Student's Solutions Manual And Supplementary Materials
For Econometric Analysis Of Cross Section And Panel
Data
Second Edition
,This File And Several Accompanying Files Contain The Solutions To The Odd-
Numbered Problems In The Book Econometric Analysis Of Cross Section And Panel
Data, By Jeffrey M. Wooldridge, MIT Press, 2002. The Empirical Examples Are
Solved Using Various Versions Of Stata, With Some Dating Back To Stata 4.0.
Partly Out Of Laziness, But Also Because It Is Useful For Students To See
Computer Output, I Have Included Stata Output In Most Cases Rather Than Type
Tables. In Some Cases, I Do More Hand Calculations Than Are Needed In Current
Versions Of Stata.
Currently, There Are Some Missing Solutions. I Will Update The Solutions
Occasionally To Fill In The Missing Solutions, And To Make Corrections. For
Some Problems I Have Given Answers Beyond What I Originally Asked. Please
Report Any Mistakes Or Discrepencies You Might Come Across By Sending Me E-
Mail At Wooldri1@Msu.Edu.
CHAPTER 2
De(Y|X1,X2) De(Y|X1,X2)
2.1. A. + B4x2 And ---------- 2 + 2b3x2 + B4x1. 1
-- - -- - -- -- - -- - -- - -- - -- - -- - -- -- - -- - -- - -- - -- - -- -- -
= b = b
Dx1 Dx2
-- - -- -- - -- - -
2
B. By Definition, E(U |X ,X ) = 0. And X X Are Just Functions
1 2 Because X2 1 2
Of (X1,X2), It Does Not Matter Whether We Also Condition On Them:
2
E(U| X1,X2,X2,X1x2) = 0.
C. All We Can Say About Var(U|X1,X2) Is That It Is Nonnegative For All X1
And X2: E(U|X1,X2) = 0 In No Way Restricts Var(U|X1,X2).
2.3. A. Y = B0 + B1x1 + B2x2 + B3x1x2 + U, Where U Has A Zero Mean Given X1
And X2: E(U|X1,X2) = 0. We Can Say Nothing Further About U.
b. De(Y|X1,X2)/Dx1 = B1 + B3x2. Because E(X2) = 0, B1 =
1
,E[De(Y|X1,X2)/Dx1]. Similarly, B2 = E[De(Y|X1,X2)/Dx2].
c. If X1 And X2 Are Independent With Zero Mean Then E(X1x2) = E(X1)E(X2)
2
= 0. Further, The Covariance Between X X And X Is E(X X W
1 2 1 1 2 X1) = E(X1x2) =
2
E(X1)E(X2) (By Independence) = 0. A Similar Argument Shows That The
Covariance Between X1x2 And X2 Is Zero. But Then The Linear Projection Of
X1x2 Onto (1,X1,X2) Is Identically Zero. Now Just Use The Law Of Iterated
Projections (Property LP.5 In Appendix 2A):
L(Y|1,X1,X2) = L(B0 + B1x1 + B2x2 + B3x1x2|1,X1,X2)
= B0 + B1x1 + B2x2 + B3l(X1x2|1,X1,X2)
= B0 + B1x1 + B2x2.
d. Equation (2.47) Is More Useful Because It Allows Us To Compute The
Partial Effects Of X1 And X2 At Any Values Of X1 And X2. Under The
Assumptions We Have Made, The Linear Projection In (2.48) Does Have As Its
Slope Coefficients On X1 And X2 The Partial Effects At The Population Average
Values Of X1 And X2 -- Zero In Both Cases -- But It Does Not Allow Us To
Obtain The Partial Effects At Any Other Values Of X1 And X2. Incidentally,
The Main Conclusions Of This Problem Go Through If We Allow X1 And X2 To Have
Any Population Means.
2.5. By Definition, Var(U1|X,Z) = Var(Y|X,Z) And Var(U2|X) = Var(Y|X). By
2 2
Assumption, These Are Constant And Necessarily Equal To S _ Var(U ) And S _
1 1 2
2 2
Var(U 2), Respectively. But Then Property CV.4 Implies That S 2 > S1. This
Simple Conclusion Means That, When Error Variances Are Constant, The Error
Variance Falls As More Explanatory Variables Are Conditioned On.
2.7. Write The Equation In Error Form As
2
, Y = G(X) + Zb + U, E(U|X,Z) = 0.
Take The Expected Value Of This Equation Conditional Only On X:
E(Y|X) = G(X) + [E(Z|X)]B,
And Subtract This From The First Equation To Get
Y - E(Y|X) = [Z - E(Z|X)]B + U
~ ~ ~ ~
Y Zb Z OR =
z) + U. Because
Is A Function Of (X,Z), E(U| = 0 (Since E(U|X,Z) =
~ ~ ~
E(Y|Z) 0
z)B, And So = . This Basic Result Is
Fundamental In The Literature On Estimating Partial Linear Models. First, One
Estimates E(Y|X) And E(Z|X) Using Very Flexible Methods, Typically, So-Called
Nonparametric Methods.
~ ^ ~
Then, After Obtaining Residuals Of The Form Yi _ Yi - E(Yi|Xi) And Zi _ Zi - -
^ ~ ~
E(Zi|Xi), B Is Estimated From An OLS Regression Yi On Zi, I = 1,...,N. Under
General Conditions, This Kind Of Nonparametric Partialling-Out Procedure Leads
To A Rn-Consistent, Asymptotically Normal Estimator Of B. See Robinson (1988)
And Powell (1994).
CHAPTER 3
3.1. To Prove Lemma 3.1, We Must Show That For All E > 0, There Exists Be < 8
And An Integer Ne Such That P[|Xn| > Be] < E, All N > Ne. We Use The
P
Following Fact: Since Xn L A, For Any E > 0 There Exists An Integer Ne Such
That P[|Xn - A| > 1] < E For All N > Ne. [The Existence Of Ne Is Implied By
Definition 3.3(1).] But |Xn| = |Xn - A + A| < |Xn - A| + |A| (By The Triangle
Inequality), And So |Xn| - |A| < |Xn - A|. It Follows That P[|Xn| - |A| > 1]
< P[|Xn - A| > 1]. Therefore, In Definition 3.3(3) We Can Take Be _ |A | + 1
(Irrespective Of The Value Of E) And Then The Existence Of Ne Follows From
Definition 3.3(1).
3
