2. The model

Let $(\Omega,{\mathcal{F}},{\bf F}=({\mathcal{F}}_t)_{t\ge 0},{\mathbb P})$ be a stochastic basis, i.e. a filtered probability space, where we are given a Wiener process $W=(W_t)_{t\ge 0}$ and $P=(P_t)_{t\ge 0}$ an independent compound Poisson process with drift c and successive jump instants $T_n$ . We denote by $p_P(\text{d} t,\text{d} x)$ the jump measure of the latter with its mean measure $\Pi_P(\text{d} x) \, \text{d} t$ , where $\Pi_P({\mathbb R})<\infty$ . We assume that $\Pi_P({\mathbb R}_+)>0$ and $\Pi_P({\mathbb R}_-)>0$ (some comments for the cases where $\Pi_P$ charges only of the half-axes will be also given).

We consider the process $X=X^{u}$ that is the solution of the non-homogeneous linear stochastic equation $X_t = u + \int_0^t X_s \, \text{d} R_s + P_t$ , where $R_t=at+\sigma W_t$ is the relative price process of the risky asset, P is a compound Poisson process with drift $c\in {\mathbb R}$ representing the business activity of the insurance company, and $u>0$ is the initial capital at time zero; we assume that $\sigma^2>0$ . The process X can be written in ‘differential’ form as $\text{d} X_t= X_t \text{d} R_t + \text{d} P_t$ , $X_0=u$ , where $\text{d} P_t = c \text{d} t + \int x {p_P}(\text{d} t,\text{d} x)$ , $P_0=0$ .

In the actuarial context $X=X^u$ represents the dynamics of the reserve of an insurance company combining both life and non-life insurance business and investing in a stock with the price given by a geometric Brownian motion $S_t = S_0\text{e}^{(a-\sigma^2/2)t+\sigma W_t}$ solving the linear stochastic differential equation $\text{d} S_t = S_t\text{d} R_t$ with initial value $S_0$ . We assume without loss of generality that $S_0=1$ . In this case we have, by the product formula,

(2.1) \begin{equation}X_t=S_t \bigg (u +\int_0^t S_s^{-1} \, \text{d} P_s\bigg).\end{equation}

In classical collective risk theory the process P is usually represented in the form

(2.2) \begin{equation}P_t=ct+\sum_{i=1}^{N_t}\xi_i,\end{equation}

where $N_t\,:\!=\,{p_P([0,t]\times {\mathbb R})}$ is a Poisson process with intensity $\alpha=\Pi_P({\mathbb R})$ independent of the random variables $\xi_n=\Delta P_{T_n}$ , where $T_n$ are successive instants of jumps of N; the independent and identically distributed (i.i.d.) random variables $\xi_n$ have the probability distribution $F(\text{d} x)=\Pi_P(\text{d} x)/\Pi_P({\mathbb R})$ . Alternatively, we can represent P using independent compound Poisson processes given by the sums in the representation $P_t=ct+\sum_{i=1}^{N^2_t}\xi^2_i-\sum_{i=1}^{N^1_t}\xi^1_i$ . The Poisson processes $N^1$ and $N^2$ with intensities $\alpha_1=\Pi_P({\mathbb R}_-)$ and $\alpha_2=\Pi_P({\mathbb R}_+)$ count, respectively, downward and upward jumps of P and have successive jump instants $T^1_n$ , $T^2_n$ . The corresponding jump sizes $\xi^1_n$ , $\xi^2_n$ are positive and have the distribution functions $F_1(x)\,:\!=\,\Pi_P\big(\mathopen{]}-x,0\mathclose{]} \big)/\alpha_1$ , $F_2(x)\,:\!=\,\Pi_P\big(\mathopen{]}0,x\mathclose{]} \big)/\alpha_2$ for $x>0$ .

Let $\tau^u\,:\!=\,\inf \{t: X^u_t\le 0\}$ (the instant of ruin), $\Psi (u)\,:\!=\,P(\tau^u<\infty)$ (the ruin probability), and $\Phi(u)\,:\!=\,1-\Psi (u)$ (the survival probability).

Note that in (2.2) N can be an independent renewal process, i.e. the counting process in which the lengths of the interarrival intervals $T_n-T_{n-1}$ form an i.i.d. sequence. In collective risk theory this corresponds to the Sparre Andersen model.

The main aims of the present work are (i) to get a result on smoothness in u of the ruin probability $\Psi$ justifying that $\Psi$ solves an IDE in the classical sense, and (ii) to deduce from the latter, in the special case of exponentially distributed jumps, an ODE and use it to obtain the exact asymptotic of the ruin probability as $u\to \infty$ .

Of course, in a model with only upward jumps (i.e. with $\Pi_P\big(\mathopen{]}-\infty,0\mathclose{[}\big) = 0$ ), if $c\ge 0$ there is no ruin; the nontrivial case where $c<0$ is studied in detail in [Reference Kabanov and Pergamenshchikov9, Reference Pergamenshchikov and Zeitouni11]. The model in [Reference Frolova, Kabanov and Pergamenshchikov7] deals with downward jumps, i.e. with the non-life insurance business, but the question of smoothness is not discussed and needs to be revisited (in [Reference Wang and Wu14] the smoothness is established under restrictions on the parameters).

Throughout the paper we use the following notation: $\beta\,:\!=\, \frac{2a}{\sigma^2}-1$ , $\kappa\,:\!=\, a - \frac{1}{2}\sigma^2 = \frac{1}{2}\sigma^2\beta$ , and $\eta_t\,:\!=\,\ln S_t=\kappa t+\sigma W_t$ .

The following simple result holds for any Sparre Andersen model with investments.

Proof.Let $\tilde \Psi(u)$ be the ruin probability for the reserve process $\tilde X^u$ corresponding to the model where the business of the company is given by $\tilde P_t=-|c|t-\sum_{i=1}^{N_t}\xi^1_i$ . Then we have $\Psi(u)\le \tilde \Psi(u)\le {\mathbb P}(Z_{\infty}>u)$ , where

\begin{equation*}Z_{\infty}\,:\!=\, -\int_0^{\infty} S_s^{-1} \, \text{d} \tilde P_s=|c|\int_0^{\infty} \text{e}^{-\kappa s - \sigma W_s} \, \text{d} s + \sum_{n=1}^\infty \text{e}^{-\kappa T_n-\sigma W_{T_n}}\xi_n^1 , \end{equation*}

and $\kappa s + \sigma W_s=\frac{1}{2}\sigma^2s\big(\beta+ \frac{2}{\sigma} \, \frac{W_s}{s}\big)$ . By the law of large numbers, for almost all $\omega$ there exists $s_0(\omega)$ such that $\beta+ \frac{2}{\sigma} \, \frac{W_s}{s} > \frac{\beta}{2}$ when $s\ge s_0(\omega)$ . Thus, the integral on the right-hand side above is a finite random variable. Also, $\text{e}^{-\kappa T_n-\sigma W_{T_n}} = \prod_{k=1}^n \zeta_k$ , where

\begin{equation*}\zeta_k \,:\!=\, \text{e}^{-\kappa (T_{k} -T_{k-1})-\sigma \big(W_{T_{k}}-W_{T_{k-1}}\big)}\end{equation*}

form an i.i.d. sequence. Note that ${\mathbb E} \zeta_1^\beta\! =\! 1$ . Hence, ${\mathbb E} \zeta_1^{\beta^{\prime}}\! <\! 1$ and ${\mathbb E} \sum_{n=1}^\infty\!(\xi^1_n)^{\beta^{\prime}}\! \prod_{k=1}^n \zeta^{\beta^{\prime}}_k\! <\infty$ . That is, $\sum_{n=1}^\infty\Big(\xi^1_n \prod_{k=1}^n \zeta_k\Big)^{\beta^{\prime}} <\infty$ almost surely (a.s.). But then $\sum_{n=1}^\infty\xi^1_n\prod_{k=1}^n \zeta_k <\infty$ (a.s.).

