# Ruin Probabilities with Investments in Random Environment: Smoothness

###### Abstract

The paper deals with the ruin problem of an insurance company investing its
capital reserve in a risky asset with the price dynamics given by a
conditional geometric Brownian motion whose parameters depend on a Markov
process describing a random variations in the economic and financial
environments. We prove smoothness of the ruin probability as a function of the
initial capital and obtain for it an integro-differential equation.

###### keywords:

ruin probabilities; risky investments; actuarial models with investments;
smoothness of ruin probabilities; stochastic volatility, regime switching

V. ANTIPOV AND Yu. KABANOV

[Lomonosov Moscow State University and “Vega” Institute]Viktor Antipov
<EMAIL_ADDRESS>
[Lomonosov Moscow State University and Université de Franche-Comté]Yuri
Kabanov

16 Route de Gray, 25030 Besançon cedex, France<EMAIL_ADDRESS>
60G51, 60G7091G05

##  1 Introduction

Ruin models with risky investments (in other words, with stochastic interest
rates) is one of the most active fields of the present day ruin theory. These
models combine classical frameworks with models of asset price dynamics
developed in mathematical finance. The goal of numerous studies is to provide
an information about asymptotic of ruin probabilities. For this there are
several approaches. One of them is based on the integro-differential equations
for the ruin probabilities. An inspection of the literature reveals that in
many cases these equations are derived assuming the smoothness, see, e.g.,
[6], [7]. To our mind, the smoothness of ruin probabilities is a rather
delicate property requiring a special (and rather involved) study but there
are very few papers on it, see [3] and [5].

In the present note we consider the price dynamics suggested by Di Masi,
Kabanov, and Runggaldier, [1], where the coefficients of a (conditional)
geometric Brownian motion depend on a Markov process with finite number of
states. Such a setting (sometimes referred to as stochastic volatility model,
regime switching, or hidden Markov chain) reflects the random dynamics of the
economic environment and seems to be adequate for a context with long term
contracts typical in insurance. It was already considered in the actuarial
literature, see, e.g., the papers [2], [4] where the analysis was based using
implicit renewal theory and the paper [7] where a second order integro-
differential equation for the ruin probability was derived using intuitive
arguments.

Our main result is the theorem asserting that if the distribution of jumps has
a density with two continuous and integrable derivatives then the ruin
probability is two times continuously differentiable.

##  2 The model

Let $(\Omega,{\cal F},{\bf F}=({\cal F}_{t})_{t\in{\mathbb{R}}_{+}},{\bf P})$
be a stochastic basis where are given three independent stochastic processes,
$W$, $P$, and $\theta^{i}$ generating the filtration ${\bf F}$:

1\. A standard Wiener process $W=(W_{t})$.

2\. A compound Poisson process $P=(P_{t})$ with drift:

$P_{t}=ct+\sum_{k=1}^{N_{t}}\xi_{k}$

where $c$ is a constant, $N=(N_{t})$ is a Poisson process with intensity
$\alpha>0$, and $(\xi_{k})$ is an i.i.d. sequence independent of $N$ with the
distribution $F_{\xi}=F_{\xi}(dx)$. Sometimes, it is more convenient the
alternative description using the notation of stochastic calculus, namely,

$P_{t}=ct+x*\pi_{t}=ct+\int_{0}^{t}\int x\pi(ds,dx)$ (2.1)

where $\pi=\pi(dt,dx)$ is the jump measure of $P$, that is a Poisson random
measure ${\mathbb{R}}_{+}\times{\mathbb{R}}$ with the deterministic
compensator $\tilde{\pi}=\tilde{\pi}(dt,dx)=\Pi(dx)dt$ (coinciding with the
mean of $\pi$) and the Lévy measure $\Pi(dx)=\alpha F_{\xi}(dx)$. We denote by
$T_{n}$the consecutive jumps of the process
$N_{t}=\pi([0,T]\times{\mathbb{R}})$ with the usual convention $T_{0}=0$.

3\. A piecewise constant right-continuous Markov process
$\theta^{i}=(\theta^{i}_{t})$ with values in the finite set
$E:=\\{0,1,...,K-1\\}$, the initial value $\theta^{i}_{0}=i$, and the
transition intensity matrix $(\lambda_{jk})$. We assume that all states are
communicating.

Recall that $\lambda_{jj}=-\sum_{k\neq j}\lambda_{jk}$ and
$\lambda_{j}:=-\lambda_{jj}>0$ for each $j$. We denote $\tau^{i}_{n}$ the
consecutive jumps of $\theta^{i}$ with the convention $\tau^{i}_{0}=0$ and
consider also the imbedded Markov chain
$\vartheta^{i}_{n}:=\theta^{i}_{\tau^{i}_{n}}$ with transition probabilities
$P_{kl}=\lambda_{kl}/\lambda_{k}$, $k\neq l$, and $P_{kk}=0$.

The conditional distribution of the length of interval
$\tau^{i}_{k+1}-\tau^{i}_{k}$ with respect to the $\sigma$-algebra ${\cal
F}_{\tau^{i}_{k}}$ is exponential with parameter
$\lambda_{\vartheta^{i}_{k}}$.

