SlowGuess/NeurIPS_2023_OCR huggingface dataset:
[60] Y. Wang, W. Huang, M. Gong, X. Geng, T. Liu, K. Zhang, and D. Tao. Identifiability and asymptotics in learning homogeneous linear ODE systems from discrete observations. arXiv preprint arXiv:2210.05955, 2022.
[61] D. J. Wilkinson. Stochastic modelling for quantitative description of heterogeneous biological systems. Nature Reviews Genetics, 10(2):122-133, 2009.
[62] M. Zakai and J. Snyders. Stationary probability measures for linear differential equations driven by white noise. Journal of Differential Equations, 8(1):27-33, 1970.
Appendix for "Generator Identification for Linear SDEs with Additive and Multiplicative Noise"
A Detailed proofs
Proof. We start by presenting the mathematical definition of a Lévy process. (cf. [51])
Definition A.1. A stochastic process  $X \coloneqq \{X_{t} : 0 \leqslant t < \infty\}$  is said to be a Lévy process if it satisfies the following properties:
1.  $X_0 = 0$  almost surely;
2. Independence of increments: For any  $0 \leqslant t_1 < t_2 < \ldots < t_n < \infty$ ,  $X_{t_2} - X_{t_1}$ ,  $X_{t_3} - X_{t_2}$ , ...,  $X_{t_n} - X_{t_{n-1}}$  are independent;
3. Stationary increments: For any  $s < t$ ,  $X_{t} - X_{s}$  is equal in distribution to  $X_{t - s}$ ;
4. Continuity in probability: For any  $\varepsilon > 0$  and  $0 \leqslant t < \infty$  it holds that  $\lim_{h \to 0} \mathbb{P}(|X_{t + h} - X_t| > \varepsilon) = 0$ .
In the following, we first show that the SDE (1) can be expressed as the form of (10), with  $Z$  being a  $p$ -dimensional Lévy process and  $a: \mathbb{R}^d \to \mathbb{R}^{d \times p}$  being Lipschitz. The first equation in the SDE (1) can be rearranged as
$$
\begin{array}{l} d X _ {t} = A X _ {t} d t + G d W _ {t} \\ = \left[ A X _ {t} \quad G \right] \left[ \begin{array}{l} d t \\ d W _ {t} \end{array} \right] \tag {18} \\ = a \left(X _ {t}\right) d Z _ {t}, \\ \end{array}
$$
$$
a (X _ {t}) = \left[ \begin{array}{c c} A X _ {t} & G \end{array} \right] \in \mathbb {R} ^ {d \times (m + 1)},
$$
$$
d Z _ {t} = \left[ \begin{array}{c} d t \\ d W _ {t} \end{array} \right] = \underbrace {\left[ \begin{array}{c} 1 \\ 0 _ {m \times 1} \end{array} \right]} _ {r} d t + \underbrace {\left[ \begin{array}{c c} 0 _ {1 \times 1} & 0 _ {1 \times m} \\ 0 _ {m \times 1} & I _ {m \times m} \end{array} \right]} _ {E} \left[ \begin{array}{c} d W _ {0, t} \\ d W _ {t} \end{array} \right], \tag {19}
$$
where  $0_{i\times j}$  denotes an  $i\times j$  zero matrix, let  $\tilde{W} := \{\tilde{W}_t:0\leqslant t < \infty \}$  with  $\tilde{W}_t = [W_{0,t},W_{1,t},\ldots ,W_{m,t}]^\top$  denote a  $(m + 1)$ -dimensional standard Brownian motion, then one can find a process  $Z := \{Z_t:0\leqslant t < \infty \}$  with
$$
Z _ {t} = r t + E \tilde {W} _ {t}, \tag {20}
$$
$$
Z _ {0} = 0,
$$
satisfying  $dZ_{t}$  described in Equation (19). Then we will show that the process  $Z$  described in (20) is a Lévy process, that is, it satisfies the four properties stated in Definition A.1.
Property 1: The first property is readily checked since  $Z_0 = 0$ .
