4.1. Preliminary results

We will need the following lemma, which bounds the probability distribution of the supremum of a Lévy process over an exponentially distributed interval of time in terms of the distribution of the Lévy process at the end of the interval. It is a variation of [Reference Willekens28, Lemma 1], where the time interval considered was fixed.

Lemma 4.1. Let X be a Lévy process, let $\tau$ be an independent exponentially distributed random variable, and suppose $0<u_0<u$. Then

(4.1) \begin{equation} \mathbb P\biggl( \sup_{0\leq s<\tau} X_s > u \biggr) \leq\dfrac{\mathbb P (X_\tau \geq u-u_0 )}{\mathbb P ( X_\tau \geq -u_0 )}.\end{equation}

Proof. Let $S_u \,:\!=\, \inf\{t\geq0\colon X_t>u\}$. Then, since $X_{S_u}\geq u$, by independence of increments of X and the memoryless property of $\tau$, we have

\begin{align*} \mathbb P(S_u<\tau ;\, X_{\tau} < u-u_0 ) &\leq \mathbb P( S_u<\tau ;\, X_{\tau} - X_{S_u} < -u_0 )\\ & =\mathbb P( S_u<\tau ;\, \tilde{X}_{\tilde{\tau}} < -u_0 )\\ &= \mathbb P(S_u<\tau)\,\mathbb P(X_{\tau}<-u_0),\end{align*}

where $\tilde{X}$ and $\tilde{\tau}$ are independent and identically distributed copies of X and $\tau$, respectively. Then (4.1) can be obtained by rearranging the inequality

\begin{align*}\mathbb P(S_u<\tau) &\leq \mathbb P (X_{\tau}\geq u-u_0 ) + \mathbb P ( S_u<\tau ;\, X_{\tau}<u-u_0 )\\ &\leq \mathbb P (X_{\tau}\geq u-u_0 ) + \mathbb P(S_u<\tau)\,\mathbb P(X_{\tau}<-u_0).\end{align*}

We now consider long-tailed distributions. Let $Q\colon \mathbb R^+\rightarrow[0,1]$ be a probability distribution and $\bar{Q}(x)\,:\!=\, 1-Q(x)$ for all $x\in\mathbb R^+$. Then Q is a long-tailed distribution if $Q(x+y)/Q(x)\rightarrow 1$ as $x\rightarrow\infty$, for any $y\in\mathbb R^+$. For any two functions $f,g\colon \mathbb R^+\rightarrow\mathbb R^+$, we will write $f\sim g$ if $\lim_{x\rightarrow\infty}f(x)/g(x) = 1$ and say that f and g are asymptotically equivalent.

For a random variable X, we define the functions

\begin{equation*} G_X(x) \,:\!=\, \int_x^\infty \mathbb P(X>u) \,{\mathrm{d}} u\quad \text{and}\quad H_X(x) \,:\!=\, \min ( 1, G_X(x) ) \end{equation*}

and refer to $H_X$ as the integrated tail of X.

In the next lemma we show that the integrated tail of a long-tailed random variable is asymptotically equivalent to an infinite series. This will be used in Lemma 4.6 to show the asymptotic equivalence of the integrated tail $G_X$ and an infinite series that we will construct.

Lemma 4.2. Suppose $K+\epsilon<0$ and that X is a long-tailed random variable, which is independent of $(T_{2n})_{n\in\mathbb N}$. Then the integrated tail of X has the asymptotics

(4.2) \begin{equation} \int_x^\infty \mathbb P(X>u)\, {\mathrm{d}} u \sim \mathbb E[T_2]|K+\epsilon| \sum_{n=0}^\infty \mathbb P( X > x- (K+\epsilon)T_{2n}).\end{equation}

Proof. By splitting the interval $(0,\infty)$ into a disjoint union, we can write

\begin{align*} G_X(x) &= \sum_{n=0}^\infty \int_{-T_{2n}(K+\epsilon)}^{-T_{2(n+1)}(K+\epsilon)} \mathbb P(X>u_1+x) \,{\mathrm{d}} u_1.\end{align*}

Then, by using the change of variables $u=-(K+\epsilon)^{-1}u_1$, we get

\begin{align*} G_X(x) &= \sum_{n=0}^\infty \int_{T_{2n}}^{T_{2(n+1)}} \mathbb P ( X > -(K+\epsilon)u + x ) \: |K+\epsilon| \,{\mathrm{d}} u.\end{align*}

By independence of $\{T_{2n}\}_{n\in\mathbb N}$ and X, we can shift the domain of integration to obtain

\begin{align*} G_X(x) &= \sum_{n=0}^\infty \int_0^{T_{2(n+1)}-T_{2n}} \mathbb P( X > x - (K+\epsilon)( u + T_{2n} ) \mid T_{2n} ) \:|K+\epsilon| \,{\mathrm{d}} u.\end{align*}

Taking expectations and noting that the left-hand side is not random, that $T_{2(n+1)}-T_{2n}\stackrel{\mathcal{L}}{=} T_2$, and that $T_{2(n+1)}-T_{2n}$ is independent of $T_{2n}$ gives

\begin{align*} G_X(x) &=|K+\epsilon|\sum_{n=0}^\infty \mathbb E\biggl[ \int_0^{\tilde{T}_2} \mathbb P( X > x-(K+\epsilon)(u + T_{2n}) \mid T_{2n} ) \,{\mathrm{d}} u \biggr],\end{align*}

where $\tilde{T}_{2}$ is an independent and identically distributed copy of $T_2$. This can be written in the integral form

(4.3) \begin{align} G_X(x)= |K+\epsilon|\sum_{n=0}^\infty \int_0^\infty \mathbb P(\tilde{T}_2 \in \,{\mathrm{d}} s ) \int_0^s \int_0^\infty \mathbb P(T_{2n}\in {\mathrm{d}} v) \,\mathbb P (X> x - (K+\epsilon)(u + v) ) \,{\mathrm{d}} u.\end{align}

Let $\delta>0$; then, since X is long-tailed, for all $s>0$ there exists $R(s)>0$ such that, whenever $z>R(s)$ and $y\in[0,-s(K+\epsilon)]$,

\begin{equation*} (1-\delta) \leq \dfrac{\mathbb P( X > z+ y )}{\mathbb P(X>z)} \leq (1+\delta),\end{equation*}

and since $-v(K+\epsilon)\geq 0$ for $v\geq0$, we have for all $x>R(s)$ and $u\in[0,s]$

\begin{align*} (1-\delta) \leq \dfrac{\mathbb P( X > x-(K+\epsilon)(v+u) )}{\mathbb P( X> x - (K+\epsilon)v )} \leq (1+\delta) .\end{align*}

To show the lower bound we use this inequality within the last two integrals of (4.3) to obtain, for $x>R(s)$,

\begin{align*}& \int_0^s \int_0^\infty \mathbb P(T_{2n}\in {\mathrm{d}} v) \,\mathbb P (X> x - (K+\epsilon)(u + v) ) \,{\mathrm{d}} u\\[5pt] &\quad \geq \int_0^s \int_0^\infty \mathbb P(T_{2n}\in {\mathrm{d}} v)(1-\delta)\,\mathbb P(X > x-(K+\epsilon)v) \,{\mathrm{d}} u.\end{align*}

Then, evaluating the integrals and noticing that the integrand is constant with respect to u gives, for $x>R(s)$,

\begin{align*} \int_0^s \int_0^\infty \mathbb P(T_{2n}\in {\mathrm{d}} v) \,\mathbb P (X> x - (K+\epsilon)(u + v) ) \,{\mathrm{d}} u \geq s(1-\delta) \,\mathbb P( X > x - (K+\epsilon)T_{2n} ).\end{align*}

Now consider some $l>0$ and suppose $x>R(l)$, so that $x>R(s)$ for any $s\in[0,l]$. Then

\begin{align*} &\int_0^\infty \mathbb P(T_2 \in {\mathrm{d}} s)\int_0^s \int_0^\infty \mathbb P(T_{2n}\in {\mathrm{d}} v)\,\mathbb P(X>x-(K+\epsilon)(u+v)) \,{\mathrm{d}} u\\ &\quad \geq \int_0^l \mathbb P(T_2 \in {\mathrm{d}} s) s(1-\delta)\,\mathbb P(X>x-(K+\epsilon)T_{2n})\\ &\quad =(1-\delta)\,\mathbb P(X>x-(K+\epsilon)T_{2n})\,\mathbb E[T_2 ;\, T_2<l].\end{align*}

Since $l>0$ and $\delta>0$ are arbitrary and $T_2$ is integrable, we can take l sufficiently large to obtain $\mathbb E[T_2 ;\, T_2<l]\geq(1-\delta)\,\mathbb E[T_2]$ and thus

\begin{align*}& \int_0^\infty \mathbb P(T_2 \in {\mathrm{d}} s)\int_0^s \int_0^\infty \mathbb P(T_{2n}\in {\mathrm{d}} v)\,\mathbb P(X>x-(K+\epsilon)(u+v)) \,{\mathrm{d}} u\\ &\quad \geq (1-\delta)^2\,\mathbb P(X>x-(K+\epsilon)T_{2n})\,\mathbb E[T_2].\end{align*}

If this is substituted into the expression for $G_X$ for $x>R(l)$, we have

\begin{align*} G_X(x) \geq (1-\delta)^2|K+\epsilon|\,\mathbb E[T_2] \sum_{n=0}^\infty \mathbb P( X>x - (K+\epsilon)T_{2n} ).\end{align*}

For the upper bound, since $-(K+\epsilon)u>0$ for $u>0$,

\begin{align*} \mathbb E\biggl[\int_0^{\tilde{T}_2}\mathbb P( X>x -(K+\epsilon)(u+T_{2n}) \mid \sigma(T_{2n})) \,{\mathrm{d}} u\biggr] \leq \mathbb E[\tilde{T_2}]\,\mathbb P ( X > x-(K+\epsilon)T_{2n} ),\end{align*}

which, substituted into the expression for $G_X$, gives, for all $x>0$,

\begin{align*} G_X(x) \leq \mathbb E[T_2]|K+\epsilon| \sum_{n=0}^\infty \mathbb P( X>x-(K+\epsilon)T_{2n} ).\end{align*}

Combining the upper and lower bounds gives (4.2).