We consider the random integer-valued measure
$p^{i}:=\sum_{n}{\varepsilon}_{(\tau^{i}_{n},\vartheta_{n}^{i})}$ where
${\varepsilon}_{x}$ is the unit mass at $x$. It can be also defined by the
counting processes

$p^{i}([0,t],k):=\sum_{\tau^{i}_{n}\leq t}{\bf
1}_{\\{\vartheta_{n}^{i}=k\\}},\qquad k\in E.$

The compensator $\tilde{p}^{i}$ (dual predictable projection) of the random
measure $p^{i}$ is given by the formula

$\tilde{p}^{i}(dt,k,)={\bf 1}_{\\{\theta^{i}_{t}\neq
k\\}}\lambda_{\theta^{i}_{t}k}dt.$ (2.2)

To alleviate formulae we shall skip the superscript $i$ (and also u) when it
will not lead to an ambiguity and use for brevity the standard “dot” notation
of the semimartingale stochastic calculus for integrals, with $L=(L_{t})$
standing for the (deterministic) process with $L_{t}:=t$.

With these conventions the Markov modulated price process $S$ can be given as
the solution of the linear integral equation

$S=1+S(a_{\theta}\cdot L+\sigma_{\theta}\cdot W).$

In the traditional symbolical form of the Ito calculus this is written as

$dS_{t}=S_{t}(a_{\theta_{t}}dt+\sigma_{\theta_{t}}dW_{t}),\qquad S_{0}=1,$

where $a_{k}\in{\mathbb{R}}$, $\sigma_{k}>0$, $k=0,...,K-1$.

Let $\kappa_{k}:=a_{k}-(1/2)\sigma^{2}_{k}$. The solution of (homogeneous)
linear equation can be represented by the formula $S={\cal E}(R)$ where ${\cal
E}$ is the stochastic exponential and $R:=a_{\theta}\cdot
L+\sigma_{\theta}\cdot W$ is the relative price process, or, in a more
explicit form, by the formula $S=e^{V}$ where

$V:=\kappa_{\theta}\cdot L+\sigma_{\theta}\cdot W$

is the logprice process.

The process $X=X^{u,i}$ is defined by the formula

$X=u+X\cdot R+P$ (2.3)

where $u>0$.

In the actuarial context the process $X=X^{u,i}$ is interpreted as the capital
reserve of an insurance company fully invested in a risky asset whose price
$S^{i}$ is a conditional geometric Brownian motion given the Markov process
$\theta^{i}$ describing the financial environment.

The process $P$ describes the business activity. There are two basic models:
of the non-life insurance ($c<0$, $F_{\xi}((0,\infty))=1$) and of the annuity
payments ($c>0$, $F_{\xi}((-\infty,0))=1$. In the literature one can find also
a mixed model where $F_{\xi}$ charges both half-axes.

We consider here the setting where only the coefficients of the price process
(i.e. the stochastic interest rate) depend on $\theta$. The extension to the
case where parameters of the business process also depends on $\theta$ is
rather straightforward and has no effect on our main results.

We assume that $P$ is not increasing (otherwise the ruin probability is zero).

Let $\tau^{u,i}:=\inf\\{t>0\colon X^{u,i}_{t}\leq 0\\}$ be the instant of ruin
corresponding to the initial capital $u$ and the initial regime $i$. Then
$\Psi_{i}(u):={\bf P}[\tau^{u,i}<\infty]$ is the ruin probability and
$\Phi_{i}(u):=1-\Psi_{i}(u)$ is the survival probability.

Our main result is a sufficient condition on the smoothness of survival (and
ruin) probabilities as functions of the initial capital.

###### Theorem 2.1

Let $\xi>0$ be a random variable with a density $f$ which is two times
continuously differentiable on $(0,\infty)$ and such that
$f^{\prime},f^{\prime\prime}\in L^{1}({\mathbb{R}}_{+})$. Then the functions
$\Phi^{i}(u)$ are two times continuously differentiable for $u>0$.

The proof is given in Section 3.

##  3 Smoothness of the survival probability

To use the theory of Markov processes we should consider not a single process
$\theta$ but a family of processes $\theta^{i}$ with initial value
$\theta^{i}_{0}=i$, $i\in E$. Automatically, other related processes also will
depend on $i$, in many cases omitted in the notations to alleviate formulae.

The basic idea to prove the smoothness of the survival probability is to use
an integral representation. Define the continuous process $Y=Y^{u,i}$ with

$Y^{u,i}_{t}:=e^{V^{i}_{t}}\left(u+c\int_{0}^{t}e^{-V^{i}_{s}}ds\right),$
(3.1)

coinciding with $X^{u,i}$ on $[0,T_{1})$.

We consider the case of non-life insurance, i.e. with $c>0$.

By virtue of the strong Markov property of $(X^{u,i},\theta^{i})$

$\Phi_{i}(u)=\sum_{j\in E}{\bf E}\left[\Phi_{j}(X^{u,i}_{T_{1}}){\bf
1}_{\\{\theta^{i}_{T_{1}}=j\\}}\right]=\sum_{j\in E}{\bf
E}\left[\Phi_{j}(X^{u,i}_{T_{1}-}+\Delta X^{u,i}_{T_{1}}){\bf
1}_{\\{\theta^{i}_{T_{1}}=j\\}}\right].$

On the interval $[0,T_{1})$ the process $X^{u,i}$ coincides with the
continuous increasing process $Y^{u,i}$ and $\Delta X^{u,i}=\xi_{1}<0$.
Putting $\alpha(dt):=\alpha e^{-\alpha t}dt$ we get that

