BS2247 Introduction to Econometrics Lecture 4: The simple regression - - PowerPoint PPT Presentation

BS2247 Introduction to Econometrics Lecture 4: The simple regression model

OLS Unbiasedness, OLS Variances, Units of measurement, Nonlinearities

• Dr. Kai Sun

Aston Business School

1 / 26

Unbiasedness of OLS

SLR for “Simple Linear Regression”

◮ Assumption SLR.1: The population model is linear in

parameters, i.e., y = β1

0 + β1 1x + u ◮ Assumption SLR.2: Use a random sample of size n,

{(xi, yi) : i = 1, 2, . . . , n}, so yi = β0 + β1xi + ui

◮ Assumption SLR.3: There is sample variation in x, i.e.,

• i(xi − ¯

x)2 > 0 (recall the formula for ˆ β1)

◮ Assumption SLR.4: Zero conditional mean, i.e., E(ui|xi) = 0

2 / 26

Unbiasedness of OLS

SLR for “Simple Linear Regression”

◮ Assumption SLR.1: The population model is linear in

parameters, i.e., y = β1

0 + β1 1x + u ◮ Assumption SLR.2: Use a random sample of size n,

{(xi, yi) : i = 1, 2, . . . , n}, so yi = β0 + β1xi + ui

◮ Assumption SLR.3: There is sample variation in x, i.e.,

• i(xi − ¯

x)2 > 0 (recall the formula for ˆ β1)

◮ Assumption SLR.4: Zero conditional mean, i.e., E(ui|xi) = 0

2 / 26

Unbiasedness of OLS

SLR for “Simple Linear Regression”

◮ Assumption SLR.1: The population model is linear in

parameters, i.e., y = β1

0 + β1 1x + u ◮ Assumption SLR.2: Use a random sample of size n,

{(xi, yi) : i = 1, 2, . . . , n}, so yi = β0 + β1xi + ui

◮ Assumption SLR.3: There is sample variation in x, i.e.,

• i(xi − ¯

x)2 > 0 (recall the formula for ˆ β1)

◮ Assumption SLR.4: Zero conditional mean, i.e., E(ui|xi) = 0

2 / 26

Unbiasedness of OLS

SLR for “Simple Linear Regression”

◮ Assumption SLR.1: The population model is linear in

parameters, i.e., y = β1

0 + β1 1x + u ◮ Assumption SLR.2: Use a random sample of size n,

{(xi, yi) : i = 1, 2, . . . , n}, so yi = β0 + β1xi + ui

◮ Assumption SLR.3: There is sample variation in x, i.e.,

• i(xi − ¯

x)2 > 0 (recall the formula for ˆ β1)

◮ Assumption SLR.4: Zero conditional mean, i.e., E(ui|xi) = 0

2 / 26

Unbiasedness of ˆ β1

In order to think about unbiasedness, we need to rewrite our estimator, ˆ β’s, in terms of the population parameter, β’s. ˆ β1 =

• i(xi − ¯

x)(yi − ¯ y)

• i(xi − ¯

x)2 =

• i(xi − ¯

x)yi

• i(xi − ¯

x)2

3 / 26

Unbiasedness of ˆ β1

Plug in yi = β0 + β1xi + ui and define SSTx =

i(xi − ¯

x)2, ˆ β1 =

• i(xi − ¯

x)(β0 + β1xi + ui) SSTx = 0 + β1 +

• i ui(xi − ¯

x) SSTx using the fact that

i(xi − ¯

x) = 0 and

• i xi(xi − ¯

x) =

i(xi − ¯

x)2 = SSTx.

4 / 26

Unbiasedness of ˆ β1

◮ Take expectation (conditional on xi) for both sides:

E(ˆ β1|xi) = β1 +

1 SSTx

• i E(ui|xi)(xi − ¯

x) = β1 (using the Assumption SLR.4 that E(ui|xi) = 0)

◮ Finally, E(ˆ

β1) = E(E(ˆ β1|xi)) = E(β1) = β1 (first equality by law of iterated expectation).

◮ The fact that E(ˆ

β1) = β1 means that ˆ β1 is unbiased for β1.

5 / 26

Unbiasedness of ˆ β1

◮ Take expectation (conditional on xi) for both sides:

E(ˆ β1|xi) = β1 +

1 SSTx

• i E(ui|xi)(xi − ¯

x) = β1 (using the Assumption SLR.4 that E(ui|xi) = 0)

◮ Finally, E(ˆ

β1) = E(E(ˆ β1|xi)) = E(β1) = β1 (first equality by law of iterated expectation).

