## 3 Detector Design
### AE-Rao Detector
\begin{split}\mathbf{I}_{\mathbf{\Theta}_{r},\mathbf{\Theta}_{s} }(\mathbf{\Theta})=&\mathrm{E}\bigg{[}\frac{\partial\ln f(\mathbf{z},\mathbf{Z})}{\partial\mathbf{\Theta}_{r}^{*}}\frac{\partial\ln f(\mathbf{z},\mathbf{Z})}{\partial\mathbf{\Theta}_{r}^{\mathrm{T}}}\bigg{]}\\ =&\mathbf{H}^{\mathrm{H}}\mathbf{R}^{-1}\mathrm{E} \Big{[}(\mathbf{z}-\mathbf{H}\boldsymbol{\xi})(\mathbf{z}-\mathbf{H} \boldsymbol{\xi})^{\mathrm{H}}\Big{]}\mathbf{R}^{-1}\mathbf{H}\ =\mathbf{H}^{\mathrm{H}}\mathbf{R}^{-1}\mathbf{H}\end{split}  \tag{18}\]  
\(\mathbf{I}_{\mathbf{\Theta}_{r},\mathbf{\Theta}_{s}}(\mathbf{\Theta})\) can be written as  
\[\mathbf{I}_{\mathbf{\Theta}_{r},\mathbf{\Theta}_{s}}(\mathbf{\Theta})\mathbf{= }\Big{[}\mathbf{I}_{\boldsymbol{\xi},\boldsymbol{\gamma}}(\mathbf{\Theta})\, \mathbf{I}_{\boldsymbol{\xi},\mathrm{vec}(\mathbf{R})}(\mathbf{\Theta})\Big{]} \tag{19}\]  
We can derive the following two equations according to (17)  
\[\mathbf{I}_{\boldsymbol{\xi},\boldsymbol{\gamma}}(\mathbf{\Theta})=-\mathrm{E} \bigg{[}\frac{\partial^{2}\ln f_{1}(\mathbf{z},\mathbf{Z})}{\partial\boldsymbol{ \xi}^{*}\partial\mathbf{\gamma}^{\mathrm{T}}}\bigg{]}=-\mathrm{E}\bigg{[} \frac{\partial}{\partial\mathbf{\gamma}^{\mathrm{T}}}\Big{(}\mathbf{H}^{ \mathrm{H}}\mathbf{R}^{-1}(\mathbf{z}-\mathbf{H}\boldsymbol{\xi})\Big{)} \bigg{]}=\mathbf{0}_{p\times L} \tag{20}\]  
and  
\[\mathbf{I}_{\boldsymbol{\xi},\mathrm{vec}(\mathbf{R})}(\mathbf{\Theta})=- \mathrm{E}\bigg{[}\frac{\partial^{2}\ln f_{1}(\mathbf{z},\mathbf{Z})}{ \partial\boldsymbol{\xi}^{*}\partial\mathrm{vec}^{\mathrm{T}}(\mathbf{R})} \bigg{]}=-\mathrm{E}\bigg{[}\frac{\partial}{\partial\mathrm{vec}^{\mathrm{T}}( \mathbf{R})}\Big{(}\mathbf{H}^{\mathrm{H}}\mathbf{R}^{-1}(\mathbf{z}-\mathbf{H }\boldsymbol{\xi})\Big{)}\bigg{]}=\mathbf{0}_{p\times N^{2}} \tag{21}\]  
Substituting (18), (19), (20), and (21) into (12) results in  
\[\Big{[}\mathbf{I}^{-1}(\mathbf{\Theta})\Big{]}_{\mathbf{\Theta}_{r}\mathbf{ \Theta}_{r}}=\Big{(}\mathbf{H}^{\mathrm{H}}\mathbf{R}^{-1}\mathbf{H}\Big{)}^{-1} \tag{22}\]  
Substituting (16), (17) and (22) into (11) yields the AE-Rao for the given \(\mathbf{R}\) as  
\[t_{\mathrm{AE-Rao}_{\mathbf{R}}}=\mathbf{z}^{\mathrm{H}}\mathbf{R}^{-1}\mathbf{H} (\mathbf{H}^{\mathrm{H}}\mathbf{R}^{-1}\mathbf{H})^{-1}\mathbf{H}^{\mathrm{H}} \mathbf{R}^{-1}\mathbf{z} \tag{23}\]  
Substituting the final alternate estimation of \(\mathbf{\gamma}_{0}\) into (7) results in the MLE of \(\mathbf{R}\) under hypothesis \(\mathbf{H}_{0}\) as  
\[\mathbf{\hat{R}}_{0}=\frac{1}{L+1}\Bigg{(}\sum_{l=1}^{L}\frac{\mathbf{z} \mathbf{z}\mathbf{z}_{l}^{\mathrm{H}}}{\gamma_{l}^{0,(l_{\mathrm{max}})}}+ \mathbf{z}\mathbf{z}^{\mathrm{H}}\Bigg{)} \tag{24}\]where the value of \(\hat{\gamma}_{l}^{0,(t_{\max})}\) can be obtained by (A9) and (A10). For convenience, we define \(\mathbf{\Sigma}_{0}=\sum\limits_{l=1}^{L}\frac{\mathbf{z}_{l}\mathbf{z}_{l}^{\mathrm{H}}}{ \hat{\gamma}_{l}^{0,(t_{\max})}}+\mathbf{z}\mathbf{z}^{\mathrm{H}}\). Substituting (24) into (23) and ignoring the constant terms results in the AE-Rao as  
\[t_{\text{AE-Rao}}=\mathbf{z}^{\mathrm{H}}\mathbf{\Sigma}_{0}^{-1}\mathbf{\mathrm{H}}\Big{(} \mathbf{\mathrm{H}}^{\mathrm{H}}\mathbf{\Sigma}_{0}^{-1}\mathbf{\mathrm{H}}\Big{)}^{-1} \mathbf{\mathrm{H}}^{\mathrm{H}}\mathbf{\Sigma}_{0}^{-1}\mathbf{z} \tag{25}\]