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A Probability measures on infinite-dimensional Hilbert spaces
In this section, we briefly present some fundamental notions related to probability measures on infinite-dimensional spaces, specifically separable Hilbert spaces  $(H,\langle \cdot ,\cdot \rangle)$ . There is abundant literature on the subject. For more details we refer to Stuart [5], Pidstrigach et al. [24], Kerrigan et al. [32], Prato [51] and references therein.
A.1 Gaussian measures on Hilbert spaces
Definition 3. Let  $(\Omega, \mathcal{F}, \mathbb{P})$  be a probability space. A measurable function  $X: \Omega \to H$  is called a Gaussian random element (GRE) if for any  $h \in H$ , the random variable  $\langle h, X \rangle$  has a scalar Gaussian distribution.
Every GRE  $X$  has a mean element  $m \in H$  defined by
$$
m = \int_ {\Omega} X (\omega) d \mathbb {P} (\omega),
$$
and a linear covariance operator  $C:H\to H$  defined by
$$
C h = \int_ {\Omega} \langle h, X (\omega) \rangle X (\omega) d \mathbb {P} (\omega) - \langle m, h \rangle m, \quad \forall h \in H.
$$
We denote  $X \sim \mathcal{N}(m, C)$  for a GRE in  $H$  with mean element  $m$  and covariance operator  $C$ . It can be shown that the covariance operator of a GRE is trace class, positive-definite and symmetric. Conversely, for any trace class, positive-definite and symmetric linear operator  $C: H \to H$  and every  $m \in H$ , there exists a GRE with  $X \sim \mathcal{N}(m, C)$ . This leads us to the following definition:
Definition 4. If  $X$  is a GRE, the pushforward of  $\mathbb{P}$  through  $X$ , denoted by  $\mathbb{P}_X$ , is called a Gaussian probability measure on  $H$ . We will write  $\mathbb{P}_X = \mathcal{N}(m, C)$ .
Let  $X\sim \mathcal{N}(m,C)$  . We can make a few remarks:
1) For any  $h\in H$  , we have  $\langle h,X\rangle \sim \mathcal{N}(\langle h,m\rangle ,\langle Ch,h\rangle)$
2)  $C$  is compact. By Mercer theorem [52] there exists  $(\lambda_j)$  and an orthonormal basis of eigenfunctions  $(v_{j})$  such that  $\lambda_{j}\geq 0$  and  $Cv_{j} = \lambda_{j}v_{j}\forall j$ . We consider the infinite-dimensional case in which  $\lambda_{j} > 0$ $\forall j$ .
3) Suppose  $m = 0$  (we call the Gaussian measure of  $X$  centered). The expected square norm of  $X$  is given by
$$
\mathbb {E} [ \| X \| _ {H} ^ {2} ] = \mathbb {E} \left[ \sum_ {j = 1} ^ {\infty} \langle v _ {j}, X \rangle^ {2} \right] = \sum_ {j = 1} ^ {\infty} \langle C v _ {j}, v _ {j} \rangle = \sum_ {j = 1} ^ {\infty} \lambda_ {j} = \operatorname {T r} (C),
$$
which is finite since  $C$  is trace class.
A.2 Absolutely continuous measures and the Feldman-Hajek theorem
Here we introduce the notion of absolute continuity for measures.
Definition 5. Let  $\mu$  and  $\nu$  be two probability measures on  $H$  equipped with its Borel  $\sigma$ -algebra  $\mathcal{B}(H)$ . Measure  $\mu$  is absolutely continuous with respect to  $\nu$  (we write  $\mu \ll \nu$ ) if  $\mu(\Sigma) = 0$  for all  $\Sigma \in \mathcal{B}(H)$  such that  $\nu(\Sigma) = 0$ .
Definition 6. If  $\mu \ll \nu$  and  $\nu \ll \mu$  then  $\mu$  and  $\nu$  are said to be equivalent and we write  $\mu \sim \nu$ . If  $\mu$  and  $\nu$  are concentrated on disjoint sets then they are called singular; in this case we write  $\mu \perp \nu$ .
Another notion that will be used throughout the paper is the Radon-Nikodym derivative.
