Solutions Manual Econometric Analysis of Cross Section and Panel Data 2nd Edition By Jeffrey M. Wooldridg

Solutions Manual
Econometric Analysis of Cross Section and Panel Data 2nd Edition

By
Jeffrey M. Wooldridge

( All Chapters Included - 100% Verified Solutions )




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, Solutions to Chapter 2 Problems
2.1. a. Simple partial differentiation gives

∂Ey|x 1 , x 2 
 1  4x2
∂x 1

and

∂Ey|x 1 , x 2 
  2  2 3 x 2   4 x 1
∂x 2

b. By definition, Eu|x 1 x 2   0. Because x 22 and x 1 x 2 are functions of x 1 , x 2 , it does not

matter whether or not we also condition on them: Eu|x 1 , x 2, x 22 , x 1 x 2   0.

c. All we can say about Varu|x 1, x 2  is that it is nonnegative for all x 1 and x 2 :

Eu|x 1 , x 2   0 in no way restricts Varu|x 1 , x 2 .

2.2. a. Because ∂Ey|x/∂x   1  2 2 x − µ, the marginal effect of x on Ey|x is a linear

function of x. If  2 is negative then the marginal effect is less than  1 when x is above its mean.

If, for example,  1  0 and  2  0, the marginal effect will eventually be negative for x far

enough above . (Whether the values for x such that ∂Ey|x/∂x  0 represents an interesting

segment of the population is a different matter.)

b. Because ∂Ey|x/∂x is a function of x, we take the expectation of ∂Ey|x/∂x over the

distribution of x: E∂Ey|x/∂x  E 1  2 2 x −    1  2 2 Ex −    1 .

c. One way to do this part is to apply Property LP.5 from Appendix 2A. We have

Ly|1,x  LEy|x   0   1 Lx − |1, x   2 Lx −  2 |1, x
  0   1 x −    2  0   1 x,

because Lx − |1, x  x −  and  0   1 x is the linear projection of x −  2 on x. By

assumption, x −  2 and x are uncorrelated, and so  1  0. It follows that



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, Ly|x   0 −  1    2  0    1 x

2.3. a. y   0   1 x 1   2 x 2   3 x 1 x 2  u, where u has a zero mean given x 1 and x 2 :

Eu|x 1 , x 2   0. We can say nothing further about u.

b. ∂Ey|x 1 , x 2 /∂x 1   1   3 x 2 . Because Ex 2   0,  1  E∂Ey|x 1 , x 2 /∂x 1 , that is,  1 is

the average partial effect of x 1 on Ey|x 1 , x 2 /∂x 1 . Similarly,  2 E∂Ey|x 1 , x 2 /∂x 2 .

c. If x 1 and x 2 are independent with zero mean then Ex 1 x 2   Ex 1 Ex 2   0. Further,

the covariance between x 1 x 2 and x 1 is Ex 1 x 2  x 1   Ex 21 x 2   Ex 21 Ex 2  (by

independence)  0. A similar argument shows that the covariance between x 1 x 2 and x 2 is zero.

But then the linear projection of x 1 x 2 onto 1, x 1 , x 2  is identically zero. Now just use the law

of iterated projections (Property LP.5 in Appendix 2A):

Ly|1, x 1 , x 2   L 0   1 x 1   2 x 2   3 x 1 x 2 |1, x 1 , x 2 
  0   1 x 1   2 x 2   3 Lx 1 x 2 |1, x 1 , x 2 
 0  1x1  2x2.

d. Equation (2.47) is more useful because it allows us to compute the partial effects of x 1

and x 2 at any values of x 1 and x 2 . Under the assumptions we have made, the linear projection

in (2.48) does have as its slope coefficients on x 1 and x 2 the partial effects at the population

average values of x 1 and x 2 – zero in both cases – but it does not allow us to obtain the partial

effects at any other values of x 1 and x 2 . Incidentally, the main conclusions of this problem go

through if we allow x 1 and x 2 to have nonzero population means.

2.4. By assumption,

Eu|x, v   0  x   1 v

for some scalars  0 ,  1 and a column vector . Now, it suffices to show that  0  0 and   0.

One way to do this is to use LP.7 in Appendix 2A, and in particular, equation (2.56). This says


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, that  0 ,  ′  ′ can be obtained by first projecting 1, x onto v, and obtaining the population

residual, r. Then, project u onto r. Now, since v has zero mean and is uncorrelated with x, the

first step projection does nothing: r  1, x. Thus, projecting u onto r is just projecting u onto

1, x. Since u has zero mean and is uncorrelated with x, this projection is identically zero,

which means that  0  0 and   0.

2.5. By definition and the zero conditional mean assumptions, Varu 1 |x, z  Vary|x, z

and Varu 2 |x  Vary|x. By assumption, these are constant and necessarily equal to

 21 ≡ Varu 1  and  22 ≡ Varu 2 , respectively. But then Property CV.4 implies that  22 ≥  21 .

This simple conclusion means that, when error variances are constant, the error variance falls

as more explanatory variables are conditioned on.

2.6. a. By linearity of the linear projection,

Lq|1, x  Lq ∗ |1, x  Le|1, x  Lq ∗ |1, x,

where the last inequality follows because Le|1, x  0 when Ee  0 and Ex ′ e  0.

Therefore, the parameters in the linear projection of q onto 1,x are the same as the linear

projection of q ∗ onto 1,x. This fact is useful for studying equations with measurement error

in the explained or explanatory variables.

b. r  q − Lq|1, x  q ∗  e − Lq|1, x  q ∗  e − Lq ∗ |1, x (from part a)

 q ∗ − Lq ∗ |1, x  e  r ∗  e.

2.7. Write the equation in error form as

y  gx  z u
Eu|x, z  0.

Take the expected value of the first equation conditional only on x:




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