The SIML Estimation of Integrated Covariance and Hedging Coefficient Under Round-off Errors, Micro-market Price Adjustments and Random Sampling

Abstract

For estimating the integrated volatility and covariance by using high frequency data, Kunitomo and Sato (Math Comput Simul 81:1272–1289, 2011; N Am J Econ Finance 26:289–309, 2013) have proposed the separating information maximum likelihood (SIML) method when there are micro-market noises. The SIML estimator has reasonable finite sample properties and asymptotic properties when the sample size is large when the hidden efficient price process follows a Brownian semi-martingale. We shall show that the SIML estimation is useful for estimating the integrated covariance and hedging coefficient when we have round-off errors, micro-market price adjustments and noises, and when the high-frequency data are randomly sampled. The SIML estimation is consistent, asymptotically normal in the stable convergence sense under a set of reasonable assumptions and it has reasonable finite sample properties with these effects.

Appendix: On Mathematical Derivations

In this Appendix, we give some details of the proof of Theorem 4.1. The method of our derivations is based on the results reported by Kunitomo and Sato (2013) and Sato and Kunitomo (2015), but we need some extra arguments. We shall use the notations $K_i\;(i\ge 1)$ as positive constants. Our proof consists of several steps.

(Step-1) The first step is to argue that the effects of the randomness of sampling and the effects of drift terms in the underlying stochastic processes are stochastically negligible under our assumptions. Our proof consists of several steps.

Although $t_i^n$ and $n^{*}$ are random variables, which are finite-valued and bounded in [0, 1] for any n. We write $y_i^a=x_i^a+v_i^a$ for $a=s$ or f ,  where $y_i^a=y_a({t_i^n})$ and $x_i^a=X_a({t_i^n})$ in the basic case. We set $y_i^n=(y^s(t_i^n),y^f(t_i^n))^{\prime },$ $x_i^n=(x^s(t_i^n),x^f(t_i^n))^{\prime }$ and also we write the underlying (unobservable) returns in the period $(t_{i-1}^n,t_i^n]$ as

$$\begin{aligned} r_i^{*}= & {} x_i^n-x_{i-1}^n =\int _{t_{i-1}^n}^{t_i^n}\mu _x (s,X(s))ds\nonumber \\&+\int _{t_{i-1}^n}^{t_i^n}\sigma _x (s,X(s))dB_s\,\,(i=1,\ldots ,n^{*}) \end{aligned}$$

(6.1)

and the martingale part as

$$\begin{aligned} r_i^n=\int _{t_{i-1}^n}^{t_i^n} \sigma _x (s,X(s)) dB_s\;\;(i=1,\ldots ,n^{*}) \end{aligned}$$

(6.2)

with $0=t_0\le t_1^n<\cdots <t_{n^{*}}^n\le 1$. By using Assumptions I and II, we have $n^{*}/n =1+o_p(1)$,

$$\begin{aligned} \mathbf{E}\left[ \left\| r_j^{*}\right\| ^2\right]= & {} \mathbf{E}\left[ \left\| \int _{t_{i-1}^n}^{t_i^n}\mu _x (s,X(s))ds \right| ^2\right] \\&+2\mathbf{E}\left[ \left( \int _{t_{i-1}^n}^{t_i^n}\mu _x (s,X(s))ds\right) ^{\prime } r_i^n\right] +\mathbf{E}\left[ \left\| r_i \right\| ^2\right] , \end{aligned}$$

and

$$\begin{aligned} \mathbf{E}\left[ \left\| \int _{t_{i-1}^n}^{t_i^n}\sigma _x(s)dB_s -\int _{t_{i-1}^n}^{t_i^n}\sigma _x\left( t_{i-1}^n\right) dB_s\right\| ^2\right] =O\left( \left( \frac{1}{n}\right) ^2\right) . \end{aligned}$$

Then we can evaluate as

$$\begin{aligned} \mathbf{E}\left[ \left\| r_i^{*} \right\| ^2-\int _{t_{i-1}^n}^{t_i^n}\mathrm{tr}(\Sigma _x(s))ds\right] =O\left( \left( \frac{1}{n}\right) ^{3/2}\right) . \end{aligned}$$

(6.3)

Hence we find that the effects of drift terms are negligible for the estimation of integrated volatility function under Assumptions I and II.

