Periodic Trawl Processes: Simulation, Statistical Inference and Applications in Energy Markets

Abstract

This article introduces the class of periodic trawl processes, which are continuous-time, infinitely divisible, stationary stochastic processes, that allow for periodicity and flexible forms of their serial correlation, including both short- and long-memory settings. We derive some of the key probabilistic properties of periodic trawl processes and present relevant examples. Moreover, we show how such processes can be simulated and establish the asymptotic theory for their sample mean and sample autocovariances. Consequently, we prove the asymptotic normality of a (generalised) method-of-moments estimator for the model parameters. We illustrate the new model and estimation methodology in an application to electricity prices.

Appendix

The appendix contains the proofs of all the technical results presented in the main paper, additional examples and a discussion of when the technical assumptions needed in our main theorems hold for periodic trawl processes.

1.1 Proof of the Second Order Properties

First, we derive the joint characteristic/cumulant function.

Proposition 8

Let$t_1<t_2$and$\theta _1, \theta _2 \in \mathbb {R}$. Then

$$\displaystyle \begin{aligned} &{\mathrm{Log}}(\mathbb{E}(\exp(i (\theta_1, \theta_2)(Y_{t_1}, Y_{t_2})^{\top})))={\mathrm{Log}}(\mathbb{E}(\exp(i \theta_1 Y_{t_1} + i\theta_2 Y_{t_2}))) \\ &= \int_{(-\infty,t_1]\times \mathbb{R}}C_{L'}( \theta_1 p(t_1-s)\mathbb{I}_{(0,g(t_1-s))}(x)+\theta_2 p(t_2-s)\mathbb{I}_{(0,g(t_2-s))}(x))dx ds \\ &+ \int_{(t_1,t_2]\times \mathbb{R}}C_{L'}(\theta_2 p(t_2-s)\mathbb{I}_{(0,g(t_2-s))}(x))dx ds, \end{aligned} $$

where$C_{L'}$denotes the cumulant function of the Lévy seed$L'$associated with the Lévy basis L.

Proof

Let $t_1<t_2$ and $\theta _1, \theta _2 \in \mathbb {R}$. Then the joint characteristic function is given by

$$\displaystyle \begin{aligned} &\mathbb{E}(\exp(i (\theta_1, \theta_2)(Y_{t_1}, Y_{t_2})^{\top}))=\mathbb{E}(\exp(i \theta_1 Y_{t_1} + i\theta_2 Y_{t_2})) \\ &=\mathbb{E}\left[\exp\left(i \theta_1 \int_{(-\infty,t_1]\times \mathbb{R}}p(t_1-s)\mathbb{I}_{(0,g(t_1-s))}(x)L(dx,ds) \right.\right.\\ & + i\theta_2 \int_{(-\infty,t_1]\times \mathbb{R}}p(t_2-s)\mathbb{I}_{(0,g(t_2-s))}(x)L(dx,ds) \\ &\quad \left.\left. +i\theta_2\int_{(t_1,t_2]\times \mathbb{R}}p(t_2-s)\mathbb{I}_{(0,g(t_2-s))}(x)L(dx,ds)\right)\right]\\ &=\mathbb{E}\left[\exp\left( i\int_{(-\infty,t_1]\times \mathbb{R}}\{ \theta_1 p(t_1-s)\mathbb{I}_{(0,g(t_1-s))}(x)\right.\right.\\ &\quad +\theta_2 p(t_2-s)\mathbb{I}_{(0,g(t_2-s))}(x)\}L(dx,ds) \\ & \left.\left.+i\theta_2\int_{(t_1,t_2]\times \mathbb{R}}p(t_2-s)\mathbb{I}_{(0,g(t_2-s))}(x)L(dx,ds)\right)\right]\\ &=\exp\left( \int_{(-\infty,t_1]\times \mathbb{R}}C( \theta_1 p(t_1-s)\mathbb{I}_{(0,g(t_1-s))}(x) \right.\\ &\quad \left.+\theta_2 p(t_2-s)\mathbb{I}_{(0,g(t_2-s))}(x); L')dx ds \right)\\ & \quad \cdot \exp\left( \int_{(t_1,t_2]\times \mathbb{R}}C(\theta_2 p(t_2-s)\mathbb{I}_{(0,g(t_2-s))}(x); L')dx ds \right). \end{aligned} $$

I.e.

$$\displaystyle \begin{aligned} &\log(\mathbb{E}(\exp(i \theta_1 Y_{t_1} + i\theta_2 Y_{t_2}))) \\ &= \int_{(-\infty,t_1]\times \mathbb{R}}C( \theta_1 p(t_1-s)\mathbb{I}_{(0,g(t_1-s))}(x)+\theta_2 p(t_2-s)\mathbb{I}_{(0,g(t_2-s))}(x); L')dx ds \\ &+ \int_{(t_1,t_2]\times \mathbb{R}}C(\theta_2 p(t_2-s)\mathbb{I}_{(0,g(t_2-s))}(x); L')dx ds. \end{aligned} $$

□

We can now easily derive the second-order properties of the periodic trawl process:

Proof (Proof of Proposition5)

For $t, t_1, t_2\in \mathbb {R}, t_1<t_2$, we have

$$\displaystyle \begin{aligned} \mathbb{E}(Y_t)&= \mathbb{E}(L')\int_{-\infty}^{t}p(t-s) g(t-s)ds=\mathbb{E}(L')\int_{0}^{\infty}p(u) g(u)du,\\ {\mathrm{Var}}(Y_{t})&= {\mathrm{Var}}(L')\int_{0}^{\infty}p^2(u)g(u)du,\\ {\mathrm{Cov}}(Y_{t_1},Y_{t_2})&=-\left.\frac{\partial^2}{\partial \theta_1 \partial \theta_2}\log(\mathbb{E}(\exp(i\theta_1 Y_{t_1}+i\theta_2 Y_{t_2})))\right|{}_{\theta_1=\theta_2=0} \\ &= {\mathrm{Var}}(L')\int_{-\infty}^{t_1}p(t_1-s)p(t_2-s)\min(g(t_1-s),g(t_2-s))ds,\\ {\mathrm{Cor}}(Y_{t_1},Y_{t_2}) &= \frac{\int_{-\infty}^{t_1}p(t_1-s)p(t_2-s)\min(g(t_1-s),g(t_2-s))ds}{\int_{0}^{\infty}p^2(u)g(u)du}. \end{aligned} $$

Recall that we assume that g is monotonically decreasing, i.e. if $x\leq y$, then $g(x)\geq g(y)$.

Since $t_1<t_2$, we have $t_1-s<t_2-s$ for $s<t_1$ and

$$\displaystyle \begin{aligned} \min(g(t_1-s),g(t_2-s))=g(t_2-s), \end{aligned} $$

hence the above expressions simplify to

$$\displaystyle \begin{aligned} {\mathrm{Cov}}(Y_{t_1},Y_{t_2}) &={\mathrm{Var}}(L')\int_{-\infty}^{t_1}p(t_1-s)p(t_2-s)g(t_2-s)ds\\ &={\mathrm{Var}}(L')\int_{0}^{\infty}p(u)p(t_2-t_1+u)g(t_2-t_1+u)du,\\ {\mathrm{Cor}}(Y_{t_1},Y_{t_2}) &= \frac{\int_{0}^{\infty}p(u)p(t_2-t_1+u)g(t_2-t_1+u)du}{\int_{0}^{\infty}p^2(u)g(u)du}. \end{aligned} $$

□

Proof (Proof of Proposition6)

Recall that

$$\displaystyle \begin{aligned} {\mathrm{Cor}}(Y_{0},Y_{t})= \frac{\int_{0}^{\infty}p(u)p(t+u)g(t+u)du}{\int_{0}^{\infty}p^2(u)g(u)du}. \end{aligned} $$

