Estimating integrated co-volatility with partially miss-ordered high frequency data

Abstract

The covariation of short-time period returns between securities plays an important role in many area of finance. Under the wide availability of high frequency financial data, realized covariation, as an ex-post measure of the covariation, can accurately estimate the quadratic covariation. However, the realized covariation fails to work when the multiple records appear. In this paper, we propose an estimator of integrated covariation, which is robust to the high frequency data containing multiple records. Consistency of the estimator and central limit theorem have been established. Moreover, several extensions which make the estimator available to different types of high frequency data are also considered. Simulation study confirms the performance of the estimator.

Appendix

We need some notations, which simplify the presentation of the proof. For the process X, we set

$$\begin{aligned}&\displaystyle \alpha _i=\frac{1}{L_i}\sum _{k=1}^{L_i} \left( \frac{k-1}{L_i}\right) ^2,\quad \alpha '_i=\frac{1}{L_{i}}\sum _{k=1}^{L_{i}} \left( 1-\frac{k-1}{L_{i}}\right) ^2, \quad (\alpha _0=\alpha _0'=0)&\end{aligned}$$

(20)

$$\begin{aligned}&\displaystyle C_t=\int _0^t\rho \sigma ^X_s\sigma ^Y_sds, \quad \Delta _{i,j}C=C_{t_{N_{i-1}+j}}-C_{t_{N_{i-1}+j-1}}&\end{aligned}$$

(21)

$$\begin{aligned}&\displaystyle c_i = \sum _{j=1}^{L_i} \left( \frac{j-1}{L_i}\right) ^2\Delta _{i,j}C, \quad c_i'=\sum _{j=1}^{L_i}\left( 1-\frac{j-1}{L_i}\right) ^2\Delta _{i,j}C&\end{aligned}$$

(22)

$$\begin{aligned}&\displaystyle \xi _i(X) = \sum _{j=1}^{L_i} \left( \frac{j-1}{L_i}\right) \Delta _{i,j}X, \quad (\xi _0=0)&\end{aligned}$$

(23)

$$\begin{aligned}&\displaystyle \xi '_i(X)=\sum _{j=1}^{L_i} \left( 1-\frac{j-1}{L_i}\right) \Delta _{i,j}X,\quad (\xi '_0=0)&\end{aligned}$$

(24)

$$\begin{aligned}&\displaystyle \kappa _{i}(X)=\sigma ^X_{i\Delta _n}\sum _{j=1}^{L_{i}} \left( \frac{j-1}{L_{i}}\right) \Delta _{i,j}W^X,\quad (\kappa _{0}=0)&\end{aligned}$$

(25)

$$\begin{aligned}&\displaystyle \kappa '_{i}(X)=\sigma ^X_{(i-1)\Delta _n}\sum _{j=1}^{L_{i}} \left( 1-\frac{j-1}{L_{i}}\right) \Delta _{i,j}W^X, \quad (\kappa '_{0}=0)&\end{aligned}$$

(26)

$$\begin{aligned}&\displaystyle \mu _i(X) := \xi _{i}(X)+\xi '_{i+1}(X), \quad \theta _{i}(X):=\kappa _{i}(X)+\kappa '_{i+1}(X), \quad {\mathcal {F}}_i:=\mathcal {F}_{i\Delta _n\vee 0}.&\end{aligned}$$

(27)

Furthermore, by a standard localization procedure as presented in Barndorff-Nielsen et al. (2006) or Jacod (2012), there is without loss of generality to impose the following assumption addition to the assumptions of theorems.

Assumption 4

$\sigma ^X$ and $\sigma ^Y$ are bounded.

Thus, to prove Theorem 1, it remains to prove that

$$\begin{aligned} \sum _{i=0}^{n-1}\frac{\mu _i(X)\mu _i(Y)}{\varpi _i}\xrightarrow {\mathbb {P}}\int _0^t\rho \sigma ^X_s\sigma ^Y_sds. \end{aligned}$$

(28)

We start from an auxiliary lemma.

