 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2021/ib/paper-1/7h/c/solution

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

h (s) = lo g (1 + e^{s})

is convex because

h^{''} (s) = \frac{e ^{s}}{( 1 + e ^{s} ) ^{2}} \geq 0.

Part (b) therefore shows that every

β \mapsto h (a_{i}^{T} β)

is convex. The absolute value function is convex, so

β \mapsto ∣ β_{j} ∣

is convex for every coordinate

j

. Finally, part (a) says that a finite sum of convex functions is convex. Therefore

Q (β) = \sum_{i = 1}^{n} lo g (1 + e^{a_{i}^{T} β}) + \sum_{j = 1}^{d} ∣ β_{j} ∣

is convex. The second sum is the L1 norm regularizer.

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