We consider a constant $M>\tau$. Since p is periodic with period $\tau$, there exist $\xi _1, \xi _2 \in [0,\tau ]$ such that,

$$\displaystyle \begin{aligned} \int_{0}^{M}p(u)p(t+u)g(t+u)du&=p(\xi_1)p(\xi_1+t)\int_{0}^{M}g(t+u)du,\\ \int_{0}^{M}(p(u))^2g(u)du&=(p(\xi_2))^2\int_{0}^{M}g(t+u)du, \end{aligned} $$

by the mean value theorem. We note that p is assumed to be continuous, and since it is also periodic, it is bounded. Also, the integrability conditions in (7) guarantee the existence of the integrals when taking the limit as $M\to \infty$. Taking the limit and setting $c(t)=p(\xi _1)p(\xi _1+t)/(p(\xi _2))^2$ leads the result; since $\mathrm {Cor}(Y_0,Y_0)=1$, we deduce that $c(0)=1$. Also, we observe that c is proportional to the $\tau$-periodic function p and is hence $\tau$-periodic itself. □

Remark 11

As mentioned in Remark 2, Barndorff-Nielsen et al. [9] proposed adding a periodic function as a multiplicative factor to g rather than as kernel function as in (8), which results in a process $(Z_t)_{t\geq 0}$ with

$$\displaystyle \begin{aligned} {} Z_t &= \int_{\mathbb{R}\times \mathbb{R}}\mathbb{I}_{(0,p(t-s)g(t-s))}(x)\mathbb{I}_{[0,\infty)}(t-s) L(dx,ds), \end{aligned} $$

(19)

compared to our earlier definition of $(Y_t)_{t\geq 0}$ with

$$\displaystyle \begin{aligned} Y_t &= \int_{\mathbb{R}\times \mathbb{R}}p(t-s)\mathbb{I}_{(0,g(t-s))}(x)\mathbb{I}_{[0,\infty)}(t-s) L(dx,ds). \end{aligned} $$

The autocorrelation function of the process Z is of the form, for $t_1<t_2$,

$$\displaystyle \begin{aligned} {\mathrm{Cor}}(Z_{t_1},Z_{t_2}) &= \frac{\int_{-\infty}^{t_1}\min(p(t_1-s)g(t_1-s), p(t_2-s)g(t_2-s))ds}{\int_{0}^{\infty}p(u)g(u)du}, \end{aligned} $$

which is potentially slightly more difficult to deal with than the autocorrelation function of our proposed periodic trawl process Y .

1.2 Proofs of the Asymptotic Theory

The following proofs extend the ideas presented in the work by Cohen and Lindner [14]. Alternatively, we could have deduced the results from the more recent work by Curato and Stelzer [16].

Proof (Proof of Theorem1)

The proof is a straightforward extension of the arguments provided in the proof of Theorem 2.1 in [14]. For the convenience of the reader and to keep this article self-contained, we will present the steps to extend the proof by Cohen and Lindner [14] to our more general setting of mixed moving average processes driven by homogeneous Lévy bases.

First of all, we continue the function $F_{\Delta }$ periodically on $\mathbb {R}$ by setting

$$\displaystyle \begin{aligned} F_{\Delta}(x,u)=\sum_{j=-\infty}^{\infty}|f(x, u+j\Delta)|, \quad u \in \mathbb{R}, \end{aligned} $$

where $F_{\Delta }(x,u)=F_{\Delta }(x,u+j\Delta )$ for all $j \in \mathbb {Z}$, $u, x \in \mathbb {R}$.

We note that the autocovariance function of Y  satisfies

$$\displaystyle \begin{aligned} |\gamma_f(j\Delta)| \leq \kappa_2\int_{\mathbb{R} \times \mathbb{R}}|f(x, -s)||f(x, j\Delta-s)|dx ds, \end{aligned} $$

for any $j \in \mathbb {Z}$ and

$$\displaystyle \begin{aligned} \begin{array}{rcl} {} \frac{1}{\kappa_2}\sum_{j=-\infty}^{\infty}|\gamma_f(j \Delta)| & \leq&\displaystyle \sum_{j=-\infty}^{\infty}\int_{\mathbb{R} \times \mathbb{R}}|f(x, -s)||f(x, j\Delta-s)|dx ds\\ & \leq&\displaystyle \int_{\mathbb{R} \times \mathbb{R}}|f(x, -s)|\sum_{j=-\infty}^{\infty}|f(x, j\Delta-s)|dx ds\\ & =&\displaystyle \int_{\mathbb{R} \times \mathbb{R}}|f(x, -s)|F_{\Delta}(x, -s)dx ds \\ & =&\displaystyle \int_{\mathbb{R} \times \mathbb{R}}|f(x, s)|F_{\Delta}(x, s)dx ds \\ & =&\displaystyle \sum_{j=-\infty}^{\infty}\int_{\mathbb{R} \times [0,\Delta]}|f(x, j\Delta +s)|F_{\Delta}(x, s)dx ds \\ & =&\displaystyle \int_{\mathbb{R} \times [0,\Delta]} \sum_{j=-\infty}^{\infty}|f(x, j\Delta +s)|F_{\Delta}(x, s)dx ds\\ & =&\displaystyle \int_{\mathbb{R} \times [0,\Delta]} F_{\Delta}^2(x, s)dx ds< \infty. \end{array} \end{aligned} $$

(20)

The above computations can be repeated without the modulus, which implies that $\sum _{j=-\infty }^{\infty }\gamma _f(j \Delta )=V_{\Delta }$.

To simplify the exposition, we shall now assume that $\mu =0$. We proceed as in [14]. Define the function $f_{m;\Delta }(x,s):=f(x,s)\mathbb {I}_{(-m \Delta , m \Delta )}(s)$, for $m \in \mathbb {N}, x, s \in \mathbb {R}$, and set

$$\displaystyle \begin{aligned} Y_{j; \Delta}^m&:=\int_{\mathbb{R} \times \mathbb{R}}f_{m; \Delta}(x,s)L(dx, ds)= \int_{\mathbb{R} \times (-m \Delta, m \Delta)}f(x, s) L(dx, ds) \\ &=\int_{\mathbb{R} \times ((-m+j)\Delta, (m+j)\Delta)}f(x, j\Delta-s) L(dx, ds). \end{aligned} $$

Since L is independently scattered, we can deduce that $(Y_{j;\Delta }^{(m)})_{j \in \mathbb {Z}}$ is a $(2m-1)$-dependent sequence, which is also strictly stationary. Hence, by Brockwell and Davis [13, Theorem 6.4.2], we know that

$$\displaystyle \begin{aligned} \sqrt{n} \; \overline{Y}_{n; \Delta}^{(m)} = n^{-1/2}\sum_{j=1}^n Y_{j;\Delta}^{(m)}\stackrel{\mathrm{d}}{\to}Z^{(m)}_{\Delta}, \quad \mathrm{as} \; n \to \infty, \end{aligned} $$

where the random variable $Z^{(m)}_{\Delta }$ satisfies $Z^{(m)}_{\Delta }\stackrel {\mathrm {d}}{=}\mathrm {N}(0, V^{(m)})$, where

$$\displaystyle \begin{aligned} V^{(m)}_{\Delta}=\sum_{j=-2m}^{2m}\gamma_{f_m}(j\Delta), \end{aligned} $$

for

$$\displaystyle \begin{aligned} \gamma_{f_m}(j\Delta) &={\mathrm{Cov}}(Y_{0;\Delta}^{(m)}, Y_{j;\Delta}^{(m)})=\kappa_2\int_{\mathbb{R} \times \mathbb{R}}f_{m;\Delta}(x, -s)f_{m;\Delta}(x, j\Delta-s)dx ds\\ &=\int_{\mathbb{R} \times ((-m+j)\Delta, (m+j)\Delta)}f(x, -s) f(x, j\Delta-s) dx ds. \end{aligned} $$

