a) (5p) What is the bias in the asymptotic IV estimator?
(b) (5p) How much correlation would have to exist between X and u before OLS has more
asymptotic bias than TSLS? ̂ ̂
(a) From equation (1) with σu = σx, plim 𝑋𝑋1,𝐼𝐼𝐼𝐼 = β1 + (.1/.2) = β1 + .5, where 𝑋𝑋1,𝐼𝐼𝐼𝐼 is the IV
estimator. So the asymptotic bias is .5.
(b) From equation (2) with σu = σx, plim 𝑋𝑋̂ = β1 + Corr(x,u), where 𝑋𝑋̂ is the OLS
7. (10p) In the discussion of the instrumental variables estimator, we showed that the least ̂̂̂
1,𝑂𝑂𝑇𝑇𝑇𝑇 1,𝑂𝑂𝑇𝑇𝑇𝑇
estimator. So we would have to have Corr(x,u) > .5 before the asymptotic bias in OLS exceeds that of IV. This is a simple illustration of how a seemingly small correlation (.1 in this case) between the IV (Z) and error (u) can still result in IV being more biased than OLS if the correlation between Z and X is weak (.2).
squares estimator 𝑋𝑋 is biased and inconsistent. Nonetheless, 𝑋𝑋 does estimate something: 𝑜𝑜𝑜𝑜𝑟𝑟 𝐸𝐸𝐵𝐵𝑖𝑖𝑚𝑚𝑋𝑋𝑜𝑜𝑜𝑜𝑟𝑟 =𝜃𝜃=𝑋𝑋+𝑄𝑄−1𝛾𝛾 𝑜𝑜𝑜𝑜𝑟𝑟
8
Derive the asymptotic covariance matrix of 𝑋𝑋̂𝑜𝑜𝑜𝑜𝑟𝑟 and show that 𝑋𝑋̂𝑜𝑜𝑜𝑜𝑟𝑟 is asymptotically normally distributed.
̂
𝑋𝑋𝑜𝑜𝑜𝑜𝑟𝑟 = 𝑋𝑋 + (𝑋𝑋′𝑋𝑋)−1𝑋𝑋′𝜀𝜀
To obtain the asymptotic distribution, write the result already in hand as
𝐸𝐸𝐵𝐵𝑖𝑖𝑚𝑚 𝑋𝑋̂𝑜𝑜𝑜𝑜𝑟𝑟 = 𝑋𝑋 +𝑄𝑄−1𝛾𝛾 ̂
For convenience, let 𝜃𝜃 ≠ 𝑋𝑋 denote 𝑋𝑋 +𝑄𝑄−1𝛾𝛾 = 𝐸𝐸𝐵𝐵𝑖𝑖𝑚𝑚 𝑋𝑋𝑜𝑜𝑜𝑜𝑟𝑟. Write the preceding in the
form 𝑋𝑋̂ − 𝜃𝜃 = �𝑋𝑋′𝑋𝑋�−1 �𝑋𝑋′𝜀𝜀� −𝑄𝑄−1𝛾𝛾 Since 𝑜𝑜𝑜𝑜𝑟𝑟 𝑛𝑛𝑋𝑋′𝑋𝑋 𝑛𝑛
𝐸𝐸𝐵𝐵𝑖𝑖𝑚𝑚� 𝑛𝑛 �=𝑄𝑄
the large sample behavior of the right-hand side is the same as that of
