Proof. The functions $\exp\!({-}\! \cdot)$ and $E_\alpha({-} \, \cdot)$ are both c.m. for positive arguments; for the latter see [Reference Pollard49]. Furthermore the function $x \rightarrow \lambda x^\alpha$ , $\alpha \in (0,1)$ , is positive with a c.m. derivative, so that its composition with $E_\alpha({-} \, \cdot)$ is also c.m. Therefore $q(\cdot, u)$ is c.m., it being a product of c.m. functions [Reference Schilling and Song55, Corollary 1.6] for all possible values of u. Observe also that $q(0+, u)=1$ , for all u. Hence, for $\gamma>0$ that $m_\gamma$ is a Lévy measure follows from [Reference Rosiński51]. To complete the proof we thus need to show that $\int_0^\infty \int_{S^{d-1}}((ru)^2 \wedge 1)m_0(dr, du) < \infty$ . From $\exp\!({-}\cdot), E_\alpha({-} \, \cdot) \leq 1 $ we have

(33) \begin{equation}\int_{S^{d-1}}\int_0 ^ 1 r^2 \frac{q(r,u;\;\alpha, \lambda, \theta )}{r}dr \sigma(du) < \sigma(S^{d-1})\int_0^1 r dr< \infty . \end{equation}

Also, by assumption (A), it follows that if $\theta_*= \min_{u \in S^{d-1}}\theta(u)=0$ , then both $\lambda_*= \min_{u \in S^{d-1}}\lambda(u),$ and $ \alpha_*= \min_{u \in S^{d-1}}\alpha(u)$ are strictly positive. Then in view of (26) it holds that

(34) \begin{align}\int_{S^{d-1}}\int_1 ^ \infty \frac{q(r,u;\;\alpha, \lambda, \theta )}{r}dr \sigma(du) & < \sigma(S^{d-1})\int_1^\infty r^{-1} E_{\alpha_*}({-}\lambda_* r^{\alpha_*} ) dr \nonumber \\ &\sim \frac{\sigma(S^{d-1})}{\lambda_* \Gamma(1-\alpha_*)} \int_1^\infty r^{-1-\alpha_*} dr < \infty. \end{align}

Similarly, if $\lambda_*, \alpha_*=0$ then $\theta_*>0$ whence

(35) \begin{equation}\int_{S^{d-1}}\int_1 ^ \infty \frac{q(r,u;\;\alpha, \lambda, \theta )}{r}dr \sigma(du) < \sigma(S^{d-1})\int_1^\infty r^{-1} e^{-r \theta_*} dr < \infty. \end{equation}

Proof. Using continuity of the Mittag-Leffler function, $E_1({-}\lambda(u) x)=\exp\!({-}\lambda(u) x)$ , $x \in \mathbb R$ , $E_\alpha(0)=1$ for all $\alpha \in (0,1]$ and $E_\alpha({-}\lambda(u)) \rightarrow 1/(1+\lambda(u))$ as $\alpha \rightarrow 0$ (see [Reference Haubold, Mathai and Saxena24]), together with dominated convergence as needed, (i)–(vi) follow operating on the Lévy measures. For (vi) write

(41) \begin{equation}m^+(x;\, \alpha, \lambda, \theta, \delta)= \delta \frac{e^{-\theta x}}{x} E_{\alpha} ( -\lambda \, x^{\alpha} )= \alpha c_1 \frac{e^{-\theta x}}{x} E_{\alpha} \left(\frac{c_2\theta^{\alpha}-1}{c_2} \, x^{\alpha} \right) \end{equation}

with $c_1=\delta /\alpha$ , $c_2=1/(\lambda +\theta^{\alpha})$ , which matches the expression of the TPL Lévy measure in [Reference Torricelli, Barabesi and Cerioli61, Proposition 2.1], given there for the Laplace exponent. Finally, (vii)–(x) are simple parameter specifications of the prior cases.

Proof. We only treat the positive part, the negative one being identical with the obvious parameter modification, and by substituting x with $|x|$ . We thus remove the subscripts for ease of notation.

Assume $\gamma=0$ . We notice that in that case $x \, m(x)$ is integrable around zero, and we can compute the Lévy–Khintchine integral without truncation. Interchanging series and integral using Fubini’s theorem, we have

(45) \begin{align} \int_0^\infty & \left(e^{i z x}-1 \right) q(x)x^{-1-\gamma} dx= \int_0^\infty \sum_{k=1}^\infty \frac{(i z x)^k}{k !} \frac{e^{-\theta x}}{x} E_{\alpha} \left( -\lambda \, x^{\alpha} \right) dx \nonumber \\ &= \sum_{k=1}^\infty \frac{(i z)^k}{k !}\int_0^\infty e^{-\theta x} x^{k-1} E_{\alpha} \left( -\lambda \, x^{\alpha} \right) dx\nonumber \\ &= \sum_{k=1}^\infty \frac{(i z)^k}{k !}\sum_{j=0}^\infty \left({-}\lambda \right)^j \frac{1}{\Gamma(1+ \alpha j)}\int_0^\infty e^{-\theta x} x^{\alpha j + k-1} dx \nonumber \\ &= \sum_{k=1}^\infty \frac{\left(\frac{i z}{\theta}\right)^k}{k !}\sum_{j=0}^\infty \left(\frac{-\lambda}{\theta^\alpha} \right)^j \frac{\Gamma( k+ \alpha j )}{\Gamma(1+ \alpha j)} \nonumber \\ &= \sum_{j=0}^\infty \left(\frac{-\lambda}{\theta^\alpha} \right)^j \frac{1}{\Gamma(1+ \alpha j)} \sum_{k=1}\frac{\left(\frac{i z}{\theta}\right)^k}{k !} \Gamma( k+ \alpha j ) . \end{align}

For $j=0$ the summations on k in (45) reduce to a logarithmic series. For $j \geq1$ , recalling the binomial series, it holds that

(46) \begin{align}\sum_{k=1}\frac{\left(\frac{i z}{\theta}\right)^k}{k !}\frac{ \Gamma( k+ \alpha j )}{\Gamma(\alpha j)}=\left(\frac{\theta}{\theta-i z}\right)^{\alpha j}-1.\end{align}

Substitute (46) in (45), and under the convergence conditions $|z| < \theta, \theta^\alpha >\lambda$ , we obtain

(47) \begin{align} & \sum_{j=1}^\infty \left(\frac{-\lambda}{ (\theta-iz)^\alpha} \right)^j \frac{1}{\alpha j} -\sum_{j=1}^\infty \left(\frac{-\lambda}{\theta^\alpha} \right)^j \frac{1}{\alpha j}+ \sum_{k=1}\frac{(i z/\theta)^k}{k} \nonumber \\&=-\frac{1}{\alpha} \left( \log\! \left( \frac{(\theta-i z)^\alpha+\lambda}{(\theta- i z)^\alpha}\right) - \log\!\left( \frac{\theta^\alpha+\lambda}{\theta^\alpha} \right) + \log\!\left(\frac{(\theta - i z)^\alpha}{\theta^\alpha} \right) \right) \nonumber \\ &= \frac{1}{\alpha} \log\! \left(\frac{\theta^\alpha +\lambda}{(\theta-i z)^\alpha+\lambda} \right).\end{align}

Here we consider the principal branch of the complex logarithm and the power function $y \mapsto y^\alpha$ for Arg $(y) \in ({-}\pi, \pi]$ .