Theorem 1. Let  $\mu$  and  $\nu$  be two measures on  $(H, \mathcal{B}(H))$  and  $\nu$  be  $\sigma$ -finite. If  $\mu \ll \nu$ , then there exists a  $\nu$ -measurable function  $f$  on  $H$  such that
$$
\mu (A ^ {\prime}) = \int_ {A ^ {\prime}} f d \nu , \quad \forall A ^ {\prime} \in \mathcal {B} (H).
$$
Furthermore,  $f$  is unique  $\nu$ -a.e. and is called the Radon-Nikodym derivative of  $\mu$  with respect to  $\nu$ . It is denoted by  $d\mu / d\nu$ .
Remark 4. In the paper, we will sometimes refer to  $f$  as the density of  $\mu$  with respect to  $\nu$ .
We are finally able to state the Feldman-Hajek theorem in its general form.
Theorem 2. The following statements hold.
1. Gaussian measures  $\mu = \mathcal{N}(m_1,C_1)$ ,  $\nu = \mathcal{N}(m_2,C_2)$  are either singular or equivalent.
2. They are equivalent if and only if the following conditions hold:
(i)  $\nu$  and  $\mu$  have the same Cameron-Martin space  $H_0 = C_1^{1/2}(H) = C_2^{1/2}(H)$ .
(ii)  $m_{1} - m_{2}\in H_{0}$
(iii) The operator  $(C_1^{-1/2} C_2^{1/2})(C_1^{-1/2} C_2^{1/2})^*$  -  $I$  is a Hilbert-Schmidt operator on the closure  $\overline{H_0}$ .
3. If  $\mu$  and  $\nu$  are equivalent and  $C_1 = C_2 = C$ , then  $\nu$ -a.s. the Radon-Nikodym derivative  $d\mu / d\nu$  is given by
$$
\frac {d \mu}{d \nu} (h) = e ^ {\Psi (h)},
$$
where  $\Psi (h) = \langle C^{-1 / 2}(m_1 - m_2),C^{-1 / 2}(h - m_2)\rangle -\frac{1}{2}\| C^{-1 / 2}(m_1 - m_2)\| _H^2\forall h\in H.$
A.3 Bayes' theorem for inverse problems
Let  $H$  and  $K$  be separable Hilbert spaces, equipped with the Borel  $\sigma$ -algebra, and  $A: H \to K$  a measurable mapping. We want to solve the inverse problem of finding  $X$  from  $Y$ , where
$$
Y = A (X) + B
$$
and  $B \in K$  denotes the noise. We adopt a Bayesian approach to this problem. We let  $(X, Y) \in H \times K$  be a random variable and compute  $X|Y$ . We first specify  $(X, Y)$  as follows:
1) Prior:  $X \sim \mu_0$  measure on  $H$ .
2) Noise:  $B \sim \eta_0$  measure on  $K$ , with  $B$  independent from  $X$ .
The random variable  $Y|X$  is then distributed according to the measure  $\eta_{x}$ , the translate of  $\eta_0$  by  $A(X)$ . We assume that  $\eta_{x} \ll \eta_{0}$ . Thus for some potential  $\Psi : H \times K \to \mathbb{R}$ ,
$$
\frac {d \eta_ {x}}{d \eta_ {0}} (y) = e ^ {- \Psi (x, y)}.
$$
The potential  $\Psi (\cdot ,y)$  satisfying the above formula is often termed the negative log likelihood of the problem. Now define  $\nu_{0}$  to be the product measure  $\nu_{0} = \mu_{0}\times \eta_{0}$ . We can finally state the following infinite-dimensional analogue of the Bayes' theorem.
Theorem 3. Assume that  $\Psi : H \times K \to \mathbb{R}$  is  $\nu_0$ -measurable and define
$$
Z (y) = \int e ^ {- \Psi (x, y)} d \mu_ {0}.
$$
$$
\frac {d \mu_ {0} (\cdot | Y = y)}{d \mu_ {0}} (x) = \frac {1}{Z (y)} e ^ {- \Psi (x, y)},
$$
where  $\mu_0(\cdot |Y = y)$  is the conditional distribution of  $X$  given  $Y = y$ .
