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# Hidden Markov model where higher noise makes smaller errors

Yury A Kutoyants

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## To cite this version:

Yury A Kutoyants. Hidden Markov model where higher noise makes smaller errors. 2025. hal-05419901

HAL Id: hal-05419901

https://univ-lemans.hal.science/hal-05419901v1

Preprint submitted on 16 Dec 2025

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Distributed under a Creative Commons Attribution 4.0 International License
# Hidden Markov model where higher noise makes smaller errors

YURY A. KUTOYANTS

Le Mans University, Le Mans, France

Tomsk State University, Tomsk, Russia

## Abstract

This work studies a model of a partially observed linear system that depends on some unknown parameters. It is supposed that the noises in the observation and state equations tend to zero but with different rates. It is shown that for some rates in the hidden component the higher noise makes the smaller error of parameter estimates. The proposed approximation of the unobserved component involves three steps. First, a method of moment estimator of unknown parameters is constructed, and second, this estimator is used to define the one-step MLE-process. Finally, the last estimator is substituted into the equations of the Kalman filter. The solution of obtained equations provides us with the desired approximation (adaptive Kalman filter). The asymptotic properties of all mentioned estimators and maximum likelihood and Bayesian estimators of the unknown parameters are detailed. The asymptotic efficiency of adaptive filtering is discussed.

Key words: Partially observed linear system, hidden Markov process, Kalman filter, parameter estimation, method of moments estimators, MLE and Bayesian estimators, One-step MLE-process, on-line approximation, adaptive Kalman filter.

# 1 Introduction

## 1.1 Model of observations

Consider the partially observed stochastic linear system

$$dX_t = f(\vartheta, t) Y_t dt + \varepsilon\sigma(t) dW_t, \quad X_0 = 0, \quad 0 \le t \le T, \quad (1)$$

$$dY_t = a(\vartheta, t) Y_t dt + \psi_{\varepsilon}b(\vartheta, t) dV_t, \quad Y_0 = y_0, \quad (2)$$

where $W_t, 0 \le t \le T$ and $V_t, 0 \le t \le T$ are independent Wiener processes, $f(\cdot, \cdot)$, $a(\cdot, \cdot)$, $b(\cdot, \cdot)$ and $\sigma(\cdot)$ are known functions, $\varepsilon \in (0, 1]$ and $\psi_{\varepsilon} \in (0, 1]$ are noise intensities.
The initial value $y_0$ is deterministic (known). The parameter $\vartheta \in \Theta \subset \mathcal{R}^d$ is unknown and the Gaussian process $Y^T = (Y_t, 0 \le t \le T)$ is unobservable (hidden). The set $\Theta \subset \mathcal{R}^d$ is open, convex and bounded.

Introduce

Conditions $\mathcal{A}_*$. The functions $f(\vartheta, t), a(\vartheta, t), b(\vartheta, t), \sigma(t); t \in [0, T], \vartheta \in \bar{\Theta}$ are bounded and $\sigma(t)^2 \ge c > 0$ for $t \in [0, T]$.

We suppose that the conditions $\mathcal{A}_*$ in this work are always fulfilled.

We have the problem of the approximation of the unobserved component $Y_t$ by observations $X^t = (X_s, 0 \le s \le t)$. It is known that the best mean squared approximation of $Y_t$ is the conditional expectation $m(\vartheta, t) = \mathbf{E}_{\vartheta}(Y_t|\mathfrak{F}_t^X)$. Here $\mathfrak{F}_t^X$ is the $\sigma$-algebra generated by the observations $X^t$. If $\vartheta$ is known then the solution of this problem is given by the following equations of Kalman-Bucy filtering [21]

$$dm(\vartheta, t) = \left[ a(\vartheta, t) - \frac{f(\vartheta, t)^2 \gamma(\vartheta, t)}{\varepsilon^2 \sigma(t)^2} \right] m(\vartheta, t) dt + \frac{f(\vartheta, t) \gamma(\vartheta, t)}{\varepsilon^2 \sigma(t)^2} dX_t, \quad (3)$$

$$\frac{d\gamma(\vartheta, t)}{dt} = 2a(\vartheta, t)\gamma(\vartheta, t) - \frac{f(\vartheta, t)^2 \gamma(\vartheta, t)^2}{\varepsilon^2\sigma(t)^2} + \psi_{\varepsilon}^2 b(\vartheta, t)^2, \quad \gamma(\vartheta, 0) = 0. \quad (4)$$

We suppose that the initial value $m(\vartheta, 0) = y_0 \neq 0$.

As the parameter $\vartheta$ is unknown, the adaptive filter is constructed in two steps: first this parameter is estimated and then the estimator is substituted in the equations (3), (4).

The adaptive filtering of the partially observed linear systems in continuous and discrete time attracted attention of many researchers, especially for the discrete time models. We can mention here some of such works [4], [7], [10], [13], [18], [41]. The main problem, of course, is the construction of the good estimators of the unknown parameters.

The likelihood ratio function for this model of observations is

$$L(\vartheta, X^T) = \exp \left( \int_0^T \frac{M(\vartheta, t)}{\varepsilon^2 \sigma(t)^2} dX_t - \int_0^T \frac{M(\vartheta, t)^2}{2\varepsilon^2 \sigma(t)^2} dt \right), \quad \vartheta \in \Theta. \quad (5)$$

Here $M(\vartheta, t) = f(\vartheta, t)m(\vartheta, t)$. Remark that the maximum likelihood estimator (MLE) $\hat{\vartheta}_{t,\varepsilon}$ or Bayes estimator (BE) $\tilde{\vartheta}_{t,\varepsilon}$ which depend on observations $X^t$ are defined by the relations

$$L(\hat{\vartheta}_{t,\varepsilon}, X^t) = \sup_{\vartheta \in \Theta} L(\vartheta, X^t), \qquad \tilde{\vartheta}_{t,\varepsilon} = \frac{\int_{\Theta} \vartheta p(\vartheta) L(\vartheta, X^t) d\vartheta}{\int_{\Theta} p(\vartheta) L(\vartheta, X^t) d\vartheta}. \quad (6)$$

In the case of BE we suppose that $\vartheta$ is a random variable and $p(\vartheta), \vartheta \in \Theta$ is the density of its distribution a priory and the loss function is quadratic. These estimators have nice
asymptotic properties, i.e., they are consistent, asymptotically normal and asymptotically efficient. But the use of them for the real approximation of $m(\vartheta_0, t)$ ($\vartheta_0$ is the true value) is almost impossible because the calculation of $\sup_{\vartheta \in \Theta} L(\vartheta, X^t)$ or the integrals in the case of BE requires many solutions (trajectories $m(\vartheta, s), \gamma(\vartheta, s), 0 \le s \le t$) of the systems (3), (4) for different $\vartheta \in \Theta$ and different $t \in (0, T]$. Of course, the numerical realization of such approximation is hard to do.

There are other approaches of parameter estimation. We mention two of them. The Expectation-Maximization (E-M) method allows to obtain an approximation of the MLE, see, e.g., [1], [6], [15] and references there in. Another possibility is to use the particle method, see, e.g., [3], [8], [9], [11], [12], [26], [42]. In the both cases the number of required solutions of (3), (4) is essentially reduced but the consistency of proposed estimators or asymptotic efficiency of the corresponding adaptive filters are often difficult to prove.

In this work the properties of estimators and adaptive filters are described in the asymptotics of small noise: $\varepsilon \to 0$ and $\psi_{\varepsilon} = \varepsilon^{\delta} \to 0$. We suppose that $\delta \in (0, 1)$. The case $\delta = 0$, i.e., $\psi_{\varepsilon} = 1$, was already studied in [30] and [32] and the case $\delta = 1$ was studied in [27], [36] (see as well [35], where similar method was used in a different problem).

Remark, that if $y_0 = 0$, then the notations $\bar{\varepsilon} = \psi_{\varepsilon}^{-1}\varepsilon$, $\bar{X}_t = \psi_{\varepsilon}^{-1}X_t$, $\bar{Y}_t = \psi_{\varepsilon}^{-1}Y_t$ allow us to re-write the system (1), (2) as

$$d\bar{X}_t = f(\vartheta, t) \bar{Y}_t dt + \bar{\varepsilon}\sigma(t) dW_t, \quad \bar{X}_0 = 0, \quad 0 \le t \le T, \quad (7)$$

$$d\bar{Y}_t = a(\vartheta, t) \bar{Y}_t dt + b(\vartheta, t) dV_t, \quad \bar{Y}_0 = 0. \quad (8)$$

To obtain consistent estimators we have to suppose that $\bar{\varepsilon} \to 0$, but in this case the system (7), (8) is a particular case (with $y_0 = 0$) of the system studied in [30] and [32]. Remark that in the case of the system (7), (8) the process $\bar{Y}_t, 0 \le t \le T$ does not depend on $\varepsilon$ and therefore the limit is the same Gaussian process. To avoid this coincidence we suppose that the initial value $y_0 \neq 0$. Then the limit $(\varepsilon = 0, \psi_{\varepsilon} = 0)$ system (1), (2) is non degenerate

$$\frac{dx_t(\vartheta)}{dt} = f(\vartheta, t) y_t(\vartheta), \quad x_0 = 0, \quad \frac{dy_t(\vartheta)}{dt} = a(\vartheta, t) y_t(\vartheta), \quad y_0, \quad 0 \le t \le T. \quad (9)$$

Recall as well the representation of the observations (1)

$$dX_t = f(\vartheta_0, t) m(\vartheta_0, t) dt + \varepsilon\sigma(t) d\bar{W}_t, \quad X_0 = 0, \quad 0 \le t \le T \quad (10)$$

where $\bar{W}_t, 0 \le t \le T$ is the innovation Wiener process [37].

The nonlinear in $Y_t$ systems similar to (1), (2) (Extended Kalman filters) with different levels of noises in the observation and hidden equations but without unknown parameters were studied in the works [2], [39], [40] and in some others. The main problem there was an approximation of the conditional expectation $E(Y_t|\mathfrak{F}_t^X)$. The problems of parameter estimation for partially observed nonlinear systems were discussed in many works too, see, e.g., [5], [16], [17], [20], [23], [24].
In the present work we study the adaptive filters constructed with the help of the recently introduced estimator, called *One-step MLE-process* $\vartheta_{t,\varepsilon}^*, 0 < t \le T$ which is a slight modification of the well known Le Cam's One-step MLE. It has the following properties which make it quite attractive in this and many other problems:

* $\vartheta_{t,\varepsilon}^*$ is $\mathfrak{F}_t^X$ measurable, i.e., it depends on the observations $X^t$,

* can be easy calculated for any $t$,

* has recursive structure,

* is asymptotically efficient.

The adaptive filter $m_{t,\varepsilon}^*$ is obtained by the substitution of $\vartheta_{t,\varepsilon}^*$ in the equations (3), (4). Moreover, the obtained adaptive filter either is asymptotically efficient or it can be easy modified to became asymptotically efficient. It is also shown that the MLE and BE are consistent, asymptotically normal

$$ \varphi_{\varepsilon}^{-1}(\hat{\vartheta}_{\varepsilon} - \vartheta_0) \implies \mathcal{N}(0, I(\vartheta_0)^{-1}), \quad \varphi_{\varepsilon}^{-1}(\tilde{\vartheta}_{\varepsilon} - \vartheta_0) \implies \mathcal{N}(0, I(\vartheta_0)^{-1}) \quad (11) $$

and asymptotically efficient. Here $\varphi_{\varepsilon} \to 0$ and $I(\vartheta_0)$ is Fisher information. The particularity of this model is in the relation between $\psi_{\varepsilon} = \varepsilon^{\delta}$ and the normalization function $\varphi_{\varepsilon} = \varepsilon^{h(\delta)}$. For the values $\delta \in (0, 1/3)$, $\delta = 1/3$ and $\delta \in (1/3, 1)$ we obtain three different statistical models with different Fisher information. Moreover, we have a curious situation in the case $\delta \in (0, 1/3)$ where for the *higher values of $\delta$ (smaller noise) we obtain the smaller values of $h(\delta)$*, i.e., *bigger error of estimation*.

Note that this approach was already applied to the construction of adaptive filters for the following models of observations: hidden ergodic Ornstein-Uhlenbeck process [34] ($T \to \infty$), hidden AR time series [33] ($T \to \infty$), hidden telegraph process [25] ($T \to \infty$), partially observed systems with unknown volatility [32] and linear system [36] the both with small noises $\varepsilon$ and $\psi_{\varepsilon} = \varepsilon$).

## 1.2 Parameter estimation

Let us first explain the relation between $\psi_{\varepsilon}$ and $\varphi_{\varepsilon}$. To study the MLE and BE we use the Ibragimov-Khasminskii general theory [19]. We suppose that all functions $f(\vartheta, t)$, $a(\vartheta, t)$, $b(\vartheta, t)$ are sufficiently smooth w.r.t. $\vartheta$ and we have regular statistical experiment. Note that according to (5) the properties of these estimators are defined by the asymptotic of the integral

$$ \int_0^T \frac{[M(\vartheta_0 + \varphi_{\varepsilon} u, t) - M(\vartheta_0, t)]^2}{\varepsilon^2 \sigma(t)^2} dt \to u^\top I(\vartheta_0) u. $$

The rate of the MLE is given by the function $\varphi_{\varepsilon}$ because in the regular statistical problems we have the convergences (11). The model (1), (2) corresponds to different statistical
problems depending on the value of $\delta$. Denote $a^*(\vartheta, t) = a(\vartheta, t) + \ell(\vartheta, t)$, where $\ell(\vartheta, t) = [\ln f(\vartheta, t)]'_t$.

Below it will be shown (see (34)) that if $\varepsilon/\psi_\varepsilon \to 0$ then the following representation is valid

$$ 
\begin{aligned}
& \frac{M(\vartheta, t) - M(\vartheta_0, t)}{\varepsilon} \\
&= \sqrt{\frac{\psi_\varepsilon}{\varepsilon} \frac{[f(\vartheta, t)b(\vartheta, t) - f(\vartheta_0, t)b(\vartheta_0, t)]}{\sqrt{2b(\vartheta, t)f(\vartheta, t)}}} \sqrt{\sigma(t)} \xi_{t,\varepsilon} (1 + o(1)) \\
&\quad + \frac{1}{\psi_\varepsilon} \frac{[a^*(\vartheta, t) - a^*(\vartheta_0, t)]}{b(\vartheta, t)f(\vartheta, t)} f(\vartheta_0, t) \sigma(t) y_t(\vartheta_0) (1 + o(1)) \\
&= \sqrt{\frac{\psi_\varepsilon}{\varepsilon}} L_1(\vartheta, \vartheta_0, t) \xi_{t,\varepsilon} (1 + o(1)) + \frac{1}{\psi_\varepsilon} L_2(\vartheta, \vartheta_0, t) (1 + o(1))
\end{aligned}
$$

with obvious notation. Here $\xi_{t,\varepsilon}, t \in (\tau_\varepsilon < t \le T)$ are Gaussian random variables, $\xi_{t,\varepsilon} \Rightarrow \xi_t \sim \mathcal{N}(0, 1)$. The random variables $\xi_t, 0 < t \le T$ are independent.

Note that we obtained 5 different problems depending on the relation between $1/\psi_\varepsilon$ and $\sqrt{\psi_\varepsilon/\varepsilon}$. Roughly speaking this can be explained as follows. Suppose that $\delta \in (0, 1)$. Then we can write

$$ 
\begin{aligned}
& \int_0^T \frac{[M(\vartheta_0 + \varphi_\varepsilon u, t) - M(\vartheta_0, t)]^2}{\varepsilon^2 \sigma(t)^2} dt \\
&= u^2 \int_0^T \left[ \sqrt{\varepsilon \psi_\varepsilon} \frac{\varphi_\varepsilon}{\varepsilon} \dot{L}_2(\vartheta_0, \vartheta_0, t) \quad \xi_{t,\varepsilon} + \frac{\varepsilon}{\psi_\varepsilon} \frac{\varphi_\varepsilon}{\varepsilon} \dot{L}_1(\vartheta_0, \vartheta_0, t) \right]^2 \frac{dt}{\sigma(t)^2} \\
&= u^2 \frac{\psi_\varepsilon \varphi_\varepsilon^2}{\varepsilon} \int_0^T \dot{L}_1(\vartheta_0, \vartheta_0, t)^2 \frac{dt}{\sigma(t)^2} + u^2 \frac{\varphi_\varepsilon^2}{\psi_\varepsilon^2} \int_0^T \dot{L}_2(\vartheta_0, \vartheta_0, t)^2 \frac{dt}{\sigma(t)^2} \\
&= u^2 \varphi_\varepsilon^2 \frac{\psi_\varepsilon}{\varepsilon} I_1(\vartheta_0) + u^2 \varphi_\varepsilon^2 \frac{1}{\psi_\varepsilon^2} I_2(\vartheta_0) \\
&= u^2 I_1(\vartheta_0) \mathbb{I}_{\{0<\delta\le 1/3\}} + u^2 I_2(\vartheta_0) \mathbb{I}_{\{1/3\le\delta\le 1\}}.
\end{aligned}
\quad (12) $$

Above we denoted $\dot{L}_i(\vartheta_0, \vartheta_0, t) = \partial L_i(\vartheta, \vartheta_0, t) / \partial \vartheta|_{\vartheta=\vartheta_0}$, $I_i(\vartheta_0), i = 1, 2$ are Fisher informations, set

$$ \varphi_\varepsilon = \varepsilon^\delta \mathbb{I}_{\{0<\delta\le 1/3\}} + \varepsilon^{\frac{1-\delta}{2}} \mathbb{I}_{\{1/3<\delta\le 1\}}, \quad (13) $$

used the relation

$$ \int_0^T \dot{L}_1(\vartheta_0, \vartheta_0, t) \dot{L}_2(\vartheta, \vartheta_0, t) \xi_{t,\varepsilon} dt = o(1) $$

and omitted once more the terms $o(1)$. Denote as well $I_0(\vartheta_0)$ the Fisher information defined in [32] ($d=1$).

The correspondence between $\psi_\varepsilon, \varphi_\varepsilon$ and $I_i(\vartheta_0), i = 0, 1, 2$ is the follows:
1. $\psi_{\varepsilon} = 1, (\delta = 0), \quad \varphi_{\varepsilon} = \sqrt{\varepsilon}, \quad I_1(\vartheta_0),$

2. $\psi_{\varepsilon} = \varepsilon^{\delta}, (0 < \delta < 1/3), \quad \varphi_{\varepsilon} = \varepsilon^{\frac{1-\delta}{2}}, \quad I_1(\vartheta_0),$

3. $\psi_{\varepsilon} = \varepsilon^{1/3}, (\delta = 1/3), \quad \varphi_{\varepsilon} = \varepsilon^{\frac{1}{3}}, \quad I_3(\vartheta_0) = I_1(\vartheta_0) + I_2(\vartheta_0),$

4. $\psi_{\varepsilon} = \varepsilon^{\delta}, (1/3 < \delta < 1), \quad \varphi_{\varepsilon} = \varepsilon^{\delta}, \quad I_2(\vartheta_0),$

5. $\psi_{\varepsilon} = \varepsilon, (\delta = 1), \quad \varphi_{\varepsilon} = \varepsilon, \quad I_0(\vartheta_0).$

The function $h(\delta)$, $\delta \in [0, 1]$ is given on Fig. 1. Thus we have 5 different problems.

<figure><img src="image_4.png" /><figcaption>Figure 1: Function $h(\delta)$</figcaption></figure>

We do not discuss the cases $\delta = 1$ and $\delta = 0$ which were already studied in the works [27], [36] and [30], [32] respectively. Remark that if $\delta = 0$ then the limit model ($\varepsilon = 0$) contains a random process.

The value of $\varphi_{\varepsilon}$ in the case $0 < \delta < 1/3$ seems to be a bit strange by the following reason. The equations (1), (2) can be called a *dynamic system perturbed by small noises*. It is natural to think that the smaller noise provides the smaller error of estimation. This is usually true but not in this case! We see that the larger values of $\delta$ (smaller noise) gives us the bigger error of estimation. The problem is not in mathematics but in the terminology. In this case the useful information about $\vartheta$ is in the functions $\psi_{\varepsilon}f(\vartheta, \cdot) b(\vartheta, \cdot)$ and therefore $\psi_{\varepsilon}$ in $\psi_{\varepsilon}b(\vartheta, \cdot)$ is no more a *noise* but can be called an *amplitude* of the signal $b(\vartheta, \cdot)$.

### 1.2.1 MDE

Below we describe the asymptotic properties of estimators in the case $\varepsilon/\psi_{\varepsilon} \to 0$ in two situations. First we consider the case with the fixed observation interval $[0, \tau]$ and then the case with the vanishing observation interval $[0, \tau_{\varepsilon}]$, where $\tau_{\varepsilon} \to 0$ as $\varepsilon \to 0$.
### 1.2.2 Fixed observation interval

The MDE $\check{\vartheta}_{\tau, \varepsilon}$ is defined by the equation

$$ \int_{0}^{\tau} [X_t - x_t(\check{\vartheta}_{\varepsilon})]^2 dt = \inf_{\vartheta \in \Theta} \int_{0}^{\tau} [X_t - x_t(\vartheta)]^2 dt. \quad (14) $$

Introduce the $d \times d$ matrix

$$ \check{\mathbf{T}}(\vartheta) = \int_{0}^{\tau} \dot{x}_{t}(\vartheta) \dot{x}_{t}^{\top}(\vartheta) dt, \quad \vartheta \in \Theta, $$

Gaussian vector

$$ \check{\zeta}(\vartheta) = \check{\mathbf{T}}(\vartheta)^{-1} \int_{0}^{\tau} \int_{0}^{t} b(\vartheta, r) \int_{r}^{t} f(\vartheta, s) \phi(\vartheta, r, s) \mathrm{d}s \mathrm{d}V_r \dot{x}_t(\vartheta) \mathrm{d}t, $$

where

$$ \phi(\vartheta, r, s) = \exp\left(\int_{r}^{s} a(\vartheta, v) \mathrm{d}v\right), $$

and the function

$$ \check{g}(\vartheta_0, \nu) \equiv \inf_{|\vartheta - \vartheta_0| > \nu} \int_{0}^{\tau} [x_t(\vartheta_0) - x_t(\vartheta)]^2 \mathrm{d}t. $$

Recall that $\dot{x}_t(\vartheta) = \partial x_t(\vartheta)/\partial \vartheta$.

Conditions $\check{\mathcal{B}}$.

$\check{\mathcal{B}}_1$. The functions $f(\vartheta, t)$, $a(\vartheta, t)$, $b(\vartheta, t)$; $\vartheta \in \Theta, t \in [0, \tau]$ have two continuous bounded derivatives w.r.t. $\vartheta \in \bar{\Theta}$.

