SlowGuess/NeurIPS_2023_OCR huggingface dataset:
$$
\begin{array}{l} \int_ {\mathbb {R} ^ {f d} \times \mathbb {R} ^ {f d}} \| x _ {0} - x _ {1} \| ^ {2} \mathrm {d} \mathbb {P} (x _ {0}, x _ {1}) - \varepsilon \mathrm {K L} (\mathbb {P} | \pi_ {0} ^ {\mathrm {u p}} \otimes \pi_ {1} ^ {\mathrm {u p}}) \\ = f ^ {2} \left(\int_ {\mathbb {R} ^ {d _ {d}} \times \mathbb {R} ^ {d _ {d}}} \| \bar {x} _ {0} - \bar {x} _ {1} \| ^ {2} \mathrm {d} \left(\operatorname {d o w n} _ {\#} \mathbb {P}\right) (\bar {x} _ {0}, \bar {x} _ {1}) - \left(\varepsilon / f ^ {2}\right) \mathrm {K L} \left(\operatorname {d o w n} _ {\#} \mathbb {P} | \pi_ {0} \otimes \pi_ {1}\right)\right). \\ \end{array}
$$
Therefore, we have the following result.
Proposition 12. Let  $\varepsilon >0$ .  $\mathbb{P}^{\star}$  is the solution of the static Schrödinger bridge with marginals  $\pi_0^{\mathrm{up}}$ ,  $\pi_1^{\mathrm{up}}$  and regularization  $\varepsilon$  if and only if  $\mathrm{down}_{\#}\mathbb{P}^{\star}$  is the solution of the static Schrödinger bridge with marginals  $\pi_0$ ,  $\pi_1$  and regularization  $\varepsilon /f^2$ .
This means in particular that the Schrödinger Bridge is not invariant via upsampling. In Appendix I.4, we confirm these results visually upon noting that for the same  $\sigma$ , running DSBM at resolution  $64 \times 64$  and  $128 \times 128$  gives different results, even after downsampling of the  $128 \times 128$  results.
D Convergence of IMF in the Gaussian setting
In this section, we study the IMF in an one-dimensional Gaussian case. We consider  $T = 1$ ,  $\Pi_0 = \Pi_1 = \mathrm{N}(0, (1/2\beta^2))$  and  $\mathbb{Q}$  associated with  $(\sigma \mathbf{B}_t)_{t \in [0,1]}$  where  $\sigma > 0$ . In what follows, we let
$$
\Sigma^ {0} = \left(1 / 2 \beta^ {2}\right) \left( \begin{array}{c c} 1 & c ^ {2} \\ c ^ {2} & 1 \end{array} \right), \tag {39}
$$
where  $c \in [0,1]$ . We also denote  $\bar{\sigma}^2 = 2\sigma^2\beta^2$ . We start with the following result which gives an explicit expression of some marginals of the reciprocal projection.
Lemma 13. Let  $\Pi_{0,1}^{0} = \mathrm{N}(0,\Sigma^{0})$  with  $\Sigma^0$  given by (39). Let  $\Pi^0 = \Pi_{0,1}^0\mathbb{Q}_{|0,1}$ . For any  $t\in [0,1]$ , we have that  $\Pi_{0,1,t}^{0} = \mathrm{N}(0,\Sigma)$  with
$$
\Sigma = (1 / 2 \beta^ {2}) \left( \begin{array}{c c c} 1 & c ^ {2} & a _ {k} ^ {t} \\ c ^ {2} & 1 & a _ {k} ^ {1 - t} \\ a _ {k} ^ {t} & a _ {k} ^ {1 - t} & b _ {k} ^ {t} \end{array} \right),
$$
where we have
$$
a _ {k} ^ {t} = 1 - t + t c ^ {2}, \quad b _ {k} ^ {t} = 1 + t (1 - t) (2 (c ^ {2} - 1) + \bar {\sigma} ^ {2}).
$$
Proof. Let  $t \in [0,1]$  and  $\mathbf{X}_t^{0,1} \sim \Pi_{t|0,1}^0 = \mathbb{Q}_{t|0,1}$ . Using (Barczy and Kern, 2013, Theorem 3.3), we get that
$$
\mathbf {X} _ {t} ^ {0, 1} = (1 - t) \mathbf {X} _ {0} + t \mathbf {X} _ {1} + \sigma (t (1 - t)) ^ {1 / 2} \mathbf {Z},
$$
with  $\mathbf{Z} \sim \mathrm{N}(0, \mathrm{Id})$  independent from  $(\mathbf{X}_0, \mathbf{X}_1)$ . Hence, we get that  $\mathbb{E}[\mathbf{X}_t^{0,1}] = 0$ , and
$$
\operatorname {C o v} \left(\mathbf {X} _ {0}, \mathbf {X} _ {t} ^ {0, 1}\right) = \mathbb {E} \left[ \mathbf {X} _ {t} ^ {0, 1} \mathbf {X} _ {0} \right] = (1 - t) / (2 \beta^ {2}) + t c ^ {2} / (2 \beta^ {2}) = a _ {k} ^ {t} / (2 \beta^ {2}).
