Limit Theory for Moderate Deviation from Integrated GARCH Processes

frontmatter\address{90 Stamford Rd, Singapore Management University} \fntext[myft1]{I would like to thank the co-editor, an associate editor and the referees for helping improve the paper. All possible errors are mine. Yubo Tao, School of Economics, Singapore Management University, 90 Stamford Road, Singapore 178903. Email: [email removed].} \begin{abstract} This paper develops the limit theory of the GARCH(1,1) process that moderately deviates from IGARCH process towards both stationary and explosive regimes. The GARCH(1,1) process is defined by equations $u_t = \sigma_t \varepsilon_t$, $\sigma_t^2 = \omega + \alpha_n u_{t-1}^2 + \beta_n\sigma_{t-1}^2$ and $\alpha_n + \beta_n$ approaches to unity as sample size goes to infinity. The asymptotic theory extends BerkesHorvathKokoszka2005 by allowing the parameters to have a slower rate of convergence. The results can be applied to unit root test for processes with mildly-integrated GARCH innovations (e.g. Boswijk2001, Cavaliere2007, Cavaliere2009) and deriving limit theory of estimators for models involving mildly-integrated GARCH processes (e.g. JensenRahbek2004, FrancqZakoian2012,FrancqZakoian2013). \end{abstract} \begin{keyword} Central Limit Theorem \sep Limiting Process \sep Localization \sep Explosive GARCH \sep Volatility Process \MSC[2010] 62M10 \sep 91B84 \end{keyword}

assumption$\lbrace \varepsilon_t\rbrace_{t=0}^n$ is an i.i.d sequence with $E\varepsilon_0^2 = 1$ and $E|\varepsilon_0|^{4+\delta} < \infty$, for some $\delta > 0$.

assumption$\alpha_n\log\log n \rightarrow 0$, $n\alpha_n \rightarrow \infty$ and $\beta_n \rightarrow 1$.

proposition[Additive Representation] Under Assumption (ref) and (ref), we have the additive representation for $\sigma_t^2$ as \begin{align*} \sigma_t^2 &= \sigma_0^2t^{t/2}e^{\sqrt{t}\gamma_n}\left(1+\dfrac{\alpha_n}{\sqrt{t}}\sum_{j=1}^{t}\xi_{t-j} + R_{t}^{(1)}\right) + \omega\left[1+\sum_{j=1}^{t}t^{j/2}e^{\frac{j\gamma_n}{\sqrt{t}}}\left(1+\dfrac{\alpha_n}{\sqrt{t}}\sum_{i=1}^{j}\xi_{t-i} + R_{t,j}^{(2)}\right)\left(1+ R_{t,j}^{(3)}\right)\right] \end{align*} where $\xi_t = \varepsilon_t^{2}-1$ and the remainder terms satisfy \begin{align*} \left\vert R_{t}^{(1)} \right\vert = O_p\left(\alpha_n^2 + \gamma_n^{2}\right), &\quad \max\limits_{1 \leq j \leq t}\left\vert R_{t,j}^{(2)} \right\vert = O_p\left(\alpha_n^2\right) \\ \max\limits_{1 \leq j \leq t}\dfrac{1}{j\log\log j}\left\vert R_{t,j}^{(2)} \right\vert = O_p\left(\dfrac{\alpha_n^2}{t}\right), &\quad \max\limits_{1 \leq j \leq t}\dfrac{1}{j}\left\vert R_{t,j}^{(3)} \right\vert = O_p\left(\dfrac{\alpha_n^2 + \gamma_n^{2}}{t}\right) \end{align*}

remarkThe key difference between our results and BerkesHorvathKokoszka2005 is the convergence rate of the approximation errors. In BerkesHorvathKokoszka2005, the approximation error $\vert R_{t}^{(p)} \vert$, $\forall p = \{1, 2, 3\}$ is of order $t(\alpha_n^2 + \gamma_n^{2})$ or $t\alpha_n^2$ asymptotically. Hence, these errors are negligible only when $\kappa \in (1/2,1)$. We relax this restrictive assumption by normalizing the original terms with $\sqrt{t}$. Under this new normalization, all the approximation errors remains negligible when $\kappa \in (0,1)$.

assumption$\dfrac{\sqrt{\lvert \gamma_n \rvert}}{\alpha_n n^{1/4}} \rightarrow \infty$, while $\dfrac{\sqrt{\lvert \gamma_n \rvert^{3}}}{\alpha_n n^{1/4}} \rightarrow 0$, as $n \rightarrow \infty$.

theorem[Near-stationary Case] Suppose $\gamma_n < 0$, then under Assumption (ref)-(ref), the random variables \begin{equation*} \dfrac{\sqrt{2\lvert \gamma_n \rvert^{3}}}{\alpha_nk(m)^{1/4}} \dfrac{1}{\sqrt{E\xi_0^2}} \left(\dfrac{\sigma_{k(m)}^2}{\omega {k(m)}^{k(m)/2}} - \sum_{j=1}^{k(m)-1}e^{\frac{j\gamma_n}{\sqrt{k(m)}}}\right) \xrightarrow{d} \mathcal{N}(0,1). \end{equation*} In addition, the random variables \begin{equation*} \left(\dfrac{\lvert\gamma_n\rvert }{\omega {k(m)}^{(k(m)+1)/2}} \right)^{1/2} u_{k(m)} \end{equation*} are asymptotically independent, each with the asymptotic distribution equals to that of $\varepsilon_0$.