Similarly, we can use that $n^{*}/n\mathop {\rightarrow }\limits ^{p} 1$ and

$$\begin{aligned} \sum _{j=1}^{n^{*}} \int _{t_{j-1}^n}^{t_j^n}\Sigma _x(s) ds -\int _0^1 \Sigma _x(s) ds \mathop {\rightarrow }\limits ^{p} O\;, \end{aligned}$$

(6.4)

and then

$$\begin{aligned} \mathbf{E}\left[ \int _{t_{i-1}^n}^{t_i^n}\Sigma _x(s) ds\right] =O\left( \frac{1}{n}\right) . \end{aligned}$$

(6.5)

We note that the volatility function $\Sigma _x(s) \;(0\le s \le 1)$ and the integrated volatility $\Sigma _x =\int _0^1 \Sigma _x (s)ds$ can be stochastic.

By above arguments we can proceed the present proof as if $n^{*}$ were fixed and was replaced by the corresponding fixed n. For instance we can use the standard method for the random sums (martingales) to the following proof of CLT (central limit theorem). (See Section 35 of Billigsley (1995) for instance.)

Then we shall use the fixed n (and $m_{n^{*}}$) as if it were $n^{*}$ (and $m_n$) in the following developments of this Appendix for the sake of the resulting simplicities. (Basically we need to replace n by $n^{*}$ in each step, and then we need to use many straightforward, but tedious arguments because $n^{*}$ is stochastic.)

(Step-2) Let $Z_{in}^{(1)}$ and $Z_{in}^{(2)}\,(i=1,\ldots ,n)$ be the i-th elements of $n\times 2$ vectors

$$\begin{aligned} \mathbf{Z}_n^{(1)} =h_n^{-1/2}\mathbf{P}_n\mathbf{C}_n^{-1}(\mathbf{X}_n-{\bar{\mathbf{Y}}}_0),\,\mathbf{Z}_n^{(2)}=h_n^{-1/2}\mathbf{P}_n\mathbf{C}_n^{-1}\mathbf{V}_n, \end{aligned}$$

(6.6)

respectively, where we denote $\mathbf{X}_n=(x_{i}^{\prime })=(x_i^s,x_i^f)$, $\mathbf{V}_n=(v_{i}^{\prime })=(v_i^s,v_i^f)$, $\mathbf{X}_n=(z_{in}^{\prime })$ are $n\times 2$ vectors with $z_{in}=z_{in}^{(1)}+z_{in}^{(2)}$ and $\mathbf{P}_n$ is defined by (2.15).

We write $z_{kn}^s$ and $z_{kn}^f$ as the first and second components of $z_{kn}$, and also we use the notations $z_{kn}^{s,j}$ and $z_{kn}^{f,j};(j=1,2)$ for the j-th components of $z_{kn}^s$ and $z_{kn}^f$. By following the proof developed by Kunitomo and Sato (2013) for the case of fixed n, we use the arguments for investigating on the asymptotic distribution of $\sqrt{m_n}[{\hat{\sigma }}_{ss}^{(x)} - \sigma _{ss}^{(x)}]$ and $\sigma _{ss}^{(x)}=\int _0^1\sigma _{ss}^{(x)} (s) ds$. (We can use the similar arguments for $\sigma _{ff}^{(x)}=\int _0^1\sigma _{ff}^{(x)} (s) ds$ and $\sigma _{sf}^{(x)}=\int _0^1\sigma _{sf}^{(x)} (s) ds$, but we omit the details.) We use the decomposition

$$\begin{aligned} \sqrt{m_n}\left[ {\hat{\sigma }}_{ss}^{(x)} - \sigma _{ss}^{(x)} \right]= & {} \sqrt{m_n}\left[ \frac{1}{m_n}\sum _{k=1}^{m_n} \left( z_{kn}^s\right) ^2 -\sigma _{ss}^{(x)}\right] \nonumber \\= & {} \sqrt{m_n}\left[ \frac{1}{m_n}\sum _{k=1}^{m_n}\left( z_{kn}^{s,1}\right) ^2-\sigma _{ss}^{(x)}\right] + \frac{1}{\sqrt{m_n}}\sum _{k=1}^{m_n} \mathbf{E}\left[ \left( z_{kn}^{s,2}\right) ^2\right] \nonumber \\&+\frac{1}{\sqrt{m_n}}\sum _{k=1}^{m_n} \left[ \left( z_{kn}^{s,2}\right) ^2-\mathbf{E}\left[ \left( z_{kn}^{s,2}\right) ^2\right] \right] \nonumber \\&+2\frac{1}{\sqrt{m_n}}\sum _{k=1}^{m_n} \left[ \left( z_{kn}^{s,1}\right) \left( z_{kn}^{s,2}\right) \right] . \end{aligned}$$