◮ The fact that E(ˆ

β1) = β1 means that ˆ β1 is unbiased for β1.

5 / 26

Unbiasedness of ˆ β1

◮ Take expectation (conditional on xi) for both sides:

E(ˆ β1|xi) = β1 +

1 SSTx

• i E(ui|xi)(xi − ¯

x) = β1 (using the Assumption SLR.4 that E(ui|xi) = 0)

◮ Finally, E(ˆ

β1) = E(E(ˆ β1|xi)) = E(β1) = β1 (first equality by law of iterated expectation).

◮ The fact that E(ˆ

β1) = β1 means that ˆ β1 is unbiased for β1.

5 / 26

Unbiasedness of ˆ β0

Recall that ˆ β0 = ¯ y − ˆ β1¯ x = (β0 + β1¯ x + ¯ u) − ˆ β1¯ x = β0 + (β1 − ˆ β1)¯ x + ¯ u

◮ Take expectation for both sides:

E(ˆ β0) = β0 + ¯ xE(β1 − ˆ β1) + E(¯ u) = β0 using the fact that E(ˆ β1) = β1 and E(¯ u) = E(1/n

i ui) = 1/n i E(ui) = 0 ◮ The fact that E(ˆ

β0) = β0 means that ˆ β0 is unbiased for β0.

6 / 26

Unbiasedness of ˆ β0

Recall that ˆ β0 = ¯ y − ˆ β1¯ x = (β0 + β1¯ x + ¯ u) − ˆ β1¯ x = β0 + (β1 − ˆ β1)¯ x + ¯ u

◮ Take expectation for both sides:

E(ˆ β0) = β0 + ¯ xE(β1 − ˆ β1) + E(¯ u) = β0 using the fact that E(ˆ β1) = β1 and E(¯ u) = E(1/n

i ui) = 1/n i E(ui) = 0 ◮ The fact that E(ˆ

β0) = β0 means that ˆ β0 is unbiased for β0.

6 / 26

Unbiasedness of ˆ β0

Recall that ˆ β0 = ¯ y − ˆ β1¯ x = (β0 + β1¯ x + ¯ u) − ˆ β1¯ x = β0 + (β1 − ˆ β1)¯ x + ¯ u

◮ Take expectation for both sides:

E(ˆ β0) = β0 + ¯ xE(β1 − ˆ β1) + E(¯ u) = β0 using the fact that E(ˆ β1) = β1 and E(¯ u) = E(1/n

i ui) = 1/n i E(ui) = 0 ◮ The fact that E(ˆ

β0) = β0 means that ˆ β0 is unbiased for β0.

6 / 26

Variance of OLS

◮ While unbiasedness means that the sampling distribution of

• ur estimate is centered around the true parameter

◮ Want to think about how spread out this distribution is ◮ Much easier to think about this variance

under an additional assumption: Assumption SLR.5: Var(u|x) = σ2 (Homoskedasticity: the conditional variance of u is a constant)

7 / 26

Variance of OLS

◮ While unbiasedness means that the sampling distribution of

• ur estimate is centered around the true parameter

◮ Want to think about how spread out this distribution is ◮ Much easier to think about this variance

under an additional assumption: Assumption SLR.5: Var(u|x) = σ2 (Homoskedasticity: the conditional variance of u is a constant)

7 / 26

Variance of OLS

◮ While unbiasedness means that the sampling distribution of

• ur estimate is centered around the true parameter

◮ Want to think about how spread out this distribution is ◮ Much easier to think about this variance

under an additional assumption: Assumption SLR.5: Var(u|x) = σ2 (Homoskedasticity: the conditional variance of u is a constant)

7 / 26

Var(u|x) = E(u2|x) − [E(u|x)]2 Since E(u|x) = 0, Var(u|x) = E(u2|x) = σ2

◮ By the law of iterated expectation,

E(u2) = E(E(u2|x)) = E(σ2) = σ2. Var(u) = E(u2) = σ2 means that the unconditional variance

• f u is a constant, too - σ2 is also called error variance.

◮ If we take the conditional variance for both sides of

y = β0 + β1x + u, Var(y|x) = Var(β0 + β1x|x) + Var(u|x) = Var(u|x) = σ2. Var(β0 + β1x|x) = 0 because when we “conditional on x”, we can view (β0 + β1x) as a constant, whose variance = 0.