theorem[Integrate Case] Suppose $\gamma_n = 0$, then under Assumption (ref) and (ref), the volatility has the asymptotic distribution \begin{equation*} \dfrac{k(m)^{1/2}}{n^{3/2}\alpha_n}\dfrac{1}{\sqrt{E\xi_0^{2}}}\left(\dfrac{\sigma_{k(m)}^2}{\omega {k(m)}^{k(m)/2}} - k(m)\right) \xrightarrow{d} \int_{0}^{t_m}xdW(x). \end{equation*} In addition, the random variables \begin{equation*} \left(\omega {k(m)}^{k(m)/2 + 1}\right)^{-1/2}u_{k(m)} \end{equation*} are asymptotically independent, each with the asymptotic distribution equals to that of $\varepsilon_0$.

assumption$ \gamma_n / \alpha_n \rightarrow 0$, as $n \rightarrow \infty$.

theorem[Near-explosive Case] Suppose $\gamma_n > 0$, then under Assumption (ref), (ref) and (ref), the volatility has the asymptotic distribution \begin{equation*} \dfrac{\gamma_n e^{-\sqrt{k(m)}\gamma_n}}{\alpha_n\sqrt{k(m)}} \dfrac{1}{\sqrt{E\xi_0^2}} \left(\dfrac{\sigma_{k(m)}^2}{\omega k(m)^{k(m)/2}} - \sum_{j=1}^{k(m)-1}e^{\frac{j\gamma_n}{\sqrt{k(m)}}}\right) \Rightarrow W(t_m). \end{equation*} In addition, the random variables \begin{equation*} \left(\dfrac{\gamma_n e^{-\sqrt{k(m)}\gamma_n}}{\omega k(m)^{(k(m)+1)/2}}\right)^{1/2}u_{k(m)} \end{equation*} are asymptotically independent, each with the asymptotic distribution equals to that of $\varepsilon_0$.

remarkAs one may notice, the rate of convergence for both volatility process and return process in all three cases decreases to 0 asymptotically. These seemingly awkward results are reasonable in the sense that the convergence rate is a part of the normalization which reflects the order of the process. In other words, when we compute a partial sum of $X$s in form of $\sum_{i=1}^{n} a_i X_i$, the normalization just plays the role of $a_i$ which is usually required to decrease to 0 for applying a central limit theorem.