The following theorem is the main result of the paper.

Property (ii), corresponding to the case where the volatility is large (with respect to the drift), also holds for a more general Sparre Andersen model with investments under minor additional assuptions on the distribution of interarrival intervals.

The proof of Theorem 2.2 is rather straightforward and will be given in the next section. Proving property (i) requires a bit more work. In Section 4 we establish a lower asymptotic bound needed to select suitable components of the fundamental solution of the ODE for the ruin probability. In Section 5 we give a sufficient condition for the smoothness of the latter, and in Section 6 we derive an integro-differential equation for it. In the case of exponential distributions of jump sizes, the solutions of this IDE also solve an ODE of the fourth order, which we analyze in the concluding Section 7.

3. Large-volatility case: Ruin with probability 1

First, we consider a general setting where the random variables $T>0$ and $\xi$ , and the Wiener process W, are independent, $\eta_t\,:\!=\,\sigma W_t+\kappa t$ , and where the objects of interest are the random variables $A \,:\!=\, \text{e}^{\eta_{T}}$ and $B\,:\!=\,\xi+c\,\int^{T}_{0} \text{e}^{-\eta_{v}} \, \text{d} v$ . In our proofs we shall use conditioning with respect to the final value of the Wiener process on a given finite interval. For convenience, we have formulated the required result as the following lemma (see, e.g., [Reference Revuz and Yor12, Chapter 1]).

Proof.Let $c<0$ . Take arbitrary $N>0$ . According to Lemma 3.1,

(3.1) \begin{equation}{\mathbb P}\bigg(\frac{B}{A} \le -N \mid (T,W_{T})=(t,x)\bigg)={\mathbb P}\left(\text{e}^{-\kappa t-\sigma x}\xi + c\zeta^x_t\le -N\right) ,\end{equation}

where the random variable

(3.2) \begin{equation}\zeta^x_t \,:\!=\, \int_0^t \exp\bigg\{{-}\kappa s-\sigma \bigg(W_s-\frac{s}{t}W_t+\frac{s}{t}x\bigg)\bigg\} \, \text{d} s >0\end{equation}

is independent of $\xi$ . The law of $\zeta^x_t$ is a probability measure charging every open interval in ${\mathbb R}_+$ . Therefore, the right-hand side of (3.1) is strictly positive for every $(t,x)\in \mathopen{]}0,\infty\mathclose{[}\times {\mathbb R}$ . Integrating with respect to the law of $(T,W_{T})$ , we get that $\frac{B}{A}$ is unbounded from below.

Let ${\mathbb P}(\xi<0)>0$ . Take ${\varepsilon}>0$ and $ u>0$ such that the probabilities ${\mathbb P}(\xi\le -{\varepsilon})$ and ${\mathbb P}(u\le T)$ are strictly positive. Then

\begin{equation*}{\mathbb P}\left(\frac{B}{A}\le -N, \xi\le -{\varepsilon} \mid (T,W_{T}) = (t,x)\right) \ge {\mathbb P}\big({-}{\varepsilon} \text{e}^{-\kappa u-\sigma x} + c\zeta^x_t\le -N\big)>0\end{equation*}

for all sufficiently small x, namely, satisfying the inequality ${\varepsilon} \text{e}^{-\kappa u-\sigma x}>N$ . Since the law of $(T,W_{T})$ is a probability measure under which every set $[0,T]\times \mathopen{]}-\infty,y\mathclose{]}$ , $y\in {\mathbb R}$ , is non-null, the result follows.

Proof.Note that $\eta_{T}-\eta_s\le \sigma (W_{T}-W_s)$ . For any $\delta\in \mathopen{]}0,1\mathclose{[}$ ,

\begin{equation*}{\mathbb E} |B|^{\delta} \le{\mathbb E} |\xi|^{\delta} + |c|^\delta{\mathbb E} \int_0^{T}\text{e}^{\eta_{T}-\eta_s} \, \text{d} s^{\delta} \le{\mathbb E} |\xi|^{\delta} + |c|^\delta {\mathbb E} T^\delta \textstyle\sup_{s\le T} \text{e}^{\delta \sigma W_s}.\end{equation*}

Let $m(\text{d} t)$ denote the distribution of T. Substituting the density of distribution of the running maximum of the Wiener process, we get

\begin{align*} {\mathbb E} T^\delta \textstyle\sup_{s\le T} \text{e}^{\delta \sigma W_s} & = \int_0^\infty t^\delta \int_0^t \sqrt{\frac 2{\pi t}}\text{e}^{\delta \sigma x}\text{e}^{-x^2/(2t)} \, \text{d} x \, m(\text{d} t) \\ & =\int_0^\infty t^\delta \text{e}^{(1/2)\delta^2\sigma^2 t}\int_0^{\sqrt t}\sqrt{\frac 2{\pi}}\text{e}^{-\big(y-\delta\sigma\sqrt t\big)^2/2} \,\text{d} y \, m(\text{d} t) \\ & \le 2{\mathbb E} T^\delta \text{e}^{(1/2)\delta^2\sigma^2 T}.\end{align*}

Our assumptions imply that ${\mathbb E}|B|^{\delta}<\infty$ for all sufficiently small $\delta>0$ .

As in [Reference Kabanov and Pergamenshchikov9], the arguments in the proof of Theorem 2.2 are based on the ergodic property of the discrete-time autoregressive process $\big(\tilde X_n^u\big)_{n\ge 1}$ with random coefficients, which is defined recursively by the relations

(3.3) \begin{equation}\tilde X_n^u=A_n\tilde X_{n-1}^u +B_n, \quad n\ge 1, \qquad \tilde X_0^u=u,\end{equation}

where $(A_n,B_n)_{n\ge 1}$ is an i.i.d. sequence in ${\mathbb R}^2$ . For the following result see [Reference Pergamenshchikov and Zeitouni11, Proposition 7.1].

