On Fractional Lévy Processes: Tempering, Sample Path Properties and Stochastic Integration

Abstract

We define two new classes of stochastic processes, called tempered fractional Lévy process of the first and second kinds (TFLP and TFLP II, respectively). TFLP and TFLP II make up very broad finite-variance, generally non-Gaussian families of transient anomalous diffusion models that are constructed by exponentially tempering the power law kernel in the moving average representation of a fractional Lévy process. Accordingly, the increment processes of TFLP and TFLP II display semi-long range dependence. We establish the sample path properties of TFLP and TFLP II. We further use a flexible framework of tempered fractional derivatives and integrals to develop the theory of stochastic integration with respect to TFLP and TFLP II, which may not be semimartingales depending on the value of the memory parameter and choice of marginal distribution.

Appendix: Proofs

Proof of Proposition 2.3

The proof of (2.7) follows by a similar argument of Proposition 2.3 in [54] and hence we omit the details. To show (2.9), apply the covariance function formula (2.7) in Proposition 2.3 for $s=t$ to arrive at

$$\begin{aligned} \mathrm{Var}\big [S^{I}_{d,\lambda }(t) \big ] = \frac{ {\mathbb {E}}( L(1)^2 ) }{ \Gamma (1+d)^2 } \Bigg [ \frac{ 2\Gamma (1+2d) }{ (2\lambda )^{1+2d} } - \frac{2\Gamma (1+d)}{ \sqrt{\pi } } \Big ( \frac{1}{ 2\lambda }\Big )^{d+\frac{1}{2} } |t|^{d+\frac{1}{2}} K_{d+\frac{1}{2}} (\lambda t)\Bigg ]. \end{aligned}$$

(A.1)

The second term inside the bracket tends to zero as $t\rightarrow \infty$, since

$$\begin{aligned} K_{d+\frac{1}{2}} (\lambda t) \sim \sqrt{\frac{\pi }{2\lambda t}} e^{-\lambda t}. \end{aligned}$$

Hence, relation (2.9) holds, as claimed. $\square$

Proof of Proposition 2.5

Starting from the definition of TFLP, we can use integration by parts (see [82, p. 1106]) to write

$$\begin{aligned} \begin{aligned} \Gamma (d+1) S^{I}_{d,\lambda }(t)&= \int _{{\mathbb {R}}}\big [ e^{-\lambda (t-x)_{+}} (t-x)_{+}^{d} - e^{-\lambda (-x)_{+}} (-x)_{+}^{d} \big ]\ dL(x)\\&= \int _{-\infty }^{t-} e^{-\lambda (t-x)} (t-x)^{d} dL(x) - \int _{-\infty }^{0-} e^{\lambda x} (-x)^{d} dL(x) \\&= \lim _{u\uparrow t}\Big ( e^{-\lambda (t-u)} (t-u)^{d} L(u) - \int _{-\infty }^{u} L(u) d(e^{-\lambda (t-u)} (t-u)^{d}) \Big ) \\&\quad - \lim _{u\uparrow 0}\Big ( e^{\lambda u} (-u)^{d} L(u) - \int _{-\infty }^{u} L(u) d(e^{\lambda u} (-u)^{d}) \Big ). \end{aligned} \end{aligned}$$

(A.2)

Using [99, Proposition 47.11], we have $e^{\lambda v}L(v)\rightarrow 0$ as $v \rightarrow 0$. Hence, for $d>0$,

$$\begin{aligned} \lim _{u\uparrow t} e^{-\lambda (t-u)} (t-u)^{d} L(u) = \lim _{u\uparrow 0} e^{\lambda u} (-u)^{d} L(u) = 0. \end{aligned}$$

Therefore, we can reexpress (A.2) as

$$\begin{aligned} \begin{aligned}&\quad \quad - \lim _{u\uparrow t} \int _{-\infty }^{u} L(u) \Big (-d e^{-\lambda (t-u)} (t-u)^{d-1} + \lambda e^{-\lambda (t-u)} (t-u)^{d} \Big ) du\\&\quad \quad + \lim _{u\uparrow 0} \int _{-\infty }^{u} L(u) \Big (-d e^{\lambda u} (-u)^{d-1} + \lambda e^{\lambda u} (-u)^{d} \Big ) du\\&= d \int _{{\mathbb {R}}} L(u) \big [ e^{-\lambda (t-u)_{+}} (t-u)_{+}^{d-1} - e^{-\lambda (-u)_{+}} (-u)_{+}^{d-1} \big ]\ du\\&\quad \quad -\lambda \int _{{\mathbb {R}}} L(u) \big [ e^{-\lambda (t-u)_{+}} (t-u)_{+}^{d} - e^{-\lambda (-u)_{+}} (-u)_{+}^{d} \big ]\ du.\\ \end{aligned} \end{aligned}$$

Hence, (2.11) holds.

To show the continuity of the process (2.11), without loss of generality fix $t\in (a,b) \subseteq {\mathbb {R}}_+$. Rewrite the first term in the expression (2.11) as

$$\begin{aligned} \Big \{\int ^{a}_{-\infty } +\int ^{t}_a \Big \}\Big ( e^{-\lambda (t-x)_{+}}{(t-x)_{+}^{d-1}}-e^{-\lambda (-x)_{+}}{(-x)_{+}^{d-1}}\Big )\ L(x)\ dx. \end{aligned}$$

