In this section, some distributions on the maximum surplus and the maximal severity of ruin are given. Before proceeding with the next Theorem, we will give a simple explanation of shift operators first. For \(t\ge 0\), let \(\theta _{t}\) be the shift operators from \(\Omega\) to itself defined by \(U(s)\circ \theta _{t}= U(s+t)\). For stopping time T, conditioning on \(\{T<\infty \}\) the map \(\theta _{T}\) is defined by \(U(t)\circ \theta _{T}=U(t+T)\) (see Revuz and Yor 1991, pp. 34, 37 and 74). Let \(G(u,a)=P^{u}(\sup \nolimits _{0\le t<T}U(t)>a,T<\infty )\) to denote the probability distribution of the supreme profit of an insurance company before the time of ruin. First we will give the explicit expression of G(u, a).
Theorem 3
For
\(u\ge 0\), we have
$$\begin{aligned} G(u,a) =\left\{ \begin{array}{ll} \Psi (u),&{}\quad a\le u,\\ \frac{\Phi (u)}{\Phi (a)}\Psi (a),&{}\quad a>u,\\ \end{array}\right. \end{aligned}$$
where
\(\Psi (u), \Phi (u)\)
are given by (7).
Proof
When \(a\le u\), it is obvious that \(G(u,a)=P^{u}(T<\infty )=\Psi (u)\). Therefore we only consider the case \(a>u\). Since \(P^{u}(\lim \nolimits _{t\rightarrow \infty }U(t)=\infty )=1\), by the strong Markov property of \(\{U(t),t\ge 0\}\) we can show that
$$\begin{aligned} G\left( u,a\right)&= {} P^{u}\left( T_{1}^{a}<T, T<\infty \right) =P^{u}\left( T_{1}^{a}<T,T\circ \theta _{T_{1}^{a}}<\infty \right) \nonumber \\&= {} P^{u}\left( T_{1}^{a}<T\right) P^{a}\left( T<\infty \right) =P^{u}\left( T_{1}^{a}<T\right) \Psi (a). \end{aligned}$$
(9)
In addition, we have
$$\begin{aligned} P^{u}\left( T_{1}^{a}<T\right)&= {} P^{u}\left( T_{1}^{a}<T, T<\infty \right) +P^{u}\left( T_{1}^{a}<T, T=\infty \right) \nonumber \\&= {} G(u,a)+\Phi (u). \end{aligned}$$
(10)
Inserting (10) into (9), we get that
$$\begin{aligned} G(u,a)=\frac{\Psi (a)}{1\Psi (a)}\Phi (u)=\frac{\Phi (u)}{\Phi (a)}\Psi (a). \end{aligned}$$
(11)
This ends the proof. \(\square\)
Corollary 2
For
\(a>u\ge 0\), we have
$$\begin{aligned} P^{u}(T_{1}^{a}<T)=\frac{\Phi (u)}{\Phi (a)}. \end{aligned}$$
Proof
This follows from (10) and (11). \(\square\)
Theorem 4
For
\(a>u\), we have
$$\begin{aligned} P^{u}\Big (\sup \limits _{0\le t<L}U(t)<a,L>0\Big )=\left\{ \begin{array}{ll} \Phi (a)\Phi (u),&{}\quad u\ge 0,\\ \Phi (a),&{}\quad u<0,\\ \end{array}\right. \end{aligned}$$
where
\(\Phi (u)\)
can be obtained by (7).
