SlowGuess/NeurIPS_2023_OCR huggingface dataset:
$$
f _ {t} ^ {\star} (x _ {t}) = \mathbb {E} _ {\Pi_ {0, T | t}} [ \bar {\alpha} _ {t} ^ {\prime} \mathbf {X} _ {0} + \bar {\beta} _ {t} ^ {\prime} \mathbf {X} _ {T} + \frac {\bar {\alpha} _ {t} ^ {\prime} \bar {\gamma} _ {t}}{\bar {\alpha} _ {t}} \mathbf {Z} \mid \mathbf {X} _ {t} = x _ {t} ].
$$
In Albergo et al. (2023), it is shown that  $\nabla \log \mathbb{M}_t^\star(x_t) = -\mathbb{E}_{\Pi_{0,T|t}}[\mathbf{Z} \mid \mathbf{X}_t = x_t] / \bar{\gamma}_t$ , where  $\mathbb{M}^\star$  is the Markovian projection. The probability flow associated with  $(\mathbf{X}_t^\star)_{t \in [0,T]}$  is given by
$$
\begin{array}{l} \mathrm {d} \mathbf {Z} _ {t} ^ {\star} = \left\{f _ {t} ^ {\star} \left(\mathbf {Z} _ {t} ^ {\star}\right) - \frac {\gamma_ {t} ^ {2}}{2} \nabla \log \mathbb {M} _ {t} ^ {\star} \left(\mathbf {Z} _ {t} ^ {\star}\right) \right\} \mathrm {d} t \\ = \left\{\mathbb {E} _ {\Pi_ {0, T | t}} \left[ \bar {\alpha} _ {t} ^ {\prime} \mathbf {X} _ {0} + \bar {\beta} _ {t} ^ {\prime} \mathbf {X} _ {T} + \left(- \alpha_ {t} \bar {\gamma} _ {t} + \frac {\gamma_ {t} ^ {2}}{2 \bar {\gamma} _ {t}}\right) \mathbf {Z} \mid \mathbf {X} _ {t} = \mathbf {Z} _ {t} ^ {\star} \right] \right\} d t \\ = \left\{\mathbb {E} _ {\Pi_ {0, T | t}} [ \bar {\alpha} _ {t} ^ {\prime} \mathbf {X} _ {0} + \bar {\beta} _ {t} ^ {\prime} \mathbf {X} _ {T} + \bar {\gamma} _ {t} ^ {\prime} \mathbf {Z} \mid \mathbf {X} _ {t} = \mathbf {Z} _ {t} ^ {\star} ] \right\} \mathrm {d} t. \\ \end{array}
$$
Hence, we recover (Albergo et al., 2023, Theorem 2.6).
Non-Markov path measures. A natural question is whether (19) arises as the bridge measure of some Markov measure. For instance, if  $\mathbb{Q}$  is associated with  $(x_0 + \mathbf{B}_t)_{t\in [0,T]}$ , then pinning the process at  $x_{T}$  at time  $T$ , we get that the associated bridge measure  $\mathbb{Q}_{|0,T}$  is given by
$$
\mathrm {d} \mathbf {X} _ {t} ^ {0, T} = (x _ {T} - \mathbf {X} _ {t}) / (T - t) \mathrm {d} t + \mathrm {d} \mathbf {B} _ {t}.
$$
Therefore, we recover (19) with  $\alpha_{t} = \beta_{t} = \frac{1}{T - t}$  and  $\gamma_{t} = 1$ . Using (20), we get that  $\bar{\alpha}_{t} = 1 - \frac{t}{T}$ ,  $\bar{\beta}_{t} = \frac{t}{T}$  and  $\bar{\gamma}_{t}^{2} = (T - t)t / T$ . We recover (4), upon noting that  $\mathbf{B}_t - \frac{t}{T}\mathbf{B}_T$  is Gaussian with zero mean and variance  $(T - t)t / T$ .
More generally, we consider a Markov measure  $\mathbb{Q}$  associated with  $(\mathbf{X}_t)_{t\in [0,T]}$  such that
$$
\mathrm {d} \mathbf {X} _ {t} = - a _ {t} \mathbf {X} _ {t} \mathrm {d} t + c _ {t} \mathrm {d} \mathbf {B} _ {t}, \quad \mathbf {X} _ {0} = x _ {0}.
