we briefly derive the Hamming distance-optimal estimator when $\x_n\in \{\mathbf{1},\mathbf{0}\}$, where $\mathbf{0}$ is an all-zero row vector. The mean Hamming distance (MHD) given the pseudo data $\q_n$ is $$\label{eq:MHD}
\begin{split}
    &\text{MHD}|\q_n=\mathbb{E}[d_{\text{MHD}}(\x_n,\widehat{\x}_n)|\q_n]=\\
    &\left\{
                \begin{array}{ll}
                 \Pr(\x_n=\mathbf{1}|\q_n),&\   \widehat{\x}_n=\mathbf{0}\ \text{and}\ \x_n=\mathbf{1},\\
                 \Pr(\x_n=\mathbf{0}|\q_n),&\   \widehat{\x}_n=\mathbf{1}\ \text{and}\ \x_n=\mathbf{0},\\
                 0,\quad\quad\quad&\ \widehat{\x}_n= \x_n.
                \end{array}
            \right.\\
\end{split}$$ Following the steps in –, the minimum MHD (MMHD) estimator is $$\label{eq:MHD_optTh}
\widetilde{\x}_n=\left\{
                \begin{array}{ll}
                 \mathbf{1},&   \sum_{j=1}^J q_n^{(j)} \geq\frac{J}{2}+\Delta_v \ln\l[\frac{1-\rho}{\rho}\r],\\
                 \mathbf{0},&   \sum_{j=1}^J q_n^{(j)} <\frac{J}{2}+\Delta_v \ln\l[\frac{1-\rho}{\rho}\r].
                \end{array}
            \right.\\$$

Mean absolute error (MAE) {#sec:MAE}
-------------------------

[**MAE-optimal estimator:**]{} The element-wise absolute error (AE) is $$\label{eq:AE}
  d_{\text{AE}}(x_n^{(j)},\widehat{x}_n^{(j)})=|x_n^{(j)}-\widehat{x}_n^{(j)}|.$$ In order to find the minimum mean absolute error (MMAE) estimate, $\widetilde{\x}_n$, we need to find the stationary point of , $$\label{eq:optimalCondition}
\frac{d\mathbb{E}[|x_n^{(j)}-\widehat{x}_n^{(j)}|\ | \q_n]}{d\widehat{x}_n^{(j)}}\bigg|_{\widehat{x}_n^{(j)}=\widetilde{x}_n^{(j)}}=0,$$ $\forall j\in\{1,\ldots,J\},n\in\{1,\ldots,N\}$. It can be proved that $\mathbb{E}[X]=\int_{0}^{\infty} \Pr(X>x)dx$, if $X\geq 0$. Therefore, $$\label{eq:expectation}
\begin{split}
  &\mathbb{E}[|x_n^{(j)}-\widehat{x}_n^{(j)}|\ | \q_n]=\int_{0}^{\infty} \Pr(|x_n^{(j)}-\widehat{x}_n^{(j)}|>t| \q_n)dt\\
  &=\int_{-\infty}^{\widehat{x}_n^{(j)}} \Pr(x_n^{(j)}<t| \q_n)dt+\int_{\widehat{x}_n^{(j)}}^{\infty} \Pr(x_n^{(j)}>t| \q_n)dt.
\end{split}$$ Using  and , we obtain $$\Pr(x_n^{(j)}<\widetilde{x}_n^{(j)} | \q_n)=\Pr(x_n^{(j)}>\widetilde{x}_n^{(j)}| \q_n).$$ That is, $$\label{eq:optimalEst}
  \int_{-\infty}^{\widetilde{x}_n^{(j)}} f(x_n^{(j)} | \q_n)dx_n^{(j)}=\int_{\widetilde{x}_n^{(j)}}^{\infty} f(x_n^{(j)} | \q_n)dx_n^{(j)}=\frac{1}{2},$$ through which we solve for the optimal estimator $\widetilde{x}_n^{(j)}$ numerically.

[**Performance limits:**]{} We calculate the MMAE as follows, $$\label{eq:MMAE}
\begin{split}
  \text{MMAE}\!&=\!\mathbb{E}[|\widetilde{x}_n^{(j)}-x_n^{(j)}|]\!=\!\int_{-\infty}^{\infty}\! \mathbb{E}[|\widetilde{x}_n^{(j)}-x_n^{(j)}|\ | \q_n] f(\q_n)d \q_n\\
  &=\int_{-\infty}^{\infty}\Bigg[\int_{-\infty}^{\widetilde{x}_n^{(j)}} -x_n^{(j)} f(x_n^{(j)}| \q_n)d x_n^{(j)}+\\
  &\quad\quad\quad\quad \int_{\widetilde{x}_n^{(j)}}^{\infty} x_n^{(j)}f(x_n^{(j)}| \q_n)d x_n^{(j)}\Bigg] f(\q_n)d \q_n,
\end{split}$$ which has 