1 Introduction and Main Results

Let \(\eta _1\), \(\eta _2,\ldots \) be independent copies of a random variable \(\eta \) with zero mean and finite variance. By Kolmogorov’s three-series theorem, the random Dirichlet series

$$\begin{aligned} S(\alpha ):=\sum _{k\ge 1}k^{-\alpha }\eta _k \end{aligned}$$

converges almost surely (a.s.) if, and only if, \(\alpha >1/2\). Throughout the paper, we additionally assume \(\eta \le b\) a.s. for some \(b>0\). Our purpose is to find a precise (as opposed to logarithmic) asymptotic behaviour of the distribution right tail of \(S(\alpha )\), that is, \(\mathbb {P}\{S(\alpha )>x\}\) as \(x\rightarrow \infty \). If \(\alpha >1\), then \(S(\alpha )\le b\zeta (\alpha )\) a.s., where \(\zeta \) is the Riemann zeta function. Thus, in what follows our attention is restricted to the nontrivial case \(\alpha \in (1/2, 1]\). At the end of Sect. 2, we discuss this case and the assumption \(\eta \le b\) in more details. The other part of Sect. 2 is concerned with the complementary situation, in which the random variable \(\eta \) is a.s. unbounded.

Put \(\psi (t):=\log \mathbb {E}[ \textrm{e}^{t\eta }]\) for \(t\in \mathbb {R}\). Under the present assumptions, the function \(\psi \) is infinitely differentiable, strictly convex and strictly increasing on \([0,\infty )\). In particular, \(\psi ^\prime \) is positive and strictly increasing on \((0,\infty )\) and \(\psi ^{\prime \prime }\) is positive on \([0,\infty )\). Recall that, for \(\rho \in (0,1]\), the Euler–Mascheroni constant \(\gamma _\rho \) is defined by

$$\begin{aligned} \gamma _\rho :=\lim _{n\rightarrow \infty }\left( \sum _{k=1}^n k^{-\rho }-(1-\rho )^{-1}n^{1-\rho }\right) \end{aligned}$$
(1)

if \(\rho \in (0,1)\), and

$$\begin{aligned} \gamma _1:=\lim _{n\rightarrow \infty }\left( \sum _{k=1}^n k^{-1}-\log n\right) \end{aligned}$$
(2)

if \(\rho =1\).

Theorem 1.1

Assume that \(\eta \le b\) a.s. with \(\mathbb {P}\{\eta =b\}=\theta \in (0,1)\), \(\mathbb {E}[\eta ]=0\) and \(\mathbb {E}[\eta ^2]<\infty \).

If \(\alpha =1\), then, as \(x\rightarrow \infty \),

$$\begin{aligned} \mathbb {P}\{S(1)>x\}~\sim ~\frac{1}{(2\pi \theta b)^{1/2}}\exp \left( -\frac{x-q}{2b}-b\textrm{e}^{(x-q)/b}\right) , \end{aligned}$$

where

$$\begin{aligned} q=b\gamma _1+\int _0^1 x^{-1}\psi ^\prime (x)\textrm{d}x+\int _1^\infty x^{-1}(\psi ^\prime (x)-b)\textrm{d}x\in \mathbb {R}. \end{aligned}$$

If \(\alpha \in (1/2, 1)\), then, as \(x\rightarrow \infty \),

$$\begin{aligned} \mathbb {P}\{S(\alpha )>x\}{} & {} \sim \frac{1}{(2\pi \theta )^{1/2}}\left( \frac{\alpha (\sigma _\alpha ^2)^\alpha }{1-\alpha }\right) ^{1/(2(1-\alpha ))}\frac{1}{x^{1/(2(1-\alpha ))}}\\{} & {} \times \exp \left( -\left( \frac{1-\alpha }{(\alpha \sigma _\alpha ^2)^\alpha }\right) ^{1/(1-\alpha )}(x-b\gamma _\alpha )^{1/(1-\alpha )}\right) , \end{aligned}$$

where

$$\begin{aligned} \sigma _\alpha ^2:=\alpha ^{-1}\int _0^\infty x^{1-1/\alpha }\psi ^{\prime \prime }(x)\textrm{d}x=(1-\alpha )\alpha ^{-3}\int _0^\infty x^{-1-1/\alpha }\psi (x)\textrm{d}x\in (0,\infty ). \end{aligned}$$

Theorem 1.2

Assume that \(\eta \le b\) a.s., \(\mathbb {P}\{b-\eta \le x\}\sim \lambda x^r\) as \(x\rightarrow 0+\) for positive \(\lambda \) and r, \(\mathbb {E}[\eta ]=0\) and \(\mathbb {E}[\eta ^2]<\infty \).

If \(\alpha =1\), then, as \(x\rightarrow \infty \),

$$\begin{aligned} \mathbb {P}\{S(1)>x\}~\sim ~\left( \frac{(2\pi )^{r-1}}{\lambda \Gamma (r+1)b}\right) ^{1/2}\exp \left( \frac{r-1}{2b}(x-q)-b\textrm{e}^{(x-q)/b}\right) \end{aligned}$$

with q as defined in Theorem 1.1.

If \(\alpha \in (1/2, 1)\), then, as \(x\rightarrow \infty \),

$$\begin{aligned} \mathbb {P}\{S(\alpha )>x\}{} & {} \sim \left( \frac{(2\pi )^{r\alpha -1}}{\lambda \Gamma (r+1)}\right) ^{1/2}\left( \frac{(1-\alpha )^{r\alpha -1}}{\alpha ^{r\alpha -1}(\sigma _\alpha ^2)^{(r-1)\alpha }}\right) ^{1/(2(1-\alpha ))}\\{} & {} \quad \times \exp \left( \frac{r}{2} \left( \left( \frac{\alpha \sigma _\alpha ^2}{1-\alpha }\right) ^{\alpha /(1-\alpha )}-1\right) \right) x^{(r\alpha -1)/(2(1-\alpha ))}\\{} & {} \quad \times \exp \left( -\left( \frac{1-\alpha }{(\alpha \sigma _\alpha ^2)^\alpha }\right) ^{1/(1-\alpha )}(x-b\gamma _\alpha )^{1/(1-\alpha )}\right) \end{aligned}$$

with \(\sigma ^2_\alpha \) as defined in Theorem 1.1.

Remark 1.3

Putting formally \(r=0\) and \(\lambda =\theta \) on the right-hand sides of the tail asymptotics of Theorem 1.2, we obtain the right-hand sides of the tail asymptotics of Theorem 1.1, as it must be. Thus, in principle the two theorems could have been combined into a single result. The same remark also concerns Theorems 1.4 and 1.5 stated below.

The random variable \(S(\alpha )\) can be seen as a special case of the series \(\sum _{k\ge 1}c_k\eta _k\), where \(c_1\), \(c_2,\ldots \) are real numbers. Perhaps, the most known representative of this family is called the Bernoulli convolution. It corresponds to \(c_k=a^k\) with some \(a\in (0,1)\) and \(\eta \) having a Rademacher distribution. The main question for the Bernoulli convolutions is whether the distribution of the series \(\sum _{k\ge 1}a^k\eta _k\) is absolutely continuous with respect to (w.r.t.) Lebesgue measure or not. It has long been known and is quite easy to prove that the distribution of \(\sum _{k\ge 1}a^k\eta _k\) is singular continuous for every \(a<1/2\). On the other hand, if \(a=1/2\), then \(\sum _{k\ge 1}a^k\eta _k\) has a uniform distribution on \([-1,1]\). Solomyak [12] has shown that the distribution of \(\sum _{k\ge 1}a^k\eta _k\) is absolutely continuous for almost all, w.r.t. Lebesgue measure, \(a\in (1/2,1)\).

It has been noticed by Yaskov [14] that the results of Reich [8] can be used to show that the distribution of \(S(\alpha )\) is absolutely continuous whenever a random variable \(\eta \) has zero mean and finite variance. Furthermore, there exists a smooth version of the density that we denote by \(g_\alpha \). Assuming that \(\eta \) has a Rademacher distribution Yaskov [14] has found an asymptotic behaviour of \(\log g_\alpha (x)\) as \(x\rightarrow \infty \). Our approach allows us to determine a precise, rather than logarithmic, asymptotic of \(g_\alpha \).

Theorem 1.4

Assume that \(\eta \le b\) a.s. with \(\mathbb {P}\{\eta =b\}=\theta \in (0,1)\), \(\mathbb {E}[\eta ]=0\) and \(\mathbb {E}[\eta ^2]<\infty \).

If \(\alpha =1\), then, as \(x\rightarrow \infty \),

$$\begin{aligned} g_1(x)~\sim ~\frac{1}{(2\pi \theta b)^{1/2}} \exp \left( \frac{x-q}{2b}-b\textrm{e}^{(x-q)/b}\right) \end{aligned}$$

with q as defined in Theorem 1.1.

If \(\alpha \in (1/2, 1)\), then, as \(x\rightarrow \infty \),

$$\begin{aligned} g_\alpha (x){} & {} \sim \frac{1}{(2\pi \theta )^{1/2}}\left( \frac{(1-\alpha )^{2\alpha -1}}{\alpha ^{2\alpha -1}(\sigma _\alpha ^2)^\alpha }\right) ^{1/(2(1-\alpha ))}x^{(2\alpha -1)/(2(1-\alpha ))}\\{} & {} \quad \times \exp \left( -\left( \frac{1-\alpha }{(\alpha \sigma _\alpha ^2)^\alpha }\right) ^{1/(1-\alpha )}(x-b\gamma _\alpha )^{1/(1-\alpha )}\right) \end{aligned}$$

with \(\sigma ^2_\alpha \) as defined in Theorem 1.1.

One of the earliest works on the distribution of \(S(\alpha )\) is the paper [9] by Rice, who has studied the case \(\alpha =1\) and \(\mathbb {P}\{\eta =\pm 1\}=1/2\). He has obtained a version of Theorem 1.4 for this particular case. It is worth mentioning that his interest on \(S(\alpha )\) was motivated by applications to digital communication systems.

Theorem 1.5

Assume that \(\eta \le b\) a.s., \(\mathbb {P}\{b-\eta \le x\}\sim \lambda x^r\) as \(x\rightarrow 0+\) for positive \(\lambda \) and r, \(\mathbb {E}[\eta ]=0\) and \(\mathbb {E}[\eta ^2]<\infty \).

If \(\alpha =1\), then

$$\begin{aligned} g_1(x)~\sim ~\left( \frac{(2\pi )^{r-1}}{\lambda \Gamma (r+1)b}\right) ^{1/2}\exp \left( \frac{r+1}{2b}(x-q)-b\textrm{e}^{(x-q)/b}\right) ,\quad x\rightarrow \infty . \end{aligned}$$

with q as defined in Theorem 1.1.

If \(\alpha \in (1/2, 1)\), then, as \(x\rightarrow \infty \),

$$\begin{aligned} g_\alpha (x){} & {} \sim ~\left( \frac{(2\pi )^{r\alpha -1}}{\lambda \Gamma (r+1)}\right) ^{1/2}\left( \frac{(1-\alpha )^{(r+2)\alpha -1}}{\alpha ^{(r+2)\alpha -1}(\sigma _\alpha ^2)^{(r+1)\alpha }}\right) ^{1/(2(1-\alpha ))}\\{} & {} \quad \times \exp \left( \frac{r}{2} \left( \left( \frac{\alpha \sigma _\alpha ^2}{1-\alpha }\right) ^{\alpha /(1-\alpha )}-1\right) \right) x^{((r+2)\alpha -1)/(2(1-\alpha ))}\\{} & {} \quad \times \exp \left( -\left( \frac{1-\alpha }{(\alpha \sigma _\alpha ^2)^\alpha }\right) ^{1/(1-\alpha )}(x-b\gamma _\alpha )^{1/(1-\alpha )}\right) \end{aligned}$$

with \(\sigma ^2_\alpha \) as defined in Theorem 1.1.

Remark 1.6

The asymptotic behaviour of the distribution tail of \(S(\alpha )\) can be deduced from the asymptotic behaviour of the corresponding density \(g_\alpha \). Indeed, integrating the asymptotic relations of Theorems 1.4 and 1.5 and applying the L’Hôpital rule we obtain the statements of Theorems 1.1 and 1.2. However, we believe that our approach used in the proofs of Theorems 1.1 and 1.2 has its own merits and may be used for investigating precise distributional tail behaviour of other infinite weighted sums of independent identically distributed random variables. One of its advantages is that absolute continuity of the distribution of a sum is not a priori required.

Remark 1.7

Our proofs can be adapted to the case \(\mathbb {P}\{b-\eta \le x\}\sim L(x) x^r\) as \(x\rightarrow 0+\), where L is a slowly varying at 0 function. While the form of the asymptotic relations in Theorems 1.2 and 1.5 remains unchanged, the parameter \(\lambda \) has to be replaced by L(1/(t(x))), where t(x) is given in Proposition 5.2.

2 A Look from a Broader Perspective

We start by discussing the situation, in which \(\mathbb {P}\{\eta>y\}>0\) for all \(y>0\). Assume first that some exponential moments of positive orders are finite, that is,

$$\begin{aligned} t_0:=\sup \{t>0:\mathbb {E}[\textrm{e}^{t\eta }]<\infty \}\in (0,\infty ). \end{aligned}$$

Set \(S'(\alpha ):=\sum _{k\ge 2}k^{-\alpha }\eta _k\). It is clear that

$$\begin{aligned} \mathbb {E}[\textrm{e}^{tS'(\alpha )}]<\infty \quad \text {for all }~ 0\le t<2^\alpha t_0. \end{aligned}$$

Therefore, the distribution tail of \(S'(\alpha )\) is lighter than that of \(\eta \). This ensures that the distribution tail of \(S(\alpha )\) is proportional to the distribution tail of \(\eta \). We illustrate this observation under the assumption that the distribution of \(\eta \) satisfies

$$\begin{aligned} \lim _{x\rightarrow \infty } \frac{\mathbb {P}\{\eta>x-y\}}{\mathbb {P}\{\eta >x\}} =\textrm{e}^{t_0y}\quad \text {for each }y\in \mathbb {R}. \end{aligned}$$
(3)

The class of distributions with this property is called in the literature \(\mathcal {L}(t_0)\), see, for example, [5]. Indeed, by the total probability formula,

$$\begin{aligned} \frac{\mathbb {P}\{S(\alpha )>x\}}{\mathbb {P}\{\eta>x\}} =\int _\mathbb {R}\frac{\mathbb {P}\{\eta>x-y\}}{\mathbb {P}\{\eta >x\}} \mathbb {P}\{S'(\alpha )\in \textrm{d}y\}. \end{aligned}$$

Having pointwise convergence in (3), one can use Lebesgue’s dominated convergence theorem to obtain

$$\begin{aligned} \lim _{x\rightarrow \infty } \frac{\mathbb {P}\{S(\alpha )>x\}}{\mathbb {P}\{\eta >x\}}=\int _\mathbb {R}\textrm{e}^{t_0y}\mathbb {P}\{S'(\alpha )\in \textrm{d}y\}\in (0,\infty ). \end{aligned}$$
(4)

To legitimate the use of Lebesgue’s theorem, we have to find an integrable majorant for the ratio \(\frac{\mathbb {P}\{\eta>x-y\}}{\mathbb {P}\{\eta >x\}}\). Fix some \(\varepsilon \in (0,(2^\alpha -1)t_0)\) and notice that (3) implies that there exists \(x_0=x_0(\varepsilon )\) such that

$$\begin{aligned} \frac{\mathbb {P}\{\eta>x-1\}}{\mathbb {P}\{\eta >x\}} \le \textrm{e}^{t_0+\varepsilon }\quad \text {for all }x\ge x_0. \end{aligned}$$

Therefore,

$$\begin{aligned} \frac{\mathbb {P}\{\eta>x-n\}}{\mathbb {P}\{\eta >x\}} \le \textrm{e}^{(t_0+\varepsilon )n}\quad \text {for all }n\le x-x_0+1. \end{aligned}$$

Now, monotonicity of \(z\mapsto \mathbb {P}\{\eta >z\}\) entails

$$\begin{aligned} \frac{\mathbb {P}\{\eta>x-y\}}{\mathbb {P}\{\eta >x\}} \le \textrm{e}^{(t_0+\varepsilon )(y+1)} \le \textrm{e}^{2^\alpha t_0}\textrm{e}^{(t_0+\varepsilon )y} \quad \text {for all }y\le x-x_0. \end{aligned}$$
(5)

Letting here \(y=x-x_0\), we also have

$$\begin{aligned} \frac{1}{\mathbb {P}\{\eta>x\}} \le \frac{\textrm{e}^{2^\alpha t_0}}{\mathbb {P}\{\eta >x_0\}} \textrm{e}^{(t_0+\varepsilon )(x-x_0)}. \end{aligned}$$

Consequently,

$$\begin{aligned} \frac{\mathbb {P}\{\eta>x-y\}}{\mathbb {P}\{\eta>x\}} \le \frac{1}{\mathbb {P}\{\eta>x\}} \le \frac{\textrm{e}^{2^\alpha t_0}}{\mathbb {P}\{\eta>x_0\}} \textrm{e}^{(t_0+\varepsilon )y}\quad \text {for all }y>x-x_0. \end{aligned}$$
(6)

Combining (5) and (6), we conclude that the function \(y\mapsto \frac{\textrm{e}^{2^\alpha t_0}}{\mathbb {P}\{\eta >x_0\}}\max \{1,\textrm{e}^{(t_0+\varepsilon )y}\}\) is an integrable majorant with respect to \(\mathbb {P}\{S'(\alpha )\in \textrm{d}y\}\). This proves (4).

