2.1 Lévy processes

We write ${\bf x}=(x_1,\ldots,x_n)\in\mathbb{R}^n$ as a row vector. For $A\subseteq\mathbb{R}^n$ , let $A_*\;:\!=\; A\backslash\{{\bf 0}\}$ and let ${\bf 1}_A$ denote the indicator function for A. Let $\mathbb{D}\;:\!=\; \{{\bf x}\in\mathbb{R}^n\colon \|{\bf x}\|\le 1\}$ be the Euclidean unit ball centred at the origin. Let $\|{\bf x}\|^2_\Sigma\;:\!=\; {\bf x}\Sigma{\bf x}'$ , where ${\bf x}\in\mathbb{R}^n$ and $\Sigma\in\mathbb{R}^{n\times n}$ . Let $I\colon [0,\infty)\to[0,\infty)$ be the identity function.

For references on Lévy processes, see [Reference Bertoin2] and [Reference Sato14]. The law of an n-dimensional Lévy process ${\bf X}=(X_1,\ldots,X_n)=({\bf X}(t))_{t\ge 0}$ is determined by its characteristic function $\Phi_{\bf X}\;:\!=\; \Phi_{{\bf X}(1)}$ with

\begin{align*} \Phi_{{\bf X}(t)}(\boldsymbol{\theta})\,\;:\!=\; \,\mathbb{E}\exp({\textrm{i}}\langle \boldsymbol{\theta},{{\bf X}(t)}\rangle) = \exp(t\Psi_{\bf X}(\boldsymbol{\theta})), \quad t\ge0,\ \boldsymbol{\theta}\in\mathbb{R}^n,\end{align*}

and characteristic exponent

\begin{align*} \Psi_{\bf X}(\boldsymbol{\theta})\;:\!=\; {\textrm{i}} \langle {\boldsymbol{\mu}},\boldsymbol{\theta}\rangle - \frac 12\|\boldsymbol{\theta}\|^2_{\Sigma} +\int_{\mathbb{R}^n_*}({\textrm{e}}^{{\textrm{i}}\langle \boldsymbol{\theta}, {\bf x}\rangle} - 1 - {\textrm{i}}\langle \boldsymbol{\theta}, {\bf x}\rangle {\bf 1}_\mathbb{D}({\bf x}))\,\mathcal{X}({\textrm{d}} {\bf x}),\end{align*}

where $\boldsymbol{\mu}\in\mathbb{R}^n$ , $\Sigma\in\mathbb{R}^{n\times n}$ is a covariance matrix, and $\mathcal{X}$ is a Lévy measure, that is, a non-negative Borel measure on $\mathbb{R}^n_*$ such that $\int_{\mathbb{R}^n_*}(1\wedge \|{\bf x}\|^2)\,\mathcal{X}({\textrm{d}} {\bf x})<\infty$ . We write ${\bf X}\sim L^n(\boldsymbol{\mu},\Sigma,\mathcal{X})$ (or ${\bf X}\sim L^n$ for short) to mean that X is an n-dimensional Lévy process with characteristic triplet $(\boldsymbol{\mu},\Sigma,\mathcal{X})$ .

An n-dimensional Lévy process T with almost surely non-decreasing sample paths is called a subordinator, and it is denoted by ${\bf T} \sim S^n({\bf d},\mathcal{T}) \;:\!=\; L^n(\boldsymbol{\mu},0,\mathcal{T})$ (or ${\bf T}\sim S^n$ for short), where ${\bf d} \;:\!=\; \boldsymbol{\mu} - \int_{\mathbb{D}_*} {\bf t} \,\mathcal{T}({\textrm{d}} {\bf t})\in[0,\infty)^n$ is the drift, and $\mathcal{T}$ is the Lévy measure. The law of T is characterised by its Laplace exponent $\Lambda_{\bf T}$ , which satisfies $\mathbb{E}[\!\exp(-\langle {\boldsymbol{\lambda}},{{\bf T}(1)}\rangle)]=\exp(-\Lambda_{\bf T}(\boldsymbol{\lambda}))$ , $\boldsymbol{\lambda}\in[0,\infty)^n$ . For ${\bf w},{\bf z} \in\mathbb{C}^n$ , let $\langle {{\bf w}},{{\bf z}}\rangle \;:\!=\; \sum_{k=1}^n w_kz_k$ , noting that there is no conjugation. The domain of $\Lambda_{\bf T}$ can be extended, giving

(2) \begin{align} \Lambda_{\bf T}({\bf z})=\langle {{\bf d}},{{\bf z}}\rangle+\int_{[0,\infty)_*^n}(1-{\textrm{e}}^{-\langle {\bf z},{\bf t}\rangle}) \,\mathcal{T}({\textrm{d}} {\bf t}),\quad\Re{\bf z}\in[0,\infty)^n\end{align}

(see the proof of [Reference Barndorff-Nielsen, Pedersen and Sato1, Theorem 3.3]).

We have the following characterisation of piecewise constant Lévy processes from [Reference Sato14, Theorem 21.2].

