2. The MMPP class of PH-distributions and properties

A phase-type distribution, to be called a PH-distribution, is defined as the distribution of the absorption time of a special state in a finite-state discrete/continuous-time Markov chain. Since there is a one-to-one correspondence between the results obtained in this paper for discrete- and continuous-time cases, we shall focus on continuous-time PH-distributions. Consider a continuous-time Markov chain (CTMC) $\{Y(t), t\geq 0\}$ with $m+1$ states and infinitesimal generator

\begin{equation*}\begin{pmatrix} T\;\;\;\;\;\; & {\bf T}^0 \\ 0\;\;\;\;\;\; & 0 \end{pmatrix}\!,\end{equation*}

where T is an $m\times m$ matrix with negative diagonal elements and non-negative off-diagonal element, ${\bf T}^0$ is a non-negative and non-zero column vector of order m, $T{\bf e}+{\bf T}^0=0$ , and ${\bf e}$ is the column vector with all elements being one. Define $X = \min\{t\colon Y(t) = m+1, t\geq0\}$ . Assume that Y(0) is distributed according to $({\boldsymbol \alpha}, \alpha_{m+1})$ . It is well known that

(2.1) \begin{align} \mathbb{P}\{X < t\} &= 1 - {\boldsymbol \alpha}\exp\{Tt\}{\bf e} \quad \text{for} \ t\geq 0, \notag \\ \mathbb{E}[X] &= -{\boldsymbol \alpha}T^{-1}{\bf e}, \notag \\ \mathbb{E}[X^2] &= 2{\boldsymbol \alpha}T^{-2}{\bf e}, \notag \\ \operatorname{SCV}(X) &= \dfrac{2{\boldsymbol \alpha}T^{-2}{\bf e}}{({\boldsymbol \alpha}T^{-1}{\bf e})^2} - 1.\end{align}

Random variable X has a PH-distribution and the pair $({\boldsymbol \alpha}, T)$ is called a PH-representation of X. We refer to Neuts [Reference Neuts16], He [Reference He8], and Buchholz, Kriege, and Felko [Reference Buchholz, Kriege and Felko4] for the general theory on phase-type distributions.

For the MMPP class of PH-distributions, we choose $T = Q-D$ , where Q is an infinitesimal generator for an irreducible CTMC and D is a non-zero diagonal matrix with non-negative diagonal elements $\{d_1, d_2, \ldots, d_m\}$ . In Section 4 we will use MMPPs to explain why T has that structure. Let ${\boldsymbol \pi} = (\pi_1, \pi_2, \ldots, \pi_m)$ be the stationary distribution of Q, i.e. ${\boldsymbol \pi}Q=0$ and ${\boldsymbol \pi}{\bf e}=1$ . Let $\lambda = {\boldsymbol \pi}D{\bf e}$ and ${\boldsymbol \pi}_{D} = {\boldsymbol \pi}D/\lambda$ . We define the set of PH-random variables X with PH-representation $({\boldsymbol \pi}, Q-D)$ and $X_{D}$ with PH-representation $({\boldsymbol \pi}_D, Q-D)$ as the MMPP class of PH-distributions. In the rest of this section we show several properties related to the PH-representations $({\boldsymbol \pi}, Q-D)$ and $({\boldsymbol \pi}_D, Q-D)$ .

Let $P_Q = I + \operatorname{diag} ({\boldsymbol \pi})Q/c_Q$ , where $\operatorname{diag} ({\boldsymbol \pi})$ is a diagonal matrix with ${\boldsymbol \pi}$ on its diagonal, and $c_Q$ is a constant satisfying $c_Q \geq \max\{-q_{i,i}, i=1, 2, \ldots, m\}$ , and $Q=(q_{i,j})$ .

Proof. Since Q is an infinitesimal generator, it is easy to see that $P_Q$ is non-negative, $P_Q{\bf e} = {\bf e}$ , and ${\bf e}'P_Q = {\bf e}' + {\boldsymbol \pi}Q/c_Q = {\bf e}'$ , where ${\bf e}'$ is the transpose of e. Thus $P_Q$ is a doubly stochastic matrix.

Proof. If $D=\lambda I$ , we have ${\boldsymbol \pi}(D-Q-\lambda{\bf e}{\boldsymbol \pi}) = {\boldsymbol \pi}(\lambda I-\lambda{\bf e}{\boldsymbol \pi}) = 0$ . Thus $D-Q-\lambda{\bf e}{\boldsymbol \pi}$ has a zero eigenvalue and so is non-invertible. This proves the sufficiency of the condition $D=\lambda I$ .