$\check{\mathcal{B}}_2$. The matrix $\check{\mathbf{T}}(\vartheta)$ is uniformly non degenerate

$$ \inf_{\vartheta \in \Theta} \inf_{\|e\|=1, e \in \mathbb{R}^d} e^{\check{\mathbf{T}}(\vartheta)} e^{\top} > 0. $$

$\check{\mathcal{B}}_3$. For any $\nu > 0$

$$ \inf_{\vartheta_0 \in \Theta} \check{g}(\vartheta_0, \nu) > 0. $$

**Theorem 1.** Let the conditions $\check{\mathcal{B}}$ hold. Then the MDE is consistent, asymptotically normal

$$ \psi_{\varepsilon}^{-1} (\check{\vartheta}_{\tau, \varepsilon} - \vartheta_0) \implies \check{\zeta}(\vartheta_0) $$

and the polynomial moments converge: for any $p > 0$

$$ \psi_{\varepsilon}^{-p} \mathbf{E}_{\vartheta_0} \|\check{\vartheta}_{\tau, \varepsilon} - \vartheta_0\|^p \to \mathbf{E}_{\vartheta_0} \|\check{\zeta}(\vartheta_0)\|^p. \quad (15) $$
Proof. This theorem is a particular case of Theorem 7.5 in [27]. Note that

$$
\begin{align}
X_t &= \int_0^t y_0 \phi(\vartheta_0, 0, s) f(\vartheta_0, s) \, ds + \varepsilon \int_0^t \sigma(s) \, dW_s \nonumber \\
&\quad + \psi_\varepsilon \int_0^t f(\vartheta_0, s) \int_0^s \phi(\vartheta_0, r, s) b(\vartheta_0, r) \, dV_r \, ds \nonumber \\
&= x_t(\vartheta_0) + \psi_\varepsilon \int_0^t f(\vartheta_0, s) \int_0^s \phi(\vartheta_0, r, s) b(\vartheta_0, r) \, dV_r \, ds \left(1 + O\left(\frac{\varepsilon}{\psi_\varepsilon}\right)\right). \tag{16}
\end{align}
$$

The cases $\psi_\varepsilon/\varepsilon \to 0$ and $\psi_\varepsilon/\varepsilon = 1$ we do not consider, but remark that in these cases the representations of the observations (16) have other asymptotics:

$$
X_t = \int_0^t y_0 \phi(\vartheta_0, 0, s) f(\vartheta_0, s) \, ds + \varepsilon \int_0^t \sigma(s) \, dW_s \left( 1 + O\left( \frac{\psi_\varepsilon}{\varepsilon} \right) \right)
$$

and

$$
X_t = \int_0^t y_0 \phi(\vartheta_0, 0, s) f(\vartheta_0, s) \, ds + \varepsilon \left[ \int_0^t \sigma(s) \, dW_s + \int_0^t L(\vartheta_0, r, t) b(\vartheta_0, r) \, dV_r \right],
$$

where

$$
L(\vartheta_0, r, t) = \int_r^t \phi(\vartheta_0, r, s) f(\vartheta_0, s) \, ds,
$$

respectively. Of course the properties of MDE in all cases can be easily described.

Remark that the derivative w.r.t. $\vartheta$ is

$$
\dot{x}_t(\vartheta) = \int_0^t y_0 \phi(\vartheta, 0, s) \dot{f}(\vartheta, s) \, ds + \int_0^t y_0 f(\vartheta, s) \phi(\vartheta, 0, s) \int_0^s \dot{a}(\vartheta, r) \, dr \, ds.
$$

<figure><img src="image_5.png" /></figure>

1.2.3 Vanishing observation interval

Let us describe the properties of MDE

$$
\check{v}_{\tau_{\varepsilon}} = \arg \inf_{\vartheta \in \Theta} \|X_t - x_t(\vartheta)\|_{\tau}, \quad \|X_t - x_t(\vartheta)\|_{\tau}^2 = \int_0^{\tau_{\varepsilon}} [X_t - x_t(\vartheta)]^2 dt
$$

constructed by the observations $X^{\tau_{\varepsilon}} = (X_t, 0 \le t \le \tau_{\varepsilon})$ on the vanishing interval $[0, \tau_{\varepsilon}]$,
where $\tau_{\varepsilon} \to 0$. We will need such estimator for the construction of One-step MLE-process
later. The parameter $\vartheta$ is one-dimensional $\vartheta \in \Theta = (\alpha, \beta)$, where $\alpha$ and $\beta$ are finite.

Conditions $\mathcal{C}^\circ$.
$\mathcal{C}_1^\circ$. The functions $f(\vartheta, t)$, $a(\vartheta, t)$, $\vartheta \in \Theta$, $t \in [0, t_0]$ have two continuous bounded derivatives w.r.t. $\vartheta$. Here $t_0 > 0$ is some (small) real from the interval $[0, T]$.

$\mathcal{C}_2^\circ$. The functions $f(\vartheta, 0)$, $\dot{f}(\vartheta, 0)$, $b(\vartheta, 0)$ and $\sigma(0)$ are separated from zero uniformly on $\vartheta \in \Theta$.

$\mathcal{C}_3^\circ$. The functions $f(\vartheta, t)$, $a(\vartheta, t)$, $b(\vartheta, t)$, $\vartheta \in \Theta$, $t \in [0, t_0]$ have continuous bounded derivatives w.r.t. $t$.

Let us denote $\kappa = \inf_{\vartheta \in \Theta} |\dot{f}(\vartheta, 0)|$ and set

$$D(\vartheta) = y_0^{-2} \dot{f}(\vartheta, 0)^{-2} \left( \frac{33}{140} f(\vartheta, 0)^2 b(\vartheta, 0)^2 + \frac{6}{5} \sigma(0)^2 \right).$$

**Proposition 1.** Let conditions $\mathcal{C}^\circ$ hold and $\tau_\varepsilon = \varepsilon/\psi_\varepsilon$. Then the MDE $\check{\vartheta}_{\tau_\varepsilon}$ is consistent and asymptotically normal

$$\sqrt{\frac{\psi_\varepsilon}{\varepsilon}} (\check{\vartheta}_{\tau_\varepsilon} - \vartheta_0) \implies \mathcal{N}(0, D(\vartheta_0)). \quad (17)$$

*Proof.* At the beginning we suppose that $\sqrt{\tau_\varepsilon}\varepsilon^{-1} \to \infty$ and then we obtain the optimal rate $\tau_\varepsilon = \varepsilon/\psi_\varepsilon$. To verify the consistency we first obtain the relation as in the proof of Theorem 2.1 in [36]

$$\mathbf{P}_{\vartheta_0} (|\check{\vartheta}_{\tau_\varepsilon} - \vartheta_0| > \nu) \le \mathbf{P}_{\vartheta_0} \left( 2 \|X_t - x_t(\vartheta_0)\|_{\tau} > \check{g}_{\tau}(\vartheta_0, \nu) \right) \\ \le \frac{4}{\check{g}_{\tau}(\vartheta_0, \nu)^2} \mathbf{E}_{\vartheta_0} \|X_t - x_t(\vartheta_0)\|_{\tau}^2, \quad (18)$$

and then consider the asymptotics of these integrals as $\tau_\varepsilon \to 0$ (see [36] where the similar problems with $\psi_\varepsilon = \varepsilon$ were considered).

Let us compare the contributions of the terms containing Wiener processes $W_t$, $0 \le t \le \tau_\varepsilon$ and $V_t$, $0 \le t \le \tau_\varepsilon$.

The asymptotic $(t \to 0)$ of one integral is

$$\begin{aligned} J_1(t) &= \int_0^t f(\vartheta_0, s) \int_0^s \phi(\vartheta_0, r, s) b(\vartheta_0, r) \, dV_r \, ds \\ &= \int_0^t \int_r^t f(\vartheta_0, s) \phi(\vartheta_0, r, s) \, ds \, b(\vartheta_0, r) \, dV_r \\ &= f(\vartheta_0, 0) b(\vartheta_0, 0) \int_0^t (t-r) \, dV_r \, (1+o(1)) \end{aligned}$$

and

$$\mathbf{E}_{\vartheta_0} \int_0^{\tau_\varepsilon} J_1(t)^2 \, dt = f(\vartheta_0, 0)^2 b(\vartheta_0, 0)^2 \int_0^{\tau_\varepsilon} \int_0^t (t-r)^2 \, dr \, dt (1+o(1))$$
$$ = \frac{f(\vartheta_0, 0)^2 b(\vartheta_0, 0)^2}{12} \tau_{\varepsilon}^4 (1 + o(1)). $$

For another integral we have

$$ \mathbf{E}_{\vartheta_0} \int_0^{\tau_{\varepsilon}} \left( \int_0^t \sigma(s) \, dW_s \right)^2 dt = \frac{\sigma(0)^2}{2} \tau_{\varepsilon}^2 (1 + o(1)). $$

The function $\check{g}_{\tau}(\vartheta_0, \nu)$ is estimated as follows. We have

$$ \begin{aligned} x_t(\vartheta) - x_t(\vartheta_0) &= y_0 \int_0^t [f(\vartheta, s) \phi(\vartheta, 0, s) - f(\vartheta_0, s) \phi(\vartheta_0, 0, s)] \, ds \\ &= y_0 t [f(\vartheta, 0) - f(\vartheta_0, 0)] (1 + o(1)). \end{aligned} $$

Thus for the small $\tau_{\varepsilon}$

$$ \begin{aligned} \|x_t(\vartheta) - x_t(\vartheta_0)\|_{\tau}^2 &= \frac{y_0^2 \tau_{\varepsilon}^3}{3} [f(\vartheta, 0) - f(\vartheta_0, 0)]^2 (1 + o(1)) \\ &= \frac{y_0^2 \tau_{\varepsilon}^3}{3} \dot{f}(\bar{\vartheta}, 0)^2 [\vartheta - \vartheta_0]^2 (1 + o(1)). \end{aligned} $$

Therefore for some $\varepsilon_0 > 0$ and $\varepsilon < \varepsilon_0$ we obtain the estimate

$$ \check{g}_{\tau}(\vartheta_0, \nu) \geq \frac{y_0 \kappa}{2} \tau_{\varepsilon}^{3/2} \nu. $$

The RHS of the inequality (18) is

$$ \check{g}_{\tau}(\vartheta_0, \nu)^{-2} \mathbf{E}_{\vartheta_0} \|X_t(\vartheta) - x_t(\vartheta_0)\|_{\tau}^2 \leq C \psi_{\varepsilon}^2 \tau_{\varepsilon} + C \frac{\varepsilon^2}{\tau_{\varepsilon}}. \quad (19) $$

Any choice of the rate $\tau_{\varepsilon} \to 0$ such that $\varepsilon^2/\tau_{\varepsilon} \to 0$ makes the MDE consistent.

The asymptotic normality of this estimator will follow from the asymptotic normality of the ratio of two integrals

$$ \begin{aligned} \check{\vartheta}_{\tau_{\varepsilon}} - \vartheta_0 &= \frac{\int_0^{\tau_{\varepsilon}} [X_t - x_t(\vartheta_0)] \dot{x}_t(\vartheta_0) \, dt}{\int_0^{\tau_{\varepsilon}} \dot{x}_t(\vartheta_0)^2 \, dt} (1 + o(1)) \\ &= \frac{3 \psi_{\varepsilon} f(\vartheta_0, 0) b(\vartheta_0, 0)}{\tau_{\varepsilon}^3 y_0 \dot{f}(\vartheta_0, 0)} \int_0^{\tau_{\varepsilon}} t \int_0^t (t-r) \, dV_r \, dt (1 + o(1)) \\ &\quad + \frac{3 \varepsilon \sigma(0)}{\tau_{\varepsilon}^3 y_0 \dot{f}(\vartheta_0, 0)} \int_0^{\tau_{\varepsilon}} t W_t \, dt (1 + o(1)). \end{aligned} $$

Here we used the asymptotics

$$ \int_0^{\tau_{\varepsilon}} \dot{x}_t (\vartheta_0)^2 \, dt = \frac{y_0^2}{3} \dot{f}(\vartheta_0, 0)^2 \tau_{\varepsilon}^3 (1 + o(1)). $$
Remark that

$$ \int_0^{\tau_{\varepsilon}} t \int_0^{\tau_{\varepsilon}} (t-r) \, dV_r \, dt = \int_0^{\tau_{\varepsilon}} \int_r^{\tau_{\varepsilon}} t(t-r) \, dt \, dV_r \sim \mathcal{N} \left( 0, \frac{11\tau_{\varepsilon}^7}{420} \right) = \sqrt{\frac{11\tau_{\varepsilon}^7}{420}} \zeta_1, \\ \int_0^{\tau_{\varepsilon}} t W_t \, dt \sim \mathcal{N} \left( 0, \frac{2\tau_{\varepsilon}^5}{15} \right) = \sqrt{\frac{2\tau_{\varepsilon}^5}{15}} \zeta_2 (1+o(1)). $$

Here $\zeta_1 \sim \mathcal{N}(0, 1)$ and $\zeta_2 \sim \mathcal{N}(0, 1)$ are independent r.v.'s.

Thus

$$ \check{\vartheta}_{\tau_{\varepsilon}} - \vartheta_0 = \frac{f(\vartheta_0, 0) b(\vartheta_0, 0)}{y_0 \dot{f}(\vartheta_0, 0)} \sqrt{\frac{33}{140}} \zeta_1 \psi_{\varepsilon} \sqrt{\tau_{\varepsilon}} (1+o(1)) \\ + \frac{\sigma(0)}{y_0 \dot{f}(\vartheta_0, 0)} \sqrt{\frac{6}{5}} \zeta_2 \frac{\varepsilon}{\sqrt{\tau_{\varepsilon}}} (1+o(1)). $$

The equation of balance

$$ \psi_{\varepsilon} \sqrt{\tau_{\varepsilon}} = \frac{\varepsilon}{\sqrt{\tau_{\varepsilon}}} $$

gives us the rate $\tau_{\varepsilon} = \varepsilon/\psi_{\varepsilon}$. Hence

$$ \sqrt{\frac{\psi_{\varepsilon}}{\varepsilon}} (\check{\vartheta}_{\tau_{\varepsilon}} - \vartheta_0) = \frac{f(\vartheta_0, 0) b(\vartheta_0, 0)}{y_0 \dot{f}(\vartheta_0, 0)} \sqrt{\frac{33}{140}} \zeta_1 + \frac{\sigma(0)}{y_0 \dot{f}(\vartheta_0, 0)} \sqrt{\frac{6}{5}} \zeta_2 + o(1). $$

Here the equality is understood in distribution.

<figure><img src="image_6.png" /></figure>

**Remark 1.** The Proposition 1 describes the properties of MDE in the case $|\dot{f}(\vartheta, 0)| > 0$. If $\dot{f}(\vartheta, 0) = 0$ and $|\dot{a}(\vartheta, 0)| > 0$, then the MDE $\check{\vartheta}_{\tau_{\varepsilon}}$ can be consistent but with the worse rate.

Note that in the case $f(\vartheta, t) = f(t)$, $a(\vartheta, t) = a(t)$ and $|\dot{b}(\vartheta, 0)| > 0$ the consistent estimation is impossible for this model of observations.

**Remark 2.** Remark that if $\sigma(t) \equiv 0$ and the observed system is

$$ \begin{aligned} \mathrm{d}X_t &= f(\vartheta_0, t) Y_t \mathrm{d}t, & X_0 &= 0, & 0 \le t \le T, \\ \mathrm{d}Y_t &= a(\vartheta_0, t) Y_t \mathrm{d}t + \psi_{\varepsilon} b(\vartheta_0, t) \mathrm{d}V_t, & Y_0 &= y_0 \ne 0, \end{aligned} $$

then the problem of estimation $\vartheta_0$ is singular. This means that even if $\psi_{\varepsilon} \equiv 1$ the value $\vartheta_0$ can be “estimated” without error.
**Remark 3.** If $\delta > 1/3$, then the rate $\varphi_{\varepsilon}^{(f)} = \psi_{\varepsilon} = \varepsilon^{\delta}$ of convergence of MDE constructed by observations on the fixed interval is better than the rate $\varphi_{\varepsilon}^{(v)} = \sqrt{\varepsilon/\psi_{\varepsilon}} = \varepsilon^{(1-\delta)/2}$ of the same estimator constructed by the observations on the vanishing interval. Indeed, the comparison of the rates gives us

$$ \frac{\varphi_{\varepsilon}^{(f)}}{\varphi_{\varepsilon}^{(v)}} = \psi_{\varepsilon} \sqrt{\frac{\psi_{\varepsilon}}{\varepsilon}} = \varepsilon^{\frac{3\delta-1}{2}} \rightarrow 0. \qquad (20) $$

If $\delta < 1/3$ the situation changes, i.e., one can think that less observations is better for the estimation of the unknown parameter! We have to note that this is not a general property of the model but the property of the particular estimator. Mention that in the next sections the properties of the MLE and BE are studied under condition $\delta > 1/3$, $\delta = 1/3$ and $\delta < 1/3$. As it was mentioned above this model of observations with $\delta < 1/3$ is in some sense more to the model studied in the Section 3.1.

**Remark 4.** Note that if $y_0 = 0$ then the MDE proposed above with fixed and vanishing intervals can not be consistent. This follows from the equality $y_t(\vartheta) \equiv 0$ which gives $x_t(\vartheta) \equiv 0$.

### 1.2.4 MLE and BE.

We consider separately three cases: $\delta \in (1/3, 1)$, $\delta \in (0, 1/3)$, $\delta = 1/3$. The detailed proofs are given in the first case only because the proofs in the two other cases follow the same lines and repeat mainly the first proof.

**Case** $\delta \in (1/3, 1)$.

The MLE $\hat{\vartheta}_{\varepsilon}$ and BE $\tilde{\vartheta}_{\varepsilon}$ are defined by the relations (6). If this equation for the MLE has more than one solution, then any one of them can be taken as MLE. Here $p(\vartheta), \vartheta \in \Theta$, $\vartheta \subset \mathbb{R}^d$ is some known continuous positive density function.

We study the properties of these estimators in the case $\psi_{\varepsilon} = \varepsilon^{\delta} \rightarrow 0$, where $\frac{1}{3} < \delta < 1$.

Denote $\ell(\vartheta, t) = \partial \ln f(\vartheta, t) / \partial t = f'_{t}(\vartheta, t) / f(\vartheta, t)$. Then the Fisher information in this problem can be written as follows

$$ I_2(\vartheta) = \int_0^T \frac{[\dot{a}(\vartheta, t) + \dot{\ell}(\vartheta, t)][\dot{a}(\vartheta, t) + \dot{\ell}(\vartheta, t)]^\top}{b(\vartheta, t)^2} y_t(\vartheta)^2 \, dt, \quad \vartheta \in \Theta. $$

Define as well the function

$$ G(\vartheta, \vartheta_0) = \int_0^T \frac{[a(\vartheta, t) - a(\vartheta_0, t) + \ell(\vartheta, t) - \ell(\vartheta_0, t)]^2 y_t(\vartheta_0)^2}{b(\vartheta, t)^2} dt. $$

The properties of estimators will be derived under the following regularity conditions.

Conditions $\hat{\mathcal{D}}$.
$\hat{\mathcal{D}}_1$. The functions $f(\vartheta, t)$, $a(\vartheta, t)$, $b(\vartheta, t)$, $t \in [0, T]$, $\vartheta \in \Theta$ and $\sigma(t)$, $t \in [0, T]$ have continuous derivatives w.r.t. $t$.

$\hat{\mathcal{D}}_2$. The functions $f(\vartheta, t)$, $b(\vartheta, t)$, $t \in [0, T]$, $\vartheta \in \Theta$ and $\sigma(t)$, $0 \le t \le T$ are bounded and separated from zero.

$\hat{\mathcal{D}}_3$. The functions $f(\vartheta, t)$, $a(\vartheta, t)$, $t \in [0, T]$, $\vartheta \in \Theta$ are two times continuously differentiable w.r.t. $\vartheta$.

$\hat{\mathcal{D}}_4$. The Fisher information matrix is uniformly non degenerate

$$ \inf_{\vartheta \in \Theta} \inf_{\|e\|=1, e \in \mathbb{R}^d} e^{\top} I_2(\vartheta) e > 0. $$

$\hat{\mathcal{D}}_5$. For any $\nu > 0$

$$ g(\nu) \equiv \inf_{\vartheta_0 \in \Theta} \inf_{\|\vartheta - \vartheta_0\| > \nu} G(\vartheta, \vartheta_0) > 0. $$

For the sake of simplicity and without loss of generality, we assume that the functions $f(\cdot), b(\cdot)$ and $\sigma(\cdot)$ are positive.

The family of measures $\{\mathbf{P}_{\vartheta}^{(\varepsilon)}, \vartheta \in \Theta\}$ induced by the observation process $X^T$ on the space $(C[0, T], \mathfrak{B}_T)$ of continuous functions on $[0, T]$ is locally asymptotically normal (LAN) (see Lemma 3 below). Therefore the mean squared estimation error satisfies the Hajek's minimax lower bound: for any $\vartheta_0 \in \Theta$ and any estimator $\bar{\vartheta}_{\varepsilon}$

$$ \lim_{\nu \to 0} \lim_{\varepsilon \to 0} \sup_{\|\vartheta - \vartheta_0\| \le \nu} \psi_{\varepsilon}^{-2} \mathbf{E}_{\vartheta} \|\bar{\vartheta}_{\varepsilon} - \vartheta\|^2 \ge \mathbf{E}_{\vartheta_0} \|\zeta(\vartheta_0)\|^2, \quad \zeta(\vartheta_0) \sim \mathcal{N}(0, I_2(\vartheta_0)^{-1}). $$

For the proof see, e.g., [19]. The estimator $\vartheta_{\varepsilon}^*$ is called asymptotically efficient if for any $\vartheta_0 \in \Theta$

$$ \lim_{\nu \to 0} \lim_{\varepsilon \to 0} \sup_{\|\vartheta - \vartheta_0\| \le \nu} \psi_{\varepsilon}^{-2} \mathbf{E}_{\vartheta} \|\vartheta_{\varepsilon}^* - \vartheta\|^2 = \mathbf{E}_{\vartheta_0} \|\zeta(\vartheta_0)\|^2. \quad (21) $$

**Theorem 2.** Let conditions $\hat{\mathcal{D}}$ hold. Then the MLE $\hat{\vartheta}_{\varepsilon}$ and BE $\tilde{\vartheta}_{\varepsilon}$ are uniformly consistent, asymptotically normal

$$ \psi_{\varepsilon}^{-1} (\hat{\vartheta}_{\varepsilon} - \vartheta_0) \implies \zeta(\vartheta_0), \quad \psi_{\varepsilon}^{-1} (\tilde{\vartheta}_{\varepsilon} - \vartheta_0) \implies \zeta(\vartheta_0), $$

the convergence of moments holds,

$$ \psi_{\varepsilon}^{-p} \mathbf{E}_{\vartheta_0} \|\hat{\vartheta}_{\varepsilon} - \vartheta_0\|^p \to \mathbf{E}_{\vartheta_0} \|\zeta(\vartheta_0)\|^p, \quad \psi_{\varepsilon}^{-p} \mathbf{E}_{\vartheta_0} \|\tilde{\vartheta}_{\varepsilon} - \vartheta_0\|^p \to \mathbf{E}_{\vartheta_0} \|\zeta(\vartheta_0)\|^p, $$

for any $p > 0$, and both estimators are asymptotically efficient.
Proof. We verify the conditions of the Theorems 3.1.1 and 3.2.1 in [19].

In the case $\vartheta = \vartheta_0$ we have

$$dm(\vartheta_0, t) = a(\vartheta_0, t)m(\vartheta_0, t) dt + \frac{\gamma(\vartheta_0, t)f(\vartheta_0, t)}{\varepsilon\sigma(t)} d\bar{W}_t, \quad m(\vartheta_0, 0) = y_0.$$

Here $\bar{W}_t, 0 \le t \le T$ is the innovation Wiener process (see (10)).

Define the normalized likelihood ratio process

$$Z_{\varepsilon}(\vartheta_0, u) = \frac{L(\vartheta_0 + \psi_{\varepsilon}u, X^T)}{L(\vartheta_0, X^T)}, \quad u \in \mathbb{U}_{\varepsilon}, \quad \mathbb{U}_{\varepsilon} = (u : \vartheta_0 + \psi_{\varepsilon}u \in \Theta).$$

The substitution of the observations (10) in $Z_{\varepsilon}(\vartheta_0, u)$ gives us the representation

$$\ln Z_{\varepsilon}(\vartheta_0, u) = \int_{0}^{T} \frac{M(\vartheta_u, t) - M(\vartheta_0, t)}{\varepsilon\sigma(t)} d\bar{W}_t - \int_{0}^{T} \frac{[M(\vartheta_u, t) - M(\vartheta_0, t)]^2}{2\varepsilon^2\sigma(t)^2} dt,$$

where $\vartheta_u = \vartheta_0 + \psi_{\varepsilon}u$. Taylor formula yields

$$M(\vartheta_0 + \psi_{\varepsilon}u, t) - M(\vartheta_0, t) = \psi_{\varepsilon}u^{\top}\dot{M}(\vartheta_0, t)(1 + O(\psi_{\varepsilon})).$$

Recall that $\dot{M}(\vartheta, t)$ is the derivative w.r.t. $\vartheta$ of the random process $M(\vartheta, t)$,

$$\dot{M}(\vartheta, t) = \dot{f}(\vartheta, t)m(\vartheta, t) + f(\vartheta, t)\dot{m}(\vartheta, t). \qquad (22)$$

Thus we need to find the asymptotics of the derivative $\dot{m}(\vartheta, t)$ of $m(\vartheta, t)$ and the related derivative $\dot{\gamma}(\vartheta, t)$ of $\gamma(\vartheta, t)$, since the equation (3) for $m(\vartheta, t)$ depends on $\gamma(\vartheta, t)$.