$\displaystyle\Phi_{i}(u)$ $\displaystyle=$ $\displaystyle\sum_{j\in E}{\bf
E}\left[{\bf
1}_{\\{\theta^{i}_{T_{1}}=j\\}}\Phi_{j}(Y^{u,i}_{T_{1}}+\xi_{1})\right]=\sum_{j\in
E}(\Upsilon^{1i}_{j}(u)+\Upsilon_{j}^{2i}(u)),$

where

$\displaystyle\Upsilon^{1i}_{j}(u)$ $\displaystyle:=$
$\displaystyle\int_{0}^{2}{\bf E}\left[{\bf
1}_{\\{\theta^{i}_{t}=j\\}}\Phi_{j}(Y^{u,i}_{t}+\xi_{1})\right]\alpha(dt),$
$\displaystyle\Upsilon^{2i}_{j}(u)$ $\displaystyle:=$
$\displaystyle\int_{2}^{\infty}{\bf E}\left[{\bf
1}_{\\{\theta^{i}_{t}=j\\}}\Phi_{j}(Y^{u,i}_{t}+\xi_{1})\right]\alpha(dt).$

Clearly,

$\displaystyle\Upsilon^{1i}_{j}(u)$ $\displaystyle:=$
$\displaystyle\int_{0}^{2}\int_{0}^{\infty}\int_{D}{\bf
E}\left[\Phi_{j}(Y_{t}^{u}(z)-w)\right]{\bf
1}_{\\{z_{t}=j\\}}m^{i}(dz)F_{|\xi|}(dw)\alpha(dt),$
$\displaystyle\Upsilon^{2i}_{j}(u)$ $\displaystyle:=$
$\displaystyle\int_{2}^{\infty}\int_{D}{\bf
E}\left[\tilde{\Phi}_{j}(Y_{t}^{u}(z))\right]{\bf
1}_{\\{z_{t}=j\\}}m^{i}(dz)(dw)\alpha(dt),$

where $m^{i}(dz)$ is the distribution of the process $\theta^{i}$ in the
Skorohod space $D$ of the càdlàg functions with a generic point $z=(z_{s})$
and $\tilde{\Phi}_{j}(y):={\bf E}\Phi_{j}(y-|\xi_{1}|)$. Recall that we
consider the case where $\xi_{1}=-|\xi_{1}|$.

###  3.1 Smoothness of $\Upsilon^{2}$.

In this case the distribution $F_{\xi}$ is not involved (and this is the
reason why we introduce the function $\tilde{\Phi}_{j}$).

###### Lemma 3.1

Let $G:{\mathbb{R}}\to[0,1]$ be a Borel function such that $G(y)=0$ for $y\leq
0$. Let $J(u,t,z):={\bf E}\left[G(Y_{t}^{u}(z))\right]$, $t\geq 2$. Then
$J(.,t,z)\in C^{\infty}((0,\infty))$ and

$\sup_{z}(\partial^{n}/\partial u^{n})J(u,t,z)\leq C(u).$

Proof. To simplify formulae we write here “local” notations $V$, $\sigma$,
etc. instead of $V(z)$, $\sigma(z)$, etc. Using the representation

$Y_{t}^{u}=e^{V_{t}-V_{1}}Y_{1}^{u}+c\int_{1}^{t}e^{V_{t}-V_{s}}ds,\qquad
t\geq 1,$

and the independence of the process $(V_{s}-V_{1})_{s\geq 1}$ and the random
variable $Y_{1}^{u}$ we get that

${\bf E}[G(Y^{u}_{t})|Y^{u}_{1}]=G(t,Y_{1}^{u}),$

where

$G(t,y):={\bf
E}\left[G\left(e^{V_{t}-V_{1}}y+c\int_{1}^{t}e^{V_{t}-V_{s}}ds\right)\right].$

Substituting the expression for $Y^{u}_{1}$ given by (3.1) and using
abbreviated notations for the integrals

$\kappa\cdot\Lambda_{t}:=\int_{0}^{t}\kappa_{s}ds,\quad\sigma\cdot
W_{t}=\int_{0}^{t}\sigma_{s}dW_{s},\quad
M_{t}:=ce^{-\kappa\cdot\Lambda}\cdot\Lambda_{t}=c\int_{0}^{t}e^{-\int_{0}^{s}\kappa_{r}dr}ds,$

we have

$\displaystyle J(u,t)$ $\displaystyle:=$ $\displaystyle{\bf
E}\left[G(Y_{t}^{u})\right]={\bf E}\left[G(t,Y_{1}^{u})\right]={\bf
E}\left[G\left(t,e^{\kappa\cdot\Lambda_{1}+\sigma\cdot
W_{1}}\big{(}u+e^{-\sigma\cdot W}\cdot M_{1}\big{)}\right)\right]$
$\displaystyle=$ $\displaystyle{\bf E}\left[\tilde{G}\left(t,e^{\sigma\cdot
W_{1}}\big{(}u+e^{-\sigma\cdot W}\cdot M_{1}\big{)}\right)\right],$

where $\tilde{G}(t,y):=G(t,e^{\kappa\cdot\Lambda_{1}}y)$.

The next step is a reduction to the case of constant coefficients. The
arguments are simple. The function $A=\sigma^{2}\cdot\Lambda$ is strictly
increasing. Let us denote $A^{-1}$ its inverse. The law of the continuous
process with independent increments $(\sigma\cdot W_{s})_{s\geq 0}$ coincides
with the law of the process $(W_{A_{s}})_{s\geq 0}$.

