2. Preliminaries

Exponential tilting, also known as the Escher transform, is a technique which transforms any probability density function f with support on $[0,\infty)$ into a new probability density function $f_\lambda$ defined by

\begin{equation*}f_\lambda(x) = \frac{e^{-\lambda x} f(x)}{\int_{0}^\infty e^{-\lambda r} f(r)\textrm{d} r}, \quad x\ge 0,\end{equation*}

where $\lambda\ge 0$ is the tilting rate. The use of exponential tilting goes back at least to [Reference Escher11], where it was used to build upon Cramér’s classical actuarial models [Reference Cramér10]. Later on, the exponential tilting method played a prominent role in the theory of option pricing [Reference Gerber and Shiu13].

The exponentially tilted version of a matrix-exponential distribution has a simple form which happens to be matrix-exponential itself. To see this, notice that if f is of the form (3), then

\begin{align*}\int_0^{\infty}e^{-\lambda r}f(r)\textrm{d} r & = \int_0^{\infty}e^{-\lambda r}(\boldsymbol{\alpha}e^{Tr}\boldsymbol{{s}})\textrm{d} r = \boldsymbol{\alpha}(\lambda I-T)^{-1}\boldsymbol{{s}},\end{align*}

where we used the fact that $T-\lambda I$ has eigenvalues with strictly negative real parts and thus $e^{(T-\lambda I) r}$ vanishes as $r\rightarrow\infty$ . Thus,

(4) \begin{align}f_\lambda(x)=\frac{e^{-\lambda x} (\boldsymbol{\alpha}e^{Tx}\boldsymbol{{s}})}{\boldsymbol{\alpha}(\lambda I-T)^{-1}\boldsymbol{{s}}}=\left(\frac{\boldsymbol{\alpha}}{\boldsymbol{\alpha}(\lambda I-T)^{-1}\boldsymbol{{s}}}\right)e^{(T-\lambda I)x}\boldsymbol{{s}},\quad x\ge 0,\end{align}

implying that $f_\lambda$ corresponds to the density function of a matrix-exponential distribution of parameters $\left(\tfrac{\boldsymbol{\alpha}}{\boldsymbol{\alpha}(\lambda I-T)^{-1}\boldsymbol{{s}}}, T-\lambda I, \boldsymbol{{s}}\right)$ .

Recall that the parameters $(\boldsymbol{\alpha},T,\boldsymbol{{s}})$ need not have a probabilistic meaning in terms of a finite-state-space Markov chain, as opposed to the parameters associated to phase-type distributions. For instance, the parameters

(5) \begin{equation}\boldsymbol{\alpha}=\left(1,0, 0\right), \quad T=\left[\begin{array}{r@{\quad}r@{\quad}r}-1 & -1 & 2/3 \\[5pt]1 & -1 & -2/3 \\[5pt]0 & 0 & -1\end{array}\right], \quad \boldsymbol{{s}}=\left[\begin{array}{r}4/3 \\[5pt]2/3 \\[5pt]1\end{array}\right]\end{equation}

yield a valid matrix-exponential distribution whose density function is given by $f(x)=\tfrac{2}{3}e^{-x}(1+\cos\!(x))$ , and where the dominant eigenvalue of T is $-1$ (see [Reference Bladt and Nielsen6, Example 4.5.21] for details). In the following section we show how to assign a probabilistic meaning to the exponentially tilted version of (5), and more generally to those having the properties A1 and A2, in terms of a finite-state Markov jump process.

3. Main results

Let $(\boldsymbol{\alpha},T,\boldsymbol{{s}})$ be parameters associated to a p-dimensional matrix-exponential distribution which have the properties A1 and A2. For $1\le i,j\le p$ denote by $t_{ij}$ the (i, j) entry of T, and denote by $s_i$ the ith entry of $\boldsymbol{{s}}$ . For $\ell\in\{+,-\}$ , define the $(p\times p)$ -dimensional matrix $T^{\ell}=\left\{t_{ij}^{\ell}\right\}_{1 \le i,j\le p}$ and the p-dimensional column vector $\boldsymbol{{s}}^{\ell}=\left(s^\ell_1,\dots, s^\ell_p\right)^\intercal$ where

\begin{align*}t_{ij}^\pm&=\max\{0,\pm t_{ij}\}\quad \forall\,i\neq j,\\[5pt]t_{ii}^\pm&=\pm \min\{0, \pm t_{ii}\}\quad\forall\, i,\mbox{ and}\\[5pt]s_i^\pm&=\max\{0,\pm s_i\}\quad\forall\, i.\end{align*}

It follows that $T^+$ has nonnegative off-diagonal elements and nonpositive diagonal elements, $T^-$ is a nonnegative matrix, $\boldsymbol{{s}}^+$ and $\boldsymbol{{s}}^-$ are nonnegative column vectors, $T=T^+-T^-$ , and $\boldsymbol{{s}}=\boldsymbol{{s}}^+-\boldsymbol{{s}}^-$ . Now, let

\begin{equation*}\lambda_0=\min\!\left\{r\ge 0\;:\; s^+_i+s^-_i+\sum_{j=1}^p (t^+_{ij} + t^-_{ij}) \le r \mbox{ for all }1\le i\le p\right\}.\end{equation*}

For some fixed $\lambda\ge\lambda_0$ , consider a (possibly) terminating Markov jump process $\{\varphi_t^\lambda\}_{t\ge 0}$ driven by the block-partitioned subintensity matrix