To show the necessity of the condition, let ${\bf u}=(u_1, u_2, \ldots, u_m)$ be a non-zero left eigenvector of $D-Q-\lambda{\bf e}{\boldsymbol \pi}$ corresponding to the eigenvalue zero. Then we obtain

(2.2) \begin{equation} 0 = {\bf u}(D-Q-\lambda{\bf e}{\boldsymbol \pi}) = {\bf u}(D-Q) - \lambda({\bf u}{\bf e}){\boldsymbol \pi}.\end{equation}

If ${\bf u}{\bf e} = 0$ , we have ${\bf u}(D-Q)=0$ . Since $D-Q$ is an M-matrix, $D-Q$ is invertible and $(D-Q)^{-1}$ is positive elementwise (see [Reference Minc13]). That leads to ${\bf u}=0$ , which is a contradiction. Thus ${\bf u}{\bf e}$ cannot be zero.

If ${\bf u}{\bf e}$ is non-zero, without loss of generality, we assume that ${\bf u}{\bf e}=1$ . By (2.2), we obtain ${\bf u}=\lambda{\boldsymbol \pi}(D-Q)^{-1}$ . Thus u is positive elementwise and, in fact, is a stochastic vector. Multiplying both sides of (2.2) by e, we obtain $\sum_{i=1}^m u_id_i = {\bf u}D{\bf e} = \lambda$ . By (2.2), we also obtain

(2.3) \begin{equation} {\bf u}(D - \lambda{\bf e}{\boldsymbol \pi}) = {\bf u}Q = {\bf u}(\!\operatorname{diag} ({\boldsymbol \pi}))^{-1}\operatorname{diag} ({\boldsymbol \pi})Q = c_Q({\bf u}(\!\operatorname{diag} ({\boldsymbol \pi}))^{-1})(P_Q-I).\end{equation}

Multiplying both sides of the above equation by $(\!\operatorname{diag} ({\boldsymbol \pi}))^{-1}{\bf u}'$ , we obtain

(2.4) \begin{equation} {\bf u}(D - \lambda{\bf e}{\boldsymbol \pi})(\!\operatorname{diag} ({\boldsymbol \pi}))^{-1}{\bf u}' = c_Q ({\bf u}(\!\operatorname{diag} ({\boldsymbol \pi}))^{-1}P_Q(\!\operatorname{diag} ({\boldsymbol \pi}))^{-1}{\bf u}' - {\bf u}(\!\operatorname{diag} ({\boldsymbol \pi}))^{-2}{\bf u}').\end{equation}

By Lemma 2.1, $P_Q$ is a doubly stochastic matrix. By the well-known Birkhoff theorem for doubly stochastic matrices (see [Reference Minc13], or [Reference Marshall, Olkin and Arnold12, A.2 Theorem]), we have

\begin{equation*} P_Q = \sum_{\sigma\in\Omega} c_{\sigma}P_{\sigma},\end{equation*}

where $\Omega$ is the set of all permutations on $\{1, 2, \ldots, m\}$ , $\sigma \in \Omega$ , $P_{\sigma}$ is the permutation matrix associated with $\sigma$ , and $\{c_\sigma, \sigma\in \Omega\}$ are non-negative numbers with unit sum. For any real row vector ${\bf x}=(x_1, x_2, \ldots, x_m)$ and permutation $\sigma$ , it is well known that ${\bf x}P_{\sigma}{\bf x}' \leq {\bf x}{\bf x}'$ , which can be shown easily as follows:

(2.5) \begin{equation} {\bf x}P_{\sigma}{\bf x}' = \sum_{i=1}^m x_ix_{\sigma(i)}=\dfrac{1}{2}\biggl(\sum_{i=1}^m 2x_ix_{\sigma(i)}\biggr)\leq \dfrac{1}{2}\biggl(\sum_{i=1}^m\bigl(x_i^2+x_{\sigma(i)}^2\bigr)\biggr) = {\bf x}{\bf x}'.\end{equation}

Consequently, the right-hand side of (2.4) is less than or equal to zero, which leads to ${\bf u}(D-\lambda{\bf e}{\boldsymbol \pi})(\!\operatorname{diag} ({\boldsymbol \pi}))^{-1}{\bf u}' \leq 0$ . This last expression can be rewritten explicitly as follows:

(2.6) \begin{equation} \sum_{i=1}^m u_i\dfrac{d_i}{\pi_i}u_i - \lambda({\bf u}{\bf e})^2 = \sum_{i=1}^m \biggl(\dfrac{1}{d_i}\biggr)\dfrac{d_i^2u_i^2}{\pi_i^2}\pi_i - \lambda \leq 0.\end{equation}

Next we consider two cases: (i) all $\{d_1, d_2, \ldots, d_m\}$ are positive, and (ii) some $\{d_1, d_2, \ldots, d_m\}$ are zero.

