# A New Proof for the Linear Filtering and Smoothing Equations, and Asymptotic Expansion of Nonlinear Filtering

Masahiro Kurisaki

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Masahiro Kurisaki. A New Proof for the Linear Filtering and Smoothing Equations, and Asymptotic Expansion of Nonlinear Filtering. 2025. hal-04876974

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# A New Proof for the Linear Filtering and Smoothing Equations, and its Application to Nonlinear Filtering

Masahiro Kurisaki*†

January 20, 2025

## Abstract

In this paper, we propose a new approach to the linear filtering and smoothing problem and demonstrate its applicability to nonlinear filtering. For the linear case, our main theorem provides an explicit expression for the conditional distribution of the hidden process given the observations, leading to a novel derivation of the linear filtering and smoothing equations. Moreover, the theorem offers an efficient framework for computing the asymptotic expansion of nonlinear filtering.

**Key words**— Liner filtering and smoothing, Kalman-Bucy filter, Rauch-Tung-Striebel smoother, nonlinear filtering, asymptotic expansion

## 1 Introduction

Estimating hidden states from noisy observations is a fundamental problem in various fields, including signal processing, control theory, and finance. One classical framework for addressing this problem is filtering, which aims to estimate the hidden states of an unobserved stochastic process {$X_t$}$_{t \ge 0}$ based on the observed data from a related stochastic process {$Y_t$}$_{t \ge 0}$. Here, {$Y_t$}$_{t \ge 0}$ provides partial information about {$X_t$}$_{t \ge 0}$ through an observation model.

In the general framework of continuous-time filtering theory, we assume that the signal process X satisfies the stochastic differential equation

$$dX_t = A(t, X_t) dt + B(t, X_t) dV_t,$$

<footnote>*Graduate School of Mathematical Sciences, the University of Tokyo
email: makurisaki@g.ecc.u-tokyo.ac.jp</footnote>

<footnote>†Japan Science and Technology Agency CREST
The author was supported by Japan Science and Technology Agency CREST JP-MJCR2115, and JSPS KAKENHI Grant Number JP24KJ0667.</footnote>
and that the observation process Y evolves according to

$$dY_t = h(t, X_t) dt + \sigma(t) dW_t,$$

where V and W are independent standard Wiener processes, and A, B, h, and σ are appropriately chosen functions.

The objective of the filtering problem is to compute the conditional expectation $E[X_t|Y_t]$ for every $t \ge 0$, where $Y_t$ denotes the σ-field generated by $\{Y_s\}_{0\le s\le t}$. This conditional expectation serves as the $L^2$-optimal estimator of $X_t$ as a function of the observation process $\{Y_s\}_{0\le s\le t}$.

In addition to filtering, the smoothing problem considers the estimation of past states of $\{X_t\}$ given future observations of $\{Y_t\}$. Specifically, the goal of smoothing is to compute $E[X_s|Y_t]$ for $0 \le s \le t$. Together, filtering and smoothing play a central role in a wide range of applications.

The development of filtering theory has a long and rich history, dating back to the seminal works of Wiener (Wiener 1949) and Kalman (Kalman 1960). While Wiener focused on optimal filtering for continuous-time stationary Gaussian processes, Kalman introduced the concept of optimal filtering for discrete-time linear systems. The continuous-time counterpart of the Kalman filter, known as the Kalman-Bucy filter (Kalman and Bucy 1961), laid the groundwork for modern filtering theory in continuous-time systems. This result was particularly influential, as it provided a systematic and efficient method for estimating the hidden states of a linear stochastic system driven by continuous-time observations. The filtering problem is solved by the well-known Kalman filter, while the smoothing problem is addressed by the Rauch-Tung-Striebel (RTS) smoother, which provides a backward smoothing method for state estimation in linear Gaussian state-space models (Rauch et al. 1965). As a result of these algorithms, filtering and smoothing problems in the linear case can be reduced to analytically solvable finite-dimensional equations.

Following these developments, significant progress was made in the realm of nonlinear filtering. The works of Stratonovich (Stratonovich 1960) and Kushner (Kushner 1964) are of particular note, as they established the fundamental equation governing continuous-time nonlinear filtering. This equation, now commonly referred to as the Stratonovich-Kushner equations, generalize the Kalman-Bucy filter to nonlinear and non-Gaussian settings. In the same period, the equivalent equation for the unconditional probability measure, which is known as the Zakai equation was derived in Mortensen (1966), Zakai (1969), Duncan (1970a,b). However, solving the Stratonovich-Kushner equation or Zakai equation analytically is often infeasible, since it is generally known to be infinite-dimensional (Hazewinkel et al. 1983).

To circumvent the infinite dimensionality, the extended Kalman filter was introduced shortly after the development of the Kalman filter and remains widely used in practical applications. The extended Kalman filter employs
a linearization approach to approximate the nonlinear model, making it heuristically straightforward to implement. However, its performance lacks rigorous mathematical guarantees except in very limited cases examined in Picard (1991). In the case where $X$ and $Y$ are one-dimensional and the observation noise is relatively small compared to the system noise, Picard (1986) proposes an asymptotically efficient filter in the limit of vanishing observation noise, overcoming the extended Kalman filter's reliance on local linearization. Kushner and Budhiraja (2000) proposes another Gaussian approximation for the conditional distribution in discrete-time nonlinear filtering, where the means and variances are iteratively estimated using the Gaussian quadrature.

Over the past few decades, considerable effort has been devoted to improving the computational efficiency and accuracy of nonlinear filtering methods. Particle filters (Gordon et al. 1993), also known as sequential Monte Carlo methods, have become a popular approach for solving high-dimensional and nonlinear filtering problems. The continuous-time counterpart of the particle filter, which addresses the problem of filtering for continuous-time stochastic processes, is discussed in detail in Crisan (2006).

Furthermore, the nonlinear filtering equation has been examined within the context of stochastic partial differential equation theory. Specifically, Lototsky (2011) applied the Wiener-Chaos expansion to the Zakai equation and expressed the solution of nonlinear filtering using solutions of numerous deterministic partial differential equations. Brigo et al. (1995) proposed the projection filter, which results in a finite-dimensional stochastic differential equation by projecting the Kushner-Stratonovich equation onto a finite-dimensional statistical manifold. Their method has been further developed in works such as Armstrong et al. (2019), and computational simulations have demonstrated its effectiveness in specific cases and over short time intervals.

In this paper, we present a new approach for both linear and nonlinear filtering. We first revisit the linear case, and analyze it through the Kallianpur-Striebel formula, Girsanov transformation, and Malliavin calculus. This leads to our main theorem for the linear model, providing an explicit formula for the conditional distribution of {$X_s$}$_{0\le s\le t}$ given $Y_t$. Consequently, well-known filtering and smoothing equations, such as the Kalman filter and the Rauch-Tung-Striebel smoother—typically presented as independent theorems in the literature (Liptser and Shiryaev 2001a,b)—are derived here as corollaries of our main theorem.

Beyond reproducing existing equations, our result establishes a stronger statement than existing approaches. In fact, while existing filtering and smoothing equations only offer information about the random variable $X_s$ at each time point, our theorem determines the distribution of the process {$X_s$}$_{0\le s\le t}$ given $Y_t$. This, for instance, allows us to obtain information about the conditional covariance $\text{Cov}(X_s, X_u | Y_t)$, and thereby we can write the
fixed-point smoother in a simpler form than the existing one (Shaked and Theodor 1992).

Furthermore, these results can be applied to the asymptotic expansion of nonlinear filtering around a linear model. This application includes not only the perturbed linear model but also the nonlinear model with small system noise. In such cases, the asymptotic expansion of the nonlinear filtering can be obtained by solving a finite-dimensional system of (ordinary) stochastic differential equations that incorporates the observations. As a result, the asymptotic expansion provides an efficient algorithm for the nonlinear filtering over a relatively wide range of systems with low computational cost.

This paper is organized as follows. In Section 2, we begin with a review of the fundamentals of nonlinear filtering and prove a new result that relaxes existing assumptions. The Kallianpur-Striebel formula plays a central role in the developments that follow.

Section 3 presents our new results for linear filtering and smoothing. We prove the main theorem in Section 3.1 and derive filtering and smoothing equations as corollaries in Section 3.2, including Kalman-Bucy filter, Rauch-Tung-Striebel smoother, and the fixed-point smoother. The heuristic idea behind the proof is outlined in Section 3.1.1.

In Section 4, we outline the application of our results to nonlinear filtering in two specific cases. This section provides an overview of the algorithm without presenting a rigorous proof, but each step will be justified in a subsequent paper. We first discuss the asymptotic expansion for a linear model with a polynomially perturbed transfer function in Section 4.1. We demonstrate that the expansion reduces to solving a stochastic differential equation and present numerical results. The simulation suggests that advancing the asymptotic expansion leads to a reduction in the integrated mean squared error of the filter. Finally, in Section 4.2, we show that the same approach can be applied to nonlinear models with small system noise.

## 2 Filtering theory

Let $(\Omega, \mathcal{F}, \{\mathcal{F}_t\}_{t \ge 0}, P)$ be a filtered probability space, and $\{X_t\}_{t \ge 0}$ be a $d_1$-dimensional, $\{\mathcal{F}_t\}$-adapted càdlàg process. Let $\{H_t\}_{t \ge 0}$ be a $d_2$-dimensional progressively measurable process, and let $\{W_t\}_{t \ge 0}$ be a $d_2$-dimensional $\{\mathcal{F}_t\}$-Brownian motion independent of both $\{X_t\}$ and $\{H_t\}$. We consider the process $\{Y_t\}_{t \ge 0}$ given by

$$ Y_t = \int_0^t H_s ds + W_t. \qquad (2.1) $$

Let $\{\mathcal{X}_t\}_{t \ge 0}$ and $\{\mathcal{Y}_t\}_{t \ge 0}$ be the usual augmentations (with null sets) of the filtrations generated by $\{X_t\}$ and $\{Y_t\}$, respectively. We define the conditional distribution $\pi_t$ of $X_t$ given the information $\{\mathcal{Y}_t\}$. More precisely,
there exists a $\mathcal{P}(\mathbb{R}^{d_1})$-valued, $\{\mathcal{Y}_t\}$-adapted process $\{\pi_t\}_{t \ge 0}$ such that, for any bounded measurable function $f: \mathbb{R}^{d_1} \to \mathbb{R}$,

$$ \pi_t(f) = E[f(X_t)|\mathcal{Y}_t] \quad \text{a.s.,} \qquad (2.2) $$

where $\mathcal{P}(\mathbb{R}^{d_1})$ denotes the space of all probability measures on $\mathbb{R}^{d_1}$ (Bain and Crisan 2009:Section 2). Equations (2.2) is still valid for measurable functions $f$ such that $E[|f(X_t)|] < \infty$ (see Remark 2.27 in Bain and Crisan (2009)).

Assume that for all $t \ge 0$,

$$ \int_{0}^{t} |H_{s}|^{2} ds < \infty \quad \text{a.s.,} \qquad (2.3) $$

and define the process $\{Z_t\}_{t \ge 0}$ by

$$ \begin{aligned} Z_t &= \exp \left( \int_0^t H_s^\top dW_s + \frac{1}{2} \int_0^t |H_s|^2 ds \right) \\ &= \exp \left( \int_0^t H_s^\top dY_s - \frac{1}{2} \int_0^t |H_s|^2 ds \right). \end{aligned} \qquad (2.4) $$

We now introduce a new probability measure $Q$ for an arbitrary fixed $T > 0$, defined by

$$ Q(A) = E[1_A Z_T^{-1}] \quad (A \in \mathcal{F}). \qquad (2.5) $$

To ensure that $Q$ is well-defined, we need to verify that

$$ E[Z_T^{-1}] = 1. \qquad (2.6) $$

In Bain and Crisan (2009), additional conditions are imposed to verify this condition:

$$ E \left[ \int_{0}^{T} |H_{s}|^{2} ds \right] < \infty \qquad (2.7) $$

and

$$ E \left[ \int_{0}^{T} Z_{s} |H_{s}|^{2} ds \right] < \infty. \qquad (2.8) $$

However, we show below that these conditions can be removed and that the local boundedness condition (2.3) alone is sufficient to ensure (2.6). To this end, we first establish the following lemma.

In what follows, for each $t \ge 0$, let $S_t$ denote the space of measurable functions $g: [0, t] \to \mathbb{R}^{d_2}$ such that $\int_0^t |g(s)|^2 ds < \infty$. Let $S_t$ be the $\sigma$-algebra generated by the projections $\varpi_s: S_t \to \mathbb{R}^{d_2}$ for $0 \le s \le t$, where $\varpi_s(g) = g(s)$ for $g \in S_t$. Then the following lemmas hold.
**Lemma 2.1.** Let {$\xi_s$}$_{0\le s\le t}$ be an $\mathbb{R}^{d_2}$-valued progressively measurable process satisfying

$$\int_0^t |\xi_s|^2 ds < \infty,$$

and $\xi$. be its path. Then the $\sigma$-algebra $\mathcal{F}_t^\xi = \sigma(\xi_s; 0 \le s \le t)$ can be represented as

$$\mathcal{F}_t^\xi = \{\xi^{-1}(A); A \in \mathcal{S}_t\}.$$

In particular, $\xi$. is an $S_t$-valued random variable.

*Proof.* Write $\mathcal{G}_t^\xi = \{\xi^{-1}(A); A \in \mathcal{S}_t\}$. Then it is clear that $\mathcal{F}_t^\xi \subset \mathcal{G}_t$ since $\xi_s^{-1}(B) = \xi^{-1}(\varpi_s^{-1}(B)) \in \mathcal{G}_t$ for any $B \in \mathcal{B}(\mathbb{R}^{d_2})$. To prove the converse inclusion, let

$$\mathcal{A} = \{A \subset S_t; \xi^{-1}(A) \in \mathcal{F}_t^\xi\}.$$

Then $\mathcal{A}$ is a $\sigma$-algebra, and $\pi_s^{-1}(B) \in \mathcal{A}$ for any $0 \le s \le t$ and $B \in \mathcal{B}(\mathbb{R}^{d_2})$ since $\xi^{-1}(\pi_s^{-1}(B)) = \xi_s^{-1}(B) \in \mathcal{F}_t^\xi$. Thus it follows $S_t \subset \mathcal{A}$ and the desired result. $\square$

**Lemma 2.2.** Let {$W_t$}$_{t\ge 0}$ be a $d_2$-dimensional Brownian motion, and {$H_t$}$_{t\ge 0}$ be a $d_2$-dimensional progressively measurable process such that for all $t \ge 0$

$$\int_0^t |H_s|^2 ds < \infty \text{ a.s.,}$$

and independent of {$W_t$}.

Then there exists a measurable function $\Phi_t : (S_t \times S_t, \mathcal{S}_t \otimes \mathcal{S}_t) \to \mathbb{R}$ such that

$$\Phi_t(H, W) = \int_0^t H_s dW_s \quad \text{a.s.,} \qquad (2.9)$$

and for $\mu_H$-a.e. $h \in S_t$

$$\Phi_t(h, W) = \int_0^t h(s) dW_s \quad \text{a.s.} \qquad (2.10)$$

Here, $H$. and $W$. denote the paths of {$H_s$}$_{0\le s\le t}$ and {$W_s$}$_{0\le s\le t}$, and $\mu_H$ is the distribution of $H$. on $S_t$.

*Proof.* (i) First assume that {$H_s$}$_{0\le s\le t}$ is continuous. Then we have

$$\sum_{i=1}^{n} H_{t/2^{n-1}}^{\top} (W_{t/2^n} - W_{t/2^{n-1}}) \xrightarrow{P} \int_{0}^{t} H_s dW_s.$$
Hence we can take a subsequence {$n(k)$}$_{k \in \mathbb{N}}$ such that

$$
\sum_{i=1}^{n(k)} H_{t/2^{n(k)}-1}^{\top} (W_{t/2^{n(k)}} - W_{t/2^{n(k)}-1}) \xrightarrow{\text{a.s.}} \int_{0}^{t} H_s dW_s. \quad (2.11)
$$

Now we introduce a sequence of measurable functions $\Phi_t^k : S_t \times S_t \to \mathbb{R}$ for $k \in \mathbb{N}$, defined by

$$
\Phi_t^k(h, w) = \sum_{i=1}^{n(k)} h \left( \frac{t}{2^{n(k)-1}} \right)^{\top} \left( w \left( \frac{t}{2^{n(k)}} \right) - w \left( \frac{t}{2^{n(k)-1}} \right) \right).
$$

We then define $\Phi_t : S_t \times S_t \to \mathbb{R}$ by

$$
\Phi_t(h, w) = \begin{cases} \lim_{k \to \infty} \Phi_t^k(h, w) & \text{if } \Phi_t^k(h, w) \text{ converges,} \\ 0 & \text{otherwise.} \end{cases}
$$

Since each $\Phi_t^k$ is measurable, $\Phi_t$ is also measurable, and (2.9) follows directly
from the construction of $\Phi_t$.

Let us verify (2.10). If we set

$$
A = \{ (h, w) \in S_t \times S_t ; \Phi_t^k(h, w) \text{ converges to } \Phi_t(h, w) \},
$$

then (2.11) means

$$
\int_{S_t \times S_t} 1_{A^c}(h, w) d\mu_{(H,W)}(h, w) = 0,
$$

where $\mu_{(H,W)}$ is the distribution of $(H, W)$. Due to the independence of $H$
and $W$, this can be rewritten as

$$
\int_{S_t} \int_{S_t} 1_{A^c}(h, w) d\mu_W(w) d\mu_H(h) = 0,
$$

and thus we have for $\mu_H$-a.e. $h \in S_t$,

$$
\int_{S_t} 1_{A^c}(h, w) d\mu_W(w) = 0.
$$

From this, it follows that for $\mu_H$-a.s. $h \in S_t$, the sequence of random variables

$$
\Phi_t^k(h, W.) = \sum_{i=1}^{n(k)} h(t/2^{n(k)-1})^{\top} (W_{t/2^{n(k)}} - W_{t/2^{n(k)-1}}) \quad (2.12)
$$

almost surely converges to $\Phi_t(h, W)$. On the other hand, since {$H_s$}$_{0\le s\le t}$
is continuous, we may assume that $h$ is continuous. For such $h$, the right-
hand side of (2.12) converges in probability to $\int_0^t h(s) dW_s$, thereby yielding
(2.10).

(ii) Next, let us assume {$H_s$}$_{0\le s\le t}$ is bounded but not necessarily continuous. For $m \in \mathbb{N}$ and $h \in S_t$, define $h^m$ by

$$h^m(s) = m \int_{(s-\frac{1}{m})\vee 0}^{s} h(u)du.$$

Then according to the Lebesgue differentiation theorem, it holds for all $\omega \in \Omega$

$$\lim_{m \to \infty} H_s^m(\omega) = H_s(\omega) \quad \text{a.e. } s \in [0, t].$$

Hence, we have $H_s^m(\omega) \to H_s(\omega)$ for almost every $(\omega, s)$ by Fubini's theorem, and it follows from the bounded convergence theorem that

$$E \left[ \left( \int_{0}^{t} H_{s}^{m} dW_{s} - \int_{0}^{t} H_{s} dW_{s} \right)^{2} \right] \le E \left[ \int_{0}^{t} (H_{s}^{m} - H_{s})^{2} ds \right] \to 0 (m \to \infty).$$

Thus, we can take a subsequence {$m(k)$}$_{k \in \mathbb{N}}$ such that

$$\int_{0}^{t} H_{s}^{m(k)} dW_{s} \xrightarrow{\text{a.s.}} \int_{0}^{t} H_{s} dW_{s} \quad (m \to \infty). \qquad (2.13)$$

On the other hand, since {$H_s^m$} is a continuous progressively measurable process, we can take measurable functions $\Phi_t^m : S_t \times S_t \to \mathbb{R}$ ($m \in \mathbb{N}$) such that

$$\Phi_t^m(H^m, W.) = \int_0^t H_s^m dW_s \quad \text{a.s.,} \qquad (2.14)$$

and for $\mu_H$-a.s. $h \in S_t$

$$\Phi_t^m(h^m, W.) = \int_0^t h^m(s) dW_s. \qquad (2.15)$$

Now we define $\Phi_t : S_t \times S_t \to \mathbb{R}$ by

$$\Phi_t(h, w) = \begin{cases} \lim_{k \to \infty} \Phi_t^{m(k)}(h^{m(k)}, w) & \text{when } \Phi_t^{m(k)}(h^{m(k)}, w) \text{ converges,} \\ 0 & \text{otherwise.} \end{cases}$$

Then (2.9) follows directly from (2.13) and (2.14). To prove (2.10), we proceed similarly to part (i), showing that for $\mu_H$-a.s. $h \in S_t$, the sequence $\Phi_t^{m(k)}(h^{m(k)}, W.)$ converges almost surely to $\Phi_t(h, W)$. By (2.15), this limit must be $\int_0^t h(s) dW_s$, thereby establishing (2.10).
(iii) Finally, consider the general case where {$H_s$} is not necessarily continuous or bounded. In this case, we define $H_s^m = H_s 1_{\{|H_s| \le m\}}$ for $m \in \mathbb{N}$. Then {$H_s^m$} is bounded, and since $|H_s^m| \le |H_s|$, we have

$$ \int_0^t H_s^m dW_s \xrightarrow{P} \int_0^t H_s dW_s, $$

so we obtain the desired result by applying the same argument as in part (ii). □

We now apply this lemma to $Z_t^{-1}$ as defined in (2.4).

**Proposition 2.3.** Under (2.3), {$Z_t^{-1}$}$_{t \ge 0}$ is a martingale.

*Proof.* First, we show that $E[Z_t^{-1}] = 1$ for all $t \ge 0$. Let $(S_t, \mathcal{S}_t)$ and $H$ be as in Lemma 2.2, and let $\Phi_t$ satisfy (2.9) and (2.10). Then $Z_t^{-1}$ can be expressed as

$$ Z_t^{-1} = \exp \left( -\Phi_t(H, W.) - \frac{1}{2} \int_0^t |H_s|^2 ds \right) \quad \text{a.s.,} $$

and we have $E[Z_t^{-1}] \le E[Z_0^{-1}] = 1$ since {$Z_t^{-1}$}$_{t \ge 0}$ is a supermartingale. Therefore, we have

$$ E[Z_t^{-1}|\mathcal{H}_t] = E \left[ \exp \left( -\Phi_t(h, W.) - \frac{1}{2} \int_0^t |h(s)|^2 ds \right) \right] \Big|_{h=H}, $$

where $\mathcal{H}_t = \sigma(H_s; 0 \le s \le t)$. Due to (2.10), the right-hand side almost surely equals

$$ E \left[ \exp \left( - \int_0^t h(s) dW_s - \frac{1}{2} \int_0^t |h(s)|^2 ds \right) \right] \Big|_{h=H} = 1, $$

and thus we obtain

$$ E[Z_t^{-1}] = E[E[Z_t^{-1}|\mathcal{H}_t]] = 1. \qquad (2.16) $$

Now the martingale property of {$Z_t^{-1}$} easily follows. In fact, noting that {$Z_t^{-1}$} is a supermartingale, we have

$$ E[Z_t^{-1} - Z_s^{-1} | \mathcal{F}_s] \le 0. $$

On the other hand, it follows from (2.16) that

$$ E[E[Z_t^{-1} - Z_s^{-1} | \mathcal{F}_s]] = E[Z_t^{-1}] - E[Z_s^{-1}] = 1 - 1 = 0. $$

Therefore we have $E[Z_t^{-1} - Z_s^{-1} | \mathcal{F}_s] = 0$. □
By Proposition 2.3, we can define the probability measure $Q$ using (2.5) under the weaker condition (2.3), instead of the stronger conditions (2.7) and (2.8). Moreover, the same results in Section 3 of Bain and Crisan (2009) still hold.

In particular, the following two results are of central importance in this paper.

**Proposition 2.4.** Under the new measure $Q$, the process $\{Y_t\}_{0\le t\le T}$ is a Brownian motion independent of $\{X_t\}_{0\le t\le T}$, and the law of $\{X_t\}_{0\le t\le T}$ is the same as its law under $P$.

More generally, if $U$ is a random variable independent of $\{W_t\}_{0\le t\le T}$, then under $Q$, $U$ is independent of $\{Y_t\}_{0\le t\le T}$ and has the same law as under $P$.

*Proof.* Although a proof of this proposition is provided in Proposition 3.13 of Bain and Crisan (2009), we present a brief proof here, utilizing Lemma 2.2.