For $\gamma \in (0,2)$ , we observe that since m decays exponentially, it holds $\int_{ \{|x| >1\}} x m(x)dx=\mu_1 < \infty$ . We can thus use the representation of the characteristic exponent with constant truncation function 1. The same calculations of (45) produce

(48) \begin{align} \int_0^\infty & \left(e^{i z x}-1 - i z x \right) q(x)x^{-1-\gamma} dx= \theta^\gamma \sum_{j=0}^\infty \left(\frac{-\lambda}{\theta^\alpha} \right)^j \frac{1}{\Gamma(1+ \alpha j)} \sum_{k=2}\frac{\left(\frac{i z}{\theta}\right)^k}{k !} \Gamma( k-\gamma+ \alpha j ).\end{align}

Now first let $\gamma \neq 1$ . Using the binomial series again we have

(49) \begin{equation} \sum_{k=2}^\infty \Gamma(k+ j \alpha- \gamma) \frac{(i z/\theta)^k}{k!} = \Gamma(j \alpha-\gamma)\left( \left(\frac{\theta - iz}{\theta}\right)^{-\alpha j +\gamma} - i z \frac{j \alpha -\gamma}{\theta} - 1\right),\end{equation}

and therefore, whenever $|z|<\theta$ and $\theta^\alpha >\lambda$ , (48) becomes, in view of the series expression (29) for the $_2R_1$ function,

(50) \begin{align}& \sum_{j=0} ^\infty \frac{\Gamma(j \alpha-\gamma) }{\Gamma(j \alpha +1) } \left((\theta-iz)^{\gamma} \left( \frac{-\lambda }{(\theta- iz)^\alpha} \right)^j - \theta^{\gamma} \left( \frac{-\lambda }{\theta^{\alpha}} \right)^j \left( iz\frac{j \alpha-\gamma}{\theta} + 1 \right) \right) \nonumber \\&=(\theta-iz)^{\gamma}\sum_{j=0} ^\infty \frac{\Gamma(j \alpha-\gamma) }{\Gamma(j \alpha +1) } \left( \frac{-\lambda }{(\theta- i z)^\alpha} \right)^j - \theta^{\gamma}\sum_{j=0} ^\infty \frac{\Gamma(j \alpha-\gamma) }{\Gamma(j \alpha +1) } \left( \frac{-\lambda }{\theta^\alpha} \right)^j\nonumber \\ &\phantom{xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx} - i z \theta^{\gamma-1}\sum_{j=0} ^\infty \frac{\Gamma(j \alpha-\gamma+1) }{\Gamma(j \alpha +1 ) } \left( \frac{-\lambda }{\theta^\alpha} \right)^j\nonumber \\ & = \Gamma({-}\gamma)\left( (\theta - iz )^\gamma \, _2R_1\left(1, -\gamma, 1, \alpha;\frac{-\lambda}{ (\theta-i z)^\alpha}\right) + i z \gamma \, \theta^{\gamma-1} \, _2R_1\left(1, 1-\gamma, 1, \alpha;-\frac{\lambda}{ \theta^\alpha} \right) \nonumber \right. \\ & \phantom{xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx} \left. -\theta^\gamma \, _2R_1\left(1, -\gamma, 1, \alpha;-\frac{\lambda}{ \theta^\alpha} \right) \right) .\end{align}

Finally, for $\gamma=1$ the term $j=0$ in (48) becomes yet another log series,

(51) \begin{align}& \sum_{k=2}^\infty \Gamma(k-1) \frac{(i z/\theta)^k}{k!} = \sum_{k=2}^\infty \frac{(i z/\theta)^k}{k(k-1)} = \sum_{k=2}^\infty \frac{(i z/\theta)^k}{k-1} -\sum_{k=2}^\infty \frac{(i z/\theta)^k}{k} \nonumber \\ & = \frac{iz}{\theta}+\log\!\left(1-\frac{iz}{\theta} \right)\left(1- \frac{iz}{\theta} \right).\end{align}

Separating this term and using (49) results in the following expression for (48):

(52) \begin{align}&\frac{\theta-iz}{\alpha}\sum_{j=1} ^\infty \frac{1 }{ j(\alpha j-1) } \left( \frac{-\lambda }{(\theta- iz)^\alpha} \right)^j - \frac{\theta}{\alpha}\sum_{j=1} ^\infty \frac{1 }{ j(\alpha j-1) } \left( \frac{-\lambda }{\theta^\alpha} \right)^j\nonumber \\ & - \frac{iz}{\alpha} \sum_{j=1} ^\infty \frac{1}{j} \left( \frac{-\lambda }{\theta^\alpha} \right)^j +i z +\log\!\left(1-\frac{iz}{\theta} \right)\left(\theta- i z \right). \end{align}

Observe that for $y \in \mathbb C$ , $|y|<1$ it holds that

(53) \begin{align}&\sum_{j=1} ^\infty \frac{y^j }{ j(\alpha j-1) }=\alpha \sum_{j=1} ^\infty \frac{y^j }{(\alpha j-1) }-\sum_{j=1} ^\infty \frac{y^j }{j } = \sum_{j=0} ^\infty \frac{y^{j+1} }{( j+\frac{\alpha-1}{\alpha}) } + \log(1-y) \nonumber \\ =& y \, \Phi \left(y,1,\frac{\alpha-1}{\alpha}\right)+ \log(1-y). \end{align}

Replacing (53) in (52), and recalling (30), the case $\gamma=1$ specializes to

(54) \begin{align} \int_0^\infty & \left(e^{i z x}-1 - i z x \right) q(x)x^{-\gamma-1} dx= \nonumber \\ &\frac{\lambda}{\alpha} \left( \frac{\Phi\left({-}\frac{\lambda}{\theta^{\alpha}}, 1, \frac{\alpha-1}{\alpha}\right) }{\theta^{\alpha-1}}- \frac{\Phi\left({-}\frac{\lambda}{(\theta-iz )^{\alpha}}, 1, \frac{\alpha-1}{\alpha}\right) }{(\theta-iz)^{\alpha-1}} \right) + \frac{\theta-i z}{\alpha} \log\!\left(\frac{(\theta-i z)^\alpha+ \lambda}{(\theta-i z)^\alpha}\right) -\nonumber \\& \frac{\theta}{\alpha}\log\! \left(\frac{\theta^\alpha+ \lambda}{\theta^\alpha}\right) + iz+ \frac{\theta-i z}{\alpha}\log\! \left(\frac{(\theta -iz)^\alpha}{\theta^\alpha}\right) + \frac{i z}{\alpha}\log\! \left(\frac{\theta^\alpha+\lambda}{\theta^\alpha}\right) \nonumber \\ =&\frac{\lambda}{\alpha} \left( \frac{\Phi\left({-}\frac{\lambda}{\theta^{\alpha}}, 1, \frac{\alpha-1}{\alpha}\right) }{\theta^{\alpha-1}} - \frac{\Phi\left({-}\frac{\lambda}{(\theta-iz)^{\alpha}}, 1, \frac{\alpha-1}{\alpha}\right) }{(\theta-iz)^{\alpha-1}} \right) + \frac{\theta-i z}{\alpha} \log\! \left(\frac{(\theta-i z)^\alpha+ \lambda}{\theta^\alpha+ \lambda }\right)+i z . \end{align}

Integrating the obtained expressions (47), (50), and (54) in $\sigma(du)$ as in (39), subtracting $i z (\mu_0-\mu)$ to (47), and adding $iz (\mu_1+\mu)$ to (54)–(50), $(i)-(iii)$ follow for all $|z|<\theta$ and $\lambda < \theta^{\alpha}$ .

