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Abstract

The SIML (abbreviation of Separating Information Maximal Likelihood) method, has been introduced by N. Kunitomo and S. Sato and their collaborators to estimate the integrated volatility of high-frequency data that is assumed to be an Itô process but with so-called microstructure noise. The SIML estimator turned out to share many properties with the estimator introduced by P. Malliavin and M. E. Mancino. The present paper establishes the consistency and the asymptotic normality under a general sampling scheme but without microstructure noise. Specifically, a fast convergence shown for Malliavin–Mancino estimator by E. Clement and A. Gloter is also established for the SIML estimator.

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Notes

  1. It has been pointed out that the “bias” \( \textbf{E} [ V^{j,j'} - \widehat{\Sigma ^{j,j'}_{n,m_n}}(0)] \) converges to zero and the mean square error \( \textbf{E} [ (V^{j,j'} - \widehat{\Sigma ^{j,j'}_{n,m_n}}(q) )^2 \) does not diverge when \( m_n = o (n) \) as \( n \rightarrow \infty \), which is not the case with the realized volatility.

  2. To be precise, they used the IBP technique to prove the convergence of \( \textrm{Res}_t \) in (3.1) and \( \langle M, W\rangle \), but used another approach to prove the convergence of \( I^2 \) for which the Malliavin differentiability for b is not required.

  3. Here we avoid using the commonly used notation D for the derivative so as not to mix it up with the Dirichlet kernel.

  4. As is remarked in Clément and Gloter (2011), the conditions (3.10) and (3.11) are not too strong. For example, the solution for a stochastic differential equation with smooth bounded coefficients naturally satisfies these conditions.

  5. Misaki and Kunitomo (2015) studied a random sampling scheme for SIML in a mild situation.

References

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Correspondence to Atsuhito Watanabe.

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A Appendices

A Appendices

1.1 A.1 A proof of Lemma 3.3

Let us put

$$\begin{aligned} D_m (x) := \frac{1}{2m} \frac{\sin m \pi x}{\sin \frac{\pi }{2}x}. \end{aligned}$$

By extending \( \varphi ^j \) and \( \varphi ^{j'} \) periodically, \( \mathcal {D}^{j,j'}_m (u,s) = D_m(\varphi ^j(u) + \varphi ^{j'} (s)) + D_m(\varphi ^j(u) - \varphi ^{j'}(s)) \) is periodic in both u and s with the period 2. Therefore, we have

$$\begin{aligned}&\sup _{s \in [0,1]} \int _0^1\big |\mathcal {D}_{m_n}^{j, j'}(u, s)\big |^p\,\textrm{d}u \\&\quad \le \sup _{c \in \textbf{R}} \left( \int _{c-1}^{c+1} \big |D_{m_n}(\varphi ^j(u)-c)\big |^p\,\textrm{d}u + \int _{-c-1}^{-c+1}\big |D_{m_n}(\varphi ^j(u)+c)\big |^p\,\textrm{d}u \right) . \end{aligned}$$

Hence, it is sufficient to show that

$$\begin{aligned} \limsup _{m_n,n \rightarrow \infty }m_{n}\sup _{c \in \mathbb {R}} \int _{c-1}^{c+1} \big |D_{m_n}(\varphi ^j(u)-c)\big |^p \textrm{d}u < \infty . \end{aligned}$$
(A.1)

We put \(a:=\limsup _{m_n, n \rightarrow \infty } m_n \rho _n \) which is in \( [ 0, \infty ) \) by assumption. Moreover, let

$$\begin{aligned} J_n^{1}(u):= \int _{c-\frac{1}{2}}^{c+\frac{1}{2}} {\varvec{1}}_{\{|u-c|>\frac{2a+2}{m_n}\}}\big |D_{m_n} ( \varphi ^j(u)-c)\big |^p \textrm{d}u \end{aligned}$$

and

$$\begin{aligned} J_n^{2}(u):= \int _{c-1}^{c+1} {\varvec{1}}_{ \{|u-c|\le \frac{2a+2}{m_n}\}}\big |D_{m_n} ( \varphi ^j(u)-c)\big |^p \textrm{d}u. \end{aligned}$$