Let $U$ be a $\sigma$-field, and let $\mathcal{H}_T$ denote the $\sigma$-field generated by $\{H_s\}_{0\le s\le T}$. Consider $\xi_1, \dots, \xi_k \in \mathbb{R}^{d_2}$ and $t_1, \dots, t_k \in [0, T]$. Using the same arguments as in Proposition 2.3 and applying the Girsanov theorem, along with the independence of $W$, $X$, $U$, and $H$, we deduce:

$$
\begin{align*}
& E_Q \left[ \exp \left( \sum_{i=1}^k \xi_i^\top Y_{t_i} \right) \middle| \mathcal{X}_T \lor \mathcal{U} \right] \\
&= E \left[ \exp \left( \sum_{i=1}^k \xi_i^\top Y_{t_i} \right) Z_T^{-1} \middle| \mathcal{X}_T \lor \mathcal{U} \right] \\
&= E \left[ \exp \left( \sum_{i=1}^k \xi_i^\top \left( \int_0^{t_i} H_s \, ds + W_{t_i} \right) \right) \right. \\
&\qquad \left. \times \exp \left( - \int_0^T H_s^\top \, dW_s - \frac{1}{2} \int_0^T |H_s|^2 \, ds \right) \middle| \mathcal{X}_T \lor \mathcal{U} \right] \\
&= E \left[ E \left[ \exp \left( \sum_{i=1}^k \xi_i^\top \left( \int_0^{t_i} H_s \, ds + W_{t_i} \right) \right) \right. \right. \\
&\qquad \left. \left. \times \exp \left( - \int_0^T H_s^\top \, dW_s - \frac{1}{2} \int_0^T |H_s|^2 \, ds \right) \middle| \mathcal{X}_T \lor \mathcal{U} \lor \mathcal{H}_T \right] \middle| \mathcal{X}_T \lor \mathcal{U} \right] \\
&= E \left[ E \left[ \exp \left( \sum_{i=1}^k \xi_i^\top \left( \int_0^{t_i} h(s) \, ds + W_{t_i} \right) \right) \right. \right. \\
&\qquad \left. \left. \times \exp \left( - \int_0^T h(s)^\top \, dW_s - \frac{1}{2} \int_0^T |h(s)|^2 \, ds \right) \right] \middle|_{h=H} \middle| \mathcal{X}_T \lor \mathcal{U} \right] \\
&= E \left[ \exp \left( \sum_{i=1}^k \xi_i^\top W_{t_i} \right) \right] = E \left[ \exp \left( \sum_{i=1}^k \xi_i^\top W_{t_i} \right) \middle| \mathcal{X}_T \lor \mathcal{U} \right].
\end{align*}
$$
Here, $h \in S_T$, and $H$ denotes the path of $\{H_s\}_{0 \le s \le T}$.

Now, let us fix $A \in \mathcal{X}_T \lor U$ arbitrarily, and define measures $\mu_1$ and $\mu_2$ on $\mathbb{R}^{d_2 \times n}$ by

$$ \mu_1(B) = E_Q[1_B(Y_{t_1}, \dots, Y_{t_k})1_A], \quad \mu_2(B) = E[1_B(W_{t_1}, \dots, W_{t_k})1_A], $$

for any $B \in \mathcal{B}(\mathbb{R}^{d_2 \times n})$. By the earlier argument, the characteristic functions of $\mu_1$ and $\mu_2$ are identical. Hence, for $t_1, \dots, t_k, s_1, \dots, s_l \in [0, T]$, the law of $(Y_{t_1}, \dots, Y_{t_k}, X_{s_1}, \dots, X_{s_l}, U)$ under $Q$ is the same as the law of $(Y_{t_1}, \dots, W_{t_k}, W_{s_1}, \dots, W_{s_l}, U)$ under $P$. This establishes the desired conclusion. $\square$

**Proposition 2.5.** (Kallianpur-Striebel formula)

For any $0 \le t \le T$ and a random variable $U$ such that $E[|U|] < \infty$, it holds that

$$ E[U|\mathcal{Y}_t] = \frac{E_Q[UZ_t|\mathcal{Y}_t]}{E_Q[Z_t|\mathcal{Y}_t]} \quad \text{a.s.,} \qquad (2.17) $$

where $E_Q$ denotes the expectation under the measure $Q$.

*Proof.* This proposition is a generalized version of Proposition 3.16 in Bain and Crisan (2009), but the proof is exactly the same as in the original.

First note that

$$ E_Q[|UZ_t|] = E[|U|Z_tZ_t^{-1}] = E[|U|] < \infty. $$

Furthermore, by

$$ P(Z_t = 0) = E[1_{\{Z_t=0\}}] = E_Q[1_{\{Z_t=0\}}Z_t] = 0, $$

and the equivalence of $P$ and $Q$, $Z_t > 0$ holds $P$ and $Q$ a.s. Hence, it is sufficient to show that

$$ E[U|\mathcal{Y}_t]E_Q[Z_t|\mathcal{Y}_t] = E_Q[UZ_t|\mathcal{Y}_t]. \qquad (2.18) $$

To prove this, let $A \in \mathcal{Y}_t$. Then we have

$$ \begin{aligned} E_Q[E[U|\mathcal{Y}_t]E_Q[Z_t|\mathcal{Y}_t]1_A] &= E_Q[E_Q[E[U|\mathcal{Y}_t]Z_t1_A|\mathcal{Y}_t]] \\ &= E_Q[E[U|\mathcal{Y}_t]Z_t1_A] \\ &= E[E[U1_A|\mathcal{Y}_t]] \\ &= E[U1_A] = E_Q[U1_AZ_t], \end{aligned} $$

which implies (2.18). $\square$

In the case where $H$ depends on $Y$ and is therefore not independent of $W$, a convenient sufficient condition for $Z_t^{-1}$ to be a martingale is provided in Kunita (2011).
### 3 Linear filtering and smoothing

In this section, we assume that the processes {$X_t$} and {$Y_t$} are solutions of the following linear equations:

$$X_t = X_0 + \int_0^t a(s)X_s ds + \int_0^t b(s) dV_s, \quad (3.1)$$

$$Y_t = \int_0^t c(s)X_s ds + \int_0^t \sigma(s) dW_s, \quad (3.2)$$

where $a, b, c$, and $\sigma$ are measurable functions taking values in $M_{d_1}(\mathbb{R})$, $M_{d_1,m_1}(\mathbb{R})$, $M_{d_2,d_1}(\mathbb{R})$, and $M_{d_2,m_2}(\mathbb{R})$, respectively. Here, {$V_t$} and {$W_t$} are independent $m_1$- and $m_2$-dimensional {$\mathcal{F}_t$}-Brownian motions. We also assume that $X_0$ is normally distributed and independent of {$V_t$} and {$W_t$}, so that {$\{(X_t, Y_t)\}$} forms a $d_1 + d_2$-dimensional Gaussian process.

Additionally, we assume that for every $t \ge 0$,

$$\int_0^t |a(s)| ds < \infty, \quad (3.3)$$

$$\int_0^t |b(s)|^2 ds < \infty, \quad (3.4)$$

$$\int_0^t |c(s)|^2 ds < \infty, \quad (3.5)$$

$$\int_0^t |\sigma(s)|^2 ds < \infty, \quad (3.6)$$

and that there exists a constant $C > 0$ such that for every $t \ge 0$,

$$\lambda_{\min}(\sigma(t)\sigma(t)^{\top}) > C, \quad (3.7)$$

where $\lambda_{\min}(\sigma(t)\sigma(t)^{\top})$ denotes the smallest eigenvalue of $\sigma(t)\sigma(t)^{\top}$. Through-out this paper, $|A|$ denotes the Frobenius norm for a matrix $A$.

Let us define

$$\overline{W}_t = \int_0^t (\sigma(s)\sigma(s)^{\top})^{-\frac{1}{2}}\sigma(s)dW_s.$$

Then {$\overline{W}_t$}$_{t \ge 0}$ is a Brownian motion since

$$\begin{align*}
\langle \overline{W} \rangle_t &= \int_0^t (\sigma(s)\sigma(s)^{\top})^{-\frac{1}{2}} \sigma(s)\sigma(s)^{\top} (\sigma(s)\sigma(s)^{\top})^{-\frac{1}{2}} ds \\
&= \int_0^t (\sigma(s)\sigma(s)^{\top})^{-\frac{1}{2}} (\sigma(s)\sigma(s)^{\top})^{\frac{1}{2}} (\sigma(s)\sigma(s)^{\top})^{\frac{1}{2}} (\sigma(s)\sigma(s)^{\top})^{-\frac{1}{2}} ds \\
&= I_{d_2}t,
\end{align*}$$
where $I_{d_2}$ is the $d_2$-dimensional identity matrix. Furthermore, let us write

$$ \bar{c}(t) = (\sigma(t)\sigma(t)^{\top})^{-\frac{1}{2}}c(t), \quad \bar{Y}_t = \int_0^t (\sigma(s)\sigma(s)^{\top})^{-\frac{1}{2}}dY_s, $$

so that

$$ \bar{Y}_t = \int_0^t \bar{c}(s) X_s ds + \bar{W}_t. $$

Then by (3.5) and (3.7), we have for $t \ge 0$

$$ \int_0^t |\bar{c}(s)X_s|^2 ds \le \int_0^t |\bar{c}(s)|^2 |X_s|^2 ds \le \frac{1}{\sqrt{C}} \int_0^t |c(s)|^2 |X_s|^2 ds < \infty, $$

where $C$ is given by (3.7). Furthermore, the filtration generated by $\{Y_t\}$ and $\{\bar{Y}_t\}$ are the same one because

$$ Y_t = \int_0^t (\sigma(s)\sigma(s)^{\top})^{\frac{1}{2}} d\bar{Y}_s. $$

Therefore, we can substitute $H_t$ with $\bar{c}(t)X_t$ and $Y_t$ with $\bar{Y}_t$ in Proposition 2.5 to obtain the following result.

**Proposition 3.1.** Let $T > 0$ and define a probability measure $Q$ by

$$ Q(A) = E \left[ 1_A Z_T^{-1} \right] $$

for $A \in \mathcal{F}$, where

$$ \begin{aligned} Z_t &= \exp \left( \int_0^t X_s^\top c(s)^\top (\sigma(s)\sigma(s)^\top)^{-1} \sigma(s) dW_s \right. \\ & \qquad \left. + \frac{1}{2} \int_0^t X_s^\top c(s)^\top (\sigma(s)\sigma(s)^\top)^{-1} c(s) X_s ds \right) \\ &= \exp \left( \int_0^t X_s^\top c(s)^\top (\sigma(s)\sigma(s)^\top)^{-1} dY_s \right. \\ & \qquad \left. - \frac{1}{2} \int_0^t X_s^\top c(s)^\top (\sigma(s)\sigma(s)^\top)^{-1} c(s) X_s ds \right). \end{aligned} \quad (3.8) $$

Then, under the new measure $Q$, the process $\{\bar{Y}_t\}_{0 \le t \le T}$ is a Brownian motion independent of $\{X_t\}_{0 \le t \le T}$, and the law of $\{X_t\}_{0 \le t \le T}$ is the same as its law under $P$.

Furthermore, for $0 \le t \le T$ and a random variable $U$ satisfying $E[|U|] < \infty$, it holds that $E[Z_t|\mathcal{Y}_t] > 0$ a.s. and

$$ E[U|\mathcal{Y}_t] = \frac{E_Q[UZ_t|\mathcal{Y}_t]}{E_Q[Z_t|\mathcal{Y}_t]}. \quad (3.9) $$
To examine the properties of the conditional distribution, we fix $t \in [0, T]$ and define the random measure $\tilde{P}_t$ on $(\Omega, \mathcal{X}_t \vee \mathcal{Y}_t)$ by

$$ \tilde{P}_t(A) = \tilde{P}_t(A|\mathcal{Y}_t) = \frac{E_Q[1_A Z_t | \mathcal{Y}_t]}{E_Q[Z_t | \mathcal{Y}_t]}, $$

and let $\tilde{E}_t$ denote the expectation with respect to $\tilde{P}_t$.

Note that this measure is well-defined. Specifically, if $C$ denotes the space of $\mathbb{R}^{d_1+d_2}$-valued continuous functions on $[0, t]$, equipped with the Borel $\sigma$-algebra associated with the supremum norm, then any $\mathcal{X}_t \vee \mathcal{Y}_t$-measurable random variable $U$ can be represented as

$$ U = \Phi(X., Y.) \quad \text{a.s.,} $$

where $\Phi: C \to \mathbb{R}$ is a measurable function, and $X$ and $Y$ denote the paths of $\{X_s\}_{0 \le s \le t}$ and $\{Y_s\}_{0 \le s \le t}$, respectively. It is well known that $C$ is a complete separable metric space, and therefore the regular conditional distribution of $(X., Y.)$ given $Y.$ is well-defined.

Let us denote this distribution by $\mu_{(X,Y)}(\cdot|Y.)$. Then for $B \in \mathcal{B}(C)$ and $A = \{(X., Y.) \in B\} \in \mathcal{X}_t \vee \mathcal{Y}_t$, we can define $\tilde{P}_t$ by

$$ \tilde{P}_t((X., Y.) \in B) = \mu_{(X,Y)}(B|Y.). $$

This $\tilde{P}_t$ is well-defined. In fact, if $A = \{(X., Y.) \in B_1\} = \{(X., Y.) \in B_2\}$ where $B_1, B_2 \in \mathcal{B}(C)$, then we have

$$ \{(X., Y.) \in B_1 \setminus B_2\} = \{(X., Y.) \in B_2 \setminus B_1\} = \emptyset $$

and thus

$$ \mu_{(X,Y)}(B_1 \times C|Y.) = \mu_{(X,Y)}((B_1 \cap B_2) \times C|Y.) = \mu_{(X,Y)}(B_2 \times C|Y.). $$

Furthermore, choose $A_i = \{(X., Y.) \in B_2 \setminus B_i\} \in \mathcal{X}_t$ where $B_i \in \mathcal{B}(C)$ for $i \in \mathbb{N}$, and assume $A_i \cap A_j = \emptyset$ ($i \neq j$). Also, set

$$ B'_1 = B_1, \quad B'_2 = B_2 \setminus B_1, \quad \dots, B'_i = B_i \setminus \bigcup_{j=1}^{i-1} B_j. $$

Then we have $A_i = \{(X., Y.) \in B'_i\}$ for every $i \in \mathbb{N}$. In fact, if there exists $\omega \in \{(X., Y.) \in B_i\} \setminus \{(X., Y.) \in B'_i\} = \{(X., Y.) \in B_i \setminus B'_i\}$ for $i \ge 2$, then $(X.(\omega), Y.(\omega)) \in B_i \cap \bigcup_{j=1}^{i-1} B_j$ and thus $\omega \in A_i \cap \bigcup_{j=1}^{i-1} A_j$. This contradicts $A_i \cap A_j = \emptyset$.

Hence, we have

$$ \begin{aligned} P\left(\bigcup_{i=1}^{\infty} A_i\right) &= P\left(\bigcup_{i=1}^{\infty} \{(X., Y.) \in B'_i\}\right) = \mu_{(X,Y)}\left(\bigcup_{i=1}^{\infty} B'_i \mid Y.\right) \\ &= \sum_{i=1}^{\infty} \mu_{(X,Y)}(B'_i) = \sum_{i=1}^{\infty} P(A_i). \end{aligned} $$
## 3.1 The Main Theorem

In this section, we prove our main theorem for the linear case, which determines the distribution of the process {$X_s$}$_{0\le s\le t}$ under the measure $\tilde{P}_t$. To describe the theorem, for each fixed $t \ge 0$, we introduce an $M_{d_1}(\mathbb{R})$-valued differentiable function $\phi(s; t)$ as the negative semidefinite symmetric solution of the matrix Riccati equation

$$ \frac{d\phi}{ds}(s; t) = -\phi(s; t)b(s)b(s)^{\top}\phi(s; t) - a(s)^{\top}\phi(s; t) - \phi(s; t)a(s) + c(s)^{\top}(\sigma(s)\sigma(s)^{\top})^{-1}c(s) \quad (3.10) $$

with the boundary condition $\phi(t; t) = 0$. This solution is well-defined. Specifically, the Riccati equation

$$ \begin{aligned} \frac{d\tilde{\phi}}{ds}(s; t) = & -\tilde{\phi}(s; t)b(t-s)b(t-s)^{\top}\tilde{\phi}(s; t) \\ & -a(t-s)^{\top}\tilde{\phi}(s; t) - \tilde{\phi}(s; t)a(t-s) \\ & +c(t-s)^{\top}(\sigma(t-s)\sigma(t-s)^{\top})^{-1}c(t-s) \end{aligned} $$

with $\tilde{\phi}(0; t) = 0$ has a positive-semidefinite solution $\tilde{\phi}(s; t)$ for $0 \le s \le t$, according to Theorems 2.1 and 2.2 in Potter (1965). It then follows immediately that $\phi(s; t) = -\tilde{\phi}(t - s; t)$ satisfies (3.10), and uniqueness can be established using Gronwall's lemma.

Also, let {$\tilde{V}_s$} be a $d_1$-dimensional Brownian motion on $(\Omega, \mathcal{F}, \{\mathcal{F}_t\}, P)$ that is independent of {$Y_s$} and $X_0$. Let {$\xi_{s;t}$}$_{0\le s\le t}$ be the solution to the stochastic differential equation

$$ d_s\xi_{s;t} = \{a(s) + b(s)b(s)^{\top}\phi(s;t)\} \xi_{s;t} ds + b(s) d\tilde{V}_s, \quad (3.11) $$

where $\xi_{0;t}$ is a normal random variable, independent of {$\tilde{V}_s$} and {$Y_s$}, with mean 0 and covariance matrix

$$ V[\xi_{0;t}] = V[X_0]^{\frac{1}{2}} \left( I_{d_1} - V[X_0]^{\frac{1}{2}}\phi(0;t)V[X_0]^{\frac{1}{2}} \right)^{-1} V[X_0]^{\frac{1}{2}}. \quad (3.12) $$

Furthermore, we define $\zeta_{s;t}$ for $0 \le s \le t$ by

$$ \zeta_{s;t} = E[X_s] + \xi_{s;t} + \int_0^t \text{Cov}(\xi_{s;t}, \xi_{u;t}) c(u)^{\top} (\sigma(u)\sigma(u)^{\top})^{-1} (dY_u - c(u)E[X_u] du). \quad (3.13) $$

Noting that it holds $E[\xi_{s;t}|\mathcal{Y}_t] = E[\xi_{s;t}] = 0$ by $E[\xi_{0;t}] = 0$ and the independence, {$\zeta_{s;t}$}$_{0\le s\le t}$ is a conditionally Gaussian process with mean

$$ \begin{aligned} E[\zeta_{s;t}|\mathcal{Y}_t] = & E[X_s] \\ & + \int_0^t \text{Cov}(\xi_{s;t}, \xi_{u;t}) c(u)^{\top} (\sigma(u)\sigma(u)^{\top})^{-1} (dY_u - c(u)E[X_u] du), \end{aligned} \quad (3.14) $$
and covariance

$$ \mathrm{Cov}(\zeta_{s;t}, \zeta_{u;t} | \mathcal{Y}_t) = \mathrm{Cov}(\xi_{s;t}, \xi_{u;t}). \qquad (3.15) $$

Then the distribution of {$X_s$}$_{0 \le s \le t}$ under $\tilde{P}_t$ is identified by the following result.

**Theorem 3.2.** Let $C_t$ be the space of all continuous functions from $[0, t]$ to $\mathbb{R}^{d_1+d_2}$, and $C_t$ be the Borel $\sigma$-filed on $C_t$.

Then the law of the process {($\zeta_{s;t}$, $Y_s$)}$_{0 \le s \le t}$ on a regular conditional probability measure $P(\cdot | \mathcal{Y}_t)$ is a version of the law of the process {($X_s$, $Y_s$)}$_{0 \le s \le t}$ on the probability measure $\tilde{P}_t(\cdot | \mathcal{Y}_t)$, in the sense that for every $A \in C_t$

$$ \tilde{P}_t((X., Y.) \in A | \mathcal{Y}_t) = P((\zeta., Y.) \in A | \mathcal{Y}_t), $$

where $X$, $Y$, and $\xi$. are paths of {$X_s$}$_{0 \le s \le t}$, {$Y_s$}$_{0 \le s \le t}$ and {$\zeta_{s;t}$}$_{0 \le s \le t}$, respectively.

### 3.1.1 Heuristic Argument

Before proving Theorem 3.2, we provide a heuristic argument to illustrate the main ideas behind the proof. The primary task in the proof is to compute the conditional distribution of to calculate conditional distribution of {$X_s$}$_{0 \le s \le t}$ given $\mathcal{Y}_t$, by directly evaluating the quotient in (3.9).

However, a significant challenge arises in applying stochastic calculus techniques, such as the Girsanov theorem, because the numerator and denominator are conditional expectations. In the following, we introduce a methodology to reduce the quotient of these conditional expectations to the computation of standard expectations.

**(I) One-dimensional case:**

Before considering the original problem, we explain our idea in a one-dimensional model given by

$$ Y = cX + \sigma Z, $$

where $X \sim N(\mu_x, \sigma_x^2)$ and $Z \sim N(0, 1)$ are independent random variables, with $\sigma_x, \sigma > 0$. Furthermore, we assume $c \neq 0$. In this model, the conditional expectation of $X$ given $Y$ can be expressed as

$$ \begin{aligned} E[X|Y=y] &= \frac{\int_{\mathbb{R}} x \exp \left( -\frac{1}{2\sigma^2} (cx - y)^2 \right) p(x) dx}{\int_{\mathbb{R}} \exp \left( -\frac{1}{2\sigma^2} (cx - y)^2 \right) p(x) dx} \\ &= \frac{\int_{\mathbb{R}} x \exp \left( -\frac{c^2 x^2}{2\sigma^2} + \frac{cxy}{\sigma^2} \right) p(x) dx}{\int_{\mathbb{R}} \exp \left( -\frac{c^2 x^2}{2\sigma^2} + \frac{cxy}{\sigma^2} \right) p(x) dx} \end{aligned} $$
$$ = \frac{E \left[ X \exp \left( -\frac{c^2 X^2}{2\sigma^2} + \frac{cXy}{\sigma^2} \right) \right]}{E \left[ \exp \left( -\frac{c^2 X^2}{2\sigma^2} + \frac{cXy}{\sigma^2} \right) \right]} \quad (3.16) $$

where $p(x)$ is the probability density function of $X$. This formula is the analogue of (3.9).

This final expression can be further rewritten as

$$ \frac{\sigma^2}{c} \frac{d}{dy} \log E \left[ \exp \left( -\frac{c^2 X^2}{2\sigma^2} + \frac{cXy}{\sigma^2} \right) \right]. \quad (3.17) $$

Here, due to the normality of $X$, it can be shown that $\log E \left[ \exp \left( -\frac{X^2}{2\sigma^2} + \frac{Xy}{\sigma^2} \right) \right]$ is a quadratic function of $y$. Therefore, we can write

$$ E[X|Y=y] = \frac{E \left[ X \exp \left( -\frac{c^2 X^2}{2\sigma^2} + \frac{cXy}{\sigma^2} \right) \right]}{E \left[ \exp \left( -\frac{c^2 X^2}{2\sigma^2} + \frac{cXy}{\sigma^2} \right) \right]} = Ay + B. $$

By taking expectation of both sides (with respect to $y$), we obtain

$$ E[X] = AE[Y] + B = AcE[X] + B. $$

Thus we have $B = E[X] - AcE[X]$ and hence

$$ E[X|Y=y] = \frac{E \left[ X \exp \left( -\frac{c^2 X^2}{2\sigma^2} + \frac{cXy}{\sigma^2} \right) \right]}{E \left[ \exp \left( -\frac{c^2 X^2}{2\sigma^2} + \frac{cXy}{\sigma^2} \right) \right]} = E[X] + A(y - cE[X]). \quad (3.18) $$

Furthermore, by differentiating (3.18), $A$ can be determined as

$$ \begin{aligned} A &= \frac{d}{dy} \frac{E \left[ X \exp \left( -\frac{c^2 X^2}{2\sigma^2} + \frac{cXy}{\sigma^2} \right) \right]}{E \left[ \exp \left( -\frac{c^2 X^2}{2\sigma^2} + \frac{cXy}{\sigma^2} \right) \right]} \\ &= \frac{c}{\sigma^2} \frac{E \left[ X^2 \exp \left( -\frac{c^2 X^2}{2\sigma^2} + \frac{cXy}{\sigma^2} \right) \right]}{E \left[ \exp \left( -\frac{c^2 X^2}{2\sigma^2} + \frac{cXy}{\sigma^2} \right) \right]} \\ &\quad - \frac{c}{\sigma^2} \left( \frac{E \left[ X \exp \left( -\frac{c^2 X^2}{2\sigma^2} + \frac{cXy}{\sigma^2} \right) \right]}{E \left[ \exp \left( -\frac{c^2 X^2}{2\sigma^2} + \frac{cXy}{\sigma^2} \right) \right]} \right)^2 \end{aligned} \quad (3.19) $$
Since A does not depend on y, we can set $y = 0$ in the final expression, which yields

$$A = \frac{c}{\sigma^2} \left\{ \frac{E\left[X^2 \exp\left(-\frac{c^2 X^2}{2\sigma^2}\right)\right]}{E\left[\exp\left(-\frac{c^2 X^2}{2\sigma^2}\right)\right]} - \left(\frac{E\left[X \exp\left(-\frac{c^2 X^2}{2\sigma^2}\right)\right]}{E\left[\exp\left(-\frac{c^2 X^2}{2\sigma^2}\right)\right]}\right)^2 \right\}. \quad (3.20)$$

Therefore, (3.18) ends up in the conditional mean formula

$$E[X|Y=y] = E[X] + \frac{c}{\sigma^2} \left\{ \frac{E\left[X^2 \exp\left(-\frac{c^2 X^2}{2\sigma^2}\right)\right]}{E\left[\exp\left(-\frac{c^2 X^2}{2\sigma^2}\right)\right]} - \left(\frac{E\left[X \exp\left(-\frac{c^2 X^2}{2\sigma^2}\right)\right]}{E\left[\exp\left(-\frac{c^2 X^2}{2\sigma^2}\right)\right]}\right)^2 \right\} (y - cE[X]). \quad (3.21)$$

Moreover, recalling the Bayesian formula (3.16), we notice that (3.19) is equivalent to

$$A = \frac{c}{\sigma^2} (E[X^2|Y=y] - E[X|Y=y]^2) = \frac{c}{\sigma^2} E[V|Y=y].$$

Since A is given by (3.20), we obtain the conditional variance formula

$$V[X|Y=y] = \frac{E\left[X^2 \exp\left(-\frac{c^2 X^2}{2\sigma^2}\right)\right]}{E\left[\exp\left(-\frac{c^2 X^2}{2\sigma^2}\right)\right]} - \left(\frac{E\left[X \exp\left(-\frac{c^2 X^2}{2\sigma^2}\right)\right]}{E\left[\exp\left(-\frac{c^2 X^2}{2\sigma^2}\right)\right]}\right)^2. \quad (3.22)$$

Further differentiating (3.19) yields the third and higher cumulants of X given Y, but they are zero since A is a constant. In this way, we can conclude that the conditional distribution of X given Y = y is normal with the mean and variance given by (3.21) and (3.22), respectively.

