On a projection least squares estimator for jump diffusion processes

Abstract

This paper deals with a projection least squares estimator of the drift function of a jump diffusion process X computed from multiple independent copies of X observed on [0, T]. Risk bounds are established on this estimator and on an associated adaptive estimator. Finally, some numerical experiments are provided.

Appendices

Appendix A. Proofs

1.1 A.1 Proof of Lemma 1

First, let us show that the symmetric matrix ${\varvec{\Psi }}_{m,\sigma }$ is positive semidefinite. Indeed, for any $y\in {\mathbb {R}}^m$,

$$\begin{aligned} y^*{\varvec{\Psi }}_{m,\sigma }y= & {} \frac{1}{NT}\sum _{j,\ell = 1}^{m}y_jy_{\ell } \sum _{i,k = 1}^{N}{\mathbb {E}}\left( \left( \int _{0}^{T}\varphi _j(X_{s}^{i})(\sigma (X_{s}^{i})dB_{s}^{i} + \gamma (X_{s}^{i})d{\mathfrak {Z}}_{s}^{i})\right) \right. \nonumber \\{} & {} \hspace{5cm}\left. \times \left( \int _{0}^{T}\varphi _{\ell }(X_{s}^{k})(\sigma (X_{s}^{k})dB_{s}^{k} + \gamma (X_{s}^{k})d{\mathfrak {Z}}_{s}^{k})\right) \right) \nonumber \\= & {} \frac{1}{NT}{\mathbb {E}}\left[ \left( \sum _{i = 1}^{N}\int _{0}^{T}\tau _y(X_{s}^{i})(\sigma (X_{s}^{i})dB_{s}^{i} + \gamma (X_{s}^{i})d{\mathfrak {Z}}_{s}^{i})\right) ^2\right] \geqslant 0 \quad \textrm{with}\quad \tau _y(.):=\sum _{j = 1}^{m}y_j\varphi _j(.). \end{aligned}$$

Moreover, by the isometry property of Itô’s integral (with respect to B), by the isometry type property of the stochastic integral with respect to ${\mathfrak {Z}}$, and since $\sigma$ and $\gamma$ are bounded, for every $j\in \{1,\dots ,m\}$,

$$\begin{aligned} y^*{\varvec{\Psi }}_{m,\sigma }y= & {} \frac{1}{T}\mathbb E\left[ \left( \int _{0}^{T}\tau _y(X_s)\sigma (X_s)dB_s\right) ^2\right] + \frac{1}{T}{\mathbb {E}}\left[ \left( \int _{0}^{T}\tau _y(X_s)\gamma (X_s)d\mathfrak Z_s\right) ^2\right] \nonumber \\= & {} \frac{1}{T}\int _{0}^{T}{\mathbb {E}}(\tau _y(X_s)^2\sigma (X_s)^2)ds +\frac{\lambda {\mathfrak {c}}_{\zeta ^2}}{T} \int _{0}^{T}\mathbb E(\tau _y(X_s)^2\gamma (X_s)^2)ds \nonumber \\\leqslant & {} (\Vert \sigma \Vert _{\infty }^{2} +\lambda \mathfrak c_{\zeta ^2}\Vert \gamma \Vert _{\infty }^{2}) \int _{-\infty }^{\infty } \left( \sum _{j = 1}^{m}y_j\varphi _j(x)\right) ^2f_T(x)dx = (\Vert \sigma \Vert _{\infty }^{2} +\lambda \mathfrak c_{\zeta ^2}\Vert \gamma \Vert _{\infty }^{2}) \Vert {\varvec{\Psi }}_{m}^{1/2}y\Vert _{2,m}^{2}. \end{aligned}$$

(5)

Therefore, since ${\varvec{\Psi }}_{m,\sigma }$ is positive semidefinite, and by Inequality (),

$$\begin{aligned} \textrm{trace}({\varvec{\Psi }}_{m}^{-1/2}{\varvec{\Psi }}_{m,\sigma }{\varvec{\Psi }}_{m}^{-1/2})\leqslant & {} m\Vert {\varvec{\Psi }}_{m}^{-1/2}{\varvec{\Psi }}_{m,\sigma }{\varvec{\Psi }}_{m}^{-1/2}\Vert _{\textrm{op}}\\= & {} m\cdot \sup \{y^*{\varvec{\Psi }}_{m,\sigma }y; y\in {\mathbb {R}}^m\text { and }\Vert {\varvec{\Psi }}_{m}^{1/2}y\Vert _{2,m} = 1\}\\\leqslant & {} (\Vert \sigma \Vert _{\infty }^{2} +\lambda \mathfrak c_{\zeta ^2}\Vert \gamma \Vert _{\infty }^{2})m. \end{aligned}$$

$\Box$

1.2 A.2 Proof of Theorem 1

The proof of Theorem 1 relies on the two following lemmas.

Lemma 2

There exists a constant $\mathfrak c_{2} > 0$, not depending on m and N, such that

$$\begin{aligned} {\mathbb {E}}(|\widehat{\textbf{E}}_{m}^{*}\widehat{\textbf{E}}_m|^2) \leqslant {\mathfrak {c}}_{2} \frac{mL(m)^2}{N^2}. \end{aligned}$$

Lemma 3

Consider the event

$$\begin{aligned} \Omega _m:= \left\{ \sup _{\tau \in {\mathcal {S}}_m} \left| \frac{\Vert \tau \Vert _{N}^{2}}{\Vert \tau \Vert _{f_T}^{2}} - 1\right| \leqslant \frac{1}{2} \right\} . \end{aligned}$$

Under Assumptions 1, there exists a constant ${\mathfrak {c}}_{3} > 0$, not depending on m and N, such that

$$\begin{aligned} {\mathbb {P}}(\Omega _{m}^{c}) \leqslant \frac{\mathfrak c_{3}}{N^7} \quad \text {and}\quad {\mathbb {P}}(\Lambda _{m}^{c}) \leqslant \frac{\mathfrak c_{3}}{N^7}. \end{aligned}$$

The proof of Lemma 2 is postponed to Subsubsection a.2.2, and the proof of Lemma 3 remains the same as the proof of Comte and Genon-Catalot (2020b), Lemma 6.1, because $(B^1,Z^1),\dots ,(B^N,Z^N)$ are independent.

