Asymptotic theory for regression models with fractional local to unity root errors

Abstract

This paper develops the asymptotic theory for parametric and nonparametric regression models when the errors have a fractional local to unity root (FLUR) model structure. FLUR models are stationary time series with semi-long range dependence property in the sense that their covariance function resembles that of a long memory model for moderate lags but eventually diminishes exponentially fast according to the presence of a decay factor governed by a an exponential tempering parameter. When this parameter is sample size dependent, the asymptotic theory for these regression models admit a wide range of stochastic processes with behavior that includes long, semi-long, and short memory processes.

Proofs

Before we prove the main results of the paper, we first state two technical lemmas upon which our results are based. Lemma 1 and Lemma 2, play an important role in establishing the asymptotic results in Sect. 4. Next, we introduce some notations that will be used in Lemma 2.

For the function f and $m\in \mathbb {N}\cup \{\infty \}$, we define the approximation

$$\begin{aligned}&f^{+}_{N,m}(y)=\sum _{j=0}^{m} f\Big (\frac{j}{N}\Big )1_{[\frac{j}{N},\frac{j+1}{N}]}(y), \qquad f^{-}_{N,m}=\sum _{j=-m}^{-1} f\Big (\frac{j}{N}\Big )1_{[\frac{j}{N},\frac{j+1}{N}]}(y), \\&f^{+}_{N}=f^{+}_{N,\infty }, \qquad f^{-}_{N}=f^{+}_{N,\infty },\qquad f_{N}=f^{+}_{N}+f^{-}_{N}. \end{aligned}$$

Lemma 1

Let $b_d(k)$ be defined as in (4). Then for any $y \in \mathbb {R}$

$$\begin{aligned} \biggl (\lambda _N + i\frac{y}{N}\biggr )^{d} \sum _{k=0}^{\infty } e^{- (\lambda _N + i\frac{y}{N})k}\, b_{d}(k) \sim 1 \end{aligned}$$

(43)

as $N\rightarrow \infty$.

proof Lemma 1

For $d>0$, since $\sum _{k=0}^Nb_d(k) \sim 1/(d\varGamma (d)) N^d$ as $N\rightarrow \infty$ according to (4) and $e^{-(\lambda _N+\frac{ix}{N})} \le 1$, then according to the Tauberian theorem for power series (Feller 1971, p. 447 Theorem 5) we have

$$\begin{aligned} \sum _{k=0}^{\infty } e^{- (\lambda _N + i\frac{y}{N})k}\, b_{d}(k) \sim (1-e^{- (\lambda _N + i\frac{y}{N})k})^{-d} \text { as } N\rightarrow \infty \end{aligned}$$

and consequently $\bigl (\lambda _N + i\frac{y}{N}\bigr )^{d}/(1-e^{- (\lambda _N + i\frac{y}{N})k})^{d}\sim 1$ for $\lambda _N \rightarrow 0$ and $N \rightarrow \infty$ as $N\rightarrow \infty$, proving (43). For $-1<d<0$, define $\widetilde{b}_d(k) = \sum _{i=k}^\infty b_d(i) \sim -1/(d\varGamma (d))k^{\widetilde{d}-1}$ with $\widetilde{d}=d+1 \in (0,1)$. Next, we have that

$$\begin{aligned} \widetilde{b}_d(0) = \sum _{i=0}^\infty b_d(i) = \sum _{i=0}^{k-1} b_d(i) + \sum _{i=k}^\infty b_d(i) = 0 \end{aligned}$$

(44)

and therefore $\sum _{i=0}^{k-1} b_d(i) = -\sum _{i=k}^\infty b_d(i)$. Using summation by parts (Giraitis et al., 2012, p. 32, Eq. 2.5.8) yields

$$\begin{aligned} \lim _{s\rightarrow \infty } \sum _{j=0}^s e^{- (\lambda _N + i\frac{y}{N})j}\, b_{d}(j)= & {} \lim _{s\rightarrow \infty } \bigl [e^{-(\lambda _N +i\frac{y}{N})s} \sum _{j=0}^s b_d(j)\\&+ \sum _{j=0}^{s-1}e^{-(\lambda _N +i\frac{y}{N})s}-e^{-(\lambda _N +i\frac{y}{N})(s+1)} \sum _{i=0}^j b_d(i)\bigr ] \\= & {} 0 + \{1- e^{- (\lambda _N + i\frac{y}{N})}\}\lim _{s\rightarrow \infty } \sum _{j=0}^{s-1}e^{-(\lambda _N +i\frac{y}{N})j} \sum _{i=0}^j b_d(i). \end{aligned}$$

By setting $j=k-1$ and using (44) we have

$$\begin{aligned}&\lim _{s\rightarrow \infty } \sum _{j=0}^s e^{- (\lambda _N + i\frac{y}{N})j}\, b_{d}(j) = \{1- e^{- (\lambda _N + i\frac{y}{N})}\}\lim _{s\rightarrow \infty } \sum _{k=1}^{s-1}e^{-(\lambda _N +i\frac{y}{N})(k-1)} \sum _{i=0}^{k-1} b_d(i) \\&\quad = -e^{(\lambda _N + i\frac{y}{N})} \{1- e^{- (\lambda _N + i\frac{y}{N})}\}\lim _{s\rightarrow \infty }\sum _{k=1}^{s-1}e^{-(\lambda _N +i\frac{y}{N})k}\sum _{i=k}^\infty b_d(i) \\&\quad = -e^{(\lambda _N + i\frac{y}{N})} \{1- e^{- (\lambda _N + i\frac{y}{N})}\}\lim _{s\rightarrow \infty }\sum _{k=1}^{s-1}e^{-(\lambda _N +i\frac{y}{N})k}\widetilde{b}_d(k). \end{aligned}$$

Application of the Tauberian theorem for power series (Feller, 1971, p. 447 Theorem 5) yields

$$\begin{aligned} \sum _{k=0}^{\infty } e^{- (\lambda _N + i\frac{y}{N})k}\, b_{d}(k) \sim \{1- e^{- (\lambda _N + i\frac{y}{N})}\}^{1-\widetilde{d}} = \{1- e^{- (\lambda _N + i\frac{y}{N})}\}^{-d} \end{aligned}$$

as $N\rightarrow \infty$, proving (43). In the general case $-j< d < -j +1$, $j = 1,2,\ldots$ (43) follows similarly using summation by parts j times. For $d=0$, it can be shown that the same result holds under an additional assumption on the sum of the $b_d(k)$’s Sabzikar and Surgailis (2018b).