We start with the asymptotics of Riccati equation. It will be convenient to introduce the function $\gamma_*(\vartheta, t) = \gamma(\vartheta, t) / (\varepsilon\psi_{\varepsilon})$. Equations (3) and (4) can be rewritten as follows

$$dm(\vartheta, t) = \left[ a(\vartheta, t) - \frac{\psi_{\varepsilon} \gamma_*(\vartheta, t) f(\vartheta, t)^2}{\varepsilon \sigma(t)^2} \right] m(\vartheta, t) dt + \frac{\psi_{\varepsilon} \gamma_*(\vartheta, t) f(\vartheta, t)}{\varepsilon \sigma(t)^2} dX_t,$$

$$\frac{\varepsilon}{\psi_{\varepsilon}} \frac{\partial \gamma_{*}(\vartheta, t)}{\partial t} = \frac{2\varepsilon}{\psi_{\varepsilon}} a(\vartheta, t) \gamma_{*}(\vartheta, t) - \frac{\gamma_{*}(\vartheta, t)^2 f(\vartheta, t)^2}{\sigma(t)^2} + b(\vartheta, t)^2,$$

subject to initial conditions $m(\vartheta, 0) = y_0, \gamma_*(\vartheta, 0) = 0$.

Let us denote

$$\gamma_0(\vartheta, t) = \frac{b(\vartheta, t) \sigma(t)}{f(\vartheta, t)}.$$

**Lemma 1.** Let the conditions $\hat{\mathcal{D}}_{1-2}$ hold and $\varepsilon\psi_{\varepsilon}^{-1} \to 0$. Then, for any $t_0 \in (0, T]$,

$$\sup_{t_0 \le t \le T} |\gamma_*(\vartheta, t) - \gamma_0(\vartheta, t)| = O\left(\frac{\varepsilon}{\psi_{\varepsilon}}\right). \qquad (23)$$
Proof. Introduce another Riccati equation

$$ \frac{\partial \hat{\gamma}(t)}{\partial t} = 2a_M \hat{\gamma}(t) - \frac{\psi_{\varepsilon} \hat{\gamma}(t)^2 f_m^2}{\varepsilon S^2} + \frac{\psi_{\varepsilon}}{\varepsilon} b_M^2, \quad \hat{\gamma}(0) = 0, \qquad (24) $$

where we denoted

$$ a_M = \sup_{0 \le t \le T} \sup_{\vartheta \in \Theta} |a(\vartheta, t)|, \quad f_m = \inf_{0 \le t \le T} \inf_{\vartheta \in \Theta} f(\vartheta, t), $$

$$ b_M = \sup_{0 \le t \le T} \sup_{\vartheta \in \Theta} b(\vartheta, t), \quad \sigma_M = \sup_{0 \le t \le T} \sigma(t). $$

Note that by condition $\hat{\mathcal{D}}_2$ we have $f_m > 0$. Due to the comparison theorem for ordinary differential equations, the estimate $\gamma_*(\vartheta, t) \le \hat{\gamma}(t)$ holds for all $t \in [0, T]$. This bound is intuitive too, i.e., as replacing the “noise coefficient” $\sigma(t)$ by its maximal value should increase the estimation error. Further, as we take smaller coefficient $f_m$ in the drift $f(\vartheta, t) Y_t$ the error increases as well. The solution of Riccati equation (24) with constant coefficients can be written explicitly

$$ \hat{\gamma}(t) = e^{-2R_{\varepsilon}t} \left[ \frac{\psi_{\varepsilon} f_m^2 (1 - e^{-2R_{\varepsilon}t})}{2\varepsilon R_{\varepsilon} \sigma_M^2} - \hat{\gamma}^{-1} \right]^{-1} + \hat{\gamma}. $$

Here

$$ R_{\varepsilon} = \left( a_m^2 + \frac{\psi_{\varepsilon}^2 b_M^2 f_m^2}{\varepsilon^2 \sigma_M^2} \right)^{1/2} = \frac{\psi_{\varepsilon}}{\varepsilon} \frac{f_m b_M}{\sigma_M} \left( 1 + O\left( \frac{\varepsilon^2}{\psi_{\varepsilon}^2} \right) \right) $$

and

$$ \hat{\gamma} = \frac{\varepsilon}{\psi_{\varepsilon}} \frac{a_m \sigma_M^2}{f_m^2} \left[ \left( 1 + \frac{\psi_{\varepsilon}^2 b_M^2 f_m^2}{\varepsilon^2 a_m^2 \sigma_M^2} \right)^{1/2} - 1 \right] = \frac{b_M \sigma_M}{f_m} \left( 1 + O\left( \frac{\varepsilon}{\psi_{\varepsilon}} \right) \right). $$

Therefore for any $t_0 \in (0, T]$

$$ \sup_{t_0 \le t \le T} \left| \hat{\gamma}(t) - \frac{b_M \sigma_M}{f_m} \right| = O\left( \frac{\varepsilon}{\psi_{\varepsilon}} \right). \qquad (25) $$

Let us write the Riccati equation in integral form

$$ \frac{\varepsilon}{\psi_{\varepsilon}} \left[ \gamma_{*}(\vartheta, t) - \gamma_{*}(\vartheta, t_0) + 2 \int_{t_0}^{t} a(\vartheta, s) \gamma_{*}(\vartheta, s) ds \right] = - \int_{t_0}^{t} \frac{\gamma_{*}(\vartheta, s)^2 f(\vartheta, s)^2}{\sigma(s)^2} ds + \int_{t_0}^{t} b(\vartheta, s)^2 ds. $$

As $\gamma_*(\vartheta, t) \le \hat{\gamma}(t)$ and $\hat{\gamma}(t)$ according to (25) is bounded, the left hand side of the integral equation tends to 0. Hence

$$ \int_{t_0}^{t} \frac{\gamma_{*}(\vartheta, s)^2 f(\vartheta, s)^2}{\sigma(s)^2} ds = \int_{t_0}^{t} b(\vartheta, s)^2 ds + O\left( \frac{\varepsilon}{\psi_{\varepsilon}} \right) $$
where $O(\frac{\varepsilon}{\psi_{\varepsilon}})$ has continuous derivative w.r.t. $t$. All functions in these integrals are continuously differentiable on $s$ and therefore it is possible to verify that

$$ \gamma_*(\vartheta, t)^2 = \frac{b(\vartheta, t)^2 \sigma(t)^2}{f(\vartheta, t)^2} + O\left(\frac{\varepsilon}{\psi_{\varepsilon}}\right) = \gamma_0(\vartheta, t)^2 + O\left(\frac{\varepsilon}{\psi_{\varepsilon}}\right). $$

<figure><img src="image_7.png" /></figure>

It can be checked that all terms of order $O(\frac{\varepsilon}{\psi_{\varepsilon}})$ above satisfy

$$ \left|O\left(\frac{\varepsilon}{\psi_{\varepsilon}}\right)\right| \leq C \frac{\varepsilon}{\psi_{\varepsilon}}, $$

where the constant $C > 0$ can be chosen independent on $\vartheta$.

Below $O_p(\frac{\varepsilon}{\psi_{\varepsilon}})$ and $o_p(\frac{\varepsilon}{\psi_{\varepsilon}})$ means that for any $p > 1$

$$ \mathbf{E}_{\vartheta_0} \left| O_p \left( \frac{\varepsilon}{\psi_{\varepsilon}} \right) \right|^p \leq C \left( \frac{\varepsilon}{\psi_{\varepsilon}} \right)^p, \quad \mathbf{E}_{\vartheta_0} \left| o_p \left( \frac{\varepsilon}{\psi_{\varepsilon}} \right) \right|^p \leq c_{\varepsilon} \left( \frac{\varepsilon}{\psi_{\varepsilon}} \right)^p, \quad c_{\varepsilon} \to 0, $$

where the constants $C > 0$ and $c_{\varepsilon} > 0$ do not depend on $\vartheta$.

**Lemma 2.** Let conditions $\hat{\mathcal{D}}_{1-3}$ hold. Then for any $u \in \mathbb{U}_{\varepsilon}$ and any $t \in (0, T]$ the relation

$$ M(\vartheta_u, t) - M(\vartheta_0, t) = \varepsilon u^{\top} \frac{[\dot{a}(\vartheta_0, t) + \dot{l}(\vartheta_0, t)]y_t(\vartheta_0) \sigma(t)}{b(\vartheta_0, t)} (1 + o(1)), \quad (26) $$

holds.

*Proof.* By Lemma 1 we can replace the function $\gamma_*(\vartheta, t)$ by the function $\gamma_0(\vartheta, t) = b(\vartheta, t) \sigma(t) f(\vartheta, t)^{-1}$. Therefore omitting the negligible terms we obtain the equation

$$ \begin{aligned} dm(\vartheta, t) &= \left[ a(\vartheta, t) - \frac{\psi_{\varepsilon} b(\vartheta, t) f(\vartheta, t)}{\sigma(t)} \right] m(\vartheta, t) dt + \frac{\psi_{\varepsilon} b(\vartheta, t)}{\sigma(t)} dX_t \\ &= \left[ a(\vartheta, t) - \frac{\psi_{\varepsilon}}{\varepsilon} B(\vartheta, t) \right] m(\vartheta, t) dt + \frac{\psi_{\varepsilon} b(\vartheta, t)}{\varepsilon \sigma(t)} f(\vartheta_0, t) m(\vartheta_0, t) dt \\ &\quad + \psi_{\varepsilon} b(\vartheta, t) d\bar{W}_t, \quad m(\vartheta, t_0), \quad t_0 \le t \le T, \end{aligned} $$

where we denoted $B(\vartheta, t) = b(\vartheta, t) f(\vartheta, t) \sigma(t)^{-1}$.

To write the equation for $M(\vartheta, t)$ we remark that

$$ dM(\vartheta, t) = f(\vartheta, t) dm(\vartheta, t) + f'_t(\vartheta, t) m(\vartheta, t) dt. $$
Thus the equations for $M(\vartheta, t), t_0 \le t \le T$ and $M(\vartheta_0, t), t_0 \le t \le T$ are

$$
\begin{aligned}
dM(\vartheta, t) &= [a(\vartheta, t) + \ell(\vartheta, t)]M(\vartheta, t)dt - \frac{\psi_{\varepsilon}}{\varepsilon}B(\vartheta, t)M(\vartheta, t)dt \\
&\quad + \frac{\psi_{\varepsilon}}{\varepsilon}B(\vartheta, t)M(\vartheta_0, t)dt + \psi_{\varepsilon}B(\vartheta, t)\sigma(t)d\bar{W}_t, \quad M(\vartheta, t_0), \\
dM(\vartheta_0, t) &= [a(\vartheta_0, t) + \ell(\vartheta_0, t)]M(\vartheta_0, t)dt + \psi_{\varepsilon}B(\vartheta_0, t)\sigma(t)d\bar{W}_t, \quad M(\vartheta_0, t_0).
\end{aligned}
$$

The particular values of $M(\vartheta, t_0)$ and $M(\vartheta_0, t_0)$ do not important because their contributions are exponentially negligible.

Let us denote $a^*(\vartheta, t) = a(\vartheta, t) + \ell(\vartheta, t)$, $\delta_M(t) = M(\vartheta, t) - M(\vartheta_0, t)$, $\delta_{a^*}(t) = a^*(\vartheta, t) - a^*(\vartheta_0, t)$ and $\delta_B(t) = B(\vartheta, t) - B(\vartheta_0, t)$. Then the equations for $M(\vartheta_0, t)$ and $\delta_M(t)$ are

$$dM(\vartheta_0, t) = a^*(\vartheta_0, t) M(\vartheta_0, t) dt + \psi_{\varepsilon} B(\vartheta_0, t) \sigma(t) d\bar{W}_t, \quad M(\vartheta_0, t_0), \quad (27)$$

$$
\begin{aligned}
d\delta_M(t) = a^*(\vartheta, t) \delta_M(t) dt &- \frac{\psi_{\varepsilon}}{\varepsilon} B(\vartheta, t) \delta_M(t) dt + \delta_{a^*}(t) M(\vartheta_0, t) dt \\
&+ \psi_{\varepsilon} \delta_B(t) \sigma(t) d\bar{W}_t, \quad \delta_M(t_0) = M(\vartheta, t_0) - M(\vartheta_0, t_0).
\end{aligned}
\quad (28)
$$

The solution of this equation is

$$
\begin{aligned}
\delta_M(t) ={}& \delta_M(0) \phi_\varepsilon(\vartheta, 0, t) + \int_0^t \phi_\varepsilon(\vartheta, s, t) \delta_{a^*}(s) M(\vartheta_0, s) ds \\
& + \int_0^t \phi_\varepsilon(\vartheta, s, t) a^*(\vartheta, s) \delta_M(s) ds + \psi_\varepsilon \int_0^t \phi_\varepsilon(\vartheta, s, t) \delta_B(s) \sigma(s) d\bar{W}_s.
\end{aligned}
\quad (29)
$$

Here

$$\phi_{\varepsilon}(\vartheta, s, t) = \exp \left( -\frac{\psi_{\varepsilon}}{\varepsilon} \int_{s}^{t} B(\vartheta, v) dv \right).$$

By Lemma 1 in [30] these integrals have the following asymptotics

$$
\begin{aligned}
& \int_{t_0}^{t} \phi_{\varepsilon}(\vartheta, s, t) \delta_{a^*}(s) M(\vartheta_0, s) ds \\
&= f(\vartheta_0, t) y_t(\vartheta_0) \int_{t_0}^{t} \phi_{\varepsilon}(\vartheta, s, t) \delta_{a^*}(s) ds \\
&\quad + [M(\vartheta_0, t) - f(\vartheta_0, t) y_t(\vartheta_0)] \int_{t_0}^{t} \phi_{\varepsilon}(\vartheta, s, t) \delta_{a^*}(s) ds \\
&\quad + \int_{t_0}^{t} \phi_{\varepsilon}(\vartheta, s, t) \delta_{a^*}(s) [M(\vartheta_0, s) - M(\vartheta_0, t)] ds \\
&= \frac{\varepsilon}{\psi_{\varepsilon}} \frac{a^*(\vartheta, t) - a^*(\vartheta_0, t)}{B(\vartheta, t)} f(\vartheta_0, t) y_t(\vartheta_0) \left(1 + O\left(\frac{\varepsilon}{\psi_{\varepsilon}}\right)\right),
\end{aligned}
\quad (30)
$$
$$ \int_{t_0}^{t} \phi_{\varepsilon}(\vartheta, s, t) a^{*}(\vartheta, s) \delta_{M}(s) \mathrm{d}s = \frac{\varepsilon}{\psi_{\varepsilon}} \frac{a^{*}(\vartheta, t)}{B(\vartheta, t)} \delta_{M}(t) \left(1 + O\left(\frac{\varepsilon}{\psi_{\varepsilon}}\right)\right), \quad (31) $$

$$ \begin{aligned} \psi_{\varepsilon} \int_{t_0}^{t} \phi_{\varepsilon}(\vartheta, s, t) \delta_{B}(s) \sigma(s) \mathrm{d}\bar{W}_{s} \\ = \sqrt{\varepsilon \psi_{\varepsilon}} \frac{[f(\vartheta, t) b(\vartheta, t) - f(\vartheta_{0}, t) b(\vartheta_{0}, t)]}{\sqrt{2B(\vartheta, t)}} \xi_{t, \varepsilon} \left(1 + O\left(\sqrt{\frac{\varepsilon}{\psi_{\varepsilon}}}\right)\right). \end{aligned} \quad (32) $$

Here

$$ \xi_{t, \varepsilon} = \sqrt{2B(\vartheta, t)} \sqrt{\frac{\varepsilon}{\psi_{\varepsilon}}} \int_{t_0}^{t} \exp\left(-\frac{\psi_{\varepsilon}}{\varepsilon}(t-s)B(\vartheta, t)\right) \mathrm{d}\bar{W}_{s} \implies \xi_{t}, \quad (33) $$

where $\xi_t, t_0 \le t \le T$ are independent r.v.'s, $\xi_t \sim \mathcal{N}(0, 1)$ (see the proof of Lemma 1 in [30] for details).

Let us clarify the order $\varepsilon/\psi_{\varepsilon}$ of the residual in (30). When we replace $M(\vartheta_0, s)$ by $M(\vartheta_0, t)$ the difference is

$$ \begin{aligned} M(\vartheta_0, t) - M(\vartheta_0, s) &= \int_s^t f(\vartheta_0, r) y_r(\vartheta_0) \mathrm{d}r + \psi_\varepsilon \int_s^t B(\vartheta_0, r) \sigma(r) \mathrm{d}\bar{W}_r \\ &= f(\vartheta_0, t) y_t(\vartheta_0) (t-s) (1 + O(t-s)) \\ &\quad + \psi_\varepsilon B(\vartheta_0, t) \sigma(t) [\bar{W}_t - \bar{W}_s] (1 + o(1)). \end{aligned} $$

Therefore

$$ \begin{aligned} & \int_{t_0}^{t} \phi_{\varepsilon}(\vartheta, s, t) \delta_{a^{*}}(s) [M(\vartheta_0, s) - M(\vartheta_0, t)] \mathrm{d}s \\ &= \psi_{\varepsilon} \delta_{a^{*}}(t) B(\vartheta_0, t) \sigma(t) \int_{t_0}^{t} \phi_{\varepsilon}(\vartheta, s, t) [\bar{W}_{s} - \bar{W}_{t}] \mathrm{d}s (1 + o(1)) \\ &= \psi_{\varepsilon} \left(\frac{\varepsilon}{\psi_{\varepsilon}}\right)^{3/2} O(1) = \left(\frac{\varepsilon^3}{\psi_{\varepsilon}}\right)^{1/2} O(1). \end{aligned} $$

Here we used Lemma 1 in [30]. We have as well the relations

$$ [M(\vartheta_0, t) - f(\vartheta_0, t) y_t(\vartheta_0)] \int_{t_0}^{t} \phi_{\varepsilon}(\vartheta, s, t) \delta_{a^{*}}(s) \mathrm{d}s = \varepsilon O(1) $$

and

$$ \begin{aligned} & f(\vartheta_0, t) y_t(\vartheta_0) \int_{t_0}^{t} \phi_{\varepsilon}(\vartheta, s, t) \delta_{a^{*}}(s) \mathrm{d}s \\ &= \frac{\varepsilon}{\psi_{\varepsilon}} \frac{[a(\vartheta, t) - a(\vartheta_0, t) + \ell(\vartheta, t) - \ell(\vartheta_0, t)] f(\vartheta_0, t) y_t(\vartheta_0)}{B(\vartheta, t)} \left(1 + O\left(\frac{\varepsilon}{\psi_{\varepsilon}}\right)\right) \end{aligned} $$
Finally we obtain for any (small) real $t_0 > 0$ the equality

$$ \delta_M(t) \left(1 - \frac{\varepsilon}{\psi_\varepsilon} \frac{a^*(\vartheta, t)}{B(\vartheta, t)}\right) = \frac{\varepsilon}{\psi_\varepsilon} \frac{a^*(\vartheta, t) - a^*(\vartheta_0, t)}{B(\vartheta, t)} f(\vartheta_0, t) y_t(\vartheta_0) (1 + O(\varepsilon^{1-\delta})) \\ + \sqrt{\varepsilon \psi_\varepsilon} \frac{[f(\vartheta, t) b(\vartheta, t) - f(\vartheta_0, t) b(\vartheta_0, t)]}{\sqrt{2B(\vartheta, t)}} \xi_{t,\varepsilon} \left(1 + O\left(\varepsilon^{\frac{1-\delta}{2}}\right)\right), \quad t_0 \le t \le T, $$

and

$$ \delta_M(t) = \frac{\varepsilon a^*(\vartheta, t) - a^*(\vartheta_0, t)}{\psi_\varepsilon B(\vartheta, t)} f(\vartheta_0, t) y_t(\vartheta_0) (1 + O(\varepsilon^{1-\delta})) \\ + \sqrt{\varepsilon \psi_\varepsilon} \frac{[f(\vartheta, t) b(\vartheta, t) - f(\vartheta_0, t) b(\vartheta_0, t)]}{\sqrt{2B(\vartheta, t)}} \xi_{t,\varepsilon} \left(1 + O\left(\varepsilon^{\frac{1-\delta}{2}}\right)\right), \quad (34) $$

where we omitted exponentially small term related with $\delta_M(t_0)$.

Thus if we put $\vartheta = \vartheta_u = \vartheta_0 + \psi_\varepsilon u$ and use Taylor expansion, then we obtain

$$ M(\vartheta_u, t) - M(\vartheta_0, t) = \frac{\varepsilon a^*(\vartheta_u, t) - a^*(\vartheta_0, t)}{\psi_\varepsilon B(\vartheta_u, t)} f(\vartheta_0, t) y_t(\vartheta_0) (1 + O(\varepsilon^{1-\delta})) \\ + \sqrt{\varepsilon \psi_\varepsilon} \frac{[f(\vartheta_u, t) b(\vartheta_u, t) - f(\vartheta_0, t) b(\vartheta_0, t)]}{\sqrt{2B(\vartheta_u, t)}} \xi_{t,\varepsilon} \left(1 + O\left(\varepsilon^{\frac{1-\delta}{2}}\right)\right) \\ = \varepsilon u^\top \frac{[\dot{a}(\vartheta_0, t) + \dot{\ell}(\vartheta_0, t)] f(\vartheta_0, t) y_t(\vartheta_0) \sigma(t)}{f(\vartheta_u, t) b(\vartheta_u, t)} (1 + O(\varepsilon^{1-\delta})) \\ = \varepsilon u^\top \frac{[\dot{a}(\vartheta_0, t) + \dot{\ell}(\vartheta_0, t)] y_t(\vartheta_0) \sigma(t)}{b(\vartheta_0, t)} (1 + O(\varepsilon^{1-\delta})). $$

<figure><img src="image_8.png" /></figure>

Remark that these expansions allow us to write (below $S(\vartheta, t) = f(\vartheta, t) b(\vartheta, t)$)

$$ \begin{aligned} & \varepsilon \psi_{\varepsilon} \int_{t_0}^{T} \frac{[f(\vartheta_u, t) b(\vartheta_u, t) - f(\vartheta_0, t) b(\vartheta_0, t)]^2}{2B(\vartheta_u, t) \varepsilon^2 \sigma(t)^2} \xi_{t, \varepsilon}^2 dt \\ &= \frac{\psi_{\varepsilon}^2 \varepsilon \psi_{\varepsilon}}{\varepsilon^2} \int_{t_0}^{T} \frac{[u^{\top} \dot{S}(\vartheta_0, t)]^2}{2B(\vartheta_0, t) \sigma(t)^2} \xi_{t, \varepsilon}^2 dt (1 + \psi_{\varepsilon} u O_p(1)) \\ &= u^2 \frac{\psi_{\varepsilon}^3}{\varepsilon} (1 + \psi_{\varepsilon} u O_p(1)) O(1) \to 0 \end{aligned} \quad (35) $$

and

$$ \frac{\varepsilon^2}{\psi_{\varepsilon}^2} \int_{t_0}^{T} \frac{[a^*(\vartheta_u, t) - a^*(\vartheta_0, t)]^2}{B(\vartheta_u, t)^2 \varepsilon^2 \sigma(t)^2} f(\vartheta_0, t)^2 y_t(\vartheta_0)^2 dt $$
$$
\begin{align}
&= \int_{t_0}^{T} \frac{[u^{\top} \dot{a}^{*}(\vartheta_0, t)]^2 y_t(\vartheta_0)^2}{b(\vartheta_0, t)^2} \mathrm{d}t (1 + \psi_{\varepsilon} u O(1)) \nonumber \\
&= \int_{t_0}^{T} \frac{[u^{\top} \dot{a}(\vartheta_0, t) + u^{\top} \dot{l}(\vartheta_0, t)]^2 y_t(\vartheta_0)^2}{b(\vartheta_0, t)^2} \mathrm{d}t (1 + \psi_{\varepsilon} u O(1)) \nonumber \\
&= u^{\top} I_2(\vartheta_0) u (1 + \psi_{\varepsilon} u O(1)) (1 + O(t_0)). \tag{36}
\end{align}
$$

Here $O_p(1)$ means that all polynomial moments of $O_p(1)$ are bounded by a constant which does not depend on $\varepsilon$.