A similar to (4) relation holds true in the case where the distribution tail of \(\eta \) is heavy, that is, \(t_0=0\). If, for example, the function \(x\mapsto \mathbb {P}\{\eta >x\}\) is regularly varying at \(\infty \) of index \(-\theta \) for \(\theta >2\), then

$$\begin{aligned} \lim _{x\rightarrow \infty } \frac{\mathbb {P}\{S(\alpha )>x\}}{\mathbb {P}\{\eta>x\}}=\lim _{x\rightarrow \infty }\frac{\mathbb {P}\{\eta _k>xk^{\alpha }\text { for some }k\ge 1\}}{\mathbb {P}\{\eta >x\}}=\sum _{k\ge 1} k^{-\alpha \theta }. \end{aligned}$$

These equalities can be proven with the help of arguments which are standard for heavy-tailed distributions. The driving force behind the aforementioned situations is the classical ‘one big jump strategy’, according to which a large value of the sum \(S(\alpha )\) is caused by a large value of a single summand.

If the moment generating function \(t\mapsto \mathbb {E}[\textrm{e}^{t\eta }]\) is finite for all \(t>0\), it is still possible that the distribution tail of \(S(\alpha )\) is of the same type as the distribution tail of \(\eta \). Here is a rather simple example. Assume that \(\eta \) is normally distributed with zero mean and unit variance. The sum \(S(\alpha )\) is then also normally distributed with zero mean and variance \(\theta ^2_\alpha :=\sum _{k\ge 1}k^{-2\alpha }\). In particular,

$$\begin{aligned} \mathbb {P}\{S(\alpha )>x\}=\mathbb {P}\{\eta >x/\theta _\alpha \}, \quad x\in \mathbb {R}. \end{aligned}$$

Actually, a weaker version of this equality holds true for a much wider class of distributions. Assume that

$$\begin{aligned} \psi (t)=\log \mathbb {E}[\textrm{e}^{t\eta }]\sim \gamma t^{p},\quad t\rightarrow \infty \end{aligned}$$
(7)

for some \(\gamma >0\) and \(p>1\). If \(\alpha >1/p\), then

$$\begin{aligned} \log \mathbb {E}[\textrm{e}^{tS(\alpha )}] =\sum _{k\ge 1}\psi (tk^{-\alpha })~\sim ~ \gamma t^p\sum _{k\ge 1}k^{-\alpha p},\quad t\rightarrow \infty . \end{aligned}$$

Applying now the Kasahara Tauberian theorem (see Theorem 4.12.7 in [2]), we conclude that

$$\begin{aligned} \log \mathbb {P}\{S(\alpha )>x\}~\sim ~\log \mathbb {P}\left\{ \eta >x/\left( \sum _{k\ge 1}k^{-\alpha p}\right) ^{1/p}\right\} , \quad x\rightarrow \infty . \end{aligned}$$
(8)

In other words, the distribution tails of \(S(\alpha )\) and \(\eta \) are proportional to each other on the logarithmic scale provided that \(\alpha >1/p\). More precise versions of (8) (with the logarithms omitted) show up in the tail asymptotics of series \(\sum _{k\ge 1}c_k\eta _k\) with \((c_k)_{k\ge 1}\) being a summable sequence. This case has been investigated by Rootzén [10, 11] and Klüppelberg and Lindner [6].

The proportionality arising in (8) disappears for \(\alpha \le 1/p\). Indeed, assume, for instance, that (7) holds and that \(\alpha <1/p\). It can be checked along the lines of the proofs of Propositions 5.3 and 5.4 that

$$\begin{aligned} \log \mathbb {E}[\textrm{e}^{tS(\alpha )}]~\sim ~\frac{t^{1/\alpha }}{\alpha }\int _0^\infty v^{-1-1/\alpha }\psi (v)\textrm{d}v,\quad t\rightarrow \infty , \end{aligned}$$

but we omit details. Since \(1/(1-\alpha )<p/(p-1)\), the Kasahara Tauberian theorem implies that

$$\begin{aligned} -\log \mathbb {P}\{S(\alpha )>x\}~\sim ~ C_1 x^{1/(1-\alpha )} \ll C_2 x^{p/(p-1)}~\sim ~- \log \mathbb {P}\{\eta >x\},\quad x\rightarrow \infty \end{aligned}$$

for appropriate positive constants \(C_1\) and \(C_2\). The first limit relation agrees with the precise (rather than logarithmic) tail behaviour exhibited in Theorems 1.1 and 1.2 for \(\alpha \in (1/2,1)\).

Plainly, relation (7) is not sufficient for obtaining a precise asymptotic. To succeed, more information on the function \(\psi \) is needed. Under our standing assumption \(\eta \le b\) a.s., \(\psi (t)\sim bt\) as \(t\rightarrow \infty \). Further terms in the asymptotic expansion for \(\psi \) depend on the behaviour of probabilities \(\mathbb {P}\{\eta \in [b-\delta ,b]\}\) as \(\delta \rightarrow 0+\). This fact justifies our assumptions in Theorems 1.1 and 1.2. It will be clear from the proofs that our argument still applies whenever

$$\begin{aligned} \psi ''(t)=t^{p-2}L(t)+o(t^{-2}),\quad t\rightarrow \infty \end{aligned}$$

for some \(p\le 1/\alpha \) and some L slowly varying at infinity. Summarizing, the assumed boundedness of \(\eta \) from the right admits a simple link between the asymptotics of \(\mathbb {P}\{\eta \in [b-\delta ,b]\}\) as \(\delta \rightarrow 0+\) and \(\psi (t)\) as \(t\rightarrow \infty \) and does not lead to a significant restriction of generality.

3 Description of Our Approach to the Tail Asymptotics

We first perform the exponential change of measure, which is standard in the area of large deviations. More precisely, for each \(t>0\), we define a new probability measure \(\mathbb {P}^{(t)}\) on the \(\sigma \)-algebra \(\sigma (\eta _1,\eta _2,\ldots )\) by

$$\begin{aligned} \mathbb {E}^{(t)}[g(\eta _1,\ldots , \eta _k)]=\frac{\mathbb {E}[\textrm{e}^{tS(\alpha )}g(\eta _1,\ldots ,\eta _k)]}{\mathbb {E}[\textrm{e}^{tS(\alpha )}]}, \end{aligned}$$
(9)

where \(\mathbb {E}^{(t)}\) denotes expectation with respect to \(\mathbb {P}^{(t)}\). The equality is assumed to hold for all \(k\in \mathbb {N}\) and each bounded Borel function \(g:\mathbb {R}^k\rightarrow \mathbb {R}\). Let \(f:\mathbb {R}\rightarrow \mathbb {R}\) be any bounded Borel function. Putting \(g(y_1,\ldots , y_k)=f\left( \sum _{j=1}^k j^{-\alpha } y_j\right) \) for \((y_1,\ldots , y_k)\in \mathbb {R}^k\) and letting \(k\rightarrow \infty \) we infer

$$\begin{aligned} \mathbb {E}^{(t)}[f(S(\alpha ))]=\frac{\mathbb {E}[\textrm{e}^{tS(\alpha )}f(S(\alpha ))]}{\mathbb {E}[\textrm{e}^{tS(\alpha )}]}. \end{aligned}$$
(10)

Equality (10) holds true for any Borel function \(f:\mathbb {R}\rightarrow \mathbb {R}\) which is not necessarily bounded, whenever the left- or right-hand side of (10) is well defined, possibly infinite.

An important property of the exponential change of measure in (9) is that the random variables \(\eta _1\), \(\eta _2,\ldots \) remain independent under \(\mathbb {P}^{(t)}\). In other words, the change of measure only affects the marginal distributions of individual random variables.

Recall the notation \(\psi (t)=\log \mathbb {E}[ \textrm{e}^{t\eta }]\) for \(t\in \mathbb {R}\). Then, \(\mathbb {E}[\textrm{e}^{tS(\alpha )}]=\exp (\sum _{k\ge 1}\psi (t/k^\alpha ))\) for \(t\in \mathbb {R}\). Using (10) with \(f(y)=y\), we obtain

$$\begin{aligned} \mathbb {E}^{(t)}[S(\alpha )]=\frac{\mathbb {E}[\textrm{e}^{tS(\alpha )}S(\alpha )]}{\mathbb {E}[\textrm{e}^{tS(\alpha )}]}=\sum _{k\ge 1}k^{-\alpha }\psi ^\prime (t/k^\alpha ). \end{aligned}$$

Fix any \(x>0\) and put \(f(y)=\textrm{e}^{-ty}\mathbbm {1}_{(x,\infty )}(y)\). Then, (10) reads

$$\begin{aligned} \mathbb {P}\{S(\alpha )>x\}= & {} \mathbb {E}[\textrm{e}^{tS(\alpha )}]\mathbb {E}^{(t)}[\textrm{e}^{-tS(\alpha )}\mathbbm {1}_{\{S(\alpha )>x\}}]\\= & {} \textrm{e}^{-tx+\sum _{k\ge 1}\psi (t/k^\alpha )}\mathbb {E}^{(t)}[\textrm{e}^{-t(S(\alpha )-x)}\mathbbm {1}_{\{S(\alpha )-x>0\}}]. \end{aligned}$$

Under \(\mathbb {P}^{(t)}\), put \(S_0^{(t)}(\alpha )=S(\alpha )-\mathbb {E}^{(t)}[S(\alpha )]\). We shall write \(S_0(\alpha )\) for \(S_0^{(t)}(\alpha )\) unless it leads to ambiguity. The function \(t\mapsto \sum _{k\ge 1}k^{-\alpha }\psi ^\prime (t/k^\alpha )\) is continuous, strictly increasing on \([0,\infty )\) and equal to 0 at 0. Hence, for each \(x\ge 0\), the equation

$$\begin{aligned} \sum _{k\ge 1}k^{-\alpha }\psi ^\prime (t/k^\alpha )=x \end{aligned}$$
(11)

has a unique solution \(t=t(x)\). We shall investigate the asymptotic behaviour of \(\mathbb {P}\{S(\alpha )>x\}\) as \(x\rightarrow \infty \) with the help of the representation

$$\begin{aligned} \mathbb {P}\{S(\alpha )>x\}{} & {} =\exp \left\{ \sum _{k\ge 1}(\psi (t(x)/k^\alpha )-(t(x)/k^\alpha )\psi ^\prime (t(x)/k^\alpha ))\right\} \nonumber \\{} & {} \quad \mathbb {E}^{(t(x))}[\textrm{e}^{-t(x)S_0(\alpha )}\mathbbm {1}_{\{S_0(\alpha )>0\}}]. \end{aligned}$$
(12)

Our analysis of the terms on the right-hand side of (12) consists of the following four steps.

Step 1. Find an asymptotic expansion of \(\mathbb {E}^{(t)}[S(\alpha )]=\sum _{k\ge 1}k^{-\alpha }\psi ^\prime (t/k^\alpha )\) as \(t\rightarrow \infty \) with a sufficient precision. It turns out that an expansion up to the term o(1/t) serves our purpose. Use the expansion to determine the asymptotic behaviour of the solution t(x) to equation (11). This step is implemented in Propositions 5.1 and 5.2.

Step 2. Find an asymptotic expansion of \(\sum _{k\ge 1}(\psi (t/k^\alpha )-(t/k^\alpha )\psi ^\prime (t/k^\alpha ))\) as \(t\rightarrow \infty \) up to the term o(1). The o(1)-precision is needed to obtain precise asymptotics for the exponential term in (12). This is done in Propositions 5.3 and 5.4.

Step 3. Find the first-order asymptotic of \(\mathbb {E}^{(t)}[\textrm{e}^{-tS_0(\alpha )}\mathbbm {1}_{\{S_0(\alpha )>0\}}]\) as \(t\rightarrow \infty \). To this end, we prove a local central limit theorem (CLT) for \(S_0(\alpha )\) under \(\mathbb {P}^{(t)}\). We note that the use of CLT-like results is also a very common tool in the study of large deviation probabilities. For example, one applies the Berry–Esseen inequality to derive an exact large deviation asymptotic for sums of independent random variables, see Chapter VIII in Petrov’s book [7]. Surprisingly, it turned out that the application of the Berry–Esseen inequality to \(S(\alpha )\) does not allow one to determine the asymptotics of \(\mathbb {E}^{(t)}[\textrm{e}^{-tS_0(\alpha )}\mathbbm {1}_{\{S_0(\alpha )>0\}}]\). A further peculiarity of \(S(\alpha )\) consists in the fact that its variance under \(\mathbb {P}^{(t)}\) goes to zero and, consequently, one has a kind of ‘superconcentration’ effect around \(\mathbb {E}^{(t)}[S(\alpha )]\). The local CLT is proven in Theorem 5.7. The expectation \(\mathbb {E}^{(t)}[\textrm{e}^{-tS_0(\alpha )}\mathbbm {1}_{\{S_0(\alpha )>0\}}]\) is analysed in Proposition 5.5.

Step 4. To conclude, replace t with t(x) in the asymptotic expansions obtained at Steps 2 and 3. In this way, both Theorems 1.1 and 1.2 follow.

4 Preparation for the Proofs

4.1 A Version of the Euler–Maclaurin Formula

There are several versions of the Euler–Maclaurin summation formula. Below we state the one that serves our needs.

Let \(m,n\in \mathbb {N}\), \(m<n\) and \(f: [m,n]\rightarrow \mathbb {R}\) be a twice continuously differentiable function. A specialization of formula (9.78) on p. 460 in [4] yields

$$\begin{aligned} \sum _{j=m}^n f(j)=\int _m^n f(x)\textrm{d}x+(f(n)+f(m))/2+(f^\prime (n)-f^\prime (m))/12+R_{m,n},\nonumber \\ \end{aligned}$$
(13)

where \(|R_{m,n}|\le (1/12)\int _m^n |f^{\prime \prime }(x)|\textrm{d}x\).