(6) \begin{align}G=\left[\begin{array}{c@{\quad}c@{\quad}c@{\quad}c}T^+-\lambda I & T^-&\boldsymbol{{s}}^+&\boldsymbol{{s}}^-\\[5pt]T^-& T^+-\lambda I & \boldsymbol{{s}}^-&\boldsymbol{{s}}^+\\[5pt]\textbf{0}&\textbf{0} & 0& 0\\[5pt]\textbf{0}&\textbf{0} & 0& 0 \end{array}\right]\end{align}

evolving on the state space $\mathcal{E}=\mathcal{E}^o\cup \mathcal{E}^a\cup\{\Delta^o\}\cup\{\Delta^a\}$ where $\mathcal{E}^o\;:\!=\;\{1^o,2^o,\dots,p^o\}$ and $\mathcal{E}^a=\{1^a,2^a,\dots,p^a\}$ . The state space $\mathcal{E}$ may be thought as the union of two sets: a collection of original states $\mathcal{E}^o\cup\{\Delta^o\}$ and a collection of anti-states $\mathcal{E}^a\cup\{\Delta^a\}$ , where both $\Delta^o$ and $\Delta^a$ are absorbing. In the case $\lambda>\lambda_0$ , the process $\{\varphi_t^\lambda\}_{t\ge 0}$ alternates between sojourn times in $\mathcal{E}^o$ and $\mathcal{E}^a$ up until one of the following happens: (a) get absorbed into $\Delta^o$ , (b) get absorbed into $\Delta^a$ , or (c) undergo termination due to the defect of (6). If $\lambda=\lambda_0$ , the states $\mathcal{E}^o\cup\mathcal{E}^a$ may or may not be transient, their status depending on the values of T.

In Theorem 3.1 we establish a link between the absorption probabilities of $\{\varphi_t^\lambda\}_{t\ge 0}$ and the vector $e^{(T-\lambda I) x}\boldsymbol{{s}}$ appearing in the exponentially tilted matrix-exponential density (4). More specifically, we express each element in $e^{(T-\lambda I) x}\boldsymbol{{s}}$ as the sum of some positive density function and some negative density function, where the positive density is associated to an absorption of $\{\varphi_t^\lambda\}_{t\ge 0}$ to $\Delta^o$ , while the negative density function corresponds to an absorption of $\{\varphi_t^\lambda\}_{t\ge 0}$ to $\Delta^a$ . To shorten notation, from now on we denote by $\mathbb{P}_j$ ( $\mathbb{E}_j$ ), $j\in\mathcal{E}$ , the probability measure (expectation) associated to $\{\varphi^\lambda_t\}_{t\ge 0}$ conditional on the event $\{\varphi^\lambda_0=j\}$ .

Theorem 3.1. Let $\lambda\ge \lambda_0$ be such that the states $\mathcal{E}^o\cup\mathcal{E}^a$ are transient. Define

(7) \begin{equation}\tau=\inf\{x\ge 0\;:\; \varphi_x^\lambda\notin \mathcal{E}^o\cup\mathcal{E}^a\}.\end{equation}

Then, for $i\in \{1,\dots, p\}$ and $x\ge 0$ ,

(8) \begin{align} \big(\boldsymbol{{e}}_i^\intercal e^{(T-\lambda I)x}\boldsymbol{{s}}\big) \textrm{d} x & = \mathbb{E}_{i^o}\left[{\unicode[Times]{x1D7D9}}\{\tau\in [x,x+\textrm{d} x]\} \beta(\varphi_\tau^\lambda)\right] \end{align}

(9) \begin{align} = (\boldsymbol{{e}}_i^\intercal,\textbf{0})\exp\!\left(\left[\begin{matrix}T^+-\lambda I & T^-\\[5pt]T^-& T^+-\lambda I\end{matrix}\right] x\right)\left[\begin{matrix}\boldsymbol{{s}}\\[5pt]-\boldsymbol{{s}}\end{matrix}\right]\textrm{d} x, \end{align}

where $\boldsymbol{{e}}_i$ denotes the column vector with 1 as its ith entry and 0 elsewhere, and $\beta(j)\;:\!=\;{\unicode[Times]{x1D7D9}}\{j=\Delta^o\}-{\unicode[Times]{x1D7D9}}\{j=\Delta^a\}$ . Moreover,

(10) \begin{align} \big(-\boldsymbol{{e}}_i^\intercal e^{(T-\lambda I)x}\boldsymbol{{s}}\big) \textrm{d} x &= \mathbb{E}_{i^a}\left[{\unicode[Times]{x1D7D9}}\{\tau\in [x,x+\textrm{d} x]\} \beta(\varphi_\tau^\lambda)\right] \end{align}

(11) \begin{align} = (\textbf{0},\boldsymbol{{e}}_i^\intercal)\exp\!\left(\left[\begin{matrix}T^+-\lambda I & T^-\\[5pt]T^-& T^+-\lambda I\end{matrix}\right] x\right)\left[\begin{matrix}\boldsymbol{{s}}\\[5pt]-\boldsymbol{{s}}\end{matrix}\right]\textrm{d} x.\end{align}