̂𝑢𝑢 √𝑛𝑛�𝑋𝑋𝑜𝑜𝑜𝑜𝑟𝑟 − 𝜃𝜃� which is the same as
𝐸𝐸𝐵𝐵𝑖𝑖𝑚𝑚�𝑋𝑋̂ − 𝜃𝜃� =𝑄𝑄−1𝐸𝐸𝐵𝐵𝑖𝑖𝑚𝑚�𝑋𝑋′𝜀𝜀� −𝑄𝑄−1𝛾𝛾
𝑜𝑜𝑜𝑜𝑟𝑟 𝑛𝑛
That is, we may replace �𝑍𝑍′𝑍𝑍� with 𝑄𝑄. Then, we seek the asymptotic distribution of
𝑋𝑋′𝜀𝜀 1𝑢𝑢
√𝑛𝑛 �𝑄𝑄−1𝐸𝐸𝐵𝐵𝑖𝑖𝑚𝑚� 𝑛𝑛 � −𝑄𝑄−1𝛾𝛾� = 𝑄𝑄−1√𝑛𝑛 �𝑛𝑛� �(𝑥𝑥𝑖𝑖𝜀𝜀𝑖𝑖 − 𝛾𝛾)
Note that 𝑋𝑋𝑖𝑖𝜀𝜀𝑖𝑖 are iid, 𝐸𝐸(𝑋𝑋𝑖𝑖𝜀𝜀𝑖𝑖) = 𝛾𝛾 and 𝑖𝑖=1
𝑉𝑉𝑊𝑊𝑟𝑟(𝑋𝑋𝑖𝑖𝜀𝜀𝑖𝑖) = 𝐸𝐸[𝑋𝑋𝑖𝑖𝑋𝑋𝑖𝑖′𝜀𝜀𝑖𝑖2] − (𝐸𝐸[𝑋𝑋𝑖𝑖𝜀𝜀𝑖𝑖])′𝐸𝐸[𝑋𝑋𝑖𝑖𝜀𝜀𝑖𝑖] = 𝐸𝐸(𝐸𝐸[𝑋𝑋𝑖𝑖𝑋𝑋𝑖𝑖′𝜀𝜀𝑖𝑖2|𝑋𝑋𝑖𝑖]) − 𝛾𝛾′𝛾𝛾
= 𝐸𝐸(𝑋𝑋𝑖𝑖𝑋𝑋𝑖𝑖′𝐸𝐸[𝜀𝜀𝑖𝑖2|𝑋𝑋𝑖𝑖]) − 𝛾𝛾′𝛾𝛾 = 𝜎𝜎2𝐸𝐸(𝑋𝑋𝑖𝑖𝑋𝑋𝑖𝑖′) − 𝛾𝛾′𝛾𝛾 = 𝜎𝜎2𝑄𝑄 − 𝛾𝛾′𝛾𝛾
Hence,byCLT,1 1 𝑢𝑢 𝑟𝑟 ̂√𝑛𝑛�𝑛𝑛𝑋𝑋′𝜀𝜀−𝛾𝛾�=√𝑛𝑛�𝑛𝑛�𝑛𝑛�(𝑥𝑥𝑖𝑖𝜀𝜀𝑖𝑖 −𝛾𝛾) → 𝑁𝑁(0,𝜎𝜎2𝑄𝑄−𝛾𝛾′𝛾𝛾)
𝑖𝑖=1
1𝑢𝑢𝑟𝑟
√𝑛𝑛�𝑋𝑋𝑂𝑂𝑇𝑇𝑇𝑇̂− 𝐸𝐸𝐵𝐵𝑖𝑖𝑚𝑚 𝑋𝑋� = 𝑄𝑄−1√𝑛𝑛 � � �(𝑥𝑥𝑖𝑖𝜀𝜀𝑖𝑖 − 𝛾𝛾) → 𝑄𝑄−1𝑁𝑁(0, 𝜎𝜎2𝑄𝑄 − 𝛾𝛾′𝛾𝛾)
And hence, by rules of convergence in distribution
𝑂𝑂𝑇𝑇𝑇𝑇 −1 2 ′ −1 2 −1 ′ −2 So,𝑆𝑆𝑊𝑊𝑦𝑦𝑉𝑉𝑊𝑊𝑟𝑟�𝑋𝑋 �=𝑄𝑄 (𝜎𝜎 𝑄𝑄−𝛾𝛾𝛾𝛾)𝑄𝑄𝑖𝑖=1=𝜎𝜎 𝑄𝑄 −𝛾𝛾𝛾𝛾𝑄𝑄
9
8. (35p) You will replicate and extend the work reported in Acemoglu, Johnson and Robin- son (2001). The authors provided an expanded set of controls when they published their 2012 extension and posted the data on the AER website. This dataset is AJR2001 on the course website. �
(a) (3p) Estimate the OLS regression
log(𝐺𝐺𝐷𝐷𝑃𝑃 𝐸𝐸𝑒𝑒𝑟𝑟 𝐶𝐶𝑊𝑊𝐸𝐸𝐶𝐶𝑆𝑆𝑊𝑊) = 0.5 𝑟𝑟𝑖𝑖𝑊𝑊𝑘𝑘 (1)
(0.06)
𝑟𝑟𝐶𝐶𝑊𝑊𝑘𝑘 = −0.61log(𝑚𝑚𝐶𝐶𝑟𝑟𝑆𝑆𝑊𝑊𝐵𝐵𝑖𝑖𝑆𝑆𝑦𝑦)+ 𝑆𝑆� (2)
the reduc�ed form regression (�0.13)
and the 2SLS regression:
𝐵𝐵𝐶𝐶𝑊𝑊(𝐺𝐺𝐷𝐷𝑃𝑃 𝐸𝐸𝑒𝑒𝑟𝑟 𝐶𝐶𝑊𝑊𝐸𝐸𝐶𝐶𝑆𝑆𝑊𝑊) = 0.94 𝑟𝑟𝑖𝑖𝑊𝑊𝑘𝑘 (3) (0.16)
(Which point estimate is different by 0.01 from the reported values? This is a
common phenomenon in empirical replication).