Let $m(s):=ce^{-\kappa\cdot\Lambda_{s}}$,
$\tilde{m}(r):=m(A^{-1}_{r})/\sigma^{2}_{A^{-1}_{r}}$. Changing the variable
we get that

$\int_{0}^{1}e^{-W_{A_{s}}}dM_{s}=\int_{0}^{1}e^{-W_{A_{s}}}m(s)ds=\int_{0}^{A_{1}}e^{-W_{r}}\tilde{m}(r)dr.$

It follows that

$J(u,t)={\bf
E}\left[\tilde{G}\left(t,e^{W_{A_{1}}}\big{(}u+e^{-W}\tilde{m}\cdot\Lambda_{A_{1}}\big{)}\right)\right].$

Let $T_{1}(s):=A^{-1}_{A_{1}s}$ and $\bar{\sigma}:=\sqrt{A_{1}}$. Then
$(\bar{\sigma}W_{t/\bar{\sigma}^{2}})_{t\geq 0}$ is a Wiener process and we
get that

$J(u,t)={\bf
E}\left[\tilde{G}\left(t,e^{\bar{\sigma}W_{1}}\big{(}u+e^{-\bar{\sigma}W}\bar{m}\cdot\Lambda_{1}\big{)}\right)\right]$

where
$\bar{m}(s):=\tilde{m}(A_{1}s)A_{1}=m(T_{1}(s))A_{1}/\sigma^{2}_{T_{1}(s)}$.

Using the conditioning with respect to $W_{1}=x$ and taking into account that
the conditional law of $(W_{t})_{t\in[0,1]}$ in this case is that of the
Brownian bridge on $[0,1]$ ending at the level $x$, that is the same as the
process $W_{s}+s(x-W_{1})$, $s\in[0,1]$, we get that

$J(u,t)=\int{\bf E}[H(t,x,u+\eta_{x})]\varphi_{0,1}(x)dx$

where $H(t,x,y)=\tilde{G}(t,e^{\bar{\sigma}x}y)$ is a function taking values
in $[0,1]$,

$\displaystyle\eta_{x}$ $\displaystyle:=$
$\displaystyle\int_{0}^{1}e^{-\bar{\sigma}(W_{s}-sW_{1})}m(s,x)ds,$ (3.2)
$\displaystyle m(s,x)$ $\displaystyle:=$ $\displaystyle
e^{-\bar{\sigma}sx}\bar{m}(s)=e^{-\bar{\sigma}sx}m(T_{1}(s))\bar{\sigma}^{2}/\sigma^{2}_{T_{1}(s)}.$
(3.3)

Let us check that $\eta_{x}$ has a density. To this end we consider on $[0,1]$
the Gaussian process $D_{s}:=W_{s}-sW_{1}-\gamma_{s}W_{1/2}$ where
$\gamma_{s}:=(2s)\wedge 1-s$. Note that

${\bf E}[D_{s}W_{1/2}]={\bf
E}[(W_{s}-sW_{1}-\gamma_{s}W_{1/2})W_{1/2}]=s\wedge 1/2-s/2-\gamma_{s}/2=0$

and, hence, $D$ and $W_{1/2}$ are independent. It follows that for any bounded
Borel function $g$

${\bf E}[g(\eta_{x})]={\bf E}\left[\int
g(B_{x}(v))\varphi_{0,1/2}(v)dv\right]$

where the strictly positive $C^{\infty}$-function

$v\mapsto B_{x}(v):=\int_{0}^{1}e^{-\bar{\sigma}(D_{s}+\gamma_{s}v)}m(s,x)ds$

(depending of $x$ and also of $\omega$, omitted as usual) is strictly
decreasing, $B_{x}(-\infty)=\infty$ and $B_{x}(\infty)=0$. The inverse
function $y\mapsto B_{x}^{-1}(y)$ is also strictly decreasing
$B_{x}^{-1}(0+)=\infty$ and $B_{x}^{-1}(\infty)=-\infty$.

After the change of variable we get that

${\bf E}[g(\eta_{x})]=\int_{0}^{\infty}g(y){\bf
E}\left[\frac{\varphi_{0,1/2}(B^{-1}_{x}(y))}{K_{x}(B^{-1}_{x}(y))}\right]dy$

where

$K_{x}(v)=B_{x}^{\prime}(v)=-\bar{\sigma}\int_{0}^{1}\
\gamma_{s}e^{-\bar{\sigma}(D_{s}+\gamma_{s}v)}m(s,x)ds.$ (3.4)

Thus, $\eta_{x}$ has the density

$\rho_{x}(y)={\bf
E}\left[\frac{\varphi_{0,1/2}(B^{-1}_{x}(y))}{K_{x}(B^{-1}_{x}(y))}\right],\qquad
y>0.$

Changing the variable we get that

$J(u,t)=\int{\bf
E}[H(t,x,u+\eta_{x})]\varphi_{0,1}(x)dx=\int_{\mathbb{R}^{2}}H(t,x,y)\rho_{x}(y-u)\varphi_{0,1}(x)dydx.$

It remains to show that $\rho_{x}(.)$ belongs to $C^{\infty}$ and its
derivatives has at most exponential growth.

Put

$Q^{(0)}(x,v):=\frac{\varphi_{0,1/2}(v)}{K_{x}(v)},\quad
Q^{(n)}(x,v):=-\frac{\partial Q^{(n-1)}(x,v)}{\partial
v}\frac{1}{K_{x}(v)},\quad n\geq 1.$

Then

$\displaystyle\frac{\partial}{\partial y}Q^{(0)}(x,B^{-1}_{x}(y))$
$\displaystyle=$ $\displaystyle\frac{\partial
Q^{(0)}(x,B^{-1}_{x}(y))}{\partial v}\frac{\partial B^{-1}_{x}(y)}{\partial
y}=-\frac{\partial Q^{(0)}(x,B^{-1}_{x}(y))}{\partial
v}\frac{1}{K_{x}(B^{-1}_{x}(y))}$ $\displaystyle=$ $\displaystyle
Q^{(1)}(x,B^{-1}_{x}(y))$

and, similarly,

$\frac{\partial}{\partial y}Q^{(n-1)}(x,B^{-1}_{x}(y))=-\frac{\partial
Q^{(n-1)}(x,B^{-1}_{x}(y))}{\partial
v}\frac{1}{K_{x}(B^{-1}_{x}(y))}=Q^{(n)}(x,B^{-1}_{x}(y)).$