4.2. Framework for an upper bound of $\log({{A}}_{\boldsymbol\infty})$

We now develop a framework for bounding $\log(A_\infty)$ whenever $\mathbb E[\xi_{T_2}]\in(-\infty,0)$. This will be used in the strong subexponential setting of Section 4.3 to obtain the right tails of $A_\infty$. This framework consists of constructing a series of stopping times between which $\xi$ is bounded from above. By studying the properties of these stopping times and the behaviour of $\xi$ at those times, we can begin to understand $\log(A_\infty)$.

For $\epsilon\in(0,-K)$ and sufficiently large $A\in\mathbb R$, define a sequence of stopping times by $\sigma_0\,:\!=\, 0$ and

\begin{equation*} \sigma_n \,:\!=\, \inf \{ t>\sigma_{n-1} \colon \xi_t-\xi_{\sigma_{n-1}} \geq (K+\epsilon)(t-\sigma_{n-1}) + A \}, \end{equation*}

for each $n\in\mathbb N$, with the convention $\inf(\varnothing)=\infty$, and setting $\sigma_n=\infty$ if $\sigma_{n-1}=\infty$. Then also define

\begin{align*}N\,:\!=\, \max\{ n\in\mathbb N_0 \mid \sigma_n<\infty \}\quad \text{and}\quad \rho_n&\,:\!=\, \mathbb P(\sigma_n<\infty \mid \sigma_{n-1}<\infty ). \end{align*}

The next lemma concerns the finiteness of N.

Proof. Define a new MAP $\{(J_t,\tilde{\xi_t})\colon t\geq0\}$ by setting $\tilde{\xi}_t \,:\!=\, \xi_t - (K+\epsilon) t$. For each $n\in\mathbb N$,

\begin{align*} 1-\rho_n &= \mathbb P\biggl( \sup_{t>\sigma_{n-1}}\tilde{\xi}_t-\tilde{\xi}_{\sigma_{n-1}} < A \mid \sigma_{n-1}<\infty \biggr),\end{align*}

then, since $\sigma_{n-1}$ is a stopping time, using the Markov additive property and summing over the events $\{J_{\sigma_{n-1}}=j\}$ for $j\in\{+,-\}$,

\begin{align*} 1-\rho_n = \sum_{j\in\{+,-\}}\mathbb P_j\biggl( \sup_{t\geq0}\tilde{\xi}_t<A \biggr)\,\mathbb P(J_{\sigma_{n-1}} = j).\end{align*}

By the strong law of large numbers,

\begin{align*} \lim_{t\rightarrow\infty}t^{-1}\tilde{\xi_t} &= \lim_{t\rightarrow\infty}t^{-1}(\xi_t - (K+\epsilon)t) = K-(K+\epsilon) = -\epsilon <0 \quad \text{a.s.}\end{align*}

and hence there a.s. exists $T>0$ such that if $t>T$ then $\tilde{\xi}_t<0$. From this we can conclude that $\sup_{t\geq0}\tilde{\xi}_t = \max (0 , \sup_{t\in[0,T]}\tilde{\xi}_t )<\infty$, since the supremum of a càdlàg process over a compact interval is bounded.

This implies that there exists an $A^*>0$ such that for all $A>A^*$ we have $\mathbb P (\sup_{t\geq0}\tilde{\xi}_t > A )<1$. From this, we conclude $N<\infty$ a.s., since

(4.4) \begin{align} \mathbb P(N>n) & = \prod_{k=1}^n\mathbb P(\sigma_{k}<\infty \mid \sigma_{k-1}<\infty ) \leq \max_{j\in\{+,-\}} \mathbb P_j\biggl( \sup_{t\geq0}\tilde{\xi}_t < A \biggr)^n.\end{align}

Since there are conditions for N to be finite, we can bound $\log(A_\infty)$ by using the stopping times $\{\sigma_n\}_{n\in\mathbb N}$ to split the process $\{\xi_t\colon t\geq0\}$ into a finite number of bounded sections.

Define the constant

\begin{align*}C\,:\!=\, \log\biggl( \dfrac{{\mathrm{e}}^A}{|K+\epsilon|} \biggr){,} \end{align*}

and by taking $A>A^*$ sufficiently large we can ensure ${\mathrm{e}}^C>2$. Then we have the following upper bound for $\log(A_\infty)$.

Lemma 4.4. If $\mathbb E[\xi_{T_2}]<0$, then

\begin{equation*} \log{A_\infty} \leq (N+1)C + \sum_{n=1}^N(\xi_{\sigma_n} - \xi_{\sigma_{n-1}})^+{,}\end{equation*}

where $(\cdot)^+$ denotes the positive part.

Proof. Following the approach of [Reference Maulik and Zwart21, Lemma 4.1], $A_\infty$ may be expanded as

\begin{align*} A_\infty &= \int_0^{\sigma_1} \,{\mathrm{e}}^{\xi_t} \,{\mathrm{d}} t + {\mathrm{e}}^{\xi_{\sigma_1}}\biggl( \int_{\sigma_1}^{\sigma_2}\,{\mathrm{e}}^{\xi_t - \xi_{\sigma_1}} \,{\mathrm{d}} t + {\mathrm{e}}^{\xi_{\sigma_2}-\xi_{\sigma_1}}\biggl( \int_{\sigma_2}^{\sigma_3}\,{\mathrm{e}}^{\xi_t - \xi_{\sigma_2}} \,{\mathrm{d}} t + \cdots \\&\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad \cdots + {\mathrm{e}}^{\xi_{\sigma_N}-\xi_{\sigma_{N-1}}}\biggl( \int_{\sigma_N}^{\sigma_{N+1}}\,{\mathrm{e}}^{\xi_t - \xi_{\sigma_N}} \,{\mathrm{d}} t \biggr) \biggr)\biggr),\end{align*}

noting that $\sigma_{N+1}=\infty$. By the definition of $\sigma_n$, we have

\begin{equation*} \int_{\sigma_n}^{\sigma_{n+1}}\,{\mathrm{e}}^{\xi_t - \xi_{\sigma_n}} \,{\mathrm{d}} t\leq \int_{\sigma_n}^{\sigma_{n+1}}\exp( (K+\epsilon)(t-\sigma_n) + A) \,{\mathrm{d}} t\leq {\mathrm{e}}^C {,}\end{equation*}

which, when substituted into the expression for $A_\infty$, gives

(4.5) \begin{equation} A_\infty \leq {\mathrm{e}}^C + {\mathrm{e}}^{\xi_{\sigma_1}} ( {\mathrm{e}}^C + {\mathrm{e}}^{\xi_{\sigma_2}-\xi_{\sigma_1}} ( {\mathrm{e}}^C + \dots + {\mathrm{e}}^{\xi_{\sigma_N}-\xi_{\sigma_{N-1}}} ( {\mathrm{e}}^C ) ) ).\end{equation}

Then, by considering the logarithms of both sides of (4.5) and repeatedly using the property $\log(A+B)\leq\log(A)+\log(B)$ whenever $A,B>2$,

\begin{align*} \log(A_\infty) \leq(N+1)C + \sum_{n=1}^N(\xi_{\sigma_n} - \xi_{\sigma_{n-1}})^+{,}\end{align*}

where we have made use of the fact that ${\mathrm{e}}^{(\xi_{\sigma_n}-\xi_{\sigma_{n-1}})^+}\geq1$ and ${\mathrm{e}}^C>2$ in order to use the log inequality.

As a consequence of this lemma, the right tails of $\log(A_\infty)$ can be studied by considering the evolution of the MAP between the stopping times $\{\sigma_n\}_{n\in\mathbb N}.$ First we consider the J component.

Let $(K_n)_{n\in\mathbb N_0}$ be the sequence of random variables, taking values in $\{+,-,\infty\}$, such that for each $n\in\mathbb N$, if $\sigma_n<\infty$ then $K_n=J_{\sigma_n}$, otherwise $K_n=\infty$. For each $\alpha,\beta\in\{+,-\}$ we will be interested in the number of times that $\{K_n\}_{n\in\mathbb N}$ transitions from $\alpha$ to $\beta$. For this purpose, define the random variable $N(\alpha,\beta)\,:\!=\, \sum_{k=1}^\infty\mathbb{1}_{\{K_{k-1}=\alpha,K_k=\beta\}}$. We also make use of the notation $f={\mathrm{o}}(g)$ for any two functions $f,g\colon \mathbb R^+\rightarrow\mathbb R^+$ such that $\lim_{x\rightarrow\infty}f(x)/g(x) = 0$.

Proposition 4.1. Suppose $\mathbb E[\xi_{T_2}]\in(-\infty,0)$. Then the sequence $(K_n)_{n\in\mathbb N_0}$ is a discrete-time homogeneous Markov chain, with $\infty$ as an absorbing state. Moreover, if $\eta$ is the stochastic matrix of $\{K_n\}_{n\in\mathbb N}$ and $\alpha,\beta,\gamma\in\{+,-\}$ with $\alpha\neq\beta$, then

\begin{align*} \eta_{\alpha,\gamma}&\rightarrow0, \quad \mathbb E_\alpha[N(\alpha,\gamma)]\sim \eta_{\alpha,\gamma} \quad {and}\quad \mathbb E_\beta[N(\alpha,\gamma)] = {\mathrm{o}}(\eta_{\alpha,\gamma})\end{align*}

as $A\rightarrow\infty$.

Proof. First we show that $\{K_n\}_{n\in\mathbb N}$ is a Markov chain. If $K_{n-1}\neq\infty$, then by the Markov additive property, since $\sigma_{n-1}$ is a stopping time, $\{\xi_{\sigma_{n-1}+t}-\xi_{\sigma_{n-1}}\mid t\geq0\}$ is independent of $\mathcal{F}_{\sigma_{n-1}}$ given $K_{n-1}$. Moreover, the random variable

\begin{align*} \Delta\sigma_n \,:\!=\, \sigma_{n} - \sigma_{n-1} &= \inf\{ t\geq0 \colon \xi_{t+\sigma_{n-1}} - \xi_{\sigma_{n-1}} \geq t(K+\epsilon) + A \}\end{align*}

is a function of $\{\xi_{t+\sigma_{n-1}}-\xi_{\sigma_{n-1}}\colon t\geq0\}$. Thus the event $\{K_n=\infty\}=\{\Delta\sigma_n=\infty\}$ is independent of $\mathcal{F}_{\sigma_{n-1}}$ given $K_{n-1}$, and has the same law as the event $\{K_1=\infty\}$ given $K_0$.