Property 2: For any  $0 \leqslant t_1 < t_2 < t_3 < \infty$ ,
$$
\begin{array}{l} Z _ {t _ {2}} - Z _ {t _ {1}} = r (t _ {2} - t _ {1}) + E (\tilde {W} _ {t _ {2}} - \tilde {W} _ {t _ {1}}) \\ = \left[ \begin{array}{c} t _ {2} - t _ {1} \\ 0 _ {m \times 1} \end{array} \right] + \left[ \begin{array}{c} 0 \\ W _ {t _ {2}} - W _ {t _ {1}} \end{array} \right]. \\ \end{array}
$$
$$
Z _ {t _ {3}} - Z _ {t _ {2}} = \left[ \begin{array}{c} t _ {3} - t _ {2} \\ 0 _ {m \times 1} \end{array} \right] + \left[ \begin{array}{c} 0 \\ W _ {t _ {3}} - W _ {t _ {2}} \end{array} \right].
$$
Since  $W_{t_2} - W_{t_1}$  and  $W_{t_3} - W_{t_2}$  are independent,  $Z_{t_3} - Z_{t_2}$  and  $Z_{t_2} - Z_{t_1}$  are independent.
Property 3: when  $s < t$ ,
$$
\begin{array}{l} Z _ {t} - Z _ {s} = \left[ \begin{array}{c} t - s \\ 0 _ {m \times 1} \end{array} \right] + \left[ \begin{array}{c} 0 \\ W _ {t} - W _ {s} \end{array} \right] \\ \sim \mathcal {N} \left(\left[ \begin{array}{c} t - s \\ 0 _ {m \times 1} \end{array} \right], \left[ \begin{array}{c c} 0 _ {1 \times 1} & 0 _ {1 \times m} \\ 0 _ {m \times 1} & (t - s) I _ {m \times m} \end{array} \right]\right). \\ \end{array}
$$
$$
\begin{array}{l} Z _ {t - s} = r (t - s) + E \tilde {W} _ {t - s} \\ = \left[ \begin{array}{c} t - s \\ 0 _ {m \times 1} \end{array} \right] + \left[ \begin{array}{c} 0 \\ W _ {t - s} \end{array} \right] \\ \sim \mathcal {N} \left(\left[ \begin{array}{c} t - s \\ 0 _ {m \times 1} \end{array} \right], \left[ \begin{array}{c c} 0 _ {1 \times 1} & 0 _ {1 \times m} \\ 0 _ {m \times 1} & (t - s) I _ {m \times m} \end{array} \right]\right). \\ \end{array}
$$
Therefore, property 3 is checked.
Property 4: Obviously, process  $Z$  described in (20) is continuous with probability one at  $t$  for all  $0 \leqslant t < \infty$ , therefore,  $Z$  has continuity in probability.
Now that we have shown that process  $Z$  is a  $p$ -dimensional Lévy process with  $p = m + 1$ . Then we will show that  $a(X_{t}) = [AX_{t} - G]$  is Lipschitz.
$$
\begin{array}{l} \left\| a \left(X _ {t}\right) - a \left(X _ {s}\right) \right\| _ {F} = \left\| \left[ A \left(X _ {t} - X _ {s}\right) \quad 0 \right] \right\| _ {F} \\ = \| A \left(X _ {t} - X _ {s}\right) \| _ {2} \\ \leqslant \| A \| _ {F} \| X _ {t} - X _ {s} \| _ {2} \\ \end{array}
$$
where  $\| M \|_F$  denotes the Frobenius norm of matrix  $M$  and  $\| v \|_2$  denotes the Euclidean norm of vector  $v$ . Now it is readily checked that function  $a: \mathbb{R}^d \to \mathbb{R}^{d \times p}$  is Lipschitz.
Similarly, we will show that the SDE (2) can also be expressed as the form of (10), with  $Z$  being a  $p$ -dimensional Lévy process and  $a: \mathbb{R}^d \to \mathbb{R}^{d \times p}$  being Lipschitz. Let us rearrange the first equation in the SDE (2):
$$
\begin{array}{l} d X _ {t} = A X _ {t} d t + \sum_ {k = 1} ^ {m} G _ {k} X _ {t} d W _ {k, t} \\ = \left[ \begin{array}{c c c c} A X _ {t} & G _ {1} X _ {t} & \ldots & G _ {m} X _ {t} \end{array} \right] \left[ \begin{array}{c} d t \\ d W _ {1, t} \\ \vdots \\ d W _ {m, t} \end{array} \right] \\ = a \left(X _ {t}\right) d Z _ {t}. \\ \end{array}
$$
Since the  $dZ_{t}$  here has the same form as that of the SDE (1), we use the same process  $Z$  described in Equation (20), which has been shown to be a Lévy process.