Lemma 2

Let $\{ X_{d,\lambda _N}(j)\}_{j\in \mathbb {Z}}$ be tempered linear process given by (3), $N\lambda _N\rightarrow \lambda _*\in (0,\infty )$ and $d>-1/2$. Let ${\mathcal {A}}_{d,\lambda _*}$ be the class of functions defined by (26) and let

$$\begin{aligned}&{\mathbf{Condition\ A}:}\ f,f^{\pm }_{N}\in { {\mathcal {A}}_{d,\lambda _N}}, \Vert f^{\pm }_{N} - f^{\pm }_{N,m}\Vert _{ {\mathcal {A}}_{d,\lambda _N} }\rightarrow 0,\ \mathrm{as}\ m\\&\quad \rightarrow \infty ,\ \Vert f-f_N\Vert _{ {\mathcal {A}}_{d,\lambda _*} }\rightarrow 0,\ \mathrm{as}\ N\rightarrow \infty \end{aligned}$$

be satisfied, then

$$\begin{aligned} \frac{1}{N^{d+1/2}}\sum _{j=-\infty }^{\infty }f\Big (\frac{j}{N}\Big )X_{d,\lambda _N}(j) {\mathop {\longrightarrow }\limits ^{\mathrm{f.d.d.}}}\int _{\mathbb {R}}f(u)\ dB^{II}_{d,\lambda _*}(u) \end{aligned}$$

(45)

as $N\rightarrow \infty$.

proof Lemma 2

We first note that $\{X_{d,\lambda _N}(j)\}_{j\in \mathbb {Z}}$ can be written as

$$\begin{aligned} X_{d,\lambda _N}(j) = \frac{1}{\sqrt{2\pi }} \int _{-\pi }^{\pi }e^{i\omega j}\sum _{k=0}^{\infty } e^{-i\omega k} e^{-\lambda _N k} b_{d}(k) \hat{B}(d\omega ), \end{aligned}$$

(46)

where $\hat{B}(d \omega )$ is complex-valued Gaussian noise with $\mathbb {E}|\hat{B}(d \omega )|^2 = d \omega$, see Brockwell and Davis (2012, Sects. 4.6–4.7). Define

$$\begin{aligned} U_{N}=\frac{1}{N^{ d+ \frac{1}{2}}} \sum _{j= -\infty }^{\infty } f\Big (\frac{j}{N}\Big ) X_{d,\lambda _N}(j), \qquad U=\int _{\mathbb {R}}\ f(u)\ dB^{II}_{d,\lambda _*}(u). \end{aligned}$$

(47)

The Wiener integral U is well-defined, since $f\in {\mathcal {A}}_{d,\lambda }$. To show that the series $U_{N}$ is well-defined in the $L^{2}(\varOmega )$, first apply the spectral representation of $\{X_{d,\lambda _N}(j)\}_{j\in \mathbb {Z}}$ given by (46)

$$\begin{aligned} \begin{aligned}&\frac{1}{ {N}^{d+ \frac{1}{2}} } \sum _{j=0}^{m} f\Big (\frac{j}{N}\Big ) X_{d,\lambda _N}(j) = \frac{1}{ {N}^{d+ \frac{1}{2}} } \int _{-\pi }^{\pi } \Bigg [\sum _{j=0}^{m} \frac{1}{\sqrt{2\pi }} f\Big (\frac{j}{N}\Big ) e^{ij\omega }\Bigg ]\\&\qquad \sum _{k=0}^{\infty } e^{ -(\lambda _N + i\omega )k } b_{d}(k)\ d\widehat{B}(\omega )\\&\quad =\frac{1}{ {N}^{d+ \frac{1}{2}} } \int _{\mathbb {R}}\Bigg [\sum _{j=0}^{m} \frac{1}{\sqrt{2\pi }} f\Big (\frac{j}{N}\Big ) e^{\frac{i j y}{N}} \Bigg ]\mathbf{1}_{[-N\pi ,N\pi ]}(y)\\&\qquad \times \sum _{k=0}^{\infty } e^{ -(\lambda _N + i\frac{y}{N} )k } b_{d}(k) d\hat{B}\big (N^{-1}y\big )\\&\quad =\frac{1}{N^{d-\frac{1}{2}}}\int _{\mathbb {R}} \Bigg [ \sum _{j=0}^{m} \frac{1}{\sqrt{2\pi }} f\Big (\frac{j}{N}\Big ) \frac{e^{\frac{i(j+1)y}{N}}-e^{\frac{ijy}{N}}}{iy} \Bigg ]\\&\qquad \times \frac{\frac{iy}{N}}{e^{\frac{iy}{N}}-1} 1_{[-N\pi ,N\pi ]}(y) \sum _{k=0}^{\infty } e^{ -(\lambda _N + i\frac{y}{N} )k } b_{d}(k) d\hat{B}\big (N^{-1}y\big )\\&\quad =\frac{1}{N^{d-\frac{1}{2}}}\int _{\mathbb {R}} \widehat{f_{N,m}}(y) \frac{\frac{iy}{N}}{e^{\frac{iy}{N}}-1} \sum _{k=0}^{\infty } e^{ -(\lambda _N + i\frac{y}{N} )k } b_{d}(k) d\hat{B}\big (N^{-1}y\big ), \end{aligned} \end{aligned}$$

(48)

where $\widehat{f_{N,m}}(y)=\sum _{j=0}^{m} f\Big (\frac{j}{N}\Big ) \frac{1}{\sqrt{2\pi }} \int _{\mathbb {R}}e^{i\omega y}\mathbf {1}_{(\frac{j}{N},\frac{j+1}{N})}(\omega )\ d\omega$ is the Fourier transform of ${f_{N,m}}$. We note

$$\begin{aligned} \sum _{k=0}^{\infty } e^{- (\lambda _N + \frac{iy}{N})k} b_{d}(k) < C \Big (\lambda _N + \frac{iy}{N}\Big )^{-d}, \end{aligned}$$