Moreover, if $\Delta\sigma_n<\infty$ then $K_n=J_{\sigma_n}=J_{\sigma_{n-1}+\Delta\sigma_n}$, hence $K_n$ is a function of

\begin{equation*}\{ (\xi_{\sigma_{n-1}+t}-\xi_{\sigma_{n-1}},J_{\sigma_{n-1}+t} )\colon t\geq0 \},\end{equation*}

and so by the Markov additive property is independent of $\mathcal{F}_{\sigma_{n-1}}$ given $K_{n-1}$ and has the same distribution as $K_1$ given $K_0$. Hence the sequence $(K_n)_{n\in\mathbb N}$ is a time homogeneous Markov chain. By definition of $\sigma_n$, $\infty$ is clearly an absorbing state for $(K_n)_{n\in\mathbb N}$.

Now we consider the limiting behaviour of $\eta$ as $A\rightarrow\infty$. Let $\alpha,\gamma\in\{+,-\}$. From the proof of Lemma 4.3 we know that $\sup_{t\geq0}\tilde{\xi}_t<\infty$ a.s., where $\tilde{\xi_t}\,:\!=\, \xi_t-(K+\epsilon)t$. Thus

\begin{align*} \lim_{A\rightarrow\infty}\mathbb P\biggl( \sup_{t\geq0} \{\xi_t -(K+\epsilon)t \} > A \biggr) = \lim_{A\rightarrow\infty}\mathbb P\biggl( \sup_{t\geq0}\tilde{\xi}_t > A \biggr) = 0{.}\end{align*}

However,

\begin{align*} \eta_{\alpha,+}+\eta_{\alpha,-} = \mathbb P_\alpha (\sigma_1<\infty ) = \mathbb P_\alpha\biggl( \sup_{t\geq0} \{\xi_t-(K+\epsilon)t \}>A\biggr)\end{align*}

and, since $\eta_{\alpha,\gamma}$ is non-negative, this implies $\lim_{A\rightarrow\infty}\eta_{\alpha,\gamma} = 0$.

Further assume that $\gamma\in\{+,-\}$ and $\alpha\neq\beta$. Then it is easily seen that

\begin{align*} \mathbb E_\theta[N(\alpha,\gamma)] &=\sum_{n=1}^\infty \mathbb P_\theta(K_n=\gamma\mid K_{n-1}=\alpha)\,\mathbb P_\theta(K_{n-1}=\alpha) =\eta_{\alpha,\gamma} ( \mathbb{1}_{\{\theta=\alpha\}} + \phi_\sigma(\alpha) ),\end{align*}

where $\phi_\theta(\alpha) \,:\!=\, \sum_{n=1}^\infty \mathbb P_\theta(K_n=\alpha)$. Since $\infty$ is an absorbing state of the Markov chain $(K_n)_{n\in\mathbb N}$ and $\alpha\neq\beta$, for each $n\in\mathbb N$,

\begin{align*} \mathbb P_\theta(K_n=\alpha) &= \mathbb P_\theta( K_n=\alpha\mid K_{n-1}=\alpha)\,\mathbb P_\theta(K_{n-1}=\alpha) \\ &\quad \, + \mathbb P_\theta( K_n=\alpha\mid K_{n-1}=\beta )\,\mathbb P_\theta(K_{n-1}=\beta)\\ &= \eta_{\alpha,\alpha}\mathbb P_\theta(K_{n-1}=\alpha) + \eta_{\beta,\alpha}\mathbb P_\theta(K_{n-1}=\beta),\end{align*}

then summing up over $n\in\mathbb N$ we have

\begin{align*} \phi_\theta(\alpha) = \eta_{\alpha,\alpha} ( \mathbb{1}_{\{\theta=\alpha\}} + \phi_\theta(\alpha) ) + \eta_{\beta,\alpha} ( \mathbb{1}_{\{\theta=\beta\}} + \phi_\theta(\beta) ),\end{align*}

and by symmetry

\begin{align*} \phi_\theta(\beta) = \eta_{\beta,\beta} ( \mathbb{1}_{\{\theta=\beta\}} + \phi_\theta(\beta) ) + \eta_{\alpha,\beta} ( \mathbb{1}_{\{\theta=\alpha\}} + \phi_\theta(\alpha) ).\end{align*}

Solving this system gives

\begin{align*} \phi_\theta(\alpha) = \dfrac{\mathbb{1}_{\{\theta=\alpha\}} (\eta_{\alpha,\alpha}(1-\eta_{\beta,\beta}) + \eta_{\beta,\alpha}\eta_{\alpha,\beta} ) + \mathbb{1}_{\{\theta=\beta\}}\eta_{\beta,\alpha} }{ (1-\eta_{\alpha,\alpha})(1-\eta_{\beta,\beta}) -\eta_{\beta,\alpha}\eta_{\alpha,\beta} },\end{align*}

and thus

\begin{align*} \mathbb E_\theta [ N(\alpha,\gamma) ] =\dfrac{\eta_{\alpha,\gamma} ( \mathbb{1}_{\{\theta=\alpha\}}(1-\eta_{\beta,\beta}) + \mathbb{1}_{\{\theta=\beta\}}\eta_{\beta,\alpha} )}{(1-\eta_{\alpha,\alpha})(1-\eta_{\beta,\beta}) -\eta_{\beta,\alpha}\eta_{\alpha,\beta}},\end{align*}

from which the two asymptotic results for $\mathbb E_\theta[N(\alpha,\gamma)]$ are a consequence of the limiting behaviour of $\eta$.

In the next lemma we consider the evolution of $\xi$ between the stopping times $\{\sigma_n\}_{n\in\mathbb N}$, conditioned on the values of $\{K_n\}_{n\in\mathbb N}$.

Lemma 4.5. Suppose that $\mathbb E[\xi_{T_2}]\in(-\infty,0)$ and $m,n\in\mathbb N$ with $m<n$. Then, conditionally on $K_{m-1}$, $K_m$, $K_{n-1}$, and $K_n$, the increments $\xi_{\sigma_m}-\xi_{\sigma_{m-1}}$ and $\xi_{\sigma_n} - \xi_{\sigma_{n-1}}$ are independent. If $\alpha,\beta\in\{+,-\}$, then conditional on the event $\{K_{n-1}=K_{m-1}=\alpha ;\, K_n=K_m=\beta\}$, the increments $\xi_{\sigma_m}-\xi_{\sigma_{m-1}}$ and $\xi_{\sigma_n} - \xi_{\sigma_{n-1}}$ are equal in distribution and independent. Furthermore, for any $l\in\mathbb N$ such that $m\neq l$ and any bounded continuous function $f\colon \mathbb R\rightarrow\mathbb R^+$, we have

\begin{align*} \mathbb E[ f(\xi_{\sigma_{m}}-\xi_{\sigma_{m-1}}) \mid \sigma(K_{m-1},K_m,K_{l-1},K_l) ] &=\mathbb E[ f(\xi_{\sigma_{m}}-\xi_{\sigma_{m-1}}) \mid \sigma(K_{m-1},K_m) ].\end{align*}

Proof. First suppose that $m<l$. Then we have

\begin{align*} &\mathbb E[f (\xi_{\sigma_m}-\xi_{\sigma_{m-1}} )\mid \sigma(K_{m-1},K_m,K_{l-1},K_l)]\\ &\quad =\sum_{\gamma,\delta\in\{+,-\}} \mathbb{1}_{\{K_{l-1}=\gamma,K_{l}=\delta\}}\dfrac{\mathbb E[f (\xi_{\sigma_m}-\xi_{\sigma_{m-1}} ) ;\, K_{l-1}=\gamma,K_l=\delta \mid \sigma(K_{m-1},K_m)]}{\mathbb P(K_{l-1}=\gamma,K_l=\delta \mid \sigma(K_{m-1},K_m))}.\end{align*}

It follows that, by using the tower property and the fact that $(K_k)_{k\in\mathbb N}$ is a Markov chain, we have

\begin{align*} &\mathbb E[f (\xi_{\sigma_m}-\xi_{\sigma_{m-1}} ) ;\, K_{l-1}=\gamma,K_l=\delta \mid \sigma(K_{m-1},K_m)]\\ &\quad =\mathbb E [ f (\xi_{\sigma_m}-\xi_{\sigma_{m-1}} ) \,\mathbb E [ \mathbb{1}_{\{K_{l-1}=\gamma,K_l=\delta\}} \mid \mathcal{F}_{\sigma_m} ] \mid \sigma(K_{m-1},K_m) ]\\ &\quad =\mathbb E [ f (\xi_{\sigma_m}-\xi_{\sigma_{m-1}} ) \,\mathbb E [ \mathbb{1}_{\{K_{l-1}=\gamma,K_l=\delta\}} \mid \sigma(K_m) ] \mid \sigma(K_{m-1},K_m) ]\\ &\quad = \mathbb P (K_{l-1}=\gamma,K_l=\delta \mid \sigma(K_{m-1},K_m) ) \,\mathbb E [ f (\xi_{\sigma_m}-\xi_{\sigma_{m-1}} ) \mid \sigma(K_{m-1},K_m) ],\end{align*}

which, when substituted into the previous equation, gives

\begin{align*} &\mathbb E[f (\xi_{\sigma_m}-\xi_{\sigma_{m-1}} ) \mid \sigma(K_{m-1},K_m,K_{l-1},K_l)]\\ &\quad = \sum_{\gamma,\delta\in\{+,-\}} \mathbb{1}_{\{K_{l-1}=\gamma,K_l=\delta\}}\,\mathbb E [ f ( \xi_{\sigma_m}-\xi_{\sigma_{m-1}} ) \mid \sigma(K_{m-1},K_m) ]\\ &\quad =\mathbb E [ f(\xi_{\sigma_m}-\xi_{\sigma_{m-1}} ) \mid \sigma(K_{m-1},K_m) ].\end{align*}

