Nonparametric estimation of the kernel function of symmetric stable moving average random functions

Abstract

We estimate the kernel function of a symmetric alpha stable ($S\alpha S$) moving average random function which is observed on a regular grid of points. The proposed estimator relies on the empirical normalized (smoothed) periodogram. It is shown to be weakly consistent for positive definite kernel functions, when the grid mesh size tends to zero and at the same time the observation horizon tends to infinity (high-frequency observations). A simulation study shows that the estimator performs well at finite sample sizes, when the integrator measure of the moving average random function is $S\alpha S$ and for some other infinitely divisible integrators.

Appendices

Appendix 1: Proofs of Theorems 1, 3

Proof of Theorem 1

We first show how (ii) follows from (i). Notice that $\left| {{\hat{g}}}(\lambda )\right| = {{\hat{g}}}(\lambda )$ for all $\lambda \in {\mathbb {R}}$, since by (F1) f is of positive type. Since $|\sqrt{a}-\sqrt{b}| \le \sqrt{ |a-b|}$ for $a, b\ge 0$, we get, using the Cauchy–Schwarz inequality,

$$\begin{aligned}&\int _{-a_n}^{a_n} \left( \sqrt{\varDelta _n I^s_{n,X}(\lambda )} - {\hat{g}}(\lambda ) \right) ^2 \hbox {d}\lambda \le \sqrt{2a_n} \cdot \\&\quad \sqrt{\int _{-a_n}^{a_n} \left| \varDelta _n I^s_{n,X}(\lambda ) - {\hat{g}}(\lambda )^2 \right| ^2 \, \hbox {d}\lambda } \overset{P}{\longrightarrow } 0,\quad n\rightarrow \infty . \end{aligned}$$

Since ${\hat{g}}\in L^2({{\mathbb {R}}})$, we then have

$$\begin{aligned} \int _{{\mathbb {R}}} \left( {\mathbf {1}}_{[-a_n,a_n]}(\lambda ) \cdot \sqrt{\varDelta _n I^s_{n,X}(\lambda )} - {\hat{g}}(\lambda )\right) ^2\, \hbox {d}\lambda \overset{P}{\longrightarrow } 0,\quad n\rightarrow \infty . \end{aligned}$$

The desired statement now follows from the Plancherel’s equality.

Now let us prove (i). Write

$$\begin{aligned} I^s_{n,X}(\lambda ) = \frac{J_{n,X}^s(\lambda )}{S_{n,X}}, \end{aligned}$$

where $J_{n,X}^s(\lambda ) = \sum _{\left| m\right| \le m_n} W_n(m)\left| \sum _{j=1}^{n} X(t_{j,n})\hbox {e}^{it_{j,n}\nu _n(m,\lambda )}\right| ^2,\quad S_{n,X} = {\sum _{j=1}^{n}X(t_{j,n})^2}.$

Let f be supported by $[-T,T]$. We will assume that $N = T/\varDelta _n$ is integer: this will simplify the exposition while not harming the rigor. The proof is rather long, so we split it into several steps for better readability. Choose $n\ge 2N+1$.

Step 1. Denominator

We start with investigating the denominator $S_{n,X}$. First, we study the behavior of a similar expression with f replaced by its discretized version. Specifically, define

$$\begin{aligned} X_{n}(t_{j,n})= & {} \sum _{k=-N}^{N-1} f(t_{k,n}) \varLambda \left( ((j-k-1)\varDelta _n,(j-k)\varDelta _n]\right) \\= & {} \int _{{\mathbb {R}}}f_n(t_{j,n} - s) \varLambda (\hbox {d}s),\ j=1,\dots ,n, \end{aligned}$$

where $f_n(x) = \sum _{k=-N}^{N-1}f(t_{k,n}){\mathbb {1}}_{[t_{k,n},t_{k+1,n})}(x)$. Denote $\varepsilon _{l,n} = \varLambda \left( ((l-1)\varDelta _n,l\varDelta _n]\right)$, $l\in {\mathbb {Z}}$. For fixed n, these variables are independent $S\alpha S$ with scale parameter $\varDelta _n^{1/\alpha }$.

Decompose

$$\begin{aligned} \sum _{j=1}^n X_{n}(t_{j,n})^2= & {} \sum _{j=1}^n \left( \sum _{l=j-N+1}^{j+N}f(t_{j-l,n})\varepsilon _{l,n} \right) ^2 \\= & {} \sum _{j=1}^n \sum _{l=j-N+1}^{j+N} f(t_{j-l,n})^2\varepsilon _{l,n}^2 \\&+\sum _{j=1}^n \sum _{\begin{array}{c} l_1,l_2=j-N+1\\ l_1\ne l_2 \end{array}}^{j+N}f(t_{j-l_1,n})f(t_{j-l_2,n})\varepsilon _{l_1,n}\varepsilon _{l_2,n}\\=\, & {} \left( \sum _{l=N+1}^{n-N}\sum _{j=l-N}^{l+N-1} + \sum _{l=2-N}^{N}\sum _{j=1}^{l+N-1} + \sum _{l=n-N+1}^{n+N}\sum _{j=l-N}^{n}\right) f(t_{j-l,n})^2\varepsilon _{l,n}^2\\&+ \sum _{j=1}^n \sum _{\begin{array}{c} l_1,l_2=j-N+1\\ l_1\ne l_2 \end{array}}^{j+N}f(t_{j-l_1,n})f(t_{j-l_2,n})\varepsilon _{l_1,n}\varepsilon _{l_2,n}\\=: & {} S_{1,n} + S_{2,n} + S_{3,n} + S_{4,n}. \end{aligned}$$

We are going to show that the last three terms are negligible. We use the shorthand $E_n = \sum _{l=N+1}^{n-N} \varepsilon _{l,n}^2$, as this will be our benchmark term. Observe that $S_{1,n} = \sum _{k=-N}^{N-1} f(t_{k,n})^2 E_n$. Thanks to the boundedness and uniform continuity of f,

$$\begin{aligned} \left| S_{1,n} - \frac{1}{\varDelta _n} \int _{-T}^T f(x)^2 \hbox {d}x\cdot E_n\right| = O\left( \varDelta _n^{-1}\omega _f(\varDelta _n) E_n \right) ,\quad n\rightarrow \infty . \end{aligned}$$

(17)