$$
We now derive the associated bridge measure  $\mathbb{Q}_{|0,T}$ :
$$
\mathbf {X} _ {T} = \exp [ - \Lambda_ {T} + \Lambda_ {t} ] \mathbf {X} _ {t} + \int_ {t} ^ {T} c _ {s} \exp [ \Lambda_ {s} - \Lambda_ {T} ] \mathrm {d} \mathbf {B} _ {s},
$$
with  $\Lambda_t = \int_0^t a_s \, \mathrm{d}s$ . We have that
$$
\begin{array}{l} c _ {t} ^ {2} \nabla_ {x _ {t}} \log \mathbb {Q} _ {T | t} (x _ {T} | x _ {t}) = (c _ {t} ^ {2} \exp [ \Lambda_ {t} - \Lambda_ {T} ] / \int_ {t} ^ {T} c _ {s} ^ {2} \exp [ 2 (\Lambda_ {s} - \Lambda_ {T}) ] \mathrm {d} s) x _ {T} \\ - \left(c _ {t} ^ {2} \exp [ 2 (\Lambda_ {t} - \Lambda_ {T}) ] / \int_ {t} ^ {T} c _ {s} ^ {2} \exp [ 2 (\Lambda_ {s} - \Lambda_ {T}) ] \mathrm {d} s\right) x _ {t}. \\ \end{array}
$$
Therefore, combining this result and (2), we get that  $\mathbb{Q}_{|0,T}$  is associated with
$$
\alpha_ {t} = a _ {t} + c _ {t} ^ {2} \exp [ - 2 \int_ {t} ^ {T} a _ {s} \mathrm {d} s ] / \int_ {t} ^ {T} c _ {s} ^ {2} \exp [ - 2 \int_ {s} ^ {T} a _ {u} \mathrm {d} u ] \mathrm {d} s,
$$
$$
\beta_ {t} = c _ {t} ^ {2} \exp \left[ - \int_ {t} ^ {T} a _ {s} \mathrm {d} s \right] / \int_ {t} ^ {T} c _ {s} ^ {2} \exp \left[ - 2 \int_ {s} ^ {T} a _ {u} \mathrm {d} u \right] \mathrm {d} s, \quad \gamma_ {t} = c _ {t}.
$$
In that case  $(a_{t},c_{t})_{t\in [0,T]}$  entirely parameterize  $(\alpha_{t},\beta_{t},\gamma_{t})_{t\in [0,T]}$ . Hence, in the Ornstein-Uhlenbeck setting, if  $\mathbb{Q}_{|0,T}$  is the bridge of a Markov measure, it is fully parameterized by two functions while in the non-Markov setting it is parameterized by three functions.
In this paper, we present our framework in the Markovian setting as the Schrödinger Bridge problem is usually defined with respect to Markov reference measures. However, our methodology could be extended in a straightforward fashion to the non-Markovian setting. This would allow for a further exploration of the design space of DSBM.
B.2 Linear SDE and Bridge Matching
In this section, we study further the diffusion bridge of linear SDEs. Arbitrary Markov measures can be chosen to build bridges; however, we want to be able to compute some representations of the bridge in an explicit way. More precisely, denoting  $(\mathbf{X}_t^{0,T})_{t\in [0,T]}$  the diffusion bridge with  $x_0,x_T$  the initial and final condition, we want to have access to the following:
- integral sampler: we want to have a formula to sample  $\mathbf{X}_t^{0,T}$  for any  $t\in [0,T]$  without having to run a stochastic process forward or backward.
- forward sampler: we want to have a forward SDE for  $\mathbf{X}_t^{0,T}$  with explicit coefficients, which might depend on  $x_{T}$ , running in a forward fashion terminating at  $x_{T}$ .
- backward sampler: we want to have a backward SDE for  $\mathbf{Y}_t^{0,T} = \mathbf{X}_{T - t}^{0,T}$  with explicit coefficients, which might depend on  $x_0$ , running in a backward fashion terminating at  $x_0$ .