(b) (3p) For the above estimates, calculate both homoskedastic and heteroskedastic-
robust standard errors. Which were used by the authors (as reported in (1)-(2)-(3)?)
(c) (3p) Calculate the 2SLS estimates by the Indirect Least Squares formula. Are they the
same?
(d) (3p) Calculate the 2SLS estimates by the two-stage approach. Are they the same?
(e) (3p) Calculate the 2SLS estimates by the control variable approach. Are they the
same?
(f) (4p) Acemoglu, Johnson and Robinson (2001) reported many specifications including
alternative regressor controls, for example latitude and africa. Estimate by least- squares the equation for logGDP adding latitude and africa as regressors. Does this regression suggest that latitude and africa are predictive of the level of GDP?
(g) (4p) Now estimate the same equation as in (f) but by 2SLS using log mortality as an instrument for risk. How does the interpretation of the effect of latitude and africa change?
(h) (4p) Return to our baseline model (without including latitude and africa ). The authors reduced form equation uses log(mortality) as the instrument, rather than, say, the level of mortality. Estimate the reduced form for risk with mortality as the instrument. (This variable is not provided in the dataset, so you need to take the exponential of the mortality variable.) Can you explain why the authors preferred the equation with log(mortality)?
10
(i) (4p)Tryanalternativereducedform,includingbothlog(mortality)andthesquareof log(mortality). Interpret the results. Re-estimate the structural equation by 2SLS using both log(mortality) and its square as instruments. How do the results change?
(j) (4p) Calculate and interpret a test for exogeneity of the instruments. Solutions:
(a) Regression result (eq 1): . reg loggdp risk
= 64
= 68.17
= 0.0000
= 0.5237
= 0.5160
= .72888
——————————————————————————
loggdp | Coef. Std. Err. t P>|t| [95% Conf. Interval]
————-+—————————————————————-
risk | .516187 .0625186 8.26 0.000 .3912141 .6411599
_cons | 4.687415 .417441 11.23 0.000 3.852962 5.521867
——————————————————————————
= 64
= 23.34
= 0.0000
= 0.2735
= 0.2618
= 1.262
——————————————————————————
risk | Coef. Std. Err. t P>|t| [95% Conf. Interval]
————-+—————————————————————-
logmort0 | -.6132892 .1269412 -4.83 0.000 -.8670411 -.3595374
_cons | 9.365895 .6105941 15.34 0.000 8.145335 10.58646
——————————————————————————
Source | SS df MS Number of obs
————-+———————————- F(1, 62)
Model | 36.2163143 1 36.2163143 Prob > F
Residual | 32.9382286 62 .531261752 R-squared
————-+———————————- Adj R-squared
Total | 69.1545429 63 1.09769116 Root MSE
Regression result (eq 2):
. reg risk logmort0
Source | SS df MS Number of obs
————-+———————————- F(1, 62)
Model | 37.1754537 1 37.1754537 Prob > F
Residual | 98.7466676 62 1.59268819 R-squared
————-+———————————- Adj R-squared
Total | 135.922121 63 2.15749399 Root MSE
Regression result (eq 3):
. predict u, residual
. ivregress 2sls loggdp (risk=logmort0)
Instrumental variables (2SLS) regression
Number of obs =
Wald chi2(1) =
Prob > chi2 =
R-squared =
Root MSE =
64
36.60
0.0000
0.1880
.93672
——————————————————————————
11
9.