On the other hand, by Feller (1966, XVII.5, Theorem 3 (i)), we have

$$\begin{aligned} \frac{E_n}{n^{2/\alpha }\varDelta _n^{2/\alpha }}\Rightarrow Z_{\alpha },\quad n\rightarrow \infty , \end{aligned}$$

(18)

where $Z_\alpha$ is some positive $\alpha /2$-stable random variable. Therefore, by Slutsky’s theorem,

$$\begin{aligned} \frac{S_{1,n}}{n^{2/\alpha }\varDelta _n^{2/\alpha -1}}\Rightarrow Z_{\alpha } \int _{-T}^T f(x)^2 \hbox {d}x,\quad n\rightarrow \infty . \end{aligned}$$

(19)

Estimating

$$\begin{aligned} S_{2,n} + S_{3,n} \le \left( \sum _{l=2-N}^{N} + \sum _{l=n-N+1}^{n+N}\right) \varepsilon _{l,n}^2\sum _{k=-N}^{N-1} f(t_{k,n})^2 =: S_{5,n}, \end{aligned}$$

similarly to (19), we get

$$\begin{aligned} \frac{S_{5,n}}{(2N)^{2/\alpha } \varDelta _n^{2/\alpha -1}}\Rightarrow Z_\alpha ^\prime \int _{-T}^T f(x)^2 \hbox {d}x,\quad n\rightarrow \infty . \end{aligned}$$

(20)

Since $N\varDelta _n=T$, we have

$$\begin{aligned} S_{2,n} + S_{3,n}= &\, {} O_P (\varDelta _n^{-1}) = O_P\left( (n\varDelta _n)^{-2/\alpha } n^{2/\alpha }\varDelta _n^{2/\alpha -1}\right) \\= &\, {} O_P \left( S_{1,n} (n\varDelta _n)^{-2/\alpha } \right) , n\rightarrow \infty . \end{aligned}$$

The term $S_{4,n}$ is estimated using Lemma 3: $S_{4,n} = O_P(N^{3/2} n^{2/\alpha -1/2}\varDelta _n^{2/\alpha }) = O_P((n\varDelta _n)^{-1/2}S_{1,n}) .$

Summing up, we have $\sum _{j=1}^{n} X_{n}(t_{j,n})^2 = S_{1,n}\left( 1+O_P((n\varDelta _n)^{-1/2})\right)$, $n\rightarrow \infty$, and $S_{1,n}$ is of order $n^{2/\alpha }\varDelta _n^{2/\alpha -1}$, in the sense of (19).

Now we get back to the denominator of $I_{n,X}(\lambda )$. For any positive vanishing sequence $\left\{ \delta _n,n\ge 1\right\},$ write the following simple estimate:

$$\begin{aligned} \left| a^2-b^2\right| \le 2\left| a(a-b)\right| + \left| a-b\right| ^2 \le \delta _n a^2 + (1+\delta _n^{-1})\left| a-b\right| ^2. \end{aligned}$$

(21)

Then, we obtain

$$\begin{aligned} \left| \sum _{j=1}^{n} X_{n}(t_{j,n})^2 - S_{n,X} \right| \le \delta _n \sum _{j=1}^{n} X_{n}(t_{j,n})^2 + (1+\delta _n^{-1})\sum _{j=1}^{n} \left( X_{n}(t_{j,n})-X(t_{j,n})\right) ^2. \end{aligned}$$

From Lemma 4 it follows that

$$\begin{aligned} \sum _{j=1}^{n} \left( X_{n}(t_{j,n})-X(t_{j,n})\right) ^2= &\, {} O_P(\Vert f_n-f\Vert ^2_\infty n^{2/\alpha } \varDelta _n^{2/\alpha -1})\\= & {}\, O_P(\omega _f(\varDelta _n)^2 S_{1,n}),\quad n\rightarrow \infty . \end{aligned}$$

Putting $\delta _n = \omega _f(\varDelta _n)$ and using (17), we conclude that

$$\begin{aligned} \varDelta _n S_{n,X} = \varDelta _n S_{1,n}\left( 1+ O_P((n\varDelta _n)^{-1/2}+ \omega _f(\varDelta _n))\right) \overset{P}{\sim } \Vert f\Vert _2^2 E_n ,\ n\rightarrow \infty . \end{aligned}$$

(22)

Step 2. Whole expression

Thanks to (22),

$$\begin{aligned}&a_n\int \nolimits _{-a_n}^{a_n}\left| \varDelta _n I_{n,X}^s(\lambda ) - \left| {{\hat{g}}}(\lambda )\right| ^2\right| ^2\hbox {d}\lambda \\&\quad = a_n\int \nolimits _{-a_n}^{a_n}\left| \frac{\varDelta _n^2 J^s_{n,X}(\lambda )}{\varDelta _nS_{n,X}} - \frac{\left| {{\hat{f}}}(\lambda )\right| ^2}{\Vert f\Vert _2^2}\right| ^2 \hbox {d}\lambda \\&\quad \le 2a_n\int \nolimits _{-a_n}^{a_n}\left| \frac{\varDelta _n^2 J^s_{n,X}(\lambda )}{\varDelta _n S_{n,X}} - \frac{\left| {\hat{f}}(\lambda )\right| ^2 E_n}{\varDelta _n S_{n,X}}\right| ^2 \hbox {d}\lambda + 2a_n\int \nolimits _{-a_n}^{a_n}\left| \frac{\left| {\hat{f}}(\lambda )\right| ^2 E_n}{\varDelta _n S_{n,X}} - \frac{\left| {{\hat{f}}}(\lambda )\right| ^2}{\Vert f\Vert _2^2}\right| ^2 \hbox {d}\lambda \\&\quad = 2a_n \left[ \int \nolimits _{-a_n}^{a_n} \left| \frac{\varDelta _n^2 J^s_{n,X}(\lambda ) - \left| {\hat{f}}(\lambda )\right| ^2 E_n}{\varDelta _n S_{n,X} }\right| ^2 \hbox {d}\lambda + \left| \frac{ \Vert f\Vert _2^2 E_n - \varDelta _n S_{n,X} }{\varDelta _n\Vert f\Vert _2^2 S_{n,X} }\right| ^2 \int \nolimits _{-a_n}^{a_n} \left| {\hat{f}}(\lambda )\right| ^4 \hbox {d}\lambda \right] \\&\quad = \frac{a_n}{\Vert f\Vert _2^4 E_n^2} \int \nolimits _{-a_n}^{a_n} \left| \varDelta _n^2 J^s_{n,X}(\lambda ) - \left| {\hat{f}}(\lambda )\right| ^2 E_n\right| ^2 \hbox {d}\lambda + o_P(1),\quad n\rightarrow \infty . \end{aligned}$$

