measure.

Until further notice, we restrict our reasoning to the case when $\xi$ is a
centred Lévy process, with Lévy measure $\Lambda$ concentrated in $[-K,K]$ for
some $K>0$, and without loss of generality we suppose that $K=1$. In
consequence, $\xi$ has finite moments of any order and we set
$\sigma^{2}_{n}:=\mathbb{E}[\xi_{1/n}^{2}]=\mathbb{E}[(\hat{X}^{(n)}_{1})^{2}]$.
Notice that under our standing hypothesis, $\hat{\xi}$ writes

$t^{p}\int_{0}^{t}s^{-p}{\mathrm{d}}B_{s}+\hat{\xi}^{(3)}_{t},\quad\quad
t\in[0,1],$

for some Brownian motion ${B}$ independent of $\hat{\xi}^{(3)}$. Further,
remark that under our restrictions, the family of discrete skeletons
$(\hat{S}^{(n)}_{k})$, for $n\in\mathbb{N}$, have typical steps centred and in
$L_{2}(\mathbb{P})$. Consequently, we can make use of Lemma 5.11.

Next, to establish Proposition 5.13 under our current restrictions, we claim
that it would be enough to show that the following convergence holds,

$\left(n^{p}a_{\lfloor nt\rfloor}\hat{S}^{(n)}_{\lfloor
nt\rfloor}\right)_{t\in[0,1]}\overset{\mathscr{L}}{\longrightarrow}\big{(}t^{-p}\hat{\xi}_{t}\big{)}_{t\in[0,1]},$
(5.35)

where now, the sequence on the left-hand side of (5.35) is a sequence of
martingales, while the process on the right hand side is the martingale
introduced in Proposition 3.2, viz.

$N_{t}:=\int_{0}^{t}s^{-p}{\mathrm{d}}B_{s}+t^{-p}\hat{\xi}^{(3)}_{t},\quad\quad
t\in[0,1].$

Indeed, for each $n$, let $M^{n}$ be the continuous time version of martingale
of Lemma 5.11 associated with the $n$-reinforced skeleton
$(\hat{S}^{(n)}_{k})_{k\in\mathbb{N}}$, i.e.

$M^{n}_{\lfloor nt\rfloor}=a_{\lfloor nt\rfloor}\hat{S}^{(n)}_{\lfloor
nt\rfloor},\quad t\geq 0,$

and remark that by Lemma 5.12, the predictable quadratic variation of
$M^{n}_{\lfloor n\cdot\rfloor}$ is given by

$\langle M^{n},M^{n}\rangle_{\lfloor nt\rfloor}=\mathds{1}_{\\{1/n\leq
t\\}}\sigma_{n}^{2}+\sum_{k=2}^{\lfloor
nt\rfloor}a^{2}_{k}\left((1-p)\sigma^{2}_{n}-p^{2}\frac{(\hat{S}_{k-1}^{(n)})^{2}}{(k-1)^{2}}+p\frac{\hat{V}_{k-1}^{(n)}}{k-1}\right).$

It follows that for each $n\in\mathbb{N}$, the following process

$N^{n}_{t}:=n^{p}M^{n}_{\lfloor nt\rfloor}=n^{p}a_{\lfloor
nt\rfloor}\hat{S}^{(n)}_{\lfloor nt\rfloor},\quad{t\geq 0},$

is also a martingale, and its predictable quadratic variation writes:

$\langle N^{n},N^{n}\rangle_{t}=n^{2p}\langle M^{n},M^{n}\rangle_{\lfloor
nt\rfloor}=\mathds{1}_{\\{1/n\leq
t\\}}n^{2p}\sigma^{2}_{n}+n^{2p}\sum_{k=2}^{\lfloor
nt\rfloor}a^{2}_{k}\left((1-p)\sigma^{2}_{n}-p^{2}\frac{(\hat{S}_{k-1}^{(n)})^{2}}{(k-1)^{2}}+p\frac{\hat{V}_{k-1}^{(n)}}{k-1}\right).$
(5.36)

Moreover, by Stirling’s formula, we have

$a_{n}=\frac{\Gamma(n)}{\Gamma(n+p)}\sim n^{-p},\quad\text{ as
}n\uparrow\infty,$

which gives:

$n^{p}a_{\lfloor nt\rfloor}\sim t^{-p},\quad\text{ as }n\uparrow\infty.$

This yields the claimed equivalence between (5.34) and (5.35) under our
current restrictions for $\xi$. For technical reasons, we shall prove first
that the convergence of the martingales $(N^{n})$ towards $N$ holds in the
interval $[\varepsilon,1]$, for any $\varepsilon>0$. This leads us to the
following lemma:

######  Lemma 5.14.

For any $\varepsilon>0$, the sequence $(N^{n}_{t})_{t\in[\varepsilon,1]}$ for
$n\in\mathbb{N}$ is tight.

###### Proof.

We denote by $(\mathscr{F}^{n}_{t})$ the natural filtration of $N^{n}$. By
Aldous’s tightness criterion (see for e.g. Kallenberg [14] Theorem 16.11), it
is enough to show that for any sequence $(\tau_{n})$ of (bounded)
$(\mathscr{F}^{n}_{t})$-stopping times in $[\varepsilon,1]$ and any sequence
of positive real numbers $(h_{n})$ converging to 0, we have

$\lim_{n\uparrow\infty}|N^{n}_{\tau_{n}+h_{n}}-N^{n}_{\tau_{n}}|=0,\quad\text{
in probability.}$

By Rebolledo’s Theorem (see e.g. Theorem 2.3.2 in Joffre and Metivier [13] )
it’s enough to show that the sequence of associated predictable quadratic
variations $(\langle N^{n},N^{n}\rangle)$ satisfies Aldous’s tightness
criterion, i.e. that

$\lim_{n\uparrow\infty}\langle N^{n},N^{n}\rangle_{\tau_{n}+h_{n}}-\langle
N^{n},N^{n}\rangle_{\tau_{n}}=0,\quad\text{ in probability}.$

In this direction, by (5.36), we have

$\displaystyle\langle N^{n},N^{n}\rangle_{\tau_{n}+h_{n}}-\langle
N^{n},N^{n}\rangle_{\tau_{n}}$ $\displaystyle=n^{2p}\sum_{k=\lfloor
n\tau_{n}\rfloor+1}^{\lfloor
n(\tau_{n}+h_{n})\rfloor}a^{2}_{k}\left((1-p)\sigma^{2}_{n}-p^{2}\frac{(\hat{S}_{k-1}^{(n)})^{2}}{(k-1)^{2}}+p\frac{\hat{V}_{k-1}^{(n)}}{k-1}\right)$
$\displaystyle\leq(1-p)n^{2p}\sum_{k=\lfloor n\tau_{n}\rfloor+1}^{\lfloor
n(\tau_{n}+h_{n})\rfloor}a^{2}_{k}\sigma^{2}_{n}+p\cdot n^{2p}\sum_{k=\lfloor
n\tau_{n}\rfloor+1}^{\lfloor
n(\tau_{n}+h_{n})\rfloor}a^{2}_{k}\frac{\hat{V}_{k-1}^{(n)}}{k-1},$ (5.37)

and it remains to show that both terms in the right hand side converge to 0 in
probability as $n\uparrow\infty$. The key now is in the asymptotic behaviour
of the series $\sum_{k=1}^{n}a^{2}_{k}$. As was already pointed out in [2],
for $p\in(0,1/2)$, we have

$\lim_{n\uparrow\infty}n^{2p-1}\sum_{k=1}^{n}a^{2}_{k}=\frac{1}{1-2p}.$ (5.38)

Furthermore, since the Lévy measure of $\xi$ is compactly supported, it holds
that

$\sigma^{2}_{n}=\mathds{E}\left[(\hat{X}^{(n)}_{1})^{2}\right]=\mathds{E}\left[\xi_{1/n}^{2}\right]=O(1/n),\quad\text{
as }n\uparrow\infty.$ (5.39)

Now, from (5.38) and (5.39) it follows that

$\lim_{n\uparrow\infty}(1-p)n^{2p-1}\sum_{k=\lfloor
n\tau_{n}\rfloor+1}^{\lfloor n(\tau_{n}+h_{n})\rfloor}a^{2}_{k}=0\quad\text{
a.s.}$

and a fortiori in probability, which entails that the first term in (5.37)
converges in probability to 0
as $n\uparrow\infty$. In order to show that the second term in (5.37) also
converges in probability to 0, we need to proceed more carefully. First, since
$\tau_{n}\in[\varepsilon,1]$, we can bound the second term in (5.37) by

$n^{2p}\sum_{k=\lfloor n\tau_{n}\rfloor+1}^{\lfloor
n(\tau_{n}+h_{n})\rfloor}a^{2}_{k}\frac{\hat{V}_{k-1}^{(n)}}{k-1}\leq
n^{2p}\frac{\sup_{\lfloor n\varepsilon\rfloor\leq k\leq
n}\hat{V}_{k}^{(n)}}{\lfloor n\varepsilon\rfloor}\sum_{k=\lfloor
n\tau_{n}\rfloor+1}^{\lfloor n(\tau_{n}+h_{n})\rfloor}a^{2}_{k}.$

Next, since $\frac{n^{2p}}{\lfloor n\varepsilon\rfloor}\sim
n^{2p-1}\varepsilon^{-1}$, in order to proceed as before we need to show that

$\sup_{\lfloor n\varepsilon\rfloor\leq k\leq
n}\hat{V}_{k}^{(n)}=O(1),\quad\text{in probability as }n\uparrow\infty,$

i.e. that the sequence is stochastically bounded. To do so we proceed as
follows: for each $n$, notice that the process

$\hat{V}^{(n)}_{k}=(\hat{X}^{(n)}_{1})^{2}+\dots+(\hat{X}^{(n)}_{k})^{2},\quad
k\geq 1,$

is the reinforced version of the random walk

${V}^{(n)}_{k}=({X}^{(n)}_{1})^{2}+\dots+({X}^{(n)}_{k})^{2},\quad k\geq 1,$

where $(\hat{X}^{(n)}_{i})^{2}$ are i.i.d. variables with law $\xi_{1/n}^{2}$.
In order to have a centred noise reinforced random walk, for $k\geq 1$ set
$\hat{Y}^{(n)}_{k}:=(\hat{X}^{(n)}_{k})^{2}-\mathbb{E}[{\xi_{1/n}^{2}}]$ and
we introduce:

$\displaystyle\hat{W}_{k}^{(n)}$
$\displaystyle:=\hat{V}^{(n)}_{k}-k\mathds{E}\left[\xi_{1/n}^{2}\right]$
$\displaystyle=\left((\hat{X}^{(n)}_{1})^{2}-\mathds{E}\left[\xi_{1/n}^{2}\right]\right)+\dots+\left((\hat{X}^{(n)}_{k})^{2}-\mathds{E}\left[\xi_{1/n}^{2}\right]\right)=\hat{Y}^{(n)}_{1}+\dots+\hat{Y}^{(n)}_{k},$

Now, the process $(\hat{W}^{(n)}_{k})_{k\in\mathbb{N}}$ is the noise
reinforced version of the centred random walk defined for $k\geq 1$ as

$\displaystyle{W}_{k}^{(n)}$
$\displaystyle={V}^{(n)}_{k}-k\mathds{E}\left[\xi_{1/n}^{2}\right]$
$\displaystyle=\left(({X}^{(n)}_{1})^{2}-\mathds{E}\left[\xi_{1/n}^{2}\right]\right)+\dots+\left(({X}^{(n)}_{k})^{2}-\mathds{E}\left[\xi_{1/n}^{2}\right]\right)={Y}^{(n)}_{1}+\dots+{Y}^{(n)}_{k},$

where $(Y_{i}^{(n)})_{i\in\mathbb{N}}$ are i.i.d. with law
$\xi_{1/n}^{2}-\mathbb{E}[{\xi_{1/n}^{2}}]$. We can now apply Corollary 4.3
from [8] to $W^{(n)}$: recalling that
$\sigma_{n}^{2}=\mathbb{E}[{\xi_{1/n}^{2}}]=O(1/n)$, we have

$\displaystyle\mathds{E}\left[\sup_{k\leq
n}\left(\hat{V}^{(n)}_{k}\right)^{2}\right]=\mathds{E}\left[\sup_{k\leq
n}\left(\hat{W}^{(n)}_{k}+k\sigma^{2}_{n}\right)^{2}\right]$
$\displaystyle\leq C\mathds{E}\left[\sup_{k\leq
n}\left(\hat{W}^{(n)}_{k}\right)^{2}\right]+O(1)$ $\displaystyle\leq
C^{\prime}\mathds{E}\left[\left(\xi_{1/n}^{2}-\mathds{E}\left[\xi_{1/n}^{2}\right]\right)^{2}\right]n+O(1).$

Once again, since $\xi$ has compactly supported Lévy measure,
$\mathbb{E}[{\xi_{1/n}^{2}}]$ and $\mathbb{E}[{\xi_{1/n}^{4}}]$ are both
$O(1/n)$ as $n\uparrow\infty$ and we deduce that

$\mathds{E}\left[\sup_{k\leq
n}\left(\hat{V}^{(n)}_{k}\right)^{2}\right]=O(1),\quad\text{ as
}n\uparrow\infty.$

Hence, by Markov’s inequality the sequence $(\sup_{k\leq
n}\hat{V}^{(n)}_{k})_{n}$ is $O(1)$ in probability and we can conclude as
before by bounding as follows for $L>0$:

$\displaystyle\mathds{P}\left(n^{2p}\frac{\sup_{\lfloor
n\varepsilon\rfloor\leq k\leq n}V_{k}^{(n)}}{\lfloor
n\varepsilon\rfloor}\sum_{k=\lfloor n\tau_{n}\rfloor+1}^{\lfloor
n(\tau_{n}+h_{n})\rfloor}a^{2}_{k}>\eta\right)$ $\displaystyle\hskip
85.35826pt\leq\mathds{P}\left(\sup_{k\leq
n}\hat{V}^{(n)}_{k}>L\right)+\mathds{P}\left(L\frac{n^{2p}}{\lfloor
n\varepsilon\rfloor}\sum_{k=\lfloor n\tau_{n}\rfloor+1}^{\lfloor
n(\tau_{n}+h_{n})\rfloor}a^{2}_{k}>\eta\right).$

∎

We shall now conclude the proof of Proportion 5.13 under our standing
assumptions, and in this direction recall our discussion prior to Lemma 5.14.
To extend the convergence to the interval $[0,1]$ we shall use a truncation
argument similar to the one employed in Section 4.3 of [8]. For each
$\varepsilon>0$, we have
$(N^{n}_{t})_{t\in[\varepsilon,1]}\overset{\mathscr{L}}{\rightarrow}(N_{t})_{t\in[\varepsilon,1]}$
and since $(N_{t+\varepsilon})_{t\in[0,1]}\rightarrow(N_{t})_{t\in[0,1]}$ by
right continuity (extending $N_{\cdot+\varepsilon}$ for
$t\in[1-\varepsilon,1]$ identically as the constant $N_{1}$ ), we deduce by
metrisability of the weak convergence that there exists some sequence
$(\varepsilon(n))_{n\in\mathbb{N}}$, converging to 0 slowly enough as
$n\uparrow\infty$ such that
$(N^{n}_{t})_{t\in[\varepsilon(n),1]}\overset{\mathscr{L}}{\rightarrow}(N_{t})_{t\in[0,1]}$
and we only need to show:

$\sup_{s\leq\varepsilon(n)}n^{p}a_{\lfloor ns\rfloor}\hat{S}^{(n)}_{\lfloor
ns\rfloor}\rightarrow 0,\quad\quad\text{ in probability as }n\uparrow\infty.$
(5.40)

In this direction, notice the inequality

$\langle N^{n},N^{n}\rangle_{s}\leq
n^{2p}\sigma^{2}_{n}+n^{2p}\sum_{k=2}^{\lfloor
ns\rfloor}a^{2}_{k}\left((1-p)\sigma^{2}_{n}+p\frac{\hat{V}_{k-1}^{(n)}}{k-1}\right).$

Since $\mathbb{E}[{\hat{V}^{(n)}_{k}}]=k\sigma^{2}_{n}$, an application of
Doob’s inequality and the previous display yield that, for any $\delta>0$, we
have

$\displaystyle\mathds{P}\left(\sup_{s\leq\varepsilon(n)}|n^{p}a_{\lfloor
ns\rfloor}\hat{S}^{(n)}_{\lfloor ns\rfloor}|\geq\delta\right)$
$\displaystyle\leq\delta^{-2}\mathds{E}\left[\langle
N^{(n)},N^{(n)}\rangle_{\varepsilon(n)}\right]$
$\displaystyle\leq\delta^{-2}n^{2p}\sigma^{2}_{n}+\delta^{-2}\mathds{E}\left[n^{2p}\sum_{k=2}^{\lfloor
nt\rfloor}a^{2}_{k}\left((1-p)\sigma^{2}_{n}+p\frac{\hat{V}_{k-1}^{(n)}}{k-1}\right)\right]$
$\displaystyle\leq\delta^{-2}\sigma_{n}^{2}n^{2p}\sum_{n=2}^{\lfloor
n\varepsilon(n)\rfloor}a^{2}_{k}.$

From the asymptotics,

$\lim_{n\uparrow\infty}n^{2p-1}\sum_{k=1}^{n}a_{k}^{2}=\frac{1}{1-2p},\quad\quad\text{and
}\quad\quad{\sigma_{n}^{2}}{}=O(1/n),$

we deduce that, as $n\uparrow\infty$, the convergence (5.40) holds and we can
conclude by an application of Lemma 3.31 - VI from Jacod and Shiryaev [12].

######  Remark 5.15.

_Before proceeding, we point out that our proof no longer works for $p\geq
1/2$: indeed, one might notice that the change in the asymptotic behaviour of
the series $\sum^{n}_{k=1}a^{2}_{k}$ for $p\geq 1/2$ makes the preceding
reasoning unfruitful. Let us be more precise: these series possess three
different asymptotic regimes depending on $p$ and are the reason behind the
different regimes appearing in the behaviour of the Elephant random walk, see
e.g. [2]. More generally, they are behind the three regimes appearing in the
invariance principles [8, 4]. When $p\geq 1/2$, there is no Brownian component
and the martingale $t^{-p}\hat{\xi}^{(3)}$ is no longer in $L_{2}(\mathbb{P})$
because $Y(t)\in L_{q}(\mathbb{P})$ for $q<1/p$. Since $N^{n}$ is converging
weakly towards $t^{-p}\hat{\xi}^{(3)}_{t}$ by (5.35), working with the
sequence of quadratic variations $\langle N^{n},N^{n}\rangle$ might not be the
right approach to obtain tightness._

#### Proof of Proposition 5.13, general case.

Let us start by introducing some notation. First, if $\hat{\mathcal{N}}$ is
the jump measure of $\hat{\xi}$, we will shorten our notation for the
compensated integrals and simply write
$\hat{\xi}_{u,v}^{(3)}(t):=(\mathds{1}_{\\{u\leq|x|<v\\}}x*\hat{\mathcal{N}}^{(sc)})_{t}$,
for $0\leq u<v$. Hence, for $K>1$, we can write

$\hat{\xi}_{0,K}^{(3)}(t)=\hat{\xi}^{(3)}_{t}+\hat{\xi}_{1,K}^{(3)}(t),\quad
t\in[0,1].$

It will also be convenient to introduce the following notation for the sums of
jumps: for fixed $0<a<b$, we write

$\Sigma_{a,b}(t):=\sum_{s\leq
t}\mathds{1}_{\\{a\leq|\Delta\hat{\xi}_{s}|<b\\}}\Delta\hat{\xi}_{s}=\sum_{i}\mathds{1}_{x_{i}\in[a,b)}x_{i}Y_{i}(t),\quad\text{
for }t\in[0,1],$ (5.41)

so that in particular we have $\hat{\xi}^{(2)}=\Sigma_{1,\infty}$. Next, if
$\xi$ can be decomposed into $\xi=L^{(1)}+L^{(2)}$, for independent Lévy
processes $L^{(1)},L^{(2)}$, we denote its reinforced skeleton by
$\hat{S}^{(n)}(\xi)=(\hat{S}^{(n)}_{\lfloor nt\rfloor}(\xi))_{t\in[0,1]}$ and
we write:

$\hat{S}^{(n)}(\xi)=\hat{S}^{(n)}(L^{(1)})+\hat{S}^{(n)}(L^{(2)}),$

for the decomposition that is naturally induced. More precisely, the two noise
reinforced random walks in the right-hand side of the previous display are
made with the same sequence of Bernoulli random variables as $\hat{S}(\xi)$,
and just result from decomposing each increment as

$\Delta^{(n)}\xi_{i}=+\Delta^{(n)}L^{(1)}_{i}+\Delta^{(n)}L^{(2)}_{i}.$

Now, we proceed by lifting progressively our restriction imposed in 5.3 as
follows:
Step 1: First, if $\xi$ satisfies that $\xi=M^{\leq K}$ where $M^{\leq K}$ is
the sum of a Brownian motion with diffusion $q$ and a compensated martingale
with jumps smaller than $K$, by 5.3 the following convergence holds in
distribution:

$\hat{S}^{(n)}\left(M^{\leq
K}\right)\overset{\mathscr{L}}{\rightarrow}q\hat{B}+\hat{\xi}_{0,K}^{(3)},\quad\text{as
}n\uparrow\infty.{}$ (End of Step 1)

Step 2: If $b$ is a deterministic constant, let $b\cdot Id:=(bt:t\geq 0)$ and
suppose now that $\xi$ can be written as $\xi=b\cdot Id+M^{\leq K}$. Then, we
can write

$\hat{S}^{(n)}(\xi)=\hat{S}^{(n)}(b\cdot Id)+\hat{S}^{(n)}(M^{\leq K}),$

where the sequence of processes $(\hat{S}^{(n)}(b\cdot Id):n\geq 1)$ is
deterministic and converges uniformly to the continuous function $b\cdot Id$.
Indeed, notice that the reinforcement doesn’t affect the drift term since
$\hat{S}^{(n)}(b\cdot Id)_{t}=b\lfloor nt\rfloor/n$. We deduce from [12, Lemma
3.33] that, as $n\uparrow\infty$, we still have

$\hat{S}^{(n)}\left(b\cdot Id+M^{\leq K}\right)=\hat{S}^{(n)}(b\cdot
Id)+\hat{S}^{(n)}\left(M^{\leq
K}\right)\overset{\mathscr{L}}{\rightarrow}b\cdot
Id+q\hat{B}+\hat{\xi}_{0,K}^{(3)}.$ (5.42)

From here, we work with the Lévy process $\xi$ with triplet
$(a,q^{2},\Lambda)$, with Lévy-Itô decomposition given by:

$\xi=a\cdot Id+M^{\leq 1}+\xi^{(3)},$

and we denote its jump measure by $\mathcal{N}$ – in particular, we have
$\xi^{(3)}=\mathds{1}_{(-1,1)^{c}}x*{\mathcal{N}}$. For any $K>1$, we can
rearrange the triplet by compensating and modifying appropriately the drift
coefficient, in such a way that we have:

$\xi=b_{K}Id+M^{\leq K}+\xi^{\geq K},$

where $\xi^{\geq K}:=\mathds{1}_{(-K,K)^{c}}x*{\mathcal{N}}$. Before moving to
Step 3, let us make the two following remarks.

$\bullet$ First, notice that for each fixed $n$, $S^{(n)}\left(\xi^{\geq
K}\right)\overset{\mathbb{P}}{\rightarrow}0$ uniformly in probability as
$K\uparrow\infty$. Indeed, we have

$\displaystyle\mathbb{P}\Big{(}\sup_{t\in[0,1]}|\hat{S}_{\lfloor
nt\rfloor}^{(n)}\left(\xi^{\geq K}\right)|>\varepsilon\Big{)}$
$\displaystyle\leq\mathds{P}\left(\Delta\xi_{t}^{\geq K}\neq 0\text{ for some
}t\in[0,1]\right),$

where the right-hand side can be written in terms of the jump process
$\mathcal{N}$ of $\xi$ as

$\mathbb{P}\Big{(}\mathcal{N}\big{(}\\{(t,x)\in[0,1]\times\mathbb{R}:|x|\geq
K\\}\big{)}\geq 1\Big{)}=1-e^{-\left((-\infty,K]\cup[K,\infty)\right)}.$
(5.42)

The right-hand side in the previous display converges to 0 as
$K\uparrow\infty$ and notice that the bound does not depend on $n$.

$\bullet$ Let $\hat{\xi}$ be the noise reinforced Lévy process of
characteristics $(a,q^{2},\Lambda,p)$ and write its jump measure by
$\hat{\mathcal{N}}$. Again, we can rewrite $\hat{\xi}$, by compensating
appropriately and modifying the drift coefficient, as follows:

$\hat{\xi}=b_{K}Id+q\hat{B}+\hat{\xi}_{0,K}^{(3)}+\Sigma_{K,\infty}.$

Arguing as before, we have the uniform convergence in probability
$b_{K}Id+q\hat{B}+\hat{\xi}_{0,K}^{(3)}\overset{\mathbb{P}}{\rightarrow}\hat{\xi}$
as $K\uparrow\infty$, since, by the description of $\hat{\mathcal{N}}$ given
in Definition 4.5, we have

$\mathbb{P}\Big{(}\sup_{t\in[0,1]}\left|\Sigma_{K,\infty}(t)\right|\geq\varepsilon\Big{)}\leq\mathds{P}\left(\hat{\mathcal{N}}\big{(}\\{(t,x)\in[0,1]\times\mathbb{R}:|x|\geq
K\\}\big{)}\geq
1\right)=1-e^{-(1-p)\Lambda\left((-\infty,K]\cup[K,\infty)\right)}.$

Step 3: To conclude, for $K>1$, we write respectively the Lévy process and the
corresponding NRLP without their jumps of size greater than $K$ as

$\xi^{\leq K}:=b_{K}Id+M^{\leq K},\hskip 28.45274pt\text{ and }\hskip
28.45274pt\hat{\xi}^{\leq K}:=b_{K}Id+q\hat{B}+\hat{\xi}_{0,K}^{(3)}.$

In (5.42), we already proved that for each fixed $K$, we have

$\hat{S}^{(n)}(\xi^{\leq K})\overset{\mathscr{L}}{\rightarrow}\hat{\xi}^{\leq
K},\text{ as }n\uparrow\infty,\hskip 14.22636pt\text{ while by our second
remark, it holds that }\hskip 14.22636pt\hat{\xi}^{\leq
K}\overset{\mathscr{L}}{\rightarrow}\hat{\xi},\text{ as }K\uparrow\infty.$

Since the convergence in distribution is metrisable, there exists an
increasing sequence $(K(n):n\geq 1)$ converging to infinity slowly enough as
$n\uparrow\infty$, such that

$\hat{S}^{(n)}\big{(}\xi^{\leq
K(n)}\big{)}\overset{\mathscr{L}}{\rightarrow}\hat{\xi},\quad\quad\text{ as
}n\uparrow\infty.$

Moreover, we can write

$\hat{S}^{(n)}\left(\xi\right)=\hat{S}^{(n)}\big{(}\xi^{\leq
K(n)}\big{)}+\hat{S}^{(n)}\big{(}\xi^{\geq K(n)}\big{)},$

where for each $\varepsilon>0$, by (5.42) we have:

$\lim_{n\uparrow\infty}\mathbb{P}\Big{(}\sup_{t\in[0,1]}|\hat{S}_{\lfloor
nt\rfloor}^{(n)}\left(\xi^{\geq
K(n)}\right)|>\varepsilon\Big{)}\leq\lim_{n\uparrow\infty}1-e^{-\Lambda((-\infty,K(n)]\cup[K(n),\infty))}=0.$

We can now apply [12, Lemma 3.31, Chapter VI] to deduce that the convergence
$\hat{S}^{(n)}(\xi)\overset{\mathscr{L}}{\rightarrow}\hat{\xi}$ holds.

