# Volatility and jump activity estimation in a stable Cox-Ingersoll-Ross model

Elise Bayraktar LAMA, Univ Gustave Eiffel, Univ Paris Est Creteil, CNRS,
F-77447 Marne-la-Vallée, France.  Emmanuelle Clément LAMA, Univ Gustave
Eiffel, Univ Paris Est Creteil, CNRS, F-77447 Marne-la-Vallée, France.

(July, 26 2024)

Abstract. We consider the parametric estimation of the volatility and jump
activity in a stable Cox-Ingersoll-Ross ($\alpha$-stable CIR) model driven by
a standard Brownian Motion and a non-symmetric stable Lévy process with jump
activity $\alpha\in(1,2)$. The main difficulties to obtain rate efficiency in
estimating these quantities arise from the superposition of the diffusion
component with jumps of infinite variation. Extending the approach proposed in
Mies [16], we address the joint estimation of the volatility, scaling and jump
activity parameters from high-frequency observations of the process and prove
that the proposed estimators are rate optimal up to a logarithmic factor.

MSC $2020$. 60G51; 60G52; 60J75; 62F12.

Key words: Cox-Ingersoll-Ross model, Lévy process; Parametric inference;
Stable process; Stochastic Differential Equation.

##  1 Introduction

Jump processes are widely studied in the literature due to their many areas of
application, especially in financial modeling where interest rates or asset
price evolutions are often described by Itô semimartingales. These processes
are characterized by the superposition of a drift term, a diffusion part and a
jump component and the estimation of the different components is the subject
of specific studies depending on the available observations. In this paper we
are interested in the estimation of the volatility and the jump activity from
high-frequency observations. The estimation of such quantities has
concentrated a lot of activity in the last decades and it is well known that
the presence of jumps, and more precisely small jumps, makes the estimation of
the volatility more difficult. Jump activity is characterized by the
Blumenthal-Getoor index $\alpha\in(0,2)$ and when jumps are of infinite
variation, i.e. $\alpha>1$, the usual statistics to estimate the volatility
such as the truncated realized quadratic variation have an asymptotic bias. On
the other hand, the estimation of the jump activity is disturbed by the
presence of a diffusion component since the process is dominated by the
Brownian Motion.

In this work, we focus on processes with infinite variation jumps. Many works
have attempted to estimate the volatility or the jump activity in this context
and we briefly present the most relevant. An efficient estimator of the
volatility, with optimal rate of convergence $\sqrt{n}$, has been proposed in
Jacod and Todorov [11] in presence of infinite variation jumps with stable-
like behavior. Their method is based on the local estimation of the volatility
by using the real part of the empirical characteristic function and the
asymptotic bias is eliminated by combining statistics with different arguments
in the characteristic function. Their estimator achieves rate and variance
optimality for $\alpha\in(1,2)$ if the jumps of the driving Lévy process are
symmetric. For the truncated realized volatility, some debiasing procedures
have been studied in Amorino and Gloter [3] and in Cooper Boniece and al. [6].
By considering different truncation levels and with a two-step debiasing
procedure, efficiency for the estimation of the volatility is obtained in [6],
with the restriction $\alpha\in(1,8/5)$. Turning to the estimation of the jump
activity, Aït-Sahalia and Jacod in [1] have proposed an estimator of $\alpha$,
in presence of a diffusion component, based on jump counting. For locally
stable jumps, their estimator is consistent with rate of convergence
$n^{\alpha/10}$ ($n^{\alpha/8}$ in the stable case) but unfortunately it fails
to achieve the optimal rate $n^{\alpha/4}\log(n)^{1-\alpha/4}$ identified in
Aït-Sahalia and Jacod [2]. This estimation method is improved in Bull [7] by
combining increments of the process with different time-scales, achieving the
near optimal rate of convergence $n^{\alpha/4-\epsilon}$ for $\epsilon$
arbitrarily small. All these methods are non-parametric and apply to
semimartingales but are not completely satisfactory. In a parametric
framework, rate optimality in estimating the jump activity has been attained
by Mies [16] for a Lévy process with characteristic triplet
$(\mu,\sigma^{2},\nu)$, assuming that the jump part is close to the
superposition of $M$ stable processes. Using a method of moments, a joint
estimator of the volatility $\sigma^{2}$, the jump activity indices
$\alpha=(\alpha_{m})_{1\leq m\leq M}$ and some scaling coefficients
$r=(r^{+}_{m},r^{-}_{m})_{1\leq m\leq M}$ is proposed. This procedure leads to
an efficient estimator of the volatility (without debiasing procedure and
without restriction on the jump activity) and to rate optimal estimators of
the jump activity and scaling coefficients.

In this paper, our aim is to study the joint parametric estimation of the
volatility and jump activity of a stochastic differential equation with
infinite variation jumps. We will focus our work on the stable Cox-Ingersoll-
Ross model (stable CIR model) described by the stochastic equation

$dX_{t}=(a-bX_{t})dt+\sigma
X_{t}^{1/2}dB_{t}+\delta^{1/\alpha}X_{t-}^{1/\alpha}dL^{\alpha}_{t},$

where $(B_{t})$ is a standard Brownian Motion and $(L_{t}^{\alpha})$ a
spectrally positive Lévy process with jump activity $\alpha\in(1,2)$. This
model is very popular in financial modeling (see Jiao et al. [12], [13]) and
also presents some specific difficulties since the coefficients of the
equation are not Lipschitz and not bounded away from zero. When $\sigma=0$
(pure-jump process), some estimation results have been obtained in Li and Ma
[15], Yang [19] and more recently in Bayraktar and Clément [5]. But when
$\sigma\neq 0$, there are almost no estimation results in the literature for
the stable CIR model and to our knowledge the only existing study concerns the
estimation of the drift term $b$ proposed in Barczy et al. [4] from continuous
observations of the process. Inspired by the work of Mies [16], we propose
here a joint estimation method of the volatility coefficient $\sigma^{2}$, the
scaling parameter $\delta$ and the jump activity index $\alpha$ from high-
frequency observations of the process that leads to an efficient estimation of
the volatility and attains rate optimality up to a factor $\log(n)$ for the
estimation of $(\delta,\alpha)$. Our main results are derived in the case
$n\Delta_{n}$ fixed, where $n$ is the number of observations and $\Delta_{n}$
the step between two consecutive observations (we recall that in this
framework and in presence of a Brownian component, the drift parameters
$(a,b)$ are not identifiable) but also in the long time asymptotics
$n\Delta_{n}\to+\infty$.

The paper is organized as follows. In Section 2, we present the model and
notation. In Section 3, we describe the estimation method and study the
asymptotic properties of estimators in both cases $n\Delta_{n}$ fixed and
$n\Delta_{n}\to+\infty$. Section 4 contains some intermediate key results that
are of intrinsic interest. Proofs are gathered in Section 5.

##  2 Setting and notation

We consider $(B_{t})_{t\geq 0}$ a Brownian motion and $(L^{\alpha}_{t})_{t\geq
0}$ an independent stable Lévy process with exponent $\alpha\in(1,2)$ defined
on a filtered probability space $(\Omega,\mathcal{F},(\mathcal{F}_{t})_{t\geq
0},\mathbb{P})$, where $(\mathcal{F}_{t})_{t\geq 0}$ is the natural augmented
filtration. We assume that $(L^{\alpha}_{t})_{t\geq 0}$ is a spectrally
positive Lévy process with characteristic function

$\mathbb{E}\left(e^{izL_{1}^{\alpha}}\right)=\exp\left(-|z|^{\alpha}\left(1-i\tan\left(\frac{\pi\alpha}{2}\right)\text{sgn}(z)\right)\right).$
(2.1)

This non-symmetric Lévy process is strictly stable (see for example Sato [18])
and admits the representation

$L^{\alpha}_{t}=\int_{0}^{t}\int_{\mathbb{R}_{+}}z\tilde{N}(ds,dz),\quad\tilde{N}=N-\overline{N},$

where $N$ is a Poisson random measure with compensator $\overline{N}$ given by

$\overline{N}(dt,dz)=dtF^{\alpha}(dz)\quad\text{for}\quad
F^{\alpha}(dz)=\frac{c_{\alpha}}{z^{\alpha+1}}{\bf 1}_{(0,+\infty)}(z)dz.$
(2.2)

From Lemma 14.11 in Sato [18], we have

$c_{\alpha}=-\frac{\alpha(\alpha-1)}{\Gamma(2-\alpha)\cos(\pi\alpha/2)},\mbox{
with }\Gamma(\alpha)=\int_{0}^{\infty}x^{\alpha-1}e^{-x}dx,\quad\alpha>0.$
(2.3)

We will study the estimation of the parameter
$\theta=(\sigma^{2},\delta,\alpha)$ of the process (stable CIR model) given by
the stochastic differential equation

$dX_{t}=(a-bX_{t})dt+\sigma
X_{t}^{1/2}dB_{t}+\delta^{1/\alpha}X_{t-}^{1/\alpha}dL^{\alpha}_{t},\quad
X_{0}=x_{0}>0.$ (2.4)

Assuming $a>0$ , $b\in\mathbb{R}$, $\delta>0$, $\sigma>0$, $x_{0}>0$ and
$2a\geq\sigma^{2}$, we have according to Proposition 3.4. in Jiao et al [12]
that $(X_{t})_{t}$ is positive.

We use the following notation. For any random variable $Z$, we write
$\mathbb{E}_{i}(Z)=\mathbb{E}(Z|\mathcal{F}_{i\Delta_{n}})$ and for $p>0$,
$\mathbb{E}^{p}(Z)=[\mathbb{E}(Z)]^{p}$ ($\mathbb{E}_{i}^{p}(Z)$ is defined
accordingly). For any process $Y$, we denote the increments with step size
$\Delta_{n}$ and the symmetrized increments by

$\Delta_{i}^{n}Y=Y_{(i+1)\Delta_{n}}-Y_{i\Delta_{n}},\quad\quad\Delta^{s,n}_{i}Y=\Delta^{n}_{i+1}Y-\Delta^{n}_{i}Y.$

We write $\Longrightarrow$ for the convergence in distribution, and
$\xRightarrow{\mathcal{L}-s}$ for the stable convergence in law. Throughout
the paper, we denote by C all positive constants (or $C_{p}$ if it depends on
an auxiliary parameter $p$), which might change from line to line.

##  3 Main results

From now on, we assume that we observe $(X_{i\Delta_{n}})_{0\leq i\leq n}$
that solves (2.4) for the parameter values $(a_{0},b_{0},\theta_{0})$
satisfying assumption H below.

H.  We assume that
$\theta_{0}=(\sigma_{0}^{2},\delta_{0},\alpha_{0})\in\Theta=(0,+\infty)\times(0,+\infty)\times(1,2)$
and $a_{0}>\sigma_{0}^{2}$.

Under H, $(X_{t})_{t\geq 0}$ is positive as previously mentioned, but also
satisfies that $\mathbb{E}(1/X_{t}^{2})<\infty$ (see Proposition 4.1). To
estimate $\theta_{0}$, we propose a moment method based on the increments of
$X$, that extends to stochastic equations the estimation method proposed in
Mies [16] for Lévy processes. For $f:\mathbb{R}\mapsto\mathbb{R}^{3}$, we
consider the estimating function $F_{n}(\theta)$ defined by

$F_{n}(\theta)=\frac{1}{n}\sum_{i=0}^{\lfloor
n/2\rfloor-1}\left[f\left(\frac{u_{n}}{\sqrt{\Delta_{n}}}\frac{\Delta^{s,n}_{2i}X}{\sqrt{X_{2i\Delta_{n}}}}\right)-P_{n}f(X_{2i\Delta_{n}},\theta)\right],$
(3.1)

where the centering term $P_{n}f(x,\theta)$, defined by (3.2), is constructed
on the following way. Considering the Euler approximation of
$\Delta^{s,n}_{2i}X$ omitting the drift term

$\Delta^{s,n}_{2i}\overline{X}=\sigma_{0}X_{2i\Delta_{n}}^{1/2}\Delta^{s,n}_{2i}B+\delta_{0}^{1/\alpha_{0}}X_{2i\Delta_{n}}^{1/\alpha_{0}}\Delta^{s,n}_{2i}L^{\alpha_{0}},$

we approximate
$\mathbb{E}_{2i}f\left(\frac{u_{n}}{\sqrt{\Delta_{n}}}\frac{\Delta^{s,n}_{2i}X}{\sqrt{X_{2i\Delta_{n}}}}\right)$
by
$\mathbb{E}_{2i}f\left(\frac{u_{n}}{\sqrt{\Delta_{n}}}\frac{\Delta^{s,n}_{2i}\overline{X}}{\sqrt{X_{2i\Delta_{n}}}}\right)$.
Now observing that $(\Delta^{s,n}_{2i}B,\Delta^{s,n}_{2i}L^{\alpha})$ has the
distribution of
$(\sqrt{2}\sqrt{\Delta_{n}}B_{1},(2\Delta_{n})^{1/\alpha}S_{1}^{\alpha})$,
where $S_{1}^{\alpha}$ is a symmetric stable random variable, independent of
$B_{1}$, with characteristic function
$\mathbb{E}e^{izS_{1}^{\alpha}}=e^{-|z|^{\alpha}}$, we have

$\displaystyle
P_{n}f(x,\theta)=\mathbb{E}_{2i}f\left(\frac{u_{n}}{\sqrt{\Delta_{n}}}\frac{\Delta^{s,n}_{2i}\overline{X}}{\sqrt{X_{2i\Delta_{n}}}}\right)=\mathbb{E}\left[f\left(\sqrt{2}u_{n}\sigma
B_{1}+\frac{(2\delta
x)^{1/\alpha}}{\sqrt{x}}u_{n}\Delta_{n}^{1/\alpha-1/2}S^{\alpha}_{1}\right)\right].$
(3.2)

At this stage, we assume that $u_{n}$ goes to zero as $n$ goes to infinity and
$u_{n}/\sqrt{\Delta_{n}}$ goes to infinity, we will precise later the rate of
convergence of the sequence $(u_{n})$. We estimate the parameter $\theta_{0}$
by solving the estimating equation $F_{n}(\theta)=0$ for $F_{n}$ defined by
(3.1) with $f=(f_{k})_{1\leq k\leq 3}$, $f_{1}(x)=\cos(x)$, $f_{2}(x)=1-K(x)$,
$f_{3}(x)=f_{2}(2x)$ and

$K(x)=\left\\{\begin{array}[]{l}1\;\text{ if}\;|x|\leq 1,\\\
(1+\exp(\frac{1}{2-|x|}-\frac{1}{1-|x|}))^{-1}\;\text{ if}\;1<|x|<2,\\\
0\;\text{ if}\;|x|\geq 2.\end{array}\right.$

The choice of $f_{1}$ is natural to estimate $\sigma^{2}$ and has been used in
Jacod and Todorov [11]. For this choice, the expression of $P_{n}f_{1}$ is
explicit, it is the real part of the characteristic function of
$u_{n}\Delta^{s,n}_{2i}\overline{X}/\sqrt{\Delta_{n}X_{2i\Delta_{n}}}$

$P_{n}f_{1}(x,\theta)=\exp(-u_{n}^{2}\sigma^{2}-2\delta
x^{1-\alpha/2}u_{n}^{\alpha}\Delta_{n}^{1-\alpha/2}).$ (3.3)

To estimate $\delta$ and $\alpha$, related to the jump component of the
process, we consider here smooth truncations of the small jumps, with
different truncation levels. $P_{n}f_{2}$ and $P_{n}f_{3}$ are not explicit
(an asymptotic expansion is given in Lemma 4.2) but easy to compute
numerically.