This procedure has two important implications:

* Differentiating the conditional expectation and substituting $y = 0$ reduces the problem to the computation of a standard expectation, which no longer depends on y.

* By analyzing the conditional expectation, we have derived the conditional distribution.
(II) The original case:

Now, let us apply the same methodology to the original model. Although not all steps are rigorously justified, the following informal computations serve to convey the intuition and underlying structure of the argument.

For simplicity, assume a, b, c, X and Y are one-dimensional, and consider the case of $\sigma = 1$. Also, we assume that $c(t) > 0$ for all $t \ge 0$.

Due to Proposition 3.1, the conditional expectation of $X_s$ ($0 \le s \le t$) given $\mathcal{Y}_t$ can be written as

$$E[X_s|\mathcal{Y}_t] = \frac{E_Q \left[ X_s \exp \left( \int_0^t c(s) X_s dY_s - \frac{1}{2} \int_0^t c(s)^2 X_s^2 ds \right) \middle| \mathcal{Y}_t \right]}{E_Q \left[ \exp \left( \int_0^t c(s) X_s dY_s - \frac{1}{2} \int_0^t c(s)^2 X_s^2 ds \right) \middle| \mathcal{Y}_t \right]} \quad (3.23)$$

(corresponding to (3.16)).

Recall that under $Q$, $Y$ is a Brownian motion independent of $X$. Hence, if $D$ denotes the Malliavin derivative with respect to $Y$, then we heuristically observe that

$$ \begin{aligned} & D_s E_Q \left[ \exp \left( \int_0^t c(s) X_s dY_s - \frac{1}{2} \int_0^t c(x)^2 X_s^2 ds \right) \middle| \mathcal{Y}_t \right] \\ &= D_s \int \exp \left( \int_0^t c(s) x(s) dY_s - \frac{1}{2} \int_0^t c(s)^2 x(s)^2 ds \right) \mu_X(dx) \\ &= \int D_s \exp \left( \int_0^t c(s) x(s) dY_s - \frac{1}{2} \int_0^t c(s)^2 x(s)^2 ds \right) \mu_X(dx) \quad (3.24) \\ &= \int \frac{c(s)}{\sigma(s)} \exp \left( \int_0^t c(s) x(s) dY_s - \frac{1}{2} \int_0^t c(s)^2 x(s)^2 ds \right) \mu_X(dx) \\ &= \frac{c(s)}{\sigma(s)} E_Q \left[ X_s \exp \left( \int_0^t c(s) X_s dY_s - \frac{1}{2} \int_0^t c(s)^2 ds \right) \middle| \mathcal{Y}_t \right] \end{aligned} $$

Therefore, by the chain rule, (3.23) can be rewritten as

$$\frac{1}{c(s)} D_s \log E_Q \left[ X_s \exp \left( \int_0^t c(s) X_s dY_s - \frac{1}{2} \int_0^t c(s)^2 X_s^2 ds \right) \middle| \mathcal{Y}_t \right] \quad (3.25)$$

(corresponding to (3.17)).

Here, recalling that $X$ is a Gaussian process independent of $Y$ under $Q$, we have the form

$$ \begin{aligned} & E_Q \left[ X_s \exp \left( \int_0^t c(s) X_s dY_s - \frac{1}{2} \int_0^t c(s)^2 X_s^2 ds \right) \middle| \mathcal{Y}_t \right] \\ &= \exp \left( h_0(t) + \int_0^t h_1(s; t) dY_s + \int_0^t \int_0^s h_2(s, u; t) dY_u dY_s \right). \end{aligned} \quad (3.26) $$
Thus (3.23) (which is equals to (3.25)) can be written in the form

$$
\begin{aligned}
E[X_s|\mathcal{Y}_t] &= \frac{E_Q \left[ X_s \exp \left( \int_0^t c(s) X_s dY_s - \frac{1}{2} \int_0^t c(s)^2 X_s^2 ds \right) \middle| \mathcal{Y}_t \right]}{E_Q \left[ \exp \left( \int_0^t c(s) X_s dY_s - \frac{1}{2} \int_0^t c(s)^2 X_s^2 ds \right) \middle| \mathcal{Y}_t \right]} \\
&= g_0(t) + \int_0^t g_1(s, u; t) dY_u.
\end{aligned}
$$

By taking the expectations of both sides, we obtain

$$
E[X_s|\mathcal{Y}_t] = g_0(t) + \int_0^t g_1(s, u; t)c(u)E[X_u]du,
$$

and thus it holds

$$
\begin{aligned}
E[X_s|\mathcal{Y}_t] &= \frac{E_Q \left[ X_s \exp \left( \int_0^t c(s) X_s dY_s - \frac{1}{2} \int_0^t c(s)^2 X_s^2 ds \right) \middle| \mathcal{Y}_t \right]}{E_Q \left[ \exp \left( \int_0^t c(s) X_s dY_s - \frac{1}{2} \int_0^t c(s)^2 X_s^2 ds \right) \middle| \mathcal{Y}_t \right]} \\
&= E[X_t] + \int_0^t g_1(s, u; t) (dY_u - c(u)E[X_u]du). \quad (3.27)
\end{aligned}
$$

(corresponding to (3.18)).

Furthermore, $g_1$ is obtained by the Malliavin derivative:

$$
\begin{aligned}
& g_1(s, u; t) \\
&= D_u \frac{E_Q \left[ X_s \exp \left( \int_0^t c(s) X_s dY_s - \frac{1}{2} \int_0^t c(s)^2 X_s^2 ds \right) \middle| \mathcal{Y}_t \right]}{E_Q \left[ \exp \left( \int_0^t c(s) X_s dY_s - \frac{1}{2} \int_0^t c(s)^2 X_s^2 ds \right) \middle| \mathcal{Y}_t \right]} \\
&= c(u) \left( \frac{E_Q \left[ X_s X_u \exp \left( \int_0^t c(s) X_s dY_s - \frac{1}{2} \int_0^t c(s)^2 X_s^2 ds \right) \middle| \mathcal{Y}_t \right]}{E_Q \left[ \exp \left( \int_0^t c(s) X_s dY_s - \frac{1}{2} \int_0^t c(s)^2 X_s^2 ds \right) \middle| \mathcal{Y}_t \right]} \right. \\
&\qquad \left. - \frac{E_Q \left[ X_s \exp \left( \int_0^t c(s) X_s dY_s - \frac{1}{2} \int_0^t c(s)^2 X_s^2 ds \right) \middle| \mathcal{Y}_t \right]}{E_Q \left[ \exp \left( \int_0^t c(s) X_s dY_s - \frac{1}{2} \int_0^t c(s)^2 X_s^2 ds \right) \middle| \mathcal{Y}_t \right]} \right. \\
&\qquad \left. \times \frac{E_Q \left[ X_u \exp \left( \int_0^t c(s) X_s dY_s - \frac{1}{2} \int_0^t c(s)^2 X_s^2 ds \right) \middle| \mathcal{Y}_t \right]}{E_Q \left[ \exp \left( \int_0^t c(s) X_s dY_s - \frac{1}{2} \int_0^t c(s)^2 X_s^2 ds \right) \middle| \mathcal{Y}_t \right]} \right) \quad (3.28)
\end{aligned}
$$
Since $g_1$ does not depend on $Y$, we can substitute 0 into $Y$, and obtain

$$g_1(s, u; t) = c(u)\gamma(s, u; t) \qquad (3.29)$$

where

$$\gamma(s, u; t) = \frac{E \left[ X_s X_u \exp \left( -\frac{1}{2} \int_0^t c(s)^2 X_s^2 ds \right) \middle| \mathcal{Y}_t \right]}{E_Q \left[ \exp \left( -\frac{1}{2} \int_0^t c(s)^2 X_s^2 ds \right) \middle| \mathcal{Y}_t \right]} - \frac{E \left[ X_s \exp \left( -\frac{1}{2} \int_0^t c(s)^2 X_s^2 ds \right) \middle| \mathcal{Y}_t \right]}{E \left[ \exp \left( -\frac{1}{2} \int_0^t c(s)^2 X_s^2 ds \right) \middle| \mathcal{Y}_t \right]} \times \frac{E \left[ X_u \exp \left( -\frac{1}{2} \int_0^t c(s)^2 X_u^2 ds \right) \middle| \mathcal{Y}_t \right]}{E_Q \left[ \exp \left( -\frac{1}{2} \int_0^t c(s)^2 X_u^2 ds \right) \middle| \mathcal{Y}_t \right]} \qquad (3.30)$$

(corresponding to (3.20)). Here, we again used the fact that $X$ has the same law under $P$ and $Q$. Substituting this expression into (3.27), we obtain the conditional expectation formula

$$E[X_s|\mathcal{Y}_t] = E[X_t] + \int_0^t c(u)\gamma(s, u; t)(dY_u - c(u)E[X_u]du). \qquad (3.31)$$

(corresponding to (3.21)). Moreover, recalling the Kallianpur-Striebel formula (3.23), we can see (3.28) as

$$E[X_s X_u | \mathcal{Y}_t] - E[X_s | \mathcal{Y}_t] E[X_u | \mathcal{Y}_t] = \text{Cov}(X_s, X_u | \mathcal{Y}_t).$$

Since this is equivalent to (3.29), we obtain the conditional covariance formula

$$\text{Cov}(X_s, X_u | \mathcal{Y}_t) = \gamma(s, u; t). \qquad (3.32)$$

(corresponding to (3.22)). The third and higher derivatives can be obtained as further Malliavin derivatives of (3.28), but they should be zero since $g_1$ does not depend on $Y$.

In this way, we have shown that the conditional distribution of $\{X_s\}_{0\le s\le t}$ given $\mathcal{Y}_t$ is a Gaussian process where the mean and covariance is given by (3.31) and (3.32). Therefore, calculation of the conditional distribution is reduced to evaluating $\gamma(s, u; t)$ given by (3.30).

In order to evaluate (3.30), let us introduce $\phi(s; t)$ and $\xi_{s;t}$ ($0 \le s \le t$) defined by (3.10) and (3.11) (with $\sigma = 1$), and assume $X_0 = 0$ for simplicity.
Then by the density formula in Liptser and Shiryaev (2001a), we have

$$ \frac{d\mu_{\xi.}}{d\mu_X}(X.) = \exp \left( \int_0^t \phi(s; t) X_s dX_s \right. \\ \left. - \frac{1}{2} \int_0^t \{2a(s) + b(s)^2\phi(s; t)\} \phi(s; t) X_s^2 ds \right). \quad (3.33) $$

where X. and ξ. are the paths of {X_s}_{0≤s≤t} and {ξ_{s;t}}_{0≤s≤t} respectively, and μ_X and μ_ξ are their distributions. By Itô's formula and (3.10), we have

$$ \begin{aligned} & \phi(t; t)X_t^2 - \phi(0; t)X_0 \\ &= 2 \int_0^t \phi(s; t)X_s dX_s + \int_0^t X_s^2 \frac{d}{ds}\phi(s; t)ds + \int_0^t \phi(s; t)b(s)^2ds \\ &= 2 \int_0^t \phi(s; t)X_s dX_s + \int_0^t X_s^2 \{-\phi(s; t)b(s)^2 - 2a(s)\phi(s; t) + c(s)^2\}ds \\ & \qquad + \int_0^t \phi(s; t)b(s)^2ds. \end{aligned} $$

Recalling $\phi(t; t) = 0$ and $X_0 = 0$, we obtain

$$ \begin{aligned} & \int_0^t \phi(s; t) X_s dX_s \\ &= \frac{1}{2} \int_0^t \{2a(s) + b(s)^2\phi(s; t)\} \phi(s; t) X_s^2 ds - \int_0^t c(s)^2 X_s^2 ds \\ & \qquad - \int_0^t \phi(s; t) b(s)^2 ds \end{aligned} $$

Therefore, (3.33) can be rewritten as

$$ \frac{d\mu_{\xi.}}{d\mu_X}(X.) = \exp \left( - \int_0^t c(s)^2 X_s^2 ds - \int_0^t \phi(s; t) b(s)^2 ds \right), $$

and (3.30) yields

$$ \begin{aligned} \gamma(s, u; t) = & \frac{E \left[ X_u X_s \frac{d\mu_{\xi.}}{d\mu_X}(X.) \right] \exp \left( - \int_0^t \phi(s; t) b(s)^2 ds \right)}{\exp \left( - \int_0^t \phi(s; t) b(s)^2 ds \right)} \\ & - \frac{E \left[ X_s \frac{d\mu_{\xi.}}{d\mu_X}(X.) \right] \exp \left( - \int_0^t \phi(s; t) b(s)^2 ds \right)}{\exp \left( - \int_0^t \phi(s; t) b(s)^2 ds \right)} \\ & \times \frac{E \left[ X_u \frac{d\mu_{\xi.}}{d\mu_X}(X.) \right] \exp \left( - \int_0^t \phi(s; t) b(s)^2 ds \right)}{\exp \left( - \int_0^t \phi(s; t) b(s)^2 ds \right)} \\ = & E[\xi_{s;t} \xi_{u;t}] - E[\xi_{s;t}] E[\xi_{u;t}] = \text{Cov}(\xi_{s;t}, \xi_{u;t}). \end{aligned} \quad (3.34) $$
Using this result, (3.31) and (3.32) can be rewritten as

$$E[X_s|\mathcal{Y}_t] = E[X_t] + \int_0^t c(u)\gamma(s, u; t)(dY_u - c(u)E[X_u]du).$$

and

$$\mathrm{Cov}(X_s, X_u|\mathcal{Y}_t) = E[\xi_{s;t}\xi_{u;t}],$$

which leads to Theorem 3.2.

### 3.1.2 Proof of the Theorem

In this section, we present the proof based on the discussion above. The core of the proof involves justifying the Malliavin derivative (3.24), which will be stated in Lemma 3.5. In this lemma, we first establish (3.24) by replacing the stochastic integral with a discrete sum, and then obtain the desired result by taking the limit of this equality. The convergence of the limit is guaranteed by Lemma 3.4.

Before proceeding with this argument, we start by verifying the expression (3.26) using the Gaussian property of $X$. For this purpose, we rewrite $Z_t$ as

$$ 
\begin{aligned}
Z_t &= \exp \left( \int_0^t X_s^\top c(s)^\top (\sigma(s)\sigma(s)^\top)^{-1} dY_s \right. \\
&\quad \left. - \frac{1}{2} \int_0^t X_s^\top c(s)^\top (\sigma(s)\sigma(s)^\top)^{-1} c(s) X_s ds \right) \\
&= \exp \left( X_t^\top \hat{Y}_t - \int_0^t \hat{Y}_s^\top dX_s \right. \\
&\quad \left. - \frac{1}{2} \int_0^t X_s^\top c(s)^\top (\sigma(s)\sigma(s)^\top)^{-1} c(s) X_s ds \right),
\end{aligned} 
\quad (3.35) $$

where

$$\hat{Y}_s = \int_0^s c(u)^\top (\sigma(u)\sigma(u)^\top)^{-1} dY_u. \quad (3.36)$$

Furthermore, let us write

$$ 
\begin{aligned}
z_t(y) = \exp \left( X_t^\top y(t) - \int_0^t y(s)^\top dX_s \right. \\
\left. - \frac{1}{2} \int_0^t X_s^\top c(s)^\top (\sigma(s)\sigma(s)^\top)^{-1} c(s) X_s ds \right)
\end{aligned} 
\quad (3.37) $$

for a bounded measurable function $y: [0, t] \to \mathbb{R}^{d_1}$.

Then the following lemma holds.
**Lemma 3.3.** Let {$\bar{\xi}_s$}$_{0\le s\le t}$ be the solution of

$$
d\bar{\xi}_{s;t} = \{a(s) + b(s)b(s)^{\top}\phi(s;t)\} \bar{\xi}_{s;t} ds + b(s)d\tilde{V}_s, \quad \bar{\xi}_{0;t} = 0, \quad (3.38)
$$

and define

$$
\alpha(s;t) = \exp \left( \int_0^s \{a(u) + b(u)b(u)^\top \phi(u;t)\} du \right). \quad (3.39)
$$

Then for any bounded measurable $y: [0,t] \to \mathbb{R}^{d_1}$ and $u_1, \dots, u_n \in [0,t]$,
it holds

$$
\begin{align}
& E_Q[z_t(y)] \nonumber \\
&= \sqrt{\det(I_{d_1} - \Sigma^{\frac{1}{2}}\phi(0;t)\Sigma^{\frac{1}{2}})} \exp\left(\frac{1}{2}\int_0^t b(s)^{\top}\phi(s;t)b(s)ds\right) \nonumber \\
&\quad \times \exp\left(\frac{1}{2}\beta(t;y)^{\top}\Sigma^{\frac{1}{2}}(I_{d_1} - \Sigma^{\frac{1}{2}}\phi(0;t)\Sigma^{\frac{1}{2}})^{-1}\Sigma^{\frac{1}{2}}\beta(t;y)\right) \nonumber \\
&\quad \times \exp\left(\mu^{\top}\alpha(t;t)^{\top}y(t)\right) \tag{3.40} \\
&\quad - \int_0^t y(s)^{\top}\{a(s) + b(s)b(s)^{\top}\phi(s;t)\}\alpha(s;t)ds\mu + \frac{1}{2}\mu^{\top}\phi(0;t)\mu \nonumber \\
&\quad \times \exp\left(\frac{1}{2}y(t)^{\top}E[\bar{\xi}_{t;t}\bar{\xi}_{t;t}^{\top}]y(t) + \frac{1}{2}\int_0^t y(s)^{\top}b(s)b(s)^{\top}y(s)ds\right. \nonumber \\
&\qquad \left. - \int_0^t y(s)b(s)b(s)^{\top}y(t)ds\right), \nonumber
\end{align}
$$

and

$$
\frac{E_Q[X_{u_1} \otimes \cdots \otimes X_{u_n} z_t(y)]}{E_Q[z_t(y)]}
\begin{align}
&= E \left[ \bigotimes_{j=1}^n \left\{ \alpha(u_j; t) \left( \mu + \Sigma^{\frac{1}{2}} (I_{d_1} - \Sigma^{\frac{1}{2}} \phi(0; t) \Sigma^{\frac{1}{2}})^{-\frac{1}{2}} (U + \beta(t; y)) \right) \right. \right. \tag{3.41} \\
&\qquad \left. \left. + \bar{\xi}_{u_j; t} + E[\bar{\xi}_{u_j; t} \eta_t(y)] \right\} \right], \nonumber
\end{align}
$$

where

$$
\mu = E[X_0], \quad \Sigma = \mathrm{Cov}(X_0, X_0),
$$

$$
\beta(t;y) = \alpha(t;t)^{\top} y(t) - \int_{0}^{t} \alpha(s;t)^{\top} \left\{ a(s) + b(s)b(s)^{\top} \phi(s;t) \right\}^{\top} y(s)ds \\
+ \phi(0;t)\mu,
$$

$$
\eta_t(y) = \bar{\xi}_{t;t}^{\top} y(t) - \int_0^t y(s)^{\top} d\bar{\xi}_{s;t},
$$

$$
E[\bar{\xi}_{s;t} \eta_t(y)] = E[\bar{\xi}_{s;t} \bar{\xi}_{t;t}^\top] y(t) - \int_0^s b(u)b(u)^\top y(u)du,
$$
and U is a $d_1$-dimensional standard normal random variable independent of $\{\tilde{\xi}_{s,t}\}_{0\le s\le t}$.

*Proof.* Let us define $\{\tilde{\xi}_{s,t}\}_{0\le s\le t}$ by

$$d\tilde{\xi}_{s,t} = \{a(s) + b(s)b(s)^{\top}\phi(s;t)\} \tilde{\xi}_{s;t}ds + b(s)d\tilde{V}_s, \quad \tilde{\xi}_{0;t} = X_0, \quad (3.42)$$

and let $X$. and $\tilde{\xi}$. be the paths of $\{X_s\}_{0\le s\le t}$ and $\{\tilde{\xi}_{s,t}\}_{0\le s\le t}$, and $\mu_X$ and $\mu_{\tilde{\xi}}$ be their distributions on $P$. Then by (7.138) in Liptser and Shiryaev (2001a), $\mu_X$ and $\mu_{\tilde{\xi}}$ are equivalent, and it holds that

$$\frac{d\mu_{\tilde{\xi}}}{d\mu_X}(X.) = \exp\left( \int_0^t X_s^\top \phi(s;t) dX_s - \frac{1}{2} \int_0^t X_s^\top \phi(s;t) (2a(s) + b(s)b(s)^\top \phi(s;t)) X_s ds \right). \quad (3.43)$$

Here, note that the condition (7.137) in the reference can be replaced by

$$\int_0^T |\alpha_t(x)|^2 dt < \infty \quad \mu_{\tilde{\xi}} \text{ and } \mu_{\eta}\text{-a.s.,}$$

and this condition can be written as

$$\int_0^t |b(s)b(s)^\top \phi(s;t) \tilde{\xi}_{s,t}|^2 ds < \infty, \quad \int_0^t |b(s)b(s)^\top \phi(s;t) X_s|^2 ds < \infty \quad \text{a.s.}$$

in our setup. This can be verified by (3.4) and the continuities of $\phi(s;t)$, $X_s$ and $\tilde{\xi}_{s,t}$.

