 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2012/iii/paper-36/3/c/solution

Codex  0 Created 2026-10-03 Updated 2026-10-07

For each

n \geq 1

f (u) = 1, g_{n} (u) = 1 + \frac{u - 1/2}{n} .

The second is at least

1/2

and integrates to one. Their expected values differ by

ψ (g_{n}) - ψ (f) = \frac{1}{n} \int_{0}^{1} u (u - 1/2) d u = \frac{1}{12 n} .

g_{n} / f = 1 + Δ_{n}

gives

P_{f} Δ_{n}^{2} = \frac{1}{n} \int_{0}^{1} (u - 1/2)^{2} d u = \frac{1}{12 n} .

The supplied product bound, also obtained from the chi-squared divergence of product measures, implies

∥ P_{f}^{(n)} - P_{g_{n}}^{(n)} ∥_{1}^{2} \leq (1 + \frac{1}{12 n})^{n} - 1 \leq e^{1/12} - 1 \leq \frac{1}{11} < 1.

For the final elementary estimate, compare the exponential series with the geometric series:

e^{x} \leq (1 - x)^{- 1}

for

0 \leq x < 1

. Thus the overlap of the product laws is at least

1/2

R_{n}^{*} \geq \frac{1}{4} \cdot \frac{1}{144 n} \cdot \frac{1}{2} = \frac{1}{1152 n} .

So one may take $C = 1/1152$ , uniformly for all $n \geq 1$ . The alternatives are allowed to depend on

n

, because the minimax risk takes a supremum over all continuous probability density functions separately at each sample size.

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