(49)

for $d>-\frac{1}{2}$ and a constant C by Lemma 1. Using (48) and (49), we have

$$\begin{aligned}&\mathbb {E}\Bigg | \frac{1}{ {N}^{d+ \frac{1}{2}} }\sum _{j=0}^{m}f\Big (\frac{j}{N}\Big ) {X}_{d,\lambda _N}(j)\Bigg |^{2}\nonumber \\&\quad =\int _{\mathbb {R}}\Big | \widehat{f_{N,m}}(y)\Big |^{2} \Bigg | \frac{\frac{iy}{N}}{e^{\frac{iy}{N}}-1} \Bigg |^{2} \frac{1}{ {N}^{2d} }\ \!\!\Big | \sum _{k=0}^{\infty } e^{- (\lambda _N + \frac{iy}{N})k} b_{d}(k) \Big |^{2}\ dy \nonumber \\&\quad \le \frac{\pi ^2}{4}\, C \int _{\mathbb {R}}\Big | \widehat{f_{N,m}}(y)\Big |^{2} \Big [ (N\lambda _N)^2 + y^2 \Big ]^{-d}\ dy \nonumber \\&\quad = C' \Vert f_{N,m}\Vert ^{2}_{{\mathcal {A}}_{3,N\lambda _N}}, \end{aligned}$$

(50)

where $C'$ is another constant. Now, for $m_2>m_1\ge 1$, we have

$$\begin{aligned} \mathbb {E}\Bigg |\frac{1}{ {N}^{d+ \frac{1}{2}} }\sum _{j=m_1+1}^{m_2}f\Big (\frac{j}{N}\Big ) X_{d,\lambda _N}(j)\Bigg |^{2}\le C'\Vert {f}^+_{N,m_2}-{f}^+_{N,m_1}\Vert ^{2}_{{\mathcal {A}}_{3,\lambda _N}}\rightarrow 0 \end{aligned}$$

as $m_1,m_2\rightarrow \infty$ and this shows the series is well-defined. The following remark illustrates the inequality in (50).

Remark 5

In (50) we used Lemma 1 and $\Big |\frac{\frac{iy}{N}}{e^{\frac{iy}{N}}-1} \Big |^{2} \le \frac{\pi ^2}{4}$ for $y \in [-N\pi ,N\pi ]$. This can be seen as follows

$$\begin{aligned} \Bigg |\frac{\frac{iy}{N}}{e^{\frac{iy}{N}}-1} \Bigg |^{2} = \frac{\frac{y^2}{N^2}}{|\cos \frac{y}{N}+i\sin \frac{y}{N}-1|^2} = \frac{\frac{y^2}{N^2}}{2(1-\cos \frac{y}{N})} = \frac{1}{4} \frac{\frac{y^2}{N^2}}{\sin ^2 \frac{y}{2N}}. \end{aligned}$$

Then taking the limit yields

$$\begin{aligned} \lim _{N\rightarrow \infty } \frac{1}{4} \frac{\frac{N^2\pi ^2}{N^2}}{\sin ^2 \frac{\pm N\pi }{2N}} = \frac{1}{4} \lim _{N\rightarrow \infty } \frac{\pi ^2}{\sin ^2 \pm \frac{\pi }{2}}= \frac{\pi ^2}{4}. \end{aligned}$$

Next, we show that $U_{N}$ converges in distribution to U as $N\rightarrow \infty$. Similar to Meerschaert and Sabzikar (2014, Theorem 3.15), the set of elementary functions are dense in ${\mathcal {A}}_{d,\lambda }$ and then there exists a sequence of elementary functions $f^{l}$ such that $\Vert f-f^{l}\Vert _{{\mathcal {A}}_{3}}\rightarrow 0$, as $l\rightarrow \infty$. Now, assume

$$\begin{aligned} U^{l}_{N}=\frac{1}{N^{ d+ \frac{1}{2}}}\sum _{j=-\infty }^{\infty } f^{l}\Big (\frac{j}{N}\Big ) X_{d,\lambda _N}(j),\quad \ U^{l}=\varGamma ^{-1}(d+1)\int _{\mathbb {R}}f^{l}(u)\ dB^{II}_{d,\lambda _*}(u). \end{aligned}$$

(51)

Observe that $U^{l}_{N}$ is well-defined, since $U^{l}_{N}$ has a finite number of terms and the elementary function $f^{l}$ is in ${\mathcal {A}}_{3}$. According to Meerschaert and Sabzikar (1968, Theorem 4.2.), the series $U_{N}$ converges in distribution to U if

Step 1:

$U^{l}{\mathop {\longrightarrow }\limits ^{d}}U$, as $l\rightarrow \infty$,

Step 2:

for all $l\in \mathbb {N}$, $U^{l}_{N}{\mathop {\longrightarrow }\limits ^{d}}U^{l}$, as $N\rightarrow \infty$,

Step 3:

$\limsup _{l\rightarrow \infty }\limsup _{N\rightarrow \infty }\mathbb {E}\Big |U^{l}_{N}-U_{N}\Big |^{2}=0$.

Step 1: The random variables $U^{l}$ and U have normal distribution with mean zero and variances $\Vert f^{l}\Vert _{{\mathcal {A}}_{3,\lambda _N}}$ and $\Vert f\Vert _{{\mathcal {A}}_{3,\lambda _N}}$, respectively, since f and $f^{l}$ are in ${{\mathcal {A}}_{3,\lambda _N}}$. Therefore $\mathbb {E}\Big |U^{l}-U\Big |^{2}=\Vert f^{l}-f\Vert _{{\mathcal {A}}_{3,\lambda _N}}\rightarrow 0$ as $l\rightarrow \infty$.

Step 2: Note that $f^{l}$ is an elementary function and hence $U^{l}_{N}$, given by (51), can be written as $U^{l}_{N}=\frac{1}{ N^{d+\frac{1}{2} } }\int _{\mathbb {R}} f^{l}(u) dS_{d,\lambda _N}(u)$. Now, apply part (iii) of Theorem 4.3 in Sabzikar and Surgailis (2018b) to see that $\frac{S_{d,\lambda _N}(u)}{ N^{d+\frac{1}{2} } }{\mathop {\longrightarrow }\limits ^{\mathrm{f.d.d.}}}\varGamma ^{-1}(d+1) B^{II}_{d,\lambda _*}(u)$, as $N\rightarrow \infty$, and this implies that $U^{l}_{N}{\mathop {\longrightarrow }\limits ^{\mathrm{f.d.d.}}}U^{l}$, as $N\rightarrow \infty$.