With this last result we conclude the proof of Proposition 5.13.

####  5.3.3 Convergence of finite-dimensional distributions

We maintain the notation and setting introduced at the beginning of Section 5.

######  Proposition 5.16.

Let $\xi$ be a Lévy process of characteristic triplet $(a,q^{2},\Lambda)$ and
denote its characteristic exponent by $\Psi$. Fix $p\in(0,1)$ an admissible
memory parameter, and for each $n$, let $(S^{(n)}_{k},\hat{S}^{(n)}_{k})$ be
the sequence of $n$-skeletons and its corresponding reinforced versions as
defined in (5.4). Then, there is the weak convergence in the sense of finite-
dimensional distributions,

$\Big{(}\big{(}S^{(n)}_{\lfloor
nt\rfloor}\big{)}_{t\in[0,1]},\big{(}\hat{S}^{(n)}_{\lfloor
nt\rfloor}\big{)}_{t\in[0,1]}\Big{)}\overset{f.d.d.}{\longrightarrow}\Big{(}(\xi_{t})_{t\in[0,1]},(\hat{\xi}_{t})_{t\in[0,1]}\Big{)},$
(5.43)

where we denoted by $(\xi,\hat{\xi})$ a pair of processes with law
characterised by (5.2).

Remark that since the convergence is in the sense of finite dimensional
distributions, the restriction $p<1/2$ is dropped. Our proof will rely on two
results taken respectively from [7] and [9]. We state them without proof for
ease of reading:
Corollary 3.7 of [7] Let $F$ be a continuous functional on counting functions
such that $F(0)=0$ where, with a slight abuse of notation we still write $0$
for the identically $0$ trajectory. Further, suppose that there exists $c>0$
and $1\leq\gamma<1/p$ such that $|F(\omega)|\leq c\omega(1)^{\gamma}$ for
every counting function $\omega:[0,1]\rightarrow\mathbb{N}$. Then, if $Y$ is a
Yule-Simon process with parameter $1/p$, the following convergence holds in
$L_{1}(\mathbb{P})$:

$\lim_{n\rightarrow\infty}\frac{1}{n}\sum_{j=1}^{n}F\left(N_{j}(\lfloor
n\cdot\rfloor)\right)=(1-p)\mathds{E}\left[F(Y)\right].$ (5.44)

The second result concerns the asymptotic behaviour of $\Psi$.
Lemma 3.1 of [9] The asymptotic behaviour of the characteristic exponent
$\Psi$ as $|z|\uparrow\infty$ is given by:

$|\Psi(z)|=\begin{cases}o(|z|^{2+\eta})\quad\hskip 11.38109pt\text{when }q\neq
0\\\ o(|z|^{\beta(\Lambda)+\eta})\quad\text{when }q=0\text{ and }\int_{|x|\leq
1}|x|\Lambda({\mathrm{d}}x)=\infty\\\ o(|z|^{1+\eta})\quad\hskip
11.38109pt\text{when }q=0\text{ and }\int_{|x|\leq
1}|x|\Lambda({\mathrm{d}}x)<\infty.\end{cases}$

Now we have all the ingredients needed for the proof of Proposition 5.16.

###### Proof.

We fix $k\geq 1$, $0<\lambda_{1}<\dots<\lambda_{k}\leq 1$, and let
$\beta_{1},\dots\beta_{k}$ be real numbers. In order to establish the finite
dimensional convergence, it suffices to show that

$\mathds{E}\left[\exp\bigg{\\{}i\sum_{j=1}^{k}\big{(}\lambda_{j}S_{\lfloor
nt_{j}\rfloor}+\beta_{j}\hat{S}_{\lfloor
nt_{j}\rfloor}\big{)}\bigg{\\}}\right],$ (5.45)

converges as $n\uparrow\infty$ towards (5.2). In this direction, for each $n$,
we write $(N_{\ell}^{(n)}(k))_{k\geq 1,\ell\geq 1}$ the counting process of
repetitions of $\hat{S}^{(n)}$ introduced in (5.12). Recalling the identity
(5.13), we can write,

$\hat{S}_{\lfloor nt\rfloor}^{(n)}=\sum_{\ell=1}^{n}N_{\ell}^{(n)}(\lfloor
nt\rfloor)X_{\ell}^{(n)}\hskip 28.45274pt\text{and}\hskip
28.45274pt{S}_{\lfloor
nt\rfloor}^{(n)}=\sum_{\ell=1}^{n}\mathds{1}_{\\{\ell\leq\lfloor
nt\rfloor\\}}X_{\ell}^{(n)},$

with $\mathbb{E}[{e^{i\lambda X^{(n)}_{\ell}}}]=e^{\frac{1}{n}\Psi(\lambda)}$
for every $\ell$. Then, by independence of the counting processes
$(N_{\ell}^{(n)}(k))_{k\geq 1,\ell\geq 1}$ from the sequence
$(X^{(n)}_{\ell})_{\ell\geq 0}$, the characteristic function (5.45) can be
written as follows

$\displaystyle\mathds{E}\left[\exp\bigg{\\{}i\sum_{j=1}^{k}\big{(}\lambda_{j}S_{\lfloor
nt_{j}\rfloor}+\beta_{j}\hat{S}_{\lfloor
nt_{j}\rfloor}\big{)}\bigg{\\}}\right]$
$\displaystyle=\mathds{E}\left[\exp\bigg{\\{}i\sum_{\ell=1}^{n}\left(\sum_{j=1}^{k}\left(\lambda_{j}N_{\ell}^{(n)}(\lfloor
nt_{j}\rfloor)+\beta_{j}\mathds{1}_{\\{\ell\leq\lfloor
nt_{j}\rfloor\\}}\right)\right)X_{\ell}^{(n)}\bigg{\\}}\right]$
$\displaystyle=\mathds{E}\left[\exp\bigg{\\{}-\frac{1}{n}\sum_{\ell=1}^{n}\Psi\left(\sum_{j=1}^{k}\lambda_{j}N_{\ell}^{(n)}(\lfloor
nt_{j}\rfloor)+\beta_{j}\mathds{1}_{\\{\ell\leq\lfloor
nt_{j}\rfloor\\}}\right)\bigg{\\}}\right].$

Remark that since the law of $(N_{\ell}^{(n)}(k))_{k\geq 1,\ell\geq 1}$
doesn’t depend on $n$, we can drop the up-script $(n)$ in the last display.
Next, recall that $N_{\ell}(\lfloor nt\rfloor)=0$ for all $t\in[0,1]$ if
$\varepsilon_{\ell}=1$ while on the other hand, if $\varepsilon_{\ell}=1$,
$N_{\ell}(\lfloor ns\rfloor)=0$ for $\lfloor ns\rfloor<l$, and
$N_{\ell}(\lfloor ns\rfloor)\geq 1$ if $\lfloor ns\rfloor\geq l$ . Hence, we
have:

$\mathds{1}_{\\{\ell\leq\lfloor
ns\rfloor\\}}=\mathds{1}_{\big{\\{}N_{\ell}\lfloor ns\rfloor\geq
1\big{\\}}},\quad\quad\text{ on }\\{\varepsilon_{\ell}=0\\}.$

By the previous observations, we can write:

$\displaystyle\frac{1}{n}\sum_{\ell=1}^{n}\Psi\left(\sum_{j=1}^{k}\lambda_{j}N_{\ell}(\lfloor
nt_{j}\rfloor)+\beta_{j}\mathds{1}_{\\{\ell\leq\lfloor
nt_{j}\rfloor\\}}\right)$ (5.46)
$\displaystyle=\frac{1}{n}\sum_{\ell=1}^{n}\Psi\left(\sum_{j=1}^{k}\lambda_{j}N_{\ell}(\lfloor
nt_{j}\rfloor)+\beta_{j}\mathds{1}_{\big{\\{}N_{\ell}\lfloor nt_{j}\rfloor\geq
1\big{\\}}}\right)\mathds{1}_{\\{\varepsilon_{\ell}=0\\}}+\frac{1}{n}\sum_{\ell=1}^{n}\Psi\left(\sum_{j=1}^{k}\beta_{j}\mathds{1}_{\\{\ell\leq\lfloor
nt_{j}\rfloor\\}}\right)\mathds{1}_{\\{\varepsilon_{\ell}=1\\}}.$

Now, let us establish the convergence in probability of both terms in the
previous display separately. Starting with the first one, we introduce the
functional $F:D[0,1]\rightarrow\mathds{C}$ defined as follows:

$F(\omega):=\Psi\Bigg{(}\sum_{j=1}^{k}\lambda_{j}\omega(t_{j})+\beta_{j}\mathds{1}_{\\{\omega(t_{j})\in[1,\infty]\\}}\Bigg{)},$
(5.47)

for $\omega:[0,1]\rightarrow\mathbb{N}$ a generic counting function. This is a
$\mathbb{Q}$ \- a.s. continuous functional, since
$\omega\mapsto\mathds{1}_{\\{\omega(s)\in[1,\infty]\\}}$ can be written as
$\omega\mapsto\omega(s)\wedge 1$, which is a composition of a
$\mathbb{Q}$-a.s. continuous functional with the continuous mapping $x\mapsto
x\wedge 1$. Moreover, we have $F(0)=0$, and notice that we can bound:

$\left|\sum_{j=1}^{k}\lambda_{j}\omega(t_{j})+\beta_{j}\mathds{1}_{\\{\omega(t_{j})\in[1,\infty]\\}}\right|\leq
w(1)\Bigg{(}\sum_{j=1}^{k}\left|\lambda_{j}\right|+\left|\beta_{j}\right|\Bigg{)},$

by monotonicity of $\omega$ and the inequality
$\mathds{1}_{\\{\omega(s)\in[1,\infty]\\}}\leq\omega(s)$. Now, by Lemma 3.1 of
[9], we deduce that $F$ satisfies the hypothesis of Corollary 3.7 from [7],
since

$|F(\omega)|\leq|\omega(1)|^{\gamma}K\Bigg{(}\sum_{j=1}^{k}|\lambda_{j}|+|\beta_{j}|\Bigg{)}^{\gamma},\quad\text{with
}\quad\begin{cases}\gamma\in(2,1/p),\quad\text{ if }q\neq 0\\\
\gamma\in(1,1/p),\quad\text{ if }q=0,\end{cases}$ (5.48)

for a constant $K$ that only depends on $\beta(\Lambda)$ and $q$. From an
application of Corollary 3.7 of [7], we obtain the following convergence:

$\displaystyle\lim_{n\rightarrow\infty}\frac{1}{n}\sum_{\ell=1}^{n}\Psi\Bigg{(}\sum_{j=1}^{k}\lambda_{j}N_{\ell}(\lfloor
nt_{j}\rfloor)+\beta_{j}\mathds{1}_{\big{\\{}N_{\ell}\lfloor nt_{j}\rfloor\geq
1\big{\\}}}\Bigg{)}\mathds{1}_{\\{\varepsilon_{\ell}=0\\}}$
$\displaystyle=\lim_{n\rightarrow\infty}\frac{1}{n}\sum_{\ell=1}^{n}\Psi\Bigg{(}\sum_{j=1}^{k}\lambda_{j}N_{\ell}(\lfloor
nt_{j}\rfloor)+\beta_{j}\mathds{1}_{\big{\\{}N_{\ell}\lfloor nt_{j}\rfloor\geq
1\big{\\}}}\Bigg{)}$
$\displaystyle=(1-p)\mathds{E}\left[\Psi\Bigg{(}\sum_{j=1}^{k}\lambda_{j}Y(t_{j})+\beta_{j}\mathds{1}_{\big{\\{}Y(t_{j})\geq
1\big{\\}}}\Bigg{)}\right].$ (5.49)

Turning our attention to the second term, similarly, we claim that:

$\lim_{n\rightarrow\infty}\frac{1}{n}\sum_{\ell=1}^{n}\Psi\Bigg{(}\sum_{j=1}^{k}\beta_{j}\mathds{1}_{\\{\ell\leq\lfloor
nt_{j}\rfloor\\}}\Bigg{)}\mathds{1}_{\\{\varepsilon_{\ell}=1\\}}=p\mathds{E}\left[\Psi\Bigg{(}\sum_{j=1}^{n}\beta_{j}\mathds{1}_{\big{\\{}U\leq
t_{j}\big{\\}}}\Bigg{)}\right].$ (5.50)