It follows that

$\frac{\partial^{n}}{\partial
y^{n}}Q^{(0)}(x,B^{-1}_{x}(y))=Q^{(n)}(x,B^{-1}_{x}(y)).$

It is easily seen that

$Q^{(n)}(x,v)=\frac{\varphi_{0,1/2}(v)}{K^{n+1}_{x}(v)}\sum_{k=0}^{n}P_{k}(v)R_{n-k}(x,v),$

where $P_{k}(v)$ is a polynomial of order $k$ and $R_{n-k}(x,v)$ is a linear
combination of products of derivatives of $K_{x}(v)$ in variable $v$.

Recall that

$m(s,x)=e^{-\bar{\sigma}sx}m(T_{1}(s))\bar{\sigma}^{2}/\sigma^{2}_{T_{1}(s)}=c\,e^{-\bar{\sigma}sx-\kappa\cdot\Lambda_{T_{1}(s)}}\bar{\sigma}^{2}/\sigma^{2}_{T_{1}(s)}.$

It follows that for all $s\in[0,1]$ and $x\in{\mathbb{R}}$ we have the bounds

$c_{m}e^{-\bar{\sigma}|x|}\leq m(s,x)\leq C_{m}e^{\bar{\sigma}|x|}$

where $c_{m},C_{m}>0$ are constants depending only on bounds on $|a|$ and
$\sigma$. Let $\sigma^{*}:=\sup_{s}\sigma_{s}$,
$\kappa^{*}:=\sup_{s}|\kappa_{s}|$, and $W_{1}^{*}:=\sup_{s\in[0,1]}|W_{s}|$.
Note that $\gamma\leq 1/2$ and

$\gamma^{n}\cdot\Lambda_{1}=2^{-(n+1)}/(n+1),\quad n\geq 1.$

Clearly,

$e^{-\sigma^{*}(3W^{*}_{1}+v)}\leq e^{-\bar{\sigma}(D_{s}+\gamma_{s}v)}\leq
e^{\sigma^{*}(3W^{*}_{1}+v)},$

Combining this with the above bounds we easily get from (3.4) that

$\frac{1}{4}\sigma_{*}c_{m}e^{-\sigma^{*}(|x|+|v|+3W_{1}^{*})}\leq|K_{x}(v)|\leq\frac{1}{4}\sigma^{*}C_{m}e^{\sigma^{*}(|x|+|v|+3W_{1}^{*})},$
$\frac{(\sigma_{*}/2)^{n+1}}{n+2}c_{m}e^{-\sigma^{*}(|x|+|v|+3W_{1}^{*})}\leq\left|\frac{\partial^{n}}{\partial
v^{n}}K_{x}(v)\right|\leq\frac{(\sigma^{*}/2)^{n+1}}{n+1}C_{m}e^{\sigma^{*}(|x|+|v|+3W_{1}^{*})}.$

It follows that there exists a constant $C_{n}>0$ such that

$|Q^{(n)}(x,v)|\leq C_{n}e^{C_{n}|x|}(1+v^{n})e^{-v^{2}}e^{C_{n}W_{1}^{*}}.$
(3.5)

Therefore,

${\bf E}\left[\sup_{y}|Q^{(n)}(x,B^{-1}_{x}(y))|\right]\leq{\bf
E}\left[\sup_{v\in\mathbb{R}}\left|Q^{(n)}(x,v)\right|\right]\leq\tilde{C}_{n}e^{C_{n}|x|}{\bf
E}\left[e^{C_{n}W_{1}^{*}}\right]<\infty$

for some constant $\tilde{C}_{n}>0$. Thus, for every $x$ the derivative
$(\partial^{n}/\partial y^{n})Q^{(0)}(x,B^{-1}_{x}(y))$ admits an integrable
bound not depending on $y$. This implies that the function
$y\mapsto\rho_{x}(y)$ is infinitely differentiable and

$\frac{\partial^{n}}{\partial y^{n}}\rho_{x}(y)={\bf
E}\left[\frac{\partial^{n}}{\partial y^{n}}Q^{(0)}(x,B^{-1}_{x}(y))\right].$

Moreover, increasing in the need the constant $C_{n}$ we obtain the inequality

$\sup_{y}\left|\frac{\partial^{n}}{\partial y^{n}}\rho_{x}(y)\right|\leq
C_{n}e^{C_{n}|x|}.$

Since $B^{-1}_{x}(0+)=\infty$, the inequality (3.5) implies that
$Q^{(n)}(x,B^{-1}_{x}(0+))=0$. Hence, $(\partial^{n}/\partial
y^{n})\rho_{x}(0)=0$.