1.2.1 A.2.1 Steps of the proof

First of all,

$$\begin{aligned} \Vert {{\widetilde{b}}}_m - b_I\Vert _{N}^{2}= & {} \Vert b_I\Vert _{N}^{2}{\textbf {1}}_{\Lambda _{m}^{c}} + \Vert {\widehat{b}}_m - b_I\Vert _{N}^{2}{{\textbf {1}}}_{\Lambda _m}\\= & {} \Vert b_I\Vert _{N}^{2}{{\textbf {1}}}_{\Lambda _{m}^{c}} + \Vert {\widehat{b}}_m - b_I\Vert _{N}^{2}{{\textbf {1}}}_{\Lambda _m\cap \Omega _m} + \Vert {\widehat{b}}_m - b_I\Vert _{N}^{2}{{\textbf {1}}}_{\Lambda _m\cap \Omega _{m}^{c}} =: U_1 + U_2 + U_3. \end{aligned}$$

Let us find suitable bounds on ${\mathbb {E}}(U_1)$, ${\mathbb {E}}(U_2)$ and ${\mathbb {E}}(U_3)$.

• Bound on ${\mathbb {E}}(U_1)$. By Cauchy-Schwarz’s inequality,

  $$\begin{aligned} {\mathbb {E}}(U_1)\leqslant & {} {\mathbb {E}}(\Vert b_I\Vert _{N}^{4})^{1/2}\mathbb P(\Lambda _{m}^{c})^{1/2} \leqslant \mathbb E\left( \frac{1}{T}\int _{0}^{T}b_I(X_t)^4dt\right) ^{1/2} {\mathbb {P}}(\Lambda _{m}^{c})^{1/2}\\\leqslant & {} {\mathfrak {c}}_1{\mathbb {P}}(\Lambda _{m}^{c})^{1/2}<\infty \quad \textrm{with}\quad {\mathfrak {c}}_1 =\left( \int _{-\infty }^{\infty }b_I(x)^4f_T(x)dx\right) ^{1/2} <\infty . \end{aligned}$$

• Bound on ${\mathbb {E}}(U_2)$. Let $\Pi _{N,m}(.)$ be the orthogonal projection from ${\mathbb {L}}^2(I,f_T(x)dx)$ onto $\mathcal S_m$ with respect to the empirical scalar product $\langle .,.\rangle _N$. Then,

  $$\begin{aligned} \Vert {\widehat{b}}_m - b_I\Vert _{N}^{2} = \Vert {\widehat{b}}_m -\Pi _{N,m}(b_I)\Vert _{N}^{2} + \min _{\tau \in {\mathcal {S}}_m}\Vert b_I -\tau \Vert _{N}^{2}. \end{aligned}$$

  (6)

  As in the proof of Comte and Genon-Catalot (2020b), Proposition 2.1, on $\Omega _m$,

  $$\begin{aligned} \Vert {\widehat{b}}_m -\Pi _{N,m}(b_I)\Vert _{N}^{2} = \widehat{\textbf{E}}_{m}^{*}{\widehat{\varvec{\Psi }}}_{m}^{-1}\widehat{\textbf{E}}_m \leqslant 2\widehat{\textbf{E}}_{m}^{*}{\varvec{\Psi }}_{m}^{-1}\widehat{\textbf{E}}_m. \end{aligned}$$

  So,

  $$\begin{aligned} {\mathbb {E}}(\Vert {\widehat{b}}_m -\Pi _{N,m}(b_I)\Vert _{N}^{2} {\textbf {1}}_{\Lambda _m\cap \Omega _m})\leqslant & {} 2\mathbb E\left( \sum _{j,\ell = 1}^{m}[\widehat{\textbf{E}}_m]_j[\widehat{\textbf{E}}_m]_{\ell } [{\varvec{\Psi }}_{m}^{-1}]_{j,\ell }\right) \\= & {} \frac{2}{NT}\sum _{j,\ell = 1}^{m} [{\varvec{\Psi }}_{m,\sigma }]_{j,\ell } [{\varvec{\Psi }}_{m}^{-1}]_{j,\ell } = \frac{2}{NT}\textrm{trace}({\varvec{\Psi }}_{m}^{-1/2} {\varvec{\Psi }}_{m,\sigma } {\varvec{\Psi }}_{m}^{-1/2}). \end{aligned}$$

  Then, by Equality (6) and Lemma 1,

  $$\begin{aligned} {\mathbb {E}}(U_2)\leqslant & {} {\mathbb {E}}\left( \min _{\tau \in \mathcal S_m}\Vert b_I -\tau \Vert _{N}^{2}\right) + \frac{2}{NT}\textrm{trace}({\varvec{\Psi }}_{m}^{-1/2} {\varvec{\Psi }}_{m,\sigma } {\varvec{\Psi }}_{m}^{-1/2})\\\leqslant & {} \min _{\tau \in {\mathcal {S}}_m}\Vert b_I -\tau \Vert _{f_T}^{2} + \frac{2m}{NT} (\Vert \sigma \Vert _{\infty }^{2} +\lambda \mathfrak c_{\zeta ^2}\Vert \gamma \Vert _{\infty }^{2}). \end{aligned}$$

• Bound on ${\mathbb {E}}(U_3)$. Since

  $$\begin{aligned} \Vert {\widehat{b}}_m -\Pi _{N,m}(b_I)\Vert _{N}^{2} = \widehat{\textbf{E}}_{m}^{*}{\widehat{\varvec{\Psi }}}_{m}^{-1}\widehat{\textbf{E}}_m, \end{aligned}$$

  by the definition of the event $\Lambda _m$, and by Lemma 2,

  $$\begin{aligned} {\mathbb {E}}(\Vert {\widehat{b}}_m -\Pi _{N,m}(b_I)\Vert _{N}^{2} {\textbf {1}}_{\Lambda _m\cap \Omega _{m}^{c}})\leqslant & {} \mathbb E(\Vert {\widehat{\varvec{\Psi }}}_{m}^{-1}\Vert _{\textrm{op}} |\widehat{\textbf{E}}_{m}^{*}\widehat{\textbf{E}}_m| {{\textbf {1}}}_{\Lambda _m\cap \Omega _{m}^{c}})\\\leqslant & {} \frac{{\mathfrak {c}}_T}{L(m)}\cdot \frac{NT}{\log (NT)} {\mathbb {E}}(|\widehat{\textbf{E}}_{m}^{*}\widehat{\textbf{E}}_m|^2)^{1/2} {\mathbb {P}}(\Omega _{m}^{c})^{1/2} \leqslant \frac{\mathfrak c_2m^{1/2}}{\log (NT)}{\mathbb {P}}(\Omega _{m}^{c})^{1/2}, \end{aligned}$$