As for the function  $a(X_{t})$
$$
\begin{array}{l} \parallel a (X _ {t}) - a (X _ {s}) \parallel_ {F} = \parallel [ A (X _ {t} - X _ {s}) G _ {1} (X _ {t} - X _ {s}) \dots G _ {m} (X _ {t} - X _ {s}) ] \parallel_ {F} \\ \leqslant \left\| A \left(X _ {t} - X _ {s}\right) \right\| _ {2} + \sum_ {k = 1} ^ {m} \left\| G _ {k} \left(X _ {t} - X _ {s}\right) \right\| _ {2} \\ \leqslant \| A \| _ {F} \| X _ {t} - X _ {s} \| _ {2} + \sum_ {k = 1} ^ {m} \| G _ {k} \| _ {F} \| X _ {t} - X _ {s} \| _ {2} \\ = \left(\| A \| _ {F} + \sum_ {k = 1} ^ {m} \| G _ {k} \| _ {F}\right) \| X _ {t} - X _ {s} \| _ {2}, \\ \end{array}
$$
it is readily checked that function  $a:\mathbb{R}^d\to \mathbb{R}^{d\times p}$  is Lipschitz.
A.2 Proof of Proposition 3.1
Proof. For the backward direction, when  $b(x) = \tilde{b}(x)$  and  $c(x) = \tilde{c}(x)$  for all  $x \in \mathbb{R}^d$ , it is obviously that  $(\mathcal{L}f)(x) = (\tilde{\mathcal{L}}f)(x)$  for all  $f \in C_b^2(\mathbb{R}^d)$  and  $x \in \mathbb{R}^d$ , that is  $\mathcal{L} = \tilde{\mathcal{L}}$ .
For the forward direction, since  $(\mathcal{L}f)(x_1) = (\tilde{\mathcal{L}} f)(x_1)$  for all  $f\in C_b^2 (\mathbb{R}^d)$  and  $x_{1}\in \mathbb{R}^{d}$ .
We first set
$$
f (x) = x _ {p},
$$
where  $x_{p}$  denotes the  $p$ -th component of variable  $x$ . It is readily checked that
$$
b _ {p} (x _ {1}) = \tilde {b} _ {p} (x _ {1}),
$$
for all  $x_{1}\in \mathbb{R}^{d}$  and  $p = 1,\ldots ,d$  .As a result,
$$
b (x) = \tilde {b} (x), \quad \text {f o r a l l} x \in \mathbb {R} ^ {d}.
$$
Then we set
$$
f (x) = \left(x _ {p} - x _ {1 p}\right) \left(x _ {q} - x _ {1 q}\right),
$$
where  $x_{1p}$  denotes the  $p$ -th component of  $x_{1}$ . It is readily checked that
$$
c _ {p q} (x _ {1}) = \tilde {c} _ {p q} (x _ {1}),
$$
for all  $x_{1}\in \mathbb{R}^{d}$  and  $p,q = 1,\ldots ,d$  . Consequently,
$$
c (x) = \tilde {c} (x), \quad \text {f o r a l l} x \in \mathbb {R} ^ {d}.
$$
Proof. For the forward direction, since
$$
X (\cdot ; x _ {0}, A, G) \stackrel {{\mathrm {d}}} {{=}} X (\cdot ; x _ {0}, \tilde {A}, \tilde {G}),
$$
one has
$$
\mathbb {E} \left[ X _ {t} \right] = \mathbb {E} \left[ \tilde {X} _ {t} \right], \quad \forall 0 \leqslant t <   \infty .
$$
$$
\left(X _ {t} - \mathbb {E} [ X _ {t} ]\right) _ {0 \leqslant t <   \infty} \stackrel {\mathrm {d}} {=} \left(\tilde {X} _ {t} - \mathbb {E} [ \tilde {X} _ {t} ]\right) _ {0 \leqslant t <   \infty},
$$
in particular, one has
$$
\mathbb {E} \left\{\left(X _ {t + h} - \mathbb {E} \left[ X _ {t + h} \right]\right) \left(X _ {t} - \mathbb {E} \left[ X _ {t} \right]\right) ^ {\top} \right\} = \mathbb {E} \left\{\left(\tilde {X} _ {t + h} - \mathbb {E} \left[ \tilde {X} _ {t + h} \right]\right) \left(\tilde {X} _ {t} - \mathbb {E} \left[ \tilde {X} _ {t} \right]\right) ^ {\top} \right\} \quad \text {f o r a l l} 0 \leqslant t, h <   \infty .
$$
For the backward direction, we know that the solution of the SDE (1) is a Gaussian process. The distribution of a Gaussian process can be fully determined by its mean and covariance functions. Therefore, the two processes have the same distribution when the mean and covariance are the same for both processes for all  $0 \leqslant t, h < \infty$ .