Proof
It follows from \(P^{u}(\lim \nolimits _{t\rightarrow \infty }U(t)=\infty )=1\) that \(P^{u}(L<\infty )=1\) and \(P^{u}(T_{1}^{a}<\infty )=1\). Note that
$$\begin{aligned} \left( \sup \limits _{0\le t<L}U(t)<a,L>0\right) =\left( L<T_{1}^{a},L>0\right) =\left( T\circ \theta _{T_{1}^{a}}=\infty ,L>0\right) . \end{aligned}$$
When \(u<0\), we have \(P^{u}(L>0)=1\). Thus we can obtain
$$\begin{aligned} P^{u}\left( L<T_{1}^{a},L>0\right)&= {} P^{u}\left( T\circ \theta _{T_{1}^{a}}=\infty \right) =P^{u}\left( T_{1}^{a}<\infty ,T\circ \theta _{T_{1}^{a}}=\infty \right) \\&= {} P^{a}(T=\infty )=\Phi (a). \end{aligned}$$
When \(u\ge 0\), we have \((L>0)=(T<\infty )\), then it is easy to see that
$$\begin{aligned}&P^{u}\left( L<T_{1}^{a},L>0\right) =P^{u}\left( T_{1}^{a}<\infty ,T<\infty ,T\circ \theta _{T_{1}^{a}}=\infty \right) \\&\quad =P^{u}\left( T_{1}^{a}<\infty ,T<T_{1}^{a},T\circ \theta _{T_{1}^{a}}=\infty \right) =P^{u}\left( T<T_{1}^{a}\right) \Phi (a). \end{aligned}$$
By Corollary 2, we have \(P^{u}(T<T_{1}^{a})=\frac{\Phi (a)\Phi (u)}{\Phi (a)}\). Hence \(P^{u}(L<T_{1}^{a},L>0)=\Phi (a)\Phi (u)\). This completes the proof. \(\square\)
In the following, we consider the maximum surplus and the maximal severity of ruin before the time of recovery. To some extent, as ‘indexes’, they can describe the ‘best’ situation and the ‘worst’ situation the company would experience before the surplus process upcrossing level zero after ruin for the first time. Their joint distributions are derived.
Theorem 5
For
\(a>u\ge 0\)
and
\(b>0\), we have
$$\begin{aligned} P^{u}\Big (\sup \limits _{0\le t<T_{1}^{0}}U(t)<a,\inf \limits _{0\le t<T_{1}^{0}}U(t)\ge b\Big )=\frac{\Phi (a)\Phi _{b}(u)\Phi (u)\Phi _{b}(a)}{\Phi (a)\Phi _{b}(0)}, \end{aligned}$$
where
\(\Phi (u), \Phi _{b}(u)\)
are presented by (7) and (8).
Proof
Let \(A= \{\sup \nolimits _{0\le t<T_{1}^{0}}U(t)\ge a,\inf \nolimits _{0\le t<T_{1}^{0}}U(t)\ge b\}\), and then we have
$$\begin{aligned} P^{u}(A\cap \{T=\infty \})&= {} P^{u}\Big (\sup \limits _{t\ge 0}U(t)\ge a,\inf \limits _{t\ge 0}U(t)\ge b,\inf \limits _{t\ge 0}U(t)\ge 0\Big )\\&= {} P^{u}\Big (\inf \limits _{t\ge 0}U(t)\ge 0\Big )=\Phi (u), \end{aligned}$$
$$\begin{aligned} P^{u}(A\cap \{T<\infty \})&= {} P^{u}\Big (\sup \limits _{0\le t<T}U(t)\ge a,\inf \limits _{T\le t<T_{1}^{0}}U(t)\ge b,T<\infty \Big )\\&= {} P^{u}\Big (T_{1}^{a}<T,\Big (\inf \limits _{T\le t<T_{1}^{0}}U(t)\ge b,T<\infty \Big )\circ \theta _{T_{1}^{a}}\Big )\\&= {} P^{u}\left( T_{1}^{a}<T\right) P^{a}\Big (\inf \limits _{T\le t<T_{1}^{0}}U(t)\ge b,T<\infty \Big ). \end{aligned}$$
Following the line of Picard (1994), we introduce \(\widetilde{M}=Max\{U(t),U(t)<0\}\) in order to obtain \(M=Max\{U(t),T\le t<T_{1}^{0}\}\). Clearly \(\widetilde{M}\le z\) means that the surplus process never goes under the level \(z\), so that
$$\begin{aligned} P^{u}(\widetilde{M}\le z)=1\Psi _{z}(u),\quad z\ge 0. \end{aligned}$$
The event \(\widetilde{M}\le z\) is equivalent to that ruin does not occur or ruin occurs and \(\widetilde{M}\le z\). By the total probability formula, we have