$$
Similarly, we get that  $\mathrm{Cov}(\mathbf{X}_1,\mathbf{X}_t^{0,1}) = a_k^{1 - t} / (2\beta^2)$ . Finally, we get that
$$
\begin{array}{l} \mathrm {V a r} (\mathbf {X} _ {t} ^ {0, 1}) = \mathbb {E} [ ((1 - t) \mathbf {X} _ {0} + t \mathbf {X} _ {1}) ^ {2} ] + \sigma^ {2} t (1 - t) \\ = (1 - t) ^ {2} / (2 \beta^ {2}) + 2 t (1 - t) c ^ {2} / (2 \beta^ {2}) + t ^ {2} / (2 \beta^ {2}) + \bar {\sigma} ^ {2} t (1 - t) / (2 \beta^ {2}) \\ = (1 - 2 t + 2 t ^ {2} + 2 t (1 - t) c ^ {2} + \bar {\sigma} ^ {2} t (1 - t)) / (2 \beta^ {2}) \\ = (1 + t (1 - t) (2 (c ^ {2} - 1) + \bar {\sigma} ^ {2})) / (2 \beta^ {2}), \\ \end{array}
$$
Leveraging Lemma 13, we can give an explicit expression of the drift term in the Markovian projection.
Lemma 14. Let  $\Pi_{0,1}^{0} = \mathrm{N}(0,\Sigma^{0})$  with  $\Sigma^0$  given by (39). Let  $\Pi^0 = \Pi_{0,1}^0\mathbb{Q}_{|0,1}$ . For any  $t\in [0,1]$  and  $x_{t}\in \mathbb{R}^{d}$ , we have that
$$
\sigma^ {2} \mathbb {E} _ {\Pi_ {1 | t} ^ {0}} [ \nabla \log \mathbb {Q} _ {1 | t} | \mathbf {X} _ {t} = x _ {t} ] = \frac {(1 - 2 t) (c ^ {2} - 1) - \bar {\sigma} ^ {2} t}{1 + t (1 - t) (2 (c ^ {2} - 1) + \bar {\sigma} ^ {2})} x _ {t}.
$$
Hence, the Markovian projection of  $\Pi^0$ , denoted  $\mathbb{M}^1$  is associated with  $(\mathbf{X}_t)_{t\in [0,1]}$  with
$$
\mathbf {X} _ {0} \sim \Pi_ {0}, \quad \mathrm {d} \mathbf {X} _ {t} = \frac {(1 - 2 t) \left(c ^ {2} - 1\right) - \bar {\sigma} ^ {2} t}{1 + t (1 - t) \left(2 \left(c ^ {2} - 1\right) + \bar {\sigma} ^ {2}\right)} \mathbf {X} _ {t} + \sigma \mathrm {d} \mathbf {B} _ {t}. \tag {40}
$$
Proof. Using (Barczy and Kern, 2013, Theorem 3.2), we get that  $\mathbb{Q}_{|0,1}$  is associated with
$$
\mathrm {d} \mathbf {X} _ {t} ^ {0, 1} = \sigma^ {2} \nabla \log \mathbb {Q} _ {1 | t} (x _ {1} | \mathbf {X} _ {t}) \mathrm {d} t + \sigma \mathrm {d} \mathbf {B} _ {t}, \qquad \mathbf {X} _ {0} ^ {0, 1} = x _ {0},
$$
where for any  $t \in [0,1)$ , we have and  $x_{t} \in \mathbb{R}^{d}$ ,  $\sigma^2 \nabla \log \mathbb{Q}_{1|t}(x_1|x_t) = (x_1 - x_t) / (1 - t)$ . Therefore, we get that  $\mathbb{M}^1$  is associated with  $(\mathbf{X}_t)_{t \in [0,1]}$  such that
$$
\mathrm {d} \mathbf {X} _ {t} = \mathbb {E} _ {\Pi_ {1 | t} ^ {0}} [ \sigma^ {2} \nabla \log \mathbb {Q} _ {1 | t} (\mathbf {X} _ {1} | \mathbf {X} _ {t}) | \mathbf {X} _ {t} ] \mathrm {d} t + \sigma \mathrm {d} \mathbf {B} _ {t}, \qquad \mathbf {X} _ {0} = x _ {0}.
