SlowGuess/NeurIPS_2023_OCR huggingface dataset:
$$
\begin{array}{l} \mathbb {E} \left[ e ^ {\sum_ {k = 1} ^ {m} G _ {k} W _ {k, t}} x _ {0} x _ {0} ^ {T} e ^ {\sum_ {k = 1} ^ {m} G _ {k} ^ {\top} W _ {k, t + h}} \right] \\ = \mathbb {E} \left[ e ^ {\sum_ {k = 1} ^ {m} G _ {k} W _ {k, t}} x _ {0} x _ {0} ^ {T} e ^ {\sum_ {k = 1} ^ {m} G _ {k} ^ {\top} W _ {k, t}} e ^ {- \sum_ {k = 1} ^ {m} G _ {k} ^ {\top} W _ {k, t}} e ^ {\sum_ {k = 1} ^ {m} G _ {k} ^ {\top} W _ {k, t + h}} \right] \tag {52} \\ = \mathbb {E} \left[ e ^ {\sum_ {k = 1} ^ {m} G _ {k} W _ {k, t}} x _ {0} x _ {0} ^ {T} e ^ {\sum_ {k = 1} ^ {m} G _ {k} ^ {\top} W _ {k, t}} e ^ {\sum_ {k = 1} ^ {m} G _ {k} ^ {\top} \left(W _ {k, t + h} - W _ {k, t}\right)} \right] \\ = \mathbb {E} \left[ e ^ {\sum_ {k = 1} ^ {m} G _ {k} W _ {k, t}} x _ {0} x _ {0} ^ {T} e ^ {\sum_ {k = 1} ^ {m} G _ {k} ^ {\top} W _ {k, t}} \right] \mathbb {E} \left[ e ^ {\sum_ {k = 1} ^ {m} G _ {k} ^ {\top} (W _ {k, t + h} - W _ {k, t})} \right], \\ \end{array}
$$
because the Brownian motion  $W_{k,t}$  has independent increments.
It is known that, for  $Z\sim \mathcal{N}(0,1)$ , we have that the  $j$ th moment is
$$
\mathbb {E} (Z ^ {j}) = \left\{ \begin{array}{c l} 0  , & j \text {i s o d d}, \\ 2 ^ {- j / 2} \frac {j !}{(j / 2) !}, & j \text {i s e v e n}. \end{array} \right.
$$
Since  $W_{k,t}\sim \mathcal{N}(0,t)$  , we have
$$
\begin{array}{l} \mathbb {E} \left[ e ^ {G _ {k} W _ {k, t}} \right] = \mathbb {E} \left[ \sum_ {j = 0} ^ {\infty} \frac {(G _ {k}) ^ {j} (W _ {k , t}) ^ {j}}{j !} \right] \\ = \sum_ {j = 0} ^ {\infty} \frac {(G _ {k}) ^ {j} \mathbb {E} [ (W _ {k , t}) ^ {j} ]}{j !} \\ = \sum_ {j = 0, 2, 4 \dots} ^ {\infty} \frac {(G _ {k}) ^ {j} (t / 2) ^ {j / 2}}{(j / 2) !} \\ = \sum_ {i = 0} ^ {\infty} \frac {\left(G _ {k} ^ {2} t / 2\right) ^ {i}}{i !} \\ = e ^ {G _ {k} ^ {2} t / 2}. \\ \end{array}
$$
$$
\mathbb {E} \big [ e ^ {G _ {k} ^ {\top} (W _ {k, t + h} - W _ {k, t})} \big ] = e ^ {(G _ {k} ^ {\top}) ^ {2} h / 2}.
$$
$$
\mathbb {E} \left[ e ^ {\sum_ {k = 1} ^ {m} G _ {k} ^ {\top} \left(W _ {k, t + h} - W _ {k, t}\right)} \right] = e ^ {\sum_ {k = 1} ^ {m} \left(G _ {k} ^ {\top}\right) ^ {2} h / 2}. \tag {53}
$$
By combining Equations (51), (52) and (53), one readily obtains that
$$
P (t, t + h) = P (t, t) e ^ {A ^ {\top} h}, \tag {54}
$$
we denote  $P(t) \coloneqq P(t, t)$ . Set  $P_i(t, t + h) = \mathbb{E}[X_t^i(X_{t + h}^i)^\top]$ , since  $X^1 \stackrel{\mathrm{d}}{=} X^2$ , it follows that
$$
P _ {1} (t, t + h) = \mathbb {E} \left[ X _ {t} ^ {1} \left(X _ {t + h} ^ {1}\right) ^ {\top} \right] = \mathbb {E} \left[ X _ {t} ^ {2} \left(X _ {t + h} ^ {2}\right) ^ {\top} \right] = P _ {2} (t, t + h) \quad \forall t, h \geqslant 0.
