Information loss in volatility measurement with flat price trading

Abstract

A model of financial asset price determination is proposed that incorporates flat trading features into an efficient price process. The model involves the superposition of a Brownian semimartingale process for the efficient price and a Bernoulli process that determines the extent of flat price trading. The approach is related to sticky price modeling and the Calvo pricing mechanism in macroeconomic dynamics. A limit theory for the conventional realized volatility (RV) measure of integrated volatility is developed. The results show that RV is still consistent but has an inflated asymptotic variance that depends on the probability of flat trading. Estimated quarticity is similarly affected, so that both the feasible central limit theorem and the inferential framework suggested in Barndorff-Nielsen and Shephard (J Royal Stat Soc Ser B (Stat Methodol) 64:253–280, 2002) remain valid under flat price trading even though there is information loss due to flat trading effects. The results are related to work by Jacod (J Financ Econom 16:526–569, 2018) and Mykland and Zhang (Ann Stat 34:1931–1963, 2006) on realized volatility measures with random and intermittent sampling, and to ACD models for irregularly spaced transactions data. Extensions are given to include models with microstructure noise. Some simulation results are reported. Empirical evaluations with tick-by-tick data indicate that the effect of flat trading on the limit theory under microstructure noise is likely to be minor in most cases, thereby affirming the relevance of existing approaches.

Appendix

Proof of Theorem 2.1

The specification of p(t) implies that

$$\begin{aligned} p_{i,m}= & {} p_{i,m}^{*}\xi _{i,m}+p_{i-1,m}(1-\xi _{i,m}), \end{aligned}$$

(1.39)

$$\begin{aligned} p_{i,m}-p_{i-1,m}= & {} (p_{i,m}^{*}-p_{i-1,m})\xi _{i,m}, \end{aligned}$$

(1.40)

and

$$\begin{aligned} p_{i,m}^{*}-p_{i,m}=(p_{i,m}^{*}-p_{i-1,m})(1-\xi _{i,m}). \end{aligned}$$

(1.41)

Taking conditional expectations of both sides of equation (1.39), we have

$$\begin{aligned} E(p_{i,m}|{\mathcal {F}}_{i-1,m})= & {} E(p_{i,m}^{*}|{\mathcal {F}}_{i-1,m})\pi +p_{i-1,m}(1-\pi ) \nonumber \\= & {} E(p_{i-1,m}^{*}|{\mathcal {F}}_{i-1,m})\pi +p_{i-1,m}(1-\pi ) \end{aligned}$$

(1.42)

To compute $E(p_{i-1,m}^{*}|{\mathcal {F}}_{i-1,m})$, note that if $p_{i-1,m}\ne p_{i-2,m}$, then $p_{i-1,m}=p_{i-1,m}^{*}$ and hence $E(p_{i-1,m}^{*}|{\mathcal {F}}_{i-1,m})=p_{i-1,m}$. If $p_{i-1,m}=p_{i-2,m}$ but $p_{i-2,m}\ne p_{i-3,m}$, then $p_{i-2,m}=p_{i-2,m}^{*}$, $p_{i-1,m}^{*}=p_{i-2,m}^{*}+\varepsilon _{i-1,m}$, and ${\mathcal {F}} _{i-1,m}={\mathcal {F}}_{i-2,m}$. Hence $E(p_{i-1,m}^{*}|{\mathcal {F}} _{i-1,m})=E(p_{i-2,m}^{*}+\varepsilon _{i-1,m}|{\mathcal {F}} _{i-2,m})=p_{i-2,m}=p_{i-1,m}$. Similarly, if $p_{i-1,m}=\cdots =p_{i-K,m}$ but $p_{i-K,m}\ne p_{i-K=1,m}$, then $p_{i-K,m}=p_{i-K,m}^{*}$, $p_{i-1,m}^{*}=p_{i-K,m}^{*}+\varepsilon _{i-K+1,m}+\cdots +\varepsilon _{i-1,m}$ and ${\mathcal {F}}_{i-1,m}=\cdots ={\mathcal {F}}_{i-K,m}$ . Hence $E(p_{i-1,m}^{*}|{\mathcal {F}}_{i-1,m})=E(p_{i-K,m}^{*}+\varepsilon _{i-K+1,m}+\cdots +\varepsilon _{i-1,m}|{\mathcal {F}} _{i-K,m})=p_{i-K,m}=p_{i-1,m}$. In general, we have $E(p_{i-1,m}^{*}| {\mathcal {F}}_{i-1,m})=p_{i-1,m},$and so $E(p_{i,m}|{\mathcal {F}} _{i-1,m})=p_{i-1,m},$ as required. $\square$

Proof of Theorem 2.2

Let $K_{i}$ be the run time of flat trading prior to $t_{i,m}.$ As discussed in the paper, the maximum run time, ${\bar{K}} _{m}$, for a sequence of identical Bernoulli draws in a sample of size m has mean $E\left( {\bar{K}}_{m}\right) =O\left( \log _{1/\pi }\left\{ m\left( 1-\pi \right) \right\} \right) =O\left( \frac{\log \left\{ m\left( 1-\pi \right) \right\} }{\log \frac{1}{\pi }}\right)$ and variance $Var({\bar{K}} _{m})=\frac{\pi ^{2}}{6\log ^{2}\left( \frac{1}{\pi }\right) }.$ It follows that

$$\begin{aligned} {\bar{K}}_{m}=O_{p}\left( \log m\right) , \end{aligned}$$

(1.43)

and so each $K_{i}$ is at most $O_{p}\left( \log m\right) .$ It follows that in equi-spaced sampling when $t_{i,m}=\frac{i}{m},$ we have a maximum grid size $h_{{\bar{K}}_{m},m}=\frac{{\bar{K}}_{m}}{m}=O_{p}\left( \frac{\log m}{m} \right) \rightarrow 0$ as $m\rightarrow \infty .$ In the case of a general grid $\left\{ t_{i,m}\right\} ,$ the maximum grid size is bounded as follows for large m

$$\begin{aligned} h_{{\bar{K}}_{m},m}=\sup _{i}\left| t_{i,m}-t_{i-{\bar{K}}_{m},m}\right| \le \sup _{i}\left| t_{i,m}-t_{i-m^{\delta },m}\right| \text { for some }\delta >0. \end{aligned}$$

Hence, $h_{{\bar{K}}_{m},m}\rightarrow 0$ if $h_{m^{\delta },m}\rightarrow 0$ as $m\rightarrow \infty$ for some $\delta >0.$ In both cases, therefore, the grid size tends to zero and standard quadratic variation theory ensures that

$$\begin{aligned} \sum _{i=1}^{m}[p_{i,m}-p_{i-1,m}]^{2}\rightarrow _{p}\int _{0}^{1}\sigma ^{2}\left( t\right) \textrm{d}t. \end{aligned}$$

(1.44)