proof[Proof of Proposition (ref)] First, note the GARCH(1,1) model can be written into the following multiplicative form: \begin{align*} \sigma_t^2 &= \sigma_0^2\prod_{i=1}^{t}\left(\beta_n + \alpha_n\varepsilon_{t-i}^2\right) + \omega\left[1 + \sum_{j=1}^{t-1}\prod_{i=1}^{j}\left(\beta_n + \alpha_n\varepsilon_{t-i}^{2}\right)\right] \\ &= \sigma_0^2 t^{t/2} \prod_{i=1}^{t}\dfrac{\left(\beta_n + \alpha_n\varepsilon_{t-i}^2\right)}{\sqrt{t}} + \omega\left[1 + t^{t/2}\sum_{j=1}^{t-1}\prod_{i=1}^{j}\dfrac{\left(\beta_n + \alpha_n\varepsilon_{t-i}^{2}\right)}{\sqrt{t}}\right]. \end{align*} Note that \begin{equation*} \max\limits_{1 \leq i \leq t} \dfrac{\left\vert \beta_n + \alpha_n\varepsilon_{t-i}^2 - 1 \right\vert}{\sqrt{t}} \leq \dfrac{\vert \gamma_n \vert}{\sqrt{t}} + \alpha_n\max\limits_{1 \leq i \leq t}\dfrac{\vert \varepsilon_{t-i}^2 - 1 \vert}{\sqrt{t}} = \dfrac{\vert \gamma_n \vert}{\sqrt{t}} + \alpha_n\max\limits_{1 \leq i \leq t-1}\dfrac{\vert \varepsilon_{i}^2 - 1 \vert}{\sqrt{t}}. \end{equation*} Then by Assumption (ref) and ChowTeicher2012, we have the almost sure convergence of \begin{equation*} \max_{1 \leq j \leq t-1}\vert \varepsilon_{i}^2 - 1 \vert = O(\sqrt{t}). \end{equation*} Therefore, the term above is \begin{equation*} \max\limits_{1 \leq i \leq t} \dfrac{\left\vert \beta_n + \alpha_n\varepsilon_{t-i}^2 - 1 \right\vert}{\sqrt{t}} = o_p(1). \end{equation*} Now consider the sequence of events $$ A_n = \left\lbrace \max\limits_{1 \leq i \leq t} \dfrac{\lvert\beta_n + \alpha_n\varepsilon_{t-i}^2 - 1 \rvert}{\sqrt{t}} \leq \dfrac{1}{2}\right\rbrace. $$ From the previous result we know $\lim\limits_{n \rightarrow \infty}P(A_n) = 1$. Then by Taylor expansion, $\lvert \log(1+x) - x \rvert \leq 2x^2$, $\lvert x\rvert \leq 1/2$ on the event $A_n$, which implies \begin{align*} \left\vert R_{t,j}^{(3)} \right\vert &= \left\vert \sum_{i=1}^{j} \log\dfrac{\left(\beta_n + \alpha_n\varepsilon_{t-i}^2\right)}{\sqrt{t}} - \sum_{i=1}^{j}\dfrac{\left(\gamma_n + \alpha_n\xi_{t-i}\right)}{\sqrt{t}} \right\vert \\ &= \left\vert \sum_{i=1}^{j} \log\dfrac{\left(\gamma_n + \alpha_n\xi_{t-i} + 1\right)}{\sqrt{t}} - \sum_{i=1}^{j}\dfrac{\left(\gamma_n + \alpha_n\xi_{t-i}\right)}{\sqrt{t}} \right\vert \\ &\leq \sum_{i=1}^{j} \left\vert \log\left(\dfrac{\gamma_n + \alpha_n\xi_{t-i}}{\sqrt{t}} + 1 \right)- \dfrac{\left(\gamma_n + \alpha_n\xi_{t-i}\right)}{\sqrt{t}} \right\vert \\ &\leq 2 \sum_{i=1}^{j} \dfrac{\left(\gamma_n + \alpha_n\xi_{t-i}\right)^2}{t} \leq \dfrac{4j\gamma_n^2}{t} + \dfrac{4\alpha_n^2\sum_{i=1}^{j}\xi_{t-i}^2}{t}. \end{align*} By Assumption (ref) and law of large numbers (LLN), we know \begin{equation*} \max\limits_{1 \leq j \leq t} \dfrac{1}{j} \left\vert \sum_{i=1}^{j} \xi_{t-i}^2\right\vert \sim \max\limits_{1 \leq j \leq t} \dfrac{1}{j} \left\vert \sum_{i=1}^{j} \xi_{i}^2\right\vert = O_p(1). \end{equation*} Then by the equation above, we have \begin{equation*} \max\limits_{1 \leq i \leq j} \dfrac{1}{j}\lvert R_{t,j}^{(3)} \rvert = O_p\left(\dfrac{\gamma_n^2 + \alpha_n^2}{t}\right). \end{equation*} Now by direct plugging into the key multiplicative term we care about, we have \begin{align*} \prod_{i=1}^{j}\dfrac{\left(\beta_n + \alpha_n\varepsilon_{t-i}^2\right)}{\sqrt{t}} &= \exp\left\lbrace \sum_{i=1}^{j} \log\left(\dfrac{\beta_n + \alpha_n\varepsilon_{t-i}^2}{\sqrt{t}}\right)\right\rbrace \\ &= \exp\left\lbrace \dfrac{j\gamma_n}{\sqrt{t}}\right\rbrace \exp\left\lbrace\dfrac{\alpha_n\sum_{i=1}^{j}\xi_{t-i}}{\sqrt{t}} \right\rbrace \exp\left\lbrace R_{t,j}^{(3)}\right\rbrace \\ &= e^{\frac{j\gamma_n}{\sqrt{t}}}\exp\left\lbrace\dfrac{\alpha_n\sum_{i=1}^{j}\xi_{t-i}}{\sqrt{t}} \right\rbrace\left(1+ R_{t,j}^{(3)}\right). \end{align*} Further, note $\lbrace \xi_t\rbrace_{t=1}^{n}$ is an i.i.d sequence with $E\xi_0^{2} < \infty$, then we know \begin{equation*} \max\limits_{1 \leq j \leq t} \left\vert \sum_{i=1}^{j} \xi_{t-i} \right\vert = O_p(\sqrt{t}), \end{equation*} which implies \begin{equation*} \max\limits_{1 \leq j \leq t} \left\vert \dfrac{\alpha_n}{\sqrt{t}}\sum_{i=1}^{j} \xi_{t-i} \right\vert = O_p(\alpha_n) = o_p(1). \end{equation*} Similarly, we define the sequence of events \begin{equation*} B_n = \left\lbrace \max\limits_{1 \leq j \leq t}\left\vert \dfrac{\alpha_n}{\sqrt{t}}\sum_{i=1}^{j} \xi_{t-i} \right\vert \leq \dfrac{1}{2} \right\rbrace, \end{equation*} which is known to have the property $\lim\limits_{n \rightarrow \infty} P(B_n) = 1$. Then by Taylor expansion, $\lvert \exp(x) - (1+x) \rvert \leq \sqrt{e}x^2/2 $ when $\lvert x \rvert \leq 1/2$, on the event $B_n$ \begin{equation*} \left\vert R_{t,j}^{(2)}\right\vert = \left\vert\exp\left\lbrace \dfrac{\alpha_n}{\sqrt{t}}\sum_{i=1}^{j} \xi_{t-i} \right\rbrace - \left(1 + \dfrac{\alpha_n}{\sqrt{t}}\sum_{i=1}^{j} \xi_{t-i}\right)\right\vert \leq \dfrac{\sqrt{e}}{2}\left(\dfrac{\alpha_n}{\sqrt{t}}\sum_{i=1}^{j} \xi_{t-i}\right)^2 = O_p\left(\alpha_n^2\right), \end{equation*} and by law of iterated logarithm, we know \begin{equation*} \max\limits_{1 \leq j \leq t}\dfrac{1}{j \log\log j}\left(\dfrac{\alpha_n}{\sqrt{t}}\sum_{i=1}^{j} \xi_{t-i}\right)^2 = O_p\left(\dfrac{\alpha_n^2}{t}\right). \end{equation*} Combining the results above, we have thus showed that \begin{equation*} \prod_{i=1}^{j}\left(\dfrac{\beta_n + \alpha_n\varepsilon_{t-i}^2}{\sqrt{t}}\right) = e^{\frac{j\gamma_n}{\sqrt{t}}}\left(1 + \dfrac{\alpha_n}{\sqrt{t}}\sum_{i=1}^{j}\xi_{t-i} + R_{t,j}^{(2)}\right)\left(1 + R_{t,j}^{(3)}\right). \end{equation*} Lastly, by the equation above, we know \begin{align*} \prod_{i=1}^{t}\left(\dfrac{\beta_n + \alpha_n\varepsilon_{t-i}^2}{\sqrt{t}}\right) &= e^{\frac{t\gamma_n}{\sqrt{t}}}\left(1 + \dfrac{\alpha_n}{\sqrt{t}}\sum_{i=1}^{t}\xi_{t-i} + O_p(\alpha_n^2)\right)\left(1 + O_p(\gamma_n^2 + \alpha_n^2)\right) \\ &= e^{\sqrt{t}\gamma_n}\left(1 + \dfrac{\alpha_n}{\sqrt{t}}\sum_{i=1}^{t}\xi_{t-i} + O_p(\gamma_n^2 + \alpha_n^2)\right), \end{align*} and this establishes $R_{t}^{(1)}$.