loggdp | Coef. Std. Err. z P>|z| [95% Conf. Interval]
————-+—————————————————————-
risk | .9294897 .1536318 6.05 0.000 .6283768 1.230603
_cons | 1.994296 1.007904 1.98 0.048 .0188405 3.969751
——————————————————————————
= 64
= 56.67
= 0.0000
= 0.6501
= 0.6386
= .62982
——————————————————————————
loggdp | Coef. Std. Err. t P>|t| [95% Conf. Interval]
————-+—————————————————————-
Instrumented: risk
Instruments: logmort0
. reg loggdp risk u
Source | SS df MS Number of obs
————-+———————————- F(2, 61)
Model | 44.9573006 2 22.4786503 Prob > F
Residual | 24.1972423 61 .396676104 R-squared
————-+———————————- Adj R-squared
Total | 69.1545429 63 1.09769116 Root MSE
risk | .9294897 .1032975 9.00 0.000 .7229335
u | -.5689 .1211919 -4.69 0.000 -.8112382
1.136046
-.3265617
3.34941
_cons | 1.994296 .6776848 2.94 0.005
—————————————————————————— (b) through (k) : see pr8.log
(25p) Using MROZ.dta that we discussed in lecture.
(a) (3p) Run the following regression and report (copy/paste) your results, interpret the
coefficient of education.
(b) (3p) Explain why the regression in part (a) may suffer from endogeneity problem?
(c) (3p) Use the variable mother’s education (motheduc) as an instrumental variable for education, report your result and interpret the coefficient of education, is it significant? Why? Write this regression in regression equation form.
(d) (4p) Run the same regression in part (c) in two separate steps using OLS (that is: run the first and second stage regressions separately) and report your results. What is the difference between your result here and the result in part (c) where you used ivregression command? Be specific. Also, write both stages in regression equation form.
(e) (4p) Use both mother’s and father’s education as instrumental variables with ivregress 2sls command, check for endogeneity. Write this regression in regression equation form.
𝐿𝐿𝑛𝑛(𝑤𝑤𝑊𝑊𝑊𝑊𝑒𝑒) = 𝑋𝑋0 + 𝑋𝑋1𝐸𝐸𝐼𝐼𝑆𝑆𝑂𝑂 + 𝑒𝑒
(f) (4p) Run the following regression using ivreg command and both motheduc and fatheduc as instrumental variables for education:
𝐿𝐿𝑛𝑛(𝑤𝑤𝑊𝑊𝑊𝑊𝑒𝑒)=𝑋𝑋0 +𝑋𝑋1𝐸𝐸𝐼𝐼𝑆𝑆𝑂𝑂+𝑋𝑋2 𝐸𝐸𝑥𝑥𝐸𝐸𝑒𝑒𝑟𝑟+𝑋𝑋3 𝐸𝐸𝑥𝑥𝐸𝐸𝑒𝑒𝑟𝑟𝑆𝑆𝐸𝐸+𝑆𝑆
.6391811
12
Report your results in equation form as well. Also write the first and second stage in general equation form (note that you will not see the first stage with ivreg command but you do know what it is)
(g) (4p) Using the regression in part (f) test exogeneity of both IV’s (motheduc and fatheduc)
Solutions: (a) Results:
reg lwage educ, r
Linear regression
• • • • • • • • • • • • •
For 428 married women (mroz.dta) the estimated return on education is approximately 10.86%
(b) There are several reasons why this regression suffers from endogeneity problem (i) Omitted variable bias, say experience is omitted and not only experience is associated with wage but also it is correlated with education so not including experience in the wage regression will cause OVB in the coefficient estimate of education. (ii) variables like “ability” cannot be measured and causes attenuation bias. Hence, educ is endogenous, 10.86% may be overstating the true value of return on education. We must use IV, such as mother’s education.
