Infinite Level GREM-Like K-Processes Existence and Convergence

Abstract

We derive the existence of infinite level GREM-like K-processes by taking the limit of a sequence of finite level versions of such processes as the number of levels diverges. The main step in the derivation is obtaining the convergence of the sequence of underlying finite level clock processes. This is accomplished by perturbing these processes so as to turn them into martingales, and resorting to martingale convergence to obtain convergence for the perturbed clock processes; nontriviality of the limit requires a specific choice of parameters of the original process; we conclude the step by showing that the perturbation washes away in the limit. The perturbation is done by inserting suitable factors into the expression of the clocks, as well as rescaling the resulting expression suitably; the existence of such factors is itself established through martingale convergence.

Appendices

Martingality Inducing Factors—Proofs

Proof of Proposition 2.1

Without loss of generality, let us assume $k = 0$, and define the following random variables. For $n > 0$ and $\lambda > 0$, let

$$\begin{aligned} Z_n(\lambda ) := \exp \left\{ - \sum _{y|_n} \left( \lambda \hat{\gamma }_n(y|_n) \right) ^{\alpha _{n+1}} \right\} . \end{aligned}$$

Let us show that these random variables form a martingale with respect to the filtration $({\mathcal {D}}_n)$, where ${\mathcal {D}}_n$ is the $\sigma$-algebra generated by $\{{\tilde{\gamma }}_j(x|_j): x|_j \in {\mathbb {N}}^j, \,j \le n \}$. Let us at this point note that it readily follows from Proposition B.2 below that

$$\begin{aligned} {\tilde{\gamma }}_k(x|_{k-1}):=\left\{ {\tilde{\gamma }}_k(x|_{k-1},x): x \in {\mathbb {N}}\right\} \end{aligned}$$

(A.1)

is a Poisson point process, with intensity measure ${\tilde{\mu }}(dt) := \frac{c^*_k}{t^{1+\alpha _k}}\,dt, t > 0$.

We may thus use Campbell’s Theorem (see e.g. [33], Section 3.2) in the following:

$$\begin{aligned} {\mathbb {E}}\left( Z_{n+1}(\lambda ) \Big | {\mathcal {D}}_{n} \right)&= {\mathbb {E}}\left[ \exp \left\{ - \sum _{y|_{n+1}} \left( \lambda \hat{\gamma }_{n+1}(y|_{n+1}) \right) ^{\alpha _{n+2}} \right\} \Big |{\mathcal {D}}_{n} \right] \nonumber \\&= \prod _{y|_n} {\mathbb {E}}\left[ \exp \left\{ - \left( \lambda \hat{\gamma }_{n}(y|_{n})\right) ^{\alpha _{n+2}} \sum _{y_{n+1}} {\tilde{\gamma }}_{n+1}(y|_{n+1}) ^{\alpha _{n+2}} \right\} \Big | {\mathcal {D}}_{n} \right] \nonumber \\&= \prod _{y|_n} \exp \left\{ - \int _0^\infty (1 - \exp \left( - \left( \lambda \hat{\gamma }_{n}(y|_{n})\right) ^{\alpha _{n+2}} t^{\alpha _{n+2}}\right) ) {\tilde{\mu }}_{n+1}(d t) \right\} \nonumber \\&= \prod _{y|_n} \exp \left\{ - \left( \lambda \hat{\gamma }_{n}(y|_{n})\right) ^{\alpha _{n+1}} \right\} = \exp \left\{ - \sum _{y|_n}\left( \lambda \hat{\gamma }_{n}(y|_{n})\right) ^{\alpha _{n+1}} \right\} \nonumber \\&= Z_n(\lambda ). \end{aligned}$$

(A.2)

Since $(Z_n(1))$ is a positive martingale, it converges a.s.; thus, its exponent must converge as well.