(i) If all $\{d_1, d_2, \ldots, d_m\}$ are positive, we choose ${\boldsymbol \pi}$ as a measure defined on $\{1, 2, \ldots, m\}$ , $p=-1$ , $q=1/2$ (so that $1/p + 1/q = 1$ ), $f(i) = 1/d_{i}$ , and $g(i) = (d_iu_i/\pi_i)^2$ . By Hölder’s inequality (see (iii) in [12, D.1.C]),

\begin{equation*} \int fg \,\mathrm{d}{\boldsymbol \pi} \geq \biggl(\int f^p \,\mathrm{d}{\boldsymbol \pi}\biggr)^{1/p} \biggl(\int g^q \,\mathrm{d}{\boldsymbol \pi}\biggr)^{1/q},\end{equation*}

we obtain

(2.7) \begin{align} \sum_{i=1}^m \biggl(\dfrac{1}{d_i}\biggr)\dfrac{d_i^2u_i^2}{\pi_i^2}\pi_i & \geq \biggl(\sum_{i=1}^m\biggl(\dfrac{1}{d_i}\biggr)^{-1}\pi_i\biggr)^{-1}\biggl(\sum_{i=1}^m\biggl(\dfrac{d_i^2u_i^2}{\pi_i^2}\biggr)^{0.5}\pi_i\biggr)^2 \notag \\* &= \biggl(\sum_{i=1}^m d_i\pi_i\biggr)^{-1}\biggl(\sum_{i=1}^m d_iu_i\biggr)^2 \notag \\* &= \lambda,\end{align}

in which we have used the facts ${\boldsymbol \pi}D{\bf e}={\bf u}D{\bf e}=\lambda$ . Combining (2.6) and (2.7), we obtain $\lambda \leq \sum_{i=1}^mu_i^2d_i/\pi_i \leq \lambda$ , which implies that $\sum_{i=1}^mu_i^2d_i/\pi_i = \lambda$ . On the other hand, the equality in (2.7) holds if and only if $f^p$ and $g^q$ are proportional, i.e. $f^p/g^q$ is constant, which implies that $d_i/(d_iu_i/\pi_i) = \pi_i/u_i$ are the same for $i=1, 2, \ldots, m$ . Since ${\bf u}{\bf e}={\boldsymbol \pi}{\bf e} = 1$ , we must have ${\boldsymbol \pi}={\bf u}$ , which leads to ${\boldsymbol \pi}D = \lambda{\boldsymbol \pi}$ by (2.2). Consequently, we must have $d_i = \lambda$ for all $i=1, 2, \ldots, m$ , i.e. $D=\lambda I$ .

(ii) Next we consider the case for which some $\{d_1, \ldots, d_m\}$ are zero. We show that this is impossible if $D-Q-\lambda{\bf e}{\boldsymbol \pi}$ is non-invertible. Without loss of generality we suppose that $d_i=0$ for $i=1, 2, \ldots, k$ , and $d_i>0$ for $i=k+1, \ldots, m$ . Accordingly, we partition vectors ${\bf u}=({\bf u}_1, {\bf u}_2)$ , ${\boldsymbol \pi} = ({\boldsymbol \pi}_1, {\boldsymbol \pi}_2)$ , and matrices

\begin{equation*} Q = \begin{pmatrix} Q_{1,1}\;\;\;\;\; & Q_{1,2} \\ Q_{2,1}\;\;\;\;\; & Q_{2,2} \end{pmatrix}\!, \quad D = \begin{pmatrix} 0\;\;\;\;\; & 0 \\ 0\;\;\;\;\; & D_2 \end{pmatrix}\!.\end{equation*}

In the above proof, since $d_i=0$ for $i=1, \ldots, k$ , we can restrict ${\boldsymbol \pi}$ to ${\boldsymbol \pi}_2$ , u to ${\bf u}_2$ , and D to $D_2$ in (2.7) to obtain ${\bf u}_2 = \gamma{\boldsymbol \pi}_2$ for some positive constant $\gamma$ . We claim that $\gamma=1$ . By ${\bf u}(D-Q-\lambda{\bf e}{\boldsymbol \pi})=0$ , we obtain

\begin{align*} {\bf u}_1(Q_{1,1}+\lambda {\bf e}{\boldsymbol \pi}_1) + {\bf u}_2(Q_{2,1}+\lambda {\bf e}{\boldsymbol \pi}_1) &= 0, \\* {\bf u}_1(Q_{1,2}+\lambda {\bf e}{\boldsymbol \pi}_2) + {\bf u}_2(-D_2 + Q_{2,2}+\lambda {\bf e}{\boldsymbol \pi}_2) &= 0. \end{align*}