We want to show that this expression is continuous as a function of t. On one hand, the mapping $t\mapsto \int _{-\infty }^a L(u) \big [ e^{-\lambda (t-u)_{+}} (t-u)_{+}^{d-1} - e^{-\lambda (-u)_{+}} (-u)_{+}^{d-1} \big ]\ du$ is continuous. This is a consequence of the dominated convergence theorem, since

$$\begin{aligned}&{\mathbf {1}}_{(-\infty ,a]}(u)| L(u)|\left| \big [ e^{-\lambda (t-u)_{+}} (t-u)_{+}^{d-1} - e^{-\lambda (-u)_{+}} (-u)_{+}^{d-1}\right| \\&\quad \le {\mathbf {1}}_{(-\infty ,a]}(u)| L(u)| \Big ( e^{-\lambda (a-u)} (b-u)_{+}^{d-1}+ e^{-\lambda (-u)_{+}} (-u)_{+}^{d-1}\Big )\in L^1({\mathbb {R}}), \end{aligned}$$

where we use the fact that L is locally bounded. On the other hand, by making the change of variable $z = t-u$,

$$\begin{aligned} \int ^t_a L(u) e^{-\lambda (t-u)_{+}} (t-u)_{+}^{d-1} \ du = \int _{{\mathbb {R}}} 1_{[0,t-a]}(z) L(t-z) e^{-\lambda z} z^{d-1} \ dz. \end{aligned}$$

(A.3)

However, the integrand in (A.3) is bounded in absolute value by

$$\begin{aligned} \sup _{w \in (a,b)}|L(w)| 1_{[0,b-a]}(z) e^{-\lambda z} z^{d-1} \in L^1({\mathbb {R}}). \end{aligned}$$

Therefore, by the dominated convergence theorem, the mapping $t \mapsto \int _a^t L(u) \big [ e^{-\lambda (t-u)_{+}} (t-u)_{+}^{d-1} - e^{-\lambda (-u)_{+}} (-u)_{+}^{d-1} \big ]\ du$ is also continuous. Hence, the first term in the expression (2.11) is continuous as a function of t, as claimed. Again by the dominated convergence theorem, the second term in the expression (2.11) is also continuous as a function of t. This establishes that the process (2.11) is continuous. $\square$

Proof of Theorem 2.6

First, we establish (a). We use the modification of $S^{I}_{d,\lambda }$ given in Theorem 2.5 to write

$$\begin{aligned} \begin{aligned} |S^{I}_{d,\lambda }(t) - S^{I}_{d,\lambda }(s)|&\le \frac{1}{\Gamma (d)} \int _{{\mathbb {R}}} \Big | e^{-\lambda (t-u)_{+}} (t-u)_{+}^{d-1} - e^{-\lambda (s-u)_{+}} (s-u)_{+}^{d-1} \Big | \Big |L(u)\Big | \ du\\&\quad + \frac{\lambda }{\Gamma (d+1)} \int _{{\mathbb {R}}} \Big | e^{-\lambda (t-u)_{+}} (t-u)_{+}^{d} - e^{-\lambda (s-u)_{+}} (s-u)_{+}^{d} \Big | \Big |L(u)\Big | \ du. \end{aligned} \end{aligned}$$

(A.4)

Recall that $0 < d \le 1/2$. For notational simplicity, consider a parameter $\beta$, which can be interpreted either as d or $d-1$, i.e. $\beta \in (-1,-1/2]\cup (0,1/2]$. Define

$$\begin{aligned} W_{\beta }(s,t) = \int _{{\mathbb {R}}} \Big | e^{-\lambda (t-u)_{+}} (t-u)_{+}^{\beta } - e^{-\lambda (s-u)_{+}} (s-u)_{+}^{\beta } \Big | \Big |L(u)\Big | \ du. \end{aligned}$$

For s, t satisfying $-T \le s \le t \le T$, we obtain

$$\begin{aligned} \begin{aligned} W_{\beta }(s,t) =&\int _{s}^{t} e^{-\lambda (t-u)} (t-u)^{\beta } \Big |L(u)\Big | \ du\\&+ \int _{-\infty }^{s} \Big | e^{-\lambda (t-u)} (t-u)^{\beta } - e^{-\lambda (s-u)} (s-u)^{\beta } \Big | \Big |L(u)\Big | \ du\\ \le&\sup _{|u|\le T} |L(u)| \int _{s}^{t} e^{-\lambda (t-u)} (t-u)^{\beta } \ du \\&+ \int _{-\infty }^{s} e^{-\lambda (t-u)} \Big | (t-u)^{\beta } - (s-u)^{\beta } \Big | \Big |L(u)\Big | \ du\\&+ \int _{-\infty }^{s} (s-u)^{\beta } \Big | e^{-\lambda (t-u)} - e^{-\lambda (s-u)} \Big | \Big |L(u)\Big | \ du. \end{aligned} \end{aligned}$$

Using the substitution $h=t-s$, we get

$$\begin{aligned} \begin{aligned} W_{\beta }(s,t)&\le \frac{h^{\beta +1}}{\beta +1}\sup _{|u|\le T} |L(u)| + e^{-\lambda h}\int _{-\infty }^{s} e^{-\lambda (s-u)} \Big | (h+s-u)^{\beta } - (s-u)^{\beta } \Big | \Big |L(u)\Big | \ du\\&\quad + \int _{-\infty }^{s} | e^{-\lambda h}-1 | (s-u)^{\beta } e^{-\lambda (s-u)} \Big |L(u)\Big | \ du\\&= \frac{h^{\beta +1}}{\beta +1}\sup _{|u|\le T} |L(u)| + e^{-\lambda h}\int _{0}^{\infty } e^{-\lambda v} \Big | (h+v)^{\beta } - v^{\beta } \Big | \Big |L(s-v)\Big | \ dv\\&\quad + ( 1- e^{-\lambda h} ) \int _{0}^{\infty } v^{\beta } e^{-\lambda v} \Big | L(s-v) \Big | \ dv\\&=: I_1 + I_2 + I_3. \end{aligned} \end{aligned}$$

(A.5)

Since L is locally bounded, then

$$\begin{aligned} I_1 \le C_{1}(\omega ) h^{\beta +1} \end{aligned}$$

(A.6)

for an almost surely finite random variable $C_1$. Next, observe that

$$\begin{aligned} \limsup _{|v|\rightarrow \infty }\quad \frac{|L(v)|}{|v|} = 0 \end{aligned}$$

(A.7)

by [99, Proposition 48.9]. In particular, the integrands appearing in $I_2$ and $I_3$ are finite almost surely (since $\lambda >0$). Since $( 1- e^{-\lambda h} ) \le \lambda h$ for $h>0$, we conclude that there is an almost sure finite continuous random variable $C_3(\omega )$ such that

$$\begin{aligned} I_{3}(\omega ) \le C_{3}(\omega ) h \end{aligned}$$

(A.8)

for all $-T\le s\le t\le T$.