Proof. The block structure of (6) implies that

\begin{align*}\mathbb{P}_{i^o}(\tau\in [x,x+\textrm{d} x], \varphi_\tau^\lambda=\Delta^o) & = (\boldsymbol{{e}}_i^\intercal,\textbf{0})\exp\!\left(\left[\begin{matrix}T^+-\lambda I & T^-\\[5pt]T^-& T^+-\lambda I\end{matrix}\right] x\right)\left[\begin{matrix}\boldsymbol{{s}}^+\\[5pt]\boldsymbol{{s}}^-\end{matrix}\right]\textrm{d} x,\\[5pt]\mathbb{P}_{i^o}(\tau\in [x,x+\textrm{d} x], \varphi_\tau^\lambda=\Delta^a) & = (\boldsymbol{{e}}_i^\intercal,\textbf{0})\exp\!\left(\left[\begin{matrix}T^+-\lambda I & T^-\\[5pt]T^-& T^+-\lambda I\end{matrix}\right] x\right)\left[\begin{matrix}\boldsymbol{{s}}^-\\[5pt]\boldsymbol{{s}}^+\end{matrix}\right]\textrm{d} x;\end{align*}

therefore, the right-hand side of (8) is equal to (9). Next, we prove that (8) holds.

Define the collection of $(p\times p)$ -dimensional matrices $\{\Phi_{oa}(x)\}_{x\ge 0}$ , $\{\Phi_{ao}(x)\}_{x\ge 0}$ , $\{\Phi_{oo}(x)\}_{x\ge 0}$ , and $\{\Phi_{aa}(x)\}_{x\ge 0}$ by

\begin{align*}(\Phi_{oa}(x))_{ij}& =\mathbb{P}_{i^o}(\tau>x, \varphi_x^\lambda=j^a) ,\qquad(\Phi_{ao}(x))_{ij} =\mathbb{P}_{i^a}(\tau>x, \varphi_x^\lambda=j^o) ,\\[5pt](\Phi_{oo}(x))_{ij}& =\mathbb{P}_{i^o}(\tau>x, \varphi_x^\lambda=j^o) ,\qquad(\Phi_{aa}(x))_{ij} =\mathbb{P}_{i^a}(\tau>x, \varphi_x^\lambda=j^a) ,\end{align*}

for all $i,j\in\{1,\dots,p\}$ . By the symmetry of the subintensity matrix G it is clear that for all $x\ge 0$ , $\Phi_{oa}(x)=\Phi_{ao}(x)$ and $\Phi_{oo}(x)=\Phi_{aa}(x)$ , even if their probabilistic interpretations differ. For all $x\ge 0$ , let $\Phi_{o}(x)\;:\!=\;\Phi_{oo}(x)-\Phi_{oa}(x)$ and $\Phi_{a}(x)\;:\!=\;\Phi_{aa}(x)-\Phi_{ao}(x)$ . Define

\begin{equation*}\gamma=\inf\{r \ge 0\;:\;\varphi_r^\lambda\notin\mathcal{E}^o\}.\end{equation*}

Then, for $i,j\in\{1,\dots,p\}$ ,

(12) \begin{align}\boldsymbol{{e}}_i^\intercal\Phi_{o}(x)\boldsymbol{{e}}_j&=\mathbb{P}_{i^o}(\tau>x, \varphi_x^\lambda=j^o) - \mathbb{P}_{i^o}(\tau>x, \varphi_x^\lambda=j^a)\nonumber\\[5pt]&=\left\{\mathbb{P}_{i^o}(\gamma>x,\tau>x, \varphi_x^\lambda=j^o) + \mathbb{P}_{i^o}(\gamma\le x,\tau>x, \varphi_x^\lambda=j^o) \right\}\nonumber\\[5pt] &\quad - \left\{\mathbb{P}_{i^o}(\gamma>x,\tau>x, \varphi_x^\lambda=j^a) + \mathbb{P}_{i^o}(\gamma\le x,\tau>x, \varphi_x^\lambda=j^a)\right\} \nonumber\\[5pt]& = \left\{\mathbb{P}_{i^o}(\gamma>x, \varphi_x^\lambda=j^o) + \int_{0}^x \mathbb{P}_{i^o}(\gamma\in[r,r+\textrm{d} r], \tau>x, \varphi_x^\lambda=j^o)\right\}\nonumber\\[5pt] &\quad -\left\{\mathbb{P}_{i^o}(\gamma>x, \varphi_x^\lambda=j^a) + \int_{0}^x \mathbb{P}_{i^o}(\gamma\in[r,r+\textrm{d} r], \tau>x, \varphi_x^\lambda=j^a)\right\}\nonumber\\[5pt]& = \mathbb{P}_{i^o}(\gamma>x, \varphi_x^\lambda=j^o)\nonumber\\[5pt]&\quad + \int_{0}^x \sum_{k=1}^p \mathbb{P}_{i^o}(\gamma\in[r,r+\textrm{d} r], \varphi_\gamma^\lambda=k^a) \mathbb{P}_{k^a}(\tau>x-r, \varphi_{x-r}^\lambda=j^o)\nonumber \\[5pt]&\quad - \int_{0}^x \sum_{k=1}^p \mathbb{P}_{i^o}(\gamma\in[r,r+\textrm{d} r], \varphi_\gamma^\lambda=k^a) \mathbb{P}_{k^a }(\tau>x-r, \varphi_{x-r}^\lambda=j^a),\end{align}

