Ruin probabilities for a Lévy-driven generalised Ornstein–Uhlenbeck process

Abstract

We study the asymptotics of the ruin probability for a process which is the solution of a linear SDE defined by a pair of independent Lévy processes. Our main interest is a model describing the evolution of the capital reserve of an insurance company selling annuities and investing in a risky asset. Let $\beta >0$ be the root of the cumulant-generating function $H$ of the increment $V_{1}$ of the log-price process. We show that the ruin probability admits the exact asymptotic $Cu^{-\beta }$ as the initial capital $u\to \infty$, assuming only that the law of $V_{T}$ is non-arithmetic without any further assumptions on the price process.

Appendix A: Tails of solutions of distributional equations

1.1 A.1 Kesten–Goldie theorem

Here we present a short account of needed results on distributional equations (random equations in the terminology of [15]) of the form

$$ Y_{\infty } \stackrel{d}{=}Q+M Y_{\infty },\qquad Y_{\infty } \mbox{ independent of } (M,Q), $$

(A.1)

where $(M,Q)$ is an ${{\mathbb{R}}}^{2}$-valued random variable such that $M>0$ and $\mathbf{P}[M\neq 1]>0$ and $\stackrel{d}{=}$ is equality in law. This is a symbolic notation which means that we are given, in fact, a two-dimensional distribution ℒ on $(0,\infty )\times {\mathbb{R}}$ not concentrated on $\{1\}\times {\mathbb{R}}$, and the problem is to find a probability space with random variables $Y_{\infty }$ and $(M,Q)$ on it such that $Y_{\infty }$ and $(M,Q)$ are independent, $\mathcal{L}(M,Q)= \mathcal{L}$ and $\mathcal{L}(Y_{\infty })= \mathcal{L}(Q+M Y _{\infty })$. Uniqueness in this problem means uniqueness of the distribution of $Y_{\infty }$.

In the sequel, $(M_{j},Q_{j})$ form an i.i.d. sequence whose generic term $(M,Q)$ has the distribution ℒ and $Z_{j}:=M_{1}\cdots M_{j}$, $Z_{n}^{*}:=\sup _{j\le n}Z_{j}$.

If there is $p>0$ such that $\mathbf{E}[ M^{p}]<1$ and $\mathbf{E}[ |Q|^{p}]< \infty$, then the solution $Y_{\infty }$ of (A.1) can be easily realised on the probability space $(\Omega ,\mathcal{F},\mathbf{P})$ where the sequence $(M_{j},Q_{j})$ is defined; in fact, we can take the limit in $L^{p}$ of the series $\sum _{j\ge 0}Z_{j-1}Q_{j}$, see the beginning of the proof of Proposition 5.1.

The following classical result from renewal theory is the Kesten–Goldie theorem; see [15, Theorem 4.1].

Theorem A.1

Suppose that$(M,Q)$is such that the distribution of$\ln M$is non-arithmetic and, for some$\beta >0$,

$$\begin{aligned} {\mathbf{E}}[M^{\beta }]=1, \qquad {\mathbf{E}}[M^{\beta }(\ln M)^{+}]< \infty , \qquad {\mathbf{E}}[|Q|^{\beta }]< \infty . \end{aligned}$$

(A.2)

Then

$$\begin{aligned} \lim _{u\to \infty } u^{\beta }{\mathbf{P}}[Y_{\infty } >u] =&C_{+}< \infty , \\ \lim _{u\to \infty } u^{\beta }{\mathbf{P}}[Y_{\infty } < -u] =&C_{-}< \infty , \end{aligned}$$

where$C_{+}+C_{-}>0$.

Theorem A.1 leaves open the question when the constant $C_{+}$ is strictly positive. The expression

$$ C_{+}=\frac{\mathbf{E}[((Q+MY_{\infty } )^{+})^{\beta }- ((MY_{\infty } )^{+})^{\beta }]}{\beta {\mathbf{E}}[M^{\beta }\ln M] } $$

(A.3)

given in [15] and involving the unknown distribution of $Y_{\infty }$ is not helpful. How to check whether the right-hand side of this formula is strictly positive? Recently, Guivarc’h and Le Page [18] showed for the above case where the distribution of $\ln M$ is non-arithmetic that $C_{+}>0$ if and only if $Y_{\infty }$ is unbounded from above; see also Buraczewski and Damek [9] for simpler arguments. Of course, this criterion is not a result formulated in terms of the given data; it involves a property of the unknown distribution of $Y_{\infty }$, namely that the support is unbounded. But this property can be checked in the model considered in the present paper.