Using ${\bf u}_2 = \gamma{\boldsymbol \pi}_2$ and ${\bf u}{\bf e}={\bf u}_1{\bf e}+{\bf u}_2{\bf e} = 1$ , the above equations become

\begin{align*} {\bf u}_1Q_{1,1} + \gamma{\boldsymbol \pi}_2Q_{2,1} + \lambda{\boldsymbol \pi}_1 & = 0, \\* {\bf u}_1Q_{1,2} + \gamma{\boldsymbol \pi}_2(Q_{2,2}-D_2) + \lambda{\boldsymbol \pi}_2 &= 0. \end{align*}

Using ${\boldsymbol \pi}_1={\boldsymbol \pi}_2Q_{2,1}(-Q_{1,1})^{-1}$ and the above two equations, we obtain

\begin{equation*} {\boldsymbol \pi}_2 (\gamma(Q_{2,2}+Q_{2,1}(-Q_{1,1})^{-1}Q_{1,2}-D_2)+\lambda(I+Q_{2,1}(-Q_{1,1})^{-2}Q_{1,2})) = 0.\end{equation*}

Using ${\boldsymbol \pi}_2(Q_{2,2}+Q_{2,1}(-Q_{1,1})^{-1}Q_{1,2}) = 0$ , the above equation can be reduced to

\begin{equation*} {\boldsymbol \pi}_2 (\gamma D_2 - \lambda(I+Q_{2,1}(-Q_{1,1})^{-2}Q_{1,2})) = 0.\end{equation*}

Note that $\lambda = {\boldsymbol \pi}D{\bf e} = {\boldsymbol \pi}_2D_2{\bf e}>0$ . Multiplying both sides of the above equation by e yields, after canceling $\lambda$ ,

\begin{align*} \gamma = &\ {\boldsymbol \pi}_2(I+Q_{2,1}(-Q_{1,1})^{-2}Q_{1,2}){\bf e}\\ = & \ {\boldsymbol \pi}_2{\bf e} + {\boldsymbol \pi}_2Q_{2,1}(-Q_{1,1})^{-1}(-Q_{1,1})^{-1}Q_{1,2} {\bf e} = {\boldsymbol \pi}_2{\bf e} + {\boldsymbol \pi}_1{\bf e} = 1,\end{align*}

which proves the claim that $\gamma=1$ . On the other hand, we rewrite (2.3) as

\begin{equation*} \dfrac{1}{c_Q}{\bf u}(D - \lambda{\bf e}{\boldsymbol \pi}) + {\bf u}(\!\operatorname{diag} ({\boldsymbol \pi}))^{-1} = ({\bf u}(\!\operatorname{diag} ({\boldsymbol \pi}))^{-1})P_Q.\end{equation*}

For two vectors ${\bf x}=(x_1, x_2, \ldots, x_m)$ and ${\bf y}=(y_1, y_2, \ldots, y_m)$ , we reorganize their elements in ascending order to obtain $(x_{(1)}, x_{(2)}, \ldots, x_{(m)})$ and $(y_{(1)}, y_{(2)}, \ldots, y_{(m)})$ . Vector x majorizes y if $\sum_{i=1}^jx_{(i)} \leq \sum_{i=1}^jy_{(i)}$ , for $j=1, 2, \ldots, m-1$ , and ${\bf xe}={\bf ye}$ (see [Reference Marshall, Olkin and Arnold12, A.1]). Since $P_Q$ is a doubly stochastic matrix, vector ${\bf u}(\!\operatorname{diag} ({\boldsymbol \pi}))^{-1}$ majorizes ${\bf u}(D - \lambda{\bf e}{\boldsymbol \pi})/c_Q + {\bf u}(\!\operatorname{diag} ({\boldsymbol \pi}))^{-1}$ (see [Reference Marshall, Olkin and Arnold12, B.2 Theorem (Hardy, Littlewood, and Pólya)]). Define

\begin{align*} i_{(1)} &= \textrm{arg min}\biggl\{\dfrac{u_i}{\pi_i}, \ i=1, 2, \ldots, m\biggr\}, \\ j_{(1)} &= \textrm{arg min}\biggl\{\dfrac{u_j}{\pi_j} + \dfrac{1}{c_Q}(u_jd_j-\lambda\pi_j), \ j=1, 2, \ldots, m\biggr\}. \end{align*}