We focus on linear SDEs of the form  $\mathrm{d}\mathbf{X}_t = -\alpha \beta_t\mathbf{X}_t\mathrm{d}t + \sigma \beta_t^{1 / 2}\mathrm{d}\mathbf{B}_t$  , which are particularly amenable, where  $(\beta_{t})_{t\in [0,T]}$  is a schedule with  $\beta \in \mathrm{C}([0,T],(0, + \infty))$
B.2.1 Brownian motion
First, we consider the Brownian motion setting and let  $(\mathbf{X}_t)_{t\in [0,T]}$  be associated with  $\mathbb{Q}$  with  $\mathrm{d}\mathbf{X}_t = \beta_t^{1 / 2}\mathrm{d}\mathbf{B}_t$ . We consider  $(\mathbf{X}_t^{0,T})_{t\in [0,T]}$  conditioned at both ends  $\mathbf{X}_0^{0,T} = x_0$  and  $\mathbf{X}_T^{0,T} = x_T$ . First, using (Barczy and Kern, 2013, Theorem 3.3), we have that for any  $t\in [0,T]$
$$
\mathbf {X} _ {t} ^ {0, T} = \frac {R (t , T)}{R (0 , T)} x _ {0} + \frac {R (0 , t)}{R (0 , T)} (x _ {T} - \mathbf {X} _ {T}) + \mathbf {X} _ {t},
$$
with  $R(s,t) = \int_s^t\beta_u\mathrm{d}u = \sigma^2 (B_t - B_s)$ , where for any  $t\in [0,T]$ ,  $B_{t} = \int_{0}^{t}\beta_{s}\mathrm{d}s$ . Therefore, we get that
$$
\mathbf {X} _ {t} ^ {0, T} = \left(1 - \frac {B (t)}{B (T)}\right) x _ {0} + \frac {B (t)}{B (T)} \left(x _ {T} - \mathbf {X} _ {T}\right) + \mathbf {X} _ {t}. \tag {22}
$$
(22) defines the integral sampler. In addition, using (Barczy and Kern, 2013, Theorem 3.2), we have
$$
\mathbf {X} _ {0} ^ {0, T} = x _ {0}, \qquad \mathrm {d} \mathbf {X} _ {t} ^ {0, T} = \{- \sigma^ {2} \beta_ {t} / \gamma (t, T) \mathbf {X} _ {t} ^ {0, T} + \sigma^ {2} \beta_ {t} / \gamma (t, T) x _ {T} \} \mathrm {d} t + \sigma \beta_ {t} ^ {1 / 2} \mathrm {d} \mathbf {B} _ {t},
$$
where  $\gamma (s,t) = \sigma^2\big(B(t) - B(s)\big)$ . Therefore, we get that
$$
\mathbf {X} _ {0} ^ {0, T} = x _ {0}, \quad \mathrm {d} \mathbf {X} _ {t} ^ {0, T} = \left\{- \frac {\beta_ {t}}{B (T) - B (t)} \mathbf {X} _ {t} ^ {0, T} + \frac {\beta_ {t}}{B (T) - B (t)} x _ {T} \right\} \mathrm {d} t + \sigma \beta_ {t} ^ {1 / 2} \mathrm {d} \mathbf {B} _ {t}. \tag {23}
$$
(23) defines the forward sampler. Finally, we derive the backward sampler by considering the time-reversal of the forward unconditional process (initialized at  $x_0$ ). Following Haussmann and Pardoux (1986),
$$
\mathbf {Y} _ {0} ^ {0, T} = x _ {T}, \quad \mathrm {d} \mathbf {Y} _ {t} ^ {0, T} = \sigma^ {2} \beta_ {T - t} \nabla \log \mathbb {Q} _ {T - t | 0} \left(\mathbf {Y} _ {t} ^ {0, T} \mid x _ {0}\right) \mathrm {d} t + \sigma \beta_ {T - t} ^ {1 / 2} \mathrm {d} \mathbf {B} _ {t}. \tag {24}
$$
In addition, we have that
$$
\mathbf {X} _ {t} = x _ {0} + \sigma B (t) ^ {1 / 2} \varepsilon_ {t}, \qquad \varepsilon_ {t} \sim \mathrm {N} (0, \mathrm {I d}).
$$
Hence, we get that for any  $t\in [0,T]$  and  $x\in \mathbb{R}^d$
$$
\nabla \log \mathbb {Q} _ {t | 0} (x | x _ {0}) = - (x - x _ {0}) / (\sigma^ {2} B (t)).
$$
Combining this result and (24), we get
$$
\mathbf {Y} _ {0} ^ {0, T} = x _ {T}, \quad \mathrm {d} \mathbf {Y} _ {t} ^ {0, T} = \left\{- \frac {\beta_ {T - t}}{B (T - t)} \mathbf {Y} _ {t} ^ {0, T} + \frac {\beta_ {T - t}}{B (T - t)} x _ {0} \right\} \mathrm {d} t + \sigma \beta_ {T - t} ^ {1 / 2} \mathrm {d} \mathbf {B} _ {t}. \tag {25}
$$
Combining (22), (23) and (25), we get
$$
\mathbf {X} _ {t} ^ {0, T} = \lambda_ {t} x _ {0} + \varphi_ {t} (x _ {T} - \mathbf {X} _ {T}) + \mathbf {X} _ {t}.