  where the constant ${\mathfrak {c}}_2 > 0$ doesn’t depend on m and N. Moreover,

  $$\begin{aligned} \min _{\tau \in {\mathcal {S}}_m}\Vert \tau - b_I\Vert _{N}^{2} \leqslant \Vert b_I\Vert _{N}^{2} \quad \textrm{because}\quad 0\in {\mathcal {S}}_m, \end{aligned}$$

  and then,

  $$\begin{aligned} \Vert {\widehat{b}}_m - b_I\Vert _{N}^{2}= & {} \Vert {\widehat{b}}_m -\Pi _{N,m}(b_I)\Vert _{N}^{2} + \min _{\tau \in {\mathcal {S}}_m}\Vert \tau - b_I\Vert _{N}^{2}\\\leqslant & {} \Vert {\widehat{b}}_m -\Pi _{N,m}(b_I)\Vert _{N}^{2} +\Vert b_I\Vert _{N}^{2}. \end{aligned}$$

  Therefore,

  $$\begin{aligned} {\mathbb {E}}(U_3)\leqslant & {} {\mathbb {E}}(\Vert {\widehat{b}}_m -\Pi _{N,m}(b_I)\Vert _{N}^{2} {{\textbf {1}}}_{\Lambda _m\cap \Omega _{m}^{c}}) + {\mathbb {E}}(\Vert b_I\Vert _{N}^{2}{{\textbf {1}}}_{\Lambda _m\cap \Omega _{m}^{c}})\\\leqslant & {} \frac{{\mathfrak {c}}_2m^{1/2}}{\log (NT)} \mathbb P(\Omega _{m}^{c})^{1/2} + {\mathfrak {c}}_1\mathbb P(\Omega _{m}^{c})^{1/2}. \end{aligned}$$

So,

$$\begin{aligned} {\mathbb {E}}(\Vert {{\widetilde{b}}}_m - b_I\Vert _{N}^{2})\leqslant & {} \min _{\tau \in {\mathcal {S}}_m}\Vert b_I -\tau \Vert _{f_T}^{2}\\{} & {} \hspace{1cm} + \frac{2m}{NT}(\Vert \sigma \Vert _{\infty }^{2} + \lambda {\mathfrak {c}}_{\zeta ^2}\Vert \gamma \Vert _{\infty }^{2}) + {\mathfrak {c}}_2 \frac{\sqrt{m{\mathbb {P}}(\Omega _{m}^{c})}}{\log (NT)} + {\mathfrak {c}}_1( {\mathbb {P}}(\Lambda _{m}^{c})^{1/2} +{\mathbb {P}}(\Omega _{m}^{c})^{1/2}). \end{aligned}$$

Therefore, by Lemma 3, there exists a constant ${\mathfrak {c}}_3 > 0$, not depending on m and N, such that

$$\begin{aligned} {\mathbb {E}}(\Vert {{\widetilde{b}}}_m - b_I\Vert _{N}^{2}) \leqslant \min _{\tau \in {\mathcal {S}}_m}\Vert b_I -\tau \Vert _{f_T}^{2} + \mathfrak c_3\left( \frac{m}{NT} +\frac{1}{N}\right) . \end{aligned}$$

$\square$

1.2.2 A.2.2 Proof of Lemma 2

In the sequel, the quadratic variation of any piecewise continuous stochastic process $(\Gamma _t)_{t\in [0,T]}$ is denoted by $(\llbracket \Gamma \rrbracket _t)_{t\in [0,T]}$. First of all, note that since B and Z are independent, for every $j\in \{1,\dots ,m\}$,

$$\left[\kern-0.15em\left[{\int_{0}^{.} {\varphi _{j} } (X_{s} )(\sigma (X_{s} )dB_{s} + \gamma (X_{s} )d{\mathfrak {Z}}_{s} )} \right]\kern-0.15em\right] _{T} = \int_{0}^{T} {\varphi _{j} } (X_{s} )^{2} \sigma (X_{s} )^{2} ds + \int_{0}^{T} {\varphi _{j} } (X_{s} )^{2} \gamma (X_{s} )^{2} dZ_{s}^{{(2)}} = \int_{0}^{T} {\varphi _{j} } (X_{s} )^{2} (\sigma (X_{s} )^{2} + {\mathfrak{c}}_{{\zeta ^{2} }} \lambda \gamma (X_{s} )^{2} )ds + \int_{0}^{T} {\varphi _{j} } (X_{s} )^{2} \gamma (X_{s} )^{2} d{\mathfrak {Z}}_{s}^{{(2)}}$$

where, for every $t\in [0,T]$,

$$\begin{aligned} Z_{t}^{(2)}:=\llbracket {\mathfrak {Z}}\rrbracket _t = \sum _{n = 1}^{\nu _t}\zeta _{n}^{2} \quad \textrm{and}\quad {\mathfrak {Z}}_{t}^{(2)}:= Z_{t}^{(2)} -{\mathfrak {c}}_{\zeta ^2}\lambda t. \end{aligned}$$

By Jensen’s inequality and Burkholder-Davis-Gundy’s inequality (see Dellacherie and Meyer, 1980, p. 303), there exists a constant ${\mathfrak {c}}_1 > 0$, not depending on m and N, such that

$$ \begin{aligned} {\mathbb{E}}(|\widehat{{\mathbf{E}}}_{m}^{*} \widehat{{\mathbf{E}}}_{m} |^{2} )\leqslant m\sum\limits_{{j = 1}}^{m} {\mathbb{E}} ([\widehat{{\mathbf{E}}}_{m} ]_{j}^{4} ) & \leqslant\frac{{{\mathfrak{c}}_{1} m}}{{N^{4} T^{4} }}\sum\limits_{{j = 1}}^{m} {\mathbb{E}} \left( {\left[\kern-0.15em\left[ {\sum\limits_{{i = 1}}^{N} {\int_{0}^{.} {\varphi _{j} } } (X_{s}^{i} )(\sigma (X_{s}^{i} )dB_{s}^{i} + \gamma (X_{s}^{i} )d3_{s}^{i} )} \right]\kern-0.15em\right]_{T}^{2} } \right) \\ & \leqslant \frac{{2{\mathfrak{c}}_{1} m}}{{N^{2} T^{4} }}\sum\limits_{{j = 1}}^{m} {\mathbb{E}} \left( {\left[\kern-0.15em\left[ {\int\limits_{0}^{.} {\varphi _{j} } (X_{s} )(\sigma (X_{s} )dB_{s} + \gamma (X_{s} )d3_{s} )} \right]\kern-0.15em\right]_{T}^{2} } \right) \\ & \leqslant \frac{{4{\mathfrak{c}}_{1} m}}{{N^{2} T^{4} }}\sum\limits_{{j = 1}}^{m} \begin{gathered} \left[ {{\mathbb{E}}\left[ {\left( {\int_{0}^{T} {\varphi _{j} } (X_{s} )^{2} (\sigma (X_{s} )^{2} + {\mathfrak{c}}_{{\zeta ^{2} }} \lambda \gamma (X_{s} )^{2} )ds} \right)^{2} } \right]} \right] \hfill \\ + {\mathbb{E}}\left[ {\left( {\int_{0}^{T} {\varphi _{j} } (X_{s} )^{2} \gamma (X_{s} )^{2} d3_{s}^{{(2)}} } \right)^{2} } \right]. \hfill \\ \end{gathered} \\ \end{aligned} $$