2.2. Weak subordination

Let ${\bf X}\sim L^n$ and ${\bf T}\sim S^n({\bf d},\mathcal{T})$ . Let ${\bf t}=(t_1,\ldots,t_n)\in[0,\infty)^n$ and $\langle{(1),\ldots,(n)}\rangle$ be a permutation of $\{1,\ldots,n\}$ such that $t_{(1)} \leq \cdots \leq t_{(n)}$ , and define $\Delta t_{(k)}\;:\!=\; t_{(k)}-t_{(k-1)}$ , $1\leq k \leq n$ , with $t_{(0)}\;:\!=\; 0$ . Further, let $\boldsymbol{\pi}_{J}\colon \mathbb{R}^n\to\mathbb{R}^n$ be the projection onto the coordinate axes in $J\subseteq\{1,\ldots,n\}$ . For all ${\bf t}\in[0,\infty)^n$ , by [Reference Buchmann, Lu and Madan5, Proposition 2.1], the random vector ${\bf X}({\bf t})\;:\!=\; (X_1(t_1),\ldots, X_n(t_n))$ is infinitely divisible with characteristic exponent

(3) \begin{align}({\bf t}\diamond \Psi_{\bf X})(\boldsymbol{\theta})\;:\!=\; \sum_{k=1}^n\Delta t_{(k)} \Psi_{\bf X}(\boldsymbol{\pi}_{\{(k),\ldots,(n)\}}(\boldsymbol{\theta})),\quad \boldsymbol{\theta}\in\mathbb{R}^n, \end{align}

and characteristic function

(4) \begin{align} \Phi_{{\bf X}({\bf t})}(\boldsymbol{\theta}) = \exp({{\bf t} \diamond \Psi_{{\bf X}}(\boldsymbol{\theta})}).\end{align}

It is convenient to consider both the subordinator and the subordinated process together as a joint 2n-dimensional Lévy process. In this form, we can define weak subordination as follows (see [Reference Buchmann, Lu and Madan5, Proposition 3.1]).

Definition 1. The weak subordination of ${\bf X}\sim L^n$ and ${\bf T}\sim S^n({\bf d},\mathcal{T})$ is the joint Lévy process ${\bf Z}\stackrel{D}{=}({\bf T},{\bf X}\odot {\bf T})$ with characteristic exponent

(5) \begin{align} \Psi_{\bf Z}(\boldsymbol{\theta}) = {\textrm{i}}\langle {{\bf d}},{\boldsymbol{\theta}_1}\rangle+({\bf d}\diamond\Psi_{{\bf X}})(\boldsymbol{\theta}_2)+\int_{[0,\infty)^n_*} (\Phi_{({\bf t},{\bf X}({\bf t}))}(\boldsymbol{\theta})-1)\,\mathcal{T}({\textrm{d}} {\bf t}),\end{align}

where $\boldsymbol{\theta}=(\boldsymbol{\theta}_1,\boldsymbol{\theta}_2)$ , $\boldsymbol{\theta}_1,\boldsymbol{\theta}_2\in\mathbb{R}^n$ .

This is a valid characteristic exponent for a Lévy process. Weak subordination can equivalently be defined in terms of a characteristic triplet. From [Reference Buchmann, Lu and Madan5, Definition 2.1], if ${\bf d} ={\bf 0}$ , then ${\bf Z}\sim L^{2n}({\bf m},\Theta,\mathcal{Z})$ , where

(6) \begin{align} {\bf m} = \int_{\mathbb{D}_*} ({\bf t},{\bf x})\,\mathcal{Z}({\textrm{d}}{\bf t},{\textrm{d}}{\bf x}), \qquad\ \ \ \end{align}

(7) \begin{align} \Theta =0,\qquad\qquad\qquad\qquad\qquad\end{align}

\begin{align*} \mathcal{Z}({\textrm{d}}{\bf t},{\textrm{d}}{\bf x}) & = {\bf 1}_{[0,\infty)^n_*\times \mathbb{R}^n}({\bf t},{\bf x}) \mathbb{P}({\bf X}({\bf t})\in{\textrm{d}}{\bf x})\mathcal{T}({\textrm{d}}{\bf t}).\end{align*}

For additional details on weak subordination, see [Reference Buchmann, Lu and Madan5].

2.3. Poisson random measures

For references on Poisson random measures and their relationship to the jumps of Lévy processes, see [Reference Bertoin2], [Reference Çinlar6], and [Reference Kingman9]. A Poisson random measure (PRM) $\mathbb{Z}$ with intensity measure $\mu$ on a measurable space $(E,\mathcal{E})$ is a random measure such that $\mathbb{Z}(A)\sim \operatorname{Poisson}(\mu(A))$ for all $A\in\mathcal{E}$ , and $\mathbb{Z}(A_1),\ldots,\mathbb{Z}(A_m)$ are independent for all disjoint $A_1,\ldots,A_m\in\mathcal{E}$ . In general, a PRM has the form $\mathbb{Z}=\sum_{i=1}^\infty \boldsymbol{\delta}_{{\bf Z}_i}$ , where $\boldsymbol{\delta}_{{\bf Z}_i}$ , $i\in\mathbb{N}$ , is the Dirac measure at the random vector ${\bf Z}_i$ taking values in $(E,\mathcal{E})$ . Define the random variable $Z_f\;:\!=\; \int_{E} f({\bf x})\,\mathbb{Z}({\textrm{d}} {\bf x})$ , where f is a non-negative, $\mathcal{E}$ -measurable real function. The Laplace functional of $\mathbb{Z}$ is

(8) \begin{align} L(f)\;:\!=\; \mathbb{E} \big[ {\textrm{e}}^{-Z_f}\big] = \mathbb{E}\biggl[\prod_{i=1}^\infty \,{\textrm{e}}^{-f({\bf Z}_i)} \biggr] = \exp\biggl(-\int_E (1-{\textrm{e}}^{-f({\bf x})})\,\mu({\textrm{d}} {\bf x}) \biggr)\end{align}

(see [Reference Kingman9, equation (3.35)]). The Laplace functional is well-defined with $L(f)\in[0,1]$ , where this equality can be interpreted as 0 if $Z_f<\infty$ a.s. fails. Two PRMs are equal if their Laplace functionals are equal [Reference Çinlar6, Chapter VI, Proposition 1.4].