proof[Proof of Theorem (ref)] First, we focus on the volatilities. Denote $k = \lfloor nt \rfloor$, $0 < t \leq 1$, \begin{align*} \sigma_k^2 &= \omega + \sigma_0^2 k^{k/2} e^{\sqrt{k}\gamma_n}\left(1+ \dfrac{\alpha_n}{\sqrt{k}}\sum_{j=1}^{k}\xi_{k-j} + R_k^{(1)}\right) + \omega k^{k/2}\sum_{j=1}^{k-1}e^{\frac{j\gamma_n}{\sqrt{k}}}\left(1+\dfrac{\alpha_n}{\sqrt{k}}\sum_{i=1}^{j}\xi_{k-i} + R_{k,j}^{(2)}\right)R_{k,j}^{(3)} \\ &\ \ \ + \omega k^{k/2} \sum_{j=1}^{k-1}e^{\frac{j\gamma_n}{\sqrt{k}}} R_{k,j}^{(2)} + \omega k^{k/2} \sum_{j=1}^{k-1}e^{\frac{j\gamma_n}{\sqrt{k}}}\left(1+\dfrac{\alpha_n}{\sqrt{k}}\sum_{i=1}^{j}\xi_{k-i}\right) \\ &= \omega + \sigma_{k,1}^{2} + \sigma_{k,2}^{2} + \sigma_{k,3}^{2} + \sigma_{k,4}^{2}. \end{align*} For $\sigma_{k,1}^{2}$, note $k^{-1/2}\sum_{j=1}^{k}\xi_{k-j}$ is asymptotically normal, then by Proposition (ref), \begin{equation*} \dfrac{\alpha_n}{\sqrt{k}}\sum_{j=1}^{k}\xi_{k-j} + R_{k}^{(1)} = o_p(1), \end{equation*} and this implies \begin{align*} \left\lvert \sigma_{k,1}^2 \right\rvert &= O_p\left(k^{k/2}e^{\sqrt{k}\gamma_n}\right). \end{align*} For $\sigma_{k,2}^{2}$, note by Lemma 4.1 in BerkesHorvathKokoszka2005, we have \begin{equation} \sum_{j=1}^{k}je^{\frac{j\gamma_n}{\sqrt{k}}} \sim \dfrac{k}{\lvert \gamma_n \rvert^2}\Gamma(2), \end{equation} and note that \begin{equation} \max\limits_{1\leq j \leq k-1} \left\lvert \dfrac{\alpha_n}{\sqrt{k}}\sum_{i=1}^{j}\xi_{k-i} + R_{k,j}^{(2)} \right\rvert = o_p(1). \end{equation} Then by equation ((ref)), ((ref)) and Proposition (ref) we have \begin{align*} \left\lvert \sigma_{k,2}^{2} \right\rvert &= \left\lvert \omega k^{k/2}\sum_{j=1}^{k-1}je^{\frac{j\gamma_n}{\sqrt{k}}}\left(1 + \dfrac{\alpha_n}{\sqrt{k}}\sum_{i=1}^{j}\xi_{k-i} + R_{k,j}^{(2)}\right)\dfrac{1}{j}R_{k,j}^{(3)} \right\rvert \\ &= O_p(1)\omega k^{k/2} \dfrac{\alpha_n^2 + \gamma_n^2}{k} \dfrac{k}{\lvert \gamma_n \rvert^2} \\ &= O_p\left(\dfrac{k^{k/2}\left(\alpha_n^2 + \gamma_n^2\right)}{\gamma_n^2}\right). \end{align*} For $\sigma_{k,3}^{2}$, similarly, by Proposition (ref) and Lemma 4.1 in BerkesHorvathKokoszka2005, we have \begin{align*} \left\lvert \sigma_{k,3}^{2} \right\rvert &= \left\lvert \omega k^{k/2}\sum_{j=1}^{k-1}e^{\frac{j\gamma_n}{\sqrt{k}}}R_{k,j}^{(2)} \right\rvert \\ &= O_p(1)\omega k^{k/2}\dfrac{\alpha_n^2}{k}\sum_{j=1}^{k-1}je^{\frac{j\gamma}{\sqrt{k}}}\log\log j \\ &= O_p\left(\dfrac{k^{k/2} \left(\alpha_n^{2}\log\log k\right) }{\gamma_n^2}\right). \end{align*} Lastly, for $\sigma_{k,4}^{2}$, by Lemma 4.1 in (ref) we have \begin{align*} \sigma_{k,4}^{2} &= \omega k^{k/2} \sum_{j=1}^{k-1}e^{\frac{j\gamma_n}{\sqrt{k}}} + \omega k^{k/2} \dfrac{\alpha_n}{\sqrt{k}} \sum_{j=1}^{k-1}e^{\frac{j\gamma_n}{\sqrt{k}}} \sum_{i=1}^{j}\xi_{k-i} \\ &= O_p\left(\dfrac{k^{k/2}k^{1/2}}{\lvert \gamma_n \rvert}\right) + \omega k^{k/2} \dfrac{\alpha_n}{\sqrt{k}} \sum_{j=1}^{k-1}e^{\frac{j\gamma_n}{\sqrt{k}}} \sum_{i=1}^{j}\xi_{k-i}. \end{align*} Therefore, we only have to consider the last term in the above equation. Define \begin{equation*} \tau_m = k(m)^{-1/4}\sum_{j=1}^{k(m)-1}e^{\frac{j\gamma_n}{\sqrt{k(m)}}} \xi_{k(m)-j}, \quad 1 \leq m \leq N, \end{equation*} and \begin{equation*} \tau_m^* = k(m)^{-1/2}\sum_{j=1}^{k(m)-1}e^{\frac{j\gamma_n}{\sqrt{k(m)}}} \sum_{i=1}^{j}\xi_{k(m)-i}, \quad 1 \leq m \leq N. \end{equation*} Then by Cramer-Wold device (Theorem 29.4 of Billingsley1995), we have \begin{align*} \sum_{m=1}^{N}\mu_m\tau_m &= \sum_{i=1}^{k(1)-1}\sum_{m=1}^{N}\dfrac{\mu_m}{k(m)^{1/4}} e^{\frac{(k(m)-i)\gamma_n}{\sqrt{k(m)}}} + \sum_{i=k(1)}^{k(2)-1}\sum_{m=2}^{N}\dfrac{\mu_m}{k(m)^{1/4}} e^{\frac{(k(m)-i)\gamma_n}{\sqrt{k(m)}}} \\ &+ \cdots + \sum_{i=k(N-1)}^{k(N)-1}\dfrac{\mu_N}{k(N)^{1/4}} e^{\frac{(k(N)-i)\gamma_n}{\sqrt{k(N)}}} \\ &= S_1 + S_2 + \cdots + S_N. \end{align*} Observe that \begin{align*} ES_1^2 &= E\xi_0^2\left( \sum_{i=1}^{k(1)-1}\sum_{m=1}^{N}k(m)^{-1/4}\mu_m e^{\frac{(k(m)-i)\gamma_n}{\sqrt{k(m)}}}\right)^2 \\ &= E\xi_0^2 \sum_{m=1}^{N}\dfrac{\mu_m^2}{\sqrt{k(m)}}\sum_{i=1}^{k(1)-1} e^{\frac{2(k(m)-i)\gamma_n}{\sqrt{k(m)}}} + E\xi_0^2 \sum_{1 \leq m \neq l \leq N}\left(k(m)k(l)\right)^{-1/4}\mu_m\mu_l\sum_{i=1}^{k(1)-1} e^{\frac{(k(m)-i)\gamma_n}{\sqrt{k(m)}}+\frac{(k(l)-i)\gamma_n}{\sqrt{k(l)}}} \\ &= E\xi_0^2 \dfrac{\mu_1^2}{\sqrt{k(1)}}\sum_{i=1}^{k(1)-1} e^{\frac{2(k(1)-i)\gamma_n}{\sqrt{k(1)}}} + E\xi_0^2 \sum_{m=2}^{N}\dfrac{\mu_m^2}{\sqrt{k(m)}}\sum_{i=1}^{k(1)-1} e^{\frac{2(k(m)-i)\gamma_n}{\sqrt{k(m)}}} \\ &\ \ \ + E\xi_0^2 \sum_{1 \leq m \neq l \leq N}\left(k(m)k(l)\right)^{-1/4}\mu_m\mu_l\sum_{i=1}^{k(1)-1} e^{\frac{(k(m)-i)\gamma_n}{\sqrt{k(m)}}+\frac{(k(l)-i)\gamma_n}{\sqrt{k(l)}}} \\ &= E\xi_0^2 \dfrac{\mu_1^2}{\sqrt{k(1)}}\sum_{i=1}^{k(1)-1} e^{\frac{2i\gamma_n}{\sqrt{k(1)}}} + E\xi_0^2 \sum_{m=2}^{N}\dfrac{\mu_m^2}{\sqrt{k(m)}}e^{\frac{2(k(m)-k(1))\gamma_n}{\sqrt{k(m)}}}\sum_{i=1}^{k(1)-1} e^{\frac{2i\gamma_n}{\sqrt{k(m)}}} \\ &\ \ \ + E\xi_0^2 \sum_{1 \leq m \neq l \leq N}(k(m)k(l))^{-1/4}\mu_m\mu_le^{\frac{(k(m)-k(1))\gamma_n}{\sqrt{k(m)}}+\frac{(k(l)-k(1))\gamma_n}{\sqrt{k(l)}}}\sum_{i=1}^{k(1)-1} e^{\frac{i\gamma_n}{\sqrt{k(m)}}+\frac{i\gamma_n}{\sqrt{k(l)}}} \\ &\sim E\xi_0^2 \mu_1^2\dfrac{1}{2\lvert \gamma_n\rvert} + E\xi_0^2 \sum_{m=2}^{N}\mu_m^2e^{\frac{2(k(m)-k(1))\gamma_n}{\sqrt{k(m)}}} \dfrac{1}{2\lvert \gamma_n \rvert} \\ &\ \ \ + E\xi_0^2 \sum_{1 \leq m \neq l \leq N}\dfrac{\mu_m\mu_l}{(\sqrt{k(m)} +\sqrt{k(l)}) \lvert\gamma_n\rvert} e^{\frac{(k(m)-k(1))\gamma_n}{\sqrt{k(m)}} + \frac{(k(l)-k(1))\gamma_n}{\sqrt{k(l)}}} \\ &= E\xi_0^2 \mu_1^2\dfrac{1}{2\lvert \gamma_n\rvert} + o\left(\dfrac{1}{\lvert \gamma_n \rvert}\right), \end{align*} we then have \begin{align*} E\left(\sum_{m=1}^{N} \mu_m \tau_m\right)^2 &= \left(\sum_{m=1}^{N} \mu_m^2\right) E\xi_0\dfrac{1}{2\lvert \gamma_n \rvert} + o\left(\dfrac{1}{\lvert \gamma_n \rvert}\right). \end{align*} Observe also that, for some $c_i$, $1 \leq i \leq k(N)-1$, we have \begin{equation*} \sum_{m=1}^{N} \mu_m \tau_m = \sum_{i=1}^{k(N)-1} c_i \xi_i, \end{equation*} and by Jensen's inequality, we know for some $\delta > 0$, \begin{align*} \lvert c_i \rvert^{2+\delta} &= \left\vert k(1)^{-1/4}\mu_1 e^{\frac{(k(1)-i) \gamma_n}{\sqrt{k(1)}}} + k(2)^{-1/4}\mu_1 e^{\frac{(k(2)-i) \gamma_n}{\sqrt{k(2)}}}+ \cdots +k(N)^{-1/4}\mu_1 e^{\frac{(k(N)-i) \gamma_n}{\sqrt{k(N)}}} \right\vert^{2+\delta} \\ &\leq C_1(N)\left[\dfrac{\lvert \mu_1\rvert^{2+\delta}}{k(1)^{1/2+\delta/4}}e^{\frac{(k(1) - i)(2+\delta)\gamma_n}{\sqrt{k(1)}}} + \cdots + \dfrac{\lvert \mu_N\rvert^{2+\delta}}{k(N)^{1/2+\delta/4}}e^{\frac{(k(N) - i)(2+\delta)\gamma_n}{\sqrt{k(N)}}}\right]. \end{align*} This implies that \begin{equation*} \sum_{i=1}^{k(N)-1} \lvert c_i \rvert^{2+\delta} \sim C_1(N)\lvert \mu_1\rvert^{2+\delta} \dfrac{1}{k(1)^{\delta/4}(2+\delta)\lvert \gamma_n \rvert} + O\left(\dfrac{1}{k(2)^{\delta/4}\lvert \gamma_n\rvert}\right) = o\left(\dfrac{1}{\lvert \gamma_n \rvert}\right). \end{equation*} Now we can easily check the Liapounov's condition, where \begin{equation*} \dfrac{\left(\sum_{i=1}^{k(N)-1}\lvert c_i \rvert^{2+\delta} E\lvert \xi_i\rvert^{2+\delta}\right)^{1/(2+\delta)}}{\left(\sum_{i=1}^{k(N)-1}c_i^2 