Lemma 1

Under the Assumptions 1–2 and 4, we have

$$\begin{aligned} \sum _{i=0}^{n-1}\frac{\big |\mu _i(X)\mu _i(Y)-\theta _{i}(X)\theta _{i}(Y)\big |}{\varpi _i} \xrightarrow {\mathbb {P}} 0,\end{aligned}$$

(29)

$$\begin{aligned} \sqrt{n}\sum _{i=0}^{n-1}\frac{\big |\mu _i(X)\mu _i(Y)-\theta _{i}(X)\theta _{i}(Y)\big |}{\varpi _i} \xrightarrow {\mathbb {P}} 0. \end{aligned}$$

(30)

Proof of Lemma 1

By Cauchy–Schwarz inequality, B urkholder inequality, Itô’s Isometry and the fundamental inequality

$$\begin{aligned} |x-y|\ge \big ||x|-|y|\big |, \end{aligned}$$

simple computations show that

$$\begin{aligned}&E\left[ \sum _{i=0}^{n-1}\frac{|\mu _i(X)\mu _i(Y)-\theta _{i}(X)\theta _{i}(Y)|}{\varpi _i}\right] \\&\qquad \le E\left[ \sum _{i=0}^{n-1}\frac{|\mu _i(X)-\theta _{i}(X)||\mu _{i}(Y)|+|\mu _{i}(Y)-\theta _i(Y)||\theta _i(X)|}{\varpi _i}\right] \\&\qquad =K\sum _{i=0}^{n-1}\left( \left( \int _{i\Delta _n}^{(i+1)\Delta _n}E[|\sigma ^X_s-\sigma ^X_{i\Delta _n}|^2]ds\Delta _n\right) ^{\frac{1}{2}}+ \left( \int _{i\Delta _n}^{(i+1)\Delta _n}E[|\sigma ^Y_s-\sigma ^Y_{i\Delta _n}|^2]ds\Delta _n\right) ^{\frac{1}{2}}\right) \\&\qquad \rightarrow 0, \end{aligned}$$

because $\{\sigma _s, s>0\}$ is cádlàg. $\square$

Proof of Theorem 1

In view of Lemma 1, it is enough to prove

$$\begin{aligned} \sum _{i=0}^{n-1}\frac{\theta _i(X)\theta _i(Y)}{\varpi _i}\xrightarrow {\mathbb {P}} \int _0^t\rho \sigma ^X_s\sigma ^Y_sds. \end{aligned}$$

(31)

We further let $\tilde{\theta }_i=\frac{\theta _i(X)\theta _i(Y)}{\varpi _i}$ and $\tilde{\theta }'_i=E[\tilde{\theta }_i|\mathcal {{F}}_{i}].$ We firstly have

$$\begin{aligned} E\left[ (\tilde{\theta }_i)^2|\mathcal {F}_i \right] \le K\Delta _n^2, \end{aligned}$$

(32)

and in particular $\tilde{\theta }'_i\le K\Delta _n$. Since $E\left[ (\tilde{\theta }_i-\tilde{\theta }'_i)(\tilde{\theta }_j-\tilde{\theta }'_j)\right] =0$ when $|i-j|\ge 2$, thus

$$\begin{aligned} E\left[ \left| \sum _{i=0}^{n-1}\tilde{\theta }_i-\sum _{i=0}^{n-1}\tilde{\theta }'_i\right| ^2\right]= & {} E\left[ \sum _{i=0}^{n-1}\sum _{j=0}^{n-1}(\tilde{\theta }_i-\tilde{\theta }'_i)(\tilde{\theta }_j-\tilde{\theta }'_j)\right] \end{aligned}$$

(33)

$$\begin{aligned}\le & {} K\sum _{i=0}^{n-1}E[(\tilde{\theta }_i-\tilde{\theta }'_i)^2]\le K\Delta _n\rightarrow 0, \end{aligned}$$

(34)

by Burkholder–Davis–Gundy’s inequality. Thus, to prove the theorem, it suffices to show

$$\begin{aligned} \sum _{i=0}^{n-1}\tilde{\theta }'_i\xrightarrow {\mathbb {P}} \int _0^t\rho \sigma _s^X\sigma ^Y_sds. \end{aligned}$$

(35)

Observe that

$$\begin{aligned}&\left| \sum _{i=0}^{n-1}\tilde{\theta }'_i-\int _0^t\rho \sigma _s^X\sigma ^Y_sds\right| \end{aligned}$$

(36)

$$\begin{aligned}&\qquad =\left| \sum _{i=0}^{n-1}\rho \sigma ^X_{(i-1)\Delta _n}\sigma ^Y_{(i-1)\Delta _n}\Delta _n-\int _0^t\rho \sigma ^X_s\sigma _s^Yds\right| . \end{aligned}$$