(6.7)

Because under Assumption II the effects of differences due to $n^{*}$ and n are small, we notice that we can ignore the effects of $\sum _{m_{n^{*}}}^n ( z_{kn}^s )^2$ and $[m_{n^{*}}-m_n]/m_n$ in the following evaluations.

Then we shall investigate the conditions that three terms except the first one of (6.7) are $o_p(1)$. If they are satisfied, we could estimate the integrated volatility consistently as if there were no noise terms because other terms can be ignored asymptotically as $n\rightarrow \infty$.

Let $\mathbf{b}_k= (b_{kj})= \mathbf{e}_k^{\prime }\mathbf{P}_n\mathbf{C}_n^{-1}=(b_{kj})$ and $\mathbf{e}_k^{\prime }=(0,\ldots ,1,0,\ldots )$ be an $n\times 1$ vector. We write $z_{kn}^{(2)}=\sum _{j=1}^n b_{kj}v_{j}$ and use the relation

$$\begin{aligned} \left( \mathbf{P}_n\mathbf{C}_n^{-1}\mathbf{C}_n^{'-1}\mathbf{P}_n^{\prime }\right) _{k,k^{\prime }} =\delta (k,k^{\prime })4 \sin ^2 \left[ \frac{\pi }{2n+1}\left( k-\frac{1}{2}\right) \right] . \end{aligned}$$

Then because we have $n\sum _{j=1}^n b_{kj}b_{k^{\prime }j}=\delta (k,k^{\prime }) a_{kn},$ $\mathbf{E}(v_{ss}^{(v)})$ are bounded and (2.4), it is straightforward to find $K_1$ such that

$$\begin{aligned} \mathbf{E}\left[ \left( z_{kn}^{s,2}\right) \right] ^2 = n \mathbf{E}\left[ \sum _{i=1}^n b_{ki}v_i\sum _{j=1}^n b_{kj} v_{j}\right] \le K_1 \times a_{kn}, \end{aligned}$$

(6.8)

where we use the notation $b_{kj}=0\;(j\le 0)$. (We take $\rho =\max \{\rho _s,\rho _f\}$ and apply (32) of Kunitomo and Sato (2011).) Also Kunitomo and Sato (2013) have shown that

$$\begin{aligned} \frac{1}{m_n}\sum _{k=1}^{m_n} a_{k n} = \frac{1}{m_n} 2n \sum _{k=1}^{m_n} \left[ 1-\cos \left( \pi \frac{2k-1}{2n+1}\right) \right] = O\left( \frac{m_n^2}{n}\right) \end{aligned}$$

(6.9)

and the second term of (6.7) becomes

$$\begin{aligned} \frac{1}{\sqrt{m_n}}\sum _{k=1}^{m_n} \mathbf{E}\left[ z_{kn}^{s,2}\right] ^2 \le K_1 \frac{1}{\sqrt{m_n}}\sum _{k=1}^{m_n} a_{kn} =O\left( \frac{m_n^{5/2}}{n}\right) . \end{aligned}$$

(6.10)

This term is o(1) if $0< \alpha <0.4$. (When we use the MSIML estimation, the effect of this term can be removed from $\sqrt{m_n}[ \hat{\sigma _{m.ss}^{(x)}} - \sigma _{ss}^{(x)}]$ and we can improve the condition such that $0< \alpha < 0.5$ in Theorem 4.1.)