We must finally prove the analyticity of $\Psi_X$ on S and that the constraint $\lambda <\theta^\alpha$ can be lifted. We know that $\Psi_X$ can be analytically continued on a horizontal strip $\{z \in \mathbb C\,:\, -a < \mbox{Im}(z) < b \}$ , $a, b \in \mathbb R \cup \{\infty\}, a \neq b $ (see [Reference Lukacs and Szász41]), if $E[e^{-c X}]<\infty$ , for all $c \in ({-}a,b)$ . In turn — by, e.g. [Reference Sato53, Corollary 25.8] — this is equivalent to $\int_{\{|x| > 1\}}e^{-c x}m(x)dx <\infty$ for all such c. In our instance we thus have $a =\theta_+,\ b= \theta_-$ . Since the expression for $\Psi_X$ has been found on the real line, that (42)–(44) also provide the analytic continuation of $\Psi_X$ on S would then follow from the identity principle of analytic functions, if we can show that such expressions are analytic on S.

Assume first $\gamma=0$ . The function $y \mapsto y^\alpha$ is analytic outside the logarithm branch cut $\mathbb C \setminus ({-}\infty, 0]$ yielding the condition $(\theta_+-iz) \notin ({-}\infty, 0]$ , i.e. Im $(z) \neq \theta_+$ . Hence we obtain, $S^+\,:\!=\{z \in \mathbb C: \mbox{Im}(z)>-\theta_+ \}$ . Furthermore, the logarithm branch cut is never crossed by the function $\frac{\theta_+^{\alpha_+}+\lambda_+}{(\theta_+-iz)^\alpha+\lambda_+}$ , so that $\log\! \left(\frac{\theta_+^{\alpha_+}+\lambda_+}{(\theta_+-iz)^{\alpha_+}+\lambda_+}\right)$ is analytic on $S^+$ . The same argument applies to $(\theta_-+iz)$ , which is analytical on $S^-\,:\!=\{z \in \mathbb C: \mbox{Im}(z)<\theta_- \}$ , and hence $\psi_X$ is analytical on S, and so is $\Psi_X$ .

For $\gamma \in (0,2)$ , $\gamma \neq 1$ , we exploit the fact that it is known (e.g. [Reference Karp and Prilepkina27]) that $_2 R_1$ can be analytically continued on $\mathbb C \setminus [1, \infty)$ . On the other hand, on the regions $S^{\pm}$ , the functions $-\lambda_\pm/(\theta_\pm \pm iz)^{\alpha_\pm}$ never attain real positive values, again yielding analiticity of $\psi_X$ on S. Lerch’s transcendent can also be analytically continued on $\mathbb C \setminus [1,\infty)$ , e.g. [Reference Guillera and Jonathan23, Lemma 2.2], and then the claim also follows for the case $\gamma=1$ . That the condition $\lambda < \theta^\alpha$ can be lifted also follows from the arguments above, since the parameter ranges of $\lambda, \theta, \alpha$ always lie in the domain of analyticity of the continued functions.

Proof. Denote $m(x)=m_+(x)+m_-(x)$ the positive and negative parts of the Lévy density. That $k^X_1=\mu+\mu_1$ for all i.d. distribution is well known (e.g [Reference Sato53, Example 25.12]). In the case $\theta_+, \theta_->0$ we apply that, for analytic distributions, cumulants and Lévy moments coincide when $n>1$ (again [Reference Sato53, Chapter 25]). In our case $k^X_n=\int_{\mathbb R} x^n m(x)dx $ , and for $n \geq 1$ , $\gamma \neq 1$ , and $\theta_+^{\alpha_+} >\lambda_+$ it holds that

(80) \begin{align}\int_0^\infty x^n m_+( dx) &= \delta_+ \int_0^\infty e^{-\theta_+ x} x^{n-\gamma_+-1} E_{\alpha_+} \left( -\lambda_+ \, x^{\alpha_+} \right) dx \nonumber \\&= \delta_+ \theta_+^{n-\gamma_+} \sum_{j=0}^\infty \left(\frac{-\lambda_+}{\theta_+^{\alpha_+}} \right)^j \frac{\Gamma(n-\gamma_++ \alpha_+ j )}{\Gamma(1+ \alpha_+ j)}\end{align}

leading, together with the analogous calculation for $m_-$ , to (76)–(77). The case for arbitrary parameters follows by analytic continuation. Because of the analyticity of $\psi_X$ , for $\gamma \neq 1$ the conclusion is also immediate from Lemma 1(i,ii), since we can differentiate $\psi$ in z and take the limit $z \rightarrow 0$ to obtain the cumulants.

Regarding Y, again because of [Reference Sato53, Corollary 25.8], the first statement follows from $\int_{\{ |x|>1 \}}x^k m_\pm(x)dx <\infty$ if and only if $k <\alpha_\pm +\gamma_\pm$ , which once again is a consequence of (26). For the variance it is possible to differentiate twice the characteristic exponent (68) and take the limit $z \rightarrow 0$ . To this end, since $2-\gamma_+<\alpha_+$ we can apply Lemma 1 with $b=\alpha_+$ , $a=-\gamma_+$ , $c=\lambda_+$ , and z replaced by $-i z$ , so that by part (i) it holds that

(81) \begin{equation}\mbox{Var}[Y^+]=\psi''(0)=\frac{\delta_+\Gamma(2-\gamma_+) }{ ({-}iz)^{2-\gamma_+}} \, _2R_1\left(1, 2-\gamma_+ , 1, \alpha_+; - \frac{ \lambda_+ }{ ({-}i z)^{\alpha_+}} \right)\end{equation}

and then, using part (ii),

(82) \begin{align}k^{Y^+}_2&= \delta_+ \lim_{z \rightarrow 0} \frac{\delta_+\Gamma(2-\gamma_+) }{ ({-}iz)^{2-\gamma_+}} \, _2R_1\left(1, 2-\gamma_+ , 1, \alpha_+; - \frac{ \lambda_+ }{ ({-}i z)^{\alpha_+}} \right)\nonumber\\&= \delta_+ \frac{\lambda_+^{\frac{\gamma_+-2}{\alpha_+}}}{\alpha_+}\frac{\pi}{\sin\left( \pi \frac{2-\gamma_+}{\alpha_+}\right)\Gamma(\gamma_+-1)} \nonumber \\ &= \delta_+ \ell(\gamma_+-2,\alpha_+, \lambda_+).\end{align}

The same arguments apply to $k^{Y^-}_2$ , and the proof is finished.

