SlowGuess/NeurIPS_2023_OCR huggingface dataset:
$$
X _ {r, 2} \rightarrow \left(X _ {1}, X _ {2}, X _ {r, 1}\right)\rightarrow X _ {3}. \tag {115}
$$
for some deterministic functions  $\eta_1(.)$  and  $\eta_2(.)$ .
Comparing the two regions  $\bar{\mathcal{C}}_{\mathrm{RDP}}$  and  $\mathcal{C}_{\mathrm{RDP}}$ , we identify the following differences. The Markov chain in (74) is more restricted comparing to (115). Moreover, the Markov chain (75) does not exist in  $\bar{\mathcal{C}}_{\mathrm{RDP}}$ . The following lemma states that  $\bar{\mathcal{C}}_{\mathrm{RDP}} = \mathcal{C}_{\mathrm{RDP}}$ . Now, for a given  $(\mathrm{D},\mathrm{P})$ , let  $\bar{\mathcal{R}} (\mathrm{D},\mathrm{P})$  denote the set of rate tuples  $\mathsf{R}$  such  $(\mathsf{R},\mathsf{D},\mathsf{P})\in \bar{\mathcal{C}}_{\mathrm{RDP}}$ , then this lemma implies that  $\bar{\mathcal{R}} (\mathrm{D},\mathrm{P}) = \mathcal{R}(\mathrm{D},\mathrm{P})$  which completes the proof of the outer bound. Moreover, notice that the above proof only deals with the statistics of the representations and reconstructions and does not depend on the choice of the PLF. So, it holds for both PLF-FMD and PLF-JD. This concludes the proof.
We conclude this section by the following lemma.
Lemma 1 For first-order Markov sources, we have
$$
\mathcal {C} _ {\mathrm {R D P}} = \bar {\mathcal {C}} _ {\mathrm {R D P}}. \tag {116}
$$
Proof: This result for the scenario without perception constraint has been similarly observed in [36, Eq. (12)]. The proof in this section is provided for completeness.
First, notice that the set of Markov chains in (74)-(76) is more restricted than the ones in (114)-(115), hence  $\mathcal{C}_{\mathrm{RDP}} \subseteq \bar{\mathcal{C}}_{\mathrm{RDP}}$ . Now, it remains to prove that  $\bar{\mathcal{C}}_{\mathrm{RDP}} \subseteq \mathcal{C}_{\mathrm{RDP}}$ . Consider the following facts
1. The distortion constraints in (111) depend only on the joint distribution of  $(X_{j},\hat{X}_{j})$ , and thus on the joint distribution of  $(X_{j},X_{r,1},\ldots ,X_{r,j})$ . So, imposing the Markov chain  $X_{r,2}\rightarrow (X_2,X_{r,1})\rightarrow X_1$  does not affect the expected distortion  $\mathbb{E}[\| X_2 - \hat{X}_2\|^2 ]$  since it does not depend on the joint distribution of  $X_{1}$  with  $(X_{r,1},X_{r,2},X_{2})$ . A similar argument holds for other frames;
2. The perception constraints in (112) depend on the joint distributions  $P_{X_1\dots X_j}$  and  $P_{\hat{X}_1,\dots ,\hat{X}_j}$  (hence on  $P_{X_{r,1}\dots X_{r,j}}$ ). Thus, imposing  $X_{r,2}\rightarrow (X_2,X_{r,1})\rightarrow X_1$  does not affect  $\phi_2(P_{X_1X_2},P_{\hat{X}_1\hat{X}_2})$  since it does not depend on the joint distribution of  $X_{1}$  with  $(X_{r,1},X_{r,2},X_2)$ . A similar argument holds for other frames;
3. Moreover, the rate constraints in (109) and (110) would be further lower bounded by
$$
R _ {2} \geq I \left(X _ {1}, X _ {2}; X _ {r, 2} \mid X _ {r, 1}\right) \geq I \left(X _ {2}; X _ {r, 2} \mid X _ {r, 1}\right), \tag {117}
$$
$$
R _ {3} \geq I \left(X _ {1}, X _ {2}, X _ {3}; \hat {X} _ {3} \mid X _ {r, 1}, X _ {r, 2}\right) \geq I \left(X _ {3}; \hat {X} _ {3} \mid X _ {r, 1}, X _ {r, 2}\right). \tag {118}
$$
Thus, the set of rate constraints is optimized by the set of Markov chains (74)-(76).
