Nonparametric estimation in a mixed-effect Ornstein–Uhlenbeck model

Abstract

Two adaptive nonparametric procedures are proposed to estimate the density of the random effects in a mixed-effect Ornstein–Uhlenbeck model. First a kernel estimator is introduced with a new bandwidth selection method developed recently by Goldenshluger and Lepski (Ann Stat 39:1608–1632, 2011). Then, we adapt an estimator from Comte et al. (Stoch Process Appl 7:2522–2551, 2013) in the framework of small time interval of observation. More precisely, we propose an estimator that uses deconvolution tools and depends on two tuning parameters to be chosen in a data-driven way. The selection of these two parameters is achieved through a two-dimensional penalized criterion. For both adaptive estimators, risk bounds are provided in terms of integrated $\mathbb {L}^2$-error. The estimators are evaluated on simulations and show good results. Finally, these nonparametric estimators are applied to neuronal data and are compared with previous parametric estimations.

Appendices

Appendix 1: Proofs

1.1 Proof of Theorem 3.1

Given $h \in {\mathcal {H}}_{N,T}$, we denote:

$$\begin{aligned} V(h)=\kappa _1\frac{\Vert K\Vert ^2_1\Vert K\Vert ^2}{Nh}+\kappa _2 \frac{\sigma ^4\Vert K\Vert ^2_1\Vert K''\Vert ^2}{T^2h^5}{=:}V_1(h)+V_2(h). \end{aligned}$$

Using the definition of A(h) and of ${\widehat{h}}$ we obtain

$$\begin{aligned} \Vert {\widehat{f}}_{{\widehat{h}}}-f\Vert ^2\le & {} 3\Vert {\widehat{f}}_{{\widehat{h}}} -{\widehat{f}}_{h,{\widehat{h}}}\Vert ^2+3\Vert {\widehat{f}}_{h,{\widehat{h}}} -{\widehat{f}}_{h}\Vert ^2+3\Vert {\widehat{f}}_{h}-f\Vert ^2\\\le & {} 3\left( A(h)+V({\widehat{h}}) \right) + 3\left( A({\widehat{h}})+V(h) \right) +3 \Vert {\widehat{f}}_h-f\Vert ^2\\\le & {} 6A(h)+6V(h)+3 \Vert {\widehat{f}}_h-f\Vert ^2. \end{aligned}$$

Thus,

$$\begin{aligned} \mathbb {E}[\Vert {\widehat{f}}_{{\widehat{h}}}-f\Vert ^2] \le 6\mathbb {E}[A(h)]+6V(h)+3 \mathbb {E}[ \Vert {\widehat{f}}_h-f\Vert ^2], \end{aligned}$$

hence, we only have to study the term $\mathbb {E}[A(h)]$. We can decompose $\Vert {\widehat{f}}_{h,h'}-{\widehat{f}}_{h'}\Vert ^2$ as follows:

$$\begin{aligned}&\Vert {\widehat{f}}_{h,h'}-{\widehat{f}}_{h'}\Vert ^2 \le 5\Vert {\widehat{f}}_{h,h'}-\mathbb {E}[{\widehat{f}}_{h,h'}]\Vert ^2+ 5\Vert \mathbb {E}[{\widehat{f}}_{h,h'}]-{f}_{h,h'}\Vert ^2+5\Vert {f}_{h,h'}\\&\quad -{f}_{h'}\Vert ^2+5\Vert {f}_{h'}-\mathbb {E}[{\widehat{f}}_{h'}]\Vert ^2+ 5\Vert \mathbb {E}[{\widehat{f}}_{h'}]-{\widehat{f}}_{h'}\Vert ^2 \end{aligned}$$

thus

$$\begin{aligned} A(h)\le 5(D_1+D_2+D_3+D_4+D_5) \end{aligned}$$

with:

$$\begin{aligned} D_1:= & {} \underset{h' \in {\mathcal {H}}_{N,T}}{\sup } \Vert {f}_{h,h'}-{f}_{h'}\Vert ^2,\\ D_2:= & {} \underset{h' \in {\mathcal {H}}_{N,T}}{\sup } \left( \Vert {\widehat{f}}_{h'}-\mathbb {E}[{\widehat{f}}_{h'}]\Vert ^2-\frac{V_1(h')}{10}\right) _+,\\ ~~D_3:= & {} \underset{h' \in {\mathcal {H}}_{N,T}}{\sup } \left( \Vert {\widehat{f}}_{h,h'}-\mathbb {E}[{\widehat{f}}_{h,h'}]\Vert ^2-\frac{V_1(h')}{10}\right) _+\\ D_4:= & {} \underset{h' \in {\mathcal {H}}_{N,T}}{\sup } \left( \Vert \mathbb {E}[{\widehat{f}}_{h'}]-{f}_{h'}\Vert ^2-\frac{V_2(h')}{10}\right) _+,\\ ~~D_5:= & {} \underset{h' \in {\mathcal {H}}_{N,T}}{\sup } \left( \Vert \mathbb {E}[{\widehat{f}}_{h,h'}]-{f}_{h,h'}\Vert ^2-\frac{V_2(h')}{10}\right) _+. \end{aligned}$$

According to Young’s inequality (see Theorem 9.1), we obtain

$$\begin{aligned} \Vert f_{h,h'}-f_{h'}\Vert ^2=\Vert K_{h'}{\star } (f_{h}-f)\Vert ^2\le \Vert K_{h'}\Vert _1^2 \Vert f_{h}-f\Vert ^2= \Vert K\Vert _1^2 \Vert f_{h}-f\Vert ^2 \end{aligned}$$

thus

$$\begin{aligned} D_1\le \Vert K\Vert _1^2 \Vert f_{h}-f\Vert ^2. \end{aligned}$$

(18)

Let us study the term $D_2$. We denote ${\mathcal {B}}(1)=\{g\in \mathbb {L}^2(\mathbb {R}), \Vert g\Vert =1\}$. We define

$$\begin{aligned} \nu _{N,h}(g):= \langle g,{\widehat{f}}_{h}-\mathbb {E}[{\widehat{f}}_{h}]\rangle \end{aligned}$$

then $|\nu _{N,h}(g)|\le \Vert g\Vert \Vert {\widehat{f}}_{h}-\mathbb {E}[{\widehat{f}}_{h}]\Vert$ thus, the estimator ${\widehat{f}}_{h}$ satisfies:

$$\begin{aligned} \Vert {\widehat{f}}_{h}-\mathbb {E}[{\widehat{f}}_{h}]\Vert ^2=\underset{g\in {\mathcal {B}}(1)}{\sup }(\nu _{N,h}(g))^2. \end{aligned}$$