Differentiating the function $u\mapsto J(u,t)$ we have:

$\frac{\partial^{n}}{\partial u^{n}}J(u,t)=\int\left[\int
H(t,x,y)\frac{\partial^{n}}{\partial
u^{n}}\rho_{x}(y-u)dy\right]\varphi_{0,1}(x)dx.$

###  3.2 Smoothness of $\Upsilon^{1}$.

Now we get a result on smoothness of the function

$\Upsilon^{1}(u)=\int_{0}^{2}\int_{D}{\bf
E}\left[\Phi(Y_{t}^{u}(z)+\xi)]\right]m(t,dz)\alpha(dt).$

Recall that $c>0$ and $\xi<0$. For us it is more convenient to work with the
random variable $\zeta:=-\xi>0$. The needed result follows from

###### Proposition 3.2

Let $Y^{u}=e^{V}(u+ce^{-V}\cdot\Lambda)$ where $V=\kappa\cdot\Lambda+\sigma
W$, $\kappa$ and $\sigma>0$ are measurable functions. Let $\zeta>0$ be a
random variable with a density $f$ two times continuously differentiable on
$(0,\infty)$ and such that $f^{\prime},f^{\prime\prime}\in
L^{1}({\mathbb{R}}_{+})$. Then the function $\phi_{t}(u):={\bf
E}\left[\Phi(Y_{t}^{u}-\zeta)]\right]$ is two times continuously
differentiable in $u$ and there is a constant $C$ depending only of
$\kappa^{*}$ and $\sigma^{*}$ and such that

$|\phi_{t}^{\prime}(u)|\leq
C||f^{\prime}||_{L^{1}},\qquad|\phi_{t}^{\prime\prime}(u)|\leq
C||f^{\prime\prime}||_{L^{1}}.$ (3.6)

for all $t\in[0,2]$, $u>0$.

Proof. Let $H:{\mathbb{R}}\to[0,1]$ be a measurable function such that $H=0$
on ${\mathbb{R}}_{-}$ and let $\gamma>0$ be a constant. Then the function

$h(y):=\int_{{\mathbb{R}}_{+}}H(y-\gamma
w)f(w)dw=(1/\gamma)\int_{{\mathbb{R}}_{+}}H(z)f((y-z)/\gamma)dz.$

It follows that $h$ is two times differentiable,

$h^{\prime}(y)=(1/\gamma^{2})\int_{\mathbb{R}}H(z)f^{\prime}((y-z)/\gamma)dz,\qquad
h^{\prime\prime}(y)=(1/\gamma^{3})\int_{\mathbb{R}}H(z)f^{\prime\prime}((y-z)/\gamma)dz,$

$|h^{\prime}(y)|\leq(1/\gamma)||f^{\prime}||_{L^{1}}$ and
$|h^{\prime\prime}(y)|\leq(1/\gamma^{2})||f^{\prime\prime}||_{L^{1}}$.

Let us transform our problem to the above elementary framework. Using the
definitions of $Y^{u}$ and $V$ we get that

$\phi_{t}(u)={\bf E}[\Phi(Y^{u}_{t}-\zeta)]={\bf
E}\big{[}\Phi_{t}\big{(}e^{\sigma\cdot
W_{t}}{\color[rgb]{1,0,0}(}u-e^{-\sigma\cdot
W}m\cdot\Lambda_{t}{\color[rgb]{1,0,0})}-e^{-\kappa\cdot\Lambda_{t}}\zeta\big{)}\big{]},$

where the functions $\Phi_{t}(y):=\Phi(e^{\kappa\cdot\Lambda_{t}}y)$,
$m:=ce^{-\kappa\cdot\Lambda_{t}}$. Thus,

$\phi_{t}(u)=\int_{{\mathbb{R}}}{\bf E}\big{[}\Phi_{t}\big{(}e^{\sigma\cdot
W_{t}}{\color[rgb]{1,0,0}(}u-e^{-\sigma\cdot
W}m\cdot\Lambda_{t}{\color[rgb]{1,0,0})}-e^{-\kappa\cdot\Lambda_{t}}w\big{)}\big{]}f(w)dw.$

To get rid of stochastic integrals in the above formula we consider on
${\mathbb{R}}_{+}$ the strictly increasing function
$A:=\sigma^{2}\cdot\Lambda$ with the inverse $A^{-1}$ and observe that the law
of the process $\sigma\cdot W$ is the same as of the process $W_{A}$. Changing
the variable in the integral with respect to $\Lambda$ we get that
$e^{W_{A}}m\cdot\Lambda_{t}=e^{W}\tilde{m}\cdot\Lambda_{A_{t}}$ where the
function $\tilde{m}(r):=m(A^{-1}_{r})/\sigma^{2}_{A^{-1}_{r}}$.

It follows that

${\bf E}\big{[}\Phi_{t}\big{(}e^{\sigma\cdot W_{t}}(u-e^{-\sigma\cdot
W}m\cdot\Lambda_{t})-e^{-\kappa\cdot\Lambda_{t}}w\big{)}\big{]}={\bf
E}\big{[}\Phi_{t}\big{(}e^{W_{A_{t}}}(u-e^{-W}\tilde{m}\cdot\Lambda_{A_{t}})-e^{-\kappa\cdot\Lambda_{t}}w\big{)}\big{]}.$

The change of variable $s\to st/A_{t}$ replace the integration over the
interval $[0,A_{t}]$ by the integration over the interval $[0,t]$ and the rhs
of the above equality is equal to

${\bf
E}\big{[}\Phi_{t}\big{(}e^{\sqrt{A_{t}/t}W_{t}}(u-e^{-\sqrt{A_{t}/t}W}\bar{m}_{t}\cdot\Lambda_{t})-e^{-\kappa\cdot\Lambda_{t}}w\big{)}\big{]}$
(3.7)

where
$\bar{m}_{t}(s):=\tilde{m}(A_{t}s)A_{t}=m(T_{t}(s))A_{t}/\sigma^{2}_{T_{t}(s)}$
with the abbreviatio $T_{t}(s):=A^{-1}_{A_{t}s}$.