$\square$

The result may be proved directly by writing the left side of (1.44) in terms of the empirical quadratic variation of the efficient price $RV^{(m)}(p^{*})=\sum _{i=1}^{m}(p_{i,m}^{*}-p_{i-1,m}^{*})^{2}$ and showing that the error converges in probability to zero. The derivation is useful in later arguments so it is given here. In particular, from Eq. (1.40), we have

$$\begin{aligned} \sum _{i=1}^{m}[p_{i,m}-p_{i-1,m}]^{2}= & {} \sum _{i=1}^{m}(p_{i,m}^{*}-p_{i-1,m})^{2}\xi _{i,m}^{2} \nonumber \\= & {} \sum _{i=1}^{m}(p_{i,m}^{*}-p_{i-1,m})^{2}E[\xi _{i,m}^{2}]+\sum _{i=1}^{m}(p_{i,m}^{*}-p_{i-1,m})^{2} \nonumber \\{} & {} \quad \times \left( \xi _{i,m}^{2}-E[\xi _{i,m}^{2}]\right) =\pi \sum _{i=1}^{m}(p_{i,m}^{*}-p_{i-1,m})^{2}\nonumber \\{} & {} \quad +\sum _{i=1}^{m}(p_{i,m}^{*}-p_{i-1,m})^{2}\left( \xi _{i,m}^{2}-E[\xi _{i,m}^{2}]\right) . \end{aligned}$$

(1.45)

Write the sum $\sum _{i=1}^{m}(p_{i,m}^{*}-p_{i-1,m})^{2}$ in the first term above as follows

$$\begin{aligned} \sum _{i=1}^{m}(p_{i,m}^{*}-p_{i-1,m})^{2}= & {} \sum _{i=1}^{m}(p_{i,m}^{*}-p_{i-1,m}^{*}+p_{i-1,m}^{*}-p_{i-1,m})^{2} \nonumber \\= & {} RV^{(m)}(p^{*})+2\sum _{i=1}^{m}(p_{i,m}^{*}-p_{i-1,m}^{*})\left( p_{i-1,m}^{*}-p_{i-1,m}\right) \nonumber \\{} & {} +\sum _{i=1}^{m}(p_{i-1,m}^{*}-p_{i-1,m})^{2} \nonumber \\= & {} RV^{(m)}(p^{*})+2\sum _{i=1}^{m}\varepsilon _{i,m}\left( p_{i-1,m}^{*}-p_{i-1,m}\right) \nonumber \\{} & {} +\sum _{i=1}^{m}(p_{i-1,m}^{*}-p_{i-2,m})^{2}(1-\xi _{i-1,m})^{2} \nonumber \\= & {} RV^{(m)}(p^{*})+2\sum _{i=1}^{m}\varepsilon _{i,m}\left( p_{i-1,m}^{*}-p_{i-1,m}\right) \nonumber \\{} & {} +\sum _{i=1}^{m}(p_{i-1,m}^{*}-p_{i-2,m})^{2}(1-\pi ) \nonumber \\{} & {} +\sum _{i=1}^{m}(p_{i-1,m}^{*}-p_{i-2,m})^{2}\left\{ (1-\xi _{i-1,m})^{2}-(1-\pi )\right\} .\nonumber \\ \end{aligned}$$

(1.46)

Set $A_{m}=\sum _{i=1}^{m}(p_{i,m}^{*}-p_{i-1,m})^{2}$. So $A_{m-1}=A_{m}-(p_{m,m}^{*}-p_{m-1,m})^{2}$. Substituting out $A_{m-1}$ in Eq. (1.46) gives

$$\begin{aligned} A_{m}= & {} RV^{(m)}(p^{*})+2\sum _{i=1}^{m}\varepsilon _{i,m}\left( p_{i-1,m}^{*}-p_{i-1,m}\right) +A_{m}(1-\pi ) \\{} & {} -(1-\pi )(p_{m,m}^{*}-p_{m-1,m})^{2}+\sum _{i=1}^{m}(p_{i-1,m}^{*}-p_{i-2,m})^{2}\\{} & {} \times \left\{ (1-\xi _{i-1,m})^{2}-(1-\pi )\right\} . \end{aligned}$$

Hence

$$\begin{aligned} A_{m}= & {} \frac{1}{\pi }RV^{(m)}(p^{*})+\frac{2}{\pi }\sum _{i=1}^{m} \varepsilon _{i,m}\left( p_{i-1,m}^{*}-p_{i-1,m}\right) \nonumber \\{} & {} -\frac{1-\pi }{\pi }(p_{m,m}^{*}-p_{m-1,m})^{2}\nonumber \\{} & {} +\frac{1}{\pi } \sum _{i=1}^{m}(p_{i-1,m}^{*}-p_{i-2,m})^{2}\left\{ (1-\xi _{i-1,m})^{2}-(1-\pi )\right\} . \end{aligned}$$

(1.47)

Substituting (1.47) into (1.45) we have

$$\begin{aligned} \sum _{i=1}^{m}[p_{i,m}-p_{i-1,m}]^{2}= & {} RV^{(m)}(p^{*})+2\sum _{i=1}^{m}\varepsilon _{i,m}\left( p_{i-1,m}^{*}-p_{i-1,m}\right) \nonumber \\{} & {} -(1-\pi )(p_{m,m}^{*}-p_{m-1,m})^{2} \nonumber \\{} & {} +\sum _{i=1}^{m}(p_{i-1,m}^{*}-p_{i-2,m})^{2}\left\{ (1-\xi _{i-1,m})^{2}-(1-\pi )\right\} \nonumber \\{} & {} +\sum _{i=1}^{m}(p_{i,m}^{*}-p_{i-1,m})^{2}\left( \xi _{i,m}^{2}-\pi \right) \nonumber \\= & {} RV^{(m)}(p^{*})+\underset{A}{\underbrace{2\sum _{i=1}^{m}\varepsilon _{i,m}\left( p_{i-1,m}^{*}-p_{i-1,m}\right) }}\nonumber \\{} & {} \underset{B}{\underbrace{ -(1-\pi )(p_{m,m}^{*}-p_{m-1,m})^{2}}} \nonumber \\{} & {} \underset{C}{\underbrace{-2\sum _{i=1}^{m}(p_{i,m}^{*}-p_{i-1,m})^{2}\xi _{i,m}(1-\xi _{i,m})}} \nonumber \\= & {} RV^{(m)}(p^{*})+A+B+C. \end{aligned}$$

(1.48)

Since $\xi _{i,m}$ is a Bernoulli variable, $\xi _{i,m}\left( 1-\xi _{i,m}\right) =0$ a.s.,  and so $C=0$. Consider term B. Note that for some duration $K_{m}\ge 1$ and for which at most $K_{m}=O_{p}(\log m)$ we have

$$\begin{aligned} p_{m,m}^{*}-p_{m-1,m}=p_{m,m}^{*}-p_{m-K_{m},m}^{*}=\int _{t_{m-K_{m},m}}^{1}\sigma (s)\textrm{d}B(s)=O_{p}\left( \sqrt{h_{{\bar{K}}_{m},m} }\right) , \end{aligned}$$

so that

$$\begin{aligned} B=-(1-\pi )(p_{m,m}^{*}-p_{m-1,m})^{2}=-(1-\pi )O_{p}\left( h_{{\bar{K}} _{m},m}\right) =o_{p}(1), \end{aligned}$$

(1.49)

For term A, note that $\varepsilon _{i,m}=\int _{t_{i-1,m}}^{t_{i,m}}\sigma (s)\textrm{d}B(s)$ and

$$\begin{aligned} p_{i-1,m}^{*}-p_{i-1,m}= & {} (p_{i-1,m}^{*}-p_{i-2,m})(1-\xi _{i-1,m})\nonumber \\= & {} (p_{i-1,m}^{*}-p_{i-K_{i-1},m}^{*})(1-\xi _{i-1,m}), \end{aligned}$$

(1.50)

for some $K_{i-1}\ge 2$ and where $K_{i-1}$ is at most of $O_{p}(\log m).$ Then

$$\begin{aligned} \sum _{i=1}^{m}\varepsilon _{i,m}\left( p_{i-1,m}^{*}-p_{i-1,m}\right) =\sum _{i=1}^{m}\varepsilon _{i,m}(p_{i-1,m}^{*}-p_{i-K_{i-1},m}^{*})(1-\xi _{i-1,m}). \end{aligned}$$

Now $\varepsilon _{i,m}$ is independent of $\xi _{i-1,m},$ and $E(\varepsilon _{i,m})=0$ and $Var(\varepsilon _{i,m})\rightarrow 0$, as $m\rightarrow \infty ,$ because $h_{{\bar{K}}_{m},m}=o\left( 1\right) ,$ while $\sum _{i=1}^{m}(p_{i-1,m}^{*}-p_{i-K_{i-1},m}^{*})^{2}(1-\xi _{i-1,m})^{2}$ is bounded as $m\rightarrow \infty .$ It follows that $A=o_{p}(1).$ Thus, $\sum _{i=1}^{m}[p_{i,m}-p_{i-1,m}]^{2}=RV^{(m)}(p^{*})+o_{p}(1),$ as required.