We can also compute the scalar product which defines $\nu _{N,h}$ and we obtain

$$\begin{aligned} \nu _{N,h}(g)=\frac{1}{N}\sum _{j=1}^N \left( g{\star } K_h^-(Z_{j,T})-\mathbb {E}[g{\star } K_h^-(Z_{j,T})] \right) \end{aligned}$$

(19)

with $K_h^-(x):=K_h(-x)$. This finally conducts to:

$$\begin{aligned} \mathbb {E}[D_2] \le \underset{h' \in {\mathcal {H}}_{N,T}}{\sum } \mathbb {E}\left[ \underset{g \in {\mathcal {B}}(1)}{\sup } (\nu _{N,h'}(g))^2-\frac{V_1(h')}{10} \right] _+. \end{aligned}$$

This bound and Eq. (19) leads to apply Talagrand’s Theorem (9.2). We have to compute 3 quantities: M, $H^2$ and v.

First:

$$\begin{aligned} \underset{g \in {\mathcal {B}}(1)}{\sup } \Vert g{\star } K_{h'}^-\Vert _{\infty }= & {} \underset{g \in {\mathcal {B}}(1)}{\sup }\underset{x\in \mathbb {R}}{\sup } \left| \int g(y)K_{h'}^-(x-y)dy\right| \nonumber \\= & {} \underset{g \in {\mathcal {B}}(1)}{\sup }\underset{x\in \mathbb {R}}{\sup } |\langle g,K_{h'}^-(.-x)\rangle | \nonumber \\\le & {} \underset{g \in {\mathcal {B}}(1)}{\sup }\Vert g\Vert \Vert K_{h'}\Vert =\frac{\Vert K\Vert }{\sqrt{h'}}:=M. \end{aligned}$$

(20)

Secondly, the bound of Proposition 3.1 gives

$$\begin{aligned} \mathbb {E}\left[ \underset{g \in {\mathcal {B}}(1)}{\sup } (\nu _{N,h}(g))^2\right] =\mathbb {E}\left[ \Vert {\widehat{f}}_{h}-\mathbb {E}[{\widehat{f}}_{h}]\Vert ^2\right] \le \frac{\Vert K\Vert ^2}{Nh}:=H^2. \end{aligned}$$

(21)

Thirdly:

$$\begin{aligned} \underset{g \in {\mathcal {B}}(1)}{\sup } \left( \mathrm {Var} ( g{\star } K_{h'}^-(Z_{1,T}) ) \right)\le & {} \underset{g \in {\mathcal {B}}(1)}{\sup } \mathbb {E}[ (g{\star } K_{h'}^-(Z_{1,T}))^2 ]\\\le & {} 2\underset{g \in {\mathcal {B}}(1)}{\sup } \mathbb {E}[ (g{\star } K_{h'}^-(\phi _1))^2 ]\\&+\,2\underset{g \in {\mathcal {B}}(1)}{\sup } \mathbb {E}[ (g{\star } (K_{h'}^-(Z_{1,T})-K_{h'}^-(\phi _1))^2 ]. \end{aligned}$$

Let us investigate the two terms separately. Young’s inequality gives:

$$\begin{aligned} \mathbb {E}\left[ (g{\star } K_{h'}^-(\phi _1))^2 \right]= & {} \int \left( g{\star } K_{h'}^-(x)\right) ^2 f(x)dx \le \Vert f\Vert \Vert g{\star } K_{h'}^-\Vert ^2_4\nonumber \\= & {} \frac{\Vert f\Vert \Vert K\Vert ^2_{4/3}}{\sqrt{h'}}:=v_1. \end{aligned}$$

(22)

Then, one can write: $K_{h'}(x-Z_{1,T})-K_{h'}(x-\phi _1)=(\phi _1-Z_{1,T}) \int _0^1 (K_{h'})'(x-\phi _1+u(\phi _1-Z_{1,T}))du$, thus

$$\begin{aligned}&(g{\star } K_{h'}^-(Z_{1,T})-g{\star } K_{h'}(\phi _1))^2\\&\qquad =(\phi _1-Z_{1,T})^2 \left( \int g(x) \int _0^1 (K_{h'})'(x-\phi _1+u(\phi _1-Z_{1,T}))dudx \right) ^2\\&\qquad \le (\phi _1-Z_{1,T})^2 \int g^2(x)\left( \int _0^1 (K_{h'})'^2(x-\phi _1+u(\phi _1-Z_{1,T}))du \right) dx\\&\qquad \le (\phi _1-Z_{1,T})^2 \Vert g\Vert ^2 \int (K_{h'})'^2(y)dy= (\phi _1-Z_{1,T})^2\Vert (K_{h'})'\Vert ^2. \end{aligned}$$

With $\mathbb {E}[(\phi _1-Z_{1,T})^2]=\frac{\sigma ^2}{T^2}\mathbb {E}[W_1(T)^2] =\frac{\sigma ^2}{T}$, the assumption $T^{-1}\le h^{5/2}$ leads to

$$\begin{aligned} \mathbb {E}\left[ (g{\star } K_{h'}^-(Z_{1,T})-g{\star } K_{h'}(\phi _1))^2\right] \le \frac{\Vert K'\Vert ^2 \sigma ^2}{h'^3T}\le \frac{\Vert K'\Vert ^2 \sigma ^2}{\sqrt{h'}}:=v_2. \end{aligned}$$

(23)

Finally $v=v_1+v_2=A_0/\sqrt{h'}$ with $A_0=\Vert f\Vert \Vert K\Vert ^2_{4/3}+\Vert K'\Vert ^2\sigma ^2$.

If $\kappa _1 \Vert K\Vert _1^2 \ge 40$, with the assumption $1/(Nh) \le 1$, Talagrand’s inequality (under the assumptions of the Theorem 3.1) gives

$$\begin{aligned} \mathbb {E}\left( \underset{g \in {\mathcal {B}}(1)}{\sup } (\nu _{N,h'}(g))^2-\frac{V_1(h')}{10}\right) _+\le & {} \frac{C_1}{N\sqrt{h'}} e^{-C_2/\sqrt{h'}}+C_3\frac{1}{h' N^2} e^{-C_4 \sqrt{N}}\\\le & {} \frac{C_5}{N} \sum _{h'\in {\mathcal {H}}_{N,T}} \frac{1}{\sqrt{h'}} e^{-C_6/\sqrt{h'}} \le \frac{C_5 S(C_6)}{N}. \end{aligned}$$

One can lead the study of $D_3$ as we have done for $D_2$, using the same steps and tools. However $K_h {\star } K_{h'}$ instead of $K_{h'}$, adds $\Vert K\Vert _1$ in M and $\Vert K\Vert ^2_1$ in $H^2$ and v.