(37)

The required result follows from the Riemann integrability. $\square$

Proof of Theorem 2

To derive the above central limit theorem, note that the summands are one-step correlated, as illustrated in following picture. One of the methodologies to deal with the correlated variables is constructing the martingale differences, then using the martingale central limit theorem. This is similar to the central limit theorem of bi-power variation, as shown in Barndorff-Nielsen et al. (2006). Here, we apply the technique of “big blocks” and “small blocks”, which is classical method proving the central limit theorem of summation of correlated random variables, in probability theory. The extension to stochastic processes, have been developed in Jacod et al. (2009) and Vetter (2010). The “big blocks” (including p terms) are used to construct independent terms in the summation, and eventually dominates the asymptotic behavior. Whereas the “small blocks” (including only one term), which is asymptotically negligible, will be removed from the summation. Following is the detail. Giving a positive integer p, we define

$$\begin{aligned}&\displaystyle a_i(p)=i(p+1)+1, \quad b_i(p)=(i+1)(p+1),&\\&\displaystyle A_i(p)=\{k\in N^+: a_i(p)\le k<b_i(p)\}, \quad B_i(p)=\{k\in N^+: b_i(p)\le k<a_{i+1}(p)\},&\\&\displaystyle i_n(p)=\max \{i: b_i(p)\le n-1\}=\lfloor \frac{n-1}{p+1}\rfloor -1,&\\&\displaystyle j_n(p)=b_{i_n(p)}(p).&\end{aligned}$$

That is, the ith “big block” consists of $\{\mu _i: i\in A_i(p)\}$, whereas the ith ”small blocks” contains $\{\mu _i: i\in B_i(p)\}$. Because $\mu _i$ is 1-dependent sequence, after removing the small blocks, we get an independent (conditional) sequence. Denote:

$$\begin{aligned} \hat{\mu }_k= & {} \frac{1}{\varpi _i}\left\{ \sum _{j=1}^{L_k}\left( 1-\frac{j-1}{L_k}\right) ^2\Delta _{k,j}X\Delta _{k,j}Y\right. \\&\left. +\sum _{j=1}^{L_{k+1}}\left( 1-\frac{j-1}{L_{k+1}}\right) ^2\Delta _{k+1,j}X\Delta _{k+1,j}Y-\left( c_k+c'_{k+1}\right) \right\} . \end{aligned}$$

We first collect all the terms in $A_i(p)$, define as

$$\begin{aligned} \varsigma _i(p,1)=\sum _{j=a_i(p)}^{b_i(p)-1}\hat{\mu }_j, \quad \varsigma '_i(p,1)=E\left[ \varsigma _i(p,1)|\mathcal {{F}}_{a_i(p)-1}\right] . \end{aligned}$$

(38)

and all the terms in $B_i(p)$, define as

$$\begin{aligned} \varsigma _i(p,2)=\sum _{j=b_i(p)}^{a_{i+1}(p)-1}\hat{\mu }_k=\hat{\mu }_{b_i(p)}, \quad \varsigma '_i(p,2)=E\left[ \varsigma _i(p,2)|\mathcal {{F}}_{b_i(p)-1}\right] . \end{aligned}$$

(39)

Notice that $\varsigma _i(p,1)$ contains p summands (“big”), whereas $\varsigma _i(p,2)$ contains only 1 summand (“small”), because finally we will let $p\rightarrow \infty$, hence the small blocks are asymptotically negligible. To realize this, we set

$$\begin{aligned}&\displaystyle M(p)=\sum _{j=0}^{i_n(p)}\left[ \varsigma _j(p,1)-\varsigma '_j(p,1)\right] , \quad M'(p)=\sum _{j=0}^{i_n(p)}\varsigma '_j(p,1);&\end{aligned}$$

(40)

$$\begin{aligned}&\displaystyle N(p)=\sum _{j=0}^{i_n(p)}\left[ \varsigma _j(p,2)-\varsigma '_j(p,2)\right] , \quad N'(p)=\sum _{j=0}^{i_n(p)}\varsigma '_j(p,2);&\end{aligned}$$

(41)

$$\begin{aligned}&\displaystyle C(p)=\sum _{j=j_n(p)}^{n-1}\hat{\mu }_{j}.&\end{aligned}$$

(42)