Thus, it remains to prove that

$$\begin{aligned} a_n\int _{-a_n}^{a_n}\left| \varDelta _n^2 J^s_{n,X}(\lambda ) - \left| {{\hat{f}}}(\lambda )\right| ^2 E_n\right| ^2 \hbox {d}\lambda = o_P(E_n^2),\quad n\rightarrow \infty . \end{aligned}$$

(23)

Step 3. Numerator

As with the denominator, we start with examining the discretized version of $J_{n,X}^s(\lambda )$:

$$\begin{aligned} R_n(\lambda )= & {} \sum _{\left| m\right| \le m_n} W_n(m) \left| \sum _{j=1}^{n} X_{n}(t_{j,n})\hbox {e}^{it_{j,n}\nu _{n}(m,\lambda )}\right| ^2\\= & {} \sum _{\left| m\right| \le m_n} W_n(m) \left| \sum _{j=1}^{n}\sum _{l=j-N+1}^{j+N}f(t_{j-l,n}) \varepsilon _{l,n}\hbox {e}^{it_{j,n}\nu _{n}(m,\lambda )}\right| ^2 . \end{aligned}$$

We proceed in three substeps, first considering the following expression

$$\begin{aligned} R_{1,n}(\lambda ) = \sum _{\left| m\right| \le m_n} W_n(m) \left| \sum _{l=N+1}^{n-N}\sum _{j=l-N}^{l+N-1}f(t_{j-l,n}) \varepsilon _{l,n}\hbox {e}^{it_{j,n}\nu _{n}(m,\lambda )}\right| ^2. \end{aligned}$$

Step 3a). We shall show

$$\begin{aligned} a_n \int _{-a_n}^{a_n}\left| \left| {{\hat{f}}}(\lambda )\right| ^2 E_n - \varDelta _n^2 R_{1,n}(\lambda )\right| ^2 \hbox {d}\lambda = o_P(E_n^2),\quad n\rightarrow \infty . \end{aligned}$$

(24)

We have for $\lambda \in [-a_n,a_n]$ that

$$\begin{aligned} R_{1,n}(\lambda )= & {} \sum _{\left| m\right| \le m_n} W_n(m) \left| \sum _{l=N+1}^{n-N}\varepsilon _{l,n} \hbox {e}^{it_{l,n}\nu _{n}(m,\lambda )}\sum _{j=l-N}^{l+N-1}f(t_{j-l,n})\hbox {e}^{it_{j-l,n}\nu _{n}(m,\lambda )}\right| ^2\\= & {} \sum _{\left| m\right| \le m_n} W_n(m) \left| \sum _{k=-N}^{N-1} f(t_{k,n})\hbox {e}^{it_{k,n}\nu _n(m,\lambda )}\right| ^2\left| \sum _{l=N+1}^{n-N}\varepsilon _{l,n}\hbox {e}^{it_{l,n}\nu _n(m,\lambda )}\right| ^2\\= &\, {} F_{n}(\lambda )\sum _{l=N+1}^{n-N}\varepsilon _{l,n}^2 + \sum _{N+1\le l_1\ne l_2\le n-N} a_{l_1,l_2,n}(\lambda )\varepsilon _{l_1,n}\varepsilon _{l_2,n} , \end{aligned}$$

where

$$\begin{aligned}&F_n(\lambda ) = \sum _{\left| m\right| \le m_n} W_n(m) \left| \sum _{k=-N}^{N-1} f(t_{k,n})\hbox {e}^{it_{k,n}\nu _{n}(m,\lambda )}\right| ^2,\\&a_{l_1,l_2,n}(\lambda ) = \sum _{\left| m\right| \le m_n} W_n(m)\left| \sum _{k=-N}^{N-1} f(t_{k,n})\hbox {e}^{it_{k,n} \nu _{n}(m,\lambda )} \right| ^2 \hbox {e}^{i(l_1-l_2)\varDelta _n\nu _{n}(m,\lambda )}{\mathbb {1}}_{[-a_n,a_n]}(\lambda ). \end{aligned}$$

With the help of Lemma 6, we obtain

$$\begin{aligned}&a_n \int _{-a_n}^{a_n}\left| \left| {{\hat{f}}}(\lambda )\right| ^2 E_n - \varDelta _n^2 R_{1,n}(\lambda )\right| ^2 \hbox {d}\lambda \\&\quad = O\left( a^2_n W_n^{(2)}(n\varDelta _n )^{-2} + a^2_n\omega _f(\varDelta _n)^2 + a_n^4 \varDelta _n^2 \right) E^2_n \\&\qquad +\, 2a_n\varDelta _n^4 \int _{{{\mathbb {R}}}} \left| \sum _{N+1\le l_1\ne l_2\le n-N} a_{l_1,l_2,n}(\lambda )\varepsilon _{l_1,n}\varepsilon _{l_2,n}\right| ^2 \hbox {d}\lambda ,\quad n\rightarrow \infty . \end{aligned}$$

By Lemma 2,

$$\begin{aligned} \int _{{{\mathbb {R}}}}\left| \sum _{N+1\le l_1\ne l_2\le n-N} a_{l_1,l_2,n}(\lambda ) \varepsilon _{l_1,n}\varepsilon _{l_2,n} \right| ^2 \hbox {d}\lambda = O_P\left( A_n n^{4/\alpha -2}\varDelta _n^{4/\alpha }\right) , \quad n\rightarrow \infty , \end{aligned}$$

where, by Lemma 7,

$$\begin{aligned} A_n = \int _{-a_n}^{a_n}\sum _{N+1\le l_1\ne l_2\le n-N} \left| a_{l_1,l_2,n}(\lambda )\right| ^2 \hbox {d}\lambda = O\left( a_n W_n^*(K_n^* n)^2\right) ,\quad n\rightarrow \infty , \end{aligned}$$

with

$$\begin{aligned} K_n^* = \sup _{\left| m\right| \le m_n}\left| \sum _{k=-N}^{N-1} f(t_{k,n})\hbox {e}^{it_{k,n} \nu _{n}(m,\lambda )}\right| ^2 \le \left( \sum _{k=-N}^{N-1} \left| f(t_{k,n})\right| \right) ^2 \sim \varDelta _n^{-2}\Vert f\Vert ^2_1,\ n\rightarrow \infty . \end{aligned}$$

Hence,

$$\begin{aligned}&a_n\varDelta _n^4\int _{{{\mathbb {R}}}} \left| \sum _{N+1\le l_1\ne l_2\le n-N} a_{l_1,l_2,n}(\lambda )\varepsilon _{l_1,n}\varepsilon _{l_2,n}\right| ^2 \hbox {d}\lambda \\&\quad = O_P(a_n^2 W_n^* (n\varDelta _n)^{4/\alpha }) = O_P\left( a_n^2 W_n^* E_n^2\right) ,\ n\rightarrow \infty . \end{aligned}$$

Combining the estimates, we get (24).

