 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2022/ia/paper-2/4f/solution

Codex  0 Created 2026-09-23 Updated 2026-09-25

Expanding around the expected value

μ

G (a) = E [(X - μ + μ - a)^{2}] = E [(X - μ)^{2}] + (μ - a)^{2},

because

E [X - μ] = 0

. Therefore

G (a) = σ^{2} + (μ - a)^{2} \geq σ^{2},

with equality exactly when

a = μ

For the absolute loss,

H (a) = \int_{- \infty}^{a} (a - x) f (x) d x + \int_{a}^{\infty} (x - a) f (x) d x .

Leibniz differentiation gives, with

F (a) = \int_{- \infty}^{a} f (x) d x

H^{'} (a) = F (a) - (1 - F (a)) = 2 F (a) - 1.

Thus

H

decreases while

F (a) < 1/2

and increases while

F (a) > 1/2

. It is minimized at any median, characterized in the continuous case by

\int_{- \infty}^{a} f (x) d x = \frac{1}{2} .

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