We then have the following decomposition:

$$\begin{aligned}&\sqrt{n}\left( RCV_t^{weight}-\int _0^t\rho \sigma _s^X\sigma _s^Yds\right) \nonumber \\&\quad =\sqrt{n}\left( M(p)+M'(p)+N(p)+N'(p)+C(p)+R(p)\right) , \end{aligned}$$

(43)

where, R(p) consists of some “residuals” from above expansions. It is easy to follow the similar steps in Jacod et al. (2009) (identity (5.14), Lemma 5.5 and Lemma 5.6), we obtain that the last four terms are asymptotically negligible, when $p\rightarrow \infty$ and $n\rightarrow \infty$, i.e.,

$$\begin{aligned} \lim _{p\rightarrow \infty }\limsup _{n\rightarrow \infty }P\left( \sqrt{n}(|M'(p)|+|N(p)|+|N'(p)|+|C(p)|+|R(p)|)>\delta \right) =0,\quad \end{aligned}$$

(44)

for any $\delta >0$. We now establish the central limit theorem for M(p) by an auxiliary lemma. $\square$

Lemma 2

Under the same assumptions and notations as in Theorem 2 and the Assumption 4 holds, we have for any fixed p,

$$\begin{aligned} \sqrt{n}M(p)\xrightarrow {{\mathcal {L}}-(s)}\int _0^t\gamma (p)_sdB_s, \end{aligned}$$

(45)

where,

$$\begin{aligned} (\gamma (p)_s)^2=\frac{1}{p+1}\left( p(\rho ^2+1)+\frac{2(p-1)}{(\alpha +\alpha ')^2}\varrho \right) (\sigma ^X_s\sigma _s^Y)^2, \end{aligned}$$

(46)

$\varrho$ is defined in (60) later.

Proof of Lemma 2

Since $L_i\equiv L$, hence we have

$$\begin{aligned} \alpha _i=\alpha:= & {} \frac{1}{L}\sum _{k=1}^{L} \Big (\frac{k-1}{L}\Big )^2;\end{aligned}$$

(47)

$$\begin{aligned} \alpha '_i=\alpha ':= & {} \frac{1}{L}\sum _{k=1}^{L}\Big (1-\frac{k-1}{L}\Big )^2. \end{aligned}$$

(48)

By a martingale central limit theorem argument as presented in Theorem IX7.28 in Jacod and Shiryayev (2003), we need to show the following conditions:

$$\begin{aligned}&n\sum _{i=0}^{i_n(p)}E\left[ \varsigma ^2_i(p,1)|\mathcal {F}_{a_i(p)-1}\right] \xrightarrow {\mathbb {P}}\int _0^t(\gamma (p))^2ds,\end{aligned}$$

(49)

$$\begin{aligned}&n^2\sum _{i=0}^{i_n(p)}E\left[ \varsigma ^4_i(p,1)|\mathcal {F}_{a_i(p)-1}\right] \xrightarrow {\mathbb {P}} 0,\end{aligned}$$

(50)

$$\begin{aligned}&\sqrt{n}\sum _{i=0}^{i_n(p)}E\left[ \varsigma _i(p,1)\Delta W^X(p)_i|\mathcal {F}_{a_i(p)-1}\right] \xrightarrow {\mathbb {P}}0,\end{aligned}$$

(51)

$$\begin{aligned}&\sqrt{n}\sum _{i=0}^{i_n(p)}E\left[ \varsigma _i(p,1)\Delta W^Y(p)_i|\mathcal {F}_{a_i(p)-1}\right] \xrightarrow {\mathbb {P}}0,\end{aligned}$$

(52)

$$\begin{aligned}&\sqrt{n}\sum _{i=0}^{i_n(p)}E\left[ \varsigma _i(p,1)\Delta N(p)_i|\mathcal {F}_{a_i(p)-1}\right] \xrightarrow {\mathbb {P}}0, \end{aligned}$$

(53)

where $\Delta V(p)_i=V_{s_{b_i(p)}}-V_{s_{a_i(p)-1}}$ for any process V, and the Eq. (53) holds for any bounded martingale N which is orthogonal to both $W^X$ and $W^Y$ and also for $N=W^X$ and $N=W^Y$.

