A copula spectral test for pairwise time reversibility

Abstract

In this paper, we propose a new frequency domain test for pairwise time reversibility at any specific couple of quantiles of two-dimensional marginal distribution. The proposed test is applicable to a very broad class of time series, regardless of the existence of moments and Markovian properties. By varying the couple of quantiles, the test can detect any violation of pairwise time reversibility. Our approach is based on an estimator of the $L^2$-distance between the imaginary part of copula spectral density kernel and its value under the null hypothesis. We show that the limiting distribution of the proposed test statistic is normal and investigate the finite sample performance by means of a simulation study. We illustrate the use of the proposed test by applying it to stock price data.

Appendix Details for the proofs in Sect. 3

This appendix contains the detailed proofs of Theorems 1 and 2. For ease of notation, we use C for any generic positive constant.

1.1 Main lemmas used in the proofs

Recall that the rth-order joint cumulant $\text {cum}(\zeta _1,\ldots , \zeta _r)$ of the random vector $(\zeta _1,\ldots ,\zeta _r)$ is defined as

$$\begin{aligned} \textrm{cum}(\zeta _1,\ldots ,\zeta _r):=\sum _{\{\nu _1,\ldots ,\nu _p\}}(-1)^{p-1} (p-1)! \big (\textbf{E}\prod _{j\in \nu _1}\zeta _j\big )\cdots \big (\textbf{E}\prod _{j\in \nu _p}\zeta _j\big ), \end{aligned}$$

with summation extending over all partitions $\{\nu _1,\ldots ,\nu _p\}$, $p=1,\ldots ,r$, of $\{1,\ldots ,r\}$ (cf. Brillinger 2001, pp. 19). If $\zeta _1=\cdots =\zeta _r=\zeta$, we use $\textrm{cum}_r(\zeta )$ to denote $\textrm{cum}(\zeta _1,\ldots ,\zeta _r)$.

Lemma 1

If assumption (M) holds, then the following assumption (C) also holds.

(C) There exist constants $\rho \in (0,1)$ and $K<\infty$ such that, for arbitrary intervals $A_1,\ldots , A_p\subset \mathbb {R}$ and arbitrary $t_1,\ldots ,t_p\in \mathbb {Z}$,

$$\begin{aligned} \big |\textrm{cum}\big (\mathbb {I}_{A_1}(x_{t_1}),\ldots ,\mathbb {I}_{A_p}(x_{t_p})\big )\big |\le K \rho ^{\max _{i,j}|t_i-t_j|}. \end{aligned}$$

(19)

Proof

Let $\alpha (n):=\sup \{\textbf{P}(A B)-\textbf{P}(A)\textbf{P}(B): \, A\in \sigma (x_k; k\le 0),\, B\in \sigma (x_k; k\ge n)\}$. According to Bradley (2005), we have $\alpha (n)\le \frac{1}{2}\beta (n)$. Then, by applying Proposition 3.1 of Kley et al. (2016), we obtain the lemma. $\square$

For $\omega \in (0,\pi ]$ and $(\tau _1,\tau _2)\in [0,1]^2$, we denote

$$\begin{aligned} \textbf{y}_{n,U}^{\tau _1, \tau _2}(\omega )=\big (\textrm{Re}\, y_{n,U}^{\tau _1}(\omega ),\textrm{Im}\, y_{n,U}^{\tau _1}(\omega ),\textrm{Re}\,y_{n,U}^{\tau _2}(\omega ), \textrm{Im}\,y_{n,U}^{\tau _2}(\omega )\big )^T, \end{aligned}$$

where $y_{n,U}^{\tau }(\omega )$ is defined in (14).

Then, according to (1.6) of Kley et al. (2015), we have the following lemma.

Lemma 2

If $\{x_t\}_{t\in \mathbb {Z}}$ is strictly stationary and satisfies assumption (C), then for every $\omega \in (0,\pi ]$, we have

$$\begin{aligned} \textbf{y}_{n,U}^{\tau _1, \tau _2}(\omega )\rightsquigarrow \mathbb {Y}^{\tau _1, \tau _2}(\omega ), \end{aligned}$$

as n goes to infinity, where $\mathbb {Y}^{\tau _1, \tau _2}(\omega )=(\mathbb {C}^{\tau _1}(\omega ),\mathbb {D}^{\tau _1}(\omega ),\mathbb {C}^{\tau _2}(\omega ),\mathbb {D}^{\tau _2}(\omega ))^T$ follows a four-dimensional zero-mean Gaussian distribution with covariance matrix $\frac{1}{2}\begin{pmatrix} \Sigma _{11} &{} \Sigma _{12}\\ \Sigma _{21} &{} \Sigma _{22} \end{pmatrix}$, with

$$\begin{aligned} \Sigma _{ij}={\left\{ \begin{array}{ll} \begin{pmatrix} f_{q_{\tau _i},q_{\tau _i}}(\omega ) &{} 0\\ 0 &{} f_{q_{\tau _i},q_{\tau _i}}(\omega )\end{pmatrix}, &{} \text {if }i=j,\\ \begin{pmatrix} \textrm{Re} f_{q_{\tau _i},q_{\tau _j}}(\omega ) &{} \textrm{Im} f_{q_{\tau _i},q_{\tau _j}}(\omega )\\ -\textrm{Im} f_{q_{\tau _i},q_{\tau _j}}(\omega ) &{} \textrm{Re} f_{q_{\tau _i},q_{\tau _j}}(\omega )\end{pmatrix},&\text {if }i\ne j. \end{array}\right. } \end{aligned}$$

(20)

Moreover, $\textbf{y}_{n,U}^{\tau _1, \tau _2}(\omega )$ is asymptotically independent for distinct $\omega$’s.