For the fourth term of (6.7),

$$\begin{aligned} \mathbf{E}\left[ \frac{1}{\sqrt{m_n}} \sum _{j=1}^{m_n} z_{kn}^{s,1}z_{kn}^{s,2} \right] ^2= & {} \frac{1}{m_n}\sum _{k,k^{\prime }=1}^{m_n} \mathbf{E} \left[ z_{kn}^{s,1}z_{k^{\prime },n}^{(1)} z_{kn}^{s,2}z_{k^{\prime },n}^{s,2} \right] \\= & {} \mathbf{E}\left[ 2\sum _{j,j^{\prime }=1}^ns_{jk}s_{j^{\prime }k^{\prime }} \mathbf{E}\left( r_jr_{j^{\prime }}\vert {\mathcal {F}}_{\min (j,j^{\prime })}\right) z_{kn}^{s,2}z_{k^{\prime }n}^{s,2} \right] \\= & {} O \left( \frac{m_n^2}{n}\right) , \end{aligned}$$

where

$$\begin{aligned} s_{jk}= \cos \left[ \frac{2\pi }{2n+1}\left( j-\frac{1}{2}\right) \left( k-\frac{1}{2}\right) \right] \end{aligned}$$

for $j,k=1,2,\ldots ,n$. (See Lemma 1.3 of Kunitomo and Sato (2008).) In the above evaluation we have used the relation

$$\begin{aligned} \left| \sum _{j=1}^n s_{jk}s_{j,k^{\prime }}\right| \le \left[ \sum _{j=1}^n s_{jk}^2\right] =n/2+1/4\,\,\mathrm{for\, any}\,k\ge 1. \end{aligned}$$

For the third term of (6.7), we need to consider the variance of

$$\begin{aligned} \left( z_{kn}^{s,2}\right) ^2-\mathbf{E}\left[ \left( z_{kn}^{s,2}\right) ^2\right] = n \sum _{j,j^{\prime }=1}^n b_{kj}b_{k,j^{\prime }} \left[ v_{j}^s v_{j^{a '}}-\mathbf{E}\left( v_{j}^s v_{j^{\prime }}^s\right) \right] \end{aligned}$$

and then it is enough to evaluate the expectation of $\left[ (z_{kn}^{s, 2})^2-\mathbf{E}[ (z_{kn}^{s, 2})^2 ]\right] \left[ (z_{k^{\prime },n}^{s, 2})^2-\mathbf{E}[ (z_{k^{\prime },n}^{s, 2})^2 ]\right]$. When $v_j^s$ are correlated and satisfy (2.4), becuse of the independence assumption on $w_i\;(i=1,\ldots ,n),$ it is possible to evaluate that there exists a positive constant $K_3$ such that

$$\begin{aligned} \mathbf{E}\left[ \frac{1}{\sqrt{m_n}}\sum _{j=1}^{m_n}\left( \left( z_{kn}^{s, 2}\right) ^2 -\mathbf{E}\left[ \left( z_{kn}^{s,2}\right) ^2\right] \right) \right] ^2\le & {} \frac{K_3}{m_n}\left( \sum _{k=1}^{m_n} a_{kn}\right) ^2 \nonumber \\&= O\left( \frac{1}{m_n}\times \left( \frac{m_n^2}{n}\right) ^2\right) .\qquad \end{aligned}$$

(6.11)

Thus the third term of (6.7) is negligible if $0< \alpha <0.4$. (see Section 5 of Kunitomo and Sato (2011). When $\{ v_j^s\}$ have the moving average (MA) representation as (2.4), the term is $O(m^5/n^2)$ because of (6.9) and then it can be negligible if if $0< \alpha <0.4$.)

When $\{ v_j^s\}$ are mutually independent, it is possible to evaluate that the left-hand side of (6.11) is less that $(K_3/m_n)\sum _{k=1}^{m_n}a_{kn}^2\;,$ which is $O(m_n^4/n^2)$. Then the third term of (6.7) is negligible if $0< \alpha <0.5$. (See (6.13) of Kunitomo and Sato (2013). This point the important step for the proof of Part (iii) because then the only concern is on (6.10).)

