ML Interview Q Series: L2 Regularization's Impact on Selecting the Optimal Logistic Regression Hyperplane
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22. We have 100 data points: 50 are POSITIVE (in quadrant Q1) and 50 are NEGATIVE (in quadrant Q3), hence 2D points. Imagine you are using Logistic Regression with regularization and obtain these hyperplanes: 0.1x + 0.1y = 0, 100x + 100y = 0, 3x + 3y = 0. Which hyperplane would be best, and why?
The decision boundary in all three cases is effectively the same line (because 0.1x + 0.1y = 0, 3x + 3y = 0, and 100x + 100y = 0 all simplify to x + y = 0, or y = -x, if we ignore any potential intercept term). Quadrant Q1 (positive quadrant where x > 0, y > 0) is separated from Quadrant Q3 (negative quadrant where x < 0, y < 0) by that line, so geometrically all three hyperplanes lead to the same decision boundary.
The question then is which of these parameter vectors would be favored by logistic regression with a regularization term (most typically, L2 regularization). L2 regularization penalizes large coefficient magnitudes by adding the squared norm of the parameter vector to the cost function. In other words, if w is our weight vector, L2 regularization adds something proportional to the sum of squares of w’s components to the loss. Hence, the model is encouraged to keep weights small in magnitude, provided that smaller weights can still classify the data effectively.
In the three lines given:
0.1x + 0.1y = 0 corresponds to the parameter vector w = (0.1, 0.1)
3x + 3y = 0 corresponds to w = (3, 3)
100x + 100y = 0 corresponds to w = (100, 100)
Although each of these parameter vectors produces the same geometric boundary, their L2 norms differ. Specifically, for w = (0.1, 0.1), the Euclidean norm is smaller compared to the norms for (3, 3) or (100, 100). Since logistic regression with L2 regularization favors the solution with the smallest norm (assuming all three solutions classify the data equally well), the hyperplane 0.1x + 0.1y = 0 is best. It achieves the same decision boundary while incurring the least penalty from the regularizer. That is exactly why, in practice, you would see the parameter magnitudes settle around smaller values when the data is perfectly separable by these directions.
The reason it is “best” is that this solution optimizes both the data fitting term (log-likelihood) and the regularization term (which penalizes large coefficients). Since all three hyperplanes yield the same classification boundary, the difference in the objective function value across these solutions comes primarily from the regularization penalty, and the one with the smallest norm (the 0.1x + 0.1y = 0 line) yields the lowest overall cost.
Logistic Regression’s learning algorithm (often by gradient descent or a variant such as LBFGS, or coordinate-based methods in some implementations) will converge to the parameter vector that minimizes the total loss (data fitting + regularization). Because scaling up the weights (for instance, using (100, 100) instead of (0.1, 0.1)) grows the regularization penalty significantly, it is not favored unless the data strongly demands large coefficients to fit more difficult boundaries—which is not the situation here. In this particular case, all three boundaries perfectly separate quadrant Q1 from quadrant Q3, so the model with the smallest norm wins out under regularization.
If the boundary is the same, why do different magnitudes matter?
Different magnitudes of the weights alter the penalty in the cost function. L2 regularization specifically imposes a cost proportional to the squared magnitude of the weight vector. Even if multiple weight vectors define the same line in the 2D plane, the logistic regression optimization procedure will pick the one with the smallest L2 norm because it minimizes the total loss. The boundary in pure geometric terms does not change among these three lines, but the cost function is not just about geometry; it also includes that penalty term to control overfitting and keep the model’s parameters in smaller ranges.
In standard logistic regression, there is also typically an intercept term. If the intercept is zero in all three hyperplanes, then we are purely looking at the weight vector’s magnitude. One can incorporate a nonzero intercept if the data demands a shift of the boundary, but here the question specifically gave lines that pass through the origin. That’s why the intercept is effectively zero in all three cases. With L2 regularization, large absolute values in any component of the model’s parameter vector are penalized; so if w is scaled up by a large factor, the cost from the regularization term significantly increases.
The scaling property that the same geometry can be described by w or by a constant multiple of w does not imply logistic regression is scale-invariant. The logistic function for a data point x is σ(w⋅x+b), and if you multiply w by a large constant without adjusting b, you alter the predicted probabilities. The only reason all three are mentioned here as effectively the “same boundary” is because they share the same ratio of coefficients and we’re ignoring intercept. When searching for an optimal solution, the combination of the data-likelihood term and the L2 penalty leads the algorithm to pick smaller coefficients that do not degrade classification accuracy.
This captures the conceptual reason why 0.1x + 0.1y = 0 is preferred under L2 regularization. It yields the same classification boundary but with the minimum penalty, and that is why it is the best among those three lines in the context of a regularized logistic regression model.
How does the regularization term look mathematically?
Below is a short conceptual form of the L2 regularized logistic regression cost (omitting intercept for illustrative purposes):
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Could a larger coefficient magnitude ever be beneficial?
In some cases, if the data is not so cleanly separable or if certain outliers heavily influence the loss term, the optimization might find that slightly larger magnitudes in some directions help reduce the misclassification or log-loss. However, here the boundary is straightforward: Q1 vs. Q3 with a line y = -x. If that line perfectly classifies the data, increasing the magnitude of the weights does not give any additional benefit on the data side of the cost function, yet it increases the penalty. Consequently, there is no reason for the algorithm to settle on the large-magnitude solutions.
An additional subtlety arises if we consider a nonzero intercept. In practice, we would let the algorithm learn that intercept, and often the final solution might be x + y + b = 0 for some b, adjusting slightly to maximize the margin or minimize misclassifications. But the key insight is that among solutions that achieve the same classification boundary, the L2 penalty heavily favors smaller coefficients.
Are these boundaries guaranteed to pass exactly through the origin?
In the question, all provided hyperplanes are written with no intercept. In actual logistic regression solutions, the optimal boundary might not necessarily pass through the origin. For quadrant Q1 vs. Q3 classification, the boundary that best separates the data might indeed pass near the origin, but typically a logistic regression model will include an intercept term to shift the decision line optimally. Still, the question’s focus is on the magnitude of the weights, and all three given weight vectors produce the same slope, so we can treat them as the same boundary for discussion’s sake.
What if there was an L1 penalty instead of L2?