Proof Theorem 2.3

From (1.48) we have

$$\begin{aligned}{} & {} \sqrt{m}\left\{ \sum _{i=1}^{m}[p_{i,m}-p_{i-1,m}]^{2}-IV\right\} \nonumber \\{} & {} \quad =\sqrt{m} \left\{ RV^{(m)}(p^{*})-IV\right\} \nonumber \\{} & {} \qquad +\sqrt{m}2\sum _{i=1}^{m}\varepsilon _{i,m}\left( p_{i-1,m}^{*}-p_{i-1,m}\right) \nonumber \\{} & {} \qquad -\sqrt{m}(1-\pi )(p_{m,m}^{*}-p_{m-1,m})^{2} \nonumber \\{} & {} \qquad +\sqrt{m}\sum _{i=1}^{m}(p_{i-1,m}^{*}-p_{i-2,m})^{2}\left\{ (1-\xi _{i-1,m})^{2}-(1-\pi )\right\} \nonumber \\{} & {} \qquad +\sqrt{m}\sum _{i=1}^{m}(p_{i,m}^{*}-p_{i-1,m})^{2}\left( \xi _{i,m}^{2}-\pi \right) \nonumber \\{} & {} \quad =\sqrt{m}\left\{ RV^{(m)}(p^{*})-IV\right\} +\sqrt{m}A+\sqrt{m}B+\sqrt{ m}C \nonumber \\{} & {} \quad =\sqrt{m}\left\{ RV^{(m)}(p^{*})-IV\right\} +\sqrt{m}A+\sqrt{m}B, \end{aligned}$$

(1.51)

since $C=0,$ a.s. . Standard theory (Barndorff-Nielsen and Shephard 2002; Barndorff-Nielsen et al. 2006; Jacod 2018) gives the CLT

$$\begin{aligned} \sqrt{m}\left\{ RV^{(m)}(p^{*})-IV\right\} \overset{d}{\rightarrow }\ {\mathcal {N}}\left( 0,2\int _{0}^{1}\sigma ^{4}(t)\textrm{d}t\right) , \end{aligned}$$

(1.52)

stably as $m\rightarrow \infty .$ We now study the asymptotic behavior of $\sqrt{m}A,$ and $\sqrt{m}B$.

For term $\sqrt{m}A$, from (1.50) we get

$$\begin{aligned} \sqrt{m}A= & {} \sqrt{m}2\sum _{i=1}^{m}\varepsilon _{i,m}(p_{i-1,m}^{*}-p_{i-K_{i-1},m}^{*})(1-\xi _{i-1,m}) \\= & {} 2\sqrt{m}\sum _{i=1}^{m}\int _{t_{i-1,m}}^{t_{i,m}}\sigma (s)\textrm{d}B(s)(p_{i-1,m}^{*}-p_{i-K_{i-1},m}^{*})(1-\xi _{i-1,m}) \\= & {} 2\sqrt{m}\sum _{i=1}^{m}\nu _{i,m}(p_{i-1,m}^{*}-p_{i-K_{i-1},m}^{*}), \end{aligned}$$

where $\nu _{i,m}=\int _{t_{i-1,m}}^{t_{i,m}}\sigma (s)\textrm{d}B(s)(1-\xi _{i-1,m})$ is uncorrelated with $(p_{i-1,m}^{*}-p_{i-K_{i-1},m}^{*}),$ because of the martingale property, and has mean 0 and conditional variance $\left( 1-\pi \right) m\int _{t_{i-1,m}}^{t_{i,m}}\sigma (s)^{2}\textrm{d}s.$ So $\sqrt{ m}A$ is a martingale with conditional variance

$$\begin{aligned} m\left( 1-\pi \right) \sum _{i=1}^{m}(p_{i-1,m}^{*}-p_{i-K_{i-1},m}^{*})^{2}\int _{t_{i-1,m}}^{t_{i,m}}\sigma (s)^{2}\textrm{d}s. \end{aligned}$$

By stochastic Taylor series expansion we have

$$\begin{aligned}{} & {} p_{i-1,m}^{*}-p_{i-K_{i-1},m}^{*}\nonumber \\{} & {} \quad =\int _{t_{i-K_{i-1},m}}^{t_{i-1,m}}\sigma (s)\textrm{d}B(s) \nonumber \\{} & {} \quad =\left\{ \sigma \left( t_{i-K_{i-1},m}\right) +O_{p}\left( \sqrt{K_{i-1}h} \right) \right\} \left( B\left( t_{i-1,m}\right) -B\left( t_{i-K_{i-1},m}\right) \right) \nonumber \\{} & {} \quad =\sigma \left( t_{i-K_{i-1},m}\right) \left( B\left( t_{i-1,m}\right) -B\left( t_{i-K_{i-1},m}\right) \right) +O_{p}\left( K_{i-1}h\right) \nonumber \\{} & {} \quad =\sigma \left( \frac{i-1}{m}\right) \left( B\left( \frac{i-1}{m}\right) -B\left( \frac{i-K_{i-1}}{m}\right) \right) +O_{p}\left( \frac{K_{i-1}}{m} \right) \end{aligned}$$

(1.53)

on the equispaced grid $\{t_{i,m}=\frac{i}{m}:i=0,\ldots ,m\}$ with $h=\frac{1}{ m}.$ Then

$$\begin{aligned}{} & {} m\sum _{i=1}^{m}(p_{i-1,m}^{*}-p_{i-K_{i-1},m}^{*})^{2}\int _{t_{i-1,m}}^{t_{i,m}}\sigma (s)^{2}\textrm{d}s \\{} & {} \quad =m\sum _{i=1}^{m}\left\{ \sigma \left( \frac{i-1}{m}\right) \left( B\left( \frac{i-1}{m}\right) -B\left( \frac{i-K_{i-1}}{m}\right) \right) +O_{p}\left( \frac{K_{i-1}}{m}\right) \right\} ^{2} \\{} & {} \qquad \times \left\{ \sigma \left( \frac{i-1}{m}\right) ^{2}+O_{p}\left( \frac{1 }{\sqrt{m}}\right) \right\} \frac{1}{m} \\{} & {} \quad =\sum _{i=1}^{m}\left\{ \sigma \left( \frac{i-1}{m}\right) ^{4}\left( B\left( \frac{i-1}{m}\right) -B\left( \frac{i-K_{i-1}}{m}\right) \right) ^{2} \left[ 1+O_{p}\left( \frac{K_{i-1}}{m}\right) \right] \right\} \\{} & {} \quad =\sum _{i=1}^{m}\left\{ \sigma \left( \frac{i-1}{m}\right) ^{4}\frac{ K_{i-1}-1}{m}\left[ 1+O_{p}\left( \frac{K_{i-1}}{m}\right) \right] \right\} \\{} & {} \quad \rightarrow _{p}\left( \int _{0}^{1}\sigma ^{4}(t)\textrm{d}t\right) E(K_{i-1}-1). \end{aligned}$$

