Christopher Dougherty EC220 - Introduction to econometrics (chapter 4) Slideshow: semilogarithmic models Original citation: Dougherty, C.

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Transcript Christopher Dougherty EC220 - Introduction to econometrics (chapter 4) Slideshow: semilogarithmic models Original citation: Dougherty, C.

Slide 1

Christopher Dougherty

EC220 - Introduction to econometrics
(chapter 4)
Slideshow: semilogarithmic models
Original citation:
Dougherty, C. (2012) EC220 - Introduction to econometrics (chapter 4). [Teaching Resource]
© 2012 The Author
This version available at: http://learningresources.lse.ac.uk/130/
Available in LSE Learning Resources Online: May 2012
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http://learningresources.lse.ac.uk/


Slide 2

SEMILOGARITHMIC MODELS

Y   1e

2X

This sequence introduces the semilogarithmic model and shows how it may be applied to
an earnings function. The dependent variable is linear but the explanatory variables,
multiplied by their coefficients, are exponents of e.
1


Slide 3

SEMILOGARITHMIC MODELS

Y   1e
dY
dX

  1  2e

2X

2X

  2Y

The differential of Y with respect to X simplifies to 2Y.

2


Slide 4

SEMILOGARITHMIC MODELS

Y   1e
dY
dX

  1  2e

dY Y
dX

2X

2X

  2Y

 2

Hence the proportional change in Y per unit change in X is equal to 2. It is therefore
independent of the value of X.
3


Slide 5

SEMILOGARITHMIC MODELS

Y   1e
Y   Y   1e
  1e
 Ye

2X

 2  X  X 
2X

e

 2X

 2X


 2 X

 Y  1   2X 
2



 2 X

Y  Y   2X 
2


2

2


 ... 



 ... 


Strictly speaking, this interpretation is valid only for small values of 2. When 2 is not
small, the interpretation may be a little more complex.
4


Slide 6

SEMILOGARITHMIC MODELS

Y   1e
Y   Y   1e
  1e
 Ye

2X

 2  X  X 
2X

e

 2X

 2X


 2 X

 Y  1   2X 
2



 2 X

Y  Y   2X 
2


2

2


 ... 



 ... 


Suppose that X increases by an amount X and that as a consequence Y increases by an
amount Y.
5


Slide 7

SEMILOGARITHMIC MODELS

Y   1e
Y   Y   1e
  1e
 Ye

2X

 2  X  X 
2X

e

 2X

 2X


 2 X

 Y  1   2X 
2



 2 X

Y  Y   2X 
2


2

2


 ... 



 ... 


We can rewrite the right side of the equation as shown.

6


Slide 8

SEMILOGARITHMIC MODELS

Y   1e
Y   Y   1e
  1e
 Ye

2X

 2  X  X 
2X

e

 2X

 2X


 2 X

 Y  1   2X 
2



 2 X

Y  Y   2X 
2


2

2


 ... 



 ... 


We can simplify the right side of the equation as shown.

7


Slide 9

SEMILOGARITHMIC MODELS

Y   1e
Y   Y   1e
  1e
 Ye

2X

 2  X  X 
2X

e

 2X

 2X


 2 X

 Y  1   2X 
2


2


 ... 


2
3





X
Z
Z
2
Z
 Y e Y 12 ZX

 ...... 
2! 23!



2

Now expand the exponential term using the standard expression for e to some power.

8


Slide 10

SEMILOGARITHMIC MODELS

Y   1e
Y   Y   1e
  1e
 Ye

2X

 2  X  X 
2X

e

 2X

 2X


 2 X

 Y  1   2X 
2



 2 X

Y  Y   2X 
2


2

2


 ... 



 ... 


Subtract Y from both sides.

9


Slide 11

SEMILOGARITHMIC MODELS


 2 X
 Y  Y   2  X 
2


  X  2 negligible

Y

 X 

not negligible


 ... 


Y  Y 2X

Y / X

2

2

Y / X
Y

 2

 2 X
2

 2 

 ...

2

2

2

 2 

2

 ...

if  X is one unit

We now consider two cases: X so small that (X)2 is negligible, and the alternative.

10


Slide 12

SEMILOGARITHMIC MODELS


 2 X
 Y  Y   2  X 
2


  X 2

Y

 X 

not negligible


 ... 


Y  Y 2X

negligible

Y / X

2

2

 2

Y / X
Y

 2 X
2

 2 

 ...