In L1-regularized logistic regression, the penalty is proportional to the absolute values of the coefficients rather than their squares. Even so, scaling up the same boundary direction from 0.1x + 0.1y = 0 to 100x + 100y = 0 would still produce a larger penalty in the L1 case, because the sum of absolute values would be larger. Hence the principle would be the same in that smaller magnitude coefficients for the same boundary are preferred. The difference is that L1 might set some coefficients to zero for sparseness, but in a 2D scenario with such a simple boundary, that distinction is less relevant here. The key idea remains that the regularization term penalizes large weights, so smaller is favored if the data fit is unchanged.
Potential edge cases or pitfalls
One potential pitfall is if we do not realize that the logistic regression objective combines both the fit (which depends on predictions for each data point) and the regularization cost (which depends on coefficient magnitudes). Someone might argue that (100, 100) or (3, 3) “is the same line,” but in reality, scaling up the weight vector changes the predictions in the logistic function unless an intercept is simultaneously scaled in a way to preserve the probabilities. The net effect on the logistic regression loss function can become quite different. Usually, to preserve exactly the same predictions, you would have to re-scale the intercept as well. If the intercept is zero in all three cases, logistic regression definitely treats the solution with smaller coefficients as better because it directly lowers the penalty.
Another subtlety is that in some frameworks or libraries, optimization methods might converge to a specific scale for the weights due to their initialization, convergence thresholds, or numeric stability. However, conceptually in L2-regularized logistic regression, if multiple sets of coefficients yield the same classification boundary, the set with the smallest L2 norm is always preferred by the cost function.
In summary
For L2-regularized logistic regression, among the three given hyperplanes 0.1x + 0.1y = 0, 3x + 3y = 0, and 100x + 100y = 0, they are all essentially the same boundary in the plane. But because L2 regularization penalizes large weights, the parameter vector (0.1, 0.1) has the smallest norm and thus leads to the lowest regularized cost. Therefore, it is the best solution among those three.
Follow-up Questions
Could all three lines produce the same predicted probabilities for logistic regression?
Strictly speaking, if each line has zero intercept, scaling up the weights changes the logits w⋅x. For a given point x, 0.1x + 0.1y is not the same numeric value as 3x + 3y, so the predicted probabilities would differ unless the intercept is adjusted accordingly. However, if the classification problem is very clean (Q1 vs. Q3 points that are well separated), then all points might be so far on the correct side of the boundary that their logistic outputs all saturate to nearly 1 or nearly 0 in exactly the same way. In such a scenario, even though the numeric logit changes, the classification decisions (above 0.5 or below 0.5) do not. Hence for classification decisions alone, it might appear the same. But from the point of view of the optimization cost, the model with smaller magnitude weights is more optimal because it penalizes the large magnitude vectors more heavily.
What is the mathematical reason that 0.1x + 0.1y = 0 has a smaller penalty?
Why is there no mention of an intercept in these hyperplanes?
The question directly gave lines of the form “some_constant * x + some_constant * y = 0.” In many real-world logistic regression scenarios, the model would look like w · x + b = 0. Possibly, the question is illustrating a simplified case with the data arranged such that the origin is the perfect place to put the decision boundary. Or it might be ignoring the intercept for simplicity. If the data truly straddles around the origin in Q1 vs. Q3, a boundary through the origin can be good enough. But in general, we allow a bias (intercept) term to shift the line up/down/left/right for a better fit.
What if the data is not perfectly separable?
If the data is not perfectly separable, logistic regression tries to find the set of coefficients that best separates positive and negative points in a probabilistic sense while also minimizing the penalty from large coefficients. The boundary will still be influenced by the geometry of the data, and the final solution might not pass through the origin, but the principle remains: L2 regularization favors smaller norms, so if multiple boundaries achieve similar predictive performance, the boundary with the smallest norm is chosen.
Could we confirm this logic by running an example in Python?
Yes, we can. For instance, you could generate 2D points in Q1 and Q3, label them, and run a logistic regression with regularization in scikit-learn or PyTorch. In scikit-learn, for instance:
import numpy as np
from sklearn.linear_model import LogisticRegression
# Generate data in Q1
n = 50
np.random.seed(0)
X_q1 = np.random.uniform(low=1.0, high=3.0, size=(n, 2))
y_q1 = np.ones(n)
# Generate data in Q3
X_q3 = np.random.uniform(low=-3.0, high=-1.0, size=(n, 2))
y_q3 = np.zeros(n)
X = np.vstack([X_q1, X_q3])
y = np.concatenate([y_q1, y_q3])
model = LogisticRegression(C=1.0, penalty='l2') # C is 1/lambda
model.fit(X, y)
print("Coefficients:", model.coef_)
print("Intercept:", model.intercept_)
You would typically see that the learned coefficients have moderate magnitudes (not extremely large), and the intercept is non-zero unless the data is perfectly symmetrical around the origin. Indeed, the solution that emerges respects the L2 regularization principle, generally preferring smaller coefficient magnitudes to reduce the overall cost.
Is there a trick question aspect that scaling the weights might make no difference?
One might imagine that because all lines x + y = 0, 3x + 3y = 0, 100x + 100y = 0 are the same geometrically, it is irrelevant. However, in the specific cost function for logistic regression with L2 penalty, it does matter. It might not matter for a purely geometric unregularized perceptron if only sign matters, but in logistic regression, the scale of the weight vector has direct implications for the penalty (and for the predicted probabilities if the intercept is not scaled accordingly). So it is not a trick question; it is simply highlighting that L2 regularization explicitly drives down the norm of the weight vector.
Conclusion
Among the three specified hyperplanes (0.1x + 0.1y = 0, 3x + 3y = 0, and 100x + 100y = 0), the best under L2-regularized logistic regression is 0.1x + 0.1y = 0, solely because it incurs the smallest L2 penalty while effectively creating the same geometric boundary for separating quadrant Q1 from quadrant Q3. This is a direct consequence of how the regularization term penalizes large coefficient magnitudes, making the minimal-norm solution optimal in cases where multiple configurations yield the same classification boundary.
Below are additional follow-up questions
How do extremely large or extremely small coefficients affect numerical stability during training, and how might this influence the final choice of hyperplane?
When training logistic regression with gradient-based optimizers (such as gradient descent, LBFGS, or others), extremely large or small coefficients can introduce numerical instability:
• Large coefficients can lead to large logits (the value of w⋅x), and for negative classes, the term in the exponential exp(−yi(w⋅xi+b)) could cause numerical overflow or underflow, depending on sign. Extremely large positive logits might push floating-point values into “infinity,” resulting in NaNs or Inf in the computation.
• Very small coefficients combined with large input feature values can have the opposite effect, making the computed probabilities appear saturated or cause the gradient to vanish in certain edge cases.