(c) Results:
——————————————————————————
| Robust
lwage | Coef. Std. Err. t P>|t| [95% Conf. Interval] ————-+—————————————————————- educ | .1086487 .0134153 8.10 0.000 .0822802 .1350171 _cons | -.185196�8 .1707481 -1.08 0.279 -.5208104 .1504168 ——————————————————————————
𝐿𝐿𝑛𝑛(𝑊𝑊𝑊𝑊𝑊𝑊𝑒𝑒) = −0.1852 + 0.1086𝐸𝐸𝐼𝐼𝑆𝑆𝑂𝑂 (0.171) (0.0134)
Number of obs
F(1, 426)
Prob > F
R-squared
Root MSE
= 428
= 65.59
= 0.0000
= 0.1179
= .68003
13
• • • • • • • • • • • • • • • •
ivreg lwage (educ = motheduc), r
Instrumental variables (2SLS) regression
Number of obs
F(1, 426)
Prob > F
R-squared
Root MSE
= 428
= 0.96
= 0.3267
= 0.0688
= .69869
——————————————————————————
| Robust
lwage | Coef. Std. Err. t P>|t| [95% Conf. Interval]
————-+—————————————————————-
educ | .0385499 .0392631 0.98 0.327 -.0386235 .1157234
_cons | .7021743 .4937329 1.42 0.156 -.2682815 1.67263
——————————————————————————
Instrumented: educ
Instruments: motheduc � ——————————————————————————
First Stage:
. reg educ motheduc if lwage!=., r
Linear regression
Number of obs
F(1, 426)
Prob > F
R-squared
Root MSE
= 428
= 70.72
= 0.0000
= 0.1498
= 2.1098
𝐿𝐿𝑛𝑛(𝑊𝑊𝑊𝑊𝑊𝑊𝑒𝑒) = 0.7022 + 0.0385𝐸𝐸𝐼𝐼𝑆𝑆𝑂𝑂 (0.493) (0.0393)
the estimated return on education is approximately 3.85% (d) Results:
——————————————————————————
| Robust
educ | Coef. Std. Err. t P>|t| [95% Conf. Interval]
————-+—————————————————————-
motheduc | .2673697 .0317937 8.41 0.000 .2048776 .3298618
_cons | 10.11449 .3192591 31.68 0.000 9.486975 10.74201
——————————————————————————
.�
𝐸𝐸𝐼𝐼𝑆𝑆𝑂𝑂 = 10.11 + 0.2674𝑀𝑀𝐶𝐶𝑆𝑆h𝑒𝑒𝑟𝑟𝐸𝐸𝐼𝐼𝑆𝑆𝑂𝑂
Second Stage:
• predict educ_hat
. reg lwage educ_hat, r
Linear regression
(0.319) (0.032)
Number of obs
F(1, 426)
= 428 = 0.89
14
——————————————————————————
| Robust
lwage | Coef. Std. Err. t P>|t| [95% Conf. Interval]
————-+—————————————————————-
educ_hat | .0385499 .04084 0.94 0.346 -.041723 .1188229
_cons | .7021746 .5�137775 1.37 ——————————————————————————
𝐿𝐿𝑛𝑛(𝑊𝑊𝑊𝑊𝑊𝑊𝑒𝑒) = 0.7022 + 0.0385𝐸𝐸𝐼𝐼𝑆𝑆𝑂𝑂 (0.514) (0.041)
This regression has WRONG STANDARD ERRORS!
First Stage: 𝐸𝐸𝐼𝐼𝑆𝑆𝑂𝑂 = 𝛾𝛾 + 𝛾𝛾 �𝑀𝑀𝐶𝐶𝑆𝑆h𝑒𝑒𝑟𝑟𝐸𝐸𝐼𝐼𝑆𝑆𝑂𝑂 + 𝛾𝛾 𝐹𝐹𝑊𝑊𝑆𝑆h𝑒𝑒𝑟𝑟 𝐸𝐸𝐼𝐼𝑆𝑆𝑂𝑂 + 𝜐𝜐
012 Second Stage: 𝐿𝐿𝑛𝑛(𝑤𝑤𝑊𝑊𝑊𝑊𝑒𝑒) = 𝑋𝑋0 + 𝑋𝑋1𝐸𝐸𝐼𝐼𝑆𝑆𝑂𝑂 + 𝑆𝑆
(e) Results:
. ivregress 2sls lwage (educ = motheduc fatheduc), r
Instrumental variables (2SLS) regression Number of obs =
Wald chi2(1) =
Prob > chi2 =
R-squared =
Root MSE =
428
2.17
0.1406
0.0841
.69131
——————————————————————————
| Robust
lwage | Coef. Std. Err. z P>|z| [95% Conf. Interval]
————-+—————————————————————-
educ | .0504905 .0342656 1.47 0.141 -.0166689 .1176498
_cons | .5510205 .4299931 1.28 0.200 -.2917506 1.393792
——————————————————————————
Instrumented: educ
Instruments: motheduc fatheduc
. estat endogenous
. estat endogenous
Tests of endogeneity
Ho: variables are exogenous
Robust score chi2(1)
Robust regression F(1,42�5)
= 3.82345 (p = 0.0505)
= 3.93747 (p = 0.0479)
Education is marginally endogenous.
