I am learning LinearRegression (specifically in sklearn; Python's SciKit library) We are making models, fitting them with training datasets, then scoring them against datasets:

model = LinearRegression()
model.fit(X_train, y_train)
score_on_train = model.score(X_train, y_train)
score_on_test = model.score(X_test, y_test)

My class materials materials say:

the model should always perform better on the training set than the testing set. This because the model was trained on the training data and not on the testing data. Intuitively, the model should perform better on data that it has seen before versus data it has not seen.

But this is not true for my datasets; the model doesn't perform better on training data;

the model.score(...) on the training dataset was lower than scoring the test dataset! score_on_train < score_on_test

But I am tempted by this "Intuitively..." explanation.

Is it always true that a model will perform better on its training data than some test data ? Why or why not ? Maybe the text I quoted is trying to describe a different phenomenon.


So far, responses suggest the model should perform better on training data most of the time. But I tried this suggestion: "Try different train/test splits and see if the problem persists." when I run 1000 trials of 1000 make_regression simulated data : the training data scores higher in only ~50% of cases; hardly most of the time.

Am I doing something wrong? How can I avoid "information leaking"?

from sklearn.preprocessing import StandardScaler
from sklearn.model_selection import train_test_split
from sklearn.datasets import make_regression
from sklearn.metrics import r2_score, mean_squared_error
import math

#~100 trials
for i in range(1,1000):

    #In each trial, generate 1000 random observations
    X, y = make_regression(n_features=1, n_samples=1000, noise = 4, random_state=i)
    y=y.reshape(-1, 1) 
    #split observations into training and testing
    X_train, X_test, y_train, y_test = train_test_split(X, y, random_state=i, train_size=0.8)#42)

    #Scale... (am I doing this properly?)
    X_scaler = StandardScaler().fit(X_train)
    y_scaler = StandardScaler().fit(y_train)

    X_train_scaled = X_scaler.transform(X_train)
    X_test_scaled = X_scaler.transform(X_test)
    y_train_scaled = y_scaler.transform(y_train)
    y_test_scaled = y_scaler.transform(y_test)

    mdl = LinearRegression()

    #Train the model to the training data
    mdl.fit(X_train_scaled, y_train_scaled)

    #But score the model on the training data, *and the test data*
        #mdl.score does R-squared coefficient, so this code is equivalent:
        r2_score(y_train_scaled, mdl.predict(X_train_scaled)),
        r2_score(y_test_scaled, mdl.predict(X_test_scaled)),
        #             mdl.score(X_train_scaled, y_train_scaled),
        #             mdl.score(X_test_scaled, y_test_scaled)

        # https://stackoverflow.com/a/18623635/1175496
        math.sqrt(mean_squared_error(y_train_scaled, mdl.predict(X_train_scaled))),
        math.sqrt(mean_squared_error(y_test_scaled, mdl.predict(X_test_scaled)))

train_vs_test_df = pd.DataFrame(results,  columns=('r2__train', 'r2__test', 'rmse__train', 'rmse__test'))

# Count how frequently the winner is the model's score on training data set
train_vs_test_df['r2__winner_is_train'] = train_vs_test_df['r2__train'] > train_vs_test_df['r2__test']
train_vs_test_df['rmse__winner_is_train'] = train_vs_test_df['rmse__train'] > train_vs_test_df['rmse__test']

the first 10 trials testing the model score of train vs model score of test show model score of train are only winners ~ half the time

And when I check how many times the training data scored better:(497, 505)


... training data scores a higher R-squared score in only 497 cases! And the training data scores a higher RMSE-score in only 507 cases! (meaning it's only better in 493 cases). In other words, roughly half! (This is very different than "always" / "almost always" which I am led to believe)

When I change the above parameters, (like changing what amount is used as training data vs amount used as test data... or changing the sample size... or changing the random_state... the test data performs better only about half the time?