We observe that $\lim _{m\to \infty }\gamma _{f_m}(j\Delta )=\gamma _{f}(j\Delta )$ for all $j \in \mathbb {Z}$; also

$$\displaystyle \begin{aligned} |\gamma_{f_m}(j\Delta)|\leq \kappa_2 \int_{\mathbb{R} \times \mathbb{R}}|f(x, -s)| |f(x, j\Delta-s)| dx ds, \end{aligned} $$

and $\sum _{j=-\infty }^{\infty } \int _{\mathbb {R} \times \mathbb {R}}|f(x, -s)| |f(x, j\Delta -s)|<\infty$ by the computations in (20). Hence, Lebesgue’s Dominated Convergence Theorem implies that $\lim _{m\to \infty }V^{(m)}_{\Delta }=V_{\Delta }$ and we get that $Z_{\Delta }^{(m)}\stackrel {\mathrm {d}}{\to } Z_{\Delta }$, where $Z \stackrel {\mathrm {d}}{=} \mathrm {N}(0, V_{\Delta })$.

It remains to control the difference $n^{1/2}(\overline {Y}_{n; \Delta }-\overline {Y}_{n; \Delta }^{(m)})$. We argue as follows. Using similar arguments as above, we note that $\lim _{m\to \infty }\sum _{j=-\infty }^{\infty }\gamma _{f-f_{m;\Delta }}(j\Delta ) =0$. Hence, we have that

$$\displaystyle \begin{aligned} &\lim_{m\to \infty} \lim_{n\to \infty} {\mathrm{Var}}(n^{1/2}(\overline{Y}_{n; \Delta}-\overline{Y}_{n; \Delta}^{(m)})) \\ &{\,=\,}\lim_{m\to \infty} \lim_{n\to \infty} n{\mathrm{Var}}\left(n^{-1}\sum_{j=1}^n\int_{\mathbb{R} \times \mathbb{R}}(f(x, j\Delta{\,-\,}s)-f_{m;\Delta}(x, j\Delta{\,-\,}s))L(dx, ds)\right)\\ &\stackrel{(\star)}{=}\lim_{m\to \infty} \sum_{j=-\infty}^{\infty} \gamma_{f-f_{m;\Delta}}(j\Delta)=0, \end{aligned} $$

where the equality $(\star )$ follows from [13, Theorem 7.1.1]. Chebyshev’s inequality allows us to conclude that, for any $\epsilon >0$,

$$\displaystyle \begin{aligned} \lim_{m\to \infty} \limsup_{n\to \infty} \mathbb{P}(n^{1/2}|\overline{Y}_{n; \Delta}-\overline{Y}_{n; \Delta}^{(m)}|>\epsilon)=0. \end{aligned} $$

As stated in [14], the final step of the proof consists of an application of a Slutsky-type theorem as presented in [13, Proposition 6.3.9]. □

Proof (Proof of Lemma1)

For $t_1, t_2, t_3, t_4 \in \mathbb {R}$, we have, for any $a_1, a_2, a_3, a_4 \in \mathbb {R}$, the following expression for the joint characteristic function

$$\displaystyle \begin{aligned} &\psi((a_1,a_2,a_3,a_4);(Y_{t_1},Y_{t_2},Y_{t_3},Y_{t_4})):= \mathbb{E}(\exp(i(a_1Y_{t_1}+a_2Y_{t_2}+a_3Y_{t_3}+a_4Y_{t_4}))\\ &=\mathbb{E}\left[\exp\left(i \int_{\mathbb{R}\times \mathbb{R}}\left(\sum_{j=1}^4a_jf(x, t_j-s)\right)L(ds, dx)\right) \right]\\ &= \exp\left[ \int_{\mathbb{R} \times \mathbb{R}} C\left(\sum_{j=1}^4a_jf(x, t_j-s); L' \right) dx ds \right], \end{aligned} $$

where $C(\cdot ; L')$ denotes the cumulant function of the Lévy seed $L'$, which we will present next.

Suppose $L'$ has characteristic triplet $(c, A, \nu )$ w.r.t. the truncation function $\tau (y)=\mathbb {I}_{[-1,1]}(y)$. I.e. we have the following representation for its characteristic function, for any $\theta \in \mathbb {R}$,

$$\displaystyle \begin{aligned} \mathbb{E}(\exp(i \theta L'))=\exp\left(i c \theta -\frac{1}{2}A \theta^2 + \int_{\mathbb{R} }(e^{iy \theta}-1-i\theta y \tau(y))\nu(dy)\right). \end{aligned} $$

We recall that $\mathbb {E}(L')=c +\int _{\mathbb {R}}y(1-\tau (y))\nu (dy)$. Since we are assuming that $\mathbb {E}(L')=0$, we get that $c=-\int _{\mathbb {R}}y(1-\tau (y))\nu (dy)=-\int _{\mathbb {R}}y\mathbb {I}_{[-1,1]^c}(y)\nu (dy)$ and, hence,

$$\displaystyle \begin{aligned} \mathbb{E}(\exp(i \theta L'))=\exp\left( -\frac{1}{2}A \theta^2 + \int_{\mathbb{R} }(e^{iy \theta}-1-i\theta y )\nu(dy)\right). \end{aligned} $$

I.e. the corresponding cumulant function is given by

$$\displaystyle \begin{aligned} C(\theta; L')= -\frac{1}{2}A \theta^2 + \int_{\mathbb{R} }(e^{iy \theta}-1-i\theta y )\nu(dy). \end{aligned} $$

Moreover,

$$\displaystyle \begin{aligned} &C((a_1,a_2,a_3,a_4);(Y_{t_1},Y_{t_2},Y_{t_3},Y_{t_4})) :=\int_{\mathbb{R} \times \mathbb{R}} C\left(\sum_{j=1}^4a_jf(x, t_j-s); L' \right) dx ds\\ &= -\frac{1}{2}A \int_{\mathbb{R} \times \mathbb{R}} \left(\sum_{j=1}^4a_jf(x, t_j-s)\right)^2 dx ds \\ & \quad +\int_{\mathbb{R} \times \mathbb{R}} \int_{\mathbb{R} }\left(e^{iy \sum_{j=1}^4a_jf(x, t_j-s)}-1-i\sum_{j=1}^4a_jf(x, t_j-s) y \right)\nu(dy) dx ds\\ &= -\frac{1}{2}A \sum_{j,k=1}^4 a_j a_k \int_{\mathbb{R} \times \mathbb{R}} f(x, t_j-s) f(x, t_k-s) dx ds \\ & \quad +\int_{\mathbb{R} \times \mathbb{R}} \int_{\mathbb{R} }\left(e^{iy \sum_{j=1}^4a_jf(x, t_j-s)}-1-i\sum_{j=1}^4a_jf(x, t_j-s) y \right)\nu(dy) dx ds\\ &= -\frac{1}{2}A \sum_{j=1}^4 a_j^2 \int_{\mathbb{R} \times \mathbb{R}} f^2(x, t_j-s) dx ds \\ & \quad -\frac{1}{2}A \sum_{j,k=1, j\neq k}^4 a_j a_k \int_{\mathbb{R} \times \mathbb{R}} f(x, t_j-s) f(x, t_k-s) dx ds \\ & \quad +\int_{\mathbb{R} \times \mathbb{R}} \int_{\mathbb{R} }\left(e^{iy \sum_{j=1}^4a_jf(x, t_j-s)}-1-i\sum_{j=1}^4a_jf(x, t_j-s) y \right)\nu(dy) dx ds. \end{aligned} $$

