3.1. First passage time distribution

Let $\Theta_w$ be the first passage time of the process M(t) through a fixed boundary $w>0$ , i.e.

(17) \begin{equation}\Theta_w\;:\!=\;\inf\{t\ge 0: M(t)=w\}.\end{equation}

Note that the random variable $\Theta_w$ is identified with the first time instant at which the process X(t) crosses a fixed level w. The relationship between the moments of $\Theta_w$ and those of the constant phase C is shown in the following proposition.

Proposition 2. For $w>0$ , the expected value and the variance of the first passage time $\Theta_w$ are given by

(18) \begin{equation}{\mathbb E}(\Theta_w)={\mathbb E}(Z(w))=\lambda w {\mathbb E}(C),\end{equation}

(19) \begin{equation}Var(\Theta_w)=2 w {\mathbb E}(Z(w))+ Var(Z(w))=2\lambda w^2 {\mathbb E}(C)+\lambda w {\mathbb E}(C^2),\end{equation}

where the compound Poisson process Z(t) is as defined in Equation (11).

Proof. By using integration by parts, we have

\begin{equation*}\begin{aligned}\psi_Z(w,-s)&=\int_0^{+\infty}e^{-sx}h_Z(x,w)\textrm{d}x=s\int_0^{+\infty}e^{-sx}\mathbb P(Z(w)\le x)\textrm{d}x\\[5pt] &=s\int_0^{+\infty}e^{-sx}\mathbb P(M(x+w)>w)\textrm{d}x=s e^{sw}\int_{w}^{+\infty}e^{-sy}\mathbb P(M(y)>w)\textrm{d}y.\end{aligned}\end{equation*}

Hence, the equality

(20) \begin{equation}\frac{e^{-sw} \psi_Z(w,-s)}{s}=\int_{w}^{+\infty}e^{-sy}\mathbb P(M(y)>w)\textrm{d}y\end{equation}

holds, so that

(21) \begin{equation}\int_{w}^{+\infty}e^{-sy}\!\left(1-\mathbb P(M(y)>w)\right)\!\textrm{d}y=\frac{e^{-sw}}{s}-\int_w^{+\infty}e^{-sy}\mathbb P(M(y)>w)\textrm{d}y=\frac{e^{-sw}}{s}(1-\psi_Z(w,-s)).\end{equation}

Since

\begin{align*}\lim_{s\to 0}\int_{w}^{+\infty}e^{-sy}\left(1-\mathbb P(M(y)>w)\right)\textrm{d}y=\int_w^{+\infty}\mathbb P(M(y)\le w)\textrm{d}y=\int_w^{+\infty}\mathbb P(\Theta_w>y)\textrm{d}y=\mathbb E(\Theta_w),\end{align*}

and if we recall that

\begin{align*}\lim_{s\to 0} \frac{e^{-sw}}{s}(1-\psi_Z(w,-s))=\lim_{s\to 0} \left.\frac{\textrm{d}}{\textrm{d}s}\psi_Z(w,-s)\right|_{s=0}=\mathbb E(Z(w)),\end{align*}

the proof follows from Equation (21) and the first of the equations in (14).

Moreover, from Equation (20), by differentiating both members twice with respect to s and evaluating the result for $s=0$ , we obtain

\begin{align*}2 w \mathbb E(Z(w))+\mathbb E(Z^2(w))=2 w \frac{\textrm{d}}{\textrm{d}s}M_Z(w,s)\Big\vert_{s=0}+\frac{\textrm{d}^2}{\textrm{d}s^2}M_Z(w,0)=\int_w^{+\infty} 2 y \mathbb{P}(M(y)\le w) \textrm{d}y,\end{align*}

so that

(22) \begin{equation}\mathbb E(\Theta_w^2)= \mathbb E(Z(w))(2w+\mathbb E(Z(w)))+Var (Z(w)).\end{equation}

Hence, Equation (19) follows from Equations (14) and (22).

4.1. Exponentially distributed downward random times

Let us assume that the random variables $D_i$ , $i\in {\mathbb N}$ , are exponentially distributed with parameter $\mu>0$ . From now on, we assume $\mu>\lambda$ so that the condition (6) is satisfied. The probability density function of the constant phase C is provided in the following theorem.

Theorem 3. In the case of exponentially distributed downward times, for $t> 0$ , the probability density function of the constant phase random variable is given by

(23) \begin{equation}f_C(t)=\frac{\mu e^{-(\lambda+\mu)t/2}}{t\sqrt{\lambda\mu}}I_1(t\sqrt{\lambda\mu}),\end{equation}

where $I_n(x)$ is the nth-order modified Bessel function of the first kind,

(24) \begin{equation}I_n(x)\;:\!=\;\sum_{m=0}^{+\infty}\frac{1}{\Gamma(m+n+1)m!}\left(\frac x2\right)^{2m+n}.\end{equation}

Proof. If $D\sim Exp(\mu)$ , then $g^{(n)}(x)=\frac{\mu^n x^{n-1}e^{-\mu x}}{(n-1)!}$ , $n\in\mathbb N$ , $x>0$ . Hence, from Equation (7), for $t\ge0$ , it follows that

\begin{eqnarray*}&&\frac{1}{t}\sum_{n=1}^{+\infty} \frac{(\lambda t)^n e^{-\lambda t}}{n!} \int_0^t \frac{x \mu^n (t-x)^{n-1}e^{-\mu (t-x)}}{(n-1)!} \mu e^{-\mu x} \textrm{d}x\\[5pt] && =\frac{\mu e^{-(\lambda+\mu)t}}{t}\sum_{n=1}^{+\infty}\frac{(\lambda\mu t)^n}{n!(n-1)!}\int_0^t x(t-x)^{n-1}\textrm{d}x\\[5pt] && =\mu e^{-(\lambda+\mu)t}\sum_{n=1}^{+\infty} \frac{(\lambda\mu t^2)^n}{n!(n+1)!}=\mu e^{-(\lambda+\mu)t}\left(-1+\frac{I_1(2t\sqrt{\lambda\mu})}{t\sqrt{\lambda\mu}}\right),\end{eqnarray*}

so that the proof follows from Equation (10).

Plots of the density $f_C(t)$ and of the corresponding distribution function $F_C(t)$ are given in Figure 2 for some choices of the parameters.

Figure 2.

The density $f_C(t)$ (left) and the distribution function $F_C(t)$ (right) for $\lambda=0.2$ (solid), $\lambda=0.5$ (dashed), $\lambda=0.8$ (dotted), and $\mu=0.5$ .

In the following proposition we provide the expression for the n-fold convolution of the density $f_C$ .

Proposition 3. In the case of exponentially distributed downward times, for $t>0$ and $n\in\mathbb N$ , the n-fold convolution of the density $f_C$ is given by

(25) \begin{equation}f_C^{(n)}(t)=\frac{n}{t}\left(\frac{\mu}{\lambda}\right)^{n/2} e^{-(\lambda+\mu)t/2}I_n(t\sqrt{\lambda\mu}).\end{equation}

Proof. The proof is by induction on n. The equality holds for $n=1$ . We suppose that Equation (25) holds for $n\in\mathbb N$ , and we show it is then true for $n+1$ . We have

\begin{equation*}\begin{aligned}f_C^{(n+1)}(t)&=\int_0^tf_C^{(n)}(x)f_C(t-x)\textrm{d}x\\[5pt] &=\int_0^t \frac{n}{x}\left(\frac{\mu}{\lambda}\right)^{n/2} e^{-(\lambda+\mu)x/2}I_n(x\sqrt{\lambda\mu})\frac{\mu e^{-(\lambda+\mu)(t-x)/2}}{(t-x)\sqrt{\lambda\mu}}I_1((t-x)\sqrt{\lambda\mu})\textrm{d}x\\[5pt] &=\frac{n+1}{t}\left(\frac{\mu}{\lambda}\right)^{(n+1)/2}e^{-(\lambda+\mu)t/2}I_{n+1}(t\sqrt{\lambda\mu}),\end{aligned}\end{equation*}

following the equation (cf. Equation 11 in Section $2.15.19$ of Prudnikov et al. [Reference Prudnikov, Brychkov and Marichev33])

\begin{equation*}\int_0^a \frac{1}{x(a-x)}I_n(ac-cx)I_m(cx)\textrm{d}x=\frac{n+m}{anm}I_{m+n}(ac),\qquad a,n,m>0.\end{equation*}

Starting from Proposition 3, we now provide an explicit expression for the density $h_Z(z,t)$ of the compound process Z(t).

Proposition 4. In the case of exponentially distributed downward times, for $t>0$ , the density function of Z(t) is given by

\begin{equation*}h_Z(z,t)\;:\!=\;\frac{\textrm{d}}{\textrm{d}z} \mathbb P(Z(t)\le z) =\frac{e^{-\lambda t}e^{-(\lambda+\mu)z/2}}{z}(t\sqrt{\lambda\mu})\left(2\frac{t}{z}+1\right)^{-1/2}I_1\left(z\sqrt{\lambda\mu\left(1+\frac{2t}{z}\right)}\right).\end{equation*}

Moreover, we have $\mathbb P(Z(t)=0)=e^{-\lambda t}$ .

Proof. The equality follows from straightforward calculations. Indeed, for any $t> 0$ ,

\begin{equation*}\begin{aligned}h_Z(z,t)&=\sum_{n=0}^{+\infty}\frac{(\lambda t)^n e^{-\lambda t}}{n!} \, \frac{\mu^n}{(\lambda\mu)^{n/2}}\, \frac{n}{z}\, e^{-(\lambda+\mu)z/2}I_n(z\sqrt{\lambda\mu})\\[5pt] &=\frac{e^{-\lambda t}e^{-(\lambda+\mu)z/2}}{z}\sum_{n=1}^{+\infty}\frac{(t\sqrt{\lambda\mu})^{n/2}}{(n-1)!}I_n(z\sqrt{\lambda\mu})\\[5pt] &=\frac{e^{-\lambda t}e^{-(\lambda+\mu)z/2}}{z}t\sqrt{\lambda\mu}\left(2\frac{t}{z}+1\right)^{-1/2}I_1\left(\sqrt{\lambda\mu z^2+2\lambda\mu tz}\right)\\[5pt] &=\frac{e^{-\lambda t}e^{-(\lambda+\mu)z/2}}{z}t\sqrt{\lambda\mu}\left(2\frac{t}{z}+1\right)^{-1/2}I_1\left(z\sqrt{\lambda\mu(1+2t/z)} \right),\end{aligned}\end{equation*}

where we have used the following relation (cf. Equation 4 in Section $5.8.3$ of Prudnikov et al. [Reference Prudnikov, Brychkov and Marichev33]):

\begin{equation*}\sum_{k=0}^{+\infty}\frac{t^k}{k!}I_{n+k}(z)=\left(2\frac{t}{z}+1\right)^{-n/2}I_n\left(\sqrt{z^2+2tz}\right).\end{equation*}

Some plots of the density $h_Z(z,t)$ are provided in Figure 3 for different choices of the parameters.

