Robust adaptive efficient estimation for semi-Markov nonparametric regression models

Abstract

We consider the nonparametric robust estimation problem for regression models in continuous time with semi-Markov noises. An adaptive model selection procedure is proposed. Under general moment conditions on the noise distribution a sharp non-asymptotic oracle inequality for the robust risks is obtained and the robust efficiency is shown. It turns out that for semi-Markov models the robust minimax convergence rate may be faster or slower than the classical one.

Appendix

1.1 A.1 Inequalities for the purely discontinuous martingales

Now let us recall the Novikov inequalities, Novikov (1975), also referred to as the Bichteler–Jacod inequalities (see Bichteler and Jacod 1983; Marinelli and Röckner 2014)

Lemma A.1

Let $\mu$ be a jump measure and $\widetilde{\mu }$ its compensator on the ${{\mathbb {R}}}_{*}={{\mathbb {R}}}\setminus \{0\}$. Then for for any $T>0$, any predictable function h and any $p\ge 2$

$$\begin{aligned} \mathbf{E}\sup _{0 \le t\le T} \left| \int _{[0,T]\times {{\mathbb {R}}}_{*}} h \;\mathrm {d}(\mu -\widetilde{\mu }) \right| ^{p} \le C^{*}_{p} \mathbf{E}_{Q}\check{J}_{p,T}(h)\,, \end{aligned}$$

(A.1)

where $C^{*}_{p}$ is some positive constant and

$$\begin{aligned} \check{J}_{p,T}(h)= \left( \int _{[0,T]\times {{\mathbb {R}}}_{*}} h^2 \; \mathrm {d}\widetilde{\mu } \right) ^{p/2} + \int _{[0,T]\times {{\mathbb {R}}}_{*}} h^p \, \mathrm {d}\widetilde{\mu } \,. \end{aligned}$$

1.2 A.2 Property of the penalty term

Lemma A.2

For any $n\ge \,1$ and $\lambda \in \Lambda$,

$$\begin{aligned} P^{0}_n(\lambda ) \le \mathbf{E}_{Q} \text{ Err }_n(\lambda )+\frac{\mathbf{C}_{1,Q,n}}{n}, \end{aligned}$$

where the coefficient $P^{0}_n(\lambda )$ was defined in (9.2).

Proof

By the definition of $\text{ Err }_n(\lambda )$ one has

$$\begin{aligned} \mathbf{E}_{Q}\text{ Err }_n(\lambda )&= \mathbf{E}_{Q}\sum _{j=1}^{n} \left( (\lambda (j)-1) \theta _{j}+ \frac{\lambda (j)}{\sqrt{n}}\xi _{j,n} \right) ^2\\&= \sum _{j=1}^{n} (\lambda (j)-1)^{2} \theta ^{2}_{j}+ \frac{1}{n} \sum _{j=1}^{n} \lambda ^{2}(j)\mathbf{E}_{Q} \xi ^{2}_{j,n} \,. \end{aligned}$$

So, denoting by $\lambda ^{2}=(\lambda ^{2}(j))_{1\le j\le n}$ we obtain that

$$\begin{aligned} \mathbf{E}_{Q}\text{ Err }_n(\lambda )\ge \frac{1}{n} \sum _{j=1}^{n} \lambda ^{2}(j)\mathbf{E}_{Q} \xi ^{2}_{j,n} = P^{0}_n(\lambda )+\mathbf{B}_{1,Q,n}(\lambda ^{2})\,. \end{aligned}$$

where $\lambda ^{2}=(\lambda ^{2}(j))_{1\le j\le n}$. Now Proposition 7.1 implies the desired result, i.e.

$$\begin{aligned} \mathbf{E}_{Q}\, \text{ Err }_n(\lambda ) \ge \, P^{0}_n(\lambda )-\frac{\mathbf{C}_{1,Q,n}}{n}\,. \end{aligned}$$