where in the last equality we used that $\{\gamma>x, \varphi_x^\lambda=j^a\}=\varnothing$ and the Markov property of $\{\varphi_{x}^\lambda\}_{x\ge 0}$ . Note that all the elements in (12) correspond to transition probabilities or intensities that can be expressed in matricial form as follows:

\begin{align*}\mathbb{P}_{i^o}(\gamma>x, \varphi_x^\lambda=j^o)&=\boldsymbol{{e}}_i^\intercal e^{(T^+-\lambda I)x}\boldsymbol{{e}}_j,\\[5pt]\mathbb{P}_{i^o}(\gamma\in[r,r+\textrm{d} r], \varphi_\gamma^\lambda=k^a)&= \boldsymbol{{e}}_i^\intercal e^{(T^+-\lambda I)r}T^-\boldsymbol{{e}}_k\textrm{d} r,\\[5pt]\mathbb{P}_{k^a}(\tau>x-r, \varphi_{x-r}^\lambda=j^o) & = \boldsymbol{{e}}_k^\intercal \Phi_{ao}(x-r)\boldsymbol{{e}}_j,\\[5pt]\mathbb{P}_{k^a}(\tau>x-r, \varphi_{x-r}^\lambda=j^a) & = \boldsymbol{{e}}_k^\intercal \Phi_{aa}(x-r)\boldsymbol{{e}}_j.\end{align*}

Substituting these expressions into (12) and using the identity $I=\sum_{k=1}^p \boldsymbol{{e}}_k\boldsymbol{{e}}_k^\intercal$ gives

\begin{align*}\boldsymbol{{e}}_i^\intercal\Phi_{o}(x)\boldsymbol{{e}}_j & = \boldsymbol{{e}}_i^\intercal e^{(T^+-\lambda I)x}\boldsymbol{{e}}_j + \int_{0}^x \sum_{k=1}^p \left(\boldsymbol{{e}}_i^\intercal e^{(T^+-\lambda I)r}T^-\boldsymbol{{e}}_k\right) \left(\boldsymbol{{e}}_k^\intercal \Phi_{ao}(x-r)\boldsymbol{{e}}_j\right) \textrm{d} r\\[5pt]& \quad- \int_{0}^x \sum_{k=1}^p \left(\boldsymbol{{e}}_i^\intercal e^{(T^+-\lambda I)r}T^-\boldsymbol{{e}}_k\right) \left(\boldsymbol{{e}}_k^\intercal \Phi_{aa}(x-r)\boldsymbol{{e}}_j\right)\textrm{d} r\\[5pt]& = \boldsymbol{{e}}_i^\intercal\!\left(e^{(T^+-\lambda I)x} + \int_{0}^xe^{(T^+-\lambda I)r}T^-\left[\Phi_{ao}(x-r)-\Phi_{aa}(x-r)\right]\textrm{d} r\right)\boldsymbol{{e}}_j\\[5pt]& = \boldsymbol{{e}}_i^\intercal\!\left(e^{(T^+-\lambda I)x} + \int_{0}^xe^{(T^+-\lambda I)r}T^-\left[\Phi_{oa}(x-r)-\Phi_{oo}(x-r)\right]\textrm{d} r\right)\boldsymbol{{e}}_j\\[5pt]& = \boldsymbol{{e}}_i^\intercal\!\left(e^{(T^+-\lambda I)x} + \int_{0}^xe^{(T^+-\lambda I)r}(\!-T^-)\Phi_{o}(x-r)\textrm{d} r\right)\boldsymbol{{e}}_j,\end{align*}

so that $\{\Phi_{o}(x)\}_{x\ge 0}$ is the bounded solution to the matrix-integral equation

\begin{equation*}\Phi_{o}(x)=e^{(T^+-\lambda I)x} + \int_{0}^xe^{(T^+-\lambda I)r}(\!-T^-)\Phi_{o}(x-r)\textrm{d} r.\end{equation*}

By [Reference Bean, Nguyen, Nielsen and Peralta3, Theorem 3.10],

\begin{equation*}\Phi_{o}(x)=e^{[(T^+-\lambda I)+(-T^-)]x}=e^{(T-\lambda I)x}.\end{equation*}

The Markov property implies that

\begin{align*}&\mathbb{P}_{i^o}(\tau\in [x,x+\textrm{d} x], \varphi_\tau^\lambda=\Delta^o)\\[5pt]&\quad = \sum_{k=1}^p \mathbb{P}_{i^o}(\tau>x, \varphi_x^\lambda=k^o)\mathbb{P}_{k^o}(\tau\in[x,x+\textrm{d} x], \varphi_\tau^\lambda=\Delta^o)\\[5pt]&\quad\quad+\sum_{k=1}^p\mathbb{P}_{i^o}(\tau>x, \varphi_x^\lambda=k^a)\mathbb{P}_{k^a}(\tau\in[x, x+\textrm{d} x], \varphi_\tau^\lambda=\Delta^o)\\[5pt]&\quad = \sum_{k=1}^p (\boldsymbol{{e}}_i^\intercal\Phi_{oo}(x)\boldsymbol{{e}}_k)(\boldsymbol{{e}}_k^\intercal \boldsymbol{{s}}^+)\textrm{d} x + \sum_{k=1}^p (\boldsymbol{{e}}_i^\intercal\Phi_{oa}(x)\boldsymbol{{e}}_k)(\boldsymbol{{e}}_k^\intercal \boldsymbol{{s}}^-\textrm{d} x)\\[5pt]&\quad = \boldsymbol{{e}}_i^\intercal (\Phi_{oo}(x)\boldsymbol{{s}}^+ + \Phi_{oa}(x)\boldsymbol{{s}}^-)\textrm{d} x.\end{align*}