4. The mutual information terms  $I(X_{1};X_{r,1})$ ,  $I(X_{2};X_{r,2}|X_{r,1})$  and  $I(X_{3};\hat{X}_{3}|X_{r,1},X_{r,2})$  depend on distributions  $P_{X_1X_{r,1}}$ ,  $P_{X_{r,1}X_{r,2}X_2}$  and  $P_{X_3\hat{X}_3X_{r,1}X_{r,2}}$ , respectively. So, these distributions should be preserved by the set of Markov chains. The first two distributions are preserved by the choice of (73)-(74). Now, since we have first-order Markov sources (see Definition 3), preserving the joint distributions of  $P_{X_{r,1}X_1}$  and  $P_{X_{r,1}X_{r,2}X_2}$  is sufficient to preserve the distribution  $P_{X_{r,1}X_{r,2}X_3}$ . So, preserving the joint distribution of  $P_{\hat{X}_3X_{r,1}X_{r,2}}$  is sufficient to keep  $I(X_3;\hat{X}_3|X_{r,1},X_{r,2})$  unchanged.
Considering the above four facts, without loss of optimality, one can impose the following Markov chains
$$
X _ {r, 1} \rightarrow X _ {1} \rightarrow \left(X _ {2}, X _ {3}\right), \tag {119}
$$
$$
X _ {r, 2} \rightarrow \left(X _ {2}, X _ {r, 1}\right)\rightarrow \left(X _ {1}, X _ {3}\right), \tag {120}
$$
$$
\hat {X} _ {3} \rightarrow \left(X _ {3}, X _ {r, 1}, X _ {r, 2}\right)\rightarrow \left(X _ {1}, X _ {2}\right). \tag {121}
$$
This concludes the proof for the PLF-JD. For the PLF-FMD, notice that the only difference is the second fact stated above. But, this also holds since the perception constraints depend only on  $P_{X_j}$  and  $P_{\hat{X}_j}$  (hence on  $P_{X_{r,1},\ldots ,X_{r,j}}$ ).
E Gauss-Markov Source Model
We first remark that the Wasserstein-2 distance can also be replaced by the KL-divergence in most of the following analysis. The common properties between these two measures are convexity and the fact that they both depend on only second-order statistics when restricted to Gaussian source model.
Theorem 5 For the Gauss-Markov source model, any tuple  $(\mathsf{R},\mathsf{D},\mathsf{P})\in \mathcal{C}_{\mathsf{RDP}}$  can be attained by a jointly Gaussian distribution over  $(X_{r,1},X_{r,2},X_{r,3})$  and identity mappings for  $\eta_j(\cdot)$  in Definition 4.
Proof: First, notice that a proof for the setting without perception constraint is provided in [37]. The following proof is different from [37] in some steps and also involves the perception constraint.
For a given tuple  $(\mathsf{R},\mathsf{D},\mathsf{P})\in \mathcal{C}_{\mathsf{RDP}}$ , let  $X_{r,1}^{*}$ ,  $X_{r,2}^{*}$ ,  $\hat{X}_1^* = \eta_1(X_{r,1}^*)$ ,  $\hat{X}_2^* = \eta_2(X_{r,1}^*,X_{r,2}^*)$  and  $\hat{X}_3^*$  be random variables satisfying (73)-(75). Let  $P_{\hat{X}_1^G|X_1}$ ,  $P_{\hat{X}_2^G|\hat{X}_1^GX_2}$  and  $P_{\hat{X}_3^G|\hat{X}_1^GX_2^G X_3}$  be jointly Gaussian distributions such that the following conditions are satisfied.
$$
\operatorname {c o v} \left(\hat {X} _ {1} ^ {G}, X _ {1}\right) = \operatorname {c o v} \left(\hat {X} _ {1} ^ {*}, X _ {1}\right), \tag {122}
$$
$$
\operatorname {c o v} \left(\hat {X} _ {1} ^ {G}, \hat {X} _ {2} ^ {G}, X _ {2}\right) = \operatorname {c o v} \left(\hat {X} _ {1} ^ {*}, \hat {X} _ {2} ^ {*}, X _ {2}\right), \tag {123}
$$
$$
\operatorname {c o v} \left(\hat {X} _ {1} ^ {G}, \hat {X} _ {2} ^ {G}, \hat {X} _ {3} ^ {G}, X _ {3}\right) = \operatorname {c o v} \left(\hat {X} _ {1} ^ {*}, \hat {X} _ {2} ^ {*}, \hat {X} _ {3} ^ {*}, X _ {3}\right), \tag {124}
$$
In general, the Gaussian random variables which satisfy the constraints in (122)-(124) can be written in the following format
$$
X _ {1} = \nu \hat {X} _ {1} ^ {G} + Z _ {1}, \tag {125}
$$
$$
\hat {X} _ {2} ^ {G} = \omega_ {1} \hat {X} _ {1} ^ {G} + \omega_ {2} X _ {2} + Z _ {2}, \tag {126}
$$
$$
\hat {X} _ {3} ^ {G} = \tau_ {1} \hat {X} _ {1} ^ {G} + \tau_ {2} \hat {X} _ {2} ^ {G} + \tau_ {3} X _ {3} + Z _ {3}, \tag {127}
$$
for some real  $\nu, \omega_1, \omega_2, \tau_1, \tau_2, \tau_3$  where  $\hat{X}_1^G \sim \mathcal{N}(0, \sigma_{\hat{X}_1^G}^2)$ ,  $\hat{X}_2^G \sim \mathcal{N}(0, \sigma_{\hat{X}_2^G}^2)$ ,  $Z_1, Z_2$  and  $Z_3$  are Gaussian random variables with zero mean and variances  $\alpha_1^2, \alpha_2^2, \alpha_3^2$ , independent of  $\hat{X}_1^G$ ,  $(\hat{X}_1^G, X_2)$  and  $(\hat{X}_1^G, \hat{X}_2^G, X_3)$ , respectively.