Then, let us study the term $D_4$. If $\kappa _2 \ge 10/(3\Vert K\Vert _1^2)$, the bound (9) leads us to

$$\begin{aligned} D_4= & {} \underset{h' \in {\mathcal {H}}_{N,T}}{\sup } \left( \Vert \mathbb {E}[{\widehat{f}}_{h'}]-{f}_{h'}\Vert ^2-\frac{V_2(h')}{10}\right) _+\\\le & {} \underset{h' \in {\mathcal {H}}_{N,T}}{\sup } \left( \frac{\Vert K''\Vert ^2\sigma ^4}{3h'^5T^2}-\frac{\kappa _2 \Vert K\Vert _1^2 \Vert K''\Vert ^2\sigma ^4}{10 T^2 h^{'5}}\right) _+=0 \end{aligned}$$

thus $D_4=0$. Finally, similarly, if $\kappa _2\ge 10/3$, we obtain

$$\begin{aligned} D_5= & {} \underset{h' \in {\mathcal {H}}_{N,T}}{\sup } \left( \Vert \mathbb {E}[{\widehat{f}}_{h,h'}]-{f}_{h,h'}\Vert ^2-\frac{V_2(h')}{10}\right) _+\\\le & {} \underset{h' \in {\mathcal {H}}_{N,T}}{\sup } \left( \frac{\Vert K''\Vert ^2 \Vert K\Vert _1^2\sigma ^4}{3h^5T^2}-\frac{\kappa _2 \Vert K\Vert _1^2 \Vert K''\Vert ^2\sigma ^4}{10 T^2 h^{'5}}\right) _+=0. \end{aligned}$$

Thus finally we obtained that:

$$\begin{aligned} \mathbb {E}[A(h)] \le 5 \left( \Vert K\Vert _1^2 \Vert f_h-f\Vert ^2 +\frac{c}{N} \right) \end{aligned}$$

(24)

with c a constant depending on $\Vert f\Vert ,\Vert K\Vert _1,\Vert K\Vert ,\Vert K\Vert _{4/3}$. Finally we have shown that for all $h \in {\mathcal {H}}_{N,T}$:

$$\begin{aligned} \mathbb {E}[\Vert {\widehat{f}}_{{\widehat{h}}}-f\Vert ^2]\le & {} 6\kappa _1 \frac{ \Vert K\Vert _1^2 \Vert K\Vert ^2}{Nh}+ 6\kappa _2 \frac{\Vert K\Vert _1^2 \Vert K''\Vert ^2 \sigma ^4}{T^2h^5}\\&+\,3 \left( 2\Vert f-f_h\Vert ^2+ \frac{ \Vert K\Vert ^2 }{Nh} + \frac{ 2\Vert K''\Vert ^2 \sigma ^4}{3T^2h^5} \right) \\&+ \,30\left( \Vert K\Vert ^2_1 \Vert f-f_h\Vert ^2+\frac{c}{N} \right) \\\le & {} \left( 6+ \frac{3}{\Vert K\Vert _1^2 \kappa _1} \right) V_1(h)+\left( 6+ \frac{9}{2\Vert K\Vert _1^2 \kappa _2} \right) V_2(h)\\&+ \,(30\Vert K\Vert _1^2+6) \Vert f_h-f\Vert ^2 +\frac{C}{N}\\\le & {} C_1 \underset{h\in {\mathcal {H}}_{N,T}}{\inf } \{\Vert f-f_h\Vert ^2+ V(h)\} +\frac{C_2}{N}. \end{aligned}$$

where $C_1= \max (7, 30 \Vert K\Vert _1^2+6)$ and $C_2$ depends on $\Vert f\Vert ,\Vert K\Vert _1,\Vert K\Vert ,\Vert K\Vert _{4/3}$. $\square$

1.2 Proof of Proposition 4.1

The bias term is $\Vert f-\mathbb {E}[{\widetilde{f}}_{m,s}]\Vert ^2$. Let us compute $\mathbb {E}[{\widetilde{f}}_{m,s}]$. As the $Z_{j,\tau }$ are i.i.d. when $\tau$ is fixed and due to the independence of $\phi _1$ and $W_1$, we obtain:

$$\begin{aligned} \mathbb {E}[{\widetilde{f}}_{m,s}(x)]= & {} \frac{1}{2\pi } \int _{-m}^{m} e^{-iux} \mathbb {E}\left[ e^{iuZ_{1,m^2/s^2}+u^2\sigma ^2s^2/(2m^2) }\right] du\\= & {} \frac{1}{2\pi } \int _{-m}^{m} e^{-iux} \mathbb {E}\left[ e^{iu\phi _1+ iu\sigma W_1(m^2/s^2)s^2/m^2 + u^2 \sigma ^2s^2/(2m^2)}\right] du\\= & {} \frac{1}{2\pi } \int _{-m}^{m} e^{-iux+u^2 \sigma ^2 s^2/(2m^2)} f^*(u) \mathbb {E}\left[ e^{iu\sigma W_1(m^2/s^2) s^2/m^2}\right] du\\= & {} \frac{1}{2\pi } \int _{-m}^{m} e^{-iux+u^2 \sigma ^2 s^2/(2m^2)} f^*(u) e^{-u^2\sigma ^2 s^2/(2m^2)} du\\= & {} \frac{1}{2\pi } \int _{-m}^{m} e^{-iux} f^*(u) du{=:} f_{m}(x). \end{aligned}$$

Therefore this gives $\mathbb {E}[{\widetilde{f}}_{m,s}(x)]={f}_{m}(x)$, and $\Vert f-\mathbb {E}[{\widetilde{f}}_{m,s}]\Vert ^2=\Vert f-{f}_{m}\Vert ^2=\frac{1}{2\pi } \int _{|u| \ge m}|f^*(u)|^2du$.