A.4 Proof of Lemma 3.3
$$
\operatorname {r a n k} \left(\left[ \gamma_ {1} \mid A \gamma_ {1} \right| \dots \left| A ^ {d - 1} \gamma_ {1} \right| \dots \left| \gamma_ {n} \right| \left| A \gamma_ {n} \right| \dots \left| A ^ {d - 1} \gamma_ {n} \right|\right) <   d,
$$
then for all  $l = [l_1,\dots ,l_n]^\top \in \mathbb{R}^n$
$$
\operatorname {r a n k} \left(\left[ \beta \mid A \beta \right| \dots \left| A ^ {d - 1} \beta \right]\right) <   d,
$$
where  $\beta \coloneqq l_1\gamma_1 + \ldots +l_n\gamma_n$  . Consequently, the corresponding ODE system
$$
\begin{array}{l} \dot {x} (t) = A x (t), \\ \begin{array}{l} x (t) - A x (t), \\ x (0) = \beta , \end{array} \tag {21} \\ \end{array}
$$
is not identifiable from  $\beta$  by [57, Theorem 2.5.], where  $\dot{x}(t)$  denotes the first derivative of  $x(t)$  with respect to time  $t$ .
$$
\tilde {\beta} := Q ^ {- 1} \beta \in \mathbb {R} ^ {d},
$$
$$
w _ {k} := \left\{ \begin{array}{l l} \tilde {\beta} _ {k} \in \mathbb {R} ^ {1}, & \text {f o r} k = 1, \ldots , K _ {1}, \\ (\tilde {\beta} _ {2 k - K _ {1} - 1}, \tilde {\beta} _ {2 k - K _ {1}}) ^ {\top} \in \mathbb {R} ^ {2}, & \text {f o r} k = K _ {1} + 1, \ldots , K. \end{array} \right.
$$
Simple calculation shows that
$$
\begin{array}{l} \tilde {\beta} = Q ^ {- 1} \beta \\ = Q ^ {- 1} \left(l _ {1} \gamma_ {1} + \dots + l _ {n} \gamma_ {n}\right) \\ = l _ {1} \tilde {\gamma} _ {1} + \ldots + l _ {n} \tilde {\gamma} _ {n}, \\ \end{array}
$$
therefore, one has
$$
w _ {k} = l _ {1} w _ {1, k} + \dots + l _ {n} w _ {n, k}, \quad \text {f o r a l l} k \in \{1, \dots , K \}. \tag {22}
$$
By [43, Theorem 2.4], we know that for any  $l \in \mathbb{R}^n$ , there always exists  $k \in \{1, \dots, K\}$  such that  $w_k = 0$  ( $\in \mathbb{R}^1$  or  $\mathbb{R}^2$ ) since the ODE (21) is not identifiable from initial state  $\beta$ . Next, we will show that this result is satisfied only when there exists a  $k$  such that  $w_{j,k} = 0$  ( $\in \mathbb{R}^1$  or  $\mathbb{R}^2$ ) for all  $j = 1, \dots, n$ . Let us rearrange the Equation (22) as
$$
\left[ \begin{array}{c c c} w _ {1, 1} & \ldots & w _ {n, 1} \\ \vdots & \ddots & \vdots \\ w _ {1, K} & \ldots & w _ {n, K} \end{array} \right] \left[ \begin{array}{c} l _ {1} \\ \vdots \\ l _ {n} \end{array} \right] = \left[ \begin{array}{c} w _ {1} \\ \vdots \\ w _ {K} \end{array} \right],
$$
assume for any  $k \in \{1, \ldots, K\}$ ,  $[w_{1,k}, \ldots, w_{n,k}]^\top \neq 0$ , then there always exists a  $l \in \mathbb{R}^n$  such that  $w_k \neq 0$  for all  $k = \{1, \ldots, K\}$ . The reason is that under this circumstance, for any  $k \in \{1, \ldots, K\}$ , the set of  $l$ 's such that  $w_k = 0$  has Lebesgue measure zero in  $\mathbb{R}^n$ . Therefore, the set of  $l$ 's such that there exists a  $k$  such that  $w_k = 0$  has Lebesgue measure zero in  $\mathbb{R}^n$ . This result creates a contradiction. Thus, there must exist a  $k$ , such that  $[w_{1,k}, \ldots, w_{n,k}]^\top = 0$ , that is  $|w_{j,k}| = 0$  for all  $j = 1, \ldots, n$ .