For \(m_n\) and n large enough, we see that

$$\begin{aligned} \begin{aligned} |u-c|&=\big |\varphi ^{j}(u)-c-(\varphi ^{j}(u)-u)| \le \big |\varphi ^{j}(u)-c\big |+ \big |\varphi ^{j}(u)-u\big |\\&\le \big |\varphi ^{j}(u)-c\big |+\rho _n. \end{aligned} \end{aligned}$$
(A.2)

Therefore, we obtain

$$\begin{aligned} \begin{aligned} |u-c|>\frac{2a+2}{m_n}&\Rightarrow \big |\varphi ^{j}(u)-c\big | \ge \frac{2}{m_n}+\frac{2a-m_n\rho _n}{m_n}>\frac{2+a}{m_n} \\&\Rightarrow \frac{1}{2 m_n \big |\varphi ^{j}(u)-c\big |} \le \frac{1}{2} \frac{1}{2+a} < 1. \end{aligned} \end{aligned}$$
(A.3)

Since it holds that

$$\begin{aligned} \big |D_{m_n}(x)\big | \le 1 \wedge \frac{1}{2m_n x} \end{aligned}$$
(A.4)

for \( x \in \textbf{R} \), it follows from (A.3) that

$$\begin{aligned} \begin{aligned} J_n^{1}(u) \le \int _{c-1}^{c+1} {\varvec{1}}_{\{|u-c|>\frac{2a+2}{m_n}\}}\left| \frac{1}{2m_n(\varphi ^j(u)-c)}\right| ^p \textrm{d}u. \end{aligned} \end{aligned}$$
(A.5)

Since (A.2) implies, for sufficiently large n, \(|u-c|>\frac{2a+2}{m_n} \Rightarrow \left| \varphi ^{j}(u)-c\right| \ge |u-c|-\frac{2a}{m_n}\), one has

$$\begin{aligned} \begin{aligned}&\int _{c-1}^{c+1} {\varvec{1}}_{\{|u-c|>\frac{2a+2}{m_n}\}}\left| \frac{1}{2m_n(\varphi ^j(u)-c)}\right| ^p \textrm{d}u \\&\quad \le \int _{c-1}^{c+1} {\varvec{1}}_{\{|u-c|>\frac{2a+2}{m_n}\}}\left| \frac{1}{2m_n\left( |u-c|-\frac{2a}{m_n}\right) }\right| ^p \textrm{d}u,\\&\quad = \int _{c-1}^{c+1} {\varvec{1}}_{\{\frac{m_n}{2}|u-c|-a>1\}}\frac{1}{\left| 4\left( \frac{m_n}{2}|u-c|-a\right) \right| ^p} \,\textrm{d}u, \\ \end{aligned} \end{aligned}$$

(by changing variables with \( w = \frac{m_n}{2} |u-c|-a\))

$$\begin{aligned} \begin{aligned}&= 4 \int _{-a}^{\frac{m_n}{2}-a} 1_{\{w > 1\}} \frac{1}{4^p m_n} w^{-p} \,\textrm{d}w \le \left( \frac{1}{4}\right) ^p\frac{4}{m_n} \int _1^{\infty }\omega ^{-p}\textrm{d}\omega . \end{aligned} \end{aligned}$$
(A.6)

This establishes (A.1) since clearly one has

$$\begin{aligned} \begin{aligned} J_n^2(u) \le \int _{c-1}^{c+1} {\varvec{1}}_{\{|u-c|\le \frac{2a+2}{m_n}\}} \textrm{d}u \le \frac{4a+4}{m_n} \end{aligned} \end{aligned}$$

by (A.4). \(\square \)

