22institutetext: Lomonosov Moscow State University, Federal Research Center
“Computer Science and Control” of the Russian Academy of Sciences Moscow,
Russia, and Université de Franche-Comté, Laboratoire de Mathématiques, UMR
CNRS 6623, 16 Route de Gray, 25030 Besançon, France
22email<EMAIL_ADDRESS>
33institutetext: Lomonosov Moscow State University, Moscow, Russia
33email<EMAIL_ADDRESS>
# Ruin Probabilities for a Sparre Andersen Model with Investments: the Case of
Annuity Payments

Yuri Kabanov  Platon Promyslov

###### Abstract

This note is a complement to the paper by Eberlein, Kabanov, and Schmidt on
the asymptotic of the ruin probability in a Sparre Andersen non-life insurance
model with investments a risky asset whose price follows a geometric Lévy
process. Using the techniques of semi-Markov processes we extend the result of
the mentioned paper to the case of annuities and models with two-sided jumps.

###### Keywords:

Ruin probabilities Sparre Andersen model Actuarial models with investments
Renewal processes Annuities Distributional equations

Dedicated to the memory of Tomas Björk.

Mathematics Subject Classification (2010) 60G44

JEL Classification G22 $\cdot$ G23

##  1 Introduction

In the classical Sparre Andersen model of insurance company the counts of
claims form a renewal process. In recent studies, see ACT , EKT  and
references therein, this model was enriched by the assumption that the capital
reserve of the insurance company is fully invested in a risky asset whose
price evolves as a geometric Lévy process. In the paper EKT  by Eberlein,
Kabanov, and Schmidt it was considered the non-life insurance version of such
a model. It was shown that under rather mild hypotheses on the business
process the asymptotic behavior of the (ultimate) ruin probability is
essentially the same as in the Cramér–Lundberg model with risky investments.
Namely, the ruin probability decays, up to a multiplicative constant, as the
function $u^{-\beta}$ where $u$, the initial capital, tends to infinity. The
decay rate $\beta$ depends only of characteristics of the price process. The
method of analysis in EKT  is based heavily on the assumption that the risk
process has only downward jumps and, therefore, crosses the zero level only by
a jump. This specific feature allows a straightforward reduction to a
discrete-time Markovian framework.

The approach of EKT  left an open question whether the results hold also in
the case of upward jumps. This is a feature of the annuity model when the risk
process crosses the zero level in a continuous way. In a less popular mixed
model with two-sided jumps the crossing may happen in both way. Of course, a
positive answer is expected here: this was already established for the
Cramér–Lundberg models with investments analyzed by Kabanov, Pergamenshchikov,
and Pukhlyakov, KP , KPukh , as well as in very general Lévy
Ornstein–Uhlenbeck models introduced and studied by Paulsen, see Paul-93 ,
Paul , Paul-98 , Paul-G , and a more recent paper KP2020  by Kabanov and
Pergamenshchikov.

Our note, based on the study EKT , gives a positive answer for a Sparre
Andersen model with investments in its annuity version, with the upward jumps.
We discuss briefly the needed changes leading to a result for a model with
upward and downward jumps used serving to describe the evolution of the
capital reserve of a company with two types of the business activity. Our
techniques is based on the imbedding of a semi-Markov process into a Markov
one by increasing the dimensionality.

In the paper we use standard notations of stochastic calculus and concepts
discussed in details in KP2020 , EKT .

##  2 The model

The Sparre Andersen model with risky investments considered contains two
ingredients:

1\. The price process of a risky financial asset $S=(S_{t})_{t\geq 0}$. It is
of the form $S={\cal E}(R)$ where ${\cal E}$ is the stochastic exponential,
$R$ is a Lévy process with the Lévy triplet $(a,\sigma^{2},\Pi)$ and such that
$\Pi((-\infty,-1])=0$. The latter condition ensures that the jumps $\Delta
R>-1$, hence, the price $S>0$. In such a case, $S=e^{V}$ where $V=\ln S$ is
again a Lévy process which can be given by the formula

$V_{t}=at-\frac{1}{2}\sigma^{2}t+\sigma
W_{t}+h*(\mu-\nu)_{t}+(\ln(1+x)-h)*\mu_{t},$ (2.1)

where $h(x):=xI_{\\{|x|\leq 1\\}}$. The Lévy triplet of $V$ is
$(a_{V},\sigma^{2},\Pi_{V})$ with

$a_{V}=a-\frac{\sigma^{2}}{2}+\Pi(h(\ln(1+x))-h)$

and $\Pi_{V}=\Pi\varphi^{-1}$, $\varphi:x\mapsto\ln(1+x)$.

It is assume that $R$ is non-deterministic, that is, at least one of the
parameters $\sigma^{2}$ or $\Pi$ is not zero.

2\. The ”business process“. It is an independent of $S$ compound renewal
process $P=(P_{t})$. Classically, it can be written in the form

$P_{t}=ct+\sum_{i=1}^{N_{t}}\xi_{i},$ (2.2)

where $N=(N_{t})$ is a counting renewal process with the interarrival times
(lengths of the inter jump intervals) $U_{i}:=T_{i}-T_{i-1}$, $i\geq 2$,
forming an i.i.d. sequence independent of the i.i.d. sequence of random
variables $\xi_{i}=\Delta P_{T_{i}}$, $i\geq 1$, with the common law
$F_{\xi}$, $F_{\xi}(\\{0\\})=0$. In the sequel, a “generic“ r.v. with such a
law is denoted by $\xi$. As usual, $T_{0}:=0$. The common law of $U_{i}$ we
denoted by $F$ and use the same character for its distribution function.

The risk process $X=X^{u}$, $u>0$, is defined as the solution of the non-
homogeneous linear stochastic equation

$X_{t}=u+\int_{0}^{t}X_{s-}dR_{s}+P_{t}.$ (2.3)

The ruin probability is the function of the initial capital $\Psi(u):={\bf
P}[\tau^{u}<\infty]$ where $\tau^{u}:=\inf\\{t:X^{u}_{t}\leq 0\\}$.