If \(f: [m,\infty )\rightarrow \mathbb {R}\) is a twice continuously differentiable function with \(\lim _{x\rightarrow \infty }f(x)=\lim _{x\rightarrow \infty }f^\prime (x)=0\) and \(\int _m^\infty |f^{\prime \prime }(x)|\textrm{d}x<\infty \), then

$$\begin{aligned} \sum _{j\ge m} f(j)=\int _m^\infty f(x)\textrm{d}x+f(m)/2-f^\prime (m)/12+R_m, \end{aligned}$$
(14)

where \(|R_{m}|\le (1/12)\int _m^\infty |f^{\prime \prime }(x)|\textrm{d}x\).

4.2 Auxiliary Results

Lemma 4.1

Assume that \(\eta \le b\) a.s. and \(\mathbb {P}\{\eta =b\}=\theta \in (0,1)\). Then, as \(t\rightarrow \infty \),

  1. (a)

    \(\psi (t)=bt+\log \theta +o(1)\);

  2. (b)

    \(t\psi ^\prime (t)=bt+o(1)\) and \(L^\prime (t)=o(1/t)\), where \(L(t):=\log \mathbb {E}[\textrm{e}^{-t(b-\eta )}]\) for \(t\ge 0\);

  3. (c)

    \(\lim _{t\rightarrow \infty } t^2 \psi ^{\prime \prime }(t)=\lim _{t\rightarrow \infty } t^3 |\psi ^{\prime \prime \prime }(t)|=0\).

Proof

(a) This is justified as follows

$$\begin{aligned} \psi (t){} & {} =bt+\log \mathbb {E}\left[ \textrm{e}^{-t(b-\eta )}\right] =bt+\log \theta +\log \left( 1+\theta ^{-1}\mathbb {E}\left[ \textrm{e}^{-t(b-\eta )}\mathbbm {1}_{\{\eta <b\}}\right] \right) \\{} & {} =bt+\log \theta +o(1),\quad t\rightarrow \infty . \end{aligned}$$

The last equality stems from the fact that \(\lim _{t\rightarrow \infty }\mathbb {E}\left[ \textrm{e}^{-t(b-\eta )}\mathbbm {1}_{\{\eta <b\}}\right] =0\).

(b) Put \(\ell (t):=\mathbb {E}[\textrm{e}^{-t(b-\eta )}]\) for \(t\ge 0\). Then \(t\psi ^\prime (t)=bt+t\ell ^\prime (t)/\ell (t)=bt+tL^\prime (t)\). By Lebesgue’s dominated convergence theorem, for \(n\in \mathbb {N}\),

$$\begin{aligned} t^n\left| \ell ^{(n)}(t)\right| =\mathbb {E}[(t(b-\eta ))^n \textrm{e}^{-t(b-\eta )}]~\rightarrow ~ 0,\quad t\rightarrow \infty , \end{aligned}$$
(15)

where \(\ell ^{(n)}\) denotes the nth derivative of the function \(\ell \). Indeed, \(\lim _{t\rightarrow \infty }(t(b-\eta ))^n \textrm{e}^{-t(b-\eta )}=0\) a.s., and the function \(x\mapsto x^n\textrm{e}^{-x}\) is bounded on \([0,\infty )\). Since

$$\begin{aligned} \lim _{t\rightarrow \infty }\ell (t)=\theta , \end{aligned}$$
(16)

the claims of part (b) follow from (15) with \(n=1\).

(c) The proof is analogous to that of part (b). We only treat the third derivative. Since

$$\begin{aligned} \psi ^{\prime \prime \prime }(t)=\frac{\ell ^{\prime \prime \prime }(t)\ell (t)-\ell ^\prime (t)\ell ^{\prime \prime }(t)}{(\ell (t))^2}-\frac{2\ell ^\prime (t)}{\ell (t)}\frac{\ell ^{\prime \prime }(t)\ell (t)-(\ell ^\prime (t))^2}{(\ell (t))^2}, \end{aligned}$$
(17)

the result is secured by (15) with \(n=1,2,3\) and (16). \(\square \)

Lemma 4.2

Assume that \(\eta \le b\) a.s. and \(\mathbb {P}\{b-\eta \le x\}\sim \lambda x^r\) as \(x\rightarrow 0+\) for positive \(\lambda \) and r. Then, as \(t\rightarrow \infty \),

  1. (a)

    \(\psi (t)=bt-r\log t+\log \left( \lambda \Gamma (r+1)\right) +o(1)\);

  2. (b)

    \(L^\prime (t)=-rt^{-1}+o(1/t)\) and \(\psi ^\prime (t)=b-rt^{-1}+o(1/t)\);

  3. (c)

    \(\psi ^{\prime \prime }(t)=L^{\prime \prime }(t)=rt^{-2}+o(1/t^2)\) and \(\psi ^{\prime \prime \prime }(t)=L^{\prime \prime \prime }(t)=-2rt^{-3}+o(1/t^3)\).

Proof

(a) By Theorem 1.7.1’ in [2],

$$\begin{aligned} \ell (t)=\mathbb {E}\textrm{e}^{-t(b-\eta )}~\sim ~\lambda \Gamma (1+r)t^{-r},\quad t\rightarrow \infty , \end{aligned}$$
(18)

where \(\Gamma \) is the Euler gamma function. This entails

$$\begin{aligned} L(t)=\log \ell (t)=-r\log t+\log \left( \lambda \Gamma (r+1)\right) +o(1), \end{aligned}$$

whence

$$\begin{aligned} \psi (t)=bt+L(t)=bt-r\log t+\log \left( \lambda \Gamma (r+1)\right) +o(1). \end{aligned}$$

(b) Using \(\ell (t)=\int _t^\infty (-\ell ^\prime (x))\textrm{d}x\), the fact that \(-\ell '\) is nonincreasing and the monotone density theorem (Theorem 1.7.2 in [2]) we infer

$$\begin{aligned} -\ell ^\prime (t)~\sim ~\lambda r\Gamma (1+r)t^{-(1+r)},\quad t\rightarrow \infty . \end{aligned}$$
(19)

Relations (18) and (19) entail

$$\begin{aligned} L^\prime (t)=rt^{-1}+o(1/t),\quad t\rightarrow \infty . \end{aligned}$$

The proof of part (c) is analogous, hence omitted. The basic observation is that, for \(n\ge 2\), \((-1)^n \ell ^{(n)}\) is a nonincreasing function. This enables us to use the monotone density theorem. \(\square \)

Lemma 4.3

Assume that \(\psi \) is finite in some right vicinity of 0, that \(\mathbb {E}[\eta ]=0\) and \(\mathbb {E}[\eta ^2]\in (0,\infty )\). Then, as \(t\rightarrow 0+\),

$$\begin{aligned} \psi (t)~\sim ~\mathbb {E}[\eta ^2]t^2/2, \ \ \psi ^\prime (t)~\sim ~\mathbb {E}[\eta ^2] t\quad \text {and} \ \ \lim _{t\rightarrow 0+}\psi ^{\prime \prime }(t)=\mathbb {E}[\eta ^2]. \end{aligned}$$

If, in addition, \(\eta \le b\) a.s., then

$$\begin{aligned} \lim _{t\rightarrow 0+}t \psi ^{\prime \prime \prime }(t)=0. \end{aligned}$$

Proof

The first three claims are standard, and we omit a proof.

As for the last limit relation, observe that, in view of (17), the limit \(\lim _{t\rightarrow 0+}\psi ^{\prime \prime \prime }(t)\) is finite provided that \(\mathbb {E}[|\eta |^3]<\infty \). Thus, the last claim holds trivially in this case. If \(\mathbb {E}[|\eta |^3]=\infty \), then, as \(t\rightarrow 0+\), the limits of all functions, except \(\ell ^{\prime \prime \prime }\), appearing in (17) are still finite. It remains to note that, by Lebesgue’s dominated convergence theorem,

$$\begin{aligned} t\left| \ell ^{\prime \prime \prime }(t)\right| =\mathbb {E}[(b-\eta )^2(t(b-\eta ))\textrm{e}^{-t(b-\eta )}]~\rightarrow ~ 0,\quad t\rightarrow 0+ \end{aligned}$$

because the function \(x\mapsto x\textrm{e}^{-x}\) is bounded on \([0,\infty )\) and \(\mathbb {E}[(b-\eta )^2]<\infty \).

\(\square \)

5 Proofs of Theorems 1.1 and 1.2

We follow the steps outlined in Sect. 3. Step 1 is realized by the following propositions.

Proposition 5.1

Assume that \(\eta \le b\) a.s. with \(\mathbb {P}\{\eta =b\}=\theta \in (0,1)\), \(\mathbb {E}[\eta ]=0\) and \(\mathbb {E}[\eta ^2]<\infty \). If \(\alpha =1\), then

$$\begin{aligned} \sum _{k\ge 1}k^{-1}\psi ^\prime (t/k)=b\log t+q+o(1/t),\quad t\rightarrow \infty , \end{aligned}$$
(20)

where \(q=b\gamma _1+\int _0^1 x^{-1}\psi ^\prime (x)\textrm{d}x+\int _1^\infty x^{-1}(\psi ^\prime (x)-b)\textrm{d}x\in \mathbb {R}\). For each \(x\ge 0\), the equation \(\sum _{k\ge 1}k^{-1}\psi ^\prime (t/k)=x\) has a unique solution \(t=t(x)\) satisfying

$$\begin{aligned} t(x)=\exp ((x-q)/b)+o(1),\quad x\rightarrow \infty . \end{aligned}$$
(21)

If \(\alpha \in (1/2, 1)\), then

$$\begin{aligned} \sum _{k\ge 1}k^{-\alpha }\psi ^\prime (t/k^\alpha )=r_\alpha t^{-1+1/\alpha }+b\gamma _\alpha +o(1/t),\quad t\rightarrow \infty , \end{aligned}$$
(22)

where \(r_\alpha :=\alpha \sigma ^2_\alpha /(1-\alpha )\in (0,\infty )\). For each \(x\ge 0\), the equation \(\sum _{k\ge 1}k^{-1}\psi ^\prime (t/k)=x\) has a unique solution \(t=t(x)\) satisfying

$$\begin{aligned} t(x)=(r_\alpha ^{-1}(x-b\gamma _\alpha )+o(x^{-\alpha /(1-\alpha )}))^{\alpha /(1-\alpha )},\quad x\rightarrow \infty . \end{aligned}$$
(23)

Proposition 5.2

Assume that \(\eta \le b\) a.s., \(\mathbb {P}\{b-\eta \le x\}\sim \lambda x^r\) as \(x\rightarrow 0+\) for positive \(\lambda \) and r, \(\mathbb {E}[\eta ]=0\) and \(\mathbb {E}[\eta ^2]<\infty \). If \(\alpha =1\), then

$$\begin{aligned} \sum _{k\ge 1}k^{-1}\psi ^\prime (t/k)=b\log t+q+rt^{-1}/2+o(1/t),\quad t\rightarrow \infty \end{aligned}$$
(24)

with the same q as in Proposition 5.1. For each \(x\ge 0\), the equation \(\sum _{k\ge 1}k^{-1}\psi ^\prime (t/k) =x\) has a unique solution \(t=t(x)\) satisfying

$$\begin{aligned} t(x)=\exp ((x-q)/b)-r/(2b)+o(1),\quad x\rightarrow \infty . \end{aligned}$$
(25)

If \(\alpha \in (1/2, 1)\), then

$$\begin{aligned} \sum _{k\ge 1}k^{-\alpha }\psi ^\prime (t/k^\alpha )=r_\alpha t^{-1+1/\alpha }+b\gamma _\alpha +rt^{-1}/2+o(1/t),\quad t\rightarrow \infty \end{aligned}$$
(26)

with the same \(r_\alpha \) as in Proposition 5.1. For each \(x\ge 0\), the equation \(\sum _{k\ge 1}k^{-1}\psi ^\prime (t/k) =x\) has a unique solution \(t=t(x)\) satisfying

$$\begin{aligned} t(x)= & {} (r_\alpha ^{-1}(x-b\gamma _\alpha )-(rr_\alpha ^{(2\alpha -1)/(1-\alpha )}/2)x^{-\alpha /(1-\alpha )}\nonumber \\{} & {} +o(x^{-\alpha /(1-\alpha )}))^{\alpha /(1-\alpha )},\quad x\rightarrow \infty . \end{aligned}$$
(27)

Now we are passing to Step 2.

Proposition 5.3

Assume that \(\eta \le b\) a.s. with \(\mathbb {P}\{\eta =b\}=\theta \in (0,1)\), \(\mathbb {E}[\eta ]=0\) and \(\mathbb {E}[\eta ^2]<\infty \). Then for \(\alpha \in (1/2, 1]\), as \(t\rightarrow \infty \),

$$\begin{aligned} \sum _{k\ge 1}\left( \psi (t/k^\alpha )-(t/k^\alpha )\psi ^\prime (t/k^\alpha )\right) =-\alpha \sigma _\alpha ^2t^{1/\alpha }-2^{-1}\log \theta +o(1) \end{aligned}$$
(28)

with \(\sigma _\alpha ^2\) as defined in Theorem 1.1 (in particular, \(\sigma _1^2=b\)).

Proposition 5.4

Assume that \(\eta \le b\) a.s., \(\mathbb {P}\{b-\eta \le x\}\sim \lambda x^r\) as \(x\rightarrow 0+\) for positive \(\lambda \) and r, \(\mathbb {E}[\eta ]=0\) and \(\mathbb {E}[\eta ^2]<\infty \). Then for \(\alpha \in (1/2, 1]\), as \(t\rightarrow \infty \),

$$\begin{aligned}&\sum _{k\ge 1}\left( \psi (t/k^\alpha )-(t/k^\alpha )\psi ^\prime (t/k^\alpha )\right) \nonumber \\&\quad =-\alpha \sigma _\alpha ^2t^{1/\alpha }+(r/2)\log t + (r/2)(\alpha \log (2\pi )-1)-(1/2)\nonumber \\&\qquad \log (\lambda \Gamma (r+1))+o(1) \end{aligned}$$
(29)

with \(\sigma _\alpha ^2\) as defined in Theorem 1.1.

Step 3 is implemented with the help of the result given next.

Proposition 5.5

Assume that \(\eta \le b\) a.s., \(\mathbb {E}[\eta ]=0\) and \(\mathbb {E}[\eta ^2]\in (0,\infty )\). Then

$$\begin{aligned} \lim _{t\rightarrow \infty } t^{1/(2\alpha )}\mathbb {E}^{(t)}\left[ \textrm{e}^{-tS_0(\alpha )}\mathbbm {1}_{\{S_0(\alpha )>0\}}\right] =(2\pi \sigma _\alpha ^2)^{-1/2} \end{aligned}$$

with \(\sigma ^2_\alpha \) as defined in Theorem 1.1.

With these propositions at hand, we are ready to prove Theorems 1.1 and 1.2.

Proof of Theorem 1.1

We only discuss the case \(\alpha \in (1/2,1)\), the case \(\alpha =1\) being simpler.