$$
$$
\mathrm {d} \mathbf {X} _ {t} = (\mathbb {E} _ {\Pi_ {1 | t} ^ {0}} [ \mathbf {X} _ {1} | \mathbf {X} _ {t} ] - x _ {t}) / (1 - t) \mathrm {d} t + \sigma \mathrm {d} \mathbf {B} _ {t}, \qquad \mathbf {X} _ {0} = x _ {0}.
$$
Using Lemma 13, we have that for any  $t \in [0,1]$  and  $x_{t} \in \mathbb{R}^{d}$ ,  $\mathbb{E}_{\Pi_{1|t}^{0}}[\mathbf{X}_{1}|\mathbf{X}_{t} = x_{t}] = a_{k}^{1 - t} / b_{k}^{t}x_{t}$ . In addition, we have for any  $t \in [0,1]$
$$
\begin{array}{l} a _ {k} ^ {1 - t} / b _ {k} ^ {t} - 1 = (t + (1 - t) c ^ {2} - 1 - t (1 - t) (2 (c ^ {2} - 1) + \bar {\sigma} ^ {2})) / (1 + t (1 - t) (2 (c ^ {2} - 1) + \bar {\sigma} ^ {2})) \\ = \left((1 - t) \left(c ^ {2} - 1\right) - t (1 - t) \left(2 \left(c ^ {2} - 1\right) + \bar {\sigma} ^ {2}\right)\right) / \left(1 + t (1 - t) \left(2 \left(c ^ {2} - 1\right) + \bar {\sigma} ^ {2}\right)\right) \\ = (1 - t) \left(\left(c ^ {2} - 1\right) - t \left(2 \left(c ^ {2} - 1\right) + \bar {\sigma} ^ {2}\right)\right) / \left(1 + t (1 - t) \left(2 \left(c ^ {2} - 1\right) + \bar {\sigma} ^ {2}\right)\right) \\ = (1 - t) \left(\left(1 - 2 t\right)\left(c ^ {2} - 1\right) - t \bar {\sigma} ^ {2}\right)\left. \right) / \left(1 + t (1 - t) \left(2 \left(c ^ {2} - 1\right) + \bar {\sigma} ^ {2}\right)\right), \\ \end{array}
$$
Note that since  $\sigma > 0$  and  $c^2 \in [0,1]$ , we get that for any  $t \in [0,1]$ ,  $1 + t(1 - t)(2(c^2 - 1) + \bar{\sigma}^2) > 0$  and therefore the drift is well-defined, smooth and sublinear. In particular, (40) admits a unique strong solution. In what follows, we denote  $G: [0,1] \times [0,1] \to \mathbb{R}$  given for any  $t \in [0,1]$  by
$$
G (t, c ^ {2}) = \int_ {0} ^ {t} \frac {(1 - 2 s) (c ^ {2} - 1) - \bar {\sigma} ^ {2} s}{1 + s (1 - s) (2 (c ^ {2} - 1) + \bar {\sigma} ^ {2})} \mathrm {d} s.
$$
We have the following useful lemma.
Lemma 15. Let  $c \in [0,1]$ ,  $\bar{\sigma} > 0$  and  $p = 2c^2 + \bar{\sigma}^2$ . We distinguish three cases:
(a) If  $p < 2$ , then
$$
G (1, c ^ {2}) = - \bar {\sigma} ^ {2} (4 - p ^ {2}) ^ {- 1 / 2} \tan^ {- 1} ((4 - p ^ {2}) ^ {1 / 2} / p).
$$
(b) If  $p = 2$ , then
$$
G (1, c ^ {2}) = - \bar {\sigma} ^ {2} / 2.
$$
(c) If  $p > 2$ , then
$$
G (1, c ^ {2}) = - \bar {\sigma} ^ {2} (p ^ {2} - 4) ^ {- 1 / 2} \tanh ^ {- 1} ((p ^ {2} - 4) ^ {1 / 2} / p).
$$
This lemma is useful combined with the following proposition, which gives the analytical update formula of Gaussian IMF covariances as a function of  $G(1,c^2)$ .
