## 3 Training Signal Optimization
### Optimal Power Allocation  
In this section, we determine the matrices \(\mathbf{D}_{P}^{*}\), \(\mathbf{D}_{T}^{*}\), and \(\mathbf{D}_{Q}^{*}\). Substituting the structures \(\mathbf{P}=\mathbf{U}_{T}\mathbf{D}_{P}\mathbf{U}_{Q^{\prime}}^{H}\mathbf{ \tilde{T}}=\mathbf{U}_{T}\mathbf{D}_{T}\mathbf{U}_{T}^{H}\), and \(\tilde{\mathbf{Q}}=\mathbf{U}_{Q}\mathbf{D}_{Q}\mathbf{U}_{Q}^{H}\) into (14), we can rewrite the estimated MSE \(J\) in (12) as  
\[\zeta(\mathbf{d}_{P},\mathbf{d}_{T},\mathbf{d}_{Q})=\beta\sum_{i=1}^{\
u} \biggl{(}\frac{1}{\lambda_{T,i}+d_{T,i}}+\frac{d_{P,i}}{\lambda_{Qi}+d_{Qi}} \biggr{)}^{-1}+\beta\sum_{i=\
u+1}^{M_{t}}(\lambda_{T,i}+d_{T,i}), \tag{17}\]  
where \(\beta=\mathrm{Tr}(\hat{\mathbf{R}})+\sqrt{M_{t}e_{R}}\) and \(\
u=\min\{M_{t},L\}\) denotes the maximum rank of \(\mathbf{P}\). Also,  
\[\mathbf{d}_{P}=[d_{P,1},\cdots,d_{P,\
u}]^{T},\]  
\[\mathbf{d}_{T}=[d_{T,1},\cdots,d_{T,M_{t}}]^{T},\]  
and  
\[\mathbf{d}_{Q}=[d_{Q,1},\cdots,d_{Q,L}]^{T}\]  
are the vectors of the diagonal elements of \(\mathbf{D}_{P}\mathbf{D}_{P}^{T}\), \(\mathbf{D}_{T}\), and \(\mathbf{D}_{Q}\), respectively. From (17), the optimal training power allocation \(\mathbf{d}_{P}^{*}\) can be obtained by solving the following problem:  
\[\min_{\mathbf{d}_{P}} \max_{\mathbf{d}_{T},\mathbf{d}_{Q}}\ \ \zeta(\mathbf{d}_{P},\mathbf{d}_{T},\mathbf{d}_{Q})\] (18a) s.t. \[\mathbf{d}_{P}\in\mathcal{D}_{\mathcal{P}},\ \ \mathbf{d}_{T}\in \mathcal{D}_{\mathcal{T}},\ \ \mathbf{d}_{Q}\in\mathcal{D}_{\mathcal{Q}}, \tag{18b}\]  
where the constraint sets \(\mathcal{D}_{P}\), \(\mathcal{D}_{\mathcal{T}}\), and \(\mathcal{D}_{\mathcal{Q}}\) are defined as  
\[\mathcal{D}_{\mathcal{P}}=\{\mathbf{d}_{P}:\sum_{i=1}^{\
u}d_{P,i}\leq\mathcal{ P}_{T},\ \ d_{P,i}\geq 0,\ i=1,\cdots,\
u\},\]  
\[\mathcal{D}_{\mathcal{T}}=\{\mathbf{d}_{T}:\|\mathbf{d}_{T}\|^{2}\leq\epsilon _{T},\ \ \lambda_{T,i}+d_{T,i}\geq 0,\ i=1,\cdots,M_{t}\}\]  
and  
\[\mathcal{D}_{\mathcal{Q}}=\{\mathbf{d}_{Q}:\|\mathbf{d}_{Q}\|^{2}\leq\epsilon _{Q},\ \ \lambda_{Qi}+d_{Q,i}\geq 0,\ i=1,\cdots,L\},\]  
respectively. In the following lemma, we show that the cost function of the problem (18) is convex-concave.  
**Definition 1** (Convex-concave function).: _[_28; 29_]_ _We say the function \(\varphi(\mathbf{a},\mathbf{b}):\mathcal{A}\times\mathcal{B}\subset\mathbb{R}^{n} \times\mathbb{R}^{m}\rightarrow\mathbb{R}\) is convex-concave if \(\varphi(\mathbf{a},\mathbf{b})\) is a convex function of \(\mathbf{a}\in\mathcal{A}\) for fixed \(\mathbf{b}\in\mathcal{B}\) and a concave function of \(\mathbf{b}\in\mathcal{B}\) for fixed \(\mathbf{a}\in\mathcal{A}\)._  
**Lemma 1**.: _The MSE \(\zeta\) in (17) is convex in \(\mathbf{d}_{P}\in\mathcal{D}_{\mathcal{P}}\) for fixed \((\mathbf{d}_{T},\mathbf{d}_{Q})\in\mathcal{D}_{\mathcal{T}}\times\mathcal{D}_ {\mathcal{Q}}\), and concave in \(\mathbf{d}_{T}\in\mathcal{D}_{\mathcal{T}}\) and \(\mathbf{d}_{Q}\in\mathcal{D}_{\mathcal{Q}}\) for fixed \(\mathbf{d}_{P}\in\mathcal{D}_{\mathcal{P}}\)._  
Proof.: See Appendix B.  
Lemma 1 implies that the power allocation problem (18) is convex-concave since the constraint sets \(\mathcal{D}_{\mathcal{P}}\), \(\mathcal{D}_{\mathcal{T}}\), and \(\mathcal{D}_{\mathcal{Q}}\) are all convex. Unfortunately, it is not possible to find aclosed-form solution in general due to the nonlinearity of the objective function and norm constraints on \(\mathbf{d}_{T}\) and \(\mathbf{d}_{Q}\). However, the optimal power allocation \(\mathbf{d}_{P}^{*}\) can be computed numerically by using well-known methods such as the interior point method or barrier method [28; 30].