# Optimal recovery of linear operators from information of random functions

K. Yu. Osipenko  Moscow State University,
Institute for Information Transmission Problems, Russian Academy of Sciences,
Moscow<EMAIL_ADDRESS>
###### Abstract

The paper concerns problems of the recovery of linear operators defined on
sets of functions from information of these functions given with stochastic
errors. The constructed optimal recovery methods, in general, do not use all
the available information. As a consequence, optimal methods are obtained for
recovering derivatives of functions from Sobolev classes by the information of
their Fourier transforms given with stochastic errors. A similar problem is
considered for solutions of the heat equation.

###### keywords:

optimal recovery , random functions , Fourier transform

###### MSC:

[2010] 41A65 , 41A46 , 49N30 , 60G35

##  1 Introduction

There are several approaches to recovery problems from inaccurate information.
One of them concerns the case when the error in the initial data is
deterministic. Quite a lot of works are devoted to this case. The main results
can be found in [8], [9], [10], [11], [13] and the literature cited there.

Another approach is related to the fact that the initial information is
considered to be given with a random error. There are also many papers
dedicated to this topic. The following are the closest to the setting under
consideration: [2], [1], [14], [15], [12], [4]. A distinctive specificity of
this paper is that the information used here is not random vectors, but random
functions.

##  2 General setting

Denote by $\mathcal{W}$ the set of functions $x(\cdot)\in L_{2}(\mathbb{R})$
for which

$\int_{\mathbb{R}}\nu(t)|x(t)|^{2}\,dt<\infty,$

where $\nu(\cdot)$ is continuous and positive almost everywhere. Put

$W=\biggl{\\{}\,x(\cdot)\in\mathcal{W}:\int_{\mathbb{R}}\nu(t)|x(t)|^{2}\,dt\leq
1\,\biggr{\\}}.$

Consider the problem of optimal recovery of the operator $\Lambda
x(\cdot)=\mu(\cdot)x(\cdot)$ on the class $W$ by functions $x(\cdot)$ given
with random errors (we assume that $\mu(\cdot)$ is continuous and such that
$\Lambda$ maps $\mathcal{W}$ into $L_{2}(\mathbb{R})$). More precisely, for a
fixed $\delta>0$ and every $x(\cdot)\in W$ we consider the set of random
functions

$Y_{\delta}(x(\cdot))=\\{\,y_{\xi}(\cdot)\in
L_{2}(\mathbb{R}):\mathbb{M}y_{\xi}(\cdot)=x(\cdot),\
\mathbb{D}y_{\xi}(\cdot)\leq\delta\,\\}.$

We will also assume that the set of these random functions and the
corresponding probability measures are such that they allow for a change in
integration so that the equalities are valid

$\mathbb{M}\int_{\mathbb{R}}p(t)y_{\xi}(t)\,dt=\int_{\mathbb{R}}p(t)\mathbb{M}y_{\xi}(t)\,dt,\quad
p(t)\in L_{2}(\mathbb{R}),$ (1)

and

$\mathbb{M}\int_{\mathbb{R}}|y_{\xi}(t)|^{2}\,dt=\int_{\mathbb{R}}\mathbb{M}|y_{\xi}(t)|^{2}\,dt.$
(2)

As recovery methods we consider all possible mappings $\varphi\colon
L_{2}(\mathbb{R})\to L_{2}(\mathbb{R})$. The error of a method $m$ is defined
as

$e(\Lambda,W,\delta,\varphi)=\left(\sup_{\begin{subarray}{c}x(\cdot)\in W\\\
y_{\xi}(\cdot)\in Y_{\delta}(x(\cdot))\end{subarray}}\mathbb{M}\left(\|\Lambda
x(\cdot)-\varphi(y_{\xi}(\cdot))(\cdot)\|^{2}_{L_{2}(\mathbb{R})}\right)\right)^{1/2}.$

The problem is to find the error of optimal recovery

$E(\Lambda,W,\delta)=\inf_{\varphi\colon L_{2}(\mathbb{R})\to
L_{2}(\mathbb{R})}e(\Lambda,W,\delta,\varphi)$ (3)

and a method on which this infimum is attained which is called optimal.

We assume that $|\mu(\cdot)|$ and $\nu(\cdot)$ are even functions,
$|\mu(t)|>0$ almost everywhere, and $|\mu(\cdot)|/\sqrt{\nu(\cdot)}$ is a
monotonically decreasing function on $\mathbb{R}_{+}=[0,+\infty)$. Put

$f(s)=\int_{|t|\leq
s}\left(\frac{\sqrt{\nu(s)}}{|\mu(s)|}\frac{|\mu(t)|}{\sqrt{\nu(t)}}-1\right)\nu(t)\,dt.$
(4)

It is easy to check that $f(\cdot)$ is a monotonically increasing function.
Assume that $f(s)\to+\infty$ as $s\to+\infty$. Then for any $\delta>0$ the
equation $f(s)=\delta^{-2}$ has a unique solution $t_{\delta}$.

######  Theorem 1.

For all $\delta>0$

$E(\Lambda,W,\delta)=\delta\biggl{(}\int_{|t|\leq
t_{\delta}}|\mu(t)|^{2}\biggl{(}1-\dfrac{\sqrt{\nu(t)}}{|\mu(t)|}\frac{|\mu(t_{\delta})|}{\sqrt{\nu(t_{\delta})}}\biggr{)}\,dt\biggr{)}^{1/2}.$
(5)

Moreover, the method

$\varphi(y_{\xi}(\cdot))(t)=\left(1-\dfrac{\sqrt{\nu(t)}}{|\mu(t)|}\frac{|\mu(t_{\delta})|}{\sqrt{\nu(t_{\delta})}}\right)_{+}\mu(t)y_{\xi}(t)$

is optimal $(a_{+}=\max\\{a,0\\})$.