When large weights are not needed to perfectly separate the data, L2 regularization naturally keeps weight magnitudes moderate or small, helping avoid these stability issues. In practice, frameworks like scikit-learn or PyTorch implement safe computations (e.g., log-sum-exp or special functions) to reduce overflow risk. Nonetheless, if the same line can be expressed by scaled versions of the weight vector, the solution with smaller magnitude is numerically safer and less prone to causing very large intermediate values.
A subtlety arises if the input features themselves are not scaled or normalized. If some features have large magnitudes, the corresponding learned coefficients might stay proportionally smaller to balance out the product w⋅x. Conversely, for a feature that is scaled to very small numerical values, the learned coefficient might become very large to ensure it has enough impact on the logit. Regularization counters these extremes by penalizing large coefficients; thus, for the same decision boundary, the training procedure will typically land on a stable, lower-norm solution.
A pitfall to watch out for is ignoring numeric warnings or overflow messages that might appear during training. Beginners sometimes do not realize that extremely large or small magnitudes in the parameter space could be causing the training to diverge or produce inaccurate results. One real-world issue is if your data has some huge outliers, which might heavily influence the gradient steps, pushing the weight vector to very large values before the regularization penalty can counterbalance it.
Therefore, from both a cost function perspective and a numerical stability perspective, the solution with smaller norm is strongly preferred—further supporting why the hyperplane with (0.1, 0.1) is a better logistic regression solution than (3, 3) or (100, 100) when all else is equal.
Does logistic regression always find a unique solution for linearly separable data, or could multiple solutions exist with the same cost?
In purely geometric terms, any scalar multiple of a given weight vector that yields the same sign for all training points can represent the same boundary. However, logistic regression’s cost function includes the predicted probabilities, and these probabilities do change if you scale the weight vector without adjusting the intercept. Therefore, from a strict optimization standpoint, the logistic regression solution in linearly separable data might still be unique once you factor in the intercept and the regularization term.
Under L2 regularization, there is generally a single global minimum. The model finds one unique w, b that balances the data fit and the regularization penalty. If we tried to scale w by a constant factor, we would have to adjust the intercept in a very specific manner to yield the same predicted probabilities on all training points. The L2 penalty grows quadratically with the magnitude of w, so that scaling approach is typically suboptimal unless the intercept is carefully tuned. In practice, even that rarely yields an exact equivalence because for it to be perfect, each data point’s logit w⋅x + b must remain unchanged. That means if we multiply w by α, we need to shift b by some offset that keeps w⋅x + b the same for all x. Because x takes on multiple values, a single shift in b typically cannot perfectly preserve w⋅x + b for all points if w is scaled by α ≠ 1.
For linearly separable data, if there were no regularization, logistic regression can theoretically push the norm of the weight vector to become arbitrarily large (maximizing the margin in a sense akin to perceptron solutions). But with regularization, you will converge to a finite-norm global optimum. Hence, in standard L2-regularized logistic regression, you typically get one unique minimum up to numerical tolerances.
A subtlety: If you have degenerate cases (e.g., perfectly overlapping data points or exact collinearity), you could have infinitely many solutions that achieve the same cost. Even in that scenario, L2 regularization typically breaks ties in favor of the smallest Euclidean norm solution. So effectively, the solution is unique once the penalty is included. Practitioners might see slight differences depending on the solver or random initialization, but in theory, the solution is effectively unique in a well-conditioned problem.
How might cross-validation be used to choose the regularization strength, and how does it affect which hyperplane is chosen?
Cross-validation (CV) is a method of splitting the training data into multiple folds, training on some folds while validating on the remaining fold, and repeating this process to assess generalization. In logistic regression, the regularization strength (often controlled by a parameter like λ or equivalently C = 1/λ) is a hyperparameter that strongly influences the magnitude of the coefficients. A large λ (or small C) indicates strong regularization, pushing weights to be smaller. A small λ (large C) means weaker regularization, allowing weights to grow larger if it helps reduce classification error.
By running logistic regression for various candidate values of λ (or C) and measuring average validation accuracy (or another suitable metric) across folds, one selects the value that best generalizes to unseen data. A stronger regularizer might lead to smaller weight norms, potentially smoothing out overfitting. A weaker regularizer can let the model overfit if it tries to perfectly classify the training set with very large coefficients.
If you have data that is linearly separable—like Q1 vs. Q3 with no overlap—very small λ (weak regularization) might push the weight vector to extreme magnitudes in an attempt to classify with near 100% confidence for each point. This can cause unstable numerical behavior or degrade generalization if any slight perturbation occurs. On the other hand, moderate or strong regularization can keep the weights more balanced, leading to a simpler decision boundary that still classifies well but remains numerically stable and robust. Through cross-validation, you can empirically see which setting yields the best performance on held-out folds.
A pitfall is forgetting to tune the regularization parameter. Some practitioners might stick to a default value that is either too large or too small for their data, resulting in suboptimal classification performance. Another real-world issue: if the data is extremely clean (like artificially separated quadrants) cross-validation might not reveal much difference in accuracy across various λ because they all achieve 100% accuracy. In that scenario, smaller-norm solutions might still be preferred for interpretability and stability. Tools such as scikit-learn automatically manage an inverse-regularization parameter C, so it’s important to confirm which convention the library uses (C or λ) and keep track accordingly.
What if the positive data (Q1) were slightly shifted, or some noise was introduced? Could the boundary still be y = -x?
If the positive data in quadrant Q1 is shifted or has some noise, or the negative data in quadrant Q3 is scattered around, the ideal boundary might no longer be perfectly y = -x. Logistic regression will then adjust w and b to best fit the “majority trend” while minimizing misclassification. The resulting line could be close to y = -x, but might shift to capture the distribution of the points more accurately.
In real-world scenarios, data is rarely perfectly separable. You might have some “positive” points appear in the region typically associated with negative points (outliers), or vice versa. With logistic regression, you do not simply draw a line halfway between Q1 and Q3. Instead, the training algorithm tries to find w, b that minimize the total log-loss plus the regularization penalty. If the data is mostly in Q1 vs. Q3 but with slight noise, the learned boundary might rotate a bit and shift away from passing directly through the origin.
An edge case arises when the data is heavily noisy or has outliers. If some of those outliers are extremely far from the cluster, they can exert strong influence on the solution, potentially pulling the boundary in ways that degrade overall accuracy or stability. This is particularly relevant for logistic regression because it is not as robust to outliers in feature space as some robust methods or SVM with certain loss modifications. Nevertheless, L2 regularization provides a stabilizing effect, preventing the weight vector from blowing up to misclassify outliers with near-perfect confidence.