Step 3b). We get

$$\begin{aligned} a_n\int _{-a_n}^{a_n}\left| \varDelta _n^2 R_{n}(\lambda ) - \left| {{\hat{f}}}(\lambda )\right| ^2E_n\right| ^2 \hbox {d}\lambda = o_P(E_n^2),\quad n\rightarrow \infty . \end{aligned}$$

(25)

Indeed, write

$$\begin{aligned}&a_n\int _{-a_n}^{a_n}\left| \varDelta _n^2 R_{n}(\lambda ) - \left| {{\hat{f}}}(\lambda )\right| ^2E_n\right| ^2 \hbox {d}\lambda \\&\quad \le 2a_n\int _{-a_n}^{a_n}\varDelta _n^4\left| R_n(\lambda ) - R_{1,n}(\lambda )\right| ^2 \hbox {d}\lambda + 2a_n \int _{-a_n}^{a_n}\left| \varDelta _n^2 R_{1,n}(\lambda ) - \left| {{\hat{f}}}(\lambda )\right| ^2E_n\right| ^2 \hbox {d}\lambda \\&\quad = 2a_n\varDelta _n^4\int _{-a_n}^{a_n}\left| R_n(\lambda ) - R_{1,n}(\lambda )\right| ^2 \hbox {d}\lambda + o_P(E_n^2),\quad n\rightarrow \infty . \end{aligned}$$

Let us estimate the first expression. Take some positive vanishing sequence $\left\{ \theta _n,n\ge 1\right\}$, which will be specified later. Using (21), we have

$$\begin{aligned}&\left| R_{1,n}(\lambda ) - R_n(\lambda )\right| \le \theta _n R_{1,n}(\lambda ) + (1+\theta _n^{-1})\\&\qquad \times \sum _{\left| m\right| \le m_n}W_n(m)\left| \left( \sum _{l=2-N}^{N}\sum _{j=1}^{l+N-1} + \sum _{l=n-N+1}^{n+N}\sum _{j=l-N}^{n}\right) \right. \nonumber \\&\quad \qquad \left. \varepsilon _{l,n}\hbox {e}^{i t_{l,n} \nu _{n}(m,\lambda )}f(t_{j-l,n})\hbox {e}^{it_{j-l,n}\nu _{n}(m,\lambda )} \right| ^2\\&\quad \le \theta _n R_{1,n}(\lambda ) + 2(1+\theta _n^{-1})\left( R_{2,n}(\lambda )+R_{3,n}(\lambda )\right) , \end{aligned}$$

where

$$\begin{aligned} R_{2,n}(\lambda )= & {} \sum _{\left| m\right| \le m_n}W_n(m)\left| \sum _{l=2-N}^{N}\varepsilon _{l,n}\hbox {e}^{i t_{l,n} \nu _{n}(m,\lambda )}\sum _{k=1-l}^{N-1}f(t_{k,n})\hbox {e}^{it_{k,n}\nu _{n}(m,\lambda )} \right| ^2,\\ R_{3,n}(\lambda )= & {} \sum _{\left| m\right| \le m_n}W_n(m)\left| \sum _{l=n-N+1}^{n+N}\varepsilon _{l,n}\hbox {e}^{i t_{l,n} \nu _{n}(m,\lambda )}\sum _{k=-N}^{n-l} f(t_{k,n})\hbox {e}^{it_{k,n}\nu _{n}(m,\lambda )} \right| ^2. \end{aligned}$$

Hence,

$$\begin{aligned}&a_n\int _{-a_n}^{a_n}\left| R_n(\lambda ) - R_{1,n}(\lambda )\right| ^2 \hbox {d}\lambda \\&\quad \le 2a_n\theta _n^2 \int _{-a_n}^{a_n} R_{1,n}(\lambda )^2 \hbox {d}\lambda + 16a_n(1+\theta _n^{-1})^2\int _{-a_n}^{a_n}\left( R_{2,n}(\lambda )^2 + R_{3,n}(\lambda )^2\right) \hbox {d}\lambda . \end{aligned}$$

Now

$$\begin{aligned} R_{2,n}(\lambda )= & {} \sum _{l=2-N}^{N} \varepsilon _{l,n}^2 \sum _{\left| m\right| \le m_n}W_n(m) \left| \sum _{k=1-l}^{N-1}f(t_{k,n})\hbox {e}^{it_{k,n}\nu _{n}(m,\lambda )} \right| ^2\\&+ \sum _{\begin{subarray}{c} l_1,l_2=2-N\\ l_1\ne l_2 \end{subarray}}^{N-1} b_{l_1,l_2,n}(\lambda ) \varepsilon _{l_1,n}\varepsilon _{l_2,n}\\\le & {} \sum _{l=2-N}^{N} \varepsilon _{l,n}^2 \left( \sum _{k=-N}^{N-1}\left| f(t_{k,n})\right| \right) ^2 + \sum _{\begin{subarray}{c} l_1,l_2=2-N\\ l_1\ne l_2 \end{subarray}}^{N} b_{l_1,l_2,n}(\lambda ) \varepsilon _{l_1,n}\varepsilon _{l_2,n} \\=: & {} R_{4,n} + R_{5,n}(\lambda ), \end{aligned}$$

where

$$\begin{aligned}&b_{l_1,l_2,n}(\lambda ) \nonumber \\&\quad = \sum _{\left| m\right| \le m_n}W_n(m)\hbox {e}^{i(l_1-l_2)\varDelta _n\nu _{n}(m,\lambda )} \sum _{k_1=1-l_1}^{N-1} \sum _{k_2=1-l_2}^{N-1}f(t_{k_1,n}) f(t_{k_2,n})\hbox {e}^{i(k_1-k_2)\varDelta _n\nu _{n}(m,\lambda )}. \end{aligned}$$