The remaining part of the appendix is a compendium of facts needed to cover also the arithmetic case.

1.2 A.2 Grincevic̆ius theorem

The theorem below is a simplified version of [17, Theorem 2(b)], but with a slightly weaker assumption on $Q$, namely $\mathbf{E}[|Q|^{\beta }]<\infty$, as used in our study. For the reader’s convenience, we give a complete proof after recalling some concepts and facts from renewal theory.

Theorem A.2

Suppose that (A.2) holds and the distribution of$\ln M$is concentrated on the lattice${{\mathbb{Z}}d}=\{0,\pm d, \pm 2d,\dots \}$, where$d>0$. Then

$$ \limsup _{u\to \infty } u^{\beta }{\mathbf{P}}[Y_{\infty } >u]< \infty . $$

We consider the convolution-type linear operator which is well defined for all positive as well as for (Lebesgue-) integrable functions by the formula

$$ \check{\psi }(x)=: \int ^{x}_{-\infty } e^{-(x-y)} \psi (y) d y. $$

Clearly, the functions $\psi$ and $\check{\psi }$ are simultaneously integrable or not and

$$ \int _{{\mathbb{R}}}\check{\psi }(x)d x= \int _{{\mathbb{R}}}\psi (x)d x. $$

Suppose that $\psi \ge 0$ is integrable. Then $\check{\psi }(x+\delta )\ge e^{-\delta }\check{\psi }(x)$ for any $\delta >0$ and

$$ \delta \inf _{x\in [j\delta , (j+1)\delta ]} \check{\psi }(x)\ge \delta e^{-\delta } \check{\psi }(j\delta )\ge e^{-2\delta } \int _{(j-1)\delta }^{j\delta }\check{\psi }(x)\,dx, $$

implying that

$$ \underline{U}(\check{\psi },\delta ):=\delta \sum _{j\in {\mathbb{Z}}} \inf _{x\in [j\delta , (j+1)\delta ]} \check{\psi }(x)\ge e^{-2\delta }\int _{{\mathbb{R}}}\check{\psi }(x)\,dx. $$

Similarly,

$$ \bar{U}(\check{\psi },\delta ):=\delta \sum _{j\in {\mathbb{Z}}} \sup _{x\in [j\delta , (j+1)\delta ]} \check{\psi }(x)\le e^{2\delta } \int _{{\mathbb{R}}}\check{\psi }(x)\,dx. $$

Thus $\bar{U}(\check{\psi },\delta )<\infty$ and $\bar{U}( \check{\psi },\delta )- \underline{U}(\check{\psi },\delta ) \to 0$ as $\delta \to \infty$. These two properties mean by definition that the function $\check{\psi }$ is directly Riemann-integrable. Arguing for the positive and negative parts, we obtain that if $\psi$ is integrable, then $\check{\psi }$ is directly Riemann-integrable.

We use in the sequel the following renewal theorem (see [19, Proposition 2.1]) for the random walk $S _{n}:=\sum _{i=1}^{n}\xi _{i}$ on a lattice.

Proposition A.3

Let$\xi _{i}$be i.i.d. random variables taking values in the lattice${\mathbb{Z}}d$, $d>0$, and having finite expectation$m:=\mathbf{E}[ \xi _{i}]>0$. Let$F:{\mathbb{R}}\to {\mathbb{R}}$be a measurable function. If$x\in {\mathbb{R}}$is such that$\sum _{j\in {\mathbb{Z}}}|F(x+jd)|< \infty$, then

$$ \lim _{n\to \infty }{\mathbf{E}}\bigg[ \sum _{k\ge 0}F(x+nd-S_{k}) \bigg]=\frac{d}{m} \sum _{j\in {\mathbb{Z}}}F(x+jd). $$