On the other hand, by Itô's formula and (3.10), we have

$$
\begin{align*}
& X_t^\top \phi(t;t) X_t - X_0^\top \phi(0;t) X_0 \\
&=2 \int_0^t X_s^\top \phi(s;t) dX_s + \int_0^t X_s^\top \frac{\partial}{\partial s} \phi(s;t) X_s ds + \int_0^t b(s)^\top \phi(s;t) b(s) ds \\
&=2 \int_0^t X_s^\top \phi(s;t) dX_s - \int_0^t X_s^\top \phi(s;t) b(s) b(s)^\top \phi(s;t) X_s ds \\
&\quad - \int_0^t X_s^\top a(s)^\top \phi(s;t) X_s ds - \int_0^t X_s^\top \phi(s;t) a(s) X_s ds \\
&\quad + \int_0^t X_s^\top c(s)^\top (\sigma(s)\sigma(s)^\top)^{-1} c(s) X_s ds + \int_0^t b(s)^\top \phi(s;t) b(s) ds \\
&=2 \int_0^t X_s^\top \phi(s;t) dX_s - \int_0^t X_s^\top \phi(s;t) b(s) b(s)^\top \phi(s;t) X_s ds \\
&\quad - 2 \int_0^t X_s^\top \phi(s;t) a(s) X_s ds \\
&\quad + \int_0^t X_s^\top c(s)^\top (\sigma(s)\sigma(s)^\top)^{-1} c(s) X_s ds + \int_0^t b(s)^\top \phi(s;t) b(s) ds.
\end{align*}
$$
Therefore, noting that $\phi(t; t) = 0$, it holds

$$ \int_{0}^{t} X_{s}^{\top} \phi(s; t) dX_{s} = \frac{1}{2} \int_{0}^{t} X_{s}^{\top} \phi(s; t) b(s) b(s)^{\top} \phi(s; t) X_{s} ds \\ + \int_{0}^{t} X_{s}^{\top} \phi(s; t) a(s) X_{s} ds \\ - \frac{1}{2} \int_{0}^{t} X_{s}^{\top} c(s)^{\top} (\sigma(s) \sigma(s)^{\top})^{-1} c(s) X_{s} ds \\ - \frac{1}{2} \int_{0}^{t} b(s)^{\top} \phi(s; t) b(s) ds - \frac{1}{2} X_{0}^{\top} \phi(0; t) X_{0}. $$

Together with (3.43), we obtain

$$ \frac{d\mu_{\tilde{\xi}}}{d\mu_X}(X.) = \exp\left( -\frac{1}{2} \int_0^t X_s^\top c(s)^\top (\sigma(s)\sigma(s)^\top)^{-1} c(s) X_s ds \right. \\ \left. - \frac{1}{2} \int_0^t b(s)^\top \phi(s; t) b(s) ds - \frac{1}{2} X_0^\top \phi(0; t) X_0 \right). \quad (3.44) $$

Hence, it follows that

$$ z_t(y) = \exp \left( X_t^\top y(t) - \int_0^t y(s)^\top dX_s \right. \\ \left. + \frac{1}{2} \int_0^t b(s)^\top \phi(s; t) b(s) ds + \frac{1}{2} X_0^\top \phi(0; t) X_0 \right) \frac{d\mu_{\tilde{\xi}}}{d\mu_X}(X.) $$

and

$$ E_Q[z_t(y)] = E \left[ \exp \left( \tilde{\xi}_{t;t}^\top y(t) - \int_0^t y(s)^\top d\tilde{\xi}_{s;t} \right) \right. \\ \left. + \frac{1}{2} \int_0^t b(s)^\top \phi(s; t) b(s) ds + \frac{1}{2} X_0^\top \phi(0; t) X_0 \right) \right]. \quad (3.45) $$

Furthermore, note that $\tilde{\xi}_{s,t}$ in (3.42) can be written as

$$ \tilde{\xi}_{s,t} = \alpha(s; t) X_0 + \bar{\xi}_{s,t} \quad (3.46) $$

(for the definitions of $\bar{\xi}$ and $\alpha$, see (3.38) and (3.39)). Substituting this
expression into (3.45), we obtain

$$
\begin{align}
& E_Q[z_t(y)] \nonumber \\
&= \exp \left( \frac{1}{2} \int_0^t b(s)^\top \phi(s; t) b(s) ds \right) \nonumber \\
&\quad \times E \left[ \exp \left( X_0^\top \alpha(t; t)^\top y(t) \right) \right. \nonumber \\
&\qquad \left. - \int_0^t y(s)^\top \left\{ a(s) + b(s) b(s)^\top \phi(s; t) \right\} \alpha(s; t) ds X_0 + \frac{1}{2} X_0^\top \phi(0; t) X_0 \right] \nonumber \\
&\quad \times E \left[ \exp \left( \bar{\xi}_{t;t}^\top y(t) - \int_0^t y(s)^\top d\bar{\xi}_{s;t} \right) \right]. \tag{3.47}
\end{align}
$$

For the first expectation on the right-hand side, let $X_0 = \mu + \Sigma^{\frac{1}{2}}U$, where $U$ is a standard normal random variable, $\mu \in \mathbb{R}^{d_1}$, and $\Sigma \in M_{d_1}(\mathbb{R})$. Then, by straightforward calculation, we obtain

$$
\begin{align}
& E \left[ \exp \left( X_0^\top \alpha(t; t)^\top y(t) \right) \right. \nonumber \\
& \qquad \left. - \int_0^t y(s)^\top \left\{ a(s) + b(s) b(s)^\top \phi(s; t) \right\} \alpha(s; t) ds X_0 + \frac{1}{2} X_0^\top \phi(0; t) X_0 \right] \nonumber \\
&= \sqrt{\det |I_{d_1} - \Sigma^{\frac{1}{2}} \phi(0; t) \Sigma^{\frac{1}{2}}|} \exp \left( \frac{1}{2} \beta(t; y)^\top \Sigma^{\frac{1}{2}} (I_{d_1} - \Sigma^{\frac{1}{2}} \phi(0; t) \Sigma^{\frac{1}{2}})^{-1} \Sigma^{\frac{1}{2}} \beta(t; y) \right) \nonumber \\
& \quad \times \exp \left( \mu^\top \alpha(t; t)^\top y(t) \right. \nonumber \\
& \qquad \left. - \int_0^t y(s)^\top \left\{ a(s) + b(s) b(s)^\top \phi(s; t) \right\} \alpha(s; t) ds \mu + \frac{1}{2} \mu^\top \phi(0; t) \mu \right). \tag{3.48}
\end{align}
$$

Here, we used the negative semi-definiteness of $\phi(0; t)$.

For the second expectation, since $\bar{\xi}_{t;t}^\top y(t) - \int_0^t y(s)^\top d\bar{\xi}_{s;t}$ is a Gaussian random variable with mean 0, it follows that

$$
\begin{align}
& E \left[ \exp \left( \bar{\xi}_{t;t}^\top y(t) - \int_0^t y(s)^\top d\bar{\xi}_{s;t} \right) \right] \nonumber \\
&= \exp \left( \frac{1}{2} E \left[ \left( \bar{\xi}_{t;t}^\top y(t) - \int_0^t y(s)^\top d\bar{\xi}_{s;t} \right)^2 \right] \right) \nonumber \\
&= \exp \left( \frac{1}{2} y(t)^\top E[\bar{\xi}_{t;t} \bar{\xi}_{t;t}^\top] y(t) + \frac{1}{2} \int_0^t y(s)^\top b(s) b(s)^\top y(s) ds \right. \nonumber \\
& \qquad \left. - \int_0^t y(s)^\top b(s) b(s)^\top y(t) ds \right). \tag{3.49}
\end{align}
$$
Putting (3.47), (3.48) and (3.49) together, we obtain (3.40).

In the same way as (3.45) and (3.47), we can show that

$$
\begin{align*}
E_Q[X_{u_1} \otimes \cdots \otimes X_{u_n} z_t(y)]
&= E\left[\check{\xi}_{u_1;t} \otimes \cdots \otimes \check{\xi}_{u_n;t} \exp\left(\check{\xi}_{t;t}^\top y(t) - \int_0^t y(s)^\top d\check{\xi}_{s;t}\right)\right. \\
&\qquad \left. + \frac{1}{2} \int_0^t b(s)^\top \phi(s;t) b(s) ds + \frac{1}{2} X_0^\top \phi(0;t) X_0\right] \\
&= \exp\left(\frac{1}{2} \int_0^t b(s)^\top \phi(s;t) b(s) ds\right) \\
&\qquad \times E\left[\left\{\alpha(u_1;t) X_0 + \bar{\xi}_{u_1;t}\right\} \otimes \cdots \otimes \left\{\alpha(u_1;t) X_0 + \bar{\xi}_{u_1;t}\right\} \times \exp\left(X_0^\top \alpha(t;t)^\top y(t)\right)\right. \\
&\qquad \left. - \int_0^t y(s)^\top \left\{a(s) + b(s) b(s)^\top \phi(s;t)\right\} \alpha(s;t) ds X_0 + \frac{1}{2} X_0^\top \phi(0;t) X_0\right) \\
&\qquad \times \exp\left(\bar{\xi}_{t;t}^\top y(t) - \int_0^t y(s)^\top d\bar{\xi}_{s;t}\right).
\end{align*}
$$

From this expression, a straightforward calculation yields (3.41) by using
the independence of $X_0$ and $\{\bar{\xi}_s\}_{0\le s\le t}$.

<figure><img src="image_2.png" /></figure>

Now we introduce discretized versions of $Z_t$ and $\{X_s\}_{0\le s\le t}$ by

$$
Z_t^{(n)} = \exp \left( X_t^\top \hat{Y}_t - \sum_{i=1}^n \hat{Y}_{s_{i-1}}^\top (X_{s_i} - X_{s_{i-1}}) \right. \\
\left. - \frac{1}{2} \int_0^t X_s^\top c(s)^\top (\sigma(s)\sigma(s)^\top)^{-1} c(s) X_s ds \right), \quad (3.50)
$$

and

$$
X_s^{(n)} = \sum_{i=1}^{n} X_{s_{i-1}} 1_{[s_{i-1}, s_i]}(s) + X_t 1\{t\}(s) = X_{\frac{t}{n}[\frac{s}{t}n]} \quad (3.51)
$$

$$
\text{where } s_i = s_i^n = \frac{i}{n} t.
$$

Then the previous lemma gives the following results.

**Lemma 3.4.** For any fixed $u_1, \dots, u_k \in [0, t]$, it holds that

$$
\frac{E_Q[X_{u_1}^{(n)} \otimes \cdots \otimes X_{u_k}^{(n)} Z_t^{(n)} | \mathcal{Y}_t]}{E_Q[Z_t^{(n)} | \mathcal{Y}_t]} \xrightarrow{L^2} \frac{E_Q[X_{u_1} \otimes \cdots \otimes X_{u_k} Z_t | \mathcal{Y}_t]}{E_Q[Z_t | \mathcal{Y}_t]} \quad (3.52)
$$

as $n \to \infty$.
Proof. First, recall (3.36), (3.35), and (3.37), allowing us to write $Z_t = z_t(\hat{Y})$. Then, as in the proof of Proposition 2.3, Lemma 2.2 yields

$$E_Q[Z_t|\mathcal{Y}_t] = E_Q[z_t(y)]|_{y=\hat{Y}}$$

and

$$E_Q[X_{u_1} \otimes \cdots \otimes X_{u_k} Z_t|\mathcal{Y}_t] = E_Q[X_{u_1} \otimes \cdots \otimes X_{u_k} z_t(y)]|_{y=\hat{Y}}$$

Furthermore, if we define

$$\hat{Y}^{(n)}(s) = \sum_{i=1}^{n} \hat{Y}_{s_{i-1}} 1_{[s_{i-1}, s_i)}(s) + \hat{Y}_t 1_{\{t\}}(s),$$

then we have $Z_t^{(n)} = z_t(\hat{Y}^{(n)})$, and it follows that

$$E_Q[Z_t^{(n)}|\mathcal{Y}_t] = E_Q[z_t(y)]|_{y=\hat{Y}^{(n)}}$$

and

$$E_Q[X_{u_1}^{(n)} \otimes \cdots \otimes X_{u_k}^{(n)} Z_t^{(n)}|\mathcal{Y}_t] = E_Q[X_{u_1}^{(n)} \otimes \cdots \otimes X_{u_k}^{(n)} z_t(y)]|_{y=\hat{Y}^{(n)}}$$

Therefore, the desired result follows from (3.41) and the convergence

$$\sup_{0 \le s \le t} E_Q \left[ \left| \hat{Y}_s - \hat{Y}_s^{(n)} \right|^p \right] \to 0 \quad (n \to \infty)$$

for $p \ge 1$.

<figure><img src="image_3.png" /></figure>

**Lemma 3.5.** For every $s, u \in [0, t]$, we have

$$\tilde{E}_t[X_s] = E[X_s] + \int_0^t \mathrm{Cov}(\xi_s, \xi_u)c(u)^\top(\sigma(u)\sigma(u)^\top)^{-1}(dY_u - c(u)E[X_u]du) \quad (3.53)$$

and

$$\tilde{E}_t[X_s \otimes X_u] - \tilde{E}_t[X_s] \otimes \tilde{E}_t[X_u] = \mathrm{Cov}(\xi_s, \xi_u). \quad (3.54)$$

Furthermore, for any $u_1, \cdots, u_k \in [0, t]$, the third and higher cumulants of $X_{u_1}, \cdots, X_{u_k}$ under $\tilde{P}_t$ are almost surely zero.

**Proof.** First, let us write

$$z_t^{(n)}(y_0, \cdots, y_n) = \exp \left( X_t^\top y_n - \sum_{i=1}^n y_{i-1}^\top (X_{s_i} - X_{s_{i-1}}) \right. \\ \left. - \frac{1}{2} \int_0^t X_s^\top c(s)^\top (\sigma(s)\sigma(s)^\top)^{-1} c(s) X_s ds \right),$$
and consider the function

$$F(y_0, \dots, y_n) = E_Q[z_t^{(n)}(y_0, \dots, y_n)]$$

for $y_0, \dots, y_n \in \mathbb{R}^{d_1}$. Then it is easy to verify that

$$\begin{aligned} \partial_{y_i} F(y_0, \dots, y_n) &= -E_Q\left[(X_{s_{i+1}} - X_{s_i}) z_t^{(n)}(y_0, \dots, y_n)\right] \quad (i = 0, 1, \dots, n), \end{aligned}$$

where $X_{s_{n+1}} = 0$ for convenience, and higher derivatives can be calculated in the same manner. Therefore, we have

$$\begin{aligned} & \sum_{i,j=0}^{n} \partial_{y_i} \otimes \partial_{y_j} \log F(y_0, \dots, y_n) 1_{[0,s_{i+1})}(s) 1_{[0,s_{j+1})}(u) \\ &= \sum_{i,j=0}^{n} \frac{E_Q[(X_{s_{i+1}} - X_{s_i}) \otimes (X_{s_{j+1}} - X_{s_j}) z_t^{(n)}(y_0, \dots, y_n)]}{E_Q[z_t^{(n)}(y_0, \dots, y_n)]} 1_{[0,s_{i+1})}(s) 1_{[0,s_{j+1})}(u) \\ & \quad - \sum_{i,j=0}^{n} \frac{E_Q[(X_{s_{i+1}} - X_{s_i}) z_t^{(n)}(y_0, \dots, y_n)]}{E_Q[z_t^{(n)}(y_0, \dots, y_n)]} 1_{[0,s_{i+1})}(s) \\ & \quad \otimes \frac{E_Q[(X_{s_{j+1}} - X_{s_j}) z_t^{(n)}(y_0, \dots, y_n)]}{E_Q[z_t^{(n)}(y_0, \dots, y_n)]} 1_{[0,s_{j+1})}(u) \\ &= \frac{E_Q[X_s^{(n)} \otimes X_u^{(n)} z_t^{(n)}(y_0, \dots, y_n)]}{E_Q[z_t^{(n)}(y_0, \dots, y_n)]} \\ & \quad - \frac{E_Q[X_s^{(n)} z_t^{(n)}(y_0, \dots, y_n)] \otimes E_Q[X_u^{(n)} z_t^{(n)}(y_0, \dots, y_n)]}{E_Q[z_t^{(n)}(y_0, \dots, y_n)]^2}, \end{aligned} \quad (3.55)$$

where $X^{(n)}$ is defined in (3.51).

On the other hand, it follows from (3.40) that $\log F(y_1, \dots, y_n)$ is a quadratic form of $y_0, \dots, y_n$, and thus $\partial_{y_i} \otimes \partial_{y_j} \log F(y_0, \dots, y_n)$ does not depend on $y_0, \dots, y_n$. Thus (3.55) does not depend on $y_0, \dots, y_n$, and we have

$$\begin{aligned} & \frac{E_Q[X_s^{(n)} \otimes X_u^{(n)} z_t^{(n)}(y_0, \dots, y_n)]}{E_Q[z_t^{(n)}(y_0, \dots, y_n)]} \\ & \quad - \frac{E_Q[X_s^{(n)} z_t^{(n)}(y_0, \dots, y_n)] \otimes E_Q[X_u^{(n)} z_t^{(n)}(y_0, \dots, y_n)]}{E_Q[z_t^{(n)}(y_0, \dots, y_n)]^2} \\ &= \frac{E_Q[X_s^{(n)} \otimes X_u^{(n)} z_t^{(n)}(0, \dots, 0)]}{E_Q[z_t^{(n)}(0, \dots, 0)]} \\ & \quad - \frac{E_Q[X_s^{(n)} z_t^{(n)}(0, \dots, 0)] \otimes E_Q[X_u^{(n)} z_t^{(n)}(0, \dots, 0)]}{E_Q[z_t^{(n)}(0, \dots, 0)]^2}. \end{aligned}$$
Substituting $(\hat{Y}_{s_0}, \dots, \hat{Y}_{s_n})$ into $(y_0, \dots, y_n)$, we obtain

$$
\begin{align}
& \frac{E_Q[X_s^{(n)} \otimes X_u^{(n)} Z_t^{(n)} | \mathcal{Y}_t]}{E_Q[Z_t^{(n)} | \mathcal{Y}_t]} \nonumber \\
& - \frac{E_Q[X_s^{(n)} Z_t^{(n)} | \mathcal{Y}_t] \otimes E_Q[X_u^{(n)} Z_t^{(n)} | \mathcal{Y}_t]}{E_Q[Z_t^{(n)} | \mathcal{Y}_t]^2} \nonumber \\
& = \frac{E_Q\left[X_s^{(n)} \otimes X_u^{(n)} \exp\left(-\frac{1}{2} \int_0^t X_s^\top c(s)^\top (\sigma(s)\sigma(s)^\top)^{-1} c(s) X_s ds\right)\right]}{E_Q\left[\exp\left(-\frac{1}{2} \int_0^t X_s^\top c(s)^\top (\sigma(s)\sigma(s)^\top)^{-1} c(s) X_s ds\right)\right]} \nonumber \\
& \quad - \frac{E_Q\left[X_s^{(n)} \exp\left(-\frac{1}{2} \int_0^t X_s^\top c(s)^\top (\sigma(s)\sigma(s)^\top)^{-1} c(s) X_s ds\right)\right]}{E_Q\left[\exp\left(-\frac{1}{2} \int_0^t X_s^\top c(s)^\top (\sigma(s)\sigma(s)^\top)^{-1} c(s) X_s ds\right)\right]} \nonumber \\
& \quad \otimes \frac{E_Q\left[X_u^{(n)} \exp\left(-\frac{1}{2} \int_0^t X_s^\top c(s)^\top (\sigma(s)\sigma(s)^\top)^{-1} c(s) X_s ds\right)\right]}{E_Q\left[\exp\left(-\frac{1}{2} \int_0^t X_s^\top c(s)^\top (\sigma(s)\sigma(s)^\top)^{-1} c(s) X_s ds\right)\right]} \tag{3.56}
\end{align}
$$

since $z_t^{(n)}(\hat{Y}_{s_0}, \dots, \hat{Y}_{s_n}) = Z_t^{(n)}$, as introduced in (3.50).

By Lemma 3.4, the left-hand side of this converges in $L^2$ to

$$
\begin{align*}
& \frac{E_Q[X_s \otimes X_u Z_t | \mathcal{Y}_t]}{E_Q[Z_t | \mathcal{Y}_t]} - \frac{E_Q[X_s Z_t | \mathcal{Y}_t] \otimes E_Q[X_u Z_t | \mathcal{Y}_t]}{E_Q[Z_t | \mathcal{Y}_t]^2} \\
& = \tilde{E}_t[X_s \otimes X_u] - \tilde{E}_t[X_s] \otimes \tilde{E}_t[X_u]
\end{align*}
$$

as $n \to \infty$. On the other hand, the $L^2$-convergence of the right-hand side directly follows from the $L^2$-convergence of $X_s^{(n)}$. Therefore, we obtain

$$
\begin{align*}
& \tilde{E}_t[X_s \otimes X_u] - \tilde{E}_t[X_s] \otimes \tilde{E}_t[X_u] \\
& = \frac{E_Q\left[X_s \otimes X_u \exp\left(-\frac{1}{2} \int_0^t X_s^\top c(s)^\top (\sigma(s)\sigma(s)^\top)^{-1} c(s) X_s ds\right) \middle| \mathcal{Y}_t\right]}{E_Q\left[\exp\left(-\frac{1}{2} \int_0^t X_s^\top c(s)^\top (\sigma(s)\sigma(s)^\top)^{-1} c(s) X_s ds\right) \middle| \mathcal{Y}_t\right]} \\
& \quad - \frac{E_Q\left[X_s \exp\left(-\frac{1}{2} \int_0^t X_s^\top c(s)^\top (\sigma(s)\sigma(s)^\top)^{-1} c(s) X_s ds\right) \middle| \mathcal{Y}_t\right]}{E_Q\left[\exp\left(-\frac{1}{2} \int_0^t X_s^\top c(s)^\top (\sigma(s)\sigma(s)^\top)^{-1} c(s) X_s ds\right) \middle| \mathcal{Y}_t\right]} \\
& \quad \otimes \frac{E_Q\left[\exp\left(-\frac{1}{2} \int_0^t X_s^\top c(s)^\top (\sigma(s)\sigma(s)^\top)^{-1} c(s) X_s ds\right) \middle| \mathcal{Y}_t\right]}{E_Q\left[\exp\left(-\frac{1}{2} \int_0^t X_s^\top c(s)^\top (\sigma(s)\sigma(s)^\top)^{-1} c(s) X_s ds\right) \middle| \mathcal{Y}_t\right]}
\end{align*}
$$
$$
\begin{align}
&= \frac{E_Q \left[ \tilde{\xi}_{s;t} \otimes \tilde{\xi}_{u;t} \exp \left( \frac{1}{2} X_0^\top \phi(0;t) X_0 \right) \right]}{E_Q \left[ \exp \left( \frac{1}{2} X_0^\top \phi(0;t) X_0 \right) \right]} \nonumber \\
&\quad - \frac{E_Q \left[ \tilde{\xi}_{s;t} \exp \left( \frac{1}{2} X_0^\top \phi(0;t) X_0 \right) \right]}{E_Q \left[ \exp \left( \frac{1}{2} X_0^\top \phi(0;t) X_0 \right) \right]} \otimes \frac{E_Q \left[ \tilde{\xi}_{u;t} \exp \left( \frac{1}{2} X_0^\top \phi(0;t) X_0 \right) \right]}{E_Q \left[ \exp \left( \frac{1}{2} X_0^\top \phi(0;t) X_0 \right) \right]}. \tag{3.57}
\end{align}
$$

Here, $\tilde{\xi}$ is introduced in (3.42), and we used the density formula (3.44) in the final equality.

Now, recall the expression (3.46):

$$
\tilde{\xi}_{s;t} = \alpha(s;t)X_0 + \bar{\xi}_{s;t},
$$

where $\bar{\xi}_{s;t}$ is independent of $X_0$ and $E[\bar{\xi}_{s;t}] = 0$. Using this expression, (3.57)
becomes equivalent to

$$
\begin{align*}
& \operatorname{Cov}(\bar{\xi}_{s;t}, \bar{\xi}_{u;t}) \\
& \quad + \frac{E \left[ \alpha(s;t) X_0 \otimes \alpha(u;t) X_0 \exp \left( \frac{1}{2} X_0^\top \phi(0;t) X_0 \right) \right]}{E \left[ \exp \left( \frac{1}{2} X_0^\top \phi(0;t) X_0 \right) \right]} \\
& \quad - \frac{E \left[ \alpha(s;t) X_0 \exp \left( \frac{1}{2} X_0^\top \phi(0;t) X_0 \right) \right]}{E \left[ \exp \left( \frac{1}{2} X_0^\top \phi(0;t) X_0 \right) \right]} \otimes \frac{E \left[ \alpha(u;t) X_0 \exp \left( \frac{1}{2} X_0^\top \phi(0;t) X_0 \right) \right]}{E \left[ \exp \left( \frac{1}{2} X_0^\top \phi(0;t) X_0 \right) \right]} \\
&= \operatorname{Cov}(\bar{\xi}_{s;t}, \bar{\xi}_{u;t}) + \alpha(s;t) \Sigma^{\frac{1}{2}} (I - \Sigma^{\frac{1}{2}} \phi(0;t) \Sigma^{\frac{1}{2}})^{-1} \Sigma^{\frac{1}{2}} \alpha(u;t)^{\top} \\
&= \operatorname{Cov}(\xi_{s;t}, \xi_{u;t}),
\end{align*}
$$

Here, we used $V[X_0] = \Sigma$ and (3.12). This implies (3.54).