Could the presence of correlated features impact the choice of the hyperplane and the magnitude of coefficients?
Yes, correlated features can lead to inflated or unstable estimates of individual coefficients if you do not have regularization. Logistic regression without regularization might produce very large (positive or negative) weights in attempts to “explain away” the correlation among features. This can create a scenario where multiple, nearly equivalent solutions exist with wildly different weight vectors.
Under L2 regularization, the presence of correlated features typically leads the solution to distribute the weights more evenly across those correlated features, instead of letting a single coefficient become extremely large. This is because the penalty for large coefficients can be significantly higher than the penalty for distributing the contribution across multiple features. When features are strongly correlated, the boundary might still effectively be the same line, but the model might express it in different ways through combinations of correlated features. L2 regularization typically encourages a more balanced distribution of weights to reduce the overall squared norm.
A subtle pitfall arises when interpreting the magnitude of coefficients in the presence of correlated features. A large coefficient might not necessarily mean that feature is overwhelmingly influential if it is part of a correlated group. Real-world examples include text classification with bag-of-words models, where synonyms or overlapping phrases can cause correlation. Or in numerical data, certain features might be correlated (e.g., height in inches vs. height in centimeters). Regularization helps keep these effects from ballooning individual weights.
In the simple 2D quadrant example, it’s presumably just x and y, so correlation is not a major factor. But if x and y were scaled or combined in certain transformations, the correlation could affect the final learned hyperplane or how the weight vector is represented.
How do feature standardization or normalization practices interact with the preference for smaller weights?
When you standardize or normalize features, each feature is transformed to have zero mean and unit variance, or at least scaled to a defined range (e.g., [0,1]). In logistic regression, scaling features can drastically impact the magnitude of the learned coefficients and reduce the risk of some coefficients becoming very large just to compensate for large input feature values. However, after standardization, a hyperplane w⋅x = 0 in the scaled feature space might not directly correspond to the same line in the original unscaled space, though the model effectively uses the standardized space for training.
Regularization is often more stable and effective when features are scaled. Without scaling, if one feature is in thousands-of-units scale, it might push the optimizer to produce extremely small or large coefficient magnitudes to manage that difference. This can also lead to slow or inconsistent convergence in gradient-based methods. After standardizing, logistic regression tends to converge faster, and the interplay between the log-loss and the L2 penalty becomes more direct and balanced. Then, among any solutions that produce the same classification performance, the model that yields a smaller norm in the standardized space is favored.
A potential pitfall is that if you forget to apply the same scaling transformation to your test set (or production data), you will get incorrect results. Also, you must keep in mind that standardization changes the interpretability of the coefficients. A weight of 2.0 in the standardized space means that a one-standard-deviation change in the feature modifies the log-odds by 2.0. In the original scale, that relationship is different. Nonetheless, from the perspective of L2 regularization, consistent feature scaling helps ensure a stable and well-defined preference for smaller weights.
Is there a scenario where weaker regularization might still be preferred, even if it implies larger coefficient magnitudes?
Yes. If the data is not perfectly separable or has a certain pattern where a steeper logit slope in one dimension can yield substantially better performance on the training set (and presumably on the validation set), then the solution might favor somewhat larger weights if they reduce classification errors for borderline points.
In other words, the best generalization might be achieved at some moderate or lower level of regularization, because the logistic model can “sharpen” the decision boundary around critical regions of the feature space. Cross-validation typically reveals whether the improvement in data fitting outweighs the increased penalty from larger weights.
A real-world example: Suppose certain borderline points are crucial to classify correctly (e.g., in medical diagnosis, false negatives are extremely costly). The model might accept a higher penalty to better separate those borderline points by shifting or rotating the decision boundary. If that results in improved recall for the positive class without excessively harming overall performance, it might be favored even though it means a bigger coefficient vector.
A subtlety is that your metric of interest (accuracy, F1-score, AUC, etc.) might weigh certain types of errors differently. If the standard logistic regression objective (log-loss plus L2 penalty) does not align well with your real-world cost of errors, you might still prefer a hyperplane that leads to better domain-specific outcomes. In that case, you could shift the regularization parameter or the decision threshold to meet your needs, acknowledging that it might come at the cost of bigger weights.
How might we verify experimentally that the small-norm solution generalizes better or is more stable over new data?
One common technique is to run repeated train/test splits or use k-fold cross-validation. For each split, fit the model with a range of regularization strengths. Observe:
• The magnitude of the coefficients (the L2 norm). • The classification accuracy (or another performance metric) on the validation set. • Potential stability metrics, such as how sensitive the boundary is to small random perturbations in the training data or how stable the predictions are across folds.
Often, you will see that as regularization is relaxed (e.g., smaller λ, bigger C), the model’s training performance might improve or remain perfect if data is easily separable. But the validation or test performance might plateau or even slightly worsen, especially if the model is overfitting borderline points. Meanwhile, the coefficient norm tends to grow dramatically. This is a practical demonstration that large coefficient values are not necessarily beneficial once you factor in real-world data noise or distribution shifts.
For instance, you could do something like:
import numpy as np
from sklearn.linear_model import LogisticRegression
from sklearn.model_selection import train_test_split
# Synthetic data
np.random.seed(42)
X_pos = np.random.randn(50, 2) + [2, 2] # Q1 cluster
X_neg = np.random.randn(50, 2) - [2, 2] # Q3 cluster
X = np.vstack([X_pos, X_neg])
y = np.array([1]*50 + [0]*50)
X_train, X_test, y_train, y_test = train_test_split(X, y, test_size=0.2)
for C_val in [0.01, 0.1, 1, 10, 100]:
model = LogisticRegression(C=C_val, penalty='l2', solver='lbfgs')
model.fit(X_train, y_train)
train_acc = model.score(X_train, y_train)
test_acc = model.score(X_test, y_test)
weight_norm = np.linalg.norm(model.coef_)
print(f"C={C_val}, Weight Norm={weight_norm:.4f}, "
f"Train Acc={train_acc:.2f}, Test Acc={test_acc:.2f}")
You might see that for extreme values of C (very weak or very strong regularization), the performance can degrade or remain roughly the same while the weight norm changes drastically. This gives empirical evidence of how smaller-norm solutions can still work just as well (or better) for classification without risking extreme coefficient magnitudes that might reduce stability or interpretability.