Proof of Theorem A.2

Let the solution of (A.1) be realised on some probability space $(\Omega ,\mathcal{F},\mathbf{P})$. We use the notation $(M,Q)$ instead of $(M_{1},Q_{1})$ and as usual define the tail function $\bar{G}(u):=\mathbf{P}[Y_{\infty }> u]$. Set $g(x):=e^{\beta x} \bar{G}(e^{x})$. Since $Y_{\infty }$ and $M$ are independent, we have $\mathbf{P}[MY_{\infty }>e^{x}]=\mathbf{E}[ \bar{G}(e^{x-\ln M})]$. Introducing the new probability measure $\tilde{\mathbf{P}}:=M^{\beta } {\mathbf{P}}$ and noting that

$$ e^{\beta x}{\mathbf{P}}[MY_{\infty }>e^{x}] =\mathbf{E}[M^{\beta }e ^{\beta (x-\ln M)} \bar{G}(e^{x-\ln M})]= \tilde{\mathbf{E}}[g(x- \ln M)], $$

we obtain the identity (called renewal equation)

$$ g(x)=D(x)+\tilde{\mathbf{E}}[g(x-\ln M)], $$

(A.4)

where $D(x):=e^{\beta x}(\mathbf{P}[Y_{\infty } >e^{x}]-\mathbf{P}[MY _{\infty }>e^{x}])$. The Jensen inequality for the convex function $x\mapsto x\ln x$ implies that $\tilde{\mathbf{E}}[\ln M]=\mathbf{E}[M ^{\beta }\ln M]>0$ and hence $\tilde{\mathbf{E}}[|\ln M|]<\infty$. Let us check that the function $x\mapsto D(x)$ is integrable. To this end, we note that for any random variables $\xi ,\eta$,

$$ \big|{\mathbf{P}}[\xi >s]-\mathbf{P}[\eta >s]\big|\le {\mathbf{P}}[ \eta ^{+}\le s< \xi ^{+}]+\mathbf{P}[\xi ^{+}\le s< \eta ^{+}]. $$

Using the Fubini theorem, we obtain that

$$\begin{aligned} \int _{0}^{\infty }{\mathbf{P}}[\eta ^{+}\le s< \xi ^{+}]s^{\beta -1}\,ds =& \mathbf{E}\bigg[ I_{\{\eta _{+}< \xi _{+}\}}\int _{\eta ^{+}}^{\xi ^{+}} s ^{\beta -1}\,ds\bigg] \\ =&\frac{1}{\beta }{\mathbf{E}}\big[\big((\xi ^{+})^{\beta }-(\eta ^{+})^{ \beta }\big)^{+}\big]. \end{aligned}$$

Applying the above bound and identity with $\xi :=Q+MY_{\infty } \stackrel{d}{=}Y_{\infty }$ and $\eta :=M Y_{\infty }$, we get that

$$ \int _{{\mathbb{R}}} \vert D(x)\vert d x=\int _{0}^{\infty }\big|{\mathbf{P}}[ \xi >s]-\mathbf{P}[\eta >s]\big|s^{\beta -1}\,ds\le \frac{1}{\beta } {\mathbf{E}}\left [\big\vert (\xi ^{+})^{\beta }-(\eta ^{+})^{\beta } \big\vert \right ], $$

and it remains to verify that

$$ {\mathbf{E}}\big[\big|\big((Q+\eta )^{+}\big)^{\beta }-(\eta ^{+})^{ \beta }\big|\big] < \infty $$

(A.5)

when $\mathbf{E}[\vert Q\vert ^{\beta }]<\infty$. But $|((Q+\eta )^{+})^{ \beta }-(\eta ^{+})^{\beta }|=\zeta _{1}+\zeta _{2}$ with positive summands

$$\begin{aligned} \zeta _{1} :=&I_{\{-Q< \eta \le 0\}}(Q+\eta )^{\beta }+I_{\{0< \eta \le -Q\}}\eta ^{\beta }\le |Q|^{\beta }, \\ \zeta _{2} :=&I_{\{Q+\eta >0, \eta >0\}}|(Q+\eta )^{\beta }-\eta ^{ \beta }|. \end{aligned}$$

If $\beta \le 1$, the random variable $\zeta _{2}$ is also dominated by the random variable $|Q|^{\beta }$. If $\beta >1$, the inequality $|x^{\beta }-y^{\beta }|\le \beta |x-y|(x\vee y)^{\beta -1}$ for $x,y\ge 0$ combined with the inequality $(|a|+|b|)^{\beta -1}\le 2^{( \beta -2)^{+}} (|a|^{\beta -1}+|b|^{\beta -1} )$ leads to the estimate

$$ \zeta _{2}\le 2^{(\beta -2)^{+}}\beta |Q| (|\eta |^{\beta -1}+|Q|^{ \beta -1}). $$