It follows by the martingale central limit theorem (Hall and Heyde 1980, Theorem 3.2) that

$$\begin{aligned} \sqrt{m}A\overset{d}{\rightarrow }A_{\infty }=_{d}2\times \mathcal{M}\mathcal{N}\left( 0,\left( 1-\pi \right) \left( \int _{0}^{1}\sigma ^{4}(t)\textrm{d}t\right) E(K_{i-1}-1)\right) , \end{aligned}$$

stably in the sense that $\left\{ Z,\sqrt{m}A\right\} \overset{d}{ \rightarrow }\left\{ Z,A_{\infty }\right\}$ for $Z=\int _{0}^{1}\sigma ^{4}(t)\textrm{d}t.$ Observe that

$$\begin{aligned} K_{i-1}-2=\left\{ \begin{array}{cl} 0 &{} \text { with probability }\pi \\ 1 &{} \text { with probability }\pi (1-\pi ) \\ 2 &{} \text { with probability }\pi (1-\pi )^{2} \\ \vdots &{} \end{array} \right. , \end{aligned}$$

so that $E(K_{i-1}-2)=\pi (1-\pi )+2\pi (1-\pi )^{2}+\cdots =\frac{1-\pi }{ \pi },$ which implies that $E(K_{i-1}-1)=\frac{1}{\pi }$. Thus

$$\begin{aligned} \sqrt{m}A\overset{d}{\rightarrow }\mathcal{M}\mathcal{N}\left( 0,4\frac{1-\pi }{\pi } \int _{0}^{1}\sigma ^{4}(t)\textrm{d}t\right) , \end{aligned}$$

(1.54)

stably.

Next consider term $\sqrt{m}B$. From (1.49) we have

$$\begin{aligned} \sqrt{m}B=-\sqrt{m}O_{p}\left( \frac{\log m}{m}\right) (1-\pi )=o_{p}(1). \end{aligned}$$

(1.55)

Observe that the components of $\sqrt{m}A$ involve the product $\varepsilon _{i,m}(p_{i-1,m}^{*}-p_{i-K_{i-1},m}^{*})(1-\xi _{i-1,m})$ whereas $\sqrt{m}\left\{ RV^{(m)}(p^{*})-IV\right\}$ is a centred quadratic in the increments $\varepsilon _{i,m}=p_{i,m}^{*}-p_{i-1,m}^{*}$, so that $\sqrt{m}A$ and $\sqrt{m}\left\{ RV^{(m)}(p^{*})-IV\right\}$ are asymptotically uncorrelated. By combining (1.54) and (1.52), it follows that

$$\begin{aligned} \sqrt{m}\left\{ RV^{(m)}(p)-IV\right\} \overset{d}{\rightarrow }\mathcal{M}\mathcal{N}\left( 0, \frac{4-2\pi }{\pi }\int _{0}^{1}\sigma ^{4}(t)\textrm{d}t\right) , \end{aligned}$$

giving the required result. $\square$

Proof of Lemma 2.4

From Eq. (1.40), we have

$$\begin{aligned} m\sum _{i=1}^{m}[p_{i,m}-p_{i-1,m}]^{4}= & {} \pi m\sum _{i=1}^{m}(p_{i,m}^{*}-p_{i-1,m})^{4}\nonumber \\{} & {} +m\sum _{i=1}^{m}(p_{i,m}^{*}-p_{i-1,m})^{4}\left( \xi _{i,m}^{4}-E[\xi _{i,m}^{4}]\right) . \end{aligned}$$

(1.56)

Consider $\sum _{i=1}^{m}(p_{i,m}^{*}-p_{i-1,m})^{4}$ in the first term, giving

$$\begin{aligned}{} & {} \sum _{i=1}^{m}(p_{i,m}^{*}-p_{i-1,m})^{4} \nonumber \\{} & {} \quad =\sum (p_{i,m}^{*}-p_{i-1,m}^{*}+p_{i-1,m}^{*}-p_{i-1,m})^{4} \nonumber \\{} & {} \quad =\sum _{i=1}^{m}\left( p_{i,m}^{*}-p_{i-1,m}^{*}\right) ^{4}+4\sum _{i=1}^{m}\left( p_{i,m}^{*}-p_{i-1,m}^{*}\right) ^{3}\left( p_{i-1,m}^{*}-p_{i-1,m}\right) \nonumber \\{} & {} \quad \quad +\,6\sum _{i=1}^{m}\left( p_{i,m}^{*}-p_{i-1,m}^{*}\right) ^{2}\left( p_{i-1,m}^{*}-p_{i-1,m}\right) ^{2} \nonumber \\{} & {} \quad \quad +\,4\sum _{i=1}^{m}\left( p_{i,m}^{*}-p_{i-1,m}^{*}\right) \left( p_{i-1,m}^{*}-p_{i-1,m}\right) ^{3}+\sum _{i=1}^{m}\left( p_{i-1,m}^{*}-p_{i-1,m}\right) ^{4} \nonumber \\{} & {} \quad =\sum _{i=1}^{m}\left( p_{i,m}^{*}-p_{i-1,m}^{*}\right) ^{4}+4\sum _{i=1}^{m}\left( p_{i,m}^{*}-p_{i-1,m}^{*}\right) ^{3}\left( p_{i-1,m}^{*}-p_{i-1,m}\right) \nonumber \\{} & {} \quad \quad +\,6\sum _{i=1}^{m}\left( p_{i,m}^{*}-p_{i-1,m}^{*}\right) ^{2}\left( p_{i-1,m}^{*}-p_{i-1,m}\right) ^{2} \nonumber \\{} & {} \quad \quad +\,4\sum _{i=1}^{m}\left( p_{i,m}^{*}-p_{i-1,m}^{*}\right) \left( p_{i-1,m}^{*}-p_{i-1,m}\right) ^{3}\nonumber \\{} & {} \quad \quad +m\sum _{i=1}^{m}\left( p_{i-1,m}^{*}-p_{i-2,m}\right) ^{4}(1-\pi ) \nonumber \\{} & {} \quad \quad +\,\sum _{i=1}^{m}\left( p_{i-1,m}^{*}-p_{i-2,m}\right) ^{4}((1-\xi _{i-1,m})^{4}-(1-\pi )). \end{aligned}$$

(1.57)