  • 3
    $\begingroup$ Your test set can always 'coincidentally' score slightly higher. Large discrepancies can mean either leaking information from the test set (perhaps through scaling?), or your test set being 'easier' than your training set, both of which are problematic. Try different train/test splits and see if the problem persists. $\endgroup$ Jul 16, 2019 at 5:28
  • $\begingroup$ Hi @FransRodenburg I tried your suggestion with different train/test splits, by changing the random_state in 100 different trials... but trial data only scores higher ~half the time; am I doing something wrong? I posted my code.... sorry it's such a code-centric question on Statistics site... $\endgroup$ Jul 16, 2019 at 6:12
  • 1
    $\begingroup$ I'd say your results are very close, which would suggest you're in the desirable middle between under- and overfitting. What does score do? Is that the RMSE or something else? Finally, $49$ is technically less than $\frac{100}{2}$, but is it really less than half? At least not significantly... $\endgroup$ Jul 16, 2019 at 6:22
  • 1
    $\begingroup$ For standard linear regression, you should try and calculate the RMSE: $\sqrt{\sum{y_i - \bar{y}}/n}$. The $\text{R}^2$ is not a measure of predictive performance and can often be misleading. The reason they're so close is (1) you're simulating data and then splitting it, assuring the train and test set come from identical populations and (2) you're using an ordinary regression model. This model has little capacity and will therefore not overfit easily on the train set. $\endgroup$ Jul 16, 2019 at 6:33
  • 1
    $\begingroup$ I should probably turn this into an answer later, but in short: In real data, you will almost always see better performance on the training set, due to small differences in the test and train set, and (hopefully) sufficiently large capacity of your model. $\endgroup$ Jul 16, 2019 at 6:37

2 Answers 2


If your training data is a very good representation of your sample space, then there will be little difference in performance measures between the training and test data. With enough coverage of the sample space, your test data is well-represented in the training set, and looks very much like something the model has "seen before". Numerically, your RMSE estimates on the training and test data look very close, I'd be interested to check if there's any significant difference between them. It's a coin flip whether training or test looks better by RMSE, which indicates that your training data is a very good representation of the test data.

Looking at the model you're fitting, it's not too hard to see why this is the case. You're building a regression model to predict an output using just one single input feature. Even with noise, it's very easy to find a linear model that fits well, especially when given 800 data points to train on. When you go to the test set, there's nothing there that wasn't adequately represented in the training, and the model is simple enough that overfitting isn't really an issue. For this simple case, your training and test data are reasonably equivalent, which is why it's a 50-50 chance of which one performs better.

  • $\begingroup$ Thanks , you're highlighting what other folks have said, I think I understand now: A model that has more capacity, i.e. if I deliberately over-fit, the over-fit model will always train better A train data set that has more differences from test data set (i.e. a training data set which is not very "representative"), and/or a smaller train data set. Can you suggest how I can " check if there's any significant difference between them" (RMSE for train vs RMSE vs test)? Is it a p-value ? Do I run a t-test of the distribution rmse__train vs rmse__test; and get the resulting p-value? $\endgroup$ Jul 16, 2019 at 16:55
  • 1
    $\begingroup$ @TheRedPea Good question - a hypothesis test of difference in RMSE may not be as straightforward as I thought, but it seems that a Diebold-Mariano test might be applicable here (although I am unfamiliar with this test). The second test you suggest would tell if train RMSE is distributed differently than test RMSE over the 1000 iterations, but wouldn't say if they're different for any one particular run. $\endgroup$ Jul 16, 2019 at 18:13

The class materials seem sensible. In general, you expect better performance on the training data. However, this is by no means always true. For example, what if your test data happened to consist of just a single observation, which the model predicted 100% correctly?

As an aside, this illustrates the difference between classrooms and real life. In the classroom, you are always going to get data which was divided randomly into training and test, and so you will almost always see better performance on the training data. But in real life, you might very well get test data which is concentrated around some particular x-value where your model happens to predict well, and see better performance on test than training ( not saying it is common, but I have seen it happen! )

  • $\begingroup$ Hi @Flounderer I tried to create a classroom-style experience: I use make_regression function to create (random?) fake data, and I use test_train_split to split (randomly?) between training and test data sets... yet the training data sets score higher only ~ half the time... I posted my code, I wonder if I'm doing something wrong... , ... sorry it's such a code-centric question on Statistics site... $\endgroup$ Jul 16, 2019 at 6:14
  • $\begingroup$ @Frans Rodenburg has me convinced me that the opposite can also happen: that the "classroom simulation" is in fact the reason that the training data scores better than test data only ~50% of time (Frans also recommends I use RMSE instead of R2) . That in fact, a "real data" situation is more likely to have higher-scoring train vs test. (i.e. "always" or "almost always") $\endgroup$ Jul 16, 2019 at 6:40

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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