Using the independence of $(M,Q)$ and $Y_{\infty }$, the Hölder inequality and taking into account that $\mathbf{E}[M^{\beta }]=1$ and $\mathbf{E}[| Y_{\infty }|^{p}]<\infty$ for $p\in [0,\beta )$, we get that

$$ \mathbf{E}[|Q||\eta |^{\beta -1} ]= \mathbf{E}[|Q|M^{\beta -1} ] {\mathbf{E}}[| Y_{\infty }|^{\beta -1} ] \le (\mathbf{E}[|Q|^{\beta }])^{1/ \beta } {\mathbf{E}}[|Y_{\infty }|^{\beta -1} ] < \infty . $$

Thus (A.5) holds. The integrability of $D$ allows us to transform (A.4) into the equality

$$ \check{g}(x)=\check{D}(x)+\tilde{\mathbf{E}}[\check{g}(x-\ln M)]. $$

Iterating, we obtain that

$$ \check{g}(x)=\sum _{n= 0}^{N-1}\tilde{\mathbf{E}}[\check{D}(x-{S}_{n})]+ \tilde{\mathbf{E}}[\check{g}(x-{S}_{N})], $$

(A.6)

where $S_{n}:=\sum _{i=1}^{n}\xi _{i}$ for $n\ge 1$ and $(\xi _{i})$ is a sequence of independent random variables on $(\Omega ,\mathcal{F}, \tilde{\mathbf{P}})$, independent of $Y_{\infty }$, such that $\mathcal{L}(\xi _{i},\tilde{\mathbf{P}})=\mathcal{L}(\ln M, \tilde{\mathbf{P}})$. In particular, $\tilde{\mathbf{E}}[e^{-\beta \xi _{i}}]=1$.

By the strong law of large numbers, we have $S_{N}/N\to \tilde{\mathbf{E}}[\ln M]>0$$\tilde{\mathbf{P}}$-a.s. as $N\to \infty$ and therefore $y-\ln S_{N}\to -\infty$$\tilde{\mathbf{P}}$-a.s. for every $y$. Since $\tilde{\mathbf{E}}[e ^{-\beta S_{N}}]=1$, we have by dominated convergence that

$$ \tilde{\mathbf{E}}[g(y-{S}_{N})]=\tilde{\mathbf{E}}[e^{\beta (y-S_{N})} \bar{G}(e^{y-S_{N}})]\longrightarrow 0. $$

It follows that the remainder term $\mathbf{E}[\check{g}(x-{S}_{N})]$ in (A.6) tends to zero so that

$$ \check{g}(x)=\sum _{k\ge 0}\tilde{\mathbf{E}}[\check{D}(x-{S}_{k})]. $$

Using Proposition A.3 (with $F=\check{D}$), we obtain that for any $x>0$,

$$ \lim _{n\to \infty }\check{g}(x+d n)= \frac{d}{\tilde{\mathbf{E}}[ \ln M]}\sum _{j\in {\mathbb{Z}}} \check{D}(x+jd) \le \bar{U}(\check{D},d) < \infty . $$

Replacing in the integral below the function $\bar{G}(e^{y})$ by its maximal value $\bar{G}(e^{x})$, we get

$$ \check{g}(x):=\int ^{x}_{-\infty } e^{-(x-y)}e^{\beta y} \bar{G}(e^{y}) d y \ge \frac{1}{\beta +1}g(x) $$

and therefore

$$ \limsup _{u\to \infty } u^{\beta }{\mathbf{P}}[Y_{\infty }> u]= \limsup _{x\to \infty } g(x) \le (\beta +1) \limsup _{x\to \infty } \check{g}(x)< \infty . $$

Theorem A.2 is proved. □

1.3 A.3 Buraczewski–Damek approach

The following result, usually formulated in terms of the supremum of the random walk $S_{n}:=\sum _{i=1}^{n}{\ln M_{i}}$, is well known (see e.g. Kesten [23, Theorem A] for a much more general setting).

