Probabilistic View of Explosion in an Inelastic Kac Model

Abstract

Let $\{\mu (\cdot ,t):t\ge 0\}$ be the family of probability measures corresponding to the solution of the inelastic Kac model introduced in Pulvirenti and Toscani (J Stat Phys 114:1453–1480, 2004). It has been proved by Gabetta and Regazzini (J Stat Phys 147:1007–1019, 2012) that the solution converges weakly to equilibrium if and only if a suitable symmetrized form of the initial data belongs to the standard domain of attraction of a specific stable law. In the present paper it is shown that, for initial data which are heavier-tailed than the aforementioned ones, the limiting distribution is improper in the sense that it has probability $1/2$ “adherent” to $-\infty$ and probability $1/2$ “adherent” to $+\infty$. It is explained in which sense this phenomenon is amenable to a sort of explosion, and the main result consists in an explicit expression of the rate of such an explosion. The presentation of these statements is preceded by a discussion about the necessity of the assumption under which their validity is proved. This gives the chance to make an adjustment to a portion of a proof contained in the above-mentioned paper by Gabetta and Regazzini.

Appendices

Appendix 1: Proof of Proposition 1

By virtue of the Skorokhod construction there is a subset ${\widehat{\Omega }}'$ of ${\widehat{\Omega }}$, with ${\widehat{{\mathcal {P}}}}$-probability $1$, such that the recursive relations (4) hold true for every point of ${\widehat{\Omega }}'$. Then, we redefine the ${\hat{\beta }^{(n)}}$’s according to (4) outside ${\widehat{\Omega }}'$. This does not change the distribution of ${\widehat{W}}_n$. Next, we introduce the compact space

$$\begin{aligned} M:= \overline{{\mathbb {N}}}^\infty \times \Big (\times _{j\ge 1}\overline{{\mathbb {N}}}_j^\infty \Big )\times \Big (\times _{j\ge 1}I_j^\infty \Big ) \end{aligned}$$

where $\overline{{\mathbb {N}}}_1,\overline{{\mathbb {N}}}_2,\dots$ are copies of $\overline{{\mathbb {N}}}:=\{1,2,\dots ,+\infty \}$ and $I_1,I_2,\dots$ are copies of $[0,2\pi ]$. We define the vector-valued function $\widehat{Y}:{\widehat{\Omega }}\rightarrow M$ such that

$$\begin{aligned} \widehat{Y}:=\Big ( ({\hat{\nu }}^{(j)})_{j\ge 1}, ({\hat{i}}^{(j)})_{j\ge 1}, ({\hat{\theta }}^{(j)})_{j\ge 1}\Big ) \end{aligned}$$

(same notation as in Sect. 2), and the mappings

$$\begin{aligned} f_i(\widehat{Y})&:= \Big ( ({\hat{\nu }}^{(1)},{\hat{i}}^{(1)}),\dots ,({\hat{\nu }}^{(i)},{\hat{i}}^{(i)}),\big (c_p({\hat{\theta }}_j^{(1)}),s_p({\hat{\theta }}_j^{(1)})\big )_{j\ge 1},\qquad \qquad (\widehat{Y}\in M)\\&\qquad \dots ,\big (c_p({\hat{\theta }}_j^{(i)}),s_p({\hat{\theta }}_j^{(i)})\big )_{j\ge 1}\Big ) \end{aligned}$$

$i=1,2,\dots$. It should be noted that $({\hat{\nu }}^{(i)},{\hat{i}}^{(i)})$ specifies a McKean tree, say ${\widehat{\mathcal {T}}}_i$. The leading idea of the proof is the construction of a decreasing sequence $(A_n)_{n\ge 1}$ of nonempty closed subsets of $M$ such that inequalities (5)–(6) are met simultaneously when $\widehat{Y}$ belongs to $A_n$, for every $n$. To show this we first exhibit, for each $n$, a distinguished McKean tree ${ \mathcal {T}}_n$ together with a specific sequence of angles $\theta ^{(n)}$ for which the desired inequalities occur. Then, we will make use of the distributional properties of $({\widehat{\mathcal {T}}}_n,{\hat{\theta }^{(n)}})$ to state the existence of suitable neighbourhoods of the above pair on which the inequalities of interest (5)–(6) are preserved. The rule we follow to construct the $A_n$’s is recursive. In the sequel, we confine ourselves to describing the first two steps since the step from $A_n$ to $A_{n+1}$ is essentially the same but with a more complicated notation.