The cases of major interest are: $c>0$ and $\xi_{i}<0$ (a non-life insurance
model, considered in EKT ) and $c<0$ and $\xi_{i}>0$ (annuities payments
model). The latter case studied here, is often interpreted as a model of
venture company paying salary and selling innovations. The case where
$F_{\xi}$ charges both half-axes can be viewed as a model of company combined
two types of activity, see, e.g., ACT . and we study also the case where
$\xi_{i}$ may take positive and negative values.

If $c\geq 0$ and $\xi>0$ the ruin never happens and this case is excluded from
considerations.

Standing assumption. The cumulant generating function $H:q\to\ln{\bf
E}\,e^{-qV_{T_{1}}}$ of the random variable $V_{T_{1}}$ has a root $\beta>0$
not laying on the boundary of the effective domain of $H$. That is, if the
${\rm int\,dom}\,H=(\underline{q},\bar{q})$, there is a unique root
$\beta\in(0,\bar{q})$.

We are looking for conditions under which

$0<\liminf_{u\to\infty}u^{\beta}\Psi(u)\leq\limsup_{u\to\infty}u^{\beta}\Psi(u)<\infty.$
(2.4)

The paper EKT  treats the case of the non-life insurance. We formulate its
main result in a more transparent form.

######  Theorem 2.1 (EKT )

Suppose that the drift $c\geq 0$, the law $F_{\xi}$ is concentrated on
$(-\infty,0)$, ${\bf E}[|\xi|^{\beta}]<\infty$, and ${\bf
E}[e^{{\varepsilon}T_{1}}]<\infty$ for some ${\varepsilon}>0$. Then (2.4)
holds if at least one of the following conditions are fulfilled:

1. $\sigma\neq 0$ or $\xi$ is unbounded from below.

2a. $\Pi((-1,0))>0$ and $\Pi((0,\infty))>0$.

2b. $\Pi((-1,0))=0$ and $\Pi(h)=\infty$.

2c. $\Pi((0,\infty))=0$ and $\Pi(|h|)=\infty$.

2d. $\Pi((-\infty,0))=0$, $0<\Pi(h)<\infty$, $F((0,t))>0$ for every $t>0$.

2e. $\Pi((0,\infty))=0$, $0<\Pi(|h|)<\infty$, $F((0,t))>0$ for every $t>0$.

The proof in EKT  used heavily the assumption that the business process has a
positive drift and negative claims corresponding to the non-life insurance
setting. In such a case the ruin may happen only at an instant of jump and,
therefore, one needs to monitor the risk process only at $T_{1}$, $T_{2}$, and
so on. Such a reduction to a discrete-time ruin model does not work if
$\xi_{i}>0$.

In our paper we consider the annuity model of the Sparre Andersen type where
the ruin occurs because exhausting resources and the risk process reaches zero
in a continuous way. The main result can be formulated as follows.

###### Theorem 2.2

Suppose that the drift $c<0$, the law $F_{\xi}$ is concentrated on
$(0,\infty)$, ${\bf E}[\xi^{\beta}]<\infty$, and ${\bf
E}[e^{{\varepsilon}T_{1}}]<\infty$ for some ${\varepsilon}>0$. Then (2.4)
holds if at least one of the following conditions are fulfilled:

1. $\sigma\neq 0$.

2a. $\Pi((-1,0))>0$ and $\Pi((0,\infty))>0$.

2b. $\Pi((-1,0))=0$ and $\Pi(h)=\infty$.

2c. $\Pi((0,\infty))=0$ and $\Pi(|h|)=\infty$.

2d. $\Pi((-1,0))=0$, $0<\Pi(h)<\infty$, $F((t,\infty))>0$ for every $t>0$.

2e. $\Pi((0,\infty))=0$, $0<\Pi(|h|)<\infty$, $F((t,\infty))>0$ for every
$t>0$.

For the mixed case we have the following result.

###### Theorem 2.3

Suppose that the drift $c\in{\bf R}$, the law $F_{\xi}$ charges both half-
lines $(-\infty,0)$ and $(0,\infty)$, ${\bf E}[|\xi|^{\beta}]<\infty$, and
${\bf E}[e^{{\varepsilon}T_{1}}]<\infty$ for some ${\varepsilon}>0$. Then
(2.4) holds if at least one of the following conditions are fulfilled:

1. $\sigma\neq 0$ or $|\xi|$ is unbounded.

2a. $\Pi((-1,0))>0$ and $\Pi((0,\infty))>0$.

2b. $\Pi((-1,0))=0$ and $\Pi(h)=\infty$.

2c. $\Pi((0,\infty))=0$ and $\Pi(|h|)=\infty$.

2d. $\Pi((-1,0))=0$, $0<\Pi(h)<\infty$, $F((t,\infty))>0$ for every $t>0$ in
the case $c<0$ and $F_{\xi}((0,{\varepsilon}))>0$ for every ${\varepsilon}>0$
in the case $c\geq 0$.

2e. $\Pi((0,\infty))=0$, $0<\Pi(|h|)<\infty$, $F((t,\infty))>0$ for every
$t>0$ in the case $c<0$ and $F_{\xi}((0,{\varepsilon}))>0$ for every
${\varepsilon}>0$ in the case $c\geq 0$.

The annuity and the mixed setting require a different approach inspired by the
theory of semi-Markov processes. Namely, we consider the business process $P$
as the component of the two-dimensional Markov process $(P,D)$ where the
second component $D=D^{r}$ is a “clock”, i.e. a process measuring the elapsed
time after the instant of last claim. We assume that the law of $U=T_{1}$ may
be different from the common law of the further interarrival times: at the
instant zero a portion $r$ of the interarrival time is already elapsed. This
feature admits obvious justifications: e.g., the venture company may change
the governance when a project was still in progress.