E\xi_i^{2}\right)^{1/2}} = o\left(\lvert \gamma_n \rvert^{1/2 - 1/(2+\delta)}\right) = o_p(1). \end{equation*} Then by Liapounov central limit theorem (Theorem 27.3, p.362 of Billingsley1995), we have \begin{equation*} \sqrt{2\lvert \gamma_n \rvert}\left[\tau_1, \tau_2, \cdots, \tau_N\right] \xrightarrow{d} \sqrt{E\xi_{0}^2}\left[\eta_1, \eta_2, \cdots, \eta_N\right], \end{equation*} where $\eta_1, \eta_2, \cdots, \eta_N$ are independent standard normal random variables. Now we have to check the relationship between $\tau_m$ and $\tau_m^*$. Note by $k^{-1/2}\left(e^{\frac{\gamma_n}{\sqrt{k}}}-1\right)^{-1} = \left(\gamma_n + o(1)\right)^{-1}$, we have \begin{align*} \dfrac{1}{\sqrt{k}}\sum_{j=i}^{k-1}e^{\frac{j\gamma_n}{\sqrt{k}}} - \lvert \gamma_n\rvert^{-1}e^{\frac{i\gamma_n}{\sqrt{k}}} &= \dfrac{1}{\sqrt{k}}\dfrac{e^{\frac{k\gamma_n}{\sqrt{k}}} - e^{\frac{i\gamma_n}{\sqrt{k}}}}{e^{\frac{\gamma_n}{\sqrt{k}}}-1} - \lvert \gamma_n\rvert^{-1}e^{\frac{i\gamma_n}{\sqrt{k}}} \\ &= \left(\gamma_n + o(1)\right)^{-1}\left(e^{\frac{k\gamma_n}{\sqrt{k}}} - e^{\frac{i\gamma_n}{\sqrt{k}}}\right) - \lvert \gamma_n \rvert^{-1} e^{\frac{i\gamma_n}{\sqrt{k}}} \\ &= \left(\gamma_n^{-1} + O(1)\right) e^{\frac{k\gamma_n}{\sqrt{k}}} - e^{\frac{i\gamma_n}{\sqrt{k}}}O(1). \end{align*} Then, we know \begin{align*} E\left[\sqrt{2\lvert \gamma_n\rvert^3}\tau_m^* - \sqrt{2\lvert \gamma_n\rvert}\tau_m\right]^2 &= \dfrac{2\lvert \gamma_n \rvert^3}{\sqrt{k}} E\left[\dfrac{1}{\sqrt{k}}\sum_{i=1}^{k-1}\left(\sum_{j=i}^{k-1}e^{\frac{j\gamma}{\sqrt{k}}}\right)\xi_{k-i} - \lvert \gamma_n\rvert^{-1}\sum_{i=1}^{k-1}e^{\frac{i\gamma_n}{\sqrt{k}}}\xi_{k-i} \right]^{2} \\ &= \dfrac{2\lvert \gamma_n \rvert^3}{\sqrt{k}} E\xi_0^2\sum_{i=1}^{k-1}\left(\dfrac{1}{\sqrt{k}}\sum_{j=i}^{k-1}e^{\frac{j\gamma_n}{\sqrt{k}}} - \lvert \gamma_n\rvert^{-1}e^{\frac{i\gamma_n}{\sqrt{k}}}\right)^2 \\ &\sim \dfrac{2\lvert \gamma_n \rvert^3}{\sqrt{k}} E\xi_0^2\left(k\gamma_n^{-2}e^{\sqrt{k}\gamma_n} + \dfrac{\sqrt{k}}{2\lvert\gamma\rvert} - 2\gamma_n^{-1}e^{\sqrt{k}\gamma_n}\dfrac{\sqrt{k}}{\lvert\gamma\rvert}\right) \\ &= 2E\xi_0^2 O\left(\sqrt{k}\lvert\gamma_n\rvert e^{\sqrt{k}\gamma_n}\right) + o_p(1) \\ &= o_p(1), \end{align*} where the last equality comes from the well known limits of $xe^{-x}$, \begin{equation*} \lim\limits_{x \rightarrow \infty} \dfrac{x}{e^{x}} = \lim\limits_{x \rightarrow \infty}\dfrac{1}{e^x} = 0 \quad and \quad \lim\limits_{x \rightarrow 0} \dfrac{x}{e^{x}} = 0. \end{equation*} Therefore, we have \begin{equation*} \sqrt{2\lvert \gamma_n \rvert^{3}}\left[\tau_1^*, \tau_2^*, \cdots, \tau_N^* \right] \xrightarrow{d} \sqrt{E\xi_0^2}\left[\eta_1, \eta_2, \cdots, \eta_N\right], \end{equation*} Now combine the results above, we have, for each $k = {\lfloor nt_m \rfloor}$, $m = 1, \cdots, N$ \begin{align*} \dfrac{\sqrt{2\lvert \gamma_n \rvert^{3}}}{\alpha_n k^{1/4}} \dfrac{1}{\sqrt{E\xi_0^2}} \left(\dfrac{\sigma_{k}^2}{\omega {k}^{k/2}} - \sum_{j=1}^{k-1}e^{\frac{j\gamma_n}{\sqrt{k}}}\right) \xrightarrow{d} \mathcal{N}(0,1). \end{align*} Now, for returns, we know from the above result that \begin{align*} \dfrac{\lvert\gamma_n\rvert\sigma_{k}^2}{\omega {k}^{(k+1)/2}} - 1 = O_p\left(\dfrac{\alpha_n n^{1/4}}{\sqrt{\lvert \gamma_n \rvert}}\right) = o_p(1). \end{align*} Therefore, by the return equation, we have \begin{equation*} \left(\dfrac{\lvert\gamma_n\rvert }{\omega {k}^{(k+1)/2}} \right)^{1/2} u_k = \left(\dfrac{\lvert\gamma_n\rvert\sigma_{k}^2}{\omega {k}^{(k+1)/2}}\right)^{1/2} \varepsilon_k \sim \varepsilon_k. \end{equation*}