Step 3: By a similar arguments of (49) and (50), we have

$$\begin{aligned} \begin{aligned} \mathbb {E}\Big |U^{l}_{N}-U_{N}\Big |^{2}&=\int _{\mathbb {R}} \Big | \widehat{f^{l}_{N}}(y)-\widehat{f_N}(y)\Big |^{2} \Big |\frac{\frac{iy}{N}}{e^{\frac{iy}{N}}-1} \Big |^{2} \frac{1}{N^{2d}}\ \Big |1-e^{-(\lambda _N+\frac{iy}{N})}\Big |^{-2d}\ dy\\&\le C \int _{\mathbb {R}} \Big | \widehat{f^{l}_{N}}(y)-\widehat{f_N}(y)\Big |^{2} \Big [ (N \lambda _N)^{2} + y^2 \Big ]^{-d}\ dy, \end{aligned} \end{aligned}$$

(52)

where $\widehat{f^{l}_{N}}(y)$ and $\widehat{f_N}(y)$ are the Fourier transforms of

$$\begin{aligned} f^{l}_{N}(u):=\sum _{j=0}^{\infty }f^{l}\Big (\frac{j}{N}\Big )\mathbf{1}_{ \Big (\frac{j}{N},\frac{Nx-j+1}{N}\Big )}(u) \end{aligned}$$

and $f_{N}:=\sum _{j=0}^{\infty }f\Big (\frac{j}{N}\Big ) \mathbf{1}_{ \big (\frac{j}{N},\frac{Nx-j+1}{N}\big ) }(u)$ respectively. Note that $f^{l}$ is an elementary function and therefore $\widehat{f^{l}_{N}}$ converges to $\widehat{f^{l}}$ at every point and $\Big |\widehat{f^{l}_{N}}(\omega )-\widehat{f^{l}}(\omega )\Big |\le \widehat{g^{l}}(\omega )$ uniformly in N, for some function $\widehat{g^{l}}(\omega )$ which is bounded by the minimum of $C_1$ and $C_2|\omega |^{-1}$ for all $\omega \in \mathbb {R}$ (see Theorem 3.2. in Pipiras and Taqqu (2000) for more details). Let $\mu _{d,\lambda }(d\omega )= (\lambda ^2+\omega ^2)^{-d}\ d\omega$ be the measure on the real line for $d >-\frac{1}{2}$, then $\widehat{g^{l}}(\omega )\in L^{2}(\mathbb {R},\mu _{d,\lambda })$. Now apply the Dominated Convergence Theorem to see that

$$\begin{aligned} \Vert f^{l}_{N}-f^{l}\Vert ^{2}_{{\mathcal {A}}_{3}}=\Vert \widehat{f^{l}_{N}}-\widehat{f^{l}}\Vert ^{2}_{L^{2}(\mathbb {R},\mu _{d,\lambda _*})}\rightarrow 0, \end{aligned}$$

(53)

as $N\rightarrow \infty$. From (48) and (53), we have

$$\begin{aligned} \begin{aligned} \mathbb {E}\Big |U^{l}_{N}-U_{N}\Big |^{2}&\le C\Vert \widehat{ f^l_N } - \widehat{f_N} \Vert ^{2}_{{\mathcal {A}}_{3}}\\&\le C\Big [\Vert \widehat{ f^l_N }- \widehat{f^l }\Vert ^{2}_{{\mathcal {A}}_{3}}+\Vert \widehat{f} - \widehat{f_N }\Vert ^{2}_{{\mathcal {A}}_{3}}+\Vert \widehat{f^l }-\widehat{f}\Vert ^{2}_{{\mathcal {A}}_{3}}\Big ]. \end{aligned} \end{aligned}$$

The first two terms tend to zero as $N\rightarrow \infty$ because of (53) and Condition A respectively, and the last term tends to zero as $l\rightarrow 0$ (see step 1) and this completes the proof of Step 3. $\square$

proof Theorem 1

We prove only part (c) and omit the proofs of parts (a) and (b) due to the similarity of proofs. We first show that

$$\begin{aligned} \frac{1}{(Nh)^{d+\frac{1}{2}}}\sum _{j=1}^{N} K\Big (\frac{Nx-j}{Nh}\Big ) X_{d,\lambda _N}(j){\mathop {\longrightarrow }\limits ^{\mathrm{f.d.d.}}}\frac{1}{\varGamma (d+1)}\int _{0}^{2}K^{'}(1-t) B^{II}_{d,\lambda _*}(t) dt, \end{aligned}$$

(54)

where $B^{II}_{d,\lambda _*}(t)$ is TFBMII. Starting from the l.h.s of (54), using Riemann sums to integrals and integration by parts

$$\begin{aligned} \begin{aligned}&\frac{1}{(Nh)^{d+\frac{1}{2}}} \sum _{j=1}^{N} K\Big (\frac{Nx-j}{Nh}\Big ) X_{d,\lambda _N}(j) = \frac{1}{(Nh)^{d+\frac{1}{2}}} \int _{0}^{1} K\Big (\frac{x-y}{h}\Big ) dS_{d,\lambda _N}(y)\\&\quad = \frac{1}{h(Nh)^{d+\frac{1}{2}}} \int _{0}^{1} K^{'}\Big (\frac{x-y}{h}\Big ) S_{d,\lambda _N}(y) dy\\&\quad = \frac{1}{(Nh)^{d+\frac{1}{2}}} \int _{-1}^{1} K^{'}(u) S_{d,\lambda _N}(x-hu) du\\&\quad = \frac{1}{(Nh)^{d+\frac{1}{2}}} \int _{-1}^{1} K^{'}(u) \sum _{j=\lfloor N(x-h)\rfloor }^{\lfloor N(x-hu)\rfloor } X_{d,\lambda _N}(j) du, \end{aligned} \end{aligned}$$

(55)

where we used

$$\begin{aligned} S_{d,\lambda _N}(x-hu) =\sum _{j=1}^{\lfloor N(x-h)\rfloor -1}X_{d,\lambda _N}(j)+ \sum _{j=\lfloor N(x-h)\rfloor }^{\lfloor N(x-hu)\rfloor } X_{d,\lambda _N}(j) \end{aligned}$$

and the assumptions on the kernel function K to see that

$$\begin{aligned} \frac{\sum _{j=1}^{\lfloor N(x-h)\rfloor -1}X_{d,\lambda _N}(j)}{(Nh)^{d+\frac{1}{2}}} \int _{-1}^{1} K^{'}(u) du=0. \end{aligned}$$

Next, by stationarity of $X_{d,\lambda _N}(j)$ and a change of variable we have

$$\begin{aligned} \begin{aligned} \sum _{j=\lfloor N(x-h)\rfloor }^{\lfloor N(x-hu)\rfloor } X_{d,\lambda _N}(j)&= \sum _{j=1}^{ \lfloor N(x-hu)\rfloor - \lfloor N(x-h)\rfloor +1} X_{d,\lambda _N}(j+ \lfloor N(x-h)\rfloor -1)\\&{\mathop {=}\limits ^{f.d.d.}}\sum _{j=1}^{ \lfloor N(x-hu)\rfloor - \lfloor N(x-h)\rfloor +1} X_{d,\lambda _N}(j)= \sum _{j=1}^{ l_{x}(u) } X_{d,\lambda _N}(j), \end{aligned} \end{aligned}$$