The asymptotic properties of estimators depend on the connections between the
number of observations $n$ and the discretization step $\Delta_{n}$ and we
study separately the two cases $n\Delta_{n}$ fixed and
$n\Delta_{n}\rightarrow+\infty$.

###  3.1 Fixed observation window

We consider that $n\Delta_{n}$ is fixed and without loss of generality, we
assume that $n\Delta_{n}=1$. We first prove the existence of a joint estimator
with non-diagonal rate of convergence $\Lambda_{n}$ defined by

$\Lambda_{n}(\theta)=\frac{1}{\sqrt{n}}\left(\begin{array}[]{ccc}1&0&0\\\
0&\frac{1}{u_{n}^{\alpha/2}\Delta_{n}^{1/2-\alpha/4}}&-\delta\ln(\frac{u_{n}}{\sqrt{\Delta_{n}}})\frac{1}{u_{n}^{\alpha/2}\Delta_{n}^{1/2-\alpha/4}}\\\
0&0&\frac{1}{u_{n}^{\alpha/2}\Delta_{n}^{1/2-\alpha/4}}\end{array}\right).$
(3.4)

######  Theorem 3.1.

We assume H, $n\Delta_{n}=1$ and $u_{n}=1/[\ln(1/\Delta_{n})]^{p}$ with
$p>1/2$. Then there exists a sequence of random vectors $(\hat{\theta}_{n})$
solving $F_{n}(\hat{\theta}_{n})=0$ with probability tending to one, i.e.
$\mathbb{P}(F_{n}(\hat{\theta}_{n})=0)\to 1$. This sequence satisfies
$\hat{\theta}_{n}\to\theta_{0}$ in probability as $n\to\infty$, and

$\Lambda_{n}(\theta_{0})^{-1}(\hat{\theta}_{n}-\theta_{0})\xRightarrow[n\to\infty]{\mathcal{L}-s}W(\theta_{0})^{-1}\Sigma(\theta_{0})^{1/2}\mathcal{N},$

where $\mathcal{N}$ is a standard Gaussian variable independent of
$\mathcal{F}_{1}$. $\Sigma$ is the symmetric definite positive matrix and $W$
is the non-singular matrix respectively defined by

$\Sigma(\theta)=\frac{1}{2}\left(\begin{array}[]{ccc}2\sigma^{4}&0&0\\\
0&c_{\alpha}\delta
I(\alpha,X)\int_{\mathbb{R}}\frac{f_{2}^{2}(v)}{|v|^{\alpha+1}}dv&c_{\alpha}\delta
I(\alpha,X)\int_{\mathbb{R}}\frac{(f_{2}f_{3})(v)}{|v|^{\alpha+1}}dv\\\
0&c_{\alpha}\delta
I(\alpha,X)\int_{\mathbb{R}}\frac{(f_{2}f_{3})(v)}{|v|^{\alpha+1}}dv&c_{\alpha}\delta
I(\alpha,X)\int_{\mathbb{R}}\frac{f_{3}^{2}(v)}{|v|^{\alpha+1}}dv\end{array}\right),$
(3.5) $W(\theta)=\frac{1}{2}\left(\begin{array}[]{ccc}1&0&0\\\
0&-\psi(\alpha)I(\alpha,X)&-\delta\partial_{\alpha}[\psi(\alpha)I(\alpha,X)]\\\
0&-2^{\alpha}\psi(\alpha)I(\alpha,X)&-\delta\partial_{\alpha}[2^{\alpha}\psi(\alpha)I(\alpha,X)]\end{array}\right),$
(3.6)

with $I(\alpha,X)=\int_{0}^{1}X_{t}^{1-\alpha/2}dt$,
$\psi(\alpha)=c_{\alpha}\int\frac{f_{2}(v)}{|v|^{\alpha+1}}dv$ and
$c_{\alpha}$ given by (2.3).

We deduce from Theorem 3.1 that
$\sqrt{n}(\hat{\sigma}^{2}_{n}-\sigma_{0}^{2})\to\mathcal{N}(0,4\sigma_{0}^{4})$,
so the estimator is rate optimal and achieves the optimal variance up to a
factor 2, which is explained by the use of the symmetrized increments
$\Delta^{s,n}_{2i}X$. Optimality for the marginal estimation of
$(\delta,\alpha)$ has been discussed in Aït-Sahalia and Jacod [2]. For the
process $X_{t}=\sigma B_{t}+\delta L_{T}^{\alpha}$ with $\sigma>0$, they show
that the optimal rates of convergence for the estimation of $\delta$ and
$\alpha$ are respectively
$1/((\log(1/\Delta_{n})^{\alpha/4}\Delta_{n}^{\alpha/4})$ and
$\log(1/\Delta_{n})/((\log(1/\Delta_{n})^{\alpha/4}\Delta_{n}^{\alpha/4})$.
But from Mies [16] a diagonal normalisation with marginal optimality for each
parameter leads to a singular information matrix, suggesting that the optimal
rate of convergence in the joint estimation of $(\delta,\alpha)$ is non-
diagonal with slower rate of convergence by a factor $\log(1/\Delta_{n})$ and
probably achieved by the choice $u_{n}=1/(\log(1/\Delta_{n}))^{1/2}$ in (3.4).
This choice is theoretically possible up to a constant depending on the true
value of the parameters since the proof of Lemma 4.2 still works with
$u_{n}=1/(4\sigma_{0}(\log(1/\Delta_{n}))^{1/2})$ (see (5.16)). This choice is
not possible in practice, this is why we assume the near optimal condition
$u_{n}=1/(\log(1/\Delta_{n}))^{p}$ with $p>1/2$ in Theorem 3.1.

We now assume that $\alpha_{0}$ is known and only consider the estimation of
$\beta=(\sigma^{2},\delta)$ by solving
$F_{n}(\sigma^{2},\delta,\alpha_{0})=0$, where $F_{n}$ is defined by (3.1)
with $f=(f_{1},f_{2})$. In that case, we can also prove a uniqueness result,
which implies that any solution to the estimating equation is consistent.

######  Theorem 3.2.

We assume that $\alpha_{0}$ is known and that $(\sigma_{0}^{2},\delta_{0})$
belongs to the interior of a compact subset of $(0,+\infty)\times(0,+\infty)$.
Then under the assumptions of Theorem 3.1 any sequence
$(\hat{\sigma}^{2}_{n},\hat{\delta}_{n})$ that solves
$F_{n}(\sigma^{2},\delta,\alpha_{0})=0$ converges in probability to
$(\sigma_{0}^{2},\delta_{0})$. Such a sequence exists and is unique in the
sense that if $(\tilde{\sigma}^{2}_{n},\tilde{\delta}_{n})$ is another
sequence that solves $F_{n}(\sigma^{2},\delta,\alpha_{0})=0$ then
$\mathbb{P}((\hat{\sigma}^{2}_{n},\hat{\delta}_{n})\neq(\tilde{\sigma}^{2}_{n},\tilde{\delta}_{n}))$
goes to zero as n goes to infinity. This sequence obviously satisfies the
asymptotic properties of Theorem 3.1.

###  3.2 Ergodic case

We consider in this section the long time asymptotics $n\Delta_{n}\to+\infty$
and we restrict our study to the case $b_{0}>0$. With this assumption, the
process $(X_{t})_{t\geq 0}$ is geometrically ergodic (we refer to Li and Ma
[15] Proposition 2.2 and Theorem 2.5) and the stationary distribution
$\pi_{0}$ has Laplace transform

$\int_{0}^{+\infty}e^{-\lambda
y}\pi_{0}(dy)=\exp\left(-\int_{0}^{\lambda}\frac{a_{0}x}{\frac{\sigma_{0}^{2}}{2}x^{2}+\frac{\overline{\delta_{0}}^{\alpha_{0}}}{\alpha_{0}}+b_{0}x}dx\right),$

with
$\overline{\delta_{0}}^{\alpha_{0}}=\delta_{0}\alpha_{0}/|\cos(\frac{\pi\alpha_{0}}{2})|$.
Since the Laplace transform is $\mathcal{C}^{1}$ at zero, we check easily that
$\int_{0}^{\infty}x\pi_{0}(dx)<+\infty$, and setting
$\overline{I}(\alpha)=\int_{0}^{\infty}x^{1-\alpha/2}\pi_{0}(dx)$ we deduce
that $\alpha\mapsto\overline{I}(\alpha)$ is derivable for $\alpha\in(1,2)$.

######  Theorem 3.3.

We assume H and $b_{0}>0$, $n\Delta_{n}\to+\infty$ ,
$u_{n}=1/[\ln(1/\Delta_{n})]^{p}$ with $p>1/2$, and the additional condition
$\sqrt{n}\Delta_{n}^{1/\alpha_{0}+\alpha_{0}/4-1/2-\epsilon_{0}}\to 0$ for
$\epsilon_{0}>0$. Then the result of Theorem 3.1 holds with

$\Lambda_{n}(\theta_{0})^{-1}(\hat{\theta}_{n}-\theta_{0})\xRightarrow[n\to\infty]{}\overline{W}(\theta_{0})^{-1}\overline{\Sigma}(\theta_{0})^{1/2}\mathcal{N},$

where $\mathcal{N}$ is a standard Gaussian variable. $\overline{\Sigma}$ is
the symmetric definite positive (deterministic) matrix and $\overline{W}$ is
the (deterministic) non-singular matrix respectively defined by (3.5) and
(3.6) replacing $I(\alpha,X)$ by
$\overline{I}(\alpha)=\int_{0}^{+\infty}x^{1-\alpha/2}\pi_{0}(dx)$.

The condition
$\sqrt{n}\Delta_{n}^{1/\alpha_{0}+\alpha_{0}/4-1/2-\epsilon_{0}}\to 0$ is
compatible with $n\Delta\to+\infty$ if $\alpha_{0}\in(1,2)$ but stronger for
$\alpha_{0}$ near $2$. We also mention that we can also prove in the ergodic
case a uniqueness result for the estimation of $(\sigma^{2},\delta)$ (similar
to Theorem 3.2) if we assume that $\alpha_{0}$ is known.

##  4 Auxiliary results

###  4.1 Moments and approximation

Since $X_{t}$ is positive but not lower bounded by some strictly positive
constant, we first establish conditions to ensure that
$\mathbb{E}(1/X^{p}_{t})$ is finite. We also give some bounds for the moments
of $(X_{t})_{t}$ and its increments.

######  Proposition 4.1.

Let $(X_{t})_{t\geq 0}$ be the solution of (2.4) with $a>0$ ,
$b\in\mathbb{R}$, $\delta>0$, $\sigma>0$, $x_{0}>0$ and $2a>\sigma^{2}$. Then
we have

$\begin{array}[]{l}(i)\quad\text{for}\;1\leq
p<\frac{2a}{\sigma^{2}},\quad\sup_{t\geq
0}\mathbb{E}(\frac{1}{X_{t}^{p}})<+\infty,\end{array}$

and for $1\leq p<\alpha$, $\exists C_{p}>0$ such that $\forall
t_{1},t_{2}\in[0,1]$ and $\forall i\in[\\![0;n-1]\\!]$

$\begin{array}[]{l}(ii)\;\mathbb{E}\left(\sup_{t_{1}\leq u\leq
t_{2}}X_{u}^{p}|\mathcal{F}_{t_{1}}\right)\leq C_{p}(1+X_{t_{1}}^{p}),\\\
(iii)\;\mathbb{E}_{i}\left(\sup_{s\in[i\Delta_{n},(i+1)\Delta_{n}]}\left|\int_{i\Delta_{n}}^{s}X_{t-}^{1/\alpha}dL^{\alpha}_{t}\right|^{p}\right)\leq
C_{p}\Delta_{n}^{p/\alpha}\left(1+X_{i\Delta_{n}}^{p/\alpha}\right),\\\
(iv)\;\mathbb{E}_{i}\left(\sup_{s\in[i\Delta_{n},(i+1)\Delta_{n}]}\left|X_{s}-X_{i\Delta_{n}}\right|^{p}\right)\leq
C_{p}\Delta_{n}^{p/2}\left(1+X_{i\Delta_{n}}^{p}\right),\\\
(v)\;\mathbb{E}_{i}\left(\left|\int_{i\Delta_{n}}^{(i+1)\Delta_{n}}(X_{s-}^{1/\alpha}-X_{i\Delta_{n}}^{1/\alpha})dL^{\alpha}_{s}\right|^{p}\right)\leq
C_{p}\Delta_{n}^{p/\alpha+p/2}\left(1+\frac{1}{X_{i\Delta_{n}}^{p(1-1/\alpha)}}+X_{i\Delta_{n}}^{p/\alpha}\right).\par\end{array}$

We next give a sharp bound for the Euler approximation (without drift) of the
symmetrized normalized increments of $X$. This result can be seen as an
extension of Lemma 11 in Jacod Todorov [11]. We obtain here a better rate of
weak approximation and this result is a key step in the proof of Proposition
4.2. To shorten the notation, we set

$\rho_{i}^{n}=\frac{u_{n}}{\sqrt{\Delta_{n}}}\frac{\Delta^{s,n}_{i}X}{\sqrt{X_{i\Delta_{n}}}}\quad\text{and}\quad\overline{\rho}_{i}^{n}=\frac{u_{n}}{\sqrt{\Delta_{n}}}\frac{\Delta^{s,n}_{i}\overline{X}}{\sqrt{X_{i\Delta_{n}}}},$
(4.1)

where
$\Delta^{s,n}_{i}\overline{X}=\sigma_{0}X_{i\Delta_{n}}^{1/2}\Delta^{s,n}_{i}B+\delta_{0}^{1/\alpha_{0}}X_{i\Delta_{n}}^{1/\alpha_{0}}\Delta^{s,n}_{i}L^{\alpha_{0}}$
and $(X_{t})_{t\geq 0}$ solves (2.4) with $a_{0}>0$ , $b_{0}\in\mathbb{R}$,
$\delta_{0}>0$, $\sigma_{0}>0$, $x_{0}>0$ and $2a_{0}\geq\sigma_{0}^{2}$.