Next, we compute the fourth moments, where we recall that

$$\displaystyle \begin{aligned} & \left.\frac{\partial^4}{\partial a_1 \partial a_2 \partial a_3 \partial a_4}\psi((a_1,a_2,a_3,a_4);(Y_{t_1},Y_{t_2},Y_{t_3},Y_{t_4}))\right|{}_{a_1=a_2=a_3=a_4=0} \\ &\quad = \mathbb{E}(Y_{t_1}Y_{t_2}Y_{t_3}Y_{t_4}). \end{aligned} $$

We now abbreviate the functions to $\psi$ and C without stating their arguments and a subscript denotes the corresponding partial derivative, e.g. $C_{a_1}=\frac {\partial }{\partial a_1}C((a_1,a_2,a_3,a_4);(Y_{t_1},Y_{t_2},Y_{t_3},Y_{t_4})$ and similarly for higher order partial derivatives. Since $\psi = \exp (C)$, we have

$$\displaystyle \begin{aligned} \psi_{a1} &= \psi C_{a_1},\\ \psi_{a_1,a_2} &= \psi [ C_{a_1, a_2}+ C_{a_1} C_{a_2}],\\ \psi_{a_1,a_2,a_3} &= \psi [ (C_{a_1, a_2}+ C_{a_1} C_{a_2}) C_{a_3} +C_{a_1, a_2, a_3} +C_{a_1, a_3}C_{a_2}+C_{a_1}C_{a_2, a_3} ]\\ &= \psi [ C_{a_1} C_{a_2}C_{a_3} +C_{a_1}C_{a_2, a_3} +C_{a_2} C_{a_1, a_3} +C_{a_3} C_{a_1, a_2} + +C_{a_1, a_2, a_3} ],\\ \psi_{a_1,a_2,a_3,a_4} &= \psi [ C_{a_1} C_{a_2}C_{a_3} C_{a_4} +C_{a_1}C_{a_2, a_3} C_{a_4} +C_{a_2} C_{a_1, a_3}C_{a_4} \\ & \quad +C_{a_3} C_{a_1, a_2} C_{a_4} +C_{a_1, a_2, a_3}C_{a_4} \\ & + C_{a_1, a_4} C_{a_2}C_{a_3} + C_{a_1} C_{a_2, a_4}C_{a_3} + C_{a_1} C_{a_2}C_{a_3, a_4}\\ & \quad + C_{a_1, a_4}C_{a_2, a_3} + C_{a_1}C_{a_2, a_3, a_4} \\ & + C_{a_2, a_4} C_{a_1, a_3} + C_{a_2} C_{a_1, a_3, a_4} + C_{a_3,a_4} C_{a_1, a_2} \\ &\quad + C_{a_3} C_{a_1, a_2, a_4} +C_{a_1, a_2, a_3, a_4} ]. \end{aligned} $$

Here we have

$$\displaystyle \begin{aligned} C_{a_i}|{}_{a_1=a_2=a_3=a_4=0}&=0, \quad \mathrm{for}\; i=1, 2, 3, 4,\\ C_{a_i,a_j}|{}_{a_1=a_2=a_3=a_4=0} &=-\left(A+ \int_{\mathbb{R}} y^2 \nu(dy)\right) \int_{\mathbb{R} \times \mathbb{R}} f(x, t_i-s) f(x, t_j-s) dx ds,\\ & \qquad \mathrm{for}\, i, j = 1, 2, 3, 4,\\ C_{a_1,a_2,a_3,a_4}&=\int_{\mathbb{R}\times \mathbb{R}}\prod_{j=1}^4 f(x, t_j-s)dxds \int_{\mathbb{R}}y^4\nu(dy). \end{aligned} $$

The above results imply that

$$\displaystyle \begin{aligned} &\mathbb{E}(Y_{t_1}Y_{t_2}Y_{t_3}Y_{t_4})\\ &= \left(A+ \int_{\mathbb{R}} y^2 \nu(dy)\right)^2 \\ &\left( \int_{\mathbb{R} \times \mathbb{R}} f(x, t_1-s) f(x, t_2-s)dx ds \int_{\mathbb{R} \times \mathbb{R}} f(x, t_3-s) f(x, t_4-s)dx ds \right .\\ & + \int_{\mathbb{R} \times \mathbb{R}} f(x, t_1-s) f(x, t_3-s)dx ds \int_{\mathbb{R} \times \mathbb{R}} f(x, t_2-s) f(x, t_4-s)dx ds \\ &\left. +\int_{\mathbb{R} \times \mathbb{R}} f(x, t_1-s) f(x, t_4-s)dx ds \int_{\mathbb{R} \times \mathbb{R}} f(x, t_2-s) f(x, t_3-s)dx ds \right)\\ & + \int_{\mathbb{R}} y^4 \nu(dy) \int_{\mathbb{R} \times \mathbb{R}} f(x, t_1-s) f(x, t_2-s) f(x, t_3-s) f(x, t_4-s) dx ds. \end{aligned} $$

We note that $\kappa _4:=\int _{\mathbb {R}}y^4\nu (dy)=(\eta -3)\kappa _2^2$ and $\kappa _2=A+\int _{\mathbb {R}}y^2\nu (dy)$. We can further simplify the above formula as follows:

$$\displaystyle \begin{aligned} &\mathbb{E}(Y_{t_1}Y_{t_2}Y_{t_3}Y_{t_4})\\ &= \left(A+ \int_{\mathbb{R}} y^2 \nu(dy)\right)^2 \\ &\left( \int_{\mathbb{R} \times \mathbb{R}} f(x, t_1-t_2+s) f(x, s)dx ds \int_{\mathbb{R} \times \mathbb{R}} f(x, t_3-t_4+s) f(x, s)dx ds \right .\\ & + \int_{\mathbb{R} \times \mathbb{R}} f(x, t_1-t_3+s) f(x, s)dx ds \int_{\mathbb{R} \times \mathbb{R}} f(x, t_2-t_4+s) f(x, s)dx ds \\ &\left. +\int_{\mathbb{R} \times \mathbb{R}} f(x, t_1-t_4+s) f(x, s)dx ds \int_{\mathbb{R} \times \mathbb{R}} f(x, t_2-t_3+s) f(x, s)dx ds \right)\\ & + \int_{\mathbb{R}} y^4 \nu(dy) \int_{\mathbb{R} \times \mathbb{R}} f(x, t_1-t_3+s) f(x, t_2-t_3) f(x, s) f(x, t_4-t_3+s) dx ds\\ &= \gamma(t_1-t_2)\gamma(t_3-t_4) + \gamma(t_1-t_3) \gamma(t_2-t_4) +\gamma(t_1-t_4)\gamma(t_2-t_3) \\ & + \kappa_4 \int_{\mathbb{R} \times \mathbb{R}}f(x, t_1-t_3+s) f(x, t_2-t_3) f(x, s) f(x, t_4-t_3+s) dx ds. \end{aligned} $$

□

Proof (Proof of Proposition7)

We first expand the covariance of the sample autocovariances as follows

$$\displaystyle \begin{aligned} {\mathrm{Cov}}(\widehat{\gamma}_{n;\Delta}^*(\Delta p), \widehat{\gamma}_{n,\Delta}^*(\Delta q)) {\,=\,}\mathbb{E}(\widehat{\gamma}_{n;\Delta}^*(\Delta p) \widehat{\gamma}_{n;\Delta}^*(\Delta q)) {\,-\,} \mathbb{E}(\widehat{\gamma}_{n;\Delta}^*(\Delta p)) \mathbb{E}(\widehat{\gamma}_{n;\Delta}^*(\Delta q)), \end{aligned} $$

where

$$\displaystyle \begin{aligned} \mathbb{E}(\widehat{\gamma}_{n;\Delta}^*(\Delta p)) &=\frac{1}{n}\sum_{j=1}^{n}\mathbb{E}(Y_{j\Delta}Y_{(j+p)\Delta})= \frac{1}{n}\sum_{j=1}^{n}\gamma(p\Delta)=\gamma(p\Delta),\\ \mathbb{E}(\widehat{\gamma}_{n;\Delta}^*(\Delta q)) &=\frac{1}{n}\sum_{k=1}^{n}\mathbb{E}(Y_{k\Delta}Y_{(k+q)\Delta})= \frac{1}{n}\sum_{k=1}^{n}\gamma(q\Delta)=\gamma(q\Delta). \end{aligned} $$