By Jensen’s inequality,

$$\begin{aligned} \left( \int _{0}^{T}\varphi _j(X_s)^2(\sigma (X_s)^2 + \mathfrak c_{\zeta ^2}\lambda \gamma (X_s)^2)ds\right) ^2= & {} T^2\left( \int _{0}^{T}\varphi _j(X_s)^2(\sigma (X_s)^2 + {\mathfrak {c}}_{\zeta ^2}\lambda \gamma (X_s)^2)\frac{ds}{T}\right) ^2\\\leqslant & {} T\int _{0}^{T}\varphi _j(X_s)^4(\sigma (X_s)^2 + {\mathfrak {c}}_{\zeta ^2}\lambda \gamma (X_s)^2)^2ds. \end{aligned}$$

Moreover, since $\Vert {{\textbf {x}}}\Vert _{4,m}\leqslant \Vert {{\textbf {x}}}\Vert _{2,m}$ for every ${{\textbf {x}}}\in {\mathbb {R}}^d$,

$$\begin{aligned} \sup _{x\in I}\sum _{j = 1}^{m}\varphi _j(x)^4 = \sup _{x\in I}\Vert (\varphi _j(x))_j\Vert _{4,m}^{4}\leqslant \sup _{x\in I}\Vert (\varphi _j(x))_j\Vert _{2,m}^{4}\leqslant L(m)^2. \end{aligned}$$

So, by applying twice the Fubini–Tonelli theorem,

$$\begin{aligned}{} & {} \sum _{j = 1}^{m} \mathbb E\left[ \left( \int _{0}^{T}\varphi _j(X_s)^2(\sigma (X_s)^2 + {\mathfrak {c}}_{\zeta ^2}\lambda \gamma (X_s)^2)ds\right) ^2\right] \\{} & {} \leqslant 2T\sum _{j = 1}^{m}\int _{0}^{T} \mathbb E(\varphi _j(X_s)^4(\sigma (X_s)^4 + {\mathfrak {c}}_{\zeta ^2}^{2}\lambda ^2\gamma (X_s)^4))ds\\{} & {} = 2T^2\int _{-\infty }^{\infty } \underbrace{\left( \sum _{j = 1}^{m}\varphi _j(x)^4\right) }_{\leqslant L(m)^2} (\sigma (x)^4 +\mathfrak c_{\zeta ^2}^{2}\lambda ^2\gamma (x)^4)f_T(x)dx, \end{aligned}$$

and by the isometry type property of the stochastic integral with respect to ${\mathfrak {Z}}^{(2)}$,

$$\begin{aligned} \sum _{j = 1}^{m}{\mathbb {E}}\left[ \left( \int _{0}^{T}\varphi _j(X_s)^2 \gamma (X_s)^2d{\mathfrak {Z}}_{s}^{(2)}\right) ^2\right]= & {} \lambda {\mathfrak {c}}_{\zeta ^4} \sum _{j = 1}^{m}\int _{0}^{T}{\mathbb {E}}(\varphi _j(X_s)^4\gamma (X_s)^4)ds\\\leqslant & {} \lambda {\mathfrak {c}}_{\zeta ^4}TL(m)^2 \int _{-\infty }^{\infty }\gamma (x)^4f_T(x)dx. \end{aligned}$$

Therefore,

$$\begin{aligned} {\mathbb {E}}(|\widehat{\textbf{E}}_{m}^{*}\widehat{\textbf{E}}_m|^2) \leqslant \frac{{\mathfrak {c}}_2}{N^2T^2}mL(m)^2 \end{aligned}$$

with

$$\begin{aligned} {\mathfrak {c}}_2 = 8{\mathfrak {c}}_1\left( \int _{-\infty }^{\infty }(\sigma (x)^4 + \mathfrak c_{\zeta ^2}^{2}\lambda ^2\gamma (x)^4)f_T(x)dx + \lambda \mathfrak c_{\zeta ^4}\int _{-\infty }^{\infty }\gamma (x)^4f_T(x)dx \right) . \end{aligned}$$

$\square$

1.3 A.3 Proof of Theorem 2

Let us consider the events

$$\begin{aligned} \Omega _N:=\bigcap _{m\in {\mathcal {M}}_{N}^{+}}\Omega _m \quad \textrm{and}\quad \Xi _N:=\{{\mathcal {M}}_N\subset \widehat{\mathcal M}_N\subset {\mathcal {M}}_{N}^{+}\}, \end{aligned}$$

where

$$\begin{aligned} {\mathcal {M}}_{N}^{+}:=\left\{ m\in \{1,\dots ,N_T\}: \mathfrak c_{\varphi }^{2}m(\Vert {\varvec{\Psi }}_m^{-1}\Vert _{\textrm{op}}^{2}\vee 1) \leqslant 4{\mathfrak {d}}_T\frac{NT}{\log (NT)}\right\} , \end{aligned}$$

and let us recall that

$$\begin{aligned} {\mathfrak {d}}_T = \min \left\{ \frac{{\mathfrak {c}}_T}{2}, \frac{1}{64{\mathfrak {c}}_{\varphi }^{2}T(\Vert f_T\Vert _{\infty } +\sqrt{{\mathfrak {c}}_T/2}/(3{\mathfrak {c}}_{\varphi }))} \right\} . \end{aligned}$$