Lemma 3.4. Suppose that ${\mathbb E} |A_1|^\delta < 1$ and ${\mathbb E} |B_1|^\delta < \infty$ for some $\delta \in \mathopen{]}0,1\mathclose{[}$ . Then, for any $u\in {\mathbb R}$ , the sequence $\tilde X_n^u$ converges in $L^\delta$ (and hence in probability) to the random variable $\tilde X_\infty^0=\sum_{k=1}^\infty B_k\prod_{j=1}^{k-1}A_j$ and, for any bounded uniformly continuous function f,

(3.4) \begin{equation}\frac 1n\sum_{k=1}^n f\big(\tilde X_k^u\big)\to {\mathbb E} f\big(\tilde X_\infty^0\big) \ \ {in\ probability\ as }\ n\to \infty . \end{equation}

Proof of Theorem 2.2. Let $\tilde X_n=\tilde X^u_n\,:\!=\,X^u_{T_n}$ , where $X^u$ is the process given by (2.1) and (2.2) where N is a renewal process. In this case,

(3.5) \begin{align}X^u_{T_n} & = \text{e}^{\eta_{T_n}}u+\sum_{k=1}^n\text{e}^{\eta_{T_n}-\eta_{T_k}}\Bigg(\xi_{k}+ c \int^{T_{k}}_{T_{k-1}} \text{e}^{\eta_{T_{k}}-\eta_{v}} \, \text{d} v\Bigg) , \nonumber \\A_{k} & \,:\!=\, \text{e}^{\eta_{T_{k}}-\eta_{T_{k-1}}},\qquad B_{k}\,:\!=\,\xi_{k} + c \int^{T_{k}}_{T_{k-1}} \text{e}^{\eta_{T_{k}}-\eta_{v}} \, \text{d} v.\end{align}

Since the Wiener process W and P (a compound renewal process with drift) are independent, $(A_k,B_k)_{k\ge 1}$ is an i.i.d. sequence.

According to (2.1) and (2.2),

(3.6) \begin{equation}\tilde X_{n}={\mathcal{E}}_n u+\sum^{n}_{k=1}B_{k}\frac {{\mathcal{E}}_n}{{\mathcal{E}}_k}, \qquad {\mathcal{E}}_n\,:\!=\,\prod^{n}_{j=1}\,A_{j}.\end{equation}

Clearly, $\tilde X_{n}=A_n\tilde X_{n-1}+B_n$ , i.e. $\tilde X_{n}$ satisfies (3.3).

In the case where $\beta < 0$ we get the result by virtue of Lemmas 3.2 and 3.3 as well as Corollary 3.1(ii), taking into account that

\begin{equation*}{\mathbb E} A_1^{-\beta} = {\mathbb E} \text{e}^{-\beta\eta_{T_1}} = \int_0^\infty {\mathbb E} \text{e}^{-\beta\sigma W_t-(1/2)(\beta\sigma)^2t} \, m(\text{d} t) = 1\end{equation*}

and, therefore, ${\mathbb E} A_1^{\delta} < 1$ for any $\delta \in \mathopen{]}0,-\beta\mathclose{[}$ .

In the case where $\beta=0$ , we consider a suitably chosen random subsequence of $\tilde X_n$ which satisfies a linear difference equation with the required properties.

Let $\widehat X_n=\widehat X^u_n\,:\!=\,\tilde X_{\theta^n}$ , where $\theta_n\,:\!=\, \inf\big\{k> \theta_{n-1}\colon {\mathcal{E}}_k<{\mathcal{E}}_{\theta_{n-1}}\big\}$ . Thus, $\theta_n$ are ladder times for the random walk $M_k=\ln {\mathcal{E}}_k$ . If $\beta=0$ , then $M_1=\sigma W_{T_1}$ and $ {\mathbb E} M_1 = 0$ , $ {\mathbb E} M_1^2 = {\mathbb E} T_1 < \infty$ . Therefore, there is a finite constant C such that

(3.7) \begin{equation}{\mathbb P}(\theta_{1} > n) \le Cn^{-1/2} \end{equation}

(see [Reference Feller5, Theorem 1a, Chapter XII.7] and the remark before it).

In particular, $\theta_n<\infty $ and the differences $\theta_n-\theta_{n-1}$ form a sequence of finite independent random variables distributed as $\theta_1$ . The discrete-time process

\begin{equation*}\widehat X^u_n={\mathcal{E}}_{\theta_n}u+\sum^{\theta_n}_{k=1}B_{k}\frac {{\mathcal{E}}_{\theta_n}}{{\mathcal{E}}_k}\end{equation*}

solves the linear equation $\widehat X_n^u=\widehat A_n\widehat X_{n-1}^u +\widehat B_n$ , $n\ge 1$ , $\widehat X_0^u=u$ , where

\begin{equation*}\widehat A_n\,:\!=\,\frac {{\mathcal{E}}_{\theta_n}}{{\mathcal{E}}_{\theta_{n-1}}}, \qquad \widehat B_n\,:\!=\,\sum_{k=\theta_{n-1}+1}^{\theta_n} B_k\frac {{\mathcal{E}}_{\theta_n}}{{\mathcal{E}}_k}.\end{equation*}

By construction, $ \widehat A_1<1$ and

\begin{equation*}\big\vert \widehat B_{1} \big\vert\le \sum_{k=\theta_{n-1}+1}^{\theta_n}\vert B_k\vert \frac {{\mathcal{E}}_{\theta_n}}{{\mathcal{E}}_k}\le \sum^{\theta_{1}}_{j=1}\,\vert {B}_{j}\vert.\end{equation*}

According to Lemma 3.3, ${\mathbb E} \vert B_{1}\vert^{\delta} < \infty$ for some $\delta \in \mathopen{]}0,1\mathclose{[}$ . Taking $r \in \mathopen{]}0,\frac{\delta}{5}\mathclose{[}$ and defining the sequence $l_{n}\,:\!=\,\big[n^{4r}\big]$ , we have, using the Chebyshev inequality and (3.7),

\begin{align*}{\mathbb E} \big\vert \widehat B_{1} \big\vert^{r} & \le 1+r\sum_{n\ge 1}{n^{r-1}}{\mathbb P}\Bigg( \sum^{\theta_{1}}_{j=1}\big\vert B_{j}\big\vert>n\Bigg)\\[2mm] &\le 1+r\sum_{n\ge 1}{n^{r-1}}\,{\mathbb P}\Bigg( \sum^{l_{n}}_{j=1}\big\vert B_{j}\big\vert>n\Bigg)+ r\sum_{n\ge 1} {n^{r-1}}{\mathbb P}\big( \theta_{1}>l_{n}\big)\\[2mm] &\le 1+r{\mathbb E} \vert Q_{1}\vert^{\delta} \sum_{n\ge 1} l_{n}n^{r-1-\delta} +rC\sum_{n\ge 1} n^{r-1}l_{n}^{-1/2}<\infty.\end{align*}

To apply Corollary 3.1(ii) it remains to check that the random variable $\frac{\widehat B_1}{\widehat A_1}$ is unbounded from below. But Lemma 3.2 asserts that this ratio coinciding with $\frac{B_1}{A_1}$ on the set $\big\{W_{T_1}<0\big\}$ of strictly positive probability is unbounded from below on this set.

4. Lower asymptotic bound

The next result we need for our asymptotic analysis indicates that the ruin probability decreases at infinity not faster than a certain power function. The proof given below also covers the more general case where P is a compound renewal process with drift given by the representation (2.2) where N is a counting renewal process.