Hence lemma A.2. $\square$

1.3 A.3 Properties of the Fourier transform

Theorem A.3

Cauchy (1825) Let U be a simply connected open subset of ${{\mathbb {C}}},$ let $g : U \rightarrow {{\mathbb {C}}}$ be a holomorphic function, and let $\gamma$ be a rectifiable path in U whose start point is equal to its end point. Then

$$\begin{aligned} \oint _{\gamma }\,g(z)\mathrm {d}z=0\,. \end{aligned}$$

Proposition A.4

Let $g : {{\mathbb {C}}}\rightarrow {{\mathbb {C}}}$ be a holomorphic function in $U=\left\{ z\in {{\mathbb {C}}}\,:\,-\beta _{1} < \text{ Im }\right.$$\left. z< \beta _{2}\right\}$ for some $0<\beta _{1}\le \infty$ and $0<\beta _{2}\le \infty$. Assume that, for any $-\beta _{1}< t\le 0,$

$$\begin{aligned} \int _{{{\mathbb {R}}}}\,\vert g(\theta +i t)\vert \,\mathrm {d}\theta <\infty \quad \text{ and }\quad \lim _{\vert \theta \vert \rightarrow \infty }\,g(\theta +i t)\,=0\,. \end{aligned}$$

(A.2)

Then, for any $x\in {{\mathbb {R}}}$ and for any $0<\beta <\beta _{1},$

$$\begin{aligned} \int _{{{\mathbb {R}}}}\,e^{i\theta x} g(\theta )\,\mathrm {d}\theta =e^{-\beta x} \int _{{{\mathbb {R}}}}\,e^{i\theta x} g(\theta -i\beta )\,\mathrm {d}\theta . \end{aligned}$$

(A.3)

Proof

First note that the conditions of this theorem imply that

$$\begin{aligned} \int _{{{\mathbb {R}}}}\,e^{i\theta x} g(\theta )\,\mathrm {d}\theta = \lim _{N\rightarrow \infty }\, \int ^{N}_{-N}\,e^{i\theta x} g(\theta )\,\mathrm {d}\theta \,. \end{aligned}$$

We fix now $0<\beta <\beta _{1}$ and we set for any $N\ge 1$

$$\begin{aligned} \gamma&=\{z\in {{\mathbb {C}}}:-N\le \text{ Re }z\le N\,,\,\text{ Im }z=0\} \cup \{z\in {{\mathbb {C}}}:-N\le \text{ Im }z\le N\,,\,\text{ Re }z=N\}\\&\cup \{z\in {{\mathbb {C}}}:-N\le \text{ Re }z\le N\,,\,\text{ Im }z=-\beta \} \cup \{z\in {{\mathbb {C}}}:-\beta \le \text{ Im }z\le 0\,,\,\text{ Re }z=-N\}\,. \end{aligned}$$

Now, in view of the Cauchy theorem, we obtain that for any $N\ge 1$

$$\begin{aligned} \oint _{\gamma }\,e^{iz x}\,g(z)\mathrm {d}z&= \int ^{N}_{-N}\,e^{i\theta x} g(\theta )\,\mathrm {d}\theta + \int ^{-\beta }_{0}\,e^{i(N+i t)x} g(N+i t)\,\mathrm {d}t \nonumber \\&\quad + \int ^{-N}_{N} e^{i(-i\beta +\theta ) x} g(-i\beta +\theta )\mathrm {d}\theta \nonumber \\&\quad + \int ^{0}_{-\beta }e^{i(-N+i t)x} g(-N+i t)\mathrm {d}t =0\,. \end{aligned}$$

(A.4)

The conditions (A.2) provide that

$$\begin{aligned} \lim _{N\rightarrow \infty }\,\int ^{-\beta }_{0}\,e^{i(N+i t)x} g(N+i t)\,\mathrm {d}t= \lim _{N\rightarrow \infty }\,\int ^{0}_{-\beta }\,e^{i(-N+i t)x} g(-N+i t)\,\mathrm {d}t=0\,. \end{aligned}$$

Therefore, letting $N\rightarrow \infty$ in (A.4) we obtain (A.3). Hence we get the desired result. $\square$

The following technical lemma is also needed in this paper.