As a reminder, the sets $\widehat{{\mathcal {M}}}_N$ and ${\mathcal {M}}_N$ introduced in Sect. 4 are, respectively, defined by

$$\begin{aligned} \widehat{{\mathcal {M}}}_N =\left\{ m\in \{1,\dots ,N_T\}: \mathfrak c_{\varphi }^{2}m(\Vert {\widehat{\varvec{\Psi }}}_{m}^{-1}\Vert _{\textrm{op}}^{2}\vee 1) \leqslant {\mathfrak {d}}_T\frac{NT}{\log (NT)}\right\} \end{aligned}$$

and

$$\begin{aligned} {\mathcal {M}}_N =\left\{ m\in \{1,\dots ,N_T\}: \mathfrak c_{\varphi }^{2}m(\Vert {\varvec{\Psi }}_{m}^{-1}\Vert _{\textrm{op}}^{2}\vee 1) \leqslant \frac{{\mathfrak {d}}_T}{4}\cdot \frac{NT}{\log (NT)}\right\} . \end{aligned}$$

The proof of Theorem 2 relies on the three following lemmas.

Lemma 4

Under Assumptions 1, 2 and 3, there exists a constant ${\mathfrak {c}}_{4} > 0$, not depending on N, such that

$$\begin{aligned} {\mathbb {P}}(\Xi _{N}^{c})\leqslant \frac{\mathfrak c_{4}}{N^6}. \end{aligned}$$

Lemma 5

(Bernstein type inequality) Consider the empirical process

$$\begin{aligned} \nu _N(\tau ):= \frac{1}{NT}\sum _{i = 1}^{N}\int _{0}^{T}\tau (X_{s}^{i})( \sigma (X_{s}^{i})dB_{s}^{i} +\gamma (X_{s}^{i})d{\mathfrak {Z}}_{s}^{i}); \tau \in \mathcal S_1\cup \dots \cup {\mathcal {S}}_{N_T}. \end{aligned}$$

Under Assumption 3, for every $\xi ,v > 0$,

$$\begin{aligned} {\mathbb {P}}(\nu _N(\tau )\geqslant \xi ,\Vert \tau \Vert _{N}^{2}\leqslant v^2) \leqslant \exp \left[ -\frac{NT\xi ^2}{4[\mathfrak c_{5} (\Vert \sigma \Vert _{\infty }^{2} + \Vert \gamma \Vert _{\infty }^{2})v^2 +\xi \Vert \tau \Vert _{\infty ,I}\Vert \gamma \Vert _{\infty }]}\right] \end{aligned}$$

with

$$\begin{aligned} {\mathfrak {c}}_{6} = \frac{1}{2}\left[ 1\vee \int _{-\infty }^{\infty }e^{\mathfrak b|z|/2}\pi _{\lambda }(dz)\right] . \end{aligned}$$

Lemma 6

Under Assumptions 1 and 3, there exists a constant ${\mathfrak {c}}_{7} > 0$, not depending on N, such that for every $m\in {\mathcal {M}}_N$,

$$\begin{aligned} {\mathbb {E}}\left[ \left( \left[ \sup _{\tau \in \mathcal B_{m,m'}}\nu _n(\tau )\right] ^2 - p(m,{\widehat{m}})\right) _{+}{\textbf {1}}_{\Xi _N\cap \Omega _N}\right] \leqslant \frac{\mathfrak c_{8}}{NT} \end{aligned}$$

where, for every $m'\in {\mathcal {M}}_N$,

$$\begin{aligned} {\mathcal {B}}_{m,m'}:= \left\{ \tau \in {\mathcal {S}}_{m\wedge m'}:\Vert \tau \Vert _{f_T} = 1\right\} \quad \textrm{and}\quad p(m,m'):=\frac{{\mathfrak {c}}_{\text {cal}}}{8}\cdot \frac{m\vee m'}{NT}. \end{aligned}$$

The proof of Lemma 5 is postponed to Subsubsection A.3.2. Lemma 6 is a consequence of Lemma 5 thanks to the $\mathbb L_{f_T}^{2}$-${\mathbb {L}}^{\infty }$ chaining technique (see Comte, 2001, Proposition 4). Finally, the proof of Lemma 4 remains the same as the proof of Comte and Genon-Catalot (2020b), Eq. (6.17), because $(B^1,{\mathfrak {Z}}^1),\dots ,(B^N,{\mathfrak {Z}}^N)$ are independent.

1.3.1 A.3.1 Steps of the proof

First of all,

$$\begin{aligned} \Vert {\widehat{b}}_{{\widehat{m}}} - b_I\Vert _{N}^{2}= & {} \Vert {\widehat{b}}_{{\widehat{m}}} - b_I\Vert _{N}^{2}{{\textbf {1}}}_{\Xi _{N}^{c}} + \Vert {\widehat{b}}_{{\widehat{m}}} - b_I\Vert _{N}^{2}{{\textbf {1}}}_{\Xi _N} \nonumber \\=: & {} U_1 + U_2. \end{aligned}$$

(7)

Let us find suitable bounds on ${\mathbb {E}}(U_1)$ and $\mathbb E(U_2)$.

• Bound on ${\mathbb {E}}(U_1)$. Since

  $$\begin{aligned} \Vert {\widehat{b}}_{{\widehat{m}}} -\Pi _{N,{\widehat{m}}}(b_I)\Vert _{N}^{2} = \widehat{\textbf{E}}_{{\widehat{m}}}^{*}{\widehat{\varvec{\Psi }}}_{{\widehat{m}}}^{-1} \widehat{\textbf{E}}_{{\widehat{m}}}, \end{aligned}$$

  by the definition of $\widehat{{\mathcal {M}}}_N$, and by Lemma 2,

  $$\begin{aligned} {\mathbb {E}}(\Vert {\widehat{b}}_{{\widehat{m}}} -\Pi _{N,{\widehat{m}}}(b_I)\Vert _{N}^{2} {{\textbf {1}}}_{\Xi _{N}^{c}})\leqslant & {} \mathbb E(\Vert {\widehat{\varvec{\Psi }}}_{{\widehat{m}}}^{-1}\Vert _{\textrm{op}} |\widehat{\textbf{E}}_{NT}^{*}\widehat{\textbf{E}}_{NT}| {{\textbf {1}}}_{\Xi _{N}^{c}})\\\leqslant & {} \left[ {\mathfrak {d}}_T \frac{NT}{\log (NT)}\right] ^{1/2} {\mathbb {E}}(|\widehat{\textbf{E}}_{NT}^{*}\widehat{\textbf{E}}_{NT}|^2)^{1/2} {\mathbb {P}}(\Xi _{N}^{c})^{1/2} \leqslant \frac{\mathfrak c_1N}{\log (NT)}{\mathbb {P}}(\Xi _{N}^{c})^{1/2}, \end{aligned}$$