(56)

where $l_{x}(u)= \lfloor N(x-hu)\rfloor - \lfloor N(x-h)\rfloor +1$. Consequently, from (55) and (56), we get

$$\begin{aligned}&\frac{1}{(Nh)^{d+\frac{1}{2}}} \int _{-1}^{1} K^{'}(u)\sum _{j=\lfloor N(x-h)\rfloor }^{\lfloor N(x-hu)\rfloor }X_{d,\lambda _N}(j)\ du\nonumber \\&\quad {\mathop {=}\limits ^{f.d.d.}}\frac{1}{(Nh)^{d+\frac{1}{2}}} \int _{-1}^{1}K^{'}(u)\sum _{j=1}^{ l_{x}(u) } X_{d,\lambda _N}(j)\ du \nonumber \\&\quad =\frac{1}{(Nh)^{d+\frac{1}{2}}} \int _{0}^{2}K^{'}(1-t)\sum _{j=1}^{ \lfloor Nht\rfloor } X_{d,\lambda _N}(j) dt + o_{p}(1) \end{aligned}$$

(57)

According to Sabzikar and Surgailis (2018b, Theorem 4.3) we have

$$\begin{aligned} \frac{1}{(Nh)^{d+\frac{1}{2}}} \sum _{j=1}^{ \lfloor Nh t\rfloor } X_{d,\lambda _N}(j){\mathop {\longrightarrow }\limits ^{\mathrm{f.d.d.}}}\frac{1}{\varGamma (d+1)}B^{II}_{d,\lambda _*}(t) \end{aligned}$$

(58)

in D[0, 2] provided $Nh\rightarrow \infty$. Now, the desired result (54) follows from (57), (58), and the continuous mapping theorem. $\square$

proof of Theorem 2

The proof of this theorem follows by Proposition 1 and Theorem 1 and hence we omit the details. $\square$

proof of Theorem 3

For brevity, we restrict the proof of the theorem to $k=2$. Moreover, we just prove part (c) of the theorem since the proofs of the other two cases are similar. We first note that for each $0<x_i<1$,

$$\begin{aligned} \widehat{A}_{N,i}= (Nh)^{\frac{1}{2}-d}[\hat{m}(x_i)-\mathbb {E}\hat{m}(x_i)]=\frac{1}{(Nh)^{d+\frac{1}{2}}}\sum _{1\le s \le N}K\Big (\frac{Nx_i -s}{Nh}\Big )X_{d,\lambda _N}(s). \end{aligned}$$

(59)

Let $j_i$ be integers such that $|Nx_i-j_i|\le 1$ for $i=1,2$ and (similar to Deo (1997)) define

$$\begin{aligned} \widetilde{A}_{Ni}= \frac{1}{(Nh)^{d+\frac{1}{2}}}\sum _{s=|j_i|-\lfloor Nh \rfloor }^{|j_i|+\lfloor Nh \rfloor } K\Big (\frac{j_i -s}{Nh}\Big )X_{d,\lambda _N}(s) \end{aligned}$$

(60)

for $i=1, 2$. Since the kernel function K vanishes in $\mathbb {R}\setminus [-1,1]$ and $|K^{'}(x)|\le C$ for all $x\in [-1,1]$, it follows that

$$\begin{aligned} \widehat{A}_{N,i}-\widetilde{A}_{N,i}= o_{p}(1) \end{aligned}$$

(61)

for $i=1,2$. By a change of variable $s=\nu +j_1 -Nh$ and $s=\nu +j_2 -Nh-\lfloor N\delta \rfloor$, with $\delta = x_2-x_1$, for $\widetilde{A}_{N,1}$ and $\widetilde{A}_{N,2}$ respectively, use the fact that $X_{d,\lambda _N}$ is stationary, and $|K^{'}(x)|\le C$ to see that

$$\begin{aligned} \Big ( \widetilde{A}_{N,1}, \widetilde{A}_{N,2} \Big ){\mathop {=}\limits ^{f.d.d.}}\Big ( A^{*}_{N,1}, A^{*}_{N,2} \Big )+ o_{p}(1), \end{aligned}$$

(62)

where

$$\begin{aligned} A^{*}_{N,1} = \sum _{\nu =1}^{2\lfloor Nh\rfloor } K\Big (\frac{\nu }{Nh}-1\Big ) X_{d,\lambda _N}(\nu ) \end{aligned}$$

(63)

and

$$\begin{aligned} A^{*}_{N,2} = \sum _{\nu =\lfloor N\delta \rfloor }^{ \lfloor N\delta \rfloor + 2\lfloor Nh\rfloor } K\Big (\frac{\nu - \lfloor N\delta \rfloor }{Nh}-1\Big ) X_{d,\lambda _N}(\nu ). \end{aligned}$$

(64)

We use the partial sums $A^{*}_{N,i}$, for $i=1,2$, to establish the functional limit theorems. Let $\{ K_m\}$ be a sequence of elementary functions such that $K_m\rightarrow K$ in $L^{2}$ as $m\rightarrow \infty$. Define $A^{*}_{m,Ni}$ be as (63) and (64) with $K_m(x)=\sum _{i=1}^{m} a_i \mathbf{1}( t_{i-1},t_{i} )(x)$, where $a_i$ are some constants and $-1\le t_i \le 1$ for $i=0, \ldots , m$. We can rewrite $A^{*}_{m,N,i}$ as

$$\begin{aligned} A^{*}_{m,N,i} = 2^{d+\frac{1}{2}} \int _{0}^{2} K_{m}(u-1) dS^{*}_{Ni}(u) + o_{p}(1), \end{aligned}$$

(65)

where

$$\begin{aligned} S^{*}_{N,1}(s) = \frac{1}{(2Nh)^{d+\frac{1}{2}}}\sum _{t=1}^{\lfloor \lfloor Nh\rfloor s\rfloor } X_{d,\lambda _N}(t) \end{aligned}$$

(66)

and

$$\begin{aligned} S^{*}_{N,2}(s) = \frac{1}{(2Nh)^{d+\frac{1}{2}}}\sum _{t=1}^{\lfloor \lfloor Nh\rfloor s\rfloor } X_{d,\lambda _N}(t+ \lfloor N\delta \rfloor ). \end{aligned}$$

(67)