###### Proof.

1\. The lower bound. Consider the segment $[-A,A]\subset\mathbb{R}$. Let us
divide it on $2N$ parts by the points $x_{j}=\pm j\dfrac{A}{N}$,
$j=0,1,\ldots,N$. Set $a=(a_{1},\ldots,a_{2N})$,

$x_{a}(t)=\begin{cases}a_{2j-1},&t\in[x_{j-1},x_{j}),\ j=1,\ldots N,\\\
a_{2N-1},&t=A,\\\ a_{2j},&t\in[x_{-j},x_{-j+1}),\ j=1,\ldots,N,\\\
0,&|t|>A.\end{cases}$

Let $\tau=(\tau_{1},\ldots,\tau_{2N})$,
$\tau_{1}\geq\tau_{2}\geq\ldots\geq\tau_{2N}>0$, and $x_{\tau}(\cdot)\in W$.
Put

$B=\\{\,x_{a}(\cdot)\in W:a_{j}=\pm\tau_{j},\quad j=1,\ldots,2N\,\\}.$

It is obvious that $B\subset W$. Set

$p_{j}=\frac{\delta^{2}}{\delta^{2}+\tau_{j}^{2}},\quad j=1,\ldots,2n.$

Due to the monotony conditions of $\tau_{j}$, we have

$0<p_{1}\leq\ldots\leq p_{2n}<1.$

Any $x(\cdot)\in B$ can be written in the form

$x(\cdot)=\sum_{j=1}^{2n}s_{j}(x)\tau_{j}e_{j}(\cdot),$

where $s_{j}(x)\in\\{-1,1\\}$, and

$e_{2j-1}(\cdot)=\chi_{[x_{j-1},x_{j})}(\cdot),\quad
e_{2j}(\cdot)=\chi_{[x_{-j},x_{-j+1})}(\cdot),\quad j=1,\ldots,N$

($\chi_{\Omega}(\cdot)$ is the characteristic function of the set $\Omega$).

For each $x(\cdot)\in B$ we define the distribution $\eta(x)(\cdot)$ in the
following way:

$\eta(x)(\cdot)=\begin{cases}0,&\mbox{with probability }p_{1},\\\
\dfrac{s_{1}(x)\tau_{1}}{1-p_{1}}e_{1}(\cdot)&\mbox{with probability
}p_{2}-p_{1},\\\
\displaystyle\sum_{j=1}^{2}\dfrac{s_{j}(x)\tau_{j}}{1-p_{j}}e_{j}(\cdot),&\mbox{with
probability }p_{3}-p_{2},\\\ \dotfill&\dotfill\textstyle\\\
\displaystyle\sum_{j=1}^{2n-1}\dfrac{s_{j}(x)\tau_{j}}{1-p_{j}}e_{j}(\cdot),&\mbox{with
probability }p_{2n}-p_{2n-1},\\\
\displaystyle\sum_{j=1}^{2n}\dfrac{s_{j}(x)\tau_{j}}{1-p_{j}}e_{j}(\cdot),&\mbox{with
probability }1-p_{2n}.\end{cases}$

Thus, we have the following distribution for $\eta(x)(t)$ if
$t\in[x_{j-1},x_{j})$:

$\eta(x)(t)=\begin{cases}0,&\mbox{with probability }p_{2j-1},\\\
\dfrac{s_{2j-1}(x)\tau_{2j-1}}{1-p_{2j-1}},&\mbox{with probability
}1-p_{2j-1},\end{cases}\\\ j=1,\ldots,N,$

and if $t\in[x_{-j},x_{-j+1})$:

$\eta(x)(t)=\begin{cases}0,&\mbox{with probability }p_{2j},\\\
\dfrac{s_{2j}(x)\tau_{2j}}{1-p_{2j}},&\mbox{with probability
}1-p_{2j},\end{cases}\quad j=1,\ldots,N.$

It is easy to verify that $\mathbb{M}(\eta(x)(\cdot))(\cdot)=x(\cdot)$.
Moreover,

$\mathbb{D}(\eta(x)(\cdot))(t)=\begin{cases}\delta^{2},&t\in[A,A],\\\
0,&|t|>A.\end{cases}$

Consequently, $\eta(x)(\cdot)\in Y_{\delta}(x(\cdot))$ for all $x(\cdot)\in
B$.

Let $\varphi$ be an arbitrary recovery method. Taking into account that the
set $B$ is finite (with $2^{2N}$ elements), we have

$e^{2}(\Lambda,W,\delta,\varphi)\geq\sup_{x(\cdot)\in B}\mathbb{M}\|\Lambda
x(\cdot)-\varphi(\eta(x)(\cdot))(\cdot)\|_{L_{2}(\mathbb{R})}^{2}\\\
=\sup_{x(\cdot)\in
B}\biggl{(}\sum_{j=1}^{2N+1}(p_{j}-p_{j-1})\biggl{\|}\Lambda
x(\cdot)-\varphi\biggl{(}\sum_{k=1}^{j-1}\dfrac{s_{k}(x)\tau_{k}}{1-p_{k}}e_{k}(\cdot)\biggr{)}(\cdot)\biggr{\|}_{L_{2}(\mathbb{R})}^{2}\biggr{)}\\\
\geq\frac{1}{2^{2N}}\sum_{x(\cdot)\in
B}\biggl{(}\sum_{j=1}^{2N+1}(p_{j}-p_{j-1})\biggl{\|}\Lambda
x(\cdot)-\varphi\biggl{(}\sum_{k=1}^{j-1}\dfrac{s_{k}(x)\tau_{k}}{1-p_{k}}e_{k}(\cdot)\biggr{)}(\cdot)\biggr{\|}_{L_{2}(\mathbb{R})}^{2}\biggr{)}\\\
=\frac{1}{2^{2N}}\sum_{j=1}^{2N+1}(p_{j}-p_{j-1})\sum_{x(\cdot)\in
B}\biggl{\|}\Lambda
x(\cdot)-\varphi\biggl{(}\sum_{k=1}^{j-1}\dfrac{s_{k}(x)\tau_{k}}{1-p_{k}}e_{k}(\cdot)\biggr{)}(\cdot)\biggr{\|}_{L_{2}(\mathbb{R})}^{2};$
(6)

here $p_{0}=0$ and $p_{2N+1}=1$. Set

$B_{s_{1},\ldots,s_{j-1}}=\\{\,x(\cdot)\in
B:s_{1}(x)=s_{1},\ldots,s_{j-1}(x)=s_{j-1}\,\\},$

$j=1,\ldots,2N+1$ (for $j=1$ this set coincides with $B$). Then

$\frac{p_{j}-p_{j-1}}{2^{2N}}\sum_{x(\cdot)\in B}\biggl{\|}\Lambda
x(\cdot)-\varphi\biggl{(}\sum_{k=1}^{j-1}\dfrac{s_{k}(x)\tau_{k}}{1-p_{k}}e_{k}(\cdot)\biggr{)}(\cdot)\biggr{\|}_{L_{2}(\mathbb{R})}^{2}=\frac{p_{j}-p_{j-1}}{2^{2N}}\\\
\times\sum_{s_{1},\ldots,s_{j-1}}\,\,\sum_{x\in
B_{s_{1},\ldots,s_{j-1}}}\biggl{\|}\Lambda
x(\cdot)-\varphi\biggl{(}\sum_{k=1}^{j-1}\dfrac{s_{k}\tau_{k}}{1-p_{k}}e_{k}(\cdot)\biggr{)}(\cdot)\biggr{\|}_{L_{2}(\mathbb{R})}^{2}.$