Indeed, if for each $n$ we denote by $u(n)$ a uniform random variable on
$\\{1,\dots,n\\}$ independent of the i.i.d. sequence $(\varepsilon_{n})_{n}$
of Bernoulli with parameter $p$, we have

$\displaystyle\mathds{E}\left[\frac{1}{n}\sum_{\ell=1}^{n}\Psi\Bigg{(}\sum_{j=1}^{k}\beta_{j}\mathds{1}_{\\{\ell\leq\lfloor
nt_{j}\rfloor\\}}\Bigg{)}\mathds{1}_{\\{\varepsilon_{l}=1\\}}\right]$
$\displaystyle=\mathds{E}\left[\Psi\Bigg{(}\sum_{j=1}^{k}\beta_{j}\mathds{1}_{\\{u(n)\leq\lfloor
nt_{j}\rfloor\\}}\Bigg{)}\mathds{1}_{\\{\varepsilon_{u(n)}=1\\}}\right]$
$\displaystyle=\mathds{E}\left[\Psi\Bigg{(}\sum_{j=1}^{k}\beta_{j}\mathds{1}_{\\{u(n)\leq\lfloor
nt_{j}\rfloor\\}}\Bigg{)}\right]p,$ (5.51)

since $\varepsilon_{u(n)}$ is independent of $u(n)$ for each $n$. Further,
since $u(n)/n$ converges in law towards a uniform random variable in $[0,1]$,
the sequence of step processes $(\mathds{1}_{\\{u(n)\leq\lfloor
n\cdot\rfloor\\}})_{n\in\mathbb{N}}$ converges weekly towards
$\mathds{1}_{\\{U\leq\cdot\\}}$. Consequently, as $n\uparrow\infty$, (5.3.3)
converges towards

$p\,\mathds{E}\left[\Psi\Bigg{(}\sum_{j=1}^{k}\beta_{j}\mathds{1}_{\\{U\leq
t_{j}\\}}\Bigg{)}\right],$

where we recall that $\mathds{1}_{\\{U\leq t\\}}$ has the same distribution as
$\mathds{1}_{\\{Y(t)\geq 1\\}}$ by the description (2.1). Finally, recall the
identity of Proposition 5.3 for characteristic function of the finite
dimensional distributions of the pair $(\xi,\hat{\xi})$. It follows from
(5.46) and the limits (5.3.3), (5.50) that as $n\uparrow\infty$, we have the
convergence towards the characteristic function of the finite-dimensional
distributions of $(\xi,\hat{\xi})$,

$\displaystyle\lim_{n\uparrow\infty}\mathds{E}\left[\exp\bigg{\\{}i\sum_{j=1}^{k}\big{(}\lambda_{j}S_{\lfloor
nt_{j}\rfloor}+\beta_{j}\hat{S}_{\lfloor
nt_{j}\rfloor}\big{)}\bigg{\\}}\right]$
$\displaystyle=\exp\bigg{\\{}p\mathds{E}\left[\Psi\Bigg{(}\sum_{j=1}^{k}\lambda_{j}\mathds{1}_{\\{Y(t_{j})\geq
1\\}}\Bigg{)}\right]+(1-p)\mathds{E}\left[\Psi\Bigg{(}\sum_{j=1}^{k}\lambda_{j}\mathds{1}_{\\{Y(t_{j})\geq
1\\}}+\beta_{i}Y(s_{i})\Bigg{)}\right]\bigg{\\}}.$

∎

This result paired with the tightness established in Proposition 5.13 proves
Theorem 5.4.

##  6 Applications

We conclude this work with two sections devoted to applications.

###  6.1 Rates of growth at the origin

In this section we turn our attention to the trajectorial behaviour of noise
reinforced Lévy processes at the origin. In this direction let us start by
recalling a well known result established by Blumenthal and Getoor [9] for
Lévy processes. Let $\xi$ be a Lévy process with characteristic triplet
$(a,q^{2},\Lambda)$ with no Gaussian component, viz. $q=0$; in particular
$\beta(\Lambda)=\beta$. Further, we make the following hypothesis:
$\bullet$ If $\int_{\\{|x|\leq 1\\}}|x|\Lambda({\mathrm{d}}x)=\infty$, the
characteristic exponent can be written as follows:

$\Psi(\lambda)=\int_{\mathbb{R}}\left(e^{i\lambda x}-1-i\lambda
x\mathds{1}_{\\{|x|\leq 1\\}}\right)\Lambda({\mathrm{d}}x).$

Observe that in this case, we have $\beta(\Lambda)\in[1,2]$.
$\bullet$ If $\int_{\\{|x|\leq 1\\}}|x|\Lambda({\mathrm{d}}x)<\infty$, which
can happen for $\beta(\Lambda)\in[0,1]$, we suppose $\Psi$ takes the following
form:

$\Psi(\lambda)=\int_{\mathbb{R}}\left(e^{i\lambda
x}-1\right)\Lambda({\mathrm{d}}x).$

This is, when $\int_{[0,1]}|x|\Lambda({\mathrm{d}}x)<\infty$ we are supposing
that the Lévy process has no linear drift, the reason being that in that case
the behaviour at 0 is dominated by the drift term. We insist in the fact that
when $\beta(\Lambda)=1$ the integral $\int_{\\{|x|\leq
1\\}}|x|\Lambda({\mathrm{d}}x)$ can be finite or infinite.

We will be working for the rest of the section under these hypothesis, and we
will refer to them as hypothesis (H). It was established by Blumenthal and
Getoor in [9] that under (H), the behaviour at zero of a Lévy process is
dictated by the Blumenthal-Getoor index of the Lévy measure $\Lambda$. More
precisely, almost surely, we have:

$\lim_{t\downarrow 0}{t^{-\gamma}}\xi_{t}=0,\quad\text{ if
}\beta(\Lambda)<1/\gamma\quad\quad\quad\text{ and
}\quad\quad\quad\limsup_{t\downarrow
0}{t^{-\gamma}}|\xi_{t}|=\infty,\quad\text{ if }\beta(\Lambda)>1/\gamma.$

We will show that the same result still holds if we replace the Lévy process
$\xi$ by its noise reinforced version. Concretely, the main result of the
section is the following:

######  Proposition 6.1.

Let $\xi$ be a Lévy process with triplet $(a,q^{2},\Lambda)$ satisfying
hypothesis _(H)_ , and consider $\hat{\xi}$ its noise reinforced version for
an admissible parameter $p$. Then, almost surely, we have

$\lim_{t\downarrow 0}{t^{-\gamma}}\hat{\xi}_{t}=0,\quad\hskip 17.07164pt\hskip
8.53581pt\text{ if }\beta(\Lambda)<1/\gamma,$ (6.1)

while

$\limsup_{t\downarrow 0}{t^{-\gamma}}|\hat{\xi}_{t}|=\infty,\quad\text{ if
}\beta(\Lambda)>1/\gamma.$ (6.2)

The rest of the section is devoted to the proof of Proposition 6.1 and it is
achieved in several steps. We start by proving the second statement (6.2), in
Lemma 6.2 we prove (6.1) for $\beta(\Lambda)\geq 1$, $\int_{|x|\leq
1}|x|\Lambda({\mathrm{d}}x)=\infty$ and the case $\beta(\Lambda)\leq 1$,
$\int_{|x|\leq 1}|x|\Lambda({\mathrm{d}}x)<\infty$ is treated separately in
Lemma 6.4.

###### Proof of (6.1)..

It suffice to prove that for some $r>0$ and $\varepsilon>0$ a.s. there exists
a sequence of jumps occurring in $[0,\varepsilon]$ at times, that we denote by
$(t_{i})$, satisfying

$|\Delta\hat{\xi}_{t_{i}}|>t_{i}^{\gamma-r}.$

Now, recall from the discussion following (4.11) that the jump measure
$\hat{\mathcal{N}}$ of $\hat{\xi}$ dominates a Poisson point process with
intensity $(1-p)({\mathrm{d}}u\otimes\Lambda)$, say $\mathcal{N}^{\prime}$. If
we denote the atoms of $\mathcal{N}^{\prime}$ by $(u_{i},x_{i})$, we deduce
that

$\\#\\{(u_{i},x_{i})\in\mathcal{N}^{\prime}:u_{i}\in[0,\varepsilon]\text{ and
}|x_{i}|>2u_{i}^{\gamma-r}\\},$

is distributed Poisson with parameter

$(1-p){\mathrm{d}}u\otimes\Lambda\left((u,x)\in[0,\varepsilon]\times\mathbb{R}:|x|^{1/(\gamma-r)}>2\cdot
u\right)=\int_{\mathbb{R}}\left(2^{-1}|x|^{1/(\gamma-r)}\wedge\varepsilon\right)\Lambda({\mathrm{d}}x)(1-p).$
(6.3)

Now, take $r>0$ small enough such that the inequality
$1/(\gamma-r)<\beta(\Lambda)$ still holds. For such a choice of $r$, the
integral (6.3) is infinite by definition of the index $\beta(\Lambda)$ and the
claim follows. ∎

Now we focus on showing that $\lim_{t\downarrow
0}t^{-\gamma}|\hat{\xi}_{t}|=0$ for $\gamma\in(0,1/\beta(\Lambda))$. In this
direction, let us start introducing some notation and with some preliminary
remarks. First, notice that since we are interested in the behaviour of
$\hat{\xi}$ at the origin, we can rely on the original construction in [7] in
terms of Poissonian sums of Yule-Simon processes that we recalled in Section
2.3. Next, under (H), $\hat{\xi}$ can be written either as a sum of a
compensated integral $\hat{\xi}^{(3)}$ and a reinforced compound Poisson
process $\hat{\xi}^{(2)}$ viz.

$\hat{\xi}=\hat{\xi}^{(3)}+\hat{\xi}^{(2)},\hskip 17.07164pt\hskip
42.67912pt\text{if }\beta(\Lambda)>1,$ (6.4)

or as an absolutely convergent series of jumps,

$\hat{\xi}=\sum_{s\leq t}\Delta\hat{\xi}_{s},\,\,t\in[0,1],\hskip
28.45274pt\text{if }\beta(\Lambda)<1.$ (6.5)

We stress that if $\beta(\Lambda)=1$, $\hat{\xi}$ takes the form (6.4) or
(6.5) depending respectively on if $\int_{\\{|x|\leq
1\\}}|x|\Lambda({\mathrm{d}}x)$ is infinite or not, and remark that $\gamma$
can be strictly larger than one only when $\beta(\Lambda)<1$. Since
$\hat{\xi}^{(2)}$ is a finite sum of weighted Yule processes and
$\hat{\xi}^{(2)}_{0}=0$, independently of the value of $\beta(\Lambda)$ it
holds that $\lim_{t\downarrow 0}t^{-\gamma}\hat{\xi}^{(2)}_{t}=0$ and we can
consequently restrict our study of (6.4) resp. (6.5) to the case where
$\hat{\xi}=\hat{\xi}^{(3)}$ resp. $\hat{\xi}$ has Lévy measure concentrated in
$[0,1]$ – and hence is a reinforced, driftless subordinator.

######  Lemma 6.2.

Suppose that $\beta(\Lambda)\geq 1$ and let $\gamma\in(0,1\wedge
1/\beta(\Lambda))$. Then,

$\lim_{T\downarrow 0}\mathds{E}\left[\sup_{s\leq
T}s^{-\gamma}|\hat{\xi}^{(3)}_{s}|\right]=0.$

In particular, if $\beta(\Lambda)\geq 1$ with $\int_{\\{|x|\leq
1\\}}|x|\Lambda({\mathrm{d}}x)=\infty$, we have $\lim_{t\downarrow
0}t^{-\gamma}|\hat{\xi}^{(3)}_{t}|=0$ a.s.

