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TABLE 7.12
(
Continued
)
guj75772_ch07.qxd 27/08/2008 10:53 AM Page 226
Chapter 7
Multiple Regression Analysis: The Problem of Estimation
227
500
450
400
350
250
300
150
200
50
100
0
1/3/95
2/27/95
4/24/95
6/19/95
8/14/95
10/9/95
12/4/95
1/29/96
3/25/96
5/20/96
7/15/96
9/9/96
11/4/96
12/30/96
2/24/97
4/21/97
6/16/97
8/11/97
10/6/97
12/1/97
1/26/98
3/23/98
5/18/98
7/13/98
9/8/98
11/2/98
12/28/98
2/22/99
4/19/99
6/14/99
8/9/99
10/4/99
11/29/99
Price
Date
FIGURE 7.4
Qualcomm stock
prices over time.
c.
Finally, fit the following cubic or
third-degree polynomial:
Y
i
=
β
0
+
β
1
X
i
+
β
2
X
2
i
+
β
3
X
3
i
+
u
i
where
Y
=
stock price and
X
=
time. Which model seems to be the best estimator
for the stock prices?
Appendix
7A
7A.1
Derivation of OLS Estimators
Given in Equations (7.4.3) to (7.4.5)
Differentiating the equation
ˆ
u
2
i
=
(
Y
i
− ˆ
β
1
− ˆ
β
2
X
2
i
− ˆ
β
3
X
3
i
)
2
(7.4.2)
partially with respect to the three unknowns and setting the resulting equations to zero, we obtain
∂
ˆ
u
2
i
∂
ˆ
β
1
=
2
(
Y
i
− ˆ
β
1
− ˆ
β
2
X
2
i
− ˆ
β
3
X
3
i
)(
−
1)
=
0
∂
ˆ
u
2
i
∂
ˆ
β
2
=
2
(
Y
i
− ˆ
β
1
− ˆ
β
2
X
2
i
− ˆ
β
3
X
3
i
)(
−
X
2
i
)
=
0
∂
ˆ
u
2
i
∂
ˆ
β
3
=
2
(
Y
i
− ˆ
β
1
− ˆ
β
2
X
2
i
− ˆ
β
3
X
3
i
)(
−
X
3
i
)
=
0
Simplifying these, we obtain Eqs. (7.4.3) to (7.4.5).
guj75772_ch07.qxd 11/08/2008 04:22 PM Page 227
228
Part One
Single-Equation Regression Models
In passing, note that the three preceding equations can also be written as
ˆ
u
i
=
0
ˆ
u
i
X
2
i
=
0
(Why?)
ˆ
u
i
X
3
i
=
0
which show the properties of the least-squares fit, namely, that the residuals sum to zero and that they
are uncorrelated with the explanatory variables
X
2
and
X
3
.
Incidentally, notice that to obtain the OLS estimators of the
k
-variable linear regression model
(7.4.20) we proceed analogously. Thus, we first write
ˆ
u
2
i
=
(
Y
i
− ˆ
β
1
− ˆ
β
2
X
2
i
− · · · − ˆ
β
k
X
ki
)
2
Differentiating this expression partially with respect to each of the
k
unknowns, setting the resulting
equations equal to zero, and rearranging, we obtain the following
k
normal equations in the
k
unknowns:
Y
i
=
n
ˆ
β
1
+ ˆ
β
2
X
2
i
+ ˆ
β
3
X
3
i
+ · · · + ˆ
β
k
X
ki
Y
i
X
2
i
= ˆ
β
1
X
2
i
+ ˆ
β
2
X
2
2
i
+ ˆ
β
3
X
2
i
X
3
i
+ · · · + ˆ
β
k
X
2
i
X
ki
Y
i
X
3
i
= ˆ
β
1
X
3
i
+ ˆ
β
2
X
2
i
X
3
i
+ ˆ
β
3
X
2
3
i
+ · · · + ˆ
β
k
X
3
i
X
ki
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Y
i
X
ki
= ˆ
β
1
X
ki
+ ˆ
β
2
X
2
i
X
ki
+ ˆ
β
3
X
3
i
X
ki
+ · · · + ˆ
β
k
X
2
ki
Or, switching to small letters, these equations can be expressed as
y
i
x
2
i
= ˆ
β
2
x
2
2
i
+ ˆ
β
3
x
2
i
x
3
i
+ · · · + ˆ
β
k
x
2
i
x
ki
y
i
x
3
i
= ˆ
β
2
x
2
i
x
3
i
+ ˆ
β
3
x
2
3
i
+ · · · + ˆ
β
k
x
3
i
x
ki
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
y
i
x
ki
= ˆ
β
2
x
2
i
x
ki
+ ˆ
β
3
x
3
i
x
ki
+ · · · + ˆ
β
k
x
2
ki
It should further be noted that the
k
-variable model also satisfies these equations:
ˆ
u
i
=
0
ˆ
u
i
X
2
i
=
ˆ
u
i
X
3
i
= · · · =
ˆ
u
i
X
ki
=
0
guj75772_ch07.qxd 11/08/2008 04:22 PM Page 228
Chapter 7
Multiple Regression Analysis: The Problem of Estimation
229
7A.2
Equality between the Coefficients of PGNP
in Equations (7.3.5) and (7.6.2)
Letting
Y
=
CM,
X
2
=
PGNP, and
X
3
=
FLR and using the deviation form, write
y
i
=
b
1 3
x
3
i
+ ˆ
u
1
i
(1)
x
2
i
=
b
2 3
x
3
i
+ ˆ
u
2
i
(2)
Now regress
ˆ
u
1
on
ˆ
u
2
to obtain:
a
1
=
ˆ
u
1
i
ˆ
u
2
i
ˆ
u
2
2
i
= −
0
.