Therefore

$$
\int_{t_0}^{T} \frac{[M(\vartheta_u, t) - M(\vartheta_0, t)]^2}{\varepsilon^2 \sigma(t)^2} \mathrm{d}t = u^{\top} I(\vartheta_0) u (1 + \psi_{\varepsilon} u O_p(1)) (1 + O_p(t_0)). \quad (37)
$$

To simplify the exposition we will write $I_2(\vartheta_0)$ instead of $I_2(\vartheta_0)(1+O(t_0))$ because the
difference between them is of order

$$
\int_0^{t_0} \frac{\| \dot{a}(\vartheta_0, t) + \dot{l}(\vartheta_0, t) \|^2 y_t(\vartheta_0)^2}{b(\vartheta_0, t)^2} \mathrm{d}t \le C t_0
$$

can be made less than any $\epsilon$, $0 < \epsilon \ll 1$ by the choice of $t_0$.

**Lemma 3.** Let the conditions $\hat{\mathcal{D}}_{1-4}$ hold. Then the family of measures $\{\mathbf{P}_{\vartheta}^{(\varepsilon)}, \vartheta \in \Theta\}$ is uniformly on compacts $\mathbb{K} \in \Theta$ locally asymptotically normal in $\Theta$, i.e., the normalized likelihood ratio $Z_{\varepsilon}(\vartheta_0, u)$ for all $\vartheta_0 \in \Theta$ admits the representation

$$
Z_{\varepsilon}(\vartheta_0, u) = \exp \left( u^{\top} \Delta_{\varepsilon}(\vartheta_0, X^T) - \frac{1}{2} u^{\top} I_2(\vartheta_0) u + r_{\varepsilon}(\vartheta_0, u, X^T) \right), \quad u \in \mathbb{U}_{\varepsilon},
$$

where

$$
\Delta_{\varepsilon}(\vartheta_0, X^T) = \int_{t_0}^{T} \frac{[\dot{a}(\vartheta_0, t) + \dot{l}(\vartheta_0, t)]y_t(\vartheta_0)}{b(\vartheta_0, t)} \mathrm{d}\bar{W}_t = \Delta(\vartheta_0) \sim \mathcal{N}(0, I_2(\vartheta_0)).
$$

Here the last equality is in distribution and $r_{\varepsilon}(\vartheta_0, u, X^T) \to 0$ uniformly on $\vartheta_0 \in \mathbb{K}$.

*Proof.* The process $Z_{\varepsilon}(\vartheta_0, u)$ by Lemma 2 and (35), (36) admits of the representation

$$
\begin{align*}
\ln Z_{\varepsilon}(\vartheta_0, u) &= \frac{\psi_{\varepsilon} u^{\top}}{\varepsilon} \int_{0}^{T} \dot{M}(\vartheta_0, t) \mathrm{d}\bar{W}_t - \frac{\psi_{\varepsilon}^2}{2\varepsilon^2} \int_{0}^{T} \left[ u^{\top} \dot{M}(\vartheta_0, t) \right]^2 \mathrm{d}t + o_p(1) \\
&= u^{\top} \int_{0}^{T} \frac{[\dot{a}(\vartheta_0, t) + \dot{l}(\vartheta_0, t)] y_t(\vartheta_0)}{b(\vartheta_0, t)} \mathrm{d}\bar{W}_t \\
&\quad - \frac{1}{2} \int_{0}^{T} \frac{[u^{\top} \dot{a}(\vartheta_0, t) + u^{\top} \dot{l}(\vartheta_0, t)]^2 y_t(\vartheta_0)^2}{b(\vartheta_0, t)^2} \mathrm{d}t + o_p(1)
\end{align*}
$$
$$ = u^{\top} \Delta(\vartheta_0) - \frac{1}{2} u^{\top} I_2(\vartheta_0) u + o(1). $$

<figure><img src="image_9.png" /></figure>

**Lemma 4.** Let conditions $\hat{\mathcal{D}}_{1-3}$ hold. Then for any compact $\mathbb{K} \subset \Theta$ and any $R > 0$ there exists a constant $C > 0$ such that for $\|u_1\| + \|u_2\| \le R$

$$ \sup_{\vartheta_0 \in \mathbb{K}} \mathbf{E}_{\vartheta_0} \left[ Z_{\varepsilon} (\vartheta_0, u_1)^{\frac{1}{2m}} - Z_{\varepsilon} (\vartheta_0, u_2)^{\frac{1}{2m}} \right]^{2m} \le C (1 + R^{2m}) \|u_2 - u_1\|^{2m}. \quad (38) $$

*Proof.* We have

$$ \mathbf{E}_{\vartheta_0} \left[ Z_{\varepsilon} (\vartheta_0, u_1)^{\frac{1}{2m}} - Z_{\varepsilon} (\vartheta_0, u_2)^{\frac{1}{2m}} \right]^{2m} \le \frac{C_1(m)}{\varepsilon^{4m}} T^{2m-1} \int_0^T \frac{\mathbf{E}_{\vartheta_{u_2}} \delta_M (u,t)^{4m}}{\sigma(t)^2} \mathrm{d}t \\ + \frac{C_2(m)}{\varepsilon^{2m}} T^{m-1} \int_0^T \frac{\mathbf{E}_{\vartheta_{u_2}} \delta_M (u,t)^{2m}}{\sigma(t)^2} \mathrm{d}t, \quad (39) $$

where we used the inequality (1.40) in [28] and denoted $\delta_M(u,t) = M(\vartheta_0 + \psi_\varepsilon u_2, t) - M(\vartheta_0 + \psi_\varepsilon u_1, t)$.

Consider the term $\mathbf{E}_{\vartheta_{u_1}} \delta_M (u,t)^{2m}$. The function $\delta_M(u, \cdot)$ with $\mathbf{P}_{\vartheta_{u_1}}$ probability one is solution of the equation

$$ \mathrm{d}\delta_M(u,t) = a^*(\vartheta_{u_2}, t) \delta_M(u,t) \mathrm{d}t - \frac{\psi_\varepsilon}{\varepsilon} B(\vartheta_{u_2}, t) \delta_M(u,t) \mathrm{d}t + \delta_{a^*}(u,t) M(\vartheta_{u_1}, t) \mathrm{d}t \\ + \psi_\varepsilon \delta_B(u,t) \sigma(t) \mathrm{d}\bar{W}_t, \quad \delta_M(u,0) = [f(\vartheta_{u_2}, 0) - f(\vartheta_{u_1}, 0)] y_0. $$

Here $\delta_{a^*}(u,t) = a^*(\vartheta_{u_2}, t) - a^*(\vartheta_{u_1}, t) = \psi_\varepsilon (u_2 - u_1)^{\top} \dot{a}^*(\vartheta_{\bar{u}_1}, t)$ and $\delta_B(u,t) = B(\vartheta_{u_2}, t) - B(\vartheta_{u_1}, t) = \psi_\varepsilon (u_2 - u_1)^{\top} \dot{B}(\vartheta_{\bar{u}_1}, t)$, $B(\vartheta, t) = \gamma_*(\vartheta, t) f(\vartheta, t) \sigma(t)^{-2}$.

We have the representation

$$ \delta_M(u,t) = \delta_M(u,0) \Phi(\vartheta_{u_2}, 0,t) + \int_0^t \Phi(\vartheta_{u_2}, s,t) \delta_{a^*}(u,s) M(\vartheta_{u_1}, s) \mathrm{d}s \\ + \psi_\varepsilon \int_0^t \Phi(\vartheta_{u_2}, s,t) \delta_B(u,s) \sigma(s) \mathrm{d}\bar{W}_s \quad (40) $$

where $\bar{W}_s, 0 \le s \le T$ is the innovation Wiener process (under the measure $\mathbf{P}_{\vartheta_{u_1}}$) and

$$ \Phi(\vartheta_{u_2}, s, t) = \exp \left( -\frac{\psi_{\varepsilon}}{\varepsilon} \int_s^t B(\vartheta_{u_2}, v) \mathrm{d}v + \int_s^t a^*(\vartheta_{u_2}, v) \mathrm{d}v \right). $$

There exist constants $c_* > 0$ and $C_* > 0$ such that for $t \ge t_0$ we have $B(\vartheta_{u_2}, v) > c_*$ and

$$ \int_0^T |a^*(\vartheta, v)| \mathrm{d}v < C_* $$
Thus

$$ \Phi(\vartheta_{u_2}, s, t) \leq \exp\left(-\frac{\psi_{\varepsilon}}{\varepsilon}c_{*}(t-s) + C_{*}\right), \quad (41) $$

where $c_* > 0$ is some constant.

Remark that for $t \in (0, T]$

$$ \begin{aligned} \varepsilon^{-1} |\delta_M(u, 0)| \Phi(\vartheta_{u_2}, 0, t) &\le C \|u_2 - u_1\| y_0 |\dot{f}(\tilde{\vartheta}, 0)| \frac{\psi_{\varepsilon}}{\varepsilon} e^{-\psi_{\varepsilon}/\varepsilon c_* t} \\ &\le C r_{\varepsilon} \|u_2 - u_1\|, \end{aligned} \quad (42) $$

where $r_{\varepsilon} \to 0$.

For any $m = 1, 2, \dots$ and $t > 0$ we have

$$ \begin{aligned} \mathbf{E}_{\vartheta_{u_1}} \delta_M (u, t)^{2m} &\le C r_{\varepsilon}^{2m} \varepsilon^{2m} \|u_2 - u_1\|^{2m} \\ &\quad + C \mathbf{E}_{\vartheta_{u_1}} \left( \int_0^t e^{-\frac{\psi_{\varepsilon}}{\varepsilon} c_*(t-s)} \delta_{a^*}(s) M(\vartheta_{u_1}, s) \mathrm{d}s \right)^{2m} \\ &\quad + C \psi_{\varepsilon}^{2m} \left( \int_0^t e^{-\frac{2\psi_{\varepsilon}}{\varepsilon} c_*(t-s)} \delta_B(s)^2 \sigma(\vartheta_{u_1}, s)^2 \mathrm{d}s \right)^m. \end{aligned} $$

Here

$$ \begin{aligned} & \mathbf{E}_{\vartheta_{u_1}} \left( \int_0^t e^{-\frac{\psi_{\varepsilon}}{\varepsilon} c_*(t-s)} \delta_{a^*}(s) M(\vartheta_{u_1}, s) \mathrm{d}s \right)^{2m} \\ &\le \left( \int_0^t e^{-\frac{\psi_{\varepsilon}}{\varepsilon} c_*(t-s)} \mathrm{d}s \right)^{2m-1} \int_0^t e^{-\frac{\psi_{\varepsilon}}{\varepsilon} c_*(t-s)} \delta_{a^*}(s)^{2m} \mathbf{E}_{\vartheta_{u_1}} M(\vartheta_{u_1}, s)^{2m} \mathrm{d}s \\ &\le C \psi_{\varepsilon}^{2m} \left( \frac{\varepsilon}{\psi_{\varepsilon}} \right)^{2m} \|u_2 - u_1\|^{2m} = C \varepsilon^{2m} \|u_2 - u_1\|^{2m}. \end{aligned} \quad (43) $$

The Gaussian random process $M(\vartheta_{u_1}, \cdot)$ satisfies (27) where $\vartheta_0$ is replaced by $\vartheta_{u_1}$ and $\bar{W}_t, 0 \le t \le T$ is the innovation Wiener process under probability $P_{\vartheta_{u_1}}$. Therefore $M(\vartheta_{u_1}, \cdot)$ has bounded polynomial moments.

For the second integral we have

$$ \begin{aligned} & \psi_{\varepsilon}^{2m} \left( \int_{t_0}^t e^{-\frac{2\psi_{\varepsilon}}{\varepsilon} c_*(t-s)} \delta_B(s)^2 \sigma(\vartheta_{u_1}, s)^2 \mathrm{d}s \right)^m \\ &\le C \psi_{\varepsilon}^{4m} \|u_2 - u_1\|^{2m} \left( \int_{t_0}^t e^{-\frac{2\psi_{\varepsilon}}{\varepsilon} c_*(t-s)} \mathrm{d}s \right)^m \\ &\le C \psi_{\varepsilon}^{4m} \|u_2 - u_1\|^{2m} \left( \frac{\varepsilon}{\psi_{\varepsilon}} \right)^m = C (\varepsilon \psi_{\varepsilon}^3)^m \|u_2 - u_1\|^{2m}. \end{aligned} \quad (44) $$

The similar estimates we have for the term $\mathbf{E}_{\vartheta_{u_2}} \delta_M (u, t)^{4m}$ too. The substitution of these estimates in (39) proves (38).

<figure><img src="image_10.png" /></figure>
**Lemma 5.** Assume that conditions $\hat{\mathcal{D}}$ hold, then for any compact $\mathbb{K}$ and any $N > 0$ there exist constants $\kappa > 0$ and $C_N > 0$ such that

$$ \sup_{\vartheta_0 \in \mathbb{K}} \mathbf{P}_{\vartheta_0} \left( Z_{\varepsilon} (\vartheta_0, u) \geq e^{-\kappa \|u\|^2} \right) \leq \frac{C_N}{\|u\|^N}. \quad (45) $$

*Proof.* Denote $\delta_M(t) = M(\vartheta_u, t) - M(\vartheta_0, t)$ and write

$$ \begin{aligned} & \mathbf{P}_{\vartheta_0} \left( \ln Z_{\varepsilon} (\vartheta_0, u) \geq -\kappa \|u\|^2 \right) \\ &= \mathbf{P}_{\vartheta_0} \left( \int_0^T \frac{\delta_M(t)^2}{2\varepsilon\sigma(t)} \mathrm{d}\bar{W}_t - \int_0^T \frac{\delta_M(t)^2}{4\varepsilon^2\sigma(t)^2} \mathrm{d}t \geq -\frac{\kappa}{2} \|u\|^2 \right) \\ &\leq \mathbf{P}_{\vartheta_0} \left( \int_0^T \frac{\delta_M(t)}{2\varepsilon\sigma(t)} \mathrm{d}\bar{W}_t - \int_0^T \frac{\delta_M(t)^2}{8\varepsilon^2\sigma(t)^2} \mathrm{d}t \geq \frac{\kappa}{2} \|u\|^2 \right) \\ &\qquad + \mathbf{P}_{\vartheta_0} \left( -\int_0^T \frac{\delta_M(t)^2}{8\varepsilon^2\sigma(t)^2} \mathrm{d}t \geq -\kappa \|u\|^2 \right) \\ &\leq e^{-\frac{\kappa}{2} u^2} + \mathbf{P}_{\vartheta_0} \left( \int_0^T \frac{\delta_M(t)^2}{\varepsilon^2\sigma(t)^2} \mathrm{d}t \leq 8\kappa \|u\|^2 \right). \end{aligned} \quad (46) $$

Here we used the equality

$$ \mathbf{E}_{\vartheta_0} \exp \left( \int_0^T \frac{\delta_M(t)}{2\varepsilon\sigma(t)} \mathrm{d}\bar{W}_t - \int_0^T \frac{\delta_M(t)^2}{8\varepsilon^2\sigma(t)^2} \mathrm{d}t \right) = 1. $$

As it follows from (34)

$$ \delta_M(t) = \delta_M^\circ(t) + O\left(\sqrt{\varepsilon\psi_\varepsilon}\right) $$

where

$$ \delta_M^\circ(t) = \frac{\varepsilon}{\psi_\varepsilon} \frac{[a(\vartheta, t) - a(\vartheta_0, t) + \ell(\vartheta, t) - \ell(\vartheta_0, t)] f(\vartheta_0, t) y_t(\vartheta_0) \sigma(t)}{b(\vartheta, t) f(\vartheta, t)}. $$

Set $\mu_s(\vartheta_0) = f(\vartheta_0, s) y_s(\vartheta_0)$ and remind the obvious equality (see (27))

$$ \begin{aligned} \mathbf{E}_{\vartheta_0} M(\vartheta_0, s) &= y_0 f(\vartheta_0, 0) \exp \left( \int_0^s \left[ a(\vartheta_0, v) + f(\vartheta_0, v)^{-1} f'_v(\vartheta_0, v) \right] \mathrm{d}v \right) \\ &= y_0 f(\vartheta_0, s) \exp \left( \int_0^s a(\vartheta_0, r) \mathrm{d}v \right) = f(\vartheta_0, s) y_s(\vartheta_0) = \mu_s(\vartheta_0). \end{aligned} $$

We have the estimate (42) and therefore for any $t \in (0, T]$

$$ |\delta_M(0)| \Phi(\vartheta_0, 0, t) \leq C e^{-\frac{\psi_\varepsilon}{\varepsilon} c_* t} \to 0. $$
That is why we have omitted this term below.

Let us return to the last probability in (46)

$$ 
\begin{aligned}
& \mathbf{P}_{\vartheta_0} \left( -\int_0^T \frac{\delta_M(t)^2}{\varepsilon^2 \sigma(t)^2} dt \geq -8\kappa \|u\|^2 \right) \\
& = \mathbf{P}_{\vartheta_0} \left( \frac{G(\vartheta_u, \vartheta_0)}{\psi_{\varepsilon}^2} - \int_0^T \frac{\delta_M(t)^2}{\varepsilon^2 \sigma(t)^2} dt \geq \frac{G(\vartheta_u, \vartheta_0)}{\psi_{\varepsilon}^2} - 8\kappa \|u\|^2 \right) \quad (47)
\end{aligned}
$$

Consider two sets $\mathbb{A} = (u : \|\vartheta_u - \vartheta_0\| \leq \nu), \nu > 0$ and $\mathbb{A}^c = \mathbb{U}_\varepsilon \setminus \mathbb{A}$.

For $u \in \mathbb{A}$ and sufficiently small $\nu$ we can write

$$ 
\begin{aligned}
\frac{G(\vartheta_u, \vartheta_0)}{\psi_{\varepsilon}^2} &= u^\top \mathrm{I}_2(\vartheta_0) u (1 + o(1)) \geq \frac{1}{2} u^\top \mathrm{I}_2(\vartheta_0) u \\
&\geq \frac{\|u\|^2}{2} \inf_{\vartheta \in \Theta} \inf_{\|e\|=1, e \in \mathbb{R}^d} e^\top \mathrm{I}_2(\vartheta) e = \frac{\|u\|^2}{2} \hat{\kappa}.
\end{aligned}
$$

Therefore

$$ \frac{G(\vartheta_u, \vartheta_0)}{\psi_{\varepsilon}^2} - 8\kappa \|u\|^2 \geq \left[ \frac{\hat{\kappa}}{2} - 8\kappa \right] \|u\|^2 = \kappa_1 \|u\|^2, $$

where $\kappa$ is such that $\kappa_1 > 0$.

Further, for any $m = 1, 2, \dots$

$$ 
\begin{aligned}
& \mathbf{P}_{\vartheta_0} \left( \frac{G(\vartheta_u, \vartheta_0)}{\psi_{\varepsilon}^2} - \int_0^T \frac{\delta_M(t)^2}{\varepsilon^2 \sigma(t)^2} dt \geq \kappa_1 \|u\|^2 \right) \\
& \leq \mathbf{P}_{\vartheta_0} \left( \int_0^T \frac{|\delta_M(t)^2 - \delta_M^\circ(t)^2|}{\varepsilon^2 \sigma(t)^2} dt \geq \kappa_1 \|u\|^2 \right) \\
& \leq \frac{T^{2m-1}}{\kappa_1^{2m} \|u\|^{4m}} \int_0^T \frac{\mathbf{E}_{\vartheta_0} |\delta_M(t)^2 - \delta_M^\circ(t)^2|^{2m}}{\varepsilon^{4m} \sigma(t)^{4m}} dt \\
& \leq \frac{T^{2m-1}}{\kappa_1^{2m} \|u\|^{4m}} \int_0^T \frac{\left( \mathbf{E}_{\vartheta_0} |\delta_M(t) + \delta_M^\circ(t)|^{4m} \mathbf{E}_{\vartheta_0} |\delta_M(t) - \delta_M^\circ(t)|^{4m} \right)^{1/2}}{\varepsilon^{4m} \sigma(t)^{4m}} dt. \quad (48)
\end{aligned}
$$

We have the estimates

$$ 
\begin{aligned}
\frac{|\delta_M^\circ(t)|^{4m}}{\varepsilon^{4m}} &\leq \frac{C}{\psi_{\varepsilon}^{4m}} \psi_{\varepsilon}^{4m} \|u\|^{4m} = C \|u\|^{4m}, \\
\frac{\mathbf{E}_{\vartheta_0} |\delta_M(t)|^{4m}}{\varepsilon^{4m}} &\leq C \|u\|^{4m} + C \left( \frac{\psi_{\varepsilon}^3}{\varepsilon} \right)^{2m} \|u\|^{4m} \leq C \|u\|^{4m},
\end{aligned}
$$
$$ \frac{\mathbf{E}_{\vartheta_0} |\delta_M(t) - \delta_M^\circ(t)|^{4m}}{\varepsilon^{4m}} \le C \frac{\psi_\varepsilon^{4m}}{\varepsilon^{4m}} \left(\frac{\varepsilon}{\psi_\varepsilon}\right)^{6m} \|u\|^{4m} = C \left(\frac{\varepsilon}{\psi_\varepsilon}\right)^{2m} \|u\|^{4m}. $$

Remind that $\psi_\varepsilon^3/\varepsilon \to 0$.

Finally we obtain for $u \in A$

$$ \mathbf{P}_{\vartheta_0} \left( - \int_0^T \frac{\delta_M(t)^2}{\varepsilon^2 \sigma(t)^2} dt \ge -8\kappa \|u\|^2 \right) \le C \frac{T^{2m}}{\kappa_1^{2m}} \left( \frac{\varepsilon}{\psi_\varepsilon} \right)^m \le C \|u\|^{-\frac{(1-\delta)m}{\delta}}, $$

where we used the relations

$$ \vartheta_0 + \psi_\varepsilon u \in \Theta, \quad \|u\| \le \frac{\mathrm{Dim}(\Theta)}{\psi_\varepsilon} = \varepsilon^{-\delta} \mathrm{Dim}(\Theta), \quad \varepsilon < \frac{\mathrm{Dim}(\Theta)^{\frac{1}{\delta}}}{|u|^{\frac{1}{\delta}}}. \quad (49) $$

Here $\mathrm{Dim}(\Theta) = \sup_{\vartheta_1, \vartheta_2 \in \Theta} \|\vartheta_1 - \vartheta_2\|$. Hence $\varepsilon^{m(1-\delta)} \le C |u|^{-\frac{(1-\delta)m}{\delta}}$. Therefore if we set $m = N\delta/(1-\delta)$, then we obtain (39) for $u \in A$.

Consider now the case $u \in A^c$. We have

$$ \frac{G(\vartheta_u, \vartheta_0)}{\psi_\varepsilon^2} \ge \frac{g(\nu)}{\psi_\varepsilon^2} = \frac{g(\nu)}{(\|u\| \psi_\varepsilon)^2} \|u\|^2 \ge \frac{g(\nu)}{\mathrm{Dim}(\Theta)^2} \|u\|^2 $$

and

$$ \frac{G(\vartheta_u, \vartheta_0)}{\psi_\varepsilon^2} - 8\kappa \|u\|^2 \ge \left( \frac{g(\nu)}{\mathrm{Dim}(\Theta)^2} - 8\kappa \right) \|u\|^2 = \kappa_2 \|u\|^2 $$

where $\kappa$ is such that $\kappa_2 > 0$ too.