For the backward direction, there exists  $k$  such that  $|w_{j,k}| = 0$  for all  $j = 1,\dots ,n$ , that is  $w_{j,k} = 0$  ( $\in \mathbb{R}^1$  or  $\mathbb{R}^2$ ) for all  $j = 1,\ldots ,n$ . Simple calculation shows that
$$
\gamma_ {j} = Q \tilde {\gamma} _ {j} = \sum_ {p = 1} ^ {k - 1} Q _ {p} w _ {j, p} + \sum_ {p = k + 1} ^ {K} Q _ {p} w _ {j, p},
$$
$$
\begin{array}{l} A ^ {q} \gamma_ {j} = Q \Lambda^ {q} Q ^ {- 1} \gamma_ {j} \\ = Q \Lambda^ {q} \tilde {\gamma} _ {j} \\ = \sum_ {p = 1} ^ {k - 1} Q _ {p} J _ {p} ^ {q} w _ {j, p} + \sum_ {p = k + 1} ^ {K} Q _ {p} J _ {p} ^ {q} w _ {j, p}, \\ \end{array}
$$
recall that
$$
J _ {k} = \left\{ \begin{array}{l l} \lambda_ {k}, & \text {i f} k = 1, \ldots , K _ {1}  , \\ \left[ \begin{array}{c c} a _ {k} & - b _ {k} \\ b _ {k} & a _ {k} \end{array} \right], & \text {i f} k = K _ {1} + 1, \ldots , K  . \end{array} \right.
$$
Then matrix
$$
\begin{array}{l} M := \left[ \gamma_ {1} \mid A \gamma_ {1} \right| \dots \left| A ^ {d - 1} \gamma_ {1} \right| \dots \left| \gamma_ {n} \right| \left| A \gamma_ {n} \right| \dots \left| A ^ {d - 1} \gamma_ {n} \right] \\ = Q _ {- k} C, \\ \end{array}
$$
$$
Q _ {- k} = \left[ Q _ {1} \right| \dots \left| Q _ {k - 1} \right| \left| Q _ {k + 1} \right| \dots \left| Q _ {K} \right],
$$
and matrix  $C$  denotes:
$$
\left[ \begin{array}{c c c c c c c c c} w _ {1, 1} & J _ {1} w _ {1, 1} & \ldots & J _ {1} ^ {d - 1} w _ {1, 1} & \ldots & w _ {n, 1} & J _ {1} w _ {n, 1} & \ldots & J _ {1} ^ {d - 1} w _ {n, 1} \\ \vdots & \vdots & \ddots & \vdots & \ddots & \vdots & \vdots & \ddots & \vdots \\ w _ {1, k - 1} & J _ {k - 1} w _ {1, k - 1} & \ldots & J _ {k - 1} ^ {d - 1} w _ {1, k - 1} & \ldots & w _ {n, k - 1} & J _ {k - 1} w _ {n, k - 1} & \ldots & J _ {k - 1} ^ {d - 1} w _ {n, k - 1} \\ w _ {1, k + 1} & J _ {k + 1} w _ {1, k + 1} & \ldots & J _ {k + 1} ^ {d - 1} w _ {1, k + 1} & \ldots & w _ {n, k + 1} & J _ {k + 1} w _ {n, k + 1} & \ldots & J _ {k + 1} ^ {d - 1} w _ {n, k + 1} \\ \vdots & \vdots & \ddots & \vdots & \ddots & \vdots & \vdots & \ddots & \vdots \\ w _ {1, K} & J _ {K} w _ {1, K} & \ldots & J _ {K} ^ {d - 1} w _ {1, K} & \ldots & w _ {n, K} & J _ {K} w _ {n, K} & \ldots & J _ {K} ^ {d - 1} w _ {n, K} \end{array} \right]
$$
$$
\operatorname {r a n k} (M) = \operatorname {r a n k} \left(Q _ {- k} C\right) \leqslant \min  \left(\operatorname {r a n k} \left(Q _ {- k}\right), \operatorname {r a n k} (C)\right).
$$
When  $k \in \{1, \dots, K_1\}$ ,  $Q_{-k} \in \mathbb{R}^{d \times (d - 1)}$ , and  $\mathrm{rank}(Q_{-k}) = d - 1$ , while when  $k \in \{K_1 + 1, \dots, K\}$ ,  $Q_{-k} \in \mathbb{R}^{d \times (d - 2)}$ , and  $\mathrm{rank}(Q_{-k}) = d - 2$ . In both cases,  $\mathrm{rank}(Q_{-k}) < d$ , thus  $\mathrm{rank}(M) < d$ .
