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# On the Markov transformation of Gaussian processes

Armand Ley

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Armand Ley. On the Markov transformation of Gaussian processes. 2024. hal-04724163

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# On the Markov transformation of Gaussian processes

Armand Ley

October 2024

**Abstract:** Given a Gaussian process $(X_t)_{t \in \mathbb{R}}$, we construct a Gaussian Markov process with the same one-dimensional marginals using sequences of transformations of $(X_t)_{t \in \mathbb{R}}$ “made Markov” at finitely many times. We prove that there exists at least such a Markov transform of $(X_t)_{t \in \mathbb{R}}$. In the case the instantaneous decorrelation rate of $(X_t)_{t \in \mathbb{R}}$ is continuous, we prove that the Markov transform is uniquely determined and characterized through the same instantaneous decorrelation rate.

# 1 Introduction

## 1.1 Context and main results

During the last decades, partly under the impulsion of mathematical finance, the question of mimicking stochastic processes has become a recurrent problem. Put in a very general way, the problem can be expressed as follows: Given a stochastic process $(X_t)_{t \in T}$, one can ask if there exists a process $(Y_t)_{t \in T}$ preserving certain properties of $(X_t)_{t \in T}$ while satisfying some additional conditions. To motivate our problem of mimicking Gaussian processes, we now present a selection of four mimicking problems:

1. The Kellerer problem of mimicking a martingale with a Markov martingale process;

2. The Gyöngy problem of mimicking an Itô process by the solution of a SDE;

3. The problem of faking Brownian motion;

4. The problem of mimicking an $\mathbb{R}^2$-valued process by an $\mathbb{R}^2$-valued order-preserving Markov process.

We shall then expose with more details the problem of Boubel and Juillet about mimicking an increasing process for the stochastic order with a Markov process that has non-decreasing trajectories. This mimicking problem is the most important one for this article, as Markov transformation, the construction method used to build its solution, is central in our article.

In a seminal article, Strassen [23] investigated if, given a coupling $(X_t)_{t \in \{1,2\}}$, there exists a coupling $(Y_t)_{t \in \{1,2\}}$ with the same 1-marginals satisfying the martingale property, i.e., $\mathbb{E}(Y_2|Y_1) = Y_1$. He proved that such a coupling exists if and only if $X_1$ is smaller than $X_2$ for the convex order, i.e., $\mathbb{E}(f(X_1)) \le \mathbb{E}(f(X_2))$ for every function $f$ that is convex<fnref n="1" />. Kellerer [16] generalized this result: A real-valued process $(X_t)_{t \in \mathbb{R}}$ can be mimicked by a Markov martingale having the same 1-marginals if and only

<footnote n="1">¹We refer to [21, 22] for more information about stochastic orders.</footnote>
if its components are increasing for the convex order (see also [1, 7, 12]). In a different vein, Gyöngi [10], using an approach suggested by Krylo [17], showed that any Itô process $(X_t)_{t \in \mathbb{R}}$ with coefficients satisfying certain conditions can be mimicked by a Markov process which is a solution of a stochastic differential equation and which has the same one-dimensional marginals. Another mimicking problem is to fake the Brownian motion, that is, to find a non-Brownian process that has as many common features with the Brownian motion as possible. It is known (see e.g. [19]) that the Brownian motion is the only continuous martingale with Brownian marginals that satisfies the strong Markov property. Beiglböck and al. [2], in the line of previous articles (see their introduction for numerous references) finally proved that there exists a non-Brownian continuous martingale with Brownian marginals that is Markov. More recently, for processes valued in product spaces, Bérard and Frénais [8] studied the case of a homogeneous Markov process $(X_t^{x_1}, X_t^{x_2})_{t \ge 0}$ starting at $x_1 \le x_2$ and whose marginal processes $(X_t^{x_i})_{t \ge 0}$ are governed by the same stochastically monotone Feller semi-group. They showed that it can be mimicked by a Feller process $(Y_t^{x_1}, Y_t^{x_2})_{t \ge 0}$ starting at $(x_1, x_2)$, satisfying Law $((X_t^{x_i})_{t \ge 0}) = \text{Law}((Y_t^{x_i})_{t \ge 0})$ for $i = 1, 2$ and $Y_t^{x_1} \le Y_t^{x_2}$ for every $t \in \mathbb{R}_+$.

Boubel and Juillet [7] studied the problem of mimicking a stochastic process $(X_t)_{t \in \mathbb{R}}$ by a Markov process with the same 1-marginals and with non-decreasing trajectories. If we ignore the Markov property, it is well known that there exists a solution if and only if the marginals are increasing for the stochastic order, i.e., $\mathbb{E}(f(X_s)) \le \mathbb{E}(f(X_t))$ for every $s < t \in \mathbb{R}^2$ and $f: \mathbb{R} \to \mathbb{R}$ non-decreasing. An explicit solution is then given by the quantile process, i.e., by $(G_t(U))_{t \in \mathbb{R}}$ where $U$ is a uniformly distributed random variable on $[0, 1]$ and $G_t: q \in [0, 1] \mapsto \inf\{x \in \mathbb{R} ; \mu_t(|-\infty, x|) \ge q\} \in \mathbb{R}$ stands for the quantile function of $\mu_t$. However, as shown in [13], the quantile process is not Markov in general. To describe how their solution to the mimicking problem with the Markov property is obtained, we have to introduce the notions of “transformation of a process made Markov at certain times” and of “Markov transforms”. Given a finite set of times $R \subset \mathbb{R}$ and a process $X = (X_t)_{t \in \mathbb{R}}$, we say that a process $X^R$ is the $2$-transformation of $X$ made Markov at times $R$ if:

* On every interval between two successive times of $R$, $X^R$ and $X$ have the same law;

* For every $r \in R$, $X^R$ is made Markov at time $r$: If one knows the value of the trajectory at time $r$, then the future $(X_t^R)_{t>r}$ of the trajectory does not depend on the past $(X_t^R)_{t<r}$ of the trajectory.

A Markov transform of $X$ is then a Markov process $X'$ obtained as the limit (for the finite-dimensional topology) of a sequence of processes $(X^{R_n})_{n \ge 1}$, with $(R_n)_{n \ge 1}$ an admissible$^3$ sequence of sets of times. Their solution to the mimicking problem, called Markov-quantile process, is then obtained as a Markov transform of the quantile process. Since $X^R$ has the same 1-marginals as $X$, a Markov transform is a Markov process that has the same 1-marginals as $X$. This draws a general construction of mimicking processes. The hope is that making a process Markov at certain times and passing to the limit preserves certain property of the original process while adding the Markov property. In the article, we study this construction in the case of Gaussian processes, confirming its interests but also highlighting its limitations (see Section 5). As expected, it turns out that Markov transforms of “regular” Gaussian processes are solutions of the mimicking problem presented hereby in Theorem A. Assume we have a Gaussian

<footnote>$^2$As all the processes made Markov at times $R$ follow the same law, we will talk about the transformation made Markov at times $R$. A rigorous definition of Markov transforms will be given in Definition 34.</footnote>

<footnote>$^3$If we denote by $\sigma_R := \sup_{x \in \mathbb{R}} d(x, R \setminus \{x\})$ the mesh of a finite set $R \subset \mathbb{R}$, a sequence of sets of times $(R_n)_{n \ge 1}$ is admissible if : $\lim_{n \to +\infty} \inf(R_n) = -\infty$, $\lim_{n \to +\infty} \sup(R_n) = +\infty$ and $\lim_{n \to +\infty} \sigma_{R_n} = 0$.</footnote>
process $X = (X_t)_{t \in \mathbb{R}}$ with a continuous covariance function $K$ and an instantaneous decorrelation rate (or instantaneous decay rate of the correlation) $\alpha^K$ given by

$$ \alpha^K(t) := \lim_{h \to 0^+} \frac{1}{h} \left( 1 - \frac{K(t, t+h)}{\sqrt{K(t,t)}\sqrt{K(t+h, t+h)}} \right), \qquad (1) $$

that is well defined<fnref n="4" /> and continuous. The mimicking problem of interest is to find a Gaussian Markov process that meets the 1-marginals of $X$ and has the same instantaneous decorrelation rate. The following result solves this problem and establishes that, under a reinforced condition, the solution of this mimicking problem is a Markov transform.

**Theorem A.** Let $X = (X_t)_{t \in \mathbb{R}}$ denote a Gaussian process with continuous covariance function $K$ and positive variance function. Assume $\alpha^K$ (recall (1)) is well defined and continuous.

1. Existence: There exists a Gaussian process $Y = (Y_t)_{t \in \mathbb{R}}$ with covariance function $K'$ satisfying:

(1) For every $t \in \mathbb{R}$, $\text{Law}(X_t) = \text{Law}(Y_t)$;

(2) The process $Y$ has the same instantaneous decorrelation rate as $X$, i.e.,

$$ \lim_{h \to 0^+} \frac{1}{h} \left( 1 - \frac{K'(t, t+h)}{\sqrt{K'(t, t)} \sqrt{K'(t+h, t+h)}} \right) = \alpha^K(t); $$

(3) $(Y_t)_{t \in \mathbb{R}}$ is a Markov process.

In the following of the article, if a Gaussian process $Y$ satisfies (1), (2) and (3), we allow ourselves to say that $Y$ is a mimicking process of $X$.

2. Uniqueness in law: Every mimicking process of $X$ with covariance function $K' : \mathbb{R}^2 \to \mathbb{R}$ has the same mean function as $(X_t)_{t \in \mathbb{R}}$ and, for every $s < t \in \mathbb{R}^2$,

$$ K'(s, t) = K(s, s)^{1/2} K(t, t)^{1/2} \exp \left( - \int_s^t \alpha^K(u) du \right). $$

3. The mimicking process is a Markov transform (under the following reinforced hypothesis): Assume

$$ \sup_{v \in [s, t]} \left| \alpha^K(v) - \frac{1}{h} \left( 1 - \frac{K(v, v+h)}{\sqrt{K(v,v)}\sqrt{K(v+h, v+h)}} \right) \right| \underset{h \to 0^+}{\longrightarrow} 0 \qquad (2) $$

for every $s < t \in \mathbb{R}^2$. Let $(R_n)_{n \ge 1} \in \mathcal{A}$ be an admissible sequence and $Y$ be the mimicking process of $X$ (see Point 1 and 2). For every $n \ge 1$, we denote by $X^{R_n}$ the transformation of $X$ made Markov at times $R_n$. Then $(X^{R_n})_{n \ge 1}$ and $Y$ almost surely have continuous paths and $X^{R_n}$ converges weakly to $Y$ on compact sets.

<footnote n="4">⁴As we will see in Section 5, the decay rate of the correlation does not always converge when $h$ goes to $0^+$.</footnote>
Note that, for every $s < t \in \mathbb{R}^2$, the correlation of $(Y_s, Y_t)$ is inversely proportional to the exponential value of the sum of the instantaneous decorrelation rate from $s$ to $t$. Hence, informally, if $X$ is highly correlated between $s$ and $t$, the instantaneous decorrelation rate of $X$ will be small between $s$ and $t$, so that the correlation coefficient of $(Y_s, Y_t)$ will be high. Theorem A will be proved in Theorem 45, where we also prove that our mimicking process is the solution of a given stochastic differential equation (SDE). According to Theorem A, under hypothesis (2), the general method of making a process Markov at certain times and passing to the limit behaves well: Our process $(X_t)_{t \in \mathbb{R}}$ admits a Markov transform and this Markov transform is a Markov process with the same 1-marginal as $X$ that preserves its instantaneous decorrelation rate. However, in general, there is no reason why a process should admit a Markov transform and, if it does, no reason why it should be unique. Given a process $X$, there is no guarantee that one can find an admissible sequence $(R_n)_{n \ge 1}$ and a Markov process $X'$ such that $\lim_{n \to +\infty} X^{R_n} = X'$, nor that it is impossible to find two admissible sequences $(R_n^1)_{n \ge 1}$, $(R_n^2)_{n \ge 1}$ and two distinct Markov processes $X'^1, X'^2$ satisfying $\lim_{n \to +\infty} X^{R_n^1} = X'^1$ and $\lim_{n \to +\infty} X^{R_n^2} = X'^2$. As we will see, it is natural to distinguish two notions of Markov transform: $X'$ is a strong Markov transform if for every admissible sequence $(R_n)_{n \ge 1}$, $\lim_{n \to +\infty} X^{R_n} = X'$, whereas $X'$ is a weak Markov transform of $X$ if there exists an admissible sequence $(R_n)_{n \ge 1}$ such that $\lim_{n \to +\infty} X^{R_n} = X'$. We shall see that it is easier to study a local version of Markov transforms. Instead of requiring that $X'$ is the limit of $X^{R_n}$ for an admissible sequence $(R_n)_{n \ge 1}$, we rather require that, for each pair of times $s < t \in \mathbb{R}^2$, there exists a sequence of partitions $(R_n^{s,t})_{n \ge 1}$ of $[s,t]$ with mesh size going to 0 and such that the laws of $(X_s^{R_n^{s,t}}, X_t^{R_n^{s,t}})_{n \ge 1}$ converge to the law of $(X_s', X_t')$. In this case, we say that $X'$ is a weak local Markov transform of $X$. If every sequence of partitions $(R_n^{s,t})_{n \ge 1}$ leads to convergence, we say that $X'$ is a strong local Markov transform of $X$. This local version of Markov transform is less stringent than the former global version of Markov transform, as we just ask for the convergence of the two-dimensional laws and, more importantly, the time sets dependence on $(s,t)$ is allowed$^5$.

Assuming condition (2), Theorem A implies that there that there exists a (unique) strong global Markov transform that is characterized as the unique solution of our mimicking problem. Similarly, Boubel and Juillet showed that every quantile process admits a unique weak (not strong) global Markov transform (the Markov-quantile process) and they give a characterization of this process in terms of stochastic orders. Hence, they asked [7, §5.5.1, Open Question a)] whether the weak local Markov transform of a process (when it exists) is always unique and, if it is, how to characterize it. The following result shows that a (stationary) Gaussian process can admit infinitely many weak local Markov transforms and undermines the hope to find a nice characterization of the set of weak local Markov transforms in general.

**Theorem B.** There exists a stationary Gaussian process $(X_t)_{t \in \mathbb{R}}$ whose set of weak local Markov transforms is the set of all the processes $(Y_t)_{t \in \mathbb{R}}$ satisfying:

1. The process $(Y_t)_{t \in \mathbb{R}}$ is centered Gaussian with constant variance function equal to 1;

2. The covariance function of $(Y_t)_{t \in \mathbb{R}}$ is non-negative;

3. The process $(Y_t)_{t \in \mathbb{R}}$ is Markov.

<footnote>$^5$We finally end up with four notions of Markov transforms: weak local Markov transform, strong local Markov transform, weak global Markov transform and strong global Markov transform.</footnote>
Hence, the set of Markov transforms of $(X_t)_{t \in \mathbb{R}}$ is not reduced to a singleton but contains a large variety of processes. In particular, Theorem B shows that a weak local Markov transform of a stationary process is not necessarily stationary. Before presenting the organization of the paper, note that all the involved concepts depend only on the law of the involved processes. Hence, we shall work directly with measures on a product space instead of stochastic processes.

## 1.2 Organization of the article

In Section 2, we introduce some notation and give some definitions. At first, we recall the operation of concatenation and composition of transport plans that will often be used in this paper. Then, we thoroughly define the notions of weak local Markov transform and strong local Markov transform of a measure and see how these notions behave relatively to some transformations.

In Section 3, we give some general results on Gaussian measures. We begin by proving the concatenation formula, which is an explicit formula of the law obtained by concatening Gaussian transport plans with non-singular marginals (Lemma 12). This will enable us to give a criterion to find out if a Gaussian measure with non-singular marginals is Markov (Proposition 13), to show that the concatenation of Gaussian measures is a continuous operation (Lemma 14) and to prove that a weak local Markov transform of a Gaussian measure remains a Gaussian measure (Proposition 15). Applying a result of Kellerer [16, Theorem 1], we show the existence of a weak local Markov transform in the case of Gaussian measures with non-singular marginals (Theorem 17). This is the same conclusion as [7, Theorem 2.26], but in the context of Gaussian measures, our kernels do not need be increasing and multi-dimensional marginals will be considered.

In Section 4, we establish a sufficient criterion to prove that a real-valued Gaussian process admits a strong local Markov transform (Theorem 23), which is also a preliminary version of Point 3 of Theorem A. If $K$ and $\alpha_K$ are defined as in Theorem A and we assume that Hypothesis (2) is satisfied, then $X$ admits a strong local Markov transform and this strong Markov transform is also its mimicking process$^6$. We also give a sufficient criterion to identify weak local Markov transforms of a *stationary* Gaussian process, by looking at the cluster points of the decay rate of its correlation function (Theorem 25): If $\alpha$ is a cluster point of this decay rate when $h \to 0^+$, the (renormalized) stationary Ornstein-Uhlenbeck process with parameter $\alpha$ is a weak local Markov transform of $X$. We finally apply our criterion on strong local Markov transforms of Gaussian processes, starting with fractional Brownian motion.

Section 5 is devoted to the proof of Theorem B. First, we use the Weierstrass’s continuous nowhere differentiable functions [11] to construct a probability measure $\mu$ on $\mathbb{R}$ whose Fourier transform has a decay rate that has infinitely many cluster points at $0^+$ (Lemma 30 and Proposition 31). Then, we apply a theorem of Bochner [5] to construct a stationary Gaussian process $(X_t)_{t \in \mathbb{R}}$ using $\mu$. Finally, we apply the identification criterion of Markov transforms of stationary processes proved in Section 4 to establish Theorem B (labelled as Theorem 33).

Finally, in Section 6, after properly defining measures made Markov at times $R$ for a finite set $R \subset \mathbb{R}$ (Definition 34), we study the link between local Markov transforms and global Markov transforms. We shall see that, in the Gaussian case, there is no difference between a strong local Markov transform and a strong global Markov transform (Proposition 38). For stationary Gaussian processes, we show that our criterion to identify weak *local* Markov transforms still holds for weak *global* Markov transforms

<footnote>$^6$Since weak convergence implies finite-dimensional convergence, this result is weaker than Point 3 of Theorem A.</footnote>
(Proposition 39). Then, we apply some standard results about convergence of processes to carry out the convergence from the finite-dimensional topology to the topology on continuous processes associated to the uniform norm (Theorem 43). We finally state and prove Theorem A, to which we add a SDE characterization of the mimicking process (Theorem 45).

## 2 Preliminaries and (weak) Markov transformation

In this section, we fix some generic notation that will be used in the rest of the article.

