Logistic Regression

Lecture 17

Josh Lim

Duke University
STA 199 Summer 2026: Session 2
Adapted from slides by Mine Çetinkaya-Rundel, Katie Solarz & John Zito

July 27, 2026

So where were we…

Thus far…

We have been studying regression:

• What combinations of data types have we seen?

• What did the picture look like?

Recap: simple linear regression

Numerical response and one numerical predictor:

Recap: simple linear regression

Numerical response and one categorical predictor (two levels):

Recap: multiple linear regression

Numerical response; numerical and categorical predictors:

Today: a binary response

y = \begin{cases} 1 & &&\text{eg. Yes, Win, True, Heads, Success}\\ 0 & &&\text{eg. No, Lose, False, Tails, Failure}. \end{cases}

Who cares?

If we can model the relationship between predictors (x) and a binary response (y), we can use the model to do a special kind of prediction called classification.

Example: is the e-mail spam or not?

\mathbf{x}: \text{word and character counts in an e-mail.}

y = \begin{cases} 1 & \text{it's spam}\\ 0 & \text{it's legit} \end{cases}

Example: is it cancer or not?

\mathbf{x}: \text{features in a medical image.}

y = \begin{cases} 1 & \text{it's cancer}\\ 0 & \text{it's healthy} \end{cases}

Example: will they default?

\mathbf{x}: \text{financial and demographic info about a loan applicant.}

y = \begin{cases} 1 & \text{applicant is at risk of defaulting on loan}\\ 0 & \text{applicant is safe} \end{cases}

Example: Who said it: Taylor Swift, or Shakespeare?

\mathbf{x}: \text{word counts (e.g., thou, love, heartbreak), stylistic features}

y = \begin{cases} 1 & \text{Taylor Swift}\\ 0 & \text{William Shakespeare} \end{cases}

How do we model this type of data?

Straight line of best fit is a little silly

Instead: S-curve of best fit

Instead of modeling y directly, we model the probability that y=1:

• “Given new email, what’s the probability that it’s spam?’’
• “Given new image, what’s the probability that it’s cancer?’’
• “Given new loan application, what’s the probability that they default?’’

Why don’t we model y directly?

• Recall regression with a numerical response:

  • Our models do not output guarantees for y, they output predictions that describe behavior on average;

• Similar when modeling a binary response:

  • Our models cannot directly guarantee that y will be zero or one. The correct analog to “on average” for a 0/1 response is “what’s the probability?”

So, what is this S-curve, anyway?

It’s the logistic function:

\text{Prob}(y = 1) = \frac{e^{\beta_0+\beta_1x}}{1+e^{\beta_0+\beta_1x}}.

If you set p = \text{Prob}(y = 1) and do some algebra, you get the simple linear model for the log-odds:

\log\left(\frac{p}{1-p}\right) = \beta_0+\beta_1x.

This is called the logistic regression model.

But first a message from the Capitol

Probability? Odds? Log-odds?

• p = \text{Prob}(y = 1) is a probability. A number between 0 and 1;

• p / (1 - p) is the odds. A number between 0 and \infty;

“The odds of this lecture going well are 10 to 1.”

• The log odds \log(p / (1 - p)) is a number between -\infty and \infty, which is suitable for the linear model.

• Why does this “transformation” work? The log function maps positive numbers (0, \infty) to all real numbers (-\infty, \infty).

Probability to odds

Zooming in

Odds to log odds

Logistic regression

\log\left(\frac{p}{1-p}\right) = \beta_0+\beta_1x.

• The logit function \log(p / (1-p)) is an example of a link function that transforms the linear model to have an appropriate range;

• This is an example of a generalized linear model

Estimation

• We estimate the parameters \beta_0,\,\beta_1 using maximum likelihood (don’t worry about it) to get the “best fitting” S-curve;

• The fitted model is

\log\left(\frac{\hat{p}}{1-\hat{p}}\right) = b_0+b_1x.

Today’s data

email |> 
  dplyr::select(c(spam, dollar, viagra, winner, password, exclaim_mess)) |> 
  glimpse()

Rows: 3,921
Columns: 6
$ spam         <fct> 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0,…
$ dollar       <dbl> 0, 0, 4, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 2, 0,…
$ viagra       <dbl> 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0,…
$ winner       <fct> no, no, no, no, no, no, no, no, no, no, no, no,…
$ password     <dbl> 0, 0, 0, 0, 2, 2, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0,…
$ exclaim_mess <dbl> 0, 1, 6, 48, 1, 1, 1, 18, 1, 0, 2, 1, 0, 10, 4,…

Fitting a logistic model

logistic_fit <- logistic_reg() |>
  fit(spam ~ exclaim_mess, data = email)

tidy(logistic_fit)

# A tibble: 2 × 5
  term          estimate std.error statistic p.value
  <chr>            <dbl>     <dbl>     <dbl>   <dbl>
1 (Intercept)  -2.27      0.0553     -41.1     0    
2 exclaim_mess  0.000272  0.000949     0.287   0.774

Fitted equation for the log-odds:

\log\left(\frac{\hat{p}}{1-\hat{p}}\right) = -2.27 + 0.000272\times exclaim~mess

Interpreting the intercept

If exclaim_mess = 0, then

\hat{p}=\widehat{P(y=1)}=\frac{e^{-2.27}}{1+e^{-2.27}}\approx 0.09.