If $x(\cdot)\in B_{s_{1},\ldots,s_{j-1}}$, then

$x(\cdot)=\sum_{k=1}^{j-1}s_{k}\tau_{k}e_{k}(\cdot)+z(x)(\cdot),\quad
z(x)(\cdot)=\sum_{k=j}^{2N}s_{k}(x)\tau_{k}e_{k}(\cdot).$

Moreover, with every element

$\sum_{k=1}^{j-1}s_{k}\tau_{k}e_{k}(\cdot)+z(x)(\cdot)\in
B_{s_{1},\ldots,s_{j-1}}$

the set $B_{s_{1},\ldots,s_{j-1}}$ contains the element

$\sum_{k=1}^{j-1}s_{k}\tau_{k}e_{k}(\cdot)-z(x)(\cdot).$

Thus,

$\frac{p_{j}-p_{j-1}}{2^{2N}}\sum_{s_{1},\ldots,s_{j-1}}\,\,\sum_{x\in
B_{s_{1},\ldots,s_{j-1}}}\biggl{\|}\Lambda
x(\cdot)-\varphi\biggl{(}\sum_{k=1}^{j-1}\dfrac{s_{k}\tau_{k}}{1-p_{k}}e_{k}(\cdot)\biggr{)}(\cdot)\biggr{\|}_{L_{2}(\mathbb{R})}^{2}\\\
=\frac{p_{j}-p_{j-1}}{2^{2N}}\sum_{s_{1},\ldots,s_{j-1}}\,\,\sum_{x\in
B_{s_{1},\ldots,s_{j-1}}}\biggl{\|}\Lambda\biggl{(}\sum_{k=1}^{j-1}s_{k}\tau_{k}e_{k}(\cdot)+z(x)(\cdot)\biggr{)}\\\
-\varphi\biggl{(}\sum_{k=1}^{j-1}\dfrac{s_{k}\tau_{k}}{1-p_{k}}e_{k}(\cdot)\biggr{)}(\cdot)\biggr{\|}_{L_{2}(\mathbb{R})}^{2}\\\
=\frac{p_{j}-p_{j-1}}{2^{2N}}\sum_{s_{1},\ldots,s_{j-1}}\,\,\sum_{x\in
B_{s_{1},\ldots,s_{j-1}}}\biggl{\|}\Lambda\biggl{(}\sum_{k=1}^{j-1}s_{k}\tau_{k}e_{k}(\cdot)\biggr{)}+\Lambda
z(x)(\cdot)\\\
-\varphi\biggl{(}\sum_{k=1}^{j-1}\dfrac{s_{k}\tau_{k}}{1-p_{k}}e_{k}(\cdot)\biggr{)}(\cdot)\biggr{\|}_{L_{2}(\mathbb{R})}^{2}\\\
=\frac{p_{j}-p_{j-1}}{2^{2N+1}}\sum_{s_{1},\ldots,s_{j-1}}\,\,\sum_{x\in
B_{s_{1},\ldots,s_{j-1}}}\biggl{(}\biggl{\|}\Lambda\biggl{(}\sum_{k=1}^{j-1}s_{k}\tau_{k}e_{k}(\cdot)\biggr{)}+\Lambda
z(x)(\cdot)\\\
-\varphi\biggl{(}\sum_{k=1}^{j-1}\dfrac{s_{k}\tau_{k}}{1-p_{k}}e_{k}(\cdot)\biggr{)}(\cdot)\biggr{\|}_{L_{2}(\mathbb{R})}^{2}+\biggl{\|}\Lambda\biggl{(}\sum_{k=1}^{j-1}s_{k}\tau_{k}e_{k}(\cdot)\biggr{)}-\Lambda
z(x)(\cdot)\\\
-\varphi\biggl{(}\sum_{k=1}^{j-1}\dfrac{s_{k}\tau_{k}}{1-p_{k}}e_{k}(\cdot)\biggr{)}(\cdot)\biggr{\|}_{L_{2}(\mathbb{R})}^{2}\biggr{)}\\\
\geq\frac{p_{j}-p_{j-1}}{2^{2N}}\sum_{s_{1},\ldots,s_{j-1}}\,\,\sum_{x\in
B_{s_{1},\ldots,s_{j-1}}}\|\Lambda z(x)(\cdot)\|_{L_{2}(\mathbb{R})}^{2}\\\
=\frac{p_{j}-p_{j-1}}{2^{2N}}\sum_{x\in B}\|\Lambda
z(x)(\cdot)\|_{L_{2}(\mathbb{R})}^{2}=(p_{j}-p_{j-1})\sum_{k=j}^{2N}\mu_{k}\tau_{k}^{2},$

where

$\mu_{2j-1}=\int_{x_{j-1}}^{x_{j}}|\mu(t)|^{2}\,dt,\quad\mu_{2j}=\int_{x_{-j}}^{x_{-j+1}}|\mu(t)|^{2}\,dt,\quad
j=1,\ldots,N.$

Substituting this estimate into (6), we get

$e^{2}(\Lambda,W,I,\delta,\varphi)\geq\sum_{j=1}^{2N+1}(p_{j}-p_{j-1})\sum_{k=j}^{2N}\mu_{k}\tau_{k}^{2}\\\
=\sum_{j=1}^{2N}\biggl{(}p_{j}\sum_{k=j}^{2N}\mu_{k}\tau_{k}^{2}-p_{j}\sum_{k=j+1}^{2N}\mu_{k}\tau_{k}^{2}\biggr{)}=\sum_{j=1}^{2N}p_{j}\mu_{j}\tau_{j}^{2}=\sum_{j=1}^{2N}\frac{\delta^{2}}{\delta^{2}+\tau_{j}^{2}}\mu_{j}\tau_{j}^{2}.$

Since the method $\varphi$ was chosen arbitrarily, we have

$E^{2}(\Lambda,W,I,\delta)\geq\sup_{\begin{subarray}{c}\tau_{1}\geq\ldots\geq\tau_{2N}>0\\\
x_{\tau}(\cdot)\in
W\end{subarray}}\sum_{j=1}^{2N}\frac{\delta^{2}}{\delta^{2}+\tau_{j}^{2}}\mu_{j}\tau_{j}^{2}.$
(7)