Proposition 4.1. Suppose that $c<0$ or the random variable $\xi_1$ is unbounded from below. Then there exists $\beta_{*}>0$ such that

(4.1) \begin{equation} \liminf_{u\to\infty}\,u^{\beta_{*}}\,\Psi(u)\,>\,0. \end{equation}

Proof.Using (3.6) we easily get that on the set $\cap^{n}_{k=1}\,\Gamma_{k}$ ,

\begin{equation*}\tilde X_{n}\le u \varrho^{n}+\frac{1}{\varrho(1-\varrho)}.\end{equation*}

From the representation $\tilde X_{n+1}=A_{n+1} \tilde X_{n} + B_{n+1}$ and the above bound we infer that on the set $\Big(\cap^{n}_{k=1} \Gamma_{k}\Big)\cap D_{n+1}$ ,

\begin{equation*}\tilde X_{n+1}\le u \varrho^{n-1}+\frac{1}{\varrho^{2}(1-\varrho)}-b= u \varrho^{n-1} -b_1 , \qquad b_1 \,:\!=\, b - \frac{1}{\big({\varrho^{2}(1-\varrho)}\big)} > 0 .\end{equation*}

Let $u>b_1$ and let $n=n(u)\,:\!=\,2+[{(1/\ln\varrho)}{\ln(b_{1}/u)}] $ , where [x] means the integer part, $x-1<[x]\le x$ . Then $u \varrho^{n-1}= u \text{e}^{(n-1)\ln \varrho} < u\text{e}^{\ln (b_1/u)} - b_1$ , and therefore

\begin{equation*} {\mathbb P}(\vartheta^u<\infty)\ge {\mathbb P}\left( \cap^{n}_{k=1}\,\Gamma_{k}\cap D_{n+1}\right)=\left({\mathbb P}(\Gamma_{1})\right)^{n}{\mathbb P}(D_{1}). \end{equation*}

Take $\beta_*\,:\!=\,\frac{\ln{\mathbb P}(\Gamma_{1})}{\ln\varrho}$ . Then

\begin{align*}u^{\beta_*} {\mathbb P}\big(\vartheta^u<\infty\big) & \ge \exp\bigg\{\frac {\ln{\mathbb P}(\Gamma_{1})}{\ln\varrho}\ln u+n\ln {\mathbb P}(\Gamma_{1})\bigg\}{\mathbb P}(D_{1}) \\& \ge \exp\bigg\{\bigg(2+\frac {\ln b_1}{\ln \varrho}\bigg)\ln {\mathbb P}(\Gamma_{1})\bigg\}{\mathbb P}(D_{1}).\end{align*}

Since ${\mathbb P}\big(\tau^u<\infty\big)\ge {\mathbb P}\big(\vartheta^u<\infty\big)$ , the lemma is proven.

It remains to show that our assumptions ensure that for any $ \varrho, b>0$ the probability ${\mathbb P} (A_{1}\le \varrho, \ B_{1}\le -b)$ is strictly positive. With this in mind, take $v_1, v_2>0$ and $x_0$ such that ${\mathbb P}(T_1\in [v_1, v_2])>0$ and $|\kappa |v_2 + \sigma x_0 \le \ln \varrho$ . Let $\Delta\,:\!=\, [v_1, v_2]\times \mathopen{]}-\infty,x_0\mathclose{]}$ . By virtue of Lemma 3.1, for any $(t,x)\in \Delta$ we have

(4.2) \begin{equation}{\mathbb P} \Big(A_{1}\le \varrho, \ B_{1}\le -b \mid \big(T_1,W_{T_1}\big)=(t,x)\Big) = {\mathbb P}\big(\xi_1+ c\Upsilon ^x_{t}\le -b\big) ,\end{equation}

where $\Upsilon^x_t \,:\!=\, \text{e}^{ \sigma x+\kappa t } \zeta_t^x$ , with $\zeta_t^x$ given by (3.2). Recall that $\zeta_t^x$ takes arbitrarily large values with strictly positive probability (regardless of x and $t>0$ ), and $\xi_1$ and $\zeta_t^x$ are independent. Therefore, the probability on the right-hand side is strictly positive if $\xi_1$ is unbounded from below or $c<0$ . Integrating (4.2) over $\Delta$ with respect to the distribution of $\big(T_1,W_{T_1}\big)$ charging $\Delta$ , we get the result.

5. Regularity of the ruin probability

Following tradition (seemingly justified by notational convenience), we shall work with the survival probability $\Phi=1-\Psi$ , which has the same regularity property and satisfies the same equations.

Proof.Define the continuous process $Y^{u}_{t}\,:\!=\,S_t\Big (u+c\int_{[0,t]} S^{-1}_s \, \text{d} s\Big)$ coinciding with $X^u$ on $\mathopen{[}0,T_1\mathclose{[}$ , and introduce the stopping time $\theta^{u}\,:\!=\,\inf\{t\ge 0 \,:\, Y^{u}_{t}\le 0\}$ . By virtue of the strong Markov property of $X ^u$ ,

(5.1) \begin{equation}\Phi(u)={\mathbb E} \Phi\Big(X^u_{\theta^{u}\wedge T_1}\Big) = {\mathbb E} \Phi\Big(Y^u_{\theta^{u}\wedge T_1}+\Delta X^u_{\theta^{u}\wedge T_1}\Big). \end{equation}

Due to the independence of W and the Poisson processes $N^1$ and $N^2$ , the values of $\theta^u(\omega) $ , $T^1_1(\omega)$ , and $T^2_1(\omega)$ are all different for almost all $\omega$ . Since $\Phi\big(X^u_{\theta^{u}}\big)\mathbf{1}_{\big\{\theta^{u}< T_1\big\}}=0$ we have the representation $\Phi=K^\downarrow +K^\uparrow $ , where

\begin{align*}K^\downarrow (u) & \,:\!=\,{\mathbb E} \mathbf{1}_{\Big\{Y^u_{T^1_1}>0\Big\}}\mathbf{1}_{\big\{T_1^1<T_1^2\big\}}\Phi\Big(Y^u_{T^1_1}-\xi^1_1\Big),\\ K^\uparrow (u) & \,:\!=\,{\mathbb E} \mathbf{1}_{\Big\{Y^u_{T_1^ 2}>0\Big\}}\mathbf{1}_{\big\{T_1^1>T_1^2\big\}}\Phi\bigg(Y^u_{T_1^2}+\xi^2_1\bigg). \end{align*}

The analysis of the smoothness of these two functions is similar, so we consider the first one. It is convenient to represent it as the sum $K^\downarrow = K_1^\downarrow +K_2^\downarrow $ , where

\begin{align*} K_1^\downarrow (u) & \,:\!=\, \int_{{\mathbb R}^3} \mathbf{1}_{\big\{0<s<t\wedge 2\big\}}{\mathbb E} G \big(Y^u_s,w\big) \, m(\text{d} s, \text{d} t)\,F_1(\text{d} w), \\ K_2^\downarrow (u) & \,:\!=\, \int_{{\mathbb R}^3} \mathbf{1}_{\big\{2<s<t\big\}}{\mathbb E} G\big(Y^u_s,w\big) \, m(\text{d} s,\text{d} t)\,F_1(\text{d} w),\end{align*}

with

\begin{align*} G(y,w) & \,:\!=\,\mathbf{1}_{\{y> 0\}}\Phi(y-w)=\mathbf{1}_{\{y-w> 0\}}\Phi(y-w)=\Phi(y-w)\ \ \text{for}\ \ w\ge 0,\\ m(\text{d} s, \text{d} t) & \,:\!=\, \alpha_1\alpha_2\text{e}^{-\big(\alpha_1s+\alpha_2 t\big)} \, \text{d} s \, \text{d} t. \end{align*}

The theorem follows from Lemmas 5.1 and 5.3 below.

