system. Evaluating the fixed points (local maxima) in the free energy expression provides the MMSE for the channel  [@Tanaka2002; @GuoVerdu2005; @Krzakala2012probabilistic; @krzakala2012statistical; @MezardMontanariBook; @Barbier2015]. [*Under the assumption of self-averaging*]{} [@Tanaka2002; @GuoVerdu2005; @Krzakala2012probabilistic; @krzakala2012statistical; @MezardMontanariBook; @Barbier2015], the free energy is defined as[^4] $$\label{eq:free_energy}
  \mathcal{F}=\lim_{N\rightarrow\infty}\frac{1}{N}\mathbb{E}_{\F,\s,\z}[\log (Z)],$$ which is difficult to evaluate. The replica method [@Tanaka2002; @GuoVerdu2005; @Krzakala2012probabilistic; @krzakala2012statistical; @MezardMontanariBook; @Barbier2015] introduces $n$ replicas of the estimate $\x$ as $\x^a,\ a\in\{1,...,n\}$, and the free energy  can be approximated by the replica trick [@Krzakala2012probabilistic; @krzakala2012statistical; @MezardMontanariBook; @Barbier2015], $$\label{eq:replicaTrick}
  \mathcal{F}=\lim_{N\rightarrow\infty}\lim_{n\rightarrow 0} \frac{\mathbb{E}_{\F,\s,\z}[ Z^n]-1}{Nn}.$$ Note that the self-averaging property that leads to  and the replica trick , as well as the replica symmetry assumptions that appear in latter parts of this paper, are assumed to be valid in this work, and their rigorous justification is still an open problem in mathematical physics [@Tanaka2002; @GuoVerdu2005; @Krzakala2012probabilistic; @krzakala2012statistical; @MezardMontanariBook; @Barbier2015].[^5]

**Evaluating the free energy**: To evaluate the free energy , we calculate $\mathbb{E}_{\F,\s,\z}\l[Z^n\r]$, where $\cdot_{\F,\s,\z}$ denotes expectation with respect to (w.r.t.) $\F,\s$, and $\z$, and $Z$ is given in : $$\label{eq:EZn1}
  \mathbb{E}_{\F,\s,\z}\l[Z^n\r]\!=\!\frac{\mathbb{E}_{\s}\!\l[\displaystyle{\int {\prod_{l=1}^{N}}{\prod_{a=1}^{n}}\! P(\x_l^a)\!{\prod_{\mu=1}^{M}}\!\mathbb{X}_{\mu}\!{\prod_{l=1}^{N}}{\prod_{a=1}^{n}}d\x_l^a}\r]}{(2\pi\Delta)^{\frac{nMJ}{2}}},$$ where $$\label{eq:Xmu}
  \mathbb{X}_{\mu}=\mathbb{E}_{\F,\z}\l[\operatorname{e}^{-\frac{1}{2\Delta}{\sum_{j=1}^{J}}{\sum_{a=1}^{n}}(v_{\mu j}^a)^2}\r],$$ $a$ is the replica index, $\x^a_l$ is the $l$-th super symbol of $\x^a$, and $$\label{eq:v_mu_a}
v_{\mu j}^a={\sum_{l=1}^{N}}\F_{\mu+M(j-1),l}(\s_l-\x_l^a)+z_{\mu+M(j-1)}.$$

\[lemma:covIsSame\] In the 