Here and throughout the paper we use the superscript $r$ to emphasize that the
law of a random variable or a process depends on $r$, skipping usually $r=0$.

Formally, the “clock”, $D^{r}=(D^{r}_{t})$, is a process with the initial
value $D^{r}_{0}=r$, $D^{r}_{t}=r+t$ on the interval $[0,T_{1})$, and
$D^{r}_{t}:=t-T^{r}_{n}$ on all other interarrival intervals
$[T^{r}_{n},T^{r}_{n+1})$, $n\geq 1$. That is, the “clock” restarts from zero
at each instant $T^{r}_{n}$. We denote by $F^{r}$ the law of the first
interarrival time $T^{r}_{1}=T^{r}_{1}-T_{0}$. In accordance with our
convention $F^{0}=F$.

Alternatively, $D^{r}$ can be representing as the solution of the linear
equation

$D^{r}_{t}=r+t-\int_{[0,t]}D^{r}_{s-}dN_{s}.$

Typically, ${\bf P}[T^{r}_{1}>t]={\bf P}[T_{i}>t+r]/{\bf P}[T_{i}>r]$, $i>1$.
In the case of exponential distribution $F^{r}=F$ for all $r\geq 0$ (“absence
of memory”).

We assume that $F^{r}\geq F$.

Recall that the assumed independence of $P^{r}$ and $R$ implies that the joint
quadratic characteristic $[P^{r},R]$ is zero and the ruin process $X^{u,r}$
can be written in the form resembling the Cauchy formula for solutions of
liner differential equations:

$X^{u,r}_{t}=e^{V_{t}}(u-Y^{r}_{t}),$ (2.5)

where

$Y^{r}_{t}:=-\int_{(0,t]}{\cal
E}^{-1}_{s-}(R)dP^{r}_{s}=-\int_{(0,t]}e^{-V_{s-}}dP^{r}_{s}.$ (2.6)

The strict positivity of the process ${\cal E}(R)=e^{V}$ implies that the ruin
time

$\tau^{u,r}:=\inf\\{t\geq 0:\ X^{u,r}_{t}\leq 0\\}=\inf\\{t\geq 0:\
Y^{r}_{t}\geq u\\}.$

The crucial element of our study is the following

###### Lemma 1

Suppose that $Y^{r}_{t}\to Y^{r}_{\infty}$ almost surely as $t\to\infty$ where
$Y^{r}_{\infty}$ is a finite random variable such that $\bar{G}(u,r):={\bf
P}[Y^{r}_{\infty}>u]>0$ for every $u>0$ and $r\geq 0$. If
$\bar{G}_{*}:=\inf_{q}\bar{G}(0,q)>0$, then

$\bar{G}(u,r)\leq\,\Psi(u,r)=\frac{\bar{G}(u,r)}{{\bf
E}\left[\bar{G}(X^{u,r}_{\tau^{u,r}},D^{r}_{\tau^{u,r}})\,|\,\tau^{u,r}<\infty\right]}\leq\frac{1}{\bar{G}_{*}}{\bar{G}(u,r)}.$
(2.7)

Proof. Let $\tau$ be an arbitrary stopping time with respect to the filtration
$({\cal F}^{P,D,R}_{t})$. As we assume that the finite limit $Y^{r}_{\infty}$
exists, the random variable

$Y^{r}_{\tau,\infty}:=\begin{cases}-\lim_{N\to\infty}\int_{(\tau,\tau+N]}\,e^{-(V_{s-}-V_{\tau})}dP^{r}_{s},&\tau<\infty,\\\
0,&\tau=\infty,\end{cases}$

is well defined. On the set $\\{\tau<\infty\\}$

$Y^{r}_{\tau,\infty}=e^{V_{\tau}}(Y^{r}_{\infty}-Y^{r}_{\tau})=X^{u,r}_{\tau}+e^{V_{\tau}}(Y^{r}_{\infty}-u).$
(2.8)

Let $\zeta$ be a ${\cal F}_{\tau}^{P,D,R}$-measurable random variable.

Using the strong Markov property, we get that

${\bf P}\left[Y^{r}_{\tau,\infty}>\zeta,\ \tau<\infty\right]={\bf
E}\left[\bar{G}(\zeta,D^{r}_{\tau})I_{\\{\tau<\infty\\}}\right]$ (2.9)

Noting that $\Psi(u,r):={\bf P}\left[\tau^{u,r}<\infty\right]\geq{\bf
P}\left[Y^{r}_{\infty}>u\right]>0$, we deduce from here using (2.8) that

$\displaystyle\bar{G}(u,r)$ $\displaystyle=$ $\displaystyle{\bf
P}\left[Y^{r}_{\infty}>u,\ \tau^{u,r}<\infty\right]={\bf
P}\left[Y^{r}_{\tau^{u,r},\infty}>X^{u,r}_{\tau^{u,r}},\
\tau^{u,r}<\infty\right]$ $\displaystyle=$ $\displaystyle{\bf
P}[\tau^{u,r}<\infty]{\bf
E}\left[\bar{G}(X^{u,r}_{\tau^{u,r}},D^{r}_{\tau^{u,r}})\,|\,\tau^{u,r}<\infty\right]$
$\displaystyle\geq$ $\displaystyle{\bf P}[\tau^{u,r}<\infty]{\bf
E}\left[\bar{G}(0,D^{r}_{\tau^{u,r}})\,|\,\tau^{u,r}<\infty\right]$
$\displaystyle\geq$ $\displaystyle{\bf
P}[\tau^{u,r}<\infty]\inf_{q}\bar{G}(0,q)$

and get the result. $\Box$

In view of the above lemma the proof of the main theorem is reduced to
establishing the existence of finite limits $Y^{r}_{\infty}$ and finding the
asymptotic of the tail of their distributions.