Also,

$$\displaystyle \begin{aligned} &\mathbb{E}(\widehat{\gamma}_{n;\Delta}^*(\Delta p) \widehat{\gamma}_{n;\Delta}^*(\Delta q)) =\frac{1}{n^2}\sum_{j=1}^{n}\sum_{k=1}^{n}\mathbb{E}( Y_{j\Delta}Y_{(j+p)\Delta} Y_{k\Delta}Y_{(k+q)\Delta} ), \end{aligned} $$

where

$$\displaystyle \begin{aligned} &\mathbb{E}( Y_{j\Delta}Y_{(j+p)\Delta} Y_{k\Delta}Y_{(k+q)\Delta} )\\ &=\gamma(p\Delta)\gamma(q\Delta) + \gamma((k-j)\Delta) \gamma((j+p-k-q)\Delta) \\ &\quad +\gamma((j-k-q)\Delta)\gamma((j+p-k)\Delta) \\ & \quad + \kappa_4 \int_{\mathbb{R} \times \mathbb{R}} f(x, (j-k)\Delta+s) f(x, (j-k+p)\Delta+s) f(x, s) \\ &\quad f(x, q\Delta+s) dx ds\\ &=\gamma(p\Delta)\gamma(q\Delta) + \gamma((j-k)\Delta) \gamma((j-k+p-q)\Delta) \\ &\quad +\gamma((j-k-q)\Delta)\gamma((j-k+p)\Delta) \\ & \quad + \kappa_4 \int_{\mathbb{R} \times \mathbb{R}} f(x, (j-k)\Delta+s) f(x, (j-k+p)\Delta+s) f(x, s)\\ &\quad f(x, q\Delta+s) dx ds\\ &\stackrel{l=j-k}{=}\gamma(p\Delta)\gamma(q\Delta) + \gamma(l\Delta) \gamma((l+p-q)\Delta) +\gamma((l-q)\Delta)\gamma((l+p)\Delta) \\ & \quad + \kappa_4 \int_{\mathbb{R} \times \mathbb{R}} f(x, l\Delta+s) f(x, (l+p)\Delta+s) f(x, s) f(x, q\Delta+s) dx ds. \end{aligned} $$

Now we subtract $\gamma (p\Delta )\gamma (q\Delta )$, we set $l=j-k$, interchange the order of summation and use the stationarity to obtain

$$\displaystyle \begin{aligned} &{\mathrm{Cov}}(\widehat{\gamma}_{n;\Delta}^*(\Delta p), \widehat{\gamma}_{n;\Delta}^*(\Delta q))\\ &=\frac{1}{n^2}\sum_{j=1}^{n}\sum_{k=1}^{n}\mathbb{E}( Y_{j\Delta}Y_{(j+p)\Delta} Y_{k\Delta}Y_{(k+q)\Delta} ) -\gamma(p\Delta)\gamma(q\Delta)\\ & =\frac{1}{n^2}\sum_{j=1}^{n}\sum_{k=1}^{n}\left( \gamma((j-k)\Delta) \gamma((j-k+p-q)\Delta) \right. \\ &\quad +\gamma((j-k-q)\Delta)\gamma((j-k+p)\Delta) \\ & \quad + \kappa_4 \int_{\mathbb{R} \times \mathbb{R}} f(x, (j-k)\Delta+s) f(x, (j-k+p)\Delta+s) \\ &\quad f(x, s) f(x, q\Delta+s) dx ds\Big)\\ & =\frac{1}{n^2}\sum_{|l|<n}\sum_{k=1}^{n-|l|}\left( \gamma(l\Delta) \gamma((l+p-q)\Delta) +\gamma((l-q)\Delta)\gamma((l+p)\Delta)\right. \\ & \quad + \left. \kappa_4 \int_{\mathbb{R} \times \mathbb{R}} f(x, l\Delta+s) f(x, (l+p)\Delta+s) f(x, s) f(x, q\Delta+s) dx ds\right)\\ & =\frac{1}{n^2}\sum_{|l|<n}(n-|l|)T_{l,p,q;\Delta} =\frac{1}{n}\sum_{|l|<n}\left(1-\frac{|l|}{n}\right)T_{l,p,q;\Delta}, \end{aligned} $$

where

$$\displaystyle \begin{aligned} T_{l,p,q;\Delta}&:= \gamma(l\Delta) \gamma((l+p-q)\Delta) +\gamma((l-q)\Delta)\gamma((l+p)\Delta) \\ & \quad + \kappa_4 \int_{\mathbb{R} \times \mathbb{R}} f(x, l\Delta+s) f(x, (l+p)\Delta+s) f(x, s) f(x, q\Delta+s) dx ds. \end{aligned} $$

Hence, we have

$$\displaystyle \begin{aligned} \lim_{n\to \infty}n{\mathrm{Cov}}(\widehat{\gamma}_{n;\Delta}^*(\Delta p), \widehat{\gamma}_{n;\Delta}^*(\Delta q)) &=\lim_{n\to \infty}\sum_{|l|<n}\left(1-\frac{|l|}{n}\right)T_{l,p,q;\Delta}\\ &=\sum_{l=-\infty}^{\infty}T_{l,p,q;\Delta}, \end{aligned} $$

where

$$\displaystyle \begin{aligned} \sum_{l=-\infty}^{\infty}T_{l,p,q;\Delta} &= \sum_{l=-\infty}^{\infty}[\gamma(l\Delta) \gamma((l+p-q)\Delta) +\gamma((l-q)\Delta)\gamma((l+p)\Delta)] \\ &\quad + \kappa_4 \int_{\mathbb{R} \times \mathbb{R}}\sum_{l=-\infty}^{\infty} f(x, l\Delta+s) f(x, (l+p)\Delta+s) \\ & \qquad f(x, s) f(x, q\Delta+s) dx ds, \end{aligned} $$

by the Dominated Convergence Theorem since (13) holds. More precisely, let us justify why $\sum _{l=-\infty }^{\infty }|T_{l,p,q;\Delta }|<\infty$. The finiteness of $\sum _{l=-\infty }^{\infty }|\gamma (l\Delta ) \gamma ((l+p-q)\Delta ) +\gamma ((l-q)\Delta )\gamma ((l+p)\Delta )|$ follows from (13). For the second term, for $q \in \mathbb {Z}$, define

$$\displaystyle \begin{aligned} & \Bigg(\widetilde{G}_{q;\Delta}:\mathbb{R}\times [0, \Delta] \to \mathbb{R}, (x, u) \mapsto \widetilde{G}_{q;\Delta}(x,u) \\ &\quad =\sum_{j=-\infty}^{\infty}|f(x, u+j\Delta)||f(x, u+(j+q)\Delta)|\Bigg), \end{aligned} $$

which is in $L^2(\mathbb {R} \times [0, \Delta ])$ due to (11). We consider the periodic continuation of $\widetilde {G}_{q;\Delta }$ and set

$$\displaystyle \begin{aligned} & \Bigg(\widetilde{G}_{q;\Delta}:\mathbb{R}\times \mathbb{R} \to \mathbb{R}, (x, u) \mapsto \widetilde{G}_{q;\Delta}(x,u)\\ &\quad =\sum_{j=-\infty}^{\infty}|f(x, u+j\Delta)||f(x, u+(j+q)\Delta)|\Bigg). \end{aligned} $$