Proof. By, for example, [Reference De Oliveira, Mainardi and Vaz13, Equation (2.16)], the density in the Bernstein representation of $E_\alpha({-}\cdot^\alpha)$ is

(88) \begin{equation}s^E(x)=\frac{x^{\alpha-1}}{\pi }\frac{ \sin\!(\alpha \pi )}{ x^{2\alpha}+ 2 x^{\alpha} \cos(\alpha \pi)+1}, \qquad x>0,\end{equation}

which is the p.d.f. of the ratio of two independent $\alpha$ -stable r.v.s (see [Reference James25, p. 9]). After an application of the Laplace transform rules we see that the tempering function $q(x)=q(x,1){\unicode{x1D7D9}}_{\{x>0\}}+q(|x|, -1){\unicode{x1D7D9}}_{\{x <0 \}} $ is given by

(89) \begin{equation}q(x)=\left\{ \begin{array}{lr}\displaystyle{\int_0^\infty e^{-y x} \lambda_+^{-1/\alpha_+}s^{E}\big( (y-\theta_+) \lambda_+^{-1/\alpha_+}\big)}dy & x > 0, \\ \\\displaystyle{\int_0^\infty e^{-y |x|} \lambda_-^{-1/\alpha_-}s^{E}\big( (y-\theta_-) \lambda_-^{-1/\alpha_-}\big)} dy & x< 0. \end{array} \right.\end{equation}

Substituting (88) in (89) and using this in (16) with $\sigma$ as in (36) we obtain (86). Moreover, writing explicitly (17), for $B \in \mathcal B(\mathbb R_+)$ we have, using the integral substitution $y=1/x$ ,

(90) \begin{align}R(B)&= \delta_+ \int_{\theta_+}^\infty {\unicode{x1D7D9}}_{B}\left(\frac{\mbox{sgn}(x)}{x} \right) \frac{x^\gamma}{\pi }\frac{ (x-\theta_+)^{\alpha_+-1} \sin\!(\pi \alpha_+ )}{ \lambda_+^{-1} (x-\theta_+)^{2\alpha_+}+ 2 (x-\theta_+)^{\alpha_+} \cos(\alpha_+ \pi)+\lambda_+} dx \nonumber \\& =\delta_+\int_0^{1/\theta_+}\! {\unicode{x1D7D9}}_{B}(y) \frac{ y^{-\gamma-2}}{\pi }\frac{(y^{-1}-\theta_+)^{\alpha_+-1} \sin\!(\pi \alpha_+ )}{ \lambda_+^{-1} (y^{-1}-\theta_+)^{2\alpha_+}+ 2 (y^{-1}-\theta_+)^{\alpha_+} \cos(\alpha_+ \pi)+\lambda_+} dy\nonumber \\ &=\delta_+\int_0^{1/\theta_+} \! {\unicode{x1D7D9}}_{B}(y) \frac{ y^{-\gamma-\alpha_+-1}}{\pi }\frac{(1{-}\theta_+ y)^{\alpha_+-1} \sin\!(\pi \alpha_+ )}{ \lambda_+^{-1} y^{-2 \alpha_+}(1{-}\theta_+ y )^{2\alpha_+}{+} 2 y^{-\alpha_+}(1{-}\theta_+ y)^{\alpha_+} \cos(\alpha_+ \pi){+}\lambda_+} dy \nonumber \\ & = \delta_+ \int_0^{1/\theta_+} \!{\unicode{x1D7D9}}_{B}(y) \frac{ y^{-\gamma+\alpha_+-1}}{\pi }\frac{(1-\theta_+ y)^{\alpha_+-1} \sin\!(\pi \alpha_+ )}{ \lambda_+^{-1} (1-\theta_+ y )^{2\alpha_+}+ 2(y(1-\theta_+ y))^{\alpha_+} \cos(\alpha_+ \pi)+\lambda y^{2 \alpha_+}} dy.\end{align}

The analogous computation holds for the negative part when $B \in \mathcal B(\mathbb R_-)$ , and (87) follows.

Proof. Let $r_\gamma$ be the Rosińsky density (87). For (i) from [Reference Rosiński51, Theorem 3.1], a sufficient condition for the statement to hold is that

(94) \begin{equation}\int_{\mathbb R} x^\gamma r_\gamma(x;\, {\boldsymbol \alpha}, {\boldsymbol \lambda},{\boldsymbol \theta},{\boldsymbol \delta})dx <\infty.\end{equation}

When $\theta_+,\theta_->0$ , the above is trivially verified since in that case $r_\gamma$ is supported on a bounded set. When $\theta_+=\theta_-=0$ , then using (91) with $p=\gamma$ , one has $x^\gamma r_\gamma(x) \sim O(|x|^{-1-\alpha})$ , as $x \rightarrow \pm \infty$ so that (94) still holds. Notice that convergence in distribution can be strengthened to (93), since such is the conclusion of [Reference Kallenberg26, Theorem 15.17], used in the proof of [Reference Rosiński51, Theorem 3.1].

Now to show $(ii)-(iii)$ we begin by proving the Gaussian limit in (iii) under the assumption $\theta_ + , \theta_->0$ . Denote by $\psi_h(z)$ the characteristic exponent of the Lévy process $X^h$ . By the mentioned result [Reference Kallenberg26, Theorem 15.17], for the claim to hold it is sufficient to show convergence in distribution which we verify on the characteristic exponents. Write the decomposition of $\psi_h$ in spectrally positive and negative parts as $\psi_h=\psi_h^++\psi_h^-$ , Now we can use the integral (48) since $h^{-1} \sim 0$ , which yields, after interchanging the summation order,

(95) \begin{align}\lim_{h \rightarrow \infty}h^{-1/2}\psi^+_h(z )&=\lim_{h \rightarrow \infty}h\psi^+(z h^{-1/2}) \nonumber \\ &=\theta_+^{\gamma} \delta_+ \lim_{h \rightarrow \infty} h \, \sum_{j=0}^\infty \left(\frac{-\lambda_+}{\theta_+^{\alpha}} \right)^j \frac{1}{\Gamma(1+ \alpha j)} \sum_{k=2}\frac{\left(\frac{i z h^{-1/2}}{\theta_+}\right)^k}{k !} \Gamma( k-\gamma + \alpha j ) \nonumber \\ &= \delta_+ \lim_{h \rightarrow \infty} h \, \theta_+^{\gamma_+-2} \sum_{j=0}^\infty \left(\frac{-\lambda_+}{\theta_+^{\alpha}} \right)^j \frac{ \Gamma( 2-\gamma + \alpha j ) }{\Gamma(1+ \alpha j)}\left( - \frac{z^2 h^{-1}}{2}\right)+o(1) \nonumber \\ &= -\frac{z^2}{2}\delta_+ \theta_+^{\gamma-2} \Gamma(2-\gamma) \, _2R_1\left(1, 2-\gamma, 1;\;\alpha, -\frac{\lambda_+}{\theta_+^{\alpha}}\right).\end{align}

After carrying out the corresponding computation for $\psi^-_h$ , recalling Proposition 2, we notice that in the final expression the factor Var $[X_1]$ appears as multiplying the characteristic exponent of the standard Brownian motion, which establishes the claim.