8 / 26

Var(u|x) = E(u2|x) − [E(u|x)]2 Since E(u|x) = 0, Var(u|x) = E(u2|x) = σ2

◮ By the law of iterated expectation,

E(u2) = E(E(u2|x)) = E(σ2) = σ2. Var(u) = E(u2) = σ2 means that the unconditional variance

• f u is a constant, too - σ2 is also called error variance.

◮ If we take the conditional variance for both sides of

y = β0 + β1x + u, Var(y|x) = Var(β0 + β1x|x) + Var(u|x) = Var(u|x) = σ2. Var(β0 + β1x|x) = 0 because when we “conditional on x”, we can view (β0 + β1x) as a constant, whose variance = 0.

8 / 26

Var(u|x) = E(u2|x) − [E(u|x)]2 Since E(u|x) = 0, Var(u|x) = E(u2|x) = σ2

◮ By the law of iterated expectation,

E(u2) = E(E(u2|x)) = E(σ2) = σ2. Var(u) = E(u2) = σ2 means that the unconditional variance

• f u is a constant, too - σ2 is also called error variance.

◮ If we take the conditional variance for both sides of

y = β0 + β1x + u, Var(y|x) = Var(β0 + β1x|x) + Var(u|x) = Var(u|x) = σ2. Var(β0 + β1x|x) = 0 because when we “conditional on x”, we can view (β0 + β1x) as a constant, whose variance = 0.

8 / 26

Var(u|x) = E(u2|x) − [E(u|x)]2 Since E(u|x) = 0, Var(u|x) = E(u2|x) = σ2

◮ By the law of iterated expectation,

E(u2) = E(E(u2|x)) = E(σ2) = σ2. Var(u) = E(u2) = σ2 means that the unconditional variance

• f u is a constant, too - σ2 is also called error variance.

◮ If we take the conditional variance for both sides of

y = β0 + β1x + u, Var(y|x) = Var(β0 + β1x|x) + Var(u|x) = Var(u|x) = σ2. Var(β0 + β1x|x) = 0 because when we “conditional on x”, we can view (β0 + β1x) as a constant, whose variance = 0.

8 / 26

Homoskedastic Case

y

f(y|x)

. .

E(y|x) = 0 + 1x

x1 x2

9 / 26

Heteroskedastic Case

f(y|x)

. . .

E(y|x) = 0 + 1x

x x1 x2 x3

10 / 26

Variance of ˆ β1

Recall that ˆ β1 = β1 +

• i ui(xi−¯

x) SSTx

Var(ˆ β1|xi) = 1 SST 2

x

Var(

• i

(xi − ¯ x)ui|xi) = 1 SST 2

x

• i

Var((xi − ¯ x)ui|xi) = 1 SST 2

x

• i

(xi − ¯ x)2Var(ui|xi) = σ2 SST 2

x

• i

(xi − ¯ x)2 = σ2 SST 2

x

SSTx = σ2 SSTx

11 / 26

Variance of ˆ β1

What happens to Var(ˆ β1|xi) if (1) σ2 increases (2) SSTx increases (3) n increases

12 / 26

Variance of ˆ β0

Recall that ˆ β0 = ¯ y − ˆ β1¯ x Var(ˆ β0|xi) = Var(¯ y|xi) + Var(ˆ β1¯ x|xi) − 2Cov(¯ y, ˆ β1¯ x|xi) = Var(¯ y|xi) + (¯ x)2Var(ˆ β1|xi) − 2¯ xCov(¯ y, ˆ β1|xi) = σ2/n + (¯ x)2 σ2 SSTx Note that Var(¯ y|xi) = Var((1/n)

i yi|xi) = (1/n2) i Var(yi|xi) =

(1/n2)

i σ2 = (1/n2) · nσ2 = σ2/n

13 / 26

Variance of ˆ β0

Recall that ˆ β0 = ¯ y − ˆ β1¯ x Var(ˆ β0|xi) = Var(¯ y|xi) + Var(ˆ β1¯ x|xi) − 2Cov(¯ y, ˆ β1¯ x|xi) = Var(¯ y|xi) + (¯ x)2Var(ˆ β1|xi) − 2¯ xCov(¯ y, ˆ β1|xi) = σ2/n + (¯ x)2 σ2 SSTx Note that Var(¯ y|xi) = Var((1/n)

i yi|xi) = (1/n2) i Var(yi|xi) =

(1/n2)

i σ2 = (1/n2) · nσ2 = σ2/n

13 / 26

Variance of ˆ β0

Simplify Var(ˆ β0|xi) = σ2/n + (¯ x)2 σ2 SSTx and get: Var(ˆ β0|xi) = σ2

i x2 i

n · SSTx (practice of summation operator)

14 / 26

σ2 is unknown

◮ σ2 appears in both Var(ˆ

β0|xi) and Var(ˆ β1|xi), but the problem is that the error variance, σ2, is generally unknown, because we don’t observe the error term, u, hence Var(u).