  where the constant ${\mathfrak {c}}_1 > 0$ does not depend on N. Then,

  $$\begin{aligned} {\mathbb {E}}(U_1)\leqslant & {} {\mathbb {E}}(\Vert {\widehat{b}}_{{\widehat{m}}} -\Pi _{N,{\widehat{m}}}(b_I)\Vert _{N}^{2} {{\textbf {1}}}_{\Xi _{N}^{c}}) + {\mathbb {E}}(\Vert b_I\Vert _{N}^{2}{{\textbf {1}}}_{\Xi _{N}^{c}})\\\leqslant & {} \frac{{\mathfrak {c}}_1N}{\log (NT)}\mathbb P(\Xi _{N}^{c})^{1/2} + {\mathfrak {c}}_2{\mathbb {P}}(\Xi _{N}^{c})^{1/2} \end{aligned}$$

  with

  $$\begin{aligned} {\mathfrak {c}}_2 =\left( \int _{-\infty }^{\infty }b_I(x)^4f_T(x)dx\right) ^{1/2}. \end{aligned}$$

  So, by Lemma 4, there exists a constant ${\mathfrak {c}}_3 > 0$, not depending on N, such that

  $$\begin{aligned} {\mathbb {E}}(U_1) \leqslant \frac{{\mathfrak {c}}_3}{N}. \end{aligned}$$

• Bound on ${\mathbb {E}}(U_2)$. Note that

  $$\begin{aligned} U_2= & {} \Vert {\widehat{b}}_{{\widehat{m}}} - b_I\Vert _{N}^{2}{\textbf {1}}_{\Xi _N\cap \Omega _{N}^{c}} + \Vert {\widehat{b}}_{{\widehat{m}}} - b_I\Vert _{N}^{2}{{\textbf {1}}}_{\Xi _N\cap \Omega _N}\\=: & {} U_{2,1} + U_{2,2}. \end{aligned}$$

  On the one hand, by Lemma 3, there exists a constant ${\mathfrak {c}}_4 > 0$, not depending on N, such that

  $$\begin{aligned} {\mathbb {P}}(\Xi _N\cap \Omega _{N}^{c}) \leqslant \sum _{m\in \mathcal M_{N}^{+}}{\mathbb {P}}(\Omega _{m}^{c}) \leqslant \frac{\mathfrak c_4}{N^6}. \end{aligned}$$

  Then, as for ${\mathbb {E}}(U_1)$, there exists a constant $\mathfrak c_5 > 0$, not depending on N, such that

  $$\begin{aligned} {\mathbb {E}}(U_{2,1}) \leqslant \frac{{\mathfrak {c}}_5}{N}. \end{aligned}$$

  On the other hand,

  $$\begin{aligned} \gamma _N(\tau ') -\gamma _N(\tau ) = \Vert \tau ' - b\Vert _{N}^{2} -\Vert \tau - b\Vert _{N}^{2} - 2\nu _N(\tau ' -\tau ) \end{aligned}$$

  for every $\tau ,\tau '\in {\mathcal {S}}_1\cup \dots \cup {\mathcal {S}}_{N_T}$. Moreover, since

  $$\begin{aligned} {\widehat{m}} = \arg \min _{m\in \widehat{{\mathcal {M}}}_N}\{-\Vert {\widehat{b}}_m\Vert _{N}^{2} + \textrm{pen}(m)\} = \arg \min _{m\in \widehat{\mathcal M}_N}\{\gamma _N({\widehat{b}}_m) + \textrm{pen}(m)\}, \end{aligned}$$

  for every $m\in \widehat{{\mathcal {M}}}_N$,

  $$\begin{aligned} \gamma _N({\widehat{b}}_{{\widehat{m}}}) + \textrm{pen}({\widehat{m}}) \leqslant \gamma _N({\widehat{b}}_m) + \textrm{pen}(m). \end{aligned}$$

  (8)

  On the event $\Xi _N =\{{\mathcal {M}}_N\subset \widehat{\mathcal M}_N\subset {\mathcal {M}}_{N}^{+}\}$, Inequality (8) remains true for every $m\in {\mathcal {M}}_N$. Then, on $\Xi _N$, for any $m\in {\mathcal {M}}_N$, since ${\mathcal {S}}_m +{\mathcal {S}}_{{\widehat{m}}}\subset \mathcal S_{m\vee {\widehat{m}}}$ under Assumption 2,

  $$\begin{aligned} \Vert {\widehat{b}}_{{\widehat{m}}} - b_I\Vert _{N}^{2}\leqslant & {} \Vert {\widehat{b}_m} - b_I\Vert _{N}^{2} + 2\Vert {\widehat{b}}_{{\widehat{m}}} -{\widehat{b}_{m}}\Vert _{f_T} \nu _N \left( \frac{{\widehat{b}}_{{\widehat{m}}} -{\widehat{b}}_m}{ \Vert {\widehat{b}}_{{\widehat{m}}} -{\widehat{b}}_m\Vert _{f_T}}\right) + \textrm{pen}(m) - \textrm{pen}({\widehat{m}})\\\leqslant & {} \Vert {\widehat{b}}_m - b_I\Vert _{N}^{2} + \frac{1}{8}\Vert {\widehat{b}}_{{\widehat{m}}} -{\widehat{b}}_m\Vert _{f_T}^{2}\\{} & {} \hspace{1.5cm} + 8\left( \left[ \sup _{\tau \in \mathcal B_{m,{\widehat{m}}}}|\nu _N(\tau )|\right] ^2 - p(m,{\widehat{m}})\right) _+ + \textrm{pen}(m) + 8p(m,{\widehat{m}}) - \textrm{pen}({\widehat{m}}). \end{aligned}$$