If we extended to higher dimensions (e.g., 100 features instead of 2D), does the same principle of preferring smaller norm solutions still hold?
Yes, absolutely. In higher-dimensional spaces, logistic regression with L2 regularization behaves similarly: it discourages large coefficients unless they are strictly necessary to improve predictive performance. Indeed, the curse of dimensionality can make classification more challenging, and overfitting is a big risk when the dimension is high relative to the number of data points.
The same principle that the solution with the smallest L2 norm is favored (among solutions that achieve similar data fit) is even more critical in high dimensions, where many weight components might be driven to near zero. However, in extremely high dimensions, L1 regularization (Lasso) is also popular because it can drive many coefficients exactly to zero, providing a level of feature selection. Yet even with L2, the fundamental preference for smaller weight magnitude remains.
A pitfall arises if one does not carefully tune the regularization in high-dimensional settings. If the regularization is too strong, it might push nearly all weights to be very close to zero, underfitting the data. If it is too weak, the model might latch onto spurious correlations and produce large weights for unimportant features. Balancing that with cross-validation is essential to achieve a stable and meaningful solution in high-dimensional contexts.
Do other models, such as a linear SVM, exhibit a similar preference for smaller coefficient magnitudes?
In soft-margin SVM (where data can overlap or be misclassified with some penalty), there is a regularization parameter that also controls the trade-off between margin size and misclassification penalty. The final SVM boundary similarly tries to keep ∥w∥ small unless forced by the data to grow to reduce misclassifications. So from a high-level perspective, linear SVM and regularized logistic regression both penalize large coefficient magnitudes, though the details of how the penalty interacts with the loss function differ.
An edge case arises if you compare logistic regression with a certain regularization parameter to SVM with a certain cost parameter. The solutions might differ in subtle ways in how they treat borderline points or outliers. But conceptually, both prefer smaller norms if it achieves good classification performance. In the scenario of Q1 vs. Q3 data, both might yield a boundary close to y = -x with moderate weight magnitudes. If the data is extremely clean, either approach can classify it perfectly, but the exact boundary might shift slightly. And in practice, logistic regression outputs probabilities, while SVM output is typically a margin-based decision score (though you can calibrate SVM scores into probabilities with additional steps).
What if the labels were heavily imbalanced? For example, 90 points in Q1 and only 10 in Q3?
Class imbalance can affect logistic regression in multiple ways. If you have 90 positive examples and 10 negative examples, the standard logistic loss function without weighting might place more emphasis on not misclassifying the majority class. As a result, the boundary might shift in a way that inadvertently increases false negatives.
One approach is to use class weights, effectively penalizing mistakes on the minority class more heavily. Many frameworks allow you to specify a “class_weight” parameter or to re-scale the training data so that the cost of misclassifying a minority example is increased. This modifies the objective function to weigh errors from each class differently. In that adjusted objective, the magnitude of the coefficients might be larger or smaller depending on how severely the model tries to push the boundary away from the minority class region. Even so, L2 regularization still penalizes the squared norm of the weights, so among solutions that satisfy the cost-sensitive constraints, the one with smaller norm tends to be preferred.
A pitfall is ignoring the imbalance entirely. The model might focus too strongly on Q1 data (the majority class) and get a trivial 90% accuracy by classifying everything as Q1, which is useless for practical purposes if you truly care about identifying Q3 points. Another potential subtlety is that if the minority examples are widely spread or not as well-defined in that quadrant, the model might struggle to place a boundary that achieves good recall for Q3. The solution might require bigger magnitude weights or a larger offset in the intercept to capture that minority region. Tuning the class weight or using a balanced approach in cross-validation can guide the model toward a more effective boundary for the minority class.
Could a small shift in the intercept dramatically change the classification outcome near the origin in borderline cases?
Yes. In logistic regression, the intercept (often denoted b) can shift the decision boundary up or down. For a 2D model where the boundary is w₁ x + w₂ y + b = 0, changing b by a small amount can translate the line in the plane. If the data points near the boundary are borderline, even a small shift might flip some from positive to negative or vice versa.
In the Q1 vs. Q3 scenario, if the data is close to perfectly separable by y = -x, but a few points lie near that boundary or slightly deviate, logistic regression might learn a small intercept to push the boundary in a way that better accounts for those borderline samples. This shift can slightly reduce misclassification or log-loss for those borderline points, even if the slope remains near -1.
In real-world data, intercept shifts can be significant if the average location of positive vs. negative points is not symmetrically centered around the origin. Regularization also applies to the intercept in many frameworks, albeit sometimes with a weaker effect or sometimes not at all (depending on the library). If the intercept is included in the penalty, then the model tries to keep both w and b small. If the intercept is excluded from regularization, b might be free to shift if that improves classification. Checking the documentation of your logistic regression implementation is important to see how the intercept is treated.
A potential pitfall is forgetting to center or scale the data. If the features are not centered, the intercept might need to be quite large or quite negative to account for an offset in the average location of data points. This can be suboptimal if the user does not realize that a simpler or more stable solution might be found by centering the data prior to logistic regression.
Could adding polynomial or interaction terms change the scaling argument?
Yes, adding polynomial or interaction terms changes how the model can shape the decision boundary:
• In plain 2D with features x and y, if you add polynomial terms like x², xy, y², etc., the decision boundary is no longer a simple line. It becomes a curve in the original x-y space. But in the transformed feature space, logistic regression is still effectively finding a linear boundary (with respect to the polynomial features).
• Because you now have higher-dimensional parameter vectors (w might include coefficients for x, y, x², xy, y², etc.), the L2 regularization penalizes the norm across all these coefficients. Even if the final boundary is still reminiscent of y = -x, the solution might achieve better classification by combining multiple polynomial features in certain data distributions.
• The scaling argument persists because each coefficient in the expanded feature space is still penalized. If multiple sets of coefficients produce the same final classification, the set with smaller overall norm in that expanded space is preferred. The presence of polynomial or interaction terms might let the model fit minor variations or curves in the data, but the principle of preferring smaller magnitudes remains.
A potential real-world pitfall is that as you increase polynomial degree, you can drastically inflate the dimensionality, increasing the risk of overfitting. Overfitting might lead to large coefficients if the model chases every minor fluctuation in the data. L2 regularization helps mitigate that but cannot entirely prevent it if the polynomial features grow too extensive. Another subtlety is that the question about “which hyperplane is best?” in the original sense might become “which hyperplane in the transformed space best?”—but the same minimal-norm principle still applies in that higher-dimensional feature representation.