◮ What we observe are the residuals, ˆ

ui.

◮ We can use the residuals

to form an estimate of the error variance: ˆ σ2 =

1 n−2

• i ˆ

u2

i = SSR n−2

n − 2: degrees of freedom (DF) for SSR. (ˆ ui is restricted by two moment restrictions.)

15 / 26

σ2 is unknown

◮ σ2 appears in both Var(ˆ

β0|xi) and Var(ˆ β1|xi), but the problem is that the error variance, σ2, is generally unknown, because we don’t observe the error term, u, hence Var(u).

◮ What we observe are the residuals, ˆ

ui.

◮ We can use the residuals

to form an estimate of the error variance: ˆ σ2 =

1 n−2

• i ˆ

u2

i = SSR n−2

n − 2: degrees of freedom (DF) for SSR. (ˆ ui is restricted by two moment restrictions.)

15 / 26

σ2 is unknown

◮ σ2 appears in both Var(ˆ

β0|xi) and Var(ˆ β1|xi), but the problem is that the error variance, σ2, is generally unknown, because we don’t observe the error term, u, hence Var(u).

◮ What we observe are the residuals, ˆ

ui.

◮ We can use the residuals

to form an estimate of the error variance: ˆ σ2 =

1 n−2

• i ˆ

u2

i = SSR n−2

n − 2: degrees of freedom (DF) for SSR. (ˆ ui is restricted by two moment restrictions.)

15 / 26

Estimated Variances

Substitute ˆ σ2 for σ2, and we can get the estimated variances:

• Var(ˆ

β1) =

ˆ σ2 SSTx

• Var(ˆ

β0) = ˆ

σ2

i x2 i

n·SSTx

and standard errors: se(ˆ β1) =

• Var(ˆ

β1) se(ˆ β0) =

• Var(ˆ

β0)

16 / 26

Estimated Variances

Substitute ˆ σ2 for σ2, and we can get the estimated variances:

• Var(ˆ

β1) =

ˆ σ2 SSTx

• Var(ˆ

β0) = ˆ

σ2

i x2 i

n·SSTx

and standard errors: se(ˆ β1) =

• Var(ˆ

β1) se(ˆ β0) =

• Var(ˆ

β0)

16 / 26

Units of measurement

◮ Suppose we have data on education and wage,

and estimated a wage equation using Stata: wage = ˆ β0 + ˆ β1educ + ˆ u = 146.95 + 60.21educ + ˆ u, R2 = 0.107

◮ Now we would like to interpret the results.

(1) educ (education) is measured as years of schooling (2) wage is measured as pounds per month

◮ So ˆ

β1 = 60.21 would mean that: if one increases his education by 1 year, then on average, his wage will increase by 60.21 pounds per month.

17 / 26

Units of measurement

◮ Suppose we have data on education and wage,

and estimated a wage equation using Stata: wage = ˆ β0 + ˆ β1educ + ˆ u = 146.95 + 60.21educ + ˆ u, R2 = 0.107

◮ Now we would like to interpret the results.

(1) educ (education) is measured as years of schooling (2) wage is measured as pounds per month

◮ So ˆ

β1 = 60.21 would mean that: if one increases his education by 1 year, then on average, his wage will increase by 60.21 pounds per month.

17 / 26

Units of measurement

◮ Suppose we have data on education and wage,

and estimated a wage equation using Stata: wage = ˆ β0 + ˆ β1educ + ˆ u = 146.95 + 60.21educ + ˆ u, R2 = 0.107

◮ Now we would like to interpret the results.

(1) educ (education) is measured as years of schooling (2) wage is measured as pounds per month

◮ So ˆ

β1 = 60.21 would mean that: if one increases his education by 1 year, then on average, his wage will increase by 60.21 pounds per month.

17 / 26

Changing units of measurement: change in scale

◮ Suppose that we change the units of measurement for educ

and wage, such that (1) educ∗ is measured as months of schooling (2) wage∗ is measured as hundred pounds per month

◮ So for the same wage equation, wage∗ = ˆ

β∗

0 + ˆ

β∗

1educ∗ + ˆ

u∗, what would be ˆ β∗

0 and ˆ

β∗

1?

