= Solution
Write $h=h_n$ and $I=[t_0,t_0+h]$, restricting to meshes small enough that $t_0+h\leq1$. 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 $q_n(s)=g_n(t_{i-1})$ on each observation cell $[t_{i-1},t_i)$. 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 $2\Delta$. Thus the <quadrature error> obeys
$$
B_n=\int_0^1(q_n-g_n)\sigma^2\,ds,\qquad
|B_n|\leq2\|\sigma^2\|_\infty\frac{\Delta}{h}.
$$
The fluctuation has <variance> at most $2S\Delta/h^2$. Its zero <expected value> gives
$$
\mathbb E\bigl(\widehat\Lambda_n(g_n)-\Lambda(g_n)\bigr)^2
\leq\frac{S}{h^2}(2\Delta+4\Delta^2)
\leq6S\frac{\Delta}{h^2},
$$
using $\Delta\leq1$.
\Image[/past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2016/iii/paper-209-window-cells.png]
{title=Only the two endpoint cells contribute to the window quadrature error}
The remaining <bias> is the difference between an interval average and the <spot variance>. By the <Hölder continuity> of $\sigma^2$,
$$
A_h=\Lambda(g_n)-\sigma^2(t_0),\qquad
|A_h|\leq\frac1h\int_0^h C s^\beta\,ds
=\frac{C}{1+\beta}h^\beta.
$$
Retaining the centered fluctuation and bounding $(B_n+A_h)^2\leq2B_n^2+2A_h^2$ yields
$$
\mathbb E\bigl(\widehat\Lambda_n(g_n)-\sigma^2(t_0)\bigr)^2
\leq10S\frac{\Delta}{h^2}+2C^2h^{2\beta}.
$$
For $h=\Delta^{1/(2+2\beta)}$, both $\Delta/h^2$ and $h^{2\beta}$ equal $\Delta^{2\beta/(2\beta+2)}$. Therefore \b[$D_1=10$ and $D_2=2C^2$ satisfy both requested bounds]:
$$
\boxed{\mathbb E\bigl(\widehat\Lambda_n(g_n)-\sigma^2(t_0)\bigr)^2
\leq(10\|\sigma^4\|_\infty+2C^2)\Delta^{2\beta/(2\beta+2)}.}
$$
The constants are independent of the observation partition, $\sigma$, $t_0$ and $\beta$. The exponent follows from a <bias-variance tradeoff> for a shrinking window.
Back to article page