B Proofs of Section 4
B.1 Proofs of Lemma 2 and Proposition 1
We assume that  $C_{\mu}$  in (9) and  $C$  in (1) have the same basis of eigenfunctions  $(v_j)$  and that  $C_{\mu} v_j = \mu_j v_j$ ,  $C v_j = \lambda_j v_j \forall j$ . We define  $X_t^{(j)} = \langle X_t, v_j \rangle$ ,  $y^{(j)} = \langle y, v_j \rangle$  and  $S^{(j)}(t, x, y) = \langle S(t, x, y), v_j \rangle$  so that in (1)  $S(t, x, y) = \sum_j S^{(j)}(t, x, y) v_j$ . We assume  $j \in \mathcal{I}^{(n)}$  so that we consider a mode corresponding to an observation. We then have
$$
d X _ {t} ^ {(j)} = - \frac {1}{2} X _ {t} ^ {(j)} d t + \sqrt {\lambda^ {(j)}} d W _ {t} ^ {(j)},
$$
with  $W^{(j)}$  standard Brownian motions which are independent for the different modes  $j$ . Note also that with  $C_{\mu}$  and  $C$  having the same basis of eigenfunctions the system of modes is diagonalized so that the  $X_{t}^{(j)}$  processes are independent with respect to mode  $j$ , both for the observed and un-observed modes. Thus we have
$$
X _ {0} ^ {(j)} = \sqrt {\mu_ {j}} \eta_ {0} ^ {(j)}, Y ^ {(j)} = X _ {0} ^ {(j)} + \sigma_ {B} \eta_ {1} ^ {(j)}, X _ {t} ^ {(j)} = X _ {0} ^ {(j)} e ^ {- t / 2} + \sqrt {\lambda_ {j} (1 - e ^ {- t})} \eta_ {2} ^ {(j)},
$$
for  $\eta_i^{(j)}$  independent standard Gaussian random variables. We then seek
$$
x _ {0} ^ {(j, y)} = \mathbb {E} [ X _ {0} ^ {(j)} \mid X _ {t} = x, Y = y ] = \mathbb {E} [ X _ {0} ^ {(j)} \mid X _ {t} ^ {(j)} = x ^ {(j)}, Y ^ {(j)} = y ^ {(j)} ],
$$
which in this Gaussian setting is the  $L^2$  projection of  $X_0^{(j)}$  onto  $X_{t}^{(j)}$  and  $Y^{(j)}$ . Thus we can write  $x_0^{(j,y)} = ax^{(j)} + by^{(j)}$  with  $(a,b)$  solving
$$
\mathbb {E} [ (a X _ {t} ^ {(j)} + b Y ^ {(j)} - X _ {0} ^ {(j)}) Y ^ {(j)} ] = 0, \quad \mathbb {E} [ (a X _ {t} ^ {(j)} + b Y ^ {(j)} - X _ {0} ^ {(j)}) X _ {t} ^ {(j)} ] = 0,
$$
$$
a = \frac {e ^ {t / 2}}{1 + (e ^ {t} - 1) p ^ {(j)} (1 + q ^ {(j)})}, \qquad b = \frac {p ^ {(j)} q ^ {(j)} (e ^ {t} - 1)}{1 + (e ^ {t} - 1) p ^ {(j)} (1 + q ^ {(j)})},
$$
for  $p^{(j)} = \lambda_j / \mu_j$ ,  $q^{(j)} = \mu_j / \sigma_B^2$ .
We then get in view of (8)
$$
S ^ {(j)} (t, x, y) = - \left(\frac {e ^ {t} p ^ {(j)} \left(1 + q ^ {(j)}\right)}{1 + \left(e ^ {t} - 1\right) p ^ {(j)} \left(1 + q ^ {(j)}\right)}\right) x ^ {(j)} + \left(\frac {e ^ {t / 2} p ^ {(j)} q ^ {(j)}}{1 + \left(e ^ {t} - 1\right) p ^ {(j)} \left(1 + q ^ {(j)}\right)}\right) y ^ {(j)}. \tag {31}
$$
Note that with some abuse of notation we then have
$$
S ^ {(j)} (t, x, y) = S ^ {(j)} (t, x ^ {(j)}, y ^ {(j)}),
$$
which is important since then also the time reversed system diagonalizes. We remark that for an unobserved mode we get by a similar, but easier, calculation
$$
S ^ {(j)} (t, x, y) = - \left(\frac {e ^ {t} p ^ {(j)}}{1 + (e ^ {t} - 1) p ^ {(j)}}\right) x ^ {(j)},
$$
which simply corresponds to setting  $\sigma_B = \infty$  in (31).