Proof for $n=1$ Our aim is to combine a tree with a sequence of angles in such a way that they yield (5)–(6). We begin by focusing on the complete tree of suitable depth $N_1$ and on angles equal to $\pi /4$. The common value of the associated ${\hat{\beta }}$’s is $[(1/\sqrt{2})^{2/\alpha }]^{N_1}=(1/2)^{N_1/\alpha }$. We choose $N_1$ and $m_1$ in such a way that these ${\hat{\beta }}$’s are all greater than $1/(x_{m_1}-\varepsilon )$, which is equivalent to $N_1\le \log _2(x_{m_1}-\varepsilon )^\alpha$. This inequality is satisfied if we select $m_1$ such that $x_{m_1}$ is greater than $(1+\varepsilon )$ and $N_1=\lfloor \log _2(x_{m_1}-\varepsilon )^\alpha \rfloor$. Now consider this complete McKean tree with $2^{N_1}$ leaves (see the construction of a generic tree at the end of Sect. 2) and assume all its leaves germinate to obtain the complete tree with $\nu ^{(1)}=2^{N_1+1}$ leaves, that is ${ \mathcal {T}}_1$. According to the usual left-to-right order, ${\hat{\theta }}^{(1)}_{2^{N_1}+k-1}$ will indicate the angle associated with the germination of the $k$th leaf of the original tree (i.e. before germination), $k=1,\dots ,2^{N_1}$. With a view to the construction of the ${\hat{\beta }}^{(1)}$’s, we start from a reference situation with $(2^{N_1}-1)$ angles $\theta ^{(1)}_1=\dots =\theta ^{(1)}_{2^{N_1}-1}=\pi /4$ and in which we choose values $\theta ^{(1)}_{2^{N_1}+k-1}, k=1,\dots ,2^{N_1}$, meeting $1/x_{m_1}\le |\cos \theta ^{(1)}_{2^{N_1}+k-1}|^{2/\alpha }(1/2)^{N_1/\alpha }\le 1/(x_{m_1}-\varepsilon )$. Values of this kind exist since $2^{N_1/\alpha }/(x_{m_1}-\varepsilon )\le 1$. With this choice of $(2^{N_1+1}-1)$ angles, inequality (5) follows immediately. Moreover, it is easy to check that $\sum _{i=1}^{\lfloor \frac{\nu ^{(1)}+1}{2}\rfloor }|\beta ^{(1)}_{2i-1,\nu ^{(1)}}|^\alpha \ge \frac{1}{x_{m_1}^\alpha }\lfloor \frac{\nu ^{(1)}+1}{2}\rfloor \ge \frac{1}{2}(\frac{x_{m_1}-\varepsilon }{x_{m_1}})^\alpha =:a>0$. Then, there are a tree, i.e. the above ${ \mathcal {T}}_1$ with $\nu ^{(1)}=2^{N_1+1}$ leaves, and a nondegenerate closed interval ${\mathcal {I}}_1$ including $(1/2,\dots ,1/2,|\cos \theta ^{(1)}_{2^{N_1}}|^2,\dots ,|\cos \theta ^{(1)}_{2^{N_1+1}-1}|^2)$ with the following property: the ${\hat{\beta }}^{(1)}$’s associated with the values of $\Big (({\hat{\nu }}^{(1)},{\hat{i}}^{(1)}),\; \big (c_p({\hat{\theta }}_j^{(1)}),s_p({\hat{\theta }}_j^{(1)})\big )_{j\ge 1}\Big )$ contained in $\{{ \mathcal {T}}_1\}\times {\mathcal {I}}_1\times [0,1]^\infty$ satisfy (5)–(6). It is of paramount importance to note that, thanks to the distributional properties of $({\widehat{\mathcal {T}}}_1,{\hat{\theta }}^{(1)})$, the probability of the event $\{ f_1(\widehat{Y})\in \{{ \mathcal {T}}_1\}\times {\mathcal {I}}_1\times [0,1]^\infty \}$ is strictly positive. Now, $A_1:=f_1^{-1}\Big (\{{ \mathcal {T}}_1\}\times {\mathcal {I}}_1\times [0,1]^\infty \Big )$ is a closed subset of $M$ thanks to the continuity of $f_1$.