Set $B_{m}=\sum _{i=1}^{m}(p_{i,m}^{*}-p_{i-1,m})^{4}$. So $B_{m-1}=B_{m}-(p_{m,m}^{*}-p_{m-1,m})^{4}$. Substituting out $B_{m-1}$ in Eq. (1.57) and solving for $B_{m}$, we get

$$\begin{aligned} B_{m}= & {} \frac{1}{\pi }\sum _{i=1}^{m}\left( p_{i,m}^{*}-p_{i-1,m}^{*}\right) ^{4}+\frac{4}{\pi }\sum _{i=1}^{m}\left( p_{i,m}^{*}-p_{i-1,m}^{*}\right) ^{3}\left( p_{i-1,m}^{*}-p_{i-1,m}\right) \nonumber \\{} & {} +\,\frac{6}{\pi }\sum _{i=1}^{m}\left( p_{i,m}^{*}-p_{i-1,m}^{*}\right) ^{2}\left( p_{i-1,m}^{*}-p_{i-1,m}\right) ^{2} \nonumber \\{} & {} +\,\frac{4}{\pi }\sum _{i=1}^{m}\left( p_{i,m}^{*}-p_{i-1,m}^{*}\right) \left( p_{i-1,m}^{*}-p_{i-1,m}\right) ^{3} \nonumber \\{} & {} -\,\frac{1-\pi }{\pi }(p_{m,m}^{*}-p_{m-1,m})^{2}\nonumber \\{} & {} +\,\frac{1}{\pi } \sum _{i=1}^{m}(p_{i-1,m}^{*}-p_{i-2,m})^{4}\left\{ (1-\xi _{i-1,m})^{4}-(1-\pi )\right\} . \end{aligned}$$

(1.58)

Substituting (1.58) into (1.56) we have

$$\begin{aligned} m\sum [p_{i,m}-p_{i-1,m}]^{4}= & {} m\sum _{i=1}^{m}\left( p_{i,m}^{*}-p_{i-1,m}^{*}\right) ^{4}\nonumber \\{} & {} +\,4m\sum _{i=1}^{m}\left( p_{i,m}^{*}-p_{i-1,m}^{*}\right) ^{3}\left( p_{i-1,m}^{*}-p_{i-1,m}\right) \nonumber \\{} & {} +\,6m\sum _{i=1}^{m}\left( p_{i,m}^{*}-p_{i-1,m}^{*}\right) ^{2}\left( p_{i-1,m}^{*}-p_{i-1,m}\right) ^{2} \nonumber \\{} & {} +\,4m\sum _{i=1}^{m}\left( p_{i,m}^{*}-p_{i-1,m}^{*}\right) \left( p_{i-1,m}^{*}-p_{i-1,m}\right) ^{3}\nonumber \\{} & {} -\,(1-\pi )(p_{m,m}^{*}-p_{m-1,m})^{2} \nonumber \\{} & {} +\,m\sum (p_{i-1,m}^{*}-p_{i-2,m})^{4}\left\{ (1-\xi _{i-1,m})^{4}-(1-\pi )\right\} \nonumber \\{} & {} +\,m\sum _{i=1}^{m}(p_{i,m}^{*}-p_{i-1,m})^{4}\left( \xi _{i,m}^{4}-\pi \right) \nonumber \\= & {} m\sum _{i=1}^{m}\left( p_{i,m}^{*}-p_{i-1,m}^{*}\right) ^{4}\nonumber \\{} & {} +\, \underset{A}{\underbrace{4m\sum _{i=1}^{m}\left( p_{i,m}^{*}-p_{i-1,m}^{*}\right) ^{3}\left( p_{i-1,m}^{*}-p_{i-1,m}\right) }} \nonumber \\{} & {} +\,\underset{B}{\underbrace{6m\sum _{i=1}^{m}\left( p_{i,m}^{*}-p_{i-1,m}^{*}\right) ^{2}\left( p_{i-1,m}^{*}-p_{i-1,m}\right) ^{2} }} \nonumber \\{} & {} +\,\underset{C}{\underbrace{4m\sum _{i=1}^{m}\left( p_{i,m}^{*}-p_{i-1,m}^{*}\right) \left( p_{i-1,m}^{*}-p_{i-1,m}\right) ^{3}}} \nonumber \\{} & {} \underset{D}{\underbrace{-(1-\pi )m(p_{m,m}^{*}-p_{m-1,m})^{4}}} \nonumber \\{} & {} +\,\underset{E}{\underbrace{m\sum (p_{i,m}^{*}-p_{i-1,m})^{4}(1-\xi _{i,m})^{2}\xi _{i,m}^{2}}} \nonumber \\= & {} m\sum _{i=1}^{m}\left( p_{i,m}^{*}-p_{i-1,m}^{*}\right) ^{4}+A+B+C+D+E. \nonumber \\ \end{aligned}$$

(1.59)

As in (1.53) we have

$$\begin{aligned} p_{i,m}^{*}-p_{i-1,m}^{*}= & {} \sigma \left( t_{i-1,m}\right) \left( B\left( t_{i,m}\right) -B\left( t_{i,m}\right) \right) +O_{p}\left( \frac{1}{ m}\right) \nonumber \\= & {} \sigma \left( t_{i-1,m}\right) \frac{\epsilon _{i,m}}{\sqrt{m}} +O_{p}\left( \frac{1}{m}\right) , \end{aligned}$$

(1.60)

where $\epsilon _{i,m}$ is iid ${\mathcal {N}}\left( 0,1\right) .$ Hence

$$\begin{aligned} m\sum _{i=1}^{m}\left( p_{i,m}^{*}-p_{i-1,m}^{*}\right) ^{4}= & {} \sum _{i=1}^{m}\sigma \left( t_{i-1,m}\right) ^{4}\frac{\epsilon _{i,m}^{4} }{m}+O_{p}\left( \frac{1}{m^{1/2}}\right) \nonumber \\= & {} \sum _{i=1}^{m}\sigma \left( t_{i-1,m}\right) ^{4}\frac{E\left( \epsilon _{i,m}^{4}\right) }{m}+O_{p}\left( \frac{1}{m^{1/2}}\right) \nonumber \\{} & {} \rightarrow _{p}\,3\int _{0}^{1}\sigma ^{4}(t)\textrm{d}t. \end{aligned}$$

(1.61)

Hence

$$\begin{aligned} \frac{2}{3}m\sum _{i=1}^{m}\left( p_{i,m}^{*}-p_{i-1,m}^{*}\right) ^{4}\rightarrow _{p}2\int _{0}^{1}\sigma ^{4}(t)\textrm{d}t. \end{aligned}$$

(1.62)

This corresponds with the result obtained in Barndorff-Nielsen and Shephard (2002).

We now consider the limit behavior of terms A, B, C, D,  and E. First, for term E, since $\left( 1-\xi _{i,m}\right) ^{2}\xi _{i,m}^{2}=0$ almost surely, $E=0$. Second, for term D, note that

$$\begin{aligned} D=-(1-\pi )m(p_{m,m}^{*}-p_{m-1,m})^{4}=-(1-\pi )m\times O_{p}\left( \frac{\log ^{2}m}{m^{2}}\right) =o_{p}(1). \end{aligned}$$

Next, consider term A, viz.,

$$\begin{aligned}{} & {} m\sum _{i=1}^{m}\left( p_{i,m}^{*}-p_{i-1,m}^{*}\right) ^{3}\left( p_{i-1,m}^{*}-p_{i-1,m}\right) \nonumber \\{} & {} \quad =m\sum _{i=1}^{m}\left( p_{i,m}^{*}-p_{i-1,m}^{*}\right) ^{3}\left( p_{i-1,m}^{*}-p_{i-2,m}\right) (1-\xi _{i-1,m}) \nonumber \\{} & {} \quad =m\sum _{i=1}^{m}\left\{ \sigma \left( t_{i-1,m}\right) \frac{\epsilon _{i,m}}{\sqrt{m}}+O_{p}\left( \frac{1}{m}\right) \right\} ^{3}\left( p_{i-1,m}^{*}-p_{i-K_{i-1},m}^{*}\right) (1-\xi _{i-1,m}) \nonumber \\{} & {} \quad \text {for some }K_{i-1}=O_{p}(\log m) \nonumber \\{} & {} \quad =\frac{1}{\sqrt{m}}\sum _{i=1}^{m}\sigma ^{3}\left( t_{i-1,m}\right) \epsilon _{i,m}^{3}\left( p_{i-1,m}^{*}-p_{i-K_{i-1},m}^{*}\right) (1-\xi _{i-1,m})+o_{p}\left( 1\right) . \end{aligned}$$