Consider next  $\mathbb{E}[S^{(j)}(t,x,y)^2 ]$  . Note first that
$$
\begin{array}{l} \mathbb {E} \left[ X _ {t} ^ {(j)} \mid Y = y \right] = \left(\frac {q ^ {(j)} e ^ {- t / 2}}{1 + q ^ {(j)}}\right) y ^ {(j)}, \\ \operatorname {V a r} \left[ X _ {t} ^ {(j)} \mid Y = y \right] = e ^ {- t} \operatorname {V a r} \left[ X _ {t} ^ {(j)} \mid Y = y \right] + \lambda_ {j} \left(1 - e ^ {- t}\right) \\ = \left(1 + \left(e ^ {t} - 1\right) p ^ {(j)} \left(1 + q ^ {(j)}\right)\right) \left(\frac {\mu_ {j} e ^ {- t}}{1 + q ^ {(j)}}\right). \\ \end{array}
$$
We can then easily check that the score is conditionally centered  $\mathbb{E}[S^{(j)}(t,X_t,Y)\mid Y = y] = 0$  and we then get
$$
\begin{array}{l} \mathbb {E} \left[ \left(S ^ {(j)} \left(t, X _ {t}, Y\right)\right) ^ {2} \mid Y = y \right] = \left(\frac {e ^ {t} p ^ {(j)} \left(1 + q ^ {(j)}\right)}{1 + \left(e ^ {t} - 1\right) p ^ {(j)} \left(1 + q ^ {(j)}\right)}\right) ^ {2} \operatorname {V a r} \left[ X _ {t} ^ {(j)} \mid Y = y \right] \\ = \frac {e ^ {t} \mu_ {j} (p ^ {(j)}) ^ {2} (1 + q ^ {(j)})}{1 + (e ^ {t} - 1) p ^ {(j)} (1 + q ^ {(j)})}, \\ \end{array}
$$
which gives Proposition 1 upon summing over the mode index  $j$ , where we define  $q^{(j)} = 0$  for the unobserved modes.
C Proofs of Section 5
C.1 Discussion about an alternative approach
The following lemma is a complementary result related to Remark 1. It shows that we can actually derive the expression of the score from the results contained in Millet et al. [38]. The result is powerful, but requires the verification of technical conditions.
Lemma 3. Under the conditions stated in Proposition 3 the score is defined by
$$
S (x, y, t) = - \left(1 - e ^ {- t}\right) ^ {- 1} \left(x - e ^ {- t / 2} \mathbb {E} [ X _ {0} | X _ {t} = x, Y = y ]\right) \tag {32}
$$
and the time reversed diffusion takes the form in (14).
$$
X ^ {(j)} = \left\langle v _ {j}, X \right\rangle \text {f o r} C v _ {j} = \lambda_ {j} v _ {j}.