Proof for $n=2$. We consider the above tree ${ \mathcal {T}}_1$, with $\nu ^{(1)}$ leaves, in combination with the sequence $(\theta ^{(1)}_1,\dots ,\theta ^{(1)}_{\nu ^{(1)}})$ of angles associated with ${ \mathcal {T}}_1$ in the previous part of the proof. We next append a suitable complete tree to each leaf of ${ \mathcal {T}}_1$, in accord with the following rule, to obtain ${ \mathcal {T}}_2$. For the actual construction of the latter, we consider each node $k=1,\dots ,2^{N_1}$ of depth $N_1$ in ${ \mathcal {T}}_1$, and we replace its “left child” (“right child”, respectively) with a complete tree of suitable depth $N^{(k)}_{2,1}$ ($N^{(k)}_{2,2}$, respectively). Arguing as in the first part of the proof, we determine $m_2, N^{(k)}_{2,1}$ and $N^{(k)}_{2,2}$ in such a way that $[(1/\sqrt{2})^{2/\alpha }]^{(N_1+N^{(k)}_{2,1})}|\cos \theta ^{(1)}_{2^{N_1}+k-1}|^{2/\alpha }\ge 1/(x_{m_2}-\varepsilon )$ and $[(1/\sqrt{2})^{2/\alpha }]^{(N_1+N^{(k)}_{2,2})}|\sin \theta ^{(1)}_{2^{N_1}+k-1}|^{2/\alpha }\ge 1/(x_{m_2}-\varepsilon )$, which are equivalent to $N^{(k)}_{2,1}\le \log _2(x_{m_2}-\varepsilon )^\alpha +\log _2|\cos \theta ^{(1)}_{2^{N_1}+k-1}|^2-N_1$ and $N^{(k)}_{2,2}\le \log _2(x_{m_2}-\varepsilon )^\alpha +\log _2|\sin \theta ^{(1)}_{2^{N_1}+k-1}|^2 -N_1$. Then, we define $m_2$ to be the first index $m>m_1$ an $x_m$ satisfies $\lfloor \log _2(x_m-\varepsilon )^\alpha +\log _2|\beta ^{(1)}_{j,\nu ^{(1)}}|^\alpha \rfloor \ge 1$ for every $j=1,\dots ,\nu ^{(1)}$. Then, we put $N^{(k)}_{2,1}:=\lfloor \log _2(x_{m_2}-\varepsilon )^\alpha +\log _2|\cos \theta ^{(1)}_{2^{N_1}+k-1}|^2 \rfloor -N_1$ and $N^{(k)}_{2,2}:=\lfloor \log _2(x_{m_2}-\varepsilon )^\alpha +\log _2|\sin \theta ^{(1)}_{2^{N_1}+k-1}|^2 \rfloor -N_1$ for $k=1,\dots ,2^{N_1}$. As to the remaining part of the argument, for the sake of notational simplicity we confine ourselves to giving a detailed description for $k=1$. We assume that each leaf of the complete tree of depth $N^{(1)}_{2,1}$, previously appended to the “left child” of the node $k=1$, germinates. We choose $\theta ^{(2)}_{2^{N_1+1}+2^{N^{(1)}_{2,1}}+j-2} -$ the angle associated with the $j$th germination ($j=1,\dots ,2^{N^{(1)}_{2,1}}-1$) $-$ in such a way that $1/x_{m_2}\le (1/2)^{(N_1+N^{(1)}_{2,1})/\alpha }|\cos \theta ^{(1)}_{2^{N_1}}|^{2/\alpha }|\cos \theta ^{(2)}_{2^{N_1+1}+2^{N^{(1)}_{2,1}}+j-2}|^{2/\alpha }\le 1/(x_{m_2}-\varepsilon )$. As to its existence, it suffices to note that $2^{(N_1+N^{(1)}_{2,1})/\alpha }/[(x_{m_2}-\varepsilon )|\cos \theta ^{(1)}_{2^{N_1}}|^{2/\alpha }]\le 1$. We now repeat the procedure for the “right child” of the node $k=1$ with $N^{(1)}_{2,2}$ in place of $N^{(1)}_{2,1}$ and $\sin \theta ^{(1)}_{2^{N_1}}$ in place of $\cos \theta ^{(1)}_{2^{N_1}}$. The argument to extend this construction to all the remaining nodes $k=2,\dots ,2^{N_1}$ is essentially the same and requires only obvious small changes in notation. The resulting tree $-$ with $\nu ^{(2)}=2\sum _{k=1}^{2^{N_1}}(2^{N^{(k)}_{2,1}}+2^{N^{(k)}_{2,2}})$ leaves $-$ is ${ \mathcal {T}}_2$. The angles to be associated with ${ \mathcal {T}}_2$ are $\theta ^{(2)}_h=\theta ^{(1)}_h$ for $h=1,\dots ,2^{N_1+1}-1$, while every angle pertaining to the complete trees appended, of depths $N^{(\cdot )}_{2,1}$ and $N^{(\cdot )}_{2,2}$ respectively, is equal to $\pi /4$. It is now easy to show that $\beta ^{(2)}$, generated by ${ \mathcal {T}}_2$ together with the aforesaid angles, satisfies (5)–(6) with $K=2$. Moreover, the complete subtree ${ \mathcal {T}}'_2$ of ${ \mathcal {T}}_2$ having $\nu ^{(1)}=2^{N_1}$ leaves coincides with ${ \mathcal {T}}_1$ and, as to the angles associated with this subtree, one has $\theta ^{(2)}_h=\theta ^{(1)}_h$ for $h=1,\dots ,2^{N_1+1}-1$. As a consequence, (5)–(6) are satisfied also for $K=1$. At this stage, we can state the existence of a tree, the same ${ \mathcal {T}}_2$ as above, and of a nondegenerate closed interval ${\mathcal {I}}_2$ such that: (a) ${\mathcal {I}}_2={\mathcal {I}}'_1\times {\mathcal {I}}_{1,2}$ for which ${\mathcal {I}}'_1\subset {\mathcal {I}}_1$; (b) the ${\hat{\beta }}^{(2)}$’s associated with the values of $f_2(\widehat{Y})$ contained in $\{{ \mathcal {T}}_1\}\times \{{ \mathcal {T}}_2\}\times {\mathcal {I}}_1\times [0,1]^\infty \times {\mathcal {I}}_2\times [0,1]^\infty$ satisfy (5)–(6). Moreover, the event $\{f_2(\widehat{Y})\in \{{ \mathcal {T}}_1\}\times \{{ \mathcal {T}}_2\}\times {\mathcal {I}}_1\times [0,1]^\infty \times {\mathcal {I}}_2\times [0,1]^\infty \}$ has strictly positive probability. The set $A_2:=f_2^{-1}(\{{ \mathcal {T}}_1\}\times \{{ \mathcal {T}}_2\}\times {\mathcal {I}}_1\times [0,1]^\infty \times {\mathcal {I}}_2\times [0,1]^\infty )$ is a closed subset of $M$ and it is easy to conclude, from the definitions of $f_i$ ($i=1,2$), that $A_2\subset A_1$.

Conclusion. Assuming that the procedure has been repeated in the same way to obtain nonempty closed sets of $M\supset A_1\supset A_2\supset \cdots \supset A_n\supset \cdots$, the finite intersection principle leads us to conclude that $\bigcap _{n\ge 1}A_n$ is nonempty. To complete the argument, it suffices to note that $\widehat{Y}^{-1}\big (\bigcap _{n\ge 1}A_n\big )\ne \emptyset$ and (5)–(6) are satisfied for every $\omega ^*$ in $\widehat{Y}^{-1}\big (\bigcap _{n\ge 1}A_n\big )$.

Appendix 2: Proof of Proposition 2

Considering the Skorokhod representation, in order that the law of (3), under ${\mathcal {P}}_t$, converges weakly, it is necessary that ${\widehat{\Lambda }^{(n)}}(\omega )$ converges weakly as $n\rightarrow +\infty$, for every $\omega$ in ${\widehat{\Omega }}$. Then, assuming such a convergence, from the central limit theorem for symmetric triangular arrays (see Theorem 24 in Sect. 16.8 of [21]), for every $\omega$ in ${\widehat{\Omega }}$ there exists a Lévy measure $\nu (\omega )$ such that

$$\begin{aligned} \nu (\omega )[x,+\infty )=\lim _{n\rightarrow +\infty }\sum _{j=1}^{{\hat{\nu }^{(n)}}(\omega )}{\hat{\lambda }^{(n)}}_j(\omega )[x,+\infty ) \end{aligned}$$