Proof. Let  $\tilde{A} \in \mathbb{R}^{d \times d}$  and  $\tilde{G} \in \mathbb{R}^{d \times m}$ , such that  $X(\cdot; x_0, A, G) \stackrel{\mathrm{d}}{=} X(\cdot; x_0, \tilde{A}, \tilde{G})$ , we denote as  $X \stackrel{\mathrm{d}}{=} \tilde{X}$ . For simplicity of notation, in the following, we denote  $A_1 := A$ ,  $A_2 := \tilde{A}$ ,  $G_1 := G$  and  $G_2 := \tilde{G}$ , and denote  $X \stackrel{\mathrm{d}}{=} \tilde{X}$  as  $X^1 \stackrel{\mathrm{d}}{=} X^2$ .
Sufficiency. We will show that under the identifiability condition (14), one has  $(A_{1},G_{1}G_{1}^{\top}) = (A_{2},G_{2}G_{2}^{\top})$ .
We first show that  $H_{1} = H_{2}$  ( $H_{i} \coloneqq G_{i}G_{i}^{T}$ ). Indeed, since  $X^{1}, X^{2}$  have the same distribution, one has
$$
\mathbb {E} \left[ f \left(X _ {t} ^ {1}\right) \right] = \mathbb {E} \left[ f \left(X _ {t} ^ {2}\right) \right] \tag {23}
$$
for all  $0 \leqslant t < \infty$  and  $f \in C^{\infty}(\mathbb{R}^{d})$ . By differentiating (23) at  $t = 0$ , one finds that
$$
\left(\mathcal {L} _ {1} f\right) \left(x _ {0}\right) = \left(\mathcal {L} _ {2} f\right) \left(x _ {0}\right), \tag {24}
$$
where  $\mathcal{L}_i$  is the generator of  $X^i$ $(i = 1,2)$ . Based on the Proposition 2.1,
$$
(\mathcal {L} _ {i} f) (x _ {0}) = \sum_ {k = 1} ^ {d} \sum_ {l = 1} ^ {d} (A _ {i}) _ {k l} x _ {0 l} \frac {\partial f}{\partial x _ {k}} (x _ {0}) + \frac {1}{2} \sum_ {k, l = 1} ^ {d} (H _ {i}) _ {k l} \frac {\partial^ {2} f}{\partial x _ {k} \partial x _ {l}} (x _ {0}),
$$
where  $(M)_{kl}$  denotes the  $kl$ -entry of matrix  $M$ , and  $x_{0l}$  is the  $l$ -th component of  $x_0$ . Since (24) is true for all  $f$ , by taking
$$
f (x) = (x _ {p} - x _ {0 p}) (x _ {q} - x _ {0 q}),
$$
it is readily checked that
$$
\left(H _ {1}\right) _ {p q} = \left(H _ {2}\right) _ {p q},
$$
for all  $p,q = 1,\ldots ,d$ . As a result,  $H_{1} = H_{2}$ . Let us call this matrix  $H$ .
Next, we show that  $A_{1} = A_{2}$ . We first show the relationship between  $A_{i}$  and  $x_{0}$ , and then show the relationship between  $A_{i}$  and  $H$ . To this end, one first recalls that
$$
X _ {t} ^ {i} = e ^ {A _ {i} t} x _ {0} + \int_ {0} ^ {t} e ^ {A _ {i} (t - s)} G _ {i} d W _ {s}.
$$
Set  $m_i(t)\coloneqq \mathbb{E}[X_t^i ]$  , we know that  $m_{i}(t)$  satisfies the ODE
$$
\begin{array}{l} \dot {m} _ {i} (t) = A _ {i} m _ {i} (t), \quad \forall 0 \leqslant t <   \infty , \\ m _ {i} (0) = x _ {0}, \end{array} \tag {25}
$$
where  $\hat{f}(t)$  denotes the first derivative of function  $f(t)$  with respect to time  $t$ .
$$
m _ {i} (t) = e ^ {A _ {i} t} x _ {0}.
$$
Since  $X^1 \stackrel{\mathrm{d}}{=} X^2$ , one has
$$
\mathbb {E} \left[ X _ {t} ^ {1} \right] = \mathbb {E} \left[ X _ {t} ^ {2} \right]
$$
for all  $0 \leqslant t < \infty$ . That is
$$
e ^ {A _ {1} t} x _ {0} = e ^ {A _ {2} t} x _ {0}, \quad \forall 0 \leqslant t <   \infty .
