6.1 Lecture #6 Studenmund(2006) Chapter 7 1. Suppressing the intercept 2. Alternative Functional forms 3. Scaling and units of measurement Objectives:

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:

6.2


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

^

6.3


Regression through the origin

The intercept term is absent or zero.i.e.,

iii XY 1

iY

iX

1

1^

ii XYSRF 1:^^

0

6.4


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

6.5


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.

6.6


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

6.7


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.

6.8


Example 1:(cont.)

fi rER

fERm

1

1

Security market line

6.9


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

6.10


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

6.11


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

6.12


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

6.13


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.

6.14


True model is nonlinear

Y

X

Income

Age6015

PRF

SRF

But run the wrong linear regression model and makes a wrong prediction

6.15


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

6.16


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

6.17


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

6.18


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

6.19


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

6.20


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

6.21


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

6.22


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

6.23


Example 2:

2

10

X

Y

transforming2

*

1

1

X

X

*

110 XY rewrite

However, X1* cannot be computed, because is unknown.

2

6.24


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.

6.25


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)

6.26


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

6.27


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

6.28


25325.1304.100 MPNG ^

(1.368) (39.20)

Linear model

6.29


GNP = -1.6329.21 + 2584.78 lnM2(-23.44) (27.48)

^

Lin-log model

6.30


lnGNP = 6.8612 + 0.00057 M2(100.38) (15.65)

^

Log-lin model

6.31


2ln9882.05529.0ln MNPG ^

(3.194) (42.29)

Log-log model

6.32


Wage(y)

unemp.(x)

SRF

10.43

wage=10.343-3.808(unemploy)(4.862) (-2.66)

^

6.33


)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

6.34


lnwage = 1.9038 - 1.175ln(unemploy)(10.375) (-2.618)

^

6.35


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

6.36


(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^

6.37


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^

6.38


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

6.39


lnY

fitted or

estimated

Step 2:Run the log-log modeland obtain

C

LNX1

LNX2

CV2 =

Y _ =

0.07481

10.09653= 0.0074

^

6.40


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

6.41


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

6.42


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:

6.43


= 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


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

==> *

6.45


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 *

6.46


0^

0*^

5

10

Y

X

25

50

6.47


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


0^

5

10

Y

X

5050

6.49


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


5

10

Y

X0^

25

6.51


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


GNPBIBDGP 1739.0001.37 ^(-0.485) (3.217)

Both in billion measure:

…B: BillionBillion of 1972 dollar

6.53


GNPMIMDGP 1739.052.37001 ^(-0.485) (3.217)

Both in million measure:

…M: MillionMillion of 1972 dollar

6.54


GNPBIMDGP 9491.17352.37001 ^(-0.485) (3.217)

6.55


GNPMIBDGP 00017.00015.37 ^(-0.485) (3.217)

6.56


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^

Documents

6.1 Lecture #6 Studenmund(2006) Chapter 7 1. Suppressing the intercept 2. Alternative Functional forms 3. Scaling and units of measurement Objectives: