• We have seen different data types in the first session.
• One of them was factor, representing categorical data:
• A person is male or female
• A plane is passenger, cargo or military
• Some good is produced in Spain, France, China or UK.

• A dummy variable is either TRUE or FALSE (or 0 or 1).
• We use dummies to mark category membership: if member, then TRUE.
• for example, \[ \begin{aligned} \text{is.male}_i &= \begin{cases} 1 & \text{if }i\text{ is male} \\ 0 & \text{if }i\text{ is not male}. \\ \end{cases} \\ \end{aligned} \]
• Notice that whether 0 corresponds to TRUE or FALSE is up to you. Just be consistent!

• We defined is.male. Let's define its pendant.
• That is, for females, \[ \begin{aligned} \text{is.female}_i &= \begin{cases} 1 & \text{if }i\text{ is female} \\ 0 & \text{if }i\text{ is not female} \\ \end{cases} \\ \end{aligned} \]
• let's all create this dataset:

df1 = data.frame(income=c(3000,5000,7500,3500),
             sex=c("male","female","male","female"))

• Let's run regression \(y = b_0 + b_1 is.male + b_2 is.female\)
• First, we create those dummy variables:

df1$is.male = df1$sex == "male"
df1$is.female = df1$sex == "female"

• and then let's run this:

lm(income ~ is.male + is.female,df1)

• What do you see? 🤔

df1$linear_comb = df1$is.male + df1$is.female
df1

##   income    sex is.male is.female linear_comb
## 1   3000   male    TRUE     FALSE           1
## 2   5000 female   FALSE      TRUE           1
## 3   7500   male    TRUE     FALSE           1
## 4   3500 female   FALSE      TRUE           1

• Oops. is.male + is.female is always equal 1!
• In other words, is.male = 1 - is.female. A perfect colinearity!
• Multiple regression fails. 😠

• Notice: Inclusion of both dummies doesn't add anything
• If someone is male they are not female.
• So we drop one of the categories. Only do \(y = b_0 + b_1 is.female\)

lm1 = lm(income ~ is.female,df1)
lm1

## 
## Call:
## lm(formula = income ~ is.female, data = df1)
## 
## Coefficients:
##   (Intercept)  is.femaleTRUE  
##          5250          -1000

• We have excluded is.male.
• So what's the effect of being male now?
• Well, male means is.female = 0. So male is subsumed in the intercept!
• At is.female = 0, i.e. \(\widehat{y} = b_0 + b_1 \cdot 0=\) 5250
• Coefficient on is.female is \(b_1=\) -1000. It measures the difference in intercept from being female.
• That means: \(\widehat{y} = b_0 + b_1 \cdot 1=\) 4250

• So, we have seen that \[ b_1 = E[Y|\text{is.female}=1] - E[Y|\text{is.female}=0] \]
• This was the meaning of the red arrow.

• Time for you to play around with the Binary Regression!
• Try to find the best line again!

library(ScPoEconometrics)
launchApp("reg_dummy")

• What if we added \(\text{exper}_i\in \mathbb{N}\) to that regression? \[ y_i = b_0 + b_1 \text{is.female}_i + b_2 \text{exper}_i + e_i \]
• As before, dummy acts as intercept shifter. We have \[ y_i = \begin{cases} b_0 + b_1 + b_2 \text{exper}_i + e_i & \text{if is.female=1} \\ b_0 + \hphantom{b_1} + b_2 \text{exper}_i + e_i & \text{if is.female=0} \end{cases} \]
• intercept is \(b_0 + b_1\) for women but \(b_0\) for men
• Slope \(b_2\) identical for both!

library(ScPoEconometrics)
launchApp("reg_dummy_example")

• Let \(\alpha\) represent the scale of the regression.
• E.g. if \(\alpha=1\), measure in Euros, \(\alpha=\frac{1}{1000}\) in thousands of Euros.
• \(\log\) transforming \(x\), we obtain \[ y = b_0 + b_1 \log(\alpha x) = b_0 + \log \alpha + b_1 \log x \] hence, scale \(\alpha\) moves to intercept, slope becomes invariant to it!

• Now transform both outcome and regressor: \[ \log y = b_0 + b_1 \log x \]
• Then the slope coefficient is \[ b_1 = \frac{d\log y}{d \log x} = \frac{dy/y}{dx/x} \]
• This represents the elasticity of \(y\) with respect to \(x\)
• \(y\) changes by \(b_1\) percent if \(x\) changes by one percent.

• Now only the outcome is log transformed.
• We have a semi-elasticity formulation. \[ \log y = b_0 + b_1 x \]
• Slope coefficient becomes \[ b_1 = \frac{d\log y}{d x} \]
• A one-unit change in \(x\) increases the logarithm of the outcome by \(b_1\) units.
• Small \(\Delta x\): exponentiating \(b_1\) approximates effect of \(x\) on level of \(y\). (not on \(\log(y)\))

• Wages \(w\) are often \(\log\)-transformed
• \(w\) are non-negative
• \(w\) are approximately log-normally distributed (so \(\log(w)\sim N\))
• Here is a typical equation: What's the return of experience? \[ \ln w_i = b_0 + b_1 exp_i + e_i \]

data(Wages,package="Ecdat")  # load data
str(Wages)

## 'data.frame':    4165 obs. of  12 variables:
##  $ exp    : int  3 4 5 6 7 8 9 30 31 32 ...
##  $ wks    : int  32 43 40 39 42 35 32 34 27 33 ...
##  $ bluecol: Factor w/ 2 levels "no","yes": 1 1 1 1 1 1 1 2 2 2 ...
##  $ ind    : int  0 0 0 0 1 1 1 0 0 1 ...
##  $ south  : Factor w/ 2 levels "no","yes": 2 2 2 2 2 2 2 1 1 1 ...
##  $ smsa   : Factor w/ 2 levels "no","yes": 1 1 1 1 1 1 1 1 1 1 ...
##  $ married: Factor w/ 2 levels "no","yes": 2 2 2 2 2 2 2 2 2 2 ...
##  $ sex    : Factor w/ 2 levels "female","male": 2 2 2 2 2 2 2 2 2 2 ...
##  $ union  : Factor w/ 2 levels "no","yes": 1 1 1 1 1 1 1 1 1 2 ...
##  $ ed     : int  9 9 9 9 9 9 9 11 11 11 ...
##  $ black  : Factor w/ 2 levels "no","yes": 1 1 1 1 1 1 1 1 1 1 ...
##  $ lwage  : num  5.56 5.72 6 6 6.06 ...

• \(E[\ln w] = 6.68\) and \(E[w] = 882.93\)
• Additional year of experience increases \(E[\ln w]\) by 0.0088.
• Approximate effect on \(E[w]=\exp(b_1)=1.009\)
• Precise effect is \(100 * (\exp(b_1)-1) = 0.885\)
• Anway: Tiny effect! 😒

• Let's see if there are important differences by gender in this: \[ \ln w_i = b_0 + b_1 exp_i + b_2 sex_i + e_i \]
• We just update our lm object as follows:

lm_sex = update(lm_w, . ~ . + sex)  
# update lm_w with same LHS, same RHS, but add sex to it