Let us introduce the notations

$Q^{r}_{k}:=-\int_{(T^{r}_{k-1},T^{r}_{k}]}e^{-(V_{s-}-V_{T^{r}_{k-1}})}dP^{r}_{s},\qquad
M^{r}_{k}:=e^{-(V_{T^{r}_{k}}-V_{T^{r}_{k-1}})}.$ (2.10)

###### Lemma 2

The random variables $Y^{r}_{\infty}$ admit the representations

$Y^{r}_{\infty}=Q^{r}_{1}+M^{r}_{1}\tilde{Y}^{r}_{\infty},$

where

$Q^{r}_{1}:=-\int_{[0,T^{r}_{1}]}e^{-V_{s-}}dP^{r}_{s},\qquad
M^{r}_{1}:=e^{-V_{T^{r}_{1}}},$ (2.11)

$(Q^{r}_{1},M^{r}_{1})$ and $\tilde{Y}^{r}_{\infty}$ are independent, and the
laws of $\tilde{Y}^{r}_{\infty}$ and $Y^{0}_{\infty}$ coincide.

Proof. Note that

$\displaystyle Y^{r}_{T^{r}_{n}}$ $\displaystyle=$
$\displaystyle-\int_{[0,T^{r}_{1}]}e^{-V_{s-}}dP^{r}_{s}-\sum_{k=2}^{n}e^{V_{T^{r}_{k-1}}}\int_{(T^{r}_{k-1},T^{r}_{k}]}e^{-(V_{s-}-V_{T^{r}_{k-1}})}dP^{r}_{s}$
$\displaystyle=$ $\displaystyle
Q^{r}_{1}+M^{r}_{1}\left(Q^{r}_{2}+\sum_{k=3}^{n}M^{r}_{2}...M^{r}_{k-1}Q^{r}_{k}\right),$

where the random variable in the parentheses is independent of
$(Q^{r}_{1},M^{r}_{1})$ and has the same distribution as $Y^{0}_{n-1}$. $\Box$

###### Lemma 3

Suppose that $Y_{\infty}$ is unbounded from above. If $c<0$, then

$\inf_{q}\bar{G}(0,q)\geq{\bf E}[\bar{G}(\xi,0)]>0.$

If $c\in{\bf R}$ and the distribution function $F^{r}\leq F$, then

$\inf_{q}\bar{G}(0,q)>0.$

Proof. Using Lemma 2 we have:

$\displaystyle\bar{G}(0,r)$ $\displaystyle=$ $\displaystyle{\bf
P}[Y^{r}_{\infty}>0]={\bf P}[Q_{1}^{r}/M_{1}^{r}+\tilde{Y}^{r}_{\infty}>0]$
$\displaystyle=$ $\displaystyle\int{\bf
P}\left[|c|e^{V_{t}}\int_{[0,t]}e^{-V_{s}}ds-\xi_{1}+\tilde{Y}^{r}_{\infty}>0\right]F_{T_{1}^{r}}(dt)$
$\displaystyle\geq$ $\displaystyle{\bf
P}[\tilde{Y}^{r}_{\infty}>\xi_{1}]=\int{\bf
P}[\tilde{Y}^{r}_{\infty}>x]F_{\xi}(dx)=\int\bar{G}(x,0)F_{\xi}(dx)>0$

since $Y_{\infty}$ is unbounded.

The inspection of the proof reveals that the majority of arguments does work
with minor changes also for the case where $c$ is of an arbitrary sign and the
law $F_{\xi}$ charges $(-\infty,0)$ and $(0,\infty)$. In particular, the proof
that the finite limit $Y_{\infty}$ exists remains the same.

Put $f_{t}:=|\xi_{1}|+|c|te^{2V^{*}_{t}}$, where $V^{*}_{t}:=\sup_{s\leq
t}|V_{s}|$. Then

$|Q_{1}^{r}|/M_{1}^{r}\leq|\xi_{1}|+|c|e^{V_{T^{r}_{1}}}\int_{[0,T_{1}^{r}]}e^{-V_{s}}ds\leq
f_{{T_{1}^{r}}}.$

It follows that

$\displaystyle\bar{G}(0,r)$ $\displaystyle=$ $\displaystyle{\bf
P}[\tilde{Y}^{r}_{\infty}>-Q_{1}^{r}/M_{1}^{r}]={\bf
E}\bar{G}(-Q_{1}^{r}/M_{1}^{r},0)\geq{\bf E}\bar{G}(|Q_{1}^{r}|/M_{1}^{r},0)$
$\displaystyle\geq$ $\displaystyle{\bf E}\int\bar{G}(f_{t},0)F^{r}(dt)=-{\bf
E}\int F^{r}(t)d\bar{G}(f_{t},0)$ $\displaystyle\geq$ $\displaystyle-{\bf
E}\int F(t)d\bar{G}(f_{t},0)\geq{\bf E}\int\bar{G}(f_{t},0)F(dt)>0,$

where we use the property $F^{r}\geq F$. Thus, $\inf_{r}\bar{G}(0,r)>0$.
$\Box$

##  3 Tails of solutions of distributional equations

As a number of results on the ruin with investments, the proof is based on the
implicit renewal theory. As in EKT  we shall use the following formulation
combining several useful facts:

###### Theorem 3.1

Suppose that for some $\beta>0$,

$\displaystyle{\bf E}[M^{\beta}]=1,\ \ \ {\bf E}[M^{\beta}\,(\ln
M)^{+}]<\infty,\ \ \ {\bf E}[|Q|^{\beta}]<\infty.$ (3.1)

Let $Y_{\infty}$ be the solution of the distributional equation
$Y_{\infty}\stackrel{{\scriptstyle d}}{{=}}Q+MY_{\infty}$ and let
$\bar{G}(u):={\bf P}[Y_{\infty}>u]$. Then $\limsup
u^{\beta}\bar{G}(u)<\infty$. If the random variable $Y_{\infty}$ is unbounded
from above, then $\liminf u^{\beta}\bar{G}(u)>0$.