The variance term is:

$$\begin{aligned} \mathbb {E}\left[ \Vert {\widetilde{f}}_{m,s}-{f}_{m}\Vert ^2 \right]= & {} \frac{1}{2\pi } \mathbb {E}\left[ \int _{-m}^{m} \left| \frac{1}{N} \sum _{j=1}^{N} e^{iuZ_{j,m^2/s^2} } e^{\frac{u^2 \sigma ^2 s^2}{2m^2}}-f^*(u)\right| ^2du \right] \\= & {} \frac{1}{2\pi N} \int _{-m}^{m} e^{\frac{u^2 \sigma ^2 s^2}{m^2}} \mathrm {Var}\left( e^{iuZ_{1,m^2/s^2}} \right) du \\\le & {} \frac{1}{2\pi N} \int _{-m}^{m} e^{\frac{u^2 \sigma ^2 s^2}{m^2}} du = \frac{m}{\pi N} \int _{0}^{1} e^{s^2 \sigma ^2 v^2} du. \end{aligned}$$

$\Box$

1.3 Proof of Theorem 4.1

Let us study the term $\Vert {\widetilde{f}}_{{\widetilde{m}},{\widetilde{s}}}-f\Vert ^2$. We decompose it into a sum of three terms and the definition of $({\widetilde{m}},{\widetilde{s}})$ (15) implies for all $(m,s) \in {\mathcal {C}}$

$$\begin{aligned} \Vert {\widetilde{f}}_{{\widetilde{m}},{\widetilde{s}}}-f\Vert ^2\le & {} 3\left( \Vert {\widetilde{f}}_{{\widetilde{m}},{\widetilde{s}}}-{\widetilde{f}}_{({\widetilde{m}},{\widetilde{s}})\wedge (m,s)}\Vert ^2+\Vert {\widetilde{f}}_{({\widetilde{m}},{\widetilde{s}})\wedge (m,s)}-{\widetilde{f}}_{m,s}\Vert ^2+\Vert {\widetilde{f}}_{m,s}-f\Vert ^2 \right) \nonumber \\\le & {} 3\left( \varGamma _{m,s}+\text {pen}({\widetilde{m}},{\widetilde{s}}) \right) + 3\left( \varGamma _{{\widetilde{m}},{\widetilde{s}}}+\text {pen}(m,s)\right) +3 \Vert {\widetilde{f}}_{m,s}-f\Vert ^2 \nonumber \\\le & {} 6\varGamma _{m,s}+6\text {pen}(m,s)+3 \Vert {\widetilde{f}}_{m,s}-f\Vert ^2 \end{aligned}$$

(25)

Now we study ${\varGamma }_{m,s}$. First:

$$\begin{aligned}&\Vert {\widetilde{f}}_{(m,s)\wedge (m',s')}-{\widetilde{f}}_{m',s'}\Vert ^2\\&\quad \le 3 \left( \Vert {\widetilde{f}}_{m',s'}-f_{m'}\Vert ^2+\Vert f_{m'}-f_{m\wedge m'}\Vert ^2 +\Vert f_{m \wedge m'}- {\widetilde{f}}_{(m',s')\wedge (m,s)}\Vert ^2 \right) . \end{aligned}$$

Thus:

$$\begin{aligned} \varGamma _{m,s}\le & {} \underset{(m',s')\in {\mathcal {C}}}{\max } \left( 3\Vert {\widetilde{f}}_{m',s'}-f_{m'}\Vert ^2 + 3\Vert f_{m'}-f_{m \wedge m'}\Vert ^2 +3\Vert f_{m\wedge m'}\right. \\&\left. - \,{\widetilde{f}}_{(m',s')\wedge (m,s)}\Vert ^2 -\text {pen}({m',s'}) \right) _+\\\le & {} 3\underset{(m',s')\in {\mathcal {C}}}{\max } \left( \Vert {\widetilde{f}}_{m',s'}-f_{m'}\Vert ^2-\frac{1}{6}\text {pen}({m',s'})\right) _+\\&+ \,3\underset{(m',s')\in {\mathcal {C}}}{\max } \left( \Vert {\widetilde{f}}_{(m',s')\wedge (m,s)}-f_{m \wedge m'} \Vert ^2 -\frac{1}{6}\text {pen}({m',s'}) \right) _+\\&+\, 3 \underset{{m'}\in {\mathcal {M}}}{\max } \Vert f_{m'}-f_{m\wedge m'}\Vert ^2. \end{aligned}$$

The last maximum can be explicit. If $m'\le m$, then $\Vert f_{m'}-f_{m \wedge m'}\Vert ^2=\Vert f_{m'}-f_{m'}\Vert ^2=0$. Otherwise,

$$\begin{aligned} \Vert f_{m'}-f_{m \wedge m'}\Vert ^2=\Vert f_{m'}-f_{m}\Vert ^2=\int _{m \le |u|\le m'} |f^*(u)|^2du \le \Vert f-f_{m}\Vert ^2. \end{aligned}$$

Finally:

$$\begin{aligned} \underset{{m'}\in {\mathcal {M}}}{\max } \Vert f_{m'}-f_{m\wedge m'}\Vert ^2 \le \Vert f-f_{m}\Vert ^2. \end{aligned}$$

We get the following bound for $\varGamma _{m,s}$:

$$\begin{aligned} \varGamma _{m,s}\le & {} 3\underset{(m',s')\in {\mathcal {C}}}{\max } \left( \Vert {\widetilde{f}}_{m',s'}-f_{m'}\Vert ^2-\frac{1}{6}\text {pen}({m',s'})\right) _+ \nonumber \\&+\, 3\underset{(m',s')\in {\mathcal {C}}}{\max } \left( \Vert {\widetilde{f}}_{(m',s')\wedge (m,s)}-f_{m \wedge m'} \Vert ^2 -\frac{1}{6}\text {pen}({m',s'}) \right) _+ \nonumber \\&+\, 3 \Vert f-f_{m}\Vert ^2. \end{aligned}$$

(26)

Then we gather Eqs. (25) and (26):

$$\begin{aligned} \Vert {\widetilde{f}}_{{\widetilde{m}},{\widetilde{s}}}-f\Vert ^2\le & {} 6\text {pen}(m,s)+3\Vert {\widetilde{f}}_{m,s}-f\Vert ^2 + 18 \Vert f-f_{m}\Vert ^2\\&+\, \underset{(m',s')\in {\mathcal {C}}}{\max } 18\left( \Vert {\widetilde{f}}_{m',s'}-f_{m'}\Vert ^2-\frac{1}{6}\text {pen}({m',s'})\right) _+\\&+\, \underset{(m',s')\in {\mathcal {C}}}{\max }18 \left( \Vert {\widetilde{f}}_{(m',s')\wedge (m,s)}-f_{m \wedge m'} \Vert ^2 -\frac{1}{6}\text {pen}({m',s'}) \right) _+ . \end{aligned}$$