Since $\widetilde {G}_{q;\Delta }$ is periodic and, restricted to $\mathbb {R}\times [0, \Delta ]$ square-integrable, we have

$$\displaystyle \begin{aligned} &\int_{\mathbb{R} \times \mathbb{R}} \sum_{l=-\infty}^{\infty} |f(x, l\Delta+s) f(x, (l+p)\Delta+s)| |f(x, s) f(x, q\Delta+s)| dx ds\\ &=\int_{\mathbb{R} \times \mathbb{R}}\widetilde{G}_p(s) |f(x, s) f(x, q\Delta+s)| dx ds \\ &=\sum_{j=-\infty}^{\infty}\int_{\mathbb{R} \times [j\Delta, (j+1)\Delta]}\widetilde{G}_p(s) |f(x, s) f(x, q\Delta+s)| dx ds\\ &=\sum_{j=-\infty}^{\infty}\int_{\mathbb{R} \times [0, \Delta]}\widetilde{G}_p(s+j\Delta) |f(x, s+j\Delta) f(x, (q+j)\Delta+s)| dx ds\\ &=\sum_{j=-\infty}^{\infty}\int_{\mathbb{R} \times [0, \Delta]}\widetilde{G}_p(s) |f(x, s+j\Delta) f(x, (q+j)\Delta+s)| dx ds\\ &=\int_{\mathbb{R} \times [0, \Delta]}\widetilde{G}_p(s) \sum_{j=-\infty}^{\infty}|f(x, s+j\Delta) f(x, (q+j)\Delta+s)| dx ds\\ &=\int_{\mathbb{R} \times [0, \Delta]}\widetilde{G}_p(s) \widetilde{G}_q(s) dx ds<\infty. \end{aligned} $$

Equation (14) follows from the same calculations as above without the modulus sign in the definition of $\widetilde {G}$. □

Proof (Proof of Theorem2)

• 1. For a function f with compact support, the result can be deduced as in [13, Proposition 7.3.2]. The general case can be handled as follows, where we adapt the proof of [14, Theorem 3.5] to our more general setting. As in the proof for the sample mean, define the function $f_{m;\Delta }(x,s):=f(x,s)\mathbb {I}_{(-m \Delta , m \Delta )}(s)$, for $m \in \mathbb {N}, x, s \in \mathbb {R}$, and set

  $$\displaystyle \begin{aligned} Y_{j; \Delta}^m &:=\int_{\mathbb{R} \times \mathbb{R}}f_{m; \Delta}(x,s)L(dx, ds) \\ & =\int_{\mathbb{R} \times ((-m+j)\Delta, (m+j)\Delta)}f(x, j\Delta-s) L(dx, ds). \end{aligned} $$

  We denote by $\gamma _m$ the autocovariance function of the process $(Y_{j; \Delta }^m)_{j\in \mathbb {Z}}$. We set

  $$\displaystyle \begin{aligned} \gamma_{n;\Delta;m}^*(p\Delta)=\sum_{j=1}^n Y_{j; \Delta}^mY_{j+p; \Delta}^m, \quad p=0, \ldots, h. \end{aligned} $$

  Then, we have

  $$\displaystyle \begin{aligned} &\sqrt{n}(\gamma_{n;\Delta;m}^*(0)-\gamma_m(0), \ldots, \gamma_{n;\Delta;m}^*(h\Delta)-\gamma_m(h\Delta))^{\top} \\ &\quad \stackrel{\mathrm{d}}{\to}Z_{\Delta;m}\sim \mathrm{N}(0,V_{\Delta;m}), \quad n \to \infty, \end{aligned} $$

  where the asymptotic covariance matrix is given by $V_{\Delta ;m}=(v_{pq; \Delta ;m})_{p,q=0,\ldots ,h} \in \mathbb {R}^{h+1,h+1}$ with $v_{pq;\Delta ;m}$ defined as

  $$\displaystyle \begin{aligned} v_{pq;\Delta;m}&:=(\eta-3)\kappa_2^2\int_{\mathbb{R}\times[0,\Delta]}G_{p;\Delta;m}(x,u)G_{q;\Delta;m}(x,u)dxdu \\ &+\sum_{l=-\infty}^{\infty}[\gamma_m(l\Delta)\gamma_m((l+p-q)\Delta)\\ &+ \gamma_m((l-q)\Delta)\gamma_m((l+p)\Delta)], \\ G_{q;\Delta;m}(x,u)&:=\sum_{j=-\infty}^{\infty}f_m(x, u+j\Delta)f_m(x, u+(j+q)\Delta), \quad u \in [0, \Delta]. \end{aligned} $$

  We would like to show that $\lim _{m\to \infty }V_{\Delta ;m}=V_{\Delta }$. For this, we note that

  $$\displaystyle \begin{aligned} G_{q;\Delta;m}(x,u)&=\sum_{j=-\infty}^{\infty}f_m(x, u+j\Delta)f_m(x, u+(j+q)\Delta)\\ & \to G_{q;\Delta}(x,u)=\sum_{j=-\infty}^{\infty}f(x, u+j\Delta)f(x, u+(j+q)\Delta), \end{aligned} $$

  uniformly in $u \in [0, \Delta ]$, as $m\to \infty$, by Lebesgue’s Dominated Convergence Theorem, since the function $(x,u)\mapsto \sum _{j=-\infty }^{\infty }|f(x, u+j\Delta )||f(x, u+(j+q)\Delta )|$ is in $L^2(\mathbb {R} \times [0, \Delta ])$ by (11) and is therefore almost surely finite. Moreover, we note that

  $$\displaystyle \begin{aligned} |G_{q;\Delta;m}(x,u)|&\leq\sum_{j=-\infty}^{\infty}|f(x, u+j\Delta)||f(x, u+(j+q)\Delta)|, \end{aligned} $$

  uniformly in u and m. Hence, an application of the Dominated Convergence Theorem leads to $G_{q;\Delta ;m}\to G_{q;\Delta }$ in $L^2(\mathbb {R} \times [0, \Delta ])$ as $m \to \infty$. We also note that

  $$\displaystyle \begin{aligned} |\gamma_{m}(j\Delta)|&\leq \int_{\mathbb{R}\times \mathbb{R}}|f(x, u)||f(x, u+j\Delta)|dxdu, \quad \forall m \in \mathbb{N}, \forall j \in \mathbb{Z}. \end{aligned} $$

  Moreover, $\lim _{m\to \infty }\gamma _m(j\Delta )=\gamma (j\Delta )$ for all $j \in \mathbb {Z}$. Assumption (15) together with the Dominated Convergence Theorem allows us to conclude that $(\gamma _m(j\Delta ))_{j \in \mathbb {Z}}$ converges in $L^2(\mathbb {Z})$ to $(\gamma (j\Delta ))_{j \in \mathbb {Z}}$. Combining this result with our earlier finding of the convergence of $G_{q;\Delta ;m}$ implies that $\lim _{m\to \infty }V_{\Delta ;m}=V_{\Delta }$ and

  $$\displaystyle \begin{aligned} Z_{\Delta;m}\stackrel{\mathrm{d}}{\to} Z_{\Delta}, \quad m \to \infty, \end{aligned} $$

  where $Z_{\Delta }\stackrel {\mathrm {d}}{=}\mathrm {N}(0, V_{\Delta })$.

