Past exam of the mathematics course of the University of Cambridge / 2017 / iii / Paper 326 / 3 / 2 / b

Table of contents
- Solution b

Solution

The indicator functional of a constraint set restricts the conjugate supremum to

[- 1, 1]

. Complete the square:

p x - \frac{x ^{2}}{2} = \frac{p ^{2}}{2} - \frac{( x - p ) ^{2}}{2} .

Its maximizer is

x = p

when

∣ p ∣ \leq 1

, and the nearer endpoint

x = sign (p)

otherwise. The convex conjugate of a constrained quadratic is therefore

E^{*} (p) = {\frac{1}{2} p^{2}, ∣ p ∣ - \frac{1}{2}, ∣ p ∣ \leq 1, ∣ p ∣ > 1.

This is the unit-threshold Huber loss; both branches and their first derivatives agree at the transition points. The original PDF constrains

∣ x ∣ \leq 1

, including both signs; the damaged TeX loses that absolute value.

 Ancestors (11)

 View article source

 Discussion (0)

There are no discussions about this article yet.

 Articles by others on the same topic (0)

There are currently no matching articles.

  See all articles in the same topic Create my own version