In the same way, since the third and higher derivatives of log F(y₀, ..., yₙ)
are zero, we have

$$
0 = \sum_{i_1, \dots, i_k = 0}^{n} \partial_{y_{i_1}} \otimes \dots \otimes \partial_{y_{i_k}} \log F(\hat{Y}_{s_0}, \dots, \hat{Y}_{s_n}) 1_{[0, s_{i_1}]}(u_1) \cdots 1_{[0, s_{i_k}]}(u_k)
$$

for $k \ge 3$. The right-hand side converges in probability to the $k$-th cumulant of $X_{u_1}, \dots, X_{u_k}$, and thus it must be almost surely zero.
Finally, to prove (3.53), let us consider

$$
\begin{align}
& \sum_{i=0}^{n} \partial_{y_i} \log F(\hat{Y}_{s_0}, \dots, \hat{Y}_{s_n}) 1_{[0, s_{i+1})}(s) \nonumber \\
&= - \sum_{i=0}^{n} \frac{E_Q[(X_{s_{i+1}} - X_{s_i}) Z_t^{(n)} | \mathcal{Y}_t]}{E_Q[Z_t^{(n)} | \mathcal{Y}_t]} 1_{[0, s_{i+1})}(s) = \frac{E_Q[X_s^{(n)} Z_t^{(n)} | \mathcal{Y}_t]}{E_Q[Z_t^{(n)} | \mathcal{Y}_t]}, \tag{3.58}
\end{align}
$$

and let $D$ be the Malliavin derivative with respect to the Brownian motion

$$
\bar{Y}_s = \int_0^s (\sigma(s)\sigma(s)^\top)^{-\frac{1}{2}} dY_s
$$

on $Q$. Since $\log F(y_0, \dots, y_n)$ is a quadratic form, $\partial_{y_i} \log F(y_0, \dots, y_n)$ is
linear. Hence, by the definition of the Malliavin derivative of smooth random
variables (Nualart 2006) and $\hat{Y}_t = \int_0^t c(s)^\top (\sigma(s)\sigma(s)^\top)^{-\frac{1}{2}} d\bar{Y}_s$, the Malliavin
derivative of (3.58) is

$$
\begin{align*}
& D_u \frac{E_Q[X_s^{(n)} Z_t^{(n)} | \mathcal{Y}_t]}{E_Q[Z_t^{(n)} | \mathcal{Y}_t]} \\
&= \sum_{i=0}^{n} (\sigma(u)\sigma(u)^{\top})^{-\frac{1}{2}} c(u) \partial_{y_i} \partial_{y_i} \log F(\hat{Y}_{s_0}, \dots, \hat{Y}_{s_n}) 1_{[0, s_i]}(u) 1_{[0, s_j]}(u) \\
&= \frac{(\sigma(u)\sigma(u)^{\top})^{-\frac{1}{2}} c(u) E[X_u^{(n)} \otimes X_s^{(n)} Z_t^{(n)} | \mathcal{Y}_t]}{E[Z_t^{(n)} | \mathcal{Y}_t]} \\
&\quad - \frac{(\sigma(u)\sigma(u)^{\top})^{-\frac{1}{2}} c(u) E[X_u^{(n)} Z_t^{(n)} | \mathcal{Y}_t] \otimes E[X_s^{(n)} Z_t^{(n)} | \mathcal{Y}_t]}{E[Z_t^{(n)} | \mathcal{Y}_t]^2}.
\end{align*}
$$

According to (3.56), this is deterministic. Thus, by the Clerk-Ocone formula
(Nualart 2006), we can write

$$
\frac{E_Q[X_s^{(n)} Z_t^{(n)} | \mathcal{Y}_t]}{E_Q[Z_t^{(n)} | \mathcal{Y}_t]} = g^{(n)}(s; t) + \int_0^t \left( \frac{E[X_u^{(n)} \otimes X_s^{(n)} Z_t^{(n)} | \mathcal{Y}_t]}{E[Z_t^{(n)} | \mathcal{Y}_t]} - \frac{E_Q[X_u^{(n)} Z_t^{(n)} | \mathcal{Y}_t] \otimes E_Q[X_s^{(n)} Z_t^{(n)} | \mathcal{Y}_t]}{E_Q[Z_t^{(n)} | \mathcal{Y}_t]^2} \right)^\top c(u)^{\top} (\sigma(u)\sigma(u)^{\top})^{-1} dY_u,
$$

where $g^{(n)}(s;t)$ is a deterministic function. Due to Lemma 3.4, for every $s, t$, $g^{(n)}(s;t)$ converges, and

$$
\left\{ \frac{E_Q[X_u^{(n)} \otimes X_s^{(n)} Z_t^{(n)} | \mathcal{Y}_t]}{E_Q[Z_t^{(n)} | \mathcal{Y}_t]} - \frac{E_Q[X_u^{(n)} Z_t^{(n)} | \mathcal{Y}_t] \otimes E_Q[X_s^{(n)} Z_t^{(n)} | \mathcal{Y}_t]}{E_Q[Z_t^{(n)} | \mathcal{Y}_t]^2} \right\}_{0 \le u \le t}
$$
converges in $L^2(\Omega; H)$ where $H = L^2([0, t])$. On the other hand, we have shown above that for each $s, u \in [0, t]$, this converges to $\text{Cov}(\xi_u, \xi_s)$ in $L^2$.

Therefore, by letting $n \to \infty$, we obtain

$$ 
\begin{aligned}
\tilde{E}_t[X_s] &= \left( \frac{E_Q[X_s Z_t | \mathcal{Y}_t]}{E_Q[Z_t | \mathcal{Y}_t]} \right) \\
&= g(s; t) + \int_0^t \text{Cov}(\xi_s, \xi_u) c(u)^\top (\sigma(u) \sigma(u)^\top)^{-1} dY_u,
\end{aligned}
$$

where $g(s; t)$ is a deterministic function. By taking expectations of both sides under $P$, it follows that

$$ 
\begin{aligned}
E[X_s](=E[E[X_s|\mathcal{Y}_t]] &= E[\tilde{E}_t[X_s]]) \\
&= g(s; t) + \int_0^t \text{Cov}(\xi_s, \xi_u) c(u)^\top (\sigma(u)\sigma(u)^\top)^{-1} c(u) E[X_u] du.
\end{aligned}
$$

This gives the expression of $g(s; t)$, which leads to the desired result.

<figure><img src="image_4.png" /></figure>

**Proof of Theorem 3.2.** Let $\varpi_t : C_t \to \mathbb{R}^{d_1+d_2}$ be the projection where $\varpi_t(w) = w(t)$, and $\mathcal{P}$ be the class of sets of the form

$$ \varpi_{u_1}^{-1}(A_1 \times B_1) \cap \cdots \cap \varpi_{u_k}^{-1}(A_k \times B_k), $$

where $A_1, \cdots, A_k \in \mathcal{B}(\mathbb{R}^{d_1})$, $B_1, \cdots, B_k \in \mathcal{B}(\mathbb{R}^{d_2})$ and $u_1, \cdots, u_k \in [0, t]$.

Furthermore, let us define $\mathcal{D}$ by

$$ \mathcal{D} = \{ A \in \mathcal{C}_t; \tilde{P}_t((X., Y.) \in A) = P((\xi., Y.) \in A | \mathcal{Y}_t) \text{ a.s.} \}. $$

Due to (3.14), (3.15) and Lemma 3.5, it follows that for fixed $u_1, \cdots, u_k \in [0, t]$, $(X_{u_1}, \cdots, X_{u_k})$ under $\tilde{P}_t$ and $(\zeta_{u_1;t}, \cdots, \zeta_{u_k;t})$ under $P(\cdot | \mathcal{Y}_t)$ have the almost surely same mean and covariance, and their third and higher cumulants are almost surely zero.

Therefore, for fixed $u_1, \cdots, u_k \in [0, t]$, there exists a null set $N$ such that the distributions of $(X_{u_1}, \cdots, X_{u_k})$ under $\tilde{P}_t(\cdot)(\omega)$ and $(\zeta_{u_1;t}, \cdots, \zeta_{u_k;t})$ under $P(\cdot | \mathcal{Y}_t)(\omega)$ are the same for $\omega \in \Omega \setminus N$. Thus we have for $A_1, \cdots, A_k \in \mathcal{B}(\mathbb{R}^{d_1})$ and $B_1, \cdots, B_k \in \mathcal{B}(\mathbb{R}^{d_2})$

$$ 
\begin{aligned}
& \tilde{P}_t(X_{u_1} \in A_1, \cdots X_{u_k} \in A_k, Y_{u_1} \in B_1, \cdots, Y_{u_k} \in B_k) \\
&= \tilde{P}_t(X_{u_1} \in A_1, \cdots X_{u_k} \in A_k) 1_{\{Y_{u_1} \in B_1, \cdots, Y_{u_k} \in B_k\}} \\
&= P(\zeta_{u_1,t} \in A_1, \cdots \zeta_{u_k,t} \in A_k | \mathcal{Y}_t) 1_{\{Y_{u_1} \in B_1, \cdots, Y_{u_k} \in B_k\}} \\
&= P(\zeta_{u_1,t} \in A_1, \cdots \zeta_{u_k,t} \in A_k, Y_{u_1} \in B_1, \cdots, Y_{u_k} \in B_k | \mathcal{Y}_t).
\end{aligned}
$$

This means $\mathcal{P} \subset \mathcal{D}$, and we obtain the desired result by the $\pi$-$\lambda$ theorem.

<figure><img src="image_5.png" /></figure>
## 3.2 Filtering and Smoothing Equations

In the previous section, we showed that {$X_s$}$_{0\le s\le t}$ is conditionally Gaussian given $Y_t$, and we obtained explicit formulas for its mean and covariance functions. In this section, we derive differential equations for these functions.

First, we consider the differential equation for $\mathrm{Cov}(\xi_{t;t}, \xi_{t;t})$, which corresponds to the covariance matrix of $X_t$ given {$Y_t$}. To this end, we define $\gamma(t)$ as the positive-semidefinite solution of the Riccati equation

$$ \frac{d}{dt}\gamma(t) = -\gamma(t)c(t)^{\top}(\sigma(t)\sigma(t)^{\top})^{-1}c(t)\gamma(t) + a(t)\gamma(t) + \gamma(t)a(t)^{\top} \quad (3.59) \\ + b(t)b(t)^{\top} $$

with $\gamma(0) = V[X_0]$. The existence of the solution is guaranteed by Theorems 2.1 and 2.2 in Potter (1965).

To avoid potential singularity issues with $\gamma(t)$, we introduce $\gamma^\epsilon(t)$ and $\xi_{s;t}^\epsilon$ as solutions to the same equations for $\gamma(t)$ and $\xi_{s;t}$, where $V[X_0]$ is replaced by $V[X_0] + \epsilon I_{d_1}$ for $\epsilon > 0$. Specifically, $\gamma^\epsilon(t)$ and $\xi_{s;t}^\epsilon$ are given as solutions to the equations

$$ \frac{d}{dt}\gamma^{\epsilon}(t) = -\gamma^{\epsilon}(t)c(t)^{\top}(\sigma(t)\sigma(t)^{\top})^{-1}c(t)\gamma^{\epsilon}(t) + a(t)\gamma^{\epsilon}(t) + \gamma^{\epsilon}(t)a(t)^{\top} \\ + b(t)b(t)^{\top} \quad (3.60) $$

and

$$ d\xi_{s;t}^{\epsilon} = \{a(s) + b(s)b(s)^{\top}\phi(s;t)\} \xi_{s;t}^{\epsilon} ds + b(s) d\tilde{V}_s, \quad (3.61) $$

where

$$ \gamma^{\epsilon}(0) = V[X_0] + \epsilon I_{d_1}, \quad (3.62) $$

and $\xi_{0;t}^\epsilon$ is a Gaussian random variable with mean zero and covariance

$$ V[\xi_{0;t}^{\epsilon}] = \{(V[X_0] + \epsilon I_{d_1})^{-1} - \phi(0;t)\}^{-1}. \quad (3.63) $$

By Corollary 1 to Theorem 2.1 in Potter (1965), $\gamma^\epsilon(t)$ is strictly positive. We can also establish the following lemma.

**Lemma 3.6.** (1) For every $t \ge 0$, $\gamma^\epsilon(s)$ is bounded for $\epsilon \in [0, 1]$ and $s \in [0, t]$.

(2) The matrix $\gamma^\epsilon(t)$ converges uniformly to $\gamma(t)$ on any compact set as $\epsilon \to +0$.
Proof. (1) Take an arbitrary $x \in \mathbb{R}^{d_1}$. Then (3.59) yields

$$
\begin{align*}
\frac{d}{dt} x^\top \gamma^\epsilon(t) x &= -x^\top \gamma^\epsilon(t) c(t)^\top (\sigma(t) \sigma(t)^\top)^{-1} c(t) \gamma^\epsilon(t) x \\
&\quad + x^\top a(t) \gamma^\epsilon(t) x + x^\top \gamma^\epsilon(t) a(t)^\top x + x^\top b(t) b(t)^\top x \\
&\le 2|x^\top a(t) \gamma^\epsilon(t)^{\frac{1}{2}} \gamma^\epsilon(t)^{\frac{1}{2}} x| + x^\top b(t) b(t)^\top x \\
&\le 2\sqrt{x^\top a(t) \gamma^\epsilon(t) a(t)^\top x} \sqrt{x^\top \gamma^\epsilon(t) x} + x^\top b(t) b(t)^\top x \\
&\le 2\|a(t)\|_2 \|\gamma^\epsilon(t)\|_2 |x|^2 + \|b(t)\|_2^2 |x|^2,
\end{align*}
$$

where $\| \cdot \|_2$ is the matrix 2-norm, and we used

$$
x^\top a(t) \gamma^\epsilon(t) a(t)^\top x = |\gamma^\epsilon(t)^{\frac{1}{2}} a(t)^\top x|^2 \le \|\gamma^\epsilon(t)\|_2 \|a(t)\|_2^2 |x|^2.
$$

From this, it follows that

$$
\|\gamma^\epsilon(t)\|_2 = \max_{|x|=1} x^\top \gamma^\epsilon(t) x \le \|\gamma^\epsilon(0)\|_2 + \int_0^t \{2\|a(s)\|_2 \|\gamma^\epsilon(s)\|_2 + \|b(s)\|_2^2\} ds.
$$

Therefore, by the Gronwall's lemma, we obtain

$$
\begin{align*}
\|\gamma^\epsilon(t)\|_2 \le & \|\gamma^\epsilon(0)\|_2 \exp \left( 2 \int_0^t \|a(r)\|_2 dr \right) \\
& + \int_0^t \exp \left( 2 \int_s^t \|a(r)\|_2 dr \right) \|b(s)\|_2^2 ds,
\end{align*}
$$

which leads to the desired result.

Statement (2) can be proven routinely using the result of (1) along with
Gronwall's lemma.
□

**Proposition 3.7.** It holds that

$$
\gamma(t) = V[\xi_{t,t}].
$$

Proof. First, assume that $\gamma(t)$ is strictly positive for every $t \ge 0$. Then we have

$$
\begin{align*}
\frac{d}{ds} (\gamma(s)^{-1}) &= - \gamma(s)^{-1} \frac{d}{ds} \gamma(s) \gamma(s)^{-1} \\
&= - \gamma(s)^{-1} b(s) b(s)^{\top} \gamma(s)^{-1} - a(s)^{\top} \gamma(s)^{-1} - \gamma(s)^{-1} a(s) \\
&\quad + c(s)^{\top} (\sigma(s) \sigma(s)^{\top})^{-1} c(s).
\end{align*}
$$

This means that $\gamma(s)^{-1}$ satisfies the same equation (3.10) as $\phi(s; t)$. Hence,
if we set $z(s; t) = \gamma(s)^{-1} - \phi(s; t)$, then it follows that

$$ 
\begin{aligned}
& \frac{d}{ds} z(s; t) \\
&= - \left\{ \gamma(s)^{-1} b(s) b(s)^{\top} \gamma(s)^{-1} - \phi(s; t)^{-1} b(s) b(s)^{\top} \phi(s; t)^{-1} \right\} \\
&\quad - a(s)^{\top} \left\{ \gamma(s)^{-1} - \phi(s; t) \right\} - \left\{ \gamma(s)^{-1} - \phi(s; t) \right\} a(s) \\
&= - z(s; t) b(s) b(s)^{\top} z(s; t) \\
&\quad - \left\{ a(s)^{\top} + \phi(s; t) b(s) b(s)^{\top} \right\} z(s; t) - z(s; t) \left\{ a(s) + b(s) b(s)^{\top} \phi(s; t) \right\}.
\end{aligned}
$$

Furthermore, if we set $w(s; t) = z(s; t)^{-1}$ (this is well-defined since $\psi(s; t)$ is negative-semidefinite), we have

$$ 
\begin{aligned}
\frac{d}{ds} w(s; t) &= - z(s; t)^{-1} \frac{d}{ds} z(s; t) z(s; t)^{-1} \\
&= b(s) b(s)^{\top} \\
&\quad + w(s; t) \left\{ a(s)^{\top} + \phi(s; t) b(s) b(s)^{\top} \right\} \\
&\quad + \left\{ a(s) + b(s) b(s)^{\top} \phi(s; t) \right\} w(s; t)
\end{aligned}
\qquad (3.64) $$

Therefore, $w(s; t)$ can be written as

$$ 
\begin{aligned}
w(s; t) = & \exp \left( \int_{0}^{s} \left\{ a(r) + b(r) b(r)^{\top} \phi(r; t) \right\} dr \right) V[\xi_{0}; t] \\
& \times \exp \left( \int_{0}^{s} \left\{ a(r)^{\top} + \phi(r; t) b(r) b(r)^{\top} \right\} dr \right) \\
& + \int_{0}^{s} \exp \left( \int_{u}^{s} \left\{ a(r) + b(r) b(r)^{\top} \phi(r; t) \right\} dr \right) \\
& \times b(u) b(u)^{\top} \exp \left( \int_{u}^{s} \left\{ a(r)^{\top} + \phi(r; t) b(r) b(r)^{\top} \right\} dr \right) du.
\end{aligned}
\qquad (3.65) $$

Here, we used $\gamma(0) = V[X_0]$ and (3.12) to establish

$$ w(0; s) = (V[X_0]^{-1} - \phi(0; t))^{-1} = V[\xi_0; t]. $$

On the other hand, since $\{\xi_{s;t}\}_{0\le s\le t}$ is the solution of (3.11), it can be expressed as

$$ 
\begin{aligned}
\xi_{s;t} = & \exp \left( \int_{0}^{s} \left\{ a(r) + b(r) b(r)^{\top} \phi(r; t) \right\} dr \right) \xi_{0;t} \\
& + \int_{0}^{s} \exp \left( \int_{u}^{s} \left\{ a(r) + b(r) b(r)^{\top} \phi(r; t) \right\} dr \right) b(u) d\tilde{V}_{u}.
\end{aligned}
\qquad (3.66) $$
Hence, it follows immediately from (3.65) that $V[\xi_{s;t}] = w(s;t)$. In particular, since

$$w(s;t) = \{\gamma(s)^{-1} - \phi(s;t)\}^{-1} \quad (3.67)$$

and $\phi(t;t) = 0$, we have $w(t;t) = \gamma(t)$, which gives the desired result.

In the case where $\gamma(t)$ becomes singular for some $t$, we replace $V[X_0]$ with $V[X_0] + \epsilon I_{d_1}$ for $\epsilon > 0$, yielding

$$\gamma^{\epsilon}(t) = V[\xi_{t;t}^{\epsilon}],$$

since $\gamma^{\epsilon}(t)$ is strictly positive. We obtain the conclusion by letting $\epsilon \to 0$. $\square$

**Proposition 3.8.** Let $0 \le u \le s \le t$. If $\gamma(r)$ is non-singular for every $r \in [0, t]$, it holds that

$$\mathrm{Cov}(\xi_{s;t}, \xi_{u;t}) = V[\xi_{s;t}] \exp \left( - \int_{u}^{s} \left\{ a(r)^{\top} + \gamma(r)^{-1} b(r) b(r)^{\top} \right\} dr \right).$$

*Proof.* Since $\xi_{s;t}$ is given by the explicit formula (3.66), we can express

$$\begin{align}
\mathrm{Cov}(\xi_{s;t}, \xi_{u;t})
&= \exp \left( \int_{0}^{s} \left\{ a(r) + b(r)b(r)^{\top}\phi(r;t) \right\} dr \right) V[\xi_{0;t}] \nonumber \\
&\quad \exp \left( \int_{0}^{u} \left\{ a(r)^{\top} + \phi(r;t)b(r)b(r)^{\top} \right\} dr \right) \nonumber \\
&\quad + \int_{0}^{u} \exp \left( \int_{v}^{s} \left\{ a(r) + b(r)b(r)^{\top}\phi(r;t) \right\} dr \right) \nonumber \\
&\qquad \times b(v)b(v)^{\top} \exp \left( \int_{v}^{u} \left\{ a(r)^{\top} + \phi(r;t)b(r)b(r)^{\top} \right\} dr \right) dv \nonumber \\
&= \exp \left( \int_{u}^{s} \left\{ a(r) + b(r)b(r)^{\top}\phi(r;t) \right\} dr \right) w(u;t), \tag{3.68}
\end{align}$$

where $w(u;t) = V[\xi_{u;t}]$ is given by (3.65). Thus, it follows from (3.64) and (3.67) that

$$\begin{align*}
d_u \mathrm{Cov}(\xi_{s;t}, \xi_{u;t})
&= \exp \left( \int_u^s \left\{ a(r) + b(r)b(r)^\top \phi(r;t) \right\} dr \right) \\
&\quad \times \left[ - \left\{ a(u) + b(u)b(u)^\top \phi(u;t) \right\} w(u;t) + \frac{\partial}{\partial u} w(u;t) \right] du \\
&= \exp \left( \int_u^s \left\{ a(r) + b(r)b(r)^\top \phi(r;t) \right\} dr \right) \\
&\quad \times \left[ - \left\{ a(u) + b(u)b(u)^\top \phi(u;t) \right\} w(u;t) + b(u)b(u)^\top \right]
\end{align*}$$
$$
\begin{align*}
&+ w(u; t) \{a(u)^{\top} + \phi(u; t)b(u)b(u)^{\top}\} \\
&+ \{a(u) + b(u)b(u)^{\top}\phi(u; t)\} w(u; t)] du \\
&= \exp\left(\int_{u}^{s} \{a(r) + b(r)b(r)^{\top}\phi(r; t)\} dr\right) w(u, t) \\
&\quad \times \{w(u; t)^{-1}b(u)b(u)^{\top} + a(u)^{\top} + \phi(u; t)b(u)b(u)^{\top}\} du \\
&= \operatorname{Cov}(\xi_{s;t}, \xi_{u;t}) \\
&\quad \times [\{\gamma(u)^{-1} - \phi(u; t)\} b(u)b(u)^{\top} + a(u)^{\top} + \phi(u; t)b(u)b(u)^{\top}] du \\
&= \operatorname{Cov}(\xi_{s;t}, \xi_{u;t}) \{a(u)^{\top} + \gamma(u)^{-1}b(u)b(u)^{\top}\} du,
\end{align*}
$$

which implies the conclusion.

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

**Proposition 3.9.** For every $s \ge 0$, it holds that

$$
d_t \mathrm{Cov}(\xi_{t;t}, \xi_{s;t}) = \{a(t) - \gamma(t)c(t)^{\top}(\sigma(t)\sigma(t)^{\top})^{-1}c(t)\} \mathrm{Cov}(\xi_{t;t}, \xi_{s;t})dt.
$$

Proof. First, assume that $\gamma(t)$ is non-singular for every $t \ge 0$. In this case, by Propositions 3.7 and 3.8, we can write

$$
\mathrm{Cov}(\xi_{t;t}, \xi_{s;t}) = \gamma(t) \exp \left( - \int_s^t \{a(r)^\top + \gamma(r)^{-1} b(r) b(r)^\top\} dr \right).
$$

Hence, it holds

$$
\begin{align*}
d_t \mathrm{Cov}(\xi_{t;t}, \xi_{s;t})
&= \left[ \frac{d}{dt} \gamma(t) - \gamma(t) \{a(t)^\top + \gamma(t)^{-1} b(t) b(t)^\top \} \right] \\
&\quad \times \exp \left( - \int_s^t \{a(r)^\top + \gamma(r)^{-1} b(r) b(r)^\top \} dr \right) dt \\
&= \left\{ -\gamma(t) c(t)^\top (\sigma(t) \sigma(t)^\top)^{-1} c(t) \gamma(t) + a(t) \gamma(t) \right\} \\
&\quad \times \exp \left( - \int_s^t \{a(r)^\top + \gamma(r)^{-1} b(r) b(r)^\top \} dr \right) dt \\
&= \left\{ -\gamma(t) c(t)^\top (\sigma(t) \sigma(t)^\top)^{-1} c(t) + a(t) \right\} \mathrm{Cov}(\xi_{t;t}, \xi_{s;t}) dt,
\end{align*}
$$

which implies the conclusion.