Lemma A.5

Let $g : [a,b]\rightarrow {{\mathbb {R}}}$ be a function from $\mathbf{L}_{1}[a,b]$. Then, for any fixed $-\infty \le a<b\le +\infty ,$

$$\begin{aligned} \lim _{N\rightarrow \infty }\,\int ^{b}_{a}\,g(x)\,\sin (Nx)\mathrm {d}x=0 \quad \text{ and }\quad \lim _{N\rightarrow \infty }\,\int ^{b}_{a}\,g(x)\,\cos (Nx)\mathrm {d}x=0 \,. \end{aligned}$$

(A.5)

Proof

Let first $-\infty< a<b< +\infty$. Assume that g is continuously differentiable, i.e. $g\in \mathbf{C}^{1}[a,b]$. Then integrating by parts gives us

$$\begin{aligned} \int ^{b}_{a}\,g(x)\,\sin (Nx)\,\mathrm {d}x =\frac{1}{N}\left( g(b)\,\sin (Nb)\ - g(a)\,\sin (Na)\ - \int ^{b}_{a}\,g^{'}(x)\,\cos (Nx)\,\mathrm {d}x \right) \,. \end{aligned}$$

So, from this we obtain that

$$\begin{aligned} \left| \int ^{b}_{a}\,g(x)\,\sin (Nx)\,\mathrm {d}x \right| \le \frac{\vert g(b)\vert +\vert g(a)\vert +(b-a)\max _{a\le x\le b}\vert g^{'}(x)\vert }{N} \,. \end{aligned}$$

This implies the first limit in (A.5) for this case. The second one is obtained similarly. Let now g be any absolutely integrated function on [a, b], i.e. $g\in \mathbf{L}_{1}[a,b]$. In this case there exists a sequence $g_{n}\in \mathbf{C}^{1}[a,b]$ such that

$$\begin{aligned} \lim _{n\rightarrow \infty }\int ^{b}_{a}\vert g(x)-g_{n}(x)\vert \mathrm {d}x=0\,. \end{aligned}$$

Therefore, taking into account that for any $n\ge 1$

$$\begin{aligned} \lim _{N\rightarrow \infty }\,\int ^{b}_{a}\,g_{n}(x)\,\sin (Nx)\mathrm {d}x=0\,, \end{aligned}$$

we obtain that

$$\begin{aligned} \limsup _{n\rightarrow \infty }\,\vert \int ^{b}_{a}\,g(x)\,\sin (Nx)\mathrm {d}x\vert \le \int ^{b}_{a}\vert g(x)-g_{n}(x)\vert \mathrm {d}x\,. \end{aligned}$$

So, letting in this inequality $n\rightarrow \infty$ we obtain the first limit in (A.5) and, similarly, we obtain the second one. Let now $b=+\infty$ and $a=-\infty$. In this case we obtain that for any $-\infty<a<b<+\infty$

$$\begin{aligned} \left| \int ^{+\infty }_{-\infty }\,g(x)\,\sin (Nx)\mathrm {d}x \right| \le&\left| \int ^{+\infty }_{-\infty }\,g(x)\,\sin (Nx)\mathrm {d}x \right| + \int ^{+\infty }_{b}\,\vert g(x)\,\vert \mathrm {d}x\\&+ \int ^{a}_{-\infty }\,\vert g(x)\,\vert \mathrm {d}x \,. \end{aligned}$$

Using here the previous results we obtain that for any $-\infty<a<b<+\infty$

$$\begin{aligned} \limsup _{N\rightarrow \infty } \left| \int ^{+\infty }_{-\infty }\,g(x)\,\sin (Nx)\mathrm {d}x \right| \le \int ^{+\infty }_{b}\,\vert g(x)\,\vert \mathrm {d}x + \int ^{a}_{-\infty }\,\vert g(x)\,\vert \mathrm {d}x \,. \end{aligned}$$

Passing here to limit as $b\rightarrow +\infty$ and $a\rightarrow -\infty$ we obtain the first limit in (A.5). Similarly, we can obtain the second one. $\square$

Let us now study the inverse Fourier transform. To this end, we need the following local Dini condition.

($\mathbf{D}$) Assume that, for some fixed$x\in {{\mathbb {R}}},$there exist the finite limits

$$\begin{aligned} g(x-)=\lim _{z\rightarrow x-}g(z) \quad \text{ and }\quad g(x+)=\lim _{z\rightarrow x+}g(z) \end{aligned}$$

and there exists$\delta =\delta (x)>0$for which

$$\begin{aligned} \int ^{\delta }_{0}\, \frac{ \vert \widetilde{g}(x,t) - \widetilde{g}_{0}(x) \vert }{t} \mathrm {d}t\,<\,\infty \,, \end{aligned}$$

where$\widetilde{g}(x,t)=g((x+t)+g(x-t))/2$and$\widetilde{g}_{0}(x)=(g(x+)+g(x-))/2$.