######  Lemma 4.1.

Let $f$ be an even bounded $\mathcal{C}^{2}$ function with bounded
derivatives. Then we have (with $\rho_{i}^{n}$ and $\overline{\rho}_{i}^{n}$
defined by (4.1)) for $\epsilon>0$, $\exists C_{\epsilon}$ such that $\forall
i\in[\\![0;n-1]\\!]$

$\displaystyle|\mathbb{E}_{i}(f(\rho_{i}^{n})-f(\overline{\rho}_{i}^{n}))|\leq$
$\displaystyle
C_{\epsilon}u_{n}\Delta_{n}^{1/\alpha_{0}-\epsilon}\left(1+X_{i\Delta_{n}}^{1/2}\frac{1}{X_{i\Delta_{n}}}\left[1+\frac{1}{\Delta_{n}}\mathbb{E}_{i}\int_{i\Delta_{n}}^{(i+2)\Delta_{n}}\frac{dv}{X_{v}}\right]\right),$
(4.2)

and for $1\leq p<\alpha_{0}$, $\exists C_{p}$ such that $\forall
i\in[\\![0;n-1]\\!]$

$\displaystyle\mathbb{E}_{i}(|f(\rho_{i}^{n})-f(\overline{\rho}_{i}^{n})|^{p})\leq
C_{p}u_{n}^{p}\Delta_{n}^{p/2}(1+\frac{1}{X_{i\Delta_{n}}^{p}}+X_{i\Delta_{n}}^{p/2}).$
(4.3)

Our estimating equations are based on the computation of the expectation
$P_{n}f(x,\theta)$ given by (3.2). For $f(x)=\cos(x)$ this expectation has an
explicit expression but not for $f(x)=1-K(x)$. The next lemma gives an
asymptotic expansion of $P_{n}f(x,\theta)$ for functions vanishing in a
neighborhood of zero. It is similar to Lemma 6.1 (i) in [16] derived from
Itô’s Formula. Here we propose a different proof, which easily allows higher
order expansion, by using the explicit form of the conditional distribution of
the Euler approximation $\Delta^{s,n}_{i}\overline{X}$ as a convolution
between a Gaussian distribution and a Stable distribution.

######  Lemma 4.2.

Let $f:\mathbb{R}\mapsto\mathbb{R}$ be a bounded even function such that $f=0$
on $[-\eta,+\eta]$ for some $\eta>0$. We assume that
$u_{n}[\ln(1/\Delta_{n})]^{1/2}\rightarrow 0$. Then we have $\forall x>0$

$\displaystyle
P_{n}f(x,\theta)=u_{n}^{\alpha}\Delta_{n}^{1-\alpha/2}c_{\alpha}\delta
x^{1-\alpha/2}\int\frac{f(v)}{|v|^{\alpha+1}}dv+u_{n}^{\alpha}\Delta_{n}^{1-\alpha/2}\varepsilon_{n}(\theta,x),\quad\text{with
$c_{\alpha}$ given by \eqref{eq:c-alpha},}$

where for any compact subset $A$ of $\Theta$, $\sup_{\theta\in
A}|\varepsilon_{n}(\theta,x)|\leq(1+x)\varepsilon_{n},$ with
$\varepsilon_{n}\rightarrow 0$.

###  4.2 Convergences

The asymptotic behavior of the estimator $\hat{\theta}_{n}$ is derived using
the theory of estimating equations (see [10]) and is based on the convergence
of $F_{n}(\theta_{0})$ defined by (3.1) and its gradient
$\nabla_{\theta}F_{n}(\theta)$. We set

$A_{n}(\theta)=\sqrt{n}\;diag\left(\frac{1}{u_{n}^{2}},\frac{1}{u_{n}^{\alpha/2}\Delta_{n}^{1/2-\alpha/4}},\frac{1}{u_{n}^{\alpha/2}\Delta_{n}^{1/2-\alpha/4}}\right).$
(4.4)

We first state a Central Limit Theorem for $F_{n}(\theta_{0})$ with rate
$A_{n}(\theta_{0})$.

######  Proposition 4.2.

We assume H and $u_{n}=1/[\ln(1/\Delta_{n})]^{p}$ with $p>1/2$.

  1. 1.

If $n\Delta_{n}=1$, then we have the stable convergence in law

$A_{n}(\theta_{0})F_{n}(\theta_{0})\xRightarrow[n\to\infty]{\mathcal{L}-s}\Sigma(\theta_{0})^{1/2}\mathcal{N},$

where $\mathcal{N}$ is a standard Gaussian variable independent of
$\mathcal{F}_{1}$ and $\Sigma(\theta_{0})$ is the symmetric definite positive
matrix defined by (3.5).

  2. 2.

If $n\Delta_{n}\to+\infty$, $b_{0}>0$ and
$\sqrt{n}\Delta_{n}^{1/\alpha_{0}+\alpha_{0}/4-1/2-\epsilon_{0}}\to 0$ for
some $\epsilon_{0}>0$, then we have the convergence in distribution

$A_{n}(\theta_{0})F_{n}(\theta_{0})\xRightarrow[n\to\infty]{}\overline{\Sigma}(\theta_{0})^{1/2}\mathcal{N},$

where $\mathcal{N}$ is a standard Gaussian variable and
$\overline{\Sigma}(\theta_{0})$ is the symmetric definite positive matrix
defined by (3.5) replacing $I(\alpha,X)$ by
$\overline{I}(\alpha)=\int_{0}^{+\infty}x^{1-\alpha/2}\pi_{0}(dx)$.

Turning to the convergence of $\nabla_{\theta}F_{n}$ and recalling that
$\Lambda_{n}(\theta)$ is defined by (3.4), we obtain the following
convergence.

######  Proposition 4.3.

Let $A$ be a compact subset of $\Theta$. We assume H and
$u_{n}=1/[\ln(1/\Delta_{n})]^{p}$ with $p>1/2$.

  1. 1.

If $n\Delta_{n}=1$, then we have the convergence in probability

$\sup_{\theta\in
A}|A_{n}(\theta)\nabla_{\theta}F_{n}(\theta)\Lambda_{n}(\theta)-W(\theta)|\to
0,$

where $W(\theta)$ is the non-singular matrix defined by (3.6).

  2. 2.

If $n\Delta_{n}\to+\infty$ and $b_{0}>0$, then we have the convergence in
probability

$\sup_{\theta\in
A}|A_{n}(\theta)\nabla_{\theta}F_{n}(\theta)\Lambda_{n}(\theta)-\overline{W}(\theta)|\to
0,$

where $\overline{W}(\theta)$ is defined by (3.6) replacing $I(\alpha,X)$ by
$\overline{I}(\alpha)=\int_{0}^{+\infty}x^{1-\alpha/2}\pi_{0}(dx)$.

##  5 Proofs

###  5.1 Auxiliary results

####  5.1.1 Proof of Proposition 4.1

We only prove $(i)$ since the other items can be deduced with similar
arguments as the ones given for the proof of Proposition 2.2 in [5]. The
Laplace transform of $X_{t}$ has the explicit expression (we refer to Li [14],
see also Section 3.1 in [12] and Proposition 2.1. in [4])

$\mathbb{E}(e^{-uX_{t}})=\exp\left(-x_{0}v_{t}(u)-\int_{v_{t}(u)}^{u}\frac{F(z)}{R(z)}dz\right),$
(5.1)

where $t\to v_{t}(u)$ is the unique locally bounded solution of

$\frac{\partial}{\partial t}v_{t}(u)=-R(v_{t}(u)),\quad v_{0}(u)=u.$ (5.2)

For the $\alpha$-CIR model, the branching mechanism and the immigration rate
are given by

$R(z)=\frac{\sigma^{2}}{2}z^{2}+\frac{\overline{\delta}^{\alpha}}{\alpha}z^{\alpha}+bz,\quad
F(z)=az,\quad z\in[0,+\infty),$

with
$\overline{\delta}=\delta^{1/\alpha}(\alpha/|\cos(\frac{\pi\alpha}{2})|)^{1/\alpha}$.
Setting $z_{0}=\inf\\{z>0,\;R(z)\geq 0\\}$, $z_{0}=0$ when $b\geq 0$ and
$z_{0}>0$ when $b<0$ with $z_{0}<+\infty$ as $\lim_{z\to+\infty}R(z)=+\infty$.
From Corollary 3.2 of Li [14], $\lim_{t\to\infty}v_{t}(u)=z_{0}$ and
$\forall\;u>z_{0}$, $t\to v_{t}(u)$ is decreasing. Now we have $\forall t\geq
0$

$\mathbb{E}(\frac{1}{X_{t}^{p}})=C_{p}\int_{\mathbb{R}_{+}}u^{p-1}\mathbb{E}(e^{-uX_{t}})du\leq
C_{p}(z_{0})+C_{p}\int_{z_{0}+1}^{+\infty}u^{p-1}\mathbb{E}(e^{-uX_{t}})du.$

We fix $\epsilon\in(0,1)$ and for $u\geq z_{0}+1$ we set
$t_{u}^{\epsilon}=\inf\\{t\geq 0,v_{t}(u)\leq u^{\epsilon}\\}$. With
$u_{0,\epsilon}=\max(z_{0}+1,z_{0}^{1/\epsilon})$

$\displaystyle\sup_{t\geq
0}\;\int_{u_{0,\epsilon}}^{+\infty}u^{p-1}\mathbb{E}(e^{-uX_{t}})du\leq\int_{u_{0,\epsilon}}^{+\infty}u^{p-1}\left[\sup_{t\in[0,t_{u}^{\epsilon}]}\;\mathbb{E}(e^{-uX_{t}})+\sup_{t\geq
t_{u}^{\epsilon}}\;\mathbb{E}(e^{-uX_{t}})\right]du.$ (5.3)

Using that $t\to v_{t}(u)$ is non-increasing for $u>z_{0}$, we obtain from
(5.1) and observing that $v_{t_{u}^{\epsilon}}(u)=u^{\epsilon}$

$\displaystyle\sup_{t\in[0,t_{u}^{\epsilon}]}\mathbb{E}(e^{-uX_{t}})$
$\displaystyle\leq\sup_{t\in[0,t_{u}^{\epsilon}]}\exp(-x_{0}v_{t}(u))\leq\exp(-x_{0}u^{\epsilon}),$
$\displaystyle\sup_{t\geq t_{u}^{\epsilon}}\mathbb{E}(e^{-uX_{t}})$
$\displaystyle\leq\sup_{t\geq
t_{u}^{\epsilon}}\exp\left(-\int_{v_{t}(u)}^{u}\frac{F(z)}{R(z)}dz\right)\leq\exp\left(-\int_{u^{\epsilon}}^{u}\frac{F(z)}{R(z)}dz\right).$

For $z\in[u^{\epsilon},u]$

$\displaystyle\frac{F(z)}{R(z)}\geq\frac{2a}{\sigma^{2}}\frac{1}{z(1+\frac{\overline{\delta}^{\alpha}2}{\alpha\sigma^{2}z^{2-\alpha}})+\frac{2b}{\sigma^{2}}}\geq\frac{2a}{\sigma^{2}}\frac{1}{zg(u)+\frac{2b}{\sigma^{2}}},\;\text{with}\;g(u)=1+\frac{\overline{\delta}^{\alpha}2}{\alpha\sigma^{2}u^{\epsilon(2-\alpha)}}.$

Hence integrating the previous inequality

$\displaystyle\int_{u^{\epsilon}}^{u}\frac{F(z)}{R(z)}dz\geq\frac{2a}{\sigma^{2}g(u)}\ln\left(\frac{ug(u)+\frac{2b}{\sigma^{2}}}{u^{\epsilon}g(u)+\frac{2b}{\sigma^{2}}}\right)=\frac{2a}{\sigma^{2}g(u)}[\ln(u^{1-\epsilon})+h(u)],$

with

$h(u)=\ln\left(\frac{g(u)+\frac{2b}{u\sigma^{2}}}{g(u)+\frac{2b}{u^{\epsilon}\sigma^{2}}}\right).$

We have $g(u)\geq 1$, $\lim_{u\to+\infty}g(u)=1$ and
$\lim_{u\to+\infty}h(u)=0$, then

$\displaystyle\exp\left(-\int_{u^{\epsilon}}^{u}\frac{F(z)}{R(z)}dz\right)$
$\displaystyle\leq C_{\epsilon}/u^{(1-\epsilon)\frac{2a}{\sigma^{2}g(u)}}.$

If $1\leq p<\frac{2a}{\sigma^{2}}$, we choose $\epsilon>0$ such that
$p-\frac{2a}{\sigma^{2}}+\epsilon\frac{2a}{\sigma^{2}}<0$, then we deduce

$\int_{u_{0,\epsilon}}^{+\infty}u^{p-1}\exp\left(-\int_{u^{\epsilon}}^{u}\frac{F(z)}{R(z)}dz\right)du<+\infty,$

and this allows us to conclude.

####  5.1.2 Proof of Lemma 4.1

We introduce a little more notation. We set

$a_{t}={\bf 1}_{\\{(i+1)\Delta_{n}\leq t\leq(i+2)\Delta_{n}\\}}-{\bf
1}_{\\{i\Delta_{n}\leq t<(i+1)\Delta_{n}\\}},\quad
t\in[i\Delta_{n},(i+2)\Delta_{n}].$

With this notation, we have

$\Delta^{s,n}_{i}X=\int_{i\Delta_{n}}^{(i+2)\Delta_{n}}a_{t}dX_{t}\quad\text{and}\quad\Delta^{s,n}_{i}\overline{X}=\sigma_{0}X_{i\Delta_{n}}^{1/2}\int_{i\Delta_{n}}^{(i+2)\Delta_{n}}a_{t}dB_{t}+\delta_{0}^{1/\alpha_{0}}X_{i\Delta_{n}}^{1/\alpha_{0}}\int_{i\Delta_{n}}^{(i+2)\Delta_{n}}a_{t}dL^{\alpha_{0}}_{t}.$

We then deduce the decomposition
$\rho_{i}^{n}=\overline{\rho}_{i}^{n}+\theta(1)_{i}^{n}+\theta(2)_{i}^{n}+\theta(3)_{i}^{n}$
with

$\displaystyle\theta(1)_{i}^{n}$
$\displaystyle=\sigma_{0}\frac{u_{n}}{\sqrt{\Delta_{n}}\sqrt{X_{i\Delta_{n}}}}\int_{i\Delta_{n}}^{(i+2)\Delta_{n}}a_{t}(X_{t}^{1/2}-X_{i\Delta_{n}}^{1/2})dB_{t},$
$\displaystyle\theta(2)_{i}^{n}$
$\displaystyle=\delta_{0}^{1/\alpha_{0}}\frac{u_{n}}{\sqrt{\Delta_{n}}\sqrt{X_{i\Delta_{n}}}}\int_{(\Delta_{n}}^{(i+2)\Delta_{n}}a_{t}(X_{t}^{1/\alpha_{0}}-X_{i\Delta_{n}}^{1/\alpha_{0}})dL^{\alpha_{0}}_{t},$
$\displaystyle\theta(3)_{i}^{n}$
$\displaystyle=-b_{0}\frac{u_{n}}{\sqrt{\Delta_{n}}\sqrt{X_{i\Delta_{n}}}}\int_{i\Delta_{n}}^{(i+2)\Delta_{n}}a_{t}(X_{t}-X_{i\Delta_{n}})dt.$

We have the following estimates

$\displaystyle\forall p\in[1,\alpha_{0}),\;$
$\displaystyle\mathbb{E}_{i}(|\theta(1)_{i}^{n}|^{p})\leq
C_{p}u_{n}^{p}\Delta_{n}^{p/2}(1+\frac{1}{X_{i\Delta_{n}}^{p}}),$ (5.4)
$\displaystyle\forall p\in[1,\alpha_{0}),\;$
$\displaystyle\mathbb{E}_{i}(|\theta(2)_{i}^{n}|^{p})\leq
C_{p}u_{n}^{p}\Delta_{n}^{p/\alpha_{0}}(1+\frac{1}{X_{i\Delta_{n}}^{p/2+p(1-1/\alpha_{0})}}+X_{i\Delta_{n}}^{p/\alpha_{0}-p/2}),$
(5.5) $\displaystyle\forall p\in[1,\alpha_{0}),\;$
$\displaystyle\mathbb{E}_{i}(|\theta(3)_{i}^{n}|^{p})\leq
C_{p}u_{n}^{p}\Delta_{n}^{p}(1+\frac{1}{X_{i\Delta_{n}}^{p/2}}+X_{i\Delta_{n}}^{p/2}).$
(5.6)

We deduce easily (5.5) and (5.6) from Proposition 4.1, (5.4) follows from
Burkholder’s inequality, Proposition 4.1 and the bound

$\forall x,y\in(0,+\infty)\;|y^{1/2}-x^{1/2}|\leq C|y-x|/x^{1/2}.$

Combining the estimates (5.4), (5.5), (5.6) and $|f(x+y)-f(x)|\leq C|y|$ as
$f^{\prime}$ is bounded, we deduce immediately (4.3).

