Logistic Regression

Dr. Mine Dogucu

(Normal) Linear Regression Response Variables

Birth weight of Babies (55 - 176 ounces)
Sale Prices ($12789 - $755,000)
Number of Species (6 - 129 mammals)

Logistic Regression Response Variables

Will it rain tomorrow? (Yes/No)
Is email spam? (Yes/No)
Does the candidate receive a callback? (Yes/No)

When the response variable is binary a logistic regression model can be utilized.

Are Emily and Greg More Employable than Lakisha and Jamal? A Field Experiment on Labor Market Discrimination.

Researchers respond to help-wanted ads in Boston and Chicago newspapers with fictitious resumes.

They randomly assign White sounding names to half the resumes and African American sounding names to the other half.

They create high quality resumes (more experience, likely to have an email address etc.) and low quality resumes.
For each job ad they send four resumes (two high quality and two low quality.)

Data

resume <- resume %>% 
  select(received_callback, race, years_experience, 
         job_city)

glimpse(resume)

Rows: 4,870
Columns: 4
$ received_callback <lgl> FALSE, FALSE, FALSE, FALSE, FALSE, FALSE, FALSE, FAL…
$ race              <chr> "white", "white", "black", "black", "white", "white"…
$ years_experience  <int> 6, 6, 6, 6, 22, 6, 5, 21, 3, 6, 8, 8, 4, 4, 5, 4, 5,…
$ job_city          <chr> "Chicago", "Chicago", "Chicago", "Chicago", "Chicago…

Response variable: received_callback

count(resume, received_callback) %>% 
  mutate(prop = n / sum(n))

# A tibble: 2 × 3
  received_callback     n   prop
  <lgl>             <int>  <dbl>
1 FALSE              4478 0.920 
2 TRUE                392 0.0805

Notation

$y_i$ = whether a (fictitious) job candidate receives a call back.

$\pi_i$ = probability that the $i$th job candidate will receive a call back.

$1-\pi_i$ = probability that the $i$th job candidate will not receive a call back.

Where is the line?

ggplot(resume, aes(x = race, y = received_callback)) +
  geom_point()

The Linear Model

We can model the probability of receiving a callback with a linear model.

$\text{transformation}(\pi_i) = \beta_0 + \beta_1x_{1i}+\beta_2x_{2i} +.... \beta_kx_{ki}$

$logit(\pi_i) = \beta_0 + \beta_1x_{1i}+\beta_2x_{2i} +.... \beta_kx_{ki}$

$logit(\pi_i) = log(\frac{\pi_i}{1-\pi_i})$

Note that log is natural log and not base 10. This is also the case for the log() function in R.

Probability $\pi_i$ Probability of receiving a callback.

Odds $\frac{\pi_i}{1-\pi_i}$ Odds of receiving a callback.

Logit $log(\frac{\pi_i}{1-\pi_i})$ Logit of receiving a callback.

When race is black (0)

resume %>% 
  filter(race == "black") %>% 
  count(received_callback) %>% 
  mutate(prop = n / sum(n))

# A tibble: 2 × 3
  received_callback     n   prop
  <lgl>             <int>  <dbl>
1 FALSE              2278 0.936 
2 TRUE                157 0.0645

Note that R assigns 0 an 1 to levels of categorical variables in alphabetical order. In this case black (0) and white(1)

When race is black (0)

p_b <- resume %>% 
  filter(race == "black") %>% 
  count(received_callback) %>% 
  mutate(prop = n / sum(n)) %>% 
  filter(received_callback == TRUE) %>% 
  select(prop) %>% 
  pull()

Probability of receiving a callback when the candidate has a Black sounding name is 0.0644764.

When race is white (1)

p_w <- resume %>% 
  filter(race == "white") %>% 
  count(received_callback) %>% 
  mutate(prop = n / sum(n)) %>% 
  filter(received_callback == TRUE) %>% 
  select(prop) %>% 
  pull()

Probability of receiving a callback when the candidate has a white sounding name is 0.0965092.

p_b

[1] 0.06447639

## Odds
odds_b <- p_b / (1 - p_b)
odds_b

[1] 0.06892011

## Logit
logit_b <- log(odds_b)
logit_b

[1] -2.674807

p_w

[1] 0.09650924

## Odds
odds_w <- p_w / (1 - p_w)
odds_w

[1] 0.1068182

## Logit
logit_w <- log(odds_w)
logit_w

[1] -2.236627

This is THE LINE of the linear model. As x increases by 1 unit, the expected change in the logit of receiving call back is 0.4381802. In this case, this is just the difference between logit for the white group and the black group.