Let $\textrm{Im}\, I^{\tau _1,\tau _2}(\omega )=\mathbb {C}^{\tau _1}(\omega )\mathbb {D}^{\tau _2}(\omega )-\mathbb {D}^{\tau _1}(\omega )\mathbb {C}^{\tau _2}(\omega )$. Then, we have:

Lemma 3

If $\{x_t\}_{t\in \mathbb {Z}}$ is strictly stationary and satisfies assumption (C), then $I^{\tau _1,\tau _2}(\omega )$s are independent among distinct $\omega$’s; moreover, for every $\omega \in (0,\pi ]$ and $\tau _1,\tau _2\in [0,1]$, it holds

$$\begin{aligned} \textbf{E}\big [\textrm{Im}\, I^{\tau _1,\tau _2}(\omega )\big ]= \textrm{Im} f_{q_{\tau _1},q_{\tau _2}}(\omega ) \end{aligned}$$

(21)

and

$$\begin{aligned} \textbf{E}\big [\big (\textrm{Im}\,I^{\tau _1,\tau _2}(\omega )\big )^2\big ]=&\frac{3}{2}\big (\textrm{Im} f_{q_{\tau _1},q_{\tau _2}}(\omega )\big )^2+\frac{1}{2}f_{q_{\tau _1},q_{\tau _1}}(\omega )f_{q_{\tau _2},q_{\tau _2}}(\omega )\nonumber \\&\quad -\frac{1}{2} \big (\textrm{Re} f_{q_{\tau _1},q_{\tau _2}}(\omega )\big )^2=: A_{\tau _1,\tau _2}(\omega ). \end{aligned}$$

(22)

Proof

To prove (21) and (22), from Lemma 2, it suffices to verify

$$\begin{aligned} \textbf{E}\big [\mathbb {C}^{\tau _1}(\omega )\mathbb {D}^{\tau _2}(\omega )-\mathbb {D}^{\tau _1}(\omega )\mathbb {C}^{\tau _2}(\omega )\big ]=\textrm{Im} f_{q_{\tau _1},q_{\tau _2}}(\omega ) \end{aligned}$$

(23)

and

$$\begin{aligned} \textbf{E}\big [\big (\mathbb {C}^{\tau _1}(\omega )\mathbb {D}^{\tau _2}(\omega )-\mathbb {D}^{\tau _1}(\omega )\mathbb {C}^{\tau _2}(\omega )\big )^2\big ]=A_{\tau _1,\tau _2}(\omega ). \end{aligned}$$

(24)

From (20), the equality (23) holds obviously. By using Lemma 2.2 of Nagao (1973) and (20), some straightforward calculations yield the equality (24).$\square$

For $p\ge 2$, $k_1,\ldots ,k_{p-1}\in \mathbb {Z}$ and the quantile levels $\tau _1,\ldots ,\tau _p\in [0,1]$, consider the copula cumulant kernel of order p

$$\begin{aligned} \gamma _{k_1,\ldots ,k_{p-1}}^U (\tau _1,\ldots ,\tau _p):=\textrm{cum}\big (\mathbb {I}_{(0,\tau _1]}(U_{0}),\mathbb {I}_{(0,\tau _1]}(U_{k_1}),\ldots ,\mathbb {I}_{(0,\tau _p]}(U_{k_{p-1}})\big ), \end{aligned}$$

where $U_t=F(x_t)$. Note that, under assumption (C), the following quantity, which we call copula spectral density kernel of order p,

$$\begin{aligned} f_{q_{\tau _1},\ldots ,q_{\tau _p}}(\omega _1,\ldots ,\omega _{p-1}):=\frac{1}{(2\pi )^{p-1}} \sum _{k_1,\ldots ,k_{p-1}=\infty }^\infty \gamma _{k_1,\ldots ,k_{p-1}}^U (\tau _1,\ldots ,\tau _p)\,\textrm{e}^{-\textrm{i}(k_1 \omega _1+\cdots +k_{p-1}\omega _{p-1})} \end{aligned}$$

exists for all $p\ge 2$ (pp. 1 of(Kley et al. 2015)).

Let

$$\begin{aligned} \varepsilon (\tau _1,\ldots ,\tau _p,\omega _1,\ldots ,\omega _p):=&\textrm{cum}\big (y_{n,U}^{\tau _1}(\omega _1),\ldots ,y_{n,U}^{\tau _p}(\omega _p)\big )\\&- (2\pi )^{p/2-1} n^{-p/2} \Delta _n\big (\sum _{i=1}^p \omega _i\big )f_{q_{\tau _1},\ldots ,q_{\tau _p}}(\omega _1,\ldots ,\omega _{p-1}), \end{aligned}$$

where $\Delta _n(\omega ):=\sum _{t=1}^n \textrm{e}^{-\textrm{i} \omega t}$.

By Theorem 1.3 of Kley et al. (2015), we obtain the following lemma directly.