**Notation 1.** We write $\llbracket a, b \rrbracket := [a, b] \cap \mathbb{N}$ for $(a, b) \in \mathbb{R}^2$ and $\{t_1 < \dots < t_m\}$ (resp. $(t_1 < \dots < t_m)$) for the set $\{t_1, \dots, t_m\}$ (resp. the sequence $(t_1, \dots, t_m)$), when $t_1 < \dots < t_m$. For each measurable space $(E, S)$, we denote by $\mathcal{P}(E)$ (resp. $\mathcal{M}_+(E)$) the set of probability measures (resp. positive finite measures on $E$). Classically, we endow every product space of measurable spaces with its cylindrical $\sigma$-algebra and every topological space with its Borel sets. For a product space $E = \prod_{t \in T} E_t$, if $T' \subset T$, we write $\text{proj}_{T'}$ the projection from $\prod_{t \in T} E_t$ to $\prod_{t \in T'} E_t$. We then denote by $f_{\#} : A \mapsto \gamma((f^{-1}(A))$ the push-forward measure of $\gamma$ by $f$ and set $P^{T'} := \text{proj}_{\#}^{T'} P \in \mathcal{P}(\prod_{t \in T'} E_t)$ for every $P \in \mathcal{P}(\prod_{t \in T} E_t)$. In case $T' = \{t_1, \dots, t_m\}$, we rather denote $P^{T'}$ by $P^{t_1, \dots, t_m}$. Furthermore, for $(\mu_t)_{t \in T} \in \prod_{t \in T} \mathcal{P}(E_t)$, we write $\text{Marg}((\mu_t)_{t \in T}) := \{P \in \mathcal{P}(\prod_{t \in T} E_t) ; \forall t \in T, P^t = \mu_t\}$.

**Notation/Definition 2** (Concatenation and composition). Let $E_1, \dots, E_d$ be Polish spaces, $(\mu_i)_{i \in \llbracket 1, d \rrbracket} \in \prod_{i=1}^d \mathcal{P}(E_i)$, and $(P_i)_{i \in \llbracket 1, d-1 \rrbracket} \in \prod_{i=1}^{d-1} \text{Marg}(\mu_i, \mu_{i+1})$. The concatenation of $P_1, \dots, P_{d-1}$ is the probability measure $P_1 \circ \dots \circ P_{d-1} \in \mathcal{P}(E_1 \times \dots \times E_d)$ defined by

$$ (P_1 \circ \dots \circ P_{d-1})(A_1 \times \dots \times A_d) = \int_{A_1 \times \dots \times A_d} \mu_1(dx_1)k^{1,2}(x_1, dx_2) \dots k^{d-1,d}(x_{d-1}, dx_d), $$

where $k^{i,i+1}$ is the probability kernel defined by the disintegration $P_i(dx_i, dx_{i+1}) = \mu_i(dx_i)k^{i,i+1}(x_i, dx_{i+1})$. Defining $k^{2,1}$ as the kernel given by the disintegration $P_1(dx_1, dx_2) = \mu_2(dx_2)k^{2,1}(x_2, dx_1)$, we leave it to the reader to verify that $(P_1 \circ P_2)(dx_1, dx_2, dx_3) = \mu_2(dx_2)[k^{2,1}(x_2, \cdot) \otimes k^{2,3}(x_2, \cdot)](dx_1, dx_3)$. This means that, conditionally to the present, the future is independent of the past. For $T \subset \mathbb{R}$ and $(E_t)_{t \in T}$ a family of Polish spaces, we say that a probability $P \in \mathcal{P}(\prod_{t \in T} E_t)$ is a Markov measure if for all subset $\{t_1 < \dots < t_m\} \subset T$, $P^{t_1, \dots, t_m} = P^{t_1, t_2} \circ \dots \circ P^{t_{m-1}, t_m}$. Of course the notion of Markov measure is related to the more usual notion of Markov process: we leave it to the reader to verify that a process is Markov (relatively to its canonical filtration) if and only if its law is a Markov measure. The composition $P_1 \circ \dots \circ P_d \in \mathcal{P}(E_1 \times E_d)$ of $P_1, \dots, P_d$ is now defined by $P_1 \circ \dots \circ P_d := \text{proj}_{\#}^{1,d}(P_1 \circ \dots \circ P_d)$. For $s < t \in \mathbb{R}^2$, we write $S_{[s,t]}$ the set of partitions of $[s, t]$, i.e., the sequences $(t_1 < \dots < t_m) \in \mathbb{R}^m$ with $t_1 = s$ and $t_m = t$. If $R = (t_1 < \dots < t_m) \in S_{[s,t]}$, we set $P_{\{R\}}^{s,t} := P^{t_1, t_2} \circ \dots \circ P^{t_{m-1}, t_m}$ and denote by $\sigma_R := \sup_{i \in \llbracket 1, d-1 \rrbracket} |t_{i+1} - t_i|$ the mesh of $R$.

We now define (weak and strong) local Markov transforms of a measure. We stress out that this notion is different from the notion of global Markov transform introduced in Definition 37.

**Definition 3** (Local Markov transform). Let us consider an interval $T \subset \mathbb{R}$, $d \ge 1$ and a measure $P \in \mathcal{P}(\mathbb{R}^d)^T)$.
1. We say that $P$ admits a weak local Markov transform if there exists a Markov measure $P'$ such that

$$ \forall s < t \in T^2, \exists (R_n)_{n \ge 1} \in (S_{[s,t]})^{\mathbb{N}^*}, \left\{ \begin{array}{l} \lim_{n \to +\infty} P_{\{R_n\}}^{s,t} = P'^{s,t} \\ \lim_{n \to +\infty} \sigma_{R_n} = 0 \end{array} \right. . $$

In this case, we say that $P'$ is a weak local Markov transform of $P$.

2. We say that $P$ admits a strong local Markov transform if there exists a Markov measure $P'$ such that

$$ \forall s < t \in T^2, \forall (R_n)_{n \ge 1} \in (S_{[s,t]})^{\mathbb{N}^*}: \lim_{n \to +\infty} \sigma_{R_n} = 0 \implies \lim_{n \to +\infty} P_{\{R_n\}}^{s,t} = P'^{s,t}. $$

In this case, we say that $P'$ is a strong Markov transform of $P$.

3. Given two $\mathbb{R}^d$-valued stochastic processes $(X_t)_{t \in T}$ and $(Y_t)_{t \in T}$, we say that $(Y_t)_{t \in T}$ is a weak (resp. strong) local Markov transform of $(X_t)_{t \in T}$ if the law of $(Y_t)_{t \in T}$ is a weak (resp. strong) Markov transform of the law of $(X_t)_{t \in T}$.

**Remark 4.** 1. If $P'$ is a strong local Markov transform of a measure $P \in \mathcal{P}((\mathbb{R}^d)^T)$ and $Q$ is a weak local Markov transform of $P$, then $P' = Q$. Indeed, for every $s < t \in T^2$, there exists $(R_n)_{n \ge 1} \in (S_{[s,t]})^{\mathbb{N}^*}$ such that $Q^{s,t} = \lim_{n \to +\infty} P_{\{R_n\}}^{s,t}$ and $\lim_{n \to +\infty} \sigma_{R_n} = 0$. As $P$ is a strong local Markov transform, we have $P'^{s,t} = \lim_{n \to +\infty} P_{\{R_n\}}^{s,t}$, hence $P'^{s,t} = Q^{s,t}$. Since a Markov measure is completely characterized by its two-dimensional laws, this implies $P' = Q$. In particular, there is at most one strong local Markov transform, and if a measure $P$ admits a strong local Markov transform, we will talk about the strong local Markov transform of $P$.

2. A strong local Markov transform of $P$ is clearly a weak local Markov transform of $P$, but the converse is false, even when the weak local Markov transform is unique. For instance, if $(Y_t)_{t \in \mathbb{R}}$ is a sequence of i.i.d. random variables with law $\mathcal{N}(0, 1)$ and $P$ is the law of the process $X = (Y_01_{t \in \mathbb{Q}} + Y_t1_{t \in \mathbb{R} \setminus \mathbb{Q}})_{t \in \mathbb{R}}$, we leave it to the reader to verify that $P' := \otimes_{t \in \mathbb{R}} \mathcal{N}(0, 1)$ is the unique weak local Markov transform of $P$, but is not a strong local Markov transform of $P$.

3. If $P$ is a Markov process, then $P_{\{R\}}^{s,t} = P'^{s,t}$ for every $R \in S_{[s,t]}$, so that $P$ is a strong local Markov transform of itself. In particular, $P$ is the only weak local Markov transform of $P$.

In [7, Theorem 2.26], Boubel and Juillet showed an existence result of a weak local Markov transform when the measure $P$ has increasing kernels in the sense given below. We denote by $\le_{st}$ the stochastic order on $\mathcal{P}(\mathbb{R})$, namely $\mu \le_{st} \nu$ if, for every non-decreasing bounded function $f$, $\int_{\mathbb{R}} f d\mu \le \int_{\mathbb{R}} f d\nu$.

**Definition 5.** 1. Consider $\mu, \nu \in \mathcal{P}(\mathbb{R})$ and $P \in \text{Marg}(\mu, \nu)$. We say that $P$ has increasing kernels for the stochastic order if there exists a disintegration $P(dx, dy) = \mu(dx)k_x(dy)$ and a Borel set $\Gamma$ such that $\mu(\Gamma) = 1$ and $k_x \le_{st} k_y$ for every $x < y \in \Gamma^2$,

2. Consider $T \subset \mathbb{R}$ and $P \in \mathcal{P}(\mathbb{R}^T)$. We say that $P$ has increasing kernels for the stochastic order if, for every $s < t \in T^2$, $P^{s,t}$ has increasing kernels for the stochastic order.

For more informations about the stochastic order and other orders on probability spaces, we refer to the monographs [21, 22]. For additional information about increasing kernels, we refer to [7, Proposition/Definition 3.11.].
**Theorem 6 (Boubel-Juillet).** Consider an interval $T \subset \mathbb{R}$ and a probability measure $P \in \mathcal{P}(\mathbb{R}^T)$. If $P$ has increasing kernels for the stochastic order, then $P$ admits a weak local Markov transform.

In [7], Theorem 6 was proved and written with $T = \mathbb{R}$, but an easy modification shows that it stays true for any interval $T \subset \mathbb{R}$. The following proposition indicates how Markov transforms behave relatively to a change in time and a transformation “component by component”. Given $S, T \subset \mathbb{R}$, $d \ge 1$, $P \in \mathcal{P}((\mathbb{R}^d)^T)$ and $\phi : S \to T$, we denote by $P^\phi$ the measure on $\mathbb{R}^S$ with finite-dimensional laws $(P^\phi)^{s_1, \dots, s_m} := P^{\phi(s_1), \dots, \phi(s_m)}$, $s_1 < \dots < s_m \in S^m$. If $(X_t)_{t \in T}$ is a stochastic process with law $P$, then $P^\phi$ is the law of the time-changed process $(X_{\phi(s)})_{s \in S}$.

**Proposition 7.** We denote by $T \subset \mathbb{R}$ an interval, we fix $d \ge 1$ and we consider $P, P' \in \mathcal{P}((\mathbb{R}^d)^T)$.

1. Let $(f_t)_{t \in T}$ be a family of continuous injective functions from $\mathbb{R}^d$ to $\mathbb{R}^d$ and set $f := (\otimes_{t \in T} f_t) : (x_t)_{t \in T} \in (\mathbb{R}^d)^T \mapsto (f_t(x_t))_{t \in T} \in (\mathbb{R}^d)^T$. If $P'$ is a weak (resp. the strong) local Markov transform of $P$, then $f_\# P'$ is a weak (resp. the strong) local Markov transform of $f_\# P$.

2. Let $S, T \subset \mathbb{R}$ be two intervals and $\phi : S \to T$ be a strictly increasing continuous function. If $P'$ is a weak (resp. the strong) local Markov transform of $P$, then $P'^\phi \in \mathcal{P}((\mathbb{R}^d)^S)$ is a weak (resp. the strong) local Markov transform of $P^\phi \in \mathcal{P}((\mathbb{R}^d)^S)$.

*Proof.* 1. Fix $s < t \in \mathbb{R}^2$. We leave it to the reader to verify that, by injectivity of the functions $(f_t)_{t \in \mathbb{R}}$, the Markov property transmits from $P'$ to $f_\# P'$ and $(f_\# Q)^{s,t}_{\{R_n\}} = (f_s \otimes f_t)_\# Q^{s,t}_{\{R_n\}}$ for every $Q \in \mathcal{P}((\mathbb{R}^d)^T)$, $R \in S_{[s,t]}$. So, if $\lim_{n \to +\infty} P^{s,t}_{\{R_n\}} = P'^{s,t}$, the continuity of $f_s \otimes f_t$ leads to $(f_\# P)^{s,t}_{\{R_n\}} = (f_s \otimes f_t)_\# P^{s,t}_{\{R_n\}} \xrightarrow{n \to +\infty} (f_s \otimes f_t)_\# P'^{s,t} = (f_\# P')^{s,t}$. This shows the result.

2. First, assume $P'$ is the strong local Markov transform of $P$. Fix $s < t \in S^2$ and $(R_n)_{n \ge 1} \in (S_{[s,t]})^{\mathbb{N}*}$ such that $\lim_{n \to +\infty} \sigma_{R_n} = 0$. Since $\phi$ is strictly increasing and uniformly continuous on $[s,t]$, $(\phi(R_n))_{n \ge 1} \in (S_{[\phi(s),\phi(t)]})^{\mathbb{N}*}$ and satisfies $\lim_{n \to +\infty} \sigma_{\phi(R_n)} = 0$. Thus $(P^\phi)^{s,t}_{\{R_n\}} = P^{\phi(s),\phi(t)}_{\{\phi(R_n)\}} \xrightarrow{n \to +\infty} (P')^{\phi(s),\phi(t)} = (P^\phi)^{s,t}$, which shows that $P^\phi$ is the strong local Markov transform of $P$. Now, assume $P'$ is a weak local Markov transform of $P$ and fix $s < t \in S^2$. We have to find a sequence $(R_n)_{n \ge 1} \in (S_{[s,t]})^{\mathbb{N}*}$ such that $\lim_{n \to +\infty} \sigma_{R_n} = 0$ and $\lim_{n \to +\infty} (P^\phi)^{s,t}_{\{R_n\}} = (P'^\phi)^{s,t}$. As $\phi(s) < \phi(t)$ and $P'$ is a weak local Markov transform of $P$, there exists a sequence $(S_n)_{n \ge 1} \in (S_{[\phi(s),\phi(t)]})^{\mathbb{N}*}$ such that $\lim_{n \to +\infty} \sigma_{S_n} = 0$ and $\lim_{n \to +\infty} P^{\phi(s),\phi(t)}_{\{S_n\}} = P'^{\phi(s),\phi(t)} = (P'^\phi)^{s,t}$. We set $R_n := \phi^{-1}(S_n)$. Since $\phi^{-1} : [\phi(s), \phi(t)] \to [s,t]$ is strictly increasing, uniformly continuous and $(P^\phi)^{s,t}_{\{R_n\}} = P^{\phi(s),\phi(t)}_{\{S_n\}}$, the sequence $(R_n)_{n \ge 1}$ meets the requirement. $\square$

**Remark 8.** • In terms of random variables, Proposition 7 tells us that if the law of $(Y_t)_{t \in T}$ is a weak (resp. the strong) local Markov transform of the law of $(X_t)_{t \in T}$, then the law of $(f_{\phi(s)}(Y_{\phi(s)}))_{s \in S}$ is a weak (resp. the strong) local Markov transform of the law of $(f_{\phi(s)}(X_{\phi(s)}))_{s \in S}$.

• Consider $S, T$ two intervals, $u : S \to \mathbb{R}^*$ and $\phi : S \to T$ a strictly injective continuous function. Assume $P \in \mathcal{P}(\mathbb{R}^T)$ is a centered Gaussian process with covariance function $K$ and $P' \in \mathcal{P}(\mathbb{R}^S)$ is
a weak (resp. the strong) Markov transform of $P$ with covariance function $K$. Applying Proposition 7, it is straightforward that the centered Gaussian process with covariance $(s,t) \in S^2 \mapsto u(s)u(t)K'(\phi(s), \phi(t))$ is a weak (resp. the strong) local Markov transform of the centered Gaussian process with covariance function $(s,t) \in S^2 \mapsto u(s)u(t)K(\phi(s), \phi(t))$.

For $\gamma \in \mathcal{P}(\mathbb{R}^d)$, we denote by $m_\gamma \in \mathbb{R}^d$ (resp. $\Sigma_\gamma \in \mathcal{M}_d(\mathbb{R})$) the expected value of $\gamma$ (resp. the covariance of $\gamma$), i.e., the expected value (resp. the covariance matrix) of a $\mathbb{R}^d$-valued random variable with law $\gamma$.

**Remark 9.** In the rest of the article, we work under the hypothesis of non-singular marginals, i.e., $\Sigma_{\mu_t} \in \text{GL}_d(\mathbb{R})$ for every $t \in T$. In this case, we can apply Proposition 7 with $f_t : x_t \mapsto \Sigma_{\mu_t}^{-1/2}(x_t - m_{\mu_t})$ and $f_t^{-1} : u_t \mapsto \Sigma_{\mu_t}^{1/2}x_t + m_{\mu_t}$, where $\Sigma_{\mu_t}^{1/2}$ stands for the symmetric and positive-definite square root of $\Sigma_{\mu_t}$. So, for a given measure $P \in \mathcal{P}((\mathbb{R}^d)^T)$, $P'$ is a weak (resp. the strong) Markov transform of $P$ if and only if $(\otimes_{t \in T} f_t)_\# P'$ is a weak (resp. the strong) Markov transform of $(\otimes_{t \in T} f_t)_\# P$. Since $(\otimes_{t \in T} f_t)_\# P' \in \text{Marg}((\tilde{\mu}_t)_{t \in T})$, where $\tilde{\mu}_t := f_t_\# \mu_t$ satisfies $m_{\tilde{\mu}_t} = 0$ and $\Sigma_{\tilde{\mu}_t} = I_d$, we can restrict our study to Markov transforms of centered processes with constant covariance function equal to the identity matrix.

### 3 Composition and Markov transformation of Gaussian measures

For every $d \ge 1$, we write $\mathcal{G}_d$ the set of centered Gaussian measures and $\mathcal{G}_d^*$ the set of centered Gaussian measures on $\mathbb{R}^d$ with invertible covariance matrix. Consider $(d_1, d_2) \in \mathbb{N}^* \times \mathbb{N}^*$, $\pi \in \mathcal{P}(\mathbb{R}^{d_1} \times \mathbb{R}^{d_2})$ and $(X_1, X_2)$ a $(\mathbb{R}^{d_1} \times \mathbb{R}^{d_2})$-valued random variable with law $\pi$. We denote by $\Sigma_\pi^{d_1, d_2} \in \mathcal{M}_{d_1, d_2}(\mathbb{R})$ the covariance between $X_1$ and $X_2$. The following lemma is well known and characterizes the weak convergence of (centered) Gaussian measures by the convergence of their covariance matrices. We refer for instance to [3, Ch. 8, Theorem 3] for a proof.

**Lemma 10.** Let consider $(\mu_n)_{n \ge 1} \in (\mathcal{G}_d)^{\mathbb{N}^*}$ and $\mu \in \mathcal{P}(\mathbb{R}^d)$. The following conditions are equivalent.

1. The measure $\mu$ is centered Gaussian and $\Sigma_\mu = \lim_{n \to +\infty} \Sigma_{\mu_n}$.

2. The sequence $(\mu_n)_{n \ge 1}$ weakly converges to $\mu$.

We now recall a standard result about conditional laws of Gaussian vectors. We refer to [24, Chapter 8, Section 9] for a proof.