Our starting point is representation (12). Invoking Proposition 5.5 in combination with \(\lim _{x\rightarrow \infty }t(x)=+\infty \) we obtain

$$\begin{aligned} \mathbb {E}^{(t(x))}\left[ \textrm{e}^{-t(x)S_0(\alpha )}\mathbbm {1}_{\{S_0(\alpha )>0\}}\right] ~\sim ~\frac{1}{(2\pi \sigma ^2_\alpha (t(x))^{1/\alpha })^{1/2}},\quad x\rightarrow \infty . \end{aligned}$$

Using (23) we infer

$$\begin{aligned}{} & {} (t(x))^{1/\alpha }-(r^{-1}_\alpha (x-b\gamma _\alpha ))^{1/(1-\alpha )}\nonumber \\{} & {} \quad =(r^{-1}_\alpha (x-b\gamma _\alpha )+o(x^{-\alpha /(1-\alpha )}))^{1/(1-\alpha )}- (r^{-1}_\alpha (x-b\gamma _\alpha ))^{1/(1-\alpha )}\nonumber \\{} & {} \quad =O(x^{1/(1-\alpha )})o(x^{-1/(1-\alpha )})=o(1),\quad x\rightarrow \infty . \end{aligned}$$
(30)

This yields

$$\begin{aligned} \mathbb {E}^{(t(x))}\left[ \textrm{e}^{-t(x)S_0(\alpha )}\mathbbm {1}_{\{S_0(\alpha )>0\}}\right] ~\sim ~\left( \frac{r^{1/(1-\alpha )}_\alpha }{2\pi \sigma ^2_\alpha x^{1/(1-\alpha )}}\right) ^{1/2},\quad x\rightarrow \infty . \end{aligned}$$
(31)

By Proposition 5.3,

$$\begin{aligned}{} & {} \exp \left( \sum _{k\ge 1}(\psi (t(x)/k^\alpha )-(t(x)/k^\alpha )\psi ^\prime (t(x)/k^\alpha ))\right) ~\sim ~ \frac{1}{\theta ^{1/2}}\exp \left( -\alpha \sigma _\alpha ^2(t(x))^{1/\alpha }\right) \\{} & {} \quad ~\sim ~ \frac{1}{\theta ^{1/2}}\exp \left( -\alpha \sigma ^2_\alpha \left( \frac{x-b\gamma _\alpha }{r_\alpha }\right) ^{1/(1-\alpha )}\right) ,\quad x\rightarrow \infty . \end{aligned}$$

Combining this with (31) proves Theorem 1.1 in the case \(\alpha \in (1/2,1)\). \(\square \)

Proof of Theorem 1.2

This proof is analogous to that of Theorem 1.1. We only give a counterpart of (30) for \(t=t(x)\) satisfying (27):

$$\begin{aligned} (t(x))^{1/\alpha }=(r^{-1}_\alpha (x-b\gamma _\alpha ))^{1/(1-\alpha )}-(rr_\alpha ^{(2\alpha -1)/(1-\alpha )})/(2(1-\alpha ))+o(1),\quad x\rightarrow \infty . \end{aligned}$$

\(\square \)

5.1 Proof of Propositions 5.1 and 5.2

Proof of Proposition 5.1

Fix any \(\beta>2/(2\alpha -1)>1\). By Lemma 4.3, \(\psi ^\prime (t)\sim \mathbb {E}[\eta ^2]t\) as \(t\rightarrow 0+\). Hence,

$$\begin{aligned} \sum _{k\ge \lfloor t^\beta \rfloor +1}k^{-\alpha }\psi ^\prime (t/k^\alpha )=O\left( t\sum _{k\ge \lfloor t^\beta \rfloor +1}k^{-2\alpha }\right) =O(t^{1-(2\alpha -1)\beta })=o(1/t),\quad t\rightarrow \infty . \end{aligned}$$

Recall that \(\psi ^\prime (t)=b+L^\prime (t)\), where \(L(t)=\log \mathbb {E}[\textrm{e}^{-t(b-\eta )}]\) for \(t\ge 0\), and write

$$\begin{aligned} \sum _{k=1}^{\lfloor t^\beta \rfloor }k^{-\alpha }\psi ^\prime (t/k^\alpha )=b\sum _{k=1}^{\lfloor t^\beta \rfloor }k^{-\alpha }+\sum _{k=1}^{\lfloor t^\beta \rfloor }k^{-\alpha }L^\prime (t/k^\alpha ):=A_\alpha (t)+B_\alpha (t). \end{aligned}$$

By Theorem 3.2 (a,b) on p. 55 in [1], in the case \(\alpha =1\),

$$\begin{aligned} A_1(t)=b\log (\lfloor t^\beta \rfloor )+b\gamma _1+O(t^{-\beta }),\quad t\rightarrow \infty , \end{aligned}$$
(32)

whereas in the case \(\alpha \in (1/2,1)\),

$$\begin{aligned} A_\alpha (t)=b(1-\alpha )^{-1}(\lfloor t^\beta \rfloor )^{1-\alpha }+b\gamma _\alpha +O(t^{-\alpha \beta }),\quad t\rightarrow \infty , \end{aligned}$$
(33)

where \(\gamma _\alpha \) is the Euler–Mascheroni constant, see (1) and (2).

We intend to use formula (13) with \(f=f_t\), \(m=1\) and \(n=\lfloor t^\beta \rfloor \), where \(f_t(x)=x^{-\alpha }L^\prime (t/x^\alpha )\). For later needs, we note that

$$\begin{aligned} f_t^\prime (x)=-\alpha \left( x^{-(\alpha +1)}L^\prime (t/x^\alpha ) +(t/x^{2\alpha +1})L^{\prime \prime }(t/x^\alpha )\right) \end{aligned}$$

and

$$\begin{aligned} f_t^{\prime \prime }(x)= & {} \alpha \Big ((\alpha +1)x^{-(\alpha +2)}L^\prime (t/x^\alpha ) +(3\alpha +1)tx^{-(2\alpha +2)}L^{\prime \prime }(t/x^\alpha )\\{} & {} +\alpha t^2 x^{-(3\alpha +2)}L^{\prime \prime \prime }(t/x^\alpha )\Big ). \end{aligned}$$

According to (13),

$$\begin{aligned} B_\alpha (t){} & {} =\int _1^{\lfloor t^\beta \rfloor }x^{-\alpha }L^\prime (t/x^\alpha )\textrm{d}x+\left( L^\prime (t)+(\lfloor t^\beta \rfloor )^{-\alpha }L^\prime (t/(\lfloor t^\beta \rfloor )^\alpha )\right) /2\nonumber \\{} & {} \quad +\alpha \Big (L^\prime (t)+tL^{\prime \prime }(t)-(\lfloor t^\beta \rfloor )^{-(\alpha +1)}L^\prime (t/(\lfloor t^\beta \rfloor )^\alpha )\nonumber \\{} & {} \quad -(t/(\lfloor t^\beta \rfloor )^{2\alpha +1})L^{\prime \prime }\left( t/(\lfloor t^\beta \rfloor )^\alpha \right) \Big )/12+R(t), \end{aligned}$$
(34)

where \(R(t)\le (1/12)\int _1^{\lfloor t^\beta \rfloor }|f_t^{\prime \prime }(x)|\textrm{d}x\).

By Lemma 4.1(b,c), \(L^\prime (t)=o(1/t)\) and \(tL^{\prime \prime }(t)=t\psi ^{\prime \prime }(t)=o(1/t)\) as \(t\rightarrow \infty \). Using \(\lim _{u\rightarrow 0+}L^\prime (u)=-b\) and the choice of \(\beta \) which ensures \(\alpha \beta \ge (2\alpha -1)\beta>2>1\) because \(\alpha \in (1/2,1]\), we infer

$$\begin{aligned} (\lfloor t^\beta \rfloor )^{-\alpha }L^\prime \left( t/(\lfloor t^\beta \rfloor )^\alpha \right) =O(t^{-\alpha \beta })=o(1/t),\quad t\rightarrow \infty . \end{aligned}$$

The latter trivially implies that \((\lfloor t^\beta \rfloor )^{-(\alpha +1)}L^\prime \left( t/(\lfloor t^\beta \rfloor )^\alpha \right) =o(1/t)\). According to Lemma 4.3,

$$\begin{aligned} (t/(\lfloor t^\beta \rfloor )^{2\alpha +1})L^{\prime \prime }\left( t/(\lfloor t^\beta \rfloor )^\alpha \right)= & {} (t/(\lfloor t^\beta \rfloor )^{2\alpha +1})\psi ^{\prime \prime }\left( t/(\lfloor t^\beta \rfloor )^\alpha \right) \\= & {} O(t^{-\beta (2\alpha +1)+1})=o(1/t). \end{aligned}$$

Further, substituting \(y=t/x^\alpha \), applying the L’Hôpital rule and Lemma 4.1(b), we obtain

$$\begin{aligned} \alpha \int _1^\infty x^{-(\alpha +2)}L^\prime (t/x^\alpha )\textrm{d}x&=t^{-(1+1/\alpha )}\int _1^\infty (t/x^\alpha )^{1/\alpha } (\alpha t/x^{\alpha +1})L^\prime (t/x^\alpha )\textrm{d}x\\&=t^{-(1+1/\alpha )}\int _0^t y^{1/\alpha }L^\prime (y)\textrm{d}y~\sim ~\alpha (\alpha +1)^{-1}L^\prime (t)\\&=o(1/t),\quad t\rightarrow \infty . \end{aligned}$$

Similarly,

$$\begin{aligned} \alpha t\int _1^\infty x^{-(2\alpha +2)}L^{\prime \prime }(t/x^\alpha )\textrm{d}x&=t^{-(1+1/\alpha )}\int _1^\infty (t/x^\alpha )^{1+1/\alpha } (\alpha t/x^{\alpha +1})L^{\prime \prime }(t/x^\alpha )\textrm{d}x\\&=t^{-(1+1/\alpha )}\int _0^t y^{1+1/\alpha }L^{\prime \prime }(y)\textrm{d}y~\sim ~\alpha (\alpha +1)^{-1}tL^{\prime \prime }(t)\\&=o(1/t),\quad t\rightarrow \infty \end{aligned}$$

and

$$\begin{aligned} \alpha t^2\int _1^\infty x^{-(3\alpha +2)}L^{\prime \prime }(t/x^\alpha )\textrm{d}x&= t^{-(1+1/\alpha )}\int _1^\infty (t/x^\alpha )^{2+1/\alpha } (\alpha t/x^{\alpha +1})L^{\prime \prime \prime }(t/x^\alpha )\textrm{d}x\\&=t^{-(1+1/\alpha )}\int _0^t y^{2+1/\alpha }L^{\prime \prime \prime }(y)\textrm{d}y~\sim ~\alpha (\alpha +1)^{-1}t^2L^{\prime \prime \prime }(t)\\&=o(1/t),\quad t\rightarrow \infty . \end{aligned}$$

This proves \(R(t)=o(1/t)\). As a result,

$$\begin{aligned} B_\alpha (t)=\int _1^{\lfloor t^\beta \rfloor }x^{-\alpha }L^\prime (t/x^\alpha )\textrm{d}x+o(1/t),\quad t\rightarrow \infty . \end{aligned}$$

Since \(\alpha \beta >1\) and \(\lim _{u\rightarrow 0+}L'(u)=-b\),

$$\begin{aligned} \int _{\lfloor t^\beta \rfloor }^{t^\beta }x^{-\alpha }L^\prime (t/x^\alpha )\textrm{d}x=O(t^{-\alpha \beta }) =o(t^{-1}). \end{aligned}$$

Consequently,

$$\begin{aligned} B_\alpha (t)=\int _1^{t^\beta }x^{-\alpha }L^\prime (t/x^\alpha )\textrm{d}x+o(1/t),\quad t\rightarrow \infty . \end{aligned}$$
(35)

By Lemma 4.3, \(\lim _{y\rightarrow 0+}y^{-1}\psi ^\prime (y)=\mathbb {E}[\eta ^2]\). Hence, the integral \(c_0(\alpha ):=\int _0^1 y^{-1/\alpha }\psi ^\prime (y)\textrm{d}y\) is well defined and

$$\begin{aligned} \int _0^{t^{1-\beta \alpha }}y^{-1/\alpha }\psi ^\prime (y)\textrm{d}y=O(t^{-(\beta \alpha -1)(2-1/\alpha )})=o(t^{-1/\alpha }),\quad t\rightarrow \infty \end{aligned}$$

(indeed, \((\beta \alpha -1)(2-1/\alpha )>1/\alpha \) by the choice of \(\beta \)). In view of \(L^\prime (y)=o(1/y)\) as \(y\rightarrow \infty \) (see Lemma 4.1(b)), we conclude that the integral \(c_1(\alpha ):=-\int _1^\infty y^{-1/\alpha }L^\prime (y)\textrm{d}y=-\int _1^\infty y^{-1/\alpha }(\psi ^\prime (y)-b)\textrm{d}y\) is also well defined and \(-\int _t^\infty y^{-1/\alpha }L^\prime (y)\textrm{d}y=o(t^{-1/\alpha })\) as \(t\rightarrow \infty \). Finally,

$$\begin{aligned}{} & {} \int _1^{t^\beta }x^{-\alpha }L^\prime (t/x^\alpha )\textrm{d}x\nonumber \\{} & {} \quad =\alpha ^{-1}t^{-1+1/\alpha }\left( \int _{t^{1-\beta \alpha }}^1 y^{-1/\alpha }L^\prime (y)\textrm{d}y+\int _1^t y^{-1/\alpha }L^\prime (y)\textrm{d}y\right) \nonumber \\{} & {} \quad =\alpha ^{-1}t^{-1+1/\alpha }\left( -b\int _{t/(\lfloor t^\beta \rfloor )^\alpha }^1 y^{-1/\alpha }\textrm{d}y+\int _0^1 y^{-1/\alpha }\psi ^\prime (y)\textrm{d}y\right. \nonumber \\{} & {} \quad -\int _0^{t^{1-\beta \alpha }}y^{-1/\alpha }\psi ^\prime (y)\textrm{d}y \nonumber \\{} & {} \quad \left. +\int _1^\infty y^{-1/\alpha }L^\prime (y)\textrm{d}y-\int _t^\infty y^{-1/\alpha }L^\prime (y)\textrm{d}y\right) . \end{aligned}$$
(36)

In the case \(\alpha =1\) we infer

$$\begin{aligned} \int _1^{t^{\beta }}x^{-1}L^\prime (t/x)\textrm{d}x=b\log t-b\log (\lfloor t^\beta \rfloor )+c_0(1)-c_1(1)+o(1/t),\quad t\rightarrow \infty . \end{aligned}$$

A combination of this, (32) and (35) yields (20). In the case \(\alpha \in (1/2,1)\) we obtain

$$\begin{aligned} \int _1^{t^{\beta }}x^{-\alpha }L^\prime (t/x^\alpha )\textrm{d}x= & {} (b(1-\alpha )^{-1}+\alpha ^{-1}(c_0(\alpha )-c_1(\alpha )))t^{-1+1/\alpha }\\{} & {} -b(1-\alpha )^{-1}(\lfloor t^\beta \rfloor )^{1-\alpha }+o(1/t) \end{aligned}$$

as \(t\rightarrow \infty \). Recall that \(\sigma ^2_\alpha =(1-\alpha )\alpha ^{-3}\int _0^\infty y^{-1-1/\alpha }\psi (y)\textrm{d}y\). This entails

$$\begin{aligned}&b(1-\alpha )^{-1}+\alpha ^{-1}(c_0(\alpha )-c_1(\alpha ))\nonumber \\&\quad =\alpha ^{-1}\int _0^\infty y^{-1/\alpha }\psi ^\prime (y)\textrm{d}y=\alpha ^{-2}\int _0^\infty y^{-1-1/\alpha }\psi (y)\textrm{d}y\nonumber \\&\qquad =\alpha \sigma _\alpha ^2/(1-\alpha )=r_\alpha . \end{aligned}$$
(37)

Using these together with (33) and (35) we arrive at (22).

It was explained in Sect. 3 that, for each \(x\ge 0\), the equation \(\sum _{k\ge 1}k^{-\alpha }\psi ^\prime (t/k^\alpha )=x\) has a unique solution \(t=t(x)\). To determine its asymptotic behaviour in the case \(\alpha =1\), write

$$\begin{aligned} b\log t+q+o(1/t)=x \end{aligned}$$
(38)

or, equivalently, \(t^b \textrm{e}^{o(1/t)}=\textrm{e}^{x-q}\). As a consequence, \(t(x)=\exp ((x-q)/b)(1+\varepsilon (x))\), where \(\varepsilon \) satisfies \(\lim _{x\rightarrow \infty }\varepsilon (x)=0\). Plugging this into (38) we infer \(\varepsilon (x)=o(\textrm{e}^{-x/b})\) as \(x\rightarrow \infty \) and thereupon (21).