The slope of the line is:

logit_w - logit_b

[1] 0.4381802

The intercept is

logit_b

[1] -2.674807

model_r <- glm(received_callback ~ race,
               data = resume,
               family = binomial)

tidy(model_r)

# A tibble: 2 × 5
  term        estimate std.error statistic   p.value
  <chr>          <dbl>     <dbl>     <dbl>     <dbl>
1 (Intercept)   -2.67     0.0825    -32.4  1.59e-230
2 racewhite      0.438    0.107       4.08 4.45e-  5

$log(\frac{\hat \pi_i}{1-\hat \pi_i}) = -2.67 + 0.438\times racewhite_i$

Scale	Range
Probability	0 to 1
Odds	0 to $\infty$
Logit	- $\infty$ to $\infty$

We will consider years of experience as an explanatory variable. Normally, we would also include race in the model and have multiple explanatory variables, however, for learning purposes, we will keep the model simple.

model_y <- glm(received_callback ~ years_experience,
               data = resume,
               family = binomial)

tidy(model_y)

# A tibble: 2 × 5
  term             estimate std.error statistic   p.value
  <chr>               <dbl>     <dbl>     <dbl>     <dbl>
1 (Intercept)       -2.76     0.0962     -28.7  5.58e-181
2 years_experience   0.0391   0.00918      4.26 2.07e-  5

model_y_summary <- tidy(model_y)

intercept <- model_y_summary %>% 
  filter(term == "(Intercept)") %>% 
  select(estimate) %>% 
  pull()

slope <- model_y_summary %>% 
  filter(term == "years_experience") %>% 
  select(estimate) %>% 
  pull()

From logit to odds

Logit for a Candidate with 1 year of experience (rounded equation)

$-2.76 + 0.0391 \times 1$

Odds for a Candidate with 1 year of experience

$odds = e^{logit}$

$\frac{\pi_i}{1-\pi_i} = e^{log(\frac{\pi_i}{1-\pi_i})}$

$\frac{\hat\pi_i}{1-\hat\pi_i} = e^{-2.76 + 0.0391 \times 1}$

From odds to probability

$\pi_i = \frac{odds}{1+odds}$

$\pi_i = \frac{\frac{\pi_i}{1-\pi_i}}{1+\frac{\pi_i}{1-\pi_i}}$

$\hat\pi_i = \frac{e^{-2.76 + 0.0391 \times 1}}{1+e^{-2.76 + 0.0391 \times 1}} = 0.0618$

Note you can use exp() function in R for exponentiating number e.

exp(1)

[1] 2.718282

Logistic Regression model

Logit form:

$log(\frac{\pi_i}{1-\pi_i}) = \beta_0 + \beta_1x_{1i}+\beta_2x_{2i} +.... \beta_kx_{ki}$

Probability form:

$\large{\pi_i = \frac{e^{\beta_0 + \beta_1x_{1i}+\beta_2x_{2i} +.... \beta_kx_{ki}}}{1+e^{\beta_0 + \beta_1x_{1i}+\beta_2x_{2i} +.... \beta_kx_{ki}}}}$

Estimated probability of a candidate with 0 years of experience receiving a callback

$\hat\pi_i = \frac{e^{-2.76 + 0.0391 \times 0}}{1+e^{-2.76 + 0.0391 \times 0}} = 0.0595$

Estimated probability of a candidate with 1 year of experience receiving a callback

$\hat\pi_i = \frac{e^{-2.76 + 0.0391 \times 1}}{1+e^{-2.76 + 0.0391 \times 1}} = 0.0618$

model_ryc <- glm(received_callback ~ race + 
                  years_experience + job_city,
               data = resume,
               family = binomial)

tidy(model_ryc)

# A tibble: 4 × 5
  term             estimate std.error statistic  p.value
  <chr>               <dbl>     <dbl>     <dbl>    <dbl>
1 (Intercept)       -2.78     0.134      -20.8  6.18e-96
2 racewhite          0.440    0.108        4.09 4.39e- 5
3 years_experience   0.0332   0.00940      3.53 4.11e- 4
4 job_cityChicago   -0.329    0.108       -3.04 2.33e- 3

The estimated probability that a Black candidate with 10 years of experience, residing in Boston, would receive a callback.