2

2

2

 2 

2

 ...

if  X is one unit

If (X)2 is negligible, we obtain the same interpretation of 2 as we did using the calculus, as
one would expect.
11


Slide 13

SEMILOGARITHMIC MODELS


 2 X
 Y  Y   2  X 
2


  X 2

Y

 X 

not negligible


 ... 


Y  Y 2X

negligible

Y / X

2

2

Y / X
Y

 2

 2 X
2

 2 

2

2

 ...

2

 2 

2

 ...

If (X)2 is not negligible, the proportional change in Y given a X change in X has an extra
term. (We are assuming that X is small enough that higher powers of X can be
neglected.)
12


Slide 14

SEMILOGARITHMIC MODELS


 2 X
 Y  Y   2  X 
2


  X 2

Y

 X 

not negligible


 ... 


Y  Y 2X

negligible

Y / X

2

2

Y / X
Y

 2

 2 X
2

 2 

2

2

 ...

2

 2 

2

 ...

if X is one unit

Usually we talk about the effect of a one-unit change in X. If X = 1, the proportional change
in Y is as shown. The issue now becomes whether 2 is so small that the second and
subsequent terms can be neglected.
13


Slide 15

SEMILOGARITHMIC MODELS

Y   1e
X 0



2X

Y   1e   1
0

1 is the value of Y when X is equal to zero (note that e0 is equal to 1).
14


Slide 16

SEMILOGARITHMIC MODELS

Y   1e

2X

log Y  log  1 e

2X

 log  1  log e

2X

  1   2 X log e
'

 1  2X
'

To fit a function of this type, you take logarithms of both sides. The right side of the
equation becomes a linear function of X (note that the logarithm of e, to base e, is 1). Hence
we can fit the model with a linear regression, regressing log Y on X.
15


Slide 17

SEMILOGARITHMIC MODELS
. reg LGEARN S
Source |
SS
df
MS
-------------+-----------------------------Model | 38.5643833
1 38.5643833
Residual |
148.14326
538 .275359219
-------------+-----------------------------Total | 186.707643
539
.34639637

Number of obs
F( 1,
538)
Prob > F
R-squared
Adj R-squared
Root MSE

=
=
=
=
=
=

540
140.05
0.0000
0.2065
0.2051
.52475

-----------------------------------------------------------------------------LGEARN |
Coef.
Std. Err.
t
P>|t|
[95% Conf. Interval]
-------------+---------------------------------------------------------------S |
.1096934
.0092691
11.83
0.000
.0914853
.1279014
_cons |
1.292241
.1287252
10.04
0.000
1.039376
1.545107
------------------------------------------------------------------------------

Here is the regression output from a wage equation regression using Data Set 21. The
estimate of 2 is 0.110. As an approximation, this implies that an extra year of schooling
increases hourly earnings by a proportion 0.110.
16


Slide 18

SEMILOGARITHMIC MODELS
. reg LGEARN S
Source |
SS
df
MS
-------------+-----------------------------Model | 38.5643833
1 38.5643833
Residual |
148.14326
538 .275359219
-------------+-----------------------------Total | 186.707643
539
.34639637

Number of obs
F( 1,
538)
Prob > F
R-squared
Adj R-squared
Root MSE

=
=
=
=
=
=

540
140.05
0.0000
0.2065
0.2051
.52475

-----------------------------------------------------------------------------LGEARN |
Coef.
Std. Err.
t
P>|t|
[95% Conf. Interval]
-------------+---------------------------------------------------------------S |
.1096934
.0092691
11.83
0.000
.0914853
.1279014
_cons |
1.292241
.1287252
10.04
0.000
1.039376
1.545107
------------------------------------------------------------------------------

In everyday language it is usually more natural to talk about percentages rather than
proportions, so we multiply the coefficient by 100. It implies that an extra year of schooling
increases hourly earnings by 11.0%.
17


Slide 19

SEMILOGARITHMIC MODELS
. reg LGEARN S
Source |
SS
df
MS
-------------+-----------------------------Model | 38.5643833
1 38.5643833
Residual |
148.14326
538 .275359219
-------------+-----------------------------Total | 186.707643
539
.34639637

Number of obs
F( 1,
538)
Prob > F
R-squared
Adj R-squared
Root MSE

=
=
=
=
=
=

540
140.05
0.0000
0.2065
0.2051
.52475

-----------------------------------------------------------------------------LGEARN |
Coef.
Std. Err.
t
P>|t|
[95% Conf. Interval]
-------------+---------------------------------------------------------------S |
.1096934
.0092691
11.83
0.000
.0914853
.1279014
_cons |
1.292241
.1287252
10.04
0.000
1.039376
1.545107
------------------------------------------------------------------------------

 X 

2

not negligible

Y / X
Y

2

2

 2 

 ...