Note that in the conventional notation

$e^{-\sqrt{A_{t}/t}W}\bar{m}_{t}\cdot\Lambda_{t}:=\int_{0}^{t}e^{-\sqrt{A_{t}/t}W_{s}}\bar{m}_{t}(s)ds.$

Using the conditioning with respect to $W_{t}=x$ and taking into account that
the conditional law of $(W_{s})_{s\in[0,t]}$ is that of the Brownian bridge on
$[0,t]$ ending at the level $x$, that is the same as of the process
$W_{s}+(s/t)(x-W_{t})$, $s\in[0,t]$, we infer that (3.7) can be written as

$\frac{1}{\sqrt{t}}\int_{{\mathbb{R}}}H(u-\gamma_{t,x}w)\varphi_{0,1}(x/\sqrt{t})dx$

where $H(y)=H_{t,x}(y):={\bf E}[\Phi_{t}(e^{\sqrt{A_{t}/t}x}(y-\eta_{t,x})]$,
$\gamma_{t,x}:=e^{-\kappa\cdot\Lambda_{t}-\sqrt{A_{t}/t}x}$,

$\eta_{t,x}:=\int_{0}^{t}e^{-\sqrt{A_{t}/t}(W_{s}-(s/t)W_{t})}m_{t,x}(s)ds,\quad
m_{t,x}(s):=e^{(s/t)x-\sqrt{A_{t}/t}x}\bar{m}_{t}(s).$

Thus,

$\phi_{t}(u)=\frac{1}{\sqrt{t}}\int_{{\mathbb{R}}}\left[\int_{{\mathbb{R}}}H(u-\gamma_{t,x}w)f(w)dw\right]\varphi_{0,1}(x/\sqrt{t})dx.$

Let denote by $h(u)=h_{t,x}(u):={\bf E}[H(u-\gamma_{t,x}\zeta)]$, i.e. the
function in parentheses above. Then $h$ is two times continuously
differentiable and there is a constant $C$ such that for all $t\in[0,2]$ and
all $x$

$\displaystyle|h^{\prime}(u)|$ $\displaystyle\leq$
$\displaystyle(1/\gamma_{t,x})||f^{\prime}||_{L^{1}}\leq
Ce^{C|x|}||f^{\prime}||_{L^{1}},$ $\displaystyle|h^{\prime\prime}(u)|$
$\displaystyle\leq$
$\displaystyle(1/\gamma^{2}_{t,x})||f^{\prime\prime}||_{L^{1}}\leq
Ce^{C|x|}||f^{\prime\prime}||_{L^{1}}.$

Since

$\frac{1}{\sqrt{t}}\int_{{\mathbb{R}}}e^{C|x|}\varphi_{0,1}(x/\sqrt{t})dx=\int_{{\mathbb{R}}}e^{2C\sqrt{t}|x|}\varphi_{0,1}(x)dx\leq\int_{{\mathbb{R}}}e^{4C|x|}\varphi_{0,1}(x)dx<\infty,$

this implies that the function $\phi_{t}(u)$ is two times continuously
differentiable in $u$ and the bounds (3.6) hold.

###### Remark 3.3

Minor changes in the above proof allows to get smoothness result for annuity
model and for the model where the distribution $F_{\xi}$ charges both half-
axes.

##  4 Integro-differential equations

Knowing that the survival probability is a $C^{2}$-function, the derivation of
the integro-differential equation is easy and we get it for all variants of
the model.

###### Proposition 4.1

Assume that $\Phi=(\Phi_{0},\Phi_{1},\dots,\Phi_{K-1})\in C^{2}$. Then $\Phi$
satisfies the system of integro-differential equations:

$\frac{1}{2}\sigma^{2}_{i}u^{2}\Phi^{\prime\prime}_{i}(u)+(a_{i}u+c)\Phi^{\prime}_{i}(u)+\mathcal{I}(\Phi_{i})(u)+\sum_{j\in
E}\lambda_{ij}\Phi_{j}(u),\quad i\in E,$ (4.1)

where

$\displaystyle\mathcal{I}(\Phi_{i})(u)=\alpha\int(\Phi_{i}(u+x)-\Phi_{i}(u))F_{\xi}(dx)=\alpha\int\Phi_{i}(u+x)F_{\xi}(dx)-\alpha\Phi_{i}(u).$

Proof. Take arbitrary $h>0$ and ${\varepsilon}>0$ such that
$u\in[{\varepsilon},{\varepsilon}^{-1}]\subset(0,1)$. Skipping, as usual, $i$
in the notations we define the stopping time

$\tau^{{\varepsilon}}_{h}:=\inf\big{\\{}t\geq 0\colon\
X_{t}\notin[{\varepsilon},,{\varepsilon}^{-1}]\big{\\}}\wedge
T_{1}\wedge\tau_{1}\wedge h.$

The Ito formula applied to the function $\Phi$ and the semimartingale
$(X,\theta)=(X^{u,i},\theta^{i})$ yields in the identity

$\Phi_{\theta_{t}}(X_{t})=\Phi_{i}(u)+\Phi^{\prime}_{\theta_{-}}(X_{-})\sigma\cdot
W_{t}+{\cal L}\Phi_{\theta}(X)\cdot\Lambda_{t}+\sum_{s\leq
t}\big{(}\Phi_{\theta_{s}}(X_{s})-\Phi_{\theta_{s-}}(X_{s-})\big{)}$ (4.2)

where

${\cal
L}\Phi_{\theta}(X):=\frac{1}{2}\sigma_{\theta}^{2}X^{2}\Phi^{\prime\prime}_{\theta}(X)+(a_{\theta}X+c)\Phi^{\prime}_{\theta}(X),\quad\Sigma_{t}:=\sum_{s\leq
t}\big{(}\Phi_{\theta_{s}}(X_{s})-\Phi_{\theta_{s-}}(X_{s-})\big{)}.$