In the previous section we introduce a process $Y=(Y_{t})$. Assuming that it
has at infinity a limit $Y_{\infty}$, which is a finite unbounded from above
random variable, we have proved that it solves the required distributional
equation and its tail function gives lower and upper bounds for the ruin
probability. It remains to check that the hypotheses of Theorem 2.2 ensure the
assumed properties and get the result applying the above theorem. We do this
in the next sections.

##  4 The existence of the limit $Y^{r}_{\infty}$

First, we recall several results from EKT .

######  Lemma 4 ( EKT , Lemma 2.1)

Let $T>0$ be a random variable independent of $R$. Suppose that ${\bf
E}[e^{{\varepsilon}T}]<\infty$ for some ${\varepsilon}>0$. Let
$\beta\in(0,\bar{q})$ be the root of the equation $H(q)=0$. If
$q\in[\beta,\bar{q})$ is such that $H(q)\leq{\varepsilon}/2$, then

${\bf E}\left[\sup_{s\leq T}e^{-qV_{s}}\right]<\infty.$ (4.1)

###### Corollary 1

Suppose that ${\bf E}[e^{{\varepsilon}T_{1}}]<\infty$ for some
${\varepsilon}>0$. Let

$\widehat{Q}_{1}:=\sup_{t\leq T_{1}}|e^{-V_{-}}\boldsymbol{\cdot}P_{t}|.$

If ${\bf E}[|\xi_{1}|^{\beta}]<\infty$, then ${\bf
E}[|\widehat{Q}_{1}^{\beta}]<\infty$.

Though the above assertion is a bit more general than Corollary 2.2 in EKT ,
the proof is exactly the same. Note also that it does not depend on the sign
of $c$ or $\xi_{1}$ and needs only the integrability of $|\xi_{1}|^{\beta}$.
It implies, in particular, that ${\bf E}[|Q_{1}^{\beta}|]<\infty$

###### Lemma 5

Suppose that ${\bf E}[e^{{\varepsilon}T_{1}}]<\infty$ and ${\bf
E}[|\xi_{1}|^{\beta\wedge{\varepsilon}\wedge 1}]<\infty$ for some
${\varepsilon}>0$. Then $Y_{t}\to Y_{\infty}$ almost surely as $t\to\infty$
where $Y_{\infty}$ is a finite random variable.

Proof. The convergence a.s. of the sequence $Y_{T_{n}}$, $n\geq 1$, to a
finite r.v. $Y_{\infty}$ has been proven in Lemma 4.1 of EKT  as well as the
fact that $\rho:={\bf E}[M_{1}^{p}]<1$ for any
$p\in(0,\beta\wedge{\varepsilon}\wedge 1)$.

Put $I_{n}:=(T_{n-1},T_{n}]$ and

$\Delta_{n}:=\sup_{v\in
I_{n}}\left|\int_{(T_{n-1},v]}e^{-V_{s-}}dP_{s}\right|=\prod_{i=1}^{n-1}M_{i}\sup_{v\in
I_{n}}\left|\int_{(T_{n-1},v]}e^{-(V_{s-}-V_{T_{n-1}})}dP_{s}\right|.$

By virtue of the Borel–Cantelli lemma, to get the announced result it is
sufficient to show that for every $\delta>0$

$\sum_{n=1}^{\infty}{\bf P}[\Delta_{n}\geq\delta]<\infty.$

But this is true because the Chebyshev inequality and the Corollary 1 imply
that ${\bf P}[\Delta_{n}\geq\delta]\leq\delta^{-p}\rho^{p}{\bf
E}[\widehat{Q}_{1}|^{p}]$. $\Box$

By Lemma 2 the sequence $Y^{r}_{T^{r}_{n}}$ converges a.s. to $Y^{r}_{\infty}$
and the same arguments as above allows us to conclude that $Y^{r}_{t}$ also
converges.

##  5  When the distribution of $Y_{\infty}$ is unbounded from above?

The question in the title of the section is studied in EKT  for the non-life
insurance case, i.e. when $c<0$ and $F_{\xi}((0,\infty))=1$. In the present
paper we provide sufficient conditions for the unboundedness from above for
all new cases using the techniques developed in the mentioned paper. It is
based on the following elementary observation: if $f:X\times Y\to{\bf R}$ is a
measurable function, r.v. $\eta$ and $\zeta$ are independent and have the laws
$F_{\eta}$ and $F_{\zeta}$, then the r.v. $f(\eta,\zeta)$ is unbounded from
above provided that there exists a measurable set $X_{0}\subseteq X$ with
$F_{\eta}(X_{0})>0$ such that the r.v. $f(x,\zeta)$ is unbounded from above
for every $x\in X_{0}$.

Let $A_{n}:=M_{1}...M_{n}$, $n\geq 1$, $A_{0}:=0$.

A tractable sufficient condition is give by the following

######  Lemma 6 (EKT , Lemma 5.1)

If there exists $n\geq 1$ such that the random variables $Q_{1}$ and
$(Q_{1}+\dots+A_{n-1}Q_{n})/A_{n}$ are unbounded from above, then $Y_{\infty}$
is unbounded from above.

It usually works already with $n=1$ but sometimes we need it with $n=2$. A
short look at the expressions

$\displaystyle Q_{1}$ $\displaystyle=$
$\displaystyle-c\int_{0}^{T_{1}}e^{-V_{r}}dr-e^{-V_{T_{1}}}\xi_{1}$ (5.1)
$\displaystyle Q_{1}/A_{1}$ $\displaystyle=$ $\displaystyle-
ce^{V_{T_{1}}}\int_{0}^{T_{1}}e^{-V_{r}}dr-\xi_{1},$ (5.2) $\displaystyle
Q_{1}/A_{2}+Q_{2}/M_{2}$ $\displaystyle=$ $\displaystyle-
ce^{V_{T_{2}}}\int_{0}^{T_{2}}e^{-V_{r}}dr-\xi_{1}e^{V_{T_{2}}-V_{T_{1}}}-\xi_{2}$
(5.3)

shows that $Y_{\infty}$ is unbounded from above when $\xi$ is unbounded from
below (of course, the latter property is not fulfilled for the annuity model).