$$\begin{aligned} P^{u}(\widetilde{M}\le z)=1\Psi (u)+P^{u}(M\le z)(1\Psi _{z}(0)). \end{aligned}$$
Hence we obtain
$$\begin{aligned} P^{u}(M\le z)=\frac{\Phi _{z}(u)\Phi (u)}{\Phi _{z}(0)}. \end{aligned}$$
Then we get
$$\begin{aligned} P^{a}\Big (\inf \limits _{T\le t<T_{1}^{0}}U(t)\ge b,T<\infty \Big )= P^{a}(M\le b)=\frac{\Phi _{b}(a)\Phi (a)}{\Phi _{b}(0)}. \end{aligned}$$
By Corollary 2, we have
$$\begin{aligned} P^{u}\Big (\sup \limits _{0\le t<T_{1}^{0}}U(t)\ge a,\inf \limits _{0\le t<T_{1}^{0}}U(t)\ge b\Big )=P^{u}(A)=\Phi (u)+\frac{\Phi (u)(\Phi _{b}(a)\Phi (a))}{\Phi (a)\Phi _{b}(0)}. \end{aligned}$$
Using similar argument, we obtain
$$\begin{aligned}&P^{u}\Big (\inf \limits _{0\le t<T_{1}^{0}}U(t)\ge b,T=\infty \Big )=\Phi (u),\\&P^{u}\Big (\inf \limits _{0\le t<T_{1}^{0}}U(t)\ge b,T<\infty \Big )=P^{u}\Big (\inf \limits _{T\le t<T_{1}^{0}}U(t)\ge b,T<\infty \Big ) =\frac{\Phi _{b}(u)\Phi (u)}{\Phi _{b}(0)}. \end{aligned}$$
Therefore, we can get
$$\begin{aligned}&P^{u}\Big (\sup \limits _{0\le t<T_{1}^{0}}U(t)<a,\inf \limits _{0\le t<T_{1}^{0}}U(t)\ge b\Big )\\&\quad =P^{u}\Big (\inf \limits _{0\le t<T_{1}^{0}}U(t)\ge b\Big )P^{u}\Big (\sup \limits _{0\le t<T_{1}^{0}}U(t)\ge a,\inf \limits _{0\le t<T_{1}^{0}}U(t)\ge b\Big )\\&\quad =\frac{\Phi (a)\Phi _{b}(u)\Phi (u)\Phi _{b}(a)}{\Phi (a)\Phi _{b}(0)}. \end{aligned}$$
This ends the proof. \(\square\)
Remark 2

1.
When \(g(x)=c\), the risk model simplifies to the classical compound Poisson risk model, Theorem 5 is the same as Lemma 3.5 in Wu et al. (2003).

2.
When \(g(x)=c, a=\infty\), Theorem 5 simplifies to
$$\begin{aligned} P^{u}\Big (\inf \limits _{0\le t<T_{1}^{0}}U(t)\ge b\Big )=\frac{\Phi (u+b)\Phi (u)}{\Phi (b)}, \end{aligned}$$
which coincides with Theorem 1 in Picard (1994).

3.
When \(g(x)=c+\delta x\) for \(x \ge 0\) and \(g(x)=c+r x\) for \(x<0\), the risk model is reduced to the risk model with credit and debit interests. Let \(a=\infty\), Theorem 5 simplifies to
$$\begin{aligned} P^{u}\Big (\inf \limits _{0\le t<T_{1}^{0}}U(t)\ge b\Big )=\frac{\Phi _{b}(u)\Phi (u)}{\Phi _{b}(0)}, \end{aligned}$$
which is the same as (6.2) in Li and Lu (2013).
Next, we consider the maximum surplus and the maximal severity of ruin before the time of the surplus process leaving zero ultimately, which describe the best situation and the worst situation the company would experience before the time of the surplus process leaving zero ultimately. We obtain their explicit expression in the following theorem.
Theorem 6
For
\(a>u\ge 0\)
and
\(b>0\), we have
$$\begin{aligned} P^{u}\Big (\sup \limits _{0\le t<L}U(t)<a,\inf \limits _{0\le t<L}U(t)\ge b,L>0\Big )=\frac{\Phi _{b}(u)}{\Phi _{b}(a)}\Phi (a)\Phi (u), \end{aligned}$$
In particular,
$$\begin{aligned}&P^{u}\Big (\inf \limits _{0\le t<L}U(t)\ge b,L>0\Big )=\Phi _{b}(u)\Phi (u),\\&P^{u}\Big (\sup \limits _{0\le t<L}U(t)<a,L>0\Big )=\Phi (a)\Phi (u), \end{aligned}$$
where
\(\Phi (u), \Phi _{b}(u)\)
are given by (7) and (8).
