From Choice Probabilities to Predictions
A multinomial logit model does not merely label a choice as “most likely.” It assigns a probability to every available alternative. Those probabilities can support different prediction tasks: selecting the most likely category, estimating how often a category will be chosen in a population, or assigning a score to the quality of a full probability forecast. These tasks are related, but they do not measure the same thing.
Suppose each observation \(i\) faces the same set of \(J\) mutually exclusive and exhaustive alternatives. The fitted multinomial logit model gives probabilities \(\hat p_{ij}\), where \(j\) indexes alternatives. For the standard multinomial logit specification, these probabilities are obtained from the alternatives’ systematic utilities \(V_{ij}\). For a new observation, the covariates and choice set must be available, and the fitted model must produce a valid probability vector: each probability is between zero and one and the probabilities sum to one.
A predicted category is commonly the alternative with the largest predicted probability. This rule is appropriate when every incorrect classification has the same cost. If mistakes have different costs, the best decision instead depends on those costs; the largest-probability category need not minimize expected loss.
Prediction quality should ordinarily be evaluated on observations not used to fit or select the model, with the same outcome definitions and a choice set appropriate to the prediction task. This is held-out evaluation. Performance measured on the fitting data can look overly favorable because the model was estimated using those outcomes. Held-out results describe performance for the population represented by the evaluation data; they do not guarantee performance under a different population or choice environment.
Category Predictions and Classification Accuracy
Classification accuracy is the proportion of observations for which the predicted category equals the observed choice. It is easy to interpret, but it uses only the largest-probability category and ignores how confident the model was. A forecast assigning probabilities \(0.51,0.49\) and one assigning \(0.99,0.01\) produce the same predicted category, even though they express very different uncertainty.
Here \(y_i\) is the observed category and \(I(\cdot)\) equals one when its argument is true and zero otherwise. A confusion table can further show which observed categories are predicted as which categories. Accuracy alone can be misleading when categories are unbalanced: a model that always predicts a very common alternative may achieve high accuracy while failing to identify less common choices. It is therefore useful to report category-specific performance or a probability-based score as well.
Worked Example: Turning Probabilities Into Choice Predictions
Suppose a hypothetical held-out sample contains three travelers, each choosing among car, transit, and bike. The fitted model gives the following probabilities, and the final column records each traveler’s observed choice.
| Traveler | Car | Transit | Bike | Observed choice | Predicted category |
|---|---|---|---|---|---|
| A | 0.60 | 0.30 | 0.10 | Car | Car |
| B | 0.25 | 0.50 | 0.25 | Bike | Transit |
| C | 0.20 | 0.35 | 0.45 | Bike | Bike |
For traveler A, car has the largest probability, \(0.60\), so car is predicted and the prediction is correct. For B, transit has probability \(0.50\), greater than either competing probability, so transit is predicted; the observed choice was bike, so this prediction is incorrect. For C, bike has the largest probability, \(0.45\), and the prediction is correct.
The accuracy is
Thus, two of these three held-out choices were classified correctly, an accuracy of about \(66.7\%\). This small sample does not establish that the model will have that accuracy in other data. Accuracy also does not show whether the probabilities were well calibrated or whether the model assigned sensible probability to the observed choices.
Calibration: Do the Probabilities Match Frequencies?
A model is calibrated when its predicted probabilities correspond to observed frequencies over repeated predictions in the population being considered. For alternative \(j\), class-specific calibration means that among observations assigned probability \(p\) to that alternative, the long-run proportion choosing it is \(p\). In a finite sample, exact equality is not expected, so calibration is assessed approximately.
A practical table groups observations by predicted probability for a particular alternative. Within each group, compare the mean predicted probability with the observed proportion choosing that alternative. Groups must be defined transparently; very wide groups can hide patterns, while very narrow groups can contain too few observations for stable proportions. Calibration tables are descriptive diagnostics, not proof that probabilities are calibrated.
Worked Example: Reading a Class-Specific Calibration Table
Consider eight hypothetical held-out trips. For each trip, the table records the model’s probability of transit and whether transit was actually chosen. The remaining probability is divided equally between car and bike, so each row still gives a valid three-alternative probability vector. The trips are grouped into low- and high-probability bins.
| Trip | Predicted transit probability | Transit chosen? | Bin |
|---|---|---|---|
| 1 | 0.10 | No | Low |
| 2 | 0.20 | No | Low |
| 3 | 0.20 | Yes | Low |
| 4 | 0.30 | No | Low |
| 5 | 0.60 | Yes | High |
| 6 | 0.70 | Yes | High |
| 7 | 0.80 | No | High |
| 8 | 0.90 | Yes | High |
In the low bin, the mean predicted probability is
One of the four trips chose transit, so the observed proportion is \(1/4=0.25\). In the high bin, the mean predicted probability is
Three of the four trips chose transit, giving an observed proportion of \(3/4=0.75\). The high-bin average matches its observed proportion, while the low-bin proportion is five percentage points higher than its mean forecast. With only four observations per bin, that difference is descriptive and imprecise; it does not establish systematic underprediction.