###### Proof.

Recall from Proposition 3.2 that $(t^{-p}\hat{\xi}^{(3)}_{t})_{t\in[0,1]}$ is
a martingale. We start by fixing $s<u$ two times in $[0,1]$ and notice that
for any $r\in(\beta(\Lambda),1/p\wedge 2)$ (or $r=2$ if $\beta(\Lambda)=2$ ),
by Doob’s inequality in $L_{r}(\mathbb{P})$ we have

$\displaystyle\mathds{E}\left[\sup_{t\in[s,u]}t^{-\gamma}|\hat{\xi}^{(3)}_{t}|\right]$
$\displaystyle\leq
s^{-(\gamma-p)}\mathds{E}\left[\sup_{t\in[s,u]}t^{-p}|\hat{\xi}^{(3)}_{t}|\right]$
$\displaystyle\leq s^{-(\gamma-p)}\mathds{E}\left[\sup_{t\in[s,u]}t^{-p\cdot
r}|\hat{\xi}_{t}^{(3)}|^{r}\right]^{1/r}\leq c\cdot
s^{-(\gamma-p)}u^{-p}\mathds{E}\left[|\hat{\xi}_{u}^{(3)}|^{r}\right]^{1/r},$
(6.6)

for some constant $c$. In order to bound the expectation on the right hand
side, we recall from the proof of Lemma 2.6 in Bertoin [7] that the following
bound holds 444The bound was first established for non-atomic Lévy measures
$\Lambda$, but it was later shown that a similar bound holds if $\Lambda$ has
atoms by an approximation argument. for some constant $C$ large enough:

$\mathds{E}\left[|\hat{\xi}^{(3)}_{u}|^{r}\right]^{1/r}\leq
C\mathds{E}\left[\sum_{j}Y_{j}(u)^{r}|x_{j}|^{r}\right]^{1/r}.$ (6.7)

Next, by Campbell’s formula we have

$\mathds{E}\left[\sum_{j}Y_{j}(u)^{r}|x_{j}|^{r}\right]=\mathds{E}\left[Y(u)^{r}\right]\int_{\\{|x|<1\\}}|x|^{r}\Lambda({\mathrm{d}}x)<\infty,$
(6.8)

and remark that
$\mathds{E}\left[Y(u)^{r}\right]=u\cdot\mathds{E}\left[\eta^{r}\right]$ where
$\eta$ stands for a Yule-Simon random variable with parameter $1/p$. It now
follows that we can bound
$\mathds{E}\left[|\hat{\xi}^{(3)}_{u}|^{r}\right]^{1/r}\leq K\cdot u^{1/r}$
for a positive constant $K$ depending only on $r$. This observation paired
with the bound we obtained in (6.1), yields:

$\mathds{E}\left[\sup_{t\in[s,u]}t^{-\gamma}|\hat{\xi}^{(3)}_{t}|\right]\leq
s^{-(\gamma-p)}u^{-p+1/r}\cdot K,$ (6.9)

for a finite constant $K$ that only depends on the choice of $r$. Now, set
$t_{0}:=1$, $t_{n}:=2^{-n},\text{ for }n\geq 1$ and fix $N\in\mathbb{N}$.
Applying the bound (6.9) to each interval $[2^{-(n+1)},2^{-n}]$, we get:

$\mathds{E}\left[\sup_{t\leq
t_{N}}t^{-\gamma}|\hat{\xi}^{(3)}_{t}|\right]\leq\sum_{n\geq
N}\mathds{E}\left[\sup_{t\in[t_{n+1},t_{n}]}t^{-\gamma}|\hat{\xi}^{(3)}_{t}|\right]\leq
2^{\gamma-p}\sum_{n\geq N}2^{n(\gamma-1/r)},$ (6.10)

and to conclude it suffices to show that, for an appropriate choice of $r$,
the inequality $\gamma-1/r<0$ is satisfied. Since
$r\in(\beta(\Lambda),1/p\wedge 2)$, we can always choose $\varepsilon$ small
enough such that $r:=\beta(\Lambda)+\varepsilon$ belongs to
$(\beta(\Lambda),1/p\wedge 2)$ and $\gamma<1/(\beta(\Lambda)+\varepsilon)$,
since we recall that $\gamma<1/\beta(\Lambda)$. For such a particular choice
of $r$, the series (6.10) converge and we obtain the desired result. In
particular, this proves the statement of Proposition 6.1 when
$\int_{\\{|x|\leq 1\\}}|x|\Lambda({\mathrm{d}}x)=\infty$, which is when
$\hat{\xi}=\hat{\xi}^{(3)}$. ∎

The statement (6.1) of Proposition 6.1 is incomplete only when the Lévy
measure fulfils the integrability condition $\int_{\\{|x|\leq
1\\}}|x|\Lambda({\mathrm{d}}x)<\infty$. Recalling the discussion prior to
Lemma 6.2, we henceforth assume that the Lévy process is a driftless
subordinator with jumps smaller than one, say $(T_{t})$, and we denote by
$(\hat{T}_{t})$ the corresponding reinforced version for a memory parameter
$p\in(0,1)$. It is then convenient to work with its Laplace transform at time
$t\in[0,1]$,

$\mathds{E}\left[e^{-\lambda\hat{T}_{t}}\right]=\exp\Big{(}-\mathds{E}\left[\Phi(Y(t)\lambda)\right]\Big{)},\quad\quad\text{
for }\lambda\geq 0,$

for
$\Phi(\lambda):=(1-p)\int_{\mathbb{R}^{+}}\left(1-e^{-x\lambda}\right)\Lambda({\mathrm{d}}x)$
and $Y$ is a Yule-Simon process with parameter $1/p$. The following result
from [9] will be needed and we state it for the reader’s convenience:

######  Theorem 6.3.

_[Blumenthal, Getoor][9]_ If $\Phi(\lambda)$ is the Laplace exponent of a
driftless subordinator with Lévy measure $\Lambda$, then for any
$\varepsilon>0$,

$\Phi(\lambda)=o(\lambda^{\beta(\Lambda)+\varepsilon}),\quad\text{ as
}\lambda\uparrow\infty.$

Let $\varepsilon>0$, fix $\lambda>0$ and observe from Theorem 6.3 that for
$t\in(0,1)$, there exists positive constants $K$ and $R$ such that

$\Phi(\eta\lambda t^{-\gamma})\leq\begin{cases}K\quad\hskip 52.63759pt\text{
if }\eta\lambda t^{-\gamma}\leq R,\\\ \left(\lambda\eta
t^{-\gamma}\right)^{\beta(\Lambda)+\varepsilon}\quad\text{ if }\eta\lambda
t^{-\gamma}>R.\\\ \end{cases}$

Consequently, for $t\in(0,1)$ the following bound holds:

$\displaystyle t\Phi(\eta\lambda t^{-\gamma})\leq t\left(K+\left(\lambda\eta
t^{-\gamma}\right)^{\beta(\Lambda)+\varepsilon}\right)=tK+(\lambda\eta)^{\beta(\Lambda)+\varepsilon}t^{1-\gamma\beta(\Lambda)-\gamma\varepsilon}.$
(6.11)

######  Lemma 6.4.

Let $\hat{T}$ be a reinforced subordinator of memory parameter $p$ and Lévy
measure $\Lambda$. Then, for any $\gamma\in\mathbb{R}^{+}$ such that
$\gamma<1/\beta(\Lambda)$,

$\lim_{t\downarrow 0}t^{-\gamma}\hat{T}_{t}=0\quad\text{a.s.}$

The proof relies on the same techniques used for subordinators, see [5,
Proposition 10 - III.4].

###### Proof.

Consider $t\in[0,1]$ and fix $a>0$. An application of Markov’s inequality for
$g(r)=1-e^{-r}$ and the inequality $g(r)\leq r$ for $r\geq 0$ yield

$\displaystyle\mathbb{P}\big{(}\hat{T}_{t}>a\big{)}$
$\displaystyle\leq(1-e^{-1})^{-1}\left(1-\exp\left\\{-\mathds{E}\left[\Phi\left(a^{-1}Y(t)\right)\right]\right\\}\right)$
$\displaystyle\leq(1-e^{-1})^{-1}\mathds{E}\left[\Phi\left(a^{-1}Y(t)\right)\right].$

Since $\Phi(0)=0$ and $Y(t)$ conditioned to $Y(t)\geq 1$ follows the Yule-
Simon distribution with parameter $1/p$, for a constant $C$ we deduce the
bound:

$\mathbb{P}\big{(}\hat{T}_{t}>a\big{)}\leq
Ct\mathds{E}\left[\Phi\left(\eta/a\right)\right],$ (6.12)

where we denoted by $\eta$ a Yule Simon random variable with parameter $1/p$.
Now, let $h$ be an increasing function with $\lim_{t\downarrow 0}h(t)=0$, and
consider $a=h(2^{-n})$, $t=2^{-(n-1)}$. Then, by (6.12) and from summing over
$n\in\mathbb{N}$, we deduce

$\sum_{n=1}^{\infty}\mathds{P}\left(\hat{T}_{2^{-(n-1)}}>h(2^{-n})\right)\leq
2C\mathds{E}\left[\sum_{n=1}^{\infty}2^{-n}\Phi\left(\eta/h(2^{-n})\right)\right].$
(6.13)

In order to apply a Borel-Cantelli argument, we specialise in our case of
interest: we set $h(t):=t^{\gamma}$ and we show that the right hand side of
(6.13) is finite. From the first inequality in (6.11) with $\lambda=1$, we get

$\sum_{n=1}^{\infty}2^{-n}\Phi\left(\eta/h(2^{-n})\right)\leq
K\sum_{n=1}^{\infty}2^{-n}+\eta^{\beta(\Lambda)+\varepsilon}\sum_{n=1}^{\infty}(2^{-n})^{1-\gamma\beta(\Lambda)-\gamma\varepsilon}.$

For $\varepsilon$ small enough, we have both
$\eta^{\beta(\Lambda)+\varepsilon}\in L_{1}(\mathbb{P})$ (since $\eta$ is in
$L_{q}(\mathbb{P})$ for any $q<1/p$ and $\beta(\Lambda)<1/p$) and
$1-\gamma\beta(\Lambda)-\gamma\varepsilon>0$, by our standing assumption
$1>\gamma\beta(\Lambda)$. Consequently, we have

$\sum_{n=1}^{\infty}\mathds{P}\left(\hat{T}_{2^{-(n-1)}}>(2^{-n})^{\gamma}\right)<\infty,$

which entails by Borel-Cantelli that $\hat{T}_{2^{-(n-1)}}<(2^{-n})^{\gamma}$
holds for all $n$ large enough, a.s. From a monotony argument, it follows that
a.s. $\hat{T}_{t}<t^{\gamma}$ for all $t$ small enough and in consequence
$\limsup_{t\downarrow 0}t^{-\gamma}\hat{T}_{t}\leq 1$. If we now take
$h(t)=\delta t^{\gamma}$ for $\delta\in(0,1)$, by the same reasoning we obtain
$\limsup_{t\downarrow 0}t^{-\gamma}\hat{T}_{t}\leq\delta$ which leads to the
desired result. ∎

Finally, our proof of Proposition 6.1 is complete.

###  6.2 Noise reinforced Lévy processes as infinitely divisible processes

As was already mentioned in Section 4.3, NRLPs are infinitely divisible
processes – abbreviated ID processes. In this final section, we study their
properties under this new scope. In this direction, we start by giving a brief
overview of the theory; our exposition mainly follows Rosinksi [21] and
Chapter 3 of Samorodnitsky [22]. Then, we identify the features of NRLPs in
this setting and more precisely, we identify the functional triplet of NRLPs,
in the sense of ID processes. The objective here is hence to put Lévy
processes and their NRLPs counterparts in the context of ID processes and
compare then through this new lens. As an application, making use of the
Isomorphism Theorem for ID processes [21, Theorem 4.4] we establish the
following result:

######  Proposition 6.5.

Let $\hat{\xi}$ be a noise reinforced Lévy process with characteristics
$(a,0,\Lambda,p)$. Let $f:\mathbb{R}\rightarrow\mathbb{R}^{+}$ be a bounded,
continuous function with $f(x)=O(x^{2})$ at $0$. Then, we have

$\lim_{h\downarrow
0}h^{-1}\mathbb{E}\big{[}{f\big{(}\hat{\xi}_{h}\big{)}}\big{]}=p^{-1}{(1-p)}\int_{\mathbb{R}}\Lambda({\mathrm{d}}x)\sum_{k=1}^{\infty}f(kx)\emph{B}(k,1/p+1).$

Note that the probability distribution appearing in the previous display is
the Yule-Simon distribution (2.2). For an analogous result in the setting of
Lévy processes, we refer to [21, Proposition 4.13] and we shall use in our
proof similar type of arguments. To simplify notation, for the rest of the
section we work with NRLPs in $[0,1]$, but our exposition can be adapted to
$\mathbb{R}^{+}$ with some slight changes. Hence, we can make use of the
construction of NRLPs from [7] in terms of Poissonian Yule-Simon series that
we recalled at the end of Section 3. This construction will be used for the
rest of the section.