We now prove (4.2). With the previous notation, we have the decomposition

$\displaystyle f(\rho_{i}^{n})-f(\overline{\rho}_{i}^{n})=$ $\displaystyle
f(\rho_{i}^{n})-f(\overline{\rho}_{i}^{n}+\theta(1)_{i}^{n}+\theta(2)_{i}^{n})+f(\overline{\rho}_{i}^{n}+\theta(1)_{i}^{n}+\theta(2)_{i}^{n})-f(\overline{\rho}_{i}^{n}+\theta(1)_{i}^{n})$
$\displaystyle+f(\overline{\rho}_{i}^{n}+\theta(1)_{i}^{n})-f(\overline{\rho}_{i}^{n}),$

and since $f^{\prime}$ is bounded it yields

$|\mathbb{E}_{i}(f(\rho_{i}^{n})-f(\overline{\rho}_{i}^{n}))|\leq
C(\mathbb{E}_{i}|\theta(3)_{i}^{n}|+\mathbb{E}_{i}|\theta(2)_{i}^{n}|+|\mathbb{E}_{i}(f(\overline{\rho}_{i}^{n}+\theta(1)_{i}^{n})-f(\overline{\rho}_{i}^{n}))|).$

Using (5.5) and (5.6) with $p=1$

$\displaystyle\mathbb{E}_{i}|\theta(3)_{i}^{n}|+\mathbb{E}_{i}|\theta(2)_{i}^{n}|\leq
Cu_{n}\Delta_{n}^{1/\alpha_{0}}(1+\frac{1}{X_{i\Delta_{n}}^{3/2-1/\alpha_{0}}}+X_{i\Delta_{n}}^{1/2}).$
(5.7)

It remains to bound
$|\mathbb{E}_{i}(f(\overline{\rho}_{i}^{n}+\theta(1)_{i}^{n})-f(\overline{\rho}_{i}^{n}))|$.
Since $f^{\prime\prime}$ is bounded, we have

$\displaystyle|\mathbb{E}_{i}(f(\overline{\rho}_{i}^{n}+\theta(1)_{i}^{n})-f(\overline{\rho}_{i}^{n}))|\leq|\mathbb{E}_{i}(\theta(1)_{i}^{n}f^{\prime}(\overline{\rho}_{i}^{n}))|+C\mathbb{E}_{i}|\theta(1)_{i}^{n}|^{2}.$

To finish the proof, we show the following estimates

$\mathbb{E}_{i}|\theta(1)_{i}^{n}|^{2}\leq
Cu_{n}^{2}\Delta_{n}\left(1+\frac{1}{X_{i\Delta_{n}}}\left[1+\frac{1}{\Delta_{n}}\mathbb{E}_{i}\int_{i\Delta_{n}}^{(i+2)\Delta_{n}}\frac{dv}{X_{v}}\right]\right),$
(5.8)

and for $\epsilon>0$

$\displaystyle|\mathbb{E}_{i}(\theta(1)_{i}^{n}f^{\prime}(\overline{\rho}_{i}^{n}))|\leq
C_{\epsilon}u_{n}\Delta_{n}^{1/\alpha_{0}-\epsilon}\left(1+\frac{1}{X_{i\Delta_{n}}^{1/2}}\left[1+\frac{1}{\Delta_{n}}\mathbb{E}_{i}\int_{i\Delta_{n}}^{(i+2)\Delta_{n}}\frac{dv}{X_{v}^{1/2}}\right]\right).$
(5.9)

Collecting (5.7), (5.8) and (5.9), we deduce (4.2).

Proof of (5.8) We write Itô’s formula for the function $x\to\sqrt{x}$ which is
$\mathcal{C}^{2}$ on $(0,+\infty)$

$\sqrt{X_{t}}-\sqrt{X_{s}}=\int_{s}^{t}\overline{b}_{v}dv+\int_{s}^{t}\frac{\sigma_{0}}{2}dB_{v}+\int_{s}^{t}\int\overline{c}_{v-}\tilde{N}(dv,dz),$

where

$\displaystyle\overline{b}_{v}$
$\displaystyle=\frac{(a_{0}-b_{0}X_{v})}{2X_{v}^{1/2}}-\frac{\sigma_{0}^{2}}{8X_{v}^{1/2}}+\int(\sqrt{X_{v-}+\delta_{0}^{1/\alpha_{0}}X_{v-}^{1/\alpha_{0}}z}-\sqrt{X_{v-}}-\frac{1}{2X_{v-}^{1/2}}\delta_{0}^{1/\alpha_{0}}X_{v-}^{1/\alpha_{0}}z)F^{\alpha_{0}}(z)dz,$
$\displaystyle\overline{c}_{v-}$
$\displaystyle=(X_{v-}+\delta_{0}^{1/\alpha_{0}}X_{v-}^{1/\alpha_{0}}z)^{1/2}-(X_{v-})^{1/2}.$

Moreover, we check that $|\overline{b}_{v}|\leq
C(1+\frac{1}{\sqrt{X_{v}}}+\sqrt{X_{v}})$ and that $|\overline{c}_{v}|\leq
C(1+\sqrt{X_{v}})(\sqrt{z}{\bf 1}_{\\{z>1\\}}+z{\bf 1}_{\\{0<z\leq 1\\}}).$ We
obtain the decomposition

$\theta(1)_{i}^{n}=\mu(1)_{i}^{n}+\mu(2)_{i}^{n}+\mu(3)_{i}^{n},$ (5.10)

with

$\displaystyle\mu(1)_{i}^{n}$
$\displaystyle=\frac{u_{n}\sigma_{0}}{\sqrt{\Delta_{n}X_{i\Delta_{n}}}}\int_{i\Delta_{n}}^{(i+2)\Delta_{n}}a_{t}(\int_{i\Delta_{n}}^{t}\overline{b}_{v}dv)dB_{t},$
$\displaystyle\mu(2)_{i}^{n}$
$\displaystyle=\frac{u_{n}\sigma_{0}}{\sqrt{\Delta_{n}X_{i\Delta_{n}}}}\frac{\sigma_{0}}{2}\int_{i\Delta_{n}}^{(i+2)\Delta_{n}}a_{t}(B_{t}-B_{i\Delta_{n}})dB_{t},$
$\displaystyle\mu(3)_{i}^{n}$
$\displaystyle=\frac{u_{n}\sigma_{0}}{\sqrt{\Delta_{n}X_{i\Delta_{n}}}}\int_{i\Delta_{n}}^{(i+2)\Delta_{n}}a_{t}\left(\int_{i\Delta_{n}}^{t}\int\overline{c}_{v-}\tilde{N}(dv,dz)\right)dB_{t}.$

From Burkholder’s inequality and Proposition 4.1 (ii), we have

$\displaystyle\forall p\in$
$\displaystyle[1,2],\,\mathbb{E}_{i}|\mu(1)_{i}^{n}|^{p}\leq
C_{p}u_{n}^{p}\Delta_{n}^{p}\left(1+\frac{1}{X_{i\Delta_{n}}^{p/2}}\left[1+\frac{1}{\Delta_{n}}\mathbb{E}_{i}\int_{i\Delta_{n}}^{(i+2)\Delta_{n}}\frac{dv}{X_{v}^{p/2}}\right]\right),$
(5.11) $\displaystyle\forall p$ $\displaystyle\geq
1,\,\mathbb{E}_{i}|\mu(2)_{i}^{n}|^{p}\leq
C_{p}\frac{u_{n}^{p}}{X_{i\Delta_{n}}^{p/2}}\Delta_{n}^{p/2},$ (5.12)
$\displaystyle\forall p$
$\displaystyle\in(\alpha_{0},2],\;\mathbb{E}_{i}|\mu(3)_{i}^{n}|^{p}\leq
C_{p}u_{n}^{p}\Delta_{n}\left(1+\frac{1}{X_{i\Delta_{n}}^{p/2}}\right).$
(5.13)

Note that to obtain (5.13), we have used

$\mathbb{E}_{i}\sup_{i\Delta_{n}\leq
t\leq(i+2)\Delta_{n}}|\int_{i\Delta_{n}}^{t}\int\overline{c}_{v-}\tilde{N}(dv,dz)|^{p}\leq
C_{p}\mathbb{E}_{i}\int_{i\Delta_{n}}^{(i+2)\Delta_{n}}\int|\overline{c}_{v-}|^{p}\overline{N}(dv,dz),$

combined with $|\overline{c}_{v}|\leq C(1+\sqrt{X_{v}})(\sqrt{z}{\bf
1}_{\\{z>1\\}}+z{\bf 1}_{\\{0<z\leq 1\\}})$. We then deduce (5.8) from the
previous bounds with $p=2$.

Proof of (5.9) Since $f^{\prime}$ is bounded, we have using (5.10) and
Hölder’s inequality with $\alpha_{0}<p\leq 2$

$|\mathbb{E}_{i}(\theta(1)_{i}^{n}f^{\prime}(\overline{\rho}_{i}^{n}))|\leq
C(\mathbb{E}_{i}|\mu(1)_{i}^{n}|+|\mathbb{E}_{i}(\mu(2)_{i}^{n}f^{\prime}(\overline{\rho}_{i}^{n}))|+\mathbb{E}^{1/p}_{i}|\mu(3)_{i}^{n}|^{p}).$
(5.14)

It remains to consider
$\mathbb{E}_{i}(\mu(2)_{i}^{n}f^{\prime}(\overline{\rho}_{i}^{n}))$. We split
$\overline{\rho}_{i}^{n}$ in two parts (Brownian part and pure-jump part)

$\overline{\rho}_{i}^{n}=\frac{\sigma_{0}u_{n}}{\sqrt{\Delta_{n}}}\Delta^{s,n}_{i}B+\frac{\delta_{0}^{1/\alpha_{0}}X_{i\Delta_{n}}^{1/\alpha_{0}}u_{n}}{\sqrt{\Delta_{n}}\sqrt{X_{i\Delta_{n}}}}\Delta^{s,n}_{i}L^{\alpha_{0}}:=\overline{\rho}_{i}^{1,n}+\overline{\rho}_{i}^{2,n},$

and write

$\mathbb{E}_{i}(\mu(2)_{i}^{n}f^{\prime}(\overline{\rho}_{i}^{n}))=\mathbb{E}_{i}(\mu(2)_{i}^{n}f^{\prime}(\overline{\rho}_{i}^{1,n}))+\mathbb{E}_{i}(\mu(2)_{i}^{n}(f^{\prime}(\overline{\rho}_{i}^{n})-f^{\prime}(\overline{\rho}_{i}^{1,n}))).$

Since $f^{\prime}$ is odd, an explicit computation gives
$\mathbb{E}_{i}(\mu(2)_{i}^{n}f^{\prime}(\overline{\rho}_{i}^{1,n}))=0$. For
the second term, we use successively that $f^{\prime\prime}$ is bounded,
Hölder’s inequality with $1<q<\alpha_{0}$ and the scaling property of
$L^{\alpha_{0}}$ to obtain

$\displaystyle\mathbb{E}_{i}|\mu(2)_{i}^{n}(f^{\prime}(\overline{\rho}_{i}^{n})-f^{\prime}(\overline{\rho}_{i}^{1,n}))|$
$\displaystyle\leq
C\mathbb{E}_{i}|\mu(2)_{i}^{n}\overline{\rho}_{i}^{2,n}|\leq
C_{q}\frac{u_{n}}{\sqrt{\Delta_{n}}}(1+X_{i\Delta_{n}}^{1/2})\mathbb{E}_{i}^{\frac{1}{q^{\prime}}}|\mu(2)_{i}^{n}|^{q^{\prime}}\mathbb{E}^{\frac{1}{q}}_{i}|\Delta^{s,n}_{i}L^{\alpha_{0}}|^{q}$
$\displaystyle\leq
C_{q}u_{n}^{2}(1+\frac{1}{X_{i\Delta_{n}}^{1/2}})\Delta_{n}^{1/\alpha_{0}}.$
(5.15)

We finally deduce (5.9) from (5.14) (using (5.11) with $p=1$ and (5.13) with
$p=\alpha_{0}+\epsilon$) and (5.15).

####  5.1.3 Proof of Lemma 4.2

We assume that $\theta=(\sigma^{2},\delta,\alpha)\in A$ a compact subset of
$\Theta$. We denote by $g_{\alpha}$ the density of $S_{1}^{\alpha}$, by $\Phi$
the density of the standard Gaussian distribution and we set
$a_{n}(x,\theta)=(2\delta)^{1/\alpha}u_{n}(x\Delta_{n})^{1/\alpha-1/2}$. Then

$\displaystyle
P_{n}f(x,\theta)=2\int_{\mathbb{R}_{+}}f(v)dv\int_{\mathbb{R}}g_{\alpha}\left(\frac{v-y}{a_{n}(x,\theta)}\right)\frac{1}{a_{n}(x,\theta)}\Phi\left(\frac{y}{\sqrt{2}u_{n}\sigma}\right)\frac{1}{\sqrt{2}u_{n}\sigma}dy.$

Using successively a change of variables and Fubini’s Theorem yields

$\displaystyle P_{n}f(x,\theta)$
$\displaystyle=2\int_{\mathbb{R}}\Phi(z)dz\int_{\mathbb{R}_{+}}f(v)g_{\alpha}\left(\frac{v-\sqrt{2}u_{n}\sigma
z}{a_{n}(x,\theta)}\right)\frac{1}{a_{n}(x,\theta)}dv.$

Recalling that $f=0$ on $[-\eta,\eta]$, we split the previous integral in two
parts $P_{n}f(x,\theta)=I_{n}^{1}+I_{n}^{2}$ where

$\displaystyle
I_{n}^{1}=2\int_{\\{|z]>\frac{\eta}{2\sqrt{2}u_{n}\sigma}\\}}\Phi(z)dz\int_{\mathbb{R}_{+}}f(v)g_{\alpha}\left(\frac{v-\sqrt{2}u_{n}\sigma
z}{a_{n}(x,\theta)}\right)\frac{1}{a_{n}(x,\theta)}dv,$ $\displaystyle
I_{n}^{2}=2\int_{\\{|z]\leq\frac{\eta}{2\sqrt{2}u_{n}\sigma}\\}}\Phi(z)dz\int_{\\{v>\eta\\}}f(v)g_{\alpha}\left(\frac{v-\sqrt{2}u_{n}\sigma
z}{a_{n}(x,\theta)}\right)\frac{1}{a_{n}(x,\theta)}dv.$

By assumption, $u_{n}^{2}=\overline{u}_{n}/\ln(1/\Delta_{n})$ with
$\overline{u}_{n}\rightarrow 0$, so we deduce immediately, since $f$ is
bounded, that

