Past exam of the mathematics course of the University of Cambridge / 2016 / iii / Paper 209 / 1

Put ${\delta }_i=t_i-t_{i-1}$, $\Delta ={\max }_i{\delta }_i$, $v_i={\int }_{t_{i-1}}^{t_i}\sigma (s{)}^{2}ds$ and $S=\Vert {\sigma }^4{\Vert }_{\infty }$. The deterministic integrand in the Itô integral implies that the increments $Z_i=X_{t_i}-X_{t_{i-1}}$ are independent random variables with normal distributions $N(0,v_i)$. The Gaussian fourth moment givesThese identities include $v_i=0$. Since the summands of $M_n$ are centered and independent, all cross terms in its second moment vanish. Consequently,Thus $D=2$ works for every observation partition. The last step uses the total interval length, rather than assuming equally spaced observations.

Use the same $M_n,v_i,S,\Delta$ as in part (a). The error of the weighted realized variance splits into its centered fluctuation and its deterministic quadrature error:The Hölder continuity of $g$ controls this bias:Since $\mathbb{E}{M}_n=0$, the bias-variance decomposition of mean squared error has no cross term. Part (a) therefore givesHence $\tilde{D}=3$ is a universal choice. This argument separately controls the statistical fluctuation and the Riemann sum approximation.

Write $h=h_n$ and $I=[t_0,t_0+h]$, restricting to meshes small enough that $t_0+h\le 1$. This discontinuous indicator function does not satisfy the hypothesis of part (b). Instead apply part (a) with $\Vert g_n{\Vert }_{\infty }=1/h$ 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 obeysThe fluctuation has variance at most $2S\Delta /h^2$. Its zero expected value givesusing $\Delta \le 1$.

The remaining bias is the difference between an interval average and the spot variance. By the Hölder continuity of ${\sigma }^2$,Retaining the centered fluctuation and bounding $(B_n+A_h{)}^2\le 2B_n^2+2A_h^2$ yieldsFor $h={\Delta }^{1/(2+2\beta )}$, both $\Delta /h^2$ and $h^{2\beta }$ equal ${\Delta }^{2\beta /(2\beta +2)}$. Therefore $D_1=10$ and $D_2=2C^2$ satisfy both requested bounds: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.