𝐿𝐿𝑛𝑛(𝑤𝑤𝑊𝑊𝑊𝑊𝑒𝑒) = 0.551 + 0.051𝐸𝐸𝐼𝐼𝑆𝑆𝑂𝑂 (0.431) (0.0343)
15
Prob > F =
R-squared =
Root MSE =
0.3457
0.0022
.72324
0.172 -.3076798 1.712029
(f) Results:
• • • • • • • • • • • • • • • • • •
ivreg lwage exper expersq (educ = motheduc fatheduc), r
Instrumental variables (2SLS) regression
Number of obs
F(3, 424)
Prob > F
R-squared
Root MSE
= 428
= 6.15
= 0.0004
= 0.1357
= .67471
——————————————————————————
| Robust
lwage | Coef. Std. Err. t P>|t| [95% Conf. Interval]
————-+—————————————————————-
.1269261
.074728
-.0000536
.8928998
——————————————————————————
educ |
exper |
expersq |
_cons |
.0613966
.0441704
-.000899
.0481003
.0333386
.0155464
.0004301
.4297977
1.84 0.066
-.0041329
.0136128
-.0017443
-.7966992
Instrumente�d: educ
Instruments: exper expersq motheduc fatheduc ——————————————————————————
𝐿𝐿𝑛𝑛(𝑤𝑤𝑊𝑊𝑊𝑊𝑒𝑒) = 0.048 + 0.061𝐸𝐸𝐼𝐼𝑆𝑆𝑂𝑂 + 0.044𝐸𝐸𝑥𝑥𝐸𝐸𝑒𝑒𝑟𝑟 − 0.0009𝐸𝐸𝑥𝑥𝐸𝐸𝑒𝑒𝑟𝑟𝑆𝑆𝐸𝐸
(0.429) (0.033) (0.016)
First Stage: 𝐸𝐸𝐼𝐼𝑆𝑆𝑂𝑂 = 𝛾𝛾0 + 𝛾𝛾1𝑀𝑀𝐶𝐶𝑆𝑆h𝑒𝑒𝑟𝑟𝐸𝐸�𝐼𝐼𝑆𝑆𝑂𝑂 + 𝛾𝛾2𝐹𝐹𝑊𝑊𝑆𝑆h𝑒𝑒𝑟𝑟 𝐸𝐸𝐼𝐼𝑆𝑆𝑂𝑂 + 𝛾𝛾2𝐸𝐸𝑥𝑥𝐸𝐸𝑒𝑒𝑟𝑟 + 𝛾𝛾3𝐸𝐸𝑥𝑥𝐸𝐸𝑒𝑒𝑟𝑟𝑆𝑆𝐸𝐸 + 𝜐𝜐
(0.0004) Second Stage: 𝐿𝐿𝑛𝑛(𝑤𝑤𝑊𝑊𝑊𝑊𝑒𝑒) = 𝑋𝑋0 + 𝑋𝑋1𝐸𝐸𝐼𝐼𝑆𝑆𝑂𝑂 + 𝑋𝑋2𝐸𝐸𝑥𝑥𝐸𝐸𝑒𝑒𝑟𝑟 + 𝑋𝑋3𝐸𝐸𝑥𝑥𝐸𝐸𝑒𝑒𝑟𝑟𝑆𝑆𝐸𝐸 + 𝑆𝑆
(g) Results:
. reg eps_hat motheduc fatheduc exper expersq
= 428
= 0.09
= 0.9845
= 0.0009
= -0.0086
= .67521
——————————————————————————
eps_hat | Coef. Std. Err. t P>|t| [95% Conf. Interval]
————-+—————————————————————-
motheduc | -.0066065 .0118864 -0.56 0.579 -.0299704 .0167573
fatheduc | .0057823 .0111786 0.52 0.605 -.0161902 .0277547
exper | -.0000183 .0133291 -0.00 0.999 -.0262179 .0261813
expersq | 7.34e-07 .0003985 0.00 0.999 -.0007825 .000784
_cons | .0109641 .1412571 0.08 0.938 -.2666892 .2886173
——————————————————————————
. test motheduc fatheduc
Source | SS df MS Number of obs
————-+———————————- F(4, 423)
Model | .170503136 4 .042625784 Prob > F
Residual | 192.84951 423 .455909007 R-squared
————-+———————————- Adj R-squared
Total | 193.020013 427 .452037502 Root MSE
2.84 0.005
-2.09 0.037
0.11 0.911
16
( 1) motheduc = 0
( 2) fatheduc = 0
F( 2, 423) = 0.19
Prob > F = 0.8295
17