####  6.2.1 Preliminaries on infinitely divisible processes

Let us introduce some standard notation mostly taken from [21]. For $T$ a
nonempty set, we denote by $\mathbb{R}^{T}$ the set of $\mathbb{R}$-valued
functions indexed by $t\in T$. If $S\subset T$ is an arbitrary subset and
$e=(e(t))_{t\in T}\in\mathbb{R}^{T}$, we write $e_{S}$ for the restriction of
$e$ to $S$. Further, let $\pi_{S}$ be the canonical projection
$\pi_{S}:\mathbb{R}^{T}\rightarrow\mathbb{R}^{S}$ from $\mathbb{R}^{T}$ into
$\mathbb{R}^{S}$, viz. the function defined as $\pi_{S}(e):=e_{S}$. For finite
subsets of $T$ of the form $I:=\\{t_{1},\dots,t_{k}\\}\subset T$, the space
$\mathbb{R}^{I}$ is identified with $\mathbb{R}^{k}$ and we write:

$e_{I}=(e({t_{1}}),\dots,e({t_{k}}))\in\mathbb{R}^{I}.$

As usual, the space $\mathbb{R}^{T}$ is equipped with the cylindrical sigma
field $\mathscr{B}^{T}:=\sigma(\pi_{t}\,:\,t\in T)$ generated by the
projection mappings. For any arbitrary $S\subset T$, we denote by $0_{S}$ the
$0$ element of $\mathbb{R}^{S}$ and we write
$\mathscr{B}_{0}^{S}:=\\{A\in\mathscr{B}^{S}:0_{S}\notin A\\}$. Consequently,

$\pi^{-1}_{S}({0}_{S})=\\{e\in\mathbb{R}^{T}:e(t)=0\text{ for all }t\in
S\\}=:\textbf{{0}}_{S}\subset\mathbb{R}^{T}.$

Notice however that this subset is not $\mathscr{B}^{T}$ measurable when $S$
is uncountable. Finally, for $x\in\mathbb{R}$ we set $\llbracket
x\rrbracket:=x\mathds{1}_{\\{|x|\leq 1\\}}$ and if
$x=(x_{1},\dots,x_{k})\in\mathbb{R}^{k}$, the term $\llbracket x\rrbracket$
should be interpreted component-wise, viz. $\llbracket
x\rrbracket:=\left(\llbracket x_{1}\rrbracket,\dots,\llbracket
x_{k}\rrbracket\right)$. Now let us start with the following definition:

######  Definition 6.6.

An $\mathbb{R}$-valued stochastic process $X=(X_{t})_{t\in T}$ is said to be
infinitely divisible (in law) if for any $n\in\mathbb{N}$, there exist
independent and identically distributed processes $Y^{(n,1)},\dots Y^{(n,n)}$
such that

$X\overset{\mathscr{L}}{=}Y^{(n,1)}+\dots Y^{(n,n)}.$

When $T=\\{1\\}$ is a singleton, this is just the definition of a real valued
infinitely-divisible random variable, in which case, the characteristic
function of $X_{1}$ takes the Lévy-Khintchine form:

$\mathds{E}\left[e^{i\theta X_{1}}\right]=\exp\Big{\\{}i\theta
b-\frac{q^{2}}{2}\theta^{2}+\int_{\mathbb{R}}\left(e^{i\theta
x}-1-i\theta\llbracket x\rrbracket\right)\nu({\mathrm{d}}x)\Big{\\}},$

for $q,b\in\mathbb{R}$, $\nu$ a Lévy measure. Further, it is well known that
the set of infinitely divisible random variables and distributions of Lévy
processes are in bijection and it is clear that if $X$ is a Lévy process with
characteristic exponent as in the previous display, we have

$X\overset{\mathscr{L}}{=}Y^{(n,1)}+\dots+Y^{(n,n)},$

where for each $i\in\\{1,\dots,n\\}$, $Y^{(n,i)}$ is an independent copy of a
Lévy process with characteristic triplet $(b/n,q/n,\nu/n)$. Said otherwise,
Lévy processes are infinitely divisible processes. Moreover, from the formula
for the characteristic function of Proposition 2.2, it is clear that NRLPs are
in turn infinitely divisible.

Now, recall that a Gaussian process $X=(X_{t})_{t\in T}$ is a $T$-indexed
process satisfying that, for any $I=\\{{t_{1}},\dots,{t_{k}}\\}\subset T$, the
vector $X_{I}=(X_{t_{1}},\dots,X_{t_{k}})$ is Gaussian. In the sequel we also
assume that the Gaussian processes we work with are centred. Gaussian
processes are characterised by their covariance function, in the sense that
the law of $X$ is completely determined by the semi-definite positive function
$\Gamma:T\times T\rightarrow\mathbb{R}$ defined by

${\Gamma}(t,s):=\mathds{E}\left[X_{t}X_{s}\right],\quad\quad\text{ for }t,s\in
T.$ (6.14)

The following characterisation of infinitely divisible stochastic processes
shows that they are the natural generalisation of Gaussian processes:

######  Proposition 6.7.

_[Proposition 3.1.3][22]_ An $\mathbb{R}$-valued stochastic process
$X=(X_{t})_{t\in T}$ is infinitely divisible if and only if for any finite
collection of indices $I=\\{t_{1},\dots t_{k}\\}\subset T$, the random vector
$X_{I}=(X_{t_{1}},\dots X_{t_{k}})$ is infinitely divisible.

Hence, if $X$ is an infinitely divisible process, by Lévy-Kintchine
representation and the previous proposition, for every
$I=\\{t_{1},\dots,t_{k}\\}$ there exists: an $\mathbb{R}^{k}$-valued measure
$\nu_{I}({\mathrm{d}}x)$ verifying

$\int_{\mathbb{R}^{k}}1\wedge|x|^{2}\nu_{I}({\mathrm{d}}x)<\infty,\hskip
42.67912pt\text{ and }\hskip 42.67912pt\nu_{I}\left(\\{0_{I}\\}\right)=0,$

a semi-definite positive $I\times I$ matrix $\Gamma_{I}$ and an
$\mathbb{R}^{k}$ vector, that we denote as ${b}({I})$, satisfying for every
$\theta\in\mathbb{R}^{I}$ the identity:

$\displaystyle\mathds{E}\left[\exp\left\\{\sum_{t\in
I}\theta_{t}X_{t}\right\\}\right]=\exp\left\\{i\langle{b}({I}),{\theta}\rangle-\frac{1}{2}\langle{\theta}\,\Gamma_{I},{\theta}\rangle+\int_{\mathbb{R}^{I}}\left(e^{i\langle{\theta},x\rangle}-1-i\langle{\theta},\llbracket
x\rrbracket\rangle\right)\nu_{I}({\mathrm{d}}x)\right\\}.$ (6.15)

It is possible to show that one can recover the collection of triplets
$((b(I),\Gamma_{I},\nu_{I}):I\subset T,\,|I|<\infty)$ from a so called
functional triplet $(b,\Gamma,\bar{\nu})$, consisting in a path
$b\in\mathbb{R}^{T}$, a covariance function $\Gamma:T\times
T\rightarrow\mathbb{R}$ and a path-valued measure $\bar{\nu}$ defined in
$\mathbb{R}^{T}$, satisfying for any finite $I\subset{T}$, that

$b(I)=\pi_{I}(b)\hskip
56.9055pt\Gamma_{I}={\left.\kern-1.2pt\Gamma\vphantom{\big{|}}\right|_{I}}\hskip
56.9055pt\nu_{I}({\mathrm{d}}x)=\bar{\nu}\circ\pi_{I}^{-1}({\mathrm{d}}x)\quad\text{
in }\mathscr{B}_{0}^{I},$

where $\bar{\nu}$ satisfies some regularity and integrability conditions that
we now introduce:

######  Definition 6.8.

A measure $\bar{\nu}$ on $\mathbb{R}^{T}$ is called a path Lévy measure if it
satisfies the following two conditions:

  1. _(i)_

$\int_{\mathbb{R}^{T}}|e(t)|^{2}\wedge
1\,\bar{\nu}({\mathrm{d}}e)<\infty\quad\text{ for all }t\in T.$

  2. _(ii)_

$\text{ For every }A\in\mathscr{B}^{T},\text{ there exists a countable subset
}T_{A}\subset T\text{ such that
}\bar{\nu}(A)=\bar{\nu}(A\setminus\pi_{T_{A}}^{-1}(0_{T_{A}}))$.

Moreover, we consider the following third condition:

  1. _(iii)_

$\text{There exists a countable subset }T_{0}\subset T\text{ such that
}\bar{\nu}(\pi_{T_{0}}^{-1}({0}_{T_{0}}))=0.$

Then, (iii) is a stronger statement than (ii) and it has been shown that a
path Lévy measure is $\sigma$-finite if and only if (iii) holds – see e.g.
[21]. Condition (ii) states roughly speaking that $\bar{\nu}$ ”does not charge
the origin”. As we already mentioned, in general $0_{T}$ is not measurable and
hence we can not state this condition as in the finite-dimensional case of
Lévy measures. One of the main results of the theory states that infinitely
divisible processes are in bijection with functional triplets
$(b,\Gamma,\bar{\nu})$, we refer to [21] for the proof:

######  Theorem 6.9.

For every infinitely divisible stochastic process $X=(X_{t})_{t\in T}$ there
exists a unique generating triplet $({b},{\Gamma},\bar{\nu})$ consisting of a
path $b\in\mathbb{R}^{T}$, a covariance function $\Gamma$ in $T\times T$ and a
path Lévy measure $\bar{\nu}$ in $\mathbb{R}^{T}$ such that for any finite
$I\subset\hat{T}$,

$\mathds{E}\left[\exp\left\\{i\sum_{t\in
I}\theta_{t}X_{t}\right\\}\right]=\exp\left\\{i\langle
b_{I},\theta\rangle-\frac{1}{2}\langle\theta\,\Gamma_{I},\theta\rangle+\int_{\mathbb{R}^{T}}\left(e^{i\langle\theta,e_{I}\rangle}-1-i\langle\theta,\llbracket
e_{I}\rrbracket\rangle\right)\bar{\nu}({\mathrm{d}}e)\right\\}.$ (6.16)

Conversely, for every generating triplet $({b},{\Gamma},\bar{\nu})$ there
exists an infinitely divisible process satisfying (6.16).

Maintaining the notation of Theorem 6.9, it follows in particular that the law
of any ID process $X$ can be written as a sum of two independent processes
$X\overset{\mathscr{L}}{=}G+P,$ where $G$ is Gaussian with covariance $\Gamma$
and $P$ is a so-called Poissonian ID process. When the equality $X=G+P$ holds
almost surely, we call respectively $G$ and $P$ the Gaussian part and the
Poissonian part of $X$. Let us conclude our presentation with the following
notion that will be of use:

######  Definition 6.10.

A process $V=(V_{t})_{t\in T}$ defined in a measure space $(S,\mathscr{S},n)$
is called a representant of a path Lévy measure $\bar{\nu}$ if for any finite
$I\subset{T}$, we have

$n\left(s\in S\,:\,V_{I}(s)\in B\right)=\bar{\nu}_{I}(B),\quad\quad\text{ for
every }B\in\mathscr{B}_{0}^{I}.$

The representation is called exact if $n\circ V^{-1}=\bar{\nu}$.

This is, if $V$ is only a representant, the measure $\nu\circ V^{-1}$ might
not be a Lévy measure since it might ”charge the origin”. In the situations we
will be interested in the representations will always be exact, and we only
enunciate the weaker definition to write the results we need in their full
generality. Representants allow to build explicitly Poissonian ID processes in
terms of Poisson random measure, for more details we refer to [21], see also
our brief discussion before the proof of Proposition 6.5 below.

####  6.2.2 The characteristic triplet of a NRLP

We can now start investigating Lévy processes and their reinforced
counterparts as ID processes, and we start with a basic analysis of the
former. More precisely, we identify the path Lévy measure of Lévy processes as
well as an exact representant. These results are known [21, Example 2.23 ] and
the statements are only included to contrast with the analogous results for
NRLPs – see Lemma 6.13 below.

######  Lemma 6.11.

The following assertions hold:

  1. _(i)_

Let $\xi$ be a Lévy process with characteristic triplet $(a,q,\Lambda)$. The
path Lévy measure $\bar{\nu}$ of $\xi$ is given by,

$\bar{\nu}({\mathrm{d}}e):=({\mathrm{d}}t\otimes\Lambda)\circ
V^{-1}({\mathrm{d}}e),$

where we denoted by $V$ the mapping
$V:\mathbb{R}^{+}\times\mathbb{R}\mapsto\mathbb{R}^{\mathbb{R}^{+}}$ defined
as $V(s,x):=x\mathds{1}_{\\{s\leq\cdot\\}}$.

  2. _(ii)_

Consider a measure $\Lambda$ on $\mathbb{R}$ with $\Lambda(0)=0$ and let $V$
be defined as in _(i)_. Then, the condition $\int
1\wedge|x|^{2}\Lambda({\mathrm{d}}x)<\infty$ holds if and only if
$\bar{\nu}:=({\mathrm{d}}t\otimes\Lambda)\circ V^{-1}$ is a path Lévy measure
in $\mathbb{R}^{\mathbb{R}^{+}}$. Moreover, if the later holds, the path Lévy
measure $\bar{\nu}$ is $\sigma$-finite.

In particular, from (i) we get that $V$ is an exact representant of
$\bar{\nu}$, on
$(S,\mathscr{S},n):=(\mathbb{R}^{+}\times\mathbb{R},\mathscr{B}(\mathbb{R}^{+})\otimes\mathscr{B}(\mathbb{R}),{\mathrm{d}}t\otimes\Lambda)$.
We now turn our attention to noise reinforced Lévy processes and we start with
the following technical lemma:

######  Lemma 6.12.

Let $\hat{\xi}$ be an NRLP of characteristic triplet $(a,0,\Lambda,p)$ and let
$T=[0,1]$. Then, for any $t\in T$, we have

$\mathds{E}\left[\int_{\mathbb{R}}\big{|}\llbracket
Y(t)x\rrbracket-Y(t)\llbracket x\rrbracket\big{|}\Lambda(dx)\right]<\infty.$
(6.17)

###### Proof.

First, recalling that $\llbracket x\rrbracket=x\mathds{1}_{\\{|x|\leq 1\\}}$,
we can write

$\mathds{E}\left[\big{|}\llbracket Y(t)x\rrbracket-Y(t)\llbracket
x\rrbracket\big{|}\right]=\mathds{E}\left[\big{|}Y(t)\llbracket
x\rrbracket\big{|}\mathds{1}_{\\{|xY(t)|>1\\}}\right]+\mathds{E}\left[\big{|}Y(t)x-Y(t)\llbracket
x\rrbracket\big{|}\mathds{1}_{\\{|xY(t)|\leq 1\\}}\right].$ (6.18)

Remark that since $Y$ takes values in $\\{0,1,2,\dots\\}$, the second term in
the right-hand side vanishes. On the other hand, for any
$q\in(\beta(\Lambda)\vee 1,1/p)$, we have

$\displaystyle\int_{\mathbb{R}}\mathds{E}\left[\big{|}Y(t)\llbracket
x\rrbracket\big{|}\mathds{1}_{\\{|x|Y(t)>1\\}}\right]\Lambda(dx)$
$\displaystyle=\int_{\\{|x|\leq
1\\}}\mathds{E}\left[Y(t)\mathds{1}_{\\{|x|Y(t)>1\\}}\right]|x|\Lambda(dx)$
$\displaystyle\leq\mathds{E}\left[Y^{q}(t)\right]^{1/q}\int_{\\{|x|\leq
1\\}}\mathds{P}\left(Y(t)>1/|x|\right)^{(q-1)/q}|x|\Lambda(dx),$ (6.19)

where we recall that $Y\in L^{q}(\mathbb{P})$ for any $q<1/p$. To conclude,
recall the asymptotic behaviour from (10) in [7],

$\mathds{P}\left(Y(t)>1/|x|\right)\sim t\Gamma(1/p+1)|x|^{1/p},\quad\text{as
}x\downarrow 0.$

It now follows that we can take $q$ close enough to $1/p$ such that the
integral in (6.2.2) is finite and we deduce (6.17). ∎

Now, we identify the path Lévy measure of NRLPs.