Figure 3.

Left: the density $h_Z(z,t)$ as a function of t for $z=2$ , $\lambda=0.2$ (solid), $\lambda=0.5$ (dashed), $\lambda=0.8$ (dotted), and $\mu=0.8$ . Right: the density $h_Z(z,t)$ as a function of z for $\mu=0.3$ , $\lambda=0.5$ , $t=1$ (solid), $t=2$ (dashed), and $t=3$ (dotted).

The following proposition provides an expression for the survival function of the supremum M(t) in terms of a series of products of modified Bessel functions and generalized Laguerre polynomials.

Proposition 5. In the case of exponentially distributed downward times, for $t>0$ and $0\le w< t$ , the survival function of the supremum M(t) can be expressed as

(26) \begin{eqnarray}&& \mathbb P (M(t)>w)= 1- \frac{2 w \sqrt{\lambda \mu} }{(\lambda+\mu) \sqrt{t^2-w^2}} \textrm{e}^{-\frac{(\lambda+\mu)t}{2}}\textrm{e}^{\frac{(\mu-\lambda)w}{2}} \nonumber \\[5pt] && \times \sum_{j=0}^{+\infty} j! \left(-\frac{2 \sqrt{\lambda\mu}}{(\lambda+\mu)^2\sqrt{t^2-w^2}}\right)^j\, I_{j+1}\left(\sqrt{\lambda\mu (t^2-w^2)}\right)\, L_{j}^{-2j-1}(t(\lambda+\mu)),\end{eqnarray}

where $L_n^k(x)$ , $n\in {\mathbb N}$ , denotes the generalized Laguerre polynomial. Moreover, $\mathbb P (M(t)=t)=e^{-\lambda t}$ .

The proof of Proposition 5 is given in Appendix A.

The following proposition provides an asymptotic result concerning the survival probability.

Proposition 6. For $w>0$ , we have

\begin{equation*}\lim_{t\to+\infty} \mathbb {P}\left(M(t)>w\right)=1.\end{equation*}

In the case $\lambda=\mu$ , the following proposition provides an expression for the survival function of the supremum M(t) in terms of the Marcum Q-function, which is a special function occurring as a complementary cumulative distribution function for a non-central chi-squared random variable.

Proposition 7. In the case of exponentially distributed downward times, for $t>0$ , $0\le w<t$ , and $\lambda=\mu$ , the survival function of the supremum M(t) is given by

(27) \begin{equation}\mathbb P\left(M(t)>w\right)= e^{-\lambda t}I_0\left(\lambda\sqrt{t-w}\sqrt{t+w}\right)+2\left[1-Q_{1}\left(\sqrt{\lambda(t+w)},\sqrt{\lambda(t-w)}\right)\right],\end{equation}

where $Q_1\left(a,b\right)$ denotes the Marcum Q-function of order 1, defined as

\begin{align*}Q_1(a,b)=\int_b^{+\infty}x\exp\left(-\frac{x^2+a^2}{2}\right)I_0(ax)\textrm dx,\qquad b>0.\end{align*}

Moreover, $\mathbb P (M(t)=t)=e^{-\lambda t}$ .

Proof. From the definition of the compound Poisson process Z(t) (see Equation (11)), and by Proposition 4, we have that, for $0\le w\le t$ ,

\begin{equation*}\begin{aligned}\mathbb P (M(t)>w)&=H_Z(t-w,w)=\int_0^{t-w}h_Z(x,w)\textrm{d}x\\[5pt] &=e^{-\lambda w}w\sqrt{\lambda\mu}\int_0^{t-w}\frac{e^{-\frac{x(\lambda+\mu)}{2}}}{x}\, \frac{1}{\sqrt{2\frac{w}{x}+1}}I_1\left(x\sqrt{\lambda\mu\left(1+2\frac{w}{x}\right)}\right)\textrm{d}x+{e^{-\lambda w}}.\end{aligned}\end{equation*}

Making use of integration by parts, we have

(28) \begin{align}&\mathbb P\left(M(t)>w\right)=e^{-\lambda t}I_0\left(\lambda \sqrt{t-w}\sqrt{t+w}\right)\nonumber\\[5pt] &+\int_0^{t-w}e^{-\lambda(x+w)}\left[\lambda I_0\left(I_0\left(\lambda\sqrt{x}\sqrt{x+2w}\right)-\frac{\lambda \sqrt{x}}{\sqrt{x+2w}}I_1\left(\lambda \sqrt{x}\sqrt{x+2w}\right)\right)\right]\textrm dx\nonumber \\[5pt] &=e^{-\lambda t}I_0\left(\lambda \sqrt{t-w}\sqrt{t+w}\right)+\lambda \int_w^t e^{-\lambda y}\!\left[I_0\left(\lambda\sqrt{y-w}\sqrt{y+w}\right)-\frac{\sqrt{y-w}}{\sqrt{y+w}}I_1\!\left(\lambda\frac{\sqrt{y-w}}{\sqrt{y+w}}\right)\!\right]\!\textrm dy.\end{align}

Considering the definition of the modified Bessel function of the first kind, we have

(29) \begin{align}&\int_w^t e^{-\lambda y/2}\left(I_0\left(\lambda\sqrt{t+w}\sqrt{y-w}\right)-\sqrt{\frac{y-w}{t+w}}I_1\left( \lambda\sqrt{t+w}\sqrt{y-w}\right)\right)\textrm dy\nonumber\\[5pt] &=\frac{2 e^{-\lambda w/2}}{\sqrt{t+w}}\int_0^{\sqrt{t-w}}z e^{-\lambda z^2/2}\left(\sqrt{t+w}I_0\left(\lambda z \sqrt{t+w}\right)-zI_1\left(\lambda z \sqrt{t+w}\right)\right)\textrm dz\nonumber\\[5pt] &=\frac{2e^{-\lambda w/2}}{\sqrt{t+w}}\int_0^{\sqrt{t-w}}z e^{-\lambda/2 z^2}\left(\sum_{n=0}^{+\infty}\frac{\left(\sqrt{t+w}\right)^{2n+1}\left(\frac{\lambda}{2}z\right)^{2n}}{n!n!}-z \sum_{n=0}^{+\infty}\frac{\left(\sqrt{t+w}\right)^{2n+1}\left(\frac{\lambda}{2}z\right)^{2n+1}}{n!(n+1)!}\right)\textrm dz\nonumber\\[5pt] &=\frac{2e^{-\lambda w/2}}{\lambda}\left[\sum_{n=0}^{+\infty}\frac{\left(\frac{\lambda}{2}(t+w)\right)^n}{n!n!}\gamma\left(n+1,\frac{\lambda}{2}(t-w)\right)-\sum_{n=0}^{+\infty}\frac{\left(\frac{\lambda}{2}(t+w)\right)^n}{n!(n+1)!}\gamma\left(n+2,\frac{\lambda}{2}(t-w)\right)\right]\!,\end{align}

where $\gamma(s,x)$ denotes the lower incomplete gamma function. Hence, recalling the recurrence relation (see, for instance, Equation (6.5.22) of [Reference Abramowitz and Stegun1])

(30) \begin{equation}\gamma(s+1,x)=s\gamma(s,x)-x^se^{-x},\end{equation}

we find that Equation (29) becomes

\begin{align*}&\frac{2e^{-\lambda w/2}}{\lambda}\left[\sum_{n=0}^{+\infty}\frac{\left(\frac{\lambda}{2}(t+w)\right)^n}{n!n!}\gamma\left(n+1,\frac{\lambda(t-w)}{2}\right)\right.\\[5pt] &\left.-\sum_{n=0}^{+\infty}\frac{\left(\frac{\lambda}{2}(t+w)\right)^n}{n!(n+1)!}\left((n+1)\gamma\left(n+1,\frac{\lambda (t-w)}{2}\right)-e^{-\lambda/2(t-w)}\left(\frac\lambda2(t-w)\right)^{n+1}\right)\right]\\[5pt] &=\frac{2e^{-\lambda w/2}}{\lambda}\left[\sum_{n=0}^{+\infty}\frac{\left(\frac{\lambda}{2}(t+w)\right)^n\left(\frac{\lambda}{2}(t-w)\right)^{n+1}}{n!(n+1)!}e^{-\lambda(t-w)/2}\right]\\[5pt] &=\frac{2e^{-\lambda t}}{\lambda}\frac{\sqrt{t-w}}{\sqrt{t+w}}I_1\left(\lambda\sqrt{t-w}\sqrt{t+w}\right).\end{align*}

Hence

\begin{align*}&\int_w^t e^{-\lambda t/2}e^{-\lambda y/2}I_0\left(\lambda \sqrt{y-w}\sqrt{t+w}\right)\textrm dy \\[5pt] &= \int_w^t e^{-\lambda y}\left[I_0\left(\lambda\sqrt{y-w}\sqrt{y+w}-\frac{\sqrt{y-w}}{\sqrt{y+w}}I_1\left(\lambda \sqrt{y-w}\sqrt{y+w}\right)\right)\right]\textrm dy.\end{align*}

Finally, from Equation (28) it follows that

\begin{equation*}\begin{aligned}\mathbb{P}\left(M(t)>w\right)&=e^{-\lambda t}I_0\left(\lambda \sqrt{t-w}\sqrt{t+w}\right)+\lambda \int_w^t e^{-\lambda t/2}e^{-\lambda y/2}I_0\left(\lambda \sqrt{y-w}\sqrt{t+w}\right)\textrm dy\\[5pt] &= e^{-\lambda t}I_0\left(\lambda \sqrt{t-w}\sqrt{t+w}\right) + \int_0^{\lambda(t-w)}e^{-\lambda/2(t+w)-x/2}I_0\left(\sqrt{\lambda x}\sqrt{t+w}\right)\textrm dx,\end{aligned}\end{equation*}

and the desired result immediately follows from recalling the definition of the Marcum Q-function.

The result obtained in Proposition 7 is in agreement with that given in Cinque and Orsingher [Reference Cinque and Orsingher6] (cf. Equation (2.26)), as proven in the following remark.

