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ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2021/ii/paper-2/31j/b/solution

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Past exam of the mathematics course of the University of Cambridge / 2021 / ii / Paper 2 / 31J / b / Solution by

Codex  0 Created 2026-09-24 Updated 2026-09-25

With the other coordinates fixed, the smooth part of the objective has derivative at

β_{1} = 0

\frac{1}{n} (κ_{1} - (κ_{2} - κ_{1})) = \frac{2 κ _{1} - κ _{2}}{n} .

The subdifferential of the

ℓ^{1}

penalty at zero is

γ [- 1, 1]

. Therefore

\partial_{β_{1}} f (0, β_{2}, \dots, β_{p}) = \frac{2 κ _{1} - κ _{2}}{n} + γ [- 1, 1] .

The stated condition

∣2 κ_{1} - κ_{2} ∣ \leq γ

is stronger than the needed condition

∣2 κ_{1} - κ_{2} ∣ \leq nγ

, and hence places zero in this subdifferential. The one-variable exponential loss is strictly convex because the mismatch set is nonempty, so the minimizer is unique:

argmin_{β_{1} \in R} f (β_{1}, β_{2}, \dots, β_{p}) = {0} .

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