Similarly,

\begin{equation*}\mathbb{P}_{i^o}(\tau\in [x,x+\textrm{d} x], \varphi_\tau^\lambda=\Delta^a) = \boldsymbol{{e}}_i^\intercal (\Phi_{oa}(x)\boldsymbol{{s}}^+ + \Phi_{oo}(x)\boldsymbol{{s}}^-)\textrm{d} x.\end{equation*}

Thus,

\begin{align*}&\mathbb{P}_{i^o}(\tau\in [x,x+\textrm{d} x], \varphi_\tau^\lambda=\Delta^o) - \mathbb{P}_{i^o}(\tau\in [x,x+\textrm{d} x], \varphi_\tau^\lambda=\Delta^a)\\[5pt]&\quad = \boldsymbol{{e}}_i^\intercal ([\Phi_{oo}(x)\boldsymbol{{s}}^+ + \Phi_{oa}(x)\boldsymbol{{s}}^-]-[\Phi_{oa}(x)\boldsymbol{{s}}^+ + \Phi_{oo}(x)\boldsymbol{{s}}^-])\textrm{d} x\\[5pt]&\quad = \boldsymbol{{e}}_i^\intercal ([\Phi_{oo}(x)-\Phi_{oa}(x)]\boldsymbol{{s}}^+ - [\Phi_{oo}(x)-\Phi_{oa}(x)]\boldsymbol{{s}}^-)\textrm{d} x\\[5pt]& \quad = \boldsymbol{{e}}_i^\intercal\Phi_{o}(x)\boldsymbol{{s}}\textrm{d} x = \boldsymbol{{e}}_i^\intercal e^{(T-\lambda I)x}\boldsymbol{{s}}\textrm{d} x,\end{align*}

so that (8) holds. Analogous arguments follow for (10) and (11), which completes the proof.

Heuristically, Equations (8) and (10) imply that initiating $\{\varphi^\lambda_t\}_{t\ge 0}$ in the anti-state $i^a$ has the opposite effect, in terms of sign, to initiating in the original state $i^o$ . In the following we exploit this fact to provide a probabilistic interpretation not only for the elements of $e^{(T-\lambda I)x}\boldsymbol{{s}}$ , but also for the exponentially tilted matrix-exponential density $\boldsymbol{\alpha} e^{(T-\lambda I)x}\boldsymbol{{s}}$ .

Define $w^+$ and $w^-$ by

\begin{equation*}w^\pm=\sum_{i=1}^p \max\{0,\pm\alpha_i\},\end{equation*}

and define $\boldsymbol{\alpha}^+=(\alpha_1^+,\dots,\alpha_p^+)$ and $\boldsymbol{\alpha}^-=(\alpha_1^-,\dots,\alpha_p^-)$ by

\begin{align*}\alpha^{\pm}_{i}=\left\{\begin{array}{c@{\quad}c@{\quad}c}\frac{1}{w^\pm}\max\{0,\pm\alpha_i\}&\mbox{if}&w^\pm>0,\\[5pt]0&\mbox{if}& w^\pm=0.\end{array}\right.\end{align*}

If $w^\pm>0$ , then $\boldsymbol{\alpha}^\pm$ is a probability vector, and in general,

(13) \begin{equation}\boldsymbol{\alpha}=w^+\boldsymbol{\alpha}^+ - w^-\boldsymbol{\alpha}^-.\end{equation}

In some sense, $(w^+ + w^-)^{-1}\boldsymbol{\alpha}$ can be thought as a mixture of the probability vectors $\boldsymbol{\alpha}^+$ and $\boldsymbol{\alpha}^-$ , with the latter contributing ‘negative mass’. Fortunately, this ‘negative mass’ in the context of $\boldsymbol{\alpha} e^{(T-\lambda I)x}\boldsymbol{{s}}$ can be given a precise probabilistic interpretation by means of anti-states as follows.

Theorem 3.2. Let $f_\lambda(x)=(\boldsymbol{\alpha}(\lambda I-T)^{-1}\boldsymbol{{s}})^{-1}\boldsymbol{\alpha}e^{(T-\lambda I)x}\boldsymbol{{s}}$ , $x\ge 0$ , be the density of the exponentially tilted matrix-exponential distribution of parameters $(\boldsymbol{\alpha},T,\boldsymbol{{s}})$ . Define the vectors

\begin{equation*}\widehat{\boldsymbol{\alpha}}^+\;:\!=\; \tfrac{w^+}{w^++w^-}\boldsymbol{\alpha}^+\quad{\textit{and}}\quad\widehat{\boldsymbol{\alpha}}^-\;:\!=\; \tfrac{w^-}{w^++w^-}\boldsymbol{\alpha}^-,\end{equation*}

and suppose $\varphi_0^\lambda\sim (\widehat{\boldsymbol{\alpha}}^+, \widehat{\boldsymbol{\alpha}}^-)$ . Then