The condition $x_{\tau}(\cdot)\in W$ means that

$\int_{\mathbb{R}}\nu(t)|x_{\tau}(t)|^{2}\,dt=\sum_{j=1}^{2N}\nu_{j}\tau_{j}^{2}\leq
1,$

where

$\nu_{2j-1}=\int_{x_{j-1}}^{x_{j}}\nu(t)\,dt,\quad\nu_{2j}=\int_{x_{-j}}^{x_{-j+1}}\nu(t)\,dt,\quad
j=1,\ldots,N.$

Hence,

$E^{2}(\Lambda,W,I,\delta)\geq\sup_{\begin{subarray}{c}\tau_{1}\geq\ldots\geq\tau_{2N}>0\\\
\sum_{j=1}^{2N}\nu_{j}\tau_{j}^{2}\leq
1\end{subarray}}\sum_{j=1}^{2N}\frac{\delta^{2}}{\delta^{2}+\tau_{j}^{2}}\mu_{j}\tau_{j}^{2}.$

Let $\tau=(\tau_{1},\ldots,\tau_{k},0,\ldots,0)$, $1\leq k<2N$,
$\tau_{1}\geq\ldots\geq\tau_{k}>0$, and

$\sum_{j=1}^{k}\nu_{j}\tau_{j}^{2}\leq 1.$

For sufficiently small $\varepsilon>0$ we put
$\tau_{\varepsilon}=(\tau_{1}(\varepsilon),\ldots,\tau_{2N}(\varepsilon))$
where

$\tau_{j}(\varepsilon)=\begin{cases}\sqrt{\tau_{j}^{2}-\varepsilon},&1\leq
j\leq k,\\\ C\sqrt{\varepsilon},&k+1\leq j\leq 2N,\end{cases}$

and

$C=\left(\frac{\sum_{j=1}^{k}\nu_{j}}{\sum_{j=k+1}^{2N}\nu_{j}}\right)^{1/2}.$

Then

$\sum_{j=1}^{2N}\nu_{j}\tau_{j}^{2}(\varepsilon)=\sum_{j=1}^{k}\nu_{j}\tau_{j}^{2}-\varepsilon\sum_{j=1}^{k}\nu_{j}+C^{2}\varepsilon\sum_{j=k+1}^{2N}\nu_{j}=\sum_{j=1}^{k}\nu_{j}\tau_{j}^{2}\leq
1.$

For $\varepsilon<\tau_{k}^{2}/(1+C^{2})$ we have

$\sqrt{\tau_{k}^{2}-\varepsilon}>C\sqrt{\varepsilon}.$

Consequently, for such $\varepsilon$

$\tau_{1}(\varepsilon)\geq\ldots\geq\tau_{2N}(\varepsilon)>0.$

It follows from (7) that

$E^{2}(\Lambda,W,I,\delta)\geq\sum_{j=1}^{2N}\frac{\delta^{2}}{\delta^{2}+\tau_{j}^{2}(\varepsilon)}\mu_{j}\tau_{j}^{2}(\varepsilon).$

Passing to the limit as $\varepsilon\to 0$, we obtain

$E^{2}(\Lambda,W,I,\delta)\geq\sum_{j=1}^{k}\frac{\delta^{2}}{\delta^{2}+\tau_{j}^{2}}\mu_{j}\tau_{j}^{2}.$

Thus,

$E^{2}(\Lambda,W,I,\delta)\geq\sup_{\begin{subarray}{c}\tau_{1}\geq\ldots\geq\tau_{2N}\geq
0\\\ \sum_{j=1}^{2N}\nu_{j}\tau_{j}^{2}\leq
1\end{subarray}}\sum_{j=1}^{2N}\frac{\delta^{2}}{\delta^{2}+\tau_{j}^{2}}\mu_{j}\tau_{j}^{2}.$
(8)

Let the piecewise continuous function $x(\cdot)$ be such that $|x(\cdot)|$ is
an even function monotonically decreasing on $\mathbb{R}_{+}$ and

$\int_{\mathbb{R}}\nu(t)|x(t)|^{2}\,dt<1.$ (9)

Consider the integral

$I=\int_{\mathbb{R}}\frac{\delta^{2}}{\delta^{2}+|x(t)|^{2}}|\mu(t)|^{2}|x(t)|^{2}\,dt.$

Let us fix $\varepsilon>0$ and find $A>0$ such that

$I_{1}=\int_{-A}^{A}\frac{\delta^{2}}{\delta^{2}+|x(t)|^{2}}|\mu(t)|^{2}|x(t)|^{2}\,dt>I-\varepsilon.$
(10)

It is obvious that

$I_{2}=\int_{-A}^{A}\nu(t)|x(t)|^{2}\,dt<1.$

We will approximate these integrals by integral sums over partitions of
$T_{N}$, representing segments $[x_{j-1},x_{j}]$, $[x_{-j},x_{-j+1}]$,
$j=1,\ldots,N$, and points $t_{2j-1}=\dfrac{2j-1}{2N}$,
$t_{2j}=-\dfrac{2j-1}{2N}$, $j=1,\ldots,2N$.

For any $\varepsilon_{1}>0$, there is such a $N_{1}$ that for all $N>N_{1}$

$\frac{1}{N}\sum_{j=1}^{2N}\frac{\delta^{2}}{\delta^{2}+\tau_{j}^{2}}|\mu(t_{j})|^{2}\tau_{j}^{2}>I_{1}-\varepsilon_{1},$
(11)

where $\tau_{j}=|x(t_{j})|$. Moreover, there is such a $N_{2}$ that for all
$N>N_{2}$ for some $\omega>0$ the inequality

$\frac{1}{N}\sum_{j=1}^{2N}\nu(t_{j})\tau_{j}^{2}<1-\omega$

holds.