(1.63)

The component $\sigma ^{3}\left( t_{i-1,m}\right) \epsilon _{i,m}^{3}\left( p_{i-1,m}^{*}-p_{i-K_{i-1},m}^{*}\right) (1-\xi _{i-1,m})$ in the sum (1.63) has mean zero and conditional variance $E\left[ \epsilon _{i,m}^{6}\right] \left( 1-\pi \right) \sigma ^{6}\left( t_{i-1,m}\right) \left( p_{i-1,m}^{*}-p_{i-K_{i-1},m}^{*}\right) ^{2}=O_{p}\left( \frac{\log m}{m}\right)$ since from (1.53)

$$\begin{aligned} p_{i-1,m}^{*}-p_{i-K_{i-1},m}^{*}= & {} \sigma \left( \frac{i-1}{m} \right) \left( B\left( \frac{i-1}{m}\right) -B\left( \frac{i-K_{i-1}}{m} \right) \right) +O_{p}\left( \frac{K_{i-1}}{m}\right) \nonumber \\= & {} \sigma \left( \frac{i-1}{m}\right) \frac{\eta _{K_{i-1}}}{\sqrt{m}} +O_{p}\left( \frac{K_{i-1}}{m}\right) =O_{p}\left( \sqrt{\frac{\log m}{m}} \right) , \end{aligned}$$

(1.64)

where

$$\begin{aligned} \eta _{K_{i-1}}:=B\left( \frac{i-1}{m}\right) -B\left( \frac{i-K_{i-1}}{m} \right) =O_{p}\left( \sqrt{\log m}\right) . \end{aligned}$$

It follows that

$$\begin{aligned} E\left\{ \frac{1}{\sqrt{m}}\sum _{i=1}^{m}\sigma ^{3}\left( t_{i-1,m}\right) \epsilon _{i,m}^{3}\left( p_{i-1,m}^{*}-p_{i-K_{i-1},m}^{*}\right) (1-\xi _{i-1,m})\right\} ^{2}=O_{p}\left( \frac{\log m}{m}\right) , \end{aligned}$$

and so $A=o_{p}(1)$. Similarly, for term C, $4m\sum \left( p_{i,m}^{*}-p_{i-1,m}^{*}\right) \left( p_{i-1,m}^{*}-p_{i-1,m}\right) ^{3}$ is $o_{p}(1)$.

Next, consider term B. Using (1.40) and (1.64), we have

$$\begin{aligned}{} & {} m\sum _{i=1}^{m}\left( p_{i,m}^{*}-p_{i-1,m}^{*}\right) ^{2}\left( p_{i-1,m}^{*}-p_{i-1,m}\right) ^{2} \nonumber \\{} & {} \quad =m\sum _{i=1}^{m}\left( p_{i,m}^{*}-p_{i-1,m}^{*}\right) ^{2}\left( p_{i-1,m}^{*}-p_{i-2,m}\right) ^{2}(1-\xi _{i-1,m})^{2} \nonumber \\{} & {} \quad =m\sum _{i=1}^{m}\left( p_{i,m}^{*}-p_{i-1,m}^{*}\right) ^{2}\left( p_{i-1,m}^{*}-p_{i-K_{i-1},m}^{*}\right) ^{2}(1-\xi _{i-1,m})^{2} \nonumber \\{} & {} \qquad \text { for some }K_{i-1}=O_{p}(\log m) \nonumber \\{} & {} \quad =m\sum _{i=1}^{m}\left\{ \sigma \left( t_{i-1,m}\right) \frac{\epsilon _{i,m}}{\sqrt{m}}+O_{p}\left( \frac{1}{m}\right) \right\} ^{2} \nonumber \\{} & {} \qquad \times \left( \sigma \left( t_{i-1,m}\right) \frac{\eta _{K_{i-1}}}{\sqrt{m}}+O_{p}\left( \frac{ \log m}{m}\right) \right) ^{2}(1-\xi _{i,m})^{2} \nonumber \\{} & {} \quad =\frac{1}{m}\sum _{i=1}^{m}\sigma ^{2}(t_{i,m})\epsilon _{i,m}^{2}\sigma \left( t_{i-1,m}\right) ^{2}\eta _{K_{i-1}}^{2}(1-\xi _{i-1,m})^{2}+O_{p}\left( \sqrt{\frac{\log m}{m}}\right) \nonumber \\{} & {} \quad =\frac{1}{m}\sum _{i=1}^{m}\sigma ^{4}(t_{i,m})E\left[ \epsilon _{i,m}^{2}\eta _{K_{i-1}}^{2}(1-\xi _{i-1,m})^{2}\right] \nonumber \\{} & {} \qquad +\frac{1}{m}\sum _{i=1}^{m}\sigma ^{4}(t_{i,m})\left\{ \epsilon _{i,m}^{2}\eta _{K_{i-1}}^{2}(1-\xi _{i-1,m})^{2}-E\left[ \epsilon _{i,m}^{2}\eta _{K_{i-1}}^{2}(1-\xi _{i-1,m})^{2}\right] \right\} \nonumber \\{} & {} \qquad +O_{p}\left( \sqrt{\frac{\log m}{m}}\right) \nonumber \\{} & {} \quad =\frac{1}{m}\sum _{i=1}^{m}\sigma ^{4}\left( t_{i,m}\right) (1-\pi )E(K_{i-1}-1)+O_{p}\left( \frac{\log m}{\sqrt{m}}\right) \nonumber \\{} & {} \quad \rightarrow _{p} \,\left( \int _{0}^{1}\sigma ^{4}(t)\textrm{d}t\right) (1-\pi )E(K_{i-1}-1)=\frac{(1-\pi )}{\pi }\int _{0}^{1}\sigma ^{4}(t)\textrm{d}t, \end{aligned}$$

(1.65)

since $\epsilon _{i,m},$ $\eta _{K_{i-1}},$ and $\xi _{i-1,m}$ are independent. Therefore,

$$\begin{aligned} m\sum \left[ p_{i,m}-p_{i-1,m}\right] ^{4}= & {} 3\int _{0}^{1}\sigma ^{4}(t)\textrm{d}t+6 \frac{(1-\pi )}{\pi }\int _{0}^{1}\sigma ^{4}(t)\textrm{d}t+o_{p}(1) \\= & {} \frac{6-3\pi }{\pi }\int _{0}^{1}\sigma ^{4}(t)\textrm{d}t+o_{p}(1), \end{aligned}$$

leading to the required result. $\square$

Proof Theorem 2.5

This follows directly from Lemma 2.4 and Theorem 2.3. $\square$

Proof Theorem 3.1

This proof is given in the Online Supplement to the paper. $\square$

Proof Theorem 4.1

As in Zhang et al. (2005) we set $K=cm^{2/3}$ for some constant $c>0.$ Then

$$\begin{aligned} \frac{K}{{\bar{m}}}=\frac{K^{2}}{m}=c^{2}m^{1/3}. \end{aligned}$$

The two time scale measures involving the efficient price series are denoted by $\left[ p^{*}\right] ^{(avg)}$ and $\left[ p^{*}\right] ^{ {\mathcal {G}}_{m}},$ and the corresponding measures using the actual price series are denoted by $\left[ p\right] ^{(avg)}$ and $\left[ p\right] ^{ {\mathcal {G}}_{m}}.$