**Lemma 11 (Conditioning of Gaussian measures).** Let $(m, n)$ be in $\mathbb{N}^* \times \mathbb{N}^*$.

1. Fix $\pi \in \mathcal{P}(\mathbb{R}^{m+n})$ a Gaussian measure and set $\mu := \text{proj}^{1, \dots, m}_{\#\pi}$, $\nu := \text{proj}^{m+1, \dots, m+n}_{\#\pi}$, $\Sigma_\pi := \Sigma_{\pi}^{m,n} \in \mathcal{M}_{m,n}(\mathbb{R})$. If $\mu \in \mathcal{G}_m^*$ and $k : \mathbb{R}^m \to \mathcal{P}(\mathbb{R}^n)$ is a probability kernel such that $\pi(dx, dy) = \mu(dx)k_x(dy)$, then

$$ \mu(dx) - \text{a.s., } k_x = \mathcal{N} \left( \Sigma_{\pi}^t \Sigma_{\mu}^{-1} x, \Sigma_{\nu} - \Sigma_{\pi}^t \Sigma_{\mu}^{-1} \Sigma_{\pi} \right). $$

2. Consider $\mu \in \mathcal{G}_m^*$, $(A, \Gamma) \in \mathcal{M}_{n,m}(\mathbb{R}) \times \mathcal{M}_n(\mathbb{R})$ and $k : x \in \mathbb{R}^m \to \mathcal{N}(Ax, \Gamma) \in \mathcal{P}(\mathbb{R}^n)$. Then, writing $\pi(dx, dy) = \mu(dx)k_x(dy)$, we have

$$ \pi = \mathcal{N} \left( 0_{\mathbb{R}^{m+n}}, \begin{pmatrix} \Sigma_{\mu} & \Sigma_{\mu} A^t \\ A \Sigma_{\mu} & \Gamma + A \Sigma_{\mu} A^t \end{pmatrix} \right) \in \mathcal{G}_{m+n}. $$
For $\mu \in \mathcal{P}(\mathbb{R}^m)$, $\nu \in \mathcal{P}(\mathbb{R}^n)$, we set $\mathcal{G}(\mu, \nu) := \mathcal{G}_{m+n} \cap \mathrm{Marg}(\mu, \nu)$. Using Lemma 11, we obtain an explicit formula for the concatenation and the composition of Gaussian measures.

**Lemma 12.** Let us consider $(\mu_1, \dots, \mu_p) \in \mathcal{G}_{d_1}^* \times \dots \times \mathcal{G}_{d_p}^*$ and $(P_i)_{i \in [1, p-1]} \in \prod_{i=1}^{p-1} \mathcal{G}(\mu_i, \mu_{i+1})$. We denote $\Sigma_{P_i}^{d_i, d_{i+1}}$ by $\Sigma_{i, i+1}$ and $\Sigma_{\mu_i}$ by $\Sigma_{i, i}$.

**Concatenation formula:** $P_1 \circ \dots \circ P_{p-1} = \mathcal{N}(0, (A_{i,j})_{1 \le i, j \le p})$, where $A_{i,j} \in \mathcal{M}_{d_i, d_j}(\mathbb{R})$ is defined by

$$A_{i,j} = \begin{cases} \Sigma_{i,i+1} \Sigma_{i+1,i+1}^{-1} \Sigma_{i+1,i+2} \dots \Sigma_{j-1,j-1}^{-1} \Sigma_{j-1,j} \in \mathcal{M}_{d_i, d_j}(\mathbb{R}) & \text{if } i < j \\ \Sigma_{i,i} & \text{if } i = j \\ A_{j,i}^t & \text{if } j < i \end{cases}$$

**Composition formula:**

$$P_1 \circ \dots \circ P_{p-1} = \mathcal{N} \left( 0, \left( \begin{array}{cc} \Sigma_{\mu_1} & \Sigma_{1,2} \Sigma_{2,2}^{-1} \Sigma_{2,3} \dots \Sigma_{p-1,p-1}^{-1} \Sigma_{p-1,p} \\ (\Sigma_{1,2} \Sigma_{2,2}^{-1} \Sigma_{2,3} \dots \Sigma_{p-1,p-1}^{-1} \Sigma_{p-1,p})^t & \Sigma_{\mu_p} \end{array} \right) \right).$$

*Proof.* The proof of the concatenation formula is based on a recursion on $p$, Lemma 11 and the decomposition $P_{1,2} \circ P_{2,3} = \mu_2(dx_2)[k^{2,1}(x_2, \cdot) \otimes k^{2,3}(x_2, \cdot)](dx_1, dx_3)$ (see Notation/Definition 2). The composition formula is then immediately implied by the concatenation formula and the projection on $\mathbb{R}^{d_1} \times \mathbb{R}^{d_p}$. $\square$

In the case where the covariance matrices are identity matrices, for every $i < j \in [1, d]^2$, $A_{i,j} = \Sigma_{i,i+1} \dots \Sigma_{j-1,j} \in \mathcal{M}_{d_i, d_j}$ is the product of the $j-i$ matrices successive matrices $\Sigma_{k,k+1}$ starting with $k=i$. Lemma 12 allows us to recover$^7$ a characterization of the Markov property for Gaussian measures.

**Proposition 13.** Let fix $d \ge 1$, $T \subset \mathbb{R}$, $(\mu_t)_{t \in T} \in (\mathcal{G}_d^*)^T$ and denote by $P \in \mathrm{Marg}((\mu_t)_{t \in T})$ a centered Gaussian measure. For $s < t \in T^2$, we set $\Sigma_{t,t} := \Sigma_{\mu_t}$ and $\Sigma_{s,t} := \Sigma_{P_{s,t}}^{d,d}$. The measure $P$ is Markov if and only if

$$\forall s < t < u \in T^3, \Sigma_{s,u} = \Sigma_{s,t} \Sigma_{t,u}^{-1} \Sigma_{t,u}. \quad (3)$$

*Proof.* If $P$ is Markov, then for every $s < t < u \in T^3$, according to the composition formula, we have $\Sigma_{s,u} = \Sigma_{P_{s,u}}^{d,d} = \Sigma_{P_{s,t}, P_{t,u}}^{d,d} = \Sigma_{\mu_t}^{d,d} \Sigma_{P_{t,u}}^{-1} = \Sigma_{s,t} \Sigma_{t,t}^{-1} \Sigma_{t,u}$. For the converse implication, we assume that Hypothesis (3) is true and we want to show that $P^{t_0, \dots, t_m} = P^{t_0, t_1} \circ \dots \circ P^{t_{m-1}, t_m}$ for all $t_0 < \dots < t_m \in T^{m+1}$. Since, for every $i < j \in [1, d-1]$, applying (3) recursively, we obtain $\Sigma_{t_i, t_{i+1}} \Sigma_{t_{i+1}, t_{i+1}}^{-1} \Sigma_{t_{i+1}, t_{i+2}} \dots \Sigma_{t_{j-1}, t_{j-1}}^{-1} \Sigma_{t_{j-1}, t_j} = \Sigma_{t_i, t_j}$. According to the concatenation formula, $P^{t_0, \dots, t_m}$ and $P^{t_0, t_1} \circ \dots \circ P^{t_{m-1}, t_m}$ are two centered Gaussian measures with the same covariance matrix, hence are equal. $\square$

Using Lemma 10 and the concatenation formula of Lemma 12, we obtain the continuity of concatenation and composition on product spaces of Gaussian transport plans.

**Lemma 14.** Let us consider $(\mu_1, \dots, \mu_p) \in \mathcal{G}_{d_1}^* \times \dots \times \mathcal{G}_{d_p}^*$, $P_i \in \mathcal{G}(\mu_i, \mu_{i+1})$ and, for $i \in [1, p-1]$, $(P_i^n)_{n \ge 1} \in \mathcal{G}(\mu_i, \mu_{i+1})^{\mathbb{N}^*}$. If $\lim_{n \to +\infty} P_i^n = P_i$ for every $i \in [1, d-1]$, then $P_1 \circ \dots \circ P_p = \lim_{n \to +\infty} P_1^n \circ \dots \circ P_p^n$ and $P_1 \circ \dots \circ P_p = \lim_{n \to +\infty} P_i^n \circ \dots \circ P_p^n$.

<footnote>$^7$This criterion seems to be known for a long time [6], but to the best of our knowledge, not its multi-dimensional version.</footnote>
**Proof.** According to Lemma 10, for every $i < j \in [1, d-1]$, the function $A_{i,j} : (P_1, \dots, P_d) \in \mathcal{G}(\mu_1, \mu_2) \times \dots \times \mathcal{G}(\mu_{p-1}, \mu_p) \mapsto \sum_{P_i}^{d_i, d_{i+1}} \sum_{\mu_i}^{-1} \dots \sum_{\mu_{j-1}}^{-1} \sum_{P_i}^{d_{j-1}, d_j}$ is continuous. According to Lemma 10 and the concatenation formula in Lemma 12, the function $\mathcal{C} : (P_1, \dots, P_{d-1}) \in \mathcal{G}(\mu_1, \mu_2) \times \dots \times \mathcal{G}(\mu_{d-1}, \mu_d) \mapsto P_1 \circ \dots \circ P_{d-1}$ is continuous. The continuity of the composition is then a consequence of the continuity of projections. $\square$

Applying Lemma 12, we obtain that a weak local Markov transform of a Gaussian measure is a Gaussian measure.

**Proposition 15.** Fix an interval $T \subset \mathbb{R}$, $d \ge 1$, $(\mu_t)_{t \in T} \in (\mathcal{G}_d^*)^T$ and denote by $P \in \mathcal{P}((\mathbb{R}^d)^T)$ a Gaussian measure. If $P' \in \mathcal{P}((\mathbb{R}^d)^T)$ is a weak Markov transform of $P$, then $P'$ is a Gaussian measure.

**Proof.** According to Remark 9, we can assume that our $(\mu_l)_{l \in T}$ are centered. Fix $s < t \in T^2$ and a sequence $(R_n)_{n \ge 1} \in (S_{[s,t]})^{\mathbb{N}^*}$ such that $\lim_{n \to +\infty} P_{\{R_n\}}^{s,t} = P^{s,t}$. According to Lemma 12, $P_{\{R_n\}}^{s,t} \in \mathcal{G}(\mu_s, \mu_t)$. It is well known that $\text{Marg}(\mu_s, \mu_t)$ is closed and according to Lemma 10, a limit of a sequence of centered Gaussian measures is a centered Gaussian measure. Thus $\mathcal{G}(\mu_s, \mu_t)$ is closed and we obtain $P^{s,t} \in \mathcal{G}(\mu_s, \mu_t)$. Hence, for each $t_1 < \dots < t_p \in T^p$, since $P'$ is a Markov measure, $P^{t_1, \dots, t_p} = P^{t_1, t_2 \circ \dots \circ t_p}$ is Gaussian, which proves the desired result. $\square$

In the context of Gaussian measures, the hypothesis of increasing kernel in Theorem 6 can be removed. For this purpose, we adapt the proof of Boubel-Juillet [7, Theorem 2.26.]. We first recall a theorem of Kellerer, main tool of the proof (see [16, Theorem 1]). This theorem is an existence result of a Markov measure satisfying certain constraints. It generalizes the standard Kolmogorov extension theorem of a Markov process fitting a consistent family of two-dimensional laws (just take $\mathcal{N}^{s,t} = \{\mu_{s,t}\}$ below).

**Theorem 16.** Let $T$ be an interval, and $(\mu_t)_{t \in T}$ a family of probability measures on some Polish space $E$. For every $s < t \in T^2$, we consider a subset $\mathcal{N}_{s,t}$ of $\mathcal{P}(E^2)$. Assume that, for every $s < t \in T^2$:

(1) $\mathcal{N}_{s,t}$ is non-empty;

(2) $\mathcal{N}_{s,t} \subset \text{Marg}(\mu_s, \mu_t)$;

(3) $\mathcal{N}_{s,t}$ is closed for the weak topology;

(4) For every $r < s < t \in T^3$ and $(P, P') \in \mathcal{N}_{r,s} \times \mathcal{N}_{s,t}$, $P \cdot P' \in \mathcal{N}_{r,t}$;

(5) For every $d \ge 1$ and $t_1 < \dots < t_d \in T^d$, if for every $i \in [1, d-1]$ the sequences $(Q_{t_i, t_{i+1}}^n)_{n \ge 1} \in (\mathcal{N}_{t_i, t_{i+1}})^{\mathbb{N}^*}$ converge weakly to $Q_{t_i, t_{i+1}}$, then the sequence $(Q_{t_1, t_2}^n \circ \dots \circ Q_{t_{d-1}, t_d}^n)_{n \ge 1}$ tends weakly to $Q_{t_1, t_2} \circ \dots \circ Q_{t_{d-1}, t_d}$.

Then, there exists a Markov measure $P \in \text{Marg}((\mu_t)_{t \in T})$ satisfying $\text{proj}_{\#}^{s,t} P \in \mathcal{N}_{s,t}$ for every $s < t \in T^2$.

If $R = (r_1, \dots, r_p)$ and $S = (s_1, \dots, s_q)$ are such that $r_p = s_1$, we denote $(r_1, \dots, r_p, s_2, \dots, s_p)$ by $R + S \in S_{[r_1, s_q]}$.

**Theorem 17.** Consider an interval $T \subset \mathbb{R}$ and $(\mu_t)_{t \in T} \in (\mathcal{G}_d^*)^T$. Then, every Gaussian measure $P \in \text{Marg}((\mu_t)_{t \in T})$ admits a weak local Markov transform.
Proof. For every $s < t \in T^2$ and $\sigma > 0$, put $\mathcal{N}_{s,t}^{\sigma} := \{ P_{\{R\}}^{s,t} ; R \in \mathcal{S}_{[s,t]} \text{ and } \sigma_R \le \sigma \}$. For each $s < t$, we set $\mathcal{N}_{s,t} := \cap_{\sigma>0} \overline{\mathcal{N}}_{s,t}^{\sigma}$. In order to apply Theorem 16, we establish that the conditions (1) to (5) are fulfilled. First recall that $\mathcal{G}(\mu_s, \mu_t) \subset \text{Marg}(\mu_s, \mu_t)$ is closed and $\text{Marg}(\mu_s, \mu_t)$ is compact, so that $\mathcal{G}(\mu_s, \mu_t)$ is compact. For every $s < t \in T^2$ and $\sigma > 0$, according to the composition formula of Lemma 12, we get $\emptyset \subsetneq \mathcal{N}_{s,t}^{\sigma} \subset \mathcal{G}(\mu_s, \mu_t)$, which implies $\overline{\mathcal{N}}_{s,t}^{\sigma} \subset \mathcal{G}(\mu_s, \mu_t)$. Hence, $\mathcal{N}_{s,t}$ is a decreasing intersection of non-empty compact subsets of $\mathcal{G}(\mu_s, \mu_t)$, thus a non-empty compact subset of $\mathcal{G}(\mu_s, \mu_t)$. This establishes (1), (2) and (3). To prove (4), consider $Q_1 \in \mathcal{N}_{s,t}$, $Q_2 \in \mathcal{N}_{t,u}$ and $\sigma > 0$. There exists two sequences of partitions $(R_n)_{n \ge 1} \in (\mathcal{S}_{[s,t]})^{\mathbb{N}^*}$ and $(S_n)_{n \ge 1} \in (\mathcal{S}_{[t,u]})^{\mathbb{N}^*}$ such that $Q_1 = \lim_{n \to +\infty} P_{\{R_n\}}^{s,t}$, $Q_2 = \lim_{n \to +\infty} P_{\{S_n\}}^{t,u}$ and $\max(\sigma_{R_n}, \sigma_{S_n}) \le \sigma$. According to Lemma 14, we get $P_{\{R_n+S_n\}}^{s,u} = P_{\{R_n\}}^{s,t} \cdot P_{\{S_n\}}^{t,u} \underset{n \to +\infty}{\longrightarrow} Q_1 \cdot Q_2$. Since $\sigma_{R_n+S_n} = \max(\sigma_{R_n}, \sigma_{S_n}) \le \sigma$, we have $Q_1 \cdot Q_2 \in \overline{\mathcal{N}}_{s,u}^{\sigma}$. This being true for all $\sigma > 0$, $Q_1 \cdot Q_2 \in \mathcal{N}_{s,u}$, so that (4) is true. For (5), just recall that $\mathcal{N}_{s,t} \subset \mathcal{G}(\mu_s, \mu_t)$ and apply Lemma 14. Thus, Theorem 16 applies and there exists a Markov measure $P' \in \mathcal{P}(\mathbb{R}^T)$ satisfying $P'^{s,t} \in \mathcal{N}_{s,t}$ for every $s < t \in T^2$. This means exactly that $P'$ is a weak local Markov transform of $P$. $\square$

Theorem 17 improves Theorem 6 for Gaussian measures. First, our result is valid for any $(\mu_t)_{t \in T} \in (\mathcal{G}_d^*)^{\mathbb{N}^*}$ with $d \ge 1$, whereas Theorem 6 only applies when $d = 1$. Moreover if $d = 1$, we do not ask that $P \in \text{Marg}((\mu_t)_{t \in T})$ has increasing kernels. For a Gaussian process, having increasing kernels means having a non-negative covariance function. Indeed, for $P \in \text{Marg}((\mu_t)_{t \in T})$ with covariance function $K$, according to the first point of Lemma 11, $P^{s,t}(dx, dy) = \mu_t(dx)k_x^{s,t}(dy)$ with $k_x^{s,t} = \mathcal{N}(K(s,t)/K(s,s)x, K(t,t) - K(s,t)^2/K(s,s))$. Recall that, for every $(x, y, \sigma) \in \mathbb{R}^2 \times \mathbb{R}_+$, $\mathcal{N}(x, \sigma) \le s$ and $\mathcal{N}(y, \sigma)$ if and only if $x \le y$. Hence, $P^{s,t}$ has increasing kernels if and only if $K(s,t) \ge 0$, which proves that $P$ has increasing kernels if and only if $K$ only takes non-negative values.

# 4 Identification criteria of Markov transform for Gaussian measure.

From now on, we restrict ourselves to real-valued Gaussian processes. The proof of the following proposition is straightforward, but the result is nevertheless crucial. It uses that, for every Gaussian process $P$, $s < t \in \mathbb{R}^2$ and $R = (t_0 < \dots < t_m)$, the correlation coefficient of $P_{\{R\}}^{s,t}$ is the product of the correlation coefficients of $P^{t_0, t_1}, \dots, P^{t_{d-1}, t_d}$.

**Proposition 18.** Fix $(\mu_t)_{t \in T} \in (\mathcal{G}_1^*)^T$ and denote by $P, P' \in \text{Marg}((\mu_t)_{t \in T})$ two Gaussian processes with covariance functions $K$ and $K'$ respectively. For every $s < t \in T^2$ and $(R_n)_{n \ge 1} = ((t_k^n)_{k \in \{1, m_n\}})_{n \ge 1} \in (\mathcal{S}_{[s,t]})^{\mathbb{N}^*}$, the following conditions are equivalent:

$$1. \quad P'^{s,t} = \lim_{n \to +\infty} P_{\{R_n\}}^{s,t}$$

$$2. \quad K'(s,t) = \lim_{n \to +\infty} \frac{\prod_{k=0}^{m_n-1} K(t_k^n, t_{k+1}^n)}{\prod_{k=1}^{m_n-1} K(t_k^n, t_k^n)}.$$

*Proof.* According to Lemma 10, $P'^{s,t} = \lim_{n \to +\infty} P_{\{R_n\}}^{s,t}$ if and only if $\Sigma_{P'^{s,t}} = \lim_{n \to +\infty} \Sigma_{P_{\{R_n\}}^{s,t}}$. Hence, writing $u_n$ the fraction appearing in the limit of Point 2., the composition formula of Lemma 12 implies that $P'^{s,t} = \lim_{n \to +\infty} P_{\{R_n\}}^{s,t}$ boils down to

$$\lim_{n \to +\infty} \begin{pmatrix} \Sigma_{\mu_s} & u_n \\ u_n & \Sigma_{\mu_t} \end{pmatrix} = \begin{pmatrix} \Sigma_{\mu_s} & K'(s,t) \\ K'(s,t) & \Sigma_{\mu_t} \end{pmatrix},$$
that is $K'(s, t) = \lim_{n \to +\infty} u_n$.