(37)

for every positive $x$ for which $\nu (\omega )\{x\}=0$. If (7) is violated, there is $(x_m)_{m\ge 1}$ satisfying (8) and, in view of Proposition 1, there are $(m_n)_{n\ge 1}$ and $\omega ^*$ in ${\widehat{\Omega }}$ for which (5)–(6) are valid for some $\varepsilon >0$. There is also a strictly positive $x_0<1$ for which $\nu (\omega ^*)\{x_0\}=0$, and (37), with $\omega =\omega ^*$ and $x=x_0$, becomes

$$\begin{aligned}&\nu (\omega ^*)[x_0,+\infty )\\&\qquad =\lim _{n\rightarrow +\infty }\sum _{j=1}^{{\hat{\nu }^{(n)}}(\omega ^*)}{\hat{\lambda }^{(n)}}_j(\omega ^*)[x_0,+\infty )\\&\qquad = \lim _{n\rightarrow +\infty }\sum _{j=1}^{{\hat{\nu }^{(n)}}(\omega ^*)}\left[ 1-F^*_0\left( \frac{x_0}{|{\hat{\beta }^{(n)}}_{j,{\hat{\nu }^{(n)}}(\omega ^*)}(\omega ^*)|}\right) \right] \\&\qquad \ge \limsup _{n\rightarrow +\infty }\sum _{j=1}^{{\hat{\nu }^{(n)}}(\omega ^*)}\left[ 1-F^*_0\left( \frac{1}{|{\hat{\beta }^{(n)}}_{j,{\hat{\nu }^{(n)}}(\omega ^*)}(\omega ^*)|}\right) \right] \quad (x_0<1)\\&\qquad \ge \limsup _{n\rightarrow +\infty }\sum _{i=1}^{\lfloor \frac{{\hat{\nu }^{(n)}}(\omega ^*)+1}{2}\rfloor }\left[ 1-F^*_0\left( \frac{1}{|{\hat{\beta }^{(n)}}_{2i-1,{\hat{\nu }^{(n)}}(\omega ^*)}(\omega ^*)|}\right) \right] \frac{|{\hat{\beta }^{(n)}}_{2i-1,{\hat{\nu }^{(n)}}(\omega ^*)}(\omega ^*)|^\alpha }{|{\hat{\beta }^{(n)}}_{2i-1,{\hat{\nu }^{(n)}}(\omega ^*)}(\omega ^*)|^\alpha }\\&\qquad \ge \limsup _{n\rightarrow +\infty }\sum _{i=1}^{\lfloor \frac{{\hat{\nu }^{(n)}}(\omega ^*)+1}{2}\rfloor }[1-F^*_0(x_{m_n})](x_{m_n}-\varepsilon )^\alpha |{\hat{\beta }^{(n)}}_{2i-1,{\hat{\nu }^{(n)}}(\omega ^*)}(\omega ^*)|^\alpha \\&\qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \hbox {(from}(5) )\\&\qquad \ge a \limsup _{n\rightarrow +\infty } x_{m_n}^\alpha [1-F^*_0(x_{m_n})]\left( \frac{x_{m_n}-\varepsilon }{x_{m_n}}\right) ^\alpha \quad \hbox {(from} (6))\\&\qquad = a \limsup _{n\rightarrow +\infty } \rho (x_{m_n}). \end{aligned}$$

This is a contradiction when (7) is violated.

Appendix 3: Proof of Proposition 3

In view of (3), for every $B$ in ${\mathcal {B}}({\mathbb {R}})$ we have

$$\begin{aligned} \mu (B,t)&= {\mathcal {P}}_t(V\in B)\\&= {\mathbb {E}}_t\Big ({\mathcal {P}}_t(V\in B{\tilde{\nu }})\Big )\\&= \sum _{n\ge 1}{\mathcal {P}}_t\{{\tilde{\nu }}=n\}{\mathcal {P}}_t\left\{ \sum _{j=1}^n{\tilde{\beta }}_{j,n}X_j\in B\right\} \end{aligned}$$

where the last equality holds since ${\tilde{\nu }}$ is stochastically independent of $(X,{\tilde{i}},{\tilde{\theta }})$. Thus, to prove the proposition, it is enough to show that

$$\begin{aligned} \sup _{n\ge 1} {\mathcal {P}}_t\left\{ \sum _{j=1}^n{\tilde{\beta }}_{j,n}X_j\in I^c_\varepsilon \right\} \le \varepsilon \end{aligned}$$

holds for every $\varepsilon >0$ and for a suitable interval $I_\varepsilon$ such as that introduced immediately before the wording of the proposition. The proof is based on inequality (18) applied to $Y=\sum _{j=1}^n{\tilde{\beta }}_{j,n}X_j$, which has the same p.d. as $\sum _{j=1}^n|{\tilde{\beta }}_{j,n}|X_j$, thanks to the symmetry of the common distribution $F^*_0$ of the $X$’s. The c.f. of the sum with absolute values of the coefficients, in view of the independence of $X$ and ${\tilde{\beta }}$ and the mutual independence of the $X$’s, is given by $u\mapsto {\mathbb {E}}_t[\prod _{j=1}^n \varphi _0(u|{\tilde{\beta }}_{j,n}|)]$, and hence

$$\begin{aligned}&{\mathcal {P}}_t\left\{ \left| \sum _{j=1}^n|{\tilde{\beta }}_{j,n}|X_j\right| >C\right\} \le \frac{1}{\Delta }\left( 1+\frac{2\pi }{C\Delta }\right) ^2\int \limits _0^\Delta \left[ 1-{\mathbb {E}}_t\left( \prod _{j=1}^n \varphi ^*_0(u|{\tilde{\beta }}_{j,n}|)\right) \right] du\nonumber \\&\quad = {\mathbb {E}}_t \left( \frac{1}{\Delta }\left( 1+\frac{2\pi }{C\Delta }\right) ^2\int \limits _0^\Delta \left[ 1-\prod _{j=1}^n \varphi ^*_0(u|{\tilde{\beta }}_{j,n}|)\right] du\right) . \end{aligned}$$

(38)

From now on, the proof proceeds, with slight changes, like in Step 4 in Sect. 3 of [25]. If

$$\begin{aligned} 0\le 1-\varphi ^*_0(u)\le \frac{3}{8}, \end{aligned}$$

(39)

then there is $\theta$ such that $0\le |\theta |\le 1$ and

$$\begin{aligned} \log \varphi ^*_0(u)=-\Big \{1-\varphi ^*_0(u)\Big \}+\frac{4\theta }{5}\Big \{1-\varphi ^*_0(u)\Big \}^2. \end{aligned}$$