Lemma 4

If $\{x_t\}_{t\in \mathbb {Z}}$ is strictly stationary and satisfies assumption (C), then

$$\begin{aligned} \sup _{\tau _1,\ldots ,\tau _p\in [0,1]} \sup _{\omega _1,\ldots ,\omega _p\in (0,\pi ]}\big |\varepsilon (\tau _1,\ldots ,\tau _p,\omega _1,\ldots ,\omega _p)\big |=O(n^{-p/2}). \end{aligned}$$

(25)

Since $|\Delta _n(\omega )|\le n$ holds, and $f_{q_{\tau _1},\ldots ,q_{\tau _p}}(\omega _1,\ldots ,\omega _{p-1})$ is bounded above uniformly for $(\omega _1,\ldots ,\omega _{p-1})\in (0,\pi ]^{p-1}$ (pp. 1 of Kley et al. 2015), we obtain:

Corollary 1

If $\{x_t\}_{t\in \mathbb {Z}}$ is strictly stationary and satisfies assumption (C), then

$$\begin{aligned} \sup _{\tau _1,\ldots ,\tau _p\in [0,1]} \sup _{\omega _1,\ldots ,\omega _p\in (0,\pi ]}\big |\textrm{cum}\big (y_{n,U}^{\tau _1}(\omega _1),\ldots ,y_{n,U}^{\tau _p}(\omega _p)\big )\big |=O(n^{1-p/2}). \end{aligned}$$

(26)

Lemma 5

If $\{x_t\}_{t\in \mathbb {Z}}$ is strictly stationary and satisfies assumption (C), then

$$\begin{aligned}&\textbf{E}\big [\textrm{Im}\, I_{n,U}^{\tau _1,\tau _2}(\omega )\big ]= \textrm{Im} f_{q_{\tau _1},q_{\tau _2}}(\omega ) + O(\frac{1}{n}) \end{aligned}$$

(27)

and

$$\begin{aligned} \textbf{E}\big [\textrm{Im}\,I_{n,U}^{\tau _1,\tau _2}(\omega _1)\, \textrm{Im}\,I_{n,U}^{\tau _1,\tau _2}(\omega _2)\big ] =\textrm{Im} f_{q_{\tau _1},q_{\tau _2}}(\omega _1)\, \textrm{Im} f_{q_{\tau _1},q_{\tau _2}}(\omega _2)+O(\frac{1}{\sqrt{n}}) \end{aligned}$$

(28)

hold uniformly for all $\tau _1,\tau _2\in [0,1]$ and for all $\omega ,\omega _1,\omega _2\in (0,\pi ]$. More generally, for each $p\in \mathbb {N}$ and $k_1,\ldots ,k_p\in \mathbb {N}$,

$$\begin{aligned} \textbf{E}\big [\big (\textrm{Im}\,I_{n,U}^{\tau _1,\tau _2}(\omega _1)\big )^{k_1}\cdots \big (\textrm{Im}\,I_{n,U}^{\tau _1,\tau _2}(\omega _p)\big )^{k_p}\big ] =\textbf{E}\big [\big (\textrm{Im}\,I^{\tau _1,\tau _2}(\omega _1)\big )^{k_1}\cdots \big (\textrm{Im}\,I^{\tau _1,\tau _2}(\omega _p)\big )^{k_p}\big ]+O(\frac{1}{\sqrt{n}}) \end{aligned}$$

(29)

hold uniformly for all $\tau _1,\tau _2\in [0,1]$ and all $\omega _1,\ldots ,\omega _2\in (0,\pi ]$.

Proof

Note that,

$$\begin{aligned} \textrm{Im}\,I_{n,U}^{\tau _1,\tau _2}(\omega )=\frac{I_{n,U}^{\tau _1,\tau _2}(\omega )-I_{n,U}^{\tau _1,\tau _2}(-\omega )}{2\,\textrm{i}} =\frac{y_{n,U}^{\tau _1}(-\omega )y_{n,U}^{\tau _2}(\omega )-y_{n,U}^{\tau _1}(\omega )y_{n,U}^{\tau _2}(-\omega )}{2\,\textrm{i}}. \end{aligned}$$

(30)

First expressing moments of the form $\textbf{E}[y_{n,U}^{\tau _1}(\omega _1)\cdots y_{n,U}^{\tau _p}(\omega _p)]$ in terms of cumulants, then using (25) for $p=2$ and (26) for $p\ge 3$ recursively, we can obtain (27)–(29).

To illustrate, we only prove (27).

Expressing $\textbf{E}[y_{n,U}^{\tau _1}(-\omega )y_{n,U}^{\tau _2}(\omega )]$ in terms of cumulants, we obtain

$$\begin{aligned} \textbf{E}[y_{n,U}^{\tau _1}(-\omega )y_{n,U}^{\tau _2}(\omega )] =\textrm{cum}(y_{n,U}^{\tau _1}(-\omega ),y_{n,U}^{\tau _2}(\omega ))+ \textbf{E}[y_{n,U}^{\tau _1}(-\omega )]\textbf{E}[y_{n,U}^{\tau _2}(\omega )]. \end{aligned}$$

(31)

According to (25), we have

$$\begin{aligned} \textrm{cum}(y_{n,U}^{\tau _1}(-\omega ),y_{n,U}^{\tau _2}(\omega ))= f_{q_{\tau _1},q_{\tau _2}}(\omega )+\varepsilon (\tau _1,\tau _2,-\omega ,\omega ) =f_{q_{\tau _1},q_{\tau _2}}(\omega )+O(\frac{1}{n}), \end{aligned}$$

(32)

where $O(\frac{1}{n})$ holds uniformly for all $\tau _1,\tau _2\in [0,1]$ and all $\omega \in (0,\pi ]$. Combining (31) with (32), we obtain

$$\begin{aligned} \textbf{E}[y_{n,U}^{\tau _1}(-\omega )y_{n,U}^{\tau _2}(\omega )] =f_{q_{\tau _1},q_{\tau _2}}(\omega )+ \textbf{E}[y_{n,U}^{\tau _1}(-\omega )]\textbf{E}[y_{n,U}^{\tau _2}(\omega )]+O(\frac{1}{n}). \end{aligned}$$