$\large{\hat\pi_i = \frac{e^{-2.78 + (0.0440 \times 0) + (0.0332\times10) + (-0.0329\times 0)}}{1+e^{-2.78 + (0.0440 \times 0) + (0.0332\times10) + (-0.0329\times 0)}} = 0.0796}$

We have used the data for educational purposes. The original study considers many other variables that may influence whether someone receives a callback or not. Read the original study for other considerations.

Are Emily and Greg More Employable than Lakisha and Jamal? A Field Experiment on Labor Market Discrimination.

Model Evaluation

babies <- babies %>% 
  mutate(low_bwt = case_when(bwt < 88 ~ TRUE,
                             bwt >= 88~ FALSE)) %>% 
  drop_na(gestation)

model_g <- glm(low_bwt ~ gestation, 
               data = babies,
               family = "binomial")

tidy(model_g)

# A tibble: 2 × 5
  term        estimate std.error statistic  p.value
  <chr>          <dbl>     <dbl>     <dbl>    <dbl>
1 (Intercept)  17.5      2.24         7.82 5.11e-15
2 gestation    -0.0758   0.00846     -8.96 3.27e-19

$\hat p = \frac{\exp(b_0 +b_1x)}{1+\exp(b_0 + b_1x)}$

$\hat p$ when gestation is 284 = $\frac{\exp(17.5 -0.0758 \cdot 284)}{1+\exp(17.5 -0.0758 \cdot 284)} = \frac{\exp(-4.0272)}{1+ \exp(-4.0272)} = 0.01751203$

babies <- babies %>% 
  add_predictions(model_g) 
select(babies, pred)

# A tibble: 1,223 × 1
     pred
    <dbl>
 1 -4.02 
 2 -3.87 
 3 -3.64 
 4 -3.87 
 5 -4.17 
 6 -0.986
 7 -1.06 
 8 -4.40 
 9 -5.15 
10 -9.10 
# ℹ 1,213 more rows

babies <- babies %>% 
  mutate(pred_p = exp(pred)/(1+exp(pred)))
select(babies, pred, pred_p)

# A tibble: 1,223 × 2
     pred   pred_p
    <dbl>    <dbl>
 1 -4.02  0.0177  
 2 -3.87  0.0205  
 3 -3.64  0.0256  
 4 -3.87  0.0205  
 5 -4.17  0.0152  
 6 -0.986 0.272   
 7 -1.06  0.257   
 8 -4.40  0.0122  
 9 -5.15  0.00574 
10 -9.10  0.000112
# ℹ 1,213 more rows

Cutoff

babies <- babies %>% 
  mutate(pred_y = case_when(pred_p < 0.5 ~ FALSE, 
                           pred_p >= 0.5 ~ TRUE))
select(babies, low_bwt, pred, pred_p, pred_y)

# A tibble: 1,223 × 4
   low_bwt   pred   pred_p pred_y
   <lgl>    <dbl>    <dbl> <lgl> 
 1 FALSE   -4.02  0.0177   FALSE 
 2 FALSE   -3.87  0.0205   FALSE 
 3 FALSE   -3.64  0.0256   FALSE 
 4 FALSE   -3.87  0.0205   FALSE 
 5 FALSE   -4.17  0.0152   FALSE 
 6 FALSE   -0.986 0.272    FALSE 
 7 FALSE   -1.06  0.257    FALSE 
 8 FALSE   -4.40  0.0122   FALSE 
 9 FALSE   -5.15  0.00574  FALSE 
10 FALSE   -9.10  0.000112 FALSE 
# ℹ 1,213 more rows

Confusion Matrix

janitor::tabyl(babies, low_bwt, pred_y) %>% 
  janitor::adorn_totals(c("row", "col"))

 low_bwt FALSE TRUE Total
   FALSE  1161    5  1166
    TRUE    53    4    57
   Total  1214    9  1223

Sensitivity (true-positive rate): 4/57 = 0.0702
Specificity (true-negative rate): 1161/1166 = 0.9957
Accuracy: (1161+4)/1223 = 0.9526