We explicitly derive the coefficients  $\nu, \omega_1, \omega_2, \tau_1, \tau_2$  and  $\tau_3$  in the following. Multiplying both sides of (125) by  $\hat{X}_1^G$  and taking an expectation, we get
$$
\mathbb {E} \left[ X _ {1} \hat {X} _ {1} ^ {G} \right] = \nu \sigma_ {\hat {X} _ {1} ^ {G}} ^ {2}. \tag {128}
$$
According to (122), the above equation can be written as follows
$$
\mathbb {E} \left[ X _ {1} \hat {X} _ {1} ^ {*} \right] = \nu \mathbb {E} \left[ \hat {X} _ {1} ^ {* 2} \right]. \tag {129}
$$
Multiplying both sides of (126) by the vector  $\left[\hat{X}_1^G X_2\right]$  and taking an expectation, we have
$$
[ \mathbb {E} [ \hat {X} _ {1} ^ {G} \hat {X} _ {2} ^ {G} ] \quad \mathbb {E} [ X _ {2} \hat {X} _ {2} ^ {G} ] ] = \left[ \begin{array}{l l} \omega_ {1} & \omega_ {2} \end{array} \right] \left( \begin{array}{c c} \sigma_ {\hat {X} _ {1} ^ {G}} ^ {2} & \mathbb {E} [ X _ {2} \hat {X} _ {1} ^ {G} ] \\ \mathbb {E} [ X _ {2} \hat {X} _ {1} ^ {G} ] & \sigma_ {2} ^ {2} \end{array} \right) \tag {130}
$$
Considering the fact that  $\mathbb{E}[X_2\hat{X}_1^G] = \rho_1\mathbb{E}[X_1\hat{X}_1^G]$  and according to (123), the above equation can be written as follows
$$
\left[ \mathbb {E} \left[ \hat {X} _ {1} ^ {*} \hat {X} _ {2} ^ {*} \right] \quad \mathbb {E} \left[ X _ {2} \hat {X} _ {2} ^ {*} \right] \right] = \left[ \omega_ {1} \quad \omega_ {2} \right] \binom {\mathbb {E} \left[ \hat {X} _ {1} ^ {* 2} \right]} {\rho_ {1} \mathbb {E} \left[ X _ {1} \hat {X} _ {1} ^ {*} \right]} \quad \begin{array}{c} \rho_ {1} \mathbb {E} \left[ X _ {1} \hat {X} _ {1} ^ {*} \right] \\ \sigma_ {2} ^ {2} \end{array} . \tag {131}
$$
Similarly, multiplying both sides of (127) by the vector  $[\hat{X}_1^G\hat{X}_2^G X_3]$ , taking an expectation and considering (124), we get
$$
\left[ \mathbb {E} \left[ \hat {X} _ {1} ^ {*} \hat {X} _ {3} ^ {*} \right] \mathbb {E} \left[ \hat {X} _ {2} ^ {*} \hat {X} _ {3} ^ {*} \right] \mathbb {E} \left[ X _ {3} \hat {X} _ {3} ^ {*} \right] \right] = \left[ \begin{array}{l l l} \tau_ {1} & \tau_ {2} & \tau_ {3} \end{array} \right] \left( \begin{array}{c c c} \mathbb {E} \left[ \hat {X} _ {1} ^ {* 2} \right] & \mathbb {E} \left[ \hat {X} _ {1} ^ {*} \hat {X} _ {2} ^ {*} \right] & \rho_ {1} \rho_ {2} \mathbb {E} \left[ X _ {1} \hat {X} _ {1} ^ {*} \right] \\ \mathbb {E} \left[ \hat {X} _ {1} ^ {*} \hat {X} _ {2} ^ {*} \right] & \mathbb {E} \left[ \hat {X} _ {2} ^ {* 2} \right] & \rho_ {2} \mathbb {E} \left[ X _ {2} \hat {X} _ {2} ^ {*} \right] \\ \rho_ {1} \rho_ {2} \mathbb {E} \left[ X _ {1} \hat {X} _ {1} ^ {*} \right] & \rho_ {2} \mathbb {E} \left[ X _ {2} \hat {X} _ {2} ^ {*} \right] & \mathbb {E} \left[ \hat {X} _ {3} ^ {* 2} \right] \end{array} \right). \tag {132}
$$
Solving equations (129), (131) and (132), we get
$$
\sigma_ {\hat {X} _ {1} ^ {G}} ^ {2} = \mathbb {E} \left[ \hat {X} _ {1} ^ {* 2} \right], \tag {133}
