lecture #6
DESCRIPTION
Lecture #6. Studenmund(2006) Chapter 7. Objectives:. 1. Suppressing the intercept 2. Alternative Functional forms 3. Scaling and units of measurement. Estimated relationship Suppressing the intercept. ^. Such an effect potentially biases the β s and inflates their t-values. Y. - PowerPoint PPT PresentationTRANSCRIPT
6.1
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Lecture #6
Studenmund(2006) Chapter 7
1. Suppressing the intercept2. Alternative Functional forms3. Scaling and units of measurement
Objectives:
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XYi 10ˆˆˆ
True relation
Y
X
XYi'
1' ˆˆ
Estimated relationshipEstimated relationshipSuppressing the interceptSuppressing the intercept
Such an effect potentially biases the βs and inflates their t-values
^
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Regression through the origin
The intercept term is absent or zero.i.e.,
iii XY 1
iY
iX
1
1^
ii XYSRF 1:^^
0
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The estimated model:
i1 XY ~~ or
iiXY ~~ 1
Regression through the origin
Applied OLS method:
21
i
ii
X
YX~
2
2
1iX
Var
^~
and
~
N -1
22 and
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Some feature of no-intercept model
1. i~ need not be zero
2. R2can be occasions turn out to be negative. may not be appropriate for the summary of statistics.
3. df does not include the constant term, i.e., (n-k)
In practice:
•1. A very strong priori or theoretical expectation, otherwise stick to the conventional intercept-present model.
•2. If intercept is included in the regression model but it turns out to be statistically insignificant, then we have to drop the intercept to re-run the regression.
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iXY 10
Regression through origin
i’X Y 1
^
2
22
n
1
22
n
’~
22
2
2
YYXX
YYXXR
or
22
2
2
yxxy
R
2
2
Y2X2
( XY )Rraw
1x2
xy^
1 X2
XY~
1
X2Var
~ ^
2
21
xVar
^ ^
N-k-1 N-k
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Example 1: Capital Asset Pricing Model (CAPM) security ii’s expect risk premium=expected market risk premium
fm1fi rERrER
expected rate of return on security i
risk free of return
expected rate of return on market portfolio
1 as a measure of systematic risk.
1 >1 ==> implies a volatile or aggressive security.
1 <1 ==> implies a defensive security.
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Example 1:(cont.)
fi rER
fERm
1
1
Security market line
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Example 2: Covered Interest Parity
International interest rate differentials equal exchange rate forward premium.
i.e., )(*
eeF
ii 1
NN f
eeF
ii )( *
*ii
eeF
1
1
Covered interest parity line
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Example 2:(Cont.)
in regression:
i10 ue
eFii )()( *
0)( 0E
If covered interest parity holds, 0 is expected to be zero.
Use the t-test to test the intercept to be zero
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y: Return on A Future Fund, %
X: Return on Fisher Index, %
XY 0899.1^(5.689)
Formal report: R2=0.714SEE=19.54
N=10
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The t-value shows that b0 is statistically insignificant different from zero
XY 0691.12797.1 ^(0.166) (4.486)
R2=0.715SEE=20.69
N=10
1.279 - 0
7.668
H0: 0 = 0
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Functional Forms of Regression
The term linear in a simple regression model means that there are linear in the parameters; variables in the regression model may or may not be linear.
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True model is nonlinear
Y
X
Income
Age6015
PRF
SRF
But run the wrong linear regression model and makes a wrong prediction
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Yi = 0 + 1Xi + i
Examples of Linear Statistical Models
ln(Yi) = 0 + 1Xi + i
Yi = 0 + 1 ln(Xi) + i
Yi = 0 + 1Xi + i2
Examples of Non-linear Statistical Models
Yi = 0 + 1Xi + i
2
Yi = 0 + 1Xi + exp(2Xi) + i
Yi = 0 + 1Xi + i
2
Linear vs. Nonlinear
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Different Functional Forms
5. Reciprocal (or inverse)
Attention to each form’s slope and elasticity
1. Linear2. Log-Log3. Semilog • Linear-Log or Log-Linear
4. Polynomial
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Functional Forms of Regression models
Transform into linear log-form:
iXlnlnYln 1
iXY **
1
*
0
* iXlnYln
1
*
0==>
==>1
*
1 where
**
*
lnln
XdX
YdY
XdYd
dXdY elasticity
coefficient
2. Log-log model:ieXY
0
This is a non-linear model
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Functional Forms of Regression modelsQ
uan
tity
Dem
and
Y
Xprice
1
0 XY
lnY
lnX
XY lnlnln 10
lnY
lnX
XY lnlnln 10 Qu
anti
ty
Dem
and
price
Y
X
1
0 XY
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Functional Forms of Regression models3. Semi log model:
Log-lin model or lin-log model:
iiiXY
10ln
iiiXY ln10
or
and
1
relative change in Y
absolute change in X YdXdY
dXY
dY
dXYd 1ln
1
absolute change in Y
relative change in X 1lnX
dXdY
XddY
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5. Reciprocal (or inverse) transformations
i
i
i X
Y )1
(10
Functional Forms of Regression models(Cont.)
