What the Intercept Represents
A logistic regression model describes a conditional probability through a linear predictor. The intercept has a precise interpretation: it is the model’s log-odds for the predictor settings that make every non-intercept term equal to zero. Those settings are often called the baseline. The baseline probability is not automatically the overall probability in the population; it is the probability for the specified baseline predictor values under the fitted model.
For a binary response \(Y\), let \(p(\mathbf{x})=P(Y=1\mid \mathbf{x})\), where \(\mathbf{x}\) denotes the predictor values. A logistic regression model uses the logit link and a linear predictor. For example, with two numerical predictors, one may write \(\eta(\mathbf{x})=\beta_0+\beta_1x_1+\beta_2x_2\). The model sets the conditional log-odds equal to this predictor.
When all predictors are numerical and recorded so that zero is meaningful, the baseline uses \(x_j=0\) for each predictor. With categorical predictors, a baseline may instead use specified reference levels, depending on the model’s coding. In either case, check the actual predictor settings before interpreting the intercept. A zero in a model term does not always mean a zero in the underlying measurement: for example, a centered predictor equals zero at its centering value.
From Log-Odds to Odds and Probability
The word log-odds is important: the intercept is not itself the odds. If \(p_0\) is the baseline conditional probability, then its odds are \(p_0/(1-p_0)\). The logit link says that the logarithm of these odds equals the linear predictor. At baseline, the predictor is \(\beta_0\); exponentiating gives the odds. Solving the odds equation for probability gives the probability formula.
This conversion is valid for the logistic link. The inverse-logit formula always gives a probability strictly between zero and one for a finite intercept. A positive intercept gives baseline odds greater than one and baseline probability greater than one-half; a zero intercept gives odds one and probability one-half; and a negative intercept gives odds less than one and probability less than one-half.
Why the Conversion Works
The conversion follows directly from the definition of odds and the logit link; it does not require an additional approximation. Start at baseline with \(p_0/(1-p_0)=\exp(\beta_0)\). Multiply both sides by \(1-p_0\), then collect the terms involving \(p_0\):
Adding \(\exp(\beta_0)p_0\) to both sides gives \(p_0[1+\exp(\beta_0)]=\exp(\beta_0)\). Dividing by the positive quantity \(1+\exp(\beta_0)\) yields \(p_0=\exp(\beta_0)/(1+\exp(\beta_0))\). Dividing numerator and denominator by \(\exp(\beta_0)\) gives the equivalent form \(1/[1+\exp(-\beta_0)]\). Both expressions are useful for checking arithmetic.
Set every non-intercept term to zero, or identify the corresponding reference-level settings in the model.
At those settings, the linear predictor is \(\beta_0\), so the baseline log-odds equal \(\beta_0\).
Compute \(\exp(\beta_0)\). Do not report \(\beta_0\) itself as the odds.
Use \(p_0=\text{odds}/(1+\text{odds})\), or use the inverse-logit formula directly.
Worked Interpretations
Worked Example: A Negative Intercept and Baseline Risk
A hypothetical insurer models whether an application receives a specified follow-up review, with \(Y=1\) indicating a review. Conditional on its predictors, each outcome is Bernoulli with probability \(p(\mathbf{x})\), and the model uses the logit link. Its fitted linear predictor is \(\eta=\beta_0+\beta_1x_1+\beta_2x_2\), with \(\beta_0=\log(1/4)\). Find the baseline log-odds, odds, and probability when \(x_1=x_2=0\).
Method. Evaluate the linear predictor at the stated baseline, then exponentiate the log-odds and convert the resulting odds to a probability.
At \(x_1=x_2=0\), \(\eta=\log(1/4)+\beta_1(0)+\beta_2(0)=\log(1/4)\). Thus the baseline log-odds are \(\log(1/4)\), approximately \(-1.3863\). The baseline odds are \(\exp(\log(1/4))=1/4=0.25\).
The baseline probability is \(p_0=(1/4)/(1+1/4)=(1/4)/(5/4)=1/5=0.20\). To four decimal places, it is \(0.2000\). Check: the odds calculated from this probability are \(0.20/(1-0.20)=0.20/0.80=0.25\), as required.
Interpretation and trap avoided. At the stated baseline, the model assigns a 20% conditional probability of review. The negative intercept is a log-odds value, not a negative probability; exponentiating it produces positive odds.