$$
$$
\mathbf {X} _ {0} ^ {0, T} = x _ {0}, \qquad \mathrm {d} \mathbf {X} _ {t} ^ {0, T} = \left\{\kappa_ {t} ^ {f} \mathbf {X} _ {t} ^ {0, T} + \Psi_ {t} ^ {f} x _ {T} \right\} \mathrm {d} t + \sigma \beta_ {t} ^ {1 / 2} \mathrm {d} \mathbf {B} _ {t},
$$
$$
\mathbf {Y} _ {0} ^ {0, T} = x _ {T}, \qquad \mathrm {d} \mathbf {Y} _ {t} ^ {0, T} = \kappa_ {T - t} ^ {b} \mathbf {Y} _ {t} ^ {0, T} + \Psi_ {T - t} ^ {b} x _ {0} \} \mathrm {d} t + \sigma \beta_ {T - t} ^ {1 / 2} \mathrm {d} \mathbf {B} _ {t},
$$
$$
\lambda_ {t} = 1 - \frac {B (t)}{B (T)}, \qquad \varphi_ {t} = \frac {B (t)}{B (T)},
$$
$$
\kappa_ {t} ^ {f} = - \frac {\beta_ {t}}{B (T) - B (t)}, \qquad \Psi_ {t} ^ {f} = \frac {\beta_ {t}}{B (T) - B (t)},
$$
$$
\kappa_ {t} ^ {b} = - \frac {\beta_ {t}}{B (t)}, \qquad \Psi_ {t} ^ {b} = \frac {\beta_ {t}}{B (t)}.
$$
B.2.2 Ornstein-Uhlenbeck
Second, we consider the Ornstein-Uhlenbeck setting and let  $(\mathbf{X}_t)_{t\in [0,T]}$  with  $\mathrm{d}\mathbf{X}_t = -\alpha \beta_t\mathbf{X}_t\mathrm{d}t + \sigma \beta_t^{1 / 2}\mathrm{d}\mathbf{B}_t$  with  $\alpha \neq 0$ . We consider  $(\mathbf{X}_t^{0,T})_{t\in [0,T]}$ , the stochastic process  $(\mathbf{X}_t)_{t\in [0,T]}$  conditioned at both ends  $\mathbf{X}_0^{0,T} = x_0$  and  $\mathbf{X}_T^{0,T} = x_T$ . First, using (Barczy and Kern, 2013, Theorem 3.3), we have that for any  $t\in [0,T]$
$$
\mathbf {X} _ {t} ^ {0, T} = \frac {R (t , T)}{R (0 , T)} x _ {0} + \frac {R (0 , t)}{R (0 , T)} \left(x _ {T} - \mathbf {X} _ {T}\right) + \mathbf {X} _ {t},
$$
with  $R(s,t) = \exp [\alpha (B(t) - B(s))] \gamma (s,t)$ , with  $\gamma (s,t) = \int_s^t\sigma^2\beta (u)\exp [-2\alpha (B(t) - B(u))]\mathrm{d}u$ . In particular, we have
$$
\gamma (s, t) = \frac {\sigma^ {2}}{2 \alpha} (1 - \exp [ - 2 \alpha (B (t) - B (s)) ], \quad R (s, t) = \frac {\sigma^ {2}}{\alpha} \sinh (\alpha (B (t) - B (s))). \tag {26}
$$
Therefore, we get that
$$
\mathbf {X} _ {t} ^ {0, T} = \frac {\sinh (\alpha (B (T) - B (t))}{\sinh (\alpha B (T))} x _ {0} + \frac {\sinh (\alpha B (t))}{\sinh (\alpha B (T))} \left(x _ {T} - \mathbf {X} _ {T}\right) + \mathbf {X} _ {t}. \tag {27}
$$
(27) defines the integral sampler. In addition, using (Barczy and Kern, 2013, Theorem 3.2) and (26), we have  $\mathbf{X}_0^{0,T} = x_0$  and
$$
\begin{array}{l} \mathrm {d} \mathbf {X} _ {t} ^ {0, T} = \left\{- \alpha \beta_ {t} \mathbf {X} _ {t} - \frac {\sigma^ {2} \beta_ {t} \exp [ - 2 \alpha (B (T) - B (t)) ]}{\gamma (t , T)} \mathbf {X} _ {t} ^ {0, T} + \frac {\sigma^ {2} \beta_ {t} \exp [ - \alpha (B (T) - B (t)) ]}{\gamma (t , T)} x _ {T} \right\} \mathrm {d} t + \sigma \beta_ {t} ^ {1 / 2} \mathrm {d} \mathbf {B} _ {t} \\ = \left\{- \alpha \beta_ {t} \mathbf {X} _ {t} - \frac {2 \alpha \beta_ {t}}{\exp [ 2 \alpha (B (T) - B (t)) ] - 1} \mathbf {X} _ {t} ^ {0, T} + \frac {2 \alpha \beta_ {t}}{\exp [ - \alpha (B (T) - B (t)) ] - \exp [ - \alpha (B (T) - B (t)) ]} x _ {T} \right\} \mathrm {d} t + \sigma \beta_ {t} ^ {1 / 2} \mathrm {d} \mathbf {B} _ {t} \\ = \left\{- \alpha \beta_ {t} \frac {\exp [ 2 \alpha (B (T) - B (t)) ] + 1}{\exp [ 2 \alpha (B (T) - B (t)) ] - 1} \mathbf {X} _ {t} ^ {0, T} + \frac {2 \alpha \beta_ {t}}{\exp [ - \alpha (B (T) - B (t)) ] - \exp [ - \alpha (B (T) - B (t)) ]} x _ {T} \right\} \mathrm {d} t + \sigma \beta_ {t} ^ {1 / 2} \mathrm {d} \mathbf {B} _ {t} \\ = \{- \alpha \beta_ {t} \coth (\alpha (B (T) - B (t))) \mathbf {X} _ {t} ^ {0, T} + \alpha \beta_ {t} \operatorname {c s c h} (\alpha (B (T) - B (t))) x _ {T} \} \mathrm {d} t + \sigma \beta_ {t} ^ {1 / 2} \mathrm {d} \mathbf {B} _ {t}. \\ \end{array}