Remark 10.1

Note that the condition $(\mathbf{H}_{1}$) is the “uniform” version of the condition $\mathbf{D}$).

Proposition A.6

Let $g : {{\mathbb {R}}}\rightarrow {{\mathbb {R}}}$ be a function from $\mathbf{L}_{1}({{\mathbb {R}}})$. If, for some $x\in {{\mathbb {R}}},$ this function satisfies Condition $\mathbf{D}$, then

$$\begin{aligned} \widetilde{g}_{0}(x)= \frac{g(x+)+g(x-)}{2} = \frac{1}{2\pi } \int _{{{\mathbb {R}}}}\,e^{-i\theta x} \widehat{g}(\theta )\,\mathrm {d}\theta \,, \end{aligned}$$

(A.6)

where $\widehat{g}(\theta )=\int _{{{\mathbb {R}}}}\,e^{i\theta t}\,g(t)\,\mathrm {d}t$.

Proof

First, for any fixed $N>0$ we set

$$\begin{aligned} J_{N}(x)=\frac{1}{2\pi }\, \int ^{N}_{-N}\,e^{-i\theta x} \widehat{g}(\theta )\,\mathrm {d}\theta =\frac{1}{\pi }\, \int _{{{\mathbb {R}}}}\,g(z)\, \int ^{N}_{0}\,\cos (\theta (z- x))\,\mathrm {d}\theta \mathrm {d}z\,. \end{aligned}$$

Here, the inner integral may be represented as

$$\begin{aligned} J_{N}(x)= \frac{1}{\pi }\, \int _{{{\mathbb {R}}}}\,g(z)\, \frac{\sin (N(z- x))}{z- x} \, \mathrm {d}z = \frac{2}{\pi }\, \int ^{\infty }_{0}\, \widetilde{g}(x,t) \, \frac{\sin (N t)}{t} \, \mathrm {d}t\,. \end{aligned}$$

Taking into account that for any $N>0$ the integral

$$\begin{aligned} \frac{2}{\pi }\, \int ^{\infty }_{0}\, \frac{\sin (N t)}{t} \, \mathrm {d}t =1 \end{aligned}$$

(A.7)

we obtain that

$$\begin{aligned} J_{N}(x)-\widetilde{g}_{0}(x) = \frac{2}{\pi }\, \int ^{\infty }_{0}\, \frac{(\widetilde{g}(x,t)-\widetilde{g}_{0}(x))\sin (N t)}{t} \, \mathrm {d}t \,. \end{aligned}$$

Now we represent the last integral as

$$\begin{aligned} \int ^{\infty }_{0}\, \frac{(\widetilde{g}(x,t)-\widetilde{g}_{0}(x))\sin (N t)}{t} \, \mathrm {d}t =I_{1,N} + I_{2,N} - \widetilde{g}_{0}(x) I_{3,N}\,, \end{aligned}$$

where $I_{1,N}=\int ^{\delta }_{0}\,(\widetilde{g}(x,t)-\widetilde{g}_{0}(x))\,\sin (N t)\,t^{-1}\,\mathrm {d}t$,

$$\begin{aligned} I_{2,N}=\int ^{\infty }_{\delta }\, t^{-1}\,\widetilde{g}(x,t)\sin (N t)\mathrm {d}t \quad \text{ and } \quad I_{3,N}=\int ^{\infty }_{\delta }\,t^{-1}\,\sin (N t)\,\mathrm {d}t \,. \end{aligned}$$

Condition $\mathbf{D}$) and Lemma A.5 imply that $I_{1,N}\rightarrow 0$ as $N\rightarrow \infty$. Now note that, since $g\in \mathbf{L}_{1}({{\mathbb {R}}}),$ then the function $t^{-1}\,\widetilde{g}(x,t)$ is absolutely integrated. Therefore, in view of Lemma A.5, $I_{2,N}\rightarrow 0$ as $N\rightarrow \infty$. As to the last integral we use the property (A.7), i.e., the changing of the variables gives

$$\begin{aligned} I_{3,N}=\int ^{\infty }_{\delta N}\, \frac{\sin t}{t}\,\mathrm {d}t \rightarrow 0 \quad \text{ as }\quad N\rightarrow \infty \,. \end{aligned}$$