The second claim can be obtained by using (A.2) to explicitly compute:

$$\begin{aligned} {\mathbb {E}}\left[ \exp \left\{ - \lambda \sum _{y|_n} \hat{\gamma }_n(y|_n)^{\alpha _{n+1}} \right\} \right]&= {\mathbb {E}}\left[ Z_n(\lambda ^{1/\alpha _{n+1}}) \right] \nonumber \\&= {\mathbb {E}}\left[ Z_1(\lambda ^{1/\alpha _{n+1}}) \right] \nonumber \\&= \exp \{ -\lambda ^{\alpha _{1}/\alpha _{n+1}}\} \xrightarrow {n \rightarrow \infty } \exp \{-\lambda ^{\alpha _{1}}\}, \end{aligned}$$

(A.3)

since $\alpha _n\rightarrow 1$ as $n\rightarrow \infty$, where the latter equality follows from Campbell’s Theorem, once we note that $\{[{\tilde{\gamma }}_1(x|_1)]^{\alpha _2},\,x|_1\in {\mathbb {N}}\}$ is a Poisson point process with intensity measure $\mu ^*_1(dt)=(c^*_1/\alpha _2)/t^{1+\alpha _1/\alpha _2}\,dt$.

To establish the final claim, let us fix a realization such that (2.2) holds for every $x|_k \in {\mathbb {N}}^k$, $k\ge 0$, and fix an arbitrary $\epsilon > 0$, then:

$$\begin{aligned} W_k(x|_k)&= \lim _{n \rightarrow \infty } \sum _{y_n \in [x|_k]_n} \left( \frac{\hat{\gamma }_n(y|_n)}{\hat{\gamma }_k(x|_k)} \right) ^{\alpha _{n+1}}\nonumber \\&= \lim _{n \rightarrow \infty } \sum _{x_{k+1}} ({\tilde{\gamma }}_{k+1}(x|_{k+1}))^{\alpha _{n+1}} \sum _{y_n \in [x|_{k+1}]_n} \left( \frac{\hat{\gamma }_n(y|_n)}{\hat{\gamma }_{k+1}(x|_{k+1})} \right) ^{\alpha _{n+1}}\nonumber \\&= \lim _{n \rightarrow \infty } \sum _{\begin{array}{c} x_{k+1}:\\ {\tilde{\gamma }}_{k+1}(x|_{k+1}) > \epsilon \end{array}} ({\tilde{\gamma }}_{k+1}(x|_{k+1}))^{\alpha _{n+1}} \sum _{y_n \in [x|_{k+1}]_n} \left( \frac{\hat{\gamma }_n(y|_n)}{\hat{\gamma }_{k+1}(x|_{k+1})} \right) ^{\alpha _{n+1}} \end{aligned}$$

(A.4)

$$\begin{aligned}&\quad + \lim _{n \rightarrow \infty } \sum _{\begin{array}{c} x_{k+1}:\\ {\tilde{\gamma }}_{k+1}(x|_{k+1}) \le \epsilon \end{array}} ({\tilde{\gamma }}_{k+1}(x|_{k+1}))^{\alpha _{n+1}} \sum _{y_n \in [x|_{k+1}]_n} \left( \frac{\hat{\gamma }_n(y|_n)}{\hat{\gamma }_{k+1}(x|_{k+1})} \right) ^{\alpha _{n+1}}. \end{aligned}$$

(A.5)

Let us treat these two terms separately, and show that their respective limits exist in probability. For (A.4), since the outermost sum is finite and $\alpha _n \rightarrow 1$ as $n \rightarrow \infty$, we have that

$$\begin{aligned} {}(\hbox {A}.4)&= \sum _{\begin{array}{c} x_{k+1}:\\ {\tilde{\gamma }}_{k+1}(x|_{k+1})> \epsilon \end{array}} {\tilde{\gamma }}_{k+1}(x|_{k+1}) \lim _{n \rightarrow \infty } \sum _{y_n \in [x|_{k+1}]_n} \left( \frac{\hat{\gamma }_n(y|_n)}{\hat{\gamma }_k(x|_{k+1})} \right) ^{\alpha _{n+1}}\\&= \sum _{\begin{array}{c} x_{k+1}:\\ {\tilde{\gamma }}_{k+1}(x|_{k+1}) > \epsilon \end{array}} {\tilde{\gamma }}_{k+1}(x|_{k+1}) W_{k+1}(x|_{k+1}) \xrightarrow {\epsilon \rightarrow 0} \sum _{x_{k+1}} {\tilde{\gamma }}_{k+1}(x|_{k+1}) W_{k+1}(x|_{k+1}). \end{aligned}$$

Let ${\mathcal {D}}'_k={\mathcal {D}}'_k(x|_k)$ be the $\sigma$-algebra generated by