Let now \(\alpha \in (1/2,1)\). Starting with

$$\begin{aligned} r_\alpha t^{-1+1/\alpha }+b\gamma _\alpha +o(1/t)=x \end{aligned}$$
(39)

we conclude that \(t(x)=(r_\alpha ^{-1}(x-b\gamma _\alpha )+\delta (x))^{\alpha /(1-\alpha )}\) for some \(\delta \) satisfying \(\lim _{x\rightarrow \infty }\delta (x)=0\). Plugging this expression into (39), we obtain \(\delta (x)=o(x^{-\alpha /(1-\alpha )})\) as \(x\rightarrow \infty \). Thus, representation (23) does indeed hold. \(\square \)

Proof of Proposition 5.2

The proof of Proposition 5.1 goes through with the exception of a few places that we now point out. The major distinction in the present setting is that \(-L^\prime (t)\sim r/t\) as \(t\rightarrow \infty \) rather than \(L^\prime (t)=o(1/t)\). By Lemma 4.2, in formula (34) \(L^\prime (t)=-rt^{-1}+o(1/t)\) and \(L^\prime (t)+tL^{\prime \prime }(t)=o(1/t)\) as \(t\rightarrow \infty \). Also, \(R(t)=o(1/t)\) as \(t\rightarrow \infty \). Indeed,

$$\begin{aligned}{} & {} \int _1^\infty |f_t^{\prime \prime }(x)|\textrm{d}x=t^{-(1+1/\alpha )}\int _0^t\bigg |(\alpha +1)y^{1/\alpha }L^\prime (y)+(3\alpha +1) y^{1+1/\alpha }L^{\prime \prime }(y)\\{} & {} \qquad +\alpha y^{2+1/\alpha }L^{\prime \prime \prime }(y)\bigg |\textrm{d}y\\{} & {} =o(1/t),\quad t\rightarrow \infty \end{aligned}$$

because the integrand is \(o(y^{-1+1/\alpha })\) by Lemma 4.2. Hence, formula (35) transforms into

$$\begin{aligned} B_\alpha (t)=\int _1^{\lfloor t^\beta \rfloor }x^{-\alpha }L^\prime (t/x^\alpha )\textrm{d}x-rt^{-1}/2+o(1/t),\quad t\rightarrow \infty . \end{aligned}$$

By Lemma 4.2(b), in formula (36)

$$\begin{aligned} -\alpha ^{-1}t^{-1+1/\alpha }\int _t^\infty y^{-1/\alpha }L^\prime (y)\textrm{d}y=rt^{-1}+o(1/t),\quad t\rightarrow \infty . \end{aligned}$$

Combining pieces together we conclude that, in comparison with the case \(\mathbb {P}\{\eta =b\}\in (0,1)\) treated in Proposition 5.1, the asymptotic expansions of \(\sum _{k\ge 1}k^{-\alpha }\psi ^\prime (t/k^\alpha )\) have the additional summand \(rt^{-1}/2\), that is, formulae (24) and (26) do indeed hold.

The argument leading to (25) and (27) is similar to that used to prove (21) and (23). For instance, to obtain (25) we represent the solution t to

$$\begin{aligned} b\log t+q+r/(2t)+o(1/t)=x \end{aligned}$$
(40)

in the form \(t(x)=\exp ((x-q)/b)(1+\varepsilon (x))\) with \(\lim _{x\rightarrow \infty }\varepsilon (x)=0\). Substituting this into (40) we obtain (25).

\(\square \)

5.2 Proof of Propositions 5.3 and 5.4

Proof of Proposition 5.3

For a fixed \(t>0\), put \(f_t(x):=\psi (t/x^\alpha )-(t/x^\alpha )\psi ^\prime (t/x^\alpha )\) for \(x>0\). Then,

$$\begin{aligned} \lim _{x\rightarrow \infty }f_t(x)=\lim _{x\rightarrow \infty }f_t^\prime (x)=0. \end{aligned}$$
(41)

The former is a consequence of \(\psi (0)=0\) and \(\lim _{y\rightarrow 0+}y\psi ^\prime (y)=0\), see Lemma 4.3. The latter follows from \(f_t^\prime (x)=\alpha (t^2/x^{2\alpha +1})\psi ^{\prime \prime }(t/x^\alpha )\) and the fact that \(\lim _{y\rightarrow 0+}\psi ^{\prime \prime }(y)=\mathbb {E}[\eta ^2]<\infty \) which holds by Lemma 4.3.

In view of (41), an application of formula (14) with \(f=f_t\) yields

$$\begin{aligned} \sum _{j\ge 1}f_t(j)=\int _1^\infty f_t(x)\textrm{d}x+ f_t(1)/2-f_t^\prime (1)/12+R_1(t), \end{aligned}$$

where \(|R_1(t)|\le (1/12)\int _1^\infty |f_t^{\prime \prime }(x)|\textrm{d}x\). By Lemma 4.1(a,b), \(f_t(1)=\psi (t)-t\psi ^\prime (t)=\log \theta +o(1)\) as \(t\rightarrow \infty \). Further, by Lemma 4.1(c), \(f_t^\prime (1)=\alpha t^2 \psi ^{\prime \prime }(t)\rightarrow 0\) as \(t\rightarrow \infty \). Now, we intend to prove that

$$\begin{aligned} \lim _{t\rightarrow \infty } R_1(t)=0. \end{aligned}$$
(42)

To this end, noting that

$$\begin{aligned} f_t^{\prime \prime }(x)=-\alpha t^2 \left( (2\alpha +1)x^{-2\alpha -2}\psi ^{\prime \prime }(t/x^\alpha )+\alpha t x^{-3\alpha -2}\psi ^{\prime \prime \prime }(t/x^\alpha )\right) , \end{aligned}$$

we obtain

$$\begin{aligned}{} & {} t^{-1/\alpha }\int _1^\infty (\alpha t/x^{\alpha +1})(t/x^\alpha )^{(\alpha +1)/\alpha }\psi ^{\prime \prime }(t/x^\alpha )\textrm{d}x\\{} & {} \quad =t^{-1/\alpha }\int _0^t x^{1+1/\alpha }\psi ^{\prime \prime }(x)\textrm{d}x~\rightarrow ~0,\quad t\rightarrow \infty \end{aligned}$$

and

$$\begin{aligned}{} & {} t^{-1/\alpha }\int _1^\infty (\alpha t/x^{\alpha +1})(t/x^\alpha )^{(2\alpha +1)/\alpha }\left| \psi ^{\prime \prime \prime }(t/x^\alpha )\right| \textrm{d}x\\{} & {} \quad =t^{-1/\alpha }\int _0^t x^{2+1/\alpha }\left| \psi ^{\prime \prime \prime }(x)\right| \textrm{d}x~\rightarrow ~0,\quad t\rightarrow \infty . \end{aligned}$$

Here, the limit relations are secured by \(\lim _{y\rightarrow \infty } y^2\psi ^{\prime \prime }(y)=0\) and \(\lim _{y\rightarrow \infty } y^3|\psi ^{\prime \prime \prime }(y)| =0\), respectively, see Lemma 4.1(c). The proof of (42) is complete.

According to (37),

$$\begin{aligned} \int _0^\infty y^{-1-1/\alpha }(\psi (y)-y\psi ^\prime (y))\textrm{d}y=-\alpha ^2 \sigma _\alpha ^2. \end{aligned}$$

In view of Lemma 4.1(a,b),

$$\begin{aligned} t^{1/\alpha }\int _t^\infty y^{-1-1/\alpha }(\psi (y)-y\psi ^\prime (y))\textrm{d}y= & {} t^{1/\alpha }\int _t^\infty y^{-1-1/\alpha }(\log \theta +o(1))\textrm{d}y\\= & {} \alpha \log \theta +o(1),\quad t\rightarrow \infty . \end{aligned}$$

With these at hand, changing the variable yields

$$\begin{aligned} \int _1^\infty f_t(x)\textrm{d}x{} & {} =\alpha ^{-1}t^{1/\alpha }\Bigg (\int _0^\infty y^{-1-1/\alpha }(\psi (y)-y\psi ^\prime (y))\textrm{d}y\\{} & {} \qquad -\int _t^\infty y^{-1-1/\alpha }(\psi (y)-y\psi ^\prime (y))\textrm{d}y\Bigg )\\{} & {} \quad =-\alpha \sigma _\alpha ^2 t^{1/\alpha }-\log \theta +o(1),\quad t\rightarrow \infty . \end{aligned}$$

Combining fragments together, we arrive at (28).

Finally, observe that \(\sigma _1^2=\int _0^\infty \psi ^{\prime \prime }(x)\textrm{d}x=\lim _{s\rightarrow \infty }\psi ^\prime (s)-\lim _{x \rightarrow 0+}\psi ^\prime (x)=b\) by Lemma 4.1(b) and Lemma 4.3. The proof of Proposition 5.3 is complete. \(\square \)

Proof of Proposition 5.4

Fix any \(\beta >2/(2\alpha -1)\). By Lemma 4.3, \(\psi (t)\sim \mathbb {E}[\eta ^2]t^2/2\) and \(\psi ^\prime (t)\sim \mathbb {E}[\eta ^2]t\) as \(t\rightarrow 0+\). Using

$$\begin{aligned} -t\sum _{k\ge \lfloor t^\beta \rfloor +1}k^{-\alpha }\psi ^\prime (t/k^\alpha )\le \sum _{k\ge \lfloor t^\beta \rfloor +1}\left( \psi (t/k^\alpha )-(t/k^\alpha )\psi ^\prime (t/k^\alpha )\right) \le \sum _{k\ge \lfloor t^\beta \rfloor +1}\psi (t/k^\alpha ) \end{aligned}$$

we conclude that

$$\begin{aligned} \sum _{k\ge \lfloor t^\beta \rfloor +1}\left( \psi (t/k^\alpha )-(t/k^\alpha )\psi ^\prime (t/k^\alpha )\right) =o(1),\quad t\rightarrow \infty . \end{aligned}$$

Indeed,

$$\begin{aligned} \sum _{k\ge \lfloor t^\beta \rfloor +1}\psi (t/k^\alpha )=O\left( t^2\sum _{k\ge \lfloor t^\beta \rfloor +1}k^{-2\alpha }\right) =O(t^{2-\beta (2\alpha -1)})=o(1),\quad t\rightarrow \infty , \end{aligned}$$

and an analogous asymptotic estimate holds true for the lower bound involving \(\psi ^\prime \).

Recalling that \(f_t(x)=\psi (t/x^\alpha )-(t/x^\alpha )\psi ^\prime (t/x^\alpha )\) and setting

$$\begin{aligned} h_t(x) :&=f_t(x)+r\log (t/x^\alpha )\\&=\psi (t/x^\alpha )-(t/x^\alpha )\psi ^\prime (t/x^\alpha )+r\log (t/x^\alpha ), \end{aligned}$$

note that

$$\begin{aligned} h_t^\prime (x)=(\alpha t^2/x^{1+2\alpha })\psi ^{\prime \prime }(t/x^\alpha )-r\alpha /x \end{aligned}$$

and

$$\begin{aligned} h_t^{\prime \prime }(x)=-\alpha (((2\alpha +1)t^2/x^{2\alpha +2})\psi ^{\prime \prime }(t/x^\alpha )+(\alpha t^3/x^{3\alpha +2})\psi ^{\prime \prime \prime }(t/x^\alpha )-r/x^2). \end{aligned}$$

We shall use a representation

$$\begin{aligned}{} & {} \sum _{k=1}^{\lfloor t^\beta \rfloor } \left( \psi (t/k^\alpha )-(t/k^\alpha )\psi ^\prime (t/k^\alpha )\right) =\left( \sum _{k=1}^{\lfloor t^\beta \rfloor }h_t(k)-r\int _1^{\lfloor t^\beta \rfloor }\log (t/x^\alpha )\textrm{d}x\right) \nonumber \\{} & {} \quad +r\left( \int _1^{\lfloor t^\beta \rfloor }\log (t/x^\alpha )\textrm{d}x-\sum _{k=1}^{\lfloor t^\beta \rfloor }\log (t/k^\alpha )\right) =:C_1(t)+C_2(t). \end{aligned}$$
(43)

By Stirling’s formula,

$$\begin{aligned} C_2(t)&=r\left( -\log t+\alpha \left( \sum _{k=1}^{\lfloor t^\beta \rfloor }\log k-\int _1^{\lfloor t^\beta \rfloor }\log x\,\textrm{d}x\right) \right) \\&=-r\log t+r\alpha \big ((\lfloor t^\beta \rfloor +1/2)\log (\lfloor t^\beta \rfloor )-\lfloor t^\beta \rfloor +(1/2)\log (2\pi )+o(1)\\&\hspace{1cm}-\lfloor t^\beta \rfloor \log (\lfloor t^\beta \rfloor )+\lfloor t^\beta \rfloor -1\big )\\&=-r\log t+(r\alpha /2)\log (\lfloor t^\beta \rfloor )+r\alpha ((1/2)\log (2\pi )-1)+o(1),\quad t\rightarrow \infty . \end{aligned}$$

Next, applying formula (13) with \(f=h_t\), \(m=1\) and \(n=\lfloor t^\beta \rfloor \) and recalling that \(h_t(x)=f_t(x)+r\log (t/x^\alpha )\), we conclude that

$$\begin{aligned} C_1(t)=\int _1^{\lfloor t^\beta \rfloor }f_t(x)\textrm{d}x+(h_t(1)+h_t(\lfloor t^\beta \rfloor ))/2+(h_t^\prime (\lfloor t^\beta \rfloor )-h_t^\prime (1))/12+R_2(t), \end{aligned}$$

where \(R_2(t)\le (1/12)\int _1^{\lfloor t^\beta \rfloor }|h_t^{\prime \prime }(x)|\textrm{d}x\). By Lemma 4.2, as \(t\rightarrow \infty \),

$$\begin{aligned} h_t(1)=\psi (t)-t\psi ^\prime (t)+r\log t =r+\log \left( \lambda \Gamma (r+1)\right) +o(1) \end{aligned}$$

and

$$\begin{aligned} h^\prime _t(1)=\alpha (t^2\psi ^{\prime \prime }(t)-r)=o(1). \end{aligned}$$

Further, by Lemma 4.3, as \(t\rightarrow \infty \),

$$\begin{aligned} h_t(\lfloor t^\beta \rfloor )&=\psi (t/(\lfloor t^\beta \rfloor )^\alpha )-(t/(\lfloor t^\beta \rfloor )^\alpha )\psi ^\prime (t/(\lfloor t^\beta \rfloor )^\alpha )+r(\log t-\alpha \log (\lfloor t^\beta \rfloor ))\\&=r(\log t-\alpha \log (\lfloor t^\beta \rfloor ))+o(1) \end{aligned}$$

and

$$\begin{aligned} h^\prime _t(\lfloor t^\beta \rfloor )=(\alpha t^2/(\lfloor t^\beta \rfloor )^{1+2\alpha })\psi ^{\prime \prime }(t/(\lfloor t^\beta \rfloor )^\alpha )-r\alpha /\lfloor t^\beta \rfloor =o(1). \end{aligned}$$

The relation \(R_2(t)=o(1)\) as \(t\rightarrow \infty \) follows from

$$\begin{aligned}{} & {} \int _1^\infty |h_t^{\prime \prime }(x)|\textrm{d}x=t^{-1/\alpha }\int _1^\infty \Big |(2\alpha +1)(t/x^\alpha )^{1+1/\alpha }(-\alpha t/x^{\alpha +1})\psi ^{\prime \prime }(t/x^\alpha )\\{} & {} \qquad +\alpha (t/x^\alpha )^{2+1/\alpha }(-\alpha t/x^{\alpha +1})\psi ^{\prime \prime \prime }(t/x^\alpha )-r(t/x^\alpha )^{-1+1/\alpha } (-\alpha /x^{\alpha +1})\Big |\textrm{d}x\\{} & {} \quad =t^{-1/\alpha }\int _0^t \left| (2\alpha +1)y^{1+1/\alpha }\psi ^{\prime \prime }(y)+\alpha y^{2+1/\alpha }\psi ^{\prime \prime \prime }(y)-ry^{-1+1/\alpha }\right| \textrm{d}y\\{} & {} \quad =o(1),\quad t\rightarrow \infty \end{aligned}$$

because the integrand is \(o(y^{-1+1/\alpha })\) by Lemma 4.2.