In regard to $I_2$, consider the decomposition

$$\begin{aligned} \Big \{\int _{0}^{1} + \int _{1}^{\infty } \Big \} \quad e^{-\lambda v} | (h+v)^{\beta } - v^{\beta } | |L(s-v)| \ dv. \end{aligned}$$

(A.9)

By the mean value theorem, for each $v>0$ there exists some $v_{h}\in [v,v+h]$ such that $(h+v)^{\beta } - v^{\beta } = h\beta {v_{h}}^{\beta -1}$. Thus, we can bound the second integral in (A.9) by

$$\begin{aligned} \begin{aligned}&\int _{1}^{\infty } e^{-\lambda v} | (h+v)^{\beta } - v^{\beta } | |L(s-v)| \ dv \\&\quad \le \int _{1}^{\infty } e^{-\lambda v} h \beta \max \{v^{\beta }, (v+h)^{\beta }\} |L(s-v)| \ dv \le C_{2,1} h, \end{aligned} \end{aligned}$$

(A.10)

$-T \le s\le t \le T$. In (A.10), $C_{2,1}$ is an almost surely finite random variable as a consequence of (A.7). On the other hand, the first integral in (A.9) can be bounded by

$$\begin{aligned} \begin{aligned}&\int _{0}^{1} e^{-\lambda v} \Big | (h+v)^{\beta } - v^{\beta } \Big | \Big |L(s-v)\Big | \ dv \\&\quad \le \sup _{v\in [-T-1,T]} |L(v)| \int _{0}^{1} \Big | (h+v)^{\beta } - v^{\beta } \Big | \ dv \\&\quad = \sup _{v\in [-T-1,T]} |L(v)| \Big | \int _{0}^{1} (h+v)^{\beta }\ dv - \int _0^1 v^{\beta } \ dv \Big | \\&\quad = \sup _{v\in [-T-1,T]} |L(v)|\quad \frac{1}{\beta +1} \Big | (1+h)^{\beta +1} - h^{\beta +1} -1 \Big | \\&\quad \le \sup _{v\in [-T-1,T]} |L(v)| \quad \frac{1}{\beta +1} \Big ( |(1+h)^{\beta +1}-1| + h^{\beta +1} \Big ). \end{aligned} \end{aligned}$$

Using a Taylor expansion, it follows that there is an almost surely finite random variable $C_{2,2}$ such that

$$\begin{aligned} \begin{aligned}&\int _{0}^{1} e^{-\lambda v} \Big | (h+v)^{\beta } - v^{\beta } \Big | \Big |L(s-v)\Big | \ dv \le C_{2,2}|h|^{\min (1,\beta +1)} \end{aligned} \end{aligned}$$

(A.11)

for $s,t\in [-T,T]$. Combining (A.5), (A.6), (A.8), (A.10), and (A.11), we see that

$$\begin{aligned} |W_{\beta }(s,t)| \le C_{\beta } h^{\min (1,\beta +1)} \end{aligned}$$

(A.12)

for $s,t\in [-T,T]$, where $C_{\beta }$ is an almost surely finite random variable. Applying (A.12) to (A.4) with $\beta = d$ and $\beta = d-1$ yields

$$\begin{aligned} |S^{I}_{d,\lambda }(t) - S^{I}_{d,\lambda }(s)| \le C_{T} |t-s|^{d}, \quad s,t\in [-T,T], \end{aligned}$$

which establishes (2.20).

To show (b), let $-\frac{1}{2}< d < 0$. In this case, the kernel function $g^{I}_{d,\lambda ,\cdot }(s)$ is not locally bounded and in fact the mapping $t\longmapsto g^{I}_{d,\lambda ,t}(s)$, $t\in {\mathbb {R}}$, is unbounded and discontinuous for all s. Therefore, Theorem 4 in [108] implies that the sample paths of $S^{I}_{d,\lambda }$ are unbounded and discontinuous with positive probability, as claimed. $\square$

Proof of Proposition 2.7

To prove (a), note that TFLN has the same covariance structure as tempered fractional Gaussian noise (TFGN), up to a constant. Expression (2.15) can be obtained by following the same argument as in Chen et al. [46, Appendix 2] for the asymptotic behavior of TFGN over large covariance lags.

To show (b), let a(t) be the time domain kernel of the moving average representation (2.14) of TFLN. Then, the spectral density is given by

$$\begin{aligned} h^{I}(\omega ) = \frac{1}{2\pi } \quad \Big | \int _{{\mathbb {R}}}e^{-i\omega t} a(t) dt\Big |^2 = \frac{1}{2\pi } \quad \Big | \frac{ e^{i\omega } -1 }{ (\lambda + i\omega )^{d+1} } \Big |^2. \end{aligned}$$

This establishes (2.16). $\square$

The next lemma is mentioned in Sect. 2.2. As a consequence of the lemma, $S^{I\!I}_{d,\lambda }(t)$ is well defined for any $t>0$.

Lemma A.1

Let $g^{I\! I}_{d,\lambda ,t}(y)$ be the function (2.8). Then,

$$\begin{aligned} g^{I\! I}_{d,\lambda ,t}(y) \in L^2({\mathbb {R}}) \end{aligned}$$

(A.13)

for any $t \in {\mathbb {R}}$ and any $\lambda >0$, $d > -\frac{1}{2}$.