Proof
Note that the event \(\{\sup \nolimits _{0\le t<L}U(t)<a,\inf \nolimits _{0\le t<L}U(t)\ge b,L>0\}\) is equivalent to the event \(\{T<T_{1}^{a}<T_{b},\inf \nolimits _{t\ge 0}U(T_{1}^{a}+t)\ge 0\}\), so we can obtain
$$\begin{aligned}&P^{u}\Big (\sup \limits _{0\le t<L}U(t)<a,\inf \limits _{0\le t<L}U(t)\ge b,L>0\Big )\\&\quad =P^{u}\Big (T<T_{1}^{a}<T_{b},\inf \limits _{t\ge 0}U(T_{1}^{a}+t)\ge 0\Big )\\&\quad = P^{u}\Big (T_{1}^{a}<T_{b},\inf \limits _{t\ge 0}U(T_{1}^{a}+t)\ge 0\Big )P^{u}\Big (T_{1}^{a}<T,\inf \limits _{t\ge 0}U(T_{1}^{a}+t)\ge 0\Big )\\&\quad =P^{u}\Big [P^{u}\Big (T_{1}^{a}<T_{b},\inf \limits _{t\ge 0}U(T_{1}^{a}+t)\ge 0{\mathscr {F}}_{T_{1}^{a}}\Big )\Big ]\\&\qquad P^{u}\Big [P^{u}\Big (T_{1}^{a}<T,\inf \limits _{t\ge 0}U(T_{1}^{a}+t)\ge 0{\mathscr {F}}_{T_{1}^{a}}\Big )\Big ]\\&\quad =P^{u}(T_{1}^{a}<T_{b})P^{a}\Big (\inf \limits _{t\ge 0}U(t)\ge 0\Big )P^{u}\left( T_{1}^{a}<T\right) P^{a}\Big (\inf \limits _{t\ge 0}U(t)\ge 0\Big ). \end{aligned}$$
Using the same argument as Theorem 3 and Corollary 2, we have
$$\begin{aligned} P^{u}(T_{1}^{a}<T_{b})=\frac{\Phi _{b}(u)}{\Phi _{b}(a)},\quad P^{u}(T_{1}^{a}<T)=\frac{\Phi (u)}{\Phi (a)}. \end{aligned}$$
Hence we can obtain
$$\begin{aligned} P^{u}\left( \sup \limits _{0\le t<L}U(t)<a,\inf \limits _{0\le t<L}U(t)\ge b,L>0\right) =\frac{\Phi _{b}(u)}{\Phi _{b}(a)}\Phi (a)\Phi (u). \end{aligned}$$
This completes the proof. \(\square\)
Example 1
Let \(g(x)=c\) and the individual claim amount distribution be exponential with mean value \(\mu\). Then
$$\begin{aligned} f^{n*}(x)=\frac{1}{\mu ^{n}\Gamma (n)}x^{n1}e^{\frac{x}{\mu }}, \quad x>0, \end{aligned}$$
and
$$\begin{aligned} p(s,u,x)&= {} e^{\lambda s}\sum \limits _{n=1}^{\infty }\frac{(\lambda s)^{n}}{n!}f^{n*}(u+csx)\\&= {} \frac{\lambda s}{\mu }e^{\frac{ux+(c+\lambda \mu )s}{\mu }} \quad \sum \limits _{n=1}^{\infty }\left[ \frac{1}{n \Gamma ^{2}(n)}\left( \frac{\lambda s(ux+cs)}{\mu }\right) ^{n1}\right] . \end{aligned}$$
Thus
$$\begin{aligned} g_{u}^{x}(s)=\frac{\lambda cs}{\mu } e^{\frac{ux+(c+\lambda \mu )s}{\mu }}\quad\sum \limits _{n=1}^{\infty }\left[ \frac{1}{n \Gamma ^{2}(n)}\left( \frac{\lambda s(ux+cs)}{\mu }\right) ^{n1}\right] . \end{aligned}$$
Hence
$$\begin{aligned} f_{u}^{x}(s)=&\frac{1}{s}e^{\frac{ux+(c+\lambda \mu )s}{\mu }}\quad\sum \limits _{n=0}^{\infty }(1)^{n}\sum \limits _{m_{1}=1}^{\infty }\cdots \sum \limits _{m_{n}=1}^{\infty }\sum \limits _{m=1}^{\infty } \sum \limits _{k=1}^{m1}\left( \frac{\lambda cs^{2}}{\mu }\right) ^{\sum \limits _{i=1}^{n}m_{i}}\\&\quad \times \prod \limits _{i=1}^{n}\frac{B(2\sum \nolimits _{k=1}^{i1}m_{k},2m_{i})}{m_{i} \Gamma ^{2}(m_{i})}\left( \frac{\lambda s}{\mu }\right) ^{m} \frac{B(2\sum \nolimits _{i=1}^{n}m_{i},m+k1)}{m \Gamma ^{2}(m)}C_{m1}^{k}(ux)^{m1k}(cs)^{k+1}, \end{aligned}$$