Now suppose $m>l$, and through a direct application of the Markov additive property we have

\begin{align*} &\mathbb E [ f(\xi_{\sigma_{m}}-\xi_{\sigma_{m-1}}) \mid \sigma(K_{l-1},K_l,K_{m-1},K_m) ]\\ &\quad =\sum_{\alpha\in\{+,-\}}\mathbb{1}_{\{K_m=\alpha\}}\dfrac{\mathbb E [ f(\xi_{\sigma_{m}}-\xi_{\sigma_{m-1}}) ;\, K_m=\alpha \mid \sigma(K_{l-1},K_l,K_{m-1}) ]}{\mathbb P(K_m=\alpha \mid \sigma(K_{l-1},K_l,K_{m-1}) ) }\\ &\quad =\sum_{\alpha\in\{+,-\}}\mathbb{1}_{\{K_m=\alpha\}} \dfrac{\mathbb E [ f(\xi_{\sigma_{m}}-\xi_{\sigma_{m-1}}) ;\, K_m=\alpha \mid \sigma( K_{m-1} ) ]}{\mathbb P(K_m=\alpha \mid \sigma (K_{m-1}) )}\\ &\quad = \mathbb E [ f(\xi_{\sigma_{m}}-\xi_{\sigma_{m-1}}) \mid \sigma(K_{m-1},K_m) ].\end{align*}

To see the independence of increments, suppose that $f,g\colon \mathbb R\rightarrow\mathbb R^+$ are bounded continuous functions, and then

\begin{align*} &\mathbb E [ f(\xi_{\sigma_n}-\xi_{\sigma_{n-1}})g(\xi_{\sigma_m} - \xi_{\sigma_{m-1}}) \mid \sigma(K_{m-1},K_m,K_{n-1},K_n) ]\\[4pt] &\quad = \sum_{\alpha\in\{+,-\}}\mathbb{1}_{\{K_n=\alpha\}}\dfrac{\mathbb E [ f(\xi_{\sigma_n}-\xi_{\sigma_{n-1}})g(\xi_{\sigma_m} - \xi_{\sigma_{m-1}}) ;\, K_n=\alpha \mid \sigma(K_{m-1},K_m,K_{n-1}) ]}{\mathbb P ( K_n=\alpha \mid \sigma(K_{m-1},K_m,K_{n-1}) )}.\end{align*}

Then, by the tower property, we get

\begin{align*} &\mathbb E [ f(\xi_{\sigma_n}-\xi_{\sigma_{n-1}})g(\xi_{\sigma_m} - \xi_{\sigma_{m-1}}) ;\, K_n=\alpha \mid \sigma(K_{m-1},K_m,K_{n-1}) ]\\ &\quad =\mathbb E [ g(\xi_{\sigma_m} - \xi_{\sigma_{m-1}}) \,\mathbb E [ f(\xi_{\sigma_n}-\xi_{\sigma_{n-1}}) ;\, K_n=\alpha \mid \mathcal{F}_{\sigma_{n-1}} ] \mid \sigma(K_{m-1},K_m,K_{n-1}) ]\\ &\quad =\mathbb E [ g(\xi_{\sigma_m} - \xi_{\sigma_{m-1}}) \,\mathbb E [ f(\xi_{\sigma_n}-\xi_{\sigma_{n-1}}) ;\, K_n=\alpha \mid \sigma(K_{\sigma_{n-1}}) ] \mid \sigma(K_{m-1},K_m,K_{n-1}) ]\\ &\quad =\mathbb E [ f(\xi_{\sigma_n}-\xi_{\sigma_{n-1}}) ;\, K_n=\alpha \mid \sigma(K_{\sigma_{n-1}}) ]\,\mathbb E [ g(\xi_{\sigma_m} - \xi_{\sigma_{m-1}}) \mid \sigma(K_{m-1},K_m) ].\end{align*}

Plugging this into the previous equation yields

\begin{align*} \qquad\qquad &\mathbb E [ f(\xi_{\sigma_n}-\xi_{\sigma_{n-1}})g(\xi_{\sigma_m} - \xi_{\sigma_{m-1}}) \mid \sigma(K_{m-1},K_m,K_{n-1},K_n) ]\\[4pt] &\quad = \sum_{\alpha\in\{+,-\}}\mathbb{1}_{\{K_n=\alpha\}}\dfrac{\mathbb E [ f(\xi_{\sigma_n}-\xi_{\sigma_{n-1}}) ;\, K_n=\alpha \mid \sigma(K_{\sigma_{n-1}}) ]}{\mathbb P ( K_n=\alpha \mid \sigma(K_{m-1},K_m,K_{n-1}) )}\\[4pt] &\quad \quad \, \times \mathbb E [ g(\xi_{\sigma_m} - \xi_{\sigma_{m-1}}) \mid \sigma(K_{m-1},K_m) ] \\[4pt] &\quad = \mathbb E [g(\xi_{\sigma_m} - \xi_{\sigma_{m-1}}) \mid \times \sigma(K_{m-1},K_m) ]\\[4pt] &\quad \quad \, \times \sum_{\alpha\in\{+,-\}}\mathbb{1}_{\{K_n=\alpha\}}\dfrac{\mathbb E [ f(\xi_{\sigma_n}-\xi_{\sigma_{n-1}}) ;\, K_n=\alpha \mid \sigma(K_{\sigma_{n-1}}) ] }{\mathbb P ( K_n=\alpha \mid \sigma(K_{n-1}) )}\\[4pt] &\quad = \mathbb E [g(\xi_{\sigma_m} - \xi_{\sigma_{m-1}}) \mid \sigma(K_{m-1},K_m) ]\,\mathbb E [ f(\xi_{\sigma_n}-\xi_{\sigma_{n-1}}) \mid \sigma(K_{\sigma_{n-1}},K_{\sigma_n}) ].\end{align*}

4.3. Lamperti–Kiu processes of strong subexponential type

Strong subexponential distributions are a widely studied class of heavy-tailed distributions, because of their mathematical tractability and their appearance in empirical data. We will use [Reference Foss, Korshunov and Zachary11] as a reference to the background theory of subexponential distributions, within which further discussion of the use of these distributions can be found.

For a probability distribution $Q\colon \mathbb R^+\rightarrow[0,1]$, define $\bar{Q}(x)\,:\!=\,q 1-Q(x)$ for all $x\in\mathbb R^+$. Then, if $\overline{Q*Q}(x)/\overline{Q}(x)\rightarrow 2$ as $x\rightarrow\infty$, we say that Q is a subexponential distribution. It can be shown (see e.g. [Reference Foss, Korshunov and Zachary11]) that all subexponential distributions are also long-tailed. The distribution Q is a strong subexponential distribution if it also satisfies the property that

\begin{align*}\lim_{x\rightarrow\infty}\dfrac{1}{\bar{Q}(x)}\int_0^x\bar{Q}(x-y)\,\bar{Q}(y) \,{\mathrm{d}} y = 2m,\end{align*}

where $m=\mathbb E[X^+]$ and X is a random variable with distribution Q. We will refer to a random variable with a (strong) subexponential distribution as a (strong) subexponential random variable.

Let $\mathcal{S}$ denote the set of real-valued subexponential random variables and let $\mathcal{S}^*$ denote the subset of $\mathcal{S}$ comprising strong subexponential random variables.

For a random variable X, recall the definitions

\begin{equation*} G_X(x) \,:\!=\, \int_x^\infty \mathbb P(X>u) \,{\mathrm{d}} u\quad \text{and}\quad H_X(x) \,:\!=\, \min ( 1, G_X(x) ). \end{equation*}

If $H_X$ is a subexponential distribution then we write $X\in\mathcal{S}_I$, and from [Reference Foss, Korshunov and Zachary11, Theorem 3.27] we have $\mathcal{S}^*\subset\mathcal{S}_I$.

For ease of notation, we also define the integrated tails $H(x) \,:\!=\, H_{\xi_{T_2}}(x)$ and, for each $\alpha\in\{+,-\}$, define

\begin{equation*} H_{\xi_\alpha}\,:\!=\, H_{\xi^{(\alpha)}_{\zeta_\alpha}}\quad \text{and}\quad H^{(\alpha)} \,:\!=\, H_{\xi_\alpha} + H_{U_{-\alpha}}. \end{equation*}

Let us introduce a subset of components of the Lamperti–Kiu decomposition given by $L\,:\!=\, \{ \xi^{(+)}_{\zeta_+} , \xi^{(-)}_{\zeta_-} , U^+, U^- \}$.

For any functions $f,g\colon \mathbb R^+\rightarrow\mathbb R^+$, we write $f={\mathrm{O}}(g)$ if $\limsup_{x\rightarrow\infty}f(x)/g(x)\in\mathbb R$.

If a Lamperti–Kiu process is of this type, there is a heaviest-tailed component of the Lamperti–Kiu decomposition and it is strong subexponential. Denote this component by $X\in L$. Let $B\subseteq\{+,-\}$ be the set of all $\beta\in\{+,-\}$ such that

\begin{equation*}\limsup_{x\rightarrow\infty}H_X(x)^{-1}H^{(\beta)}(x)\neq0.\end{equation*}

Then, for any $b\in B$ and $\beta\in\{+,-\}\setminus B$, we have $H^{(\beta)}(x) = {\mathrm{o}}( H^{(b)}(x) )$ as $x\rightarrow\infty$.