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

If one writes $c_K : (s,t) \mapsto K(s,t)/\sqrt{K(s,s)K(t,t)}$ for the correlation function of $P$ and $c_{K'} : (s,t) \mapsto K'(s,t)/\sqrt{K'(s,s)K'(t,t)}$ for the correlation function of $K'$, the second point becomes

$$ \lim_{n \to +\infty} c_K(t_0^n, t_1^n) \cdots c_K(t_{m_n-1}^n, t_{m_n}^n) = c_{K'}(s,t). $$

According to Proposition 18, being a local Markov transform of a Gaussian measure is a property depending only on the covariance functions of the involved measures. Thus, in order to find a criteria to identify Markov transforms of Gaussian measures, we focus on covariance functions of Gaussian measures, i.e., positive semi-definite kernels. For $T \subset \mathbb{R}$ and $K: T \times T \to \mathbb{R}$, we recall that $K$ is said to be a positive semi-definite kernel if for all $(t_1, \dots, t_m) \in T^m$, the matrix $(K(t_i, t_j))_{1 \le i,j \le m}$ is symmetric and positive semi-definite. Moreover, this kernel is said to be stationary if there exists $\tilde{K}: \mathbb{R} \to \mathbb{R}$ such that, for every $(s,t) \in T \times T$, we have $K(s,t) = \tilde{K}(t-s)$. It is a standard fact that a function $K: T \times T \to \mathbb{R}$ is a positive semi-definite kernel (resp. stationary positive semi-definite kernel) if and only if there exists a Gaussian measure (resp. a stationary Gaussian measure) on $\mathbb{R}^T$ with covariance function $K$. For a proof, one can e.g. refer to [24, Chapter 3]. Before stating our criteria to identify Markov transform of Gaussian measure, we define the variance, correlation and instantaneous decorrelation rate associated with a positive semi-definite kernel.

**Definition 19.** Let $K: T \times T \to \mathbb{R}$ be a positive semi-definite kernel. We denote by $v_K: t \in T \mapsto K(t,t) \in \mathbb{R}$ the variance function of the kernel $K$ and by $\sigma_K := \sqrt{v_K}$ its standard deviation function.

1. The kernel $K$ is said non-singular if its variance function of $K$ takes positive values.

2. If $K$ is non-singular, we denote by $c_K : (s,t) \mapsto \sigma_K^{-1}(s)\sigma_K^{-1}(t)K(s,t)$ its correlation function. Moreover, for every point $t \in T \setminus \{\sup T\}$, we define the decay rate of the correlation of $K$ at point $t$ as the function $L_t^K: h \in (T-t) \cap \mathbb{R}_+^* \mapsto h^{-1}(1-c_K(t,t+h)) \in \mathbb{R}_+$.

3. If for every $t \in T \setminus \{\sup T\}$, $\alpha^K(t) := \lim_{h \to 0+} L_t^K(h)$ exists, then $\alpha^K: t \in T \setminus \{\sup T\} \mapsto \alpha^K(t) \in \mathbb{R}_+$ is well-defined and we call it the instantaneous decorrelation rate of $K$.

If $X = (X_t)_{t \in T}$ is a centered real-valued random process with covariance function $K$, then $v_K$ is the variance function of $X$, $c_K$ is the correlation function of $X$ and $K$ is non-singular if and only if $X_t = 0$ a.s. never happens. Denote by $\mathbb{C}(X_s, X_t)$ the correlation between $X_s$ and $X_t$. Then, the map $L_t^K: h \mapsto L_t^K(h) = h^{-1}(\mathbb{C}(X_t, X_t) - \mathbb{C}(X_s, X_t))$ is the decay rate of the function $s \mapsto \mathbb{C}(X_s, X_t)$, that is the decay rate of the correlation to $X_t$. Hence, $\alpha^K(t)$ is the instantaneous decay of the correlation to $X_t$, i.e., the instantaneous decay rate of $X$ at time $t$.

**Proposition 20.** Consider an interval $T \subset \mathbb{R}$ and a non-negative measurable map $\alpha: T \to [0, +\infty]$. Then $K_\alpha : (s,t) \in T \times T \mapsto \exp(-\int_{\min(s,t)}^{\max(s,t)} \alpha(u)du)$ is a positive semi-definite kernel and the centered Gaussian measure with covariance function $K_\alpha$ is Markov. We denote this process $P_\alpha \in \mathcal{P}(\mathbb{R}^T)$,

*Proof.* First, we prove that $K_\alpha$ is a positive semi-definite kernel. Fix $s < t \in T^2$ and set $A_{s,t} := \begin{pmatrix} K_\alpha(s,s) & K_\alpha(s,t) \\ K_\alpha(t,s) & K_\alpha(t,t) \end{pmatrix}$. Since $\mathrm{Tr}(A_{s,t}) = K_\alpha(s,s) + K_\alpha(t,t) = 1+1=2 > 0$ and $\det(A_{s,t}) = 1-$
$\exp(-2 \int_t^s \alpha(v)dv) \ge 0$, the matrix $A_{s,t}$ is positive semi-definite and the probability measure $\mu_{s,t} := \mathcal{N}(0, A_{s,t})$ is well defined. For every $s < t < u \in T^3$, the composition formula in Lemma 12 implies $\mu_{s,t} \cdot \mu_{t,u} = \mathcal{N}(0, A)$, where

$$A = \begin{pmatrix} K_{\alpha}(s, s) & K_{\alpha}(s, t)K_{\alpha}(t, u)/K_{\alpha}(t, t) \\ K_{\alpha}(s, t)K_{\alpha}(t, u)/K_{\alpha}(t, t) & K_{\alpha}(u, u) \end{pmatrix}.$$

Since

$$\frac{K_{\alpha}(s, t)K_{\alpha}(t, u)}{K_{\alpha}(t, t)} = \exp \left( - \int_{s}^{t} \alpha(u)du \right) \exp \left( - \int_{t}^{u} \alpha(x)dx \right) / 1 = K_{\alpha}(s, u), \quad (4)$$

we have $A = A_{s,u}$, which implies $\mu_{s,t} \cdot \mu_{t,u} = \mu_{s,u}$. According to the Kolmogorov extension theorem, there exists a unique Markov measure $P_{\alpha} \in \mathcal{P}(\mathbb{R}^T)$ such that $P_{\alpha}^{s,t} = \mu_{s,t}$ for every $s < t \in T^2$. According to the composition formula in Lemma 12, for every $t_1 < \dots < t_m \in T^m$, we have $P_{\alpha}^{t_1, \dots, t_m} = P_{\alpha}^{t_1, t_2} \circ \dots \circ P_{\alpha}^{t_m-1, t_m} \in \mathcal{G}_m$. Hence $P_{\alpha}$ is a Gaussian process and its covariance function is $K_{\alpha}$. In particular, $K_{\alpha}$ is a positive semi-definite kernel. Finally, according to Proposition 13 and Equation (4), $P_{\alpha}$ is Markov. $\square$

**Remark 21.** 1. If $\alpha$ is constant equal to 0, then $K_{\alpha}(s, t) = 1$ for every $s < t \in \mathbb{R}^2$. Thus $P_{\alpha}$ is the law of a completely correlated process $(Z)_{t \in \mathbb{R}}$, where $Z \sim \mathcal{N}(0, 1)$.

2. If $\alpha$ is constant equal to $+\infty$, then $K_{\alpha}(s, t) = 0$ for every $s < t \in \mathbb{R}^2$. Hence, $P_{\alpha}$ is the law of a sequence $(X_t)_{t \in \mathbb{R}}$ of i.i.d. random variables with law $\mathcal{N}(0, 1)$.

3. The stationary Ornstein-Uhlenbeck process with parameter $\alpha \in \mathbb{R}_+^*$ is defined as the solution to the stochastic differential equation

$$\begin{cases} dX_t = -\alpha X_t dt + dB_t, \\ X_0 = Z \end{cases}, \quad (5)$$

where $(B_t)_{t \ge 0}$ is a standard Brownian motion and $Z$ is a random variable independent from $(B_t)_{t \ge 0}$ with law $\mathcal{N}(0, 1/2\alpha)$. It is well known that a stationary Ornstein-Uhlenbeck is a centered Gaussian process with covariance function $(s, t) \mapsto \frac{1}{2\alpha} \exp(-\alpha|t-s|)$. Thus $P_{\alpha}$ is the law of $(\sqrt{2\alpha}X_t)_{t \ge 0}$, where $(X_t)_{t \ge 0}$ is an Ornstein-Uhlenbeck process.

We can now state our criterion to identify the strong local Markov transforms. Since the expansion of $\ln(1+x)$ at point 0 will often be used, we fix a notation.

**Notation 22.** Let $\varepsilon : ] -1, +\infty[ \to \mathbb{R}$ be the function defined by

$$\varepsilon(x) = \begin{cases} \frac{\ln(1+x)-x}{x} & \text{if } x \in ]-1, +\infty[ \setminus \{0\} \\ 0 & \text{if } x = 0 \end{cases}.$$

This function is continuous, $\varepsilon(0) = 0$ and $\ln(1+x) = x + x\varepsilon(x)$ for every $x \in ]-1, +\infty[$.

**Theorem 23.** Consider an interval $T \subset \mathbb{R}$ and a continuous positive semi-definite kernel $K : T \times T \to \mathbb{R}$ with constant variance function equal to 1. We denote by $P$ the centered Gaussian process with covariance function $K$. For every $(s, t, h^*) \in T \times T \times \mathbb{R}_+^*$, we set

$$C_{s,t}(h^*) := \{(v, h); v \in [s, t, v + h \in [s, t], h \in ]0, h^*]\}.$$
1. Assume $\alpha^K$ is well defined, continuous and for every $s < t \in T \times T$

$$ \sup_{(v,h) \in C_{s,t}(h^*)} |L_v^K(h) - \alpha^K(v)| \xrightarrow{h^* \to 0^+} 0. \qquad (6) $$

Then $P_{\alpha^K}$ is the strong local Markov transform of P.

2. Assume, for every $s < t \in T \times T$

$$ \inf_{(v,h) \in C_{s,t}(h^*)} L_v^K(h) \xrightarrow{h^* \to 0^+} +\infty. \qquad (7) $$

Then $P_{+\infty}$ is the strong local Markov transform of P.

**Proof.** In order to prove the first point, let fix $s < t \in T^2$ and $(R_n)_{n \ge 1} \in (S_{[s,t]})^{\mathbb{N}^*}$ such that $\lim_{n \to +\infty} \sigma_{R_n} = 0$. Put $L := L^K$, $\alpha := \alpha^K$, $R_n = (t_0^n, \dots, t_{m_n}^n)$ for $n \ge 1$ and $h_k^n := t_{k+1}^n - t_k^n$ for $k \in [0, m_n - 1]$. According to Proposition 18, we have to establish the limit

$$ \prod_{k=0}^{m_n-1} K(t_k^n, t_{k+1}^n) \xrightarrow{n \to +\infty} \exp\left(-\int_s^t \alpha(u) du\right), $$

that is,

$$ \sum_{k=0}^{m_n-1} \ln(K(t_k^n, t_k^n + h_k^n)) \xrightarrow{n \to +\infty} -\int_s^t \alpha(u) du. \qquad (8) $$

Defining $\varepsilon$ as in Notation 22, for every $(v, h) \in T \times \mathbb{R}^*$ such that $v+h \in T$ and $K(v, v+h) > 0$, we get $\ln(K(v, v+h)) = (K(v, v+h) - 1)[1 + \varepsilon(K(v, v+h) - 1)] = -hL_v(h)[1 + \varepsilon(-hL_v(h))]$. Thus,

$$ \sum_{k=0}^{m_n-1} \ln(K(t_k^n, t_k^n + h_k^n)) = - \sum_{k=0}^{m_n-1} h_k^n L_{t_k^n}^n(h_k^n) [1 + \varepsilon(-h_k^n L_{t_k^n}^n(h_k^n))], \qquad (9) $$

$$ \begin{aligned} \text{which implies that } u_n &:= \left| \underbrace{\sum_{k=0}^{m_n-1} \ln(K(t_k^n, t_k^n + h_k^n))}_{\le 0} - \underbrace{\left(-\int_s^t \alpha(u) du\right)}_{\le 0} \right| \text{satisfies} \\ u_n &= \left| \sum_{k=0}^{m_n-1} -h_k^n L_{t_k^n}^n(h_k^n) [1 + \varepsilon(-h_k^n L_{t_k^n}^n(h_k^n))] - \left(-\int_s^t \alpha(u) du\right) \right| \\ &= \left| \sum_{k=0}^{m_n-1} h_k^n L_{t_k^n}^n(h_k^n) - \left(\int_s^t \alpha(u) du\right) + \sum_{k=0}^{m_n-1} h_k^n L_{t_k^n}^n(h_k^n) \varepsilon(-h_k^n L_{t_k^n}^n(h_k^n)) \right| \\ &= \left| \left(\sum_{k=0}^{m_n-1} h_k^n L_{t_k^n}^n(h_k^n) - \sum_{k=0}^{m_n-1} h_k^n \alpha(t_k^n)\right) + \left(\sum_{k=0}^{m_n-1} h_k^n \alpha(t_k^n) - \int_s^t \alpha(u) du\right) + \sum_{k=0}^{m_n-1} h_k^n L_{t_k^n}^n(h_k^n) \varepsilon(-h_k^n L_{t_k^n}^n(h_k^n)) \right| \\ &\le \underbrace{\left| \sum_{k=0}^{m_n-1} h_k^n L_{t_k^n}^n(h_k^n) - \sum_{k=0}^{m_n-1} h_k^n \alpha(t_k^n) \right|}_{:= a_n} + \underbrace{\left| \sum_{k=0}^{m_n-1} h_k^n \alpha(t_k^n) - \int_s^t \alpha(u) du \right|}_{:= b_n} + \underbrace{\left| \sum_{k=0}^{m_n-1} h_k^n L_{t_k^n}^n(h_k^n) \varepsilon(-h_k^n L_{t_k^n}^n(h_k^n)) \right|}_{:= c_n} \end{aligned} $$
We are now left to prove that $\lim_{n \to +\infty} a_n = \lim_{n \to +\infty} b_n = \lim_{n \to +\infty} c_n = 0$. For every $n \ge 1$, we have

$$0 \le a_n \le \sum_{k=0}^{m_n-1} h_k^n |L_{t_k^n}(h_k^n) - \alpha(t_k^n)| \le |t-s| \sup_{(v,h) \in C_{s,t}(\sigma_{R_n})} |L_v(h) - \alpha(v)|.$$

According to Hypothesis (6), $\lim_{n \to +\infty} a_n = 0$. Since $\alpha$ is continuous and continuous functions are Riemann-integrable, we immediately get $\lim_{n \to +\infty} b_n = 0$. In order to prove that $\lim_{n \to +\infty} c_n = 0$, put $C_{s,t} := \{(v,h); v \in [s,t], v+h \in [s,t], 0 \le h\}$ and $L_v(0) := \alpha(v)$ for every $v \in [s,t]$. We want to prove that $L$ is continuous on the compact set $C_{s,t}$. Since $K$ is continuous, we know that $L$ is continuous on $C_{s,t} \setminus ([s,t] \times \{0\})$ and we are left proving the continuity on $[s,t] \times \{0\}$. Let us fix $\varepsilon > 0$, $v^* \in [s,t]$ and consider $\alpha_\varepsilon \in \mathbb{R}_+$ such that

$$\left\{ \begin{array}{l} \sup_{(v,h) \in C_{s,t}(\alpha_\varepsilon)} |L_v(h) - \alpha(v)| \le \varepsilon/2 \\ \sup_{v \in [s,t]; |v-v^*| \le \alpha_\varepsilon} |\alpha(v) - \alpha(v^*)| \le \varepsilon/2 \end{array} \right. .$$

For every $(v,h) \in C_{s,t} \cap ([v^* - \alpha_\varepsilon, v^* + \alpha_\varepsilon] \times [-\alpha_\varepsilon, \alpha_\varepsilon]) \subset C_{s,t}(\alpha_\varepsilon) \cup ([v^* - \alpha_\varepsilon, v^* + \alpha_\varepsilon] \times \{0\})$, we have

$$\begin{aligned} |L_v(h) - L_{v^*}(0)| &= |L_v(h) - \alpha(v^*)| \\ &\le |L_v(h) - \alpha(v)| + |\alpha(v) - \alpha(v^*)| \\ &\le \sup_{(v,h) \in C_{s,t}(\alpha_\varepsilon)} |L_v(h) - \alpha(v)| + \sup_{v \in [s,t]; |v-v^*| \le \alpha_\varepsilon} |\alpha(v) - \alpha(v^*)| \\ &\le \varepsilon/2 + \varepsilon/2 = \varepsilon, \end{aligned}$$

which shows the desired continuity. Hence $M := \sup_{(v,h) \in C_{s,t}} L_v(h)$ is finite and we have

$$0 \le c_n \le \sum_{k=0}^{m_n-1} h_k^n L_{t_k^n}(h_k^n) |\varepsilon(-h_k^n L_{t_k^n}(h_k^n))| \le |t-s|M \sup_{|x| \le M\sigma_{R_n}} |\varepsilon(x)|,$$

which implies $\lim_{n \to +\infty} c_n = 0$ and finishes the proof of the first point. We now prove the second point. Using the same notation as before, we are left to prove (8) with $\alpha$ constant equal to $+\infty$, that is $\lim_{n \to +\infty} \sum_{k=0}^{m_n-1} \ln (K(t_k^n, t_{k+1}^n)) = -\infty$. As for (9), this amounts to show

$$\sum_{k=0}^{m_n-1} h_k^n L_{t_k^n}(h_k^n) [1 + \varepsilon (-h_k^n L_{t_k^n}(h_k^n))] \xrightarrow{n \to +\infty} +\infty.$$

One can find a rank $N \ge 1$ such that for every $n \ge N$ and $k \in [0, m_n - 1]$, we have $\varepsilon(-h_k^n L_{t_k^n}(h_k^n)) = \varepsilon(K(t_k^n, t_k^n + h_k^n) - 1) \ge -1/2$. Thus, for every $n \ge N$,

$$\sum_{k=0}^{m_n-1} h_k^n L_{t_k^n}(h_k^n) [1 + \varepsilon (-h_k^n L_{t_k^n}(h_k^n))] \ge \sum_{k=0}^{m_n-1} h_k^n L_{t_k^n}(h_k^n) 2^{-1} \ge \frac{t-s}{2} v_n,$$

where $v_n := \inf_{(v,h) \in C_{s,t}(\sigma_{R_n})} L_v(h)$. According to Hypothesis (7) we get $\lim_{n \to +\infty} v_n = +\infty$, which finishes the proof. $\square$
**Remark 24.** 1. We ask that the variance of $K$ is constant equal to 1 and $P$ is centered only to simplify the statement of our result and the notation used in the proof. If we just assume that $K$ is non-singular, i.e., the variance function of $K$ is positive, we obtain that the Gaussian process with covariance $K' : (s,t) \mapsto \sqrt{K(s,s)K(t,t)}K_\alpha(s,t)$ and same mean function as $P$ is the strong local Markov transform of $P$. This is a straightforward consequence of Remark 9.