We first notice that our penalty function is increasing in s and m, thus we get the following bound for the last term:

$$\begin{aligned}&\mathbb {E}\left[ \underset{(m',s')\in {\mathcal {C}}}{\max } \left( \Vert {\widetilde{f}}_{(m',s')\wedge (m,s)}-f_{m \wedge m'} \Vert ^2 -\frac{1}{6}\text {pen}((m',s')\wedge (m,s)) \right) _+\right] \\&\quad \le \mathbb {E}\left[ \underset{m' \le m,s'\le s}{\max } \left( \Vert {\widetilde{f}}_{m',s'}-f_{m'} \Vert ^2 -\frac{1}{6}\text {pen}(m',s') \right) _+\right] \\&\qquad +\,\mathbb {E}\left[ \underset{ m \le m',s\le s' }{\max } \left( \Vert {\widetilde{f}}_{m,s}-f_{m} \Vert ^2 -\frac{1}{6}\text {pen}(m,s) \right) _+\right] \\&\qquad +\,\mathbb {E}\left[ \underset{ m \le m' ,s'\le s }{\max } \left( \Vert {\widetilde{f}}_{m,s'}-f_{m} \Vert ^2 -\frac{1}{6}\text {pen}(m,s') \right) _+\right] \\&\qquad +\,\mathbb {E}\left[ \underset{ m' \le m, s\le s'}{\max } \left( \Vert {\widetilde{f}}_{m',s}-f_{ m'} \Vert ^2 -\frac{1}{6}\text {pen}(m',s) \right) _+\right] \\&\quad \le 4 \sum _{m' \in {\mathcal {M}}} \sum _{s' \in {\mathcal {S}}}\mathbb {E}\left[ \Vert {\widetilde{f}}_{m',s'}-f_{m'}\Vert ^2 -\frac{1}{6}\text {pen}({m',s'})\right] _+. \end{aligned}$$

Moreover, according to Proposition 4.1 and using the inequality $\int _0^1 e^{\sigma ^2s^2v^2}dv \le e^{\sigma ^2s^2}$, we obtain, for all $(m,s) \in {\mathcal {C}}$,

$$\begin{aligned} \mathbb {E}\left[ \Vert {\widetilde{f}}_{{\widetilde{m}},{\widetilde{s}}}-f\Vert ^2\right]\le & {} 5\times 18 \sum _{m' \in {\mathcal {M}}}\sum _{s' \in {\mathcal {S}}} \mathbb {E}\left[ \Vert {\widetilde{f}}_{m',s'}-f_{m'}\Vert ^2-\frac{1}{6}\text {pen}({m',s'})\right] _+ + 6\text {pen}(m,s)\\&+ \,3\frac{m}{\pi N} e^{\sigma ^2s^2} + 21 \Vert f-f_{m}\Vert ^2 . \end{aligned}$$

Then we obtain the announced result with the following Lemma.

Lemma 8.1

There exists a constant $C'>0$ such that for $\text {pen}(m,s)$ defined by $\text {pen}(m,s)=\kappa \frac{m}{ N} e^{\sigma ^2s^2}$,

$$\begin{aligned} \sum _{m' \in {\mathcal {M}}} \sum _{s' \in {\mathcal {S}}}\mathbb {E}\left[ \Vert {\widetilde{f}}_{m',s'}-f_{m'}\Vert ^2 -\frac{1}{6}\text {pen}({m',s'})\right] _+ \le \frac{C'(P+1)}{N}. \end{aligned}$$

According to Lemma 8.1, to be proved next, we choose $\text {pen}(m,s)=\kappa \frac{m}{ N} e^{\sigma ^2s^2}$, thus, there exist two constants $C=145,~C'>0$ such that,

$$\begin{aligned} \mathbb {E}\left[ \Vert {\widetilde{f}}_{{\widetilde{m}},{\widetilde{s}}}-f\Vert ^2\right]\le & {} 5\times 18 \sum _{m' \in {\mathcal {M}}} \sum _{s' \in {\mathcal {S}}}\mathbb {E}\left[ \Vert {\widetilde{f}}_{m',s'}-f_{m'}\Vert ^2 -\frac{1}{6}\text {pen}({m',s'})\right] _+\\&+ \left( 6\kappa +\frac{3}{\pi }\right) \frac{m}{ N} e^{\sigma ^2s^2} + 21 \Vert f-f_{m}\Vert ^2 \\\le & {} C \underset{(m,s)\in {\mathcal {C}}}{\inf }\left\{ \Vert f-f_{m}\Vert ^2 +\frac{m}{N} e^{\sigma ^2 s^2 } \right\} + \frac{C'}{N}.~~ \end{aligned}$$

$\Box$

Proof of Lemma 8.1

For a couple $(m,s)\in {\mathcal {C}}$ fixed, let us consider the subset $S_{m}:= \{t \in \mathbb {L}^1(\mathbb {R})\cap \mathbb {L}^2(\mathbb {R}), \text {supp}(t^*)=[-m,m] \}$. For $t \in S_{m}$,

$$\begin{aligned} \nu _N(t)=\frac{1}{N} \sum _{j=1}^{N} \left( \varphi _t(Z_{j,m^2/s^2})-\mathbb {E}\left[ \varphi _t\left( Z_{j,m^2/s^2}\right) \right] \right) \end{aligned}$$

with $\varphi _t(x):=\frac{1}{2\pi } \int {\overline{t^*(u)}}e^{iux+\sigma ^2u^2s^2/(2m^2)}du$, then $\nu _N(t)=\frac{1}{2\pi } \langle t^*,({\widetilde{f}}_{m,s}-f_{m})^*\rangle$. This leads to

$$\begin{aligned} \Vert {\widetilde{f}}_{m,s}-f_{m}\Vert ^2= \underset{t \in S_{m},~ \Vert t\Vert =1}{\sup } |\nu _N(t)|^2. \end{aligned}$$

(27)