1.2 A.2 A proof of Proposition 3.1

Under the condition that \( \rho _n m_n^2 \rightarrow 0 \), we have

$$\begin{aligned} m_n \int _0^1 \int _0^u \left( \mathcal {D}^{j,j'} (s,u) \mathcal {D}^{k,k'} (s,u) -(\mathcal {D} (s,u))^2 \right) f(s,u) \,\textrm{d}u \textrm{d}s \rightarrow 0 \quad (n \rightarrow \infty ). \end{aligned}$$

by a similar argument as the one we did for the proof of Lemma 2.1. Therefore, it suffices to prove that

$$\begin{aligned}&m \int _0^1 \int _0^u (\mathcal {D} (s,u))^2 f(s,u) \,\textrm{d}u \textrm{d}s -\frac{1}{2}\int _0^1 f(s,s) \, \textrm{d}s \rightarrow 0 \quad (n \rightarrow \infty ). \end{aligned}$$

We note that

$$\begin{aligned} \begin{aligned} \int _0^1 \int _0^u (\mathcal {D} (s,u))^2 f(s,u) \,\textrm{d}u \textrm{d}s = \frac{1}{2} \int _0^1 \int _0^1 |\mathcal {D} (s,u)|^2 f(s,u) \,\textrm{d}u \textrm{d}s \end{aligned} \end{aligned}$$

holds by extending f(su) from \( \{ (s,u): s\le u \} \) to [0, 1] symmetrically. Then, by letting

$$\begin{aligned} g(s) :=m\int _0^1 |\mathcal {D} (u,s)|^2\,\textrm{d}u, \end{aligned}$$

we have

$$\begin{aligned}&m \int _0^1 \int _0^u (\mathcal {D} (s,u))^2 f(s,u) \,\textrm{d}u \textrm{d}s -\frac{1}{2}\int _0^1 f(s,s) \, \textrm{d}s \nonumber \\&\quad = \frac{m}{2} \int _0^1 \int _0^1 |\mathcal {D} (s,u)|^2 ((f(s,u) -f(s,s))\,\textrm{d}u \textrm{d}s -\frac{1}{2}\int _0^1 (1-g(s))f(s,s) \, \textrm{d}s. \end{aligned}$$
(A.7)

On the other hand, it follows from the expression (2.9) that

$$\begin{aligned} \begin{aligned} g(u)&= \frac{4}{m}\sum _{l=1}^{m} \sum _{l'=1}^{m}\cos \left( l-\frac{1}{2}\right) \pi u \cos \left( l'-\frac{1}{2}\right) \pi u \int _0^1 \cos \\&\quad \times \left( l-\frac{1}{2}\right) \pi s \cos \left( l'-\frac{1}{2}\right) \pi s \,\textrm{d}s. \end{aligned} \end{aligned}$$

Since it holds that

$$\begin{aligned} \begin{aligned}&\int _0^1 \cos \left( l-\frac{1}{2}\right) \pi s \cos \left( l'-\frac{1}{2}\right) \pi s \,\textrm{d}s \\&\quad = \frac{1}{2} \int _0^1 \left( \cos (l+l'-1)\pi s + \cos (l-l')\pi s \right) \,\textrm{d}s \\&\quad = {\left\{ \begin{array}{ll} 1/2 &{} l=l' \\ 0 &{} l \ne l' \end{array}\right. }, \end{aligned} \end{aligned}$$

we have

$$\begin{aligned} \begin{aligned} g(u)&= \frac{2}{m}\sum _{l=1}^{m}\cos ^2\left( l-\frac{1}{2}\right) \pi u \\&= 1 + \frac{1}{m} \sum _{l=1}^m \cos (2l-1) \pi u \\&= 1+ \frac{1}{2m} \frac{\sin 2 m \pi u}{\sin 2 \pi u} = 1 + D_m (2u). \end{aligned} \end{aligned}$$

Then, by Lemma 3.3, we see that

$$\begin{aligned} \frac{1}{2}\int _0^1 (1-g(s))f(s,s) \, \textrm{d}s \rightarrow 0 \quad (m \rightarrow \infty ). \end{aligned}$$

Finally we shall prove the convergence of the first term in (A.7). Recalling (2.5), we have

$$\begin{aligned} \begin{aligned}&m \left| \int _0^1 \int _0^1 |\mathcal {D} (s,u)|^2 (f(s,u) -f(s,s))\,\textrm{d}u \textrm{d}s \right| \\&\quad \le \int _{[0,1]^2} 2m ((D_m (u+s))^2 + (D_m (u-s))^2) | f(s,u) -f(s,s) | \,\textrm{d}u \textrm{d}s. \end{aligned} \end{aligned}$$
(A.8)