Using Lemma 2, we obtain

$$\begin{aligned} \int _{-a_n}^{a_n}R_{5,n}(\lambda )^2 \hbox {d}\lambda = O_P(a_n\varDelta _n^{-4}),\quad n\rightarrow \infty . \end{aligned}$$

Further, using (20), $R_{4,n}\sim \varDelta _n^{-2} \Vert f\Vert _1^2\sum _{l=2-N}^{N} \varepsilon _{l,n}^2= O_P(\varDelta _n^{-2}),\ n\rightarrow \infty ,$ hence

$$\begin{aligned} \int _{-a_n}^{a_n} R_{2,n}(\lambda )^2 \, \hbox {d}\lambda = O_P(a_n\varDelta _n^{-4}), n\rightarrow \infty . \end{aligned}$$

Similarly, $\int _{-a_n}^{a_n}R_{3,n}(\lambda )^2 \, \hbox {d}\lambda = O_P(a_n\varDelta _n^{-4})$, $n\rightarrow \infty$.

Setting $\theta _n = a_n^{1/4}(n\varDelta _n)^{-1/\alpha }$, we get by (A3) that

$$\begin{aligned} a_n(1+\theta _n^{-1})^2 \int _{-a_n}^{a_n}\left( R_{2,n}(\lambda )^2 + R_{3,n}(\lambda )^2\right) \hbox {d}\lambda= & {} O_P\left( a_n^{3/2}\varDelta _n^{-4}(n\varDelta _n)^{2/\alpha }\right) \\= & {} o_P\left( n^{4/\alpha }\varDelta _n^{4/\alpha -4}\right) \end{aligned}$$

as $n\rightarrow \infty$. Therefore, we arrive at

$$\begin{aligned}&a_n\int _{-a_n}^{a_n}\left| \varDelta _n^2 R_n(\lambda ) - \varDelta _n^2 R_{1,n}(\lambda )\right| ^2 \hbox {d}\lambda \\&\quad \le 2a_n^{3/2}(n\varDelta _n)^{-2/\alpha }\varDelta _n^4 \int _{-a_n}^{a_n} R_{1,n}(\lambda )^2 \hbox {d}\lambda + o_P(n^{4/\alpha }\varDelta _n^{4/\alpha }),\quad n\rightarrow \infty . \end{aligned}$$

Noting that $o_P(n^{4/\alpha }\varDelta _n^{4/\alpha }) = o_P(E_n^2)$, $n\rightarrow \infty$, and by Step 3a)

$$\begin{aligned} \varDelta _n^4 \int _{-a_n}^{a_n} R_{1,n}(\lambda )^2 \hbox {d}\lambda\le &\, {} 2\left( \int _{-a_n}^{a_n} \left| {{\hat{f}}}(\lambda )\right| ^4 \hbox {d}\lambda \cdot E_n^2 + \int _{-a_n}^{a_n}\left| \left| {{\hat{f}}}(\lambda )\right| ^2 E_n - \varDelta _n^2 R_{1,n}(\lambda )\right| ^2 \hbox {d}\lambda \right) \\= & {} \left( 2\int _{-a_n}^{a_n} \left| {{\hat{f}}}(\lambda )\right| ^4 \hbox {d}\lambda + o_P(1) \right) E_n^2,\quad n\rightarrow \infty , \end{aligned}$$

we get by (A3) $a_n\int _{-a_n}^{a_n}\left| \varDelta _n^2 R_n(\lambda ) - \varDelta _n^2 R_{1,n}(\lambda )\right| ^2\hbox {d}\lambda = o_P(E_n^2),\quad n\rightarrow \infty ,$ whence (25) follows from (24).

Step 3c). Finally, we have

$$\begin{aligned} a_n\varDelta _n^4 \int _{-a_n}^{a_n}\left| J^s_{n,X}(\lambda ) - R_{n}(\lambda )\right| ^2\hbox {d}\lambda = o_P(E_n^2),\quad n\rightarrow \infty . \end{aligned}$$

(26)

Using (21) again, write

$$\begin{aligned} \left| J^s_{n,X}(\lambda ) - R_{n}(\lambda )\right|= &\, {} \left| R_n(\lambda )-\sum _{\left| m\right| \le m_n}W_n(m)\left| \sum _{j=1}^{n} X(t_{j,n})\hbox {e}^{it_{j,n}\nu _{n}(m,\lambda )}\right| ^2\right| \nonumber \\\le &\, {} \delta _n R_n(\lambda ) + (1+\delta _n^{-1}) \sum _{\left| m\right| \le m_n}W_n(m)\nonumber \\&\left| \sum _{j=1}^{n} \left( X_{n}(t_{j,n})-X(t_{j,n})\right) \hbox {e}^{it_{j,n}\nu _{n}(m,\lambda )}\right| ^2\nonumber \\= &\, {} \delta _n R_n(\lambda ) + (1+\delta _n^{-1}) \sum _{\left| m\right| \le m_n}W_n(m)\nonumber \\&\left| \int _{{\mathbb {R}}}\sum _{j=1}^{n}\left( f_n(t_{j,n}-s)- f(t_{j,n}-s)\right) \hbox {e}^{it_{j,n}\nu _{n}(m,\lambda )}\varLambda (ds)\right| ^2 \nonumber \\=: & {} \delta _n R_n(\lambda ) + (1+\delta _n^{-1})R_{6,n}(\lambda ) \end{aligned}$$

(27)

for $\delta _n = a_n^{1/4}\omega _f(\varDelta _n)$. Hence,

$$\begin{aligned} a_n\int _{-a_n}^{a_n}\left| J^s_{n,X}(\lambda ) - R_{n}(\lambda )\right| ^2\hbox {d}\lambda \le 2a_n \delta _n^2 \int _{-a_n}^{a_n}R_n(\lambda )^2\hbox {d}\lambda + 2a_n(1+\delta _n^{-1})^2 \int _{-a_n}^{a_n}R_{6,n}(\lambda )^2\hbox {d}\lambda . \end{aligned}$$