Let ${\bf X}\sim L^n(\boldsymbol{\mu},\Sigma,\mathcal{X})$ with $\mathcal{X}\neq 0$ . For a fixed sample path, a time t is a jumping time of X if the jump $\Delta {\bf X}(t)\;:\!=\; {\bf X}(t)-{\bf X}(t-)\neq {\bf 0}$ . The following result is [Reference Sato14, Theorem 21.3].

For the Lévy process X, the countable sequence of random vectors giving the time and size of the jumps, $({\bf Z}_{i})_{i\in\mathbb{N}} \;:\!=\; (t,\Delta {\bf X}(t))_{t>0,\Delta {\bf X}(t)\neq {\bf 0}}$ , is a Poisson point process. Consequently, $\mathbb{Z}=\sum_{i=1}^\infty \boldsymbol{\delta}_{{\bf Z}_i}$ is the PRM of X (or of the jumps of X), defined on the Borel space $([0,\infty)\times\mathbb{R}^n_*,\mathcal{B}([0,\infty)\times\mathbb{R}^n_*))$ with intensity measure ${\textrm{d}} t \otimes \mathcal{X}$ [Reference Bertoin2, Chapter I, Theorem 1].

3.1. Consistency of weak subordination for stacked univariate subordination

Here we show that the laws of weak and strong subordination coincide when the latter satisfies the stacked univariate subordination property in condition (). The proof follows along the lines of [Reference Buchmann, Lu and Madan5, Proposition 3.3].

Proof. Recall that $n_1+\cdots+n_d=n$ and let $\boldsymbol{\theta}=(\boldsymbol{\theta}_1,\boldsymbol{\theta}_2) = (\boldsymbol{\theta}_{11},\ldots, \boldsymbol{\theta}_{1d},\boldsymbol{\theta}_{21},\ldots, \boldsymbol{\theta}_{2d})$ , $\boldsymbol{\theta}_1,\boldsymbol{\theta}_2\in\mathbb{R}^n$ , $\boldsymbol{\theta}_{1m},\boldsymbol{\theta}_{2m}\in\mathbb{R}^{n_m}$ for all $1\leq m \leq d$ . Since T and X are independent processes, using (4) and conditioning on T, we get

(9) \begin{align} \Phi_{({\bf T},{\bf X}\circ {\bf T})}(\boldsymbol{\theta})=\mathbb{E}[\!\exp({\textrm{i}} \langle {\boldsymbol{\theta}_1},{{\bf T}(1)}\rangle+({\bf T}(1)\diamond\Psi_{{\bf X}})(\boldsymbol{\theta}_2))]. \end{align}

Let ${\bf e}=(1,\ldots,1)\in\mathbb{R}^n$ . Since T and X satisfy the stacked univariate subordination condition for n-dimensional processes, the subordinator $({\bf T},{\bf T})$ and the subordinate $(I{\bf e},{\bf X})$ satisfy this condition for 2n-dimensional processes. Thus $(I{\bf e},{\bf X})\circ ({\bf T},{\bf T})= ({\bf T},{\bf X}\circ {\bf T})$ is a Lévy process by [Reference Barndorff-Nielsen, Pedersen and Sato1, Theorem 3.3], so it suffices to show that $\Psi_{({\bf T},{\bf X}\circ {\bf T})}=\Psi_{({\bf T},{\bf X}\odot {\bf T})}$ .

Noting that ${\bf X} = ({\bf Y}_1,\ldots, {\bf Y}_d)$ , where ${\bf Y}_1\sim L^{n_1},\ldots, {\bf Y}_d\sim L^{n_d}$ are independent Lévy processes, Kac’s theorem gives

\begin{align*} \Psi_{\bf X}(\boldsymbol{\theta}_2) =\sum_{m=1}^d \Psi_{{\bf Y}_m}(\boldsymbol{\theta}_{2m}). \end{align*}

Form the partition $\{1,\ldots,n\}= J_1\cup\cdots \cup J_d$ , where

\begin{equation*} J_1 \;:\!=\; \{1,\ldots,n_1\}, J_2 \;:\!=\; \{n_1+1,\ldots,n_1+n_2\},\ldots, J_d \;:\!=\; \{n_1+\cdots+n_{d-1}+1, \ldots, n\}. \end{equation*}

Let ${\bf r}=(r_1,\ldots,r_d)\in[0,\infty)^d$ and $\langle(1),\ldots,(d)\rangle$ be a permutation of $\{1,\ldots, d\}$ such that $r_{(1)}\leq \cdots \leq r_{(d)}$ . Define the projections $\boldsymbol{\pi}_m\;:\!=\; \boldsymbol{\pi}_{J_{(m)} \cup\cdots \cup J_{(d)}}$ , $1\leq m \leq d$ . Thus, for all $1\leq m\leq d$ ,

(10) \begin{align} \Psi_{\bf X}(\boldsymbol{\pi}_m(\boldsymbol{\theta}_2)) =\sum_{k=m}^d \Psi_{{\bf Y}_{(k)}}(\boldsymbol{\theta}_{2(k)}). \end{align}

