Which of the following is a reason for using Jensen-Shannon (JS) distance over a Kolmogorov- Smirnov (KS) test for numeric feature drift detection?

Question

Options

• AAll of these reasons
• BJS is not normalized or smoothed
• CNone of these reasons
• DJS is more robust when working with large datasets
• EJS does not require any manual threshold or cutoff determinations

• B
  17% (5)
• C
  10% (3)
• D
  70% (21)
• E
  3% (1)

Explanation

Jensen-Shannon distance handles large datasets more gracefully than the KS test because the KS test becomes hypersensitive at scale - with very large sample sizes, even trivially small distribution differences yield statistically significant results, flooding you with false drift alerts. JS distance measures the magnitude of distributional difference on a normalized 0-to-1 scale, so its interpretation stays stable regardless of dataset size.

Why the distractors fail:

• B is wrong because JS is normalized (bounded 0–1) and smoothing can be applied to it - the claim is the opposite of reality.
• E is wrong because JS distance still requires you to manually set a threshold to decide what magnitude of divergence counts as "drift." The KS test actually has a built-in significance framework (p-values) that arguably reduces arbitrary cutoff decisions.
• A is wrong because it depends on B and E being correct, which they aren't.
• C is wrong because D is a valid reason.

Memory tip: Think of the KS test as a microscope - powerful, but at high magnification (large N) it makes every speck look like a problem. JS distance is more like a ruler: it gives you a consistent, bounded measurement no matter how closely you look.

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