Hence we have the desired result. $\square$

1.4 A.4 Properties of the analytic functions

We need the following identity theorem (see, for example, Sveshnikov and Tikhonov 1978, Corollary 1, p. 78)

Theorem A.7

Let function $f(z)\not \equiv 0$ and be analytic in the domain D. In any closed bounded subdomain $D'\subset D$ it has only a finite numbers of zeros.

Lemma A.8

Assume that the condition $(\mathbf{H}_{3}$)–$(\mathbf{H}_{4}$) hold. Then there exists some $0<\beta _{*}<\beta$ such that the function $\check{G}$ defined in (5.9) is holomorphic in

$$\begin{aligned} U= \{\theta \in {{\mathbb {C}}}\,:\, \text{ Im }(\theta )\ge -\beta _{*}\} \end{aligned}$$

and for any $x\ge -\beta _{*}$

$$\begin{aligned} \sup _{\theta \in {{\mathbb {R}}}}\, \vert \check{G}(\theta +i x) \vert <\infty \,. \end{aligned}$$

(A.8)

Proof

Firstly, we remind that thanks to the condition $(\mathbf{H}_{3}$) the function $\widehat{g}(\theta )$ is holomorphic in $D=\{\theta \in {{\mathbb {C}}}\,:\,\text{ Im }(\theta )>-\beta \}$. Therefore, the function $\check{G}(\theta )$ is holomorphic for any $\theta$ from $D\cap \{\theta \in {{\mathbb {C}}}\,:\,\widehat{g}(\theta )\ne 1\}$. In this case its derivative can be calculated as

$$\begin{aligned} \check{G}^{\prime }(\theta )=\frac{\widehat{g}^{\prime }(\theta )}{(1-\widehat{g}(\theta ))^{2}} -\frac{1}{i{\check{\tau }}\theta ^{2}} \,, \end{aligned}$$

where $\widehat{g}^{\prime }(\theta )$ is the derivative of the function $\widehat{g}(\theta )$. Moreover, using the Taylor expansion for $e^{z}$, the function $\widehat{g}(\theta )$ and its derivative $\widehat{g}^{\prime }(\theta )$ can be represent as

$$\begin{aligned} \widehat{g}(\theta )&= \mathbf{E}_{Q}e^{i \tau \theta }\, = 1+i {\check{\tau }} \theta -\frac{{\check{\tau }}_{1} \theta ^{2}}{2} +\theta ^3 A_{1}(\theta )\,,\nonumber \\ \widehat{g}^{\prime }(\theta )&= i\mathbf{E}_{Q}\tau \,e^{i \tau \theta }\, = i{\check{\tau }} -{\check{\tau }}_{1} \theta + \theta ^2 A_{2}(\theta )\,, \end{aligned}$$

(A.9)

where ${\check{\tau }}_{1}=\mathbf{E}_{Q}\tau ^{2}_{1}$ and the terms $A_{1}(\theta )$ and $A_{2}(\theta )$ are such that for any $L>0$

$$\begin{aligned} \sup _{\vert \theta \vert \le L}\vert A_{1}(\theta )\vert<\infty \quad \text{ and }\quad \sup _{\vert \theta \vert \le L}\vert A_{2}(\theta )\vert <\infty \,. \end{aligned}$$

Therefore, there exists $0<\delta <1$ for which

$$\begin{aligned} \inf _{\vert \theta \vert \le \delta }\, \frac{\vert 1 - \widehat{g}(\theta ) \vert }{\vert \theta \vert } >0\,. \end{aligned}$$

(A.10)

So, the expansion (A.9) implies, that the function $\check{G}(\theta )$ is holomorphic at the point $\theta =0$ and for such $\delta >0$

$$\begin{aligned} \sup _{\vert \theta \vert \le \delta }\, \vert \check{G}(\theta ) \vert<\infty \quad \text{ and }\quad \sup _{\vert \theta \vert \le \delta }\, \vert \check{G}^{\prime }(\theta ) \vert <\infty \,. \end{aligned}$$