(Step-3) The third step is to give the asymptotic variance of the first term of (6.7), that is,

$$\begin{aligned} \sqrt{m_n}\left[ \frac{1}{m_n}\sum _{k=1}^{m_n}\left( z_{kn}^{s,(1)}\right) ^2 - \sigma _{ss}^{(x)}\right] \end{aligned}$$

(6.12)

because it is of the order $O_p(1)$. We write

$$\begin{aligned} z_{kn}^{s,(1)}= \sqrt{ \frac{n}{2n+1} }\sum _{j=1}^n r_j^{s} \left( e^{i \theta _{kj}}+e^{-i \theta _{kj}}\right) , \end{aligned}$$

(6.13)

where $r_j^{s}$ is the first component of $r_j^{*}\;(= x_j^n-x_{j-1}^n)$ in (6.1) and $\theta _{kj}=[2\pi /(2n+1)](k-1/2)(j-1/2)$. By using the relation that $(e^{i \theta _{kj}}+e^{-i \theta _{k j}})^2 =2+e^{2i \theta _{kj}}+e^{-2 i \theta _{kj}}$, we represent

$$\begin{aligned}&\left[ \frac{2n+1}{2n}\right] \left[ \frac{1}{m_n}\sum _{k=1}^{m_n} \left( z_{kn}^{s,(1)}\right) ^2 -\int _{0}^{1}\sigma _{ss}^{(x)}(s) ds \right] \nonumber \\&\quad =\frac{1}{m_n}\sum _{k=1}^{m_n} \left\{ \frac{1}{2}\left[ \sum _{j=1}^n \left( r_j^s\right) ^2 \left( e^{i \theta _{kj}}+e^{-i \theta _{kj}} \right) ^2 -2\left( 1+\frac{1}{2n}\right) \int _{0}^{1}\sigma _{ss}^{(x)}(s)ds\right] \right. \nonumber \\&\quad \quad \left. +\left[ \sum _{j\ne j^{\prime }=1}^{n} r_j^s r_{j^{\prime }}^s \left( e^{i \theta _{kj}}+e^{-i \theta _{kj}}\right) \left( e^{i \theta _{kj^{\prime }}}+e^{-i \theta _{kj^{\prime }}}\right) \right] \right\} \nonumber \\&\quad = 2\sum _{j>j^{\prime }=1}^{n} r_j^s r_{j^{\prime }}^s \left[ \frac{1}{m_n}\sum _{k=1}^{m_n}\left( e^{i \theta _{kj}}+e^{-i \theta _{kj}}\right) \left( e^{i \theta _{kj^{\prime }}}+e^{-i \theta _{kj^{\prime }}}\right) \right] \nonumber \\&\quad \quad +\sum _{j=1}^{n}\left[ \left( r_j^s\right) ^2-\int _{t_{j-1}^n}^{t_j^n}\sigma _{ss}^{(x)}(s)ds\right] \nonumber \\&\quad \quad + \frac{1}{2}\sum _{j=1}^{n}\left[ \left( r_j^s\right) ^2 -\int _{t_{j-1}}^{t_j^n}\sigma _{ss}^{(x)}(s)ds\right] \left[ \sum _{k=1}^{m_n}\left( e^{2i\theta _{kj}}+e^{-2i\theta _{kj}}\right) \right] \nonumber \\&\quad \quad -\frac{1}{2n} \sum _{j=1}^{n}\left[ \int _{t_{j-1}}^{t_j^n}\sigma _{ss}^{(x)}(s) ds\right] \nonumber \\&\quad =(A)+(B)+(C)+(D),(\mathrm{say}). \end{aligned}$$

(6.14)

Then by using the derivations in Kunitomo and Sato (2013) (also Sato and Kunitomo (2015)), it is possible to show that except the first term (A) $\sqrt{m_n} (B) \mathop {\rightarrow }\limits ^{p} 0,$ $\sqrt{m_n} (C) \mathop {\rightarrow }\limits ^{p} 0,$ and $\sqrt{m_n} (D) \mathop {\rightarrow }\limits ^{p} 0$ as $m_n \rightarrow \infty \;(n \rightarrow \infty )$. Also by using the simple relation

$$\begin{aligned} \left( e^{i\theta _{kj}}+e^{-i\theta _{k j}}\right) \left( e^{i\theta _{kj^{\prime } }}+e^{-i \theta _{kj^{\prime }}}\right)= & {} \left( e^{i\left( \theta _{kj}+\theta _{kj^{\prime }}\right) }+e^{-i\left( \theta _{kj}+\theta _{kj^{\prime }}\right) }\right) \\&+\left( e^{i\left( \theta _{kj}-\theta _{kj^{\prime }}\right) }+e^{-i\left( \theta _{kj}-\theta _{kj^{\prime }}\right) }\right) , \end{aligned}$$

the random quantity $\sqrt{m_n}\left[ \frac{1}{m_n}\sum _{k=1}^{m_n} (z_{kn}^{s,(1)})^2 -\sigma _{ss}^{(x)} \right]$ is asymptotically equivalent to

$$\begin{aligned} (A)^{\prime }=2\sum _{j> j^{\prime }=1}^{n} r_j^s r_{j^{\prime }}^s h_m(j,j^{\prime }), \end{aligned}$$