Proof.Using the representation $Y_s^u= \text{e}^{\eta_s-\eta_1}Y_1^u + \int_{[1,s]} \text{e}^{\eta_s-\eta_r} \, \text{d} r$ , $s\ge 2$ , and noticing that the random variable $Y_1^u$ is independent of the process $(\eta_s-\eta_1)_{s\ge 1}$ , we get

\begin{align*} {\mathbb E} G\big(Y^u_s,w\big) \mid Y^u_1 = \tilde G\big(s,Y_1^u,w\big),\end{align*}

where $\tilde G(s,y,w)\,:\!=\,{\mathbb E} G\Big(\text{e}^{\eta_s-\eta_1}y+\int_{[1,s]} \text{e}^{\eta_s-\eta_r} \text{d} r,w\Big) $ . Substituting the expression for $Y_1^u$ , we obtain ${\mathbb E} G\big(Y^u_s,w\big) = {\mathbb E} \tilde G\big(s,Y^u_1,w\big) = {\mathbb E} \tilde G\big(s, \text{e}^{\kappa+\sigma W_1}(u +c R_1),w\big)$ with $R_1\,:\!=\,\int_{[0,1]} \text{e}^{-\kappa r -\sigma W_r} \, \text{d} r$ . Conditioning on $W_1=x$ and Lemma 3.1 lead to the representation ${\mathbb E} G\big(Y^u_s,w\big) =\int_{{\mathbb R}} {\mathbb E} \tilde G\big(s, \text{e}^{\kappa +\sigma x}(u+ \zeta^x),w\big) \varphi_{0,1}(x) \, \text{d} x$ , where $\varphi_{0,1}(x)$ is the standard Gaussian density, and the random variable $\zeta^x \,:\!=\, c\int_{[0,1]} \text{e}^{-\kappa r-\sigma (W_r+r(x-W_1))} \, \text{d} r$ .

The case $c= 0$ is easy. By the change of variable $z=\kappa +\sigma x+\ln u$ we get

\begin{equation*}{\mathbb E} G\big(Y^u_s,w\big) = \frac 1\sigma \int_{{\mathbb R}} \tilde G(s,\text{e}^z,w)\varphi_{0,1}\bigg(\frac{z - \kappa- \ln u}{\sigma}\bigg) \, \text{d} z.\end{equation*}

The function $u\mapsto \varphi_{0,1}\big(\frac{1}{\sigma}(z - \kappa- \ln u)\big)$ belongs to $C^{\infty}(\mathopen{]}0,\infty\mathclose{[})$ , and its derivatives on any interval $[u_1,u_2]\subset \mathopen{]}0,\infty\mathclose{[}$ are dominated by integrable functions. It follows that the function $u\mapsto {\mathbb E} G\big(Y^u_s,w\big) $ belongs to $C^{\infty}(\mathopen{]}0,\infty\mathclose{[})$ , and its derivatives are locally bounded.

Now, let $c\neq 0$ . Lemma 5.2 below asserts that, for every x, the random variable $\zeta^x$ has a density $\rho (x,\cdot)$ such that ${\mathbb E} G\big(Y^u_s,w\big) = \int_{{\mathbb R}^2}\tilde G\big(s,\text{e}^{\kappa +\sigma x}z,w\big)\rho(x,z-u)\varphi_{0,1}(x) \, \text{d} x \, \text{d} z$ . Since this density belongs to $C^{\infty}$ and its derivatives are of subexponential growth in x, the function $y\mapsto {\mathbb E} G\big(Y^u_s,w\big) $ also belongs to $C^{\infty}$ and has bounded derivatives. So, the function $u\mapsto K_2^\downarrow (u)$ has the same property and the lemma is proved.

The required smoothness property of $K_1^\downarrow$ follows from the next lemma.

Proof.First, we are able to observe that $h(y)= \int_{\mathbb R} G(y-x)f(x) \, \text{d} x = \int_{\mathbb R} G(z)f(y-z) \, \text{d} z$ and $h^{\prime}(y)=\int_{\mathbb R} G(z)f^{\prime}(y-z) \, \text{d} z$ . It follows that $|h^{\prime}(y)|\le ||f^{\prime}||_{L^1}$ and $|h^{\prime\prime}(y)|\le ||f^{\prime\prime}||_{L^1}$ .

Using the representation $Y^u_s=\text{e}^{\eta_s}u+\int_{[0,s]}\text{e}^{\eta_s-\eta_r} \, \text{d} r$ , and arguing in the same spirit as above but conditioning this time on the random variable $W_s\sim {\mathcal{N}}(0,s)$ and considering the Brownian bridge on [0,s], we obtain

\begin{equation*}{\mathbb E} G\big(Y_s^u-\xi\big) = \frac{1}{\sqrt{s}\,}\int_{{\mathbb R}}{\mathbb E} h\big(\text{e}^{\kappa s+\sigma x }(u+\zeta^{s,x})\big) \varphi_{0,1}\left(\frac{x}{\sqrt{s}\,}\right) \, \text{d} x,\end{equation*}

where $\zeta^{s,x} \,:\!=\, c\int_{[0,s]}\exp\big\{{-} \big(\frac{\sigma r x}{s} + \kappa r+\sigma (W_{r}-\frac{r}{s}W_{s}\big)\big\} \, \text{d} r$ . If $c=0$ , then

\begin{equation*}{\mathbb E} G\big(Y_s^u-\xi\big) = \int_{{\mathbb R}} h\Big(\text{e}^{\kappa s+\sigma {\sqrt s}x }u\Big)\varphi_{0,1}(x) \, \text{d} x\end{equation*}

and the required property is obvious.

Let $c\neq 0$ . It is easily seen that the random variable $\zeta^{s,x}$ has a $C^\infty$ density (the same as $\zeta^x$ but with the parameters cs, $\kappa s$ , and $\sigma s^{1/2}$ ). Unfortunately, the derivatives of this density have non-integrable singularities at $s=0$ . For this reason, the arguments used above do not work.