(33)

With arguments similar to prove (33), we have

$$\begin{aligned} \textbf{E}[y_{n,U}^{\tau _1}(\omega )y_{n,U}^{\tau _2}(-\omega )] =f_{q_{\tau _1},q_{\tau _2}}(-\omega )+ \textbf{E}[y_{n,U}^{\tau _1}(\omega )]\textbf{E}[y_{n,U}^{\tau _2}(-\omega )]+O(\frac{1}{n}). \end{aligned}$$

(34)

Noting that $\textbf{E}[y_{n,U}^{\tau }(\omega )]=(\sqrt{2\pi n})^{-1} \tau \sum _{t=1}^n \textrm{e}^{\textrm{i}\omega t}$, we find

$$\begin{aligned} \textbf{E}[y_{n,U}^{\tau _1}(-\omega )]\textbf{E}[y_{n,U}^{\tau _2}(\omega )]=\textbf{E}[y_{n,U}^{\tau _1}(\omega )]\textbf{E}[y_{n,U}^{\tau _2}(-\omega )]. \end{aligned}$$

(35)

Combining (30) and (33)–(35) yields

$$\begin{aligned} \textbf{E}\big [ \textrm{Im}\,I_{n,U}^{\tau _1,\tau _2}(\omega )\big ] =\frac{f_{q_{\tau _1},q_{\tau _2}}(\omega )-f_{q_{\tau _1},q_{\tau _2}}(-\omega )}{2\,\textrm{i}}+O(\frac{1}{n}) =\textrm{Im}\,f_{q_{\tau _1},q_{\tau _2}}(\omega )+O(\frac{1}{n}). \end{aligned}$$

$\square$

Lemma 6

If assumption (M) is satisfied, then we have for each $\tau \in [0,1]$,

$$\begin{aligned} \sup _{\omega \in (0,2\pi ]}|y_{n,R}^{\tau }(\omega )-y_{n,U}^{\tau }(\omega )|=o_P(n^{-1/4+\delta /2}) \end{aligned}$$

(36)

holds, where $\delta$ can be taken to any constant satisfying $0<\delta <1/2$, as in Condition 1.

Proof

If assumption (M) is satisfied, then for every $0<\delta <1/2$, there exists a $\theta >1/\delta -1$ such that $\beta (n)=O(n^{-\theta })$. It follows from (A.13) of Dette et al. (2015) that

$$\begin{aligned} \sup _{\omega \in (0,2\pi ]}|y_{n,R}^{\tau }(\omega )-y_{n,U}^{\tau }(\omega )|&=n^{-1/4} (n^{\frac{1}{1+\theta }}\log n)^{1/2} \log n \,O_P(1)\\&=n^{-1/4+\delta /2} \frac{(\log n)^{3/2}}{n^{\delta /2-1/2(1+\theta )}} \,O_P(1)=n^{-1/4+\delta /2} \,o_P(1), \end{aligned}$$

since $\delta >1/(1+\theta )$ holds. This proves the lemma.$\square$

Lemma 7

If assumption (M) is satisfied, then

$$\begin{aligned}&\sup _{\omega \in (0,2\pi ]}|I_{n,U}^{\tau _1,\tau _2}(\omega )|=O_P(n^{1/k}), \end{aligned}$$

(37)

$$\begin{aligned}&\sup _{\omega \in (0,2\pi ]}|I_{n,R}^{\tau _1,\tau _2}(\omega )|=O_P(n^{1/k}) \end{aligned}$$

(38)

and

$$\begin{aligned} \sup _{\omega \in (0,2\pi ]}|I_{n,R}^{\tau _1,\tau _2}(\omega )-I_{n,U}^{\tau _1,\tau _2}(\omega )|=o_P(n^{-1/4+\delta /2}) \end{aligned}$$

(39)

hold for any $k\in \mathbb {N}$ and any $\delta \in (0,1/2)$.

Proof

By Lemma 1, assumption (M) implies assumption (C). Since $y_{n,U}^{\tau }(\omega )$ is defined as a piecewise constant function extended from (14), by Lemma A.6 of Kley et al. (2016), we have $\sup _{\omega \in (0,2\pi ]}|y_{n,U}^{\tau }(\omega )|=O_P(n^{1/(2k)})$ for any $k\in \mathbb {N}$. Since the inequality $\sup _{\omega \in (0,2\pi ]}|I_{n,U}^{\tau _1,\tau _2}(\omega )|\le \big (\sup _{\omega \in (0,2\pi ]}|y_{n,U}^{\tau }(\omega )|\big )^2$ holds, we obtain (37).