$|I_{n}^{1}|\leq C\exp(-\frac{\eta^{2}}{16u_{n}^{2}\sigma^{2}})\leq
C\Delta_{n}^{C^{\prime}/\overline{u}_{n}},$ (5.16)

where $C$ and $C^{\prime}$ do not depend on $\theta$. Turning to $I_{n}^{2}$,
we observe that for $|z|\leq\frac{\eta}{2\sqrt{2}u_{n}\sigma}$ and $v>\eta$,

$\frac{v-\sqrt{2}u_{n}\sigma
z}{a_{n}(x,\theta)}\geq\frac{\eta}{2a_{n}(x,\theta)}>0.$

Using the asymptotic expansion of $g_{\alpha}$ (see Sato [18]), we have

$\forall
y>0\quad|g_{\alpha}(y)-\frac{c_{\alpha}}{2y^{\alpha+1}}|\leq\frac{C}{y^{2\alpha+1}},\;\text{with}\;c_{\alpha}\;\text{
given by \eqref{eq:c-alpha}}$

and we deduce

$g_{\alpha}\left(\frac{v-\sqrt{2}u_{n}\sigma
z}{a_{n}(x,\theta)}\right)=\frac{c_{\alpha}a_{n}(x,\theta)^{\alpha+1}}{2v^{\alpha+1}}\frac{1}{(1-\sqrt{2}\sigma
u_{n}z/v)^{\alpha+1}}(1+R^{0}_{n}(z,v)),$

with

$|R^{0}_{n}(z,v)|\leq
C\frac{a_{n}(x,\theta)^{\alpha}}{v^{\alpha}}\frac{1}{(1-\sqrt{2}\sigma
u_{n}z/v)^{\alpha}}\leq C\frac{a_{n}(x,\theta)^{\alpha}}{v^{\alpha}},$

where we used $|\sqrt{2}\sigma u_{n}z/v|\leq 1/2$ for the last inequality.
Moreover, we have the first order expansion

$\frac{1}{(1-\sqrt{2}\sigma
u_{n}z/v)^{\alpha+1}}=1+R_{n}^{1}(z,v),\;\text{with}\;|R_{n}^{1}(z,v)|\leq
Cu_{n}z/v.$

Reporting these results in $I_{n}^{2}$, we obtain

$\displaystyle I_{n}^{2}=$
$\displaystyle\int_{\\{|z]\leq\frac{\eta}{2\sqrt{2}u_{n}\sigma}\\}}\Phi(z)dz\int_{\\{v>\eta\\}}f(v)\frac{c_{\alpha}a_{n}(x,\theta)^{\alpha}}{v^{\alpha+1}}(1+R_{n}^{1}(z,v))(1+R^{0}_{n}(z,v))dv$
$\displaystyle=$ $\displaystyle
c_{\alpha}a_{n}(x,\theta)^{\alpha}\left(\int_{\\{v>\eta\\}}\frac{f(v)}{v^{\alpha+1}}dv+\int_{\\{|z]>\frac{\eta}{2\sqrt{2}u_{n}\sigma}\\}}\Phi(z)dz\int_{\\{v>\eta\\}}\frac{f(v)}{v^{\alpha+1}}dv\right.$
$\displaystyle\left.+\int_{\\{|z]\leq\frac{\eta}{2\sqrt{2}u_{n}\sigma}\\}}\Phi(z)dz\int_{\\{v>\eta\\}}\frac{f(v)}{v^{\alpha+1}}[R_{n}^{1}(z,v)(1+R^{0}_{n}(z,v))+R^{0}_{n}(z,v)]dv\right).$

This finally gives

$I_{n}^{2}=c_{\alpha}a_{n}(x,\theta)^{\alpha}\left(\int_{\\{v>\eta\\}}\frac{f(v)}{v^{\alpha+1}}dv+R_{n}\right),$
(5.17)

with

$|R_{n}|\leq
C(\Delta_{n}^{C^{\prime}/\overline{u}_{n}}+u_{n}+u_{n}^{\alpha}(x\Delta_{n})^{1-\alpha/2})\leq
C(1+\sqrt{x})(\Delta_{n}^{C^{\prime}/\overline{u}_{n}}+u_{n}).$

From (5.16) and (5.17), we deduce the result of Lemma 4.2.

####  5.1.4 Proof of Proposition 4.2

The proof consists first to replace, in the expression of $F_{n}(\theta_{0})$,
$\rho_{2i}^{n}$ by $\overline{\rho}_{2i}^{n}$ (defined by (4.1)) by using
Lemma 4.1 and next to study the convergence in distribution of

$A_{n}(\theta_{0})\frac{1}{n}\sum_{i=0}^{\lfloor\frac{n}{2}\rfloor-1}[f(\overline{\rho}_{2i}^{n})-P_{n}f(X_{2i\Delta_{n}},\theta_{0})].$

We set

$\zeta_{2i}^{n}=A_{n}(\theta_{0})\frac{1}{n}[f(\overline{\rho}_{2i}^{n})-P_{n}f(X_{2i\Delta_{n}},\theta_{0})],$
(5.18)

hence $\zeta_{2i}^{n}=(\zeta_{2i}^{k,n})_{1\leq k\leq 3}$ with

$\displaystyle\zeta_{2i}^{1,n}$
$\displaystyle=\frac{1}{u_{n}^{2}\sqrt{n}}[f_{1}(\overline{\rho}_{2i}^{n})-P_{n}f_{1}(X_{2i\Delta_{n}},\theta_{0})],$
$\displaystyle\zeta_{2i}^{k,n}$
$\displaystyle=\frac{1}{u_{n}^{\alpha_{0}/2}\Delta_{n}^{1/2-\alpha_{0}/4}\sqrt{n}}[f_{k}(\overline{\rho}_{2i}^{n})-P_{n}f_{k}(X_{2i\Delta_{n}},\theta_{0})],\quad\text{for}\quad
k\in\\{2,3\\}.$

1\.  Case $n\Delta_{n}=1$. We first remark that for any $0<p\leq 1$ we have
the convergence in probability

$\frac{1}{n}\sum_{i=0}^{\lfloor\frac{n}{2}\rfloor-1}X_{2i\Delta_{n}}^{p}\xrightarrow[n\to\infty]{\mathbb{P}}\frac{1}{2}\int_{0}^{1}X_{t}^{p}dt.$

Moreover from Proposition 4.1 and recalling $a_{0}>\sigma_{0}^{2}$, we have
for $1\leq p\leq 2$

$\displaystyle\mathbb{E}\left(\frac{1}{n}\sum_{i=0}^{\lfloor\frac{n}{2}\rfloor-1}\frac{1}{X_{i\Delta_{n}}^{p}}\right)\leq\sup_{t\geq
0}\mathbb{E}(\frac{1}{X_{t}^{p}})<+\infty,$ (5.19)

and using $1/(X_{i\Delta_{n}}X_{v})\leq
1/(2X_{i\Delta_{n}}^{2})+1/(2X_{v}^{2})$

$\displaystyle\mathbb{E}\left(\frac{1}{n}\sum_{i=0}^{\lfloor\frac{n}{2}\rfloor-1}\frac{1}{\Delta_{n}}\mathbb{E}_{i}\int_{i\Delta_{n}}^{(i+2)\Delta_{n}}\frac{dv}{X_{i\Delta_{n}}X_{v}}\right)\leq\sup_{t\geq
0}\mathbb{E}(\frac{1}{X_{t}^{2}})<+\infty.$ (5.20)

Next from Lemma 4.1 and choosing $\epsilon>0$ such that
$1/\alpha_{0}-\epsilon>1/2$ and $1/\alpha_{0}-\epsilon>1-\alpha_{0}/4$, which
is possible since $1<\alpha_{0}<2$, we have the convergence in probability

$A_{n}(\theta_{0})\frac{1}{n}\sum_{i=0}^{\lfloor\frac{n}{2}\rfloor-1}|\mathbb{E}_{2i}(f(\rho_{2i}^{n})-f(\overline{\rho}_{2i}^{n}))|\to
0.$

Observing that for $f$ bounded
$\mathbb{E}_{2i}|f(\overline{\rho}_{2i}^{n})-P_{n}f(X_{2i\Delta_{n}},\theta_{0})|^{2}\leq
C\mathbb{E}_{2i}|f(\overline{\rho}_{2i}^{n})-P_{n}f(X_{2i\Delta_{n}},\theta_{0})|^{p}$
with $p<\alpha_{0}$, we check with Lemma 4.1

$\displaystyle\frac{n}{u_{n}^{4}}\frac{1}{n^{2}}$
$\displaystyle\sum_{i=0}^{\lfloor\frac{n}{2}\rfloor-1}\mathbb{E}_{2i}|f_{1}(\rho_{2i}^{n})-f_{1}(\overline{\rho}_{2i}^{n})|^{2}\to
0,$ $\displaystyle\frac{n}{u_{n}^{\alpha_{0}}\Delta_{n}^{1-\alpha_{0}/2}}$
$\displaystyle\frac{1}{n^{2}}\sum_{i=0}^{\lfloor\frac{n}{2}\rfloor-1}\mathbb{E}_{2i}|f_{k}(\rho_{2i}^{n})-f_{k}(\overline{\rho}_{2i}^{n})|^{2}\to
0\quad k\in\\{2,3\\}.$

Consequently, we have the convergence in probability

$\displaystyle
A_{n}(\theta_{0})\frac{1}{n}\sum_{i=0}^{\lfloor\frac{n}{2}\rfloor-1}[f(\rho_{2i}^{n})-f(\overline{\rho}_{2i}^{n})]\to
0,$ (5.21)

and the proof of Proposition 4.2 reduces to show the stable convergence

$A_{n}(\theta_{0})\frac{1}{n}\sum_{i=0}^{\lfloor\frac{n}{2}\rfloor-1}[f(\overline{\rho}_{2i}^{n})-P_{n}f(X_{2i\Delta_{n}},\theta_{0})]\xRightarrow[n\to\infty]{\mathcal{L}-s}\Sigma(\theta_{0})^{1/2}\mathcal{N},$

for $\mathcal{N}$ a standard Gaussian variable independent of
$\Sigma(\theta_{0})$ defined by (3.5). According to Theorem IX.7.26 in Jacod
and Shiryaev [9], it is sufficient to show the following convergences in
probability (with $\zeta_{2i}^{n}$ defined by (5.18))

$\displaystyle\sum_{i=0}^{\lfloor\frac{n}{2}\rfloor-1}\mathbb{E}_{2i}(\zeta_{2i}^{n})\to
0,$ (5.22)
$\displaystyle\sum_{i=0}^{\lfloor\frac{n}{2}\rfloor-1}\mathbb{E}_{2i}(|\zeta_{2i}^{k,n}|^{p})\to
0\quad\text{for some}\;p>2,\quad k=1,2,3,$ (5.23)
$\displaystyle\sum_{i=0}^{\lfloor\frac{n}{2}\rfloor-1}\mathbb{E}_{2i}(\zeta_{2i}^{k,n}\zeta_{2i}^{l,n})\to\Sigma_{k,l}(\theta_{0})\quad\text{for}\quad
k,l=1,2,3,$ (5.24)
$\displaystyle\sum_{i=0}^{\lfloor\frac{n}{2}\rfloor-1}\mathbb{E}_{2i}(\zeta_{2i}^{k,n}(M_{2(i+1)\Delta_{n}}-M_{2i\Delta_{n}}))\to
0\quad\text{for}\quad k=1,2,3,$ (5.25)

where the last convergence holds for any square integrable martingale $M$. The
convergence (5.22) is immediate since
$\mathbb{E}_{2i}f(\overline{\rho}_{2i}^{n})=P_{n}f(X_{2i\Delta_{n}},\theta_{0})$
and (5.25) is addressed in Lemma 5.1 below.

We deduce (5.23) from Lemma 4.2 for $k=2$ and $k=3$. For $k=1$, we recall that
using the notation (4.1) with $f_{1}(x)=\cos(x)$

$P_{n}f_{1}(X_{2i\Delta_{n}},\theta_{0})=\mathbb{E}_{2i}f_{1}(\overline{\rho}_{2i}^{n})=\exp(-u_{n}^{2}\sigma_{0}^{2}-2\delta_{0}X_{2i\Delta_{n}}^{1-1/\alpha_{0}}u_{n}^{\alpha_{0}}\Delta_{n}^{1-1/\alpha_{0}}),$
(5.26)

and

$\mathbb{E}_{2i}f_{1}(2\overline{\rho}_{2i}^{n})=\exp(-4u_{n}^{2}\sigma_{0}^{2}-2^{1+\alpha_{0}}\delta_{0}X_{2i\Delta_{n}}^{1-1/\alpha_{0}}u_{n}^{\alpha_{0}}\Delta_{n}^{1-1/\alpha_{0}}).$

Using $\cos^{2}(a)=(1+\cos(2a))/2$, we deduce

$\mathbb{E}_{2i}(f_{1}(\overline{\rho}_{2i}^{n})^{2})=(1+\mathbb{E}_{2i}f_{1}(2\overline{\rho}_{2i}^{n}))/2,$

and for $q\geq 2$, since $f_{1}$ is bounded

$\mathbb{E}_{2i}(|f_{1}(\overline{\rho}_{2i}^{n})-P_{n}f_{1}(X_{2i\Delta_{n}},\theta_{0})|^{q})\leq\mathbb{E}_{2i}(f_{1}(\rho_{2i}^{n})-\mathbb{E}_{2i}f_{1}(\overline{\rho}_{2i}^{n}))^{2}=\frac{\mathbb{E}_{2i}f_{1}(2\overline{\rho}_{2i}^{n})-2\mathbb{E}^{2}_{2i}f_{1}(\overline{\rho}_{2i}^{n})+1}{2}.$
(5.27)

We now expand $\mathbb{E}_{2i}f_{1}(\overline{\rho}_{2i}^{n})$ and
$\mathbb{E}_{2i}f_{1}(2\overline{\rho}_{2i}^{n})$ around 0 and get that

$\mathbb{E}_{2i}f_{1}(2\overline{\rho}_{2i}^{n})-2\mathbb{E}^{2}_{2i}f_{1}(\overline{\rho}_{2i}^{n})+1=4u_{n}^{4}\sigma_{0}^{4}+R_{n}(x,\theta_{0}),\;\text{with}\;|R_{n}(x,\theta_{0})|\leq
C(u_{n}^{6}+u_{n}^{\alpha_{0}}\Delta_{n}^{1-\alpha_{0}/2})(1+x).$ (5.28)

This gives (5.23) for $k=1$.