See Lemma 3 of Sect. 3 in [16]. Obviously, there is $\Delta _0>0$ so that (39) holds true if $|u|\le \Delta \le \Delta _0$. For any $u$ of this kind, since $|{\tilde{\beta }}_{j,n}|\le 1$ for every $j$ and $n$, it turns out that $u|{\tilde{\beta }}_{j,n}|\le \Delta$ and

$$\begin{aligned} \log \varphi ^*_0(u|{\tilde{\beta }}_{j,n}|)=-\Big \{1-\varphi ^*_0(u|{\tilde{\beta }}_{j,n}|)\Big \}+\frac{4\theta }{5}\Big \{1-\varphi ^*_0(u|{\tilde{\beta }}_{j,n}|)\Big \}^2 \end{aligned}$$

is valid for every $j$ and $n$. Thus, the last integrand in (38) can be developed as follows:

$$\begin{aligned} 1-\prod _{j=1}^n \varphi ^*_0(u|{\tilde{\beta }}_{j,n}|)&= 1-\exp \left( \sum _{j=1}^n\log \varphi ^*_0(u|{\tilde{\beta }}_{j,n}|)\right) \\&= 1-\exp \left( \sum _{j=1}^n\left[ -(1-\varphi ^*_0(u|{\tilde{\beta }}_{j,n}|))+\frac{4}{5}\theta (1-\varphi ^*_0(u|{\tilde{\beta }}_{j,n}|))^2 \right] \right) \\&\le 1-\exp \left( \sum _{j=1}^n\left[ -(1-\varphi ^*_0(u|{\tilde{\beta }}_{j,n}|))-\frac{3}{10}(1-\varphi ^*_0(u|{\tilde{\beta }}_{j,n}|))\right] \right) \\&= 1-\exp \left( -\frac{13}{10}\sum _{j=1}^n(1-\varphi ^*_0(u|{\tilde{\beta }}_{j,n}|))\right) \le \frac{13}{10}\sum _{j=1}^n(1-\varphi ^*_0(u|{\tilde{\beta }}_{j,n}|)). \end{aligned}$$

This entails

$$\begin{aligned} {\mathcal {P}}_t\left\{ \left| \sum _{j=1}^n|{\tilde{\beta }}_{j,n}|X_j\right| >C\right\} \le \frac{13}{10}\frac{1}{\Delta }\left( 1+\frac{2\pi }{C\Delta }\right) ^2{\mathbb {E}}_t\left( \sum _{j=1}^n\int \limits _0^\Delta \left[ 1-\varphi ^*_0(u|{\tilde{\beta }}_{j,n}|)\right] du\right) . \end{aligned}$$

Now, proceeding as in Step 4 of [25], with $|{\tilde{\beta }}_{j,n}|$ in place of $1/n^{1/\alpha }$,

$$\begin{aligned}&\int \limits _0^\Delta \Big [1-\varphi ^*_0(u|{\tilde{\beta }}_{j,n}|)\Big ]du\\&\qquad \le 2\Delta ^{\alpha +1}|{\tilde{\beta }}_{j,n}|^\alpha M \int \limits _0^{+\infty }\left( {\mathbb {I}}_{\{(0,\varepsilon )\}}(u)\frac{y^{1-\alpha }}{6}+{\mathbb {I}}_{\{(\varepsilon ,+\infty )\}}(u)\frac{1+u}{u^{\alpha +2}}\right) du \end{aligned}$$

where $M:=\sup _{x\ge 0}\rho (x)$ which is finite by hypothesis. After noticing that the integral in the right-hand-side is finite, and denoting its value by $A$, we have

$$\begin{aligned} {\mathcal {P}}_t\left\{ \left| \sum _{j=1}^n|{\tilde{\beta }}_{j,n}|X_j\right| >C\right\} \le \frac{13}{5}MA\Delta ^\alpha \left( 1+\frac{2\pi }{C\Delta }\right) ^2 \end{aligned}$$

and hence, taking $C:=C_\varepsilon =\Delta ^{-1}\ge \Big (\frac{13}{5}\frac{MA(1+2\pi )^2}{\varepsilon }\Big )^{1/\alpha }$,

$$\begin{aligned} {\mathcal {P}}_t\left\{ \Bigg |\sum _{j=1}^n|{\tilde{\beta }}_{j,n}|X_j\Bigg |>C\right\} \le \varepsilon \end{aligned}$$

for every $n$, i.e. $\sup _{n\ge 1} {\mathcal {P}}_t\Big \{\sum _{j=1}^n|{\tilde{\beta }}_{j,n}|X_j\in I^c_\varepsilon \Big \}\le \varepsilon$ with $I_\varepsilon :=[-C_\varepsilon ,C_\varepsilon ]$.

Appendix 4: Rates of Explosion in Example 2

In the probabilistic setting at the beginning of Sect. 2, let us replace each ${\mathcal {P}}_t$ by ${\bar{\mathcal {P}}}_t$, with the proviso that, under ${\bar{\mathcal {P}}}_t$, all the random elements maintain the previous p.d.’s, with the exception of the $X_j$’s whose common p.d.f. becomes ${\bar{F}}^*_{0,t}(u):= F^*_0(u \cdot \xi (t))$ for every $u>0$ and for $\xi (t):=x(t)/\varepsilon _1(t)$ where $t\mapsto \varepsilon _1(t)$ is a strictly positive function on $(0,+\infty )$ such that $\varepsilon _1(t)\searrow 0$, as $t\rightarrow +\infty$, and $t\mapsto x(t)$ is the same as the one specified in each of the examples we are dealing with. Then, for every $x>0$,

$$\begin{aligned} \mu \Big ((-x\cdot \xi (t), x\cdot \xi (t)),t\Big )=\bar{\mathcal {P}}_t\left\{ \sum _{j=1}^{{\tilde{\nu }}}|{\tilde{\beta }}_{j,{\tilde{\nu }}}|X_j\in (-x,x)\right\} . \end{aligned}$$

