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Logarithmic functions of Y and/or X
Last class, we saw that we could approximate a model in which X has a
non-linear effect on Y by using a polynomial population model:
𝑌𝑖 = 𝛽0 + 𝛽1𝑋𝑖 + 𝛽2𝑋𝑖2 + 𝛽3𝑋𝑖
3 + ⋯ + 𝛽𝑟𝑋𝑖𝑟 + 𝑢𝑖
Other regressors may be added as usual
This is a polynomial of degree r
Often a quadratic or cubic term is enough
r can be determined by a series of t or F tests
The validity of this approximation is based on a Taylor series
expansion
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For example, take the quadratic form:
𝑌𝑖 = 𝛽0 + 𝛽1𝑋𝑖 + 𝛽2𝑋𝑖2 + 𝑢𝑖
𝜕𝑌
𝜕𝑋1= 𝛽1 + 2𝛽2𝑋
This polynomial model is capable of capturing
diminishing/increasing marginal effects
The effect of 𝑋 on 𝑌 depends on the value of 𝑋
Consider again the model:
TestScorei = 0 + 1Incomei + 2(Incomei)2 + ui
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OLS estimates this model as:
𝑇𝑒𝑠𝑡𝑆𝑐𝑜𝑟𝑒̂𝑖 = 607.3 + 3.85𝐼𝑛𝑐𝑜𝑚𝑒𝑖 − 0.0423𝐼𝑛𝑐𝑜𝑚𝑒𝑖
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(2.9) (0.27) (0.0048)
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Exercises:
i) Using the above regression line, calculate the effect on test scores of an
increase in the budget of $1000, when school income is $5000 and when
school income is $20,000.
ii) Test the hypothesis that Income has a linear effect on TestScore.
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Another way (besides using polynomials) to express a non-linear
relationship is in percentages. For example:
Wage gap across professions/time
Elasticity of demand – a 1% increase in P leads to a Q% decrease in
quantity demanded
Test scores and district income
The effect of age or yrseduc on earnings.
What kind of model should we estimate so that (for example) a change in
X causes a percentage change in Y?
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Logarithms can be used to approximate percentage changes
Percentage change:
∆𝑋
𝑋× 100 =
𝑥2 − 𝑥1
𝑥1× 100
𝑥1 is the initial value of X, 𝑥2 is the final value of X.
The approximation:
[ln(𝑋 + ∆𝑋) − ln 𝑋] × 100 ≅∆𝑋
𝑋× 100
(ln 𝑥2 − ln 𝑥1) × 100 ≅𝑥2 − 𝑥1
𝑥1× 100
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The approximation is better the smaller ∆𝑥.
Change in 𝑥 Percentage change:
𝑥2 − 𝑥1
𝑥1× 100
Approximated
percentage change
(ln 𝑥2 − ln 𝑥1) × 100
𝑥1 = 1 , 𝑥2 = 2 100% 69.32%
𝑥1 = 1 , 𝑥2 = 1.1 10% 9.53%
𝑥1 = 1 , 𝑥2 = 1.01 1% 0.995%
𝑥1 = 5 , 𝑥2 = 6 20% 18.23%
𝑥1 = 11 , 𝑥2 = 12 9.09% 8.70%
𝑥1 = 11 , 𝑥2 = 11.1 0.91% 0.91%
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So how is this helpful?
Three log regression specifications
Case Population regression function
I. linear-log Yi = 0 + 1ln(Xi) + ui
II. log-linear ln(Yi) = 0 + 1Xi + ui
III. log-log ln(Yi) = 0 + 1ln(Xi) + ui
The interpretation of the slope coefficient differs in each case.
The interpretation can be found by figuring out the change in Y for a
given change in X.
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I. linear-log
Y = 0 + 1ln(X) (1)
Now change X: Y + Y = 0 + 1ln(X + X) (2)
Subtract (2) – (1): Y = 1[ln(X + X) – ln(X)]
Use the approximation: ln(X + X) – ln(X) ∆𝑋
𝑋,
So Y 1∆𝑋
𝑋
Or 1 ∆𝑌
∆𝑋 𝑋⁄ (small X)
1 is the change in Y due to a proportional (percentage) change in X.
1% change in X is associated with a 0.01𝛽1 change in Y.
