Solution Homework 8
8.7 (i) This follows from the simple fact that, for uncorrelated random variables, the variance of
the sum is the sum of the variances: Var( fi  vi ,e )  Var( fi )  Var(vi ,e )   2f   v2 .
(ii) We compute the covariance between any two of the composite errors as
Cov(ui ,e , ui , g )  Cov( fi  vi ,e , fi  vi , g )  Cov( fi , fi )  Cov( fi , vi , g )  Cov( vi ,e , fi )  Cov( vi ,e , vi , g )
 Var( fi )  0  0  0   2f ,
where we use the fact that the covariance of a random variable with itself is its variance and the
assumptions that fi , vi ,e , and vi , g are pairwise uncorrelated.
(iii) This is most easily solved by writing
mi1 ei 1 ui ,e  mi1 ei 1 ( fi  ui ,e )  fi  mi1 ei 1 vi ,e .
m
m
m
Now, by assumption, fi is uncorrelated with each term in the last sum; therefore, fi is uncorrelated
m
with mi1 ei 1 vi ,e . It follows that



Var fi  mi1  ei 1 vi ,e  Var  fi   Var mi1  ei 1 vi ,e
m
m

  2f   v2 / mi ,
where we use the fact that the variance of an average of mi uncorrelated random variables with
common variance (  v2 in this case) is simply the common variance divided by mi – the usual
formula for a sample average from a random sample.
(iv) The standard weighting ignores the variance of the firm effect,  2f . Thus, the
(incorrect) weight function used is 1/ hi  mi . A valid weighting function is obtained by writing
the variance from (iii) as Var(ui )   2f [1  ( v2 /  2f ) / mi ]   2f hi . But obtaining the proper
weights requires us to know (or be able to estimate) the ratio  v2 /  2f . Estimation is possible, but
we do not discuss that here. In any event, the usual weight is incorrect. When the mi are large or
the ratio  v2 /  2f is small – so that the firm effect is more important than the individual-specific
effect – the correct weights are close to being constant. Thus, attaching large weights to large
firms may be quite inappropriate.
SOLUTIONS TO COMPUTER EXERCISES
C8.11 (i) The usual OLS standard errors are in (), the heteroskedasticity-robust standard errors
are in []:
nettfa = 1.50
(15.31)
[19.09]
+ .774 inc  1.60 age + .029 age2 + 2.47 male + 6.98 e401k
(.062)
(0.77)
(.009)
(2.05)
(2.13)
[.102]
[1.08]
[.014]
[2.06]
[2.19]
n = 2,017, R2 = .128.
(ii) The smallest hˆi is about 12.83 and the largest is about 58,059.74. Thus, there is wide
variation in the estimated conditional variances.
(iii) The usual WLS standard errors are in (), the standard errors robust to misspecified
variance are in []:
nettfa = 2.58
(9.94)
[8.19]
+ .456 inc  .613 age + .013 age2 + 1.42 male + 4.26 e401k
(.058)
(.541)
(.007)
(1.03)
(1.23)
[.062]
[.408]
[.005]
[0.82]
[1.14]
n = 2,017, R2 = .062.
Interestingly, except for the income coefficient, the robust standard errors are actually smaller
than the usual standard error. This could just be sampling variation, or it could be that the
variance function is misspecified in such a way that, when it is used in WLS, the usual standard
errors overestimate the actual sampling variation.
(iv) The robust standard error for the e401k coefficient is 2.19, while that for WLS is 1.14.
Thus, the WLS standard error is just over half as large as the OLS standard error. Assuming that
the zero conditional mean assumption actually holds – something that is not clear given some
nontrivial changes in the WLS estimates as compared with OLS – the smaller robust standard
error for WLS suggests it is the more efficient procedure, whether or not we have properly
specified the “skedastic” function.