Remark 2. The survival function of M(t), obtained by considering the distribution function given in Equation (2.26) of [Reference Cinque and Orsingher6] for $c_1=c_2=1$ , can be identified with Equation (27). Indeed,

\begin{align*} &\mathbb P\left(M(t)>w\right)=1-e^{-\lambda t}\sum_{j=0}^{+\infty}I_{j+1}\left(\lambda\sqrt{(t-w)(t+w)}\right)\left[\left(\sqrt{\frac{t+w}{t-w}}\right)^{j+1}-\left(\sqrt{\frac{t-w}{t+w}}\right)^{j+1}\right]\\[5pt] &=1-e^{-\lambda t}\sum_{m=1}^{+\infty}\frac{\left(\frac{\lambda}{2}\right)^m}{m!}(t+w)^m\sum_{j=0}^{m-1}\frac{\left(\frac\lambda 2\right)^j}{j!}(t-w)^j+e^{-\lambda t}\sum_{j=0}^{+\infty}\frac{\left(\frac{\lambda}{2}\right)^j}{j!}(t+w)^j\sum_{h=j+1}^{+\infty}\frac{\left(\frac{\lambda}{2}\right)^h}{h!}(t-w)^h\\[5pt] &=1+e^{-\lambda t}\left(-1+e^{\lambda/2(t-w)}\right)+e^{\lambda/2(t-w)}e^{-\lambda t}\sum_{m=1}^{+\infty}\frac{\left(\frac{\lambda}{2}\right)^m}{m!}(t+w)^m\\[5pt] & \times \left[\frac{\gamma(m+1,\frac\lambda 2(t-w))}{\Gamma\left(m+1\right)}+\frac{\gamma(m,\frac\lambda 2(t-w))}{\Gamma\left(m\right)}-1\right]. \end{align*}

Hence, by Equation (30), the survival function can be expressed as

\begin{equation*}\begin{aligned}\mathbb P\left(M(t)>w\right)=&-2e^{-\lambda t}+2e^{-\frac{\lambda}{2}(t+w)}+e^{-\lambda t}I_0\left(\lambda\sqrt{t-w}\sqrt{t+w}\right)\\[5pt] &+2e^{-\frac{\lambda}{2}(t+w)}\sum_{m=1}^{+\infty}\frac{\left(\frac{\lambda}{2}(t+w)\right)^m}{m!}\gamma\left(m+1,\frac{\lambda}{2}(t-w)\right)\\[5pt] &=e^{-\lambda t}I_0\left(\lambda\sqrt{t-w}\sqrt{t+w}\right)+2\left[1-Q_1\left(\sqrt{\lambda(t+w)},\sqrt{\lambda(t-w)}\right)\right],\end{aligned}\end{equation*}

where the last equality holds by virtue of Equation (61.2.5) of Hansen [Reference Hansen20].

Figure 4 shows the behavior of the distribution function $\mathbb P(M(t)\le w)$ for some choices of the parameters, obtained from (26) as the complementary function.

In the following proposition, we provide the expression for the expected value of M(t).

Proposition 8. In the case of exponentially distributed downward times, the expected value of the supremum M(t) is given by

(31) \begin{equation}\begin{aligned}\mathbb E\left(M(t)\right)&=\frac{1-e^{-\lambda t}}{\lambda}+\frac{\mu}{2\lambda} \sum_{h=0}^{+\infty}\frac{\left(\frac{\lambda\mu}{4}\right)^h }{h!(h+1)!}\sum_{r=0}^{h} \binom{h}{r}\left(\frac{2}{\lambda}\right)^r (r+1)!\\[5pt] &\times \left[\left(\frac{2}{\lambda+\mu}\right)^{1+2h-r}\gamma\left(1+2h-r,\frac{(\lambda+\mu)t}{2}\right)-e^{-\lambda t}t^{1+2h-r} \right.\\[5pt] &\times \left.\sum_{k=0}^{r+1} \frac{(\lambda t)^{k}}{k!}\beta\left(2h-r+1,k+1\right) {_1}F_1\left(1+2h-r;\;2+2h+k-r;\;\frac{(\lambda-\mu)t}{2}\right)\right],\end{aligned}\end{equation}

where ${_1}F_1(a;\;b;\;z)$ is the confluent hypergeometric function

(32) \begin{equation}{_1}F_1(a,b;\;z)\;:\!=\;\sum_{k=0}^{+\infty}\frac{(a)_kz^k}{(b)_kk!},\end{equation}

$\gamma(s,x)$ is the lower incomplete gamma function, and $\beta(x,y)$ denotes the beta function.

Figure 4.

The distribution function $\mathbb P(M(t)\le w)$ , in the case of exponentially distributed downward times, when $t=30$ . Left: $\mu=0.5$ , $\lambda=0.5$ (solid), $\lambda=0.3$ (dashed), and $\lambda=0.2$ (dotted). Right: $\lambda=0.5$ , $\mu=0.5$ (solid), $\mu=0.7$ (dashed), and $\mu=0.9$ (dotted).

The proof of Proposition 8 is given in Appendix B.

Remark 3. In the case $\lambda=\mu$ , the expected value of M(t) is given by

(33) \begin{equation}\mathbb E\left(M(t)\right)=e^{-\lambda t}t\left(I_0(\lambda t)+I_1\left(\lambda t\right)\right),\qquad t\ge 0.\end{equation}

The result follows from Equation (31) if we note that for $\mu=\lambda$ ,

\begin{align*}&\sum_{k=0}^{r+1}\frac{(\lambda t)^k}{k!}\beta\left(2h-r+1,k+1\right) {_1F_1}\left(1+2h-r;\;2+2h+k-r;\;0\right)\\[5pt] &\qquad =\frac{(\lambda t)^{1-2(h+1)+r}\left[(2h-r)!\Gamma\left(3+h,\lambda t\right)+(2h+2)!\Gamma(1+2h-r,\lambda t)\right]}{(2h+2)!}.\end{align*}

We remark that Equation (33) coincides with Equation (5.10) of Cinque and Orsingher [Reference Cinque and Orsingher5].

4.2. Erlang distributed downward random times

Let us now consider the case $D\sim Erlang(\mu,k)$ with $k\in \mathbb N$ , $\mu>0$ . In the sequel we shall assume $k/\mu<1/\lambda$ , which ensures the validity of the condition (6). The density of D is given by

\begin{align*}g(t)=\frac{\mu^k t^{k-1}e^{-\mu t}}{(k-1)!}, \quad t\ge 0.\end{align*}

In the following proposition, we provide the expression for the density of the random variable C.

Proposition 9. In the case of Erlang distributed downward random times, for any $t>0$ , the density of the constant phase C is given by

(34) \begin{equation}f_C(t)=\frac{1}{2}\mu^k k \left(\frac{t}{2}\right)^{k-1} e^{-(\lambda+\mu)t/2} W_{k,k+1}\Big[\lambda \mu^k \left(\frac{t}{2}\right)^{k+1}\Big],\end{equation}

where

(35) \begin{equation}W_{\rho,\beta}(z)\;:\!=\;\sum_{j=0}^{+\infty} \frac{z^j}{j! \Gamma(\rho j+ \beta)},\quad \rho>-1, \beta\in {\mathbb C},\end{equation}

is the Wright function.

Proof. From Equation (3), the density of Y(t) is

\begin{align*}h(u,t)=\frac{e^{-\lambda t -\mu u}}{u} \sum_{n=1}^{+\infty}\frac{(\lambda t \mu^k u^k)^n}{n! (nk-1)!},\quad 0<u<t.\end{align*}

Hence, taking into account Equation (7), the probability density of T, for any $t\ge 0$ , is

\begin{eqnarray*}&& f_T(t)=\textrm{e}^{-\lambda t} g(t)+\int_0^t \frac{x}{t}h(t-x,t)g(x)\textrm{d} x\\[5pt] && =\textrm{e}^{-\lambda t} g(t) + \int_0^t\frac{x}{t}\left(\frac{e^{-\lambda t -\mu (t-x)}}{t-x} \sum_{n=1}^{+\infty}\frac{(\lambda t \mu^k (t-x)^k)^n}{n! (nk-1)!} \right) \frac{\mu^kx^{k-1}e^{-\mu x}}{(k-1)!}\textrm{d}x \\[5pt] && =\textrm{e}^{-\lambda t} g(t) + \frac{\mu^k}{t(k-1)!}e^{-\lambda t-\mu t}\sum_{n=1}^{+\infty}\frac{(\lambda t \mu^k)^n}{n! (nk-1)!}\int_0^t x^k(t-x)^{kn-1}\textrm{d}x\\[5pt] && =\textrm{e}^{-\lambda t} g(t) + \frac{\mu^k}{t(k-1)!}e^{-\lambda t-\mu t}\sum_{n=1}^{+\infty}\frac{(\lambda t \mu^k)^n}{n! (nk-1)!}\cdot\frac{t^{k(n+1)}k!(nk-1)!}{(k+kn)!}\\[5pt] && =\textrm{e}^{-\lambda t} g(t)+ \mu^k k t^{k-1} e^{-t(\lambda+\mu)}\sum_{n=1}^{+\infty}\frac{(\lambda\mu^k t^{k+1})^n}{n!(k+kn)!}.\end{eqnarray*}

Finally, by Equation (10), we have for $t\ge 0$

\begin{eqnarray*}&& f_C(t)=\frac{1}{2} e^{-(\lambda+\mu)t/2} \frac{\mu^k ({t}/{2})^{k-1}}{(k-1)!}+\frac{1}{2}\mu^k k \left(\frac{t}{2}\right)^{k-1}e^{-(\lambda+\mu)t/2}\sum_{n=1}^{+\infty} \frac{(\lambda \mu^k \left({t}/{2}\right)^{k+1})^n}{n!(k+kn)!}\\[5pt] && =\frac{1}{2}\mu^k k \left(\frac{t}{2}\right)^{k-1}e^{-(\lambda+\mu)t/2} \sum_{n=0}^{+\infty} \frac{(\lambda \mu^k \left({t}/{2}\right)^{k+1})^n}{n!(k+kn)!},\end{eqnarray*}

so that the proof immediately follows from Equation (35).

Remark 4. Note that for $k=1$ Equation (34) becomes

\begin{align*}\begin{aligned}f_C(t)\big|_{k=1}=&\frac12\mu e^{-(\lambda+\mu)t/2}\,W_{1,2}\left(\lambda\mu\left(\frac t2\right)^2\right)=\frac12\mu e^{-(\lambda+\mu)t/2}\,\sum_{m=0}^{+\infty}\frac{\left(\lambda\mu\left(\frac t2\right)^2\right)^m}{m!(m+1)!}\\[5pt] =&\mu e^{-(\lambda+\mu)t/2}\frac{I_1\left(\sqrt{\lambda\mu}t\right)}{\sqrt{\lambda\mu}t},\quad t\ge 0,\end{aligned}\end{align*}

which is the density of C in the case of exponentially distributed downward random times.