$$
Taking  $k$ -th derivative of  $e^{A_i t} x_0$  with respect to  $t$ , one finds that
$$
\left. \frac {d ^ {k}}{d t ^ {k}} \right| _ {t = 0} e ^ {A _ {i} t} x _ {0} = A _ {i} ^ {k} x _ {0},
$$
for all  $k = 1,2,\ldots$  . Consequently,
$$
A _ {1} ^ {k} x _ {0} = A _ {2} ^ {k} x _ {0}.
$$
Let us denote this vector  $A^k x_0$ . Obviously, one gets
$$
A _ {1} A ^ {k - 1} x _ {0} = A _ {2} A ^ {k - 1} x _ {0} \quad \text {f o r a l l} k = 1, 2, \dots . \tag {26}
$$
In the following, we show the relationship between  $A_{i}$  and  $H$ . Let us denote
$$
Y _ {t} ^ {i} := \int_ {0} ^ {t} e ^ {A _ {i} (t - s)} G _ {i} d W _ {s} = X _ {t} ^ {i} - \mathbb {E} [ X _ {t} ^ {i} ]
$$
$$
V _ {i} (t, t + h) := \mathbb {E} \left[ Y _ {t + h} ^ {i} \cdot \left(Y _ {t} ^ {i}\right) ^ {T} \right].
$$
$$
\begin{array}{l} V _ {i} (t, t + h) = e ^ {A _ {i} h} \int_ {0} ^ {t} e ^ {A _ {i} (t - s)} H e ^ {A _ {i} ^ {\top} (t - s)} d s \tag {27} \\ = e ^ {A _ {i} h} V _ {i} (t), \\ \end{array}
$$
where  $V_{i}(t)\coloneqq V_{i}(t,t)$
Since  $X^1 \stackrel{\mathrm{d}}{=} X^2$ , by Lemma 3.2, one has
$$
V _ {1} (t, t + h) = V _ {2} (t, t + h), \forall 0 \leqslant t, h <   \infty .
$$
To obtain information about  $A_{i}$ , let us fix  $t$  for now and take  $k$ -th derivative of (27) with respect to  $h$ . One finds that
$$
\left. \frac {d ^ {k}}{d h ^ {k}} \right| _ {h = 0} V _ {i} (t, t + h) = A _ {i} ^ {k} V _ {i} (t), \tag {28}
$$
for all  $k = 1,2,\ldots$
On the other hand, the function  $V_{i}(t)$  satisfies the ODE [56]
$$
\dot {V} _ {i} (t) = A _ {i} V _ {i} (t) + V _ {i} (t) A _ {i} ^ {\top} + H, \quad 0 \leqslant t <   \infty ,
$$
$$
V _ {i} (0) = 0.
$$
$$
\dot {V} _ {i} (0) = A _ {i} V _ {i} (0) + V _ {i} (0) A _ {i} + H = H.
$$
By differentiating (28) at  $t = 0$ , it follows that
$$
\left. \frac {d}{d t} \right| _ {t = 0} \frac {d ^ {k}}{d h ^ {k}} \Big | _ {h = 0} V _ {i} (t, t + h) = A _ {i} ^ {k} H,
$$
$$
A _ {1} ^ {k} H = A _ {2} ^ {k} H.
$$
Let us denote this matrix  $A^k H$ . Obviously, by rearranging this matrix, one gets
$$
A _ {1} A ^ {k - 1} H = A _ {2} A ^ {k - 1} H \quad \text {f o r a l l} k = 1, 2, \dots . \tag {29}
$$
Recall our identifiability condition is that  $\mathrm{rank}(M) = d$  with
$$
M := \left[ x _ {0} \mid A x _ {0} \right| \dots \left. \mid A ^ {d - 1} x _ {0} \mid H _ {\cdot 1} \mid A H _ {\cdot 1} \right| \dots \left. \mid A ^ {d - 1} H _ {\cdot 1} \right| \dots \left. \mid H _ {\cdot d} \mid A H _ {\cdot d} \right| \dots \left. \mid A ^ {d - 1} H _ {\cdot d} \right].
$$
If we denote the  $j$ -th column in  $M$  as  $M_{.j}$ , one gets
$$
A _ {1} M. _ {j} = A _ {2} M. _ {j},
$$
for all  $j = 1, \dots, d + d^2$  by Equations (26) and (29).
This means one can find a full-rank matrix  $B \in \mathbb{R}^{d \times d}$  by horizontally stacking  $d$  linearly independent columns of matrix  $M$ , such that  $A_{1}B = A_{2}B$ . Since  $B$  is invertible, one thus concludes that  $A_{1} = A_{2}$ . Hence, the sufficiency of the condition is proved.