$$
$$
\nu = \frac {\mathbb {E} \left[ X _ {1} \hat {X} _ {1} ^ {*} \right]}{\mathbb {E} \left[ \hat {X} _ {1} ^ {* 2} \right]}, \tag {134}
$$
$$
\alpha_ {1} ^ {2} = \sigma_ {1} ^ {2} - \frac {\mathbb {E} \left[ X _ {1} \hat {X} _ {1} ^ {*} \right]}{\mathbb {E} \left[ \hat {X} _ {1} ^ {* 2} \right]}, \tag {135}
$$
$$
\omega_ {1} = \frac {\nu \rho_ {1} \mathbb {E} \left[ \hat {X} _ {1} ^ {*} \hat {X} _ {2} ^ {*} \right] - \mathbb {E} \left[ X _ {2} \hat {X} _ {2} ^ {*} \right]}{\nu^ {2} \rho_ {1} ^ {2} \sigma_ {\hat {X} _ {1} ^ {G}} ^ {2} - \sigma_ {2} ^ {2}}, \tag {136}
$$
$$
\omega_ {2} = \frac {\nu \rho_ {1} \sigma_ {\hat {X} _ {1} ^ {G}} ^ {2} \mathbb {E} \left[ X _ {2} \hat {X} _ {2} ^ {*} \right] - \sigma_ {2} ^ {2} \mathbb {E} \left[ \hat {X} _ {1} ^ {*} \hat {X} _ {2} ^ {*} \right]}{\nu^ {2} \rho_ {1} ^ {2} \sigma_ {\hat {X} _ {1} ^ {G}} ^ {4} - \sigma_ {2} ^ {2} \sigma_ {\hat {X} _ {1} ^ {G}} ^ {2}}, \tag {137}
$$
$$
\alpha_ {2} ^ {2} = \mathbb {E} \left[ \hat {X} _ {2} ^ {* 2} \right] - \alpha_ {2} ^ {2} \sigma_ {\hat {X} _ {1} ^ {G}} ^ {2} - \omega_ {2} ^ {2} \sigma_ {2} ^ {2} - 2 \omega_ {1} \omega_ {2} \rho_ {1} \nu \sigma_ {\hat {X} _ {1} ^ {G}} ^ {2}. \tag {138}
$$
For the third step, the coefficients and noise variance of (127) are given as follows
$$
\left[ \begin{array}{c c c} \tau_ {1} & \tau_ {2} & \tau_ {3} \end{array} \right]
$$
$$
= \left[ \mathbb {E} \left[ \hat {X} _ {1} ^ {*} \hat {X} _ {3} ^ {*} \right] \mathbb {E} \left[ \hat {X} _ {2} ^ {*} \hat {X} _ {3} ^ {*} \right] \mathbb {E} \left[ X _ {3} \hat {X} _ {3} ^ {*} \right] \right] \binom {\mathbb {E} \left[ \hat {X} _ {1} ^ {* 2} \right]} {\mathbb {E} \left[ \hat {X} _ {1} ^ {*} \hat {X} _ {2} ^ {*} \right]} \begin{array}{c c c} \mathbb {E} \left[ \hat {X} _ {1} ^ {*} \hat {X} _ {2} ^ {*} \right] & \mathbb {E} \left[ \hat {X} _ {1} ^ {*} \hat {X} _ {2} ^ {*} \right] & \rho_ {1} \rho_ {2} \mathbb {E} \left[ X _ {1} \hat {X} _ {1} ^ {*} \right] \\ \rho_ {1} \rho_ {2} \mathbb {E} \left[ X _ {1} \hat {X} _ {1} ^ {*} \right] & \mathbb {E} \left[ \hat {X} _ {2} ^ {* 2} \right] & \rho_ {2} \mathbb {E} \left[ X _ {2} \hat {X} _ {2} ^ {*} \right] \\ & \rho_ {2} \mathbb {E} \left[ X _ {2} \hat {X} _ {2} ^ {*} \right] & \mathbb {E} \left[ \hat {X} _ {3} ^ {* 2} \right] \end{array} } ^ {- 1},
$$
$$
\begin{array}{l} \alpha_ {3} ^ {2} = \mathbb {E} [ \hat {X} _ {3} ^ {* 2} ] - \tau_ {1} ^ {2} \mathbb {E} [ \hat {X} _ {1} ^ {* 2} ] - \tau_ {2} ^ {2} \mathbb {E} [ \hat {X} _ {2} ^ {* 2} ] - \tau_ {3} ^ {2} \mathbb {E} [ X _ {3} ^ {2} ] \\ - 2 \tau_ {1} \tau_ {2} \mathbb {E} \left[ \hat {X} _ {1} ^ {*} \hat {X} _ {2} ^ {*} \right] - 2 \tau_ {1} \tau_ {3} \rho_ {1} \rho_ {2} \mathbb {E} \left[ X _ {1} \hat {X} _ {1} ^ {*} \right] - 2 \tau_ {2} \tau_ {3} \rho_ {2} \mathbb {E} \left[ X _ {2} \hat {X} _ {2} ^ {*} \right], \tag {140} \\ \end{array}
$$