$\{{\tilde{\gamma }}_{k+1}(x|_{k+1}) : x_{k+1} \in {\mathbb {N}}\}$. Using (A.3), we can compute the Laplace transform of (A.5) as:

$$\begin{aligned}&{\mathbb {E}}\left[ \exp \left\{ - \lambda \sum _{\begin{array}{c} x_{k+1}:\\ {\tilde{\gamma }}_{k+1}(x|_{k+1}) \le \epsilon \end{array}} ({\tilde{\gamma }}_{k+1}(x|_{k+1}))^{\alpha _{n+1}} \sum _{y_n \in [x|_{k+1}]_n} \left( \frac{\hat{\gamma }_n(y|_n)}{\hat{\gamma }_{k+1}(x|_{k+1})} \right) ^{\alpha _{n+1}} \right\} \right] \\&\quad =\, {\mathbb {E}}\left[ \prod _{\begin{array}{c} x_{k+1}:\\ {\tilde{\gamma }}_{k+1}(x|_{k+1}) \le \epsilon \end{array}} {\mathbb {E}}\left[ \exp \left\{ - \lambda ({\tilde{\gamma }}_{k+1}(x|_{k+1}))^{\alpha _{n+1}}\right. \right. \right. \\&\qquad \left. \left. \left. \sum _{y_n \in [x|_{k+1}]_n} \left( \frac{\hat{\gamma }_n(y|_n)}{\hat{\gamma }_{k+1}(x|_{k+1})} \right) ^{\alpha _{n+1}} \right\} \Big | {\mathcal {D}}'_k \right] \right] \\&\quad =\, {\mathbb {E}}\left[ \prod _{\begin{array}{c} x_{k+1}:\\ {\tilde{\gamma }}_{k+1}(x|_{k+1}) \le \epsilon \end{array}} \exp \left\{ - \left( \lambda ({\tilde{\gamma }}_{k+1}(x|_{k+1}))^{\alpha _{n+1}} \right) ^{\frac{\alpha _{k+2}}{\alpha _{n+1}}} \right\} \right] \xrightarrow {n \rightarrow \infty }\\&{\mathbb {E}}\left[ \exp \left\{ - \lambda ^{\alpha _{k+2}} \!\!\!\!\!\!\!\sum _{\begin{array}{c} x_{k+1}:\\ {\tilde{\gamma }}_{k+1}(x|_{k+1}) \le \epsilon \end{array}}\!\!\!\!\!\!\! ({\tilde{\gamma }}_{k+1}(x|_{k+1}))^{\alpha _{k+2}} \right\} \right] \!= \exp \left\{ \!-\! \int _0^\epsilon \left( 1 - e^{-{(\lambda x)}^{\alpha _{k+2}}} \right) {\tilde{\mu }}_{k+1}(d x) \right\} \\&\quad \xrightarrow {\epsilon \rightarrow 0} 1. \end{aligned}$$

We used Campbell’s Theorem in the last equality. With this we can conclude that the expression in (A.5) converges in probability to 0 as $\epsilon \rightarrow 0$.

Finally, note that we have shown that (A.4) $+$ (A.5) converges in probability to both $W_{k}(x|_k)$ and $\sum _{x_{k+1}} {\tilde{\gamma }}_{k+1}(x|_{k+1}) W_{k+1}(x|_{k+1})$ as $\epsilon \rightarrow 0$, which implies that these latter two quantities are almost surely equal. $\square$

Proof of Corollary 3.1

Let $A_n := \{ x|_n \in {\mathbb {N}}^n: \hat{\gamma }_n(x|_n) > 1 \}$ and $m_n := \max \{ \hat{\gamma }_n(x|_n): x|_n \in {\mathbb {N}}^n\}$. Using Proposition 2.1, we get that almost surely

$$\begin{aligned} W_0(\emptyset )&= \lim _{n \rightarrow \infty } \sum _{x|_n \in {\mathbb {N}}^n} \left( \hat{\gamma }_n(x|_n) \right) ^{\alpha _{n+1}} \ge \limsup _{n \rightarrow \infty } \sum _{x|_n \in A_n} \left( \hat{\gamma }_n(x|_n) \right) ^{\alpha _{n+1}} \ge \limsup _{n \rightarrow \infty } |A_n|;\\ W_0(\emptyset )&= \lim _{n \rightarrow \infty } \sum _{x|_n \in {\mathbb {N}}^n} \left( \hat{\gamma }_n(x|_n) \right) ^{\alpha _{n+1}} \ge \limsup _{n \rightarrow \infty } \left( m_n \right) ^{\alpha _{n+1}} = \limsup _{n \rightarrow \infty } m_n. \end{aligned}$$