Define $h_{n,m}(s,\lambda ) = \sum _{j=1}^n \left( f_n(t_{j,n}-s)- f(t_{j,n}-s)\right) \hbox {e}^{it_{j,n}\nu _{n}(m,\lambda )} {\mathbb {1}}_{[-a_n,a_n]}(\lambda ).$ Note that the summands do not exceed $\omega _f(\varDelta _n)$, and at most 2N of them are not zero. Hence, $\Vert h_{n,m}(\cdot ,\lambda )\Vert _\infty \le 2N\omega _f(\varDelta _n)$. Applying Lemma 5, we get

$$\begin{aligned} a_n\int _{-a_n}^{a_n} R_{6,n}(\lambda )^2 \hbox {d}\lambda= &\, {} O_P(a_n^2 N^4 \omega _f(\varDelta _n)^4 (n\varDelta _n)^{4/\alpha }) \\= &\, {} O_P(a_n^2\omega _f(\varDelta _n)^4 n^{4/\alpha }\varDelta _n^{4/\alpha -4}),\ n\rightarrow \infty . \end{aligned}$$

Recalling that $\delta _n\rightarrow 0$ and $a_n^{3/2}\omega _f(\varDelta _n)^2\rightarrow 0$ as $n\rightarrow \infty$ and using (25), we ultimately obtain

$$\begin{aligned}&a_n\varDelta _n^4\int _{-a_n}^{a_n} \left| J^s_{n,X}(\lambda ) - R_n(\lambda )\right| ^2 \, \hbox {d}\lambda \\&\quad =O_P \left( a_n^{3/2}(\omega _f(\varDelta _n))^{2} \left( \int _{{\mathbb {R}}} \left| {\hat{f}}(\lambda )\right| ^4 \, \hbox {d}\lambda +o_P(1)\right) E_n^2 \right. \\&\qquad \left. +\, \varDelta _n^4 a_n^{-1/2}(\omega _f(\varDelta _n))^{-2}a_n^{2}\omega _f(\varDelta _n)^4 n^{4/\alpha }\varDelta _n^{4/\alpha -4} \right) \\&\quad =O_p\left( a_n^{3/2}(\omega _f(\varDelta _n))^{2}(E_n^2+n^{4/\alpha } \varDelta _n^{4/\alpha })\right) =o_p(E_n^2),\quad n\rightarrow \infty . \end{aligned}$$

Combining (25) and (26), we come to (23).$\square$

Proof of Theorem 3

Consider first the case $\alpha \in [1,2)$. (i) Using the triangle inequality and the Hölder inequality, we get

$$\begin{aligned} \left| \widetilde{\Vert g\Vert }_{\alpha ,T} - \Vert g\Vert _\alpha \right|\le & {} \left( \int _{-T}^{T} \left| {{\tilde{g}}}(t) - g(t)\right| ^\alpha \hbox {d}t\right) ^{1/\alpha }\\\le & {} \left( (2T)^{2/\alpha -1}\int _{-T}^{T} \left| {{\tilde{g}}}(t) - g(t)\right| ^2 \hbox {d}t\right) ^{1/2} \overset{P}{\longrightarrow }0, \quad n\rightarrow \infty . \end{aligned}$$

(ii) Similarly to (i),

$$\begin{aligned} \left| \widetilde{\Vert g\Vert }_{\alpha ,b_n} - \Vert g\Vert _\alpha \right|\le & {} \left( \int _{-b_n}^{b_n} \left| {{\tilde{g}}}(t) - g(t)\right| ^\alpha \hbox {d}t\right) ^{1/\alpha } + \left( \int _{\{t: |t|>b_n\}} \left| g(t)\right| ^\alpha \hbox {d}t\right) ^{1/\alpha }\\\le & {} \left( (2b_n)^{2/\alpha -1}\int _{-b_n}^{b_n} \left| {{\tilde{g}}}(t) - g(t)\right| ^2 \hbox {d}t\right) ^{1/2}\\&+ \left( \int _{\{t: |t|>b_n\}} \left| g(t)\right| ^\alpha \hbox {d}t\right) ^{1/\alpha } \overset{P}{\longrightarrow }0, \quad n\rightarrow \infty , \end{aligned}$$

in view of (8).

The proof of (iii) uses the same ideas and is based on (9).

For $\alpha \in (0,1)$, the proof goes in a similar manner through the triangle inequality for $\Vert \cdot \Vert _\alpha ^\alpha$. $\square$

Appendix 2: Auxiliary statements

Lemma 1

Let $(E,{\mathcal {E}},\nu )$ be a $\sigma$-finite measure space, $\varLambda$ be an independently scattered S$\alpha$S random measure on E with the control measure $\nu$, and $\{f_t, t\in {\mathbf {T}}\}\subset L^\alpha (E,{\mathcal {E}},\nu )$ be a family of functions indexed by some parameter set ${\mathbf {T}}$, $\varphi$ be a positive probability density on E. Then

$$\begin{aligned} X_t = \int _E f_t(x) \varLambda ({\mathrm{d}}x),\quad t\in {\mathbf {T}}, \end{aligned}$$

has the same finite-dimensional distributions as the almost surely convergent series

$$\begin{aligned} X'_t = C_\alpha ^{1/\alpha } \sum _{k=1}^{\infty } \varGamma _k^{-1/\alpha } \varphi (\xi _k)^{-1/\alpha } f_t(\xi _k) \zeta _k,\quad t\in {\mathbf {T}}, \end{aligned}$$

where $\left\{ \zeta _k,k\ge 1\right\}$ are iid standard Gaussian random variables, $\left\{ \xi _k,k\ge 1\right\}$ are iid random elements of E with density $\varphi$, $\varGamma _k=\eta _1+\dots +\eta _k$, $\left\{ \eta _k,k\ge 1\right\}$ are iid ${\text {Exp}}(1)$-distributed random variables, and these three sequences are independent;

$$\begin{aligned} C_\alpha = \left( {\mathsf {E}}\left[ \,|g_1|^{\alpha }\,\right] \int _0^\infty x^{-\alpha }\sin x\, {\mathrm{d}}x\right) ^{-1} = {\left\{ \begin{array}{ll} \frac{(1-\alpha )\sqrt{\pi }}{2^{\alpha /2}\varGamma ((\alpha +1)/2)\varGamma (2-\alpha )\cos (\pi \alpha /2)}, &{} \alpha \ne 1,\\ \sqrt{2/\pi }, &{} \alpha = 1. \end{array}\right. } \end{aligned}$$