Would the shape of the logistic regression loss allow for local minima or is it guaranteed to have one global minimum?
The logistic regression loss function (the negative log-likelihood, often referred to as log-loss) combined with an L2 penalty is convex in the parameters w and b, assuming we are talking about a standard linear model. Convexity implies there is only one global minimum and no local minima. Hence, gradient-based methods should converge to the global optimum, provided that the learning rate and optimization steps are chosen well.
A potential edge case or pitfall is when the data set is extremely large or has certain pathological conditions (like extreme collinearity or extremely poorly scaled features) that make the solver’s path to convergence very slow or ill-conditioned. But the function itself is convex for standard logistic regression with L2 regularization, so the solution is unique in that sense.
In more complex scenarios—such as neural networks, or logistic regression with certain non-linear transformations or constraints—non-convexities can arise. But for the standard logistic regression scenario, we do not have local minima. That’s one reason logistic regression is a popular baseline method: it’s relatively straightforward and consistent to train without getting stuck in spurious local minima.
What if a domain constraint forces the coefficients to be non-negative? Does the scaling argument still apply?
Sometimes, domain knowledge might suggest that certain coefficients cannot be negative (e.g., in a medical risk model, we might only allow non-negative contributions). This changes the feasible region of the parameter space. The standard logistic regression objective remains convex, but now it has box constraints on the parameters (w ≥ 0). The solution is still found via specialized optimization routines for constrained convex problems.
The preference for smaller norms still applies within that feasible region. If multiple feasible weight vectors classify the data equally well, the L2 penalty will guide the model to pick the one with the smallest squared norm. So among any sets of non-negative solutions that achieve the same boundary, the solution with minimal norm is favored. However, the geometry is more restricted, so it’s possible that some directions in the parameter space are disallowed, and the final boundary might differ from the unconstrained solution.
A pitfall is ignoring constraints or imposing them incorrectly. If you require w ≥ 0 but the data strongly demands negative coefficients for good separation, the model performance will degrade. Conversely, if you do trust your domain constraint, imposing w ≥ 0 can enhance interpretability and keep the model aligned with real-world knowledge while still leveraging the beneficial effect of L2 regularization to avoid large coefficients.
Does the fact that all three hyperplanes pass through the origin imply something special about the data distribution in the context of logistic regression?
If all candidate hyperplanes pass through the origin, that usually implies no intercept term (b = 0). In practice, logistic regression typically learns an intercept automatically. For quadrant-based data, it might happen that the best dividing line indeed goes through the origin if the data is symmetric around that origin.
When the intercept is strictly zero, it means the boundary is w₁ x + w₂ y = 0. The preference for smaller norms remains. If there was an intercept, the boundary would be w₁ x + w₂ y + b = 0, potentially shifting the line. The presence of that intercept can help if the data is not centered or if the distributions do not exactly revolve around the origin.
A pitfall is forgetting that real-world data rarely is so nicely symmetrical that the origin is the natural dividing point. People sometimes artificially remove an intercept in experiments or academic examples because it simplifies the math. But in production scenarios, we almost always keep the intercept to accommodate offsets in the data. The essential point is that even if you allow an intercept, the solution with smaller weight magnitude is still preferred when the classification outcome is equivalent.
Could adding a random initialization lead to different final solutions for the same data?
Logistic regression with a convex loss typically converges to the same global minimum, so in theory, random initialization should not matter. In practice, if you use gradient descent with a poorly chosen learning rate or early stopping, you might end up near a suboptimal region. But with well-tuned optimization, random initialization is typically irrelevant because the objective is convex.
A subtlety arises with extremely large or complicated feature sets, or with regularization methods like L1 that can produce corner solutions (some weights are exactly zero). In certain borderline cases, the path the solver takes might cause different sets of weights to be zeroed out. That can lead to solutions that have the same cost but differ in which features are active. For standard L2 logistic regression in 2D, though, you get one unique solution (plus or minus numerical tolerance), so random initialization has no significant effect.
A real-world pitfall is incorrectly concluding your model is non-convex because you see different results from repeated runs. Often, it’s an issue of data ordering, different default hyperparameters, or the solver not converging fully. If you run it to convergence with the same conditions, it should yield the same solution.
How might we interpret the coefficients if the final boundary is y = -x, but with a small or large magnitude?
When the final boundary is effectively y = -x, it suggests that an increase in y by 1 can be offset by an increase in x by 1 to stay on the boundary. In other words, x and y have the same “weight” in magnitude but differ in sign. If the logistic regression solution is (w₁, w₂) = (0.1, 0.1), that implies the logit is 0.1⋅x + 0.1⋅y. With no intercept, a data point satisfies 0 = 0.1⋅x + 0.1⋅y => x + y = 0.
If the magnitude of the coefficients was (100, 100), the boundary is still x + y = 0, but the interpretation for the log-odds changes drastically for points not exactly on the boundary. For example, if x + y = 1, then w⋅x = 100, leading to a very high predicted probability for the positive class—arguably an overconfident outcome for just a small shift from the boundary. Conversely, if x + y = -1, the logit becomes -100, yielding a near-zero predicted probability for the positive class. The smaller magnitude solution is typically more moderate in these transitions.
In practice, we often examine the logistic function σ(w⋅x+b) at certain reference points to see how the probability changes. A large norm for w can cause the model to saturate near 1 or 0 probability for even small changes in x, which might be undesirable if we want more gradual probability transitions.
A potential pitfall is misinterpreting the scale of these coefficients. If you see an extremely large coefficient, you might think that feature is highly significant, but it could be a side-effect of how the data is scaled or of an under-regularized solution. Checking the magnitude of w₁, w₂, the intercept, and the effect on actual predictions is crucial to avoid misinterpretation.
Could the optimization algorithm converge to a suboptimal line if the learning rate is poorly set?
While the logistic regression objective is convex, a poorly set learning rate (too large) can cause gradient descent to diverge, or to oscillate in parameter space and fail to converge. If the learning rate is too small, it might take an extremely long time to get close to the global optimum, or it might get stuck near a suboptimal region due to numerical precision issues.
In many libraries, robust solvers like LBFGS or Newton’s method-based algorithms handle logistic regression well, automatically adjusting step sizes. However, if you implement logistic regression from scratch using vanilla gradient descent, you must carefully tune the learning rate. The main difference from a truly non-convex problem is that you do not have local minima to get trapped in, but you can have suboptimal behavior if your steps are so large that you skip over the valley of the global minimum or cause numerical overflows.