Next, due to (1), we can write ${\bf T}= {\bf R} A$ for some $A\in\mathbb{R}^{d\times n}$ , where ${\bf R}= (R_1,\ldots,R_d)\sim S^d({\bf d},\mathcal{R})$ . Then (3) and (10) give

(11) \begin{align} ({\bf r} A)\diamond \Psi_{{\bf X}}(\boldsymbol{\theta}_2) &= \sum_{m=1}^d (r_{(m)}-r_{(m-1)})\Psi_{{\bf X}}(\boldsymbol{\pi}_{m}(\boldsymbol{\theta}_2))\notag\\* &= \sum_{m=1}^{d-1} r_{(m)}(\Psi_{{\bf X}}(\boldsymbol{\pi}_{m}(\boldsymbol{\theta}_2)) - \Psi_{{\bf X}}(\boldsymbol{\pi}_{m+1}(\boldsymbol{\theta}_2))) + r_{(d)}\Psi_{{\bf X}}(\boldsymbol{\pi}_{d}(\boldsymbol{\theta}_2))\notag\\ &= \sum_{m=1}^d r_{(m)}\Psi_{{\bf Y}_{(m)}}(\boldsymbol{\theta}_{2(m)})\notag\\* &=\sum_{m=1}^d r_{m}\Psi_{{\bf Y}_m}(\boldsymbol{\theta}_{2m}). \end{align}

Let ${\bf z}\;:\!=\; (z_1,\ldots,z_d)\in\mathbb{C}^d$ , where $z_m = -{\textrm{i}}\langle {\boldsymbol{\theta}_{1m}},{{\bf e}_m}\rangle-\Psi_{{\bf Y}_m}(\boldsymbol{\theta}_{2m})$ , $1 \leq m \leq d$ . Using (11), we have

(12) \begin{align} -\langle {{\bf z}},{{\bf r}}\rangle&=\sum_{m=1}^d {\textrm{i}}\langle {\boldsymbol{\theta}_{1m}},{r_m{\bf e}_m}\rangle+r_m\Psi_{{\bf Y}_m}(\boldsymbol{\theta}_{2m})\notag\\ &={\textrm{i}}\langle {\boldsymbol{\theta}_1},{{\bf r} A}\rangle + ({\bf r} A)\diamond \Psi_{{\bf X}}(\boldsymbol{\theta}_2). \end{align}

Thus (9) becomes $\Phi_{({\bf T},{\bf X}\circ {\bf T})}(\boldsymbol{\theta})=\mathbb{E}[\!\exp(-\langle {{\bf z}},{{\bf R}(1)}\rangle)]$ . By noting that $\Re{\bf z}\in[0,\infty)^d$ and using (2) to obtain the Laplace exponent of R, we have $\Psi_{({\bf T},{\bf X}\circ {\bf T})}(\boldsymbol{\theta})=-\Lambda_{\bf R}({\bf z})$ , where

\begin{align*} \Lambda_{\bf R}({\bf z})=\langle {{\bf d}},{{\bf z}}\rangle+\int_{[0,\infty)_*^d}(1-{\textrm{e}}^{-\langle {{\bf z}},{\bf r}\rangle })\,\mathcal{R}({\textrm{d}} {\bf r}). \end{align*}

Using (4) and (12), we have ${\textrm{e}}^{-\langle {{\bf z}},{\bf r}\rangle }=\Phi_{({\bf r} A,{\bf X}({\bf r} A))}(\boldsymbol{\theta})$ for ${\bf r}\in[0,\infty)^d_*$ . Thus

\begin{align*} \Psi_{({\bf T},{\bf X}\circ {\bf T})}(\boldsymbol{\theta})& = {\textrm{i}}\langle {\boldsymbol{\theta}_1},{{\bf d} A}\rangle + ({\bf d} A)\diamond \Psi_{{\bf X}}(\boldsymbol{\theta}_2)+\int_{[0,\infty)^d_*} (\Phi_{({\bf r} A ,{\bf X}({\bf r} A))}(\boldsymbol{\theta})-1)\,\mathcal{R}({\textrm{d}} {\bf r})\\* &= {\textrm{i}}\langle {\boldsymbol{\theta}_1},{{\bf d} A}\rangle + ({\bf d} A)\diamond \Psi_{{\bf X}}(\boldsymbol{\theta}_2) + \int_{[0,\infty)^n_*} (\Phi_{({\bf t},{\bf X}({\bf t}))}(\boldsymbol{\theta})-1)\,(\mathcal{R}\circ A^{-1}) ({\textrm{d}} {\bf t}) \end{align*}

by the transformation theorem. This matches the right-hand side of (5) because ${\bf T} \sim S^n({\bf d} A, \mathcal{R}\circ A^{-1})$ . Therefore $({\bf T},{\bf X}\circ {\bf T})\stackrel{D}{=}({\bf T},{\bf X}\odot {\bf T})$ .

3.2. Necessity of weak subordination for deterministic and pure-jump, finite activity subordinators

In this subsection we assume that $({\bf T},{\bf X}\circ{\bf T})\sim L^{2n}$ , and under some conditions we show that it is equal in law to $({\bf T},{\bf X}\odot{\bf T})$ . This is in contrast to Theorem 1, where we proved, under a condition for which it is known that $({\bf T},{\bf X}\circ{\bf T})\sim L^{2n}$ , that it is equal in law to $({\bf T},{\bf X}\odot{\bf T})$ . The conditions we consider in this subsection are that the subordinator T is deterministic, or that T is pure-jump with finite activity, which are dealt with in Proposition 1 and Theorem 2, respectively. In the former case we use the weaker assumption ${\bf X}\circ {\bf T}\sim L^{n}$ .