$$
In the formula, coth is the hyperbolic cotangent function defined as  $\coth(x) = \frac{1}{\tanh(x)} = \frac{\cosh(x)}{\sinh(x)}$  and  $\operatorname{csch}$  is the hyperbolic cosecant function defined as  $\operatorname{csch}(x) = \frac{1}{\sinh(x)}$ . Combining this result and (26), we get that
$$
\mathbf {X} _ {0} ^ {0, T} = x _ {0}, \mathrm {d} \mathbf {X} _ {t} ^ {0, T} = \left\{- \alpha \beta_ {t} \coth (\alpha (B (T) - B (t))) \mathbf {X} _ {t} ^ {0, T} + \alpha \beta_ {t} \operatorname {c s c h} (\alpha (B (T) - B (t))) x _ {T} \right\} \mathrm {d} t + \sigma \beta_ {t} ^ {1 / 2} \mathrm {d} \mathbf {B} _ {t}. \tag {28}
$$
(28) defines the forward sampler. Finally, we derive the backward sampler by considering the time-reversal of the forward unconditional process (initialized at  $x_0$ ). Following Haussmann and Pardoux (1986),
$$
\mathbf {Y} _ {0} ^ {0, T} = x _ {T}, \quad \mathrm {d} \mathbf {Y} _ {t} ^ {0, T} = \left\{\alpha \beta_ {T - t} \mathbf {Y} _ {t} ^ {0, T} + \sigma^ {2} \beta_ {T - t} \nabla \log \mathbb {Q} _ {T - t | 0} \left(\mathbf {Y} _ {t} ^ {0, T} | x _ {0}\right) \right\} \mathrm {d} t + \sigma \beta_ {T - t} ^ {1 / 2} \mathrm {d} \mathbf {B} _ {t}. \tag {29}
$$
$$
\mathbf {X} _ {t} = \exp [ - \alpha B (t) ] x _ {0} + \frac {\sigma}{\sqrt {2 \alpha}} \big (1 - \exp [ - 2 \alpha B (t) ] \big) ^ {1 / 2} \varepsilon_ {t}, \qquad \varepsilon_ {t} \sim \mathrm {N} (0, \mathrm {I d}).
$$
$$
\nabla \log \mathbb {Q} _ {t | 0} (x | x _ {0}) = - 2 \alpha (x - \exp [ - \alpha B (t) ] x _ {0}) / (\sigma^ {2} (1 - \exp [ - 2 \alpha B (t) ])).
$$
Combining this result and (29), we get  $\mathbf{Y}_0^{0,T} = x_T$  and
$$
\begin{array}{l} \mathrm {d} \mathbf {Y} _ {t} ^ {0, T} = \{\alpha \beta_ {T - t} \mathbf {Y} _ {t} ^ {0, T} - \frac {2 \alpha \beta_ {T - t}}{1 - \exp [ - \alpha B (T - t) ]} \mathbf {Y} _ {t} ^ {0, T} + \frac {2 \alpha \beta_ {T - t} \exp [ - \alpha B (T - t) ]}{1 - \exp [ - 2 \alpha B (T - t) ]} x _ {0} \} \mathrm {d} t + \sigma \beta_ {T - t} ^ {1 / 2} \mathrm {d} \mathbf {B} _ {t} \\ = \left\{- \alpha \beta_ {T - t} \frac {1 + \exp [ - 2 \alpha B (T - t) ]}{1 - \exp [ - 2 \alpha B (T - t) ]} \mathbf {Y} _ {t} ^ {0, T} + \frac {2 \alpha \beta_ {T - t}}{\exp [ \alpha B (T - t) ] - \exp [ - \alpha B (T - t) ]} x _ {0} \right\} \mathrm {d} t + \sigma \beta_ {T - t} ^ {1 / 2} \mathrm {d} \mathbf {B} _ {t} \\ = \{- \alpha \beta_ {T - t} \coth (\alpha B (T - t)) \mathbf {Y} _ {t} ^ {0, T} + \alpha \beta_ {T - t} \operatorname {c s c h} (\alpha B (T - t)) x _ {T} \} \mathrm {d} t + \sigma \beta_ {T - t} ^ {1 / 2} \mathrm {d} \mathbf {B} _ {t}. \\ \end{array}
$$
$$
\mathbf {Y} _ {0} ^ {0, T} = x _ {T}, \mathrm {d} \mathbf {Y} _ {t} ^ {0, T} = \left\{- \alpha \beta_ {T - t} \coth (\alpha B (T - t)) \mathbf {Y} _ {t} ^ {0, T} + \alpha \beta_ {T - t} \operatorname {c s c h} (\alpha B (T - t)) x _ {T} \right\} \mathrm {d} t + \sigma \beta_ {T - t} ^ {1 / 2} \mathrm {d} \mathbf {B} _ {t}.