Proposition 10. In the case of Erlang distributed downward times, the upper bound U(t, w) provided in Proposition 1 has the following expression: for any $t>w$ ,

(36) \begin{equation}U(t,w)=\exp\left\{-\lambda w\left[1-k\left(\frac{\mu}{\lambda+\mu}\right)^k\sum_{j=0}^{+\infty}\frac{(\lambda\mu^k)^j\gamma\left(j+jk+k,\frac{(\lambda+\mu)(t-w)}{2}\right)}{j!(\lambda+\mu)^{j+jk}\Gamma\left(jk+k+1\right)} \right]\right\},\end{equation}

where $\gamma(s,x)$ is the lower incomplete gamma function.

Proof. From Equation (34), the distribution function of C is given by

(37) \begin{equation}F_C(t)=\int_0^t f_C(s)\textrm{d}s=k\left(\frac{\mu}{\lambda+\mu}\right)^k\sum_{j=0}^{+\infty}\frac{(\lambda\mu^k)^j\gamma\left(j+jk+k,\frac{(\lambda+\mu)t}{2}\right)}{j!(\lambda+\mu)^{j+jk}\Gamma\left(jk+k+1\right)}.\end{equation}

Hence, the result immediately follows from Proposition 1.

In Figure 5 we provide plots of the lower bound L(t, w) defined in Equation (16) with $F_C(w)$ as given in Equation (37) and U(t, w) as given in Equation (36), for different choices of parameters.

Figure 5.

The lower bound L(t, w) and the upper bound U(t, w) of the survival function of M(t), in the case of Erlang distributed downward times, for $\mu=9$ , $k=2$ , $t=3$ , $\lambda=2$ (left), and $\lambda=3$ (right).

4.3. Weighted exponentially distributed downward random times

Let us assume that $D\sim WE(\alpha,\mu)$ , with $\alpha,\mu>0$ , so that the density of D is given by

(38) \begin{equation}g(t)=\frac{\alpha+1}{\alpha}\mu e^{-\mu t} (1-e^{-\alpha\mu t}), \quad t>0.\end{equation}

A random variable having such a density is said to have a weighted exponential distribution (see Gupta and Kundu [Reference Gupta and Kundu19] and Das and Kundu [Reference Das and Kundu8]). Note that the exponential distribution can be obtained from Equation (38) by letting $\alpha\to +\infty$ . We assume $0<\lambda\le \frac{\mu}{2}$ , $\alpha>0$ (or $\frac{\mu}{2}<\lambda<\mu$ , $\alpha>\frac{\mu-2\lambda}{\lambda-\mu}$ ), so that the condition (6) is fulfilled. In the following proposition we obtain the expression for the density of the constant phase C.

Theorem 4. In the case of weighted exponentially distributed downward times, the probability density function of the random variable C is given by

(39) \begin{equation}f_C(t)=\frac{(1+\alpha)\mu}{2\alpha}e^{-\frac{(\lambda+\mu)t}{2}-\frac{\alpha\mu t}{4}}\sqrt{\pi\alpha\mu\frac{t}{2}}\sum_{n=0}^{+\infty}\frac{\left(\lambda\frac{t^2}{4}\right)^n}{n!(n+1)!}\left(\frac{(1+\alpha)\mu}{\alpha}\right)^nI_{n+1/2}\left(\frac{\alpha\mu t}{4}\right),\quad t\ge 0.\end{equation}

Proof. From Condition 4 of [Reference Gupta and Kundu19], by considering two independent exponentially distributed random variables, i.e. $U\sim Exp(\mu)$ and $V\sim Exp(\mu(1+\alpha))$ , we can easily see that

\begin{align*}U+V\overset{d}{=}D \sim WE(\alpha,\mu).\end{align*}

Hence, if $\{D_1,\ldots,D_n\}$ is a collection of n independent random variables such that $D_i\sim WE(\alpha,\mu)$ for any $i=1,\ldots,n$ , then

\begin{equation*}\sum_{i=1}^n D_i\overset{d}{=}\widetilde U +\widetilde V,\end{equation*}

where $\widetilde U\sim Erlang(\mu(1+\alpha),n)$ and $\widetilde V\sim Erlang (\mu,n)$ . Moreover, denoting by $f_{\widetilde U}(x)$ and $f_{\widetilde V}(x)$ the densities of $\widetilde U$ and $\widetilde V$ respectively, we have

\begin{equation*}\begin{aligned}g^{(n)}(t)&=\int_{-\infty}^{+\infty} f_{\widetilde U}(x)f_{\widetilde V}(t-x)\textrm{d}x=\int_0^t \frac{\mu^n (1+\alpha)^nx^{n-1}e^{-\mu(1+\alpha)x}}{(n-1)!}\cdot \frac{\mu^n(t-x)^{n-1}e^{-\mu(t-x)}}{(n-1)!}\textrm{d}x\\[5pt] &=\frac{\mu^{2n}(1+\alpha)^n}{[(n-1)!]^2}\int_0^t x^{n-1}e^{-\mu(1+\alpha)x}(t-x)^{n-1}e^{-\mu(t-x)}\textrm{d}x\\[5pt] &=\left[\frac{\mu(1+\alpha)}{\alpha}\right]^n\frac{e^{-\frac{1}{2}(2+\alpha)\mu t}}{(n-1)!}\sqrt{\frac{\alpha\mu \pi }{t}} t^n I_{n-1/2}\left(\frac{\alpha\mu t}{2}\right).\end{aligned}\end{equation*}

Hence, from Equation (7), it follows that

\begin{equation*}\begin{aligned}f_T(t)&=\mu e^{-(\lambda +\mu)t}\frac{\alpha+1}{\alpha}(1-e^{-\alpha\mu t})-\frac{1}{t}\sum_{n=1}^{+\infty}\frac{(\lambda t)^n e^{-\lambda t}}{n!}\int_0^te^{-\mu t-\alpha\mu t-1/2\alpha\mu x}\left(-1+e^{\alpha\mu(t-x)}\right)\\[5pt] &\times \sqrt{\frac{\alpha\mu\pi}{x}}\left(\frac{(1+\alpha)\mu x}{\alpha}\right)^{n+1}(x-t)I_{n-1/2}\left(\frac{\alpha\mu x}{2}\right)\frac{1}{n! \, x}\textrm{d}x\\[5pt] &=\mu e^{-(\lambda +\mu)t}\frac{\alpha+1}{\alpha}(1-e^{-\alpha\mu t}) -\frac{e^{-\mu t-\lambda t -\alpha\mu t}}{t}\sqrt{\alpha\mu \pi}\\[5pt] &\times \sum_{n=1}^{+\infty}\frac{(\lambda t)^n}{n!(n-1)!}\left(\frac{(1+\alpha)\mu}{\alpha}\right)^{n+1}\int_0^t e^{\frac{\alpha\mu x}{2}}x^{n-1/2}(t-x) I_{n-1/2}\left(\frac{\alpha\mu x}{2}\right)\textrm{d}x\\[5pt] &+\frac{e^{-\mu t-\lambda t}}{t}\sqrt{\alpha\mu \pi}\sum_{n=1}^{+\infty}\frac{(\lambda t)^n}{n!(n-1)!}\left(\frac{(1+\alpha)\mu}{\alpha}\right)^{n+1}\!\int_0^t e^{-\frac{\alpha\mu x}{2}}x^{n-1/2}(t-x) I_{n-1/2}\!\left(\frac{\alpha\mu x}{2}\right)\!\textrm{d}x\\[5pt] &=\mu e^{-(\lambda +\mu)t}\frac{\alpha+1}{\alpha}(1-e^{-\alpha\mu t})+e^{-\mu t-\lambda t-\frac{\alpha\mu t}{2}}\sqrt{\alpha\mu\pi t}\sum_{n=1}^{+\infty}\frac{(\lambda t^2)^n}{n!(n+1)!}\left(\frac{(1+\alpha)\mu}{\alpha}\right)^{n+1}\\[5pt] &\times I_{n+1/2}\left(\frac{\alpha\mu t}{2}\right) =e^{-\mu t-\lambda t-\frac{\alpha\mu t}{2}}\sqrt{\alpha\mu\pi t}\sum_{n=0}^{+\infty}\frac{(\lambda t^2)^n}{n!(n+1)!}\left(\frac{(1+\alpha)\mu}{\alpha}\right)^{n+1}I_{n+1/2}\left(\frac{\alpha\mu t}{2}\right).\end{aligned}\end{equation*}

Finally, the result follows from taking into account Equation (10).

Remark 5. For $\alpha\to +\infty$ , since (see, for instance, Equation (9.7.1) of [Reference Abramowitz and Stegun1])

\begin{align*}I_{n+1/2}\left(\frac{\alpha \mu t}{4}\right)\propto \frac{e^{\alpha\mu t/4}}{\sqrt{\frac{\pi\alpha\mu t}{2}}},\end{align*}

we have that

\begin{align*}\lim_{\alpha\to+\infty} f_C(t)=\frac 12\mu e^{-(\lambda+\mu)t/2} \frac{I_1\left(t\sqrt{\lambda\mu}\right)}{\sqrt{\lambda\mu}t},\quad t\ge 0,\end{align*}

which is identical to the density function of C in the case of exponentially distributed downward random times.

The following proposition provides the expression for the upper bound for the survival function of M(t).

Proposition 11. In the case of weighted exponentially distributed downward times, the upper bound U(t, w) (see Proposition 1) has the following expression:

(40) \begin{equation}\begin{aligned}U(t,w)&=\exp\left\{-\lambda w \left[1- \frac{(1+\alpha)\mu}{2\alpha}\frac{\sqrt{2\pi}\alpha\mu}{(\lambda+\mu+\alpha\mu)^2}\sum_{n=0}^{+\infty}\frac{1}{n!(n+1)!}\left(\frac{(1+\alpha)\mu\lambda}{\alpha}\right)^n\right. \right.\\[5pt] &\times \left. \left. \frac{1}{(\lambda+\mu+\alpha\mu)^{3n}}\left(\frac{\alpha\mu}{2}\right)^n\sum_{k=0}^{+\infty}\left(\frac{\alpha\mu/2}{\lambda+\mu+\alpha\mu}\right)^{2k}\frac{1}{\Gamma\left(k+n+3/2\right)k!} \right.\right.\\[5pt] &\left.\left. \times\gamma\left(2+2k+3n,(\lambda+\alpha\mu+\mu)\frac{t-w}{2}\right)\right]\right\},\quad t>w>0,\end{aligned}\end{equation}

where $\gamma(s,x)$ is the lower incomplete gamma function.