Proposition A.4

If$M$satisfies (A.2), then

$$ \liminf _{u\to \infty }u^{\beta } {\mathbf{P}}[Z^{*}_{\infty }>u]>0. $$

Proof

Let $F(x):= \mathbf{P}[\ln M\le x]$, $\bar{F}(x):=1-F(x)$ and $S_{n}:=\sum _{i=1}^{n}{\xi _{i}}$, where $\xi _{i}:=\ln M_{i}$. The function $\bar{H}(x):= \mathbf{P}[ \sup _{n\in {\mathbb{N}}} S_{n}>x]$ admits the representation

Putting $Z(x):=e^{\beta x}\bar{H}(x)$, $z(x):=e^{\beta x}\bar{F}(x)$ and $\tilde{\mathbf{P}}:=e^{\beta \xi _{1}} {\mathbf{P}}$, we obtain from here that

$$ Z(x)=z(x)+ \tilde{\mathbf{E}}[Z(x-\xi _{1})I_{\{\xi _{1}\le x\}}]. $$

The same arguments as were used in deriving (A.6) lead to the representation

$$ Z(x)=\tilde{\mathbf{E}}\bigg[\sum _{k\ge 0}z(x-S_{k})I_{\{S_{k}\le x\}} \bigg]. $$

The function $\hat{z}(x):=z(x)I_{\{x\ge 0\}}$ is directly Riemann-integrable. Indeed, for $j\ge 0$, we have that

$$ \sup _{x\in [j\delta , (j+1)\delta ]} z(x)\le e^{\beta (j+1)\delta } \bar{F}(j\delta )\le e^{2\beta \delta }\int ^{j\delta }_{(j-1)\delta } e^{\beta v} \bar{F}(v)\,d v $$

and therefore

$$ \bar{U}(\hat{z},\delta )=\delta z(0)+\delta \sum _{j\ge 0} \sup _{x\in [j\delta , (j+1)\delta ]} z(x)\le \delta z(0)+ e^{2\beta \delta }\int _{-\delta }^{\infty }e^{\beta v} \bar{F}(v)\,d v. $$

In the same spirit, we get

$$ \inf _{x\in [j\delta , (j+1)\delta ]} z(x)\ge e^{\beta j\delta } \bar{F}\big((j+1)\delta \big)\ge e^{-2\beta \delta }\int ^{(j+2)\delta }_{(j+1)\delta } e^{\beta v} \bar{F}(v)\,d v $$

and

$$ \underline{U}(\hat{z},\delta )=\delta \sum _{j\ge 0} \sup _{x\in [j\delta , (j+1)\delta ]} z(x)\ge e^{-2\beta \delta } \int _{\delta }^{\infty }e^{\beta v} \bar{F}(v)\,d v. $$

Taking into account that

$$ \int _{{\mathbb{R}}}e^{\beta v} \bar{F}(v)\,d v = \frac{1}{\beta } {\mathbf{E}}[e ^{\beta \xi _{1}}] =\frac{1}{\beta }< \infty , $$

we get from here that $\bar{U}(\hat{z},\delta )<\infty$ and $\bar{U}(\hat{z},\delta )-\underline{U}(\hat{z},\delta )\to 0$ as $\delta \to 0$. Using renewal theory, we obtain that if the law of $\xi$ is non-arithmetic,

$$ \lim _{x\to \infty } e^{\beta x}\bar{H}(x)=\frac{1}{\tilde{\mathbf{E}}[ \xi ]} \int ^{\infty }_{0} z(v)\,dv; $$

(A.7)

see e.g. [13, Chap. XI, 9]. If the law of $\xi$ is arithmetic with step $d>0$, then according to Proposition A.3, for any $x>0$, we have

$$ \lim _{n\to \infty } e^{\beta (x+nd)}\bar{H}(x+nd)=\frac{d}{ \tilde{\mathbf{E}}[\xi ]}\sum _{j\in {\mathbb{Z}}}z(x+jd) I_{\{x+jd \ge 0\}}. $$

(A.8)

The equalities (A.7) and (A.8) imply the statement. □

The proof of the result below, formulated in a form to cover our needs, follows the same lines as in Lemma 2.6 of the Buraczewski–Damek paper [9] with minor changes to include also the arithmetic case.