$$
$$
d X ^ {(j)} = - \frac {1}{2} X ^ {(j)} d t + \sqrt {\lambda_ {j}} d W ^ {(j)} \text {f o r} W ^ {(j)} = \langle v _ {j}, W \rangle , \tag {33}
$$
and where we assume that  $C$  is of trace class. This is then an infinite dimensional system of the type considered in Millet et al. [38]. We proceed to verify some conditions stated in Millet et al. [38]: (i) the coefficients of the system (33) satisfy standard growth and Lipschitz continuity conditions (assumption  $(H1, H4)$  satisfied); (ii) the coefficients depend on finitely many coordinates (assumption  $(H2)$  satisfied); the system is time independent and diagonal (assumption  $(H5)$  satisfied). Moreover define  $\check{x}^{(j)} = (x_1, \ldots, x_{j-1}, x_{j+1}, \ldots)$ , then the law of  $X_t^{(j)}$  given  $\check{X}_t^{(j)}$  has for  $t > 0$  density  $p_t(x^{(j)}|\check{X}_t^{(j)} = \check{x}^{(j)}, Y = y)$  with respect to Lebesgue measure and so that for  $t_0 > 0$  and each  $j$ :  $\int_{t_0}^T \mathbb{E}[|\partial_{x^{(j)}}\log(p_t(x^{(j)})|Y = y)dt < \infty$ . Then it follows from Theorems 3.1 and 4.3 in Millet et al. [38] that the time reversed problem is associated with the well-posed martingale problem defined by the coefficients in (14) for the score being:
$$
\langle v _ {j}, S (x, y, t) \rangle = \frac {\lambda_ {j} \frac {\partial}{\partial x ^ {j}} \left(p _ {t} (x ^ {(j)} | \check {X} _ {t} ^ {(j)} = \check {x} ^ {(j)} , Y = y)\right)}{p _ {t} (x ^ {(j)} | \check {X} _ {t} ^ {(j)} = \check {x} ^ {(j)} , Y = y)},
$$
with the convention that the right hand side is null on the set  $\{p_t(x^{(j)}|\check{X}_t^{(j)} = \check{x}^{(j)},Y = y) = 0\}$ . It then follows for  $t > 0$
$$
\begin{array}{l} \langle v _ {j}, S (x, y, t) \rangle \\ = \int_ {\mathbb {R}} d \mu_ {0} (x _ {0} ^ {(j)} | \check {X} _ {t} ^ {(j)} = \check {x} ^ {(j)}, Y = y) \frac {\lambda_ {j} \frac {\partial}{\partial x ^ {j}} \left(p _ {t} (x ^ {(j)} | \check {X} _ {t} ^ {(j)} = \check {x} ^ {(j)} , X _ {0} ^ {(j)} = x _ {0} ^ {(j)} , Y = y)\right)}{p _ {t} (x ^ {(j)} | \check {X} _ {t} ^ {(j)} = \check {x} ^ {(j)} , Y = y)}. \\ \end{array}
$$
We then get
$$
\begin{array}{l} \langle v _ {j}, S (x, y, t) \rangle = - \int_ {\mathbb {R}} d \mu_ {0} \left(x _ {0} ^ {(j)} \mid \check {X} _ {t} ^ {(j)} = \check {x} ^ {(j)}, Y = y\right) \\ \times \left(\frac {x ^ {j} - e ^ {- t / 2} x _ {0} ^ {(j)}}{1 - e ^ {- t}}\right) \frac {\left(p _ {t} \left(x ^ {(j)} \mid \check {X} _ {t} ^ {(j)} = \check {x} ^ {(j)} , X _ {0} ^ {(j)} = x _ {0} ^ {(j)} , Y = y\right)\right)}{p _ {t} \left(x ^ {(j)} \mid \check {X} _ {t} ^ {(j)} = \check {x} ^ {(j)} , Y = y\right)} \\ = - \int_ {\mathbb {R}} d \mu_ {0} (x _ {0} ^ {(j)} | X _ {t} = x, Y = y) \left(\frac {x ^ {j} - e ^ {- t / 2} x _ {0} ^ {(j)}}{1 - e ^ {- t}}\right). \\ \end{array}
$$
C.2 A preliminary lemma
The following lemma is the equivalent of [24, Lemma 3]. It is used in the forthcoming proof of Proposition 2.
Lemma 4. In the finite-dimensional setting  $x \in \mathbb{R}^D$ , we have for any  $0 \leq s \leq t \leq T$ :
$$
\nabla \log p _ {t, y} (x _ {t}) = e ^ {(t - s) / 2} \mathbb {E} _ {y} \left[ \nabla \log p _ {s, y} (X _ {s}) | X _ {t} = x _ {t} \right],
$$
where  $\mathbb{E}_y$  is the expectation with respect to the distribution of  $X_0$  and  $W$  given  $Y = y$  and  $p_{t,y}$  is the pdf of  $X_t$  under this distribution.