By the closure properties of $\mathcal{S}^*$ [Reference Foss, Korshunov and Zachary11, Corollary 3.16], it follows that $\xi_{T_2}$ is also strong subexponential with tails and integrated tails, respectively, given, as $x\rightarrow\infty$, by

\begin{align*}\mathbb P (\xi_{T_2} > x )\sim \sum_{\beta\in\{+,-\}}\bigl(\mathbb P\bigl( \xi^{(\beta)}_{\zeta_\beta} > x\bigr) + \mathbb P( U^\beta > x ) \bigr) \end{align*}

and

\begin{align*}H(x)\sim \sum_{\beta\in\{+,-\}} H^{(\beta)}(x)\sim \sum_{\beta\in B} H^{(\beta)}(x). \end{align*}

Recall from Section 2 that $K\,:\!=\, \mathbb E[\xi_{T_2}]/\mathbb E[T_2]$, and hence, when $\mathbb E[\xi_{T_2}]\neq0$, it follows that $K\neq0$. The main result of this section, which extends [Reference Maulik and Zwart21, Section 4, p. 166] to Lamperti–Kiu processes, is the following result.

Theorem 4.1. Suppose that Y is a Lamperti–Kiu process of strong subexponential type such that $\mathbb E[\xi_{T_2}]\in(-\infty,0)$. Then

(4.6) \begin{align} \mathbb P(A_\infty > x) \sim \dfrac{H(\log(x))}{\mathbb E[T_2]|K|}, \quad as \ x\rightarrow\infty.\end{align}

Furthermore, $A_\infty$ is long-tailed and $\log(A_\infty)$ is subexponential.

The proof of Theorem 4.1 requires a number of intermediary lemmas. These lemmas are stated with proofs below, followed by the proof of Theorem 4.1 at the end of this section.

Set $Z_n \,:\!=\, \xi_{\sigma_n}-\xi_{\sigma_{n-1}}$ for each $n\geq1$. We are now in a position to consider the asymptotic behaviour of the survival function of $Z_n$ conditioned on $K_{n-1}$ and $K_n$, under the assumptions of Theorem 4.1. Recall that for $\alpha\in\{+,-\}$, $H^{(\alpha)} = H_{\xi_\alpha}(x) + H_{U_{-\alpha}}(x)$.

Lemma 4.6. Suppose that Y is a Lamperti–Kiu process of strong subexponential type such that $\mathbb E[\xi_{T_2}]\in(-\infty,0)$, and fix $\alpha,\beta\in\{+,-\}$. Then, if $\beta\in B$, for each $n\in\mathbb N$,

\begin{align*} \limsup_{x\rightarrow\infty}\dfrac{\mathbb P(Z_n >x \mid K_{n-1}=\alpha,K_n=\beta) }{H^{(\beta)}(x)} \leq \dfrac{1}{\eta^{(\alpha,\beta)}|K+\epsilon|\,\mathbb E[T_2]}.\end{align*}

Furthermore, if $\beta\in\{+,-\}\setminus B$ and $b\in B$, then, for each $n\in\mathbb N$,

\begin{align*} \limsup_{x\rightarrow\infty}\dfrac{\mathbb P(Z_n >x \mid K_{n-1}=\alpha,K_n=\beta) }{H^{(b)}(x)} = 0.\end{align*}

Proof. Suppose that $x>A$, let $u_0\in(0,x-A)$, and fix $\alpha,\beta\in\{+,-\}$. For ease of notation, let $\sigma\,:\!=\, \sigma_1$. Then, since

\begin{equation*} \mathbb P(Z_n >x \mid K_{n-1}=\alpha,K_n=\beta ) = \mathbb P_\alpha(\xi_{\sigma} > x \mid \sigma < \infty, \: J_{\sigma}=\beta), \end{equation*}

we have

\begin{align*}& \mathbb P(Z_n >x \mid K_{n-1}=\alpha,K_n=\beta ) = \dfrac{1}{\eta^{(\alpha,\beta)}}\sum_{m=0}^\infty \mathbb P_\alpha (\xi_{\sigma}>x ;\, T_{m}\leq \sigma < T_{m+1} ;\, J_\sigma=\beta ).\end{align*}

To bound the elements of the sum, first consider the strict inequality $T_{m}<\sigma<T_{m+1}$ for some $m\in\mathbb N$; then

\begin{align*} \mathbb P_\alpha ( \xi_\sigma>x ;\, T_{m}<\sigma<T_{m+1} ;\, J_\sigma=\beta ) \leq \mathbb P_\alpha\biggl(\sup_{T_{m}< u<T_{m+1}} \xi_u >x ;\, \xi_{T_{m}}<(K+\epsilon)T_{m}+A ;\, J_{T_m}=\beta \biggr){,}\end{align*}

and using the Lamperti–Kiu decomposition followed by Lemma 4.1, we have

\begin{align*} \mathbb P_\alpha ( \xi_\sigma>x ;\, T_{m}<\sigma<T_{m+1} ;\, J_\sigma=\beta ) &\leq \mathbb P_\alpha\biggl( \sup_{0< u<\tilde{\zeta}_\beta} \tilde{\xi}^{(\beta)}_u > x - (K+\epsilon)T_m - A ;\, J_{T_m}=\beta \biggr)\\[3pt] & \leq \dfrac{\mathbb P_\alpha\bigl( \tilde{\xi}^{(\beta)}_{\tilde{\zeta}_\beta} \geq x-(K+\epsilon)T_m-A-u_0 ;\, J_{T_m}=\beta \bigr)}{ \mathbb P\bigl( \xi^{(\beta)}_{\zeta_\beta} \geq -u_0 \bigr) }\end{align*}

where $\tilde{\xi}^{(\beta)}$ and $\tilde{\zeta}_\beta$ are independent copies of the Lévy process $\xi^{(\beta)}$ and the exponential random variable $\zeta_\beta$, respectively.

In the case when $\sigma=T_{m}$,

\begin{align*} \mathbb P_\alpha (\xi_\sigma>x ;\, T_{m}=\sigma ;\, J_\sigma=\beta ) &\leq\mathbb P_\alpha ( \xi_{T_{m}}>x ;\, \xi_{T_m-} \leq (K+\epsilon)T_{m}+A ;\, J_{T_m}=\beta )\\ &\leq\mathbb P_\alpha ( \xi_{T_{m}} - \xi_{T_m-} > x - (K+\epsilon)T_{m}-A ;\, J_{T_m}=\beta ) \\&= \mathbb P_\alpha ( U^{(-\beta)}>x-(K+\epsilon)T_{m} - A ;\, J_{T_m}=\beta ).\end{align*}

If $\alpha=\beta$ then there must be an even number of changes of J before $\sigma$, so there exists $m\in\mathbb N$ such that $\sigma\in[T_{2m},T_{2m+1})$. Hence combining the two results above gives

\begin{align*} \mathbb P ( Z_n >x \mid K_{n-1}=\alpha,K_n=\beta ) \leq& \dfrac{1}{\eta^{(\alpha,\beta)}} \biggl(\sum_{m=0}^\infty \dfrac{\mathbb P_\alpha\bigl(\tilde{\xi}_{\tilde{\zeta}_\beta}^{(\beta)} \geq x-(K+\epsilon)T_{2m}-A-u_0\bigr)}{\mathbb P\bigl(\xi^{(\beta)}_{\zeta_\beta}>-u_0\bigr)} \\* & + \sum_{m=0}^\infty\mathbb P_\alpha ( U^{(-\beta)} \geq x- (K+\epsilon)T_{2m} - A -u_0 ) \biggr).\end{align*}

If $\alpha\neq\beta$ then there is an odd number of changes in J before time $\sigma$. However, $T_{2m+1}\geq T_{2m}$, so the inequalities can be weakened to give the same result as the $\alpha=\beta$ case.

For ease of notation define

\begin{align*} Q_\beta(u_0) \,:\!=\, \mathbb P\bigl( \xi_{\zeta_\beta}^{(\beta)} \geq -u_0 \bigr)\leq 1.\end{align*}

If $\xi^{(\beta)}_{\zeta_\beta}$ is long-tailed, we can use Lemma 4.2 to obtain the asymptotic approximation

\begin{align*} \sum_{m=0}^\infty \mathbb P_\alpha\bigl( \tilde{\xi}^{(\beta)}_{\tilde{\zeta}_\beta} \geq x-(K+\epsilon)T_{2m}-A-u_0 \bigr) &\sim \dfrac{G_{\xi_\beta}(x)}{|K+\epsilon|\,\mathbb E[T_2]},\end{align*}

as $x\rightarrow\infty$. Similarly, if $U^{(-\beta)}$ is long-tailed we have

\begin{align*} \sum_{m=0}^\infty\mathbb P_{\alpha} ( U^{(-\beta)}> x-(K+\epsilon)T_{2m} -A -u_0 ) &\sim \dfrac{G_{U^{(-\beta)}}(x)}{|K+\epsilon|\,\mathbb E[T_2]},\end{align*}

as $x\rightarrow\infty$. We will consider separately the cases where both of the asymptotics hold, exactly one holds, or neither hold.