We also have by Cauchy–Schwarz inequality

$$\begin{aligned} \Vert \varphi _t\Vert _{\infty }\le & {} \frac{1}{2\pi } \int |t^*(u)|e^{\sigma ^2u^2s^2/(2m^2)}du \le \frac{1}{2\pi } \left( \int _{-m}^{m} |t^*(u)|^2du\right) ^{1/2} \left( \int _{-m}^{m} e^{\sigma ^2u^2s^2/m^2}du \right) ^{1/2} \\\le & {} \frac{\sqrt{2m}}{\sqrt{2\pi }}e^{\sigma ^2{s}^2/2} \end{aligned}$$

thus

$$\begin{aligned} \underset{t\in S_{m},\Vert t\Vert =1}{\sup } \Vert \varphi _t\Vert _{\infty } \le \frac{\sqrt{m}}{\sqrt{\pi }}e^{\sigma ^2{s}^2/2}:=M. \end{aligned}$$

Then, by Proposition 4.1,

$$\begin{aligned} \mathbb {E}\left[ \underset{t\in S_{m},\Vert t\Vert =1}{\sup } |\nu _N(t)|^2\right] = \mathbb {E}\left[ \Vert {\widetilde{f}}_{m,s}-f_{m}\Vert ^2\right] \le \frac{m}{\pi N} \int _0^1 e^{\sigma ^2s^2v^2}dv\le \frac{m}{\pi N} e^{\sigma ^2s^2}:=H^2. \end{aligned}$$

Using Fubini and Cauchy–Schwarz inequalities we obtain for all $(m,s)\in {\mathcal {C}}$:

$$\begin{aligned} 4\pi \underset{t\in S_{m},~\Vert t\Vert =1}{\sup } \mathrm {Var}(\varphi _t(Z_{j,m^2/s^2}))\le & {} \underset{t\in S_{m},\Vert t\Vert =1}{\sup } \iint t^*(u)t^*(-v)\mathbb {E}\left[ e^{i(u-v)Z_{j,m^2/s^2}}\right] \\&\qquad \times e^{(u^2+v^2)\sigma ^2s^2/(2m^2)} dudv\\\le & {} 2\pi \left( \iint _{[-m,m]^2} |f^*(u-v)|^2 e^{(u^2+v^2)\sigma ^2s^2/m^2} dudv \right) ^{1/2}\\\le & {} 2\pi \left( e^{2\sigma ^2s^2} \iint _{[-m,m]^2} |f^*(u-v)|^2 dudv \right) ^{1/2}\\\le & {} 2\pi e^{\sigma ^2s^2} \sqrt{2m} \left( \int _{-2m}^{2m} |f^*(z)|^2dz\right) ^{1/2}\\\le & {} 2\sqrt{2m}\sqrt{2}\pi \sqrt{\pi } e^{\sigma ^2 s^2} \Vert f\Vert {=:} 4\pi ^2v,\\ v:= & {} \frac{\sqrt{m}e^{\sigma ^2s^2} \Vert f\Vert }{\sqrt{\pi }}. \end{aligned}$$

Finally using that $m \le N$, $s \le 2/\sigma$ and $\sum _{s \in {\mathcal {S}}} s=(4/\sigma )(1-(1/2)^{P+1})< 4/\sigma$, the Talagrand’s inequality with $\alpha =1/2$ if $4H^2 \le \text {pen}(m,s)/6$ implies,

$$\begin{aligned}&\sum _{s\in {\mathcal {S}}} \sum _{m \in {\mathcal {M}}} \mathbb {E} \left[ \Vert {\widetilde{f}}_{m,s}-f_{m}\Vert ^2-\frac{1}{6}\text {pen}({s,m})\right] _+\\&\quad \le \sum _{s\in {\mathcal {S}}} \sum _{m \in {\mathcal {M}}} \left( \frac{C_1 \Vert f\Vert }{N} e^{\sigma ^2 s^2} \sqrt{m} e^{-C_2 \frac{\sqrt{m}}{\Vert f\Vert }} + C_3\frac{m}{N^2} e^{\sigma ^2 s^2} e^{-C_4\sqrt{N}} \right) \\&\quad \le \sum _{s\in {\mathcal {S}}} \frac{C_1 \Vert f\Vert }{N} e^{\sigma ^2 s^2} \left( \sum _{m \in {\mathcal {M}}} \sqrt{m} e^{-C_2 \frac{\sqrt{m}}{\Vert f\Vert }} \right) + \sum _{s\in {\mathcal {S}}} \sum _{m \in {\mathcal {M}}} C_3 e^{4} \frac{1}{N} e^{-C_4\sqrt{m}}\\&\quad \le \frac{C_1 \Vert f\Vert (P+1) e^{4}}{N} \left( \sum _{m \in {\mathcal {M}}} \sqrt{m} e^{-C_2 \frac{\sqrt{m}}{\Vert f\Vert }} \right) + C_3 e^{4}\frac{P+1}{N} \sum _{m \in {\mathcal {M}}} e^{-C_4\sqrt{m}}\\&\quad \le \frac{C'(P+1)}{N} \end{aligned}$$

because with the definition of ${\mathcal {M}}$, $\sum _{m \in {\mathcal {M}}} \sqrt{m} e^{-C_2 \frac{\sqrt{m}}{\Vert f\Vert }}\le a_1 \sum _{k \in \mathbb {N}} k^{1/4} e^{-a_2 k^{1/4}} < +\infty$, and $\sum _{m \in {\mathcal {M}}} e^{-C_4 m^{1/2}}\le \sum _{k\in \mathbb {N}} e^{-a_3 k^{1/4}} <+\infty$, with $a_1, a_2, a_3$ three positive constants. Notice that $C'>0$ depends on $\sigma , \Vert f\Vert$, $\varDelta$.

We choose $\text {pen}(m,s)=\kappa m e^{\sigma ^2 s^2} /N$ with $\kappa \ge 24$. $\square$

Appendix 2

1.1 Young inequality

This inequality can be found in Briane and Pagès (2006) for example.

Theorem 9.1

Let f be a function belonging to $\mathbb {L}^p(\mathbb {R})$ and g belonging to $\mathbb {L}^q(\mathbb {R})$, let p, q, r be real numbers in $[1, +\infty ]$ and such that

$$\begin{aligned} \frac{1}{p}+\frac{1}{q}=\frac{1}{r}+1. \end{aligned}$$

Then,

$$\begin{aligned} \Vert f{\star } g\Vert _r \le \Vert f\Vert _p \Vert g\Vert _q. \end{aligned}$$

1.2 Talagrand’s inequality

The following result is a consequence of the Talagrand concentration inequality Talagrand (1996) given in Birgé and Massart (1997).