Proof. We can write
$$
p _ {t, y} (x _ {t}) = \int_ {\mathbb {R} ^ {D}} p _ {s, y} (x _ {s}) p _ {t | s, y} (x _ {t} | x _ {s}) d x _ {s},
$$
where  $p_{t|s,y}(\cdot |x_s)$  is the pdf of  $X_t$  given  $Y = y$  and  $X_s = x_s$ . It is, in fact, equal to the pdf of  $X_t$  given  $X_s = x_s$ , which is the pdf of the multivariate Gaussian distribution with mean  $\exp(-(t - s)/2)x_s$  and covariance  $(1 - \exp(-(t - s)))C$ . Therefore
$$
p _ {t, y} (x _ {t}) = \int_ {\mathbb {R} ^ {D}} p _ {s, y} (x _ {s}) p _ {t | s} (x _ {t} | x _ {s}) d x _ {s}.
$$
We can then deduce that
$$
\begin{array}{l} \nabla p _ {t, y} (x _ {t}) = \frac {1}{(2 \pi) ^ {D / 2} (1 - e ^ {- (t - s)}) ^ {1 / 2} \det (C) ^ {1 / 2}} \int_ {\mathbb {R} ^ {D}} d x _ {s} p _ {s, y} (x _ {s}) \\ \times \nabla_ {x _ {t}} \exp \Big (- \frac {1}{2 (1 - e ^ {- (t - s)})} (x _ {t} - \exp (- (t - s) / 2) x _ {s}) ^ {T} C ^ {- 1} (x _ {t} - \exp (- (t - s) / 2) x _ {s}) \Big) \\ = - \frac {e ^ {(t - s) / 2}}{(2 \pi) ^ {D / 2} (1 - e ^ {- (t - s)}) ^ {1 / 2} \det (C) ^ {1 / 2}} \int_ {\mathbb {R} ^ {D}} d x _ {s} p _ {s, y} (x _ {s}) \\ \times \nabla_ {x _ {s}} \exp \Big (- \frac {1}{2 (1 - e ^ {- (t - s)})} (x _ {t} - \exp (- (t - s) / 2) x _ {s}) ^ {T} C ^ {- 1} (x _ {t} - \exp (- (t - s) / 2) x _ {s}) \Big) \\ = \frac {e ^ {(t - s) / 2}}{(2 \pi) ^ {D / 2} (1 - e ^ {- (t - s)}) ^ {1 / 2} \det  (C) ^ {1 / 2}} \int_ {\mathbb {R} ^ {D}} d x _ {s} \nabla_ {x _ {s}} \left(p _ {s, y} \left(x _ {s}\right)\right) \\ \times \exp \Big (- \frac {1}{2 (1 - e ^ {- (t - s)})} (x _ {t} - \exp (- (t - s) / 2) x _ {s}) ^ {T} C ^ {- 1} (x _ {t} - \exp (- (t - s) / 2) x _ {s}) \Big) \\ = e ^ {(t - s) / 2} \int_ {\mathbb {R} ^ {D}} d x _ {s} \nabla_ {x _ {s}} \big (p _ {s, y} (x _ {s}) \big) p _ {t | s} (x _ {t} | x _ {s}), \\ \end{array}
$$
$$
\nabla p _ {t, y} (x _ {t}) = e ^ {(t - s) / 2} \int_ {\mathbb {R} ^ {D}} p _ {t | s} (x _ {t} | x _ {s}) p _ {s, y} (x _ {s}) \nabla \log p _ {s, y} (x _ {s}) d x _ {s}.