The two time scale estimator is $\widehat{\left[ p^{*}\right] }=\left[ p \right] ^{(avg)}-\frac{{\bar{m}}}{J_{m}}\left[ p\right] ^{{\mathcal {G}}_{m}}.$ Decompose the estimation error as

$$\begin{aligned} \widehat{\left[ p^{*}\right] }-\int _{0}^{1}\sigma ^{2}(t)\textrm{d}t=\left( \widehat{\left[ p^{*}\right] }-\left[ p^{*}\right] ^{(avg)}\right) +\left( \left[ p^{*}\right] ^{(avg)}-\int _{0}^{1}\sigma ^{2}(t)\textrm{d}t\right) . \end{aligned}$$

(1.66)

From the proof of Theorem A.1 and equation (56) in Zhang et al. (2005), we have

$$\begin{aligned} \sqrt{\frac{K}{{\bar{m}}}}\left\{ \widehat{\left[ p^{*}\right] }-\left[ p^{*}\right] ^{(avg)}\right\} \Rightarrow {\mathcal {N}}\left( 0,8\sigma _{u}^{4}\right) , \end{aligned}$$

(1.67)

and their proof applies in the present case with only minor notational changes to account for random $\tau _{j}$ sampling under Assumption D2 (ii).

The second component of (1.66) is the discretization effect and is the same as that in Zhang et al. (2005), after allowing for the fact that we have random stopping times and adjusting accordingly, as we do below. Theorems 2 and 3 of Zhang et al. (2005) give the following limit theory

$$\begin{aligned} \left( \frac{m}{K}\right) ^{1/2}\left( \left[ p^{*}\right] ^{(avg)}-\int _{0}^{1}\sigma ^{2}(t)\textrm{d}t\right) \Rightarrow \mathcal{M}\mathcal{N}\left( 0,\eta ^{2}\right) , \end{aligned}$$

(1.68)

where $\eta ^{2}$ is the probability limit of the quantity

$$\begin{aligned} \eta _{m}^{2}=\sum _{j=1}^{J_{m}}\sigma ^{4}\left( \tau _{j}\right) h_{j}\left( \tau _{j}-\tau _{j-1}\right) , \end{aligned}$$

(1.69)

where

$$\begin{aligned} h_{j}=\frac{4}{K\overline{\Delta \tau }}\sum _{i=1}^{\left( K-1\right) \wedge j}\left( 1-\frac{i}{K}\right) ^{2}\left( \tau _{j-i}-\tau _{j-i-1}\right) , \end{aligned}$$

(1.70)

and in the present case $m\overline{\Delta \tau }=J_{m}^{-1} \sum _{j=1}^{J_{m}}D_{m,j}.$ The limiting forms of $h_{j}$ and the random (variance) quantity $\eta ^{2}$ are evaluated below.

We first observe that result (1.68) relies on the following representation, given in the Proof of Theorem 2 of Zhang et al. (2005),

$$\begin{aligned} \left[ p^{*}\right] ^{(avg)}-\left[ p^{*}\right] ^{{\mathcal {G}} _{m}}= & {} 2\sum _{j=1}^{J_{m}-1}\left( p^{*}\left( \tau _{j}\right) -p^{*}\left( \tau _{j-1}\right) \right) \\{} & {} \sum _{i=1}^{\left( K-1\right) \wedge j}\left( 1-\frac{i}{K}\right) \left( p^{*}\left( \tau _{j-i}\right) -p^{*}\left( \tau _{j-i-1}\right) \right) +O_{p}\left( \frac{K}{m}\right) \end{aligned}$$

whose leading term is

$$\begin{aligned}{} & {} 2\sum _{j=1}^{J_{m}-1}\sigma \left( \tau _{j-1}\right) \left( B\left( \tau _{j}\right) -B\left( \tau _{j-1}\right) \right) \\{} & {} \quad \sum _{i=1}^{\left( K-1\right) \wedge j}\left( 1-\frac{i}{K}\right) \sigma \left( \tau _{j-i}\right) \left( B\left( \tau _{j-i}\right) -B\left( \tau _{j-i-1}\right) \right) \\{} & {} \qquad \times \left\{ 1+O_{p}\left( \frac{\max _{j\le m}D_{m,j}}{m}\right) \right\} . \end{aligned}$$

Now

$$\begin{aligned}{} & {} 2\sum _{j=1}^{J_{m}-1}\sigma \left( \tau _{j-1}\right) \left( B\left( \tau _{j}\right) -B\left( \tau _{j-1}\right) \right) \\{} & {} \sum _{i=1}^{\left( K-1\right) \wedge j}\left( 1-\frac{i}{K}\right) \sigma \left( \tau _{j-i}\right) \left( B\left( \tau _{j-i}\right) -B\left( \tau _{j-i-1}\right) \right) \end{aligned}$$

is a local martingale with conditional variance whose leading term, as in the proof of Theorem 2 of Zhang et al. (2005), is

$$\begin{aligned}{} & {} 4\sum _{j=1}^{J_{m}-1}\sigma \left( \tau _{j-1}\right) ^{2}\left( \tau _{j}-\tau _{j-1}\right) \chi _{1j}^{2}\\{} & {} \quad \sum _{i=1}^{\left( K-1\right) \wedge j}\left( 1-\frac{i}{K}\right) ^{2}\sigma \left( \tau _{j-i}\right) ^{2}\left( \tau _{j-i}-\tau _{j-i-1}\right) \chi _{1j-i}^{2} \\{} & {} \quad =4\sum _{j=1}^{J_{m}-1}\sigma \left( \tau _{j-1}\right) ^{2}\left( \tau _{j}-\tau _{j-1}\right) \\{} & {} \quad \sum _{i=1}^{\left( K-1\right) \wedge j}\left( 1- \frac{i}{K}\right) ^{2}\sigma \left( \tau _{j-i}\right) ^{2}\left( \tau _{j-i}-\tau _{j-i-1}\right) \textrm{E}\left( \chi _{1j}^{2}\right) \textrm{E} \left( \chi _{1j-1}^{2}\right) +o_{p}\left( 1\right) \\{} & {} \quad =4\sum _{j=1}^{J_{m}-1}\sigma \left( \tau _{j-1}\right) ^{4}\left( \tau _{j}-\tau _{j-1}\right) \\{} & {} \quad \sum _{i=1}^{\left( K-1\right) \wedge j}\left( 1- \frac{i}{K}\right) ^{2}\left( \tau _{j-i}-\tau _{j-i-1}\right) \left\{ 1+O_{p}\left( \frac{K\max _{j\le m}D_{m,j}}{m}\right) \right\} +o_{p}\left( 1\right) \\{} & {} \quad =K\overline{\Delta \tau }\sum _{j=1}^{J_{m}}\sigma ^{4}\left( \tau _{j}\right) h_{j}\left( \tau _{j}-\tau _{j-1}\right) \left\{ 1+o_{p}\left( 1\right) \right\} =K\overline{\Delta \tau }\left\{ \eta _{m}^{2}+o_{p}\left( 1\right) \right\} , \end{aligned}$$

thereby giving (1.69) by martingale central limit theory since, under D2(ii)$^{\prime }$ and $K=cm^{2/3},$

$$\begin{aligned} \frac{K\max _{j\le m}D_{m,j}}{m}=o_{p}\left( 1\right) . \end{aligned}$$