######  Lemma 6.13.

The following assertions hold:

  1. _(i)_

Let $\hat{\xi}$ be a NRLP with characteristic triplet $(a,q^{2},\Lambda,p)$.
The path Lévy measure $\bar{\nu}$ of $\hat{\xi}$ is given by,

$\bar{\nu}:=(1-p)(\Lambda\otimes\mathbb{Q})\circ V^{-1},$

where $V:D[0,1]\times\mathbb{R}\rightarrow\mathbb{R}^{[0,1]}$ is defined by
$V(x,y):=xy$.

  2. _(ii)_

Let $(a,0,\Lambda)$ be the characteristic triplet of a Lévy Process and let
$V$ be defined as in _(i)_. Then, if a memory parameter $p\in(0,1)$ is
admissible for the triplet $(a,0,\Lambda)$, the measure
$\bar{\nu}:=(\Lambda\otimes\mathbb{Q})\circ V^{-1}$ is a $\sigma$-finite path
Lévy measure in $\mathbb{R}^{[0,1]}$. On the other hand, if
$1/p<\beta(\Lambda)$, then the integrability condition 6.8 \- _(i)_ fails.

In particular, from (i) we get that $V$ is an exact representant of
$\bar{\nu}$, in
$(S,\mathscr{S},n)=(D[0,1]\times\mathbb{R},\mathscr{B}(D[0,1])\otimes\mathscr{B}(\mathbb{R}),\mathbb{Q}\otimes\Lambda(1-p))$.
On other hand, (ii) gives a natural interpretation in the terminology of ID
processes for the admissibility of $p$ for $\Lambda$.

###### Proof.

To identify the Lévy measure, let us write the characteristic function of the
finite dimensional distributions of $\hat{\xi}$ in the form (6.16) and to
simplify notation, we suppose that $a,q=0$. In this direction, consider a
finite $I\subset T$,
$\theta=(\theta_{t_{1}},\dots,\theta_{t_{k}})\in\mathbb{R}^{I}$, and denote by
$y=(y(t))_{t\in[0,1]}$ an arbitrary counting function. Recall the formula for
the finite dimensional distributions of $\hat{\xi}$ from Proposition 2.2, for
$t=1$. It now follows by Lemma 6.12 and the triangle inequality that we have:

$\int_{\mathbb{R}}\Lambda({\mathrm{d}}x)\mathds{E}\left[\left|e^{i\langle\theta,Y_{I}\rangle
x}-1-i\langle\theta,\llbracket xY_{I}\rrbracket\rangle\right|\right]<\infty.$
(6.20)

Now, we can write

$\displaystyle\mathds{E}\left[\exp\left\\{i\sum_{t\in
I}\theta_{t}\hat{\xi}_{t}\right\\}\right]$
$\displaystyle=\exp\left\\{(1-p)\int_{\mathbb{R}}\Lambda({\mathrm{d}}x)\mathds{E}\left[e^{\langle\theta,Y_{I}\rangle
x}-1-i\langle\theta,Y_{I}\rangle\llbracket x\rrbracket\right]\right\\}$
$\displaystyle=\exp\bigg{\\{}(1-p)\int_{\mathbb{R}\times
D[0,1]}\left(e^{\langle\theta,(xy)_{I}\rangle}-1-i\langle\theta,\llbracket
xy_{I}\rrbracket\rangle\right)\Lambda\otimes\mathbb{Q}({\mathrm{d}}x,{\mathrm{d}}y)$
$\displaystyle\hskip 71.13188pt+i\int_{\mathbb{R}\times
D[0,1]}\langle\theta,\llbracket
xy_{I}\rrbracket\rangle-\langle\theta,y_{I}\rangle\llbracket
x\rrbracket\Lambda\otimes\mathbb{Q}({\mathrm{d}}x,{\mathrm{d}}y)(1-p)\bigg{\\}},$

where all the terms in the previous expression are well defined by Lemma 6.12
and (6.20). Since $(xy)_{I}=\pi_{I}(V(x,y))$ and
$\bar{\nu}=(1-p)(\Lambda\otimes Q)\circ V^{-1}$, we obtain for a clear choice
of $b_{I}$ that

$\mathds{E}\left[\exp\left\\{\sum_{t\in
I}\theta_{t}\hat{\xi}_{t}\right\\}\right]=\exp\left\\{\int_{\mathbb{R}^{T}}\left(e^{\langle{\theta},e_{I}\rangle}-1-i\langle{{\theta}},\llbracket
e_{I}\rrbracket\rangle\right)\bar{\nu}({\mathrm{d}}e)+\langle\theta,b_{I}\rangle\right\\}.$

Next, notice that condition (iii) of Definition 6.8 is satisfied by
$(\Lambda\otimes\mathbb{Q})\circ V^{-1}$. Indeed, if we let $T_{0}:=\\{1\\}$
and $\bm{0}_{T_{0}}:=\\{e\in\mathbb{R}^{T}\,:\,e(1)=0\\}$, recalling that
$Y(1)\geq 1$ a.s., we deduce that

$\Lambda\otimes\mathbb{Q}\left((x,y):xy\in\\{\bm{0}_{T_{0}}\\}\right)=\Lambda(\\{0\\})=0.$

To conclude, let us show that $\bar{\nu}$ satisfies the integrability
condition (i) of Definition 6.8 if $p$ is an admissible memory parameter for
$\Lambda$, viz. if $\beta(\Lambda)<1/p$, while when $\beta(\Lambda)>1/p$, the
condition fails. By definition of $\bar{\nu}$, we have

$\int_{\mathbb{R}^{T}}\left(|e(t)|^{2}\wedge
1\right)\bar{\nu}({\mathrm{d}}e)=\int_{\mathbb{R}}\mathds{E}\left[|xY_{t}|^{2}\wedge
1\right]\Lambda({\mathrm{d}}x),$ (6.21)

and write:

$\mathds{E}\left[|xY_{t}|^{2}\wedge
1\right]=|x|^{2}\mathds{E}\left[|Y_{t}|^{2}\mathds{1}_{\\{Y_{t}\leq
1/|x|\\}}\right]+\mathds{P}\left(Y_{t}>1/|x|\right).$ (6.22)

Now, recalling from (10) of [7] the asymptotic behaviour,

$\mathds{P}\left(Y_{t}>1/|x|\right)\sim t\Gamma(p^{-1}+1)|x|^{1/p},\quad\text{
as }|x|\downarrow 0,$

it follows that if $\beta(\Lambda)<1/p$, the term
$\mathds{P}\left(Y_{t}>1/|x|\right)$ is integrable with respect to $\Lambda$
and infinite if $\beta(\Lambda)>1/p$. Let us now show that the same holds for

$\int_{[0,1]}|x|^{2}\mathds{E}\left[|Y_{t}|^{2}\mathds{1}_{\\{Y_{t}\leq
1/|x|\\}}\right]\Lambda({\mathrm{d}}x).$ (6.23)

Recalling Lemma 2.1, we get:

$\mathds{E}\left[|Y_{t}|^{2}\mathds{1}_{\\{Y_{t}\leq
1/|x|\\}}\right]=\sum_{n=1}^{\lfloor
1/|x|\rfloor}n^{2}\mathds{P}\left(Y_{t}=n\right)=tp^{-1}\sum_{n=1}^{\lfloor
1/|x|\rfloor}n^{2}\text{B}\left(n,p^{-1}+1\right),$

where we denoted by B the Beta function. Now, from the asymptotic behaviour

$\text{B}(n,p^{-1}+1)\sim n^{-(1+p)/p}\Gamma(p^{-1}+1),\quad\text{ as
}n\uparrow\infty,$

it follows that (6.23) is finite if $\beta(\Lambda)<1/p$ and infinite if
$\beta(\Lambda)>1/p$. ∎

Let us state the two last result that we need for the proof of Proposition
6.5. First, the Poissonian part of ID processes consists, roughly speaking, in
Poissonian sums of i.i.d. trajectories – for instance, remark that for NRLPs
those trajectories are the weighted Yule-Simon processes – for more examples
see e.g. [21, Section 3]. More precisely, let $X=(X_{t})_{t\in T}$ be an
infinitely divisible process with characteristic triplet
$(b,\Sigma,\bar{\nu})$ and suppose that $V=(V_{t})_{t\in T}$ is a representant
of $\bar{\nu}$ defined on a $\sigma$-finite measure space $(S,\mathscr{S},n)$.
To simplify notation, set $\chi(u):=\mathds{1}_{\\{|u|\leq 1\\}}$ and consider
$\mathcal{M}$ a Poisson random measure in $(S,\mathscr{S})$ with intensity
$n$. Then, the following process has the same distribution as $X$,

$b_{t}+G_{t}+\int_{S}V_{t}(s)\big{(}\mathcal{M}({\mathrm{d}}s)-\chi(V_{t}(s))n({\mathrm{d}}s)\big{)},\quad
t\in T,$ (6.24)

where $G=(G_{t})_{t\in T}$ is an independent Gaussian process with covariance
$\Sigma$. The integration in the previous display should read as a compensated
integral, and for a detailed statement we refer to [21, Proposition 3.1]. For
example, notice that if $X$ is a Lévy process, $\mathcal{M}$ is Poisson with
intensity ${\mathrm{d}}t\otimes\Lambda({\mathrm{d}}x)$ and replacing $V$ by
$x\mathds{1}_{\\{s\leq\cdot\\}}$ yields a Lévy-Itô representation. Finally, we
give one of the statements that we use of the Isomorphism Theorem of
infinitely divisible processes needed for our proof.

######  Theorem 6.14.

_[Isomorphism Theorem][21, 4.4]_ Let $X=(X_{t})_{t\in T}$ be an infinitely
divisible process given by (6.24). Choose an arbitrary measurable function
$\emph{q}:S\mapsto\mathbb{R}^{+}$ such that
$\int_{S}\emph{q}(s)n({\mathrm{d}}s)=1$ and set
$N(\emph{q}):=\int_{S}\emph{q}(s)\mathcal{M}({\mathrm{d}}s)$. Then, for any
measurable functional $F:\mathbb{R}^{T}\mapsto\mathbb{R}$, we have

$\displaystyle\mathds{E}\left[F((X_{t})_{t\in
T})\mathds{1}_{\\{N(\emph{q})>0\\}}\right]=\int_{S}\mathds{E}\left[F\left((X_{t}+V_{t}(s))_{t\in
T}\right)\left(N(\emph{q})+\emph{q}(s)\right)^{-1}\right]\emph{q}(s)n({\mathrm{d}}s).$

This allows for instance to study the law of $X$ under different
conditionings, for appropriate elections of q. This will be used in our
reasoning below. Now, let us conclude the proof of Proposition 6.5.