By applying the triangular inequality, we have

$$\begin{aligned} |I_{n,R}^{\tau _1,\tau _2}(\omega )-I_{n,U}^{\tau _1,\tau _2}(\omega )|&=|y_{n,R}^{\tau _1}(\omega )| |y_{n,R}^{\tau _2}(\omega )-y_{n,U}^{\tau _2}(\omega )|+|y_{n,U}^{\tau _2}(\omega )| |y_{n,R}^{\tau _1}(\omega )-y_{n,U}^{\tau _1}(\omega )|. \end{aligned}$$

For any $\delta \in (0,1/2)$, choosing any $\delta _0\in (0,\delta )$, taking $k=[1/(\delta -\delta _0)]$ and applying (36), we obtain

$$\begin{aligned}&\sup _{\omega \in (0,2\pi ]} |y_{n,R}^{\tau _1}(\omega )| |y_{n,R}^{\tau _2}(\omega )-y_{n,U}^{\tau _2}(\omega )|\le \sup _{\omega \in (0,2\pi ]} |y_{n,R}^{\tau _1}(\omega )| \sup _{\omega \in (0,2\pi ]} |y_{n,R}^{\tau _2}(\omega )-y_{n,U}^{\tau _2}(\omega )|\\&\qquad =O_P(n^{1/(2k)})\, o_P(n^{-1/4+\delta _0/2})=o_P(n^{-1/4+\delta /2}), \end{aligned}$$

since $\sup _{\omega \in (0,2\pi ]} |y_{n,R}^{\tau }(\omega )|\le \sup _{\omega \in (0,2\pi ]} |y_{n,U}^{\tau }(\omega )|+\sup _{\omega \in (0,2\pi ]} |y_{n,R}^{\tau }(\omega )-y_{n,U}^{\tau }(\omega )|=O_P(n^{1/(2k)})$ holds for any $k\in \mathbb {N}$. Similarly, we obtain $\sup _{\omega \in (0,2\pi ]} |y_{n,U}^{\tau _2}(\omega )| |y_{n,R}^{\tau _1}(\omega )-y_{n,U}^{\tau _1}(\omega )|=o_P(n^{-1/4+\delta /2})$. This proves (39).

Since we have (37) and (39), using the triangular inequality straightforwardly produces (38).$\square$

Lemma 8

If assumption (C) is satisfied, then the CSDK is uniformly Hölder continuous, i.e.,

$$\begin{aligned} |f_{q_{\tau _1},q_{\tau _2}}(\omega _1)- f_{q_{\tau _1},q_{\tau _2}}(\omega _2)|\le C |\omega _1-\omega _2| \end{aligned}$$

(40)

holds uniformly for all $(\tau _1,\tau _2)\in [0,1]^2$.

Proof

The assumption (C) implies that $|\textbf{Cov}(\mathbb {I}_{A_1}(x_{t_1}),\mathbb {I}_{A_2}(x_{t_2}))|\le K \rho ^{|t_1-t_2|}$ holds for arbitrary intervals $A_1, A_2\subset \mathbb {R}$ and arbitrary $t_1,t_2\in \mathbb {Z}$, so does the $\gamma _k^U(\tau _1,\tau _2)$. By the triangular inequality, we obtain

$$\begin{aligned}&|f_{q_{\tau _1},q_{\tau _2}}(\omega _1)-f_{q_{\tau _1},q_{\tau _2}}(\omega _2)| =\Big |\frac{1}{2\pi }\sum _{k=-\infty }^\infty \gamma _k^U(\tau _1,\tau _2) (\textrm{e}^{-\textrm{i}k\omega _1}-\textrm{e}^{-\textrm{i}k\omega _2})\Big |\\&\quad \le \frac{1}{2\pi }\sum _{k=-\infty }^\infty | \gamma _k^U(\tau _1,\tau _2) | \, |\textrm{e}^{-\textrm{i}k\omega _1}| \, |1-\textrm{e}^{\textrm{i}k(\omega _1-\omega _2)}|\le \frac{K}{2\pi }(\sum _{k=-\infty }^\infty k \rho ^k) \, |\omega _1-\omega _2|=C|\omega _1-\omega _2|, \end{aligned}$$

where the last inequality follows from the inequality $|\textrm{e}^{\textrm{i} a}-1|\le |a|$.$\square$

1.2 Proofs of Theorems 1 and 2

Proof of Theorem 1. Let

$$\begin{aligned} \widetilde{T}_M^{(\tau _1,\tau _2)}=\frac{\pi }{M} \sum _{m=1}^M \text {Im} f_{q_{\tau _1},q_{\tau _2}}\big (\frac{m}{M}\pi \big ) f_{q_{\tau _1},q_{\tau _2}}\big (\frac{m-1}{M}\pi \big ), \end{aligned}$$

and let

$$\begin{aligned} T_M^{(\tau _1,\tau _2)}=\frac{\pi }{M} \sum _{m=1}^M \Big (\text {Im} f_{q_{\tau _1},q_{\tau _2}}\big (\frac{m}{M}\pi \big )\Big )^2. \end{aligned}$$

We have the decomposition

$$\begin{aligned} \sqrt{M}\big (T_{n,M}^{(\tau _1,\tau _2)}-T^{(\tau _1,\tau _2)}\big ) =&\sqrt{M}\big (T_{n,M}^{(\tau _1,\tau _2)}-\widetilde{T}_{n,M}^{(\tau _1,\tau _2)}\big ) +\sqrt{M}\big (\widetilde{T}_{n,M}^{(\tau _1,\tau _2)}-\widetilde{T}_M^{(\tau _1,\tau _2)}\big ) \\&+\sqrt{M}\big (\widetilde{T}_M^{(\tau _1,\tau _2)}-T_M^{(\tau _1,\tau _2)}\big ) +\sqrt{M}\big (T_M^{(\tau _1,\tau _2)}-T^{(\tau _1,\tau _2)}\big ), \end{aligned}$$

where $\widetilde{T}_{n,M}^{(\tau _1,\tau _2)}$ is given by (12).

