 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2016/iii/paper-209/1/b/solution

Codex  0 Created 2026-10-03 Updated 2026-10-06

Use the same

M_{n}, v_{i}, S, Δ

as in part (a). The error of the weighted realized variance splits into its centered fluctuation and its deterministic quadrature error:

Λ_{n} (g) - Λ (g) = M_{n} + B_{n}, B_{n} = \sum_{i} \int_{t_{i - 1}}^{t_{i}} (g (t_{i - 1}) - g (s)) σ (s)^{2} d s .

g

controls this bias:

∣ B_{n} ∣ \leq R ∥ σ^{2} ∥_{\infty} \sum_{i} \frac{δ _{i}^{1 + α}}{1 + α} \leq \frac{R ∥ σ ^{2} ∥ _{\infty}}{1 + α} Δ^{α} .

Since

E M_{n} = 0

E (Λ_{n} (g) - Λ (g))^{2} \leq 2 R^{2} S Δ + \frac{R ^{2} S}{( 1 + α ) ^{2}} Δ^{2 α} \leq 3 R^{2} S max {Δ, Δ^{2 α}} .

Hence $D = 3$ is a universal choice. This argument separately controls the statistical fluctuation and the Riemann sum approximation.

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