Since $W_0(\emptyset ) < \infty$, we have that $\limsup _n |A_n| < \infty$ and $\limsup _n m_n < \infty$ a.s. Thus,

$$\begin{aligned} \sum _{x|_n} \hat{\gamma }_n(x|_n) = \sum _{x|_n \in A_n} \hat{\gamma }_n(x|_n) + \sum _{x|_n\ \not \in A_n} \hat{\gamma }_n(x|_n) \le m_n |A_n| + \sum _{x|_n\ \not \in A_n} \left( \hat{\gamma }_n(x|_n) \right) ^{\alpha _{n+1}}. \end{aligned}$$

From the previous remark, the limsup of the first term of this last sum is bounded a.s. by $W^2_0(\emptyset )$, while the $\limsup$ of the second term is bounded by $W_0(\emptyset )$. $\square$

Proof of Lemma 3.2

Recall from Proposition 2.1 that $W_k(x|_k)$ has a stable distribution, which may be denoted, in the notation of [37, Sections 1.1. and 1.2], as follows: $W_k(x|_k)\sim S_{\alpha _{k+1}}(\sigma ,0,1)$, with $\sigma ^{\alpha _{k+1}}= \cos \left( \frac{\pi \alpha _{k+1} }{2}\right)$. We may thus write the characteristic function of $W_k(x|_k) - 1$ as follows:

$$\begin{aligned} \varphi _k (u)&:= {\mathbb {E}}\left[ e^{i u (W(x|_k) - 1)} \right] \nonumber \\&= \exp \left\{ - |u|^{\alpha _{k+1}} \left[ \cos \left( \frac{\pi \alpha _{k+1}}{2} \right) -i {{\,\mathrm{sgn}\,}}(u) \sin \left( \frac{\pi \alpha _{k+1}}{2} \right) \right] - i u \right\} \nonumber \\&= \exp \left\{ - |u|^{\alpha _{k+1}} \cos \left( \frac{\pi \alpha _{k+1}}{2} \right) -i \left[ u - |u|^{\alpha _{k+1}} {{\,\mathrm{sgn}\,}}(u) \sin \left( \frac{\pi \alpha _{k+1}}{2} \right) \right] \right\} . \end{aligned}$$

(A.6)

Theorem 2.2 from [34] states that

$$\begin{aligned} {\mathbb {E}}\left[ \left| W(x|_k) - 1 \right| ^{\alpha _k} \right]&= \frac{1}{\cos \left( \frac{\pi \alpha _k}{2} \right) } {{\,\mathrm{Re}\,}}\left[ \frac{\alpha _k}{\Gamma (1-\alpha _k)} \int _0^\infty \frac{1 - \varphi _k(-u)}{u^{1+\alpha _k}} d u \right] . \end{aligned}$$

Note that $\frac{\alpha _k}{\cos \left( \frac{\pi \alpha _k}{2}\right) \Gamma (1-\alpha _k)}$ converges to $\frac{2}{\pi }$ as $k\rightarrow \infty$. So we are left with showing that the real part of the above integral vanishes as $k\rightarrow \infty$. Fixing an arbitrary $\epsilon > 0$, we write

$$\begin{aligned}&{{\,\mathrm{Re}\,}}\left[ \int _0^\infty \frac{1 - \varphi _k(-u)}{u^{1+\alpha _k}} d u \right] \nonumber \\&\quad = \int _0^\infty \frac{1}{u^{1+\alpha _{k}}} \left[ 1 - \exp \left\{ -u^{\alpha _{k+1}} \cos \left( \frac{\pi \alpha _{k+1}}{2} \right) \right\} \cos \left( u - u^{\alpha _{k+1}} \sin \left( \frac{\pi \alpha _{k+1}}{2} \right) \right) \right] d u \nonumber \\&\quad = \int _0^\epsilon \frac{1}{u^{1+\alpha _{k}}} \left[ 1 - \exp \left\{ -u^{\alpha _{k+1}} \cos \left( \frac{\pi \alpha _{k+1}}{2} \right) \right\} \cos \left( u - u^{\alpha _{k+1}} \sin \left( \frac{\pi \alpha _{k+1}}{2} \right) \right) \right] d u \end{aligned}$$