To prove (9), it suffices to verify

$$\begin{aligned}&\sqrt{M}\big (T_{n,M}^{(\tau _1,\tau _2)}-\widetilde{T}_{n,M}^{(\tau _1,\tau _2)}\big ){\mathop {\longrightarrow }\limits ^{P}}0, \end{aligned}$$

(41)

$$\begin{aligned}&\sqrt{M}\big (\widetilde{T}_{n,M}^{(\tau _1,\tau _2)}-\widetilde{T}_M^{(\tau _1,\tau _2)}\big )\rightsquigarrow N(0,V^{(\tau _1,\tau _2)}), \end{aligned}$$

(42)

$$\begin{aligned}&\sqrt{M}\big (\widetilde{T}_M^{(\tau _1,\tau _2)}-T_M^{(\tau _1,\tau _2)}\big )\longrightarrow 0 \end{aligned}$$

(43)

and

$$\begin{aligned} \sqrt{M}\big (T_M^{(\tau _1,\tau _2)}-T^{(\tau _1,\tau _2)}\big )\rightarrow 0, \end{aligned}$$

(44)

as n goes to infinity, where “${\mathop {\longrightarrow }\limits ^{P}}$” denotes convergence in probability.

Proof of (41). By the triangular inequality, we have

$$\begin{aligned}&\big |\sqrt{M}\big (T_{n,M}^{(\tau _1,\tau _2)}-\widetilde{T}_{n,M}^{(\tau _1,\tau _2)}\big )\big |\\&\qquad \le \frac{\pi }{\sqrt{M}}\sum _{m=1}^M \big |\text {Im}\,I_{n,R}^{\tau _1,\tau _2}\big (\frac{m-1}{M}\pi \big )\big | \big |\text {Im}\,I_{n,R}^{\tau _1,\tau _2}\big (\frac{m}{M}\pi \big )- \text {Im}\,I_{n,U}^{\tau _1,\tau _2}\big (\frac{m}{M}\pi \big ) \big |\\&\qquad \qquad + \frac{\pi }{\sqrt{M}}\sum _{m=1}^M \big |\text {Im}\,I_{n,U}^{\tau _1,\tau _2}\big (\frac{m}{M}\pi \big )\big | \big |\text {Im}\,I_{n,R}^{\tau _1,\tau _2}\big (\frac{m-1}{M}\pi \big )- \text {Im}\,I_{n,U}^{\tau _1,\tau _2}\big (\frac{m-1}{M}\pi \big ) \big |\\&\qquad =: B_{1,n,M}+B_{2,n,M}. \end{aligned}$$

For any $\delta \in (0,1/2)$, choosing a $\delta _0\in (0,\delta )$ and taking $k=[2/(\delta -\delta _0)]$, it follows from Lemma 7 that

$$\begin{aligned} B_{1,n,M}&\le \pi \sqrt{M} \sup _{\omega \in (0,2\pi ]}|I_{n,R}^{\tau _1,\tau _2}(\omega )| \sup _{\omega \in (0,2\pi ]}|I_{n,R}^{\tau _1,\tau _2}(\omega )-I_{n,U}^{\tau _1,\tau _2}(\omega )|\\&=o_P(M^{1/2}/n^{1/4-\delta _0/2-1/k})=o_P\Big (\sqrt{\frac{M}{n^{1/2-\delta }}}\Big ) \end{aligned}$$

holds. Similarly, we also have $B_{2,n,M}=o_P\Big (\sqrt{\frac{M}{n^{1/2-\delta }}}\Big )$. This proves (41).

Proof of (42). According to Lemma P4.5 of Brillinger (2001), to prove (42), it is required to check that the q-th-order cumulant of $\sqrt{M}\big (\widetilde{T}_{n,M}^{(\tau _1,\tau _2)}-\widetilde{T}_M^{(\tau _1,\tau _2)}\big )$ behaves in the manner required by the $N(0,V^{(\tau _1,\tau _2)})$, for each $q\ge 1$.

By (28), we know $\textbf{E}\big [\sqrt{M}\big (\widetilde{T}_{n,M}^{(\tau _1,\tau _2)}-\widetilde{T}_M^{(\tau _1,\tau _2)}\big )\big ]=O(\sqrt{M/n})$. This indicates that the first cumulant behaves in the manner required.

From (29), we obtain that for each $k\in \mathbb {N}$,

$$\begin{aligned}&\textbf{E}\Big [\Big (\sum _{m=1}^M \text {Im}\,I_{n,M}^{\tau _1,\tau _2}\big (\frac{m}{M}\pi \big ) \, \text {Im}\,I_{n,M}^{\tau _1,\tau _2}\big (\frac{m-1}{M}\pi \big )\Big )^2\Big ]\\&\qquad =\textbf{E}\Big [\Big (\sum _{m=1}^M \text {Im}\,I^{\tau _1,\tau _2}\big (\frac{m}{M}\pi \big ) \, \text {Im}\,I^{\tau _1,\tau _2}\big (\frac{m-1}{M}\pi \big )\Big )^2\Big ]+O(\frac{M^2}{\sqrt{n}}) \end{aligned}$$

holds. This implies that

$$\begin{aligned}&\textbf{Var}\big (\sqrt{M}\widetilde{T}_{n,M}^{(\tau _1,\tau _2)}\big ) =V_M^{(\tau _1,\tau _2)}+O\big (\frac{M}{n^{1/2}}\big ), \end{aligned}$$