To prove (ii) and (iii) when $\theta_+=\theta_-=0$ , we proceed by setting $\kappa>0$ and analyze $\psi_h(z h^{-1/k})=h \psi(z h^{-1/\kappa})$ . We must expand (67) and (68) around $h^{-1} \sim 0$ ; in order to do this we can use [Reference Kilbas, Saxena, Saigo and Trujillo28, Theorem 5.2], providing a series representation for the $_2 R_1$ function of complex argument outside the unit circle. Under our parameter specification it holds (recall also footnote 1)

(96) \begin{align}&\Gamma({-}\gamma)\, _2 R_1\left(1, -\gamma, 1, \alpha;\; w \right)= \nonumber \\ &- \sum_{k=0}^{\infty} \frac{\Gamma({-}\gamma-\alpha(k+1))}{\Gamma(1-\alpha(k+1))} (w^{-k-1})\nonumber\\&+ \frac{({-}w^{\gamma/\alpha})}{\alpha }\sum_{k=0}^\infty \frac{\Gamma( (k-\gamma)/\alpha) \Gamma(1-(k-\gamma)/\alpha)}{k! \Gamma(1+\gamma-k)}({-}1)^k ({-}w)^{-k/\alpha} \nonumber \\ &= - \sum_{k=0}^{\infty} \frac{\Gamma({-}\gamma-\alpha(k+1))}{\Gamma(1-\alpha(k+1))} (w^{-k-1})+ \sum_{k=0}^\infty \frac{({-}1)^k}{k!} \ell(\gamma -k , \alpha , -w)\end{align}

after using Euler’s summation and recalling that $\ell(\cdot, \cdot, \cdot)$ is given by (69). Setting $w=\frac{-\lambda_+}{h^{-\alpha/\kappa}|z|^{\alpha}\Theta_+(z)}$ , we obtain further

(97) \begin{align}&\Gamma({-}\gamma)h^{-\gamma/\kappa}\, _2 R_1\left(1, -\gamma, 1, \alpha; \frac{-\lambda_+h^{\alpha/\kappa} }{|z|^{\alpha}\Theta_+(z)} \right)\nonumber \\ &= - \sum_{k=0}^{\infty} \frac{\Gamma({-}\gamma-\alpha(k+1))}{\Gamma(1-\alpha(k+1))} \left(\frac{ ({-} iz)^{\alpha}}{-\lambda_+ }\right)^{k+1}h^{-(\gamma+\alpha(k+1))/\kappa}\nonumber\\&\quad+ \sum_{k=0}^\infty \frac{({-}1)^k}{k!} h^{(\gamma-k)/\kappa} \ell\left(\gamma -k , \alpha , \frac{\lambda_+}{({-}iz)^{\alpha}} \right).\end{align}

If $\alpha + \gamma<1$ , observing that $\ell\left(\gamma-k , \alpha , \frac{\lambda_+}{({-}iz)^{\alpha}} \right)=({-}iz)^{-\gamma}\ell(\gamma-k, \alpha, \lambda_+)$ in the positive part of (67) with $\mu=-\mu_0$ we obtain,

(98) \begin{align}&\delta_+\Bigg( \Gamma({-}\gamma)|z|^\gamma \Theta_+(z)h^{-\gamma/\kappa}\, _2 R_1\left(1, -\gamma, 1, \alpha; \frac{-\lambda_+h^{\alpha/\kappa} }{|z|^{\alpha}\Theta_+(z)} \right)- \ell(\gamma,\alpha, \lambda_+) \Bigg) \nonumber \\ &= \delta_+\Bigg( - ({-}iz)^{\gamma} \sum_{k=0}^{\infty} \frac{\Gamma({-}\gamma-\alpha(k+1))}{\Gamma(1-\alpha(k+1))} \left(\frac{ ({-} iz)^{\alpha}}{-\lambda_+ }\right)^{k+1}h^{-(\gamma+\alpha(k+1))/\kappa} \nonumber \\ & \phantom{xxxxxxxxxxxxxxxxxxxxxxxxxxx}+ \sum_{k=1}^\infty \frac{({-}1)^k}{k!} h^{-k/\kappa} \ell\left(\gamma -k , \alpha , \frac{\lambda_+}{({-}iz)^{\alpha}} \right) \Bigg). \end{align}

The leading order in (98) corresponds to the term $k=0$ of the first series. We thus have

(99) \begin{align}& \lim_{h \rightarrow \infty} h \, \psi^+(h^{-1/\kappa})= \delta_+ \lim_{h\rightarrow \infty} h \frac{\Gamma({-}\gamma-\alpha)}{\Gamma(1-\alpha)\lambda_+ }({-}i z)^{\alpha +\gamma} h^{-(\gamma +\alpha)/\kappa} = \delta_+ \frac{\Gamma({-}\gamma-\alpha)}{\Gamma(1-\alpha)\lambda_+ }({-}i z)^{\alpha + \gamma},\end{align}

with the last equality holding if and only if $\kappa=\alpha +\gamma$ . Now it is well known that (e.g. [Reference Sato53, Lemma 14.11])

(100) \begin{equation}({-}iz)^{\alpha+\gamma}\Gamma({-}\alpha-\gamma)=\int_0^{\infty} \frac{(e^{izx}-1)}{x^{1+\gamma+\alpha}}dx \qquad \alpha+\gamma \in (0,1), \, z \in \mathbb R,\end{equation}

which is the characteristic exponent of a spectrally positive $\alpha+\gamma$ stable r.v. with unit $\delta_+$ and $\mu=\mu^*_0$ . Substituting in (99) produces the positive part of the characteristic exponent of Z. Repeating the above for $\psi^-_h$ yields (iii) when $\alpha+\gamma \in (0,1)$ .