iii XY )(*
10==> Where
i
iX
X1*
4. Polynomial: Quadratic term to capture the nonlinear pattern
Yi= 0 + 1 Xi +2X2i + i
Yi
Xi
1>0, 2<0
Yi
Xi
1<0, 2>0
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Some features of reciprocal model
XY
1 Y
0X
0
0 and 01
Y
X
0
0
+
-
XY
1
00 and 01
Y
0
X0 01 /
00 and 01
Y
0
X0
01 /
00 and 01
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Two conditions for nonlinear, non-additive equation transformation.
1. Exist a transformation of the variable.
2. Sample must provide sufficient information.
Example 1:Suppose
213
2
12110 XXXXY
transforming X2* = X1
2
X3* = X1X2
rewrite *
33
*
22110 XXXY
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Example 2:
2
10
X
Y
transforming2
*
1
1
X
X
*
110 XY rewrite
However, X1* cannot be computed, because is unknown.
2
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Application of functional form regression
1. Cobb-Douglas Production function:
eKLY 0
Transforming:
KLY
KLY
lnlnln
lnlnlnln
210
210
==>
1lnln
LdYd
2lnln
KdYd
: elasticity of output w.r.t. labor input
: elasticity of output w.r.t. capital input.
121 ><
Information about the scale of returns.
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2. Polynomial regression model: Marginal cost function or total cost function
costs
y
MC
i.e.
costs
y
XXY 2
210 (MC)
orcosts
y
TCXXXY 3
3
2
210 (TC)
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linear XY 10
1
dXdY X
)(1Y
XY lnln 10 Log-log
1ln
ln
XdXY
dY
Xd
Yd1
)(1 X
Y
dX
dY ==>
Slope ElasticitySummary
Model Equation)(
dXdY )(
XdXY
dY
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Summary(Count.)
ReciprocalX
Y1
10 1
2 )1
(1
dX
X
dY
Xd
dY
X2dXdY -1
1
)-1
(1XY
==>
Lin-log XY ln10 1
ln
XdXdY
XddY
Y1
1
XdXdY 1
1==>
YdXdY
1==>
XY 10ln Log-linX1
1
ln dXY
dY
dXYd
Slope Elasticity
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25325.1304.100 MPNG ^
(1.368) (39.20)
Linear model
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GNP = -1.6329.21 + 2584.78 lnM2(-23.44) (27.48)
^
Lin-log model
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lnGNP = 6.8612 + 0.00057 M2(100.38) (15.65)
^
Log-lin model
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2ln9882.05529.0ln MNPG ^
(3.194) (42.29)
Log-log model
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Wage(y)
unemp.(x)
SRF
10.43
wage=10.343-3.808(unemploy)(4.862) (-2.66)
^
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)1
(x
y
SRF-1.428
uN
uN: natural rate of unemployment
Reciprocal Model
(1/unemploy)
Wage = -1.4282+8.7243 )1
(x
(-.0690) (3.063)
^
The 0 is statistically insignificantTherefore, -1.428 is not reliable
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lnwage = 1.9038 - 1.175ln(unemploy)(10.375) (-2.618)
^
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Lnwage = 1.9038 + 1.175 ln )1
(X
(10.37) (2.618)
^
Antilog(1.9038) = 6.7113, therefore it is a more meaningful and statistically significant bottom line for min. wage
Antilog(1.175) = 3.238, therefore it means that one unit X increase will have 3.238 unit decrease in wage
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(MacKinnon, White, Davidson)MWD Test for the functional form (Wooldridge, pp.203)
Procedures:
1. Run OLS on the linear model, obtain Y ^
Y = 0 + 1 X1 + 2 X2 ^ ^ ^ ^
2. Run OLS on the log-log model and obtain lnY
lnY = 0 + 1 ln X1 + 2 ln X2^ ^ ^ ^
3. Compute Z1 = ln(Y) - lnY ^ ^4. Run OLS on the linear model by adding z1
Y = 0’ + 1’ X1 + 2’ X2 + 3’ Z1
^ ^ ^ ^ ^
and check t-statistic of 3’If t*
3 > tc ==> reject H0 : linear model^
If t*3
< tc ==> not reject H0 : linear model^
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MWD test for the functional form (Cont.)