Using the above sufficient condition of the unboundedness, we examine various
cases.

1. Let $\sigma\neq 0$. In this case the following lemma is helpful:

###### Lemma 7

Let $K>0$, $\sigma\neq 0$ and $0\leq s<t$. Then the random variables

$\zeta:=Ke^{\sigma W_{t}}-\int_{0}^{t}e^{\sigma
W_{r}}dr,\qquad\tilde{\zeta}:=Ke^{\sigma(W_{t}-W_{s})}-e^{\sigma
W_{t}}\int_{0}^{t}e^{\sigma W_{r}}dr,$ (5.4)

are unbounded from below and from above.

The property that $\zeta$ and $\tilde{\zeta}$ are unbounded from above has
been proven in EKT , Lemma 5.2. The unboundedness from below can be
established by similar arguments. It is also clear, that if $K=0$, then
$\zeta$ and $\tilde{\zeta}$ are unbounded from below.

The process $\bar{V}:=V-\sigma W$ is independent of the Wiener process $W$. If
$c<0$, then

$Q_{1}\geq|c|\inf_{r\leq T_{1}}e^{-\bar{V}_{r}}\int_{0}^{T_{1}}e^{-\sigma
W_{r}}dr-\xi_{1}e^{-\bar{V}_{T_{1}}}e^{-\sigma W_{T_{1}}}.$

Using the conditioning with respect to $\bar{V}$, $\xi_{1}$, $T_{1}$ and the
previous lemma, we get that $Q_{1}$ is unbounded from above. Since

$Q_{1}/A_{1}\geq|c|e^{\bar{V}_{T_{1}}}\inf_{r\leq
T_{1}}e^{-\bar{V}_{r}}e^{\sigma W_{T_{1}}}\int_{0}^{T_{1}}e^{-\sigma
W_{r}}dr-\xi_{1},$

we conclude in the same way that $Q_{1}/A_{1}$ is unbounded from above. If
$c\geq 0$, then necessary $F_{\xi}(-\infty,0))>0$ (recall that we exclude the
case $c\geq 0$, $\xi>0$ when the ruin is impossible). Lemma 5.2 implies that
the random variables $Q_{1}$ and $Q_{1}/A_{2}+Q_{2}/M_{2}$ are unbounded from
above.

2. Let $\sigma=0$ and let $\xi$ be bounded from below. We treat separately several subcases.

2a. $\Pi((-1,0))>0$ and $\Pi((0,\infty))>0$. Fix ${\varepsilon}>0$ such that
$\Pi((-1,-{\varepsilon}))>0$ and $\Pi(({\varepsilon},\infty))>0$ and put

$V^{(1)}:=I_{\\{-1<x<-{\varepsilon}\\}}\ln(1+x)*\mu+I_{\\{x>{\varepsilon}\\}}\ln(1+x)*\mu.$

Then the processes $V^{(1)}$ and $V^{(2)}:=V-V^{(1)}$ are independent.

Note that $V^{(1)}$ is the sum of two independent compound Poisson processes
with negative and positive jumps, respectively, and the absolute values of
jumps are larger than some constant $c_{\varepsilon}>0$.

###### Lemma 8

Let $K>0$, $t>0$. Then the random variable

$\zeta:=Ke^{-V^{(1)}_{t}}-\int_{0}^{t}e^{-V^{(1)}_{r}}dr$ (5.5)

is unbounded from above and from below, the random variable

$\widehat{\zeta}:=e^{-V^{(1)}_{t}}\int_{0}^{t}e^{-V^{(1)}_{r}}dr.$

is unbounded from above.

Proof. The arguments are simple and we explain only the idea. One can consider
trajectories where $V^{(1)}$ has a lot of negative jumps in a neighborhood of
zero while all positive jumps are concentrate in a neighborhood of $t$.
Choosing suitable parameters and using the independence of processes with
positive and negative jumps we obtain that, with a strictly positive
probability, the first term in the definition of $\zeta$ is arbitrary close to
zero while the integral is arbitrary large. Thus, $\zeta$ is unbounded from
below. Symmetric arguments lead to the conclusion that $\zeta$ is unbounded
from above. $\Box$

Let $c<0$. Then the following bounds are obvious:

$Q_{1}\geq|c|\inf_{r\leq
T_{1}}e^{-V^{(2)}_{r}}\int_{0}^{T_{1}}e^{-V^{(1)}_{r}}dr-\xi_{1}e^{-\bar{V}^{(2)}_{T_{1}}}e^{-V^{(1)}_{T_{1}}},$
$Q_{1}/A_{1}\geq|c|e^{V^{(2)}_{T_{1}}}\inf_{r\leq
T_{1}}e^{-V^{(2)}_{r}}e^{V^{(1)}_{T_{1}}}\int_{0}^{T_{1}}e^{-V^{(1)}_{r}}dr-\xi_{1}.$

By conditioning with respect to the random variables $V^{(2)}$, $T_{1}$,
$\xi_{1}$, which are independent of $V^{(1)}$, and using Lemma 8 we easily
obtain that the random variables $Q_{1}$ and $Q_{1}/A_{1}$ are unbounded from
above and, by Lemma 6, so is $Y_{\infty}$.

Let $c\geq 0$. The same arguments as in EKT  show that $Q_{1}$ and
$Q_{1}/A_{2}+Q_{2}/M_{2}$ are unbounded from above on the non-null set
$\\{\xi_{1}<0,\;\xi_{2}<0\\}$.

2b. $\Pi((-1,0))=0$, $\Pi(h)=\infty$.