Proof. Since T is deterministic, ${\bf T}={\bf d} I$ . The stationary and independent increment property of the Lévy process ${\bf X}\circ{\bf T}\sim L^{n}$ is equivalent to, for all $0\le t_1<\cdots <t_{m+1}$ , $m\geq1$ ,

\begin{align*} & \mathbb{E} \biggl[ \prod_{k=1}^{m} \exp({\textrm{i}}\langle {\boldsymbol{\theta}_{2k}},{{\bf X}(t_{k+1}{\bf d} )-{\bf X}(t_k{\bf d})}\rangle)\biggr] =\prod_{k=1}^{m} \mathbb{E} [ \!\exp({\textrm{i}}\langle {\boldsymbol{\theta}_{2k}},{{\bf X}((t_{k+1}-t_k){\bf d})}\rangle)], \end{align*}

for $\boldsymbol{\theta}_{21},\ldots, \boldsymbol{\theta}_{2m}\in\mathbb{R}^n.$ Multiplying both sides by $\prod_{k=1}^{m} \exp({\textrm{i}}\langle {\boldsymbol{\theta}_{1k}},{(t_{k+1}-t_k){\bf d}}\rangle)$ , $\boldsymbol{\theta}_{11},\ldots, \boldsymbol{\theta}_{1m}\in\mathbb{R}^n$ , shows that $({\bf T},{\bf X}\circ{\bf T})$ also has stationary and independent increments. Further, $({\bf T},{\bf X}\circ{\bf T})(0)={\bf 0}$ a.s. and the sample paths of $({\bf T},{\bf X}\circ{\bf T})$ are a.s. càdlàg. Thus $({\bf T},{\bf X}\circ{\bf T})\sim L^{2n}$ , so we just need to verify $\Psi_{({\bf T},{\bf X}\circ {\bf T})}=\Psi_{({\bf T},{\bf X}\odot {\bf T})}$ .

Let $\boldsymbol{\theta}=(\boldsymbol{\theta}_1,\boldsymbol{\theta}_2)$ , $\boldsymbol{\theta}_1,\boldsymbol{\theta}_2\in\mathbb{R}^n$ . Noting that ${\bf T}={\bf d} I$ and using (4), we have

\begin{align*} \Psi_{({\bf T},{\bf X}\circ {\bf T})}(\boldsymbol{\theta})={\textrm{i}} \langle {\boldsymbol{\theta}_1},{{\bf d}}\rangle+({\bf d}\diamond\Psi_{{\bf X}})(\boldsymbol{\theta}_2), \end{align*}

which is the same as $\Psi_{({\bf T},{\bf X}\odot {\bf T})}(\boldsymbol{\theta})$ from (5).

We now consider the case where T is a pure-jump subordinator. The following lemma establishes the Laplace functional corresponding to the PRM of the weakly subordinated process $({\bf T},{\bf X}\odot{\bf T})$ without the finite activity assumption on T. It is based on the marked Poisson point process of jumps of $({\bf T},{\bf X}\odot{\bf T})$ given in the proof of [Reference Buchmann, Lu and Madan5, Theorem 2.1 (ii)], and closely follows the arguments in the proof of [Reference Çinlar6, Chapter VI, Theorem 3.2].

Throughout the rest of this section, we let $E\;:\!=\; [0,\infty)\times[0,\infty)^n_*\times \mathbb{R}^n$ , and let Q be the mapping $({\bf t}^*,B)\mapsto \mathbb{P}({\bf X}({\bf t})\in B)$ for ${\bf t}^*\;:\!=\; (t,{\bf t})\in [0,\infty)\times [0,\infty)^n_*$ and Borel sets $B\subseteq\mathbb{R}^n$ .

Lemma 3. Assume ${\bf T}\sim S^n({\bf 0},\mathcal{T})$ , $\mathcal{T}\neq 0$ , ${\bf X}\sim L^n$ . The PRM of the Lévy process $({\bf T},{\bf X}\odot{\bf T})$ , denoted $\mathbb{Z}_{\odot}$ , has Laplace functional

(13) \begin{align} \mathbb{E}\biggl[\!\exp \biggl(-\int_{E} f({\bf t}^*,{\bf y})\,\mathbb{Z}_{\odot}({\textrm{d}} {\bf t}^*, {\textrm{d}} {\bf y}) \biggr)\biggr]= \mathbb{E} \biggl[\prod_{i=1}^\infty \,{\textrm{e}}^{-g({\bf T}^*_i)} \biggr], \end{align}

where f is a non-negative, measurable real function,

\begin{align*} {\textrm{e}}^{-g({\bf t}^*)}&=\int_{\mathbb{R}^n} \,{\textrm{e}}^{-f({\bf t}^*,{\bf y})} \,Q({\bf t}^*,{\textrm{d}}{\bf y}),\\* ({\bf T}^*_i)_{i\in\mathbb{N}} &\;:\!=\; (t,\Delta {\bf T} (t))_{t>0,\Delta {\bf T} (t)\neq{\bf 0}}. \end{align*}

Proof. Noting that the Lévy process $({\bf T},{\bf X}\odot{\bf T})$ has only countably many jumps by Lemma 2, the Poisson point process of jumps can be written as

\begin{equation*} (t,\Delta {\bf T} (t), \Delta({\bf X}\odot {\bf T})(t))_{t>0,\Delta {\bf T} (t)\neq{\bf 0}} =\!: \, ({\bf T}^*_i,{\bf Y}_i)_{i\in\mathbb{N}}. \end{equation*}