We now prove (5.24). From (5.27) and (5.28) we have

$\displaystyle\sum_{i=0}^{\lfloor\frac{n}{2}\rfloor-1}\mathbb{E}_{2i}(\zeta_{2i}^{1,n})^{2}=\frac{1}{n}\frac{1}{u_{n}^{4}}\sum_{i=0}^{\lfloor\frac{n}{2}\rfloor-1}\frac{\mathbb{E}_{2i}f_{1}(2\overline{\rho}_{2i}^{n})-2\mathbb{E}^{2}_{2i}f_{1}(\overline{\rho}_{2i}^{n})+1}{2}\to\sigma_{0}^{4}.$

We then apply Lemma 4.2 to $f_{k}$, $f_{l}$ and $f_{k}f_{l}$ with
$k,l\in\\{2,3\\}$. It yields

$\displaystyle\mathbb{E}_{2i}$
$\displaystyle[f_{k}(\overline{\rho}_{2i}^{n})-P_{n}f_{k}(X_{2i\Delta_{n}},\theta_{0})][f_{l}(\overline{\rho}_{2i}^{n})-P_{n}f_{l}(X_{2i\Delta_{n}},\theta_{0})]$
$\displaystyle=P_{n}(f_{k}f_{l})(X_{2i\Delta_{n}},\theta_{0})-P_{n}f_{k}(X_{2i\Delta_{n}},\theta_{0})P_{n}f_{l}(X_{2i\Delta_{n}},\theta_{0})$
$\displaystyle=u_{n}^{\alpha_{0}}\Delta_{n}^{1-\alpha_{0}/2}c_{\alpha_{0}}\delta_{0}X_{2i\Delta_{n}}^{1-\alpha_{0}/2}\int\frac{f_{k}f_{l}(v)}{|v|^{\alpha_{0}+1}}dv+u_{n}^{\alpha_{0}}\Delta_{n}^{1-\alpha_{0}/2}\varepsilon_{n}(\theta_{0},X_{2i\Delta_{n}}),$

where

$|\varepsilon_{n}(\theta_{0},X_{2i\Delta_{n}})|\leq(1+X_{2i\Delta_{n}})\varepsilon_{n},\quad\text{with}\quad\varepsilon_{n}\rightarrow
0.$

Consequently, we deduce for $k,l\in\\{2,3\\}$

$\displaystyle\sum_{i=0}^{\lfloor\frac{n}{2}\rfloor-1}\mathbb{E}_{2i}(\zeta_{2i}^{k,n}\zeta_{2i}^{l,n})\to\frac{1}{2}c_{\alpha_{0}}\delta_{0}\int_{0}^{1}X_{s}^{1-\alpha_{0}/2}ds\int\frac{f_{k}f_{l}(v)}{|v|^{\alpha_{0}+1}}dv.$

We finally address the cross terms
$\sum_{i=0}^{\lfloor\frac{n}{2}\rfloor-1}\mathbb{E}_{2i}(\zeta_{2i}^{1,n}\zeta_{2i}^{k,n})$
for $k\in\\{2,3\\}$. We have

$\displaystyle\mathbb{E}_{2i}$
$\displaystyle[(f_{1}(\overline{\rho}_{2i}^{n})-P_{n}f_{1}(X_{2i\Delta_{n}},\theta_{0})][f_{k}(\overline{\rho}_{2i}^{n})-P_{n}f_{k}(X_{2i\Delta_{n}},\theta_{0}))]$
$\displaystyle=P_{n}(f_{1}f_{k})(X_{2i\Delta_{n}},\theta_{0})-P_{n}f_{1}(X_{2i\Delta_{n}},\theta_{0})P_{n}f_{k}(X_{2i\Delta_{n}},\theta_{0}).$

Using again Lemma 4.2 for $f_{1}f_{k}$ and $f_{k}$ with $k\in\\{2,3\\}$ and
observing that $P_{n}f_{1}(X_{2i\Delta_{n}},\theta_{0})$ is bounded we have

$|P_{n}(f_{1}f_{k})(X_{2i\Delta_{n}},\theta_{0})-P_{n}f_{1}(X_{2i\Delta_{n}},\theta_{0})P_{n}f_{k}(X_{2i\Delta_{n}},\theta_{0})|\leq
Cu_{n}^{\alpha_{0}}\Delta_{n}^{1-\alpha_{0}/2}(1+X_{2i\Delta_{n}}),$

and we obtain for $k\in\\{2,3\\}$,

$\sum_{i=0}^{\lfloor\frac{n}{2}\rfloor-1}\mathbb{E}_{2i}(\zeta_{2i}^{1,n}\zeta_{2i}^{k,n})\to
0.$

We finally conclude that

$\sum_{i=0}^{\lfloor\frac{n}{2}\rfloor-1}\mathbb{E}_{2i}(\zeta_{2i}^{n}(\zeta_{2i}^{n})^{T})\to\Sigma(\theta_{0}).$

We have proved equation (5.24).

We check easily that $\Sigma(\theta_{0})$ is definite positive and we finish
the proof of Proposition 4.2 in the case $n\Delta_{n}=1$ with Lemma 5.1 below.

######  Lemma 5.1.

Let $M$ be a square-integrable martingale. Let $f_{1}$ and $f_{2}$ be even
bounded functions. We assume moreover that $f_{1}$ is $\mathcal{C}^{1}$ with
bounded derivative, $f_{2}=0$ on $[-\eta,\eta]$, for $\eta>0$ and that
$\Delta_{n}=1/n$ and $u_{n}=1/[\ln(1/\Delta_{n})]^{p}$ with $p>1/2$. Then we
have the convergence in probability (where $\overline{\rho}_{2i}^{n}$ is
defined by (4.1) and
$\Delta_{2i}^{2,n}M=M_{2(i+1)\Delta_{n}}-M_{2i\Delta_{n}}$)

$(i)\;\frac{1}{u_{n}^{2}\sqrt{n}}\sum_{i=0}^{\lfloor\frac{n}{2}\rfloor-1}\mathbb{E}_{2i}[(f_{1}(\overline{\rho}_{2i}^{n})-P_{n}f_{1}(X_{2i\Delta_{n}},\theta_{0}))\Delta_{2i}^{2,n}M]\to
0,$
$(ii)\;\frac{1}{u_{n}^{\alpha_{0}/2}\Delta_{n}^{1/2-\alpha_{0}/4}\sqrt{n}}\sum_{i=0}^{\lfloor\frac{n}{2}\rfloor-1}\mathbb{E}_{2i}[(f_{2}(\overline{\rho}_{2i}^{n})-P_{n}f_{2}(X_{2i\Delta_{n}},\theta_{0}))\Delta_{2i}^{2,n}M]\to
0.$

###### Proof.

From Theorem 2.2.15 in [8], we only consider the cases where $M$ is the
Brownian Motion $B$ or $M$ is orthogonal to $B$.

(i) Proceeding as in the proof of (5.9), we separate the Brownian component
and the pure-jump component
$\overline{\rho}_{2i}^{n}=\overline{\rho}_{2i}^{1,n}+\overline{\rho}_{2i}^{2,n}$
and since $f_{1}$ and $f_{1}^{\prime}$ are bounded, for $1\leq p<\alpha_{0}$

$\mathbb{E}_{2i}|f_{1}(\overline{\rho}_{2i}^{n})-f_{1}(\overline{\rho}_{2i}^{1,n})|^{2}\leq
C_{p}u_{n}^{p}\Delta_{n}^{p(1/\alpha_{0}-1/2)}X_{2i\Delta_{n}}^{p(1/\alpha_{0}-1/2)}.$

Choosing $p$ close to $\alpha_{0}$ and using Proposition 4.1 (ii), it yields
for $\epsilon>0$

$\mathbb{E}|f_{1}(\overline{\rho}_{2i}^{n})-f_{1}(\overline{\rho}_{2i}^{1,n})|^{2}\leq
C_{\epsilon}u_{n}^{\alpha_{0}-\epsilon}\Delta_{n}^{1-\alpha_{0}/2-\epsilon}.$

Setting

$S_{n}^{1}=\frac{1}{u_{n}^{2}\sqrt{n}}\sum_{i=0}^{\lfloor\frac{n}{2}\rfloor-1}\mathbb{E}_{2i}[(f_{1}(\overline{\rho}_{2i}^{n})-f_{1}(\overline{\rho}_{2i}^{1,n}))\Delta_{2i}^{2,n}M],$

we deduce from Cauchy-Schwarz inequality and the previous bound with
$\epsilon$ such that $1-\alpha_{0}/2-\epsilon>0$

$\displaystyle\mathbb{E}|S_{n}^{1}|$
$\displaystyle\leq\frac{C_{\epsilon}}{u_{n}^{2}\sqrt{n}}\left(\sum_{i=0}^{\lfloor\frac{n}{2}\rfloor-1}u_{n}^{\alpha_{0}-\epsilon}\Delta_{n}^{1-\alpha_{0}/2-\epsilon}\sum_{i=0}^{\lfloor\frac{n}{2}\rfloor-1}\mathbb{E}(\Delta_{2i}^{2,n}M)^{2}\right)^{1/2}\to
0,$

observing that the second sum is bounded since $M$ is a square-integrable
martingale. So the proof of (i) reduces to $S_{n}^{2}\to 0$ where

$S_{n}^{2}=\frac{1}{u_{n}^{2}\sqrt{n}}\sum_{i=0}^{\lfloor\frac{n}{2}\rfloor-1}\mathbb{E}_{2i}[f_{1}(\overline{\rho}_{2i}^{1,n})\Delta_{2i}^{2,n}M].$

If $M=B$, using that $f_{1}$ is even and recalling that
$\overline{\rho}_{2i}^{1,n}=\frac{\sigma_{0}u_{n}}{\sqrt{\Delta_{n}}}\Delta^{s,n}_{2i}B$,
an explicit calculus gives
$\mathbb{E}_{2i}[f_{1}(\overline{\rho}_{2i}^{1,n})\Delta_{2i}^{2,n}M]=0$. If
$M$ is a square integrable martingale orthogonal to $B$, we also have
$\mathbb{E}_{2i}[f_{1}(\overline{\rho}_{2i}^{1,n})\Delta_{2i}^{2,n}M]=0$ from
the representation Theorem and Itô’s Formula.

(ii) We follow the proof of Lemma 8 in [7]. We first assume that $M=B$. From
Hölder’s inequality with $p>1$, we have

$|\mathbb{E}_{2i}[f_{2}(\overline{\rho}_{2i}^{n})\Delta_{2i}^{2,n}B]|\leq
C\mathbb{E}^{1/p}_{2i}(f_{2}(\overline{\rho}_{2i}^{n}))^{p}\sqrt{\Delta_{n}}\leq
C\Delta_{n}^{(1-\alpha_{0}/2)/p}\sqrt{\Delta_{n}}(1+X_{2i\Delta_{n}}),$

where for the second inequality we applied Lemma 4.2 to $f_{2}^{p}$. We deduce
then (ii) in that case choosing $p<2$.

We now assume that $M$ is orthogonal to $B$. From the representation Theorem
(we refer to Chapter III.4 of [9]), we have for some predictable process
$(G_{t})$

$M_{t}=\int_{0}^{t}\int
G_{s}\tilde{N}(ds,dz)\;\text{with}\;\mathbb{E}\int_{0}^{1}\int
G_{s}^{2}\overline{N}(ds,dz)<\infty,$

and also with $(h_{t})$ and $(H_{t})$ predictable and $(H_{t})$ bounded (since
$f_{2}$ is bounded)

$f_{2}(\overline{\rho}_{2i}^{n})-P_{n}f_{2}(X_{2i\Delta_{n}},\theta_{0})=\int_{2i\Delta_{n}}^{2(i+1)\Delta_{n}}h_{s}dB_{s}+\int_{2i\Delta_{n}}^{2(i+1)\Delta_{n}}\int
H_{s}\tilde{N}(ds,dz).$

We then deduce from Itô’s Formula

$\mathbb{E}_{2i}[(f_{2}(\overline{\rho}_{2i}^{n})-P_{n}f_{2}(X_{2i\Delta_{n}},\theta_{0}))\Delta_{2i}^{2,n}M]=\mathbb{E}_{2i}\int_{2i\Delta_{n}}^{2(i+1)\Delta_{n}}\int
H_{s}G_{s}\overline{N}(ds,dz).$

With the decomposition $G_{s}=G_{s}{\bf 1}_{|G_{s}|>\epsilon_{n}}+G_{s}{\bf
1}_{|G_{s}|\leq\epsilon_{n}}$, the proof reduces to study the convergence of

$\displaystyle
T^{1}_{n}=\frac{1}{u_{n}^{\alpha_{0}/2}\Delta_{n}^{1/2-\alpha_{0}/4}\sqrt{n}}\sum_{i=0}^{\lfloor\frac{n}{2}\rfloor-1}\mathbb{E}_{2i}\int_{2i\Delta_{n}}^{2(i+1)\Delta_{n}}\int
H_{s}G_{s}{\bf 1}_{|G_{s}|>\epsilon_{n}}\overline{N}(ds,dz),$ $\displaystyle
T^{2}_{n}=\frac{1}{u_{n}^{\alpha_{0}/2}\Delta_{n}^{1/2-\alpha_{0}/4}\sqrt{n}}\sum_{i=0}^{\lfloor\frac{n}{2}\rfloor-1}\mathbb{E}_{2i}\int_{2i\Delta_{n}}^{2(i+1)\Delta_{n}}\int
H_{s}G_{s}{\bf 1}_{|G_{s}|\leq\epsilon_{n}}\overline{N}(ds,dz).$

Since $H$ is bounded, we have immediately

$\displaystyle\mathbb{E}|T^{1}_{n}|$
$\displaystyle\leq\frac{C}{u_{n}^{\alpha_{0}/2}\Delta_{n}^{1/2-\alpha_{0}/4}\sqrt{n}}\sum_{i=0}^{\lfloor\frac{n}{2}\rfloor-1}\mathbb{E}\int_{2i\Delta_{n}}^{2(i+1)\Delta_{n}}\int\frac{G^{2}_{s}}{\epsilon_{n}}\overline{N}(ds,dz),$
$\displaystyle\leq\frac{C}{\epsilon_{n}u_{n}^{\alpha_{0}/2}\Delta_{n}^{1/2-\alpha_{0}/4}\sqrt{n}}\mathbb{E}\int_{0}^{1}\int
G_{s}^{2}\overline{N}(ds,dz)\leq\frac{C}{\epsilon_{n}u_{n}^{\alpha_{0}/2}\Delta_{n}^{1/2-\alpha_{0}/4}\sqrt{n}}.$
(5.29)

With $\epsilon_{n}=\Delta_{n}^{\alpha_{0}/8}$, we deduce
$\mathbb{E}|T_{n}^{1}|\to 0$.