If $u_{i_{(1)}}/\pi_{i_{(1)}}\geq 1$ , we have $u_i/\pi_i\geq 1$ for all $i=1, 2, \ldots, m$ . By ${\bf u}{\bf e}={\boldsymbol \pi}{\bf e}=1$ , we must have ${\bf u}={\boldsymbol \pi}$ , which leads to ${\boldsymbol \pi}D=\lambda {\boldsymbol \pi}$ and $D=\lambda I$ by (2.2) and contradicts to $d_i=0$ for $i=1, 2, \ldots, k$ . Now we must have $u_{i_{(1)}}/\pi_{i_{(1)}} < 1$ . Then we must have $i_{(1)}\leq k$ and $d_{i_{(1)}} = 0$ , since $u_i/\pi_i=\gamma = 1$ for $i=k+1, \ldots, m$ . By the above definition of vector majorization, we must have

\begin{equation*} \dfrac{u_{i_{(1)}}}{\pi_{i_{(1)}}} + \dfrac{1}{c_Q}(u_{i_{(1)}}d_{i_{(1)}} - \lambda\pi_{i_{(1)}}) \geq \dfrac{u_{j_{(1)}}}{\pi_{j_{(1)}}} + \dfrac{1}{c_Q}(u_{j_{(1)}}d_{j_{(1)}} - \lambda\pi_{j_{(1)}}) \geq \dfrac{u_{i_{(1)}}}{\pi_{i_{(1)}}},\end{equation*}

which leads to $-\lambda\pi_{i_{(1)}} \geq 0$ , which is a contradiction. Consequently, if $D-Q-\lambda{\bf e}{\boldsymbol \pi}$ is non-invertible, then $\{d_1, \ldots, d_m\}$ must all be positive and equal to $\lambda$ , i.e. $D=\lambda I$ . This completes the proof.

Proof. For convenience, let $\Phi(x) = (D-xQ)^{-1}$ , for $x>0$ . Since $D-xQ$ is an M-matrix for $x>0$ , $\Phi(x)$ exists and is positive elementwise. Then we have $(D-xQ)\Phi(x) = I$ . Multiplying both sides of the expression by ${\boldsymbol \pi}$ , we obtain ${\boldsymbol \pi}D\Phi(x) = {\boldsymbol \pi}$ . Since ${\boldsymbol \pi}$ is positive elementwise, $\lim_{x\to\infty}D\Phi(x)$ exists. Since $xQ\Phi(x) = I - D\Phi(x)$ , $\lim_{x\to\infty}Q\Phi(x) = \lim_{x\to\infty}(I-D\Phi(x))/x = 0$ . Since Q is an irreducible infinitesimal generator, its left and right eigenvectors of eigenvalue zero (a.k.a. Perron–Frobenius eigenvectors [Reference Minc13]) are unique up to a constant. Thus we must have $\lim_{x\to\infty}\Phi(x) = {\bf e}{\bf y}$ , where y is a positive vector. In a similar way, we can show that $\lim_{x\to\infty}\Phi(x)Q = 0$ . Thus we must have $\lim_{x\to\infty}\Phi(x) = \rho {\bf e}{\boldsymbol \pi}$ , where $\rho$ is a positive constant. Letting x go to infinity in ${\boldsymbol \pi}D\Phi(x) = {\boldsymbol \pi}$ , we obtain ${\boldsymbol \pi}D\rho{\bf e}{\boldsymbol \pi} = {\boldsymbol \pi}$ , which leads to $\rho = 1/({\boldsymbol \pi}D{\bf e}) = 1/\lambda$ .

3. Main results for the MMPP class of PH-distributions

First, we show that $1/\lambda$ is a lower bound of the mean of X. We state the result in a more general manner, which is also needed in the proof of the result. Let $X_x$ be the PH-random variable with PH-representation $({\boldsymbol \pi}, xQ-D)$ for all $x > 0$ . We recall that D is diagonal, non-zero, and non-negative, and Q is irreducible throughout this paper.

Theorem 3.1. For $\{D, Q, {\boldsymbol \pi}\}$ defined in Section 2, we have

\begin{equation*} \mathbb{E}[X_x] = {\boldsymbol \pi}(D-xQ)^{-1}{\bf e} \geq \dfrac{1}{\lambda} \quad \textit{for} \ x>0.\end{equation*}

The equality holds in the equation if and only if all diagonal elements of D are equal to each other (i.e. $X_x$ has an exponential distribution with parameter $\lambda$ ). In addition, we have $\lim_{x\to0}\mathbb{E}[X_x] = {\boldsymbol \pi}D^{-1}{\bf e}$ and $\lim_{x\to\infty}\mathbb{E}[X_x] = 1/\lambda$ .