2

 0.110 

if X is one unit

 0.110  2
2

 0 . 116

If we take account of the fact that a year of schooling is not a marginal change, and work
out the effect exactly, the proportional increase is 0.116 and the percentage increase 11.6%.
18


Slide 20

SEMILOGARITHMIC MODELS
. reg LGEARN S
Source |
SS
df
MS
-------------+-----------------------------Model | 38.5643833
1 38.5643833
Residual |
148.14326
538 .275359219
-------------+-----------------------------Total | 186.707643
539
.34639637

Number of obs
F( 1,
538)
Prob > F
R-squared
Adj R-squared
Root MSE

=
=
=
=
=
=

540
140.05
0.0000
0.2065
0.2051
.52475

-----------------------------------------------------------------------------LGEARN |
Coef.
Std. Err.
t
P>|t|
[95% Conf. Interval]
-------------+---------------------------------------------------------------S |
.1096934
.0092691
11.83
0.000
.0914853
.1279014
_cons |
1.292241
.1287252
10.04
0.000
1.039376
1.545107
------------------------------------------------------------------------------

 X 

2

not negligible

Y / X
Y

2

2

 2 

 ...

2

 0.110 

if X is one unit

 0.110  2
2

 0 . 116

In general, if a unit change in X is genuinely marginal, the estimate of 2 will be small and
one can interpret it directly as an estimate of the proportional change in Y per unit change
in X.
19


Slide 21

SEMILOGARITHMIC MODELS
. reg LGEARN S
Source |
SS
df
MS
-------------+-----------------------------Model | 38.5643833
1 38.5643833
Residual |
148.14326
538 .275359219
-------------+-----------------------------Total | 186.707643
539
.34639637

Number of obs
F( 1,
538)
Prob > F
R-squared
Adj R-squared
Root MSE

=
=
=
=
=
=

540
140.05
0.0000
0.2065
0.2051
.52475

-----------------------------------------------------------------------------LGEARN |
Coef.
Std. Err.
t
P>|t|
[95% Conf. Interval]
-------------+---------------------------------------------------------------S |
.1096934
.0092691
11.83
0.000
.0914853
.1279014
_cons |
1.292241
.1287252
10.04
0.000
1.039376
1.545107
------------------------------------------------------------------------------

 X 

2

not negligible

Y / X
Y

2

2

 2 

 ...

2

 0.110 

if X is one unit

 0.110  2
2

 0 . 116

However if a unit change in X is not small, the coefficient may be large and the second term
might not be negligible. In the present case, a year of schooling is not marginal and the
refinement does make a small difference.
20


Slide 22

SEMILOGARITHMIC MODELS
. reg LGEARN S
Source |
SS
df
MS
-------------+-----------------------------Model | 38.5643833
1 38.5643833
Residual |
148.14326
538 .275359219
-------------+-----------------------------Total | 186.707643
539
.34639637

Number of obs
F( 1,
538)
Prob > F
R-squared
Adj R-squared
Root MSE

=
=
=
=
=
=

540
140.05
0.0000
0.2065
0.2051
.52475

-----------------------------------------------------------------------------LGEARN |
Coef.
Std. Err.
t
P>|t|
[95% Conf. Interval]
-------------+---------------------------------------------------------------S |
.1096934
.0092691
11.83
0.000
.0914853
.1279014
_cons |
1.292241
.1287252
10.04
0.000
1.039376
1.545107
------------------------------------------------------------------------------

 X 

2

not negligible

Y / X
Y

2

2

 2 

 ...