Proof. Recalling Equation (24), from Equation (39) we have that the distribution of C, for $t>w>0$ , is given by

(41) \begin{equation}\begin{aligned}F_C(w)&=\frac{(1+\alpha)\mu}{2\alpha}\sqrt{\pi\alpha\mu} \sum_{n=0}^{+\infty} \frac{1}{n!(n+1)!}\left(\frac{(1+\alpha)\mu\lambda}{\alpha}\right)^n\\[5pt] & \times\int_0^w \exp\left(-\frac{(\lambda+\mu)t}{2}-\frac{\alpha\mu t}{4}\right)\left(\frac{t}{2}\right)^{2n+1/2} I_{n+1/2}\left(\frac{\alpha\mu t}{2}\right)\textrm dt\\[5pt] &=\frac{(1+\alpha)\mu}{2\alpha}\sqrt{\pi\alpha\mu} \sum_{n=0}^{+\infty} \frac{1}{n!(n+1)!}\left(\frac{(1+\alpha)\mu\lambda}{\alpha}\right)^n\\[5pt] &\times \sum_{k=0}^{+\infty}\frac{1}{\Gamma(k+n+3/2)k!}\int_0^w \exp\left(-\frac{(\lambda+\mu)t}{2}-\frac{\alpha\mu t}{4}\right)\left(\frac{t}{2}\right)^{2n+1/2}\left(\frac{\alpha\mu t}{4}\right)^{2k+n+1/2}\textrm dt\\[5pt] &=\frac{(1+\alpha)\mu}{2\alpha}\sqrt{\pi\alpha\mu} \sum_{n=0}^{+\infty} \frac{1}{n!(n+1)!}\left(\frac{(1+\alpha)\mu\lambda}{\alpha}\right)^n\\[5pt] &\times \sum_{k=0}^{+\infty}\frac{2^{1/2-2k-n}}{\Gamma\left(k+n+3/2\right)k!}\cdot\frac{(\alpha\mu)^{1/2+2k+n}}{(\lambda+\mu+\alpha\mu)^{2+2k+3n}} \,\gamma\left(2+2k+3n,(\lambda+\alpha\mu+\mu)\frac{w}{2}\right).\end{aligned}\end{equation}

Hence, the result follows from Proposition 1.

Figure 6 shows plots of the lower bound L(t, w) defined in Equation (16) with $F_C(w)$ as given in Equation (41) and the upper bound U(t, w) given in Equation (40), in the case of weighted exponentially distributed downward times, for various choices of the parameters.

Figure 6.

The lower bound L(t, w) and the upper bound U(t, w) of the survival function of M(t), in the case of weighted exponentially distributed downward times, for $\alpha=1$ , $\lambda=4$ , $t=3$ , $\mu=10$ (left), and $\mu=12$ (right).

4.4. Mixed exponential downward random times

This section is devoted to the case in which the downward time D is distributed as a mixture of two exponential distributions, so that the density of D is given by

(42) \begin{equation}f_D(t)=b_1\mu_1e^{-\mu_1 t}+b_2\mu_2e^{-\mu_2 t},\quad t\ge 0,\end{equation}

with $b_i\ge 0$ , for $i=1,2$ and $b_1+b_2=1$ . Throughout the section we assume that

\begin{align*}\frac{b1}{\mu_1}+\frac{b_2}{\mu_2}<\frac{1}{\lambda},\end{align*}

so that the condition (6) is fulfilled.

The following theorem provides the expression for the density of the constant phase C.

Theorem 5. In the case of mixed exponential downward times (see Equation (42)), for $t\geq 0$ , the probability density function of the constant phase C is given by

(43) \begin{align}&f_C(t)=\frac{1}{2}\sum_{n=0}^{+\infty}\left(\frac{\lambda t^2}{4}\right)^n \frac{e^{-\lambda t/2}}{(n+1)!}\sum_{k=0}^n \frac{\binom{n}{k}}{n!}b_2^n\mu_2^n\left(\frac{b_1\mu_1}{b_2\mu_2}\right)^k\nonumber \\[5pt]&\times \left[b_1\mu_1e^{-\mu_1 t/2}{_1F_1}\left(n-k,n+2,(\mu_1-\mu_2)\frac{t}{2}\right) +b_2\mu_2e^{-\mu_2 t/2}{_1F_1}\left(k,n+2,(\mu_2-\mu_1)\frac{t}{2}\right)\right]\!,\end{align}

where ${_1}F_1(a,b;\;z)$ is as defined in Equation (32).

Proof. Let us start with the computation of the n-fold convolution of $f_D$ . Setting $f_i(t)\;:\!=\;\mu_ie^{-\mu_i t}$ for any $i=1,2$ , we have

\begin{equation*}\begin{aligned}f_D^{(n)}(t)=\sum_{k=0}^{n} \binom{n}{k}b_1^kb_2^{n-k}f_1^{(k)}(t)\star f_2^{(n-k)}(t), \quad t\ge 0,\end{aligned}\end{equation*}

where $f\star g$ denotes the convolution of f and g. In particular, we have that

\begin{eqnarray*}&& f_1^{(k)}(t)\star f_2^{(n-k)}(t)=\int_{-\infty}^{+\infty}f_1^{(k)}(x)f_2^{(n-k)}(t-x) \textrm{d}x=\frac{\mu_1^k\mu_2^{n-k}e^{-\mu_2 t}}{(k-1)!(n-k-1)!}\\[5pt] && \times \int_0^t x^{k-1}(t-x)^{n-k-1}e^{-\mu_1 x}e^{\mu_2 x}\textrm{d}x=\frac{\mu_1^k\mu_2^{n-k}e^{-\mu_2 t}t^{n-1} {_1F_1(k,n,(\mu_2-\mu_1)t)}}{(n-1)!}.\end{eqnarray*}

Hence, $f_D^{(n)}(t)$ can be expressed as follows:

\begin{align*}f_D^{(n)}(t)=\sum_{k=0}^n\binom{n}{k}b_1^kb_2^{n-k}\mu_1^k\mu_2^{n-k}e^{-\mu_2 t} t^{n-1} \frac{{_1F_1(k,n,(\mu_2-\mu_1)t)}}{(n-1)!}.\end{align*}

From Equation (7), we have

\begin{equation*}\begin{aligned}f_T(t)&=\frac{1}{t}\sum_{n=0}^{+\infty}p(n,\lambda t)\int_0^t x f_D^{(n)}(t-x)f_D(x)\textrm{d}x=\frac{1}{t}\sum_{n=0}^{+\infty}p(n,\lambda t)\sum_{k=0}^n\frac{\binom{n}{k}}{(n-1)!}b_2^n\mu_2^n \left(\frac{b_1\mu_1}{b_2\mu_2}\right)^k\\[5pt] &\times \left[b_1\mu_1\int_0^t xe^{-\mu_2(t-x)}(t-x)^{n-1}e^{-\mu_1 x}{_1F_1(k,n,(\mu_2-\mu_1)(t-x))}\textrm{d}x\right.\\[5pt] &+\left. b_2\mu_2\int_0^t xe^{-\mu_2(t-x)}(t-x)^{n-1}e^{-\mu_2 x}{_1F_1(k,n,(\mu_2-\mu_1)(t-x))}\textrm{d}x \right].\end{aligned}\end{equation*}

It is not hard to show that

\begin{equation*}\begin{aligned}&\int_0^t x e^{-\mu_2(t-x)}(t-x)^{n-1}e^{-\mu_1x}{_1F_1(k,n,(\mu_2-\mu_1)(t-x))}\textrm{d}x\\[5pt] &=e^{-\mu_1t}\int_0^t x(t-x)^{n-1}{_1F_1(n-k,n,(\mu_1-\mu_2)(t-x))}\textrm{d}x \\[5pt] &=e^{-\mu_1 t}\frac{t^{n+1}}{n(n+1)}{_1F_1(n-k,n+2,(\mu_1-\mu_2)t)},\end{aligned}\end{equation*}

and

\begin{align*}&\int_0^t x e^{-\mu_2(t-x)}(t-x)^{n-1}e^{-\mu_2x}{_1F_1(k,n,(\mu_2-\mu_1)(t-x))}\textrm{d}x\\[5pt] & = e^{-\mu_2t}\int_0^t x(t-x)^{n-1}{_1F_1(k,n,(\mu_2-\mu_1)(t-x))}\textrm{d}x\\[5pt] & = e^{-\mu_2 t}\frac{t^{n+1}}{n(n+1)}{_1F_1(k,n+2,(\mu_2-\mu_1)t)}.\end{align*}

Hence, Equation (43) follows from Equation (5).

Remark 6. For $b_1=b_2=\frac 12$ and $\mu_1=\mu_2=\mu$ , since $_1F_1(n-k,n+2,0)=1$ , we have that

\begin{align*}\begin{aligned}f_C(t)\Big|_{b_1=b_2=\frac{1}{2},\, \mu_1=\mu_2=\mu}&=\frac{\mu e^{-(\lambda+\mu) t/2}}{2}\sum_{n=0}^{+\infty}\left(\frac{\lambda t^2}{4}\right)^n\frac{\left(\frac\mu 2\right)^n}{n!(n+1)!}\,\sum_{k=0}^n\binom{n}{k}\\[5pt] &=\frac{\mu e^{-(\lambda+\mu) t/2}I_1\left(\sqrt{\lambda\mu}t\right)}{\sqrt{\lambda\mu}\,t},\quad t\ge 0,\end{aligned}\end{align*}

which is the same as the corresponding result in the case of exponentially distributed downward random times.