Proof. If $D=\lambda I$ , it is easy to verify that $\mathbb{E}[X_x] = 1/\lambda$ for all $x>0$ . Thus all results hold.

Now we suppose that $\{d_1, \ldots, d_m\}$ are not all equal to each other. First, we show that $\lim_{x\to0}\mathbb{E}[X_x] > 1/\lambda$ . If one of the diagonal elements of D is zero, $\lim_{x\to0}\mathbb{E}[X_x] = \infty > 1/\lambda$ . If all diagonal elements of D are positive, $\lim_{x\to0}\mathbb{E}[X_x] = \mathbb{E}[X_0] = {\boldsymbol \pi}D^{-1}{\bf e} = \sum_{i=1}^m \pi_i/d_i$ . Then $\lim_{x\to0}\mathbb{E}[X_x] > 1/\lambda$ is equivalent to

\begin{equation*}\biggl(\sum_{i=1}^m \pi_i/d_i\biggr)\biggl(\sum_{i=1}^m \pi_id_i\biggr) > 1,\end{equation*}

which can be verified easily as follows:

\begin{align*} \biggl(\sum_{i=1}^m \pi_i/d_i\biggr)\biggl(\sum_{i=1}^m \pi_id_i\biggr) &=\sum_{i=1}^{m}\pi_i^2 + \sum_{i< j}\pi_i\pi_j\biggl(\dfrac{d_i}{d_j} + \dfrac{d_j}{d_i}\biggr) \\ & > \sum_{i=1}^{m}\pi_i^2 + 2\sum_{i< j}\pi_i\pi_j \\ &= \biggl(\sum_{i=1}^m\pi_i\biggr)^2\\* &= 1.\end{align*}

Thus we must have $\mathbb{E}[X_x] > 1/\lambda$ , if x is close to zero (recall that $\{d_1, \ldots, d_m\}$ are not all equal to each other). If $\mathbb{E}[X_x] > 1/\lambda$ holds for all $x>0$ , the desired result is obtained. Otherwise, there must be $x>0$ such that $\mathbb{E}[X_x] = 1/\lambda$ , i.e. ${\boldsymbol \pi}(D-xQ)^{-1}{\bf e} = 1/\lambda$ , since $\mathbb{E}[X_x]$ is continuous in x for $x>0$ .

Next we show that ${\boldsymbol \pi}(D-xQ)^{-1}{\bf e} = 1/\lambda$ if and only if matrix $D-xQ-\lambda{\bf e}{\boldsymbol \pi}$ is non-invertible. If the matrix is non-invertible, there exists a non-zero row vector u such that ${\bf u}(D-xQ-\lambda{\bf e}{\boldsymbol \pi}) = 0$ . By the proof of Lemma 2.2, ${\bf u}{\bf e}$ cannot be zero, and without loss of generality we suppose that ${\bf u}{\bf e}=1$ . By (2.2) we obtain ${\bf u} = \lambda {\boldsymbol \pi}(D-xQ)^{-1}$ . Multiplying both sides of the expression by e, we obtain $1 = {\bf u}{\bf e} = \lambda {\boldsymbol \pi}(D-xQ)^{-1}{\bf e}$ . So we have shown the sufficiency of the condition. On the other hand, if $1 = \lambda {\boldsymbol \pi}(D-xQ)^{-1}{\bf e}$ , we can verify that $\lambda{\boldsymbol \pi}(D-xQ)^{-1}(D-xQ-\lambda{\bf e}{\boldsymbol \pi}) = 0$ , which implies that $D-xQ-\lambda{\bf e}{\boldsymbol \pi}$ has an eigenvalue zero. Thus $D-xQ-\lambda{\bf e}{\boldsymbol \pi}$ is non-invertible. This proves the necessity of the condition.

Since we suppose that $\{d_1, \ldots, d_m\}$ are not all equal to each other, by Lemma 2.2, matrix $D-xQ-\lambda{\bf e}{\boldsymbol \pi}$ is invertible. Thus ${\lambda\boldsymbol \pi}(D-xQ)^{-1}{\bf e} = 1$ cannot be true, which is a contradiction. Consequently, $\mathbb{E}[X_x]$ has to be above $1/\lambda$ for all $x>0$ for this case.

Combining results for the two cases (i.e. $\{d_1, \ldots, d_m\}$ are all equal or not), we have shown ${\boldsymbol \pi}(D-xQ)^{-1}{\bf e} \geq 1/\lambda$ for $x>0$ .