Proposition 16. Let  $\Pi_{0,1}^{0} = \mathbb{N}(0,\Sigma^{0})$  with  $\Sigma^0$  given by (39). Then  $\Pi_{0,1}^{1} = \mathbb{N}(0,\Sigma^{1})$  with
$$
\Sigma^ {1} = \left(1 / 2 \beta^ {2}\right) \left( \begin{array}{c c} 1 & c _ {1} ^ {2} \\ c _ {1} ^ {2} & 1 \end{array} \right), \qquad c _ {1} ^ {2} = f (c _ {0} ^ {2}),
$$
with  $f:[0,1]\to [0,1]$  given for any  $c\in [0,1]$  by
$$
f (c ^ {2}) = \exp [ G (1, c ^ {2}) ].
$$
Proof. We have that  $\mathbf{X}_1 = \exp [G(1,c^2)]\mathbf{X}_0 + \mathbf{M}_1$ , where  $\mathbf{M}_1$  is a Gaussian random variable with zero mean independent from  $\mathbf{X}_0$ . Therefore, we get that  $\mathrm{Cov}(\mathbf{X}_0,\mathbf{X}_1) = \exp [G(1,c^2)] / (2\beta^2)$ . In addition, we have that  $\mathbb{E}[\mathbf{X}_1^2 ] = \mathbb{E}[\mathbf{X}_0^2 ] = 1 / 2\beta^2$ , which concludes the proof.
Iterating the procedure in Proposition 16, we obtain a sequence of IMF covariances  $(c_n^2)_{n\in \mathbb{N}}$  satisfying  $c_{n + 1}^2 = f(c_n^2)$ . Finally, we show that this iterative procedure recovers the true SB coupling  $\Pi_{0,1}^{\mathrm{SB}} = \mathrm{N}(0,\Sigma^{\mathrm{SB}})$  as a fixed point. The formula of  $\Sigma^{\mathrm{SB}}$  is given e.g. in (Bunne et al., 2023, Equation (2)) which we use below.
Proposition 17. Let  $\Pi_{0,1}^{\mathrm{SB}} = \mathrm{N}(0,\Sigma^{\mathrm{SB}})$  be the true static SB solution, with
$$
\Sigma^ {\mathrm {S B}} = (1 / 2 \beta^ {2}) \left( \begin{array}{c c} 1 & c _ {\mathrm {S B}} ^ {2} \\ c _ {\mathrm {S B}} ^ {2} & 1 \end{array} \right), \qquad c _ {\mathrm {S B}} ^ {2} = \frac {1}{2} (\sqrt {4 + \bar {\sigma} ^ {4}} - \bar {\sigma} ^ {2}).
$$
Then  $\Pi_{0,1}^{\mathrm{SB}}$  is a fixed point of the iterative procedure in Proposition 16, i.e.  $f(c_{\mathrm{SB}}^{2}) = c_{\mathrm{SB}}^{2}$ .
Proof. By straightforward calculations,  $p_{\mathrm{SB}} = 2c_{\mathrm{SB}}^2 + \bar{\sigma}^2 = \sqrt{4 + \bar{\sigma}^4}$ . If  $\bar{\sigma} = 0$ ,  $p_{\mathrm{SB}} = 2$  and thus  $G(1, c_{\mathrm{SB}}^2) = -\bar{\sigma}^2 / 2 = 0$ . Hence  $f(c_{\mathrm{SB}}^2) = \exp(G(1, c_{\mathrm{SB}}^2)) = 1 = c_{\mathrm{SB}}^2$ . If  $\bar{\sigma} > 0$ , we have  $p_{\mathrm{SB}} > 2$  and  $G(1, c_{\mathrm{SB}}^2) = -\tanh^{-1}(\bar{\sigma}^2 / \sqrt{4 + \bar{\sigma}^4})$ . Hence,  $f(c_{\mathrm{SB}}^2) = \exp(G(1, c_{\mathrm{SB}}^2)) = \frac{1}{2} (\sqrt{4 + \bar{\sigma}^4} - \bar{\sigma}^2) = c_{\mathrm{SB}}^2$ . We thus correctly recover the true static SB solution  $\Pi_{0,1}^{\mathrm{SB}}$  as fixed point of IMF.
We visualize the convergence of this fixed point procedure for a variety of parameter settings in Figure 12. The convergence appears to be very fast in only two or three iterations.
(a)  $\bar{\sigma} = 0.1, c_0^2 = 0$
(b)  $\bar{\sigma} = 1, c_0^2 = 0$
(c)  $\bar{\sigma} = 1, c_0^2 = 1$
Figure 12: Convergence of IMF in the analytic case given by Proposition 16.