$$
Combining (27), (28) and (30), we get
$$
\begin{array}{l} \mathbf {X} _ {t} ^ {0, T} = \lambda_ {t} x _ {0} + \varphi_ {t} \left(x _ {T} - \mathbf {X} _ {T}\right) + \mathbf {X} _ {t}. \\ \mathbf {X} _ {0} ^ {0, T} = x _ {0}, \qquad \mathrm {d} \mathbf {X} _ {t} ^ {0, T} = \left\{\kappa_ {t} ^ {f} \mathbf {X} _ {t} ^ {0, T} + \Psi_ {t} ^ {f} x _ {T} \right\} \mathrm {d} t + \sigma \beta_ {t} ^ {1 / 2} \mathrm {d} \mathbf {B} _ {t}, \\ \mathbf {Y} _ {0} ^ {0, T} = x _ {T}, \qquad \mathrm {d} \mathbf {Y} _ {t} ^ {0, T} = \left\{\kappa_ {T - t} ^ {b} \mathbf {Y} _ {t} ^ {0, T} + \Psi_ {T - t} ^ {b} x _ {0} \right\} \mathrm {d} t + \sigma \beta_ {T - t} ^ {1 / 2} \mathrm {d} \mathbf {B} _ {t}, \\ \end{array}
$$
$$
\begin{array}{l} \lambda_ {t} = \frac {\sinh (\alpha (B (T) - B (t))}{\sinh (\alpha B (T))}, \qquad \varphi_ {t} = \frac {\sinh (\alpha B (t))}{\sinh (\alpha B (T))}, \\ \kappa_ {t} ^ {f} = - \alpha \beta_ {t} \coth (\alpha (B (T) - B (t))), \quad \Psi_ {t} ^ {f} = \alpha \beta_ {t} \operatorname {c s c h} (\alpha (B (T) - B (t))), \\ \kappa_ {t} ^ {b} = - \alpha \beta_ {t} \coth (\alpha B (t)), \quad \Psi_ {t} ^ {b} = \alpha \beta_ {t} \operatorname {c s c h} (\alpha B (t)). \\ \end{array}
$$
Using that  $\tanh (x)\sim x$  and  $\sinh (x)\sim x$  for  $x\to 0$ , we recover the Brownian motion setting by letting  $\alpha \rightarrow 0$ . Note that Albergo et al. (2023) show that given an integral sampler, which does not necessarily comes from a Markovian process, a forward sampler with the same marginals can be defined, although it does not necessarily satisfy the fact that the paths have the same distribution.
We refer the reader to Chung and Walsh (2006); Rogers and Williams (2000) for an introduction to Doob  $h$ -transform. Our theoretical treatment of the Doob  $h$ -transform closely follows Palmowski and Rolski (2002).
First, we introduce the infinitesimal generator  $\mathcal{A}$  given for any  $f\in C_c^\infty ([0,T]\times \mathbb{R}^d,\mathbb{R}),t\in [0,T]$  and  $x\in \mathbb{R}^d$  by
$$
\mathcal {A} f (t, x) = \left\langle f _ {t} (x), \nabla f (t, x) \right\rangle + \frac {\sigma_ {t} ^ {2}}{2} \Delta f (t, x) + \partial_ {t} f (t, x). \tag {31}
$$
The following assumption ensures that the diffusion associated with  $\mathbb{Q}$  as well as its Markovian projections are well-defined.
A1.  $f, \sigma$  and  $(t,x_{t})\mapsto \mathbb{E}_{\Pi_{T|t}}[\nabla \log \mathbb{Q}_{T|t}(\mathbf{X}_{T}|\mathbf{X}_{t})\mid \mathbf{X}_{t} = x_{t}]$  are locally Lipschitz and there exist  $C > 0$ ,  $\psi \in \mathrm{C}([0,T],\mathbb{R}_+)$  such that for any  $t\in [0,T]$  and  $x_0,x_t\in \mathbb{R}^d$ , we have
$$
\begin{array}{l} \left\| f _ {t} \left(x _ {t}\right) \right\| \leq C \left(1 + \left\| x _ {t} \right\|\right), \quad C \geq \sigma_ {t} \geq 1 / C, \\ \left\| \mathbb {E} _ {\Pi_ {T | t}} [ \nabla \log \mathbb {Q} _ {T | t} (\mathbf {X} _ {T} | \mathbf {X} _ {t}) \mid \mathbf {X} _ {t} = x _ {t} ] \right\| \leq C \psi (t) (1 + \| x _ {t} \)). \\ \end{array}
$$
We consider the following assumption, which will ensure that we can apply Doob  $h$ -transform techniques.