where  $(.)^{-1}$  denotes the inverse of a matrix.
Now, we look at the rate constraints.
Rate Constraints:
Consider the rate constraint of the first step as follows
$$
\begin{array}{l} R _ {1} \geq I \left(X _ {1}; X _ {r, 1} ^ {*}\right) (141) \\ = H \left(X _ {1}\right) - H \left(X _ {1} \mid X _ {r, 1} ^ {*}\right) (142) \\ \geq H \left(X _ {1}\right) - H \left(X _ {1} \mid \hat {X} _ {1} ^ {*}\right) (143) \\ = H \left(X _ {1}\right) - H \left(X _ {1} - \mathbb {E} \left[ X _ {1} \mid \hat {X} _ {1} ^ {*} \right] \mid \hat {X} _ {1} ^ {*}\right) (144) \\ \geq H \left(X _ {1}\right) - H \left(X _ {1} - \mathbb {E} \left[ X _ {1} \mid \hat {X} _ {1} ^ {*} \right]\right) (145) \\ \geq H \left(X _ {1}\right) - H \left(X _ {1} - \mathbb {E} \left[ X _ {1} \mid \hat {X} _ {1} ^ {G} \right]\right) (146) \\ = H \left(X _ {1}\right) - H \left(X _ {1} - \mathbb {E} \left[ X _ {1} \mid \hat {X} _ {1} ^ {G} \right] \mid \hat {X} _ {1} ^ {G}\right) (147) \\ = I \left(X _ {1}; \hat {X} _ {1} ^ {G}\right) (148) \\ \end{array}
$$
(143) follows because  $\hat{X}_1^*$  is a function of  $X_{r,1}^{*}$
(146) follows because for a given covariance matrix in (122), the Gaussian distribution maximizes the differential entropy;
(147) follows because the MMSE is uncorrelated from the data and since the random variables are Gaussian, the MMSE would be independent of the data.
Next, consider the rate constraint of the second step as the following
$$
\begin{array}{l} R _ {2} \geq I \left(X _ {2}; X _ {r, 2} ^ {*} \mid X _ {r, 1} ^ {*}\right) (149) \\ = H \left(X _ {2} \mid X _ {r, 1} ^ {*}\right) - H \left(X _ {2} \mid X _ {r, 1} ^ {*}, X _ {r, 2} ^ {*}\right) (150) \\ \geq H \left(X _ {2} \mid X _ {r, 1} ^ {*}\right) - H \left(X _ {2} \mid \hat {X} _ {1} ^ {*}, \hat {X} _ {2} ^ {*}\right) (151) \\ \geq H \left(X _ {2} \mid X _ {r, 1} ^ {*}\right) - H \left(X _ {2} \mid \hat {X} _ {1} ^ {G}, \hat {X} _ {2} ^ {G}\right) (152) \\ = H \left(\rho_ {1} X _ {1} + N _ {1} \mid X _ {r, 1} ^ {*}\right) - H \left(X _ {2} \mid \hat {X} _ {1} ^ {G}, \hat {X} _ {2} ^ {G}\right) (153) \\ \geq \frac {1}{2} \log \left(\rho_ {1} ^ {2} 2 ^ {2 H \left(X _ {1} \mid X _ {r, 1} ^ {*}\right)} + 2 ^ {2 H (N _ {1})}\right) - H \left(X _ {2} \mid \hat {X} _ {1} ^ {G}, \hat {X} _ {2} ^ {G}\right) (154) \\ \geq \frac {1}{2} \log \left(\rho_ {1} ^ {2} 2 ^ {- 2 R _ {1}} 2 ^ {2 H \left(X _ {1}\right)} + 2 ^ {2 H \left(N _ {1}\right)}\right) - H \left(X _ {2} \mid \tilde {X} _ {1} ^ {G}, \tilde {X} _ {2} ^ {G}\right), (155) \\ \end{array}
$$
(151) follows because  $\hat{X}_1^*$  and  $\hat{X}_2^*$  are deterministic functions of  $X_{r,1}^*$  and  $(X_{r,1}^*, X_{r,2}^*)$ , respectively;
(152) follows because for a given covariance matrix in (123), the Gaussian distribution maximizes the differential entropy;
(154) follows from entropy power inequality (EPI) [38, pp. 22];
(155) follows from (142).
