Write

h = h_{n}

and

I = [t_{0}, t_{0} + h]

, restricting to meshes small enough that

t_{0} + h \leq 1

. This discontinuous indicator function does not satisfy the hypothesis of part (b). Instead apply part (a) with

∥ g_{n} ∥_{\infty} = 1/ h

and handle its two boundary cells directly.

Let

q_{n} (s) = g_{n} (t_{i - 1})

[t_{i - 1}, t_{i})

. Away from cells containing the two ends of

I

,

q_{n} = g_{n}

almost everywhere. Even if an endpoint is itself an observation time, at most two cells contribute, and their total length is at most

2Δ

. Thus the quadrature error obeys

B_{n} = \int_{0}^{1} (q_{n} - g_{n}) σ^{2} d s, ∣ B_{n} ∣ \leq 2∥ σ^{2} ∥_{\infty} \frac{Δ}{h} .

The fluctuation has variance at most

2 S Δ/ h^{2}

. Its zero expected value gives

E (Λ_{n} (g_{n}) - Λ (g_{n}))^{2} \leq \frac{S}{h ^{2}} (2Δ + 4 Δ^{2}) \leq 6 S \frac{Δ}{h ^{2}},

using

Δ \leq 1

.

The remaining bias is the difference between an interval average and the spot variance. By the Hölder continuity of

σ^{2}

,

A_{h} = Λ (g_{n}) - σ^{2} (t_{0}), ∣ A_{h} ∣ \leq \frac{1}{h} \int_{0}^{h} C s^{β} d s = \frac{C}{1 + β} h^{β} .

Retaining the centered fluctuation and bounding

(B_{n} + A_{h})^{2} \leq 2 B_{n}^{2} + 2 A_{h}^{2}

yields

E (Λ_{n} (g_{n}) - σ^{2} (t_{0}))^{2} \leq 10 S \frac{Δ}{h ^{2}} + 2 C^{2} h^{2 β} .

For

h = Δ^{1/ (2 + 2 β)}

, both

Δ/ h^{2}

and

h^{2 β}

equal

Δ^{2 β / (2 β + 2)}

. Therefore $D_{1} = 10$ and $D_{2} = 2 C^{2}$ satisfy both requested bounds:

E (Λ_{n} (g_{n}) - σ^{2} (t_{0}))^{2} \leq (10∥ σ^{4} ∥_{\infty} + 2 C^{2}) Δ^{2 β / (2 β + 2)} .

The constants are independent of the observation partition,

σ

,

t_{0}

and

β

. The exponent follows from a bias-variance tradeoff for a shrinking window.

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