Proof of Lemma A.1

Let $t >0$. By applying Minkowski’s inequality to (2.8), we arrive at

$$\begin{aligned}&\Vert g^{I\! I}_{d,\lambda ,t}(\cdot )\Vert _{2} \le \Big (\int _{\mathbb {R}}(t-y)_+^{2d} \ e^{-2 \lambda (t-y)_+} \ dy\Big )^{1/2} + \Big (\int _{\mathbb {R}}(-y)_+^{2d} \ e^{-2 \lambda (-y)_+} \ dy\Big )^{1/2}\\&\quad + \, \lambda \Big ( \int _{\mathbb {R}}\ \Big \{\int ^t_0 (s-y)_+^{d} \ e^{-\lambda (s-y)_+} \ ds \Big \}^{2}dy \Big )^{1/2} < \infty , \end{aligned}$$

where finiteness is a consequence of the facts that $2d+1>0$ and $\lambda >0$. Since $g^{I\! I}_{d,\lambda ,-t}(y) = -g^{I\! I}_{d,\lambda ,t}(y+t)$ for any $t, y \in {\mathbb {R}}$, (A.13) holds. $\square$

Proof of Proposition 2.9

We first note that $g^{I\! I}_{d,\lambda ,t}(y) = d\int _{0}^{t} (s-y)_+^{d-1} e^{-\lambda (s-y)_+} ds$, where $g^{I\! I}_{d,\lambda ,t}(y)$ is the function given by (2.8). Hence,

$$\begin{aligned} S^{I\!I}_{d,\lambda }(t)=\frac{1}{\Gamma (d+1)} \int _{{\mathbb {R}}} g^{I\! I}_{d,\lambda ,t}(y)\ dL(y) = \frac{1}{\Gamma (d)} \int _{{\mathbb {R}}}\int _{0}^{t} (s-y)_+^{d-1} e^{-\lambda (s-y)_+} ds \ dL(y) \end{aligned}$$

(A.14)

From Proposition 2.1,

$$\begin{aligned} {\text{ Cov }}\Big (\int _{{\mathbb {R}}}f(y)\ dL(y), \int _{{\mathbb {R}}}g(y)\ dL(y)\Big )={\mathbb {E}}[L(1)^2] \int _{{\mathbb {R}}} f(y)g(y) dy \end{aligned}$$

(A.15)

Now, by Lemma A.1, we can apply (A.15) to TFLP $I\!I$ in (A.14) to write

$$\begin{aligned} \begin{aligned}&{\text {Cov}}\Big (S^{I\!I}_{d,\lambda }(t),S^{I\!I}_{d,\lambda }(s)\Big ) \\&\quad \quad =\frac{{\mathbb {E}}[L(1)^2]}{(\Gamma (d))^{2}}\int _{{\mathbb {R}}} g^{I\! I}_{d,\lambda ,t}(y)g^{I\! I}_{d,\lambda ,s}(y) dy\\&\quad \quad =\frac{{\mathbb {E}}[L(1)^2]}{(\Gamma (d))^{2}}\int _{{\mathbb {R}}}\Big (\int _{0}^{t}\int _{0}^{s}(u-y)_{+}^{d-1}(v-y)_{+}^{d-1}e^{-\lambda (u-y)_{+}}e^{-\lambda (v-y)_{+}}dv\ du\Big )dy\\&\quad \quad =\frac{{\mathbb {E}}[L(1)^2]}{(\Gamma (d))^{2}}\int _{0}^{t}\int _{0}^{s}\Bigg [\int _{-\infty }^{\min (u,v)} (u-y)^{d-1}(v-y)^{d-1}e^{-\lambda (u-y)}e^{-\lambda (v-y)}dy\Bigg ]dv\ du. \end{aligned} \end{aligned}$$

(A.16)

Using the relation

$$\begin{aligned} \int _{0}^{\infty }x^{\nu -1}(x+\beta )^{\nu -1}e^{-\mu x}dx=\frac{1}{\sqrt{\pi }}\left( \frac{\beta }{\mu }\right) ^{\nu -\frac{1}{2}}e^{\frac{\beta \mu }{2}}\ \Gamma (\nu )K_{\frac{1}{2}-\nu }\Big (\frac{\beta \mu }{2}\Big ), \end{aligned}$$

(A.17)

(see [109, p. 348]), we have

$$\begin{aligned} \int _{-\infty }^{\min (u,v)}(u-y)^{d-1}(v-y)^{d-1}e^{-\lambda (u-y)}e^{-\lambda (v-y)}dy =\frac{\Gamma (d)}{\sqrt{\pi }}\Big (\frac{|u-v|}{2\lambda }\Big )^{d-\frac{1}{2}}K_{d-\frac{1}{2}}(\lambda |u-v|). \end{aligned}$$

(A.18)

Therefore, from (A.16) and (A.18), we have

$$\begin{aligned} {\text{ Cov }}\Big (S^{I\!I}_{d,\lambda }(t),S^{I\!I}_{d,\lambda }(s)\Big )=\frac{{\mathbb {E}}[L(1)^2]}{\sqrt{\pi }\Gamma (d)(2\lambda )^{d-\frac{1}{2}}}\int _{0}^{t}\int _{0}^{s} |u-v|^{d-\frac{1}{2}}K_{d-\frac{1}{2}}(\lambda |u-v|)dv\ du \end{aligned}$$

for any $d>0$ and $\lambda >0$, as claimed. $\square$

Proof of Proposition 2.11

The proof follows the similar technique that was employed in Theorem 2.5 and hence we omit it. $\square$

Proof of Theorem 2.12

We use the Kolmogorov-$\breve{\text {C}}$entsov theorem (e.g., [105, p. 53]) to establish the claim. Since $\lambda >0$ is fixed, we can assume $\lambda =1$ without loss of generality. Since the increments of $S^{I\!I}_{d,1}(t)$ are stationary, it suffices to show that

$$\begin{aligned} {\mathbb {E}}|S^{I\!I}_{d,1}(t)|^2 \le C t^{1+\beta } \end{aligned}$$