Theorem 9.2

Consider $n \in \mathbb {N}^*$, ${\mathcal {F}}$ a class at most countable of measurable functions, and $(X_i)_{i\in \{1,\ldots ,N\}}$ a family of real independent random variables. One defines, for all $f\in {\mathcal {F}}$,

$$\begin{aligned} \nu _N(f) =\frac{1}{N}\sum _{i=1}^{N} (f(X_i)-\mathbb {E}[f(X_i)]). \end{aligned}$$

Supposing there are three positive constants M, H and v such that $\underset{f\in {\mathcal {F}}}{\sup } \Vert f\Vert _{\infty } \le M$,

$\mathbb {E}[\underset{f\in {\mathcal {F}}}{\sup } |\nu _Nf| ] \le H$, and $\underset{f\in {\mathcal {F}}}{\sup } ({1}/{N})\sum _{i=1}^{N} \mathrm {Var}(f(X_i)) \le v$, then for all $\alpha >0$,

$$\begin{aligned} \mathbb {E}\left[ \left( \underset{f\in {\mathcal {F}}}{\sup } |\nu _N(f)|^2-2(1+2\alpha )H^2 \right) _+ \right]\le & {} \frac{4}{a} \left( \frac{v}{N} \exp \left( -a \alpha \frac{N H^2}{v} \right) \right. \\&+ \left. \frac{49M^2}{a C^2(\alpha )N^2} \exp \left( -\frac{\sqrt{2}a C(\alpha )\sqrt{\alpha }}{7}\frac{NH}{M} \right) \right) \end{aligned}$$

with $C(\alpha )=(\sqrt{1+\alpha }-1) \wedge 1$, and $a=\frac{1}{6}$.

1.3 Discretization

Indeed, if we assume that the times of observations are the $t_k=k\delta , k=1,\ldots ,N$ and $0<\delta <1$, we must study the error applied by discretization of the $Z_{j,\tau }$. Then, for any $0<m^2/s^2 \le T$ we use:

$$\begin{aligned} {\widehat{Z}}_{j,m^2/s^2}=\frac{s^2}{m^2}\left[ X_j(\delta [ m^2/(s^2\delta )])-X_j(0)+\frac{\delta }{\alpha }\sum _{k=1}^{[ m^2/(s^2\delta )]} X_j((k-1)\delta ) \right] \end{aligned}$$

(28)

to approximate $Z_{j,m^2/s^2}$ given by (2). The corresponding estimator of f is

$$\begin{aligned} {\widehat{{\widetilde{f}}}}_{m,s}(x)=\frac{1}{2\pi } \int _{-m}^{m} e^{-iux} \frac{1}{N} \sum _{j=1}^N e^{iu{\widehat{Z}}_{j,m^2/s^2}} e^{\frac{u^2\sigma ^2s^2}{2m^2}} du. \end{aligned}$$

(29)

We investigate the error:

$$\begin{aligned} \mathbb {E}[\Vert {\widehat{{\widetilde{f}}}}_{m,s}-f\Vert ^2] \le 2\mathbb {E}[\Vert {\widehat{{\widetilde{f}}}}_{m,s}-{\widetilde{f}}_{m,s}\Vert ^2] +2 \mathbb {E}\left[ \Vert {\widetilde{f}}_{m,s}-f\Vert ^2\right] \end{aligned}$$

where the second term of the right hand side is bounded by Proposition 4.1. Then, Plancherel–Parseval’s Theorem implies:

$$\begin{aligned} \mathbb {E}[\Vert {\widehat{{\widetilde{f}}}}_{m,s}-{\widetilde{f}}_{m,s}\Vert ^2]\le & {} \frac{1}{2\pi }\mathbb {E}\left[ \int _{-m}^{m} \frac{1}{N}\sum _{j=1}^N e^{u^2\sigma ^2s^2/m^2} \left| e^{iu{\widehat{Z}}_{j,m^2/s^2}}-e^{iu{Z}_{j,m^2/s^2}}\right| ^2du\right] \\\le & {} \frac{1}{2\pi }\int _{-m}^{m} e^{u^2\sigma ^2s^2/m^2} \mathbb {E}\left[ \left| e^{iu{\widehat{Z}}_{1,m^2/s^2}} -e^{iu{Z}_{1,m^2/s^2}}\right| ^2\right] du \end{aligned}$$

and

$$\begin{aligned} \mathbb {E}\left[ \left| e^{iu{\widehat{Z}}_{1,m^2/s^2}} -e^{iu{Z}_{1,m^2/s^2}}\right| ^2\right] \le |u|^2\mathbb {E}\left[ \left| {\widehat{Z}}_{1,m^2/s^2}-{Z}_{1,m^2/s^2}\right| ^2\right] \end{aligned}$$

thus we study the last term. For all $(m,s) \in {\mathcal {C}}$, $m^2/s^2 \le T$,

$$\begin{aligned} {Z}_{1,m^2/s^2}-{\widehat{Z}}_{1,m^2/s^2}= & {} \frac{s^2}{m^2}\left( X_j(m^2/s^2) -X_j(\delta [ m^2/(s^2\delta ) ]) \right) \\&+\frac{s^2}{\alpha m^2} \sum _{k=1}^{[ m^2/(s^2\delta ) ]} \int _{(k-1)\delta }^{k\delta }(X_j(s)-X_j((k-1)\delta ))ds \end{aligned}$$

then by Cauchy–Schwarz’s inequality we obtain

$$\begin{aligned} ({Z}_{1,m^2/s^2}-{\widehat{Z}}_{1,m^2/s^2})^2\le & {} \frac{2 s^4}{m^4}\left( X_j(m^2/s^2)-X_j(\delta [ m^2/(s^2\delta ) ]) \right) ^2\\&+\frac{2s^4}{\alpha ^2 m^4} \left[ \sum _{k=1}^{[ m^2/(s^2\delta ) ]}\int _{(k-1)\delta }^{k\delta }(X_j(s)-X_j((k-1)\delta ))ds\right] ^2. \end{aligned}$$

Höder’s inequality yields

$$\begin{aligned} \left[ \sum _{k=1}^{\left[ \frac{m^2}{s^2\delta } \right] }\int _{(k-1)\delta }^{k\delta }(X_j(s)-X_j((k-1)\delta ))ds\right] ^2\le & {} \sum _{k=1}^{\left[ \frac{m^2}{s^2\delta } \right] }\left[ \int _{(k-1)\delta }^{k\delta }(X_j(s)-X_j((k-1)\delta ))ds\right] ^2 \left[ \frac{m^2}{s^2\delta } \right] \\\le & {} \left[ \frac{m^2}{s^2\delta } \right] \delta \sum _{k=1}^{\left[ \frac{m^2}{s^2\delta } \right] }\int _{(k-1)\delta }^{k\delta }(X_j(s)-X_j((k-1)\delta ))^2 ds . \end{aligned}$$