For the case where $\gamma(t)$ is singular, we have

$$
d_t \mathrm{Cov}(\xi_{t;t}^{\epsilon}, \xi_{s;t}^{\epsilon}) = \{-\gamma^{\epsilon}(t)c(t)^{\top}(\sigma(t)\sigma(t)^{\top})^{-1}c(t) + a(t)\} \mathrm{Cov}(\xi_{t;t}^{\epsilon}, \xi_{s;t}^{\epsilon})dt
$$

and thus

$$
\mathrm{Cov}(\xi_{t;t}^{\epsilon}, \xi_{s;t}^{\epsilon}) = \exp \left( \int_s^t \{a(r) - \gamma^{\epsilon}(r)c(r)^{\top}(\sigma(r)\sigma(r)^{\top})^{-1}c(r)\} dr \right) \gamma^{\epsilon}(s). \tag{3.69}
$$
By the uniform convergence of $\gamma^\epsilon$ on any compact set, we can take the limit $\epsilon \to 0$ to obtain

$$ \mathrm{Cov}(\xi_{t;t}, \xi_{s;t}) = \exp \left( \int_s^t \left\{ a(r) - \gamma(r)c(r)^\top (\sigma(r)\sigma(r)^\top)^{-1}c(r) \right\} dr \right) \gamma(s), $$

which leads to the desired result. □

Now, let us define

$$ \mu_{s;t} = E[X_s] + \int_0^t \mathrm{Cov}(\xi_{s;t}, \xi_{u;t}) c(u)^\top (\sigma(u)\sigma(u)^\top)^{-1} (dY_u - c(u)E[X_u]du). \quad (3.70) $$

Recalling Theorem 3.2, we find that for every $0 \le s \le t$,

$$ \mu_{s;t} = E[\zeta_{s;t} | \mathcal{Y}_t] = E[X_s | \mathcal{Y}_t] \quad \text{a.s.} $$

The previous proposition provides the stochastic differential equation for $\mu_{t,t}$, which is known as the Kalman-Bucy filter.

**Proposition 3.10. (Kalman-Bucy filter)**

The process $\{\mu_{t,t}\}_{t \ge 0}$ satisfies the stochastic differential equation

$$ d\mu_{t,t} = a(t)\mu_{t,t}dt + \gamma(t)c(t)^\top(\sigma(t)\sigma(t)^\top)^{-1}\{dY_t - c(t)\mu_{t,t}dt\}. $$

*Proof.* Set

$$ \tilde{\mu}_{t;t} = \mu_{t;t} - E[X_t], \quad \tilde{Y}_t = Y_t - \int_0^t c(u)E[X_u]du. \quad (3.71) $$

Then according to Proposition 3.9 and (3.70), we have

$$ \begin{aligned} \tilde{\mu}_{t,t} &= \int_0^t \mathrm{Cov}(\xi_{t;t}, \xi_{u;t}) c(u)^\top (\sigma(u)\sigma(u)^\top)^{-1} d\tilde{Y}_u \\ &= \int_0^t \left[ \gamma(u) + \int_u^t \left\{ -\gamma(s)c(s)^\top (\sigma(s)\sigma(s)^\top)^{-1} c(s) + a(s) \right\} \mathrm{Cov}(\xi_{s;s}, \xi_{u;s}) ds \right] \\ &\quad \times c(u)^\top (\sigma(u)\sigma(u)^\top)^{-1} d\tilde{Y}_u \\ &= \int_0^t a(s) \int_0^s \mathrm{Cov}(\xi_{s;s}, \xi_{u;s}) c(u)^\top (\sigma(u)\sigma(u)^\top)^{-1} d\tilde{Y}_u ds \\ &\quad + \int_0^t \gamma(u)c(u)^\top (\sigma(u)\sigma(u)^\top)^{-1} d\tilde{Y}_u \\ &\quad - \int_0^t \gamma(s)c(s)^\top (\sigma(s)\sigma(s)^\top)^{-1} c(s) \int_0^s \mathrm{Cov}(\xi_{s;s}, \xi_{u;s}) c(u)^\top (\sigma(u)\sigma(u)^\top)^{-1} d\tilde{Y}_u ds \\ &= \int_0^t a(s)\tilde{\mu}_{s,s}ds + \int_0^t \gamma(u)c(u)^\top (\sigma(u)\sigma(u)^\top)^{-1} \{d\tilde{Y}_u - c(s)\tilde{\mu}_{s,s}ds\}, \end{aligned} $$
which gives the equation

$$d\tilde{\mu}_{t,t} = a(t)\tilde{\mu}_{t,t}dt + \gamma(t)c(t)^{\top}(\sigma(t)\sigma(t)^{\top})^{-1}\{d\tilde{Y}_t - c(t)\tilde{\mu}_{t,t}dt\}.$$

By (3.71) and $dE[X_t] = a(t)E[X_t]dt$, this is equivalent to the desired result.

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

We can also derive the backward algorithm for fixed-interval smoothing, commonly referred to as the Rauch-Tung-Striebel smoother.

**Proposition 3.11. (Rauch-Tung-Striebel smoother)**

If $\gamma(r)$ is strictly positive for every $r \in [0, t]$, it holds

$$\mu_{s;t} = \mu_{t;t} - \int_{s}^{t} \{a(r)\mu_{r;t} + b(r)b(r)^{\top}\gamma(r)^{-1}(\mu_{r;t} - \mu_{r;r})\} dr$$

for every $0 \le s \le t$.

*Proof.* Set

$$\tilde{\mu}_{t;t} = \mu_{t;t} - E[X_t], \quad \hat{Y}_t = \int_{0}^{t} c(u)^{\top}(\sigma(u)\sigma(u)^{\top})^{-1}(dY_u - c(u)E[X_u]du).$$

According to Proposition 3.8, for $s \le u$ we have

$$d_s \mathrm{Cov}(\xi_{s;t}, \xi_{u;t}) = \{a(s) + b(s)b(s)^{\top}\gamma(s)^{-1}\} \mathrm{Cov}(\xi_{s;t}, \xi_{u;t})ds$$

On the other hand, it follows from (3.68) that for $s \ge u$

$$d_s \mathrm{Cov}(\xi_{s;t}, \xi_{u;t}) = \{a(s) + b(s)b(s)^{\top}\phi(s;t)\} \mathrm{Cov}(\xi_{s;t}, \xi_{u;t})ds. \quad (3.72)$$

Therefore, it holds

$$
\begin{align*}
\tilde{\mu}_{s;t} &= \int_0^t \left\{ \mathrm{Cov}(\xi_{t;t}, \xi_{u;t}) - \int_s^t \frac{\partial}{\partial r} \mathrm{Cov}(\xi_{r;t}, \xi_{u;t}) dr \right\} d\hat{Y}_u \\
&= \tilde{\mu}_{t;t} - \int_s^t \int_0^t \frac{\partial}{\partial r} \mathrm{Cov}(\xi_{r;t}, \xi_{u;t}) d\hat{Y}_u dr \\
&= \tilde{\mu}_{t;t} - \int_s^t \int_0^r \{a(r) + b(r)b(r)^{\top}\phi(r;t)\} \mathrm{Cov}(\xi_{r;t}, \xi_{u;t}) d\hat{Y}_u dr \\
&\quad - \int_s^t \int_r^t \{a(r) + b(r)b(r)^{\top}\gamma(r)^{-1}\} \mathrm{Cov}(\xi_{r;t}, \xi_{u;t}) d\hat{Y}_u dr \\
&= \tilde{\mu}_{t;t} - \int_s^t \{a(r) + b(r)b(r)^{\top}\gamma(r)^{-1}\} \int_0^r \mathrm{Cov}(\xi_{r;t}, \xi_{u;t}) d\hat{Y}_u dr \\
&\quad + \int_s^t \int_0^r b(r)b(r)^{\top}\{\gamma(r)^{-1} - \phi(r;t)\} \mathrm{Cov}(\xi_{r;t}, \xi_{u;t}) d\hat{Y}_u dr \\
&= \tilde{\mu}_{t;t} - \int_s^t \{a(r) + b(r)b(r)^{\top}\gamma(r)^{-1}\} \tilde{\mu}_{r;t} dr \\
&\quad + \int_s^t b(r)b(r)^{\top} \int_0^r V[\xi_{r;t}]^{-1} \mathrm{Cov}(\xi_{r;t}, \xi_{u;t}) d\hat{Y}_u dr.
\end{align*}
$$
In the last equation, we utilized the relation (3.67).

Furthermore, according to Proposition 3.8, $V[\xi_{r;t}]^{-1}\text{Cov}(\xi_{r;t}, \xi_{u;t})$ does not depend on $t$. Hence, we have

$$ \int_{0}^{r} V[\xi_{r;t}]^{-1} \text{Cov}(\xi_{r;t}, \xi_{u;t}) d\hat{Y}_{u} \\ = \int_{0}^{r} V[\xi_{r;r}]^{-1} \text{Cov}(\xi_{r;r}, \xi_{u;r}) d\hat{Y}_{u} = \gamma(r)^{-1} \tilde{\mu}_{r,r}. $$

Therefore, we obtain

$$ \tilde{\mu}_{s;t} = \tilde{\mu}_{t;t} - \int_{s}^{t} \{a(r) + b(r)b(r)^{\top}\gamma(r)^{-1}\} \tilde{\mu}_{r;t} dr \\ + \int_{s}^{t} b(r)b(r)^{\top}\gamma(r)^{-1}\tilde{\mu}_{r,r} dr. $$

Finally, we obtain the desired result by using $\tilde{\mu}_{t;t} = \mu_{t;t} - E[X_t]$ and $dE[X_t] = a(t)E[X_t]dt$. $\square$

Next, we derive the differential equations for $\text{Cov}(\xi_{s;t}, \xi_{u;t})$ and $\mu_{s;t}$ with respect to $t$.

**Proposition 3.12.** It holds for $s, u \ge 0$ and $t \ge s \lor u$ that

$$ d_t \text{Cov}(\xi_{s;t}, \xi_{u;t}) = - \text{Cov}(\xi_{t;t}, \xi_{s;t})^\top c(t)^\top (\sigma(t)\sigma(t)^\top)^{-1} c(t) \text{Cov}(\xi_{t;t}, \xi_{u;t}) dt. $$

*Proof.* First assume that $\gamma(t)$ is positive for every $t \ge 0$. By exchanging $\gamma(s)^{-1}$ and $\phi(s;t)$ in (3.64), and recalling $w(s;t) = V[\xi_{s;t}]$, it holds

$$ \frac{d}{ds} V[\xi_{s;t}] = -b(s)b(s)^{\top} \\ + V[\xi_{s;t}] \{a(s)^{\top} + \gamma(s)^{-1}b(s)b(s)^{\top}\} \\ + \{a(s)^{\top} + \gamma(s)^{-1}b(s)b(s)^{\top}\} V[\xi_{s;t}]. $$

Thus, we obtain for $s \le t$

$$ V[\xi_{s;t}] = \exp \left( - \int_{s}^{t} \{a(r) + b(r)b(r)^{\top}\gamma(r)^{-1}\} dr \right) \gamma(t) \\ \times \exp \left( - \int_{s}^{t} \{a(r)^{\top} + \gamma(r)^{-1}b(r)b(r)^{\top}\} dr \right) \\ + \int_{s}^{t} \exp \left( \int_{u}^{s} \{a(r) + b(r)b(r)^{\top}\gamma(r)^{-1}\} dr \right) \\ \times b(u)b(u)^{\top} \exp \left( \int_{u}^{s} \{a(r)^{\top} + \gamma(r)^{-1}b(r)b(r)^{\top}\} dr \right) du, $$
where we used $V[\xi_{t;t}] = \gamma(t)$. Differentiating this yields

$$
\begin{align*}
d_t V[\xi_{s;t}]
&= \exp \left( - \int_s^t \left\{ a(r) + b(r)b(r)^\top \gamma(r)^{-1} \right\} dr \right) \\
&\quad \times \left[ - \left\{ a(t) + b(t)b(t)^\top \gamma(t)^{-1} \right\} \gamma(t) \right. \\
&\qquad \left. - \gamma(t) \left\{ a(t)^\top + \gamma(t)^{-1}b(t)b(t)^\top \right\} + \frac{d}{dt}\gamma(t) + b(t)b(t)^\top \right] \\
&\quad \times \exp \left( - \int_s^t \left\{ a(r)^\top + \gamma(r)^{-1}b(r)b(r)^\top \right\} dr \right) dt \\
&= - \exp \left( - \int_s^t \left\{ a(r) + b(r)b(r)^\top \gamma(r)^{-1} \right\} dr \right) \\
&\quad \times \gamma(t)c(t)^\top (\sigma(t)\sigma(t)^\top)^{-1} c(t)\gamma(t) \\
&\quad \times \exp \left( - \int_s^t \left\{ a(r)^\top + \gamma(r)^{-1}b(r)b(r)^\top \right\} dr \right) dt
\end{align*}
$$

for $t \ge s$.

Thus, from Proposition 3.8, we obtain

$$
\begin{align}
d_t \mathrm{Cov}(\xi_{s;t}, \xi_{u;t})
&= d_t V[\xi_{s;t}] \exp \left( - \int_u^s \left\{ a(r)^\top + \gamma(r)^{-1} b(r) b(r)^\top \right\} dr \right) \nonumber \\
&= - \exp \left( - \int_s^t \left\{ a(r) + b(r) b(r)^\top \gamma(r)^{-1} \right\} dr \right) \nonumber \\
&\qquad \times \gamma(t) c(t)^\top (\sigma(t) \sigma(t)^\top)^{-1} c(t) \gamma(t) \nonumber \\
&\qquad \times \exp \left( - \int_u^t \left\{ a(r)^\top + \gamma(r)^{-1} b(r) b(r)^\top \right\} dr \right) dt \tag{3.73} \\
&= - \mathrm{Cov}(\xi_{t;t}, \xi_{s;t})^\top c(t)^\top (\sigma(t) \sigma(t)^\top)^{-1} c(t) \mathrm{Cov}(\xi_{t;t}, \xi_{u;t}) dt \nonumber
\end{align}
$$

for $0 \le u \le s \le t$.

In the case where $s \le u$, we can derive the same formula by noting that

$$
\frac{\partial}{\partial t} \mathrm{Cov}(\xi_{s;t}, \xi_{u;t}) = \frac{\partial}{\partial t} \mathrm{Cov}(\xi_{u;t}, \xi_{s;t})^{\top}.
$$

Finally, assume that $\gamma(t)$ may be singular. In this case, we have

$$
d_t \mathrm{Cov}(\xi_{s;t}^{\epsilon}, \xi_{u;t}^{\epsilon}) = - \mathrm{Cov}(\xi_{t;t}^{\epsilon}, \xi_{s;t}^{\epsilon})^{\top} c(t)^{\top} (\sigma(t)\sigma(t)^{\top})^{-1} c(t) \mathrm{Cov}(\xi_{t;t}^{\epsilon}, \xi_{u;t}^{\epsilon}) dt,
$$

where $\gamma(s)^\epsilon$ and $\xi_{s;t}^\epsilon$ are defined in (3.60) to (3.63). By (3.69), $\mathrm{Cov}(\xi_{t;t}^\epsilon, \xi_{s;t}^\epsilon)$ converges uniformly to $\mathrm{Cov}(\xi_{t;t}, \xi_{s;t})$ on any compact set as $\epsilon \to 0$, leading to the desired result.

$\square$

<figure><img src="image_8.png" /></figure>
**Corollary 3.13.** For $0 \le u \le s \le t$, $\text{Cov}(\xi_{s;t}, \xi_{u;t})$ has the expression

$$ \begin{aligned} \text{Cov}(\xi_{s;t}, \xi_{u;t}) &= \exp \left( \int_{u}^{s} \left\{ a(r) - \gamma(r)c(r)^{\top}(\sigma(r)\sigma(r)^{\top})^{-1}c(r) \right\} dr \right) \gamma(u) \\ &\quad - \gamma(s) \int_{s}^{t} \exp \left( \int_{s}^{v} \left\{ a(r)^{\top} - c(r)^{\top}(\sigma(r)\sigma(r)^{\top})^{-1}c(r)\gamma(r) \right\} dr \right) \\ &\quad \times c(v)^{\top}(\sigma(v)\sigma(v)^{\top})^{-1}c(v) \\ &\quad \times \exp \left( \int_{u}^{v} \left\{ a(r) - \gamma(r)c(r)^{\top}(\sigma(r)\sigma(r)^{\top})^{-1}c(r) \right\} dr \right) dv\gamma(u) \end{aligned} $$

*Proof.* By Proposition 3.12, we have

$$ \begin{aligned} \text{Cov}(\xi_{s;t}, \xi_{u;t}) &= \text{Cov}(\xi_{s;s}, \xi_{u;s}) \\ &\quad - \int_{s}^{t} \text{Cov}(\xi_{v;v}, \xi_{s;v})^{\top} c(v)^{\top} (\sigma(v)\sigma(v)^{\top})^{-1} c(v) \text{Cov}(\xi_{v;v}, \xi_{u;v}) dv. \end{aligned} $$

Then the desired result is immediately obtained, recalling that it follows from Proposition 3.9 that

$$ \text{Cov}(\xi_{t;t}, \xi_{s;t}) = \exp \left( \int_{s}^{t} \left\{ a(r) - \gamma(r)c(r)^{\top}(\sigma(r)\sigma(r)^{\top})^{-1}c(r) \right\} dr \right) \gamma(s) $$

for $0 \le s \le t$.

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

The equation for fixed-point smoothing is presented in a simpler form than the literature (Shaked and Theodor 1992).

**Proposition 3.14. (Fixed-point smoother)**

For any $s \ge 0$, the process $\{\mu_{s;t}\}_{t \ge s}$ satisfies the stochastic differential equation

$$ d_t \mu_{s;t} = \text{Cov}(\xi_{s;t}, \xi_{t;t}) c(t)^{\top} (\sigma(t)\sigma(t)^{\top})^{-1} (dY_t - c(t)\mu_{t;t}dt). $$

*Proof.* By the definition of $\mu_{s;t}$ and Proposition 3.12, we have

$$ \begin{aligned} d_t \mu_{s;t} &= \text{Cov}(\xi_{s;t}, \xi_{t;t}) c(t)^{\top} (\sigma(t)\sigma(t)^{\top})^{-1} (dY_t - c(t)E[X_t]dt) \\ &\quad - \int_0^t \text{Cov}(\xi_{t;t}, \xi_{s;t})^{\top} c(t)^{\top} (\sigma(t)\sigma(t)^{\top})^{-1} c(t) \text{Cov}(\xi_{t;t}, \xi_{u;t}) \\ &\quad \times c(u)^{\top} (\sigma(u)\sigma(u)^{\top})^{-1} (dY_u - c(u)E[X_u]du)dt \\ &= \text{Cov}(\xi_{s;t}, \xi_{t;t}) c(t)^{\top} (\sigma(t)\sigma(t)^{\top})^{-1} (dY_t - c(t)E[X_t]dt) \\ &\quad - \text{Cov}(\xi_{t;t}, \xi_{s;t})^{\top} c(t)^{\top} (\sigma(t)\sigma(t)^{\top})^{-1} c(t) (\mu_{t;t} - E[X_t])dt \\ &= \text{Cov}(\xi_{s;t}, \xi_{t;t}) c(t)^{\top} (\sigma(t)\sigma(t)^{\top})^{-1} (dY_t - c(t)\mu_{t;t}dt). \end{aligned} $$

<figure><img src="image_10.png" /></figure>
# 4 Application to nonlinear filtering

In this section, we will imply that the results derived in Section 3 can be applied to calculation of nonlinear filtering in two specific models: the linear model with a polynomially perturbed transfer function, and the nonlinear model with small system noise. For the first example, we provide the results of numerical simulations.

Since this work is already enough complicated and long, we only present a heuristic outline of how the linear results can be applied to nonlinear filtering. As a result, many procedures below lack mathematical rigor, but all steps here can be mathematically justified, which will be shown in a subsequent paper. We added this section in order to exhibit that our research is not limited to the linear model, and actually improves the accuracy of nonlinear filtering.

## 4.1 Linear model with polynomially perturbed transfer function

### 4.1.1 Algorithm

In this section, we assume that $X$, $Y$, $V$, and $W$ are one-dimensional, and consider the following model:

$$dX_t = aX_t dt + b dV_t, \qquad X_0 = 0, \qquad (4.1)$$

$$dY_t = (cX_t + \epsilon g(X_t)) dt + \sigma dW_t, \qquad Y_0 = 0, \qquad (4.2)$$

where $a, b, c \in \mathbb{R}$, $0 < \epsilon < 1$, $\sigma > 0$, and $g$ is a polynomial. In this model, we aim to calculate the conditional expectation $E[X_t|Y_t]$.

Fix $T > 0$, and introduce a new probability measure $Q$ by

$$Q(A) = E \left[ 1_A \exp \left( - \int_0^T \frac{cX_t + \epsilon g(X_t)}{\sigma^2} dW_t - \frac{1}{2} \int_0^T \frac{\{cX_t + \epsilon g(X_t)\}^2}{\sigma^2} dt \right) \right]$$

for $A \in \mathcal{F}$. This measure is well-defined due to Proposition 2.3.

Furthermore, by Proposition 2.5, we can write

$$E[X_t|Y_t] = \frac{E_Q \left[ X_t \exp \left( \int_0^t \frac{cX_s + \epsilon g(X_s)}{\sigma^2} dY_s - \frac{1}{2} \int_0^t \frac{\{cX_s + \epsilon g(X_s)\}^2}{\sigma^2} ds \right) \middle| Y_t \right]}{E_Q \left[ \exp \left( \int_0^t \frac{cX_s + \epsilon g(X_s)}{\sigma^2} dY_s - \frac{1}{2} \int_0^t \frac{\{cX_s + \epsilon g(X_s)\}^2}{\sigma^2} ds \right) \middle| Y_t \right]} \quad (4.3)$$

for $0 \le t \le T$, where $E_Q$ denotes the expectation with respect to $Q$. Let us denote this conditional expectation by $m_t^\epsilon$, and consider expanding $m_t^\epsilon$ with respect to $\epsilon$.
For this purpose, define a conditional measure $\tilde{P}_t$ by

$$ \tilde{P}_t(A) = \frac{E_Q \left[ 1_A \exp \left( \int_0^t \frac{cX_s}{\sigma^2} dY_s - \frac{1}{2} \int_0^t \frac{(cX_s)^2}{\sigma^2} ds \right) \middle| \mathcal{Y}_t \right]}{E_Q \left[ \exp \left( \int_0^t \frac{cX_s}{\sigma^2} dY_s - \frac{1}{2} \int_0^t \frac{(cX_s)^2}{\sigma^2} ds \right) \middle| \mathcal{Y}_t \right]} $$

for $A \in \mathcal{X}_t \vee \mathcal{Y}_t$, and let $\tilde{E}_t$ be the expectation with respect to $\tilde{P}_t$ (see (??)).
Then the coefficient of $\epsilon^0$ in $m_t^\epsilon$ can be written as

$$ m_t^0 = \frac{E_Q \left[ X_t \exp \left( \int_0^t \frac{cX_s}{\sigma^2} dY_s - \frac{1}{2} \int_0^t \frac{(cX_s)^2}{\sigma^2} ds \right) \middle| \mathcal{Y}_t \right]}{E_Q \left[ \exp \left( \int_0^t \frac{cX_s}{\sigma^2} dY_s - \frac{1}{2} \int_0^t \frac{(cX_s)^2}{\sigma^2} ds \right) \middle| \mathcal{Y}_t \right]} = \tilde{E}_t[X_t]. $$

Furthermore, differentiating (4.3) yields

$$ \begin{aligned} & \frac{d}{d\epsilon} m_t^\epsilon \Big|_{\epsilon=0} \\ &= \frac{E_Q \left[ X_t \int_0^t g(X_s)(dY_s - cX_s ds) \exp \left( \int_0^t \frac{cX_s}{\sigma^2} dY_s - \frac{1}{2} \int_0^t \frac{(cX_s)^2}{\sigma^2} ds \right) \middle| \mathcal{Y}_t \right]}{E_Q \left[ \exp \left( \int_0^t \frac{cX_s}{\sigma^2} dY_s - \frac{1}{2} \int_0^t \frac{(cX_s)^2}{\sigma^2} ds \right) \middle| \mathcal{Y}_t \right]} \\ & \quad - \frac{E_Q \left[ X_t \exp \left( \int_0^t \frac{cX_s}{\sigma^2} dY_s - \frac{1}{2} \int_0^t \frac{(cX_s)^2}{\sigma^2} ds \right) \middle| \mathcal{Y}_t \right]}{E_Q \left[ \exp \left( \int_0^t \frac{cX_s}{\sigma^2} dY_s - \frac{1}{2} \int_0^t \frac{(cX_s)^2}{\sigma^2} ds \right) \middle| \mathcal{Y}_t \right]} \\ & \quad \times \frac{E_Q \left[ \int_0^t g(X_s)(dY_s - cX_s ds) \exp \left( \int_0^t \frac{cX_s}{\sigma^2} dY_s - \frac{1}{2} \int_0^t \frac{(cX_s)^2}{\sigma^2} ds \right) \middle| \mathcal{Y}_t \right]}{E_Q \left[ \exp \left( \int_0^t \frac{cX_s}{\sigma^2} dY_s - \frac{1}{2} \int_0^t \frac{(cX_s)^2}{\sigma^2} ds \right) \middle| \mathcal{Y}_t \right]} \\ &= \tilde{E}_t \left[ X_t \int_0^t g(X_s)(dY_s - cX_s ds) \right] - \tilde{E}_t [X_t] \tilde{E}_t \left[ \int_0^t g(X_s)(dY_s - cX_s ds) \right], \end{aligned} $$

which gives the coefficient of $\epsilon^1$.