We use the decomposition of $V$ depending on the choice of
${\varepsilon}\in(0,1)$. Namely, put

$\displaystyle V^{\varepsilon}$ $\displaystyle:=$ $\displaystyle
I_{\\{x\leq{\varepsilon}\\}}h*(\mu-\nu)+I_{\\{x\leq{\varepsilon}\\}}(\ln(1+x)-h)*\mu,$
(5.6) $\displaystyle\tilde{V}^{\varepsilon}$ $\displaystyle:=$ $\displaystyle
I_{\\{x>{\varepsilon}\\}}h*(\mu-\nu)+I_{\\{x>{\varepsilon}\\}}(\ln(1+x)-h)*\mu.$
(5.7)

Note that $V_{t}=at+V^{\varepsilon}_{t}+\tilde{V}^{\varepsilon}_{t}$ and

$\tilde{V}^{\varepsilon}=I_{\\{x>{\varepsilon}\\}}\ln(1+x)*\mu-
I_{\\{x>{\varepsilon}\\}}h*\nu.$

###### Lemma 9

Let $K>0$, $t>0$. Then the random variables

$\eta:=\int_{0}^{t}e^{-V_{r}}dr-
Ke^{-V_{t}},\qquad\eta^{\prime}:=e^{V_{t}}\int_{0}^{t}e^{-V_{r}}dr$ (5.8)

are unbounded from above.

Proof. Without loss of generality we assume that $a=0$. Fix $N>0$ and choose
${\varepsilon}>0$ small enough to ensure that
$\Pi(I_{\\{x>{\varepsilon}\\}}h)\geq N$. Let
$\Gamma_{\varepsilon}:=\\{\sup_{s\leq t}|V^{\varepsilon}_{s}|\leq 1\\}$. Let
denoting by $J^{\varepsilon}$ and $\bar{J}^{\varepsilon}$ the processes in the
lhs of (5.6). Using the Doob inequality and the elementary bound
$x-\ln(1+x)\leq x^{2}/2$ for $x>0$, we get that

$\displaystyle{\bf P}\left[\sup_{s\leq t}|V^{\varepsilon}_{s}|>1\right]$
$\displaystyle\leq$ $\displaystyle{\bf P}\left[\sup_{s\leq
t}|J^{\varepsilon}_{s}|>1/2\right]+{\bf
P}\left[|\bar{J}^{\varepsilon}_{t}|>1/2\right]$ $\displaystyle\leq$
$\displaystyle 2{\bf E}\left[\sup_{s\leq t}|J^{\varepsilon}_{s}|\right]+2{\bf
E}\left[|\bar{J}^{\varepsilon}_{t}|\right]$ $\displaystyle\leq$ $\displaystyle
2(I_{\\{x\leq{\varepsilon}\\}}h^{2}*\nu_{t})^{1/2}+I_{\\{x\leq{\varepsilon}\\}}h^{2}*\nu_{t}\to
0,\quad{\varepsilon}\to 0.$

Thus, the set $\Gamma_{\varepsilon}$ is non-null, at least, for sufficiently
small ${\varepsilon}$, and on this set

$\eta\geq\frac{1}{e}\int_{0}^{t}e^{-\tilde{V}^{\varepsilon}_{r}}dr-
Ke^{-\tilde{V}^{\varepsilon}_{t}+1}.$ (5.9)

On the intersection
$\Gamma_{\varepsilon}\cap\\{I_{\\{x>{\varepsilon}\\}}h*\mu_{t/2}=0,\
\ln(1+{\varepsilon})\mu((t/2,t]\times({\varepsilon},1])\geq Nt+1\\}$ we have

$\eta\geq\frac{1}{e}\int_{0}^{t/2}e^{Nr}dr-Ke=\frac{1}{eN}(e^{Nt/2}-1)-Ke.$

Due to the independence of $V^{\varepsilon}$ and $\tilde{V}^{\varepsilon}$
this intersection is a non-null set. Since $N$ is arbitrary large, the
required property of $\eta$ holds.

The analysis of $\eta^{\prime}$ follows the same line. At the first stage we
replace $V$ by $\tilde{V}$ and compensate the linear decrease of $V$ by a
large number of positive jumps on the second half of the interval $[0,t]$.
$\Box$

Let $c<0$. Then the random variables

$\displaystyle Q_{1}$ $\displaystyle=$
$\displaystyle|c|\int_{0}^{T_{1}}e^{-V_{r}}dr-e^{-V_{T_{1}}}\xi_{1},$
$\displaystyle Q_{1}/A_{1}$ $\displaystyle=$
$\displaystyle|c|e^{V_{T_{1}}}\int_{0}^{T_{1}}e^{-V_{r}}dr-\xi_{1}$

are unbounded from above by virtue of the lemma.

Let $c\geq 0$. The same arguments as in EKT  lead to the conclusion that that
$Q_{1}$ and $Q_{1}/A_{2}+Q_{2}/M_{2}$ are unbounded from above on the non-null
set $\\{\xi_{1}<0,\;\xi_{2}<0\\}$.

2c. $\Pi((0,\infty))=0$, $\Pi(|h|)=\infty$.

Let $c<0$. Again we use the decomposition of $V$ depending on the choice of
${\varepsilon}\in(0,1)$. Put

$\displaystyle V^{\varepsilon}$ $\displaystyle:=$ $\displaystyle
I_{\\{x\geq-{\varepsilon}\\}}h*(\mu-\nu)+I_{\\{x\geq-{\varepsilon}\\}}(\ln(1+x)-h)*\mu,$
(5.10) $\displaystyle\tilde{V}^{\varepsilon}$ $\displaystyle:=$ $\displaystyle
I_{\\{x<-{\varepsilon}\\}}h*(\mu-\nu)+I_{\\{x<-{\varepsilon}\\}}(\ln(1+x)-h)*\mu,$
(5.11)

$L:=I_{\\{x<-{\varepsilon}\\}}\ln(1+x)*\mu$ and
$\Pi_{\varepsilon}:=\Pi(I_{\\{x<-{\varepsilon}\\}}|h|)\uparrow\infty$ as
${\varepsilon}\to 0$. Then

