Between \(L_1\) and \(L_2\) norms, which one of the following is CORRECT in the treatment of outliers in the data?
Show Hint
- \(L_1\) 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
The Correct Option is B
Solution and Explanation
The \(L_1\) and \(L_2\) norms of a residual vector \(e_i = d_i^{obs}-d_i^{pred}\) are defined as:
\[ \lVert e\rVert_1 = \sum_i |e_i|, \qquad \lVert e\rVert_2 = \left(\sum_i e_i^2\right)^{1/2} \]
Consider how the contribution of a single residual to the misfit changes as that residual grows large (i.e., becomes an outlier). For the \(L_1\) norm, the contribution of a data point to the total misfit is \(|e_i|\), which grows only linearly with the size of the residual. For the \(L_2\) (least-squares) norm, the contribution is \(e_i^2\), which grows quadratically.
Concretely, if one residual is 10 times larger than the rest, its contribution to the \(L_1\) misfit is only 10× a typical term, but its contribution to the \(L_2\) misfit is 100× a typical term. When an inversion or regression minimizes an \(L_2\)-based objective function, a single large outlier can therefore dominate the total misfit and pull the fitted model strongly toward itself (the model "tries hard" to reduce that one huge squared term). An \(L_1\)-based (least-absolute-deviations) fit is comparatively insensitive to the same outlier, since its influence only scales linearly — this is exactly why \(L_1\)-norm (robust) inversion is preferred in geophysical data processing when the data contain occasional bad picks or spikes.
Hence the \(L_2\) norm gives higher weightage to outliers than the \(L_1\) norm.
\(\boxed{\text{Option (B)}}\)
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