In the case when both $\xi^{(\beta)}_{\zeta_\beta}$ and $U^{(-\beta)}$ are subexponential (and hence are long-tailed) for all $\delta>0$, there exists an $R>0$ such that, for all $x>R$,

\begin{align*} \dfrac{\mathbb P (Z^{(\alpha,\beta)} > x )}{G_{\xi_\beta}(x) + G_{U^{(-\beta)}}(x)} &\leq \dfrac{1}{ (G_{\xi_\beta}(x)+G_{U^{(-\beta)}}(x) )\eta^{(\alpha,\beta)}} \\[3pt] &\quad \, \times \biggl(\dfrac{(1+\delta)G_{\xi_\beta}(x)}{Q_\beta(u_0)|K+\epsilon|\,\mathbb E[T_2]} + \dfrac{(1+\delta)G_{U^{(-\beta)}}(x)}{|K+\epsilon|\,\mathbb E[T_2]} \biggr)\\[3pt] &\leq \dfrac{(1+\delta)}{\eta^{(\alpha,\beta)}Q_\beta(u_0)|K+\epsilon|\,\mathbb E[T_2]}{,}\end{align*}

where the second inequality holds since $Q_\beta(u_0)<1$. Since $\delta$ was arbitrary, taking the $\limsup$ as $x\rightarrow\infty$ yields

\begin{align*} \limsup_{x\rightarrow\infty} \dfrac{\mathbb P (Z_n >x \mid K_{n-1}=\alpha,K_n=\beta )}{G_{\xi_\beta}(x) + G_{U^{(-\beta)}}(x)} \leq \dfrac{1}{\eta^{(\alpha,\beta)}Q_\beta(u_0)|K+\epsilon|\,\mathbb E[T_2]}.\end{align*}

In the case when exactly one of $\xi^{(\beta)}_{\zeta_\beta}$ and $U_{-\beta}$ is subexponential, it asymptotically dominates the other as $x\rightarrow\infty$, since Y is of strong subexponential type. Suppose that it is $\xi^{(\beta)}_{\zeta_\beta}$ that is subexponential and note that the following argument is symmetric in $\xi^{(\beta)}_{\zeta_\beta}$ and $U^{(-\beta)}$. For all $\delta>0$, there exists $\hat{\delta}>0$ such that $\hat{\delta}(1+\hat{\delta})<\delta/2$, and an $R>0$ such that for all $x>R$ and $n\in\mathbb N$,

\begin{align*} \mathbb P_{\alpha} ( U^{(-\beta)}> x-(K+\epsilon)T_{2n} -A -u_0 ) \leq \hat{\delta}\mathbb P_\alpha\bigl( \tilde{\xi}^{(\beta)}_{\tilde{\zeta}_\beta} \geq x-(K+\epsilon)T_{2n}-A-u_0 \bigr).\end{align*}

Thus, for all $x>R$, for suitably large R,

\begin{align*} \sum_{m=0}^\infty \mathbb P_{\alpha} ( U^{(-\beta)}> x-(K+\epsilon)T_{2m} -A -u_0 ) \leq \dfrac{\hat{\delta}(1+\hat{\delta})G_{\xi_\beta}(x)}{|K+\epsilon|\,\mathbb E[T_2]},\end{align*}

which gives for $x>R$

\begin{align*} \dfrac{\mathbb P (Z_n >x \mid K_{n-1}=\alpha,K_n=\beta )}{G_{\xi_\beta}(x) + G_{U^{(-\beta)}}(x)} &\leq \dfrac{1}{\eta^{(\alpha,\beta)}G_{\xi_\beta}(x)} \biggl(\dfrac{(1+{\delta}/{2})G_{\xi_\beta}(x)}{|K+\epsilon|\,\mathbb E[T_2]Q_\beta(u_0)} + \dfrac{({\delta}/{2}) G_{\xi_\beta}}{|K+\epsilon|\,\mathbb E[T_2]} \biggr)\\[3pt] &\leq \dfrac{1+\delta}{\eta^{(\alpha,\beta)}|K+\epsilon|\,\mathbb E[T_2]Q_\beta(u_0)}.\end{align*}

Therefore, since $\delta>0$ was arbitrary, we may take the $\limsup$ first as $\delta\rightarrow0$ and then as $x\rightarrow\infty$ to obtain

\begin{align*} \limsup_{x\rightarrow\infty} \dfrac{\mathbb P (Z_n >x \mid K_{n-1}=\alpha,K_n=\beta )}{G_{\xi_\beta}(x) + G_{U^{(-\beta)}}(x)} \leq \dfrac{1}{\eta^{(\alpha,\beta)}|K+\epsilon|\,\mathbb E[T_2]Q_\beta(u_0)}.\end{align*}

Finally, we consider the case when neither is subexponential. Since Y is of strong subexponential type, the tails of $\tilde{\xi}_{\tilde{\zeta}_\beta}^{(\beta)}$ and $U^{(-\beta)}$ are dominated by the tails of at least one of either $\tilde{\xi}_{\tilde{\zeta}_{b}}^{(b)}$ or $U^{(-b)}$. Denote the dominating random variable by X and let $W\in\bigl\{ \tilde{\xi}_{\tilde{\zeta}_\beta}^{(\beta)} , U^{(-\beta)} \bigr\}$. Following the above calculation, for all $\delta>0$ there exists an $R>0$ such that for any $x>R$ and $n\in\mathbb N$,

\begin{align*} \mathbb P_\alpha ( W \geq x-(K+\epsilon)T_{2n}-A-u_0 ) \leq \delta\mathbb P_\alpha ( X>x-(K+\epsilon)T_{2n}-A-u_0 ).\end{align*}

Then, by using the results of the previous two cases, for suitably large $R>0$ we have

\begin{align*} \sum_{n=0}^\infty \mathbb P_\alpha (W\geq x-(K+\epsilon)T_{2n}-A-u_0 ) &\leq \dfrac{\delta(1+\delta)G_X(x)}{|K+\epsilon|\,\mathbb E[T_2]H_b(u_0)}{.}\end{align*}

Hence, for $x>R$,

\begin{align*} \dfrac{\mathbb P ( Z_n >x \mid K_{n-1}=\alpha,K_n=\beta )}{G_{\xi_{-b}}(x)+G_{U^{(b)}}(x)} &\leq \dfrac{2\delta(1+\delta)}{\eta^{(\alpha,\beta)}|K+\epsilon|\,\mathbb E[T_2]Q_b(u_0)}\end{align*}

so, as $x\rightarrow\infty$, since $\delta>0$ was arbitrary,

\begin{align*} \mathbb P \big( Z^{(\alpha,\beta)}>x \big) = {\mathrm{o}}( G_{\xi_{b}}(x)+G_{U^{(-b)}}(x) ).\end{align*}

Since all the components of the Lamperti–Kiu decomposition are finite, for sufficently large x we have $G_{(\cdot)}(x) = H_{(\cdot)}(x)$. Hence, in the first two cases, we obtain

\begin{align*} \limsup_{x\rightarrow\infty}\dfrac{\mathbb P(Z_n >x \mid K_{n-1}=\alpha,K_n=\beta )}{H_{\xi_\beta}(x) + H_{U^{(-\beta)}}(x)} \leq \dfrac{1}{\eta^{(\alpha,\beta)} Q(u_0)|K+\epsilon|\,\mathbb E[T_2]}.\end{align*}

As we are taking the limit $x\rightarrow\infty$, we may also take $u_0\rightarrow\infty$ and use the fact that $Q(u_0)\rightarrow1$ to obtain

\begin{align*} \limsup_{x\rightarrow\infty}\dfrac{\mathbb P(Z_n >x \mid K_{n-1}=\alpha,K_n=\beta) }{H_{\xi_\beta}(x) + H_{U^{(-\beta)}}(x)} \leq \dfrac{1}{\eta^{(\alpha,\beta)}|K+\epsilon|\,\mathbb E[T_2]}{,}\end{align*}

whilst in the third case

\begin{align*} \limsup_{x\rightarrow\infty}\dfrac{\mathbb P(Z_n >x \mid K_{n-1}=\alpha,K_n=\beta) }{H_{\xi_{-\beta}}(x) + H_{U^{(\beta)}}(x)} = 0.\end{align*}

This completes the proof.

The following lemma provides a way to use a bound on a random variables distribution function to produce an inequality between random variables.

Proof. Let $P(x) \,:\!=\, \mathbb P(Z\geq x)$ and $V\sim \text{Unif}(0,1)$ be independent of Z. We will use the notation $P(x^+)\,:\!=\, \lim_{y\downarrow x}P(y)$, which exists for all $x\geq0$ since P is non-increasing and bounded from below. Define the random function $U\colon \mathbb R^+\rightarrow[0,1]$ by setting $ U(x) \,:\!=\, P(x) - V(P(x)-P(x^+)).$ Let $x_1<x_2$; then since P is non-increasing, $P(x^{+})\leq U(x)\leq P(x)$ for all $x\in\mathbb R$ and

\begin{align*} U(x_2) \leq P(x_2) \leq P(x_1^+) = P(x_1) -1(P(x_1) - P(x_1^+)) \leq U(x_1),\end{align*}

hence U is also non-increasing.

Furthermore, suppose $U(x_1)=U(x_2)$ for some $x_1<x_2$. Then $ P(x_1^+) \leq U(x_1) = U(x_2) \leq P(x_2) \leq P(x_1^+) ,$ where the last inequality is because P is non-increasing and so $P(x_1^+)=P(x_2)$. If $P(x_1^+)=P(x_1)$, then we have $P(x_2)=P(x_1)$. Otherwise we have

\begin{align*} U(x_1) > P(x_1^+) \geq P(x_2) > U(x_2).\end{align*}

This is a contradiction. Hence, if $x_1<x_2$ and $U(x_1)=U(x_2)$, then $P(x_1^+)=P(x_1)=P(x_2)$ a.s., so $\mathbb P( x_1 \leq Z < x_2) = 0$.

From this we can now conclude that, for all $x\in\mathbb R$, we have

\begin{align*} \mathbb P( U(z) \leq U(x) ) &=\mathbb P(Z \geq x ) + \mathbb P( Z<x ;\, U(Z)=U(x) )\\[4pt] &=P(x) + \mathbb P(Z<x ;\, U(Z)=U(x)).\end{align*}

However, by the above calculation, we get

\begin{align*} \mathbb P(Z<x ;\, U(Z)=U(x)) \leq \mathbb P(Z<x ;\, P(Z)=P(x) ) =0\end{align*}

and hence, for all $x\in\mathbb R$, we have $ \mathbb P(U(z) \leq U(x) ) = P(x).$

Now let $q\in[0,1]$ and suppose that there exists $x\in\mathbb R^+$ with $P(x^+)=P(x)=q$, so

\begin{align*} \mathbb P( U(Z) \leq q) &= \mathbb P( U(Z) \leq P(x) )\\ &= \mathbb P( U(Z) \leq U(x) ) = q.\end{align*}

If there is not such an x, then, since $\lim_{x\rightarrow-\infty}P(x)=1$ and $\lim_{x\rightarrow\infty}P(x)=0$, there exists $x\in\mathbb R^+$ such that $q\in[P(x^+),P(x))$. For any $y>x$, we have $ U(y) \leq P(y) \leq P(x^+),$ so $U(y)\notin(P(x^+),P(x))$ and similarly, for any $y<x$, we have $ U(y) \geq U(y^+) \geq P(x) ,$ so $U(y)\notin(P(x^+),P(x))$. Hence $U(Z)\in(q,P(x))\subset(P(x^+),P(x))$ implies $Z=x$.