Moreover, note that in view of lemma A.5 for any $0<r<\beta$

$$\begin{aligned} \lim _{\vert \text{ Re }(\theta )\vert \rightarrow \infty ,\vert \text{ Im }(\theta )\vert \le r}\,\widehat{g}(\theta )\, =0\,. \end{aligned}$$

(A.11)

Taking into account here that, thanks to (5.6), the module $\vert \widehat{g}(\theta ) \vert < 1$ for any $\theta \in {{\mathbb {R}}}\setminus \{ 0\}$, we obtain

$$\begin{aligned} \mathbf{g}_{\delta }= \sup _{\vert \theta \vert> \delta } \vert \widehat{g}(\theta ) \vert < 1 \quad \text{ for } \text{ any }\quad \delta >0 \,. \end{aligned}$$

(A.12)

Therefore, the function $\check{G}(\theta )$ is holomorphic at the line $\{z\in {{\mathbb {C}}}\,:\,\text{ Im }(z)=0\}$ and

$$\begin{aligned} \sup _{\theta \in {{\mathbb {R}}}}\, \vert \check{G}(\theta ) \vert <\infty \,. \end{aligned}$$

(A.13)

In the general case note that, due to Lemma 5.1, the function $1-\widehat{g}(\theta )$ has no zeros on the line $\left\{ z\in {{\mathbb {C}}}\,:\,\text{ Im }(z)=-\beta _{1} \right\}$ for some $0<\beta _{1}<\beta$. Moreover, in view of Theorem A.7 for any $N>1$ there can be only finitely many zeros of the function $1-\widehat{g}(\theta )$ in

$$\begin{aligned} \left\{ z\in {{\mathbb {C}}}\,:\,-\beta _{1}\le \text{ Im }(z)\le 0\,,\,\vert \text{ Re }(z)\vert \le N \right\} \subset D\,. \end{aligned}$$

The limiting equality (A.11) implies that there exists $N>0$ such that the function $1-\widehat{g}(\theta )\ne 0$ on the set

$$\begin{aligned} \left\{ z\in {{\mathbb {C}}}\,:\,-\beta _{1}\le \text{ Im }(z)\le 0\,,\,\vert \text{ Re }(z)\vert > N \right\} \,. \end{aligned}$$

So, there can be only finitely many zeros of the function $1-\widehat{g}(\theta )$ in

$$\begin{aligned} \left\{ z\in {{\mathbb {C}}}\,:\,-\beta _{1}\le \text{ Im }(z)\le 0 \right\} \,. \end{aligned}$$

Moreover, taking into account the Taylor expansion (A.9) one can check that $\theta =0$ is an isolated zero of the $1-\widehat{g}(\theta )$. Therefore, there exists some $\beta _{*}>0$ for which the function $1-\widehat{g}(\theta )$ has no zeros in $\left\{ z\in {{\mathbb {C}}}\,:\,-\beta _{*}\le Im(z)<0 \right\}$. Note also that the function $1-\widehat{g}(\theta )$ has no zeros in $\left\{ z\in {{\mathbb {C}}}\,:\,\text{ Im }(z)>0 \right\}$. Since we already shown that the function $\check{G}(\theta )$ is holomorphic at the line $\{z\in {{\mathbb {C}}}\,:\,\text{ Im }(z)=0\}$ we obtain that it is holomorphic for any $\theta \in {{\mathbb {C}}}$ with $\text{ Im }(\theta )\ge -\beta _{*}$ and in view of (A.13) we obtain the upper bound (A.8). Hence Lemma A.8. $\square$

1.5 A.5 Proof of (5.8)

Fist note, that for any $\vert \theta \vert > \delta$ the inequality (A.12) implies

$$\begin{aligned} \vert G_{\epsilon }(\theta ) \vert \le \, \frac{1}{\vert 1-(1-\epsilon )\widehat{g}(\theta ) \vert } + \frac{\vert 1-i{\check{\tau }}\theta \vert }{\vert \epsilon -i{\check{\tau }}\theta \vert } \, \le \frac{1}{1-\mathbf{g}_{\delta }} +\frac{1}{{\check{\tau }} \delta } \,, \end{aligned}$$

i.e.