where \(B(x,y)=\int _{0}^{1}t^{x1}(1t)^{y1}dt,x>0,y>0\) is the Betafunction and \(C_{n}^{m}=\frac{n!}{m!(nm)!}\) is a combinatorial number. It follows that
$$\begin{aligned} \Psi _{x}(u)=\frac{\lambda \mu }{c}e^{\frac{(c\lambda \mu)(ux)}{c\mu}},\quad \Psi (u)=\frac{\lambda \mu }{c}e^{\frac{(c\lambda \mu )u}{c\mu }}, \end{aligned}$$
and
$$\begin{aligned} \Phi _{x}(u)&= {} 1\Psi _{x}(u)=1\frac{\lambda \mu }{c}e^{\frac{(c\lambda \mu )(ux)}{c\mu }},\\ \Phi (u)&= {} 1\Psi (u)=1\frac{\lambda \mu }{c}e^{\frac{(c\lambda \mu )u}{c\mu }}. \end{aligned}$$
By Theorem 3, we have
$$\begin{aligned} G(u,a) =\left\{ \begin{array}{ll} \frac{\lambda \mu }{c}e^{\frac{(c\lambda \mu )u}{c\mu }},&{}\quad a\le u,\\ \frac{\lambda \mu e^{\frac{(c\lambda \mu )a}{c\mu }}\quad\left( c\lambda \mu e^{\frac{(c\lambda \mu )u}{c\mu }}\right ) }{c \left( c\lambda \mu e^{\frac{(c\lambda \mu )a}{c\mu }}\right) }, &{}\quad a>u.\\ \end{array}\right. \end{aligned}$$
By Theorem 4, we get
$$\begin{aligned} P^{u}\Big (\sup \limits _{0\le t<L}U(t)<a,L>0\Big )=\left\{ \begin{array}{ll} \frac{\lambda \mu }{c}\left( e^{\frac{(c\lambda \mu )u}{c\mu }}e^{\frac{(c\lambda \mu )a}{c\mu }}\right) ,&{}\quad u\ge 0,\\ 1\frac{\lambda \mu }{c}e^{\frac{(c\lambda \mu )a}{c\mu }},&{}\quad u<0.\\ \end{array}\right. \end{aligned}$$
By Theorem 5, we have
$$\begin{aligned}&P^{u}\Big (\sup \limits _{0\le t<T_{1}^{0}}U(t)<a,\inf \limits _{0\le t<T_{1}^{0}}U(t)\ge b\Big )\\&\quad =\frac{\left( c\lambda \mu e^{\frac{(c\lambda \mu )(u+b)}{c\mu }}\right) }{\left( c\lambda \mu e^{\frac{(c\lambda \mu )b}{c\mu }}\right) } \frac{\left( c\lambda \mu e^{\frac{(c\lambda \mu )u}{c\mu }}\right) \left( c\lambda \mu e^{\frac{(c\lambda \mu )(a+b)}{c\mu }}\right) }{ \left( c\lambda \mu e^{\frac{(c\lambda \mu )a}{c\mu }}\right) \left( c\lambda \mu e^{\frac{(c\lambda \mu )b}{c\mu } }\right) }. \end{aligned}$$
When \(u=0\), the result is the same as W(0, a, b) in Wu et al. (2003). By Theorem 6, we have
$$\begin{aligned}&P^{u}\Big (\sup \limits _{0\le t<L}U(t)<a,\inf \limits _{0\le t<L}U(t)\ge b,L>0\Big )\\&\quad =\frac{\left( c\lambda \mu e^{\frac{(c\lambda \mu )(u+b)}{c\mu } }\right) }{\left( c\lambda \mu e^{\frac{(c\lambda \mu )(a+b)}{c\mu }}\right) }\left( 1\frac{\lambda \mu }{c}e^{\frac{(c\lambda \mu )a}{c\mu }}\right) \left( 1\frac{\lambda \mu }{c}e^{\frac{(c\lambda \mu )u}{c\mu }}\right) . \end{aligned}$$