(d)  $\bar{\sigma} = 5, c_0^2 = 1$
E Discrete-Time Markovian Projection
We derive in this section a discrete-time version of the Markovian projection and show that, in some limiting case, we recover the continuous-time projection. In the discrete case, we let
$$
\begin{array}{l} \pi (x _ {0: N}) = \pi (x _ {0}, x _ {N}) \prod_ {k = 0} ^ {N - 1} q _ {k + 1 | 0, k, N} (x _ {k + 1} | x _ {0}, x _ {k}, x _ {N}) \\ = \pi (x _ {0}, x _ {N}) \prod_ {k = 0} ^ {N - 2} q _ {k + 1 | k, N} (x _ {k + 1} | x _ {k}, x _ {N}). \\ \end{array}
$$
We consider a Markovian measure  $p$  given by  $p(x_{0:N}) = p(x_0)\prod_{k = 0}^{N - 1}p_{k + 1|k}(x_{k + 1}|x_k)$ . Now let us compute  $\mathrm{KL}(\pi |p)$ . We have
$$
\begin{array}{l} \mathrm {K L} (\pi (x _ {0: N}) | p (x _ {0: N})) = \sum_ {k = 0} ^ {N - 2} \int_ {\mathbb {R} ^ {d} \times \mathbb {R} ^ {d}} \mathrm {K L} (q _ {k + 1 | k, N} | p _ {k + 1 | k}) \pi_ {k, N} (x _ {k}, x _ {N}) \mathrm {d} x _ {k} \mathrm {d} x _ {N} \\ + \mathrm {K L} (\pi_ {0} | p _ {0}) + \int_ {\mathbb {R} ^ {d}} \mathrm {K L} (\pi (x _ {N} | x _ {0}) | p (x _ {N} | x _ {N - 1})) \pi (x _ {0}, x _ {N - 1}) \mathrm {d} x _ {0} \mathrm {d} x _ {N - 1}. \\ \end{array}
$$
In what follows, we denote
$$
\begin{array}{l} \mathcal {L} _ {0} = \mathrm {K L} (\pi_ {0} | p _ {0}), \mathcal {L} _ {N} = \int_ {\mathbb {R} ^ {d}} \mathrm {K L} (\pi (x _ {N} | x _ {0}) | p (x _ {N} | x _ {N - 1})) \pi (x _ {0}, x _ {N - 1}) \mathrm {d} x _ {0} \mathrm {d} x _ {N - 1}, \\ \mathcal {L} _ {k + 1} = \int_ {\mathbb {R} ^ {d} \times \mathbb {R} ^ {d}} \mathrm {K L} (q _ {k + 1 | k, N} | p _ {k + 1 | k}) \pi_ {k, N} (x _ {k}, x _ {N}) \mathrm {d} x _ {k} \mathrm {d} x _ {N}, \\ \end{array}
$$
We have the following proposition.
Proposition 18. The minimizer  $p_{k + 1|k}$  of  $\mathcal{L}_{k + 1}$  is given by
$$
p _ {k + 1 \mid k} (x _ {k + 1} \mid x _ {k}) = \int_ {\mathbb {R} ^ {d}} q _ {k + 1 \mid k, N} (x _ {k + 1} \mid x _ {k}, x _ {N}) \pi_ {N \mid k} (x _ {N} \mid x _ {k}) \mathrm {d} x _ {N}. \tag {41}
$$
If  $p_0 = q_0$ , then for any  $k \in \{0, \dots, N - 1\}$ ,  $p_k = \pi_k$ . In addition, assume that  $p_{k + 1|k}(x_{k + 1}|x_k) = \exp[-\| x_{k + 1} - x_k - \gamma f(x_k)\|^2/(2\gamma)]/(2\pi\gamma)^{-d/2}$  and  $q_{k + 1|k,N}(x_{k + 1}|x_k,x_N) = \exp[-\| x_{k + 1} - x_k - \gamma f(x_k,x_N)\|^2/(2\gamma)]/(2\pi\gamma)^{d/2}$ . Finally, assume that  $\| x_{k + 1} - x_k\| \leq \gamma^{1/2}$ . Then, we have that
$$
f \left(x _ {k}\right) = \int_ {\mathbb {R} ^ {d}} f \left(x _ {k}, x _ {N}\right) \pi \left(x _ {N} \mid x _ {k}\right) \mathrm {d} x _ {N} + o \left(\gamma^ {1 / 2}\right). \tag {42}
$$
Proof. The proofs of (41) and (42) are straightforward and left to the reader. We now prove that if  $p_0 = \pi_0$ , then for any  $k \in \{1, \dots, N\}$ ,  $p_k = \pi_k$ . First, we have that for any  $k \in \{0, \dots, N - 1\}$ ,
$$
\pi (x _ {k}, x _ {k + 1}, x _ {N}) = \pi (x _ {k}, x _ {N}) q (x _ {k + 1} | x _ {k}, x _ {N}).
