# Accurate Performance Analysis of Coded Large-Scale Multiuser MIMO Systems with MMSE Receivers
## 3 Distribution of the SINR  
As shown, in Refs. [11, 18, 19], the problem encountered in performance analysis of MMSE-detected MIMO systems is to explore an efficient approach to evaluate Equation (6). In this section, we develop a simple and efficient way to derive the distribution expression for SINR in massive multiuser MIMO scenarios. To do that, firstly, we consider the expression for the SINR in Equation (6). The term \(\sum_{r=1,r\
eq k}^{K}\zeta_{r}\mathbf{h}_{r}\mathbf{h}_{r}^{H}\) is a \(R\times R\) Hermitian matrix that can be reformulated as \(\mathbf{H}_{k}\mathbf{D}_{kk}\mathbf{H}_{k}^{H}\), where \(\mathbf{H}_{k}\) is a \(R\times(K-1)\) matrix by removing the \(k\)-th column of \(\mathbf{H}\) and \(\mathbf{D}_{kk}\) corresponds to \(\mathbf{D}\) with the \(k\)th row and the \(k\)th column deleted. We note that Equation (6) contains \(K-1\) different nonzero eigenvalues to be solved. The matrix \(\mathbf{H}_{k}\mathbf{D}_{kk}\mathbf{H}_{k}^{H}\) is a complex Wishart matrix; let \(\xi_{1},\cdots,\xi_{K-1}\) be the ordered eigenvalues of the diagonal matrix \(\mathbf{D}_{kk}\) such that \(\xi_{1}>\cdots>\xi_{K-1}>0\). The joint PDF of the ordered positive eigenvalues of \(\mathbf{H}_{k}\mathbf{D}_{kk}\mathbf{H}_{k}^{H}\), \(\lambda_{1},\cdots,\lambda_{K-1}\), equals [20]  
\[f_{\lambda}=L\det[\mathbf{E}]\prod_{l=1}^{K-1}\lambda_{l}^{R-K+1}\prod_{k<l}^ {K-1}(\lambda_{k}-\lambda_{l}), \tag{7}\]  
where \(\mathbf{E}=\{e^{-\lambda_{k}/\xi_{l}}\}_{k,l=1,\cdots,K-1}\),  
\[L=\prod_{l=1}^{K-1}\frac{1}{\xi_{l}^{R}(R-l)!}\prod_{k<l}^{K-1}\frac{\xi_{k}\xi_ {l}}{\xi_{k}-\xi_{l}}. \tag{8}\]As illustrated in [19], the joint distribution of non-zero eigenvalues is directly used without approximating, and, even if the dimension of the matrix is less than 4, it is difficult to obtain a closed solution of the SINR distribution. Influenced by the large-scale fading factor, the eigenvalue joint distribution function of \(\mathbf{H}\mathbf{D}\mathbf{H}^{H}\) is more complicated than the eigenvalue joint distribution function of \(\mathbf{H}\mathbf{H}^{H}\). Therefore, method [16] is not suitable for Equation (6). Here, we employ the approximate marginal PDF of the eigenvalues to evaluate Equation (6). The marginal PDF of an unordered eigenvalue of \(\mathbf{H}_{k}\mathbf{D}_{kk}\mathbf{H}_{k}^{H}\) is [21]  
\[f(\lambda)=\frac{L}{K-1}\sum_{i=1}^{K-1}\sum_{j=1}^{K-1}\mathcal{D}_{ij}\lambda ^{R-K+j}e^{-\lambda/\xi_{i}}, \tag{9}\]  
where \(\mathcal{D}_{ij}\) is the \((i,j)\)-th cofactor of the matrix  
\[\mathcal{D}=\left\{\frac{(R-K+k)!}{\xi_{l}^{-R+K-k-1}}\right\}_{l,k=1,\cdots,K -1}. \tag{10}\]  
Then, the PDF of \(\gamma_{k}\) conditioned on \(\mathbf{H}\) can be written as follows:  
\[f_{\gamma_{k}}(x|\mathbf{H})=\zeta_{k}\underbrace{\mathbb{E}_{\lambda}\left[ \frac{1}{\lambda_{i,k}+\rho}\right]}_{\xi}\sum_{i=1}^{K-1}|\tilde{n}_{i,k}|^{ 2}+\frac{\zeta_{k}}{\rho}\sum_{j=K}^{R}|\tilde{n}_{j,k}|^{2}, \tag{11}\]  
where \(\mathcal{G}\) can be calculated as follows:  
\[\mathcal{G} =\mathbb{E}\bigg{[}\frac{1}{\lambda+\rho}\bigg{]}=\int_{0}^{ \infty}\frac{1}{\lambda+\rho}f(\lambda)\mathbf{d}\lambda\] \[=\frac{L}{K-1}\sum_{i=1}^{K-1}\sum_{j=1}^{K-1}\mathcal{D}_{ij} \int_{0}^{\infty}\frac{1}{\lambda+\rho}\lambda^{R-K+j}e^{-\lambda/\xi_{i}} \mathbf{d}\lambda\] \[=\frac{L}{K-1}\sum_{i=1}^{K-1}\sum_{j=1}^{K-1}\mathcal{D}_{ij} \Gamma(R-K+j+1)\int_{0}^{\infty}e^{-\rho t}(t+\frac{1}{\xi_{i}})^{-R+K-j-1} \mathbf{d}t \tag{12}\] \[=\frac{L}{K-1}\sum_{i=1}^{K-1}\sum_{j=1}^{K-1}\mathcal{D}_{ij} \Gamma(R-K+j+1)\] \[\times\bigg{[}\frac{1}{(-R+K-j-2)!}\sum_{t=1}^{-R+K-j-2}(t-1)!(- \rho)^{-R+k-j-2-t}(\frac{1}{\xi_{i}})^{-t}-\frac{(-\rho)^{-R+K-j-2}}{(-R+K-j- 2)!}e^{\frac{\rho}{\xi_{i}}}\mathbf{E}(-\frac{\rho}{\xi_{i}})\bigg{]},\]  
where we have used [22] (\[L=\prod_{l=1}^{K-1}\frac{1}{\xi_{l}^{R}(R-l)!}\prod_{k<l}^{K-1}\frac{\xi_{k}\xi_ {l}}{\xi_{k}-\xi_{l}}. \tag{8}\]0353) for the final equality. \(\mathbf{E}\mathbf{i}(\mathbf{x})\) represents the exponential integral \(\int_{-\infty}^{x}\frac{\rho}{\tau}\mathrm{d}t\).