The smooth function $x\to \zeta ^{s,x}$ is strictly decreasing and maps ${\mathbb R}$ onto ${\mathbb R}_+$ (when $c>0$ ) or is strictly increasing and maps ${\mathbb R}$ onto ${\mathbb R}_-$ (when $c<0$ ). Let $z(s,\cdot)$ denote its inverse, which is a function decreasing from $+\infty$ to $-\infty$ (when $c>0$ ) and increasing from $-\infty$ to $+\infty$ (when $c<0$ ). The partial derivative in x is given by

\begin{equation*}z_x(s,x)=-\frac{s}{L(s,z(s,x))},\end{equation*}

where

\begin{equation*} L(s,x) = c \sigma \int_{[0,s]} r \exp\left\{{-} \left(\frac{\sigma r x}{s} + \kappa r +\sigma \left(W_{r} - \frac{r}{s}W_{s}\right) \right) \right\} \, \text{d} r. \end{equation*}

In both cases, $z_{xx}(s,x)>0$ for $s>0$ .

Changing the variable, we obtain that ${\mathbb E} h\big(Y_s^u\big) = {\mathbb E} H(s,u)$ , where

\begin{equation*}H(s,u)\,:\!=\,\frac{1}{\sqrt{s}\,} \int^{\infty}_{0}h\big(\text{e}^{\kappa s+\sigma z(s,x) }(u+x)\big){\varphi_{0,1}\left(\frac{z(s,x)}{\sqrt{s}}\right)}z_x(s,x) \, \text{d} x, \qquad c>0.\end{equation*}

For $s\in \mathopen{[}0,2\mathclose{]}$ ,

\begin{align*}\frac{1}{\sqrt{s}\,} \int^{\infty}_{0} \text{e}^{\sigma z(s,x) }\varphi_{0,1}\left(\frac{z(s,x)}{\sqrt{s}}\right) z_x(s,x) \, \text{d} x = \int \text{e}^{\sigma \sqrt{s}z}\varphi_{0,1}(z) \, \text{d} z \le \int \text{e}^{\sigma \sqrt{2}z}\varphi_{0,1}(z) \, \text{d} z ,\end{align*}

where the last integral is a finite constant. This estimate and the boundedness of h ^′ legitimate the differentiation under the sign of the integrals. In the case $c>0$ we have

\begin{equation*}H_u(s,u) = \frac{1}{\sqrt{s}\,} \int^{\infty}_{0} \text{e}^{\kappa s+\sigma z(s,x) }h^{\prime}\big(\text{e}^{\kappa s+\sigma z(s,x) }(u+x)\big){\varphi_{0,1}\left(\frac{z(s,x)}{\sqrt{s}}\right)}z_x(s,x) \, \text{d} x ,\end{equation*}

and the bound $|H_u(s,u)|\le C$ for all $s\in \mathopen{]}0,2\mathclose{]}$ .

Repeating the arguments, we get

\begin{equation*}H_{uu}(s,u) = \frac{1}{\sqrt{s}\,} \int^{\infty}_{0} \text{e}^{2\kappa s+2\sigma z(s,x) }h^{\prime\prime}\big(\text{e}^{\kappa s+\sigma z(s,x) }(u+x)\big){\varphi_{0,1}\left(\frac{z(s,x)}{\sqrt{s}}\right)}z_x(s,x) \, \text{d} x , \end{equation*}

and $|H_u(s,u)|\le C$ for all $s \in \mathopen{]}0,2\mathclose{]}$ .

Similar arguments are applied in the case $c<0$ . The lemma is proven.

6. The integro-differential equation for the survival probability

Proposition 6.1. Suppose that $\Psi\in C^2$ (see the sufficient conditions for this in Theorem 5.1). Then the function $\Phi$ on $\mathopen{]}0,\infty\mathclose{[}$ satisfies the following equation:

(6.1) \begin{equation} \frac 12 \sigma^2u^2 \Phi^{\prime\prime}(u)+ (au+c)\Phi^{\prime}(u) +\int (\Phi(u+y)-\Phi(u)) \, \Pi_P(\text{d} y) = 0.\end{equation}

ProofFor $h>0$ and $\epsilon>0$ small enough to ensure that $u\in \mathopen{]}\epsilon,\epsilon^{-1}\mathclose{[}$ , put

\begin{equation*}\tau^{\epsilon}_h : = \inf\big\{t\ge 0 : X^{u}_{t} \notin [\epsilon, \epsilon^{-1}] \big\}\wedge h\wedge T_1.\end{equation*}

Let ${\mathcal{L}}^0\Phi(u)\,:\!=\, \frac12\sigma^2u^2 \Phi^{\prime\prime}(u)+ (au+c)\Phi^{\prime}(u)$ . By the Itô formula,

\begin{multline*} \Phi\Big(X^u_{\tau^{\epsilon}_h}\Big) = \Phi(u) + \sigma \int_{0}^{\tau^{\epsilon}_h}X^u_{s}\Phi^{\prime}\big(X^u_{s}\big)\,\text{d} W_s + \int_{0}^{\tau^{\epsilon}_h}{\mathcal{L}}^0\Phi\big(X^u_{s}\big) \, \text{d} s \\ + \int_{0}^{\tau^{\epsilon}_h} \int\big(\Phi\big(X^u_{s-}+x\big) - \Phi\big(X^u_{s-}\big)\big) \, p_P(\text{d} s, \text{d} x).\end{multline*}

Due to the strong Markov property, $\Phi(u) = {\mathbb E} \Phi\Big(X^u_{\tau^{\epsilon}_h}\Big)$ (since $\Phi(u)=0$ for $u\le 0$ ). For every $\epsilon>0$ the integrands above are bounded by constants, and hence the expectation of the stochastic integral with respect to the Wiener process is zero. The expectation of the integral with respect to the integer-valued measure $p_P(\text{d} s, \text{d} x)$ is equal to the integral with respect to the compensator of the latter, i.e. to ${\mathbb E} \int_{0}^{\tau^{\epsilon}_h}\int\big(\Phi\big(X^u_{s-}+x\big) - \Phi\big(X^u_{s-}\big)\big) \, \text{d} s \, \Pi_P(\text{d} x)$ . Moreover, $\tau^\epsilon_h=h$ when h is sufficiently small (the threshold below which the equality holds, of course, depends on $\omega$ ).

It follows that, independently of $\epsilon$ ,

\begin{equation*}\frac 1h {\mathbb E}\int_{0}^{\tau^{\epsilon}_h}\Big(\frac 12{\sigma^2} \big(X^u_{s}\big)^2\Phi^{\prime\prime}\big(X^u_{s}\big)+\big(aX^u_{s}+c\big)\Phi^{\prime}\big(X^u_{s}\big)\Big) \, \text{d} s \to {\mathcal{L}}^0 \Phi(u)\end{equation*}

as $h\to 0$ . Finally,

\begin{equation*}\frac 1h {\mathbb E} \int_{0}^{\tau^{\epsilon}_h} \int\big(\Phi\big(X^u_{s-}+x\big) - \Phi\big(X^u_{s-}\big)\big) \, \text{d} s \, \Pi_P(\text{d} x) \to \int (\Phi(u+y)-\Phi(u)) \, \Pi_P(\text{d} y).\end{equation*}

It follows that $\Phi$ satisfies (6.1).