Necessity. In the following, we will show that when  $A$  has distinct eigenvalues. The condition (14) stated in Theorem 3.4 is also necessary. Specifically, we will show that when the identifiability condition (14) is not satisfied, one can always find a  $\tilde{A}$  with  $(A, GG^{\top}) \neq (\tilde{A}, GG^{\top})$  such that  $X \stackrel{\mathrm{d}}{=} \tilde{X}$ . Recall that for simplicity of notation, we denote  $A_1 \coloneqq A$ ,  $A_2 \coloneqq \tilde{A}$ , and denote  $X \stackrel{\mathrm{d}}{=} \tilde{X}$  as  $X^1 \stackrel{\mathrm{d}}{=} X^2$ , where process  $X^i = \{X_t^i : 0 \leqslant t < \infty\}$ , and  $X_t^i = X(t; x_0, A_i, G)$  following the form described in the solution process (3). In the following, we may use both  $A$  and  $A_1$  interchangeably according to the context.
By Lemma 3.2, to guarantee  $X^1 \stackrel{\mathrm{d}}{=} X^2$  one only needs to show that
$$
\mathbb {E} \left[ X _ {t} ^ {1} \right] = \mathbb {E} \left[ X _ {t} ^ {2} \right], \quad \forall 0 \leqslant t <   \infty ,
$$
$$
e ^ {A _ {1} t} x _ {0} = e ^ {A _ {2} t} x _ {0}, \quad \forall 0 \leqslant t <   \infty ,
$$
$$
e ^ {A _ {1} h} V (t) = e ^ {A _ {2} h} V (t), \quad \forall 0 \leqslant t, h <   \infty , \tag {30}
$$
$$
V _ {1} (t) = V _ {2} (t), \forall 0 \leqslant t <   \infty ,
$$
where  $V(t)\coloneqq V_1(t) = V_2(t)$
Recall that  $H = GG^{\top}$ . For simplicity of notation, abusing notation a bit, we denote  $H_{0} \coloneqq x_{0}$ . Let
$$
\tilde {H} _ {\cdot j} := Q ^ {- 1} H _ {\cdot j}, \quad \text {f o r a l l} j = 0, \dots , d,
$$
$$
w _ {j, k} := \left\{ \begin{array}{l l} \tilde {H} _ {\cdot j, k} \in \mathbb {R} ^ {1}, & \text {f o r} k = 1, \ldots , K _ {1}  , \\ (\tilde {H} _ {\cdot j, 2 k - K _ {1} - 1}, \tilde {H} _ {\cdot j, 2 k - K _ {1}}) ^ {\top} \in \mathbb {R} ^ {2}, & \text {f o r} k = K _ {1} + 1, \ldots , K  . \end{array} \right.
$$
When the identifiability condition (14) is not satisfied, that is
$$
\operatorname {r a n k} \left( \right.\left[ \right. H _ {0} \mid A H _ {0} \left. \right| \dots \mid A ^ {d - 1} H _ {0} \mid H _ {1} \mid A H _ {1} \mid \dots \mid A ^ {d - 1} H _ {1} \mid \dots \mid H _ {d} \mid A H _ {d} \mid \dots \mid A ^ {d - 1} H _ {d} \left. \right]\left. \right) <   d,
$$
by Lemma 3.3, there exists  $k$  such that  $|w_{j,k}| = 0$ , i.e.,  $w_{j,k} = 0$  ( $\in \mathbb{R}^1$  or  $\mathbb{R}^2$ ), for all  $j = 0, \ldots, d$ . Recall that
$$
\begin{array}{l} V (t) = V _ {1} (t) = \int_ {0} ^ {t} e ^ {A (t - s)} H e ^ {A ^ {\top} (t - s)} d s \\ = \int_ {0} ^ {t} Q e ^ {\Lambda (t - s)} Q ^ {- 1} Q [ \tilde {H}. _ {1} | \dots | \tilde {H}. _ {d} ] e ^ {A ^ {\top} (t - s)} d s \tag {31} \\ = Q \int_ {0} ^ {t} e ^ {\Lambda (t - s)} [ \tilde {H}. _ {1} | \dots | \tilde {H}. _ {d} ] e ^ {A ^ {\top} (t - s)} d s \\ := Q \int_ {0} ^ {t} W e ^ {A ^ {\top} (t - s)} d s, \\ \end{array}
$$
$$
W = e ^ {\Lambda (t - s)} [ \tilde {H}. _ {1} | \dots | \tilde {H}. _ {d} ],
$$
and some calculation shows that