(45)

with

$$\begin{aligned}&V_M^{(\tau _1,\tau _2)}=\frac{\pi ^2}{M}\sum _{m=1}^M \textbf{Var}\big (\text {Im}\,I^{\tau _1,\tau _2}(\frac{m}{M}\pi ) \text {Im}\, I^{\tau _1,\tau _2}(\frac{m-1}{M}\pi )\big ) \nonumber \\&\quad +2\frac{\pi ^2}{M}\sum _{m=2}^M \textbf{Cov}\big (\text {Im}\, I^{\tau _1,\tau _2}(\frac{m}{M}\pi ) \text {Im}\, I^{\tau _1,\tau _2}(\frac{m-1}{M}\pi ), \text {Im}\, I^{\tau _1,\tau _2}(\frac{m-1}{M}\pi ) \text {Im}\, I^{\tau _1,\tau _2}(\frac{m-2}{M}\pi )\big ). \end{aligned}$$

Employing Lemma 3 in combination with straightforward derivations yields

$$\begin{aligned} V_M^{(\tau _1,\tau _2)}=&\frac{\pi ^2}{M}\sum _{m=1}^M \Big \{A_{\tau _1,\tau _2}\big (\frac{m}{M}\pi \big )A_{\tau _1,\tau _2}\big (\frac{m-1}{M}\pi \big )\nonumber \\&\quad -\big (\textrm{Im} f_{q_{\tau _1},q_{\tau _2}}\big (\frac{m}{M}\pi \big )\big )^2 \big (\textrm{Im} f_{q_{\tau _1},q_{\tau _2}}\big (\frac{m-1}{M}\pi \big )\big )^2\Big \} \nonumber \\&+2\frac{\pi ^2}{M}\sum _{m=2}^M \Big \{\textrm{Im} f_{q_{\tau _1},q_{\tau _2}}\big (\frac{m}{M}\pi \big ) A_{\tau _1,\tau _2}\big (\frac{m-1}{M}\pi \big )\textrm{Im} f_{q_{\tau _1},q_{\tau _2}}\big (\frac{m-2}{M}\pi \big ) \nonumber \\&\quad -\textrm{Im} f_{q_{\tau _1},q_{\tau _2}}\big (\frac{m}{M}\pi \big )\big (\textrm{Im} f_{q_{\tau _1},q_{\tau _2}}\big (\frac{m-1}{M}\pi \big )\big )^2\textrm{Im} f_{q_{\tau _1},q_{\tau _2}}\big (\frac{m-2}{M}\pi \big )\Big \}, \end{aligned}$$

(46)

where $A_{\tau _1,\tau _2}(\cdot )$ is defined in (22). As M goes to infinity, we have an alternative expression of (46) as

$$\begin{aligned}&V_M^{(\tau _1,\tau _2)}=\frac{\pi ^2}{M}\sum _{m=1}^M \Big \{\Big (A_{\tau _1,\tau _2}\big (\frac{m}{M}\pi \big )\Big )^2 -\Big (\textrm{Im} f_{q_{\tau _1},q_{\tau _2}}\big (\frac{m}{M}\pi \big )\Big )^4\Big \} \nonumber \\&\qquad +2\frac{\pi ^2}{M}\sum _{m=2}^M \Big \{\Big (\textrm{Im} f_{q_{\tau _1},q_{\tau _2}}\big (\frac{m}{M}\pi \big ) \Big )^2 A_{\tau _1,\tau _2}\big (\frac{m}{M}\pi \big ) -\Big (\textrm{Im} f_{q_{\tau _1},q_{\tau _2}}\big (\frac{m}{M}\pi \big )\Big )^4\Big \}+o(1) \nonumber \\&\quad = V^{(\tau _1,\tau _2)}+o(1). \end{aligned}$$

(47)

Combining (45) with (47) shows that the second-order cumulant of $\sqrt{M}\big (\widetilde{T}_{n,M}^{(\tau _1,\tau _2)}-\widetilde{T}_M^{(\tau _1,\tau _2)}\big )$ behaves in the manner required.

Finally, we prove the equality

$$\begin{aligned} \textrm{cum}_q(\sqrt{M}\widetilde{T}_{n,M}^{(\tau _1,\tau _2)})=O\big (\frac{M^{q/2}}{n^{2q-4}}\big ) \end{aligned}$$

(48)

holds for general $q\ge 3$.

Since the equality

$$\begin{aligned} \textrm{cum}_q(\sqrt{M}\widetilde{T}_{n,M}^{(\tau _1,\tau _2)})=&\frac{\pi ^q}{M^{q/2}}\sum _{m_1=1}^M \cdots \sum _{m_q=1}^M \Big \{\textrm{cum} \Big (\text {Im}\,I_{n,U}^{\tau _1,\tau _2}\big (\frac{m_1}{M}\pi \big ) \, \text {Im}\,I_{n,U}^{\tau _1,\tau _2}\big (\frac{m_1-1}{M}\pi \big ), \\&\quad \cdots , \text {Im}\,I_{n,U}^{\tau _1,\tau _2}\big (\frac{m_q}{M}\pi \big ) \, \text {Im}\,I_{n,U}^{\tau _1,\tau _2}\big (\frac{m_q-1}{M}\pi \big )\Big )\Big \} \end{aligned}$$

holds, to prove (48), it suffices to check

$$\begin{aligned} \textrm{cum} \big (\text {Im}\,I_{n,U}^{\tau _1,\tau _2}(\omega _1) \, \text {Im}\,I_{n,U}^{\tau _1,\tau _2}(\omega _2), \ldots , \text {Im}\,I_{n,U}^{\tau _1,\tau _2}(\omega _{2q-1}) \, \text {Im}\,I_{n,U}^{\tau _1,\tau _2}(\omega _{2q})\big )=O\big (n^{4-2q}\big ) \end{aligned}$$