2

 0.110 

if X is one unit

 0.110  2
2

 0 . 116

In general, when 2 is less than 0.1, there is little to be gained by working out the effect
exactly.
21


Slide 23

SEMILOGARITHMIC MODELS
. reg LGEARN S
Source |
SS
df
MS
-------------+-----------------------------Model | 38.5643833
1 38.5643833
Residual |
148.14326
538 .275359219
-------------+-----------------------------Total | 186.707643
539
.34639637

Number of obs
F( 1,
538)
Prob > F
R-squared
Adj R-squared
Root MSE

=
=
=
=
=
=

540
140.05
0.0000
0.2065
0.2051
.52475

-----------------------------------------------------------------------------LGEARN |
Coef.
Std. Err.
t
P>|t|
[95% Conf. Interval]
-------------+---------------------------------------------------------------S |
.1096934
.0092691
11.83
0.000
.0914853
.1279014
_cons |
1.292241
.1287252
10.04
0.000
1.039376
1.545107
------------------------------------------------------------------------------

log b1  1 . 292

b1  e

1 . 292

 3 . 64

The intercept in the regression is an estimate of log 1. From it, we obtain an estimate of 1
equal to e1.29, which is 3.64.
22


Slide 24

SEMILOGARITHMIC MODELS
. reg LGEARN S
Source |
SS
df
MS
-------------+-----------------------------Model | 38.5643833
1 38.5643833
Residual |
148.14326
538 .275359219
-------------+-----------------------------Total | 186.707643
539
.34639637

Number of obs
F( 1,
538)
Prob > F
R-squared
Adj R-squared
Root MSE

=
=
=
=
=
=

540
140.05
0.0000
0.2065
0.2051
.52475

-----------------------------------------------------------------------------LGEARN |
Coef.
Std. Err.
t
P>|t|
[95% Conf. Interval]
-------------+---------------------------------------------------------------S |
.1096934
.0092691
11.83
0.000
.0914853
.1279014
_cons |
1.292241
.1287252
10.04
0.000
1.039376
1.545107
------------------------------------------------------------------------------

log b1  1 . 292

b1  e

1 . 292

 3 . 64

Literally this implies that a person with no schooling would earn $3.64 per hour. However it
is dangerous to extrapolate so far from the range for which we have data.
23


Slide 25

Logarithm of hourly earnings

SEMILOGARITHMIC MODELS

5

4

3

2

1

0
0

1

2

3

4

5

6

7

8

9

10 11 12 13 14 15 16 17 18 19 20

Years of schooling
Here is the scatter diagram with the semilogarithmic regression.

24


Slide 26

SEMILOGARITHMIC MODELS

120

Hourly earnings ($)

100
80
60
40
20
0
0

1

2

3

4

5

6

7

8

9

10 11 12 13 14 15 16 17 18 19 20

-20

Years of schooling
Here is the semilogarithmic regression line plotted in a scatter diagram with the
untransformed data, with the linear regression shown for comparison.
25


Slide 27

SEMILOGARITHMIC MODELS

120

Hourly earnings ($)

100
80
60
40
20
0
0

1

2

3

4

5

6

7

8

9

10 11 12 13 14 15 16 17 18 19 20

-20

Years of schooling
There is not much difference in the fit of the regression lines, but the semilogarithmic
regression is more satisfactory in two respects.
26


Slide 28

SEMILOGARITHMIC MODELS

120

Hourly earnings ($)

100
80
60
40
20
0
0

1

2

3

4

5

6

7

8

9

10 11 12 13 14 15 16 17 18 19 20

-20

Years of schooling
The linear specification predicts that earnings will increase by about $2 per hour with each
additional year of schooling, which is implausible for high levels of education. The semilogarithmic specification allows the increment to increase with level of education.
27


Slide 29

SEMILOGARITHMIC MODELS

120

Hourly earnings ($)

100
80
60
40
20
0
0

1

2

3

4

5

6

7

8

9

10 11 12 13 14 15 16 17 18 19 20

-20

Years of schooling
Second, the linear specification predicts negative earnings for an individual with no
schooling. The semilogarithmic specification predicts hourly earnings of $3.64, which at
least is not obvious nonsense.
28


Slide 30

Copyright Christopher Dougherty 2011.
These slideshows may be downloaded by anyone, anywhere for personal use.
Subject to respect for copyright and, where appropriate, attribution, they may be
used as a resource for teaching an econometrics course. There is no need to
refer to the author.
The content of this slideshow comes from Section 4.2 of C. Dougherty,
Introduction to Econometrics, fourth edition 2011, Oxford University Press.
Additional (free) resources for both students and instructors may be
downloaded from the OUP Online Resource Centre
http://www.oup.com/uk/orc/bin/9780199567089/.
Individuals studying econometrics on their own and who feel that they might
benefit from participation in a formal course should consider the London School
of Economics summer school course
EC212 Introduction to Econometrics
http://www2.lse.ac.uk/study/summerSchools/summerSchool/Home.aspx
or the University of London International Programmes distance learning course
20 Elements of Econometrics
www.londoninternational.ac.uk/lse.

11.07.25