All else being equal, the decision boundary of an SVM model gets smoother (simpler; refer to the image below) as the regularization parameter (a.k.a. a cost parameter) increases.

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BlackTom AI · Accepted Answer

When evaluating how the SVM decision boundary behaves as the regularization parameter C changes, the intuition can be tricky. 
Option 1: True — This would claim that increasing C makes the boundary smoother or simpler. In practice, a larger C reduces the penalty for misclassifications and allows the model to fit the training data more closely, which tends to produce a more complex, wiggly boundary that captures more intricacies of the data. Therefore this option contradicts typical SVM behavior.
Option 2: False — This aligns with the standard understanding: increasing C reduces regularization strength, which often leads to a more flexible and potentially overfitted boundary, not a smoother one. The margin can become narrower and the decision surface can bend around individual data points to reduce training errors, resulting in a boundary that is less smooth and more complicated.
Option 3 (if considered): To discuss the generic effect, one might recall that with stronger regularization (smaller C), the model prefers a simpler, smoother decision surface to maximize margin, at the cost of more training errors. Conversely, weakening regularization (larger C) typically relaxes that simplicity constraint, enabling more complex boundaries.
Option 4 (edge case considerations): In some synthetic or highly constrained datasets, changes in C might not visibly alter the boundary much, but the general trend remains: higher C tends to reduce smoothness rather than increase it, so the statement claiming smoother boundaries with larger C is not generally correct.

BU.330.780.T1.SP25 Quiz 2- Requires Respondus LockDown Browser

All else being equal, the decision boundary of an SVM model gets smoother (simpler; refer to the image below) as the regularization parameter (a.k.a. a cost parameter) increases.

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