To show these convergences, we use the following approximations:

$$\begin{aligned} \kappa _{i,l}= & {} \sigma ^X_{l\Delta _n}\sigma ^Y_{l\Delta _n}\sum _{j=1}^{L_{i}}\left( \frac{j-1}{L_{i}}\right) ^2\Delta _{i,j}W^X\Delta _{i,j}W^Y, \quad (\kappa _{0}=0)\end{aligned}$$

(54)

$$\begin{aligned} \kappa '_{i,l}= & {} \sigma ^X_{l\Delta _n}\sigma ^Y_{l\Delta _n}\sum _{j=1}^{L_{i}}\left( 1-\frac{j-1}{L_{i}}\right) ^2\Delta _{i,j}W^X\Delta _{i,j}W^Y, \quad (\kappa '_{0}=0), \end{aligned}$$

(55)

for $0\le i\le n-1$ and integer l. Hence, we have the following new notations:

$$\begin{aligned} \theta _{i,l}:=\frac{1}{\varpi _i}\left\{ \kappa _{i,l}+\kappa '_{i+1,l}-(c_i+c'_{i+1})\right\} ; \quad \tilde{\theta }_{i,l}:=\theta _{i,l}-E\left[ \theta _{i,l}|\mathcal {{F}}_{l}\right] . \end{aligned}$$

and the following approximations are used for different kinds of “blocks”:

$$\begin{aligned} \bar{\mu }_k= \left\{ \begin{array}{c@{\quad }c} \tilde{\theta }_{k,a_i(p)-1},&{} k\in A_i(p)\\ \tilde{\theta }_{k,b_i(p)-1},&{} k\in B_i(p)\\ \tilde{\theta }_{k,j_n(p)-1},&{} k\ge j_n(p) \end{array} \right. . \end{aligned}$$

(56)

Using the approximations of Jacod et al. (2009) (Lemma 5.2 and Lemma 5.3), we only need to show (49)–(53) by redefining

$$\begin{aligned} \varsigma _i(p,1)=\sum _{j=a_i(p)}^{b_i(p)-1}\bar{\mu }_j, \quad \varsigma '_i(p,1)=E\left[ \varsigma _i(p,1)|\mathcal {{F}}_{a_i(p)-1}\right] . \end{aligned}$$

(57)

Direct calculation can show (50) and it is easy to see that

$$\begin{aligned} E\left[ \varsigma _i(p,1)\Delta W^X(p)_i|\mathcal {F}_{s_{a_i(p)}}\right] =0, \quad E\left[ \varsigma _i(p,1)\Delta W^Y(p)_i|\mathcal {F}_{s_{a_i(p)}}\right] =0. \end{aligned}$$

(58)

The proof of (53) is the same as Jacod et al. (2009) (Lemma 5.7) or Barndorff-Nielsen et al. (2006). We hence left to proving the Equation (49). By (47), we have

$$\begin{aligned} \alpha _{i}+\alpha '_{i}\equiv \alpha +\alpha '. \end{aligned}$$

(59)

Hence, recall (56), when $k,r\in A_i(p)$, we have

$$\begin{aligned}&E\left[ \bar{\mu }_{k}^2|\mathcal {F}_{a_i(p)-1}\right] =\left( \sigma ^X_{s_{a_i(p)-1}}\sigma ^Y_{s_{a_i(p)-1}}\right) ^2\left( \rho ^2+1\right) \Delta ^2, \\&E\left[ \bar{\mu }_k\bar{\mu }_{r}|\mathcal {F}_{a_i(p)-1}\right] = \left\{ \begin{array}{l@{\quad }l} 0, &{} |k-r|>1 \\ \frac{(\sigma ^X_{s_{a_i(p)-1}}\sigma ^Y_{s_{a_i(p)-1}})^2}{(\alpha +\alpha ')^2}\big (\rho ^2(A+B)+A-\rho ^2\alpha \alpha '\big )\Delta ^2, &{} |k-r|=1 \end{array} \right. \end{aligned}$$

where,

$$\begin{aligned} \left\{ \begin{array}{l} A=\frac{1}{L^6}\Big (\sum _{k=1}^L\big (k-1\big )\big (L-k+1\big )\Big )^2,\\ B=\frac{1}{L^6}\sum _{k=1}^L\big (k-1\big )^2\cdot \sum _{j=1}^L\big (L-j+1\big )^2=\alpha \alpha ',\\ \end{array} \right. \end{aligned}$$