If, instead, $\alpha +\gamma \in (1,2)$ we must use (68) with $\mu=-\mu_1$ and in (97) we have, from the relation $\ell\left(\gamma-1 , \alpha , \frac{\lambda_+}{({-}iz)^{\alpha}} \right)=({-}iz)^{1-\gamma}\ell(\gamma-1, \alpha, \lambda_+)$ , that

(101) \begin{align}&\delta_+ \Bigg( \Gamma({-}\gamma)|z|^\gamma \Theta_+(z)h^{-\gamma/\kappa}\, _2 R_1\left(1, -\gamma, 1, \alpha; \frac{-\lambda_+h^{\alpha/\kappa} }{|z|^{\alpha}\Theta_+(z)} \right)- \ell(\gamma,\alpha, \lambda_+)\nonumber \\ &\phantom{xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx}- i z h^{-1/\kappa}\ell(\gamma-1,\alpha, \lambda_+) \Bigg) \nonumber \\ &= \delta_+ \Bigg({-} ({-}iz)^{\gamma} \sum_{k=0}^{\infty} \frac{\Gamma({-}\gamma-\alpha(k+1))}{\Gamma(1-\alpha(k+1))} \left(\frac{ ({-} iz)^{\alpha}}{-\lambda_+ }\right)^{k+1}h^{-(\gamma+\alpha(k+1))/\kappa}+ \nonumber \\& \phantom{xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx} \sum_{k=2}^\infty \frac{({-}1)^k}{k!} h^{-k/\kappa} \ell\left(\gamma -k , \alpha , \frac{\lambda_+}{({-}iz)^{\alpha}} \right)\Bigg).\end{align}

Now, when $\alpha + \gamma \in (0,2)$ , again the $k=0$ term in the first series leads, and the statement follows as in (99), this time observing that

(102) \begin{equation}({-}iz)^{\alpha+\gamma} \Gamma({-}\alpha-\gamma)=\int_0^{\infty} \frac{(e^{izx}-1-izx)}{x^{1+\gamma+\alpha}}dx \qquad \alpha+\gamma \in (1,2), \, z \in \mathbb R,\end{equation}

which is again the characteristic exponent of a spectrally positive $\alpha+\gamma$ stable r.v. and unit $\delta_+$ , but this time with $\mu=-\mu^*_1$ . This concludes the proof of (ii).

Finally, if $\alpha +\gamma>2$ in (97), then the leading order corresponds to the term $k=2$ in the second series, so that

(103) \begin{align}\lim_{h \rightarrow \infty} h \, \psi^+(h^{-1/\kappa})&= \delta_+ \lim_{h \rightarrow \infty} h \frac{h^{-2/\kappa}}{2} ({-}iz)^{\gamma} \ell\left(\gamma-2, \alpha,\lambda_+ \right)({-}iz)^{2-\gamma}\nonumber \\&=-\frac{ z^2}{2} \delta_+ \ell \left(\gamma-2, \alpha,\lambda_+ \right)\end{align}

provided that $\kappa=2$ . Together with the analogous computation for $\psi^-_h$ , this completes the proof of (iii) and of the proposition.

Proof. Indicating q the GRGTS tempering function in Cartesian coordinates, from [Reference Rosiński51, Theorem 4.1], we have that a necessary and sufficient condition for the absolute continuity of a TS $_\gamma$ Lévy process with respect to the given stable one is

(106) \begin{equation}\int_{\{|x|<1\}} (1-q(x))^2 x^{-\gamma-1} dx <\infty.\end{equation}

That is, for $x>0$

(107) \begin{equation}\int_0^1 \left(1- e^{-\theta_+ x}E_{\alpha_+} ( -\lambda_+ \, x^{\alpha_+}) \right)^2 dx < \infty.\end{equation}

We have, when $x \sim 0$ and using (27),

(108) \begin{equation}e^{-\theta_+ x} \sim 1-\theta_+ x , \qquad E_{\alpha_+} ( - \lambda_+ \, x^{\alpha_+}) \sim 1 - \frac{\lambda_+ x^{\alpha_+}}{\Gamma(\alpha_+ +1)},\end{equation}

and hence

(109) \begin{equation}1- e^{-\theta_+ x}E_{\alpha_+} ( -\lambda_+ \, x^{\alpha_+}) \sim \theta_+ x + \frac{\lambda_+ x^{\alpha_+}}{\Gamma(\alpha_+ +1)}, \qquad x \sim 0.\end{equation}

Since $\alpha_+ <1$ this implies the following leading order for $x \sim 0$ ,

(110) \begin{equation}\left(1- e^{-\theta_+ x}E_{\alpha_+} ( -\lambda_+ \, x^{\alpha_+}) \right)^2 \sim \frac{\lambda_+^2 x^{2 \alpha_+ }}{\Gamma(\alpha_+ +1)^2}.\end{equation}

Comparing with (106) we see that the condition for convergence is $\alpha_+> \gamma /2$ . The same calculation on $-1<x<0$ in (106) establishes the claim.

Proof. According to [Reference Sato53, Theorems 33.1, 33.2], $P \sim P'$ if and only if the given relation between $\mu$ and $\mu'$ holds, and the Hellinger distance between the absolutely continuous Lévy measures m(x)dx and $m^{\prime}$ (x)dx is finite, that is

(115) \begin{equation}\int_{\mathbb R} \left(\sqrt{m(x)}-\sqrt{m'(x)} \right)^2 dx <\infty.\end{equation}

Furthermore, $l(x)=\log(m'(x)/m(x))$ , and then (114) is clear by (39), once the first assertion is proved.

Now letting $I(x)=\left(\sqrt{m(x)}-\sqrt{m'(x)} \right)^2$ and using (26), for large x we have

(116) \begin{align}I(x) \sim \left( \sqrt{\frac{\delta_+}{\lambda_+\Gamma(1-\alpha_+) x^{\gamma_+ +\alpha_++1}} }e^{-x\theta_+/2} - \sqrt{\frac{\delta'_+}{\lambda'_+ \Gamma(1-\alpha'_+) x^{\gamma'_+ +\alpha'_++1}} }e^{-x\theta'_+/2} \right)^2,\end{align}

which is always integrable at $+\infty$ , whatever the value of $\theta_+$ by (26). The corresponding convergence holds for large negative x.

In a right neighborhood of 0 we have the condition

(117) \begin{align}\int_0^1 &\left( \sqrt{\delta_+ \frac{E_{\alpha_+}({-}\lambda_+ x^{\alpha_+})}{x^{1+\gamma_+}}}e^{-x\theta_+/2} - \sqrt{\delta'_+ \frac{E_{\alpha'_+}({-}\lambda'_+ x^{\alpha'_+})}{x^{1+\gamma'_+}}}e^{-x\theta'_+/2}\right)^2 dx < \infty.\end{align}

We write

(118) \begin{equation}I(x) = \delta'_+ \frac{e^{-\theta_+' x}E_{\alpha_+'}({-}\lambda_+' x^{\alpha_+'})}{ x^{\alpha_+'+1}}\left(x^{\frac{\gamma_+'-\gamma_+}{2}} \sqrt{\frac{\delta_+ E_{\alpha_+}({-}\lambda_+ x^{\alpha_+}) }{\delta'_+ E_{\alpha_+'}({-}\lambda_+' x^{\alpha_+'})} } e^{\frac{-\theta_+-\theta_+'}{2}x} -1 \right)^2, \quad x>0. \end{equation}

Assuming $\gamma'_+<\gamma_+$ and $\delta_+ \neq \delta_-$ , since $\exp\!(\!\cdot\!)\sim E_\alpha(\!\cdot\!) \sim 1$ as $x \rightarrow 0^+$ , we have that $I(x) \sim \delta_+ x^{-1-\alpha_+'-\gamma_+ +\gamma'_+}$ , which diverges. The case $\gamma_+ >\gamma'_+$ is similarly excluded. Therefore, for convergence we must have $\gamma_+=\gamma_+'$ . Once this is further assumed, if we allow $\delta_+ \neq \delta_+'$ then $I(x)\sim (\sqrt{\delta_+}-\sqrt{\delta_+'})^2x^{-1-\alpha'_+}$ , again a divergent integrand. Therefore, for convergence it is necessary that both $\gamma_+=\gamma_+'$ and $\delta_+=\delta_+'$ . Expanding the right-hand term in (118) in its McLaurin series using (27) we obtain, with $\alpha_*=\min\{\alpha_+, \alpha_+' \}$ ,

