7
$\begingroup$

At scikit-learn website they have a very nice picture showing the need to calibrate [some] classifiers to correct bias in predicted probabilities:

enter image description here

And they have a very nice explanation of why one would want to calibrate bagged and boosted trees (they have explanation for other classifiers as well):

Methods such as bagging and random forests that average predictions from a base set of models can have difficulty making predictions near 0 and 1 because variance in the underlying base models will bias predictions that should be near zero or one away from these values. Because predictions are restricted to the interval [0,1], errors caused by variance tend to be one- sided near zero and one. For example, if a model should predict p = 0 for a case, the only way bagging can achieve this is if all bagged trees predict zero. If we add noise to the trees that bagging is averaging over, this noise will cause some trees to predict values larger than 0 for this case, thus moving the average prediction of the bagged ensemble away from 0. We observe this effect most strongly with random forests because the base-level trees trained with random forests have relatively high variance due to feature subsetting

It might be interesting to see the quote from the source they are referring:

Other models such as neural nets and bagged trees do not have these biases and predict well calibrated probabilities

which states bagged trees are "well calibrated".

First of all, I doubt averaging predictions near 0 or 1 works the way as presented above in the first quote. To my knowledge binary RF classifier counts number of 0/1 bins where a datapoint ends up over the overall forest, not one-sided probabilities. So if we have a certain datapont passed through an ensemble of 100 decision trees, and that datapoint ends up in bin #1 99 times, then the prob of 1 is .99. Again, no issue with one-sidedness.

Though, calibrating different classifiers does have a significant, sometimes non-monotonic impact on logloss:

   Samples          Model  LogLoss Before  LogLoss After    Gain
0     1000       Logistic          0.3941         0.3854  1.0226
0    10000       Logistic          0.1340         0.1345  0.9959
0   100000       Logistic          0.1645         0.1645  0.9999
0     1000    Naive Bayes          0.3025         0.2291  1.3206
0    10000    Naive Bayes          0.4094         0.3055  1.3403
0   100000    Naive Bayes          0.4119         0.2594  1.5881
0     1000  Random Forest          0.4438         0.3137  1.4146
0    10000  Random Forest          0.3450         0.2776  1.2427
0   100000  Random Forest          0.3104         0.1642  1.8902
0     1000      Light GBM          0.2993         0.2219  1.3490
0    10000      Light GBM          0.2074         0.2182  0.9507
0   100000      Light GBM          0.2397         0.2534  0.9459
0     1000        Xgboost          0.1870         0.1638  1.1414
0    10000        Xgboost          0.3072         0.2967  1.0351
0   100000        Xgboost          0.1136         0.1186  0.9575
0     1000       Catboost          0.1834         0.1901  0.9649
0    10000       Catboost          0.1251         0.1377  0.9085
0   100000       Catboost          0.1600         0.1727  0.9264

Question

I would be grateful if somebody shares their opinion under what condition a certain classifier will produce [un]biased probability estimates, and how to test if probability predictions are truly unbiased.

Reproducible code for the table and plot:

from sklearn.datasets import make_classification
from sklearn.metrics import log_loss
from sklearn.naive_bayes import GaussianNB
from sklearn.linear_model import LogisticRegression
from sklearn.ensemble import RandomForestClassifier
from sklearn.svm import SVC
from sklearn.calibration import calibration_curve, CalibratedClassifierCV
from xgboost import XGBClassifier
from lightgbm import LGBMClassifier
from catboost import CatBoostClassifier

np.random.seed(42)

# Create classifiers
lrc = LogisticRegression(n_jobs=-1)
gnb = GaussianNB()
svc = SVC(C=1.0, probability=True,)
rfc = RandomForestClassifier(n_estimators=300, max_depth=3,n_jobs=-1)
xgb = XGBClassifier(
    n_estimators=300,
    max_depth=3,
    objective="binary:logistic",
    eval_metric="logloss",
    use_label_encoder=False,
)
lgb = LGBMClassifier(n_estimators=300, objective="binary", max_depth=3)
cat = CatBoostClassifier(n_estimators=300, max_depth=3, objective="Logloss", verbose=0)

df = pd.DataFrame()

plt.figure(figsize=(10, 10))
ax1 = plt.subplot2grid((3, 1), (0, 0), rowspan=2)
ax2 = plt.subplot2grid((3, 1), (2, 0))
ax1.plot([0, 1], [0, 1], "k:", label="Perfectly calibrated")

for clf, name in [
    (lrc, "Logistic"),
    (gnb, "Naive Bayes"),
#     (svc, "Support Vector Classification"),
    (rfc, "Random Forest"),
    (lgb, "Light GBM"),
    (xgb, "Xgboost"),
    (cat, "Catboost"),
]:
    print(name)
    for nsamples in [1000,10000,100000]:
        train_samples = 0.75 
        X, y = make_classification(
            n_samples=nsamples, n_features=20, n_informative=2, n_redundant=2
        )
        i = int(train_samples * nsamples)
        X_train = X[:i]
        X_test  = X[i:]
        y_train = y[:i]
        y_test  = y[i:]
        clf.fit(X_train, y_train)
        prob_pos = clf.predict_proba(X_test)[:, 1]
        fraction_of_positives, mean_predicted_value = calibration_curve(
            y_test, prob_pos, n_bins=10
        )