(14) \begin{align} f_\lambda(x) \textrm{d} x &= \frac{(w^++w^-)}{\boldsymbol{\alpha}(\lambda I-T)^{-1}\boldsymbol{{s}}}\mathbb{E}\left[{\unicode[Times]{x1D7D9}}\{\tau\in [x,x+\textrm{d} x]\} \beta(\varphi_\tau^\lambda)\right]\end{align}

(15) \begin{align} = \frac{(w^++w^-)}{\boldsymbol{\alpha}(\lambda I-T)^{-1}\boldsymbol{{s}}}\left((\widehat{\boldsymbol{\alpha}}^+,\widehat{\boldsymbol{\alpha}}^-)\exp\!\left(\left[\begin{matrix}T^+-\lambda I & T^-\\[5pt]T^-& T^+-\lambda I\end{matrix}\right] x\right)\left[\begin{matrix}\boldsymbol{{s}}\\[5pt]-\boldsymbol{{s}}\end{matrix}\right]\right)\textrm{d} x,\end{align}

where $\tau$ and $\beta(\!\cdot\!)$ are defined as in Theorem 3.1.

Proof. Equation (13) implies that

(16) \begin{align}f_\lambda(x)&= \frac{1}{\boldsymbol{\alpha}(\lambda I-T)^{-1}\boldsymbol{{s}}}\left(\sum_{i=1}^p w^+\alpha^+_{i}\!\left(\boldsymbol{{e}}_i^\intercal e^{(T-\lambda I)x}\boldsymbol{{s}}\right) + \sum_{i=1}^p w^-\alpha^-_i\left(-\boldsymbol{{e}}_i^\intercal e^{(T-\lambda I)x}\boldsymbol{{s}}\right)\right)\nonumber\\[5pt]& = \frac{(w^++w^-)}{\boldsymbol{\alpha}(\lambda I-T)^{-1}\boldsymbol{{s}}}\left(\sum_{i=1}^p \tfrac{w^+}{w^++w^-}\alpha^+_i \left(\boldsymbol{{e}}_i^\intercal e^{(T-\lambda I)x}\boldsymbol{{s}}\right) + \sum_{i=1}^p \tfrac{w^-}{w^++w^-}\alpha^-_i\left(-\boldsymbol{{e}}_i^\intercal e^{(T-\lambda I)x}\boldsymbol{{s}}\right)\right).\end{align}

Equality (14) follows from (16), (8), and (10). Equality (15) follows from (16), (9), and (11).

Example 3.1. Let $(\boldsymbol{\alpha},T,\boldsymbol{{s}})$ be the matrix-exponential parameters corresponding to (5). As noted previously, these parameters by themselves lack a probabilistic interpretation, so we apply Theorem 3.1 to construct one. For such parameters we take the tilting parameter $\lambda\;:\!=\;\lambda_0=2$ , leading to the block-partitioned matrices

and $w^+=1$ , $w^-=0$ . We can then verify that

\begin{align*}&\frac{(w^++w^-)}{\boldsymbol{\alpha}(\lambda I-T)^{-1}\boldsymbol{{s}}}\left((\widehat{\boldsymbol{\alpha}}^+,\widehat{\boldsymbol{\alpha}}^-)\exp\!\left(\left[\begin{matrix}T^+-\lambda I & T^-\\[5pt]T^-& T^+-\lambda I\end{matrix}\right] x\right)\left[\begin{matrix}\boldsymbol{{s}}\\[5pt]-\boldsymbol{{s}}\end{matrix}\right]\right)\\[5pt]&\qquad=\frac{1}{\boldsymbol{\alpha}(\lambda I-T)^{-1}\boldsymbol{{s}}} \left(\frac{2}{3}e^{-3x}(1+\cos\!(x))\right)=\frac{e^{-2x}}{\boldsymbol{\alpha}(\lambda I-T)^{-1}\boldsymbol{{s}}}\left(\frac{2}{3} e^{-x}(1+\cos\!(x))\right),\end{align*}

the latter corresponding to the exponentially tilted matrix-exponential density function $f(x)=\tfrac{2}{3} e^{-x}(1+\cos\!(x))$ .

A probabilistic interpretation of $f_\lambda$ alternative to that of (14) is the following.

Corollary 3.1. Define $\boldsymbol{{d}}= (d_1,\dots,d_p)^\intercal\;:\!=\;-(T^+-\lambda I)\textbf{1} - T^-\textbf{1}$ to be the termination intensities vector from $\mathcal{E}^o$ or $\mathcal{E}^a$ , and define $\boldsymbol{{q}}^\pm=(q_1^\pm,\dots,q_p^\pm)^\intercal$ by

\begin{equation*}q^\pm_i=\left\{\begin{array}{c@{\quad}c@{\quad}c}\frac{s^\pm_i}{d_i}&{\textit{if}} &d_i>0,\\[5pt] 0& {\textit{if}}&d_i=0.\end{array}\right.\end{equation*}

Let $\bar{q}\;:\;\mathcal{E}^o\cup\mathcal{E}^a\mapsto \mathbb{R}$ be defined by

\begin{align*}\bar{q}(i^o)=q^+_i - q^-_i\quad{\textit{and}}\quad\bar{q}(i^a)=q^-_i - q^+_i\quad\mbox{ for }\quad i\in\{1,\dots, p\}.\end{align*}

Then

(17) \begin{align}f_\lambda(x) \textrm{d} x &= \left(\frac{w^++w^-}{\boldsymbol{\alpha}(\lambda I-T)^{-1}\boldsymbol{{s}}}\right)\mathbb{E}\left[{\unicode[Times]{x1D7D9}}\{\tau\in [x,x+\textrm{d} x]\} \bar{q}\big(\varphi^\lambda_{\tau^-}\big)\right], \end{align}

where $\{\varphi^\lambda_t\}_{t\ge 0}$ and $\tau$ are defined as in Theorem 3.2.