5. Compute Z2 = antilog (lnY) - Y^ ^
6. Run OLS on the log-log model by adding Z2
lnY = 0’ + 1’ ln X1 + 2’ ln X2 + 3’ Z2^ ^ ^ ^ ^
If t*3
> tc ==> reject H0 : log-log model^
If t*3
< tc ==> not reject H0 : log-log model^
and check t-statistic of ’3^
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MWD TEST: TESTING the Functional form of regression
CV1 =
Y _ =
1583.279
24735.33
= 0.064
^
Y
Example:(Table 7.3)Step 1:Run the linear modeland obtain
C
X1
X2
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lnY
fitted or
estimated
Step 2:Run the log-log modeland obtain
C
LNX1
LNX2
CV2 =
Y _ =
0.07481
10.09653= 0.0074
^
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MWD TEST
tc0.05, 11 = 1.796
tc0.10, 11 = 1.363
t* < tc at 5%=> not reject H0
t* > tc at 10%=> reject H0
Step 4:H0 : true model is linear
C
X1
X2
Z1
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MWD Test
tc0.025, 11 = 2.201
tc0.05, 11 = 1.796
tc0.10, 11 = 1.363
Since t* < tc
=> not reject H0
Comparing the C.V. =C.V.1
C.V.2
=0.064
0.0074
Step 6:
H0 : true model is
log-log model
CLNX1LNX2Z2
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Y
^The coefficient of variationcoefficient of variation:
C.V. =
It measures the average error of the sample regression function relative to the mean of Y.
Linear, log-linear, and log-log equations can be meaningfully compared.
The smaller C.Vsmaller C.V. of the model, the more preferredmore preferred equationequation (functional model).
Criterion for comparing two different functional models:
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= 4.916 means that model 2 is better
Coefficient Variation (C.V.)
/ Y of model 1 ^
/ Y of model 2 ^
= 2.1225/89.612
0.0217/4.4891=
0.0236
0.0048
Compare two different functional form models:
Model 1linear model
Model 2log-log model
6.44
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Scaling and units of measurement
X + iY 10
1 : the slope of the regression line.
1 =
Units of change of y
Units of change of x=
dXdY
orXY
if Y* = 1000Y
X* = 1000X
then
*
10
* XY ^^^ *
iXY ^^^100010001000 10
1000
==> *
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Changing the scale of X and Y
Yi/k = (0/k)+(1)Xi/k + i/k
Yi = 0 + 1Xi + iR2 and thet-statistics are no changein regression
results for 1
but all otherstatistics are change. 0 = 0/k
* and
Yi = 0 + 1Xi+ i * * * *
Xi = Xi/k *
*i = i/kYi = Yi/kwhere *
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0^
0*^
5
10
Y
X
25
50
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Changing the scale of x
Yi = 0 + (k1)(Xi/k) + i
Yi = 0 + 1Xi + i
Yi = 0 + 1Xi+ i * *
1 = k1* Xi = Xi/k
*
where
and
The estimatedcoefficient andstandard errorchange but theother statisticsare unchanged.
6.48
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0^
5
10
Y
X
5050
6.49
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Changing the scale of Y
Yi/k = (1/k) + (1/k)Xi + i/k
Yi = 0 + 1Xi + i
All statistics are changed except for the t-statistics and R2.
0 = 0/k * and
Yi = 0 + 1Xi + i * * **
1 = 1/k *
*i = i/kYi = Yi/kwhere *
6.50
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5
10
Y
X0^
25
6.51
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Effects of scaling and units change
But t-statistic
F-statistic will not be affected.
R2
All properties of OLS estimations are also unaffected.
The values of i, SEE, RSS will be affected.
6.52
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GNPBIBDGP 1739.0001.37 ^(-0.485) (3.217)
Both in billion measure:
…B: BillionBillion of 1972 dollar
6.53
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GNPMIMDGP 1739.052.37001 ^(-0.485) (3.217)
Both in million measure:
…M: MillionMillion of 1972 dollar
6.54
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GNPBIMDGP 9491.17352.37001 ^(-0.485) (3.217)
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GNPMIBDGP 00017.00015.37 ^(-0.485) (3.217)
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The “ex-post” and “ex ante” forecasting:For example: Suppose you have data of C and Y from 1947–1999.And the estimated consumption expenditures for 1947-1995 is
Given values of Y96 = 10,419; Y97 = 10,625; … Y99 = 11,286
The calculated predictions or the “ex postex post” forecasts are:
1996: C96 = 238.4 + 0.87(10,149) = 9.355
1997: C97 = 238.4 + 0.87(10,625) = 9.535.50
…..
1999: C99 = 238.4 + 0.87(11285) = 10,113.70^
^^
Ct = 238.4 + 0.87Yt^1947 – 1995:
The calculated predictions or the “ex anteex ante” forecasts base on the assumed value of Y2000=12000:
2000: C2000 = 238.4 + 0.87(12,000) = 10678.4^