In order to find higher order coefficients, write

$$ \begin{aligned} & \exp \left( \int_0^t \frac{cX_s + \epsilon g(X_s)}{\sigma^2} dY_s - \frac{1}{2} \int_0^t \frac{\{cX_s + \epsilon g(X_s)\}^2}{\sigma^2} ds \right) \\ &= \exp \left( \int_0^t \frac{cX_s}{\sigma^2} dY_s - \frac{1}{2} \int_0^t \frac{(cX_s)^2}{\sigma^2} ds \right) \\ & \quad \times \exp \left( \epsilon \int_0^t \frac{g(X_s)}{\sigma^2} (dY_s - cX_s ds) - \frac{1}{2} \epsilon^2 \int_0^t \frac{g(X_s)^2}{\sigma^2} ds \right). \end{aligned} $$
Here, set

$$K_t^\epsilon = \exp \left( \epsilon \int_0^t \frac{g(X_s)}{\sigma^2} (dY_s - cX_s ds) - \frac{1}{2} \epsilon^2 \int_0^t \frac{g(X_s)^2}{\sigma^2} ds \right).$$

Then $K_t^\epsilon$ is an exponential local martingale, and we have the expansion

$$
\begin{align*}
K_t^\epsilon &= 1 + \frac{\epsilon}{\sigma^2} \int_0^t K_s^\epsilon g(X_s) (dY_s - cX_s ds) \\
&= 1 + \frac{\epsilon}{\sigma^2} \int_0^t g(X_s) (dY_s - cX_s ds) \\
&\quad + \frac{\epsilon^2}{\sigma^4} \int_0^t K_u^\epsilon g(X_u) (dY_u - cX_u du) g(X_s) (dY_s - cX_s ds) \\
&= \dots \\
&= 1 + \sum_{i=1}^n \frac{\epsilon^i}{\sigma^{2i}} \int_0^t \int_0^{t_i} \dots \int_0^{t_2} g(X_{t_1}) \dots g(X_{t_n}) \\
&\quad (dY_{t_1} - cX_{t_1} dt_1) \dots (dY_{t_i} - cX_{t_i} dt_i) \\
&\quad + \frac{\epsilon^{n+1}}{\sigma^{2n+2}} \int_0^t \int_0^{t_{n+1}} \dots \int_0^{t_2} K_{t_1}^\epsilon g(X_{t_1}) \dots g(X_{t_n}) \\
&\quad (dY_{t_1} - cX_{t_1} dt_1) \dots (dY_{t_n} - cX_{t_n} dt_n)
\end{align*}
$$

for any $n \in \mathbb{N}$. The last term is shown to be $O(\epsilon^{n+1})$ in some sense. Therefore, (4.3) can be expanded as

$$
\begin{align*}
E[X_t | \mathcal{Y}_t] &= \frac{E_Q \left[ X_t K_t^\epsilon \exp \left( \int_0^t \frac{cX_s}{\sigma^2} dY_s - \frac{1}{2} \int_0^t \frac{(cX_s)^2}{\sigma^2} ds \right) \middle| \mathcal{Y}_t \right]}{E_Q \left[ K_t^\epsilon \exp \left( \int_0^t \frac{cX_s}{\sigma^2} dY_s - \frac{1}{2} \int_0^t \frac{(cX_s)^2}{\sigma^2} ds \right) \middle| \mathcal{Y}_t \right]} \\
&= \frac{E_Q \left[ X_t K_t^\epsilon \exp \left( \int_0^t \frac{cX_s}{\sigma^2} dY_s - \frac{1}{2} \int_0^t \frac{(cX_s)^2}{\sigma^2} ds \right) \middle| \mathcal{Y}_t \right]}{E_Q \left[ K_t^\epsilon \exp \left( \int_0^t \frac{cX_s}{\sigma^2} dY_s - \frac{1}{2} \int_0^t \frac{(cX_s)^2}{\sigma^2} ds \right) \middle| \mathcal{Y}_t \right]} \\
&= \frac{E_Q \left[ \exp \left( \int_0^t \frac{cX_s}{\sigma^2} dY_s - \frac{1}{2} \int_0^t \frac{(cX_s)^2}{\sigma^2} ds \right) \middle| \mathcal{Y}_t \right]}{E_Q \left[ \exp \left( \int_0^t \frac{cX_s}{\sigma^2} dY_s - \frac{1}{2} \int_0^t \frac{(cX_s)^2}{\sigma^2} ds \right) \middle| \mathcal{Y}_t \right]}
\end{align*}
$$
$$
\begin{align}
&= \frac{\tilde{E}_t[X_t K_t^\epsilon]}{\tilde{E}_t[K_t^\epsilon]} \nonumber \\
&= J_t^0(X_t) + \{J_t^1(X_t) - J_t^0(X_t)J_t^1(1)\} \epsilon \nonumber \\
&\quad + \{J_t^2(X_t) - J_t^0(X_t)J_t^2(1) - J_t^1(X_t)J_t^1(1) + J_t^0(X_t)J_t^1(1)^2\} \epsilon^2 \nonumber \\
&\quad + \dots + \sum_{k=0}^n \sum_{p=1}^{n-k} \sum_{\substack{i_1, \dots, i_p \in \mathbb{N} \\ i_1 + \dots + i_p = n-k}} (-1)^p J_t^k(X_t) J_t^{i_1}(1) \dots J_t^{i_p}(1) \epsilon^n \nonumber \\
&\quad + O(\epsilon^{n+1}), \tag{4.4}
\end{align}
$$

where

$$
J_t^i(U) = \frac{1}{\sigma^{2n}} \tilde{E}_t \left[ U \int_0^t \int_0^{t_i} \dots \int_0^{t_2} g(X_{t_1}) \dots g(X_{t_i}) \right. \\
\left. \times (dY_{t_1} - cX_{t_1}dt_1) \dots (dY_{t_i} - cX_{t_i}dt_i) \right] \quad (4.5)
$$

for a random variable $U$. Thus, the expansion of $E[X_t|\mathcal{Y}_t]$ is reduced to computing the (conditional) expectation (4.5).

According to Theorem 3.2, the law of {$X_s$}$_{0\le s\le t}$ under $\tilde{E}_t$ is the same as the conditional law of {$\zeta_{s;t}$}$_{0\le s\le t}$ given $\mathcal{Y}_t$, as introduced in (3.13). Hence, we obtain

$$
J_t^i(f(X_t)) = \frac{1}{\sigma^{2n}} E \left[ f(\zeta_{t;t}) \int_0^t \int_0^{t_i} \dots \int_0^{t_2} g(\zeta_{t_1;t}) \dots g(\zeta_{t_i;t}) \right. \\
\left. \times (dY_{t_1} - c\zeta_{t_1;t}dt_1) \dots (dY_{t_i} - c\zeta_{t_i;t}dt_i) \right] \Big| \mathcal{Y}_t . \quad (4.6)
$$

Recall that {$\zeta_{s;t}$}$_{0\le s\le t}$ is a conditionally Gaussian process given $\mathcal{Y}_t$ whose mean and covariance is given by

$$
E[\zeta_{s;t}|\mathcal{Y}_t] = \mu_{s;t}, \quad \text{Cov}(\zeta_{s;t}, \zeta_{u;t}|\mathcal{Y}_t) = \text{Cov}(\zeta_{s;t}, \xi_{u;t}),
$$

where $\mu$ and $\xi$ were introduced in (3.70) and (3.11), respectively. Thus, if $g$
is a polynomial, (4.6) can be expressed using $\mu_{s;t}$ and $\text{Cov}(\zeta_{s;t}, \xi_{u;t})$.

For instance, when $g(x) = x^3$, we have

$$
J_t^0(X_t) = E[\zeta_{t;t} | \mathcal{Y}_t] = \mu_{t;t},
$$

$$
\begin{align*}
J_t^1(X_t) &= E \left[ \zeta_{t;t} \int_0^t \zeta_{s;t}^3 (dY_s^\epsilon - c\zeta_{s;t}ds) \middle| \mathcal{Y}_t \right] \\
&= \int_0^t \left( E \left[ \zeta_{t;t} \zeta_{s;t}^3 \middle| \mathcal{Y}_t \right] dY_s - cE \left[ \zeta_{t;t} \zeta_{s;t}^4 \middle| \mathcal{Y}_t \right] ds \right)
\end{align*}
$$
$$
\begin{align*}
&= \mu_{t;t} \int_0^t \mu_{s;t}^3 (dY_s - c\mu_{s;t}ds) + 3\mu_{t;t} \int_0^t \mu_{s;t}V[\xi_{s;t}](dY_s - c\mu_{s;t}ds) \\
&\quad + 3 \int_0^t \mu_{s;t}^2 \mathrm{Cov}(\xi_{t;t}, \xi_{s;t})(dY_s - c\mu_{s;t}ds) \\
&\quad - c \int_0^t \mu_{s;t}^3 \mathrm{Cov}(\xi_{t;t}, \xi_{s;t})ds - 3c\mu_{t;t} \int_0^t \mu_{s;t}^2 \mathrm{Cov}(\xi_{t;t}, \xi_{s;t})ds \\
&\quad + 3 \int_0^t \mathrm{Cov}(\xi_{t;t}, \xi_{s;t})V[\xi_{s;t}](dY_s - c\mu_{s;t}ds) \\
&\quad - 6c \int_0^t \mu_{s;t} \mathrm{Cov}(\xi_{t;t}, \xi_{s;t})V[\xi_{s;t}]ds - 3c\mu_{t;t} \int_0^t \mathrm{Cov}(\xi_{t;t}, \xi_{s;t})V[\xi_{s;t}]ds, \\
J_t^0(1) &= E \left[ \left. \int_0^t \zeta_{s;t}^3 (dY_s - c\zeta_{s;t}ds) \right| \mathcal{Y}_t \right] \\
&= \int_0^t (E \left[ \zeta_{s;t}^3 | \mathcal{Y}_t \right] dY_s - cE \left[ \zeta_{s;t}^4 | \mathcal{Y}_t \right] ds) \\
&= \int_0^t \mu_{s;t}^3 (dY_s - c\mu_{s;t}ds) + 3 \int_0^t \mu_{s;t}V[\xi_{s;t}](dY_s - c\mu_{s;t}ds) \\
&\quad - 3c \int_0^t \mu_{s;t}^2 \mathrm{Cov}(\xi_{t;t}, \xi_{s;t})ds - 3c \int_0^t \mathrm{Cov}(\xi_{t;t}, \xi_{s;t})V[\xi_{s;t}]ds.
\end{align*}
$$

Using this, the coefficient of $\epsilon^1$ in (4.4) is represented as

$$
\begin{align}
&J_t^1(X_t) - J_t^0(X_t)J_t^1(1) \nonumber \\
&=3 \int_0^t \mu_{s;t}^2 \mathrm{Cov}(\xi_{t;t}, \xi_{s;t})(dY_s - c\mu_{s;t}ds) - c \int_0^t \mu_{s;t}^3 \mathrm{Cov}(\xi_{t;t}, \xi_{s;t}) ds \nonumber \\
&\quad + 3 \int_0^t \mathrm{Cov}(\xi_{t;t}, \xi_{s;t})V[\xi_{s;t}](dY_s - c\mu_{s;t}ds) \tag{4.7} \\
&\quad - 6c \int_0^t \mu_{s;t} \mathrm{Cov}(\xi_{t;t}, \xi_{s;t})V[\xi_{s;t}] ds. \nonumber
\end{align}
$$

Here, for numerical stability, we preserved $dY_s^\epsilon - c\mu_{s;t}ds$ without fully expanding integrals.

Note also that some of the stochastic integrals above are not well-defined, since $\{\zeta_{s;t}\}_{0\le s\le t}$ and $\{\mu_{s;t}\}_{0\le s\le t}$ are not adapted (they depend on the full history of $\{Y_s\}_{0\le s\le t}$). However, we can justify the procedure by appropriately interpreting the integrals.

Anyway, $J_t^i(X_t)$ and $J_t^i(1)$ are expanded into a sum of terms of the form

$$
A_t(p, q, r, \alpha; n) = \int_0^t \int_0^{t_n} \cdots \int_0^{t_2} \prod_{i=1}^n (\mu_{t_i;t})^{p_i} \prod_{i=1}^n \mathrm{Cov}(\xi_{t;t}, \xi_{t_i;t})^{q_i} \\
\times \prod_{1 \le i \le j \le n} \mathrm{Cov}(\xi_{t_i;t}, \xi_{t_j;t})^{r_{ij}} d\lambda_{t_1;t}^{(\alpha_1)} \cdots d\lambda_{t_n;t}^{(\alpha_n)},
$$
where $n \in \mathbb{N}$, $p = (p_i)_{i=1,\dots,n} \in \mathbb{Z}_+^n$, $q = (q_i)_{i=1,\dots,n} \in \mathbb{Z}_+^n$, $r = (r_{ij})_{i,j=1,\dots,n} \in M_n(\mathbb{Z}_+)$, $\alpha = (\alpha_i)_{i=1,\dots,n} \in \{0,1\}^n$, and

$$d\lambda_{s;t}^{(i)} = \begin{cases} ds & (i = 0) \\ dY_s - c\mu_{s;t}ds & (i = 1). \end{cases}$$

For example, (4.7) can be written as

$$3A_t(2, 1, 0, 1; 1) - cA_t(3, 1, 0, 0; 1) + 3A_t(0, 1, 1, 1; 1) - 6cA_t(1, 1, 1, 0; 1).$$

To derive a recursive formula for $A_t(p, q, r, \alpha; n)$, recall the equations derived in Section 3.2:

$$d_t \mu_{t;t} = a \mu_{t;t} dt + \frac{c\gamma(t)}{\sigma^2} (dY_t - c\mu_{t;t} dt), \quad (4.8)$$

$$d_t \mu_{s;t} = \frac{c}{\sigma^2} \mathrm{Cov}(\xi_{s;t}, \xi_{t;t}) (dY_t - c\mu_{t;t} dt), \quad (4.9)$$

$$d_t \mathrm{Cov}(\xi_{t;t}, \xi_{s;t}) = \left\{ a - \frac{c^2 \gamma(t)}{\sigma^2} \right\} \mathrm{Cov}(\xi_{t;t}, \xi_{s;t}) dt, \quad (4.10)$$

$$d_t \mathrm{Cov}(\xi_{s;t}, \xi_{u;t}) = -\frac{c^2}{\sigma^2} \mathrm{Cov}(\xi_{t;t}, \xi_{s;t}) \mathrm{Cov}(\xi_{t;t}, \xi_{u;t}) dt, \quad (4.11)$$

where $\gamma(t) = \mathrm{Cov}(\xi_{t;t}, \xi_{t;t})$ is the solution of

$$\frac{d}{dt}\gamma(t) = -\frac{c^2}{\sigma^2}\gamma(t)^2 + 2a\gamma(t) + b^2. \quad (4.12)$$

As mentioned above, the stochastic integral defining $A_t(p, q, r, \alpha; n)$ may not be interpreted as an Itô integral, but it can be differentiated in accordance with the following rule, which is similar to Itô's rule:

**Rule:** If $\alpha_{s;t}$ satisfies

$$d_t \alpha_{s;t} = p_{s;t} dt + q_{s;t} dY_t,$$

then

$$d_t \int_0^t \alpha_{s;t} dY_s = \alpha_{t;t} dY_t + \int_0^t p_{s;t} dY_s dt + \int_0^t q_{s;t} dY_s dY_t + \sigma^2 q_{t;t} dt. \quad (4.13)$$

The final term results from the Itô correction term $q_{t;t}dY_t \times dY_t$.

For example, let us consider the derivative of

$$A_t(1, 0, 0, 1; 1) = \int_0^t \mu_{s;t}(dY_s - c\mu_{s;t}ds) = \int_0^t \mu_{s;t}(dY_s - c\mu_{s;t}ds).$$

By (4.9) and applying (4.13) yields

$$d_t \int_0^t \mu_{s;t} dY_s = \mu_{t;t} dY_t + \frac{c}{\sigma^2} \int_0^t \mathrm{Cov}(\xi_{s;t}, \xi_{t;t}) dY_s (dY_t - c\mu_{t;t} dt) + c\gamma(t) dt.$$
Thus, we obtain the equation

$$
\begin{align*}
dA_t(1, 0, 0, 1; 1) &= \mu_{t;t}(dY_t - c\mu_{t;t}dt) \\
&\quad + \frac{c}{\sigma^2} \int_0^t \operatorname{Cov}(\xi_{s;t}, \xi_{t;t})(dY_s - c\mu_{s;t}ds) (dY_t - c\mu_{t;t}dt) \\
&\quad - \frac{c^2}{\sigma^2} \int_0^t \mu_{s;t}\operatorname{Cov}(\xi_{s;t}, \xi_{t;t})ds (dY_t - c\mu_{t;t}dt) \\
&\quad - \frac{c^2}{\sigma^2} \int_0^t \operatorname{Cov}(\xi_{s;t}, \xi_{t;t})^2 ds dt + c\gamma(t)dt \\
&= \mu_{t;t}(dY_t - c\mu_{t;t}dt) \\
&\quad + \frac{c}{\sigma^2} A_t(0, 1, 0, 1; 1)(dY_t - c\mu_{t;t}dt) \\
&\quad - \frac{c^2}{\sigma^2} A_t(1, 1, 0, 0; 1)(dY_t - c\mu_{t;t}dt) \\
&\quad - \frac{c^2}{\sigma^2} A_t(0, 2, 0, 0; 1)dt + c\gamma(t)dt.
\end{align*}
$$

In the same way, we have

$$
\begin{align*}
dA_t(0, 1, 0, 1; 1) &= \left\{ a - \frac{c^2 \gamma(t)}{\sigma^2} \right\} \int_0^t \operatorname{Cov}(\xi_{s;t}, \xi_{t;t})(dY_s - c\mu_{s;t}ds) \, dt \\
&\quad - \frac{c^2}{\sigma^2} \int_0^t \operatorname{Cov}(\xi_{s;t}, \xi_{t;t})^2 ds (dY_t - c\mu_{t;t}dt), \\
&= \left\{ a - \frac{c^2 \gamma(t)}{\sigma^2} \right\} A_t(0, 1, 0, 1; 1) dt \\
&\quad - \frac{c^2}{\sigma^2} A_t(0, 2, 0, 0; 1) (dY_t - c\mu_{t;t}dt),
\end{align*}
$$

$$
\begin{align*}
dA_t(1, 1, 0, 0; 1) &= \left\{ a - \frac{c^2 \gamma(t)}{\sigma^2} \right\} A_t(1, 1, 0, 0; 1) \\
&\quad + \frac{c}{\sigma^2} \int_0^t \operatorname{Cov}(\xi_{s;t}, \xi_{t;t})^2 ds (dY_t - c\mu_{t;t}dt), \\
&= \left\{ a - \frac{c^2 \gamma(t)}{\sigma^2} \right\} dA_t(1, 1, 0, 0; 1) \\
&\quad + \frac{c}{\sigma^2} A_t(0, 2, 0, 0; 1) (dY_t - c\mu_{t;t}dt),
\end{align*}
$$

$$
\begin{align*}
dA_t(0, 2, 0, 0; 1) &= 2 \left\{ a - \frac{c^2 \gamma(t)}{\sigma^2} \right\} \int_0^t \operatorname{Cov}(\xi_{s;t}, \xi_{t;t})^2 ds \, dt \\
&= 2 \left\{ a - \frac{c^2 \gamma(t)}{\sigma^2} \right\} A_t(0, 2, 0, 0; 1) dt.
\end{align*}
$$

Therefore, $A_t(1, 0, 0, 1; 1)$ can be computed by solving the four-dimensional
stochastic differential equation

$$
\begin{align*}
dA_t(1,0,0,1;1) ={}& \mu_{t;t}(dY_t - c\mu_{t;t}dt) + \frac{c}{\sigma^2}A_t(0,1,0,1;1)(dY_t - c\mu_{t;t}dt) \\
& - \frac{c^2}{\sigma^2}A_t(1,1,0,0;1)(dY_t - c\mu_{t;t}dt) \\
& - \frac{c^2}{\sigma^2}A_t(0,2,0,0;1)dt + c\gamma(t)dt,
\end{align*}
$$

$$
\begin{align*}
dA_t(0,1,0,1;1) ={}& \left\{ a - \frac{c^2 \gamma(t)}{\sigma^2} \right\} A_t(0,1,0,1;1) dt \\
& - \frac{c^2}{\sigma^2} A_t(0,2,0,0;1) (dY_t - c\mu_{t;t}dt),
\end{align*}
$$

$$
\begin{align*}
dA_t(1,1,0,0;1) ={}& \left\{ a - \frac{c^2\gamma(t)}{\sigma^2} \right\} dA_t(1,1,0,0;1) \\
& + \frac{c}{\sigma^2} A_t(0,2,0,0;1) (dY_t - c\mu_{t;t}dt),
\end{align*}
$$

$$
dA_t(0,2,0,0;1) = 2 \left\{ a - \frac{c^2 \gamma(t)}{\sigma^2} \right\} A_t(0,2,0,0;1) dt,
$$

where $\mu_{t;t}$ and $\gamma(t)$ is given by (4.8) and (4.12). A numerical solution to this equation can be recursively computed using the Euler-Maruyama method.

Similarly, $A_t(p, q, r, \alpha; n)$ is eventually reduced to the case of $p = r = \alpha = 0$ by repeatedly differentiating it using (4.9)–(4.11). A closed system of equations can then be obtained due to (4.10).

The algorithm can be summarized in the following steps:

1. Expand the conditional expectation as in (4.4).

2. Due to Theorem 3.2, replace $X$ with $\zeta$ as in (4.6).

3. Express (4.6) in terms of the mean $\mu_{s;t}$ and covariance $\text{Cov}(\xi_{s;t}, \xi_{u;t})$, resulting in an expression like (4.7).

4. Differentiate each term from the previous step according to (4.9)–(4.11) and the rule (4.13), and derive a finite-dimensional stochastic differential equation to calculate the term.

4.1.2 Numerical simulation 1: Linear case

In the first simulation, we set $g(x) = x$ and consider the following model

$$
dX_t = aX_t dt + b dV_t, \quad X_0 = 0,
$$

$$
dY_t = (cX_t + \epsilon X_t) dt + \sigma dW_t, \quad Y_0 = 0,
$$

In this case, the true value of the conditional expectation $E[X_t|Y_t]$ can be
calculated through the Kalman-Bucy filter, which serves as a benchmark for
evaluating the accuracy of our asymptotic expansion.
We simulated a path of $X_t$ and $Y_t$ over the interval $0 \le t \le 10$ using the parameters

$$a = -0.4, \quad b = 0.5, \quad c = 1, \quad \sigma = 0.3, \quad \epsilon = 0.2.$$

For the simulated path, we expanded the conditional expectation $E[X_t|Y_t]$ with respect to $\epsilon$ following the algorithm described in the previous section. Figure 1 shows the plots of this expansion and the true conditional expansion calculated by the Kalman-Bucy filter. presents the plots of this expansion alongside the true conditional expectation computed using the Kalman-Bucy filter. As shown in the figure, the expanded filters converge toward the true line as the order of expansion increases, demonstrating the validity of our algorithm.

<figure><img src="image_11.png" /><figcaption>Figure 1: Plots of the expanded filter and the true filter.</figcaption></figure>

### 4.1.3 Numerical simulation: cubic sensor

In the second simulation, we consider the case of $g(x) = x^3$:

$$dX_t = aX_t dt + b dV_t, \qquad X_0 = 0,$$

$$dY_t = (cX_t + \epsilon X_t^3) dt + \sigma dW_t, \qquad Y_0 = 0,$$
This model is referred to as the cubic sensor, which serves as a benchmark in several studies (Armstrong et al. 2019).

In this case, the true filter cannot be computed to directly evaluate the performance of the expansion. Instead, we generated 1,000 sample paths of $X_t$ and $Y_t$ on the interval $0 \le t \le 100$, and computed the asymptotic expansion for each path. To assess the accuracy of the expansion, we evaluated the integrated squared error for each path, defined as

$$ \int_{0}^{100} (X_t - N_{t}^{[n], \epsilon})^2 dt, $$

where

$$ N_{t}^{[n], \epsilon} = J_{t}^{0}(X_{t}) + \{J_{t}^{1}(X_{t}) - J_{t}^{0}(X_{t})J_{t}^{1}(1)\} \epsilon + \{J_{t}^{2}(X_{t}) - J_{t}^{0}(X_{t})J_{t}^{2}(1) - J_{t}^{1}(X_{t})J_{t}^{1}(1) + J_{t}^{0}(X_{t})J_{t}^{1}(1)^{2}\} \epsilon^{2} + \dots + \sum_{k=0}^{n} \sum_{p=1}^{n-k} \sum_{i_{1}, \dots, i_{p} \in \mathbb{N}} (-1)^{p} J_{t}^{k}(X_{t}) J_{t}^{i_{1}}(1) \dots J_{t}^{i_{p}}(1) \epsilon^{n} $$

is the n-th order approximation (see (4.4)). Due to the $L^2$-optimality of the conditional expectation, the true filter is expected to minimize the mean of integrated squared error.