Proof

The statement follows from Samorodnitsky and Taqqu (1994, Section 3.11) by noting that

$$\begin{aligned} X_t = \int _E f_t(x) \varphi (x)^{-1/\alpha }M(\hbox {d}x), \end{aligned}$$

where M is an independently scattered S$\alpha$S random measure on E defined by

$$\begin{aligned} M(A) = \int _{A} \varphi ^{1/\alpha }(x) \varLambda (x),\quad A\in {\mathcal {E}}, \end{aligned}$$

so that the control measure of M has $\nu$-density $\varphi$.$\square$

Lemma 2

Let, for each $n\ge 1$, $\left\{ \varepsilon _{m,n},m=1,\dots ,n\right\}$ be iid $S\alpha S$ random variables with scale parameter $\sigma _n$. Let also $\left\{ a_{j,l,n},\ 1\le j<l\le n\right\}$ be a collection of measurable functions $a_{j,l,n}:{{\mathbb {R}}}\rightarrow {\mathbb {C}}$ such that

$$\begin{aligned} A_n = \int _{{{\mathbb {R}}}}\sum _{1\le j<l\le n}\left| a_{j,l,n}(\lambda )\right| ^2 {\mathrm{d}}\lambda <\infty . \end{aligned}$$

Then

$$\begin{aligned} \int _{{{\mathbb {R}}}}\left| \sum _{1\le j<l\le n} a_{j,l,n}(\lambda )\varepsilon _{j,n}\varepsilon _{l,n}\right| ^2{\mathrm{d}}\lambda = O_P(A_n\sigma _n^4 n^{4/\alpha -2}),\quad n\rightarrow \infty . \end{aligned}$$

Proof

W.l.o.g. we can assume that $\sigma _n = 1$. We shall use the LePage series representation. For each $n\ge 1$, the variables $\left\{ \varepsilon _{m,n},m=1,\dots ,n\right\}$ have the same joint distribution as $\left\{ \varLambda ([m-1,m]),m=1,\dots ,n\right\}$, where $\varLambda$ is an independently scattered S$\alpha$S random measure on [0, n] with the Lebesgue control measure. By Lemma 1, this distribution coincides with that of

$$\begin{aligned} {\tilde{\varepsilon }}_{m,n} = n^{1/\alpha }C_\alpha ^{1/\alpha } \sum _{k=1}^\infty \varGamma _k^{-1/\alpha } {\mathbb {1}}_{[m-1,m]}(\xi _k)\zeta _k,\quad m=1,\dots ,n, \end{aligned}$$

where $\left\{ \varGamma _k,k\ge 1\right\}$ and $\left\{ \zeta _k,k\ge 1\right\}$ are as in Lemma 1, $\left\{ \xi _k,k\ge 1\right\}$ are iid U([0, n]). Since the boundedness in probability relies only on marginal distributions (for fixed n), we can assume that $\varepsilon _{m,n} = {\tilde{\varepsilon }}_{m,n}$. Let $\varXi _n(\lambda ) = \sum _{1\le j<l\le n} a_{j,l,n}(\lambda )\varepsilon _{j,n}\varepsilon _{l,n}$. A generic term in the expansion of $|\varXi _n(\lambda )|^2$ has, up to a non-random constant, the form

$$\begin{aligned} \varGamma _{k_1}^{-1/\alpha }\varGamma _{k_1'}^{-1/\alpha } \varGamma _{k_2}^{-1/\alpha }\varGamma _{k_2'}^{-1/\alpha } {\mathbb {1}}_{[j_1-1,j_1]}(\xi _{k_1}){\mathbb {1}}_{[l_1-1,l_1]}(\xi _{k_1'}) {\mathbb {1}}_{[j_2-1,j_2]}(\xi _{k_2}){\mathbb {1}}_{[l_2-1,l_2]}(\xi _{k_2'}) \zeta _{k_1}\zeta _{k_1'}\zeta _{k_2}\zeta _{k_2'}. \end{aligned}$$

Recall that $\left\{ \zeta _k,k\ge 1\right\}$ are independent and centered. Then, given $\varGamma$’s and $\xi$’s, such term has a non-zero expectation only if $k_1=k_2$, $k_1'=k_2'$ or $k_1=k_2'$, $k_2=k_1'$ (for $k_1=k_1'$ it is zero since $j_1\ne l_1$), so we must also have $j_1=j_2$, $l_1=l_2$ or $j_1 = l_2$, $j_2 =l_1$ respectively so that the product of indicators is not zero. The latter, however, is impossible, since $j_1<l_1$ and $j_2<l_2$. Consequently, the lemma of Fatou implies

$$\begin{aligned}&{\mathsf {E}}\left[ \,|\varXi _n(\lambda )|^2\mid \varGamma \,\right] \\&\quad \le C_\alpha ^{4/\alpha } n^{4/\alpha } \sum _{k\ne k'}^\infty \varGamma _{k}^{-2/\alpha }\varGamma _{k'}^{-2/\alpha }\sum _{1\le j<l\le n}^{n}\left| a_{j,l,n}(\lambda )\right| ^2{\mathsf {E}}\left[ \,{\mathbb {1}}_{[j-1,j]}(\xi _{k}){\mathbb {1}}_{[l-1,l]}(\xi _{k'})\,\right] \\&\quad = C_\alpha ^{4/\alpha }n^{4/\alpha } \sum _{k\ne k'}^\infty \varGamma _{k}^{-2/\alpha }\varGamma _{k'}^{-2/\alpha }\sum _{1\le j<l\le n}^{n}\left| a_{j,l,n}(\lambda )\right| ^2 P\left( \xi _k\in [j-1,j]\right) P\left( \xi _{k'}\in [l-1,l]\right) \\&\quad \le C_\alpha ^{4/\alpha }n^{4/\alpha -2}\sum _{1\le j<l\le n}\left| a_{j,l,n}(\lambda )\right| ^2 \left( \sum _{k=1}^\infty \varGamma _{k}^{-2/\alpha }\right) ^2. \end{aligned}$$

Integrating over $\lambda$, we get

$$\begin{aligned} {\mathsf {E}}\left[ \,\int _{{{\mathbb {R}}}}|\varXi _n(\lambda )|^2\hbox {d}\lambda \,\left| \, \right. \varGamma \,\right] \le C_\alpha ^{4/\alpha }n^{4/\alpha -2}A_n \left( \sum _{k=1}^\infty \varGamma _{k}^{-2/\alpha }\right) ^2. \end{aligned}$$

By the strong law of large numbers, $\varGamma _k \sim k$, $k\rightarrow \infty$, a.s. Therefore, given $\varGamma$’s, $\int _{{{\mathbb {R}}}}|\varXi _n(\lambda )|^2\hbox {d}\lambda = O_P(A_n n^{4/\alpha -2})$, $n\rightarrow \infty$, whence the required statement follows.$\square$

The following lemma is an immediate corollary of the proof of Lemma 2.