Next we want to consider $\mathbb P( P(x)=U(Z) ;\, Z\neq x)$. Notice that if $z>x$ then $U(z)\leq P(x^+)<P(x)$ and so $\mathbb P( P(x)=U(z) ;\, Z>x)=0$. If $z<x$ and $P(x)=U(z)$, then $P(x)=U(z)\geq P(z^+)\geq P(x)$ and so $P(x)=P(z^+)$. However, if there is a discontinuity point $y\in[z,x)$, then $U(z)\geq P(z^+)\geq P(y)>P(y^+)\geq P(x)$ so $\mathbb P(P(x)=U(z) ;\, Z<x)=0$, and we conclude that $P(z)=P(z^+)=P(x)$ and hence $ \mathbb P( P(x)=U(Z) ;\, Z\neq x ) = 0.$

From this and since V is uniformly distributed on [0,1], we have

\begin{align*} \mathbb P(U(Z)\in(q,P(x)]) &= \mathbb P(Z=x)\,\mathbb P(U(x)\in(q,P(x)]) + \mathbb P(Z\neq x ;\, U(Z)=P(x) )\\ &= \mathbb P(Z=x)\,\mathbb P\biggl( \dfrac{q-P(x^+)}{P(x)-P(x^+)} < V \leq \dfrac{P(x)-P(x^+)}{P(x)-P(x^+)}\biggr) + 0\\ &= (P(x)-P(x^+) )\dfrac{q-P(x^+)}{P(x)-P(x^+)},\end{align*}

and hence

\begin{align*} \mathbb P( U(Z) \leq q ) &= \mathbb P( U(Z) \leq P(x) ) - \mathbb P( U(Z) \in (q,P(x)] )\\ &= P(x) - (P(x) - P(x^+))\dfrac{(P(x)-q)}{P(x) - P(x^+)} =q.\end{align*}

Hence U(Z) is uniformly distributed on [0,1]. Using [Reference Embrechts and Hofert9, Proposition 3.1], the random variable defined by $X\,:\!=\, U^{-}(Z)\,:\!=\, \inf \{ x\in\mathbb R^+ \mid F(x) < U(Z) \}$ has the distribution F. Moreover, since $F(x)\geq P(x)$ for all $x\in\mathbb R^+$, we have

\begin{align*} F(Z) \geq P(Z) \geq P(Z) - V(P(Z)-P(Z^+)) = U(Z)\end{align*}

and, since F is non-increasing, we have $ X = \inf \{ x\in\mathbb R^+ \mid F(x) < U(Z) \} \geq Z,$ as required.

We are now able to derive the upper bound of Theorem 4.1.

Lemma 4.8. If Y is a Lamperti–Kiu process of strong subexponential type, then the right tail of $A_\infty$ satisfies

(4.7) \begin{equation} \limsup_{x\rightarrow\infty}\dfrac{\mathbb P(\log(A_\infty)>x)}{H(x)} \leq \dfrac{1}{|K|\,\mathbb E[T_2]},\end{equation}

where H is the integrated tail from Theorem 4.1.

Proof. Fix $\sigma\in\{+,-\}$ and let $\delta_2>0$. For sufficiently large $A>0$, by Proposition 4.1, we know $\mathbb E_\sigma[N(\alpha,\beta)]/\eta^{(\alpha,\beta)}\leq \mathbb{1}_{\{\sigma=\alpha\}} + \delta_2$. Now fix such an $A>0$ and let $\delta_1>0$.

From Lemma 4.4 we have $ \log(A_\infty) \leq (N+1)C + \sum_{i=1}^NZ_i^+.$

For each $i\in\mathbb N$, we have a tail estimate for $Z_i^+$, given $(K_n)_{n\in\mathbb N_0}$, from Lemma 4.6, which, used in conjunction with Lemma 4.7, gives the existence of random variables $X_i(k)$ for each $k\in\cup_{n\in\mathbb N}\{+,-\}^n$ with $k_0=\sigma$ such that:

1. (1) each $X_i(k)$ is a function of $Z_i$ and a random variable independent of the rest of the system,

2. (2) $X_i(k)\geq (Z_i^+ +C)\mathbb{1}_{\{N=n ;\, (K_0,\ldots,K_n)=k\}}$,

3. (3) $X_i(k)$ has tails given by $\min (1,H^{(k_i)}(x)(\eta^{(k_{i-1},k_i)}|K+\epsilon|\,\mathbb E[T_2]|)^{-1} )$ if $k_i\in B$ and tails which are ${\mathrm{o}}(\min(1,H^{(b)}(x)))$ for $b\in B$ if $k_i\notin B$.

Then, summing up over the sample paths of $(K_n)_{n\in\mathbb N_0}$, we have the upper bound

\begin{align*} \mathbb P_\sigma\biggl( \sum_{i=1}^N Z_i^+ +C>x \biggr) &\leq\sum_{n\in\mathbb N}\sum_{\substack{ k\in\{+,-\}^{n+1} \\ k_0=\sigma } }\mathbb P_\sigma\biggl( \sum_{i=1}^n X_i(k)>x ;\, N=n ;\, (K_0,\ldots,K_n)=k \biggr).\end{align*}

For ease of notation, let $\bar{\eta}=\max_{\alpha,\beta\in\{+,-\}}\eta_{\alpha,\beta}$, $\underline{\eta}=\min_{\alpha,\beta\in\{+,-\}}\eta_{\alpha,\beta}$, and $d\in(0,(1-2\bar{\eta})/2\bar{\eta})$.

For each $\alpha,\beta\in\{+,-\}$, $n\in\mathbb N$ and $k\in\{+,-\}^{n+1}$ such that $k_0=\sigma$, let

\begin{equation*}n_{\alpha,\beta}(k)\,:\!=\, \sum_{i=0}^n\mathbb{1}_{\{k_{i-1}=\alpha,k_i=\beta\}}.\end{equation*}

Then, by Lemma 4.5, given the event $\{N=n ;\, (K_0,\ldots,K_n)=k\}$, the sum

\begin{equation*}Y_{\alpha,\beta}(k)\,:\!=\, \sum_{i=1}^nX_i\mathbb{1}_{\{k_{i-1}=\alpha,k_i=\beta\}}\end{equation*}

is a sum of $n_{\alpha,\beta}(k)$ i.i.d. random variables, and hence, in the case $\beta\in B$, from Kesten’s bound [Reference Foss, Korshunov and Zachary11, Theorem 3.34],

\begin{align*} & \mathbb P_\sigma\biggl( \sum_{i=1}^nX_i(k)\mathbb{1}_{\{k_{i-1}=\alpha,k_i=\beta\}} >x \mid N=n ;\, (K_0,\ldots,K_n)=k \biggr)\\ &\quad \leq c(d)(1+d)^{n_{\alpha,\beta}(k)}\mathbb P_\sigma(X_1((\alpha,\beta))>x)\\ &\quad \leq \dfrac{c(d)(1+d)^nF(x)}{\underline{\eta}|K+\epsilon|\,\mathbb E[T_2]}.\end{align*}

In the case $\beta\notin B$, since Y is of strong subexponential type, for any $b\in B$,

\begin{align*} & \mathbb P_\sigma\biggl( \sum_{i=1}^nX_i(k)\mathbb{1}_{\{k_{i-1}=\alpha,k_i=\beta\}} >x \mid N=n ;\, (K_0,\ldots,K_n)=k \biggr)\\ &\quad \leq \mathbb P_\sigma\biggl( \sum_{i=1}^nW_i\mathbb{1}_{\{k_{i-1}=\alpha,k_i=\beta\}} >x \mid N=n ;\, (K_0,\ldots,K_n)=k \biggr)\\&\quad \leq \dfrac{c(d)(1+d)^nH(x)}{\underline{\eta}|K+\epsilon|\,\mathbb E[T_2]},\end{align*}

where for each $i\in\mathbb N$ the random variable $W_i$ depends only on $X_i(k)$ and has the distribution of $X_1((\alpha,b))$. We can now use Corollaries 3.16 and 3.18 in [Reference Foss, Korshunov and Zachary11, Chapter 3, p. 52] to sum the $Y_{\alpha,\beta}$ and obtain the bound

\begin{align*} &\mathbb P_\sigma\biggl( \sum_{i=1}^nX_i(k) >x \mid N=n ;\, (K_0,\ldots,K_n)=k \biggr)\\[4pt] &\quad =\mathbb P_\sigma\biggl( \sum_{\alpha,\beta\in\{+,-\}}\sum_{i=1}^nX_i(k)\mathbb{1}_{\{k_{i-1}=\alpha,k_i=\beta\}} >x \mid N=n ;\, (K_0,\ldots,K_n)=k \biggr)\\[4pt] &\quad \leq \dfrac{4c(d)(1+d)^nH(x)}{\underline{\eta}|K+\epsilon|\,\mathbb E[T_2]}.\end{align*}

Using the bound on the distribution of N from (4.4), we see that there is an $M\in\mathbb N$ such that $\mathbb E[\underline{\eta}^{-1}4c(d)(1+d)^N ;\, N>M]\leq \delta_1$ for sufficiently small d. This then gives

\begin{align*} \mathbb P_\sigma\biggl( \sum_{i=1}^N X_i((K_n)_{n\in\mathbb N})>x ;\, N>M \biggr) \leq \dfrac{\delta_1 H(x)}{|K+\epsilon|\,\mathbb E[T_2]}.\end{align*}