By the mean value theorem for integrals, there are $\xi_{j}$ such that
$\xi_{2j-1}\in[x_{j-1},x_{j}]$, $\xi_{2j}\in[x_{-j},x_{-j+1}]$,
$j=1,\ldots,N$, and $\mu_{j}=|\mu(\xi_{j})|^{2}/N$. Using the same arguments
we obtain that there are $\eta_{j}$ such that $\eta_{2j-1}\in[x_{j-1},x_{j}]$,
$\eta_{2j}\in[x_{-j},x_{-j+1}]$, $j=1,\ldots,N$, and
$\nu_{j}=\nu(\eta_{j})/N$. Due to the uniform continuity of the functions
$|\mu(\cdot)|$ and $\nu(\cdot)$ on the segment $[-A,A]$ for any
$\varepsilon_{2}>0$, there is $N_{3}$ such that for all
$s_{1},s_{2}\in[-A,A]$, $|s_{1}-s_{2}|<1/N_{3}$, inequalities

$||\mu(s_{1})|^{2}-|\mu(s_{2})|^{2}|<\varepsilon_{2},\quad|\nu(s_{1})-\nu(s_{2})|<\varepsilon_{2}$
(12)

hold.

Put

$M=\operatorname*{vraisup}_{t\in[0,1]}|x(t)|^{2}$

(due to the monotonous decrease of $|x(\cdot)|$ on $\mathbb{R}_{+}$, instead
of the segment $[0,1]$, we may take any segment $[0,b]$, $b>0$). Choose
$\varepsilon_{2}<\omega/(2M)$. Let $N>\max\\{N_{1},N_{2},N_{3}\\}$. Then
$\nu_{j}=\nu(\eta_{j})/N<\nu(t_{j})/N+\varepsilon_{2}/N$. Consequently,

$\sum_{j=1}^{2N}\nu_{j}\tau_{j}^{2}\leq\frac{1}{N}\sum_{j=1}^{2N}(\nu(t_{j})+\varepsilon_{2})\tau_{j}^{2}\leq\frac{1}{N}\sum_{j=1}^{2N}\nu(t_{j})\tau_{j}^{2}+2M\varepsilon_{2}<1.$
(13)

It follows from (12) that
$\mu_{j}=|\mu(\xi_{j})|^{2}/N>|\mu(t_{j})|^{2}/N-\varepsilon_{2}/N$.
Therefore,

$\sum_{j=1}^{2N}\frac{\delta^{2}}{\delta^{2}+\tau_{j}^{2}}\mu_{j}\tau_{j}^{2}\geq\frac{1}{N}\sum_{j=1}^{2N}\frac{\delta^{2}}{\delta^{2}+\tau_{j}^{2}}(|\mu(t_{j})|^{2}-\varepsilon_{2})\tau_{j}^{2}\\\
\geq\frac{1}{N}\sum_{j=1}^{2N}\frac{\delta^{2}}{\delta^{2}+\tau_{j}^{2}}|\mu(t_{j})|^{2}\tau_{j}^{2}-2M\varepsilon_{2}.$

Taking into account (13), (11), and (10), it follows from (7) that

$E^{2}(\Lambda,W,I,\delta)\geq\sum_{j=1}^{2N}\frac{\delta^{2}}{\delta^{2}+\tau_{j}^{2}}\mu_{j}\tau_{j}^{2}\geq\frac{1}{N}\sum_{j=1}^{2N}\frac{\delta^{2}}{\delta^{2}+\tau_{j}^{2}}|\mu(t_{j})|^{2}\tau_{j}^{2}-2M\varepsilon_{2}\\\
\geq I_{1}-\varepsilon_{1}-2M\varepsilon_{2}\geq
I-\varepsilon-\varepsilon_{1}-2M\varepsilon_{2}.$

Due to the fact that $\varepsilon$, $\varepsilon_{1}$ and $\varepsilon_{2}$
can be chosen arbitrarily small, we get

$E^{2}(\Lambda,W,I,\delta)\geq\sup_{\begin{subarray}{c}x(\cdot)\in W_{0}\\\
\int_{\mathbb{R}}\nu(t)|x(t)|^{2}\,dt<1\end{subarray}}\int_{\mathbb{R}}\frac{\delta^{2}}{\delta^{2}+|x(t)|^{2}}|\mu(t)|^{2}|x(t)|^{2}\,dt,$
(14)

where $W_{0}$ is the set of piecewise continuous functions $x(\cdot)$ such
that $|x(\cdot)|$ is an even function monotonically decreasing on
$\mathbb{R}_{+}$.

We show that the strict inequality on the right side of (14) can be replaced
by a non-strict one. Let $x(\cdot)\in W_{0}$ and

$\int_{\mathbb{R}}\nu(t)|x(t)|^{2}\,dt=1.$

Consider the function $y(\cdot)=(1+\varepsilon)^{-1/2}x(\cdot)$,
$\varepsilon>0$. Then it follows from (14) that

$E^{2}(\Lambda,W,I,\delta)\geq\frac{1}{1+\varepsilon}\int_{\mathbb{R}}\frac{\delta^{2}}{\delta^{2}+\dfrac{|x(t)|^{2}}{1+\varepsilon}}|\mu(t)|^{2}|x(t)|^{2}\,dt.$

Since

$\frac{1}{\delta^{2}+\dfrac{|x(t)|^{2}}{1+\varepsilon}}\geq\frac{1}{\delta^{2}+|x(t)|^{2}},$

we have

$E^{2}(\Lambda,W,I,\delta)\geq\frac{1}{1+\varepsilon}\int_{\mathbb{R}}\frac{\delta^{2}}{\delta^{2}+|x(t)|^{2}}|\mu(t)|^{2}|x(t)|^{2}\,dt.$

Passing $\varepsilon$ to zero, we get

$E^{2}(\Lambda,W,I,\delta)\geq\int_{\mathbb{R}}\frac{\delta^{2}}{\delta^{2}+|x(t)|^{2}}|\mu(t)|^{2}|x(t)|^{2}\,dt.$

Thus,

$E^{2}(\Lambda,W,I,\delta)\geq\sup_{\begin{subarray}{c}x(\cdot)\in W_{0}\\\
\int_{\mathbb{R}}\nu(t)|x(t)|^{2}\,dt\leq
1\end{subarray}}\int_{\mathbb{R}}\frac{\delta^{2}}{\delta^{2}+|x(t)|^{2}}|\mu(t)|^{2}|x(t)|^{2}\,dt.$
(15)

2\. The upper bound. Let us find the error of the methods having the form

$\varphi(y_{\xi}(\cdot))(\cdot)=\alpha(\cdot)\mu(\cdot)y_{\xi}(\cdot).$

Put $z_{\xi}(\cdot)=y_{\xi}(\cdot)-x(\cdot)$. Then
$\mathbb{M}z_{\xi}(\cdot)=0$, $\mathbb{D}z_{\xi}(\cdot)\leq\delta^{2}$. We
have