(49)

holds uniformly for $\omega _1,\ldots ,\omega _{2q}\in (0,\pi ]$. According to (30), to check (49), we need only to verify

$$\begin{aligned}&\textrm{cum} \big (y_{n,U}^{\tau _1}(\omega _1)y_{n,U}^{\tau _2}(\omega _1) \, y_{n,U}^{\tau _1}(\omega _2)y_{n,U}^{\tau _2}(\omega _2), \ldots , y_{n,U}^{\tau _1}(\omega _{2q-1})y_{n,U}^{\tau _2}(\omega _{2q-1}) \, y_{n,U}^{\tau _1}(\omega _{2q})y_{n,U}^{\tau _2}(\omega _{2q})\big )\nonumber \\&=O\big (n^{4-2q}\big ) \end{aligned}$$

(50)

holds uniformly for $\omega _1,\ldots ,\omega _{2q}\in (0,\pi ]$. Since we have established (26), using Theorem 2.3.2 of Brillinger (2001) yields (50). This proves (48) and indicates that the q-th-order ($q\ge 3$) cumulant of $\sqrt{M}\big (\widetilde{T}_{n,M}^{(\tau _1,\tau _2)}-\widetilde{T}_M^{(\tau _1,\tau _2)}\big )$ behaves in the manner required.

Now, all the cumulants of $\sqrt{M}\big (\widetilde{T}_{n,M}^{(\tau _1,\tau _2)}-\widetilde{T}_M^{(\tau _1,\tau _2)}\big )$ behave in the manner required by the $N(0,V^{(\tau _1,\tau _2)})$. This proves (42).

Proof of (43). Employing Lemmas 1 and 8 in combination with the triangular inequality, we have that

$$\begin{aligned}&\sqrt{M}\big |\widetilde{T}_M^{(\tau _1,\tau _2)}-T_M^{(\tau _1,\tau _2)}\big | \nonumber \\&\qquad \le \frac{\pi }{\sqrt{M}} \sum _{m=1}^M \big |\text {Im} f_{q_{\tau _1},q_{\tau _2}}\big (\frac{m}{M}\pi \big )\big |\,\Big |\text {Im} f_{q_{\tau _1},q_{\tau _2}}\big (\frac{m}{M}\pi \big )-\text {Im} f_{q_{\tau _1},q_{\tau _2}}\big (\frac{m-1}{M}\pi \big )\Big |\nonumber \\&\qquad \le C \frac{1}{\sqrt{M}}\longrightarrow 0, \end{aligned}$$

(51)

as M goes to infinity.

Proof of (44). According to the first mean value theorem for definite integrals (e.g., Comenetz 2002, pp. 159), we obtain

$$\begin{aligned}&\sqrt{M}\big (T_M^{(\tau _1,\tau _2)}-T^{(\tau _1,\tau _2)}\big )\nonumber \\&\qquad = \sqrt{M} \sum _{m=1}^M \Big \{\big (\text {Im} f_{q_{\tau _1},q_{\tau _2}}\big (\frac{m}{M}\pi \big )\big )^2 \frac{\pi }{M}-\int _{\frac{m-1}{M}\pi }^{\frac{m}{M}\pi } \big (\text {Im} f_{q_{\tau _1},q_{\tau _2}}(\omega )\big )^2\,\text {d}\omega \Big \}\nonumber \\&\qquad =\frac{\pi }{\sqrt{M}} \sum _{m=1}^M \Big (\big (\text {Im} f_{q_{\tau _1},q_{\tau _2}}\big (\frac{m}{M}\pi \big )\big )^2 -\big (\text {Im} f_{q_{\tau _1},q_{\tau _2}}\big (\xi _m\big )\big )^2 \Big ), \end{aligned}$$

where $\xi _m\in (\frac{m-1}{M}\pi ,\frac{m}{M}\pi )$, $m=1,\ldots ,M$. Then, by the triangular inequality and the Hölder continuity (40), we have

$$\begin{aligned} \sqrt{M}\big |T_M^{(\tau _1,\tau _2)}-T^{(\tau _1,\tau _2)}\big | \le&C \frac{1}{\sqrt{M}} \sum _{m=1}^M \Big \{\Big (\big |\text {Im} f_{q_{\tau _1},q_{\tau _2}}\big (\frac{m}{M}\pi \big )\big |+\big |\text {Im} f_{q_{\tau _1},q_{\tau _2}}\big (\xi _m\big )\big |\Big )\nonumber \\&\qquad \times \Big |\text {Im} f_{q_{\tau _1},q_{\tau _2}}\big (\frac{m}{M}\pi \big )-\text {Im} f_{q_{\tau _1},q_{\tau _2}}\big (\xi _m\big )\Big |\Big \} \nonumber \\ \le&C \frac{\pi }{\sqrt{M}} \sum _{m=1}^M \Big |\frac{m}{M}\pi -\xi _m\Big | \le C \frac{1}{\sqrt{M}}\longrightarrow 0, \end{aligned}$$

(52)

as M goes to infinity. This completes the proof.

Proof of Theorem 2. Under the assumptions of Theorem 2, it follows from Theorem 3.5 of Kley et al. (2016) that $\hat{G}_{n,R}(\tau _1,\tau _2;\omega )$ is a consistent estimator of $f_{q_{\tau _1},q_{\tau _2}}(\omega )$ for each pair of levels $(\tau _1, \tau _2)$. Then, Combining Theorem 1, the continuous mapping theorem and the definition of definite integral, we obtain the results of Theorem 2.