$$
Assume now that  $p_k = \pi_k$ , then we have
$$
\begin{array}{l} p _ {k + 1} \left(x _ {k + 1}\right) = \int_ {\mathbb {R} ^ {d}} p _ {k} \left(x _ {k}\right) p _ {k + 1 \mid k} \left(x _ {k + 1} \mid x _ {k}\right) \mathrm {d} x _ {k} \\ = \int_ {\mathbb {R} ^ {d} \times \mathbb {R} ^ {d}} p _ {k} (x _ {k}) q _ {k + 1 | k, N} (x _ {k + 1} | x _ {k}, x _ {N}) \pi_ {N | k} (x _ {N} | x _ {k}) \mathrm {d} x _ {k} \mathrm {d} x _ {N} \\ = \int_ {\mathbb {R} ^ {d} \times \mathbb {R} ^ {d}} \pi_ {k} (x _ {k}) q _ {k + 1 | k, N} (x _ {k + 1} | x _ {k}, x _ {N}) \pi_ {N | k} (x _ {N} | x _ {k}) \mathrm {d} x _ {k} \mathrm {d} x _ {N} \\ = \int_ {\mathbb {R} ^ {d} \times \mathbb {R} ^ {d}} \pi_ {k, k + 1, N} \left(x _ {k}, x _ {k + 1}, x _ {N}\right) \mathrm {d} x _ {k} \mathrm {d} x _ {N} = \pi_ {k + 1} \left(x _ {k + 1}\right), \\ \end{array}
$$
In particular, in the previous proposition, if  $f(x_{k},x_{N}) = \nabla \log q(x_{N}|x_{k})$ , i.e. we have a discretization of the bridge then  $f(x_{k}) = \int_{\mathbb{R}^{d}}\nabla \log q(x_{N}|x_{K})\pi (x_{N}|x_{k})\mathrm{d}x_{N}$ , which recovers the Markovian projection in continuous-time.
F Comparing DSBM-IPF and DSB
We analyze further the differences between DSBM-IPF proposed here and DSB proposed in De Bortoli et al. (2021) and related algorithms in Vargas et al. (2021); Chen et al. (2022). All algorithms solve the SB problem using the IPF iterates. However, DSB-type algorithms solve for the IPF iterates using time-reversals, whereas DSBM solves for the iterates using Markovian and reciprocal projections. A comparison between these two methodologies is made in Section 4.
We investigate here further benefit (iii) of DSBM in Section 4, i.e. the benefit of explicitly projecting onto the reciprocal class of  $\mathbb{Q}$ . Intuitively speaking, we directly incorporate the reference measure  $\mathbb{Q}$  in the training procedure as our inductive bias. More formally, suppose we have at the current
IPF iteration  $\mathbb{M}^{2n}$  (Markov diffusion in the forward direction) and want to learn  $\mathbb{M}^{2n + 1}$  (Markov diffusion in the backward direction). Due to training error and the forgetting issue (Fernandes et al., 2021), however,  $\mathbb{M}^{2n}$  no longer has the correct bridge  $\mathbb{Q}_{|0,T}$ . Now suppose we first perform IMF for  $\mathbb{M}^{2n}$  and learn  $\mathbb{M}^{2n,\star}$  in the forward direction. That is to say, we repeat alternative reciprocal and Markovian projections and obtain a sequence  $(\mathbb{M}^{2n,m})_{m\in \mathbb{N}}$  in the forward direction converging to  $\mathbb{M}^{2n,\star}$ . Then  $\mathbb{M}^{2n,\star}$  now has the correct bridge  $\mathbb{Q}_{|0,T}$  by Proposition 7, since  $\mathbb{M}^{2n,\star}$  is the SB between  $\mathbb{M}_0^{2n}$  and  $\mathbb{M}_T^{2n}$ . Theoretically,  $\mathbb{M}_t^{2n} = \mathbb{M}_t^{2n,\star}$  for  $t = 0,T$ , but due to training error accumulating it may be that  $\mathbb{M}_T^{2n}\neq \mathbb{M}_T^{2n,\star}$ . However,  $\mathbb{M}_0^{2n} = \mathbb{M}_0^{2n,\star}$ , since  $\mathbb{M}^{2n,\star}$  is in the forward direction and starting from samples from  $\pi_0$ . As a result, we can obtain a Markov forward diffusion  $\mathbb{M}^{2n,\star}$ , which has  $\mathbb{M}_0^{2n,\star} = \pi_0$  and the correct bridge  $\mathbb{Q}_{|0,T}$ . These are the same set of properties that the reference measure  $\mathbb{Q}$  has. As a result, replacing  $\mathbb{Q}$  with  $\mathbb{M}^{2n,\star}$  in (6) results in the same SB solution. Consequently, now continuing the IPF iterations from  $\mathbb{M}_0^{2n,\star}$ , it is as if we restart IPF afresh using a modified SB problem
$$
\mathbb {P} ^ {\mathrm {S B}} = \operatorname * {a r g m i n} \{\mathrm {K L} (\mathbb {P} | \mathbb {M} ^ {2 n, \star}): \mathbb {P} _ {0} = \pi_ {0}, \mathbb {P} _ {T} = \pi_ {T} \}.
