## 3 The New Proposed Expression for the Conditional Expectation
: According to Reference [20] (Appendix B), the approximated MSE for the noisy case, valid at the latter stages of the deconvolutional process, may be given as  
\[E\left[x|z\right]=\frac{\frac{z+\tilde{\sigma}_{\tilde{p}}^{\prime}\left(z \right)}{2\tilde{\sigma}\left(z\right)}\sigma_{\tilde{p}}^{2}}{1+\frac{\tilde {\sigma}_{\tilde{p}}^{\prime}\left(z\right)}{2\tilde{\sigma}\left(z\right)} \sigma_{\tilde{p}}^{2}} \tag{24}\]1where  
\[E\left[x|z\right]=\frac{\frac{z+\tilde{\sigma}_{\tilde{p}}^{\prime}\left(z \right)}{2\tilde{\sigma}\left(z\right)}\sigma_{\tilde{p}}^{2}}{1+\frac{\tilde {\sigma}_{\tilde{p}}^{\prime}\left(z\right)}{2\tilde{\sigma}\left(z\right)} \sigma_{\tilde{p}}^{2}} \tag{24}\]2  
Thus, according to (19), we obtain the following equations for the Lagrange multipliers:  
\[ \begin{array}{c}2+2\lambda_{2}\tilde{m}_{2}4+2\lambda_{4}\tilde{m}_{4}8=0\\ \\ \tilde{m}_{2}12+2\lambda_{4}\tilde{m}_{6}16+2\lambda_{2}\tilde{m}_{4}8=0\end{array}  \tag{30}\]  
where  
\[\tilde{m}_{G}=E[\tilde{x}^{G}]. \tag{31}\]  
The solution of (30) for \(\lambda_{2}\) and \(\lambda_{4}\) as a function of \(\tilde{m}_{2}\), \(\tilde{m}_{4}\), and \(\tilde{m}_{6}\) is given in (26). Next, I find closed-form approximated expressions for \(\tilde{m}_{2}\), \(\tilde{m}_{4}\), and \(\tilde{m}_{6}\). When the deconvolutional process has converged and leaves the system with a convolutional noise that can be considered as very low, we may write, according to Reference [52],  
\[\sigma_{\widetilde{w}}^{2}=\frac{\sigma_{w}^{2}}{\sum_{k=0}^{R-1}\left|h_{k} \right|^{2}}. \tag{32}\]  
Since \(x\) and \(w\) are independent, by using (32) and the expression for \(SNR\) (27) we have  
\[ \begin{array}{c}\widetilde{m}_{2}=E\left[\left(x+\widetilde{w}\right)^{2} \right]=m_{2}+\sigma_{\widetilde{w}}^{2}=m_{2}+\frac{\sigma_{w}^{2}}{\sum_{k =0}^{R-1}\left|h_{k}\right|^{2}}=m_{2}\left(1+\frac{\sigma_{w}^{2}}{m_{2}\sum _{k=0}^{R-1}\left|h_{k}\right|^{2}}\right)=\\ \\ m_{2}\left(1+\frac{1}{SNR\sum_{k=0}^{R-1}\left|h_{k}\right|^{2}}\right)\\ \\ \widetilde{m}_{4}=E\left[\left(x+\widetilde{w}\right)^{4}\right]=3\sigma_{ \widetilde{w}}^{4}+6m_{2}\sigma_{\widetilde{w}}^{2}+m_{4}=3\left(\frac{\sigma _{w}^{2}}{\sum_{k=0}^{R-1}\left|h_{k}\right|^{2}}\right)^{2}+6m_{2}\frac{ \sigma_{w}^{2}}{\sum_{k=0}^{R-1}\left|h_{k}\right|^{2}}+m_{4}=\\ \\ m_{2}^{2}\left(\frac{3}{\left(SNR\sum_{k=0}^{R-1}\left|h_{k}\right|^{2} \right)^{2}}+\frac{6}{SNR\sum_{k=0}^{R-1}\left|h_{k}\right|^{2}}+\frac{m_{4}} {m_{2}^{2}}\right)\\ \\ \widetilde{m}_{6}=E\left[\left(x+\widetilde{w}\right)^{6}\right]=15\sigma_{ \widetilde{w}}^{6}+45\sigma_{\widetilde{w}}^{4}m_{2}+15\sigma_{\widetilde{w}} ^{2}m_{4}+m_{6}=\\ \\ m_{2}^{3}\left(\frac{15}{\left(SNR\sum_{k=0}^{R-1}\left|h_{k}\right|^{2} \right)^{3}}+\frac{45}{\left(SNR\sum_{k=0}^{R-1}\left|h_{k}\right|^{2} \right)^{2}}+\frac{15}{SNR\sum_{k=0}^{R-1}\left|h_{k}\right|^{2}}\frac{m_{4}} {m_{2}^{2}}+\frac{m_{6}}{m_{2}^{2}}\right)\end{array}  \tag{33}\]  
This completes our proof. Next, I turn to the 16 QAM source input. For this case, according to Reference [44], I have that  
\[E\left[x|z\right]=E\left[x_{1}|z_{1}\right]+jE\left[x_{2}|z_{2}\right]. \tag{34}\]