(A.7)

$$\begin{aligned}&\qquad + \int _\epsilon ^\infty \frac{1}{u^{1+\alpha _{k}}} \left[ 1 - \exp \left\{ -u^{\alpha _{k+1}} \cos \left( \frac{\pi \alpha _{k+1}}{2} \right) \right\} \cos \left( u - u^{\alpha _{k+1}} \sin \left( \frac{\pi \alpha _{k+1}}{2} \right) \right) \right] d u. \end{aligned}$$

(A.8)

To control (A.8), note that the integrand converges to zero as $k \rightarrow \infty$ and can be bounded by $2/u^{1+\alpha _k} \le 2/u^{3/2}$ for large enough k. Therefore, by dominated convergence, (A.8) vanishes as $k \rightarrow \infty$ for any $\epsilon > 0$.

We control (A.7), using $1-e^{-x} \le x$, as follows. It is bounded above by

$$\begin{aligned}&\int _0^\epsilon \frac{1}{u^{1+\alpha _{k}}} \left[ u^{\alpha _{k+1}} \cos \left( \frac{\pi \alpha _{k+1}}{2} \right) - \log \cos \left( u - u^{\alpha _{k+1}} \sin \left( \frac{\pi \alpha _{k+1}}{2} \right) \right) \right] d u\nonumber \\&\quad = \int _0^\epsilon \frac{1}{u^{1+\alpha _{k}}} u^{\alpha _{k+1}} \cos \left( \frac{\pi \alpha _{k+1}}{2} \right) d u - \int _0^\epsilon \frac{1}{u^{1+\alpha _{k}}} \log \cos \left( u - u^{\alpha _{k+1}} \sin \left( \frac{\pi \alpha _{k+1}}{2} \right) \right) d u\nonumber \\&\quad = \frac{\epsilon ^{\alpha _{k+1} - \alpha _k}}{\alpha _{k+1} - \alpha _k} \cos \left( \frac{\pi \alpha _{k+1}}{2} \right) - \int _0^\epsilon \frac{1}{u^{1+\alpha _{k}}} \log \cos \left( u - u^{\alpha _{k+1}} \sin \left( \frac{\pi \alpha _{k+1}}{2} \right) \right) d u. \end{aligned}$$

(A.9)

Since $\frac{1-\alpha _{k+1}}{1-\alpha _k} \xrightarrow {k \rightarrow \infty } 0$ by hypothesis and $\cos (\pi x /2)/(1-x) \xrightarrow {x \rightarrow 1} \frac{\pi }{2}$, we can rewrite the leftmost term from (A.9) as

$$\begin{aligned}&\frac{\epsilon ^{\alpha _{k+1} - \alpha _k}}{\alpha _{k+1} - \alpha _k} \cos \left( \frac{\pi \alpha _{k+1}}{2} \right) = \epsilon ^{\alpha _{k+1} - \alpha _k} \frac{1- \alpha _{k+1}}{\alpha _{k+1} - \alpha _k} \frac{\cos \left( \frac{\pi \alpha _{k+1}}{2} \right) }{1-\alpha _{k+1}}\\&\quad = \epsilon ^{\alpha _{k+1} - \alpha _k} \left( \frac{1 - \alpha _k}{1- \alpha _{k+1}} - 1 \right) ^{-1} \frac{\cos \left( \frac{\pi \alpha _{k+1}}{2} \right) }{1-\alpha _{k+1}} \xrightarrow {k \rightarrow \infty } 0. \end{aligned}$$

To control the integral in (A.9), let us first remark that there exists an $\epsilon _0 > 0$ such that if $|x| < \epsilon _0$, then $-\log \cos x < x^2$. Thus, for small enough $\epsilon > 0$ we can write