$$
If  $\mathbb{M}^{2n,\star}$  is closer than  $\mathbb{Q}$  to  $\mathbb{M}^{\star}$  in the sense of KL divergence, then we obtain a better initialization of the IPF procedure. As proposed in Algorithm 1, DSBM performs the Markovian and reciprocal projection only once before switching between the forward and backward directions. However, it is still beneficial compared to DSB with less bias accumulation in the bridge.
Alge. rically, one main difference between DSB-type algorithms and DSBM due to the above distinction occurs in the trajectory caching step. In DSB, a fixed discretization of SDE needs to be chosen, and all intermediate samples from the discretized Euler-Maruyama simulation of the SDE need to be saved. Furthermore, a second set of drift evaluation needs to be performed for all datapoints in the trajectory (De Bortoli et al., 2021, Equations (12), (13)). The IPML algorithm in Vargas et al. (2021) is also similar to DSB, but Gaussian processes are used to fit the drifts of forward and backward SDEs instead of neural networks. In Chen et al. (2022), the implicit score matching loss is used instead, but all intermediate points in the SDE trajectory also need to be saved. On the contrary, DSBM does not require intermediate samples during trajectory caching and only retains the joint samples at times  $0,T$ . Then, the intermediate trajectories are reconstructed using the reference bridge  $\mathbb{Q}_{|0,T}$ .
G Joint Training of Forward and Backward Processes
We recall below the DSBM algorithm given in Algorithm 1.
Algorithm 2 Diffusion Schrödinger Bridge Matching
1: Input: Joint distribution  $\Pi_{0,T}^{0}$ , tractable bridge  $\mathbb{Q}_{|0,T}$ , number of outer iterations  $N \in \mathbb{N}$ .  
2: Let  $\Pi^{0} = \Pi_{0,T}^{0}\mathbb{Q}_{|0,T}$ .  
3: for  $n \in \{0, \dots, N-1\}$  do  
4: Learn  $v_{\phi^*}$  using (14) with  $\Pi = \Pi^{2n}$ .  
5: Let  $\mathbb{M}^{2n+1}$  be given by (13).  
6: Let  $\Pi^{2n+1} = \mathbb{M}_{0,T}^{2n+1}\mathbb{Q}_{|0,T}$ .  
7: Learn  $v_{\theta^*}$  using (10) with  $\Pi = \Pi^{2n+1}$ .  
8: Let  $\mathbb{M}^{2n+2}$  be given by (9).  
9: Let  $\Pi^{2n+2} = \mathbb{M}_{0,T}^{2n+2}\mathbb{Q}_{|0,T}$ .  
10: end for  
11: Output:  $v_{\theta^*}, v_{\phi^*}$
Our main observation comes from Proposition 9. In particular, under mild assumptions, we have that the Markovian projection  $\mathbb{M} = \mathrm{proj}_{\mathcal{M}}(\Pi)$  is associated with
$$
\mathrm {d} \mathbf {X} _ {t} = \left\{f _ {t} \left(\mathbf {X} _ {t}\right) + \sigma_ {t} ^ {2} \mathbb {E} _ {\Pi_ {T | t}} [ \nabla \log \mathbb {Q} _ {T | t} (\mathbf {X} _ {T} | \mathbf {X} _ {t}) \mid \mathbf {X} _ {t} ] \right\} \mathrm {d} t + \sigma_ {t} \mathrm {d} \mathbf {B} _ {t}, \quad \mathbf {X} _ {0} \sim \pi_ {0},
$$
$$
\mathrm {d} \mathbf {Y} _ {t} = \left\{- f _ {T - t} (\mathbf {Y} _ {t}) + \sigma_ {T - t} ^ {2} \mathbb {E} _ {\Pi_ {0 | T - t}} [ \nabla \log \mathbb {Q} _ {T - t | 0} (\mathbf {Y} _ {t} | \mathbf {Y} _ {T}) | \mathbf {Y} _ {t} ] \right\} \mathrm {d} t + \sigma_ {T - t} \mathrm {d} \mathbf {B} _ {t}, \quad \mathbf {Y} _ {0} \sim \pi_ {T}.