2. In the case of a stationary kernel $K : (s,t) \in T^2 \mapsto \tilde{K}(t-s)$, the value of the map $L_t^K : h \mapsto h^{-1}(1-\tilde{K}(0)^{-1}\tilde{K}(h))$ does not depend on $t$ and Equation (6) and Equation (7) become $\lim_{h \to 0^+} h^{-1}(1-\tilde{K}(h)) = \alpha \in [0, +\infty]$. In this case, Theorem 23 states that, if $K$ is continuous, $\tilde{K}(0) = 1$ and $\lim_{h \to 0^+} h^{-1}(1-\tilde{K}(h)) = \alpha \in [0, +\infty]$, then $P_\alpha$ is the strong Markov transform of $P$.

**Theorem 25.** Consider an interval $T$ and a stationary positive semi-definite kernel $K : (s,t) \in T^2 \mapsto \tilde{K}(t-s) \in \mathbb{R}$ that is continuous and satisfies $\tilde{K}(0) = 1$. We denote by $P$ a centered Gaussian measure with covariance function $K$ and set $L^K : h \in \mathbb{R}_+^* \mapsto h^{-1}(1-\tilde{K}(h))$.

1. If $\lim_{h \to 0^+} L^K(h) = \alpha \in [0, +\infty]$, then $P_\alpha$ is the strong local Markov transform of $P$.

2. Assume $\alpha \in [0, +\infty]$ is a cluster point of $L^K$ at $0^+$ and consider a sequence of positive numbers $(s_n)_{n \ge 1}$ converging to zero that satisfies $\lim_{n \to +\infty} L^K(s_n) = \alpha$. For every $s < t \in T^2$, writing $R_n^{s,t} := ((s_n\mathbb{Z})\cap[s, t]) \cup \{s, t\}$, we obtain $\lim_{n \to +\infty} P_{\{R_n^{s,t}\}} = P_\alpha^{s,t}$. In particular, $P_\alpha$ is a weak local Markov transform of $P$.

*Proof.* The first point is only a restatement of the Point 2 in Remark 24 and we are left with the proof of the second point. We set $L := L^K$ and fix $s < t \in T^2$. Let $(s_n)_{n \ge 1}$ and $R_n^{s,t}$ be as in the statement. For every $n \ge 1$, we write $R_n^{s,t} := (t_0^n, \dots, t_{m_n}^n) \in S_{[s,t]}$. For every $k \in \{1, \dots, m_n - 2\}$, we have

$$ \ln(\tilde{K}(s_n)) = \ln(1+[\tilde{K}(s_n)-1]) = -s_n L(s_n) [1 + \varepsilon (\tilde{K}(s_n)-1)] = -(t_{k+1}^n - t_k^n)L(s_n) [1 + \varepsilon (\tilde{K}(s_n)-1)], $$

where $\varepsilon$ is defined in Notation 22. Hence,

$$ \begin{align*} \sum_{k=0}^{m_n-1} \ln (K(t_k^n, t_{k+1}^n)) &= \sum_{k=0}^{m_n-1} \ln (\tilde{K}(t_{k+1}^n - t_k^n)) \\ &= \ln(\tilde{K}(t_1^n - s)) + \ln(\tilde{K}(t - t_{m_n-1}^n)) + \sum_{k=1}^{m_n-2} \ln(\tilde{K}(s_n)) \\ &= \ln(\tilde{K}(t_1^n - s)) + \ln(\tilde{K}(t - t_{m_n-1}^n)) - \sum_{k=1}^{m_n-2} (t_{k+1}^n - t_k^n)L(s_n) [1 + \varepsilon (\tilde{K}(s_n)-1)] \\ &= \ln(\tilde{K}(t_1^n - s)) + \ln(\tilde{K}(t - t_{m_n-1}^n)) - (t_{m_n-1}^n - t_1^n)L(s_n) [1 + \varepsilon (\tilde{K}(s_n)-1)]. \end{align*} $$

Since $\tilde{K}(0) = 1$ and $K$ is continuous, we obtain $\lim_{n \to +\infty} \sum_{k=0}^{m_n-1} \ln (K(t_k^n, t_{k+1}^n)) = 0 + 0 - \alpha(t-s) = -\alpha(t-s)$, which implies the result according to Proposition 18. $\square$
We now apply Theorem 23 to two different examples: We shall prove that they admit a strong local Markov transform and then compute it. We begin with the fractional Brownian motion.

**Proposition 26.** Consider $H \in ]0, 1[$ and denote by $\mathcal{B}_H$ the fractional Brownian motion with Hurst parameter $H$, i.e., the centered Gaussian measure with covariance function $K_H : (s,t) \in \mathbb{R}_+^* \times \mathbb{R}_+^* \mapsto \frac{1}{2}(|t|^{2H} + |s|^{2H} - |t-s|^{2H})$. The centered Gaussian measure $\mathcal{B}_H'$ with covariance matrix

$$ K_H' : (s,t) \in \mathbb{R}_+^* \times \mathbb{R}_+^* \mapsto \begin{cases} s^H t^H \mathbb{1}_{s=t} + 0 \mathbb{1}_{s \neq t} & \text{if } H < 1/2 \\ \min(s,t) & \text{if } H = 1/2 \\ s^H t^H & \text{if } H > 1/2 \end{cases} $$

is the strong local Markov transform of $\mathcal{B}_H$.

*Proof.* For $(s,t) \in \mathbb{R}_+^* \times \mathbb{R}_+^*$,

$$ K_H(s,t) = s^H t^H \cdot \frac{1}{2} \left( \left( \frac{t}{s} \right)^H + \left( \frac{s}{t} \right)^H - \left| \left( \frac{t}{s} \right)^{1/2} - \left( \frac{s}{t} \right)^{1/2} \right|^{2H} \right) = u(s)u(t)\tilde{K}(\phi(t) - \phi(s)), \quad (10) $$

where $u : t \in \mathbb{R}_+^* \mapsto t^H \in \mathbb{R}$, $\phi : s \in \mathbb{R}_+^* \mapsto \frac{\ln(s)}{2}$ and $\tilde{K} : x \mapsto \frac{1}{2}(e^{2Hx} + e^{-2Hx} - |e^x - e^{-x}|^{2H})$. Let define $P$ as the stationary centered Gaussian process with covariance function $K : (s,t) \mapsto \tilde{K}(|t-s|)$. Using the Taylor expansion of the exponential function, we get

$$ h^{-1}(1 - \tilde{K}(h)) = o(1) + (2h)^{2H-1}[(1 + o(1))^{2H} \xrightarrow{h \to 0^+} \begin{cases} +\infty & \text{if } H < 1/2 \\ 1 & \text{if } H = 1/2 =: \beta. \\ 0 & \text{if } H > 1/2 \end{cases} $$

Thus, according to Theorem 25, $P_\beta$ is the strong local Markov transform of $P$. According to Remark Equation (10) and Remark 8, the centered Gaussian process with kernel $(s,t) \mapsto u(s)u(t) \exp(-\beta|\phi(t)-\phi(s)|)$ is the strong local Markov transform of $\mathcal{B}_H$. Finally, it is straightforward that $u(s)u(t) \exp(-\beta|\phi(t)-\phi(s)|) = K'_H(s,t)$, which gives the wanted result. $\square$

We now continue our sequence of examples with a class of non-stationary processes.

**Proposition 27.** Consider two intervals $T \subset \mathbb{R}$, $J \subset \mathbb{R}_+^*$, a Brownian motion $(B_u)_{u \ge 0}$ on a probability space $(\Omega, \mathcal{F}, \mathbb{P})$ and a family $(k_t)_{t \in T}$ of elements of $L^2(J)$ such that $\int_J k_t(u)^2 du = 1$ for every $t \in T$. Set $X_t := \int_J k_t(u) dB_u$ for every $t \in T$. The, the law $P \in \mathcal{P}(\mathbb{R}^T)$ of $(X_t)_{t \in T}$ is a Gaussian measure with covariance function $K : (s,t) \in T \times T \mapsto \int_J k_s(u)k_t(u) du$. Moreover, set $C_{s,t}(h^*) := \{(v,h) \in [s,t[ \times ]0,h^*]; v+h \in [s,t]\}$ for every $(s,t,h^*) \in T \times T \times \mathbb{R}_+^*$ and assume:

$$ (H_1) \quad \exists f : T \times J \to \mathbb{R}, \forall (s,t,u) \in T \times T \times J, \\ \sup_{(v,h) \in C_{s,t}(h^*)} \left| h^{-1}(k_v(u) - k_{v+h}(u)) - f(v,u) \right| \xrightarrow{h^* \to 0^+} 0; \quad (11) $$

$$ (H_2) \quad \exists g \in L^1(J), \exists h_0 \in \mathbb{R}_+^*, \forall (h,t,u) \in ]0, h_0] \times T \times J, \left| h^{-1}(k_t(u) - k_{t+h}(u)) - f(t,u) \right| \le g(u); $$
$(H_3) \quad \alpha : t \in T \mapsto \int_J k_t(u)f(t, u)du \text{ is continuous.}$

Then $P_\alpha$ is the strong Markov transform of $P$.

*Proof.* The hypothesis $k_s \in L^2(J)$ ensures that $(X_t)_{t \in T}$ is well defined. The process $P$ is clearly centered, Gaussian and for every $(s, t) \in T^2$,

$$K(s,t) = \mathbb{E}(X_s X_t) = \mathbb{E} \left( \int_J k_s(u) dB_u \int_J k_t(u) dB_u \right) = \mathbb{E} \left( \int_J k_s(u) k_t(u) d\langle B, B \rangle_u \right) = \int_J k_s(u) k_t(u) du.$$

Moreover, for $(v, h) \in C_{s,t}(h^*)$, $L_v^K(h) = h^{-1}(1 - K(v, v + h)) = \int_J h^{-1}(k_v(u) - k_{v+h}(u))k_v(u)du$. Hence,

$$\begin{aligned} |L_v^K(h) - \alpha(v)| &= \left| \int_J h^{-1}(k_v(u) - k_{v+h}(u)) k_v(u) du - \int_J f(v, u) k_v(u) du \right| \\ &\le \int_J |h^{-1}(k_v(u) - k_{v+h}(u)) - f_v(u)| |k_v(u)| du \\ &\le \left( \sup_{w \in [s, t]} |k_w(u)| \right) \int_J \sup_{(w,h) \in C_{s,t}(h^*)} \left\{ |h^{-1}(k_w(u) - k_{w+h}(u)) - f(w, u)| \right\} du, \end{aligned}$$

which implies

$$\sup_{(v,h) \in C_{s,t}(h^*)} |L_v^K(h) - \alpha(v)| \le \left( \sup_{w \in [s,t]} |k_w(u)| \right) \int_J \sup_{(w,h) \in C_{s,t}(h^*)} \left\{ |h^{-1}(k_w(u) - k_{w+h}(u)) - f(w, u)| \right\} du. \quad (12)$$

According to $(H_1)$ and $(H_2)$, the Lebesgue's dominated convergence theorem applies to the right-hand side of Equation (12) so that $\lim_{h^* \to 0^+} \sup_{(v,h) \in C_{s,t}(h^*)} |L_v^K(h) - \alpha(v)| = 0$. According to $(H_3)$, $\alpha$ is continuous and we have $K(t,t) = \int_J k_t(u)^2 du = 1$. Hence, Theorem 23 applies, which proves the result. $\square$

**Example 28.** Let us apply Proposition 27 to the law $P$ of $(X_s)_{s>0} := (\int_0^{+\infty} \sqrt{s}e^{-\frac{su}{2}} dB_u)_{s>0}$. In order to satisfy $(H_2)$, we rather work on $(X_t)_{t \in [a, b]}$ for a fixed couple $a < b \in (\mathbb{R}_+^*)^2$. Fix $s < t \in [a, b]^2$ and set $k_v(u) := \sqrt{v}e^{-vu/2}$ for every $(v, u) \in [s, t] \times \mathbb{R}_+^*$. Using the Taylor expansion of the exponential map and of $x \in ]-1, +\infty[ \mapsto \sqrt{1+x}$, we get

$$\begin{aligned} h^{-1}(k_v(u) - k_{v+h}(u)) &= k_v(u) \left[ h^{-1} \left( 1 - \sqrt{1 + \frac{h}{v} e^{-\frac{hu}{2}}} \right) \right] \\ &= \frac{1}{2} k_v(u) \left( u - \frac{1}{v} + \theta(h) \right), \end{aligned}$$

where $\theta: \mathbb{R} \to \mathbb{R}$ satisfies $\lim_{h \to 0^+} \theta(h) = 0$. Setting $f(v, u) := \frac{1}{2}k_v(u)(u - 1/v)$, we get

$$|h^{-1}(k_v(u) - k_{v+h}(u)) - f(v, u)| = \left| \frac{1}{2} k_v(u) \theta(h) \right| \le \frac{\sqrt{t}}{2} |\theta(h)|,$$
for every $(v, u, h) \in [s, t] \times \mathbb{R}_+ \times \mathbb{R}_+^*$. Hence, Hypothesis $(H_1)$ is satisfied. Let $h_0 \in \mathbb{R}_+^*$ be such that $\sup_{h \in [0, h_0]} |\theta(h)| \le 1$. For every $h \in [0, h_0]$ and $(v, u) \in [a, b] \times \mathbb{R}_+^*$, we have

$$|h^{-1}(k_v(u) - k_{v+h}(u)) - f(v, u)| = \left| \frac{1}{2} k_v(u) \theta(h) \right| \le \frac{1}{2} \sqrt{be^{-\frac{au}{2}}} =: g(u),$$

which shows $(H_2)$. Since $f(v, u)k_v(u) = \frac{1}{2}(uve^{-vu} - e^{-vu})$, we have $\alpha(v) = \int_0^{+\infty} f(v, u)k_v(u)du = \int_0^{+\infty} (uve^{-vu} - e^{-vu})du = 1 - 1 = 0$ for every $v \in [a, b]$. Since $(H_3)$ is obviously true, Proposition 27 applies and the Markov transform of the law of $(X_t)_{t \in [a, b]}$ is the law of the completely correlated process $(Z)_{t \in [a, b]}$, where $Z \sim \mathcal{N}(0, 1)$. This being true for every $a < b \in (\mathbb{R}_+^*)^2$, the Markov transform of the law of $(X_t)_{t \in t>0}$ is $(Z)_{t>0}$. Applying Remark 9, the strong local Markov transform of the law of $(\int_0^{+\infty} e^{-tu} dB_u)_{t>0}$ is the law of $(Z/\sqrt{2t})_{t>0}$.

# 5 Default of uniqueness for a weak Markov transform.

As already noticed in Remark 4, strong local Markov transforms are unique. However, we shall see now that this fails in the case of weak local Markov transforms. In this section, our aim is to construct a (stationary) Gaussian measure which has several weak local Markov transforms. The guiding result for our construction is a theorem of Bochner which characterizes the continuous positive semi-definite stationary kernels. For a proof, we refer to [5, Paragraph 8] (or [9, Page 208] for a more recent presentation). We denote by $\hat{\mu} : t \in \mathbb{R} \mapsto \int_{\mathbb{R}} \exp(itx)\mu(dx)$ the Fourier transform of a positive finite measure $\mu \in \mathcal{M}_+(\mathbb{R})$.

**Theorem 29 (Bochner<fnref n="8" />).** Let $K : \mathbb{R}^2 \to \mathbb{R}$ be a symmetric map. The following conditions are equivalent:

1. There exists $\mu \in \mathcal{M}_+(\mathbb{R})$ such that $K(s, t) = \hat{\mu}(t-s)$ for every $(s, t) \in \mathbb{R}^2$.

2. The function $K$ is a continuous positive semi-definite stationary kernel.

The measure $\mu$ is then unique and is called the spectral measure associated with $K$.

We now claim that, in order to obtain a stationary Gaussian measure that has not a unique weak local Markov transform, it is sufficient to find a symmetric probability measure $\mu \in \mathcal{P}(\mathbb{R})$ with a Fourier transform whose growth rate at $0^+$ admits several cluster points. To justify this, notice that if $\mu$ is a real-valued symmetric probability measure, then the function $K : (s, t) \in \mathbb{R}^2 \mapsto \hat{\mu}(t-s)$ is real valued and symmetric. According to Theorem 29, this implies that $K$ is a stationary positive semi-definite kernel, with constant variance equal to $\hat{\mu}(0) = \mu(\mathbb{R}) = 1$. Thus, for every cluster point $\alpha \in [0, +\infty]$ of the decay rate of $\hat{\mu}$ at point $0^+$, according to Theorem 25, the centered Gaussian process $P$ with covariance function $K$ admits $P_\alpha$ as weak local Markov transform. In our construction, the set of cluster points of the decay rate of the Fourier transform of $\mu$ at point $0^+$ will be $[0, +\infty]$, which guarantees an infinity of weak local Markov transforms. Denoting by sgn : $\mathbb{R} \to \{-1, 0, 1\}$ the usual sign function, our strategy is to consider a symmetric probability measure of the form $\gamma := \sum_{|k| \ge 2} a_{|k|} \delta_{\text{sgn}(k)b_{|k|}}$ whose Fourier transform has an infinite decay rate at $0^+$. Then, to obtain our measure $\mu$, we “mix” $\gamma$ with $\delta_0$, whose Fourier

<footnote n="8">⁸In general, Theorem 29 is stated for $\mathbb{C}$-valued kernels, but it is straightforward that this implies our reformulation with $\mathbb{R}$-valued kernels.</footnote>
transform has a decay rate converging to 0 at point $0^+$. More precisely, we will recursively construct a set $S \subset (\mathbb{N} \cap [2, +\infty[)$ and put

$$y_k := \begin{cases} b_k & \text{if } k \in S \\ 0 & \text{otherwise,} \end{cases}$$

in order to define our measure $\mu$ by

$$\mu := \sum_{|k| \ge 2} a_{|k|} \delta_{\text{sgn}(k) y_{|k|}} = \sum_{|k| \in S} a_{|k|} \delta_{\text{sgn}(k) b_{|k|}} + \left( \sum_{|k| \notin S} a_{|k|} \right) \delta_0.$$

Since

$$\begin{cases} \hat{\gamma}(t) = \sum_{k \ge 2} a_k (e^{itb_k} + e^{-itb_k}) = 2 \sum_{k \ge 2} a_k \cos(tb_k) \\ \hat{\mu}(t) = \sum_{k \ge 2} a_k (e^{ity_k} + e^{-ity_k}) = 2 \sum_{k \ge 2} a_k \cos(ty_k) \end{cases}, \quad (13)$$

the decay rates of $\hat{\gamma}$ and $\hat{\mu}$ at 0 are given by

$$\left\{ \begin{aligned} t^{-1}(1 - \hat{\gamma}(t)) &= 2t^{-1} \sum_{k \ge 2} a_k (1 - \cos(tb_k)) \\ t^{-1}(1 - \hat{\mu}(t)) &= 2t^{-1} \sum_{k \in S} a_k (1 - \cos(tb_k)) \end{aligned} \right. . \quad (14)$$

Put $a := 1/2$, $b := 3$ and $a_k := a^k$, $b_k := b^k$ for every $k \ge 2$. We have $ab > 1$ and (13) shows that $\hat{\gamma}/2$ is the well-known continuous nowhere differentiable Weierstrass function [25]. In Point 1 of Lemma 30 below, we rely on an article of Hardy [11] to prove $\lim_{t \to 0^+} t^{-1}(1 - \hat{\gamma}(t)) = +\infty$. We are left to find a recursive construction of $S$ such that the lacunary series $t^{-1}(1 - \hat{\mu}(t))$ of $t^{-1}(1 - \hat{\gamma}(t))$ admits the elements of $[0, +\infty]$ as cluster points. Points 2-5 of Lemma 30 are useful to define the sequence $(y_k)_{k \ge 1}$ and to show that the resulting measure $\mu$ has the wanted property. The construction itself is done in Proposition 31, using the tools of Lemma 30.