Finally, the result $\lim_{x\to\infty}\mathbb{E}[X_x] = 1/\lambda$ can be obtained directly from Lemma 2.3.

Let $X_{D,x}$ be the PH-random variable with PH-representation $({\boldsymbol \pi}_D, xQ-D)$ for all $x> 0$ . The squared coefficient of variation of $X_{D,x}$ is defined as

\begin{equation*}\operatorname{SCV}(X_{D,x}) = \mathbb{E}[(X_{D,x}- \mathbb{E}[X_{D,x}])^2]/(\mathbb{E}[X_{D,x}])^2.\end{equation*}

Theorem 3.2. For $\{D, Q, {\boldsymbol \pi}\}$ defined in Section 2, we have $\mathbb{E}[X_{D,x}]=1/\lambda$ and

(3.1) \begin{equation} \operatorname{SCV}(X_{D,x}) = 2\lambda \mathbb{E}[X_x] -1 \geq 1 \quad {for} \ x > 0.\end{equation}

The equality holds if and only if $D=\lambda I$ .

Proof. Since ${\boldsymbol \pi}D(D-xQ)^{-1}={\boldsymbol \pi}$ , we have

\begin{equation*} \mathbb{E}[X_{D,x}] = {\boldsymbol \pi}_D(D-xQ)^{-1}{\bf e} = \dfrac{{\boldsymbol \pi}D}{\lambda}(D-xQ)^{-1}{\bf e} = \dfrac{{\boldsymbol \pi}{\bf e}}{\lambda} = \dfrac{1}{\lambda}.\end{equation*}

By the definition of SCV, we have

\begin{equation*}\operatorname{SCV}(X_{D,x}) = \mathbb{E}[X^2_{D,x}]/(\mathbb{E}[X_{D,x}])^2 - 1.\end{equation*}

Further, by Theorem 3.1, we obtain

(3.2) \begin{align} \operatorname{SCV}(X_{D,x}) &= \dfrac{\mathbb{E}[X^2_{D,x}]}{(\mathbb{E}[X_{D,x}])^2} - 1 \notag \\ &= \dfrac{2{\boldsymbol \pi}_D(D-xQ)^{-2}{\bf e}}{(\mathbb{E}[X_{D,x}])^2} - 1 \notag \\ &= 2\lambda{\boldsymbol \pi}(D-xQ)^{-1}{\bf e} - 1 \notag \\ &\geq 2 - 1 \notag \\ &= 1.\end{align}

By Theorem 3.1, the equality holds in (3.2) if and only if $D=\lambda I$ , i.e. when $X_x$ and $X_{D,x}$ are exponential random variables with parameter $\lambda$ .

Define $\mathbb{E}[X_0] = {\boldsymbol \pi}D^{-1}{\bf e}$ if $d_i>0$ for $i=1, 2, \ldots, m$ , and $\infty$ otherwise. We note that, with the definition of $\mathbb{E}[X_0]$ , Theorems 3.1 and 3.2 hold for $x\geq0$ .

Proof. If at least one of $\{d_1, d_2, \ldots, d_m\}$ is zero, then $\mathbb{E}[X_0]=\infty$ and the desired result is obtained.

Now suppose that all $\{d_1, d_2, \ldots, d_m\}$ are positive. Then $\mathbb{E}[X_x]\leq \mathbb{E}[X_0]$ is equivalent to

(3.3) \begin{align} {\boldsymbol \pi}(D-xQ)^{-1}{\bf e}-{\boldsymbol \pi}D^{-1}{\bf e}& = {\boldsymbol \pi}(D-xQ)^{-1}(D - (D-xQ))D^{-1}{\bf e}\notag \\ & = x{\boldsymbol \pi}(D-xQ)^{-1}QD^{-1}{\bf e} \notag \\* &\leq 0.\end{align}

Let $Q_1 = D\operatorname{diag} ({\boldsymbol \pi})(D-xQ)^{-1}Q$ . It can be verified that $Q_1{\bf e}=0$ and ${\bf e}'Q_1=0$ , noting that $D\operatorname{diag} ({\boldsymbol \pi})=\operatorname{diag} ({\boldsymbol \pi})D$ . Then $P_1 = I + Q_1/c_1$ is a doubly stochastic matrix for $c_1 \geq \max\{-(Q_1)_{i,i}, i=1, 2, \ldots, m\}$ . Then (3.3) can be rewritten as

\begin{align*} x{\bf e}'D^{-1}D\operatorname{diag} ({\boldsymbol \pi})(D-xQ)^{-1}QD^{-1}{\bf e} &= xc_1(D^{-1}{\bf e})'(P_1-I)D^{-1}{\bf e} \\ &\leq xc_1(D^{-1}{\bf e})'(I-I)D^{-1}{\bf e} \\ &= 0,\end{align*}

where the inequality can be proved by using (2.5). The proof is complete.