Let

$$\begin{aligned} \varrho =\rho ^2(A+B)+A-\rho ^2\alpha \alpha '=(\rho ^2+1)A. \end{aligned}$$

(60)

Therefore, we obtain

$$\begin{aligned}&E\left[ \varsigma ^2_i(p,1)|\mathcal {F}_{s_{a_i(p)-1}}\right] =E\left[ \left( \sum _{k=a_i(p)}^{b_i(p)-1}\bar{\mu }_{k}\right) ^2\big |\mathcal {F}_{s_{a_i(p)-1}}\right] \\&\quad =E\left[ \sum _{k,r=a_i(p)}^{b_i(p)-1}\bar{\mu }_{k}\bar{\mu }_{r}\big |\mathcal {F}_{s_{a_i(p)-1}}\right] =\sum _{k,r=a_i(p)}^{b_i(p)-1}E\left[ \bar{\mu }_{k}\bar{\mu }_{r}\big |\mathcal {F}_{s_{a_i(p)-1}}\right] \\&\quad =\sum _{k=a_i(p)}^{b_i(p)-1}E\left[ \bar{\mu }_{k}^2|\mathcal {F}_{s_{a_i(p)-1}}\right] +\sum _{k\ne r, k,r=a_i(p)}^{b_i(p)-1}E\left[ \bar{\mu }_{k}\bar{\mu }_{r}\big |\mathcal {F}_{s_{a_i(p)-1}}\right] \\&\quad =p\left( \sigma ^X_{s_{a_i(p)-1}}\sigma ^Y_{s_{a_i(p)-1}}\right) ^2\left( \rho ^2+1\right) \Delta ^2 +2(p-1)\frac{(\sigma ^X_{s_{a_i(p)-1}}\sigma ^Y_{s_{a_i(p)-1}})2}{(\alpha +\alpha ')^2}\varrho \Delta ^2\\&\quad =\left( p(\rho ^2+1)+\frac{2(p-1)}{(\alpha +\alpha ')^2}\varrho \right) (\sigma ^X_{s_{a_i(p)-1}}\sigma ^Y_{s_{a_i(p)-1}})^2\Delta ^2. \end{aligned}$$

Hence by Riemann integrability, we obtain

$$\begin{aligned} n\sum _{i=0}^{i_n(p)}E\left[ \varsigma ^2_i(p,1)|\mathcal {F}_{s_{a_i(p)}}\right] \rightarrow ^P\int _0^t(\gamma (p)_s)^2ds, \end{aligned}$$

(61)

where,

$$\begin{aligned} (\gamma (p)_s)^2:=\frac{1}{p+1}\left( p(\rho ^2+1)+\frac{2(p-1)}{(\alpha +\alpha ')^2}\varrho \right) \sigma _s^4. \end{aligned}$$

(62)

$\square$

We now return to the proof of Theorem 2. In view of

$$\begin{aligned} (\gamma (p)_s)^2\rightarrow \gamma _s^2, \quad \text {when} \quad p\rightarrow \infty , \end{aligned}$$

we obtain the required result of Theorem 2.$\square$

Sketched Proof of Theorem 4:

For any process Z, let

$$\begin{aligned} \overline{\xi }_{k}(Z)=\sum _{j=1}^{k_n-1}g\left( \frac{j}{k_n}\right) \xi _{k+j-1}(Z), \quad \overline{\xi }'_k(Z)=\sum _{j=1}^{k_n-1}g\left( \frac{j}{k_n}\right) \xi '_{k+j-1}(Z). \end{aligned}$$

(63)

Then we have $\Delta _{i,k_n}X=\overline{\xi }_{i}(X)+\overline{\xi }'_{i+1}(X)$. Further let $\overline{\overline{\xi }}_{i}(Z)=\sqrt{k_n}\left( \overline{\xi }_{i}(Z)+\overline{\xi }'_{i+1}(Z)\right)$. Thus