(A.19)

for some $\beta > 0$ and all $0< t < 1$. Consider $g^{I\! I}_{d,1,t}$ as in (2.17). By (A.15),

$$\begin{aligned} {\mathbb {E}}|S^{I\!I}_{d,1}(t)|^2 = C \int _{-\infty }^t (g^{I\! I}_{d,1,t}(y))^2 \ dy =: C (I_1 + I_2), \end{aligned}$$

where

$$\begin{aligned} I_1= & {} \int _{-t}^t (g^{I\! I}_{d,1,t}(y))^2 \ dy= \frac{1}{\Gamma (d)}\int _{-t}^t\left( \int _{0}^{t} (s-x)^{d-1} e^{-(s-x)} ds\right) ^2dx\\\le & {} C\int _{-t}^{t} (t - y)^{2d} \ dy \le C t^{2d+1} \end{aligned}$$

and

$$\begin{aligned} \begin{aligned} I_2 =&\int _{-\infty }^{-t} (g^{I\! I}_{d,1,t}(y))^2 \ dy \le C\int _t^\infty ((t + y)^{d} \ e^{-t-y} - y^{d} \ e^{-y})^2 \ dy\\&\ + C\int _t^\infty \Big \{\int _0^t (s+y)^{d} \ e^{-s-y} \ ds \Big \}^2 \ dy = C(I_2' + I_2''). \end{aligned} \end{aligned}$$

Using $|(t + y)^{d} \ e^{-(t-y)} - y^{d} \ e^{-y}| \le |\ e^{-t}-1|\, \ e^{-y} (t + y)^{d} + \ e^{-y}\, |(t + y)^{d} - y^{d} | \le C t \, \ e^{-y} (t + y)^{d} + C t \, \ e^{-y} y^{d}$ we obtain $I'_2 \le C t^2$ and, similarly, $I''_2 \le C t^2$, implying $I_1 + I_2 \le C(t^{2d+1} + t^2)$ and ${\mathbb {E}}|S^{I\!I}_{d,1}(t)|^2 \le C(t^{2d+1} + t^2)\le t^{2d+1}$ since $d\in (0,1/2]$ and $0< t < 1$. Hence, (A.19) is satisfied with $\beta = 2d$. This completes the proof. To show (b), note that when $-\frac{1}{2}< d < 0$$g^{I\! I}_{d,\lambda ,\cdot }(s)$ is not locally bounded and $t\longmapsto g^{I\! I}_{d,\lambda ,t}(s)$, $t\in {\mathbb {R}}$ is unbounded and discontinuous for all s, and so the same proof in part (b) of Theorem 2.6 applies. $\square$

Proof of Proposition 2.13

To show (a), note that the autocovariance function of a TFGN $I\! I$ satisfies $\gamma (h) \asymp e^{-\lambda h } h^{ d -1 }$ as $h\rightarrow \infty$ (see [55]). From (2.3), TFBM $I\! I$ and TFLP $I\! I$ have the same second order structure up to constants. Hence, (2.23) holds.

To show (b), let $\Big ({\mathbb {I}}^{d,\lambda }_{-} f\Big )(x)$ be as in (3.4) with $\kappa = d$. Note that the process $X^{I\!I}_{d,\lambda }$ as in (2.14) has the integral representation

$$\begin{aligned} X^{I\!I}_{d,\lambda }(t) = \int _{{\mathbb {R}}} \Big ( {\mathbb {I}}^{d,\lambda }_{-}{} \mathbf{1}_{[t,t+1]}(x) \Big ) \ dL(x). \end{aligned}$$

Therefore, its spectral density is given by

$$\begin{aligned} \begin{aligned} h^{I\!I}(\omega )&= \frac{1}{2\pi } \Big |\int _{{\mathbb {R}}} e^{-i\omega t} \Big ( {\mathbb {I}}^{d,\lambda }_{-}{} \mathbf{1}_{[t,t+1]}(\omega ) \Big ) dt \Big |^2\\&= \frac{1}{2\pi } \Big | (\lambda + i\omega )^{-d}\int _{t}^{t+1} e^{-i\omega x} dx \Big |^2 = \frac{1}{2\pi } \frac{ 2( 1- \cos (\omega ) ) }{(\lambda ^2 + \omega ^2)^{d}\ \omega ^2}, \end{aligned} \end{aligned}$$

as claimed. $\square$

Proof of Proposition 2.14

Write

$$\begin{aligned} \phi ^I(x)= x_+^de^{-\lambda x_+}, \quad \phi ^{II}(x)= x_+^d e^{-\lambda x_+} + \lambda \int _0^{x} u_+^d e^{-\lambda u_+} du, \end{aligned}$$

(A.20)

and note that

$$\begin{aligned} S^I_{d,\lambda }(t)= & {} \frac{1}{\Gamma (d+1)}\int _{{\mathbb {R}}} \{\phi ^I(t-x)-\phi ^I(-x)\}dL(x),\nonumber \\ S^{I\! I}_{d,\lambda }(t)= & {} \frac{1}{\Gamma (d+1)}\int _{{\mathbb {R}}} \{\phi ^{II}(t-x)-\phi ^{II}(-x)\}dL(x). \end{aligned}$$

(A.21)