So, our model predicts that an email with no exclamation marks has a 9% probability of being spam.

Interpreting the slope is tricky

Recall:

\log\left(\frac{\hat{p}}{1-\hat{p}}\right) = b_0+b_1x.

Alternatively:

\frac{\hat{p}}{1-\hat{p}} = e^{b_0+b_1x} = \color{blue}{e^{b_0}e^{b_1x}} .

If we increase x by one unit, we have:

\frac{\hat{p}}{1-\hat{p}} = e^{b_0}e^{b_1(x+1)} = e^{b_0}e^{b_1x+b_1} = {\color{blue}{e^{b_0}e^{b_1x}}}{\color{red}{e^{b_1}}} .

A one unit increase in x is associated with a change in odds by a factor of e^{b_1}. Gross!

Back to the example…

\log\left(\frac{\hat{p}}{1-\hat{p}}\right) = -2.27 + 0.000272\times exclaim~mess

If the email has an additional exclamation mark, we predict the odds of an email being spam to be higher by a multiplicative factor of e^{0.000272}\approx 1.000272 on average.

Logistic regression -> classification?

Step 0: fit the model

Select a number 0 < p^* < 1:

• if \text{Prob}(y=1)\leq p^*, then predict \widehat{y}=0;
• if \text{Prob}(y=1)> p^*, then predict \widehat{y}=1.

Step 1: pick a threshold

Select a number 0 < p^* < 1:

• if \text{Prob}(y=1)\leq p^*, then predict \widehat{y}=0;
• if \text{Prob}(y=1)> p^*, then predict \widehat{y}=1.

Step 2: find the “decision boundary”

Solve for the x-value that matches the threshold:

• if \text{Prob}(y=1)\leq p^*, then predict \widehat{y}=0;
• if \text{Prob}(y=1)> p^*, then predict \widehat{y}=1.

Step 3: classify a new arrival

A new person shows up with x_{\text{new}}. Which side of the boundary are they on?

• if x_{\text{new}} \leq x^\star, then \text{Prob}(y=1)\leq p^*, so predict \widehat{y}=0 for the new person;
• if x_{\text{new}} > x^\star, then \text{Prob}(y=1)> p^*, so predict \widehat{y}=1 for the new person.

Let’s change the threshold

A new person shows up with x_{\text{new}}. Which side of the boundary are they on?

• if x_{\text{new}} \leq x^\star, then \text{Prob}(y=1)\leq p^*, so predict \widehat{y}=0 for the new person;
• if x_{\text{new}} > x^\star, then \text{Prob}(y=1)> p^*, so predict \widehat{y}=1 for the new person.

Let’s change the threshold

A new person shows up with x_{\text{new}}. Which side of the boundary are they on?

• if x_{\text{new}} \leq x^\star, then \text{Prob}(y=1)\leq p^*, so predict \widehat{y}=0 for the new person;
• if x_{\text{new}} > x^\star, then \text{Prob}(y=1)> p^*, so predict \widehat{y}=1 for the new person.

Nothing special about one predictor…

Two numerical predictors and one binary response:

“Multiple” logistic regression

For the log-odds, a multiple linear regression:

\log\left(\frac{p}{1-p}\right) = \beta_0+\beta_1x_1+\beta_2x_2+...+\beta_mx_m. On the probability scale:

\text{Prob}(y = 1) = \frac{e^{\beta_0+\beta_1x_1+\beta_2x_2+...+\beta_mx_m}}{1+e^{\beta_0+\beta_1x_1+\beta_2x_2+...+\beta_mx_m}}.

Decision boundary, again

It’s linear! Consider two numerical predictors:

• if new (x_1,\,x_2) below, \text{Prob}(y=1)\leq p^*. Predict \widehat{y}=0 for the new person;
• if new (x_1,\,x_2) above, \text{Prob}(y=1)> p^*. Predict \widehat{y}=1 for the new person.

Decision boundary, again

It’s linear! Consider two numerical predictors:

• if new (x_1,\,x_2) below, \text{Prob}(y=1)\leq p^*. Predict \widehat{y}=0 for the new person;
• if new (x_1,\,x_2) above, \text{Prob}(y=1)> p^*. Predict \widehat{y}=1 for the new person.

Decision boundary, again

It’s linear! Consider two numerical predictors:

• if new (x_1,\,x_2) below, \text{Prob}(y=1)\leq p^*. Predict \widehat{y}=0 for the new person;
• if new (x_1,\,x_2) above, \text{Prob}(y=1)> p^*. Predict \widehat{y}=1 for the new person.

Note: the classifier isn’t perfect

• There are blue points in the orange region: spam (1) emails misclassified as legit (0);
• There are orange points in the blue region: legit (0) emails misclassified as spam (1).

How do you pick the threshold?

To balance out the two kinds of errors:

• High threshold >> Hard to classify as 1 >> FP less likely; FN more likely
• Low threshold >> Easy to classify as 1 >> FP more likely; FN less likely

Silly examples

• Set p* = 0

  • Classify every email as spam (1);
  • No false negatives, but a lot of false positives;

• Set p* = 1

  • Classify every email as legit (0);
  • No false positives, but a lot of false negatives.

You pick a threshold in between to strike a balance. The exact number depends on context.

AE