José António Ferreira Machado
Paulo Fagandini
Marlon Francisco
1
Nova SBE
Econometrics
Spring 2024-25
The Simple Linear Regression - Solutions
1.1 Exercise done in the practical class.
1.2 Exercise done in the practical class.
1.3 a) β1 : If distance from a garbage incinerator increases by 1%, housing price increases by 0.3%, on
average, ceteris paribus.
β0 : If distance is 1 mile (log(1)=0) , price is exp(9.4) ≈ 12.088.
b) Incinerator may have been located purposely away from more expensive neighborhoods, i.e., distance is positively correlated with housing quality which is in ht error term and this consists in a
violation of SLR.4 - the OLS estimation is biased.
c) Quality factors are in u: house size, amenities(e.g. central heating, alarm), age, neighborhood
features (e.g. supermarket, school), ... These influence the price and can be correlated with
distance.
1.4 a) α = 0, β = 0.6, γ = 0, δ = 1
b) No. Correlation indicates the strength of a linear relationship between two variables but it tells
us nothing about causality. Although Corr(y, x) = Corr(x, y) the impact of children’s height
in parent’s height is not the same as the parent’s height impact on children’s height. The first
regression makes more sense (i.e. parent’s height impacts children’s height).
1.5 Exercise done in the practical class.
1.6 (a) Let yi = GP Ai , xi = ACTi and n=8. Then x̄ = 25.875, ȳ = 3.2125,
and
n
P
n
P
(xi − x̄)(yi − ȳ) = 5.8125,
i=1
(xi − x̄)2 = 56.875
i=1
β1 ≈ 0.1022 and β0 ≈ 0.5681
The intercept does not have a useful interpretation because ACT is not close to zero for the
population of interest. If ACT is 5 points higher, GPA increases on average, ceteris paribus,
0.1022(5)=0.51.
(b) The fitted values and residuals – rounded to four decimal places – are given with the observation
number and GPA in the following table:
1
José António Ferreira Machado
Paulo Fagandini
Marlon Francisco
Nova SBE
Econometrics
Spring 2024-25
Student
GPA
\
GP
Ai
ubi
1
2.8
2.7143
0.0857
2
3.4
3.0209
0.3791
3
3.0
3.2253
-0.2253
4
3.5
3.3275
0.1725
5
3.6
3.5319
0.0681
6
3.0
3.1231
-0.1231
7
2.7
3.1231
-0.4231
8
3.7
3.6341
0.0659
You can verify that the residuals, as reported in the table, sum to -0.0002, which is pretty close
to zero givn the inherent rounding error.
[
(c) When ACT = 20, GP
A = 0.5681 + 0.1022(20) ≈ 2.61.
(d) The SSR is 0.4347 and the SST is about 1.0288. So the R2 from the regression is 0.577. Therefore,
about 57.7% of the variation in GPA is explained by the variation in ACT in this small sample
of students.
1.7 a) (10) Linear Relationship:
Even if net public assets (NPA) are zero, the predicted CPI is 44.484
1 billion of escudos increase in NPA is estimated to increase the CPI in 5.686 points
(11) Power Relationship:
When NPA are equal to 1 billion of escudos, the predicted CPI is exp(3.289) ≈ 26.82
1 percent increase in NPA is estimated to increase the CPI by 0.578%
(12) Exponential Relationship:
Even if NPA are zero, the predicted CPI is exp(4.094) ≈ 59.98
1 billion of escudos increase in NPA is estimated to increase the CPI by 5.2%
(13) Logarithmic Relationship:
When NPA are equal to 1 billion of escudos, the predicted CPI is -42.635
1 percent increase in NPA is estimated to increase the CPI by 0.6263 points
b) As the dependent variables are not the same in the 4 models, we can only compare the R2 be2
José António Ferreira Machado
Paulo Fagandini
Marlon Francisco
Nova SBE
Econometrics
Spring 2024-25
tween models (10) and (13), and between models (11) and (12). 96.6% of the variation in CPI is
explained by L, while the logarithm of L only explains 95.4%, so the linear model is better than
the logarithmic one. Similarly, while L explains 97.3% of the variation in ln(CP I), ln(L) explains
only 96.4%, so the exponential relationship is better that the power relationship.
c) Considering the monetarist theory, we should choose the power relationship model once it assumes
a constant relationship between the growth rate of prices (CPI) and the growth rate of money (L).
\ = −41.623 + 7.399 ln(inc)
1.8 a) (14) dairy
R2 = 0.4567
R2 = 0.5190
\ = −2.556 + 0.6866 ln(inc)
(15) ln(dairy)
1
b) (14) the marginal propensity to expenditure in this model is β1 × inc
1
the elasticity expenditure/income is β1 × dairy
1
(15) the marginal propensity to expenditure in this model is β1 × inc
× dairy
the elasticity expenditure/income is simply β1
Interpretation: In both models, the marginal propensity to expenditure is inversely proportional
to the level of income, but in the log-log model this increases with the level of expenditure. While
in the linear-log model the elasticity is inversely proportional to the level of expenditure on dairy
products, in the log-log model we assume that this elasticity is constant. This means that a 1%
increase in income leads to an increase in the demand of dairy products by 0.07399 euros in the
first model and by 0.68% in the log-log model.
c) No, because the dependent variable is not the same in both models.
1.9 Exercise done in the practical class. Take special note of the following:
a) β˜1 =
Pn
i=1 yi xi
n
P
comes from the minimization of the sum of squared residuals from a model without
xi 2
i=1
intercept: minβ̃1
n
X
(yi − β̃1 xi )2 . Derive with respect to β̃1 , equal the expression to zero, and solve
i=1
for β̃1 .
e) Write down the expressions of bias and variance of β̃1 and β̂1 . For a given sample size n and
n
P
holding
xi 2 fixed, (1) if β0 close to zero, the bias of β̃1 vanishes, and as shown in d) the variance
i=1
of β̃1 is lower than that of β̂1 , so β̃1 may be preferred; (2) if x̄ close to zero, the bias of β̃1 also
vanishes, and the variance of β̂1 approximates that of β̃1 , so we become indifferent. You may also
3
José António Ferreira Machado
Paulo Fagandini
Marlon Francisco
Nova SBE
Econometrics
Spring 2024-25
calculate the mean squared error of the two estimators (M SE = Bias2 + V ariance) and reach
the same conclusions.
1.10 a) E[β˜1 |xi ] = E
b) V ar(β˜1 ) =
h P n
yi
1
σ2
n2
i=1
n
xi
i
h P i
i
P h
|xi = E n1 ni=1 β1 xxii+ui |xi = β1 + n1 ni=1 x1i E[ui |xi ] = β1
Pn
1
i=1 x2i
2
c) V ar(βˆ1 ) = Pnσ x2
i=1
i
P
P
P
Cauchy-Schwarz inequality: ( ni=1 ai bi )2 ≤ ( ni=1 a2i )( ni=1 b2i ). Denote ai = xi and bi = 1/xi :
n
X
xi .
i=1
n
n
X
X
1 2
1
≤
x2i
xi
x2i
n2 .σ 2 ≤
σ2
Pn
2
i=1 xi
≤
i=1
i=1
n
X
n
X
1
2
xi
i=1
n
2
σ X
n2
1
2
x
i=1 i
V ar(βˆ1 ) ≤ V ar(β˜1 )
1.11 No solution is provided for demonstrations.
1.12 No solution is provided for demonstrations.
4
x2
i=1 i
.σ 2