  Now, using the same arguments as in the proof of [13, Equation (7.3.9)], we can show that

  $$\displaystyle \begin{aligned} \lim_{m\to \infty}\limsup_{n\to \infty}\mathbb{P}(n^{1/2}|\gamma_{n;\Delta;m}^*(q\Delta)-\gamma_m(q\Delta)-\gamma^*(q \Delta)+\gamma(q\Delta)|>\epsilon)=0, \end{aligned} $$

  for all $\epsilon >0, q\in \{0, \ldots , h\}$. An application of a variant of Slutsky’s theorem, see [13, Proposition 6.3.9] completes the proof.

• 2. This part can be proven in similar way as the proof of [13, Proposition 7.3.4]. Also, as in the proof of [14, Theorem 3.5 b)], we observe that $\sqrt {n} \overline {Y}_{n;\Delta }$ converges to a Gaussian random variable as $n \to \infty$ due to Theorem 1 and $\overline {Y}_{n;\Delta }$ converges to 0 in probability as $n \to \infty$ (since we assume here that $\mu =0).$

• 3. For the final part of the theorem, we can argue as in the proof of [13, Theorem 7.2.1], where the $w_{pq;\Delta }$ are obtained via the Bartlett formula

  $$\displaystyle \begin{aligned} w_{pq;\Delta}=(v_{pq;\Delta}-\rho(p\Delta)v_{0q;\Delta} -\rho(q\Delta)v_{p0;\Delta} +\rho(p\Delta)\rho(q\Delta)v_{00;\Delta})/\gamma^2(0). \end{aligned} $$

  We can simplify the above formula and write

  $$\displaystyle \begin{aligned} w_{pq;\Delta}=w_{pq;\Delta}^{(1)}+w_{pq;\Delta}^{(2)}, \end{aligned} $$

  where

  $$\displaystyle \begin{aligned} w_{pq;\Delta}^{(1)}&:=\frac{(\eta-3)\kappa_2^2}{\gamma^2(0)} \int_{\mathbb{R} \times [0,\Delta]} (G_{p;\Delta}(x,u)-G_{0;\Delta}(x,u)\rho(p\Delta))\\ &\quad \cdot (G_{q;\Delta}(x,u)-G_{0;\Delta}(x,u)\rho(q\Delta))dxdu, \end{aligned} $$

  and

  $$\displaystyle \begin{aligned} w_{pq;\Delta}^{(2)}&:= \sum_{l=-\infty}^{\infty} \left[ \rho(l\Delta)\rho((l+p-q)\Delta)+\rho((l-q)\Delta)\rho((l+p)\Delta) \right.\\ & -2\rho(l\Delta)\rho((l-q)\Delta)\rho(p\Delta) -2\rho(l\Delta)\rho((l+p)\Delta)\rho(q\Delta)\\ &\left. +2\rho(p\Delta)\rho(q\Delta)\rho^2(l\Delta)\right]. \end{aligned} $$

  Note that

  $$\displaystyle \begin{aligned} \sum_{l=-\infty}^{\infty} \rho(l\Delta)\rho((l+p-q)\Delta) &= \sum_{l=-\infty}^{\infty} \rho((l+q)\Delta)\rho((l+p)\Delta),\\ \sum_{l=-\infty}^{\infty} \rho(l\Delta)\rho((l-q)\Delta)\rho(p\Delta) &= \sum_{l=-\infty}^{\infty} \rho((l+q)\Delta)\rho(l\Delta)\rho(p\Delta). \end{aligned} $$

  Hence,

  $$\displaystyle \begin{aligned} w_{pq;\Delta}^{(2)}&= \sum_{l=-\infty}^{\infty} \left[ \rho((l+q)\Delta)\rho((l+p)\Delta)+\rho((l-q)\Delta)\rho((l+p)\Delta) \right.\\ & -2\rho((l+q)\Delta)\rho(l\Delta)\rho(p\Delta) -2\rho(l\Delta)\rho((l+p)\Delta)\rho(q\Delta)\\ &\left. +2\rho(p\Delta)\rho(q\Delta)\rho^2(l\Delta)\right]. \end{aligned} $$

  Hence,

  $$\displaystyle \begin{aligned} w_{pq;\Delta}&= \frac{(\eta-3)\kappa_2^2}{\gamma^2(0)} \int_{\mathbb{R} \times [0,\Delta]} (G_{p;\Delta}(x,u)-G_{0;\Delta}(x,u)\rho(p\Delta))\\ &\quad \cdot (G_{q;\Delta}(x,u)-G_{0;\Delta}(x,u)\rho(q\Delta))dxdu\\ &+\sum_{l=-\infty}^{\infty} \left[ \rho((l+q)\Delta)\rho((l+p)\Delta)+\rho((l-q)\Delta)\rho((l+p)\Delta) \right.\\ & -2\rho((l+q)\Delta)\rho(l\Delta)\rho(p\Delta) -2\rho(l\Delta)\rho((l+p)\Delta)\rho(q\Delta)\\ &\left. +2\rho(p\Delta)\rho(q\Delta)\rho^2(l\Delta)\right]. \end{aligned} $$

□

1.3 Examples

In this subsection, we show how the asymptotic variances appearing in the asymptotic theory for the sample mean can be computed for trawl processes with either an exponential or a supGamma trawl function.

Note that, in the case when $p\equiv 1$, i.e. for the (non-periodic) trawl process, we get

$$\displaystyle \begin{aligned} \gamma(h)={\mathrm{Cov}}(Y_t, Y_{t+h})=\int_{|h|}^{\infty}g(u)du, \end{aligned} $$

for $t, h \in \mathbb {R}$.

1.3.1 Exponential Trawl

Consider the case of an exponential trawl function with $g(x)=\exp (-\lambda x)$, for $x \geq 0$. The autocovariance function is given by

$$\displaystyle \begin{aligned} \gamma(t)={\mathrm{Cov}}(Y_t, Y_0)=\frac{1}{\lambda}e^{-\lambda |t|}, \end{aligned} $$

for $t\in \mathbb {R}$, and the autocorrelation function is given by

$$\displaystyle \begin{aligned} \rho(t)={\mathrm{Cor}}(Y_t, Y_0)=e^{-\lambda |t|}, \end{aligned} $$

for $t\in \mathbb {R}$.

For the sample mean, we have the following result. Suppose that $\mathbb {E}(L')=0, \kappa _2={\mathrm {Var}}(L')<\infty , \mu \in \mathbb {R}$ and $\Delta >0$. Then

$$\displaystyle \begin{aligned} & \left(F_{\Delta}:\mathbb{R}\times [0, \Delta] \to [0, \infty], (x, u) \mapsto F_{\Delta}(x,u)=\sum_{j=-\infty}^{\infty}|f(x, u+j\Delta)|\right)\\ & \quad \in L^2(\mathbb{R} \times [0, \Delta]) \end{aligned} $$

since

$$\displaystyle \begin{aligned} F_{\Delta}(x,u)=\sum_{j=-\infty}^{\infty}|f(x, u+j\Delta)| =\sum_{j=-\infty}^{\infty}\mathbb{I}_{(0, g(u+j\Delta)}(x), \end{aligned} $$

and we have that

$$\displaystyle \begin{aligned} & \int_{\mathbb{R} \times [0, \Delta]}|F_{\Delta}(x,u)|{}^2dx du \\ &\quad =\int_{\mathbb{R} \times [0, \Delta]} \sum_{j=-\infty}^{\infty}\sum_{k=-\infty}^{\infty}\mathbb{I}_{(0, g(u+j\Delta)}(x) \mathbb{I}_{(0, g(u+k\Delta)}(x)dx du \\ &\quad =\sum_{j=-\infty}^{\infty}\sum_{k=-\infty}^{\infty}\int_{\mathbb{R} \times [0, \Delta]} \mathbb{I}_{(0, g(u+j\Delta)}(x) \mathbb{I}_{(0, g(u+k\Delta)}(x)dx du\\ &\quad =\sum_{j=0}^{\infty}\sum_{k=0}^{\infty}\int_{ [0, \Delta]} \min\{g(u+j\Delta),g(u+k\Delta)\} du\\ &\quad =\sum_{j=0}^{\infty}\int_{ [0, \Delta]} g(u+j\Delta) du +2 \sum_{j=0}^{\infty}\sum_{k=j+1}^{\infty}\int_{ [0, \Delta]} g(u+k\Delta) du, \end{aligned} $$

where we applied Tonelli’s theorem. Hence, we deduce that $\sum _{j=-\infty }^{\infty } |\gamma (\Delta j)| < \infty$,

$$\displaystyle \begin{aligned} {} V_{\Delta}:=\sum_{j=-\infty}^{\infty}\gamma( \Delta j) = \kappa_2 \int_{\mathbb{R} \times [0, \Delta]} \left(\sum_{j=1}^{\infty}f(x, u+j\Delta)\right)^2 dx du, \end{aligned} $$