$\tilde{V}^{\varepsilon}_{t}=I_{\\{x<-{\varepsilon}\\}}\ln(1+x)*\mu_{t}-I_{\\{x<-{\varepsilon}\\}}h*\nu=L_{t}+\Pi_{\varepsilon}t.$

To prove that $Q_{1}$ is inbounded from above, we argue as follows. As in the
previous subcase 2b. we reduce the problem to checking that the random
variable

$\tilde{\eta}=\int_{0}^{t}e^{-\tilde{V}^{\varepsilon}_{r}}dr-
Ke^{-\tilde{V}^{\varepsilon}_{r}}$

is unbounded from above. Let $t_{1}:=t_{2}/2$ where
$t_{2}:=1/\Pi_{\varepsilon}$. Note that $t_{2}\leq t$ when ${\varepsilon}>0$
is sufficiently small. On the set $\\{L_{t}=L_{t_{1}}\\}$ we have that

$\displaystyle\tilde{\eta}$ $\displaystyle\geq$
$\displaystyle(t_{2}-t_{1})e^{|\bar{L}_{t_{1}}|}e^{-\Pi_{\varepsilon}t_{2}}-Ke^{\bar{L}_{t_{1}}}e^{-\Pi_{\varepsilon}t}$
$\displaystyle=$ $\displaystyle
e^{|\bar{L}_{t_{1}}|}(1/(2e\Pi_{\varepsilon})-Kee^{-\Pi_{\varepsilon}t})\geq
e^{|\bar{L}_{t_{1}}|}/(4e\Pi_{\varepsilon})$

for sufficiently small ${\varepsilon}$. Since the r.v. $|\bar{L}_{t_{1}}|$ is
unbounded from above, so is $Q_{1}$.

On the set $\\{L_{t}=0,\ \xi_{1}\leq K\\}$ non-null for any ${\varepsilon}>0$
and sufficiently large $K$, we have the bound

$Q_{1}/A_{1}\geq(|c|/\Pi_{\varepsilon})(e^{\Pi_{\varepsilon}t}-1)-K.$

It follows that $Q_{1}/A_{1}$ is unbounded from above.

The case $c\geq 0$ is treated as in EKT .

2d. $\Pi((-\infty,0))=0$, $0<\Pi(h)<\infty$, and $F((t,\infty))>0$ for every
$t>0$.

In this subcase $V_{t}=L_{t}-bt$, where $L_{t}:=\ln(1+x)*\mu_{t}$ is an
increasing process and $b:=\Pi(h)-a$. Note that $b>0$, otherwise we get a
contradiction with the existence of $\beta>0$ such that $\ln{\bf E}e^{-\beta
V_{1}}=0$.

Let $c<0$. Take arbitrary $N>0$. Let $s>0$ be the solution of the equation
$(|c|/b)(e^{bs}-1)=N$. Take $K$ large enough to ensure that ${\bf P}[\xi\leq
K]>0$. Chose $t>s$ such that $F((s,t))>0$. On the non-null set

$\\{L_{s}=0,T_{1}\in(s,t),\ e^{L_{t}-L_{s}}\geq Ke^{bt},\ \xi<K\\}$

we have that

$Q_{1}\geq(|c|/b)(e^{bs}-1)-e^{bt-L_{t}}\xi\geq N-1.$

Thus, $Q_{1}$ is unbounded from above.

To prove that $Q_{1}/A_{1}$ is unbounded from above we take ${\varepsilon}>0$
and $K>0$ such that $F((t,\infty))>0$ and $F_{\xi}((0,K))>0$. Setting
$c_{\varepsilon}:=(|c|/b)(e^{-b{\varepsilon}}-e^{-2b{\varepsilon}})$ we have
that on the non-null set

$\\{T_{1}>{\varepsilon},\
L_{T_{1}-{\varepsilon}}=0,L_{T_{1}}\geq\ln((N+K)/c_{\varepsilon}),\xi<K\\}$

we have

$Q_{1}/A_{1}\geq(|c|/b)e^{L_{T_{1}}}e^{-bT_{1}}(e^{b(T_{1}-{\varepsilon})}-1)-\xi\geq
c_{\varepsilon}e^{L_{t}}-K\geq N.$

So, by Lemma 6 $Y_{\infty}$ is unbounded.

If $c\geq 0$, we proceed as in EKT , using the assumption that $F$ charges any
neighborhood of zero.

2e. $\Pi((0,\infty))=0$, $0<\Pi(|h|)<\infty$, and $F((t,\infty))>0$ for every
$t>0$.

We have again $V=L_{t}-bt$ but now the jump process $L$ is decreasing and the
constant $b<0$.

Let $c<0$. Fix $N>0$. Let $s,t>0$ be such that $F((s,t))>0$. On the non-null
set

$\\{T_{1}\in(s,t),\ |L_{s/2}|\geq N,\ L_{s/2}=L_{t},\ \xi<e^{|L_{s/2}|}\\}$

we have

$Q_{1}\geq|c|(T_{1}/2)e^{|L_{(1/2)T_{1}}|-|b|T_{1}}-e^{|L_{(1/2)T_{1}}|-|b|T_{1}}\xi\geq|c|(s/2)e^{N-|b|t}-1.$

Since $N$ is arbitrary, $Q_{1}$ is unbounded from above.

For any $t>0$ and $K>0$ on the non-null set $\\{T_{1}\geq t,\ L_{t}=0,\
\xi\leq K\\}$

$Q_{1}/A_{1}\leq|c/b|(e^{|b|t}-1)-K.$

This implies that $Q_{1}/A_{1}$ is unbounded from above.

If $c\geq 0$, we again proceed as in EKT .

## Acknowledgements

The research is funded by the grant of RSF $n^{\circ}$ 20-68-47030
”Econometric and probabilistic methods for the analysis of financial markets
with complex structure“.

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