## 3 Random Antenna Selection (RAS)
### Ergodic Capacity of \(x_{1}\)  
Ergodic Capacity of \(x_{1}\) while being assisted by the \(k\)-th antenna is given as  
\[E_{x_{1}}^{RAS}(k)= \mathbb{E}\left[\log_{2}\left(1+\frac{\theta_{1}X_{1,k}}{\theta_ {2}X_{1,k}+Z_{1}}\right)\right] \tag{12}\] \[\overset{(a)}{=} \frac{1}{\ln 2}\int_{0}^{\frac{\theta_{1}}{2}}\frac{1}{1+x} \bigg{(}1-F_{G_{1}^{(b)}}\left(\frac{x}{\theta_{1}-x\theta_{2}}\right)\bigg{)}dx\] (13) \[\overset{(b)}{=} \frac{1}{\ln 2}\frac{\theta_{1}}{\theta_{2}-1}\bigg{(}\int_{0}^{ \infty}\frac{1}{t+\frac{1}{\theta_{2}}}\bar{F}_{G_{1}}(t)dt-\int_{0}^{\infty }\frac{1}{1+t}\bar{F}_{G_{1}}(t)dt\bigg{)}, \tag{14}\]  
in which \((a)\) is obtained by performing integral-by-part while \((b)\) is given via the variable change \(t\leftarrow\frac{x}{\theta_{1}-x\theta_{2}}\) and then performs some necessary manipulations.  
To derive (14) into closed-form, we consider the following well-known transformations as  
\[e^{-pt^{2}}=G_{0,1}^{1,0}\bigg{(}pt^{2}\bigg{|}_{0}^{-}\bigg{)}, \ p>0,v>0 \tag{15a}\] \[\frac{1}{t+1}=G_{1,1}^{1,1}\bigg{(}t\bigg{|}_{0}^{0}\bigg{)},\] (15b) \[\frac{1}{(t+1)^{2}}=G_{1,1}^{1,1}\bigg{(}t\bigg{|}_{0}^{-1} \bigg{)},\] (15c) \[\bigg{(}t+\frac{1}{\theta_{2}}\bigg{)}^{-1}=\theta_{2}G_{1,1}^{1,1 }\bigg{(}\theta_{2}t\bigg{|}_{0}^{0}\bigg{)}, \tag{15d}\]  
in which \(G_{p,q}^{m,n}(\cdot|\cdot)\) denotes the Meijer's G function, thus the first integral part can be derived as  
\[\begin{split}\int_{0}^{\infty}\frac{1}{1+t}\bar{F}_{G_{1}}(t)dt &=\int_{0}^{\infty}G_{1,1}^{1,1}\bigg{(}t\bigg{|}_{0}^{0} \bigg{)}G_{0,1}^{1,0}\bigg{(}C_{1}\left(\frac{t}{\mu_{1}a_{1}}\right)^{\frac{ 2}{q}}+C_{2}\left(\frac{t}{\mu_{2}a_{2}}\right)^{\frac{2}{q}}\bigg{|}_{0}^{-} \bigg{)}dt\\ &=H_{1,2}^{2,1}\bigg{(}C_{1}\left(\frac{1}{\mu_{1}a_{1}}\right)^ {\frac{2}{q}}+C_{2}\left(\frac{1}{\mu_{2}a_{2}}\right)^{\frac{2}{q}}\bigg{|} _{0}^{\frac{2}{q}}\bigg{|}(0,2\eta^{-1})\bigg{)},\end{split} \tag{16}\]  
in which the last equation is obtained via computation of \(H_{p,q}^{m,n}(\cdot|\cdot)\) which specifies the Fox H-function. The remaining integrals can be expressed in a closed-form with similar manner. Consequently, the analytical form of the Ergodic Capacity of \(x_{1}\) can be given as  
\[\begin{split} E_{x_{1}}^{RAS}(k)=\frac{\theta_{1}}{(\theta_{2}- 1)\ln 2}\left\{H_{1,2}^{2,1}\bigg{(}C_{1}\left(\frac{1}{\mu_{1}a_{1}\theta_{2}} \right)^{\frac{2}{q}}+C_{2}\left(\frac{1}{\mu_{2}a_{2}}\right)^{\frac{2}{q}} \bigg{|}_{0}^{\frac{2}{q}}\bigg{|}_{0}^{\frac{2}{q}}\bigg{|}_{0}  
\[\overset{(c)}{=} \frac{1}{\text{ln}2}\int_{0}^{\infty}\frac{1}{1+x}\bigg{(}1-F_{G_{2 }^{(k)}}\left(\frac{x}{\theta_{2}}\right)\bigg{)}dx \tag{19}\] \[\overset{(d)}{=} \frac{1}{\text{ln}2}\int_{0}^{\infty}\frac{1}{1+x}\bar{F}_{\bar{G} _{1}}\left(\frac{x}{\theta_{2}}\right)dx+\frac{1}{\text{ln}2}\int_{0}^{\infty} \frac{1}{1+x}\bar{F}_{\bar{G}_{2}}\left(\frac{x}{\theta_{2}}\right)dx\] \[-\frac{1}{\text{ln}2}\int_{0}^{\infty}\frac{1}{1+x}\bar{F}_{G_{1} }\left(\frac{x}{\theta_{2}}\right)\bar{F}_{G_{2}}\left(\frac{x}{\theta_{2}} \right)dx. \tag{20}\]  
in which \((c)\) is obtained via integration-by-part and \((d)\) is via substituting (10) into (19). Furthermore, by applying (11), we found that the above integrals are similar to (16) and thus can be solved similarly. Therefore, \(E_{x_{2}}^{(k)}\) can be derived in closed-form as  
\[\begin{split} E_{x_{2}}^{RAS}(k)=&\frac{1}{\text{ ln}2}\left\{H_{1,2}^{2,1}\bigg{(}C_{1}\left(\frac{1}{\mu_{1}\alpha_{1}\theta_{2}} \right)^{\frac{2}{q}}\bigg{|}\begin{matrix}(0,2\eta^{-1})\\ (0,1),(0,2\eta^{-1})\end{matrix}\right)+H_{1,2}^{2,1}\bigg{(}C_{2}\left(\frac{ 1}{\mu_{2}\alpha_{2}\theta_{2}}\right)^{\frac{2}{q}}\bigg{|}\begin{matrix}(0, 2\eta^{-1})\\ (0,1),(0,2\eta^{-1})\end{matrix}\right)\\ &-H_{1,2}^{2,1}\bigg{(}C_{1}\left(\frac{1}{\mu_{1}\alpha_{1} \theta_{2}}\right)^{\frac{2}{q}}+C_{2}\left(\frac{1}{\mu_{2}\alpha_{2}\theta_{ 2}}\right)^{\frac{2}{q}}\bigg{|}\begin{matrix}(0,2\eta^{-1})\\ (0,1),(0,2\eta^{-1})\end{matrix}\bigg{)}\right\}.\end{split} \tag{21}\]