First we repeat the calculations as in (48) and obtain

$$ \mathbf{P}_{\vartheta_0} \left( - \int_0^T \frac{\delta_M(t)^2}{\varepsilon^2 \sigma(t)^2} dt \ge -8\kappa \|u\|^2 \right) \le \frac{T^{2m-1}}{\kappa_2^{2m} \|u\|^{4m}} \int_0^T \frac{\left( \mathbf{E}_{\vartheta_0} |\delta_M(t) + \delta_M^\circ(t)|^{4m} \mathbf{E}_{\vartheta_0} |\delta_M(t) - \delta_M^\circ(t)|^{4m} \right)^{1/2}}{\varepsilon^{4m} \sigma(t)^{4m}} dt. $$

Then the relations (47) and (48) allow us to write the corresponding asymptotics

$$ \frac{|\delta_M^\circ(t)|^{4m}}{\varepsilon^{4m}} \le \frac{C}{\psi_\varepsilon^{4m}}, $$

$$ \frac{\mathbf{E}_{\vartheta_0} |\delta_M(t)|^{4m}}{\varepsilon^{4m}} \le \frac{C}{\psi_\varepsilon^{4m}}, $$

$$ \frac{\mathbf{E}_{\vartheta_0} |\delta_M(t) - \delta_M^\circ(t)|^{4m}}{\varepsilon^{4m}} \le C \frac{1}{\varepsilon^{4m}} \left( \frac{\varepsilon}{\psi_\varepsilon} \right)^{6m} = C \left( \frac{\varepsilon}{\psi_\varepsilon^3} \right)^{2m}. $$
Hence

$$ 
\begin{aligned}
\mathbf{P}_{\vartheta_0} \left( - \int_0^T \frac{\delta_M(t)^2}{\varepsilon^2 \sigma(t)^2} dt \geq -8\kappa \|u\|^2 \right) &\leq C \frac{T^{2m}}{\kappa_2^{2m} u^{4m}} \frac{\varepsilon^m}{\psi_{\varepsilon}^{5m}} \leq C \frac{T^{2m}}{\kappa_2^{2m} \nu^{4m}} \left( \frac{\varepsilon}{\psi_{\varepsilon}} \right)^m \\
&\leq C \varepsilon^{m(1-\delta)} \leq \frac{C}{\|u\|^N},
\end{aligned}
$$

where we used the estimate $\psi_{\varepsilon} \|u\| \geq \nu$ and once more the relations (49).

These properties of the LR allow us to apply the Theorems 3.1.1 and 3.2.1 in [19].

<figure><img src="image_11.png" /></figure>

Since the convergence of moments is uniform on compacts of $\Theta$, we can prove the asymptotic efficiency of estimators as follows. The uniform convergence of moments for the MLE implies

$$ \lim_{\varepsilon \to 0} \sup_{\|\vartheta - \vartheta_0\| \leq \nu} \psi_{\varepsilon}^{-1} \mathbf{E}_{\vartheta} \|\hat{\vartheta}_{\varepsilon} - \vartheta\|^2 = \sup_{\|\vartheta - \vartheta_0\| \leq \nu} \mathbf{E}_{\vartheta} \|\zeta(\vartheta)\|^2 \xrightarrow{\nu \to 0} \mathbf{E}_{\vartheta_0} \|\zeta(\vartheta_0)\|^2. $$

The same convergence holds for $\mathrm{BE} \tilde{\vartheta}_{\varepsilon}$.

<figure><img src="image_12.png" /></figure>

**Remark 5.** The condition of identifiability requires that $G(\vartheta, \vartheta_0) \neq 0$ for any $\vartheta \neq \vartheta_0$ and any $\vartheta_0$. This condition is not fulfilled if, for example, the functions $a(\vartheta, t)$ and $f(\vartheta, t)$ satisfy the equation

$$ \frac{f'_{t}(\vartheta, t)}{f(\vartheta, t)} = -a(\vartheta, t) + h(t), \quad 0 \leq t \leq T, $$

where $h(t)$ is some continuous function which does not depend on $\vartheta$. This means that

$$ f(\vartheta, t) = H(t) \exp \left( - \int_{0}^{t} a(\vartheta, s) \mathrm{d}s \right), \quad H(t) = \exp \int_{0}^{t} h(s) \mathrm{d}s. \quad (50) $$

Therefore for such functions $f(\vartheta, \cdot)$ and $a(\vartheta, \cdot)$ we have $G(\vartheta, \vartheta_0) = 0$ for all $\vartheta \in \Theta$.

If the functions $f(\vartheta, \cdot)$ and $a(\vartheta, \cdot)$ satisfy the relation (50) then under the corresponding conditions we can obtain the representation

$$ \frac{M(\vartheta_u, t) - M(\vartheta_0, t)}{\varepsilon \sigma(t)} = \sqrt{\frac{\psi_{\varepsilon}}{\varepsilon}} \frac{[f(\vartheta_u, t) b(\vartheta_u, t) - f(\vartheta_0, t) b(\vartheta_0, t)]}{\sqrt{2B(\vartheta_u, t)} \sigma(t)} \xi_{t, \varepsilon} (1 + o(1)). $$

This case is in some sense close to the considered below case with $\delta \in (0, 1/3)$ and probably needs a special study.
**Case** $\delta \in (0, 1/3)$.

Let us return to the key expression (34). If $\delta \in (0, 1/3)$, then

$$ 
\begin{aligned}
& \frac{M(\vartheta_u, t) - M(\vartheta_0, t)}{\varepsilon\sigma(t)} \\
&= \frac{1}{\psi_\varepsilon} \frac{a(\vartheta_u, t) - a(\vartheta_0, t) + \ell(\vartheta_u, t) - \ell(\vartheta_0, t)}{B(\vartheta_u, t)\sigma(t)} f(\vartheta_0, t) y_t(\vartheta_0) (1 + o(1)) \\
&\quad + \sqrt{\frac{\psi_\varepsilon}{\varepsilon}} \frac{[f(\vartheta_u, t)b(\vartheta_u, t) - f(\vartheta_0, t)b(\vartheta_0, t)]}{\sqrt{2B(\vartheta_u, t)}\sigma(t)} \xi_{t,\varepsilon} (1 + o(1)) \\
&= \sqrt{\frac{\psi_\varepsilon}{\varepsilon}} \frac{[f(\vartheta_u, t)b(\vartheta_u, t) - f(\vartheta_0, t)b(\vartheta_0, t)]}{\sqrt{2B(\vartheta_u, t)}\sigma(t)} \xi_{t,\varepsilon} (1 + o(1)) \\
&= \varphi_\varepsilon \sqrt{\frac{\psi_\varepsilon}{\varepsilon}} \frac{[u^\top \dot{f}(\vartheta_0, t)b(\vartheta_0, t) + f(\vartheta_0, t)u^\top \dot{b}(\vartheta_0, t)]}{\sqrt{2B(\vartheta_0, t)}\sigma(t)} \xi_{t,\varepsilon} (1 + o(1)).
\end{aligned} 
\quad (51) $$

Therefore we have to take $\varphi_\varepsilon = \sqrt{\varepsilon/\psi_\varepsilon} = \varepsilon^{(1-\delta)/2}$. The Fisher information in this case is

$$ I_1(\vartheta_0) = \int_0^T \frac{[\dot{f}(\vartheta_0, t)b(\vartheta_0, t) + f(\vartheta_0, t)\dot{b}(\vartheta_0, t)][\dot{f}(\vartheta_0, t)b(\vartheta_0, t) + f(\vartheta_0, t)\dot{b}(\vartheta_0, t)]^\top}{2b(\vartheta_0, t)f(\vartheta_0, t)\sigma(t)} dt. $$

Therefore we obtained the same Fisher information as in the case $\psi_\varepsilon \equiv 1$ studied in [30]. Of course the normalization of the MLE is different.

Define the function

$$ G_1(\vartheta, \vartheta_0) = \int_0^T \frac{[f(\vartheta, t)b(\vartheta, t) - f(\vartheta_0, t)b(\vartheta_0, t)]^2}{2f(\vartheta, t)b(\vartheta, t)\sigma(t)} dt $$

and introduce Conditions $\hat{\mathcal{E}}^*$

$\hat{\mathcal{E}}_1^*$. The functions $f(\vartheta, t)$, $b(\vartheta, t)$, $\vartheta \in \Theta, t \in [0, T]$ and $\sigma(t)$, $t \in [0, T]$ have continuous derivatives w.r.t. $t$.

$\hat{\mathcal{E}}_2^*$. The functions $f(\vartheta, t)$, $b(\vartheta, t)$, $t \in [0, T]$, $\vartheta \in \Theta$ and $\sigma(t)$, $0 \le t \le T$ are bounded and separated from zero.

$\hat{\mathcal{E}}_3^*$. The functions $f(\vartheta, t)$, $b(\vartheta, t)$, $t \in [0, T]$, $\vartheta \in \Theta$ are two times continuously differentiable w.r.t. $\vartheta$.

$\hat{\mathcal{E}}_4^*$. The Fisher information is uniformly positive

$$ \inf_{\vartheta \in \Theta} \inf_{\|e\|=1, e \in \mathbb{R}^d} e^\top I_1(\vartheta) e > 0. $$

$\hat{\mathcal{E}}_5^*$. For any $\nu > 0$

$$ g(\nu) = \inf_{\vartheta_0 \in \Theta} \inf_{\|\vartheta - \vartheta_0\| > \nu} G_1(\vartheta, \vartheta_0) > 0. $$
**Theorem 3.** Let the conditions $\hat{\mathcal{E}}^*$ hold and $\delta \in (0, 1/3)$. Then the MLE and BE are uniformly on compacts consistent, asymptotically normal

$$ \sqrt{\frac{\psi_{\varepsilon}}{\varepsilon}} (\hat{\vartheta}_{\varepsilon} - \vartheta_0) \implies \zeta^*(\vartheta_0) \sim \mathcal{N} (0, I_1 (\vartheta_0)^{-1}), \quad \sqrt{\frac{\psi_{\varepsilon}}{\varepsilon}} (\tilde{\vartheta}_{\varepsilon} - \vartheta_0) \implies \zeta^*(\vartheta_0), $$

the moments converge and the both estimators are asymptotically efficient.

*Proof.* We give a sketch of the proof because it is quite close to the proof of Theorem 1 in [30]. Denote $S(\vartheta, t) = f(\vartheta, t)b(\vartheta, t)$, $\varphi_{\varepsilon} = (\varepsilon/\psi_{\varepsilon})^{1/2}$ and $\vartheta_u = \vartheta_0 + \varphi_{\varepsilon}u$. The normalized likelihood ratio $Z_{\varepsilon}(\vartheta_0, u) = L(\vartheta_0 + \varphi_{\varepsilon}u, X^T)/L(\vartheta_0, X^T)$, $u \in \mathbb{U}_{\varepsilon}$ according to (12) admits the representation

$$ \begin{aligned} \ln Z_{\varepsilon}(\vartheta_0, u) &= \int_0^T \frac{M(\vartheta_u, t) - M(\vartheta_0, t)}{\varepsilon\sigma(t)} d\bar{W}_t - \int_0^T \frac{[M(\vartheta_u, t) - M(\vartheta_0, t)]^2}{2\varepsilon^2\sigma(t)^2} dt \\ &= \sqrt{\frac{\psi_{\varepsilon}}{\varepsilon}} \int_0^T \frac{S(\vartheta_u, t) - S(\vartheta_0, t)}{\sqrt{2S(\vartheta_u, t)\sigma(t)}} \xi_{t,\varepsilon} d\bar{W}_t (1+o(1)) \\ &\quad - \frac{\psi_{\varepsilon}}{\varepsilon} \int_0^T \frac{[S(\vartheta_u, t) - S(\vartheta_0, t)]^2}{4S(\vartheta_u, t)\sigma(t)} \xi_{t,\varepsilon}^2 dt (1+o(1)) \\ &= \int_0^T \frac{u^{\top}\dot{S}(\vartheta_0, t)}{\sqrt{2S(\vartheta_0, t)\sigma(t)}} \xi_{t,\varepsilon} d\bar{W}_t (1+o(1)) \\ &\quad - \frac{1}{2} \int_0^T \frac{[u^{\top}\dot{S}(\vartheta_0, t)]^2}{2S(\vartheta_0, t)\sigma(t)} \xi_{t,\varepsilon}^2 dt (1+o(1)). \end{aligned} $$

The Gaussian random variables

$$ \xi_{t,\varepsilon} = \sqrt{2B(\vartheta_0, t) \frac{\psi_{\varepsilon}}{\varepsilon}} \int_0^t \exp\left(-\frac{\psi_{\varepsilon}}{\varepsilon}(t-s)B(\vartheta_0, t)\right) d\bar{W}_s, \quad 0 < t \le T $$

are asymptotically independent and for any continuous function $R(t)$, $0 \le t \le T$ the relations

$$ \int_0^T R(t) \xi_{t,\varepsilon} dt \to 0, \qquad \int_0^T R(t) \xi_{t,\varepsilon}^2 dt \to \int_0^T R(t) dt $$

hold (see Lemma 4 in [30]).

Therefore we have the limits:

$$ \int_0^T \frac{\dot{S}(\vartheta_0, t) \dot{S}(\vartheta_0, t)^\top}{2S(\vartheta_0, t) \sigma(t)} \xi_{t,\varepsilon}^2 dt \to \int_0^T \frac{\dot{S}(\vartheta_0, t) \dot{S}(\vartheta_0, t)^\top}{2S(\vartheta_0, t) \sigma(t)} dt = I_1(\vartheta_0) $$

and by the CLT

$$ \Delta_{\varepsilon}(\vartheta_0, X^T) \equiv \int_0^T \frac{\dot{S}(\vartheta_0, t)}{\sqrt{2S(\vartheta_0, t)\sigma(t)}} \xi_{t,\varepsilon} d\bar{W}_t \implies \int_0^T \frac{\dot{S}(\vartheta_0, t)}{\sqrt{2S(\vartheta_0, t)\sigma(t)}} dw(t) $$
$$= \Delta(\vartheta_0) \sim \mathcal{N}(0, I_1(\vartheta_0)).$$

Here $w(t)$, $0 \le t \le T$ is some Wiener process. Moreover, these convergences are uniform on compacts $\mathbb{K} \in \Theta$. Therefore we verified that the LR $Z_{\varepsilon}(\vartheta_0, u)$, $u \in \mathbb{U}_{\varepsilon}$ have the representation

$$Z_{\varepsilon}(\vartheta_0, u) = \exp \left( u^{\top} \Delta_{\varepsilon} (\vartheta_0, X^T) - \frac{1}{2} u^{\top} I_1 (\vartheta_0) u + r_{\varepsilon} \right)$$

where $r_{\varepsilon} \to 0$.

The proof of the estimates

$$\sup_{\vartheta_0 \in \mathbb{K}} \mathbf{E}_{\vartheta_0} \left[ Z_{\varepsilon}(\vartheta_0, u_2)^{\frac{1}{2m}} - Z_{\varepsilon}(\vartheta_0, u_1)^{\frac{1}{2m}} \right]^{2m} \le C (1 + R^{2m}) \|u_2 - u_1\|^{2m} \quad (52)$$

and

$$\sup_{\vartheta_0 \in \mathbb{K}} \mathbf{P}_{\vartheta_0} \left( Z_{\varepsilon}(\vartheta_0, u) > e^{-\kappa u^2} \right) \le \frac{C_N}{|u|^N} \quad (53)$$

follows the same lines as the proofs of Lemmas 4 and 5.

□

**Case** $\delta = 1/3$.

In this case we take $\varphi_{\varepsilon} = \varepsilon^{1/3}$ and the representation (12) is

$$\begin{aligned} & \frac{M(\vartheta_u, t) - M(\vartheta_0, t)}{\varepsilon\sigma(t)} \\ &= \frac{1}{\varepsilon^{1/3}} \frac{a(\vartheta_u, t) - a(\vartheta_0, t) + \ell(\vartheta_u, t) - \ell(\vartheta_0, t)}{B(\vartheta_u, t)\sigma(t)} f(\vartheta_0, t) y_t(\vartheta_0) (1 + o(1)) \\ & \quad + \frac{1}{\varepsilon^{1/3}} \frac{[f(\vartheta_u, t)b(\vartheta_u, t) - f(\vartheta_0, t)b(\vartheta_0, t)]}{\sqrt{2B(\vartheta_u, t)}\sigma(t)} \xi_{t,\varepsilon} (1 + o(1)) \\ &= \left[ \frac{u^\top \dot{a}^*(\vartheta_0, t) y_t(\vartheta_0)}{b(\vartheta_0, t)} + \frac{u^\top \dot{S}(\vartheta_0, t)}{\sqrt{2S(\vartheta_0, t)}\sigma(t)} \xi_{t,\varepsilon} \right] (1 + o(1)). \end{aligned}$$

Therefore if we suppose that the corresponding version of the conditions $\hat{\mathcal{D}}$ and $\hat{\mathcal{E}}^*$ holds, then

$$\begin{aligned} \ln Z_{\varepsilon}(\vartheta_0, u) &= \int_0^T \frac{M(\vartheta_u, t) - M(\vartheta_0, t)}{\varepsilon\sigma(t)} d\bar{W}_t - \int_0^T \frac{[M(\vartheta_u, t) - M(\vartheta_0, t)]^2}{\varepsilon^2\sigma(t)^2} dt \\ &= u^\top \Delta_{\varepsilon}(\vartheta_0, X^T) - \frac{1}{2} u^\top I_3(\vartheta_0) u + r_{\varepsilon} \end{aligned} \quad (54)$$
where $r_{\varepsilon} \to 0$,

$$
\begin{align*}
\Delta_{\varepsilon}(\vartheta_0, X^T) &= \int_0^T \left[ \frac{\dot{a}^*(\vartheta_0, t) y_t(\vartheta_0)}{b(\vartheta_0, t)} + \frac{\dot{S}(\vartheta_0, t)}{\sqrt{2S(\vartheta_0, t) \sigma(t)}} \xi_{t,\varepsilon} \right] d\bar{W}_t \\
&\implies \Delta(\vartheta_0) \sim \mathcal{N}(0, I_3(\vartheta_0)) \\
I_3(\vartheta_0) &= \int_0^T \left[ \frac{\dot{a}^*(\vartheta_0, t) \dot{a}^*(\vartheta_0, t)^\top y_t(\vartheta_0)^2}{b(\vartheta_0, t)^2} + \frac{\dot{S}(\vartheta_0, t) \dot{S}(\vartheta_0, t)^\top}{2S(\vartheta_0, t) \sigma(t)} \right] dt.
\end{align*}
$$

Remind that

$$
\int_0^T \frac{\| \dot{a}^*(\vartheta_0, t) \| + \| \dot{S}(\vartheta_0, t) \|}{b(\vartheta_0, t) \sqrt{2S(\vartheta_0, t) \sigma(t)}} y_t(\vartheta_0) \xi_{t,\varepsilon} \, dt \to 0.
$$

The proofs of the estimates like (52), (53) once more follows the proofs of Lemmas 4,5.
Just note that the function $G_3(\vartheta, \vartheta_0)$ in this case has the following form

$$
G_3(\vartheta, \vartheta_0) = \int_0^T \frac{[a(\vartheta, t) - a(\vartheta_0, t) + \ell(\vartheta, t) - \ell(\vartheta_0, t)]^2}{f(\vartheta, t)^2 b(\vartheta, t)^2} f(\vartheta_0, t)^2 y_t(\vartheta_0)^2 dt \\
+ \int_0^T \frac{[f(\vartheta, t) b(\vartheta, t) - f(\vartheta_0, t) b(\vartheta_0, t)]^2}{2f(\vartheta, t) b(\vartheta, t) \sigma(t)} dt
$$

Having (54) and these estimates we obtain the consistency, asymptotic normality

$$
\varepsilon^{-1/3} (\hat{\vartheta}_{\varepsilon} - \vartheta_0) \implies \hat{\zeta} (\vartheta_0) \sim \mathcal{N} (0, I_3 (\vartheta_0)^{-1}), \quad \varepsilon^{-1/3} (\tilde{\vartheta}_{\varepsilon} - \vartheta_0) \implies \hat{\zeta} (\vartheta_0),
$$

convergence of moments and asymptotic efficiency of the MLE and BE.

1.3 One-step MLE-processes

1.3.1 Case $\delta \in (1/3, 1)$.

The partially observed system is (1),(2) with $\psi_{\varepsilon} = \varepsilon^{\delta}$, $\delta \in (1/3, 1)$. The learning interval is $[0, \tau]$ and the preliminary MDE is

$$
\check{\vartheta}_{\tau, \varepsilon} = \arg \inf_{\vartheta \in \Theta} \int_{0}^{\tau} [X_t - x_t(\vartheta)]^2 dt.
$$

Note that if the conditions $\check{\mathcal{B}}$ hold then by Theorem 1 the MDE is consistent and asymp-
totically normal

$$
\psi_{\varepsilon}^{-1} (\check{\vartheta}_{\tau, \varepsilon} - \vartheta_0) \implies \check{\zeta} (\vartheta_0) \sim \mathcal{N} (0, \check{D} (\vartheta_0)).
$$
Introduce the random process $M(\check{\vartheta}_{\tau,\varepsilon}, t)$, $\tau \le t \le T$ by the equation

$$dM(\check{\vartheta}_{\tau,\varepsilon}, t) = A^*(\check{\vartheta}_{\tau,\varepsilon}, t) M(\check{\vartheta}_{\tau,\varepsilon}, t) dt + \frac{\psi_{\varepsilon}}{\varepsilon} \tilde{B}(\check{\vartheta}_{\tau,\varepsilon}, t) dX_t \quad (55)$$

subject to initial value $M(\check{\vartheta}_{\tau,\varepsilon}, \tau)$. Here

$$A^*(\check{\vartheta}_{\tau,\varepsilon}, t) = a^*(\check{\vartheta}_{\tau,\varepsilon}, t) - \frac{\psi_{\varepsilon}}{\varepsilon} \tilde{B}^*(\check{\vartheta}_{\tau,\varepsilon}, t), \quad \tilde{B}(\check{\vartheta}_{\tau,\varepsilon}, t) = \frac{\gamma_*(\check{\vartheta}_{\tau,\varepsilon}, t) f(\check{\vartheta}_{\tau,\varepsilon}, t)^2}{\sigma(t)^2}.$$

The initial value can be calculated as follows. The function $M(\vartheta, t)$, $0 \le t \le \tau$ has the representation

$$M(\vartheta, \tau) = M(\vartheta, 0) \Phi(\vartheta, 0, \tau) + \frac{\psi_{\varepsilon}}{\varepsilon} \tilde{B}(\vartheta, \tau) X_{\tau} \\ - \frac{\psi_{\varepsilon}}{\varepsilon} \int_{0}^{\tau} \Phi(\vartheta, s, \tau) X_s \left[ \tilde{B}'_{s}(\vartheta, s) - A^*(\vartheta, s) \right] ds,$$

where

$$\Phi(\vartheta, s, \tau) = \exp \left( \int_{s}^{t} A^*(\vartheta, v) dv \right).$$

Therefore we can write

$$M(\check{\vartheta}_{\tau,\varepsilon}, \tau) = M(\check{\vartheta}_{\tau,\varepsilon}, 0) \Phi(\check{\vartheta}_{\tau,\varepsilon}, 0, \tau_{\varepsilon}) + \frac{\psi_{\varepsilon}}{\varepsilon} \tilde{B}(\check{\vartheta}_{\tau,\varepsilon}, \tau) X_{\tau,\varepsilon} \\ - \frac{\psi_{\varepsilon}}{\varepsilon} \int_{0}^{\tau_{\varepsilon}} \Phi(\check{\vartheta}_{\tau,\varepsilon}, s, \tau) X_s \left[ \tilde{B}'_{s}(\check{\vartheta}_{\tau,\varepsilon}, s) - A^*(\check{\vartheta}_{\tau,\varepsilon}, s) \right] ds. \quad (56)$$

Denote

$$F(\vartheta, t) = \frac{[\dot{a}(\vartheta, t) + \dot{l}(\vartheta, t)]y_t(\vartheta)}{b(\vartheta, t) \sigma(t)}, \quad Q(\vartheta, t) = \frac{[\dot{a}(\vartheta, t) + \dot{l}(\vartheta, t)]y_t(\vartheta)}{b(\vartheta, t)}$$

and Fisher information function

$$I_0^t(\vartheta) = \int_0^t \frac{[\dot{a}(\vartheta, t) + \dot{l}(\vartheta, t)][\dot{a}(\vartheta, t) + \dot{l}(\vartheta, t)]^\top y_s(\vartheta)^2}{b(\vartheta, s)^2} ds \\ = \int_0^t Q(\vartheta, t) Q(\vartheta, t)^\top dt, \quad 0 \le t \le T.$$

We need Conditions $\mathcal{F}^*$.