**Lemma 30.** Put $a := 1/2$, $b := 3$ and $a_k := a^k$, $b_k := b^k$ for every $k \ge 2$.

1. Then $\lim_{x \to +\infty} x \sum_{k \ge 2} a_k \left(1 - \cos\left(\frac{b_k}{x}\right)\right) = +\infty$.

2. For any given sequence $(x_k)_{k \ge 1} \in \mathbb{R}^{\mathbb{N}^*}$, $\lim_{x \to +\infty} x \sum_{k \ge \lfloor x \rfloor + 1} a_k \left(1 - \cos\left(\frac{x_k}{x}\right)\right) = 0$.

3. For $p \ge 1$ and $((n_i, m_i))_{i \in [1, p]} \in (\mathbb{N}^2)^p$, the map

$$g(n_1, m_1), \dots, (n_p, m_p) : x \in \mathbb{R}_+^* \mapsto x \sum_{l=1}^p \sum_{k=n_l}^{m_l} a_k \left(1 - \cos\left(\frac{b_k}{x}\right)\right) \in \mathbb{R}_+$$

satisfies $\lim_{x \to +\infty} g(n_1, m_1), \dots, (n_p, m_p)(x) = 0$.

4. For $n \ge 2$, the map

$$f_n : x \in \mathbb{R} \mapsto x \sum_{k=n}^{\lfloor x \rfloor} a_k \left(1 - \cos\left(\frac{b_k}{x}\right)\right) \in \mathbb{R}_+$$

satisfies $\lim_{x \to +\infty} f_n(x) = +\infty$.
5. Consider a increasing sequence of integers $(n_k)_{k \ge 1}$ such that $n_0 = 2$ and define the sequence $(y_k)_{k \ge 0}$ by

$$y_k = \begin{cases} b_k & \text{if } k \in \bigcup_{i \ge 0} \llbracket n_{2i}, n_{2i+1} \rrbracket \\ 0 & \text{if } k \in \bigcup_{i \ge 0} \llbracket n_{2i+1}, n_{2(i+1)} \rrbracket. \end{cases}$$

Then the map

$$f: x \in \mathbb{R}_+^* \mapsto x \sum_{k=2}^{|x|} a_k \left(1 - \cos\left(\frac{y_k}{x}\right)\right)$$

satisfies

$$f(x) = \begin{cases} g(n_0, n_1-1), \dots, (n_{2(i-1)}, n_{2(i-1)-1})(x) + f_{n_{2i}}(x) & \text{if } x \in [n_{2i}, n_{2i+1}[ \\ g(n_0, n_1-1), \dots, (n_{2i}, n_{2i+1}-1)(x) & \text{if } x \in [n_{2i+1}, n_{2(i+1)}[. \end{cases}$$

*Proof.* 1. According to [11, Points 2.41-2.42], $\lim_{h \to 0^+} h^{-1} \sum_{k \ge 0} a^k(1 - \cos(b^k \pi h)) = +\infty$. Since $\lim_{h \to 0^+} h^{-1} a^0(1 - \cos(b^0 \pi h)) = \lim_{h \to 0^+} h^{-1} a^1(1 - \cos(b^1 \pi h)) = 0$, the change of variable $x^{-1} = \pi h$ gives the wanted result.

2. For $x \ge 2$, $0 \le x \sum_{k \ge |x|+1} a_k \left(1 - \cos\left(\frac{x_k}{x}\right)\right) \le 2x \sum_{k \ge |x|+1} a^k = 2x \frac{a^{|x|+1}}{1-a}$, which proves the result.

3. For $x \in \mathbb{R}_+^*$, $0 \le x \sum_{k=1}^p \sum_{l=n_k}^{m_k} a_l \left(1 - \cos\left(\frac{b_l}{x}\right)\right) \le x \sum_{k=1}^p \sum_{l=n_k}^{m_k} a_l \frac{b_l^2/x^2}{2} = \frac{1}{x} \sum_{k=1}^p \sum_{l=n_k}^{m_k} \frac{a_l b_l^2}{2}$, which proves the result.

4. According to Points 1, 2 and 3, the equality

$$f_n(x) = x \sum_{k \ge 2}^{\infty} a_k \left(1 - \cos\left(\frac{b_k}{x}\right)\right) - x \sum_{k=2}^{n-1} a_k \left(1 - \cos\left(\frac{b_k}{x}\right)\right) - x \sum_{k \ge |x|+1} a_k \left(1 - \cos\left(\frac{b_k}{x}\right)\right)$$

implies $\lim_{n \to +\infty} f_n(x) = +\infty - 0 - 0 = +\infty$.

5. The computation is straightforward.

□

**Proposition 31.** There exists a sequence $(y_k)_{k \ge 1} \in \mathbb{R}_+^{N^*}$ such that the cluster points of the decay rate at point $0^+$ of the Fourier transform of $\mu := \sum_{|k| \ge 2} 2^{-|k|} \delta_{\operatorname{sgn}(k)y_{|k|}}$ are the elements of $[0, +\infty]$.

*Proof.* We define the functions $f_n$ and $g_{(n_1, m_1), \dots, (n_p, m_p)}$ as in Lemma 30. According to Points 3 and 4 of Lemma 30, we can define a sequence $(n_i)_{i \ge 2}$ by

$$\left\{ \begin{array}{l} n_0 := 2 \\ \forall i \ge 0, n_{2i+1} := \inf \{n > n_{2i} ; f_{n_{2i}}(n-1) > i\} \\ \forall i \ge 0, n_{2(i+1)} := \inf \{n > n_{2i+1} ; g_{(n_0, n_1-1), \dots, (n_{2i}, n_{2i+1}-1)}(n-1) < 1/i\} \end{array} \right. .$$

As in Point 5 of Lemma 30, we associate a sequence $(y_k)_{k \ge 2}$ and a function $f$ to our sequence $(n_i)_{i \ge 2}$. Since $n_{2i+1} - 1 \in [n_{2i}, n_{2i+1}]$ and $n_{2(i+1)} - 1 \in [n_{2i+1}, n_{2(i+1)}]$, according to this same point,

$$f(n_{2i+1} - 1) = g_{(n_0, n_1-1), \dots, (n_{2(i-1)}, n_{2(i-1)-1})}(n_{2i+1} - 1) + f_{n_{2i}}(n_{2i+1} - 1) \ge f_{n_{2i}}(n_{2i+1} - 1) > i$$
and

$$f(n_{2(i+1)} - 1) = g_{(n_0, n_1-1), \dots, (n_{2i}, n_{2i+1}-1)}(n_{2(i+1)} - 1) < 1/i$$

for $i \ge 1$. Set $\mu := \sum_{|k| \ge 2} 2^{-|k|} \delta_{\text{sgn}(k)y_{|k|}}$ and denote by $L: t \in \mathbb{R}_+^* \to t^{-1}(1-\hat{\mu}(t))$ the decay rate of $\hat{\mu}$ at $0^+$. As in Equation (14),

$$L(1/x) = 2x \sum_{k \ge 2} a_k (1 - \cos(y_k/x)) = f(x) + g(x),$$

where $g(x) := \sum_{k \ge |x|+1} a_k (1 - \cos(y_k/x))$. According to Point 2 of Lemma 30, $\lim_{x \to +\infty} g(x) = 0$. Thus, writing $s_i := \frac{1}{n_{2i+1}-1}$ and $t_i := \frac{1}{n_{2(i+1)}-1}$, we get $L(s_i) = f(n_{2i+1}-1) + g(n_{2i+1}-1) \xrightarrow{i \to +\infty} +\infty + 0 = +\infty$ and $L(t_i) = f(n_{2(i+1)}-1) + g(n_{2(i+1)}-1) \xrightarrow{i \to +\infty} 0$, so that 0 and $+\infty$ are cluster points of $L$ at $0^+$. Since $L$ is continuous, according to the intermediate value theorem, any elements of $[0, +\infty]$ is a cluster points of $L$. Since $L$ is non-negative, all its cluster points are elements of $[0, +\infty]$ which finishes the proof. $\square$

The following Remark will be used in the construction and further in the article.

**Remark 32.** Let $T$ be an interval and $P \in \mathcal{P}(\mathbb{R}^T)$ a Gaussian measure with non-singular covariance function $K: T \times T \to \mathbb{R}$. In [20, Theorem 1], the authors proved that if $P$ satisfies the Markov property and $K$ is continuous, then there exists two functions $f, g$ satisfying $K(s, t) = f(s)g(t)$ for every $s < t \in \mathbb{R}^2$. Since, for every $t \in \mathbb{R}$, $K(t, t) = f(t)g(t) > 0$, both $f$ and $g$ do not vanish. Since $f$ (resp. $g$) is continuous and does not vanish $f$ (resp. $g$) is either positive or negative. As $f(t)g(t) > 0$ for every $t \in T$, either $f$ and $g$ are both positive or $f$ and $g$ are both negative. This implies $K$ is positive. Hence, every non-singular continuous covariance function of a Gaussian measure that satisfies the Markov property is positive.

We can now construct a stationary Gaussian measure with an infinity of weak local Markov transforms.

**Theorem 33.** There exists a centered stationary Gaussian measure $P \in \mathcal{P}(\mathbb{R}^\mathbb{R})$ whose set of weak local Markov transforms is the set of all the measures $P'$ satisfying:

1. The measure $P'$ is centered Gaussian with constant variance function equal to 1;

2. The covariance function of $P'$ is non-negative;

3. The measure $P'$ is Markov.

*Proof.* According to Proposition 31, we can find a symmetric probability measure $\mu$ such that the cluster points of the decay rate of $\hat{\mu}$ at $0^+$ are the elements of $[0, +\infty]$. Since $\mu$ is symmetric, the function $K: (s, t) \in \mathbb{R}^2 \mapsto \hat{\mu}(t-s)$ is real valued and symmetric. Hence, according to Theorem 29, $K$ is a stationary positive semi-definite kernel, with constant variance equal to $\hat{\mu}(0) = \mu(\mathbb{R}) = 1$. Based on Theorem 29, the Gaussian measure $P$ with covariance function $K: (s, t) \in \mathbb{R} \times \mathbb{R} \mapsto \hat{\mu}(t-s)$ is a well defined stationary Gaussian measure such that $K(t, t) = \hat{\mu}(0) = 1$ for every $t \in \mathbb{R}$. We denote by $P$ the centered Gaussian measure with covariance function $K$. Let $P'$ be a Gaussian and Markov measure with non-negative covariance function $K'$ and constant variance equal to 1. In order to show that $P'$ is a weak local Markov transform of $P$, let fix $s < t \in \mathbb{R}^2$. Since $0 \le K'(s, t) \le K'(s, s)^{1/2}K'(t, t)^{1/2} = 1$, $\alpha := -|t-s|^{-1} \ln(K'(s, t))$ is well defined and we have $\alpha \in [0, +\infty]$. Moreover $K'(s, t) = \exp(-\alpha|t-s|) = K_\alpha(s, t)$. Since $\alpha$ is a cluster point of the decay rate of $\hat{\mu}$ at $0^+$, according to Point 2 of Theorem 25,
$P_\alpha$ is a weak local Markov transform of $P$. Hence, there exists $(R_n)_{n \ge 1} = ((t_k^n)_{n \in \{1, m_n\}}) \in (S_{[s,t]})^{\mathbb{N}*}$ such that $\lim_{n \to +\infty} \sigma_{R_n} = 0$ and $\lim_{n \to +\infty} K(s, t_1^n) \dots K(t_{m_{n-1}}^n, t) = K_\alpha(s, t) = K'(s, t)$. According to Proposition 18, $P'$ is a weak local Markov transform of $P$. Conversely, let $P'$ be a weak local Markov transform of $P$ with covariance function $K'$. We want to show that $P'$ is a Gaussian and Markov measure with non-negative covariance function and constant variance function equal to 1. First $P'$ is Markov by definition and Gaussian according to Proposition 15. As $P'$ has the same variance function as $P$, its variance is constant equal to 1. Since $P'$ is a weak local Markov transform of $P$, there exists $(R_n)_{n \ge 1} = ((t_k^n)_{k \in \{1, m_n\}}) \in (S_{[s,t]})^{\mathbb{N}*}$ such that $\lim_{n \to +\infty} K(t_0^n, t_1^n) \dots K(t_{m_{n-1}}^n, t_{m_n}^n) = K'(s, t)$. According to Remark 32, $K$ is positive, which implies $K'(s, t) \ge 0$ and finishes the proof. $\square$

In particular for a measure $P$ as in Theorem 33 and any measurable function $\alpha: \mathbb{R} \to [0, +\infty]$, $P_\alpha$ is a weak local Markov transform of $P$. As one can see taking $\alpha: t \mapsto 0 \cdot 1_{t \le 1} + (+\infty) \cdot 1_{t > 1}$, this also proves that a weak Markov transform of a stationary measure is not always a stationary measure.

# 6 Global Markov transform, weak convergence of the transformations and SDE characterization of the mimicking process

In the past sections, we gave some results local Markov transforms. In this section, we shall study global Markov transform and show how these results can be used to get results about global Markov transforms. As said in the introduction, a global Markov transform is a law $P'$ of a process $X'$ obtained as limit of transformations of $X$ made Markov at certain times. We recall that, given a finite subset $R$ of $\mathbb{R}$, the transformation of $X$ made Markov at times $R$ is a process $X^R$ satisfying:

* On every interval between two successive times of $R$, $X^R$ and $X$ have the same law;

* For every $r \in R$, $X^R$ is made Markov at time $r$: if one knows the value of the trajectory at time $r$, then the future $(X_t^R)_{t>r}$ of the trajectory does not depend on the past $(X_t^R)_{t<r}$ of the trajectory.

It is possible to give a rigorous definition of the law $P_{[R]}$ of this process $X^R$. This has been done by Boubel and Juillet [7, Definition 4.18] using the Kolmogorov extension theorem.

**Definition/Proposition 34** (Measure made Markov at times $R$). Let $T \subset \mathbb{R}$ be a set, $E$ a Polish space, $P \in \mathcal{P}(E^T)$ a probability measure and $R = \{r_1 < \dots < r_m\} \subset T$ a finite set of times. For each finite set $S \subset T$, we write

$$S \cup R = \{s_1^0 < \dots < s_{k_0}^0 < r_1 < s_1^1 < \dots < s_{k_1}^1 < r_2 < \dots < s_1^{m-1} < \dots < s_{k_{m-1}}^{m-1} < r_m < s_1^m < \dots < s_{k_m}^m\}$$
and

$$\mu_S := \mathrm{proj}_{\#}^S[P^{s_1^0, \dots, s_{k_0}^0, r_1} \circ P^{r_1, s_1^1, \dots, s_{k_1}^1, r_2} \circ \dots \circ P^{r_m, s_1^m, \dots, s_{k_m}^m}].$$

Denoting by $S$ the class of finite subsets of $T$, it is readily verified that $(\mu_S)_{S \in \mathcal{S}}$ is a consistent family of measures. According to the Kolmogorov extension Theorem, there exists a unique probability $P_{[R]}$ on $E^T$ such that $\mathrm{proj}_{\#}^S P_{[R]} = \mu_S$ for every $S \in \mathcal{S}$. We say that $P_{[R]} \in \mathcal{P}(E^T)$ is the measure $P$ made Markov at times $R$. Given a process $X$ with law $P$, we say that $X^R$ is the $^9$ transformation of $X$ made Markov at times $R$ if $X^R$ has law $P_{[R]}$.

<footnote n="9">⁹We speak about the transformation, even if we just have uniqueness in law</footnote>
In order to simplify the following definitions and notation, we will assume that $T = \mathbb{R}$, but the results remain true for any interval $T \subset \mathbb{R}$. A global Markov transform will be defined as a Markov limit of a sequence $P_{[R_n]}$ for an admissible sequence $(R_n)_{n \ge 1}$ of set of times (see Notation 36). The topology that we will consider first is the topology of finite-dimensional convergence, that is to say the weak convergence on $\mathcal{P}(E^{\mathbb{R}})$, where $E^{\mathbb{R}}$ is endowed with the product topology. More explicitly, a sequence $(P_n)_{n \ge 1} \in \mathcal{P}(E^{\mathbb{R}})^{\mathbb{N}^*}$ converges to $P \in \mathcal{P}(E^{\mathbb{R}})$ for this topology if for all $s_1 < \cdots < s_m \in \mathbb{R}^m$, $(P_n^{s_1, \cdots, s_m})_{n \ge 1}$ converges to $P^{s_1, \cdots, s_m}$ for the weak topology. We denote this convergence by $P_n \xrightarrow[n \to +\infty]{\text{f.d.}} P$. The following remark states that for Gaussian measures, this convergence is equivalent to the two-dimensional convergence.

**Remark 35.** Let $\{P_n\}_{n \ge 1} \cup \{P'\} \subset \mathcal{P}(\mathbb{R}^{\mathbb{R}})$ be a set of centered Gaussian processes. Then $P_n \xrightarrow[n \to +\infty]{\text{f.d.}} P'$ if and only if, for every $s < t \in \mathbb{R}^2$, $\lim_{n \to +\infty} (P_n)^{s,t} = (P')^{s,t}$. The direct implication is trivial. Conversely, assume, for every $s < t \in \mathbb{R}^2$, $\lim_{n \to +\infty} (P_n)^{s,t} = (P')^{s,t}$ and fix $t_1 < \cdots < t_m \in \mathbb{R}^m$. According to Lemma 10, $\lim_{n \to +\infty} P_n^{t_1, \cdots, t_m} = P'^{t_1, \cdots, t_m}$ if and only if $\Sigma_{P_n^{t_1, \cdots, t_m}} \xrightarrow[n \to +\infty]{\text{f.d.}} \Sigma_{(P')^{t_1, \cdots, t_m}}$. As for every Gaussian process $Q \in \mathcal{P}(\mathbb{R}^{\mathbb{R}})$ and $i < j \in [1, m]^2$, we have $\Sigma_{Q^{t_1, \cdots, t_m}}(i, j) = \Sigma_{Q^{t_i, t_j}}(1, 2)$ we get the converse implication.

**Notation 36.** Fix $\mathbb{R}_\uparrow^k := \{(t_1, \dots, t_k) \in \mathbb{R}^k ; t_1 < \cdots < t_k\}$, $\mathbb{R}_\uparrow := \cup_{k \ge 1} \mathbb{R}_\uparrow^k$ and denote by

$$ \mathcal{A} := \left\{ (R_n)_{n \ge 1} \in (\mathbb{R}_\uparrow)^{\mathbb{N}^*} ; \lim_{n \to +\infty} \sigma_{R_n} = 0, \lim_{n \to +\infty} \inf(R_n) = -\infty \text{ and } \lim_{n \to +\infty} \sup(R_n) = +\infty \right\} $$

the set of admissible sequences.