We now turn to $T_{n}^{2}$. We first apply Lemma 4.2 to $f_{2}$ to obtain

$\displaystyle\mathbb{E}_{2i}\int_{2i\Delta_{n}}^{2(i+1)\Delta_{n}}\int
H^{2}_{s}\overline{N}(ds,dz)\leq$
$\displaystyle\mathbb{E}_{2i}(f_{2}(\overline{\rho}_{2i}^{n})-P_{n}f_{2}(X_{2i\Delta_{n}},\theta_{0}))^{2}\leq
Cu_{n}^{\alpha_{0}}\Delta_{n}^{1-\alpha_{0}/2}(1+X_{2i\Delta_{n}}).$

From Cauchy-Schwarz inequality it yields

$\displaystyle|\mathbb{E}_{2i}$
$\displaystyle\int_{2i\Delta_{n}}^{2(i+1)\Delta_{n}}\int H_{s}G_{s}{\bf
1}_{|G_{s}|\leq\epsilon_{n}}\overline{N}(ds,dz)|\leq$ $\displaystyle
Cu_{n}^{\alpha_{0}/2}\Delta_{n}^{1/2-\alpha_{0}/4}(1+\sqrt{X_{2i\Delta_{n}}})\mathbb{E}^{1/2}_{2i}\int_{2i\Delta_{n}}^{2(i+1)\Delta_{n}}\int
G^{2}_{s}{\bf 1}_{|G_{s}|\leq\epsilon_{n}}\overline{N}(ds,dz),$

and applying once again Cauchy-Schwarz inequality

$\displaystyle\mathbb{E}|T_{n}^{2}|\leq
C\left(\mathbb{E}\frac{1}{n}\sum_{i=0}^{\lfloor\frac{n}{2}\rfloor-1}(1+X_{2i\Delta_{n}})\times\mathbb{E}\sum_{i=0}^{\lfloor\frac{n}{2}\rfloor-1}\int_{2i\Delta_{n}}^{2(i+1)\Delta_{n}}\int
G^{2}_{s}{\bf 1}_{|G_{s}|\leq\epsilon_{n}}\overline{N}(ds,dz)\right)^{1/2}.$

The first expectation is bounded from Proposition 4.1, consequently

$\displaystyle\mathbb{E}|T_{n}^{2}|\leq C\mathbb{E}^{1/2}\int_{0}^{1}\int
G^{2}_{s}{\bf 1}_{|G_{s}|\leq\epsilon_{n}}\overline{N}(ds,dz).$ (5.30)

As $\epsilon_{n}$ goes to zero, we conclude $\mathbb{E}|T_{n}^{2}|\to 0$ by
dominated convergence and the proof of (ii) is finished. ∎

2\.  Case $n\Delta_{n}\to+\infty$. The proof follows essentially the same
lines as in the previous case and we only indicate the main changes. The
bounds (5.19) and (5.20) still hold. From the ergodic Theorem (if $b_{0}>0$,
the process $(X_{t})$ is geometrically ergodic, see [15]), we have the
convergence in probability for $0<p\leq 1$

$\frac{1}{n\Delta_{n}}\int_{0}^{n\Delta_{n}}X_{t}^{p}dt\xrightarrow[n\to\infty]{}\int_{0}^{+\infty}x^{p}\pi_{0}(dx),$

since $\int_{0}^{+\infty}x\pi_{0}(dx)<+\infty$. Moreover we know that
$\mathbb{E}(X_{t})\rightarrow a_{0}/b_{0}$ (see [4]), then
$\sup_{t}\mathbb{E}(X_{t})<+\infty$ and combining this with Proposition 4.1
(iv) we can prove

$\mathbb{E}|\frac{1}{n}\sum_{i=0}^{\lfloor\frac{n}{2}\rfloor-1}X_{2i\Delta_{n}}^{p}-\frac{1}{2n\Delta_{n}}\int_{0}^{n\Delta_{n}}X_{t}^{p}dt|\xrightarrow[n\to\infty]{}0.$

We deduce the convergence in probability

$\frac{1}{n}\sum_{i=0}^{\lfloor\frac{n}{2}\rfloor-1}X_{2i\Delta_{n}}^{p}\xrightarrow[n\to\infty]{}\frac{1}{2}\int_{0}^{+\infty}x^{p}\pi_{0}(dx).$
(5.31)

Now replacing $\rho_{2i}^{n}$ by $\overline{\rho}_{2i}^{n}$, we obtain (5.21)
under the additional assumption
$\sqrt{n}\Delta_{n}^{1/\alpha+\alpha/4-1/2-\epsilon_{0}}\rightarrow 0$. It
remains to prove the convergence in distribution of

$A_{n}(\theta_{0})\frac{1}{n}\sum_{i=0}^{\lfloor\frac{n}{2}\rfloor-1}[f(\overline{\rho}_{2i}^{n})-P_{n}f(X_{2i\Delta_{n}},\theta_{0})].$

We check (5.22), (5.23), (5.24) (with $\overline{\Sigma}$) by proceeding as in
the case $n\Delta_{n}$ fixed but using (5.31) and
$\sup_{t}\mathbb{E}(X_{t})<+\infty$ to conclude.

####  5.1.5 Proof of Proposition 4.3

By definition of $F_{n}(\theta)$, we have

$A_{n}(\theta)\nabla_{\theta}F_{n}(\theta)\Lambda_{n}(\theta)=-\frac{1}{n}\sum_{i=0}^{\lfloor
n/2\rfloor-1}A_{n}(\theta)\nabla_{\theta}P_{n}f(X_{2i\Delta_{n}},\theta)\Lambda_{n}(\theta).$

We start by computing

$\nabla_{\theta}P_{n}f(x,\theta)=\left(\begin{array}[]{ccc}\partial_{\sigma^{2}}P_{n}f_{1}(x,\theta)&\partial_{\delta}P_{n}f_{1}(x,\theta)&\partial_{\alpha}P_{n}f_{1}(x,\theta)\\\
\partial_{\sigma^{2}}P_{n}f_{2}(x,\theta)&\partial_{\delta}P_{n}f_{2}(x,\theta)&\partial_{\alpha}P_{n}f_{2}(x,\theta)\\\
\partial_{\sigma^{2}}P_{n}f_{3}(x,\theta)&\partial_{\delta}P_{n}f_{3}(x,\theta)&\partial_{\alpha}P_{n}f_{3}(x,\theta)\end{array}\right).$

As $P_{n}f_{1}(x,\theta)=\exp(-u_{n}^{2}\sigma^{2}-2\delta
x^{1-\alpha/2}u_{n}^{\alpha}\Delta_{n}^{1-\alpha/2})$, we have

$\displaystyle\partial_{\sigma^{2}}P_{n}f_{1}(x,\theta)$
$\displaystyle=-u_{n}^{2}P_{n}f_{1}(x,\theta),$
$\displaystyle\partial_{\delta}P_{n}f_{1}(x,\theta)$
$\displaystyle=-2x^{1-\alpha/2}u_{n}^{\alpha}\Delta_{n}^{1-\alpha/2}P_{n}f_{1}(x,\theta),$
(5.32) $\displaystyle\partial_{\alpha}P_{n}f_{1}(x,\theta)$
$\displaystyle=-2\delta
x^{1-\alpha/2}u_{n}^{\alpha}\Delta_{n}^{1-\alpha/2}\ln(u_{n}/\sqrt{x\Delta_{n}})P_{n}f_{1}(x,\theta).$

Next observing that
$\nabla_{\theta}P_{n}f_{2}(x,\theta)=-\nabla_{\theta}P_{n}K(x,\theta)$ (the
same connection holds for $f_{3}$ replacing $K(x)$ by $K(2x)$), we evaluate
$\nabla_{\theta}P_{n}K(x,\theta)$. We recall that from (3.2)

$P_{n}K(x,\theta)=\mathbb{E}K(\sqrt{2}u_{n}\sigma B_{1}+\frac{(2\delta
x)^{1/\alpha}}{\sqrt{x}}u_{n}\Delta_{n}^{1/\alpha-1/2}S^{\alpha}_{1}),$

and we denote by $g_{n,\theta,x}$ the density of $\sqrt{2}u_{n}\sigma
B_{1}+\frac{(2\delta
x)^{1/\alpha}}{\sqrt{x}}u_{n}\Delta_{n}^{1/\alpha-1/2}S^{\alpha}_{1}$. We
obtain the following result where $\mathcal{F}f$ is the Fourier transform of
$f$, defined by $\mathcal{F}f(y)=\int_{\mathbb{R}}e^{-iyz}f(z)dz$.

######  Lemma 5.2.

Let $K$ be defined by (3). Then we have

$\displaystyle\partial_{\sigma^{2}}P_{n}K(x,\theta)$
$\displaystyle=u_{n}^{2}P_{n}K^{\prime\prime}(x,\theta),$
$\displaystyle\partial_{\delta}P_{n}K(x,\theta)$
$\displaystyle=-x^{1-\alpha/2}u_{n}^{\alpha}\Delta_{n}^{1-\alpha/2}\frac{2}{2\pi}\int\mathcal{F}K(y)|y|^{\alpha}\mathcal{F}g_{n,\theta,x}(y)dy,$
$\displaystyle\partial_{\alpha}P_{n}K(x,\theta)$ $\displaystyle=-\delta
x^{1-\alpha/2}u_{n}^{\alpha}\Delta_{n}^{1-\alpha/2}\left(\frac{2}{2\pi}\int\mathcal{F}K(y)|y|^{\alpha}\ln(|y|)\mathcal{F}g_{n,\theta,x}(y)dy\right.$
$\displaystyle\left.+(\ln(\frac{u_{n}}{\sqrt{\Delta_{n}}})-\ln(\sqrt{x}))\frac{2}{2\pi}\int\mathcal{F}K(y)|y|^{\alpha}\mathcal{F}g_{n,\theta,x}(y)dy\right).$

###### Proof.

Since $K$ is compactly supported and even, its Fourier transform is a Schwartz
function and $\mathcal{F}\mathcal{F}K=2\pi K$. We also have
$-y^{2}\mathcal{F}K(y)=\mathcal{F}K^{\prime\prime}(y)$. With the previous
notation it yields

$\displaystyle\nabla_{\theta}P_{n}K(x,\theta)$
$\displaystyle=\nabla_{\theta}\int_{\mathbb{R}}K(y)g_{n,\theta,x}(y)dy=\frac{1}{2\pi}\nabla_{\theta}\int_{\mathbb{R}}(\mathcal{F}\mathcal{F}K)(y)g_{n,\theta,x}(y)dy$
$\displaystyle=$
$\displaystyle\frac{1}{2\pi}\nabla_{\theta}\int_{\mathbb{R}}\mathcal{F}K(y)\mathcal{F}g_{n,\theta,x}(y)dy=\frac{1}{2\pi}\int_{\mathbb{R}}\mathcal{F}K(y)\nabla_{\theta}\mathcal{F}g_{n,\theta,x}(y)dy,$

where we successively used duality and Lebesgue’s Theorem. By definition of
$g_{n,\theta,x}$, we have

$\mathcal{F}g_{n,\theta,x}(y)=\exp(-u_{n}^{2}\sigma^{2}y^{2}-2|y|^{\alpha}\delta
x^{1-\alpha/2}u_{n}^{\alpha}\Delta_{n}^{1-\alpha/2}).$ (5.33)

The computation of $\nabla_{\theta}\mathcal{F}g_{n,\theta,x}$ is
straightforward and the expression of $\nabla_{\theta}P_{n}K$ follows. ∎

Coming back to $\nabla_{\theta}P_{n}f_{2}(x,\theta)$ we obtain the following
result.

######  Lemma 5.3.

We assume that $u_{n}[\ln(1/\Delta_{n})]^{1/2}\to 0$. Then we have $\forall
x>0$

$\displaystyle(i)$
$\displaystyle\;\left|\frac{1}{u_{n}^{\alpha/2}\Delta_{n}^{1/2-\alpha/4}}\partial_{\sigma^{2}}P_{n}f_{2}(x,\theta)\right|\leq\varepsilon_{n}^{1}(x,\theta),$
$\displaystyle(ii)$
$\displaystyle\;\frac{1}{u_{n}^{\alpha}\Delta_{n}^{1-\alpha/2}}\partial_{\delta}P_{n}f_{2}(x,\theta)=x^{1-\alpha/2}\psi(\alpha)+x^{1-\alpha/2}\varepsilon_{n}^{2}(x,\theta),$
$\displaystyle(iii)$
$\displaystyle\frac{1}{u_{n}^{\alpha}\Delta_{n}^{1-\alpha/2}}\left(\partial_{\alpha}P_{n}f_{2}(x,\theta)-\delta\ln(\frac{u_{n}}{\sqrt{\Delta_{n}}})\partial_{\delta}P_{n}f_{2}(x,\theta)\right)=\delta
x^{1-\alpha/2}[\partial_{\alpha}\psi(\alpha)-\ln(\sqrt{x})\psi(\alpha)]$
$\displaystyle\quad\quad+x^{1-\alpha/2}\varepsilon_{n}^{3}(x,\theta)-x^{1-\alpha/2}\ln(\sqrt{x})\varepsilon_{n}^{2}(x,\theta),$

where
$\psi(\alpha)=c_{\alpha}\int\frac{f_{2}(y)}{|y|^{\alpha+1}}dy=c_{\alpha}\int\frac{1-K(y)}{|y|^{\alpha+1}}dy$
and for any compact subset $A\subset\Theta$

$\sup_{\theta\in
A}(|\varepsilon_{n}^{1}(x,\theta)|+|\varepsilon_{n}^{2}(x,\theta)|+|\varepsilon_{n}^{3}(x,\theta)|)\leq
Cu_{n}(1+\sqrt{x}).$

The same result holds for $f_{3}$ replacing $\psi(\alpha)$ by
$2^{\alpha}\psi(\alpha)$.

###### Proof.

We first prove that

$\frac{2}{2\pi}\int\mathcal{F}K(y)|y|^{\alpha}dy=c_{\alpha}\int\frac{(1-K(y))}{|y|^{\alpha+1}}dy=\psi(\alpha).$
(5.34)

By definition of $c_{\alpha}$ we have
$2|y|^{\alpha}=c_{\alpha}\int_{\mathbb{R}}\frac{1-\cos(uy)}{|u|^{\alpha+1}}du.$
Then for $\epsilon>0$

$\displaystyle\frac{2}{2\pi}\int\mathcal{F}K(y)|y|^{\alpha}dy$
$\displaystyle=\frac{c_{\alpha}}{2\pi}\int\mathcal{F}K(y)\left(\int_{\mathbb{R}}\frac{1-\cos(uy)}{|u|^{\alpha+1}}du\right)dy$
$\displaystyle=\frac{c_{\alpha}}{2\pi}\int\mathcal{F}K(y)\left(\int_{\mathbb{R}}\frac{1-\cos(uy)}{|u|^{\alpha+1}}({\bf
1}_{|u|\leq\epsilon}+{\bf 1}_{|u|>\epsilon})du\right)dy.$

Since $1-\cos(uy)=h(uy)u^{2}y^{2}$ with $h$ bounded, we can apply Fubini’s
Theorem

$\displaystyle\frac{2}{2\pi}\int\mathcal{F}K(y)|y|^{\alpha}dy=$
$\displaystyle\frac{c_{\alpha}}{2\pi}\int_{\mathbb{R}}\frac{u^{2}}{|u|^{\alpha+1}}{\bf
1}_{|u|\leq\epsilon}\left(\int\mathcal{F}K(y)h(uy)y^{2}dy\right)du$
$\displaystyle+\frac{c_{\alpha}}{2\pi}\int_{\mathbb{R}}\frac{1}{|u|^{\alpha+1}}{\bf
1}_{|u|>\epsilon}\left(\int\mathcal{F}K(y)(1-\cos(uy))dy\right)du.$

Moreover, we have

$\int\mathcal{F}K(y)dy=2\pi K(0)=2\pi,\;\
\int\mathcal{F}K(y)\cos(uy)dy=\mathcal{F}(\mathcal{F}K)(y)=2\pi K(u).$

Letting $\epsilon$ go to zero, we obtain (5.34).