$e^{2}(\Lambda,W,I,\delta,\varphi)=\sup_{\begin{subarray}{c}x(\cdot)\in W\\\
y_{\xi}(\cdot)\in Y_{\delta}(x(\cdot))\end{subarray}}\mathbb{M}\left(\|\Lambda
x(\cdot)-\varphi(y_{\xi}(\cdot))(\cdot)\|^{2}_{L_{2}(\mathbb{R})}\right)\\\
=\sup_{\begin{subarray}{c}x(\cdot)\in W\\\ y_{\xi}(\cdot)\in
Y_{\delta}(x(\cdot))\end{subarray}}\mathbb{M}\left(\|\Lambda
x(\cdot)-\varphi(x(\cdot))(\cdot)-\varphi(z_{\xi}(\cdot))(\cdot)\|^{2}_{L_{2}(\mathbb{R})}\right)\\\
=\sup_{\begin{subarray}{c}x(\cdot)\in W\\\ y_{\xi}(\cdot)\in
Y_{\delta}(x(\cdot))\end{subarray}}\left(\|\Lambda
x(\cdot)-\varphi(x(\cdot))(\cdot)\|^{2}_{L_{2}(\mathbb{R})}+\mathbb{M}(\|\varphi(z_{\xi}(\cdot))(\cdot)\|^{2}_{L_{2}(\mathbb{R})})\right.\\\
\left.-2\mathbb{M}\operatorname*{Re}(\varphi(z_{\xi}(\cdot))(\cdot),\overline{\Lambda
x(\cdot)-\varphi(x(\cdot))(\cdot)})\right);$

here $(\cdot,\cdot)$ is the standard scalar product in $L_{2}(\mathbb{R})$. It
follows from the form of $\varphi$ that

$\mathbb{M}\operatorname*{Re}(\varphi(z_{\xi}(\cdot))(\cdot),\Lambda
x(\cdot)-\varphi(x(\cdot))(\cdot))\\\
=\operatorname*{Re}\mathbb{M}\left(\alpha(\cdot)\mu(\cdot)z_{\xi}(\cdot),\overline{\Lambda
x(\cdot)-\varphi(x(\cdot))(\cdot)}\right)\\\
=\operatorname*{Re}\left(\alpha(\cdot)\mu(\cdot)\mathbb{M}z_{\xi}(\cdot),\overline{\Lambda
x(\cdot)-\varphi(x(\cdot))(\cdot)}\right)=0.$

Due to the fact that
$\mathbb{M}(|z_{\xi}(\cdot)|^{2})=\mathbb{D}(y_{\xi}(\cdot))$, we have

$e^{2}(\Lambda,W,I,\delta,\varphi)=\sup_{\begin{subarray}{c}x(\cdot)\in W\\\
y_{\xi}(\cdot)\in
Y_{\delta}(x(\cdot))\end{subarray}}\left(\int_{\mathbb{R}}|\mu(t)|^{2}|1-\alpha(t)|^{2}|x(t)|^{2}\,dt\right.\\\
\left.+\int_{\mathbb{R}}|\mu(t)|^{2}|\alpha(t)|^{2}\mathbb{D}(y_{\xi}(t))\,dt\right)=\sup_{x(\cdot)\in
W}\int_{\mathbb{R}}|\mu(t)|^{2}|1-\alpha(t)|^{2}|x(t)|^{2}\,dt\\\
+\delta^{2}\int_{\mathbb{R}}|\mu(t)|^{2}|\alpha(t)|^{2}\,dt$

Since

$\int_{\mathbb{R}}|\mu(t)|^{2}|1-\alpha(t)|^{2}|x(t)|^{2}\,dt=\int_{\mathbb{R}}\frac{|\mu(t)|^{2}}{\nu(t)}|1-\alpha(t)|^{2}\nu(t)|x(t)|^{2}\,dt\\\
\leq\operatorname*{vraisup}_{t\in\mathbb{R}}\left(\frac{|\mu(t)|^{2}}{\nu(t)}|1-\alpha(t)|^{2}\right),$

we obtain

$e^{2}(\Lambda,W,I,\delta,\varphi)\leq\operatorname*{vraisup}_{t\in\mathbb{R}}\left(\frac{|\mu(t)|^{2}}{\nu(t)}|1-\alpha(t)|^{2}\right)+\delta^{2}\int_{\mathbb{R}}|\mu(t)|^{2}|\alpha(t)|^{2}\,dt.$

Put

$\alpha(t)=\left(1-\dfrac{\sqrt{\nu(t)}}{|\mu(t)|}\frac{|\mu(t_{\delta})|}{\sqrt{\nu(t_{\delta})}}\right)_{+}.$

Due to the monotonous decreasing of the function
$|\mu(\cdot)|/\sqrt{\nu(\cdot)}$ we get

$\operatorname*{vraisup}_{t\in\mathbb{R}}\left(\frac{|\mu(t)|^{2}}{\nu(t)}|1-\alpha(t)|^{2}\right)=\frac{|\mu(t_{\delta})|^{2}}{\nu(t_{\delta})}.$

Consequently,

$e^{2}(\Lambda,W,I,\delta,\varphi)\leq\frac{|\mu(t_{\delta})|^{2}}{\nu(t_{\delta})}\\\
+\delta^{2}\int_{|t|\leq
t_{\delta}}|\mu(t)|^{2}\biggl{(}1-\dfrac{\sqrt{\nu(t)}}{|\mu(t)|}\frac{|\mu(t_{\delta})|}{\sqrt{\nu(t_{\delta})}}\biggr{)}^{2}\,dt=\frac{|\mu(t_{\delta})|^{2}}{\nu(t_{\delta})}\\\
+\delta^{2}\int_{|t|\leq
t_{\delta}}|\mu(t)|^{2}\biggl{(}\biggl{(}1-\dfrac{\sqrt{\nu(t)}}{|\mu(t)|}\frac{|\mu(t_{\delta})|}{\sqrt{\nu(t_{\delta})}}\biggr{)}-\dfrac{\sqrt{\nu(t)}}{|\mu(t)|}\frac{|\mu(t_{\delta})|}{\sqrt{\nu(t_{\delta})}}\\\
+\dfrac{\nu(t)}{|\mu(t)|^{2}}\frac{|\mu(t_{\delta})|^{2}}{\nu(t_{\delta})}\biggr{)}\,dt=\delta^{2}\int_{|t|\leq
t_{\delta}}|\mu(t)|^{2}\biggl{(}1-\dfrac{\sqrt{\nu(t)}}{|\mu(t)|}\frac{|\mu(t_{\delta})|}{\sqrt{\nu(t_{\delta})}}\biggr{)}\,dt\\\
+\frac{|\mu(t_{\delta})|}{\sqrt{\nu(t_{\delta})}}\biggl{(}\frac{|\mu(t_{\delta})|}{\sqrt{\nu(t_{\delta})}}\biggl{(}1+\delta^{2}\int_{|t|\leq
t_{\delta}}\nu(t)\,dt\biggr{)}-\delta^{2}\int_{|t|\leq
t_{\delta}}|\mu(t)|\sqrt{\nu(t)}\,dt\biggr{)}.$