Similarly, consider the rate constraint of the third frame as the following,
$$
\begin{array}{l} R _ {3} \geq I \left(X _ {3}; \hat {X} _ {3} ^ {*} \mid X _ {r, 1} ^ {*}, X _ {r, 2} ^ {*}\right) (156) \\ = H \left(X _ {3} \mid X _ {r, 1} ^ {*}, X _ {r, 2} ^ {*}\right) - H \left(X _ {3} \mid X _ {r, 1} ^ {*}, X _ {r, 2} ^ {*}, \hat {X} _ {3} ^ {*}\right) (157) \\ \geq H \left(X _ {3} \mid X _ {r, 1} ^ {*}, X _ {r, 2} ^ {*}\right) - H \left(X _ {3} \mid \hat {X} _ {1} ^ {*}, \hat {X} _ {2} ^ {*}, \hat {X} _ {3} ^ {*}\right) (158) \\ \geq H \left(X _ {3} \mid X _ {r, 1} ^ {*}, X _ {r, 2} ^ {*}\right) - H \left(X _ {3} \mid \hat {X} _ {1} ^ {G}, \hat {X} _ {2} ^ {G}, \hat {X} _ {3} ^ {G}\right) (159) \\ = H \left(\rho_ {2} X _ {2} + N _ {2} \mid X _ {r, 1} ^ {*}, X _ {r, 2} ^ {*}\right) - H \left(X _ {3} \mid \hat {X} _ {1} ^ {G}, \hat {X} _ {2} ^ {G}, \hat {X} _ {3} ^ {G}\right) (160) \\ \geq \frac {1}{2} \log \left(\rho_ {2} ^ {2} 2 ^ {2 H \left(X _ {2} \mid X _ {r, 1} ^ {*}, X _ {r, 2} ^ {*}\right)} + 2 ^ {2 H \left(N _ {2}\right)}\right) - H \left(X _ {3} \mid \hat {X} _ {1} ^ {G}, \hat {X} _ {2} ^ {G}, \hat {X} _ {3} ^ {G}\right) (161) \\ \geq \frac {1}{2} \log \left(\rho_ {2} ^ {2} 2 ^ {- 2 R _ {2}} 2 ^ {2 H \left(X _ {2} \mid X _ {r, 1} ^ {*}\right)} + 2 ^ {2 H \left(N _ {2}\right)}\right) - H \left(X _ {3} \mid \hat {X} _ {1} ^ {G}, \hat {X} _ {2} ^ {G}, \hat {X} _ {3} ^ {G}\right) (162) \\ \geq \frac {1}{2} \log \left(\rho_ {1} ^ {2} \rho_ {2} ^ {2} 2 ^ {- 2 R _ {1} - 2 R _ {2}} 2 ^ {2 H (X _ {1})} + \rho_ {2} ^ {2} 2 ^ {- 2 R _ {2}} 2 ^ {2 H (N _ {1})} + 2 ^ {2 H (N _ {2})}\right) - H \left(X _ {3} | \hat {X} _ {1} ^ {G}, \hat {X} _ {2} ^ {G}, \hat {X} _ {3} ^ {G}\right) (163) \\ \end{array}
$$
Next, we look at the distortion constraint.