Moreover, by [Reference Foss, Korshunov and Zachary11, Corollary 3.16], for all $n\leq M$ and $k\in\{+,-\}^{n+1}$ with $k_0=\sigma$ there exists $R_{n,k}>0$ such that, for all $x>R_{n,k}$,

\begin{align*} & \mathbb P_\sigma\biggl( \sum_{i=1}^n X_i(k)>x \mid N=n ;\, (K_0,\ldots,K_n)=k \biggr) \\ &\quad \leq \biggl( \biggl(1+\dfrac{\delta_1}{2}\biggr)\sum_{\alpha\in\{+,-\}}\sum_{\beta\in B} n_{\alpha,\beta}(k)\dfrac{H^{\beta}(x)}{\eta^{(\alpha,\beta)}|K+\epsilon|\,\mathbb E[T_2]}\biggr) \bar{*} \mathbb P_\sigma\biggl( \sum_{i=1}^n \mathbb{1}_{\{k_i \notin B\}} X_i(k) > x \biggr)(x),\end{align*}

where for two survival functions $\bar{F}$ and $\bar{G}$ we write $\bar{F}\bar{*}\bar{G}$ for the survival function of the convolution of the distribution functions $F \,:\!=\, 1-\bar{F}$ and $G\,:\!=\, 1-\bar{G}$. Then, since

\begin{equation*}\mathbb P_\sigma ( \mathbb{1}_{\{k_i\notin B\}}X_i(k))>x ) = {\mathrm{o}}( H^{\beta}(x) )\quad \text{for any $\beta\in B$,}\end{equation*}

by [Reference Foss, Korshunov and Zachary11, Corollary 3.18] we have

\begin{align*}& \mathbb P_\sigma\biggl( \sum_{i=1}^n X_i(k)>x \mid N=n ;\, (K_0,\ldots,K_n)=k \biggr) \\ &\quad \leq (1+\delta_1 )\sum_{\alpha\in\{+,-\}}\sum_{\beta\in B} n_{\alpha,\beta}(k)\dfrac{H^{\beta}(x)}{\eta^{(\alpha,\beta)}|K+\epsilon|\,\mathbb E[T_2]}.\end{align*}

Since there are finitely many such pairs (n,k) with $n\leq M$, we can take $R\,:\!=\, \max_{n\leq M}R_{n,k}$ so that for all $x>R$ we get

\begin{align*} &\sum_{n=1}^{M}\sum_{ \substack{ k\in\{+,-\}^{n+1} \\ k_0=\sigma }} \mathbb P_\sigma\biggl( \sum_{i=1}^nX_i(k)>x ;\, N=n ;\, (K_0,\ldots,K_n)=k \biggr)\\ &\quad \leq (1+\delta_1)\sum_{n=1}^{M}\sum_{\substack{ k\in\{+,-\}^{n+1} \\ k_0=\sigma }}\sum_{\alpha\in\{+,-\}}\sum_{\beta\in B}n_{\alpha,\beta}(k)\dfrac{H^{(\beta)}(x)}{\eta^{(\alpha,\beta)}|K+\epsilon|\,\mathbb E[T_2]}\\ &\quad \quad \, \times \mathbb P_\sigma(N=n ;\, (K_0,\ldots,K_n)=k) \\ &\quad = (1+\delta_1)\sum_{\alpha\in\{+,-\}}\sum_{\beta\in B}\,\mathbb E_\sigma[n_{\alpha,\beta}(k) ;\, N\leq M]\dfrac{H^{(\beta)}(x)}{\eta^{(\alpha,\beta)}|K+\epsilon|\,\mathbb E[T_2]}\\ &\quad \leq \dfrac{(1+\delta_1)}{|K+\epsilon|\,\mathbb E[T_2]}\sum_{\alpha\in\{+,-\}}\sum_{\beta\in B}\dfrac{\mathbb E_\sigma[n_{\alpha,\beta}(k)]H^{(\beta)}(x)}{\eta^{(\alpha,\beta)}}.\end{align*}

Next, by our choice of A we have $\mathbb E_\sigma[n(\alpha,\beta)]/\eta^{(\alpha,\beta)}\leq \mathbb{1}_{\sigma=\alpha} + \delta_2$ and thus

\begin{align*} &\sum_{n=1}^{M}\sum_{\substack{ k\in\{+,-\}^{n+1} \\ K_0=\sigma }} \mathbb P_\sigma\biggl( \sum_{i=1}^nX_i(k)>x ;\, N=n ;\, (K_0,\ldots,K_n)=k \biggr)\\ &\quad \leq \dfrac{(1+\delta_1)}{|K+\epsilon|\,\mathbb E[T_2]}\sum_{\alpha\in\{+,-\}}\sum_{\beta\in B} ( \mathbb{1}_{\{\sigma=\alpha\}} + \delta_2 )H^{(\beta)}(x)\\ &\quad = \dfrac{(1+\delta_1)}{|K+\epsilon|\,\mathbb E[T_2]}(1+2\delta_2)\sum_{\beta\in B}H^{(\beta)}(x)\\ &\quad \leq\dfrac{(1+\delta_1)^2}{|K+\epsilon|\,\mathbb E[T_2]}(1+2\delta_2)H(x),\end{align*}

where the last inequality holds for sufficiently large x since Y is of strong subexponential type. Hence, for all $x>R$, we have

\begin{align*} \mathbb P_\sigma\biggl( \sum_{i=1}^N ( Z_i^+ +C ) >x \biggr) \leq \dfrac{(1+\delta_1)^2(1+2\delta_2)}{|K+\epsilon|\,\mathbb E[T_2]}F(x) +\dfrac{\delta_1 H(x)}{|K+\epsilon|\,\mathbb E[T_2]},\end{align*}

and so from the definition of $\limsup$,

\begin{align*} \limsup_{x\rightarrow\infty}\dfrac{\mathbb P_\sigma ( C + \sum_{i=1}^N (Z_i^++C )>x )}{H(x-C)} \leq \dfrac{1+2\delta_2}{|K+\epsilon|\,\mathbb E[T_2]}.\end{align*}

However, because H is long-tailed, $\lim_{x\rightarrow\infty}H(x-C)/H(x) =1$ and therefore

\begin{align*} \limsup_{x\rightarrow\infty}\dfrac{\mathbb P_\sigma ( C + \sum_{i=1}^N (Z_i^++C )>x )}{H(x)} \leq \dfrac{1+2\delta_2}{|K+\epsilon|\,\mathbb E[T_2]}.\end{align*}

Then, by comparison,

\begin{align*} \limsup_{x\rightarrow\infty} \dfrac{\mathbb P_\sigma(\log(A_\infty)>x)}{H(x)} &\leq \dfrac{1+2\delta_2}{|K+\epsilon|\,\mathbb E[T_2]}{,}\end{align*}

and since both $\epsilon$ and $\delta_2$ were arbitrary, the result follows.

It remains to show that the lower bound for $\liminf_{x\rightarrow\infty}\mathbb P(A_\infty>x)$ also holds. To this end, for each $x\in\mathbb R$, define the stopping time

\begin{equation*}\tau_d(x) \,:\!=\, \inf \{ T_{2n} \mid n\in\mathbb N, \xi_{T_{2n}} \geq x \} \end{equation*}

and notice that $\tau_d(x)<\infty$ if and only if $\sup_{n\in\mathbb N}\xi_{T_{2n}}\geq x$. Furthermore, $J_{\tau_d(x)}=J_0$ whenever $\tau_d(x)<\infty$. Since $\xi_{T_2}$ is strong subexponential, its integrated tail, H, is subexponential and thus, by [Reference Zachary29, Theorem 1(ii)], $\mathbb P(\tau_d(x)<\infty)$ is also subexponential. Then, by considering the random walk $ (\xi_{T_{2n}} )_{n\in\mathbb N}$ in the place of $ ( \xi_n )_{n\in\mathbb N}$ in the proof of [Reference Maulik and Zwart21, Lemma 4.3], we find that for every $y>0$

(4.8) \begin{equation}\lim_{x\rightarrow\infty} \mathbb P (\xi_{\tau_d(x)}-x>y \mid \tau_d(x)<\infty ) = 1. \end{equation}

We are now able to prove Theorem 4.1.

Proof of Theorem 4.1. Equation (4.6) of Theorem 4.1 follows from inequality (4.7) of Lemma 4.8 and the inequality

\begin{equation*}\liminf_{x\rightarrow\infty} \dfrac{\mathbb P (\log (A_\infty ) > x )}{H(x)} \geq \dfrac{1}{|\mathbb E[\xi_{T_2}]|} = \dfrac{1}{\mathbb E[T_2]|K|},\end{equation*}

which we will now prove.

The following inequality is immediate:

\begin{align*} \mathbb P(\log A_\infty> x) &\geq \mathbb P\biggl( \log\biggl( \int_{\tau_d(x)}^\infty |Y_t| \,{\mathrm{d}} t \biggr) > x ;\, \tau_d(x)<\infty \biggr).\end{align*}

Applying the Markov additive property and recalling $J_{\tau_d(x)}=J_0$ gives

\begin{align*} \mathbb P(\log A_\infty> x) &\geq\mathbb P ( \xi_{\tau_d(x)} + \log(\hat{A}_{\infty,J_0}) >x \mid \tau_d(x)<\infty )\,\mathbb P(\tau_d(x)<\infty),\end{align*}

where $\hat{A}_{\infty,j}$ is an independent and identically distributed copy of $A_\infty$ with $\hat{J}_0=j$. Then, by applying (4.8), we have

\begin{align*} \liminf_{x\rightarrow\infty} \dfrac{\mathbb P ( \log A_{\infty} > x )}{H(x)} &\geq\liminf_{x\rightarrow\infty} \dfrac{\mathbb P(\tau_d(x)<\infty )}{H(x)} =\liminf_{x\rightarrow\infty} \dfrac{\mathbb P(\sup_{n\in\mathbb N}\xi(T_{2n})\geq x)}{H(x)}.\end{align*}

However, $\xi_{T_{2n}}$ is a sum of the random variables $\xi_{T_{2m}}-\xi_{T_{2(m-1)}}$, which are i.i.d. copies of $\xi_{T_2}$. Since Y is of strong subexponential type, the integrated tail, H, of $\xi_{T_2}$ is long-tailed, and we can apply Veraverbeke’s theorem [Reference Zachary29, Theorem 1(i)] to conclude that

\begin{align*} \qquad\quad\qquad\qquad\liminf_{x\rightarrow\infty}\dfrac{\mathbb P (\sup_{n\in\mathbb N}\xi(T_{2n}) \geq x )}{H(x)} \geq \dfrac{1}{|\mathbb E[\xi_{T_2}]|} = \dfrac{1}{|K|\,\mathbb E[T_2]}.\end{align*}