$\mathcal{F}^*$. The functions $\dot{a}(\vartheta, t)$, $\dot{l}(\vartheta, t)$, $b(\vartheta, t)$, $\sigma(t)$; $\vartheta \in \Theta, t \in [0, T]$ have continuous bounded derivatives w.r.t. $t$.
$\mathcal{F}_2^*$. The Fisher information function $I_0^t(\vartheta)$, $0 < \tau < t \le T$ is uniformly on $\vartheta \in \Theta$ non degenerate for all $t \in (\tau, T]$.

Introduce the One-step MLE-process $\vartheta_{t,\varepsilon}^*, \tau \le t \le T$

$$ \vartheta_{t,\varepsilon}^* = \check{\vartheta}_{\tau,\varepsilon} + I_0^t (\check{\vartheta}_{\tau,\varepsilon})^{-1} \frac{\psi_\varepsilon}{\varepsilon} \left\{ F(\check{\vartheta}_{\tau,\varepsilon}, \tau) X_\tau - \int_0^\tau X_s F'_s(\check{\vartheta}_{\tau,\varepsilon}, s) ds - \int_0^\tau F(\check{\vartheta}_{\tau,\varepsilon}, s) M(\check{\vartheta}_{\tau,\varepsilon}, s) ds + \int_\tau^t F(\check{\vartheta}_{\tau,\varepsilon}, s) [dX_s - M(\check{\vartheta}_{\tau,\varepsilon}, s) ds] \right\}. \quad (57) $$

**Theorem 4.** Let the conditions $\tilde{\mathcal{B}}, \hat{\mathcal{D}}_{1-3}$ and $\mathcal{F}^*$ hold and $\delta \in (1/3, 1)$. Then the estimator process $\vartheta_{t,\varepsilon}^*, \tau < t \le T$ is uniformly consistent, asymptotically normal

$$ \psi_\varepsilon^{-1} (\vartheta_{t,\varepsilon}^* - \vartheta_0) \implies \zeta_t (\vartheta_0) \sim \mathcal{N} (0, I_0^t (\vartheta_0)^{-1}), $$

the moments converge and this estimator is asymptotically efficient.

*Proof.* In the definition of the One-step estimator-process we used the relation between the stochastic and ordinary integrals

$$ \int_0^\tau F(\vartheta, s) dX_s = F(\vartheta, \tau) X_\tau - \int_0^\tau X_s F'_s(\vartheta, s) ds. $$

Applying Taylor expansion at the point $\vartheta_0$ we have

$$ \begin{aligned} & F(\check{\vartheta}_{\tau,\varepsilon}, \tau) X_\tau - \int_0^\tau X_s F'_s(\check{\vartheta}_{\tau,\varepsilon}, s) ds - \int_0^\tau F(\check{\vartheta}_{\tau,\varepsilon}, s) M(\check{\vartheta}_{\tau,\varepsilon}, s) ds \\ &= F(\vartheta_0, \tau) X_\tau - \int_0^\tau X_s F'_s(\vartheta_0, s) ds - \int_0^\tau F(\vartheta_0, s) M(\check{\vartheta}_{\tau,\varepsilon}, s) ds \\ &\quad + (\check{\vartheta}_{\tau,\varepsilon} - \vartheta_0)^\top \left[ \dot{F}(\vartheta_0, \tau) X_\tau - \int_0^\tau X_s \dot{F}'_s(\vartheta_0, s) ds - \int_0^\tau \dot{F}(\vartheta_0, s) M(\check{\vartheta}_{\tau,\varepsilon}, s) ds \right] \\ &\quad + \|\check{\vartheta}_{\tau,\varepsilon} - \vartheta_0\|^2 O(1) \\ &= \int_0^\tau F(\vartheta_0, s) [dX_s - M(\check{\vartheta}_{\tau,\varepsilon}, s) ds] \\ &\quad + (\check{\vartheta}_{\tau,\varepsilon} - \vartheta_0)^\top \int_0^\tau \dot{F}(\vartheta_0, s) [dX_s - M(\check{\vartheta}_{\tau,\varepsilon}, s) ds] + \|\check{\vartheta}_{\tau,\varepsilon} - \vartheta_0\|^2 O(1) \\ &= \varepsilon \int_0^\tau F(\vartheta_0, s) \sigma(s) d\bar{W}_s + (\vartheta_0 - \check{\vartheta}_{\tau,\varepsilon})^\top \int_0^\tau F(\vartheta_0, s) \dot{M}(\bar{\vartheta}_\varepsilon, s) ds \\ &\quad + \varepsilon (\check{\vartheta}_{\tau,\varepsilon} - \vartheta_0)^\top \int_0^\tau \dot{F}(\vartheta_0, s) \sigma(s) d\bar{W}_s + \frac{\varepsilon}{\psi_\varepsilon} \|\check{\vartheta}_{\tau,\varepsilon} - \vartheta_0\|^2 O(1) \\ &= \varepsilon \int_0^\tau F(\vartheta_0, s) \sigma(s) d\bar{W}_s + \frac{\varepsilon}{\psi_\varepsilon} (\vartheta_0 - \check{\vartheta}_{\tau,\varepsilon})^\top I_0^\tau (\vartheta_0) + \|\check{\vartheta}_{\tau,\varepsilon} - \vartheta_0\|^2 O(1). \end{aligned} $$
Here we used (26). For the second stochastic integral we obtain the similar expansion directly

$$
\begin{align*}
\int_{\tau}^{t} F(\check{\vartheta}_{\tau, \varepsilon}, s) [dX_s - M(\check{\vartheta}_{\tau, \varepsilon}, s) ds] &= \int_{\tau}^{t} F(\vartheta_0, s) [dX_s - M(\check{\vartheta}_{\tau, \varepsilon}, s) ds] \\
&\quad + (\check{\vartheta}_{\tau, \varepsilon} - \vartheta_0)^{\top} \int_{\tau}^{t} \dot{F}(\vartheta_0, s) [dX_s - M(\check{\vartheta}_{\tau, \varepsilon}, s) ds] + \|\check{\vartheta}_{\tau, \varepsilon} - \vartheta_0\|^2 O(1) \\
&= \varepsilon \int_{\tau}^{t} F(\vartheta_0, s) \sigma(s) d\bar{W}_s + \int_{\tau}^{t} F(\vartheta_0, s) [M(\vartheta_0, s) - M(\check{\vartheta}_{\tau, \varepsilon}, s)] ds \\
&\quad + \|\check{\vartheta}_{\tau, \varepsilon} - \vartheta_0\|^2 O(1) \\
&= \varepsilon \int_{\tau}^{t} F(\vartheta_0, s) \sigma(s) d\bar{W}_s + \frac{\varepsilon}{\psi_{\varepsilon}} (\vartheta_0 - \check{\vartheta}_{\tau, \varepsilon})^{\top} \int_{\tau}^{t} F(\vartheta_0, s)^2 ds + \frac{\varepsilon}{\psi_{\varepsilon}} \|\check{\vartheta}_{\tau, \varepsilon} - \vartheta_0\|^2 O(1).
\end{align*}
$$

This implies

$$
\begin{align*}
\frac{\vartheta_{t, \varepsilon}^* - \vartheta_0}{\psi_\varepsilon} &= \frac{\check{\vartheta}_{\tau, \varepsilon} - \vartheta_0}{\psi_\varepsilon} + I_\tau^t (\check{\vartheta}_{\tau, \varepsilon})^{-1} \int_0^t \frac{\dot{a}^*(\vartheta_0, s) y_s(\vartheta_0)}{b(\vartheta_0, s)} d\bar{W}_s \\
&\quad + \frac{1}{\psi_\varepsilon} (\vartheta_0 - \check{\vartheta}_{\tau, \varepsilon})^\top I_0^t (\check{\vartheta}_{\tau, \varepsilon})^{-1} I_0^t (\vartheta_0) + \frac{\|\check{\vartheta}_{\tau, \varepsilon} - \vartheta_0\|^2}{\psi_\varepsilon} O(1) \\
&= I_\tau^t (\vartheta_0)^{-1} \int_0^t \frac{\dot{a}^*(\vartheta_0, s) y_s(\vartheta_0)}{b(\vartheta_0, s)} d\bar{W}_s (1+o(1)) + \frac{\|\check{\vartheta}_{\tau, \varepsilon} - \vartheta_0\|^2}{\psi_\varepsilon} O(1) \\
&\implies \eta_t^*(\vartheta_0) = I_\tau^t (\vartheta_0)^{-1} \int_0^t \frac{\dot{a}^*(\vartheta_0, s) y_s(\vartheta_0)}{b(\vartheta_0, s)} dw(s).
\end{align*}
$$

The convergence of moments follows from the convergence of moments of $\check{\vartheta}_{\tau, \varepsilon}$ and conditions of Theorem 4. The uniform convergence of the second moments and the continuity of the Fisher information allows to verify the asymptotic efficiency too.

<figure><img src="image_13.png" /></figure>

**Remark 6.** This One-step MLE-process can be written in the recursive form too. Denote

$$
P(\vartheta, \tau) = F(\vartheta, \tau) X_{\tau} - \int_{0}^{\tau} X_{s} F'_{s}(\vartheta, s) ds - \int_{0}^{\tau} F(\vartheta, s) M(\vartheta, s) ds
$$

and write

$$
I_0^t (\check{\vartheta}_{\tau, \varepsilon}) \vartheta_{t, \varepsilon}^* = I_0^t (\check{\vartheta}_{\tau, \varepsilon}) \check{\vartheta}_{\tau, \varepsilon} + \frac{\psi_{\varepsilon}}{\varepsilon} \left\{ P(\check{\vartheta}_{\tau, \varepsilon}, \tau) + \int_{\tau}^{t} F(\check{\vartheta}_{\tau, \varepsilon}, s) [dX_s - M(\check{\vartheta}_{\tau, \varepsilon}, s) ds] \right\}.
$$

Therefore the stochastic differential of $\vartheta_{t, \varepsilon}^*$ is

$$
d\vartheta_{t, \varepsilon}^* = I_0^t (\check{\vartheta}_{\tau, \varepsilon})^{-1} Q (\check{\vartheta}_{\tau, \varepsilon}, t) [\check{\vartheta}_{\tau, \varepsilon} - \vartheta_{t, \varepsilon}^*] dt
$$
$$ + \frac{\psi_{\varepsilon}}{\varepsilon} I_0^t (\check{\vartheta}_{\tau, \varepsilon})^{-1} F (\check{\vartheta}_{\tau, \varepsilon}, t) [dX_t - M (\check{\vartheta}_{\tau, \varepsilon}, t) dt], \quad \tau < t \le T, \quad (58) $$

$$ \vartheta_{\tau, \varepsilon}^{*} = \check{\vartheta}_{\tau, \varepsilon} + \frac{\psi_{\varepsilon}}{\varepsilon} I_{0}^{\tau} (\check{\vartheta}_{\tau, \varepsilon})^{-1} P (\check{\vartheta}_{\tau, \varepsilon}, \tau). $$

The representation (58) allows relatively simple recursive numerical calculation of the estimator $\vartheta_{t, \varepsilon}^*, \tau < t \le T$.

### 1.3.2 Case $\delta \in (0, 1/3)$.

Remind (see (34)) that in this case $\varphi_{\varepsilon} = \varepsilon^{\frac{1-\delta}{2}}$

$$ \frac{M(\vartheta_0 + \varphi_{\varepsilon} u, t) - M(\vartheta_0, t)}{\varepsilon \sigma(t)} = \frac{u^{\top} \dot{S}(\vartheta_0, t)}{\sqrt{2B(\vartheta_0, t)} \sigma(t)} \xi_{t, \varepsilon} (1 + O(\varphi_{\varepsilon})) $$

where $S(\vartheta, t) = f(\vartheta, t)b(\vartheta, t)$ and $B(\vartheta, t) = f(\vartheta, t)b(\vartheta, t)\sigma(t)^{-1}$. The Fisher information function is

$$ I_{\tau}^{t}(\vartheta) = \int_{\tau}^{t} \frac{\dot{S}(\vartheta, s) \dot{S}(\vartheta, s)^{\top}}{2S(\vartheta, s) \sigma(s)} ds, \quad \tau < t \le T. \quad (59) $$

We have two preliminary MDEs. The first one $\check{\vartheta}_{\tau, \varepsilon}$ was constructed by the observations on the fixed learning interval $[0, \tau]$ and the second $\check{\vartheta}_{\tau, \varepsilon}$ - by the observations on the vanishing learning interval $[0, \tau_{\varepsilon}]$, $\tau_{\varepsilon} = \varepsilon^{1-\delta} \to 0$. The rates of convergence are $\varepsilon^{\delta}$ for the first and $\varepsilon^{\frac{1-\delta}{2}}$ for the second estimator respectively. The comparison of the rates (20) shows that the MDE $\check{\vartheta}_{\tau_{\varepsilon}}$ has better rate than $\check{\vartheta}_{\tau, \varepsilon}$. This leads to the following consequence: the One-step MLE-process constructed with the help of $\check{\vartheta}_{\tau, \varepsilon}$ is consistent for the values $\delta \in (1/5, 1/3)$ and if we use $\check{\vartheta}_{\tau_{\varepsilon}}$, then it is consistent for all $\delta \in (0, 1/3)$. This is a bit unusual situation (see Remark 3) and below we show why it happens.

#### Fixed learning interval

Define the MDE $\check{\vartheta}_{\tau, \varepsilon}$ by the relation (14). Suppose as well that the conditions $\mathcal{B}$ hold. Then by Theorem 1 we have

$$ \psi_{\varepsilon}^{-1} (\check{\vartheta}_{\tau, \varepsilon} - \vartheta_0) \implies \mathcal{N} (0, D(\vartheta_0)). $$

It is shown below that if we define the One-step MLE-process by the usual relation

$$ \vartheta_{t, \varepsilon}^{*} = \check{\vartheta}_{\tau, \varepsilon} + I_{\tau}^{t} (\check{\vartheta}_{\tau, \varepsilon})^{-1} \int_{\tau}^{t} \frac{\dot{M}(\check{\vartheta}_{\tau, \varepsilon}, s)}{\varepsilon \psi_{\varepsilon} \sigma(s)^2} [dX_s - M(\check{\vartheta}_{\tau, \varepsilon}, s) ds], $$

then the last term in the expansion has the expression

$$ \left[ \frac{(\check{\vartheta}_{\tau, \varepsilon} - \vartheta_0)}{\psi_{\varepsilon}} \right]^2 \varepsilon^{\frac{5\delta-1}{2}}. $$
It tends to 0 for the values $\delta > 1/5$. That is why we consider the cases $\delta \in (1/5, 1/3)$ and $\delta \in (0, 1/5)$ separately.

**Case** $\delta \in (1/5, 1/3)$.

Let us fix the learning interval $[0, \tau]$ and take $\delta \in (1/5, 1/3)$. For One-step MLE-process we need two stochastic processes $M(\check{\vartheta}_{\tau, \varepsilon}, t)$, $\tau \le t \le T$ and $\dot{M}(\check{\vartheta}_{\tau, \varepsilon}, t)$, $\tau \le t \le T$. These processes are defined by the equations

$$dM(\check{\vartheta}_{\tau, \varepsilon}, t) = \left[ a^*(\check{\vartheta}_{\tau, \varepsilon}, t) - \frac{\psi_{\varepsilon}}{\varepsilon} B(\check{\vartheta}_{\tau, \varepsilon}, t) \right] M(\check{\vartheta}_{\tau, \varepsilon}, t) dt + \frac{\psi_{\varepsilon}}{\varepsilon} B(\check{\vartheta}_{\tau, \varepsilon}, t) dX_t, \quad (60)$$

$$d\dot{M}(\check{\vartheta}_{\tau, \varepsilon}, t) = \left[ a^*(\check{\vartheta}_{\tau, \varepsilon}, t) - \frac{\psi_{\varepsilon}}{\varepsilon} B(\check{\vartheta}_{\tau, \varepsilon}, t) \right] \dot{M}(\check{\vartheta}_{\tau, \varepsilon}, t) dt + \frac{\psi_{\varepsilon}}{\varepsilon} \dot{B}(\check{\vartheta}_{\tau, \varepsilon}, t) dX_t \\ + \left[ \dot{a}^*(\check{\vartheta}_{\tau, \varepsilon}, t) - \frac{\psi_{\varepsilon}}{\varepsilon} \dot{B}(\check{\vartheta}_{\tau, \varepsilon}, t) \right] M(\check{\vartheta}_{\tau, \varepsilon}, t) dt. \quad (61)$$

The initial conditions are $M(\check{\vartheta}_{\tau, \varepsilon}, \tau) = 0$, $\dot{M}(\check{\vartheta}_{\tau, \varepsilon}, \tau) = 0$.

The One-step MLE-process is

$$\vartheta_{t, \varepsilon}^* = \check{\vartheta}_{\tau, \varepsilon} + I_{\tau}^t (\check{\vartheta}_{\tau, \varepsilon})^{-1} \int_{\tau}^{t} \frac{\dot{M}(\check{\vartheta}_{\tau, \varepsilon}, s)}{\varepsilon \psi_{\varepsilon} \sigma(s)^2} [dX_s - M(\check{\vartheta}_{\tau, \varepsilon}, s) ds].$$

We need the additional Conditions $\mathcal{F}^*$.

$\mathcal{F}_3^*$. The functions $\dot{b}(\vartheta, t)$, $\dot{f}(\vartheta, t)$, $f(\vartheta, t)$, $\sigma(t)$; $\vartheta \in \Theta, t \in [0, T]$ have continuous bounded derivatives w.r.t. $t$.

$\mathcal{F}_4^*$. The Fisher information function $I_{\tau}^t(\vartheta)$, $0 < \tau < t \le T$ is uniformly on $\vartheta \in \Theta$ positive for all $t \in (\tau, T]$.

**Theorem 5.** Let the conditions $\tilde{\mathcal{B}}$, $\hat{\mathcal{D}}_{1-3}$ and $\mathcal{F}_{3-4}^*$ hold, $\varphi_{\varepsilon} = \varepsilon^{\frac{1-\delta}{2}}$ and $\delta \in (1/5, 1/3)$. Then the estimator process $\vartheta_{t, \varepsilon}^*, \tau < t \le T$ is uniformly consistent, asymptotically normal

$$\varphi_{\varepsilon}^{-1} (\vartheta_{t, \varepsilon}^* - \vartheta_0) \implies \mathcal{N} (0, I_0^t (\vartheta_0)^{-1}),$$

the moments converge and this estimator is asymptotically efficient.

*Proof.* Consider the asymptotic behavior of this estimator. We have

$$\frac{\vartheta_{t, \varepsilon}^* - \vartheta_0}{\varphi_{\varepsilon}} = \frac{\check{\vartheta}_{\tau, \varepsilon} - \vartheta_0}{\varphi_{\varepsilon}} + I_{\tau}^t (\check{\vartheta}_{\tau, \varepsilon})^{-1} \int_{\tau}^{t} \frac{\dot{M}(\check{\vartheta}_{\tau, \varepsilon}, s)}{\varphi_{\varepsilon} \psi_{\varepsilon} \sigma(s)} d\bar{W}_s \\ + \frac{I_{\tau}^t (\check{\vartheta}_{\tau, \varepsilon})^{-1}}{\varepsilon \varphi_{\varepsilon} \psi_{\varepsilon}} \int_{\tau}^{t} \frac{\dot{M}(\check{\vartheta}_{\tau, \varepsilon}, s)}{\sigma(s)^2} [M(\vartheta_0, s) - M(\check{\vartheta}_{\tau, \varepsilon}, s)] ds.$$
The stochastic integral is asymptotically normal

$$
\begin{align*}
\int_{\tau}^{t} \frac{\dot{M}(\check{\vartheta}_{\tau, \varepsilon}, s)}{\varphi_{\varepsilon} \psi_{\varepsilon} \sigma(s)} d\bar{W}_{s} &= \int_{\tau}^{t} \frac{\dot{S}(\vartheta_{0}, s) \xi_{s, \varepsilon}}{\sqrt{2S(\vartheta_{0}, s) \sigma(s)}} d\bar{W}_{s} (1+o(1)) \\
&\implies \int_{\tau}^{t} \frac{\dot{S}(\vartheta_{0}, s)}{\sqrt{2S(\vartheta_{0}, s) \sigma(s)}} dw(s) \sim \mathcal{N}(0, I_{\tau}^{t}(\vartheta_{0})) .
\end{align*}
$$

For the ordinary integral we have

$$
\begin{align*}
& \frac{1}{\varepsilon \varphi_{\varepsilon} \psi_{\varepsilon}} \int_{\tau}^{t} \frac{\dot{M}(\check{\vartheta}_{\tau, \varepsilon}, s)}{\sigma(s)^2} [M(\vartheta_0, s) - M(\check{\vartheta}_{\tau, \varepsilon}, s)] \, ds \\
&= \frac{\vartheta_0 - \check{\vartheta}_{\tau, \varepsilon}}{\varphi_{\varepsilon}} \frac{1}{\varepsilon \psi_{\varepsilon}} \int_{\tau}^{t} \frac{\dot{M}(\check{\vartheta}_{\tau, \varepsilon}, s)^2}{\sigma(s)^2} \, ds \\
&\quad + \frac{(\vartheta_0 - \check{\vartheta}_{\tau, \varepsilon})^2}{\varphi_{\varepsilon}} \frac{1}{\varepsilon \psi_{\varepsilon}} \int_{\tau}^{t} \frac{\dot{M}(\check{\vartheta}_{\tau, \varepsilon}, s) \ddot{M}(\bar{\vartheta}_{\tau, \varepsilon}, s)}{\sigma(s)^2} \, ds \\
&= \frac{\vartheta_0 - \check{\vartheta}_{\tau, \varepsilon}}{\varphi_{\varepsilon}} \int_{\tau}^{t} \frac{\dot{S}(\vartheta_0, s)^2 \xi_{s, \varepsilon}^2}{2S(\vartheta_0, s) \sigma(s)} \, ds + (\vartheta_0 - \check{\vartheta}_{\varepsilon}) O(1) + \frac{(\vartheta_0 - \check{\vartheta}_{\tau, \varepsilon})^2}{\varphi_{\varepsilon}} O(1) \\
&= \frac{\vartheta_0 - \check{\vartheta}_{\tau, \varepsilon}}{\varphi_{\varepsilon}} [I_{\tau}^{t}(\vartheta_0) + (\vartheta_0 - \check{\vartheta}_{\tau, \varepsilon}) O(1)] + (\vartheta_0 - \check{\vartheta}_{\tau, \varepsilon}) O(1) + \frac{(\vartheta_0 - \check{\vartheta}_{\tau, \varepsilon})^2}{\varphi_{\varepsilon}} O(1).
\end{align*}
$$

For the last term we have the estimate

$$
\frac{(\vartheta_0 - \check{\vartheta}_{\tau, \varepsilon})^2}{\varphi_{\varepsilon}} = \left[ \frac{(\vartheta_0 - \check{\vartheta}_{\tau, \varepsilon})^2}{\psi_{\varepsilon}^2} \right] \frac{\psi_{\varepsilon}^2}{\varphi_{\varepsilon}} = \left[ \frac{(\vartheta_0 - \check{\vartheta}_{\tau, \varepsilon})^2}{\psi_{\varepsilon}} \right]^2 \varepsilon^{\frac{5\delta-1}{2}} \to 0
$$

because $\delta > 1/5$. Therefore

$$
\begin{align}
\frac{\vartheta_{t, \varepsilon}^* - \vartheta_0}{\varphi_{\varepsilon}} &= \frac{\check{\vartheta}_{\tau, \varepsilon} - \vartheta_0}{\varphi_{\varepsilon}} + \frac{\vartheta_0 - \check{\vartheta}_{\tau, \varepsilon}}{\varphi_{\varepsilon}} + o(1) \nonumber \\
&\quad + I_{\tau}^{t} (\vartheta_0)^{-1} \int_{\tau}^{t} \frac{\dot{S}(\vartheta_0, s) \xi_{s, \varepsilon}}{\sqrt{2S(\vartheta_0, s) \sigma(s)}} d\bar{W}_s (1+o(1)) \nonumber \\
&\implies \eta_t^*(\vartheta_0) = I_{\tau}^{t} (\vartheta_0)^{-1} \int_{\tau}^{t} \frac{\dot{S}(\vartheta_0, s)}{\sqrt{2S(\vartheta_0, s) \sigma(s)}} dw(s) \tag{62} \\
&\hspace{2.5em} \square \nonumber
\end{align}
$$

**Remark 7.** Note that the One-step MLE-process

$$
\vartheta_{t, \varepsilon}^{*} = \check{\vartheta}_{\tau, \varepsilon} + \frac{I_{\tau}^{t} (\check{\vartheta}_{\tau, \varepsilon})^{-1}}{\sqrt{\varepsilon \psi_{\varepsilon}}} \int_{\tau}^{t} \frac{\dot{S}(\check{\vartheta}_{\tau, \varepsilon}, s)}{\sqrt{2S(\check{\vartheta}_{\tau, \varepsilon}, s) \sigma(s)^{3/2}}} [dX_s - M(\check{\vartheta}_{\tau, \varepsilon}, s) \, ds]
$$

is not consistent because in the corresponding ordinary integral we obtain the term $\xi_{s,\varepsilon}$ and not $\xi_{s,\varepsilon}^2$.
**Case** $\delta \in (1/9, 1/5]$.