7. Exponentially distributed jumps: From IDE to ODE

In the case of exponentially distributed jumps the IDE can be written as

\begin{equation*} \frac {\alpha_1} {\mu_1}\int_{0} ^\infty\Phi(u-y)\text{e}^{-y/\mu_1} \, \text{d} y+\frac {\alpha_2} {\mu_2}\int_{0} ^\infty\Phi(u+y)\text{e}^{-y/\mu_2} \, \text{d} y = 0, \end{equation*}

where

\begin{align*}{\mathcal{L}}\Phi(u)\,:\!=\, \frac 12 \sigma^2u^2 \Phi^{\prime\prime}(u)+ (au+c) \Phi^{\prime}(u) -(\alpha_1+\alpha_2)\Phi(u).\end{align*}

Changing variables in the integrals, we get

(7.1) \begin{equation}{\mathcal{L}}\Phi(u)+\frac{\alpha_1} {\mu_1}I_1(u) + \frac{\alpha_2} {\mu_2}I_2(u)=0 ,\end{equation}

where

\begin{align*} I_1(u)\,:\!=\,\int_{-\infty} ^u\Phi(z)\text{e}^{-(u-z)/\mu_1} \, \text{d} z,\ I_2(u)\,:\!=\,\int_{u} ^\infty\Phi(z)\text{e}^{-(z-u)/\mu_2} \, \text{d} z. \end{align*}

Note that we have $I^{\prime}_1 =$ $\Phi-\frac{1}{\mu_1}I_1$ and $I^{\prime}_2 =-\Phi+\frac{1}{\mu_2}I_2$ .

Put ${\mathcal{T}} f\,:\!=\,\mu_1\mu_2f^{\prime\prime}+(\mu_2-\mu_1)f^{\prime}-f$ . It is easily seen that ${\mathcal{T}} I_1 = \mu_1\mu_2 \Phi^{\prime} - \mu_1\Phi$ and ${\mathcal{T}} I_2 = -\mu_1\mu_2 \Phi^{\prime} - \mu_2\Phi$ . Applying the operator ${\mathcal{T}}$ to both sides of (7.1) we see that the survival probability $\Phi$ (as well as the ruin probability $\Psi\,:\!=\,1-\Phi$ ) solves the differential equation ${\mathcal{D}}\Phi=0$ , where ${\mathcal{D}}$ is the differential operator of the fourth order:

\begin{align*} {\mathcal{D}}\Phi & \,:\!=\, {\mathcal{T}}{\mathcal{L}} \Phi+(\alpha_1\mu_2 -\alpha_2\mu_1)\Phi^{\prime}-(\alpha_1+\alpha_2)\Phi \\& = \mu_1\mu_2({\mathcal{L}} \Phi)^{\prime\prime}+(\mu_2-\mu_1)({\mathcal{L}} \Phi)^{\prime}-{\mathcal{L}}\Phi+(\alpha_1\mu_2 -\alpha_2\mu_1)\Phi^{\prime}-(\alpha_1+\alpha_2)\Phi .\end{align*}

After some simple calculation we get that the fourth-order equation obtained is in fact the following third-order differential equation for G:

\begin{align*} G\,:\!=\,\Phi^{\prime}:\ \tilde g_3(u)G^{\prime\prime\prime}+\tilde g_2(u)G^{\prime\prime}+\tilde g_1(u)G^{\prime}+\tilde g_0(u)G=0, \end{align*}

where the coefficients (depending on u) are:

\begin{align*} \tilde g_3(u)&\,:\!=\,\frac 12 \sigma^2 \mu^2 u^2,\\ \tilde g_2(u)&\,:\!=\, \mu^2 \big(\big(a+2\sigma^2\big)u+c\big)+\frac 12 \Delta \mu \sigma^2 u^2,\\ \tilde g_1(u)&\,:\!=\,\mu^2 \big(2a+\sigma^2-\alpha_1-\alpha_2\big)+ \Delta \mu\big(\sigma^2u+au+c\big) -\frac 12 \sigma^2 u^2,\\ \tilde g_0(u)&\,:\!=\, -au-c + \Delta \mu\big(a-\alpha_1-\alpha_2\big)+\big(\alpha_1\mu_2 -\alpha_2\mu_1\big) ,\end{align*}

with $\Delta \mu\,:\!=\,\mu_2-\mu_1$ , $\mu^2\,:\!=\,\mu_1\mu_2$ . Since $\mu_1 , \mu_2>0$ we get from here the equation with unit coefficient at the third derivative:

(7.2) \begin{equation}G^{\prime\prime\prime}+q_2(u) G^{\prime\prime}+q_1(u) G^{\prime}+q_0(u) G(u)=0 ,\end{equation}

where

\begin{align*} q_2(u) & \,:\!=\, \frac {\Delta \mu}{\mu^2 } +\frac{2\big(a+2\sigma^2\big)}{ \sigma^2}\frac {1}{u}+\frac{2c}{ \sigma^2} \frac{1}{u^2}, \\ q_1(u) & \,:\!=\, -\frac 1{\mu^2 }+\frac{2\big(a+\sigma^2\big)\Delta \mu}{\sigma^2\mu^2 }\frac{1}{u} +\frac{2\big(\Delta \mu\ c +\mu^2 \big(2a+\sigma^2-\alpha_1-\alpha_2\big)\big)}{\mu^2 \sigma^2}\frac{1}{u^2}, \\ q_0(u) & \,:\!=\, -\frac{2a}{\mu^2 \sigma^2}\frac{1}{u}+\frac{2\big(\Delta\mu\big(a-\alpha_1-\alpha_2\big)+\big(\alpha_1\mu_2 -\alpha_2\mu_1\big)-c\big)}{\mu^2 \sigma^2}\frac{1}{u^2}.\end{align*}

Let us denote by ${\mathcal{J}}$ the operator on the left-hand side of the basic IDE acting in the space of sufficiently smooth functions. Then ${\mathcal{T}}{\mathcal{J}}={\mathcal{D}}$ . Note that $\text{dim}\,{\text{Ker}}\, {\mathcal{D}}=4$ . Also, we have the inclusion ${\text{Ker}}\, {\mathcal{J}}\subseteq {\text{Ker}}\, {\mathcal{D}}$ . The kernel of the second-order differential operator ${\mathcal{T}}$ is a two-dimensional linear subspace generated by the functions $h_1(u)\,:\!=\,\text{e}^{-u/\mu_1}$ , $h_2(u)\,:\!=\,\text{e}^{u/\mu_2}$ . Let $f_1$ and $f_2$ be solutions of the equations ${\mathcal{J}} f_j=h_j$ . Then $f_1$ and $f_2$ are linearly independent and belong to ${\text{Ker}}\, {\mathcal{D}}$ . The four functions the identity, the survival probability $\Phi$ (assumed not to be a constant), $f_1$ , and $f_2$ form a basis in ${\text{Ker}}\, {\mathcal{D}}$ . Indeed, if their linear combination

\begin{align*} a_1f_1+a_2f_2+a_3 1+a_4\Phi=0, \end{align*}

then $a_1{\mathcal{J}} f_1+a_2{\mathcal{J}} f_2=0$ . That is, $a_1 h_1+a_2 h_2=0$ and $a_1=a_2=0$ . But the equality $a_3+a_4\Phi=0$ holds only if $a_3=a_4=0$ .