A real-world pitfall is mixing up the roles of learning rate and regularization strength. Beginners sometimes see poor results and think they must change the regularization parameter, when the real issue is that the optimizer’s learning rate is either too large or too small. Checking the cost function over epochs can help diagnose whether the algorithm is converging steadily or diverging.
Are there scenarios in which the decision boundary might appear the same as y = -x, but the actual logistic regression solution is not just (w₁, w₂) with w₁ = w₂?
If you allow an intercept, you could have w₁ = 2, w₂ = 2, b = -2, for instance, which would yield the boundary 2x + 2y - 2 = 0 => x + y - 1 = 0 => y = 1 - x. That boundary is parallel to y = -x but shifted. So from a purely visual standpoint, it might look like y = -x if you do not look closely at the axis intercept. Meanwhile, if w₁ and w₂ are not exactly equal but have a ratio close to 1, you can still get a boundary that is roughly y = -x, but not precisely.
Additionally, in higher dimensions, you might see a boundary that projects into the x-y plane looking like y = -x, but the full boundary in the higher-dimensional space includes other features, effectively rotating or shifting the boundary. The presence of those other features can cause the classification region in 2D to appear as though it is y = -x, but the full model includes more complexity.
A subtle pitfall is focusing only on the x-y plane if there are additional features. In practice, you rarely train logistic regression on just x and y. You might have more input variables, in which case the boundary in a 2D plot is only a slice or projection of the true boundary in the higher-dimensional space. The model could have a large weight on a third feature z that drastically changes how the classification boundary works when you consider all variables together.
If the data is separable and the solution tries to perfectly classify all points, do we risk overfitting?
Yes, a fundamental principle is that when data is perfectly separable, an unregularized logistic regression model tends to push the coefficients toward infinity to maximize the likelihood. This can cause extreme overfitting in real-world scenarios where data is noisy or limited. L2 regularization prevents the parameters from growing arbitrarily large, thus avoiding a pathological solution that is extremely overconfident.
This can be illustrated by the fact that in a perfectly separable dataset, the gradient of the log-likelihood might not vanish until coefficients become very large. The limit is that the log-likelihood goes to 0 misclassification error, so the partial derivatives keep pushing the magnitude of w up. Introducing L2 regularization ensures the cost function has a finite optimum, balancing perfect classification with coefficient control. This typically leads to better generalization performance, especially if new data might contain slight variations or noise.
A pitfall is ignoring signs of overfitting. One might see a training log-likelihood that is extremely high (very close to zero error) and assume that’s good. But if you do not check validation performance, you might be inadvertently building a model that is too sensitive to the training set, expecting every point to adhere exactly to the boundary. With real data, that rarely generalizes well.
What happens if we drastically change the scale of x vs. y? For instance, if x is measured in thousands and y in ones?
If x is in thousands while y is in single-digit values, then a line y = -x might look extremely steep or shallow depending on how you plot it. The actual numeric relationship is that the coefficient for x might be quite small compared to y, or vice versa, to compensate for the scale difference in the features. Without feature scaling, the solver might produce a parameter vector that has very large magnitude for y and a smaller one for x to achieve the same ratio. Alternatively, it might produce moderate coefficients if the intercept helps offset the scale difference.
The L2 regularization still penalizes large weights, so the model tries to keep them as small as possible consistent with good classification. If the scale difference is huge, you might see the solver converge slowly or produce somewhat awkward coefficient values, especially if the library does not automatically standardize. The decision boundary might still conceptually separate Q1 from Q3, but the numeric scale can distort your interpretation of the coefficients.
A real-world pitfall is failing to realize that an unscaled variable with large numeric range can overshadow other variables in gradient steps. It can also produce extremely small or extremely large learned coefficients that degrade numerical stability. Typically, you mitigate this by standardizing or at least normalizing each feature to a comparable range, making the training procedure more robust and the coefficients more interpretable.
How do we visualize or interpret the impact of the regularization parameter on the geometry of the boundary?
To visualize, one can train logistic regression on the same dataset with a grid of regularization strengths, then plot the resulting decision boundaries over the 2D space. As you vary λ from small (weak regularization) to large (strong regularization), you can see:
• Whether the boundary moves or rotates. • Whether the intercept changes or remains near zero. • How the margin of “transition” from high probability to low probability changes.
If the data is perfectly separable, you might see that for very weak regularization, the boundary starts to push more extreme slopes or intercepts, basically “pinching” the classification region to try and produce near-certain classifications. As you increase regularization strength, the boundary tends to become more moderate and stable, with smaller magnitude coefficients.
In practice, you can do something like:
import matplotlib.pyplot as plt
import numpy as np
from sklearn.linear_model import LogisticRegression
np.random.seed(0)
X_q1 = np.random.uniform(low=1.0, high=3.0, size=(50, 2))
X_q3 = np.random.uniform(low=-3.0, high=-1.0, size=(50, 2))
X = np.vstack([X_q1, X_q3])
y = np.concatenate([np.ones(50), np.zeros(50)])
C_values = [0.01, 0.1, 1, 10, 100]
colors = ['r', 'g', 'b', 'c', 'm']
plt.scatter(X_q1[:,0], X_q1[:,1], label='Positive')
plt.scatter(X_q3[:,0], X_q3[:,1], label='Negative')
x_line = np.linspace(-4, 4, 100)
for C_val, col in zip(C_values, colors):
model = LogisticRegression(C=C_val, penalty='l2', solver='lbfgs')
model.fit(X, y)
w0, w1 = model.coef_[0]
b = model.intercept_[0]
# Solve w0*x + w1*y + b = 0 => y = -(w0/w1)*x - b/w1
y_line = -(w0/w1)*x_line - b/w1
plt.plot(x_line, y_line, col, label=f'C={C_val}')
plt.legend()
plt.show()
You can see how the boundary changes. If the data is easily separable, you’ll often notice that for strong regularization (small C), the line is less “extreme,” the slope is gentler, and the intercept might be moderate. For weak regularization (large C), the boundary can become more “confident,” possibly with steeper slopes. This direct visualization can clarify how regularization shapes the geometry of the classification boundary in logistic regression.
If we introduced outliers in Q1 or Q3, could it cause the solution to shift drastically?
Yes. Logistic regression, like many models that optimize a continuous loss function, can be sensitive to outliers. A single outlier that is very far from the normal data cluster can push the boundary in an attempt to classify that outlier correctly (if the outlier’s label is the same as the rest of the cluster). The model might thus adopt a larger magnitude weight vector to try to keep that outlier on the correct side. In addition, if the outlier is extremely far away, the log-loss might punish misclassification severely, forcing the model to adapt.