It is clear that for every $\varepsilon _1(\cdot )$ such that $\mu \Big ((-x\cdot \xi (t), x\cdot \xi (t)),t\Big ) \rightarrow 1$, as $t\rightarrow +\infty , \xi (t)$ represents an upper bound to the rate of explosion $a_t$. To find such a kind of $\varepsilon _1(\cdot )$’s, we can resort to the degenerate convergence criterion provided by the central limit theorem (see, e.g. [27]). Accordingly, if

$$\begin{aligned} \bar{\mathbb {E}}_t\left( \sum _{j=1}^{{\tilde{\nu }}}\int \limits _{|x|\ge \varepsilon }d\bar{F}^*_{0,t}\left( \dfrac{x}{|{\tilde{\beta }}_{j,{\tilde{\nu }}}|}\right) \right) \rightarrow 0 \end{aligned}$$

(40)

and

$$\begin{aligned} \bar{\mathbb {E}}_t\left( \sum _{j=1}^{{\tilde{\nu }}}\int \limits _{-\tau }^\tau x^2 d\bar{F}^*_{0,t}\left( \dfrac{x}{|{\tilde{\beta }}_{j,{\tilde{\nu }}}|}\right) \right) \rightarrow 0 \end{aligned}$$

(41)

hold, as $t\rightarrow +\infty$, for every $\varepsilon >0$ and for some $\tau >0$, then there are a strictly increasing and divergent sequence of times $t_n$ and a version of the conditional distribution of $\sum _{j=1}^{{\tilde{\nu }}}|{\tilde{\beta }}_{j,{\tilde{\nu }}}|X_j$, given $({\tilde{\nu }},{\tilde{i}},{\tilde{\theta }})$, which converges weakly to the unit mass $\delta _0$ at $0$. (It is worth noting that, for each $t_n$, the above-mentioned conditional distribution is derived assuming that the reference p.d. on $(\Omega ,{\mathcal {F}})$ is just $\bar{\mathcal {P}}_{t_n}$.) Then, $\mu ((-x\cdot \xi (t_n), x\cdot \xi (t_n)),t_n)\rightarrow 1$ for every $x>0$. Since the same argument can be used to prove that every strictly increasing and divergent sequence of times $\tau _n$ includes a suitable subsequence $\tau '_n$ of the same type such that $\mu ((-x\cdot \xi (\tau '_n), x\cdot \xi (\tau '_n)),t_n)\rightarrow 1$, we get: If (40)–(41) hold, then $\mu ((-x\cdot \xi (t), x\cdot \xi (t)),t)\rightarrow 1$ for every $x>0$.

Applying this criterion to the first case considered in Example 2, we have, for sufficiently large values of $t$,

$$\begin{aligned} \bar{\mathbb {E}}_t\left( \sum _{j=1}^{{\tilde{\nu }}}\int \limits _{|x|\ge \varepsilon }d\bar{F}^*_{0,t}\left( \dfrac{x}{|{\tilde{\beta }}_{j,{\tilde{\nu }}}|}\right) \right) =\frac{2}{\varepsilon ^\beta \xi (t)^\beta }\bar{\mathbb {E}}_t\sum _{j=1}^{{\tilde{\nu }}}|{\tilde{\beta }}_{j,{\tilde{\nu }}}|^\beta =\frac{2}{\varepsilon ^\beta \xi (t)^\beta }\exp [t(2R_{2\beta /\alpha }-1)] \end{aligned}$$

where the last equality follows from Proposition 8 in [22]. Moreover, analogously,

$$\begin{aligned} \bar{\mathbb {E}}_t\left( \sum _{j=1}^{{\tilde{\nu }}}\int \limits _{-\tau }^\tau x^2 d\bar{F}^*_{0,t}\left( \dfrac{x}{|{\tilde{\beta }}_{j,{\tilde{\nu }}}|}\right) \right) =\dfrac{\tau ^{2-\beta }\beta }{(2-\beta )\xi (t)^\beta }\exp [t(2R_{2\beta /\alpha }-1)]. \end{aligned}$$

At this stage, it is easy to verify that, for $\beta \rightarrow 0^+$ of $\beta \rightarrow \alpha ^-$, in order that (40)–(41) be satisfied, it is necessary to take $\xi (t)=\dfrac{C'}{\varepsilon _1(t)^{1/\beta }}e^{t[2R_{2\beta /\alpha }-1+B(\alpha -\beta )\beta \alpha ]/\beta }$. Moreover, with such a choice of $\xi (\cdot )$, (40)–(41) turn out to be valid for every $\beta$ in $(0,\alpha )$.

As far as the second case in Example 2 is concerned, according to the same way of reasoning, we observe that

$$\begin{aligned} \bar{\mathbb {E}}_t\left( \sum _{j=1}^{{\tilde{\nu }}}\int \limits _{|x|\ge \varepsilon }d\bar{F}^*_{0,t}\left( \dfrac{x}{|{\tilde{\beta }}_{j,{\tilde{\nu }}}|}\right) \right) \le \dfrac{2\log (\varepsilon \xi (t))}{\varepsilon ^\alpha \xi (t)^\alpha }+\dfrac{2}{\varepsilon ^\alpha \xi (t)^\alpha }\bar{\mathbb {E}}_t\left( \sum _{j=1}^{{\tilde{\nu }}}|{\tilde{\beta }}_{j,{\tilde{\nu }}}|^\alpha \log \dfrac{1}{|{\tilde{\beta }}_{j,\nu }|}\right) . \end{aligned}$$