$$
Using again that  $p_{t|s,y}(\cdot |x_s) = p_{t|s}(\cdot |x_s)$  and  $p_{t|s,y}(x_t|x_s) = \frac{p_{(s,t),y}(x_s,x_t)}{p_{s,y}(x_s)}$ , we get
$$
\begin{array}{l} \nabla p _ {t, y} (x _ {t}) = e ^ {(t - s) / 2} \int_ {\mathbb {R} ^ {D}} p _ {t | s, y} (x _ {t} | x _ {s}) p _ {s, y} (x _ {s}) \nabla \log p _ {s, y} (x _ {s}) d x _ {s} \\ = e ^ {(t - s) / 2} \int_ {\mathbb {R} ^ {D}} p _ {(s, t), y} (x _ {s}, x _ {t}) \nabla \log p _ {s, y} (x _ {s}) d x _ {s}. \\ \end{array}
$$
Since  $\nabla \log p_{t,y}(x_t) = \frac{\nabla p_{t,y}(x_t)}{p_{t,y}(x_t)}$  and  $p_{s|t,y}(x_s|x_t) = \frac{p_{(s,t),y}(x_s,x_t)}{p_{t,y}(x_t)}$  we get that
$$
\begin{array}{l} \nabla \log p _ {t, y} (x _ {t}) = e ^ {(t - s) / 2} \int_ {\mathbb {R} ^ {D}} p _ {s | t, y} (x _ {s} | x _ {t}) \nabla \log p _ {s, y} (x _ {s}) d x _ {s} \\ = e ^ {(t - s) / 2} \mathbb {E} _ {y} \big [ \nabla \log p _ {s, y} (X _ {s}) | X _ {t} = x _ {t} \big ]. \\ \end{array}
$$
C.3 Proof of Proposition 2
The proof adapts the one of [24] to the conditional setting. The only difference is that the expectation is  $\mathbb{E}_y$ , which affects the distribution of  $X_0$  but not the one of  $W$ . Moreover, Lemma 4 shows that the key to the proof (the reverse-time martingale property of the finite-dimensional score) is still valid. Here  $\mathbb{E}_y$  is the expectation with respect to the distribution of  $X_0$  and  $W$  given  $Y = y$ .
To prove Proposition 2, we are left to show that the solution of the reverse-time SDE
$$
d Z _ {t} = \frac {1}{2} Z _ {t} d t + S (T - t, Z _ {t}, y) d t + \sqrt {C} d W _ {t}, \quad Z _ {0} \sim X _ {T} | Y = y \tag {34}
$$
satisfies  $Z_{T} \sim X_{0}|Y = y$ . We recall that  $X_{t}$  is the solution to the SDE
$$
d X _ {t} = - \frac {1}{2} X _ {t} d t + \sqrt {C} d W _ {t}, \quad X _ {0} \sim \mu_ {0}.
$$
We first notice that  $X_{t}$  is given by the following stochastic convolution:
$$
X _ {t} = e ^ {- t / 2} X _ {0} + \int_ {0} ^ {t} e ^ {- (t - s) / 2} \sqrt {C} d W _ {s}.
$$
For  $P^D$  the orthogonal projection on the subspace of  $H$  spanned by  $v_1, \ldots, v_D$  (the eigenfunctions of  $C$ ),  $X_t^D = P^D(X_t)$  are solutions to
$$
d X _ {t} ^ {D} = - \frac {1}{2} X _ {t} ^ {D} d t + \sqrt {(C ^ {D})} d W _ {t} ^ {D},
$$
$$
C ^ {D} = P ^ {D} C P ^ {D}, \quad W _ {t} ^ {D} = P ^ {D} W _ {t}.
$$
We define  $X_{t}^{D:M} = X_{t}^{M} - X_{t}^{D}$ . Then
$$
X _ {t} ^ {D: M} = e ^ {- t / 2} X _ {0} ^ {D: M} + \int_ {0} ^ {t} e ^ {- (t - s) / 2} \sqrt {(C ^ {D : M})} d W _ {s} ^ {D: M},
$$
where the superscript  $D:M$  indicates the projection onto  $\operatorname {span}\{v_{D + 1},\ldots ,v_M\}$  . It holds that
$$
\mathbb {E} _ {y} \big [ \sup _ {t \leq T} \| X _ {t} ^ {D: M} \| _ {H} ^ {2} \big ] \leq 2 e ^ {- t} \mathbb {E} _ {y} [ \| X _ {0} ^ {D: \infty} \| _ {H} ^ {2} ] + 2 (1 - e ^ {- t}) \sum_ {i = D + 1} ^ {\infty} \lambda_ {i} \to 0
$$
as  $D\to \infty$  , where we used Doob's  $L^2$  inequality to bound the stochastic integral. Therefore  $(X_{t}^{N})$  is a Cauchy sequence and converges to  $X_{t}$  in  $L^2 (\mathbb{P}_y)$  . Consequently, the distribution of  $X_{t}^{N}$  given  $Y = y$  converges to the distribution of  $X_{t}$  given  $Y = y$  as  $N\rightarrow +\infty$
$$
S (t, X _ {t}, y) = - (1 - e ^ {- t}) ^ {- 1} \mathbb {E} _ {y} [ X _ {t} - e ^ {- t / 2} X _ {0} \mid X _ {t} ],
$$
and recall that
$$
C ^ {D} \nabla \log p _ {t, y} ^ {D} (X _ {t} ^ {D}) = - (1 - e ^ {- t}) ^ {- 1} P ^ {D} \mathbb {E} _ {y} [ X _ {t} - e ^ {- t / 2} X _ {0} \mid X _ {t} ^ {D} ].