A2. For any  $x_0 \in \mathbb{R}^d$ ,  $\Pi_{T|0}$  is absolutely continuous w.r.t.  $\mathbb{Q}_{T|0}$ . For any  $x_0 \in \mathbb{R}^d$ , let  $\varphi_{T|0}$  be given for any  $x_T \in \mathbb{R}^d$  by  $\varphi_{T|0}(x_T|x_0) = \mathrm{d}\Pi_{T|0}(x_T|x_0) / \mathrm{d}\mathbb{Q}_{T|0}(x_T|x_0)$  and assume that for any  $x_0 \in \mathbb{R}^d$ ,  $x_T \mapsto \varphi_{T|0}(x_T|x_0)$  is bounded. For any  $x_0 \in \mathbb{R}^d$ , let  $\varphi_{t|0}$  given for any  $x_t \in \mathbb{R}^d$  and  $t \in [0,T]$  by
$$
\varphi_ {t \mid 0} \left(x _ {t} \mid x _ {0}\right) = \int_ {\mathbb {R} ^ {d}} \varphi_ {T \mid 0} \left(x _ {T} \mid x _ {0}\right) \mathrm {d} \mathbb {Q} _ {T \mid t} \left(x _ {T} \mid x _ {t}\right). \tag {32}
$$
Finally, we assume that for any  $x_0 \in \mathbb{R}^d$ ,  $(t,x_t) \mapsto 1 / \varphi_{t|0}(x_t|x_0)$  and  $(t,x_t) \mapsto \mathcal{A}\varphi_{t|0}(x_t|x_0)$  are bounded.
This means that for any  $x_0 \in \mathbb{R}^d$ ,  $(t, x_t) \mapsto \varphi_t(x_t | x_0)$  is a good function in the sense of (Palmowski and Rolski, 2002, Proposition 3.2). Note here that these assumptions could be relaxed on a case-by-case basis. We leave this study for future work.
The following lemma is a direct consequence of A2 and (32). It ensures that the  $h$ -function  $\varphi_{t|0}$  satisfies the backward Kolmogorov equation.
Lemma 11. Assume A2. Then,  $\varphi \in \mathrm{C}^{1,2}([0,T)\times \mathbb{R}^d,\mathbb{R})$  and  $\mathcal{A}\varphi_{|0} = 0$ .
Using (31), we have that for any  $x_0 \in \mathbb{R}^d$  and  $f \in \mathrm{C}_c^\infty([0, T] \times \mathbb{R}^d, \mathbb{R})$ ,  $t \in [0, T]$  and  $x_t \in \mathbb{R}^d$
$$
\left. \right.\left(\mathcal {A} \left(f \varphi_ {| 0}\right) - f \mathcal {A} \varphi_ {| 0}\right)\left(t, x _ {t}\right) / \varphi_ {| 0} (t, x _ {t}) = \mathcal {A} f \left(t, x _ {t}\right) + \sigma_ {t} ^ {2} \left\langle \nabla f (t, x _ {t}), \nabla \log \varphi_ {t | 0} \left(x _ {t} \mid x _ {0}\right)\right\rangle .
$$
Finally, we consider the following assumption, which will ensure that the Doob  $h$ -transform is well-defined.
A3. For any  $x_0 \in \mathbb{R}^d$ , there exists  $C \geq 0$  such that for any  $t \in [0, T]$  and  $x_t \in \mathbb{R}^d$ ,  $\| \nabla \log \varphi_{t|0}(x_t | x_0) \| \leq C (1 + \| x_0 \| + \| x_t \|)$ .
We are now ready to state and prove Proposition 2. Note that the Markovian projection is defined in Definition 1. Finally, we define  $\mathcal{M}$  the space of path measures such that  $\mathbb{P} \in \mathcal{M}$  if  $\mathbb{P}$  is associated with  $\mathrm{d}\mathbf{X}_t = \{f_t(\mathbf{X}_t) + v_t(\mathbf{X}_t)\} \mathrm{d}t + \sigma_t \mathrm{d}\mathbf{B}_t$ , with  $\sigma, v$  locally Lipschitz. This restriction of Markov measures allows us to apply the entropic version of the Girsanov theorem (Léonard, 2012). It has no impact on our methodology.