Using Sabzikar and Surgailis (2018b, Theorem 4.3) and the continuous mapping theorem yields

$$\begin{aligned} A^{*}_{m,N,i} {\mathop {\longrightarrow }\limits ^{\mathrm{f.d.d.}}}A^{*}_{m} = \int _{0}^{2} K_{m}(u-1) dB^{II}_{d,\lambda _*}(u), \end{aligned}$$

(68)

as $N\rightarrow \infty$ and hence

$$\begin{aligned} \begin{aligned} \sigma ^{2}_{ii}&= \int _{0}^{2} \int _{0}^{2} K_{m}(u-1) K_{m}(v-1) \mathrm{Cov}\Big ( B^{II}_{d,\lambda _*}(u), B^{II}_{d,\lambda _*}(v) \Big ) du\ dv\\&= \int _{0}^{2} \int _{0}^{2} K_{m}(u-1) K_{m}(v-1) |u-v|^{d-\frac{1}{2}} K_{d-\frac{1}{2}}(\lambda _* |u-v|) du\ dv. \end{aligned} \end{aligned}$$

(69)

Next, we need to show that $A^{*}_{m,N,1}$ and $A^{*}_{m,N,2}$ are asymptotically independent (i.e. $\sigma ^{2}_{12} = \sigma ^{2}_{21} =0$ ). Observe that

$$\begin{aligned} A^{*}_{m,N,1} = 2^{d+\frac{1}{2}} \sum _{j=1}^{m} a_{j} [ S^{*}_{N1}(t_{j}) - S^{*}_{N1}(t_{j-1})] + o_{P}(1) \end{aligned}$$

(70)

and

$$\begin{aligned} S^{*}_{N,1}(t_j) - S^{*}_{N,1}(t_{j-1}) = \frac{1}{(2Nh)^{d+\frac{1}{2}}}\sum _{s=\lfloor \lfloor Nh \rfloor t_{j-1}\rfloor }^{\lfloor \lfloor Nh\rfloor t_j\rfloor } X_{d,\lambda _N}(s) =\sum _{p=-\infty }^{\infty } d_{pN}\zeta (p), \end{aligned}$$

(71)

where

$$\begin{aligned} d_{pN}=\frac{1}{(2Nh)^{d+\frac{1}{2}}} \sum _{t= \lfloor \lfloor Nh\rfloor t_{j-1}\rfloor }^{\lfloor \lfloor Nh\rfloor t_j\rfloor } b_{d}(p-t)e^{-\lambda _N (p-t)}. \end{aligned}$$

(72)

Since $b_{d}(j)e^{-\lambda _N j}\sim C j^{d-1} e^{-\lambda _N j}$ for large lag j, see (4), then for $p> {\lfloor \lfloor Nh\rfloor t_j\rfloor }$, we have

$$\begin{aligned} |d_{pN}| \le C |Nh|^{d+\frac{1}{2}} Nh (p-{\lfloor \lfloor Nh\rfloor t_j\rfloor })^{d-1} e^{-\lambda _N (p-{\lfloor \lfloor Nh\rfloor t_j\rfloor })}, \end{aligned}$$

(73)

where C is a constant. Therefore, we get

$$\begin{aligned} \lim _{N \rightarrow \infty }\sum _{|p|>M} d^2_{pN}= 0, \end{aligned}$$

(74)

since $h\log (Nh) \rightarrow 0$ and $M=Nh\log (Nh)$. Consequently,

$$\begin{aligned} \lim _{N\rightarrow \infty }\mathbb {E}\Bigg [ S^{*}_{N1}(t_j) - S^{*}_{N1}(t_{j-1}) - \sum _{|p|\le M} d_{pN} \zeta (p) \Bigg ]^{2} = 0 \end{aligned}$$

(75)

and by a similar argument

$$\begin{aligned} \lim _{N\rightarrow \infty }\mathbb {E}\Bigg [ S^{*}_{N2}(t_j) - S^{*}_{N2}(t_{j-1}) - \sum _{|p|\le M} d_{pN} \zeta (p+\lfloor N\delta \rfloor ) \Bigg ]^{2} = 0 \end{aligned}$$

(76)

From (75), (76), and $\lfloor N\delta \rfloor - 2N\rightarrow \infty$, we conclude that $S^{*}_{N,1}(t_j) - S^{*}_{N,1}(t_{j-1})$ and $S^{*}_{N,2}(t_{j^{'}}) - S^{*}_{N,2}(t_{{ j^{'} } -1})$ are asymptotically independent for all $j, j^{'}$ and this implies that $A^{*}_{m,N,1}$ and $A^{*}_{m,N,2}$ are asymptotically independent. Thus

$$\begin{aligned} \Big (A^{*}_{m,N,1}, A^{*}_{m,N,2}\Big ){\mathop {\longrightarrow }\limits ^{\mathrm{f.d.d.}}}N_{2}\Big ( 0, \varSigma \Big ), \end{aligned}$$

(77)

where $\sigma ^{2}_{ii}$ is given by (69) and $\sigma ^{2}_{12}=\sigma ^{2}_{21}=0$. Since

$$\begin{aligned} \lim _{m\rightarrow \infty }\sigma ^{2}_{ii} = \int _{0}^{2} \int _{0}^{2} K(u-1) K(v-1) |u-v|^{d-\frac{1}{2}} K_{d-\frac{1}{2}}(\lambda _* |u-v|) du\ dv \end{aligned}$$

(78)

and by Pipiras and Taqqu (1997, Theorem 2), one obtains

$$\begin{aligned} \lim _{m\rightarrow \infty }\lim _{N\rightarrow \infty } \mathrm{Var}(A^{*}_{m,N,1}- A^{*}_{N,1})=0,\ \mathrm{for}\ i=1, 2, \end{aligned}$$

(79)

then the desired results follows by (61), (62), (78), and (79). $\square$

proof of Theorem 4

The proofs of part (a) and part (b) are similar and we just prove part (b) for $\lambda _*\in (0,\infty )$. Using the triangle inequality yields

$$\begin{aligned}&\mathbb {P}\Big ( N^{\frac{1}{2} - d}\big \Vert \hat{\theta }-\theta - ({{\,\mathrm{\mathbf {M}}\,}}_{N+}^T{{\,\mathrm{\mathbf {M}}\,}}_{N+})^{-1} {{\,\mathrm{\mathbf {M}}\,}}_{N+}^T{{\,\mathrm{\mathbf {e}}\,}}_{N}\big \Vert>\varDelta \Big ) \\&\quad \le \mathbb {P}\Big ( N^{\frac{1}{2} - d}\Big \Vert \hat{\theta }-\theta \Big \Vert>\frac{\varDelta }{2} \Big ) +\mathbb {P}\Big ( N^{\frac{1}{2} - d}\big \Vert ({{\,\mathrm{\mathbf {M}}\,}}_{N+}^T{{\,\mathrm{\mathbf {M}}\,}}_{N+})^{-1} {{\,\mathrm{\mathbf {M}}\,}}_{N+}^T{{\,\mathrm{\mathbf {e}}\,}}_{N}\big \Vert >\frac{\varDelta }{2} \Big ). \end{aligned}$$