18 / 26

Changing units of measurement: change in scale

◮ Suppose that we change the units of measurement for educ

and wage, such that (1) educ∗ is measured as months of schooling (2) wage∗ is measured as hundred pounds per month

◮ So for the same wage equation, wage∗ = ˆ

β∗

0 + ˆ

β∗

1educ∗ + ˆ

u∗, what would be ˆ β∗

0 and ˆ

β∗

1?

18 / 26

◮ First, we can say that

(1) educ∗ = educ × 12 (2) wage∗ = wage/100

◮ Second, re-write the original equation

wage = ˆ β0 + ˆ β1educ + ˆ u as: (wage/100) · 100 = ˆ β0 + ˆ β1(educ · 12)/12 + ˆ u or wage∗ · 100 = ˆ β0 + ˆ β1educ∗/12 + ˆ u This equation is essentially the same as the original one, but includes variables with changed units of measurement, wage∗ and educ∗.

19 / 26

◮ First, we can say that

(1) educ∗ = educ × 12 (2) wage∗ = wage/100

◮ Second, re-write the original equation

wage = ˆ β0 + ˆ β1educ + ˆ u as: (wage/100) · 100 = ˆ β0 + ˆ β1(educ · 12)/12 + ˆ u or wage∗ · 100 = ˆ β0 + ˆ β1educ∗/12 + ˆ u This equation is essentially the same as the original one, but includes variables with changed units of measurement, wage∗ and educ∗.

19 / 26

◮ First, we can say that

(1) educ∗ = educ × 12 (2) wage∗ = wage/100

◮ Second, re-write the original equation

wage = ˆ β0 + ˆ β1educ + ˆ u as: (wage/100) · 100 = ˆ β0 + ˆ β1(educ · 12)/12 + ˆ u or wage∗ · 100 = ˆ β0 + ˆ β1educ∗/12 + ˆ u This equation is essentially the same as the original one, but includes variables with changed units of measurement, wage∗ and educ∗.

19 / 26

◮ First, we can say that

(1) educ∗ = educ × 12 (2) wage∗ = wage/100

◮ Second, re-write the original equation

wage = ˆ β0 + ˆ β1educ + ˆ u as: (wage/100) · 100 = ˆ β0 + ˆ β1(educ · 12)/12 + ˆ u or wage∗ · 100 = ˆ β0 + ˆ β1educ∗/12 + ˆ u This equation is essentially the same as the original one, but includes variables with changed units of measurement, wage∗ and educ∗.

19 / 26

◮ Third, re-arrange the equation in the second step,

i.e., divide both sides by 100, and get wage∗ = (ˆ β0/100) + (ˆ β1/1200)educ∗ + ˆ u/100 So ˆ β∗

0 = ˆ

β0/100 = 1.4695, and ˆ β∗

1 = ˆ

β1/1200 = 60.21/1200 = 0.05.

◮ This is an example of change in scale.

Can you interpret ˆ β∗

0 and ˆ

β∗

1?

20 / 26

◮ Third, re-arrange the equation in the second step,

i.e., divide both sides by 100, and get wage∗ = (ˆ β0/100) + (ˆ β1/1200)educ∗ + ˆ u/100 So ˆ β∗

0 = ˆ

β0/100 = 1.4695, and ˆ β∗

1 = ˆ

β1/1200 = 60.21/1200 = 0.05.

◮ This is an example of change in scale.

Can you interpret ˆ β∗

0 and ˆ

β∗

1?

20 / 26

◮ Third, re-arrange the equation in the second step,

i.e., divide both sides by 100, and get wage∗ = (ˆ β0/100) + (ˆ β1/1200)educ∗ + ˆ u/100 So ˆ β∗

0 = ˆ

β0/100 = 1.4695, and ˆ β∗

1 = ˆ

β1/1200 = 60.21/1200 = 0.05.

◮ This is an example of change in scale.

Can you interpret ˆ β∗

0 and ˆ

β∗

1?

20 / 26

Changing units of measurement: Change in origin

Suppose that we change the units of measurement for educ such that educ∗ is measured as years of schooling after primary school (assume it takes 6 years to finish primary school). We’d like to know how the coefficients would change from the original equation.