Distortion Constraint: The choices in (122)-(124) imply that
$$
D _ {j} \geq \mathbb {E} [ \| X _ {j} - \hat {X} _ {j} ^ {*} \| ^ {2} ] = \mathbb {E} [ \| X _ {j} - \hat {X} _ {j} ^ {G} \| ^ {2} ], \quad j = 1, 2, 3. \tag {164}
$$
Finally, we look at the perception constraint
Perception Constraint:
Define the following distribution
$$
\begin{array}{l} P _ {U ^ {*} V ^ {*}} := \arg \quad \inf  \quad \mathbb {E} _ {\tilde {P}} [ \| U - V \| ^ {2} ]. \tag {165} \\ \tilde {P} _ {U V}: \\ P _ {U} = P _ {X _ {1}} \\ \tilde {P} _ {V} = P _ {\hat {X} _ {1} ^ {*}} \\ \end{array}
$$
Now, define  $P_{U^G V^G}$  to be a Gaussian joint distribution with the following covariance matrix
$$
\operatorname {c o v} \left(U ^ {G}, V ^ {G}\right) = \operatorname {c o v} \left(U ^ {*}, V ^ {*}\right). \tag {166}
$$
Then, we have the following set of inequalities:
$$
P _ {1} \geq W _ {2} ^ {2} \left(P _ {X _ {1}}, P _ {\hat {X} _ {1} ^ {*}}\right) = \inf  _ {\tilde {P} _ {U V}:} \mathbb {E} _ {\tilde {P}} [ \| U - V \| ^ {2} ] \tag {167}
$$
$$
\tilde {P} _ {U} = P _ {X _ {1}}
$$
$$
\tilde {P} _ {V} = P _ {\hat {X} _ {1} ^ {*}}
$$
$$
= \mathbb {E} \left[ \left\| U ^ {*} - V ^ {*} \right\| ^ {2} \right] \tag {168}
$$
$$
= \mathbb {E} \left[ \| U ^ {G} - V ^ {G} \| ^ {2} \right] \tag {169}
$$
$$
\geq W _ {2} ^ {2} \left(P _ {U ^ {\mathcal {G}}}, P _ {V ^ {\mathcal {G}}}\right) \tag {170}
$$
$$
= \inf  _ {\hat {P} _ {U V}:} \mathbb {E} _ {\hat {P}} [ \| U - V \| ^ {2} ] \tag {171}
$$
$$
\hat {P} _ {U} = P _ {U G}
$$
$$
\hat {P} _ {V} = P _ {V G}
$$
$$
= \inf  _ {\hat {P} _ {U V}:} \mathbb {E} _ {\hat {P}} [ \| U - V \| ^ {2} ] \tag {172}
$$
$$
\hat {P} _ {U} = P _ {X _ {1}}
$$
$$
\hat {P} _ {V} = P _ {\hat {X} _ {1} G}
$$
$$
= W _ {2} ^ {2} \left(P _ {X _ {1}}, P _ {\hat {X} _ {1} ^ {G}}\right), \tag {173}
$$
(168) follows from the definition in (165);
(169) follows from (166) which implies that  $(U^{*},V^{*})$  and  $(U^{G},V^{G})$  have the same second-order statistics;
(172) follows because  $P_{V^G} = P_{\hat{X}_1^G}$  which is justified in the following. First, notice that both  $P_{V^G}$  and  $P_{\hat{X}_1^G}$  are Gaussian distributions. Denote the variance of  $V^G$  by  $\sigma_{V^G}^2$  and recall that the variance of  $\hat{X}_1^G$  is denoted by  $\sigma_{\hat{X}_1^G}^2$ . According to (166),  $\sigma_{V^G}^2$  is equal to the variance of  $V^*$ . Also, from (165), we know that  $P_{V^*} = P_{\hat{X}_1^*}$ , hence the variances of  $V^*$  and  $\hat{X}_1^*$  are the same. On the other side, according to (122), we know that the variance of  $\hat{X}_1^*$  is equal to  $\sigma_{\hat{X}_1^G}^2$ . Thus, we conclude that  $\sigma_{\hat{X}_1^G}^2 = \sigma_{V^G}^2$ , which yields  $P_{V^G} = P_{\hat{X}_1^G}$ . A similar argument shows that  $P_{U^G} = P_{X_1}$ .
A similar argument holds for the perception constraint of the second and third steps for both PLFs.