Lemma 3

Let, for each $n\ge 1$, $\left\{ \varepsilon _{m,n},m=1,\dots ,n\right\}$ be iid $S\alpha S$ random variables with scale parameter $\sigma _n$. Let also $\left\{ b_{j,l,n},\ 1\le j<l\le n\right\}$ be a set of complex numbers with

$$\begin{aligned} B_n = \sum _{1\le j<l\le n}\left| b_{j,l,n}\right| ^2. \end{aligned}$$

Then

$$\begin{aligned} \sum _{1\le j<l\le n} b_{j,l,n}\varepsilon _{j,n}\varepsilon _{l,n} = O_P(B_n^{1/2}\sigma _n^2 n^{2/\alpha -1}),\quad n\rightarrow \infty . \end{aligned}$$

In the next two lemmas, $\left\{ \varDelta _n,n\ge 1\right\}$ is some vanishing sequence, $\left\{ N_n,n\ge 1\right\}$ is a sequence of positive integers such that $N_n\rightarrow \infty$, $n\rightarrow \infty$, and $N_n = o(n)$, $n\rightarrow \infty$. We denote $t_{k,n} = k\varDelta _n$, $k\in {{\mathbb {Z}}}$, $T_n = N_n\varDelta _n$, $n\ge 1$. The proofs of these lemmas are similar to the proof of Lemma 2 and thus omitted. They can be found in the arXiv version of the present paper Kampf et al. (2019).

Lemma 4

Let $\left\{ h_n,n\ge 1\right\}$ be a sequence of compactly supported bounded functions such that the bounds of both the function values and the support are uniform in n. Then

$$\begin{aligned} \sum _{j=1}^{n} Y_{t_{j,n},n}^2 = O_P\left( \Vert h_n\Vert ^2_\infty n^{2/\alpha }\varDelta _n^{2/\alpha -1}\right) ,\quad n\rightarrow \infty . \end{aligned}$$

Lemma 5

Let $\left\{ m_n,n\ge 1\right\}$ be a sequence of positive integers such that $m_n\rightarrow \infty$ as $n\rightarrow \infty$. For a deterministic sequence $\left\{ W_n(m),n\ge 1, m=-m_n,\dots ,m_n\right\}$ satisfying (W1)–(W2) and continuous functions $h_{n,m}:[-T_n,n\varDelta _n+T_n]\times {{\mathbb {R}}}\rightarrow {\mathbb {C}}$, $n\ge 1, m= -m_n,\dots ,m_n$, define

$$\begin{aligned} R_{n}(\lambda ) = \sum _{\left| m\right| \le m_n}W_n(m) \left| \int _{-T_n}^{n\varDelta _n+T_n} h_{n,m}(t,\lambda )\varLambda ({\mathrm{d}}t)\right| ^2. \end{aligned}$$

Then

$$\begin{aligned} \int _{{{\mathbb {R}}}} R_{n}(\lambda )^2 {\mathrm{d}}\lambda = O_P\left( H_n^* (n\varDelta _n)^{4/\alpha }\right) ,\quad n\rightarrow \infty , \end{aligned}$$

where $H_n^*=\int _{\mathbb {R}} H(\lambda )\, {\mathrm{d}}\lambda$ for $H(\lambda ) = \sum _{\left| m\right| \le m_n} W_n(m)\Vert h_{n,m}(\cdot ,\lambda )\Vert _\infty ^4.$

Lemma 6

Let a bounded uniformly continuous function $f:{{\mathbb {R}}}\rightarrow {\mathbb {R}}$ with compact support $[-T, T]$ and let $\varDelta _n$, $m_n$, $W_n(m)$ and $\nu _n(m,\lambda )$ be as defined in Sect. 1or 3fulfilling (W1), (W2) and (W4). Choose a sequence of integers $(N_n)_{n\in {\mathbb {N}}}$ with $N_n\cdot \varDelta _n\sim T$. Put

$$\begin{aligned} F_n(\lambda ) = \sum _{\left| m\right| \le m_n} W_n(m) \left| \sum _{k=-N_n}^{N_n-1} f(t_{k,n}){\mathrm{e}}^{it_{k,n}\nu _{n}(m,\lambda )}\right| ^2. \end{aligned}$$

Then

$$\begin{aligned} \left| \left| {{\hat{f}}}(\lambda )\right| ^2 - \varDelta _n^2 F_{n}(\lambda )\right| = O\left( \left( W_n^{(2)}\right) ^{1/2}(n\varDelta _n)^{-1} + \omega _f(\varDelta _n) + \left| \lambda \right| \varDelta _n \right) ,\quad n\rightarrow \infty . \end{aligned}$$

We also omit the proof of this and the following lemma and refer the interested reader to Kampf et al. (2019).

Lemma 7

Let $\left\{ m_n,n\ge 1\right\}$ be a sequence of positive integers such that $m_n\rightarrow \infty$, $m_n = o(n)$, $n\rightarrow \infty$, and let $\left\{ K_n(m),n\ge 1, m=-m_n,\dots ,m_n\right\}$ be a sequence in ${\mathbb {R}}$, and let $\left\{ W_n(m),n\ge 1, m=-m_n,\dots ,m_n\right\}$ be a sequence of filters satisfying (W1)–(W2). Then

$$\begin{aligned} S_n = \sum _{j_1,j_2=1}^{n}\left| \sum _{\left| m\right| \le m_n} W_n(m)K_n(m) {\mathrm{e}}^{i (j_1-j_2)m/n} \right| ^2= O(W_n^* (K_n^* n)^2),\ n\rightarrow \infty \end{aligned}$$

with $W_n^* = \max _{\left| m\right| \le m_n} W_n(m)$, $K_n^* = \max _{\left| m\right| \le m_n} |K_n(m)|$.