Let us introduce Two-step MLE-process. The interval $[0, \tau_\ast]$ we write as two intervals: $[0, \tau]$ and $[\tau, \tau_\ast]$. Remind that we have the MDE (14) $\check{\vartheta}_{\tau, \varepsilon}$ with the rate $\psi_\varepsilon$. Introduce the first One-step estimator

$$ \bar{\vartheta}_{\tau_*, \varepsilon} = \check{\vartheta}_{\tau, \varepsilon} + \frac{\mathrm{I}_{\tau}^{\tau_*(}\check{\vartheta}_{\tau, \varepsilon})^{-1}}{\varepsilon \psi_{\varepsilon}} \int_{\tau}^{\tau_*} \frac{\dot{M}(\check{\vartheta}_{\tau, \varepsilon}, s)}{\sigma(s)^2} [dX_s - M(\check{\vartheta}_{\tau, \varepsilon}, s) ds] $$

and consider the normalized difference

$$ \frac{\bar{\vartheta}_{\tau_*, \varepsilon} - \vartheta_0}{\varepsilon^\mu} = \frac{\check{\vartheta}_{\tau, \varepsilon} - \vartheta_0}{\varepsilon^\mu \psi_\varepsilon} + \frac{\mathrm{I}_\tau^{\tau_*(}\check{\vartheta}_{\tau, \varepsilon})^{-1}}{\varepsilon^{1+\mu} \psi_\varepsilon} \int_\tau^{\tau_*} \frac{\dot{M}(\check{\vartheta}_{\tau, \varepsilon}, s)}{\sigma(s)} d\bar{W}_s \\ + \frac{\mathrm{I}_\tau^{\tau_*(}\check{\vartheta}_{\tau, \varepsilon})^{-1}}{\varepsilon^{1+\mu} \psi_\varepsilon} \int_\tau^{\tau_*} \frac{\dot{M}(\check{\vartheta}_{\tau, \varepsilon}, s)}{\sigma(s)^2} [M(\vartheta_0, s) - M(\check{\vartheta}_{\tau, \varepsilon}, s)] ds \\ = \frac{\check{\vartheta}_{\tau, \varepsilon} - \vartheta_0}{\varepsilon^\mu} + \mathrm{I}_\tau^{\tau_*(}\check{\vartheta}_{\tau, \varepsilon})^{-1} \sqrt{\frac{\varepsilon^{1-2\mu}}{\psi_\varepsilon}} \int_\tau^{\tau_*} \frac{\dot{S}(\check{\vartheta}_{\tau, \varepsilon}, s) \xi_{t, \varepsilon}}{\sqrt{2S(\check{\vartheta}_{\tau, \varepsilon}, s)\sigma(s)}} d\bar{W}_s \\ - \frac{\check{\vartheta}_{\tau, \varepsilon} - \vartheta_0}{\varepsilon^\mu} \mathrm{I}_\tau^{\tau_*(}\check{\vartheta}_{\tau, \varepsilon})^{-1} \int_\tau^{\tau_*} \frac{\dot{S}(\check{\vartheta}_{\tau, \varepsilon}, s) \dot{S}(\check{\vartheta}_{\tau, \varepsilon}, s)^\top \xi_{t, \varepsilon}^2}{2S(\check{\vartheta}_{\tau, \varepsilon}, s)\sigma(s)} ds + \frac{\|\check{\vartheta}_{\tau, \varepsilon} - \vartheta_0\|^2}{\varepsilon^\mu} O(1). $$

We have

$$ \frac{\|\check{\vartheta}_{\tau, \varepsilon} - \vartheta_0\|^2}{\varepsilon^\mu} = \frac{\|\check{\vartheta}_{\tau, \varepsilon} - \vartheta_0\|^2}{\varepsilon^{2\delta}} \varepsilon^{2\delta-\mu}. $$

For the choice of $\mu$ we have two conditions:

1. $\varepsilon^{1-2\mu-\delta}$ is bounded,

2. $\varepsilon^{2\delta-\mu} \to 0$.

Or $\mu < \min((1-\delta)/2, 2\delta) = 2\delta$.

If this condition holds then $\varepsilon^{-\mu}(\bar{\vartheta}_{\tau_*, \varepsilon} - \vartheta_0)$ is bounded in probability.

Then we introduce the second estimator, which can be called Two-steps MLE by the relation

$$ \vartheta_{t, \varepsilon}^{**} = \bar{\vartheta}_{\tau_*, \varepsilon} + \frac{\mathrm{I}_{\tau_*}^t (\bar{\vartheta}_{\tau, \varepsilon})^{-1}}{\varepsilon \psi_\varepsilon} \int_{\tau_*}^t \frac{\dot{M}(\bar{\vartheta}_{\tau, \varepsilon}, s)}{\sigma(s)^2} [dX_s - M(\bar{\vartheta}_{\tau, \varepsilon}, s) ds], \quad \tau_* < t \le T. $$

The corresponding representation of the normalized difference is

$$ \frac{\vartheta_{t, \varepsilon}^{**} - \vartheta_0}{\varphi_\varepsilon} = \frac{\bar{\vartheta}_{\tau_*, \varepsilon} - \vartheta_0}{\varphi_\varepsilon} + \frac{\mathrm{I}_{\tau_*}^t (\bar{\vartheta}_{\tau_*, \varepsilon})^{-1}}{\varphi_\varepsilon \psi_\varepsilon} \int_{\tau_*}^t \frac{\dot{M}(\bar{\vartheta}_{\tau_*, \varepsilon}, s)}{\sigma(s)} d\bar{W}_s $$
$$
\begin{align*}
&+ \frac{I_{\tau_*}^t (\bar{\vartheta}_{\tau_*, \varepsilon})^{-1}}{\varepsilon \varphi_\varepsilon \psi_\varepsilon} \int_{\tau}^t \frac{\dot{M} (\bar{\vartheta}_{\tau_*, \varepsilon}, s)}{\sigma(s)^2} [M(\vartheta_0, s) - M(\bar{\vartheta}_{\tau_*, \varepsilon}, s)] \mathrm{d}s \\
&= \frac{\bar{\vartheta}_{\tau_*, \varepsilon} - \vartheta_0}{\varphi_\varepsilon} + I_{\tau}^t (\vartheta_0)^{-1} \int_{\tau}^t \frac{\dot{S}(\vartheta_0, s) \xi_{t, \varepsilon}}{\sqrt{2S(\vartheta_0, s)\sigma(s)}} \mathrm{d}\bar{W}_s \\
&\quad - \frac{\bar{\vartheta}_{\tau_*, \varepsilon} - \vartheta_0}{\varphi_\varepsilon} I_{\tau_*}^t (\vartheta_0)^{-1} \int_{\tau}^t \frac{\dot{S}(\vartheta_0, s) \dot{S}(\vartheta_0, s)^\top \xi_{t, \varepsilon}^2}{2S(\vartheta_0, s)\sigma(s)} \mathrm{d}s \\
&\quad + \frac{\|\bar{\vartheta}_{\tau_*, \varepsilon} - \vartheta_0\|^2}{\varepsilon^{2\mu}} \frac{\varepsilon^{2\mu}}{\varphi_\varepsilon} O(1).
\end{align*}
$$

Therefore the last condition on $\mu$ is $2\mu > (1 - \delta)/2$. If $\delta \le 1/5$ the inequalities

$$ \frac{1 - \delta}{4} < \mu \le 2\delta \qquad (63) $$

hold for the values $\delta \in (1/9, 1/5)$. Hence for the $\mu$ satisfying (63) we obtain the convergence

$$ \frac{\vartheta_{t, \varepsilon}^{\star\star} - \vartheta_0}{\varphi_\varepsilon} \implies \mathcal{N} (0, I_{\tau_*}^t (\vartheta_0)^{-1}). $$

Of course, this estimator can be modified (see (57)) to obtain the full Fisher information in the following sense

$$ \frac{\tilde{\vartheta}_{t, \varepsilon}^{\star\star} - \vartheta_0}{\varphi_\varepsilon} \implies \mathcal{N} (0, I_0^t (\vartheta_0)^{-1}). $$

The born 1/9 can be replaced by another smaller one if we use Three-steps MLE-process. The details of such construction can be found in the works [22] or [29].

**Vanishing learning interval**, $\delta \in (0, 1/3)$.

Suppose now that $d = 1$ and the MDE $\tilde{\vartheta}_{\tau_\varepsilon}$ is constructed by observations on the vanishing interval $[0, \tau_\varepsilon]$, where $\tau_\varepsilon = \varepsilon^{1-\delta}$. Then under conditions $\mathcal{C}^\circ$

$$ \varepsilon^{-\frac{1-\delta}{2}} (\tilde{\vartheta}_{\tau_\varepsilon} - \vartheta_0) \implies \mathcal{N} (0, D(\vartheta_0)). $$

The One-step MLE-process is

$$ \vartheta_{t, \varepsilon}^{\star} = \tilde{\vartheta}_{\tau_{\varepsilon}} + \frac{I_{\tau_{\varepsilon}}^{t} (\tilde{\vartheta}_{\tau_{\varepsilon}})^{-1}}{\varepsilon \psi_{\varepsilon}} \int_{\tau_{\varepsilon}}^{t} \frac{\dot{M} (\tilde{\vartheta}_{\tau_{\varepsilon}}, s)}{\sigma(s)^2} [dX_s - M (\tilde{\vartheta}_{\tau_{\varepsilon}}, s) \mathrm{d}s], \quad \tau < t \le T. \quad (64) $$

The random processes $M(\tilde{\vartheta}_{\tau_\varepsilon}, s)$ and $\dot{M}(\tilde{\vartheta}_{\tau_\varepsilon}, s)$ are defined by the same equations (60), (61).
The repetition of the same calculations gives us the convergence $(\varphi_{\varepsilon} = \varepsilon^{\frac{1-\delta}{2}})$

$$ 
\begin{aligned}
\frac{\vartheta_{t,\varepsilon}^* - \vartheta_0}{\varphi_{\varepsilon}} &= \frac{\check{\vartheta}_{\tau_{\varepsilon}} - \vartheta_0}{\varphi_{\varepsilon}} + \frac{\vartheta_0 - \check{\vartheta}_{\tau_{\varepsilon}}}{\varphi_{\varepsilon}} + o(1) + \frac{(\vartheta_0 - \check{\vartheta}_{\tau_{\varepsilon}})^2}{\varphi_{\varepsilon}} O(1) \\
&\quad + I_{\tau_{\varepsilon}}^t (\vartheta_0)^{-1} \int_{\tau_{\varepsilon}}^t \frac{\dot{S}(\vartheta_0, s) \xi_{s,\varepsilon}}{\sqrt{2S(\vartheta_0, s) \sigma(s)}} d\bar{W}_s (1 + o(1)) \\
\implies \eta_t^*(\vartheta_0) &= I_0^t (\vartheta_0)^{-1} \int_0^t \frac{\dot{S}(\vartheta_0, s)}{\sqrt{2S(\vartheta_0, s) \sigma(s)}} dw(s) \quad (65)
\end{aligned}
$$

because

$$ \frac{(\vartheta_0 - \check{\vartheta}_{\tau_{\varepsilon}})^2}{\varphi_{\varepsilon}} = \left[ \sqrt{\frac{\psi_{\varepsilon}}{\varepsilon}} (\vartheta_0 - \check{\vartheta}_{\tau_{\varepsilon}}) \right]^2 \frac{\varepsilon}{\varphi_{\varepsilon} \psi_{\varepsilon}} = \left[ \sqrt{\frac{\psi_{\varepsilon}}{\varepsilon}} (\vartheta_0 - \check{\vartheta}_{\tau_{\varepsilon}}) \right]^2 \varepsilon^{\frac{1-\delta}{2}} \to 0. \quad (66) $$

**Remark 8.** The estimator-process (64) can be written in recursive form too. Denote

$$ R(\vartheta, t) = \frac{\dot{S}(\vartheta, t)}{\sqrt{2S(\vartheta, t) \sigma(t)}}. $$

Then as in (58) we obtain for the estimator $\vartheta_{t,\varepsilon}^*$ the recursive representation

$$ 
\begin{aligned}
d\vartheta_{t,\varepsilon}^* = I_{\tau_{\varepsilon}}^t (\check{\vartheta}_{\tau_{\varepsilon}})^{-1} [\check{\vartheta}_{\tau_{\varepsilon}} - \vartheta_{t,\varepsilon}^*] R(\check{\vartheta}_{\tau_{\varepsilon}}, t) dt \\
+ \frac{I_{\tau_{\varepsilon}}^t (\check{\vartheta}_{\tau_{\varepsilon}})^{-1} \dot{M}(\check{\vartheta}_{\tau_{\varepsilon}}, t)}{\varepsilon \psi_{\varepsilon} \sigma(t)^2} [dX_t - M(\check{\vartheta}_{\tau_{\varepsilon}}, t) dt], \quad \vartheta_{\tau_{\varepsilon},\varepsilon}^* = \check{\vartheta}_{\tau_{\varepsilon}}, \quad \tau_{\varepsilon} < t \le T, \quad (67)
\end{aligned}
$$

### 1.3.3 Case $\delta = 1/3$.

This case can be treated as above. The learning interval is $[0, \tau]$ and

$$ \varepsilon^{-1/3} (\check{\vartheta}_{\tau,\varepsilon} - \vartheta_0) \implies \mathcal{N}(0, D(\vartheta_0)) $$

with the corresponding limit variance $D(\vartheta_0)$. The Fisher information matrix function is

$$ 
\begin{aligned}
I_{\tau}^{t}(\vartheta) = \int_{\tau}^{t} \frac{[\dot{f}(\vartheta, s)b(\vartheta, s) + f(\vartheta, s)\dot{b}(\vartheta, s)][\dot{f}(\vartheta, s)b(\vartheta, s) + f(\vartheta, s)\dot{b}(\vartheta, s)]^{\top}}{2f(\vartheta, s)b(\vartheta, s)\sigma(s)} ds \\
+ \int_{\tau}^{t} \frac{[\dot{a}(\vartheta, s) + \dot{l}(\vartheta, s)][\dot{a}(\vartheta, s) + \dot{l}(\vartheta, s)]^{\top} y_{s}(\vartheta)^{2}}{b(\vartheta, s)^{2}} ds, \quad \tau \le t \le T.
\end{aligned}
$$

The One-step MLE-process is defined by (64)

$$ \vartheta_{t,\varepsilon}^* = \check{\vartheta}_{\tau,\varepsilon} + \frac{I_{\tau}^t (\check{\vartheta}_{\tau,\varepsilon})^{-1}}{\varepsilon^{4/3}} \int_{\tau}^t \frac{\dot{M}(\check{\vartheta}_{\tau,\varepsilon}, s)}{\sigma(s)^2} [dX_s - M(\check{\vartheta}_{\tau,\varepsilon}, s) ds] \quad (68) $$
where the random processes $M(\check{\vartheta}_{\tau,\varepsilon}, s)$, $\tau \le s \le T$ and $\dot{M}(\check{\vartheta}_{\tau,\varepsilon}, s)$, $\tau \le s \le T$ are solutions of the equations like (60) and (61).

The estimator is asymptotically normal

$$ \varepsilon^{-1/3} (\vartheta_{t,\varepsilon}^* - \vartheta_0) \implies \eta_t^*(\vartheta_0) \sim \mathcal{N}(0, I_{\tau}^t (\vartheta_0)^{-1}) \quad (69) $$

but there is a question: how to write the limit variable $\eta_t^*(\vartheta_0)$? Recall the representation before the limit

$$ \varepsilon^{-1/3} (\vartheta_{t,\varepsilon}^* - \vartheta_0) = I_{\tau}^t (\vartheta_0)^{-1} \left\{ \int_{\tau}^t \frac{[\dot{f}(\vartheta_0, s)b(\vartheta_0, s) + f(\vartheta_0, s)\dot{b}(\vartheta_0, s)]}{\sqrt{2f(\vartheta_0, s)b(\vartheta_0, s)\sigma(s)}} d\bar{W}_s \right. \\ \left. + \int_{\tau}^t \frac{[\dot{a}(\vartheta_0, s) + \dot{l}(\vartheta_0, s)]y_s(\vartheta_0)}{b(\vartheta_0, s)} \xi_{s,\varepsilon} d\bar{W}_s \right\} + o(1), $$

where the limit of $\xi_{s,\varepsilon}$, $0 < s \le T$ is not a stochastic process and in the both integrals there we have the same innovation Wiener process. To describe the limit we consider for simplicity the one-dimensional case ($d=1$). Let us denote these integrals as follows

$$ J_{\varepsilon}^{(1)} = \int_{\tau}^{t} N(\vartheta_{0}, s) d\bar{W}_{s}, \qquad J_{\varepsilon}^{(2)} = \int_{\tau}^{t} K(\vartheta_{0}, s) \xi_{s,\varepsilon} d\bar{W}_{s}. $$

We have

$$ \int_{\tau}^{t} [N(\vartheta_{0}, s) + K(\vartheta_{0}, s) \xi_{s,\varepsilon}]^2 ds \rightarrow \int_{\tau}^{t} [N(\vartheta_{0}, s)^2 + K(\vartheta_{0}, s)^2] ds. $$

Hence by the CLT

$$ \varepsilon^{-1/3} (\vartheta_{t,\varepsilon}^* - \vartheta_0) \implies \eta_t^*(\vartheta_0) = \frac{1}{I_{\tau}^t(\vartheta_0)} \int_{\tau}^{t} \sqrt{N(\vartheta_0, s)^2 + K(\vartheta_0, s)^2} dw(s). $$

This limit can be written with the help of two independent Wiener processes too

$$ \eta_t^*(\vartheta_0) = I_0^t(\vartheta_0)^{-1} \left\{ \int_{\tau}^{t} \frac{[\dot{f}(\vartheta_0, s)b(\vartheta_0, s) + f(\vartheta_0, s)\dot{b}(\vartheta_0, s)]}{\sqrt{2f(\vartheta_0, s)b(\vartheta_0, s)\sigma(s)}} dw_1(s) \right. \\ \left. + \int_{\tau}^{t} \frac{[\dot{a}(\vartheta_0, s) + \dot{l}(\vartheta_0, s)]y_s(\vartheta_0)}{b(\vartheta_0, s)} dw_2(s) \right\}. \quad (70) $$

The One-step MLE-process in this case too admits extension of the limit Fisher information on the whole interval (like (57)) and a recursive representation similar to (58) and (67).
1.4 Adaptive filters.

We present the detailed proof in the case $\delta \in (1/3, 1)$ and in the other cases we just describe the results. The proofs follow the same arguments and are omitted.

1.4.1 Case $\delta \in (1/3, 1)$.

Suppose that the learning interval is $[0, \tau]$, $0 < \tau < T$ and we have the MDE $\check{\vartheta}_{\tau, \varepsilon}$ constructed by the observations $X^\tau = (X_t, 0 \le t \le \tau)$. This estimator by Theorem 1 is asymptotically normal with the rate $\psi_\varepsilon$. This allows us to admit the values $\delta \in (1/3, 1)$ for the construction of the One-step MLE-process $\vartheta_{t, \varepsilon}^*, \tau < t \le T$ (see (57)). Recall that this estimator is asymptotically normal

$$ \eta_{t, \varepsilon}^* = \frac{\vartheta_{t, \varepsilon}^* - \vartheta_0}{\psi_\varepsilon} \implies \eta_t^*(\vartheta_0) = I_0^t(\vartheta_0)^{-1} \int_0^t \frac{[\dot{a}(\vartheta_0, s) + \dot{l}(\vartheta_0, s)]y_s(\vartheta_0)}{b(\vartheta_0, s)} dw(s). $$

The adaptive filter is defined by the equation

$$ dm_{t, \varepsilon}^* = \left[ a(\vartheta_{t, \varepsilon}^*, t) - \frac{\psi_\varepsilon b(\vartheta_{t, \varepsilon}^*, t) f(\vartheta_{t, \varepsilon}^*, t)}{\sigma(t)} \right] m_{t, \varepsilon}^* dt + \frac{\psi_\varepsilon b(\vartheta_{t, \varepsilon}^*, t)}{\sigma(t)} dX_t \quad (71) $$

for the values $\tau_* \le t \le T$, where $\tau_* \in (\tau, T)$. The initial value $m_{\tau^*, \varepsilon}^*$ can be calculated similar to (56).