It follows from the definition of $t_{\delta}$ that

$\frac{|\mu(t_{\delta})|}{\sqrt{\nu(t_{\delta})}}\biggl{(}1+\delta^{2}\int_{|t|\leq
t_{\delta}}\nu(t)\,dt\biggr{)}-\delta^{2}\int_{|t|\leq
t_{\delta}}|\mu(t)|\sqrt{\nu(t)}\,dt=0.$

Thus,

$e^{2}(\Lambda,W,I,\delta,\varphi)\leq\delta^{2}\int_{|t|\leq
t_{\delta}}|\mu(t)|^{2}\biggl{(}1-\dfrac{\sqrt{\nu(t)}}{|\mu(t)|}\frac{|\mu(t_{\delta})|}{\sqrt{\nu(t_{\delta})}}\biggr{)}\,dt.$
(16)

Consider the function

$\widehat{x}(t)=\delta\biggl{(}\biggl{(}\frac{\sqrt{\nu(t_{\delta})}}{|\mu(t_{\delta})|}\frac{|\mu(t)|}{\sqrt{\nu(t)}}-1\biggr{)}_{+}\biggr{)}^{1/2}.$

It is obvious that $\widehat{x}(\cdot)\in W_{0}$. Moreover,

$\int_{\mathbb{R}}\nu(t)|\widehat{x}(t)|^{2}\,dt=\delta^{2}\int_{|t|\leq
t_{\delta}}\nu(t)\biggl{(}\frac{\sqrt{\nu(t_{\delta})}}{|\mu(t_{\delta})|}\frac{|\mu(t)|}{\sqrt{\nu(t)}}-1\biggr{)}\,dt=1.$

Taking into account (16), it follows from (15) that

$E^{2}(\Lambda,W,I,\delta)\geq\int_{\mathbb{R}}\frac{\delta^{2}}{\delta^{2}+|\widehat{x}(t)|^{2}}|\mu(t)|^{2}|\widehat{x}(t)|^{2}\,dt\\\
=\delta^{2}\int_{|t|\leq
t_{\delta}}|\mu(t)|^{2}\biggl{(}1-\dfrac{\sqrt{\nu(t)}}{|\mu(t)|}\frac{|\mu(t_{\delta})|}{\sqrt{\nu(t_{\delta})}}\biggr{)}\,dt\\\
\geq e^{2}(\Lambda,W,I,\delta,\varphi)\geq E^{2}(\Lambda,W,I,\delta).$

This implies (5) and the optimality of the method $\varphi$. ∎

Now we consider some examples of the application of Theorem 1.

##  3 Recovery of functions and their derivatives from the Fourier transform
given with random error

Denote by $\mathcal{W}_{2}^{r}(\mathbb{R})$ the set of functions $x(\cdot)\in
L_{2}(\mathbb{R})$ for which $x^{(r-1)}(\cdot)$ is locally absolutely
continuous and $x^{(r)}(\cdot)\in L_{2}(\mathbb{R})$. Put

$W_{2}^{r}(\mathbb{R})=\\{\,x(\cdot)\in\mathcal{W}_{2}^{r}(\mathbb{R}):\|x^{(r)}(\cdot)\|_{L_{2}(\mathbb{R})}\leq
1\,\\}.$

Suppose that the Fourier transform $Fx(\cdot)$ of the function $x(\cdot)\in
W_{2}^{r}(\mathbb{R})$ is given with a random error. We assume that instead of
the function $Fx(\cdot)$ we know a random function $y_{\xi}(\cdot)\in
L_{2}(\mathbb{R})$ such that $\mathbb{M}y_{\xi}(\cdot)=Fx(\cdot)$ and
$\mathbb{D}y_{\xi}(\cdot)\leq\delta$. Using this information, it is required
to recover the function $x^{(k)}(\cdot)$, $0\leq k<r$, in the
$L_{2}(\mathbb{R})$-metric.

The exact setting of the problem is as follows. For every $x(\cdot)\in
W_{2}^{r}(\mathbb{R})$ we consider the set of random functions

$Y_{\delta}(x(\cdot))=\\{\,y_{\xi}(\cdot)\in
L_{2}(\mathbb{R}):\mathbb{M}y_{\xi}(\cdot)=Fx(\cdot),\
\mathbb{D}y_{\xi}(\cdot)\leq\delta\,\\}.$ (17)

This set requires additional conditions given in the general setting (the
validity of the equalities (1) and (2)). Next, we define the error of the
recovery method $\varphi\colon L_{2}(\mathbb{R})\to L_{2}(\mathbb{R})$ as
follows

$e(D^{k},W_{2}^{r}(\mathbb{R}),\delta,\varphi)=\left(\sup_{\begin{subarray}{c}x(\cdot)\in
W_{2}^{r}(\mathbb{R})\\\ y_{\xi}(\cdot)\in
Y_{\delta}(x(\cdot))\end{subarray}}\mathbb{M}\left(\|x^{(k)}(\cdot)-\varphi(y_{\xi}(\cdot))(\cdot)\|^{2}_{L_{2}(\mathbb{R})}\right)\right)^{1/2}.$

The problem is to find the error of optimal recovery

$E(D^{k},W_{2}^{r}(\mathbb{R}),\delta)=\inf_{\varphi\colon
L_{2}(\mathbb{R})\to
L_{2}(\mathbb{R})}e(D^{k},W_{2}^{r}(\mathbb{R}),\delta,\varphi)$

and a method on which this infimum is attained.