Proposition 12. In the case of mixed exponential downward times (Equation (42)), the upper bound U(t, w) given in Proposition 1 has the following expression:

(44) \begin{equation}U(t,w)=\exp\left[-\lambda w \left(1-F_C(t-w)\right) \right],\quad t>w>0,\end{equation}

where

(45) \begin{align}& F_C(w)= b_1\mu_1 \sum_{n=0}^{+\infty} \frac{(b_2\mu_2\lambda)^n}{n!(n+1)!}\left[ \sum_{m=0}^{+\infty}\frac{(m+n-1)!\,{_2}F_1\left(1-n,-n;1-m-n;-\frac{b_1\mu_1}{b_2\mu_2}\right)}{(n-1)! (n+2)_m m! (\lambda+\mu_1)^{m+2n+1}}\right.\nonumber\\[5pt] & \times (\mu_1-\mu_2)^m \,\gamma\left(1+m+2n,\frac{(\lambda+\mu_1)w}{2}\right)+n \sum_{m=0}^{+\infty}\frac{m! {_2}F_1\left(m+1,1-n;\;2;\;-\frac{b_1\mu_1}{b_2\mu_2}\right)}{(n+2)_m m! (\lambda+\mu_2)^{m+2n+1}}\nonumber\\[5pt] & \left.\times (\mu_1-\mu_2)^{m}\,\gamma\left(1+m+2n,\frac{(\lambda+\mu_2)w}{2}\right)\right],\end{align}

$\gamma(s,x)$ is the lower incomplete gamma function, and ${_2}F_1(a,b,c;\;z)$ is the Gauss hypergeometric function, defined as

(46) \begin{equation}{_2}F_1(a,b,c;\;z)\;:\!=\;\sum_{k=0}^{+\infty}\frac{(a)_k(b)_k z^k}{(c)_k k!}.\end{equation}

Proof. From Equation (43), the distribution of C is given by

\begin{eqnarray*}&& F_C(w)=\frac{1}{2}\sum_{n=0}^{+\infty} \frac{b_2^n\mu_2^n}{n!(n+1)!}\sum_{k=0}^n\binom{n}{k}\left(\frac{b_1\mu_1}{b_2\mu_2}\right)^k \\[5pt] && \times \left[b_1\mu_1\int_0^w e^{-(\lambda +\mu_1) t/2}{_1F_1}\left(n-k,n+2,\frac{(\mu_1-\mu_2)t}{2}\right)\left(\frac{\lambda t^2}{4}\right)^n\textrm dt\right.\\[5pt] && +\left.b_2\mu_2\int_0^w e^{-(\lambda +\mu_2) t/2}{_1F_1}\left(k,n+2,\frac{(\mu_1-\mu_2)t}{2}\right)\left(\frac{\lambda t^2}{4}\right)^n\textrm dt\right].\end{eqnarray*}

Recalling Equation (32), one has

\begin{align*}&\int_0^w e^{-(\lambda+\mu_1)t/2}{_1F_1\left(n-k,n+2,\frac{(\mu_1-\mu_2)t}{2}\right)\left(\frac{\lambda t^2}{4}\right)^n}\textrm dt\\[5pt] &=\sum_{m=0}^{+\infty}\frac{(n-k)_m}{(n+2)_m m!}\int_0^w e^{-(\lambda+\mu_1)t/2}\left(\frac{(\mu_1-\mu_2)t}{2}\right)^m \left(\frac{\lambda t^2}{4}\right)^n \textrm dt\\[5pt] &=\sum_{m=0}^{+\infty}\frac{(n-k)_m}{(n+2)_m m!} 2\lambda^n \frac{(\mu_1-\mu_2)^m}{(\lambda+\mu_1)^{1+m+2n}}\gamma\left(1+m+2n,\frac{(\lambda+\mu_1)w}{2}\right).\end{align*}

Similarly, we have

\begin{align*}&\int_0^w e^{-(\lambda+\mu_2)t/2}{_1F_1\left(k,n+2,\frac{(\mu_1-\mu_2)t}{2}\right)\left(\frac{\lambda t^2}{4}\right)^n}\textrm dt\\[5pt] &=\sum_{m=0}^{+\infty}\frac{(k)_m}{(n+2)_m m!} 2\lambda^n\frac{(\mu_1-\mu_2)^m}{(\lambda+\mu_2)^{1+m+2n}}\gamma\left(1+m+2n,\frac{(\lambda+\mu_2)w}{2}\right).\end{align*}

The result immediately follows from Proposition 1 and Theorem 5 if we consider that

\begin{align*}&\sum_{k=0}^n\binom{n}{k}\left(\frac{b_1\mu_1}{b_2\mu_2}\right)^k (n-k)_m= \frac{(m+n-1)!\, {_2}F_1\left(1-n,-n;\;1-m-n; -\frac{b_1\mu_1}{b_2\mu_2}\right)}{(n-1)!},\\[5pt] &\sum_{k=0}^n\binom{n}{k}\left(\frac{b_1\mu_1}{b_2\mu_2}\right)^k (k)_m= \frac{n b_1\mu_1\,m!\, {_2}F_1\left(1+m,1-n;\;2; -\frac{b_1\mu_1}{b_2\mu_2}\right)}{b_2\mu_2}.\end{align*}

In Figure 7, we provide plots of the lower bound L(t, w) defined in Equation (16) with $F_C(w)$ as given in Equation (45) and the upper bound U(t, w) given in Equation (44), for different choices of the parameters.

Figure 7.

The lower bound L(t, w) and the upper bound U(t, w) of the survival function of M(t), in the case of mixed exponential downward times, for $b_1=b_2=0.5$ , $t=3$ , $\mu_1=2$ , $\mu_2=3$ , $\lambda=1$ (left), and $\lambda=1.5$ (right).

5.1. Exponentially distributed downward random times

Let us assume that the random variables $D_i$ , $i\in {\mathbb N}$ , are exponentially distributed with parameter $\mu>\lambda>0$ . We collect some results concerning the first passage time of X(t) through a fixed level $w>0$ .

Theorem 6. In the case of exponentially distributed downward times, the expected value of the first passage time $\Theta_w$ , defined in Equation (17), is given by

(47) \begin{equation}\mathbb E(\Theta_w)=\frac{2\lambda}{\mu-\lambda}w, \quad w\geq 0,\end{equation}

for $\mu>\lambda>0$ .

Proof. Recalling Equation (23), we have

\begin{align*}\psi_C(-s)=\frac{1}{\lambda}\left[\frac{\lambda+\mu}{2}+s-\sqrt{-\lambda\mu+\frac{(\lambda+\mu+2s)^2}{4}}\right],\end{align*}

and

(48) \begin{equation}\psi_{Z}(w,-s)=\exp\left[-\frac{w}{2}\left(\lambda-\mu-2s+\sqrt{(\lambda-\mu)^2+4(\lambda+\mu)s+4s^2}\right)\right].\end{equation}

Hence, for $\lambda<\mu$ , we have

\begin{align*}\mathbb E(Z(w))=-\left.\frac{\textrm{d}}{\textrm{d}s}\psi_Z(w,-s)\right|_{s=0}=\frac{2\lambda}{\mu-\lambda}w,\end{align*}

so that the proof follows from recalling Equation (18).

Remark 7. Note that for $\lambda\to\mu^-$ ,

\begin{align*}\lim_{\lambda\to\mu^-}\mathbb E(\Theta_w)=+\infty.\end{align*}

This result is in agreement with the one obtained by Orsingher [Reference Orsingher30] in the case of a standard symmetric telegraph process.

Theorem 7. In the case of exponentially distributed downward times, the second-order moment of the first passage time $\Theta_w$ , defined in Equation (17), is given by

\begin{align*}\mathbb E(\Theta^2_w)=\frac{4\lambda\mu w \left[2+(\mu-\lambda)w\right]}{(\mu-\lambda)^3}, \quad w\geq 0,\end{align*}

whereas for the variance of $\Theta_w$ we have

(49) \begin{equation}Var (\Theta_w)=\frac{4\lambda w \left[2\mu+(\mu^2-\lambda^2) w\right]}{(\mu-\lambda)^3},\quad w\geq 0.\end{equation}

Proof. Recalling Equations (22) and (48), we have

\begin{align*}\mathbb E(\Theta^2_w)=\left.2w\frac{\textrm{d}}{\textrm{d}s}\psi_Z(w,s)\right|_{s=0}+\left.\frac{\textrm{d}^2}{\textrm{d}s^2}\psi_Z(w,s)\right|_{s=0}=\frac{4\lambda\mu w \left[2+(\mu-\lambda)w\right]}{(\mu-\lambda)^3}.\end{align*}

Hence, Equation (49) follows from recalling Equation (47).

In Figure 8, we provide plots of the expected value (47) and the variance (49) of the first passage time $\Theta_w$ for suitable choices of the parameters.

Figure 8.

The expected value (47) (left) and the variance (49) (right) of the first passage time $\Theta_w$ for $\mu=5$ .

5.2. Erlang distributed downward random times

Let us now consider the case $D\sim Erlang(\mu,k)$ with $k\in \mathbb N$ , $\mu>0$ , and $k/\mu<1/\lambda$ . In the following propositions we provide explicit expressions for the moment generating function of the constant phase C and of the compound process Z(w).

Proposition 13. In the case of $Erlang(\mu,k)$ distributed downward random times, the moment generating function of C is given by

(50) \begin{equation}\psi_C(s)=\frac{\mu^k k}{(\lambda+\mu-2s)^k}{_1\Psi_1\begin{bmatrix}\begin{matrix}(k,k+1)\\[5pt] (k+1,k)\end{matrix}; \displaystyle\frac{\lambda\mu^k}{(\lambda+\mu-2s)^{k+1}}\end{bmatrix}},\end{equation}

where

\begin{align*}{_1\Psi_1}\begin{bmatrix}\begin{matrix}(a_1,A_1)\\[5pt] (b_1,B_1)\end{matrix};\; z\end{bmatrix}\;:\!=\;\sum_{n=0}^{+\infty}\frac{\Gamma(a_1+A_1n)}{\Gamma(b_1+B_1n)}\cdot \frac{z^n}{n!}\end{align*}

denotes the Fox–Wright function.

Proof. We have

(51) \begin{equation}\psi_C(s)=\frac{1}{2}\mu^k k \sum_{n=0}^{+\infty}\frac{(\lambda\mu^k)^n}{n! (k+kn)!}\int_0^{+\infty}e^{st-(\lambda+\mu)t/2}\left(\frac{t}{2}\right)^{(k+1)n+k-1}\textrm{d}t.\end{equation}

Taking into account that

\begin{align*}\int_0^{+\infty}e^{st-(\lambda+\mu)t/2}\left(\frac{t}{2}\right)^{(k+1)n+k-1}\textrm{d}t=\frac{2(k+n+kn-1)!}{(\lambda+\mu-2s)^{n+k(n+1)}},\end{align*}

from Equation (51) we have

\begin{equation*}\begin{aligned}\psi_C(s)&=\frac{\mu^kk}{(\lambda+\mu-2s)^k}\sum_{n=0}^{+\infty}\left(\frac{\lambda \mu^k}{(\lambda+\mu-2s)^{k+1}}\right)^n \frac{(k+n+kn-1)!}{n!(k+kn)!}\\[5pt] &=\frac{\mu^kk}{(\lambda+\mu-2s)^k} {_1\Psi_1}\begin{bmatrix}\begin{matrix}(k,k+1)\\[5pt] (k+1,k)\end{matrix}; \displaystyle\frac{\lambda\mu^k}{(\lambda+\mu-2s)^{k+1}}\end{bmatrix}.\end{aligned}\end{equation*}

Proposition 14. In the case of $Erlang(\mu,k)$ distributed downward random times, the moment generating function of Z(t) is given by

(52) \begin{equation}\psi_{Z}(t,s)=\exp\left[-\lambda t\left(1-\frac{\mu^k k }{(\lambda+\mu-2s)^k}{_1\Psi_1}\begin{bmatrix}\begin{matrix}(k,k+1)\\[5pt] (k+1,k)\end{matrix}; \displaystyle\frac{\lambda\mu^k}{(\lambda+\mu-2s)^{k+1}}\end{bmatrix}\right)\right].\end{equation}

The expected value and the variance of the process Z(t) are given in the following proposition.