Mimicking the argument used at the beginning of the proof for the sum, one can show that \(\int _{\lfloor t^\beta \rfloor }^\infty f_t(x)\textrm{d}x=o(1)\), whence

$$\begin{aligned} \int _1^{\lfloor t^\beta \rfloor } f_t(x)\textrm{d}x=\int _1^\infty f_t(x)\textrm{d}x+o(1),\quad t\rightarrow \infty . \end{aligned}$$

From the proof of Proposition 5.3, we know that

$$\begin{aligned} \int _1^\infty f_t(x)\textrm{d}x=-\alpha \sigma _\alpha ^2 t^{1/\alpha }-\alpha ^{-1}t^{1/\alpha }\int _t^\infty y^{-1-1/\alpha }(\psi (y)-y\psi ^\prime (y))\textrm{d}y. \end{aligned}$$

By Lemma 4.2(a,b), as \(t\rightarrow \infty \),

$$\begin{aligned}{} & {} t^{1/\alpha }\int _t^\infty y^{-1-1/\alpha }(\psi (y)-y\psi ^\prime (y))\textrm{d}y\\{} & {} \quad =t^{1/\alpha }\int _t^\infty y^{-1-1/\alpha }(-r\log y+r+\log (\lambda \Gamma (r+1))+o(1))\textrm{d}y\\{} & {} \quad =-r\alpha \log t+r\alpha (1-\alpha )+\alpha \log (\lambda \Gamma (r+1))+o(1). \end{aligned}$$

To calculate the integral involving \(\log \), we have used the fact that \(z\mapsto \alpha ^{-2}\textrm{e}^{-z/\alpha }z\), \(z>0\) is a density of the gamma distribution with parameters \(1/\alpha \) and 2. The corresponding distribution tail is \(z\mapsto \textrm{e}^{-z/\alpha }(z/\alpha +1)\), \(z\ge 0\) and thereupon

$$\begin{aligned} \int _t^\infty y^{-1-1/\alpha }\log y\,\textrm{d}y=\int _{\log t}^\infty \textrm{e}^{-z/\alpha }z\,\textrm{d}z=\alpha ^2 t^{-1/\alpha }(\alpha ^{-1}\log t+1). \end{aligned}$$

Thus, we have proved that

$$\begin{aligned} \int _1^{\lfloor t^\beta \rfloor }f_t(x)\textrm{d}x=-\alpha \sigma _\alpha ^2 t^{1/\alpha }+r\log t-r(1-\alpha )-\log (\lambda \Gamma (r+1))+o(1),\quad t\rightarrow \infty . \end{aligned}$$

Collecting pieces together, we arrive at (29). \(\square \)

5.3 Proof of Proposition 5.5

Here is a slight extension of formula (9): for any bounded measurable \(g:\mathbb {R}\rightarrow \mathbb {C}\)

$$\begin{aligned} \mathbb {E}^{(t)}[g(\eta _k)]= & {} \frac{\mathbb {E}[\textrm{e}^{tS(\alpha )}g(\eta _k)]}{\mathbb {E}[\textrm{e}^{tS(\alpha )}]}=\frac{\mathbb {E}[\textrm{e}^{t\eta _k/k^\alpha }g(\eta _k)]\prod _{j\ne k}\mathbb {E}[\textrm{e}^{t\eta _j/j^\alpha }]}{\mathbb {E}[\textrm{e}^{t\eta _k/k^\alpha }]\prod _{j\ne k}\mathbb {E}[\textrm{e}^{t\eta _j/j^\alpha }]}\nonumber \\= & {} \frac{\mathbb {E}[\textrm{e}^{t\eta _k/k^\alpha }g(\eta _k)]}{\mathbb {E}[\textrm{e}^{t\eta _k/k^\alpha }]}, \end{aligned}$$
(44)

where the second equality is justified by independence of \(\eta _1\), \(\eta _2,\ldots \) We recall that \(S(\alpha )=\sum _{k\ge 1}k^{-\alpha }\eta _k\) and that, under \(\mathbb {P}^{(t)}\), \(S^{(t)}_0(\alpha )=S_0(\alpha )-\mathbb {E}^{(t)}[S(\alpha )]\).

Lemma 5.6

Under the assumptions of either Theorem 1.1 or Theorem 1.2,

$$\begin{aligned} \lim _{t\rightarrow \infty }\mathbb {E}^{(t)}[\textrm{e}^{\textrm{i}ut^{1-1/(2\alpha )}S^{(t)}_0(\alpha )}]=\textrm{e}^{-\sigma ^2_\alpha u^2/2},\quad u\in \mathbb {R}\end{aligned}$$
(45)

with \(\sigma ^2_\alpha \) as defined in Theorem 1.1.

Proof

It is enough to prove that, for each \(u\in \mathbb {R}\),

$$\begin{aligned} \lim _{t\rightarrow \infty }\mathbb {E}^{(t)}[\textrm{e}^{ut^{1-1/(2\alpha )}S^{(t)}_0(\alpha )}]=\textrm{e}^{\sigma ^2_\alpha u^2/2}. \end{aligned}$$
(46)

Indeed, (46) ensures that the \(\mathbb {P}^{(t)}\)-distributions of the variables \(t^{1-1/(2\alpha )}S_0^{(t)}(\alpha )\) converge weakly as \(t\rightarrow \infty \) to the centred normal distribution with variance \(\sigma ^2_\alpha \). Relation (45) then follows by the Lévy continuity theorem for characteristic functions.

Under \(\mathbb {P}^{(t)}\), the random variables \(\eta _1\), \(\eta _2,\ldots \) are still independent but not identically distributed. The former follows from (9), and the latter follows from (44). Anyway, under \(\mathbb {P}^{(t)}\), the variable \(S_0(\alpha )=S(\alpha )-\mathbb {E}^{(t)}S(\alpha )\) is an infinite sum of independent centred random variables with finite second moments. Using this in combination with (44) yields

$$\begin{aligned}{} & {} \mathbb {E}^{(t)}[\textrm{e}^{ut^{1-1/(2\alpha )}S_0(\alpha )}]=\prod _{k\ge 1}\mathbb {E}^{(t)}[\exp ((ut^{1-1/(2\alpha )}/k^\alpha )(\eta _k-\mathbb {E}^{(t)}[\eta _k]))]\\{} & {} \quad =\prod _{k\ge 1}\frac{\mathbb {E}[\exp (((t+ut^{1-1/(2\alpha )})/k^\alpha )\eta -(ut^{1-1/(2\alpha )}/k^\alpha ) \psi ^\prime (t/k^\alpha ))]}{\mathbb {E}[\textrm{e}^{t\eta /k^\alpha }]}\\{} & {} \quad =\exp \left( \sum _{k\ge 1}\left( \psi ((t+ut^{1-1/(2\alpha )})/k^\alpha )-\psi (t/k^\alpha )-(ut^{1-1/(2\alpha )}/k^\alpha ) \psi ^\prime (t/k^\alpha )\right) \right) . \end{aligned}$$

By the mean value theorem for twice differentiable function, we further obtain, for some \(\theta _k=\theta _k(t,u)\in (0,1)\),

$$\begin{aligned}{} & {} \mathbb {E}^{(t)}[\textrm{e}^{ut^{1-1/(2\alpha )}S_0(\alpha )}]=\exp \left( (t^{2-1/\alpha }u^2/2)\sum _{k\ge 1}k^{-2\alpha }\psi ^{\prime \prime }((t+\theta _k ut^{1-1/(2\alpha )})/k^\alpha )\right) \\{} & {} \quad =\exp \left( (t^{2-1/\alpha }u^2/2)\sum _{k\ge 1}k^{-2\alpha }\psi ^{\prime \prime }(t/k^\alpha )\right) \\{} & {} \quad \times \exp \left( (t^{2-1/\alpha }u^2/2)\sum _{k\ge 1}k^{-2\alpha }(\psi ^{\prime \prime }((t+\theta _k ut^{1-1/(2\alpha )})/k^\alpha )-\psi ^{\prime \prime }(t/k^\alpha ))\right) . \end{aligned}$$

By the mean value theorem for differentiable functions, for some \(\vartheta _k{=}\vartheta _k(t,u){\in }(0,1)\),

$$\begin{aligned}{} & {} \sum _{k\ge 1}k^{-2\alpha }(\psi ^{\prime \prime }((t+\theta _k ut^{1-1/(2\alpha )})/k^\alpha )-\psi ^{\prime \prime }(t/k^\alpha ))\\{} & {} \quad =ut^{1-1/(2\alpha )}\sum _{k\ge 1}k^{-3\alpha }\psi ^{\prime \prime \prime }((t+\vartheta _k ut^{1-1/(2\alpha )})/k^\alpha ). \end{aligned}$$

Thus, (46) follows if we can prove that

$$\begin{aligned} \lim _{t\rightarrow \infty } t^{2-1/\alpha }\sum _{k\ge 1}k^{-2\alpha }\psi ^{\prime \prime }(t/k^\alpha )=\sigma _\alpha ^2\in (0,\infty ) \end{aligned}$$
(47)

and that, for each fixed \(u\in \mathbb {R}\),

$$\begin{aligned} \lim _{t\rightarrow \infty } t^{3-3/(2\alpha )}\sum _{k\ge 1}k^{-3\alpha }|\psi ^{\prime \prime \prime }((t+\vartheta _kut^{1-1/(2\alpha )})/k^\alpha )|=0. \end{aligned}$$
(48)

Proof of (47). We intend to show that the function h defined by \(h(x):=x^{-2\alpha }\psi ^{\prime \prime }(x^{-\alpha })\) is directly Riemann integrable (dRi) on \([0,\infty )\). The function \(\psi ^{\prime \prime }\) is continuous on \([0,\infty )\). In view of Lemma 4.1(c), for some \(C_1>C_2>0\), \(\psi ^{\prime \prime }(x^{-\alpha })\le C_1x^{2\alpha }\) for \(x\in (0,1]\) and \(\psi ^{\prime \prime }(x^{-\alpha })\le C_2\) for \(x>0\). Hence, the function h is continuous and bounded on \([0,\infty )\) and \(0\le h(x)\le h_1(x)\) for \(x>0\), where \(h_1(x):=C_1\mathbbm {1}_{[0,1]}(x)+C_2x^{-2\alpha }\mathbbm {1}_{(1,\infty )}(x)\) for \(x\ge 0\). Being a Lebesgue integrable nonincreasing function on \([0,\infty )\), \(h_1\) is dRi on \([0,\infty )\), hence, so is h.

As a consequence,

$$\begin{aligned}{} & {} \lim _{t\rightarrow \infty }t^{-1/\alpha }\sum _{k\ge 1}t^2k^{-2\alpha }\psi ^{\prime \prime }(tk^{-\alpha })\\{} & {} \quad =\int _0^\infty x^{-2\alpha }\psi ^{\prime \prime }(x^{-\alpha })\textrm{d}x=\alpha ^{-1}\int _0^\infty x^{1-1/\alpha }\psi ^{\prime \prime }(x)\textrm{d}x\\{} & {} \quad =\sigma _\alpha ^2\in (0,\infty ), \end{aligned}$$

which proves (47).

Proof of (48). Fix any \(u\ge 0\). The proof for fixed \(u<0\) is analogous. By Lemmas 4.1(c) and 4.2(c), for large enough \(t_1>0\) there exists \(c_1>0\) such that \(|\psi ^{\prime \prime \prime }(t)|\le c_1 t^{-3}\) whenever \(t\ge t_1\). Hence, for positive integer \(k\le (t/t_1)^{1/\alpha }\le ((t+\vartheta _kut^{1-1/(2\alpha )})/t_1)^{1/\alpha }\)

$$\begin{aligned} |\psi ^{\prime \prime \prime }((t+\vartheta _k ut^{1-1/(2\alpha )})/k^\alpha )|\le c_1k^{3\alpha }/(t+\vartheta _kut^{1-1/(2\alpha )})^3 \le c_1k^{3\alpha }t^{-3}. \end{aligned}$$

This entails

$$\begin{aligned}{} & {} t^{3-3/(2\alpha )}\sum _{k=1}^{(t/t_1)^{1/\alpha }} k^{-3\alpha }|\psi ^{\prime \prime \prime }((t+\vartheta _kut^{1-1/(2\alpha )})/k^\alpha )|\\{} & {} \quad \le c_1 t^{-3/(2\alpha )}(t/t_1)^{1/\alpha }~\rightarrow ~0,\quad t\rightarrow \infty . \end{aligned}$$

Put \(t_2:=2t_1\). By Lemma 4.3, there exists \(c_2>0\) such that \(|\psi ^{\prime \prime \prime }(t)|\le c_2 t^{-1}\) whenever \(t\in (0, t_2]\). For large enough t and any \(k\in \mathbb {N}\), \(t+\vartheta _kut^{1-1/(2\alpha )}\le 2t\). Hence, for such t and \(k\ge (2t/t_2)^{1/\alpha }\ge ((t+\vartheta _kut^{1-1/(2\alpha )})/t_2)^{1/\alpha }\)

$$\begin{aligned} |\psi ^{\prime \prime \prime }((t+\vartheta _kut^{1-1/(2\alpha )})/k^\alpha )|\le c_2k^\alpha /(t+\vartheta _kut^{1-1/(2\alpha )}) \le c_2k^{\alpha }t^{-1}. \end{aligned}$$

As a consequence,

$$\begin{aligned} t^{3-3/(2\alpha )}\sum _{k\ge (t/t_1)^{1/\alpha }} k^{-3\alpha }|\psi ^{\prime \prime \prime }((t+\vartheta _kut^{1-1/(2\alpha )})/k^\alpha )|\le c_2 t^{2-3/(2\alpha )}\sum _{k\ge (t/t_1)^{1/\alpha }}k^{-2\alpha }\\=t^{2-3/(2\alpha )}O(t^{1/\alpha -2})~\rightarrow ~0,\quad t\rightarrow \infty . \end{aligned}$$

This finishes the proof of (48).

The proof of Lemma 5.6 is complete. \(\square \)

Given next is a local limit theorem. The result and its proof resemble Stone’s [13] local limit theorem and its proof for standard random walks with the distribution of jumps belonging to the domain of attraction of a stable distribution. While Stone’s theorem deals with intervals of fixed length h, we treat intervals of length \(ht^{-1/2}\) as \(t\rightarrow \infty \).

Theorem 5.7

Assume that \(\eta \le b\) a.s., \(\mathbb {E}[\eta ]=0\) and \(\mathbb {E}[\eta ^2] \in (0,\infty )\). Then, for each \(h>0\),

$$\begin{aligned} \lim _{t\rightarrow \infty }\sup _{x\in \mathbb {R}} \left| t^{1/(2\alpha )}\mathbb {P}^{(t)}\{S_0(\alpha )\in (xt^{-1+1/(2\alpha )}, xt^{-1+1/(2\alpha )}+ht^{-1}]\}-hn_\alpha (x)\right| =0, \end{aligned}$$

where \(n_\alpha (x):=(2\pi \sigma _\alpha ^2)^{-1/2}\textrm{e}^{-x^2/(2\sigma _\alpha ^2)}\) for \(x\in \mathbb {R}\).

We first prove an auxiliary result.

Lemma 5.8

Under the assumptions of Theorem 5.7, for all \(t>0\), there exist positive constants c and \(\rho \) such that

$$\begin{aligned} \sup _{k\ge t^{1/\alpha }}\left| \mathbb {E}^{(t)}[\textrm{e}^{\textrm{i}u\eta _k}]\right| \le \textrm{e}^{-cu^2} \end{aligned}$$

for all \(u\in \mathbb {R}\) satisfying \(|u|\le \rho \).