$$\begin{aligned}&- \int _0^\epsilon \frac{du}{u^{1+\alpha _{k}}} \log \cos \left( u - u^{\alpha _{k+1}} \sin \left( \frac{\pi \alpha _{k+1}}{2} \right) \right) \le \int _0^\epsilon \frac{du}{u^{1+\alpha _{k}}} \left( u - u^{\alpha _{k+1}} \sin \left( \frac{\pi \alpha _{k+1}}{2} \right) \right) ^2 \\&\quad = \int _0^\epsilon \left[ u^{1-\alpha _k} - 2 u^{\alpha _{k+1} - \alpha _k} \sin \left( \frac{\pi \alpha _{k+1}}{2} \right) + u^{2 \alpha _{k+1} - \alpha _k - 1} \sin ^2\left( \frac{\pi \alpha _{k+1}}{2} \right) \right] d u\\&\quad = \frac{\epsilon ^{2-\alpha _k}}{2-\alpha _k} - 2 \frac{\epsilon ^{\alpha _{k+1} - \alpha _k + 1}}{\alpha _{k+1} - \alpha _k + 1} \sin \left( \frac{\pi \alpha _{k+1}}{2} \right) + \frac{\epsilon ^{2 \alpha _{k+1} - \alpha _k}}{2\alpha _{k+1} - \alpha _k} \sin ^2\left( \frac{\pi \alpha _{k+1}}{2} \right) \xrightarrow {k \rightarrow \infty } 0. \end{aligned}$$

$\square$

Auxiliary Results; Proof of Lemma 2.2

We collect here a few elementary results used in the main part of the text, and provide a proof of Lemma 2.2.

Lemma B.1

Recall (1.4,1.4b). Almost surely $X_{k-1}(\sigma _i^{k, x}) \in {\mathbb {N}}^{k-1}$, $k,x\ge 1$.

Proof

This is obvious for $k=1$; let then $k\ge 2$. We first note that for $1\le j < k$, $X_{k-1, j}(s) = \infty$ only if s belongs to the range of $\theta _j^{k-1}$, and this set has null Lebesgue measure, since $\theta _j^{k-1}$ increases only by jumps (at the points of the Poisson processes $N^{j,x}$, $x \in {\mathbb {N}}$). Combining the latter point with that fact that the Poisson processes $N^{k,x}$, $x \in {\mathbb {N}}$, are independent from $\theta _j^{k-1}$, we readily find that the probability that $\sigma _i^{k, x}$ belongs to the range of $\theta _j^{k-1}$ for some i, x vanishes, and the result follows. $\square$

Proposition B.2

Let $\{\xi _i: i \in {\mathbb {N}}\}$ be the marks of a Poisson point process on ${\mathbb {R}}^+$ with intensity measure $\mu (d t) = \frac{c}{t^{1+\alpha }}\,dt$, $t >0$, with some $\alpha \in (0, 1)$ and $c > 0$. Let $\{ R_i : i \in {\mathbb {N}}\}$ be i.i.d. positive random variables, independent from the Poisson process, such that ${\mathbb {E}}(R_1^\alpha ) <\infty$. Then $\{ R_i\xi _i : i \in {\mathbb {N}}\}$ is also a Poisson point process, with intensity measure ${\mathbb {E}}(R_1^\alpha ) \mu$.

Proof

Define $S := \{ (\xi _i, R_i): i \in {\mathbb {N}}\}$ and note that S is the set of the marks of a Poisson point process on the first quadrant of ${\mathbb {R}}^2$ with intensity measure $\pi = \mu \times \nu$, where $\nu$ is the probability measure of $R_1$.

Note that $T(x, y) = xy$ is a continuous transformation with no accumulation point except 0, so $\{ T(s): s \in S\} = \{R_i\xi _i : i \in {\mathbb {N}}\}$ is a Poisson point process (see eg. [33], Mapping Theorem, Section 2.3); its intensity measure is readily determined to be ${\mathbb {E}}(R_1^\alpha ) \mu$. $\square$

Proposition B.3

Let R be a positive random variable with Laplace transform $\phi (\lambda ) := {\mathbb {E}}(e^{-\lambda R}) = e^{-c \lambda ^\alpha }$, $\lambda >0$, for some $c > 0$, $\alpha \in (0, 1)$. Then for $0< \beta < \alpha$:

$$\begin{aligned} {\mathbb {E}}(R^\beta ) = c^{\beta /\alpha } \frac{\Gamma \left( 1 - \beta /\alpha \right) }{\Gamma (1-\beta )}. \end{aligned}$$

Proof

By Fubini and the fact that $\phi ^\prime (\lambda ) = - {\mathbb {E}}(R e^{-\lambda R})$, we have that

$$\begin{aligned} - \int _0^\infty \!\phi ^\prime (\lambda ) \lambda ^{-\beta } d \lambda= & {} {\mathbb {E}}\!\left( \int _0^\infty Re^{-\lambda R}\lambda ^{-\beta } d \lambda \right) \! \\= & {} {\mathbb {E}}\!\left( R^\beta \int _0^\infty e^{-\lambda }\lambda ^{-\beta } d \lambda \right) \! ={\mathbb {E}}(R^\beta )\Gamma (1-\beta ). \end{aligned}$$

It follows that ${\mathbb {E}}(R^\beta ) = - \frac{1}{\Gamma (1-\beta )} \int _0^\infty \phi ^\prime (\lambda ) \lambda ^{-\beta } d \lambda ,$ and differentiating $\phi (\lambda ) = e^{-c \lambda ^\alpha }$ and substituting in the latter expression yields the result. $\square$

We now justify a claim made at the beginning.