###### Proof of Proposition 6.5..

To simplify notation, we will perform a slight abuse of notation by writing
${\Lambda}$ instead of $(1-p)\Lambda$. We start by fixing $\delta\in(0,1)$
small enough such that $m=\Lambda(|x|>\delta)>0$. Now, let $h>0$ and as usual,
write $y=(y(t))_{t\in[0,1]}$ for a generic counting trajectory in $D[0,1]$.
Recall the result of Lemma 6.13 and consider a Poisson random measure
$\mathcal{M}=\sum\delta_{(x_{i},Y_{i})}$ with intensity
$\Lambda\otimes\mathbb{Q}$. Next, we set

$\text{q}(y,x):=\frac{1}{mh}\mathds{1}_{\\{|x|\geq\delta\\}}\mathds{1}_{\\{y(h)\geq
1\\}},$

and take

$N(\text{q})=\frac{1}{mh}\int_{D\times\\{|x|\geq\delta\\}}\mathds{1}_{\\{y(h)\geq
1\\}}{\mathcal{M}}({\mathrm{d}}x,{\mathrm{d}}y)=:\frac{1}{mh}S_{h}.$

Then, from the definition of $S_{h}$ we have

$S_{h}=\\#\\{(x_{i},Y_{i})\,:\,|x_{i}|\geq\delta\text{ and }Y_{i}(h)\geq
1\\}\leq\\#\\{|\Delta\hat{\xi}_{s}|\geq\delta\text{ for }s\leq h\\},$

where the inequality stands since a jump $x_{i}$ is repeated at each jump time
of its respective $Y_{i}$, and consequently might be repeated multiple times
in $[0,h]$. However, we do have $\\#\\{|\Delta\hat{\xi}_{s}|\geq\delta\text{
for }s\leq h\\}=0$ when $S_{h}=0$. Finally, we consider the functional
$F(e):=f(e(h))$ for $e\in D[0,1]$. An application of Theorem 6.14 yields:

$\displaystyle\mathds{E}\left[f(\hat{\xi}_{h})\mathds{1}_{\\{S_{h}>0\\}}\right]$
$\displaystyle=\int_{D\times\\{|x|\geq\delta\\}}\mathds{E}\left[f\big{(}\hat{\xi}_{h}+xy(h)\big{)}\frac{1}{S_{h}+1}\right]\mathds{1}_{\\{y(h)\geq
1\\}}\mathbb{Q}(dy)\Lambda({\mathrm{d}}x)$
$\displaystyle=\int_{D\times\\{|x|\geq\delta\\}}G_{h}(xy(h))\mathds{1}_{\\{y(h)\geq
1\\}}\mathbb{Q}({\mathrm{d}}y)\Lambda({\mathrm{d}}x),$

for $G_{h}(z)=\mathds{E}\left[f(\hat{\xi}_{h}+z)\frac{1}{S_{h}+1}\right]$ and
notice that $\lim_{h\downarrow 0}G_{h}(z)=f(z)$ by right-continuity – remark
that the previous display can be interpreted as the law of $\hat{\xi}_{h}$
conditioned at having at least one jump before time $h$ of size greater than
$\delta$. If we let $\eta$ be a random variable distributed Yule-Simon with
parameter $1/p$ under $\mathbb{P}$, this entails that we can write:

$\displaystyle\frac{1}{h}\mathds{E}\left[f(\hat{\xi}_{h})\right]$
$\displaystyle=\frac{1}{h}\mathds{E}\left[f(\hat{\xi}_{h})\mathds{1}_{\\{S_{h}=0\\}}\right]+\frac{1}{h}\mathds{E}\left[f(\hat{\xi}_{h})\mathds{1}_{\\{S_{h}>0\\}}\right]$
$\displaystyle=\frac{1}{h}\mathds{E}\left[f(\hat{\xi}_{h})\mathds{1}_{\\{S_{h}=0\\}}\right]+\int_{\\{|x|\geq\delta\\}}\mathds{E}\left[G_{h}(x\eta)\right]\Lambda({\mathrm{d}}x),$

where in the last equality we used that the law of $y(h)$ under

$\frac{\mathds{1}_{\\{y(h)\geq 1\\}}}{\mathbb{Q}(y(h)\geq
1)}\mathbb{Q}({\mathrm{d}}y),$

is the Yule-Simon distribution with parameter $1/p$ by Lemma 2.1 and that
$\mathbb{Q}(y(h)\geq 1)=h$. Consequently, we deduce that

$\displaystyle\Big{|}h^{-1}\mathbb{E}\big{[}{f(\hat{\xi}_{h})}{}\big{]}-\int_{\mathbb{R}}\mathds{E}\left[f(x\eta)\right]\Lambda({\mathrm{d}}x)\Big{|}$
$\displaystyle\leq
h^{-1}\mathds{E}\left[|f(\hat{\xi}_{h})|\mathds{1}_{\\{S_{h}=0\\}}\right]+\int_{\\{|x|\leq\delta\\}}\mathbb{E}\big{[}|f(x\eta)|\big{]}\Lambda({\mathrm{d}}x)+\Big{|}\int_{\\{|x|\geq\delta\\}}\mathds{E}\left[G_{h}(x\eta)\right]-\mathds{E}\left[f(x\eta)\right]\Lambda({\mathrm{d}}x)\Big{|}$
$\displaystyle=:K_{1}(h,\delta)+K_{2}(\delta)+K_{3}(h,\delta).$

Now, we study the limit as $h\downarrow 0$ of these three terms separately and
we start with $K_{1}(h,\delta)$. Recall the notation introduced in 5.3 for the
compensated integrals as well as
$\Sigma_{\delta,\infty}:=\mathds{1}_{(-\delta,\delta)^{c}}x*\hat{\mathcal{N}}$
for the process obtained by adding jumps of size greater that $\delta>0$.
Recall that on $\\{S_{h}=0\\}$, the process $\hat{\xi}$ doesn’t have jumps of
seize greater than $\delta$ before time $h$. It now follows that, restricted
to $\\{S_{h}=0\\}$, the following equality holds:

$\displaystyle\hat{\xi}_{h}=\hat{\xi}_{h}-\sum_{i}x_{i}Y_{i}(h)\mathds{1}_{\\{|x_{i}|\geq\delta\\}}=a\cdot
h+\hat{\xi}_{0,1}^{(3)}(h)+\Sigma_{1,\infty}(h)-\Sigma_{\delta,\infty}(h)=\hat{\xi}_{0,\delta}^{(3)}(h)-c_{\delta}\cdot
h,$ (6.25)

for $c_{\delta}:=-a+(1-p)^{-1}\int_{\\{\delta\leq|x|\leq
1\\}}x\Lambda({\mathrm{d}}x)$ and denote the right hand side of (6.25) by
$\hat{\xi}^{\delta}_{h}$. Now let us first consider the case
$\beta(\Lambda)<2$. Since $f$ is bounded and $O(|x|^{2})$ at the origin, for
any $q\in(\beta(\Lambda)\vee 1,1/p\wedge 2)$ satisfying $q<r$ we can bound
$|f(x)|\leq C|x|^{q}$ for all $x\in\mathbb{R}$, for some constant $C$ large
enough. Then, for a constant $C^{\prime}$ that only depends on $q$ we have

$\displaystyle K_{1}(h,\delta)$
$\displaystyle=h^{-1}\mathds{E}\left[|f(\hat{\xi}_{h})|\mathds{1}_{\\{S_{h}=0\\}}\right]$
$\displaystyle\leq
Ch^{-1}\mathds{E}\left[|\hat{\xi}_{h}^{\delta}|^{q}\right]\leq
C^{\prime}h^{-1}\mathds{E}\left[|\hat{\xi}_{0,\delta}^{(3)}(h)|^{q}\right]+C^{\prime}h^{q-1}|c^{\delta}|^{q}.$

Now, arguing as in (6.7), (6.8), recall that for $q\in(\beta(\Lambda)\vee
1,1/p\wedge 2)$ we have the following bound for the compensated sum of Yule-
Simon processes:

$\mathds{E}\left[|\hat{\xi}_{0,\delta}^{(3)}(h)|^{q}\right]\leq\mathds{E}\left[Y(h)^{q}\right]\int_{\\{|x|\leq\delta\\}}|x|^{q}\Lambda({\mathrm{d}}x)=h\cdot\mathds{E}\left[\eta^{q}\right]\int_{\\{|x|\leq\delta\\}}|x|^{q}\Lambda({\mathrm{d}}x)<\infty.$
(6.26)

Since $q-1>0$, we have $\limsup_{h\downarrow
0}K_{1}(h,\delta)\leq\mathds{E}\left[\eta^{q}\right]\int_{\\{|x|\leq\delta\\}}|x|^{q}\Lambda({\mathrm{d}}x)$
which can be made arbitrarily small for an appropriate choice of $\delta$.
Remark that the same reasoning applies for $K_{2}(\delta)$, by making use once
again of the bound $|f(x)|\leq C|x|^{q}$. Finally, since for any choice of
$\delta$, $K_{3}(h,\delta)\downarrow 0$ as $h\downarrow 0$, we obtain the
desired result.

If $\beta(\Lambda)=2$, we set $q=2$ and once again recall from page 9 of
Bertoin [7] that the inequality (6.26) still holds. In this case, since
$p\beta(\Lambda)<1$, $p$ must be smaller than $1/2$ and consequently
$\mathds{E}\left[\eta^{2}\right]<\infty$, while of course
$\int_{\\{|x|\leq\delta\\}}|x|^{2}\Lambda({\mathrm{d}}x)<\infty$ by definition
of a Lévy measure. We can then proceed as before. ∎

##  7 Appendix

This short section is devoted to proving a technical identity needed for the
proof of Lemma 4.2. The proof was omitted from the main discussion for
readability purposes.
Fix a Lévy measure $\Lambda$ in $\mathbb{R}$, $p\in(0,1)$ and denote the law
of the standard Yule process $Z=(Z(t))_{t\in\mathbb{R}^{+}}$ started at
$Z_{0}=1$ by Z. We write $D[0,\infty)$ the space of $\mathbb{R}^{+}$ indexed,
$\mathbb{R}$-valued rcll functions. Since Z is supported on the subset of
$D[0,\infty)$ of counting functions, $z=(z_{t})_{t\in\mathbb{R}^{+}}$ in the
sequel stands for a generic counting function. Moreover, if
$F:\mathbb{R}^{+}\times D[0,\infty)\mapsto\mathbb{R}^{+}$ is a measurable
function, we write $\textbf{{Z}}^{\bullet}$ for the measure in
$\mathbb{R}^{+}\times D[0,\infty)$ defined as
$\textbf{{Z}}^{\bullet}(F):=\int_{\mathbb{R^{+}}}{\mathrm{d}}u\,\mathbb{E}[F(u,Z)]$.
Roughly speaking, the objective is to describe the law of the following
”process”:

$(u,z)\mapsto\left(\mathds{1}_{\\{u\leq
t\\}}z_{p(\ln(t)-\ln(u))}\,:\,t\in\mathbb{R}^{+}\right)\quad\in D[0,\infty)$
(7.1)

defined on the measure space $(\mathbb{R}\times
D[0,\infty),\textbf{{Z}}^{\bullet})$, under different restrictions of the
measure $\textbf{{Z}}^{\bullet}$. In this direction, for $T>0$ we write

$\textbf{{Z}}^{\bullet}(\cdot\,|u\leq T):=\frac{\mathds{1}_{\\{u\leq
T\\}}}{T}{\mathrm{d}}u\,\textbf{{Z}}({\mathrm{d}}z),$

which is now a probability measure on $\mathbb{R}^{+}\times D[0,\infty)$. The
main properties of interest are stated in the following lemma, and shares
obvious similarities with Lemma 2.1.

######  Lemma 7.1.

The following properties hold:

  1. _(i)_

For each fixed $t>0$, the random variable

$(u,z)\mapsto\mathds{1}_{\\{u\leq t\\}}z_{p\ln(t/u)}\quad\text{ under
}\quad\textbf{{Z}}^{\bullet}(\cdot\,|u\leq t),$

has the same distribution as the Yule Simon random variable $\eta$ with
parameter $1/p$.

  2. _(ii)_

For every $T>0$, the process

$(u,z)\mapsto\left(\mathds{1}_{\\{u\leq
Tt\\}}z_{p\ln(Tt/u)}\,:\,t\in[0,1]\right)\quad\text{
under}\quad\textbf{{Z}}^{\bullet}(\cdot\,|u\leq T),$

has the same law as the Yule-Simon process $(Y(t))_{t\in[0,1]}$ with parameter
$1/p$.

Notice that the conditioning $\\{u\leq t\\}$ is playing the exact same role as
the conditioning on $\\{Y(t)\geq 1\\}$ in Lemma 2.1. Heuristically, (7.1) is
then a Yule-Simon process started at a time chosen according to
${\mathrm{d}}u$ in $\mathbb{R}^{+}$.

###### Proof.

(i) Since for each fixed $t$, ${\mathrm{d}}u\otimes\textbf{{Z}}(u\leq t)=t$,
for every bounded measurable function $f:\mathbb{R}\rightarrow\mathbb{R}$, we
have

$\displaystyle\textbf{{Z}}^{\bullet}\left(f\left(\mathds{1}_{\\{u\leq
t\\}}z_{p\ln(t/u)}\right)|u\leq
t\right)=t^{-1}\int_{0}^{t}{\mathrm{d}}u\,\mathbb{E}\Big{[}f\Big{(}Z\big{(}p(\ln(t)-\ln(u))\big{)}\Big{)}\Big{]},$
(7.2)

where we denoted by $Z$ a standard Yule process. Since $Z_{r}$ is distributed
geometric with parameter $e^{-r}$, it follows from the change of variable
$y=(u/t)^{p}$ and (2.2) that (7.2) equals

$\displaystyle p^{-1}\sum_{k\geq 1}f(k)\text{B}(k,1+1/p),$

and the claim follows from (2.2).

(ii) In order to show the second claim, we fix an arbitrary collection of
bounded measurable functions $(f_{i})_{i\leq k}$ with
$f_{i}:\mathbb{R}\mapsto\mathbb{R}$, and an increasing sequence of times
$0\leq t_{1}<\dots<t_{k}\leq 1$, and notice that

$\displaystyle\textbf{{Z}}^{\bullet}\left(\prod_{i=1}^{k}f_{i}\left(\mathds{1}_{\\{u\leq
Tt_{i}\\}}z_{p\ln(Tt_{i}/u)}\right)|u\leq T\right)$
$\displaystyle=\int_{0}^{1}{\mathrm{d}}u\,\mathds{E}\left[\prod_{i=1}^{k}f_{i}\left(\mathds{1}_{\\{u\leq
t_{i}\\}}Z\left(p(\ln(t_{i})-\ln(u))\right)\right)\right].$

The claim now follows from the description (2.1) by independence between $U$
and $Z$. ∎

Acknowledgement: I warmly thank Jean Bertoin for the discussions and attention
provided through the making of this work, as well as for introducing me to
noise reinforced Lévy processes.

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