$$
In particular, due to the tower property of the conditional expectations,
$$
C ^ {D} \nabla \log p _ {t, y} ^ {D} (X _ {t} ^ {D}) = \mathbb {E} _ {y} [ S (t, X _ {t}, y) \mid X _ {t} ^ {D} ].
$$
Since, by Assumption 1,
$$
\mathbb {E} _ {y} [ \| S (t, X _ {t}, y) \| _ {H} ^ {2} ] <   \infty ,
$$
the quantities  $\mathbb{E}_y[X_t - e^{-t / 2}X_0\mid X_t^D ]$  are bounded in  $L^2 (\mathbb{P}_y)$  and will converge to the limit,  $\mathbb{E}_y[S(t,X_t,y)\mid X_t] = S(t,X_t,y)$ , by the Martingale convergence theorem. We get rid of the projection  $P^{D}$  by
$$
\begin{array}{l} (1 - e ^ {- t}) \mathbb {E} _ {y} [ \| C ^ {D} \nabla \log p _ {t, y} ^ {D} (X _ {t} ^ {D}) - S (t, X _ {t}, y) \| _ {H} ^ {2} ] \\ = \mathbb {E} _ {y} [ \| P ^ {D} \mathbb {E} _ {y} [ X _ {t} - e ^ {- t / 2} X _ {0} \mid X _ {t} ^ {D} ] - \mathbb {E} _ {y} [ X _ {t} - e ^ {- t / 2} X _ {0} \mid X _ {t} ] \| _ {H} ^ {2} ] \\ \leq \mathbb {E} _ {y} [ \| \mathbb {E} _ {y} [ X _ {t} - e ^ {- t / 2} X _ {0} \mid X _ {t} ^ {D} ] - \mathbb {E} _ {y} [ X _ {t} - e ^ {- t / 2} X _ {0} \mid X _ {t} ] \| _ {H} ^ {2} ] \\ + \mathbb {E} _ {y} [ \| (I - P ^ {D}) \mathbb {E} _ {y} [ X _ {t} - e ^ {- t / 2} X _ {0} | X _ {t} ^ {D} ] \| _ {H} ^ {2} ] \\ \leq \mathbb {E} _ {y} [ \| \mathbb {E} [ X _ {t} - e ^ {- t / 2} X _ {0} \mid X _ {t} ^ {D} ] - \mathbb {E} _ {y} [ X _ {t} - e ^ {- t / 2} X _ {0} \mid X _ {t} ] \| _ {H} ^ {2} ] \\ + \mathbb {E} _ {y} [ \| (I - P ^ {D}) (X _ {t} - e ^ {- t / 2} X _ {0}) \| _ {H} ^ {2} ]. \\ \end{array}
$$
The first term vanishes due to our previous discussion. The second term vanishes since
$$
\begin{array}{l} \mathbb {E} _ {y} [ \| (I - P ^ {D}) (X _ {t} - e ^ {- t / 2} X _ {0}) \| _ {H} ^ {2} ] = \mathbb {E} _ {y} [ \| (I - P ^ {D}) \int_ {0} ^ {t} e ^ {- (t - s) / 2} \sqrt {C} d W _ {s} \| _ {H} ^ {2} ] \\ = \mathbb {E} [ \| (I - P ^ {D}) \int_ {0} ^ {t} e ^ {- (t - s) / 2} \sqrt {C} d W _ {s} \| _ {H} ^ {2} ] \\ = (1 - e ^ {- t}) \sum_ {i = D + 1} ^ {\infty} \lambda_ {j} \rightarrow 0 \\ \end{array}
$$
as  $D\to \infty$