Corollary B.4

Claim (1.3) holds true.

Proof

By the definition of $\gamma _\cdot (\cdot )$, we have that ${\mathcal {R}}_k(x|_{k-1}):=\sum _{x_k\in {\mathbb {N}}}\gamma _k(x|_{k-1},x_k)$ is an $\alpha _k$-stable random variable for every $x|_{k-1}\in {\mathbb {N}}^{k-1}$. It follows from Proposition B.2 that for every $x|_{k-2}\in {\mathbb {N}}^{k-2}$, $\left\{ \gamma _k(x|_{k-2},x_{k-1})\sum _{x_k\in {\mathbb {N}}}\gamma _k(x|_{k-2},x_{k-1},x_k),\,x_{k-1}\in {\mathbb {N}}\right\}$ is a Poisson point process with intensity measure const $\times \mu _{k-1}$, with const $={\mathbb {E}}({\mathcal {R}}_k(1,\ldots ,1)^{\alpha _{k-1}})$ finite by Proposition B.3. The result follows by iteration. $\square$

Proof of Lemma 2.2

(2.11) is established in the proof of Lemma 4.6 in [21]. We present an explicit argument for ${{\theta }}_1^n$; the case of ${\hat{\theta }}_1^n$ is similar. We proceed by induction on n. The case $n = 1$ is a direct consequence of the independence and stationarity of increments of Poisson point processes, as well as of the fact that the variables used to construct $\theta _1^1 = {\Xi }_1$ are independent.

Assuming $n \ge 2$, note that for $t, s > 0$:

$$\begin{aligned} \theta _1^n(t+s) - \theta _1^n(t)&= \sum _{x} \sum _{i = N^{n, x}(\theta _1^{n-1}(t)) + 1}^{N^{n, x}(\theta _1^{n-1}(t+s))} \gamma _n(X_{n-1}(\sigma _i^{n, x}) x) T^{n, x}_i. \end{aligned}$$

Fix $0 = t_0< t_1< \cdots < t_k$ and let us look at the joint law of $(\theta _1^n(t_i) - \theta _1^{n}(t_{i-1}))_{i = 1, \cdots , k}$.

$\theta _1^n(t_i) - \theta _1^{n}(t_{i-1})$, for varying values of i, depends on disjoint intervals of the Poisson processes $N^{n, x}$, each one with length $\theta _1^{n-1}(t_{i}) - \theta _1^{n-1}(t_{i-1})$. Because of the induction hypothesis these lengths are independent and each one has the same law as $\theta _1^{n-1}(t_i - t_{i-1})$.

Note that, as follows from Lemma B.1, we have that for fixed $t > 0$, $X_{n-1}(\theta _1^{n-1}(t)) = (\infty , \cdots , \infty )$ a.s. The instant $\theta _1^{n-1}(t)$ is thus a renewal time for $X_{n-1}$, and, by construction, the law of $X_{n-1}$ at such a renewal is the same as at time 0.

$\theta _1^n(t_i) - \theta _1^n(t_{i-1})$ also depends on the values of $X_{n-1}(r)$ for $r \in (\theta _1^{n-1}(t_{i-1}), \theta _1^{n-1}(t_i))$. Both ends of this interval are renewals, so what happens to $X_{n-1}$ inside such an interval is independent of what happens on other such intervals (namely, $(\theta _1^{n-1}(t_{j-1}),\theta _1^{n-1}(t_j))$, $j\ne i$).

Therefore we have that $\theta _1^n(t_i) - \theta _1^n(t_{i-1})$, $i\ge 1$, are independent, and the law of each $\theta _1^n(t_i) - \theta _1^n(t_{i-1})$ depends only on $t_i - t_{i-1}$. That is, $\theta _1^n$ has independent and stationary increments. $\square$