For $x>0$, the derivatives $\eta ^I(x):=\frac{d}{dx}\phi ^I(x)$, $\eta ^{II}(x):=\frac{d}{dx}\phi ^{II}(x)$ exist and satisfy $\eta ^I(x)\sim \eta ^{II}(x)\sim d x^{d-1}$, $x \rightarrow 0^+$. Hence

$$\begin{aligned} \int _a^b \left| \eta ^I(x)\right| ^\alpha dx =\infty ,\quad \int _a^b \left| \eta ^{II}(x)\right| ^\alpha dx=\infty \end{aligned}$$

for any interval [a, b) containing 0 whenever $\alpha (d-1)+1<0$, i.e., whenever $d+\frac{1}{\alpha }<1$. Hence, by Corollary 3.4 in [110], the processes

$$\begin{aligned} \int _0^t\phi ^I(t-x)dL(x), \quad \int _0^t\phi ^{II}(t-x)dL(x), \quad t\ge 0 \end{aligned}$$

are not $({\mathcal {F}}^{L}_t)_{t\ge 0}$-semimartingales, where $({\mathcal {F}}^{L}_t)_{t\ge 0}=\sigma \{L(s); 0\le s \le t \}$. Thus, since L is symmetric, in view of the representations (A.21), by Lemma 5.2 of [110] $S^I_{d,\lambda }$ and $S^{I\! I}_{d,\lambda }$ are not $({\mathcal {F}}^{L,\infty }_t)_{t\ge 0}$-semimartingales. $\square$

Proof of Proposition 3.1

The proof is similar to that of Theorem 3.9 in [111]. Write $\eta ^I(x)=\frac{d}{dx}\phi ^I(x)$ where $\phi ^I$ is given in (A.20). Note since $d>1/2$, $\eta ^I\in L^2({\mathbb {R}})$, and hence the integral $\int _{\mathbb {R}}\eta ^I(x) dL(x)$ is well-defined. Now,

$$\begin{aligned} \Gamma (d+1)S^I_{d,\lambda }(t)= & {} \int _{-\infty }^t \{\phi ^I(t-x) - \phi ^I(-x) \}dL(x)\\= & {} \int _{-\infty }^0 \{\phi ^I(t-x) - \phi ^I(-x)\} dL(x) + \int _{0}^t \phi ^I(t-x) dL(x)\\= & {} \int _{-\infty }^0 \int _0^t \eta ^I(s-x) ds dL(x) + \int _{0}^t \int _{x}^t\eta ^I(s-x)ds dL(x). \end{aligned}$$

Hence, by a stochastic version of the Fubini theorem (e.g. [112, Theorem 65]), the above process has a version that is equal to

$$\begin{aligned} \int _0^t\int _{-\infty }^0 \eta ^I(s-x) dL(x)ds + \int _{0}^t \int _{0}^s\eta ^I(s-x) dL(x)ds = \int _0^t\int _{-\infty }^s \eta ^I(s-x) dL(x)ds. \end{aligned}$$

This establishes (i).

We now turn to (ii). First note that

$$\begin{aligned} S^{I\! I}_{d,\lambda }(t) = \frac{1}{\Gamma (d+1)}\int _{{\mathbb {R}}}\{ \phi ^{II}(t-x)-\phi ^{II}(-x)\}dL(x), \end{aligned}$$

where $\phi ^{II}$ is given in (A.20). Since $d>1/2$, $\frac{d}{dx}\phi ^{II}(x)\in L^2({\mathbb {R}})$, and the rest of the proof can be done similarly to that of part (i). $\square$

The following lemma is used in Sect. 3.

Lemma A.2

Let $S^{I}_{d,\lambda }$ and $S^{I\!I}_{d,\lambda }(t)$ be a TFLP and TFLP $I\!I$ given by (2.6) and (2.17), respectively. Then, for every $t\in {\mathbb {R}}$,

• (a) when $d>0$, expressions (3.9) and (3.10) hold;

• (b) when $-\frac{1}{2}<d<0$, expressions (3.11) and (3.12) hold.

Proof of Lemma A.2

The proofs can be developed along the same lines of that of Lemma 3.4 in [61] for TFLP, and of Proposition 2.5 in [55] for TFLP $I\!I$. $\square$

Proof of Theorem 3.5

To show that ${{\mathcal {A}}}_{1}$ is an inner product space, it suffices to establish that ${\langle f,f \rangle }_{{{\mathcal {A}}}_{1}}=0$ implies $f=0$dx–a.e. If ${\langle f,f \rangle }_{{{\mathcal {A}}}_{1}}=0$, then in view of (3.17) and (3.18) we have ${\langle F,F\rangle }_{2}=0$, so $F(x)=\Big ({{\mathbb {I}}}^{d,\lambda }_{-}f\Big )(x)=0$dx–a.e. Then,

$$\begin{aligned} \Big ({{\mathbb {I}}}^{d,\lambda }_{-}f\Big )(x)=0\quad dx-\text {a.e.} \end{aligned}$$

(A.22)

Apply ${{\mathbb {D}}}^{d,\lambda }_{-}$ to both sides of Eq. (A.22) and use Lemma 2.14 in [61] to get $f(x)=0$dx–a.e. Hence, ${{\mathcal {A}}}_{1}$ is an inner product space, as claimed. $\square$

Next, we want to show that the set of elementary functions ${{\mathcal {E}}}$ is dense in ${{\mathcal {A}}}_{1} \subseteq L^2({\mathbb {R}})$. For any $f\in {{\mathcal {A}}}_{1}$, we also have $f\in {L}^{2}({\mathbb {R}})$, and hence there exists a sequence of elementary functions $(f_n)_{n \in {\mathbb {N}}}$ in $L^2({\mathbb {R}})$ such that $\Vert f-f_n\Vert _2\rightarrow 0$ as $n \rightarrow \infty$. However,

$$\begin{aligned} \Vert f-f_n\Vert ^2_{{{\mathcal {A}}}_{1}} ={\langle f-f_n,f-f_n \rangle }_{{{\mathcal {A}}}_{1}}={\langle F-F_n,F-F_n\rangle }_2= \Vert F-F_n\Vert ^2_2, \end{aligned}$$

where $F_n(x)=\Big ({{\mathbb {I}}}^{d,\lambda }_{-}{f_n}\Big )(x)$ and F(x) is given by (3.18). It can be further shown that $\Vert {{\mathbb {I}}}^{\kappa ,\lambda }_{-}(f)\Vert _{2}\le C\Vert f\Vert _{2}$ for some constant C. Then,

$$\begin{aligned} \Vert f-f_n\Vert _{{{\mathcal {A}}}_{1}}=\left\| F-F_n\right\| _{2}=\Vert {{\mathbb {I}}}^{d,\lambda }_{-}(f-f_n)\Vert _{2}\le C\Vert f-f_n\Vert _{2}. \end{aligned}$$

Since $\Vert f-f_n\Vert _2\rightarrow 0$ as $n \rightarrow \infty$, it follows that the set of elementary functions is dense in ${{{\mathcal {A}}}_{1}}$. Finally, using the example provided in the [93, Theorem 3.1], one can show that ${{\mathcal {A}}}_{1}$ is not complete.