proof[Proof of Theorem (ref)] Similar to Theorem (ref), when $\gamma_n = 0$, the volatility admits the decomposition. Then, for $\sigma_{k,1}^{2}$, by central limit theorem, we know \begin{equation*} \dfrac{\alpha_n}{\sqrt{k}}\sum_{j=1}^{k}\xi_{k-j} = O_p(\alpha_n) = o_p(1) \end{equation*} which, combining with Proposition (ref), implies that \begin{equation*} \left\lvert \sigma_{k,1}^2 \right\rvert = O_p\left(k^{k/2}\right) \end{equation*} For $\sigma_{k,2}^{2}$, note that we have established equation ((ref)), then by Proposition (ref), we have \begin{equation*} \left\lvert \sigma_{k,2}^{2} \right\rvert = O_p\left(k^{k/2}\alpha_n^{2}\right). \end{equation*} For $\sigma_{k,3}^{2}$, by Proposition (ref) we have \begin{equation*} \left\lvert \sigma_{k,3}^{2} \right\rvert = O_p(k^{k/2}\alpha_n^{2}). \end{equation*} Lastly, for $\sigma_{k,4}^2$, note by Lemma 5.1 in BerkesHorvathKokoszka2005, for $k = \lfloor nt \rfloor$, $t \in (0, 1)$ , we have \begin{equation*} \dfrac{1}{n^{3/2}}\sum_{j=1}^{\lfloor nt \rfloor - 1}\sum_{i=1}^{j}\xi_{k-i} \xrightarrow{d} \sqrt{E\xi_{0}^2}\int_{0}^{t}xdW(x), \end{equation*} where $W(x)$ is a Wiener process. Therefore, for $k(m) = \lfloor nt_m \rfloor$, $m = 1, \cdots, N$, we have \begin{align*} \dfrac{k(m)^{1/2}}{n^{3/2}\alpha_n}\left(\dfrac{\sigma_{k(m)}^2}{\omega {k(m)}^{k(m)/2}} - k(m)\right) = \dfrac{1}{n^{3/2}}\sum_{j=1}^{\lfloor nt_m \rfloor - 1}\sum_{i=1}^{j}\xi_{k(m)-i} + o_p(1) \xrightarrow{d} \sqrt{E\xi_0^{2}}\int_{0}^{t_m}xdW(x). \end{align*} Further, note the results above implies that \begin{equation*} \dfrac{\sigma_{k(m)}^2}{\omega {k(m)}^{k(m)/2 + 1}} - 1 = O_p\left(\left(\dfrac{n}{k}\right)^{3/2}\alpha_n\right) = o_p(1). \end{equation*} Hence, by return equation, we obtain \begin{equation*} \left(\dfrac{1}{\omega {k(m)}^{k(m)/2 + 1}}\right)^{1/2}u_{k(m)} = \left(\dfrac{\sigma_{k(m)}^2}{\omega {k(m)}^{k(m)/2 + 1}}\right)^{1/2}\varepsilon_{k(m)} \xrightarrow{d} \varepsilon_{k(m)}. \end{equation*}

