Question:medium

Between \(L_1\) and \(L_2\) norms, which one of the following is CORRECT in the treatment of outliers in the data?

Show Hint

L1 cost grows linearly with residual size, L2 cost grows quadratically — so a single large outlier dominates an L2 (least-squares) misfit far more than an L1 misfit.
Updated On: Jul 21, 2026
  • \(L_1\) norm gives higher weightage to outliers than \(L_2\) norm
  • \(L_2\) norm gives higher weightage to outliers than \(L_1\) norm
  • Both \(L_1\) and \(L_2\) norms give equal weightage to outliers
  • \(L_1\) norm occasionally gives higher weightage to outliers than \(L_2\) norm
Show Solution

The Correct Option is B

Solution and Explanation

Look at it through the influence function (the derivative of each norm's per-datum cost with respect to the residual), which measures how strongly a single data point pulls on the fitted parameters. For the \(L_1\) cost \(|e_i|\), the derivative is \(\text{sgn}(e_i)=\pm1\) — constant in magnitude no matter how large the residual becomes, so every outlier, however extreme, tugs on the solution with the same, bounded force. For the \(L_2\) cost \(e_i^2\), the derivative is \(2e_i\), which grows without bound as the residual grows.

An unbounded influence function is precisely the mathematical definition of non-robustness: as \(e_i\to\infty\), the \(L_2\) fit is dragged arbitrarily far to try to shrink that one term, whereas the \(L_1\) fit's response saturates. This robustness-theory view (bounded vs. unbounded influence function) gives the same conclusion as the direct linear-vs-quadratic growth argument: the \(L_2\) norm weights outliers more heavily than the \(L_1\) norm, confirming option (B).

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