$$\begin{aligned}&\frac{1}{k_n}\sum _{i=0}^{n-k_n}\Delta _{i,k_n}X(\epsilon )\Delta _{i,k_n}Y(\delta )=\frac{1}{n}\sum _{i=0}^{n-k_n}\left[ \overline{\overline{\xi }}_{i}(X)+\overline{\overline{\xi }}_{i}(\epsilon )\right] \left[ \overline{\overline{\xi }}_{i}(Y)+\overline{\overline{\xi }}_{i}(\delta )\right] \nonumber \\&\quad =\frac{1}{n}\sum _{i=0}^{n-k_n}\overline{\overline{\xi }}_{i}(X)\overline{\overline{\xi }}_{i}(Y)+\frac{1}{n}\sum _{i=0}^{n-k_n}\overline{\overline{\xi }}_{i}(X)\overline{\overline{\xi }}_{i}(\delta ) +\frac{2}{n}\sum _{i=0}^{n-k_n}\overline{\overline{\xi }}_{i}(\epsilon )\overline{\overline{\xi }}_{i}(Y)+\frac{2}{n}\sum _{i=0}^{n-k_n}\overline{\overline{\xi }}_{i}(\epsilon )\overline{\overline{\xi }}_{i}(\delta )\nonumber \\&\quad =:I_{1n}+I_{2n}+I_{3n}+I_{4n}. \end{aligned}$$

(64)

Routine techniques can show that

$$\begin{aligned} I_{2n}\xrightarrow {P}0, \qquad I_{3n}\xrightarrow {P}0. \end{aligned}$$

(65)

Firstly, using similar approximations as previous theorem, some computation yields

$$\begin{aligned} \left\{ \begin{array}{l} E\left[ \overline{\xi }_{i}(X)\overline{\xi }_{i}(Y)|{\mathcal {F}}_{s_i}\right] =k_n\bar{g}(2)\left( \frac{\Delta _n}{L}\right) \rho \sigma ^X_{s_i}\sigma ^Y_{s_i}\sum _{k=1}^L\left( \frac{k-1}{L}\right) ^2+e_{1n};\\ E\left[ \overline{\xi }_{i}(X)\overline{\xi }'_{i+1}(Y)|{\mathcal {F}}_{s_i}\right] =k_n\bar{g}(2)\left( \frac{\Delta _n}{L}\right) \rho \sigma ^X_{s_i}\sigma ^Y_{s_i}\sum _{k=1}^L\left( \frac{k-1}{L}\right) \left( 1-\frac{k-1}{L}\right) +e_{2n};\\ E\left[ \overline{\xi }'_{i+1}(X)\overline{\xi }_{i}(Y)|{\mathcal {F}}_{s_i}\right] =k_n\bar{g}(2)\left( \frac{\Delta _n}{L}\right) \rho \sigma ^X_{s_i}\sigma ^Y_{s_i}\sum _{k=1}^L\left( \frac{k-1}{L}\right) \left( 1-\frac{k-1}{L}\right) +e_{3n};\\ E\left[ \overline{\xi }_{i+1}'(X)\overline{\xi }_{i+1}(Y)|{\mathcal {F}}_{s_i}\right] =k_n\bar{g}(2)\left( \frac{\Delta _n}{L}\right) \rho \sigma ^X_{s_i}\sigma ^Y_{s_i}\sum _{k=1}^L\left( 1-\frac{k-1}{L}\right) ^2+e_{4n}. \end{array} \right. \end{aligned}$$

Notice that $\frac{1}{L}\sum _{k=1}^L\left[ \left( \frac{k-1}{L}\right) +\left( 1-\frac{k-1}{L}\right) \right] ^2=1$, thus

$$\begin{aligned} E\left[ \overline{\overline{\xi }}_{i}(X)\overline{\overline{\xi }}_{i}(Y)|{\mathcal {F}}_{s_i}\right] =\bar{g}(2)\rho \sigma ^X_{s_i}\sigma ^Y_{s_i}+e_n. \end{aligned}$$

(66)

Following the proof of Lemma 5.4 in Jacod et al. (2009), one can show that $\frac{1}{n}\sum _{i=0}^{n-k_n}e_n\xrightarrow {P} 0$. Hence

$$\begin{aligned} I_{1n}\xrightarrow {P}\bar{g}(2)\int _0^t\rho \sigma ^X_s\sigma ^Y_sds. \end{aligned}$$

(67)

Secondly, note that

$$\begin{aligned} \frac{1}{n}\sum _{i=0}^{n-k_n}\overline{\overline{\xi }}_{i}(\epsilon )\overline{\overline{\xi }}_{i}(\delta )= & {} \frac{ k_n}{n} \sum _{i=0}^{n-k_n}\Bigg \{\Big (\frac{1}{L}\sum _{j=0}^{k_n-1}\Big (g\Big (\frac{j}{k_n}\Big )-g\Big (\frac{j+1}{k_n}\Big )\Big )\sum _{k=1}^L\epsilon _{N_{i+j-1}+k}\Big )\end{aligned}$$