The former summand, in the RHS of the above inequality, goes to $0$ as $t\rightarrow +\infty$. As to the latter summand,

$$\begin{aligned}&\dfrac{2}{\varepsilon ^\alpha \xi (t)^\alpha }\bar{\mathbb {E}}_t\left( \sum _{j=1}^{{\tilde{\nu }}}|{\tilde{\beta }}_{j,{\tilde{\nu }}}|^\alpha \log \dfrac{1}{|{\tilde{\beta }}_{j,\nu }|}\right) \\&\quad \le \dfrac{4}{\alpha \varepsilon ^\alpha \xi (t)^\alpha }\bar{\mathbb {E}}_t\left( \log \sum _{j=1}^{{\tilde{\nu }}}|{\tilde{\beta }}_{j,{\tilde{\nu }}}|^\alpha \dfrac{1}{|{\tilde{\beta }}_{j,\nu }|^{\alpha /2}}\right) \qquad \left( \hbox {in view of the Jensen inequality,}\right. \\&\left. \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \hbox {since} \sum _{j=1}^{{\tilde{\nu }}}|{\tilde{\beta }}_{j,{\tilde{\nu }}}|^\alpha =1\right) \\&\quad \le \dfrac{4}{\alpha \varepsilon ^\alpha \xi (t)^\alpha }\log \bar{\mathbb {E}}_t\left( \sum _{j=1}^{{\tilde{\nu }}}|{\tilde{\beta }}_{j,{\tilde{\nu }}}|^{\alpha /2}\right) \qquad \quad \qquad \hbox {(from the Jensen inequality)}\\&\quad \le \dfrac{4\varepsilon _1(t)^\alpha }{\alpha \varepsilon ^\alpha c^\alpha }\left( \dfrac{4}{\pi }-1\right) \qquad \qquad \qquad \qquad \qquad \quad \hbox {(from Proposition 8 in} [22]) \end{aligned}$$

which completes the argument to prove (40). Finally, the validity of (41) can be verified, after an integration by parts, in the same way.

Appendix 5: Proof of Lemma 1

The first statement is an immediate consequence of well-known properties of the domain of attraction described, for example, in Sect. 2.6 of [26]. Accordingly, it turns out that $a_m=(\pi (c_1+c_2))/(2\Gamma (\alpha )\sin (\pi \alpha /2))$ where $c_1=c_2=\dfrac{K_{1,m}\theta _m}{\alpha }$ and hence

$$\begin{aligned} a_m&= \dfrac{K_{1,m}\theta _m}{\alpha }\frac{\pi }{\Gamma (\alpha )\sin (\pi \alpha /2)}\\&= \dfrac{K_{1,m}\theta _m}{\alpha }\frac{\Gamma (1-\alpha )\sin (\pi \alpha )}{\sin (\pi \alpha /2)}\\&\qquad \qquad \qquad \qquad \qquad \hbox {(by the reflection property of the} \Gamma \hbox {function)}\\&= \dfrac{2}{\alpha }K_{1,m}\theta _m \Gamma (1-\alpha )\cos (\pi \alpha /2)\\&= \dfrac{2}{\alpha }K_{1,m}\theta _m\int \limits _{(0,+\infty )}\frac{\sin x}{x^\alpha }dx \qquad \hbox {(by Euler's formula).} \end{aligned}$$

To prove that $1-\hat{g}_{1,m}(\xi )=(a_m+v_m(\xi ))\xi ^\alpha$, set $F_m(x):=1-G_{1,m}(x)=G_{1,m}(-x)$ ($x>0$) to write

$$\begin{aligned} 1-\hat{g}_{1,m}(\xi )=\int \limits _0^{+\infty }(e^{i\xi x}-1)dF_m(x)+\int \limits _0^{+\infty }(e^{-i\xi x}-1)dF_m(x). \end{aligned}$$

As to the first integral,

$$\begin{aligned}&\int \limits _0^{+\infty }(e^{i\xi x}-1)dF_m(x)\nonumber \\&\quad = -i\xi \int \limits _0^{+\infty }\Big (\cos (\xi x)+ i \sin (\xi x)\Big )F_m(x)dx\quad \hbox {(by integration by parts)}\nonumber \\&\quad = -i\xi ^\alpha \int \limits _{(0,+\infty )} \cos (x)\dfrac{h_1(x/\xi )}{x^\alpha }dx+\xi ^\alpha \int \limits _{(0,+\infty )} \sin (x)\dfrac{h_1(x/\xi )}{x^\alpha }dx \end{aligned}$$

(42)

where

$$\begin{aligned} h_1(x):= 1-K_{1,m}\left( F^*_0(x)+F^*_0(x_m)-1+\dfrac{\theta _m}{\alpha x_m^\alpha }\right) x^\alpha {\mathbb {I}}_{\{0<x< x_m\}} + \dfrac{K_{1,m}\theta _m}{\alpha }{\mathbb {I}}_{\{x\ge x_m\}}.\nonumber \\ \end{aligned}$$

(43)

Analogously,

$$\begin{aligned}&\int \limits _0^{+\infty }(e^{-i\xi x}-1)dF_m(x)\nonumber \\&\qquad = i\xi ^\alpha \int \limits _{[0,+\infty )} \cos (x)\dfrac{h_1(x/\xi )}{x^\alpha }dx+\xi ^\alpha \int \limits _{[0,+\infty )} \sin (x)\dfrac{h_1(x/\xi )}{x^\alpha }dx. \end{aligned}$$

(44)

Combination of (42), (44) and (43) gives

$$\begin{aligned} \dfrac{1-\hat{g}_{1,m}(\xi )}{\xi ^\alpha }=a_m+v_m(\xi ) \end{aligned}$$

where $a_m$ is the same as in (13) and $v_m$ is the same as in the wording of the lemma, i.e.

$$\begin{aligned} v_m(\xi )=\dfrac{2}{\xi ^\alpha }\int \limits _0^{\xi x_m} \sin x\; \omega _{m,\xi }(x)dx-\dfrac{2}{\alpha }K_{1,m}\theta _m\int \limits _0^{\xi x_m}\dfrac{\sin x}{x^\alpha }dx \end{aligned}$$

where $\omega _{m,\xi }:=1-K_{1,m}\Big [F^*_0\Big (\frac{x}{\xi }\Big )+F^*_0(x_m)-1+\dfrac{\theta _m}{\alpha x_m^\alpha }\Big )$ for $x<\xi x_m$. Since $0<\alpha <2$, we have $\int _0^{\xi x_m}(\sin x)/x^\alpha dx\rightarrow 0$ as $\xi \rightarrow 0^+$ and, then, the latter summand is $o(1)$ as $\xi \rightarrow 0^+$. As to the former,

$$\begin{aligned} \left| \dfrac{2}{\xi ^\alpha }\int \limits _0^{\xi x_m} \sin x \;\omega _{m,\xi }(x)dx\right| \le \dfrac{2}{\xi ^\alpha }\int \limits _0^{\xi x_m} |\sin x| \sup _{t\in [0,x_m]} \omega _{m,\xi }(x)dx. \end{aligned}$$

