Answers
Chapter 14
GaussMarkovUnivariate.xls Answers
1) Prove that the residuals produced by the Sample Average estimator will always
sum to zero.
A) The intuition for this is that the residuals are the deviations of the original
observations from their mean and, because the mean is in the middle the deviations will
n
sum to zero. More formally, Y 
Y
i 1
n
i
is the Sample Average estimate.
The residuals are defined as Re sidual i  Yi  Y . The sum of the residuals, then, is
 Residual i   Yi  Y 
n
n
i 1
i 1
n
  Yi  nY
i 1
n
n
i 1
i 1
  Yi   Yi
 0.
2) Produce your own, new estimator of the population average in which the CEM
describes the data generation process. Is your estimator unbiased? Is it linear?
Provide reasons for your answers.
The answers will vary.
One possibility is the Even estimator. Each even-numbered observation is weighted 0.2,
and the odd-numbered observations are given 0 weight.
This is linear because it can be written as a weighted sum of the Y values.
This is unbiased because the weights sum to 1. You can also run a Monte Carlo to
support this claim.
GaussMarkovUnivariateAns.doc
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Answers
Chapter 14
3) (Requires use of algebra of expectations.) Show that the 0.9 Estimator is indeed
biased, and compute its expected value assuming it follows the CEM.
A) We can compute the expected value of the 0.9 Estimator in exactly the same way as
n
we did for the Sample Average. The difference is that the sum of the weights,
w
i 1
0.9, not 1:
 n

 n

E  wi Yi   E  wi    i  
 i 1

i
 i 1
 n

 n

 E  wi    E  wi  i 

i
 i 1
 i 1
n
n
   wi   wi E  i 
i 1
i 1
n
   0.9   wi  0
i 1
 0.9   0
 0.9  .
Inasmuch as the expected value does not equal β, this estimator is biased.
GaussMarkovUnivariateAns.doc
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i
, is
Answers
Chapter 14
4) (Requires use of algebra of expectations.) Assume that the data generation
process is as described in Model 1 with SD(i)= 5.
Someone proposes an estimator with the following weights:
Obs
Weight
1
2
3
4
5
6
7
8
9
10
1
1
1
1
1
1
1
1
1
-8
(a) Is this an unbiased estimator of the population average? Compute its
expected value.
A) It is unbiased because the sum of the weights is 1. The computation of the expected
value looks just like that for the sample average:
 n

 n

E  wi Yi   E  wi    i  
 i 1

i
 i 1
 n

 n

 E  wi    E  wi  i 

i
 i 1
 i 1
n
n
   wi   wi E  i 
i 1
i 1
n
   1   wi  0
i 1
  0

(b) What is the SE of this estimator?
A) You can use the ComputingSEs sheet to find that its SE is huge: 42.
SD()
Variance()
5
25
Weights (w i )
wi2SD()2
Sample
Estimator Sample
Estimator for
Observation Average
for Q6
Average
Q6
1
0.1
1
0.2500
25.0000
2
0.1
1
0.2500
25.0000
3
0.1
1
0.2500
25.0000
4
0.1
1
0.2500
25.0000
5
0.1
1
0.2500
25.0000
6
0.1
1
0.2500
25.0000
7
0.1
1
0.2500
25.0000
8
0.1
1
0.2500
25.0000
9
0.1
1
0.2500
25.0000
10
0.1
-8
0.2500 1600.0000
Sum
1
1
2.500
1825.000
SquareRoot
1.581
42.720
GaussMarkovUnivariateAns.doc
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Answers
Chapter 14
5) Provide a recipe for obtaining an estimate of the slope in a bivariate version of the
standard econometric model:
Yi = Intercept + Slope*Xi.
Imagine that there are five observations; thus, n = 5.
Your recipe must involve the Y's and the X's.
To be even more concrete, you might show how the estimator works with the
following data set:
Xi Y i
10 60.33
20 112.55
30 165.03
40 218.45
50 255.03,
or this data set:
Xi Y i
10 64.93
20 110.95
30 163.12
40 215.91
50 264.88,
or another sample. Give us an estimator (a recipe) for obtaining the sample slope
that we can use with data like this.
Your recipe must be one we could use on any data set with five observations on X
and Y, not just a particular sample.
Also tell us whether your estimator is Linear and make a guess as to whether it is
Unbiased.
A) Answers will vary. However, we obtain many wrong answers.
Wrong answer A: “The answer is 0.2.”
To obtain this wrong answer, you probably tried to find the slope and have given us
an estimate, not an estimator. An estimator is a recipe that must work with any data
set, not just one of the data sets used as an example. That recipe should tell us how to
combine the X’s and Y’s. If it is a linear estimator, it should look like this:
n
Slope   wi  Yi , where you need to specify the weights as functions of X1, X2,
i 1
through X5 (on the assumption that n = 5). Notice we do not specify it in terms of 10,
20, 30, through 50 because those are specific X values and the recipe needs to be
general.
The most charitable thing we can say about 0.2 is that it is an estimator in the sense
that you could just say 0.2 no matter what the X’s and Y’s are. It is linear in the sense
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Answers
Chapter 14
n
that it looks like Slope  0.2   wi  Yi , where all the weights are 0. It is clearly
i 1
biased unless the Slope ( 1 ) magically happens to be 0.2.
Wrong answer B: Some combination of the two data sets we gave you as examples.
Again, you are using specific numbers and not giving us a general recipe. Also, note
that, in practice, you typically have one data set and will not be able to combine
observations from two data sets.
Wrong answer C: You give us an estimator, but it has the Y’s raised to some power
other than 1 and you say the estimator is linear. No, in that case it is nonlinear.
Right, but suspicious answer D: You give us the Average Slopes or Extreme Points
estimators. Perhaps you are really sharp; perhaps you read ahead and forgot to cite us
as the source for your answer!
Bottom line: This is a very hard question, which is designed to get you to think about
what the terms estimator and linear mean. You need to distinguish between estimates
and estimators, linear and nonlinear, and unbiased and biased. Now that Section 14.7
has been covered, you have the tools to determine whether linear estimators are
unbiased. Determining whether a nonlinear estimator is unbiased is really hard,
though the default answer should be “probably not.” For example, in the univariate
case, the median may or may not be unbiased for the population average. It depends
on the distribution of the tickets in the box, whereas the Sample Average is
guaranteed to be unbiased provided that the population average exists.
GaussMarkovUnivariateAns.doc
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Answers
Chapter 14
6) We set the SD of the errors to 200 in the UnivariateSample sheet, drew a sample,
and got the result pictured below.
Evaluate the following statement: "Holy cow! The Sample Average estimator is
biased when the SD of the errors gets large enough. That estimate is way low! It’s
off the chart!"
Population Parameters