The following proposition can be established by a direct adaptation of the proof of Proposition 2.1 in [93].

Proposition A.3

For $d > -1/2$, $\lambda >0$, let ${{\mathcal {E}}}$ be the set of elementary functions, let ${{\mathcal {I}}}^{d,\lambda }(f)$ be an integral (3.19) of $f \in {{\mathcal {E}}}$ with respect to the Lévy process L as in (2.1). Suppose ${{\mathcal {D}}}$ is a set of deterministic functions on ${\mathbb {R}}$ such that: (i) ${\mathcal D}$ is an inner product space with an inner product $\langle f,g\rangle _{{{\mathcal {D}}}}$ for $f,g \in {{\mathcal {D}}}$; (ii) ${{\mathcal {E}}} \subseteq {{\mathcal {D}}}$ and $\langle f,g\rangle _{{{\mathcal {D}}}} = \langle {\mathcal I}^{d,\lambda }(f),{{\mathcal {I}}}^{d,\lambda }(g)\rangle _{L^2(\Omega )}$, $f,g \in {{\mathcal {E}}}$; (iii) the set is dense in ${{\mathcal {D}}}$. Then,

• (a) there is an isometry between the space ${{\mathcal {D}}}$ and a linear subspace of $\overline{\text {Sp}}(S^{I\!I}_{d,\lambda } )$ which is an extension of the mapping $f \mapsto {{\mathcal {I}}}^{d,\lambda }(f)$, $f \in {{\mathcal {E}}}$;

• (b) ${{\mathcal {D}}}$ is isometric to $\overline{\text {Sp}}(S^{I\!I}_{d,\lambda } )$ itself if and only if ${{\mathcal {D}}}$ is complete.

We are now in a position to prove Theorem 3.8.

Proof of Theorem 3.8 :

Since $\Vert {{\mathbb {I}}}^{\kappa ,\lambda }_{-}(f)\Vert _{2}\le C\Vert f\Vert _{2}$ then the stochastic integral (3.19) is well-defined for any $f\in {{\mathcal {A}}}_{1}$. By using the isometry (2.3) and expression (3.19), it follows from Proposition A.3 and (3.17) that, for any $f,g\in {{{\mathcal {A}}}_{1}}$,

$$\begin{aligned} {\langle f,g \rangle }_{{{\mathcal {A}}}_{1}}={\langle F,G\rangle }_{L^{2}({\mathbb {R}})} ={\langle {{{\mathcal {I}}}}^{d,\lambda }(f),{{{\mathcal {I}}}}^{d,\lambda }(g)\rangle }_{L^2(\Omega )}. \end{aligned}$$

Then, Theorem 3.5 implies that ${{\mathcal {A}}}_{1}$ is isometric to a subset of $\overline{\mathrm{Sp}}(S^{I\!I}_{d,\lambda })$, as claimed. However, again by Theorem 3.5, ${{\mathcal {A}}}_{1}$ is not complete. Therefore, ${{\mathcal {A}}}_{1}$ is isometric to a strict subset of $\overline{\mathrm{Sp}}(S^{I\!I}_{d,\lambda })$. $\square$

Lemmas A.4 and A.5, stated and proved next, are used in the proof of Theorem 3.9.

Lemma A.4

Under the assumptions of Theorem 3.9, every $f\in W^{-d,2}({\mathbb {R}})$ is an element of ${{\mathcal {A}}}_{2}$ for $-\frac{1}{2}< d <0$ and $\lambda >0$, i.e., as sets, $W^{-d,2}({\mathbb {R}})={\mathcal {A}}_2$.

Proof of Lemma A.4

Given $f\in W^{-d,2}({\mathbb {R}})$, we need to show that

$$\begin{aligned} \varphi _f={\mathbb {D}}^{-d,\lambda }_{-} f \end{aligned}$$

(A.23)

for some $\varphi _f\in L^{2}( {\mathbb {R}})$. From the definition (3.8) we see that $\int (\lambda ^2+\omega ^2)^{-d}|\hat{f}(\omega )|^2\ d\omega <\infty$. Define $h_1(\omega )=(\lambda -i\omega )^{-d}\hat{f}(\omega )$ and note that $h_1$ is the Fourier transform of some function $\varphi _1\in L^{2}( {\mathbb {R}})$. Define $\varphi _f:=\varphi _{1}$ so that

$$\begin{aligned} \widehat{\varphi _f}(\omega ) = \widehat{\varphi _1}(\omega ) = {\widehat{f}}(\omega )(\lambda -i\omega )^{-d}. \end{aligned}$$

(A.24)

Since $f\in W^{-d,2}({\mathbb {R}})\subset L^2({\mathbb {R}})$, we can apply Definition 3.4 to get the desired result. $\square$

We state the following lemma that will be used to proof Theorem 3.9. We refer the reader to [61, Lemma 3.12] for the proof of the Lemma.