◮ First, we can say that educ∗ = educ − 6. ◮ Second, re-write the original equation

wage = ˆ β0 + ˆ β1educ + ˆ u as: wage = ˆ β0 + ˆ β1(educ − 6 + 6) + ˆ u or wage = ˆ β0 + ˆ β1(educ∗ + 6) + ˆ u

21 / 26

Changing units of measurement: Change in origin

Suppose that we change the units of measurement for educ such that educ∗ is measured as years of schooling after primary school (assume it takes 6 years to finish primary school). We’d like to know how the coefficients would change from the original equation.

◮ First, we can say that educ∗ = educ − 6. ◮ Second, re-write the original equation

wage = ˆ β0 + ˆ β1educ + ˆ u as: wage = ˆ β0 + ˆ β1(educ − 6 + 6) + ˆ u or wage = ˆ β0 + ˆ β1(educ∗ + 6) + ˆ u

21 / 26

Changing units of measurement: Change in origin

Suppose that we change the units of measurement for educ such that educ∗ is measured as years of schooling after primary school (assume it takes 6 years to finish primary school). We’d like to know how the coefficients would change from the original equation.

◮ First, we can say that educ∗ = educ − 6. ◮ Second, re-write the original equation

wage = ˆ β0 + ˆ β1educ + ˆ u as: wage = ˆ β0 + ˆ β1(educ − 6 + 6) + ˆ u or wage = ˆ β0 + ˆ β1(educ∗ + 6) + ˆ u

21 / 26

◮ Third, re-arrange the equation in the second step,

wage = (ˆ β0 + 6ˆ β1) + ˆ β1educ∗ + ˆ u So ˆ β∗

0 = ˆ

β0 + 6ˆ β1 = 146.95 + 6 × 60.21 = 508.21, and ˆ β∗

1 does not change from the original equation.

22 / 26

Changing units of measurement does not change the goodness-of-fit, R2 = SSE/SST. This is because (1) When there is a change in scale, SSE and SST change proportionally, and the ratio SSE/SST does not change. (2) When there is a change in origin, neither SSE nor SST changes, and the ratio SSE/SST does not change.

23 / 26

Changing units of measurement does not change the goodness-of-fit, R2 = SSE/SST. This is because (1) When there is a change in scale, SSE and SST change proportionally, and the ratio SSE/SST does not change. (2) When there is a change in origin, neither SSE nor SST changes, and the ratio SSE/SST does not change.

23 / 26

Nonlinearities in simple regression

◮ What if we estimate another wage equation,

where wage is logged:

• log(wage) = ˆ

β0 + ˆ β1educ = 0.62 + 0.08educ, R2 = 0.2 This is a log-level model.

◮ So ˆ

β1 = 0.08 would mean that: if one increases his education by 1 year, then on average, his wage will increase by 8% per month.

◮ Advantage of the log-level model:

the slope coefficient is unit-free (i.e., it’s a percentage).

24 / 26

Nonlinearities in simple regression

◮ What if we estimate another wage equation,

where wage is logged:

• log(wage) = ˆ

β0 + ˆ β1educ = 0.62 + 0.08educ, R2 = 0.2 This is a log-level model.

◮ So ˆ

β1 = 0.08 would mean that: if one increases his education by 1 year, then on average, his wage will increase by 8% per month.

◮ Advantage of the log-level model:

the slope coefficient is unit-free (i.e., it’s a percentage).

24 / 26

Nonlinearities in simple regression

◮ What if we estimate another wage equation,

where wage is logged:

• log(wage) = ˆ

β0 + ˆ β1educ = 0.62 + 0.08educ, R2 = 0.2 This is a log-level model.

◮ So ˆ

β1 = 0.08 would mean that: if one increases his education by 1 year, then on average, his wage will increase by 8% per month.

◮ Advantage of the log-level model:

the slope coefficient is unit-free (i.e., it’s a percentage).

24 / 26

Nonlinearities in simple regression

◮ Is the regression log(wage) = β0 + β1educ + u linear?

wage is non-linear in educ (because wage is exponential function of educ). However, the regression is linear in parameters!

◮ Examples:

Linear in parameter: y = β0 + β1 √x + u Non-linear in parameter: y = xβ + u

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Nonlinearities in simple regression

◮ Is the regression log(wage) = β0 + β1educ + u linear?

wage is non-linear in educ (because wage is exponential function of educ). However, the regression is linear in parameters!

◮ Examples:

Linear in parameter: y = β0 + β1 √x + u Non-linear in parameter: y = xβ + u

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Reading

Chapter 2, Introductory Econometrics - A Modern Approach, 4th Edition, J. Wooldridge

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