We rely on the uniformly continuity of f. For arbitrary sufficiently small \( \varepsilon > 0 \), we can take \( \delta > 0 \) such that

$$\begin{aligned} |s-u|<\delta \Rightarrow |f(s,u)-f(s,s)| < \varepsilon . \end{aligned}$$

Let

$$\begin{aligned} A^+_\delta := \{ (s,u) \in [0,1]^2: \delta /2< s+u < \delta /2 \}, \end{aligned}$$

and

$$\begin{aligned} A^-_\delta := \{ (s,u) \in [0,1]^2: |s-u| < \delta \}. \end{aligned}$$

Then clearly \( (s,u) \in A^{\pm }_\delta \) satisfies \( |s-u|< \delta \) and therefore \( |f(s,u)-f(s,s)| < \varepsilon \). Then, we can bound the right-hand-side of (A.8) by

$$\begin{aligned} \begin{aligned} 4 \Vert f\Vert _\infty \left( \int _{A^+_\delta } \frac{1}{2m} \frac{\sin ^2 m \pi (u+s)}{\sin ^2 \pi (u+s)/2}\,\textrm{d}u \textrm{d}s + \int _{A^-_\delta } \frac{1}{2m} \frac{\sin ^2 m \pi (u-s)}{\sin ^2 \pi (u-s)/2}\,\textrm{d}u \textrm{d}s \right) \\ + \varepsilon \left( \int _{[0,1]^2} \frac{1}{2m} \left( \frac{\sin ^2 m \pi (u+s)}{\sin ^2 \pi (u+s)/2} + \frac{\sin ^2 m \pi (u-s)}{\sin ^2 \pi (u-s)/2} \right) \,\textrm{d}u \textrm{d}s \right) . \end{aligned} \end{aligned}$$

Since

$$\begin{aligned} \begin{aligned}&\int _{A^+_\delta } \frac{1}{2m} \frac{\sin ^2 m \pi (u+s)}{\sin ^2 \pi (u+s)/2}\,\textrm{d}u \textrm{d}s + \int _{A^-_\delta } \frac{1}{2m} \frac{\sin ^2 m \pi (u-s)}{\sin ^2 \pi (u-s)/2}\,\textrm{d}u \textrm{d}s\\&\quad \le 2 \int _\delta ^{1} \frac{1}{2my^2}\,\textrm{d}y \le \frac{1}{m\delta } \end{aligned} \end{aligned}$$

and

$$\begin{aligned} \begin{aligned}&\int _{[0,1]^2} \frac{1}{2m} \left( \frac{\sin ^2 m \pi (u+s)}{\sin ^2 \pi (u+s)/2} + \frac{\sin ^2 m \pi (u-s)}{\sin ^2 \pi (u-s)/2} \right) \,\textrm{d}u \textrm{d}s \\&\quad = \frac{1}{2m} \int _{[0,1]^2} \left( \left( \sum _{l=-m+1}^m e^{(l-\frac{1}{2})\pi (s+u)} \right) ^2 + \left( \sum _{l=-m+1}^m e^{(l-\frac{1}{2})\pi (s-u)} \right) ^2\right) \,\textrm{d}u \textrm{d}s = \frac{1}{2}, \end{aligned} \end{aligned}$$

we have

$$\begin{aligned} \begin{aligned}&m \left| \int _0^1 \int _0^1 |\mathcal {D} (s,u)|^2 (f(s,u) -f(s,s))\,\textrm{d}u \textrm{d}s \right| \le \frac{4 \Vert f\Vert _\infty }{m\delta } + \frac{\varepsilon }{2}, \end{aligned} \end{aligned}$$

which shows the convergence to zero (as \( m \rightarrow \infty \)) of the first term in (A.7). \(\square \)

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Akahori, J., Namba, R. & Watanabe, A. The SIML method without microstructure noise. Jpn J Stat Data Sci (2024). https://doi.org/10.1007/s42081-024-00249-y

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