Worked Example: Zero Intercept Does Not Mean Zero Probability
In a hypothetical model of whether a small business renews a service contract, \(Y=1\) indicates renewal and \(Y\mid\mathbf{x}\) is Bernoulli with probability \(p(\mathbf{x})\). The logit model has intercept \(\beta_0=0\). All other terms vanish when the numerical predictors are zero and the categorical predictor is at its designated reference level. Find baseline log-odds, odds, and probability.
Method. Evaluate the model at the described baseline, then distinguish the log-odds from the odds by exponentiating.
At baseline, the linear predictor is \(\eta=0\), so the log-odds are \(0\). The odds are \(\exp(0)=1\). Therefore \(p_0=1/(1+1)=1/2=0.5000\), to four decimal places.
Check the result in the other direction: \(p_0/(1-p_0)=0.5/0.5=1\), and \(\log(1)=0\). The conversion is consistent. If the model were applied to 80 baseline businesses, the sum of their modeled renewal probabilities would be \(80(0.5)=40\); this is an expected count, not a guarantee that exactly 40 renew.
Interpretation and trap avoided. A zero intercept means baseline odds of one, not odds of zero. It therefore corresponds to probability one-half.
Worked Example: Recovering the Intercept From a Baseline Probability
A hypothetical model predicts whether a submitted equipment claim is accepted, with \(Y=1\) for acceptance. Assume a Bernoulli conditional response and a logistic model. At the baseline predictor settings, the modeled acceptance probability is \(0.08\). Find the baseline odds and the intercept.
Method. First convert probability to odds using \(p_0/(1-p_0)\); then take the natural logarithm because the intercept is the baseline log-odds.
The probability of non-acceptance at baseline is \(1-0.08=0.92\). The odds of acceptance are \(0.08/0.92=8/92=2/23\), approximately \(0.0870\). The intercept is therefore \(\beta_0=\log(2/23)\approx -2.4423\).
Check by converting back: \(\exp(\beta_0)=2/23\), and \((2/23)/(1+2/23)=(2/23)/(25/23)=2/25=0.08\). The recovered probability matches the stated value.
Interpretation and trap avoided. The intercept is negative because the baseline odds are below one. Do not take the logarithm of the probability alone: the logit is the logarithm of \(p_0/(1-p_0)\), not \(\log(p_0)\).
Common Mistakes and Interpretation Limits
A reliable interpretation begins by separating three quantities: the intercept \(\beta_0\), the baseline odds \(\exp(\beta_0)\), and the baseline probability \(\exp(\beta_0)/[1+\exp(\beta_0)]\). They are related, but they are not interchangeable. In particular, probability is not obtained by dividing the log-odds by one plus the log-odds.
The intercept also depends on how predictors are represented. If an age variable is measured in years, zero may describe an implausible person; if it is centered at 40 years, zero corresponds to age 40. The intercept is still the log-odds at predictor values that make the non-intercept terms zero, but the context of that baseline changes with the model specification. Likewise, for categorical predictors, use the specified reference level rather than assuming that a category has a numerical value of zero.
Finally, the baseline probability is conditional on those predictor values. It need not equal the proportion of positive outcomes in the full data set, since that data set may contain many different predictor combinations. A baseline interpretation should therefore name the predictor settings, the outcome coded as \(Y=1\), and the modeled conditional probability. The interpretation is aligned with the SOA Exam SRM (Statistics for Risk Modeling) syllabus.
Check Your Understanding
For each problem, identify the baseline settings described and keep log-odds, odds, and probability distinct.
- A hypothetical model for timely delivery has \(Y=1\) for delivery on time, a Bernoulli conditional response, a logit link, and \(\beta_0=\log(2)\). At baseline, find the log-odds, odds, and probability. (Answer: log-odds \(\log(2)\), odds \(2\), probability \(0.6667\).)
- A logistic model for whether a hypothetical repair request is approved has \(\beta_0=-\log(5)\). Find baseline odds and probability. (Answer: odds \(0.2\), probability \(0.1667\).)
- A fitted baseline model for a hypothetical device return has \(\beta_0=0\). For 120 applications all at baseline, find the modeled probability per application and the expected number of returns. (Answer: probability \(0.5000\), expected count \(60\).)
- A hypothetical logistic model gives baseline acceptance probability \(0.25\). Find the baseline odds and intercept, rounding the intercept to four decimal places. (Answer: odds \(1/3\), intercept \(\log(1/3)\approx-1.0986\).)
- A hypothetical Bernoulli logistic model for claim escalation has baseline intercept \(-2\). Compute baseline odds and probability to four decimal places, and verify by converting the probability back to odds. (Answer: odds \(0.1353\), probability \(0.1192\); \(0.1192/(1-0.1192)\approx0.1353\).)