Observe that $x\mapsto \omega _{m,\xi }(x)$ is nonnegative, monotonically nonincreasing on $[0,\xi x_m]$ and its value at $0$ is less or equal to $1/2$. Then, for $\xi \le \pi /x_m$

$$\begin{aligned} \left| \dfrac{2}{\xi ^\alpha }\int \limits _0^{\xi x_m} \sin x \;\omega _{m,\xi }(x) dx\right| \le \dfrac{1}{\xi ^\alpha }\int \limits _0^{\xi x_m}\sin x dx= \dfrac{1}{2}x_m^2\xi ^{2-\alpha }+ o(|\xi |^{3-\alpha }) \end{aligned}$$

which completes the proof that $v_m$ is $o(1)$ as $\xi \rightarrow 0^+$. As to the supremum norm of $v_m$,

$$\begin{aligned} \left\| v_m\right\|&:= \sup _{\xi \in {\mathbb {R}}^+} |v_m(\xi )|\le \sup _{\xi \in {\mathbb {R}}^+}\left| \dfrac{1}{\xi ^\alpha }\int \limits _0^{\xi x_m} \sin x \;\omega _{m,\xi }(x)dx\right| \nonumber \\&+ \sup _{\xi \in {\mathbb {R}}^+}\left| \dfrac{2}{\alpha }K_{1,m}\theta _m\int \limits _0^{\xi x_m}\dfrac{\sin x}{x^\alpha }dx\right| . \end{aligned}$$

(45)

By the second mean value theorem, there is $a$ in $[0,\xi x_m)$ such that

$$\begin{aligned} \left| \int \limits _0^{\xi x_m} \omega _{m,\xi }(x)\sin x dx\right| =\left| \omega _{m,\xi }(0)\int \limits _0^a \sin xdx\right| \le 2. \end{aligned}$$

Moreover, from $|\sin x|\le x$ for any $x\ge 0$,

$$\begin{aligned} \left| \int \limits _{0}^{\xi x_m} \omega _{m,\xi }(x)\sin xdx\right| \le \omega _{m,\xi }(0)\int \limits _{0}^{\xi x_m}|\sin x|dx\le \dfrac{x_m^2\xi ^2}{2} \end{aligned}$$

and combination of the above bounds gives

$$\begin{aligned} \left| \dfrac{1}{\xi ^\alpha } \int \limits _{0}^{\xi x_m} \omega _{m,\xi }(x)\sin x\;dx\right| \le \min \left\{ \dfrac{x_m^2\xi ^{2-\alpha }}{2},2\xi ^{-\alpha }\right\} \le 2^{1-\alpha }x_m^\alpha . \end{aligned}$$

To bound the latter summand in the RHS of (45), we begin by recalling the obvious inequalities

$$\begin{aligned} \dfrac{2}{((k+1)\pi )^\alpha }\le \left| \int \limits _{k\pi }^{(k+1)\pi } \dfrac{\sin x}{x^\alpha }dx\right| \le \dfrac{2}{(k\pi )^\alpha }\quad (k=1,2,\dots ) \end{aligned}$$

and by setting $k^*=k^*(\xi ):=\max \{k\in {\mathbb {N}}:\; k\pi \le \xi x_m\}$ for every $\xi >0$. These inequalities, combined with

$$\begin{aligned} \int \limits _0^{\xi x_m}\dfrac{\sin x}{x^\alpha }dx&= \left| \int \limits _0^{\pi }\dfrac{\sin x}{x^\alpha }dx\right| -\left| \int \limits _{\pi }^{2\pi }\dfrac{\sin x}{x^\alpha }dx\right| + \cdots \\&+ (-1)^{k^*-1}\left| \int \limits _{(k^*-1)\pi }^{k^*\pi }\dfrac{\sin x}{x^\alpha }dx\right| +(-1)^{k^*}\left| \int \limits _{k^*\pi }^{x_m\xi }\dfrac{\sin x}{x^\alpha }dx\right| \end{aligned}$$

yield

$$\begin{aligned} \int \limits _0^{\xi x_m}\dfrac{\sin x}{x^\alpha }dx&\le \int \limits _0^\pi \dfrac{\sin x}{x^\alpha }dx-\dfrac{2}{(2\pi )^\alpha } +\dfrac{2}{(2\pi )^\alpha }+\cdots -\dfrac{2}{(x_m\xi )^\alpha }{\mathbb {I}}_{\lbrace \hbox {odd numbers}\rbrace }(k^*)\\&\le \int \limits _0^\pi \dfrac{\sin x}{x^\alpha }dx \end{aligned}$$

and

$$\begin{aligned} \int \limits _0^{\xi x_m}\dfrac{\sin x}{x^\alpha }dx\ge \dfrac{2}{\pi ^\alpha } -\dfrac{2}{\pi ^\alpha }+\cdots +\dfrac{2}{(x_m\xi )^\alpha }{\mathbb {I}}_{\lbrace \hbox {even numbers}\rbrace }(k^*)\ge 0. \end{aligned}$$

This entails

$$\begin{aligned} \sup _{\xi \in {\mathbb {R}}^+}\left| \dfrac{2}{\alpha }K_{1,m}\theta _m\int \limits _0^{\xi x_m}\dfrac{\sin x}{x^\alpha }dx\right| \le \dfrac{2}{\alpha }K_{1,m}\theta _m\int \limits _0^\pi \dfrac{\sin x}{x^\alpha }dx. \end{aligned}$$