Proposition 15. In the case of $Erlang(\mu,k)$ distributed downward random times, for $t>0$ , the expected value of the process Z(t) is given by

(53) \begin{align}\mathbb E(Z(t)) = \frac{\lambda t\mu^k}{(\lambda+\mu)^{k+1}}&\left\{2k \left[k{_1\Psi_1}\begin{bmatrix}\begin{matrix}(k,k+1)\nonumber \\[5pt] (k+1,k)\end{matrix}; \displaystyle\frac{\lambda\mu^k}{(\lambda+\mu)^{k+1}}\end{bmatrix}\right.\right.\\[5pt] & \left.\left.+ {\lambda (k+1)}{_1\Psi_1}\begin{bmatrix}\begin{matrix}(2k+1,k+1)\\[5pt] (2k+1,k)\end{matrix}; \displaystyle\frac{\lambda\mu^k}{(\lambda+\mu)^{k+1}}\end{bmatrix}\right] \right\},\end{align}

and the variance is given by

(54) \begin{align}Var(Z(t))&=\lambda t \frac{4k\mu^k(k+1)}{(\lambda+\mu)^{k+2}}\Bigg\{k {_1\Psi_1}\begin{bmatrix}\begin{matrix}(k,k+1)\\[5pt] (k+1,k)\end{matrix};\displaystyle \frac{\lambda\mu^k}{(\lambda+\mu)^{k+1}}\end{bmatrix}\nonumber\\[5pt] &\quad+ \frac{\lambda\mu^k (3k+2)}{(\lambda+\mu)^{k+1}}{_1\Psi_1}\begin{bmatrix}\begin{matrix}(2k+1,k+1)\\[5pt] (2k+1,k)\end{matrix};\displaystyle \frac{\lambda\mu^k}{(\lambda+\mu)^{k+1}}\end{bmatrix}\nonumber\\[5pt] &\quad+\frac{\lambda^2\mu^{2k}(k+1)}{(\lambda+\mu)^{2k+2}}{_1\Psi_1}\begin{bmatrix}\begin{matrix}(3k+2,k+1)\\[5pt] (3k+1,k)\end{matrix};\displaystyle \frac{\lambda\mu^k}{(\lambda+\mu)^{k+1}}\end{bmatrix}\Bigg\}.\end{align}

The proof of Proposition 15 is given in Appendix C.

Finally, we provide some results concerning the expected value of the first passage time.

Proposition 16. In the case of $Erlang(\mu,k)$ distributed downward times, the expected value of the first passage time $\Theta_w$ is given by

\begin{align*}\mathbb E(\Theta_w)&= \lambda w\Bigg\{\frac{2\mu^k k}{(\lambda+\mu)^{k+1}} \Bigg[k{_1\Psi_1}\begin{bmatrix}\begin{matrix}(k,k+1)\\[5pt] (k+1,k)\end{matrix}; \displaystyle\frac{\lambda\mu^k}{(\lambda+\mu)^{k+1}}\end{bmatrix}\\[5pt] &+ \frac{\lambda\mu^k (k+1)}{(\lambda+\mu)^{k+1}}{_1\Psi_1}\begin{bmatrix}\begin{matrix}(2k+1,k+1)\\ (2k+1,k)\end{matrix}; \displaystyle\frac{\lambda\mu^k}{(\lambda+\mu)^{k+1}}\end{bmatrix}\Bigg] \Bigg\}. \end{align*}

Remark 8. When $k=1$ , we have

\begin{align*}&\mathbb E(\Theta_w)\Big|_{k=1}=\lambda w\left\{\frac{2\mu}{(\lambda+\mu)^{2}} \left[{_1\Psi_1}\begin{bmatrix}\begin{matrix}(1,2)\\[5pt] (2,1)\end{matrix}; \displaystyle\frac{\lambda\mu}{(\lambda+\mu)^{2}}\end{bmatrix}+ \frac{2\lambda\mu }{(\lambda+\mu)^{2}}{_1\Psi_1}\begin{bmatrix}\begin{matrix}(3,2)\\[5pt] (3,1)\end{matrix}; \displaystyle\frac{\lambda\mu}{(\lambda+\mu)^{2}}\end{bmatrix}\right] \right\}\\[5pt] &=\lambda w\left[\frac{2\mu}{(\lambda+\mu)^2}\left(\frac{\lambda+\mu}{\mu}+\frac{2\lambda(\lambda+\mu)}{\mu(\mu-\lambda)}\right)\right]=\frac{2\lambda}{\mu-\lambda}w,\end{align*}

which is the same as the expected value in the case of exponential downward random times.

Remark 9. From Equation (53), we can see that the mean of $\Theta_w$ depends linearly on w. Moreover, when $\lambda\to\mu^-$ , as in the exponential case, one has

\begin{align*}\mathbb E(\Theta_w)\to+\infty.\end{align*}

From Equations (53)–(54) and Equation (22), it is possible to get an explicit formula for the second-order moment of the first passage time $\Theta_w$ . The resulting expression is cumbersome, so for brevity we omit it.

In Figure 9, we provide plots of the expected value and the variance of the first passage time $\Theta_w$ for suitable choices of the parameters.

Figure 9.

The expected value obtained in Proposition 16 (left) and the variance (right) of the first passage time $\Theta_w$ for $\mu=9$ and $k=2$ .

5.3. Weighted exponentially distributed downward random times

Let us assume that $D\sim WE(\alpha,\mu)$ (see Equation (38)), with $0<\lambda\le \frac{\mu}{2}$ , $\alpha>0$ (or $\frac{\mu}{2}<\lambda<\mu$ , $\alpha>\frac{\mu-2\lambda}{\lambda-\mu}$ ).

Proposition 17. In the case of weighted exponentially distributed downward times, the moment generating function of C is given by

(55) \begin{equation}\begin{aligned}\psi_C(s)&=\frac{{4(1+\alpha)\mu^2}}{\left(2\lambda+(2+\alpha)\mu-4s\right)^{2}}\sum_{n=0}^{+\infty}\left(\frac{{8(1+\alpha)\lambda\mu^2}}{\left(2\lambda+(2+\alpha )\mu-4s\right)^3}\right)^n\frac{(3n+1)!}{(n+1)!(2n+1)!}\\[5pt] &\times {_2 F_1}\left(\frac{3(1+n)}{2},1+\frac{3n}{2},\frac 32 +n, \frac{\alpha^2\mu^2}{(2\lambda+(2+\alpha)\mu -4s)^2}\right),\end{aligned}\end{equation}

where ${_2}F_1(a,b,c;\;z)$ is the Gauss hypergeometric function (see Equation (46)).

Proof. Since

\begin{align*} &\int_0^{+\infty}e^{-\left(\frac{\lambda+\mu}{2}+\frac{\alpha\mu}{4}-s\right)t} t^{1/2+2n} I_{n+1/2}\left(\frac{\alpha\mu t}{2}\right)\textrm{d}t\\[5pt] &=\frac{2^{5/2+3n}(\alpha\mu)^{n+1/2}}{(2\lambda+(2+\alpha)\mu-4s)^{2+3n}}\frac{\Gamma\left(3n+2\right)}{\Gamma\left(\frac32+n\right)}\\[5pt] &\quad \times {_2}F_1\left(\frac{3(1+n)}{2}, 1+\frac{3}{2}n,\frac32+n,\frac{\alpha^2\mu^2}{(2\lambda+(2+\alpha)\mu-4s)^2}\right), \end{align*}

the proof immediately follows from Equation (13).

In the following proposition, we provide the expected value of the first passage time $\Theta_w$ .

Proposition 18. In the case of weighted exponentially distributed downward times, the expected value of the first passage time $\Theta_w$ is given by

(56) \begin{equation}\begin{aligned}\mathbb E(\Theta_w)&=\lambda w \left\{\frac{16(1+\alpha)\mu^2}{(2\lambda+(2+\alpha)\mu)^5}\sum_{n=0}^{+\infty}\left(\frac{8(1+\alpha)\lambda\mu^2}{(2\lambda+(2+\alpha)\mu)^3}\right)^n \frac{(2+3n)!}{n!(n+1)!}\right.\\[5pt] &\left.\left[{{_2F_1}\left(\frac{3(1+n)}{2},1+\frac{3}{2}n,\frac{3}{2}+n,\frac{\alpha^2\mu^2}{(2\lambda+(2+\alpha)\mu)^2}\right)}\right.\right.\\[5pt] &\left.\left.+\frac{3\alpha^2\mu^2(1+n)}{(3+2n)(2\lambda+(2+\alpha)\mu)^2}{_2F_1}{\left(\frac 32 n+2,\frac{5+3n}{2},\frac 52+n, \frac{\alpha^2\mu^2}{(2\lambda+(2+\alpha)\mu)^2}\right)}\right]\right\},\end{aligned}\end{equation}

where ${_2}F_1(a_1,a_2;\;b;\;z)$ is as defined in Equation (46).

Proof. Recalling Equation (55), by evaluating the first derivative with respect to s of $\psi_C(s)$ at $s=0$ , we get the expected value of C, which is given by

\begin{equation*}\begin{aligned}\mathbb E(C)&=\left.\frac{\textrm{d}}{\textrm{d}s}\psi_C(s)\right|_{s=0}=\frac{16(1+\alpha)\mu^2}{(2\lambda+(2+\alpha)\mu)^5}\sum_{n=0}^{+\infty}\left(\frac{8(1+\alpha)\lambda\mu^2}{(2\lambda+(2+\alpha)\mu)^3}\right)^n \frac{(2+3n)!}{n!(n+1)!}\\[5pt] &\left[{{_2F_1}\left(\frac{3(1+n)}{2},1+\frac{3}{2}n,\frac{3}{2}+n,\frac{\alpha^2\mu^2}{(2\lambda+(2+\alpha)\mu)^2}\right)}\right.\\[5pt] &\left.+\frac{3\alpha^2\mu^2(1+n)}{(3+2n)(2\lambda+(2+\alpha)\mu)^2}{_2F_1}{\left(\frac 32 n+2,\frac{5+3n}{2},\frac 52+n, \frac{\alpha^2\mu^2}{(2\lambda+(2+\alpha)\mu)^2}\right)}\right].\end{aligned}\end{equation*}

Hence, the result follows easily from Equation (18).