Here ${\bf T}^*_i$ and ${\bf Y}_i$ are random vectors that take values in $[0,\infty)\times [0,\infty)^n_*$ and $\mathbb{R}^n$ , respectively. So the PRM of $({\bf T},{\bf X}\odot{\bf T})$ is

\begin{align*} \mathbb{Z}_{\odot}({\textrm{d}} {\bf t}^*, {\textrm{d}}{\bf y})=\sum_{i=1}^\infty \boldsymbol{\delta}_{({\bf T}^*_i,{\bf Y}_i)}({\textrm{d}}{\bf t}^*, {\textrm{d}}{\bf y}). \end{align*}

Now Q is a probability kernel, and $\mathbb{Z}_{\odot}$ corresponds to a marked Poisson point process with marks $({\bf Y}_i)_{i\in\mathbb{N}}$ , and ${\bf Y}_i$ , $i\in\mathbb{N}$ , are conditionally independent given ${\bf T}^*\;:\!=\; ({\bf T}^*_i)_{i\in\mathbb{N}}$ with probability distribution $Q({\bf T}^*_i,{\textrm{d}} {\bf y})$ (see the proof of [Reference Buchmann, Lu and Madan5, Theorem 2.1 (ii)]). Consequently, the Laplace functional of $\mathbb{Z}_{\odot}$ is

\begin{align*} \mathbb{E}\biggl[\!\exp \biggl(-\int_{E} f({\bf t}^*,{\bf y})\,\mathbb{Z}_{\odot}({\textrm{d}} {\bf t}^*, {\textrm{d}} {\bf y}) \biggr)\biggr] &= \mathbb{E} \biggl[ \mathbb{E}\biggl[\prod_{i=1}^\infty \,{\textrm{e}}^{-f({\bf T}^*_i,{\bf Y}_i)} \Bigm| {\bf T}^* \biggr]\biggr] \notag\\* &= \mathbb{E} \biggl[\prod_{i=1}^\infty \mathbb{E}[ {\textrm{e}}^{-f({\bf T}^*_i,{\bf Y}_i)}{\,\vert\,} {\bf T}^* ]\biggr]\notag \\ &= \mathbb{E} \biggl[\prod_{i=1}^\infty\int_{\mathbb{R}^n} \,{\textrm{e}}^{-f({\bf T}^*_i,{\bf y})} \,Q({\bf T}^*_i,{\textrm{d}}{\bf y}) \biggr] \notag\\* &= \mathbb{E} \biggl[\prod_{i=1}^\infty \,{\textrm{e}}^{-g({\bf T}^*_i)} \biggr], \end{align*}

as required.

Proof. Let

\begin{equation*} (t,\Delta {\bf T} (t),\Delta({\bf X}\circ {\bf T})(t))_{t>0,\Delta ({\bf T},{\bf X}\circ {\bf T}) (t)\neq{\bf 0}} \;=\!:\; ({\bf T}^*_i,{\bf Y}^*_i)_{i\in\mathbb{N}} \end{equation*}

with ${\bf T}_i^*\, :\!=(S_i,\Delta {\bf T}(S_i))$ and ${\bf S}^*\;:\!=\; (S_i)_{i\in\mathbb{N}}$ . The PRM of $({\bf T},{\bf X}\circ{\bf T})$ is

\begin{align*} \mathbb{Z}_{\circ}({\textrm{d}} {\bf t}^*, {\textrm{d}}{\bf y})=\sum_{i=1}^\infty \boldsymbol{\delta}_{({\bf T}^*_i,{\bf Y}^*_i)}({\textrm{d}}{\bf t}^*, {\textrm{d}}{\bf y}). \end{align*}

Since $\mathcal{T}([0,\infty)^n_*)<\infty$ by assumption, the jumps of T are countable in increasing order by Lemma 2 (i). Therefore the sample paths $t\mapsto({\bf T},{\bf X}\circ{\bf T})(t)$ are piecewise constant a.s., which implies by Lemma 1 (ii) that the Lévy process $({\bf T},{\bf X}\circ{\bf T})\sim L^{2n}({\bf m},\Theta,\mathcal{Z})$ must have ${\bf m} = \int_{\mathbb{D}_*} ({\bf t},{\bf x})\,\mathcal{Z}({\textrm{d}}{\bf t},{\textrm{d}}{\bf x})$ and $\Theta = 0$ . These are the same m and $\Theta$ as $({\bf T},{\bf X}\odot{\bf T})$ in (6)–(7) provided that $({\bf T},{\bf X}\circ{\bf T})$ and $({\bf T},{\bf X}\odot{\bf T})$ have the same Lévy measure, which we now verify by showing they have the same PRM.