200
SD( )
200

200
200
200
200
200
200
200
200
200
200
Sum
i
212.14
-5.57
-141.16
-138.00
21.33
-386.52
-312.53
-94.92
-216.46
-8.18
Yi
412.14
194.43
58.84
62.00
221.33
-186.52
-112.53
105.08
-16.46
191.82
-1069.87
930.13
wi
Sample Average Estimator
wi
w i i
w i Yi
0.1
20
21.214
41.214
0.1
20
-0.557
19.443
0.1
20
-14.116
5.884
0.1
20
-13.800
6.200
0.1
20
2.133
22.133
0.1
20
-38.652
-18.652
0.1
20
-31.253
-11.253
0.1
20
-9.492
10.508
0.1
20
-21.646
-1.646
0.1
20
-0.818
19.182
1.0
200
-106.987
93.013
Observed Y
215
210
205
200
195
190
185
1
2
3
4
5
6
7
Observation Number
8
9
10
A) From the table above we can tell that the estimator is unbiased—the sum of the
weights is 1. Because the spread of the errors (SD()) is quite high, it is not surprising
that individual estimates will be very far from the true value of . The estimator is
unbiased because the long-run average value of the estimates will equal the true value of
.
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Answers
Chapter 14
7) Is the exact SE of the Diminishing Weights estimator always a constant multiple
of the exact SE of the Sample Average estimator no matter what the value of SD( )
is? If so, why? If not, why not?
A) Yes. The reason can be found in the formula for the Exact SE of a linear estimator in
this case. Both the Sample Average and Diminishing Weights estimators are linear
estimators whose SE can be easily computed. In each case the SE depends on the sum of
the squares of the weights and the SD of the errors.
 n

 n

SE  wiYi   Var  wi    i 
 i 1

 i 1



 Var  wi   wi i 
 i 1

n
 n

 Var  wi i 
 i 1


n
w
i 1
2
i
 SD 
2
n
 SD  
w
2
i 1
 SD  
2
i
Because for this DGP, the SE of every linear
estimator is directly proportional to the SD of the
error terms, the ratio of the SEs of two linear
estimators does not change when the SD of the
error terms changes.
This derivation can be verified for the special case
we considered. See the ComputingSEs sheet. We
chose different values of SD() (you must set
those values on the UnivariateSample sheet) and
obtained the following table:
n
w
i 1
2
i
SD()
10
1
5
SE(Sample
Average
Estimator)
3.162
0.316
1.581
SE(Diminishing
Weights
Estimator)
5.774
0.577
2.887
Ratio
1.826
1.826
1.826
The ratio 1.826 is simply the ratio of the sum of the squared weights for the two
estimators:
n
 Diminishin
i 1
n
g wi2

0.5 2  0.25 2    0.00195312
 Sample Average wi2
0.12  0.12    0.12
i 1
 1.826.
GaussMarkovUnivariateAns.doc
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Answers
Chapter 14
8) Scroll down to row 100 of the UnivariateSample sheet. You are offered four
estimators. Use the Monte Carlo simulation add-in to determine where to place
them in the grid below. The grid is actually composed of four separate text boxes.
Click on a box and type in your answer. Select cell range A93:I116, hold down the
Shift key, and execute Edit: Copy Picture. Paste your picture in a Word document.
biased
linear
nonlinear
unbiased
linear
arithmetic mean
(aka, average)
geometric mean
harmonic mean
median
biased
nonlinear
unbiased
The Monte Carlo results that support our answer can be accessed from
GaussMarkovUnivariate.xls. Execute Format: Sheet: Unhide and select
MCRawCompare.
The harmonic and geometric means are nonlinear estimators because they cannot be
written as a linear function of the Y’s. The harmonic mean uses reciprocals of the Y’s in
its formula and the geometric mean takes the Y’s multiplied by each other to the power
1/n, where n is the number of observations in the sample (10 in our case). For more on
various means and how they are related to each other, see
http://mathworld.wolfram.com/PowerMean.html
More specific information on the geometric mean is at
http://mathworld.wolfram.com/GeometricMean.html
More specific information for the harmonic mean can be found at
http://mathworld.wolfram.com/HarmonicMean.html.
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