$$\begin{aligned} \sup _{0<\epsilon<1}\, \sup _{\vert \theta \vert > \delta } \vert G_{\epsilon }(\theta ) \vert <\infty \,. \end{aligned}$$

Moreover, from (A.9) we can obtain that

$$\begin{aligned} G_{\epsilon }(\theta ) =\frac{-\epsilon _{1}+\epsilon _{1}(1-i{\check{\tau }}\theta ) \widehat{g}(\theta )}{(1-\epsilon _{1}\widehat{g}(\theta ))(\epsilon -i{\check{\tau }}\theta )} =\frac{\epsilon _{1}{\check{\tau }}^2 \theta ^2+\epsilon _{1}\theta ^2 (1-i{\check{\tau }}\theta )A_{0}(\theta ) }{(1-\epsilon _{1}\widehat{g}(\theta ))(\epsilon -i{\check{\tau }}\theta )} \,, \end{aligned}$$

where $\epsilon _{1}=1-\epsilon$ and $A_{0}(\theta )=-{\check{\tau }}_{1}/2+\theta A_{1}(\theta )$. Note that $G_{\epsilon }(0)=0$ for any $\epsilon >0$. Let now $0<\vert \theta \vert \le \delta$ for some $\delta$ for which the inequality (A.10) holds. In this case for some $0<C_{*}<\infty$

$$\begin{aligned} \vert G_{\epsilon }(\theta ) \vert = \frac{\vert \epsilon _{1}{\check{\tau }}^2 \theta ^2+\epsilon _{1} \theta ^2(1-i{\check{\tau }}\theta ) A_{0}(\theta )\vert }{\vert 1-\epsilon _{1} \widehat{g}(\theta )\vert \sqrt{\epsilon ^{2}+{\check{\tau }}^{2}\theta ^{2}}} \le \, \frac{C_{*}\vert \theta \vert }{\vert 1-\epsilon _{1} \widehat{g}(\theta )\vert }\,. \end{aligned}$$

Moreover, taking into account here the lower bound (A.10) and that

$$\begin{aligned} \vert 1-\epsilon _{1} \widehat{g}(\theta )\vert =\sqrt{(1-\epsilon _{1} \text{ Re }(\widehat{g}(\theta )) )^2+\epsilon _{1}^2 ( \text{ Im }(\widehat{g}(\theta )))^2} \ge \epsilon _{1} \left| \text{ Im }(\widehat{g}(\theta )) \right| \,, \end{aligned}$$

we obtain that

$$\begin{aligned} \sup _{0<\epsilon<1}\, \sup _{\vert \theta \vert \le \delta } \vert G_{\epsilon }(\theta ) \vert < \infty \,. \end{aligned}$$

Hence the inequality (5.8). $\square$

1.6 A.6 Lower bound for the robust risks

In this section we give the lower bound obtained in Konev and Pergamenshchikov (2009b) (Theorem 6.1) for the robust risks (1.3) for the “signal+white noise” model, i.e. the model (1.1) in which the noise process $(\xi _{t})_{0\le t\le n}$ is defined as $\xi _{t}=\varsigma ^{*}w_{t}$ for $0\le t\le n$. To study the efficiency properties we need to obtain the lower bound for the mean square accuracy in the set of all possible estimators $\widehat{S}_{n}$ denoted by $\Pi _{n}$.

Theorem A.9

Under Conditions (2.12)

$$\begin{aligned} \liminf _{n\rightarrow \infty }\, \left( \frac{n}{\varsigma ^{*}}\right) ^{2k/(2k+1)} \, \inf _{\widehat{S}_{n}\in \Pi _{n}}\, \sup _{S\in W^{k}_{\mathbf{r}}} \,\mathbf{E}_{Q_{0},S}\Vert \widehat{S}_{n}-S\Vert ^{2} \ge \mathbf{r}^{*}_{k}\,, \end{aligned}$$

(A.14)

where $Q_{0}$ is the distribution of the noise process $(\varsigma ^{*}w_{t})_{0\le t\le n}$ and $\mathbf{r}^{*}_{k}$ is given in (4.13).