(6.15)

where for $j,j^{\prime }=1,\ldots ,n$

$$\begin{aligned} h_m(j,j^{\prime })= & {} \frac{1}{2\sqrt{m_n}} \frac{\sin 2m\left( \theta _j+\theta _{j^{\prime }}\right) /2}{\sin \left( \theta _j+\theta _{j^{\prime }}\right) /2} +\frac{1}{2\sqrt{m_n}} \frac{ \sin 2m\left( \theta _j-\theta _{j^{\prime }}\right) /2}{ \sin \left( \theta _j-\theta _{j^{\prime }}\right) /2} \end{aligned}$$

and $\theta _j=[2\pi /(2n+1)](j-1/2)$.

By extending the evaluation method on F$\acute{e}$je-kernel (see Chapter 8 of Anderson (1971)), we can derive the variance of the asymptotic distribution of $(A)^{\prime }$, which is asymptotically equivalent to (6.12).

Lemma 6.1

Under Assumption II, as $n \rightarrow \infty$ the asymptotic variance of $(A)^{\prime }$ is given by

$$\begin{aligned} V_{ss} = 2 \int _0^1 \left[ \sigma _{ss}^{(x)}( \tau (s)) \right] ^2 d(s)^2 ds. \end{aligned}$$

(6.16)

Proof of Lemma 6.1:

From the representation of (6.15), given the sampling process $t_j^n\;(j=1,\ldots ,n^{*})$ we have the conditional expectation

$$\begin{aligned}&\mathbf{E}[ ((A)^{\prime })^2 \vert \{ t_j^s\} ]\\&\quad = 2\sum _{j> j^{\prime }=1}^{n^{*}} \sigma _{ss.j}^{(x)}\sigma _{ss.j^{\prime }}^{(x)}\left[ \frac{1}{4m}\right] \\&\quad \quad \times \left[ \frac{ \sin 2m\pi \left( j+j^{\prime }-1\right) /\left( 2^{n^{*}}+1\right) }{ \sin \pi \left( j+j^{\prime }\right) /\left( 2^{n^{*}}+1\right) } +\frac{ \sin 2m\pi \left( j-j^{\prime }-1\right) /\left( 2^{n^{*}}+1\right) }{ \sin \pi \left( j-j^{\prime }\right) /\left( 2^{n^{*}}+1\right) }\right] ^2, \end{aligned}$$

where we use the notation

$$\begin{aligned} \sigma _{ss.j}^{(x)} =\mathbf{E}\left[ \int _{t_{j-1}^n}^{t_j^n}\sigma _{ss}^{(x)}(s) ds \vert t_{j-1}^n,t_j^n\right] . \end{aligned}$$

(6.17)

Under Assumption I by using the standard approximation argument on integrations we find that

$$\begin{aligned} \mathbf{E}\left[ \int _{t_{j-1}^n}^{t_j^n}\sigma _{ss}^{(x)}(s) ds -\int _{t_{j-1}^n}^{t_j^n}\sigma _{ss}^{(x)}\left( t_{j-1}^n\right) ds\right] =o\left( \frac{1}{n}\right) \end{aligned}$$

(6.18)

and

$$\begin{aligned} \int _{t_{j-1}^n}^{t_j^n}\sigma _{ss}^{(x)}(t_{j-1}) ds =\left( t_n^n - t_{j-1}^n\right) \sigma _{ss}^{(x)}\left( t_{j-1}^n\right) =O_p\left( \frac{1}{n}\right) . \end{aligned}$$

(6.19)