Proof

Fix any \(u\in \mathbb {R}\). Using formula (44) with \(g(x)=\textrm{e}^{\textrm{i}ux}\), we infer

$$\begin{aligned} \left| \mathbb {E}^{(t)}[\textrm{e}^{\textrm{i}u\eta _k}]\right| =\frac{\left| \mathbb {E}[\textrm{e}^{(\textrm{i}u+t/k^\alpha )\eta _k}]\right| }{\mathbb {E}[\textrm{e}^{t\eta _k/k^\alpha }]} \end{aligned}$$

and thereupon

$$\begin{aligned} \sup _{k\ge t^{1/\alpha }}\left| \mathbb {E}^{(t)}[\textrm{e}^{\textrm{i}u\eta _k}]\right| \le \sup _{s\in [0,1]}\frac{\left| \mathbb {E}[\textrm{e}^{(\textrm{i}u+s)\eta }]\right| }{\mathbb {E}[\textrm{e}^{s\eta }]}. \end{aligned}$$

Write

$$\begin{aligned} \mathbb {E}[\textrm{e}^{(\textrm{i}u+s)\eta }]=\mathbb {E}[\textrm{e}^{s\eta }]+\textrm{i}u\mathbb {E}[\eta \textrm{e}^{s\eta }]-2^{-1}u^2\mathbb {E}[\eta ^2\textrm{e}^{s\eta }]+\mathbb {E}[\textrm{e}^{s\eta }(\textrm{e}^{\textrm{i}u\eta }-1-\textrm{i}u\eta +2^{-1}u^2\eta ^2)]. \end{aligned}$$

By Lemma 3.3.7 on p. 115 in [3],

$$\begin{aligned} \left| \textrm{e}^{\textrm{i}x}-1-\textrm{i}x+ 2^{-1}x^2\right| \le \min (|x|^3/6, |x|^2),\quad x\in \mathbb {R}. \end{aligned}$$
(49)

As a consequence, for any \(\kappa \in (0,1)\),

$$\begin{aligned}{} & {} \left| \mathbb {E}[\textrm{e}^{s\eta }(\textrm{e}^{\textrm{i}u\eta }-1-\textrm{i}u\eta +2^{-1}u^2\eta ^2)]\right| \\{} & {} \quad \le 6^{-1}|u|^3\mathbb {E}[|\eta |^3\textrm{e}^{s\eta }\mathbbm {1}_{\{|\eta |\le |u|^{-\kappa }\}}]+u^2 \mathbb {E}[\eta ^2\textrm{e}^{s\eta }\mathbbm {1}_{\{|\eta |>|u|^{-\kappa }\}}]\\{} & {} \quad \le 6^{-1}|u|^{3-\kappa }\mathbb {E}[\eta ^2\textrm{e}^{s\eta }]+u^2 \mathbb {E}[\eta ^2\textrm{e}^{s\eta }\mathbbm {1}_{\{|\eta |>|u|^{-\kappa }\}}]. \end{aligned}$$

Observe that \(1/A:=\inf _{s\in [0,1]}\mathbb {E}[\textrm{e}^{s\eta }]\in (0,\infty )\) and that \(\textrm{e}^{s\eta }\le \textrm{e}^{sb}\) a.s. Hence,

$$\begin{aligned}{} & {} \sup _{s\in [0,1]}\frac{\left| \mathbb {E}[\textrm{e}^{s\eta }(\textrm{e}^{\textrm{i}u\eta }-1-\textrm{i}u\eta +2^{-1}u^2\eta ^2)]\right| }{\mathbb {E}[\textrm{e}^{s\eta }]}\\{} & {} \quad \le A\textrm{e}^b (6^{-1}|u|^{3-\kappa }\mathbb {E}[\eta ^2]+u^2 \mathbb {E}[\eta ^2\mathbbm {1}_{\{|\eta |>|u|^{-\kappa }\}}]). \end{aligned}$$

This proves

$$\begin{aligned} \lim _{u\rightarrow 0} u^{-2}\sup _{s\in [0,1]}\,\frac{\left| \mathbb {E}[\textrm{e}^{s\eta }(\textrm{e}^{\textrm{i}u\eta }-1-\textrm{i}u\eta +2^{-1}u^2\eta ^2)]\right| }{\mathbb {E}[\textrm{e}^{s\eta }]}=0. \end{aligned}$$
(50)

Further, with \(\varphi (s):=\mathbb {E}[\textrm{e}^{s\eta }]\),

$$\begin{aligned}{} & {} \frac{\mathbb {E}[\textrm{e}^{s\eta }]+\textrm{i}u\mathbb {E}[\eta \textrm{e}^{s\eta }]-2^{-1}u^2\mathbb {E}[\eta ^2\textrm{e}^{s\eta }]}{\mathbb {E}[\textrm{e}^{s\eta }]}\textrm{e}^{-\textrm{i}u \psi ^\prime (s)}\\{} & {} \quad =\left( 1+\textrm{i}u\psi ^\prime (s)-2^{-1}u^2\frac{\varphi ^{\prime \prime }(s)}{\varphi (s)}\right) \left( 1-\textrm{i}u\psi ^\prime (s)-2^{-1}u^2 (\psi ^\prime (s))^2\right) \\{} & {} \qquad +\left( 1+\textrm{i}u\psi ^\prime (s)-2^{-1}u^2\frac{\varphi ^{\prime \prime }(s)}{\varphi (s)}\right) (\textrm{e}^{-\textrm{i}u\psi ^\prime (s)}-1+\textrm{i}u\psi ^\prime (s)+2^{-1}u^2 (\psi ^\prime (s))^2)\\{} & {} \quad =:K_1(s,u)+K_2(s,u). \end{aligned}$$

Since \(\mathbb {E}[\eta ]=0\), the function \(\psi \) is nondecreasing on \([0,\infty )\), whence \(\psi ^\prime (s)\ge 0\) for \(s\ge 0\). In view of (49),

$$\begin{aligned} \left| \textrm{e}^{-\textrm{i}u\psi ^\prime (s)}-1+\textrm{i}u\psi ^\prime (s)+2^{-1}u^2 (\psi ^\prime (s))^2\right| \le 6^{-1}|u|^3 (\psi ^\prime (s))^3 \end{aligned}$$

and thereupon

$$\begin{aligned} \lim _{u\rightarrow 0}u^{-2}\sup _{s\in [0,1]}|K_2(s,u)|=0. \end{aligned}$$

that

$$\begin{aligned} \frac{\varphi ^{\prime \prime }(s)}{\varphi (s)}-(\psi ^\prime (s))^2=\psi ^{\prime \prime }(s) \end{aligned}$$

and that \(\psi ^{\prime \prime }(s)\ge 0\) by convexity of \(\psi \). With this at hand, we conclude that

$$\begin{aligned} K_1(s,u)=1-2^{-1}u^2 \psi ^{\prime \prime }(s)+2^{-1}\textrm{i}u^3\psi ^\prime (s)\psi ^{\prime \prime }(s)+4^{-1}u^4 (\psi ^\prime (s))^2\frac{\varphi ^{\prime \prime }(s)}{\varphi (s)} \end{aligned}$$

and that

$$\begin{aligned} \lim _{u\rightarrow 0}u^{-2}\sup _{s\in [0,1]}|K_1(s,u)-1+2^{-1}u^2\psi ^{\prime \prime }(s)|=0. \end{aligned}$$

The function \(\psi \) is strictly log-convex on \([0,\infty )\) with \(\psi ^{\prime \prime }(0)=\mathbb {E}[\eta ^2]\). This entails \(1/B:=\inf _{s\in [0,1]}\psi ^{\prime \prime }(s)\in (0,\infty )\). Thus,

$$\begin{aligned}{} & {} \sup _{s\in [0,1]}\left| \frac{\mathbb {E}[\textrm{e}^{s\eta }]+\textrm{i}u\mathbb {E}[\eta \textrm{e}^{s\eta }]-2^{-1}u^2\mathbb {E}[\eta ^2\textrm{e}^{s\eta }]}{\mathbb {E}[\textrm{e}^{s\eta }]}\right| \\{} & {} \quad =\sup _{s\in [0,1]}\left| \frac{\mathbb {E}[\textrm{e}^{s\eta ]}+\textrm{i}u\mathbb {E}[\eta \textrm{e}^{s\eta }]-2^{-1}u^2\mathbb {E}[\eta ^2\textrm{e}^{s\eta }]}{\mathbb {E}[\textrm{e}^{s\eta }]}\textrm{e}^{-\textrm{i}u \psi ^\prime (s)}\right| \\{} & {} \quad \le \sup _{s\in [0,1]}|K_1(s,u)-1+2^{-1}u^2\psi ^{\prime \prime }(s)|+\sup _{s\in [0,1]} |1-2^{-1}u^2\psi ^{\prime \prime }(s)|+\sup _{s\in [0,1]}|K_2(s,u)|\\{} & {} \quad \le 1-2^{-1}u^2 \inf _{s\in [0,1]}\psi ^{\prime \prime }(s)+o(u^2),\quad u\rightarrow 0. \end{aligned}$$

This in combination with (50) shows that

$$\begin{aligned} \sup _{s\in [0,1]}\frac{\left| \mathbb {E}[\textrm{e}^{(\textrm{i}u+s)\eta }]\right| }{\mathbb {E}[\textrm{e}^{s\eta }]}\le 1-(2B)^{-1}u^2+o(u^2),\quad u\rightarrow 0, \end{aligned}$$

where the term \(o(u^2)\) is uniform in \(s\in [0,1]\). In particular, there exists \(u_0>0\) such that \(|o(u^2)|\le (4B)^{-1}u^2\) whenever \(|u|\le u_0\). Hence,

$$\begin{aligned} \sup _{s\in [0,1]}\frac{\left| \mathbb {E}[\textrm{e}^{(\textrm{i}u+s)\eta }]\right| }{\mathbb {E}[\textrm{e}^{s\eta }]}\le 1-(4B)^{-1}u^2\le \textrm{e}^{-u^2/(4B)} \end{aligned}$$

whenever \(|u|\le \min (u_0, (2B)^{1/2})\). \(\square \)

We are ready to prove Theorem 5.7.

Proof of Theorem 5.7

For \(x\in \mathbb {R}\), \(\delta >0\) and \(t>0\), put

$$\begin{aligned} v_{t,\,\delta }(x):=\delta ^{-1}\mathbb {P}^{(t)}\{S_0(\alpha )\in (x, x+\delta ]\}. \end{aligned}$$

Observe that \(v_{t,\,\delta }\) is a density of the \(\mathbb {P}^{(t)}\)-distribution of \(S_0(\alpha )-U_\delta \), where \(S_0(\alpha )\) and \(U_\delta \) are \(\mathbb {P}^{(t)}\)-independent, and \(U_\delta \) has a uniform distribution on \((0,\delta )\). The corresponding characteristic function is

$$\begin{aligned} \int _\mathbb {R}\textrm{e}^{\textrm{i}zx}v_{t,\,\delta }(x)\textrm{d}x=\mathbb {E}^{(t)}\left[ \textrm{e}^{\textrm{i}zS_0(\alpha )}\right] \frac{1-\textrm{e}^{-\textrm{i}\delta z}}{\textrm{i}\delta z},\quad z\in \mathbb {R}. \end{aligned}$$
(51)

Later in the proof we shall show that this characteristic function is absolutely integrable on \(\mathbb {R}\). This entails

$$\begin{aligned} v_{t,\,\delta }(x)=\frac{1}{2\pi }\int _\mathbb {R}\textrm{e}^{-\textrm{i}zx}\mathbb {E}^{(t)}\left[ \textrm{e}^{\textrm{i}zS_0(\alpha )}\right] \frac{1-\textrm{e}^{-\textrm{i}\delta z}}{\textrm{i}\delta z}\textrm{d}z,\quad x\in \mathbb {R}. \end{aligned}$$
(52)

Changing the variable \(z=ut^{1-1/(2\alpha )}\), we obtain

$$\begin{aligned} v_{t,\,\delta }(xt^{1/(2\alpha )-1})=\frac{t^{1-1/(2\alpha )}}{2\pi }\int _\mathbb {R}\textrm{e}^{-\textrm{i}ux}\mathbb {E}^{(t)}\left[ \textrm{e}^{\textrm{i}ut^{1-1/(2\alpha )}S_0(\alpha )}\right] \frac{1-\textrm{e}^{-\textrm{i}\delta ut^{1-1/(2\alpha )}}}{\textrm{i}\delta ut^{1-1/(2\alpha )}}\textrm{d}u. \end{aligned}$$

This yields

$$\begin{aligned}{} & {} t^{1/(2\alpha )}\mathbb {P}^{(t)}\{S_0(\alpha )\in (xt^{-1+1/(2\alpha )}, xt^{-1+1/(2\alpha )}+ht^{-1}]\}\\{} & {} \quad =ht^{-1+1/(2\alpha )}v_{t,\,ht^{-1}}(xt^{-1+1/(2\alpha )})\\{} & {} \quad =\frac{h}{2\pi }\int _\mathbb {R}\textrm{e}^{-\textrm{i}ux}\mathbb {E}^{(t)}\left[ \textrm{e}^{\textrm{i}ut^{1-1/(2\alpha )}S_0(\alpha )}\right] \frac{1-\textrm{e}^{-\textrm{i}hut^{-1/(2\alpha )}}}{\textrm{i}hut^{-1/(2\alpha )}}\textrm{d}u. \end{aligned}$$

Noting that \(n_\alpha (x)=(2\pi )^{-1}\int _\mathbb {R}\textrm{e}^{-\textrm{i}xu}\textrm{e}^{-\sigma _\alpha ^2 u^2/2}\textrm{d}u\), it suffices to prove that, for any \(A>0\),

$$\begin{aligned} \lim _{t\rightarrow \infty }\sup _{x\in \mathbb {R}}\left| \int _{-A}^A \textrm{e}^{-\textrm{i}ux}\left( \mathbb {E}^{(t)}\left[ \textrm{e}^{\textrm{i}ut^{1-1/(2\alpha )}S_0(\alpha )}\right] \frac{1-\textrm{e}^{-\textrm{i}hut^{-1/(2\alpha )}}}{\textrm{i}hut^{-1/(2\alpha )}}-\textrm{e}^{-\sigma _\alpha ^2 u^2/2}\right) \textrm{d}u\right| =0 \nonumber \\ \end{aligned}$$
(53)

and that

$$\begin{aligned}{} & {} \lim _{A\rightarrow \infty }\limsup _{t\rightarrow \infty }\sup _{x\in \mathbb {R}}\left| \int _{|u|>A} \textrm{e}^{-\textrm{i}ux}\mathbb {E}^{(t)}\left[ \textrm{e}^{\textrm{i}ut^{1-1/(2\alpha )}S_0(\alpha )}\right] \frac{1-\textrm{e}^{-\textrm{i}hut^{-1/(2\alpha )}}}{\textrm{i}hut^{-1/(2\alpha )}}\textrm{d}u\right| =0;\nonumber \\ \end{aligned}$$
(54)
$$\begin{aligned}{} & {} \lim _{A\rightarrow \infty }\sup _{x\in \mathbb {R}}\left| \int _{|u|>A} \textrm{e}^{-\textrm{i}ux} \textrm{e}^{-\sigma _\alpha ^2 u^2/2}\textrm{d}u\right| =0. \end{aligned}$$
(55)

Proof of (53). The supremum in (53) does not exceed

$$\begin{aligned} \int _{-A}^A \left| \mathbb {E}^{(t)}\left[ \textrm{e}^{\textrm{i}ut^{1-1/(2\alpha )}S_0(\alpha )}\right] \frac{1-\textrm{e}^{-\textrm{i}hut^{-1/(2\alpha )}}}{\textrm{i}hut^{-1/(2\alpha )}}-\textrm{e}^{-\sigma _\alpha ^2 u^2/2}\right| \textrm{d}u. \end{aligned}$$

By Lemma 5.6, \(\lim _{t\rightarrow \infty } \mathbb {E}^{(t)}\left[ \textrm{e}^{\textrm{i}ut^{1-1/(2\alpha )}S_0(\alpha )}\right] =\textrm{e}^{-\sigma _\alpha ^2 u^2/2}\) for \(u\in \mathbb {R}\). The characteristic functions \(u\mapsto (1-\textrm{e}^{-\textrm{i}hut^{-1/(2\alpha )}})/(\textrm{i}hut^{-1/(2\alpha )})\) converge as \(t\rightarrow \infty \) to r the characteristic function of degenerate at 0 distribution (\(r(u)=1\) for \(u\in \mathbb {R}\)). Since the convergence of characteristic functions is locally uniform, the latter integral converges to 0 as \(t\rightarrow \infty \).