        if nsamples in [10000]:
            ax1.plot(
                mean_predicted_value,
                fraction_of_positives,
                "s-",
                label="%s" % (name + " nsamples " + str(nsamples),),
            )

            ax2.hist(
                prob_pos,
                bins=10,
                label="%s" % (name + " nsamples " + str(nsamples),),
                histtype="step",
                lw=2,
            )

        preds = clf.predict_proba(X_test)
        ll_before = log_loss(y_test, preds)
        preds = (
            CalibratedClassifierCV(clf, cv=5)
            .fit(X_train, y_train)
            .predict_proba(X_test)
        )
        ll_after = log_loss(y_test, preds)
        df = df.append(pd.DataFrame({
            "Samples": [nsamples],
            "Model": name,
            "LogLoss Before": round(ll_before,4),
            "LogLoss After": round(ll_after,4),
            "Gain": round(ll_before/ll_after,4)
        }))

ax1.set_ylabel("Fraction of positives")
ax1.set_ylim([-0.05, 1.05])
ax1.legend(loc="lower right")
ax1.set_title("Calibration plots  (reliability curve)")

ax2.set_xlabel("Mean predicted value")
ax2.set_ylabel("Count")
ax2.legend(loc="upper center", ncol=2)

plt.tight_layout()
print(df)

Disclaimer. I do understand LogisticRegression is a regression.

$\endgroup$
8
  • $\begingroup$ The first question is answered in: stats.stackexchange.com/questions/390487/… $\endgroup$
    – Tim
    Commented Jan 12, 2021 at 20:06
  • 1
    $\begingroup$ @Tim Thanks for pointing that out. Unfortunately, the accepted answer states [RF] ... will end up with a better-calibrated model in practice than a logistic regression model, which is contrary to what is advised by sklearn, in the literature, and seen in the simulation. A comment by @whuber scikit-learn quotation about "optimizes log loss" isn't an effective explanation, because there is no necessary connection between this and being unbiased is more interesting, and I'd appreciate if somebody can elaborate on that. $\endgroup$ Commented Jan 12, 2021 at 20:26
  • 1
    $\begingroup$ If I may suggest something, it'd be probably better if you could make the question more focused, since you are about five things at once. Questions that are overtly broad are less likely to be answered from my experience, better ask few separate questions than all-in-one. $\endgroup$
    – Tim
    Commented Jan 12, 2021 at 20:29
  • $\begingroup$ @Tim. Thanks, expected. Will edit. $\endgroup$ Commented Jan 12, 2021 at 20:30
  • 1
    $\begingroup$ catboost etc are typically trained to optimise log loss - are you saying that's the wrong objective function?for RF choosing hyperparameters by accuracy is wrong, see arxiv.org/pdf/1812.05792.pdf eg fig 11, where it is shown that using the defaults for RF regression give accurate probability estimates $\endgroup$
    – seanv507
    Commented Jan 12, 2021 at 22:53

1 Answer 1

6
$\begingroup$

There are several possible scenarios when one would think about calibrating probabilities:

  1. The model is misspecified or not optimally trained. That will be the case when non-linear relationships are modeled with a linear learner; or model is too rigid due to excessive regularization (model underfits); or to the contrary, the model is too flexible (overfit or data memorization). Under/over-fit may also be due to having too few/many learning epochs or bagged trees.

  2. A wrong objective function for predicting probabilities was chosen. That will be the case for distorted probabilities predicted by sklearn RandomForest, where they use "gini" or "entropy" for objective function. Classifiers with logloss as objective function are supposed to produce non-biased probability estimates given they have enough data to learn from. Onesidedness of probabilities may explain probability distortions near interval end, but that would not explain distortions in the middle.

  3. Using optimized objective function instead of exact one. This is the case with XGBoost Random Forest implementation:

XGBoost uses 2nd order approximation to the objective function. This can lead to results that differ from a random forest implementation that uses the exact value of the objective function.

In all three scenarios (including having too little data, where it's unclear if calibration results will generalize well when new data arrives), calibration time is better spent on (1) correct model specification (2) choosing right metric (objective function) to optimize (3) collecting more data.

Case 1. Right objective, enough data (Logistic Regression from sklearn)

enter image description here

Case 2. Wrong objective (Random Forest from sklearn) enter image description here

Case 3. Right objective, needs more data (Random Forest from XGBoost) enter image description here

Case 4. Right objective, needs more data

enter image description here

enter image description here

enter image description here

PS

  1. There is nothing wrong with the "wrong" metric as all the classifiers, including sklearn's Random Forest, Naïve Bayes, or SVC, perform very well for certain tasks. The expectation these will behave well for predicting probabilities is wrong and misspecified.
  2. Calibrating well specified and well trained models, even though may show better in-sample result, most probably is fitting to test and will lead to worse generalization to new data.
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.