We can now give the definition of both weak and strong global Markov transform.

**Definition 37.** [Global Markov transform] Let $E$ be a Polish space and $P$ a probability measure on $E^{\mathbb{R}}$.

1. We say that $P$ admits a weak global Markov transform if there exists a Markov measure $P'$ on $E^{\mathbb{R}}$ and $(R_n)_{n \ge 1} \in \mathcal{A}$ such that $P_{[R_n]} \xrightarrow[n \to +\infty]{\text{f.d.}} P'$. We say that $P'$ is a weak global Markov transform of $P$.

2. We say that $P$ admits a strong global Markov transform if there exists a Markov measure $P'$ on $E^{\mathbb{R}}$ such that for every $(R_n)_{n \ge 1} \in \mathcal{A}$, we have $P_{[R_n]} \xrightarrow[n \to +\infty]{\text{f.d.}} P'$. We say that $P'$ is the strong global Markov transform of $P$.

The two-dimensional laws at time $(s, t)$ of the measure made Markov at times $R$ can be obtained by composing the transition kernel passing trough the times of $R$. More explicitly, one can readily check that $(P_{[R]})^{s,t} = P_{\{(s,t) \cup (R \cap [s,t])\}}^{s,t}$ (see Definition 2 for the right-hand side of the equality). In the following proposition, we verify that this implies that a weak (resp. strong) global Markov transforms of $P$ is a weak (resp. strong) local Markov transform of $P$. A natural question to ask is if local Markov transforms are also global Markov transform. This is less obvious and false in general. However, using Remark 35, we shall prove that it is true for strong Markov transform of Gaussian measures: If a Gaussian measure $P'$ is a strong local Markov transform of $P$, then $P'$ is also its strong global Markov transform<fnref n="10" />. In the case of weak Markov transforms, it stays true for stationary Gaussian measures under the hypothesis of Theorem 25.

<footnote n="10">¹⁰According to Remark 4, this justifies that we talk about the strong global Markov transform of $P$.</footnote>
**Proposition 38.** Let $P \in \mathcal{P}(\mathbb{R}^\mathbb{R})$ be a Gaussian measure and $P' \in \mathcal{P}(\mathbb{R}^\mathbb{R})$ be a Markov measure.

1. The measure $P'$ is the strong global Markov transform of $P$ if and only if it is the strong local Markov transform of $P$.

2. If $P'$ is a weak global Markov transform of $P$, then $P'$ is a weak local Markov transform of $P$.

*Proof.* 1. Assume $P'$ is the strong global Markov transform of $P$. In order to show that $P'$ is the strong local Markov transform of $P$, fix $s < t \in \mathbb{R}^2$ and $(\tilde{R}_n)_{n \ge 1} \in (S_{[s,t]})^{\mathbb{N}^*}$ satisfying $\lim_{n \to +\infty} \sigma_{\tilde{R}_n} = 0$. For every $n \ge 1$, we set $R_n := \tilde{R}_n \cup (([-n, s[\cup t, n]) \cap (2^{-n}\mathbb{Z}))$. We have $(R_n)_{n \ge 1} \in \mathcal{A}$, $(R_n \cap]s,t[) \cup \{s,t\} = \tilde{R}_n$ and $P_{[R_n]} \xrightarrow{n \to +\infty} P$. Hence, we get $P_{\{\tilde{R}_n\}}^{s,t} = P_{(R_n \cap]s,t[) \cup \{s,t\}}^{s,t} = (P_{[R_n]})^{s,t} \xrightarrow{n \to +\infty} P'^{s,t}$. Conversely, assume $P'$ is the strong local Markov transform of $P$. According to Remark 35, it is sufficient to show that, for every $s < t \in \mathbb{R}^2$, $\lim_{n \to +\infty} (P_{[R_n]})^{s,t} = (P')^{s,t}$, i.e., $\lim_{n \to +\infty} P_{\{\tilde{R}_n\}}^{s,t} = (P')^{s,t}$ where $\tilde{R}_n := \{s,t\} \cup (R_n \cap]s,t[)$. Since $(\tilde{R}_n)_{n \ge 1} \in (S_{[s,t]})^{\mathbb{N}^*}$, $\lim_{n \to +\infty} \sigma_{\tilde{R}_n} = 0$ and $P'$ is a strong local Markov transform of $P$, we get $(P_{[R_n]})^{s,t} = P_{\{\tilde{R}_n\}}^{s,t} \xrightarrow{n \to +\infty} (P')^{s,t}$. Thus $P'$ is the strong global Markov transform of $P$.

2. Fix $(R_n)_{n \ge 1} \in \mathcal{A}$ such that $P_{[R_n]} \xrightarrow{n \to +\infty} P'$. Given $s < t \in \mathbb{R}^2$, put $\tilde{R}_n := \{s,t\} \cup (R_n \cap [s,t])$. We have $(\tilde{R}_n)_{n \ge 1} \in (S_{[s,t]})^{\mathbb{N}^*}$, $\lim_{n \to +\infty} \sigma_{\tilde{R}_n} = 0$ and $P_{\{\tilde{R}_n\}}^{s,t} = P_{(R_n \cap]s,t[) \cup \{s,t\}}^{s,t} = (P_{[R_n]})^{s,t} \xrightarrow{n \to +\infty} (P')^{s,t}$. As this is true for every $s < t \in \mathbb{R}^2$, $P'$ is a weak local Markov transform of $P$. $\square$

Notice that we did not use the fact that $P$ is Gaussian to prove that every weak (resp. strong) global Markov transforms of $P$ is a weak (resp. strong) local Markov transform of $P$, but used it to prove that every strong local Markov transforms of $P$ is the strong global Markov transform of $P$. Indeed, we used the Remark 35, which is only valid in the Gaussian case.

**Proposition 39.** Let $K : (s,t) \in T^2 \mapsto \tilde{K}(t-s) \in \mathbb{R}$ be a continuous stationary positive semi-definite kernel that satisfies $\tilde{K}(0) = 1$. Let $P$ be a centered Gaussian measure with covariance function $K$ and set $L^K : h \mapsto h^{-1}(1-\tilde{K}(h))$. Assume $\alpha \in [0, +\infty]$ is a cluster point of $L^K$ at $0^+$ and let $(s_n)_{n \ge 1}$ be a positive sequence converging to zero such that $\lim_{n \to +\infty} L^K(s_n) = \alpha$. Then, writing $R_n := (s_n \cdot \mathbb{Z}) \cap [-n, n]$, we have $P_{[R_n]} \xrightarrow{n \to +\infty} P_\alpha$. In particular, the measure $P_\alpha$ is a weak global Markov transform of $P$.

*Proof.* Let $(s_n)_{n \ge 1}$ and $(R_n)_{n \ge 1}$ be as in the statement. According to Remark 35, we are left to prove that $\lim_{n \to +\infty} P_{[R_n]}^{s,t} = P_\alpha^{s,t}$ for every $s < t \in \mathbb{R}^2$. Since for every $n \ge \max(t, -s)$, we have $P_{[R_n]}^{s,t} = P_{\{\{s,t\} \cup \{\lfloor s, t \rfloor \cap R_n\}\}}^{s,t} = P_{\{\{s,t\} \cup \{\lfloor s, t \rfloor \cap \lfloor s_n \mathbb{Z} \rfloor \rfloor - n, n\}\}}^{s,t} = P_{\{\{s,t\} \cup \{\lfloor s, t \rfloor \cap \lfloor s_n \mathbb{Z} \rfloor\}\}}^{s,t}$, this corresponds exactly to the second point of Theorem 25. $\square$

**Corollary 40.** Let $P$ be the stationary Gaussian process appearing in Theorem 33. Then, for every $\alpha \in [0, +\infty]$, $P_\alpha$ is weak global Markov transform of $P$.

*Proof.* By construction of $P$, every $\alpha \in [0, +\infty]$ is a cluster point of the decay rate of the correlation function of $P$. According to Proposition 39, for every $\alpha \in [0, +\infty]$, the process $P_\alpha$ is a weak global Markov transform of $P$. $\square$
Until now, we only considered the topology of finite-dimensional convergence. In the case where $a < b \in \mathbb{R}^2$, we can endow $\mathcal{C}([a, b], \mathbb{R})$ with the topology of the uniform convergence, inducing a topology on $\mathcal{P}(\mathcal{C}([a, b], \mathbb{R}))$, that we simply call weak topology. We shall write $P_n \xrightarrow[n \to +\infty]{w} P$ to express that a sequence $(P_n)_{n \ge 1} \in \mathcal{P}(\mathcal{C}([a, b], \mathbb{R}))^{\mathbb{N}^*}$ converges to $P \in \mathcal{P}(\mathcal{C}([a, b], \mathbb{R}))$ for this topology, i.e., $\int fdP_n \xrightarrow[n \to +\infty]{} \int fdP$ for every continuous bounded function $f: (\mathcal{C}([a, b], \mathbb{R}), \|\cdot\|_{\infty}) \to \mathbb{R}$. Weak convergence implies convergence for the finite-dimensional topology, but in general the converse is false. However, if the family $(P_n)_{n \ge 1}$ is tight$^{11}$, then both convergences are equivalent. To carry out our convergence result from the finite-dimensional topology to the weak convergence topology, we need the Kolmogorov-Chenstov criterion to ensure that our measures $P$ and $P_{[R_n]}$ are concentrated on $\mathcal{C}([a, b], \mathbb{R})$ and the Kolmogorov tightness criterion to ensure that $\{P\} \cup \{P_{[R_n]}; n \ge 1\}$ is tight. For a proof of the Kolmogorov continuity criterion, we refer to [18, Theorem 2.9], whereas for the Kolmogorov tightness criteria we refer to [14, Theorem 23.7].

**Theorem 41.** Let $T \subset \mathbb{R}$ be a compact set.

**Kolmogorov-Chenstov criteria:** If $P \in \mathcal{P}(\mathbb{R}^T)$ and

$$ \exists a, b, C \in \mathbb{R}_+^*, \forall (s, t) \in T^2, \int_{\mathbb{R}^2} |x-y|^b P^{s,t}(dx, dy) \le C|t-s|^{1+a}, $$

then $P$ is concentrated on continuous paths.

**Kolmogorov tightness criteria:** If $(P_n)_{n \ge 1} \in \mathcal{P}(\mathcal{C}(T, \mathbb{R}))^{\mathbb{N}^*}$ and

$$ \exists a, b, C \in \mathbb{R}_+^*, \forall (s, t) \in T^2, \sup_{n \ge 1} \int_{\mathbb{R}^2} |x-y|^b P_n^{s,t}(dx, dy) \le C|t-s|^{1+a}, $$

then $(P_n)_{n \ge 1}$ is tight for the weak convergence.

In order to prove the uniform bounds of Theorem 41, we first establish a technical lemma.

**Lemma 42.** Consider a continuous semi-definite positive kernel $K: \mathbb{R}^2 \to \mathbb{R}$ such that $K(t, t) = 1$ for every $t \in \mathbb{R}$. Fix $a < b \in \mathbb{R}^2$ and denote by $P$ a centered Gaussian measure with covariance function $K$.

1. Assume, for every $s < t \in \mathbb{R}^2$,

$$ \sup_{v \in [s, t]} |L_v^K(h) - \alpha^K(v)| \xrightarrow{h \to 0^+} 0. \qquad (15) $$

Then

$$ \forall (R_n)_{n \ge 1} \in \mathcal{A}, \exists M \in \mathbb{R}_+, \forall (s, t) \in [a, b]^2, \forall n \ge 1, 1 - K_n(s, t) \le M|s - t|, $$

where $K_n$ stands for the covariance function of $P_{[R_n]}$.

<footnote>$^{11}$In this paper we only use tightness criteria not the definition of tightness. For completeness, we refer to [4].</footnote>
2. Assume $K: (s,t) \mapsto \tilde{K}(t-s)$ is a stationary kernel, $\alpha \in \mathbb{R}_+$ is a cluster point of $L^K: h \in \mathbb{R}_+^* \mapsto h^{-1}(1-\tilde{K}(h))$ and $L^K$ is bounded. Fix a positive sequence $(s_n)_{n \ge 1}$ converging to zero such that $\lim_{n \to +\infty} L^K(s_n) = \alpha$ and put $(R_n)_{n \ge 1} := ((s_n \cdot \mathbb{Z}) \cap [-n, n])_{n \ge 1} \in \mathcal{A}$. Then,

$$ \exists M \in \mathbb{R}_+, \forall (s,t) \in [a,b]^2, \forall n \ge 1, 1 - K_n(s,t) \le M|s-t|, $$

where $K_n$ stands for the covariance function of $P_{[R_n]}$.

*Proof.* We successively prove both statement.

1. Fix $(R_n)_{n \ge 1} \in \mathcal{A}$ and $s < t \in [a,b]^2$. We write $L_t$ instead of $L_t^K$, $\alpha$ instead of $\alpha^K$ and, for every $v \in \mathbb{R}$, we put $L_v(0) := \alpha(v)$. Set $\sigma^* := \sup_{n \ge 1} \sigma_{R_n}$, $M_1 := \sup_{0 \le h \le \sigma^*, v \in [a,b]} |L_v(h) - \alpha(v)|$, $M_2 := \sup_{x,y \in [a,b]} |\alpha(x) - \alpha(y)|$, $M_3 := \sup_{0 \le h \le \sigma^*, v \in [a,b]} L_v(h)$, $M_4 := \sup_{|x| \le \sigma^* M_3} |\varepsilon(x)|$ and $M_5 := M_1 + M_2 + M_3 M_4$, where $\varepsilon$ is defined in Notation 22. As in Theorem 23, hypothesis (15) implies $L$ that is continuous on $[a,b] \times [0, \sigma^*]$ and thus $M_3 < +\infty$. Since $K$, $\alpha$ and $\varepsilon$ are continuous, $M_5 < +\infty$. For a fixed $n \ge 1$, we write $\{s,t\} \cup (R_n \cap [s,t]) = \{t_0 < \dots < t_m\}$ and $h_k := t_{k+1} - t_k$ for every $k \in \{0, \dots, m-1\}$. Since, $(P_{[R_n]})^{s,t} = P_{\{(s,t) \cup (R_n \cap [s,t])\}}^{s,t}$, the composition formula in Lemma 12 gives $K_n(s,t) = K(t_0, t_1) \cdots K(t_{m-1}, t_m)$. As in the proof of Theorem 23,

$$ \begin{aligned} \left| \ln(K_n(s,t)) - \left(-\int_s^t \alpha(u)du\right) \right| & \le \left| \sum_{k=0}^{m-1} h_k L_{t_k}(h_k) - \sum_{k=0}^{m-1} h_k \alpha(t_k) \right| + \left| \sum_{k=0}^{m-1} h_k \alpha(t_k) - \int_s^t \alpha(u)du \right| \\ & \qquad + \left| \sum_{k=0}^{m-1} h_k L_{t_k}(h_k) \varepsilon(-h_k L_{t_k}(h_k)) \right| \\ & \le \sum_{k=0}^{m-1} h_k^n |L_{t_k}(h_k) - \alpha(t_k)| + \sum_{k=0}^{m-1} \int_{t_k}^{t_{k+1}} |\alpha(t_k) - \alpha(u)|du \\ & \qquad + \sum_{k=0}^{m-1} h_k L_{t_k}(h_k) |\varepsilon(-h_k L_{t_k}(h_k))| \\ & \le |t-s|M_1 + |t-s|M_2 + |t-s|M_3 M_4 \\ & = |t-s|M_5. \end{aligned} $$

Thus $\ln(K_n(s,t)) \ge -M_5|t-s| - \int_s^t \alpha(u)du \ge -M|t-s|$, where $M = M_5 + \sup_{u \in [a,b]} |\alpha(u)|$. Hence, for every $s < t \in [a,b]^2$, $K_n(s,t) \ge \exp(-M|t-s|) \ge 1 - M|s-t|$ which proves our result.

2. Let $(s_n)_{n \ge 1}$ and $R_n$ be as in the statement. Put $R_n = \{t_0 < \dots < t_m\}$, $\sigma^* := \sup_{n \ge 1} \sigma_{R_n}$ and, for every $k \in [0, m-1]$, $h_k := t_{k+1} - t_k$. Defining $\varepsilon$ as in Notation 22, for every $(v,h) \in \mathbb{R} \times \mathbb{R}_+^*$, $\ln(K(v, v+h)) = -hL^K(h) [1 + \varepsilon(\tilde{K}(h))]$. Hence,

$$ \begin{aligned} \left| \ln(K_n(s,t)) \right| & = \left| \sum_{k=0}^{m-1} h_k L^K(h_k) [1 + \varepsilon(\tilde{K}(h_k) - 1)] \right| \\ & \le |t-s|M, \end{aligned} $$
where $M := \sup_{0 \le h \le \sigma^*} |L^K(h)| (1 + \sup_{|x| \le \sigma^*} |\varepsilon(\tilde{K}(x) - 1)|)$. Since $L^K$ in bounded and $\varepsilon, \tilde{K}$ are continuous, $M$ is finite. So $\ln(K_n(s, t)) \ge -M|t-s|$, which implies $K_n(s, t) \ge e^{-M|t-s|} \ge 1 - M|t-s|$ and shows the result.
□

We can now apply our criteria to prove weak convergence.

**Theorem 43.** Let us consider a continuous positive semi-definite kernel $K : \mathbb{R} \times \mathbb{R} \to \mathbb{R}$ such that $K(t, t) = 1$ for every $t \in \mathbb{R}$. We denote by $P$ a centered Gaussian measure with covariance function $K$.