Let $S_0\;:\!=\; 0$ . Recalling that $({\bf T},{\bf X}\circ{\bf T})$ has càdlàg and piecewise constant sample paths a.s., and that $(S_i)_{i\in\mathbb{N}}$ is countable in increasing order, we have

(14) \begin{align} (\Delta {\bf T}(S_i),{\bf Y}_i^*)&=({\bf T}(S_i)-{\bf T}(S_i-), {\bf X}\circ {\bf T}(S_i)-{\bf X}\circ {\bf T}(S_i-))\notag\\* &=({\bf T},{\bf X}\circ {\bf T})(S_i)-({\bf T},{\bf X}\circ {\bf T})(S_{i-1}),\quad i\in\mathbb{N} . \end{align}

For $i\in\mathbb{N}$ , let $\mathbb{P}((\Delta {\bf T}(S_i),{\bf Y}_i^*)\in({\textrm{d}}{\bf t},{\textrm{d}}{\bf y}) {\,\vert\,} {\bf S}^*)$ denote the conditional distribution of $(\Delta {\bf T}(S_i),{\bf Y}_i^*)$ given ${\bf S}^*$ . Since $(\Delta {\bf T}(S_i),{\bf Y}_i^*)$ takes values on the Borel space $([0,\infty)^n_*\times\mathbb{R}^n,\mathcal{B}([0,\infty)^n_*\times\mathbb{R}^n))$ , there exists a regular version of the conditional distribution (see [Reference Kallenberg8, Theorem 5.3]), so we can assume that $\mathbb{P}((\Delta {\bf T}(S_i),{\bf Y}_i^*)\in({\textrm{d}}{\bf t},{\textrm{d}}{\bf y}) {\,\vert\,} {\bf S}^*)$ is a probability kernel.

Consequently, $\mathbb{P}((\Delta {\bf T}(S_i),{\bf Y}_i^*)\in({\textrm{d}}{\bf t},{\textrm{d}}{\bf y}) {\,\vert\,} {\bf S}^*)$ is a probability measure for each value of ${\bf S}^*$ , so it is determined by its characteristic function, which by the disintegration theorem (see [Reference Kallenberg8, Theorem 5.4]) is

(15) \begin{align}& \mathbb{E}[\! \exp({\textrm{i}}\langle { (\Delta {\bf T}(S_i),{\bf Y}_i^*) },{\boldsymbol{\theta}}\rangle) {\,\vert\,} {\bf S}^*] \notag\\* &\quad = \mathbb{E}[ \!\exp({\textrm{i}}\langle { ({\bf T},{\bf X}\circ {\bf T})(S_i-S_{i-1})},{\boldsymbol{\theta}}\rangle){\,\vert\,} {\bf S}^*] \notag\\&\quad = \mathbb{E}[ \!\exp({\textrm{i}}\langle { {\bf T}(S_i-S_{i-1})},{\boldsymbol{\theta}_1}\rangle) \exp({\bf T}(S_i-S_{i-1}) \diamond \Psi_{{\bf X}}(\boldsymbol{\theta}_2) ) {\,\vert\,} {\bf S}^*]\notag \\*&\quad = \mathbb{E}[ \! \exp({\textrm{i}}\langle { \Delta {\bf T}(S_i) },{\boldsymbol{\theta}_1}\rangle) \exp(\Delta {\bf T}(S_i) \diamond \Psi_{{\bf X}}(\boldsymbol{\theta}_2) ) {\,\vert\,} {\bf S}^*], \end{align}

using the stationary increment property of $({\bf T},{\bf X}\circ{\bf T})$ , (4), and the stationary increment property of T, where $\boldsymbol{\theta}=(\boldsymbol{\theta}_1,\boldsymbol{\theta}_2)$ , $\boldsymbol{\theta}_1, \boldsymbol{\theta}_2\in\mathbb{R}^n$ . Similarly, the conditional distribution $\mathbb{P}((\Delta {\bf T}(S_i), {\bf X}(\Delta {\bf T}(S_i)))\in({\textrm{d}}{\bf t},{\textrm{d}}{\bf y}) {\,\vert\,} {\bf S}^*)$ is also a probability kernel, and using the tower law, its characteristic function is

\begin{align*} & \mathbb{E}[ \!\exp({\textrm{i}}\langle { (\Delta {\bf T}(S_i),{\bf X}(\Delta {\bf T}(S_i))) },{\boldsymbol{\theta}}\rangle) {\,\vert\,} {\bf S}^*] \\* &\quad = \mathbb{E}[ \mathbb{E}[\! \exp({\textrm{i}}\langle { (\Delta {\bf T}(S_i),{\bf X}(\Delta {\bf T}(S_i))) },{\boldsymbol{\theta}}\rangle) {\,\vert\,} \Delta {\bf T}(S_i), {\bf S}^*]{\,\vert\,} {\bf S}^*], \end{align*}

which matches (15). Thus we have

(16) \begin{align} &\mathbb{P}( (\Delta {\bf T}(S_i),{\bf Y}_i^*)\in ({\textrm{d}} {\bf t}, {\textrm{d}} {\bf y}) {\,\vert\,} {\bf S}^*) \notag\\* &\quad = \mathbb{P}((\Delta {\bf T}(S_i),{\bf X}(\Delta {\bf T}(S_i))) \in ({\textrm{d}} {\bf t}, {\textrm{d}} {\bf y}){\,\vert\,} {\bf S}^*)\notag\\ &\quad = \mathbb{P}((\Delta {\bf T}(S_i),{\bf X}(\Delta {\bf T}(S_i))) \in ({\textrm{d}} {\bf t}, {\textrm{d}} {\bf y}))\notag\\ &\quad = \mathbb{P}({\bf X}(\Delta {\bf T}(S_i)) \in{\textrm{d}} {\bf y}{\,\vert\,} \Delta {\bf T}(S_i) = {\bf t})\mathbb{P}(\Delta {\bf T}(S_i) \in {\textrm{d}} {\bf t})\notag\\ &\quad = \mathbb{P}({\bf X}({\bf t})\in{\textrm{d}} {\bf y})\mathbb{P}(\Delta {\bf T}(S_i) \in {\textrm{d}} {\bf t}) \notag\\* &\quad = Q((t,{\bf t}), {\textrm{d}} {\bf y})\mathbb{P}(\Delta {\bf T}(S_i) \in {\textrm{d}} {\bf t}),\quad i\in\mathbb{N},\ t>0, \end{align}