$$
To obtain information about  $A$ , let us fix  $t$  for now and take  $j$ -th derivative of (54) with respect to  $h$ . One finds that
$$
\left. \frac {d ^ {j}}{d h ^ {j}} \right| _ {h = 0} P (t, t + h) = P (t) \left(A ^ {\top}\right) ^ {j}, \tag {55}
$$
for all  $j = 1,2,\ldots$ . It is readily checked that
$$
P _ {1} (t) \left(A _ {1} ^ {\top}\right) ^ {j} = P _ {2} (t) \left(A _ {2} ^ {\top}\right) ^ {j} \quad \forall 0 \leqslant t <   \infty . \tag {56}
$$
We know the function  $P_{i}(t)$  satisfies the ODE
$$
\dot {P} _ {i} (t) = A _ {i} P _ {i} (t) + P _ {i} (t) A _ {i} ^ {\top} + \sum_ {k = 1} ^ {m} G _ {i, k} P _ {i} (t) G _ {i, k} ^ {\top}, \quad \forall 0 \leqslant t <   \infty , \tag {57}
$$
$$
\dot {P} _ {i} (0) = A _ {i} x _ {0} x _ {0} ^ {\top} + x _ {0} x _ {0} ^ {\top} A _ {i} ^ {\top} + \sum_ {k = 1} ^ {m} G _ {i, k} x _ {0} x _ {0} ^ {\top} G _ {i, k} ^ {\top}.
$$
By differentiating (56) with respect to  $t$  at  $t = 0$ , it follows that
$$
\begin{array}{l} A _ {1} x _ {0} x _ {0} ^ {\top} \left(A _ {1} ^ {\top}\right) ^ {j} + x _ {0} x _ {0} ^ {\top} \left(A _ {1} ^ {\top}\right) ^ {j + 1} + \left(\sum_ {k = 1} ^ {m} G _ {1, k} x _ {0} x _ {0} ^ {\top} G _ {1, k} ^ {\top}\right) \left(A _ {1} ^ {\top}\right) ^ {j} \\ = A _ {2} x _ {0} x _ {0} ^ {\top} \left(A _ {2} ^ {\top}\right) ^ {j} + x _ {0} x _ {0} ^ {\top} \left(A _ {2} ^ {\top}\right) ^ {j + 1} + \left(\sum_ {k = 1} ^ {m} G _ {2, k} x _ {0} x _ {0} ^ {\top} G _ {2, k} ^ {\top}\right) \left(A _ {2} ^ {\top}\right) ^ {j}. \\ \end{array}
$$
Since we have known that  $A_1^j x_0 = A_2^j x_0$  for all  $j = 1, 2, \ldots$ , it is readily checked that
$$
\left(\sum_ {k = 1} ^ {m} G _ {1, k} x _ {0} x _ {0} ^ {\top} G _ {1, k} ^ {\top}\right) (A _ {1} ^ {\top}) ^ {j} = \left(\sum_ {k = 1} ^ {m} G _ {2, k} x _ {0} x _ {0} ^ {\top} G _ {2, k} ^ {\top}\right) (A _ {2} ^ {\top}) ^ {j},
$$
that is  $A_1^j H = A_2^j H$  for all  $j = 1, 2, \ldots$ . Let us denote this matrix  $A^j H$ . Obviously, by rearranging this matrix, one gets
$$
A _ {1} A ^ {j - 1} H = A _ {2} A ^ {j - 1} H \quad \text {f o r a l l} j = 1, 2, \dots
$$
Therefore, under condition C1, that is  $\mathrm{rank}(M) = d$  with
$$
M := \left[ x _ {0} \mid A x _ {0} \right| \dots | A ^ {d - 1} x _ {0} | H _ {. 1} | A H _ {. 1} | \dots | A ^ {d - 1} H _ {. 1} | \dots | H _ {. d} | A H _ {. d} | \dots | A ^ {d - 1} H _ {. d} ]. \tag {58}
$$
if we denote the  $j$ -th column in  $M$  as  $M_{.j}$ , one gets  $A_{1}M_{.j} = A_{2}M_{.j}$  for all  $j = 1, \ldots, d + d^{2}$  by equations (49) and (58).
This means one can find a full-rank matrix  $B \in \mathbb{R}^{d \times d}$  by horizontally stacking  $d$  linearly independent columns from matrix  $M$ , such that  $A_{1}B = A_{2}B$ . Since  $B$  is invertible, one thus concludes that  $A_{1} = A_{2}$ .
In the proof of Theorem 3.5, we have shown that when  $A_{1} = A_{2}$ , under condition C2, for all  $x \in \mathbb{R}^d$ ,
Thus the proposition is proved.
It is noteworthy that Proposition E.1 is established on the explicit solution assumption of the SDE (2), which requires both sets of vectors  $\{A, \{G_k\}_{k=1}^m\}$  and  $\{\tilde{A}, \{\tilde{G}_k\}_{k=1}^m\}$  to satisfy condition C3. As aforementioned, condition C3 is very restrictive and impractical, rendering the identifiability condition derived in this proposition unsatisfactory. Nonetheless, this condition is presented to illustrate that condition C1 is more relaxed compared to condition A1 stated in Theorem 3.5 when identifying  $A$  with the incorporation of  $G_k$ 's information.