Proof of (54). Let \(\rho \) be as in Lemma 5.8. Write, for large t,

$$\begin{aligned}{} & {} \sup _{x\in \mathbb {R}}\left| \int _{|u|>A} \textrm{e}^{-\textrm{i}ux}\mathbb {E}^{(t)}\left[ \textrm{e}^{\textrm{i}ut^{1-1/(2\alpha )}S_0(\alpha )}\right] \frac{1-\textrm{e}^{-\textrm{i}hut^{-1/(2\alpha )}}}{\textrm{i}hut^{-1/(2\alpha )}}\textrm{d}u\right| \\{} & {} \quad \le \int _{A<|u|\le \rho t^{1/(2\alpha )}}\left| \mathbb {E}^{(t)}\left[ \textrm{e}^{\textrm{i}ut^{1-1/(2\alpha )}S_0(\alpha )}\right] \right| \textrm{d}u+\int _{|u|>\rho t^{1/(2\alpha )}}\left| \mathbb {E}^{(t)}\left[ \textrm{e}^{\textrm{i}ut^{1-1/(2\alpha )}S_0(\alpha )}\right] \right| \textrm{d}u\\{} & {} \quad =:I_1(t,A)+I_2(t). \end{aligned}$$

Observe that

$$\begin{aligned} \left| \mathbb {E}^{(t)}\left[ \textrm{e}^{\textrm{i}ut^{1-1/(2\alpha )}S_0(\alpha )}\right] \right|= & {} \prod _{k\ge 1}\left| \mathbb {E}^{(t)}\left[ \textrm{e}^{\textrm{i}ut^{1-1/(2\alpha )}k^{-\alpha }(\eta _k-\mathbb {E}^{(t)}\eta _k)}\right] \right| \\= & {} \prod _{k\ge 1}\left| \mathbb {E}^{(t)}\left[ \textrm{e}^{\textrm{i}ut^{1-1/(2\alpha )}k^{-\alpha }\eta _k}\right] \right| . \end{aligned}$$

Analysis of \(I_1\). Put \(c_2:=\inf _{t\ge 1} t^{2-1/\alpha }\sum _{k\ge t^{1/\alpha }}k^{-2\alpha }\) and note that \(c_2>0\) because

$$\begin{aligned} \lim _{t\rightarrow \infty } t^{2-1/\alpha }\sum _{k\ge t^{1/\alpha }}k^{-2\alpha }=(2\alpha -1)^{-1}. \end{aligned}$$

If \(|u|\le \rho t^{1/(2\alpha )}\), then \(|u|t^{1-1/(2\alpha )}/k^\alpha \le \rho \) for all \(k\ge t^{1/\alpha }\) and, according to Lemma 5.8, for \(t\ge 1\),

$$\begin{aligned} \left| \mathbb {E}^{(t)}\left[ \textrm{e}^{\textrm{i}ut^{1-1/(2\alpha )}S_0(\alpha )}\right] \right|\le & {} \prod _{k\ge t^{1/\alpha }}\left| \mathbb {E}^{(t)}\left[ \textrm{e}^{\textrm{i}ut^{1-1/(2\alpha )}k^{-\alpha }\eta _k}\right] \right| \\\le & {} \exp (-cu^2 t^{2-1/\alpha }\sum _{k\ge t^{1/\alpha }}k^{-2\alpha })\le \textrm{e}^{-cc_2 u^2}. \end{aligned}$$

Since the function \(u\mapsto \textrm{e}^{-cc_2 u^2}\) is integrable on \(\mathbb {R}\) and, for large t, \(I_1(t,A)\le \int _{|u|>A}\textrm{e}^{-cc_2 u^2}\textrm{d}u\), we conclude that \(\lim _{A\rightarrow \infty }\limsup _{t\rightarrow \infty }I_1(t,A)=0\).

Analysis of \(I_2\). If \(|u|>\rho t^{1/(2\alpha )}\), then \(|u|t^{1-1/(2\alpha )}/k^\alpha \le \rho \) whenever \(k^\alpha \ge \rho ^{-1}|u|t^{1-1/(2\alpha )}>t\). Invoking Lemma 5.8 once again we obtain, for \(t\ge 1\),

$$\begin{aligned}{} & {} \left| \mathbb {E}^{(t)}\left[ \textrm{e}^{\textrm{i}ut^{1-1/(2\alpha )}S_0(\alpha )}\right] \right| \le \prod _{k\ge (\rho ^{-1}|u|t^{1-1/(2\alpha )})^{1/\alpha }}\left| \mathbb {E}^{(t)}\left[ \textrm{e}^{\textrm{i}ut^{1-1/(2\alpha )}k^{-\alpha }\eta _k}\right] \right| \\{} & {} \quad \le \exp \left( -cu^2 t^{2-1/\alpha }\sum _{k\ge (\rho ^{-1}|u|t^{1-1/(2\alpha )})^{1/\alpha }}k^{-2\alpha }\right) \\{} & {} \quad \le \textrm{e}^{-cc_2\rho ^{2-1/\alpha }|u|^{1/\alpha }t^{1/\alpha -1/(2\alpha ^2)}}\le \textrm{e}^{-cc_2\rho ^{2-1/\alpha }|u|^{1/\alpha }}. \end{aligned}$$

Thus, for \(t\ge 1\), \(I_2(t)\le \int _{|u|> \rho t^{1/(2\alpha )}}\textrm{e}^{-cc_2\rho ^{2-1/\alpha }|u|^{1/\alpha }}\textrm{d}u\) and thereupon \(\lim _{t\rightarrow \infty } I_2(t)=0\).

The proof of relation (55) is trivial, hence omitted. The proof of (54) is complete. We note in passing that while dealing with (54), we have shown that the characteristic function given in (51) is absolutely integrable, thereby justifying (52).

The proof of Theorem 5.7 is complete. \(\square \)

Proof of Proposition 5.5

Fix any \(h\in (0,1)\) and write

$$\begin{aligned}{} & {} t^{1/(2\alpha )}\mathbb {E}^{(t)}\left[ \textrm{e}^{-tS_0(\alpha )}\mathbbm {1}_{\{S_0(\alpha )>0\}}\right] =t^{1/(2\alpha )}\int _{(0, \lfloor t^{1/(4\alpha )}\rfloor ht^{-1}]}\textrm{e}^{-tx}\textrm{d}\mathbb {P}^{(t)}\{S_0(\alpha )\le x\}\\{} & {} \quad +t^{1/(2\alpha )}\int _{(\lfloor t^{1/(4\alpha )}\rfloor ht^{-1}, \infty )}\textrm{e}^{-tx}\textrm{d}\mathbb {P}^{(t)}\{S_0(\alpha )\le x\}. \end{aligned}$$

The second term is dominated by \(t^{1/(2\alpha )}\textrm{e}^{-\lfloor t^{1/(4\alpha )}\rfloor h}\) and as such is negligible as \(t\rightarrow \infty \). The first term is equal to

$$\begin{aligned}{} & {} t^{1/(2\alpha )}\sum _{k=0}^{\lfloor t^{1/(4\alpha )}\rfloor -1}\int _{(kh/t, (k+1)h/t]}\textrm{e}^{-tx}\textrm{d}\mathbb {P}^{(t)}\{S_0\le x\}\\{} & {} \le \sum _{k=0}^{\lfloor t^{1/(4\alpha )}\rfloor -1} \textrm{e}^{-kh}\left( t^{1/(2\alpha )}\mathbb {P}^{(t)}\{S_0(\alpha )\in (kh/t, (k+1)h/t]\}-hn_\alpha (kh t^{-1/(2\alpha )})\right) \\{} & {} \quad +h\sum _{k=0}^{\lfloor t^{1/(4\alpha )}\rfloor -1} \textrm{e}^{-kh}n_\alpha (kh t^{-1/(2\alpha )}). \end{aligned}$$

By Theorem 5.7 with \(x=kht^{-1/(2\alpha )}\), the first term on the right-hand side vanishes as \(t\rightarrow \infty \). Given \(\varepsilon >0\), \(|n_\alpha (kh t^{-1/(2\alpha )})-n_\alpha (0)|\le \varepsilon \) for all positive integers \(k\le \lfloor t^{1/(4\alpha )}\rfloor -1\) and large enough t. By a standard reasoning, it follows that \(\lim _{t\rightarrow \infty } \sum _{k=0}^{\lfloor t^{1/(4\alpha )}\rfloor -1} \textrm{e}^{-kh}(n_\alpha (kh t^{-1/(2\alpha )})-n_\alpha (0))=0\). Thus, we have proved that

$$\begin{aligned} \limsup _{t\rightarrow \infty } t^{1/(2\alpha )}\mathbb {E}^{(t)}\left[ \textrm{e}^{-tS_0(\alpha )}\mathbbm {1}_{\{S_0(\alpha )>0\}}\right] \le n_\alpha (0) h(1-\textrm{e}^{-h})^{-1}. \end{aligned}$$

Letting h tend to \(0+\) we infer

$$\begin{aligned} \limsup _{t\rightarrow \infty } t^{1/(2\alpha )}\mathbb {E}^{(t)}\left[ \textrm{e}^{-tS_0(\alpha )}\mathbbm {1}_{\{S_0(\alpha )>0\}}\right] \le n_\alpha (0)=(2\pi \sigma _\alpha ^2)^{-1/2}. \end{aligned}$$

The converse inequality for the limit inferior follows analogously. The proof of Proposition 5.5 is complete. \(\square \)

6 Proofs of Theorems 1.4 and 1.5

As we have already mentioned in the introduction, the distribution of \(S(\alpha )\) is absolutely continuous with a smooth density \(g_\alpha \). For \(t>0\), define \(g_\alpha ^{(t)}\) by

$$\begin{aligned} g_\alpha ^{(t)}(x)=\frac{\textrm{e}^{tx}g_\alpha (x)}{\mathbb {E}[\textrm{e}^{tS(\alpha )}]}, \quad x\in \mathbb {R} \end{aligned}$$

or equivalently

$$\begin{aligned} g_\alpha (x)=\mathbb {E}[\textrm{e}^{tS(\alpha )}]\textrm{e}^{-tx}g^{(t)}_\alpha (x),\quad x\in \mathbb {R}. \end{aligned}$$
(56)

The so defined \(g_\alpha ^{(t)}\) is a density of the \(\mathbb {P}^{(t)}\)-distribution of \(S(\alpha )\). It follows from the proof of Theorem 5.7 that the characteristic function \(z\mapsto \mathbb {E}^{(t)}[\textrm{e}^{\textrm{i}zS_0(\alpha )}]\), \(x\in \mathbb {R}\) is absolutely integrable. Hence, an application of the Fourier inversion formula yields

$$\begin{aligned} g_\alpha ^{(t)}(\mathbb {E}^{(t)}[S(\alpha )]) =\frac{1}{2\pi } \int _\mathbb {R}\mathbb {E}^{(t)}[\textrm{e}^{\textrm{i}zS_0(\alpha )}]\textrm{d}z. \end{aligned}$$

Substituting \(z=ut^{1-1/(2\alpha )}\), we then have

$$\begin{aligned} g_\alpha ^{(t)}(\mathbb {E}^{(t)}[S(\alpha )]) =\frac{t^{1-1/(2\alpha )}}{2\pi } \int _\mathbb {R}\mathbb {E}^{(t)}[\textrm{e}^{\textrm{i}ut^{1-1/(2\alpha )}S_0(\alpha )}]\textrm{d}u. \end{aligned}$$

Combining this representation with the equality

$$\begin{aligned} \frac{1}{\sqrt{2\pi \sigma _\alpha ^2}} =\frac{1}{2\pi }\int _\mathbb {R}\textrm{e}^{-\sigma _\alpha ^2u^2/2}\textrm{d}u, \end{aligned}$$

we infer

$$\begin{aligned} t^{1/(2\alpha )-1}g_\alpha ^{(t)}(\mathbb {E}^{(t)}[S(\alpha )]) =\frac{1}{\sqrt{2\pi \sigma _\alpha ^2}} +\frac{1}{2\pi }\int _\mathbb {R}\left( \mathbb {E}^{(t)}[\textrm{e}^{\textrm{i}ut^{1-1/(2\alpha )}S_0(\alpha )}] -\textrm{e}^{-\sigma _\alpha ^2u^2/2}\right) \textrm{d}u. \end{aligned}$$

Arguing as in the proof of Theorem 5.7, we obtain

$$\begin{aligned} \lim _{t\rightarrow \infty } \int _\mathbb {R}\left( \mathbb {E}^{(t)}[\textrm{e}^{\textrm{i}ut^{1-1/(2\alpha )}S_0(\alpha )}] -\textrm{e}^{-\sigma _\alpha ^2u^2/2}\right) \textrm{d}u=0 \end{aligned}$$

and thereupon

$$\begin{aligned} g_\alpha ^{(t)}(\mathbb {E}^{(t)}[S(\alpha )])~\sim ~\frac{t^{1-1/(2\alpha )}}{\sqrt{2\pi \sigma _\alpha ^2}},\quad t\rightarrow \infty . \end{aligned}$$

Plugging this into (56) yields

$$\begin{aligned} g_\alpha (\mathbb {E}^{(t)}[S(\alpha )])~\sim ~\mathbb {E}[\textrm{e}^{tS(\alpha )}]\textrm{e}^{-t\mathbb {E}^{(t)}[S(\alpha )]}\frac{t^{1-1/(2\alpha )}}{\sqrt{2\pi \sigma _\alpha ^2}},\quad t\rightarrow \infty . \end{aligned}$$

Using the same \(t=t(x)\) as before, that is, a unique solution to \(\mathbb {E}^{(t)}[S(\alpha )]=x\), see (11), we obtain

$$\begin{aligned} g_\alpha (x)~\sim ~\mathbb {E}[\textrm{e}^{t(x)S(\alpha )}]\textrm{e}^{-xt(x)}\frac{(t(x))^{1-1/(2\alpha )}}{\sqrt{2\pi \sigma _\alpha ^2}},\quad x\rightarrow \infty . \end{aligned}$$

By Proposition 5.5 and the argument given in Sect. 3,

$$\begin{aligned} \mathbb {P}\{S(\alpha )>x\}~\sim ~\mathbb {E}[\textrm{e}^{t(x)S(\alpha )}]\textrm{e}^{-xt(x)} \frac{(t(x))^{-1/(2\alpha )}}{\sqrt{2\pi \sigma _\alpha ^2}},\quad x\rightarrow \infty , \end{aligned}$$

whence

$$\begin{aligned} g_\alpha (x)~\sim ~ t(x)\mathbb {P}\{S(\alpha )>x\},\quad x\rightarrow \infty . \end{aligned}$$

With this at hand, the desired result is now secured by the already known asymptotics of \(\mathbb {P}\{S(\alpha )>x\}\) given in Theorems 1.1 and 1.2 and the asymptotic of t(x) given in Propositions 5.1 and 5.2.