Thus, we have proved the set of Gaussian auxiliary random variables  $(\hat{X}_1^G,\hat{X}_2^G,\hat{X}_3^G)$  given in (125)-(127) where the coefficients are chosen according to distortion-perception constraints, provides an outer bound to  $\mathcal{C}_{\mathrm{RDP}}$  which is the set of all tuples  $(\mathsf{R},\mathsf{D},\mathsf{P})$  such that
$$
R _ {1} \geq I \left(X _ {1}; \hat {X} _ {1} ^ {G}\right), \tag {174}
$$
$$
R _ {2} \geq \frac {1}{2} \log \left(\rho_ {1} ^ {2} 2 ^ {- 2 R _ {1}} 2 ^ {2 H \left(X _ {1}\right)} + 2 ^ {2 H \left(N _ {1}\right)}\right) - H \left(X _ {2} \mid \hat {X} _ {1} ^ {G}, \hat {X} _ {2} ^ {G}\right), \tag {175}
$$
$$
R _ {3} \geq \frac {1}{2} \log \left(\rho_ {1} ^ {2} \rho_ {2} ^ {2} 2 ^ {- 2 R _ {1} - 2 R _ {2}} 2 ^ {2 H \left(X _ {1}\right)} + \rho_ {2} ^ {2} 2 ^ {- 2 R _ {2}} 2 ^ {2 H \left(N _ {1}\right)} + 2 ^ {2 H \left(N _ {2}\right)}\right) - H \left(X _ {3} | \hat {X} _ {1} ^ {G}, \hat {X} _ {2} ^ {G}, \hat {X} _ {3} ^ {G}\right), \tag {176}
$$
$$
D _ {j} \geq \mathbb {E} [ \| X _ {j} - \hat {X} _ {j} ^ {G} \| ^ {2} ], \quad j = 1, 2, 3 \tag {177}
$$
$$
P _ {j} \geq W _ {2} ^ {2} \left(P _ {X _ {1} \dots X _ {j}}, P _ {\hat {X} _ {1} ^ {G} \dots \hat {X} _ {j} ^ {G}}\right). \tag {178}
$$
Now, we need to show that the above RDP region is also an inner bound to  $\mathcal{C}_{\mathrm{RDP}}$ . This is simply verified by the following choice. In iRDP region of (68)-(76), choose the following:
$$
X _ {r, j} = \hat {X} _ {j} = \hat {X} _ {j} ^ {G}, \quad j = 1, 2, 3, \tag {179}
$$
where  $(\hat{X}_1^G, \hat{X}_2^G, \hat{X}_3^G)$  satisfy (125)-(127) with coefficients chosen according to distortion-perception constraints. The lower bounds on distortion and perception constraints in (177) and (178) are immediately achieved by this choice. Now, we will look at the rate constraints. The achievable rate constraint of the first step can be written as follows
$$
R _ {1} \geq I \left(X _ {1}; \hat {X} _ {1} ^ {G}\right), \tag {180}
$$
which immediately coincides with (174). The achievable rate of the second step can be written as follows
$$
\begin{array}{l} R _ {2} \geq I \left(X _ {2}; \hat {X} _ {2} ^ {G} \mid \hat {X} _ {1} ^ {G}\right) (181) \\ = H \left(X _ {2} \mid \hat {X} _ {1} ^ {G}\right) - H \left(X _ {2} \mid \hat {X} _ {1} ^ {G}, \hat {X} _ {2} ^ {G}\right) (182) \\ = H \left(\rho_ {1} X _ {1} + N _ {1} \mid \hat {X} _ {1} ^ {G}\right) - H \left(X _ {2} \mid \hat {X} _ {1} ^ {G}, \hat {X} _ {2} ^ {G}\right) (183) \\ = \frac {1}{2} \log \left(\rho_ {1} ^ {2} 2 ^ {2 H \left(X _ {1} \mid \hat {X} _ {1} ^ {G}\right)} + 2 ^ {2 H \left(N _ {1}\right)}\right) - H \left(X _ {2} \mid \hat {X} _ {1} ^ {G}, \hat {X} _ {2} ^ {G}\right) (184) \\ \geq \frac {1}{2} \log \left(\rho_ {1} ^ {2} 2 ^ {- 2 R _ {1}} 2 ^ {2 H \left(X _ {1}\right)} + 2 ^ {2 H \left(N _ {1}\right)}\right) - H \left(X _ {2} \mid \hat {X} _ {1} ^ {G}, \hat {X} _ {2} ^ {G}\right), (185) \\ \end{array}
$$
(184) follows because EPI holds with "equality" for jointly Gaussian distributions [38, pp. 22];
(185) follows from (175).
Thus, the bound in (185) coincides with (155). A similar argument holds for the achievable rate of the third frame.
Notice that the above proof (both converse and achievability) can be extended to  $T$  frames using the sequential analysis that was presented. Thus, without loss of optimality, one can restrict to the jointly Gaussian distributions and identity functions  $\eta_{1}(.)$  and  $\eta_{2}(.,.)$  in iRDP region  $\mathcal{C}_{\mathrm{RDP}}$ .