Proof. First, notice that the jump mechanism of $\{\varphi^\lambda_t\}_{t\ge 0}$ described in (6) implies that for $i\in\{1,\dots,p\}$ ,

\begin{align*}\mathbb{P}(\varphi^\lambda_\tau=\Delta^o\mid \tau, \varphi^\lambda_{\tau^-}=i^o)&=q_i^+,\\[5pt]\mathbb{P}(\varphi^\lambda_\tau=\Delta^a\mid \tau, \varphi^\lambda_{\tau^-}=i^o)&=q_i^-,\\[5pt]\mathbb{P}(\varphi^\lambda_\tau=\Delta^o\mid \tau, \varphi^\lambda_{\tau^-}=i^a)&=q_i^-,\\[5pt]\mathbb{P}(\varphi^\lambda_\tau=\Delta^a\mid \tau, \varphi^\lambda_{\tau^-}=i^a)&=q_i^+,\end{align*}

which in turn implies that

\begin{align*}\mathbb{E}\!\left[\beta(\varphi_\tau^\lambda)\mid \tau,\varphi^\lambda_{\tau^-}\right]=\mathbb{P}\!\left(\varphi_\tau^\lambda=\Delta^o\mid \tau,\varphi^\lambda_{\tau^-}\right) - \mathbb{P}\!\left(\varphi_\tau^\lambda=\Delta^a\mid \tau,\varphi^\lambda_{\tau^-}\right)=\bar{q}(\varphi^\lambda_{\tau^-}).\end{align*}

Consequently,

\begin{align*}\mathbb{E}\!\left[{\unicode[Times]{x1D7D9}}\{\tau\in [x,x+\textrm{d} x]\} \beta(\varphi_\tau^\lambda)\right]&=\mathbb{E}\!\left[\mathbb{E}\left[{\unicode[Times]{x1D7D9}}\{\tau\in [x,x+\textrm{d} x]\} \beta(\varphi_\tau^\lambda)\mid \tau,\varphi^\lambda_{\tau^-}\right]\right]\\[5pt]& = \mathbb{E}\!\left[{\unicode[Times]{x1D7D9}}\{\tau\in [x,x+\textrm{d} x]\}\mathbb{E}\!\left[\beta(\varphi_\tau^\lambda)\mid \tau,\varphi^\lambda_{\tau^-}\right]\right]\\[5pt]& =\mathbb{E}\!\left[{\unicode[Times]{x1D7D9}}\{\tau\in [x,x+\textrm{d} x]\} \bar{q}\big(\varphi^\lambda_{\tau^-}\big)\right],\end{align*}

and the result follows from (14).

Though closely related, the interpretation provided by Corollary 3.1 is more suitable than that of Theorem 3.2 for Monte Carlo applications. Indeed, a realization of $\{\varphi^\lambda_t\}_{t\ge 0}$ may get absorbed in $\Delta^o$ , $\Delta^a$ or terminated. If termination occurs, such a realization contributes nothing to the term in the right-hand side of (14). In contrast, by observing the process until its exit time from $\mathcal{E}^o\cup\mathcal{E}^a$ and disregarding its landing point as in Corollary 3.1, we make sure that each realization contributes towards the mass in the right-hand side of (17).

4. Recovering the untilted distribution

Once the exponentially tilted density $f_\lambda$ of a matrix-exponential distribution of parameters $(\boldsymbol{\alpha},T,\boldsymbol{{s}})$ has a tractable known form, say as in (15), in principle it is straightforward to recover the original untilted density f by taking

(18) \begin{align}f(x)& =(\boldsymbol{\alpha}(\lambda I-T)\boldsymbol{{s}})e^{\lambda x} f_{\lambda}(x)\nonumber\\[5pt]& = (w^++w^-)(\widehat{\boldsymbol{\alpha}}^+,\widehat{\boldsymbol{\alpha}}^-)\exp\!\left(\left[\begin{matrix}T^+ \;\;\;\;\;\;& T^-\\[5pt]T^- \;\;\;\;\;\;& T^+\end{matrix}\right] x\right)\left[\begin{matrix}\boldsymbol{{s}}\\[5pt]-\boldsymbol{{s}}\end{matrix}\right],\qquad x\ge 0.\end{align}

While (18) is a legitimate matrix-exponential representation of f, it has two drawbacks:

1. 1. The matrix $\left[\begin{smallmatrix}T^+ & T^-\\[5pt]T^-& T^+\end{smallmatrix}\right]$ may no longer be a subintensity matrix.

2. 2. The dominant eigenvalue of $\left[\begin{smallmatrix}T^+ & T^-\\[5pt]T^-& T^+\end{smallmatrix}\right]$ may be nonnegative.