(21)

and the sample mean of $Y_{\Delta i}$, for $i=1, \ldots , n$, is asymptotically Gaussian as $n \to \infty$, i.e.

$$\displaystyle \begin{aligned} \sqrt{n}\left(\overline{Y}_{n; \Delta} - \mu \right) \stackrel{\mathrm{d}}{\to} \mathrm{N}\left(0, V_{\Delta}\right), \quad \mathrm{as} \, n \to \infty. \end{aligned} $$

For the case of an exponential trawl function, we get

$$\displaystyle \begin{aligned} V_{\Delta}&=\sum_{j=-\infty}^{\infty}\gamma(\Delta j) =\sum_{j=-\infty}^{\infty}\frac{1}{\lambda}e^{-\lambda |j|\Delta}\\ & =\frac{1}{\lambda}\left(1+2\sum_{j=1}^{\infty}e^{-\lambda j \Delta}\right) =\frac{1}{\lambda}\left(1+2\frac{e^{-\lambda \Delta}}{1-e^{-\lambda \Delta}}\right)\\ &=\frac{1+e^{-\lambda \Delta}}{\lambda(1-e^{-\lambda \Delta})}=\frac{1+e^{\lambda \Delta}}{\lambda(e^{\lambda \Delta}-1)}. \end{aligned} $$

1.3.2 SupGamma Trawl

In the case when $g(x)=(1+x/\alpha )^{-H}$, $\alpha >0, H>2$, i.e. we require a short-memory setting, $x \geq 0$, we have

$$\displaystyle \begin{aligned} \gamma(h)=\int_{|h|}^{\infty}g(x)dx=\frac{\alpha}{H-1}\left(1+\frac{|h|}{\alpha}\right)^{1-H}. \end{aligned} $$

Then

$$\displaystyle \begin{aligned} V_{\Delta}&=\sum_{j=-\infty}^{\infty}\gamma(j \Delta) =\frac{\alpha}{H-1}\sum_{j=-\infty}^{\infty} \left(1+\frac{|j|\Delta}{\alpha}\right)^{1-H} \\ &=\frac{\alpha}{H-1}\left(\frac{\Delta}{\alpha}\right)^{1-H} \sum_{j=-\infty}^{\infty} \left(\frac{\alpha}{\Delta}+|j|\right)^{1-H} \\ &=\frac{\alpha}{H-1}\left(\frac{\Delta}{\alpha}\right)^{1-H} \left(\frac{\alpha}{\Delta} \right)^{1-H} \left[ 2 \left(\frac{\alpha}{\Delta} \right)^{H-1}\zeta(H-1, \alpha/\Delta)-1\right]\\ &=\frac{\alpha}{H-1} \left[ 2 \left(\frac{\alpha}{\Delta} \right)^{H-1}\zeta(H-1, \alpha/\Delta)-1\right], \end{aligned} $$

where $\zeta$ denotes the Hurwitz Zeta function defined by $\zeta (s, a)=\sum _{k=0}^{\infty }\frac {1}{(k+a)^s}$, for $\mathrm {Re}(s)>1$.

1.4 Verifying the Assumptions of Theorem 2 for Selected Periodic Trawl Processes

For the applications discussed in Sect. 5, we need to verify the condition (11) from Proposition 7 and Assumption (15) from Theorem 2 assuming that the corresponding moment assumptions for the Lévy seed hold.

For both conditions, it is sufficient to check that a (non-periodic) trawl process satisfies the stated conditions since the periodic function is bounded. Hence, in the following, we shall set $p\equiv 1$.

1.4.1 Verifying Condition (11) from Proposition 7

We need to check that

$$\displaystyle \begin{aligned} \left(\mathbb{R}\times [0, \Delta] \to \mathbb{R}, (x, u) \mapsto \sum_{j=-\infty}^{\infty}f^2(x, u+j\Delta)\right) \in L^2(\mathbb{R} \times [0, \Delta]). \end{aligned} $$

This condition holds for trawl processes if, for $(x, u)\in \mathbb {R}\times [0, \Delta ]$,

$$\displaystyle \begin{aligned} & \sum_{j=-\infty}^{\infty}f^2(x, u+j\Delta) =\sum_{j=-\infty}^{\infty}f(x, u+j\Delta) \\ &\quad =\sum_{j=-\infty}^{\infty}\mathbb{I}_{(0, g(u+j\Delta))}(x) \mathbb{I}_{[0, \infty)}(u+j\Delta) =F_{\Delta}(x,u)\in L^2(\mathbb{R} \times [0, \Delta]). \end{aligned} $$

This is equivalent to checking that

$$\displaystyle \begin{aligned} \begin{aligned}{} & \int_{\mathbb{R} \times [0, \Delta]}|F_{\Delta}(x,u)|{}^2dxdu \\ &= \int_{\mathbb{R} \times [0, \Delta]} \sum_{j=-\infty}^{\infty}\sum_{k=-\infty}^{\infty}\mathbb{I}_{(0, g(u+j\Delta))}(x) \mathbb{I}_{[0, \infty)}(u+j\Delta) \mathbb{I}_{(0, g(u+k\Delta))}(x)\\ &\qquad \mathbb{I}_{[0, \infty)}(u+k\Delta) dxdu\\ &\propto \sum_{j=-\infty}^{\infty}\gamma(j \Delta) < \infty.\end{aligned} \end{aligned} $$

(22)

This condition is satisfied both for an exponential trawl function and also for a supGamma trawl function with short memory. In the latter case, we have that $\gamma (x) \propto (1+|x|/\alpha )^{1-H}$ for $\alpha >0, H>2$. Then, the finiteness of (22) follows using the $\zeta$-function representation.

1.4.2 Verifying Assumption (15) from Theorem 2

We need to verify

$$\displaystyle \begin{aligned} \sum_{j=-\infty}^{\infty}\left(\int_{\mathbb{R} \times \mathbb{R} } |f(x, u)||f(x, u+j\Delta)| dxdu\right)^2<\infty. \end{aligned} $$

Using very similar computations as before, we find that the above condition is equivalent to

$$\displaystyle \begin{aligned} &\sum_{j=-\infty}^{\infty}\left(\int_{\mathbb{R} \times \mathbb{R} } |f(x, u)||f(x, u+j\Delta)| dxdu\right)^2\\ &=\sum_{j=-\infty}^{\infty}\left(\int_{\mathbb{R} \times \mathbb{R} } \mathbb{I}_{(0, g(u)}(x) \mathbb{I}_{[0, \infty)}(u) \mathbb{I}_{(0, g(u+j\Delta))}(x) \mathbb{I}_{[0, \infty)}(u+j\Delta) dxdu\right)^2\\ & \propto\sum_{j=-\infty}^{\infty}\gamma^2(j\Delta)<\infty, \end{aligned} $$

which is satisfied by the exponential trawl function and the supGamma trawl functions with $H>3/2$, which includes some long-memory settings.