Proper Scoring Rules for Probability Forecasts
A scoring rule evaluates a probability forecast against the outcome that occurs. Unlike accuracy, a probability score uses the entire forecast vector. A scoring rule is proper if, in expectation, a forecaster minimizes expected loss by reporting the true probabilities. It is strictly proper if those probabilities are the unique minimizer. Proper scores discourage forecasts that distort probabilities merely to improve the score.
Two standard choices for multinomial outcomes are the log loss and the multiclass Brier score. Both are losses, so smaller values indicate better performance on the same set of observations. The log loss penalizes assigning very small probability to the category that occurs. The Brier score measures squared differences between each predicted probability and the corresponding zero-or-one outcome indicator.
For a multiclass outcome, the Brier score ranges from zero to at most two. Zero occurs when every probability vector assigns probability one to the observed category. A perfect log score is also zero, but log loss is unbounded above if a forecast assigns probability approaching zero to the category that occurs. In theory, finite-utility multinomial logit probabilities are positive; numerical rounding or other prediction procedures can nevertheless produce zeros, which require careful handling rather than silently treating the log loss as finite.
For the log-score justification, nonnegativity follows from \(\log x\leq x-1\): applying this inequality to \(x=p_j/q_j\) for categories with \(q_j>0\) gives \(\sum_j q_j\log(p_j/q_j)\leq\sum_j(p_j-q_j)\leq 0\). Equivalently, \(\sum_jq_j\log(q_j/p_j)\geq 0\); equality requires the forecast to match the true distribution. The Brier derivation shows the same principle directly through squared distance from the true probability vector.
Worked Example: Calculating Log Loss and the Brier Score
For three hypothetical held-out choices among A, B, and C, suppose the predictions and observed outcomes are:
| Observation | Predicted probabilities (A, B, C) | Observed choice |
|---|---|---|
| 1 | (0.70, 0.20, 0.10) | A |
| 2 | (0.20, 0.50, 0.30) | B |
| 3 | (0.10, 0.30, 0.60) | C |
The probabilities assigned to the observed categories are \(0.70\), \(0.50\), and \(0.60\). The mean log loss is therefore
For observation 1, the Brier loss is \((0.70-1)^2+(0.20-0)^2+(0.10-0)^2=0.14\). For observation 2 it is \((0.20-0)^2+(0.50-1)^2+(0.30-0)^2=0.38\). For observation 3 it is \((0.10-0)^2+(0.30-0)^2+(0.60-1)^2=0.26\). Thus the average Brier score is
These are summaries for this particular three-observation sample, not universal benchmarks. Comparing models with these scores is most informative when they are evaluated on the same observations and alternatives. The numerical scales of log loss and Brier score differ, so their values should not be compared directly with each other.
Prediction Is Not Structural Interpretation
Prediction evaluation asks how well a model forecasts outcomes for observations from a target setting. Structural interpretation asks what the model’s parameters and choice assumptions imply about behavior and substitution. A model can predict well in a particular evaluation sample without its structural assumptions being correct. Conversely, a theoretically motivated model can be useful for interpretation yet forecast imperfectly. Accuracy, calibration, and proper scores do not by themselves establish a causal effect, validate the model’s behavioral assumptions, or show that predicted probabilities will transport to a different choice environment.
The decision rule also matters. If a decision incurs zero loss for a correct category and equal loss for every incorrect one, choosing the largest probability minimizes expected classification error. With a general cost \(L(a,j)\) for taking action \(a\) when category \(j\) occurs, the appropriate action minimizes expected cost:
This distinction is important in applications where, for example, incorrectly predicting one alternative is more consequential than confusing two others. The model supplies probabilities; the decision-maker’s loss function determines how those probabilities should be converted into actions.
Common Mistakes and Practical Reporting
- Treating accuracy as a probability assessment. Accuracy records whether the most likely category was correct; it does not assess the quality of the full probability vector. Report a proper score or calibration assessment when probability quality matters.
- Calling a small calibration table proof of calibration. Grouped proportions fluctuate, and bin choices affect the display. Treat the table as a diagnostic summary, with group sizes and predicted means visible.
- Choosing the largest probability regardless of the decision. That rule is optimal only under equal misclassification costs. For unequal costs, minimize expected loss using the stated cost function.
- Interpreting predictive performance as structural validation. Good held-out predictions do not prove that behavioral assumptions or coefficient interpretations are correct.
- Evaluating on the fitting sample without qualification. In-sample performance can be optimistic. State how observations were held out and which population or setting the evaluation is intended to represent.
Check Your Understanding
Answer these questions using the definitions and calculations in this tutorial.
- A model predicts probabilities \(0.45\), \(0.40\), and \(0.15\) for three alternatives. What category does the maximum-probability rule predict, and what information is discarded by reporting only that category?
- In a calibration bin of 50 observations, the mean predicted probability for bike is \(0.30\), and 18 observations chose bike. Calculate the observed proportion and describe the difference from the mean forecast.
- Why can a model have high classification accuracy but poor probability forecasts?
- Explain why the log loss becomes large when a model assigns very low probability to the category that occurs.
- When would minimizing expected decision cost lead to a different action from choosing the alternative with the largest predicted probability?