$$
Considering the following losses,
$$
\theta^ {\star} = \operatorname {a r g m i n} _ {\theta} \left\{\int_ {0} ^ {T} \mathbb {E} _ {\Pi_ {t, T}} \left[ \left\| \sigma_ {t} ^ {2} \nabla \log \mathbb {Q} _ {T | t} (\mathbf {X} _ {T} | \mathbf {X} _ {t}) - v _ {\theta} (t, \mathbf {X} _ {t}) \right\| ^ {2} \right] / \sigma_ {t} ^ {2} \mathrm {d} t: \theta \in \Theta \right\}, \tag {43}
$$
$$
\phi^ {\star} = \operatorname {a r g m i n} _ {\phi} \left\{\int_ {0} ^ {T} \mathbb {E} _ {\Pi_ {t, 0}} \left[ \left\| \sigma_ {t} ^ {2} \nabla \log \mathbb {Q} _ {t | 0} \left(\mathbf {X} _ {t} \mid \mathbf {X} _ {0}\right) - v _ {\phi} (t, \mathbf {X} _ {t}) \right\| ^ {2} \right] / \sigma_ {t} ^ {2} \mathrm {d} t: \phi \in \Phi \right\}. \tag {44}
$$
If the families of functions  $\{v_{\theta} : \theta \in \Theta\}$  and  $\{v_{\phi} : \theta \in \Phi\}$  are rich enough, we have for any  $t \in [0, T]$  and  $x_t \in \mathbb{R}^d$ ,  $v_{\theta^{\star}}(t, x_t) = \sigma_t^2 \mathbb{E}_{\Pi_{T|t}}[\nabla \log \mathbb{Q}_{T|t}(\mathbf{X}_T | \mathbf{X}_t) \mid \mathbf{X}_t = x_t]$  and  $v_{\phi^{\star}}(t, x_t) = \sigma_t^2 \mathbb{E}_{\Pi_{0|t}}[\nabla \log \mathbb{Q}_{t|0}(\mathbf{X}_t | \mathbf{X}_0) \mid \mathbf{X}_t = x_t]$ . In practice, this means that the Markovian projection can be computed in a forward or backward fashion equivalently.
Therefore, given a coupling  $\Pi = \Pi^{2n}$ , we can update both  $v_{\theta}$  and  $v_{\phi}$ . This means that we train the forward and backward model jointly. We then consider  $\mathbb{M}_b^{2n + 1}$  associated with (13) and  $\mathbb{M}_f^{2n + 1}$  associated with (9). Note that if the families of functions  $\{v_{\theta} : \theta \in \Theta\}$  and  $\{v_{\phi} : \theta \in \Phi\}$  are rich enough then  $\mathbb{M}_f^{2n + 1} = \mathbb{M}_b^{2n + 1}$ .
Mixture from forward and backward. Once we have obtained both the forward update and the backward update, our next task is to define the new mixture of bridge  $\Pi^{2n+1}$ . In Algorithm 1, since we train only the backward model  $\mathbb{M}^{2n+1} = \mathbb{M}_b^{2n+1}$ , we define  $\Pi^{2n+1} = \mathbb{M}_{0,T}^{2n+1}\mathbb{Q}_{|0,T}$ . In the case of joint training, we have access to  $\mathbb{M}_b^{2n+1}$  and  $\mathbb{M}_f^{2n+1}$ . One way to define a new mixture of bridge is to compute  $\Pi^{2n+1} = \frac{1}{2} (\mathbb{M}_{b,0,T}^{2n+1}\mathbb{Q}_{|0,T} + \mathbb{M}_{f,0,T}^{2n+1})\mathbb{Q}_{|0,T}$ . This choice ensures that in the case where  $\mathbb{M}_f^{2n+1} = \mathbb{M}_b^{2n+1}$  we have
$$
\Pi^ {2 n + 1} = \mathbb {M} _ {f, 0, T} ^ {2 n + 1} \mathbb {Q} _ {| 0, T} = \mathbb {M} _ {b, 0, T} ^ {2 n + 1} \mathbb {Q} _ {| 0, T}.
$$
It also ensures that all the steps in the joint DSBM training algorithms are symmetric. We leave the study of an optimal combination of  $\mathbb{M}_f^{2n + 1}$  and  $\mathbb{M}_b^{2n + 1}$  for future work.