1. Assume, for every $s < t \in \mathbb{R}^2$,

$$ \sup_{v \in [s, t]} |L_v^K(h) - \alpha^K(v)| \xrightarrow{h \to 0^+} 0. $$

Then

$$ \forall (R_n)_{n \ge 1} \in \mathcal{A}, \forall a < b \in \mathbb{R}^2, \mathrm{proj}_{\#}^{[a,b]} P_{[R_n]} \xrightarrow{n \to +\infty} \mathrm{proj}_{\#}^{[a,b]} P_{\alpha^K}. $$

2. Assume $K : (s, t) \mapsto \tilde{K}(t-s)$ is a stationary positive semi-definite kernel, $\alpha \in \mathbb{R}_+$ is a cluster point of $L : h \in \mathbb{R}_+^* \mapsto h^{-1}(1 - \tilde{K}(h))$ and $L$ is bounded. Fix $(s_n)_{n \ge 1}$ a positive sequence converging to zero such that $\lim_{n \to +\infty} L^K(s_n) = \alpha$ and set $(R_n)_{n \ge 1} := ((s_n \cdot \mathbb{Z}) \cap [-n, n])_{n \ge 1} \in \mathcal{A}$. Then

$$ \forall a < b \in \mathbb{R}^2, \mathrm{proj}_{\#}^{[a,b]} P_{[R_n]} \xrightarrow{n \to +\infty} \mathrm{proj}_{\#}^{[a,b]} P_{\alpha}. $$

*Proof.* 1. Applying Lemma 42, there exists $M \in \mathbb{R}_+$ such that for every $(s, t, n) \in [a, b]^2 \times \mathbb{N}^*$, we have $1 - K_n(s, t) \le M|t-s|$. Hence,

$$ \int_{\mathbb{R}^2} |x - y|^2 dP_{[R_n]}^{s,t}(x, y) = K_n(s, s) + K_n(t, t) - 2K_n(s, t) = 2(1 - K_n(s, t)) \le 2M \cdot |s - t|. $$

Since for a centered Gaussian random variable $X$, one has $\mathbb{E}(X^4) = 3\mathbb{E}(X^2)^2$,

$$ \int_{\mathbb{R}^2} |x - y|^4 dP_{[R_n]}^{s,t}(x, y) = 3 \left( \int_{\mathbb{R}^2} |x - y|^2 dP_{[R_n]}^{s,t}(x, y) \right)^2 \le 12M^2 |s - t|^2. \quad (16) $$

As $\lim_{n \to +\infty} P_{[R_n]}^{s,t} = P^{s,t}$ and $(x, y) \mapsto |x - y|^4$ is lower semicontinuous and bounded by below, the Portmanteau theorem gives $\int_{\mathbb{R}^2} |x - y|^4 dP^{s,t}(x, y) \le \liminf_{n \to +\infty} \int_{\mathbb{R}^2} |x - y|^4 dP_{[R_n]}^{s,t}(x, y) \le 12M^2|s-t|^2$. According to the Kolmogorov-Chenstov criterion, we obtain that $\mathrm{proj}_{\#}^{[a,b]} P'$ is concentrated on continuous paths and $(\mathrm{proj}_{\#}^{[a,b]} P_n)_{n \ge 1}$ is a sequence of measures concentrated on continuous paths. Since our measures are concentrated on continuous paths, according to Inequality (16), the Kolmogorov tightness criteria applies and $(\mathrm{proj}_{\#}^{[a,b]} P_{[R_n]})_{n \ge 1}$ is tight. According to Point 1. of Proposition 38, we have $\mathrm{proj}_{\#}^{[a,b]} P_{[R_n]} \xrightarrow{n \to +\infty} \mathrm{proj}_{\#}^{[a,b]} P_{\alpha^K}$. Combined with tightness, this implies $\mathrm{proj}_{\#}^{[a,b]} P_{[R_n]} \xrightarrow{n \to +\infty} \mathrm{proj}_{\#}^{[a,b]} P_{\alpha^K}$, i.e., the wanted result.
2. Same proof as the first point, but applied to the sequence $(R_n)_{n \ge 1} \in \mathcal{A}$ defined by $R_n := s_n \mathbb{Z} \cap [-n, n]$. According to the second Point of Lemma 42, we also obtain Inequality (16), whereas the finite-dimensional convergence is given by Point 2. of Proposition 39. $\square$

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

**Remark 44.** In Theorem 43, $P$ is centered and $K(t,t) = 1$ for every $t \in \mathbb{R}$. However, both hypotheses are only there to simplify the statement and the proof. Assume instead that $P$ is non-centered with mean function $m$ and that, for every $t \in \mathbb{R}$, we have $K(t,t) > 0$. For every $a < b$, the map $f: (x_t)_{t \in [a,b]} \mapsto (\sqrt{K(t,t)}x_t + m(t))_{t \in [a,b]}$ is $(\sup_{t \in [a,b]} \sqrt{K(t,t)})$-Lipschitz, hence continuous for the uniform norm. Hence, one can apply Theorem 44 to the normalized measure $f_{\#}^{-1}P$ and push forward the obtained convergence results by $f$, which is continuous$^{12}$. At the end, we obtain

$$ \forall (R_n)_{n \ge 1} \in \mathcal{A}, \forall a < b \in \mathbb{R}^2, \mathrm{proj}_{\#}^{[a,b]} P_{[R_n]} \xrightarrow[n \to +\infty]{w} \mathrm{proj}_{\#}^{[a,b]} Q, $$

where $Q$ is the Gaussian process with mean function $m$ and with covariance function $K'$ given by

$$ K'(s,t) = K(s,s)^{1/2} K(t,t)^{1/2} \exp \left( - \int_s^t \alpha^K(u) du \right). $$

We now prove Theorem A of page 3 and add to it a result about the underlying dynamics the mimicking process. In Point 1 and 2, we prove that our mimicking problem has a unique solution. In Point 3, under a regularity assumption on the mean function and the variance function, we characterize the solution of this mimicking problem as the solution of a SDE. In Point 4, we show that the solution of the mimicking problem is obtained as the strong global Markov transform of the initial process.

**Theorem 45.** Let $X = (X_t)_{t \in \mathbb{R}}$ be a Gaussian process with continuous covariance function $K$ and positive variance function. Assume $\alpha^K$, the instantaneous decay rate of $K$, is well-defined and continuous.

1. Existence: There exists a Gaussian process $Y = (Y_t)_{t \in \mathbb{R}}$ with covariance $K'$ satisfying:

(1) For every $t \in \mathbb{R}$, $\mathrm{Law}(X_t) = \mathrm{Law}(Y_t)$;

(2) $\alpha^{K'} = \alpha^K$;

(3) $(Y_t)_{t \in \mathbb{R}}$ is a Markov process.

Moreover, for every $s < t \in \mathbb{R}^2$,

$$ \sup_{v \in [s,t]} \left| \alpha^K(v) - L_v^{K'}(h) \right| \xrightarrow{h \to 0^+} 0. \qquad (17) $$

If a Gaussian process $Y$ satisfies (1), (2) and (3), we allow ourselves to say that $Y$ is a mimicking process of $X$.

2. Uniqueness in law: Every mimicking process of $X$ with covariance function $K' : \mathbb{R}^2 \to \mathbb{R}$ has the same mean function as $(X_t)_{t \in \mathbb{R}}$ and, for every $s < t \in \mathbb{R}^2$,

$$ K'(s,t) = K(s,s)^{1/2} K(t,t)^{1/2} \exp \left( - \int_s^t \alpha^K(u) du \right). \qquad (18) $$

<footnote>$^{12}$To apply Theorem 43, notice that $L^K$ and $\alpha^K$ are invariant by renormalization, i.e., $L^{cK} = L^K$ and $\alpha^{cK} = \alpha^K$.</footnote>
3. Underlying dynamic of the mimicking process: Assume $m: t \mapsto \mathbb{E}(X_t)$ and $\sigma: t \mapsto K(t, t)^{1/2}$ are continuously differentiable. Then strong existence and strong uniqueness <fnref n="13" /> hold for the SDE

$$ \begin{cases} Z_t = [m'(t) + (\sigma'(t) - \alpha^K(t))Z_t] dt + \sigma(t)\sqrt{2\alpha^K(t)}dB_t \\ Z_0 \sim \mathcal{N}(0, 1) \end{cases} \qquad (19) $$

Moreover, the<fnref n="14" /> law of its solution is the law of the mimicking process of X (restricted to $\mathbb{R}_+$).

4. The mimicking process is a Markov transform (under a reinforced condition): Assume (2) is verified, i.e.,

$$ \sup_{v \in [s, t]} \left| \alpha^K(v) - \frac{1}{h} \left( 1 - \frac{K(v, v+h)}{\sqrt{K(v, v)} \sqrt{K(v+h, v+h)}} \right) \right| \underset{h \to 0^+}{\longrightarrow} 0, $$

for every $s < t \in \mathbb{R}^2$. Let $(R_n)_{n \ge 1} \in \mathcal{A}$ be an admissible sequence and $Y$ be a mimicking process. For every $n \ge 1$, we denote by $X^{R_n}$ the transformation of $X$ made Markov at times $R_n$. Then $(X^{R_n})_{n \ge 1}$ and $Y$ almost surely have continuous paths and $X^{R_n}$ converges weakly to $Y$ on compact sets<fnref n="15" />. In particular, the strong global Markov transform of $X$ is the mimicking process of $X$.

*Proof.* We write $\alpha$ instead of $\alpha^K$. To prove Point 1, we denote by $P$ the law of $X$. According to Proposition 20, the Gaussian measure $Q$ with same mean function as $P$ and covariance function $K'$ given by Formula (18) is well defined and Markov. We fix a process $Y = (Y_t)_{t \in \mathbb{R}}$ with law $Q$. For every $t \in \mathbb{R}$, $\text{Law}(Y_t) = \mathcal{N}(\mathbb{E}(Y_t), K'(t, t)) = \mathcal{N}(\mathbb{E}(X_t), K(t, t)) = \text{Law}(X_t)$, so $Y$ satisfies Condition (1). Let $\theta$ be defined by

$$ \theta: x \in \mathbb{R} \mapsto \begin{cases} \frac{e^x - 1 - x}{x} & \text{if } x \neq 0 \\ 0 & \text{if } x = 0 \end{cases}. $$

It is continuous and satisfies $e^x = 1 + x + x\theta(x)$ for every $x \in \mathbb{R}$. Hence, for every $v \in \mathbb{R}$,

$$ \begin{aligned} h^{-1}(1 - c_{K'}(v, v+h)) &= h^{-1}\left(1 - \exp\left(-\int_v^{v+h} \alpha(u)du\right)\right) \\ &= h^{-1}\int_v^{v+h} \alpha(u)dx + h^{-1}\left(\int_v^{v+h} \alpha(u)du\right)\theta\left(\int_v^{v+h} \alpha(u)du\right). \end{aligned} $$

To show (17), fix $s < t \in \mathbb{R}^2$. For every $h \in [0, 1]$, set

$$ M_h := \sup \left\{ \left| \alpha(v) - h^{-1} \int_v^{v+h} \alpha(u) du \right| ; v \in [s, t], 0 < h \le h \right\}. $$

Since

$$ \begin{aligned} M_h &\le \sup_{v \in [s, t]} h^{-1} \int_v^{v+h} |\alpha(v) - \alpha(u)| du \\ &\le \sup \left\{ |\alpha(w) - \alpha(u)| ; (w, u) \in [s, t+1]^2, |w - u| \le h \right\} \end{aligned} $$

<footnote n="13">¹³Strong uniqueness is meant in the sense of [15, Chapter 5, Definition 2.3]. By strong existence, we mean existence of a strong solution for any given brownian motion and independent condition. We refer to [15, Chapter 5, Definition 2.1] for the definition of a strong solution.</footnote>

<footnote n="14">¹⁴Since strong uniqueness holds, this process is unique up to indistinguishability.</footnote>

<footnote n="15">¹⁵This means that, for every $a < b \in \mathbb{R}^2$, $\text{Law}((X_t^{R_n})_{t \in [a, b]}) \xrightarrow{w} \text{Law}((Y_t)_{t \in [a, b]})$.</footnote>
and $\alpha$ is uniformly continuous on $[s, t+1]$, we get $\lim_{h \to 0^+} M_h = 0$. Hence, for every $v \in [s, t]$,

$$ 
\begin{aligned}
|\alpha(v) - L^{K'}(v, v+h)| &= |\alpha(v) - h^{-1} \int_{v}^{v+h} \alpha(u)du - h^{-1} \left( \int_{v}^{v+h} \alpha(u)du \right) \theta\left(-\int_{v}^{v+h} \alpha(u)du\right)| \\
&\leq |\alpha(v) - h^{-1} \int_{v}^{v+h} \alpha(u)du| + |h^{-1} \left( \int_{v}^{v+h} \alpha(u)du \right) \theta\left(-\int_{v}^{v+h} \alpha(u)du\right)| \\
&\leq M_h + \left( M_h + \sup_{u \in [s,t]} |\alpha(u)| \right) \sup \left\{ |\theta(x)| ; |x| \leq h \left( M_h + \sup_{u \in [s,t]} |\alpha(u)| \right) \right\} \\
&=: N_h,
\end{aligned}
$$

where we used

$$ \left| h^{-1} \int_{v}^{v+h} \alpha(u)du \right| \leq \left| h^{-1} \int_{v}^{v+h} \alpha(u)du - \alpha(v) \right| + |\alpha(v)| \leq M_{h^*} + \sup_{u \in [s,t]} |\alpha(u)|. $$

Hence,

$$ \sup_{v \in [s,t]} \left| \alpha^K(v) - L_v^{K'}(h) \right| \leq N_h. $$

Since $\lim_{h \to 0^+} N_h = 0$, this proves Equation 17 and in particular Condition (2). As $Q$ is Markov and Law($Y$) = $Q$, the process $Y$ is Markov, i.e., condition (3). To prove Point 2, we consider a Gaussian process $Y$ with covariance function $K'$ satisfying Conditions (1), (2) and (3). According to hypothesis (1), $Y$ clearly has the same mean function as $X$ and we are left to prove Formula (18). According to Remark 32, we can set $g_s : t \in \mathbb{R} \mapsto -\log(c_{K'}(s, t)) \in \mathbb{R}_+$ for every $s \in \mathbb{R}$. We have $g_s(s) = 0$ and $c_{K'}(s, t+h) = c_{K'}(s, t)c_{K'}(t, t+h)$ for every $(t, h) \in \mathbb{R} \times \mathbb{R}_^+$. By definition of $\alpha$, we have:

$$ 
\begin{aligned}
h^{-1}(g_s(t+h) - g_s(t)) &= -h^{-1} \left( \ln(c_{K'}(s, t+h)) - \ln(c_{K'}(s, t)) \right) \\
&= -h^{-1} \ln(c_{K'}(s, t+h)/c_{K'}(s, t)) \\
&= -h^{-1} \ln(c_{K'}(t, t+h)) \\
&= -h^{-1} (c_{K'}(t, t+h) - 1) [1 + \varepsilon (c_{K'}(t, t+h) - 1)] \\
\xrightarrow{h \to 0^+} & \alpha(t),
\end{aligned}
$$

where $\varepsilon$ is defined in Notation 22 and the convergence is obtained using Condition (3). This implies that $g_s(t) = g_s(s) + \int_s^t \alpha(u)du = \int_s^t \alpha(u)du$, i.e., Formula (18). To prove Point 3, put $b(t, x) := m'(t) + (\sigma'(t) - \alpha(t))x$ and $\sigma(t, x) := \sigma(t)\sqrt{2\alpha(t)}$. We now prove strong uniqueness holds for $(b, \sigma)$. According to [15, Chapter 5, Theorem 2.5], it is sufficient to prove that for every $T > 0$ and $n \geq 0$, there exists $K_{T,n} \in \mathbb{R}_+$ such that:

$$ \forall x, y \in [-n, n], \forall t \in [0, T], |b(t, x) - b(t, y)| + |\sigma(t, x) - \sigma(t, y)| \leq K_{T,n} |x - y|. \quad (20) $$

By continuity of $\alpha$, the constant $K_{T,n} := \sup_{t \in [0,T]} |\sigma'(t) - \alpha(t)|$ satisfies (20) and strong uniqueness holds for $(b, \sigma)$. For strong existence, we have to prove the existence of a strong solution for any probability
space $(\Omega, \mathcal{F}, \mathbb{P})$ endowed with a brownian motion $(B_t)_{t \ge 0}$ and an initial condition $Z$ independant from the brownian motion. First, prove the existence of a strong solution for the SDE

$$ \begin{cases} d\tilde{Z}_t = (-\alpha(t)\tilde{Z}_t) dt + \sqrt{2\alpha(t)}dB_t \\ \tilde{Z}_0 = Z \end{cases}. \qquad (21) $$

According to [15, Chapter 5, Section 6, Page 354], a strong solution to the SDE (21) is the process $(\tilde{Z}_t)_{t \ge 0}$ defined by

$$ \tilde{Z}_t := \Phi(t)Z + \int_0^t \frac{\Phi(t)}{\Phi(u)} \sqrt{2\alpha(u)}dB_u, $$

where $\Phi(t) := \exp(-\int_0^t \alpha(u)du)$. Hence, the process $(\tilde{Z}_t)_{t \ge 0}$ is centered Gaussian and its covariance function $K'' : (s,t) \mapsto \mathbb{E}(\tilde{Z}_s\tilde{Z}_t)$ is computed as follows. For every $s < t \in \mathbb{R}^2$,

$$ \begin{aligned} K''(s,t) &= \mathbb{E}(\Phi(s)\Phi(t)Z^2) + \mathbb{E}\left(\int_0^s \frac{\Phi(s)}{\Phi(u)}\sqrt{2\alpha(u)}dB_u \int_0^t \frac{\Phi(t)}{\Phi(v)}\sqrt{2\alpha(v)}dB_v\right) \\ &= \Phi(s)\Phi(t) + \Phi(s)\Phi(t)\int_0^s \frac{2\alpha(u)}{\Phi(u)^2}du \\ &= \Phi(s)\Phi(t)\left(1 + \int_0^s 2\alpha(u)\exp\left(\int_0^u 2\alpha(v)dv\right)du\right) \\ &= \Phi(s)\Phi(t)\left(1 + \left[\exp\left(\int_0^u 2\alpha(v)dv\right)\right]_0^s\right) \\ &= \Phi(s)\Phi(t)(1 + \Phi(s)^{-2} - 1) = \Phi(t)/\Phi(s) = K_\alpha(s,t). \end{aligned} $$

Hence $P_\alpha$ and Law($\tilde{Z}_t)_{t \ge 0}$) are two Gaussian measures with same mean and covariance function, thus are equal. Consider now the process $Z_t := m(t) + \sigma(t)\tilde{Z}_t$. This process is Gaussian, has mean function $m$ and covariance function given by (18). To obtain the fact that $(Z_t)_{t \ge 0}$ is solution of the SDE (19), we just apply the Itô formula with $f(t,x) = m(t) + \sigma(t)x$ and use the fact that $(\tilde{Z}_t)_{t \ge 0}$ is a solution of the SDE (21). We are left to prove Point 4. Since Law($X^{R_n}$) = $P_{[R_n]}$, we have to prove $\text{proj}_{\#}^{[a,b]} P_{[R_n]} \xrightarrow{n \to +\infty} \text{proj}_{\#}^{[a,b]} P'$. This follows from Theorem 43 and Remark 44. $\square$

The following numerical simulation of trajectories of a Gaussian process with law $P_\alpha$ and trajectories of solutions to the SDE (21) give an illustration of Point 4 of Theorem 45. The Gaussian process is simulated by discretizing time and using the Choleski decomposition, while our SDE is simulated with the Euler-Maruyama algorithm.

**Remark 46.** Suppose $(X_t)_{t \in \mathbb{R}}$ is a standard stationary Ornstein-Uhlenbeck process with parameter 1, $\alpha$ is a continuous non-negative function and set $\Phi : t \in \mathbb{R} \mapsto \int_0^t \alpha(u)du$. Then $(X_{\Phi(t)})_{t \in \mathbb{R}}$ has law $P_\alpha$. Indeed, $(X_{\Phi(t)})_{t \in \mathbb{R}}$ is a centered Gaussian process satisfying $\mathbb{E}(X_\Phi(s)X_{\Phi(t)}) = \exp(-|\Phi(t) - \Phi(s)|) = \exp(-\int_s^t \alpha(u)du)$.
<figure><img src="image_5.png" /><figcaption>Figure 1: Comparison of the trajectories of the SDE (21) with those of the Gaussian measure $P_α$.</figcaption></figure>
**Acknowledgment.** I would like to express my deep gratitude to Nicolas Juillet for introducing me to this research problem and for his constant support and valuable suggestions throughout the writing process.

**Copyright notice:**

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

This research was funded, in whole or in part, by the Agence

néationale de la recherche (ANR), Grant ANR-23-CE40-0017. A CC-BY public copyright license has been applied by the authors to the present document and will be applied to all subsequent versions up to the Author Accepted Manuscript arising from this submission, in accordance with the grant's open access conditions.

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Armand Ley – IRIMAS, UR 7499, Université de Haute-Alsace
18 rue des Frères Lumière, 68 093 Mulhouse, France
Email: armand.ley@uha.fr