(119) \begin{align}I(x)\sim&\frac{ \delta_+}{x^{\gamma_++1}}\left( \left( \frac{1- \frac{\lambda_+ x^{\alpha_+}}{\Gamma(1+\alpha_+)}}{1-\frac{\lambda'_+ x^{\alpha'_+}}{\Gamma(1+\alpha'_+)}} \right)^{1/2} -1 \right)^2 \nonumber \\&\sim \frac{ \delta_+}{x^{\gamma_++1}}\left( \left(1- \frac{ \lambda_+ x^{\alpha_+}}{2\Gamma(1+\alpha_+)} \right)\left(1+ \frac{\lambda'_+ x^{\alpha'_+}}{2 \Gamma(1+\alpha'_+)} \right) -1 \right)^2 \nonumber \\ & \sim \frac{ \delta_+}{x^{\gamma_++1}} \left( c x^{\alpha_*} \right)^2 = \frac{ c^2 \delta_+}{x^{\gamma_++1 - 2\alpha_*}}\end{align}

for some $ c \neq 0$ . The convergence condition is thus $\gamma_+ +1 - 2\alpha_*<1$ , i.e. $\alpha_*>\gamma_+/2$ . Repeating in a left neighborhood of 0 proves the result.

Proof. According to the theory in [Reference Sato and Yamazato54, Reference Wolfe65, Reference Zolotarev68], the regularity/boundedness of the density depends on the values of the integral part of the critical parameter $\lambda=k(0+)+|k(0-)|$ , whose value coincides with the maximum differentiability order plus one of the p.d.f. on its support. Recall that k is given by (40).

First, when $\gamma_+\delta_+>0$ we have $k(0+)=\infty$ and hence by [Reference Sato and Yamazato54, Theorem 1.2], the smoothness of the density follows. Similarly, if $\delta_-\gamma_->0$ , noticing $k(0-)=\infty$ . This proves (i).

For (ii), observe first that in such case $\mu_0<\infty$ . Assume $\gamma_+=\gamma_-=0$ and $\delta_+, \delta_->0$ ; then $k(0+)=\delta_+$ , $|k(0-)|=\delta_-$ , so that $\lambda=\delta_++\delta_-$ and the first claim follows again by [Reference Sato and Yamazato54, Theorem 1.2]. The case in which either $\delta_+$ or $\delta_-$ is 0 is analogous. For the second statement, observe that in [Reference Sato and Yamazato54] classification X is now of type III. By [Reference Sato and Yamazato54, Theorem 1.5] then $f_X$ is unbounded around $\mu_*$ if and only if the law of X is of subtype III $_1$ or III $_2$ , which happens if and only if $\delta_++\delta_-=\lambda \leq 1$ . This shows (ii).

To show strict unimodality, denote $k^-(x)=k({-}x)$ the canonical density of $-X$ . According to [Reference Sato and Yamazato54, Theorem 1.4], for $f_X$ to fail to be strictly unimodal it is necessary and sufficent that both

(121) \begin{equation}k(0+)+|k(0-)| =k^-(0-)+|k^-(0+)|=\delta_+ +\delta_-= 1\end{equation}

and that for some $\xi, \xi_- >0$

(122) \begin{equation}k(\xi)=1, \qquad k^-(\xi_-)=1.\end{equation}

But in case (i) we know $k(0+)+|k(0-)|=\lambda=\infty$ so that (121) is not satisfied. In cases (ii) assume (121) so that both $k(0+), k^-(0+) \leq 1 $ . Since k and $k^-$ from (40) are strictly decreasing on the positive half line, no strictly positive roots as in (122) can exist. Therefore (121) and (122) cannot hold simultaneously, and (iii) follows.

Proof. Part (i) follows by inspection of equation (120), after the location shift $x \mapsto x-\mu_*$ , since $E^c_{a,b}(0)=\exp\!(0)=1$ . Moreover, observe as $x \sim0$ , $k(x) \sim c x^{-\gamma}$ , $c>0$ , which is integrable around zero since $\gamma<1$ . This implies that $f_X$ is of type I $_6$ in the classification in [Reference Sato and Yamazato54], and thus part (ii) follows from an application of [Reference Sato and Yamazato54, Theorem 5.2].

Proof. We drop the parameter subscripts throughout. Let m be the Lévy density of some given RGTS $_{\boldsymbol \alpha}({\boldsymbol \lambda}, {\boldsymbol \theta}, {\boldsymbol \delta};\, \mu )$ law and consider the normalization

(127) \begin{equation}m_1(dx)\,:\!=\frac{m(x) }{\overline m(1)}{\unicode{x1D7D9}}_{\{x>1\}}dx=\frac{\delta}{\overline m(1) x}e^{-\theta x}E_\alpha({-}\lambda x^{\alpha}) {\unicode{x1D7D9}}_{\{x>1\}}dx,\end{equation}

which is a probability law. According to [Reference Watanabe63, Theorem B, (1)–(2)], the stated convolution equivalence holds if and only if so does convolution equivalence of $m_1$ of same order. Using de l’Hopital’s rule, the fundamental theorem of calculus, and (26), for all $a \in \mathbb R$ it holds that

(128) \begin{align}& \lim_{x \rightarrow \infty} \frac{\overline{ m_1}(x+a)}{\overline{ m_1}(x)}= \lim_{x \rightarrow \infty} \frac{ \displaystyle{\int_{x+a}^\infty \frac{e^{-y \theta}}{y} E_{\alpha}({-}\lambda y^\alpha)dy} }{\displaystyle{\int_{x}^\infty \frac{e^{-y \theta}}{y} E_{\alpha}({-}\lambda y^\alpha)dy}}=\frac{e^{-(x+a) \theta} x}{(x+a)} \frac{E_{\alpha}({-}\lambda (x+a)^\alpha)}{ e^{-\theta x} E_{\alpha}({-}\lambda x^\alpha)} \nonumber \\=&e^{-\theta a}\lim_{x\rightarrow \infty}\left( \frac{x}{x+a} \right) \frac{E_{\alpha}({-}\lambda (x+a)^\alpha)}{ E_{\alpha}({-}\lambda x^\alpha)}=e^{-\theta a}\lim_{x\rightarrow \infty} \left(\frac{x+a}{x} \right)^{\alpha-1}=e^{-\theta a}\end{align}

verifying (125) with $\beta=\theta$ . If $\eta$ is the law of X, and $\theta>0$ , that $\hat \eta(\theta)=e^{\psi({-}i \theta)}<\infty$ , with $\psi$ as in Theorem 1, (ii) or (iii) can be verified directly, and if $\theta=0$ there is nothing to prove.