To find an explicit expression for the limit quantity $\eta ^{2}$ in (1.68) we proceed as follows. First observe that for the present model

$$\begin{aligned} m\overline{\Delta \tau }=J_{m}^{-1}\sum _{j=1}^{J_{m}}D_{m,j}=J_{m}^{-1} \sum _{j=1}^{J_{m}}\textrm{E}\left( D_{m,j}|{\mathcal {F}}_{\tau _{j-1}}\right) +o_{p}\left( 1\right) \rightarrow _{p}\text { }\int _{0}^{1}\mu _{D}\left( s\right) \textrm{d}s, \end{aligned}$$

so that for $r\ge 0$

$$\begin{aligned} h_{j=\left[ Kr\right] }= & {} \frac{4m}{K\int _{0}^{1}\mu _{D}\left( s\right) \textrm{d}s} \sum _{i=1}^{\left( K-1\right) \wedge \left[ Kr\right] }\left( 1-\frac{i}{K} \right) ^{2}\left( \tau _{\left[ Kr\right] -i}-\tau _{\left[ Kr\right] -i-1}\right) \left\{ 1+o_{p}\left( 1\right) \right\} \\= & {} \frac{4}{K\int _{0}^{1}\mu _{D}\left( s\right) \textrm{d}s}\sum _{i=1}^{\left( K-1\right) \wedge \left[ Kr\right] }\left( 1-\frac{i}{K}\right) ^{2}D_{m, \left[ Kr\right] -i}\left\{ 1+o_{p}\left( 1\right) \right\} \\= & {} \frac{4}{K\int _{0}^{1}\mu _{D}\left( s\right) \textrm{d}s}\sum _{i=1}^{\left( K-1\right) \wedge \left[ Kr\right] }\left( 1-\frac{i}{K}\right) ^{2}\textrm{E }\left( D_{m,\left[ Kr\right] -i}|{\mathcal {F}}_{\tau _{\left[ Kr\right] -i}}\right) \left\{ 1+o_{p}\left( 1\right) \right\} \\{} & {} \rightarrow _{p}\,\frac{4\left\{ \int _{0}^{1\wedge r}\left( 1-s\right) ^{2}\mu _{D}\left( 1\wedge r-s\right) \textrm{d}s\right\} }{ \int _{0}^{1}\mu _{D}\left( s\right) \textrm{d}s}\text {.} \end{aligned}$$

Since $K/J_{m}\rightarrow 0,$ it follows that $\left[ J_{m}t\right] >K$ as $m\rightarrow \infty$ for all $t\in (0,1].$ Hence, for all $t\in (0,1],$

$$\begin{aligned} h_{j=\left[ J_{m}t\right] }= & {} \frac{4}{K\overline{\Delta \tau }} \sum _{i=1}^{\left( K-1\right) \wedge \left[ J_{m}t\right] }\left( 1-\frac{i}{ K}\right) ^{2}\left( \tau _{\left[ J_{m}t\right] -i}-\tau _{\left[ J_{m}t \right] -i-1}\right) \\{} & {} \rightarrow _{p}\,\frac{4\left\{ \int _{0}^{1}\left( 1-s\right) ^{2}\mu _{D}\left( 1-s\right) \textrm{d}s\right\} }{ \int _{0}^{1}\mu _{D}\left( s\right) \textrm{d}s}, \end{aligned}$$

and then

$$\begin{aligned} \eta _{m}^{2}=\sum _{j=1}^{J_{m}}\sigma ^{4}\left( \tau _{j}\right) h_{j}\left( \tau _{j}-\tau _{j-1}\right) \rightarrow _{p}\, \frac{4\left\{ \int _{0}^{1}\left( 1-s\right) ^{2}\mu _{D}\left( 1-s\right) \textrm{d}s\right\} }{\int _{0}^{1}\mu _{D}\left( s\right) \textrm{d}s}\int _{0}^{1}\sigma ^{4}\left( t\right) \textrm{d}t. \end{aligned}$$

Observe that when $\mu _{D}\left( s\right) =\mu _{D}$ $\ a.s.,$ the right side of the last expression simplifies to

$$\begin{aligned} 4\left\{ \int _{0}^{1}\left( 1-s\right) ^{2}\textrm{d}s\right\} \int _{0}^{1}\sigma ^{4}\left( t\right) \textrm{d}t=\frac{4}{3}\int _{0}^{1}\sigma ^{4}\left( t\right) \textrm{d}t, \end{aligned}$$

as given in Zhang et al. (2005) equation (5). It therefore follows that

$$\begin{aligned}{} & {} \left( \frac{m}{K}\right) ^{1/2}\left( \left[ p^{*}\right] ^{(avg)}-\int _{0}^{1}\sigma ^{2}(t)\textrm{d}t\right) \nonumber \\{} & {} \quad \Rightarrow \mathcal{M}\mathcal{N}\left( 0,\frac{ 4\left\{ \int _{0}^{1}\left( 1-s\right) ^{2}\mu _{D}\left( 1-s\right) \textrm{d}s\right\} }{\int _{0}^{1}\mu _{D}\left( s\right) \textrm{d}s}\int _{0}^{1}\sigma ^{4}\left( t\right) \textrm{d}t\right) . \end{aligned}$$

(1.71)

Next, consider (1.67). Since $\widehat{\left[ p^{*}\right] }=\left[ p\right] ^{(avg)}-\frac{{\bar{m}}}{J_{m}}\left[ p\right] ^{{\mathcal {G}}_{m}}$, we have

$$\begin{aligned}{} & {} \sqrt{\frac{K}{{\bar{m}}}}\left\{ \widehat{\left[ p^{*}\right] }-\left[ p^{*}\right] ^{(avg)}\right\} \nonumber \\{} & {} \quad =\sqrt{\frac{K}{{\bar{m}}}}\left\{ \left[ p\right] ^{(avg)}-\left[ p^{*} \right] ^{(avg)}-\frac{{\bar{m}}}{J_{m}}\left[ p\right] ^{{\mathcal {G}} _{m}}\right\} \nonumber \\{} & {} \quad =\sqrt{\frac{K}{{\bar{m}}}}\left\{ \left[ p\right] ^{(avg)}-\left[ p^{*} \right] ^{(avg)}-2{\bar{m}}\sigma _{u}^{2}\right\} \nonumber \\{} & {} \qquad -2\sqrt{K{\bar{m}}}\left\{ J_{m}^{-1}\left[ p\right] ^{{\mathcal {G}} _{m}}-\sigma _{u}^{2}\right\} \nonumber \\{} & {} \quad \Rightarrow {\mathcal {N}}\left( 0,8\sigma _{u}^{4}\right) , \end{aligned}$$

(1.72)

as in Theorem A.1 and equation (56) of Zhang et al. (2005). Combining (1.71) and (1.72) as in Theorem 4 of Zhang et al. (2005) and using $K=cm^{2/3}$ for some constant $c>0,$ so that $\sqrt{K/{\bar{m}}}=K/m^{1/2}=cm^{1/6}$ and $\sqrt{m/K} =m^{1/6}/c^{1/2},$ we then have the limit theory

$$\begin{aligned} m^{1/6}\left\{ \widehat{\left[ p^{*}\right] }-\int _{0}^{1}\sigma ^{2}(t)\textrm{d}t\right\} \Rightarrow \left\{ \frac{8}{c^{2}}\sigma _{u}^{4}+c\eta ^{2}\right\} ^{1/2}{\mathcal {N}}\left( 0,1\right) , \end{aligned}$$

as required. $\square$