Lemma A.5

Suppose the assumptions of Theorem 3.9 hold. If $f\in W^{-d,2}({\mathbb {R}})$, then there exists a sequence of functions $(f_n)_{n \in {\mathbb {N}}} \subseteq {{\mathcal {E}}}$ such that $\Vert f_n - f\Vert _{L^2({\mathbb {R}})}$. Moreover, when $-\frac{1}{2}< d <0$,

$$\begin{aligned} \int _{\mathbb {R}}|{\widehat{f}}_n(\omega )-\widehat{f}(\omega )|^2|\omega |^{-2d}d\omega \rightarrow 0, \quad n\rightarrow \infty . \end{aligned}$$

(A.25)

Proof of Theorem 3.9

For $f\in {{\mathcal {A}}}_{2}$ we define

$$\begin{aligned} \Vert f\Vert _{{{\mathcal {A}}}_{2}}=\sqrt{{\langle f,f\rangle }_{{{\mathcal {A}}}_{2}}}=\sqrt{{\langle \varphi _f,\varphi _f\rangle }_2}=\Vert {\varphi _f}\Vert _{2}, \end{aligned}$$

(A.26)

where $\varphi _f$ is given by (A.23). Next, use (A.24) to see that

$$\begin{aligned} \widehat{\varphi _f}(\omega ) = (\lambda -i\omega )^{-d}{\widehat{f}}(\omega ) . \end{aligned}$$

(A.27)

To verify that (3.23) is an inner product, it suffices to show that, if ${\langle f,f \rangle }_{{{\mathcal {A}}}_{2}}=0$, then

$$\begin{aligned} f=0 \quad dx-\text {a.e.} \end{aligned}$$

(A.28)

In fact,

$$\begin{aligned} 0 = \Vert f\Vert _{{{\mathcal {A}}}_{2}}^2=\Vert {\varphi _f}\Vert _{2}^2=\Vert {\widehat{\varphi _f}}\Vert ^{2}_{2} =\int _{\mathbb {R}}|{\widehat{f}}(\omega )|^{2} (\lambda ^2+\omega ^2)^{-d}\ d\omega \end{aligned}$$

(A.29)

implies that ${\widehat{f}}(\omega )=0$$d\omega$—a.e. Hence, (A.28) holds.

We now show that ${{\mathcal {E}}}$ is dense in ${{\mathcal {A}}}_2$. By Lemma A.5, there is a sequence $(f_n)_{n \in {\mathbb {N}}} \subseteq {{\mathcal {E}} }$ such that

$$\begin{aligned} \Vert f_n-f\Vert _2\rightarrow 0, \quad n \rightarrow \infty , \end{aligned}$$

(A.30)

and (A.25) holds. On the other hand, by Lemma A.4, ${{\mathcal {E}}} \subseteq W^{-d,2}({\mathbb {R}}) \subseteq {{\mathcal {A}}}_2$. By (A.29), we can write

$$\begin{aligned} \Vert f_n-f\Vert ^{2}_{{{\mathcal {A}}}_2} =\int _{\mathbb {R}}\Big |{\widehat{f}}_n(\omega )-{\widehat{f}}(\omega )\Big |^{2}(\lambda ^2 + \omega ^2)^{-d}\ d\omega =: I_{1} + I_{2}, \end{aligned}$$

where

$$\begin{aligned} I_1 = \int _{|\omega | < \lambda }\Big |{\widehat{f}}_n(\omega )-{\widehat{f}}(\omega )\Big |^{2} (\lambda ^2+\omega ^2)^{-d}\ d\omega , \quad I_2 = \int _{|\omega | \ge \lambda }\Big |{\widehat{f}}_n(\omega )-{\widehat{f}}(\omega )\Big |^{2} (\lambda ^2+\omega ^2)^{-d}\ d\omega . \end{aligned}$$

Since $|\omega | < \lambda$, then $I_1 \le 2\lambda ^{-2d} \int _{{\mathbb {R}}} |{\widehat{f}}_n(\omega )-{\widehat{f}}(\omega )\Big |^{2} \ d\omega \rightarrow 0$ as $n\rightarrow \infty$, where convergence is a consequence of (A.30). Moreover, by (A.25), $I_2 \le 2^{-d} \int _{{\mathbb {R}}} \Big |{\widehat{f}}_n(\omega )-{\widehat{f}}(\omega )\Big | |\omega |^{-2d}\ d\omega \rightarrow 0$ as $n\rightarrow \infty$. Hence, $\Vert f_n-f\Vert ^{2}_{{{\mathcal {A}}}_2} \rightarrow 0$ as $n\rightarrow \infty$, namely, ${{\mathcal {E}}}$ is dense in ${{\mathcal {A}}}_2$.

It only remains to show that ${{\mathcal {A}}}_{2}$ is complete. In fact, let $\big ( f_n \big )_{n \in {\mathbb {N}}}$ be a Cauchy sequence in ${{\mathcal {A}}}_{2}$. Then, by using the inner product (3.23), the corresponding sequence $\big ( { \varphi _{f_n} } \big )_{n \in {\mathbb {N}}}$ is Cauchy in $L^2({\mathbb {R}})$. Again by the inner product (3.23), and since $L^2({\mathbb {R}})$ is complete, there exists $\varphi _{f^*}$ such that $\Vert f_n- f^*\Vert _{{{\mathcal {A}}}_{2}} = \Vert {\varphi _{f_n} } - {\varphi _{f^*} }\Vert _{2}\rightarrow 0$, $n\rightarrow \infty$. Hence, $f^* \in {{\mathcal {A}}}_2$ and ${{\mathcal {A}}}_{2}$ is complete.

$\square$

Proof of Theorem 3.11

By Lemma A.4, the stochastic integral (3.24) is well-defined for any $f\in {{\mathcal {A}}}_{2}$. Since ${{\mathcal {A}}}_{2}$ is a complete space with inner product (3.23) and ${\mathcal {E}}$ is dense, then Proposition A.3 implies that ${{\mathcal {A}}}_{2}$ is isometric to $\overline{\mathrm{Sp}}( S^{I\!I}_{d,\lambda } )$. This completes the proof. $\square$