  Since $\Vert .\Vert _{f_T}^{2}{{\textbf {1}}}_{\Omega _N}\leqslant 2\Vert .\Vert _{N}^{2}{{\textbf {1}}}_{\Omega _N}$ on $\mathcal S_1\cup \dots \cup {\mathcal {S}}_{\max ({\mathcal {M}}_{N}^{+})}$, and since $8p(m,{\widehat{m}})\leqslant \textrm{pen}(m) + \textrm{pen}({\widehat{m}})$, on $\Xi _N\cap \Omega _N$,

  $$\begin{aligned} \Vert {\widehat{b}}_{{\widehat{m}}} - b_I\Vert _{N}^{2} \leqslant 3\Vert {\widehat{b}}_m - b_I\Vert _{N}^{2} + 4\textrm{pen}(m) + 16\left( \left[ \sup _\tau \in {\mathcal {B}}_m,{\widehat{m}}|\nu _N(\tau )|\right] ^2 - p(m,{\widehat{m}})\right) _+. \end{aligned}$$

  So, by Lemma 6,

  $$\begin{aligned} {\mathbb {E}}(U_{2,2})\leqslant & {} \min _{m\in {\mathcal {M}}_N} \{\mathbb E(3\Vert {\widehat{b}}_m - b_I\Vert _{N}^{2}{{\textbf {1}}}_{\Xi _N}) + 4\textrm{pen}(m)\} + \frac{16{\mathfrak {c}}_{6}}{NT}\\\leqslant & {} {\mathfrak {c}}_6\min _{m\in {\mathcal {M}}_N} \left\{ \inf _{\tau \in {\mathcal {S}}_m} \Vert \tau - b_I\Vert _{f_T}^{2} +\frac{m}{NT} \right\} + \frac{{\mathfrak {c}}_6}{N} \end{aligned}$$

  where ${\mathfrak {c}}_6 > 0$ is a deterministic constant not depending on N.

$\square$

1.3.2 A.2.3.2 Proof of Lemma 5

Consider $\tau \in {\mathcal {S}}_1\cup \dots \cup {\mathcal {S}}_{N_T}$ and, for any $i\in \{1,\dots ,N\}$, let $M^i(\tau ) = (M^i(\tau )_t)_{t\in [0,T]}$ be the martingale defined by

$$\begin{aligned} M^i(\tau )_t:= \int _{0}^{t}\tau (X_{s}^{i})(\sigma (X_{s}^{i})dB_{s}^{i} + \gamma (X_{s}^{i})d{\mathfrak {Z}}_{s}^{i}); \;\forall t\in [0,T]. \end{aligned}$$

Moreover, for every $\varepsilon > 0$, consider

$$\begin{aligned} Y_{\varepsilon }^{i}(\tau ):= \varepsilon M^i(\tau ) - A_{\varepsilon }^{i}(\tau ) - B_{\varepsilon }^{i}(\tau ), \end{aligned}$$

where $A_{\varepsilon }^{i}(\tau ) = (A_{\varepsilon }^{i}(\tau )_t)_{t\in [0,T]}$ and $B_{\varepsilon }^{i}(\tau ) = (B_{\varepsilon }^{i}(\tau )_t)_{t\in [0,T]}$ are the stochastic processes defined by

$$\begin{aligned} A_{\varepsilon }^{i}(\tau )_t:= & {} \frac{\varepsilon ^2}{2}\int _{0}^{t}\tau (X_{s}^{i})^2\sigma (X_{s}^{i})^2ds\\{} & {} \hspace{1cm} \textrm{and}\quad B_{\varepsilon }^{i}(\tau )_t:= \int _{0}^{t}\left[ \int _{-\infty }^{\infty }( e^{\varepsilon z\tau (X_{s}^{i})\gamma (X_{s}^{i})} - \varepsilon z\tau (X_{s}^{i})\gamma (X_{s}^{i}) - 1)\pi _{\lambda }(dz)\right] ds \end{aligned}$$

for every $t\in [0,T]$. The proof is dissected in three steps.

Step 1. Note that for any $i\in \{1,\dots ,N\}$ and $t\in [0,T]$,

$$\begin{aligned} |\tau (X_{t}^{i})\gamma (X_{t}^{i})|\leqslant \Vert \tau \Vert _{\infty ,I}\Vert \gamma \Vert _{\infty } \end{aligned}$$

and then, by Assumption 3,

$$\begin{aligned} {\mathbb {E}}\left( \int _{0}^{t}\int _{|z| > 1} |e^{\varepsilon z\tau (X_{s}^{i})\gamma (X_{s}^{i})} - 1|\pi _{\lambda }(dz)ds\right) <\infty \end{aligned}$$

for any $\varepsilon \in (0,\varepsilon ^*)$ with $\varepsilon ^* = ({\mathfrak {b}}\wedge 1)/(2\Vert \tau \Vert _{\infty ,I}\Vert \gamma \Vert _{\infty })$. So, $(\exp (Y_{\varepsilon }^{i}(\tau )_t))_{t\in [0,T]}$ is a local martingale by Applebaum (2009), Corollary 5.2.2. In other words, there exists an increasing sequence of stopping times $(T_{n}^{i})_{n\in {\mathbb {N}}}$ such that $\lim _{n\rightarrow \infty }T_{n}^{i} =\infty$ a.s. and $(\exp (Y_{\varepsilon }^{i}(\tau )_{t\wedge T_{n}^{i}})_{t\in [0,T]}$ is a martingale. Therefore, by Lebesgue’s theorem and Markov’s inequality, for every $\rho > 0$, the stochastic process $Y_{N,\varepsilon }(\tau ):= Y_{\varepsilon }^{1}(\tau ) +\dots + Y_{\varepsilon }^{N}(\tau )$ satisfies

$$\begin{aligned} {\mathbb {P}}(e^{Y_{N,\varepsilon }(\tau )_T}>\rho )= & {} \lim _{n\rightarrow \infty }{\mathbb {P}}\left( \exp \left[ \sum _{i = 1}^{N}Y_{\varepsilon }^{i}(\tau )_{T\wedge T_{n}^{i}}\right] >\rho \right) \\\leqslant & {} \frac{1}{\rho }\lim _{n\rightarrow \infty } \mathbb E(\exp (Y_{\varepsilon }^{1}(\tau )_{T\wedge T_{n}^{1}}))^N = \frac{1}{\rho }{\mathbb {E}}(\exp (Y_{\varepsilon }^{1}(\tau )_0))^N =\frac{1}{\rho }. \end{aligned}$$

Step 2. For any $\varepsilon \in (0,\varepsilon ^*)$ and $t\in [0,T]$, let us find suitable bounds on

$$\begin{aligned} A_{N,\varepsilon }(\tau )_t:=\sum _{i = 1}^{N}A_{\varepsilon }^{i}(\tau )_t \quad \textrm{and}\quad B_{N,\varepsilon }(\tau )_t:=\sum _{i = 1}^{N}B_{\varepsilon }^{i}(\tau )_t. \end{aligned}$$