proof[Proof of Theorem (ref)] Similar to proof of Theorem (ref), when $\gamma_n > 0$, the volatility admits the additive representation. For $\sigma_{k,1}^{2}$, similar to that in Theorem (ref), \begin{equation*} \left\lvert \sigma_{k,1}^2 \right\rvert = O_p\left(k^{k/2}e^{\sqrt{k}\gamma_n}\right). \end{equation*} For $\sigma_{k,2}^2$, by Proposition (ref) and equation ((ref)), we have the relation \begin{align*} \left\lvert \sigma_{k,2}^{2} \right\rvert &= \left\lvert \omega k^{k/2}\sum_{j=1}^{k-1} je^{\frac{j\gamma_n}{\sqrt{k}}}(1+o_p(1))\dfrac{1}{j}R_{k,j}^{(3)} \right\rvert \\ &=O_p(1)\omega k^{k/2}(\alpha_n^{2} + \gamma_n^{2})\dfrac{e^{\frac{k\gamma_n}{\sqrt{k}}} - e^{\frac{\gamma_n}{\sqrt{k}}}}{e^{\frac{\gamma_n}{\sqrt{k}}}-1} \\ &< O_p\left(k^{k/2}\left(\alpha_n^2 + \gamma_n^2\right) \dfrac{\sqrt{k}e^{\sqrt{k}\gamma_n}}{\gamma_n}\right), \end{align*} where the last inequality comes from the fact that \begin{equation*} \dfrac{e^{\frac{k\gamma_n}{\sqrt{k}}} - e^{\frac{\gamma_n}{\sqrt{k}}}}{e^{\frac{\gamma_n}{\sqrt{k}}}-1} < \dfrac{e^{\sqrt{k}\gamma_n}}{\gamma_n/\sqrt{k}}. \end{equation*} For $\sigma_{k,3}^2$, by Proposition (ref), we have \begin{align*} \left\lvert \sigma_{k,3}^2 \right\rvert &= \left\lvert \omega k^{k/2} \sum_{j=1}^{k-1}e^{\frac{j\gamma_n}{\sqrt{k}}} \left(j\log\log j\right) \dfrac{1}{j\log\log j} R_{k,j}^{(2)}\right\rvert \\ &= O_p(1)\omega k^{k/2} \left(k\log\log k\right) \dfrac{\alpha_n^{2}}{k}\dfrac{e^{\frac{k\gamma_n}{\sqrt{k}}} - e^{\frac{\gamma_n}{\sqrt{k}}}}{e^{\frac{\gamma_n}{\sqrt{k}}}-1} \\ &< O_p\left(k^{k/2}\left(\alpha_n^2\log\log k\right)\dfrac{\sqrt{k}e^{\sqrt{k}\gamma_n}}{\gamma_n} \right). \end{align*} Lastly, for $\sigma_{k,4}^{2}$, we have \begin{equation*} \sigma_{k,4}^2 = \omega k^{k/2} \sum_{j=1}^{k-1}e^{\frac{j\gamma_n}{\sqrt{k}}} + \omega k^{k/2} \dfrac{\alpha_n}{\sqrt{k}} \sum_{j=1}^{k-1}e^{\frac{j\gamma_n}{\sqrt{k}}} \sum_{i=1}^{j}\xi_{k-i}. \end{equation*} Now, we introduce the following lemma to assist the proof. \begin{lemma} If Assumption (ref) and (ref) hold, then \begin{equation*} \dfrac{\gamma_n^2}{k}e^{-2\sqrt{k}\gamma_n}E\left(\dfrac{1}{\sqrt{k}}\sum_{j=1}^{k-1}e^{\frac{j\gamma_n}{\sqrt{k}}}\sum_{i=1}^{j}\xi_{k-i} - \dfrac{e^{\sqrt{k}\gamma_n}}{\gamma_n}\sum_{i=1}^{k-1}\xi_i\right)^2 \rightarrow 0. \end{equation*} \end{lemma} Then by Lemma (ref), we have \begin{align*} \dfrac{\gamma_n e^{-\sqrt{k}\gamma_n}}{\sqrt{k}\alpha_n} \left(\dfrac{\sigma_{k,4}^2}{\omega k^{k/2}} - \sum_{j=1}^{k-1}e^{\frac{j\gamma_n}{\sqrt{k}}}\right) &= \dfrac{\gamma_n e^{-\sqrt{k}\gamma_n}}{\sqrt{k}}\dfrac{1}{\sqrt{k}}\sum_{j=1}^{k-1}e^{\frac{j\gamma_n}{\sqrt{k}}}\sum_{i=1}^{j}\xi_{k-i} + o_p(1) = \dfrac{1}{\sqrt{k}}\sum_{i=1}^{k-1}\xi_i + o_p(1). \end{align*} Therefore, by Donsker's theorem, we obtain that, for $k(m) = \lfloor nt_m \rfloor$, $t_m \in (0, 1)$ and $m = 1, 2, \cdots, N$, \begin{equation*} \dfrac{\gamma_n e^{-\sqrt{k(m)}\gamma_n}}{\sqrt{k(m)}\alpha_n} \dfrac{1}{\sqrt{E\xi_0^2}} \left(\dfrac{\sigma_{k(m)}^2}{\omega k(m)^{k(m)/2}} - \sum_{j=1}^{k(m)-1}e^{\frac{j\gamma_n}{\sqrt{k(m)}}}\right) \Rightarrow W(t_m), \end{equation*} where $W(t)$ is a finite dimensional Wiener process. Further, note that \begin{equation*} \dfrac{\gamma_n}{\sqrt{k}}e^{-\sqrt{k}\gamma_n}\left(\sum_{j=1}^{k-1}e^{\frac{j\gamma_n}{\sqrt{k}}} - \dfrac{\sqrt{k}e^{\sqrt{k}\gamma_n}}{\gamma_n}\right) = o(1), \end{equation*} then by the result above we know \begin{equation*} \dfrac{\gamma_n e^{-\sqrt{k(m)}\gamma_n}}{\sqrt{k(m)}} \left(\dfrac{\sigma_{k(m)}^2}{\omega k(m)^{k(m)/2}} - \sum_{j=1}^{k(m)-1}e^{\frac{j\gamma_n}{\sqrt{k(m)}}}\right) = O_p(\alpha_n) = o_p(1). \end{equation*} Hence, by return equantion, we derive \begin{equation*} \left(\dfrac{\gamma_n e^{-\sqrt{k}\gamma_n}}{\omega k^{(k+1)/2}}\right)^{1/2}u_{k} = \left(\dfrac{\gamma_n e^{-\sqrt{k}\gamma_n}}{\omega k^{(k+1)/2}}\sigma_k^2\right)^{1/2}\varepsilon_{k} \sim \varepsilon_k. \end{equation*}