Next from (5.33), we have

$\sup_{{\theta}\in A}|\mathcal{F}g_{n,\theta,x}(y)-1|\leq
Cu_{n}(1+y^{2})(1+\sqrt{x}),$

and consequently

$\displaystyle\sup_{\theta\in
A}|\frac{2}{2\pi}\int\mathcal{F}K(y)|y|^{\alpha}\mathcal{F}g_{n,\theta,x}(y)dy-\psi(\alpha)|\leq
Cu_{n}(1+\sqrt{x}),$ $\displaystyle\sup_{\theta\in
A}|\frac{2}{2\pi}\int\mathcal{F}K(y)|y|^{\alpha}\ln(|y|)\mathcal{F}g_{n,\theta,x}(y)dy-\partial_{\alpha}\psi(\alpha)|\leq
Cu_{n}(1+\sqrt{x}).$

Since $f_{2}=1-K$, we obtain (i) combining Lemma 5.2 with Lemma 4.2 (applied
to $K^{\prime\prime}$), (ii) and (iii) follow from Lemma 5.2, (5.34) and the
previous uniform bounds. ∎

With this background, we can finish the proof of Proposition 4.3. An explicit
calculus leads to
$A_{n}(\theta)\nabla_{\theta}P_{n}f(x,\theta)\Lambda_{n}(\theta)=W_{n}(x,\theta)$
with

$W_{n}=\left(\begin{array}[]{ccc}\frac{1}{u_{n}^{2}}\partial_{\sigma^{2}}P_{n}f_{1}&\frac{v_{n}}{u_{n}^{2}}\partial_{\delta}P_{n}f_{1}&\frac{v_{n}}{u_{n}^{2}}[\partial_{\alpha}P_{n}f_{1}-\delta\ln(\frac{u_{n}}{\sqrt{\Delta_{n}}})\partial_{\delta}P_{n}f_{1}]\\\
v_{n}\partial_{\sigma^{2}}P_{n}f_{2}&v_{n}^{2}\partial_{\delta}P_{n}f_{2}&v_{n}^{2}[\partial_{\alpha}P_{n}f_{2}-\delta\ln(\frac{u_{n}}{\sqrt{\Delta_{n}}})\partial_{\delta}P_{n}f_{2}]\\\
v_{n}\partial_{\sigma^{2}}P_{n}f_{3}&v_{n}^{2}\partial_{\delta}P_{n}f_{3}&v_{n}^{2}[\partial_{\alpha}P_{n}f_{3}-\delta\ln(\frac{u_{n}}{\sqrt{\Delta_{n}}})\partial_{\delta}P_{n}f_{3}]\end{array}\right),\quad
v_{n}=\frac{1}{u_{n}^{\alpha/2}\Delta_{n}^{1/2-\alpha/4}}.$

1\. Case $n\Delta_{n}=1$. We check the uniform convergences in probability for
$\alpha\in[\underline{a},\overline{a}]$, with $1<\underline{a}<\overline{a}<2$

$\sup_{\alpha\in[\underline{a},\overline{a}]}|\frac{1}{n}\sum_{i=0}^{\lfloor
n/2\rfloor-1}X_{2i\Delta_{n}}^{1-\alpha/2}-\frac{1}{2}\int_{0}^{1}X_{t}^{1-\alpha/2}dt|\to
0,$ (5.35)
$\sup_{\alpha\in[\underline{a},\overline{a}]}|\frac{1}{n}\sum_{i=0}^{\lfloor
n/2\rfloor-1}X_{2i\Delta_{n}}^{1-\alpha/2}\ln\sqrt{X_{2i\Delta_{n}}}-\frac{1}{2}\int_{0}^{1}X_{t}^{1-\alpha/2}\ln\sqrt{X_{t}}dt|\to
0.$ (5.36)

We just detail (5.36). Let $h(\alpha,x)=x^{1-\alpha/2}\ln(\sqrt{x})$. We have
for $\epsilon>0$,

$\forall
x>0,\;|h^{\prime}(\alpha,x)|\leq\frac{1}{x^{\alpha/2}}(1+|\ln(x)|)\leq
C(1+\frac{1}{x^{\alpha/2+\epsilon}}),$

and from Taylor’s formula for $x,y>0$

$|h(\alpha,y)-h(\alpha,x)|\leq
C|y-x|\int_{0}^{1}(1+\frac{1}{((1-c)x+cy)^{\alpha/2+\epsilon}})dc.$

Choosing $\epsilon$ such that $\overline{a}/2+\epsilon<1$, we deduce

$\sup_{\alpha\in[\underline{a},\overline{a}]}|h(\alpha,y)-h(\alpha,x)|\leq
C|y-x|(1+\frac{1}{x}).$

From Proposition 4.1 it yields

$\mathbb{E}_{2i}\sup_{t\in[2i\Delta_{n},2(i+1)\Delta_{n}]}\sup_{\alpha\in[\underline{a},\overline{a}]}|h(\alpha,X_{t})-h(\alpha,X_{2i\Delta_{n}})|\leq
C\sqrt{\Delta_{n}}(1+X_{2i\Delta_{n}}+\frac{1}{X_{2i\Delta_{n}}}),$

and we obtain (5.36).

Furthermore, from Lebesgue’s Theorem $\alpha\mapsto
I(\alpha,X)=\int_{0}^{1}X_{t}^{1-\alpha/2}dt$ admits a derivative and
$\partial_{\alpha}I(\alpha,X)=-\int_{0}^{1}X_{t}^{1-\alpha/2}\ln(\sqrt{X_{t}})dt$,
so in the expression of $W(\theta)$ (3.6), we have

$\partial_{\alpha}(\psi(\alpha)I(\alpha,X))=\partial_{\alpha}\psi(\alpha)\int_{0}^{1}X_{t}^{1-\alpha/2}dt-\psi(\alpha)\int_{0}^{1}X_{t}^{1-\alpha/2}\ln(\sqrt{X_{t}})dt.$

Consequently, combining (5.1.5) and Lemma 5.3 with (5.35) and (5.36), we
obtain

$\sup_{\theta\in A}|\frac{1}{n}\sum_{i=0}^{\lfloor
n/2\rfloor-1}W_{n}(X_{2i\Delta_{n}},\theta)+W(\theta)|\to 0,$

and we finally conclude from
$A_{n}(\theta)\nabla_{\theta}F_{n}(\theta)\Lambda_{n}(\theta)=-\frac{1}{n}\sum_{i=0}^{\lfloor
n/2\rfloor-1}W_{n}(X_{2i\Delta_{n}},\theta)$. We see easily that $W$ is non-
singular since

$\det
W(\theta)=\frac{1}{8}\delta\psi(\alpha)^{2}I(\alpha,X)^{2}2^{\alpha}\ln(2)\neq
0.$

2\. Case $n\Delta_{n}\rightarrow\infty$. When $n\Delta_{n}\to+\infty$, we
first prove the convergences

$\sup_{\alpha\in[\underline{a},\overline{a}]}|\frac{1}{n}\sum_{i=0}^{\lfloor
n/2\rfloor-1}X_{2i\Delta_{n}}^{1-\alpha/2}-\frac{1}{2}\int_{0}^{+\infty}x^{1-\alpha/2}\pi_{0}(dx)|\to
0,$ (5.37)
$\sup_{\alpha\in[\underline{a},\overline{a}]}|\frac{1}{n}\sum_{i=0}^{\lfloor
n/2\rfloor-1}X_{2i\Delta_{n}}^{1-\alpha/2}\ln\sqrt{X_{2i\Delta_{n}}}-\frac{1}{2}\int_{0}^{+\infty}x^{1-\alpha/2}\ln\sqrt{x}\pi_{0}(dx)|\to
0.$ (5.38)

We give the details for $\eqref{E:uni2ergo}$ and study separately the uniform
convergences in probability

$\sup_{\alpha\in[\underline{a},\overline{a}]}|\frac{1}{n}\sum_{i=0}^{\lfloor
n/2\rfloor-1}X_{2i\Delta_{n}}^{1-\alpha/2}\ln\sqrt{X_{2i\Delta_{n}}}-\frac{1}{2n\Delta_{n}}\int_{0}^{n\Delta_{n}}X_{t}^{1-\alpha/2}\ln\sqrt{X_{t}}dt|\to
0,$ (5.39)
$\sup_{\alpha\in[\underline{a},\overline{a}]}|\frac{1}{n\Delta_{n}}\int_{0}^{n\Delta_{n}}X_{t}^{1-\alpha/2}\ln\sqrt{X_{t}}dt-\int_{0}^{+\infty}x^{1-\alpha/2}\ln\sqrt{x}\pi_{0}(dx)|\to
0.$ (5.40)

The proof of (5.39) is similar to the one of (5.36) and we omit it. It remains
to check (5.40). We set
$H_{n}(\alpha)=\frac{1}{n\Delta_{n}}\int_{0}^{n\Delta_{n}}X_{t}^{1-\alpha/2}\ln\sqrt{X_{t}}dt$
and $H(\alpha)=\int_{0}^{+\infty}x^{1-\alpha/2}\ln\sqrt{x}\pi_{0}(dx)$. The
map $\alpha\to H_{n}(\alpha)$ is continuous on $[\underline{a},\overline{a}]$
and $H_{n}(\alpha)\to H(\alpha)$,
$\forall\alpha\in[\underline{a},\overline{a}]$. Moreover for
$\alpha_{1},\alpha_{2}\in[\underline{a},\overline{a}]$ and $x>0$, $\exists
c\in(\alpha_{1},\alpha_{2})$ such that

$\displaystyle|x^{1-\alpha_{1}/2}\ln(\sqrt{x})-x^{1-\alpha_{2}/2}\ln(\sqrt{x})|\leq|\alpha_{1}-\alpha_{2}|(\ln(x))^{2}x^{1-c/2}\leq
C|\alpha_{1}-\alpha_{2}|(1+x).$

With $\sup_{t}\mathbb{E}X_{t}<+\infty$, it follows that $\forall\delta>0$,
$\sup_{n}\mathbb{E}\sup_{|\alpha_{1}-\alpha_{2}|\leq\delta}|H_{n}(\alpha_{1})-H_{n}(\alpha_{2})|\leq
C\delta,$ so the sequence
$(H_{n}(\alpha)_{\alpha\in[\underline{a},\overline{a}]})$ is tight and we
obtain the uniform convergence (5.40).

Observing that
$\partial_{\alpha}\overline{I}(\alpha)=-\int_{0}^{\infty}x^{1-\alpha/2}\ln(\sqrt{x})\pi_{0}(dx)$
and using the same arguments as in the case $n\Delta_{n}$ fixed ,we obtain
from (5.1.5), Lemma 5.3, (5.37) and (5.38)

$\sup_{\theta\in A}|\frac{1}{n}\sum_{i=0}^{\lfloor
n/2\rfloor-1}W_{n}(X_{2i\Delta_{n}},\theta)+\overline{W}(\theta)|\to 0.$

This achieves the proof of Proposition 4.3.

###  5.2 Main results

####  5.2.1 Proof of Theorem 3.1 and Theorem 3.3

In both cases, $n\Delta_{n}$ fixed and $n\Delta_{n}\rightarrow+\infty$, we
apply Theorem A.2. in Mies and Podolskij [17], that extends the results of
Jacod and Sørensen [10]. We just detail the first case, where we have
strengthened condition (E.1) in [17] by assuming the stable convergence in
law. With our notation, the result of Theorem 3.1 is a consequence of the
following conditions:

(E.1)
$A_{n}(\theta_{0})F_{n}(\theta_{0})\xRightarrow[n\to\infty]{\mathcal{L}-s}\Sigma(\theta_{0})^{1/2}\mathcal{N}$,

(E.2) $\theta\to F_{n}(\theta)$ is $\mathcal{C}^{1}$ and for $r_{n}$ a
sequence of real numbers we have the convergences in probability

$\sup_{\theta\in\mathcal{B}_{r_{n}}(\theta_{0})}||A_{n}(\theta)\nabla_{\theta}F_{n}(\theta)\Lambda_{n}(\theta)-W(\theta)||\xrightarrow[n\to\infty]{}0,\;\sup_{\theta\in\mathcal{B}_{r_{n}}(\theta_{0})}\frac{||\Lambda_{n}(\theta)||\;||A_{n}(\theta)A_{n}(\theta_{0})^{-1}||}{r_{n}}\xrightarrow[n\to\infty]{}0,$
$\displaystyle\text{(E.3)}\;\sup_{\theta\in\mathcal{B}_{r_{n}}(\theta_{0})}(||A_{n}(\theta)A_{n}(\theta_{0})^{-1}-Id||+||\Lambda_{n}(\theta)\Lambda_{n}(\theta_{0})^{-1}-Id||+||W(\theta)-W(\theta_{0})||)\xrightarrow[n\to\infty]{}0,$

where $\mathcal{B}_{r_{n}}(\theta_{0})$ is the ball with center $\theta_{0}$
and radius $r_{n}$, $||\cdot||$ a matrix norm and $Id$ the identity matrix.
Condition (E.1) is proven in Proposition 4.2 and the first condition of (E.2)
is proven in Proposition 4.3. The second condition in (E.2) and condition
(E.3) are easily checked with $r_{n}=1/\ln(\Delta_{n})^{2}$.

####  5.2.2 Proof of Theorem 3.2

The existence is a consequence of Theorem 3.1. To prove the global uniqueness
result we consider the renormalized estimating function (restricted to
$\beta=(\sigma^{2},\delta)$ with $f=(f_{k})_{1\leq k\leq 2}$)

$\tilde{F}_{n}(\sigma^{2},\delta)=\tilde{A}_{n}(\alpha_{0})F_{n}(\sigma^{2},\delta,\alpha_{0}),\text{with}\;\tilde{A}_{n}(\alpha_{0})=diag\left(\frac{1}{u_{n}^{2}},\frac{1}{u_{n}^{\alpha_{0}}\Delta_{n}^{1-\alpha_{0}/2}}\right).$

From Proposition 4.2, we deduce $\tilde{F}_{n}(\sigma_{0}^{2},\delta_{0})\to
0$. Next, an explicit calculus gives

$\nabla_{\beta}\tilde{F}_{n}(\sigma^{2},\delta)=-\frac{1}{n}\sum_{i=0}^{\lfloor
n/2\rfloor-1}\left(\begin{array}[]{cc}\frac{1}{u_{n}^{2}}\partial_{\sigma^{2}}P_{n}f_{1}&\frac{1}{u_{n}^{2}}\partial_{\delta}P_{n}f_{1}\\\
\frac{1}{u_{n}^{\alpha_{0}}\Delta_{n}^{1-\alpha_{0}/2}}\partial_{\sigma^{2}}P_{n}f_{2}&\frac{1}{u_{n}^{\alpha_{0}}\Delta_{n}^{1-\alpha_{0}/2}}\partial_{\delta}P_{n}f_{2}\end{array}\right),$

and from Proposition 4.3, we have the convergence in probability for any
compact subset A of $(0,+\infty)\times(0,+\infty)$

$\sup_{(\sigma^{2},\delta)\in
A}\left|\left|\nabla\tilde{F}_{n}(\sigma^{2},\delta)-\nabla\tilde{F}(\sigma^{2},\delta)\right|\right|\to
0,\;\text{where}\;\tilde{F}(\sigma^{2},\delta)=\left(\begin{array}[]{c}\frac{1}{2}(\sigma^{2}-\sigma_{0}^{2})\\\
\frac{1}{2}\psi(\alpha_{0})\int_{0}^{1}X_{t}^{1-\alpha_{0}/2}dt(\delta_{0}-\delta)\end{array}\right).$

The global uniqueness result follows then from Theorem 2.7.a) of Jacod and
Sørensen [10], since $(\sigma_{0}^{2},\delta_{0})$ is the unique root of
$\tilde{F}(\sigma^{2},\delta)=0$.

Funding. This work was supported by french ANR reference ANR-21-CE40-0021.

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