**Proposition 2.** Let the conditions $\tilde{\mathcal{B}}, \tilde{\mathcal{D}}_{1-3}$ and $\mathcal{F}^*$ hold. Then

$$ \frac{m_{t, \varepsilon}^* - m(\vartheta_0, t)}{\psi_\varepsilon} \implies - \frac{\dot{f}(\vartheta_0, t) y_t(\vartheta_0)}{f(\vartheta_0, t)} \eta_t^*(\vartheta_0). \quad (72) $$

*Proof.* Note that from the equations for $m(\vartheta_0, t)$, $0 \le t \le T$ and $y_t(\vartheta_0)$, $0 \le t \le T$ we obtain the relation

$$ m(\vartheta_0, t) - y_t(\vartheta_0) = \psi_\varepsilon \int_0^t \exp \left( \int_s^t a(\vartheta_0, v) \, dv \right) d\bar{W}_s = \psi_\varepsilon O(1). \quad (73) $$

To study of the error $\delta_{m^*}(t) = m_{t, \varepsilon}^* - m(\vartheta_0, t)$ we remind the equation for $m(\vartheta_0, t)$

$$ dm(\vartheta_0, t) = \left[ a(\vartheta_0, t) - \frac{\psi_\varepsilon b(\vartheta_0, t) f(\vartheta_0, t)}{\sigma(t)} \right] m(\vartheta_0, t) dt + \frac{\psi_\varepsilon b(\vartheta_0, t)}{\sigma(t)} dX_t $$

and consider the equation for difference $\delta_{m^*}(t)$

$$ \begin{aligned} d\delta_{m^*}(t) = & A(\vartheta_0, t) \delta_{m^*}(t) dt + [A(\vartheta_{t, \varepsilon}^*, t) - A(\vartheta_0, t)] m_{t, \varepsilon}^* dt \\ & + \frac{\psi_\varepsilon}{\varepsilon} [b(\vartheta_{t, \varepsilon}^*, t) - b(\vartheta_0, t)] \frac{f(\vartheta_0, t)}{\sigma(t)} m(\vartheta_0, t) dt \end{aligned} $$
$$ + \psi_{\varepsilon} [b(\vartheta_{t,\varepsilon}^{*}, t) - b(\vartheta_{0}, t)] \mathrm{d}\bar{W}_{t} \\ = A(\vartheta_{0}, t) \delta_{m^{*}}(t) \mathrm{d}t + \delta_{A}(t) m_{t,\varepsilon}^{*} \mathrm{d}t + \frac{\psi_{\varepsilon}}{\varepsilon} \delta_{b}(t) \frac{f(\vartheta_{0}, t)}{\sigma(t)} m(\vartheta_{0}, t) \mathrm{d}t + \psi_{\varepsilon} \delta_{b}(t) \mathrm{d}\bar{W}_{t}. $$

Here $\delta_A(t) = A(\vartheta_{t,\varepsilon}^*, t) - A(\vartheta_0, t)$, $\delta_b(t) = b(\vartheta_{t,\varepsilon}^*, t) - b(\vartheta_0, t)$ and

$$ A(\vartheta, t) = a(\vartheta, t) - \frac{\psi_{\varepsilon} b(\vartheta, t) f(\vartheta, t)}{\varepsilon \sigma(t)} $$

Denote as well

$$ \Phi(\vartheta, s, t) = \exp \left( \int_s^t A(\vartheta, v) \mathrm{d}v \right), \quad B(\vartheta, t) = \frac{b(\vartheta, t) f(\vartheta, t)}{\sigma(t)}. $$

Then

$$ \begin{aligned} \delta_{m^*}(t) &= \delta_{m^*}(\tau) \Phi(\vartheta_0, 0, t) + \int_{\tau}^{t} \Phi(\vartheta_0, s, t) \delta_A(s) m_{s,\varepsilon}^{*} \mathrm{d}s \\ &\quad + \frac{\psi_{\varepsilon}}{\varepsilon} \int_{\tau}^{t} \Phi(\vartheta_0, s, t) \delta_b(t) \frac{f(\vartheta_0, s)}{\sigma(s)} m(\vartheta_0, s) \mathrm{d}s + \psi_{\varepsilon} \int_{\tau}^{t} \Phi(\vartheta_0, s, t) \delta_b(t) \mathrm{d}\bar{W}_s \\ &= \delta_m(\tau) \Phi(\vartheta_0, 0, t) - \frac{\psi_{\varepsilon}}{\varepsilon} \int_{\tau}^{t} \Phi(\vartheta_0, s, t) \dot{B}(\vartheta_0, s) [\vartheta_{s,\varepsilon}^{*} - \vartheta_0] m_{s,\varepsilon}^{*} \mathrm{d}s (1+o(1)) \\ &\quad + \frac{\psi_{\varepsilon}}{\varepsilon} \int_{\tau}^{t} \Phi(\vartheta_0, s, t) \dot{b}(\vartheta_0, s) [\vartheta_{s,\varepsilon}^{*} - \vartheta_0] \frac{f(\vartheta_0, s)}{\sigma(s)} m(\vartheta_0, s) \mathrm{d}s (1+o(1)) \\ &\quad + \psi_{\varepsilon} \int_{\tau}^{t} \Phi(\vartheta_0, s, t) \dot{b}(\vartheta_0, s) [\vartheta_{s,\varepsilon}^{*} - \vartheta_0] \mathrm{d}\bar{W}_s (1+o(1)) \\ &= \delta_{m^*}(\tau) \Phi(\vartheta_0, 0, t) - \frac{\psi_{\varepsilon}^2}{\varepsilon} \int_{\tau}^{t} \Phi(\vartheta_0, s, t) \dot{B}(\vartheta_0, s) \eta_{s,\varepsilon}^{*} m_{s,\varepsilon}^{*} \mathrm{d}s (1+o(1)) \\ &\quad + \frac{\psi_{\varepsilon}^2}{\varepsilon} \int_{\tau}^{t} \Phi(\vartheta_0, s, t) \dot{b}(\vartheta_0, s) \eta_{s,\varepsilon}^{*} \frac{f(\vartheta_0, s)}{\sigma(s)} m(\vartheta_0, s) \mathrm{d}s (1+o(1)) \\ &\quad + \psi_{\varepsilon}^2 \int_{\tau}^{t} \Phi(\vartheta_0, s, t) \dot{b}(\vartheta_0, s) \eta_{s,\varepsilon}^{*} \mathrm{d}\bar{W}_s (1+o(1)). \end{aligned} \quad (74) $$

The term $\delta_{m^*}(\tau) \Phi(\vartheta_0, 0, t)$ can be omitted. The Lemma 1 in [30] allow us to write

$$ \delta_{m^*}(t) = -\psi_{\varepsilon} \frac{\dot{f}(\vartheta_0, t)}{f(\vartheta_0, t)} \eta_{t,\varepsilon}^{*} y_t(\vartheta_0) (1+o(1)). $$

Here we used the convergences $m_{t,\varepsilon}^* \to y_t(\vartheta_0)$ and $m(\vartheta_0, t) \to y_t(\vartheta_0)$ (73). To verify $m_{t,\varepsilon}^* \to y_t(\vartheta_0)$ it is sufficient to prove that $E_{\vartheta_0} \delta_{m^*}(t)^2 \to 0$. We have

$$ d\delta_{m^*}(t)^2 = 2A(\vartheta_0, t) \delta_{m^*}(t)^2 \mathrm{d}t + 2 \frac{\psi_{\varepsilon}}{\varepsilon} \delta_b(t) \frac{f(\vartheta_0, t)}{\sigma(t)} m(\vartheta_0, t) \delta_{m^*}(t) \mathrm{d}t $$
$$+ 2\delta_A(t) m_{t,\epsilon}^* \delta_{m^*}(t) dt + \psi_{\epsilon}^2 \delta_b(t)^2 dt + 2\psi_{\epsilon} \delta_b(t) \delta_{m^*}(t) d\bar{W}_t.$$

Then we replace $A(\vartheta_0, t)$ by $-\frac{\psi_{\epsilon}}{\epsilon} \kappa$ with $\kappa > 0$, write the equation-inequality for $\mathbf{E}_{\vartheta_0} \delta_{m^*}(t)^2$ and apply Grönwall lemma. The terms like $\delta_b(t) m(\vartheta_0, t) \delta_{m^*}(t)$ we replace as follows

$$2 |\mathbf{E}_{\vartheta_0} \delta_b(t) m(\vartheta_0, t) \delta_{m^*}(t)| \le C \mathbf{E}_{\vartheta_0} \delta_b(t)^2 m(\vartheta_0, t)^2 + \frac{1}{C} \mathbf{E}_{\vartheta_0} \delta_{m^*}(t)^2$$

with sufficiently large constant $C > 0$ such that $\kappa > 1/C$.

Finally we obtain

$$\psi_{\epsilon}^{-1} (m_{t,\epsilon}^{*} - m(\vartheta_0, t)) \implies - \frac{\dot{f}(\vartheta_0, t) y_t(\vartheta_0)}{f(\vartheta_0, t)} \eta_t^*(\vartheta_0),$$

where

$$\eta_t^*(\vartheta_0) = I_0^t(\vartheta_0)^{-1} \int_0^t \frac{[\dot{a}(\vartheta_0, s) + \dot{l}(\vartheta_0, s)]y_s(\vartheta_0)}{b(\vartheta_0, s)} dw(s).$$

<figure><img src="image_14.png" /></figure>

**Remark 9.** It is possible to prove the asymptotic efficiency of the adaptive filter with the lower bound: for any estimator $\bar{m}_{t,\epsilon}$ and any $\vartheta_0 \in \Theta$ we have

$$\lim_{\nu \to 0} \lim_{\epsilon \to 0} \sup_{|\vartheta - \vartheta_0| < \nu} \psi_{\epsilon}^{-2} \mathbf{E}_{\vartheta} [\bar{m}_{t,\epsilon} - m(\vartheta, t)]^2 \ge \mathcal{S}_t^*(\vartheta_0)^2$$

where

$$\mathcal{S}_t^*(\vartheta_0)^2 = \frac{\dot{f}(\vartheta_0, t)^2 y_t(\vartheta_0)^2}{f(\vartheta_0, t)^2} I_0^t(\vartheta_0)^{-1}.$$

The estimator $\bar{m}_{t,\epsilon}$ of $m(\vartheta, t)$ satisfying the equality

$$\liminf_{\nu \to 0} \sup_{\epsilon \to 0} \sup_{|\vartheta - \vartheta_0| < \nu} \psi_{\epsilon}^{-2} \mathbf{E}_{\vartheta} [\bar{m}_{t,\epsilon} - m(\vartheta, t)]^2 = \mathcal{S}_t^*(\vartheta_0)^2$$

is called asymptotically efficient.

Remind that such bound can be obtained with the help of the BE as follows (see, e.g., [32], [34], [36]). We have

$$\sup_{|\vartheta - \vartheta_0| < \nu} \mathbf{E}_{\vartheta} (\bar{m}_{t,\epsilon} - m(\vartheta, t))^2 \ge \int_{\vartheta_0 - \nu}^{\vartheta_0 + \nu} \mathbf{E}_{\vartheta} (\bar{m}_{t,\epsilon} - m(\vartheta, t))^2 p(\vartheta) d\vartheta \\ \ge \int_{\vartheta_0 - \nu}^{\vartheta_0 + \nu} \mathbf{E}_{\vartheta} (\tilde{m}_{t,\epsilon} - m(\vartheta, t))^2 p(\vartheta) d\vartheta$$
where $p(\vartheta)$, $\vartheta_0 - \nu < \vartheta < \vartheta_0 + \nu$ is a density and $\tilde{m}_\varepsilon(t) = \mathbf{E}(m(\theta, t) | \mathfrak{F}_t^X)$ is a conditional expectation

$$ \tilde{m}_{\varepsilon}(t) = \int_{\vartheta_{0-\nu}}^{\vartheta_{0+\nu}} m(\theta, t) p(\theta | X^t) \, d\theta = \frac{\int_{\vartheta_{0-\nu}}^{\vartheta_{0+\nu}} m(\theta, t) p(\theta) L(\theta, X^t) \, d\theta}{\int_{\vartheta_{0-\nu}}^{\vartheta_{0+\nu}} p(\theta) L(\theta, X^t) \, d\theta}. $$

The function $p(\cdot)$ is continuous and positive.

Let us change the variables $\theta = \vartheta + \psi_{\varepsilon}u$ and recall the definition of the normalized likelihood ratio $Z_{t,\varepsilon}(\vartheta, u) = L(\vartheta + \psi_{\varepsilon}u, X^t)/L(\vartheta, X^t)$. Then using Taylor expansion we obtain

$$ \begin{aligned} \tilde{m}_{\varepsilon}(t) &= \frac{\int_{\mathbb{U}_{\varepsilon}} m(\vartheta + \psi_{\varepsilon}u, t) p(\vartheta + \psi_{\varepsilon}u) Z_{t,\varepsilon}(\vartheta, u) \, du}{\int_{\mathbb{U}_{\varepsilon}} p(\vartheta + \psi_{\varepsilon}u) Z_{t,\varepsilon}(\vartheta, u) \, du} \\ &= m(\vartheta, t) + \psi_{\varepsilon} \dot{m}(\vartheta, t) \frac{\int_{\mathbb{U}_{\varepsilon}} u Z_{t,\varepsilon}(\vartheta, u) \, du}{\int_{\mathbb{U}_{\varepsilon}} Z_{t,\varepsilon}(\vartheta, u) \, du} (1 + o(1)) \\ &= m(\vartheta, t) + \psi_{\varepsilon} \dot{m}(\vartheta, t) \tilde{u}_{t,\varepsilon} (1 + o(1)). \end{aligned} $$

Here $\tilde{u}_{t,\varepsilon}$ asymptotically equivalent to the BE $\psi_{\varepsilon}^{-1}(\tilde{\vartheta}_{t,\varepsilon} - \vartheta)$. Recall that by the Theorem 2 we have $\mathbf{E}_{\vartheta}\tilde{u}_{t,\varepsilon}^2 \to I_0^t(\vartheta)^{-1}$. Hence

$$ \begin{aligned} \psi_{\varepsilon}^{-2} \int_{\vartheta_{0-\nu}}^{\vartheta_{0+\nu}} \mathbf{E}_{\vartheta} (\tilde{m}_{t,\varepsilon} - m(\vartheta, t))^2 p(\vartheta) \, d\vartheta &= \int_{\vartheta_{0-\nu}}^{\vartheta_{0+\nu}} \mathbf{E}_{\vartheta} [\dot{m}(\vartheta, t)^2 \tilde{u}_{t,\varepsilon}^2] p(\vartheta) \, d\vartheta (1 + o(1)) \\ &\to \int_{\vartheta_{0-\nu}}^{\vartheta_{0+\nu}} \frac{\dot{f}(\vartheta, t)^2 y_t(\vartheta)^2}{f(\vartheta, t)^2} I_0^t(\vartheta)^{-1} p(\vartheta) \, d\vartheta \xrightarrow{\nu \to 0} S_t^*(\vartheta_0)^2. \end{aligned} $$

The technical details for the proof of these expansions and convergences can be found in the proof of the Theorem 3.2.1 in [19].

**Remark 10.** In the case $d > 1$ the limits are

$$ \psi_{\varepsilon}^{-1} (m_{t,\varepsilon}^{*} - m(\vartheta_0, t)) \implies - \frac{y_t(\vartheta_0)}{f(\vartheta_0, t)} \dot{f}(\vartheta_0, t)^{\top} \eta_t^{*}(\vartheta_0), $$

where

$$ \eta_t^*(\vartheta_0) = I_0^t(\vartheta_0)^{-1} \int_0^t \frac{[\dot{a}(\vartheta_0, s) + \dot{l}(\vartheta_0, s)]y_s(\vartheta_0)}{b(\vartheta_0, s)} \, dw(s) $$

and

$$ S_t^*(\vartheta_0)^2 = \frac{y_t(\vartheta_0)^2}{f(\vartheta_0, t)^2} \dot{f}(\vartheta_0, t)^{\top} I_0^t(\vartheta_0)^{-1} \dot{f}(\vartheta_0, t). $$
### 1.4.2 Case $\delta \in (1/5, 1/3)$.

Consider the fixed learning interval $[0, \tau]$ and the MDE $\check{\vartheta}_{\tau, \varepsilon}$. Recall that the asymptotic of the derivative $\dot{M}(\vartheta, t)$ contains the random processes $\xi_{t, \varepsilon}$ and we can not use this approximation for the construction of One-step MLE-process. The process $\dot{M}(\vartheta, t)$, $\tau \le t \le T$ satisfies the equation (see (61))

$$d\dot{M}(\vartheta, t) = A^*(\vartheta, t) \dot{M}(\vartheta, t) dt + \dot{A}^*(\vartheta, t) M(\vartheta, t) dt + \frac{\psi_\varepsilon}{\varepsilon} \dot{B}(\vartheta, t) dX_t. \quad (75)$$

The One-step MLE-process is

$$\vartheta_{t, \varepsilon}^* = \check{\vartheta}_{\tau, \varepsilon} + I_{\tau}^t (\check{\vartheta}_{\tau, \varepsilon})^{-1} \int_{\tau}^{t} \frac{\dot{M}(\check{\vartheta}_{\tau, \varepsilon}, s)}{\varepsilon^{1+\delta} \sigma(s)^2} [dX_s - M(\check{\vartheta}_{\tau, \varepsilon}, s) ds], \quad \tau < t \le T$$

where $M(\check{\vartheta}_{\tau, \varepsilon}, s)$, $\tau \le s \le T$ is solution of the equation (55), the Fisher information function is given in (59). The adaptive filter is the same as (71)

$$d m_{t, \varepsilon}^* = [a(\vartheta_{t, \varepsilon}^*, t) - \varepsilon^{-1+\delta} B(\vartheta_{t, \varepsilon}^*, t)] m_{t, \varepsilon}^* dt + \varepsilon^{-1+\delta} \frac{b(\vartheta_{t, \varepsilon}^*, t)}{\sigma(t)} dX_t, \quad \tau < t \le T,$$

but $\delta_{m^*}(t) = m_{t, \varepsilon}^* - m(\vartheta_0, t)$ has a different asymptotic.

**Proposition 3.** Let the conditions of Theorem 5 hold. Then

$$\frac{m_{t, \varepsilon}^* - m(\vartheta_0, t)}{\varphi_\varepsilon} \implies -\frac{\dot{f}(\vartheta_0, t)}{f(\vartheta_0, t)} \eta_t^*(\vartheta_0) y_t(\vartheta_0).$$

*Proof.* The representation (74) allows to write

$$\begin{align*}
\delta_{m^*}(t) &= -\frac{\psi_\varepsilon \varphi_\varepsilon}{\varepsilon} \int_{\tau}^{t} \Phi(\vartheta_0, s, t) \dot{B}(\vartheta_0, s) \eta_{s, \varepsilon}^* m_{s, \varepsilon}^* ds (1+o(1)) \\
&\quad + \frac{\psi_\varepsilon \varphi_\varepsilon}{\varepsilon} \int_{\tau}^{t} \Phi(\vartheta_0, s, t) \dot{b}(\vartheta_0, s) \eta_{s, \varepsilon}^* \frac{f(\vartheta_0, s)}{\sigma(s)} m(\vartheta_0, s) ds (1+o(1)) \\
&\quad + \psi_\varepsilon \varphi_\varepsilon \int_{\tau}^{t} \Phi(\vartheta_0, s, t) \dot{b}(\vartheta_0, s) \eta_{s, \varepsilon}^* d\bar{W}_s (1+o(1)) \\
&= -\varphi_\varepsilon \frac{\dot{f}(\vartheta_0, t)}{f(\vartheta_0, t)} \eta_{t, \varepsilon}^* m(\vartheta_0, t) (1+o(1)) + \varepsilon \frac{\dot{b}(\vartheta_0, t) \sqrt{\sigma(t)}}{\sqrt{2b(\vartheta_0, t) f(\vartheta_0, t)}} \eta_{t, \varepsilon}^* \xi_{t, \varepsilon} (1+o(1)).
\end{align*}$$

This implies the convergence

$$\frac{m_{t, \varepsilon}^* - m(\vartheta_0, t)}{\varphi_\varepsilon} \implies -\frac{\dot{f}(\vartheta_0, t)}{f(\vartheta_0, t)} \eta_t^*(\vartheta_0) y_t(\vartheta_0),$$
where

$$ \eta_t^*(\vartheta_0) = I_\tau^t(\vartheta_0)^{-1} \int_\tau^t \frac{\dot{f}(\vartheta_0, s) b(\vartheta_0, s) + f(\vartheta_0, s) \dot{b}(\vartheta_0, s)}{\sqrt{2f(\vartheta_0, s) b(\vartheta_0, s) \sigma(s)}} dw(s). $$

Recall that the random variables $\xi_t$ and $\eta_t^*(\vartheta_0)$ are independent. □

Note that in the one-dimensional case ($d = 1$) and vanishing learning interval the function $S_t^*(\vartheta)^2$ in this case is the same as above

$$ S_t^*(\vartheta)^2 = \frac{\dot{f}(\vartheta, t)^2 y_t(\vartheta_0)^2}{f(\vartheta_0, t)^2} I_\tau^t(\vartheta)^{-1}, \quad \vartheta \in \Theta. $$

### 1.4.3 Case $\delta = 1/3$.

The preliminary MDE $\tilde{\vartheta}_{\tau_{\varepsilon}}$ constructed by the observations on the vanishing learning interval $[0, \tau_{\varepsilon}]$, $\tau_{\varepsilon} = \varepsilon^{2/3}$ is asymptotically normal (see Proposition 1)

$$ \varepsilon^{-1/3} (\tilde{\vartheta}_{\tau, \varepsilon} - \vartheta_0) \implies \mathcal{N}(0, D(\vartheta_0)). $$

The corresponding One-step MLE-process $\vartheta_{t, \varepsilon}^*, \tau_{\varepsilon} < t \le T$ defined in (68) is asymptotically normal (69).

The adaptive filter is

$$ dm_{t, \varepsilon}^* = [a^*(\vartheta_{t, \varepsilon}^*, t) - \varepsilon^{-2/3} B(\vartheta_{t, \varepsilon}^*, t)] m_{t, \varepsilon}^* dt + \varepsilon^{-2/3} B(\vartheta_{t, \varepsilon}^*, t) dX_t, \quad \tau < t \le T, $$

where $B(\vartheta, t) = f(\vartheta, t) b(\vartheta, t) \sigma(t)^{-1}$ and $\tau \in (0, T)$.

For the error of approximation $\delta_{m^*}(t)$ we have the representation

$$ \begin{aligned} \delta_{m^*}(t) = & -\varepsilon^{-1/3} \int_{\tau}^{t} \Phi(\vartheta_0, s, t) \dot{B}(\vartheta_0, s) \eta_{s, \varepsilon}^* m_{s, \varepsilon}^* ds (1 + o(1)) \\ & + \varepsilon^{-1/3} \int_{\tau}^{t} \Phi(\vartheta_0, s, t) \frac{\dot{b}(\vartheta_0, s)}{\sigma(s)} \eta_{s, \varepsilon}^* f(\vartheta_0, s) m(\vartheta_0, s) ds (1 + o(1)) \\ & + \varepsilon^{2/3} \int_{\tau}^{t} \Phi(\vartheta_0, s, t) \dot{b}(\vartheta_0, s) \eta_{s, \varepsilon}^* d\bar{W}_s (1 + o(1)) \\ = & -\varepsilon^{1/3} \frac{\dot{B}(\vartheta_0, t)}{B(\vartheta_0, t)} \eta_{t, \varepsilon}^* m_{t, \varepsilon}^* (1 + o(1)) \\ & + \varepsilon^{1/3} \frac{\dot{b}(\vartheta_0, t)}{B(\vartheta_0, t) \sigma(s)} \eta_{t, \varepsilon}^* f(\vartheta_0, t) m(\vartheta_0, t) (1 + o(1)) \\ & + \varepsilon^{1/3} \frac{\dot{b}(\vartheta_0, t) \sqrt{\sigma(t)}}{\sqrt{2b(\vartheta_0, t) f(\vartheta_0, t)}} \eta_{t, \varepsilon}^* \xi_{t, \varepsilon} (1 + o(1)) \end{aligned} $$
$$ = -\varepsilon^{1/3} \frac{\dot{f}(\vartheta_0, t)}{f(\vartheta_0, t)} \eta_{t,\varepsilon}^{*} y_t(\vartheta_0) + \varepsilon^{1/3} \frac{\dot{b}(\vartheta_0, t) \sqrt{\sigma(t)}}{\sqrt{2b(\vartheta_0, t)} f(\vartheta_0, t)} \eta_{t,\varepsilon}^{*} \xi_{t,\varepsilon} (1+o(1)). $$

Finally

$$ \frac{m_{t,\varepsilon}^{*} - m(\vartheta_0, t)}{\varepsilon^{1/3}} = \left[ \frac{\dot{b}(\vartheta_0, t) \sqrt{\sigma(t)}}{\sqrt{2b(\vartheta_0, t)} f(\vartheta_0, t)} \xi_{t,\varepsilon} - \frac{\dot{f}(\vartheta_0, t) y_t(\vartheta_0)}{f(\vartheta_0, t)} \right] \eta_{t,\varepsilon}^{*} (1+o(1)) \\ \implies \left[ \frac{\dot{b}(\vartheta_0, t) \sqrt{\sigma(t)}}{\sqrt{2b(\vartheta_0, t)} f(\vartheta_0, t)} \xi_t - \frac{\dot{f}(\vartheta_0, t) y_t(\vartheta_0)}{f(\vartheta_0, t)} \right] \eta_t^{*}(\vartheta_0), $$

where $\eta_t(\vartheta_0)$ is defined in (70).

**Acknowledgements.** This research was financially supported by the Russian Sciences Foundation research project No 24-11-00191.

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