8. Asymptotic analysis of the differential equation for the survival probability

We analyze the asymptotic behavior of solutions of (7.2) using a result on systems with asymptotically constant coefficients. To this end, we put $y=(y_1,y_2,y_3)\,:\!=\,(G,G^{\prime},G^{\prime\prime})$ . Using matrix notation where the vectors are columns, we get from (7.2) that $y^{\prime}=A(u)y$ where

\begin{equation*}A(u)\,:\!=\,\begin{bmatrix}0 & \quad 1 &\quad 0\\[3pt] 0 & \quad 0 &\quad 1\\[3pt] - q_0(u) & \quad - q_1(u) &\quad - q_2(u)\end{bmatrix}, \qquad A(\infty)\,:\!=\,\begin{bmatrix}0 & \quad 1 &\quad 0\\[3pt] 0 & \quad 0 &\quad 1\\[3pt] 0 & \quad 1/\mu^2 &\quad - (\Delta \mu)/\mu^2\end{bmatrix}.\end{equation*}

Then, $A(u)=A(\infty)+V(u)$ , where the matrix $V(u)\,:\!=\,A(u)-A(\infty)$ is such that its norm $|V^{\prime}(u)|$ (the Euclidean or any other) is integrable on $[1,\infty[$ .

Let $\lambda_j=\lambda_j(u)$ , $j=1,2,3$ , be the roots of the characteristic equation

\begin{equation*}\lambda^3+q_2(u) \lambda^2+q_1(u) \lambda+q_0(u)=0.\end{equation*}

Note that

(8.1) \begin{equation} \lambda_1+\lambda_2+\lambda_3 = -q_2(u),\quad {\lambda_1\lambda_2}+{\lambda_2 \lambda_3}+{\lambda_1\lambda_3} = q_1(u), \quad \lambda_1\lambda_2 \lambda_3 = -q_0(u).\end{equation}

Recall that we are working under the assumptions $a> \frac12 \sigma^2 > 0$ and $\mu_1,\mu_2>0$ . Then $q_0(u)\to 0$ , $q_2(u) \to \frac {\Delta \mu}{\mu^2 }\neq 0$ , $q_1(u) \to -\frac {1}{\mu^2 } < 0$ , and $u q_0(u) \to -\frac{2a}{\mu^2 \sigma^2} < 0$ as $u\to \infty$ . The Cardano formulae imply that the roots $\lambda_j(u)$ are continuous functions having finite limits as $u\to \infty$ . According to the last equation in (8.1), at least one root, say $\lambda_3(u)$ , tends to zero as $u\to \infty$ . Since $q_1(\infty)\neq 0$ , two other roots have non-zero limits satisfying the system $\lambda_1(\infty)+\lambda_2(\infty)=-q_2(\infty)$ , $\lambda_1(\infty)\lambda_2(\infty)=q_1(\infty)$ , i.e. $\lambda_1(\infty) = \frac{1}{\mu_2}$ , $\lambda_2(\infty) = -\frac{1}{\mu_1}$ . Thus, we obtain

\begin{equation*}\lambda_1(u)=\frac 1{\mu_2}+o(1), \quad \lambda_2(u)=- \frac 1{\mu_1}+o(1), \quad \lambda_3(u)=-\frac {2a}{\sigma^2}\frac 1u+o\big(u^{-1}\big).\end{equation*}

Applying the implicit function theorem, we conclude that the function $\lambda_3(u)$ can be expanded in powers of $u^{-1}$ . In particular,

\begin{equation*}\lambda_3(u)=-\frac {2a}{\sigma^2}\frac 1u+O\big(u^{-2}\big), \qquad u\to \infty.\end{equation*}

The numbers $\lambda_1(\infty) = \frac{1}{\mu_2}$ , $\lambda_2(\infty) = -\frac{1}{\mu_1}$ , and $\lambda_3(\infty) = 0$ are eigenvalues of the matrix $A(\infty)$ .

The conditions of [Reference Hsieh and Sibuya8, Theorem VII-5-3] are fulfilled, and therefore the fundamental matrix of the equation $y^{\prime}=A(u)y$ has the form $P_0(u)(I+H(u))\exp\left\{\int_1 ^u \Lambda (s) \, \text{d} s \right\}$ , where the matrix-valued functions $P_0(u)$ and H(u) are continuous, $P_0(u)\to Q$ , $H(u)\to 0$ as $u\to \infty$ , $\Lambda (s)\,:\!=\,{\text{diag}}\,(\lambda_1(s),\lambda_2(s),\lambda_3(s))$ , I is the identity matrix, and the columns of Q are eigenvectors of $A(\infty)$ corresponding to the eigenvalues $\lambda_1(\infty)$ , $\lambda_2(\infty)$ , and $\lambda_3(\infty)$ , i.e.

\begin{equation*}Q=\begin{bmatrix}1 & \quad 1 & \quad 1\\[3pt] 1/\mu_2 & \quad -1/\mu_1 &\quad 0\\[3pt] 1/\mu_2^2 & \quad 1/\mu_1^2 &\quad 0\end{bmatrix}.\end{equation*}

A general solution G of (7.2) is a linear combination of the functions

\begin{align*} h_1(u) & \,:\!=\, (1+\theta_1(u)) \exp\left \{\int_1^u \left (\frac 1{\mu_2}+\gamma_1(s)\right)\text{d} s\right\}, \\[3pt] h_2(u) & \,:\!=\, (1+\theta_2(u)) \exp\left \{\int_1^u \left (-\frac 1{\mu_1}+\gamma_2(s)\right)\text{d} s\right\}, \\[3pt] h_3(u) & \,:\!=\, (1+\theta_3(u)) \exp\left \{-\frac {2a}{\sigma^2}\ln u\right\} \exp\left \{\int_1^u \gamma_3(s) \, \text{d} s\right\},\end{align*}

where the continuous functions $\theta_j$ and $\gamma_j$ vanish at infinity for $j=1,2,3$ , and the function $\gamma_3$ is integrable. Thus, the general solution of the fourth-order differential equation is a linear combination of a constant function and the three functions above. In particular, the ruin probability $\Psi$ is given by a certain linear combination. Clearly, the latter cannot involve the unbounded function corresponding to the integral of first function. The integrals of the others are bounded. and therefore $\Psi(u)= c_0+ c_2H_2(u)+c_3H_3(u)$ , where $H_j(u)\,:\!=\,\int_u^\infty h_j(s) \, \text{d} s$ , $j=2,3$ . Note that $c_0=0$ , the integral $H_2(u)$ converges to zero exponentially fast, and

\begin{equation*}\frac {H_3(u)}{u^{1-2a/\sigma^2}}\to \frac 1{2a/\sigma^2-1}\exp\left \{\int_1^\infty \gamma_3(s) \, \text{d} s\right\}\neq 0\end{equation*}

as $u\to \infty$ . By virtue of Proposition 4.1 $c_3\neq 0$ , and we easily obtain that $\Psi(u)\sim K u^{-\beta}$ , where $K>0$ and $\beta = \frac{2a}{\sigma^2}-1$ .