Due to independence, the processes $P$ and $\theta$ have no common jumps and

$\displaystyle\Sigma_{t}$ $\displaystyle=$
$\displaystyle\int_{0}^{t}\int_{{\mathbb{R}}}(\Phi_{\theta_{s-}}(X_{s-}+x)-\Phi_{\theta_{s-}}(X_{s}-))\pi(ds,dx)$
$\displaystyle+\int_{0}^{t}\sum_{k\in
E}(\Phi_{k}(X_{s-})-\Phi_{\theta_{s-}}(X_{s}-))p(ds,k).$

Now we replace in the formula (4.2) $t$ by $\tau^{{\varepsilon}}_{h}$ and take
the expectation from both sides using the following observations:

– In virtue of the strong Markov property $\Phi_{i}(u)={\bf
E}\Phi_{\tau^{{\varepsilon}}_{h}}(X_{\tau^{{\varepsilon}}_{h}})$ (recall that
$\Phi_{i}(u)=0$ for $u\leq 0$).

– For any ${\varepsilon}>0$, the integrands on
$[0,\tau^{{\varepsilon}}_{h}(\omega)]$ are bounded from above and, therefore,
the expectation of the stochastic integral over the Wiener process is zero.

– As ${\tau^{{\varepsilon}}_{h}}\leq\tau_{1}$, then we can replace
$\theta_{s-}$ in the integrals by $i$. In addition, $\tau^{\varepsilon}_{h}=h$
when $h$ is small enough.

– According to the definition of dual predictable projections

${\bf
E}\int_{0}^{\tau^{{\varepsilon}}_{h}}\int(\Phi_{i}(X_{s-}+x)-\Phi_{i}(X_{s-}))\pi(ds,dx)={\bf
E}\int_{0}^{\tau^{{\varepsilon}}_{h}}\int(\Phi_{i}(X_{s-}+x)-\Phi_{i}(X_{s-}))ds\Pi(dx),$
${\bf E}\int_{0}^{\tau^{{\varepsilon}}_{h}}\sum_{j\in
E\setminus\\{i\\}}\Phi_{j}(X_{s-}-\Phi_{i}(X_{s-}))p(ds,j)={\bf
E}\int_{0}^{\tau^{{\varepsilon}}_{h}}\sum_{j\in
E}\Phi_{j}(X_{s-}))\lambda_{ij}ds.$

As a result we get that

${\bf E}\int_{0}^{\tau^{{\varepsilon}}_{h}}\Bigg{(}{\cal
L}\Phi_{i}(X_{s})+\int(\Phi_{i}(X_{s}+x)-\Phi_{i}(X_{s}))F_{\xi}(dx)+\sum_{j\in
E}(\Phi_{j}(X_{s})-\Phi_{i}(X_{s}))\lambda_{ij}\Bigg{)}ds=0.$

Dividing both sides of this equality by $h$ and letting $h\downarrow 0$ we get
the result. $\Box$

###### Remark 4.2

The obtained equation is homogeneous and holds as well for the ruin
probability $\Psi(u)=1-\Phi(u)$.

##  5 Exponential distribution and differential equations

If $F_{\xi}(dx)=\mu e^{-x/\mu}dx$, then the ruin probability $\Psi\in C^{2}$.
Note that

$\frac{d}{du}\int_{0}^{\infty}\Psi_{i}(u+y)e^{-y/\mu}dy=-\Psi_{i}(u)+\frac{1}{\mu}\int_{0}^{\infty}\Psi_{i}(u+y)e^{-y/\mu}dy$

and therefore,

$\frac{d}{du}{\cal I}\Psi_{i}(u)=\frac{1}{\mu}{\cal
I}\Psi_{i}(u)-\alpha\Psi^{\prime}_{i}(u).$

Differentiating the integro-differential equation for $\Psi_{i}$ (it is same
as for $\Phi_{i}$) and subtracting from the resulting identity multiplied by
$\mu$ the equation (4.1) obtain an ODE of the third order which can be
transformed to the form

$\Psi^{\prime\prime\prime}_{i}-p_{i}(u)\Psi^{\prime\prime}_{i}+q_{i}(u)\Psi^{\prime}_{i}+\frac{2}{\mu\sigma_{i}^{2}}\frac{1}{u^{2}}\sum_{j\in
E}\lambda_{ij}\Psi^{\prime}_{j}(u)-\frac{2}{\mu\sigma_{i}^{2}}\frac{1}{u^{2}}\sum_{j\in
E}\lambda_{ij}\Psi_{j}(u)=0,$

where

$p_{i}(u):=\frac{1}{\mu}-2\Big{(}1+\frac{a_{i}}{\sigma_{i}^{2}}\Big{)}\frac{1}{u}-\frac{2c}{\sigma_{i}^{2}}\frac{1}{u^{2}},\qquad
q_{i}(u):=-\frac{2a_{i}}{\mu\sigma_{i}^{2}}\frac{1}{u}+(a_{i}-\alpha-c/\mu)\frac{2}{\sigma_{i}^{2}}\frac{1}{u^{2}}.$

Acknowledgement. This work was supported by the Russian Science Foundation
associated grants 20-68-47030 and 20-61-47043.

The authors express their thanks to Paul Embrechts who encouraged them to do a
research on the smoothness of the ruin probabilities.

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