Remark 10. For $\alpha\to+\infty$ , since

\begin{align*}&{_2F_1}\left(\frac 32 +\frac32 n, 1+\frac{3}{2}n,\frac 32+n,\frac{\alpha^2\mu^2}{(2\lambda+(2+\alpha)\mu)^2}\right)\propto \frac{(2n)!\Gamma\left(\frac{3}{2}+n\right)}{\Gamma\left(\frac{3}{2}+\frac 32 n\right)\Gamma\left(1+\frac{3}{2}n\right)}\\[5pt] &\qquad \times\left(\frac{(2\lambda +(2+\alpha)\mu)^2-\alpha^2\mu^2}{(2\lambda+(2+\alpha)\mu)^2}\right)^{-2n-1}\end{align*}

(see, for instance, Equation (9) of [Reference Erdélyi16]), we have

\begin{align*}\lim_{\alpha\to+\infty}\psi_C(s)=-\frac{\lambda+\mu-2s}{2\lambda}\left(-1+\sqrt{1-\frac{4\lambda\mu}{(\lambda+\mu-2s)^2}}\right).\end{align*}

Hence

\begin{align*}\lim_{\alpha\to+\infty}\mathbb E(\Theta_w)=\frac{2}{\mu-\lambda}\lambda ,\end{align*}

which corresponds to the expected value of $\Theta_w$ in the case of exponentially distributed downward random times.

Remark 11. From Equation (55) it is possible to get an expression for the second-order moment of $\Theta_w$ . In particular, after some calculations we have that

\begin{align*}\begin{aligned}\mathbb E(C^2)&=\left.\frac{d^2}{d s^2} \psi_C(s)\right|_{s=0}=\frac{8 (1+\alpha)\mu^2\sqrt{\pi}}{(2\lambda+(2+\alpha)\mu)^8}\sum_{n=0}^{+\infty} \left(\frac{(1+\alpha)\lambda \mu^2}{2(2\lambda+(2+\alpha)\mu)^3}\right)^n \frac{(3n+3)!}{n!(n+1)!}\\[5pt] &\times \left[\frac{4(2\lambda+(2+\alpha)\mu)^4}{\Gamma\left(\frac 32 +n\right)} {{_2F_1 \left(\frac{3(1+n)}{2},1+\frac 32 n, \frac 32+n, \frac{\alpha^2\mu^2}{(2\lambda +(2+\alpha)\mu)^2}\right)}}\right.\\[5pt] &+\frac{\alpha^4\mu^4 (4+3n)(5+3n)}{{\Gamma\left(\frac 72 +n\right)}} {{_2F_1 \left(\frac{3(2+n)}{2},\frac 72 +\frac 32 n, \frac 72+n, \frac{\alpha^2\mu^2}{(2\lambda +(2+\alpha)\mu)^2}\right)}}\\[5pt] &\left.+\frac{2\alpha^2\mu^2(2\lambda+(2+\alpha)\mu)^2(7+6n)}{{\Gamma\left(\frac 52 +n\right)}}{{_2F_1 \left(2+\frac 32 n,\frac 52 +\frac 32 n, \frac 52+n, \frac{\alpha^2\mu^2}{(2\lambda +(2+\alpha)\mu)^2}\right)}}\!\right]\!,\end{aligned}\end{align*}

where ${_2}F_1(a_1,a_2;\;b;\;z)$ is as defined in Equation (46).

The resulting expression for $\mathbb E(\Theta_w^2)$ is omitted for brevity. In Figure 10, we provide plots of the expected value and the variance of $\Theta_w$ for some choices of the parameters.

Figure 10.

The expected value (56) (left) and the variance (right) of $\Theta_w$ (obtained from Remark 11) for $\alpha=1$ and $\lambda=4$ .

5.4. Mixed exponential downward random times

This section is devoted to the case in which the downward time D is distributed as a mixture of two exponential distributions (42), with $b_i\ge 0$ , for $i=1,2$ , $b_1+b_2=1$ , and $\frac{b1}{\mu_1}+\frac{b_2}{\mu_2}<\frac{1}{\lambda}$ .

Proposition 19. In the case of mixed exponential distributed downward times, the moment generating function of C is given by

(57) \begin{equation}\psi_C(s)= \mathcal C(b_1,b_2,\mu_1,\mu_2,\lambda,s)+\mathcal D(b_1,b_2,\mu_1,\mu_2,\lambda,s),\end{equation}

where

\begin{equation*}\begin{aligned}\!\mathcal C(b_1,b_2,\mu_1,\mu_2,\lambda,s)&\;:\!=\;\frac{b_1\mu_1}{(\lambda+\mu_1-2s)}\!\sum_{n=0}^{+\infty}\frac{(2n)!}{(n+1)!n!}\!\left(\frac{b_2\mu_2\lambda}{(\lambda+\mu_1-2s)^2}\right)^n \!\sum_{k=0}^n\!\binom{n}{k}\!\left(\frac{b_1\mu_1}{b_2\mu_2}\right)^k\\[5pt] &\times {_2F_1\left(n-k,2n+1,2+n,\frac{\mu_1-\mu_2}{\lambda+\mu_1-2s}\right)},\end{aligned}\end{equation*}

with ${_2}F_1(a_1,a_2;\;b;\;z)$ as defined in Equation (46) and

\begin{equation*}\begin{aligned}\!\mathcal D(b_1,b_2,\mu_1,\mu_2,\lambda,s)&\;:\!=\;\frac{b_2\mu_2}{(\lambda+\mu_2-2s)}\sum_{n=0}^{+\infty}\frac{(2n)!}{(n+1)!n!}\!\left(\!\frac{b_2\mu_2\lambda}{(\lambda+\mu_2-2s)^2}\!\right)^n \sum_{k=0}^n\!\binom{n}{k}\!\left(\frac{b_1\mu_1}{b_2\mu_2}\!\right)^k\\[5pt] &\times{_2}F_1\left(k,2n+1,2+n,\frac{\mu_2-\mu_1}{\lambda+\mu_2-2s}\right).\end{aligned}\end{equation*}

Proof. The proof immediately follows if we recall that

\begin{equation*}\begin{aligned}&\int_0^{+\infty}e^{st}\left(\frac{\lambda t^2}{4}\right)^n e^{-\lambda t/2-\mu_1t/2}{_1F_1\left(n-k,n+2,\frac{(\mu_1-\mu_2)t}{2}\right)}\textrm{d}t\\[5pt] &=\frac{2(2n)! \lambda^n}{(\lambda+\mu_1-2s)^{2n+1}}{_2F_1\left(n-k,2n+1,2+n,\frac{\mu_1-\mu_2}{\lambda+\mu_1-2s}\right)},\end{aligned}\end{equation*}

and

\begin{equation*}\begin{aligned}&\int_0^{+\infty}e^{st}\left(\frac{\lambda t^2}{4}\right)^n e^{-\lambda t/2-\mu_2t/2}{_1F_1\left(k,n+2,\frac{(\mu_2-\mu_1)t}{2}\right)}\textrm{d}t\\[5pt] &=\frac{2(2n)! \lambda^n}{(\lambda+\mu_2-2s)^{2n+1}}{_2}F_1\left(k,2n+1,2+n,\frac{\mu_2-\mu_1}{\lambda+\mu_2-2s}\right).\end{aligned}\end{equation*}

The expected value of the first passage time $\Theta_w$ is obtained in the following proposition.

Proposition 20. In the case of downward times distributed as in Equation (42), the expected value of the first passage time $\Theta_w$ is given by

\begin{equation*}\mathbb E(\Theta_w)=\lambda w \left(\mathcal{A}(b_1,b_2,\mu_1,\mu_2,\lambda)+\mathcal B (b_1,b_2,\mu_1,\mu_2,\lambda)\right),\end{equation*}

where

\begin{equation*}\begin{aligned}&\mathcal{A}(b_1,b_2,\mu_1,\mu_2,\lambda)\;:\!=\;b_1\mu_1\sum_{n=0}^{+\infty}\frac{(2n)!}{(n+1)!n!}(b_2\mu_2\lambda)^n\sum_{k=0}^n \binom{n}{k}\left(\frac{b_1\mu_1}{b_2\mu_2}\right)^k\left(\frac{2(2n+1)}{(\lambda+\mu_1)^{2n+2}}\right.\\[5pt] &\times \left.{_2F_1\left(n-k,2n+1,2+n,\frac{\mu_1-\mu_2}{\lambda+\mu_1}\right)}+\frac{2(\mu_1-\mu_2)(n-k)(2n+1)}{(\lambda+\mu_1)^{2n+3}(2+n)}\right.\\[5pt] &\times\left.{_2F_1\left(n-k+1,2n+2,3+n,\frac{\mu_1-\mu_2}{\lambda+\mu_1}\right)}\right),\end{aligned}\end{equation*}

and

\begin{equation*}\begin{aligned}&\mathcal{B}(b_1,b_2,\mu_1,\mu_2,\lambda)\;:\!=\;b_2\mu_2\sum_{n=0}^{+\infty}\frac{(2n)!}{(n+1)!n!}(b_2\mu_2\lambda)^n\sum_{k=0}^n \binom{n}{k}\left(\frac{b_1\mu_1}{b_2\mu_2}\right)^k\left(\frac{2(2n+1)}{(\lambda+\mu_2)^{2n+2}}\right.\\[5pt] &\times \left.{_2F_1\left(k,2n+1,2+n,\frac{\mu_2-\mu_1}{\lambda+\mu_2}\right)}\right.\\[5pt]&\left. +\frac{2k(\mu_2-\mu_1)(2n+1)}{(\lambda+\mu_2)^{2n+3}(2+n)}{_2F_1\left(k+1,2n+2,3+n,\frac{\mu_2-\mu_1}{\lambda+\mu_2}\right)}\right).\end{aligned}\end{equation*}

Proof. The proof follows from Equation (18), if we recall Equation (57) and note that

\begin{equation*}\mathbb E(C)=\left.\frac{\textrm{d}}{\textrm{d}s}\psi_C(s)\right|_{s=0}=\mathcal{A}(b_1,b_2,\mu_1,\mu_2,\lambda)+\mathcal B (b_1,b_2,\mu_1,\mu_2,\lambda).\end{equation*}

Some plots of $\mathbb E(\Theta_w)$ are provided in Figure 11.

Figure 11.

The expected value of $\Theta_w$ obtained in Proposition 20 for $b_1=b_2=0.5$ , $\mu_1=2$ , and $\lambda=1$ .

Remark 12. Note that when $b_1=b_2=\frac 12$ and $\mu_1=\mu_2=\mu$ , we recover the case of exponentially distributed downward random times. Indeed,

\begin{align*}\mathcal A\left(\frac12,\frac12,\mu,\mu,\lambda\right)=\mathcal B\left(\frac12,\frac12,\mu,\mu,\lambda\right)=\frac{1}{\mu-\lambda}.\end{align*}

Hence,

\begin{align*}\mathbb E(\Theta_w)\Big|_{b_1=b_2=\frac12,\,\mu_1=\mu_2=\mu}=\frac{2\lambda}{\mu-\lambda}w,\end{align*}

which corresponds to the expected value of the first passage time $\Theta_w$ in the exponential case.