On the one hand,

$$\begin{aligned} A_{N,\varepsilon }(\tau )_t\leqslant \frac{\varepsilon ^2\Vert \sigma \Vert _{\infty }^{2}}{2} \sum _{i = 1}^{N}\int _{0}^{t}\tau (X_{s}^{i})^2ds \leqslant \frac{\varepsilon ^2\Vert \sigma \Vert _{\infty }^{2}\Vert \tau \Vert _{N}^{2}NT}{2}. \end{aligned}$$

(9)

On the other hand, for every $\beta \in (-{\mathfrak {b}}/2,\mathfrak b/2)$, by Taylor’s formula and Assumption 3,

$$\begin{aligned} \int _{-\infty }^{\infty }(e^{\beta z} -\beta z - 1)\pi _{\lambda }(dz)= & {} \beta ^2\int _{-\infty }^{\infty } \left( \int _{0}^{1}(1 -\theta )e^{\theta \beta z}d\theta \right) \pi _{\lambda }(dz)\\\leqslant & {} \frac{{\mathfrak {c}}_1}{2}\beta ^2 \quad \textrm{with}\quad {\mathfrak {c}}_1 = \int _{-\infty }^{\infty }e^{\mathfrak b|z|/2}\pi _{\lambda }(dz) <\infty . \end{aligned}$$

Since $\varepsilon \in (0,\varepsilon ^*)$, one can take $\beta =\varepsilon \tau (X_{s}^{i})\gamma (X_{s}^{i})$ for any $s\in [0,t]$ and $i\in \{1,\dots ,N\}$, and then

$$\begin{aligned} B_{N,\varepsilon }(\tau )_t\leqslant \frac{\mathfrak c_1\varepsilon ^2}{2}\sum _{i = 1}^{N}\int _{0}^{t} \tau (X_{s}^{i})^2\gamma (X_{s}^{i})^2ds \leqslant \frac{\mathfrak c_1\varepsilon ^{2}\Vert \gamma \Vert _{\infty }^{2}\Vert \tau \Vert _{N}^{2}NT}{2}. \end{aligned}$$

(10)

Therefore, Inequalities (9) and (10) lead to

$$\begin{aligned} A_{N,\varepsilon }(\tau )_t + B_{N,\varepsilon }(\tau )_t\leqslant {\mathfrak {c}}_2\varepsilon ^2(\Vert \sigma \Vert _{\infty }^{2} + \Vert \gamma \Vert _{\infty }^{2})\Vert \tau \Vert _{N}^{2}NT \quad \textrm{with}\quad {\mathfrak {c}}_2 = \frac{1}{2}(1\vee {\mathfrak {c}}_1). \end{aligned}$$

Step 3 (conclusion). Consider $M_N(\tau ):= M^1(\tau ) +\dots + M^N(\tau )$. For any $\varepsilon \in (0,\varepsilon ^*)$ and $\xi ,v > 0$, thanks to Step 2,

$$\begin{aligned} {\mathbb {P}}(\nu _N(\tau )\geqslant \xi ,\Vert \tau \Vert _{N}^{2}\leqslant v^2)\leqslant & {} {\mathbb {P}}(e^{\varepsilon M_N(\tau )_T}\geqslant e^{NT\varepsilon \xi }, A_{N,\varepsilon }(\tau )_T + B_{N,\varepsilon }(\tau )_T\leqslant {\mathfrak {c}}_2\varepsilon ^2(\Vert \sigma \Vert _{\infty }^{2} +\Vert \gamma \Vert _{\infty }^{2})NTv^2)\\\leqslant & {} {\mathbb {P}}(e^{Y_{N,\varepsilon }(\tau )_T}\geqslant \exp (NT\varepsilon \xi -\mathfrak c_2\varepsilon ^2(\Vert \sigma \Vert _{\infty }^{2} +\Vert \gamma \Vert _{\infty }^{2})NTv^2)). \end{aligned}$$

Moreover, taking

$$\begin{aligned} \varepsilon = \frac{\xi }{2{\mathfrak {c}}_2(\Vert \sigma \Vert _{\infty }^{2} +\Vert \gamma \Vert _{\infty }^{2})v^2 +\xi /\varepsilon ^*} <\varepsilon ^* \end{aligned}$$

leads to

$$\begin{aligned} NT\varepsilon \xi -{\mathfrak {c}}_2\varepsilon ^2(\Vert \sigma \Vert _{\infty }^{2} + \Vert \gamma \Vert _{\infty }^{2})NTv^2= & {} \frac{NT\xi ^2[\mathfrak c_2(\Vert \sigma \Vert _{\infty }^{2} +\Vert \gamma \Vert _{\infty }^{2})v^2 + \xi /\varepsilon ^*]}{[2{\mathfrak {c}}_2(\Vert \sigma \Vert _{\infty }^{2} +\Vert \gamma \Vert _{\infty }^{2})v^2 + \xi /\varepsilon ^*]^2}\\\geqslant & {} \frac{NT\xi ^2}{4[{\mathfrak {c}}_2(\Vert \sigma \Vert _{\infty }^{2} + \Vert \gamma \Vert _{\infty }^{2})v^2 +\xi /\varepsilon ^*]}. \end{aligned}$$

Therefore, by Step 1,

$$\begin{aligned} {\mathbb {P}}(\nu _N(\tau )\geqslant \xi ,\Vert \tau \Vert _{N}^{2}\leqslant v^2) \leqslant \exp \left( -\frac{NT\xi ^2}{4[\mathfrak c_2(\Vert \sigma \Vert _{\infty }^{2} + \Vert \gamma \Vert _{\infty }^{2})v^2 +\xi \Vert \tau \Vert _{\infty ,I}\Vert \gamma \Vert _{\infty }]}\right) . \end{aligned}$$

$\square$

Appendix B

See Table 1 and Figs. (1, 2 and 3)

Table 1 Means and StD of the MISE of ${\widehat{b}}_{{\widehat{m}}}$ (100 repetitions)

Fig. 1

Plots of b and of 10 adaptive estimations for Model 1 ($\overline{{\widehat{m}}} = 5.3$)

Fig. 2

Plots of b and of 10 adaptive estimations for Model 2 ($\overline{{\widehat{m}}} = 4.2$)

Fig. 3

Plots of b and of 10 adaptive estimations for Model 3 ($\overline{{\widehat{m}}} = 4.1$)