Theorem A.5

Suppose that (A.2) holds. If the support of the distribution of$Y_{\infty }$is unbounded from above, then

$$ \liminf _{u\to \infty } u^{\beta } {\mathbf{P}}[Y_{\infty }>u]>0. $$

Proof

Let

$$ \bar{Y}_{n}:=-\sum _{j=1}^{n} Q_{j}^{-} Z_{j-1}, \qquad Y_{n,\infty }:=\sum ^{\infty }_{j=n+1} Q_{j} \prod ^{j-1}_{\ell =n+1} M_{\ell } $$

and $Z^{*}_{n}:=\sup _{j\le n}Z_{j}$. Theorems A.1 and A.2 imply that $\mathbf{P}[\bar{Y}_{\infty } <-u]\le C _{1}u^{-\beta }$ with $C_{1}>0$. On the other hand, by Proposition A.4, $\mathbf{P}[Z^{*}_{\infty } >u]\ge C_{2}u^{-\beta }$ with $C_{2}>0$. Of course, in both cases the inequalities hold when $u$ is sufficiently large. Put $U_{n}:=\{Z_{n}>u, \bar{Y}_{n}> -Cu\}$, where $C^{\beta }:=4C_{1}/C_{2}$. The process $\bar{Y}$ decreases. Therefore, we have the inclusion $\{Z_{n}>u\}\subseteq \{\bar{Y}_{ \infty }\le -Cu \}\cup U_{n}$. It follows that for sufficiently large $u>0$, we have

$$\begin{aligned} (3/4)C_{2}u^{-\beta }\le {\mathbf{P}}[Z^{*}_{\infty }>u]=\mathbf{P} \bigg[\bigcup _{n\in {\mathbb{N}}} \{Z_{n}>u\}\bigg] \le & \mathbf{P}[ \bar{Y}_{\infty }\le -Cu]+\mathbf{P}\bigg[\bigcup _{n\in {\mathbb{N}}} U_{n}\bigg] \\ \le & 2C_{1}C^{-\beta } u^{-\beta }+\mathbf{P}\bigg[ \bigcup _{n\in {\mathbb{N}}} U_{n}\bigg] \end{aligned}$$

so that $\mathbf{P}[\bigcup _{n\in {\mathbb{N}}} U_{n}]\ge (1/4)C_{2}u ^{-\beta }$. Since $\bar{Y}_{n}+Z_{n}Y_{n,\infty }\le Y_{n}+Z _{n}Y_{n,\infty }=Y_{\infty }$, we have that

$$ \{Y_{n,\infty }>C+1\}\cap U_{n}\subseteq \{\bar{Y}_{n}+Z_{n}Y_{n, \infty }>u\}\cap U_{n}\subseteq \{Y_{\infty }>u\}\cap U_{n}. $$

Note that $\mathbf{P}[Y_{\infty }>C+1]=\mathbf{P}[Y_{n,\infty }>C+1]$ because $\mathcal{L}(Y_{n,\infty }) =\mathcal{L}(Y_{\infty })$. Using the independence of $Y_{n,\infty }$ and the sets $W_{n}:=U_{n}\cap ( \bigcup _{k=1}^{n-1}U_{k})^{c}$ forming a disjoint partition of $\bigcup _{n\in {\mathbb{N}}}U_{n}$, we get that

$$\begin{aligned} {\mathbf{P}}[Y_{\infty }>C+1]{\mathbf{P}}\bigg[\bigcup _{n\in {\mathbb{N}}} W_{n}\bigg] =&\sum _{n}{\mathbf{P}}[\{Y_{n,\infty }>C+1\}\cap W_{n}] \\ \le & \sum _{n}{\mathbf{P}}[\{Y_{\infty }>u\}\cap W_{n}] \le {\mathbf{P}}[Y _{\infty }>u]. \end{aligned}$$

Thus $\mathbf{P}[Y_{\infty }>u]\ge (1/4)bC_{2}u^{-\beta }$, where $b:=\mathbf{P}[Y_{\infty }>C+1]>0$ by the assumption that the support of $\mathcal{L}(Y_{\infty })$ is unbounded from above. The obtained asymptotic bound implies that $C_{+}>0$. □

Summarising the above results, we get for the function $\bar{G}(u)= \mathbf{P}[Y_{\infty }>u]$ the following asymptotic properties when $u\to \infty$.

Theorem A.6

Suppose that (A.2) holds. Then$\limsup _{u \to \infty } u^{ \beta }\bar{G}(u)<\infty$. If$Y_{\infty }$is unbounded from above, then$\liminf _{u \to \infty } u^{\beta }\bar{G}(u)>0$, and in the case where$\mathcal{L}(\ln M)$is non-arithmetic, $\bar{G}(u)\sim C_{+}u ^{-\beta }$with$C_{+}>0$.