(68)

$$\begin{aligned}&\times \Big (\frac{1}{L}\sum _{j=0}^{k_n-1}\Big (g\Big (\frac{j}{k_n}\Big )-g\Big (\frac{j+1}{k_n}\Big )\Big )\sum _{k=1}^L\delta _{N_{i+j-1}+k}\Big )\Bigg \}. \end{aligned}$$

(69)

Recalling $k_n=\frac{1}{\sqrt{\Delta _n}}$ and $n=\lfloor \frac{t}{\Delta _n}\rfloor$, by Law of Large Numbers,

$$\begin{aligned} I_{4n}\xrightarrow {P}\frac{\bar{g'}(2)\varphi t}{L}. \end{aligned}$$

(70)

Finally, observe that

$$\begin{aligned}&\frac{1}{2n}\sum _{i=0}^{n-1}\left( \overline{X(\epsilon )}_{s_{i+1}}-\overline{X(\epsilon )}_{s_i})(\overline{Y(\delta )}_{s_{i+1}}-\overline{Y(\delta )}_{s_i}\right) \nonumber \\&\quad =\frac{1}{2n}\sum _{i=0}^{n-1}\left( \bar{X}_{s_{i+1}}-\bar{X}_{s_{i}}\right) \left( \bar{Y}_{s_{i+1}}-\bar{Y}_{s_{i}}\right) +\frac{1}{2n}\sum _{i=0}^{n-1}\left( \bar{X}_{s_{i+1}}-\bar{X}_{s_{i}}\right) \left( \bar{\delta }_{s_{i+1}}-\bar{\delta }_{s_{i}}\right) \nonumber \\&\qquad +\frac{1}{2n}\sum _{i=0}^{n-1}\left( \bar{\epsilon }_{s_{i+1}}-\bar{\epsilon }_{s_{i}}\right) \left( \bar{Y}_{s_{i+1}}-\bar{Y}_{s_{i}}\right) +\frac{1}{2n}\sum _{i=0}^{n-1}\left( \bar{\epsilon }_{s_{i+1}}-\bar{\epsilon }_{s_{i}}\right) \left( \bar{\delta }_{s_{i+1}}-\bar{\delta }_{s_{i}}\right) \nonumber \\&\quad = J_{1n}+J_{2n}+J_{3n}+J_{4n}. \end{aligned}$$

(71)

It is easy to show that $J_{kn}\xrightarrow {P}0$ for $k=1,2,3$ under Assumption 3. And

$$\begin{aligned} J_{4n}= & {} \frac{1}{2n}\sum _{i=0}^{n-1}\left( \bar{\epsilon }_{s_{i+1}}-\bar{\epsilon }_{s_{i}}\right) \left( \bar{\delta }_{s_{i+1}}-\bar{\delta }_{s_{i}}\right) \\= & {} \frac{1}{2n}\sum _{i=1}^{n-1}\bar{\epsilon }_{s_{i+1}}\bar{\delta }_{s_{i+1}}+\frac{1}{2n}\sum _{i=1}^{n-1}\bar{\epsilon }_{s_{i+1}}\bar{\delta }_{s_{i}} +\frac{1}{2n}\sum _{i=1}^{n-1}\bar{\epsilon }_{s_{i}}\bar{\delta }_{s_{i+1}}+\frac{1}{2n}\sum _{i=1}^{n-1}\bar{\epsilon }_{s_{i}}\bar{\delta }_{s_{i}}, \end{aligned}$$

where $\bar{\epsilon }_{s_i}=\frac{1}{L}\sum _{k=1}^L\epsilon _{t^N_{N_{i-1}+k}}$ with $\bar{\epsilon }_0=0$. Since $E[\bar{\epsilon }_{s_i}\bar{\delta }_{s_i}]=\frac{\varphi }{L}$ and $E[\bar{\epsilon }_{s^n_{i-1}}\bar{\epsilon }_{s^n_{i-1}}]=0$ by Assumption 3, we get

$$\begin{aligned} J_{4n}\xrightarrow {P}\frac{\varphi }{L}, \end{aligned}$$

(72)

by Law of Large Numbers. Combining (64)–(67), (70)–(72), we obtain the result of Theorem 4. $\square$