Combining Theorems 3.1 and 3.3, we obtain $1/\lambda \leq \mathbb{E}[X_x] \leq \mathbb{E}[X_0]$ , for $x\geq 0$ .

For some special cases, we can further show that $\mathbb{E}[X_x]$ and $\operatorname{SCV}(X_{D,x})$ are decreasing in x, and $\operatorname{SCV}(X_x)\geq 1$ . A CTMC is called time-reversible if its infinitesimal generator Q satisfies $Q = (\!\operatorname{diag} ({\boldsymbol \pi}))^{-1}Q'\operatorname{diag} ({\boldsymbol \pi})$ (see [Reference Ross20, Section 6.6]).

Proof. To prove part (i), we show that the derivative of $\mathbb{E}[X_x]$ is non-positive for $x>0$ . By routine calculations, we can obtain

\begin{equation*} \dfrac{\mathrm{d}\mathbb{E}[X_x]}{\mathrm{d}x} = {\boldsymbol \pi}(D-xQ)^{-1}Q(D-xQ)^{-1}{\bf e} \quad \text{for} \ x>0.\end{equation*}

We rewrite the above expression as follows:

\begin{equation*} \dfrac{\mathrm{d}\mathbb{E}[X_x]}{\mathrm{d}x} = {\bf e}'(D-x\ \operatorname{diag} ({\boldsymbol \pi})Q(\!\operatorname{diag} ({\boldsymbol \pi}))^{-1})^{-1}(\!\operatorname{diag} ({\boldsymbol \pi})Q)(D-xQ)^{-1}{\bf e}.\end{equation*}

Note that ${\boldsymbol \pi}={\bf e}'\operatorname{diag} ({\boldsymbol \pi})$ and $\operatorname{diag} ({\boldsymbol \pi})D(\!\operatorname{diag} ({\boldsymbol \pi}))^{-1} = D$ . Let

\begin{equation*} {\bf u} = {\boldsymbol \pi}(D-xQ)^{-1}(\!\operatorname{diag} ({\boldsymbol \pi}))^{-1} = {\bf e}'(D-x\operatorname{diag} ({\boldsymbol \pi})Q(\!\operatorname{diag} ({\boldsymbol \pi}))^{-1})^{-1} \end{equation*}

and ${\bf v} = (D-xQ)^{-1}{\bf e}$ . Since Q is time-reversible, we obtain ${\bf v} = {\bf u}'$ . Using Lemma 2.1, similar to the proof of Lemma 2.2, we obtain

\begin{equation*} \dfrac{\mathrm{d}\mathbb{E}[X_x]}{\mathrm{d}x} = c_Q{\bf u}(P_Q-I){\bf u}' \leq 0 \quad \text{for} \ x>0.\end{equation*}

Therefore $\mathbb{E}[X_x]$ is non-increasing in x, and so is $\operatorname{SCV}(X_{D,x})$ by (3.1).

To prove part (ii), by (2.1), it is sufficient to show ${\boldsymbol \pi}(D-xQ)^{-2}{\bf e} \geq ({\boldsymbol \pi}(D-xQ)^{-1}{\bf e})^2$ . Using vectors u and v defined above, the inequality can be rewritten as

\begin{equation*} {\boldsymbol \pi}(D-xQ)^{-2}{\bf e} = {\bf u}\operatorname{diag} ({\boldsymbol \pi}){\bf v} \geq {\bf u}{\boldsymbol \pi}'{\boldsymbol \pi}{\bf v} = ({\boldsymbol \pi}(D-xQ)^{-1}{\bf e})^2.\end{equation*}

If Q is time-reversible, then ${\bf v}={\bf u}'$ , and the above inequality is equivalent to $\sum_{i=1}^m \pi_iu_i^2 \geq (\sum_{i=1}^m u_i\pi_i)^2$ . This last inequality can be proved as follows:

\begin{equation*} \biggl(\sum_{i=1}^m u_i\pi_i\biggr)^2 = \sum_{i=1}^mu_i^2\pi_i^2 + \sum_{i<j}\pi_i\pi_j(u_iu_j+u_ju_i) \leq \sum_{i=1}^mu_i^2\pi_i^2 + \sum_{i<j}\pi_i\pi_j(u_i^2+u_j^2)=\sum_{i=1}^m \pi_iu_i^2.\end{equation*}

This proves the theorem.

4. Applications, extensions, and future research