Let us study $\mathbb {E}[(X_j(s)-X_j((k-1)\delta ))^2]$, for $(k-1)\delta \le s \le k\delta$:

$$\begin{aligned} X_j(s)-X_j((k-1)\delta )=\int _{(k-1)\delta }^s \left( \phi _j-\frac{X_j(u)}{\alpha }\right) du+\int _{(k-1)\delta }^s \sigma dW_j(u) \end{aligned}$$

and Cauchy–Schwarz’s inequality gives

$$\begin{aligned} \mathbb {E}[(X_j(s)-X_j((k-1)\delta ))^2]\le & {} 2\mathbb {E}\left[ \left( \int _{(k-1)\delta }^s \left( \phi _j-\frac{X_j(u)}{\alpha }\right) du\right) ^2\right] + 2\mathbb {E}\left[ \left( \int _{(k-1)\delta }^s \sigma dW_j(u)\right) ^2\right] \nonumber \\\le & {} 2\mathbb {E}\left[ \int _{(k-1)\delta }^s \left( \phi _j-\frac{X_j(u)}{\alpha } \right) ^2du \right] + 2\delta \sigma ^2 \nonumber \\\le & {} 4\delta ^2 \left( \mathbb {E}(\phi _j^2)+\frac{1}{\alpha ^2} \underset{s\ge 0}{\sup }\mathbb {E}[ X_j(s)^2]\right) +2\delta \sigma ^2. \end{aligned}$$

(30)

Finally, after simplification and using for all $x\in \mathbb {R}^+$, $[ x] \le x$,

$$\begin{aligned} \mathbb {E}\left[ ({Z}_{1,m^2/s^2}-{\widehat{Z}}_{1,m^2/s^2})^2\right]\le & {} \frac{2s^4}{m^4}\mathbb {E}[\left( X_j(m^2/s^2)-X_j(\delta [ m^2/(s^2\delta )]) \right) ^2]\\&+\frac{2}{\alpha ^2} \left( 4\delta ^2 \left( \mathbb {E}(\phi _j^2)+\frac{1}{\alpha ^2} \underset{s\ge 0}{\sup }\mathbb {E}[ X_j(s)^2]\right) +2\delta \sigma ^2\right) \end{aligned}$$

and we can deal with the term $\mathbb {E}[\left( X_j(m^2/s^2)-X_j(\delta [ m^2/(s^2\delta )]) \right) ^2]$ using formula (30) and $m^2/s^2-\delta [ m^2/(s^2\delta )] \le \delta$. Thus:

$$\begin{aligned} \mathbb {E}\left[ ({Z}_{1,m^2/s^2}-{\widehat{Z}}_{1,m^2/s^2})^2\right]\le & {} \left( \frac{2s^4}{m^4}+\frac{2}{\alpha ^2}\right) \left( 4\delta ^2 \left( \mathbb {E}(\phi _j^2)+\frac{1}{\alpha ^2} \underset{s\ge 0}{\sup }\mathbb {E}[ X_j(s)^2]\right) +2\delta \sigma ^2\right) . \end{aligned}$$

Besides, for model (1), Eq. (17) implies $\mathbb {E}[X_j(s)^2]\le 3x_j^2+3\alpha ^2\mathbb {E}[\phi _j^2]+3\sigma ^2$, and $0<\delta <1$ implies

$$\begin{aligned} \mathbb {E}\left[ ({Z}_{1,m^2/s^2}-{\widehat{Z}}_{1,m^2/s^2})^2\right]\le & {} C \delta \left( \frac{2s^4}{m^4}+\frac{2}{\alpha ^2}\right) \end{aligned}$$

with C a positive constant which does not depend on $\delta$ or $m^2/s^2$. Finally,

$$\begin{aligned} \mathbb {E}[\Vert {\widehat{{\widetilde{f}}}}_{m,s}-{\widetilde{f}}_{m,s}\Vert ^2]\le & {} C \delta \left( \frac{2s^4}{m^4}+\frac{2}{\alpha ^2}\right) \frac{1}{2\pi }\int _{-m}^{m} u^2 e^{u^2\sigma ^2s^2/m^2} du\\\le & {} C'\delta \left( \int _{0}^{1} v^2 e^{v^2\sigma ^2s^2} dv \right) \left( \frac{s^4}{m}+\frac{ m^3}{\alpha ^2}\right) . \end{aligned}$$

But $s \le 2/\sigma$ and $m=\sqrt{k\varDelta }/\sigma$, with $k\in \mathbb {N}^*$ and $0<\varDelta <1$, thus we obtain

$$\begin{aligned} \mathbb {E}[\Vert {\widehat{{\widetilde{f}}}}_{m,s}-{\widetilde{f}}_{m,s}\Vert ^2]\le & {} \frac{C'}{\sigma ^3}\left( \int _{0}^{1} v^2 e^{v^2\sigma ^2s^2} dv \right) \left( 2^4\sqrt{k} \left( \frac{\delta }{\sqrt{\varDelta }}\right) +\frac{k^{3/2}}{ \alpha ^2}\left( \delta \varDelta ^{3/2}\right) \right) . \end{aligned}$$

Proposition 9.1

Under (A), assuming $\mathbb {E}[\phi _j^2]<+\infty$, the estimator ${\widehat{{\widetilde{f}}}}_{m,s}$ given by (29) satisfies

$$\begin{aligned} \mathbb {E}\left[ \Vert {\widetilde{f}}_{m,s}-f\Vert ^2 \right] \le \Vert {f}_{m}-f\Vert ^2+\frac{\sqrt{k\varDelta }}{\sigma \pi N} e^{\sigma ^2s^2} +\frac{C'}{\sigma ^3} \frac{e^{\sigma ^2s^2}}{2\sigma ^2 s^2} \left( 2^4\sqrt{k} \left( \frac{\delta }{\sqrt{\varDelta }}\right) +\frac{k^{3/2}}{ \alpha ^2}\left( \delta \varDelta ^{3/2}\right) \right) . \end{aligned}$$

Finally if $\varDelta$ is fixed and $\delta$ is small, the error is acceptable. For example if $\delta =\varDelta$ the error is of order $\sqrt{\delta }$.

For study on the kernel estimator we refer to Comte et al. (2013).