Here, we used the following parameters:

$$ a = -0.4, \quad b = 0.5, \quad c = 1, \quad \sigma = 0.3, \quad \epsilon = 0.2, $$

and simulations were performed with a step size of 0.01.

Table 1 shows the minimum, median, mean, and maximum of the integrated squared error for the 1,000 simulated paths. Note that $N_{t}^{[0], \epsilon} = \mu_{t,t}$ corresponds to the extended Kalman filter (linear approximation). We

<table><caption>Table 1: Integrated squared errors of each filter.</caption><thead><tr><th>Filter</th><th>μ<sub>t,t</sub></th><th>N<sub>t</sub><sup>[1],ε</sup></th><th>N<sub>t</sub><sup>[2],ε</sup></th></tr></thead><tbody><tr><td>Min.</td><td>8.200</td><td>7.957</td><td>7.912</td></tr><tr><td>Median</td><td>10.91</td><td>10.73</td><td>11.19</td></tr><tr><td>Mean</td><td>10.98</td><td>10.76</td><td>25.01</td></tr><tr><td>Max.</td><td>16.87</td><td>35.60</td><td>9215</td></tr></tbody></table>

can observe the superiority of the asymptotic expansion over the extended Kalman filter in terms of the minimum and median of $N_{t}^{[1], \epsilon}$, but it does not perform as well in terms of the mean and maximum.
<figure><img src="image_12.png" /><figcaption>**Figure 2:** Plots of the filtering results and the perturbation term for a single simulated path.</figcaption></figure>
This result can be explained by Figure 2, which shows the plots of $X_t$, $\mu_{t;t}$, $N_t^{[1],\epsilon}$, $N_t^{[2],\epsilon}$, and $\epsilon|X_t|^3$ for a single simulated path for the interval $0 \le t \le 10$. From the figure, we observe that $N_t^{[1],\epsilon}$ and $N_t^{[2],\epsilon}$ improve upon the extended Kalman filter $\mu_{t;t}$ over much of the interval, but $N_t^{[2],\epsilon}$ deviates from the true process around $t = 3.5$ and $t = 8$.

Looking at the lower panel, we find that this deviation aligns with the moments when the absolute value of the perturbation term $\epsilon|X_t|^3$ exceeds 1. This behavior is expected, as our expansion is performed with respect to $\epsilon X_t^3$, rather than just $\epsilon$. Consequently, the expansion diverges when $\epsilon|X_t|^3 > 1$. However, the upper panel suggests that this effect is not persistent, as the path of $N_t^{[2],\epsilon}$ returns closer to the true path as soon as $\epsilon|X_t|^3$ falls below 1. Hence, the minimum and median values in Table 1 capture the convergence of the asymptotic expansion when $\epsilon|X_t|^3 < 1$, whereas the maximum and mean values reflect its divergence when $\epsilon|X_t|^3 > 1$.

To manage the divergence of the expansion, we propose a modification to the coefficients. Let us write $N_t^{[n],\epsilon}$ as

$$N_t^{[n],\epsilon} = n_t^{[0]} + n_t^{[1]}\epsilon + n_t^{[2]}\epsilon^2 + \dots + n_t^{[n]}\epsilon^n,$$

where the coefficients are given by:

$$n_t^{[0]} = J_t^0(X_t),$$

$$n_t^{[1]} = J_t^1(X_t) - J_t^0(X_t)J_t^1(1),$$

$$n_t^{[2]} = J_t^2(X_t) - J_t^0(X_t)J_t^2(1) - J_t^1(X_t)J_t^1(1) + J_t^0(X_t)J_t^1(1)^2.$$

As discussed above, the term $n_t^{[i]}\epsilon^i$ either converges or diverges at an exponential rate, depending on the value of $\epsilon|X_t|^3$. To suppress the divergence and ensure exponential decay, we fix a parameter $r > 0$ and define the modified coefficients $\tilde{n}_t^{[i]}$ as follows:

$$\tilde{n}_t^{[0]} = n_t^{[0]}, \quad \tilde{n}_t^{[i]} = \begin{cases} n_t^{[i]} & \text{if } |n_t^{[i]}\epsilon^i| \le r|\tilde{n}_t^{[i-1]}\epsilon^{i-1}|, \\ r|\tilde{n}_t^{[i-1]}\epsilon^{i-1}| \frac{n_t^{[i]}}{|n_t^{[i]}\epsilon^i|} & \text{if } |n_t^{[i]}\epsilon^i| > r|\tilde{n}_t^{[i-1]}\epsilon^{i-1}|, \end{cases}$$

for $i = 1, 2, \dots$. Using these coefficients, we define a modified filter by

$$\tilde{N}_t^{[i],\epsilon} = \tilde{n}_t^{[0]} + \tilde{n}_t^{[1]}\epsilon + \tilde{n}_t^{[2]}\epsilon^2 + \dots + \tilde{n}_t^{[i]}\epsilon^i.$$

By construction, we have $|\tilde{n}_t^{[i]}\epsilon^i| \le r|\tilde{n}_t^{[i-1]}\epsilon^{i-1}|$, which guarantees the convergence of $\tilde{N}_t^{[i],\epsilon}$ for all $t \ge 0$ if $0 < r < 1$.

Figure 3 shows the mean integrated squared error of $\tilde{N}_t^{[1],\epsilon}$ and $\tilde{N}_t^{[2],\epsilon}$ for $r = 0.1, 0.2, \dots, 1$ and $r = \infty$, based on the 1,000 simulated paths.
<figure><img src="image_13.png" /><figcaption>Figure 3: Mean integrated squared errors of the modified filters.</figcaption></figure>

<figure><img src="image_14.png" /><figcaption>Figure 4: The modified filter of $r = 0.2$ for the same simulated path as Figure 2.</figcaption></figure>
Note that $\tilde{N}_t^{[i],\epsilon} = N_t^{[i],\epsilon}$ when $r = \infty$. From this result, we observe that the accuracy improves with the modification, and the error decreases as $r$ becomes smaller. The errors of $\tilde{N}_t^{[1],\epsilon}$ and $\tilde{N}_t^{[2],\epsilon}$ both attain their minimum at $r = 0.2$. Figure 4 shows the modified expansions with $r = 0.2$ for the same path as in Figure 2. From this figure, we observe that $\tilde{N}_t^{[1],\epsilon}$ is closer to the true path than $\mu_{t;t}$, and $\tilde{N}_t^{[2],\epsilon}$ slightly improves upon $\tilde{N}_t^{[1],\epsilon}$.

The optimal value of $r$ coincides with $\epsilon = 0.2$ in this case, but they are not necessarily related. In fact, as shown in Figure 5, the error is minimized around $r = 0.3$ regardless of the value of $\epsilon$. For each $\epsilon$, we observe that $\tilde{N}_t^{[1],\epsilon}$ and $\tilde{N}_t^{[2],\epsilon}$ can improve upon the linear approximation for sufficiently small $r$, and $\tilde{N}_t^{[2],\epsilon}$ attains a smaller error than $\tilde{N}_t^{[1],\epsilon}$.

In summary, although the raw asymptotic expansion does not necessarily improve the accuracy of the filter over the entire time horizon due to the divergence of the perturbation term, we can control this divergence by choosing an appropriate parameter $r$ and modifying the coefficients of the expansion. With this modification, we observe an overall improvement in the filter's accuracy. While it is difficult to theoretically determine the optimal value of the parameter $r$, a relatively good value of $r$ can be obtained by conducting simulations in advance.

## 4.2 Nonlinear model with small system noise

Let us consider a nonlinear unobserved process

$$d\theta_t^\epsilon = \alpha(\theta_t^\epsilon) dt + \epsilon \beta(\theta_t^\epsilon) dV_t, \quad \theta_0 = 0, \qquad (4.14)$$

and an observed process

$$dY_t = h(\theta_t^\epsilon) dt + \sigma dW_t, \quad Y_0 = 0,$$

where $0 < \epsilon < 1$ and $\sigma > 0$ are constants, $\alpha, \beta$, and $h$ are nonlinear functions of class $C^\infty$, and $V$ and $W$ are independent Wiener processes. Here, we assume $\theta^\epsilon, Y, V$, and $W$ are one-dimensional for simplicity.

In this system, we consider the asymptotic expansion of $E[\theta_t^\epsilon | Y_t]$ with respect to $\epsilon$. As in the previous example, fix $T > 0$, and define a probability measure $Q$ by

$$Q(A) = E \left[ 1_A \exp \left( - \int_0^T \frac{h(\theta_t^\epsilon)}{\sigma^2} dW_t - \frac{1}{2} \int_0^T \frac{h(\theta_t^\epsilon)^2}{\sigma^2} dt \right) \right]$$

for $A \in \mathcal{F}$. Then, by Proposition 2.5, we can write

$$E[\theta_t^\epsilon | \mathcal{Y}_t] = \frac{E_Q \left[ \theta_t^\epsilon \exp \left( \int_0^t \frac{h(\theta_t^\epsilon)}{\sigma^2} dY_s - \frac{1}{2} \int_0^t \frac{h(\theta_t^\epsilon)^2}{\sigma^2} ds \right) \middle| \mathcal{Y}_t \right]}{E_Q \left[ \exp \left( \int_0^t \frac{h(\theta_t^\epsilon)}{\sigma^2} dY_s - \frac{1}{2} \int_0^t \frac{h(\theta_t^\epsilon)^2}{\sigma^2} ds \right) \middle| \mathcal{Y}_t \right]} \quad (4.15)$$
<figure><img src="image_15.png" /><figcaption>Figure 5: Mean integrated squared errors for $\epsilon = 0.8, 0.5, 0.3, 0.1$.</figcaption></figure>

for $0 \le t \le T$, where $E_Q$ is the expectation under $Q$.

In order to expand this quotient, we consider the derivative of $\theta_t^\epsilon$ with respect to $\epsilon$. From now on, we write $\frac{\partial^i}{\partial \epsilon^i} \theta_t^\epsilon = \theta_t^{[i], \epsilon}$ and $\theta_t^{[i]} = \theta_t^{[i], \epsilon}$ for $i = 0, 1, 2, \dots$. Then, $\theta_t^{[i], \epsilon}$ and $\theta_t^{[i]}$ satisfy the equations obtained by differentiating both sides of (4.14). In particular, we have

$$ 
\begin{align*}
d\theta_t^{[1], \epsilon} &= \alpha'(\theta_t^\epsilon)\theta_t^{[1], \epsilon}dt + \beta(\theta_t^\epsilon)dV_t + \epsilon\beta'(\theta_t^{[0]})\theta_t^{[1]}dV_t, \qquad \theta_0^{[1], \epsilon} = 0 \\
d\theta_t^{[2], \epsilon} &= \alpha''(\theta_t^\epsilon)(\theta_t^{[1], \epsilon})^2 dt + \alpha'(\theta_t^\epsilon)\theta_t^{[2], \epsilon}dt + 2\beta'(\theta_t^\epsilon)\theta_t^{[1], \epsilon}dV_t \\
&\quad + \epsilon\beta''(\theta_t^\epsilon)(\theta_t^{[1], \epsilon})^2 dV_t + \epsilon\beta'(\theta_t^\epsilon)\theta_t^{[2], \epsilon}, \qquad \theta_0^{[2], \epsilon} = 0, \\
&\vdots
\end{align*}
$$

and

$$ d\theta_t^{[0]} = a(\theta_t^{[0]})dt, \qquad \theta_0^{[0]} = 0, \quad (4.16) $$
$$d\theta_t^{[1]} = \alpha'(\theta_t^{[0]})\theta_t^{[1]}dt + \beta(\theta_t^{[0]})dV_t, \qquad \theta_t^{[1]} = 0, \quad (4.17)$$

$$d\theta_t^{[2]} = \alpha''(\theta_t^{[0]})(\theta_t^{[1]})^2 dt + \alpha'(\theta_t^{[0]})\theta_t^{[2]} dt + 2\beta'(\theta_t^{[0]})\theta_t^{[1]} dV_t, \quad \theta_t^{[2]} = 0, \quad (4.18)$$

∴

(see Bichteler et al. (1987) for details).

From these equations, we see that $\theta_t^{[0]}$ is a deterministic process, and $\theta_t^{[1]}$ is a solution of the linear equation (4.17). Using these derivatives, $h(\theta_t^\epsilon)$ can be expanded as

$$h(\theta_t^\epsilon) = h(\theta_t^{[0]}) + h'(\theta_t^{[0]})\theta_t^{[1]}\epsilon + \frac{1}{2}\{h''(\theta_t^{[0]})(\theta_t^{[1]})^2 + h'(\theta_t^{[0]})\theta_t^{[2]}\}\epsilon^2 + \dots \quad (4.19)$$

Furthermore, if we assume $\beta(\theta_t^{[0]}) \neq 0$ for every $t \ge 0$, then (4.17) and (4.18) yields

$$\begin{aligned} \theta_t^{[2]} &= \int_0^t \exp \left( \int_s^t \alpha'(\theta_u^{[0]}) du \right) \alpha''(\theta_s^{[0]})(\theta_s^{[1]})^2 ds \\ &\quad + 2 \int_0^t \exp \left( \int_s^t \alpha'(\theta_u^{[0]}) du \right) \beta'(\theta_t^{[0]}) \theta_t^{[1]} dV_s \\ &= \int_0^t \exp \left( \int_s^t \alpha'(\theta_u^{[0]}) du \right) \alpha''(\theta_s^{[0]})(\theta_s^{[1]})^2 ds \\ &\quad + 2 \int_0^t \exp \left( \int_s^t \alpha'(\theta_u^{[0]}) du \right) \beta'(\theta_t^{[0]}) \theta_t^{[1]} \frac{d\theta_t^{[1]} - \alpha'(\theta_t^{[0]}) dt}{\beta(\theta_t^{[0]})} \end{aligned}$$

In virtue of Itô's formula, this can be transformed into

$$\begin{aligned} \theta_t^{[2]} &= \frac{b'(\theta_t^{[0]})}{b(\theta_t^{[0]})} (\theta_t^{[1]})^2 \\ &\quad + \int_0^t \exp \left( \int_s^t \alpha'(\theta_u^{[0]}) du \right) \\ &\quad \times \left\{ a''(\theta_s^{[0]}) - \frac{\alpha'(\theta_s^{[0]})\beta'(\theta_s^{[0]}) + \beta''(\theta_s^{[0]})}{\beta(\theta_s^{[0]})} + \frac{\beta'(\theta_s^{[0]})}{\beta(\theta_s^{[0]})} \right\} (\theta_s^{[1]})^2 ds \\ &\quad - \int_0^t \exp \left( \int_s^t \alpha'(\theta_u^{[0]}) du \right) \beta'(\theta_s^{[1]}) \beta(\theta_s^{[1]}) ds. \end{aligned}$$

In this way, $\theta_t^{[p]}$ ($p \ge 2$) can be written using polynomials of $\theta_t^{[1]}$ and integrals of polynomials of $\{\theta_s^{[1]}\}_{0 \le s \le t}$. Therefore, $h(\theta_t^\epsilon)$ can be written in the same form. Since $\{\theta_t^{[1]}\}_{t \ge 0}$ is a Gaussian process, this structure is analogous to the previous example, where $X_t$ and $c$ in (4.1) and (4.2) correspond to $\theta_t^{[1]}$ and $\epsilon h'(\theta_t^{[0]})$, respectively.
In what follows, we assume $h(\theta_t^{[0]}) = 0$ (or replace $Y_t$ with $Y_t - \int_0^t h(\theta_s^{[0]})ds$) for simplicity. Based on above arguments, we define the probability measure $\tilde{P}_t$ by

$$ \tilde{P}_t(A) = \frac{E_Q \left[ 1_A \exp \left( \int_0^t \frac{h'(\theta_t^{[0]})\theta_t^{[1]}\epsilon}{\sigma^2} dY_s - \frac{1}{2} \int_0^t \frac{h'(\theta_t^{[0]})^2(\theta_t^{[1]})^2\epsilon^2}{\sigma^2} ds \right) \middle| \mathcal{Y}_t \right]}{E_Q \left[ \exp \left( \int_0^t \frac{h'(\theta_t^{[0]})\theta_t^{[1]}\epsilon}{\sigma^2} dY_s - \frac{1}{2} \int_0^t \frac{h'(\theta_t^{[0]})^2(\theta_t^{[1]})^2\epsilon^2}{\sigma^2} ds \right) \middle| \mathcal{Y}_t \right]}, $$

and write

$$ \begin{aligned} r_t^\epsilon &= h(\theta_t^\epsilon) - h'(\theta_t^{[0]})\theta_t^{[1]}\epsilon \\ &= \frac{1}{2} \left\{ h''(\theta_t^{[0]})(\theta_t^{[1]})^2 + h'(\theta_t^{[0]})\theta_t^{[2]} \right\} \epsilon^2 \\ &\quad + \frac{1}{6} \left\{ h'''(\theta_t^{[0]})(\theta_t^{[1]})^3 + 3h''(\theta_t^{[0]})\theta_t^{[1]}\theta_t^{[2]} + h'(\theta_t^{[0]})\theta_t^{[3]} \right\} \epsilon^3 + \dots \end{aligned} $$

Then in the same way as (4.4), $E[\theta_t^{[p]} | \mathcal{Y}_t] (p \ge 0)$ is expanded as

$$ \begin{aligned} E[\theta_t^{[p]} | \mathcal{Y}_t] = &J_t^0(\theta_t^{[p]}) + \{J_t^1(\theta_t^{[p]}) - J_t^0(\theta_t^{[p]})J_t^1(1)\} \\ &+ \{J_t^2(\theta_t^{[p]}) - J_t^0(\theta_t^{[p]})J_t^2(1) - J_t^1(\theta_t^{[p]})J_t^1(1) + J_t^0(\theta_t^{[p]})J_t^1(1)^2\} \\ &+ \dots + \sum_{k=0}^{n} \sum_{p=1}^{n-k} \sum_{\substack{i_1, \dots, i_p \in \mathbb{N}, \\ i_1 + \dots + i_p = n-k}} (-1)^p J_t^k(\theta_t^{[p]}) J_t^{i_1}(1) \dots J_t^{i_p}(1) \\ &+ O(\epsilon^{2(n+1)}), \end{aligned} $$

where

$$ J_t^i(U) = \frac{1}{\sigma^{2n}} \tilde{E}_t^\epsilon \left[ U \int_0^t \int_0^{t_i} \dots \int_0^{t_2} r_{t_1}^\epsilon \dots r_{t_i}^\epsilon \times (dY_{t_1} - \epsilon h'(\theta_{t_1}^{[0]}) \theta_{t_1}^{[1]} dt_1) \dots (dY_{t_i} - \epsilon h'(\theta_{t_i}^{[0]}) \theta_{t_i}^{[1]} dt_i) \right]. $$

Note that $J_t^i(U) = O(\epsilon^{2i})$ since $r_t^\epsilon = O(\epsilon^2)$. Therefore, (4.15) is expanded as

$$ \begin{aligned} E[\theta_t^\epsilon | \mathcal{Y}_t] &= \theta_t^{[0]} + E[\theta_t^{[1]} | \mathcal{Y}_t] \epsilon + \frac{1}{2} E[\theta_t^{[2]} | \mathcal{Y}_t] \epsilon^2 + \dots \\ &= \theta_t^{[0]} + \tilde{E}_t[\theta_t^{[1]}] \epsilon + \frac{1}{2} \tilde{E}_t[\theta_t^{[2]}] \epsilon^2 \end{aligned} $$
$$
+ \left\{ \frac{1}{6} \tilde{E}_t[\theta_t^{[3]}] \right. \\
+ \frac{1}{2\sigma^2} \tilde{E}_t^{\epsilon} \left[ \theta_t^{[1]} \int_0^t \left\{ h''(\theta_t^{[0]})(\theta_t^{[1]})^2 + h'(\theta_t^{[0]})\theta_t^{[2]} \right\} (dY_t^{\epsilon} - \epsilon h'(\theta_s^{[0]})\theta_s^{[1]} ds) \right] \\
- \left. \frac{1}{2\sigma^2} \tilde{E}_t^{\epsilon} \left[ \theta_t^{[1]} \right] \tilde{E}_t^{\epsilon} \left[ \int_0^t \left\{ h''(\theta_t^{[0]})(\theta_t^{[1]})^2 + h'(\theta_t^{[0]})\theta_t^{[2]} \right\} (dY_t^{\epsilon} - \epsilon h'(\theta_s^{[0]})\theta_s^{[1]} ds) \right] \right\} \epsilon^3 \\
+ \dots
$$

Now, for every fixed $t \ge 0$, let us define $\phi(s; t)$, $\xi_{s;t}$, $\zeta_{s;t}$ ($0 \le s \le t$) by

$$
\begin{align*}
\frac{d\phi}{ds}(s; t) &= -\beta(\theta_s^{[0]})^2 \phi(s; t)^2 - 2\alpha'(\theta_s^{[0]}) \phi(s; t) + \frac{\epsilon h'(\theta_t^{[0]})^2}{\sigma^2}, \quad \phi(t; t) = 0 \\
d_s \xi_{s;t} &= \left\{ \alpha'(\theta_s^{[0]}) + \beta(\theta_s^{[0]})^2 \phi(s; t) \right\} \xi_{s;t} ds + \beta(\theta_s^{[0]}) d\tilde{V}_s, \quad \xi_{s;t} = 0, \\
\zeta_{s;t} &= \xi_{s;t} + \int_0^t \frac{\epsilon h'(\theta_u^{[0]})}{\sigma^2} \mathrm{Cov}(\xi_{s;t}, \xi_{u;t}) dY_u,
\end{align*}
$$

where $\tilde{V}$ is a Brownian motion independent of $Y$ (under $P$). Then the conditional mean and covariance of $\{\zeta_{s;t}\}_{0\le s\le t}$ is given by $E[\zeta_{s;t}|Y_t] = \mu_{s;t}$ and $\mathrm{Cov}(\zeta_{s;t}, \zeta_{u;t}|Y_t) = \mathrm{Cov}(\xi_{s;t}, \xi_{u;t})$, where

$$
\begin{align*}
d_t \mu_{t;t} &= \alpha'(\theta_t^{[0]}) \mu_{t;t} dt + \frac{\epsilon h'(\theta_t^{[0]}) \gamma(t)}{\sigma^2} (dY_t - \epsilon h'(\theta_t^{[0]}) \mu_{t;t} dt), \\
d_t \mu_{s;t} &= \frac{\epsilon h'(\theta_t^{[0]})}{\sigma^2} \mathrm{Cov}(\xi_{s;t}, \xi_{t;t}) (dY_t - \epsilon h'(\theta_t^{[0]}) \mu_{t;t} dt), \\
d_t \mathrm{Cov}(\xi_{t;t}, \xi_{s;t}) &= \left\{ \alpha'(\theta_t^{[0]}) - \frac{\epsilon^2 h'(\theta_t^{[0]})^2 \gamma(t)}{\sigma^2} \right\} \mathrm{Cov}(\xi_{t;t}, \xi_{s;t}) dt, \\
d_t \mathrm{Cov}(\xi_{s;t}, \xi_{u;t}) &= -\frac{\epsilon^2 h'(\theta_t^{[0]})^2}{\sigma^2} \mathrm{Cov}(\xi_{t;t}, \xi_{s;t}) \mathrm{Cov}(\xi_{t;t}, \xi_{u;t}) dt,
\end{align*}
$$

and $\gamma(t) = \mathrm{Cov}(\xi_{t;t}, \xi_{t;t})$ is the solution of

$$
\frac{d}{dt}\gamma(t) = -\frac{\epsilon^2 h'(\theta_t^{[0]})^2}{\sigma^2}\gamma(t)^2 + 2\alpha'(\theta_t^{[0]})\gamma(t) + \beta(\theta_t^{[0]})^2.
$$

Furthermore, the law of {$\theta_s^{[1]}$}$_{0\le s\le t}$ under $\tilde{E}_t$ is equivalent to the law of
{$\zeta_{s;t}$}$_{0\le s\le t}$ given $\mathcal{Y}_t$ under $P$. Therefore, we can calculate the coefficients of
the expansion in the same way as in the previous example, since $\theta_t^{[p]}$ ($p \ge 2$)
is represented by polynomials of {$\theta_s^{[1]}$}$_{0\le s\le t}$ as mentioned above.

The algorithm is more complicated than the first one and has yet to be
implemented on a computer. However, we expect that the modification of
coefficients will not be necessary in this case, since the expansion of $h(\theta_t^\epsilon)$ in
(4.19) is stable over the entire time horizon and divergence will not occur.
A more general case, where we cannot assume $\beta(\theta_t^{[0]}) = 0$, and X and Y are
multi-dimensional, is covered in a subsequent paper.

Acknowledgement

The author is deeply grateful to N. Yoshida for his valuable advice and
insightful discussions.

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