Also for any continuous functions a(s) and b(t) in [0, 1]

$$\begin{aligned}&2\int _0^1\int _0^1 \frac{1}{4m} \left[ \frac{\sin m\pi (s-t)}{\sin (\pi /2)(s-t)}+\frac{\sin m\pi (s+t)}{\sin (\pi /2)(s+t)}\right] ^2 a(s)b(t) ds dt\nonumber \\&\quad = 2\int _0^1\int _0^1 \frac{1}{4m}\left\{ \left[ \frac{\sin m\pi (s-t)}{ \sin (\pi /2)(s-t)}\right] ^2+\left[ \frac{ \sin m\pi (s+t)}{ \sin (\pi /2) (s+t)}\right] ^2\right. \nonumber \\&\quad \quad \left. +\left[ \frac{\sin m\pi (s-t)}{\sin (\pi /2)(s-t)}\right] \left[ \frac{\sin m\pi (s+t)}{\sin (\pi /2)(s+t)}\right] \right\} a(s)b(t) ds dt\nonumber \\&\quad =(E)+(F)+(G),(\mathrm{say}). \end{aligned}$$

(6.20)

By changing the order of integrations, as $m \rightarrow \infty$ we can evaluate the first term as

$$\begin{aligned}&2\int _0^1 \frac{1}{2m}\left[ \frac{\sin ^2(2m)\frac{\pi }{2}u}{\sin ^2(\pi /2)u}\right] \left[ \int _{0}^{1-u}a(u+t)b(t)dt\right] du\nonumber \\&\quad \longrightarrow 2 \lim _{u\rightarrow 0}\int _{0}^{1-u} a(u+t)b(t)dt\nonumber \\&=2 \int _0^1 a(t)b(t) dt. \end{aligned}$$

(6.21)

(See Lemma 8.3.3 of Anderson (1971).) Also by changing the order of integrations, we have found that the second term is negligible as

$$\begin{aligned} (F)= 2\int _0^1 \frac{1}{2m}\left[ \frac{\sin ^2(2m)\frac{\pi }{2}u}{ \sin ^2(\pi /2)u}\right] \left[ \int _0^u a(s)b(u-s)ds\right] du \longrightarrow 0\qquad \end{aligned}$$

(6.22)

as $m \rightarrow \infty$. By applying the similar argument to the third term and we can find that the third term (G) is also negligible when m is large. Under Assumption I and Assumption II, as $(i(n)-1)/n \rightarrow s$ and $(j(n)-1)/n \rightarrow t$ for n being large while a fixed m, we have $t_i^n \mathop {\rightarrow }\limits ^{p}\tau (s)$, $t_j^n\rightarrow \tau (t)$, $n(t_i^n-t_{i-1}^n) \mathop {\rightarrow }\limits ^{p} d(s)$, and $n(t_j^n-t_{j-1}^n) \mathop {\rightarrow }\limits ^{p} d(t)$. Then the only non-negligible term in (6.17) corresponds to

$$\begin{aligned} V^{\prime }= \int _0^1\int _0^1 \frac{1}{4m}\left[ \frac{\sin m\pi (s-t)}{\sin (\pi /2)(s-t)}\right] ^2 \sigma _{ss}^{(x)}(\tau (s))\sigma _{ss}^{(x)}(\tau (t))d(s)d(t) ds dt\;.\nonumber \\ \end{aligned}$$

(6.23)

Then by letting $m\rightarrow \infty ,$ we have the desired result as (6.16). $\square$

(Step-4) As the next step we need to show the stable convergence in law of (6.12), but the arguments are quite similar to (Step 4) of the proof of Theorem 3.2 in Sato and Kunitomo (2015), which is based on the method explained by Chapter VIII of Jacod and Shiryaev (2003) or Jacod and Protter (2012). Thus we have omitted the details of our arguments in this version.

(Step-5) Finally we need to deal with the integrated covariance. By modifying the derivations for the proof of the integrated covariance, we use $\hat{\sigma }_{ss}^{(x)} , \hat{\sigma }_{ff}^{(x)}$ and $\hat{\sigma }_{sf}^{(x)}$. (It is straightforward to develop the similar arguments as for $\hat{\sigma }_{ss}^{(x)}$ but they are tedious. Then we have omitted the details.) Then we can evaluate the variance of the asymptotic distributions of the integrated covariance SIML estimator. Then the resulting variance formula becomes

$$\begin{aligned} V_{sf} = \int _0^1 \left[ \sigma _{ss}^{(x)}(\tau (s)) \sigma _{ff}^{(x)}(\tau (s))+\left( \sigma _{sf}^{(x)}(\tau (s))\right) ^2 \right] d(s)^2 ds. \end{aligned}$$

(6.24)

$\square$