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Example: TestScore vs. ln(Income)
First define the new regressor: lnincome = log(income)
𝑇𝑒𝑠𝑡𝑆𝑐𝑜𝑟𝑒̂
𝑖 = 557.8 + 36.42𝑙𝑛𝑖𝑛𝑐𝑜𝑚𝑒𝑖
(3.8) (1.40)
a 1% increase in Income is associated with an increase in TestScore
of 0.36 points on the test
Standard errors, confidence intervals, R2 – all the usual tools of
regression apply here
How does this compare to the cubic model?
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II. log-linear
ln(Y) = 0 + 1X (1)
Now change X: ln(Y + Y) = 0 + 1(X + X) (2)
Subtract (2) – (1): ln(Y + Y) – ln(Y) = 1X
Use the approximation: ∆𝑌
𝑌 1X
or 1 ∆𝑌 𝑌⁄
∆𝑋 (small X)
iii) What is the interpretation of 1 here?
iv) What is R2 measuring here?
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III. log-log
ln(Yi) = 0 + 1ln(Xi) + ui (1)
Now change X: ln(Y + Y) = 0 + 1ln(X + X) (2)
Subtract: ln(Y + Y) – ln(Y) = 1[ln(X + X) – ln(X)]
Approximation: ∆𝑌
𝑌 1
∆𝑋
𝑋
or 1 ∆𝑌 𝑌⁄
∆𝑋 𝑋⁄ (small X)
A 1% change in X is associated with a 1% change in Y
1 can be interpreted as an elasticity
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Example: ln(TestScore) vs. ln(Income)
Define the new dependent variable: lnscore =
log(TestScore)
𝑙𝑛𝑠𝑐𝑜𝑟𝑒̂
𝑖 = 6.336 + 0.0554𝑙𝑛𝑖𝑛𝑐𝑜𝑚𝑒𝑖
(0.006) (1.40)
v) What is the interpretation of the estimated coefficient on lnincome?
vi) How do the predictions from this model compare to the predictions in
(i)?
Let’s see how the log-linear and log-log models fit the data:
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Summary: Logarithmic transformations
Three cases, differing in whether Y and/or X is transformed by taking
logs
Hypothesis tests and confidence intervals are now implemented and
interpreted “as usual”
The interpretation of 1 differs from case to case
Choice of specification should be guided by judgment (which
interpretation makes the most sense in your application?), tests, and
plotting predicted values
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Extensions: Non-linear effects
We can still use OLS as long as the model we are trying to estimate is
linear in the parameters (𝛽𝑠). That is,
𝑌 = 𝛽0 + 𝛽1∎ + 𝛽2∎ + ⋯
is still linear in the parameters, even though the stuff in the boxes may
be non-linear transformations of the X variables.
OLS is surprisingly capable of capturing non-linear effects. Consider the
Cobb-Douglas production function:
𝑌 = 𝐴𝐾𝛼𝐿𝛽
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vii) What parameters would you be trying to estimate?
viii) Is the model linear in these parameters?
ix) If not, can you make a transformation?
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cobbdata=read.csv("http://home.cc.umanitoba.ca/~godwinrt/7010/cobb.
csv")
attach(cobbdata)
summary(lm(log(y) ~ log(k) + log(l)))
Call:
lm(formula = log(y) ~ log(k) + log(l))
Coefficients:
Estimate Std. Error t value Pr(>|t|)
(Intercept) 1.8444 0.2336 7.896 7.33e-08 ***
log(k) 0.2454 0.1069 2.297 0.0315 *
log(l) 0.8052 0.1263 6.373 2.06e-06 ***
---
Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1
Residual standard error: 0.2357 on 22 degrees of freedom
Multiple R-squared: 0.9731, Adjusted R-squared: 0.9706
F-statistic: 397.5 on 2 and 22 DF, p-value: < 2.2e-16
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The estimated model is:
�̂� = 6.324𝐾0.25𝐿0.81
x) How would you test the hypothesis of constant returns to scale?
There are also many situations where the theoretical model is non-linear
in the parameters, and there is no way to transform it into a linear model.
In this case, OLS can’t be used. A different estimator, such as Non-
Linear Least Squares (NLS), may be used instead.
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