For the first term, we have by Weiershäuser (2012, p. 95 Theorem 5.1.9), (3) and Markov’s inequality,

$$\begin{aligned} \mathbb {P}\big ( N^{\frac{1}{2} - d}\big \Vert \hat{\theta }-\theta \big \Vert >\frac{\varDelta }{2} \Big ) \le \frac{4\mathbb {E}\big \Vert \hat{\theta }-\theta \big \Vert ^2}{\varDelta ^2 N^{1-2d}} = \frac{O[\min (N^{-\beta }N^{2d-1},N^{4d-2})]}{\varDelta ^2}. \end{aligned}$$

Since $\varDelta$ is arbitrary, $\lim _{N\rightarrow \infty }\mathbb {P}\big ( N^{\frac{1}{2} - d}\big \Vert \hat{\theta }-\theta \big \Vert >\frac{\varDelta }{2} \Big ) = 0$ for $d < 1/2$. For the second term and again by Markov’s inequality

$$\begin{aligned} \mathbb {P}\Big ( N^{\frac{1}{2} - d}\big \Vert ({{\,\mathrm{\mathbf {M}}\,}}_{N+}^T{{\,\mathrm{\mathbf {M}}\,}}_{N+})^{-1} {{\,\mathrm{\mathbf {M}}\,}}_{N+}^T{{\,\mathrm{\mathbf {e}}\,}}_{N}\big \Vert >\frac{\varDelta }{2} \Big )\le & {} \frac{4\mathbb {E}\big \Vert ({{\,\mathrm{\mathbf {M}}\,}}_{N+}^T{{\,\mathrm{\mathbf {M}}\,}}_{N+})^{-1} {{\,\mathrm{\mathbf {M}}\,}}_{N+}^T{{\,\mathrm{\mathbf {e}}\,}}_{N}\big \Vert ^2}{\varDelta ^2 N^{1-2d}}. \nonumber \\ \end{aligned}$$

(80)

Let $\varOmega = ({{\,\mathrm{\mathbf {M}}\,}}_{N+}^T{{\,\mathrm{\mathbf {M}}\,}}_{N+})^{-1} {{\,\mathrm{\mathbf {M}}\,}}_{N+}^T$, then $\mathbb {E}\Vert \varOmega {{\,\mathrm{\mathbf {e}}\,}}_N\Vert ^2 = {{\,\mathrm{\text {tr}}\,}}(\varOmega \varSigma _{{{\,\mathrm{\mathbf {e}}\,}}_N{{\,\mathrm{\mathbf {e}}\,}}_N}) + \{\mathbb {E}({{\,\mathrm{\mathbf {e}}\,}}_N)\}^T\varSigma _{{{\,\mathrm{\mathbf {e}}\,}}_N{{\,\mathrm{\mathbf {e}}\,}}_N}\mathbb {E}({{\,\mathrm{\mathbf {e}}\,}}_N)$ where ${{\,\mathrm{\text {tr}}\,}}(A)$ and $\varSigma _{{{\,\mathrm{\mathbf {e}}\,}}_N{{\,\mathrm{\mathbf {e}}\,}}_N}$ denote the trace of the matrix A and the variance-covariance matrix of ${{\,\mathrm{\mathbf {e}}\,}}_N$ respectively. Since $\{ X_{d,\lambda }(t) \}_{t\in \mathbb {Z}}$ is a tempered mean zero linear process we have $\mathbb {E}\Vert \varOmega {{\,\mathrm{\mathbf {e}}\,}}_N\Vert ^2 = {{\,\mathrm{\text {tr}}\,}}(\varOmega \varSigma _{{{\,\mathrm{\mathbf {e}}\,}}_N{{\,\mathrm{\mathbf {e}}\,}}_N})$. Further, the variance-covariance matrix of ${{\,\mathrm{\mathbf {e}}\,}}_N$ is finite, see (17). Consequently, the numerator of (80) is finite since ${{\,\mathrm{\mathbf {M}}\,}}_{N+}^T{{\,\mathrm{\mathbf {M}}\,}}_{N+}$ is full rank for $N\rightarrow \infty$ and hence the second term goes to zero for $d<1/2$. $\square$

proof of Theorem 5

The proofs of part (a) and part (b) are similar and hence we just give the proof for part (b). We first note that $\int _\mathbb {R}\mu _{(i+)}(u) dB^{II}_{d,\lambda _*}(u) = \int _{\mathbb {R}} {\mathbb {I}}^{d,\lambda _*}_{-}\mu _{(i+)}(u) dB(u)$. Observe that $\mu _{(i+)}\in L^{p}(\mathbb {R})$ for $p\ge 1$ and hence ${\mathbb {I}}^{d,\lambda _*}_{-}\mu _{(i+)}\in L^{p}$. In particular, let $p=2$ and apply the Ito-isometry to conclude that $\int _{\mathbb {R}} {\mathbb {I}}^{d,\lambda _*}_{-}\mu _{(i+)}(u) dB(u)$ is well-defined. Because of Theorem 4, we need to show that

$$\begin{aligned} N^{\frac{1}{2} - d} ({{\,\mathrm{\mathbf {M}}\,}}_{N+}^T{{\,\mathrm{\mathbf {M}}\,}}_{N+})^{-1} {{\,\mathrm{\mathbf {M}}\,}}_{N+}^T {{\,\mathrm{\mathbf {e}}\,}}_N {\mathop {\longrightarrow }\limits ^{\mathrm{f.d.d.}}}{{\varvec{\Lambda }}} \Big [ \int _\mathbb {R}\mu _{(i+)}(u)\ dB^{II}_{d,\lambda _*}(u) \Big ]_{i= 1, \ldots , p+1} \end{aligned}$$