To show (126), proceeding as in [Reference Torricelli60, Equations (2.11)–(2.13)], we arrive at the equality

(129) \begin{equation} \lim_{x \rightarrow \infty} \frac{\overline{ m_1 *m_1}(x)}{\overline{ m_1}(x)}=\lim_{x \rightarrow \infty} \frac{ \delta}{\overline m(1)} \int_1^{x/2}\left( \frac{x} {(x-z)z}\right) \frac{E_\alpha({-}\lambda(x-z)^\alpha) E_\alpha({-}\lambda z^\alpha) }{E_\alpha({-}\lambda x^\alpha)} dz.\end{equation}

Again, referring to [Reference Torricelli60, Theorem 1], convolution equivalence of $m_1$ of order $\theta$ —and hence, by [Reference Watanabe63, Theorem B, (1)–(2)], that of order $\theta$ of the given RGTS r.v.—would follow by a dominated convergence argument if the last integrand is dominated by an integrable function of z for any parameter choice. Now, for $z \in (1,x/2)$ we have $x/(x-z) \leq 2$ . Let $k(y)=E_{\alpha}({-}\lambda y^\alpha)$ . Since k is c.m., as a consequence of [Reference Torricelli60, Proposition 1(ii)], for all fixed $ z \geq 1$ the ratio $k(x-z)/k(x)$ is decreasing in x for all $x >z$ so that $k(x-z)/k(x){\unicode{x1D7D9}}_{\{1 \leq z \leq x/2\}}$ has a maximum in $x=2 z$ . Combining with the usual asymptotic estimate (26) this results in

(130) \begin{equation}\left( \frac{x} {(x-z)z}\right) \frac{E_\alpha({-}\lambda(x-z)^\alpha) E_\alpha({-}\lambda z^\alpha) }{E_\alpha({-}\lambda x^\alpha)}{\unicode{x1D7D9}}_{\{1 \leq z \leq x/2\}} < \frac{2}{z} \frac{E_\alpha({-}\lambda z^\alpha)^2 }{E_\alpha({-} \lambda (2 z)^\alpha)} \sim \frac{2^{1-\alpha}}{\lambda \Gamma(1-\alpha)} {z ^{-1-\alpha}}\end{equation}

as $z \rightarrow \infty$ . Hence the integrand in (129) is bounded by an integrable function on $[1, \infty)$ for all values of $\lambda$ and $\alpha$ .

Proof. We assume $x \rightarrow \infty$ first and drop the subscript $+$ for notational convenience. From [Reference Torricelli60, Theorem 1] and Proposition 9 above, we know that the law of X is convolution equivalent of order $\theta$ for all $\gamma \in [0,2)$ . Denote by $\overline{F_X}$ the survival function of X and by $f_X$ and m respectively its p.d.f. and Lévy measure. Set $X_0$ be the GRGS $_{{\boldsymbol \gamma}}({\boldsymbol \alpha}, {\boldsymbol \lambda}, {\boldsymbol \delta};\,\mu)$ r.v. with Lévy density $m_0(x)=e^{\theta x} m(x)$ (possibly, $\theta=0$ , and $m_0=m$ ), and corresponding survival function $\overline{F_{0,X}}$ having tilted p.d.f. $f_{0,X}(x)=e^{\theta x -\psi({-}i \theta)} f_X(x)$ , with $\psi$ as in Theorem 1 if $\theta>0$ , replaced by 0, if $\theta=0$ . It is shown in [Reference Watanabe63, Theorem B, (3)] that convolution equivalence ensures that

(132) \begin{equation}\overline{F_{0,X}}(x) \sim \overline {m_0}(x) e^{\psi({-}i \theta)}= \overline {m_0}(x) , \qquad x \rightarrow +\infty,\end{equation}

where $\psi$ is that of Theorem 2 (but in view of the second equality, this information is irrelevant). Then for all $c>0$ , using (26) we obtain

(133) \begin{equation}\lim_{x \rightarrow \infty} \frac{m_0(c x)}{m_0(x)}=c^{-1-\alpha-\gamma},\end{equation}

that is, $m_0$ is regularly varying of order $-1-\alpha-\gamma$ . Therefore,

(134) \begin{equation} m_0(x)=x^{-1-\alpha-\gamma} \ell(x)\end{equation}

for some slowly varying function $\ell$ . By [Reference Feller15, VIII.9, Theorem 1(a)], applied with $p=0$ , and (134) we see that

(135) \begin{equation} \overline {m_0}(x) \sim \frac{x \, m_0(x)}{\gamma+\alpha}=\frac{ x^{-\alpha - \gamma}\ell(x)}{\gamma+\alpha}, \qquad x \rightarrow +\infty,\end{equation}

which replaced in (132) leads to

(136) \begin{equation}\overline{F_{0,X}}(x) \sim \frac{ x^{-\alpha - \gamma}\ell(x)}{\gamma+\alpha}, \qquad x \rightarrow +\infty.\end{equation}

Now by the monotone density theorem (MDT) in its version for the tail function (see [Reference Bingham, Goldie and Teugels5, Theorems 1.7.2, 1.7.2b], and the related discussion of the alternative versions), we have that if $U(x)=\int_x^\infty u(y)dy$ with u ultimately monotone, and $U(x)\sim c x^\rho \ell(x)$ , $x \rightarrow \infty$ , $\ell(x)$ slowly varying, $\rho <0$ , then $u(x) \sim -c \rho x^{\rho-1} \ell(x)$ , $x \rightarrow \infty$ . In the present case, based on (136), and because by Proposition 7 (iii), $f_{0,X}$ is ultimately monotone, we can apply the MDT with $U=\overline{F_{0,X}}$ , $u=f_{0,X}$ , $c=(\gamma+\alpha)^{-1}$ , $\rho=-\alpha-\gamma$ . Hence

(137) \begin{equation}f_{0,X}(x) \sim x^{-1-\alpha -\gamma} \ell(x), \qquad x \rightarrow \infty.\end{equation}

Again, in view of (26) and (39) one has

(138) \begin{equation}\ell(x) \sim \frac{\delta}{\Gamma(1-\alpha)}, \qquad x \rightarrow \infty, \end{equation}

which used in (137) proves (131) for $f_{0,X}$ .

Letting then $\theta>0$ , recalling the definition of $f_{0,X}$ , by what just proved it also follows

(139) \begin{equation}f_X(x)=f_{0,X}(x)e^{-\theta x +\psi({-}i \theta)}\sim \frac{\delta}{\Gamma(1-\alpha)} x^{-1-\alpha -\gamma} e^{-\theta x +\psi({-}i \theta)}\end{equation}

and we have shown (131) in its generality. The statement for $x \rightarrow -\infty$ follows by considering the convolution equivalent GRGTS $_{{\boldsymbol \gamma}}({\boldsymbol \alpha}, {\boldsymbol \lambda}, {\boldsymbol \theta}, {\boldsymbol \delta};\,\mu)$ r.v. $-X$ of order $\theta_-$ at $x \rightarrow + \infty$ .