Proposition. Assume A1, A2, A3. Let  $\mathbb{M}^{\star} = \mathrm{proj}_{\mathcal{M}}(\Pi)$ . Then,
$$
\mathbb {M} ^ {\star} = \operatorname {a r g m i n} _ {\mathbb {M}} \{\mathrm {K L} (\Pi | \mathbb {M}): \mathbb {M} \in \mathcal {M} \},
$$
$$
\mathrm {K L} (\Pi | \mathbb {M} ^ {\star}) = \frac {1}{2} \int_ {0} ^ {T} \mathbb {E} _ {\Pi_ {0, t}} [ \| \sigma_ {t} ^ {2} \mathbb {E} _ {\Pi_ {T | 0, t}} [ \nabla \log \mathbb {Q} _ {T | t} (\mathbf {X} _ {T} | \mathbf {X} _ {t}) \mid \mathbf {X} _ {0}, \mathbf {X} _ {t} ] - v _ {t} ^ {\star} \| ^ {2} ] / \sigma_ {t} ^ {2} \mathrm {d} t.
$$
In addition, we have that for any  $t \in [0,T]$ ,  $\mathbb{M}_t^\star = \Pi_t$ . In particular,  $\mathbb{M}_T^\star = \Pi_T$ .
Proof. First, we recall that  $\Pi$  is given by  $\Pi = \mathbb{Q}\varphi_{0,T}$  with  $\varphi_{0,T} = \frac{\mathrm{d}\Pi_{0,T}}{\mathrm{d}\mathbb{Q}_{0,T}}$ . In particular, we have  $\Pi_{|0} = \mathbb{Q}_{|0}\varphi_{T|0}$ , where  $\varphi_{T|0} = \frac{\mathrm{d}\Pi_{T|0}}{\mathrm{d}\mathbb{Q}_{T|0}}$ . Therefore, using Lemma 11, (Palmowski and Rolski, 2002, Lemma 3.1, Lemma 4.1), the remark following (Palmowski and Rolski, 2002, Lemma 4.1), A1, A2 and A3, we get that  $\Pi_{|0}$  is Markov and associated with the distribution of  $(\mathbf{X}_t)_{t\in [0,T]}$  given for any  $t\in [0,T]$  by
$$
\mathbf {X} _ {t} = \int_ {0} ^ {t} \left\{f _ {s} \left(\mathbf {X} _ {s}\right) + \sigma_ {s} ^ {2} \nabla \log \varphi_ {s | 0} \left(\mathbf {X} _ {s} \mid \mathbf {X} _ {0}\right) \right\} d s + \int_ {0} ^ {t} \sigma_ {s} d \mathbf {B} _ {s}, \tag {33}
$$
where for any  $t\in [0,T],x_0,x_t\in \mathbb{R}^d$  we recall that
$$
\varphi_ {t \mid 0} (x _ {t} \mid x _ {0}) = \int_ {\mathbb {R} ^ {d}} \varphi_ {T \mid 0} (x _ {T} \mid x _ {0}) \mathrm {d} \mathbb {Q} _ {T \mid t} (x _ {T} \mid x _ {t}). \tag {34}
$$
First, we have that for any  $t\in [0,T],x_t,x_0\in \mathbb{R}^d$
$$
\mathbb {Q} _ {t | 0} (x _ {t} | x _ {0}) \varphi_ {t | 0} (x _ {t} | x _ {0}) = \int_ {\mathbb {R} ^ {d}} \mathbb {Q} _ {t | 0, T} (x _ {t} | x _ {T}, x _ {0}) \mathrm {d} \Pi_ {T | 0} (x _ {T} | x _ {0}) = \Pi_ {t | 0} (x _ {t} | x _ {0}).
$$
Therefore, we get that for any  $t \in [0,T]$  and  $x_{t}, x_{0} \in \mathbb{R}^{d}$
$$
\varphi_ {t | 0} \left(x _ {t} \mid x _ {0}\right) = \frac {\mathrm {d} \Pi_ {t | 0} \left(x _ {t} \mid x _ {0}\right)}{\mathrm {d} \mathbb {Q} _ {t | 0} \left(x _ {t} \mid x _ {0}\right)}. \tag {35}
$$
In addition, we have the following identity for any  $t \in [0,T]$ ,  $x_0,x_t,x_T \in \mathbb{R}^d$
$$
\mathbb {Q} _ {T \mid 0} (x _ {T} \mid x _ {0}) \mathbb {Q} _ {t \mid 0, T} (x _ {t} \mid x _ {0}, x _ {T}) = \mathbb {Q} _ {t \mid 0} (x _ {t} \mid x _ {0}) \mathbb {Q} _ {T \mid t} (x _ {T} \mid x _ {t}).
$$
Using (34), this result and (35), we get that for any  $t \in [0, T]$  and  $x_0, x_t \in \mathbb{R}^d$