0056
(for our example)
(3)
Note that because the
ˆ
u
’s are residuals, their mean values are zero. Using (1) and (2), we can write
(3) as
a
1
=
(
y
i
−
b
1 3
x
3
i
)(
x
2
i
−
b
2 3
x
3
i
)
(
x
2
i
−
b
2 3
x
3
i
)
2
(4)
Expand the preceding expression, and note that
b
2 3
=
x
2
i
x
3
i
x
2
3
i
(5)
and
b
1 3
=
y
i
x
3
i
x
2
3
i
(6)
Making these substitutions into (4), we get
ˆ
β
2
=
y
i
x
2
i
x
2
3
i
−
y
i
x
3
i
x
2
i
x
3
i
x
2
2
i
x
2
3
i
−
x
2
i
x
3
i
2
(7.4.19)'>(7.4.7)
−
0.0056
(for our example)
7A.3
Derivation of Equation (7.4.19)
Recall that
ˆ
u
i
=
Y
i
− ˆ
β
1
− ˆ
β
2
X
2
i
− ˆ
β
3
X
3
i
which can also be written as
ˆ
u
i
=
y
i
− ˆ
β
2
x
2
i
− ˆ
β
3
x
3
i
where small letters, as usual, indicate deviations from mean values.
Now
ˆ
u
2
i
=
(
ˆ
u
i
ˆ
u
i
)
=
ˆ
u
i
(
y
i
− ˆ
β
2
x
2
i
− ˆ
β
3
x
3
i
)
=
ˆ
u
i
y
i
guj75772_ch07.qxd 11/08/2008 04:22 PM Page 229
230
Part One
Single-Equation Regression Models
where use is made of the fact that
ˆ
u
i
x
2
i
=
ˆ
u
i
x
3
i
=
0
.
(Why?) Also
ˆ
u
i
y
i
=
y
i
ˆ
u
i
=
y
i
(
y
i
− ˆ
β
2
x
2
i
− ˆ
β
3
x
3
i
)
that is,
ˆ
u
2
i
=
y
2
i
− ˆ
β
2
y
i
x
2
i
− ˆ
β
3
y
i
x
3
i
(7.4.19)
which is the required result.
7A.4
Maximum Likelihood Estimation
of the Multiple Regression Model
Extending the ideas introduced in Chapter 4, Appendix 4A, we can write the log-likelihood function
for the
k
-variable linear regression model (7.4.20) as
ln
L
= −
n
2
ln
σ
2
−
n
2
ln (2
π
)
−
1
2
(
Y
i
−
β
1
−
β
2
X
2
i
− · · · −
β
k
X
ki
)
2
σ
2
Differentiating this function partially with respect to
β
1
,
β
2
,
. . .
,
β
k
and
σ
2
, we obtain the following
(
K
+
1) equations:
∂
ln
L
∂β
1
= −
1
σ
2
(
Y
i
−
β
1
−
β
2
X
2
i
− · · · −
β
k
X
ki
)(
−
1)
(1)
∂
ln
L
∂β
2
= −
1
σ
2
(
Y
i
−
β
1
−
β
2
X
2
i
− · · · −
β
k
X
ki
)(
−
X
2
i
)
(2)
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
∂
ln
L
∂β
k
= −
1
σ
2
(
Y
i
−
β
1
−
β
2
X
2
i
− · · · −
β
k
X
ki
)(
−
X
ki
)
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