The first item may affect the probabilistic interpretation of f, while the second one may make integration of certain functions (with respect to the density f) more difficult to handle. For instance, in the context of Example 3.1, the matrix

is not a subintensity matrix since some row sums are strictly positive, and it has 0 as its dominant eigenvalue. Having 0 as an eigenvalue implies that some entries of $\exp\!\left(\left(\begin{smallmatrix}T^+ & T^-\\[5pt]T^-& T^+\end{smallmatrix}\right)x\right)$ may potentially be of order $e^{0\cdot x}=1$ , meaning that the matrix integral

(19) \begin{equation}\int_{0}^\infty h(x) \exp\!\left(\left[\begin{smallmatrix}T^+ & T^-\\[5pt]T^-& T^+\end{smallmatrix}\right]x\right)\textrm{d} x\end{equation}

may only be well-defined for functions $h\;:\;\mathbb{R}_+\mapsto\mathbb{R}_+$ that decrease to 0 fast enough. In comparison, $\exp\!(Tx)$ with T as in (5) has entries of order $e^{\sigma_0 x}=e^{-x}$ or less, so that

(20) \begin{equation}\int_{0}^\infty h(x) \exp\!\left(Tx\right)\textrm{d} x\end{equation}

is well-defined and finite for every function $h\;:\;\mathbb{R}_+\mapsto\mathbb{R}_+$ of order $e^{-\rho x}$ for any $\rho>-1$ . This apparent disagreement between the applicability of (19) and (20) vanishes when we multiply $\exp\!\left(\left[\begin{smallmatrix}T^+ & T^-\\[5pt]T^-& T^+\end{smallmatrix}\right]x\right)$ by $\left[\begin{smallmatrix}\boldsymbol{{s}}\\[5pt] -\boldsymbol{{s}}\end{smallmatrix}\right]$ . Indeed, in the context of Example 3.1 it can be verified that the elements of the vector $\exp\!\left(\left[\begin{smallmatrix}T^+ & T^-\\[5pt]T^-& T^+\end{smallmatrix}\right]x\right)\left[\begin{smallmatrix}\boldsymbol{{s}}\\[5pt] -\boldsymbol{{s}}\end{smallmatrix}\right]$ are at most of order $e^{-x}$ , with the higher-order terms of $\exp\!\left(\left[\begin{smallmatrix}T^+ & T^-\\[5pt]T^-& T^+\end{smallmatrix}\right]x\right)$ cancelling each other when we multiply the matrix function by $\left[\begin{smallmatrix}\boldsymbol{{s}}\\[5pt] -\boldsymbol{{s}}\end{smallmatrix}\right]$ .

In the general case, this cancellation of higher-order terms is implied by Theorem 3.1 via the following arguments. If $\sigma_0$ is the dominant eigenvalue of T and has multiplicity $m_0$ , then the order of $\boldsymbol{{e}}_i^\intercal e^{(T-\lambda I)x}\boldsymbol{{s}}$ is at most $x^{m_0}e^{(\sigma_0-\lambda) x}$ . By (9) and (11), the elements of $\exp\!\left(\left[\begin{smallmatrix}T^+ -\lambda I& T^-\\[5pt]T^-& T^+-\lambda I\end{smallmatrix}\right]x\right)\left[\begin{smallmatrix}\boldsymbol{{s}}\\[5pt] -\boldsymbol{{s}}\end{smallmatrix}\right]$ are also of order less than or equal to $x^{m_0}e^{(\sigma_0-\lambda) x}$ . Finally, if we multiply the previous by $e^{\lambda x}$ , then we get that $\exp\!\left(\left[\begin{smallmatrix}T^+ & T^-\\[5pt]T^-& T^+\end{smallmatrix}\right]x\right)\left[\begin{smallmatrix}\boldsymbol{{s}}\\[5pt] -\boldsymbol{{s}}\end{smallmatrix}\right]$ is of order at most $x^{m_0}e^{\sigma_0 x}$ , which coincides with the order of $e^{Tx}$ .

In terms of expectations, (14) and (17) provide alternative ways to recover properties of any matrix-exponential density f of parameters $(\boldsymbol{\alpha},T,\boldsymbol{{s}})$ in terms of the exponentially tilted density $f_\lambda$ . Indeed, for any function $h\;:\;\mathbb{R}_+\mapsto\mathbb{R}_+$ of order $e^{-\rho x}$ or less, $\rho>\sigma_0$ , we have that

(21) \begin{align} \int_0^\infty h(x)f(x)\textrm{d} x & = (\boldsymbol{\alpha}(\lambda I-T)\boldsymbol{{s}})\int_0^\infty h(x)e^{\lambda x}f_\lambda(x)\textrm{d} x \end{align}

(22) \begin{align} = (w^++w^-)\mathbb{E}\left[ h(\tau)e^{\lambda \tau} \beta(\varphi^\lambda_\tau)\right]\end{align}

(23) \begin{align} = (w^++w^-)\mathbb{E}\left[ h(\tau)e^{\lambda \tau} \bar{q}(\varphi^\lambda_{\tau^-})\right],\end{align}

where $\{\varphi^\lambda_t\}_{t\ge 0}$ and $\tau$ are as in Theorem 3.2. Existence and finiteness of the first moment of $h(\tau)e^{\lambda \tau} \beta(\varphi^\lambda_\tau)$ in (22) is guaranteed by noting that the order $e^{-\rho+\lambda}$ of $h(x)e^{\lambda x}$ is dominated by that of $f_\lambda$ . Notice that, as opposed to the formula in (18), the representations (22) and (23) still have probabilistic interpretations in terms of the Markov jump process $\{\varphi^\lambda_t\}_{t\ge 0}$ .