It follows from the Parseval equality that

$\displaystyle\|x^{(r)}(\cdot)\|_{L_{2}(\mathbb{R})}^{2}=\frac{1}{2\pi}\int_{\mathbb{R}}|t|^{2r}|Fx(t)|^{2}\,dt,$
$\displaystyle\|x^{(k)}(\cdot)-\varphi(y_{\xi}(\cdot))(\cdot)\|^{2}_{L_{2}(\mathbb{R})}=\frac{1}{2\pi}\int_{\mathbb{R}}|(it)^{k}F(t)-F\varphi(y_{\xi}(\cdot))(t)|^{2}\,dt.$

Thus, the problem is reduced to problem (3) with $\nu(t)=t^{2r}$ and
$\mu(t)=(it)^{k}$. The function $f(\cdot)$ which was defined by (4), has the
form

$f(s)=\int_{|t|\leq
s}\left(\frac{s^{r-k}}{|t|^{r-k}}-1\right)t^{2r}\,dt=2s^{2r+1}\frac{r-k}{(2r+1)(r+k+1)}.$

The equation $f(s)=\delta^{-2}$ has the unique solution

$t_{\delta}=\left(\frac{(2r+1)(r+k+1)}{2\delta^{2}(r-k)}\right)^{\frac{1}{2r+1}}.$

It follows from Theorem 1

######  Theorem 2.

For all $\delta>0$ and $0\leq k<r$

$E(D^{k},W_{2}^{r}(\mathbb{R}),\delta)=\frac{(2r+1)^{\frac{2k+1}{2(2r+1)}}}{\sqrt{2k+1}}\left(2\delta^{2}\frac{r-k}{r+k+1}\right)^{\frac{r-k}{2(2r+1)}}.$

Moreover, the method

$\varphi(y_{\xi}(\cdot))(t)=F^{-1}\left((it)^{k}\alpha(t)y_{\xi}(t)\right)(t),$

where

$\alpha(t)=\left(1-t^{r-k}\left(\frac{2\delta^{2}(r-k)}{(2r+1)(r+k+1)}\right)^{\frac{r-k}{2r+1}}\right)_{+},$

is optimal.

Note that the optimal recovery method does not use all the information about
random functions $y_{\xi}(\cdot)$, but only the information contained in the
segment $[-t_{\delta},t_{\delta}]$. Moreover, the more accurate the
measurements (the smaller the variance $\delta^{2}$), the larger this segment
becomes.

The deterministic case of this problem was considered in [5] (see also [7]).

##  4 Recovery of the solution of the heat equation

The temperature distribution in an infinite rod is described by the equation

$\frac{\partial u}{\partial t}=\frac{\partial^{2}u}{\partial t^{2}},$

where $u(\cdot,\cdot)$ is the function on $[0,\infty)\times\mathbb{R}$) with a
given initial temperature distribution

$u(0,\cdot)=u_{0}(\cdot).$

Consider the problem of recovering the temperature distribution at an instant
of time $T$ from information about the Fourier transform of the initial
temperature distribution $u_{0}(\cdot)$, given with a random error. We assume
that the functions $u_{0}(\cdot)$, given the initial temperature distribution,
belong to the class $W_{2}^{r}(\mathbb{R})$. We are again considering a set of
random functions similar to the set of $\eqref{Yd}$

$Y_{\delta}(u_{0}(\cdot))=\\{\,y_{\xi}(\cdot)\in
L_{2}(\mathbb{R}):\mathbb{M}y_{\xi}(\cdot)=Fu_{0}(\cdot),\
\mathbb{D}y_{\xi}(\cdot)\leq\delta\,\\}$

with the same additional conditions regarding the change in the order of
integration. We define the error of the recovery method $\varphi\colon
L_{2}(\mathbb{R})\to L_{2}(\mathbb{R})$ as follows

$e(T,W_{2}^{r}(\mathbb{R}),\delta,\varphi)=\left(\sup_{\begin{subarray}{c}u_{0}(\cdot)\in
W_{2}^{r}(\mathbb{R})\\\ y_{\xi}(\cdot)\in
Y_{\delta}(u_{0}(\cdot))\end{subarray}}\mathbb{M}\left(\|u(T,\cdot)-\varphi(y_{\xi}(\cdot))(\cdot)\|^{2}_{L_{2}(\mathbb{R})}\right)\right)^{1/2}.$

The problem is to find the error of optimal recovery

$E(T,W_{2}^{r}(\mathbb{R}),\delta)=\inf_{\varphi\colon L_{2}(\mathbb{R})\to
L_{2}(\mathbb{R})}e(T,W_{2}^{r}(\mathbb{R}),\delta,\varphi)$

and a method on which this infimum is attained.

It is well known (see, for example, [3]) that for all $t\geq 0$ the equality

$F(u(t,\cdot))(\lambda)=e^{-\lambda^{2}t}Fu_{0}(\lambda)$

holds. It follows from the Parseval equality that

$\|u(T,\cdot)-\varphi(y_{\xi}(\cdot))(\cdot)\|^{2}_{L_{2}(\mathbb{R})}=\frac{1}{2\pi}\int_{\mathbb{R}}|e^{-\lambda^{2}T}Fu_{0}(\lambda)-F\varphi(y_{\xi}(\cdot))(\lambda)|^{2}\,d\lambda.$

Thus, the problem is reduced to problem (3) with $\nu(t)=t^{2r}$ and
$\mu(t)=e^{-t^{2}T}$. The function $f(\cdot)$ which was defined by (4), has
the form

$f(s)=\int_{|t|\leq
s}\left(\frac{s^{r}e^{s^{2}T}}{t^{r}e^{t^{2}T}}-1\right)t^{2r}\,dt=2s^{r}e^{s^{2}T}\int_{0}^{s}t^{r}e^{-t^{2}T}\,dt-\frac{s^{2r+1}}{2r+1}.$

It is easy to verify that $f(s)\to+\infty$ as $s\to+\infty$ (the monotonous
increase of $f(\cdot)$ was noted in the general case). Therefore, the equation
$f(s)=\delta^{-2}$ has a unique solution, which we denote by $t_{\delta}$.

From Theorem 1 we obtain the following result:

######  Theorem 3.

Put

$\alpha(t)=\left(1-\frac{t^{r}e^{t^{2}T}}{t_{\delta}^{r}e^{t_{\delta}^{2}T}}\right)_{+}.$

Then the equality

$E(T,W_{2}^{r}(\mathbb{R}),,\delta)=\delta\biggl{(}\int_{|t|\leq
t_{\delta}}e^{-2t^{2}T}\alpha(t)\,dt\biggr{)}^{1/2}$

holds. Moreover, the method

$\varphi(y_{\xi}(\cdot))(t)=F^{-1}\left(e^{-t^{2}T}\alpha(t)y_{\xi}(t)\right)(t)$

is optimal.

The deterministic case of this problem was considered in [6].

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