L2 regularization mitigates this somewhat by penalizing large coefficients. The model might decide it’s too costly to move the boundary significantly just to accommodate that outlier, especially if it means inflating the magnitude of w. But if the outlier is not truly an erroneous data point, ignoring it might degrade performance. The final solution is a trade-off. If there are multiple outliers on the boundary or beyond it, the boundary might shift more dramatically or the coefficients might grow larger.
A pitfall is not diagnosing or handling outliers properly. In real data, you might remove outliers if they are genuine errors, or you might apply a robust form of regression. Alternatively, you might keep them if they represent critical but unusual cases you want to classify correctly. Understanding the domain context is key. For the Q1 vs. Q3 toy problem, if you artificially inject a handful of points far out in Q3, you might see the boundary bend or shift more drastically, especially if the logistic model tries to not misclassify them at the cost of larger weights.
Could an ensemble of logistic regression models yield a better boundary?
An ensemble of logistic regression models is less common than, say, ensembles of decision trees (random forests, gradient boosting). However, you can still train multiple logistic regressions on different subsets of data (e.g., bagging) and average their predictions. This ensemble can sometimes reduce variance, improving generalization. Each individual logistic regression might produce a line close to y = -x if the data is in Q1 vs. Q3, but small variations in training subsets might yield slightly different slopes or intercepts. Averaging their probability predictions tends to smooth out extremes, leading to a more moderate overall boundary when you threshold at 0.5.
The principle that each individual model is still subject to an L2 penalty remains. Each is likely to converge to a smaller-norm solution than it would without regularization. But by ensembling, you combine multiple “small-norm” boundaries, possibly getting a more robust classifier to noise or random sampling. A pitfall is that if the data is trivially separable, an ensemble might be overkill. Another subtlety is that ensembles are less straightforward to interpret than a single logistic regression line. If interpretability is a key requirement, a single logistic regression might be preferable.
Could we end up with a scenario where the best hyperplane is not exactly y = -x but close to it, and all scaled versions of that vector also appear to work?
Yes, in realistic data, the true boundary might not be precisely y = -x. Instead, logistic regression might converge to w = (1.1, 0.9) with some intercept b. That boundary is close to y = -x but not identical. Since it’s not a pure ratio of w₁ = w₂, scaling that vector by a constant α simply modifies the boundary in a way that might not remain optimal. L2 regularization enforces a specific magnitude for w in balance with the data fit. If you forcibly scale up or scale down w, you need to retune the intercept to keep the same classification performance. Even then, you might lose some optimal alignment with the data because they might not be perfect multiples.
Hence, in a real-world distribution, “best” means the solution that best balances the log-loss and the regularization cost. In a perfect quadrant separation scenario, y = -x might be a neat geometric boundary, but actual data or slight noise might produce something close to that, plus a nonzero intercept. So while scaled versions might look similar, only one unique combination truly minimizes the cost. In practice, you interpret the final fitted model from your solver, which includes both w and b as found by the convex optimizer.
If the solver includes early stopping or a maximum iteration limit, could it pick a suboptimal boundary?
Yes, if the solver halts before convergence (either due to early stopping or hitting a max iteration count that is too low), you might end up with a parameter vector that does not fully minimize the cost function. This can result in a suboptimal boundary that is not quite the minimal-norm solution. The logistic regression might have partial improvements left undone.
This is especially relevant if the dataset is large or if the gradient steps are small. Checking convergence criteria is essential: many libraries provide a “tol” (tolerance) parameter that determines when to stop iterating if the improvement in the objective is below a certain threshold. If tol is too large, the solver might stop prematurely. If max_iter is too small, the solver halts as well, potentially leaving you with a solution that’s bigger in norm or less accurate than it should be. A real-world pitfall is ignoring solver warnings about convergence. Many times scikit-learn prints a “ConvergenceWarning” if it does not converge, yet people overlook it.
Hence, the best practice is to ensure the solver truly converged by either verifying that the iteration count is sufficient or adjusting tolerance parameters as needed. Otherwise, you risk an artificially inflated or deflated weight vector that does not reflect the globally optimal solution under L2 regularization.
Could the data in Q1 and Q3 still be misclassified if the logistic regression finds a weird local optimum?
Standard logistic regression with an L2 penalty is convex, meaning weird local optima should not exist in theory. Therefore, if the data is truly in Q1 vs. Q3 (with no noise or overlap), you’d expect near-perfect classification with a line close to y = -x. However, “weird” results might appear if:
• The solver is incorrectly implemented or has numeric issues. • The data was incorrectly labeled or includes anomalies that appear in Q1 but labeled negative, or in Q3 but labeled positive. • There is a mismatch in the cost function (for example, if you used a custom-coded logistic regression that introduced non-convexities).
In typical well-tested libraries, you do not get a local optimum but rather the global optimum. So any misclassification in a perfect Q1 vs. Q3 scenario typically stems from data anomalies or from a boundary shift introduced by the intercept if the data distribution is not purely symmetrical.
An additional subtlety is if you combined logistic regression with some weird data transformation or if you used a neural network classifier that has logistic output but is non-linear in the hidden layers. That is not standard logistic regression and can lead to local minima. But that goes beyond the scope of the simple linear logistic regression scenario.
What if we wanted to interpret the model in terms of odds ratios? Would the smaller-norm solution still be the most interpretable?
Odds ratios for logistic regression correspond to exponentiated coefficients: for a coefficient wᵢ, the odds ratio is exp(wᵢ). A smaller absolute wᵢ implies an odds ratio closer to 1, meaning the feature has a more moderate effect on the log-odds. A very large positive wᵢ yields an odds ratio far greater than 1, signifying a strong positive association with the positive class.
From an interpretation standpoint, extremely large or small odds ratios can be suspicious, because they suggest that small changes in the feature produce massive changes in probability. If the data truly justifies that, it can be valid, but often it is a sign of either a correlated feature or an overfitted model. So a solution with moderate coefficient magnitudes can be more interpretable and plausible for domain experts. They can see that a slight increase in x (holding y constant) multiplies the odds by a certain factor, which is not astronomically large.
Hence, from an interpretability perspective, L2 regularization not only helps avoid overfitting but also yields more believable odds ratios. A pitfall is neglecting domain knowledge. Sometimes, if you truly know a certain feature has a huge effect, a large odds ratio might be correct. But typically, extremely large ratios are suspicious or arise from data issues. Checking domain relevance is part of the interpretability process.