where the third line is obtained by noting that the conditioning on ${\bf S}^*$ can be dropped because, by applying Lemma 1 (iii) to $({\bf T},{\bf X}\circ {\bf T})$ , we see that ${\bf S}^*$ are the times of the jumps of a compound Poisson process, which are independent of the size of the jumps with distribution $(\Delta {\bf T}(S_i),{\bf X}(\Delta {\bf T}(S_i)))$ , the fourth line is obtained by [Reference Kallenberg8, Chapter 5, equation (7)], and the fifth line is obtained by the independence of T and X.

By definition, ${\bf S}^*$ are the jumping times of $({\bf T},{\bf X}\circ {\bf T})$ . Since T is a pure-jump subordinator with finite activity, by examining the sample paths, ${\bf X}\circ{\bf T}$ cannot jump unless T does. So almost surely,

\begin{equation*} \{t>0\colon \Delta({\bf X}\circ{\bf T})(t)\neq {\bf 0} \}\subseteq \{t>0\colon \Delta{\bf T}(t)\neq {\bf 0} \}, \end{equation*}

which implies ${\bf S}^*=\{t>0\colon \Delta{\bf T}(t)\neq {\bf 0} \}$ . Thus ${\bf S}^*$ are also the jumping times of T.

Next, for any non-negative, measurable real function f, we have

(17) \begin{align} \mathbb{E}\big[ {\textrm{e}}^{-f(S_i,\Delta {\bf T}(S_i),{\bf Y}^*_i)} {\,\big|\,} {\bf S}^* \big] &= \int_{[0,\infty)^n_*\times\mathbb{R}^n} \,{\textrm{e}}^{-f(S_i,{\bf t},{\bf y})} \,Q((t,{\bf t}),{\textrm{d}} {\bf y}) \mathbb{P}(\Delta {\bf T}(S_i) \in {\textrm{d}} {\bf t})\notag\\* &= \int_{[0,\infty)^n_*} \,{\textrm{e}}^{-g(S_i,{\bf t})}\,\mathbb{P}(\Delta {\bf T}(S_i) \in {\textrm{d}} {\bf t})\notag\\ &= \int_{[0,\infty)^n_*} \,{\textrm{e}}^{-g(S_i,{\bf t})}\,\mathbb{P}(\Delta {\bf T}(S_i) \in {\textrm{d}} {\bf t}{\,\vert\,} {\bf S}^*)\notag\\* &=\mathbb{E} \big[{\textrm{e}}^{-g(S_i,\Delta{\bf T}(S_i))}{\,\big|\,} {\bf S}^* \big], \quad i\in\mathbb{N},\ t>0, \end{align}

where the first line follows from the disintegration theorem (see [Reference Kallenberg8, Theorem 5.4]) and using (16), the third line follows from a similar argument to that above, T being a compound Poisson process means the time and size of the jumps are independent, and the final line follows from another application of the disintegration theorem.

Putting this together, we compute the Laplace functional of $\mathbb{Z}_{\circ}$ ,

\begin{align*} \mathbb{E}\biggl[ \!\exp\biggl(-\int_{E} f({\bf t}^*,{\bf y})\,\mathbb{Z}_{\circ}({\textrm{d}} {\bf t}^*, {\textrm{d}} {\bf y}) \biggr)\biggr] &= \mathbb{E} \biggl[ \mathbb{E}\biggl[\prod_{i=1}^\infty \,{\textrm{e}}^{-f(S_i,\Delta {\bf T}(S_i),{\bf Y}^*_i)} \Bigm| {\bf S}^* \biggr]\biggr] \\* &= \mathbb{E} \biggl[\prod_{i=1}^\infty \mathbb{E} [ {\textrm{e}}^{-f(S_i,\Delta {\bf T}(S_i),{\bf Y}^*_i)} {\,\vert\,} {\bf S}^* ]\biggr]\\ &= \mathbb{E} \biggl[\prod_{i=1}^\infty \mathbb{E} [{\textrm{e}}^{-g(S_i,\Delta{\bf T}(S_i))} {\,\vert\,} {\bf S}^* ]\biggr]\\ &= \mathbb{E} \biggl[ \mathbb{E}\biggl[\prod_{i=1}^\infty \,{\textrm{e}}^{-g(S_i,\Delta {\bf T}(S_i))} \Bigm| {\bf S}^* \biggr]\biggr]\\* &= \mathbb{E}\biggl[\prod_{i=1}^\infty \,{\textrm{e}}^{-g({\bf T}_i^*)} \biggr], \end{align*}

where the second line follows since $(\Delta {\bf T}(S_i),{\bf Y}_i^*)$ , $i\in\mathbb{N}$ , are conditionally independent given ${\bf S}^*$ because of (14) and the independent increment property of the Lévy process $({\bf T},{\bf X}\circ {\bf T})$ , the third line follows from (17), and the fourth line follows from a similar argument to the second line but applied to T. Thus we have proved $\mathbb{Z}_{\circ}=\mathbb{Z}_{\odot}$ from (13), and hence $({\bf T},{\bf X}\circ{\bf T})$ and $({\bf T},{\bf X}\odot{\bf T})$ have the same Lévy measure and hence the same characteristic triplet.