(81)

as $N\rightarrow \infty$. Observe that $N ({{\,\mathrm{\mathbf {M}}\,}}_{N+}^T{{\,\mathrm{\mathbf {M}}\,}}_{N+})^{-1} \rightarrow \varLambda$ as $N\rightarrow \infty$. Therefore the RHS of (81) follows if we show

$$\begin{aligned} \frac{1}{N^{ d+ \frac{1}{2} } } {{\,\mathrm{\mathbf {M}}\,}}_{N+}^T {{\,\mathrm{\mathbf {e}}\,}}_N {\mathop {\longrightarrow }\limits ^{\mathrm{f.d.d.}}}\Big [\int _{\mathbb {R}} \mu _{(i+)}(s)\ dB^{II}_{d,\lambda _*}(s)\Big ]_{i=1,\ldots , p+1}. \end{aligned}$$

But this is equivalent to show that

$$\begin{aligned} \frac{1}{N^{ d+ \frac{1}{2} } } {\langle \alpha , {{\,\mathrm{\mathbf {M}}\,}}_{N+}^T {{\,\mathrm{\mathbf {e}}\,}}_N \rangle } \rightarrow {\Bigl \langle \alpha , \Big [ \int _\mathbb {R}\mu _{(i+)}(u)\ dB^{II}_{d,\lambda _*}(u) \Big ]_{i= 1, \ldots , p+1} \Bigr \rangle }, \alpha \in \mathbb {R}^{p+1}. \end{aligned}$$

(82)

Note that

$$\begin{aligned} {\langle \alpha , {{\,\mathrm{\mathbf {M}}\,}}_{N+}^T {{\,\mathrm{\mathbf {e}}\,}}_N \rangle } = \sum _{i=1}^{p+1}\sum _{j=1}^{n} \alpha _i \mu _{(i+)}\big (\frac{j}{N}\big ) X_{d,\lambda _N}(j) \end{aligned}$$

(83)

and by Lemma 2

$$\begin{aligned} N^{-(d+1/2)}\sum _{i=1}^{p+1}\sum _{j=1}^{N} \alpha _i \mu _{(i+)}(\frac{j}{N}) X_{d,\lambda _N}(j){\mathop {\longrightarrow }\limits ^{\mathrm{f.d.d.}}}\int _{\mathbb {R}} m_{\alpha }(u) dB^{II}_{d,\lambda _*}(u) \end{aligned}$$

where $m_{\alpha }(u) := \sum _{i=1}^{p+1} \alpha _i \mu _{(i+)}(u)$ and this completes the proof. $\square$

proof of Theorem 6

We only proof part (c) since the other parts follow a similar procedure. Let $\varXi$ be the random vector

$$\begin{aligned} \varXi = \Big [\int _{\mathbb {R}} \mu _{(i+)}(s) dB^{II}_{d,\lambda _*}(s)\Big ]_{i=1,\ldots , p+1}. \end{aligned}$$

(84)

Then we can write

$$\begin{aligned} \int _\mathbb {R}\mu _{(i+)}(s) dB^{II}_{d,\lambda _*}(s) = \int _\mathbb {R}\Big ( {\mathbb {I}}^{d,\lambda _*}_{-}\mu _{(i+)} \big )(s) dB(s) \end{aligned}$$

(85)

for $i=1,\ldots , p+1$. We observe that $\int _\mathbb {R}\big ( {\mathbb {I}}^{d,\lambda _*}_{-}\mu _{(i+)} \big )(s) dB(s)$ is a Gaussian stochastic process with mean zero and finite variance $\int _\mathbb {R}\big | {\mathbb {I}}^{d,\lambda _*}_{-}\mu _{(i+)}(s) \big |^2 \ ds$. Using the Ito-isometry for the Wiener integrals, one can see $\varXi$ has the covariance matrix

$$\begin{aligned} \varSigma _{0}=\Bigg [\int _{\mathbb {R}} \Big ( {\mathbb {I}}^{d,\lambda _*}_{-} \mu _{(i+)} \Big )(s) \Big ( {\mathbb {I}}^{d,\lambda _*}_{-} \mu _{(k+)} \Big )(s)\ ds\Bigg ]_{i,k=1,\ldots ,p+1} \end{aligned}$$

(86)

and consequently $\varLambda \varXi$ has normal distribution with covariance matrix $\varLambda \varSigma _0 \varLambda$ and this completes the proof of the first part. Next, we have

$$\begin{aligned} \begin{aligned}&\int _{\mathbb {R}} \Big ( {\mathbb {I}}^{d,\lambda _*}_{-} \mu _{(i+)} \Big )(s) \Big ( {\mathbb {I}}^{d,\lambda _*}_{-} \mu _{(k+)} \Big )(s)\ ds= \int _{\mathbb {R}} \mathcal {F}[{\mathbb {I}}^{d,\lambda _*}_{-} \mu _{(i+)}](\omega ) \overline{\mathcal {F}[{\mathbb {I}}^{d,\lambda _*}_{-} \mu _{(i+)}](\omega )} \ d\omega \\&\quad = \int _\mathbb {R}\widehat{\mu _{(i+)}}(\omega ) \overline{\widehat{\mu _{(k+)}}(\omega )} (\lambda _*^2 + \omega ^2)^{-d}d\omega \\&\quad =\int _\mathbb {R}\int _\mathbb {R}\mu _{(i+)}(t) \mu _{(k+)}(s) \int _{\mathbb {R}} e^{i\omega (t-s)} (\lambda _*^2 + \omega ^2)^{-d} \ d\omega ds\ dt\\&\quad =2\int _\mathbb {R}\int _\mathbb {R}\mu _{(i+)}(t) \mu _{(k+)}(s) \int _{0}^{\infty } \cos (\omega (t-s)) (\lambda _*^2 + \omega ^2)^{-d} d\omega \ ds\ dt\\&\quad =C \int _\mathbb {R}\int _\mathbb {R}\mu _{(i+)}(t) \mu _{(k+)}(s) |t-s|^{d-\frac{1}{2}} K_{d-\frac{1}{2}}(\lambda _*|t-s|) ds \ dt, \end{aligned} \end{aligned}$$

(87)

where $C = \frac{2}{\varGamma (d) \sqrt{\pi } (2\lambda )^{d-\frac{1}{2}}}$ and we used

$$\begin{aligned} \int _{0}^{\infty } \frac{\cos (\omega x)}{(\lambda ^2+x^2)^{\nu +\frac{1}{2}} } \ dx = \frac{\sqrt{\pi }}{\varGamma (\nu +\frac{1}{2})} \Big ( \frac{|x|}{2\lambda } \Big )^{\nu } K_{\nu }(\lambda |x|) \end{aligned}$$

(88)

for $\nu > -\frac{1}{2}$ and $\lambda >0$ in (87). This completes the proof of the second part and Theorem. $\square$